DESIGN OF MACHINERY -5th Ed SOLUTION MANUAL

DESIGN OF MACHINERY - 5th Ed

SOLUTION MANUAL 2-1-1

PROBLEM 2-1 Statement:

Find three (or other number as assigned) of the following common devices. Sketch careful kinematic diagrams and find their total degrees of freedom. a. An automobile hood hinge mechanism b. An automobile hatchback lift mechanism c. An electric can opener d. A folding ironing board e. A folding card table f. A folding beach chair g. A baby swing h. A folding baby walker i. A fancy corkscrew as shown in Figure P2-9 j. A windshield wiper mechanism k. A dump-truck dump mechanism l. A trash truck dumpster mechanism m. A pickup tailgate mechanism n. An automobile jack o. A collapsible auto radio antenna

Solution:

See Mathcad file P0201.

Equation 2.1c is used to calculate the mobility (DOF) of each of the models below. a.

An automobile hood hinge mechanism. The hood (3) is linked to the body (1) through two rocker links (2 and 4). Number of links

L  4

Number of full joints

J1  4

Number of half joints

J2  0

M  3  ( L  1 )  2  J1  J2 M1 b.

HOOD 3 2

4

1 BODY

An automobile hatchback lift mechanism. The hatch (2) is pivoted on the body (1) and is linked to the body by the lift arm, which can be modeled as two links (3 and 4) connected through a translating slider joint. HATCH Number of links L  4 Number of full joints

J1  4


J2  0

2 3 1

M  3  ( L  1 )  2  J1  J2

4

M1

1 BODY

c.

An electric can opener has 2 DOF.

d.

A folding ironing board. The board (1) itself has one pivot (full) joint and one pin-in-slot sliding (half) joint. The two legs (2 and 3) hav a common pivot. One leg connects to the pivot joint on the board and the other to the slider joint.

DESIGN OF MACHINERY - 5th Ed


Number of links

L  3


J1  2


J2  1

1 3

2

M  3  ( L  1 )  2  J1  J2 M1 e.

A folding card table has 7 DOF: One for each leg, 2 for location in xy space, and one for angular orientation.

f.

A folding beach chair. The seat (3) and the arms (6) are ternary links. The seat is linked to the front leg(2), the back (5) and a coupling link (4). The arms are linked to the front leg (2), the rear leg (1), and the back (5). Links 1, 2, 4, and 5 are binar links. The analysis below is appropriate when the chair is not fully opened. When fully opened, one or more links are prevented from moving by a stop. Subtract 1 DOF when forced against the stop. Number of links

L  6


J1  7


J2  0

5 6 4

1

M  3  ( L  1 )  2  J1  J2

2 3

M1 g.

A baby swing has 4 DOF: One for the angular orientation of the swing with respect to the frame, and 3 for the location and orientation of the frame with respect to a 2-D frame.

h.

A folding baby walker has 4 DOF: One for the degree to which it is unfolded, and 3 for the location and orientation of the walker with respect to a 2-D frame.

i.

A fancy corkscrew has 2 DOF: The screw can be rotated and the arms rotate to translate the screw.

j.

A windshield wiper mechanism has 1 DOF: The position of the wiper blades is defined by a single input.

k.

A dump-truck dump mechanism has 1 DOF: The angle of the dump body is determined by the length of the hydraulic cylinder that links it to the body of the truck.

l.

A trash truck dumpster mechanism has 2 DOF: These are generally a rotation and a translation.

m. A pickup tailgate mechanism has 1 DOF: n.

An automobile jack has 4 DOF: One is the height of the jack and the other 3 are the position and orientation o the jack with respect to a 2-D frame.

o.

A collapsible auto radio antenna has as many DOF as there are sections, less one.

DESIGN OF MACHINERY - 5th Ed.



How many DOF do you have in your wrist and hand combined?

Solution:


1.

Holding the palm of the hand level and facing toward the floor, the hand can be rotated about an axis through the wrist that is parallel to the floor (and perpendicular to the forearm axis) and one perpendicular to the floor (2 DOF). The wrist can rotate about the forearm axis (1 DOF).

2.

Each finger (and thumb) can rotate up and down and side-to-side about the first joint. Additionally, each finger can rotate about each of the two remaining joints for a total of 4 DOF for each finger (and thumb).

3.

Adding all DOF, the total is Wrist Hand Thumb Fingers 4x4

1 2 4 16

TOTAL

23



How many DOF do the following joints have? a. Your knee b. Your ankle c. Your shoulder d. Your hip e. Your knuckle

Solution:


a.

Your knee. 1 DOF: A rotation about an axis parallel to the ground.

b.

Your ankle. 3 DOF: Three rotations about mutually perpendicular axes.

c.

Your shoulder. 3 DOF: Three rotations about mutually perpendicular axes.

d.

Your hip. 3 DOF: Three rotations about mutually perpendicular axes.

e

Your knuckle. 2 DOF: Two rotations about mutually perpendicular axes.





How many DOF do the following have in their normal environment? a. A submerged submarine b. An earth-orbit satellite c. A surface ship d. A motorcycle (road bike) e. A two-button mouse f. A computer joy stick.

Solution:


a.

A submerged submarine. Using a coordinate frame attached to earth, or an inertial coordinate frame, a submarine has 6 DOF: 3 linear coordinates and 3 angles.

b.

An earth-orbit satellite. If the satellite was just a particle it would have 3 DOF. But, since it probably needs to be oriented with respect to the earth, sun, etc., it has 6 DOF.

c.

A surface ship. There is no difference between a submerged submarine and a surface ship, both have 6 DOF. One might argue that, for an earth-centered frame, the depth of the ship with respect to mean sea level is constant, however that is not strictly true. A ship's position is generally given by two coordinates (longitude and latitude). For a given position, a ship can also have pitch, yaw, and roll angles. Thus, for all practical purposes, a surface ship has 5 DOF.

d.

A motorcycle. At an intersection, the motorcycle's position is given by two coordinates. In addition, it will have some heading angle (turning a corner) and roll angle (if turning). Thus, there are 4 DOF.

e.

A two-button mouse. A two-button mouse has 4 DOF. It can move in the x and y directions and each button has 1 DOF.

f.

A computer joy stick. The joy stick has 2 DOF (x and y) and orientation, for a total of 3 DOF.




Are the joints in Problem 2-3 force closed or form closed?

Solution:


They are force closed by ligaments that hold them together. None are geometrically closed.




Describe the motion of the following items as pure rotation, pure translation, or complex planar motion. a. A windmill b. A bicycle (in the vertical plane, not turning) c. A conventional "double-hung" window d. The keys on a computer keyboard e. The hand of a clock f. A hockey puck on the ice g. A "casement" window

Solution:


a.

A windmill. Pure rotation.

b.

A bicycle (in the vertical plane, not turning). Pure translation for the frame, complex planar motion for the wheels.

c.

A conventional "double-hung" window. Pure translation.

d.

The keys on a computer keyboard. Pure translation.

e.

The hand of a clock. Pure rotation.

f.

A hockey puck on the ice. Complex planar motion.

g.

A "casement" window. Pure rotation.




Calculate the mobility of the linkages assigned from Figure P2-1 part 1 and part 2.

Solution:

See Figure P2-1 and Mathcad file P0207.

1.

Use equation 2.1c (Kutzbach's modification) to calculate the mobility.

a.

Number of links

L  6


J1  7


J2  1

6 3 5 2

M  3  ( L  1 )  2  J1  J2

4 1

M0

(a)

1 3

b.

Number of links

L  3


J1  2


J2  1

1

M  3  ( L  1 )  2  J1  J2 M1

2 1

(b) 4

c.

Number of links

L  4


J1  4


J2  0

1

3

M  3  ( L  1 )  2  J1  J2 2

M1 (c)

1

7

d.

Number of links

L  7


J1  7


J2  1

1

6 5

M  3  ( L  1 )  2  J1  J2 M3

1

2 3

4

(d)



8

5

8 1

5 1

9 10

6

1

1 7

4

4

1

2

2 3

3

1

5 6

2 1

(e)

1

e.

g.

Number of links

L  10

Number of full joints Number of half joints

(f)

Number of links

L  6

J1  13


J1  6

J2  0


J2  2

f.

M  3  ( L  1 )  2  J1  J2

M  3  ( L  1 )  2  J1  J2

M1

M1

Number of links

L  8


J1  9


J2  2

M  3  ( L  1 )  2  J1  J2

4 1

4

7

6

3

7 1

5

8 1

2

2 1

1

1

M1 (g)

2 h.

Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2 M1

1 3 1 4 (h)




Identify the items in Figure P2-1 as mechanisms, structures, or preloaded structures.

Solution:


1.

Use equation 2.1c (Kutzbach's modification) to calculate the mobility and the definitions in Section 2.5 of the text to classify the linkages.

a.

Number of links

L  6


J1  7


J2  1

6 3 5 2

M  3  ( L  1 )  2  J1  J2 M0

4 1

Structure

(a)

1 3

b.

Number of links

L  3


J1  2


J2  1

1

M  3  ( L  1 )  2  J1  J2 M1

Mechanism

2 1

(b) 4

c.

Number of links

L  4


J1  4


J2  0

1

3

M  3  ( L  1 )  2  J1  J2 M1

2

Mechanism (c)

1

7

d.

Number of links

L  7


J1  7


J2  1

1

6 5

M  3  ( L  1 )  2  J1  J2 M3

Mechanism

1

2 3

4

(d)




Use linkage transformation on the linkage of Figure P2-1a to make it a 1-DOF mechanism.

Solution:

See Figure P2-1a and Mathcad file P0209.

1.

The mechanism in Figure P2-1a has mobility: Number of links

L  6


J1  7


J2  1

6 3 5 2

M  3  ( L  1 )  2  J1  J2 M0

4 1 1

2.

Use rule 2, which states: "Any full joint can be replaced by a half joint, but this will increase the DOF by one." One way to do this is to replace one of the pin joints with a pin-in-slot joint such as that shown in Figure 2-3c. Choosing the joint between links 2 and 4, we now have mobility: Number of links

L  6


J1  6


J2  2

6 3 5

M  3  ( L  1 )  2  J1  J2

2 4

M1

1 1




Use linkage transformation on the linkage of Figure P2-1d to make it a 2-DOF mechanism.

Solution:

See Figure P2-1d and Mathcad file P0210.

1.

7

The mechanism in Figure P2-1d has mobility: Number of links

L  7


J1  7


J2  1

M  3  ( L  1 )  2  J1  J2

1

6 5 1

2 3

4

M3 2.

Use rule 3, which states: "Removal of a link will reduce the DOF by one." One way to do this is to remove link 7 such that link 6 pivots on the fixed pin attached to the ground link (1). We now have mobility: Number of links

L  6


J1  6


J2  1

M  3  ( L  1 )  2  J1  J2

1 6 5 1

2 3

M2

4




Use number synthesis to find all the possible link combinations for 2-DOF, up to 9 links, to hexagonal order, using only revolute joints.

Solution:


1.

Use equations 2.4a and 2.6 with DOF = 2 and iterate the solution for valid combinations. Note that the number of links must be odd to have an even DOF (see Eq. 2.4). The smallest possible 2-DOF mechanism is then 5 links since three will give a structure (the delta triplet, see Figure 2-7). L  B  T  Q  P  H

L  3  M  T  2  Q  3  P  4  H L  5  T  2  Q  3  P  4  H

2.

3.

For L  5 0  T  2  Q  3  P  4  H

0=T =Q=P=H

2  T  2  Q  3  P  4  H

H  0

For L  7

Case 1:

Case 2:

4.

B  5

Q  0

Q  1

P  0

T  2  2  Q  3  P  4  H

T2

B  L  T  Q  P  H

B5

T  2  2  Q  3  P  4  H

T0

B  L  T  Q  P  H

B6

T  0

P  0

For L  9 4  T  2  Q  3  P  4  H Case 1:

H  1

Q  0

B  L  T  Q  P  H Case 2a:

H  0

B8

4  T  2  Q  3  P 9  B  T  Q  P

Case 2b:

P  1

1  T  2  Q

Q  0

B  L  T  Q  P  H Case 2c:

P  0

T  1 B7

4  T  2  Q 9  B  T  Q

Case 2c1:

Case 2c2:

Case 2c3:

Q  2 Q  1 Q  0

T  4  2  Q

T0

B  9  T  Q

B7

T  4  2  Q

T2

B  9  T  Q

B6

T  4  2  Q

T4

B  9  T  Q

B5

M  2




Find all of the valid isomers of the eightbar 1-DOF link combinations in Table 2-2 (p. 38) having a. Four binary and four ternary links. b. Five binaries, two ternaries, and one quaternary link. c. Six binaries and two quaternary links. d. Six binaries, one ternary, and one pentagonal link.

Solution:


1.

2.

a.

Table 2-3 lists 16 possible isomers for an eightbar chain. However, Table 2-2 shows that there are five possible link sets, four of which are listed above. Therefore, we expect that the 16 valid isomers are distributed among the five link sets and that there will be fewer than 16 isomers among the four link sets listed above. One method that is helpful in finding isomers is to represent the linkage in terms of molecules as defined in Franke's Condensed Notations for Structural Synthesis. A summary of the rules for obtaining Franke's molecules follows: (1) The links of order greater than 2 are represented by circles. (2) A number is placed within each circle (the "valence" number) to describe the type (ternary, quaternary, etc.) of link. (3) The circles are connected using straight lines. The number of straight lines emanating from a circle must be equal to its valence number. (4) Numbers (0, 1, 2, etc.) are placed on the straight lines to correspond to the number of binary links used in connecting the higher order links. (5) There is one-to-one correspondence between the molecule and the kinematic chain that it represents. Four binary and four ternary links. Draw 4 circles with valence numbers of 3 in each. Then find all unique combinations of straight lines that can be drawn that connect the circles such that there are exactly three lines emanating from each circle and the total of the numbers written on the lines is exactly equal to 4. In this case, there are three valid isomers as depicted by Franke's molecules and kinematic chains below.

8

1 3

3

5

1 0

0

1 3

6

3

3

1

4

2 1

7



8 0 3

3 2

0

5

0

1 3

7

6

3 1

3

4

1 2

8 5

0 3

3

4

2 0

6

0

2 3

3

3 0

7

1 2

The mechanism shown in Figure P2-5b is the same eightbar isomer as that depicted schematically above. b.

Five binaries, two ternaries, and one quaternary link. Draw 2 circles with valence numbers of 3 in each and one with a valence number of 4. Then find all unique combinations of straight lines that can be drawn that connect the circles such that there are exactly three lines emanating from each circle with valence of three and four lines from the circle with valence of four; and the total of the numbers written on the lines is exactly equal to 5. In this case, there are five valid isomers as depicted by Franke's molecules and kinematic chains below.

0

3 2

4

7

0

1

5

3

3

2

6 4

8

1

2



5 1

3 0

3 0

2

6 4

7

3

2 4

8

1

2

5 0

3 1

3

3

7

1

2

4

6

1

2

8

4

1

5 1

3 1

6

3

4

0

1

3 2

7

8 4

1

2

5 1

3 1

6

3

3

1

1

1

8

4

7

2

4

1

c.

Six binaries and two quaternary links. Draw 2 circles with valence numbers of 4 in each. Then find all unique combinations of straight lines that can be drawn that connect the circles such that there are exactly four lines emanating from each circle and the total of the numbers written on the lines is exactly equal to 6. In this case, there are two valid isomers as depicted by Franke's molecules and kinematic chains below.



0 2

7 4

4

4

5

3

6

2

8

2

1

2

1

4

1

7 4

4

d.

3

6

2 2

5

8

2 1

Six binaries, one ternary, and one pentagonal link. There are no valid implementations of 6 binary links with 1 pentagonal link.




Use linkage transformation to create a 1-DOF mechanism with two sliding full joints from a Stephenson's sixbar linkage as shown in Figure 2-14a (p. 47).

Solution:

See Figure 2-14a and Mathcad file P0213.

1.

The mechanism in Figure 2-14a has mobility: Number of links

L  6


J1  7


J2  0

A 4

3

5

B

2

6

M  3  ( L  1 )  2  J1  J2 M1

2.

1

Use rule 1, which states: "Revolute joints in any loop can be replaced by prismatic joints with no change in DOF of the mechanism, provided that at least two revolute joints remain in the loop." One way to do this is to replace pin joints at A and B with translating full slider joints such as that shown in Figure 2-3b. Note that the sliders are attached to links 3 and 5 in such a way that they can not rotate relative to the links. The number of links and 1-DOF joints remains the same. There are no 2-DOF joints in either mechanism.

A 4

3

5 2

1

6 B




Use linkage transformation to create a 1-DOF mechanism with one sliding full joint a a half joint from a Stephenson's sixbar linkage as shown in Figure 2-14b (p. 48).

Solution:


1.

The mechanism in Figure 2-14b has mobility: Number of links

L  6


J1  7


J2  0

3 5 4 2

6

M  3  ( L  1 )  2  J1  J2 1

M1

2.

To get the sliding full joint, use rule 1, which states: "Revolute joints in any loop can be replaced by prismati joints with no change in DOF of the mechanism, provided that at least two revolute joints remain in the loop." One way to do this is to replace pin joint links 3 and 5 with a translating full slider joint such as that shown in Figure 2-3b. Note that the slider is attached to link 3 in such a way that it can not rotate relative to the link. The number of links and 1-DOF joints remains the same.

3

5 4

2

6

1

3.

To get the half joint, use rule 4 on page 42, which states: "The combination of rules 2 and 3 above will keep the original DOF unchanged." One way to do this is to remove link 6 (and its two nodes) and insert a half joint between links 5 and 1. Number of links

L  5


J1  5


3 5 4

J2  1 2

M  3  ( L  1 )  2  J1  J2

1

M1 1




Calculate the Grashof condition of the fourbar mechanisms defined below. Build cardboard models of the linkages and describe the motions of each inversion. Link lengths are in inches (or double given numbers for centimeters). Part 1. a. b. c.

2 2 2

4.5 3.5 4.0

7 7 6

9 9 8

Part 2. d. e. f.

2 2 2

4.5 4.0 3.5

7 7 7

9 9 9

Solution: 1.

See Mathcad file P0215

Use inequality 2.8 to determine the Grashof condition. Condition( a b c d ) 

S  min ( a b c d ) L  max( a b c d ) SL  S  L PQ  a  b  c  d  SL return "Grashof" if SL  PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise

a.

Condition( 2 4.5 7 9 )  "Grashof"

b.

Condition( 2 3.5 7 9 )  "non-Grashof"

c.

Condition( 2 4.0 6 8 )  "Special Grashof" This is a special case Grashof since the sum of the shortest and longest is equal to the sum of the other two link lengths.

d.


e.


f.

Condition( 2 3.5 7 9 )  "non-Grashof"




Which type(s) of electric motor would you specify a. b. c.

Solution:

To drive a load with large inertia. To minimize variation of speed with load variation. To maintain accurate constant speed regardless of load variations.


a.

Motors with high starting torque are suited to drive large inertia loads. Those with this characteristic include series-wound, compound-wound, and shunt-wound DC motors, and capacitor-start AC motors.

b.

Motors with flat torque-speed curves (in the operating range) will minimize variation of speed with load variation. Those with this characteristic include shunt-wound DC motors, and synchronous and capacitor-start AC motors.

b.

Speed-controlled DC motors will maintain accurate constant speed regardless of load variations.




Describe the difference between a cam-follower (half) joint and a pin joint.

Solution:


1.

A pin joint has one rotational DOF. A cam-follower joint has 2 DOF, rotation and translation. The pin joint also captures its lubricant in the annulus between pin and bushing while the cam-follower joint squeezes its lubricant out of the joint.




Examine an automobile hood hinge mechanism of the type described in Section 2.14. Sketch it carefully. Calculate its DOF and Grashof condition. Make a cardboard model. Analyze it with a free-body diagram. Describe how it keeps the hood up.

Solution:

Solution of this problem will depend upon the specific mechanism modeled by the student.




Find an adjustable arm desk lamp of the type shown in Figure P2-2. Sketch it carefully. Measure it and sketch it to scale. Calculate its DOF and Grashof condition. Make a cardboard model. Analyze it with a free-body diagram. Describe how it keeps itself stable. Are there any positions in which it loses stability? Why?

Solution:

Solution of this problem will depend upon the specific mechanism modeled by the student.




The torque-speed curve for a 1/8 hp permanent magnet (PM) DC motor is shown in Figure P2-3. The rated speed for this fractional horsepower motor is 2500 rpm at a rated voltage of 130V. Determine: a) The rated torque in oz-in (ounce-inches, the industry standard for fractional hp motors) b) The no-load speed c) Plot the power-torque curve and determine the maximum power that the motor can deliver.

Given:

Rated speed, N

NR  2500 rpm

R

HR 

Rated power, H

R

1 8

 hp

4000 3500

Speed, rpm

3000 2500 2000 1500 1000 500 0

0

50

100

150

200

250

300

Torque, oz-in

Figure P2-3 Torque-speed Characteristic of a 1/8 HP, 2500 rpm PM DC Motor

Solution: a.


The rated torque is found by dividing the rated power by the rated speed: TR 

Rated torque, TR

HR NR

TR  50 ozf  in

b.

The no-load speed occurs at T = 0. From the graph this is 3000 rpm.

c.

The power is the product of the speed and the torque. From the graph the equation for the torque-speed curve is: 3000 rpm N ( T )    T  3000 rpm 300  ozf  in and the power, therefore, is: H ( T )  10

rpm ozf  in

2

 T  3000 rpm T

Plotting the power as a function of torque over the range T  0  ozf  in 10 ozf  in  300  ozf  in



0.25 0.225 0.2

Power, hp

0.175 0.15 0.125 0.1 0.075 0.05 0.025 0

0

50

100

150

200

250

300

Torque, oz-in

Maximum power occurs when dH/dT = 0. The value of T at maximum power is: ozf  in

Value of T at Hmax

THmax  3000 rpm

Maximum power

Hmax  H  THmax

Hmax  0.223  hp

Speed at max power

NHmax  N  THmax

NHmax  1500 rpm

2  10 rpm

THmax  150  ozf  in

Note that the curve goes through the rated power point of 0.125 hp at the rated torque of 50 oz-in.




Find the mobility of the mechanisms in Figure P2-4.

Solution:


1.


a.

This is a basic fourbar linkage. The input is link 2 and the output is link 4. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1).

Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2

A 2 3 O2

M1

b.

4

C

O4

This is a fourbar linkage. The input is link 2, which in this case is the wheel 2 with a pin at A, and the output is link 4. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1).

Number of links

L  4


J1  4


J2  0

A

2

O2

3

M  3  ( L  1 )  2  J1  J2

4 B

M1

c.

O4

This is a 3-cylinder, rotary, internal combustion engine. The pistons (sliders) 6, 7, and 8 drive the output crank (2) through piston rods (couplers 3, 4, and 5). There are 3 full joints at the crank where rods 3, 4and 5 are pinned to crank 2. The cross-hatched crank-shaft at O2 is supported by the ground link (1) through bearings. Number of links

L  8


J1  10


J2  0

M  3  ( L  1 )  2  J1  J2

6

3

2

4

M1 7

5

8


d.


This is a fourbar linkage. The input is link 2, which in this case is a wheel with a pin at A, and the output is the vertical member on the coupler, link 3. Since the lengths of links 2 and 4 (O2A and O4B) are the same, the coupler link (3) has curvilinear motion and AB remains parallel to O2O4 throughout the cycle. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1).

Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2

B O4

O2

M1

e.

3

A 2

4

This is a fourbar linkage with an output dyad. The input (rocker) is link 2 and the output (rocker) is link 8. Links 5 and 6 are redundant, i.e. the mechanism will have the same motion if they are removed. The input fourbar consists of links 1, 2, 3, and 4. The output dyad consists of links 7 and 8. The cross-hatched pivot pins at O2, O4 and O8 are attached to the ground link (1). In the calculation below, the redundant links and their joints are not counted (subtract 2 links and 4 joints from the totals). A

Number of links

L  6


J1  7


J2  0

O2

4 O4 G

E

3

2 D

5

C

6

7

M  3  ( L  1 )  2  J1  J2 O8

M1 F

H 8

f.

This is a fourbar offset slider-crank linkage. The input is link 2 (crank) and the output is link 4 (slider block). The cross-hatched pivot pin at O2 is attached to the ground link (1).

Number of links

L  4


J1  4


J2  0

4

B

3

M  3  ( L  1 )  2  J1  J2 A

M1 2 O2


g.


This is a fourbar linkage with an alternate output dyad. The input (rocker) is link 2 and the outputs (rockers) are links 4 and 6. The input fourbar consists of links 1, 2, 3, and 4. The alternate output dyad consists of links 5 and 6. The cross-hatched pivot pins at O2, O4 and O6 are attached to the ground link (1). Number of links

L  6


J1  7


J2  0

O6 3

B

A

2

M  3  ( L  1 )  2  J1  J2

4

C

6

O2

M1

5 D O4

h.

This is a ninebar mechanism with three redundant links, which reduces it to a sixbar. Since this mechanism is symmetrical about a vertical centerline, we can split it into two mirrored mechanisms to analyze it. Either links 2, 3 and 5 or links 7, 8 and 9 are redundant. To analyze it, consider 7, 8 and 9 as the redundant links. Analyzing the ninebar, there are two full joints at the pins A, B and C for a total of 12 joints. Number of links

L  9


J1  12


J2  0

6

O2 2

8

7

5

C

B

M  3  ( L  1 )  2  J1  J2

O8

A

9

3

M0 4

D

E

The result is that this mechanism seems to be a structure. By splitting it into mirror halves about the vertical centerline the mobility is found to be 1. Subtract the 3 redundant links and their 5 (6 minus the joint at A) associated joints to determine the mobility of the mechanism. Number of links

L  9  3


J1  12  5


J2  0

6 O2 2

5 B

M  3  ( L  1 )  2  J1  J2

3

M1 D

4

A



PROBLEM 2-22 Statement: Solution: 1.

Find the Grashof condition and Barker classifications of the mechanisms in Figure P2-4a, b, and d. See Figure P2-4 and Mathcad file P0222.

Use inequality 2.8 to determine the Grashof condition and Table 2-4 to determine the Barker classification. Condition( a b c d ) 


a.

This is a basic fourbar linkage. The input is link 2 and the output is link 4. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1). L1  174

L2  116

L3  108

L4  110

A 2 3

Condition L1 L2 L3 L4  "non-Grashof"

O2

4

C

This is a Barker Type 5 RRR1 (non-Grashof, longest link grounded). b.

O4

This is a fourbar linkage. The input is link 2, which in this case is the wheel with a pin at A, and the output is link 4. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1). L1  162

L2  40

L3  96

L4  122

B

A

2

3

O2 4

Condition L1 L2 L3 L4  "Grashof" This is a Barker Type 2 GCRR (Grashof, shortest link is input). d.

This is a fourbar linkage. The input is link 2, which in this case is a wheel with a pin at A, and the output is the vertical member on the coupler, link 3. Since the lengths of links 2 and 4 (O2A and O4B) are the same, the coupler link (3) has curvilinear motion and AB remains parallel to O2O4 throughout the cycle. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1). L1  150 L2  30 L3  150

L4  30

Condition L1 L2 L3 L4  "Special Grashof" This is a Barker Type 13 S2X (special case Grashof, two equal pairs, parallelogram).

O4

A

3

2 O2

B O4

4




Find the rotability of each loop of the mechanisms in Figure P2-4e, f, and g.

Solution:


1.

Use inequality 2.15 to determine the rotability of each loop in the given mechanisms.

e.

This is a fourbar linkage with an output dyad. The input (rocker) is link 2 and the output (rocker) is link 8. Links 5 and 6 are redundant, i.e. the mechanism will have the same motion if they are removed. The input fourbar consists of links 1, 2, 3, and 4. The output dyad consists of links 7 and 8. The cross-hatched pivot pins at O2, O4 and O8 are attached to the ground link (1). In the calculation below, the redundant links and their joints are not counted (subtract 2 links and 4 joints from the totals).

B

A O2

4 O4 G

E

3

2 D

5

C

6

7

O8

There are two loops in this mechanism. The first loop consists of links 1, 2, 3 (or 5), and 4. The second consists of links 1, 4, 7 (or 6), and 8. By inspection, we see that the sum of the shortest and longest in each loop is equal to the sum of the other two. Thus, both loops are Class III. f.

8

This is a fourbar offset slider-crank linkage. The input is link 2 (crank) and the output is link 4 (slider block). The cross-hatched pivot pin at O2 is attached to the ground link (1).

4

A 2 O2

O6

This is a fourbar linkage with an alternate output dyad. The input (rocker) is link 2 and the outputs (rockers) are links 4 and 6. The input fourbar consists of links 1, 2, 3, and 4. The alternate output dyad consists of links 5 and 6. The cross-hatched pivot pins at O2, O4 and O6 are attached to the ground link (1). r1  87

r2  49

r3  100

r4  153

B

3

We can analyze this linkage if we replace the slider ( 4) with an infinitely long binary link that is pinned at B to link 3 and pinned to ground (1). Then links 1 and 4 for are both infinitely long. Since these two links are equal in length and, if we say they are finite in length but very long, the rotability of the mechanism will be determined by the relative lengths of 2 and 3. Thus, this is a Class I linkage since link 2 is shorter than link 3.

g.

H

F

3

B

2 4

C

A 6

O2 5 D

Using the notation of inequality 2.15, N  4 LN  r4 L1  r2 L2  r1 LN  L1  202

O4

L3  r3 L2  L3  187

Since LN  L1  L2  L3, this is a a class II mechanism.




Find the mobility of the mechanisms in Figure P2-5.

Solution:


1.

Use equation 2.1c (Kutzbach's modification) to calculate the mobility. In the kinematic representations of the linkages below, binary links are depicted as single lines with nodes at their end points whereas higher order links are depicted as 2-D bars.

a.

This is a sixbar linkage with 4 binary (1, 2, 5, and 6) and 2 ternary (3 and 4) links. The inverted U-shaped link at the top of Figure P2-5a is represented here as the binary link 6. Number of links

L  6


J1  7


J2  0

3

5

4

2

M  3  ( L  1 )  2  J1  J2 O2

M1

b.

6

O4

This is an eightbar linkage with 4 binary (1, 4, 7, and 8) and 4 ternary (2, 3, 5, and 6) links. The inverted U-shaped link at the top of Figure P2-5b is represented here as the binary link 8. Number of links

L  8


J1  10


J2  0

M  3  ( L  1 )  2  J1  J2 M1

5

6

2

3

7

8 2 O2

4 O4




Find the mobility of the ice tongs in Figure P2-6. a. When operating them to grab the ice block. b. When clamped to the ice block but before it is picked up (ice grounded). c. When the person is carrying the ice block with the tongs.

Solution:


1.


a.

In this case there are two links and one full joint and 1 DOF. Number of links

L  2


J1  1


J2  0

M  3  ( L  1 )  2  J1  J2 b.

When the block is clamped in the tongs another link and two more full joints are added reducing the DOF to zero (the tongs and ice block form a structure). Number of links

L  2  1


J1  1  2


J2  0

M  3  ( L  1 )  2  J1  J2

c.

M1

M0

When the block is being carried the system has at least 4 DOF: x, y, and z position and orientation about a vertical axis.




Find the mobility of the automotive throttle mechanism shown in Figure P2-7.

Solution:


1.

This is an eightbar linkage with 8 binary links. It is assumed that the joint between the gas pedal (2) and the roller (3) that pivots on link 4 is a full joint, i.e. the roller rolls without slipping. The pivot pins at O2, O4, O6, and O8 are attached to the ground link (1). Use equation 2.1c (Kutzbach's modification) to calculate the mobility. Number of links

L  8


J1  10


J2  0

M  3  ( L  1 )  2  J1  J2

M1

7 6 O6

8

FULL JOINT 5 4 O4 3 2

O2

O8




Sketch a kinematic diagram of the scissors jack shown in Figure P2-8 and determine its mobility. Describe how it works.

Solution:


1.

The scissors jack depicted is a seven link mechanism with eight full and two half joints (see kinematic diagram below). Link 7 is a variable length link. Its length is changed by rotating the screw with the jack handle (not shown). The two blocks at either end of link 7 are an integral part of the link. The block on the left is threaded and acts like a nut. The block on the right is not threaded and acts as a bearing. Both blocks have pins that engage the holes in links 2, 3, 5, and 6. Joints A and B have 2 full joints apiece. For any given length of link 7 the jack is a structure (DOF = 0). When the screw is turned to give the jack a different height the jack has 1 DOF.

4

3

5

7

A

B 2

6 1

Number of links

L  7


J1  8


J2  2

M  3  ( L  1 )  2  J1  J2

M0




Find the mobility of the corkscrew in Figure P2-9.

Solution:


1.

The corkscrew is made from 4 pieces: the body (1), the screw (2), and two arms with teeth (3), one of which is redundant. The second arm is present to balance the forces on the assembly but is not necessary from a kinematic standpoint. So, kinematically, there are 3 links (body, screw, and arm), 2 full joints (sliding joint between the screw and the body, and pin joint where the arm rotates on the body), and 1 half joint where the arm teeth engage the screw "teeth". Using equation 2.1c, the DOF (mobility) is Number of links

L  3


J1  2


J2  1

M  3  ( L  1 )  2  J1  J2

M1




Figure P2-10 shows Watt's sun and planet drive that he used in his steam engine. The beam 2 is driven in oscillation by the piston of the engine. The planet gear is fixed rigidly to link 3 and its center is guided in the fixed track 1. The output rotation is taken from the sun gear 4. Sketch a kinematic diagram of this mechanism and determine its DOF. Can it be classified by the Barker scheme? If so, what Barker class and subclass is it?

Solution:


1.

Sketch a kinematic diagram of the mechanism. The mechanism is shown on the left and a kinematic model of it is sketched on the right. It is a fourbar linkage with 1 DOF (see below).

A

2

2

1

3

3 1

4

B

4

1

2.

C

Use equation 2.1c to determine the DOF (mobility). There are 4 links, 3 full pin joints, 1 half pin-in-slot joint (at B), and 1 half joint (at the interface C between the two gears, shown above by their pitch circles). Links 1 and 3 are ternary. Kutzbach's mobility equation (2.1c) Number of links

L  4


J1  3


J2  2

M  3  ( L  1 )  2  J1  J2 3.

M1

The Barker classification scheme requires that we have 4 link lengths. The motion of link 3 can be modeled by a basic fourbar if the half joint at B is replaced with a full pin joint and a link is added to connect B and the fixed pivot that is coincident with the center of curvature of the slot that guides pin B. L1  2.15

L2  1.25

L3  1.80

L4  0.54

This is a Grashof linkage and the Barker classification is I-4 (type 4) because the shortest link is the output.




Figure P2-11 shows a bicycle hand brake lever assembly. Sketch a kinematic diagram of this device and draw its equivalent linkage. Determine its mobility. Hint: Consider the flexible cable to be a link.

Solution:


1.

The motion of the flexible cable is along a straight line as it leaves the guide provided by the handle bar so it can be modeled as a translating full slider that is supported by the handlebar (link 1). The brake lever is a binary link that pivots on the ground link. Its other node is attached through a full pin joint to a third link, which drives the slider (link 4). Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2 M1

CABLE BRAKE LEVER 3 2

4 1 1




Figure P2-12 shows a bicycle brake caliper assembly. Sketch a kinematic diagram of this device and draw its equivalent linkage. Determine its mobility under two conditions. a. b.

Brake pads not contacting the wheel rim. Brake pads contacting the wheel rim.

Hint: Consider the flexible cable to be replaced by forces in this case. Solution: 1.


The rigging of the cable requires that there be two brake arms. However, kinematically they operate independently and can be analyzed that way. Therefore, we only need to look at one brake arm. When the brake pads are not contacting the wheel rim there is a single lever (link 2) that is pivoted on a full pin joint that is attached to the ground link (1). Thus, there are two links (frame and brake arm) and one full pin joint. Number of links

L  2


J1  1


J2  0

BRAKE ARM

FRAME

2

M  3  ( L  1 )  2  J1  J2 M1

2.

1

When the brake pad contacts the wheel rim we could consider the joint between the pad, which is rigidly attached to the brake arm and is, therefore, a part of link 2, to be a half joint. The brake arm (with pad), wheel (which is constrained from moving laterally by the frame), and the frame constitute a structure. Number of links

L  2


J1  1


J2  1

BRAKE ARM

FRAME

2

M  3  ( L  1 )  2  J1  J2 M0

1 1 HALF JOINT




Find the mobility, the Grashof condition, and the Barker classifications of the mechanism in Figure P2-13.

Solution:


1.

Use equation 2.1c (Kutzbach's modification) to calculate the mobility. When there is no cable in the jaw or before the cable is crimped this is a basic fourbar mechanism with with 4 full pin joints: Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2

M1

When there is a cable in the jaw this is a threebar mechanism with with 3 full pin joints. While the cable is clamped the jaws are stationary with respect to each other so that link 4 is grounded along with link 1, leaving only three operational links.

2.

Number of links

L  3


J1  3


J2  0

M  3  ( L  1 )  2  J1  J2

M0



L1  0.92

L2  0.27

L3  0.50

L4  0.60

Condition L1 L2 L3 L4  "non-Grashof" The Barker classification is II-1 (Type 5) RRR1 (non-Grashof, longest link grounded).




The approximate torque-speed curve and its equation for a 1/4 hp shunt-wound DC motor are shown in Figure P2-14. The rated speed for this fractional horsepower motor is 10000 rpm at a rated voltage of 130V. Determine: a) The rated torque in oz-in (ounce-inches, the industry standard for fractional hp motors) b) The no-load speed c) The operating speed range d) Plot the power-torque curve in the operating range and determine the maximum power that the motor can deliver in the that range.

Given:

Rated speed, N N ( T ) 

NR  10000  rpm

R

0.1 1.7

NR TR NR TR

HR 

Rated power, H

R

1 4

 hp

 T  1.1 NR if T  62.5 ozf  in  T  5.1 NR otherwise

T  0  ozf  in 2.5 ozf  in  75 ozf  in

12000

10000

Speed, rpm

8000

6000

4000

2000

0

0

25

50

75

100

Torque, oz-in

Figure P2-14 Torque-speed Characteristic of a 1/4 HP, 10000 rpm DC Motor Solution: a.


The rated torque is found by dividing the rated power by the rated speed: Rated torque, TR

TR 

HR NR

TR  25 ozf  in

b.

The no-load speed occurs at T = 0. From the graph this is 11000 rpm.

c.

The operating speed range for a shunt-wound DC motor is the speed at which the motor begins to stall up to the no-load speed. For the approximate torque-speed curve given in this problem the minimum speed is defined as the speed at the knee of the curve. Nopmin  N ( 62.5 ozf  in)

Nopmin  8500 rpm



Nopmax  N ( 0  ozf  in)

The power is the product of the speed and the torque. From the graph the equation for the torque-speed curve over the operating range is: N ( T )  

40 rpm ozf  in

 T  11000  rpm

and the power, therefore, is: H ( T )  N ( T )  T Plotting the power as a function of torque over the range T  0  ozf  in 2.5 ozf  in  62.5 ozf  in 0.750 0.700 0.650 0.600 0.550 0.500 Power, hp

d.

Nopmax  11000 rpm

0.450 0.400 0.350 0.300 0.250 0.200 0.150 0.100 0.050 0.000 0.0

12.5

25.0

37.5

50.0

62.5

75.0

Torque, oz-in

Maximum power occurs at the maximum torque in the operating range. The value of T at maximum power is: Value of T at Hmax

THmax  62.5 ozf  in

THmax  62.5 ozf  in

Maximum power

Hmax  H  THmax

Hmax  0.527  hp

Speed at max power

NHmax  N  THmax

NHmax  8500 rpm

Note that the curve goes through the rated power point of 0.25 hp at the rated torque of 25 oz-in.




Figure P2-15 shows a power hacksaw, used to cut metal. Link 5 pivots at O5 and its weight forces the sawblade against the workpiece while the linkage moves the blade (link 4) back and forth within link 5 to cut the part. Sketch its kinematic diagram, determine its mobility and its type (i.e., is it a fourbar, a Watt's sixbar, a Stephenson's sixbar, an eightbar, or what?) Use reverse linkage transformation to determine its pure revolute-jointed equivalent linkage.

Solution:


1.

Sketch a kinematic diagram of the mechanism. The mechanism is shown on the left and a kinematic model of it is sketched on the right. It is a fivebar linkage with 1 DOF (see below). 5

3

5 3

4

4

2

2

2.

1

1

1

Use equation 2.1c to determine the DOF (mobility). There are 5 links, 4 full pin joints, 1 full sliding joint, and 1 half joint (at the interface between the hacksaw blade and the pipe being cut). Kutzbach's mobility equation (2.1c) Number of links

L  5


J1  5


J2  1

M  3  ( L  1 )  2  J1  J2 3.

M1

Use rule 1 to transform the full sliding joint to a full pin joint for no change in DOF. Then use rules 2 and 3 by changing the half joint to a full pin joint and adding a link for no change in DOF. The resulting kinematically equivalent linkage has 6 links, 7 full pin joints, no half joints, and is shown below. Kutzbach's mobility equation (2.1c) Number of links

L  6


J1  7


J2  0

5 4

3 2

M  3  ( L  1 )  2  J1  J2

1 6

M1

1




Figure P2-16 shows a manual press used to compact powdered materials. Sketch its kinematic diagram, determine its mobility and its type (i.e., is it a fourbar, a Watt's sixbar, a Stephenson's sixbar, an eightbar, or what?) Use reverse linkage transformation to determine its pure revolute-jointed equivalent linkage.

Solution:


1.

Sketch a kinematic diagram of the mechanism. The mechanism is shown on the left and a kinematic model of it is sketched on the right. It is a fourbar linkage with 1 DOF (see below).

4 3

4

3 2

2

O2

O2

2.

Use equation 2.1c to determine the DOF (mobility). There are 4 links, 3 full pin joints, 1 full sliding joint, and 0 half joints. This is a fourbar slider-crank. Kutzbach's mobility equation (2.1c) Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2 3.

M1

Use rule 1 to transform the full sliding joint to a full pin joint for no change in DOF. The resulting kinematically equivalent linkage has 4 links, 4 full pin joints, no half joints, and is shown below. 4 O4 3

2 O2




Sketch the equivalent linkage for the cam and follower mechanism in Figure P2-17 in the position shown. Show that it has the same DOF as the original mechanism.

Solution:


1.

The cam follower mechanism is shown on the left and a kinematically equivalent model of it is sketched on the right.

4 1

1 4

3

3

2

2

INSTANTANEOUS CENTER OF CURVATURE OF CAM SURFACE

1

1

2.

Use equation 2.1c to determine the DOF (mobility) of the original mechanism. There are 4 links, 2 full pin joints, 1 full sliding joint, 1 pure rolling joint and 0 half joints. Kutzbach's mobility equation (2.1c) Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2 3.

M1

Use equation 2.1c to determine the DOF (mobility) of the equivalent mechanism. There are 4 links, 3 full pin joints, 1 full sliding joint, and 0 half joints. This is a fourbar slider-crank. Kutzbach's mobility equation (2.1c) Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2

M1


SOLUTIONS MANUAL 2-37-1


Describe the motion of the following rides, commonly found at an amusement park, as pure rotation, pure translation, or complex planar motion. a. A Ferris wheel b. A "bumper" car c. A drag racer ride d. A roller coaster whose foundation is laid out in a straight line e. A boat ride through a maze f. A pendulum ride g. A train ride

Solution:


a.

A Ferris wheel Pure rotation.

b.

A "bumper car" Complex planar motion.

c.

A drag racer ride Pure translation.

d.

A roller coaster whose foundation is laid out in a straight line Complex planar motion.

e.

A boat ride through a maze Complex planar motion.

f.

A pendulum ride Pure rotation.

g.

A train ride Complex planar motion.




Figure P2-1a is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.

Solution:


1.

Label the link numbers and joint letters for Figure P2-1a.

G B

3 C

5

2 A

4

1

1

6

F

E

D 1

a.

Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4 5 6

Link Order Ternary Ternary Binary Ternary Binary Binary

b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G

Joint Order 1 1 1 1 1 2 1

Half/Full Full Full Full Half Full Full Full




Figure P2-1b is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.

Solution:

See Figure P2-1b and Mathcad file P0239.

1.

Label the link numbers and joint letters for Figure P2-1b.

3 C 1 B

2 A 1

a.

Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3

Link Order Binary Binary Binary

b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C

Joint Order 1 1 1

Half/Full Full Half Full




Figure P2-1c is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.

Solution:

See Figure P2-1c and Mathcad file P0240.

1.

Label the link numbers and joint letters for Figure P2-1c. 4 1 C

D

3

2 B

A 1

a.

Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4

Link Order Binary Binary Binary Binary

b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D

Joint Order 1 1 1 1

Half/Full Full Full Full Full




Figure P2-1d is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.

Solution:


1.

Label the link numbers and joint letters for Figure P2-1d. H 1

7 G

6

E 5

F

A 1

D

2 3 B

a.

4 C

Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4 5 6 7

Link Order Binary Binary Ternary Binary Binary Ternary Binary

b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H

Joint Order 1 1 1 1 1 1 1 1

Half/Full Full Full Half Full Full Full Full Full




Find the mobility, Grashof condition and Barker classification of the oil field pump shown in Figure P2-18.

Solution:


1.



4 O4 3

O2

2

This is a basic fourbar linkage. The input is the 14-in-long crank (link 2) and the output is the top beam (link 4). The mobility (DOF) is found using equation 2.1c (Kutzbach's modification): Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2

M1

The link lengths and Grashof condition are L1 

2

( 76  12)  47.5

2

L1  79.701

Condition L1 L2 L3 L4  "Grashof" This is a Barker Type 2 GCRR (Grashof, shortest link is input).

L2  14

L3  80

L4  51.26




Find the mobility, Grashof condition and Barker classification of the aircraft overhead bin shown in Figure P2-19.

Solution:


1.



2.79 O2

6.95

B

2

9.17

9.17 4 3 O4

9.57

A

9.17

This is a basic fourbar linkage. The input is the link 2 and the output is link 4. The mobility (DOF) is found using equation 2.1c (Kutzbach's modification): Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2

The link lengths and Grashof condition are 2

2

L1  7.489

L2  9.17

2

2

L3  12.968

L4  9.57

L1 

2.79  6.95

L3 

9.17  9.17

Condition L1 L2 L3 L4  "non-Grashof" This is a Barker Type 7 RRR3 (non-Grashof, longest link is coupler).

M1




Figure P2-20 shows a "Rube Goldberg" mechanism that turns a light switch on when a room door is opened and off when the door is closed. The pivot at O2 goes through the wall. There are two spring-loaded piston-in cylinder devices in the assembly. An arrangement of ropes and pulleys inside the room transfers the door swing into a rotation of link 2. Door opening rotates link 2 CW, pushing the switch up as shown in the figure, and door closing rotates link 2 CCW, pulling the switch down. Find the mobility of the linkage.

Solution:


1.

2.

Examination of the figure shows 20 links (including the the switch) and 28 full joints. The second piston-in cylinder that actuates the switch is counted as a single binary link of variable length with joints at its ends. The other cylinder consists of two binary links, each link having one pin joint and one slider joint. There are no half joints. Use equation 2.1c to determine the DOF (mobility). Kutzbach's mobility equation (2.1c) Number of links

L  20


J1  28


J2  0

M  3  ( L  1 )  2  J1  J2 3.

M1

An alternative is to ignore the the first piston-in cylinder that acts on the third bellcrank from O2 since it does not affect the the motion of the linkage (it acts only as a damper.) In that case, subtract two links and three full joints, giving L = 18, J1 = 25 and M = 1.




All of the eightbar linkages in Figure 2-11 part 2 have eight possible inversions. Some of these will give motions similar to others. Those that have distinct motions are called distinct inversions. How many distinct inversions does the linkage in row 4, column 1 have?

Solution:

See Figure 2-11, part 2 and Mathcad file P0245.

1.

This isomer has one quaternary, two ternary, and five binary links arranged in a symetrical fashion. Due to this symmetry, grounding link 2 or 7 gives the same inversion, as do grounding 3 or 6 and 4 or 5. This makes 3 of the possible 8 inversions the same leaving 5 distinct inversions. Distinct inversions are obtained by grounding link 1, 2, 3, 4, or 8 (or 1, 5, 6, 7, or 8) for a total of 5 distinct inversions.





Solution:


1.

This isomer has four ternary, and four binary links arranged in a symetrical fashion. Due to this symmetry, grounding link 1 or 5 gives the same inversion, as do grounding 2 or 8, 4 or 6, and 3 or 7. This makes 4 of the possible 8 inversions the same leaving 4 distinct inversions. Distinct inversions are obtained by grounding link 1, 2, 3, or 4 (or 5, 6, 7, or 8) for a total of 4 distinct inversions.





Solution:


1.

This isomer has four ternary, and four binary links arranged in a symetrical fashion. Due to this symmetry, grounding link 2 or 4 gives the same inversion, as does grounding 5 or 7. This makes 2 of the possible 8 inversions the same leaving 6 distinct inversions. Distinct inversions are obtained by grounding link 1, 2, 3, 5, 6 or 8 (or 1, 3, 4, 6, 7, or 8) for a total of 6 distinct inversions.




Find the mobility of the mechanism shown in Figure 3-33.

Solution:

See Figure 3-33 and Mathcad file P0248.

1.

Use equation 2.1c to determine the DOF (mobility). There are 6 links, 7 full pin joints (two at B), and no half-joints.

Kutzbach's mobility equation (2.1c) Number of links

L  6


J1  7


J2  0

M  3  ( L  1 )  2  J1  J2

M1





Solution:


1.

Use equation 2.1c to determine the DOF (mobility). There are 6 links, 7 full pin joints, and no half-joints. Kutzbach's mobility equation (2.1c) Number of links

L  6


J1  7


J2  0

M  3  ( L  1 )  2  J1  J2

M1





Solution:


1.

Use equation 2.1c to determine the DOF (mobility). There are 6 links, 7 full pin joints, and no half-joints. Kutzbach's mobility equation (2.1c) Number of links

L  6


J1  7


J2  0

M  3  ( L  1 )  2  J1  J2

M1





Solution:


1.

Use equation 2.1c to determine the DOF (mobility). There are 8 links, 10 full pin joints (two at O4), and no half-joints. Kutzbach's mobility equation (2.1c) Number of links

L  8


J1  10


J2  0

M  3  ( L  1 )  2  J1  J2

M1





Solution:


1.

Use equation 2.1c to determine the DOF (mobility). There are 6 links, 7 full pin joints (two at O4), and no half-joints. Kutzbach's mobility equation (2.1c) Number of links

L  6


J1  7


J2  0

M  3  ( L  1 )  2  J1  J2

M1




Figure P2-1e is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.

Solution:

See Figure P2-1e and Mathcad file P0253.

1.

Label the link numbers and joint letters for Figure P2-1e. J

K

8

I

8 1

9

1

L 10

M

7

a.

4 D C

A 1

2

2 3

E

H

B 5

F


Link Order 5 nodes Quaternary Binary Binary Binary Binary

Link No. 7 8 9 10

6

G 1

b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H I J K L M

Joint Order 1 1 1 1 1 1 1 1 1 1 1 1 1

Half/Full Full Full Full Full Full Full Full Full Full Full Full Full Full

Joint Classification Grounded rotating joint Moving rotating joint Pure rolling joint Grounded rotating joint Moving rotating joint Moving translating joint Grounded rotating joint Moving rotating joint Moving rotating joint Grounded rotating joint Moving translating joint Moving rotating joint Grounded translating joint

Link Order Binary Ternary Binary Binary




Figure P2-1f is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.

Solution:

See Figure P2-1f and Mathcad file P0254.

1.

Label the link numbers and joint letters for Figure P2-1f.

F 5

E

5 1 6

1

G

H

4

a.

C 3

1

B A 1

2


Link Order Quaternary Binary Ternary Binary Ternary Binary

b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H

Joint Order 1 1 1 1 1 1 1 1

Half/Full Full Full Full Full Full Full Full Full

Joint Classification Grounded rotating joint Moving half joint Grounded translating joint Moving rotating joint Moving rotating joint Grounded rotating joint Moving half joint Grounded translating joint




Figure P2-1g is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.

Solution:

See Figure P2-1g and Mathcad file P0255.

1.

Label the link numbers and joint letters for Figure P2-1g.

I

D 4 E

5

1

4 C 2

1

a.

B

G

A

A

1

F

7

6

3

H

7 1 J

2 1

8 1

K

Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4

Link Order 5 nodes Binary Binary Ternary

Link No. 5 6 7 8

Link Order Binary Binary Ternary Binary

b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H I J K

Joint Order 1 1 1 1 1 1 1 1 1 1 1

Half/Full Full Full Full Full Half Full Full Full Full Half Full

Joint Classification Grounded rotating joint Moving rolling joint Moving rotating joint Grounded rotating joint Moving sliding joint Grounded translating joint Moving rolling joint Moving rotating joint Grounded rotating joint Moving sliding joint Grounded translating joint




For the example linkage shown in Figure 2-4 find the number of links and their respective link orders, the number of joints and their respective orders, and the mobility of the linkage.

Solution:


1.

Label the link numbers and joint letters for Figure 2-4 example.

K

1

9 8

I G

J

6

7 H

D 1

4

E

3 B A 1

C 2

1

5 F 1

2.

Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4 5

3.

Link No. 6 7 8 9

Link Order Ternary Binary Binary Binary

Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H I J K

4.

Link Order 5 nodes Binary Ternary Binary Binary

Joint Order 1 1 1 2 1 1 1 1 1 1 1

Half/Full Full Half Full Full Full Full Full Full Full Full Full

Joint Classification Grounded rotating joint Moving sliding joint Grounded rotating joint Moving rotating joint Moving translating joint Grounded rotating joint Moving rotating joint Grounded rotating joint Moving rotating joint Moving rotating joint Grounded translating joint

Use equation 2.1c to calculate the DOF (mobility). Kutzbach's mobility equation (2.1c) Number of links

L  9

M  3  ( L  1 )  2  J1  J2

Number of full joints M1

J1  11


J2  1




For the linkage shown in Figure 2-5b find the number of joints and their respective orders, and mobility for: a) The condition of a finite load W in the direction shown and a zero F b) The condition of a finite load W and a finite load F both in the directions shown after link 6 is off the stop.

Solution:

See Figure 2-5b and Mathcad file P0257.

1.

Label the link numbers and joint letters for Figure 2-5b. 1

6

O6

W

D 3

A

B

1

F 5

4 O4

2 O2

1

C

1

a)

The condition of a finite load W in the direction shown and a zero F: Using the joint letters, determine each joint's order and whether each is a half or full joint. Link 6 is grounded so joint D is a grounded rotating joint and O6 is not a joint. For this condition there is a total of 6 full joints and no half joints. Joint Letter O2 B C D O4 A

Joint Order 1 1 1 1 1 1

Half/Full Full Full Full Full Full Full

Joint Classification Grounded rotating joint Moving rotating joint Moving rotating joint Grounded rotating joint Grounded rotating joint Moving rotating joint

Use equation 2.1c to calculate the DOF (mobility). Kutzbach's mobility equation (2.1c) Number of links L  5


M  3  ( L  1 )  2  J1  J2

J1  6


J2  0

M0

b) The condition of a finite load W and a finite load F both in the directions shown after link 6 is off the stop. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D O2 O4 O6

Joint Order 1 1 1 1 1 1 1

Half/Full Full Full Full Full Full Full Full

Joint Classification Moving rotating joint Moving rotating joint Moving rotating joint Moving rotating joint Grounded rotating joint Grounded rotating joint Grounded rotating joint

Use equation 2.1c to calculate the DOF (mobility). Kutzbach's mobility equation (2.1c) Number of links L  6 M  3  ( L  1 )  2  J1  J2

Number of full joints M1

J1  7


J2  0




Figure P2-21a shows a "Nuremberg scissors" mechanism. Find its mobility.

Solution:


1.


L  10


J1  13


J2  0

M  3  ( L  1 )  2  J1  J2

M1




Figure P2-21b shows a mechanism. Find its mobility and classify its isomer type.

Solution:


1.


L  6


J1  7


J2  0

M  3  ( L  1 )  2  J1  J2

2.

M1

Using Figure 2-9, we see that the mechanism is a Stephenson's sixbar isomer ( the two ternary links are connected with two binary links and one dyad).




Figure P2-21c shows a circular saw mounted on the coupler of a fourbar linkage. The centerline of the saw blade is at a coupler point that moves in an approximate straight line. Draw its kinematic diagram and determine its mobility.

Solution:


1.

Draw a kinematic diagram of the mechanism. The saw's rotation axis is at point P and the saw is attached to link 3.

A

B 3 4

2 O2

O4

P

1

1 2.


L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2

M1




Figure P2-21d shows a log transporter. Draw a kinematic diagram of the mechanism, specify the number of links and joints, and then determine its mobility: a) For the transporter wheels locked and no log in the "claw" of the mechanism b) For the transporter wheels locked with it lifting a log c) For the transporter moving a log to a destination in a straight line.

Solution:


1.

Draw a kinematic diagram of the mechanism. Link 1 is the frame of the transporter. Joint B is of order 3. Actuators E and F provide two inputs (to get x-y motion) and actuator H provides an additional input for clamping logs. G

D

9 12 H

9

C

I 9

11

8

J 10

4 A

3

2 O2

B

5

7

E O4

1

F

1

6 O6 1

a)

Wheels locked, no log in "claw." Kutzbach's mobility equation (2.1c) Number of links

L  12


J1  15


J2  0

M  3  ( L  1 )  2  J1  J2

M3

b) Wheels locked, log held tighly in the "claw." With a log held tightly between links 9 and 10 a structure will be formed by links 9 through 12 and the log so that there will only be 9 links and 11 joints active. Number of links

L  9


J1  11


J2  0

M  3  ( L  1 )  2  J1  J2 c)

M2

Transporter moving in a straight line with the log holding mechanism inactive. There are two tires, the transporter frame, and the ground, making 4 links and two points of contact with the ground and two axels, making 4 joints.



Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2

M1




Figure P2-21d shows a plow mechanism attached to a tractor. Draw its kinematic diagram and find its mobility including the earth as a "link." a) When the tractor is stopped and the turnbuckle is fixed. (Hint: Consider the tractor and the wheel to be one with the earth.) b) When the tractor is stopped and the turnbuckle is being adjusted. (Same hint.) c) When the tractor is moving and the turnbuckle is fixed. (Hint: Add the moving tractor's DOF to those found in part a.)

Solution:


1.

Draw a kinematic diagram of the mechanism with the ground, tractor wheels, and tractor frame as link 1. Joint O4 is of order 2 and joint F is a half joint. The plow and its truss structure attach at joints D and E. Since the turnbuckle is fixed it can be modeled as a single binary link (6).

C

6

5 O4 1

D

B

4

7 7

3 A 2

7

2 O2

E 1

F

7 1

a)

Tractor stopped and turnbuckle fixed. Kutzbach's mobility equation (2.1c) Number of links

L  7


J1  8


J2  1

M  3  ( L  1 )  2  J1  J2

M1

b) When the tractor is stopped and the turnbuckle is being adjusted. Between joints C and D we now have 2 net links (2 links threaded LH and RH on one end and the turnbuckle body) and 1 additional helical full joint. Number of links L  8 Number of full joints

J1  9


J2  1

M  3  ( L  1 )  2  J1  J2 c)

M2

When the tractor is moving and the turnbuckle is fixed. If the tractor moved only in a straight line we would add 1 DOF to the 1 DOF that we got in part a for a total of M = 2. More realistically, the tractor can turn and move up and down hills so that we would add 3 DOF to the 1 DOF of part a to get a total of 4 DOF.




Figure P2-22 shows a Hart's inversor sixbar linkage. a) Is it a Watt or Stephenson linkage? b) Determine its inversion, i.e. is it a type I, II, or III?

Solution:

See Figure P2-22, Figure 2-14, and Mathcad file P0263.

1.

From Figure 2-14 we see that the Watt's sixbar has the two ternary links connected with a common joint while the Stephenson's sixbar has the two ternary links connected by binary links. Thus, Hart's inversor is a Watt's sixbar (links 1 and 2, the ternary links, are connected at a common joint). Further, the Hart's linkage is a Watt's sixbar inversion I since neither of the ternary links is grounded.




Figure P2-23 shows the top view of the partially open doors on one side of an entertainment center cabinet. The wooden doors are hinged to each other and one door is hinged to the cabinet. There is also a ternary, metal link attached to the cabinet and door through pin joints. A spring-loaded piston-in cylinder device attaches to the ternary link and the cabinet through pin joints. Draw a kinematic diagram of the door system and find the mobility of this mechanism.

Solution:


1.

Draw the kinematic diagram of this sixbar mechanism. The spring-loaded piston is just an in-line sliding joint (links 5 and 6, and joint F). The doors are binary links (3 and 4), and the metal ternary link (2) has nodes at A, B, and C. Link 1 is the cabinet.

A Cabinet

1

2

4

E Cabinet

B

D

Door

1

5 Cylinder F 6

Door

2

3

Link

G Cabinet 1 C

2.

Use equation 2.1c (Kutzbach's modification) to calculate the mobility. Number of links

L  6


J1  7


J2  0 M  3  ( L  1 )  2  J1  J2

M1




Figure P2-24a shows the seat and seat-back of a reclining chair with the linkage that connects them to the chair frame. Draw its kinematic diagram and determine its mobility with respect to the frame of the chair.

Solution:


1.

Draw a kinematic diagram of the mechanism. The chair-back attaches to link 2 and the seat with the attached slider slot is link 3. The node at at C is a half-joint as it allows two degrees of freedom.

A 1 2

3

B

C 2.

Determine the mobility of the mechanism. Kutzbach's mobility equation (2.1c) Number of links

L  3


J1  2


J2  1

M  3  ( L  1 )  2  J1  J2

M1




Figure P2-24b shows the mechanism used to extend the foot support on a reclining chair. Draw its kinematic diagram and determine its mobility with respect to the frame of the chair.

Solution:


1.

Draw a kinematic diagram of the mechanism. Link 1 is the frame.

D E 1

J

4 6

3

C 4

5

G

5 6

A 3 1

7

F

H

2

B 2.

Determine the mobility of the mechanism. Kutzbach's mobility equation (2.1c) Number of links

L  8


J1  10


J2  0

M  3  ( L  1 )  2  J1  J2

8

M1

K




Figure P2-24b shows the mechanism used to extend the foot support on a reclining chair. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.

Solution:


1.

Label the link numbers and joint letters for Figure P2-24b.

D E 1

J

4 6

3

C 4

5

G

5 6

A 3 1

F

2

8

7

H

B a.

Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4 5 6 7 8

Link Order Binary Binary Ternary Ternary Ternary Ternary Binary Binary

b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H J K

Joint Order 1 1 1 1 1 1 1 1 1 1

Half/Full Full Full Full Half Full Full Full Full Full Full

K




Figure P2-24 shows a sixbar linkage. a) Is it a Watt or Stephenson linkage? b) Determine its inversion, i.e. is it a type I, II, or III?

Solution:

See Figure P2-24, Figure 2-14, and Mathcad file P0268.

1.

From Figure 2-14 we see that the Watt's sixbar has the two ternary links connected with a common joint while the Stephenson's sixbar has the two ternary links connected by binary links. Thus, the sixbar linkage shown is a Watt's sixbar (links 3 and 4, the ternary links, are connected at a common joint). Further, the linkage shown is a Watt's sixbar inversion I since neither of the ternary links is grounded.




Define the following examples as path, motion, or function generation cases. a. b. c. d. e.

Solution:

A telescope aiming (star tracking) mechanism A backhoe bucket control mechanism A thermostat adjusting mechanism A computer printing head moving mechanism An XY plotter pen control mechanism


a.

Path generation. A star follows a 2D path in the sky.

b.

Motion generation. To dig a trench, say, the position and orientation of the bucket must be controlled.

c.

Function generation. The output is some desired function of the input over some range of the input.

d.

Path generation. The head must be at some point on a path.

e.

Path generation. The pen follows a straight line from point to point.




Design a fourbar Grashof crank-rocker for 90 deg of output rocker motion with no quick return. (See Example 3-1.) Build a cardboard model and determine the toggle positions and the minimum transmission angle.

Given:

Output angle

Solution:

See Example 3-1 and Mathcad file P0302.

Design choices: 1.

θ  90 deg

Link lengths:

L3  6.000

Link 3

L4  2.500

Link 4

2.

Draw the output link O4B in both extreme positions, B1 and B2, in any convenient location such that the desired angle of motion 4 is subtended. In this solution, link 4 is drawn such that the two extreme positions each make an angle of 45 deg to the vertical. Draw the chord B1B2 and extend it in any convenient direction. In this solution it was extended to the left.

3.

Layout the distance A1B1 along extended line B1B2 equal to the length of link 3. Mark the point A1.

4.

Bisect the line segment B1B2 and layout the length of that radius from point A1 along extended line B1B2. Mark the resulting point O2 and draw a circle of radius O2A1 with center at O2.

5.

Label the other intersection of the circle and extended line B1B2, A2.

6.

Measure the length of the crank (link 2) as O2A1 or O2A2. From the graphical solution, L2  1.76775

7.

Measure the length of the ground link (link 1) as O2O4. From the graphical solution, L1  6.2550 1.7677

6.0000

3.5355

2

A2

A1

3

B2

B1

O2 90.00° 1 4 6.2550

8.

O4

Find the Grashof condition. Condition( a b c d ) 


Condition L1 L2 L3 L4  "Grashof"




Design a fourbar mechanism to give the two positions shown in Figure P3-1 of output rocker motion with no quick-return. (See Example 3-2.) Build a cardboard model and determine the toggle positions and the minimum transmission angle.

Given:

Coordinates of A1, B1, A2, and B2 (with respect to A1):

Solution:

xA1  0.00

xB1  1.721

xA2  2.656

xB2  5.065

yA1  0.00

yB1  1.750

yA2  0.751

yB2  0.281


Design choices:

Link length:

Link 3

L3  5.000

Link 4

L4  2.000

1.

Following the notation used in Example 3-2 and Figure 3-5, change the labels on points A and B in Figure P3-1 to C and D, respectively. Draw the link CD in its two desired positions, C1D1 and C2D2, using the given coordinates.

2.

Draw construction lines from C1 to C2 and D1 to D2.

3.

Bisect line C1C2 and line D1D2 and extend their perpendicular bisectors to intersect at O4.

4.

Using the length of link 4 (design choice) as a radius, draw an arc about O4 to intersect both lines O4C1 and O4C2. Label the intersections B1 and B2.

5.

Draw the chord B1B2 and extend it in any convenient direction. In this solution it was extended to the left.

6.

Layout the distance A1B1 along extended line B1B2 equal to the length of link 3. Mark the point A1.

7.

Bisect the line segment B1B2 and layout the length of that radius from point A1 along extended line B1B2. Mark the resulting point O2 and draw a circle of radius O2A1 with center at O2.

8.

Label the other intersection of the circle and extended line B1B2, A2.

9.

Measure the length of the crank (link 2) as O2A1 or O2A2. From the graphical solution, L2  0.9469

10. Measure the length of the ground link (link 1) as O2O4. From the graphical solution, L1  5.3013

5.3013 5.0000 0.9469 A1 2

O2

O4

1

A2

3

B1

C1

4

R2.000

B2 C2 D1

D2



11. Find the Grashof condition. Condition( a b c d ) 






Design a fourbar mechanism to give the two positions shown in Figure P3-1 of coupler motion. (See Example 3-3.) Build a cardboard model and determine the toggle positions and the minimum transmission angle. Add a driver dyad. (See Example 3-4.)

Given:

Position 1 offsets:

Solution:

See figure below for one possible solution. Input file P0304.mcd from the solutions manual disk to the Mathcad program for this solution, file P03-04.4br to the program FOURBAR to see the fourbar solution linkage, and file P03-04.6br into program SIXBAR to see the complete sixbar with the driver dyad included.

xA1B1  1.721  in

yA1B1  1.750  in

1.

Connect the end points of the two given positions of the line AB with construction lines, i.e., lines from A1 to A2 and B1 to B2.

2.

Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of A1A2 was extended downward and the bisector of B1B2 was extended upward.

3.

Select one point on each bisector and label them O4 and O6, respectively. In the solution below the distances O4A and O6B were each selected to be 4.000 in. This resulted in a ground-link-length O4O6 for the fourbar of 6.457 in.

4.

The fourbar stage is now defined as O4ABO6 with link lengths Link 5 (coupler) L5 

2

xA1B1  yA1B1

Link 4 (input)

L4  4.000  in

Ground link 1b

L1b  6.457  in

2

L5  2.454 in Link 6 (output)

L6  4.000  in

5.

Select a point on link 4 (O4A) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it D. (Note that link 4 is now a ternary link with nodes at O4, D, and A.) In the solution below the distance O4D was selected to be 2.000 in.

6.

Draw a construction line through D1D2 and extend it to the left.

7.

Select a point on this line and call it O2. In the solution below the distance CD was selected to be 4.000 in.

8.

Draw a circle about O2 with a radius of one-half the length D1D2 and label the intersections of the circle with the extended line as C1 and C2. In the solution below the radius was measured as 0.6895 in.

9.

The driver fourbar is now defined as O2CDO4 with link lengths Link 2 (crank)

L2  0.6895 in

Link 4a (rocker) L4a  2.000  in

Link 3 (coupler) L3  4.000  in Link 1a (ground) L1a  4.418  in

10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Shortest link

S  L2

S  0.6895 in

Longest link

L  L1a

L  4.4180 in

Other links

P  L3

P  4.0000 in

Q  L4a

Q  2.0000 in


Condition( a b c d ) 



Condition( S L P Q)  "Grashof" O6

6

Ground Link 1b

A1

6 50.231°

5 C1

A2

5

B2

47.893°

2 O2 3

4

C2

B1 D1

4

3 D2

Ground Link 1a

O4

11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4ABO6 is non-Grashoff with toggle positions at 2 = -71.9 deg and +71.9 deg. The minimum transmission angle is 35.5 deg. The fourbar operates between 2 = +21.106 deg and -19.297 deg.




Design a fourbar mechanism to give the three positions of coupler motion with no quick return shown in Figure P3-2. (See also Example 3-5.) Ignore the points O2 and O4 shown. Build a cardboard model and determine the toggle positions and the minimum transmission angle. Add a driver dyad.

Solution:


Design choices: L5  4.250

Length of link 5:

L4b  1.375

Length of link 4b:

1.

Draw link CD in its three design positions C1D1, C2D2, C3D3 in the plane as shown.

2.

Draw construction lines from point C1 to C2 and from point C2 to C3.

3.

Bisect line C1C2 and line C2C3 and extend their perpendicular bisectors until they intersect. Label their intersection O2.

4.

Repeat steps 2 and 3 for lines D1D2 and D2D3. Label the intersection O4.

5.

Connect O2 with C1 and call it link 2. Connect O4 with D1 and call it link 4.

6.

Line C1D1 is link 3. Line O2O4 is link 1 (ground link for the fourbar). The fourbar is now defined as O2CDO4 and has link lengths of Ground link 1a

L1a  0.718

Link 2

L2  2.197

Link 3

L3  2.496

Link 4

L4  3.704 1.230

O6 0.718 1b

2.197 O4 O2

2

C1

6 A

5

1a

4.328 B

2.496

D3

C3

3 4 C2 D1

D2 3.704

7.

Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case.





Condition L1a L2 L3 L4  "Grashof" 8.

9.

Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution above the distance O4B was selected to be L4b  1.375 . Draw a construction line through B1B3 and extend it up to the right.

10. Layout the length of link 5 (design choice) along the extended line. Label the other end A. 11. Draw a circle about O6 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L6  1.230. 12. The driver fourbar is now defined as O4BAO6 with link lengths Link 6 (crank)

L6  1.230

Link 5 (coupler) L5  4.250 Link 1b (ground) L1b  4.328 Link 4b (rocker) L4b  1.375 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L4b L5  "Grashof"




Design a fourbar mechanism to give the three positions shown in Figure P3-2 using the fixed pivots O2 and O4 shown. Build a cardboard model and determine the toggle positions and the minimum transmission angle. Add a driver dyad.

Solution:


Design choices: Length of link 5:

L5  5.000

L2b  2.000

Length of link 2b:

1.


2.

Draw the ground link O2O4 in its desired position in the plane with respect to the first coupler position C1D1.

3.

Draw construction arcs from point C2 to O2 and from point D2 to O2 whose radii define the sides of triangle C2O2D2. This defines the relationship of the fixed pivot O2 to the coupler line CD in the second coupler position.

4.


5.

Transfer this relationship back to the first coupler position C1D1 so that the ground plane position O2'O4' bears the same relationship to C1D1 as O2O4 bore to the second coupler position C2D2.

6.

Repeat the process for the third coupler position and transfer the third relative ground link position to the first, or reference, position.

7.

The three inverted positions of the ground link that correspond to the three desired coupler positions are labeled O2O4, O2'O4', and O2"O4" in the first layout below and are renamed E1F1, E2F2, and E3F3, respectively, in the second layout, which is used to find the points G and H.

O2''

C1

D3

O2'

C3 C2

O4'' D1

O4'

8.

O2

Draw construction lines from point E1 to E2 and from point E2 to E3.

D2

O4


9.


Bisect line E1E2 and line E2E3 and extend their perpendicular bisectors until they intersect. Label their intersection G.

10. Repeat steps 2 and 3 for lines F1F2 and F2F3. Label the intersection H. 11. Connect E1 with G and label it link 2. Connect F1 with H and label it link 4. Reinverting, E1 and F1 are the original fixed pivots O2 and O4, respectively. 12. Line GH is link 3. Line O2O4 is link 1a (ground link for the fourbar). The fourbar is now defined as O2GHO4 and has link lengths of Ground link 1a

L1a  4.303

Link 2

L2  8.597

Link 3

L3  1.711

Link 4

L4  7.921

E3

G

3

H

E2 F3

4

2

F2 E1

1a O2

F1 O4

13. Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case. Condition( a b c d ) 


Condition L1a L2 L3 L4  "Grashof" The fourbar that will provide the desired motion is now defined as a Grashof double crank in the crossed configuration. It now remains to add the original points C1 and D1 to the coupler GH and to define the driving dyad.



14. Select a point on link 2 (O2G) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 2 is now a ternary link with nodes at O2, B, and G.) In the solution below, the distance O2B was selected to be L2b  2.000 . 15. Draw a construction line through B1B3 and extend it up to the right. 16. Layout the length of link 5 (design choice) along the extended line. Label the other end A. 17. Draw a circle about O6 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L6  0.412. 18. The driver fourbar is now defined as O2BAO6 with link lengths Link 6 (crank)

L6  0.412

Link 5 (coupler) L5  5.000 Link 1b (ground) L1b  5.369 Link 2b (rocker) L2b  2.000 19. Use the link lengths in step 18 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L2b L5  "Grashof"

G2

G3

H1

G1 3

H2 H3

2

C1

A3 O6

6

D3

C3

A1 5

C2

D1 B3

O2

D2 B1

4

1a

O4




Given: Solution: 1.

Repeat Problem 3-2 with a quick-return time ratio of 1:1.4. (See Example 3.9). Design a fourbar Grashof crank-rocker for 90 degrees of output rocker motion with a quick-return time ratio of 1:1.4. 1 Time ratio Tr  1.4 See figure below for one possible solution. Also see Mathcad file P0307.

Determine the crank rotation angles  and , and the construction angle  from equations 3.1 and 3.2. Tr = Solving for , and 

β 

α

α  β = 360  deg

β 360  deg

β  210 deg

1  Tr

α  360  deg  β

α  150 deg

δ  β  180  deg

δ  30 deg

2.

Start the layout by arbitrarily establishing the point O4 and from it layoff two lines of equal length, 90 deg apart. Label one B1 and the other B2. In the solution below, each line makes an angle of 45 deg with the horizontal and has a length of 2.000 in.

3.

Layoff a line through B1 at an arbitrary angle (but not zero deg). In the solution below, the line is 30 deg to the horizontal.

4.

Layoff a line through B2 that makes an angle  with the line in step 3 (60 deg to the horizontal in this case). The intersection of these two lines establishes the point O2.

5.

From O2 draw an arc that goes through B1. Extend O2B2 to meet this arc. Erect a perpendicular bisector to the extended portion of the line and transfer one half of the line to O2 as the length of the input crank.

3.8637 = b

90.0000°

B2 B1

B2

2.0000 = c

1.0353 = a

LAYOUT

B1

4 3 O4

O4 A1 2 O2

O2

3.0119 = d

A2 LINKAGE DEFINITION 


6.


For this solution, the link lengths are: Ground link (1)

d  3.0119 in

Crank (2)

a  1.0353 in

Coupler (3)

b  3.8637 in

Rocker (4)

c  2.000  in




Design a sixbar drag link quick-return linkage for a time ratio of 1:2, and output rocker motion of 60 degrees. (See Example 3-10.)

Given:

Time ratio

Solution: 1.

Tr 

1 2

See figure below for one possible solution. Also see Mathcad file P0308.

Determine the crank rotation angles  and  from equation 3.1. Tr = Solving for and 

β 

α

α  β = 360  deg

β 360  deg 1  Tr

α  360  deg  β

β  240 deg α  120 deg

2.

Draw a line of centers XX at any convenient location.

3.

Choose a crank pivot location O2 on line XX and draw an axis YY perpendicular to XX through O2.

4.

Draw a circle of convenient radius O2A about center O2. In the solution below, the length of O2A is a  1.000  in.

5.

Lay out angle  with vertex at O2, symmetrical about quadrant one.

6.

Label points A1 and A2 at the intersections of the lines subtending angle  and the circle of radius O2A.

7.

8.

Set the compass to a convenient radius AC long enough to cut XX in two places on either side of O2 when swung from both A1 and A2. Label the intersections C1 and C2. In the solution below, the length of AC is b  1.800  in. The line O2A is the driver crank, link 2, and the line AC is the coupler, link 3.

9.

The distance C1C2 is twice the driven (dragged) crank length. Bisect it to locate the fixed pivot O4.

10. The line O2O4 now defines the ground link. Line O4C is the driven crank, link 4. In the solution below, O4C measures c  2.262  in and O2O4 measures d  0.484  in. 11. Calculate the Grashoff condition. If non-Grashoff, repeat steps 7 through 11 with a shorter radius in step 7. Condition( a b c d ) 


Condition( a b c d )  "Grashof" 12. Invert the method of Example 3-1 to create the output dyad using XX as the chord and O4C1 as the driving crank. The points B1 and B2 will lie on line XX and be spaced apart a distance that is twice the length of O4C (link 4). The pivot point O6 will lie on the perpendicular bisector of B1B2 at a distance from XX which subtends the specified output rocker angle, which is 60 degrees in this problem. In the solution below, the length BC was chosen to be e  5.250  in.



LAYOUT OF SIXBAR DRAG LINK QUICK RETURN WITH TIME RATIO OF 1:2 a = 1.000 b = 1.800 c = 2.262 d = 0.484 e = 5.250 f = 4.524

13. For the design choices made (lengths of links 2, 3 and 5), the length of the output rocker (link 6) was measured as f  4.524  in.




Design a crank-shaper quick-return mechanism for a time ratio of 1:3 (Figure 3-14, p. 112).

Given:

Time ratio

Solution:


TR 

1 3

Design choices:

1.

Length of link 2 (crank)

L2  1.000

Length of link 5 (coupler)

L5  5.000

S  4.000

Length of stroke

Calculate  from equations 3.1. TR 

α β

α  β  360  deg

α 

360  deg 1

α  90.000 deg

1 TR

2.

Draw a vertical line and mark the center of rotation of the crank, O2, on it.

3.

Layout two construction lines from O2, each making an angle /2 to the vertical line through O2.

4.

Using the chosen crank length (see Design Choices), draw a circle with center at O2 and radius equal to the crank length. Label the intersections of the circle and the two lines drawn in step 3 as A1 and A2.

5.

Draw lines through points A1 and A2 that are also tangent to the crank circle (step 2). These two lines will simultaneously intersect the vertical line drawn in step 2. Label the point of intersection as the fixed pivot center O4.

6.

Draw a vertical construction line, parallel and to the right of O2O4, a distance S/2 (one-half of the output stroke length) from the line O2O4.

7.

Extend line O4A1 until it intersects the construction line drawn in step 6. Label the intersection B1.

8.

Draw a horizontal construction line from point B1, either to the left or right. Using point B1 as center, draw an arc of radius equal to the length of link 5 (see Design Choices) to intersect the horizontal construction line. Label the intersection as C1.

9.

Draw the slider blocks at points A1 and C1 and finish by drawing the mechanism in its other extreme position.

STROKE 4.000 C2

2.000

6

C1

B2

B1

5 O2 4

2 A2

3 O4

A1




Find the two cognates of the linkage in Figure 3-17 (p. 116). Draw the Cayley and Roberts diagrams. Check your results with program FOURBAR.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  2

Crank

L2  1

A1P  1.800

δ  34.000 deg

Coupler

L3  3

Rocker

L4  3.5

B1P  1.813

γ  33.727 deg


Draw the original fourbar linkage, which will be cognate #1, and align links 2 and 4 with the coupler. A1

B1

3

2

A1

3

OA

P

B1

4

2 OA P 4 1

OB

OB

2.

Construct lines parallel to all sides of the aligned fourbar linkage to create the Cayley diagram (see Figure 3-24) OA

2

A1

B1

3

OB

4

10

5 A2

B3

P

9 B2

6 8 7

A3

OC A1

3.

4.

Return links 2 and 4 to their fixed pivots OA and OB and establish OC as a fixed pivot by making triangle OAOBOC similar to A1B1P. Separate the three cognates. Point P has the same path motion in each cognate.

2 4

OA

10

P

Calculate the cognate link lengths based on the geometry of the Cayley diagram (Figure 3-24c, p. 114). L5  B1P L6 

L4 L3

 B1P

9

8

6

A2 OC OB

L5  1.813 L6  2.115

B2

A3 7

5.

B1

3

5 B3



P

P

OA

B2

10 A3

9 OC

7

8 A2 OC

6 OB 5

Cognate #2

B3

Cognate #3

L10  A1P

L10  1.800

L7  L9

B1P

L8  L6

A1P

L9 

L2 L3

 A1P

L9  0.600

L7  0.604

A1P

L8  2.100

B1P

From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC  L1AC 

L1 L3 L1 L3

 B1P

L1BC  1.209

 A1P

L1AC  1.200

Calculate the coupler point data for cognates #2 and #3 A3P  L8

A3P  2.100





δ  180  deg  δ  γ

δ  247.727 deg

A2P  L2

A2P  1.000

δ  δ

δ  34.000 deg

SUMMARY OF COGNATE SPECIFICATIONS:

6.

Cognate #1

Cognate #2

Cognate #3

Ground link length

L1  2.000

L1AC  1.200

L1BC  1.209

Crank length

L2  1.000

L10  1.800

L7  0.604

Coupler length

L3  3.000

L9  0.600

L6  2.115

Rocker length

L4  3.500

L8  2.100

L5  1.813

Coupler point

A1P  1.800

A2P  1.000

A3P  2.100

Coupler angle

δ  34.000 deg

δ  34.000 deg

δ  247.727 deg

Verify that the three cognates yield the same coupler curve by entering the original link lengths in program FOURBAR and letting it calculate the cognates.





Note that cognate #2 is a Grashof double rocker and, therefore, cannot trace out the entire coupler curve.




Find the three equivalent geared fivebar linkages for the three fourbar cognates in Figure 3-25a (p. 125). Check your results by comparing the coupler curves with programs FOURBAR and FIVEBAR.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  39.5

Crank

L2  15.5

Coupler

L3  14.0

Rocker

L4  20.0

A1P  26.0


Calculate the length BP and the angle using the law of cosines on the triangle APB. B1P   L3  A1P  2  L3 A1P  cos δ 2

 

2

0.5

B1P  23.270

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

δ  63.000 deg

γ  84.5843 deg

Use the Cayley diagram (see Figure 3-24) to calculate the link lengths of the two cognates. Note that the diagram is made up of three parallelograms and three similar triangles L4

L5  B1P

L5  23.270

L6 

L10  A1P

L10  26.000

L9 

L7  25.763

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  33.243

 A1P

L9  28.786

A1P

L8  37.143

B1P


A3P  20.000

A2P  L2

A2P  15.500

δ  γ

δ  84.584 deg

δ  δ

δ  63.000 deg

From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC 

3.

L1 L3

 B1P

L1BC  65.6548

L1AC 

L1 L3

 A1P

L1AC  73.3571

Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1

Cognate #2

Cognate #3

Ground link length

L1  39.500

L1AC  73.357

L1BC  65.655

Crank length

L2  15.500

L10  26.000

L7  25.763

Coupler length

L3  14.000

L9  28.786

L6  33.243



Rocker length

L4  20.000

L8  37.143

L5  23.270

Coupler point

A1P  26.000

A2P  15.500

A3P  20.000

Coupler angle

δ  63.000 deg

δ  63.000 deg

δ  84.584 deg

OC 8 B2

7

B3 9

P

6 A2 A3 10 3

A1

5

B1 4

2 1

OA

OB

4.

The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAA2PA3OB, OAA1PB3OC, and OBB1PB2OC. They are shown individually below with their associated gears. P

A2 A3 10 5

OA OB

OC



OC

7

B3

OD P

A1 2 OA

OC 8 B2

P

OE

B1 4

OB



SUMMARY OF GEARED FIVEBAR COGNATE SPECIFICATIONS: Cognate #1

Cognate #2

Cognate #3

Ground link length

L1  39.500

L1AC  73.357

L1BC  65.655

Crank length

L10  26.000

L2  15.500

L4  20.000

Coupler length

A2P  15.500

A1P  26.000

L5  23.270

Rocker length

A3P  20.000

L8  37.143

L7  25.763

Crank length

L5  23.270

L7  25.763

L8  37.143

Coupler point

A2P  15.500

A1P  26.000

B1P  23.270

Coupler angle

δ  0.00 deg

δ  0.00 deg

δ  0.00 deg

5.

Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path.

6.

Enter the geared fivebar cognate #1 specifications into program FIVEBAR to get a trace of the coupler path for the geared fivebar (see next page).






Design a sixbar, single-dwell linkage for a dwell of 90 deg of crank motion, with an output rocker motion of 45 deg.

Given:

Crank dwell period: 90 deg. Output rocker motion: 45 deg.

Solution:

See Figures 3-20, 3-21, and Mathcad file P0312.

Design choices: Ground link ratio, L1/L2 = 2.0: GLR  2.0 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.5: CLR  2.5 Coupler angle, γ  72 deg Crank length, L2  2.000 1.

For the given design choices, determine the remaining link lengths and coupler point specification. Coupler link (3) length

L3  CLR L2

L3  5.000

Rocker link (4) length

L4  CLR L2

L4  5.000

Ground link (1) length

L1  GLR  L2

L1  4.000

Angle PAB

δ 

Length AP on coupler 2.

180  deg  γ 2

AP  2  L3 cos δ

δ  54.000 deg AP  5.878

Enter the above data into program FOURBAR, plot the coupler curve, and determine the coordinates of the coupler curve in the selected range of crank motion, which in this case will be from 135 to 225 deg..



FOURBAR for Windows Angle Step Deg 135 140 145 150 155 160 165 170 175 180 185 190 195 200 205 210 215 220 225 3.

File P03-12.DAT

Coupler Pt X

Coupler Pt Y

Coupler Pt Mag

-1.961 -2.178 -2.393 -2.603 -2.809 -3.008 -3.201 -3.386 -3.563 -3.731 -3.890 -4.038 -4.176 -4.302 -4.417 -4.520 -4.610 -4.688 -4.753

7.267 7.128 6.977 6.813 6.638 6.453 6.257 6.052 5.839 5.617 5.389 5.155 4.915 4.671 4.424 4.175 3.924 3.673 3.424

7.527 7.453 7.375 7.293 7.208 7.119 7.028 6.935 6.840 6.744 6.646 6.548 6.450 6.351 6.252 6.153 6.054 5.956 5.858

Coupler Pt Ang

105.099 106.992 108.930 110.911 112.933 114.994 117.093 119.228 121.396 123.595 125.822 128.075 130.351 132.646 134.955 137.274 139.598 141.921 144.235

Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Use the points at crank angles of 135, 180, and 225 deg to define the pseudo-arc. Find the center of the pseudo-arc erecting perpendicular bisectors to the chords defined by the selected coupler curve points. The center will lie at the intersection of the perpendicular bisectors, label this point D. The radius of this circle is the length of link 5.

y 135 P

PSEUDO-ARC

180 B

225 3 D A

4

2 x O2

O4


4.


The position of the end of link 5 at point D will remain nearly stationary while the crank moves from 135 to 225 deg. As the crank motion causes the coupler point to move around the coupler curve there will be another extreme position of the end of link 5 that was originally at D. Since a symmetrical linkage was chosen, the other extreme position will be located along a line through the axis of symmetry (see Figure 3-20) a distance equal to the length of link 5 measured from the point where the axis of symmetry intersects the coupler curve near the 0 deg coupler point. Establish this point and label it E. FOURBAR for Windows Angle Step Deg 300 310 320 330 340 350 0 10 20 30 40 50 60

File P03-12.DAT

Coupler Pt X

Coupler Pt Y

Coupler Pt Mag

-4.271 -4.054 -3.811 -3.526 -3.159 -2.651 -1.968 -1.181 -0.441 0.126 0.478 0.631 0.617

0.869 0.926 1.165 1.628 2.343 3.286 4.336 5.310 6.085 6.654 7.068 7.373 7.598

4.359 4.158 3.985 3.883 3.933 4.222 4.762 5.440 6.101 6.656 7.085 7.400 7.623

Coupler Pt Ang

168.495 167.133 162.998 155.215 143.437 128.892 114.414 102.534 94.142 88.914 86.129 85.111 85.354

y 135 P

PSEUDO-ARC

180 B

5

4

225 3

AXIS OF SYMMETRY

D A

E

2

x O2 5.

O4

The line segment DE represents the maximum displacement that a link of the length equal to link 5, attached at P, will reach along the axis of symmetry. Construct a perpendicular bisector of the line segment DE and extend it to the right (or left, which ever is convenient). Locate fixed pivot O6 on the bisector of DE such that the lines O6D and O6E subtend the desired output angle, in this case 45 deg. Draw link 6 from D through O6 and extend it to any convenient length. This is the output link that will dwell during the specified motion of the crank. See next page for the completed layout and further linkage specifications.



y 135 P

PSEUDO-ARC

180

45.000° B

5

4

225

O6 BISECTOR

3 D A

E

2

x O4

O2

SUMMARY OF LINKAGE SPECIFICATIONS Original fourbar: Ground link

L1  4.000

Crank

L2  2.000

Coupler

L3  5.000

Rocker

L4  5.000

Coupler point

AP  5.878

δ  54.000 deg

Added dyad: Coupler

L5  6.363

Output

L6  2.855

Pivot O6

x  3.833

y  3.375




Design a sixbar double-dwell linkage for a dwell of 90 deg of crank motion, with an output of rocker motion of 60 deg, followed by a second dwell of about 60 deg of crank motion.

Given:

Initial crank dwell period: 90 deg Final crank dwell period: 60 deg (approx.) Output rocker motion between dwells: 60 deg

Solution:


Design choices:

1.

Ground link length

L1  5.000

Crank length

L2  2.000

Coupler link length

L3  5.000

Rocker length

L2  5.500

Coupler point data:

AP  8.750

δ  50 deg

In the absence of a linkage atlas it is difficult to find a coupler curve that meets the specifications. One approach is to start with a symmetrical linkage, using the data in Figure 3-21. Then, using program FOURBAR and by trial-and-error, adjust the link lengths and coupler point data until a satisfactory coupler curve is found. The link lengths and coupler point data given above were found this way. The resulting coupler curve is shown below and a printout of the coupler curve coordinates taken from FOURBAR is also printed below.



FOURBAR for Windows Angle Step Deg 0.000 10.000 20.000 30.000 40.000 50.000 60.000 70.000 80.000 90.000 100.000 110.000 120.000 130.000 140.000 150.000 160.000 170.000 180.000 190.000 200.000 210.000 220.000 230.000 240.000 250.000 260.000 270.000 280.000 290.000 300.000 310.000 320.000 330.000 340.000 350.000 360.000

File P03-13.DAT

Cpler Pt Cpler Pt Cpler Pt Cpler Pt X Y Mag Ang

9.353 9.846 10.167 10.286 10.226 10.031 9.746 9.406 9.039 8.665 8.301 7.958 7.647 7.376 7.151 6.977 6.853 6.778 6.748 6.755 6.792 6.847 6.912 6.976 7.031 7.073 7.099 7.112 7.120 7.137 7.184 7.288 7.481 7.792 8.233 8.779 9.353

4.742 4.159 3.491 2.840 2.274 1.815 1.457 1.180 0.963 0.787 0.637 0.507 0.391 0.291 0.209 0.151 0.126 0.140 0.201 0.316 0.488 0.719 1.008 1.351 1.741 2.170 2.626 3.098 3.570 4.030 4.458 4.834 5.131 5.312 5.332 5.147 4.742

10.487 10.688 10.750 10.671 10.476 10.194 9.854 9.480 9.090 8.701 8.325 7.974 7.657 7.382 7.154 6.978 6.854 6.779 6.751 6.763 6.809 6.885 6.985 7.105 7.243 7.398 7.569 7.757 7.965 8.196 8.455 8.746 9.072 9.430 9.809 10.177 10.487

26.886 22.900 18.951 15.437 12.537 10.257 8.503 7.152 6.081 5.187 4.391 3.644 2.928 2.256 1.671 1.242 1.051 1.182 1.708 2.678 4.110 5.996 8.300 10.963 13.911 17.057 20.302 23.536 26.632 29.448 31.819 33.555 34.446 34.286 32.931 30.384 26.886

2.

Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Fit tangent lines to the nearly straight portions of the curve. Label their intersection O6. The coordinates of O6 are (6.729, 0.046).

3.

Design link 6 to lie along these straight tangents, pivoted at O6. Provide a slot in link 6 to accommodate slider block 5, which pivots on the coupler point P. (See next page).

4.

The beginning and ending crank angles for the dwell portions of the motion are indicated on the layout and in the table above by boldface entries.



y 6 B 60.000°

260 5 4

P

3

90 O2 2

A

170 x O4

150 O6




Figure P3-3 shows a treadle-operated grinding wheel driven by a fourbar linkage. Make a cardboard model of the linkage to any convenient scale. Determine its minimum transmission angles. Comment on its operation. Will it work? If so, explain how it does.

Given:

Link lengths:

Link 2

L2  0.60 m

Link 3

L3  0.75 m

Link 4

L4  0.13 m

Link 1

L1  0.90 m

Grashof condition function: Condition( a b c d ) 


Solution: 1.


Determine the Grashof condition of the mechanism from inequality 2.8 and its Barker classification from Table 2-4. Grashof condition: Barker classification:

Condition L1 L2 L3 L4  "Grashof" Class I-4, Grashof rocker-rocker-crank, GRRC, since the shortest link is the output link.

2.

As a Grashof rocker-crank, the minimum transmission angle will be 0 deg, twice per revolution of the output (link 4) crank.

3.

Despite having transmission angles of 0 deg twice per revolution, the mechanism will work. That is, one will be able to drive the grinding wheel from the treadle (link 2). The reason is that the grinding wheel will act as a flywheel and will carry the linkage through the periods when the transmission angle is low. Typically, the operator will start the motion by rotating the wheel by hand.




Figure P3-4 shows a non-Grashof fourbar linkage that is driven from link O2A. All dimensions are in centimeters (cm). (a) (b) (c) (d)

Given:

Solution: 1.

Find the transmission angle at the position shown. Find the toggle positions in terms of angle AO2O4. Find the maximum and minimum transmission angles over its range of motion. Draw the coupler curve of point P over its range of motion.

Link lengths: Link 1 (ground)

L1  95 mm

Link 2 (driver)

L2  50 mm

Link 3 (coupler)

L3  44 mm

Link 4 (driven)

L4  50 mm


To find the transmission angle at the position shown, draw the linkage to scale in the position shown and measure the transmission angle ABO4. P

y 77.097°

B

3 A 2

4

O4

50.000° 1

x

O2

The measured transmission angle at the position shown is 77.097 deg. 2.

The toggle positions will be symmetric with respect to the O2O4 axis and will occur when links 3 and 4 are colinear. Use the law of cosines to calculate the angle of link 2 when links 3 and 4 are in toggle.

 L3  L42  L12  L22  2 L1 L2 cosθ where 2 is the angle AO2O4. Solving for 2,

 L12  L22   L3  L4 2 θ  acos   2  L1 L2   The other toggle position occurs at θ  73.558 deg

θ  73.558 deg


3.


Use the program FOURBAR to find the maximum and minimum transmission angles.

FOURBAR for Windows

File P03-15

Design #

1

Angle Step Deg

Theta2 Mag degrees

Theta3 Mag degrees

Theta4 Mag degrees

Trans Ang Mag degrees

-73.557 -58.846 -44.134 -29.423 -14.711 0.000 14.711 29.423 44.134 58.846 73.557

-73.557 -58.846 -44.134 -29.423 -14.711 0.000 14.711 29.423 44.134 58.846 73.557

30.861 64.075 77.168 83.147 80.604 68.350 50.145 32.106 16.173 0.566 -30.486

-149.490 -176.312 170.696 157.514 142.103 125.123 111.644 106.473 109.701 120.179 149.159

0.352 60.387 86.472 74.367 61.499 56.773 61.499 74.367 86.472 60.387 0.355

A partial output from FOURBAR is shown above. From it, we see that the maximum transmission angle is approximately 86.5 deg and the minimum is zero deg. 4.

Use program FOURBAR to draw the coupler curve with respect to a coordinate frame through O2O4.




Draw the Roberts diagram for the linkage in Figure P3-4 and find its two cognates. Are they Grashof or non-Grashof?

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  9.5

Crank

L2  5

Coupler

L3  4.4

Rocker

L4  5

A1P  8.90



 

2

0.5

B1P  7.401

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

δ  56.000 deg

γ  94.4701 deg


L5  B1P

L5  7.401

L6 

L10  A1P

L10  8.900

L9 

L7  8.410

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  8.410

 A1P

L9  10.114

A1P

L8  10.114

B1P


A3P  5.000

A2P  L2

A2P  5.000

δ  γ

δ  94.470 deg

δ  δ

δ  56.000 deg


3.

L1 L3

 B1P

L1BC  15.9793

L1AC 

L1 L3

 A1P

L1AC  19.2159


Cognate #2

Cognate #3

Ground link length

L1  9.500

L1AC  19.216

L1BC  15.979

Crank length

L2  5.000

L10  8.900

L7  8.410

Coupler length

L3  4.400

L9  10.114

L6  8.410



Rocker length

L4  5.000

L8  10.114

L5  7.401

Coupler point

A1P  8.900

A2P  5.000

A3P  5.000

Coupler angle

δ  56.000 deg

δ  56.000 deg

δ  94.470 deg

B2

OC

8

7

B3

P 9 6 A3

A2

5

3

B1

10

4

A1 2

OB 1

OA

6.

Determine the Grashof condition of each of the two additional cognates. Condition( a b c d ) 


Cognate #2:

Condition L10 L1AC L8 L9  "non-Grashof"

Cognate #3:

Condition L5 L1BC L6 L7  "non-Grashof"




Design a Watt-I sixbar to give parallel motion that follows the coupler path of point P of the linkage in Figure P3-4.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  9.5

Crank

L2  5

Coupler

L3  4.4

Rocker

L4  5

A1P  8.90



 

2

0.5

B1P  7.401

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

δ  56.000 deg

γ  94.4701 deg


L5  B1P

L5  7.401

L6 

L10  A1P

L10  8.900

L9 

L7  8.410

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  8.410

 A1P

L9  10.114

A1P

L8  10.114

B1P


A3P  5.000

A2P  L2

A2P  5.000

δ  γ

δ  94.470 deg

δ  δ

δ  56.000 deg


3.

L1 L3

 B1P

L1BC  15.9793

L1AC 

L1 L3

 A1P

L1AC  19.2159


Cognate #2

Cognate #3

Ground link length

L1  9.500

L1AC  19.216

L1BC  15.979

Crank length

L2  5.000

L10  8.900

L7  8.410

Coupler length

L3  4.400

L9  10.114

L6  8.410


B2


Rocker length

L4  5.000

L8  10.114

L5  7.401

Coupler point

A1P  8.900

A2P  5.000

A3P  5.000

Coupler angle

δ  56.000 deg

δ  56.000 deg

δ  94.470 deg

OC

8

7

B3

P 9 6 A3

A2

P

5

3 10

B1 4

A1 2

OB 1 3

OA

4

A1

4.

5.

All three of these cognates are non-Grashof and will, therefore, have limited motion. However, following Example 3-11, discard cognate #2 and retain cognates #1 and #3. Draw line qq parallel to line OAOC and through point OB. Without allowing links 5, 6, and 7 to rotate, slide them as an assembly along lines OAOC and qq until the free end of link 7 is at OA. The free end of link 5 will then be at point O'B and point P on link 6 will be at P'. Add a new link of length OAOC between P and P'. This is the new output link 8 and all points on it describe the original coupler curve. Join links 2 and 7, making one ternary link. Remove link 5 and reduce link 6 to a binary link. The result is a Watt-I sixbar with links numbered 1, 2, 3, 4, 6, and 8 (see next page). Link 8 is in curvilinear translation and follows the coupler path of the original point P.

B1 q

2

OB 1

OA 7

B3 P'

6 A3 5

q

O'B



P

B1 3

4

A1 8 2

OB

1

OA

B3 P'

6




Design a Watt-I sixbar to give parallel motion that follows the coupler path of point P of the linkage in Figure P3-4 and add a driver dyad to drive it over its possible range of motion with no quick return. (The result will be an 8-bar linkage).

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  9.5

Crank

L2  5

Coupler

L3  4.4

Rocker

L4  5

A1P  8.90



 

2

0.5

B1P  7.401

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

δ  56.000 deg

γ  94.4701 deg


L5  B1P

L5  7.401

L6 

L10  A1P

L10  8.900

L9 

L7  8.410

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  8.410

 A1P

L9  10.114

A1P

L8  10.114

B1P


A3P  5.000

A2P  L2

A2P  5.000

δ  γ

δ  94.470 deg

δ  δ

δ  56.000 deg

From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC  3.

L1 L3

 B1P

L1BC  15.9793

L1AC 

L1 L3

 A1P

L1AC  19.2159


Cognate #2

Cognate #3

Ground link length

L1  9.500

L1AC  19.216

L1BC  15.979

Crank length

L2  5.000

L10  8.900

L7  8.410

Coupler length

L3  4.400

L9  10.114

L6  8.410


B2


Rocker length

L4  5.000

L8  10.114

L5  7.401

Coupler point

A1P  8.900

A2P  5.000

A3P  5.000

Coupler angle

δ  56.000 deg

δ  56.000 deg

δ  94.470 deg

OC

8

7

B3

P 9 6 A3

A2

5

3 10

P

B1 4

A1 2

OB 1

OA

3

4

A1

4.

5.

6.

All three of these cognates are non-Grashof and will, therefore, have limited motion. However, following Example 3-11, discard cognate #2 and retain cognates #1 and #3. Draw line qq parallel to line OAOC and through point OB. Without allowing links 5, 6, and 7 to rotate, slide them as an assembly along lines OAOC and qq until the free end of link 7 is at OA. The free end of link 5 will then be at point O'B and point P on link 6 will be at P'. Add a new link of length OAOC between P P' and P'. This is the new output link 8 and all points on it describe the original coupler curve. Join links 2 and 7, making one ternary link. Remove link 5 and reduce link 6 to a binary link. The result is a Watt-I sixbar with links numbered 1, 2, 3, 4, 6, and 8 (see next page). Link 8 is in curvilinear translation and follows the coupler path of the original point P.

B1 q

2

OB 1

OA 7

B3

6 A3 5

Add a driver dyad following Example 3-4.

q

O'B



P

B1 3

4

A1 8 2

OB

1

OA

B3 P'

6

P

B1 3

4

A1 8 2

OB

1

OA

P'

6

B3

OC




Design a pin-jointed linkage that will guide the forks of the fork lift truck in Figure P3-5 up and down in an approximate straight line over the range of motion shown. Arrange the fixed pivots so they are close to some part of the existing frame or body of the truck.

Given:

Length of straight line motion of the forks: Δx  1800 mm

Solution:


Design choices: Use a Hoeken-type straight line mechanism optimized for straightness. Maximum allowable error in straightness of line: ΔCy  0.096  % 1.

Using Table 3-1 and the required length of straight-line motion, determine the link lengths. Link ratios from Table 3-1 for ΔCy  0.096 %: L1overL2  2.200 L3overL2  2.800

ΔxoverL2  4.181

Link lengths:

2.

L2 

Coupler

L3  L3overL2  L2

L3  1205.5 mm

Ground link


L1  947.1 mm

Rocker

L4  L3

L4  1205.5 mm

Coupler point

BP  L3

BP  1205.5 mm

L2  430.5 mm

ΔxoverL2

Calculate the distance from point P to pivot O4 (Cy). Cy 

3.

Δx

Crank

2 L32   L1  L22

Cy  1978.5 mm

Draw the fork lift truck to scale with the mechanism defined in step 1 superimposed on it..

1978.5mm

O4

P

620.0mm

4 947.1mm B

900.0mm

487.1mm 3

O2 2 A




Figure P3-6 shows a "V-link" off-loading mechanism for a paper roll conveyor. Design a pinjointed linkage to replace the air cylinder driver that will rotate the rocker arm and V-link through the 90 deg motion shown. Keep the fixed pivots as close to the existing frame as possible. Your fourbar linkage should be Grashof and be in toggle at each extreme position of the rocker arm.

Given:

Dimensions scaled from Figure P3-6: Rocker arm (link 4) distance between pin centers:

Solution:

L4  320  mm


Design choices: 1. Use the same rocker arm that was used with the air cylinder driver. 2. Place the pivot O2 80 mm to the right of the right leg and on a horizontal line with the center of the pin on the rocker arm. 3. Design for two-position, 90 deg of output rocker motion with no quick return, similar to Example 3-2. 1.

Draw the rocker arm (link 4) O4B in both extreme positions, B1 and B2, in any convenient location such that the desired angle of motion 4 is subtended. In this solution, link 4 is drawn such that the two extreme positions each make an angle of 45 deg to the vertical.

2.

Draw the chord B1B2 and extend it in any convenient direction. In this solution it was extended horizontally to the left.

3.

Mark the center O2 on the extended line such that it is 80 mm to the right of the right leg. This will allow sufficient space for a supporting pillow block bearing.

4.

Bisect the line segment B1B2 and draw a circle of that radius about O2.

5.

Label the two intersections of the circle and extended line B1B2, A1 and A2.

6.

Measure the length of the coupler (link 3) as A1B1 or A2B2. From the graphical solution, L3  1045 mm

7.

Measure the length of the crank (link 2) as O2A1 or O2A2. From the graphical solution, L2  226.274  mm

8.

Measure the length of the ground link (link 1) as O2O4. From the graphical solution, L1  1069.217 mm 1045.000 80.000 1069.217 320.000

1 4 3

1045.000

9.

Find the Grashof condition.

2

226.274









Figure P3-7 shows a walking-beam transport mechanism that uses a fourbar coupler curve, replicated with a parallelogram linkage for parallel motion. Note the duplicate crank and coupler shown ghosted in the right half of the mechanism - they are redundant and have been removed from the duplicate fourbar linkage. Using the same fourbar driving stage (links 1, 2, 3, 4 with coupler point P), design a Watt-I sixbar linkage that will drive link 8 in the same parallel motion using two fewer links.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  2.22

Crank

L2  1

Coupler

L3  2.06

Rocker

L4  2.33

A1P  3.06



 

2

0.5

B1P  1.674

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

δ  31.000 deg

γ  109.6560 deg


L5  B1P

L5  1.674

L6 

L10  A1P

L10  3.060

L9 

L7  0.812

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  1.893

 A1P

L9  1.485

A1P

L8  3.461

B1P


A3P  3.461





δ  180  deg  δ  γ 

A2P  L2

A2P  1.000

δ  δ

δ  31.000 deg

δ  39.344 deg From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC 

3.

L1 L3

 B1P

L1BC  1.8035

L1AC 

L1 L3

 A1P

L1AC  3.2977

Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1 Ground link length

L1  2.220

Cognate #2

Cognate #3

L1AC  3.298

L1BC  1.804



Crank length

L2  1.000

L10  3.060

L7  0.812

Coupler length

L3  2.060

L9  1.485

L6  1.893

Rocker length

L4  2.330

L8  3.461

L5  1.674

Coupler point

A1P  3.060

A2P  1.000

A3P  3.461

Coupler angle

δ  31.000 deg

δ  31.000 deg

δ  39.344 deg

OC 7 6

B3

A3

8

5

OB

9

4

1

B2

A2

10

P

OA 2 A1 3 B1

4.

Determine the Grashof condition of each of the two additional cognates. Condition( a b c d ) 


5.

Cognate #2:

Condition L8 L9 L10 L1AC  "Grashof"

Cognate #3:

Condition L5 L6 L7 L1BC  "Grashof"

Both of these cognates are Grashof but cognate #3 is a crank rocker. Following Example 3-11, discard cognate #2 and retain cognates #1 and #3. Draw line qq parallel to line OAOC and through point OB. Without allowing links 5, 6, and 7 to rotate, slide them as an assembly along lines OAOC and qq until the free end of link 7 is at OA. The free end of link 5 will then be at point O'B and point P on link 6 will be at P'. Add a new link of length OAOC between P and P'. This is the new output link 8 and all points on it describe the original coupler curve.



q OB 1 4 P

OA

2

7

B3

6 A3 A1

3 B1 8

5 O'B q P'

6.

Join links 2 and 7, making one ternary link. Remove link 5 and reduce link 6 to a binary link. The result is a Watt-I sixbar with links numbered 1, 2, 3, 4, 6, and 8 (see next page). Link 8 is in curvilinear translation and follows the coupler path of the original point P. The walking-beam (link 8 in Figure P3-7) is rigidly attached to link 8 below.

OB 1 4 P

OA

2

A3 A1 3 B1

6

8

P'




Find the maximum and minimum transmission angles of the fourbar driving stage (links L1, L2, L3, L4) in Figure P3-7 (to graphical accuracy).

Given:

Link lengths:

Link 2

L2  1.00

Link 3

L3  2.06

Link 4

L4  2.33

Link 1

L1  2.22



Solution: 1.



2.

Condition L1 L2 L3 L4  "Grashof" Class I-2, Grashof crank-rocker-rocker, GCRR, since the shortest link is the input link.

It can be shown (see Section 4.10) that the minimum transmission angle for a fourbar GCRR linkage occurs when links 2 and 1 (ground link) are colinear. Draw the linkage in these two positions and measure the transmission angles. O4

O4

A O2

31.510°

O2 A 85.843°

B

3.

B

As measured from the layout, the minimum transmission angle is 31.5 deg. The maximum is 90 deg.




Figure P3-8 shows a fourbar linkage used in a power loom to drive a comb-like reed against the thread, "beating it up" into the cloth. Determine its Grashof condition and its minimum and maximum transmission angles to graphical accuracy.

Given:

Link lengths:

Link 2

L2  2.00 in

Link 3

L3  8.375  in

Link 4

L4  7.187  in

Link 1

L1  9.625  in



Solution: 1.



2.

Condition L1 L2 L3 L4  "Grashof" Class I-2, Grashof crank-rocker-rocker, GCRR, since the shortest link is the input link.

It can be shown (see Section 4.10) that the minimum transmission angle for a fourbar GCRR linkage occurs when links 2 and 1 (ground link) are colinear. Draw the linkage in these two positions and measure the transmission angles.

83.634°

58.078°

3.

As measured from the layout, the minimum transmission angle is 58.1 deg. The maximum is 90.0 deg.




Draw the Roberts diagram and find the cognates for the linkage in Figure P3-9.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  2.22

Crank

L2  1.0

Coupler

L3  2.06

Rocker

L4  2.33

A1P  3.06



 

2

0.5

B1P  1.674

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

δ  31.00  deg

γ  109.6560 deg


L5  B1P

L5  1.674

L6 

L10  A1P

L10  3.060

L9 

L7  0.812

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  1.893

 A1P

L9  1.485

A1P

L8  3.461

B1P


A3P  3.461

δ  180  deg  δ  γ

δ  39.344 deg

A2P  L2

A2P  1.000

δ  δ

δ  31.000 deg


L1 L3

 B1P

L1BC  1.8035

L1AC 

L1 L3

 A1P

L1AC  3.2977


Cognate #2

Cognate #3

Ground link length

L1  2.220

L1AC  3.298

L1BC  1.804

Crank length

L2  1.000

L10  3.060

L7  0.812

Coupler length

L3  2.060

L9  1.485

L6  1.893

Rocker length

L4  2.330

L8  3.461

L5  1.674

Coupler point

A1P  3.060

A2P  1.000

A3P  3.461

Coupler angle

δ  31.000 deg

δ  31.000 deg

δ  39.344 deg



B1 P B2 3

2

9

4

A1

A2

10

8 OB

5

1

OA

1BC 1AC

B3 6

A3

7

OC




Find the equivalent geared fivebar mechanism cognate of the linkage in Figure P3-9.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  2.22

Crank

L2  1.0

Coupler

L3  2.06

Rocker

L4  2.33

A1P  3.06



 

2

0.5

B1P  1.674

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

δ  31.00  deg

γ  109.6560 deg


L5  B1P

L5  1.674

L6 

L10  A1P

L10  3.060

L9 

L7  0.812

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  1.893

 A1P

L9  1.485

A1P

L8  3.461

B1P


A3P  3.461

δ  180  deg  δ  γ

δ  39.344 deg

A2P  L2

A2P  1.000

δ  δ

δ  31.000 deg


3.

L1 L3

 B1P

L1BC  1.8035

L1AC 

L1 L3

 A1P

L1AC  3.2977


Cognate #2

Cognate #3

Ground link length

L1  2.220

L1AC  3.298

L1BC  1.804

Crank length

L2  1.000

L10  3.060

L7  0.812

Coupler length

L3  2.060

L9  1.485

L6  1.893

Rocker length

L4  2.330

L8  3.461

L5  1.674



Coupler point

A1P  3.060

A2P  1.000

A3P  3.461

Coupler angle

δ  31.000 deg

δ  31.000 deg

δ  39.344 deg

B1 P B2 3

2

9

4

A1

A2

10

8 OB

5

1

OA

1BC 1AC

B3 6

A3

4.

7

OC

The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAA2PB3OB, OAA1PA3OC, and OBB1PB2OC. The three geared fivebar cognates are summarized in the table below. SUMMARY OF GEARED FIVEBAR COGNATE SPECIFICATIONS: Cognate #1

Cognate #2

Cognate #3

Ground link length

L1  2.220

L1AC  3.298

L1BC  1.804

Crank length

L10  3.060

L2  1.000

L4  2.330

Coupler length

A2P  1.000

A1P  3.060

L5  1.674

Rocker length

L4  2.330

L8  3.461

L7  0.812

Crank length

L5  1.674

L7  0.812

L8  3.461

Coupler point

A2P  1.000

A1P  3.060

B1P  1.674

Coupler angle

δ  0.00 deg

δ  0.00 deg

δ  0.00 deg

5.

Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path (see next page)

6.







Use the linkage in Figure P3-9 to design an eightbar double-dwell mechanism that has a rocker output through 45 deg.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  2.22

Crank

L2  1.0

Coupler

L3  2.06

Rocker

L4  2.33

A1P  3.06

δ  31.00  deg


Enter the given data into program FOURBAR and print out the resulting coupler point coordinates (see table below). FOURBAR for Windows

File P03-26.DAT

Angle Step Deg


0.000 10.000 20.000 30.000 40.000 50.000 60.000 70.000 80.000 90.000 100.000 110.000 120.000 130.000 140.000 150.000 160.000 170.000 180.000 190.000 200.000 210.000 220.000 230.000 240.000 250.000 260.000 270.000 280.000 290.000 300.000 310.000 320.000 330.000 340.000 350.000 360.000

2.731 3.077 3.350 3.515 3.576 3.554 3.473 3.350 3.203 3.040 2.872 2.706 2.548 2.403 2.274 2.164 2.075 2.005 1.953 1.917 1.892 1.875 1.862 1.848 1.832 1.810 1.784 1.754 1.723 1.698 1.687 1.702 1.761 1.883 2.088 2.380 2.731

2.523 2.407 2.228 2.032 1.855 1.708 1.592 1.499 1.420 1.348 1.278 1.207 1.135 1.062 0.990 0.925 0.869 0.826 0.802 0.798 0.817 0.860 0.925 1.011 1.115 1.235 1.367 1.508 1.654 1.804 1.955 2.105 2.251 2.386 2.494 2.550 2.523

3.718 3.906 4.023 4.060 4.028 3.943 3.820 3.671 3.503 3.326 3.144 2.963 2.789 2.627 2.480 2.354 2.249 2.168 2.111 2.076 2.061 2.063 2.079 2.107 2.145 2.192 2.248 2.313 2.388 2.477 2.582 2.707 2.858 3.040 3.253 3.488 3.718

42.731 38.029 33.626 30.035 27.412 25.672 24.635 24.107 23.915 23.915 23.988 24.039 24.001 23.834 23.533 23.134 22.719 22.404 22.326 22.614 23.365 24.632 26.417 28.678 31.340 34.306 37.463 40.683 43.826 46.730 49.207 51.038 51.965 51.715 50.064 46.967 42.731



2.

Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Fit tangent lines to the nearly straight portions of the curve. Label their intersection O6.

3.

Design link 6 to lie along these straight tangents, pivoted at O6. Provide a guide on link 6 to accommodate slider block 5, which pivots on the coupler point P. O8

8 45.000° F

E D

7

B 70.140°

C 6

3 P 5 A O6 2

4

O4

O2

4.

Extend link 6 a convenient distance to point C. Draw an arc through point C with center at O6. Label the intersection of the arc with the other tangent line as point D. Attach link 7 to the pivot at C. The length of link 7 is CE, a design choice. Extend line CDE from point E a distance equal to CD. Label the end point F. Layout two intersecting lines through E and F such that they subtend an angle of 45 deg. Label their intersection O8. The link joining O8 and point E is link 8. The link lengths and locations of O6 and O8 are: Link 6

L6  2.330

Fixed pivot O6:

Link 7

x  1.892 y  0.762

L7  3.000 Fixed pivot O8:

Link 8 x  1.379 y  6.690

L8  3.498




Use the linkage in Figure P3-9 to design an eightbar double-dwell mechanism that has a slider output stroke of 5 crank units.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  2.22

Crank

L2  1.0

Coupler

L3  2.06

Rocker

L4  2.33

A1P  3.06

δ  31.00  deg


Enter the given data into program FOURBAR and print out the resulting coupler point coordinates (see table below). FOURBAR for Windows

File P03-26.DAT

Angle Step Deg


0.000 10.000 20.000 30.000 40.000 50.000 60.000 70.000 80.000 90.000 100.000 110.000 120.000 130.000 140.000 150.000 160.000 170.000 180.000 190.000 200.000 210.000 220.000 230.000 240.000 250.000 260.000 270.000 280.000 290.000 300.000 310.000 320.000 330.000 340.000 350.000 360.000

2.731 3.077 3.350 3.515 3.576 3.554 3.473 3.350 3.203 3.040 2.872 2.706 2.548 2.403 2.274 2.164 2.075 2.005 1.953 1.917 1.892 1.875 1.862 1.848 1.832 1.810 1.784 1.754 1.723 1.698 1.687 1.702 1.761 1.883 2.088 2.380 2.731

2.523 2.407 2.228 2.032 1.855 1.708 1.592 1.499 1.420 1.348 1.278 1.207 1.135 1.062 0.990 0.925 0.869 0.826 0.802 0.798 0.817 0.860 0.925 1.011 1.115 1.235 1.367 1.508 1.654 1.804 1.955 2.105 2.251 2.386 2.494 2.550 2.523

3.718 3.906 4.023 4.060 4.028 3.943 3.820 3.671 3.503 3.326 3.144 2.963 2.789 2.627 2.480 2.354 2.249 2.168 2.111 2.076 2.061 2.063 2.079 2.107 2.145 2.192 2.248 2.313 2.388 2.477 2.582 2.707 2.858 3.040 3.253 3.488 3.718

42.731 38.029 33.626 30.035 27.412 25.672 24.635 24.107 23.915 23.915 23.988 24.039 24.001 23.834 23.533 23.134 22.719 22.404 22.326 22.614 23.365 24.632 26.417 28.678 31.340 34.306 37.463 40.683 43.826 46.730 49.207 51.038 51.965 51.715 50.064 46.967 42.731



2.

Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Fit tangent lines to the nearly straight portions of the curve. Label their intersection O6.

3.

Design link 6 to lie along these straight tangents, pivoted at O6. Provide a guide on link 6 to accommodate slider block 5, which pivots on the coupler point P. F

E

8

D 7

C

B 6 70.140° 3 P 5 A O6 2

4

O4

O2

4.

Extend link 6 and the other tangent line until points C and E are 5 units apart. Attach link 7 to the pivot at C. The length of link 7 is CD, a design choice. Extend line CDE from point D a distance equal to CE. Label the end point F. As link 6 travels from C to E, slider block 8 will travel from D to F, a distance of 5 units. The link lengths and location of O6: Link 6

L6  4.351

Fixed pivot O6:

Link 7

x  1.892 y  0.762

L7  2.000




Use two of the cognates in Figure 3-26 (p. 126) to design a Watt-I sixbar parallel motion mechanism that carries a link through the same coupler curve at all points. Comment on its similarities to the original Roberts diagram.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  45

Crank

L2  56

Coupler

L3  22.5

Rocker

L4  56

A1P  11.25 δ  0.000  deg



 

2

0.5

B1P  11.250

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

γ  0.0000 deg


L5  B1P

L5  11.250

L6 

L10  A1P

L10  11.250

L9 

L7  28.000

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  28.000

 A1P

L9  28.000

A1P

L8  28.000

B1P


A3P  56.000

A2P  L2

A2P  56.000

δ  δ

δ  0.000 deg

δ  δ

δ  0.000 deg


3.

L1 L3

 B1P

L1BC  22.5000

L1AC 

L1 L3

 A1P

L1AC  22.5000


Cognate #2

Cognate #3

Ground link length

L1  45.000

L1AC  22.500

L1BC  22.500

Crank length

L2  56.000

L10  11.250

L7  28.000

Coupler length

L3  22.500

L9  28.000

L6  28.000



Rocker length

L4  56.000

L8  28.000

L5  11.250

Coupler point

A1P  11.250

A2P  56.000

A3P  56.000

Coupler angle

δ  0.000 deg

δ  0.000 deg

δ  0.000 deg

P

3

B1

A1

B2

B3

6 9

8

2

A2

7

5

10 OA

4.

4

A3

OB

OC

Both of these cognates are identical. Following Example 3-11, discard cognate #2 and retain cognates #1 and #3. Without allowing links 5, 6, and 7 to rotate, slide them as an assembly along line OAOC until the free end of link 7 is at OA. The free end of link 5 will then be at point O'B and point P on link 6 will be at P'. Add a new link of length OAOC between P and P'. This is the new output link 8 and all points on it describe the original coupler curve. P'

B1

8

3

P

A1

B3

6

7 2

4

5 OA

O'B

A3 OB


5.


Join links 2 and 7, making one ternary link. Remove link 5 and reduce link 6 to a binary link. The result is a Watt-I sixbar with links numbered 1, 2, 3, 4, 6, and 8. Link 8 is in curvilinear translation and follows the coupler path of the original point P. Link 8 is a binary link with nodes at P and P'. It does not attach to link 4 at B1.

P'

8

B1

3

P

A1

6

B3

2

OA

4

OB




Find the cognates of the Watt straight-line mechanism in Figure 3-29a (p. 131).

Given:

Link lengths:

Solution:

Coupler point data:

Ground link

L1  4

Crank

L2  2

A1P  0.500

δ  0.00 deg

Coupler

L3  1

Rocker

L4  2

B1P  0.500

γ  0.00 deg


1.

Input the link dimensions and coupler point data into program FOURBAR.

2.

Use the Cognate pull-down menu to get the link lengths for cognates #2 and #3 (see next page). Note that, for this mechanism, cognates #2 and #3 are identical. All three mechanisms are non-Grashof with limited crank angles.






Find the cognates of the Roberts straight-line mechanism in Figure 3-29b.

Given:

Link lengths:

Solution:

Coupler point data:

Ground link

L1  2

Crank

L2  1

A1P  1.000

δ  60.0 deg

Coupler

L3  1

Rocker

L4  1

B1P  1.000

γ  60.0 deg


1.

Input the link dimensions and coupler point data into program FOURBAR.

2.

Note that, for this mechanism, cognates #2 and #3 are identical with cognate #1 because of the symmetry of the linkage (draw the Cayley diagram to see this). All three mechanisms are non-Grashof with limited crank angles.




Design a Hoeken straight-line linkage to give minimum error in velocity over 22% of the cycle for a 15-cm-long straight line motion. Specify all linkage parameters.

Given:

Length of straight line motion: Δx  150  mm Percentage of cycle over which straight line motion takes place: 22%

Solution:


1.

Using Table 3-1 and the required length of straight-line motion, determine the link lengths. Link ratios from Table 3-1 for 22% cycle: L1overL2  1.975

L3overL2  2.463


Link lengths:

2.

L2 

Coupler


L3  200.24 mm

Ground link


L1  160.57 mm

Rocker

L4  L3

L4  200.24 mm

Coupler point

AP  2  L3

AP  400.49 mm

L2  81.30 mm

ΔxoverL2

Calculate the distance from point P to pivot O4 (Cy) when crank angle is 180 deg. Cy 

3.

Δx

Crank

2 L32   L1  L22

Cy  319.20 mm

Enter the link lengths into program FOURBAR to verify the design (see next page for coupler point curve). Using the PRINT facility, determine the x,y coordinates of the coupler curve and the x,y components of the coupler point velocity in the straight line region. A table of these values is printed below. Notice the small deviations over the range of crank angles from the y-coordinate and the x-velocity at a crank angle of 180 deg.

FOURBAR for Windows

File P03-31.DOC

Angle Step Deg

Cpler Pt X mm

Cpler Pt Y mm

Veloc CP X mm/sec

Veloc CP Y mm/sec

140 150 160 170 180 190 200 210 220

235.60 216.84 198.06 179.31 160.58 141.85 123.09 104.31 85.55

319.95 319.72 319.46 319.27 319.20 319.27 319.47 319.72 319.95

-1,072.61 -1,076.20 -1,075.51 -1,073.75 -1,072.93 -1,073.75 -1,075.52 -1,076.22 -1,072.63

-14.74 -13.54 -7.99 0.02 8.03 13.58 14.78 10.76

-10.73






Design a Hoeken straight-line linkage to give minimum error in straightness over 39% of the cycle for a 20-cm-long straight line motion. Specify all linkage parameters.

Given:

Length of straight line motion: Δx  200  mm Percentage of cycle over which straight line motion takes place: 39%

Solution:


1.

Using Table 3-1 and the required length of straight-line motion, determine the link lengths. Link ratios from Table 3-1 for 39% cycle: L1overL2  2.500

L3overL2  3.250


Δx ΔxoverL2

L2  55.20 mm

Link lengths:

2.

Crank

L2 

Coupler


L3  179.41 mm

Ground link


L1  138.01 mm

Rocker

L4  L3

L4  179.41 mm

Coupler point

AP  2  L3

AP  358.82 mm

Calculate the distance from point P to pivot O4 (Cy) when crank angle is 180 deg. Cy 

3.

2 L32   L1  L22

Cy  302.36 mm

Enter the link lengths into program FOURBAR to verify the design (see next page for coupler point curve). Using the PRINT facility, determine the x,y coordinates of the coupler curve and the x,y components of the coupler point velocity in the straight line region. A table of these values is printed below. Notice the small deviations over the range of crank angles from the y-coordinate and the x-velocity from a crank angle of 180 deg. FOURBAR for Windows

File P03-32.DAT

Angle Step Deg

Coupler Pt X mm

Coupler Pt Y mm

Veloc CP X mm/sec

110 120 130 140 150 160 170 180 190 200 210 220 230 240 250

237.992 225.289 211.710 197.521 182.927 168.076 153.076 138.010 122.944 107.944 93.093 78.499 64.311 50.731 38.028

302.408 302.361 302.378 302.398 302.399 302.385 302.368 302.360 302.368 302.385 302.399 302.398 302.378 302.361 302.408

-696.591 -755.847 -797.695 -826.217 -844.774 -856.043 -861.994 -863.841 -861.994 -856.043 -844.774 -826.217 -797.695 -755.847 -696.591

Veloc CP Y mm/sec -6.416 -0.019 1.426 0.664 -0.483 -1.052 -0.800 0.000 0.800 1.052 0.483 -0.664 -1.426 0.019 6.416






Design a linkage that will give a symmetrical "kidney bean" shaped coupler curve as shown in Figure 3-16 (p. 114 and 115). Use the data in Figure 3-21 (p. 120) to determine the required link ratios and generate the coupler curve with program FOURBAR.

Solution:


Design choices: Ground link ratio, L1/L2 = 2.0: GLR  2.0 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.5: CLR  2.5 Coupler angle, γ  72 deg Crank length, L2  2.000 1.


L3  CLR L2

L3  5.000


L4  CLR L2

L4  5.000


L1  GLR  L2

L1  4.000

Angle PAB

δ 


180  deg  γ 2

AP  2  L3 cos δ

δ  54.000 deg AP  5.878

Enter the above data into program FOURBAR and plot the coupler curve.




Design a linkage that will give a symmetrical "double straight" shaped coupler curve as shown in Figure 3-16. Use the data in Figure 3-21 to determine the required link ratios and generate the coupler curve with program FOURBAR.

Solution:


Design choices: Ground link ratio, L1/L2 = 2.5: GLR  2.5 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.5: CLR  2.5 Coupler angle, γ  252  deg Crank length, L2  2.000 1.


L3  CLR L2

L3  5.000


L4  CLR L2

L4  5.000


L1  GLR  L2

L1  5.000

Angle PAB

δ 


180  deg  γ 2

AP  2  L3 cos δ

δ  36.000 deg AP  8.090





Design a linkage that will give a symmetrical "scimitar" shaped coupler curve as shown in Figure 3-16. Use the data in Figure 3-21 to determine the required link ratios and generate the coupler curve with program FOURBAR. Show that there are (or are not) true cusps on the curve.

Solution:


Design choices: Ground link ratio, L1/L2 = 2.0: GLR  2.0 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.5: CLR  2.5 Coupler angle, γ  144  deg Crank length, L2  2.000 1.


L3  CLR L2

L3  5.000


L4  CLR L2

L4  5.000


L1  GLR  L2

L1  4.000

Angle PAB

δ 


180  deg  γ 2

AP  2  L3 cos δ

δ  18.000 deg AP  9.511



3.


The points at the ends of the "scimitar" will be true cusps if the velocity of the coupler point is zero at these points. Using FOURBAR's plotting utility, plot the magnitude and angle of the coupler point velocity vector. As seen below for the range of crank angle from 50 to 70 degrees, the magnitude of the velocity does not quite reach zero. Therefore, these are not true cusps.




Find the Grashof condition, inversion, any limit positions, and the extreme values of the transmission angle (to graphical accuracy) of the linkage in Figure P3-10.

Given:

Link lengths:

Link 2

L2  0.785

Link 3

L3  0.356

Link 4

L4  0.950

Link 1

L1  0.544


S  min ( a b c d ) L  max( a b c d ) SL  S  L PQ  a  b  c  d  SL return "Grashof" if SL  PQ return "Special Grashof" if SL = PQ

Solution: 1.

return "non-Grashof" otherwise See Figure P3-10 and Mathcad file P0336.

Determine the Grashof condition of the mechanism from inequality 2.8 and its Barker classification from Table 2-4. Condition L1 L2 L3 L4  "Grashof"

Grashof condition: Barker classification:

2.

Class I-3, Grashof rocker-crank-rocker, GRCR, since the shortest link is the coupler link.

A GRCR linkage will have two toggle positions. Draw the linkage in these two positions and measure the input link angles. A

B

158.286° O4

O2

O4

O2 B A

158.286°

3.

As measured from the layout, the input link angles at the toggle positions are: +158.3 and -158.3 deg.

4.

Since the coupler link in a GRCR linkage can make a full rotation with respect to the input and output rockers, the minimum transmission angle is 0 deg and the maximum is 90 deg.





Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  0.544 Crank

L2  0.785

Coupler

L3  0.356 Rocker

L4  0.950

δ  0.00 deg



 

2

0.5

B1P  0.734

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

A1P  1.09

γ  180.0000 deg


L5  B1P

L5  0.734

L6 

L10  A1P

L10  1.090

L9 

L7  1.619

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  1.959

 A1P

L9  2.404

A1P

L8  2.909

B1P


A3P  0.950

A2P  L2

A2P  0.785

δ  180  deg  δ

δ  180.000 deg

δ  δ

δ  0.000 deg


3.

L1 L3

 B1P

L1BC  1.1216

L1AC 

L1 L3

 A1P

L1AC  1.6656


Cognate #2

Cognate #3

Ground link length

L1  0.544

L1AC  1.666

L1BC  1.122

Crank length

L2  0.785

L10  1.090

L7  1.619

Coupler length

L3  0.356

L9  2.404

L6  1.959



Rocker length

L4  0.950

L8  2.909

L5  0.734

Coupler point

A1P  1.090

A2P  0.785

A3P  0.950

Coupler angle

δ  0.000 deg

δ  0.000 deg

δ  180.000 deg

B2

9

8

P B1 A1

3 4

A2

2

5

10 OA

1

A3 OC

OB 6

7

B3




Find the three geared fivebar cognates of the linkage in Figure P3-10.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  0.544 Crank

L2  0.785

Coupler

L3  0.356 Rocker

L4  0.950

δ  0.00 deg



 

2

0.5

B1P  0.734

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

A1P  1.09

γ  180.0000 deg


L5  B1P

L5  0.734

L6 

L10  A1P

L10  1.090

L9 

L7  1.619

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  1.959

 A1P

L9  2.404

A1P

L8  2.909

B1P


A3P  0.950

A2P  L2

A2P  0.785

δ  180  deg  δ

δ  180.000 deg

δ  δ

δ  0.000 deg


L1 L3

 B1P

L1BC  1.1216

L1AC 

L1 L3

 A1P

L1AC  1.6656


Cognate #2

Cognate #3

Ground link length

L1  0.544

L1AC  1.666

L1BC  1.122

Crank length

L2  0.785

L10  1.090

L7  1.619

Coupler length

L3  0.356

L9  2.404

L6  1.959

Rocker length

L4  0.950

L8  2.909

L5  0.734

Coupler point

A1P  1.090

A2P  0.785

A3P  0.950



δ  0.000 deg

Coupler angle

δ  0.000 deg

δ  180.000 deg

B2

9

8

P B1 A1

3 4

A2

2

5

10 OA

1

A3 OC

OB 6

7

B3 4.

The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAA2PA3OB, OAA1PB3OC, and OBB1PB2OC. They are specified in the summary table below. SUMMARY OF GEARED FIVEBAR COGNATE SPECIFICATIONS: Cognate #1

Cognate #2

Cognate #3

Ground link length

L1  0.544

L1AC  1.666

L1BC  1.122

Crank length

L10  1.090

L2  0.785

L4  0.950

Coupler length

A2P  0.785

A1P  1.090

L5  0.734

Rocker length

A3P  0.950

L8  2.909

L7  1.619

Crank length

L5  0.734

L7  1.619

L8  2.909

Coupler point

A2P  0.785

A1P  1.090

B1P  0.734

Coupler angle

δ  0.00 deg

δ  0.00 deg

δ  0.00 deg

5.

Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path (see next page).

6.







Find the Grashof condition, any limit positions, and the extreme values of the transmission angle (to graphical accuracy) of the linkage in Figure P3-11.

Given:

Link lengths:

Link 2

L2  0.86

Link 3

L3  1.85

Link 4

L4  0.86

Link 1

L1  2.22



Solution: 1.


Determine the Grashof condition of the mechanism from inequality 2.8 and its Barker classification from Table 2-4. Condition L1 L2 L3 L4  "non-Grashof"


2.

Class II-1, non-Grashof triple rocker, RRR1, since the longest link is the ground link.

An RRR1 linkage will have two toggle positions. Draw the linkage in these two positions and measure the input link angles. 116.037° A B

O4

O2

O4

O2 B A 116.037°

88.2° B

A 67.3° O2

O4

3.

As measured from the layout, the input link angles at the toggle positions are: +116 and -116 deg.

4.

Since the coupler link in an RRR1 linkage cannot make a full rotation with respect to the input and output rockers, the minimum transmission angle is 0 deg and the maximum is 88 deg.





Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  2.22

Crank

L2  0.86

Coupler

L3  1.85

Rocker

L4  0.86

δ  0.00 deg



 

2

0.5

B1P  0.520

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

A1P  1.33

γ  0.0000 deg


L5  B1P

L5  0.520

L6 

L10  A1P

L10  1.330

L9 

L7  0.242

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  0.242

 A1P

L9  0.618

A1P

L8  0.618

B1P


A3P  0.618

A2P  L7

A2P  0.242

δ  180  deg

δ  180.000 deg

δ  180  deg

δ  180.000 deg


L1 L3

 B1P

L1BC  0.6240

L1AC 

L1 L3

 A1P

L1AC  1.5960


Cognate #2

Cognate #3

Ground link length

L1  2.220

L1AC  1.596

L1BC  0.624

Crank length

L2  0.860

L10  1.330

L7  0.242

Coupler length

L3  1.850

L9  0.618

L6  0.242

Rocker length

L4  0.860

L8  0.618

L5  0.520

Coupler point

A1P  1.330

A2P  0.242

A3P  0.618

Coupler angle

δ  0.000 deg

δ  180.000 deg

δ  180.000 deg



B2

B1

9 A2

3

10

P

4

8

OC

OA

A3

7 6

2 A1

OB 5

B3





Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  2.22

Crank

L2  0.86

Coupler

L3  1.85

Rocker

L4  0.86

δ  0.00 deg



 

2

0.5

B1P  0.520

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

A1P  1.33

γ  0.0000 deg


L5  B1P

L5  0.520

L6 

L10  A1P

L10  1.330

L9 

L7  0.242

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  0.242

 A1P

L9  0.618

A1P

L8  0.618

B1P


A3P  0.618

A2P  L7

A2P  0.242

δ  180  deg

δ  180.000 deg

δ  180  deg

δ  180.000 deg


3.

L1 L3

 B1P

L1BC  0.6240

L1AC 

L1 L3

 A1P

L1AC  1.5960


Cognate #2

Cognate #3

Ground link length

L1  2.220

L1AC  1.596

L1BC  0.624

Crank length

L2  0.860

L10  1.330

L7  0.242

Coupler length

L3  1.850

L9  0.618

L6  0.242

Rocker length

L4  0.860

L8  0.618

L5  0.520



Coupler point

A1P  1.330

A2P  0.242

A3P  0.618

Coupler angle

δ  0.000 deg

δ  180.000 deg

δ  180.000 deg

B2

B1

9 A2

3

10

P

4

8

OC

OA

A3

7 6

2

5 B3

A1 4.

OB

The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAB2PB3OB, OAA1PA3OC, and OBB1PA2OC. The three geared fivebar cognates are summarized in the table below. SUMMARY OF GEARED FIVEBAR COGNATE SPECIFICATIONS: Cognate #1

Cognate #2

Cognate #3

Ground link length

L1  2.220

L1AC  1.596

L1BC  0.624

Crank length

L10  1.330

L2  0.860

L4  0.860

Coupler length

L2  0.860

A1P  1.330

L5  0.520

Rocker length

L4  0.860

L8  0.618

L7  0.242

Crank length

L5  0.520

L7  0.242

L8  0.618

Coupler point

L2  0.860

A1P  1.330

B1P  0.520

Coupler angle

δ  0.00 deg

δ  0.00 deg

δ  0.00 deg

5.

Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path (see next page)

6.







Find the Grashof condition, any limit positions, and the extreme values of the transmission angle (to graphical accuracy) of the linkage in Figure P3-12.

Given:

Link lengths:

Link 2

L2  0.72

Link 3

L3  0.68

Link 4

L4  0.85

Link 1

L1  1.82



Solution: 1.


Determine the Grashof condition of the mechanism from inequality 2.8 and its Barker classification from Table 2-4. Condition L1 L2 L3 L4  "non-Grashof"


2.

Class II-1, non-Grashof triple rocker, RRR1, since the longest link is the ground link.

An RRR1 linkage will have two toggle positions. Draw the linkage in these two positions and measure the input link angles.

A

55.4° B

O4

O2

O4

O2

B A

55.4°

B 88.8° O4 O2

A

3.

As measured from the layout, the input link angles at the toggle positions are: +55.4 and -55.4 deg.

4.

Since the coupler link in an RRR1 linkage it cannot make a full rotation with respect to the input and output rockers, the minimum transmission angle is 0 deg and the maximum is 88.8 deg.





Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  1.82

Crank

L2  0.72

Coupler

L3  0.68

Rocker

L4  0.85

δ  54.0 deg



 

2

0.5

B1P  0.792

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

A1P  0.97

γ  82.0315 deg


L5  B1P

L5  0.792

L6 

L10  A1P

L10  0.970

L9 

L7  0.839

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  0.990

 A1P

L9  1.027

A1P

L8  1.212

B1P


A3P  0.850

A2P  L2

A2P  0.720

δ  γ

δ  82.032 deg

δ  δ

δ  54.000 deg


L1 L3

 B1P

L1BC  2.1208

L1AC 

L1 L3

 A1P

L1AC  2.5962


Cognate #2

Cognate #3

Ground link length

L1  1.820

L1AC  2.596

L1BC  2.121

Crank length

L2  0.720

L10  0.970

L7  0.839

Coupler length

L3  0.680

L9  1.027

L6  0.990

Rocker length

L4  0.850

L8  1.212

L5  0.792

Coupler point

A1P  0.970

A2P  0.720

A3P  0.850



δ  54.000 deg

Coupler angle

δ  54.000 deg

OC

8 B2

7

9

B3

P 6 A2 1AC 10 3

OA

B1

5 4

A1 2

1BC

A3

1 OB

δ  82.032 deg





Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  1.82

Crank

L2  0.72

Coupler

L3  0.68

Rocker

L4  0.85

δ  54.0 deg



 

2

0.5

B1P  0.792

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

A1P  0.97

γ  82.0315 deg


L5  B1P

L5  0.792

L6 

L10  A1P

L10  0.970

L9 

L7  0.839

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  0.990

 A1P

L9  1.027

A1P

L8  1.212

B1P


A3P  0.850

A2P  L2

A2P  0.720

δ  γ

δ  82.032 deg

δ  δ

δ  54.000 deg


3.

L1 L3

 B1P

L1BC  2.1208

L1AC 

L1 L3

 A1P

L1AC  2.5962


Cognate #2

Cognate #3

Ground link length

L1  1.820

L1AC  2.596

L1BC  2.121

Crank length

L2  0.720

L10  0.970

L7  0.839

Coupler length

L3  0.680

L9  1.027

L6  0.990

Rocker length

L4  0.850

L8  1.212

L5  0.792

Coupler point

A1P  0.970

A2P  0.720

A3P  0.850



δ  54.000 deg

Coupler angle

δ  54.000 deg

δ  82.032 deg

OC

8 B2

7

9

B3

P 6 A2 1AC 10 3

B1

5 4

A1 2

1BC

A3

1 OB

OA

4.

The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAA2PA3OB, OAA1PB3OC, and OBB1PB2OC. SUMMARY OF GEARED FIVEBAR COGNATE SPECIFICATIONS: Cognate #1

Cognate #2

Cognate #3

Ground link length

L1  1.820

L1AC  2.596

L1BC  2.121

Crank length

L10  0.970

L2  0.720

L4  0.850

Coupler length

A2P  0.720

A1P  0.970

L5  0.792

Rocker length

A3P  0.850

L8  1.212

L7  0.839

Crank length

L5  0.792

L7  0.839

L8  1.212

Coupler point

A2P  0.720

A1P  0.970

B1P  0.792

Coupler angle

δ  0.00 deg

δ  0.00 deg

δ  0.00 deg

5.

Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path (see next page).

6.







Prove that the relationships between the angular velocities of various links in the Roberts diagram as shown in Figure 3-25 (p. 125) are true.

Given:

OAA1PA2, OCB2PB3, and OBB1PA3 are parallelograms for any position of link 2..

Proof: 1.

OAA1 and A2P are opposite sides of a parallelogram and are, therefore, always parallel.

2.

Any change in the angle of OAA1 (link 2) will result in an identical change in the angle of A2P.

3.

Angular velocity is the change in angle per unit time.

4.

Since OAA1 and A2P have identical changes in angle, their angular velocities are identical.

5.

A2P is a line on link 9 and all lines on a rigid body have the same angular velocity. Therefore, link 9 has the same angular velocity as link 2.

6.

OCB3 (link 7) and B2P are opposite sides of a parallelogram and are, therefore, always parallel.

7.

B2P is a line on link 9 and all lines on a rigid body have the same angular velocity. Therefore, link 7 has the same angular velocity as links 9 and 2.

8.

The same argument holds for links 3, 5, and 10; and links 4, 6, and 8.




Design a fourbar linkage to move the object in Figure P3-13 from position 1 to 2 using points A and B for attachment. Add a driver dyad to limit its motion to the range of positions shown, making it a sixbar. All fixed pivots should be on the base.

Given:

Length of coupler link: L3  52.000

Solution:


Design choices: Length of link 2

L2  130

Length of link 2b

L2b  40

L4  110

Length of link 4

1.

Connect the end points of the two given positions of the line AB with construction limes, i.e., lines from A1 to A2 and B1 to B2.

2.

Bisect these lines and extend their perpendicular bisectors into the base.

3.

Select one point on each bisector and label them O2 and O4, respectively. In the solution below the distances O2A was selected to be L2  130.000 and O4B to be L4  110.000 . This resulted in a ground-link-length O2O4 for the fourbar of 27.080.

4.

The fourbar stage is now defined as O2ABO4 with link lengths Ground link 1a

L1a  27.080

Link 3 (coupler)

L3  52.000

Link 2 (input)

L2  130.000

Link 4 (output)

L4  110.000

5.

Select a point on link 2 (O2A) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 2 is now a ternary link with nodes at O2, C, and A.) In the solution below the distance O2C was selected to be L2b  40.000 .

6.

Draw a construction line through C1C2 and extend it to the left.

7.

Select a point on this line and call it O6. In the solution below O6 was placed 20 units from the left edge of the base.

8.

Draw a circle about O6 with a radius of one-half the length C1C2 and label the intersections of the circle with the extended line as D1 and D2. In the solution below the radius was measured as 23.003 units.

A1

B1

3

A2

6

D1

O6

2 D2

C1 5

9.


L6  23.003

Link 5 (coupler) L5  106.866 Link 1b (ground) L1b  111.764 Link 2b (rocker) L2b  40.000

23.003 106.866 111.764

B2

4 C2

40.000 O2

O4 27.080



10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition( a b c d ) 


Condition L1b L2b L5 L6  "Grashof" min  L1b L2b L5 L6  23.003





Given:


Solution:



L2  130

Length of link 4b

L4b  40

L4  225

Length of link 4

1.


2.


3.


4.


L1a  111.758

Link 2 (input)

L2  130.000

Link 3 (coupler)

L3  52.000

Link 4 (output)

L4  225.000

5.

Select a point on link 4 (O4B) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution below the distance O4C was selected to be L4b  40.000 .

6.

Draw a construction line through C2C3 and extend it downward.

7.

Select a point on this line and call it O6. In the solution below O6 was placed 20 units from the bottom of the base.

8.

9.


A

2

B 111.758

C3

O4 83.977

Link 5 (coupler) L5  83.977

1b

6 O6

92.425

A

5 D2

10.480

Link 4b (rocker) L4b  40.000

O2

1a

L6  10.480

Link 1b (ground) L1b  92.425

4

C2


3

D3

B









Design a fourbar linkage to move the object in Figure P3-13 through the three positions shown using points A and B for attachment. Add a driver dyad to limit its motion to the range of positions shown, making it a sixbar. All fixed pivots should be on the base.

Given:


Solution:


Design choices: Length of link 4b

L4b  50

1.

Draw link AB in its three design positions A1B1, A2B2, A3B3 in the plane as shown.

2.

Draw construction lines from point A1 to A2 and from point A2 to A3.

3.

Bisect line A1A2 and line A2A3 and extend their perpendicular bisectors until they intersect. Label their intersection O2.

4.

Repeat steps 2 and 3 for lines B1B2 and B2B3. Label the intersection O4.

5.

Connect O2 with A1 and call it link 2. Connect O4 with B1 and call it link 4.

6.

Line A1B1 is link 3. Line O2O4 is link 1 (ground link for the fourbar). The fourbar is now defined as O2ABO4 and has link lengths of Ground link 1a

L1a  20.736

Link 2

L2  127.287

Link 3

L3  52.000

Link 4

L4  120.254

B1

A1 3

A2 4

D3

B2

2

O6

D1

6

5 A3 C3 O2 C1

C2 O4

7.

B3







Select a point on link 4 (O4B) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution above the distance O4C was selected to be L4b  50.000 .

9.


10. Select a point on this line and call it O6. In the solution above O6 was placed 20 units from the left edge of the base. 11. Draw a circle about O6 with a radius of one-half the length C1C3 and label the intersections of the circle with the extended line as D1 and D3. In the solution below the radius was measured as L6  45.719. 12. The driver fourbar is now defined as O4CDO6 with link lengths Link 6 (crank)

L6  45.719

Link 5 (coupler) L5  126.875 Link 1b (ground) L1b  128.545 Link 4b (rocker) L4b  50.000 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L4b L5  "Grashof" min  L6 L1b L4b L5  45.719





Given:


Solution:



L2  125

Length of link 2b

L4b  50

Length of link 4

L4  140

1.


2.


3.


4.


L1a  97.195

Link 3 (coupler)

L3  86.000

Link 2 (input)

L2  125.000

Link 4 (output)

L4  140.000

5.


6.


7.


8.

9.


A1

3 2 B1 B2

6 D1

O6

L6  25.808


4

D2

O2 1a

97.195

5


A2

1b

C1

C2

25.808 130.479

137.327

O4



10. Use the link lengths in step 9 to find the Grashof condition of the driving fourbar (it must be Grashof and the shortest link must be link 6). Condition( a b c d ) 







Given:


Solution:



L2  130

Length of link 2b

L2b  50

L4  130

Length of link 4

1.


2.


3.


4.

The fourbar stage is now defined as O2ABO4 with link lengths Ground link 1a Link 3 (coupler)

5.

6. 7.

8.

9.

L1a  67.395 L3  86.000

Link 2 (input)

L2  130.000

Link 4 (output)

L4  130.000

Select a point on link 2 (O2A) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O2, C, and A.) In the solution below the distance O2C was selected to be L2b  50.000 and the link was extended away from A to give a better position for the driving dyad. Draw a construction line through C2C3 and extend it downward. Select a point on this line and call it O6. In the solution below O6 was placed 35 units from the bottom of the base. A2

Draw a circle about O6 with a radius of one-half the length C1C2 and label the intersections of the circle with the extended line as D2 and D3. In the solution below the radius was measured as 24.647 units. The driver fourbar is now defined as O2CDO6 with link lengths Link 6 (crank)

3 C3

155°

107.974

O2

C2

1a

4

67.395

1b 5


Link 2b (rocker) L2b  50.000

B2 A3

L6  24.647

Link 1b (ground) L1b  107.974

2

D3

O4 6

98.822

B3

O6

24.647 D2










Given:


Solution:


Design choices: Length of link 4b

L4b  50

1.


2.


3.


4.


5.


6.


L1a  61.667

Link 2

L2  142.357

Link 3

L3  86.000

Link 4

L4  124.668

A1 A2

3

2 B1

D3

B2 O6

O2 6

D1

4

1b

A3

1a

5 C3

O4

7.

B3 C2

C1







Select a point on link 4 (O4B) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution above the distance O4C was selected to be L4b  50.000 .

9.



L6  45.178

Link 5 (coupler) L5  140.583 Link 1b (ground) L1b  142.205 Link 4b (rocker) L4b  50.000 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L4b L5  "Grashof" 14. Unfortunately, although the solution presented appears to meet the design specification, a simple cardboard model will quickly demonstrate that it has a branch defect. That is, in the first position shown, the linkage is in the "open" configuration, but in the 2nd and 3rd positions it is in the "crossed" configuration. The linkage cannot get from one circuit to the other without removing a pin and reassembling after moving the linkage. The remedy is to attach the points A and B to the coupler, but not at the joints between links 2 and 3 and links 3 and 4.





Given:


Solution:



L2  100

Length of link 4b

L4b  40

L4  160

Length of link 4

1.


2.


3.


4.


L1a  81.463

Link 3 (coupler)

L3  52.000

Link 2 (input)

L2  100.000

Link 4 (output)

L4  160.000

5.


6.


7.


8.


A2

B1

A1 3

B2

2 14.351

9.


L6  14.351


1a

O2 C2

5

O6

1b

C1

6 132.962 O4

Link 1b (ground) L1b  138.105 Link 4b (rocker) L4b  40.000

138.105

81.463

4

D2

D1










Given:


Solution:



L2  150

Length of link 4b

L4b  50

L4  200

Length of link 4

1.


2.


3.

Select one point on each bisector and label them O2 and O4, respectively. In the solution below the distances O2A was selected to be L2  150.000 and O4B to be L4  200.000 . This resulted in a ground-link-length O2O4 for the fourbar of L1a  80.864.

4.

The fourbar stage is now defined as O2ABO4 with link lengths Ground link 1a Link 3 (coupler)

L1a  80.864 L3  52.000

Link 2 (input)

L2  150.000

Link 4 (output)

L4  200.000

5.


6.

Draw a construction line through C2C3 and extend it downward.

7.

Select a point on this line and call it O6. In the solution below O6 was placed 25 units from the bottom of the base.

8.

9.

Draw a circle about O6 with a radius of one-half the length C1C2 and label the intersections of the circle with the extended line as D2 and D3. In the solution below the radius was measured as L6  12.763. The driver fourbar is now defined as O4CDO6 with link lengths

A2 3

C2

B2

A3

4 2

O4 C3 1a

Link 6 (crank)

L6  12.763

B3


5

1b

O2 112.498

80.864

D2

122.445 O6

D3

12.763










Given:


Solution:


Design choices: L2b  40

Length of link 2b 1.


2.


3.


4.


5.


6.


L1a  53.439

Link 2

L2  134.341

Link 3

L3  52.000

Link 4

L4  90.203

A1

A2

B1 3

B2

6 D1

4

O6 2

D3 5

C1

C2

A3

O4

1b C3 O2

7.

1a B3






Condition L1a L2 L3 L4  "non-Grashof" Although this fourbar is non-Grashof, there are no toggle points within the required range of motion. 8.

9.

Select a point on link 2 (O2A) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 2 is now a ternary link with nodes at O2, C, and A.) In the solution above the distance O2C was selected to be L2b  40.000 . Draw a construction line through C1C3 and extend it to the left.


L6  29.760

Link 5 (coupler) L5  119.665 Link 1b (ground) L1b  122.613 Link 2b (rocker) L2b  40.000 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L2b L5  "Grashof"




Design a fourbar mechanism to move the link shown in Figure P3-16 from position 1 to position 2. Ignore the third position and the fixed pivots O2 and O4 shown. Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.

Given:

Position 1 offsets:

Solution:


xC1D1  3.744  in

yC1D1  2.497  in

1.

Connect the end points of the two given positions of the line CD with construction lines, i.e., lines from C1 to C2 and D1 to D2.

2.

Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of C1C2 was extended downward and the bisector of D1D2 was extended upward.

3.

Select one point on each bisector and label them O4 and O6, respectively. In the solution below the distances O4D and O6C were each selected to be 7.500 in. This resulted in a ground-link-length O4O6 for the fourbar of 15.366 in.

4.

The fourbar stage is now defined as O4CDO6 with link lengths Link 5 (coupler) L5 

2

xC1D1  yC1D1

Link 4 (input)

L4  7.500  in

Ground link 1b

L1b  15.366 in

2


L6  7.500  in

5.

Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, B, and D.) In the solution below the distance O4B was selected to be 4.000 in.

6.

Draw a construction line through B1B2 and extend it to the right.

7.

Select a point on this line and call it O2. In the solution below the distance AB was selected to be 6.000 in.

8.

Draw a circle about O2 with a radius of one-half the length B1B2 and label the intersections of the circle with the extended line as A1 and A2. In the solution below the radius was measured as 1.370 in.

9.

The driver fourbar is now defined as O2ABO4 with link lengths Link 2 (crank)

L2  1.370  in







Condition L1a L2 L3 L4a  "Grashof" min  L1a L2 L3 L4a  1.370 in

O4 4.000 7.500

B2

B1

A1 2 O2

4

4 D1 5 5

A2

3

C2

D2 15.366

C1 6 6 7.500 O6

11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4DCO6 is non-Grashoff with toggle positions at 4 = -49.9 deg and +49.9 deg. The fourbar operates between 4 = +28.104 deg and -11.968 deg.





Given:

Position 2 offsets:

Solution:


xC2D2  4.355  in

yC2D2  1.134  in

1.


2.


3.


4.

The fourbar stage is now defined as O4DCO6 with link lengths Link 5 (coupler) L5 

2

xC2D2  yC2D2

Link 4 (input)

L4  6.000  in

Ground link 1b

L1b  14.200 in

2


L6  6.000  in

5.


6.


7.


8.


9.


L2  1.271  in








6.000 O4

4.000

7.099

4 4

B1 D2 C2

B2 3

5

A1

5 C3 6.000

6

D3

2

O2

A2

6

O6

14.200





Design a fourbar mechanism to give the three positions shown in Figure P3-16. Ignore the points O2 and O4 shown. Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.

Solution:



Length of link 3:

Length of link 4b:

L4b  4.500

1.


2.


3.


4.


5.


6.

Line C1D1 is link 5. Line O6O4 is link 1a (ground link for the fourbar). The fourbar is now defined as O6CDO4 and has link lengths of Ground link 1a

L1a  2.616

Link 6

L6  6.080

Link 5

L5  4.500

Link 4

L4  6.901

D2

D1 5 5 C1

C3

C2

D3

B2

B1 6

2.765

5

4

3

4 4

6

B3

6 O4

6.080

A1 O2

2

6.901

O6 10.611

2.616

7.

Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case. Condition( a b c d ) 


A3




9.

Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, D, and B.) In the solution above the distance O4B was selected to be L4b  4.500 . Draw a construction line through B1B3 and extend it up to the right.

10. Layout the length of link 3 (design choice) along the extended line. Label the other end A. 11. Draw a circle about O2 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L2  2.765. 12. The driver fourbar is now defined as O4BAO2 with link lengths Link 2 (crank)

L2  2.765

Link 3 (coupler) L3  10.000 Link 1b (ground) L1b  10.611 Link 4b (rocker) L4b  4.500 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition L2 L3 L1b L4b  "Grashof" min  L2 L3 L1b L4b  2.765




Design a fourbar mechanism to give the three positions shown in Figure P3-16 using the fixed pivots O2 and O4 shown. (See Example 3-7.) Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.

Solution:


Design choices:

L5  5.000

Length of link 5:

L2b  2.500

Length of link 2b:

1.


2.


3.


4.


5.


6.


7.

The three inverted positions of the ground link that correspond to the three desired coupler positions are labeled O2O4, O2'O4', and O2"O4" in the first layout below and are renamed E1F1, E2F2, and E3F3, respectively, in the second layout, which is used to find the points G and H. D2 D1

C2 C3

C1

O2

O4''

O2''

O2'

O4

O4'

D3



8.


9.



L1a  3.000

Link 2

L2  8.597

Link 3

L3  1.711

Link 4

L4  7.921

G 3

H

2

4

F1

E 1 O2 1a

F3

E3

O4

F2

E2



Condition L1a L2 L3 L4  "Grashof" The fourbar that will provide the desired motion is now defined as a Grashof double crank in the crossed configuration. It now remains to add the original points C1 and D1 to the coupler GH and to define the driving dyad.



14. Select a point on link 2 (O2G) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 2 is now a ternary link with nodes at O2, B, and G.) In the solution below, the distance O2B was selected to be L2b  2.500 . 15. Draw a construction line through B1B3 and extend it up to the left. 16. Layout the length of link 5 (design choice) along the extended line. Label the other end A. 17. Draw a circle about O6 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L6  1.541. 18. The driver fourbar is now defined as O2BAO6 with link lengths Link 6 (crank)

L6  1.541

Link 5 (coupler) L5  5.000 Link 1b (ground) L1b  5.374 Link 2b (rocker) L2b  2.500 19. Use the link lengths in step 18 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L5 L1b L2b  "Grashof"

D2

D1

C2

3

D3

C3 G3

G2 C1

3

H1

3

H2

2

G1 2

B3

2

4

H3

4 4

5 O6

B1

A3 O2

A1

6

1a

O4





Given:

Position 1 offsets:

Solution:


xC1D1  1.896  in

yC1D1  1.212  in

1.


2.


3.

Select one point on each bisector and label them O4 and O6, respectively. In the solution below the distances O6C and O4D were each selected to be 6.500 in. This resulted in a ground-link-length O4O6 for the fourbar of 14.722 in.

4.


2

xC1D1  yC1D1

Link 4 (input)

L4  6.500  in

Ground link 1b

L1b  14.722 in

2


L6  6.500  in

5.


6.


7.


8.


9.


L2  0.645  in



10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition( a b c d )  S  min ( a b c d ) L  max( a b c d ) SL  S  L PQ  a  b  c  d  SL return "Grashof" if SL  PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise




6.500 4.500 B2

4 4

D2

B1

5

6.500

C2 5 C1

O6

7.472

3

6 6

O4

D1 14.722

A2 O2

2 A1

0.645

11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4CDO6 is non-Grashoff with toggle positions at 4 = -17.1 deg and +17.1 deg. The fourbar operates between 4 = +5.216 deg and -11.273 deg.





Given:

Position 2 offsets:

Solution:


xC2D2  0.834  in

yC2D2  2.090  in

1.


2.


3.


4.


2

xC2D2  yC2D2

Link 4 (input)

L4  5.000  in

Ground link 1b

L1b  12.933 in

2


L6  5.000  in

5.


6.


7.


8.


9.


L2  0.741  in







Condition L1a L2 L3 L4  "Grashof"

O4 5.500

4.000 4 B3

7.173 4 B2

D3 D2

5 C3

5

3

A3

O2 2

A2

C2 6 6

12.933

O6






Solution:



Length of link 3:

L4b  2.500

Length of link 4b:

1.


2.


3.


4.


5.


6.


L1a  1.835

Link 6

L6  2.967

Link 5

L5  2.250

Link 4

L4  3.323

D3 B3 3.323

D2

B2

4 5

4

5

C3 1.835

4 C2

O4 6 O6

6

D1

B1 3 5 C1

6

1.403 A3 O2 2

2.967

A1

6.347

7.







Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, D, and B.) In the solution above the distance O4B was selected to be L4b  2.500 .

9.

Draw a construction line through B1B3 and extend it up to the right.

10. Layout the length of link 3 (design choice) along the extended line. Label the other end A. 11. Draw a circle about O2 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L2  1.403. 12. The driver fourbar is now defined as O2ABO4 with link lengths Link 2 (crank)

L2  1.403

Link 3 (coupler) L3  6.000 Link 1b (ground) L1b  6.347 Link 4b (rocker) L4b  2.500 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition L1b L2 L3 L4b  "Grashof" min  L1b L2 L3 L4b  1.403





Solution:


Design choices:

L5  4.000

Length of link 5:

Length of link 2b:

L2b  0.791

1.


2.


3.


4.


5.


6.


7.


D3 D2

C3

D1 C2

C1 O2

O4 O2'' O2'

O4' O4''



8.


9.



L1a  3.000

Link 2

L2  0.791

Link 3

L3  1.222

Link 4

L4  1.950

E1 O2

F1 O4

1a 2

G

4

3 E3 E2

H

F2 F3



Condition L1a L2 L3 L4  "non-Grashof" The fourbar that will provide the desired motion is now defined as a non-Grashof double rocker in the crossed configuration. It now remains to add the original points C1 and D1 to the coupler GH and to define the driving dyad, which in this case will drive link 4 rather than link 2. 14. Select a point on link 2 (O2G) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 2 is now a ternary link with nodes at O2, B, and G.) In the solution below, the distance O2B was selected to be L2b  0.791 . Thus, in this case B and G coincide. 15. Draw a construction line through B1B3 and extend it up to the left. 16. Layout the length of link 5 (design choice) along the extended line. Label the other end A. 17. Draw a circle about O6 with a radius of one-half the length B1B3 and label the intersections of the circle



with the extended line as A1 and A3. In the solution below the radius was measured as L6  0.727.

18. The driver fourbar is now defined as O2BAO6 with link lengths Link 6 (crank)

L6  0.727

Link 5 (coupler) L5  4.000 Link 1b (ground) L1b  4.012 Link 2b (rocker) L2b  0.791 19. Use the link lengths in step 18 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L5 L1b L2b  "Grashof"

D3 D2 A3 O6

D1

6 C3

A1

C2 3

5

3

1b

3

C1

4 O2

O4

4

2 G

H

4





Given:

Position 1 offsets:

Solution:


xC1D1  1.591  in

yC1D1  1.591  in

1.


2.


3.

Select one point on each bisector and label them O4 and O6, respectively. In the solution below the distances O4C and O6D were each selected to be 5.000 in. This resulted in a ground-link-length O4O6 for the fourbar of 10.457 in.

4.


2

xC1D1  yC1D1

Link 4 (input)

L4  5.000  in

Ground link 1b

L1b  10.457 in

2


L6  5.000  in

5.

Select a point on link 4 (O4C) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, B, and C.) In the solution below the distance O4B was selected to be 3.750 in.

6.


7.


8.


9.


L2  0.882  in





Condition L1a L2 L3 L4a  "Grashof"



O6

6 6

10.457

C2

5

5

A2 O2 2

C1

B2

B1

A1

3

4

4 5.000

D1

D2

3.750

7.020 O4

11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4CDO6 is non-Grashoff with toggle positions at 4 = -38.5 deg and +38.5 deg. The fourbar operates between 4 = +15.206 deg and -12.009 deg.





Given:

Position 2 offsets:

Solution:


xC2D2  2.053  in

yC2D2  0.920  in

1.


2.


3.


4.


2

xC2D2  yC2D2

Link 4 (input)

L4  5.000  in

Ground link 1b

L1b  8.773  in

2


L6  5.000  in

5.


6.


7.


8.


9.


L2  0.892  in







Condition L1a L2 L3 L4a  "Grashof"

7.019

O4

5.000

4

3.750 4 B3

B2 D2

D3 5 8.773

C3

A3 O2 2

A2

3

5 C2

6 6

O6

11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4DCO6 is non-Grashoff with toggle positions at 4 = -55.7 deg and +55.7 deg. The fourbar operates between 4 = -7.688 deg and -35.202 deg.





Solution:



Length of link 3:

Length of link 4b:

L4b  5.000

1.


2.


3.


4.


5.


6.


L1a  8.869

Link 6

L6  1.831

Link 5

L5  2.250

Link 4

L4  6.953

7.646

O4 4

O2 A1

6.953 8.869

B3

4 B1

2 1.593

D3 D2

D 1 C2 5 C1

3

A3

5 C3

6 O6

6 1.831

7.






Condition L6 L1a L4 L5  "non-Grashof" 8.

9.

Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, D, and B.) In the solution above the distance O4B was selected to be L4b  5.000 . Draw a construction line through B1B3 and extend it up to the right.

10. Layout the length of link 3 (design choice) along the extended line. Label the other end A. 11. Draw a circle about O2 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L2  1.593. 12. The driver fourbar is now defined as O2ABO4 with link lengths Link 2 (crank)

L2  1.593

Link 3 (coupler) L3  6.000 Link 1b (ground) L1b  7.646 Link 4b (rocker) L4b  5.000 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition L1b L2 L3 L4b  "Grashof" min  L1b L2 L3 L4b  1.593





Solution:


Design choices:

Length of link 5:

L5  4.000

Length of link 2b:

L2b  2.000

1.


2.


3.


4.


5.


6.


7.


D1 D2

D3

C1 C2

O2''

O4'

C3 O2

O4 O2'

O4''

8.


9.


10. Repeat steps 2 and 3 for lines F1F2 and F2F3. Label the intersection H.



11. Connect E1 with G and label it link 2. Connect F1 with H and label it link 4. Reinverting, E1 and F1 are the original fixed pivots O2 and O4, respectively. 12. Line GH is link 3. Line O2O4 is link 1a (ground link for the fourbar). The fourbar is now defined as O2GHO4 and has link lengths of Ground link 1a

L1a  4.000

Link 2

L2  2.000

Link 3

L3  6.002

Link 4

L4  7.002

H

3

E3

4

F2

G 2 O2

O4 E 2 F1

E1

F3



Condition L1a L2 L3 L4  "Grashof" The fourbar that will provide the desired motion is now defined as a non-Grashof crank rocker in the open configuration. It now remains to add the original points C1 and D1 to the coupler GH and to define the driving dyad, which in this case will drive link 4 rather than link 2.



14. Select a point on link 2 (O2G) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 2 is now a ternary link with nodes at O2, B, and G.) In the solution below, the distance O2B was selected to be L2b  2.000 . Thus, in this case B and G coincide. 15. Draw a construction line through B1B3 and extend it up to the left. 16. Layout the length of link 5 (design choice) along the extended line. Label the other end A. 17. Draw a circle about O6 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L6  1.399. 18. The driver fourbar is now defined as O2BAO6 with link lengths L6  1.399

Link 6 (crank)

Link 5 (coupler) L5  4.000 Link 1b (ground) L1b  4.257 Link 2b (rocker) L2b  2.000 19. Use the link lengths in step 18 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L2b L5  "Grashof" H1

H2

3 D1

3 D2

H3

D3 C1 C2 4 4 G2

2

3

C3 2

G1

O2

1a 2

1b

G3 5

A1 6 O6

A3

4 O4




Design a fourbar Grashof crank-rocker for 120 degrees of output rocker motion with a quick-return time ratio of 1:1.2. (See Example 3-9.)

Given:

Time ratio

Solution: 1.

Tr 

1 1.2



β 

α

α  β = 360  deg

β 360  deg

β  196  deg

1  Tr

α  360  deg  β

α  164  deg

δ  β  180  deg

δ  16 deg

2.


3.

Layoff a line through B1 at an arbitrary angle (but not zero deg). In the solution below the line is 60 deg to the horizontal.

0. 95 3

=

c

90.00°

B2 B2

B1

B1 4 O4

d

O4 3

4.4 91 =

b

3.8 33 =

0° .0 16

LAYOUT

A2

A1 2

O2

a

LINKAGE DEFINITION 0.2 55 =

O2



4.


5.


6.


d  3.833  in

Coupler (3)

b  4.491  in

Crank (2)

a  0.255  in

Rocker (4)

c  0.953  in





Given:

Time ratio

Solution: 1.

Tr 

1 1.5



β 

α

α  β = 360  deg

β 360  deg

β  216 deg

1  Tr

α  360  deg  β

α  144 deg

δ  β  180  deg

δ  36 deg

2.


3.

Layoff a line through B1 at an arbitrary angle (but not zero deg). In the solution below the line is 20 deg to the horizontal.

4.


5.


B1

B2

B2

B1

3.0524 = b O4

3

O2

4 1.2694 = a

2

A1

LAYOUT O4

O2

2.0000 = c

A2 LINKAGE DEFINITION 2.5364 = d


6.



d  2.5364 in

Coupler (3)

b  3.0524 in

Crank (2)

a  1.2694 in

Rocker (4)

c  2.000  in





Given:

Time ratio

Solution: 1.

Tr 

1 1.33



β 

α

α  β = 360  deg

β 360  deg

β  205  deg

1  Tr

α  360  deg  β

α  155  deg

δ  β  180  deg

δ  25 deg

2.


3.

Layoff a line through B1 at an arbitrary angle (but not zero deg). In the solution below the line is 150 deg to the horizontal. 2. 00 0= c

90.00°

B2

B2

B1

B1 4

O4 25.0 0°

O4 3

6.2 32 =

b

4.7 63 =

d

LAYOUT

A2

A1 2

LINKAGE DEFINITION

a

O2

0.4 35 =

O2



4.


5.


6.


d  4.763  in

Coupler (3)

b  6.232  in

Crank (2)

a  0.435  in

Rocker (4)

c  2.000  in




Design a sixbar drag link quick-return linkage for a time ratio of 1:4 and output rocker motion of 50 degrees. (See Example 3-10.)

Given:

Time ratio

Solution: 1.

Tr 

1 4


Determine the crank rotation angles  and  from equation 3.1. Tr = Solving for and 

β 

α

α  β = 360  deg

β 360  deg 1  Tr

α  360  deg  β

β  288 deg α  72 deg

2.

Draw a line of centers XX at any convenient location.

3.

Choose a crank pivot location O2 on line XX and draw an axis YY perpendicular to XX through O2.

4.

Draw a circle of convenient radius O2A about center O2. In the solution below, the length of O2A is a  1.000  in.

5.

Lay out angle  with vertex at O2, symmetrical about quadrant one.

6.

Label points A1 and A2 at the intersections of the lines subtending angle  and the circle of radius O2A.

7.

8.

Set the compass to a convenient radius AC long enough to cut XX in two places on either side of O2 when swung from both A1 and A2. Label the intersections C1 and C2. In the solution below, the length of AC is b  2.000  in. The line O2A is the driver crank, link 2, and the line AC is the coupler, link 3.

9.

The distance C1C2 is twice the driven (dragged) crank length. Bisect it to locate the fixed pivot O4.

10. The line O2O4 now defines the ground link. Line O4C is the driven crank, link 4. In the solution below, O4C measures c  2.282  in and O2O4 measures d  0.699  in. 11. Calculate the Grashoff condition. If non-Grashoff, repeat steps 7 through 11 with a shorter radius in step 7. Condition( a b c d ) 


Condition( a b c d )  "Grashof" 12. Invert the method of Example 3-1 to create the output dyad using XX as the chord and O4C1 as the driving crank. The points B1 and B2 will lie on line XX and be spaced apart a distance that is twice the length of O4C (link 4). The pivot point O6 will lie on the perpendicular bisector of B1B2 at a distance from XX which subtends the specified output rocker angle, which is 50 degrees in this problem. In the solution below, the length BC was chosen to be e  5.250  in.



9.000°

72.000°

LAYOUT OF SIXBAR DRAG LINK QUICK RETURN WITH TIME RATIO OF 1:4 a = 1.000 b = 2.000 c = 2.282 d = 0.699 e = 5.250 f = 5.400

13. For the design choices made (lengths of links 2, 3 and 5), the length of the output rocker (link 6) was measured as f  5.400  in.




Design a crank-shaper quick-return mechanism for a time ratio of 1:2.5 (Figure 3-14, p. 112).

Given:

Time ratio

Solution:


TR 

1 2.5

Design choices:

1.

Length of link 2 (crank)

L2  1.000

Length of link 5 (coupler)

L5  5.000

S  4.000

Length of stroke

Calculate  from equations 3.1. TR 

α β

α  β  360  deg

α 

360  deg 1

α  102.86 deg

1 TR

2.

Draw a vertical line and mark the center of rotation of the crank, O2, on it.

3.

Layout two construction lines from O2, each making an angle /2 to the vertical line through O2.

4.

Using the chosen crank length (see Design Choices), draw a circle with center at O2 and radius equal to the crank length. Label the intersections of the circle and the two lines drawn in step 3 as A1 and A2.

5.

Draw lines through points A1 and A2 that are also tangent to the crank circle (step 2). These two lines will simultaneously intersect the vertical line drawn in step 2. Label the point of intersection as the fixed pivot center O4.

6.

Draw a vertical construction line, parallel and to the right of O2O4, a distance S/2 (one-half of the output stroke length) from the line O2O4.

7.

Extend line O4A1 until it intersects the construction line drawn in step 6. Label the intersection B1.

8.

Draw a horizontal construction line from point B1, either to the left or right. Using point B1 as center, draw an arc of radius equal to the length of link 5 (see Design Choices) to intersect the horizontal construction line. Label the intersection as C1.

9.

Draw the slider blocks at points A1 and C1 and finish by drawing the mechanism in its other extreme position.

STROKE 4.000 2.000 C2

6

C1

B2

B1

5

O2 4

2 A2

3 O4

A1




Design a sixbar, single-dwell linkage for a dwell of 70 deg of crank motion, with an output rocker motion of 30 deg using a symmetrical fourbar linkage with the following parameter values: ground link ratio = 2.0, common link ratio = 2.0, and coupler angle  = 40 deg. (See Example 3-13.)

Given:

Crank dwell period: 70 deg. Output rocker motion: 30 deg. Ground link ratio, L1/L2 = 2.0: GLR  2.0 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.0: CLR  2.0 Coupler angle, γ  40 deg

Design choice: Crank length, L2  2.000 Solution: 1.

See Figures 3-20 and 3-21 and Mathcad file P0372.

For the given design choice, determine the remaining link lengths and coupler point specification. Coupler link (3) length

L3  CLR L2

L3  4.000


L4  CLR L2

L4  4.000


L1  GLR  L2

L1  4.000

Angle PAB

δ 


180  deg  γ 2

AP  2  L3 cos δ

δ  70.000 deg AP  2.736

Enter the above data into program FOURBAR, plot the coupler curve, and determine the coordinates of the coupler curve in the selected range of crank motion, which in this case will be from 145 to 215 deg.



FOURBAR for Windows

3.

File

P03-72

Angle Coupler Pt Step X Deg in

Coupler Pt Y in

Coupler Pt Mag in

Coupler Pt Ang in

145 150 155 160 165 170 175 180 185 190 195 200 205 210 215

3.818 3.661 3.494 3.319 3.135 2.945 2.749 2.547 2.342 2.133 1.923 1.711 1.499 1.289 1.080

4.422 4.360 4.295 4.226 4.156 4.083 4.009 3.935 3.859 3.783 3.707 3.631 3.555 3.479 3.403

120.297 122.895 125.549 128.259 131.025 133.846 136.723 139.655 142.639 145.674 148.757 151.886 155.055 158.261 161.498

-2.231 -2.368 -2.497 -2.617 -2.728 -2.829 -2.919 -2.999 -3.067 -3.124 -3.169 -3.202 -3.223 -3.232 -3.227


y

145 P

B

180

3 4 D

215 A 2

x

PSEUDO-ARC O2

4.

O4

The position of the end of link 5 at point D will remain nearly stationary while the crank moves from 145 to 215 deg. As the crank motion causes the coupler point to move around the coupler curve there will be another extreme position of the end of link 5 that was originally at D. Since a symmetrical linkage was chosen, the other extreme position will be located along a line through the axis of symmetry (see Figure 3-20) a distance equal to the length of link 5 measured from the point where the axis of symmetry intersects the coupler curve near the 0 deg coupler point. Establish this point and label it E.



FOURBAR for Windows

File

P03-72

Angle Step Deg

Coupler Pt X in

Coupler Pt Y in

Coupler Pt Mag n

Coupler Pt Ang in

340.000 345.000 350.000 355.000 0.000 5.000 10.000 15.000

-0.718 -0.615 -0.506 -0.386 -0.255 -0.117 0.022 0.155

0.175 0.481 0.818 1.178 1.549 1.917 2.269 2.598

0.739 0.781 0.962 1.240 1.570 1.920 2.269 2.603

166.325 142.001 121.717 108.135 99.365 93.499 89.434 86.581

y 145 P

B 5

180

3 5

AXIS OF SYMMETRY

4 D

215

355 A

E

2

x

PSEUDO-ARC

O4

O2

5.

The line segment DE represents the maximum displacement that a link of the length equal to link 5, attached at P, will reach along the axis of symmetry. Construct a perpendicular bisector of the line segment DE and extend it to the right (or left, which ever is convenient). Locate fixed pivot O6 on the bisector of DE such that the lines O6D and O6E subtend the desired output angle, in this case 30 deg. Draw link 6 from D through O6 and extend it to any convenient length. This is the output link that will dwell during the specified motion of the crank. SUMMARY OF LINKAGE SPECIFICATIONS Original fourbar: O6

y 6

145 P

B 5

180

30.000° 3 5

L2  2.000

Coupler

L3  4.000

Rocker

L4  4.000

Coupler point

AP  2.736 δ  70.000 deg

E 2

x O2

Crank

Added dyad:

355 A

L1  4.000

4 D

215

Ground link

O4

Coupler

L5  3.840

Output

L6  5.595

Pivot O6

x  3.841 y  5.809





Given:





L3  CLR L2

L3  5.000


L4  CLR L2

L4  5.000


L1  GLR  L2

L1  4.000

Angle PAB

δ 


180  deg  γ 2

AP  2  L3 cos δ

δ  60.000 deg AP  5.000




FOURBAR for Windows

3.

File

P03-73


Coupler Pt Y in

Coupler Pt Mag in

Coupler Pt Ang in

130 140 150 160 170 180 190 200 210 220 230

6.449 6.171 5.840 5.464 5.047 4.598 4.123 3.631 3.130 2.629 2.138

6.812 6.695 6.559 6.408 6.244 6.071 5.892 5.709 5.523 5.336 5.146

108.774 112.833 117.078 121.493 126.060 130.765 135.588 140.504 145.482 150.482 155.454

-2.192 -2.598 -2.986 -3.347 -3.675 -3.964 -4.209 -4.405 -4.551 -4.643 -4.681


y 130 P

B

180

D

230

4

3

PSEUDO-ARC

A 2

x O2

4.

O4




FOURBAR for Windows

File

P03-73


Coupler Pt Y in

Coupler Pt Mag in

Coupler Pt Ang in

340 350 0 10 20

1.429 2.316 3.316 4.265 5.047

3.013 3.237 3.746 4.414 5.078

151.688 134.332 117.727 104.920 96.371

-2.652 -2.262 -1.743 -1.137 -0.564

y 130 P 20 180

B

10 5

AXIS OF SYMMETRY

0

D

350

230

4

3 340 A

PSEUDO-ARC

E

2

x O4

O2

5.

The line segment DE represents the maximum displacement that a link of the length equal to link 5, attached at P, will reach along the axis of symmetry. Construct a perpendicular bisector of the line segment DE and extend it to the right (or left, which ever is convenient). Locate fixed pivot O6 on the bisector of DE such that the lines O6D and O6E subtend the desired output angle, in this case 30 deg. Draw link 6 from D through O6 and extend it to any convenient length. This is the output link that will dwell during the specified motion of the crank. SUMMARY OF LINKAGE y SPECIFICATIONS 130

Original fourbar: P

Ground link

L1  4.000

Crank

L2  2.000

Coupler

L3  5.000

Rocker

L4  5.000

Coupler point

AP  5.000

20 180

50.000°

B

10 5 0

O6 6

230

δ  60.000 deg

D

350

Added dyad:

4

3 340 PSEUDO-ARC

A

E

2

x O2

O4

Coupler

L5  5.395

Output

L6  2.998

Pivot O6

x  3.166 y  3.656





Given:





L3  CLR L2

L3  3.500


L4  CLR L2

L4  3.500


L1  GLR  L2

L1  4.000

Angle PAB

δ 


180  deg  γ 2

AP  2  L3 cos δ

δ  55.000 deg AP  4.015




FOURBAR for Windows

3.

File

P03-74


Coupler Pt Y in

Coupler Pt Mag in

Coupler Pt Ang in

140 150 160 170 180 190 200 210 220

5.208 4.940 4.645 4.332 4.005 3.668 3.322 2.969 2.613

5.252 5.032 4.804 4.578 4.359 4.152 3.958 3.779 3.612

97.395 100.971 104.781 108.860 113.242 117.942 122.946 128.210 133.663

-0.676 -0.958 -1.226 -1.480 -1.720 -1.945 -2.153 -2.337 -2.493


y

140 P 180

220

B

PSEUDO-ARC

4

3 A

O4

2 O2

4.

x D




FOURBAR for Windows

File

P03-74


Coupler Pt Y in

Coupler Pt Mag in

Coupler Pt Ang in

340 350 0 10 20

1.658 2.360 3.147 3.886 4.490

2.158 2.562 3.185 3.887 4.530

129.810 112.856 98.919 88.916 82.372

-1.382 -0.995 -0.494 0.074 0.601

y

140 P 20 180

10

0

220

AXIS OF SYMMETRY

B

350 PSEUDO-ARC

A

4

340 3

O4

2 O2

x

D E

5.

The line segment DE represents the maximum displacement that a link of the length equal to link 5, attached at P, will reach along the axis of symmetry. Construct a perpendicular bisector of the line segment DE and extend it to the right (or left, which ever is convenient). Locate fixed pivot O6 on the bisector of DE such that the lines O6D and O6E subtend the desired output angle, in this case 30 deg. Draw link 6 from D through O6 and extend it to any convenient length. This is the output link that will dwell during the specified motion of the crank.



y

SUMMARY OF LINKAGE SPECIFICATIONS Original fourbar:

140 P 20 180

10

0

220

Ground link

L1  4.000

Crank

L2  2.000

Coupler

L3  3.500

Rocker

L4  3.500

Coupler point

AP  4.015

B

δ  55.000 deg

350 PSEUDO-ARC

A

45.000°

340 3

4

O4

2 O2

O6

x

Added dyad:

D E

Coupler

L5  7.676

Output

L6  1.979

Pivot O6

x  6.217 y  0.653




Using the method of Example 3-11, show that the sixbar Chebychev straight-line linkage of Figure P2-5 is a combination of the fourbar Chebychev straight-line linkage of Figure 3-29d and its Hoeken's cognate of Figure 3-29e. See also Figure 3-26 for additional information useful to this solution. Graphically construct the Chebychev sixbar parallel motion linkage of Figure P2-5a from its two fourbar linkage constituents and build a physical or computer model of the result.

Solution:

See Figures P2-5, 3-29d, 3-29e, and 3-26 and Mathcad file P0375.

1.

Following Example 3-11and Figure 3-26 for the Chebyschev linkage of Figure 3-29d, the fixed pivot OC is found by laying out the triangle OAOBOC, which is similar to A1B1P. In this case, A1B1P is a striaght line with P halfway between A1 and B1 and therefore OAOBOC is also a straightline with OC halfway between OA and OB. As shown below and in Figure 3-26, cognate #1 is made up of links numbered 1, 2, 3, and 4. Cognate #2 is links numbered 1, 5, 6, and 7. Cognate #3 is links numbered 1, 8, 9, and 10.

3

P

3 3

B1

2

4

9

OC

B2

4

6

5 1

1

6

Links Removed

10 OA

2.

2

6

7

8

A3

4

B2

P2

A1

P1

6

9 B3

B1

A1

OB

A2

5 1

A2 OA

OC

OB

Discard cognate #3 and shift link 5 from the fixed pivot OB to OC and shift link 7 from OC to OB. Note that due to the symmetry of the figure above, L5 = 0.5 L3, L6 = L2, L7 = 0.5 L2 and OCOB = 0.5 OAOB. Thus, cognate #2 is, in fact, the Hoeken straight-line linkage. The original Chebyschev linkage with the Hoeken linkage superimposed is shown above right with the link 5 rotated to 180 deg. Links 2 and 6 will now have the same velocity as will 7 and 4. Thus, link 5 can be removed and link 6 can be reduced to a binary link supported and constrained by link 4. The resulting sixbar is the linkage shown in Figure P2-5.




Design a driver dyad to drive link 2 of the Evans straigh-line linkage in Figure 3-29f from 150 deg to 210 deg. Make a model of the resulting sixbar linkage and trace the couple curve.

Given:

Output angle

Solution:

See Figjre 3-29f, Example 3-1, and Mathcad file P0376.

Design choices:

θ  60 deg

Link lengths:

L2  2.000

Link 2

Link 5

L5  3.000

1.

Draw the input link O2A in both extreme positions, A1 and A2, at the specified angles such that the desired angle of motion 2 is subtended.

2.

Draw the chord A1A2 and extend it in any convenient direction. In this solution it was extended downward.

3.

Layout the distance A1C1 along extended line A1A2 equal to the length of link 5. Mark the point C1.

4.

Bisect the line segment A1A2 and layout the length of that radius from point C1 along extended line A1A2. Mark the resulting point O6 and draw a circle of radius O6C1 with center at O6.

5.

Label the other intersection of the circle and extended line A1A2, C2.

6.

7.

A1

Measure the length of the crank (link 6) as O6C1 or O6C2. From the graphical solution, L6  1.000 Measure the length of the ground link (link 1) as O2O6. From the graphical solution, L1  3.073

P2 3 B1 , B 2

2 O2

4 A2

P1

1

5 C1

O4

6 O6

3.073" C2

8.


2.932"



0.922" L1 = 2.4 L2 = 2 L3 = 3.2 L4 = 2.078 L5 = 3.00 L6 = 1.00 AP = 5.38




Design a driver dyad to drive link 2 of the Evans straigh-line linkage in Figure 3-29g from -40 deg to 40 deg. Make a model of the resulting sixbar linkage and trace the couple curve.

Given:

Output angle

Solution:

See Figjre 3-29G, Example 3-1, and Mathcad file P0377.

Design choices:

θ  80 deg

Link lengths:

L2  2.000

Link 2

Link 5

L5  3.000

1.

Draw the input link O2A in both extreme positions, A1 and A2, at the specified angles such that the desired angle of motion 2 is subtended.

2.

Draw the line A1C1 and extend it in any convenient direction. In this solution it was extended at a 30-deg angle from A1O2 (see note below) .

3.

Layout the distance A1C1 along extended line A1C1 equal to the length of link 5. Mark the point C1.

4.

Bisect the line segment A1A2 and layout the length of that radius from point C1 along extended line A1C1. Mark the resulting point O6 and draw a circle of radius O6C1 with center at O6.

5.

6.

7.

C2

O6 P2

Extend a line from A2 through O6. Label the other intersection of the circle and extended line A2O6, C2. Measure the length of the crank (link 6) as O6C1 or O6C2. From the graphical solution, L6  1.735

6

3.165"

C1

Measure the length of the ground link (link 1) as O2O6. From the graphical solution, L1  3.165


O2 B2

B1

3 2

4 A1

1

O4

P1



A2 5

Note: If the angle between link 2 and link 5 is zero the resulting driving fourbar will be a special Grashof. For angles greater than zero but less than 33.68 degrees it is a Grashof crank-rocker. For angles greater than 33.68 it is a non-Grashof double rocker. 8.

L1 = 4.61 L2 = 2 L3 = 2.4 L4 = 2.334 L5 = 3.00 L6 = 1.735 AP = 3.00




Figure 6 on page ix of the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD) shows a 50-point coupler that was used to generate the curves in the atlas. Using the definition of the vector R given in Figure 3-17b of the text, determine the 10 possible pairs of values of  and R for the first row of points above the horizontal axis if the gridpoint spacing is one half the length of the unit crank.

Given:

Grid module g  0.5

Solution:

See Figure 6 H&N Atlas, Figure 3-17b, and Mathcad file P0378.

1.

The moving pivot point is located on the 3rd grid line from the bottom and the third grid line from the left when the crank angle is  radians. Let the number of horizontal grid spaces from the left end of the coupler to the coupler point be n  2 1  7 and the number of vertical grid spaces from the coupler to the coupler point be m  2 1  2

2.

For the first row of points above the horizontal axis shown in Figure 6, n  2 1  7 and m  1.

3.

The angle, , between the coupler and the line from the coupler/crank pivot to the coupler point is

π π ϕ( m n )  if  n  0 atan2( n m) if  m = 0 0 if  m  0     



4.



2

2



The distance, R, from the pivot to the coupler point along the same line is 2

R( m n )  g  m  n

2

ϕ( m n ) n 

5.



deg



R( m n ) 

-2.000

153.435

1.118

-1.000

135.000

0.707

0.000

90.000

0.500

1.000

45.000

0.707

2.000

26.565

1.118

3.000

18.435

1.581

4.000

14.036

2.062

5.000

11.310

2.550

6.000

9.462

3.041

7.000

8.130

3.536

The coupler point distance, R, like the link lengths A, B, and C is a ratio of the given length to the the length of the driving crank.




The set of coupler curves in the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD, page 16 of the PDF file) has A = B = C = 1.5. Model this linkage with program FOURBAR using the coupler point fartherest to the left in the row shown on page 1 and plot the resulting coupler curve.

Given:

A  1.5

Solution:

See Figure on page 1 H&N Atlas, Figure 3-17b, and Mathcad file P0379.

B  1.5

C  1.5

1.


2.

For the second column of points to the left of the coupler pivot and the second row of points above the horizontal axis n  2 and m  2. The grid spacing is g  0.5

3.





4.



2

6.

2

2



R( m n )  1.414

Determine the values needed for input to FOURBAR. Link 2 (Crank)

a  1

Link 3 (Coupler)

b  A  a

b  1.500

Link 4 (Rocker)

c  B a

c  1.500

Link 1 (Ground)

d  C a

d  1.500

Distance to coupler point

R( m n )  1.414

Angle from link 3 to coupler point

ϕ( m n )  135.000 deg

Calculate the coordinates of O4. Let the angle between links 2 and 3 be  , then

 A 2  ( 1  C) 2  B2   2  A  ( 1  C) 

7.

2

The distance from the pivot to the coupler point, R, along the same line is R( m n )  g  m  n

5.



α  acos

α  33.557 deg

xO4  C cos α

xO4  1.250

yO4  C sin α

yO4  0.829

Enter this data into FOURBAR and then plot the coupler curve. (See next page)

ϕ( m n )  135.000 deg






The set of coupler curves on page 17 in the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD, page 32 of the PDF file) has A = 1.5, B = C = 3.0. Model this linkage with program FOURBAR using the coupler point fartherest to the right in the row shown and plot the resulting coupler curve.

Given:

A  1.5

Solution:


B  3.0

C  3.0

1.


2.

For the fifth column of points to the right of the coupler pivot and the first row of points above the horizontal axis n  5 and m  1. The grid spacing is g  0.5

3.





4.



2

6.

2

2



R( m n )  2.550


a  1

Link 3 (Coupler)

b  A  a

b  1.500

Link 4 (Rocker)

c  B a

c  3.000

Link 1 (Ground)

d  C a

d  3.000


R( m n )  2.550


ϕ( m n )  11.310 deg


 A 2  ( 1  C) 2  B2   2  A  ( 1  C) 

7.

2


5.



α  acos

α  39.571 deg


xO4  2.313

yO4  C sin α

yO4  1.911


ϕ( m n )  11.310 deg






The set of coupler curves on page 21 in the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD, page 36 of the PDF file) has A = 1.5, B = C = 3.5. Model this linkage with program FOURBAR using the coupler point fartherest to the right in the row shown and plot the resulting coupler curve.

Given:

A  1.5

Solution:


B  3.5

C  3.5

1.


2.

For the fourth column of points to the right of the coupler pivot and the second row of points above the horizontal axis n  4 and m  2. The grid spacing is g  0.5

3.





4.



2

6.

2

2



R( m n )  2.236


a  1

Link 3 (Coupler)

b  A  a

b  1.500

Link 4 (Rocker)

c  B a

c  3.500

Link 1 (Ground)

d  C a

d  3.500


R( m n )  2.236


ϕ( m n )  26.565 deg


 A 2  ( 1  C) 2  B2   2  A  ( 1  C) 

7.

2


5.



α  acos

α  40.601 deg


xO4  2.657

yO4  C sin α

yO4  2.278


ϕ( m n )  26.565 deg






The set of coupler curves on page 34 in the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD, page 49 of the PDF file) has A = 2.0, B = 1.5, C = 2.0. Model this linkage with program FOURBAR using the coupler point fartherest to the right in the row shown and plot the resulting coupler curve.

Given:

A  2.0

Solution:


B  1.5

C  2.0

1.


2.

For the sixth column of points to the right of the coupler pivot and the first row of points below the horizontal axis n  6 and m  1. The grid spacing is g  0.5

3.





4.



2

6.

2

2



R( m n )  3.041


a  1

Link 3 (Coupler)

b  A  a

b  2.000

Link 4 (Rocker)

c  B a

c  1.500

Link 1 (Ground)

d  C a

d  2.000


R( m n )  3.041


ϕ( m n )  9.462 deg


 A 2  ( 1  C) 2  B2   2  A  ( 1  C) 

7.

2


5.



α  acos

α  26.384 deg


xO4  1.792

yO4  C sin α

yO4  0.889


ϕ( m n )  9.462 deg






The set of coupler curves on page 115 in the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD, page 130 of the PDF file) has A = 2.5, B = 1.5, C = 2.5. Model this linkage with program FOURBAR using the coupler point fartherest to the right in the row shown and plot the resulting coupler curve.

Given:

A  2.5

Solution:


B  1.5

C  2.5

1.


2.

For the second column of points to the right of the coupler pivot and the second row of points below the horizontal axis n  2 and m  2. The grid spacing is g  0.5

3.





4.



2

6.

2

2



R( m n )  1.414


a  1

Link 3 (Coupler)

b  A  a

b  2.500

Link 4 (Rocker)

c  B a

c  1.500

Link 1 (Ground)

d  C a

d  2.500


R( m n )  1.414


ϕ( m n )  45.000 deg


 A 2  ( 1  C) 2  B2   2  A  ( 1  C) 

7.

2


5.



α  acos

α  21.787 deg


xO4  2.321

yO4  C sin α

yO4  0.928


ϕ( m n )  45.000 deg






Design a fourbar mechanism to move the link shown in Figure P3-19 from position 1 to position 2. Ignore the third position and the fixed pivots O2 and O4 shown. Build a cardboard model that demonstrates the required movement.

Given:

Position 1 offsets:

Solution:

See figure below and Mathcad file P0384 for one possible solution.

xC1D1  17.186 in

yC1D1  0.604  in

1.


2.

Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of C1C2 was extended upward and the bisector of D1D2 was also extended upward.

3.

Select one point on each bisector and label them O2 and O4, respectively. In the solution below the distances O2C and O4D were selected to be 15.000 in. and 8.625 in, respectively. This resulted in a ground-link-length O2O4 for the fourbar of 9.351 in.

4.

The fourbar is now defined as O2CDO4 with link lengths Link 3 (coupler) L3 

2

xC1D1  yC1D1

Link 2 (input)

L2  14.000 in

Ground link 1

L1  9.351  in

2

L3  17.197 in Link 4 (output)

L4  7.000  in

9.35 1

15 .00 0

O2

O4

17.197

8.6 25

D2

C1

D1

C2




Design a fourbar mechanism to move the link shown in Figure P3-19 from position 2 to position 3. Ignore the first position and the fixed pivots O2 and O4 shown. Build a cardboard model that demonstrates the required movement.

Given:

Position 2 offsets:

Solution:

See figure below and Mathcad file P0385 for one possible solution.

xC2D2  15.524 in

yC2D2  7.397  in

1.


2.

Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of C2C3 was extended upward and the bisector of D2D3 was also extended upward.

3.

Select one point on each bisector and label them O2 and O4, respectively. In the solution below the distances O2C and O4D were selected to be 15.000 in and 8.625 in, respectively. This resulted in a ground-link-length O2O4 for the fourbar of 9.470 in.

4.


2

xC2D2  yC2D2

Link 2 (input)

L2  15.000 in

Ground link 1b

L1b  9.470  in

2

L3  17.196 in Link 4 (output)

L6  8.625  in

8.625

D3

9.47 0

O2 O4

15.000

D2 96 17.1

C3

C2







Design a fourbar mechanism to give the three positions shown in Figure P3-19. Ignore the points O2 and O4 shown. Build a cardboard model that has stops to limit its motion to the range of positions designed.

Solution:


1.


2.


3.


4.


5.


6.

Line C1D1 is link 3. Line O2O4 is link 1 (ground link for the fourbar). The fourbar is now defined as O2CDO4 an has link lengths of Ground link 1

L1  9.187

Link 2

L2  14.973

Link 3

L3  17.197

Link 4

L4  8.815 D3

8.815

9.18 7 O2

14 .97 3

O4

2 4

D2

17.197 C1

D1

3

C3 C2




Design a fourbar mechanism to give the three positions shown in Figure P3-17 using the fixed pivots O2 and O4 shown. (See Example 3-7.) Build a cardboard model that has stops to limit its motion to the range of positions designed.

Solution:


1.


2.


3.


4.


5.


6.


7.

The three inverted positions of the ground link that correspond to the three desired coupler positions are labeled O2O4, O2'O4', and O2"O4" in the first layout below and are renamed E1F1, E2F2, and E3F3, respectively, in the second layout, which is used to find the points G and H. D3

O'2 O2

O"2

O4

O'4

D2

C1

O" 4

D1 C3

C2

First layout for steps 1 through 7



E2 O'2 E1 O2 E3 O" 2

F1 O4

O'4 F2 2 4 F3 O"4 3

G

H

Second layout for steps 8 through 12 8.


9.



L1a  9.216

Link 2

L2  16.385

Link 3

L3  18.017

Link 4

L4  8.786





Condition L1a L2 L3 L4  "non-Grashof" The fourbar that will provide the desired motion is now defined as a non-Grashof double rocker in the open configuration. It now remains to add the original points C1 and D1 to the coupler GH.

9.21 6

O2

16 .38 5

O4

4

C1

3

D1 H

G

18.017

8.786

2


SOLUTION MANUAL 4-7a-1

PROBLEM 4-7a Statement:

Given:

The link lengths and value of 2 for some fourbar linkages are defined in Table P4-1. The linkage configuration and terminology are shown in Figure P4-1. For row a, find all possible solutions (both open and crossed) for angles 3 and 4 using the vector loop method. Determine the Grashof condition. Link 2 Link 1 d  6  in a  2  in b  7  in

Link 3 Solution: 1.

c  9  in

Link 4

See Mathcad file P0407a.

Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1 

d

K2 

a

K1  3.0000

2

d

K3 

c

K2  0.6667

 

 

 

2 a c

B  1.0000

 

C  K1   K2  1   cos θ  K3

C  3.5566

Use equation 4.10b to find values of 4 for the open and crossed circuits. Open:





2

θ  2  atan2 2  A B 

B  4 A  C



θ  242.714  deg

θ  θ  360  deg



θ  602.714  deg



2

Crossed: θ  2  atan2 2  A B 

B  4 A  C



θ  216.340  deg

Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4 

2

d

K5 

b

2

2

c d a b

 

2 a b

2

K4  0.8571 K5  0.2857

 

D  cos θ  K1  K4 cos θ  K5

D  1.6774

 

E  2  sin θ

E  1.0000

 

F  K1   K4  1   cos θ  K5 4.

2

A  0.7113

B  2  sin θ

3.

2

a b c d

K3  2.0000

A  cos θ  K1  K2 cos θ  K3

2.

θ  30 deg

F  2.5906

Use equation 4.13 to find values of 3 for the open and crossed circuits. Open:





θ  2  atan2 2  D E 

2

E  4  D F



θ  θ  360  deg





Crossed: θ  2  atan2 2  D E 

θ  271.163  deg θ  631.163  deg

2

E  4  D F



θ  244.789  deg

2


5.

Check the Grashof condition. Condition( a b c d ) 


Condition( a b c d )  "Grashof"





A position vector is defined as having length equal to your height in inches (or centimeters). The tangent of its angle is defined as your weight in lbs (or kg) divided by your age in years. Calculate the data for this vector and: a. Draw the position vector to scale on Cartesian axes. b. Write an expression for the position vector using unit vector notation. c. Write an expression for the position vector using complex number notation, in both polar and Cartesian forms.

Assumptions: Height  70, weight  160, age  20 Solution:

The magnitude of the vector is R  Height. The angle that the vector makes with the x-axis is θ  atan

weight 

  age 

a.

θ  82.875 deg

θ  1.446 rad

Draw the position vector to scale on Cartesian axes. y 100 80

R

60 70.000

1.


40

82.875°

20 0

b.

x 20

40

60

80

100

Write an expression for the position vector using unit vector notation.

 cos θ    sin θ 

R  R 

R

 8.682     69.459 

R = 8.682 i + 69.459 j

c. Write an expression for the position vector using complex number notation, in both polar and Cartesian forms. j  1.446

Polar form:

R  68 e

Cartesian form:

R  8.682  j  69.459




A particle is traveling along an arc of 6.5 inch radius. The arc center is at the origin of a coordinate system. When the particle is at position A, its position vector makes a 45-deg angle with the X axis. At position B, its vector makes a 75-deg angle with the X axis. Draw this system to some convenient scale and: a. b. c.

d.

Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. Write a vector equation for the position difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. Check the result of part c with a graphical method.

Given:

Circle radius and vector magnitude, R  6.5 in; vector angles: θA  45 deg

Solution:


θB  75 deg

1.

Establish an X-Y coordinate frame and draw a circle with center at the origin and radius R.

2.

Draw lines from the origin that make angles of 45 and 75 deg with respect to the X axis. Label the intersections of the lines with the circles as A and B, respectively. Make the line segment OA a vector by putting an arrowhead at A, pointing away from the origin. Label the vector RA. Repeat for the line segment OB, labeling it RB.

Y 8 B 6 A 4

RB RA

2

0 a.

b.

X 2

4

6

8

Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. j  θA

Polar form:

RA  R e

Cartesian form:

RA  R  cos θA  j  sin θA 

j

RA  6.5 e

π 4

RA  ( 4.596  4.596j) in

Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms.


c.


j

j  θB

Polar form:

RB  R e

RB  6.5 e

Cartesian form:

RB  R  cos θB  j  sin θB 

180

RB  ( 1.682  6.279j) in

Write a vector equation for the position difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. RBA  RB  RA

d.

75 π

RBA  ( 2.914  1.682j) in

Check the result of part c with a graphical method.

Y 8

3.365 B

6

RBA

1.682

A 4

RB

2.914

2 RA 0

X 2

4

6

8

On the layout above the X and Y components of RBA are equal to the real and imaginary components calculated, confirming that the calculation is correct.




Two particles are traveling along an arc of 6.5 inch radius. The arc center is at the origin of a coordinate system. When one particle is at position A, its position vector makes a 45-deg angle with the X axis. Simultaneously, the other particle is at position B, where its vector makes a 75deg angle with the X axis. Draw this system to some convenient scale and: a. b. c.

d.

Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. Write an exp ession for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. Write a vector equation for the relative position of the particle at B with respect to the particle at A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. Check the result of part c with a graphical method.

Given:

Circle radius and vector magnitude, R  6.5 in; vector angles: θA  45 deg

Solution:


θB  75 deg

1.


2.


Y 8 B 6 A 4

RB RA

2

0 a.

b.

X 2

4

6

8


Polar form:

RA  R e

Cartesian form:

RA  R  cos θA  j  sin θA 

j

RA  6.5 e

π 4

RA  ( 4.596  4.596j) in

Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms.


c.


j

j  θB

Polar form:

RB  R e

RB  6.5 e

Cartesian form:

RB  R  cos θB  j  sin θB 

180

RB  ( 1.682  6.279j) in

Write a vector equation for the relative position of the particle at B with respect to the particle at A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. RBA  RB  RA

d.

75 π

RBA  ( 2.914  1.682j) in


Y 8

3.365 B

6

RBA

1.682

A 4

RB

2.914

2 RA 0

X 2

4

6

8

On the layout above the X and Y components of RBA are equal to the real and imaginary components calculated, confirming that the calculation is correct.




A particle is traveling along the line y = -2x + 10. When the particle is at position A, its position vector makes a 45-deg angle with the X axis. At position B, its vector makes a 75-deg angle with the X axis. Draw this system to some convenient scale and: a. b. c.

d.

Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. Write a vector equation for the position difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. Check the result of part c with a graphical method.

Given:

Vector angles: θA  45 deg

Solution:


θB  75 deg

1.

Establish an X-Y coordinate frame and draw the line y = -2x + 10.

2.

Draw lines from the origin that make angles of 45 and 75 deg with respect to the X axis. Label the intersections of the lines with the line in step 1 as A and B, respectively. Make the line segment OA a vector by putting an arrowhead at A, pointing away from the origin. Label the vector RA. Repeat for the line segment OB, labeling it RB.

Y 10 y = -2x + 10 8 B 6

4

RB

2

0

3.

A

RA X 2

4

6

8

Calculate the coordinates of points A and B. xA tan  θA = 2  xA  10

xA 

10

2  tan  θA

yA  xA tan  θA xB tan  θB = 2  xB  10

xB 

10

2  tan  θB

xA  3.333 yA  3.333 xB  1.745



yB  xB tan  θB 4.

a.

b.

yB  6.511

Calculate the distances of points A and B from the origin. 2

2

RA  4.714

2

2

RB  6.741

RA 

xA  yA

RB 

xB  yB


j

Polar form:

RA  RA e

Cartesian form:

RA  RA  cos θA  j  sin θA 

RA  4.714  e

π 4

RA  3.333  3.333j

Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. j

j  θB

Polar form:

RB  RB e

RB  6.741  e

Cartesian form:

RB  RB  cos θB  j  sin θB 

75 π 180

RB  1.745  6.511j

Y c.

10

Write a vector equation for the position difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically.

y = -2x + 10 8 B

RBA  RB  RA RBA  1.589  3.178j d.

Check the result of part c with a graphical method. On the layout above the X and Y components of RBA are equal to the real and imaginary components calculated, confirming that the calculation is correct.

6 3.178 4

3.553

RBA RB A

2

0

RA X 2

4

6 1.589

8




Two particles are traveling along the line y = -2x2 - 2x +10. When one particle is at position A, its position vector makes a 45-deg angle with the X axis. Simultaneously, the other particle is at position B, where its vector makes a 75-deg angle with the X axis. Draw this system to some convenient scale and: a. b. c.

d.

Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. Write a vector equation for the relative position of the particle at B with respect to the particle at A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. Check the result of part c with a graphical method.

Given:

Vector angles: θA  45 deg

Solution:


θB  75 deg

1.

Establish an X-Y coordinate frame and draw the line y = -2x2 - 2x +10.

2.

Draw lines from the origin that make angles of 45 and 75 deg with respect to the X axis. Label the intersections of the lines with the line drawn in step 1 as A and B, respectively. Make the line segment OA a vector by putting an arrowhead at A, pointing away from the origin. Label the vector RA. Repeat for the line segment OB, labeling it RB.

Y 10 y = -2x^2 - 2x + 10 8

6 B 4 RB 2

A RA

0

3.

X 2

4

6

8

Calculate the coordinates of points A and B. xA tan  θA = 2  xA  2  xA  10 2

2  tan  θA  tan  θA    xA    1    1    2  2  2   

1

 20 

xA  1.608



yA  xA tan  θA

yA  1.608

xB tan  θB = 2  xB  2  xB  10 2

2  tan  θB  tan  θB    xB    1    1    2  2  2   

1

yB  xB tan  θB 4.

a.

b.

c.

xB  1.223

yB  4.564

Calculate the distances of points A and B from the origin. 2

2

RA  2.275

2

2

RB  4.725

RA 

xA  yA

RB 

xB  yB


Polar form:

RA  R e

Cartesian form:

RA  RA  cos θA  j  sin θA 

j

RA  2.275  e

π 4

RA  1.608  1.608j

Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. j  θB

Polar form:

RB  R e

Cartesian form:

RB  RB  cos θB  j  sin θB 

j

RB  4.725  e

75 π 180

RB  1.223  4.564j

Write a vector equation for the relative position of the particle at B with respect to the particle at A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. RBA  RB  RA

d.

 20 

RBA  0.386  2.955j


On the layout on the next page the X and Y components of RBA are equal to the real and imaginary components calculated, confirming that the calculation is correct.



Y 10 y = -2x^2 - 2x + 10 8

6 B 4 RB

2.955

RBA

2.980

2 A RA 0

X 4

2 0.386

6

8




The link lengths and value of 2 for some fourbar linkages are defined in Table P4-1. The linkage configuration and terminology are shown in Figure P4-1. For row a, draw the linkage to scale and graphically find all possible solutions (both open and crossed) for angles 3 and 4. Determine the Grashoff condition.

Given:

Link 1

d  6  in

Link 2

a  2  in

Link 3

b  7  in

Link 4

c  9  in

Solution:

θ2  30 deg

See figure below for one possible solution. Also see Mathcad file P0406a.

1.

Lay out an xy-axis system. Its origin will be the link 2 pivot, O2.

2.

Draw link 2 to some convenient scale at its given angle.

3.

Draw a circle with center at the free end of link 2 and a radius equal to the given length of link 3.

4.

Locate pivot O4 on the x-axis at a distance from the origin equal to the given length of link 1.

5.

Draw a circle with center at O4 and a radius equal to the given length of link 4.

6.

The two intersections of the circles (if any) are the two solutions to the position analysis problem, crossed and open. If the circles don't intersect, there is no solution.

7.

Draw links 3 and 4 in their two possible positions (shown as solid for open and dashed for crossed in the figure) and measure their angles 3 and 4 with respect to the x-axis. From the solution below, OPEN

θ θ

CROSSED

θ θ

8.

31 41 32 42

 88.84  deg  117.29 deg  360  deg  115.21 deg

θ

 360  deg  143.66 deg

θ

42

 244.790 deg  216.340 deg

y


32

B

S  min ( a b c d )

OPEN

L  max( a b c d ) SL  S  L

3

PQ  a  b  c  d  SL

4

return "Grashof" if SL  PQ 88.837°

return "Special Grashof" if SL = PQ

117.286°

A

return "non-Grashof" otherwise

2 O2


115.211°

O4 143.660°

CROSSED B'

x




Given:

The link lengths and value of 2 for some fourbar linkages are defined in Table P4-1. The linkage configuration and terminology are shown in Figure P4-1. For row a, find all possible solutions (both open and crossed) for angles 3 and 4 using the vector loop method. Determine the Grashof condition. Link 2 Link 1 d  6  in a  2  in b  7  in

Link 3

c  9  in

Link 4

θ  30 deg

Two argument inverse tangent atan2( x y ) 

return 0.5 π if x = 0  y  0 return 1.5 π if x = 0  y  0 return atan 

y 

  if x  0  x 

atan 

y 

   π otherwise  x 

Solution: 1.



d

K2 

a

K1  3.0000

2

d

K3 

c

K2  0.6667

 

 

2 a c

A  0.7113

 

B  2  sin θ

B  1.0000

 

C  K1   K2  1   cos θ  K3

C  3.5566

Use equation 4.10b to find values of 4 for the open and crossed circuits.

Open:





2

θ  2  atan2 2  A B 

B  4 A  C



θ  477.286 deg

θ  θ  360  deg



θ  117.286 deg



2

Crossed: θ  2  atan2 2  A B  3.

2

K3  2.0000

A  cos θ  K1  K2 cos θ  K3

2.

2

a b c d

B  4 A  C



θ  216.340 deg


2

d

K5 

b

2

2

c d a b

 

2 a b

 

D  cos θ  K1  K4 cos θ  K5

 

E  2  sin θ

2

K4  0.8571 K5  0.2857 D  1.6774 E  1.0000

 

F  K1   K4  1   cos θ  K5

F  2.5906

2


4.







θ  2  atan2 2  D E 

2

E  4  D F



θ  θ  360  deg





Crossed: θ  2  atan2 2  D E  5.

θ  88.837 deg 2

E  4  D F






θ  448.837 deg

θ  244.789 deg




Expand equation 4.7b and prove that it reduces to equation 4.7c (p. 157).

Solution:


1.

Write equation 4.7b and expand the two terms that are squared. 2



 

 2  a cosθ  c cosθ  d2

b  a  sin θ  c sin θ

2.

(4.7b)

a sinθ  c sinθ2  a2 sinθ2  2 a c sinθ sinθ  c2 sinθ2

(a)

a cosθ  c cosθ  d2  a2 cosθ2  2 a c cosθ cosθ  2 a d cosθ  2 2 2  2  c d  cos θ  c  cos θ  d

(b)

Add the two expanded terms, equations a and b, noting the identity sin 2x + cos2x = 1. 2

2

2

2

 

 

    

 

 

b  a  c  d  2  a  d  cos θ  2  c d  cos θ  2  a  c sin θ  sin θ  cos θ  cos θ This is equation 4.7c.




The link lengths, value of 2, and offset for some fourbar slider-crank linkages are defined in Table P4-2. The linkage configuration and terminology are shown in Figure P4-2. For row a, draw the linkage to scale and graphically find all possible solutions (both open and crossed) for angles 3 and slider position d.

Given:

Link 2

a  1.4 in

Link 3

Offset

c  1  in

θ  45 deg

Solution:

b  4  in


1.


2.


3.


4.

Draw a horizontal line through y = c (the offset).

5.

The two intersections of the circle with the horizontal line (if any) are the two solutions to the position analysis problem, crossed and open. If the circle and line don't intersect, there is no solution.

6.

Draw link 3 and the slider block in their two possible positions (shown as solid for open and dashed for crossed in the figure) and measure the angle 3 and length d for each circuit. From the solution below, θ31  360  deg  179.856  deg

θ31  180.144 deg

θ32  0.144  deg

d  3.010  in

d  4.990  in 1

2

Y d2 = 3.010

d1 = 4.990

3(CROSSED)

B'

A 2

0.144° O2

45.000°

3 (OPEN)

B 179.856° 1.000 X




Given:

Solution:

The link lengths, value of 2, and offset for some fourbar slider-crank linkages are defined in Table P4-2. The linkage configuration and terminology are shown in Figure P4-2. For row a, using the vector loop method, find all possible solutions (both open and crossed) for angles 3 and slider position d. Link 2 Offset

a  1.4 in c  1  in

Link 3 b  4  in θ  45 deg

See Figure P4-2 and Mathcad file P0410a. Y d2 = 3.010

d1 = 4.990

3(CROSSED)

B'

A 2

0.144°

45.000°

3 (OPEN)

B 179.856° 1.000 X

O2

1.

Determine 3 and d using equations 4.16 and 4.17. Crossed:

 a sin θ  c   b  

θ  0.144 deg

 

d 2  3.010 in

θ  asin

 

d 2  a  cos θ  b  cos θ Open:

 a  sin θ  c  π b  

θ  180.144 deg

 

d 1  4.990 in

θ  asin 

 

d 1  a  cos θ  b  cos θ




The link lengths and the value of 2 and  for some inverted fourbar slider-crank linkages are defined in Table P4-3. The linkage configuration and terminology are shown in Figure P4-3. For row a, draw the linkage to scale and graphically find both open and closed solutions for 3 and 4 and vector RB.

Given:

Link 1

d  6  in

Link 2

Link 4

c  4  in

γ  90 deg

Solution:

a  2  in θ  30 deg


1.


2.


3a. If  = 90 deg, locate O4 on the x-axis at a distance equal the length of link 1 (d) from the origin. Draw a circle with center at O4 and radius equal to the length of link 4 (c). From point A, draw two lines that are tangent to the circle. The points of tangency define the location of the points B for the open and crossed circuits. 3b. When  is not 90 deg there are two approaches to a graphical solution for link 3 and the location of point B: 1) establish the position of link 4 and the angle  by trial and error, or 2) calculate the distance from point A to point B (the instantaneous length of link 3). Using the second approach, from triangle O2AO4

y

B



b

 c

A a 2 d

x 04

02

2

2

 

2

AO4 = a  d  2  a  d  cos θ

and, from triangle AO4B (for the open circuit)

AO4 = b  c  2  b  c cos π  γ 2

2

2

where a, b, c, and d are the lengths of links 2, 3, 4, and 1, respectively. Eliminating AO4 and solving for the unknown distance b for the open branch, b 1 

1 2



 2  c cos π  γ 

2 c cos π  γ 2  4  c2  a2  d2  2 a d cos θ

b 1  1.7932 in 2

2

2

For the closed branch: AO4 = b  c  2  b  c cos( γ) b 2 

1 2



 2  c cos γ 

and

 2 c cosγ 2  4  c2  a2  d2  2 a d cos θ



b 2  1.7932 in Draw a circle with center at point A and radius b 1. Draw a circle with center at O4 and radius equal to the length of link 4 (c). The intersections of these two circles is the solution for the open and crossed locations of the point B. 4.

Draw the complete linkage for the open and crossed circuits, including the slider. The results from the graphical solution below are: θ  127.333  deg

OPEN

CROSSED

θ  100.959  deg

θ  142.666  deg

θ  169.040  deg

RB1  3.719 at 40.708 deg

RB2  2.208 at -20.146 deg

B y 90.0°

b

127.333° c

A

142.666°

a 30.000°

d

x 04

02 B'

169.040° 79.041°




Given:

The link lengths and the value of 2 and  for some inverted fourbar slider-crank linkages are defined in Table P4-3. The linkage configuration and terminology are shown in Figure P4-3. For row a, using the vector loop method, find both open and closed solutions for 3 and 4 and vector RB. Link 1 Link 2 d  6  in a  2  in Link 4

Solution: 1.

c  4  in

γ  90 deg

θ  30 deg


Determine the values of the constants needed for finding 4 from equations 4.25 and 4.26.

 



 



P  a  sin θ  sin γ  a  cos θ  d  cos γ

 

2.

3.

4.

5.



 

P  1.000 in



Q  a  sin θ  cos γ  a  cos θ  d  sin γ

Q  4.268 in

R  c sin γ

R  4.000 in

T  2  P

T  2.000 in

S  R  Q

S  0.268 in

U  Q  R

U  8.268 in

Use equation 4.26c to find values of 4 for the open and crossed circuits.



T  4 S U



T  4 S U

OPEN

θ  2  atan2 2  S T 

CROSSED

θ  2  atan2 2  S T 

2



θ  142.667 deg

2



θ  169.041 deg

Use equation 4.22 to find values of 3 for the open and crossed circuits. OPEN

θ  θ  γ

θ  232.667 deg

CROSSED

θ  θ  γ

θ  79.041 deg

Determine the magnitude of the instantaneous "length" of link 3 from equation 4.24a. OPEN

b 1 

CROSSED

b 2 

    sin θ  γ

a  sin θ  c sin θ

b 1  1.793 in

    sin θ  γ

a  sin θ  c sin θ

b 2  1.793 in

Find the position vector RB from the definition given in the text. OPEN

  

   b1 cosθ  j  sinθ

RB1  a  cos θ  j  sin θ RB1  RB1

RB1  3.719 in

θ  arg RB1

θ  40.707 deg


CROSSED


  

   b2 cosθ  j  sinθ

RB2  a  cos θ  j  sin θ RB2  RB2

RB2  3.091 in

θ  arg RB2

θ  63.254 deg

DESIGN OF MACHINERY

SOLUTION MANUAL 4-12c-1

PROBLEM 4-12c Statement:

Given:

The link lengths and the value of 2 and for some inverted fourbar slider-crank linkages are defined in Table P4-3. The linkage configuration and terminology are shown in Figure P4-3. For row c, using the vector loop method, find both open and closed solutions for 3 and 4 and vector RB . Link 1

d  3 in

Link 2

Link 4

c  6 in

 45 deg

a 10 in  45 deg

Two argument inverse tangent atan2 (x  y)   return 0.5  if x = 0 y 0 return 1.5  if x = 0 y 0

 y if x 0 return atan  x     y  atan  x    otherwise    Solution: 1.

See Mathcad file P0412c.

Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a.



  

P a  sin  sin   a cos  d  cos 



2.

P 7.879 in

  

Q a  sin  cos   a cos  d  sin 

Q 2.121 in

R c sin 

R 4.243 in

T  2 P

T 15.757 in

S R Q

S 2.121 in

U Q R

U 6.364 in

Use equation 4.22c to find values of 4 for the open and crossed circuits. OPEN



2



46.400 deg



2



163.739 deg

  2 atan2 2  S T  T 4 S U 360 deg

CROSSED   2 atan2 2  S T  T 4 S U 360 deg 3.

4.


 

91.400 deg

CROSSED

 

118.739 deg

Determine the magnitude of the instantaneous "length" of link 3 from equation 4.20a. OPEN

b1  

CROSSED

b2  

   sin   

a sin  c sin 

   sin 

a sin  c  sin 

b1 2.727 in

b2 11.212 in

DESIGN OF MACHINERY

5.


Find the position vector RB from the definition given on page 162 of the text. OPEN

CROSSED

 

 

  

 

RB1  acos  j  sin  b 1cos  j  sin  RB1   RB1

RB1 8.356in

 arg  RB1

31.331 deg

 

 

  

 

RB2  acos  j  sin  b 2cos  j  sin  RB2   RB2

RB2 12.764 in

 arg  RB2

12.488 deg




Find the transmission angles of the linkage in row a of Table P4-1.

Given:

Link 1

d  6  in

Link 2

a  2  in

Link 3

b  7  in

Link 4

c  9  in

Solution: 1.



d a

2

d

K2 

K1  3.0000

K3 

c

K2  0.6667

 

 

   

2 a c

C  3.5566

Use equation 4.10b to find 4 for the open circuit.





2

θ  2  atan2 2  A B 

B  4 A  C



θ  242.714 deg


2

d

K5 

b

2

2

c d a b

 

2 a b

2

K4  0.8571 K5  0.2857

 

D  cos θ  K1  K4 cos θ  K5

D  1.6774

 

E  2  sin θ

E  1.0000

 

F  K1   K4  1   cos θ  K5

F  2.5906

Use equation 4.13 to find 3 for the open circuit.





θ  2  atan2 2  D E 

2

E  4  D F

θ  θ  360  deg 5.

2

B  1.0000

C  K1   K2  1   cos θ  K3

4.

2

A  0.7113

B  2  sin θ

3.

2

a b c d

K3  2.0000

A  cos θ  K1  K2 cos θ  K3

2.

θ  30 deg



θ  271.163 deg θ  631.163 deg

Use equations 4.32 to find the transmission angle.

θtrans θ θ 

t  θ  θ

θtrans θ θ  208.449 deg

return t if t  0.5 π π  t otherwise 6.

It can be shown that the triangle ABO4 in Figure 4-17 is symmetric with respect to the line AO4 for the crossed branch and, therefore, the transmission angle for the crossed branch is identical to that for the open branch.




Find the minimum and maximum values of the transmission angle for all the Grashof crankrocker linkages in Table P4-1.

Given:

Table P4-1 data:

i  1 2  14 Row  i

"a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" Solution: 1.

d 

a 

b 

c 

6 7 3 8 8 5 6 20 4 20 4 9 9 9

2 9 10 5 5 8 8 10 5 10 6 7 7 7

7 3 6 7 8 8 8 10 2 5 10 10 11 11

9 8 8 6 6 9 9 10 5 10 7 7 8 6

i

i

i

i

See Table P4-1 and Mathcad file P0414.

Determine which of the linkages in Table P4-1 are Grashof. Condition( a b c d ) 


Row a

Condition( 6 2 7 9 )  "Grashof"

Row b


Row c


Row d

Condition( 8 5 7 6 )  "Special Grashof"

Row e


Row f


Row g


Row h

Condition( 20 10 10 10)  "non-Grashof"

Row i


Row j


Row k

Condition( 4 6 10 7 )  "non-Grashof"

Row l




Row m


Row n


2.

Determine which of the Grashof linkages are crank-rockers. To be a Grashof crank-rocker, the linkage must be Grashof and the shortest link is either 2 or 4. This is true of rows a, d, and e.

3.

Use equations 4.32 and 4.33 to calculate the maximum and minimum transmission angles.

Row a

i  1

 b 2  c 2  d  a 2   i  i  i i  μ  acos   2 b  c i i    

μ  if  μ 

π 2

 

π  μ μ

 b 2  c 2  d  a 2   i  i  i i  μ  acos   2 b  c i i  

Row d

i  4

 

π 2

 

π  μ μ


i  5

μ  25.209 deg

 b 2  c 2  d  a 2   i  i  i i  μ  acos   2 b  c i i   μ  if  μ 

Row e

μ  58.412 deg

μ  0.000 deg

μ  25.209 deg

 b 2  c 2  d  a 2   i  i  i i  μ  acos   2 b  c i i    

μ  if  μ 

π 2

 

π  μ μ


μ  44.049 deg

μ  18.573 deg




Find the input angles corresponding to the toggle positions of the non-Grashof linkages in Table P4-1.

Given:

Table P4-1 data:

i  1 2  14 Row  i

"a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" Solution: 1.

d 

a 

b 

c 

6 7 3 8 8 5 6 20 4 20 4 9 9 9

2 9 10 5 5 8 8 10 5 10 6 7 7 7

7 3 6 7 8 8 8 10 2 5 10 10 11 11

9 8 8 6 6 9 9 10 5 10 7 7 8 6

i

i

i

i

See Table P4-1 and Mathcad file P0415.

Determine which of the linkages in Table P4-1 are Grashof. Condition( a b c d ) 


Row a


Row b


Row c


Row d

Condition( 8 5 7 6 )  "Special Grashof"

Row e


Row f


Row g


Row h


Row i


Row j


Row k


Row l



2.


Row m


Row n


There are six non-Grashof rows in the Table: Rows h, and j through n. For each row there are two possible arguments to the arccos function given in equation (4.37). They are: i  8

Row  "h"

j  1

i

ai  di  bi  ci  2

arg

j 1

2

2

2

b c 

2 a  d

i i

arg

j 2

i i

ai  di  bi  ci  2

2

2

2

2 a  d

Row  "j"

i i

ai  di  bi  ci  2

j 1

2

2

2

b c 

2 a  d

i i

j 2

2

2

2

2 a  d

Row  "k"

2

2

2

b c 

2 a  d

i i

j 2

2

2

2

b c 

2 a  d

i i

i  12

Row  "l"

i i

ai  di  bi  ci  2

j 1

2

2

2

b c 

2 a  d

i i

j 2

2

2

2

2 a  d

Row  "m"

2

2

2

b c 

2 a  d

i i

j 2

2

2

2 a  d

i i

i i

a d

i i

ai  di  bi  ci  2

arg

a d

i i

ai  di  bi  ci  2

j 1

i i

j  5

i

arg

a d b c



i i

i  13

i i i i

ai  di  bi  ci  2

arg

i i

a d

j  4

i

arg

i i

a d

i i

ai  di  bi  ci  2

arg

a d

i i

ai  di  bi  ci  2

j 1

i i

j  3

i

arg

a d b c



i i

i  11

i i i i

ai  di  bi  ci  2

arg

i i

a d

j  2

i

arg

b c 

i i

i  10

i i

a d

2

b c 

i i

a d

i i


i  14


Row  "n"

j  6

i

ai  di  bi  ci  2

arg

j 1

2

2

2

b c 

2 a  d

i i

arg

j 2

2

2

2 a  d

2

b c 

i i

 1.250  1.188  0.896 arg    0.960  0.960   0.833 3.

a d

i i

ai  di  bi  ci  2

i i

i i

a d

i i

 0.688   4.938  1.262   1.833  1.262  0.250

Choose the argument values that lie between plus and minus 1,



1 2



θ2h  75.5 deg



2 2



θ2j  46.6 deg



3 1



θ2k  26.4 deg



4 1



θ2l  16.2 deg

θ2h  acos arg θ2j  acos arg

θ2k  acos arg θ2l  acos arg



5 1



θ2m  16.2 deg



6 1



θ2n  33.6 deg

θ2m  acos arg θ2n  acos arg




The link lengths, gear ratio, phase angle, and the value of 2 for some geared fivebar linkages are defined in Table P4-4. The linkage configuration and terminology are shown in Figure P4-4. For row a, draw the linkage to scale and graphically find all possible solutions for angles 3 and 4.

Given:

Link 1

d  4  in

Link 2

a  1  in

Link 3

b  7  in

Link 4

c  9  in

Link 5

f  6  in

Gear ratio

λ  2.0

Phase angle

ϕ  30 deg

Input angle

θ  60 deg

Solution: 1.


Determine whether or not an idler is required. idler 

"required" if λ  0 "not-required" otherwise

idler  "required" 2.

Choose radii for gears 2 and 5 by making a design choice for their center distance (which must be increased if an idler is required). Let the standard center distance when no idler is required be C  0.5 c then C = r2  r5

and

λ =

r2 r5

Solving for r2 and r5, r5 

C λ 1

r2  r5 λ

r5  1.500 in r2  3.000 in

If an idler is required, increase the center distance. C  if  idler = "required" C  r5 C

C  6.000 in

Note that the amount by which C is increased if an idler is required is a design choice that is made based on the size of the gears and the space available. 3.

Using equation 4.27c, determine the angular position of link 5 corresponding to the position of link 2. θ  λ θ  ϕ

θ  150 deg

4.


5.


6.


7.

Locate pivot O4 on the x-axis at a distance from the origin equal to the given length of link 1.

8.

Draw link 5 to some convenient scale at its calculated angle.

9.


10. The two intersections of the circles (if any) are the two solutions to the position analysis problem, crossed and open. If the circles don't intersect, there is no solution. 11. Draw links 3 and 4 in their two possible positions (shown as solid for open and dashed for crossed in the figure) and measure their angles 3 and 4 with respect to the x-axis. From the solution below,



θ  173.64 deg

OPEN

θ  360  deg  177.715  deg θ  182.285 deg

CROSSED

θ  360  deg  115.407  deg

θ  360  deg  124.050  deg

θ  244.593 deg

θ  235.950 deg

12. Draw gears 2 and 5 schematically at their calculated radii. If an idler is required, draw it tangent to gears 2 and 5. Its diameter is a design choice that will be made on strength and space requirements. It does not affect the gear ratio.

y C

4

B

177.7152°

173.6421° 3

5 2

B`

124.0501° x

O2

3

150.0000°

115.4074°

4

O5

DESIGN OF MACHINERY - 5th ed.



Given:

Solution: 1.

The link lengths, gear ratio, phase angle, and the value of 2 for some geared fivebar linkages are defined in Table P4-4. The linkage configuration and terminology are shown in Figure P4-4. For row a, using the vector loop method, find all possible solutions for angles 3 and 4. Link 1

d  4  in

Link 2

a  1  in

Link 3

b  7  in

Link 4

c  9  in

Link 5

f  6  in

Gear ratio

λ  2.0

Phase angle

ϕ  30 deg

Input angle

θ  60 deg


Determine the values of the constants needed for finding 3 and 4 from equations 4.27h and 4.27i.







 







 

A  2  c d  cos λ θ  ϕ  a  cos θ  f



2

A  36.6462 in 2

B  2  c d  sin λ θ  ϕ  a  sin θ

2

2

2

2



2

B  20.412 in

  

C  a  b  c  d  f  2  a  f  cos θ   2  d  a  cos θ  f  cos λ θ  ϕ    2  a  d  sin θ  sin λ θ  ϕ



     





2

C  37.4308 in

2

D  C  A

D  0.78461 in

E  2  B

E  40.823 in

F  A  C

F  74.077 in

2 2





  a cosθ  f 

G  28.503 in





  a sinθ

H  15.876 in

2

G  2  b   d  cos λ θ  ϕ

2

H  2  b   d  sin λ θ  ϕ

2

2.

2

2

2



2

  

K  a  b  c  d  f  2  a  f  cos θ   2  d  a  cos θ  f  cos λ θ  ϕ    2  a  d  sin θ  sin λ θ  ϕ

K  26.569 in

L  K  G

L  1.933 in

M  2  H

M  31.751 in

N  G  K

N  55.072 in



     





2

2 2

2

Use equations 4.28h and 4.28i to find values of 3 and 4 for the open and crossed circuits. OPEN





M  4  L N





E  4  D F





M  4  L N





E  4  D F

θ  2  atan2 2  L M  θ  2  atan2 2  D E 

CROSSED

θ  2  atan2 2  L M  θ  2  atan2 2  D E 

2

2

2

2



 



θ  173.642 deg θ  177.715 deg θ  115.407 deg θ  124.050 deg




The angle between the X and x axes is 25 deg. Find the angular displacement of link 4 when link 2 rotates clockwise from the position shown (+37 deg) to horizontal (0 deg). How does the transmission angle vary and what is its minimum between those two positions? Find the toggle positions of this linkage in terms of the angle of link 2.

Given:

Link lengths: Crank

L2  116

Coupler

L3  108

Rocker

L4  110

Ground link

L1  174

Crank angle for position shown (relative to O2O4):

θ  62 deg

Y y

A

2

Crank rotation angle from position shown to horizontal:

37°

Δθ  37 deg 3 X O2

25°

B

O4

4

x

Solution: 1.

See Figure P4-5a and Mathcad file P0418a.

Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the crossed circuit). K1 

L1

K2 

L2

K1  1.5000

 

2

L1

K3 

L3

K2  1.6111

 

2

2

L2  L3  L4  L1

2

2  L2 L4

K3  1.7307

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ  2.

Determine 4 for the position shown and after the crank has moved to the horizontal position.

 

θ  θ θ



θ  θ θ  Δθ 3.

θ  183.5 deg



θ  212.8 deg

Subtract the two values of 4 to find the angular displacement of link 3 when link 2 rotates clockwise from the position shown to the horizontal.   θ  θ

4.

 2  4 A θ Cθ 

B θ

  29.2 deg


L1 L3

2

K5 

2

2

L4  L1  L2  L3 2  L2 L3

2

K4  1.6111

K5  1.7280


 


 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

5.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

Use equation 4.13 to find values of 3 for the crossed circuit.



 



 

 

θ θ  2   atan2 2  D θ E θ  6.

Determine 3 for the position shown and after the crank has moved to the horizontal position.

 

θ  θ θ

θ  275.1 deg



θ  θ θ  Δθ 7.

 2  4 Dθ F θ 

E θ



θ  256.1 deg

Use equations 4.28 to find the transmission angles. μ  π  θ  θ

μ  88.4 deg

μ  θ  θ

μ  43.4 deg

The transmission angle is smaller when the crank is in the horizontal position. 8.

Check the Grashof condition of the linkage. Condition( a b c d ) 


Condition L1 L2 L3 L4  "non-Grashof" 9.

Using equations 4.37, determine the crank angles (relative to the XY axes) at which links 3 and 4 are in toggle. 2

arg1 

2

2

2  L2 L1 2

arg2 

2

L2  L1  L3  L4

2

2

2

L2  L1  L3  L4 2  L2 L1





L3 L4 L2 L1 L3 L4 L2 L1

θ2toggle  acos arg2 The other toggle angle is the negative of this.

arg1  1.083

arg2  0.094

θ2toggle  95.4 deg


SOLUTION MANUAL 4-18b-1

PROBLEM 4-18b Statement:

Find and plot the angular position of links 3 and 4 and the transmission angle as a function of the angle of link 2 as it rotates through one revolution.

Given:

Link lengths: Wheel (crank)

L2  40

a  L2

Coupler

L3  96

b  L3

Rocker

L4  122

c  L4

Ground link

L1  162

d  L1

Two argument inverse tangent atan2( x y ) 

Y

return 0.5 π if x = 0  y  0 return atan 

y 

B

A

2

return 1.5 π if x = 0  y  0

y 3 X

  if x  0  x 

O2 4

y atan     π otherwise  x  Solution: 1.

See Figure P4-5b and Mathcad file P0418b. O4


x


Condition( a b c d )  "Grashof" 2.

Define one cycle of the input crank: θ  0  deg 1  deg  360  deg

3.

Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). K1 

d

K2 

a

K1  4.0500

 

2

d

K3 

c

K2  1.3279

 

2

2

a b c d

2

2 a c

K3  3.4336

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ  4.

 2  4 A θ Cθ 

B θ

If the calculated value of 4 is greater than 2, subtract 2 from it. If it is negative, make it positive.

 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ


5.



2

d

K5 

b

 

2

2

c d a b

2

K4  1.6875

2 a b

 

K5  2.8875

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

6.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ


 





 

 

θ θ  2   atan2 2  D θ E θ  7.

 2  4 Dθ F θ 

E θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ

8.

Plot 3 and 4 as functions of the crank angle 2 (measured from the ground link). Angular Displacement of Coupler & Rocker

Coupler or Rocker angle, deg

200

 

150

θ θ  deg

  100

θ θ  deg

50

0

0

45

90

135

180

225

270

θ deg Crank angle, deg

9.

Use equations 4.32 to find the transmission angle.

 

 

 

Tran θ  θ θ  θ θ

 

 

 

Trans θ  if  Tran θ 

π 2

 

 

 π  Tran θ Tran θ



315

360



10. Plot the transmission angle.

Transmission Angle 90

Transmission Angle, deg

80

 

Trans θ

70

deg 60

50

40

0

45

90

135

180 θ deg

Wheel angle, deg

225

270

315

360




Find and plot the position of any one piston as a function of the angle of crank 2 as it rotates through one revolution. Once one piston's motion is defined, find the motions of the other two pistons and their phase relationship to the first piston. Y

Given: L2  19

a  L2

Piston-rod length L3  70

b  L3

Crank length

6

3

2

5

8 X

c  0

Offset

4

Solution:

See Figure P4-5c and Mathcad file P0418c. 7

1.

Let pistons 1, 2, and 3 be links 7, 6, and 8, respectively.

2.

Solve first for piston 6. Establish 2 as a range variable: θ  0  deg 2  deg  360  deg

3.

Determine 3 and d using equations 4.16 and 4.17.

 

 a sin θ  b 

θ θ  asin 

4.

c

 

  

d 1 θ  a  cos θ  b  cos θ θ

For each piston (slider) the crank angle is measured counter-clock-wise from the centerline of the piston, which goes through the O2 in all cases. Thus, when the crank angle for piston 1 is 0 deg, it is 120 deg for piston 2 and 240 deg for piston 3. Thus, the crank angles for pistons 2 and 3 are

 

 

θ θ  θ  120  deg 5.

 

π 

θ θ  θ  240  deg

Determine 3 and d for pistons 2 and 3.

 

 a sin θ θ   b 

θ θ  asin 

 

c

π 

    b cosθθ

d 2 θ  a  cos θ θ

 

 a sin θ θ   c  π b  

θ θ  asin 

 

    b cosθθ

d 3 θ  a  cos θ θ

6.

Plot the piston displacements as a function of crank angle (referenced to line AC (see next page).



Piston Displacement (1, 2, and 3) 90

Piston displacement, mm

80

  d2 θ 70 d3 θ d1 θ

60

50

0

60

120

180

240

300

θ deg Piston 1 crank angle, deg.

The solid line is piston 1, the dotted line is piston 2, and the dashed line is piston 3.

360


SOLUTION MANUAL 4-18d-1

PROBLEM 4-18d Statement:

Find the total angular displacement of link 3 and the total stroke of the box as link 2 makes a complete revolution.

Given:

Ground link

L1  150

Input crank

L2  30

Coupler link

L3  150

Output crank

L4  30

Solution:

See Figure P4-5d and Mathcad file P0418d.

Y 3 2

B

A

O2

O4

X A 4

1.

This is a special-case Grashof mechanism in the parallelogram form (see Figure 2-17 in the text). As such, the coupler link 3 executes curvilinear motion and is always parallel to the ground link 1. Thus, the total angular motion of link 3 as crank 2 makes one complete revolution is zero degrees.

2.

The stroke of the box will be equal to twice the length of the crank link in one complete revolution of the crank stroke  2  L2

stroke  60


SOLUTION MANUAL 4-18e-1

PROBLEM 4-18e Statement:

Determine the ratio of angular displacement between links 8 and 2 as a function of angular displacement of input crank 2. Plot the transmission angle at point B for one revolution of crank 2. Comment on the behavior of this linkage. Can it make a full revolution as shown?

Given:

Link lengths:

Crank (O2A)

a 1  20

Coupler (L3)

b 1  160

Crank (O4B)

c1  20

B

A

Ground link (O2O4) d 1  160

O2

3

4 O4 G

E

2 D

5

C

6

7

Ground link (O4O8) d 2  120

Solution:

Crank (O4G)

a 2  30

Coupler (L6)

b 2  120

Crank (O8F)

c2  30

O8 H

F 8

See Figure P4-5e and Mathcad file P0418e.

1.

This is an eightbar, 1-DOF linkage with two redundant links (3 and 6 or 5 and 7) making it, effectively, a sixbar. It is composed of a fourbar (1, 2, 3, and 4) with an output dyad (7 and 8). The input fourbar is a special-case Grashof in the parallelogram configuration. Thus, the output angle is equal to the input angle and the couplers execute curvilinear motion with links 3 and 5 always parallel to the horizontal. The output dyad also behaves like a special-case Grashof with parallelogram configuration so that the angular motion of link 8 is equal to that of link 4. Therefore, the ratio of angular displacement between links 8 and 2 is unity. The mechanism is not capable of making a full revolution. The couplers 3 and 5 (also 6 and 7) cannot pass by each other near 2 = 0 and 180 deg because of interference with the pins that connect them to their cranks.

2.

Define the approximate range of motion of the input crank: θ  0  deg 2  deg  180  deg

3.

Define 3 and 4.

 

θ  0.deg

Use equations 4.32 to find and plot the transmission angle.

 

 

 

tran θ  θ  θ θ

 

 



 

 

 

Tran θ  if tran θ  π tran θ  π tran θ

 

Trans θ  if  Tran θ 

π 2

 

 

 π  Tran θ Tran θ



Transmission Angle at B Transmission Angle, deg

4.

θ θ  θ

90 80 70 60 Trans θ 50 40 deg 30 20 10 0

 

0

45

90 θ deg Crank angle, deg

135

180


SOLUTION MANUAL 4-18f-1

PROBLEM 4-18f Statement:

Find and plot the displacement of piston 4 and the angular displacement of link 3 as a function of the angular displacement of crank 2.

Given:

Link lengths: Crank length, L2

Solution:

a  63

Piston-rod length, L3

b  130

Offset

c  52

4

B Y, x

3

See Figure P4-5f and Mathcad file P0418f.

1.

Establish 2 as a range variable: θ  0  deg 1  deg  360  deg

2.

Determine 3 and d in global XY coord using equations 4.16 and 4.17.

A

 a sin θ  90 deg  c  θ θ  asin  π b  

 





2 y

X

O2

  

d θ  a  cos θ  90 deg  b  cos θ θ

Plot the piston displacement (directly below) and rod angle (next page) as functions of crank angle in the global XY coordinate frame. Piston Displacement 200

150 Piston displacement, mm

3.

 

d θ 100

50

0

0

60

120

180 θ deg Crank angle, deg.

240

300

360



Piston-Rod Angular Displacement 260

Angular displacement, deg

240

 

220

θ θ  deg

200

180

160

0

60

120


240

300

360


SOLUTION MANUAL 4-18g-1

PROBLEM 4-18g Statement:

Find and plot the angular displacement of link 6 versus the angle of input link 2 as it is rotated from the position shown (+30 deg) to a vertical position (+90 deg). Find the toggle positions of this linkage in terms of the angle of link 2.

Given:

Link lengths:

Y

Input (L2)

a  49

Rocker (L4)

c  153

B

Coupler (L3)

b  100

Ground link (L1)

d  87

3 30° A

2 4

Angle from x axis to X axis:

α  121  deg

Starting angle:

θ  30 deg

Crank rotation angle from position shown to vertical:

Solution:

O6

X

6

C O2 5

D

y

O4

Δθ  60 deg

x

121°

See Figure P4-5g and Mathcad file P0418g.

1.

Define one cycle of the input crank in global coord: θ  θ θ  1  deg  θ  Δθ

2.


d

K2 

a

K1  1.7755

 

2

d

K3 

c

K2  0.5686







2

2

a b c d

2

2 a c

K3  1.5592



A θ  cos θ  α  K1  K2 cos θ  α  K3

 





 

 





 

 

θ θ  2   atan2 2  A θ B θ  3.





C θ  K1   K2  1   cos θ  α  K3

B θ  2  sin θ  α

 2  4 A θ Cθ   α

B θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ

4.

Plot 4 as a function of the crank angle 2 (measured from the X-axis) as it rotates from the position shown to the vertical position.



Angular Displacement of Rocker Link 4 120

Rocker angle, deg

110

 

θ θ 

100

deg

90

80 30

40

50

60

70

80

90


4.



Condition( a b c d )  "non-Grashof" 5.

Using equations 4.37, determine the crank angles (relative to the x-axis) at which links 3 and 4 are in toggle. 2

arg1 

2

2



2 a d 2

arg2 

2

a d b c

2

2

a d b c 2 a d

θ2toggle  acos arg1

2



b c a d b c a d

arg1  0.840

arg2  6.338


The other toggle angle is the negative of this. Thus, in the global XY frame the toggle positions are:

θ2XYtoggle  θ2toggle  α

θ2XYtoggle  88.130 deg

θ2XYtoggle  θ2toggle  α

θ2XYtoggle  153.870 deg


SOLUTION MANUAL 4-18h-1

PROBLEM 4-18h Statement:

Find link 4's maximum displacement vertically downward from the position shown. What will the angle of input link 2 be at that position?

Given:

Link lengths:

c2

a  19.8 mm

Crank length, L2 or L8

Solution:

Coupler length, L3 or L5

b1  19.4 mm

Offset of 1, 2, 3, 4

c1  4.5 mm

Distance from O2 to O8

L1  45.8 mm

Coupler length, L5 or L7

b2  13.3 mm

Offset of 1, 2, 5, 6

c2  22.9 mm

Angle of link 2 as shown

θ  47 deg

6

O2

O8

A

2

8

7

5

C

B

9

3

4

D

E

c1

See Figure P4-5h and Mathcad file P0418h.

1.

Links 1, 2, 3, 4, 5, and 6 make up two offset slider-cranks with a common crank, link 2. Links 7, 8, and 9 are kinematically redundant and contribute only to equalizing the forces in the mirror image links. Slider-crank 1, 2, 3, 4 is in the open circuit, and slider-crank 1, 2, 5, 6 is in the crossed circuit.

2.

Calculate the displacement of link 4 with respect to link 2 angle for the position shown in Figure P4-5h using equations 4.17 and 4.16b.

 a sin θ  c1  π b1  

θ  149.038 deg

 

d 10  30.14 mm

θ  asin 





d 10  a  cos θ  b1 cos θ 3.

Link 4 will reach its maximum downward displacement when links 8 and 9 and links 2 and 3 are in the toggle position. However, it is possible that they may not be able to reach this position because links 5 and 7 may be too short to allow links 2 and 8 to rotate far enough to reach toggle with 3 and 9, respectively.

4.

Using equation 4.16a, determine the angle that the crank will make with the x axis (see layout below) when links 5 and 7 are horizontal (5 = -90 deg). This will be the least value of the angle 2.

 a sin θ  c2  θ  asin  b2  

22.9 = c2 O2

y

θ  90 deg

A

a b2

 

29.0°

sin θ  1.000

 

a  sin θ  c2 b2

B

47°

A'

D' b1

 1.000

c2  b2  θ  asin   a 

D 5.90 D'

θ  29.00 deg 4.5= c1


5.

SOLUTION MANUAL 4-18h-2

Use equations 4.17 and 4.16b to determine the displacement of D' with respect to O2.

 a  sin θ  c1  π b1  

θ  164.759 deg

 

d 1  36.03 mm

θ  asin 

 

d 1  a  cos θ  b1 cos θ 6.

The maximum displacement of link 4 from the position shown in Figure P4-5h is the difference between the displacement found in step 5 and that found in step 2.

Δdmax  d 1  d 10

Δdmax  5.90 mm




For one revolution of the driving link 2 of the walking-beam indexing and pick-and-place mechanism in Figure P4-6, find the horizontal stroke of link 3 for the portion of their motion where their tips are above the platen. Express the stroke as a percentage of the crank length O2B. What portion of a revolution of link 2 does this stroke correspond to? Also find the total angular displacement of link 6 over one revolution of link 2.

Given:

Measured lengths: Input crank length (O2A)

a  40

Coupler length (L3)

b  108

Output crank length (L4)

c  40

95

Q

3

A

64

2

4

O4

p  119.81

Coupler data (finger at Q) Distance from O2 to the platen surface

1.

D

C

Ground link length (O2O4) d  108

Solution:

73

E

6

δ  37.54  deg e  64

B

O2

7

O6 O5 185


Links 1, 2, 3 and 4 are a special-case Grashof linkage in the parallelogram form. The tip of the finger at point Q (left end of the coupler) is used as the coupler point. The distance from the tip to the platen is . Top platen surface Q p

 

b D

A a

c d

x

O2

O4

y

2.

Define the crank angle as a range variable and define 3 ,which is constant because the coupler has curvilinear motion.. θ  0  deg 1  deg  360  deg θ  0  deg

3.

82

5

Use equations 4.27 to define the y-component of the vector RP. RP  RA  RPA

  

 

RA  a  cos θ  j  sin θ

 







RPA  p  cos θ  δ  j  sin θ  δ

 

 





RPy θ  a  sin θ  p  sin θ  δ


4.


Define the distance of point Q above the platen (note the direction of the positive y axis in the figure above).

 

 

ε θ  e  RPy θ 5.

Plot  as a function of crank angle 2. Height of Q Above Platen 60

14

168

Height Above Platen

40

 

20

ε θ

0

 20

 40

0

60

120

180

240

300

360

θ deg

6.

From the graph we see that the coupler point Q is above the platen when the crank angle is greater than 168 deg and less than 14 deg. To find the horizontal stroke during that range of 2, calculate the x-components of any point on the coupler, say point A, for those two crank angles and subtract them. Ax1  a  cos( 14 deg)

Ax1  38.812

Ax2  a  cos( 168  deg)

Ax2  39.126

Horizontal stroke when above the platen normalized by dividing by the crank length Stroke  7.

Ax1  Ax2

Stroke  1.95

a

times the crank length

Links 1, 4, 5, and 6 constitute a Grashoff crank-rocker-rocker. The extreme positions of the output rocker (link 6) occur when links 4 and 5 are in extended and overlapping toggle positions (see Figure 3-1b in the text for example, but in this case the mechanism is in the crossed circuit). C2

29.609°

C1

6

O6

B1

O5 5

B2



Given link lengths: LO5B  13

L7  193

LO6C  92

LO5O6  128

In the first position (links 5 and 7 extended), the angle between link 6 and the ground link is:

 L 2  L 2   L  L  2  O6C O5O6 O5B 7  α  acos     L 2 L O6C O5O6  

α  138.312 deg

In the second position (links 5 and 7 overlapping), the angle between link 6 and the ground link is:

 LO6C2  LO5O62   L7  LO5B 2 α  acos   2  LO6C LO5O6  

α  108.702 deg

The total angular displacement of link 6 is the difference between these two angles.

Δ12  α  α

Δ12  29.609 deg




Figure P4-7 shows a power hacksaw, used to cut metal. Link 5 pivots at O5 and its weight forces the saw blade against the workpiece while the linkage moves the blade (link 4) back and forth on link 5 to cut the part. It is an offset slider-crank mechanism. The dimensions are shown in the figure. For one revolution of the driving link 2 of the hacksaw mechanism on the cutting stroke, find and plot the horizontal stroke of the saw blade as a function of the angle of link 2.

Given:

Link Lengths:

3

Crank length, L2

a  75 mm

Coupler length, L3

b  170  mm

Offset

c  45 mm

B

A 4

5

2 O2 O5

1

Assumptions: The arm that guides the slider (hacksaw blade carrier) remains horizontal throughout the stroke. Solution:


1.

This is a slider-crank mechanism in the crossed circuit. The offset is the vertical distance from the horizontal centerline through O2 to point B.

2.

Establish 2 as a range variable: θ  0  deg 2  deg  360  deg

3.

Determine 3 and d using equations 4.16a and 4.17.

 

 a  sin θ  c   b  

θ θ  asin

 

 

  

d θ  a  cos θ  b  cos θ θ

Plot the blade (point B) displacement as a function of crank angle. Hacksaw Blade Stroke  50

 100 Blade displacement, mm

4.

 

d θ

 150

mm

 200

 250

0

60

120


240

300

360




Given:

For the linkage in Figure P4-8, find its limit (toggle) positions in terms of the angle of link O2A referenced to the line of centers O2O4 when driven from link O2A. Then calculate and plot the xy coordinates of coupler point P between those limits, referenced to the line of centers O2O4. P Link lengths: Input (O2A)

a  5.00 in

Coupler (AB)

b  4.40 in

Rocker (O4B)

c  5.00 in

Ground link

d  9.50 in

y

Y

Coupler point data:

B

p  8.90 in δ  56 deg

3 A

4

Coordinate transformation angle: x

2

α  14 deg

O4 1

14.000° X

O2

See Figure P4-8 and Mathcad file P0421. Solution: 1. Define the coordinate systems. The local frame has origin at O2 with the positive x axis going through O4. Let the global frame also have its origin at O2 with the positive X axis to the right. 2.




Using equations 4.37, determine the crank angles (relative to the line AD) at which links 3 and 4 are in toggle. 2

arg1 

2

2



2 a d 2

arg2 

2

a d b c 2

2

a d b c 2 a d

2





arg1  1.209 arg2  0.283

θ2toggle  73.6 deg

The other toggle angle is the negative of this. 4.

Define one cycle of the input crank between limit positions: θ  θ2toggle θ2toggle  1  deg  θ2toggle

5.

Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12.


K1 


d a

K1  1.9000

 

2

d

K4 

K5 

b

K4  2.1591

 

2

2

c d a b

2

2 a b

K5  2.4911

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

6.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

Use equation 4.13 to find values of 3 for the open circuit.

 





 

 2  4 Dθ F θ 

 

θ θ  2   atan2 2  D θ E θ  7.

E θ

Use equations 4.31 to define the x- and y-components of the vector RP. RP  RA  RPA

  

 


 








 

 

  



 

RPx θ  a  cos θ  p  cos θ θ  δ 8.

 



Transform the coupler point coordinates in the local frame to the global frame using coordinate transformation equations.

 

 

 

 

 

 

XP θ  RPx θ  cos α  RPy θ  sin α YP θ  RPx θ  sin α  RPy θ  cos α Plot the coordinates of the coupler point in the global system.

COUPLER CURVE 1.2

1

0.8

Y

9.

  

RPy θ  a  sin θ  p  sin θ θ  δ

0.6

0.4

0.2

0  0.4

 0.2

0

0.2 X

0.4

0.6




For the walking beam mechanism of Figure P4-9, calculate and plot the x and y components of the position of the coupler point P for one complete revolution of the crank O2A. Hint: Calculate them first with respect to the ground link O2O4 and then transform them into the global XY coordinate system (i.e., horizontal and vertical in the figure).

Given:

Link lengths:

Coupler point data:

Ground link

d  2.22

Crank

a  1

Coupler

b  2.06

Rocker

c  2.33

1.

δ  31.000 deg

α  26.5 deg

Coordinate transformation angle: Solution:

p  3.06


Define the coordinate systems. The local frame has origin at O2 with the positive x axis going through O4. Let the global frame also have its origin at O2 with the positive X axis to the right. Y

x

y O4 4

1

26.500° X

O2

P

2 A 3 B

2.

Define one revolution of the input crank: θ  0  deg 2  deg  360  deg

3.


d

K4 

a

K1  2.2200

 

2

d

K5 

b

K4  1.0777

 

2

2 a b

K5  1.1512

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

4.


 





 

 

θ θ  2   atan2 2  D θ E θ  5.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

 2  4 Dθ F θ 

E θ


2

c d a b

2



  

 


 








 

 

  



 


 

  

Transform the coupler point coordinates in the local frame to the global frame using coordinate transformation equations.

 

 

 

 

 

 

XP θ  RPx θ  cos α  RPy θ  sin α YP θ  RPx θ  sin α  RPy θ  cos α Plot the coordinates of the coupler point in the global system.

COUPLER CURVE 0.5

0

Y

7.




 0.5

1

 1.5

2

2.5

3

3.5 X

4

4.5




For the linkage in Figure P4-10, calculate and plot the angular displacement of links 3 and 4 and the path coordinates of point P with respect to the angle of the input crank O2A for one revolution.

Given:

Link lengths:

B

y 3

Ground link

d  2.22

Crank

a  1.0

Coupler

b  2.06

Rocker

c  2.33

b

Coupler point data: p  3.06

P p

A

δ  31.00  deg

2

a

4

2

4

c d

Solution:

x

1

O2

O4


1.


2.


d a

K1  2.2200

 

d

K2 

2

K3 

c

K2  0.9528

 

2

2

a b c d

2

2 a c

K3  1.5265

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 

 



 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  3.

B θ

If the calculated value of 4 is greater than 2, subtract 2 from it.

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ 4.


2

d

K5 

b

 

2

2

c d a b

2

K4  1.0777

2 a b

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

5.


 





 

 

θ θ  2   atan2 2  D θ E θ  6.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

 2  4 Dθ F θ 

E θ


 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

K5  1.1512


7.


Plot 3 and 4 as functions of the crank angle 2 (measured from the ground link). Angular Displacement of Coupler  240

Coupler angle, deg

 260

   280

θ θ  deg

 300  320  340

0

60

120

180

240

300

360

300

360


Angular Displacement of Rocker  200

Rocker angle, deg

 220

 

θ θ 

 240

deg  260

 280

0

60

120

180

240


8.


  

 


 








 

 

  




 

 

  




Plot the coordinates of the coupler point in the local xy coordinate system.



COUPLER POINT CURVE 3

2.5

Y

2

1.5

1

0.5 1.5

2

2.5

3 X

3.5

4




For the linkage in Figure P4-11, calculate and plot the angular displacement of links 3 and 4 with respect to the angle of the input crank O2A for one revolution.

Given:

Link lengths: Link 2

a  2.00 in

Link 3

b  8.375  in

Link 4

c  7.187  in

Link 1

d  9.625  in

A 3 2

B

2

O2 4

1

O4

Solution:


1.


2.


d a

K1  4.8125

 

2

d

K2 

K3 

c

K2  1.3392

 

2

2

a b c d

2

2 a c

K3  2.7186

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 

 



 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  3.

B θ


 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ 4.


2

d

K5 

b

 

2

2

c d a b

2

K4  1.1493

2 a b

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

5.


 





 

 

θ θ  2   atan2 2  D θ E θ  6.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

 2  4 Dθ F θ 

E θ


 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

K5  3.4367


Plot 3 and 4 as functions of the crank angle 2 (measured from the ground link).

Angular Displacement of Coupler

Coupler angle, deg

 290

 300

 310

 320

 330

0

60

120

180

240

300

360

300

360

Crank angle, deg

Angular Displacement of Rocker  220

 230 Rocker angle, deg

7.


 240

 250

 260

0

60

120

180 Crank angle, deg

240




For the linkage in Figure P4-12, find its limit (toggle) positions in terms of the angle of link O2A referenced to the line of centers O2O4 when driven from link O2A. Then calculate and plot the angular displacement of links 3 and 4 and the path coordinates of point P with respect to the angle of the input crank O2A over its possible range of motion referenced to the line of centers O2O4.

Given:

Link lengths: Input (O2A)

a  0.785

Coupler (AB)

b  0.356

Rocker (O4B)

c  0.950

Ground link

d  0.544

A

1.

B

158.286° c

a O2

Coupler point data: p  1.09 Solution:

b

O4

d

δ  0  deg





double rocker

Using the geometry defined in Figure 3-1a in the text, determine the input crank angles (relative to the line O2O4) at which links 2 and 3, and 3 and 4 are in toggle.

 d2  ( a  b ) 2  c2  θ  acos  2 d ( a  b) 

θ  55.937 deg

 a2  d 2  ( b  c) 2  2 a d  

θ  acos 3.

θ  158.286 deg

Define one cycle of the input crank between limit positions: θ  θ θ  1  deg  θ

4.


d

K2 

a

K1  0.6930

 

2

d

K3 

c

K2  0.5726

 

K3  1.1317

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 

 



 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ 

 2  4 A θ Cθ 

B θ

2

2

a b c d 2 a c

2


5.



 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 6.

Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. 2

d

K4 

K5 

b

 

2

2

c d a b

2

K4  1.5281

2 a b

 

K5  0.2440

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

7.




 



 

 

θ θ  2   atan2 2  D θ E θ  8.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

 2  4 Dθ F θ 

E θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 9.



360 300 240 180 120 60 0 50

60

70

80

90

100

110

120

130

Crank angle, deg Coupler Rocker

10. Use equations 4.31 to define the x- and y-components of the vector RP. RP  RA  RPA

  

 


140

150

160


 









 

 

  



RPx θ  a  cos θ  p  cos θ θ  δ

 

 

11. Plot the coordinates of the coupler point in the local xy coordinate system.

COUPLER POINT PATH

Coupler Point Coordinate - y

2

1.5

1

0.5

0

0

0.5

  




1

Coupler Point Coordinate - x

1.5

2





Given:

Link lengths: a  0.86

Input (O2A) Coupler (AB)

b  1.85

Rocker (O4B)

c  0.86

Ground link

d  2.22

1.

B c

a O2

Coupler point data: p  1.33 Solution:

116.037° b

A

O4

d

δ  0  deg






arg1 

2

2



2 a d 2

arg2 

2

a d b c

2

2

a d b c 2 a d

2



b c

arg1  1.228

a d b c

arg2  0.439

a d




Define one cycle of the input crank between limit positions: θ  θ2toggle θ2toggle  1  deg  θ2toggle

4.


d

K2 

a

K1  2.5814

 

d c

K2  2.5814

 

2

K3 

K3  2.0181

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

2

2

a b c d

C θ  K1   K2  1   cos θ  K3

2 a c

2


 






 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 6.


d

K4 

K5 

b

 

2

2

c d a b

2

K4  1.2000

2 a b

 

K5  2.6244

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

7.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ


 





 

 

θ θ  2   atan2 2  D θ E θ  8.

 2  4 Dθ F θ 

E θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 9.



360 300 240 180 120 60 0  120

 80

 40

0

40



  

 


80

120


 









 

 

  



 


 

  


COUPLER POINT PATH 1

Y

0.5

0

 0.5 0.5

1




1.5 X

2

2.5




For the linkage in Figure P4-13, find its limit (toggle) positions in terms of the angle of link O4B referenced to the line of centers O4O2 when driven from link O4B. Then calculate and plot the angular displacement of links 2 and 3 and the path coordinates of point P with respect to the angle of the input crank O4B over its possible range of motion referenced to the line of centers O4O2.

Given:

Link lengths:

116.037° b 3 4

a  0.86

Input (O4B)

b  1.85

Coupler (AB) Rocker (O2A)

c  0.86

Ground link

d  2.22

x


c A

O2

B a O4

d

δ  0  deg y

Solution: 1.





Using equations 4.33, determine the crank angles (relative to the line O4O2) at which links 2 and 3 are in toggle. 2

arg1 

2

2



2 a d 2

arg2 

2

a d b c

2

2

a d b c 2 a d

2





arg1  1.228

arg2  0.439



Define one cycle of the input crank between limit positions: θ  θ4toggle θ4toggle  1  deg  θ4toggle

4.


d a

K2 

d c

2

K3 

K1  2.5814

K2  2.5814

    B θ  2  sin θ

  C θ  K1   K2  1   cos θ  K3

A θ  cos θ  K1  K2 cos θ  K3

2

2

a b c d

K3  2.0181

2 a c

2


 






 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 6.


d

K4 

 

b

K5 

2

2

c d a b

2

K4  1.2000

2 a b

 

K5  2.6244

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

7.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ


 





 

 

θ θ  2   atan2 2  D θ E θ  8.

 2  4 Dθ F θ 

E θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 9.



360 300 240 180 120 60 0  120

 80

 40

0

40


10. Use equations 4.31 to define the x- and y-components of the vector RP. RP  RB  RPB

  

 

RB  a  cos θ  j  sin θ

80

120


 








RPB  p  cos θ  δ  j  sin θ  δ

 

 

  



 

RPx θ  a  cos θ  p  cos θ θ  δ

 

  

11. Plot the coordinates of the coupler point in the local xy coordinate system. COUPLER POINT PATH 1

Y

0.5

0

 0.5

1

0

0.25

0.5



RPy θ  a  sin θ  p  sin θ θ  δ

0.75 X

1

1.25

1.5




For the rocker-crank linkage in Figure P4-14, find the maximum angular displacement possible for the treadle link (to which force F is applied). Determine the toggle positions. How does this work? Explain why the grinding wheel is able to fully rotate despite the presence of toggle positions when driven from the treadle. How would you get it started if it was in a toggle position?

Given:

Link lengths:

x

Input (O2A)

a  600  mm

Coupler (AB)

b  750  mm

Rocker (O4B)

c  130  mm

Ground link

d  900  mm

B

B''

c

O4

B'

b

d

43.331° A'' 25.182° a

A

O2

A'

y

Solution: 1.


Use Figure 3-1(b) in the text to calculate the angles that link O2A makes with the ground link in the toggle positions.

 a2  d 2  ( b  c) 2  θ  acos 2 a d  

θ  43.331 deg

 a2  d 2  ( b  c) 2  2 a d  

θ  68.513 deg

θ  acos 2.

Subtract these two angles to get the maximum angular displacement of the treadle.   θ  θ

3.

  25.182 deg

Despite having transmission angles of 0 deg twice per revolution, the mechanism will work. That is, one will be able to drive the grinding wheel from the treadle (link 2). The reason is that the grinding wheel will act as a flywheel and will carry the linkage through the periods when the transmission angle is low. Typically, the operator will start the motion by rotating the wheel by hand if it is in or near a toggle position.





Given:


a  0.72

Coupler (AB)

b  0.68

Rocker (O4B)

c  0.85

Ground link

d  1.82

P A B


1.

O4

O2

δ  54 deg Solution:

55.355°






arg1 

2

2



2 a d 2

arg2 

2

a d b c 2

2

a d b c 2 a d

2



b c

arg1  1.451

a d b c

arg2  0.568

a d




Define one cycle of the input crank between limit positions: θ  θ2toggle θ2toggle  0.5 deg  θ2toggle

4.


d a

K1  2.5278

 

K2 

 

d c

K2  2.1412

 

2

K3 

K3  3.3422

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

2

2

a b c d

C θ  K1   K2  1   cos θ  K3

2 a c

2


 






 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 6.


2

d

K5 

b

 

2

2

c d a b

2

K4  2.6765

2 a b

 

K5  3.6465

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

7.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

Use equation 4.13 to find values of θ3 for the open circuit.

 





 

 

θ θ  2   atan2 2  D θ E θ  8.

 2  4 Dθ F θ 

E θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 9.

Plot 3 and 4 as functions of the crank angle 2 (measured from the ground link).

Angular Displacement of Coupler & Rocker


360 300 240 180 120 60 0  60

 45

 30

 15

0

15

30



45

60



  

 


 








 

 

  



 


 

  


COUPLER POINT PATH 1.4

1.2

Y

1

0.8

0.6

0.4

0.2 0.2

0.4

0.6




0.8 X

1

1.2

1.4




For the linkage in Figure P4-15, find its limit (toggle) positions in terms of the angle of link O4B referenced to the line of centers O4O2 when driven from link O4B. Then calculate and plot the angular displacement of links 2 and 3 and the path coordinates of point P with respect to the angle of the input crank O4B over its possible range of motion referenced to the line of centers O4O2.

Given:

Link lengths: Input (O4B)

a  0.85

Coupler (AB)

b  0.68

Rocker (O2A)

c  0.72

Ground link

d  1.82

P 47.885° c

Coupler point data: p  0.792 δ  82.032 deg Solution: 1.

B A

a

b

O4

O2





Using equations 4.37, determine the crank angles (relative to the line O4O2) at which links 2 and 3 are in toggle. 2

arg1 

2

2



2 a d 2

arg2 

2

a d b c 2

2

a d b c 2 a d


2



b c

arg1  1.304

a d b c

arg2  0.671

a d



Define one cycle of the input crank between limit positions: θ  θ4toggle θ4toggle  0.5 deg  θ4toggle

4.


d

K2 

a

K1  2.1412

 

d c

K2  2.5278

 

2

K3 

K3  3.3422

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

2

2

a b c d

C θ  K1   K2  1   cos θ  K3

2 a c

2


 






 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 6.


2

d

K5 

b

 

2

2

c d a b

2

K4  2.6765

2 a b

 

K5  3.4420

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

7.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ


 





 

 

θ θ  2   atan2 2  D θ E θ  8.

 2  4 Dθ F θ 

E θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 9.



360 300 240 180 120 60 0  60

 45

 30

 15

0

15

30


10. Use equations 4.27 to define the x- and y-components of the vector RP. RP  RB  RPB

45

60



  

 

RB  a  cos θ  j  sin θ

 







RPB  p  cos θ  δ  j  sin θ  δ

 

 

  



 

RPx θ  a  cos θ  p  cos θ θ  δ

 

  


COUPLER POINT PATH 1.5

Y

1

0.5

0

0

0.2

0.4

0.6 X



RPy θ  a  sin θ  p  sin θ θ  δ

0.8

1




Write a computer program (or use an equation solver such as Mathcad, Matlab, or TKSolver) to find the roots of y = 9x2 + 50x - 40. Hint: Plot the function to determine good guess values.

Solution:


1.

Plot the function. 2

x  10 9.5  10

f ( x)  9  x  50 x  40

200

100

f ( x)

0

 100

 200  10

8

6

4

2

0

2

x

2.

From the graph, make guesses of x1  6 , x2  1

3.

Define the program using the pseudo code in the text. nroot( f df x) 

y  f ( x) y  TOL

return x if

while y  TOL xx

y df ( x)

y  f ( x) x

where,

3

TOL  1.000  10

4.

Define the derivative of the given function. df ( x)  18 x  50

5.

Use the program to find the roots. r1  nroot f df x1

r1  6.265

r2  nroot f df x2

r2  0.709

4

6

8

10




Write a computer program (or use an equation solver such as Mathcad, Matlab, or TKSolver) to find the roots of y = -x3 - 4x2 + 80x - 40. Hint: Plot the function to determine good guess values.

Solution:


1.

Plot the function. 3

x  15 14.5  10

2

f ( x)  x  4  x  80 x  40

200

100

f ( x)

0

 100

 200  20

 15

 10

5

0

x

2.

From the graph, make guesses of x1  11 , x2  0 , x3  7

3.

Define the program using the pseudo code in the text. nroot( f df x) 

y  f ( x) y  TOL

return x if


y df ( x)

y  f ( x) x

where,

3

TOL  1.000  10

2

4.

Define the derivative of the given function. df ( x)  3  x  8  x  80

5.

Use the program to find the roots. r1  nroot f df x1

r1  11.355


r2  0.515


r3  6.840

5

10




Figure 4-18 (p. 193) plots the cubic function from equation 4.34. Write a computer program (or use an equation solver such as Mathcad, Matlab, or TKSolver) to investigate the behavior of the Newton-Raphson algorithm as the initial guess value is varied from x = 1.8 to 2.5 in steps of 0.1. Determine the guess value at which the convergence switches roots. Explain this root-switching phenomenon based on your observations from this exercise.

Solution:


1.

Define the range of the guess value, the function, and the derivative of the function. xguess  1.8 1.9  2.5 3

2

2

f ( x)  x  2  x  50 x  60 2.

df ( x)  3  x  4  x  50

Define the root-finding program using the pseudo code in the text. nroot( f df x) 

y  f ( x) y  TOL

return x if


y df ( x)

y  f ( x) x 3.

Find the roots that correspond to the guess values. r( xguess)  nroot( f df xguess) 1

1

f ( xguess) df ( xguess)

1

1

1

1.800

1

33.080

1

-2.362

1

-1.177

2

1.900

2

31.570

2

-2.564

2

-1.177

3

2.000

3

30.000

3

-2.800

3

-1.177

2.100 df ( xguess)  4

28.370

nextx( xguess)  4

-3.079

r( xguess)  4

-1.177

xguess  4

4.

nextx( xguess)  xguess 

5

2.200

5

26.680

5

-3.410

5

-1.177

6

2.300

6

24.930

6

-3.807

6

-1.177

7

2.400

7

23.120

7

-4.289

7

6.740

8

2.500

8

21.250

8

-4.882

8

-7.562

Find the roots of the derivative (values of x where the slope is zero). ddf ( x)  6  x  4

5.

xz1  nroot( df ddf 5 )

xz1  4.803

xz2  nroot( df ddf 4 )

xz2  3.470

For guess values up to 2.3, the root found is that whose slope is nearly the same as the slope of the function at the guess value. At 2.4, the value of x that is calculated next results in a slope that throws the next x-value to the right of the extreme function value at x = 3.470. Subsequent estimates of x then follow down the slope to x = 6.740. At a guess value of 2.5, the value of x that is calculated next is to the left of the extreme function value at x = -4.803. Subsequent estimates of x follow up the slope to x = -7.562.




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular position of link 4 and the position of slider 6 in Figure 3-33 as a function of the angle of input link 2.

Given:

Link lengths: Input crank (L2)

a  2.170

Fourbar coupler (L3)

b  2.067

Output crank (L4)

c  2.310

Sllider coupler (L5)

e  5.40

Fourbar ground link (L1) Solution:

d  1.000


1.

This sixbar drag-link mechanism can be analyzed as a fourbar Grashof double crank in series with a slidercrank mechanism using the output of the fourbar, link 4, as the input to the slider-crank.

2.


3.

Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit) in the global XY system. K1 

d

K2 

a

K1  0.4608

 

2

d

K3 

c

K2  0.4329

 

2

2

a b c d

2

2 a c

K3  0.6755

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 

 





 

 

θ θ  2   atan2 2  A θ B θ  4.

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ

 2  4 A θ Cθ   102 deg

B θ

If the calculated value of 4 is greater than 2, subtract 2 from it and if it is negative, make it positive.

 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 5.

Determine the slider-crank motion using equations 4.16 and 4.17 with 4 as the input angle.

 

 c sin θ θ   π e  

 

    e cosθθ

θ θ  asin

f θ  c cos θ θ 6.

Plot the angular position of link 4 and the position of link 6 as functions of the angle of input link 2. See next page.



Angular Position of Link 4 360 315 270

 

θ  θ

225 180

deg 135 90 45 0

0

45

90

135

180

225

270

315

360

315

360

θ deg

Position of Slider 6 With Respect to O4 8 7.167 6.333

 

f θ

5.5 4.667 3.833 3

0

45

90

135

180 θ deg

225

270




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angles at points B and C of the linkage in Figure 3-33 as a function of the angle of input link 2.

Given:


a  2.170


b  2.067

Output crank (L4)

c  2.310


e  5.40

d  1.000

Fourbar ground link (L1) Solution:


1.

This sixbar drag-link mechanism can be analyzed as a fourbar Grashof double crank in series with a slidercrank mechanism using the output of the fourbar, link 4, as the input to the slider-crank.

2.


3.


d a

K1  0.4608

 

2

d

K2 

K3 

c

K2  0.4329

 

2

2

a b c d

2

2 a c

K3  0.6755

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 

 



 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  4.

B θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 5.


 

2

d b

K5 

2

2

c d a b

2

K4  0.4838

2 a b

 

K5  0.5178

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 6.

Use equation 4.13 to find 3 for the open circuit.

 





 

 

θ θ  2   atan2 2  D θ E θ  7.

 2  4 Dθ F θ 

E θ


 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ


 


  

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 8.

Calculate (using equations 4.32) and plot the transmission angle at B.

θtransB1 θ  θ θ  θ θ

 

θtransB θ  if  θtransB1 θ 

π 2

 

 

π  θtransB1 θ θtransB1 θ



Transmission Angle at B 40 35 30

θtransB θ

25 20

deg 15 10 5 0

0

45

90

135

180

225

270

315

360

θ deg

9.


 c sin θ θ   π e  

 


10. Calculate (using equations 4.32) and plot the transmission angle at C.

θtransC1 θ  θ θ

 

θtransC θ  if  θtransC1 θ 

π 2

 

 

π  θtransC1 θ θtransC1 θ



Transmission Angle at C 30 25

θtransC θ

20 15

deg 10 5 0

0

45

90

135

180 θ deg

225

270

315

360




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of the coupler point of the approximate straight-line linkage shown in Figure 3-29f (p. 142). Use program Fourbar to check your result.

Given:


a  1.000

Coupler (AB)

b  1.600

Rocker (O4B)

c  1.039

Ground link

d  1.200

p  2.690

Coupler point data:

α  60 deg

Coordinate rotation angle: Solution: 1.

δ  0  deg

See Figure 3-29f and Mathcad file P0436.

Check the Grashof condition of the linkage and determine its Baker classification. Condition( a b c d ) 


Condition( a b c d )  "non-Grashof" Since b (link 3) is the longest link and the linkage is non-Grashof, this is a Class 3 triple rocker. Using Figure 3-1a as a guide, determine the limiting values of 2 at the toggle positions. For links 2 and 3 colinear:

 ( b  a) 2  d2  c2 π  2 d ( b  a) 

θ  acos

θ  240 deg

For links 3 and 4 colinear:

 a2  d 2  ( b  c) 2  θ  acos 2 d a   2.

θ  27.683 deg

θ  θ

Define one cycle of the input crank (driving through the links 2-3 toggle position): θ  θ θ  1  deg  360  deg  θ

3.


d

K4 

a

K1  1.2000

 

2

d

K5 

b

K4  0.7500

 

2

2 a b

K5  1.2251

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ 4.

 

 

F θ  K1   K4  1   cos θ  K5


2

c d a b

2




 




 

 

θ θ  2   atan2 2  D θ E θ  5.

 2  4 Dθ F θ 

E θ


     RPA  p   cos θ  δ  j  sin θ  δ  RA  a  cos θ  j  sin θ

 

 

  



 


  



Plot the coordinates of the coupler point in the global X,Y coordinate system using equations 4.0b to rotate the local coordinates to a global frame.

 

 

 

 

 

 

PX θ  RPx θ  cos( α)  RPy θ  sin( α) PY θ  RPx θ  sin( α)  RPy θ  cos( α) COUPLER POINT PATH 2

1

Coupler Point Coordinate - Y

6.

 


0

1

2

3

4

0

1

2 Coupler Point Coordinate - X

3

4




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular position of link 6 in Figure 3-34 as a function of the angle of input link 2.

Given:

Link lengths:

Solution: 1.

Input crank (L2)

g  1.556

First coupler (L3)

f  4.248

First rocker (L4)

c  2.125

Third coupler (CD)

b  2.158

Output rocker (L6)

a  1.542

Second ground link (O4O6) d  1.000

Angle CDB

δ  36 deg

Distance (BD)

O2O4 ground link offsets:

h X  3.259

h Y  2.905


Calculate the length of the O2O4 ground link and the angle that it makes with the global XY system. h 

2.

p  3.274

2

hX  hY

2

 hY    hX 

γ  atan

h  4.366

γ  41.713 deg

Calculate the distance BC on link 5. This is the length of vector R51. Also, calculate the angle between vectors R51 and R52 e 

b  p  2  b  p  cos δ 2

2

e  1.986

Second coupler (BC)

 b 2  e2  p 2    2 b e 

α  acos

α  104.305 deg

β  π  α

β  75.695 deg

3.

This is a Stephenson's sixbar linkage similar to the one shown in Figure 4-13. Since the output link 6 is known to rotate 180 deg and return for a full revolution of link 2 we can use links 6, 5, and 4 as a first-stage fourbar with known input (link 6) and then solve for vector loop equations to get the corresponding motion of link 2.

4.

Define the rotation of the output crank: θ  90 deg 91 deg  270  deg

5.

Use equations 4.8a and 4.10 to calculate 4 in the local xy coordinate system as a function of 6 (for the crossed circuit). K1 

d

K2 

a

K1  0.6485

 

2

d

K3 

c

K2  0.4706

 

2

2

a b c d

2

2 a c

K3  0.4938

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ  6.

 2  4 A θ Cθ 

B θ

Use equations 4.12 and 4.13 to calculate 5 in the local xy coordinate system as a function of 6 (for the crossed circuit). K4 

d b

K4  0.463

2

K5 

2

2

c d a b

K5  0.529

2 a b

2


 


 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5



 



 

 

θ θ  2   atan2 2  D θ E θ  7.

 2  4 Dθ F θ 

E θ

Transform the angles for 4 and 52 into the global XY system and define 51 in the global system.

 

 

θ θ  θ θ  90 deg

 

 

 

 

θ θ  θ θ  90 deg θ θ  θ θ  β 8.

Define a vector loop for the remaining links and solve the resulting vector equation by separating it into real and imaginary parts using the method of section 4.5 and the identities of equations 4.9.

Y X

O2 2 R2

R1

6

A

O6

y

D R 5 R52 3 C R4

3

O4 4 x

R51 B R1 + R4 + R51 + R3 - R2 = 0. In this equation the unknowns are 3 and 2. Following the method of Section 4.5, substitute the complex number notation for each position vector and separate the resulting equations into real and imaginary parts:

 

 

 

 

f  cos θ = g  cos θ  G1 f  sin θ = g  sin θ  G2 where

 

    e cosθθ

G1 θ  h  cos γ  c cos θ θ

 



    e sinθθ

G2 θ  h  sin γ  c sin θ θ 9.

Solve these equations in the manner of equations 4.11 and 4.12 using the identities of equations 4.9 gives:

 2  G2θ2  f 2

2

 

g  G1 θ

 

 

G3 θ 

2 g

 

A' θ  G1 θ  G3 θ

 

 

B' θ  2  G2 θ

 

 

 

C' θ  G1 θ  G3 θ



 θ  B'  B'2  4 A'  C' = 2  A' 2

tan 



 



 

 

θ θ  2   atan2 2  A' θ B' θ 

 2  4 A' θ C'θ 

B' θ

10. Plot 6 vs 2 in global XY coordinates: Rotation of Link 6 vs Link 2 320 300 280 260

 

θ  θ

240 220

deg 200 180 160 140 120

0

20

40

60

80 θ deg

100  90

120

140

160

180




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angles at points B, C, and D of the linkage in Figure 3-34 as a function of the angle of input link 2.

Given:

Link lengths:

Solution: 1.

Input crank (L2)

g  1.556

First coupler (L3)

f  4.248

First rocker (L4)

c  2.125

Third coupler (CD)

b  2.158

Output rocker (L6)

a  1.542

Second ground link (O4O6) d  1.000

Angle CDB

δ  36 deg

Distance (BD)

O2O4 ground link offsets:

h X  3.259

h Y  2.905


Calculate the length of the O2O4 ground link and the angle that it makes with the global XY system. h 

2.

p  3.274

2

hX  hY

2

 hY    hX 

γ  atan

h  4.366

γ  41.713 deg

Calculate the distance BC on link 5. This is the length of vector R51. Also, calculate the angle between vectors R51 and R52 e 

b  p  2  b  p  cos δ 2

2

e  1.986

Second coupler (BC)

 b 2  e2  p 2    2 b e 

α  acos

α  104.305 deg

β  π  α

β  75.695 deg

3.

This is a Stephenson's sixbar linkage similar to the one shown in Figure 4-13. Since the output link 6 is known to rotate 180 deg and return for a full revolution of link 2 we can use links 6, 5, and 4 as a first-stage fourbar with known input (link 6) and then solve for vector loop equations to get the corresponding motion of link 2.

4.

Define the rotation of the output crank: θ  90 deg 91 deg  270  deg

5.

Use equations 4.8a and 4.10 to calculate 4 in the local xy coordinate system as a function of 6 (for the crossed circuit). K1 

d

K2 

a

K1  0.6485

 

2

d

K3 

c

K2  0.4706

 

2

2

a b c d

2

2 a c

K3  0.4938

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ  6.

 2  4 A θ Cθ 

B θ

Use equations 4.12 and 4.13 to calculate 5 in the local xy coordinate system as a function of 6 (for the crossed circuit). K4 

d b

K4  0.463

2

K5 

2

2

c d a b

K5  0.529

2 a b

2


 


 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5



 



 

 2  4 Dθ F θ 

 

θ θ  2   atan2 2  D θ E θ  7.

E θ

Transform the angles for 4 and 52 into the global XY system and define 51 in the global system.

 

 

θ θ  θ θ  90 deg

 

 

 

 

θ θ  θ θ  90 deg θ θ  θ θ  β 8.

Define a vector loop for the remaining links and solve the resulting vector equation by separating it into real and imaginary parts using the method of section 4.5 and the identities of equations 4.9.

Y X

O2 2 R2

R1

6

A

O6

y

D R 5 R52 3 C R4

3

O4 4 x

R51 B R1 + R4 + R51 + R3 - R2 = 0. In this equation the unknowns are 3 and 2. Following the method of Section 4.5, substitute the complex number notation for each position vector and separate the resulting equations into real and imaginary parts:

 

 

 

 

g  cos θ = f  cos θ  G1 g  sin θ = f  sin θ  G2 where

G1 θ  h  cos γ  c cos θ θ

 

    e cosθθ

G2 θ  h  sin γ  c sin θ θ

 

9.



    e sinθθ

Solve these equations for 2 in the manner of equations 4.11 and 4.12 using the identities of equations 4.9 gives:

 2  G2θ2  f 2

2

 

g  G1 θ

 

 

G3 θ 

2 g

 

A' θ  G1 θ  G3 θ

 

 

B' θ  2  G2 θ

 

 

 

C' θ  G1 θ  G3 θ



 θ  B'  B'2  4 A'  C' = 2  A' 2

tan 



 



 

 

θ θ  2   atan2 2  A' θ B' θ 

 2  4 A' θ C'θ 

B' θ

10. Solve these equations for 3 in the manner of equations 4.11 and 4.12 using the identities of equations 4.9 gives:

 2  G2θ2  f 2

2

 

g  G1 θ

 

 

G4 θ 

2 f

 

 

D' θ  G1 θ  G4 θ

 

E' θ  2  G2 θ

 

 

 

F' θ  G1 θ  G4 θ

 θ  E'  E'2  4 D' F' tan   = 2  D' 2

 





 

 

θ θ  2   atan2 2  D' θ E' θ 

 2  4 D'θ F'θ 

E' θ

11. Calculate (using equations 4.32) and plot the transmission angle at B.

θtransB1 θ  θ θ  θ θ

 

θtransB θ  if  θtransB1 θ 

π 2

 

 

π  θtransB1 θ θtransB1 θ



Transmission Angle at B 90 80 70 60

θtransB θ 50 deg

40 30 20 10 0 125

150

175

200

225

 

θ θ  deg


θtransC1 θ  θ θ  θ θ

250

275

300

325



 

θtransC θ  if  θtransC1 θ 

π 2

 

 

π  θtransC1 θ θtransC1 θ



Transmission Angle at C 60 55 50

θtransC θ

45

deg 40 35 30 125

150

175

200

225

 

250

275

300

325

θ θ  deg

13. Calculate (using equations 4.32) and plot the transmission angle at D.

θtransD1 θ  θ  θ θ

 

θtransD θ  if  θtransD1 θ 

π 2

 

 

π  θtransD1 θ θtransD1 θ



Transmission Angle at D 60 50 40

θtransD θ

30

deg 20 10 0 125

150

175

200

225

 

θ  θ deg

250

275

300

325




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of the coupler point of the approximate straight-line linkage shown in Figure 3-29g (p. 142). Use program Fourbar to check your result.

Given:


a  1.000

Coupler (AB)

b  1.200

Rocker (O4B)

c  1.167

Ground link

d  2.305

p  1.5

Coupler point data:

α  30 deg


δ  180  deg

See Figure 3-29g and Mathcad file P0439.



Condition( a b c d )  "non-Grashof" Since d (link 1) is the longest link and the linkage is non-Grashof, this is a Class 1 triple rocker. Using Figure 3-1a as a guide, determine the limiting values of 2 at the toggle positions. For links 3 and 4 colinear:

 a2  d 2  ( b  c) 2  2 d a  

θ  acos 1.

θ  81.136 deg

θ  θ

Define one cycle of the input crank: θ  θ θ  0.5 deg  θ

2.


d

K4 

a

K1  2.3050

 

2

d

K5 

b

K4  1.9208

 

2

2 a b

K5  2.6630

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

3.


 





 

 

θ θ  2   atan2 2  D θ E θ  4.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

 2  4 Dθ F θ 

E θ

Use equations 4.31 to define the x- and y-components of the vector RP.

2

c d a b

2



RP  RA  RPA


 

 

  



 


  



Plot the coordinates of the coupler point in the global X,Y coordinate system using equations 4.0b to rotate the local coordinates to a global frame.

 

 

 

 

 

 

PX θ  RPx θ  cos( α)  RPy θ  sin( α) PY θ  RPx θ  sin( α)  RPy θ  cos( α) COUPLER POINT PATH 2

1


5.

 


0

1

2 2

 1.5

1 Coupler Point Coordinate - X

 0.5

0





Given:

Link lengths:

Solution:

Input crank (L2)

a  1.00

First coupler (L3)

b  3.80

Common rocker (O4B)

c  1.29

Second coupler (L5)

b'  1.29

First ground link (O2O4)

d  3.86

Common rocker (O4C)

a'  1.43

Output rocker (L6)

c'  0.77

Second ground link (O4O6) d'  0.78

Angle BO4C

α  157  deg


1.

This sixbar drag-link mechanism can be analyzed as two fourbar linkages in series that use the output of the first fourbar, link 4, as the input to the second fourbar.

2.


3.


d

K2 

a

K1  3.8600

 

2

d

K3 

c

K2  2.9922

 

2

2

a b c d

2

2 a c

K3  1.2107

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  4.

B θ

Use equations 4.8a and 4.10 to calculate 6 as a function of 2 (for the open circuit).

 

K' 1 

d'

K' 2 

a'

K' 1  0.5455

 

 

θ θ  θ θ  α

Input angle to second fourbar:

2

d'

K' 3 

c'

K' 2  1.0130

2

2

2  a' c'

K' 3  0.7184

    K'1  K'2 cosθθ  K'3

A' θ  cos θ θ

 

  

 

 





 

 

θ θ  2   atan2 2  A' θ B' θ 

5.

    K'3

C' θ  K' 1   K' 2  1   cos θ θ

B' θ  2  sin θ θ

 2  4 A' θ C'θ 

B' θ

Plot the angular position of link 6 as a function of the angle of input link 2.

2

a'  b'  c'  d'



Angular Position of Link 6 100 75 50 25

 

θ  θ

0  25

deg  50  75  100  125  150

0

45

90

135

180 θ deg

225

270

315

360




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angles at points B, C, and D of the linkage in Figure 3-35 as a function of the angle of input link 2.

Given:

Link lengths:

Solution:

Input crank (L2)

a  1.00

First coupler (L3)

b  3.80

Common rocker (O4B)

c  1.29

Second coupler (L5)

b'  1.29


d  3.86

Common rocker (O4C)

a'  1.43

Output rocker (L6)

c'  0.77


Angle BO4C

α  157  deg


1.

This sixbar drag-link mechanism can be analyzed as two fourbar linkages in series that use the output of the first fourbar, link 4, as the input to the second fourbar.

2.


3.


d

K2 

a

K1  3.8600

 

2

d

K3 

c

K2  2.9922

 

2

2 a c

K3  1.2107

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ  4.

 2  4 A θ Cθ 

B θ

Use equations 4.12 and 4.13 to calculate 3 as a function of 2 (for the crossed circuit). K4 

2

d

K5 

b

K4  1.016

 

2

2

c d a b

2

2 a b

K5  3.773

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5

 





 

 

θ θ  2   atan2 2  D θ E θ  5.

 2  4 Dθ F θ 

E θ


θtransB1 θ  θ θ  θ θ

 


π 2

 

2

a b c d

 




2



Transmission Angle at B 90 82.5 75 67.5


60

deg 52.5 45 37.5 30

0

45

90

135

180

225

270

315

360

θ deg

6.


 

K' 1 

d'

K' 2 

a'

K' 1  0.5455

 

 

θ θ  θ θ  α


2

d'

K' 3 

c'

K' 2  1.0130

2

2

2

a'  b'  c'  d' 2  a' c'

K' 3  0.7184

    K'1  K'2 cosθθ  K'3

A' θ  cos θ θ

 

  

 



 



 

 2  4 A' θ C'θ 

 

θ θ  2   atan2 2  A' θ B' θ  7.

    K'3

C' θ  K' 1   K' 2  1   cos θ θ

B' θ  2  sin θ θ

B' θ

Use equations 4.12 and 4.13 to calculate 5 as a function of 2 (for the crossed circuit). K' 4 

2

d'

K' 5 

b'

K' 4  0.605

 

2

2

2

c'  d'  a'  b' 2  a' b'

K' 5  1.010

    K'1  K'4 cosθθ  K'5

D' θ  cos θ θ

 

 

    K'5

F' θ  K' 1   K' 4  1   cos θ θ

 





 

 

θ θ  2   atan2 2  D' θ E' θ  8.

 2  4 D'θ F'θ 

E' θ

Calculate (using equations 4.32) and plot the transmission angle at C.

θtransC1 θ  θ θ  θ θ

 


π 2

 

  

E' θ  2  sin θ θ

 






Transmission Angle at C 30

24


18

deg 12

6

0

0

45

90

135

180

225

270

315

360

θ deg

9.

Calculate (using equations 4.32) and plot the transmission angle at D.

θtransD1 θ  θ θ  θ θ

 

θtransD θ  if  θtransD1 θ 

π 2

 

 

π  θtransD1 θ θtransD1 θ



Transmission Angle at D 120 100 80

θtransD θ

60

deg 40 20 0

0

45

90

135

180 θ deg

225

270

315

360




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of the coupler point of the approximate straight-line linkage shown in Figure 3-29h (p. 142). Use program Fourbar to check your result.

Given:


a  1.000

Coupler (AB)

b  1.000

Rocker (O4B)

c  1.000

Ground link

d  2.000

p  2.0

Coupler point data: Solution: 1.

δ  0  deg

See Figure 3-29h and Mathcad file P0442.




 a2  d 2  ( b  c) 2  2 d a  

θ  acos 1.

θ  75.522 deg

θ  θ

Define one cycle of the input crank: θ  θ θ  0.25 deg  θ

2.

Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. d

K1 

K4 

a

K1  2.0000

 

2

d

K5 

b

K4  2.0000

 

2

2 a b

K5  2.5000

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

3.


 





 

 

θ θ  2   atan2 2  D θ E θ  4.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

 2  4 Dθ F θ 

E θ


  

 


2

c d a b

2


 









 

 

  



 


 

  

COUPLER POINT PATH 2


1.5

1

0.5

0

1

1.5




2 Coupler Point Coordinate - X

2.5

3





Given:

Link lengths:

Solution:

Input crank (L2)

a  0.450

First coupler (L3)

b  0.990

Common rocker (O4B)

c  0.590


d  1.000

Common rocker (O4C)

a'  0.590

Second coupler (CD)

b'  0.325

Output rocker (L6)

c'  0.325


Link 7 (L7)

e  0.938

Link 8 (L8)

f  0.572

Link 5 extension (DE)

p  0.823

Angle DCE

δ  7.0 deg

Angle BO4C

α  128.6  deg


1.

This eightbar can be analyzed as a fourbar (links 1, 2, 3, and 4) with its output (link 4) as the input to another fourbar (links 1, 4, 5, and 6). Since links 1 and 4 are common to both, we have an eightbar linkage with links 7 & 8 included. Start by analyzing the input fourbar.

2.


3.

Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). d

K1 

K2 

a

K1  2.2222

 

2

d

K3 

c

K2  1.6949

 

2

2

a b c d

2

2 a c

K3  1.0744

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ  4.

 2  4 A θ Cθ 

B θ


 

K' 1 

d'

K' 2 

a'

K' 1  0.7102

 

 

θ θ  θ θ  α


2

d'

K' 3 

c'

K' 2  1.2892

2

2

2  a' c'

K' 3  1.3655

    K'1  K'2 cosθθ  K'3

A' θ  cos θ θ

 

  

 

 





 

 

θ θ  2   atan2 2  A' θ B' θ  5.

    K'3

C' θ  K' 1   K' 2  1   cos θ θ

B' θ  2  sin θ θ

 2  4 A' θ C'θ 

B' θ

Use equations 4.11b, 4.12 and 4.13 to calculate 5 as a function of 2.

2

a'  b'  c'  d'


K' 4 

SOLUTION MANUAL 4-43-2 2

d'

K' 5 

b'

K' 4  1.289

 

2

2

2

c'  d'  a'  b' 2  a' b'

K' 5  1.365

    K'1  K'4 cosθθ  K'5

D' θ  cos θ θ

 

  

E' θ  2  sin θ θ

 

    K'5

F' θ  K' 1   K' 4  1   cos θ θ

 





 

 

θ θ  2   atan2 2  D' θ E' θ  6.

 2  4 D'θ F'θ 

E' θ

Define a vector loop for links 1, 5, 6, 7, and 8 as shown below and write the vector loop equation.

Y C

4 R12

O4

5 O6

X D

R6 6

8

R8 RDE

F

R7 7

E

R12 + R6 +RDE + R7 - R8 = 0. Solving for R7 gives R7 = R8 - R12 - R6 - RDE. In this equation the only unknowns are 7 and 8. Following the method of Section 4.5, substitute the complex number notation for each position vector and separate the resulting equation into real and imaginary parts:

 

 

 

 

e cos θ = f  cos θ  D1 e sin θ = f  sin θ  D2 where

 

    p  cosθθ  δ

D1 θ  d'  c'  cos θ θ

 

    p  sinθθ  δ

D2 θ  c'  sin θ θ 7.

Solve these equations in the manner of equations 4.11 and 4.12 using the identities of equations 4.9 gives:


 

D3 θ 

 

f

2


 2  D2θ2  e2

 D1 θ

2 f

 

 

 

A' θ  D1 θ  D3 θ

 

 

B' θ  2  D2 θ

 

 

C' θ  D1 θ  D3 θ

 θ  B'  B'2  4 A'  C' = 2  A' 2

tan 

 





 

 

θ θ  2   atan2 2  A' θ B' θ  8.

 2  4 A' θ C'θ 

B' θ

Plot the angular position of link 8 as a function of the angle of input link 2. If 81 is greater than 360 deg, subtra 2 from it.

 

 

θ θ  if  θ 

π

    θ θ  0  θ θ  2 π θ θ 

4

Angular Position of Link 8 360 315 270 225

  180

θ  θ

135

deg 90 45 0  45  90

0

45

90

135

180

225

270

θ deg

The graph shows that link 8 rotates 360 deg between 2 = 19 deg and 2 = 209 deg. θ( 19 deg)  42 deg θ( 209  deg)  θ( 19 deg)  360.0 deg

θ( 209  deg)  2  π  42 deg

315

360




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angles at points B, C, D, E, and F of the linkage in Figure 3-36 as a function of the angle of input link 2.

Given:

Link lengths:

Solution:

Input crank (L2)

a  0.450

First coupler (L3)

b  0.990

Common rocker (O4B)

c  0.590


d  1.000

Common rocker (O4C)

a'  0.590

Second coupler (CD)

b'  0.325

Output rocker (L6)

c'  0.325


Link 7 (L7)

e  0.938

Link 8 (L8)

f  0.572


p  0.823

Angle DCE

δ  7.0 deg

Angle BO4C

α  128.6  deg


1.

This eightbar can be analyzed as a fourbar (links 1, 2, 3, and 4) with its output (link 4) as the input to another fourbar (links 1, 4, 5, and 6). Since links 1 and 4 are common to both, we have an eightbar linkage with links 7 & 8 included. Start by analyzing the input fourbar.

2.


3.

Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). d

K1 

K2 

a

K1  2.2222

 

2

d

K3 

c

K2  1.6949

 

2

2

a b c d 2 a c

K3  1.0744

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ  4.

 2  4 A θ Cθ 

B θ

Use equations 4.12 and 4.13 to calculate 3 as a function of 2 (for the open circuit). K4 

2

d

K5 

b

K4  1.010

 

2

2

c d a b

2

2 a b

K5  2.059

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

F θ  K1   K4  1   cos θ  K5

 





 

 

θ θ  2   atan2 2  D θ E θ  5.

 2  4 Dθ F θ 

E θ

Use equations 4.8a and 4.10 to calculate 6 as a function of 2 (for the open circuit). Input angle to second fourbar:

 

E θ  2  sin θ

 

 

θ θ  θ θ  α

2


K' 1 


d'

K' 2 

a'

K' 1  0.7102

 

2

d'

K' 3 

c'

K' 2  1.2892

2

2

2

a'  b'  c'  d' 2  a' c'

K' 3  1.3655

    K'1  K'2 cosθθ  K'3

A' θ  cos θ θ

 

  

 



 



 

 

θ θ  2   atan2 2  A' θ B' θ  6.

    K'3

C' θ  K' 1   K' 2  1   cos θ θ

B' θ  2  sin θ θ

 2  4 A' θ C'θ 

B' θ

Use equations 4.11b, 4.12 and 4.13 to calculate 5 as a function of 2. K' 4 

2

d' b'

K' 4  1.289

 

2

2

2

c'  d'  a'  b'

K' 5 

2  a' b'

K' 5  1.365

    K'1  K'4 cosθθ  K'5

D' θ  cos θ θ

 

  

E' θ  2  sin θ θ

 

    K'5

F' θ  K' 1   K' 4  1   cos θ θ

 





 

  

 



 

 

θ θ  2   atan2 2  D' θ E' θ 

 

 2  4 D'θ F'θ 

E' θ

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

   



 

 

θ θ  if  105  deg  θ  322  deg  θ θ  0 θ θ  2  π θ θ  7.

Define a vector loop for links 1, 5, 6, 7, and 8 as shown on the next page and write the vector loop equation. R12 + R6 +RDE + R7 - R8 = 0. Solving for R7 gives R7 = R8 - R12 - R6 - RDE. In this equation the only unknowns are 7 and 8. Following the method of Section 4.5, substitute the complex number notation for each position vector and separate the resulting equation into real and imaginary parts:

 

 

 

 

e cos θ = f  cos θ  D1 e sin θ = f  sin θ  D2 where

 

    p  cosθθ  δ

D1 θ  d'  c'  cos θ θ

 

    p  sinθθ  δ

D2 θ  c'  sin θ θ



Y C

4 R12

O4

5 X

O6

D

R6 6

R8

8

RDE

R7

F

E

7 8.

Solve these equations in the manner of equations 4.10 using the identities of equations 4.9 gives:

 

D3 θ 

f

2

 2  D2θ2  e2

 D1 θ

2 f

 

 

 

A' θ  D1 θ  D3 θ

 

 

B' θ  2  D2 θ

 

 

 

C' θ  D1 θ  D3 θ

 θ  B'  B'2  4 A'  C' = 2  A' 2

tan 



 



 

 

θ θ  2   atan2 2  A' θ B' θ 

 

 

θ θ  if  θ  9.

 2  4 A' θ C'θ 

B' θ

π

    θ θ  0  θ θ  2 π θ θ 4 

Similarly, solve these equations in the manner of equations 4.11, 4.12 and 4.13 using the identities of equations 4.9 gives:

 

D4 θ 

 

f

2

 2  D2θ2  e2

 D1 θ

2 e

 

 

D' θ  D1 θ  D4 θ

 θ  E'  E'2  4 D' F' = 2  D' 2

tan 

 

 

E' θ  2  D2 θ

 

 

 

F' θ  D1 θ  D4 θ




 




 

 2  4 D'θ F'θ 

 

θ θ  2   atan2 2  D' θ E' θ 

E' θ

10. Calculate (using equations 4.32) and plot the transmission angle at B.

θtransB1 θ  θ θ  θ θ

 


π 2

 

 





θtransB θ 60 deg

50 40 30 20

0

45

90

135

180

225

270

315

360

θ deg


θtransC1 θ  θ θ  θ θ θtransC2 θ  if 

π

2

 

 

 

 θtransC1 θ  2  π π  θtransC1 θ θtransC1 θ




100

θtransC θ deg 50

0

0

45

90

135

180 θ deg

225

270

315

360



12. Calculate (using equations 4.32) and plot the transmission angle at D.

θtransD1 θ  θ θ  θ θ  π θtransD2 θ  if 

π

2

 

 

 

 θtransD1 θ  π π  θtransD1 θ θtransD1 θ



Transmission Angle at D 90 80 70 60

θtransD θ 50 deg

40 30 20 10 0

0

45

90

135

180

225

270

315

360

315

360

θ deg

13. Calculate (using equations 4.32) and plot the transmission angle at E.

θtransE1 θ  θ θ  θ θ

 

θtransE θ  if  θtransE1 θ 

π 2

 

 

π  θtransE1 θ θtransE1 θ



Transmission Angle at E 90 80 70

θtransE θ

60

deg 50 40 30

0

45

90

135

180 θ deg

225

270




Model the linkage shown in Figure 3-37a in Fourbar. Export the coupler curve coordinates to Excel and calculate the error function versus a true circle.

Given:


a  0.136

Coupler (AB)

b  1.000

Rocker (O4B)

c  1.000

Ground link

d  1.414


p  2.000

δ  0  deg


Model the linkage in Fourbar.

2.

Write coupler point coordinates to a data file.

3.

Import the data file into Excell and add columns for the true circle coordinates and radius. Analyze the coupler point radius to determine its mean, maximum deviation from mean, and average absolute deviation from its mean (See next two pages, note that the last four columns were added to the imported data).


FOURBAR P0445 Tom Cook


Design #

Selected Linkage Parameters a = 0.136 b = 1.000 Angle Step Deg

2

8/9/2006

c = 1.000

d = 1.414

Coupler Pt Coupler Pt Coupler Pt Coupler Pt True Circ True Circ True Circ Coupler Pt X Y Mag Ang X Y R R in in in in in in in in 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120 125 130 135 140 145 150 155 160 165 170 175 180 185 190 195 200 205 210

1.414 1.4283 1.4423 1.4561 1.4693 1.4819 1.4937 1.5046 1.5145 1.5233 1.5309 1.5373 1.5425 1.5465 1.5493 1.5508 1.5512 1.5505 1.5488 1.546 1.5424 1.5379 1.5326 1.5266 1.52 1.5128 1.5052 1.4971 1.4887 1.4799 1.471 1.4618 1.4524 1.4429 1.4333 1.4237 1.414 1.4043 1.3947 1.3851 1.3756 1.3662 1.357

1.5384 1.5379 1.5363 1.5336 1.5299 1.5251 1.5195 1.5129 1.5055 1.4974 1.4885 1.479 1.469 1.4585 1.4475 1.4363 1.4248 1.4131 1.4014 1.3897 1.378 1.3665 1.3552 1.3442 1.3336 1.3235 1.314 1.3051 1.2969 1.2895 1.2829 1.2772 1.2725 1.2688 1.2661 1.2645 1.2639 1.2645 1.2661 1.2688 1.2725 1.2772 1.2829

2.0895 2.0988 2.1072 2.1147 2.1212 2.1265 2.1307 2.1337 2.1355 2.136 2.1353 2.1333 2.1301 2.1258 2.1203 2.1138 2.1063 2.0979 2.0887 2.0788 2.0683 2.0572 2.0458 2.0341 2.0221 2.0101 1.998 1.9861 1.9744 1.9629 1.9518 1.9411 1.931 1.9214 1.9124 1.9041 1.8965 1.8897 1.8836 1.8784 1.8739 1.8703 1.8674

47.413 47.1164 46.8058 46.4846 46.1562 45.8237 45.4901 45.1581 44.8303 44.5088 44.1957 43.8925 43.6007 43.3213 43.0555 42.8038 42.567 42.3456 42.1401 41.9507 41.7781 41.6225 41.4846 41.3647 41.2636 41.1819 41.1204 41.0799 41.0612 41.0654 41.0931 41.1453 41.2227 41.3257 41.455 41.6105 41.7924 42.0001 42.2329 42.49 42.7699 43.0708 43.3907

1.4140 1.4021 1.3904 1.3788 1.3675 1.3565 1.3460 1.3360 1.3266 1.3178 1.3098 1.3026 1.2962 1.2907 1.2862 1.2826 1.2801 1.2785 1.2780 1.2785 1.2801 1.2826 1.2862 1.2907 1.2962 1.3026 1.3098 1.3178 1.3266 1.3360 1.3460 1.3565 1.3675 1.3788 1.3904 1.4021 1.4140 1.4259 1.4376 1.4492 1.4605 1.4715 1.4820

1.5500 1.5495 1.5479 1.5454 1.5418 1.5373 1.5318 1.5254 1.5182 1.5102 1.5014 1.4920 1.4820 1.4715 1.4605 1.4492 1.4376 1.4259 1.4140 1.4021 1.3904 1.3788 1.3675 1.3565 1.3460 1.3360 1.3266 1.3178 1.3098 1.3026 1.2962 1.2907 1.2862 1.2826 1.2801 1.2785 1.2780 1.2785 1.2801 1.2826 1.2862 1.2907 1.2962

0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360

0.1244 0.1247 0.1255 0.1268 0.1284 0.1302 0.1322 0.1341 0.1359 0.1375 0.1386 0.1394 0.1398 0.1398 0.1394 0.1386 0.1376 0.1365 0.1354 0.1342 0.1334 0.1327 0.1324 0.1325 0.1330 0.1340 0.1353 0.1370 0.1389 0.1409 0.1430 0.1449 0.1466 0.1480 0.1492 0.1498 0.1501 0.1498 0.1492 0.1480 0.1466 0.1449 0.1430


215 220 225 230 235 240 245 250 255 260 265 270 275 280 285 290 295 300 305 310 315 320 325 330 335 340 345 350 355 360

1.3481 1.3393 1.3309 1.3228 1.3152 1.308 1.3014 1.2954 1.2901 1.2856 1.282 1.2792 1.2775 1.2768 1.2772 1.2787 1.2815 1.2855 1.2907 1.2971 1.3047 1.3135 1.3234 1.3343 1.3461 1.3587 1.3719 1.3857 1.3997 1.414

1.2895 1.2969 1.3051 1.314 1.3235 1.3336 1.3442 1.3552 1.3665 1.378 1.3897 1.4014 1.4131 1.4248 1.4363 1.4475 1.4585 1.469 1.479 1.4885 1.4974 1.5055 1.5129 1.5195 1.5251 1.5299 1.5336 1.5363 1.5379 1.5384


1.8655 1.8643 1.864 1.8645 1.8659 1.868 1.871 1.8747 1.8793 1.8846 1.8907 1.8975 1.905 1.9132 1.922 1.9314 1.9415 1.952 1.963 1.9744 1.9861 1.998 2.0101 2.0222 2.0342 2.0461 2.0577 2.0688 2.0795 2.0895

43.7272 44.0777 44.439 44.8081 45.1815 45.5556 45.9269 46.2915 46.6458 46.986 47.3085 47.61 47.8871 48.1368 48.3563 48.5433 48.6956 48.8117 48.8904 48.9308 48.9328 48.8966 48.823 48.713 48.5685 48.3915 48.1846 47.9504 47.6921 47.413

1.4920 1.5014 1.5102 1.5182 1.5254 1.5318 1.5373 1.5418 1.5454 1.5479 1.5495 1.5500 1.5495 1.5479 1.5454 1.5418 1.5373 1.5318 1.5254 1.5182 1.5102 1.5014 1.4920 1.4820 1.4715 1.4605 1.4492 1.4376 1.4259 1.4140

1.3026 1.3098 1.3178 1.3266 1.3360 1.3460 1.3565 1.3675 1.3788 1.3904 1.4021 1.4140 1.4259 1.4376 1.4492 1.4605 1.4715 1.4820 1.4920 1.5014 1.5102 1.5182 1.5254 1.5318 1.5373 1.5418 1.5454 1.5479 1.5495 1.5500

The mean value of the coupler point radius is

0.1366

The maximum deviation from the mean is

0.0135

The average absolute deviation from the mean is

0.005127

0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360

0.1409 0.1389 0.1370 0.1353 0.1340 0.1330 0.1325 0.1324 0.1327 0.1334 0.1342 0.1354 0.1365 0.1376 0.1386 0.1394 0.1398 0.1398 0.1394 0.1386 0.1375 0.1359 0.1341 0.1322 0.1302 0.1284 0.1268 0.1255 0.1247 0.1244




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of point P in Figure 3-37a as a function of the angle of input link 2. Also plot the variation (error) in the path of point P versus that of point A.

Given:


a  0.136

Coupler (AB)

b  1.000

Rocker (O4B)

c  1.000

Ground link

d  1.414

p  2.000


δ  0  deg



S  min ( a b c d ) L  max( a b c d ) SL  S  L PQ  a  b  c  d  SL return "Grashof" if SL  PQ return "Special Grashof" if SL = PQ

return "non-Grashof" otherwise crank rocker Condition( a b c d )  "Grashof" 1.


2.

Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. d

K1 

K4 

a

K1  10.3971

 

2

d

K5 

b

K4  1.4140

 

2

2

c d a b

2

2 a b

K5  7.4187

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

3.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ




 



 

 

θ θ  2   atan2 2  D θ E θ  4.

 2  4 Dθ F θ 

E θ


  

 


 








 

 

  




 

 

  




Plot the coordinates of the coupler point in the local xy coordinate system.



COUPLER POINT PATH

Coupler Point Coordinate - y

 1.2

1.414

 1.3

 1.4

 1.414

 1.5

 1.6 1.2

1.3

1.4

1.5

1.6

Coupler Point Coordinate - x

6.

Replot, transforming the coupler path to 0,0 and plot the path of point A.

 

 

 

XA θ  a  cos θ

 

YA θ  a  sin θ

PATHS OF POINTS A &P 0.2

0.1

0

 0.1

 0.2  0.2

 0.1 Point P Point A

0

0.1

0.2




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angle at point B of the linkage in Figure 3-37a as a function of the angle of input link 2.

Given:


a  0.136

Coupler (AB)

b  1.000

Rocker (O4B)

c  1.000

Ground link

d  1.414

p  2.000


δ  0  deg





crank rocker


2.


d

K2 

a

K1  10.3971

 

2

d

K3 

c

K2  1.4140

 

2

2 a c

K3  7.4187

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ  3.

 2  4 A θ Cθ 

B θ


2

d

K5 

b

K4  1.414

 

2

2

c d a b

2

2 a b

K5  7.419

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 





 

 

 

E θ  2  sin θ

F θ  K1   K4  1   cos θ  K5 θ θ  2   atan2 2  D θ E θ 

 2  4 Dθ F θ 

E θ

2

a b c d

2


4.



θtransB1 θ  θ θ  θ θ

 


π 2

 

 




Transmission Angle at B 90

85

θtransB θ deg 80

75

0

45

90

135

180 θ deg

225

270

315

360

DESIGN OF MACHINERY - 5thEd.



Figure 3-29f shows Evan's approximate straight-line linkage #1. Determine the range of motion of link 2 for which the point P varies no more than 0.0025 from the straight line X = 1.690 (assuming that O2 is the origin of a global coordinate frame whose positive X axis is rotated 60 deg from O2O4).

Given:


a  1.000

Coupler (AB)

b  1.600

Rocker (O4B)

c  1.039

Ground link

d  1.200

p  2.690

Coupler point data:

α  60 deg


δ  0  deg

See Figure 3-29f and Mathcad file P0448.



Condition( a b c d )  "non-Grashof" Since b (link 3) is the longest link and the linkage is non-Grashof, this is a Class 3 triple rocker. Using Figure 3-1a as a guide, determine the limiting values of 2 at the toggle positions. For links 2 and 3 colinear:

 ( b  a) 2  d2  c2 π  2 d ( b  a) 


θ  240 deg

For links 3 and 4 colinear:

 a2  d 2  ( b  c) 2  2 d a  

θ  acos 1.

θ  27.683 deg

θ  θ

Define one cycle of the input crank (driving through the links 2-3 toggle position): θ  θ θ  1  deg  360  deg  θ

2.


d

K4 

a

K1  1.2000

 

2

d

K5 

b

K4  0.7500

 

2

2 a b

K5  1.2251

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ 3.

 

 

F θ  K1   K4  1   cos θ  K5


2

c d a b

2

DESIGN OF MACHINERY - 5thEd.

 






 

 

θ θ  2   atan2 2  D θ E θ  4.

 2  4 Dθ F θ 

E θ

Use equations 4.31 to define the x-components of the vector RP. RP  RA  RPA


 

 

  



 


 

  

Plot the X coordinate of the coupler point in the global X,Y coordinate system using equations 4.0b to rotate the local coordinates to a global frame.

 

 

 

PX θ  RPx θ  cos( α)  RPy θ  sin( α) X COORDINATE 1.7 1.698 Coupler Point Coordinate - X

1.696 1.694 1.692 1.69 1.688 1.686 1.684 1.682 1.68 180

210

240

270

300

Input Angle - Theta2

6.




Using the graph for guess values, solve by trial and error to find 2 for X = 1.6900 +/-0.0025. PX ( 201.525  deg)  1.69250

θ2min  201.525  deg

PX ( 273.450  deg)  1.69250

θ2max  273.450  deg




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of point P in Figure 3-37b as a function of the angle of input link 2.

Given:

Link lengths:

Solution: 1.

Input crank (L2)

a  0.50

First coupler (AB)

b  1.00

Rocker 4 (O4B)

c  1.00

Rocker 5 (L5)

c'  1.00

Ground link (O2O4)

d  0.75

Second coupler 6 (CD)

b'  1.00

Coupler point (DP)

p  1.00

Distance to OP (O2OP)

d'  1.50


Links 4, 5, BC, and CD form a parallelogram whose opposite sides remain parallel throughout the motion of the fourbar 1, 2, AB, 4. Define a position vector whose tail is at point D and whose tip is at point P and another whose tail is at O4 and whose tip is at point D. Then, since R5 = RAB and RDP = -R4, the position vector from O2 to P is P = R1 + RAB - R4. Separating this vector equation into real and imaginary parts gives the equations for the X and Y coordinates of the coupler point P.

 

 

 

XP = d  b  cos θ  c cos θ

 

YP = b  sin θ  c sin θ

2.


3.


d a

K1  1.5000

 

2

d

K2 

K3 

c

K2  0.7500

 

2

2

a b c d

2

2 a c

K3  0.8125

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  4.

B θ


 

2

d

K5 

b

2

2

c d a b

2

2 a b

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

K4  0.7500

K5  0.8125

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5 5.


 





 

 

θ θ  2   atan2 2  D θ E θ  6.

 2  4 Dθ F θ 

E θ

Define a local xy coordinate system with origin at OP and with the positive x axis to the right. The coordinates of P are transformed to xP = XP - d', yP = YP.

 

    c cosθθ  d'

xP θ  d  b  cos θ θ

 

    c sinθθ

yP θ  b  sin θ θ


7.


Plot the path of P as a function of the angle of link 2.

Path of Coupler Point P About OP 0.6 0.5 0.4 0.3 0.2 0.1

 

yP θ

0

 0.1  0.2  0.3  0.4  0.5  0.6  0.6  0.5

 0.4  0.3

 0.2  0.1

0

 

xP θ

0.1

0.2

0.3

0.4

0.5

0.6




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angles at points B, C, and D of the linkage in Figure 3-37b as a function of the angle of input link 2.

Given:

Link lengths:

Solution:

Input crank (L2)

a  0.50

First coupler (AB)

b  1.00

Rocker 4 (O4B)

c  1.00

Rocker 5 (L5)

c'  1.00

Ground link (O2O4)

d  0.75


b'  1.00

Coupler point (DP)

p  1.00


d'  1.50


1.

Links 4, 5, BC, and CD form a parallelogram whose opposite sides remain parallel throughout the motion of the fourbar 1, 2, AB, 4. Therefore, the transmission angles at points B and D will be the same and the transmission angle at point C will be the complement of the angle at B.

2.


3.


d a

K1  1.5000

 

2

d

K2 

K3 

c

K2  0.7500

 

2

2

a b c d

2

2 a c

K3  0.8125

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 



 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ

 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  4.

B θ


 

2

d

K5 

b

2

2

c d a b

2

K4  0.7500

2 a b

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5 5.


 





 

 

θ θ  2   atan2 2  D θ E θ  6.

 2  4 Dθ F θ 

E θ

Calculate (using equations 4.32) and plot the transmission angles at B and D.

θtransB1 θ  θ θ  θ θ

 


π 2

K5  0.8125

 

 






Transmission Angles at B and D 80

60


40

deg

20

0

0

45

90

135

180

225

270

315

360

θ deg

6.

Calculate and plot the transmission angle at C.

θtransC1 θ  180  deg  θtransB θ

 


π 2

 

 





60


40

deg

20

0

0

45

90

135

180 θ deg

225

270

315

360




Figure 3-29g shows Evan's approximate straight-line linkage #2. Determine the range of motion of link 2 for which the point P varies no more than 0.005 from the straight line X = -0.500 (assuming that O2 is the origin of a global coordinate frame whose positive X axis is rotated 30 deg from O2O4).

Given:


a  1.000

Coupler (AB)

b  1.200

Rocker (O4B)

c  1.167

Ground link

d  2.305

p  1.50

Coupler point data:

α  30 deg


δ  180  deg

See Figure 3-29g and Mathcad file P0451.




 a2  d 2  ( b  c) 2  2 d a  

θ  acos 1.

θ  81.136 deg

θ  θ

Define one cycle of the input crank (driving through the links 2-3 toggle position): θ  θ θ  0.5 deg  θ

2.


d

K4 

a

K1  2.3050

 

2

d

K5 

b

K4  1.9208

 

2

2 a b

K5  2.6630

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

3.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ


 





 

 

θ θ  2   atan2 2  D θ E θ 

 2  4 Dθ F θ 

E θ

2

c d a b

2


4.


Use equations 4.31 to define the x and y-components of the vector RP. RP  RA  RPA


 

 

  



 


 

  

Plot the X coordinate of the coupler point in the global X,Y coordinate system using equations 4.0b to rotate the local coordinates to a global frame.

 

 

 

PX θ  RPx θ  cos( α)  RPy θ  sin( α) X COORDINATE  0.48

Coupler Point Coordinate - X

 0.485  0.49  0.495

 

PX θ

 0.5  0.505  0.51  0.515  0.52

0

15

30

45

60

θ deg Input Angle - Theta2

6.




Using the graph for guess values, solve by trial and error to find 2 for X = -0.500 +/-0.005 PX ( 11.59  deg)  0.49500

θ2min  11.59  deg

PX ( 57.80  deg)  0.50500

θ2max  57.80  deg

75




For the linkage in Figure P4-16, what are the angles that link 2 makes with the positive X-axis when links 2 and 3 are in toggle positions?

Given:

Link lengths:

Solution: 1.

Input (O2A)

a  14

Rocker (O4B)

c  51.26

b  80

Coupler (AB)

O4 offset in XY coordinates:

O4X  47.5

Ground link:

d 

2

O4X  O4Y

O4Y  76  12 2

O4Y  64.000

d  79.701





Define the coordinate frame transformation angle:

 O4Y    O4X 

δ  π  atan 3.

crank rocker

δ  126.582 deg

Calculate the angle of link 2 in the XY system when links 2 and 3 are in the overlapped toggle position.

 ( b  a) 2  d2  c2 δ  2 ( b  a)  d 

θ21XY  acos 4.

θ21XY  86.765 deg

Calculate the angle of link 2 in the XY system when links 2 and 3 are in the extended toggle position.

 ( b  a) 2  d2  c2 δ  2 ( b  a)  d 

θ22XY  acos

θ22XY  93.542 deg




The coordinates of the point P1 on link 4 in Figure P4-16 are (114.68, 33.19) with respect to the xy coordinate system when link 2 is in the position shown. When link 2 is in another position the coordinates of P2 with respect to the xy system are (100.41, 43.78). Calculate the coordinates of P1 and P2 in the XY system for the two positions of link 2. What is the salient feature of the coordinates of P1 and P2 in the XY system?

Given:

Vertical and horizontal offsets from O2 to O4. O2O4X  47.5 in

O2O4Y  64 in

Coordinates of P1 and P2 in the local system

Solution: 1.

P1x  114.68 in

P1y  33.19  in

P2x  100.41 in

P2y  43.78  in


Calculate the angle from the global X axis to the local x axis.

 O2O4Y    O2O4X 

δ  180  deg  atan 2.

3.

δ  126.582 deg

Use equations 4.0b to transform the given coordinates from the local to the global system. P1X  P1x cos( δ)  P1y sin( δ)

P1X  95.00 in

P1Y  P1x sin( δ)  P1y cos( δ)

P1Y  72.31 in

P2X  P2x cos( δ)  P2y sin( δ)

P2X  95.00 in

P2Y  P2x sin( δ)  P2y cos( δ)

P2Y  54.54 in

In the global XY system the X-coordinates are the same for each point, which indicates that the head on the end of the rocker beam 4 is designed such that its tangent is always parallel to the Y-axis.




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular position of link 4 with respect to the XY coordinate frame and the transmission angle at point B of the linkage in Figure P4-16 as a function of the angle of input link 2 with respect to the XY frame.

Given:

Link lengths:

Solution: 1.

Input (O2A)

a  14

Coupler (AB)

b  80

Rocker (O4B)

c  51.26

Ground link

d  79.70





Define the coordinate frame transformation angle: δ  90 deg  atan

47.5 

  64 

3.

crank rocker

δ  126.582 deg

Define one cycle of the input crank with respect to the XY frame:

θ2 θ2XY   θ2XY  δ

θ2XY  0  deg 1  deg  360  deg 4.


d

K2 

a

K1  5.6929

2

d

K3 

c

K2  1.5548

2

2

a b c d

2

2 a c

K3  1.9339

A  θ2XY   cos θ2 θ2XY    K1  K2 cos θ2 θ2XY    K3 B θ2XY   2  sin θ2 θ2XY  



C θ2XY   K1   K2  1   cos θ2 θ2XY    K3



θ θ2XY   2   atan2 2  A  θ2XY  B θ2XY  

2

θ θ2XY   θ θ2XY   δ  2  π 5.


d b

K4  0.996

2

K5 

2

2

c d a b

K5  4.607

2 a b



B θ2XY   4  A  θ2XY   C θ2XY  

2



D θ2XY   cos θ2 θ2XY    K1  K4 cos θ2 θ2XY    K5 E θ2XY   2  sin θ2 θ2XY   F  θ2XY   K1   K4  1   cos θ2 θ2XY    K5





θ θ2XY   2   atan2 2  D θ2XY  E θ2XY   6.



E θ2XY   4  D θ2XY   F  θ2XY    2

Plot the angular position of link 4 as a function of the input angle of link 2 with respect to the XY frame.

Angular Position of Link 4 40 35 30 θ θ2XY 25 20 deg 15 10 5





0

0

45

90

135

180

225

270

315

360

θ2XY deg

7.


θtransB1 θ2XY   θ θ2XY   θ θ2XY 

 

θtransB θ2XY   if  θtransB1 θ2XY  

π 2

 

π  θtransB1 θ2XY  θtransB1 θ2XY  


θtransB θ2XY 

75 70

deg 65 60 55 50

0

45

90

135

180

θ2XY deg

225

270

315

360




For the linkage in Figure P4-17, calculate the maximum CW rotation of link 2 from the position shown, which is -20.60 deg with respect to the local xy system. What angles do link 3 and link 4 rotate through for that excursion of link 2?

Given:


a  9.17

Coupler (AB)

b  12.97

Rocker (O4B)

c  9.57

Ground link

d  7.49

Initial position of link 2: θ  26.00  deg  2  π (with respect to xy system) Solution: 1.




Condition( d b a c)  "non-Grashof" 2.

Using equations 4.37, determine the crank angles (relative to the line O2O4) at which links 3 and 4 are in toggle 2

arg1 

2

2



2 a d 2

arg2 

2

a d b c

2

2

a d b c

2



2 a d

b c

arg1  0.936

a d b c

arg2  2.678

a d

θ  acos arg1

θ  20.55 deg

The other toggle angle is the negative of this. θ  θ  2  π 3.

θ  339.45 deg

Calculate the CW rotation of link 2 from the initial position to the toggle position. Δ  θ  θ

4.

Δ  313.45 deg


d

K4 

a

K1  0.8168

 

2

d

K5 

b

K4  0.5775

 

2

2 a b

K5  0.9115

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

6.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ


 





 

 

θ θ  2   atan2 2  D θ E θ 

 2  4 Dθ F θ 

E θ

2

c d a b

2



 

Initial angular position of link 3:

θ θ  2  π  647.755 deg

Final angular position of link 3:

θ θ  0.001  deg  250.764 deg







   

  θ θ  0.001  deg  θ θ  2  π 7.



  898.518 deg


d

K2 

a

K1  0.8168

 

2

d

K3 

c

K2  0.7827

 

2

2 a c

K3  0.3621

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 



 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ

 

 

θ θ  2   atan2 2  A θ B θ 

 2  4 A θ Cθ 

B θ

 

Initial angular position of link 4:

θ θ  2  π  659.462 deg

Final angular position of link 4:

θ θ  0.001  deg  250.615 deg





   

  θ θ  0.001  deg  θ θ  2  π





2

a b c d

  910.077 deg

2




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of point P in Figure P4-17 with respect to the XY coordinate system as a function of the angle of input link 2 with respect to the XY coordinate system.

Given:


a  9.174

Coupler (AB)

b  12.971

Rocker (O4B)

c  9.573

Ground link

d  7.487

p  15.00


δ  0  deg





Using equations 4.37, determine the crank angles (relative to the line O2O4) at which links 3 and 4 are in toggle 2

arg1 

2

2



2 a d 2

arg2 

2

a d b c

2

2

a d b c

2



2 a d

b c

arg1  0.937

a d b c

arg2  2.679

a d




Define the coordinate transformation angle. Transformation angle:

4.

α  atan

6.95 

  2.79 

α  68.128 deg

Define one cycle of the input crank between limit positions: θ  θ2toggle θ2toggle  1  deg  2  π  θ2toggle

5.


d

K4 

a

K1  0.8161

 

2

d

K5 

b

K4  0.5772

 

 

D θ  cos θ  K1  K4 cos θ  K5

2

2

c d a b 2 a b

K5  0.9110

2


 


 

 

6.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ


 





 

 

θ θ  2   atan2 2  D θ E θ  7.

 2  4 Dθ F θ 

E θ



 

 

  



 


 

  

Transform these local xy coordinates to the global XY coordinate system using equations 4.0b.

 

 

 

 

 

 

RPX θ  RPx θ  cos α  RPy θ  sin α RPY θ  RPx θ  sin α  RPy θ  cos α Plot the coordinates of the coupler point in the global XY coordinate system. COUPLER POINT PATH 5

0


9.

5

 10

 15

 20  10

5

0




5

10

Coupler Point Coordinate - X

15

20




For the linkage in Figure P4-17, calculate the coordinates of the point P in the XY coordinate syste if its coordinates in the xy system are (2.71, 10.54).

Given:

Vertical and horizontal offsets from O2 to O4. O2O4X  2.790  in

O2O4Y  6.948  in

Coordinates of P in the local system Px  12.816 in Solution: 1.


Calculate the angle from the global X axis to the local x axis.

 O2O4Y    O2O4X 

δ  atan 2.

Py  10.234 in

δ  68.122 deg

Use equations 4.0b to transform the given coordinates from the local to the global system. PX  Px cos( δ)  Py sin( δ)

PX  14.273 in

PY  Px sin( δ)  Py cos( δ)

PY  8.079 in




The elliptical trammel in Figure P4-18 must be driven by rotating link 3 in a full circle. Derive analytical expressions for the positions of points A, B, and a point C on link 3 midway between A and B as a function of 3 and the length AB of link 3. Use a vector loop equation. (Hint: Place the global origin off the mechanism, preferably below and to the left and use a total of 5 vectors.) Code your solution in an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of point C for one revolution of link 3.

Solution:


1.

Establish the global XY system such that the coordinates of the intersection of the slot centerlines is at (d X,d Y ). Then, define position vectors R1X, R1Y , R2, R3, and R4 as shown below. Y

4 B

3 R3



3

C

R2 2

R4

A 1

R1Y

X R1X

2.

Write the vector loop equation: R1Y + R2 + R3 - R1X - R4 = 0 then substitute the complex number notation for each position vector. The equation then becomes:

 π  π j   j  θ 3 2 j ( 0) j ( 0) 2 dY  e    a e  c e  dX  e  b e   = 0 j

3.

Substituting the Euler identity into this equation gives: d Y  j  a  c  cos θ3  j  sin θ3   d X  b  j = 0

4.

Separate this equation into its real (x component) and imaginary (y component) parts, setting each equal to zero. a  c cos θ3  d X = 0

5.

Solve for the two unknowns a and b in terms of the constants d X and d Y and the independent variable 3. Where (a,d Y ) and (d X,b) are the coordinates of points A and B, respectively, and c is the length of link 3. With no loss of generality, let d X = d Y = d. Then, a = d  c cos θ3

6.

b = d  c sin θ3

The coordinates of the point C are: CX = d  0.5 cos θ3

7.

d Y  c sin θ3  b = 0

CY = d  0.5 c sin θ3

Using a local coordinate system whose origin is located at the intersection of the centerlines of the two slots and transforming the above functions to the local xy system:


8.


a x = c cos θ3

ay = 0

bx = 0

b y = c sin θ3

To plot the path of point C as a function of 3, let c  1 and define a range function for 3

θ3  0  deg 1  deg  360  deg Cx θ3  0.5 c cos θ3

Cy θ3  0.5 sin θ3 Path of Point C

1

0.5

 

Cy θ3

0

 0.5

1 1

 0.5

0

 

Cx θ3

0.5

1




Calculate and plot the angular position of link 6 in Figure P4-19 as a function of the angle of input link 2.

Given:

Link lengths:

Solution: 1.

Input crank (L2)

a  1.75

First coupler (AB)

b  1.00

First rocker (O4B)

c  1.75

Ground link (O2O4)

d  1.00

Second input (BC)

e  1.00

Second coupler (L5)

f  1.75

Output rocker (L6)

g  1.00

Third coupler (BE)

h  1.75

Ternary link (O4C)

i  2.60


Because the linkage is symmetrical and composed of two parallelograms the analysis can be done with simple trigonometry.

39.582°

C B

5

D

3

3

A

2

4

6 E

O4

2.

1

O2

Calculate the fixed angle that line BC makes with the extension of line O4B using the law of cosines.

 c2  e2  i 2    2  c e 

δ  π  acos

δ  39.582 deg

3.


4.

Because links 1, 2, 3, and 4 are a parallelogram, link 4 will have the same angle as link 2 and AB will always be parallel to O2O4. And because links BC, 5, 6, and BE are also a parallelogram, link 6 will have the same angle as link BC. Thus,

 

θ θ  θ  δ 5.

Plot the angular position of link 6 as a function of the angle of input link 2 (see next page).



Angular Position of Link 6 400 360 320 280 240

 

θ  θ

200

deg 160 120 80 40 0

0

45

90

135

180 θ deg

225

270

315

360



PROBLEM 4-60a The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row a, draw the linkage to scale and graphically find all possible solutions (both open and crossed) for angles 2 and θ3.

Statement:

Given:

Link 2

a  1.4 in

Link 3

b  4  in

Offset

c  1  in

Slider position

d  2.5 in

See Figure P4-2, Table P4-5, and Mathcad file P0460a.

Solution: 1.


2.

Draw a circle centered at the origin with radius equal to a at some convenient scale.

3.

Draw construction lines to define the point (d,c).

4.

From the point (d,c) draw an arc with radius equal to b.

5.

The two intersections of the circle and arc are the two solutions to the position analysis problem, crossed and open. If the circle and line don't intersect, there is no solution.

6.

Draw links 2 and 3 in their two possible positions (shown as solid for branch 1 and dashed for branch 2 in the figure) and measure the angles θ2 and 3 for each branch. From the solution below, Branch 1:

θ21  176.041  deg

θ31  ( 180  13.052)  deg

θ31  193.052 deg

Branch 2:

θ22  132.439  deg

θ32  ( 180  30.551)  deg

θ32  210.551 deg

13.052° 30.551°

176.041° 000 b = 4.

c = 1.000"

a = 1.400

132.439°

d = 2.500"




Given:

Solution: 1.

The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row a, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2

a  1.4 in

Link 3

Offset

c  1  in

Slider position

d  2.5 in


Determine both values of 2 using equations 4.20 and 4.21. 2

2

2

K1  a  b  c  d

2

2

K1  6.790  in

2

K2  2  a  c

K2  2.800  in

K3  2  a  d

K3  7.000  in

A  K1  K3

A  0.210  in

B  2  K2

B  5.600  in

C  K1  K3

C  13.790 in

2

2 2 2

  2  atan2 2  A B 

θ21  2  atan2 2  A B  θ22 2.

b  4  in

 2 B  4  A  C 2

B  4 A  C

θ21  176.041  deg θ22  132.439  deg

Determine both values of 3 using equation 4.16a or 4.17.

β ( a b c d α) 

θ  asin



a  sin( α)  c 

 

b

d 1  a  cos( α)  b  cos( θ ) return θ if d 1 = d asin 



a  sin( α)  c  b

  π otherwise 

θ31  β  a b c d θ21

θ31  193.052 deg

θ32  β  a b c d θ22

θ32  210.551 deg




Given:

Solution: 1.

The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row b, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2

a  2  in

Link 3

Offset

c  3  in

Slider position

d  5  in

See Figure P4-2, Table P4-5, and Mathcad file P0461b.


2

2

K1  a  b  c  d

2

2

K1  2.000  in

2

K2  2  a  c

K2  12.000 in

K3  2  a  d

K3  20.000 in

A  K1  K3

A  22.000 in

B  2  K2

B  24.000 in

C  K1  K3

C  18.000 in

2

2

2 2

  2  atan2 2  A B 

θ21  2  atan2 2  A B  θ22 2.

b  6  in

 2 B  4  A  C 2

B  4 A  C

θ21  54.117 deg θ22  116.045  deg


β ( a b c d α) 

θ  asin



a  sin( α)  c 

 

b




a  sin( α)  c  b


θ31  β  a b c d θ21

θ31  129.640 deg

θ32  β  a b c d θ22

θ32  168.433 deg




Given:

Solution: 1.

The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row c, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2

a  3  in

Link 3

Offset

c  2  in

Slider position

d  8  in

See Figure P4-2, Table P4-5, and Mathcad file P0461c.


2

2

K1  a  b  c  d

2

2

K1  13.000 in

2

K2  2  a  c

K2  12.000 in

K3  2  a  d

K3  48.000 in

A  K1  K3

A  61.000 in

B  2  K2

B  24.000 in

C  K1  K3

C  35.000 in

2

2 2 2

  2  atan2 2  A B 

θ21  2  atan2 2  A B  θ22 2.

b  8  in

 2 B  4  A  C 2

B  4 A  C

θ21  88.803 deg θ22  60.731 deg


β ( a b c d α) 

θ  asin



a  sin( α)  c 

 

b




a  sin( α)  c  b


θ31  β  a b c d θ21

θ31  172.824  deg

θ32  β  a b c d θ22

θ32  215.249  deg


SOLUTION MANUAL 4-61d-1

PROBLEM 4-61d Statement:

Given:

Solution: 1.

The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row d, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2

a  3.5 in

Link 3

Offset

c  1  in

Slider position

d  8  in

See Figure P4-2, Table P4-5, and Mathcad file P0461d.


2

2

K1  a  b  c  d

2

2

K1  22.750 in 2

K2  2  a  c

K2  7.000  in

K3  2  a  d

K3  56.000 in

A  K1  K3

A  78.750 in

B  2  K2

B  14.000 in

C  K1  K3

C  33.250 in

2 2

2

2

  2  atan2 2  A B 

θ21  2  atan2 2  A B  θ22 2.

b  10 in

 2 B  4  A  C 2

B  4 A  C

θ21  286.648  deg θ22  300.898  deg


β ( a b c d α) 

θ  asin



a  sin( α)  c 

 

b




a  sin( α)  c  b


θ31  β  a b c d θ21

θ31  25.806 deg

θ32  β  a b c d θ22

θ32  11.556 deg


SOLUTION MANUAL 4-61e-1

PROBLEM 4-61e Statement:

Given:

Solution: 1.

The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row e, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2

a  5  in

Link 3

Offset

c  5  in

Slider position

d  15 in

See Figure P4-2, Table P4-5, and Mathcad file P0461e.


2

2

K1  a  b  c  d

2

2

K1  125.000  in 2

K2  2  a  c

K2  50.000 in

K3  2  a  d

K3  150.000  in

A  K1  K3

A  25.000 in

B  2  K2

B  100.000  in

C  K1  K3

C  275.000  in

2

2 2 2

  2  atan2 2  A B 

θ21  2  atan2 2  A B  θ22 2.

b  20 in

 2 B  4  A  C 2

B  4 A  C

θ21  123.804  deg θ22  160.674  deg


β ( a b c d α) 

θ  asin



a  sin( α)  c 

 

b




a  sin( α)  c  b


θ31  β  a b c d θ21

θ31  152.759  deg

θ32  β  a b c d θ22

θ32  170.371  deg



PROBLEM 4-61f Statement:

Given:

Solution: 1.

The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row f, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2

a  3  in

Link 3

Offset

c  0  in

Slider position

d  12 in

See Figure P4-2, Table P4-5, and Mathcad file P0461f.


2

2

K1  a  b  c  d

2

2

K1  16.000 in 2

K2  2  a  c

K2  0.000  in

K3  2  a  d

K3  72.000 in

A  K1  K3

A  88.000 in

B  2  K2

B  0.000  in

C  K1  K3

C  56.000 in

2 2

2 2

  2  atan2 2  A B 

θ21  2  atan2 2  A B  θ22 2.

b  13 in

 2 B  4  A  C 2

B  4 A  C

θ21  282.840  deg θ22  282.840  deg


β ( a b c d α) 

θ  asin



a  sin( α)  c 

 

b




a  sin( α)  c  b


θ31  β  a b c d θ21

θ31  13.003 deg

θ32  β  a b c d θ22

θ32  13.003 deg



PROBLEM 4-61g Statement:

Given:

Solution: 1.

The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row g, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2

a  7  in

Link 3

Offset

c  10 in

Slider position

d  25 in

See Figure P4-2, Table P4-5, and Mathcad file P0461g.


2

2

K1  a  b  c  d

2

2

K1  149.000  in

2

K2  2  a  c

K2  140.000  in

K3  2  a  d

K3  350.000  in

A  K1  K3

A  499.000  in

B  2  K2

B  280.000  in

C  K1  K3

C  201.000  in

2

2 2 2

  2  atan2 2  A B 

θ21  2  atan2 2  A B  θ22 2.

b  25 in

 2 B  4  A  C 2

B  4 A  C

θ21  88.519 deg θ22  44.916 deg


β ( a b c d α) 

θ  asin



a  sin( α)  c 

 

b




a  sin( α)  c  b


θ31  β  a b c d θ21

θ31  186.898  deg

θ32  β  a b c d θ22

θ32  216.705  deg




Design a fourbar mechanism to give the two positions shown in Figure P3-1 of output rocker motion with no quick-return. (See Problem 3-3).

Given:

Coordinates of the points C1, D1, C2, and D2 with respect to C1: C1x  0.0

D1x  1.721

C2x  2.656

D2x  5.065

C1y  0.0

D1y  1.750

C2y  0.751

D2y  0.281

Assumptions: Use the pivot point between links 3 and 4 (C1 and C2 )as the precision points P1 and P2. Define position vectors in the global frame whose origin is at C1. Solution:

See solution to Problem 3-3 and Mathcad file P0501.

1.

Note that this is a two-position function generation (FG) problem because the output is specified as an angular displacement of the rocker, link 4. See Section 5.13 which details the 3-position FG solution. See also Section 5.3 in which the equations for the two-position motion generation problem are derived. These are really the same problem and have the same solution. The method of Section 5.3 will be used here.

2.

Two solution methods are derived in Section 5.3 and are presented in equations 5.7 and 5.8 for the left dyad and equations 5.11 and 5.12 for the right dyad. The first method (equations 5.7 and 5.11) looks like the better one to use in this case since it allows us to choose the link's angular positions and excursions and solve for the lengths of links 2 and 3 (w and z). Unfortunately, this method fails in this problem because of the requirement for a non-quick-return Grashof linkage, which requires the angular displacement of link 3 in going from position 1 to position 2 to be zero (2 = 0) causing a divide-by-zero error in equations 5.7d.

3.

Method 2 (equations 5.8 and 5.12) requires the choosing of two angles and a length for each dyad.

4.

To obtain the same solution as was done graphically in Problem 3-4, we need to know the location of the fixed pivot O4 with respect to the given CD. While we could take the results from Problem 3-3 and use them here to establish the location of O4, that won't be done. Instead, we will use the point C as the joint between links 3 and 4 as well as the precision point P.

5.

Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1. R1 

 C1x     C1y 

R2 

 C2x     C2y 

 P21x     R2  R1  P21y 

P21x  2.656 P21y  0.751

p 21  6.

2

2

P21x  P21y

p 21  2.760

From the trigonometric relationships given in Figure 5-1, determine 2. From the requirement for a non-quick-return, α  0. δ  atan2 P21x P21y

7.

δ  15.789 deg

From a graphical solution (see figure below), determine the values necessary for input to equations 5.8. z  5.000

β  180  deg

ϕ  δ



3.483 5.0000

1.380

2.027 5.621

O4

1

A1 O2

2

4 2.095

0.989

3

56.519°

A2 C1

D2 C2 2.922

D1

8.

Solve for the WZ dyad using equations 5.8. Z1x  z cos ϕ

Z1x  4.811

 

A  2.000

D  sin α

 

B  0.000

E  p 21  cos δ

 

C  0.000

F  p 21  sin δ

A  cos β  1 B  sin β

C  cos α  1

W1x 

W1y  w 

Z1y  z sin ϕ

 

D  0.000

 

E  2.656

 

F  0.751

A   C Z1x  D Z1y  E  B  C Z1y  D Z1x  F 

W1x  1.328

2  A A   C Z1y  D Z1x  F   B  C Z1x  D Z1y  E

W1y  0.375

2  A 2

Z1y  1.360

2

W1x  W1y

w  1.380

θ  atan2 W1x W1y

θ  164.211  deg

This is the expected value of w (one half of p 21) based on the design choices made in the graphical solution and the assumptions made in this problem. 9.

From the graphical solution (see figure above), determine the values necessary for input to equations 5.12. s  0

γ  56.519 deg

ψ  0  deg

10. Solve for the US dyad using equations 5.12. S 1x  s cos ψ

S 1x  0.000

S 1y  s sin ψ

 

A  0.448

D  sin α

 

B  0.834

E  p 21  cos δ

A  cos γ  1 B  sin γ

 

S 1y  0.000 D  0.000

 

E  2.656



 

C  cos α  1

C  0.000

 

F  p 21  sin δ

A   C S 1x  D S 1y  E  B  C S 1y  D S 1x  F 

U1x 

2  A A   C S 1y  D S 1x  F   B  C S 1x  D S 1y  E

U1y 

2

U1x  2.027

U1y  2.095

2  A

u 

F  0.751

2

U1x  U1y

u  2.915

σ  atan2 U1x U1y

σ  134.048  deg

This is the expected value of u based on the design choices made in the graphical solution and the assumptions made in this problem. 11. Solve for links 3 and 1 using the vector definitions of V and G. Link 3:

V1x  z cos ϕ  s cos ψ

V1x  4.811

V1y  z sin ϕ  s sin ψ

V1y  1.360

θ  atan2 V1x V1y

θ  15.789 deg

v  Link 1:

2

2

V1x  V1y

v  5.000

 

G1x  w cos θ  v cos θ  u  cos σ

 

G1x  5.510

G1y  w sin θ  v sin θ  u  sin σ

G1y  1.110

θ  atan2 G1x G1y

θ  11.391 deg

g 

2

2

G1x  G1y

g  5.621

12. Determine the initial and final values of the input crank with respect to the vector G.

θ2i  θ  θ

θ2i  152.820  deg

θ2f  θ2i  β

θ2f  332.820  deg

13. Define the coupler point with respect to point A and the vector V. rp  z

δp  ϕ  θ

rp  5.000

δp  0.000  deg

which is correct for the assumption that the precision point is at C. 14. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  0  deg R1 

2

ρ  0.000  deg 2

C1x  C1y

R1  0.000



 

O2x  3.483

 

O2y  0.985

 

O4x  2.027

 

O4y  2.095

O2x  R1 cos ρ  z cos ϕ  w cos θ O2y  R1 sin ρ  z sin ϕ  w sin θ O4x  R1 cos ρ  s cos ψ  u  cos σ O4y  R1 sin ρ  s sin ψ  u  sin σ

15. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis the line O2O4.

θrot  atan2 O4x  O2x  O4y  O2y

θrot  11.391 deg

16. Determine the Grashof condition. Condition( a b c d ) 


Condition( g u v w)  "Grashof" 17. DESIGN SUMMARY Link 2:

w  1.380

θ  164.211  deg

Link 3:

v  5.000

θ  15.789 deg

Link 4:

u  2.915

σ  134.048  deg

Link 1:

g  5.621

θ  11.391 deg

Coupler:

rp  5.000

δp  0.000  deg

Crank angles:

θ2i  152.820  deg θ2f  332.820  deg




Design a fourbar mechanism to give the two positions shown in Figure P3-1 of coupler motion.

Given:

Coordinates of the points A1, B1, A2, and B2 with respect to A1: A1x  0.0

B1x  1.721

A2x  2.656

B2x  5.065

A1y  0.0

B1y  1.750

A2y  0.751

B2y  0.281

Assumptions: Use the points A1 and A2 as the precision points P1 and P2. Define position vectors in the global frame whose origin is at A1. Solution:

See solution to Problem 3-4 and Mathcad file P0502.

1.

Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. See Section 5.3 in which the equations for the two-position motion generation problem are derived.

2.

Two solution methods are derived in Section 5.3 and are presented in equations 5.7 and 5.8 for the left dyad and equations 5.11 and 5.12 for the right dyad.

3.

Method 1 (equations 5.7 and 5.11) requires the choosing of three angles for each dyad. Method 2 (equations 5.8 and 5.12) requires the choosing of two angles and a length for each dyad. Method 1 is used in this solution.

4.

In order to obtain the same solution as was done graphically in Problem 3-4, the necessary assumed values were taken from that solution as shown below.

5.


 A1x     A1y 

R2 

 A2x     A2y 

 P21x     R2  R1  P21y 

P21x  2.656 P21y  0.751

p 21  6.

7.

2

2

P21x  P21y

p 21  2.760

From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α  atan2 A2x  B2x  A2y  B2y   atan2 A1x  B1x  A1y  B1y

α  303.481  deg

δ  atan2 P21x P21y

δ  15.789 deg

From the graphical solution (see figure below), determine the values necessary for input to equations 5.7. θ  94.394 deg

β  40.366 deg

ϕ  45.479 deg



O4 93.449° jY 54.330° 2.760 u

A1 X 45.479°

0.281

P 21

15.789°

B2

0.751

v 1.750

A2 134.521° 2.656

2.409

B1 w

40.366°

94.394° 75.124° O2

8.

Solve for the WZ dyad using equations 5.7.

  



 

  



 

  



 

  



 

A  cos θ  cos β  1  sin θ  sin β

B  cos ϕ  cos α  1  sin ϕ  sin α

 

C  p 21  cos δ

D  sin θ  cos β  1  cos θ  sin β

E  sin ϕ  cos α  1  cos ϕ  sin α w 

C  E  B F

 

F  p 21  sin δ z 

A  E  B D

w  4.000

A  F  C D A  E  B D

z  0.000

These are the expected values of w and z based on the design choices made in the graphical solution and the assumptions made in this problem. 9.

From the graphical solution (see figure above), determine the values necessary for input to equations 5.11. σ  93.449 deg

γ  54.330 deg

ψ  134.521  deg

10. Solve for the US dyad using equations 5.11.

  



 

  



 

A'  cos σ  cos γ  1  sin σ  sin γ

B'  cos ψ  cos α  1  sin ψ  sin α

 

C  p 21  cos δ



  



 

  



 

D'  sin σ  cos γ  1  cos σ  sin γ

E'  sin ψ  cos α  1  cos ψ  sin α u 

C E'  B' F

 

F  p 21  sin δ s 

A'  E'  B' D'

u  4.000

A'  F  C D' A'  E'  B' D'

s  2.455

These are the expected values of u and s based on the design choices made in the graphical solution and the assumptions made in this problem. 11. Solve for the links 3 and 1 using the vector definitions of V and G. Link 3:


V1x  1.721


V1y  1.750

θ  atan2 V1x V1y

θ  45.479 deg

v 

Link 1:

2

2

V1x  V1y

v  2.455

 


 

G1x  1.655


G1y  6.231

θ  atan2 G1x G1y

θ  75.123 deg

g 

2

2

G1x  G1y

g  6.447

12. Determine the initial and final values of the input crank with respect to the vector G.

θ2i  θ  θ

θ2i  19.271 deg

θ2f  θ2i  β

θ2f  21.095 deg

13. Define the coupler point with respect to point A and the vector V. rp  z

δp  ϕ  θ

rp  0.000

δp  0.000  deg

which is correct for the assumption that the precision point is at A. 14. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  0  deg R1 

ρ  0.000  deg

2

A1x  A1y

2

R1  0.000

 

O2x  0.306

 

O2y  3.988

O2x  R1 cos ρ  z cos ϕ  w cos θ O2y  R1 sin ρ  z sin ϕ  w sin θ



 

O4x  1.962

 

O4y  2.243

O4x  R1 cos ρ  s cos ψ  u  cos σ O4y  R1 sin ρ  s sin ψ  u  sin σ

15. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.


θrot  75.123 deg



Condition( g u v w)  "non-Grashof" 17. DESIGN SUMMARY Link 2:

w  4.000

θ  94.394 deg

Link 3:

v  2.455

θ  45.479 deg

Link 4:

u  4.000

σ  93.449 deg

Link 1:

g  6.447

θ  75.123 deg

Coupler:

rp  0.000

δp  0.000  deg

Crank angles:

θ2i  19.271 deg θ2f  21.095 deg




Design a fourbar mechanism to give the three positions of coupler motion with no quick return shown in Figure P3-2. (See Problem 3-5). Ignore the fixed pivot points in the Figure.

Given:

Coordinates of points A and B with respect to point A1: A1x  0.0

A1y  0.0

B1x  0.741

B1y  2.383

A2x  2.019

A2y  1.905

B2x  4.428

B2y  2.557

A3x  3.933

A3y  1.035

B3x  6.304

B3y  0.256

Assumptions: Let points A1, A2, and A3 be the precision points P1, P2, and P3, respectively. Solution: 1.

2.

3.

See Figure P3-2 Mathcad file P0503.

Determine the magnitudes and orientation of the position difference vectors. 2

2

p 21  2.776

δ  atan2 A2x A2y 

δ  43.336 deg

2

2

p 31  4.067

δ  atan2 A3x A3y 

δ  14.744 deg

p 21 

A2x  A2y

p 31 

A3x  A3y

Determine the angle changes of the coupler between precision points.

θP1  atan2 A1x  B1x  A1y  B1y

θP1  107.273  deg

θP2  atan2 A2x  B2x  A2y  B2y

θP2  164.856  deg

θP3  atan2 A3x  B3x  A3y  B3y

θP3  161.812  deg

α  θP2  θP1

α  57.582 deg

α  θP3  θP1

α  269.085  deg

The free choices for this linkage are (from the graphical solution to Problem 3-5): β  78.375 deg

4.

β  135.560  deg

γ  59.771 deg

γ  107.023  deg

Evaluate terms in the WZ coefficient matrix and constant vector from equations 5.26 and form the matrix and vector:

  D  sin α G  sin β L  p 31  cos δ

A  cos β  1

 A F AA   B G 

B C D 

 G H K  A D C   F K H 

 

B  sin β

  H  cos α  1 M  p 21  sin δ E  p 21  cos δ

 E  L CC    M  N   

  F  cos β  1 K  sin α N  p 31  sin δ C  cos α  1

 W1x     W1y   AA  1 CC  Z1x     Z1y 



The components of the W and Z vectors are: W1x  2.178

W1y  0.286

Z1x  0.000

Z1y  0.000


θ  172.523  deg

ϕ  atan2 Z1x Z1y

ϕ  174.344  deg

 W1x2  W1y2 , w  2.197  

The length of link 2 is: w 

 Z1x2  Z1y2 , z  0.000  

The length of vector Z is: z  5.

Evaluate terms in the US coefficient matrix and constant vector from equations 5.31 and form the matrix and vector:

 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

F'  cos γ  1

 

M  p 21  sin δ

B' C D 

 E  L CC    M  N   



G' H K 

  F' K H 

A'

D C

N  p 31  sin δ

 U1x  U   1y   AA  1 CC  S1x     S1y 

The components of the U and S vectors are: U1x  1.995

U1y  3.121

S 1x  0.741

S 1y  2.383


σ  122.591  deg

ψ  atan2 S 1x S 1y

ψ  107.267  deg

The length of link 4 is: u 

 U 2  U 2 , u  3.704 1y   1x

The length of vector S is: s  6.

 S 1x2  S 1y2 , s  2.495  

Solve for links 3 and 1 using the vector definitions of V and G. Link 3:

V1x  Z1x  S 1x

V1x  0.741

V1y  Z1y  S 1y

V1y  2.383

θ  atan2 V1x V1y

θ  72.734 deg

v  Link 1:

2

2

V1x  V1y

G1x  W1x  V1x  U1x

v  2.495 G1x  0.558


G1y  W1y  V1y  U1y

G1y  0.452

θ  atan2 G1x G1y

θ  39.009 deg

g  7.

8.

9.


2

2

G1x  G1y

g  0.718

Determine the initial and final values of the input crank with respect to the vector G.

θ2i  θ  θ

θ2i  211.532  deg

θ2f  θ2i  β

θ2f  75.972 deg

Define the coupler point with respect to point A and the vector V. rp  z

δp  ϕ  θ

rp  0.000

δp  247.078  deg

Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x  z cos ϕ  w cos θ

O2x  2.178

O2y  z sin ϕ  w sin θ

O2y  0.286

O4x  s cos ψ  u  cos σ

O4x  2.736

O4y  s sin ψ  u  sin σ

O4y  0.738



θrot  39.009 deg




w  2.197

θ  172.523  deg

Link 3:

v  2.495

θ  72.734 deg

Link 4:

u  3.704

σ  122.591  deg

Link 1:

g  0.718

θ  39.009 deg

Coupler:

rp  0.000

δp  247.078  deg



Crank angles:

θ2i  211.532  deg θ2f  75.972 deg 13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.

Y

0.718 2.197 O4 O2

2

A1

1a

B3

2.496

A3

3 4 A2 B1

B2 3.704

This is the same result as that found in Problem 3-5.

X




Design a fourbar mechanism to give the three positions shown in Figure P3-2 (see Problem 3-6). Use analytical synthesis and design it for the fixed pivots shown.

Given:

Link end points (with respect to A1): A1x  0.0

B1x  0.741

A2x  2.019

B2x  4.428

A3x  3.933

B3x  6.304

A1y  0.0

B1y  2.383

A2y  1.905

B2y  2.557

A3y  1.035

B3y  0.256

Fixed pivot points (with respect to A1): O2x  0.995 Solution: 1.

2.

3.

O2y  5.086

O4x  5.298

O4y  5.086


Determine the angle changes between precision points from the body angles given.

θP1  atan2 A1x  B1x A1y  B1y

θP1  107.273  deg

θP2  atan2 A2x  B2x A2y  B2y

θP2  164.856  deg

θP3  atan2 A3x  B3x A3y  B3y

θP3  161.812  deg

α  θP2  θP1

α  57.582 deg

α  θP3  θP1

α  269.085  deg

Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. P21x  A2x

P21x  2.019

P31x  A3x

P31x  3.933

P21y  A2y

P21y  1.905

P31y  A3y

P31y  1.035

R1x  O2x

R1x  0.995

R1y  O2y

R1y  5.086

R2x  R1x  P21x

R2x  1.024

R2y  R1y  P21y

R2y  3.181

R3x  R1x  P31x

R3x  2.938

R3y  R1y  P31y

R3y  4.051

2

2

R1  5.182

2

2

R2  3.342

2

2

R3  5.004

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ  atan2 R1x R1y

ζ  101.069  deg

ζ  atan2 R2x R2y

ζ  72.156 deg



ζ  atan2 R3x R3y 4.

ζ  54.048 deg

Solve for 2 and 3 using equations 5.34





















 

C3  8.007

 

C4  5.127

 

C5  5.851

 

C6  1.294

C1  R3 cos α  ζ  R2 cos α  ζ C2  R3 sin α  ζ  R2 sin α  ζ C3  R1 cos α  ζ  R3 cos ζ





C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C2  3.679

A1  C3  C4

A1  90.406

A2  C3 C6  C4 C5

A2  19.633

A3  C4 C6  C3 C5

A3  53.487

A4  C2 C3  C1 C4

A4  22.524

A5  C4 C5  C3 C6

A5  19.633

A6  C1 C3  C2 C4

A6  29.689

K1  A2  A4  A3  A6

K1  1.146  10

K2  A3  A4  A5  A6

K2  1.788  10

2

K3 

3 3

2

2

2

A1  A2  A3  A4  A6

2

2

3

K3  1.769  10

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  90.915 deg

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  23.770 deg

The first value is the same as 3, so use the second value

β  β

 A5  sin β  A3  cos β  A6   A1  

β  16.790 deg

 A3  sin β  A2  cos β  A4   A1  

β  16.790 deg

β  acos

β  asin

Since both values are the same, 5.

C1  1.352

Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2.

β  β



R1x  O4x

6.

7.

R1x  5.298

R1y  O4y

R2x  R1x  P21x

R2x  3.279

R2y  R1y  P21y

R2y  3.181

R3x  R1x  P31x

R3x  1.365

R3y  R1y  P31y

R3y  4.051

2

2

R1  7.344

2

2

R2  4.568

2

2

R3  4.275

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  5.086


ζ  136.170  deg

ζ  atan2 R2x R2y

ζ  135.869  deg

ζ  atan2 R3x R3y

ζ  108.621  deg

Solve for 2 and 3 using equations 5.34





















 

C3  3.636

 

C4  9.430

 

C5  3.855

 

C6  4.927






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C1  1.023 C2  4.349

A1  C3  C4

A1  102.135

A2  C3 C6  C4 C5

A2  18.434

A3  C4 C6  C3 C5

A3  60.472

A4  C2 C3  C1 C4

A4  25.460

A5  C4 C5  C3 C6

A5  18.434

A6  C1 C3  C2 C4

A6  37.287

K1  A2  A4  A3  A6

K1  1.785  10

K2  A3  A4  A5  A6

K2  2.227  10

3 3


2

K3 


2

2

2

A1  A2  A3  A4  A6

2 3

K3  2.198  10

2

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  90.915 deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  11.643 deg


γ  

 A5  sin γ  A3  cos γ  A6   A1  

  11.069 deg

 A3  sin γ  A2  cos γ  A4   A1  

  11.069 deg

  acos

  asin

γ  

Since both angles are the same, 8.

Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. p 21 

2

2

P21x  P21y

p 21  2.776

δ  atan2 P21x P21y p 31 

2

δ  43.336 deg

2

P31x  P31y

p 31  4.067

δ  atan2 P31x P31y 9.

δ  14.744 deg

Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector: A  cos β  1 B  sin β C  cos α  1

 

 

 

 

E  p 21  cos δ

 

H  cos α  1

D  sin α G  sin β

 

L  p 31  cos δ

 A F AA   B G 

B C D 

 G H K  A D C   F K H 

10. The components of the W and Z vectors are:

 

 

 

M  p 21  sin δ

 E  L CC    M  N   

 

F  cos β  1

 

K  sin α

 

N  p 31  sin δ

 W1x   W1y   AA  1 CC  Z1x   Z1y   



W1x  3.594

W1y  7.810 2

w 

11. The length of link 2 is:

Z1x  4.589

2

W1x  W1y

Z1y  2.724

w  8.597

12. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:

 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

K  sin α

 

L  p 31  cos δ

 A'  F' AA    B'  G' 

 

F'  cos γ  1

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

E   L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x     U1y   AA  1 CC  S1x   S1y   

13. The components of the W and Z vectors are: U1x  2.400


U1y  7.549 2

u 

U1x  U1y

S1x  2.898

2

S1y  2.463

u  7.921

15. Solving for links 3 and 1 from equations 5.2a and 5.2b. V1x  Z1x  S1x

V1x  1.691

V1y  Z1y  S1y

V1y  0.261

The length of link 3 is:

v 

2

2

V1x  V1y

v  1.711

G1x  W1x  V1x  U1x

G1x  4.303

G1y  W1y  V1y  U1y

G1y  5.329  10


g 

2

G1x  G1y

2

 14

g  4.303

16. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1. O2x  Z1x  W1x

O2x  0.995

O2y  Z1y  W1y

O2y  5.086

O4x  S1x  U1x

O4x  5.298

O4y  S1y  U1y

O4y  5.086

These check with Figure P3-2.



17. Determine the location of the coupler point with respect to point A and line AB. 2

2

z  5.336

2

2

s  3.803

Distance from A to P

z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y

rP  z

ψ  atan2( S1x S1y)

ψ  139.639  deg

ϕ  atan2( Z1x Z1y )

ϕ  149.309  deg

θ  atan2 z cos ϕ  s cos ψ z sin ϕ  s sin ψ  θ  171.233  deg

δp  ϕ  θ

δp  21.924 deg

18. DESIGN SUMMARY Link 1:

g  4.303

Link 2:

w  8.597

Link 3:

v  1.711

Link 4:

u  7.921

Coupler point:

rP  5.336

δp  21.924 deg

19. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct.




See Project P3-8. Define three positions of the boat and analytically synthesize a linkage to move through them.

Assumptions: Launch ramp angle is 15 deg to the horizontal. Solution: 1.

See Project P3-8 and Mathcad file P0505.

This is an open-ended design problem that has many valid solutions. First define the problem more completely than is stated by deciding on three positions for the boat to move through. The figure below shows one such set of positions (dimensions are in mm). Y 3539 1453 1 deg. P2

15 deg.

725 X

P1 1179

1261

1431

0 deg. P3

O4

O2

WATER LEVEL 331 2694 RAMP

2.

From the figure, the design choices are: P21x  1453

P21y  725

P31x  3539

P31y  1261

O2x  331

O2y  1179

O4x  2694

O4y  1431

Body angles:

θP1  15 deg

θP2  1  deg

θP3  0  deg

3.

The methods of Section 5.8 are used to get a solution for this problem. The solution is sensitive to small changes in the design choices so a trial-and-error approach is warranted.

4.


5.

α  θP2  θP1

α  14.000 deg

α  θP3  θP1

α  15.000 deg

Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. R1x  O2x

R1x  331.000

R1y  O2y 3

R2x  R1x  P21x

R2x  1.122  10

R2y  R1y  P21y

R2y  1.904  10

R3x  R1x  P31x

R3x  3.208  10

R3y  R1y  P31y

R3y  82.000

3 3

R1y  1.179  10

3


6.

7.


2

2

R1  1.225  10

3

2

2

R2  2.210  10

3

2

2

R3  3.209  10

3

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y


ζ  74.318 deg

ζ  atan2 R2x R2y

ζ  120.510  deg

ζ  atan2 R3x R3y

ζ  178.536  deg






















 

C3  3.833  10

 

C4  1.135  10

 

C5  1.728  10

 

C6  840.097

C1  R3 cos α  ζ  R2 cos α  ζ C2  R3 sin α  ζ  R2 sin α  ζ



C4  R1 sin α  ζ  R3 sin ζ







C6  R1 sin α  ζ  R2 sin ζ 2

C2  1.433  10

3

3

3

C5  R1 cos α  ζ  R2 cos ζ



3

3

C3  R1 cos α  ζ  R3 cos ζ



C1  2.542  10

2

7

A1  C3  C4

A1  1.598  10

A2  C3 C6  C4 C5

A2  5.182  10

A3  C4 C6  C3 C5

A3  5.671  10

6

A4  C2 C3  C1 C4

A4  2.607  10

6

A5  C4 C5  C3 C6

A5  5.182  10

6

A6  C1 C3  C2 C4

A6  1.137  10

7

K1  A2  A4  A3  A6

K1  5.096  10

K2  A3  A4  A5  A6

K2  7.370  10

2

K3 

6

13 13

2

2

2

A1  A2  A3  A4  A6

2

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

13

K3  3.015  10

β  125.675  deg



 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  15.000 deg

The second value is the same as 3, so use the first value

β  β

 A5  sin β  A3  cos β  A6   A1  

β  39.836 deg

 A3  sin β  A2  cos β  A4   A1  

β  39.836 deg



β  β


Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2. R1x  O4x

9.

R1x  2.694  10

3

R1y  O4y

R2x  R1x  P21x

R2x  1.241  10

3

R2y  R1y  P21y

R2y  2.156  10

3

R3x  R1x  P31x

R3x  845.000

R3y  R1y  P31y

R3y  170.000

2

2

R1  3.050  10

3

2

2

R2  2.488  10

3

2

2

R3  861.931

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  1.431  10


ζ  27.976 deg

ζ  atan2 R2x R2y

ζ  60.075 deg

ζ  atan2 R3x R3y

ζ  168.625  deg

10. Solve for 2 and 3 using equations 5.34





















 

C3  3.818  10

 

C4  514.981

 

C5  1.719  10

 

C6  1.419  10






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ

C1  2.536  10

3

C2  1.392  10

3

3

3 3

3

DESIGN OF MACHINERY - 5th Ed. 2


2

7

A1  C3  C4

A1  1.484  10

A2  C3 C6  C4 C5

A2  6.303  10

A3  C4 C6  C3 C5

A3  5.832  10

6

A4  C2 C3  C1 C4

A4  4.008  10

6

A5  C4 C5  C3 C6

A5  6.303  10

6

A6  C1 C3  C2 C4

A6  1.040  10

7

K1  A2  A4  A3  A6

K1  3.537  10

K2  A3  A4  A5  A6

K2  8.891  10

2

K3 

6

13 13

2

2

2

A1  A2  A3  A4  A6

2

2

13

K3  1.115  10

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  151.615  deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  15.000 deg


γ  γ

 A5  sin γ  A3  cos γ  A6   A1  

  56.167 deg

 A3  sin γ  A2  cos γ  A4   A1  

  56.167 deg

  acos

  asin

γ  

Since both angles are the same,

11. Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. p 21 

2

2

P21x  P21y

δ  atan2 P21x P21y

p 31 

2

2

P31x  P31y

δ  atan2 P31x P31y

3

p 21  1.624  10

δ  153.482  deg

3

p 31  3.757  10

δ  160.388  deg



12. Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:

 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 

M  p 21  sin δ

B C D 

 A F AA   B G 

 

C  cos α  1

N  p 31  sin δ

 E  L CC    M  N   

 G H K  A D C   F K H 

 W1x   W1y   AA  1 CC  Z1x   Z1y   

13. The components of the W and Z vectors are: W1x  1.331  10

3

w 


W1y  1.653  10 2

3

Z1x  1.000  10

2

W1x  W1y

3

Z1y  474.187

w  2122.473


 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

F'  cos γ  1

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   

16. The components of the W and Z vectors are: 3

U1x  1.690  10


u 

U1y  951.273 2

U1x  U1y

2

S1x  1.004  10

3

S1y  479.727

u  1939.291


V1x  2.004  10

V1y  Z1y  S1y

V1y  953.914


v 

2

2

V1x  V1y

v  2219.601

3



G1x  W1x  V1x  U1x

G1x  2.363  10

G1y  W1y  V1y  U1y

G1y  252.000


2

g 

G1x  G1y

2

3

g  2376.399


O2x  331.000

O2y  Z1y  W1y

O2y  1179.000

O4x  S1x  U1x

O4x  2694.000

O4y  S1y  U1y

O4y  1431.000

These check with the design choices shown in the figure above. 20. Determine the location of the coupler point with respect to point A and line AB. 2

2

z  1106.834

2

2

s  1112.770


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y

rP  z


ψ  25.538 deg

ϕ  atan2( Z1x Z1y )

ϕ  154.633  deg


δp  ϕ  θ

δp  0.086  deg


g  2376.4

Link 2:

w  2122.5

Link 3:

v  2219.6

Link 4:

u  1939.3

Coupler point:

rP  1106.8

δp  0.086  deg

22. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct. The solution is drawn below to show the locations of the moving pivots for the three positions chosen (see next page). 23. The design needs to be checked for the presence of toggle positions within its desired range of motion. This design has none, but is close to toggle at position 1. This could be used as a locking feature. 24. The transmission angles need to be checked also. This design has poor transmission angles, especially in and near positions 1 and 3. Unfortunately, this is where a large overturning moment is created by the mass of the boat. A large mechanical advantage input device will need to be used here, such as a hydraulic cylinder or geared drive. Note that the drive mechanism must also resist being overdriven (back driven) by the load as the boat descends from its high point onto the trailer.



A2 A1

B2

B1 A3 B3 WATER LEVEL

RAMP

O4

O2




See Project P3-20. Define three positions of the dumpster and analytically synthesize a linkage to move through them. The fixed pivots must be located on the existing truck.

Solution:


1.

This is an open-ended design problem that has many valid solutions. First define the problem more completely than it is stated by deciding on three positions for the dumpster box to move through. The figure below shows one such set of positions (dimensions are in mm). 1999 59.1 deg. 590

30.3 deg.

P3 P2 1817 0 deg.

1202

226 311

P1 O2 O4

2036 2094

2.

From the figure, the design choices are: P21x  590

P21y  1202

P31x  1999

P31y  1817

O2x  2094

O2y  226

O4x  2036

O4y  311

Body angles:

θP1  0  deg

θP2  30.3 deg

θP3  59.1 deg

3.


4.


5.

α  θP2  θP1

α  30.300 deg

α  θP3  θP1

α  59.100 deg


R1x  2.094  10

3

R1y  O2y

R2x  R1x  P21x

R2x  1.504  10

3

R2y  R1y  P21y

R2y  1.428  10

3

R3x  R1x  P31x

R3x  95.000

R1y  226.000



R3y  R1y  P31y

6.

7.

R3y  2.043  10

3

2

2

R1  2.106  10

3

2

2

R2  2.074  10

3

2

2

R3  2.045  10

3

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y


ζ  6.160  deg

ζ  atan2 R2x R2y

ζ  43.515 deg

ζ  atan2 R3x R3y

ζ  87.338 deg






















 

C3  786.433

 

C4  130.152

 

C5  189.927

 

C6  176.392






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C1  495.777 C2  212.019

5

A1  C3  C4

A1  6.354  10

A2  C3 C6  C4 C5

A2  1.140  10

A3  C4 C6  C3 C5

A3  1.723  10

5

A4  C2 C3  C1 C4

A4  2.313  10

5

A5  C4 C5  C3 C6

A5  1.140  10

5

A6  C1 C3  C2 C4

A6  3.623  10

5

K1  A2  A4  A3  A6

K1  3.607  10

K2  A3  A4  A5  A6

K2  8.115  10

2

K3 

5

10 10

2

2

2

A1  A2  A3  A4  A6

2

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

10

K3  8.816  10

β  72.976 deg



 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  59.100 deg


β  β

 A5  sin β  A3  cos β  A6   A1  

β  34.802 deg

 A3  sin β  A2  cos β  A4   A1  

β  34.802 deg



β  β



9.

R1x  2.036  10

3

R1y  O4y

R2x  R1x  P21x

R2x  1.446  10

3

R2y  R1y  P21y

R2y  1.513  10

3

R3x  R1x  P31x

R3x  37.000

R3y  R1y  P31y

R3y  2.128  10

3

2

2

R1  2.060  10

3

2

2

R2  2.093  10

3

2

2

R3  2.128  10

3

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  311.000


ζ  8.685  deg

ζ  atan2 R2x R2y

ζ  46.297 deg

ζ  atan2 R3x R3y

ζ  89.004 deg






















 

C3  741.712

 

C4  221.269

 

C5  154.965

 

C6  217.266






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ

C1  486.018 C2  161.777



2

5

A1  C3  C4

A1  5.991  10

A2  C3 C6  C4 C5

A2  1.269  10

A3  C4 C6  C3 C5

A3  1.630  10

5

A4  C2 C3  C1 C4

A4  2.275  10

5

A5  C4 C5  C3 C6

A5  1.269  10

5

A6  C1 C3  C2 C4

A6  3.247  10

5

K1  A2  A4  A3  A6

K1  2.406  10

K2  A3  A4  A5  A6

K2  7.828  10

2

K3 

5

10 10

2

2

2

A1  A2  A3  A4  A6

2

2

10

K3  7.953  10

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  86.724 deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  59.100 deg


γ  γ

 A5  sin γ  A3  cos γ  A6   A1  

  39.743 deg

 A3  sin γ  A2  cos γ  A4   A1  

  39.743 deg

  acos

  asin

γ  



2

2

P21x  P21y

δ  atan2 P21x P21y p 31 

2

2

P31x  P31y

δ  atan2 P31x P31y

p 21  1338.994 δ  116.144  deg p 31  2701.387 δ  137.731  deg




 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 

M  p 21  sin δ

B C D 

 A F AA   B G 

 

C  cos α  1

 E  L CC    M  N   

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x   W1y   AA  1 CC  Z1x   Z1y   

13. The components of the W and Z vectors are: W1x  3.194  10

3

W1y  829.763

Z1x  1.100  10

3

Z1y  603.763

14. The length of link 2 is: w 

2

2

W1x  W1y

w  3299.543


 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

F'  cos γ  1

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   

16. The components of the W and Z vectors are: 3

U1x  1.621  10

U1y  13.492

S1x  415.016

S1y  297.508

17. The length of link 4 is: u 

2

U1x  U1y

2

u  1621.040


V1x  1.515  10

V1y  Z1y  S1y

V1y  901.272

3




v 

2

V1x  V1y

v  1762.404

G1x  W1x  V1x  U1x

G1x  58.000

G1y  W1y  V1y  U1y

G1y  85.000


2

g 

G1x  G1y

2

g  102.903


O2x  2094.000

O2y  Z1y  W1y

O2y  226.000

O4x  S1x  U1x

O4x  2036.000

O4y  S1y  U1y

O4y  311.000


2

z  1254.370

2

2

s  510.637


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y

rP  z


ψ  35.635 deg

ϕ  atan2( Z1x Z1y )

ϕ  151.228  deg


δp  ϕ  θ

δp  1.984  deg


g  102.9

Link 2:

w  3299.5

Link 3:

v  1762.4

Link 4:

u  1621.0

Coupler point:

rP  1254.4

δp  1.984  deg

22. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct. The solution is drawn below to show the locations of the moving pivots for the three positions chosen (see next page). 23. The design needs to be checked for the presence of toggle positions within its desired range of motion. This design has none, but is close to toggle at position 3.



24. The transmission angles need to be checked also. A large mechanical advantage input device will need to be used here, such as a hydraulic cylinder. Note that the drive mechanism must also resist being overdriven (back driven) by the load as the dumpster descends from its high point onto the truck.

A3

A2

B3

B2

O2

B1 O4

A1




See Project P3-7. Define three positions of the computer monitor and analytically synthesize a linkage to move through them. The fixed pivots must be located on the floor or wall.

Solution:


1.

This is an open-ended design problem that has many valid solutions. First define the problem more completely than it is stated by deciding on three positions for the computer monitor to move through. The figure below shows one such set of positions (dimensions are in inches). Y 33.816

O2 97 deg. 11.580 P1

O4

X

90 deg. 7.812

1.272

14.472

P2

WALL 85 deg.

P3

2.148

5.736

2.

From the figure, the design choices are: P21x  2.148

P21y  7.812

P31x  5.736

P31y  14.472

O2x  33.816

O2y  11.580

O4x  33.816

O4y  1.272

Body angles:

θP1  97 deg

θP2  90 deg

θP3  85 deg

3.


4.


5.

α  θP2  θP1

α  7.000  deg

α  θP3  θP1

α  12.000 deg


R1x  33.816

R1y  O2y

R2x  R1x  P21x

R2x  31.668

R2y  R1y  P21y

R2y  19.392

R3x  R1x  P31x

R3x  28.080

R3y  R1y  P31y

R3y  26.052

R1y  11.580


6.

7.


2

2

R1  35.744

2

2

R2  37.134

2

2

R3  38.304

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y


ζ  161.097  deg

ζ  atan2 R2x R2y

ζ  148.519  deg

ζ  atan2 R3x R3y

ζ  137.146  deg






















 

C3  7.405

 

C4  21.756

 

C5  3.307

 

C6  12.019






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C1  3.962 C2  10.052

A1  C3  C4

A1  528.143

A2  C3 C6  C4 C5

A2  17.049

A3  C4 C6  C3 C5

A3  285.981

A4  C2 C3  C1 C4

A4  11.771

A5  C4 C5  C3 C6

A5  17.049

A6  C1 C3  C2 C4

A6  248.020

K1  A2  A4  A3  A6

K1  7.073  10

K2  A3  A4  A5  A6

K2  7.595  10

2

K3 

4 3

2

2

2

A1  A2  A3  A4  A6

2

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

4

K3  6.760  10

β  24.258 deg



 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  12.000 deg


β  β

 A5  sin β  A3  cos β  A6   A1  

β  12.435 deg

 A3  sin β  A2  cos β  A4   A1  

β  12.435 deg



β  β



9.

R1x  33.816

R1y  O4y

R2x  R1x  P21x

R2x  31.668

R2y  R1y  P21y

R2y  6.540

R3x  R1x  P31x

R3x  28.080

R3y  R1y  P31y

R3y  13.200

2

2

R1  33.840

2

2

R2  32.336

2

2

R3  31.028

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  1.272


ζ  177.846  deg

ζ  atan2 R2x R2y

ζ  168.331  deg

ζ  atan2 R3x R3y

ζ  154.822  deg






















 

C3  4.733

 

C4  21.475

 

C5  1.741

 

C6  11.924






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ

C1  2.856 C2  9.867



2

A1  C3  C4

A1  483.571

A2  C3 C6  C4 C5

A2  19.043

A3  C4 C6  C3 C5

A3  264.299

A4  C2 C3  C1 C4

A4  14.646

A5  C4 C5  C3 C6

A5  19.043

A6  C1 C3  C2 C4

A6  225.402

K1  A2  A4  A3  A6

K1  5.929  10

K2  A3  A4  A5  A6

K2  8.163  10

2

K3 

4 3

2

2

2

A1  A2  A3  A4  A6

2

2

4

K3  5.630  10

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  27.678 deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  12.000 deg


γ  γ

 A5  sin γ  A3  cos γ  A6   A1  

  14.435 deg

 A3  sin γ  A2  cos γ  A4   A1  

  14.435 deg

  acos

  asin

γ  



2

2

P21x  P21y

δ  atan2 P21x P21y

p 31 

2

2

P31x  P31y

δ  atan2 P31x P31y

p 21  8.102 δ  74.626 deg

p 31  15.567 δ  68.379 deg




 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 

M  p 21  sin δ

B C D 

 A F AA   B G 

 

C  cos α  1

N  p 31  sin δ

 E  L CC    M  N   

 G H K  A D C   F K H 

 W1x   W1y   AA  1 CC  Z1x   Z1y   

13. The components of the W and Z vectors are: W1x  36.030

W1y  8.098

Z1x  2.214

Z1y  3.482

14. The length of link 2 is: w 

2

2

W1x  W1y

w  36.929


 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

F'  cos γ  1

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   


U1y  2.592

S1x  1.522

17. The length of link 4 is: u 

2

U1x  U1y

2

u  32.398


V1x  3.736

V1y  Z1y  S1y

V1y  7.346

S1y  3.864




v 

2

V1x  V1y

v  8.241

G1x  W1x  V1x  U1x

G1x  0.000

G1y  W1y  V1y  U1y

G1y  12.852


2

g 

G1x  G1y

2

g  12.852


O2x  33.816

O2y  Z1y  W1y

O2y  11.580

O4x  S1x  U1x

O4x  33.816

O4y  S1y  U1y

O4y  1.272


2

z  4.126

2

2

s  4.153


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y


ψ  111.494  deg

ϕ  atan2( Z1x Z1y )

ϕ  57.551 deg

rP  z

θ  atan2 z cos ϕ  s cos ψ z sin ϕ  s sin ψ  θ  63.046 deg

δp  ϕ  θ

δp  5.495  deg


g  12.852

Link 2:

w  36.929

Link 3:

v  8.241

Link 4:

u  32.398

Coupler point:

rP  4.126

δp  5.495  deg

22. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct. The solution is drawn below to show the locations of the moving pivots for the three positions chosen (see next page). 23. The design needs to be checked for the presence of toggle positions within its desired range of motion. This design has none.



24. The transmission angles need to be checked also. A means to support the weight of the monitor must be provided. The figure below shows a spring placed between links to provide the balancing moment. Further design and analysis needs to be done to optimize the spring placement in order to compensate for its change in force with deflection and the change in moment arm as the linkage moves.

SPRING O2

A1

A2 O4

B1

WALL A3 B2

B3




Design a linkage to carry the body in Figure P5-1 through the two positions P1 and P2 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown. Use the free choices given below.

Given:

Coordinates of the points P1 and P2 with respect to P1: P1x  0.0

P1y  0.0

P2x  1.236

P2y  2.138

Angles made by the body in positions 1 and 2:

θP1  210  deg

θP2  147.5  deg

Free choices for the WZ dyad : z  1.075

β  27.0 deg

ϕ  204.4  deg

γ  40.0 deg

ψ  74.0 deg

Free choices for the US dyad : s  1.240 Solution:


1.

Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. Because of the data given in the hint, the second method of Section 5.3 will be used here.

2.


 P1x     P1y 

R2 

 P2x     P2y 

 P21x     R2  R1  P21y 

P21x  1.236 P21y  2.138

p 21  3.

4.

2

2

P21x  P21y

p 21  2.470

From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α  θP2  θP1

α  62.500 deg

δ  atan2 P21x P21y

δ  120.033  deg



 

A  0.109

D  sin α

 

B  0.454

E  p 21  cos δ

 

C  0.538

F  p 21  sin δ

A  cos β  1 B  sin β

C  cos α  1

W1x 

Z1x  0.979

 

A   C Z1x  D Z1y  E  B  C Z1y  D Z1x  F  2  A

Z1y  0.444 D  0.887

 

E  1.236

 

F  2.138

W1x  1.462


W1y  w 


A   C Z1y  D Z1x  F   B  C Z1x  D Z1y  E

W1y  3.367

2  A 2

2

W1x  W1y

w  3.670

θ  atan2 W1x W1y 5.

θ  113.472  deg

Solve for the US dyad using equations 5.12. S 1x  s cos ψ

S 1x  0.342

 

A  0.234

D  sin α

 

B  0.643

E  p 21  cos δ

 

C  0.538

F  p 21  sin δ

A  cos γ  1 B  sin γ

C  cos α  1

U1x 

U1y  u 

S 1y  s sin ψ

 

D  0.887

 

E  1.236

 

F  2.138

A   C S 1x  D S 1y  E  B  C S 1y  D S 1x  F  2  A A   C S 1y  D S 1x  F   B  C S 1x  D S 1y  E

2

U1x  U1y

u  5.461

σ  atan2 U1x U1y 6.

σ  125.619  deg



V1x  1.321


V1y  1.636

θ  atan2 V1x V1y

θ  128.914  deg

v  Link 1:

2

2

V1x  V1y

v  2.103

 


 

G1x  0.398


G1y  0.564

θ  atan2 G1x G1y

θ  54.796 deg

g  7.

U1x  3.180

U1y  4.439

2  A 2

S 1y  1.192

2

2

G1x  G1y

g  0.690


θ2i  θ  θ

θ2i  58.677 deg



θ2f  θ2i  β 8.

9.

θ2f  85.677 deg


δp  ϕ  θ

rp  1.075

δp  333.314  deg

Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  0  deg R1 

ρ  0.000  deg

2

2

P1x  P1y

R1  0.000

 

O2x  2.441

 

O2y  3.811

 

O4x  2.838

 

O4y  3.247



θrot  atan2 O4x  O2x  O4y  O2y

θrot  54.796 deg




w  3.670

θ  113.472  deg

Link 3:

v  2.103

θ  128.914  deg

Link 4:

u  5.461

σ  125.619  deg

Link 1:

g  0.690

θ  54.796 deg

Coupler:

rp  1.075

δp  333.314  deg

Crank angles:

θ2i  58.677 deg θ2f  85.677 deg



13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.

O2

1.236

G1 Y

O4 U2

P2 S2

B2

V2 2.138

W2

U1

Z2 A2

Z1 62.5°

W1

A1 X

P1 V1

S1 B1




Design a linkage to carry the body in Figure P5-1 through the two positions P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown. Hint: First try a rough graphical solution to create realistic values for free choices.

Given:

Coordinates of the points P2 and P3 with respect to P1: P2x  1.236

P2y  2.138

P3x  2.500

P3y  2.931


θP2  147.5  deg Solution:

θP3  110.2  deg


1.

Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. The second method of Section 5.3 will be used here.

2.

Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1.

 P2x     P2y 

R1 

R2 

 P3x     P3y 

 P21x     R2  R1  P21y 

P21x  1.264 P21y  0.793

p 21  3.

2

2

P21x  P21y

p 21  1.492


α  37.300 deg

δ  atan2 P21x P21y

δ  147.897  deg O4

4.

From a graphical solution (see figure at right), determine the values necessary for input to equations 5.8.

Y

1.250

P3 43.806°

z  0.0 B3

β  43.806 deg

P2 32.500°

57.012° B2

O2

ϕ  32.500 deg

P1

5.


Z1x  0.000


 

A  0.278

D  sin α

 

B  0.692

E  p 21  cos δ

 

C  0.205

F  p 21  sin δ

A  cos β  1 B  sin β

C  cos α  1

 

Z1y  0.000 D  0.606

 

E  1.264

 

F  0.793

X


W1x 

W1y  w 



W1x  1.618


W1y  1.175

2  A 2

2

W1x  W1y

w  2.000

θ  atan2 W1x W1y 5.

θ  35.994 deg

From the graphical solution (see figure above), determine the values necessary for input to equations 5.12. s  1.250

6.

γ  57.012 deg

ψ  147.5  deg


S 1x  1.054

 

A  0.456

D  sin α

 

B  0.839

E  p 21  cos δ

 

C  0.205

F  p 21  sin δ

A  cos γ  1 B  sin γ

C  cos α  1

U1x 

U1y  u 

S 1y  s sin ψ

 

A   C S 1x  D S 1y  E  B  C S 1y  D S 1x  F  2  A A   C S 1y  D S 1x  F   B  C S 1x  D S 1y  E 2  A 2

2

U1x  U1y

 

E  1.264

 

F  0.793

U1x  0.675

U1y  1.883

σ  70.278 deg



V1x  1.054


V1y  0.672

θ  atan2 V1x V1y

θ  32.500 deg

v  Link 1:

2

2

V1x  V1y

v  1.250

 


 

G1x  1.997


G1y  2.386

θ  atan2 G1x G1y

θ  50.070 deg

g  8.

D  0.606

u  2.000

σ  atan2 U1x U1y 7.

S 1y  0.672

2

2

G1x  G1y

g  3.112



9.


θ2i  θ  θ

θ2i  14.077 deg

θ2f  θ2i  β

θ2f  29.729 deg


δp  ϕ  θ

rp  0.000

δp  0.000  deg

which is correct for the assumption that the precision point is at C. 10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  atan2 P2x P2y R1 

2

ρ  120.033  deg

2

P2x  P2y

R1  2.470

 

O2x  2.854

 

O2y  0.963

 

O4x  0.857

 

O4y  3.349



θrot  atan2 O4x  O2x  O4y  O2y 12. Determine the Grashof condition. Condition( a b c d ) 

θrot  50.070 deg



w  2.000

θ  35.994 deg

Link 3:

v  1.250

θ  32.500 deg

Link 4:

u  2.000

σ  70.278 deg

Link 1:

g  3.112

θ  50.070 deg

Coupler:

rp  0.000

δp  0.000  deg

Crank angles:

θ2i  14.077 deg θ2f  29.729 deg




Design a linkage to carry the body in Figure P5-1 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown. Use the free choices given below.

Given:


P1y  0.0

P2x  1.236

P3x  2.500

P3y  2.931

P2y  2.138

Angles made by the body in positions 1, 2 and 3:

θP1  210  deg

θP2  147.5  deg

θP3  110.2  deg

Free choices for the WZ dyad : β  30.0 deg

β  60.0 deg

Free choices for the US dyad : γ  10.0 deg Solution: 1.

2.

3.

γ  25.0 deg



2

p 21  2.470

δ  atan2 P2x P2y

δ  120.033  deg

2

2

p 31  3.852

δ  atan2 P3x P3y

δ  130.463  deg

p 21 

P2x  P2y

p 31 

P3x  P3y

Determine the angle changes of the coupler between precision points. α  θP2  θP1

α  62.500 deg

α  θP3  θP1

α  99.800 deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

 

K  sin α

 

L  p 31  cos δ

 

M  p 21  sin δ

B C D 

 E  L CC    M  N   



G H K 

  F K H  A

 

F  cos β  1

 

G  sin β

 A F AA   B G 

 

C  cos α  1

D C

N  p 31  sin δ

 W1x  W   1y   AA  1 CC  Z1x     Z1y 

The components of the W and Z vectors are: W1x  2.920

W1y  1.720

Z1x  0.756

Z1y  0.442




ϕ  149.697  deg

 W1x2  W1y2 , w  3.389  


 Z1x2  Z1y2 , z  0.876  



θ  30.493 deg


 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

F'  cos γ  1

 

M  p 21  sin δ

B' C D 

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

N  p 31  sin δ

 U1x  U   1y   AA  1 CC  S1x     S1y 


U1y  2.693

S 1x  0.792

S 1y  2.418


σ  110.545  deg

ψ  atan2 S 1x S 1y

ψ  108.125  deg


 U1x2  U1y2 , u  2.875  


 S 1x2  S 1y2 , s  2.544  


V1x  Z1x  S 1x

V1x  0.036

V1y  Z1y  S 1y

V1y  1.976

θ  atan2 V1x V1y

θ  88.968 deg

v  Link 1:

2

2

V1x  V1y

v  1.977

G1x  W1x  V1x  U1x

G1x  3.965

G1y  W1y  V1y  U1y

G1y  1.003

θ  atan2 G1x G1y

θ  14.202 deg


g  6.

7.

8.

9.

2


2

G1x  G1y

g  4.090


θ2i  θ  θ

θ2i  16.291 deg

θ2f  θ2i  β

θ2f  76.291 deg


δp  ϕ  θ

rp  0.876

δp  238.665  deg


O2x  2.164

O2y  z sin ϕ  w sin θ

O2y  1.278

O4x  s cos ψ  u  cos σ

O4x  1.801

O4y  s sin ψ  u  sin σ

O4y  0.274

Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.


θrot  14.202 deg




w  3.389

θ  30.493 deg

Link 3:

v  1.977

θ  88.968 deg

Link 4:

u  2.875

σ  110.545  deg

Link 1:

g  4.090

θ  14.202 deg

Coupler:

rp  0.876

δp  238.665  deg

Crank angles:

θ2i  16.291 deg θ2f  76.291 deg




Y

P3 S3

Z3 A3

B1

S2 P2

B3

V3

V2 S1

A2

W3

60.0°

B2

10.0° 25.0° V1

Z1

30.0°

A1

U2 U1 X

W2

P1 W1 G1

O2

O4




Given:

Solution: 1.

2.

Design a linkage to carry the body in Figure P5-1 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis and design it for the fixed pivots shown. P21x  1.236 O2x  2.164

P21y  2.138 O2y  1.260

P31x  2.500 O4x  2.190

P31y  2.931 O4y  1.260

Body angles:

θP1  210  deg

θP2  147.5  deg

θP3  110.2  deg


Determine the angle changes between precision points from the body angles given. α  θP2  θP1

α  62.500 deg

α  θP3  θP1

α  99.800 deg


3.

4.

R1x  2.164

R1y  O2y

R2x  R1x  P21x

R2x  0.928

R2y  R1y  P21y

R2y  3.398

R3x  R1x  P31x

R3x  0.336

R3y  R1y  P31y

R3y  4.191

2

2

R1  2.504

2

2

R2  3.522

2

2

R3  4.204

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  1.260


ζ  30.210 deg

ζ  atan2 R2x R2y

ζ  74.725 deg

ζ  atan2 R3x R3y

ζ  94.584 deg






















 

C3  1.209

 

C4  6.538

 

C5  1.189






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ

C1  0.372 C2  3.726







 

C6  R1 sin α  ζ  R2 sin ζ 2

C6  4.736

2

A1  C3  C4

A1  44.206

A2  C3 C6  C4 C5

A2  2.046

A3  C4 C6  C3 C5

A3  32.399

A4  C2 C3  C1 C4

A4  6.937

A5  C4 C5  C3 C6

A5  2.046

A6  C1 C3  C2 C4

A6  23.911

K1  A2  A4  A3  A6

K1  760.497

K2  A3  A4  A5  A6

K2  273.669

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

K3  140.232

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  60.217 deg

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  99.800 deg


β  β

 A5  sin β  A3  cos β  A6   A1  

β  30.143 deg

 A3  sin β  A2  cos β  A4   A1  

β  30.143 deg



β  β



R1x  2.190

R1y  O4y

R2x  R1x  P21x

R2x  3.426

R2y  R1y  P21y

R2y  3.398

R3x  R1x  P31x

R3x  4.690

R3y  R1y  P31y

R3y  4.191

R1 

2

2

R1x  R1y

R1  2.527

R1y  1.260


6.

7.


2

2

R2  4.825

2

2

R3  6.290

R2 

R2x  R2y

R3 

R3x  R3y


ζ  150.086  deg

ζ  atan2 R2x R2y

ζ  135.235  deg

ζ  atan2 R3x R3y

ζ  138.216  deg






















 

C3  6.304

 

C4  2.247

 

C5  3.532

 

C6  0.874






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C1  2.380 C2  3.298

A1  C3  C4

A1  44.796

A2  C3 C6  C4 C5

A2  2.431

A3  C4 C6  C3 C5

A3  24.233

A4  C2 C3  C1 C4

A4  15.441

A5  C4 C5  C3 C6

A5  2.431

A6  C1 C3  C2 C4

A6  22.414

K1  A2  A4  A3  A6

K1  505.612

K2  A3  A4  A5  A6

K2  428.679

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

2

K3  336.363

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  19.215 deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  99.800 deg




γ  γ

 A5  sin γ  A3  cos γ  A6   A1  

  6.628  deg

 A3  sin γ  A2  cos γ  A4   A1  

  6.628  deg

  acos

  asin

Since 2 is not in the first quadrant , 8.

γ  

Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. 2

p 21 

2

P21x  P21y

p 21  2.470

δ  atan2 P21x P21y 2

p 31 

δ  120.033  deg

2

P31x  P31y

p 31  3.852

δ  atan2 P31x P31y 9.

δ  130.463  deg

Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:

 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

A  F AA   B G 

 

C  cos α  1

 

M  p 21  sin δ

B C D 

E   L CC    M  N   

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x     W1y   AA  1 CC  Z1x   Z1y   

10. The components of the W and Z vectors are: W1x  2.915 11. The length of link 2 is:

w 

W1y  1.702 2

Z1x  0.751

2

W1x  W1y

Z1y  0.442

w  3.376


 

A'  cos γ  1

 

D  sin α

 

B'  sin γ

 

E  p 21  cos δ

 

C  cos α  1

 

F'  cos γ  1



 

 

G'  sin γ

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

H  cos α  1

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   



U1y  3.634 2

u 

U1x  U1y

S1x  0.819

2

S1y  2.374

u  3.884


V1x  0.068

V1y  Z1y  S1y

V1y  1.932 v 


2

2

V1x  V1y

v  1.933

G1x  W1x  V1x  U1x

G1x  4.354

G1y  W1y  V1y  U1y

G1y  2.220  10

g 


2

G1x  G1y

2

 15

g  4.354


O2x  2.164

O2y  Z1y  W1y

O2y  1.260

O4x  S1x  U1x

O4x  2.190

O4y  S1y  U1y

O4y  1.260

These check with Figure P5-1. 17. Determine the location of the coupler point with respect to point A and line AB. 2

2

z  0.871

2

2

s  2.511


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y

rP  z


ψ  109.037  deg

ϕ  atan2( Z1x Z1y )

ϕ  149.555  deg

θ  atan2 z cos ϕ  s cos ψ z sin ϕ  s sin ψ 



θ  87.994 deg

δp  ϕ  θ

δp  237.549  deg


g  4.354

Link 2:

w  3.376

Link 3:

v  1.933

Link 4:

u  3.884

Coupler point:

rP  0.871

δp  237.549  deg





Design a linkage to carry the body in Figure P5-2 through the two positions P1 and P2 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown. Use the free choices given below.

Given:


P1y  0.0

P2x  1.903

P2y  1.347


θP1  101.0  deg

θP2  62.0 deg


β  30.0 deg

ϕ  150.0  deg

γ  40.0 deg

ψ  50.0 deg



1.

Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. Because of the data given in the hint, the second method of Section 5.3 will be used here.

2.


 P1x     P1y 

R2 

 P2x     P2y 

 P21x     R2  R1  P21y 

P21x  1.903 P21y  1.347

p 21  3.

4.

2

2

P21x  P21y

p 21  2.331


α  39.000 deg

δ  atan2 P21x P21y

δ  35.292 deg



A  0.134

D  sin α

 

B  0.500

E  p 21  cos δ

 

E  1.903

 

C  0.223

F  p 21  sin δ

 

F  1.347

B  sin β

C  cos α  1

 

Z1y  1.000

 

A  cos β  1

W1x 

Z1x  1.732

A   C Z1x  D Z1y  E  B  C Z1y  D Z1x  F  2  A

D  0.629

W1x  0.452


W1y  w 


A   C Z1y  D Z1x  F   B  C Z1x  D Z1y  E

W1y  1.896

2  A 2

2

W1x  W1y

w  1.949

θ  atan2 W1x W1y 5.

θ  76.607 deg


S 1x  1.928 A  0.234

D  sin α

 

B  0.643

E  p 21  cos δ

 

E  1.903

 

C  0.223

F  p 21  sin δ

 

F  1.347

B  sin γ

C  cos α  1

U1y  u 

 

D  0.629


2

2

U1x  U1y

u  6.284 σ  81.540 deg



V1x  3.660


V1y  3.298

θ  atan2 V1x V1y

θ  137.980  deg

v  Link 1:

2

2

V1x  V1y

v  4.927

 


 

G1x  4.133


G1y  7.617

θ  atan2 G1x G1y

θ  118.485  deg

g  7.

U1x  0.924

U1y  6.216

2  A

σ  atan2 U1x U1y 6.

S 1y  2.298

 

A  cos γ  1

U1x 

S 1y  s sin ψ

2

2

G1x  G1y

g  8.667


θ2i  θ  θ

θ2i  195.092  deg



θ2f  θ2i  β 8.

9.

θ2f  165.092  deg


δp  ϕ  θ

rp  2.000

δp  12.020 deg


ρ  0.000  deg

2

2

P1x  P1y

R1  0.000

 

O2x  1.281

 

O2y  0.896

 

O4x  2.853

 

O4y  8.514




θrot  118.485  deg




w  1.949

θ  76.607 deg

Link 3:

v  4.927

θ  137.980  deg

Link 4:

u  6.284

σ  81.540 deg

Link 1:

g  8.667

θ  118.485  deg

Coupler:

rp  2.000

δp  12.020 deg

Crank angles:

θ2i  195.092  deg θ2f  165.092  deg




O4

Y

1.903

U2 G1

U1

B2

S2

39.0° B1

V2 P2

V1 O2 S1

Z2 1.347

W1

W2

P1

X Z1 A1

A2




Design a linkage to carry the body in Figure P5-2 through the two positions P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown. Hint: First try a rough graphical solution to create realistic values for free choices.

Given:


P2y  1.347

P3x  1.389

P3y  1.830


θP2  62.0 deg Solution:

θP3  39.0 deg


1.


2.

Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1.

 P2x     P2y 

R1 

R2 

 P3x     P3y 

 P21x     R2  R1  P21y 

P21x  0.514 P21y  0.483

p 21  3.

4.

2

2

P21x  P21y


α  23.000 deg

δ  atan2 P21x P21y

δ  136.781  deg

From a graphical solution (see figure next page), determine the values necessary for input to equations 5.8. z  3

5.

p 21  0.705

β  45.0 deg

ϕ  100  deg


Z1x  0.521


 

A  0.293

D  sin α

 

B  0.707

E  p 21  cos δ

A  cos β  1 B  sin β

 

W1x 

W1y  w 

D  0.391

 

C  0.079 F  p 21  sin δ   A   C Z1x  D Z1y  E  B  C Z1y  D Z1x  F 

C  cos α  1

2  A A   C Z1y  D Z1x  F   B  C Z1x  D Z1y  E 2  A 2

2

W1x  W1y


Z1y  2.954

E  0.514 F  0.483 W1x  1.476

W1y  1.807 w  2.333 θ  50.759 deg



Y

B3 2.000

O4

20.000°

P3

B2

P2

20.0°

3.000 X

P1

100.0°

23.0° A3 45.000° A2

O2

5.

From the graphical solution (see figure above), determine the values necessary for input to equations 5.12. s  2.000

6.

γ  20.0 deg

ψ  20.0 deg


S 1x  1.879

 

A  0.060

D  sin α

 

B  0.342

E  p 21  cos δ

 

C  0.079

F  p 21  sin δ

A  cos γ  1 B  sin γ

C  cos α  1

U1x 

U1y 

u 

S 1y  s sin ψ

 

A   C S 1x  D S 1y  E  B  C S 1y  D S 1x  F  2  A A   C S 1y  D S 1x  F   B  C S 1x  D S 1y  E 2  A 2

2

U1x  U1y

σ  atan2 U1x U1y 7.

S 1y  0.684 D  0.391

 

E  0.514

 

F  0.483

U1x  3.346

U1y  0.306

u  3.360 σ  5.216  deg



V1x  2.400



V1y  z sin ϕ  s sin ψ θ  atan2 V1x V1y v  Link 1:

2

θ  123.413  deg

2

V1x  V1y

v  4.359

 


 

9.

G1x  4.271


G1y  5.751

θ  atan2 G1x G1y

θ  126.601  deg

g  8.

V1y  3.638

2

2

G1x  G1y

g  7.163


θ2i  θ  θ

θ2i  75.842 deg

θ2f  θ2i  β

θ2f  30.842 deg


δp  ϕ  θ

rp  3.000

δp  23.413 deg

10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  atan2 P2x P2y R1 

2

ρ  35.292 deg

2

P2x  P2y

R1  2.331

 

O2x  0.948

 

O2y  3.414

 

O4x  3.323

 

O4y  2.337




θrot  126.601  deg



Condition( g u v w)  "non-Grashof"




w  2.333

θ  50.759 deg

Link 3:

v  4.359

θ  123.413  deg

Link 4:

u  3.360

σ  5.216  deg

Link 1:

g  7.163

θ  126.601  deg

Coupler:

rp  3.000

δp  23.413 deg

Crank angles:

θ2i  75.842 deg θ2f  30.842 deg




Design a linkage to carry the body in Figure P5-2 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown.

Given:


P1y  0.0

P2x  1.903

P3x  1.389

P3y  1.830

P2y  1.347


θP1  101  deg

θP2  62.0 deg

θP3  39.0 deg


β  75.0 deg

Free choices for the US dyad : γ  0.0 deg Solution: 1.

2.

3.

γ  30.0 deg



2

p 21  2.331

δ  atan2 P2x P2y

δ  35.292 deg

2

2

p 31  2.297

δ  atan2 P3x P3y

δ  52.801 deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  39.000 deg

α  θP3  θP1

α  62.000 deg


 

B  sin β

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1 D  sin α G  sin β

 

L  p 31  cos δ

 A F AA   B G 

B C D 

 G H K  A D C   F K H 

The components of the W and Z vectors are:

 

 

C  cos α  1

 

 

 

M  p 21  sin δ

 E  L CC    M  N   

 

F  cos β  1

 

K  sin α

 

N  p 31  sin δ

 W1x  W   1y   AA  1 CC  Z1x     Z1y 



W1x  3.110

W1y  1.061

Z1x  0.297

Z1y  3.201


θ  18.843 deg


ϕ  84.698 deg

 W1x2  W1y2 , w  3.286  


 Z1x2  Z1y2 , z  3.215  



 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

F'  cos γ  1

 

M  p 21  sin δ

B' C D 

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

N  p 31  sin δ

 U1x  U   1y   AA  1 CC  S1x     S1y 

The components of the U and S vectors are: U1x  1.658

U1y  3.361

S 1x  2.853

S 1y  2.013


σ  63.740 deg

ψ  atan2 S 1x S 1y

ψ  144.792  deg


 U1x2  U1y2 , u  3.748  


 S 1x2  S 1y2 , s  3.492  


V1x  Z1x  S 1x

V1x  3.150

V1y  Z1y  S 1y

V1y  1.188

θ  atan2 V1x V1y

θ  20.657 deg

v  Link 1:

2

2

V1x  V1y

v  3.367

G1x  W1x  V1x  U1x

G1x  4.602

G1y  W1y  V1y  U1y

G1y  3.234



θ  atan2 G1x G1y

g  6.

7.

8.

9.

2

θ  35.099 deg

2

G1x  G1y

g  5.625


θ2i  θ  θ

θ2i  16.256 deg

θ2f  θ2i  β

θ2f  91.256 deg


δp  ϕ  θ

rp  3.215

δp  64.041 deg


O2x  3.407

O2y  z sin ϕ  w sin θ

O2y  2.140

O4x  s cos ψ  u  cos σ

O4x  1.195

O4y  s sin ψ  u  sin σ

O4y  5.374



θrot  35.099 deg




w  3.286

θ  18.843 deg

Link 3:

v  3.367

θ  20.657 deg

Link 4:

u  3.748

σ  63.740 deg

Link 1:

g  5.625

θ  35.099 deg

Coupler:

rp  3.215

δp  64.041 deg



Crank angles:

θ2i  16.256 deg

θ2f  91.256 deg


Y

P3 P2

Z3 A3 Z2 P1

V3 W3

X

S3

A2

S2

S1

75.0°

W2

B3

Z1

V2

B1 , B2

40.0° O2

W1 V1 30.0°

A1 G1

U1 U2

U3

O4




Given:

Solution: 1.

2.

Design a linkage to carry the body in Figure P5-2 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis and design it for the fixed pivots shown. P21x  1.903

P21y  1.347

P31x  1.389

P31y  1.830

O2x  0.884

O2y  1.251

O4x  3.062

O4y  1.251

Body angles:

θP1  101  deg

θP2  62 deg

θP3  39 deg



α  39.000 deg

α  θP3  θP1

α  62.000 deg


3.

4.

R1x  0.884

R1y  O2y

R2x  R1x  P21x

R2x  2.787

R2y  R1y  P21y

R2y  2.598

R3x  R1x  P31x

R3x  2.273

R3y  R1y  P31y

R3y  3.081

2

2

R1  1.532

2

2

R2  3.810

2

2

R3  3.829

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  1.251


ζ  54.754 deg

ζ  atan2 R2x R2y

ζ  42.990 deg

ζ  atan2 R3x R3y

ζ  53.582 deg






















 

C3  0.753

 

C4  3.274

 

C5  1.313






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ

C1  0.103 C2  2.205







 

C6  R1 sin α  ζ  R2 sin ζ 2

C6  2.182

2

A1  C3  C4

A1  11.288

A2  C3 C6  C4 C5

A2  2.654

A3  C4 C6  C3 C5

A3  8.134

A4  C2 C3  C1 C4

A4  1.324

A5  C4 C5  C3 C6

A5  2.654

A6  C1 C3  C2 C4

A6  7.297

K1  A2  A4  A3  A6

K1  55.842

K2  A3  A4  A5  A6

K2  30.136

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

K3  0.392

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  118.708  deg

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  62.000 deg


β  β

 A5  sin β  A3  cos β  A6   A1  

β  59.564 deg

 A3  sin β  A2  cos β  A4   A1  

β  59.564 deg



β  β



R1x  3.062

R1y  O4y

R2x  R1x  P21x

R2x  1.159

R2y  R1y  P21y

R2y  2.598

R3x  R1x  P31x

R3x  1.673

R3y  R1y  P31y

R3y  3.081

2

2

R1  3.308

2

2

R2  2.845

R1 

R1x  R1y

R2 

R2x  R2y

R1y  1.251


R3  6.

7.


2

2

R3x  R3y

R3  3.506


ζ  157.777  deg

ζ  atan2 R2x R2y

ζ  114.042  deg

ζ  atan2 R3x R3y

ζ  118.502  deg






















 

C3  1.340

 

C4  0.210

 

C5  0.433

 

C6  0.301






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C1  1.111 C2  1.204

A1  C3  C4

A1  1.840

A2  C3 C6  C4 C5

A2  0.495

A3  C4 C6  C3 C5

A3  0.517

A4  C2 C3  C1 C4

A4  1.847

A5  C4 C5  C3 C6

A5  0.495

A6  C1 C3  C2 C4

A6  1.236

K1  A2  A4  A3  A6

K1  1.553

K2  A3  A4  A5  A6

K2  0.344

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

2

K3  1.033

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  62.000 deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  36.991 deg


γ  



 A5  sin γ  A3  cos γ  A6   A1  

  73.415 deg

 A3  sin γ  A2  cos γ  A4   A1  

  73.415 deg

  acos

  asin


γ  


p 21 

2

P21x  P21y

p 21  2.331

δ  atan2 P21x P21y 2

p 31 

δ  35.292 deg

2

P31x  P31y

p 31  2.297

δ  atan2 P31x P31y 9.

δ  52.801 deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 

M  p 21  sin δ

B C D 

 A F AA   B G 

 

C  cos α  1

 E  L CC    M  N   

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x   W1y   AA  1 CC  Z1x   Z1y   

10. The components of the W and Z vectors are: W1x  1.262 w 


W1y  1.109 2

Z1x  0.378

2

W1x  W1y

Z1y  2.360

w  1.680


 

A'  cos γ  1

 

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

H  cos α  1

D  sin α

G'  sin γ

 

L  p 31  cos δ

 

 

 

M  p 21  sin δ

 

F'  cos γ  1

 

K  sin α

 

N  p 31  sin δ


 A' F' AA    B'  G' 


B' C D 

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   



U1y  0.830 2

u 

U1x  U1y

S1x  2.736

2

S1y  0.421

u  0.892


V1x  2.359

V1y  Z1y  S1y

V1y  1.939 v 


2

2

V1x  V1y

v  3.054

G1x  W1x  V1x  U1x

G1x  3.946

G1y  W1y  V1y  U1y

G1y  1.110  10

g 


2

G1x  G1y

2

 15

g  3.946


O2x  0.884

O2y  Z1y  W1y

O2y  1.251

O4x  S1x  U1x

O4x  3.062

O4y  S1y  U1y

O4y  1.251


2

z  2.390

2

2

s  2.769


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y


ψ  171.262  deg

ϕ  atan2( Z1x Z1y )

ϕ  99.095 deg

rP  z

θ  atan2 z cos ϕ  s cos ψ z sin ϕ  s sin ψ  θ  39.430 deg



δp  ϕ  θ

δp  59.666 deg


g  3.946

Link 2:

w  1.680

Link 3:

v  3.054

Link 4:

u  0.892

Coupler point:

rP  2.390

δp  59.666 deg





Design a linkage to carry the body in Figure P5-3 through the two positions P1 and P2 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown.

Given:


P1y  0.0

P2x  0.907

P2y  0.0


θP1  111.8  deg

θP2  191.1  deg


β  44.0 deg

ϕ  50.0 deg

γ  55.0 deg

ψ  20.0 deg



1.


2.


 P1x     P1y 

R2 

 P2x     P2y 

 P21x     R2  R1  P21y 

P21x  0.907 P21y  0.000

p 21  3.

4.

2

2

P21x  P21y

p 21  0.907


α  79.300 deg

δ  atan2 P21x P21y

δ  180.000  deg


  B  sin β

A  cos β  1

 

C  cos α  1 W1x 

W1y 

Z1x  0.964


 

A  0.281

D  sin α

B  0.695

E  p 21  cos δ

C  0.814

F  p 21  sin δ

A   C Z1x  D Z1y  E  B  C Z1y  D Z1x  F  2  A A   C Z1y  D Z1x  F   B  C Z1x  D Z1y  E 2  A

Z1y  1.149 D  0.983

 

E  0.907

 

F  0.000 W1x  1.705

W1y  2.490



2

w 

2

W1x  W1y

w  3.018

θ  atan2 W1x W1y 5.

θ  124.405  deg


S 1x  2.349

S 1y  s sin ψ

 

A  0.426

D  sin α

 

B  0.819

E  p 21  cos δ

 

C  0.814

F  p 21  sin δ

A  cos γ  1 B  sin γ

C  cos α  1

 

D  0.983

 

E  0.907

 

F  0.000


U1x 


U1y 

2

2

U1x  U1y

u  2.654

σ  atan2 U1x U1y 6.

σ  76.373 deg



V1x  1.385


V1y  2.004

θ  atan2 V1x V1y

θ  124.648  deg

v  Link 1:

2

2

V1x  V1y

v  2.436

 


 

8.

G1x  3.715


G1y  2.094

θ  atan2 G1x G1y

θ  150.596  deg

g  7.

U1x  0.625

U1y  2.579

2  A

u 

S 1y  0.855

2

2

G1x  G1y

g  4.265


θ2i  θ  θ

θ2i  275.001  deg

θ2f  θ2i  β

θ2f  319.001  deg


δp  ϕ  θ

rp  1.500

δp  74.648 deg


9.



ρ  0.000  deg

2

2

P1x  P1y

R1  0.000

 

O2x  0.741

 

O2y  1.341

 

O4x  2.975

 

O4y  3.434




θrot  150.596  deg




w  3.018

θ  124.405  deg

Link 3:

v  2.436

θ  124.648  deg

Link 4:

u  2.654

σ  76.373 deg

Link 1:

g  4.265

θ  150.596  deg

Coupler:

rp  1.500

δp  74.648 deg

Crank angles:

θ2i  275.001  deg

θ2f  319.001  deg




Y

A1 Z1 P2

V1

P1

Z2 A2

X S1

44.0°

B1

W2 V2

S2

55.0° U1

G1

B2 U2

O4

W1 O2




Design a linkage to carry the body in Figure P5-3 through the two positions P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown.

Given:

Coordinates of the points P2 and P3 with respect to P1: P2x  0.907

P2y  0.0

P3x  1.447

P3y  0.0


θP2  191.1  deg

θP3  237.4  deg


β  66.0 deg

ϕ  60.0 deg

γ  44.0 deg

ψ  30.0 deg



1.


2.


 P2x     P2y 

R2 

 P3x     P3y 

 P21x     R2  R1  P21y 

P21x  0.540 P21y  0.000

p 21  3.

4.

2

2

P21x  P21y

p 21  0.540


α  46.300 deg

δ  atan2 P21x P21y

δ  180.000  deg


A  0.593

D  sin α

 

B  0.914

E  p 21  cos δ

 

C  0.309

F  p 21  sin δ

B  sin β

C  cos α  1

W1y 


 

A  cos β  1

W1x 

Z1x  1.000

 

A   C Z1x  D Z1y  E  B  C Z1y  D Z1x  F  2  A A   C Z1y  D Z1x  F   B  C Z1x  D Z1y  E 2  A

Z1y  1.732 D  0.723

 

E  0.540

 

F  0.000 W1x  0.227

W1y  1.771



2

w 

2

W1x  W1y

w  1.786

θ  atan2 W1x W1y 5.

θ  97.314 deg


S 1x  2.598

S 1y  s sin ψ

 

A  0.281

D  sin α

 

B  0.695

E  p 21  cos δ

 

C  0.309

F  p 21  sin δ

A  cos γ  1 B  sin γ

C  cos α  1

 

D  0.723

 

E  0.540

 

F  0.000


U1x 


U1y 

2

2

U1x  U1y

u  2.608

σ  atan2 U1x U1y 6.

σ  65.609 deg



V1x  1.598


V1y  3.232

θ  atan2 V1x V1y

θ  116.310  deg

v  Link 1:

2

2

V1x  V1y

v  3.606

 


 

8.

G1x  2.902


G1y  3.836

θ  atan2 G1x G1y

θ  127.111  deg

g  7.

U1x  1.077

U1y  2.375

2  A

u 

S 1y  1.500

2

2

G1x  G1y

g  4.810


θ2i  θ  θ

θ2i  224.425  deg

θ2f  θ2i  β

θ2f  290.425  deg


δp  ϕ  θ



rp  2.000 9.

δp  56.310 deg

Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  atan2 P2x P2y R1 

2

ρ  180.000  deg

2

P2x  P2y

R1  0.907

 

O2x  1.680

 

O2y  0.039

 

O4x  4.582

 

O4y  3.875




θrot  127.111  deg




w  1.786

θ  97.314 deg

Link 3:

v  3.606

θ  116.310  deg

Link 4:

u  2.608

σ  65.609 deg

Link 1:

g  4.810

θ  127.111  deg

Coupler:

rp  2.000

δp  56.310 deg

Crank angles:

θ2i  224.425  deg

θ2f  290.425  deg




Y

A2 66.0° V1

Z1

A3

W1 Z2 W2

P3

O2

P2

S1

B2

U1

S2

V2

G1 44.0° B1 U2

O4

P1 X




Design a linkage to carry the body in Figure P5-3 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown.

Given:


P1y  0.0

P3x  1.447

P3y  0.0

P2x  0.907

P2y  0.0


θP1  111.8  deg

θP2  191.1  deg

θP3  237.4  deg


β  80.0 deg

Free choices for the US dyad : γ  20.0 deg Solution: 1.

2.

3.

γ  50.0 deg



2

p 21  0.907

δ  atan2 P2x P2y

δ  180.000  deg

2

2

p 31  1.447

δ  atan2 P3x P3y

δ  180.000  deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  79.300 deg

α  θP3  θP1

α  125.600  deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

 

K  sin α

 

L  p 31  cos δ

 

M  p 21  sin δ

B C D 

 E  L CC    M  N   



G H K 

  F K H  A

 

F  cos β  1

 

G  sin β

 A F AA   B G 

 

C  cos α  1

D C

N  p 31  sin δ

 W1x  W   1y   AA  1 CC  Z1x     Z1y 

The components of the W and Z vectors are: W1x  1.696

W1y  0.038

Z1x  0.396

Z1y  0.872




ϕ  114.406  deg

 W1x2  W1y2 , w  1.696  


 Z1x2  Z1y2 , z  0.958  



θ  1.280  deg


 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

M  p 21  sin δ

B' C D 

 E  L CC    M  N   



G' H K 

  F' K H 

A'

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

F'  cos γ  1

D C

N  p 31  sin δ

 U1x  U   1y   AA  1 CC  S1x     S1y 


U1y  0.409

S 1x  0.048

S 1y  0.650


σ  15.412 deg

ψ  atan2 S 1x S 1y

ψ  85.790 deg


 U1x2  U1y2 , u  1.538  


 S 1x2  S 1y2 , s  0.652  


V1x  Z1x  S 1x

V1x  0.444

V1y  Z1y  S 1y

V1y  0.222

θ  atan2 V1x V1y

θ  153.426  deg

v  Link 1:

2

2

V1x  V1y

v  0.496

G1x  W1x  V1x  U1x

G1x  0.230

G1y  W1y  V1y  U1y

G1y  0.225

θ  atan2 G1x G1y

θ  135.700  deg


g  6.

7.

8.

9.

2


2

G1x  G1y

g  0.322


θ2i  θ  θ

θ2i  134.419  deg

θ2f  θ2i  β

θ2f  214.419  deg


δp  ϕ  θ

rp  0.958

δp  39.020 deg


O2x  1.300

O2y  z sin ϕ  w sin θ

O2y  0.834

O4x  s cos ψ  u  cos σ

O4x  1.530

O4y  s sin ψ  u  sin σ

O4y  1.059



θrot  135.700  deg




w  1.696

θ  1.280  deg

Link 3:

v  0.496

θ  153.426  deg

Link 4:

u  1.538

σ  15.412 deg

Link 1:

g  0.322

θ  135.700  deg

Coupler:

rp  0.958

δp  39.020 deg

Crank angles:

θ2i  134.419  deg θ2f  214.419  deg




Y A3 B3

A2

P3 P2 W2 O2 O4

U1

B2 U2

B1 W1

X

P1 S1 A1



PROBLEM 5-19 Statement: Given:

Solution: 1.

2.

Design a linkage to carry the body in Figure P5-3 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis and design it for the fixed pivots shown. P21x  0.907

P21y  0.0

P31x  1.447

P31y  0.0

O2x  1.788

O2y  1.994

O4x  0.212

O4y  1.994

Body angles:

θP1  111.8  deg

θP2  191.1  deg

θP3  237.4  deg



α  79.300 deg

α  θP3  θP1

α  125.600  deg


3.

4.

R1x  1.788

R1y  O2y

R2x  R1x  P21x

R2x  0.881

R2y  R1y  P21y

R2y  1.994

R3x  R1x  P31x

R3x  0.341

R3y  R1y  P31y

R3y  1.994

2

2

R1  2.678

2

2

R2  2.180

2

2

R3  2.023

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  1.994


ζ  48.118 deg

ζ  atan2 R2x R2y

ζ  66.163 deg

ζ  atan2 R3x R3y

ζ  80.296 deg






















 

C3  3.003

 

C4  1.701

 

C5  2.508






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ

C1  0.238 C2  1.150







 

C6  R1 sin α  ζ  R2 sin ζ 2

C6  0.133

2

A1  C3  C4

A1  11.912

A2  C3 C6  C4 C5

A2  4.666

A3  C4 C6  C3 C5

A3  7.307

A4  C2 C3  C1 C4

A4  3.048

A5  C4 C5  C3 C6

A5  4.666

A6  C1 C3  C2 C4

A6  2.671

K1  A2  A4  A3  A6

K1  5.293

K2  A3  A4  A5  A6

K2  34.731

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

K3  25.158

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  125.600  deg

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  37.070 deg


β  β

 A5  sin β  A3  cos β  A6   A1  

β  18.241 deg

 A3  sin β  A2  cos β  A4   A1  

β  18.241 deg



β  β



R1x  0.212

R1y  O4y

R2x  R1x  P21x

R2x  1.119

R2y  R1y  P21y

R2y  1.994

R3x  R1x  P31x

R3x  1.659

R3y  R1y  P31y

R3y  1.994

R1 

2

2

R1x  R1y

R1  2.005

R1y  1.994


6.

7.


2

2

R2  2.287

2

2

R3  2.594

R2 

R2x  R2y

R3 

R3x  R3y


ζ  96.069 deg

ζ  atan2 R2x R2y

ζ  119.300  deg

ζ  atan2 R3x R3y

ζ  129.760  deg






















 

C3  0.161

 

C4  3.327

 

C5  0.880

 

C6  1.832






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C1  1.297 C2  0.811

A1  C3  C4

A1  11.096

A2  C3 C6  C4 C5

A2  3.222

A3  C4 C6  C3 C5

A3  5.954

A4  C2 C3  C1 C4

A4  4.186

A5  C4 C5  C3 C6

A5  3.222

A6  C1 C3  C2 C4

A6  2.906

K1  A2  A4  A3  A6

K1  3.816

K2  A3  A4  A5  A6

K2  34.288

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

2

K3  25.658

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  125.600  deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  41.699 deg




γ  

 A5  sin γ  A3  cos γ  A6   A1  

  31.159 deg

 A3  sin γ  A2  cos γ  A4   A1  

  31.159 deg

  acos

  asin

γ  

Since both values are the same , 8.


p 21 

2

P21x  P21y

p 21  0.907

δ  atan2 P21x P21y 2

p 31 

δ  180.000  deg

2

P31x  P31y

p 31  1.447

δ  atan2 P31x P31y 9.

δ  180.000  deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 A F AA   B G 

 

C  cos α  1

 

M  p 21  sin δ

 E  L CC    M  N   

B C D 

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x   W1y   AA  1 CC  Z1x   Z1y   

10. The components of the W and Z vectors are: W1x  1.943


w 

W1y  1.529 2

Z1x  0.155

2

W1x  W1y

Z1y  0.465

w  2.472


 

A'  cos γ  1

 

D  sin α

 

B'  sin γ

 

E  p 21  cos δ

 

C  cos α  1

 

F'  cos γ  1



 

 

G'  sin γ

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

H  cos α  1

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   

13. The components of the W and Z vectors are: U1x  0.395 14. The length of link 4 is:

U1y  2.460 2

u 

U1x  U1y

S1x  0.183

2

S1y  0.466

u  2.491


V1x  0.338

V1y  Z1y  S1y

V1y  0.931 v 


2

2

V1x  V1y

v  0.991

G1x  W1x  V1x  U1x

G1x  2.000

G1y  W1y  V1y  U1y

G1y  0.000

g 


2

G1x  G1y

2

g  2.000


O2x  1.788

O2y  Z1y  W1y

O2y  1.994

O4x  S1x  U1x

O4x  0.212

O4y  S1y  U1y

O4y  1.994


2

z  0.491

2

2

s  0.501


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y


ψ  68.558 deg

ϕ  atan2( Z1x Z1y )

ϕ  108.446  deg

rP  z

θ  atan2 z cos ϕ  s cos ψ z sin ϕ  s sin ψ  θ  109.959  deg



δp  ϕ  θ

δp  1.513  deg


g  2.000

Link 2:

w  2.472

Link 3:

v  0.991

Link 4:

u  2.491

Coupler point:

rP  0.491

δp  1.513  deg





Write a program to generate and plot the circle-point and center-point circles for Problem 5-19 using an equation solver or any program language.

Given: P21x  0.907

P21y  0.0

P31x  1.447

P31y  0.0

O2x  1.788

O2y  1.994

O4x  0.212

O4y  1.994

Body angles:

θP1  111.8  deg

θP2  191.1  deg

θP3  237.4  deg

Assumptions: Let the position 1 to position 2 rotation angles be: β  18.241 deg and γ  31.159 deg Let the position 1 to position 2 coupler rotation angle be: α  79.3 deg Solution: 1.



2

2

P21x  P21y

p 21  0.907

δ  atan2 P21x P21y p 31 

2

δ  180.000  deg

2

P31x  P31y

p 31  1.447

δ  atan2 P31x P31y 2.

δ  180.000  deg

Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α  125.6  deg

β  0  deg 1  deg  360  deg

 

B  sin β

 

 

E  p 21  cos δ

A  cos β  1 D  sin α

 

 

 

C  cos α  1

 

F β  cos β  1

 

K α  sin α

 

N  p 31  sin δ

 

H α  cos α  1

 

M  p 21  sin δ

G β  sin β L  p 31  cos δ

 

 

 

   

B C D   A  F β G β H α K α           AA  α β    B A D C  G β F β K α H α           

 E  L CC    M  N   

 1 CC















W1y α β  DD α β 2









Z1y α β  DD α β 4

DD α β  AA α β

W1x α β  DD α β 1 Z1x α β  DD α β 3


















3.


Check this against the solutions in Problem 5-19:





W1y α 37.070 deg  1.529





Z1y α 37.070 deg  0.465

W1x α 37.070 deg  1.943 Z1x α 37.070 deg  0.155









These are the same as the values calculated in Problem 5-19. 4.

Form the vector N, whose tip describes the center-point circle for the WZ dyad.

























Nx α β  W1x α β  Z1x α β Ny α β  W1y α β  Z1y α β 5.

Plot the center-point circle for the WZ dyad. Center-Point Circle for WZ Dyad 0

1





Ny α β   2

3

4 2

1



0

1



2

Nx α β 

4.

Form the vector Z, whose tip describes the center-point circle for the WZ dyad. β  37.070 deg





α  0  deg 1  deg  360  deg





Zx α β  Z1x α β

5.









Zy α β  Z1y α β

Plot the circle-point circle for the WZ dyad (see next page).



Circle-Point Circle for WZ Dyad  0.1

 0.2





Zy α β   0.3

 0.4

 0.5  0.2

 0.1

0





0.1

0.2

0.3

Zx α β 

6.

Evaluate terms in the US coefficient matrix and constant vector from equations (5.31) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α  125.6  deg

γ  0  deg 1  deg  360  deg

 

B  sin γ

 

 

E  p 21  cos δ

A  cos γ  1 D  sin α

 

 

 

C  cos α  1

 

F γ  cos γ  1

 


 

N  p 31  sin δ

 

H α  cos α  1

 

M  p 21  sin δ

G γ  sin γ L  p 31  cos δ

 

 

 

   

B C D   A  F γ G γ H α K α           AA  α γ    B A D C  G γ F γ K α H α           

 E  L CC    M  N   

 1 CC















U1y α γ  DD α γ 2









S1y α γ  DD α γ 4

DD α γ  AA α γ

U1x α γ  DD α γ 1 S1x α γ  DD α γ 3 7.






















U1x α 41.699 deg  0.395





U1y α 41.699 deg  2.460









S1x α 41.699 deg  0.183



S1y α 41.699 deg  0.466


Form the vector M, whose tip describes the center-point circle for the US dyad.

























Mx α γ  U1x α γ  S1x α γ My α γ  U1y α γ  S1y α γ 9.

Plot the center-point circle for the US dyad. Center-Point Circle for US Dyad 0

 0.5

1



My α γ

  1.5

2

 2.5  1.5

1

 0.5



Mx α γ

0



0.5

10. Form the vector S, whose tip describes the center-point circle for the US dyad. γ  41.699 deg





α  0  deg 1  deg  360  deg





Sx α γ  S1x α γ

11. Plot the circle-point circle for the WZ dyad (see next page).









Sy α γ  S1y α γ



Circle-Point Circle for the US Dyad 1.5

1



Sy α γ



0.5

0  0.5

0





Sx α γ

0.5

1




Design a fourbar linkage to carry the box in Figure P5-4 from position 1 to 2 without regard for the fixed pivots shown. Use points A and B for your attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base.

Given:


P1y  0.0

P2x  184.0

P2y  17.0


θP1  90.0 deg

θP2  45.0 deg

Coordinates of the points A1 and B1 with respect to P1: A1x  17.0

A1y  43.0

B1x  69.0

B1y  43.0

Free choice for the WZ dyad : β  44.0 deg Free choice for the US dyad : γ  55.0 deg Solution:


1.


2.


 P1x     P1y 

R2 

 P2x     P2y 

 P21x     R2  R1  P21y 

P21x  184.000 P21y  17.000

p 21  3.

4.

2

2

P21x  P21y


α  45.000 deg

δ  atan2 P21x P21y

δ  5.279  deg

Using Figure P5-4, the given data, and the law of cosines, determine z, s, , and . z 

P1x  A1x2   P1y  A1y 2

z  46.239

s 

P1x  B1x 2  P1y  B1y 2

s  81.302

v 

 A1x  B1x 2  A1y  B1y2

v  52.000

 v2  z2  s2    2  v z 

ϕ  acos

 v2  s2  z2    2  v s 

ψ  π  acos 5.

p 21  184.784


ϕ  111.571  deg

ψ  148.069  deg



Z1x  z cos ϕ

Z1x  17.000 A  0.281

D  sin α

 

B  0.695

E  p 21  cos δ

 

E  184.000

 

C  0.293

F  p 21  sin δ

 

F  17.000

B  sin β

C  cos α  1

W1y  w 

 

D  0.707


W1x  53.979


W1y  192.131

2  A 2

2

W1x  W1y

w  199.570

θ  atan2 W1x W1y 6.

θ  105.693  deg


S 1x  69.000

D  sin α

 

B  0.819

E  p 21  cos δ

 

E  184.000

 

C  0.293

F  p 21  sin δ

 

F  17.000

C  cos α  1

U1y  u 

S 1y  43.000

A  0.426

B  sin γ

U1x 

S 1y  s sin ψ

 

A  cos γ  1

 

D  0.707


2

U1x  15.598

U1y  154.713

2  A 2

U1x  U1y

u  155.497

σ  atan2 U1x U1y 7.

Z1y  43.000

 

A  cos β  1

W1x 


σ  95.757 deg



V1x  52.000


V1y  0.000

θ  atan2 V1x V1y

θ  0.000  deg

v  Link 1:

2

2

V1x  V1y

v  52.000

 


G1x  13.619



 


G1y  37.418

θ  atan2 G1x G1y

θ  70.000 deg

g  8.

9.

2

2

G1x  G1y

g  39.819


θ2i  θ  θ

θ2i  35.692 deg

θ2f  θ2i  β

θ2f  8.308  deg


δp  ϕ  θ

rp  46.239

δp  111.571  deg

10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  0  deg R1 

ρ  0.000  deg

2

2

P1x  P1y

R1  0.000

 

O2x  70.979

 

O2y  235.131

 

O4x  84.598

 

O4y  197.713


These fixed pivot points fall on the base and are, therefore, acceptable. 11. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.


θrot  70.000 deg







w  199.570

θ  105.693  deg

Link 3:

v  52.000

θ  0.000  deg

Link 4:

u  155.497

σ  95.757 deg

Link 1:

g  39.819

θ  70.000 deg

Coupler:

rp  46.239

δp  111.571  deg

Crank angles:

θ2i  35.692 deg

θ2f  8.308  deg


Y

P1

X Z1 A1

P2

S1 B1

Z2

V1 A2

V2 55.0°

U1 44.0° W1

W2

O4

O2

U2

S2

B2





Given:


P1y  0.0

P3x  211.0

P3y  180.0


θP1  90.0 deg

θP3  0.0 deg


A1y  43.0

B1x  69.0

B1y  43.0



1.


2.


p 21  3.

4.

 P1x     P1y 

R2 

2

 P3x     P3y 

 P21x     R2  R1  P21y 

2

P21x  P21y

P21y  180.000 p 21  277.346


α  90.000 deg

δ  atan2 P21x P21y

δ  40.467 deg


P1x  A1x2   P1y  A1y 2

z  46.239

s 

P1x  B1x 2  P1y  B1y 2

s  81.302

v 

 A1x  B1x 2  A1y  B1y2

v  52.000

 v2  z2  s2    2  v z 

ϕ  acos

ϕ  111.571  deg

 v2  s2  z2    2  v s 

ψ  π  acos 5.

P21x  211.000

ψ  148.069  deg


Z1x  17.000


Z1y  43.000



 

A  0.658

D  sin α

 

B  0.940

E  p 21  cos δ

 

C  1.000

F  p 21  sin δ

A  cos β  1 B  sin β

C  cos α  1 W1x 

W1y  w 

 

D  1.000

 

E  211.000

 

F  180.000


W1x  34.467


W1y  184.825

2  A 2

2

W1x  W1y

w  188.012

θ  atan2 W1x W1y 6.

θ  79.436 deg


S 1x  69.000

 

A  1.087

D  sin α

 

B  0.996

E  p 21  cos δ

 

C  1.000

F  p 21  sin δ

A  cos γ  1 B  sin γ

C  cos α  1 U1x 

U1y  u 

S 1y  s sin ψ

 

D  1.000

 

E  211.000

 

F  180.000


2

U1x  U1y

u  154.999

σ  atan2 U1x U1y 7.

U1x  44.882

U1y  148.358

2  A 2

S 1y  43.000

σ  73.168 deg



V1x  52.000


V1y  0.000

θ  atan2 V1x V1y

θ  0.000  deg

v  Link 1:

2

2

V1x  V1y

v  52.000


G1x  41.585


G1y  36.467

θ  atan2 G1x G1y

θ  41.248 deg

 

 


g 

8.

9.

2


G1x  G1y

g  55.310


θ2i  θ  θ

θ2i  38.188 deg

θ2f  θ2i  β

θ2f  31.812 deg


δp  ϕ  θ

rp  46.239

δp  111.571  deg

10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  0  deg R1 

ρ  0.000  deg

2

2

P1x  P1y

R1  0.000

 

O2x  17.467

 

O2y  227.825

 

O4x  24.118

 

O4y  191.358




θrot  41.248 deg




w  188.012

θ  79.436 deg

Link 3:

v  52.000

θ  0.000  deg

Link 4:

u  154.999

σ  73.168 deg



Link 1:

g  55.310

θ  41.248 deg

Coupler:

rp  46.239

δp  111.571  deg

Crank angles:

θ2i  38.188 deg θ2f  31.812 deg 14. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design. Y

P1

X Z1

S1 B1

A1

V1

U1 W1 70.0°

95.0° P3

O4

O2

W2

A3

U2

Z2

V2

S2 B3





Given:


P2y  17.0

P3x  211.0

P3y  180.0


θP2  45.0 deg

θP3  0.0 deg


A1y  43.0

B1x  69.0

B1y  43.0



1.


2.


p 21  3.

4.

 P2x     P2y 

R2 

2

 P3x     P3y 

 P21x     R2  R1  P21y 

2

P21x  P21y

P21y  163.000 p 21  165.221


α  45.000 deg

δ  atan2 P21x P21y

δ  80.595 deg


0.0  A1x 2  0.0  A1y 2

z  46.239

s 

0.0  B1x2   0.0  B1y 2

s  81.302

v 

 A1x  B1x 2  A1y  B1y2

v  52.000

 v2  z2  s2    45 deg  2  v z 

ϕ  acos

ϕ  66.571 deg

 v2  s2  z2    45 deg  2  v s 

ψ  π  acos 5.

P21x  27.000

ψ  103.069  deg


Z1x  18.385


Z1y  42.426



 

A  0.500

D  sin α

 

B  0.866

E  p 21  cos δ

 

C  0.293

F  p 21  sin δ

A  cos β  1 B  sin β

C  cos α  1

W1x 

W1y  w 

 

D  0.707

 

E  27.000

 

F  163.000


W1x  117.950


W1y  70.852

2  A 2

2

W1x  W1y

w  137.594

θ  atan2 W1x W1y 6.

θ  30.993 deg


S 1x  18.385

 

A  0.293

D  sin α

 

B  0.707

E  p 21  cos δ

 

C  0.293

F  p 21  sin δ

A  cos γ  1 B  sin γ

C  cos α  1 U1x 

U1y  u 

S 1y  s sin ψ

 

D  0.707

 

E  27.000

 

F  163.000


2

U1x  U1y

u  204.640

σ  atan2 U1x U1y 7.

U1x  201.643

U1y  34.896

2  A 2

S 1y  79.196

σ  9.818  deg



V1x  36.770


V1y  36.770

θ  atan2 V1x V1y v  Link 1:

2

θ  45.000 deg

2

V1x  V1y

v  52.000

 


 


G1x  46.924 G1y  0.813



θ  atan2 G1x G1y g  8.

9.

2

θ  179.007  deg

2

G1x  G1y

g  46.931


θ2i  θ  θ

θ2i  210.000  deg

θ2f  θ2i  β

θ2f  150.000  deg


δp  ϕ  θ

rp  46.239

δp  111.571  deg


2

ρ  5.279  deg

2

P2x  P2y

R1  184.784

 

O2x  47.665

 

O2y  130.278

 

O4x  0.742

 

O4y  131.092




θrot  179.007  deg




w  137.594

θ  30.993 deg

Link 3:

v  52.000

θ  45.000 deg

Link 4:

u  204.640

σ  9.818  deg



Link 1:

g  46.931

θ  179.007  deg

Coupler:

rp  46.239

δp  111.571  deg

Crank angles:

θ2i  210.000  deg θ2f  150.000  deg 14. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design. Y

P1

X P2

A2

Z2

S2

V2

B2

W1 O4

U1 O2

W2 P3 U2 A3 V2 B3

Z2 S2




Given:

Design a fourbar linkage to carry the box in Figure P5-4 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use any points on the object as attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base. Coordinates of the points P1 , P2 and P3 with respect to P1: P1x  0.0

P1y  0.0

P2x  184.0

P3x  211.0

P3y  180.0

P2y  17.0


θP1  90.0 deg

θP2  45.0 deg

θP3  0.0 deg

Free choices for the WZ dyad : β  80.0 deg

β  160.0  deg


2.

3.

γ  170.0  deg



2

p 21  184.784

δ  atan2 P2x P2y

δ  5.279  deg

2

2

p 31  277.346

δ  atan2 P3x P3y

δ  40.467 deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  45.000 deg

α  θP3  θP1

α  90.000 deg

Evaluate terms in the WZ coefficient matrix and constant vector from equations 5.26 and form the matrix and vector: A  cos β  1 B  sin β C  cos α  1

  D  sin α G  sin β L  p 31  cos δ  A F AA   B G 

 

  H  cos α  1 M  p 21  sin δ E  p 21  cos δ

B C D 

 E  L CC    M  N   



G H K 

  F K H  A

  F  cos β  1 K  sin α N  p 31  sin δ

D C

 W1x  W   1y   AA  1 CC  Z1x     Z1y 


W1y  109.176

Z1x  43.555

Z1y  29.523


θ  115.406  deg


ϕ  145.869  deg


 W1x2  W1y2 , w  120.864  



 Z1x2  Z1y2 , z  52.618  



  D  sin α

  E  p 21  cos δ

A'  cos γ  1

 

C  cos α  1

 

G'  sin γ

 

H  cos α  1

 

 

M  p 21  sin δ

B' C D 

 E  L CC    M  N   



G' H K 

  F' K H 

A'

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

  F'  cos γ  1

B'  sin γ

D C

N  p 31  sin δ

 U1x  U   1y   AA  1 CC  S1x     S1y 


U1y  94.023

S 1x  62.812

S 1y  62.282


σ  110.448  deg

ψ  atan2 S 1x S 1y

ψ  135.242  deg


 U1x2  U1y2 , u  100.345  



V1x  Z1x  S 1x

V1x  19.256

V1y  Z1y  S 1y

V1y  32.759

θ  atan2 V1x V1y

θ  59.552 deg

v  Link 1:

2

2

V1x  V1y

v  38.000

G1x  W1x  V1x  U1x

G1x  2.458

G1y  W1y  V1y  U1y

G1y  17.606

θ  atan2 G1x G1y

θ  82.052 deg

g  7.

 S 1x2  S 1y2 , s  88.456  

2

2

G1x  G1y

g  17.777


θ2i  θ  θ

θ2i  197.458  deg

θ2f  θ2i  β

θ2f  37.458 deg


8.

9.



δp  ϕ  θ

rp  52.618

δp  205.422  deg


O2x  95.410

O2y  z sin ϕ  w sin θ

O2y  138.699

O4x  s cos ψ  u  cos σ

O4x  97.868

O4y  s sin ψ  u  sin σ

O4y  156.305



θrot  82.052 deg




w  120.864

θ  115.406  deg

Link 3:

v  38.000

θ  59.552 deg

Link 4:

u  100.345

σ  110.448  deg

Link 1:

g  17.777

θ  82.052 deg

Coupler:

rp  52.618

δp  205.422  deg

Crank angles:

θ2i  197.458  deg

θ2f  37.458 deg




Y

P1

X Z1 S1

P2

A1 V1 B1

Z2 S2

W1

A2 V2 B2

W2

U1

U2

O2 W3

O4

P3

U3

S3

V3 B3

Z3 A3




Design a fourbar linkage to carry the box in Figure P5-4 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use points A and B for your attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base.

Given:

Coordinates of the points P1 , P2 and P3 with respect to P1: P1x  0.0

P1y  0.0

P2x  184.0

P3x  211.0

P3y  180.0

P2y  17.0


θP1  90.0 deg

θP2  45.0 deg

θP3  0.0 deg

Coordinates of the points A1 and B1 with respect to P1: A1x  17.0 Solution: 1.

2.

3.

A1y  43.0

B1y  43.0



2

p 21  184.784

δ  atan2 P2x P2y

δ  5.279  deg

2

2

p 31  277.346

δ  atan2 P3x P3y

δ  40.467 deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  45.000 deg

α  θP3  θP1

α  90.000 deg


P1x  A1x2   P1y  A1y 2

z  46.239

s 

P1x  B1x 2  P1y  B1y 2

s  81.302

v 

 A1x  B1x 2  A1y  B1y2

v  52.000

 v2  z2  s2    2  v z 

ϕ  acos

 v2  s2  z2    2  v s 

4.

B1x  69.0

ϕ  111.571  deg

ψ  π  acos

ψ  148.069  deg


Z1x  17.000


Z1y  43.000

S 1x  s cos ψ

S 1x  69.000

S 1y  s sin ψ

S 1y  43.000

Use equations 5.24 to solve for w, , 2, and 3. Since the points A and B are to be used as pivots, z and  are known from the calculations above.



Guess:

W1x  50

W1y  200

β  80 deg

Given

W1x cos β  1  W1y sin β  = p 21  cos δ  Z1x cos α  1  Z1y sin α

β  160  deg

       

   

 

       

   

 

       

   

 

       

   

 

W1x cos β  1  W1y sin β  = p 31  cos δ  Z1x cos α  1  Z1y sin α W1y cos β  1  W1x sin β  = p 21  sin δ  Z1y cos α  1  Z1x sin α W1y cos β  1  W1x sin β  = p 31  sin δ  Z1y cos α  1  Z1x sin α

 W1x   W1y     Find  W1x W1y β β  β   β  β  86.887 deg

β  165.399  deg

The components of the W vector are: W1x  65.636 The length of link 2 is: w  5.


W1y  86.672

θ  127.136  deg

 W1x2  W1y2 , w  108.720  

Use equations 5.28 to solve for u, , 2, and 3. Since the points A and B are to be used as pivots, s and  are known from the calculations above. Guess:

U1x  30

U1y  100

γ  80 deg

Given

U1x cos γ  1  U1y sin γ  = p 21  cos δ  S 1x cos α  1  S 1y sin α

       

   

 

       

   

 

       

   

 

       

   

 

U1x cos γ  1  U1y sin γ  = p 31  cos δ  S 1x cos α  1  S 1y sin α U1y cos γ  1  U1x sin γ  = p 21  sin δ  S 1y cos α  1  S 1x sin α U1y cos γ  1  U1x sin γ  = p 31  sin δ  S 1y cos α  1  S 1x sin α

γ  160  deg



 U1x   U1y     Find  U1x U1y γ γ  γ   γ  γ  76.700 deg

γ  161.878  deg

The components of the U vector are: U1x  33.074

U1y  110.894

The length of link 4 is: u  6.

Link 1:

V1x  52.000

V1y  Z1y  S 1y

V1y  0.000

θ  atan2 V1x V1y

θ  0.000  deg

2

2

V1x  V1y

v  52.000

G1x  W1x  V1x  U1x

G1x  19.438

G1y  W1y  V1y  U1y

G1y  24.222

θ  atan2 G1x G1y

θ  51.253 deg

g 

9.

 U1x2  U1y2 , u  115.721  

V1x  Z1x  S 1x

v 

8.

σ  106.607  deg


7.


2

2

G1x  G1y

g  31.057


θ2i  θ  θ

θ2i  178.389  deg

θ2f  θ2i  β

θ2f  12.990 deg


δp  ϕ  θ

rp  46.239

δp  111.571  deg


O2x  82.636

O2y  z sin ϕ  w sin θ

O2y  129.672

O4x  s cos ψ  u  cos σ

O4x  102.074

O4y  s sin ψ  u  sin σ

O4y  153.894

10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis



to the line O2O4.


θrot  51.253 deg

11. Determine the Grashof condition.




w  108.720

θ  127.136  deg

Link 3:

v  52.000

θ  0.000  deg

Link 4:

u  115.721

σ  106.607  deg

Link 1:

g  31.057

θ  51.253 deg

Coupler:

rp  46.239

δp  111.571  deg

Crank angles:

θ2i  178.389  deg

Y

θ2f  12.990 deg P1


X Z1 A1

P2

S1 B1 V1

A2

U1

14. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.

Z2

S2

V2

B2

W2

W1

U2

O2 O4

W3

P3 A3

U3

Z3

V3

S3 B3




Given:

Solution: 1.

2.

Design a fourbar linkage to carry the box in Figure P5-4 through the three positions shown in their numbered order using the fixed pivots shown. Determine the range of the transmission angle. P21x  184.0

P21y  17.0

P31x  211.0

P31y  180.0

O2x  86.0

O2y  132.0

O4x  104.0

O4y  155.0

Body angles:

θP1  90 deg

θP2  45 deg

θP3  0  deg



α  45.000 deg

α  θP3  θP1

α  90.000 deg


3.

4.

R1x  86.000

R1y  O2y

R2x  R1x  P21x

R2x  98.000

R2y  R1y  P21y

R2y  115.000

R3x  R1x  P31x

R3x  125.000

R3y  R1y  P31y

R3y  48.000

2

2

R1  157.544

2

2

R2  151.093

2

2

R3  133.899

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  132.000


ζ  123.085  deg

ζ  atan2 R2x R2y

ζ  49.563 deg

ζ  atan2 R3x R3y

ζ  21.007 deg


    C2  R3 sin α  ζ  R2 sin α  ζ C1  R3 cos α  ζ  R2 cos α  ζ







C3  7.000

 

C4  134.000

 

C5  65.473

 

C6  39.149

C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C2  24.329

 

C3  R1 cos α  ζ  R3 cos ζ



C1  60.553

C6  R1 sin α  ζ  R2 sin ζ



A1  C3  C4

2

A1  1.800  10

4

A2  C3 C6  C4 C5

A2  9.047  10

3

A3  C4 C6  C3 C5

A3  4.788  10

3

A4  C2 C3  C1 C4

A4  7.944  10

A5  C4 C5  C3 C6

A5  9.047  10

A6  C1 C3  C2 C4

A6  3.684  10

3

K1  A2  A4  A3  A6

K1  5.423  10

7

K2  A3  A4  A5  A6

K2  7.136  10

7

2

K3 

3 3

2

2

2

A1  A2  A3  A4  A6

2 7

K3  7.136  10

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  90.000 deg

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  164.466  deg


β  β

 A5  sin β  A3  cos β  A6   A1  

β  85.240 deg

 A3  sin β  A2  cos β  A4   A1  

β  85.240 deg



Since 2 is not in the first quadrant, 5.

β  β


R1x  104.000

R1y  O4y

R2x  R1x  P21x

R2x  80.000

R2y  R1y  P21y

R2y  138.000

R3x  R1x  P31x

R3x  107.000

R3y  R1y  P31y

R3y  25.000

2

2

R1  186.657

2

2

R2  159.512

R1 

R1x  R1y

R2 

R2x  R2y

R1y  155.000


R3  6.

7.


2

2

R3x  R3y

R3  109.882


ζ  123.860  deg

ζ  atan2 R2x R2y

ζ  59.899 deg

ζ  atan2 R3x R3y

ζ  13.151 deg






















 

C3  48.000

 

C4  129.000

 

C5  43.938

 

C6  45.141






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

C2  13.338

A1  C3  C4

A1  1.894  10

4

A2  C3 C6  C4 C5

A2  7.835  10

3

A3  C4 C6  C3 C5

A3  3.714  10

3

A4  C2 C3  C1 C4

A4  9.682  10

A5  C4 C5  C3 C6

A5  7.835  10

A6  C1 C3  C2 C4

A6  5.561  10

3

K1  A2  A4  A3  A6

K1  5.520  10

7

K2  A3  A4  A5  A6

K2  7.953  10

7

2

K3 

2

C1  80.017

3 3

2

2

2

A1  A2  A3  A4  A6

2

2

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3    K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3   The first value is the same as 3, so use the second value

7

K3  7.953  10

γ  90.000 deg

  159.525  deg γ  



 A5  sin γ  A3  cos γ  A6   A1  

  75.253 deg

 A3  sin γ  A2  cos γ  A4   A1  

  75.253 deg

  acos

  asin


γ  


p 21 

2

P21x  P21y

p 21  184.784

δ  atan2 P21x P21y 2

p 31 

δ  5.279  deg

2

P31x  P31y

p 31  277.346

δ  atan2 P31x P31y 9.

δ  40.467 deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 A F AA   B G 

 

C  cos α  1

 

M  p 21  sin δ

B C D 

 E  L CC    M  N   

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x   W1y   AA  1 CC  Z1x   Z1y   

10. The components of the W and Z vectors are: W1x  62.394


w 

W1y  91.663 2

Z1x  23.606

2

W1x  W1y

Z1y  40.337

w  110.884


 

A'  cos γ  1

 

B'  sin γ

 

E  p 21  cos δ

 

H  cos α  1

D  sin α

G'  sin γ

 

 

 

C  cos α  1

 

F'  cos γ  1

 

K  sin α



 

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   

13. The components of the U and S vectors are: U1x  29.920


U1y  116.933 2

u 

U1x  U1y

2

S1x  74.080

S1y  38.067

u  120.700


V1x  50.474

V1y  Z1y  S1y

V1y  2.270 v 


2

2

V1x  V1y

v  50.525

G1x  W1x  V1x  U1x

G1x  18.000

G1y  W1y  V1y  U1y

G1y  23.000

g 


2

G1x  G1y

2

g  29.206


O2x  86.000

O2y  Z1y  W1y

O2y  132.000

O4x  S1x  U1x

O4x  104.000

O4y  S1y  U1y

O4y  155.000


2

z  46.736

2

2

s  83.288


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y


ψ  152.803  deg

ϕ  atan2( Z1x Z1y )

ϕ  120.337  deg

rP  z

θ  atan2 z cos ϕ  s cos ψ z sin ϕ  s sin ψ  θ  2.575  deg



δp  ϕ  θ

δp  117.762  deg


g  29.206

Link 2:

w  110.884

Link 3:

v  50.525

Link 4:

u  120.700

Coupler point:

rP  46.736

δp  117.762  deg





Design a fourbar linkage to carry the box in Figure P5-5 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use any points on the object as attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base.

Given:


P1y  0.0

P2x  421.0

P3x  184.0

P3y  1400.0

P2y  963.0


θP1  0.0 deg

θP2  27.0 deg

θP3  88.0 deg


β  100.0  deg


2.

3.

γ  80.0 deg



δ  atan2 P2x P2y

δ  66.386 deg

3

δ  atan2 P3x P3y

δ  82.513 deg

2

2

p 21  1.051  10

2

2

p 31  1.412  10

p 21 

P2x  P2y

p 31 

P3x  P3y


α  27.000 deg

α  θP3  θP1

α  88.000 deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

 

K  sin α

 

L  p 31  cos δ

 

M  p 21  sin δ

B C D 

 E  L CC    M  N   



G H K 

  F K H  A

 

F  cos β  1

 

G  sin β

 A F AA   B G 

 

C  cos α  1

D C

N  p 31  sin δ

 W1x  W   1y   AA  1 CC  Z1x     Z1y 

The components of the W and Z vectors are: 3

W1x  784.602

W1y  362.803

Z1x  1.092  10

Z1y  39.947


θ  155.184  deg


ϕ  2.094  deg



 W1x2  W1y2 , w  864.423  


 Z1x2  Z1y2 , z  1093.069  



 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

M  p 21  sin δ

B' C D 

E   L CC    M  N   



G' H K 

  F' K H 

A'

 

K  sin α

 

L  p 31  cos δ

 A'  F' AA    B'  G' 

 

F'  cos γ  1

D C

N  p 31  sin δ

 U1x  U   1y   AA  1 CC  S1x     S1y 


U1y  281.738

σ  atan2 U1x U1y The length of link 4 is: u 

σ  163.052  deg

 S 1x2  S 1y2 , s  806.978  


V1x  Z1x  S 1x

V1x  289.619

V1y  Z1y  S 1y

V1y  42.852

θ  atan2 V1x V1y

θ  8.416  deg

v  Link 1:

2

2

V1x  V1y

v  292.772

G1x  W1x  V1x  U1x

G1x  429.556

G1y  W1y  V1y  U1y

G1y  38.214

θ  atan2 G1x G1y

θ  5.084  deg

g  7.

ψ  atan2 S 1x S 1y

 U1x2  U1y2 , u  966.514  


S 1x  802.719

2

2

G1x  G1y

g  431.252


S 1y  82.798 ψ  5.889  deg


8.

9.


θ2i  θ  θ

θ2i  150.100  deg

θ2f  θ2i  β

θ2f  50.100 deg


δp  ϕ  θ

rp  1093.069

δp  10.511 deg


O2x  307.736

O2y  z sin ϕ  w sin θ

O2y  402.750

O4x  s cos ψ  u  cos σ

O4x  121.820

O4y  s sin ψ  u  sin σ

O4y  364.536



θrot  5.084  deg




w  864.423

θ  155.184  deg

Link 3:

v  292.772

θ  8.416  deg

Link 4:

u  966.514

σ  163.052  deg

Link 1:

g  431.252

θ  5.084  deg

Coupler:

rp  1093.069

δp  10.511 deg

Crank angles:

θ2i  150.100  deg θ2f  50.100 deg 13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.



Y

P3

P2

B2

B3

A2 U2 A1

B1

A3

W2

U3 P1

W3 W1 O2

U1

O4

X





Given:


P1y  0.0

P3x  184.0

P3y  1400.0

P2x  421.0

P2y  963.0


θP1  0.0 deg

θP2  27.0 deg

θP3  88.0 deg

Coordinates of the points A1 and B1 with respect to P1: A1x  1080 Solution: 1.

2.

3.

A1y  0.0

B1y  60



δ  atan2 P2x P2y

δ  66.386 deg

3

δ  atan2 P3x P3y

δ  82.513 deg

2

2

p 21  1.051  10

2

2

p 31  1.412  10

p 21 

P2x  P2y

p 31 

P3x  P3y


α  27.000 deg

α  θP3  θP1

α  88.000 deg


P1x  A1x2   P1y  A1y 2

z  1080

s 

P1x  B1x 2  P1y  B1y 2

s  742.428

v 

 A1x  B1x 2  A1y  B1y2

v  345.254

ϕ  0

ϕ  0.000  deg

 z2  s2  v2    2 z s 

4.

B1x  740

ψ  acos

ψ  4.635  deg


Z1x  1080


Z1y  0.000

S 1x  s cos ψ

S 1x  740.000

S 1y  s sin ψ

S 1y  60.000




Guess:

W1x  50

W1y  200

β  80 deg

Given


β  160  deg

       

   

 

       

   

 

       

   

 

       

   

 


 W1x   W1y     Find  W1x W1y β β  β   β 

β  54.243 deg

β  107.466  deg



W1y  289.533

θ  158.387  deg

 W1x2  W1y2 , w  786.051  


U1x  30

U1y  100

γ  80 deg

Given


       

   

 

       

   

 

       

   

 

       

   

 

γ  160  deg


 U1x   U1y     Find  U1x U1y γ γ  γ   γ 

γ  411.378  deg

γ  437.949  deg




U1y  232.957


Link 1:

V1x  340.000

V1y  Z1y  S 1y

V1y  60.000

θ  atan2 V1x V1y

θ  10.008 deg

2

2

V1x  V1y

v  345.254

G1x  W1x  V1x  U1x

G1x  532.233

G1y  W1y  V1y  U1y

G1y  3.424

θ  atan2 G1x G1y

θ  0.369  deg

2

g 

9.

 U1x2  U1y2 , u  951.962  

V1x  Z1x  S 1x

v 

8.

σ  165.835  deg


7.


2

G1x  G1y

g  532.244


θ2i  θ  θ

θ2i  158.755  deg

θ2f  θ2i  β

θ2f  51.289 deg


δp  ϕ  θ

rp  1080.000

δp  10.008 deg


O2x  349.215

O2y  z sin ϕ  w sin θ

O2y  289.533

O4x  s cos ψ  u  cos σ

O4x  183.018

O4y  s sin ψ  u  sin σ

O4y  292.957


θrot  atan2 O4x  O2x  O4y  O2y 11. Determine the Grashof condition.

θrot  0.369  deg






w  786.051

θ  158.387  deg

Link 3:

v  345.254

θ  10.008 deg

Link 4:

u  951.962

σ  165.835  deg

Link 1:

g  532.244

θ  0.369  deg

Coupler:

rp  1080.000

δp  10.008 deg

Crank angles:


Y

P3

P2

B3

A2

B2 U2

A1

B1

A3

W2 W 3

U3 P1

W1 O2

U1

O4

X




Solution: 1.

2.

Design a linkage to carry the object in Figure P5-5 through the three positions shown in their numbered order using the fixed pivots shown. Determine the range of the transmission angle. P21x  421.0

P21y  963.0

P31x  184.0

P31y  1400.0

O2x  362.0

O2y  291.0

O4x  182.0

O4y  291.0

Body angles:

θP1  0  deg

θP2  27 deg

θP3  88 deg



α  27.000 deg

α  θP3  θP1

α  88.000 deg


3.

4.

R1x  362.000

R1y  O2y

R2x  R1x  P21x

R2x  783.000

R2y  R1y  P21y

R2y  1.254  10

R3x  R1x  P31x

R3x  546.000

R3y  R1y  P31y

R3y  1.691  10

3

3

2

2

R1  464.462

2

2

R2  1.478  10

3

2

2

R3  1.777  10

3

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  291.000


ζ  38.795 deg

ζ  atan2 R2x R2y

ζ  58.019 deg

ζ  atan2 R3x R3y

ζ  72.105 deg






















 

C3  824.189

 

C4  1.319  10

 

C5  592.567

 

C6  830.373






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ

C1  944.701 C2  928.284

3



A1  2.419  10

A2  C3 C6  C4 C5

A2  9.725  10

A3  C4 C6  C3 C5

A3  1.584  10

A4  C2 C3  C1 C4

A4  4.810  10

A5  C4 C5  C3 C6

A5  9.725  10

4

A6  C1 C3  C2 C4

A6  2.003  10

6

K1  A2  A4  A3  A6

K1  3.219  10

K2  A3  A4  A5  A6

K2  5.670  10

11

K3  4.543  10

11

2

K3 

4 6

5

12

2

2

2

A1  A2  A3  A4  A6

2

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  88.000 deg

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  107.980  deg


β  β

 A5  sin β  A3  cos β  A6   A1  

β  54.008 deg

 A3  sin β  A2  cos β  A4   A1  

β  54.008 deg




6

A1  C3  C4

β  β


R1x  182.000

R1y  O4y

R2x  R1x  P21x

R2x  239.000

R2y  R1y  P21y

R2y  1.254  10

R3x  R1x  P31x

R3x  2.000

R3y  R1y  P31y

R3y  1.691  10

2

2

R1  343.227

2

2

R2  1.277  10

R1 

R1x  R1y

R2 

R2x  R2y

3

3

3

R1y  291.000


R3  6.

7.


2

2

R3x  R3y

R3  1.691  10

3


ζ  122.023  deg

ζ  atan2 R2x R2y

ζ  79.209 deg

ζ  atan2 R3x R3y

ζ  89.932 deg






















 

C3  299.174

 

C4  1.863  10

 

C5  533.274

 

C6  1.077  10








C5  R1 cos α  ζ  R2 cos ζ





3

C6  R1 sin α  ζ  R2 sin ζ 2

3

C2  1.225  10

3

C4  R1 sin α  ζ  R3 sin ζ



C1  478.979

2

6

A1  C3  C4

A1  3.559  10

A2  C3 C6  C4 C5

A2  6.710  10

A3  C4 C6  C3 C5

A3  2.166  10

A4  C2 C3  C1 C4

A4  5.257  10

A5  C4 C5  C3 C6

A5  6.710  10

5

A6  C1 C3  C2 C4

A6  2.425  10

6

K1  A2  A4  A3  A6

K1  5.606  10

K2  A3  A4  A5  A6

K2  4.884  10

2

K3 

5 6

5

12 11

2

2

2

A1  A2  A3  A4  A6

2

2

11

K3  6.838  10

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  88.000 deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  78.042 deg


γ  



 A5  sin γ  A3  cos γ  A6   A1  

  51.463 deg

 A3  sin γ  A2  cos γ  A4   A1  

  51.463 deg

  acos

  asin


γ  


p 21 

2

P21x  P21y

p 21  1051.004

δ  atan2 P21x P21y 2

p 31 

δ  66.386 deg

2

P31x  P31y

p 31  1412.040

δ  atan2 P31x P31y 9.

δ  82.513 deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 A F AA   B G 

 

C  cos α  1

 

M  p 21  sin δ

B C D 

 E  L CC    M  N   

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x   W1y   AA  1 CC  Z1x   Z1y   

10. The components of the W and Z vectors are: W1x  728.089 11. The length of link 2 is:

w 

3

W1y  294.291 2

Z1x  1.090  10

2

W1x  W1y

Z1y  3.291

w  785.316


 

A'  cos γ  1

 

B'  sin γ

 

E  p 21  cos δ

 

H  cos α  1

D  sin α

G'  sin γ

 

 

 

C  cos α  1

 

F'  cos γ  1

 

K  sin α



 

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   


U1y  231.572 2

u 

U1x  U1y

2

S1x  739.699

S1y  59.428

u  950.344


V1x  350.390

V1y  Z1y  S1y

V1y  62.718 v 


2

2

V1x  V1y

v  355.959

G1x  W1x  V1x  U1x

G1x  544.000

G1y  W1y  V1y  U1y

G1y  5.116  10

g 


2

G1x  G1y

2

 13

g  544.000


O2x  362.000

O2y  Z1y  W1y

O2y  291.000

O4x  S1x  U1x

O4x  182.000

O4y  S1y  U1y

O4y  291.000


2

z  1.090  10

2

2

s  742.082


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y

3


ψ  4.593  deg

ϕ  atan2( Z1x Z1y )

ϕ  0.173  deg

rP  z


δp  ϕ  θ

δp  9.975  deg




g  544.000

Link 2:

w  785.316

Link 3:

v  355.959

Link 4:

u  950.344

Coupler point:

rP  1090.094

δp  9.975  deg





To the linkage solution from Problem 5-29, add a driver dyad with a crank to control the motion of your fourbar so that it cannot move beyond positions one and three.

Given:

Solution to Problem 5-29:

Solution: 1.

Length of link 2

w  785.316

Angle of link 2 in first position

θ  157.992  deg

Rotation angles for link 2

β  54.008 deg

β  107.980  deg

Coordinates of O2

O2x  362.0

O2y  291.0


Link 2 of the solution to Problem 5-29 will become the driven link for the driver dyad. The driver dyad will be links 5 and 6 and the fixed pivot for the dyad will be at O6. Select a point on link 2 of Problem 5-29 and label it C. Let the distance O2C be R2  200. The solution that follows uses the algorithm presented in Section 5.2 with changes in nomenclature to account for the fact that the driven link is link 2 and the points A and B are already defined on the fourbar of Problem 5-29..

2.

Determine the coordinates of the points C1 and C3 using equations 5.0a. Determine the vector M using 5.0b. C1x  O2x  R2 cos θ

C1x  547.426

C1y  O2y  R2 sin θ

C1y  216.053





C3x  233.475





C3y  137.764

C3x  O2x  R2 cos θ  β C3y  O2y  R2 sin θ  β RC1 

 C1x     C1y 

RC3 

 C3x     C3y 

M  RC3  RC1

M

 313.952     78.289 

3.

Select a suitable value for the multiplier, K, in equation 5.0d say K  3.0.

4.

Determine the coordinates of the crank pivot, O6 using equation 5.0d. Place the pivot to the left of O2 (by subtracting KM from RC3) so that it will be on the base below and to the left of O2. RO6  RC3  K M

O6x  RO6

1

O6x  1175.330 5.

2

O6y  372.630

Determine the length of the driving crank using equation 5.0e.





R6  R2 sin 0.5 β 6.

O6y  RO6

R6  161.783

Determine the length of the driver dyad coupler, link 5, and the ground link from eqauation 5.0f. R5  RC3  RO6  R6 RO2 

R5  808.914

 O2x     O2y 

R1  RO2  RO6

R1  817.416


7.


Determine the Grashof condition. R1  817.416

R2  200.000

R5  808.914

R6  161.783



Condition R1 R2 R5 R6  "Grashof" 8.

Draw the linkage using the link lengths and fixed pivot coordinates calculated above to verify that the driver dyad will perform as required.

Y

P3

P2

A2

B3

B2 A3

A1

B1

C2

D2 D1

O6

D3

C1

X

P1 C3

O2

O4





Given:


P1y  0.0

P3x  148.0

P3y  187.0

P2x  130.0

P2y  29.0


θP1  90.0 deg

θP2  65.0 deg

θP3  11.0 deg

Coordinates of the points A1 and B1 with respect to P1: A1x  69.0 Solution: 1.

2.

3.

A1y  43.0

B1x  17.0

B1y  43.0



2

p 21  133.195

δ  atan2 P2x P2y

δ  12.575 deg

2

2

p 31  238.481

δ  atan2 P3x P3y

δ  51.640 deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  25.000 deg

α  θP3  θP1

α  101.000  deg


P1x  A1x2   P1y  A1y 2

z  81.302

s 

P1x  B1x 2  P1y  B1y 2

s  46.239

v 

 A1x  B1x 2  A1y  B1y2

v  52.000

 v2  z2  s2    2  v z 

ϕ  acos

 v2  s2  z2    2  v s 

ϕ  31.931 deg


ψ  68.429 deg


Z1x  69.000


Z1y  43.000

S 1x  s cos ψ

S 1x  17.000

S 1y  s sin ψ

S 1y  43.000


4.


Use equations 5.24 to solve for w, , 2, and 3. Since the points A and B are to be used as pivots, z and  are known from the calculations above. Guess:

W1x  50

W1y  200

β  80 deg

Given


       

   

 

       

   

 

       

   

 

       

   

 

β  160  deg



β  52.277 deg

β  96.147 deg



W1y  118.432

θ  118.167  deg

 W1x2  W1y2 , w  134.341  


U1x  30

U1y  100

γ  80 deg

Given


       

   

 

       

   

 

U1x cos γ  1  U1y sin γ  = p 31  cos δ  S 1x cos α  1  S 1y sin α

              U1y  cos γ  1   U1x sin γ  = p 31  sin δ  S 1y  cos α  1   S 1x sin α U1y cos γ  1  U1x sin γ  = p 21  sin δ  S 1y cos α  1  S 1x sin α

γ  160  deg




γ  147.982  deg


U1y  77.634


Link 1:

V1x  52.000

V1y  Z1y  S 1y

V1y  0.000

θ  atan2 V1x V1y

θ  0.000  deg

2

2

V1x  V1y

v  52.000

G1x  W1x  V1x  U1x

G1x  34.515

G1y  W1y  V1y  U1y

G1y  40.798

θ  atan2 G1x G1y

θ  49.768 deg

g 

9.

 U1x2  U1y2 , u  90.203  

V1x  Z1x  S 1x

v 

8.

σ  120.609  deg


7.


2

2

G1x  G1y

g  53.439


θ2i  θ  θ

θ2i  68.399 deg

θ2f  θ2i  β

θ2f  27.748 deg


δp  ϕ  θ

rp  81.302

δp  31.931 deg


O2x  5.585

O2y  z sin ϕ  w sin θ

O2y  161.432

O4x  s cos ψ  u  cos σ

O4x  28.930

O4y  s sin ψ  u  sin σ

O4y  120.634





θrot  49.768 deg

11. Determine the Grashof condition.




w  134.341

θ  118.167  deg

Link 3:

v  52.000

θ  0.000  deg

Link 4:

u  90.203

σ  120.609  deg

Link 1:

g  53.439

θ  49.768 deg

Coupler:

rp  81.302

δp  31.931 deg

Crank angles:

θ2i  68.399 deg θ2f  27.748 deg 13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design. 14. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.


SOLUTION MANUAL 5-31-5 Y P1 Z1

A1

X

S1

S2

B1

V2 B2

U1 W1

U2

W2 O4 W3 O2

P2

Z2

A2

V1

A3

U3 V3 B3

Z3 S3 P3




Design a fourbar linkage to carry the box in Figure P5-6 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use any points on the object as attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base.

Given:


P1y  0.0

P2x  130.0

P3x  148.0

P3y  187.0

P2y  29.0


θP1  90.0 deg

θP2  65.0 deg

θP3  11.0 deg


β  95.0 deg


2.

3.

γ  145.0  deg



2

p 21  133.195

δ  atan2 P2x P2y

δ  12.575 deg

2

2

p 31  238.481

δ  atan2 P3x P3y

δ  51.640 deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  25.000 deg

α  θP3  θP1

α  101.000  deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 A F AA   B G 

 

C  cos α  1

 

M  p 21  sin δ

B C D 

 E  L CC    M  N   

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x  W   1y   AA  1 CC  Z1x     Z1y 


W1y  118.196

Z1x  69.575

Z1y  44.896


θ  118.246  deg


ϕ  32.834 deg



 W1x2  W1y2 , w  134.173  


 Z1x2  Z1y2 , z  82.803  



 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

F'  cos γ  1

 

M  p 21  sin δ

B' C D 

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

N  p 31  sin δ

 U1x     U1y   AA  1 CC  S1x     S1y 


U1y  82.956

S 1x  17.754

S 1y  37.985


σ  119.180  deg

ψ  atan2 S 1x S 1y

ψ  64.949 deg


 U1x2  U1y2 , u  95.014  



V1x  Z1x  S 1x

V1x  51.820

V1y  Z1y  S 1y

V1y  6.910

θ  atan2 V1x V1y

θ  7.596  deg

v  Link 1:

2

2

V1x  V1y

v  52.279

G1x  W1x  V1x  U1x

G1x  34.647

G1y  W1y  V1y  U1y

G1y  42.151

θ  atan2 G1x G1y

θ  50.580 deg

g  7.

 S 1x2  S 1y2 , s  41.930  

2

2

G1x  G1y

g  54.563


θ2i  θ  θ

θ2i  67.666 deg



θ2f  θ2i  β 8.

9.

θ2f  27.334 deg


δp  ϕ  θ

rp  82.803

δp  25.238 deg


O2x  6.076

O2y  z sin ϕ  w sin θ

O2y  163.092

O4x  s cos ψ  u  cos σ

O4x  28.571

O4y  s sin ψ  u  sin σ

O4y  120.941



θrot  50.580 deg




w  134.173

θ  118.246  deg

Link 3:

v  52.279

θ  7.596  deg

Link 4:

u  95.014

σ  119.180  deg

Link 1:

g  54.563

θ  50.580 deg

Coupler:

rp  82.803

δp  25.238 deg

Crank angles:

θ2i  67.666 deg θ2f  27.334 deg 13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design. 14. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.



Y P1 Z1 A1 V1

X

S1 B1

S2 V2 U1

W1

B2

U2

W2 O4 W3 O2

P2

Z2

A2

A3

U3 V3

Z3

B 3 S3 P3




Solution: 1.

2.

Design a linkage to carry the object in Figure P5-6 through the three positions shown in their numbered order using the fixed pivots shown. Determine the range of the transmission angle. P21x  130.0

P21y  29.0

P31x  148.0

P31y  187.0

O2x  6.2

O2y  164.0

O4x  28.0

O4y  121.0

Body angles:

θP1  90 deg

θP2  65 deg

θP3  11 deg



α  25.000 deg

α  θP3  θP1

α  101.000  deg


3.

4.

R1x  6.200

R1y  O2y

R2x  R1x  P21x

R2x  136.200

R2y  R1y  P21y

R2y  135.000

R3x  R1x  P31x

R3x  154.200

R3y  R1y  P31y

R3y  23.000

2

2

R1  164.117

2

2

R2  191.769

2

2

R3  155.906

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  164.000


ζ  87.835 deg

ζ  atan2 R2x R2y

ζ  44.746 deg

ζ  atan2 R3x R3y

ζ  8.484  deg






















 

C3  5.604

 

C4  14.379

 

C5  61.271

 

C6  11.014






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ

C1  23.501 C2  73.444



A1  C3  C4

A1  238.152

A2  C3 C6  C4 C5

A2  819.286

A3  C4 C6  C3 C5

A3  501.727

A4  C2 C3  C1 C4

A4  749.483

A5  C4 C5  C3 C6

A5  819.286

A6  C1 C3  C2 C4

A6  924.339

K1  A2  A4  A3  A6

K1  1.503  10

K2  A3  A4  A5  A6

K2  1.133  10

2

K3 

5 6

2

2

2

A1  A2  A3  A4  A6

2

K3  1.141  10

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  101.000  deg

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  94.106 deg


β  β

 A5  sin β  A3  cos β  A6   A1  

β  53.072 deg

 A3  sin β  A2  cos β  A4   A1  

β  53.072 deg




6

β  β


R1x  28.000

R1y  O4y

R2x  R1x  P21x

R2x  102.000

R2y  R1y  P21y

R2y  92.000

R3x  R1x  P31x

R3x  120.000

R3y  R1y  P31y

R3y  66.000

2

2

R1  124.197

2

2

R2  137.361

R1 

R1x  R1y

R2 

R2x  R2y

R1y  121.000


R3  6.

7.


2

2

R3x  R3y

R3  136.953


ζ  103.029  deg

ζ  atan2 R2x R2y

ζ  42.049 deg

ζ  atan2 R3x R3y

ζ  28.811 deg






















 

C3  4.120

 

C4  70.398

 

C5  76.240

 

C6  29.497






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

C2  7.150

A1  C3  C4

A1  4.973  10

3

A2  C3 C6  C4 C5

A2  5.489  10

3

A3  C4 C6  C3 C5

A3  1.762  10

3

A4  C2 C3  C1 C4

A4  675.715

A5  C4 C5  C3 C6

A5  5.489  10

A6  C1 C3  C2 C4

A6  544.601

K1  A2  A4  A3  A6

K1  2.749  10

K2  A3  A4  A5  A6

K2  4.180  10

2

K3 

2

C1  10.017

3

6 6

2

2

2

A1  A2  A3  A4  A6

2

2

K3  4.628  10

6

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  145.661  deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  101.000  deg


γ  γ



 A5  sin γ  A3  cos γ  A6   A1  

  77.265 deg

 A3  sin γ  A2  cos γ  A4   A1  

  77.265 deg

  acos

  asin


γ  


p 21 

2

P21x  P21y

p 21  133.195

δ  atan2 P21x P21y 2

p 31 

δ  12.575 deg

2

P31x  P31y

p 31  238.481

δ  atan2 P31x P31y 9.

δ  51.640 deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 

M  p 21  sin δ

B C D 

 A F AA   B G 

 

C  cos α  1

 E  L CC    M  N   

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x   W1y   AA  1 CC  Z1x   Z1y   

10. The components of the W and Z vectors are: W1x  61.917 w 


W1y  112.415 2

Z1x  68.117

2

W1x  W1y

Z1y  51.585

w  128.339


 

A'  cos γ  1

 

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

H  cos α  1

D  sin α

G'  sin γ

 

L  p 31  cos δ

 

 

 

M  p 21  sin δ

 

F'  cos γ  1

 

K  sin α

 

N  p 31  sin δ


 A' F' AA    B'  G' 


B' C D 

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   



U1y  80.382 2

u 

U1x  U1y

S1x  18.374

2

S1y  40.618

u  92.800


V1x  49.743

V1y  Z1y  S1y

V1y  10.967 v 


2

2

V1x  V1y

v  50.938

G1x  W1x  V1x  U1x

G1x  34.200

G1y  W1y  V1y  U1y

G1y  43.000

g 


2

G1x  G1y

2

g  54.942


O2x  6.200

O2y  Z1y  W1y

O2y  164.000

O4x  S1x  U1x

O4x  28.000

O4y  S1y  U1y

O4y  121.000


2

z  85.446

2

2

s  44.581


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y


ψ  65.660 deg

ϕ  atan2( Z1x Z1y )

ϕ  37.136 deg

rP  z

θ  atan2 z cos ϕ  s cos ψ z sin ϕ  s sin ψ  θ  12.433 deg

δp  ϕ  θ

δp  24.704 deg




g  54.942

Link 2:

w  128.339

Link 3:

v  50.938

Link 4:

u  92.800

Coupler point:

rP  85.446

δp  24.704 deg

19. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct. 20. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.




Design a fourbar linkage to carry the bolt in Figure P5-7 from positions 1 to 2 to 3 without regard for the fixed pivots shown. The bolt is fed into the gripper in the z direction (into the paper). The gripper grabs the bolt, and your linkage moves it to position 3 to be inserted into the hole. A second degree of freedom within the gripper assembly (not shown) pushes the bolt into the hole. The moving pivots should be on, or close to, the gripper assembly, and the fixed pivots should be on the base. Use the free choices given below.

Given:


P1y  0.0

P2x  99.0

P3x  111.3

P3y  151.8

P2y  13.0


θP1  272.3  deg

θP2  301.7  deg

θP3  270.0  deg

Free choices for the WZ dyad : β  70 deg

β  140  deg

Free choices for the US dyad : γ  5  deg Solution: 1.

2.

3.

γ  49 deg



2

p 21  99.850

δ  atan2 P2x P2y

δ  7.481  deg

2

2

p 31  188.231

δ  atan2 P3x P3y

δ  53.751 deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  29.400 deg

α  θP3  θP1

α  2.300  deg


 

B  sin β

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1 D  sin α G  sin β

 

L  p 31  cos δ

 A F AA   B G 

B C D 

 G H K  A D C   F K H 

The components of the W and Z vectors are:

 

 

C  cos α  1

 

 

 

M  p 21  sin δ

 E  L CC    M  N   

 

F  cos β  1

 

K  sin α

 

N  p 31  sin δ

 W1x  W   1y   AA  1 CC  Z1x     Z1y 



W1x  86.624

W1y  50.030

Z1x  198.147

Z1y  233.314


θ  149.991  deg


ϕ  49.660 deg

 W1x2  W1y2 , w  100.033  


 Z1x2  Z1y2 , z  306.100  



  D  sin α

  E  p 21  cos δ

A'  cos γ  1

 

C  cos α  1

 

G'  sin γ

 

H  cos α  1

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

  F'  cos γ  1

B'  sin γ

 

M  p 21  sin δ

B' C D 

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

N  p 31  sin δ

 U1x  U   1y   AA  1 CC  S1x     S1y 


U1y  205.365

S 1x  3.375

S 1y  166.927


σ  62.360 deg

ψ  atan2 S 1x S 1y

ψ  88.842 deg


 U1x2  U1y2 , u  231.821  


 S 1x2  S 1y2 , s  166.961  


V1x  Z1x  S 1x

V1x  194.772

V1y  Z1y  S 1y

V1y  66.387

θ  atan2 V1x V1y

θ  18.821 deg

v  Link 1:

2

2

V1x  V1y

v  205.775

G1x  W1x  V1x  U1x

G1x  0.602

G1y  W1y  V1y  U1y

G1y  221.722

θ  atan2 G1x G1y

θ  89.844 deg

g 

2

2

G1x  G1y

g  221.723


7.

8.

9.



θ2i  θ  θ

θ2i  239.836  deg

θ2f  θ2i  β  2  π

θ2f  19.836 deg


δp  ϕ  θ

rp  306.100

δp  30.838 deg


O2x  111.523

O2y  z sin ϕ  w sin θ

O2y  183.284

O4x  s cos ψ  u  cos σ

O4x  110.920

O4y  s sin ψ  u  sin σ

O4y  38.438



θrot  89.844 deg




w  100.033

θ  149.991  deg

Link 3:

v  205.775

θ  18.821 deg

Link 4:

u  231.821

σ  62.360 deg

Link 1:

g  221.723

θ  89.844 deg

Coupler:

rp  306.100

δp  30.838 deg

Crank angles:




Y A1 O2

B1

B2

A2 A3 P2

B3 X

P1 O4

P3




Solution: 1.

2.

Design a linkage to carry the bolt in Figure P5-7 from positions 1 to 2 to 3 using the fixed pivots shown. See Problem 5-34 for more details. P21x  99.0

P21y  13.0

P31x  111.3

P31y  151.8

O2x  111.5

O2y  183.2

O4x  111.5

O4y  38.8

Body angles:

θP1  272.3  deg

θP2  301.7  deg

θP3  270  deg



α  29.400 deg

α  θP3  θP1

α  2.300  deg


3.

4.

R1x  111.500

R1y  O2y

R2x  R1x  P21x

R2x  210.500

R2y  R1y  P21y

R2y  170.200

R3x  R1x  P31x

R3x  222.800

R3y  R1y  P31y

R3y  335.000

2

2

R1  214.463

2

2

R2  270.700

2

2

R3  402.324

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  183.200


ζ  58.674 deg

ζ  atan2 R2x R2y

ζ  38.957 deg

ζ  atan2 R3x R3y

ζ  56.373 deg






















 

C3  118.742

 

C4  147.473

 

C5  23.426

 

C6  65.329






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ

C1  155.059 C2  3.973


2


2

A1  3.585  10

A2  C3 C6  C4 C5

A2  4.303  10

A3  C4 C6  C3 C5

A3  1.242  10

4

A4  C2 C3  C1 C4

A4  2.240  10

4

A5  C4 C5  C3 C6

A5  4.303  10

3

A6  C1 C3  C2 C4

A6  1.900  10

4

K1  A2  A4  A3  A6

K1  1.395  10

K2  A3  A4  A5  A6

K2  3.598  10

2

K3 

3

8 8

2

2

2

A1  A2  A3  A4  A6

2 8

K3  1.250  10

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  139.911  deg

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  2.300  deg


β  β

 A5  sin β  A3  cos β  A6   A1  

β  69.984 deg

 A3  sin β  A2  cos β  A4   A1  

β  69.984 deg



β  β

Since both angles are the same, 5.

4

A1  C3  C4


R1x  111.500

R1y  O4y

R2x  R1x  P21x

R2x  210.500

R2y  R1y  P21y

R2y  51.800

R3x  R1x  P31x

R3x  222.800

R3y  R1y  P31y

R3y  113.000

2

2

R1  118.058

2

2

R2  216.780

R1 

R1x  R1y

R2 

R2x  R2y

R1y  38.800


R3  6.

7.


2

2

R3x  R3y

R3  249.818


ζ  19.187 deg

ζ  atan2 R2x R2y

ζ  13.825 deg

ζ  atan2 R3x R3y

ζ  26.893 deg






















 

C3  109.833

 

C4  147.294

 

C5  132.407

 

C6  36.739






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

C2  32.384

A1  C3  C4

A1  3.376  10

4

A2  C3 C6  C4 C5

A2  1.547  10

4

A3  C4 C6  C3 C5

A3  1.995  10

4

A4  C2 C3  C1 C4

A4  1.918  10

3

A5  C4 C5  C3 C6

A5  1.547  10

A6  C1 C3  C2 C4

A6  8.852  10

K1  A2  A4  A3  A6

K1  2.063  10

K2  A3  A4  A5  A6

K2  9.865  10

2

K3 

2

C1  37.169

4 3

8

2

2

2

A1  A2  A3  A4  A6

7

2

2

8

K3  2.101  10

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  2.300  deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  48.814 deg


γ  



 A5  sin γ  A3  cos γ  A6   A1  

  4.951  deg

 A3  sin γ  A2  cos γ  A4   A1  

  4.951  deg

  acos

  asin


γ  


p 21 

2

P21x  P21y

p 21  99.850

δ  atan2 P21x P21y 2

p 31 

δ  7.481  deg

2

P31x  P31y

p 31  188.231

δ  atan2 P31x P31y 9.

δ  53.751 deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 

M  p 21  sin δ

B C D 

 A F AA   B G 

 

C  cos α  1

 E  L CC    M  N   

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x   W1y   AA  1 CC  Z1x   Z1y   

10. The components of the W and Z vectors are: W1x  86.684 w 


W1y  49.977 2

Z1x  198.184

2

W1x  W1y

Z1y  233.177

w  100.059


 

A'  cos γ  1

 

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

H  cos α  1

D  sin α

G'  sin γ

 

L  p 31  cos δ

 

 

 

M  p 21  sin δ

 

F'  cos γ  1

 

K  sin α

 

N  p 31  sin δ


 A' F' AA    B'  G' 


B' C D 

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   

13. The components of the W and Z vectors are: U1x  108.268


U1y  205.938

S1x  3.232

2

u  232.664

2

u 

U1x  U1y

S1y  167.138


V1x  194.953

V1y  Z1y  S1y

V1y  66.039 v 


2

2

V1x  V1y

v  205.834

G1x  W1x  V1x  U1x

G1x  1.990  10

G1y  W1y  V1y  U1y

G1y  222.000

g 


2

G1x  G1y

2

 13

g  222.000


O2x  111.500

O2y  Z1y  W1y

O2y  183.200

O4x  S1x  U1x

O4x  111.500

O4y  S1y  U1y

O4y  38.800


2

z  306.020

2

2

s  167.169


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y


ψ  88.892 deg

ϕ  atan2( Z1x Z1y )

ϕ  49.638 deg

rP  z


δp  ϕ  θ

δp  30.924 deg




g  222.000

Link 2:

w  100.059

Link 3:

v  205.834

Link 4:

u  232.664

Coupler point:

rP  306.020

δp  30.924 deg





To the linkage solution from Problem 5-35, add a driver dyad with a crank to control the motion of your fourbar so that it cannot move beyond positions one and three.

Given:

Solution to Problem 5-35:

Solution: 1.

Length of link 4

u  232.664


θ  62.268 deg

Rotation angles for link 2

β  4.951  deg

β  48.814 deg

Coordinates of O4

O4x  111.5

O4y  38.8


Link 4 of the solution to Problem 5-35 will become the driven link for the driver dyad. The driver dyad will be links 5 and 6 and the fixed pivot for the dyad will be at O6. Select a point on link 4 of Problem 5-35 and label it C. Let the distance O4C be R4  60. The solution that follows uses the algorithm presented in Section 5.2 with changes in nomenclature to account for the fact that the driven link is link 2 and the points A and B are already defined on the fourbar of Problem 5-35.

2.


C1x  83.580

C1y  O4y  R4 sin θ

C1y  14.308





C3x  53.147





C3y  24.840


3. 4.

 C1x     C1y 

RC3 

 C3x     C3y 

M  RC3  RC1

Determine the coordinates of the crank pivot, O6 using equation 5.0d. Place the pivot to the left of O4 (by subtracting KM from RC3) so that it will be on the base above and to the left of O2. O6x  RO6

1

O6x  144.446

O6y  RO6

2

O6y  92.604

Determine the length of the driving crank using equation 5.0e.





R6  R4 sin 0.5 β 6.

 30.433     39.148 


RO6  RC3  K M

5.

M

R6  24.793


R5  123.965

 O4x     O4y 

R1  RO2  RO6

R1  135.472


7.


Determine the Grashof condition. R1  135.472

R4  60.000

R5  123.965

R6  24.793




Draw the linkage using the link lengths and fixed pivot coordinates calculated above to verify that the driver dyad will perform as required.

Y A1 D2

D1 O6

O2

B1

B2

D3

A2 A3 C1

P2

C2

B3 X

C3

P1

O4

P3




Figure P5-8 shows an off-loading mechanism for paper rolls. The V-link is rotated through 90 deg by an air-driven fourbar slider-crank linkage. Design a pin-jointed fourbar linkage to replace the existing off-loading station and perform essentially the same function. Choose three positions of the roll including its two end positions and synthesize a substitute mechanism. Use a link similar to the existing V-link as one of your links.

Given:


P1y  0.0

P3x  1000.0

P3y  1000.0

P2x  450

P2y  140


θP1  90.0 deg

θP2  65.0 deg

θP3  0.0 deg

Coordinates of the points A1 and B1 with respect to P1: A1x  400.0 Solution: 1.

2.

3.

4.

A1y  1035.0

B1x  35.0

B1y  600.0



2

p 21  471.275

2

2

p 31  1.414  10

p 21 

P2x  P2y

p 31 

P3x  P3y

3

δ  atan2 P2x P2y

δ  17.281 deg

δ  atan2 P3x P3y

δ  45.000 deg


α  25.000 deg

α  θP3  θP1

α  90.000 deg


P1x  A1x2   P1y  A1y 2

z  1109.606

s 

P1x  B1x 2  P1y  B1y 2

s  601.020

v 

 A1x  B1x 2  A1y  B1y2

v  615.183

ϕ  atan2 A1x A1y   π

ϕ  291.130  deg

ψ  π  atan2 B1x B1y

ψ  93.338 deg


Z1x  400.000


Z1y  1035.000

S 1x  s cos ψ

S 1x  35.000

S 1y  s sin ψ

S 1y  600.000




Guess:

W1x  50

W1y  200

β  80 deg

Given


β  160  deg

       

   

 

       

   

 

       

   

 

       

   

 



β  47.757 deg



W1y  194.748

θ  163.879  deg

 W1x2  W1y2 , w  701.388  


U1x  30

U1y  100

γ  80 deg

Given


       

   

 

       

   

 

       

   

 

       

   

 

γ  160  deg



γ  25.881 deg

The components of the U vector are:

γ  71.807 deg



U1x  117.927

U1y  469.583


Link 1:

V1x  435.000

V1y  Z1y  S 1y

V1y  435.000

θ  atan2 V1x V1y

θ  45.000 deg

2

2

V1x  V1y

v  615.183

G1x  W1x  V1x  U1x

G1x  356.736

G1y  W1y  V1y  U1y

G1y  160.165

θ  atan2 G1x G1y

θ  155.821  deg

2

g 

9.

 U1x2  U1y2 , u  484.164  

V1x  Z1x  S 1x

v 

8.

σ  75.903 deg


7.


2

G1x  G1y

g  391.042


θ2i  θ  θ

θ2i  8.058  deg

θ2f  θ2i  β

θ2f  39.699 deg


δp  ϕ  θ

rp  1109.606

δp  336.130  deg

Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x  z cos ϕ  w cos θ O2y  z sin ϕ  w sin θ

O4x  s cos ψ  u  cos σ O4y  s sin ψ  u  sin σ

O2x  273.809 O2y  1229.748 O4x  82.927 O4y  1069.583


θrot  atan2 O4x  O2x  O4y  O2y 11. Determine the Grashof condition. Condition( a b c d ) 


θrot  155.821  deg



Condition( g w u v)  "Grashof" 12. DESIGN SUMMARY Link 2:

w  701.388

θ  163.879  deg

Link 3:

v  615.183

θ  45.000 deg

Link 4:

u  484.164

σ  75.903 deg

Link 1:

g  391.042

θ  155.821  deg

Coupler:

rp  1109.606

δp  336.130  deg

Crank angles:

θ2i  8.058  deg θ2f  39.699 deg 13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design. 14. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.

1000.0 450.0 P1

140.0 P2

90 deg 65 deg

1000.0

A3 B1

B3

A2 A1

B2

O4 O2

0 deg

P3




Design a fourbar linkage to carry the object in Figure P5-9 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use points C and D for your attachment points. Determine the range of the transmission angle.

Given:

Coordinates of the points P1 , P2 and P3 with respect to C1: P1x  0.0

P1y  0.0

P3x  7.600

P3y  1.000

P2x  4.500

P2y  1.900


θP1  33.70  deg

θP2  14.60  deg

θP3  0.0 deg

Coordinates of the points C1 and D1 with respect to P1: C1x  0.0 Solution: 1.

2.

3.

4.

C1y  0.0

D1x  3.744

D1y  2.497



2

p 21  4.885

δ  atan2 P2x P2y

δ  22.891 deg

2

2

p 31  7.666

δ  atan2 P3x P3y

δ  7.496  deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  19.100 deg

α  θP3  θP1

α  33.700 deg


P1x  C1x2   P1y  C1y 2

z  0.000

s 

P1x  D1x2   P1y  D1y 2

s  4.500

v 

 C1x  D1x2   C1y  D1y2

v  4.500

ϕ  θP1

ϕ  33.700 deg

ψ  θP1  π

ψ  213.700  deg


Z1x  0.000


Z1y  0.000

S 1x  s cos θP1

S 1x  3.744

S 1y  s sin θP1

S 1y  2.497

Use equations 5.24 to solve for w, , 2, and 3. Since the points C and D are to be used as pivots, z and  are known from the calculations above. Use guess values from a graphical solution such as that for Problem 3-57. Guess:

W1x  4

W1y  4

β  50 deg

β  80 deg


Given


       

   

 

       

   

 

       

   

 

       

   

 

W1x cos β  1  W1y sin β  = p 21  cos δ  Z1x cos α  1  Z1y sin α W1x cos β  1  W1y sin β  = p 31  cos δ  Z1x cos α  1  Z1y sin α W1y cos β  1  W1x sin β  = p 21  sin δ  Z1y cos α  1  Z1x sin α W1y cos β  1  W1x sin β  = p 31  sin δ  Z1y cos α  1  Z1x sin α


β  47.370 deg

β  78.160 deg



W1y  4.179

θ  136.576  deg

 W1x2  W1y2 , w  6.080  

Use equations 5.28 to solve for u, , 2, and 3. Since the points C and D are to be used as pivots, s and  are known from the calculations above. Guess:

U1x  4

U1y  4

γ  44 deg

Given


       

   

 

       

   

 

       

   

 

       

   

 

γ  76 deg



γ  76.170 deg


U1y  6.080


σ  117.929  deg




 U1x2  U1y2 , u  6.881  


V1x  Z1x  S 1x

V1x  3.744

V1y  Z1y  S 1y

V1y  2.497

θ  atan2 V1x V1y v  Link 1:

2

8.

9.

2

V1x  V1y

v  4.500

G1x  W1x  V1x  U1x

G1x  2.551

G1y  W1y  V1y  U1y

G1y  0.596

θ  atan2 G1x G1y

θ  13.155 deg

2

g  7.

θ  33.700 deg

2

G1x  G1y

g  2.620


θ2i  θ  θ

θ2i  123.420  deg

θ2f  θ2i  β

θ2f  45.261 deg

Define the coupler point with respect to point C and the vector V. rp  z

δp  ϕ  θ

rp  0.000

δp  0.000  deg


O2x  4.416

O2y  z sin ϕ  w sin θ

O2y  4.179

O4x  s cos ψ  u  cos σ

O4x  6.967

O4y  s sin ψ  u  sin σ

O4y  3.583


θrot  atan2 O4x  O2x  O4y  O2y 11. Determine the Grashof condition. Condition( a b c d ) 


Condition( g u v w)  "Grashof"

θrot  13.155 deg




w  6.080

θ  136.576  deg

Link 3:

v  4.500

θ  33.700 deg

Link 4:

u  6.881

σ  117.929  deg

Link 1:

g  2.620

θ  13.155 deg

Coupler:

rp  0.000

δp  0.000  deg

Crank angles:

θ2i  123.420  deg

7.600

θ2f  45.261 deg

4.500 Y



4.500

D2

D1 C3

C2

D3

78.160° C1 47.370° y

X 43.852° 76.170° x 13.150°

6.080

2.620




Design a fourbar linkage to carry the object in Figure P5-9 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use any points on the object as attachment points. Determine the range of the transmission angle.

Given:


P1y  0.0

P3x  7.600

P3y  1.000

P2x  4.500

P2y  1.900


θP1  33.70  deg

θP2  14.60  deg

θP3  0.0 deg

Coordinates of the points C1 and E1 (used for attachment) with respect to P1: C1x  0.0 Solution: 1.

2.

3.

4.

C1y  0.0

E1x  3.744

E1y  0.000



2

p 21  4.885

δ  atan2 P2x P2y

δ  22.891 deg

2

2

p 31  7.666

δ  atan2 P3x P3y

δ  7.496  deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  19.100 deg

α  θP3  θP1

α  33.700 deg


P1x  C1x2   P1y  C1y 2

z  0.000

s 

P1x  E1x 2  P1y  E1y 2

s  3.744

v 

 C1x  E1x 2  C1y  E1y2

v  3.744

ϕ  atan2 E1x  C1x E1y  C1y

ϕ  0.000  deg

ψ  ϕ  π

ψ  180.000  deg


Z1x  0.000


Z1y  0.000

S 1x  s cos ψ

S 1x  3.744

S 1y  s sin ψ

S 1y  0.000


W1x  4

W1y  4

β  50 deg

β  80 deg


Given


       

   

 

       

   

 

       

   

 

       

   

 



β  78.160 deg



W1y  4.179

θ  136.576  deg

 W1x2  W1y2 , w  6.080  


U1x  4

U1y  4

γ  44 deg

Given


       

   

 

       

   

 

       

   

 

       

   

 

γ  76 deg



γ  84.280 deg


U1y  4.391


σ  123.354  deg



V1x  Z1x  S 1x

V1x  3.744

V1y  Z1y  S 1y

V1y  0.000

θ  atan2 V1x V1y

θ  0.000  deg

v  Link 1:

2

9.

2

V1x  V1y

v  3.744

G1x  W1x  V1x  U1x

G1x  2.218

G1y  W1y  V1y  U1y

G1y  0.211

θ  atan2 G1x G1y

θ  5.444  deg

2

g 

8.

 U1x2  U1y2 , u  5.256  


7.


2

G1x  G1y

g  2.228


θ2i  θ  θ

θ2i  142.019  deg

θ2f  θ2i  β

θ2f  63.860 deg


δp  ϕ  θ

rp  0.000

δp  0.000  deg

Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x  z cos ϕ  w cos θ O2y  z sin ϕ  w sin θ

O4x  s cos ψ  u  cos σ O4y  s sin ψ  u  sin σ

O2x  4.416 O2y  4.179 O4x  6.634 O4y  4.391



θrot  5.444  deg






w  6.080

θ  136.576  deg

Link 3:

v  3.744

θ  0.000  deg

Link 4:

u  5.256

σ  123.354  deg

Link 1:

g  2.228

θ  5.444  deg

Coupler:

rp  0.000

δp  0.000  deg

Crank angles:

θ2i  142.019  deg θ2f  63.860 deg

13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design. 14. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.

7.600 4.500 Y D2

D1

4.500

C2

D3

C3 E2

C1

X

E1

E3

6.080 O2 O4

5.257

2.228




Given:

Solution: 1.

2.

Design a fourbar linkage to carry the object in Figure P5-9 through the three positions shown in their numbered order using the fixed pivots shown. Determine the range of the transmission angle. P21x  4.500

P21y  1.900

P31x  7.600

P31y  1.000

O2x  2.900

O2y  5.100

O4x  5.900

O4y  5.100

Body angles:

θP1  33.70  deg

θP2  14.60  deg

θP3  0.0 deg



α  19.100 deg

α  θP3  θP1

α  33.700 deg


3.

4.

R1x  2.900

R1y  O2y

R2x  R1x  P21x

R2x  1.600

R2y  R1y  P21y

R2y  7.000

R3x  R1x  P31x

R3x  4.700

R3y  R1y  P31y

R3y  6.100

2

2

R1  5.867

2

2

R2  7.181

2

2

R3  7.701

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  5.100


ζ  119.624  deg

ζ  atan2 R2x R2y

ζ  77.125 deg

ζ  atan2 R3x R3y

ζ  52.386 deg






















 

C3  4.283

 

C4  0.248

 

C5  2.672






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ

C1  1.222 C2  0.710







 

C6  R1 sin α  ζ  R2 sin ζ 2

C6  1.232

2

A1  C3  C4

A1  18.405

A2  C3 C6  C4 C5

A2  4.613

A3  C4 C6  C3 C5

A3  11.748

A4  C2 C3  C1 C4

A4  3.343

A5  C4 C5  C3 C6

A5  4.613

A6  C1 C3  C2 C4

A6  5.059

K1  A2  A4  A3  A6

K1  44.009

K2  A3  A4  A5  A6

K2  62.605

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

K3  71.350

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  33.700 deg

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  76.089 deg


β  β

 A5  sin β  A3  cos β  A6   A1  

β  47.808 deg

 A3  sin β  A2  cos β  A4   A1  

β  47.808 deg



β  β

Use the negative value, 5.


R1x  5.900

R1y  O4y

R2x  R1x  P21x

R2x  1.400

R2y  R1y  P21y

R2y  7.000

R3x  R1x  P31x

R3x  1.700

R3y  R1y  P31y

R3y  6.100

R1 

2

2

R1x  R1y

R1  7.799

R1y  5.100


6.

7.


2

2

R2  7.139

2

2

R3  6.332

R2 

R2x  R2y

R3 

R3x  R3y


ζ  139.160  deg

ζ  atan2 R2x R2y

ζ  101.310  deg

ζ  atan2 R3x R3y

ζ  74.427 deg






















 

C3  3.779

 

C4  1.417

 

C5  2.506

 

C6  0.250






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C1  0.883 C2  1.393

A1  C3  C4

A1  16.286

A2  C3 C6  C4 C5

A2  4.496

A3  C4 C6  C3 C5

A3  9.117

A4  C2 C3  C1 C4

A4  4.011

A5  C4 C5  C3 C6

A5  4.496

A6  C1 C3  C2 C4

A6  5.310

K1  A2  A4  A3  A6

K1  30.381

K2  A3  A4  A5  A6

K2  60.441

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

2

K3  58.811

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  33.700 deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  92.928 deg




γ  

 A5  sin γ  A3  cos γ  A6   A1  

  55.029 deg

 A3  sin γ  A2  cos γ  A4   A1  

  55.029 deg

  acos

  asin

γ  

Use the negative value , 8.


p 21 

2

P21x  P21y

p 21  4.885

δ  atan2 P21x P21y 2

p 31 

δ  22.891 deg

2

P31x  P31y

p 31  7.666

δ  atan2 P31x P31y 9.

δ  7.496  deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 A F AA   B G 

 

C  cos α  1

 

M  p 21  sin δ

B C D 

 E  L CC    M  N   

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x   W1y   AA  1 CC  Z1x   Z1y   

10. The components of the W and Z vectors are: W1x  5.043 11. The length of link 2 is:

w 

W1y  3.126 2

Z1x  2.143

2

W1x  W1y

Z1y  1.974

w  5.933


 

A'  cos γ  1

 

D  sin α

 

B'  sin γ

 

E  p 21  cos δ

 

C  cos α  1

 

F'  cos γ  1



 

 

G'  sin γ

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

H  cos α  1

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   


U1y  2.998 2

u 

U1x  U1y

S1x  3.127

2

S1y  2.102

u  4.083


V1x  5.271

V1y  Z1y  S1y

V1y  0.128 v 


2

2

V1x  V1y

v  5.272

G1x  W1x  V1x  U1x

G1x  3.000

G1y  W1y  V1y  U1y

G1y  2.176  10

g 


2

G1x  G1y

2

 14

g  3.000


O2x  2.900

O2y  Z1y  W1y

O2y  5.100

O4x  S1x  U1x

O4x  5.900

O4y  S1y  U1y

O4y  5.100


2

z  2.914

2

2

s  3.768


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y


ψ  146.092  deg

ϕ  atan2( Z1x Z1y )

ϕ  42.651 deg

rP  z




θ  1.392  deg

δp  ϕ  θ

δp  44.042 deg


g  3.000

Link 2:

w  5.933

Link 3:

v  5.272

Link 4:

u  4.083

Coupler point:

rP  2.914

δp  44.042 deg



DESIGN OF MACHINERY - 5th Ed,




Given:


P1y  0.0

P3x  1.750

P3y  2.228

P2x  0.743

P2y  1.514


θP1  62.59  deg

θP2  68.25  deg

θP3  90.0 deg


2.

3.

4.

C1y  0.0

D1x  1.036

D1y  1.998



2

p 21  1.686

δ  atan2 P2x P2y

δ  116.140 deg

2

2

p 31  2.833

δ  atan2 P3x P3y

δ  128.148 deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  5.660 deg

α  θP3  θP1

α  27.410 deg


P1x  C1x2   P1y  C1y 2

z  0.000

s 

P1x  D1x2   P1y  D1y 2

s  2.251

v 

 C1x  D1x2   C1y  D1y2

v  2.251

ϕ  θP1

ϕ  62.590 deg

ψ  θP1  π

ψ  242.590 deg


Z1x  0.000


Z1y  0.000

S 1x  s cos θP1

S 1x  1.036

S 1y  s sin θP1

S 1y  1.998

Use equations 5.24 to solve for w, , 2, and 3. Since the points C and D are to be used as pivots, z and  are known from the calculations above. Use guess values from a graphical solution such as that for Problem 3-61.



Guess:

W1x  3

W1y  0.5

β  33.3 deg

Given


β  57.0 deg

       

   

 

       

   

 

       

   

 

       

   

 


 W1x   W1y     Find  W1x W1y β β  β   β  β  33.028 deg

β  57.045 deg

The components of the W vector are: W1x  2.925 The length of link 2 is: w  5.


W1y  0.496

θ  9.626 deg

 W1x2  W1y2 , w  2.967  


U1x  3.2

U1y  0.8

γ  32.3 deg

Given


       

   

 

       

   

 

       

   

 

       

   

 

γ  68.4 deg


 U1x   U1y     Find  U1x U1y γ γ  γ   γ    γ  32.559 deg

γ  68.259 deg



The components of the U vector are: U1x  3.223

U1y  0.815


Link 1:

V1x  1.036

V1y  Z1y  S 1y

V1y  1.998

θ  atan2 V1x V1y

θ  62.590 deg

2

2

V1x  V1y

v  2.251

G1x  W1x  V1x  U1x

G1x  0.738

G1y  W1y  V1y  U1y

G1y  1.679

θ  atan2 G1x G1y

θ  66.275 deg

2

g 

9.

 U1x2  U1y2 , u  3.324  

V1x  Z1x  S 1x

v 

8.

σ  14.189 deg


7.


2

G1x  G1y

g  1.834


θ2i  θ  θ

θ2i  56.650 deg

θ2f  θ2i  β

θ2f  0.395 deg


δp  ϕ  θ

rp  0.000

δp  0.000 deg


O2x  2.925

O2y  z sin ϕ  w sin θ

O2y  0.496

O4x  s cos ψ  u  cos σ

O4x  2.187

O4y  s sin ψ  u  sin σ

O4y  1.183


θrot  atan2 O4x  O2x  O4y  O2y 11. Determine the Grashof condition.

θrot  66.275 deg






w  2.967

θ  9.626 deg

Link 3:

v  2.251

θ  62.590 deg

Link 4:

u  3.324

σ  14.189 deg

Link 1:

g  1.834

θ  66.275 deg

Coupler:

rp  0.000

δp  0.000 deg

Crank angles:

θ2i  56.650 deg

D3

θ2f  0.395 deg

B3 3.323

D2

B2

4 5

4

5

C3 1.835

O6


C2

O4 6


4

6

D1

B1 3 5 C1

6

1.403 A3 O2 2

2.967

6.347

A1





Given:


P1y  0.0

P3x  1.750

P3y  2.228

P2x  0.743

P2y  1.514


θP1  62.59  deg

θP2  68.25  deg

θP3  90.0 deg


2.

3.

4.

C1y  0.0

E1x  1.036

E1y  0.000



2

p 21  1.686

δ  atan2 P2x P2y

δ  116.140 deg

2

2

p 31  2.833

δ  atan2 P3x P3y

δ  128.148 deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  5.660 deg

α  θP3  θP1

α  27.410 deg


P1x  C1x2   P1y  C1y 2

z  0.000

s 

P1x  E1x 2  P1y  E1y 2

s  1.036

v 

 C1x  E1x 2  C1y  E1y2

v  1.036

ϕ  atan2 E1x  C1x E1y  C1y

ϕ  0.000 deg

ψ  ϕ  π

ψ  180.000 deg


Z1x  0.000


Z1y  0.000

S 1x  s cos ψ

S 1x  1.036

S 1y  s sin ψ

S 1y  0.000


W1x  3

W1y  0.5

β  33.3 deg

β  57.0 deg


Given


       

   

 

       

   

 

       

   

 

       

   

 



β  57.045 deg



W1y  0.496

θ  9.626 deg

 W1x2  W1y2 , w  2.967  


U1x  3.2

U1y  0.8

γ  32.3 deg

Given


       

   

 

       

   

 

       

   

 

       

   

 

γ  68.4 deg


 U1x   U1y     Find  U1x U1y γ γ  γ   γ  γ  22.323 deg

γ  41.857 deg


U1y  1.088


σ  13.676 deg



V1x  Z1x  S 1x

V1x  1.036

V1y  Z1y  S 1y

V1y  0.000

θ  atan2 V1x V1y

θ  0.000 deg

v  Link 1:

2

9.

2

V1x  V1y

v  1.036

G1x  W1x  V1x  U1x

G1x  0.509

G1y  W1y  V1y  U1y

G1y  0.591

θ  atan2 G1x G1y

θ  130.699 deg

2

g 

8.

 U1x2  U1y2 , u  4.600  


7.


2

G1x  G1y

g  0.780


θ2i  θ  θ

θ2i  140.325 deg

θ2f  θ2i  β

θ2f  197.370 deg


δp  ϕ  θ

rp  0.000

δp  0.000 deg


O2x  2.925

O2y  z sin ϕ  w sin θ

O2y  0.496

O4x  s cos ψ  u  cos σ

O4x  3.434

O4y  s sin ψ  u  sin σ

O4y  1.088



θrot  130.699 deg






w  2.967

θ  9.626 deg

Link 3:

v  1.036

θ  0.000 deg

Link 4:

u  4.600

σ  13.676 deg

Link 1:

g  0.780

θ  130.699 deg

Coupler:

rp  0.000

δp  0.000 deg

Crank angles:

D3

θ2i  140.325 deg θ2f  197.370 deg

D2 E3


C3

E2

D1

C2 0.780


C1 O2

E1

O4 2.967 4.600




Given:

Solution: 1.

2.

Design a fourbar linkage to carry the object in Figure P5-10 through the three positions shown in their numbered order using the fixed pivots shown. Determine the range of the transmission angle. Add a driver dyad with a crank to control the motion of your fourbar so that it cannot move beyond positions 1 and 3. P21x  0.743

P21y  1.514

P31x  1.750

P31y  2.228

O2x  3.100

O2y  1.200

O4x  0.100

O4y  1.200

Body angles:

θP1  62.59  deg

θP2  68.25  deg

θP3  90.0 deg



α  5.660 deg

α  θP3  θP1

α  27.410 deg


3.

4.

R1x  3.100

R1y  O2y

R2x  R1x  P21x

R2x  2.357

R2y  R1y  P21y

R2y  2.714

R3x  R1x  P31x

R3x  1.350

R3y  R1y  P31y

R3y  3.428

2

2

R1  3.324

2

2

R2  3.595

2

2

R3  3.684

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  1.200


ζ  21.161 deg

ζ  atan2 R2x R2y

ζ  49.027 deg

ζ  atan2 R3x R3y

ζ  68.505 deg






















 

C3  0.850

 

C4  0.936

 

C5  0.610






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ

C1  0.162 C2  0.050







 

C6  R1 sin α  ζ  R2 sin ζ 2

C6  1.214

2

A1  C3  C4

A1  1.597

A2  C3 C6  C4 C5

A2  0.461

A3  C4 C6  C3 C5

A3  1.654

A4  C2 C3  C1 C4

A4  0.194

A5  C4 C5  C3 C6

A5  0.461

A6  C1 C3  C2 C4

A6  0.091

K1  A2  A4  A3  A6

K1  0.061

K2  A3  A4  A5  A6

K2  0.364

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

K3  0.221

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  27.410 deg

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  133.549 deg

The first value is the same as 3 so use the second value

β  β

 A5  sin β  A3  cos β  A6   A1  

β  124.137 deg

 A3  sin β  A2  cos β  A4   A1  

β  55.863 deg



β  β

Use the first value, 5.


R1x  0.100

R1y  O4y

R2x  R1x  P21x

R2x  0.643

R2y  R1y  P21y

R2y  2.714

R3x  R1x  P31x

R3x  1.650

R3y  R1y  P31y

R3y  3.428

R1 

2

2

R1x  R1y

R1  1.204

R1y  1.200


6.

7.


2

2

R2  2.789

2

2

R3  3.804

R2 

R2x  R2y

R3 

R3x  R3y


ζ  85.236 deg

ζ  atan2 R2x R2y

ζ  103.329 deg

ζ  atan2 R3x R3y

ζ  115.703 deg






















 

C3  1.186

 

C4  2.317

 

C5  0.624

 

C6  1.510






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C1  0.160 C2  1.135

A1  C3  C4

A1  6.774

A2  C3 C6  C4 C5

A2  0.345

A3  C4 C6  C3 C5

A3  4.239

A4  C2 C3  C1 C4

A4  0.977

A5  C4 C5  C3 C6

A5  0.345

A6  C1 C3  C2 C4

A6  2.820

K1  A2  A4  A3  A6

K1  12.289

K2  A3  A4  A5  A6

K2  3.165

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

2

K3  9.452

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  27.410 deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  56.299 deg




γ  

 A5  sin γ  A3  cos γ  A6   A1  

  43.866 deg

 A3  sin γ  A2  cos γ  A4   A1  

  43.866 deg

  acos

  asin

γ  

Use the negative value , 8.


p 21 

2

P21x  P21y

p 21  1.686

δ  atan2 P21x P21y 2

p 31 

δ  116.140 deg

2

P31x  P31y

p 31  2.833

δ  atan2 P31x P31y 9.

δ  128.148 deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 A F AA   B G 

 

C  cos α  1

 

M  p 21  sin δ

 E  L CC    M  N   

B C D 

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x   W1y   AA  1 CC  Z1x   Z1y   



w 

W1y  0.489 2

Z1x  2.479

2

W1x  W1y

Z1y  1.689

w  0.790


 

A'  cos γ  1

 

D  sin α

 

B'  sin γ

 

E  p 21  cos δ

 

C  cos α  1

 

F'  cos γ  1



 

 

G'  sin γ

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

H  cos α  1

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   



U1y  1.294 2

u 

U1x  U1y

S1x  1.559

2

S1y  2.494

u  1.950


V1x  0.920

V1y  Z1y  S1y

V1y  0.805 v 


2

2

V1x  V1y

v  1.222

G1x  W1x  V1x  U1x

G1x  3.000

G1y  W1y  V1y  U1y

G1y  0.000

g 


2

G1x  G1y

2

g  3.000


O2x  3.100

O2y  Z1y  W1y

O2y  1.200

O4x  S1x  U1x

O4x  0.100

O4y  S1y  U1y

O4y  1.200


2

z  3.000

2

2

s  2.941


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y


ψ  57.983 deg

ϕ  atan2( Z1x Z1y )

ϕ  34.270 deg

rP  z




θ  41.182 deg

δp  ϕ  θ

δp  75.452 deg


g  3.000

Link 2:

w  0.790

Link 3:

v  1.222

Link 4:

u  1.950

Coupler point:

rP  3.000

δp  75.452 deg

19. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct. However, this mechanism as designed has a branch defect. The three positions can only be reached by changing the links from a crossed circuit to an open circuit.





Given:


P1y  0.0

P3x  2.751

P3y  2.015

P2x  2.332

P2y  0.311


θP1  45.0 deg

θP2  24.14  deg

θP3  86.84  deg


2.

3.

4.

C1y  0.0

D1x  1.591

D1y  1.591



2

p 21  2.353

δ  atan2 P2x P2y

δ  172.404 deg

2

2

p 31  3.410

δ  atan2 P3x P3y

δ  143.779 deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  20.860 deg

α  θP3  θP1

α  41.840 deg


P1x  C1x2   P1y  C1y 2

z  0.000

s 

P1x  D1x2   P1y  D1y 2

s  2.250

v 

 C1x  D1x2   C1y  D1y2

v  2.250

ϕ  θP1

ϕ  45.000 deg

ψ  θP1  π

ψ  225.000 deg


Z1x  0.000


Z1y  0.000

S 1x  s cos θP1

S 1x  1.591

S 1y  s sin θP1

S 1y  1.591

Use equations 5.24 to solve for w, , 2, and 3. Since the points C and D are to be used as pivots, z and  are known from the calculations above. Use guess values from a graphical solution such as that for Problem 3-65.



Guess:

W1x  1.75 W1y  0.47

β  81 deg

β  138  deg

Given


       

   

 

       

   

 

       

   

 

       

   

 



β  137.178 deg



W1y  1.547

θ  57.632 deg

 W1x2  W1y2 , w  1.831  


U1x  0.1

U1y  7.0

γ  17.5 deg

Given


       

   

 

       

   

 

       

   

 

       

   

 

γ  37 deg



γ  17.457 deg

γ  37.139 deg




U1y  5.598


Link 1:

V1x  1.591

V1y  Z1y  S 1y

V1y  1.591

θ  atan2 V1x V1y

θ  45.000 deg

2

2

V1x  V1y

v  2.250

G1x  W1x  V1x  U1x

G1x  1.561

G1y  W1y  V1y  U1y

G1y  8.736

θ  atan2 G1x G1y

θ  100.130 deg

2

g 

9.

 U1x2  U1y2 , u  6.958  

V1x  Z1x  S 1x

v 

8.

σ  53.567 deg


7.


2

G1x  G1y

g  8.874


θ2i  θ  θ

θ2i  42.498 deg

θ2f  θ2i  β

θ2f  94.681 deg


δp  ϕ  θ

rp  0.000

δp  0.000 deg


O2x  0.980

O2y  z sin ϕ  w sin θ

O2y  1.547

O4x  s cos ψ  u  cos σ

O4x  2.541

O4y  s sin ψ  u  sin σ

O4y  7.189



θrot  100.130 deg






w  1.831

θ  57.632 deg

Link 3:

v  2.250

θ  45.000 deg

Link 4:

u  6.958

σ  53.567 deg

Link 1:

g  8.874

θ  100.130 deg

Coupler:

rp  0.000

δp  0.000 deg

Crank angles:

θ2i  42.498 deg

7.646

θ2f  94.681 deg

O4 4

O2 A3

6.958 8.874

B1

4 B3

2 1.593

D1 D2


3

A1

D3 5

C2

C3

6

5 C1

6 O6 1.831






Given:

Coordinates of the points P1 , P2 and P3 with respect to C1: P1x  0.0 P1y  0.0 P2x  2.332 P3x  2.751

P2y  0.311

P3y  2.015


θP1  45.0 deg

θP2  24.14  deg

θP3  86.84  deg


2.

3.

4.

C1y  0.0

E1x  1.591

E1y  0.000



2

p 21  2.353

δ  atan2 P2x P2y

δ  172.404 deg

2

2

p 31  3.410

δ  atan2 P3x P3y

δ  143.779 deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  20.860 deg

α  θP3  θP1

α  41.840 deg


P1x  C1x2   P1y  C1y 2

z  0.000

s 

P1x  E1x 2  P1y  E1y 2

s  1.591

v 

 C1x  E1x 2  C1y  E1y2

v  1.591

ϕ  atan2 E1x  C1x E1y  C1y

ϕ  0.000 deg

ψ  ϕ  π

ψ  180.000 deg


Z1x  0.000


Z1y  0.000

S 1x  s cos ψ

S 1x  1.591

S 1y  s sin ψ

S 1y  0.000


W1x  1.75 W1y  0.47

β  81 deg

β  138  deg


Given


       

   

 

       

   

 

       

   

 

       

   

 



β  137.178 deg



W1y  1.547

θ  57.632 deg

 W1x2  W1y2 , w  1.831  


U1x  1

U1y  5.0

γ  17.5 deg

Given


       

   

 

       

   

 

       

   

 

       

   

 

γ  37 deg


 U1x   U1y     Find  U1x U1y γ γ  γ   γ  γ  21.548 deg

γ  27.543 deg


U1y  5.963


σ  59.417 deg



V1x  Z1x  S 1x

V1x  1.591

V1y  Z1y  S 1y

V1y  0.000

θ  atan2 V1x V1y

θ  0.000 deg

v  Link 1:

2

9.

2

V1x  V1y

v  1.591

G1x  W1x  V1x  U1x

G1x  0.952

G1y  W1y  V1y  U1y

G1y  7.510

θ  atan2 G1x G1y

θ  97.228 deg

2

g 

8.

 U1x2  U1y2 , u  6.926  


7.


2

G1x  G1y

g  7.570


θ2i  θ  θ

θ2i  39.596 deg

θ2f  θ2i  β

θ2f  97.582 deg


δp  ϕ  θ

rp  0.000

δp  0.000 deg


O2x  0.980

O2y  z sin ϕ  w sin θ

O2y  1.547

O4x  s cos ψ  u  cos σ

O4x  1.933

O4y  s sin ψ  u  sin σ

O4y  5.963


θrot  atan2 O4x  O2x  O4y  O2y 11. Determine the Grashof condition. Condition( a b c d ) 


θrot  97.228 deg




w  1.831

θ  57.632 deg

Link 3:

v  1.591

θ  0.000 deg

Link 4:

u  6.926

σ  59.417 deg

Link 1:

g  7.570

θ  97.228 deg

Coupler:

rp  0.000

δp  0.000 deg

Crank angles:

θ2i  39.596 deg θ2f  97.582 deg

O4 6.926



7.570 D1 D2

D3 C2 C3

C1

E1

O2 1.831




Given:

Solution: 1.

2.

Design a fourbar linkage to carry the object in Figure P5-11 through the three positions shown in their numbered order using the fixed pivots shown. Determine the range of the transmission angle. P21x  2.332

P21y  0.311

P31x  2.751

P31y  2.015

O2x  3.679

O2y  3.282

O4x  0.321

O4y  3.282

Body angles:

θP1  45.0 deg

θP2  24.14  deg

θP3  86.84  deg



α  20.860 deg

α  θP3  θP1

α  41.840 deg


3.

4.

R1x  3.679

R1y  O2y

R2x  R1x  P21x

R2x  1.347

R2y  R1y  P21y

R2y  2.971

R3x  R1x  P31x

R3x  0.928

R3y  R1y  P31y

R3y  1.267

2

2

R1  4.930

2

2

R2  3.262

2

2

R3  1.571

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  3.282


ζ  41.736 deg

ζ  atan2 R2x R2y

ζ  65.611 deg

ζ  atan2 R3x R3y

ζ  53.780 deg






















 

C3  0.376

 

C4  3.632

 

C5  3.260

 

C6  1.214






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ

C1  2.297 C2  2.258



A1  C3  C4

A1  13.335

A2  C3 C6  C4 C5

A2  11.382

A3  C4 C6  C3 C5

A3  5.637

A4  C2 C3  C1 C4

A4  7.492

A5  C4 C5  C3 C6

A5  11.382

A6  C1 C3  C2 C4

A6  9.068

K1  A2  A4  A3  A6

K1  136.387

K2  A3  A4  A5  A6

K2  60.979

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

K3  60.933

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  90.019 deg

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  41.840 deg

The second value is the same as 3 so use the first value

β  β

 A5  sin β  A3  cos β  A6   A1  

β  99.989 deg

 A3  sin β  A2  cos β  A4   A1  

β  80.011 deg



β  β

Use the first value, 5.


R1x  0.321

R1y  O4y

R2x  R1x  P21x

R2x  2.653

R2y  R1y  P21y

R2y  2.971

R3x  R1x  P31x

R3x  3.072

R3y  R1y  P31y

R3y  1.267

2

2

R1  3.298

2

2

R2  3.983

R1 

R1x  R1y

R2 

R2x  R2y

R1y  3.282


R3  6.

7.


2

2

R3x  R3y

R3  3.323


ζ  95.586 deg

ζ  atan2 R2x R2y

ζ  131.764 deg

ζ  atan2 R3x R3y

ζ  157.587 deg






















 

C3  0.644

 

C4  0.964

 

C5  3.522

 

C6  0.210






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C1  1.539 C2  1.834

A1  C3  C4

A1  1.343

A2  C3 C6  C4 C5

A2  3.260

A3  C4 C6  C3 C5

A3  2.469

A4  C2 C3  C1 C4

A4  0.303

A5  C4 C5  C3 C6

A5  3.260

A6  C1 C3  C2 C4

A6  2.758

K1  A2  A4  A3  A6

K1  7.799

K2  A3  A4  A5  A6

K2  8.243

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

2

K3  11.309

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  41.840 deg

 K  K 2  K 2  K 2  2 1 2 3   K1  K3  

  51.335 deg

  2  atan


γ  



 A5  sin γ  A3  cos γ  A6   A1  

  8.321 deg

 A3  sin γ  A2  cos γ  A4   A1  

  8.321 deg

  acos

  asin

γ  

Both are the same so use the first value , 8.


p 21 

2

P21x  P21y

p 21  2.353

δ  atan2 P21x P21y 2

p 31 

δ  172.404 deg

2

P31x  P31y

p 31  3.410

δ  atan2 P31x P31y 9.

δ  143.779 deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 A F AA   B G 

 

C  cos α  1

 

M  p 21  sin δ

B C D 

 E  L CC    M  N   

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x   W1y   AA  1 CC  Z1x   Z1y   



w 

W1y  1.000 2

Z1x  1.947

2

W1x  W1y

Z1y  2.282

w  2.000


 

A'  cos γ  1

 

B'  sin γ

 

E  p 21  cos δ

 

H  cos α  1

D  sin α

G'  sin γ

 

 

 

C  cos α  1

 

F'  cos γ  1

 

K  sin α



 

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   


U1y  6.903 2

u 

U1x  U1y

S1x  0.857

2

S1y  3.621

u  7.002


V1x  1.090

V1y  Z1y  S1y

V1y  5.903 v 


2

2

V1x  V1y

v  6.002

G1x  W1x  V1x  U1x

G1x  4.000

G1y  W1y  V1y  U1y

G1y  1.776  10

g 


2

G1x  G1y

2

 15

g  4.000


O2x  3.679

O2y  Z1y  W1y

O2y  3.282

O4x  S1x  U1x

O4x  0.321

O4y  S1y  U1y

O4y  3.282


2

z  3.000

2

2

s  3.721


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y


ψ  76.683 deg

ϕ  atan2( Z1x Z1y )

ϕ  49.525 deg

rP  z

θ  atan2 z cos ϕ  s cos ψ z sin ϕ  s sin ψ  θ  79.535 deg

δp  ϕ  θ

δp  30.010 deg




g  4.000

Link 2:

w  2.000

Link 3:

v  6.002

Link 4:

u  7.002

Coupler point:

rP  3.000

δp  30.010 deg






Write a program to generate and plot the circle-point and center-point circles for Problem 5-40 using an equation solver or any program language. P21x  4.500

P21y  1.900

P31x  7.600

P31y  1.000

O2x  2.900

O2y  5.100

O4x  5.900

O4y  5.100

Body angles:

θP1  33.70  deg

θP2  14.60  deg

θP3  0.0 deg

Assumptions: Let the position 1 to position 2 rotation angles be: β  47.808 deg and γ  55.029 deg Let the position 1 to position 2 coupler rotation angle be: α  19.100 deg Solution: 1.



2

2

P21x  P21y

p 21  4.885

δ  atan2 P21x P21y p 31 

2

δ  22.891 deg

2

P31x  P31y

p 31  7.666

δ  atan2 P31x P31y 2.

δ  7.496 deg

Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α  33.700 deg

β  0  deg 0.5 deg  360  deg

 

B  sin β

 

 

E  p 21  cos δ

A  cos β  1 D  sin α

 

 

 

C  cos α  1

 

F β  cos β  1

 


 

N  p 31  sin δ

 

H α  cos α  1

 

M  p 21  sin δ

G β  sin β L  p 31  cos δ

 

 

 

   

B C D   A   F  β G β H  α K α   AA  α β   B D A C     G β F  β K α H  α 

 E  L CC    M  N   

 1 CC















W1y α β  DD α β 2









Z1y α β  DD α β 4

DD α β  AA α β

W1x α β  DD α β 1 Z1x α β  DD α β 3 3.






















W1x α 76.089 deg  5.043





W1y α 76.089 deg  3.126









Z1x α 76.089 deg  2.143



Z1y α 76.089 deg  1.974




























Plot the center-point circle for the WZ dyad. Center-Point Circle for WZ Dyad 4

6

8





Ny α β 

 10

 12

 14

0

2

4





6

8

10

Nx α β 

4.

Form the vector Z, whose tip describes the circle-point circle for the WZ dyad. β  76.089 deg





α  0  deg 1  deg  360  deg





Zx α β  Z1x α β

5.









Zy α β  Z1y α β




Circle-Point Circle for WZ Dyad 2

1

0



Zy α β 

 1

2

3 4

3

2



Zx α β 

6.



1

0

1

Evaluate terms in the US coefficient matrix and constant vector from equations (5.31) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α  33.700 deg

γ  0  deg 1  deg  360  deg

 

B  sin γ

 

 

E  p 21  cos δ

A  cos γ  1 D  sin α

 

 

 

C  cos α  1

 

F γ  cos γ  1

 


 

N  p 31  sin δ

 

H α  cos α  1

 

M  p 21  sin δ

G γ  sin γ L  p 31  cos δ

 

 

 

   

B C D   A   F  γ G γ H  α K α   AA  α γ   B D A C     G γ F  γ K α H  α 

 E  L CC    M  N   

 1 CC















U1y α γ  DD α γ 2









S1y α γ  DD α γ 4

DD α γ  AA α γ























U1x α 92.928 deg  2.773





U1y α 92.928 deg  2.998









S1x α 92.928 deg  3.128



S1y α 92.928 deg  2.102




























Plot the center-point circle for the US dyad. Center-Point Circle for US Dyad 4

6

8




  10

 12

 14

0

2

4



6



8

10


10. Form the vector S, whose tip describes the circle-point circle for the US dyad. γ  92.928 deg





α  0  deg 1  deg  360  deg





Sx α γ  S1x α γ










Sy α γ  S1y α γ



Circle-Point Circle for the US Dyad 1

0

1




 2

3

4 1

0

1






2

3

4




Write a program to generate and plot the circle-point and center-point circles for Problem 5-43 using an equation solver or any program language. P21x  0.743

P21y  1.514

P31x  1.750

P31y  2.228

O2x  3.100

O2y  1.200

O4x  0.100

O4y  1.200

Body angles:

θP1  62.59  deg

θP2  68.25  deg

θP3  90.0 deg

Given:

Assumptions: Let the position 1 to position 2 rotation angles be: β  124.137  deg and γ  43.866 deg Let the position 1 to position 2 coupler rotation angle be: α  5.660  deg Solution: 1.



2

2

P21x  P21y

p 21  1.686

δ  atan2 P21x P21y p 31 

2

δ  116.140 deg

2

P31x  P31y

p 31  2.833

δ  atan2 P31x P31y 2.

δ  128.148 deg

Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α  27.410 deg

β  0  deg 0.5 deg  360  deg

 

B  sin β

 

 

E  p 21  cos δ

A  cos β  1 D  sin α

 

 

 

C  cos α  1

 

F β  cos β  1

 


 

N  p 31  sin δ

 

H α  cos α  1

 

M  p 21  sin δ

G β  sin β L  p 31  cos δ

 

 

 

   


 E  L CC    M  N   

 1 CC















W1y α β  DD α β 2









Z1y α β  DD α β 4

DD α β  AA α β

W1x α β  DD α β 1 Z1x α β  DD α β 3 3.






















W1x α 133.549  deg  0.621





W1y α 133.549  deg  0.489









Z1x α 133.549  deg  2.479



Z1y α 133.549  deg  1.689





























0





Ny α β   1

2

3 7

6



5

4



3

Nx α β 

4.

Form the vector Z, whose tip describes the circle-point circle for the WZ dyad. β  133.549  deg





α  8  deg 9  deg  364  deg



Zx α β3  Z1x α β

5.



Plot the circle-point arc for the WZ dyad (see next page).









Zy α β  Z1y α β




10

0



Zy α β 

  10

 20

 30  20

 10

0



Zx α β 

6.



10

20

30

Evaluate terms in the US coefficient matrix and constant vector from equations (5.31) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α  27.410 deg

γ  0  deg 1  deg  360  deg

 

B  sin γ

 

 

E  p 21  cos δ

A  cos γ  1 D  sin α

 

 

 

C  cos α  1

 

F γ  cos γ  1

 


 

N  p 31  sin δ

 

H α  cos α  1

 

M  p 21  sin δ

G γ  sin γ L  p 31  cos δ

 

 

 

   


 E  L CC    M  N   

 1 CC















U1y α γ  DD α γ 2









S1y α γ  DD α γ 4

DD α γ  AA α γ























U1x α 56.299 deg  1.459





U1y α 56.299 deg  1.294









S1x α 56.299 deg  1.559



S1y α 56.299 deg  2.494





























2

4




 6

8

 10 8

6

4





2

0

2


10. Form the vector S, whose tip describes the circle-point circle for the US dyad. γ  56.299 deg





α  10 deg 11 deg  363  deg





Sx α γ  S1x α γ

11. Plot the circle-point arc for the WZ dyad (see next page).









Sy α γ  S1y α γ




10






0

 10

 20  15

 10



5




0

5




Write a program to generate and plot the circle-point and center-point circles for Problem 5-46 using an equation solver or any program language.

Given: P21x  2.332

P21y  0.311

P31x  2.751

P31y  2.015

O2x  3.679

O2y  3.282

O4x  0.321

O4y  3.282

Body angles:

θP1  45.0 deg

θP2  24.14  deg

θP3  86.84  deg

Assumptions: Let the position 1 to position 2 rotation angles be: β  99.989 deg and γ  8.321  deg Let the position 1 to position 2 coupler rotation angle be: α  20.860 deg Solution: 1.



2

2

P21x  P21y

p 21  2.353

δ  atan2 P21x P21y p 31 

2

δ  172.404 deg

2

P31x  P31y

p 31  3.410

δ  atan2 P31x P31y 2.

δ  143.779 deg

Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α  41.840 deg

β  0  deg 1  deg  360  deg

 

B  sin β

 

 

E  p 21  cos δ

A  cos β  1 D  sin α

 

 

 

C  cos α  1

 

F β  cos β  1

 


 

N  p 31  sin δ

 

H α  cos α  1

 

M  p 21  sin δ

G β  sin β L  p 31  cos δ

 

 

 

   


 E  L CC    M  N   

 1 CC















W1y α β  DD α β 2









Z1y α β  DD α β 4

DD α β  AA α β



















3.







W1y α 90.019 deg  1.000





Z1y α 90.019 deg  2.282

W1x α 90.019 deg  1.732 Z1x α 90.019 deg  1.947





































2





Ny α β   4

6

8 4

2



0



2

Nx α β 

4.

Form the vector Z, whose tip describes the circle-point circle for the WZ dyad. β  90.019 deg





α  0  deg 1  deg  360  deg





Zx α β  Z1x α β

5.









Zy α β  Z1y α β





2



Zy α β 



0

2

4  10

8

6



Zx α β 

6.



4

2

0


γ  0  deg 1  deg  360  deg

 

B  sin γ

 

 

E  p 21  cos δ

A  cos γ  1 D  sin α

 

 

 

C  cos α  1

 

F γ  cos γ  1

 


 

N  p 31  sin δ

 

H α  cos α  1

 

M  p 21  sin δ

G γ  sin γ L  p 31  cos δ

 

 

 

   


 E  L CC    M  N   

 1 CC















U1y α γ  DD α γ 2









S1y α γ  DD α γ 4

DD α γ  AA α γ























U1x α 51.335 deg  1.178





U1y α 51.335 deg  6.903









S1x α 51.335 deg  0.857



S1y α 51.335 deg  3.621





























0





My α γ  20

 40

 60  60

 40

 20



0



20


10. Form the vector S, whose tip describes the circle-point circle for the US dyad. γ  51.335 deg





α  0  deg 1  deg  360  deg





Sx α γ  S1x α γ










Sy α γ  S1y α γ




6

5




 4

3

2 2

1

0






1

2

3




In Example 5-2 the precision points and rotation angles are specified while the input and output rotation angles  and  are free choices. Using the choices given for 2 and 2, determine the radii and center coordinates of the center-point circles for O2 and O4. Plot those circles (or portions of them) and show that the choices of 3 and 3 give a solution that falls on the center-point circles.

Given: P21x  2.394

P21y  1.449

P31x  3.761

P31y  1.103

O2x  1.234

O2y  7.772

O4x  2.737

O4y  0.338

Body angles:

θP1  38.565 deg

θP2  6.435  deg

θP3  47.865 deg

Assumptions: Let the position 1 to position 2 rotation angles be: β  342.3  deg and γ  30.9 deg Let the position 1 to position 2 coupler rotation angle be: α  45.0 deg Solution: 1.



2

2

P21x  P21y

p 21  2.798

δ  atan2 P21x P21y p 31 

2

δ  31.185 deg

2

P31x  P31y

p 31  3.919

δ  atan2 P31x P31y 2.

δ  16.345 deg

Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. Since this is a non-Grashof linkage, the range of 3 is limited to approximately -56 deg < 3 < -3 deg. α  9.3 deg

β  56 deg 55 deg  3  deg

 

B  sin β

 

 

E  p 21  cos δ

A  cos β  1 D  sin α

 

 

 

C  cos α  1

 

F β  cos β  1

 


 

N  p 31  sin δ

 

H α  cos α  1

 

M  p 21  sin δ

G β  sin β L  p 31  cos δ

 

 

 

   

B C D   A  F β G β H α K α           AA  α β    B A D C  G β F β K α H α           

 E  L CC    M  N   

 1 CC















W1y α β  DD α β 2









Z1y α β  DD α β 4

DD α β  AA α β



















3.


Check this against the solutions in Example 5-2:





W1y α 324.8  deg  6.832





Z1y α 324.8  deg  0.940

W1x α 324.8  deg  0.055 Z1x α 324.8  deg  1.179









These are the same as the values calculated in Example 5-2. 4.



























Plot the center-point circle for the WZ dyad. Portion Center-Point Circle for WZ Dyad 5

 10





Ny α β   15

 20

 25  15

 10





5

0

Nx α β 

4.

Find the center and radius of this arc by taking three points on it and, from them, determine the radius and center point of the circle. Three points: x1  Nx α 56 deg x1  0.470

  y1  Ny α 56 deg x2  Nx α 10 deg y2  Ny α 10 deg x3  Nx α 3  deg y3  Ny α 3  deg

y1  6.430 x2  5.506 y2  14.302 x3  13.631 y3  23.695



2

a 

Form four ratios:

b 

2

d 

4.

2

a  12.278

2   y2  y1 

x2  x1

b  0.640

y2  y1 2

c 

2

y2  y1  x2  x1

2

2

y3  y2  x3  x2

2

c  32.547

2   y2  y1 

x3  x2

d  0.865

y3  y2 ac

Circle center x coordinate:

h 

h  89.999

Circle center y coordinate:

k  a  b  h

Center-point circle radius:

R 

bd

k  45.303

 x1  h  2   y1  k 2

R  103.40

Show that the point determined by 3 = 324.8 deg falls on this circle.

  yb3  Ny α 324.8  deg xb3  Nx α 324.8  deg

Coordinates of this point:

xb3  1.234 yb3  7.772

Substitute into radius equation: R 

 xb3  h 2  yb3  k2

R  103.42

Since the radius is the same, this point does fall on the center=point circle for the WZ dyad. 6.


γ  0  deg 1  deg  360  deg

 

B  sin γ

 

 

E  p 21  cos δ

A  cos γ  1 D  sin α

 

 

 

C  cos α  1

 

F γ  cos γ  1

 


 

N  p 31  sin δ

 

H α  cos α  1

 

M  p 21  sin δ

G γ  sin γ L  p 31  cos δ







 

 

   

B C D   A  F γ G γ H α K α           AA  α γ    B D A C  G γ     F  γ K α H  α 



 

 E  L CC    M  N   

 1 CC



DD α γ  AA α γ





U1x α γ  DD α γ 1









U1y α γ  DD α γ 2













S1x α γ  DD α γ 3 7.







S1y α γ  DD α γ 4

Check this against the solutions in Example 5-2:





U1y α 80.6 deg  1.825





S1y α 80.6 deg  1.487

U1x α 80.6 deg  2.628 S1x α 80.6 deg  0.109









These are the same as the values calculated in Example 5-2. 8.




























40

20






0

 20

 40

 60

0

20

40





60

80

100


10. Find the center and radius of this circle by taking three points on it and, from them, determine the radius and center point of the circle. Three points:

  y1  My α 0  deg x1  Mx α 0  deg

x1  42.228 y1  44.088



  y2  My α 90 deg x3  Mx  α 350  deg y3  My α 350  deg x2  Mx α 90 deg

2

Form four ratios:

a  b 

2

d 

y2  0.214 x3  39.862 y3  42.592 2

y2  y1  x2  x1 2   y2  y1 

x2  x1

a  41.977 b  0.891

y2  y1 2

c 

2

x2  2.744

2

2

y3  y2  x3  x2

2

2   y2  y1 

x3  x2

c  38.321 d  0.876

y3  y2 ac

Circle center x coordinate:

h 

Circle center y coordinate:

k  a  b  h

Center-point circle radius:

R 

bd

h  45.441 k  1.479

 x1  h  2   y1  k 2

R  42.73

11. Show that the point determined by 3 = 80.6 deg falls on this circle. Coordinates of this point:

  yg3  My  α 80.6 deg xg3  Mx α 80.6 deg

xg3  2.738 yg3  0.338

Substitute into radius equation: R 

 xg3  h 2  yg3  k2

R  42.72

Since the radius is the same, this point does fall on the centerpoint circle for the US dyad.




Design a driver dyad to move link 2 of Example 5-1 from position 1 to position 2 and return.

Given:

Solution to Example 5-1: Length of link 2

w  2.467


θ  71.6 deg

Rotation angle for link 2

β  38.4 deg

Coordinates of O2

O2x  0.00

Design Choice:

Solution: 1.

O2y  0.00

See Example 5-1, Figure 5-3 and Mathcad file P0551.

Link 2 of the solution to Example 5-1 will become the driven link for the driver dyad. The driver dyad will be links 5 and 6 and the fixed pivot for the dyad will be at O6. Select a point on link 2 of Example 5-1 and label it C. Let the distance O2C be R2  1.200. The solution that follows uses the algorithm presented in Section 5.2 with changes in nomenclature to account for the fact that the driven link is link 2 and the points A and B are already defined on the fourbar of Example 5-1.

2.


C1x  0.379

C1y  O2y  R2 sin θ

C1y  1.139





C2x  0.410





C2y  1.128

C2x  O2x  R2 cos θ  β C2y  O2y  R2 sin θ  β RC1 

 C1x     C1y 

RC2 

 C2x     C2y 

M  RC2  RC1

3.


4.

Determine the coordinates of the crank pivot, O6 using equation 5.0d. RO6  RC1  K M

O6x  RO6

1

O6x  1.989 5.





2

O6y  1.106

R6  0.395


R5  1.973

 O2x     O2y 

R1  RO2  RO6 7.

 0.789     0.011 

Determine the length of the driving crank using equation 5.0e. R6  R2 sin 0.5 β

6.

O6y  RO6

M

Determine the Grashof condition.

R1  2.275



R1  2.275

R2  1.200

R5  1.973

R6  0.395




Draw the linkage using the link lengths and fixed pivot coordinates calculated above to verify that the driver dyad will perform as required. Solve for the coordinates of O4 using equations 5.2: Given:

z  1.298

ϕ  26.5 deg

u  1.486

σ  15.4 deg

Z1x  z cos ϕ S 1x  s cos ψ

W1x  w cos θ U1x  u  cos σ Z1 

 Z1x     Z1y 

ψ  104.1  deg


Z1x  1.162

Z1y  0.579

S 1y  s sin ψ

S 1x  0.252

S 1y  1.004

W1y  w sin θ

W1x  0.779

W1y  2.341

U1y  u  sin σ

U1x  1.433 S1 

s  1.035

 S1x     S1y 

V1  Z1  S1

G1  W1  V1  U1

g  G1

g  1.701

U1y  0.395

 W1x     W1y   0.760  G1     1.522 

U1 

v  V1

v  1.476

P2

P1

B2 A1 A2

38.40°

B1

O6

C2

O4

° .60 71

C1

D1

D2

y

O2

 U1x     U1y 

W1 

x





Given:

Solution to Example 5-2:

Solution: 1.

Length of link 2

w  6.832


θ  atan


β  35.20  deg

Coordinates of O2

O2x  1.234

6.832 

  0.055 

θ  89.539 deg

O2y  7.772

See Example 5-2, Figure 5-5, Table 5-1, and Mathcad file P0552.


2.


C1x  1.202

C1y  O2y  R2 sin θ

C1y  3.772





C3x  1.098





C3y  4.522


 C1x     C1y 

RC3 

 C3x     C3y 

M  RC3  RC1

3.


4.


O6x  RO6

1

O6x  5.698 5.





2

O6y  6.022

R6  1.209


R5  6.047

 O2x     O2y 

R1  RO2  RO6 7.

 2.300     0.750 

Determine the length of the driving crank using equation 5.0e. R6  R2 sin 0.5 β

6.

O6y  RO6

M


R1  7.149



R1  7.149

R2  4.000

R5  6.047

R6  1.209




Draw the linkage using the link lengths and fixed pivot coordinates calculated above to verify that the driver dyad will perform as required. Solve for the vectors W, Z, U, and S. Given:

y

W1x  0.055

O4 P1

x

W1y  6.832 Z1x  1.179

A1 P3

Z1y  0.940

B1

U1x  2.628

A3

U1y  1.825

B3

S 1x  0.109 S 1y  1.487

C1

 W1x  W1     W1y 

D1

 S1x     S1y 

w  W1

° 39 .5 89

 Z1x  Z1     Z1y   U1x  U1     U1y  S1 

C3

35.2 00°

O6 D3

O2

w  6.832

u  U1

u  3.200


θ  89.539 deg


σ  145.222 deg

z  Z1

z  1.508

s  S1

s  1.491


ϕ  38.565 deg

ψ  atan2 S 1x S 1y

ψ  94.192 deg





Given:

Solution to Example 5-3:

Solution: 1.

Length of link 2

w  1.000


θ  atan


β  23.96  deg

Coordinates of O2

O2x  1.712

0.500 

  0.866 

θ  30.001 deg

O2y  0.033

See Example 5-3, Figure 5-7, Table 5-2, and Mathcad file P0553.


2.


C1x  1.279

C1y  O2y  R2 sin θ

C1y  0.283





C3x  1.418





C3y  0.437


 C1x     C1y 

RC3 

 C3x     C3y 

M  RC3  RC1

3.


4.


O6x  RO6

1

O6x  1.696 5.





2

O6y  0.746

R6  0.104


R5  0.519

 O2x     O2y 

R1  RO2  RO6 7.

 0.139     0.154 

Determine the length of the driving crank using equation 5.0e. R6  R2 sin 0.5 β

6.

O6y  RO6

M


R1  0.713



R1  0.713

R2  0.500

R5  0.519

R6  0.104




Draw the linkage using the link lengths and fixed pivot coordinates calculated above to verify that the driver dyad will perform as required. Solve for the vectors W, Z, U, and S. Given: W1x  0.866

Z1x  0.846

U1x  0.253

S 1x  0.035

W1y  0.500

Z1y  0.533

U1y  0.973

S 1y  1.006

W1 

 W1x     W1y 

Z1 

w  W1

 Z1x     Z1y 

U1 

w  1.000

 U1x     U1y 

S1 

u  U1

 S1x     S1y 

u  1.005


θ  30.001 deg


σ  104.575 deg

z  Z1

z  1.000

s  S1

s  1.007


ψ  atan2 S 1x S 1y

ϕ  32.212 deg

ψ  91.993 deg

y

B1

B3 A3

D3 O6

D1

A1

C3 C1 O2

P3

P1

O4

x




Design a fourbar linkage to carry the object in Figure P5-12 from position 1 to 2 using points C and D for your attachment points. The fixed pivots should be within the indicated area.

Given:

Coordinates of the points P1 (C1) and P2 (C2) : P1x  0.0

P1y  0.0

P2x  13.871

P2y  3.299

v  12.387

Length of the coupler (link 3):


θP1  0.0 deg

θP2  24.0 deg

Coordinates of the points C1 and D1 with respect to P1: C1y  0.0 D1x  C1x  v cos θP1

C1x  0.0

Free choice for the WZ dyad : β  45 deg

D1x  12.387 D1y  0.000

Free choice for the US dyad : γ  70 deg Solution:

D1y  C1y  v sin θP1


1.


2.


 P1x     P1y 

R2 

 P2x     P2y 

 P21x     R2  R1  P21y 

P21x  13.871 P21y  3.299

p 21  3.

4.

2

2

P21x  P21y


α  24.000 deg

δ  atan2 P21x P21y

δ  13.378 deg


P1x  C1x2   P1y  C1y 2

z  0.000

s 

P1x  D1x2   P1y  D1y 2

s  12.387

v 

 C1x  D1x2   C1y  D1y2

v  12.387

 v2  z2  s2    θP1  2  v z 

5.

p 21  14.258

ϕ  acos

ϕ  90.000 deg

 v2  s2  z2    θP1 ψ  π  acos  2  v s 

ψ  180.000  deg





Z1x  0.000 A  0.293

D  sin α

 

B  0.707

E  p 21  cos δ

 

E  13.871

 

C  0.086

F  p 21  sin δ

 

F  3.299

B  sin β

C  cos α  1

W1y  w 

 

D  0.407


W1x  10.918


W1y  15.094

2  A 2

2

W1x  W1y

w  18.629

θ  atan2 W1x W1y 6.

θ  125.878  deg


S 1x  12.387

D  sin α

 

B  0.940

E  p 21  cos δ

 

E  13.871

 

C  0.086

F  p 21  sin δ

 

F  3.299

C  cos α  1

U1y  u 

S 1y  0.000

A  0.658

B  sin γ

U1x 

S 1y  s sin ψ

 

A  cos γ  1

 

D  0.407


2

U1x  5.158

U1y  10.010

2  A 2

U1x  U1y

u  11.261

σ  atan2 U1x U1y 7.

Z1y  0.000

 

A  cos β  1

W1x 


σ  117.262  deg



V1x  12.387


V1y  0.000

θ  atan2 V1x V1y

θ  0.000  deg

v  Link 1:

2

2

V1x  V1y

v  12.387

 


G1x  6.627



 


G1y  5.084

θ  atan2 G1x G1y

θ  37.495 deg

g  8.

9.

2

2

G1x  G1y

g  8.353


θ2i  θ  θ

θ2i  88.383 deg

θ2f  θ2i  β

θ2f  43.383 deg


δp  ϕ  θ

rp  0.000

δp  90.000 deg

10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  0.0 deg R1 

ρ  0.000  deg

2

2

P1x  P1y

R1  0.000

 

O2x  10.918

 

O2y  15.094

 

O4x  17.545

 

O4y  10.010




θrot  37.495 deg







w  18.629

θ  125.878  deg

Link 3:

v  12.387

θ  0.000  deg

Link 4:

u  11.261

σ  117.262  deg

Link 1:

g  8.353

θ  37.495 deg

Coupler:

rp  0.000

δp  90.000 deg

Crank angles:

θ2i  88.383 deg

θ2f  43.383 deg


8.353 O2

O4

88.383

18.629

43.383 11.261

Y

45.000°

70.000°

D2

D1 X

C1

24.000

12.387

C2





Given:


P1y  0.0

P2x  19.544

P2y  0.373

v  12.387

Length of the coupler (link 3)


θP1  00.0 deg

θP2  90.0 deg

Coordinates of the points C1 and D1 : C1x  0.0

Solution:

C1y  0.0

D1x  C1x  v cos θP1



D1x  12.387

Free choice for the US dyad : γ  180  deg

D1y  0.000


1.


2.


p 21  3.

4.

 P1x     P1y 

R2 

2

 P2x     P2y 

 P21x     R2  R1  P21y 

2

P21x  P21y

P21y  0.373 p 21  19.548


α  90.000 deg

δ  atan2 P21x P21y

δ  1.093  deg


P1x  C1x2   P1y  C1y 2

z  0.000

s 

P1x  D1x2   P1y  D1y 2

s  12.387

v 

 C1x  D1x2   C1y  D1y2

v  12.387

 v2  z2  s2    θP1  2  v z 

ϕ  acos

ϕ  90.000 deg

 v2  s2  z2    θP1  2  v s 

ψ  π  acos 5.

P21x  19.544

ψ  180.000  deg


Z1x  0.000


Z1y  0.000



 

A  0.826

D  sin α

 

B  0.985

E  p 21  cos δ

 

E  19.544

 

C  1.000

F  p 21  sin δ

 

F  0.373

A  cos β  1 B  sin β

C  cos α  1 W1x 

W1y  w 

 

D  1.000


W1x  9.994


W1y  11.459

2  A 2

2

W1x  W1y

w  15.205

θ  atan2 W1x W1y 6.

θ  131.093  deg


S 1x  12.387 A  2.000

D  sin α

 

B  0.000

E  p 21  cos δ

 

E  19.544

 

C  1.000

F  p 21  sin δ

 

F  0.373

B  sin γ

C  cos α  1

U1y  u 

 

D  1.000


2

U1x  3.579

U1y  6.007

2  A 2

U1x  U1y

u  6.992

σ  atan2 U1x U1y 7.

S 1y  0.000

 

A  cos γ  1

U1x 

S 1y  s sin ψ

σ  120.783  deg



V1x  12.387


V1y  0.000

θ  atan2 V1x V1y

θ  0.000  deg

v  Link 1:

2

2

V1x  V1y

v  12.387

 


 

G1x  5.971


G1y  5.452

θ  atan2 G1x G1y

θ  42.399 deg


g  8.

9.

2


G1x  G1y

g  8.086


θ2i  θ  θ

θ2i  88.694 deg

θ2f  θ2i  β

θ2f  8.694  deg


δp  ϕ  θ

rp  0.000

δp  90.000 deg

10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  0.0 deg R1 

ρ  0.000  deg

2

2

P1x  P1y

R1  0.000

 

O2x  9.994

 

O2y  11.459

 

O4x  15.966

 

O4y  6.007




θrot  42.399 deg




w  15.205

θ  131.093  deg

Link 3:

v  12.387

θ  0.000  deg

Link 4:

u  6.992

σ  120.783  deg

Link 1:

g  8.086

θ  42.399 deg

Coupler:

rp  0.000

δp  90.000 deg



Crank angles:

θ2i  88.694 deg θ2f  8.694  deg 14. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.

O2

Y

8.086

D3

15.205 O4

80.000 88.694

D1 X

C1

C3

12.387

6.992





Given:


P1y  3.299

P2x  19.544

P2y  0.373

v  12.387

Length of the coupler (link 3):


θP1  24.0 deg

θP2  90.0 deg

Coordinates of the points C2 and D2 : C2x  P1x

Solution:

C2y  P1y

D2x  C2x  v cos θP1



D2x  25.187

Free choice for the US dyad : γ  90 deg

D2y  1.739


1.


2.


p 21  3.

4.

 P1x     P1y 

R2 

2

 P2x     P2y 

 P21x     R2  R1  P21y 

2

P21x  P21y

P21y  2.926 p 21  6.383


α  66.000 deg

δ  atan2 P21x P21y

δ  27.284 deg


P1x  C2x2   P1y  C2y 2

z  0.000

s 

P1x  D2x2   P1y  D2y 2

s  12.387

v 

 C2x  D2x2   C2y  D2y2

v  12.387

 v2  z2  s2    θP1  2  v z 

ϕ  acos

ϕ  114.000  deg

 v2  s2  z2    θP1  2  v s 

ψ  π  acos 5.

P21x  5.673

ψ  204.000  deg


Z1x  0.000


Z1y  0.000



 

A  0.181

D  sin α

 

B  0.574

E  p 21  cos δ

 

E  5.673

 

C  0.593

F  p 21  sin δ

 

F  2.926

A  cos β  1 B  sin β

C  cos α  1

W1x 

W1y  w 

 

D  0.914


W1x  1.804


W1y  10.459

2  A 2

2

W1x  W1y

w  10.614

θ  atan2 W1x W1y 6.

θ  80.216 deg


S 1x  11.316 A  1.000

D  sin α

 

B  1.000

E  p 21  cos δ

 

E  5.673

 

C  0.593

F  p 21  sin δ

 

F  2.926

B  sin γ

C  cos α  1

U1y  u 

 

D  0.914


2

U1x  7.959

U1y  2.316

2  A 2

U1x  U1y

u  8.289

σ  atan2 U1x U1y 7.

S 1y  5.038

 

A  cos γ  1

U1x 

S 1y  s sin ψ

σ  16.224 deg



V1x  11.316


V1y  5.038

θ  atan2 V1x V1y v  Link 1:

2

θ  24.000 deg

2

V1x  V1y

v  12.387

 


 

G1x  5.161


G1y  3.105

θ  atan2 G1x G1y

θ  31.035 deg


g  8.

9.

2


2

G1x  G1y

g  6.023


θ2i  θ  θ

θ2i  49.181 deg

θ2f  θ2i  β

θ2f  14.181 deg


δp  ϕ  θ

rp  0.000

δp  90.000 deg


2

ρ  13.378 deg

2

P1x  P1y

R1  14.258

 

O2x  12.067

 

O2y  7.160

 

O4x  17.228

 

O4y  4.055




θrot  31.035 deg




w  10.614

θ  80.216 deg

Link 3:

v  12.387

θ  24.000 deg

Link 4:

u  8.289

σ  16.224 deg

Link 1:

g  6.023

θ  31.035 deg

Coupler:

rp  0.000

δp  90.000 deg



Crank angles:

θ2i  49.181 deg θ2f  14.181 deg 14. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.

D3

8.289

Y O2

90.000 O4

D2 10.614 C3 35.000

12.387 C2

X




Design a fourbar linkage to carry the object in Figure P5-12 through the three positions shown in their numbered order using points C and D for your attachment points. The fixed pivots should be within the indicated area.

Given:

Coordinates of the points P1 , P2 and P3 : P1x  0.0

P1y  0.0

P2x  13.871

P3x  19.544

P3y  0.373

P2y  3.299


θP1  0.0 deg

θP2  24.0 deg

θP3  90.0 deg


2.

3.

C1y  0.0

D1y  0.0



2

p 21  14.258

δ  atan2 P2x P2y

δ  13.378 deg

2

2

p 31  19.548

δ  atan2 P3x P3y

δ  1.093  deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  24.000 deg

α  θP3  θP1

α  90.000 deg


P1x  C1x2   P1y  C1y 2

z  0.000

s 

P1x  D1x2   P1y  D1y 2

s  12.387

v 

 C1x  D1x2   C1y  D1y2

v  12.387

 v2  z2  s2    θP1  2  v z 

ϕ  acos

 v2  s2  z2    θP1  2  v s 

4.

D1x  12.387

ϕ  90.000 deg


ψ  180.000  deg


Z1x  0.000


Z1y  0.000

S 1x  s cos ψ

S 1x  12.387

S 1y  s sin ψ

S 1y  0.000

Use equations 5.24 to solve for w, , 2, and 3. Since the points C and D are to be used as pivots, z and  are known from the calculations above.



Guess:

W1x  2

W1y  15

β  45 deg

β  80 deg

Given


       

   

 

       

   

 

       

   

 

       

   

 


 W1x   W1y     Find  W1x W1y β β  β   β  β  56.754 deg

β  81.324 deg



W1y  11.190

θ  131.755  deg

 W1x2  W1y2 , w  15.000  


U1x  3

U1y  4

γ  70 deg

Given


       

   

 

       

   

 

       

   

 

       

   

 


γ  90 deg



 U1x   U1y     Find  U1x U1y γ γ  γ   γ  γ  119.119  deg

γ  222.077  deg


U1y  4.631


Link 1:

V1x  12.387

V1y  Z1y  S 1y

V1y  0.000

θ  atan2 V1x V1y

θ  0.000  deg

2

2

V1x  V1y

v  12.387

G1x  W1x  V1x  U1x

G1x  8.287

G1y  W1y  V1y  U1y

G1y  6.559

θ  atan2 G1x G1y

θ  38.362 deg

g 

9.

 U1x2  U1y2 , u  7.492  

V1x  Z1x  S 1x

v 

8.

σ  141.822  deg


7.


2

2

G1x  G1y

g  10.569


θ2i  θ  θ

θ2i  93.393 deg

θ2f  θ2i  β

θ2f  12.069 deg


δp  ϕ  θ

rp  0.000

δp  90.000 deg


O2x  9.989

O2y  z sin ϕ  w sin θ

O2y  11.190

O4x  s cos ψ  u  cos σ

O4x  18.276

O4y  s sin ψ  u  sin σ

O4y  4.631





θrot  38.362 deg




w  15.000

θ  131.755  deg

Link 3:

v  12.387

θ  0.000  deg

Link 4:

u  7.492

σ  141.822  deg

Link 1:

g  10.569

θ  38.362 deg

Coupler:

rp  0.000

δp  90.000 deg

Crank angles:

θ2i  93.393 deg θ2f  12.069 deg 13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.

D3

O2 10.569 Y

15.000 7.492 O4

93.393 56.754

12.069 D2

81.324 D1

C3

C1

C2

12.387

X




A ship is steaming due north at 20 knots (nautical miles per hour). A submarine is laying in wait 1/2 mile due west of the ship. The sub fires a torpedo on a course of 85 degrees. The torpedo travels at a constant speed of 30 knots. Will it strike the ship? If not, by how many nautical miles will it miss? Hint: Use the relative velocity equation and solve graphically or analytically.

Units:

naut_mile  1

knots 

Given:

Speed of ship

Vs  20 knots

naut_mile

Vt  30 knots

Speed of torpedo

hr

θs  90 deg

θt  15 deg

Initial distance between ship and torpedo

d i  0.5 naut_mile

Note that, for compass headings, due north is 0 degrees, due east 90 degrees, and the angle increases clockwise. However, in a right-handed Cartesian system, due north is 90 degrees (up) and due east is 0 degrees (to the right). The Cartesian system has been used above to define the ship and torpedo headings. Solution:


1.

The key to this solution is to recognize that the only information of interest is the relative velocity of one vessel to the other. The ship captain wants to know the relative velocity of the torpedo versus the ship, Vts = Vt - Vs. In effect, we want to resolve the situation with respect to a moving coordinate system attached to the ship.

2.

The figure below shows the initial positions of the torpedo and ship and their velocities. Vs = 20 knots Vt = 30 knots

torpedo

0.5 n. mi.

ship

3.

The figure below shows the vector diagram that solves the relative velocity equation Vts = Vt - Vs. For the torpedo to hit the moving ship, the relative velocity vector has to be perpendicular to the ship's velocity vector (if you were on the ship observing the torpedo, it would appear to be headed directly for you). As the velocity diagram shows, the relative velocity vector is not perpendicular to the ship's velocity vector so it will miss and pass behind the ship.

4.

Determine the distance by which the torpedo will miss the ship.

Vt

Time required for torpedo to travel 0.5 nautical miles due east tt_east 

di

Vt cos θt

tt_east  62.117 s

Distance traveled by the torpedo due north in that time d t_north  Vt sin θt  tt_east

d t_north  0.134 naut_mile

Distance traveled by the ship due north in that time d s_north  Vs tt_east

d s_north  0.345 naut_mile

Distance by which the torpedo will miss the ship d miss  d s_north  d t_north

d miss  0.211 naut_mile

-Vs V ts

22.89°




A plane is flying due south at 500 mph at 35,000 feet altitude, straight and level. A second plane is initially 40 miles due east of the first plane, also at 35,000 feet altitude, flying straight and level at 550 mph. Determine the compass angle at which the second plane would be on a collision course with the first. How long will it take for the second plane to catch the first? Hint: Use the relative velocity equation and solve graphically or analytically.

Given: Speed of first plane

V1  500  mph

Speed of second plane

V2  550  mph

Initial distance between planes

d i  40 mi

θ1  270  deg

Note that, for compass headings, due north is 0 degrees, due east 90 degrees, and the angle increases clockwise. However, in a right-handed Cartesian system, due north is 90 degrees (up) and due east is 0 degrees (to the right). The Cartesian system has been used above to define the plane headings. Solution:

See Mathcad file P0601b.

1.

The key to this solution is to recognize that the only information of interest is the relative velocity of one plane to the other. For a collision to occur, the relative velocity of the second plane with respect to the first must be perpendicular to the velocity vector of the first.

2.

The figure below shows the initial positions of the two planes and their velocities.

plane 1

40 mi.

plane 2



V1 = 500 mph

3.

4.

5.

V2 = 550 mph

The figure below shows the vector diagram that solves the relative velocity equation V2 = V1 + V21. To construct this diagram, chose a convenient velocity scale and draw V1 to its correct length with the arrow head pointing straight down (indicating due south). From the tip of the vector, layoff a horizontal construction line to the left (due west) an undetermined length. From the tail of the V1 vector, construct a circle whose radius is equal to the scaled length of vector V2. The intersection of the circle and the horizontal construction line determines the length of V21. Draw the arrowheads for V2 and V21 pointing toward the intersection of the circle and construction line. Label the horizontal vector V21 and the vector that joins the tail of V1 with the head of V21 as V2. The angle  between V1 and V2 is the required direction for V2 in order that plane 2 collides with plane 1.  The angle  can also be determined analytically from the velocity triangle as 24.620° follows. V1  V2  V1  θ  acos  θ  24.620 deg   V2  V 21 The time it will take for the second plane to catch the first is the time that it will take plane 2 to travel the 40 miles to the west. t 

di

V2 sin θ

t  628.468 s

t  10.474 min




A point is at a 6.5-in radius on a body in pure rotation with  = 100 rad/sec. The rotation center is at the origin of a coordinate system. When the point is at position A, its position vector makes a 45 deg angle with the X axis. At position B, its position vector makes a 75 deg angle with the X axis. Draw this system to some convenient scale and: a. Write an expression for the particle's velocity vector in position A using complex number notation, in both polar and Cartesian forms. b. Write an expression for the particle's velocity vector in position B using complex number notation, in both polar and Cartesian forms. c. Write a vector equation for the velocity difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the velocity difference numerically. d. Check the result of part c with a graphical method.

Given: ω  100 

Rotation speed

Solution: 1.

rad sec

Vector angles

θA  45 deg

Vector magnitude

R  6.5 in

θB  75 deg


Calculate the magnitude of the velocity at points A and B using equation 6.3. V  R ω

V  650.000

in sec

2.


3.


4.

Choose a convenient velocity scale and draw the two velocity vectors VA and VB at the tips of RA and RB, respectively. The velocity vectors will be perpendicular to their respective position vectors.

Y 0

8

1

2 in

Distance scale:

VB B

0

VA

500 in/sec

Velocity scale:

6 A 4

RB RA

2

0 a.

X 2

4

6

8

Write an expression for the particle's velocity vector in position A using complex number notation, in both polar and Cartesian forms.


Polar form:


RA  R e

VA  R j  ω e Cartesian form:

j

j  θA

j

j  θA

4

VA  650  j  e

VA  R j  ω  cos θA  j  sin θA 

RB  R e

in sec

Cartesian form:

j

j  θB

VB  R j  ω e

RA  6.5 e

π 4 j

j  θB

VB  650  j  e

75 π 180

VB  R j  ω  cos θB  j  sin θB  VB  ( 627.852  168.232j)

in sec

Write a vector equation for the velocity difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. VBA  VB  VA

d.

π

Write an expression for the particle's velocity vector in position B using complex number notation, in both polar and Cartesian forms. Polar form:

c.

4

RA  6.5 e

VA  ( 459.619  459.619j) b.

π

VBA  ( 168.232  291.387j)

in sec

Check the result of part c with a graphical method. Solve the equation VB = VA + VBA using a velocity scale of 250 in/sec per drawing unit.

Y

Velocity scale factor kv  250 

in 0

sec

1.166

Horizontal component

VA VBA

VBAx  0.673  kv VBAx  168.3

500 in/sec

Velocity scale:

VB

in

Ov

X

0.673

sec

Vertical component VBAy  1.166  kv

VBAy  291.5

in sec

On the layout above the X and Y components of VBA are equal to the real and imaginary components calculated, confirming that the calculation is correct.




Given:

A point A is at a 6.5-in radius on a body in pure rotation with  = -50 rad/sec. The rotation center is at the origin of a coordinate system. At the instant considered its position vector makes a 45 deg angle with the X axis. A point B is at a 6.5-in radius on another body in pure rotation with  = +75 rad/sec. Its position vector makes a 75 deg angle with the X axis. Draw this system to some convenient scale and: a. Write an expression for the particle's velocity vector in position A using complex number notation, in both polar and Cartesian forms. b. Write an expression for the particle's velocity vector in position B using complex number notation, in both polar and Cartesian forms. c. Write a vector equation for the velocity difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. d. Check the result of part c with a graphical method. ω  50

Rotation speeds

Solution: 1.

rad

ω  75

sec

Vector angles

θA  45 deg

Vector magnitude

R  6.5 in

rad sec

θB  75 deg


Calculate the magnitude of the velocity at points A and B using equation 6.3. VA  R ω

VA  325.000

VB  R ω

VB  487.500

in sec in

sec

1.


2.


3.

Choose a convenient velocity scale and draw the two velocity vectors VA and VB at the tips of RA and RB, respectively. The velocity vectors will be perpendicular to their respective position vectors. Y 8 VB

0

1

2 in

Distance scale: B

0

500 in/sec

Velocity scale:

6 A 4

RB RA

2

0

a.

VA

X 2

4

6

8

Write an expression for the particle's velocity vector on body A using complex number notation, in both polar and Cartesian forms.


Polar form:


RA  R e

j

j  θA

Cartesian form:

j

j  θA

VA  R j  ω  cos θA  j  sin θA  in sec

Write an expression for the particle's velocity vector on body B using complex number notation, in both polar and Cartesian forms. Polar form:

RB  R e

j

j  θB

RA  6.5 e

VB  R j  ω e Cartesian form:

j  θB

4

VB  487.5  j  e

75 π 180

VB  R j  ω  cos θB  j  sin θB  in sec

Write a vector equation for the velocity difference between the points on bodies B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. VBA  VB  VA

d.

π

j

VB  ( 470.889  126.174j) c.

π 4

VA  325  j  e

VA  ( 229.810  229.810j) b.

4

RA  6.5 e

VA  R j  ω e

π

VBA  ( 700.699  355.984j)

in sec

Check the result of part c with a graphical method. Solve the equation VB = VA + VBA using a velocity scale of 250 in/sec per drawing unit.

Y

Velocity scale factor kv  250 

in

0

Horizontal component VBAx  2.803  kv VBAx  700.7

500 in/sec

Velocity scale:

sec

VB 1.424

Ov

VBA

VA

X

in sec

2.803

Vertical component VBAy  1.424  kv

VBAy  356.0

in sec

On the layout above the X and Y components of VBA are equal to the real and imaginary components calculated, confirming that the calculation is correct.




For the fourbar defined in Table P6-1, line a, find the velocities of the pin joints A and B, and of the instant centers I1,3 and I2,4. Then calculate 3 and 4 and find the velocity of point P. Use a graphical method.

Given:


d  6  in

Link 2

a  2  in

Link 3

b  7  in

Link 4

c  9  in

θ  30 deg

Crank angle:

ω  10 rad sec

Crank velocity:

1

Coupler point data: Rpa  6  in Solution: 1.

δ  30 deg

See Figure P6-1 and Mathcad file P0604a.

Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. B 0

1 IN

SCALE

P 5.9966 6.8067

148.2007°

3.3384 I1,3

A

117.2861°

88.8372° O2

O4

I 2,4 1.7118

4.2882

From the layout above:

2.

O2I24  1.7118 in

O4I24  4.2882 in

AI13  3.3384 in

BI13  5.9966 in

θ  117.2861 deg

θ  88.8372  deg

PI13  6.8067 in

Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point A.


3.


in

VA  a  ω

VA  20.0

θVA  θ  90 deg

θVA  120.0 deg

sec

Determine the angular velocity of link 3 using equation 6.9a. ω 

VA AI13

ω  5.991

rad

CW

sec

4.

Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection. VB  BI13 ω in VB  35.925 sec θVB  θ  90 deg θVB  27.286 deg

5.

Use equation 6.9c to determine the angular velocity of link 4. ω 

6.

VB c

ω  3.99

rad

CW

sec

Use equation 6.9d and inspection of the layout to determine the magnitude and direction of the velocity at point P. in

VP  PI13 ω

VP  40.778

θVP  148.2007 deg  90 deg

θVP  58.201 deg

sec




A general fourbar linkage configuration and its notation are shown in Figure P6-1. The link lengths, coupler point location, and the values of 2 and 2 for the same fourbar linkages as used for position analysis in Chapter 4 are redefined in Table P6-1, which is the same as Table P4-1. For row a, find the velocities of the pin joints A and B, and coupler point P. Calculate 3 and 4. Draw the linkage to scale and label it before setting up the equations.

Given:

Link lengths: d  6

Link 1

a  2

Link 2

Rpa  6

Coupler point:

c  9

Link 4

δ  30 deg

Link 2 position and velocity: θ  30 deg Solution:

b  7

Link 3

ω  10

See Mathcad file P0605a. y

1.

Draw the linkage to scale and label it.

2.


d

K2 

a

K1  3.0000 2

K3 

B

OPEN

3

d

4

c

K2  0.6667 2

2

a b c d

88.837°

2

K3  2.0000

2 a c

2 O2

 

117.286°

A

O4

115.211°

 

A  cos θ  K1  K2 cos θ  K3

143.660°

 

B  2  sin θ

 

C  K1   K2  1   cos θ  K3 A  0.7113 3.

B  1.0000

CROSSED

C  3.5566

B'






2

θ  2  atan2 2  A B 

B  4 A  C



θ  242.714 deg

θ  θ  360  deg Crossed: 4.



θ  602.714 deg



2

θ  2  atan2 2  A B 

B  4 A  C



θ  216.340 deg


2

d

K5 

b

2

2

c d a b

 

2 a b

 

D  cos θ  K1  K4 cos θ  K5

 

E  2  sin θ

2

K4  0.8571 D  1.6774 E  1.0000

 

F  K1   K4  1   cos θ  K5

F  2.5906

K5  0.2857

x


5.







2

θ  2  atan2 2  D E 

E  4  D F



θ  271.163 deg

θ  θ  360  deg Crossed: 6.

7.



θ  631.163 deg



2

θ  2  atan2 2  D E 

E  4  D F



θ  244.789 deg

Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18. ω 

a  ω sin θ  θ  b sin θ  θ

 

 

ω  5.991

ω 

a  ω sin θ  θ  c sin θ  θ

 

 

ω  3.992

Determine the velocity of points A and B for the open circuit using equations 6.19.



 

 

VA  a  ω sin θ  j  cos θ VA  10.000  17.321j



arg VA  120 deg

VA  20

 

 

VB  c ω sin θ  j  cos θ VB  31.928  16.470j 8.

arg VB  27.286 deg

VB  35.926

Determine the velocity of the coupler point P for the open circuit using equations 6.36.











VPA  Rpa ω sin θ  δ  j  cos θ  δ VPA  31.488  17.337j VP  VA  VPA VP  21.488  34.658j 9.

arg VP  58.201 deg

VP  40.779

Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18. ω 

a  ω sin θ  θ  b sin θ  θ

 

 

ω  0.662

ω 

a  ω sin θ  θ  c sin θ  θ

 

 

ω  2.662

10. Determine the velocity of point B for the crossed circuit using equations 6.19.



 

 

VB  c ω sin θ  j  cos θ VB  14.195  19.295j

arg VB  126.340 deg

VB  23.954

11. Determine the velocity of the coupler point P for the crossed circuit using equations 6.36.











VPA  Rpa ω sin θ  δ  j  cos θ  δ



VPA  3.960  0.332j VP  VA  VPA VP  13.960  16.989j

VP  21.989

arg VP  129.411 deg




The general linkage configuration and terminology for an offset fourbar slider-crank linkage are shown in Fig P6-2. The link lengths and the values of 2 and 2 are defined in Table P6-2. For row a, find the velocities of the pin joints A and B and the velocity of slip at the sliding joint using a graphical method.

Given:

Link lengths:

Solution: 1.

Link 2

a  1.4 in

Link 3

b  4  in

Offset

c  1  in

θ  45 deg

ω  10 rad sec

1


Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest. Direction of VBA Y

Axis of transmission

Direction of VA

A 2

45.000°

Axis of slip and Direction of VB

B

3

179.856°

1.000 X

O2

2.

3.

Use equation 6.7 to calculate the magnitude of the velocity at point A. in VA  a  ω VA  14.000 θVA  θ  90 deg sec

θVA  135 deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.

4.

0

5 units/sec

VA Y

135.000° V BA X

VB

From the velocity triangle we have: 1.975


Velocity scale factor:

VB  1.975  in kv

5.


kv 

5  in sec

1

in

VB  9.875

in sec

θVB  180  deg

Since the slip axis and the direction of the velocity of point B are parallel, Vslip = VB.




The general linkage configuration and terminology for an offset fourbar slider-crank linkage are shown in Fig P6-2. The link lengths and the values of 2 and 2 are defined in Table P6-2. For row a, find the velocities of the pin joints A and B and the velocity of slip at the sliding joint using the analytic method. Draw the linkage to scale and label it before setting up the equations.

Given:

Link lengths: Link 2 (O2 A) Crank angle

Solution: 1.

a  1.4

Link 3 (AB)

b  4

θ  45 deg

Crank angular velocity

Offset (yB) ω  10


Draw the linkage to scale and label it. Y d1 = 4.990

d2 = 3.010

3(CROSSED)

B'

A 2

0.144°

45.000°

B

3 (OPEN)

179.856° 1.000 X

O2

2.

Determine 3 and d using equations 4.16 and 4.17. Crossed:

 a sin θ  b 

θ  asin

 

c

 

θ  0.144 deg

 

d 2  a  cos θ  b  cos θ

Open:

d 2  3.010

 a  sin θ  c  π b  

θ  180.144 deg

 

d 1  4.990

θ  asin 

 

d 1  a  cos θ  b  cos θ 3.

4.

Determine the angular velocity of link 3 using equation 6.22a:

   

ω  2.475

   

ω  2.475

Open

ω 

a cos θ   ω b cos θ

Crossed

ω 

a cos θ   ω b cos θ

Determine the velocity of pin A using equation 6.23a:



 

 

VA  a  ω sin θ  j  cos θ VA  9.899  9.899i

VA  14.000

arg VA  135.000 deg

c  1


5.


Determine the velocity of pin B using equation 6.22b: Open

VB1  a  ω sin θ  b  ω sin θ

 

 

VB1  9.875

Crossed

VB2  a  ω sin θ  b  ω sin θ

 

 

VB2  9.924

The angle of VB is 0 deg if VB is positive and 180 deg if VB negative. 6.

The velocity of slip is the same as the velocity of pin B.




The general linkage configuration and terminology for an inverted fourbar slider-crank linkage are shown in Fig P6-3. The link lengths and the values of 2 and 2 and  are defined in Table P6-3. For row a, using a graphical method, find the velocities of the pin joints A and B and the velocity of slip at the sliding joint. Draw the linkage to scale.

Given:

Link lengths: Link 1 d  6  in c  4  in

Link 4 Solution: 1.

Link 2

a  2  in

γ  90 deg

θ  30 deg

ω  10 rad sec

1


Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest. Direction of VA Axis of transmission and direction of VBA Axis of slip and Direction of VB

B y

1.793 90.0°

b a

127.333° c 142.666°

A

30.000°

d

x 04

02

2.

3.

Use equation 6.7 to calculate the magnitude of the velocity at point A. in VA  a  ω VA  20.000 θVA  θ  90 deg sec

θVA  120 deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B on link 3. The equation to be solved graphically is VB3 = VA3 + VBA3 a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.

4.

0

VA

V BA 52.667° VB 0.771

From the velocity triangle we have: Velocity scale factor:

kv 

10 in sec in

10 in/sec

1.846 Y

1

X


5.


in

VB3  0.771  in kv

VB3  7.710

VBA3  1.846  in kv

VBA3  18.460

sec

θVB3  52.667 deg

in sec

Determine the angular velocity of link 3 using equation 6.7. From the linkage layout above:b  1.793  in and ω 

VBA3 b

ω  10.296

rad

θ  142.666  deg CW

sec

The way in which link 3 slides in link 4 requires that ω  ω 6.

7.

Determine the magnitude and sense of the vector VB4 using equation 6.7. in

VB4  c ω

VB4  41.182

θVB4  θ  90 deg

θVB4  52.666 deg

sec

Note that VB3 and VB4 are in the same direction in this case. The velocity of slip is in

Vslip  VB3  VB4

Vslip  33.472

θslip  θVB4  180  deg

θslip  232.666 deg

sec




The general linkage configuration and terminology for an inverted fourbar slider-crank linkage are shown in Fig P6-3. The link lengths and the values of 2 and 2 and  are defined in Table P6-3. For row a, using an analytic method, find the velocities of the pin joints A and B and the velocity of slip at the sliding joint. Draw the linkage to scale and label it before setting up the equations.

Given:

Link lengths: Link 1 d  6 c  4

Link 4 Solution: 1.

Link 2

a  2

γ  90 deg

θ  30 deg

ω  10


Draw the linkage to scale and label it. B y 90.0°

127.333°

b

c

A

142.666°

a 30.000°

d

x 04

02 B'

169.040° 79.041°

2.

Determine the values of the constants needed for finding 4 from equations 4.25 and 4.26. P  a  sin θ  sin γ  a  cos θ  d  cos γ

      Q  a  sin θ  cos γ   a  cos θ  d   sin γ

3.

4.

5.

P  1.000 Q  4.268

R  c sin γ

R  4.000

T  2  P

T  2.000

S  R  Q

S  0.268

U  Q  R

U  8.268

Use equation 4.26 to find values of 4 for the open and crossed circuits.

  2  atan2 2  S T 

OPEN

θ  2  atan2 2  S T 

CROSSED

θ

 2 T  4 S U  2

T  4 S U

θ  142.667 deg θ  169.041 deg


θ  θ  γ

θ  232.667 deg

CROSSED

θ  θ  γ

θ  79.041 deg

Determine the magnitude of the instantaneous "length" of link 3 from equation 4.20a.


6.

7.

OPEN

b 1 

CROSSED

b 2 


    sin θ  γ

a  sin θ  c sin θ

b 1  1.793

    sin θ  γ

a  sin θ  c sin θ

b 2  1.793

Determine the angular velocity of link 4 using equation 6.30c: OPEN

ω 

CROSSED

ω 

a  ω cos θ  θ









b 1  c cos γ

a  ω cos θ  θ b 2  c cos γ

ω  10.292

ω  3.639




 

 

VA  a  ω sin θ  j  cos θ VA  10.000  17.321i 8.

 

VA  20.000

arg VA  120.000 deg

Determine the velocity of point B on link 4 using equation 6.31: OPEN

 

VB4x1  c ω sin θ

VB4x1  24.966

 

VB4y1  32.734

VB4y1  c ω cos θ VB41 

2

VB4x1  VB4y1

2

VB41  41.168

θVB1  atan2 VB4x1 VB4y1 CROSSED

θVB1  52.667 deg

 

VB4x2  c ω sin θ

VB4x2  2.767

 

VB4y2  14.289

VB4y2  c ω cos θ VB42 

2

VB4x2  VB4y2

2

VB42  14.555

θVB2  atan2 VB4x2 VB4y2 9.

θVB2  100.959 deg

Determine the slip velocity using equation 6.30a: OPEN

CROSSED

Vslip1 

Vslip2 

 



 

 

 

 

a  ω sin θ  ω b 1 sin θ  c sin θ

 

cos θ

 



a  ω sin θ  ω b 2 sin θ  c sin θ

 

cos θ

Vslip1  33.461

Vslip2  4.351




The link lengths, gear ratio (), phase angle (), and the values of 2 and 2 for a geared fivebar from row a of Table P6-4 are given below. Draw the linkage to scale and graphically find 3 and 4, using a graphical method.

Given:


f  6  in

Link 3

b  7  in

Link 2

a  1  in

Link 4

c  9  in

Link 5

d  4  in

Gear ratio, phase angle, and crank angle: λ  2 Solution: 1.

ϕ  30 deg

θ  150 deg

Choose the pitch radii of the gears. Since the gear ratio is positive, an idler must be used between gear 2 and gear 5. Let the idler be the same diameter as gear 5, and let all three gears be in line.

r5 

λ 

f

r2 r5

r5  1.200 in

λ3

r2  λ r5

r2  2.400 in

Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Direction of VBC Direction of VBA

Direction of VC

Direction of VA

C

4 B 3

A

2

4.

1

Determine the angle of link 5 using the equation in Figure P6-4.

f  r2  3  r5

3.

ω  10 rad sec


θ  λ θ  ϕ 2.

θ  60 deg

5 O5

O2

Use equation 6.7 to calculate the magnitude of the velocity at points A and C. in

VA  a  ω

VA  10.00

ω  λ ω

ω  20.000

rad

VC  80.00

in

VC  d  ω

sec

θ  60 deg  90 deg

sec

sec

θ  150  deg  90 deg


5.


Use equation 6.5 to (graphically) determine the magnitudes of the relative velocity vectors VBA and VBC. The equation to be solved graphically is the last of the following three. VB = VA + VBA

VB = VC + VBC

VA + VBA = VC + VBC

a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, layout the known vector VC. d. From the tip of VC, draw a construction line with the direction of VBC, magnitude unknown. e. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VBC construction line and drawing VBC from the tip of VC to the intersection of the VBA construction line. 6.


0

kv 

50 in sec

50 in/sec

1

Y VA

in

X

7.

VBA  4.562  in kv

VBA  228.100

VBC  3.051  in kv

VBC  152.550

in sec VC

in sec

Determine the angular velocity of links 3 and 4 using equation 6.7.

ω 

VBA b

ω  32.586

4.562

rad

3.051

sec V BA

ω 

VBC c

Both links are rotating CCW.

ω  16.950

rad sec

V BC




Given:

Solution: 1.

The general linkage configuration and terminology for a geared fivebar linkage are shown in Figure P6-4. The link lengths, gear ratio (), phase angle (), and the values of 2 and 2 are defined in Table P6-4. For row a, find 3 and 4, using an analytic method. Draw the linkage to scale and label it before setting up the equations. Link lengths: Link 1

d  4

Link 2

a  1

Link 3

b  7

Link 4

c  9

Link 5

f  6

Input angle

θ  60 deg

Gear ratio

λ  2.0

Phase angle

ϕ  30 deg

ω  10


Draw the linkage to scale and label it. y C

4

B

177.7152°

173.6421° 3

5 124.0501°

2

x

O2

3

150.0000°

115.4074°

O5

4

B`

2.


     B  2  c  d  sin λ θ  ϕ  a  sin θ 


2

2

2

2



2



A  36.646 B  20.412

  


C  37.431

D  C  A

D  0.785

E  2  B

E  40.823

F  A  C

F  74.077



     





    a cosθ  f  H  2  b   d  sin λ θ  ϕ   a  sin θ G  2  b   d  cos λ θ  ϕ

2

2

2

2



2

  

K  a  b  c  d  f  2  a  f  cos θ   2  d  a  cos θ  f  cos λ θ  ϕ    2  a  d  sin θ  sin λ θ  ϕ L  K  G



     





G  28.503 H  15.876

K  26.569 L  1.933


3.

M  2  H

M  31.751

N  G  K

N  55.072


CROSSED

   2   atan2 2  D E   2   atan2 2  L M   2   atan2 2  D E 

θ  2  atan2 2  L M  θ θ θ

4.


 2 E  4  D F   2 M  4  L N   2 E  4  D F   2

M  4  L N

θ  173.642 deg θ  177.715 deg θ  115.407 deg θ  124.050 deg

Determine the position and angular velocity of gear 5 from equations 4.27c and 6.32c θ  λ θ  ϕ

θ  150.000 deg

ω  λ ω

ω  20.000

Angular velocity of links 3 and 4 from equations 6.33 OPEN

ω  

 

  b   cos θ  2  θ  cos θ 

ω  32.585 ω 

 

  c sin θ

 

a  ω sin θ  b  ω sin θ  d  ω sin θ

ω  

CCW

 



  b   cos θ  2  θ  cos θ 



2  sin θ  a  ω sin θ  θ  d  ω sin θ  θ

ω  75.191 ω 



CCW

ω  16.948

CROSSED



2  sin θ  a  ω sin θ  θ  d  ω sin θ  θ

CW

 

  c sin θ

 

a  ω sin θ  b  ω sin θ  d  ω sin θ

ω  59.554

CW

CCW




Find all of the instant centers of the linkages shown in Figure P6-5.

Solution:


a.

This is a fourbar slider-crank with n  4.

1.

Determine the number of instant centers for this mechanism using equation 6.8a. C 

2.

n ( n  1)

C6

2

Draw the linkage to scale and identify those ICs that can be found by inspection. C

2,3 Y

A

1,2

2

3

3,4

X O2

B 4

1,4 at infinity

3.

Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4 I1,3: I1,2-I2,3 and I1,4-I3,4

I2,4: I1,2-I1,4 and I2,3-I3,4 1,3 1 4

2,4

C

2 3

A 1,2

2

3

3,4

2,3

O2

4 B 1,4 at infinity 1,4 at infinity

b.

This is a fourbar with planetary motion (roller 3), n  4.

1.


n ( n  1) 2

C6

2,3 at contact point (behind link 4) 1,3

2.

Draw the linkage to scale and identify those ICs that can be found by inspection, which in this case, is all of them. c. This is a fourbar with n  4. 1.

3 A 4


2.

1

n ( n  1) 2

2 O2

C6

Draw the linkage to scale and identify those ICs that can be found by inspection.

1,2 and 2,4 and 1,4


SOLUTION MANUAL 6-12-2 C 2,3

3,4 3 A

B 4

2

O2

O4 1,2

3.

1,4


I2,4: I1,2-I1,4 and I2,3-I3,4 1,3 at infinity

1,3 at infinity

1

C 2,3

4

3,4

2

3 3 2,4 at infinity A

2

2,4 at infinity

O2

2,4 at infinity

B

4

2,4 at infinity

O4 1,2

1,4

1,3 at infinity

1,3 at infinity

d.

This is a fourbar with n  4.

1.


2.

n ( n  1)

C6

2

Draw the linkage to scale and identify those ICs that can be found by inspection. 2,3

3,4

1,2 3 O2

4

A

2 1,4 at infinity

3.


I2,4: I1,2-I1,4 and I2,3-I3,4 2,3

3,4

1,2 1 3 O2

4

A

2,4

4

2 3

2

1,3

1,4 at infinity 1,4 at infinity

e.

This is a threebar with n  3.

1.

Determine the number of instant centers for this mechanism using equation 6.8a.


C  2.


n ( n  1)

C3

2

Draw the linkage to scale and identify those ICs that can be found by inspection. 1,2 2

1,3 3

O2

3.

O3

Use Kennedy's Rule and a linear graph to find the remaining IC, I2,3 I2,3: I1,2-I1,3 Common normal 1 2,3 1,2 2

2

3

1,3 3

O2

O3

f.

This is a fourbar with n  4.

1.


2.

n ( n  1)

C6

2


1,4 at infinity

3.

C

B

Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4

4 3

I1,3: I1,2-I2,3 and I1,4-I3,4

I2,4: I1,2-I1,4 and I2,3-I3,4 2,3 VA

A

2

2,4 at infinity 3,4

1,4 at infinity

1,2 at infinity C

B

1

4 3 1,3

4

2 3

2,3 VA

A

2

1,2 at infinity





Solution:


a.

This is a fourbar inverted slider-crank with n  4.

1.


2.

n ( n  1)

C6

2


2

3 3,4 at infinity 1 1,2 4

1 1,4

3.


I2,4: I1,2-I1,4 and I2,3-I3,4 3,4 at infinity 2,3

2,4 2

1 4

3

2

3,4 at infinity 1

3

1,2 4

1 1,4

1,3

b.

This is a sixbar with slider, n  6.

1.


n ( n  1) 2

C  15

2.


3.

Use Kennedy's Rule and a linear graph to find the remaining 7 ICs. I1,3: I1,2-I2,3 and I1,4-I3,4

5,6 6 2,3 2

1,6 at infinity 3

1

5

I1,5: I1,6-I5,6 and I1,4-I4,5

3,4; 3,5; 4,5

I2,5: I1,2-I1,5 and I2,4-I4,5

1 4

1,2

I3,6: I1,6-I1,3 and I3,4-I4,6

I4,6: I1,6-I1,4 and I4,5-I5,6

I2,4: I1,2-I1,4 and I2,3-I3,4

I2,6: I1,2-I1,6 and I2,5-I5,6

1,4

1



3,6

1,3

1,6 at infinity 2,5

to 2,4

5,6 6 to 2,4

1,5

2,3

1 1,6 at infinity

2 3

6

2

1

5

5

1,6 at infinity

3 4

3,4; 3,5; and 4,5 1 4

1,2

1,4

2,6 4,6

1,6 at infinity 1

c.

This is a sixbar with slider and roller with n  6.

1.


2.

n ( n  1)

C  15

2

Draw the linkage to scale and identify those ICs that can be found by inspection. 2,5 5,6

3.

Use Kennedy's Rule and a linear graph to find the remaining 8 ICs.

1,2

5

6

2,3

2

1

I2,6: I1,2-I1,6 and I2,5-I5,6 I1,5: I1,6-I5,6 and I1,2-I2,5

1

1,6

3

I4,5: I1,4-I1,5 and I2,4-I2,5 I3,6: I3,6-I5,6 and I2,3-I2,6 I1,3: I1,2-I2,3 and I1,4-I3,4

I3,5: I3,4-I4,5 and I2,5-I2,3

I2,4: I1,2-I1,4 and I2,5-I2,4

I4,6: I4,5-I5,6 and I3,4-I3,6

4 3,4

3,5 2,5 4,5

1,4 at infinity

1,5

5,6

1,2

5

1 6

2,6

1

2,3

2

6

2,4

2

1,4 at infinity 5

1

1,6

3

4,6 3,6 4 3,4

1,4 at infinity

1,3

1

d.

This is a sixbar with slider and roller with n  6.

1.


2.

n ( n  1) 2

3 4

C  15


1

1,4 at infinity



3,4 4 1 1,4 3

5,6 5

1,2

2

6 2,3; 2,5; and 3,5

1

1

1,6 at infinit

3.


I3,6: I1,6-I1,3 and I3,5-I5,6

I2,6: I1,2-I1,6 and I2,5-I5,6

I1,5: I1,6-I5,6 and I1,2-I2,5

I4 ,5: I1,4-I1,5 and I3,5-I3,4

I2,4: I1,2-I1,4 and I2,3-I3,4

I4,6: I1,6-I1,4 and I4,5-I5,6 2,4 4,6 3,4 4,5

4

1,3

1 1

1,4 6 3

1,5

1,6 at infinity

2

5

3 4

2,6

5,6 5

1,2

2 1

6 2,3; 2,5; and 3,5

3,6

1

1,6 at infinity 1,6 at infinity





Solution:


a.

This is a pin-jointed fourbar with n  4.

1.


n ( n  1)

C6

2

2.


3.


3,4

2,3

3

I1,3: I1,2-I2,3 and I1,4-I3,4

I2,4: I1,2-I1,4 and I2,3-I3,4

2

4

1,2

1,4

3,4 1 2,3

3 2

4

1 2

4 3

1,2 2,4 1

1,4

1,3

b.

This is a fourbar inverted slider-crank, n  4.

1.


n ( n  1)

C6

2

2.

Draw the linkage to scale and identify those ICs that can be found by inspection, which in this case, is all of them.

3.


2,3 3 1,4

4 2

I2,4: I1,2-I1,4 and I2,3-I3,4 1

3,4 at infinity 1,2

2,3 1

3 1,4

4

4

2

2

2,4 3

1

1,2

3,4 at infinity

1,3 3,4 at infinity

1,4 at infinity

c.

This is a fourbar double slider with n  4.

1.


4 3,4 3

C  2.

n ( n  1) 2

C6

2,3 2 1

Draw the linkage to scale and identify those ICs that can be found by inspection. 1,2 at infinity


3.


Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4 1,4 at infinity

I1,3: I1,2-I2,3 and I1,4-I3,4

2,4 at infinity

1,3

I2,4: I1,2-I1,4 and I2,3-I3,4 1

4 4

3,4

2

3 3

2,3 2 1 1,2 at infinity

d.


1.


n ( n  1)

C6

2

2.


3.


2,3

3,4

1,2

I2,4: I1,2-I1,4 and I2,3-I3,4

3

2

4 1,4

1 2,3

1,3

1

3,4 4

3

2

2

4 1,4

3

2,4 1 1,2

e.

This is a fourbar effective slider-crank with n  4.

1.


n ( n  1)

C6

2

2.


3.


2,3

3,4

I2,4: I1,2-I1,4 and I2,3-I3,4

3

2

4

1,3

2,3 1 1,2

1

2,4 4

3,4 3

2

2 3

4 1 1,2 1,4 at infinity

1,4 at infinity

1,4 at infinity



f.

This is a fourbar cam-follower with n  4.

1.


n ( n  1)

C6

2

2.


3.


3,4 2,3 3

1,4

I2,4: I1,2-I1,4 and I2,3-I3,4

4 2

2,4

1

3,4 1,3

1

1,2 at infinity

2,3 4

3

1,4

2

4 3

2 1 1,2 at infinity 1,2 at infinity

g.

This is a fourbar slider-crank with n  4.

1.


n ( n  1)

C6

2

2.


3.


2,3

2 3,4

I1,3: I1,2-I2,3 and I1,4-I3,4

3

4

1,2

I2,4: I1,2-I1,4 and I2,3-I3,4

1,4 at infinity

1 1,3 2,3

1,4 at infinity

1

2

2,4

3,4 3

4

2

4 3

1,2

2

2,3 1

1

h.

This is a fourbar inverted slider-crank with n  4.

1.


2.

n ( n  1) 2

1,2

1,4 at infinity

3

C6

3,4 at infinity


4 1,4

1


3.



3,4 at infinity

2,4

I2,4: I1,2-I1,4 and I2,3-I3,4

2

2,3

1

1,2 4

2

1 3

3,4 at infinity

3 3,4 at infinity

1,3 4 1

1,4

i.

This is a fourbar slider-crank (hand pump) with n  4.

1.


2.

n ( n  1)

C6

2


3

3,4 4 1,2 at infinity

2

1

1,4

3.


I2,4: I1,2-I1,4 and I2,3-I3,4 2,3 3 1,3

1,2 at infinity

3,4

1 4

1,2 at infinity

2

4

1

2,4

1,2 at infinity 1,4

2 3





Solution:


a.


1.


n ( n  1)

C6

2

2.


3.


2,3 B

I1,3: I1,2-I2,3 and I1,4-I3,4 2 3

I2,4: I1,2-I1,4 and I2,3-I3,4 O2

1,2

B 4

O4

3,4 1,4

2,3

1

2

1,2

4

2

3 3 4 1,3 3,4 and 2,4 1,4

b.


1.


2.

n ( n  1) 2

2,3

C6 2

Draw the linkage to scale and identify those ICs that can be found by inspection, which in this case, is all of them.

O2

B

A

3

3,4 4

3.


1,4

1,2

O4



I2,4: I1,2-I1,4 and I2,3-I3,4 1,3

1

2,3 B

A

2,4

4

3

2 3

2

O2

3,4 4 1,4

1,2

O4

c.

This is an eightbar (three slider-cranks with a common crank) with n  8.

1.


2.

3.

n ( n  1)

C  28

2


7


1,7 at infinity

4

2

I2,7: I1,2-I1,7 and I2,4-I4,7

3,6

8

1 3

1,6 at infinity

5

1,8 at infinity 2,3; 2,4; 2,5; 3,4; 3,5; and 4,5


5,8

1,2

4,7

6

I3,7: I1,3-I1,7 and I3,4-I4,7

I6,8: I2,7-I2,8 and I3,7-I3,8

I3,8: I1,3-I3,8 and I3,5-I5,8

I4,6: I1,6-I1,4 and I3,4-I3,6

I2,8: I1,2-I1,8 and I2,5-I5,8

I4,8: I3,8-I3,4 and I4,5-I5,8

I5,6: I2,5-I2,6 and I3,5-I3,6

I1,4: I1,7-I4,7 and I1,2-I2,4

I5,7: I2,5-I2,7 and I3,5-I3,7

I6,7: I2,6-I2,7 and I3,6-I3,7

I1,3: I1,2-I2,3 and I1,6-I3,6

I6,8: I4,6-I4,8 and I3,6-I3,8

4,7

7

4

2

3 2,8

1,6 at infinity 3,6

Note that, for clarity, not all ICs are shown.

5

8 1,8 at infinity 2,6 3,8 2,3; 2,4; 2,5; 3,4; 3,5; and 4,5 3,7

1,7 at infinity 2,7

5,8

1,2 4,6

1,4

6 1

1,3

To 1,5 To 1,5



d.


1.


2.

n ( n  1)

C6

2

E

Draw the linkage to scale and identify those ICs that can be found by inspection. 1

2,3

3.

3,4


3 2

I1,3: I1,2-I2,3 and I1,4-I3,4

4 1,2

I2,4: I1,2-I1,4 and I2,3-I3,4

1,4

E

1,3 at infinity

1

1

2,3

4

3,4

2

3 3

2 2,4 at infinity 4 1,2

1,4

e.

This is an eightbar with n  8.

1.


2.

n ( n  1)

C  28

2

3,4

2,3 1,2


4

4,6 2

4,7

5 4,5

2,5

3.

1,4

3

The remaining 17 ICs are at infinity.

6

7

7,8

6,8 8 4 1,4 at infinity 3

1

f.

This is an offset crank-slider with n  4.

1.


2,3

C  2 1,2

1,8

3,4

2.

n ( n  1) 2

C6



3.



I2,4: I1,2-I1,4 and I2,3-I3,4 3,4

4

1,3

1,4 at infinity

1 3

1

4

2

2,3

2

3

2,4

1,4 at infinity 1,2

g.

This is a sixbar with n  6.

1.


3,4

n ( n  1)

1,6

C  15

2

2,3

3 1,4 2

2.


3.


6

4

5

2,5

5,6

1,4


4,5

I3,5: I1,3-I1,5 and I3,2-I2,5 I2,4: I1,2-I1,4 and I3,4-I2,3 I2,6: I1,6-I1,2 and I2,5-I5,6 I4,6: I1,4-I1,6 and I2,4-I2,6

3,5 at infinity 3,4

1,6

I4,5: I4,6-I5,6 and I3,4-I3,5

2,4 3

I3,6: I3,4-I4,6 and I3,5-I5,6

3,5 at infinity

1,2 2

2,3 and 1,5

6

4 5 4,5 and 1,3

1,4 4,6 at infinity

3,6

2,6

5,6




The linkage in Figure P6-5a has the dimensions and crank angle given below. Find 3, VA, VB, and VC for the position shown for 2 = 15 rad/sec in the direction shown. Use the velocity difference graphical method.

Given:

Link lengths: Link 2 (O2 to A)

a  0.80 in

1.

p  1.33 in

Link 3 (A to B)

b  1.93 in

Angle BAC

δ  38.6 deg

Offset

c  0.38 in

Crank angle:

θ  34.3 deg

Input crank angular velocity Solution:

Coupler point data: Distance from A to C

ω  15 rad sec

1

CCW


Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Direction of VCA Direction of VA

C

Y

A

38.600°

34.300° 154.502°

X O2

Direction of VB B

Direction of VBA

2.

Use equation 6.7 to calculate the magnitude of the velocity at point A. VA  a  ω

3.

VA  12.00

in

θ  34.3 deg  90 deg

sec

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.

0

VA

5 in/sec

Y

2.197

124.300° V BA X

VB 2.298


4.


5.

kv 

5  in sec

1

in in

VB  2.298  in kv

VB  11.490

VBA  2.197  in kv

VBA  10.985

VBA

in sec

b

ω  5.692

rad sec

Determine the magnitude and sense of the vector VCA using equation 6.7. VCA  p  ω

VCA  7.570

in sec

θCA  ( 154.502  180  38.6  90)  deg 7.

θB  180  deg

sec

Determine the angular velocity of link 3 using equation 6.7. ω 

6.


θCA  76.898 deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VA + VCA

VA

a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, layout the (now) known vector VCA. c. Complete the vector triangle by drawing VC from the tail of VA to the tip of the VCA vector.

Y 0

V CA

5 in/sec

153.280° VC X

8.

1.130


VC  1.130  in kv

kv 

5  in sec

1

in

VC  5.650

in sec

θC  153.28 deg




The linkage in Figure P6-5a has the dimensions and crank angle given below. Find 3, VA, VB,

Given:

and VC for the position shown for 2 = 15 rad/sec in the direction shown. Use the instant center graphical method. Link lengths: Coupler point data: Link 2 (O2 to A)

a  0.80 in

Distance from A to C

p  1.33 in

Link 3 (A to B)

b  1.93 in

Angle BAC

δ  38.6 deg

Offset

c  0.38 in

Input crank angular velocity Solution: 1.

θ  34.3 deg

Crank angle:

ω  15 rad sec

1

CCW

See Figure P6-5a and Mathcad file P0616b.

Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. From the layout: AI13  2.109  in

1,3 2.109

BI13  2.019  in

0.993 116.732°

CI13  0.993  in 2,4

θC  ( 360  116.732 )  deg 2.


C 2.019

A 1,2

2

3

O2

4

VA  a  ω VA  12.000

B

in

1,4 at infinity

θVA  124.3 deg


4.

1,4 at infinity

sec

θVA  θ  90 deg 3.

3,4

2,3

VA AI13

ω  5.690

rad

CW

sec

Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection. VB  BI13 ω

VB  11.488

in sec

θVC  180  deg 5.

Determine the magnitude of the velocity at point C using equation 6.9b. Determine its direction by inspection. in

VC  CI13 ω

VC  5.650

θVC  θC  90 deg

θVC  153.268 deg

sec




The linkage in Figure P6-5a has the dimensions and crank angle given below. Find 3, VA, VB,

Given:

and VC for the position shown for 2 = 15 rad/sec in the direction shown. Use an analytical method. Link lengths: Coupler point data: Link 2 (O2 to A)

a  0.80 in

Distance from A to C

Rca  1.33 in

Link 3 (A to B)

b  1.93 in

Angle BAC

δ  38.6 deg

Offset

c  0.38 in


ω  15 rad sec

1

CCW

See Figure P6-5a and Mathcad file P0616c.

C

Draw the linkage to scale and label it. Y

2.

θ  34.3 deg

Crank angle:

Determine 3 and d using equation 4.17.

A

38.600°

34.300° 154.502°

X O2

0.380"

B

 a  sin θ  c  π b  

θ  asin 

θ  154.502 deg

 

 

d 1  a  cos θ  b  cos θ 3.

Determine the angular velocity of link 3 using equation 6.22a: ω 

4.

d 1  2.403 in

   


ω  5.691

rad sec




 

 

VA  a  ω sin θ  j  cos θ VA  ( 6.762  9.913j) 5.

in sec

VA  12.000

arg VA  124.300 deg

in sec

Determine the velocity of pin B using equation 6.22b:

 

 

VB  a  ω sin θ  b  ω sin θ VB  11.490 6.

in

VB  11.490

sec

arg VB  180.000 deg

in sec

Determine the velocity of the coupler point C for the open circuit using equations 6.36.











VCA  Rca ω sin π  θ  δ  j  cos π  θ  δ VCA  ( 1.716  7.371j)

in sec

VC  VA  VCA VC  ( 5.047  2.542j)

in sec

VC  5.651

in sec

arg VC  153.268 deg

Note that 3 is defined at point B for the slider-crank and at point A for the pin-jointed fourbar. Thus, to use equation 6.36a for the slider-crank, 180 deg must be added to the calculated value of 3.




The linkage in Figure P6-5c has the dimensions and effective crank angle given below. Find 3, 4, VA, VB, and VC for the position shown for 2 = 15 rad/sec in the direction shown. Use the velocity difference graphical method.

Given:

Link lengths:

Coupler point:

Link 2 (point of contact to A)

a  0.75 in

Distance A to C

p  1.2 in

Link 3 (A to B)

b  1.5 in

Angle BAC

δ  30 deg

Link 4 (point of contact to B)

c  0.75 in

Link 1 (between contact points)

d  1.5 in

ω  15 rad sec


θ  77 deg

Crank angle: 1

See Figure P6-5c and Mathcad file P0617a.

Although the mechanism shown in Figure P6-5c is not entirely pin-jointed, it can be analyzed for the position shown by its effective pin-jointed fourbar, which is shown below. Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. 0

0.5

1 in

Direction of VCA Y

C Direction of VA Direction of VBA

30.000°

Direction of VB

3

A

B b 4

2 a

77.000°

c d

X

O2

O4 Effective link 2

2.

Effective link 4


3.

77.000°

VA  11.250

in sec

θ  77 deg  90 deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relati velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.



Direction of V B

Y

Direction of V BA

167.000° VA

VB X

0

4.

From the velocity triangle we have: VB  VA VBA  0 

5.

ω 

sec

θ  167  deg

in sec

VBA b VB c

ω  0.000

rad sec rad

ω  15.000

sec

Determine the magnitude of the vector VCA using equation 6.7. VCA  p  ω

7.

in

VB  11.250

Determine the angular velocity of links 3 and 4 using equation 6.7. ω 

6.

5 in/sec

VCA  0.000

in sec

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VA + VCA Normally, we would draw the velocity triangle represented by this equation. However, since VCA is zero, in

VC  VA

VC  11.250

θC  θ

θC  167.000 deg

sec




The linkage in Figure P6-5c has the dimensions and effective crank angle given below. Find 3,

Given:

4, VA, VB, and VC for the position shown for 2 = 15 rad/sec in the direction shown. Use the instant center graphical method. Link lengths: Coupler point: Link 2 (point of contact to A)

a  0.75 in

Link 3 (A to B)

b  1.5 in


c  0.75 in


d  1.5 in

Solution: 1.

p  1.2 in

Angle BAC

δ  30 deg

Crank angle:

ω  15 rad sec

Input crank angular velocity

Distance A to C

1

θ  77 deg

CCW

See Figure P6-5c and Mathcad file P0617b.

Although the mechanism shown in Figure P6-5c is not entirely pin-jointed, it can be analyzed for the position shown by its effective pin-jointed fourbar, which is shown below. Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. 1,3 at infinity

1,3 at infinity C

2,3

3,4 30.000°

3

2,4 at infinity A

2

77.000° 2,4 at infinity

O2

77.000° 2,4 at infinity

O4 1,2

1,3 at infinity

Since I

1,3

2,4 at infi

B

4

1,4

1,3 at infinity

is at infinity, link 3 is not rotating ( ω  0  rad sec

1

). Thus, the velocity of every point on

link 3 is the same. From the layout above: θ  77.000 deg 2.

Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point A. in VA  a  ω VA  11.250 sec

θVA  θ  90 deg 3.

θ  0.000  deg

θVA  167.0 deg

Determine the magnitude of the velocity at point B knowing that it is the same as that of point A. in

VB  VA

VB  11.250

θVB  θ  90 deg

θVB  167.000 deg

sec


4.


5.


VB c

ω  15

rad

CCW

sec

Determine the magnitude of the velocity at point C knowing that it is the same as that of point A. in

VC  VA

VC  11.250

θVC  θVA

θVC  167.000 deg

sec




The linkage in Figure P6-5c has the dimensions and effective crank angle given below. Find 3,

Given:

4, VA, VB, and VC for the position shown for 2 = 15 rad/sec in the direction shown. Use an analytical method. Link lengths: Coupler point: Link 2 (point of contact to A)

a  0.75 in

Distance A to C

Rca  1.2 in

Link 3 (A to B)

b  1.5 in

Angle BAC

δ  30 deg


c  0.75 in


d  1.5 in

Crank angle:

θ  77 deg

ω  15 rad sec


1

See Figure P6-5c and Mathcad file P0617c.


C

0

0.5

1 in

3

30.000° A

B b 4

2 a

77.000°

c

77.000°

d

X O4

O2 Effective link 2

2.

Effective link 4


d

K2 

a

K1  2.0000

2

d

K3 

c

K2  2.0000

2

2

a b c d

2

2 a c

K3  1.0000

    B  2  sin θ C  K1   K2  1   cos θ  K3

A  cos θ  K1  K2 cos θ  K3

A  1.2250 3.

B  1.9487

C  2.3251

Use equation 4.10b to find values of 4 for the open circuit.





θ  2  atan2 2  A B 

2

B  4 A  C



θ  283.000 deg

θ  θ  360  deg 4.

θ  643.000 deg


d b

2

K5 

2

2

c d a b 2 a b

2

K4  1.0000



K5  2.0000

    E  2  sin θ F  K1   K4  1   cos θ  K5

D  cos θ  K1  K4 cos θ  K5

5.

D  3.5501 E  1.9487 F  0.0000






2

θ  2  atan2 2  D E 

E  4  D F



θ  360.000 deg

θ  θ  360  deg 6.

Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18. ω  ω 

7.

θ  0.000 deg

    a  ω sin θ  θ  c sin θ  θ a  ω sin θ  θ  b sin θ  θ

ω  0.000

rad sec

ω  15.000

rad sec




 

 

VA  a  ω sin θ  j  cos θ VA  ( 10.962  2.531j)



 

in sec

in

VA  11.250

sec

 

arg VA  167.000 deg

VB  c ω sin θ  j  cos θ VB  ( 10.962  2.531j) 8.

in

in

VB  11.250

sec

sec

arg VB  167.000 deg

Determine the velocity of the coupler point C for the open circuit using equations 6.36.











VCA  Rca ω sin θ  δ  j  cos θ  δ VCA  0.000

in sec

VC  VA  VCA VC  ( 10.962  2.531j)

in sec

VC  11.250

in sec

arg VC  167.000 deg




Given:

Solution: 1.

The linkage in Figure P6-5f has the dimensions and coupler angle given below. Find 3, VA, VB, and VC for the position shown for VA = 10 in/sec in the direction shown. Use the velocity difference graphical method. Link lengths and angles: Coupler point: Link 3 (A to B)

b  1.8 in

Distance A to C

p  1.44 in

Coupler angle

θ  128  deg

Angle BAC

δ  49 deg

Slider 4 angle

θ  59 deg

Input slider velocity

VA  10 in sec

See Figure P6-5f and Mathcad file P0618a.

Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest.

Direction of VB 0

0.5

1 in

Y

Direction of VBA

C

B 4

Direction of VCA

3 b

49.000°

128.000°

59.000°

VA

X A

2.

The magnitude and sense of the velocity at point A. VA  10.000

3.

2

in sec

θ  180  deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relative velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.

0

5 in/sec

Y

V BA 4.784 VB

3.436

38.000°

59.000° X

VA

1


4.


5.

kv 

5  in sec

1

in in

VB  3.438  in kv

VB  17.190

VBA  4.784  in kv

VBA  23.920

sec in sec

VBA

ω  13.289

b

θ  38 deg

rad sec

Determine the magnitude and sense of the vector VCA using equation 6.7. VCA  p  ω

VCA  19.136

in sec

θCA  ( 128  49  90)  deg 7.

θ  59 deg


6.


θCA  11.000 deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VA + VCA a. b. c.

Choose a convenient velocity scale and layout the known vector VA. From the tip of VA, layout the (now) known vector VCA. Complete the vector triangle by drawing VC from the tail of VA to the tip of the VCA vector. 0

5 in/sec

Y 1.902 VA X 11.000°

22.572° VC V CA

3.827

8.

From the velocity triangle we have:


VC  1.902  in kv

kv 

5  in sec

1

in

VC  9.510

in sec

θC  22.572 deg




Given:

Solution: 1.

The linkage in Figure P6-5f has the dimensions and coupler angle given below. Find 3, VA, VB, and VC for the position shown for VA = 10 in/sec in the direction shown. Use the instant center graphical method. Link lengths and angles: Coupler point: Link 3 (A to B)

b  1.8 in

Distance A to C

p  1.44 in

Coupler angle

θ  128  deg

Angle BAC

δ  49 deg

Slider 4 angle

θ  59 deg

VA  10 in sec


See Figure P6-5f and Mathcad file P0618b.

Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. 2,4 at infinity

From the layout:

3,4

1,4 at infinity

C

B

AI13  0.753  in

BI13  1.293  in

CI13  0.716  in

θC  67.428 deg

4

0.716 67.428° 3

2.

4.

VA AI13

ω  13.280

1,3

1.293


3.

1

0.753

rad

VA

CW

sec

A

2 2,3

Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection.

1,2 at infinity

in

VB  BI13 ω

VB  17.17

θVB  θ

θVB  59.000 deg

sec

Determine the magnitude of the velocity at point C using equation 6.9b. Determine its direction by inspection. in

VC  CI13 ω

VC  9.509

θVC  θC  90 deg

θVC  22.572 deg

sec




Given:

Solution: 1.

The linkage in Figure P6-5f has the dimensions and coupler angle given below. Find 3, VA, VB, and VC for the position shown for VA = 10 in/sec in the direction shown. Use an analytical method. Link lengths and angles: Coupler point: Link 3 (A to B)

b  1.8 in

Distance A to C

Rca  1.44 in

Coupler angle

θ  128  deg

Angle BAC

δ  49 deg

Slider 4 angle

θ  59 deg

VA  10 in sec


1

See Figure P6-5f and Mathcad file P0618c.

Draw the mechanism to scale and define a vector loop using the fourbar slider-crank derivation in Section 6.7 as a model. 0

0.5

1 in

Y

C

B 4 R3 R4 3 b

49.000°

128.000°

59.000° VA

X

R2

2.

A

2

Write the vector loop equation, differentiate it, expand the result and separate into real and imaginary parts to solve for 3 and VB. R2  R3  R4 a e

j  θ

 b e

j  θ

 c e

j  θ

where a is the distance from the origin to point A, a variable; b is the distance from A to B, a constant; and c is the distance from the origin to point B, a variable. Angle 2 is zero, 3 is the angle that AB makes with the x axis, and 4 is the constant angle that slider 4 makes with the x axis. Differentiating, j  θ d  d  j  θ a  j  b  ω e   c   e dt  dt 

Substituting the Euler equivalents, d d  a  b  ω sin θ  j  cos θ   c   cos θ  j  sin θ dt  dt  Separating into real and imaginary components and solving for 3 and VB. Note that dc/dt = VB and da/dt = VA



ω 

 

 

  b   sin θ  tan  θ  cos θ  VA tan θ

  

 

ω  13.288

rad sec


VB 


 

VA  b  ω sin θ

 

VB  17.180

cos θ

  

 

sec

arg VB  59.000 deg

VB  VB cos θ  j  sin θ 3.

in

Determine the velocity of the coupler point C using equations 6.36.









VCA  Rca ω sin θ  δ  j  cos θ  δ VCA  ( 18.783  3.651j)



in sec

VA  VA VC  VA  VCA VC  ( 8.783  3.651j)

in sec

VC  9.512

in sec

arg VC  22.572 deg




The cam-follower in Figure P6-5d has O2A = 0.853 in. Find V4, Vtrans, and Vslip for the position shown with 2 = 20 rad/sec in the direction (CCW) shown.

Given:

ω  20 rad sec

1

O A a  0.853  in 2

Assumptions: Rolling contact (no sliding) Solution: 1.


Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest. Direction of VA Axis of transmission

Direction of V4 3 O2

4

B A 2

Axis of slip 0.853

2.


3.

VA  17.060

in sec

Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A. The equation to be solved graphically is VA = Vtrans + Vslip a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA and drawing Vtrans from the tail of VA to the intersection of the Vslip construction line.

Y 0.768 0

VA

10 in/sec

1.523

V slip

V trans

V4

X

0.870

4.


Vslip  1.523  in kv

kv 

10 in sec

1

in

Vslip  15.230

in sec

Vtrans  0.768  in kv

Vtrans  7.680

V4  0.870  in kv

V4  8.700

in sec

in sec




The cam-follower in Figure P6-5e has O2A = 0.980 in and O3A = 1.344 in. Find 3, Vtrans, and Vslip for the position shown with 2 = 10 rad/sec in the direction (CW) shown.

Given:

ω  10 rad sec

1

Distance from O to A: a  0.980  in 2

Distance from O3 to A: b  1.344  in Assumptions: Roll-slide contact Solution:


1.

Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest.

2.

Use equation 6.7 to calculate the magnitude of the velocity at point A.

Direction of VA2 Axis of transmission Direction of VA3

VA  a  ω

VA  9.800

in

Axis of slip

sec 2

3.

O2

Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A. The equation to be solved graphically is

3

A

O3

1.344

VA2 = Vtrans + VA2slip

0.980

a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA and drawing Vtrans from the tail of VA to the intersection of the Vslip construction line. Y

0

4.

From the velocity triangle we have: Velocity scale factor: VA2slip  1.134  in kv Vtrans  1.599  in kv VA3slip  2.048  in kv VA3  2.598  in kv

5.

5  in sec

1

1.599

in

VA2slip  5.670 Vtrans  7.995

sec

2.598

in sec

VA3slip  10.240 VA3  12.990

V trans

in

V A2

in

V A3

VA3 b

The relative slip velocity is

ω  9.665 Vslip  VA3slip  VA2slip

V A2slip

1.134

sec V A3slip

in sec


6.

kv 

rad

CCW

sec Vslip  4.570

in sec

2.048

X

5 in/sec




The linkage in Figure P6-6b has L1 = 61.9, L2 = 15, L3 = 45.8, L4 = 18.1, L5 = 23.1 mm. 2 is 68.3 deg in the xy coordinate system, which is at -23.3 deg in the XY coordinate system. The X component of O2C is 59.2 mm. Find, for the position shown, the velocity ratio VI5,6/VI2,3 and the mechanical advantage from link 2 to link 6. Use the velocity difference graphical method.

Given:

Link lengths:

Solution:

Link 1

d  61.9 mm

Link 2

a  15.0 mm

Link 3

b  45.8 mm

Link 4

c  18.1 mm

Link 5

e  23.1 mm

Offset

f  59.2 mm

from O2

Crank angle:

θ  45 deg

Coordinate rotation angle

α  23.3 deg Global XY system to local xy system

Global XY system

See Figure P6-6b and Mathcad file P0621a.

1.


2.

Choose an arbitrary value for the magnitude of the velocity at I2,3 (point A). Let link 2 rotate CCW. VA  10

y 5,6 6 2,3

mm

A

sec

Direction of VCB C

45.0° 3

2

θVC  45 deg  90 deg

1

5

X

O2 23.3°

B Direction of VBA

1

θVC  135.000 deg 3.

Direction of VC

Direction of VA Y

4

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relative velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is

O4 x

1 Direction of VB

VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line. 4.

From the velocity triangle we have: VA


kv 

5  mm sec

1

0

5 mm/sec

2.053

in

VB  2.248  in kv VB  11.2

Y

mm sec

θVB  ( 360  167.558 )  deg

VBA

X 167.553°

VB 2.248


5.


Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VB + VCB a. Choose a convenient velocity scale and layout the (now) known vector VB. b. From the tip of VB, draw a construction line with the direction of VCB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VC, magnitude unknown. d. Complete the vector triangle by drawing VCB from the tip of VB to the intersection of the VC construction line and drawing VC from the tail of VB to the intersection of the VCB construction line. VA

Y 0

V BA

5 mm/sec

X VB 1.128 VC V CB

6.

From the velocity triangle we have: kv 


VC  1.128  in kv

7.

The ratio V

/V I5,6

8.

is I2,3

VC

5  mm sec

1

in

VC  5.6

mm sec

θVC  270  deg

 0.564

VA

Use equations 6.12 and 6.13 to derive an expression for the mechanical advantage for this linkage where the input is a rotating crank and the output is a slider. Fin =

Fout =

mA =

Tin rin

Pout Vout Fout Fin

Pin

=

rin ωin =

=

=

Pin VA

Pout VC Pout VA  VC Pin

mA 

VA VC

mA  1.773




The linkage in Figure P6-6b has L1 = 61.9, L2 = 15, L3 = 45.8, L4 = 18.1, L5 = 23.1 mm. 2 is 68.3 deg in the xy coordinate system, which is at -23.3 deg in the XY coordinate system. The X component of O2C is 59.2 mm. Find, for the position shown, the velocity ratio VI5,6/VI2,3 and the mechanical advantage from link 2 to link 6. Use the instant center graphical method.

Given:

Link lengths:

Solution: 1.

Link 1

d  61.9 mm

Link 2

a  15.0 mm

Link 3

b  45.8 mm

Link 4

c  18.1 mm

Link 5

e  23.1 mm

Offset

f  59.2 mm

from O2

Crank angle:

θ  45 deg



Global XY system

See Figure P6-6b and Mathcad file P0621b.

Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. From the layout:

2.

AI13  44.594 mm

BI13  50.121 mm

BI15  22.683 mm

CI15  11.377 mm

1,3

5,6


6 1,5 2,3 A

mm 2

sec

50.121

C

3

1

5

22.683

X

O2

θVC  θ  90 deg

B 1

θVC  135.000 deg 3.

11.377

44.594 Y

4

Determine the angular velocity of link 3 using equation 6.9a.

O4 1

ω 

VA

ω  0.224

AI13

rad

CW

sec

4.

Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection. mm VB  BI13 ω VB  11.239 sec

5.

Determine the angular velocity of link 5 using equation 6.9a. ω 

6.

VB

ω  0.495

BI15

The ratio V

/V I5,6

8.

CW

sec

Determine the magnitude of the velocity at point C using equation 6.9b. Determine its direction by inspection. VC  CI15 ω

7.

rad

is I2,3

VC VA

VC  5.637

mm

downward

sec

 0.56

Use equations 6.12 and 6.13 to derive an expression for the mechanical advantage for this linkage where the input is a rotating crank and the output is a slider.


Fin =

Fout =

mA =

Tin rin

Pout Vout Fout Fin

Pin

=

rin ωin =

=


=

Pin VA

Pout VC Pout VA  VC Pin

mA 

VA VC

mA  1.77




Given:

Solution: 1.

The linkage in Figure P6-6d has L2 = 15, L3 = 40.9, L5 = 44.7 mm. 2 is 24.2 deg in the XY coordinate system. Find, for the position shown, the velocity ratio VI5,6/VI2,3 and the mechanical advantage from link 2 to link 6. Use the velocity difference graphical method. Link lengths: Link 2

a  15.0 mm

Link 3

b  40.9 mm

Link 5

c  44.7 mm

Offset

f  0  mm

Crank angle:

θ  24.2 deg

from O2

See Figure P6-6d and Mathcad file P0622a.


4 1

C

Direction of VBA

3

Direction of VA

5,6 A

5

2

O2

6 B 1

2,3

1

Direction of VB

2.


3.

mm

θVC  θ  90 deg

sec

θVC  114.200 deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relati velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line. VA Y 0

5 mm/sec

V BA X

VB 1.073


4.




VB  1.073  in kv

5.

The ratio V

/V I5,6

6.

is I2,3

VB

5  mm sec

1

in

VB  5.4

mm sec

θVB  180  deg

 0.54

VA


Fout =

mA =

Tin rin

Pout Vout

Fout Fin

Pin

=

rin ωin

=

=

=

Pin VA

Pout VB

Pout VA  VB Pin

mA 

VA VB

mA  1.86




Given:

Solution: 1.

The linkage in Figure P6-6d has L2 = 15, L3 = 40.9, L5 = 44.7 mm. 2 is 24.2 deg in the XY coordinate system. Find, for the position shown, the velocity ratio VI5,6/VI2,3 and the mechanical advantage from link 2 to link 6. Use the instant center graphical method. Link lengths: Link 2 Link 3 a  15.0 mm b  40.9 mm Link 5

c  44.7 mm

Crank angle:

θ  24.2 deg

f  0  mm

Offset

from O2

See Figure P6-6d and Mathcad file P0622b.

Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. 4 C

1

3

1,5

48.561

26.075

5,6 A

5

2

O2

6 B 1

2,3

1

From the layout: AI15  48.561 mm 2.


3.

mm

θVC  θ  90 deg

sec

VA

ω  0.206

AI15

The ratio V

/V I5,6

6.

rad

CW

sec

Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection. VB  BI15 ω

5.

θVC  114.200 deg


4.

BI15  26.075 mm

is I2,3

VB VA

VB  5.370

mm

to the left

sec

 0.54


mA =

Tin rin Fout Fin

=

=

Pin rin ωin

=

Pout VA  VC Pin

Pin

Fout =

VA mA 

VA VB

Pout Vout

mA  1.86

=

Pout VC




Generate and draw the fixed and moving centrodes of links 1 and 3 for the linkage in Figure P6-7a

Solution:


1.

Draw the linkage to scale, find the instant center I1,3, and repeat for several positions of the linkage. The locus of points I1,3 is the fixed centrode.

FIXED CENTRODE

2.

Invert the linkage, grounding link 3. Draw the linkage to scale, find the instant center I1,3, and repeat for several positions of the linkage. The locus of points I1,3 is the moving centrode. (See next page.)



MOVING CENTRODE




The linkage in Figure P6-8a has the dimensions and crank angle given below. Find f, VA, and VB for the position shown for  = 15 rad/sec clockwise (CW). Use the velocity difference graphical method.

Given:

Link lengths:

Crank angle:

Link 2 (O2 to A)

a  116  mm

Link 3 (A to B)

b  108  mm

Link 4 (B to O4)

c  110  mm

Link 1 (O2 to O4)

d  174  mm

Coordinate rotation angle Solution: 1.

θ  37 deg

Global XY system

Input crank angular velocity ω  15 rad sec

α  25 deg

1

Global XY system to local xy system


Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Y

Direction of VA A

37.000° 70.133°

2 3

X

O2 d

22.319° B

Direction of VBA 4

Direction of VB

2.


3.

O4

VA  1740.0

mm sec

θA  37 deg  90 deg




1000 mm/sec

0

Y

X 0.952 53.000° VB VA

VBA

1.498 112.319°

4.


5.

1000 mm sec

1

in mm

VB  0.952  in kv

VB  952.0

VBA  1.498  in kv

VBA  1498.0

sec mm sec


6.

kv 

VBA b

ω  13.87

rad sec

Determine the angular velocity of link 4 using equation 6.7. ω 

VB c

ω  8.655

rad sec

θB  112.319  deg




The linkage in Figure P6-8a has the dimensions and crank angle given below. Find f, VA, and VB for the position shown for  = 15 rad/sec clockwise (CW). Use the instant center graphical method.

Given:


a  116  mm

Link 3 (A to B)

b  108  mm

Link 4 (B to O4)

c  110  mm

Link 1 (O2 to O4)

d  174  mm

Coordinate rotation angle Solution: 1.

Crank angle:

α  25 deg

θ  37 deg

Global XY system


1



Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. Y

From the layout:

2,3

A

AI13  125.463  mm

125.463

BI13  68.638 mm θ  157.681  deg

2

3

O2 1,2

X 2,4 157.681°

1,3 B

2.

3.


5.

4

O4

3,4 1,4

mm

VA  a  ω

VA  1740.0

θVA  θ  90 deg

θVA  53.0 deg

sec


4.

68.638

VA AI13

ω  13.869

rad

CW

sec

Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection. mm

VB  BI13 ω

VB  951.915

θVB  θ  90 deg

θVB  247.681 deg

sec


VB c

ω  8.654

rad sec

CCW




The linkage in Figure P6-8a has the dimensions and crank angle given below. Find 4, VA, and VB for the position shown for  = 15 rad/sec clockwise (CW). Use an analytical method.

Given:

Solution:

Link lengths:

Crank angle:

Link 2 (O2 to A)

a  116  mm

Link 3 (A to B)

b  108  mm

Link 4 (B to O4)

c  110  mm

Link 1 (O2 to O4)

d  174  mm

θ  62 deg

Global XY system

Input crank angular velocity ω  15 rad sec

1

CW


1.


2.


y A

K1  K2 

d

K1  1.5000

a

3 O2

d

62.000°

K2  1.5818

c 2

K3 

2

2

2

a b c d

2

2

K3  1.7307

2 a c

B

    B  2  sin θ C  K1   K2  1   cos θ  K3

4

A  cos θ  K1  K2 cos θ  K3

A  0.0424 3.

B  1.7659

4.

x

C  2.0186

Use equation 4.10b to find values of 4 for the crossed circuit.





2

θ  2  atan2 2  A B 

B  4 A  C



θ  182.681 deg


2

d b

2

2

c d a b

K5 

2

2 a b

    E  2  sin θ F  K1   K4  1   cos θ  K5

D  cos θ  K1  K4 cos θ  K5

5.



6.

K4  1.6111

K5  1.7280

D  2.0021 E  1.7659 F  0.0589




2

θ  2  atan2 2  D E 

E  4  D F



θ  275.133 deg


 

 

a  ω sin θ  θ  b sin θ  θ

O4

ω  13.869

rad sec


ω 

7.


 

 

a  ω sin θ  θ  c sin θ  θ

ω  8.654

rad sec

Determine the velocity of points A and B for the crossed circuit using equations 6.19.



 

 

VA  a  ω sin θ  j  cos θ VA  ( 1536.329  816.881i)



 

mm sec

VA  1740.000

 

mm sec

arg VA  28.000 deg

VB  c ω sin θ  j  cos θ VB  ( 44.524  950.875j)

mm sec

VB  951.917

mm sec

arg VB  87.319 deg




The linkage in Figure P6-8a has the dimensions given below. Find and plot 4, VA, and VB in

Given:

the local coordinate system for the maximum range of motion that this linkage allows if 2 = 15 rad/sec clockwise (CW). Link lengths: Link 2 (O2 to A)

a  116  mm

Link 3 (A to B)

b  108  mm

Link 4 (B to O4)

c  110  mm

Link 1 (O2 to O4)

d  174  mm

ω  15 rad sec


1

CW


1.


2.

Determine the range of motion for this non-Grashof triple rocker using equations 4.33.

y A

2

arg1 

2

2



2 a d 2

arg2 

2

a d b c

2

2

a d b c

2

2 a d

arg1  1.083



b c

2 3

a d

O2

b c

2

a d

62.000°

B

arg2  0.094


4

O4


x

The other toggle angle is the negative of this. Thus, θ  θ2toggle  1  deg θ2toggle  2  deg  θ2toggle  1  deg 3.


d

K2 

a

K1  1.5000 2

K3 

d c

K2  1.5818 2

2

a b c d

2

K3  1.7307

2 a c

      B θ  2  sin θ C θ  K1   K2  1   cos θ  K3

A θ  cos θ  K1  K2 cos θ  K3

4.


 





 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


 

2

d

K5 

b

 

2

2

c d a b 2 a b

 

D θ  cos θ  K1  K4 cos θ  K5

2

K4  1.6111

K5  1.7280


 


 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 6.




 



 

 

θ θ  2   atan2 2  D θ E θ  7.

Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.

 

a  ω

 

a  ω

ω θ 

ω θ  8.

 2  4 Dθ F θ 

E θ

b

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 

sin θ θ  θ



sin θ  θ θ


 



 

 

VA θ  a  ω sin θ  j  cos θ

 

  

 

 

 

  

 

VAx θ  Re VA θ

  

VAy θ  Im VA θ

    j  cosθθ

VB θ  c ω θ  sin θ θ

 

VBx θ  Re VB θ

Plot the angular velocity of the output link, 4, and the x and y components of the velocities at points A and B. ANGULAR VELOCITY OF LINK 4 60 40 Angular Velocity, rad/sec

9.

  

VBy θ  Im VB θ

20 0

 

ω  θ 

sec rad

 20  40  60  80  100  100

 75

 50

 25

0 θ deg

25

50

75

100



VELOCITY COMPONENTS, POINT A 0.5

Joint Velocity, m/sec

0

 

VAy θ 

 0.5 sec m 1

 1.5

2 2

 1.5

1

 0.5

0

 

0.5

VAx θ 

1

1.5

2

sec m

VELOCITY COMPONENTS, POINT B 4

Joint Velocity, m/sec

3 2

 

VBy θ 

sec 1 m

0 1 2 3 4

3

2

1

0

 

VBx θ 

1 sec m

2

3

4




The linkage in Figure P6-8b has the dimensions and crank angle given below. Find 4, VA, and

Given:

VB for the position shown for 2 = 20 rad/sec counterclockwise (CCW). Use the velocity difference graphical method. Link lengths: Crank angle: Link 2 (O2 to A)

a  40 mm

Link 3 (A to B)

b  96 mm

Link 4 (B to O4)

c  122  mm

Link 1 (O2 to O4)

d  162  mm

θ  57 deg


α  36 deg


Global XY system

1


See Figure P6-8b and Mathcad file P0628. Solution: 1. Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. 2.

Direction of VA

Use equation 6.7 to calculate the magnitude of the velocity at point A.

VA  800.000

mm sec

2

y

B 3

A

2

X O2

θA  θ  90 deg 3.

Direction of VB

Y

VA  a  ω

Direction of VBA

4

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relative velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is VB = VA + VBA

O4 x

a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line. 0

4.

400 mm/sec

From the velocity triangle we have: 147.000°

Velocity scale factor: kv 

400  mm sec

1

VA Y

in

VB  1.790  in kv

VB  716.0

mm

1.293" V BA

sec X

θB  186.406  deg VBA  1.293  in kv 5.

mm sec

6.406°

1.790" 85.486°


6.

VBA  517.2

VB

VBA b

ω  5.388

rad sec


VB c

ω  5.869

rad sec




Given:

The linkage in Figure P6-8b has the dimensions and crank angle given below. Find 4, VA, and VB for the position shown for 2 = 20 rad/sec counterclockwise (CCW). Use the instant center graphical method. Link lengths: Crank angle: a  40 mm θ  57 deg Global XY system Link 2 (O2 to A) Input crank angular velocity b  96 mm Link 3 (A to B) 1 c  122  mm ω  20 rad sec Link 4 (B to O4) d  162  mm Link 1 (O2 to O4) Coordinate rotation angle

Solution: 1.

α  36 deg



Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. From the layout: AI13  148.700  mm

1,3

BI13  133.192  mm

θ  96.406 deg 2.

133.192

Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point A. mm VA  a  ω VA  800.0 sec

θVA  θ  90 deg 3.

148.700

Y 2

A

2

θVA  147.0 deg

y

B 3 X

O2 4


VA AI13

ω  5.380

rad

96.406°

CW

sec

O4 x

4.

5.


VB  BI13 ω

VB  716.6

θVB  θ  90 deg

θVB  186.406 deg

sec


VB c

ω  5.874

rad sec

CCW




The linkage in Figure P6-8b has the dimensions and crank angle given below. Find 4, VA, and

Given:

VB for the position shown for 2 = 20 rad/sec counterclockwise (CCW). Use an analytical method. Link lengths: Crank angle: Link 2 (O2 to A)

a  40 mm

Link 3 (A to B)

b  96 mm

Link 4 (B to O4)

c  122  mm

Link 1 (O2 to O4)

d  162  mm α  36 deg

Coordinate rotation angle Solution:


2.


d

K2 

a

K1  4.0500 2

K3 

1


Y

y

2

c

B

3

A

X 93.000 4

K2  1.3279 2

2

a b c d

2

K3  3.4336

2 a c

A  0.5992

B  1.9973

θ  θ  α

O4 x

 

B  2  sin θ

C  7.6054






θ  2  atan2 2  A B 

2

B  4 A  C

  2 π

θ  587.614 deg


2

d

K5 

b

2

2

c d a b

    E  2  sin θ F  K1   K4  1   cos θ  K5

2

2 a b

D  cos θ  K1  K4 cos θ  K5

K4  1.6875

K5  2.8875

D  7.0782 E  1.9973 F  1.1265






θ  2  atan2 2  D E  6.

ω  20 rad sec

O2

    C  K1   K2  1   cos θ  K3

5.


2

d

A  cos θ  K1  K2 cos θ  K3

4.

Global XY system


1.

3.

θ  57 deg

2

E  4  D F

  2 π

θ  688.496 deg


ω 

a  ω sin θ  θ  b sin θ  θ

 

 

ω  5.385

ω 

a  ω sin θ  θ  c sin θ  θ

 

 

ω  5.868

rad sec

rad sec


7.





 

 

VA  a  ω sin θ  j  cos θ VA  ( 798.904  41.869i )



 

mm sec

VA  800.000

mm

VB  715.900

mm

 

sec

arg VA  177.000 deg

VB  c ω sin θ  j  cos θ VB  ( 528.774  482.608j)

mm sec

sec

arg VB  137.614 deg




The linkage in Figure P6-8b has the dimensions and crank angle given below. Find and plot 4, VA, and VB in the local coordinate system for the maximum range of motion that this linkage allows if 2 = 20 rad/sec counterclockwise (CCW).

Given:


a  40 mm

Link 3 (A to B)

b  96 mm

Link 4 (B to O4)

c  122  mm

Link 1 (O2 to O4)

d  162  mm

ω  20 rad sec


1

CCW


1.


2.

Determine Grashof condition. Condition( S L P Q) 

Y 2

SL  S  L

y B

3

A

93.000 X

2

PQ  P  Q

O2 4

return "Grashof" if SL  PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise O4

Condition( a d b c)  "Grashof"

x

Crank-rocker

θ  0  deg 1  deg  360  deg 3.


d

K2 

a

K1  4.0500

 

2

d

K3 

c

K2  1.3279

 

2

2

2 a c

K3  3.4336

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

2

a b c d

 

B θ  2  sin θ

 

C θ  K1   K2  1   cos θ  K3 4.




 



 

 

θ θ  2   atan2 2  A θ B θ  5.

 2  4 A θ Cθ 

B θ


2

d

K5 

b

2

2

c d a b

2

K4  1.6875

2 a b

K5  2.8875

      E θ  2  sin θ F  θ  K1   K4  1   cos θ  K5

D θ  cos θ  K1  K4 cos θ  K5

6.


 





 

 

θ θ  2   atan2 2  D θ E θ  7.

 2  4 Dθ F θ 

E θ



 

a  ω

 

a  ω

ω θ 

ω θ  8.

b

c




    sin θ θ  θ θ 



  sin θ θ  θ θ 

sin θ θ  θ



sin θ  θ θ

Determine the velocity of points B and C for the open circuit using equations 6.19.

 



 

 

VA θ  a  ω sin θ  j  cos θ

 

  

 

 

 

  

 


  


    j  cosθθ

VB θ  c ω θ  sin θ θ

 

VBx θ  Re VB θ

Plot the angular velocity of the output link, 4, and the magnitudes of the velocities at points B and C. ANGULAR VELOCITY OF LINK 4

Angular Velocity, rad/sec

10

5

 

ω  θ 

sec 0

rad

5

 10

0

45

90

135

180

225

270

315

360

θ deg

VELOCITY COMPONENTS, POINT A

3

1 10 Joint Velocity, mm/sec

9.

  


 

VAx θ 

 

VAy θ 

sec

500

mm sec mm

0  500 3

 1 10

0

45

90

135

180 θ deg

225

270

315

360



VELOCITY COMPONENTS, POINT B

3

Joint Velocity, mm/sec

1 10

600

 

VBx θ 

 

VBy θ 

sec mm

200

sec

 200

mm  600  1 10

3

0

45

90

135

180 θ deg

225

270

315

360




The offset slider-crank linkage in Figure P6-8f has the dimensions and crank angle given below. Find VA, and VB for the position shown if 2 = 25 rad/sec CW. Use the velocity difference graphical method.

Given:

Link lengths:

Solution:

Link 2

a  63 mm

Crank angle:

θ  51 deg

Link 3

b  130  mm


Offset

c  52 mm

ω  25 rad sec

1

CW


1.


2.


VA  1575.0

Direction of VB 4 B

mm

Direction of VBA

1

sec

3

θA  51 deg  90 deg 3.

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B. The equation to be solved graphically is

Direction of VA A 2 51.000°

VB = VA + VBA O2

a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line. 4.

0


kv 

VB  2.208  in kv

θVB  270  deg

1000 mm sec

1000 mm/sec

Y

1 X

in VB  2208

mm VA

sec 2.208

VB V BA




The offset crank-slider linkage in Figure P6-8f has the dimensions and crank angle given below. Find VA, and VB for the position shown for 2 = 25 rad/sec CW. Use the instant center graphical method.

Given:


a  63 mm

Crank angle:

θ  51 deg

Link 3

b  130  mm

Offset

c  52 mm

ω  25 rad sec


1

CW


Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis. 166.309 4 B

1,3

1 3

A

118.639

2

O2

From the layout above: AI13  118.639  mm 2.


θVA  θ  90 deg 3.



4.

BI13  166.309  mm

VA AI13

ω  13.276

rad

CCW

sec

Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection. VB  BI13 ω

θVB  270  deg

VB  2207.8

mm sec




The offset slider-crank linkage in Figure P6-8f has the dimensions and crank angle given below. Find VA, and VB for the position shown for 2 = 25 rad/sec CW. Use an analytical method.

Given:


a  63 mm

Link 3

b  130  mm

Offset

c  52 mm

θ  141  deg

Crank angle:

Local xy coordinate system Input crank angular velocity ω  25 rad sec

Solution:


Y

1.

Draw the linkage to a convenient scale.

2.

Determine 3 and d using equations 4.16 for the crossed circuit.

4

 a  sin θ  b 

θ  asin

 

c

 

1 3

 

d  141.160 mm

A

52.000 2

Determine the angular velocity of link 3 using equation 6.22a: ω 

4.

B

θ  44.828 deg

d  a  cos θ  b  cos θ 3.

1

   

a cos θ   ω b cos θ

ω  13.276

rad

 

141.000°

sec

x




X, y

O2

 

VA  a  ω sin θ  j  cos θ VA  ( 991.180  1224.005i )

mm sec

In the global coordinate system, 5.

VA  1575.000

mm sec

θVA  arg VA  90 deg

arg VA  51.000 deg

θVA  39.000 deg


 

 

VB  a  ω sin θ  b  ω sin θ

VB  2207.849

mm sec

VB  VB In the global coordinate system,

θVB  arg VB  90 deg

θVB  90.000 deg




The offset crank-slider linkage in Figure P6-8f has the dimensions and crank angle given below. Find and plot VA, and VB in the global coordinate system for the maximum range of motion that this linkage allows if 2 = 25 rad/sec CW.

Given:


a  63 mm

Link 3

b  130  mm ω  25 rad sec


c  52 mm

Offset 1


1.

Draw the linkage to a convenient scale. The coordinate rotation angle is α  90 deg

2.

Determine the range of motion for this slider-crank linkage. 4

θ  0  deg 2  deg  360  deg 3.

 a  sin θ  b 

 

3

c

 

 

a b



2

   ω cos θ θ  cos θ

O2

Determine the x and y components of the velocity of pin A using equation 6.23a:

 



 

 

VA θ  a  ω sin θ  j  cos θ

 

  


 

  

 

 


In the global coordinate system,

 

 

VAX θ  VAy θ 6.

VAY θ  VAx θ


 

 

    

VBx θ  a  ω sin θ  b  ω θ  sin θ θ In the global coordinate system,

 

 

VBY θ  VBx θ 7.

A

52.000

Determine the angular velocity of link 3 using equation 6.22a: ω θ 

5.

B

1

Determine 3 using equations 4.16 for the crossed circuit. θ θ  asin

4.

Y

Plot the x and y components of the velocity of A. (See next page.)

2 x

X, y



VELOCITY OF POINT A 2000

Velocity, mm/sec

1000

 

VAX θ 

 

VAY θ 

sec mm 0

sec mm

 1000

 2000

0

60

120

180

240

300

360

θ deg

Plot the velocity of point B. VELOCITY OF POINT B 2000

1000 Velocity, mm/sec

7.

0

 

VBY θ 

sec mm  1000

 2000

 3000

0

60

120

180 θ deg

240

300

360




Given:

The linkage in Figure P6-8d has the dimensions and crank angle given below. Find VA, VB, and Vbox for the position shown for 2 = 30 rad/sec clockwise (CW). Use the velocity difference graphical method. Link lengths: Link 2 Link 3 a  30 mm b  150  mm Link 4

c  30 mm

Crank angle:

θ  58 deg

Global XY system ω  30 rad sec


d  150  mm

Link 1 1

CW


1.


2.

Use equation 6.7 to calculate the magnitude of the velocity at point A. mm VA  a  ω VA  900.000 sec

Vbox 1

Direction of VBA

θA  58 deg  90 deg 3.

Direction of VA

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relative velocity VBA, and the angular velocity of link 3. The equation to be solved graphically is

B

3

A

Direction of VB

2 O4

O2

4

VB = VA + VBA a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line. Y

4.


Direction of VBA

VB  VA

X 0

VB  900.000

θB  32 deg 5.

VB

sec VBA  0 

in

VA

sec


6.

500 mm/sec

32.000°

mm

VBA b

ω  0.000

rad sec

ω 

VB c

ω  30.000

rad sec

Determine the magnitude of the vector Vbox . This is a special case Grashof mechanism in the parallelogram configuration. Link 3 does not rotate, therefore all points on link 3 have the same velocity. The velocity Vbox is the horizontal component of VA. mm Vbox  VA cos θA Vbox  763.243 sec




The linkage in Figure P6-8d has the dimensions and effective crank angle given below. Find VA,

Given:

VB, and Vbox in the global coordinate system for the position shown for 2 = 30 rad/sec CW. Use an analytical method. Link lengths:

Solution: 1.

Link 2

a  30 mm

Link 3

b  150  mm

Link 4

c  30 mm

Link 1

d  150  mm

Crank angle:

θ  58 deg


ω  30 rad sec

1

See Figure P6-8d and Mathcad file P0637. Vbox

Draw the linkage to scale and label it. 1

2.


d

K2 

a

K1  5.0000 2

K3 

2

c

2

2

2

K3  1.0000

2 a c

    C  K1   K2  1   cos θ  K3

 

A  cos θ  K1  K2 cos θ  K3 A  6.1197 3.

B  1.6961

B  2  sin θ

C  2.8205






2

θ  2  atan2 2  A B 

B  4 A  C



θ  302.000 deg

θ  θ  360  deg 4.

θ  662.000 deg


2

d

K5 

b

2

2

c d a b 2 a b

    E  2  sin θ F  K1   K4  1   cos θ  K5

D  cos θ  K1  K4 cos θ  K5

5.

2

K4  1.0000

K5  5.0000

D  8.9402 E  1.6961 F  0.0000






2

θ  2  atan2 2  D E 

E  4  D F

θ  θ  360  deg 6.

O4

O2

K2  5.0000

a b c d

B

3

A

d



θ  360.000 deg θ  0.000 deg


 

 

a  ω sin θ  θ  b sin θ  θ

ω  0.000

rad sec

4


ω  7.


 

 

a  ω sin θ  θ  c sin θ  θ

ω  30.000

rad sec




 

 

VA  a  ω sin θ  j  cos θ VA  ( 763.243  476.927j)



 

mm sec

VA  900.000

mm

VB  900.000

mm

 

sec

arg VA  32.000 deg

VB  c ω sin θ  j  cos θ VB  ( 763.243  476.927j) 8.

mm sec

sec

arg VB  32.000 deg

Determine the velocity Vbox. Since link 3 does not rotate (this is a special case Grashof linkage in the parallelogram mode), all points on it have the same velocity. Therefore, Vbox  VA

Vbox  900.000

mm sec




The linkage in Figure P6-8d has the dimensions and effective crank angle given below. Find and plot VA, VB, and Vbox in the global coordinate system for the maximum range of motion that this linkage allows if 2 = 30 rad/sec CW.

Given:


a  30 mm

Link 3

b  150  mm

Link 4

c  30 mm

Link 1

d  150  mm

ω  30 rad sec


1


1.


2.

Determine the range of motion for this special-case Grashof double crank.

Vbox 1

θ  0  deg 2  deg  360  deg 3.


d

K2 

a

K1  5.0000 2

K3 

B

3

2 O4

O2

c

K2  5.0000 2

2

a b c d

2

K3  1.0000

2 a c

 

d

A

 

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

C θ  K1   K2  1   cos θ  K3 4.


 





 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


2

d

K5 

b

2

2

c d a b

2

2 a b

K4  1.0000

K5  5.0000


D θ  cos θ  K1  K4 cos θ  K5

6.


 





 

 

θ θ  2   atan2 2  D θ E θ  7.

 2  4 Dθ F θ 

E θ

Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.

 

a  ω

 

a  ω

ω θ 

ω θ 

b

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 

sin θ θ  θ



sin θ  θ θ

4


8.



 



 

 

 

 

VA θ  a  ω sin θ  j  cos θ VA θ  VA θ

 

 

    j  cosθθ

VB θ  c ω θ  sin θ θ

 

 

VB θ  VB θ 9.

Plot the angular velocity of the output link, 4, and the magnitudes of the velocities at points A and B. Since this is a special-case Grashof linkage in the parallelogram configuration, 3 = 0 and 4 = 2 for all values of 2. Similarly, VA, VB, and Vbox all have the same constant magnitude through all values of 2. ω( 5  deg)  30.000

rad sec

VA( 5  deg)  900.000

mm

VB( 5  deg)  900.000

mm

sec

sec

ω( 135  deg)  30.000

rad sec

VA( 135  deg)  900.000

mm

VB( 135  deg)  900.000

mm

sec

sec




The linkage in Figure P6-8g has the dimensions and crank angle given below. Find 4, VA, and

Given:

VB for the position shown for 2 = 15 rad/sec clockwise (CW). Use the velocity difference graphical method. Link lengths: Link 2 (O2 to A)

a  49 mm

Link 2 (O2 to C)

a'  49 mm

Link 3 (A to B)

b  100  mm

Link 5 (C to D)

b'  100  mm

Link 4 (B to O4)

c  153  mm

Link 6 (D to O6)

c'  153  mm

Link 1 (O2 to O4)

d  87 mm

Link 1 (O2 to O6)

d'  87 mm

θ  29 deg

Crank angle:

Solution: 1.

Global XY system


ω  15 rad sec


α  119  deg

1

CW




Y

Direction of VB

Direction of VBA O6

B 3 29.000°

A

2

4 C

6

O2

X

Direction of VA

5

D

O4 2.

Use equation 6.7 to calculate the magnitude of the velocity at point B. VB  a  ω

3.

VB  735.0

mm sec

θB  29 deg  90 deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point A, the magnitude of the relati velocity VAB, and the angular velocity of link 3. The equation to be solved graphically is VA = VB + VAB a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VAB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VA, magnitude unknown. d. Complete the vector triangle by drawing VAB from the tip of VB to the intersection of the VA construction line and drawing VA from the tail of VB to the intersection of the VAB construction line.


0


500 mm/sec

Y 0.124°

1.517 VB

X V BA

VA

4.


VA  1.517  in kv

5.

kv 

500  mm sec

VA  758.5

1

in mm sec


VA c

ω  4.958

rad sec

θA  0.124  deg




The linkage in Figure P6-8g has the dimensions and crank angle given below. Find 4, VA, and

Given:

VB for the position shown for 2 = 15 rad/sec clockwise (CW). Use the instant center graphical method. Link lengths: Link 2 (O2 to A)

a  49 mm

Link 2 (O2 to C)

a'  49 mm

Link 3 (A to B)

b  100  mm

Link 5 (C to D)

b'  100  mm

Link 4 (B to O4)

c  153  mm

Link 6 (D to O6)

c'  153  mm

Link 1 (O2 to O4)

d  87 mm

Link 1 (O2 to O6)

d'  87 mm

θ  29 deg

Crank angle:

Solution: 1.

Global XY system


ω  15 rad sec


α  119  deg

1

CW



Y

Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and 4 make with the x axis.

97.094 O6 B 3

From the layout: AI13  97.094 mm

BI13  100.224  mm

1,3

100.224

θ  89.876 deg 2.

3.

5.

6

89.876° 5

X

D

O4



4.

A

O2

C


θVA  θ  90 deg

2

4

VA AI13

ω  7.570

rad

CW

sec


VB  BI13 ω

VB  758.694

θVB  θ  90 deg

θVB  0.124 deg

sec


VB c

ω  4.959

rad sec

CW




The linkage in Figure P6-8g has the dimensions and crank angle given below. Find 4, VA, and VB for the position shown for 2 = 15 rad/sec clockwise (CW). Use an analytical method.

Given:


a  49 mm

Link 2 (O2 to C)

a'  49 mm

Link 3 (A to B)

b  100  mm

Link 5 (C to D)

b'  100  mm

Link 4 (B to O4)

c  153  mm

Link 6 (D to O6)

c'  153  mm

Link 1 (O2 to O4)

d  87 mm

Link 1 (O2 to O6)

d'  87 mm

θ  148  deg Local xy system

Crank angle:

ω  15 rad sec


1


1.


2.


Y

d

K2 

a

K1  1.7755 2

K3 

O6

d

B 3

c

K2  0.5686 2

2

a b c d

2

K3  1.5592

2 a c

2

4 C

    B  2  sin θ C  K1   K2  1   cos θ  K3 3.

B  1.0598

D O4 x






2

θ  2  atan2 2  A B  4.

5

C  4.6650

B  4 A  C



θ  208.876 deg


2

d

K5 

b

2

2

c d a b

2

2 a b

    E  2  sin θ F  K1   K4  1   cos θ  K5

D  cos θ  K1  K4 cos θ  K5

5.

K5  0.3509

D  3.0104 E  1.0598 F  2.2367






2

θ  2  atan2 2  D E  6.

K4  0.8700

E  4  D F



θ  266.892 deg


 

 

a  ω sin θ  θ  b sin θ  θ

ω  7.570

rad sec

X

6

O2 148.000°

A  cos θ  K1  K2 cos θ  K3

A  0.5821

A

y


ω  7.


 

 

a  ω sin θ  θ  c sin θ  θ

ω  4.959

rad sec

Determine the velocity of points B and A for the crossed circuit using equations 6.19.



 

 

VA  a  ω sin θ  j  cos θ VA  ( 389.491  623.315i)



 

mm sec

VA  735.000

mm

VB  758.694

mm

 

sec

arg VA  58.000 deg

VB  c ω sin θ  j  cos θ VB  ( 366.389  664.362j)

mm sec

sec

arg VB  118.876 deg




The linkage in Figure P6-8g has the dimensions and crank angle given below. Find and plot 4, VA, and VB in the local coordinate system for the maximum range of motion that this linkage allows if 2 = 15 rad/sec clockwise (CW).

Given:


a  49 mm

Link 2 (O2 to C)

a'  49 mm

Link 3 (A to B)

b  100  mm

Link 5 (C to D)

b'  100  mm

Link 4 (B to O4)

c  153  mm

Link 6 (D to O6)

c'  153  mm

Link 1 (O2 to O4)

d  87 mm

Link 1 (O2 to O6)

d'  87 mm

ω  15 rad sec


1

Y


1.


2.

Determine the range of motion for this non-Grashof triple rocker using equations 4.37. 2

arg1 

2

2

a d b c

2



2 a d 2

arg2 

O6

2

2

a d b c

2

2 a d




b c

B 3

arg1  0.840

a d

C

b c

O2

arg2  6.338

a d

A

2

4

2

5 D


O4 x

The other toggle angle is the negative of this. Thus,

θ  θ2toggle  1  deg θ2toggle  2  deg  359  deg  θ2toggle

3.


d

K2 

a

K1  1.7755

 

2

d

K3 

c

2

a b c d

K3  1.5592

 

B θ  2  sin θ

 

2

2 a c

K2  0.5686

 

A θ  cos θ  K1  K2 cos θ  K3

 

2

 

 

C θ  K1   K2  1   cos θ  K3 4.


 





 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


 

2

d

K5 

b

2

2

c d a b 2 a b

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

F θ  K1   K4  1   cos θ  K5

2

K4  0.8700

 

K5  0.3509

 

E θ  2  sin θ

X

6

y


6.





 



 

 

θ θ  2   atan2 2  D θ E θ  7.


 

a  ω

 

a  ω

ω θ 

ω θ 

8.

 2  4 Dθ F θ 

E θ

b

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 

sin θ θ  θ



sin θ  θ θ


     VBx θ  Re VB θ 

 

VB θ  a  ω sin θ  j  cos θ

 

 

 

  

 

  


    j  cosθθ

VA θ  c ω θ  sin θ θ

 


Plot the angular velocity of the output link, 4, and the x and y components of the velocities at points B and A. ANGULAR VELOCITY OF LINK 4 60 40 Angular Velocity, rad/sec

9.

  


20 0

 

ω  θ 

sec rad

 20  40  60  80  100

0

45

90

135

180 θ deg

225

270

315

360



VELOCITY COMPONENTS, POINT B 1000


500

 

VBx θ 

 

VBy θ 

sec mm 0

sec mm

 500

 1000

0

45

90

135

180

225

270

315

360

315

360

θ deg

VELOCITY COMPONENTS, POINT A 4000


3000 2000

 

VAx θ 

 

VAy θ 

sec mm 1000 sec mm

0

 1000  2000  3000

0

45

90

135

180 θ deg

225

270




The 3-cylinder radial compressor in Figure P6-8c has the dimensions and crank angle given below Find piston velocities V6, V7, and V8 for the position shown for 2 = 15 rad/sec clockwise (CW). Use the velocity difference graphical method.

Given:

Link lengths:

Solution: 1. 2.

Link 2

a  19 mm

Links 3, 4, and 5

b  70 mm

c  0  mm

Offset

Crank angle:

θ  53 deg Global XY system


ω  15 rad sec

Cylinder angular spacing

α  120  deg

1

CW


Direction of V62 Direction of V8

Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Use equation 6.7 to calculate the magnitude of the velocity at rod pin on link 2.

8 6

V2  a  ω

V2  285.0

3

mm

5 X

Direction of V6

1 4

Use equation 6.5 to (graphically) determine the magnitude of the velocity at pistons 6, 7, and 8.. The equations to be solved graphically are V7 = V2 + V72

2

sec

θ2  53 deg  90 deg 3.

Direction of V82

Y

V6 = V2 + V62

Direction of V72 7

V8 = V2 + V82 Direction of V7

a. Choose a convenient velocity scale and layout the known vector V2. b. From the tip of V2, draw a construction line with the direction of V72, magnitude unknown. c. From the tail of V2, draw a construction line with the direction of V7, magnitude unknown. d. Complete the vector triangle by drawing V72 from the tip of V2 to the intersection of the V7 construction line and drawing V7 from the tail of V2 to the intersection of the V72 construction line. e. Repeat for V6 and V8. 0

4.

200 mm/sec


kv 

V7  1.046  in kv

200  mm sec

1

Y

in

V62

V7  209.2

1.046

V82

V6  83.4

V7

mm sec

θV6  150  deg V8  292.6

X

V8

sec

V2

V8  1.463  in kv

V6

mm

θV7  270  deg V6  0.417  in kv

0.417

1.463

mm sec

θV8  210  deg

V72




The 3-cylinder radial compressor in Figure P6-8c has the dimensions and crank angle given below. Find V6, V7, and V8 for the position shown for 2 = 15 rad/sec clockwise (CW). Use an analytical method.

Given:


a  19 mm b  70 mm

Links 3, 4, and 5

Solution: 1. 2.


ω  15 rad sec


α  120  deg


Crank angle:

θ  37 deg

CW

Local x'y' system

Y

x''

θ  170.599 deg

 

x'''

y'''

Determine 4 and d' using equation 4.17.

 a sin θ  b 

8 6

c

3

π 

 

2 1

d'  3.316 in

4 y''


   

5 X, y'

(x'y' system)

d'  a  cos θ  b  cos θ

7

a cos θ ω    ω b cos θ 4.

c  0  mm


θ  asin 

3.

1

Offset

ω  3.296

rad sec 37.000°

Determine the velocity of the rod pin on link 2 using equation 6.23a:



 

x'

 

V2  a  ω sin θ  j  cos θ V2  ( 171.517  227.611i)

mm sec


V2  285.000

mm sec

arg V2  53.000 deg

θV2  arg V2  90 deg

θV2  143.000 deg

Determine the velocity of piston 7 using equation 6.22b:

 

 

V7  a  ω sin θ  b  ω sin θ V7  209.204

mm

V7  209.204

sec


sec

arg V7  0.000 deg

θV7  arg V7  90 deg

θV7  90.000 deg

Determine 3 and d'' using equation 4.17. θ  θ  120  deg

θ  157.000 deg

 a sin θ  c  π b  

θ  asin 

 

 

d''  a  cos θ  b  cos θ 7.

mm

θ  173.912 deg d''  2.052 in

Determine the angular velocity of link 5 using equation 6.22a: ω 

   


ω  3.769

rad sec

(x''y'' system)


8.


Determine the velocity of the rod pin on link 2 using equation 6.23a:



 

 

V2  a  ω sin θ  j  cos θ V2  ( 111.358  262.344i)

mm sec


V2  285.000

arg V2  67.000 deg

mm sec

θV2  arg V2  150  deg

θV2  217.000 deg


 

 

V6  a  ω sin θ  b  ω sin θ mm

V6  83.378

V6  83.378

sec


arg V6  0.000 deg

mm sec

θV6  arg V6  150  deg

θV6  150.000 deg

10. Determine 5 and d''' using equation 4.17. θ  θ  120  deg

θ  277.000 deg

 a sin θ  b 

θ  asin 

 

c

π 

(x'''y''' system)

θ  195.629 deg

 

d'''  a  cos θ  b  cos θ

d'''  2.745 in

11. Determine the angular velocity of link 5 using equation 6.22a: ω 

   

a cos θ   ω b cos θ

ω  0.515

rad sec

12. Determine the velocity of the rod pin on link 2 using equation 6.23a:



 

 

V2  a  ω sin θ  j  cos θ V2  ( 282.876  34.733i )

mm sec


V2  285.000

mm sec

θV2  arg V2  30 deg

arg V2  173.000 deg

θV2  143.000 deg

13. Determine the velocity of piston 8 using equation 6.22b:

 

 

V8  a  ω sin θ  b  ω sin θ V8  292.592

mm sec


V8  292.592

mm sec

θV8  arg V8  30 deg

arg V8  180.000 deg

θV8  210.000 deg




The 3-cylinder radial compressor in Figure P6-8c has the dimensions and crank angle given below. Find and plot V6, V7, and V8 for one revolution of the crank if 2 = 15 rad/sec clockwise (CW).

Given:

Link lengths:

Solution: 1. 2.

Link 2

a  19 mm

Links 3, 4, and 5

b  70 mm

c  0  mm

Offset


ω  15 rad sec


α  120  deg

1

CW


Draw the linkage to scale and label it. Note that there are three local coordinate systems.

Y

x''

x'''

y''' 8

Determine the range of motion for this slider-crank linkage. This will be the same in each coordinate frame.

6

3

5 X, y'

θ  0  deg 2  deg  360  deg 3.

1

 a sin θ  θ θ  asin  b  4.

c

4

π 

(x'y' system)

y'' 7


 

ω θ  5.

2

Determine 4 using equation 4.17.

a b



   ω cos θ θ  cos θ

x'


 

 

    

V7 θ  a  ω sin θ  b  ω θ  sin θ θ 6.


 

 a sin θ  α  c  π b  

θ θ  asin  7.


 

ω θ  8.

(x''y'' system)

    

a cos θ  α   ω b cos θ θ


 





    

V6 θ  a  ω sin θ  α  b  ω θ  sin θ θ 9.

Determine 5 and d''' using equation 4.17.

 

 a sin θ  2  α  c  π b  

θ θ  asin 

10. Determine the angular velocity of link 5 using equation 6.22a:

 

ω θ 





a cos θ  2  α   ω b cos θ θ

  

(x'''y''' system)



11. Determine the velocity of piston 8 using equation 6.22b:

 





    

V8 θ  a  ω sin θ  2  α  b  ω θ  sin θ θ 12. Plot the velocities of pistons 6, 7, and 8.

VELOCITY OF PISTONS 6, 7, AND 8 400

 

Velocity, mm/sec

V6 θ 

 

V7 θ 

 

V8 θ 

sec 200 mm sec mm

0

sec mm  200

 400

0

60

120

180 θ deg Crank Angle, deg

240

300

360




Figure P6-9 shows a linkage in one position. Find the instantaneous velocities of points A, B, and P if link O2A is rotating CW at 40 rad/sec.

Given:


a  5.00 in

Link 3

b  4.40 in

Link 4

c  5.00 in

Link 1

d  9.50 in

Rpa  8.90 in

δ  56 deg

Coupler point:

θ  50 deg

Crank angle and speed: Solution:

ω  40 rad sec

1


1.

Draw the linkage to scale and label it. All calculated angles are in the local xy system.

2.


d

K2 

a

K1  1.9000 2

K3 

d c

K2  1.9000 2

2

a b c d

 

P

y

Y B

2

K3  2.4178

2 a c

3

4

A

 

A  cos θ  K1  K2 cos θ  K3 2

 

B  2  sin θ

14.000°

 

A  0.0607

B  1.5321

C  2.4537





2

B  4 A  C



θ  246.992 deg

θ  θ  360  deg

θ  606.992 deg


2

d b

2

2

c d a b

K5 

    E  2  sin θ F  K1   K4  1   cos θ  K5

2 a b

D  cos θ  K1  K4 cos θ  K5

5.

2

K4  2.1591

K5  2.4911

D  2.3605 E  1.5321 F  0.1539






2

θ  2  atan2 2  D E 

E  4  D F

θ  θ  360  deg 6.

X

O2

Use equation 4.10b to find values of 4 for the open circuit. θ  2  atan2 2  A B 

4.

O4

1

C  K1   K2  1   cos θ  K3 3.

x 50.000°



θ  349.895 deg θ  709.895 deg


 

 

a  ω sin θ  θ  b sin θ  θ

ω  41.552

rad sec


ω  7.


 

 

a  ω sin θ  θ  c sin θ  θ

ω  26.320

rad sec




 

 

VA  a  ω sin θ  j  cos θ VA  ( 153.209  128.558i)



 

in sec

VA  200.000

 

in sec

arg VA  40.000 deg

VB  c ω sin θ  j  cos θ VB  ( 121.130  51.437i ) 8.

in

VB  131.598

sec

in sec

arg VB  23.008 deg

Determine the velocity of the coupler point P for the open circuit using equations 6.36.











VPA  Rpa ω sin θ  δ  j  cos θ  δ VPA  ( 338.121  149.797i)

in sec

VP  VA  VPA VP  ( 184.912  21.239i )

in sec

VP  186.128

in sec

arg VP  173.448 deg




Figure P6-10 shows a linkage and its coupler curve. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the velocity of the coupler point P at 2-deg increments of crank angle for ω 2 = 100 rpm. Check your results with program FOURBAR.

Given:

Link lengths:

Solution:

Link 2 (O2 to A)

a  1.00 in

Link 3 (A to B)

b  2.06 in

Link 4 (B to O4)

c  2.33 in

Link 1 (O2 to O4)

d  2.22 in

Coupler point:

Rpa  3.06 in

  31 deg

Crank speed:

  100  rpm


1.


2.

Determine the range of motion for this Grashof crank rocker.

B

y 3

θ  0  deg 0.5 deg  360  deg 3.

b


d

K2 

a

K1  2.2200 2

K3 

2

a b c d

d c

2

 

4

c

x

1

 

A θ  cos θ  K1  K2 cos θ  K3

 

2

O2

K3  1.5265

 

a

4

d

2

2 a c

 

p A

K2  0.9528 2

P

O4

 

B θ  2  sin θ

 

C θ  K1   K2  1   cos θ  K3 4.


 





 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


 

2

d b

K5 

2

2

c d a b

2

2 a b

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

K4  1.0777

K5  1.1512

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5 6.


 





 

 

θ θ  2   atan2 2  D θ E θ  7.

 2  4 Dθ F θ 

E θ


 

ω θ 

a   b



    sin θ θ  θ θ  sin θ θ  θ


 

ω θ  8.

a   c






  sin θ θ  θ θ  sin θ  θ θ

Determine the velocity of point A using equations 6.19.

 



 

 

 

VA θ  a   sin θ  j  cos θ 9.

 

VA θ  VA θ

Determine the velocity of the coupler point P using equations 6.36.

       VP θ  VA θ  VPA θ



  



VPA θ  Rpa ω θ  sin θ θ    j  cos θ θ   10. Plot the magnitude and direction of the velocity at coupler point P.

 

 

 

VP θ  VP θ

Magnitude:

  

Direction: θVP1 θ  arg VP θ

θVP θ  if  θVP1 θ  0 θVP1 θ θVP1 θ  2  π MAGNITUDE

Velocity, in/sec

30

20

 

VP θ 

s in 10

0

0

45

90

135

180

225

270

315

360

θ deg

DIRECTION

Vector Angle, deg

360

270

θVP θ

180

deg 90

0

0

45

90

135

180 θ deg

225

270

315

360




Figure P6-11 shows a linkage that operates at 500 crank rpm. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the velocity of point B at 2-deg increments of crank angle. Check your result with program FOURBAR.

Units:

rpm  2  π rad min

Given:

Link lengths:

1

Link 2 (O2 to A)

a  2.000  in

Link 3 (A to B)

Link 4 (B to O4)

c  7.187  in

Link 1 (O2 to O4)

ω  500  rpm


ω  52.360 rad sec


2.


A 3 2

θ  0  deg 0.5 deg  360  deg

d

K2 

a

K1  4.8125 2

K3 

1

d

2

2

O4

2

K3  2.7186

2 a c

      C θ  K1   K2  1   cos θ  K3

 

 

B θ  2  sin θ


 





 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ 

B θ


2

d

K5 

b

 

2

2

c d a b

2

2 a b

 

 

K4  1.1493

 

D θ  cos θ  K1  K4 cos θ  K5

 

K5  3.4367

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5 6.


 





 

 

θ θ  2   atan2 2  D θ E θ  7.

 2  4 Dθ F θ 

E θ


 

ω θ 

a  ω b



    sin θ θ  θ θ  sin θ θ  θ

4

c

A θ  cos θ  K1  K2 cos θ  K3

5.

2

K2  1.3392

a b c d

B

O2


4.

d  9.625  in 1


1.

3.

b  8.375  in


 

ω θ  8.

a  ω c






  sin θ θ  θ θ  sin θ  θ θ

Determine the velocity of point B using equations 6.19.

 

 

    j  cosθθ

VB θ  c ω θ  sin θ θ

 

 

θVB1 θ  arg VB θ 

VB θ  VB θ

Plot the magnitude and angle of the velocity at point B. MAGNITUDE OF VELOCITY AT B

Velocity, in/sec

150

100

 

VB θ 

sec in 50

0

0

60

120

180

240

300

360

θ deg

θVB θ  if  θVB1 θ  0 θVB1 θ  π θVB1 θ  DIRECTION OF VELOCITY AT B 60

50

40 Angle, deg

9.

θVB θ

30

deg 20

10

0

0

60

120

180 θ deg

240

300

360




Figure P6-12 shows a linkage and its coupler curve. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the velocity of the coupler point P at 2-deg increments of crank angle over the maximum range of motion possible. Check your results with program FOURBAR.

Given:

Link lengths:

Solution:

Link 2 (O2 to A)

a  0.785  in

Link 3 (A to B)

b  0.356  in

Link 4 (B to O4)

c  0.950  in

Link 1 (O2 to O4)

d  0.544  in

Coupler point:

Rpa  1.09 in

δ  0  deg

Crank speed:

ω  20 rpm

See Figure P6-12 and Mathcad file P0649. y

1.


2.


 a2  d 2  ( b  c) 2  2 a d  

P 3 B 3

A


4 2

θ  158.286 deg

2

θ  θ  1  deg θ  2  deg  θ  1  deg 3.


d

K2 

a

K1  0.6930

 

O4

d c

2

2

2

a b c d

K2  0.5726

K3 

 

B θ  2  sin θ

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

x O2

2

2 a c

K3  1.1317

 

 

C θ  K1   K2  1   cos θ  K3 4.


 





 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


2

d

K5 

b

2

2

c d a b

2

2 a b

      F  θ  K1   K4  1   cos θ  K5

 

D θ  cos θ  K1  K4 cos θ  K5

6.

K5  0.2440

 

E θ  2  sin θ


 





 

 

θ θ  2   atan2 2  D θ E θ  7.

K4  1.5281

 2  4 Dθ F θ 

E θ



 

ω θ 

 

ω θ  8.


    b sin θ θ  θ θ  a  ω sin θ  θ θ   c sin θ θ  θ θ  a  ω





 



 

 

VA θ  a  ω sin θ  j  cos θ 9.


       VP θ  VA θ  VPA θ



  



VPA θ  Rpa ω θ  sin θ θ  δ  j  cos θ θ  δ

10. Plot the magnitude and direction of the coupler point P.

 

 

 


Magnitude:

  

Direction: θVP θ  arg VP θ

Velocity, mm/sec

MAGNITUDE

60

 

VP θ 

sec in

40 20 0  200

 150

 100

 50

0

50

100

50

100

150

200

θ deg

DIRECTION 200

Vector Angle, deg

100

θVP θ

0

deg  100

 200  200

 150

 100

 50

0 θ deg

150

200




Given:

Solution: 1.

Figure P6-13 shows a linkage and its coupler curve. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the velocity of the coupler point P at 2-deg increments of crank angle over the maximum range of motion possible. Check your results with program FOURBAR. Link lengths: Link 2 (O2 to A)

a  0.86 in

Link 3 (A to B)

b  1.85 in

Link 4 (B to O4)

c  0.86 in

Link 1 (O2 to O4)

d  2.22 in

Coupler point:

Rpa  1.33 in

δ  0  deg

Crank speed:

ω  80 rpm


Draw the linkage to scale and label it. y

B

3 4 P

x

O4

O2 3 2

A

2.


arg1 

2

2



2 a d 2

arg2 

2

a d b c 2

2

a d b c 2 a d


2



b c

arg1  1.228

a d b c

arg2  0.439

a d




d

K2 

a

K1  2.5814

 

 

d c

2

K3 

2

2

a b c d 2 a c

K2  2.5814

K3  2.0181

 

B θ  2  sin θ

A θ  cos θ  K1  K2 cos θ  K3

 

 

2



 

 

C θ  K1   K2  1   cos θ  K3 4.




 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


2

d

K5 

b

 

2

2

c d a b

2

K4  1.2000

2 a b

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

K5  2.6244

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5 6.




 



 

 

θ θ  2   atan2 2  D θ E θ  7.


 

ω θ 

8.

 2  4 Dθ F θ 

E θ

a  ω b

    sin θ θ  θ θ 




 

ω θ 

a  ω c





  sin θ θ  θ θ  sin θ  θ θ


 



 

 

VA θ  a  ω sin θ  j  cos θ 9.


 

 

  



  




 

 

 

VP θ  VA θ  VPA θ


 

 


Magnitude:

Direction:

θVP θ  arg VP θ 

MAGNITUDE 20

Velocity, mm/sec

15

 

VP θ 

sec in

10

5

0  120

 90

 60

 30

0 θ deg

30

60

90

120



DIRECTION 270 240

Vector Angle, deg

210 180

θ' VP θ 150 deg

120 90 60 30 0  120

 90

 60

 30

0 θ deg

30

60

90

120




Figure P6-14 shows a linkage and its coupler curve. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the velocity of the coupler point P at 2-deg increments of crank angle over the maximum range of motion possible. Check your results with program FOURBAR.

Given:

Link lengths:

Solution: 1.

Link 2 (O2 to A)

a  0.72 in

Link 3 (A to B)

b  0.68 in

Link 4 (B to O4)

c  0.85 in

Link 1 (O2 to O4)

d  1.82 in

Coupler point:

Rpa  0.97 in

δ  54 deg

Crank speed:

ω  80 rpm


Draw the linkage to scale and label it. P y

B

3 4

A 2

x O2

2.

O4


arg1 

2

2



2 a d 2

arg2 

2

a d b c 2

2

a d b c 2 a d

2



b c

arg1  1.451

a d b c

arg2  0.568

a d





d

K2 

a

K1  2.5278 2

K3 

d c

K2  2.1412 2

2

a b c d 2 a c

2

K3  3.3422


 


 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

C θ  K1   K2  1   cos θ  K3 4.


 





 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


2

d

K5 

b

 

2

2

c d a b

2

K4  2.6765

2 a b

 

K5  3.6465

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 6.


 





 

 

θ θ  2   atan2 2  D θ E θ  7.


 

a  ω

 

a  ω

ω θ 

ω θ  8.

 2  4 Dθ F θ 

E θ

b

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 






 



 

 

VA θ  a  ω sin θ  j  cos θ 9.


 

 

  



  




 

 

 



 

 

Magnitude:


Direction:

θVP θ  arg VP θ 



MAGNITUDE 20

Velocity, in/sec

15

 

VP θ 

sec in

10

5

0  60

 30

0

30

60

θ deg

DIRECTION 200

Vector Angle, deg

150 100

θVP θ

50

deg 0  50  100  60

 30


30

60




Figure P6-15 shows a power hacksaw that is an offset crank-slider mechanism that has the dimensions given below. Draw an equivalent linkage diagram; then calculate and plot the velocity of the saw blade with respect to the piece being cut over one revolution of the crank, which rotates at 50 rpm.

Given:


a  75 mm

Link 3

b  170  mm ω  50 rpm


c  45 mm

Offset


Draw the equivalent linkage to a convenient scale and label it. y

B

3

b

A

4 a 2

c

2

O2

2.

Determine the range of motion for this crank-slider linkage. θ  0  deg 2  deg  360  deg

3.

Determine 3 using equations 4.16 for the crossed circuit.

 a  sin θ  b 

 

θ θ  asin 4.

 


 

ω θ 

5.

c

a b



   ω cos θ θ  cos θ


 

 

    

VB θ  a  ω sin θ  b  ω θ  sin θ θ

7.

Plot the magnitude of the velocity of B. (See next page.)

x



VELOCITY OF POINT B 600

Velocity, mm/sec

400

200

 

VB θ 

sec mm 0

 200

 400

0

60

120

180 θ deg

240

300

360




Given:

Figure P6-16 shows a walking-beam indexing and pick-and-place mechanism that can be analyzed as two fourbar linkages driven by a common crank. Calculate and plot the absolute velocities of points E and P and the relative velocity between points E and P for one revolution of gear 2. Link lengths ( walking-beam linkage): Link 2 (O2 to A)

a'  40 mm

Link 3 (A to D)

b'  108  mm

Link 4 (O4 to D)

c'  40 mm

Link 1 (O2 to O4)

d'  108  mm

Link lengths (pick and place linkage): Link 5 (O5 to B)

a  13 mm

Link 7 (B to C)

b  193  mm

Link 6 (C to O6)

c  92 mm

Link 1 (O5 to O6)

d  128  mm

u  164  mm

Crank speed:

ω  10 rpm

Rocker point E:

ϕ  143  deg

Gears 4 & 5 phase angle Solution: 1.


Draw the walking-beam linkage to scale and label it. 30 mm Y

P

58 mm

80°

A

D c'

b'

a' d'

x'

O2

O4

2.

Determine the range of motion for this mechanism. θ  0  deg 2  deg  360  deg

3.

X

y'

(local x'y' coordinate system)

This part of the mechanism is a special-case Grashof in the parallelogram configuration. As such, the coupler does not rotate, but has curvilinear motion with ever point on it having the same velocity. Therefore, it is only necessary to calculate the X-component of the velocity at point A in order to determine the velocity of the cylinder center, P.

 



 

 

VA θ  a' ω sin θ  j  cos θ

 

  


In the global X-Y coordinate frame,

 

 

VP θ  VAx θ


4.


Draw the pick and place linkage to scale and label it.

E

C

7 6

b

c

O6 5.

B

a O5

Establish the relationship between 5 and 2. Note that gear 5 is driven by gear 4 and that their ratio is -1 (i.e., they rotate in opposite directions with the same speed). Also, because the walking beam fourbar is special Grashof, 4 = 2. Thus,

 

θ θ  θ  ϕ 6.

5

d 1

ω  ω

and


d

K1  9.8462

a 2

K3 

2

2

a b c d

d c

K2  1.3913

2

K3  5.1137

2 a c

 

K2 

    K1  K2 cosθθ  K3

A θ  cos θ θ

 

  

B θ  2  sin θ θ

 

     K3

C θ  K1   K2  1   cos θ θ 7.




 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  8.

B θ


d b

2

K5 

2

2

c d a b 2 a b

2

K4  0.6632

K5  9.0351


 


    K1  K4 cosθθ  K5

D θ  cos θ θ

 

  

E θ  2  sin θ θ

 

     K5

F θ  K1   K4  1   cos θ θ 9.




 



 

 

θ θ  2   atan2 2  D θ E θ 

 2  4 Dθ F θ 

E θ

10. Determine the angular velocity of links 6 and 7 using equations 6.18.

         

ω θ 

 

a  ω sinθ θ  θ θ  b sin θ θ  θ θ

 

a  ω sin θ θ  θ θ  c sin θ θ  θ θ

ω θ 

     

   

10. Determine the velocity of the rocker point E using equations 6.34.

 

 

    j  cosθθ

VE θ  u  ω θ  sin θ θ 11. Transform this into the global XY system.

 

  

 

 

 

VEx θ  Re VE θ

 

VEX θ  VEx θ

 

  

VEy θ  Im VE θ

 

VEY θ  VEx θ

 

 

VEXY θ  VEX θ  j  VEY θ

12. Calculate and plot the velocity of E relative to P.

 

 

 

VEP θ  VEXY θ  VP θ

RELATIVE VELOCITY 100

Velocity, mm/sec

80

 

V EP θ 

sec

60

mm 40

20

0

0

30

60

90

120

150

180

210

θ deg Crank Angle, deg

240

270

300

330

360




Figure P6-17 shows a paper roll off-loading mechanism driven by an air cylinder. In the position shown, it has the dimensions given below. The V-links are rigidly attached to O4A. The air cylinder is retracted at a constant velocity of 0.2 m/sec. Draw a kinematic diagram of the mechanism, write the necessary equations, and calculate and plot the angular velocity of the paper roll and the linear velocity of its center as it rotates through 90 deg CCW from the position shown.

Given:

Link lengths and angles:

Paper roll location from O4:

Link 4 (O4 to A)

c  300  mm

u  707.1  mm

Link 1 (O2 to O4)

d  930  mm

δ  181  deg

Link 4 initial angle

θ  62.8 deg

adot  200  mm sec

Input cylinder velocity Solution: 1.

with respect to local x axis 1


Draw the mechanism to scale and define a vector loop using the fourbar derivation in Section 6.7 as a model. V-Link

Roll Center

707.107

45.000° x

O4 c

4

46.000° A

1

O4

d

4

3

O2

2 b

R1

R4 2 A

R3

R2

O2

a f y

2.

Write the vector loop equation, differentiate it, expand the result and separate into real and imaginary parts to solve for f, 2, and 4. R1  R4  R2  R3 d e

j  θ

 c e

j  θ

(a)

 a  e

j  θ

 b e

j  θ

(b)

where a is the distance from the origin to the cylinder piston, a variable; b is the distance from the cylinder piston to A, a constant; and c is the distance from 4 to point A, a constant. Angle 1 is zero, 3 = 2, and 4 is the variable angle that the rocker arm makes with the x axis. Solving the position equations: Let

f  a  b

then, making this substitution and substituting the Euler equivalents,

  

 

  

 

d  c cos θ  j  sin θ  f  cos θ  j  sin θ

(c)

Separating into real and imaginary components and solving for 2 and f,

 



 

 

θ θ  atan2 d  c cos θ c sin θ

(d)



  sin θ θ  c sin θ

 

f θ 

(e)

Differentiate equation b. j  c ω e

j  θ

 d f   ej  θ   j  f  ω  ej  θ     dt 



(f)

Substituting the Euler equivalents,



 

 

c ω sin θ  j  cos θ 

 d f    cos θ   j  sin θ    f  ω   sin θ   j  cos θ           dt 

(g)

Separating into real and imaginary components and solving for 4 . Note that df/dt = adot.

 

ω θ 

3.

adot

  

c sin θ θ  θ

(h)



Plot 4 over a range of 4 of θ  θ θ  1  deg  θ  90 deg

ANGULAR VELOCITY, LINK 4 1.2


1 0.8

 

ω  θ 

sec rad

0.6 0.4 0.2 0 60

80

100

120

140

160

θ deg Link 4 Angle, deg

4.

Determine the velocity of the center of the paper roll using equation 6.35. The direction is in the local xy coordinate system.

 

 

 

 









VU θ  u  ω θ  sin θ  δ  j  cos θ  δ VU θ  VU θ 5.

θVU  θ  arg VU θ 

Plot the magnitude and direction of the velocity of the paper roll center. (See next page.)



MAGNITUDE OF PAPER CENTER VELOCITY

Velocity, mm/sec

600

 

VU θ 

sec mm

400

200

0 60

80

100

120

140

160

θ deg Rocker Arm Position, deg

DIRECTION OF PAPER CENTER VELOCITY 80

Vector Angle, deg

60

40

θVU  θ

20

deg 0

 20

 40 60

80

100

120 θ deg

Rocker Arm Position, deg

140

160




Figure P6-18 shows a powder compaction mechanism. Calculate its mechanical advantage for the position shown.

Given:

Link lengths: Link 2 (A to B)

a  105  mm

Link 3 (B to D)

b  172  mm

c  27 mm

Offset

Distance to force application: rin  301  mm

Link 2 (AC)

θ  44 deg

Position of link 2: Solution: 1.

Let

ω  1  rad sec

1


Draw the linkage to scale and label it. X

2.


4 D

 a sin θ  c  π b  

θ  asin 

C

θ  164.509 deg 3.

3

Determine the angular velocity of link 3 using equation 6.22a: 44.000°

   

a cos θ ω    ω b cos θ 4.

B 2

Determine the velocity of pin D using equation 6.22b:

 

 

VD  a  ω sin θ  b  ω sin θ 5.

Y

Positive upward

Calculate the velocity of point C using equation 6.23a:



 

 

VC  rin ω sin θ  j  cos θ VC  VC 6.

Calculate the mechanical advantage using equation 6.13. mA 

VC VD

mA  3.206

A




Figure P6-18 shows a powder compaction mechanism. Calculate and plot its mechanical advantage as a function of the angle of link AC as it rotates from 15 to 60 deg.

Given:


a  105  mm

Link 3 (B to D)

b  172  mm

Offset

c  27 mm


Link 2 (AC)

Initial and final positions of link 2: θ  15 deg Solution: 1.

θ  60 deg

Let ω  1  rad sec

See Figure P6-18 and Mathcad file P0655b.

Draw the linkage to scale and label it. X

4 D C

3

2 B 2

Y

2.

A

Determine the range of motion for this slider-crank linkage. θ  θ θ  1  deg  θ

3.


 a sin θ  b 

 

θ θ  asin  4.

π 


 

ω θ 

5.

c

a b



   ω cos θ θ  cos θ


 

 

    

VD θ  a  ω sin θ  b  ω θ  sin θ θ 6.

Calculate the velocity of point C using equation 6.23a:

Positive upward

1



 



 

 

 

 

VC θ  rin ω sin θ  j  cos θ VC θ  VC θ

Calculate the mechanical advantage using equation 6.13.

 

mA θ 

  VD θ VC θ

MECHANICAL ADVANTAGE 12 11 10 9 Mechanical Advantage

7.

8 7

 

m A θ 6 5 4 3 2 1 0 15

20

25

30

35

40 θ deg

Crank Angle, deg

45

50

55

60




Figure P6-19 shows a walking beam mechanism. Calculate and plot the velocity Vout for one revolution of the input crank 2 rotating at 100 rpm.

Given:

Link lengths:

Solution: 1.

Link 2 (O2 to A)

a  1.00 in

Link 3 (A to B)

b  2.06 in

Link 4 (B to O4)

c  2.33 in

Link 1 (O2 to O4)

d  2.22 in

Coupler point:

Rpa  3.06 in

δ  31 deg

Crank speed:

ω  100  rpm



x

y O4 4

1 26.00°

X

O2

P

2 A 3 B

2.

Determine the range of motion for this Grashof crank rocker. θ  0  deg 1  deg  360  deg

3.


d

K2 

a

K1  2.2200 2

K3 

c

K2  0.9528 2

2

a b c d

 

d

2

K3  1.5265

2 a c

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

C θ  K1   K2  1   cos θ  K3 4.


 





 

 

θ θ  2   atan2 2  A θ B θ 

 2  4 A θ Cθ 

B θ


5.



2

d

K5 

b

 

2

2

c d a b

2

K4  1.0777

2 a b

 

K5  1.1512

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 6.


 





 

 

θ θ  2   atan2 2  D θ E θ  7.


 

a  ω

 

a  ω

ω θ 

ω θ  8.

 2  4 Dθ F θ 

E θ

b

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 






 



 

 

VA θ  a  ω sin θ  j  cos θ 9.


       VP θ  VA θ  VPA θ



  




10. Plot the X-component (global coordinate system) of the velocity of the coupler point P. Coordinate rotation angle: α  26 deg

 

  

  

Vout θ  Re VP θ  cos α  Im VP θ  sin α Vout

Velocity, in/sec

20

10

0

 10

0

45

90

135

180

Crank Angle, deg

225

270

315

360




Figure P6-20 shows a crimping tool. For the dimensions given below, calculate its mechanical advantage for the position shown.

Given:

Link lengths: Link 2 (AB)

a  0.80 in

Link 3 (BC)

b  1.23 in

Link 4 (CD)

c  1.55 in

Link 1 (AD)

d  2.40 in

Link 4 (CD)

rout  1.00 in

Distance to force application: rin  4.26 in

Link 2 (AB)

θ  49 deg

Initial position of link 2: Solution: 1.


Draw the mechanism to scale and label it. A

2

B

2

3

C Fout

1

Fin

4

49.000°

D

2.


d

K2 

a

K1  3.0000 2

K3 

d c

K2  1.5484 2

2

a b c d

2

K3  2.9394

2 a c

 

 

A  cos θ  K1  K2 cos θ  K3

 

B  2  sin θ

 

C  K1   K2  1   cos θ  K3 A  0.4204 3.

B  1.5094

Use equation 4.10b to find value of 4 for the open circuit.





θ  2  atan2 2  A B  4.

C  4.2675

2

B  4 A  C



θ  236.482 deg


d b

2

K5 

2

2

c d a b 2 a b

2

K4  1.9512

K5  2.8000



 

 

D  cos θ  K1  K4 cos θ  K5

D  3.8638

 

E  2  sin θ

E  1.5094

 

F  K1   K4  1   cos θ  K5 5.






θ  2  atan2 2  D E  6.

F  0.8241

2

E  4  D F



θ  325.961 deg

Referring to Figure 6-10, calculate the values of the angles  and . ν  θ  θ

ν  374.961 deg

If  > 360 deg, subtract 360 deg from it. ν  if  ν  360  deg ν  360  deg ν

ν  14.961 deg

μ  θ  θ

μ  89.479 deg

If  > 90 deg, subtract it from 180 deg. μ  if  μ  90 deg 180  deg  μ μ 7.

μ  89.479 deg

Using equation 6.13e, calculate the mechanical advantage of the linkage in the position shown. mA 

c sin μ rin  a  sin ν rout

mA  31.969




Figure P6-20 shows a crimping tool. For the dimensions given below, calculate and plot its mechanical advantage as a function of the angle of link AB as it rotates from 60 to 45 deg.

Given:

Link lengths: Link 2 (AB)

a  0.80 in

Link 3 (BC)

b  1.23 in

Link 4 (CD)

c  1.55 in

Link 1 (AD)

d  2.40 in

Link 4 (CD)

rout  1.00 in

Distance to force application: rin  4.26 in

Link 2 (AB)

θ  60 deg

Range of positions of link 2: Solution: 1.

θ  45 deg


Draw the mechanism to scale and label it.

A

2

B

2

3

C Fout

1

Fin

4

2

D

2.

Calculate the range of 2 in the local coordinate system (required to calculate 3 and 4). θ  θ θ  1  deg  θ

3.


d

K2 

a

K1  3.0000

 

c

K2  1.5484

2

K3 

d

2

2

a b c d

2

K3  2.9394

2 a c

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

C θ  K1   K2  1   cos θ  K3 4.


 





 

 

θ θ  2   atan2 2  A θ B θ  5.

 2  4 A θ Cθ 

B θ



K4 


2

d

K5 

b

 

2

2

c d a b

2

K4  1.9512

2 a b

 

K5  2.8000

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 6.




 



 

 

θ θ  2   atan2 2  D θ E θ  7.

 2  4 Dθ F θ 

E θ

Referring to Figure 6-10, calculate the values of the angles  and .

 

 

ν θ  θ  θ θ

 

 

 

μ θ  θ θ  θ θ

Using equation 6.13e, calculate and plot the mechanical advantage of the linkage over the given range.

 

mA θ 

   rin  a  sin ν θ  rout

c sin μ θ

MECHANICAL ADVANTAGE vs HANDLE ANGLE 60

50 Mechanical Advantage

8.

40

 

m A θ

30

20

10 45

48

51

54 θ deg

Handle Angle, deg

57

60




Figure P6-21 shows a locking pliers. Calculate its mechanical advantage for the position shown. Scale any dimensions needed from the diagram.

Solution:


1.

Draw the linkage to scale in the position given and find the instant centers.

F

1,4

1,2

P 1

O4

O2 2 3

B

A F

2.

2,3

4

P

3,4 and 1,3

Note that the linkage is in a toggle position (links 2 and 3 are in line) and the angle between links 2 and 3 is 0 deg. From the discussion below equation 6.13e in the text, we see that the mechanical advantage for this linkage in this position is theoretically infinite.




Given:

Figure P6-22 shows a fourbar toggle clamp used to hold a workpiece in place by clamping it at D. The linkage will toggle when link 2 reaches 90 deg. For the dimensions given below, calculate its mechanical advantage for the position shown. Link lengths: Link 2 (O2A) a  70 mm c  34 mm

Link 4 (O4B)

Link 3 (AB)

b  35 mm

Link 1 (O2O4)

d  48 mm

Link 4 (O4D)

rout  82 mm


Link 2 (O2C)

θ  104  deg

Initial position of link 2: Solution: 1.

Global XY system


Draw the mechanism to scale and label it. To establish the position of O4 with respect to O2 (in the global coordinate frame), draw the linkage in the toggle position with 2 = 90 deg. The fixed pivot O4 is then 48 mm fr O2 and 34 mm from B' (see layout). Y C' C

Linkage in toggle position

A'

A 3 2

x

B

D

4

D'

B' 135.069°

O4

X O2 y

2.

Calculate the value of 2 in the local coordinate system (required to calculate 3 and 4). Rotation angle of local xy system to global XY system: θ  θ  α

3.

α  135.069  deg

θ  31.069 deg


d

K1  0.6857

a 2

K3 

2

2

a b c d 2 a c

K2 

d c

2

K3  1.4989

K2  1.4118



 

 

A  cos θ  K1  K2 cos θ  K3

 

B  2  sin θ

 

C  K1   K2  1   cos θ  K3 A  0.4605 4.

B  1.0321






2

θ  2  atan2 2  A B  5.

C  0.1189

B  4 A  C



θ  129.480 deg


2

d

K5 

b

2

2

c d a b

 

2

2 a b

K5  1.4843

 

D  cos θ  K1  K4 cos θ  K5

D  0.1388

 

E  2  sin θ

E  1.0321

 

F  K1   K4  1   cos θ  K5 6.

F  0.4804






θ  2  atan2 2  D E  7.

K4  1.3714

2

E  4  D F



θ  196.400 deg

Referring to Figure 6-10, calculate the values of the angles  and . ν  θ  θ

ν  165.331 deg

If  > 90 deg, subtract it from 180 deg. ν  if  ν  90 deg 180  deg  ν ν

ν  14.669 deg

μ  θ  θ

μ  66.920 deg

If  > 90 deg, subtract it from 180 deg. μ  if  μ  90 deg 180  deg  μ μ 8.

μ  66.920 deg

Using equation 6.13e, calculate the mechanical advantage of the linkage in the position shown. mA 

c sin μ rin  a  sin ν rout

mA  2.970




Given:

Figure P6-22 shows a fourbar toggle clamp used to hold a workpiece in place by clamping it at D. The linkage will toggle when link 2 reaches 90 deg. For the dimensions given below, calculate and plot its mechanical advantage as a function of the angle of link AB as link 2 rotates from 120 to 90 deg (in the global coordinate system). Link lengths: Link 2 (O2A)

a  70 mm

Link 3 (AB)

b  35 mm

Link 4 (O4B)

c  34 mm

Link 1 (O2O4)

d  48 mm

Link 4 (O4D)

rout  82 mm


Link 2 (O2C)

θ  120  deg

Range of positions of link 2: Solution: 1.

θ  90.1 deg Global XY system


Draw the mechanism to scale and label it. To establish the position of O4 with respect to O2 (in the global coordinate frame), draw the linkage in the toggle position with 2 = 90 deg. The fixed pivot O4 is then 48 mm fr O2 and 34 mm from B' (see layout). Linkage in initial position Y C'

C Linkage in final position

A' A

3

x

4

B

2

D

B' 135.069°

O4

D'

X O2 y

2.

Calculate the range of 2 in the local coordinate system (required to calculate 3 and 4). Rotation angle of local xy system to global XY system:

α  135.069  deg

θ  θ  α θ  α  1  deg  θ  α 3.


d

K1  0.6857

a 2

K3 

2

2

a b c d 2 a c

K2 

d c

2

K3  1.4989

K2  1.4118


 


 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

C θ  K1   K2  1   cos θ  K3 4.




 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


2

d

K5 

b

 

2

2

c d a b

2

K4  1.3714

2 a b

 

K5  1.4843

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 6.




 



 

 

θ θ  2   atan2 2  D θ E θ  7.

 2  4 Dθ F θ 

E θ

Referring to Figure 6-10, calculate the values of the angles  and .

    μ θ  θ θ  θ θ ν θ  θ  θ θ

8.

Using equation 6.13e, calculate and plot the mechanical advantage of the linkage over the given range.

 

mA θ 

   rin  a  sin ν θ  rout

c sin μ θ

MECHANICAL ADVANTAGE 50

40

 

30

m A θ

20

10

0 90

95

100

105 θ  α deg

110

115

120




Given:

Figure P6-23 shows a surface grinder. The workpiece is oscillated under the spinning grinding wheel by the slider-crank linkage that has the dimensions given below. Calculate and plot the velocity of the grinding wheel contact point relative to the workpiece over one revolution of the crank. Link lengths: Link 2 (O2 to A)

a  22 mm

Link 3 (A to B)

b  157  mm

Grinding wheel diameter

d  90 mm


ω  120  rpm

Grinding wheel angular velocity ω  3450 rpm Solution: 1.

Offset

c  40 mm

CCW CCW



5

4 2 3

A

B

c

2 O2

2.

Determine the range of motion for this slider-crank linkage. θ  0  deg 1  deg  360  deg

3.


 a sin θ  c  π b  

 

θ θ  asin  4.


 

ω θ 

5.

a b



   ω cos θ θ  cos θ


 

 

    

VB θ  a  ω sin θ  b  ω θ  sin θ θ 6.

Positive to the right

Calculate the velocity of the grinding wheel contact point using equation 6.7: VG 

d 2

 ω

VG  16.258

m sec

Directed to the right


The velocity of the grinding wheel contact point relative to the workpiece, which has velocity VB, is

 

 

VGB θ  VG  VB θ

RELATIVE VELOCITY AT CONTACT POINT 16.6

16.4

Velocity, m/sec

7.


 

VGB θ 

sec m

16.2

16

15.8

0

60

120


240

300

360




Figure P6-24 shows an inverted slider-crank mechanism. Given the dimensions below, find 2, 3, 4, VA4, Vtrans, and Vslip for the position shown with VA2 = 20 in/sec in the direction shown.

Given:

Link lengths: a  2.5 in

Link 2 (O2A)

Link 4 (O4A)

c  4.1 in

Link 1 (O2O4)

d  3.9 in

Measured angles: θ  75.5 deg

θtrans  26.5 deg

Velocity of point A on links 2 and 3: Solution: 1.

θslip  116.5  deg

VA2  20 in sec

1

VA3  VA2


Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest. Y

Axis of slip A

O2

X

11.500° 2 3

Direction of VA4

1

Axis of transmission 4 Direction of VA2 O4

2.

Use equation 6.7 to calculate the angular velocity of link 2. ω 

3.

VA2 a

ω  8.000

rad

CW

sec

Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on links 2 and 3.. The equation to be solved graphically is VA3 = Vtrans + VA3slip a. Choose a convenient velocity scale and layout the known vector VA3. b. From the tip of VA3, draw a construction line with the direction of VA3slip, magnitude unknown. c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing VA3slip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the VA3slip construction line.



Y

0

10 in/sec

1.686

X

Axis of slip

V A4

V trans

VA4slip 1.509

VA3

Axis of transmission V A3slip

2.064

4.


5.

kv 

10 in sec

1

in in

VA4  1.686  in kv

VA4  16.9

Vslip  2.064  in kv

Vslip  20.6


Vtrans  15.1

sec in sec in sec


VA4 c

ω  4.1

Because link 3 slides within link 4, 3 = 4.

rad sec

CW




Figure P6-25 shows a drag-link mechanism with dimensions. Write the necessary equations and solve them to calculate and plot the angular velocity of link 4 for an input of 2 = 1 rad/sec. Comment on the uses for this mechanism.

Given:

Link lengths: Link 2 (L2)

a  1.38 in

Link 3 (L3)

b  1.22 in

Link 4 (L4)

c  1.62 in

Link 1 (L1)

d  0.68 in

ω  1  rad sec


1

CW



A

3 B

2 2 4 x O2

2.

O4

Determine the range of motion for this Grashof double crank. θ  0  deg 2  deg  360  deg

3.


d

K2 

a

K1  0.4928 2

K3 

c

K2  0.4198 2

2

a b c d

 

d

2

K3  0.7834

2 a c

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

C θ  K1   K2  1   cos θ  K3 4.


 





 

 

θ θ  2   atan2 2  A θ B θ  5.

 2  4 A θ Cθ 

B θ



K4 


2

d

K5 

b

 

2

2

2

c d a b

K4  0.5574

2 a b

 

K5  0.3655

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 6.




 



 

 

θ θ  2   atan2 2  D θ E θ  7.


 

a  ω

 

a  ω

ω θ 

ω θ 

b

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 





Plot the angular velocity of link 4.

ANGULAR VELOCITY, LINK 4  0.5


9.

 2  4 Dθ F θ 

E θ

1

 

ω  θ 

sec rad

 1.5

2

 2.5

0

60

120


240

300

360




Figure P6-25 shows a drag-link mechanism with dimensions. Write the necessary equations and solve them to calculate and plot the centrodes of instant center I2,4.

Given:

Measured link lengths: Link 2 (L2)

L2  1.38 in

Link 3 (L3)

L3  1.22 in

Link 4 (L4)

L4  1.62 in

Link 1 (L1)

L1  0.68 in

ω  1  rad sec


1


Draw the linkage to scale and label it. Instant center I2,4 is at the intersection of line AB with line O2O4. To get the first centrode, ground link 2 and let link 3 be the input. Then we have a  L3

b  L4

c  L1 A

d  L2 3 B

2 y 2 4 O2

O4 4

x

2.

Determine the range of motion for this Grashof double rocker. From Figure 3-1a on page 80, one toggle angle is

 a2  d 2  ( b  c) 2  2 a d  


θ  124.294 deg

The other toggle angle is the negative of this. The range of motion is θ  θ  1  deg θ  2  deg  θ  1  deg 3.


d

K2 

a

K1  1.1311 2

K3 

 

d c

K2  2.0294 2

2

a b c d 2 a c

 

2

K3  0.7418

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ



 

 

C θ  K1   K2  1   cos θ  K3 4.




 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ

Calculate the coordinates of the intersection of line BC with line AD. This will be the instant center I2,4.

 

Line BC:

y  x tan θ

Line AD:

y  0  ( x  d )  tan θ

 

Eliminating y and solving for the x- and y-coordinates of the intersection,

 

x24 θ 

6.

   tan θ θ   tan θ d  tan θ θ

 

 

 

y24 θ  x24 θ  tan θ

Plot the fixed centrode.

FIXED CENTRODE 10

y-Coordinate, in

5

 

y24 θ

0

in 5

 10  10

5

0

 

x24 θ in

x-Coordinate, in

7.

Invert the linkage, making C and D the fixed pivots. Then, a  L1

8.

b  L2

c  L3

Determine the range of motion for this Grashof crank rocker. θ  0  deg 0.5 deg  360  deg

d  L4

5

10


SOLUTION MANUAL 6-63-3 4 A

x

3 B

y

2

2

4 O2

9.

O4


d a

K1  2.3824 2

K3 

d

K2 

K2  1.3279 2

2

a b c d

 

c

2

K3  1.6097

2 a c

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

C θ  K1   K2  1   cos θ  K3 10. Use equation 4.10b to find values of 4 for the open circuit.

 





 

 

θ θ  2   atan2 2  A θ B θ 

 2  4 A θ Cθ 

B θ

11. Calculate the coordinates of the intersection of line BC with line AD. This will be the instant center I2,4.

 

Line AD:

y  x tan θ

Line BC:

y  0  ( x  d )  tan θ

 

Eliminating y and solving for the x- and y-coordinates of the intersection,

 

x24 θ 

6.

   tan θ θ   tan θ d  tan θ θ

Plot the moving centrode. (See next page.)

 

 

 

y24 θ  x24 θ  tan θ



MOVING CENTRODE 10

y-Coordinate, in

5

 

y24 θ

0

in 5

 10  10

5

0

 

x24 θ in

x-Coordinate, in

5

10




Figure P6-26 shows a mechanism with dimensions. Use a graphical method to calculate the velocities of points A, B, and C and the velocity of slip for the position shown.

Given:

Link lengths and angles: Link 1 (O2O4)

d  1.22 in

Angle O2O4 makes with X axis

θ  56.5 deg

Link 2 (O2A)

a  1.35 in

Angle 2 makes with X axis

θ  14 deg

Link 4 (O4B)

e  1.36 in

Link 5 (BC)

f  2.69 in

Link 6 (O6C)

g  1.80 in

Angle O6C makes with X axis

θ  88 deg

Angular velocity of link 2 Solution: 1.

ω  20 rad sec

1

CW



B

Axis of slip Y

Axis of transmission O4

4

0.939

132.661° A 3 2 X O2

Direction of VA3

2.

Use equation 6.7 to calculate the magnitude of the velocity at point A on links 2 and 3. VA3  a  ω

3.

VA3  27.000

in sec

θVA3  θ  90 deg

θVA3  76.0 deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on link 3. The equation to be solved graphically is VA3 = Vtrans + Vslip a. Choose a convenient velocity scale and layout the known vector VA3. b. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the Vslip construction line.



Y 0

12 in/sec

1.295" X Vtrans

VA3 2.369"

4.


5.

kv 

in in


Vtrans  15.540

sec in sec

The true velocity of point A on link 4 is Vtrans, VA4  15.54 

in sec

Determine the angular velocity of link 4 using equation 6.7.

ω 

VA4 c

c  0.939  in and

ω  16.550

rad

θ  132.661  deg

CW

sec

Determine the magnitude and sense of the vector VB using equation 6.7. VB  e ω

θVA4  θ  90 deg 8.

1

Vslip  28.428

From the linkage layout above:

7.

12 in sec

Vslip  2.369  in kv

VA4  Vtrans 6.

Vslip

VB  22.507

in sec

θVA4  42.661 deg

Draw links 1, 4, 5, and 6 to a convenient scale. Indicate the directions of the velocity vectors of interest. (See next page.)



Direction of VCB

Direction of VB

B

Y O4

4

C A 3

Direction of VC 2

X O2

O6

9.

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VB + VCB a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VCB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VC, magnitude unknown. d. Complete the vector triangle by drawing VCB from the tip of VB to the intersection of the VC construction line and drawing VC from the tail of VB to the intersection of the VCB construction line. VB

0

12 in/sec

Y

VCB X VC 2.088"

10. From the velocity triangle we have: Velocity scale factor: VC  2.088  in kv

kv 

12 in sec

VC  25.1

in in sec

1

θVC  θ  90 deg θVC  2.0 deg




Figure P6-27 shows a cam and follower. Distances are given below. Find the velocities of points A and B, the velocity of transmission, velocity of slip, and 3 if 2 = 50 rad/sec (CW). Use a graphical method.

Given:

ω  50 rad sec

1

Distance from O2 to A:

a  1.890  in

Distance from O3 to B:

b  1.645  in

Assumptions: Roll-slide contact Solution: 1.


Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities of interest. Axis of slip Direction of VB

1.890

Axis of transmission A

B

3

2

O2

O3

1.645

Direction of VA

2.


3.

VA  94.500

in sec

Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A. The equation to be solved graphically is VA = Vtrans + VAslip a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA and drawing Vtrans from the tail of VA to the intersection of the Vslip construction line.



50 in/sec

0

VB 2.304

V Bslip

Y 1.317 X

3.256

Vtrans

1.890 VA

4.


5.

V A2slip

kv 

50 in sec

1

in in

VA  1.890  in kv

VA  94.5

VB  2.304  in kv

VB  115.2

Vslip  3.256  in kv

Vslip  162.8


Vtrans  65.8

sec in sec in sec in sec


VB b

ω  70.0

rad sec

CW




Figure P6-28 shows a quick-return mechanism with dimensions. Use a graphical method to calcula the velocities of points A, B, and C and the velocity of slip for the position shown.

Given:


d  1.69 in


Link 2 (L2)

a  1.00 in

Angle link 2 makes with X axis θ  99 deg

Link 4 (L4)

e  4.76 in

Link 5 (L5)

f  4.55 in

Offset (O2C)

g  2.86 in


ω  10 rad sec

1

θ  15.5 deg

CCW



B


Direction of VA3

4

Y Axis of slip A 3 2.068

2 44.228° O2

X O4

2.


3.

VA2  10.000

in sec

θVA2  θ  90 deg

θVA2  189.0 deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on link 3. The equation to be solved graphically is VA3 = Vtrans + Vslip



a. Choose a convenient velocity scale and layout the known vector VA3. b. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the Vslip construction line. Y 0

5 in/sec

X

VA3 V trans

1.154

4.

kv 

in in


Vtrans  5.770

sec in sec

The true velocity of point A on link 4 is Vtrans, VA4  5.77

in sec


ω 

VA4 c

c  2.068  in and

ω  2.790

rad

θ  44.228 deg

CCW

sec


θVB  θ  90 deg 8.

1

Vslip  8.170


7.

5  in sec

Vslip  1.634  in kv


1.634


5.

Vslip

VB  13.281

in sec

θVB  134.228 deg




Direction of VCB Direction of VB B 5 5.805°

C 6

4 Direction of VC

Y

A 3

2 44.228° O2

X O4

9.

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VB + VCB a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VCB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VC, magnitude unknown. d. Complete the vector triangle by drawing VCB from the tip of VB to the intersection of the VC construction line and drawing VC from the tail of VB to the intersection of the VCB construction line.

10. From the velocity triangle we have:

Velocity scale factor: VC  1.659  in kv

kv 

5  in sec

VC  8.30

VB

1

0

5 in/sec

in in sec

Y

θVC  180  deg V CB X

VC 1.659




Given:

Figure P6-29 shows a drum pedal mechanism . For the dimensions given below, find and plot the mechanical advantage and the velocity ratio of the linkage over its range of motion. If the input velocity Vin is a constant and Fin is constant, find the output velocity, output force, and power in over the range of motion. Link lengths: Link 2 (O2A)

a  100  mm

Link 3 (AB)

b  28 mm

Link 4 (O4B)

c  64 mm

Link 1 (O2O4)

d  56 mm

Link 3 (AP)

rout  124  mm

Distance to force application: rin  48 mm

Link 2

Solution: 1.

1

Input force and velocity:

Fin  50 N

Vin  3  m sec

Range of positions of link 2:

θ  162  deg

θ  171  deg


Draw the mechanism to scale and label it. P

3

B 3

Fin

4

Vin

A 2

2

x

1

O4

O2 y

2.




α  180  deg



θ  θ  α  θ  α  1  deg  θ  α 3.


d

K2 

a

K1  0.5600 2

K3 

 

d c

K2  0.8750 2

2

a b c d 2 a c

 

2

K3  1.2850

 

A θ  cos θ  K1  K2 cos θ  K3


 


 

B θ  2  sin θ

 

 

C θ  K1   K2  1   cos θ  K3 4.




 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


d

K4 

K5 

b

 

2

2

c d a b

2

K4  2.0000

2 a b

 

K5  1.7543

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 6.


 





 

 

θ θ  2   atan2 2  D θ E θ 

Using equations 6.13d and 6.18a, where out is 3, calculate and plot the mechanical advantage of the linkage over the given range.

 

mA θ 

     rin  rout a  sin θ θ  θ

b  sin θ θ  θ θ

MECHANICAL ADVANTAGE 0.14

Mechanical Advantage

7.

 2  4 Dθ F θ 

E θ

0.13

 

mA θ 0.12

0.11

0.1 162

164

166

168 θ α deg

Pedal Angle, deg

170

172


8.


Calculate and plot the velocity ratio using equation 6.13d,

 

1

mV θ 

 

mA θ

VELOCITY RATIO 10

Velocity Ratio

8

 

6

mV θ

4 2 0 162

164

166

168

170

172

θ  α deg Pedal Angle, deg

Calculate and plot the output velocity using equation 6.13a.

 

 

Vout θ  Vin mV θ

OUTPUT VELOCITY

Velocity, m/sec

9.

20

 

Vout θ 

sec m 10

0 162

164

166

168 θ α deg

Pedal Angle, deg

170

172



10. Calculate and plot the output force using equation 6.13a.

 

 

Fout θ  Fin mA θ

OUTPUT FORCE 8

Force, N

6

 

Fout θ

4

N 2

0 162

164

166

168 θ  α deg

Pedal Angle, deg

170

172




Figure 3-33 shows a sixbar slider crank linkage. Find all of its instant centers in the position shown:

Given:

Number of links n  6

Solution:


1.


2.

n ( n  1)

C  15

2

Draw the linkage to scale and identify those ICs that can be found by inspection (8).

Y 2,3 3

3,4; 3,5; 4,5 1,6 at infinity

2 4 1,2 O2

5 X 6

1,4 O4 2.

1,5

5,6

Use Kennedy's Rule and a linear graph to find the remaining 7 ICs: I1,3; I1,5; I2,4; I2,5; I2,6; I3,6; and I4,6

1

I1,5: I1,6-I5,6 and I1,4-I4,5

6

2

5

3

I2,5: I1,2-I1,5 and I2,3-I3,5 I1,3: I1,2-I2,3 and I1,5-I3,5 I3,6: I1,6-I1,3 and I3,5-I5,6 I2,4: I2,3-I3,4 and I2,5-I4,5

4

2,5

I2,6: I1,2-I1,6 and I2,5-I5,6

2,6

I4,6: I1,4-I1,6 and I4,5-I5,6

Y 2,3

4,6 3,6

3 3,4; 3,5; 4,5; 2,4

2 4 1,2 O2

1,6 at infinity

5 X 6

1,4 O4 1,3

5,6




Calculate and plot the centrodes of instant center I24 of the linkage in Figure 3-33 so that a pair of noncircular gears can be made to replace the driver dyad 23.

Given:

Link lengths:

Solution: 1.

Input crank (L2)

L2  2.170


L3  2.067

Output crank (L4)

L4  2.310

Fourbar ground link (L1)

L1  1.000


Invert the linkage, grounding link 2 such that the input link is 3, the coupler is 4, and the output link is 1. a  L3

b  L4

c  L1

d  L2

2.

Define the input crank motion for this inversion: θ  47 deg 47.5 deg  102.5  deg

3.


d

K2 

a

K1  1.0498

 

2

d

K3 

c

K2  2.1700

 

2

2

a b c d

2

2 a c

K3  1.1237

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ 

 

  

 

 2  4 A θ Cθ 

B θ

 

θ θ  if θ θ  2  π θ θ  2  π θ θ 5.

Calculate the coordinates of the intersection of links 1 and 3 in the xy coordinate system.

 

x242 θ  

6.

   tan θ  tan  θ θ 

 

 

 

y242 θ  x242 θ  tan θ

Invert the linkage, grounding link 4 such that the input link is 1, the coupler is 2, and the output link is 3. a  L1

8.

d  tan θ θ

b  L2

c  L3

d  L4


d

K2 

a

K1  2.3100

 

2

d

K3 

c

K2  1.1176

 

2

2

a b c d 2 a c

K3  1.4271

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 

 





 

 

θ θ  2   atan2 2  A θ B θ 

 

  

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ

 

 2  4 A θ Cθ 

B θ

 

θ θ  if θ θ  2  π θ θ  2  π θ θ 5.

Calculate the coordinates of the intersection of links 1 and 3 in the xy coordinate system.

2


 

x244 θ  


   tan θ  tan  θ θ  d  tan θ θ

 

 

 

y244 θ  x244 θ  tan θ

LINK 2 GROUNDED 5 4 3 2

 

1

y242 θ 0 1 2 3 4 5 4

3

2

1

0

 

1

x242 θ

7.

Define the input crank motion for this inversion: θ  41 deg 42 deg  241  deg LINK 4 GROUNDED 10 8 6 4 2

 

y244 θ

0 2 4 6 8

 10  10  8  6  4  2

0

 

x244 θ

2

4

6

8

10




Find the velocity of the slider in Figure 3-33 for 2 = 110 deg with respect to the global X axis assuming 2 = 1 rad/sec CW. Use a graphical method.

Given:


a  2.170  in

Link 3 (A to B)

b  2.067  in

Link 4 (O4 to B)

c  2.310  in

Link 1 (O2 to O4)

d  1.000  in

Link 5 (B to C)

e  5.400 Crank angle:

θ2  110  deg

  102  deg

Coordinate angle

ω  1  rad sec


1

CW



Direction of VA Y Direction of VBA A

147.635° 3 Direction of VB

B

2 4 O2

5 58.950°

158.818°

X

6

O4 C Direction of VC Direction of VCB 2.

3.

Use equation 6.7 to calculate the magnitude of the velocity at point A. in

VA  a  ω

VA  2.170

θVA  θ2  90 deg

θVA  20.000 deg

sec



0


VA

1 in/sec

Y

X V BA

VB 1.325

4.


VB  1.325  in kv 5.

kv 

1  in sec

1

in

VB  1.325

in

θVB  31.050 deg

sec

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C, the magnitude of the relative velocity VCB, and the angular velocity of link 3. The equation to be solved graphically is VC = VB + VCB a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VCB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VC, magnitude unknown. d. Complete the vector triangle by drawing VCB from the tip of VB to the intersection of the VC construction line and drawing VC from the tail of VB to the intersection of the VCB construction line.

0

1.400

1 in/sec

Y VC X

VB 4.

V CB


VC  1.400  in kv

kv 

1  in sec

1

in

VC  1.400

in sec

θVC  0.0 deg




Find the velocity of the slider in Figure 3-33 for 2 = 110 deg with respect to the global X axis assuming 2 = 1 rad/sec CW. Use the method of instant centers.

Given:


a  2.170  in

Link 3 (A to B)

b  2.067  in

Link 4 (O4 to B)

c  2.310  in

Link 1 (O2 to O4)

d  1.000  in

Link 5 (B to C)

e  5.400 Crank angle:

θ2  110  deg

  102  deg

Coordinate angle

ω  1  rad sec


1

CW

See Figure 3-33 and Mathcad file P0670b.

Draw the linkage to scale in the position given, find instant centers I1,3 and I1,5, and the distances from the pin joints to the instant centers. See Problem 6-68 for the determination of IC locations.

1,5

Y A 3 B

2 4

5 X

O2

C

6

O4 1,3 From the layout above: AI13  2.609  in 2.

BI13  1.641  in

BI15  9.406  in

CI15  9.896  in



3.

VA  2.170

θVA  θ2  90 deg

θVA  20.0 deg

VA AI13

ω  0.832

rad

CW

sec

Determine the magnitude of the velocity at point B using equation 6.9b. The direction of VB is down and to the right VB  1.365

in sec

Use equation 6.9c to determine the angular velocity of link 5. ω 

6.

sec


VB  BI13 ω 5.

in

VA  a  ω

ω  4.


VB BI15

ω  0.145

rad

CCW

sec

Determine the magnitude of the velocity at point C using equation 6.9b. VC  CI15 ω

VC  1.436

in sec

to the right




Find the velocity of the slider in Figure 3-33 for 2 = 110 deg with respect to the global X axis assuming 2 = 1 rad/sec CW. Use an analytical method.

Given:


a  2.170  in

Link 3 (A to B)

b  2.067  in

Link 4 (O4 to B)

c  2.310  in

Link 1 (O2 to O4)

d  1.000  in

Link 5 (B to C)

e  5.400  in Crank angle:

θ2XY  110  deg

  102  deg

Coordinate angle

ω2  1  rad sec


1

CW

See Figure P6-33 and Mathcad file P0670c.


Y A 3 B

2 4

5 X

O2

y 6

O4 x 2.

C

102°

Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. Transform crank angle to the local xy coordinate system:

θ2  θ2XY   K1 

d

K1  0.4608

a 2

K3 

θ2  212.000 deg

2

2

a b c d

K2 

d c

2

2 a c

K3  0.6755

A  cos θ2  K1  K2 cos θ2  K3 B  2  sin θ2 C  K1   K2  1   cos θ2  K3 A  0.2662 3.

B  1.0598

C  2.3515


K2  0.4329







2

θ4xy  2  atan2 2  A B  4.

2

d b

2

2

K4  0.4838

2 a b

E  2  sin θ2

E  1.0598

F  K1   K4  1   cos θ2  K5

F  0.3808






2

E  4  D F

  2 π

a  ω2 sin θ2  θ3xy  c sin θ4xy  θ3xy

ω4  0.591

rad sec

Transform 4 back to the global XY system.

θ4  662.365 deg

Determine 5 and d, with respect to O4, using equation 4.17. cc  0  in

 c sin θ4  cc  π e  

θ5  asin 

θ5  158.818 deg

dd  c cos θ4  e cos θ5

dd  6.272 in


ω5  7.

θ3xy  649.050 deg

Determine the angular velocity of link 4 for the open circuit using equations 6.18.

Offset:

6.

K5  0.5178

D  2.2370

θ4  θ4xy   5.

θ4xy  560.365 deg

D  cos θ2  K1  K4 cos θ2  K5

ω4  7.

2

c d a b

K5 

θ3xy  2  atan2 2  D E  6.

  2 π


5.

B  4 A  C

c cos θ4   ω4 e cos θ5

ω5  0.145

rad sec

Determine the velocity of pin C using equation 6.22b: VC  c ω4 sin θ4  e ω5 sin θ5 VC  1.436

in sec

VC  1.436

in sec

 

arg VC  0.000 deg




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular velocity of link 4 and the linear velocity of slider 6 in the sixbar slider-crank linkage of Figure 3-33 as a function of the angle of input link 2 for a constant 2 = 1 rad/sec CW. Plot Vc both as a function of 2 and separately as a function of slider position as shown in the figure. What is the percent deviation from constant velocity over 240 deg < 2 < 270 deg and over 190 < 2 < 315 deg?

Given:


a  2.170


b  2.067

Output crank (L4)

c  2.310


e  5.40

d  1.000


  1 

Crank velocity: Solution:

rad sec


1.

This sixbar drag-link mechanism can be analyzed as a fourbar Grashof double crank in series with a crankslider mechanism using the output of the fourbar, link 4, as the input to the crank-slider.

2.


3.

Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit) in the global XY coordinate system. 2

K1 

d

K2 

a

K1  0.4608

 

K3 

d c

K2  0.4329

 

2

2

a b c d

2

2 a c

K3  0.6755

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 

 



 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 2  4 A θ Cθ   102 deg

 

θ θ  2   atan2 2  A θ B θ  4.

B θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 5.


 

 c sin θ θ   π e  

 

    e cosθθ


f θ  c cos θ θ 5.


d b

2

K5 

2

2

c d a b 2 a b

2

K4  0.4838

K5  0.5178


 


 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 6.




 



 

 2  4 Dθ F θ 

 

θ θ  2   atan2 2  D θ E θ  7.

E θ


 

 θ 

a   c





 


  

 

sin θ θ  102  deg  θ θ

 0.5  0.75 1

 

 θ  1.25  1.5  1.75 2

0

45

90

135

180

225

270

θ deg

8.


 

ω θ 

9.

       

c cos θ θ    θ e cos θ θ

Determine the velocity of pin C using equation 6.22b:

 

      e ωθ sinθθ

VC θ  c  θ  sin θ θ

315

360



2 1 0

 

VC θ  1 2 3 4

0

45

90

135

180

225

270

315

360

θ deg

2

1

0

 

VC θ  1 2 3 4

3

4

5

 

f θ

6

7

8




Figure 3-34 shows a Stephenson's sixbar mechanism. Find all of its instant centers in the position shown:

Given:


Solution:


1.


n ( n  1)

C  15

2

a.

In part (a) of the figure.

1.


1,2 O2

2 4,6

1,6

5,6

2,3

6 O6 3

5

1,4 O4

4,5 4 5 2.

3,5

Use Kennedy's Rule and a linear graph to find the remaining 8 ICs: I1,3; I1,5; I2,4; I2,5; I2,6; I3,4; I3,6; and I4,6 I1,5: I1,6-I5,6 and I1,4-I4,5 I2,5: I1,2-I1,5 and I2,3-I3,5 I1,3: I1,2-I2,3 and I1,5-I3,5 I3,4: I1,4-I1,3 and I4,5-I3,5 I2,8: I2,3-I3,4 and I2,5-I4,5 I2,6: I1,2-I1,6 and I2,5-I5,6 I3,6: I1,3-I1,6 and I3,5-I5,6

1,2; 2,5; 2,4; 2,6

I4,6: I1,4-I1,6 and I4,5-I5,6 O2

1 6

2

2 6 3

5

3 4

1,6

5,6

2,3

5

O6

1,5

1,4 O4

4,5 4 5 3,5; 1,3; 3,4; 3,6



b.

In part (b) of the figure.

1.


1,2 5,6 2

O2

2,3

5

6

4,5 3

4

5

1,6 O6 1,4 O4

3,5

2.

Use Kennedy's Rule and a linear graph to find the remaining 8 ICs: I1,3; I1,5; I2,4; I2,5; I2,6; I3,4; I3,6; and I4,6 I1,5: I1,6-I5,6 and I1,4-I4,5

3,6

1,2

I2,5: I1,2-I1,5 and I2,3-I3,5

5,6; 4,6 2 2,3

I1,3: I1,2-I2,3 and I1,5-I3,5

O2

5 4,5

3

5

I3,4: I1,4-I1,3 and I4,5-I3,5

6 1,6 O6 1,4; 1,5 O4

4


2,6


3,5; 3,4 To 1,3 1 6

2

5

3 4

2,5; 2,4; 2,8



c.

In part (c) of the figure.

1.


2,3 1,6

2 1,2

4,5

O2 5 5

3

6 O6 1,4 O4

4 3,5

2.

5,6


I2,4: I2,3-I3,4 and I2,5-I4,5

I2,5: I1,2-I1,5 and I2,3-I3,5

I2,6: I1,2-I1,6 and I2,5-I5,6

I1,3: I1,2-I2,3 and I1,5-I3,5

I3,6: I1,3-I1,6 and I3,5-I5,6

I3,4: I1,4-I1,3 and I4,5-I3,5

I4,6: I1,4-I1,6 and I4,5-I5,6

2,3 4,6

1

1,6

2 6

2

5

3

1,2; 2,5; 2,4: 2,6 4,5 O2

4

5 5

3

4 3,5; 1,3; 3,4; 3,6

1,5

6 O6 1,4 O4

5,6




Find the angular velocity of link 6 in Figure 3-34b for 6 = 90 deg with respect to the global X axis assuming 2 = 10 rad/sec CW. Use a graphical method.

Given:


g  1.556  in

Link 3 (A to B)

f  4.248  in

Link 4 (O4 to C)

c  2.125  in

Link 5 (C to D)

b  2.158  in

Link 6 (O6 to D)

a  1.542  in

Link 5 (B to D)

p  3.274  in

Link 1 X-offset

d X  3.259  in

Link 1 Y-offset

d Y  2.905  in

Angle CDB

δ5  36.0 deg

Output rocker angle:

θ  90 deg Global XY system ω  10 rad sec


1

CW

See Figure 3-34b and Mathcad file P0673a.


Direction of VCD Y

Direction of VD

Direction of VA X 2 A

D

O2

5

6

C Direction of VAB

O6 3

5

Direction of VC

4 O4 B

Direction of VBD 2.

Since this linkage is a Stephenson's II sixbar, we will have to start at link 6 and work back to link 2. We will assume a value for 6 and eventually find a value for 2. We will then multiply the magnitudes of all velocities by the ratio of the actual 2 to the found 2. Use equation 6.7 to calculate the magnitude of the velocity at point D. 1 Assume: CW ω  1  rad sec VD  a  ω

3.

VD  1.542

in sec

θVD  0  deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C, the magnitude of the relative velocity VCD, and the angular velocity of link 5. The equation to be solved graphically is VC = VD + VCD a. b.

Choose a convenient velocity scale and layout the known vector VD. From the tip of VD, draw a construction line with the direction of VCD, magnitude unknown.



c. From the tail of VD, draw a construction line with the direction of VC, magnitude unknown. d. Complete the vector triangle by drawing VCD from the tip of VD to the intersection of the VC construction line and drawing VC from the tail of VD to the intersection of the VCD construction line. 0

1 in/sec

1.289

VC

VCD

1.309 127.003°

54.195° VD

4.



5.

1

in in

VC  1.289  in kv

VC  1.289

VCD  1.309  in kv

VCD  1.309

θVC  54.195 deg

sec in

θVCD  127.003  deg

sec

Determine the angular velocity of links 5 and 4 using equation 6.7. ω  ω 

6.

1  in sec

VCD b VC c

ω  0.607

rad

ω  0.607

rad

sec

sec

Determine the magnitude and sense of the vector VBD using equation 6.7. VBD  p  ω

VBD  1.986

in sec

θVBD  θVCD  δ5 7.

θVBD  163.003 deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B. The equation to be solved graphically is VB = VD + VBD a. b. c.

Choose a convenient velocity scale and layout the known vector VD. From the tip of VD, layout the (now) known vector VBD. Complete the vector triangle by drawing VB from the tail of VD to the tip of the VBD vector. 0

VBD VB 0.682 121.607°

VD

1 in/sec


8.



kv 


VB  0.682  in kv 9.

1  in sec

1

in

VB  0.682

in

θVB  121.607  deg

sec

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point A, the magnitude of the relati velocity VAB, and the angular velocity of link 2. The equation to be solved graphically is VA = VB + VAB a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VAB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VA, magnitude unknown. d. Complete the vector triangle by drawing VAB from the tip of VB to the intersection of the VA construction line and drawing VA from the tail of VB to the intersection of the VAB construction line. VAB

0

VB

1 in/sec

VA 0.645

131.690°

10. From the velocity triangle we have:


VA  0.645  in kv

kv 

1  in sec

1

in

VA  0.645

in

θVA  131.690  deg

sec

11. Determine the angular velocity of link 2 with respect to the assumed value of 6 using equation 6.7.

ω 

VA g

ω  0.415

rad sec

12. Calculate the actual value of the angular velocity of link 6.  

ω rad  ω sec

  24.124

rad sec

CW




Find the angular velocity of link 6 in Figure 3-34b for 6 = 90 deg with respect to the global X axis assuming 2 = 10 rad/sec CW. Use the method of instant centers.

Given:


g  1.556  in

Link 3 (A to B)

f  4.248  in

Link 4 (O4 to C)

c  2.125  in

Link 5 (C to D)

b  2.158  in

Link 6 (O6 to D)

a  1.542  in

Link 5 (B to D)

p  3.274  in

Link 1 X-offset

d X  3.259  in

Link 1 Y-offset

d Y  2.905  in

Angle CDB

δ5  36.0 deg


θ  90 deg Global XY system


  10 rad sec

1

CW

See Figure 3-34b and Mathcad file P0673b.

Draw the linkage to scale in the position given, find instant centers I1,3 and I1,5, and the distances from the pin joints to the instant centers. See Problem 6-72b for determination of IC locations. D A

2

O2

5

6

C 3

5

4 B

To 1,3

From the layout above: AI13  22.334 in

BI13  23.650 in

BI15  1.124  in

DI15  2.542  in

O6 1,5 O4


2.


Start from point D with an assumed value for 6 and work to find 2. Then, use the ratio of the actual value of 2 to the found value to calculate the actual value of 6. Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at point D. rad   1  sec in VD  a   VD  1.542 sec

θVD  θ  90 deg 3.


4.

VD DI15

VB BI13

VB  0.682

in sec

ω  0.029

rad

CCW

sec

VA  0.644

in

to the left

sec

Use equation 6.9c to determine the angular velocity of link 2 based on the assumed value of 6.  

8.

CW

sec

Determine the magnitude of the velocity at point A using equation 6.9b. VA  AI13  ω

7.

rad

Use equation 6.9c to determine the angular velocity of link 3. ω 

6.

ω  0.607

Determine the magnitude of the velocity at point B using equation 6.9b. VB  BI15 ω

5.

θVD  0.0 deg

VA g

  0.414

rad

CW

sec

Multiply the assumed vaue of 6 by the ratio of 21 over 22 to get the value of 6 for 2 = 10 rad/sec.  

 rad   sec

  24.166

rad sec

CW




Find the angular velocity of link 6 in Figure 3-34b for 6 = 90 deg with respect to the global X axis assuming 2 = 10 rad/sec CW. Use an analytic method.

Given:

Solution: 1.


g  1.556  in

Link 3 (A to B)

f  4.248  in

Link 4 (O4 to C)

c  2.125  in

Link 5 (C to D)

b  2.158  in

Link 6 (O6 to D)

a  1.542  in

Link 5 (B to D)

p  3.274  in

Link 1 X-offset

d X  3.259  in

Link 1 Y-offset

d Y  2.905  in

Angle CDB

δ5  36.0 deg

Link 1 (O4 to O6)

d  1.000  in


θ6XY  90 deg

Global XY system


ω  10 rad sec

Coordinate rotationm angle

δ  90 deg

1

CW

See Figure 3-34b and Mathcad file P0673c.

Transform the crank angle to the local coordinate system. Draw the linkage to scale and label it.

θ6  θ6XY  δ

θ6  180.000 deg

Y

2 A

D

O2

5

X

6

C

y O6

3

5

4 O4 B x

2.


d a

K1  0.6485

3.

K2 

d c

K2  0.4706

2

K3 

K3  0.4938

A  cos θ6  K1  K2 cos θ6  K3

A  0.6841

B  2  sin θ6

B  0.0000

C  K1   K2  1   cos θ6  K3

C  2.6129


2

2

a b c d 2 a c

2







θ4  2  atan2 2  A B  4.

2

d

K5 

b

θ4  234.195 deg

2

2

c d a b

2

2 a b

K4  0.4634

D  cos θ6  K1  K4 cos θ6  K5

D  2.6407

E  2  sin θ6

E  0.0000

F  K1   K4  1   cos θ6  K5

F  0.6563

K5  0.5288






θ51  2  atan2 2  D E  6.




5.

2

B  4 A  C

2

E  4  D F



θ51  307.003 deg

Determine the angular velocity of links 4 and 5 for the open circuit using equations 6.18. Initially assume 6 = 1 rad/sec. then by trial and error, change it to make 2 = 10 rad/ sec CW.

ω6  1  rad sec

1

ω5 

a  ω6 sin θ4  θ6  b sin θ51  θ4

ω5  0.607

rad

ω4 

a  ω6 sin θ6  θ51  c sin θ4  θ51

ω4  0.607

rad

sec

sec




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular velocity of link 6 in the sixbar linkage of Figure 3-34 as a function of 2 for a constant 2 = 1 rad/sec CW.

Given:

Link lengths:

Solution: 1.

Link 2 (O2 to A)

g  1.556  in

Link 3 (A to B)

f  4.248  in

Link 4 (O4 to C)

c  2.125  in

Link 5 (C to D)

b  2.158  in

Link 6 (O6 to D)

a  1.542  in

Link 5 (B to D)

p  3.274  in

Link 1 X-offset

d X  3.259  in

Link 1 Y-offset

d Y  2.905  in

Angle CDB

δ5  36.0 deg

Link 1 (O4 to O6)

d  1.000  in



Global XY system


ω  10 rad sec

Coordinate rotationm angle

δ  90 deg

1

CW


This problem is long and may be more appropriate for a project assignment. The solution involves defining vector loops and solving the resulting equations using a method such as Newton-Raphson.




Figure 3-35 shows a Stephenson's sixbar mechanism. Find all of its instant centers in the position shown:

Given:


Solution:


1.


n ( n  1)

C  15

2

a.

In part (a) of the figure.

1.

Draw the linkage to scale and identify those ICs that can be found by inspection (7). 4,5

5 5,6

4

2 1,2 O2

O4

3

2,3

1,4

6

1,6 O6

3,4

2.

Use Kennedy's Rule and a linear graph to find the remaining 8 ICs: I1,3; I1,5; I2,4; I2,5; I2,6; I3,5; I3,6; and I4,6 4,5

5 4

2 2,3

1,2; 2,5; 2,4; 2,6 O2

3

O4

5,6 1,4

6

1,6 O6

4,6 1,3; 3,5; 3,6

1

3,4

I1,5: I1,6-I5,6 and I1,4-I4,5 I1,3: I1,2-I2,3 and I1,4-I3,4 I3,5: I1,5-I1,3 and I3,4-I4,5 I2,5: I1,2-I1,5 and I2,3-I3,5 I2,4: I2,3-I3,4 and I2,5-I4,5 I2,6: I1,2-I1,6 and I2,5-I5,6 I3,6: I1,3-I1,6 and I3,5-I5,6 I4,6: I1,4-I1,6 and I4,5-I5,6

1,5

6

2

5

3 4



b.

In part (b) of the figure.

1.

Draw the linkage to scale and identify those ICs that can be found by inspection (7). 1,6 2 2,3

1,4

1,2

3

O2

5

O4 5,6

3,4

2.

4,5 6

4


I2,4: I2,3-I3,4 and I2,5-I4,5

I1,3: I1,2-I2,3 and I1,4-I3,4

I2,6: I1,2-I1,6 and I2,5-I5,6

I3,5: I1,5-I1,3 and I3,4-I4,5

I3,6: I1,3-I1,6 and I3,5-I5,6

I2,5: I1,2-I1,5 and I2,3-I3,5

I4,6: I1,4-I1,6 and I4,5-I5,6 1,5

1,2

1

2 6

2

2,3 3

5

3

4,5

1,4 O2

6

4 O4

5,6 4

1,2; 2,5; 2,4; 2,6

3,4; 1,3; 3,5; 3,6

4,6 5

1,6




Use a compass and straightedge to draw the linkage in Figure 3-35 with link 2 at 90 deg and find the angular velocity of link 6 assuming 2 = 10 rad/sec CCW. Use a graphical method.

Given:


a  1.000  in

Link 3 (A to B)

b  3.800  in

Link 4 (O4 to B)

c  1.286  in

Link 1 (O2 to O4)

d  3.857  in

Link 4 (O4 to D)

e  1.429  in

Link 5 (D to E)

f  1.286

Link 6 (O6 to E)

g  0.771  in

Crank angle:

θ2  90 deg

ω  10 rad sec


1

CCW

See Figure 3-35 and Mathcad file P0676a.

Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Direction of VA Direction of VD

Y A

D

122.085°

100.938°

33.359°

2

5

3 4 O2

X

6 O4 O6 E

35.228°

B Direction of VE Direction of VBA

Direction of VED

Direction of VB

2.

3.


VA  a  ω

VA  10.000

θVA  θ2  90 deg

θVA  180.000 deg

sec




Y

1.671 1.063 0

5 in/sec

VB

VBA

X VA

4.



VBA  1.063  in kv

VBA  5.315   6.497

c

θVB  147.915  deg

sec in

θVBA  56.641 deg

sec

rad

CW

sec

Calculate the magnitude and direction of VD. VD  e 

6.

in

VB  8.355

VB

1

in

VB  1.671  in kv

  5.

5  in sec

VD  9.284

in

θVD  55.085 deg

sec

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point E, the magnitude of the relati velocity VED, and the angular velocity of link 3. The equation to be solved graphically is VE = VD + VED a. Choose a convenient velocity scale and layout the known vector VD. b. From the tip of VD, draw a construction line with the direction of VED, magnitude unknown. c. From the tail of VD, draw a construction line with the direction of VE, magnitude unknown. d. Complete the vector triangle by drawing VED from the tip of VD to the intersection of the VE construction line and drawing VE from the tail of VD to the intersection of the VED construction line.

4.

Y


VE  2.450  in kv ω 

kv 

5  in sec

5 in/sec

1

in

VE  12.250

0

X

in sec

2.450

VE VD

g

ω  15.888

rad

CW

VE

sec V DE




Use a compass and straightedge to draw the linkage in Figure 3-35 with link 2 at 90 deg and find the angular velocity of link 6 assuming 2 = 10 rad/sec CCW. Use the method of instant centers.

Given:


a  1.000  in

Link 3 (A to B)

b  3.800  in

Link 4 (O4 to B)

c  1.286  in

Link 1 (O2 to O4)

d  3.857  in

Link 4 (O4 to D)

e  1.429  in

Link 5 (D to E)

f  1.286

Link 6 (O6 to E)

g  0.771  in

Crank angle:

θ2  90 deg


ω  10 rad sec

1

CCW


Draw the linkage to scale in the position given, find instant centers I1,3 and I1,5, and the distances from the pin joints to the instant centers. See Problem 6-68 for the determination of IC locations.

1,5 A 2

D 5

3 4 O2

6 O4 O6 E

B

1,3 From the layout above: AI13  7.152  in 2.

BI13  5.975  in

DI15  0.947  in

EI15  1.249  in


θVA  θ2  90 deg 3.

BI15  1.740  in

θVA  180.0 deg


VA AI13

ω  1.398

rad sec

CCW


4.

Determine the magnitude of the velocity at point B using equation 6.9b. VB  BI13 ω

5.

VB c

VD DI15

ω  6.496

rad

CW

sec

VD  9.283

in sec

ω  9.803

rad

CW

sec

Determine the magnitude of the velocity at point E using equation 6.9b. VE  EI15 ω

9.

sec


8.

in

Determine the magnitude of the velocity at point D using equation 6.9b. VD  e ω

7.

VB  8.354


6.


VE  12.244

in sec

Use equation 6.9c to determine the angular velocity of link 6. ω 

VE g

ω  15.88

rad sec

CW




Use a compass and straightedge to draw the linkage in Figure 3-35 with link 2 at 90 deg and find the angular velocity of link 6 assuming 2 = 10 rad/sec CCW. Use an analytical method.

Given:


a  1.000  in

Link 3 (A to B)

b  3.800  in

Link 4 (O4 to B)

c  1.286  in

Link 1 (O2 to O4)

d  3.857  in

Link 4 (O4 to D)

e  1.429  in

Link 5 (D to E)

f  1.286  in

Link 6 (O6 to E)

g  0.771  in

Link 1 (O4 to O6)

h  0.786  in

θ2  90 deg

Crank angle:

ω2  10 rad sec


1

CCW

See Figure 3-35 and Mathcad file P0676c.


Y A

D

122.085° 33.359°

2

100.938° 5

3 4 O2

X

6 O4 O6 E

35.228°

B 2.


d

K2 

a 2

K3 

2

2

a b c d

d

K1  3.8570

c

K2  2.9992

2

K3  1.2015

2 a c


B  2.0000






θ41  2  atan2 2  A B  4.

C  5.0585

2

B  4 A  C



θ41  237.915 deg


d b

2

K5 

2

2

c d a b 2 a b

D  cos θ2  K1  K4 cos θ2  K5

2

K4  1.0150 D  7.6284

K5  3.7714


5.


E  2  sin θ2

E  2.0000

F  K1   K4  1   cos θ2  K5

F  0.0856






θ3  2  atan2 2  D E  6.

7.

2

E  4  D F



θ3  326.641 deg


ω3 

a  ω2 sin θ41  θ2  b sin θ3  θ41

ω3  1.398

ω4 

a  ω2 sin θ2  θ3  c sin θ41  θ3

ω4  6.496

sec rad sec

Link 4 will be the input to the second fourbar (links 4, 5, 6, and 1). Let the angle that link 4 makes in the second fourbar be

θ42  θ41  157  deg  360  deg 8.

rad

θ42  34.915 deg


h

K2 

e 2

K3 

e f

2

2

g h

h

K1  0.5500

g

K2  1.0195

2

K3  0.7263

2  e g

A  cos θ42  K1  K2 cos θ42  K3 B  2  sin θ42

C  K1   K2  1   cos θ42  K3 A  0.1603 9.

B  1.1447

C  0.3796






θ6  2  atan2 2  A B 

2

B  4 A  C



θ6  35.228 deg

10. Determine the values of the constants needed for finding 5 from equations 4.11b and 4.12. K4 

2

h

K5 

f

2

2

g h e f 2  e f

2

K4  0.6112

D  cos θ42  K1  K4 cos θ42  K5

D  0.2408

E  2  sin θ42

E  1.1447

F  K1   K4  1   cos θ42  K5

F  0.7807

K5  1.0119

11. Use equation 4.13 to find values of 5 for the open circuit.





θ5  2  atan2 2  D E 

2

E  4  D F



θ5  280.938 deg

12. Determine the angular velocity of link6 for the open circuit using equations 6.18.

 e ω4  sin θ42  θ5   g  sin θ6  θ5

ω6  

ω6  15.885

rad sec




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular velocity of link 6 in the sixbar linkage of Figure 3-35 as a function of 2 for a constant 2 = 1 rad/sec CCW.

Given:


a  1.000  in

Link 3 (A to B)

b  3.800  in

Link 4 (O4 to B)

c  1.286  in

Link 1 (O2 to O4)

d  3.857  in

Link 4 (O4 to D)

e  1.429  in

Link 5 (D to E)

f  1.286  in

Link 6 (O6 to E)

g  0.771  in

Link 1 (O4 to O6)

h  0.786  in

  1  rad sec


1

CCW

See Figure 3-35 and Mathcad file P0677.   0  deg 1  deg  360  deg

1.

Define the range of the input angle:

2.


d a 2

K3 

K2 

d

2

2

2

a b c d

K1  3.8570

c

K3  1.2015

2 a c

 

K2  2.9992

 

 

 

A   cos   K1  K2 cos   K3

 

 

B   2  sin 

 

C   K1   K2  1   cos   K3 3.




 



 

 

   2   atan2 2  A  B   4.

 2  4 A  C 

B 


2

d

K5 

b

 

2

2

c d a b

2

K4  1.0150

2 a b

 

 

 

D   cos   K1  K4 cos   K5

 

K5  3.7714

 

E   2  sin 

 

F   K1   K4  1   cos   K5 5.


 





 

 

   2   atan2 2  D  E   6.


 

  

 

   7.

 2  4 D F  

E 

    b sin       a   sin       c sin       a  



sin    



 


 

      157  deg  360  deg 8.

Determine the values of the constants needed for finding 6 from equations 4.8a and 4.10a. h

K1 

K2 

e 2

2

e f

K3 

2

g h

h

K1  0.5500

g 2

K3  0.7263

2  e g

 

K2  1.0195

    K1  K2 cos  K3 B'  2 sin

A'   cos  

 

     K3

C'   K1   K2  1   cos   9.




 



 2  4 A'  C' 

   

   2   atan2 2  A'  B'  

B' 

10. Determine the values of the constants needed for finding 5 from equations 4.11b and 4.12. 2

h

K4 

K5 

f

 

2

2

g h e f

2

K4  0.6112

2  e f

K5  1.0119

    K1  K4 cos  K5 E'  2 sin

D'   cos  

 

     K5

F'   K1   K4  1   cos  




 



 

 

   2   atan2 2  D'  E'  

 2  4 D' F' 

E' 

12. Determine the angular velocity of link6 for the open circuit using equations 6.18.

 e    sin          g  sin      

 

  

ANGULAR VELOCITY - LINK 6

Angular velocity, rad/sec

3 2 1 0 1 2

0

45

90

135

180

225

Crank angle, deg

270

315

360




Figure 3-36 shows an eightbar mechanism. Find all of its instant centers in the position shown in part (a) of the figure:

Given:


Solution:


1.


1.

n ( n  1)

C  28

2

Draw the linkage to scale and identify those ICs that can be found by inspection (11). 1,4; 1,8; 4,8 3,4

4,5

3 4 1,2 2

O4

O2

8

5,6

6

1,6

5

O6

2,3 7

7,8

5,7

2.


1,4; 1,8; 4,8; 5,8; 1,5

1,7


3,4; 1,3

4,5; 3,5

3

I2,8: I2,4-I4,8 and I1,2-I1,8

4 2

O2

1,2; 2,4

6

1,6

O4

8

O6 7,8

2,6; 2,8; 6,8; 2,7 at infinity

1 8

I2,6: I1,2-I1,6 and I2,4-I4,6

3,6; 3,7; 3,8 6,7

7

3

I2,5: I2,6-I5,6 and I2,3-I4,5

6

4 5

I3,5: I3,4-I4,5 and I2,3-I2,5 I3,6: I2,3-I2,6 and I3,5-I5,6 I3,7: I3,6-I6,7 and I2,3-I2,7

I5,8: I5,6-I6,8 and I4,5-I4,8

I3,8: I3,7-I7,8 and I2,3-I2,8

I1,5: I1,6-I5,6 and I1,4-I4,5

I4,7: I4,6-I6,7 and I3,4-I3,7

I1,7: I1,8-I7,8 and I1,6-I6,7

7 4,7

2

I6,8: I2,8-I2,6 and I1,8-I1,6

I2,7: I2,8-I7,8 and I2,6-I6,7

5

2,5

2,3

I6,7: I5,6-I5,7 and I6,8-I7,8

5,6; 4,6

5,7




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular velocity of link 8 in the linkage of Figure 3-36 as a function of 2 for a constant 2 = 1 rad/sec CCW.

Given:


a  0.450

First coupler (L3)

b  0.990

Common rocker (O4B)

c  0.590


d  1.000

Common rocker (O4C)

a'  0.590

Second coupler (CD)

b'  0.325

Output rocker (L6)

c'  0.325


Link 7 (L7)

e  0.938

Link 8 (L8)

f  0.572


p  0.823

Angle DCE

δ  7.0 deg

Angle BO4C

α  128.6  deg


  1  rad sec

1

CCW


See problem 4-43 for the position solution. The velocity solution will use the same vector loop equations for links 5, 6, 7, and 8, differentiated with respect to time. This problem is suitable for a project assignment and is probably too long for an overnight assignment.




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the magnitude and direction of the velocity of point P in Figure 3-37a as a function of 2 for a constant 2 = 1 rad/sec CCW. Also calculate and plot the velocity of point P versus point A.

Given:

Link lengths:

Solution: 1.

Link 2 (O2 to A)

a  0.136

Link 3 (A to B)

b  1.000

Link 4 (B to O4)

c  1.000

Link 1 (O2 to O4)

d  1.414

Coupler point:

Rpa  2.000

  0  deg

Crank speed:

  1  rad sec

1



2.


d

K2 

a

K1  10.3971 2

K3 

c

K2  1.4140

2

2

a b c d

2

K3  7.4187

2 a c

 

d

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

C θ  K1   K2  1   cos θ  K3 3.


 





 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  4.

B θ


2

d

K5 

b

2

2

c d a b

2

2 a b

K4  1.4140

K5  7.4187


D θ  cos θ  K1  K4 cos θ  K5

5.


 





 

 

θ θ  2   atan2 2  D θ E θ  6.

 2  4 Dθ F θ 

E θ



 

a  

 

a  

ω θ 

ω θ  7.

b

c




    sin θ θ  θ θ 



  sin θ θ  θ θ 






 



 

 

 

VA θ  a   sin θ  j  cos θ 8.

 



 

 

  



  



VPA θ  Rpa ω θ  sin θ θ    j  cos θ θ  

 

 

 


Plot the magnitude and direction of the velocity at coupler point P.

 

 

Magnitude:


Direction:

θVP1 θ  arg VP θ  θVP θ  if  θVP1 θ  0 θVP1 θ  2  π θVP1 θ  MAGNITUDE

0.18

0.16 Velocity, mm/sec

9.

0.14

0.12

0.1

0

45

90

135

180

Crank Angle, deg

225

270

315

360



DIRECTION

Vector Angle, deg

300

200

100

0

0

45

90

135

180

Crank Angle, deg

225

270

315

360




Calculate the percent error of the deviation from constant velocity magnitude of point P in Figure 3-37a.

Given:

Link lengths:

Solution: 1.

Link 2 (O2 to A)

a  0.136

Link 3 (A to B)

b  1.000

Link 4 (B to O4)

c  1.000

Link 1 (O2 to O4)

d  1.414

Coupler point:

Rpa  2.000

  0  deg

Crank speed:

  1  rad sec

1



2.


d

K2 

a

K1  10.3971 2

K3 

c

K2  1.4140

2

2

a b c d

2

K3  7.4187

2 a c

 

d

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

C θ  K1   K2  1   cos θ  K3 3.


 





 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  4.

B θ


2

d

K5 

b

2

2

c d a b

2

2 a b

K4  1.4140

K5  7.4187


D θ  cos θ  K1  K4 cos θ  K5

5.


 





 

 

θ θ  2   atan2 2  D θ E θ  6.

 2  4 Dθ F θ 

E θ


 

ω θ 

a   b



    sin θ θ  θ θ  sin θ θ  θ


 

ω θ  7.

a   c






  sin θ θ  θ θ  sin θ  θ θ


 



 

 

 

VA θ  a   sin θ  j  cos θ 8.

 



 

 

  



  



VPA θ  Rpa ω θ  sin θ θ    j  cos θ θ  

 

 

 


Calculate the magnitude of the velocity at coupler point P.

 

 


Magnitude:

 

err θ 

Deviation from constant speed:

 

 

VP θ  VA θ

 

VA θ

DEVIATION FROM CONSTANT SPEED 30

20 Percent Error

9.

10

0

 10

 20

0

45

90

135

180

Crank Angle, deg

225

270

315

360




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the magnitude and the direction of the velocity of point P in Figure 3-37b as a function of 2. Also calculate and plot the velocity of point P versus point A.

Given:


a  0.50

First coupler (AB)

b  1.00

Rocker 4 (O4B)

c  1.00

Rocker 5 (L5)

c'  1.00

Ground link (O2O4)

d  0.75


b'  1.00

Coupler point (DP)

p  1.00


d'  1.50

  1 

Crank speed: Solution: 1.

rad sec



 

 

 


 


2.

Define one revolution of the input crank: θ  0  deg 0.5 deg  360  deg

3.


d a

K1  1.5000

 

2

d

K2 

K3 

c

K2  0.7500

 

2

2

a b c d

2

2 a c

K3  0.8125

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 



 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ

 

 2  4 A θ Cθ   2 π

 

θ θ  2   atan2 2  A θ B θ  4.

B θ


 

2

d

K5 

b

2

2

c d a b

2

2 a b

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

K4  0.7500

K5  0.8125

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5 5.


 





 

 

θ θ  2   atan2 2  D θ E θ  6.

 2  4 Dθ F θ   2 π

E θ




 

    c cosθθ  d'

xP θ  d  b  cos θ θ

 

    c sinθθ

yP θ  b  sin θ θ 7.

Use equations 6.18 to calculate 3 and 4.

 

a  

 

a  

 θ 

b

 θ 

    sin θ θ  θ θ 



  sin θ θ  θ θ 





Differentiate the position equations with respect to time to get the velocity components.

 

      c θ sinθθ

VPx θ  b   θ sin θ θ

 

 

    c θ cosθθ

VPy θ  b   θ  cos θ θ

 

 2  VPyθ2

VP θ 

VPx θ

MAGNITUDE 0.6 0.4 Velocity, mm/sec

8.

c



0.2 0  0.2  0.4  0.6

0

45

90

135

180

Crank Angle, deg x Component y Component

225

270

315

360




Find all instant centers of the linkage in Figure P6-30 in the position shown.

Given:


Solution:


1.


1.

n ( n  1)

C6

2

Draw the linkage to scale and identify those ICs that can be found by inspection (4). 3,4

4

P1

1,4

O4

3

P2

Y

2,3 2 1,2

X O2

1,3

2.


I2,4: I1,2-I1,4 and I2,3-I3,4

3,4

4

P1 O4 P2

1,4 3 Y

2,3 2 1,2

X O2 2,4




Find the angular velocities of links 3 and 4 and the linear velocity of points A, B and P1 in the XY coordinate system for the linkage in Figure P6-30 in the position shown. Assume that 2 = 45 deg in the XY coordinate system and 2 = 10 rad/sec. The coordinates of the point P1 on link 4 are (114.68, 33.19) with respect to the xy coordinate system. Use a graphical method.

Given:


a  14.00  in

Link 3 (A to B)

b  80.00  in

Link 4 (O4 to B)

c  51.26  in

Link 1 X-offset

d X  47.5 in

Link 1 Y-offset

d Y  76.00  in  12.00  in

Coupler point x-offset

p x  114.68 in

Coupler point y-offset p y  33.19  in

Crank angle:

θ  45 deg ω  10 rad sec


1

CCW

See Figure P6-30 and Mathcad file P0684a. Solution: 1. Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Direction of VB Direction of VBA B

x

4

P1 O4

29.063°

P2

3

Y

99.055° 45.000° A 2

X O2

Direction of VA

y

2.


3.

VA  140.000

in sec

θVA  45 deg  90 deg



0


100 in/sec

0.409" Y 1.206" 119.063° VBA VB

VA

135.000°

9.055°

4.

X



5.

VB  120.600

VBA  0.409  in kv

VBA  40.900

in sec

θVB  119.063  deg

in sec


ω 

VBA

ω  0.511

b VB

ω  2.353

c

rad sec

rad sec

Transform the xy coordinates of point P1 into the XY system using equations 4.0b. Coordinate transformation angle

7.

1

in

VB  1.206  in kv

ω 

6.

100  in sec

δ  atan2 d X d Y 

δ  126.582 deg

PX  p x  cos δ  p y  sin δ

PX  94.998 in

PY  p x  sin δ  p y  cos δ

PY  72.308 in

Calculate the distance from O4 to P1 and use equation 6.7 to calculate the velocity at P1. Distance from O4 to P1: VP1  e ω

e 

 PX  d X  2   PY  d Y  2 VP1  113.446

in sec

e  48.219 in




Find the angular velocities of links 3 and 4 and the linear velocity of points A, B and P1 in the XY coordinate system for the linkage in Figure P6-30 in the position shown. Assume that 2 = 45 deg in the XY coordinate system and 2 = 10 rad/sec. The coordinates of the point P1 on link 4 are (114.68, 33.19) with respect to the xy coordinate system. Use the method of instant centers.

Given:


a  14.00  in

Link 3 (A to B)

b  80.00  in

Link 4 (O4 to B)

c  51.26  in

Link 1 X-offset

d X  47.5 in

Link 1 Y-offset

d Y  76.00  in  12.00  in


p x  114.68 in


Crank angle:

θ  45 deg


ω  10 rad sec

1

CCW


Draw the linkage to scale in the position given, find instant center I1,3 and the distance from the pin joints to the instant center.

1,3

B

4

P1 O4 P2

3 Y

A 2

X O2

From the layout above: AI13  273.768  in 2.

BI13  235.874  in



3.

VA  140.000

θVA  θ  90 deg

θVA  135.0 deg

VA

ω  0.511

AI13

CW

sec

VB  120.622

in sec

Use equation 6.9c to determine the angular velocity of link 4. VB

ω  2.353

c

rad

CCW

sec

Transform the xy coordinates of point P1 into the XY system using equations 4.0b. Coordinate transformation angle

7.

rad

Determine the magnitude of the velocity at point B using equation 6.9b.

ω  6.

sec


VB  BI13 ω 5.

in

VA  a  ω

ω  4.


δ  atan2 d X d Y 

PX  p x  cos δ  p y  sin δ

PX  94.998 in

PY  p x  sin δ  p y  cos δ

PY  72.308 in

δ  126.582 deg

Calculate the distance from O4 to P1 and use equation 6.7 to calculate the velocity at P1. Distance from O4 to P1: VP1  e ω

e 

 PX  d X  2   PY  d Y  2

VP1  113.467

in sec

e  48.219 in




Find the angular velocities of links 3 and 4 and the linear velocity of points A, B and P1 in the XY coordinate system for the linkage in Figure P6-30 in the position shown. Assume that 2 = 45 deg in the XY coordinate system and 2 = 10 rad/sec. The coordinates of the point P1 on link 4 are (114.68, 33.19) with respect to the xy coordinate system. Use an analytical method.

Given:


a  14.00  in

Link 3 (A to B)

b  80.00  in

Link 4 (O4 to B)

c  51.26  in

Link 1 X-offset

d X  47.5 in

Link 1 Y-offset

d Y  76.00  in  12.00  in


p x  114.68 in


Crank angle:


δ  126.582  deg

Coordinate transformation angle: Input crank angular velocity Solution: 1.

XY coord system

ω2  10 rad sec

1

CCW


Draw the linkage to scale and label it. x

B

97.519°

4

P1 O4

27.528°

P2

3 81.582°

Y

126.582° 45.000° A 2

X O2

y

Transform the crank angle from global to local coordinate system:

θ2  θ2XY  δ

2.

2

d 

Distance O2O4:

θ2  81.582 deg dX  dY

2

d  79.701 in


d a 2

K3 

K2 

d

2

2

2

a b c d 2 a c

c

K1  5.6929

K2  1.5548

K3  1.9340

A  cos θ2  K1  K2 cos θ2  K3

B  2  sin θ2



C  K1   K2  1   cos θ2  K3 A  3.8401 3.

B  1.9785






2

θ4  2  atan2 2  A B  4.

C  7.2529

B  4 A  C



θ4  262.482 deg


2

d

K5 

b

2

2

c d a b

2

K4  0.9963

2 a b

D  cos θ2  K1  K4 cos θ2  K5

D  10.0081

E  2  sin θ2

E  1.9785

F  K1   K4  1   cos θ2  K5 5.

7.

F  1.0849






2

θ3  2  atan2 2  D E  6.

K5  4.6074

E  4  D F



θ3  332.475 deg


ω3 

a  ω2 sin θ4  θ2  b sin θ3  θ4

ω3  0.511

ω4 

a  ω2 sin θ2  θ3  c sin θ4  θ3

ω4  2.353

rad sec

rad sec

Determine the velocity of points A and B for the crossed circuit using equations 6.19. VA  a  ω2  sin θ2  j  cos θ2  VA  ( 138.492  20.495i )

θVAXY  arg VA  δ

in sec

VA  140.000

in sec

θVAXY  135.000 deg

VB  c ω4  sin θ4  j  cos θ4  VB  ( 119.588  15.781j )

in sec

θVBXY  arg VB  δ 8.

in sec

θVBXY  119.064 deg

Calculate the distance from O4 to P1 and the angle BO4P1. e 

Distance from O4 to P1:

 py    px  d 

δ4  180  deg  atan 9.

VB  120.624

 px  d 2  py 2

δ4  136.503 deg

Determine the velocity of point P1 using equations 6.35. VP1  e ω4  sin θ4  δ  j  cos θ4  δ  VP1  ( 55.122  99.181j )

θVPXY  arg VP1  δ

in sec

VP1  113.469

in sec

θVPXY  245.646 deg

e  48.219 in




Given:

Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the absolute velocity of point P1 in Figure P6-30 as a function of 2 for 2 = 10 rad/sec. The coordinates of the point P1 on link 4 are (114.68, 33.19) with respect to the xy coordinate system. Link lengths: Link 2 (O2 to A)

a  14.00  in

Link 3 (A to B)

b  80.00  in

Link 4 (O4 to B)

c  51.26  in

Link 1 X-offset

d X  47.5 in

Link 1 Y-offset

d Y  76.00  in  12.00  in


p x  114.68 in


Crank angle:

θ2XY  0  deg 1  deg  360  deg   126.582  deg

Coordinate transformation angle: Input crank angular velocity Solution: 1.

XY coord system

  10 rad sec

1

CCW


Draw the linkage to scale and label it. x

B

97.519°

4

P1 O4

27.528°

P2

3 81.582°

Y

126.582° 45.000° A 2

X O2

y

Transform the crank angle from global to local coordinate system:  θ2XY   θ2XY   d 

Distance O2O4: 2.

2

dX  dY

2

d  79.701 in


d a 2

K3 

K2 

d

2

2

2

a b c d

c

K2  1.5548

K3  1.9340

2 a c



K1  5.6929







A  θ2XY   cos  θ2XY   K1  K2 cos  θ2XY   K3

  C θ2XY   K1   K2  1   cos  θ2XY    K3 B θ2XY   2  sin  θ2XY 


3.







 θ2XY   2   atan2 2  A  θ2XY  B θ2XY   4.



B θ2XY   4  A  θ2XY   C θ2XY    2


2

d

K5 

b



2

2

c d a b

2

K4  0.9963

2 a b





K5  4.6074



D θ2XY   cos  θ2XY   K1  K4 cos  θ2XY   K5

  F  θ2XY   K1   K4  1   cos  θ2XY    K5 E θ2XY   2  sin  θ2XY  5.






 θ2XY   2   atan2 2  D θ2XY  E θ2XY   6.

 

a   sin  θ2XY    θ2XY   c sin  θ2XY    θ2XY 

 

Calculate the distance from O4 to P1 and the angle BO4P1. e 

Distance from O4 to P1:

 px  d 2  py 2

e  48.219 in

 py    px  d 

  180  deg  atan

  136.503 deg

Determine and plot the velocity of point P1 using equations 6.35.









VP1 θ2XY   e  θ2XY   sin  θ2XY     j  cos  θ2XY    VP1 θ2XY   VP1 θ2XY 



θVPXY  θ2XY   arg VP1 θ2XY    

P1 VELOCITY MAGNITUDE 200

150 Velocity - in/sec

8.

2

Determine the angular velocity of link 4 for the crossed circuit using equations 6.18.  θ2XY  

7.



E θ2XY   4  D θ2XY   F  θ2XY   

100

50

0

0

45

90

135

180

225

270

Crank Angle - Global XY System

315

360



P1 VELOCITY DIRECTION

Velocity angle - Global XY System

300

200

100

0

0

45

90

135

180

225

270

Crank Angle - Global XY System

315

360




Find all instant centers of the linkage in Figure P6-31 in the position shown.

Given:


Solution:


1.


1.

n ( n  1)

C6

2


Y y

O2 1,2

X 2 1 O4 1,4 3,4

4

P

2,3 x 3 2.


I2,4: I1,2-I1,4 and I2,3-I3,4

Y y

O2 1,2

X 2 1 O4 1,4

1,3

3,4

4

P

2,3 x 3 2,4




Find the angular velocities of links 3 and 4 and the linear velocity of point P in the XY coordinate system for the linkage in Figure P6-31 in the position shown. Assume that 2 = -94.121 deg in the XY coordinate system and 2 = 1 rad/sec. The position of the coupler point P on link 3 with respect to point A is: p = 15.00, 3 = 0 deg. Use a graphical method.

Given:


a  9.17 in

Link 3 (A to B)

b  12.97  in

Link 4 (O4 to B)

c  9.57 in

Link 1 X-offset

d X  2.79 in

Link 1 Y-offset

d Y  6.95 in

Coupler point data:

p  15.00  in

δ  0  deg

Crank angle:

θ  94.121 deg ω  1  rad sec


1

CW


Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Y y

O2

X Direction VBA 2 1

68.121°

Direction VPA

26.000° O4 B

4 Direction VA

P

A 72.224° 3 60.472°

Direction VB

x

2.

3.


VA  a  ω

VA  9.170

θVA  θ  90 deg

θVA  184.121 deg

sec




Y 0

25 in/sec VA X

1.798"

1.783" VBA

4.


5.

kv 

25 in sec

1

in in

VB  1.783  in kv

VB  44.575

VBA  1.798  in kv

VBA  44.950

ω 

VBA b VB

in

θVBA  274.103  deg

sec

c

ω  3.466

rad

ω  4.658

rad

sec

sec

Determine the magnitude and sense of the vector VPA using equation 6.7. VPA  p  ω

VPA  51.985

in sec

θVPA  θVBA 7.

θVB  262.352  deg

sec


6.

VB

θVPA  274.103 deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point P. The equation to be solved graphically is VP = VA + VPA a. b. c.

8.

Choose a convenient velocity scale and layout the known vector VA. From the tip of VA, layout the (now) known vector VPA. Complete the vector triangle by drawing VP from the tail of VA to the tip of the VPA vector. Y


0

25 in/sec VA


VP  2.059  in kv

θP  96.051 deg

kv 

25 in sec

1

X

in

VP  51.475

96.051°

in 2.059"

sec VPA

VP




Find the angular velocities of links 3 and 4 and the linear velocity of point P in the XY coordinate system for the linkage in Figure P6-31 in the position shown. Assume that 2 = -94.121 deg in the XY coordinate system and 2 = 1 rad/sec. The position of the coupler point P on link 3 with respect to point A is: p = 15.00, 3 = 0 deg. Use the method of instant centers.

Given:


a  9.174  in

Link 3 (A to B)

b  12.971 in

Link 4 (O4 to B)

c  9.573  in

Link 1 X-offset

d X  2.790  in

Link 1 Y-offset

d Y  6.948  in

Coupler point data:

p  15.00  in

δ  0  deg

Crank angle:

θ  94.121 deg ω  1  rad sec


1

CW


Draw the linkage to scale in the position given, find instant center I1,3 and the distance from the pin joints to the instant center.

Y y

O2

X 2 1 O4

1,3

B

4

P

A x 3 From the layout above: AI13  2.647  in 2.

θVA  184.121 deg


4.

PI13  14.825 in


θVA  θ  90 deg 3.

BI13  12.862 in

VA AI13

ω  3.466

rad

CW

sec

Determine the magnitude of the velocity at point B using equation 6.9b. in VB  BI13 ω VB  44.577 sec


5.


6.


VB c

ω  4.657

rad

CW

sec

Determine the magnitude of the velocity at point P using equation 6.9b. VP  PI13 ω

VP  51.381

in sec




Find the angular velocities of links 3 and 4 and the linear velocity of point P in the XY coordinate system for the linkage in Figure P6-31 in the position shown. Assume that 2 = -94.121 deg in the XY coordinate system and 2 = 1 rad/sec. The position of the coupler point P on link 3 with respect to point A is: p = 15.00, 3 = 0 deg. Use an analytical method.

Given:


a  9.174  in

Link 3 (A to B)

b  12.971 in

Link 4 (O4 to B)

c  9.573  in

Link 1 X-offset

d X  2.790  in

Link 1 Y-offset

d Y  6.948  in

Coupler point data:

p  15.00  in

δ  0  deg

Crank angle:

θ2XY  94.121 deg Global XY coord system δ  68.121 deg

Coordinate transformation angle:

ω2  1  rad sec


1

CW



Y y

O2

X 2 1

68.121°

26.000° O4 B

4

P

A 72.224° 3 60.472° Transform crank angle into local xy coordinate system:

θ2  θ2XY  δ

θ2  26.000 deg d 

Calculate distance O2O4: 2.

x

2

dX  dY

2

d  7.487 in


d

K2 

a 2

K3 

2

2

d c

a b c d

K1  0.8161

2

2 a c

K3  0.3622

A  cos θ2  K1  K2 cos θ2  K3 C  K1   K2  1   cos θ2  K3 A  0.2581

K2  0.7821

B  0.8767

C  0.4234

B  2  sin θ2


3.







2

θ4  2  atan2 2  A B  4.

B  4 A  C

  2 π

θ4  60.488 deg


2

d b

2

2

c d a b

K5 

2

K4  0.5772

2 a b

D  cos θ2  K1  K4 cos θ2  K5

D  0.3096

E  2  sin θ2

E  0.8767

F  K1   K4  1   cos θ2  K5 5.

7.

F  0.4749






2

θ3  2  atan2 2  D E  6.

K5  0.9111

E  4  D F

  2 π

θ3  72.236 deg

Determine the angular velocity of links 3 and 4 using equations 6.18.

ω3 

a  ω2 sin θ4  θ2  b sin θ3  θ4

ω3  3.467

rad

ω4 

a  ω2 sin θ2  θ3  c sin θ4  θ3

ω4  4.658

rad

sec

sec

Determine the velocity of points A and B using equations 6.19. VA  a  ω2  sin θ2  j  cos θ2  VA  ( 4.022  8.246i)

in

VA  9.174

sec

θVAXY  arg VA  δ

in sec

θVAXY  184.121 deg

VB  c ω4  sin θ4  j  cos θ4  VB  ( 38.807  21.967i )

in

VB  44.593

sec

θVBXY  arg VB  δ 8.

in sec

θVBXY  97.633 deg

Determine the velocity of the coupler point P for the crossed circuit using equations 6.36.









VPA  p  ω3 sin θ3  δ  j  cos θ3  δ VPA  ( 49.529  15.867i )



in s

VP  VA  VPA

VP  ( 45.507  24.113i )

θVPXY  arg VP  δ

θVPXY  96.039 deg

in sec

VP  51.501

in sec




Figure P6-32 shows a fourbar double slider known as an elliptical trammel. F ind all its instant centers in the position shown.

Given:


Solution:


1.


1.

n ( n  1)

C6

2

Draw the linkage to scale and identify those ICs that can be found by inspection (4). Y

3,4 4

2,3

To 1,4 at infinity

3 2

1

X

To 1,2 at infinity

2.


I2,4: I1,2-I1,4 and I2,3-I3,4

Y

2,3

3,4 4 1,3

To 1,4 at infinity

3 2

1

To 2,4 at infinity

X

To 1,2 at infinity




The elliptical trammel in Figure P6-32 must be driven by rotating link 3 in a full circle. Points on line AB describe ellipses. Find and draw (mannually or with a computer) the fixed and moving centrodes of instant center I13. (Hint: These are called the Cardan circles.)

Solution:


1.

The instant center I13 is shown as the point P in the diagram below. The length of link 3, c, is constant and it is a diagonal of the rectangle OBPA. Therefore, the other diagonal, OP, has a constant length, c, regardless of the current lengths a and b. Thus, the point P (I13) travels on a circle of radius c. This circle is the fixed centrode.

c 4 B

P 3

b O

2 A

1

a

2.

Now, invert the mechanism by holding link 3 fixed and allowing the slots to move, which can only be in a circle about O. The locus of points P will then be a circle of radius 0.5c with center at O'. This is the moving centrode.

c 4 B

P 3 O'

b O

2 A

1

a




Derive analytical expressions for the velocities of points A and B in Figure P6-32 as a function of 3, 3 and the length AB of link 3. Use a vector loop equation.

Solution:


1.

Establish the global XY system such that the coordinates of the intersection of the slot centerlines is at (d X,d Y ). Then, define position vectors R1X, R1Y , R2, R3, and R4 as shown below. Y 4 3 R3 R2

B

3

C 2

R4

A 1

R 1Y

X R 1X

2.


 π  π j   j  θ 3 2 j ( 0) j ( 0) 2 dY  e    a e  c e  dX  e  b e   = 0 j

3.

Differentiate this equation with respect to time. j   d a  j  c  e dt

4.

 j  π   2  d    b e    = 0 dt

Substituting the Euler identity into this equation gives: Va  jc    cos θ3  j  sin θ3   Vb j = 0

5.

Separate this equation into its real (x component) and imaginary (y component) parts, setting each equal to zero. Va  c  sin θ3 = 0

6.

c  cos θ3  Vb = 0

Solve for the two unknowns Va and Vb in terms of the independent variables 3 and 3 Va = c  sin θ3

Vb = c  cos θ3




The linkage in Figure P6-33a has link 2 at 120 deg in the global XY coordinate system. Find 6

Given:

and VD in the global XY coordinate system for the position shown if 2 = 10 rad/sec CCW. Use the velocity difference graphical method. Link lengths: Link 2 (O2 to A)

a  6.20 in

Link 3 (A to B)

b  4.50 in

Link 4 (O4 to B)

c  3.00 in

Link 5 (C to D)

e  5.60 in

Link 3 (A to C)

p  2.25 in

Link 4 (O4 to D)

f  3.382  in

Link 1 X-offset

d X  7.80 in

Link 1 Y-offset

d Y  0.62 in

Angle ACB

δ3  0.0 deg

Angle BO4D

δ4  110.0  deg

Input rocker angle:

θ  120  deg

Global XY system

ω  10 rad sec


1

CW


Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Direction of VA Direction of VCA Axis of slip

A Y

87.972° 113.057° 5

D

C

110°

2

3

120°

Axis of transmission Direction of VDC

411.709° B O4

O2 Direction of VB

2.

Direction of VBA


3.

X

VA  62.000

in sec

θVA  120  deg  90 deg




Y 0

50 in/sec

X VA

1.206 VB VBA 78.291°

1.433 4.


5.

8.

1

in in

VB  60.300

VBA  1.433  in kv

VBA  71.650

θVB  78.291 deg

sec in

θVBA  23.057 deg

sec


ω 

7.

50 in sec

VB  1.206  in kv

ω 

6.

kv 

VBA b VB c

ω  15.922

rad

CCW

sec

ω  20.100

rad

CW

sec

Determine the magnitude and sense of the vector VCA using equation 6.7. in

VCA  p  ω

VCA  35.825

θVCA  θVBA  δ3

θVCA  23.057 deg

sec

Determine the magnitude and sense of the vector Vtrans using equation 6.7. in

Vtrans  f  ω

Vtrans  67.978

θtrans  θVB  δ4

θtrans  31.709 deg

sec

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VA + VCA a. b. c.

Choose a convenient velocity scale and layout the known vector VA. From the tip of VA, layout the (now) known vector VCA. Complete the vector triangle by drawing VC from the tail of VA to the tip of the VCA vector.



Y 0

50 in/sec

X VA VC

0.991

VCA 114.720° 9.



VC  0.991  in kv 9.

kv 

50 in sec

1

in

VC  49.550

in

θVC  114.720  deg

sec

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point D, the magnitude of the relative velocity VDC. The equations to be solved (simultaneously) graphically are VD = VC + VDC

and

VD = Vtrans + Vslip

a. Choose a convenient velocity scale and layout the known vector VC. b. From the tip of VC, draw a construction line with the direction of VDC, magnitude unknown. c. Repeat steps a and b with Vtrans and Vslip. c. From the tail of VC, draw a construction line to the intersection of the two construction lines. d. Complete the vector polygon by drawing VDC from the tip of VC to the intersection of the VD construction line and drawing VD from the tail of VC to the intersection of the VDC construction line. 10. From the velocity polygon we have:

Y VD Vslip

VDC


50 in sec

3.045

95.189°

1

in

Vtrans VD  3.045  in kv

VD  152.250

in

X

sec

θVD  95.189 deg

0 VC

  ω

  20.100

rad sec

50 in/sec





Given:

and VD in the global XY coordinate system for the position shown if 2 = 10 rad/sec CCW. Use the instant center graphical method. Link lengths: Link 2 (O2 to A)

a  6.20 in

Link 3 (A to B)

b  4.50 in

Link 4 (O4 to B)

c  3.00 in

Link 5 (C to D)

e  5.60 in

Link 3 (A to C)

p  2.25 in

Link 4 (O4 to D)

f  3.382  in

Link 1 X-offset

d X  7.80 in

Link 1 Y-offset

d Y  0.62 in

Angle ACB

δ3  0.0 deg

Angle BO4D

δ4  110.0  deg

Input rocker angle:

θ  120  deg

Global XY system

ω  10 rad sec


1

CW


Draw the linkage to scale in the position given, find instant centers I1,3 and I1,5, and the distances from the pin joints to the instant centers. 2,3 A 6

5,6 5

D

Y

1,6

3

1,5

3,5 C

3,6 2 1,3

To 4,6 at infinity

B 3,4

4 O4

1,2 1,4

O2

X

From the layout above: AI13  3.893  in 2.

BI13  3.788  in

θVA  210.0 deg


4.

CI15  1.411  in DI15  4.333  in DI16  7.573  in


θVA  θ  90 deg 3.

CI13  3.113  in

VA AI13

ω  15.926

rad sec

Determine the magnitude of the velocity at point B using equation 6.9b. VB  BI13 ω

VB  60.328

in sec

CCW


5.

Use equation 6.9c to determine the angular velocity of links 4 and 6.  

VB c

w6   6.

rad

w6  20.109

rad

CW

sec CW

sec

VC  49.578

in sec


8.

  20.109

Determine the magnitude of the velocity at point C using equation 6.9b. VC  CI13 ω

7.


VC CI15

ω  35.137

rad

CW

sec

Determine the magnitude of the velocity at point D using equation 6.9b. VD  DI15 ω

VD  152.247

in sec

@ 95.185 deg





Given:

and VD in the global XY coordinate system for the position shown if 2 = 10 rad/sec CCW. Use an analytical method. Link lengths: Link 2 (O2 to A)

a  6.20 in

Link 3 (A to B)

b  4.50 in

Link 4 (O4 to B)

c  3.00 in

Link 5 (C to D)

e  5.60 in

Link 3 (A to C)

p  2.25 in

Link 4 (O4 to D)

f  3.382  in

Link 1 X-offset

d X  7.80 in

Link 1 Y-offset

d Y  0.62 in

Angle ACB

δ3  0.0 deg

Angle BO4D

δ4  110.0  deg

Input rocker angle:

θ2XY  120  deg

Global XY system

δ  175.455  deg


ω2  10 rad sec


1

CCW


Draw the linkage to scale and label it. A

6

Y

71.487° D

5

C

110°

120°

55.455° B 16.254°

4

x

2

3

O4

175.455°

O2 y

Calculate the distance O2O4:

2.

d 

2

dX  dY

2

d  7.825 in

Transform 2XY into the local coordinate θ2  θ2XY  δ θ2  55.455 deg system: Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1 

d a 2

K3 

K2 

d

2

2

2

a b c d

c

2 a c

K1  1.2620 K3  2.3767

A  cos θ2  K1  K2 cos θ2  K3 C  K1   K2  1   cos θ2  K3 A  0.2028 3.

B  1.6474

K2  2.6082

C  1.5927


B  2  sin θ2

X







θ41  2  atan2 2  A B  4.

2

B  4 A  C




2

d

K5 

b

2

2

c d a b

2

2 a b

D  cos θ2  K1  K4 cos θ2  K5

K5  1.9877

E  1.6474

F  K1   K4  1   cos θ2  K5

F  0.3067






θ3  2  atan2 2  D E  6.

K4  1.7388 D  1.6967

E  2  sin θ2 5.

θ41  163.746 deg

2

E  4  D F

  2 π

θ3  648.512 deg

At this point write vector loop equations for links 3, 4, 5, and 6 to solve for the position and velocty of link 6.




The linkage in Figure P6-33b has link 3 perpendicular to the X-axis and links 2 and 4 are parallel to each other. Find 3, VA, VB, and VP if 2 = 15 rad/sec CW. Use the velocity difference graphical method.

Given:


a  2.75 in

Link 3 (A to B)

b  3.26 in

Link 4 (O4 to B)

c  2.75 in

Link 1 (O4 to B)

d  4.43 in

Coupler point data:

p  1.63 in

δ  0  deg ω  15 rad sec


1

CW



Direction of VB Direction of VBA B

O4

O2 36.351° A

Direction of VA θ  36.352 deg 2.

3.


VA  a  ω

VA  41.250

θVA  θ  90 deg

θVA  126.352 deg

sec


4.

Without actually drawing the vector polygon, we see that VA and VB have the same angle with respect to the X axis. Therefore,


VB  VA 5.

ω 

7.

and

VBA  0 

in sec

θVB  θVA

θVBA  0  deg


6.


VBA b VB c

ω  0.000

rad sec

ω  15.000

rad sec

Determine the magnitude and sense of the vector VPA using equation 6.7. in

VPA  p  ω

VPA  0.000

θVPA  θVBA

θVPA  0.000 deg

sec

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point P. The equation to be solved graphically is VP = VA + VPA a. b. c.

8.

Choose a convenient velocity scale and layout the known vector VA. From the tip of VA, layout the (now) known vector VPA. Complete the vector triangle by drawing VP from the tail of VA to the tip of the VPA vector.

Again, without drawing the vector polygon we see that, VP  VA At this instant, link 3 is in pure translation with every point on it having the same velocity.




The linkage in Figure P6-33b has link 3 perpendicular to the X-axis and links 2 and 4 are parallel to each other. Find 3, VA, VB, and VP if 2 = 15 rad/sec CW. Use the instant center graphical method.

Given:


a  2.75 in

Link 3 (A to B)

b  3.26 in

Link 4 (O4 to B)

c  2.75 in

Link 1 (O4 to B)

d  4.43 in

Coupler point data:

p  1.63 in

δ  0  deg ω  15 rad sec


1

CW

See Figure P6-33b and Mathcad file P0695. Solution: 1. Draw the linkage to scale in the position given, find instant center I1,3 and the distance from the pin joints to the instant center.

Y

B

4 P

O2 2

O4

X

3 36.351° A

Since O2A and O4B are parallel, I1,3 is at infinity and link 3 does not rotate for this instantaneous position of the linkage. Thus, link 3 is in pure translation and all points on it will have the same velocity. 2.

Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at points A, B, and P, which will be the same. VA  a  ω

VA  41.250

in sec

  36.351 deg

θVA    90 deg

θVA  126.351 deg

The velocity vectors for points B and P are identical to VA.




The linkage in Figure P6-33b has link 3 perpendicular to the X-axis and links 2 and 4 are parallel to each other. Find 3, VA, VB, and VP if 2 = 15 rad/sec CW. Use an analytical method.

Given:


a  2.75 in

Link 3 (A to B)

b  3.26 in

Link 4 (O4 to B)

c  2.75 in

Link 1 (O4 to B)

d  4.43 in

Coupler point data:

p  1.63 in

δ  0  deg w2  15 rad sec


1

CW


Draw the linkage to a convenient scale and label it.

B

P

O2

O4 36.351°

A θ  36.351 deg 2.


d a 2

K3 

K2 

d

2

2

2

a b c d

K1  1.6109

c

K3  1.5949

2 a c

    B  2  sin θ C  K1   K2  1   cos θ  K3

A  cos θ  K1  K2 cos θ  K3

3.

A  0.5081 B  1.1855 C  1.1029






2

  2  atan2 2  A B  4.

B  4 A  C

  2 π

  143.649 deg


d b

2

K5 

2

2

c d a b

2

2 a b

    E  2  sin θ F  K1   K4  1   cos θ  K5

D  cos θ  K1  K4 cos θ  K5

5.

K2  1.6109


K4  1.3589 D  1.3983 E  1.1855 F  0.2127

K5  1.6873







2

  2  atan2 2  D E  6.

7.

E  4  D F

  2 π

  89.995 deg

Determine the angular velocity of links 3 and 4 using equations 6.18.  

a  w2 sin   θ  b sin   

 

 

  0.000

rad

 

a  w2 sin θ    c sin   

 

 

  15.000

rad

sec

sec

Determine the velocity of points A and B using equations 6.19.



 

 

VA  a  w2 sin θ  j  cos θ VA  ( 24.450  33.223i )

in

VA  41.250

sec

θVAXY  arg VA



in sec

θVAXY  126.351 deg

 

 

VB  c  sin   j  cos  VB  ( 24.450  33.223i )

in

VB  41.250

sec

θVBXY  arg VB 8.

in sec

θVBXY  126.351 deg










VPA  p   sin   δ  j  cos   δ



VPA  1.936  10 VP  VA  VPA

θVPXY  arg VP

4



in  sec

8

 1.676i  10

VP  41.250

in sec

θVPXY  126.351 deg




The crosshead linkage shown in Figure P6-33c has 2 DOF with inputs at crossheads 2 and 5. Find instant centers I1,3 and I1,4.

Given:


Solution:


1.


1.

n ( n  1)

C  10

2

Draw the linkage to scale and identify those ICs that can be found by inspection (4). 2,5 at infinity 2 3,4

1,2 at infinity

3 2,3 1

4 5

4,5 1,5 at infinity

2.

Use Kennedy's Rule and a linear graph to find the remaining ICs required to find I1,3 and I1,4. I2,4: I1,2-I1,4 and I2,3-I3,4

2,5 at infinity 3,4

I3,5: I2,3-I2,5 and I4,5-I3,4

2


1,3; 1,4

1,2 at infinity

3 2,3; 2,4 1

4 5

4,5; 3,5 1,5 at infinity




The crosshead linkage shown in Figure P6-33c has 2 DOF with inputs at cross heads 2 and 5. Find VB, VP3, and VP4 if the crossheads are each moving toward the origin of the XY coordinate system with a speed of 20 in/sec. Use a graphical method.

Given:


b  34.32  in

Link 4 (B to C)

c  50.4 in

Link 2 Y-offset

a  59.5 in

Link 5 X-offset

d  57 in

AP3  31.5 in

BP3  22.2 in

BP4  41.52  in

CP 4  27.0 in

Coupler point data:

V2Y  20 in sec

Input velocities: Solution: 1.

1

V5X  20 in sec

1



Y

Direction of VP3A Direction of VBA Direction of VBC

2 P3

A Direction of VA

3

Direction of VP4C B

P4

4

5 X C Direction of VC 2.

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relative velocities VBA, VBC and the angular velocities of links 3 and 4. The equations to be solved (simultaneously) graphically are VB = VA + VBA

VC = VB + VCB

a. Choose a convenient velocity scale and layout the known vectors VA and VC b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tip of VC, draw a construction line with the direction of VCB, magnitude unknown. d. From the origin, draw a construction line to the intersection of the two construction lines in steps b and c. e. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line. And then do the same with the VCB vector (See next page)



5.932 Y 0

VBC

10 in/sec

VB VBA

60.123° 32.718°

46.990°

X

VC

6.004

4.385

VA

3.

From the velocity triangles we have: Velocity scale factor:

4.

10 in sec

1

in in

VB  4.385  in kv

VB  43.850

VBA  6.004  in kv

VBA  60.040

VBC  5.932  in kv

VBC  59.320

θVB  46.990 deg

sec in

θVBA  60.123 deg

sec in

θVBC  32.718 deg

sec

Determine the angular velocity of links 3 and 4 using equation 6.7. ω  ω 

6.

kv 

VBA b VBC c

ω  1.749

rad

CCW

sec

ω  1.177

rad

CW

sec

Determine the magnitudes of the vectors VP3A and VP4C using equation 6.7. VP3A  AP3  ω

VP3A  55.107

in sec



VP4C  CP4 ω

7.

in

VP4C  47.234

sec

Use equation 6.5 to (graphically) determine the magnitude of the velocities at points P3 and P4. The equations be solved graphically are VP3 = VA + VP3A a. b. c. d.

VP4 = VC + VP4C

Choose a convenient velocity scale and layout the known vector VA. From the tip of VA, layout the (now) known vector VP3A. Complete the vector triangle by drawing VP3 from the tail of VA to the tip of the VP3A vector. Repeat for VP4.

V P3A

V P3 Y

0

3.537

10 in/sec

V P4 164.011°

V P4C

104.444°

X

VC 6.598

VA

8.

From the velocity triangles we have:


kv 

10 in sec

1

in

VP3  3.537  in kv

VP3  35.370

VP4  6.598  in kv

VP4  65.980

in sec in sec

θP3  104.444  deg θP4  164.011  deg




The linkage in Figure P6-33d has the path of slider 6 perpendicular to the global X-axis and link 2 aligned with the global X-axis. Find VA in the position shown if the velocity of the slider is 20 in/sec downward. Use the velocity difference graphical method.

Given:


a  12 in

Link 3 (A to B)

b  24 in

Link 4 (O4 to B)

c  18 in

Link 5 (C to D)

f  24 in

Link 4 (O4 to C)

e  18 in

Link 1 X-offset

d X  19 in

Link 1 Y-offset

d Y  28 in

δ4  0.0 deg

Angle BO4C

Solution: 1.

Output crank angle:

θ  0  deg


VD  20 in sec

Global XY system 1

θVD  90.0 deg


Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest. Direction of VC

Direction of VCD C

Direction of VB O4 19.963°

4

116.161°

B 5

Y

D 3

O2

2

114.410°

A

6 X

Direction of VD

Direction of VAB Direction of VA

2.

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solved graphically is VC = VD + VCD a. Choose a convenient velocity scale and layout the known vector VD. b. From the tip of VD, draw a construction line with the direction of VCD, magnitude unknown. c. From the tail of VD, draw a construction line with the direction of VC, magnitude unknown. d. Complete the vector triangle by drawing VCD from the tip of VD to the intersection of the VC construction line and drawing VC from the tail of VD to the intersection of the VCD construction line.



Y

X 1.806 0

10 in/sec 70.037°

VC VD V CD

3.


VC  1.806  in kv 4.

kv 

10 in sec

1

in

VC  18.060

in

θVC  70.037 deg

sec

Determine the magnitude and sense of the vector VB . VB  VC

VB  18.060

in sec

θVB  θVC  180  deg 5.

θVB  109.963 deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point A. The equation to be solved graphically is VA = VB + VAB a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VAB, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VA, magnitude unknown. d. Complete the vector triangle by drawing VAB from the tip of VB to the intersection of the VA construction line and drawing VA from the tail of VB to the intersection of the VAB construction line.

6.


Y


10 in sec

1

VA VB

in

VA  1.977  in kv VA  19.770

V AB

1.977

0

in sec

θVA  90.0 deg

X

10 in/sec




The linkage in Figure P6-33a has the path of slider 6 perpendicular to the global X-axis and link 2 aligned with the global X-axis. Find VA in the position shown if the velocity of the slider is 20 in/sec downward. Use the instant center graphical method.

Given:

Link lengths:

Solution: 1.

Link 2 (O2 to A)

a  12 in

Link 3 (A to B)

b  24 in

Link 4 (O4 to B)

c  18 in

Link 5 (C to D)

f  24 in

Link 3 (A to C)

p  2.25 in

Link 4 (O4 to C)

e  18 in

Link 1 X-offset

d X  19 in

Link 1 Y-offset

d Y  28 in

Angle BO4C

δ4  0.0 deg

Output crank angle:

θ  0  deg




θVD  90.0 deg


Draw the linkage to scale in the position given, find instant centers I1,3 and I1,5, and the distances from the pin joints to the instant centers.

C

O4

4

5

B 1,5

D 6

3

From the layout above:

1,3

AI13  70.085 in 2.

VD DI15

VC e

CW

sec

VC  18.057

ω  1.003

in sec

rad

CW

sec

VB  18.057


7.

rad

Determine the magnitude of the velocity at point B using equation 6.9b. VB  c ω

6.

ω  0.286


5.

CI15  63.095 in

Determine the magnitude of the velocity at point C using equation 6.9b. VC  CI15 ω

4.

BI13  64.013 in

2


3.

O2

VB BI13

ω  0.282

in sec

rad

CCW

sec

Determine the magnitude of the velocity at point A using equation 6.9b. in Upward VA  AI13  ω VA  19.769 sec

A

DI15  69.886 in




For the linkage of Figure P6-33e, write a computer program or use an equation solver to find and plot VD in the global coordinate system for one revolution of link 2 if 2 = 10 rad/sec CW.

Given:

Link lengths: a  5.00

Input crank (L2)

Output crank (O4B) c  6.00 d  2.500


  10

Crank velocity: Solution:


b  5.00

Slider coupler (L5)

e  15.00

Angle BO4C

  83.621 deg

rad sec


1.

This sixbar drag-link mechanism can be analyzed as a fourbar Grashof double crank in series with a crankslider mechanism using the output of the fourbar, link 4, as the input to the crank-slider.

2.


3.

Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit) in the global XY coordinate system. K1 

d

K2 

a

K1  0.5000

 

2

d

K3 

c

K2  0.4167

 

2

2

a b c d

2

2 a c

K3  0.7042

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  4.

B θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 5.


 

 c sin θ θ    π e  

 

  




  

f θ  c cos θ θ    e cos θ θ 5.


2

d

K5 

b

K4  0.5000

 

2

2

c d a b

2

2 a b

K5  0.4050

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5


6.





 



 

 2  4 Dθ F θ 

 

θ θ  2   atan2 2  D θ E θ  7.


 

 θ  8.

a   c





  sin θ θ  θ θ  sin θ  θ θ


 

ω θ 

        

c cos θ θ      θ e cos θ θ


 

    



    

VD θ  c  θ  sin θ θ    e ω θ  sin θ θ

VELOCITY OF PIN D 50

25

0 Velocity

9.

E θ

 25  50  75  100

0

45

90

135

180

Crank angle, deg

225

270

315

360




For the linkage of Figure P6-33f, locate and identify all instant centers.

Given:


Solution:


1.


1.

n ( n  1)

C  28

2

To 1,8 at infinity 7,8

8


7 5,6 6

2.


1,6 O6

5,7

I1,3: I1,2-I2,3 and I1,4-I3,4

5

I1,5: I1,6-I5,6 and I1,3-I3,5 2,3


3,4

I2,5: I1,2-I1,5 and I2,4-I4,5

4

I2,6: I1,2-I1,6 and I2,5-I5,6

3

2

3,5

1,4

O4

1,2

O2

I3,6: I2,3-I2,6 and I1,2-I1,6 5,8


To 1,8 at infinity


1,5

8

7,8

I6,8: I5,8-I5,6 and I2,8-I2,6

6,7 1,7

I2,7: I2,8-I7,8 and I2,5-I5,7

7 5,6

1,3; 3,8

I6,7: I5,6-I5,7 and I6,8-I7,8

6

I3,7: I3,6-I6,7 and I2,3-I2,7

1,6 O6

5,7

I4,6: I1,4-I1,6 and I4,5-I5,6 5

I4,7: I4,6-I6,7 and I3,4-I3,7

3,7

6,8 2,7

I1,7: I1,8-I7,8 and I1,6-I6,7

2,3 4,5 4,7

3 1,4 O4

2,4

3,5

2,5

2

4 3,4 O2

1,2

2,8 4,8

To 2,6 and 3,6 4,6




The linkage of Figure P6-33f has link 2 at 130 deg in the global XY coordinate system. Find VD in the global coordinate system for the position shown if 2 = 15 rad/sec CW. Use any graphical method.

Given:


a  5.0 in

Link 3 (A to B)

b  8.4 in

Link 4 (O4 to B)

c  2.5 in

Link 1 (O2 to O4)

d 1  12.5 in

Link 5 (C to E)

e  8.9 in

Link 5 (C to D)

h  5.9 in

Link 6 (O6 to E)

f  3.2 in

Link 7 (D to F)

k  6.4 in

Link 3 (A to C)

g  2.4 in

Link 1 (O2 to O6)

d 2  10.5 in

θ2  130  deg

Crank angle:

Slider axis offset and angle:

s  11.7 in

ω  15 rad sec


  150  deg 1

CW



Direction of VF 8

Y

F

Direction of VFD

Direction of VE

7 E

Direction of VEC

6 Direction of VDC

5

Direction of VCA Direction VBA

C

Direction of VB

O6

D

Direction of VA

A

3

2

B 4 O4

X

O2

Angles measured from layout:

2.

θVB  24.351 deg

θVBA  79.348 deg

θVCA  θVBA

θVE  64.594 deg

θVEC  162.461  deg

θVDC  θVEC

θVF  150.00 deg

θVFD  198.690  deg


VA  a  ω

VA  75.000

θVA  θ2  90 deg

θVA  40.000 deg

sec


3.


Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relative velocity VBA, and the angular velocity of link 3. The equations to be solved graphically are VB = VA + VBA

and

VC = VA + VCA

a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.

2.668

VA 0.943 V CA

0

VC

25 in/sec

3.302

Y 22.053° X

V BA VB 3.192

4.

From the velocity polygon we have: Velocity scale factor:

25 in sec

in

VB  79.800

VBA  3.302  in kv

VBA  82.550

VBA b

1

in

VB  3.192  in kv

  5.

kv 

  9.827

θVB  24.351 deg

sec in

θVBA  79.348 deg

sec

rad

CCW

sec

Calculate the magnitude and direction of VCA and determine the magnitude and velocity of VC from the velocity polygon above. VCA  g  

VCA  23.586

Length of VCA on velocity polygon:vCA  VC  2.668  in kv

in sec

VCA

vCA  0.943 in

kv

VC  66.700

θVCA  79.348 deg

in sec

θVC  22.053deg


6.


Use equation 6.5 to (graphically) determine the magnitude of the velocity at point E, the magnitude of the relative velocity VED, and the angular velocity of link 5. The equations to be solved graphically are VE = VC + VEC

and

VD = VC + VDC

a. Choose a convenient velocity scale and layout the known vector VC. b. From the tip of VC, draw a construction line with the direction of VEC, magnitude unknown. c. From the tail of VC, draw a construction line with the direction of VE, magnitude unknown. d. Complete the vector triangle by drawing VEC from the tip of VC to the intersection of the VE construction line and drawing VE from the tail of VC to the intersection of the VEC construction line. 1.821

1.207 V EC VE

1.716 0

VD

V DC

25 in/sec VC Y

45.934°

X 1.900

7.

From the velocity polygon we have: Velocity scale factor:

25 in sec

VEC  1.821  in kv

VEC  45.525

e

  5.115

θVE  64.594 deg

sec in

θVEC  162.461 deg

sec

rad

CCW

sec

Calculate the magnitude and direction of VDE and determine the magnitude and velocity of VD from the velocity polygon above. VDC  h  

VDC  30.179

Length of VDC on velocity polygon: vDC  VD  1.900  in kv 9.

in

VE  42.900

VEC

1

in

VE  1.716  in kv

  8.

kv 

in sec

VDC

vDC  1.207 in

kv

VD  47.500

θVDC  162.461 deg

in sec

θVD  45.934deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point F, the magnitude of the relative velocity VFD, and the angular velocity of link 5. The equation to be solved graphically is VF = VD + VFD



a. Choose a convenient velocity scale and layout the known vector VD. b. From the tip of VD, draw a construction line with the direction of VFD, magnitude unknown. c. From the tail of VD, draw a construction line with the direction of VF, magnitude unknown. d. Complete the vector triangle by drawing VFD from the tip of VD to the intersection of the VF construction line and drawing VF from the tail of VD to the intersection of the VFD construction line. Y 0

VD

25 in/sec

V FD 150.000° VF

45.934°

X 1.158

10. From the velocity polygon we have: Velocity scale factor:

VF  1.158  in kv

kv 

25 in sec

1

in

VF  28.950

in sec

θVF  150.000 deg




For the linkage of Figure P6-34, locate and identify all instant centers.

Given:


Solution:


1.


2.

n ( n  1)

C  15

2

Draw the linkage to scale and identify those ICs that can be found by inspection (7). 4,5 C

3,4

2,3

B

5

A

3

5,6 3

D

2

4

6 E 3,6 O4

1,4

2.

1,2 O2

1


I3,5: I2,3-I2,5 and I3,4-I4,5

I4,6: I3,4-I3,6 and I4,5-I5,6

I1,3: I1,2-I2,3 and I1,4-I3,4

I2,6: I1,4-I4,6 and I2,3-I3,6

I1,5: I1,2-I2,5 and I1,3-I3,5

I2,5: I2,6-I5,6 and I2,4-I4,5

I1,6: I1,3-I3,6 and I1,4-I4,6

1 6

2

5

3 4

1,3

1,5 3,5

4,6

4,5

2,5

C

2,3

B 3,4

5

2,6

A

3

5,6 D

3

2

4

2,4

6 E 3,6 1,4

1,6

O4

1

1,2 O2




For the linkage of Figure P6-34, show that I1,6 is stationary for all positions of link 2.

Given:

Link lengths:

Solution: 1.

Input crank (L2)

a  1.75

First coupler (AB)

b  1.00

First rocker (O4B)

c  1.75

Ground link (O2O4)

d  1.00

Second input (BC)

e  1.00

Second coupler (L5)

f  1.75

Output rocker (L6)

g  1.00

Third coupler (BE)

h  1.75

Ternary link (AE)

k  2.60


Because the linkage is symmetrical and composed of two parallelograms the analysis can be done with simple trigonometry. We will show that the path of point E on link 6 is a circle and that the center of the circle is the instant center I1,6.

C e

f 5

 = 25.396°

3

h

B

RP k

R2

4

D g 6

A

b 3

2 a

c E

RE

x

O4

1 d

O2 y

2.

Calculate the fixed angle that line AB (b) makes with line AE (k) using the law of cosines. This is the angle that the coupler point, E, makes with link 3 in the fourbar made up of links 1, 2, 3, and 4. Because links 1, 2, 3, and 4 are a parallelogram, link 4 will have the same angle as link 2 and AB will always be parallel to O2O4.

 b 2  k2  h 2    2 b k 

δ  acos 3.

δ  25.396 deg

Define the vectors R2, RP , and RE as shown above. Then, RE = R2 + RP or,

 

  

   k cosδ  jsin δ

RE θ = a  cos θ  jsin θ

Separating into real and imaginary parts, the coordinates of point E for all positions of link 2 are:

 

 

 

 

xE θ  a  cos θ  k cos δ yE θ  a  sin θ  k sin δ



This pair of equations represent the parametric equation of a circle with radius a and center at xEC  k cos δ

xEC  2.349

yEC  k sin δ

yEC  1.115

4.


5.

Plot the position of point E as a function of the angle of input link 2.

PATH OF POINT E 4

2.349

3

2

 

yE θ

1.115

1

0

1

0

1

2

 

3

4

5

xE θ

6.

From Problem 6-104, the instant center I1,6 is located at the intersection of a line through O4 that is parallel to BE (angle ) with a line through E that is parallel to O2A (angle 2). (See figure on next page). The coordinates of the intersection of these two lines for the position shown ( θ  64.036 deg ) are:

 b 2  h 2  k2    2 b h 

Angle BAE:

α  π  acos

Coordinates of point E:

xE  a  cos θ  k cos δ

α  39.582 deg

 

xE  3.115

 

yE  0.458

yE  a  sin θ  k sin δ Line through E with angle 2:

y ( x)   x  xE  tan θ  yE

Coordinates of point O4:

xO4  d

 

yO4  0



Line through O4 with angle :

y ( x)   x  xO4  tan  α

Solving for the intersection (point F, the instant center I1,6) of these two lines: xF 

  tan  θ  tan α

xE tan θ  yE  d  tan α

xF  2.349

yF   xF  d   tan α

yF  1.115

These are the coordinates of the instant center I1,6 and the center of the circle through which point E passes. Thus, the instant center I1,6 is stationary and is a hinge point that link 6 rotates about with respect to link 1.

1,3

1,5 3,5

4,6

4,5

2,5

C

2,3

B 3,4

5 5,6

2,6

A

3

39.582° D

3

2

4

6

2,4

64.036°

E 3,6

1,4

O4

1

1,2 O2 1.115"

1.750" 1,6

F

2.349"




Figure P6-26 shows a mechanism with dimensions. Use a graphical method to determine the velocities of points A and B and the velocity of slip for the position shown. Ignore links 5 and 6. ω 2 = 24 rad/s clock-wise.

Given:


d  1.22 in


θ  56.5 deg

Link 2 (O2A)

a  1.35 in


θ  14 deg

Link 4 (O4B)

e  1.36 in


ω  24 rad sec

1

CW



B

Axis of slip Y


4

0.939

132.661° A 3 2 X O2

Direction of VA3

2.


3.

VA3  32.400

in sec

θVA3  θ  90 deg

θVA3  76.0 deg




Y 0

10 in/sec

1.295 X V trans

V A3 2.369

4.


5.

kv 

1

in in

Vslip  28.43 


Vtrans  15.54 

sec in sec


in sec

Determine the angular velocity of link 4 using equation 6.7. From the linkage layout above: ω 

7.

12 in sec

Vslip  2.369  in kv


V slip

VA4 c

c  0.939  in and

ω  16.55 

rad

θ  132.661  deg

CW

sec


θVA4  θ  90 deg

VB  22.51 

in sec

θVA4  42.66  deg




Figure P6-26 shows a mechanism with dimensions. Use an analytic method to calculate the velocities of points A and B and the velocity of slip for the position shown. Ignore links 5 and 6. ω 2 = 24 rad/s clock-wise.

Given:


d  1.22 in


θ  56.5 deg

Link 2 (O2A)

a  1.35 in


θ  14 deg

Link 4 (O4B)

e  1.36 in ω  24 rad sec


1

CW



B

Axis of slip Y


4

0.939

132.661° A 3 2 X O2

Direction of VA3

2.

Establish an x-y frame with origin at O2 and positive x-axis through O4. Using the given data, calculate b, θ2, θ3, and θ4 with respect to the x-y frame. From the vector loop equation for this geometry, θ  θ  θ

θ  42.500 deg

 a  sin θ    d  a cos θ 

θ  180  deg  atan

b  a  3.

  sin θ sin θ

θ  256.161  deg

b  0.939 in

Using Example 6-5, determine VA2, Vtrans, and Vslip. Velocity on link 2 at A: VA2  a  ω

at an angle of

θVA2  θ  sign ω  90 deg

θ  θ


VA2  32.400

in s


θVA2  132.500  deg

Angle between VA2 vector and link 4 axis:

4.

α  θVA2  θ  2  π

α  28.661 deg

Vslip  VA2 cos( α)

Vslip  28.430

Vtrans  VA2 sin( α)

Vtrans  15.540

in s in s

Calculate the angular velocity of link 4 and the velocity at point B. Angular velocity of link 4: Velocity at point B:

ω 

Vtrans b

VB  e ω

ω  16.544

rad

VB  22.500

in

s s




Figure P6-28 shows a quick-return mechanism with dimensions. Use a graphical method to determine the velocities of points A and B and the velocity of slip for the position shown. Ignore links 5 and 6. ω 2 = 16 rad/s.

Given:


d  1.69 in


Link 2 (L2)

a  1.00 in


Link 4 (L4)

e  4.76 in


ω  16 rad sec

1

θ  15.5 deg

CCW



B


Direction of VA3

4


2 44.228° O2

X O4

2.


3.

VA2  16.000

in sec

θVA2  θ  90 deg

θVA2  189.0  deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on link 3. The equation to be solved graphically is VA3 = Vtrans + Vslip a. b.

Choose a convenient velocity scale and layout the known vector VA3. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown.



c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the Vslip construction line. Y 0

5 in/sec

X

VA3 V trans

1.154

4.

kv 

1

in in

Vslip  8.17


Vtrans  5.77

sec in sec

The true velocity of point A on link 4 is Vtrans, VA4  5.77

in sec


7.

5  in sec

Vslip  1.634  in kv


1.634


5.

Vslip

VA4 c

c  2.068  in and

ω  2.790 

rad

θ  44.228 deg

CCW

sec


θVB  θ  90 deg

VB  13.28 

in sec

θVB  134.228  deg




Figure P6-28 shows a quick-return mechanism with dimensions. Use an analytic method to calculate the velocities of points A and B and the velocity of slip for the position shown. Ignore links 5 and 6. ω 2 = 16 rad/s.

Given:


d  1.69 in


Link 2 (L2)

a  1.00 in

Angle link 2 makes with X axis θ  99 deg

Link 4 (L4)

e  4.76 in


ω  16 rad sec

1

θ  15.5 deg

CCW



B

Axis of transmission Direction of VA3

4

Y

Axis of slip 99.00°

A 3 2.06"

2 X

O2 15.5 x

O4 y

2.

Establish an x-y frame with origin at O2 and positive x-axis through O4. Using the given data, calculate b, θ2, θ3, and θ4 with respect to the x-y frame. From the vector loop equation for this geometry,





θ  θ  θ  180  deg

 a  sin θ    d  a cos θ 

θ  180  deg  atan

b  a 

  sin θ sin θ

b  2.059 in

θ  96.500 deg θ  208.855  deg

θ  θ


3.


Using Example 6-5, determine VA2, Vtrans, and Vslip. Velocity on link 2 at A: VA2  a  ω VA2  16.000

at an angle of in s

θVA2  θ  sign ω  90 deg θVA2  6.500  deg

Angle between VA2 vector and link 4 axis:

α  θVA2  θ  2  π

α  144.645  deg

Vslip  VA2 cos( α)

Vslip  13.049

Vtrans  VA2 sin( α) 4.

Vtrans  9.258

in s

in s

Calculate the angular velocity of link 4 and the velocity at point B. Angular velocity of link 4: Velocity at point B:

ω 

Vtrans b

VB  e ω

ω  4.497 

rad

VB  21.405

in

s s




Given:

Solution: 1.

The general linkage configuration and terminology for an offset fourbar slider-crank linkage are shown in Fig P6-2. The link lengths and the values of d and d-dot are defined in Table P6-5. For row a, find the velocity of the pin joint A and the angular velocity of the crank using a graphical method. Link lengths: Link 2

a  1.4 in

Link 3

b  4  in

Offset

c  1  in

θ  176.041  deg

ddot  10 in sec

1



Direction of VBA 13.052° B

. d = VB

176.041° A

000 b = 4. a = 1.400

c = 1.000"

O2

d = 2.500"

0

10 inches/sec

Direction of VA 2.

Use equation 6.6.24b to (graphically) determine the magnitude of the velocity at point A. The equation to be solved graphically is

VA

VA = VB + VBA a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VA, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VB to the intersection of the VA construction line and drawing VA from the tail of VB to the intersection of the VBA construction line. 3.

VBA

3.414" 3.330"

Y

86.041°

103.052°

VB

From the velocity triangle we have: 1.000"

X



VA  3.330  in kv

4.


kv 

10 in sec

1

in

VA  33.300

in

θVB  86.041 deg

sec

Use equation 6.7 to find the angular velocity of the crank (link 2).

ω2 

VA a

ω2  23.786

rad sec

CW




Given:

Solution: 1.

The general linkage configuration and terminology for an offset fourbar slider-crank linkage are shown in Fig P6-2. The link lengths and the values of d and d-dot are defined in Table P6-5. For row a, find the velocity of pin joint A and the angular velocity of the crank using the analytic method. Draw the linkage to scale and label it before setting up the equations. Link lengths: Link 2 (O2 A)

a  1.4 in

Link 3 (AB)

b  4  in

Slider position

d  2.5 in

Slider velocity

ddot  10

Offset (yB)

c  1  in

in s



Y

13.052° . d

B

A

c = 1.000

000 b = 4.

176.041°

O2

X

a = 1.400 2.

Determine the open value of 2 using equations 4.20 and 4.21. 2

2

2

K1  a  b  c  d

2

2

K2  2  a  c

K2  2.8 in

K3  2  a  d

K3  7 in

A  K1  K3

A  0.21 in

B  2  K2

B  5.6 in

C  K1  K3

C  13.79 in

2 2

2 2



2

θ2  2  atan2 2  A B  3.

2

K1  6.79 in

B  4 A  C



θ2  176.041  deg

Determine the value of 3 using equation 4.16a or 4.17.

β ( a b c d α) 

θ  asin



a  sin( α)  c 

 

b




θ3  β  a b c d θ2

a  sin( α)  c  b


θ3  193.052  deg


4.

Determine the value of ω 2 using equation 6.24b.

ω2  5.


ddot cos θ3

ω2  23.785

a   cos θ2  sin θ3  sin θ2  cos θ3 

Determine the value of VA using equation 6.7. in

VA  a  ω2

VA  33.299

θVA  θ2  sign ω2  90 deg

θVA  86.041 deg

s

rad s




A point at a 6.5-in radius is on a body that is in pure rotation with  = 100 rad/sec and a constant  = -500 rad/sec2 at point A. The rotation center is at the origin of a coordinate system. When the point is at position A, its position vector makes a 45 deg angle with the X axis. It takes 0.01 sec to reach point B. Draw this system to some convenient scale, calculate the  and  of position B, and: a. Write an expression for the particle's acceleration vector in position A using complex number notation, in both polar and Cartesian forms. b. Write an expression for the particle's acceleration vector in position B using complex number notation, in both polar and Cartesian forms. c. Write a vector equation for the acceleration difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the acceleration difference numerically. d. Check the result of part c with a graphical method.

Given: Initial rotation speed Rotation acceleration

ω  100 

rad

α  500 

rad

sec

sec

Solution: 1.

Time to reach point B

t  0.01 sec

Vector angle

θ  45 deg

Vector magnitude

R  6.5 in


Calculate the position and angular velocity at point B using equations 6.1 and 7.1. ω  α t  ω θ 

2.

2

α t

ω  95.000

rad sec

2

2

 ω t  θ

θ  100.863 deg

Calculate the magnitudes and directions of the normal and tangential components of acceleration at points A and B using equations 7.2. 2

a An  R ω

a An  65000

in sec

a At  R α

2

sec

θAn  225.000 deg

θAt  θ  sign α  90 deg

in

a At  3250

θAn  θ  180  deg

2

θAt  45.000 deg a Bn  R ω

2

a Bn  58663

in sec

a Bt  R α

a Bt  3250

in sec

2

2

θBn  θ  180  deg θBt  θ  sign α  90 deg θBt  10.863 deg

3.


θBn  280.863 deg



4.

Draw lines from the origin that make angles of 45 and 100.863 deg with respect to the X axis. Label the intersections of the lines with the circles as A and B, respectively.

5.

Choose a convenient acceleration scale and draw the two acceleration component vectors at points A and B.

Y 0

8 B

1

2 in

Distance scale:

ABt

0

10000 in/sec^2

Acceleration scale:

6 A 4

AAt

2 ABn

AAn

0

a.

X 2

4

8

6

Write an expression for the particle's acceleration vector in position A using complex number notation, in both polar and Cartesian forms.

Polar form:

RA  R e

j

j  θ

VA  R j  ω e AA  R j  α e

AA  6.50 j  α e



j

j  θ

j

Cartesian form:

π 4

VA  650  j  e 2 j  θ

 R j  ω  e π 4

2

 6.50 ω  e

 

j

π 4

   R ω2 cosθ  j  sinθ

AA  R α sin θ  j  cos θ AA  ( 43664  48260i)

in sec

b.

4

RA  6.5 e

j  θ

π

2

Write an expression for the particle's velocity vector in position B using complex number notation, in both polar and Cartesian forms.

Polar form:

RB  R e

j  θ

VB  R j  ω e

π

j

RB  6.5 e j  θ

 100.863

180 j

VA  650  j  e

π 180

 100.863



AB  R j  α e

j  θ j

AB  6.50 j  α e

Cartesian form:



2 j  θ

 R j  ω  e π

 100.863

180

 6.50 ω  e

 

π

   R ω2 cosθ  j  sinθ

in sec

2

Write a vector equation for the velocity difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. ABA  AB  AA

ABA  ( 57912  8739i)

in sec

d.

 100.863

180

AB  R α sin θ  j  cos θ AB  ( 14248  56999i)

c.

2

j

2

Check the result of part c with a graphical method. Solve the equation ABt + ABn = AAt + AAn + ABA using an acceleration scale of 10000 in/sec2 per drawing unit. Acceleration scale factor

ka  10000 

in sec

Horizontal component

2

ABAx  5.791  ka

ABAx  57910

in sec

ABAy  0.874  ka

Vertical component

ABAy  8740

in sec

AAt

ABt

0

2

2

10000 IN/S/S

Acceleration Scale

AAn ABA

0.874

ABn

5.791

On the layout above the X and Y components of ABA are equal (to three significant figures) to the real and imaginary components calculated, confirming that the calculation is correct.




In Problem 7-1 let A and B represent points on separate, rotating bodies both having the given  and  at t = 0, A = 45 deg, and B = 120 deg. Find their relative acceleration.

Given: ω  100 

Initial rotation speed

rad sec

α  500 

Rotation acceleration

rad sec

Solution: 1.

Vector angles

θ  45 deg

Vector magnitude

R  6.5 in

2

θ  120  deg


Calculate the acceleration of points A and B using equation 7.3.



 

   R ω2 cosθ  j  sinθ

AA  R α sin θ  j  cos θ AA  ( 43664  48260i)

in sec



 

2

   R ω2 cosθ  j  sinθ

AB  R α sin θ  j  cos θ AB  ( 35315  54667i)

in sec

2.

2

Calculate the relative acceleration of point B with respect to A using equation 7.4. ABA  AB  AA

ABA  ( 78978  6407i)

in sec

Magnitude:

ABA  ABA

ABA  79238

in sec

Direction:

θ  arg ABA

2

θ  4.638 deg

2




The link lengths, coupler point location, and the values of 2, 2, and 2 for the same fourbar linkages as used for position and velocity analysis in Chapters 4 and 6 are redefined in Table P7-1, which is the same as Table P6-1. The general linkage configuration and terminology are shown in Figure P7-1. For row a, draw the linkage to scale and graphically find the accelerations of points A and B. Then, calculate 3 and 4 and the acceleration of point P.

Given:

Link lengths:

Link 1

d  6  in

Link 2

a  2  in

Link 3

b  7  in

Link 4

c  9  in

RPA  6  in

Coupler point:

δ  30 deg θ  30 deg

Link 2 position, velocity, and acceleration: Solution: 1.

ω  10

rad

α  0 

sec

rad sec

SeeFigure P7-1 and Mathcad file P0703a.

In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution below, θ  88.837 deg y

θ  117.286  deg Using equation (6.18), ω 

OPEN

P

 

 

a  ω sin θ  θ  b sin θ  θ

ω  5.991 rad sec ω 

B

3 4

1 88.837°

 

 

a  ω sin θ  θ  c sin θ  θ

ω  3.992 rad sec

2 O4

O2

x

1

2.

The graphical solution for accelerations uses equation 7.4:

3.

For point B, this becomes: ABn  c ω

117.286°

A

(APt + APn) = (AAt + AAn) + (APAt + APAn)

(ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where

2

ABn  143.403 in sec

θABn  θ  180  deg 2

θABn  297.286 deg

AAn  a  ω

AAn  200.000 in sec

θAAn  θ  180  deg

θAAn  210.000 deg

AAt  a  α

AAt  0.000 in sec 2

2

2

2

ABAn  b  ω

ABAn  251.239 in sec

θABAn  θ  180  deg

θABAn  268.837 deg

2

2


4.


Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of 4 + 180 deg and AAn at an angle of 2 + 180 deg. From the tip of AAn, draw ABAn at an angle of 3 + 180 deg. Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABAt, and AB. 0

100 IN/S/S

Acceleration Scale AA

136.080° n

AB

5.009

AB t

AB

n

ABA

t ABA

4.800 1.826

5.

From the graphical solution above, Acceleration scale factor

ka  100 

in sec

6.

2

ABAt  1.826  ka

ABAt  182.6 in sec

ABt  4.800  ka

ABt  480.0 in sec

AB  5.009  ka

AB  500.9 in sec

2

2

2

at an angle of -136.08 deg

Calculate 3 and 4using equation 7.6. α  α 

ABAt

α  26.086 rad sec

b ABt

α  53.333 rad sec

c

2

2

CCW

CCW

From the graphical solution we see that both of the tangential acceleration vectors for links 3 and 4 point to the left, so both of the angular accelerations are positive. 7.

For point P, equation 7.4 becomes: APAn  RPA ω

2

AP = (AAt + AAn) + (APAt + APAn) , where APAn  215.348 in sec

θAPAn  θ  δ  180  deg

θAPAn  298.837 deg

APAt  RPA α

APAt  156.514 in sec

2

2



θAPAt  θ  δ  90 deg 8.

θAPAt  208.837 deg

Repeat proceedure of step 4 for the equation in step 7. 0

100 IN/S/S

Acceleration Scale AA 119.550°

n

APA 4.186 AP t

APA

9.


ka  100 

in sec

AP  4.186  ka

2

AP  418.6 in sec

2





The link lengths, coupler point location, and the values of 2, 2, and 2 for the same fourbar linkages as used for position and velocity analysis in Chapters 4 and 6 are redefined in Table P7-1, which is the same as Table P6-1. The general linkage configuration and terminology are shown in Figure P7-1. For row a, find the accelerations of points A and B using the analytic method. Then, calculate 3 and 4 and the acceleration of point P.

Given:


d  6

Link 2

a  2

Link 3

b  7

Link 4

c  9

RPA  6

Coupler point:

δ  30 deg θ  30 deg


Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. d

K2 

a

K1  3.0000

2

d

K3 

c

K2  0.6667

 

 

2

2

2 a c

A  0.7113

 

B  2  sin θ

B  1.0000

 

C  K1   K2  1   cos θ  K3

C  3.5566






2

θ  2  atan2 2  A B 

B  4 A  C



θ  242.714 deg

θ  θ  360  deg



θ  602.714 deg



2

Crossed: θ  2  atan2 2  A B 

B  4 A  C



θ  216.340 deg


2

d

K5 

b

2

2

c d a b

 

2

2 a b

 

D  cos θ  K1  K4 cos θ  K5

K4  0.8571 D  1.6774

 

E  2  sin θ

E  1.0000

 

F  K1   K4  1   cos θ  K5 4.

2

a b c d

K3  2.0000

A  cos θ  K1  K2 cos θ  K3

3.

α  0


K1 

2.

ω  10

F  2.5906






θ  2  atan2 2  D E  θ  θ  360  deg

2

E  4  D F



θ  271.163 deg θ  631.163 deg

K5  0.2857


Crossed: 5.



2

θ  2  atan2 2  D E 

E  4  D F



θ  244.789 deg

Determine the angular velocity of links 3 and 4 for the open and crossed circuits using equations 6.18. ω 

a  ω sin θ  θ  b sin θ  θ

 

 

ω  5.991

ω 

a  ω sin θ  θ  c sin θ  θ

 

 

ω  3.992

CROSSED ω 

a  ω sin θ  θ  b sin θ  θ

 

 

ω  0.662

ω 

a  ω sin θ  θ  c sin θ  θ

 

 

ω  2.662

OPEN

6.




Using the Euler identity to expand equation 7.13a for AA. Separate into real and imaginary parts to give the x and y components



 

   a ω2 cosθ  j  sinθ

 

2

AA = a  α sin θ  j  cos θ

 

AAx  a  α sin θ  a  ω  cos θ

AAx  173.205

 

AAy  100.000

2

 

AAy  a  α cos θ  a  ω  sin θ AA 

2

AAx  AAy

θAA  atan2 AAx AAy 

2

The acceleration of pin A is 7.

AA  200.000

θAA  150 deg

at

Use equations 7.12 to determine the angular accelerations of links 3 and 4 for the open and crossed circuits.

 

OPEN A  c sin θ A  7.999

 

 

 

B  b  sin θ

D  c cos θ

E  b  cos θ

B  6.999

D  4.126

E  0.142

 

2

 

2

 

2

 

2

 

2

 

2

 

C  a  α sin θ  a  ω  cos θ  b  ω  cos θ  c ω  cos θ C  244.045

 

F  a  α cos θ  a  ω  sin θ  b  ω  sin θ  c ω  sin θ F  223.741 α 

C D  A  F A  E  B D

 

CROSSED A  c sin θ A  5.333

α  26.080

 

α 

C  E  B F A  E  B D

 

α  53.331

 

B  b  sin θ

D  c cos θ

E  b  cos θ

B  6.333

D  7.250

E  2.982



 

2

 

2

 

2

 

2

 

2

 

2

 

C  a  α sin θ  a  ω  cos θ  b  ω  cos θ  c ω  cos θ C  223.253

 

F  a  α cos θ  a  ω  sin θ  b  ω  sin θ  c ω  sin θ F  135.002 α  8.

C D  A  F

α  77.920

A  E  B D

α 

C  E  B F

Use equation 7.13c to determine the acceleration of point B for the open and crossed circuits. OPEN ABx1  c  α sin θ  ω  cos θ

 

2

 

ABx1  360.826

ABy1  c  α cos θ  ω  sin θ

 

AB1 

2

ABx1  ABy1

2

 

ABy1  347.485

2

AB1  500.941

θAB1  atan2 ABx1 ABy1 

θAB1  136.1 deg

CROSSED ABx2  c  α sin θ  ω  cos θ

 

2

 

ABx2  321.587

ABy2  c  α cos θ  ω  sin θ

 

AB2 

2

ABx2  ABy2

2

 

ABy2  329.551

2

AB2  460.459

θAB2  atan2 ABx2 ABy2  9.

α  50.669

A  E  B D

θAB2  45.7 deg

Use equations 7.32 to find the acceleration of the point P for the open and crossed circuits.



 

   a ω2 cosθ  j  sinθ

OPEN AA  a  α sin θ  j  cos θ



      RPA ω   cos θ  δ  j  sin θ  δ 

APA1  RPA α sin θ  δ  j  cos θ  δ 2

AP1  AA  APA1 AP1  AP1



AP1  418.556

 

arg AP1  119.548 deg

   a ω2 cosθ  j  sinθ

CROSSED AA  a  α sin θ  j  cos θ



      RPA ω   cos θ  δ  j  sin θ  δ 

APA2  RPA α sin θ  δ  j  cos θ  δ 2

AP2  AA  APA2 AP2  AP2

AP2  298.225

arg AP2  11.282 deg




The link lengths and the values of 2, 2, and 2 for the some non inverted offset fourbar slider-crank linkages are defined in Table P7-2. The general linkage configuration and termin- olo are shown in Figure P7-2. For row a, draw the linkage to scale and graphically find the accelerations of the pin joints A and B the acceleration of slip at the sliding joint.

Given: a  1.4 in

Link lengths:

Link 2

Offset:

c  1  in θ  45 deg


b  4  in

Link 3

ω  10

rad sec

α  0 

rad sec

2


In order to solve for the accelerations at point B, we will need 3, and 3. From the graphical position solution below, θ  360  deg  179.856  deg ω 

Using equation 6.22a,

θ  180.144 deg

   

a  ω cos θ  b cos θ

ω  2.475 rad sec

1

Direction of ABAt Y


Direction of AAt

A 2

45.000°

Axis of slip and Direction of AB

B

3

179.856°

1.000 X

O2

2.


3.

For point B, this becomes:

(APt + APn) = (AAt + AAn) + (APAt + A

AB = (AAt + AAn) + (ABAt + ABAn) , where

2

AAn  a  ω

AAn  140.000 in sec

θAAn  θ  180  deg

θAAn  225.000 deg

AAt  a  α

AAt  0.000 in sec 2

2

2



θABAn  θ


2


4.


Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw AAn at an angle of AAn. From the tip of AAn, draw ABAn at an angle of ABAn. Now that the vectors with known magnitudes are drawn, from the origin and the tip of ABAn, draw construction lines in the directions of AB (horizontal) and ABAt, respectively. The intersection of these two lines are the tips of ABAt, and AB. 4.950 Y AB X t BA

A

0

25 IN/S/S

Acceleration Scale AA

n

ABA

5.


ka  25

in sec

AB  4.950  ka 6.

2

AB  123.8 in sec

2

The acceleration of slip is equal to AB since link 1 is stationary.

at an angle of 180 deg




The link lengths and the values of 2, 2, and 2 for the some non inverted offset fourbar slider-crank linkages are defined in Table P7-2. The general linkage configuration and terminology are shown in Figure P7-2. For row a, draw the linkage to scale and find the accelerations of the pin joints A and B the acceleration of slip at the sliding joint using an analytical method.

Given:


a  1.4 in

b  4  in

Link 3

c  1  in

Offset:


θ  45 deg

ω  10

rad sec

α  0 

sec


Draw the linkage to scale and label it. Y d2 = 3.010

d1 = 4.990

3(CROSSED)

B'

A 2

0.144°

45.000°

3 (OPEN)

B 179.856° 1.000 X

O2

2.

Determine 3 and d using equations 4.16 and 4.17. Open:

 a  sin θ  c  π b  

θ  180.144 deg

 

d 2  4.990 in

θ  asin 

 

d 2  a  cos θ  b  cos θ Crossed:

 a sin θ  c   b  

θ  0.144 deg

 

d 1  3.010 in

θ  asin

 

d 1  a  cos θ  b  cos θ 3.

Determine the angular velocity of link 3 using equation 6.22a. Open

Crossed

4.

   

ω  2.475

   

ω  2.475

ω 

a cos θ   ω b cos θ

ω 


rad

rad sec

rad sec

Using the Euler identity to expand equation 7.15b for AA, determine its magnitude, and direction.

2





 

   a ω2 cosθ  j  sinθ

AA  a  α sin θ  j  cos θ in

AA  ( 98.995  98.995i )

sec

AA  140.0

The acceleration of pin A is

in sec

5.

at 2

θAA  135.0 deg

Determine the angular acceleration of link 3 using equation 7.16d.

α 

Open

 

α 

2

 

  b  cos θ

2

 

rad sec

Crossed

  b  cos θ 2

a  α cos θ  a  ω  sin θ  b  ω  sin θ

α  24.764

2

 

2

a  α cos θ  a  ω  sin θ  b  ω  sin θ

α  24.764

rad sec

6.

θAA  arg AA

AA  AA

2

2

Use equation 7.16e for the acceleration of pin B. Open:

 

2

 

 

2

 

AB2  a  α sin θ  a  ω  cos θ  b  α sin θ  b  ω  cos θ AB2  123.7

in sec

2

A negative sign means that AB is to the left

Crossed:

 

2

 

 

2

 

AB1  a  α sin θ  a  ω  cos θ  b  α sin θ  b  ω  cos θ AB1  74.2

in sec

2





The link lengths and the values of 2, 2, and  for an inverted fourbar slider-crank linkage are defined in Table P7-3, row a, and are given below. Find the acceleration of the pin joints A and B and the acceleration of slip at the sliding joint. Solve by the analytic vector loop method of Section 7.3 for the open configuration of the linkage.

Given:

Link lengths: a  2  in

Link 2

Link 2 position, velocity, and accel.

1.

Link 1

d  6  in

γ  90 deg

Angle between links 3 and 4

Solution:

c  4  in

Link 4

θ  30 deg

ω  10

rad sec

α  25

sec


Draw the linkage to scale and label it. B y 90.0°

127.333°

b

c 142.666°

A

a 30.000°

d

x 04

02

2.

Use the equations in Section 4.7 to solve for the positions of links 3 and 4 and for the length b.

 



 




 



 

P  1.000  in




Q  4.268  in

R  c sin γ

S  R  Q

T  2  P

U  Q  R

R  4.000  in

S  0.268  in

T  2.000  in

U  8.268  in





θ  2  atan2( 2  S )  T  θ  θ  γ b 

 

2

T  4 S U



θ  142.667  deg θ  232.667  deg

 

a  sin θ  c sin θ

 

b  1.793  in

sin θ

3.

Calculate the angular velocity of links 3 and 4 and the slip velocity using equations 6.30.

rad 2


ω 






a  ω cos θ  θ b  c cos γ

bdot 

 

   cos θ

 

a  ω sin θ  ω b  sin θ  c sin θ

ω  10.292

rad

bdot  33.461

in

sec

sec

ω  ω 4.

Solve for the accelerations using equations (7.26) and (7.27).



P  a  α cos θ  θ



P  46.138 in sec

2





Q  77.075 in sec

2





R  423.705  in sec

Q  a  ω  sin θ  θ R  c ω  sin θ  θ S  2  bdot ω





α  130.56 rad sec

T

















K  639.230  in  sec



2



N  2  bdot c ω sin θ  θ bddot  

2

L  150.000  in  sec



M  2.035  10  in  sec

2

M  ω   b  c  2  b  c cos θ  θ 2

2



L  a  α b  sin θ  θ  c sin θ  θ 2

2

2

K  a  ω  b  cos θ  θ  c cos θ  θ



2

T  1.793  in

P Q R S 2

2

S  688.757  in sec

T  b  c cos θ  θ α 

2

2

3



2

3

2

bddot  128.48

in

N  2.755  10  in  sec

K L M  N T

sec

2

2

2

The acceleration of slip is bddot. It is directed along link 3, positive inward from B towards A, so that its angle is 3. For the acceleration of the pin joints A and B,



 

   a ω2 cosθ  j  sinθ

AA  a  α sin θ  j  cos θ AA  AA

AA  206.155 

in sec

  

2


θAA  135.964  deg

   c ω2 cosθ  j  sinθ

AB  c α sin θ  j  cos θ AB  AB

AB  672.505 

in sec

2

θAB  arg AB

θAB  88.280 deg




The link lengths and the values of 2, 2, and  for an inverted fourbar slider-crank linkage are defined in Table P7-3, row a, and are given below. Find the acceleration of the pin joints A and B and the acceleration of slip at the sliding joint. Solve by the analytic vector loop method of Section 7.3 for the crossed configuration of the linkage.

Given:

Link lengths: a  2  in

Link 2

θ  30 deg

Link 2 position, velocity, and accel.

1.

Link 1

d  6  in

γ  90 deg


Solution:

c  4  in

Link 4

ω  10

rad

rad

α  25

sec

sec



A a b

30.000°

d

x 04

02 c B

169.040° 79.041°

2.


 



 




 



 

P  1.000 in




Q  4.268 in

R  c sin γ

S  R  Q

T  2  P

U  Q  R

R  4.000 in

S  0.268 in

T  2.000 in

U  8.268 in



θ  2  atan2( 2  S ) T  θ  θ  γ b 

 

2



T  4  S  U

θ  169.041 deg θ  79.041 deg

 

a  sin θ  c sin θ

 

b  1.793 in

sin θ 3.

Calculate the angular velocity of links 3 and 4 and the slip velocity using equations 6.30.

2


ω 






a  ω cos θ  θ b  c cos γ

bdot 

 

ω  3.639

   cos θ

 

a  ω sin θ  ω b  sin θ  c sin θ

rad sec

bdot  33.461

in sec

ω  ω 4.

Solve for the accelerations using equations (7.26) and (7.27).



P  a  α cos θ  θ



P  16.312 in sec

2





Q  189.057 in sec

2





R  52.961 in sec

Q  a  ω  sin θ  θ R  c ω  sin θ  θ S  2  bdot ω





α  9.93 rad sec

T

















K  639.230 in  sec L  46.352 in  sec



M  254.418 in  sec

2

M  ω   b  c  2  b  c cos θ  θ 2



2



N  2  bdot c ω sin θ  θ bddot  

2



L  a  α b  sin θ  θ  c sin θ  θ 2

2

2

K  a  ω  b  cos θ  θ  c cos θ  θ



2

T  1.793 in

P Q R S 2

2

2

S  243.508 in sec

T  b  c cos θ  θ α 

2

2

2



2

N  974.033 in  sec

K L M  N

bddot  18.98

T

2

2

in sec

2

The acceleration of slip is bddot. It is directed along link 3, positive inward from B towards A, so that its angle is 3. For the acceleration of the pin joints A and B,



 

   a ω2 cosθ  j  sinθ


AA  206.155

in sec

  

2


θAA  135.964 deg

   c ω2 cosθ  j  sinθ

AB  c α sin θ  j  cos θ AB  AB

AB  66.195

in sec

2


θAB  47.822 deg




The link lengths, gear ratio (), phase angle (), and the values of 2, 2, and 2 for a geared fivebar from row a of Table P7-4 are given below. Find 3 and 4 and the linear acceleration of point P.

Given:

Link lengths: Link 1 f  6  in

Link 3

b  7  in

a  1  in

Link 4

c  9  in

Link 2

Link 5

d  4  in

Gear ratio, phase angle, and crank angle:

Solution: 1.

λ  2

ϕ  30 deg

θ  60 deg

Coupler data:

Rpa  6  in

δ  30 deg

α  0  rad sec

θ  150 deg

Choose the pitch radii of the gears. Since the gear ratio is positive, an idler must be used between gear 2 and gear 5. Let the idler be the same diameter as gear 5, and let all three gears be in line.

r5 

λ 

f

r2 r5

r5  1.200 in

λ3

r2  λ r5

r2  2.400 in

Draw the linkage to scale and label it. 4

C

B 3

A

5 O5

O2

2 P

4.

2


f  r2  3  r5

3.

1


θ  λ θ  ϕ 2.

ω  10 rad sec








 







 




2

B  2  c d  sin λ θ  ϕ  a  sin θ

2

2

2

2



2

B  20.412 in

  




     





2

A  36.6462 in

2

C  37.4308 in

2

D  C  A

D  0.78461 in

E  2  B

E  40.823 in

2



2

F  A  C

F  74.077 in





  a cosθ  f 

G  28.503 in





  a sinθ

H  15.876 in

2

G  2  b   d  cos λ θ  ϕ

2

H  2  b   d  sin λ θ  ϕ

2

5.

2

2

2



  

2

K  a  b  c  d  f  2  a  f  cos θ   2  d  a  cos θ  f  cos λ θ  ϕ    2  a  d  sin θ  sin λ θ  ϕ

K  26.569 in

L  K  G

L  1.933 in

M  2  H

M  31.751 in

N  G  K

N  55.072 in



     





2

2 2

2






M  4  L N





E  4  D F





M  4  L N





E  4  D F

2

θ  2  atan2 2  L M 

2

θ  2  atan2 2  D E 

CROSSED

2

θ  2  atan2 2  D E  6.

Use equation 6.32c to find 5. ω  λ ω

7.

Calculate 3 and 4 using equations 6.33.

OPEN

ω 

 

ω 





θ  177.715 deg θ  115.407 deg

θ  124.050 deg



 



 



b  cos θ  2  θ  cos θ rad sec

 

  c sin θ

 

a  ω sin θ  b  ω sin θ  d  ω sin θ

ω  16.948

CROSSED





θ  173.642 deg

2  sin θ  a  ω sin θ  θ  d  ω sin θ  θ

ω  32.585

ω 



2

θ  2  atan2 2  L M 



rad sec

 







2  sin θ  a  ω sin θ  θ  d  ω sin θ  θ

ω  75.191

 



 

b  cos θ  2  θ  cos θ rad sec




ω 


 

  c sin θ

 

a  ω sin θ  b  ω sin θ  d  ω sin θ

ω  59.554 8.

Use equation 7.28c to find 5. α  λ α

9.

Calculate 3 and 4 using equations 7.29.

rad sec







2



a  α sin θ  θ  a  ω  cos θ  θ 



2



2





 b  ω  cos θ  θ  d  ω  cos θ  θ  OPEN

α 





2

 d  α sin θ  θ  c ω





b  sin θ  θ

α  3191

rad sec

2







2



a  α sin θ  θ  a  ω  cos θ  θ 



2



2





 c ω  cos θ  θ  d  ω  cos θ  θ  α 





2

 d  α sin θ  θ  b  ω





c sin θ  θ

α  2492

rad sec

2







2



a  α sin θ  θ  a  ω  cos θ  θ 



2



2





 b  ω  cos θ  θ  d  ω  cos θ  θ  CROSSED

α 





2

 d  α sin θ  θ  c ω





b  sin θ  θ

α  6648

rad sec

2







2



a  α sin θ  θ  a  ω  cos θ  θ 



2



2





 c ω  cos θ  θ  d  ω  cos θ  θ  α 





2

 d  α sin θ  θ  b  ω

α  5950





c sin θ  θ rad sec

2




An automobile driver took a curve too fast. The car spun out of control about its CG and slid off the road in a northeasterly direction. The friction of the skidding tires provided a 0.25g linear deceleration. The car rotated at 100 rpm. When the car hit the tree head-on at 30 mph, it took 0.1 sec to come to rest.

Given:

V  30 mph

Solution:


a.

Ac  0.25 g

ω  100  rpm

What was the acceleration experienced by the child seated on the middle of the rear seat, 2 ft behind the car's CG, just prior to impact? r  2  ft

ω  10.472 rad sec

1

Ac  8.044 ft  sec

2

We want to solve the vector equation Ap = Ac + Apcn + Apct where P is the position of the child and C is the CG. Since  is zero, Apct is zero. Ac and Apcn both have the same direction, but are opposite in sense (see diagram). 2

Apcn  r ω

Apcn  6.817 g Ac

Ap  Ac  Apcn

V 

Ap  6.567 g

n

A pc

This acceleration is toward the tree. Ap  211.3 ft  sec Ap  2535 in sec b.

t

A pc

2

2

What force did the 100 lb child exert on her seat belt harness as a result of the acceleration just prior to impact? W  100  lbf The seat back must provide a force of

c.

2'

W

F 

g

 Ap

F  657 lbf

Assuming a constant deceleration during the impact, what was the magnitude of the average deceleration felt by the passengers in that interval? t  0.1 sec Assume that the rotation has ceased. Then, the average velocity during impact is Vavg 

V

Vavg  22.0

2

and the average deceleration is

Aavg 

Adding the deceleration due to skidding, Adec  7.088 g

Vavg t

ft sec

Aavg  6.838 g

Aavg  220.0 ft  sec

Adec  Ac  Aavg Adec  228 ft  sec

2

Adec  2737 in sec

2

2




For row a in Table P7-1, find the angular jerk of links 3 and 4 and the linear jerk of the pin between links 3 and 4 (point B). Assume an angular jerk of zero on link 2. The linkage configuration and terminology are shown in Figure P7-1.

Given:


d  6

Link 2

a  2

Link 3

b  7

Link 4

c  9

RPA  6

Coupler point:

δ  30 deg

Link 2 position, velocity, and acceleration:θ  30 deg Solution: 1.

d

K2 

a

2

d

K3 

c

K2  0.6667

 

2

2

a b c d

2

2 a c

K3  2.0000

 

A  cos θ  K1  K2 cos θ  K3

A  0.7113

 

B  2  sin θ

B  1.0000

 

C  K1   K2  1   cos θ  K3

C  3.5566






2

θ  2  atan2 2  A B 

B  4 A  C



θ  242.714 deg

θ  θ  360  deg



θ  602.714 deg



2

Crossed: θ  2  atan2 2  A B 

B  4 A  C



θ  216.340 deg


2

d

K5 

b

2

2

c d a b

 

2

2 a b

 

D  cos θ  K1  K4 cos θ  K5

K4  0.8571 D  1.6774

 

E  2  sin θ

E  1.0000

 

F  K1   K4  1   cos θ  K5 4.

ϕ  0


K1  3.0000

3.

α  0


K1 

2.

ω  10

F  2.5906






θ  2  atan2 2  D E 

2

E  4  D F



θ  θ  360  deg Crossed:





θ  2  atan2 2  D E 

θ  271.163 deg θ  631.163 deg

2

E  4  D F



θ  244.789 deg

K5  0.2857


5.

6.


Determine the angular velocity of links 3 and 4 for the open and crossed circuits using equations 6.18. OPEN ω 

a  ω sin θ  θ  b sin θ  θ

 

 

ω  5.991

ω 

a  ω sin θ  θ  c sin θ  θ

 

 

ω  3.992

CROSSED ω 

a  ω sin θ  θ  b sin θ  θ

 

 

ω  0.662

ω 

a  ω sin θ  θ  c sin θ  θ

 

 

ω  2.662

Use equations 7.12 to determine the angular accelerations of links 3 and 4 for the open and crossed circuits.

 

 

OPEN A  c sin θ A  7.999

 

 

B  b  sin θ

D  c cos θ

E  b  cos θ

B  6.999

D  4.126

E  0.142

 

2

 

2

 

2

 

2

 

2

 

2

 

C  a  α sin θ  a  ω  cos θ  b  ω  cos θ  c ω  cos θ C  244.045

 

F  a  α cos θ  a  ω  sin θ  b  ω  sin θ  c ω  sin θ F  223.741 α 

C D  A  F

α  26.080

A  E  B D

 

α 

 

CROSSED A  c sin θ A  5.333

C  E  B F A  E  B D

 

α  53.331

 

B  b  sin θ

D  c cos θ

E  b  cos θ

B  6.333

D  7.250

E  2.982

 

2

 

2

 

2

 

2

 

2

 

2

 

C  a  α sin θ  a  ω  cos θ  b  ω  cos θ  c ω  cos θ C  223.253

 

F  a  α cos θ  a  ω  sin θ  b  ω  sin θ  c ω  sin θ F  135.002 α  7.

C D  A  F

α  77.920

A  E  B D

α 

C  E  B F A  E  B D

α  50.669

Use equations 7.36 and 7.37 to determine the angular jerk of links 3 and 4 for the open and crossed circuits. OPEN

3

 

3

A  a  ω  sin θ

 

B  3  a  ω α cos θ

 

C  a  ϕ sin θ

 

 

D  b  ω  sin θ

G  3  c ω α cos θ

 

E  3  b  ω α cos θ 3

 

F  c ω  sin θ

 

H  c sin θ

 

K  b  sin θ



 

3

3

L  a  ω  cos θ

 

Q  3  b  ω α sin θ

 

 

N  a  ϕ cos θ

ϕ 

 

T  3  c ω α sin θ U  c cos θ

ϕ  749.012

K U  H  R A  B  C  D  E  F  G  H  ϕ

ϕ  1242.6

K

     

 

 

R  b  cos θ

K ( N  L  M  P  Q  S  T )   R ( A  B  C  D  E  F  G)

3

3

S  c ω  cos θ

 

M  3  a  ω α sin θ

ϕ 

 

P  b  ω  cos θ

   3 a ω α cosθ  j  sinθ   

J A  a  ω  sin θ  j  cos θ  a  ϕ sin θ  j  cos θ 3

     

   3 b ω α cosθ  j  sinθ   

J BA1  b  ω  sin θ  j  cos θ  b  ϕ sin θ  j  cos θ J B1  J A  J BA1 JB1  J B1

JB1x  Re J B1

JB1y  Im J B1

JB1  9301.9

JB1x  9134.7

JB1y  1755.5

3

 

3

CROSSED A  a  ω  sin θ

 

E  3  b  ω α cos θ

 

C  a  ϕ sin θ

   

K  b  sin θ

3

 

S  c ω  cos θ

3

 

 

 

R  b  cos θ

K ( N  L  M  P  Q  S  T )   R ( A  B  C  D  E  F  G) K U  H  R A  B  C  D  E  F  G  H  ϕ K 3

 

Q  3  b  ω α sin θ

N  a  ϕ cos θ

ϕ 

 

P  b  ω  cos θ

M  3  a  ω α sin θ

     

 

H  c sin θ

3

F  c ω  sin θ

L  a  ω  cos θ

ϕ 

 

G  3  c ω α cos θ

 

B  3  a  ω α cos θ

3

 

D  b  ω  sin θ

   

T  3  c ω α sin θ

 

U  c cos θ

ϕ  246.639 ϕ  740.2

   3 b ω α cosθ  j  sinθ   

J BA2  b  ω  sin θ  j  cos θ  b  ϕ sin θ  j  cos θ J B2  J A  J BA2 JB2  J B2

JB2x  Re J B2

JB2y  Im J B2

JB2  4178.7

JB2x  4147.9

JB2y  506.4




You are riding on a carousel that is rotating at a constant 12 rpm. It has an inside radius of 4 ft and an outside radius of 12 ft. You begin to run from the inside to the outside along a radius. Your peak velocity with respect to the carousel is 4 mph and occurs at a radius of 8 ft. What is your maximum Coriolis acceleration magnitude and its direction with respect to the carousel?

Given:

Carousel angular velocity Peak velocity Radius at peak velocity

Solution: 1.

rad

ω  12 rpm

ω  1.257 

Vslip  4  mph

Vslip  5.867 

sec ft sec

r  8  ft


Draw a plan view of the carousel floor showing your position at peak velocity and the velocity and Coriolis acceleration vectors. Axis of Transmission

Axis of Slip

Ac V slip

 r

2.

The direction of your path defines the axis of slip. The transmission axis is perpendicular to the axis of slip and positive in the direction of the tangential velocity of the carousel. Thus, the direction of the Coriolis acceleration vector is along the positive transmission axis.

3.

Use equation 7.19 to calculate the magnitude of the Coriolis component of your acceleration. Ac  2  Vslip  ω

Ac  14.745

ft sec

2

Ac  176.93

in sec

2




The linkage in Figure P7-5a has the dimensions and crank angle given below. Find 3, AA, AB, and AC for the position shown for 2 = 15 rad/sec and 2 = 10 rad/sec2 in the direction shown. Use the acceleration difference graphical method.

Given:

a  0.8 in

Link lengths:

Link 2

Offset:

c  0.38 in

Coupler point data:

p  1.33 in

b  1.93 in

Link 3

δ  38.6 deg

Link 2 position, velocity, and acceleration:θ  34.3 deg Solution: 1.

ω  15

rad sec

α  10

rad sec

2


In order to solve for the accelerations at points A, B and C, we will need 3, and 3. From the graphical position solution below, θ  154.502  deg ω 


   


ω  5.691 rad sec

1

Direction of ACAt Direction of AAt

C

Y

A

38.600°

34.300° 154.502°

X O2

Direction of AB B

Direction of ABAt

2.


3.

For point B, this becomes: AB = (AAt + AAn) + (ABAt + ABAn) , where 2


AAn  a  ω

AAn  180.000 in sec

θAAn  θ  180  deg

θAAn  214.300 deg

AAt  a  α

AAt  8.000 in sec

θAAt  θ  90 deg

θAAt  124.300 deg

2

2

2



θABAn  θ


2


4.


Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw AAn at an angle of AAn. From the tip of AAn, draw AAt at an angle of AAt. From the tip of AAt, draw ABAn at an angle of ABAn. Now that the vectors with known magnitudes are drawn, from the origin and the tip of ABAn, draw construction lines in the directions of AB (horizontal) and ABAt, respectively. The intersection of these two lines are the tips of ABAt, and AB. 7.089 Y AB X t BA

A 3.010

0

25 IN/S/S

Acceleration Scale n BA

A

n

t

AA

5.

AA


ka  25

in sec

6.

AB  177.2 in sec

ABAt  3.010  ka



2

Calculate 3 using equation 7.6. ABAt

α  38.990 rad sec

b

2

CCW

For point C, equation 7.4 becomes: AC = (AAt + AAn) + (ACAt + ACAn) , where 2

8.

2

AB  7.089  ka

α  7.

2

ACAn  p  ω

ACAn  43.070 in sec

θAPAn  θ  δ


ACAt  p  α

ACAt  51.856 in sec

θAPAt   θ  δ  90 deg


Repeat procedure of step 4 for the equation in step 7.

2

2



Y

8.554

X

165.351°

AC

0 t ACA

25 IN/S/S

Acceleration Scale

n

t

n CA

A

9.

AA

AA


ka  25

in sec

AC  8.554  ka

2

AC  213.9 in sec

2





The linkage in Figure P7-5a has the dimensions and crank angle given below. Find 3, AA, AB, and AC for the position shown for 2 = 15 rad/sec and 2 = 10 rad/sec2 in the direction shown. Use an analytical method.

Given:

Link lengths: Link 2 Offset:

a  0.8 in

Link 3

b  1.93 in

c  0.38 in

Coupler point data:

p  1.33 in

δ  38.6 deg

Link 2 position, velocity, and acceleration:θ  34.3 deg Solution: 1.

ω  15

rad sec

α  10

See Figure P7-5a and Mathcad file P0713b.

Draw the linkage to scale and label it. C

Y

A

38.600°

34.300° 154.502°

X O2

B

2.

Determine 3 and d using equations 4.16 and 4.17.

 a sin θ  b 

θ  asin 

 

c

π 

θ  154.502 deg

 

d  a  cos θ  b  cos θ 3.


4.

d  2.403 in

   


ω  5.691

rad sec




 

   a ω2 cosθ  j  sinθ

AA  a  α sin θ  j  cos θ AA  ( 153.206  94.826i )

in sec


2

AA  180

in sec

5.


AA  AA

at 2


θAA  148.2 deg

rad sec

2


α 

6.

 


  b  cos θ 2

2

 

a  α cos θ  a  ω  sin θ  b  ω  sin θ

α  38.990

rad sec

2

Use equation 7.16e for the acceleration of pin B.

 

 

2

 

2

 

AB  a  α sin θ  a  ω  cos θ  b  α sin θ  b  ω  cos θ in

AB  177.2

sec 7.

2


Determine the acceleration of the coupler point C using equations 7.32. Note that 3 is defined from point B in Figure 7-6 and from point A in Figure 7-9. To use equation 7.32 for a slider-crank we must redefine 3. θ  θ  180  deg

θ  25.498 deg



      p  ω   cos θ  δ  j  sin θ  δ 

ACA  p  α sin θ  δ  j  cos θ  δ 2

AC  AA  ACA AC  ( 206.910  54.083i )

in sec

2

θAC  arg AC

AC  AC

The acceleration of point C is AC  213.861

in sec

at 2

θAC  165.352 deg




The linkage in Figure P7-5b has the dimensions and effective crank angle given below. Find 3, AA, AB, and AC for the position shown for 2 = 15 rad/sec and 2 = 10 rad/sec2 in the directions shown. Use the acceleration difference graphical method.

Given:

Solution: 1.

Link lengths: Link 2 (point of contact to A)

a  0.75 in

Link 3 (A to B)

b  1.5 in


c  0.75 in


d  1.5 in

Coupler point: Distance A to C

p  1.2 in

Angle BAC

δ  30 deg

Crank angle:

θ  77 deg


ω  15 rad sec

1

α  10 rad sec

2

See Figure P7-5b and Mathcad file P0714a.

Although the mechanism shown in Figure P6-5b is not entirely pin-jointed, it can be analyzed for the position shown by its effective pin-jointed fourbar, which is shown below. Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest. 0

0.5

1 in

Direction of ACAt Y

C Direction of AAt Direction of ABAt

30.000°

Direction of ABt

3

A

B b 4

2 a

77.000°

c d

O2

X O4

Effective link 2

2.

77.000°

Effective link 4

In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above, θ  0.0 deg

θ  77.0 deg

This is a special-case Grashof in the parallelogram configuration. Therefore, ω  0.0 rad sec

1

ω  ω

3.


4.

For point B, this becomes:


(ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where


ABn  c ω

2

ABn  168.750 in sec

θABn  θ  180  deg 2

AAn  168.750 in sec

θAAn  θ  180  deg

θAAn  257.000 deg

AAt  a  α


θAt  θ  90 deg

θAt  167.000 deg

2

2

θABn  257.000 deg

AAn  a  ω

ABAn  b  ω 5.


2

2


2

Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of 4 + 180 deg and AAn at an angle of 2 + 180 deg. From the tip of AAn, draw AAt at an angle of 2 + 90 deg. Now that the vectors with known magnitudes are drawn, from the tips of ABn and AAt, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABAt, and AB. 0

25 IN/S/S

Acceleration Scale 105.545°

From the layout ABt = AAt and ABAt = 0. Also, AB = AA. Acceleration scale factor ka  25

in sec

6.757

2

AB  6.757  ka AB  168.9 in sec

2


AB n

AA

AB n

AA

AAt ABt

6.

Since ABAt = 0, 3 = 0 and, since AB = AA, 4 = 2.

7.

For point P, equation 7.4 becomes: AP = (AAt + AAn) + (APAt + APAn) , where APAt and APAn are both zero since  and 3 are zero. Therefore, AP = AA.




The linkage in Figure P7-5b has the dimensions and effective crank angle given below. Find

Given:

3, AA, AB, and AC for the position shown for 2 = 15 rad/sec and 2 = 10 rad/sec2 in the directions shown. Use an analytical method. Link lengths: Link 2 (point of contact to A)

a  0.75 in

Link 3 (A to B)

b  1.5 in


c  0.75 in


d  1.5 in

Coupler point:

Solution: 1.

Rca  1.2 in

Crank angle:

θ  77 deg

α  10 rad sec

Angle BAC

δ  30 deg


ω  15 rad sec



C

30.000°

0

0.5

1 in

3

A

B b 4

2 77.000°

a

c d

77.000°

X O4

O2 Effective link 2

2.

Effective link 4


d

K2 

a

K1  2.0000 2

K3 

d c

K2  2.0000 2

2

a b c d

2

K3  1.0000

2 a c

 

 

A  cos θ  K1  K2 cos θ  K3

 

B  2  sin θ

 

C  K1   K2  1   cos θ  K3 A  1.2250 3.

2

Distance A to C

B  1.9487

C  2.3251






θ  2  atan2 2  A B  θ  θ  360  deg

2

B  4 A  C



θ  283.000 deg θ  77.000 deg

1


4.



2

d

K5 

b

2

2

c d a b

 

2

K4  1.0000

2 a b

 

D  cos θ  K1  K4 cos θ  K5

D  3.5501

 

E  2  sin θ

E  1.9487

 

F  K1   K4  1   cos θ  K5 5.

F  0.0000

Use equation 4.13 to find values of 3 .





2

θ  2  atan2 2  D E 

E  4  D F



θ  360.000 deg

θ  θ  360  deg 6.

7.

K5  2.0000

θ  0.000 deg

Determine the angular velocity of links 3 and 4 using equations 6.18. ω 

a  ω sin θ  θ  b sin θ  θ

 

 

ω  0.000

ω 

a  ω sin θ  θ  c sin θ  θ

 

 

ω  15.000

rad sec rad sec

Using the Euler identity to expand equation 7.13a for AA. Determine the magnitude, and direction.



 

   a ω2 cosθ  j  sinθ

AA  a  α sin θ  j  cos θ AA  ( 45.268  162.738i)

in sec


in

AA  168.917

sec 8.


AA  AA

2

at 2

θAA  105.5 deg

Use equations 7.12 to determine the angular accelerations of links 3 and 4.

 

 

 

 

A  c sin θ

B  b  sin θ

D  c cos θ

E  b  cos θ

A  0.731 in

B  0.000 in

D  0.169 in

E  1.500 in

 

2

 

2

 

2

2

 

2

 

2

 

C  a  α sin θ  a  ω  cos θ  b  ω  cos θ  c ω  cos θ C  7.308 in sec

2

 

 

F  a  α cos θ  a  ω  sin θ  b  ω  sin θ  c ω  sin θ F  1.687 in sec α 

9.

2

C D  A  F A  E  B D

α  0.000

rad sec

2

Use equation 7.13c to determine the acceleration of point B.

α 

C  E  B F A  E  B D

α  10.000

rad sec

2





 

   c ω2 cosθ  j  sinθ

AB  c α sin θ  j  cos θ AB  ( 45.268  162.738i)

in sec

The acceleration of pin B is

2


AB  AB

AB  168.917

in sec

at 2

θAB  105.5 deg

10. Use equations 7.32 to find the acceleration of the point C.



     2  Rca ω   cos θ  δ  j  sin θ  δ 

ACA  Rca α sin θ  δ  j  cos θ  δ

AC  AA  ACA AC  AC

AC  168.917

in sec

2

arg AC  105.545 deg




The linkage in Figure P7-5c has the dimensions and coupler angle given below. Find 3, AB, and AC for the position shown for VA = 10 in/sec and AA = 15 in/sec2 in the directions shown. Use the acceleration difference graphical method.

Given:

Link lengths and angles: Link 3 (A to B)

b  1.8 in

Coupler angle

θ  128  deg

Slider 4 angle

θ  59 deg

p  1.44 in

Angle BAC

δ  49 deg

Coupler point: Distance A to C Input slider motion Solution: 1.

VA  10 in sec

1

AA  15 in sec

2

See Figure P7-5c and Mathcad file P0715a.

In order to solve for the accelerations at points B and C, we will need 3. From the layout below and Problem 6-18, rad θ  128  deg ω  13.288 sec Direction of AB 0

0.5

1 in

Y

Direction of ABAt

C

B 4

Direction of ACAt

3 b

49.000°

128.000°

59.000°

AA VA

X A

2

2.


3.

For point B, this becomes: AB = AA + (ABAt + ABAn) , where AA  15.000 in sec

2

2

4.

(APt + APn) = (AAt + AAn) + (APAt + A

θAA  0  deg



θABAn  θ  180  deg


2

Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw AA at an angle of AA. From the tip of AA, draw ABAn at an angle of ABAn. Now that the vectors with known magnitudes are drawn, from the origin and the tip of ABAn, draw construction lines in the directions of AB and ABAt, respectively. The intersection of these two lines are the tips of ABAt, and AB.


SOLUTION MANUAL 7-15a-2 Y AA X

0

100 IN/S/S

Acceleration Scale n

ABA

9.126

8.638 AB t ABA

5.


ka  100 

in sec

6.

2

AB  9.126  ka

AB  912.6 in sec

ABAt  8.638  ka



2

Calculate 3 using equation 7.6. α 

7.

2

ABAt

α  479.9 rad sec

b

2

CW

For point C, equation 7.4 becomes: AC = AA + (ACAt + ACAn) , where 2



θAPAn  θ  δ  180  deg


ACAt  p  α


θAPAt   θ  δ  90 deg


2

2


8.


Repeat procedure of step 4 for the equation in step 7.

Y 7.215 AA

0

100 IN/S/S

X

Acceleration Scale

AC 170.609° t ACA

n

ACA

9.


ka  100 

in sec

AC  7.215  ka

2

AC  721.5 in sec

2





The linkage in Figure P7-5c has the dimensions and coupler angle given below. Find 3, AB, and AC for the position shown for VA = 10 in/sec and AA = 15 in/sec2 in the directions shown. Use an analytical method.

Given:

Link lengths and angles: Link 3 (A to B)

b  1.8 in

Coupler angle

θ  128  deg

Slider 4 angle

θ  59 deg

Coupler point: Distance A to C

Rca  1.44 in

Angle BAC

δ  49 deg VA  10 in sec

Input slider motion Solution: 1.

1

AA  15 in sec

2

See Figure P7-5c and Mathcad file P0715b.

Draw the mechanism to scale and define a vector loop using the fourbar slider-crank derivation in Section 7.3 as a model. 0

0.5

1 in

Y

C

B 4 R3 R4 3 b

49.000°

128.000°

59.000° VA

X

R2

2.

A

2

Write the vector loop equation, differentiate it, expand the result and separate into real and imaginary parts to solve for 3 and VB. R2  R3  R4

a e

j  θ

 b e

j  θ

 c e

j  θ

where a is the distance from the origin to point A, a variable; b is the distance from A to B, a constant; and c is the distance from the origin to point B, a variable. Angle 2 is zero, 3 is the angle that AB makes with the x axis, and 4 is the constant angle that slider 4 makes with the x axis. Differentiating, j  θ d  d  j  θ a  j  b  ω e   c   e dt  dt 




 

    d c   cosθ  j  sinθ

d a  b  ω sin θ  j  cos θ dt

 dt 



Separating into real and imaginary components and solving for 3. Note that dc/dt = VB and da/dt = VA ω 

3.

  b   sin θ  tan  θ  cos θ  VA tan θ

ω  13.288

rad sec

Differentiate the velocity equation, expand it and solve for 3 and AB. d

 a  b α  j  ej  θ  b ω 2 j  ej  θ  d2 c  ej  θ    2 2

2

dt

dt

Substituting the Euler equivalents, d

2

dt

2



 

  

a  b  α sin θ  j  cos θ 2

  

  

d

 b  ω  cos θ  j  sin θ

2

dt

2

 0

  

 

c  cos θ  j  sin θ

Separating into real and imaginary components and solving for 3 and AB. Note that d 2c/dt2 = AB and d 2a/dt2 = AA

α 

AB 

4.

 

 b  cos θ  θ 2

AA  sin θ  b  ω  sin θ  θ

 



α  479.924

sec

 

2

b  α cos θ  b  ω  sin θ

 

rad

AB  912.662

sin θ

2

in sec

2

Determine the acceleration of the coupler point C using equations 7.32.









  Rca ω2 cosθ  δ  j  sinθ  δ

ACA  Rca α sin θ  δ  j  cos θ  δ

ACA  ( 726.910  117.729j)

in sec

AA  AA AC  ( 711.910  117.729j)

2

AC  AA  ACA in sec

2

AC  721.579

in sec

2

arg AC  170.610 deg




For the linkage shown in Figure P7-6a, write the vector loop equations; differentiate them, and do a complete position, velocity, and acceleration analysis of the linkage. Measure the linkage geometry from the figure. Assume 2 = 10 rad/sec and 2 = 20 rad/sec2.

Given:

Link lengths and angles: Link 2 (O2 to A)

a  5.6 mm

Link 4 (O4 to C)

c  9.5 mm

Link 3 offset (A to B)

f  9.5 mm

Link 1 (O2 to O4)

d  38.8 mm

Link 2 position, velocity and acceleration Solution: 1.

θ  135  deg

ω  10

rad

α  20

sec

rad sec

2


Draw the mechanism to scale and define a vector loop using the fourbar inverted slider-crank derivation in Section 7.3 as a model. 9.5 y

B

B 135.00°

A

2

2

5.6 O2

2

3 R3

O2

3

1

R5 A R2

R1 4

4 R4

C

38.8 O4

O4

1

C 4 x

9.5

2.

Write the vector loop equation, expand the result and separate into real and imaginary parts to solve for 3 and b. R2  R5  R1  R4  R3 a e

j  θ

 f e

j  θ

 d  e

j  θ

 c e

j  θ

 b e

j  θ

where a is the distance from pivot O2 to point A, a constant; f is the distance from A to B, a constant; b is the distance from B to C, a variable; and c is the distance from pivot O4 to point C, a constant. Angle 2 is the input crank angle, 3 is the angle that BC makes with the x axis (measured from C), and 4 is the variable angle that link 4 makes with the x axis. The angular relationships are θ  θ

and

θ  θ  90 deg


  

   f  cosθ  j  sinθ  d  c cosθ  j  sinθ   b   cos θ  j  sin θ 

a  cos θ  j  sin θ

If we stipulate that f = c, this reduces to



  

   d  b cosθ  j  sinθ


Separating into real and imaginary components and solving for 3 and b.



 

θ  174.7 deg

θ  θ  90 deg

θ  84.7 deg

b 

3.

 

θ  atan2 a  cos θ  d a  sin θ

  sin θ

a  sin θ

b  42.9 mm

Differentiate the position equation to get the velocity equation. a e

Position (f and c terms cancel) a  j  ω e

Velocity

j  θ

 bdot e

j  θ

j  θ

 d  b  e

 b  j  ω e

j  θ

j  θ

Separating into real and imaginary components and solving for 3 and bdot. ω 





a  cos θ  θ b

 

 ω

ω  1.00

 

b  ω sin θ  a  ω sin θ

bdot 

 

rad sec mm

bdot  35.8

sec

cos θ 4.

Differentiate the velocity equation to get the acceleration equation. Velocity

a  j  ω e

Acceleration

a  j  α e

j  θ

j  θ

 bdot e

j  θ

 b  j  ω e

2 j  θ

2

 a  j  ω  e

j  θ

 bddot e

j  θ

 b  j  α e

 bdot j  ω e

j  θ

1

2

 b  j  ω  e

b

  a  α cos θ  θ  a  ω  sin θ  θ  bdot ω



α  9.50



2



rad sec

bddot 



2

     2    bdot ω cos θ  b   α cos θ  ω  sin θ

1

sin θ

bddot  316.0

 a   α cos θ  ω  sin θ

 

mm sec

2

2



2 j  θ

Separating into real and imaginary components and solving for 3 and bddot. α 

j  θ

   




For the linkage shown in Figure P7-6b, write the vector loop equations; differentiate them, and do a complete position, velocity, and acceleration analysis of the linkage. Use the linkage geometry given below. Assume 2 = 10 rad/sec and 2 = 20 rad/sec2.

Given:

Measured lengths and angles: Link 1

d  61.9 mm

Link 2

a  15.0 mm

Link 3

b  45.8 mm

Link 4

c  18.1 mm

Link 5

e  23.1 mm

Offset

f  2.6 mm

from x' axis

Crank angle:

θ  45 deg

Coordinate rotation angles


Global XY system

β  113.3  deg Local xy system to local x'y' system γ  90 deg Link 2 position, velocity and acceleration Solution: 1.

Global XY system to local x'y' system θ  68.3 deg

ω  10

rad sec

α  20

rad sec

2


Draw the mechanism to scale and label it. x'

Y

y

6 A 2 O2

C 3

1

5

2

X 23.300°

B

1

113.300° 4

4 y' O4 1

x

2.

This mechanism can be analyzed as a pin-jointed fourbar (links 1, 2, 3, and 4) and a fourbar slider-crank (links 1, 4, 5, and 6). Link 4 is the common, non-stationary link in the two branches and is the input to the slider-crank. Since the vector loops are the same as those defined for the pin-jointed and slider-crank linkages, we can use the equations derived in the text with slight modification of variable names for the slider-crank. There are three coordinate systems: the global XY frame, the local xy frame for the pin-jointed fourbar, and the local x'y' frame for the slider crank. Inputs to the fourbars must be in their respective local coordinates. All output will be given in local coordinate system.

3.


d a

K1  4.1267

K2 

d c

K2  3.4199


2

K3 

2


2

a b c d

2

K3  4.2110

2 a c

 

 

A  cos θ  K1  K2 cos θ  K3

A  0.8104

 

B  2  sin θ

B  1.8583

 

C  K1   K2  1   cos θ  K3 4.






2

θ  2  atan2 2  A B  5.

C  6.7034

B  4  A  C  360  deg

θ  125.692 deg


2

d

K5 

b

2

2

c d a b

 

2

K4  1.3515

2 a b

 

D  cos θ  K1  K4 cos θ  K5

D  7.4978

 

E  2  sin θ

E  1.8583

 

F  K1   K4  1   cos θ  K5 6.



8.

F  0.0160




2

θ  2  atan2 2  D E  7.

K5  4.2406

E  4  D F

  360deg

θ  0.955 deg


a  ω sin θ  θ  b sin θ  θ

 

 

ω  3.357

ω 

a  ω sin θ  θ  c sin θ  θ

 

 

ω  9.307

rad sec

rad sec

Using the Euler identity to expand equation 7.13a for AB. Determine the magnitude, and direction (in the local coordinate system).



 

   a ω2 cosθ  j  sinθ

AB  a  α sin θ  j  cos θ AB  ( 833  1283i)

mm sec


AB  1530

mm sec

9.


AB  AB

2

2

at

θAB  123.01 deg

Use equations 7.12 to determine the angular accelerations of links 3 and 4 for the crossed circuit.

 

 

 

 





A  14.700 mm

B  0.763 mm

D  10.560 mm

E  45.794 mm



 

2

 

2

 

2

 

2

 

2

 

C  a  α sin θ  a  ω  cos θ  b  ω  cos θ  c ω  cos θ 3

2

 

2

C  2.264  10 mm sec

 

F  a  α cos θ  a  ω  sin θ  b  ω  sin θ  c ω  sin θ F  18.160 mm sec α 

9.

2

C D  A  F

α  34.706

A  E  B D

rad sec

α 

2

C  E  B F

α  152.221

A  E  B D

rad sec

Use equation 7.13c to determine the acceleration of point C.



 

   c ω2 cosθ  j  sinθ

AC  c α sin θ  j  cos θ AC  ( 1323  2881i)

mm sec


AC  AC

2

AC  3170

The acceleration of pin C is

mm sec

at

2

θAC  114.7 deg

10. Determine 5 and the vertical distance from O4 to C (d') for the slider-crank using equations 4.16 and 4.17.

 c sin θ  β  f  π e  

θ  163.7 deg



d'  39.8 mm

θ  asin 



 

d'  c cos θ  β  e cos θ

11. Determine the angular velocity of link 5 using equation 6.22a.

ω 





c cos θ  β   ω e cos θ

 

ω  7.421

rad sec

12. Determine the angular acceleration of link 5 using equation 7.16d.

α 







2



2

 

c α cos θ  β  c ω  sin θ  β  e ω  sin θ

 

e cos θ

α  122.305

sec

13. Use equation 7.16e for the acceleration of pin C.





2





 

2

 

AC  c α sin θ  β  c ω  cos θ  β  e α sin θ  e ω  cos θ AC  162.9

in sec

2

rad

A negative sign means that AC is downward

2

2




For the linkage shown in Figure P7-6c, write the vector loop equations; differentiate them, and do a complete position, velocity, and acceleration analysis of the linkage. Use the linkage geometry given below. Assume 2 = 10 rad/sec and 2 = 20 rad/sec2.

Given:


a1  11.7 mm

Link 22

a2  20.0 mm

Link 3

b1  25.0 mm

Link 5

b2  25.9 mm

Offset

c1  3.7 mm

Offset'

c2  24.7 mm from x' axis

θ  61.3 deg

Crank angle:

Link 2 velocity and acceleration Solution: 1.

θ  13.3 deg ω  10

rad

α  20

sec

rad sec

2


Draw the mechanism to scale and label it. x'

24.7

Y 25.9 20.0 C

6 1

5 11.7

2

D 13.3°

X, y O2

A

1 3 61.3° 25.0 B

y'

1 4

x 3.7

2.

This mechanism can be analyzed as a fourbar slider-crank (links 1, 2, 3, and 4) and a fourbar slider-crank (links 1, 2, 5, and 6). Link 2 is the common, non-stationary link in the two branches and is the input to both slider-cranks. Since the vector loops are the same as those defined for the slider-crank linkage, we can use the equations derived in the text with slight modification of variable names. There are three coordinate systems: the global XY frame, the local xy frame for the first slider-crank, and the local x'y' frame for the second slider crank. Inputs to the fourbars must be in their respective local coordinates. All output will be given in local coordinate system.

3.

Determine  and the vertical distance from O2 to B (d1) for the first slider-crank using equations 4.16 and 4.17.



 a1 sin θ  c1  π b1  

θ  asin 

 

θ  164.8 deg

 

d1  a1 cos θ  b1 cos θ 4.


5.

d1  29.7 mm

   

a1 cos θ   ω b1 cos θ

ω  2.329

rad sec




 

   a1 ω2 cosθ  j  sinθ

AA  a1 α sin θ  j  cos θ AA  ( 767.114  913.889i)

mm sec

2

AA  1193


mm sec

6.

θAA  130.0 deg

at

2


α 

7.


AA  AA

 

  b1 cos θ 2

 

2

a1 α cos θ  a1 ω  sin θ  b1 ω  sin θ

α  36.4

rad sec

2


 

 

2

 

2

 

AB  a1 α sin θ  a1 ω  cos θ  b1 α sin θ  b1 ω  cos θ AB  659

mm sec

8.

A negative sign means that AB is upward

2

Determine  and the distance perpendicular to the x' axis from O2 to BD (d2) for the second slider-crank using equations 4.16 and 4.17.

 a2 sin θ  c2  π b2  

θ  asin 

 

θ  230.9 deg

 

d2  a2 cos θ  b2 cos θ 9.

d2  35.8 mm


   

a2 cos θ   ω b2 cos θ

ω  11.915

rad sec

10. Using the Euler identity to expand equation 7.15b for AC, determine its magnitude, and direction.



 

   a2 ω2 cosθ  j  sinθ

AC  a2 α sin θ  j  cos θ AC  ( 2038.378  70.828i )

mm sec

2

AC  AC




AC  2040

The acceleration of pin C is

mm sec

at deg θAC  178.0

2


α 

 

  b2 cos θ 2

2

 

a2 α cos θ  a2 ω  sin θ  b2 ω  sin θ

α  179.0

rad sec

12. Use equation 7.16e for the acceleration of pin D.

 

2

 

 

2

 

AD  a2 α sin θ  a2 ω  cos θ  b2 α sin θ  b2 ω  cos θ AD  7956

mm sec

2

A negative sign means that AD is downward and to the right

2




For the linkage shown in Figure P7-6d, write the vector loop equations; differentiate them, and do a complete position, velocity, and acceleration analysis of the linkage. Use the linkage geometry given below. Assume 2 = 10 rad/sec and 2 = 20 rad/sec2.

Given:


a1  15.0 mm

Link 22

a2  15 mm

Link 3

b1  40.9 mm

Link 5

b2  44.7 mm

Offset

c1  1.0 mm

Offset'

c2  0.0 mm

θ  65.8 deg

Crank angle:

Solution: 1.

θ  24.2 deg ω  10

Link 2 velocity and acceleration

from x' axis

rad

α  20

sec

rad sec

2


Draw the mechanism to scale and label it. Y, x 1.0

1

4

B 40.9 65.8°

3

44.7 A 2

y

24.2°

5

6 C

O2

X

1

1 15.0

2.

This mechanism can be analyzed as a fourbar slider-crank (links 1, 2, 3, and 4) and a fourbar slider-crank (links 1, 2, 5, and 6). Link 2 is the common, non-stationary link in the two branches and is the input to both slider-cranks. Since the vector loops are the same as those defined for the slider-crank linkage, we can use the equations derived in the text with slight modification of variable names. There are two coordinate systems: the global XY frame and the local xy frame for the first slider-crank. Inputs to the fourbars must be in their respective local coordinates. All output will be given in local coordinate system.

3.

Determine  and the vertical distance from O2 to B (d1) for the first slider-crank using equations 4.16 and 4.17.

 a1 sin θ  c1  π b1  

θ  asin 

 

 

d1  a1 cos θ  b1 cos θ

θ  198.1 deg d1  45.0 mm


4.


5.


   

a1 cos θ   ω b1 cos θ

ω  1.581

rad sec




 

   a1 ω2 cosθ  j  sinθ

AA  a1 α sin θ  j  cos θ mm

AA  ( 341.249  1491.157i )

sec

2

AA  1530


mm sec

6.

θAA  102.9 deg

at

2


α 

7.


AA  AA

 

  b1 cos θ 2

 

2

a1 α cos θ  a1 ω  sin θ  b1 ω  sin θ

α  37.5

rad sec

2


 

2

 

 

2

 

AB  a1 α sin θ  a1 ω  cos θ  b1 α sin θ  b1 ω  cos θ AB  37.5

mm sec

8.

A positive sign means that AB is upward

2

Determine  and the distance perpendicular to the x' axis from O2 to BD (d2) for the second slider-crank using equations 4.16 and 4.17.

 a2 sin θ  c2  π b2  

θ  asin 

 

θ  172.1 deg

 

d2  a2 cos θ  b2 cos θ 9.

d2  58.0 mm


   

a2 cos θ   ω b2 cos θ

ω  3.090

rad sec

10. Determine the angular acceleration of link 5 using equation 7.16d. α 

 

  b2 cos θ 2

2

 

a2 α cos θ  a2 ω  sin θ  b2 ω  sin θ

α  6.4

sec


 

2

 

 

2

 

AC  a2 α sin θ  a2 ω  cos θ  b2 α sin θ  b2 ω  cos θ AC  1875

mm sec

2

rad

A negative sign means that AC is to the left

2




Figure P7-7 shows a sixbar linkage with the dimensions and crank angle given below. Find the angular acceleration of link 6 if 2 is a constant 1 rad/sec.

Given:


a  1.00 in

Link 3

b  5.00 in

Link 6

d  3.00 in

Distance DB

LDB  1.50 in

Link 2 position, velocity, and acceleration:θ  45 deg Solution: 1.

c  0.0 in

Offset:

ω  1 

rad

α  0 

sec

rad sec

2


Links 1, 2, 3, and 5 constitute a fourbar slider-crank. In order to solve for the accelerations at points B, C and D, we will need 3, and 3. From the graphical position solution below, θ  171.870  deg ω 


   


ω  0.143 rad sec

1

CW

Axis of Slip Axis of Transmission 15.566°

Y

O6 6

B

171.870°

2

4

D

3

C 5

X

O2

2.

The graphical solution for accelerations of pin-jointed linkages uses equation 7.4: (APt + APn) = (AAt + AAn) + (APAt + APAn)

3.

For point C, this becomes: AC = (ABt + ABn) + (ACBt + ACBn) , where 2

2

ABn  a  ω

ABn  1.000 in sec

θABn  θ  180  deg

θABn  225.000 deg

ABt  a  α


ACBn  b  ω

θACBn  θ

2

2

ACBn  0.102 in sec

2

θACBn  171.870 deg


4.


Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn. From the tip of ABn, draw ACBn at an angle of ACBn. Now that the vectors with known magnitudes are drawn, from the origin and the tip of ACBn, draw construction lines in the directions of AC (horizontal) and ACBt, respectively. The intersection of these two lines are the tips of ACBt, and AC. 2.837 Y AC X t ACB

2.799

0

0.25 IN/S/S


AB n

ACB

5.


in

ka  0.25

sec

6.

2

AC  2.837  ka

AC  0.709 in sec

ACBt  2.799  ka

ACBt  0.700 in sec


2

Calculate 3 using equation 7.6. α 

7.

2

ACBt

α  0.140 rad sec

b

2

CCW

In order to solve for the acceleration at points D, we will need 6, and 6. From the graphical position solution above, θ  15.566 deg  180  deg

θ  195.566 deg

From the vector diagram in Figure P7-7, VD  0.40

in sec

Vslip  0.65 8.

and

ω 

VD

ω  0.133

d

in sec

For point D, use equation 7.19, referenced to point B instead of O2: (ADt + ADn)= AB + (ADBt + ADBn + ADBcor + ADBslip) , where ADn  d  ω

2

θADn  θ  180  deg

ADn  0.053 in sec

θADn  15.566 deg

2

rad sec


AB  a  ω


2

AB  1.000 in sec

θAB  θ  180  deg

θAB  225.000 deg

ADBt  LDB α

ADBt  0.210 in sec

θADBt  θ  90 deg

θADBt  81.870 deg

2

9.

2

2

2

ADBn  LDB ω

ADBn  0.031 in sec

θADBn  θ

θADBn  171.870 deg

ADBcor  2  Vslip  ω

ADBcor  0.186 in sec

θADBcor  θ  90 deg

θADBcor  261.870 deg

2

Repeat procedure of step 4 for the equation in step 8. Y ADn 0

X

0.25 IN/S/S

Acceleration Scale 3.483 n DB

A

t ADB cor ADB

AB

ADt

slip ADB

10. From the graphical solution above, Acceleration scale factor

ka  0.25

in sec

ADt  3.483  ka

2

ADt  0.871 in sec

2


The angular acceleration of link 6 is α 

ADt d

α  0.290

rad sec

2

CCW




The linkage in Figure P7-8a has the dimensions and crank angle given below. Find 4, AA, and AB in the global coordinate system for the position shown for 2 = 15 rad/sec clockwise (CW) and 2 = 25 rad/sec2 CCW. Use the acceleration difference graphical method.

Given:

Solution: 1.


a  116  mm

Link 3 (A to B)

b  108  mm

Link 4 (B to O4)

c  110  mm

Link 1 (O2 to O4)

d  174  mm

Crank angle:

θ  62 deg


ω  15 rad sec


α  25 deg

Local xy system 1

α  25 rad sec

2



Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest. Y

Direction of AAt

y A

37.000° 70.133°

2

3

O2 X 2

22.319° B

Direction of ABAt 4

Direction of ABt

O4

x

2.

In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above (in the local xy coordinate system), θ  180  deg  70.133 deg  α

θ  275.133 deg

θ  180  deg  22.319 deg  α

θ  182.681 deg

Using equation (6.18),

3.

 

 

ω  13.869 rad sec

 

 

ω  8.654 rad sec

ω 

a  ω sin θ  θ  b sin θ  θ

ω 

a  ω sin θ  θ  c sin θ  θ


1

1



4.


For point B, this becomes: (ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where ABn  c ω

2

ABn  8237.9 mm sec

θABn  θ  180  deg

θABn  2.681 deg

2

AAn  a  ω

AAn  26100.0 mm sec

θAAn  θ  180  deg

θAAn  242.000 deg

AAt  a  α

AAt  2900.0 mm sec

θAAt  θ  90 deg

θAAt  152.000 deg

2

5.

2

2

2


ABAn  20773 mm sec

θABAn  θ  180  deg


2

Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn +  and AAn at an angle of AAn + . From the tip of AAn, draw AAt at an angle of AAt + . From the tip of AAt, draw ABAn at an angle of BAn + . Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABt, and ABAt.

n

ABA

41.409

Y

y

t

ABA

X 131.308

AB x 149.340°

16.401° n

AB t

AB

AA

t

AA

6.

0

200 mm/s/s


AA


ka  200 

mm sec

2

AA  131.308  ka

AA  26262 mm sec

AB  41.409 ka

AB  8282 mm sec

2

2

at an angle of -149.34 deg at an angle of -16.40 deg



PROBLEM 7-22 The linkage in Figure P7-8a has the dimensions and crank angle given below. Find 4, AA, and

Statement:

AB in the global coordinate system for the position shown for 2 = 15 rad/sec clockwise (CW) and 2 = 25 rad/sec2 CCW. Use an analytical method. Given:

Link lengths:

Solution: 1. 2.

Link 2 (O2 to A)

a  116  mm

Link 3 (A to B)

b  108  mm

Link 4 (B to O4)

c  110  mm

Link 1 (O2 to O4)

d  174  mm

Crank angle:

θ  62 deg


ω  15 rad sec


β  25 deg


α  25 rad sec

2





d

K2 

a

K1  1.5000 2

K3 

y A

d c 2

K2  1.5818 2

2

a b c d

3 2

O2

2

X

K3  1.7307

2 a c

 

2

 

A  cos θ  K1  K2 cos θ  K3

B

 

B  2  sin θ

O4

4

 

C  K1   K2  1   cos θ  K3 A  0.0424 3.

B  1.7659






2

θ  2  atan2 2  A B  4.

x

C  2.0186

B  4 A  C



θ  182.681 deg


2

d

K5 

b

2

2

c d a b

 

2 a b

 

D  cos θ  K1  K4 cos θ  K5

 

E  2  sin θ

2

K4  1.6111 D  2.0021 E  1.7659

 

F  K1   K4  1   cos θ  K5

F  0.0589

K5  1.7280


5.







2

θ  2  atan2 2  D E  6.

7.

E  4  D F



θ  275.133 deg


 

 

ω  13.869

 

 

ω  8.654

ω 

a  ω sin θ  θ  b sin θ  θ

ω 

a  ω sin θ  θ  c sin θ  θ

rad sec

rad sec

Using the Euler identity to expand equation 7.13a for AA. Determine the magnitude, and direction (in the global coordinate system).



 

   a ω2 cosθ  j  sinθ

AA  a  α sin θ  j  cos θ mm

AA  ( 14814  21683i)

sec

AA  26261


mm sec

8.

θAA  arg AA  β

AA  AA

2

at

2

θAA  149.3 deg


 

 

 

 





A  5.145 mm

B  107.567 mm

D  109.880 mm

E  9.662 mm

 

2

 

2

 

2

 

2

 

2

 


2

 

2

C  2.490  10 mm sec

 

F  a  α cos θ  a  ω  sin θ  b  ω  sin θ  c ω  sin θ 3

F  1.380  10 mm sec α 

9.

C D  A  F

2

α  231.119

A  E  B D

rad sec

α 

2

C  E  B F A  E  B D

α  7.768

sec

Use equation 7.13c to determine the acceleration of point B for the crossed circuit.



 

   c ω2 cosθ  j  sinθ

AB  c α sin θ  j  cos θ AB  ( 8189  1239i)

mm sec

2


θAB  arg AB  β

AB  AB AB  8282

mm sec

2

at

θAB  16.4 deg

rad

(Global)

2




The linkage in Figure P7-8a has the dimensions and crank angle given below. Find and plot 4, AA, and AB in the local coordinate system for the maximum range of motion that this linkage allows if 2 = 15 rad/sec CW and 2 = 25 rad/sec2 CCW.

Given:


a  116  mm

Link 4 (B to O4)

c  110  mm

Solution: 1.

d  174  mm

Link 1 (O2 to O4)

ω  15 rad sec


b  108  mm

Link 3 (A to B) 1

α  25 rad sec

2


Draw the linkage to scale and label it. Y y A

2 3 2

O2

X 2

B O4

4

x

2.


arg1 

2

2



2 a d 2

arg2 

2

a d b c

2

2

a d b c 2 a d

2



b c

arg1  1.083

a d b c

arg2  0.094

a d



The other toggle angle is the negative of this. Thus, θ  θ2toggle  0.5 deg θ2toggle  1  deg  θ2toggle  0.5 deg 3.


d a

K1  1.5000

K2 

d c

K2  1.5818



2

2

a b c d

K3 

2

K3  1.7307

2 a c

 

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

C θ  K1   K2  1   cos θ  K3 4.




 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


d

K4 

K5 

b

 

2

2

c d a b

2

K4  1.6111

2 a b

 

K5  1.7280

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 6.


 





 

 

θ θ  2   atan2 2  D θ E θ  7.


 

a  ω

 

a  ω

ω θ 

ω θ 

8.

 2  4 Dθ F θ 

E θ

b

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 





Using the Euler identity to expand equation 7.13a for AA. Determine the x and y components in the local coordinate system.

 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ

 

  

AAx θ  Re AA θ 9.

 

  

AAy θ  Im AA θ

Use equations 7.12 to determine the angular acceleration of link 4.

 

  

B θ  b  sin θ θ

 

  

E θ  b  cos θ θ

A θ  c sin θ θ

D θ  c cos θ θ

 

  

 

  

 

 

2

 

 2

    c ωθ2 cosθθ

 

 

2

 

 2     c ωθ2 sinθθ

C θ  a  α sin θ  a  ω  cos θ  b  ω θ  cos θ θ F θ  a  α cos θ  a  ω  sin θ  b  ω θ  sin θ θ


 

α θ 


        A  θ  E θ  B θ  D θ C θ  E θ  B θ  F θ

10. Use equation 7.13c to determine the acceleration of point B. Determine the x and y components in the local coordinate system.

 

 

    j  cosθθ  c ωθ2 cosθθ  j  sinθθ

AB θ  c α θ  sin θ θ

 

  

ABx θ  Re AB θ

 

  

ABy θ  Im AB θ

11. Plot the angular acceleration for link 4 and the acceleration components for pins B and C.

Angular Accel., rad/sec^2

ANGULAR ACCELERATION OF LINK 4

0  1000  2000  3000  4000  100

 50

0

50

100

Crank Angle, deg

ACCELERATION OF POINT A 30

Acceleration, m/sec^2

20 10 0  10  20  30  100

 50

0 Crank Angle, deg

x component y component

50

100



ACCELERATION OF POINT B 300


200 100 0  100  200  300  100

 50

0 Crank Angle, deg

x component y component

50

100




The linkage in Figure P7-8b has the dimensions and crank angle given below. Find 4, AA, and

Given:

AB in the global coordinate system for the position shown for  = 20 rad/sec CCW. Use the acceleration difference graphical method. Link lengths:

Solution: 1.

Link 2 (A to B)

a  40 mm

Link 3 (B to C)

b  96 mm

Link 4 (C to D)

c  122  mm

Link 1 (A to D)

d  162  mm

Crank angle:

θ  93 deg


ω  20 rad sec


α  36 deg


α  0  rad sec

2



Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest. Direction of AAt

Direction of ABt

Y 2

Direction of ABAt

y

B

57.0° 3

A

2

X O2

36.0° 4

O4 x

2.

In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above (in the local xy coordinate system), θ  31.486 deg

θ  132.406  deg

Using equation (6.18), ω 

a  ω sin θ  θ  b sin θ  θ

 

 

ω  5.388 rad sec

ω 

a  ω sin θ  θ  c sin θ  θ

 

 

ω  5.870 rad sec

1

1


3.


4.


2

ABn  4203 mm sec

2



θABn  θ  180  deg

θABn  47.594 deg

2

AAn  a  ω

AAn  16000 mm sec

θAAn  θ  180  deg

θAAn  273.000 deg

AAt  a  α

AAt  0.000 mm sec 2

5.

2

2 2



θABAn  θ  180  deg

θABAn  148.514 deg

Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn and AAn at an angle of AAn . From the tip of AAn, draw AAt at an angle of AAt . From the tip of AAt, draw ABAn at an angle of BAn . Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABt, and ABAt.

Y

0

500 mm/s/s y

Acceleration Scale X 24.340

n

AB

AB x

t

AB

AA

t ABA n

22.244

ABA 6.


ka  500 

mm sec

7.

2 2

AA  AAn

AA  16000 mm sec

AB  24.340 ka

AB  12170 mm sec

ABt  22.244 ka

ABt  11122 mm sec

2

at an angle of 273.0 deg (Global) at an angle of 242.610 deg

2

Calculate 4 using equation 7.6. α 

ABt c

α  91.2 rad sec

2

CCW




The linkage in Figure P7-8b has the dimensions and crank angle given below. Find , AA, and

Given:

AB in the global coordinate system for the position shown for  = 20 rad/sec CCW constant. Use an analytical method. Link lengths:

Solution: 1.

Link 2 (A to B)

a  40 mm

Link 3 (B to C)

b  96 mm

Link 4 (C to D)

c  122  mm

Link 1 (A to D)

d  162  mm

Crank angle:

θ  93 deg


ω  20 rad sec


β  36 deg


α  0  rad sec




y

Y 2.

2


d

K2 

a

K1  4.0500 2

K3 

2

d

2

2

57.0°

2

c

B

3 X

O2

K2  1.3279

a b c d

A

36.0° 4

2

K3  3.4336

2 a c

 

 

A  cos θ  K1  K2 cos θ  K3

 

O4

B  2  sin θ

x

 

C  K1   K2  1   cos θ  K3 A  0.5992 3.

B  1.9973

C  7.6054

Use equation 4.10b to find values of 4 for the open circuit in the local xy coordinate system.





2

θ  2  atan2 2  A B  4.

B  4 A  C

θ  132.386 deg


2

d

K5 

b

2

2

c d a b

 

2 a b

 

D  cos θ  K1  K4 cos θ  K5

 

E  2  sin θ

2

K4  1.6875

K5  2.8875

D  7.0782 E  1.9973

 

F  K1   K4  1   cos θ  K5 5.

  2 π

F  1.1265

Use equation 4.13 to find values of 3 for the open circuit in the local xy coordinate system.







2

θ  2  atan2 2  D E  6.

7.

E  4  D F

  2 π

θ  31.504 deg


a  ω sin θ  θ  b sin θ  θ

 

 

ω  5.385

ω 

a  ω sin θ  θ  c sin θ  θ

 

 

ω  5.868

rad sec

rad sec




 

   a ω2 cosθ  j  sinθ

AA  a  α sin θ  j  cos θ AA  ( 837  15978i)

mm sec

AA  16000


mm sec

8.

θAA  arg AA  β

AA  AA

2

at

2

θAA  123.0 deg


 

 

 

 





A  90.111 mm

B  50.166 mm

D  82.244 mm

E  81.850 mm

 

2

 

2

 

2

 

2

 

2

 


2

 

2

C  4.368  10 mm sec

 


F  1.433  10 mm sec α 

9.

C D  A  F A  E  B D

2

α  81.037

rad sec

α 

2

C  E  B F A  E  B D

α  93.586

sec

Use equation 7.13c to determine the acceleration of point B for the open circuit.



 

   c ω2 cosθ  j  sinθ

AB  c α sin θ  j  cos θ

AB  ( 5601  10800i)

mm sec


2

θAB  arg AB  β

AB  AB AB  12166

mm sec

2

at

θAB  153.4 deg

rad

(Global)

2




Given:

The linkage in Figure P7-8b has the dimensions and crank angle given below. Find and plot , AA, and AB in the local coordinate system for the maximum range of motion that this linkage allows if  = 20 rad/sec CCW constant. Link lengths: Link 2 (A to B)

a  40 mm

Link 3 (B to C)

b  96 mm

Link 4 (C to D)

c  122  mm

Link 1 (A to D)

d  162  mm

ω  20 rad sec

Input crank angular velocity Solution: 1. 2.

1

α  0  rad sec



y

Y

The range of 2 for this Grashof crank-rocker is:

2

θ  0  deg 1  deg  360  deg

A

57.0°

2 3.

2


d

K2 

a

K1  4.0500 2

2

a b c d

O2

36.0° 4

d c

2

K3  3.4336

2 a c

 

X

K2  1.3279

2

K3 

B

3

 

O4 x

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 

4.

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ




 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


 

2

d

K5 

b

2

2

c d a b

2

2 a b

 

K4  1.6875

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 6.


 





 

 

θ θ  2   atan2 2  D θ E θ 

 2  4 Dθ F θ 

E θ

K5  2.8875


7.


 

a  ω

 

a  ω

ω θ 

ω θ 

8.


b

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 





Using the Euler identity to expand equation 7.13a for AA. Determine the x and y components in the local coordinate system.

 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ

 

  

 

AAx θ  Re AA θ 9.

  



 

  

B θ  b  sin θ θ

 

  

E θ  b  cos θ θ



 

  

 

  

 

 

2

 

 2

    c ωθ2 cosθθ

 

 

2

 

 2     c ωθ2 sinθθ


 

α θ 


10. Use equation 7.13c to determine the acceleration of point B. Determine the x and y components in the local coordinate system.

 

 


AB θ  c α θ  sin θ θ

 

  

 

ABx θ  Re AB θ

  

ABy θ  Im AB θ

11. Plot the angular acceleration for link 4 and the acceleration components for pins A and B. ANGULAR ACCELERATION OF LINK 4 Angular Accel., rad/sec^2

200 100 0  100  200

0

45

90

135

180

225

Crank Angle, deg

270

315

360



ACCELERATION OF POINT A


20

10

0

 10

 20

0

45

90

135

180

225

270

315

360

Crank Angle, deg x component y component



10

0

 10

 20

0

45

90

135

180


225

270

315

360




The offset crank-slider linkage in Figure P7-8f has the dimensions and crank angle given below. Find AA, and AB in the global coordinate system for the position shown for  = 25 rad/sec CW, constant. Use the acceleration difference graphical method.

Given:

Link lengths: Link 2 (D to E)

a  63 mm

Offset

c  52 mm

Link 2 position, velocity, and acceleration: θ  141  deg

1.

ω  25

rad sec

α  0 

α  90 deg

Coordinate rotation angle: Solution:

b  130  mm

Link 3 (E to F)

rad sec


In order to solve for the accelerations at point F, we will need 3, and 3. From the graphical position solution below (in the local coordinate system), θ  44.828 deg ω 


   


ω  13.276 rad sec

1

Direction of AB 4 B

Direction of ABAt

1 3 44.828°

Direction of AAt

Y

A 2 51.000°

O2

X, y

x

2.

The graphical solution for accelerations uses equation 7.4: (APt + APn) = (AAt + AAn) + (APAt + APAn)

3.

For point F, this becomes: AF = (AEt + AEn) + (AFEt + AFEn) , where (angles in the global coordinate system) AEn  a  ω

2

AEn  39375  mm sec

θAEn  θ  α  180  deg

θAEn  231.000  deg

AEt  a  α

AEt  0.000  mm sec 2

2

2

AFEn  b  ω

AFEn  22911  mm sec

θAFEn  θ  α

θAFEn  45.172 deg

2

2


4.


Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw AEn at an angle of AEn. From the tip of AEn, draw AFEn at an angle of AFEn. Now that the vectors with known magnitudes are drawn, from the origin and the tip of AFEn, draw construction lines in the directions of AF (vertical) and AFEt, respectively. The intersection of these two lines are the tips of AFEt, and AF. Y

X,y

0

500 mm/s/s

Acceleration Scale x 76.546 n

AE AF t AFE n

AFE

5.


ka  500 

mm sec

AF  76.546 ka

2

AF  38273  mm sec

2

at an angle of -90 deg (global)




The offset crank-slider linkage in Figure P7-8f has the dimensions and crank angle given below. Find AA, and AB in the global coordinate system for the position shown for  = 25 rad/sec CW, constant. Use an analytical method.

Given:


a  63 mm

Offset

c  52 mm

Link 2 position, velocity, and acceleration: θ  141  deg

1.

ω  25

rad

rad

α  0 

sec

sec

α  90 deg


b  130  mm

Link 3 (A to B)

2


Draw the linkage to a convenient scale. 4

2.

Determine 3 (in the local coordinate system) and d using equations 4.16 for the crossed circuit.

 a  sin θ  b 

θ  asin

 

c

 

 

d  141.160  mm

Y 52.000


A 2 51.000°

   

a cos θ ω    ω b cos θ 4.

3

θ  44.828 deg

d  a  cos θ  b  cos θ 3.

B 1

ω  13.276

rad O2

sec

Using the Euler identity to expand equation 7.15b for AB, determine its magnitude, and direction (global).



 

X, y

x

   a ω2 cosθ  j  sinθ

AA  a  α sin θ  j  cos θ

AA  ( 30600.122  24779.490i) 

mm sec

AA  39375 


mm sec

5.

at

2

θAA  129.0  deg


α 

6.

θAA  arg AA  α

AA  AA

2

 

  b  cos θ 2

2

 


α  93.574

sec


 

2

 

 

2

 

AB  a  α sin θ  a  ω  cos θ  b  α sin θ  b  ω  cos θ AB  38274 

mm sec

2

rad

A positive sign means that AB is downward

2

DESIGN OF MACHINERY



The offset crank-slider linkage in Figure P7-8f has the dimensions and crank angle given below. Find and plot AA, and AB in the global coordinate system for the maximum range of motion that this linkage allows if  = 25 rad/sec CW, constant..

Given:


a  63 mm

Offset

c  52 mm ω  25

Link 2 velocity, and acceleration:

1. 2.

rad

rad

α  0 

sec

sec

2

α  90 deg



Y

Draw the linkage to a convenient scale.

4

Determine the range of motion for this slider-crank linkage.

B

1 3

θ  0  deg 2  deg  360  deg 3.


 

 a  sin θ  c   b  

2


 

ω θ 

5.

A

52.000

θ θ  asin 4.

b  130  mm

Link 3 (A to B)

  a   ω b cos θ θ 

O2

cos θ

2

X, y

x

Using the Euler identity to expand equation 7.15b for AA, determine its X and Y components in the global coordinate system.

 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ

 

  

  

 

  

  

AAX θ  Re AA θ  cos α  Im AA θ  sin α AAY θ  Re AA θ  sin α  Im AA θ  cos α 6.


 

α θ  7.

 

   2    b  cos θ θ  2

a  α cos θ  a  ω  sin θ  b  ω θ  sin θ θ


 

2     2  b  α θ  sin θ θ   b  ω θ  cos θ θ 

AB θ  a  α sin θ  a  ω  cos θ 

8.

Plot the components of AA and the magnitude of AB. A positive value for AB is downward, a negative value, upward.

DESIGN OF MACHINERY


PIN A ACCELERATION COMPONENTS


40

20

0

 20

 40

0

45

90

135

180

225

270

315

360


PIN B ACCELERATION MAGNITUDE 100


50

0

 50

 100

0

45

90

135

180

Crank Angle, deg

225

270

315

360




The linkage in Figure P7-8d has the dimensions and crank angle given below. Find AA, AB, and

Given:

Abox in the global coordinate system for the position shown for  = 30 rad/sec CW, constant . Use the acceleration difference graphical method. Link lengths:

Solution: 1.

Link 2 (O2 to A)

a  30 mm

Link 3 (A to B)

b  150  mm

Link 4 (O4 to B)

c  30 mm

Link 1 (O2 to O4)

d  150  mm

Crank angle:

θ  58 deg


ω  30 rad sec


α  0  rad sec

2


Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest.

Vbox 1

Direction of ABAt

A 2

58°

B

3

58° Direction of ABt

O4

O2

4

Direction of AAt 2.

In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above (in the global XY coordinate system), θ  0  deg

θ  0.000 deg

θ  θ

θ  58.000 deg

This is a special-case Grashof in the parallelogram configuration. Therefore, ω  0.0 rad sec

1

ω  ω (APt + APn) = (AAt + AAn) + (APAt + APAn)

3.


4.


2


θABn  θ  180  deg 2

θABn  238.000 deg

AAn  a  ω


θAAn  θ  180  deg

θAAn  238.000 deg

AAt  a  α

AAt  0.0 mm sec 2


2


2

2

2


5.


In this case, there is no need to draw an acceleration diagram since AAt and ABAn are zero. This means that ABAt and ABt will also be zero and AA = AAn, AB = ABn.

6.

Since ABAt = 0, 3 = 0 and, since AB = AA, 4 = 2= 0.

7.

For the box, equation 7.4 becomes: Abox = (AAt + AAn) + (AboxAt + AboxAn) , where AboxAt and AboxAn are both zero since 3 and 3 are zero. Therefore, Abox = AAx.




The linkage in Figure P7-8d has the dimensions and crank angle given below. Find AA, AB, and

Given:

Abox in the global coordinate system for the position shown for  = 30 rad/sec CW, constant. Use an analytical method. Link lengths:

Solution: 1.

Link 2 (O2 to A)

a  30 mm

Link 3 (A to B)

b  150  mm

Link 4 (O4 to B)

c  30 mm

Link 1 (O2 to O4)

d  150  mm

Crank angle:

θ  58 deg


ω  30 rad sec

1

α  0  rad sec

2



E Vbox 1

A

58°

2

3

58° B

O2

2.

O4

4


d

K2 

a

K1  5.0000 2

K3 

d c

K2  5.0000 2

2

a b c d

2

K3  1.0000

2 a c

 

 

A  cos θ  K1  K2 cos θ  K3

 

B  2  sin θ

 

C  K1   K2  1   cos θ  K3 A  6.1197 3.

B  1.6961






θ  2  atan2 2  A B  θ  θ  360  deg 4.

C  2.8205

2

B  4 A  C



θ  302.000 deg θ  58.000 deg



K4 


2

d b

2

2

c d a b

K5 

2

K4  1.0000

2 a b

 

 

D  cos θ  K1  K4 cos θ  K5

D  8.9402

 

E  2  sin θ

E  1.6961

 

F  K1   K4  1   cos θ  K5 5.

F  0.0000






2

θ  2  atan2 2  D E 

E  4  D F



θ  360.000 deg

θ  θ  360  deg 6.

7.

K5  5.0000

θ  0.000 deg


 

 

ω  0.000

 

 

ω  30.000

ω 

a  ω sin θ  θ  b sin θ  θ

ω 

a  ω sin θ  θ  c sin θ  θ

rad sec rad sec




 

   a ω2 cosθ  j  sinθ

AA  a  α sin θ  j  cos θ AA  ( 14308  22897i)

mm sec


AA  27000

mm sec

8.


AA  AA

2

θAA  122.0 deg

at

2


 

 

 

 





A  1.002 in

B  0.000 in

D  0.626 in

E  5.906 in

 

2

 

2

 

2

2

 

2

 

2

 

C  a  α sin θ  a  ω  cos θ  b  ω  cos θ  c ω  cos θ C  0.000 mm sec

2

 

 

F  a  α cos θ  a  ω  sin θ  b  ω  sin θ  c ω  sin θ F  0.000 mm sec α  9.

C D  A  F A  E  B D

2

α  0.000

rad sec

2


α 

C  E  B F A  E  B D

α  0.000

rad sec

2





 

   c ω2 cosθ  j  sinθ

AB  c α sin θ  j  cos θ AB  ( 14308  22897i)

mm sec


2


AB  AB AB  27000

mm sec

at

2

θAB  122.0 deg

10. This is a special case Grashof in the parallelogram configuration. All points on link 3 have the same velocity and acceleration. The acceleration of the box will be equal to the X-component of the acceleration of any point on link 3. Abox  Re AA The acceleration of the box is Abox  14308

mm sec

2

a negative sign means Abox is to the left




The linkage in Figure P7-8d has the dimensions and crank angle given below. Write a computer program or use an equation solver to find and plot AA, AB, and Abox in the global coordinate system for the maximum range of motion that this linkage allows if  = 30 rad/sec CW, constant .

Given:


a  30 mm

Link 3 (A to B)

Link 4 (O4 to B)

c  30 mm

Link 1 (O2 to O4)

ω  30 rad sec


1

α  0  rad sec

b  150  mm d  150  mm 2



Vbox 1

A

3

58°

2

58° B

O2

2.

O4

4


d

K2 

a

K1  5.0000 2

2

a b c d

2

K3  1.0000

2 a c

 

c

K2  5.0000

2

K3 

d

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

C θ  K1   K2  1   cos θ  K3 3.




 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  4.

B θ


 

2

d

K5 

b

 

2

2

c d a b 2 a b

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

2

K4  1.0000

K5  5.0000



 

 

F θ  K1   K4  1   cos θ  K5 5.




 



 

 2  4 Dθ F θ 

 

θ θ  2   atan2 2  D θ E θ  6.


 

a  ω

 

a  ω

ω θ 

ω θ 

7.

E θ

b

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 





Using the Euler identity to expand equation 7.13a for AA. Determine the XY components.

 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ

 

  

AAX θ  Re AA θ 8.

 

  

AAY θ  Im AA θ


 

  

B' θ  b  sin θ θ

 

  

E' θ  b  cos θ θ

A' θ  c sin θ θ

D' θ  c cos θ θ

 

  

 

  

 

 

2

 

 2     c ωθ2 cosθθ

 

 

2

 

 2     c ωθ2 sinθθ

C' θ  a  α sin θ  a  ω  cos θ  b  ω θ  cos θ θ F' θ  a  α cos θ  a  ω  sin θ  b  ω θ  sin θ θ

 

α θ 

9.

        A'  θ  E' θ  B' θ  D' θ C' θ  D' θ  A' θ  F' θ

 

α θ 

        A'  θ  E' θ  B' θ  D' θ C' θ  E' θ  B' θ  F' θ


 

       j  cosθθ  2  c ω θ   cos θ θ   j  sin θ θ  

AB θ  c α θ  sin θ θ

 

  

ABX θ  Re AB θ

 

  

ABY θ  Im AB θ

10. This is a special case Grashof in the parallelogram configuration. All points on link 3 have the same velocity an acceleration. The acceleration of the box will be equal to the X-component of the acceleration of any point on link 3.

 

  

Abox θ  Re AA θ

11. Plot the accelerations over the range: θ  0  deg 2  deg  360  deg



ACCELERATION OF PINS A & B


40

20

0

 20

 40

0

45

90

135

180

225

270

315

360

Input Angle, deg x component y component

ACCELERATION OF THE BOX 40


20

0

 20

 40

0

45

90

135

180

Input Angle, deg

225

270

315

360




The linkage in Figure P7-8g has the dimensions and crank angle given below. Find 4, AA, and

Given:

AB in the global coordinate system for the position shown if  = 15 rad/sec CW, constant. Use the acceleration difference graphical method. Link lengths:

Solution: 1.

Link 2 (O2 to A)

a  49 mm

Link 3 (A to B)

b  100  mm

Link 4 (O4 to B)

c  153  mm

Link 1 (O2 to O4)

d  87 mm

Crank angle:

θ  148  deg Local xy system


ω  15 rad sec


β  119  deg Global XY system to local xy system

1

α  0  rad sec

2


Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest. Y

Direction of ABt

Direction of ABAt O6

B 3 29.000°

2

4 C

A

5

X

6

O2

Direction of A y

D

O4 119.000° x 2.

In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above (in the local xy coordinate system), θ  266.812  deg

θ  208.876  deg


a  ω sin θ  θ  b sin θ  θ

 

 

ω  7.576 rad sec

ω 

a  ω sin θ  θ  c sin θ  θ

 

 

ω  4.967 rad sec

1

1


3.


4.

For point B, this becomes: (ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where (angles in the global coordinate system)

DESIGN OF MACHINERY - 5th Ed. 2

2


θABn  θ  180  deg  β

θABn  90.124 deg

AAn  a  ω

2


θAAn  θ  180  deg  β

θAAn  209.000 deg

AAt  a  α

AAt  0.000 mm sec


5.


2

2


2


θAABn  θ  180  deg  β

θAABn  32.188 deg

2

Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn and AAn at an angle of AAn . From the tip of AAn, draw AAt at an angle of AAt. From the tip of AAt, draw ABAn at an angle of ABAn . Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABt, and ABAt. Y

X 42.029 y x 116.604° AB A

n A

t BA

A

n

0

AB

100 mm/s/s

t B

A

Acceleration Scale 18.740mm n

ABA

6.

From the graphical solution above, Acceleration scale factor ka  100 

mm sec

7.

2

AA  AAn

AA  11025 mm sec

AB  42.029 ka

AB  4203 mm sec

ABt  18.740 ka

ABt  1874 mm sec

2

2

at an angle of 209 deg at an angle of -116.6 deg

2

Calculate 4 using equation 7.6. α 

ABt c

α  12.2 rad sec

2

CCW



PROBLEM 7-34 The linkage in Figure P7-8g has the dimensions and crank angle given below. Find 4, AA, and

Statement:

AB in the global coordinate system for the position shown if  = 15 rad/sec CW and 2 = 10 rad/sec CCW, constant. Use an analytical method. Link lengths: Link 2 (O2 to A) Link 3 (A to B) a  49 mm b  100  mm

Given:

c  153  mm

Link 4 (O4 to B)

Solution: 1. 2.

Crank angle:

θ  148  deg Local xy system (see layout below)

Input crank angular velocity Coordinate rotation angle

ω  15 rad sec α  10 rad sec   119  deg Global XY system to local xy system

1

2

See Figure P7-8g and Mathcad file P0734. Y

Draw the linkage to scale and label it. Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1 

d

K2 

a

K1  1.7755 2

K3 

O6 B

d

3

c 29.000°

K2  0.5686 2

2

a b c d

C

K3  1.5592

2 a c

 

5

   

y

B  1.0598

119.000° x

C  4.6650






2

θ  2  atan2 2  A B 

B  4 A  C



θ  208.876 deg


2

d

K5 

b

2

2

c d a b

 

2

2 a b

 

D  cos θ  K1  K4 cos θ  K5

K4  0.8700 D  3.0104

 

E  2  sin θ

E  1.0598

 

F  K1   K4  1   cos θ  K5 5.

D

O4

C  K1   K2  1   cos θ  K3

4.

X

6

O2

 

B  2  sin θ

A  0.5821

A

2

4

2

A  cos θ  K1  K2 cos θ  K3

3.

d  87 mm

Link 1 (O2 to O4)

F  2.2367






θ  2  atan2 2  D E 

2

E  4  D F



θ  266.892 deg

K5  0.3509


6.

7.


Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18. ω 

a  ω sin θ  θ  b sin θ  θ

 

 

ω  7.570

rad

ω 

a  ω sin θ  θ  c sin θ  θ

 

 

ω  4.959

rad

sec

sec

Using the Euler identity to expand equation 7.13a for AA. Determine the magnitude, and direction (in the local coordinate system).



 

   a ω2 cosθ  j  sinθ

AA  a  α sin θ  j  cos θ AA  ( 9090  6258i)

mm sec

AA  11036


mm sec

8.

θAA  arg AA  

AA  AA

2

at

2

θAA  84.46 deg


 

 

 

 





A  73.887 mm

B  99.853 mm

D  133.977 mm

E  5.422 mm

 

2

 

2

 

2

 

2

 

2

 


C  6.106  10 mm sec

 

2

2

 


F  2.353  10 mm sec α 

9.

C D  A  F

2

α  49.646

A  E  B D

rad sec

α 

2

C  E  B F A  E  B D

α  15.552




 

   c ω2 cosθ  j  sinθ

AB  c α sin θ  j  cos θ AB  ( 4444  267i)

mm sec

2


θAB  arg AB  

AB  AB AB  4452

mm sec

2

at

θAB  115.6 deg

rad sec

2




At t = 0, the non-Grashof linkage in Figure P7-8g has the local axis at -119 deg and O2A at 29 deg in the global XY coordinate system and 2 = 0. Write a computer program or use an equation solver to find and plot 4, VA, AA, VB, and AB in the local coordinate system for the maximum range of motion that this linkage allows if 2 = 15 rad/sec CCW, constant. Link lengths:

Given:

Solution: 1. 2.

Link 2 (O2 to A)

a  49 mm

Link 3 (A to B)

Link 4 (O4 to B)

c  153  mm

Link 1 (O2 to O4)

1

b  100  mm d  87 mm

α  15 rad sec

2


ω  0  rad sec

Initial crank angle:

  148  deg Local xy system (see layout below)

See Figure P7-8g and Mathcad file P0735. Y

Draw the linkage to scale and label it. Determine the range of motion for this non-Grashof triple rocker using equations 4.37. 2

arg1 

2

a d b c

2

2

a d b c

O6

2



2 a d 2

arg2 

2

2

2 a d





B

arg1  0.840

3

arg2  6.338

2

4

A


O2

2

5

The other toggle angle is the negative of this. Thus,

y

θ    0.5 deg   1  deg  360  deg  θ2toggle  0.5 deg

O4

D

x

3.


d

K2 

a

K1  1.7755

 

d c

K2  0.5686

2

K3 

2

2

a b c d

2

K3  1.5592

2 a c

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

C θ  K1   K2  1   cos θ  K3 4.


 





 

 

θ θ  2   atan2 2  A θ B θ  5.

X

6

C

 2  4 A θ Cθ 

B θ




d

K4 

b

 

2

2

c d a b

K5 

2

K4  0.8700

2 a b

 

K5  0.3509

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 6.


 





 

 

θ θ  2   atan2 2  D θ E θ  7.

Determine the angular velocity of links 2, 3, and 4 using equations 6.18.

 

ω θ 





2  θ    α

 

 

a  ω θ

 

a  ω θ

ω θ 

ω θ 

8.

 2  4 Dθ F θ 

E θ

b

 

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 





Using the Euler identity to expand equation 7.13a for AA. and determine its magnitude.

 



 

 

 

   a ωθ2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ AA θ  AA θ 9.


 

  

B θ  b  sin θ θ

 

  

E θ  b  cos θ θ



 

 

 

 

 2

 

 

  

 

  

 2

    c ωθ2 cosθθ

C θ  a  α sin θ  a  ω θ  cos θ  b  ω θ  cos θ θ

 2    2     c ωθ2 sinθθ C θ  E θ  B θ  F  θ A  θ  E θ  B θ  D θ

F θ  a  α cos θ  a  ω θ  sin θ  b  ω θ  sin θ θ

 

α θ 

10. Use equation 7.13c to determine the acceleration of point B and determine its magnitude.

 

 


AB θ  c α θ  sin θ θ

 

 

AB θ  AB θ

11. Plot the angular velocity and acceleration for link 4 and the velocity and acceleration components for pins A and B.



ANGULAR VELOCITY OF LINK 4

Angular Accel., rad/sec^2

20

0

 20

 40 100

150

200

250

300

350

300

350

300

350

Crank Angle, deg

ANGULAR ACCELERATION OF LINK 4 Angular Accel., rad/sec^2

100

50

0  50  100 100

150

200

250

Crank Angle, deg

VELOCITY OF POINT A 500

Velocity, mm/sec

400 300 200 100 0 100

150

200

250

Crank Angle, deg



ACCELERATION OF POINT A

Acceleration, mm/sec^2

5 4 3 2 1 0 100

150

200

250

300

350

300

350

300

350

Crank Angle, deg

VELOCITY OF POINT B

Velocity, m/sec

2

0

2

4 100

150

200

250

Crank Angle, deg

ACCELERATION OF POINT B


20

15

10

5

0 100

150

200

250

Crank Angle, deg




The 3-cylinder radial compressor in Figure P7-8c has the dimensions and crank angle given below. Find the piston accelerations A6, A7, and A8 for the position shown for  = 15 rad/sec CW, constant. Use the acceleration difference graphical method.

Given:


a  19 mm

Offset

c  0  mm

Link 2 position, velocity, and acceleration: θ  53 deg

1.

ω  15

rad sec

α  0 

β  120  deg

Cylinder angular spacing: Solution:

b  70 mm

Link 3

rad sec


In order to solve for the accelerations at piston 7, we will need 4, and 4. From the graphical position solution below (in the local coordinate system), α  90 deg

Coordinate system rotation angle: θ  θ  α

θ  37.000 deg ω 


θ  180  deg  9.401  deg

   

a  ω cos θ  b cos θ

ω  3.296 rad sec

θ  170.599 deg

1

Direction of ACBt Direction of A8 Direction of AEB

Y

6.088°

6

E 8

C 3

2

O2 Direction of A6

5

15.629°

X

B

1 4

Direction of ADBt 7 D

9.401°

Direction of A7


2.


3.

For point D, this becomes: AD = (ABt + ABn) + (ADBt + ADBn) , where (angles in the global coordinate system) 2

2

ABn  a  ω


θABn  θ  α  180  deg

θABn  127.000 deg

ABt  a  α

ABt  0.000 mm sec

ADBn  b  ω

2

θADBn  θ  α

ADBn  760 mm sec

θADBn  80.599 deg

2

2

2


4.


Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn. From the tip of ABn, draw ADBn at an angle of ADBn. Now that the vectors with known magnitudes are drawn, from the origin and the tip of ADBn, draw construction lines in the directions of AD (vertical) and ADBt, respectively. The intersection of these two lines are the tips of ADBt, and AD. t ADB

n

ADB

AD n

0

AB Y

75.171

50 mm/s/s

Acceleration Scale

X,y

x

5.

From the graphical solution above, ka  50

Acceleration scale factor

mm sec

AD  75.171 ka 6.

AD  3759 mm sec

2

at an angle of 90 deg (global)

In order to solve for the accelerations at piston 6, we will need 3, and 3. From the graphical position solution above (in the local coordinate system), α  90 deg  β

Coordinate system rotation angle: θ  θ  α


7.

2

θ  157.000 deg ω 

θ  180  deg  6.088  deg

   

a  ω cos θ  b cos θ

ω  3.769 rad sec

1

For point C, equation 7.4 becomes: AC = (ABt + ABn) + (ACBt + ACBn) , where (angles in the global coordinate system) ACBn  b  ω

n

2

ACBn  994.4 mm sec

ACB A

n B

Y

2

θACBn  θ  α

x

t ACB

0

Acceleration Scale

X

62.266

Repeat the procedure of step 4 for the equation in step 7 using ABn and ABt from step 3.

50 mm/s/s

150.000°

AC

θACBn  36.088 deg 8.

θ  173.912 deg

y


9.


From the graphical solution above, mm

ka  50


sec AC  62.266 ka

2

AC  3113 mm sec

2


10. In order to solve for the accelerations at piston 8, we will need 5, and 5. From the graphical position solution above (in the local coordinate system), α  360  deg   90 deg  2  β

Coordinate system rotation angle: θ  θ  α

θ  83.000 deg ω 


θ  180  deg  15.629 deg

   

a  ω cos θ  b cos θ

ω  0.515 rad sec

θ  195.629 deg

1

11. For point E, equation 7.4 becomes: AE = (ABt + ABn) + (AEBt + AEBn) , where (angles in the global coordinate system) 2

AEBn  b  ω

AEBn  18.583 mm sec

θAEBn  θ  α

θAEBn  225.629 deg

2

12. Repeat the procedure of step 4 for the equation in step 11 using ABn and ABt from step 3. n

AEB

0 A

50 mm/s/s

n B

Acceleration Scale

Y

x

y t EB

A

30.000° AE

X 12.934

Note that, at the acceleration scale chosen, AEBn is so small that it can hardly be seen in the layout above. 13. From the graphical solution above, Acceleration scale factor

ka  50

mm sec

AE  12.934 ka

2

AE  647 mm sec

2





The 3-cylinder radial compressor in Figure P7-8c has the dimensions and crank angle given below. Find the piston accelerations A6, A7, and A8 for the position shown for  = 15 rad/sec CW, constant. Use an analytical method.

Given:


a  19 mm

Offset

c  0  mm

b  70 mm

Link 3

Link 2 position, velocity, and acceleration: θ  37 deg

1.

α  0 

sec

rad sec

See Figure P7-8c and Mathcad file P0737. Y

Draw the linkage to scale and label it. x''

2.

rad

β  120  deg


ω  15

Determine 4 and d' using equation 4.17. Coordinate rotation angle α  90 deg (x'y' system)

 a sin θ  θ4  asin  b 

c

π 

 

d'  a  cos θ  b  cos θ4

x'''

y''' 6

E 8

C 3

2

5 X, y'

θ4  170.599 deg

B 1

d'  3.316 in

4 y''

3.


ω4  4.

 

a cos θ   ω b cos θ4

ω4  3.296

7 D

rad

37.000°

sec

x'

Using the Euler identity to expand equation 7.15b for AB, determine its magnitude, and direction (global).



 

   a ω2 cosθ  j  sinθ

AB  a  α sin θ  j  cos θ AB  ( 3414.167  2572.759i )

mm sec

2

AB  4275


mm sec

5.

at

2

θAB  127 deg


α4  6.

θAB  arg AB  α  360  deg

AB  AB

 

  b  cos θ4

a  α cos θ  a  ω  sin θ  b  ω4  sin θ4 2

2

α4  35.456

sec

Use equation 7.16e for the acceleration of pin D.

 

 

AD  a  α sin θ  a  ω  cos θ  b  α4 sin θ4  b  ω4  cos θ4 2

rad

2

2

2


mm

AD  3759

sec 7.


A negative sign means that AD is inward

2

Determine 3 and d'' using equation 4.17. θ  θ  120  deg

θ  157.000 deg

 a sin θ  c  π b  

  asin 

 

  173.912 deg

 

d''  a  cos θ  b  cos  8.

d''  2.052 in

Determine the angular velocity of link 3 using equation 6.22a:  

9.

(x''y'' system)

   

a cos θ   ω b cos 

  3.769

rad sec


 

 

  b  cos  2

2

 

a  α cos θ  a  ω  sin θ  b    sin 

  22.483

rad sec

2


 

2

 

 

 

2

AC  a  α sin θ  a  ω  cos θ  b   sin   b    cos  AC  3113

mm sec

A positive sign means that AC is outward

2

11. Determine 5 and d''' using equation 4.17. θ  θ  120  deg

θ  277.000 deg

 a sin θ  c  π b  

θ5  asin 

(x'''y''' system)

θ5  195.629 deg

 

d'''  a  cos θ  b  cos θ5

d'''  2.745 in


ω5 

 

a cos θ   ω b cos θ5

ω5  0.515

rad sec


α5 

 

  b  cos θ5

a  α cos θ  a  ω  sin θ  b  ω5  sin θ5 2

2

α5  62.869

sec

14. Use equation 7.16e for the acceleration of pin E.

 

 

AE  a  α sin θ  a  ω  cos θ  b  α5 sin θ5  b  ω5  cos θ5 AE  646.72

mm sec

2

2

rad

2

A positive sign means that AE is outward

2




The 3-cylinder radial compressor in Figure P7-8c has the dimensions and crank angle given below. Find and plot the piston accelerations A6, A7, and A8 for one revolution of the crank if  = 15 rad/sec CW, constant.

Given:


a  19 mm

Offset

c  0  mm

Link 2 velocity, and acceleration: ω  15

1. 2.

rad sec

sec

2

Draw the linkage to scale and label it. Note that there are three local coordinate systems. Determine the range of motion for this slider-crank linkage. This will be the same in each coordinate frame.

Y

x'' 6

E 8

C 3

a b



2

5 X, y'

(x'y' system)


ω4 θ 

x'''

y'''


 a sin θ  c  θ4 θ  asin  π b   4.

rad


θ  0  deg 2  deg  360  deg 3.

α  0 

α  120  deg


b  70 mm

Link 3

B 1 4 y''

   ω cos θ4 θ 

7 D

cos θ

37.000° 5.


α4 θ  6.

 

x'

   2    b  cos θ4 θ  2

a  α cos θ  a  ω  sin θ  b  ω4 θ  sin θ4 θ

Use equation 7.16e for the acceleration of pin D.

 

2     2  b  α4 θ  sin θ4 θ   b  ω4 θ  cos θ4 θ 

AD θ  a  α sin θ  a  ω  cos θ 

7.


 a sin θ  α  c  π b  

θ3 θ  asin  8.


9.

a cos θ  α   ω b cos θ3 θ Determine the angular acceleration of link 3 using equation 7.16d.

ω3 θ 

    

(x''y'' system)


α3 θ 






  b  cos θ3 θ 

 2   

2

a  α cos θ  α  a  ω  sin θ  α  b  ω3 θ  sin θ3 θ


 


AC θ  a  α sin θ  α  a  ω  cos θ  α 

11. Determine 5 and d''' using equation 4.17.

 a sin θ  2  α  b 

θ5 θ  asin 

c

π 

(x'''y''' system)


ω5 θ 





a cos θ  2  α   ω b cos θ5 θ

  


α5 θ 





  b  cos θ5 θ  2

 2   

a  α cos θ  2  α  a  ω  sin θ  2  α  b  ω5 θ  sin θ5 θ

14. Use equation 7.16e for the acceleration of pin E.

 


AE θ  a  α sin θ  2  α  a  ω  cos θ  2  α 

15. Plot the piston accelerations. PISTON ACCELERATIONS 4


2

0 2 4 6

0

30

60

90

120

150

180

210

Piston D Crank Angle, deg Piston C Piston D Piston E

240

270

300

330

360




Figure P7-9 shows a linkage in one position. Find the instantaneous accelerations of points A, B, and P if link O2A is rotating CW at 40 rad/sec.

Given:


a  5.00 in

Link 3 (B to C)

b  4.40 in

Link 4 (C to D)

c  5.00 in

Link 1 (A to D)

d  9.50 in

Rpa  8.90 in

δ  56 deg

Coupler point:

Crank angle and motion: θ  50 deg Solution:

ω  40 rad sec

1

α  0  rad sec

2


1.


2.


P

K1 

d

K1  1.9000 2

K3 

d

K2 

a

c

K2  1.9000 2

2

a b c d

y

Y

2

K3  2.4178

2 a c

 

B

 

3

A  cos θ  K1  K2 cos θ  K3

 

B  2  sin θ

2

x 50.000°

3.

B  1.5321

14.000°

C  2.4537





2

B  4 A  C



θ  246.992 deg

θ  θ  360  deg

θ  113.008 deg


2

d

K5 

b

2

2

c d a b

 

2 a b

 

D  cos θ  K1  K4 cos θ  K5

   




θ  2  atan2 2  D E  θ  θ  360  deg

K4  2.1591

E  1.5321

F  K1   K4  1   cos θ  K5



2

D  2.3605

E  2  sin θ

5.

X

O2

Use equation 4.10b to find values of 4 for the open circuit. θ  2  atan2 2  A B 

4.

O4

1

 

C  K1   K2  1   cos θ  K3 A  0.0607

4

A

2

E  4  D F

F  0.1539



θ  349.895 deg θ  10.105 deg

K5  2.4911


6.

7.



a  ω sin θ  θ  b sin θ  θ

 

 

ω  41.552

ω 

a  ω sin θ  θ  c sin θ  θ

 

 

ω  26.320

rad sec rad sec




 

   a ω2 cosθ  j  sinθ

AA  a  α sin θ  j  cos θ in

AA  ( 5142.3  6128.4i )

sec The acceleration of pin A is

in

AA  8000

sec 8.


AA  AA

2

at 2

θAA  130.0 deg


 

 

 

 





A  4.602 in

B  0.772 in

D  1.954 in

E  4.332 in

 

2

 

2

 

2

4

2

 

2

 

2

α  356.538

rad

 

C  a  α sin θ  a  ω  cos θ  b  ω  cos θ  c ω  cos θ C  1.398  10 in sec

 

2

3

2

 

F  a  α cos θ  a  ω  sin θ  b  ω  sin θ  c ω  sin θ F  4.273  10 in sec α  9.

C D  A  F A  E  B D

sec

α 

2

C  E  B F A  E  B D

α  2.977  10




 

   c ω2 cosθ  j  sinθ

AB  c α sin θ  j  cos θ AB  ( 12346.3  9005.8i )

in sec


2


AB  AB in

AB  15282

sec

at 2

θAB  143.9 deg

10. Use equations 7.32 to find the acceleration of the point C.



      Rpa ω   cos θ  δ  j  sin θ  δ 

ACA  Rpa α sin θ  δ  j  cos θ  δ 2

AC  AA  ACA AC  AC

AC  23073

in sec

2

arg AC  111.525 deg

3 rad

sec

2




Figure P7-10 shows a linkage and its coupler curve. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the acceleration of the coupler point P at 2-deg increments of crank angle. Check your results with program FOURBAR.

Units:


Given:

Link lengths:

Solution: 1. 2.

1

Link 2 (O2 to A)

a  10 mm

Link 3 (A to B)

b  20.6 mm

Link 4 (B to O4)

c  23.3 mm

Link 1 (O2 to O4)

d  22.2 mm

Coupler point:

Rpa  30.6 mm

δ  31 deg

Crank motion:

ω  100  rpm

α  0  rad sec


Draw the linkage to scale and label it. Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1 

d

K2 

a

K1  2.2200 2

K3 

 

 

B

y

d

3

c

K2  0.9528 2

2

a b c d

b p

K3  1.5265

2 a c

 

A

 

2

a

 

B θ  2  sin θ

4

2

 

 

1

O2



 



 

 2  4 A θ Cθ 

 

B θ


 

2

d

K5 

b

2

2

c d a b

2

2 a b

 

K4  1.0777

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 Use equation 4.13 to find values of 3 .

 





 

 

θ θ  2   atan2 2  D θ E θ  6.

x O4

Use equation 4.10b to find values of 4 for the open circuit. θ θ  2   atan2 2  A θ B θ 

5.

4

c d

C θ  K1   K2  1   cos θ  K3

4.

P

2

A θ  cos θ  K1  K2 cos θ  K3

3.

2

 2  4 Dθ F θ 

E θ


K5  1.1512


 

a  ω

 

a  ω

ω θ 

b

ω θ  7.

c




    sin θ θ  θ θ 



  sin θ θ  θ θ 





Using the Euler identity to expand equation 7.13a for AA.

 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ 8.

Use equations 7.12 to determine the angular accelerations of link 3.

 

  

B' θ  b  sin θ θ

 

  

E' θ  b  cos θ θ

A' θ  c sin θ θ

D' θ  c cos θ θ

 

  

 

  

 

 

2

 

 2     c ωθ2 cosθθ

 

 

2

 

 2     c ωθ2 sinθθ


 

α θ  9.


Use equations 7.32 to find the acceleration of the point P.

 

           2  Rpa ω θ   cos θ θ  δ  j  sin θ θ  δ  AP θ  AA θ  APA θ APA θ  Rpa α θ  sin θ θ  δ  j  cos θ θ  δ

 

 

θAP θ  arg AP θ 

AP θ  AP θ

10. Plot the magnitude and direction of the coupler point P. (See next page.) θ  0  deg 1  deg  360  deg MAGNITUDE OF COUPLER POINT ACCELERATION

Acceleration, in/sec^2

200

150

100

50

0

0

45

90

135

180

Crank Angle, deg

225

270

315

360



DIRECTION OF COUPLER POINT ACCELERATION 200

Direction Angle, deg

100

0

 100

 200

0

45

90

135

180

Crank Angle, deg

225

270

315

360




Figure P7-11 shows a linkage that operates at 500 crank rpm. Find and plot the magnitude and direction of the acceleration of point B at 2-deg increments of crank angle. Check your result with program FOURBAR.

Given:


a  2.000  in

Link 3 (A to B)

b  8.375  in

Link 4 (B to O4)

c  7.187  in

Link 1 (O2 to O4)

d  9.625  in

ω  500  rpm


α  0  rad sec Solution:

2

1


1.


2.


A


d

K2 

a

K1  4.8125 2

2

a b c d

O2 d c

4

1

2

K3  2.7186

2 a c

 

 

O4

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ 4.




 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


 

2

d

K5 

b

2

2

c d a b

2

2 a b

 

K4  1.1493

K5  3.4367

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 6.


 





 

 

θ θ  2   atan2 2  D θ E θ  7.

B

2

K2  1.3392

2

K3 

3 2

θ  0  deg 2  deg  360  deg 3.

ω  52.360 rad sec

 2  4 Dθ F θ 

E θ



 

ω θ 

 

ω θ  8.


    b sin θ θ  θ θ  a  ω sin θ  θ θ   c sin θ θ  θ θ  a  ω





     D θ  c cos θ θ 

     E θ  b  cos θ θ 


B θ  b  sin θ θ

 

 

2

 

 2

    c ωθ2 cosθθ

 

 

2

 

 2     c ωθ2 sinθθ


 

α θ  9.


Determine the acceleration of point B using equations 7.13c.

 

 


AB θ  c α θ  sin θ θ

 

 

θAB θ  arg AB θ 

AB θ  AB θ

10. Plot the magnitude and angle of the acceleration at point B. MAGNITUDE OF ACCELERATION AT B Acceleration, in/sec^2

8000 6000 4000 2000 0

0

60

120

180

240

300

360

300

360

DIRECTION OF ACCELERATION AT B

Angle, deg

100

0

 100

 200

0

60

120

180

240




Figure P7-12 shows a linkage and its coupler curve. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the acceleration of the coupler point P at 2-deg increments of crank angle over the maximum range of motion possible. Check your results with program FOURBAR. Link lengths:

Given:

Solution:

Link 2 (O2 to A)

a  0.785  in

Link 3 (A to B)

b  0.356  in

Link 4 (B to O4)

c  0.950  in

Link 1 (O2 to O4)

d  0.544  in

Coupler point:

Rpa  1.09 in

δ  0  deg

Crank speed:

ω  20 rpm

α  0  rad sec

2


1.


2.


y

 a2  d 2  ( b  c) 2  θ  acos 2 a d  

3 B 3 A

θ  158.286 deg 4

θ  θ  0.5 deg θ  1  deg  θ  0.5 deg 3.

d

K2 

a

K1  0.6930 2

K3 

2

 

d

x O2

c

2

2

a b c d

2

 

K3  1.1317

2 a c

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 

C θ  K1   K2  1   cos θ  K3




 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

O4

K2  0.5726

B θ  2  sin θ 4.

2


P

B θ


2

d

K5 

b

2

2

c d a b

2

2 a b

K4  1.5281


D θ  cos θ  K1  K4 cos θ  K5

6.


 





 

 

θ θ  2   atan2 2  D θ E θ 

 2  4 Dθ F θ 

E θ

K5  0.2440


7.


 

a  ω

 

a  ω

ω θ 

ω θ  8.


b

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 






 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ 9.


     D' θ  c cos θ θ 

     E' θ  b  cos θ θ 

A' θ  c sin θ θ

 

 

 

 

2

B' θ  b  sin θ θ

 

 2     c ωθ2 cosθθ

C' θ  a  α sin θ  a  ω  cos θ  b  ω θ  cos θ θ

   2     c ωθ2 sinθθ C'  θ  D' θ  A'  θ  F' θ A'  θ  E' θ  B' θ  D' θ 2

F' θ  a  α cos θ  a  ω  sin θ  b  ω θ  sin θ θ

 

α θ 

10. Use equations 7.32 to find the acceleration of the point P.

 


 

 

θAP θ  arg AP θ 


11. Plot the magnitude and direction of the coupler point P. MAGNITUDE OF COUPLER POINT ACCELERATION


40

30

20

10

0  200

 150

 100

 50

0

50

100

150

200





100

0

 100

 200  200

 150

 100

 50

0

Crank Angle, deg

50

100

150

200




Figure P7-13 shows a linkage and its coupler curve. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the acceleration of the coupler point P at 2-deg increments of crank angle over the maximum range of motion possible. Check your results with program FOURBAR.

Given:

Link lengths:

Solution: 1.

Link 2 (O2 to A)

a  0.86 in

Link 3 (A to B)

b  1.85 in

Link 4 (B to O4)

c  0.86 in

Link 1 (O2 to O4)

d  2.22 in

Coupler point:

Rpa  1.33 in

δ  0  deg

Crank motion:

ω  80 rpm

α  0  rad sec

2



y

B

3 4 P

O4

O2 3 2

A

2.


arg1 

2

2



2 a d 2

arg2 

2

a d b c

2

2

a d b c 2 a d

2





arg1  1.228

arg2  0.439




x



d

K1 

K2 

a

K1  2.5814 2

2

2

2

K3  2.0181

2 a c

 

c

K2  2.5814

a b c d

K3 

d

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

C θ  K1   K2  1   cos θ  K3 4.




 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


d

K4 

K5 

b

 

2

2

c d a b

2

K4  1.2000

2 a b

 

K5  2.6244

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 6.


 





 

 

θ θ  2   atan2 2  D θ E θ  7.


 

a  ω

 

a  ω

ω θ 

ω θ  8.

 2  4 Dθ F θ 

E θ

b

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 






 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ 9.


 

  

B' θ  b  sin θ θ

 

  

E' θ  b  cos θ θ

A' θ  c sin θ θ

D' θ  c cos θ θ

 

  

 

  

 

 

2

 

 2     c ωθ2 cosθθ

 

 

2

 

 2     c ωθ2 sinθθ



 

α θ 




 

           2  Rpa ω θ   cos θ θ  δ  j  sin θ θ  δ 

APA θ  Rpa α θ  sin θ θ  δ  j  cos θ θ  δ

 

 

 

 

 

AP θ  AA θ  APA θ

θAP θ  arg AP θ 




MAGNITUDE OF COUPLER POINT ACCELERATION

150

100

50

0  200

 100

0

100

200



100

0

 100

 200  200

 100

0

100

200




Figure P7-14 shows a linkage and its coupler curve. Write a computer program or use an equation solver to calculate and plot the magnitude and direction of the acceleration of the coupler point P at 2-deg increments of crank angle over the maximum range of motion possible. Check your results with program FOURBAR.

Given:

Link lengths:

Solution: 1.

Link 2 (O2 to A)

a  0.72 in

Link 3 (A to B)

b  0.68 in

Link 4 (B to O4)

c  0.85 in

Link 1 (O2 to O4)

d  1.82 in

Coupler point:

Rpa  0.97 in

δ  54 deg

Crank motion:

ω  80 rpm

α  0  rad sec

2


Draw the linkage to scale and label it. P y

B

3 4

A 2

x O2

2.

O4


arg1 

2

2



2 a d 2

arg2 

2

a d b c

2

2

a d b c 2 a d

2



b c

arg1  1.451

a d b c

arg2  0.568

a d





d a

K1  2.5278

K2 

d c

K2  2.1412



2

2

a b c d

K3 

2

K3  3.3422

2 a c

 

 

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

C θ  K1   K2  1   cos θ  K3 4.




 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


d

K4 

K5 

b

 

2

2

c d a b

2

K4  2.6765

2 a b

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

K5  3.6465

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5 6.




 



 

 

θ θ  2   atan2 2  D θ E θ  7.


 

a  ω

 

a  ω

ω θ 

ω θ  8.

 2  4 Dθ F θ 

E θ

b

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 






 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ 9.


 

  

B' θ  b  sin θ θ

 

  

E' θ  b  cos θ θ

A' θ  c sin θ θ

D' θ  c cos θ θ

 

  

 

  

 

 

2

 

 2     c ωθ2 cosθθ

 

 

2

 

 2     c ωθ2 sinθθ


 

α θ 



 



 


 

θAP θ  arg AP θ 




MAGNITUDE OF COUPLER POINT ACCELERATION

200

100

0  60

 40

 20

0

20

40

60

Crank Angle, deg

DIRECTION OF COUPLER POINT ACCELERATION


50

0

 50

 100  60

 40

 20

0 Crank Angle, deg

20

40

60




Given:

Figure P7-15 shows a power hacksaw that is an offset crank-slider mechanism that has the dimensions given below. Draw an equivalent linkage diagram and then calculate and plot the acceleration of the saw blade with respect to the piece being cut over one revolution of the crank, which rotates at 50 rpm. Link lengths: Link 2 (O2 to A)

a  75 mm

Link 3 (A to B)

b  170  mm ω  50 rpm


c  45 mm

Offset

α  0  rad sec

2


Draw the equivalent linkage to a convenient scale and label it.

y

B

3

b

A

4 a 2

c

2

O2 2.


3.


 a  sin θ  c   b  

 

θ θ  asin 4.


 

ω θ  5.

b



   ω cos θ θ  cos θ


 

α θ  6.

a

 

   2    b  cos θ θ  2



 



x





2

0

2

4

0

60

120

180 Crank Angle, deg

240

300

360




Figure P7-16 shows a walking beam indexing and pick-and-place mechanism that can be analyzed as two fourbar linkages driven by a common crank. Calculate and plot the relative acceleration between points E and P for one revolution of gear 2.

Given:

Link lengths (walking-beam linkage): Link 2 (O2 to A)

a'  40 mm

Link 3 (A to D)

b'  108  mm

Link 4 (O4 to D)

c'  40 mm

Link 1 (O2 to O4)

d'  108  mm

Link lengths (pick and place linkage): Link 2 (O5 to B)

a  13 mm

Link 7 (B to C)

b  193  mm

Link 6 (C to O6)

c  92 mm

Link 1 (O5 to O6)

d  128  mm

Rocker point E:

u  164  mm

Crank speed:

ω  10 rpm

1.

2

ϕ  143  deg

Gears 4 & 5 phase angle Solution:

α  0  rad sec


Draw the walking-beam linkage to scale and label it. 30 mm Y

P

58 mm

80°

A

D b'

c'

a' d'

x'

O2

O4

X

y'

2.

Determine the range of motion for this mechanism. θ  0  deg 2  deg  360  deg

3.

(local x'y' coordinate system)

This part of the mechanism is a special-case Grashof in the parallelogram configuration. As such, the coupler does not rotate, but has curvilinear motion with every point on it having the same velocity and acceleration. Therefore, it is only necessary to calculate the X-component of the acceleration at point A in order to determine the acceleration of the cylinder center, P.

 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ

 

  

AAx θ  Re AA θ In the global coordinate frame,

 

 

AP θ  AAx θ 4.

Draw the pick-and-place linkage to scale and label it.



E

C

7 6

b

c

O6 5.

B

a O5

Establish the relationship between 5 and 2. Note that gear 5 is driven by gear 4 and that their ratio is -1 (i.e., they rotate in opposite directions with the same speed). Also, because the walking beam fourbar is a special Grashof, 4 = 2. Thus,

 

θ θ  θ  ϕ 6.

5

d 1

ω  ω

and

α  α


d

K2 

a

K1  9.8462 2

2

a b c d

2

K3  5.1137

2 a c

 

c

K2  1.3913

2

K3 

d

    K1  K2 cosθθ  K3

A θ  cos θ θ

 

  

B θ  2  sin θ θ

 

     K3

C θ  K1   K2  1   cos θ θ 7.




 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  8.

B θ


 

2

d

K5 

b

2

2

2 a b

    K1  K4 cosθθ  K5

D θ  cos θ θ

 

2

c d a b

  

E θ  2  sin θ θ

K4  0.6632

K5  9.0351



 

     K5

F θ  K1   K4  1   cos θ θ 9.


 





 

 

θ θ  2   atan2 2  D θ E θ 

 2  4 Dθ F θ 

E θ

10. Determine the angular velocity of links 7 and 6 using equations 6.18.

         

ω θ 

 

a  ω sinθ θ  θ θ  b sin θ θ  θ θ

 

a  ω sin θ θ  θ θ  c sin θ θ  θ θ

ω θ 

     

   

11. Use equations 7.12 to determine the angular acceleration of link 6.

 

  

B θ  b  sin θ θ

 

  

E θ  b  cos θ θ

A θ  c sin θ θ

D θ  c cos θ θ

 

  

 

  

 

    a ω2 cosθθ  2 2  b  ω θ θ   cos θ θ   c ω θ  cos θ θ 

 

    a ω2 sinθθ  2 2  b  ω θ θ   sin θ θ   c ω θ  sin θ θ 

C θ  a  α sin θ θ

F θ  a  α cos θ θ

 

α θ 


12. Determine the acceleration of the rocker point E using equations 7.13c (substituting the distance to point E from O6 for the distance c).

 

 

    j  cosθθ  u ωθ2 cosθθ  j  sinθθ

AE θ  u  α θ  sin θ θ

13. Transform this into the global XY system.

     AEX  θ  AEx  θ

     AEY  θ  AEx  θ

AEx θ  Re AE θ

 

 

AEy θ  Im AE θ

 

AEXY θ  AEX θ  j  AEY θ

14. Calculate and plot the acceleration of E relative to P. (See next page).

 

 

 

AEP θ  AEXY θ  AP θ

 



 

AEPX θ  Re AEP θ

 



 

AEPY θ  Im AEP θ



RELATIVE ACCELERATION of E with respect to P 80 70


60 50 40 30 20 10 0

0

30

60

90

120

150

180

210

Crank Angle, deg

240

270

300

330

360




Figure P6-17 shows a paper roll off-loading mechanism driven by an air cylinder. In the position shown, it has the dimensions given below. The V-links are rigidly attached to O4A. The air cylinder is retracted at a constant acceleration of 0.1 m/sec2. Draw a kinematic diagram of the mechanism, write the necessary equations, and calculate and plot the angular acceleration of the paper roll and the linear acceleration of its center as it rotates through 90 deg CCW from the position shown.

Given:


Paper roll location from O4:

Link 4 (O4 to A)

c  300  mm

u  707.1  mm

Link 1 (O2 to O4)

d  930  mm

δ  181  deg

Link 4 initial angle

θ  62.8 deg

addot  100  mm sec

Input cylinder acceleration Solution: 1.

with respect to local x axis 2


Draw the mechanism to scale and define a vector loop using the fourbar derivation in Section 6.7 as a model. V-Link

Roll Center

707.107

45.000° x

O4 4

c 46.000° A

1

O4

d

4

3

O2

2 b

R1

R4 2 A

R3

R2

O2

a f y

2.

Write the vector loop equation, differentiate it, expand the result and separate into real and imaginary parts to solve for f, 2, and 4. R1  R4  R2  R3 d e

j  θ

 c e

j  θ

(a)

 a  e

j  θ

 b e

j  θ

(b)

where a is the distance from the origin to the cylinder piston, a variable; b is the distance from the cylinder piston to A, a constant; and c is the distance from 4 to point A, a constant. Angle 1 is zero, 3 = 2, and 4 is the variable angle that the rocker arm makes with the x axis. Solving the position equations: Let

f  a  b

then, making this substitution and substituting the Euler equivalents,

  

 

  

 

d  c cos θ  j  sin θ  f  cos θ  j  sin θ

(c)

Separating into real and imaginary components and solving for 2 and f,

 



 

 

θ θ  atan2 d  c cos θ c sin θ

(d)



  sin θ θ  c sin θ

 

f θ 

(e)

For constant deceleration of the hydraulic cylinder rod, the velocity of the rod is

 

adot θ  addot

  

 

2  f θ  f θ addot

Differentiate equation b. j  c ω e

j  θ



 d f   ej  θ   j  f  ω  ej  θ     dt 

(f)




 

 

c ω sin θ  j  cos θ 

d f   dt

   cos θ   j  sin θ    f  ω   sin θ   j  cos θ         

(g)

Separating into real and imaginary components and solving for 4 and 2. Note that df/dt = adot.

 

ω θ 

 

ω θ  3.

  c sin θ θ  θ adot θ

(h)

 

     f  θ  sin θ θ 

c ω θ  adot θ  cos θ θ

(i)

Differentiate equation f, expand the result and separate into real and imaginary parts to solve for 4. j  c α e

j  θ

2

2 j  θ

 j  c ω  e

 addot e

j  θ

 j  adot ω e

 j  adot ω e

j  θ

j  θ

 j  f  α e



j  θ

(f) 2

Separating into real and imaginary components and solving for 4.

 addot  2 adot θ   ω  θ   f  θ   ω  θ  2  sin 2  θ  θ             

 

α θ 





 

c sin θ  θ θ

Plot 4 over a range of 4 of θ  θ θ  1  deg  θ  90 deg ANGULAR ACCELERATION, LINK 4 Angular Acceleration, rad/sec^2

4.

 2    

 c ω θ  sin θ θ  θ

1 0 1 2 3 4 60

80

100

120

Link 4 Angle, deg

140

2 j  θ

 j  f  ω  e

160


5.


Determine the acceleration of the center of the paper roll using equation 7.31. The direction is in the local xy coordinate system.

 

 







AU θ  u  α θ  sin θ  δ  j  cos θ  δ

 

 

θAU  θ  arg AU θ 

AU θ  AU θ

Plot the magnitude and direction of the acceleration of the paper roll center.


MAGNITUDE OF PAPER CENTER ACCELERATION

2

1

0 60

80

100

120

140

160


DIRECTION OF PAPER CENTER ACCELERATION 200

100 Vector Angle, deg

6.

  u ωθ2 cosθ  δ  j  sinθ  δ

0

 100

 200 60

80

100

120


140

160




Figure P7-18 shows a mechanism and its dimensions. Find the accelerations of points A, B, and C for the position shown.

Given:

Lengths and angles:

Solution: 1.

Link 2 (O2 to A)

a  0.80 in

Link 1 (O2 to O4)

d  1.85 in

Link 4 (O4 to B)

a'  2.97 in

Link 5 (B to C)

b'  2.61 in

Crank angle

θ  37.5 deg


β  81.5 deg

Input crank motion

ω  40

Local xy system c'  3.25 in

Slider-crank offset

rad

rad

α  1500

min

min

2


Draw the mechanism to scale and define a vector loop for the input portion (links 1, 2, 3, and 4) using the fourbar crank-slider derivation in Section 7.3 as a model. 6 C

5 B

Y 4

O2

y R2

2 A

3

y

O2

A

R1 R3

37.5° X

O4 x

2.

O4 x

81.5°

Write the vector loop equation, expand the result and separate into real and imaginary parts to solve for 3 and b (distance from O4 to A). R2  R1  R3 a e

j  θ

 d  b  e

j  θ


  

   d  bcosθ  j  sinθ





 

 

θ  atan2 a  cos θ  d a  sin θ

θ  158.2 deg


b  3.


  sin θ

a  sin θ

b  1.31 in

Differentiate the position equation, expand it and solve for 3 and bdot. a  j  ω e ω 

j  θ

a  ω b

bdot 

 b  j  ω e



j  θ

 bdot e

j  θ



 cos θ  θ

 

ω  0.208

 

a  ω cos θ  b  ω cos θ

 

rad sec

bdot  0.459

sin θ 4.

sec

Differentiate the velocity equation, expand it and solve for 3. a  j   e

j  θ

2 j  θ

2

 a  j  ω  e

 bdot j  ω e

j  θ

 b  j   e

2 j  θ

2

 b  j  ω  e α 

1 b

j  θ

 bddot e



j  θ


j  θ

  a  α cos θ  θ  a  ω  sin θ  θ  2  bdot ω





2





rad

α  0.249

sec 5.

in

2

Determine the acceleration of point A on link 2 (in the local xy coordinate system) using equations 7.13a.



 

   a ω2 cosθ  j  sinθ


in

AA  0.487

sec 6.

2


θAA  174.348 deg

Determine the acceleration of points A and B on link 3 (in the Global XY coordinate system) using equations 7.13a. Transform 3 to the global coordinate frame:



 

θ  θ  β  360  deg

θ  120.337 deg

   b ω2 cosθ  j  sinθ

AA3  b  α sin θ  j  cos θ AA3  AA3

AA3  0.331

in sec



 

2

θAA3  arg AA3

θAA3  20.518 deg

   a' ω2 cosθ  j  sinθ

AB  a' α sin θ  j  cos θ AB  AB

AB  0.752

in sec

7.

2



 a' sin θ  c'  π b'  

θ  asin 

θ  195.254 deg

θAB  20.518 deg


8.


9.


   

a' cos θ   ω b' cos θ

ω  0.124

rad sec


α 

 

  b' cos θ 2

2

 

a' α cos θ  a' ω  sin θ  b' ω  sin θ

α  0.100

sec


 

2

 

 

2

 

AC  a' α sin θ  a' ω  cos θ  b' α sin θ  b' ω  cos θ AC  0.73

in sec

2

rad

A positive sign means that AC is to the right

2




Figure P7-19 shows a walking beam mechanism. Calculate and plot the acceleration Aout for one revolution of the input crank 2 rotating at 100 rpm.

Given:

Link lengths:

Solution: 1.

Link 2 (O2 to A)

a  1.00 in

Link 3 (A to B)

b  2.06 in

Link 4 (B to O4)

c  2.33 in

Link 1 (O2 to O4)

d  2.22 in

Coupler point:

p  3.06 in

δ  31 deg

Crank speed:

ω  100  rpm

α  0  rad sec

2



x

y O4 4

1 26.00°

X

O2

P

2 A 3 B

2.


3.


d

K2 

a

K1  2.2200 2

K3 

 

d c

K2  0.9528 2

2

a b c d 2 a c

2

K3  1.5265

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

C θ  K1   K2  1   cos θ  K3 4.






 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


d

K4 

K5 

b

 

2

2

c d a b

2

K4  1.0777

2 a b

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

K5  1.1512

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5 6.


 





 

 

θ θ  2   atan2 2  D θ E θ  7.


 

a  ω

 

a  ω

ω θ 

ω θ  8.

 2  4 Dθ F θ 

E θ

b

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 





Determine the acceleration of point A using equation 7.13a.

 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ 9.


 

  

B θ  b  sin θ θ

 

  

E θ  b  cos θ θ



 

  

 

  

 

 

2

 

 2

    c ωθ2 cosθθ

 

 

2

 

 2     c ωθ2 sinθθ


 

α θ 

        A  θ  E θ  B θ  D θ

C θ  D θ  A θ  F θ

10. Determine the acceleration of the coupler point P using equations 7.32.

 

           2  p  ω θ   cos θ θ  δ  j  sin θ θ  δ 

APA θ  p  α θ  sin θ θ  δ  j  cos θ θ  δ

 

 

 

AP θ  AA θ  APA θ

11. Plot the X-component (global coordinate system) of the acceleration of the coupler point P. (See next page). Coordinate rotation angle:

 

  

α  26 deg

  

Aout θ  Re AP θ  cos α  Im AP θ  sin α



Aout


400

200

0

 200

 400

0

45

90

135

180

Crank Angle, deg

225

270

315

360




Given:

Figure P7-20 shows a surface grinder. The workpiece is oscillated under the spinning grinding wheel by the slider-crank linkage that has the dimensions given below. Calculate and plot the acceleration of the grinding wheel contact point relative to the workpiece over one revolution of the crank. Link lengths: Link 2 (O2 to A)

a  22 mm

Link 3 (A to B)

b  157  mm

Grinding wheel diameter

d  90 mm


ω  30 rpm

Offset

CCW

Grinding wheel angular velocity ω  3450 rpm Solution: 1.

CCW



5

4 2 3

A

B

2

c

O2 2.


3.


 a sin θ  b 

 

θ θ  asin  4.

 

a b



   ω cos θ θ  cos θ


 

α θ  6.

π 


ω θ  5.

c

 

   2    b  cos θ θ  2



 



c  40 mm

α  0  rad sec

2


The acceleration of the grinding wheel contact point relative to the workpiece, which has zero tangential acceleration due to constant angular velocity, is AB (positive to the right). RELATIVE ACCELERATION AT CONTACT POINT 200

100


7.


0

 100

 200

 300

0

60

120


240

300

360




Figure P7-21 shows a drag-link mechanism with dimensions. Write the necessary equations and solve them to calculate and plot the angular acceleration of link 4 for an input of 2 = 1 rad/sec. Comment on the uses for this mechanism.

Given:

Link lengths: Link 2 (L2)

a  1.38 in

Link 3 (L3)

b  1.22 in

Link 4 (L4)

c  1.62 in

Link 1 (L1)

d  0.68 in

ω  1  rad sec


1

α  0  rad sec

CW

2


1.


2.

Determine the range of motion for this Grashof double crank.

y

A

3 B

θ  0  deg 2  deg  360  deg 3.


d

K2 

a

K1  0.4928 2

K3 

 

2 2

d

4

c

K2  0.4198 2

2

a b c d

x O2

2

K3  0.7834

2 a c

 

 

 

A θ  cos θ  K1  K2 cos θ  K3 4.

O4

 



 



 

 2  4 A θ Cθ 

 

B θ


2

d

K5 

b

 

2

2

c d a b

2

2 a b

 

K4  0.5574

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 6.


 





 

 

θ θ  2   atan2 2  D θ E θ  7.

 

C θ  K1   K2  1   cos θ  K3

Use equation 4.10b to find values of 4 for the open circuit. θ θ  2   atan2 2  A θ B θ 

5.

 

B θ  2  sin θ

 2  4 Dθ F θ 

E θ


 

ω θ 

a  ω b



    sin θ θ  θ θ  sin θ θ  θ

K5  0.3655


 

ω θ  8.

a  ω c






  sin θ θ  θ θ  sin θ  θ θ


 

  

B θ  b  sin θ θ

 

  

E θ  b  cos θ θ



 

 

 

 

2

 

 

  

 

  

 2

    c ωθ2 cosθθ

C θ  a  α sin θ  a  ω  cos θ  b  ω θ  cos θ θ

   2     c ωθ2 sinθθ C θ  E θ  B θ  F  θ A  θ  E θ  B θ  D θ 2

F θ  a  α cos θ  a  ω  sin θ  b  ω θ  sin θ θ

 

α θ 

Plot the angular acceleration of link 4.

ANGULAR ACCELERATION, LINK 4 2

Angular Acceleration, rad/sec^2

9.

1

0

1

2

0

60

120


240

300

360




Figure P7-22 shows a mechanism with dimensions. Use a graphical method to calculate the accelerations of points A, B, and C for the position shown.

Given:


d  1.22 in


θ  56.5 deg

Link 2 (O2A)

a  1.35 in

Angle O2A makes with X axis

θ  14 deg

Link 4 (O4B)

e  1.36 in

Link 5 (BC)

f  2.69 in

Link 6 (CO6)

g  1.80 in

Angle CO6 makes with X axis

θ  88 deg

ω  20 rad sec

Motion of link 2 Solution: 1.

1

α  0  rad sec

CW

2



B

Axis of slip Y


4

0.939

132.661° A 3 2 X O2

Direction of VA3

2.


3.

VA3  27.000

in sec

θVA3  θ  90 deg

θVA3  76.0 deg




Y 0

10 in/sec

1.295 X V trans

V A3 2.369

4.


5.

kv 

10 in sec

in


Vtrans  12.950

sec in sec

The true velocity of point A on link 4 is Vtrans, VA4  12.95

in sec


ω 

VA4 c

c  0.939  in and

ω  13.791

rad

θ  132.661  deg

CW

sec


θVA4  θ  90 deg 8.

in

Vslip  23.690


7.

1

Vslip  2.369  in kv


V slip

VB  18.756

in sec

θVA4  42.661 deg




Draw links 1, 4, 5, and 6 to a convenient scale. Indicate the directions of the velocity vectors of interest.

Direction of VCB

Direction of VB

B

Y O4

4

C A 3

Direction of VC 2

X O2

O6

VB

0

10 in/sec

Y

V CB X VC 2.088

9.


kv 

10 in sec

1

θVC  θ  90 deg

in in

VC  2.088  in kv

VC  20.9

VCB  1.519  in kv

VCB  15.2

θVC  2.0 deg

sec in sec

10. Determine the angular velocity of links 5 and 6 using equation 6.7. ω 

ω 

VCB f VC g

ω  5.647

rad

CCW

sec

ω  11.600

rad sec

CCW



11. For the acceleration of point A, use equation 7.19: (AA2t + AA2n) = AA4t + AA4n + AAcor + AAslip , where AA2n  a  ω

2

AA2n  540.000 in sec

θAA  θ  180  deg

θAA  194.000 deg

AA2t  a  α

AA2t  0.000 in sec 2

2

2 2

AA4n  c ω

AA4n  178.6 in sec

θAA4n  θ

θAA4n  132.661 deg

AAcor  2  Vslip  ω

AAcor  653.430 in sec

θAAcor  θ  90 deg

θAAcor  222.661 deg

2

12. Repeat procedure of steps 3 and 7 for the equation in step 11.

1.751 slip

AA t

AA4 AA4

0

2.501 n A4

A

Y

100 IN/S/S

Acceleration Scale

88.220°

cor

AA

X

AA2

13. From the acceleration polygon above, Acceleration scale factor

in

ka  100 

sec

2 2

AA4  2.501  ka

AA4  250.1 in sec

AA4t  1.751  ka

AA4t  175.1 in sec

at an angle of 88.22 deg

2

The angular acceleration of link 4 is α 

AA4t c

α  186.5

rad sec

CCW

2

14. The graphical solution for accelerations in pin-jointed fourbars uses equation 7.4: (APt + APn) = (AAt + AAn) + (APAt + APAn)



15. For point C, this becomes: (ACt + ACn) = (ABt + ABn) + (ACBt + ACBn) , where ACn  g  ω

2

ACn  242.2 in sec

θACn  θ  180  deg ABn  e ω

2

θACn  268.000 deg

2


2

θABn  θ  180  deg

θABn  312.661 deg

ABt  e α


θABt  θ  90 deg

θABt  222.661 deg

2

ACBn  f  ω

2


2

θ  27.5 deg

θACBn  θ


16. Repeat procedure of step 12 for the equation in step 15. Y

Y X

X 84.453°

91.769°

2.443 A

n B

0

100 IN/S/S

n B

A

Acceleration Scale

3.623

ACn

ABt

AC ACt

t ACB

ABt

AB n ACB


ka  100 

in sec

2 2

AB  3.623  ka

AB  362.3 in sec

AC  2.443  ka

AC  244.3 in sec

2

at an angle of -91.8 deg at an angle of -84.5 deg




Figure P7-23 shows a quick-return mechanism with dimensions. Use a graphical method to calculate the accelerations of points A, B, and C for the position shown.

Given:


d  1.69 in


Link 2 (L2)

a  1.00 in


Link 4 (L4)

e  4.76 in

Link 5 (L5)

f  4.55 in

Offset (O2C)

g  2.86 in


ω  10 rad sec

1

CCW

θ  195.5  deg

α  0  rad sec

2



B


Direction of VA3

4


2 44.228° O2

X O4

2.


3.

VA3  10.000

in sec

θVA3  θ  90 deg

θVA3  189.0 deg




a. Choose a convenient velocity scale and layout the known vector VA3. b. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the Vslip construction line. Y 0

5 in/sec

X

VA3 V trans

1.154

4.

kv 

5  in sec

1

in in

Vslip  1.634  in kv

Vslip  8.170


Vtrans  5.770

sec in sec

The true velocity of point A on link 4 is Vtrans, VA4  Vtrans

6.

1.634


5.

Vslip

VA4  5.77

in sec


VA4 c

c  2.068  in and

ω  2.790

rad

θ  44.228 deg

CCW

sec

7.


8.




Direction of VCB Direction of VB B 5 5.805°

C 6

4 Direction of VC

Y

A 3

2 44.228° O2

X O4

9.

From the velocity triangle we have: VB

kv 


5  in sec

VC  8.30

0

1

5 in/sec

in in

Y

sec V CB

θVC  180  deg VCB  1.913  in kv

VCB  9.57

X

VC

in

1.659

sec

10. Determine the angular velocity of link 5 using equation 6.7. From the linkage layout above: ω 

VCB

θ  5.805  deg

ω  2.102

f

rad

CCW

sec

11. For the acceleration of point A, use equation 7.19: (AA2t + AA2n) = AA4t + AA4n + AAcor + AAslip , where AA2n  a  ω

2

AA2n  100.000 in sec

2



θAA  θ  180  deg

θAA  279.000 deg

AA2t  a  α

AA2t  0.000 in sec 2

2 2

AA4n  c ω

AA4n  16.099 in sec

θAA4n  θ  180  deg

θAA4n  224.228 deg



θAAcor  θ  90 deg

θAAcor  45.772 deg

2

(Vslip is negative)

12. Repeat procedure of steps 3 and 7 for the equation in step 11. Y

X

n

AA4

1.581 69.810° AA4

t AA4

0

25 IN/S/S

slip

AA

Acceleration Scale

1.444

AA2 cor

AA


in

ka  25

sec

2 2

AA4  1.581  ka

AA4  39.5 in sec

AA4t  1.444  ka



2


AA4t

α  17.5

c

rad sec

CW

2

14. The graphical solution for accelerations in pin-jointed fourbars uses equation 7.4: (APt + APn) = (AAt + AAn) + (APAt + APAn) 15. For point C, this becomes: AC = (ABt + ABn) + (ACBt + ACBn) , where ABn  e ω

2

θABn  θ  180  deg


θABn  224.228 deg

2


SOLUTION MANUAL 7-53-5 2



θABt  θ  90 deg

θABt  45.772 deg

2

ACBn  f  ω


θACBn  θ


2

16. Repeat procedure of step 12 for the equation in step 15. 1.717

Y

AC

n

AB

X

69.808°

t ACB

0

25 IN/S/S

Acceleration Scale 3.639

AB t B

A

n

ACB


ka  25

in sec

2 2

AB  3.639  ka

AB  91.0 in sec

AC  1.717  ka

AC  42.9 in sec

2

at an angle of -69.81 deg at an angle of 0 deg




Figure P7-23 shows a quick-return mechanism with dimensions. Use an analytical method to calculate the accelerations of points A, B, and C for one revolution of the input link.

Given:


Solution: 1.

Link 1 (O2O4)

d  1.69 in

Link 4 (L4)

a'  4.76 in

Link 2 (L2)

a  1.00 in

Link 5 (L5)

b'  4.55 in

Input crank motion

ω  10 rad sec


β  164.5  deg

1

α  0  rad sec

CCW Offset (AE)

2

c'  2.86 in


Draw the mechanism to scale and define a vector loop for the input portion (links 1, 2, 3, and 4) using the fourbar crank-slider derivation in Section 7.3 as a model. θ  0  deg 0.5 deg  360  deg B 5 C 6

4

Y

A 3

2 O2

15.5°

X x

O4 y

R2

R3

R1 x y


2.



j  θ

 d  b  e

j  θ


  

   d  bcosθ  j  sinθ



 



 

 

  sin θ θ 

 

θ θ  atan2 a  cos θ  d a  sin θ a  sin θ

b θ 

3.

Differentiate the position equation, expand it and solve for 3 and bdot. a  j  ω e

j  θ

 

ω θ 

a  ω

 

b θ

 

bdot θ 

4.

 b  j  ω e

j  θ



 bdot e

j  θ

 

 cos θ  θ θ

 

    sin θ θ 

  

a  ω cos θ  b θ  ω θ  cos θ θ


j  θ

2

2 j  θ

 a  j  ω  e

 bdot j  ω e 2

j  θ

2 j  θ

 b  j  ω  e

 

α θ  5.

1

 

b θ

 b  j   e

j  θ

 bddot e

j  θ

  bdot j  ω e

j  θ

  a  α cos θ  θ θ

   a ω2 sinθθ  θ  2 bdotθ ωθ



Determine the acceleration of point A on link 2 (in the global XY coordinate system) using equations 7.13a.

 



 

 

 

   a ω2 cosθ  j  sinθ

AA2 θ  a  α sin θ  j  cos θ AA2 θ  AA2 θ 6.

θAA2 θ  arg AA2 θ   β

Determine the acceleration of points A and B on link 4 (in the Global XY coordinate system) using equations 7.13a. Transform and rename 3 to the global coordinate frame:

θ θ  θ θ  β

Rename 3 to 4:

α θ  α θ

 

 

 

 

 

 

ω θ  ω θ


 


         j  cosθθ  2  b  θ  ω θ   cos θ θ   j  sin θ θ  

AA4 θ  b θ  α θ  sin θ θ

 

 

θAA4 θ  arg AA4 θ 

AA4 θ  AA4 θ

 

       j  cosθθ  2  a' ω θ   cos θ θ   j  sin θ θ  

AB θ  a' α θ  sin θ θ

 

 

θAB θ  arg AB θ 

AB θ  AB θ 7.


 a' sin θ θ   c'   b'  

 

θ θ  asin 8.


 

ω θ 

9.

       

a' cos θ θ   ω θ b' cos θ θ


 

α θ 

 

    a' ωθ2 sinθθ  b' ωθ2 sinθθ b' cos θ θ 

a' α θ  cos θ θ


 

      a' ωθ2 cosθθ  2  b' α θ  sin θ θ   b' ω θ  cos θ θ 

AC θ  a' α θ  sin θ θ

ACCELERATION OF POINT A 250


200

150

100

50

0

0

60

120


240

300

360




ACCELERATION OF POINT B

1500

1000

500

0

0

60

120

180

240

300

360

300

360

Crank Angle, deg

ACCELERATION OF POINT C


1000

500

0

 500

 1000

0

60

120


240




Figure P7-24 shows a drum pedal mechanism . For the dimensions given below, find and plot the output acceleration of the linkage over its range of motion if the input velocity Vin is a constant.

Given:

Link lengths: Link 2 (O2A)

a  100  mm

Link 3 (AB)

b  28 mm

Link 4 (O4B)

c  64 mm

Link 1 (O2O4)

d  56 mm

Link 3 (AP)

rout  124  mm

Distance to force application: rin  48 mm

Link 2

Vin  3  m sec

Input velocity:

1

α  0  rad sec

θ  162  deg

Range of positions of link 2:

2

δ  0  deg

θ  171  deg

See Figure P7-24 and Mathcad file P0755. Solution: 1. Draw the mechanism to scale and label it. 2.

P


3

α  180  deg





θ  θ  α  θ  α  1  deg  θ  α 3.


d

K1  0.5600 2

K3 

 

3

2

2

2

2

 

 

 

1

O4

 

A θ  cos θ  K1  K2 cos θ  K3

2

x

K3  1.2850

2 a c

Vin

A

c

K2  0.8750

a b c d

Fin

4

d

K2 

a

B

y

 

B θ  2  sin θ

 

C θ  K1   K2  1   cos θ  K3 4.




 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


 

2

d

K5 

b

2

2

c d a b

2

2 a b

 

 

 

K4  2.0000

 

D θ  cos θ  K1  K4 cos θ  K5

 


 





 

 

θ θ  2   atan2 2  D θ E θ 

K5  1.7543

 

E θ  2  sin θ

F θ  K1   K4  1   cos θ  K5 6.

 2  4 Dθ F θ 

E θ

O2


7.

Determine the angular velocity of link 2 using equation 6.7 and links 3 and 4 using equations 6.18. Vin

ω 

ω  62.500

rin

 

a  ω

 

a  ω

ω θ 

ω θ 

8.


b

c

rad sec



    sin θ θ  θ θ 



  sin θ θ  θ θ 





Determine the acceleration of point A using equation 7.13a.

 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ 9.


 

  

B θ  b  sin θ θ

 

  

E θ  b  cos θ θ



 

  

 

  

 

 

2

 

 2

    c ωθ2 cosθθ

 

 

2

 

 2     c ωθ2 sinθθ


 

α θ 

        A  θ  E θ  B θ  D θ

C θ  D θ  A θ  F θ

10. Determine the acceleration of the coupler point P using equations 7.32.

 

           2  rout ω θ   cos θ θ  δ  j  sin θ θ  δ  AP θ  AA θ  APA θ APA θ  rout α θ  sin θ θ  δ  j  cos θ θ  δ

11. Plot the magnitude of the acceleration of the coupler point P. (See next page).

 

 

Aout θ  AP θ



Aout 222000


220000

218000

216000

214000

212000  18

 16.3

 14.7

 13

 11.3

 9.7

8

Crank Angle, deg

Note the acceleration scale values. If the scale went to zero, Aout would be a horizontal (constant) line over the range of 2 plotted.




A tractor-trailer tipped over while negotiating an on-ramp to the New York Thruway. The road has a 50-ft radius at that point and tilts 3 deg toward the outside of the curve. The 45-ft-long by 8-ft-wide by 8.5-ft-high trailer box (13 ft from ground to top) was loaded with 44 415 lb of paper rolls in two rows by two high as shown in Figure P7-25. The rolls are 40-in-dia by 38-in-long and weigh about 900 lb each. They are wedged against rolling backward but not against sliding sidewards. The empty trailer weighed 14 000 lb. The driver claims that he was traveling at less than 15 mph and that the load of paper shifted inside the trailer, struck the trailer sidewall, and tipped the truck. The paper company that loaded the truck claims the load was properly stowed and would not shift at that speed. Independent test of the coefficient of friction between similar paper rolls and a similar trailer floor give a value of 0.43 +/- 0.08. The composite center of gravity of the loaded trailer is estimated to be 7.5 ft above the road. Determine the truck speed that would cause the truck to just begin to tip and the speed at which the rolls will just begin to slide sidways. What do you think caused the accident?

Given:

Weight of paper

Wp  44415  lbf

Weight of trailer

Wt  14000  lbf

Radius of curve

r  50 ft

Nominal coefficient of friction

μnom  0.43

Coefficient of friction uncertainty

u μ  0.08

Trailer width

w  8  ft

Height of CG from pavement

h  7.5 ft

Assumptions: 1. The paper rolls act as a monolith since they are tightly strapped together with steel bands. 2. The tractor has about 15 deg of potential roll freedom versus the trailer due to their relative angle in plan during the turn combined with the substantial pitch freedom in the fifth wheel. So the trailer can tip independently of the tractor. 3. The outside track width of the trailer tires is equal to the width of the trailer. Solution: 1.


First, calculate the location of the trailer's CG with respect to the outside wheel when it is on the reverse-banked curve. From the figure at right, Tilt angle

θ  3  deg

a  h  tan( θ )

a  0.393 ft

w

b  3.607 ft

b 

2

a

xbar  b  cos( θ )

xbar  3.602 ft

ybar  b  sin( θ ) 

h

3° 7.500'

ybar

cos( θ )

ybar  7.699 ft The coordinates of the CG of the loaded trailer with respect to the lower outside corner of the tires are: xbar  3.602 ft

a

b

ybar  7.699 ft

xbar 4.000'


2.


The trailer is on the verge of tipping over when the couple due to centrifugal force is equal to the couple formed by the weight of the loaded trailer acting through its CG and the vertical reaction at the outside edge of the tires. At this instant, it is assumed that the entire weight of the trailer is reacted at the outside tires with the inside tires carrying none of the weight. Any increase in tangential velocity of the tractor and trailer will result in tipping. Summing moments about the tire edge (see figure below),

M

Fw xbar  Fc ybar = 0

(1)

where Fc is the centrifugal force due to the normal acceleration as the tractor and trailer go through the curve. The normal acceleration is a tip =

vtip

Fc

2

(2)

r

and the force necessary to keep the tractor trailer following a circular path is Fc = mtot  a tip

Fw ybar (3)

where mtot is the total mass of the trailer and its payload. Combining equations (2) and (3) and solving for vtip, we have vtip =

Fc r

Rx Ry xbar

(4)

mtot

or, vtip =

Fc r g

(5)

Fw

3. Calculate the minimum tipping velocity of the tractor/trailer. From equations (1) and (5), Total weight

Fw  Wt  Wp

Centrifugal force required to tip the trailer

Fc 

Minimum tipping speed

vtip 

xbar ybar

 Fw

Fc r g Fw

Fw  58415 lbf Fc  27329 lbf

vtip  18.7 mph

Thus, with the assumptions that we have made, the trailer would not begin to tip over until it reached a speed of vtip  18.7 mph 4.

The load will slip when the friction force between the paper rolls and the trailer floor is no longer sufficient to react the centrifugal force on the paper rolls. Looking at the FBD of the paper rolls below, we see that Normal force between paper and floor

Fn = Wp cos( θ )  Fcp sin( θ )

(6)



Tangential force tending to slide the paper Ft = Wp sin( θ )  Fcp cos( θ )

Fcp

(7)

Wp

Centrifugal force on the paper 2

Wp vs Fcp =  as =  g g r Wp

But, the maximum friction force is

Ft

(8)

Fn

Ff = μ  Fn = Ft

(9)

Substituting equation (9) into (7), then combining (6) and (7) to eliminate Fn, and solving for Fcp yields

Fcp =

Wp ( μ  cos( θ )  sin( θ ) )

(10)

μ  sin( θ )  cos( θ )

Substituting equation (10) into (8), to eliminate Fcp, and solving for vs yields

vs =

5.

 μ  cos( θ )  sin( θ )   r g  μ  sin( θ )  cos( θ )   

(11)

Use the upper and lower limit on the coefficient of friction to determine an upper and lower limit on the speed necessary to cause sliding. Maximum coefficient

μmax  μnom  u μ

μmax  0.510

Minimum coefficient

μmin  μnom  u μ

μmin  0.350

Maximum velocity to cause sliding

vsmax 

 μmax  cos( θ )  sin( θ )   μ  sin( θ )  cos( θ )   r g  max 

vsmax  18.3 mph

Minimum velocity to cause sliding

vsmin 

6.

 μmin  cos( θ )  sin( θ )   μ  sin( θ )  cos( θ )   r g  min 

vsmin  14.8 mph

This very rough analysis shows that , if the coefficient of friction was at or near the low end of its measured value, the paper load could slide at a tractor/trailer speed of 15 mph, which would lead to the trailer tipping over. In any case, it appears that the paper load would slide before the truck would tip with the load in place.




Figure P7-26 shows a V-belt drive. The sheaves (pulleys) have pitch diameters of 150 and 300 mm, respectively. The smaller sheave is driven at a constant 1750 rpm. For a cross-sectional, differential element of the belt, write the equations of its acceleration for one complete trip around both pulleys including its travel between the pulleys. Compute and plot the acceleration of the differential element versus time for one circuit around the belt path. What does your analysis tell you about the dynamic behavior of the belt?

Given:

Sheave radii and speed: r2  75 mm

r4  150  mm

ω  1750 rpm

Assumptions: Center distance: C  450  mm Solution: 1.


Draw a schematic representation of the V-belt drive to scale and label it.

443.706 B 3 9.594°

A 4

2

R150.000 D

R75.000

C

From the layout,

2.

Angle to point of tangency

β  9.594  deg

Distance from A to B

LAB  443.706  mm

Distance from B to C

LBC   π  2  β  r4

Distance from C to D

LCD  443.706  mm

Distance from D to A

LDA   π  2  β  r2

LDA  210.502 mm

Total path length

L  LAB  LBC  LCD  LDA

L  1619.387 mm

Calculate the angular velocity of each sheave. Sheave 2 Sheave 4

3.

LBC  521.473 mm

ω  183.260 ω 

r2 r4

 ω

rad sec ω  91.630

rad sec

Calculate the acceleration of an element on the belt for each portion of the path, starting at A.



mm

AAB  0 

From A to B the acceleration is zero since the belt velocity is constant

sec

2

From B to C the tangential acceleration is zero since the belt velocity is constant The normal component is ABC  r4 ω

6 mm

2

ABC  1.259  10

sec

2

ACD  0 

From C to D the acceleration is zero since the belt velocity is constant

mm sec

2

From D to A the tangential acceleration is zero since the belt velocity is constant The normal component is ADA  r2 ω

6 mm

2

ADA  2.519  10

sec 4.

2

Plot the acceleration over the entire path using a path variable of s. s  0  mm 5  mm  L Define a range function that will plot a variable only between two limits using the Heaviside step function . R( s a b )  Φ ( s  a )  Φ ( s  b ) The acceleration function for the entire path is. Let

s1  LAB

s2  LAB  LBC

s3  LAB  LBC  LCD

A ( s)  AAB  R s 0  mm s1  ABC  R s s1 s2  ACD  R s s2 s3  ADA  R s s3 L

BELT ACCELERATION ALONG PATH


3000

2000

1000

0

0

0.25

0.5

0.75

1

1.25

1.5

1.75

2

Distance Along Path, m

5.

The graph shows the sudden change in acceleration as the belt enters and leaves a sheave. This results in infinite jerk at these points. This results in belt "hop" at these points, a condition that will cause eventual fatigue failure and wear in the belt. Because of the belt hop caused by infinite jerk at the initial belt/sheave tangency points the span of the belt is continuously vibrating.




Write a program using an equation solver or any computer language to solve for the displacements, velocities, and accelerations in an offset crank-slider linkage as shown in Figure P7-2. Plot the variation in all link's angular and pin's linear positions, velocities, and accelerations with a constant angular velocity input to the crank over one revolution for both open and crossed configurations of the linkage. To test the program, use data from row a of Table P7-2. Check your results with program SLIDER.

Enter:

Link lengths: a  1.4 in

Link 2

Link 3

c  1  in

Offset:

ω  10

Link 2 velocity and acceleration: Crank position range: Solution: 1.

b  4  in rad

α  0 

sec

rad sec

2

θ  0  deg 2  deg  360  deg


Draw the linkage and label it.

Y 4 B 3 b

A

c 2 a

2 X

O2 d 2.

Determine 3 and d using equations 4.16 and 4.17. Open:

 a  sin θ  b 

 

θ θ  asin 

 

 

c

π 

  

d 2 θ  a  cos θ  b  cos θ θ

Crossed:

 a sin θ  b 

 

θ θ  asin

 

 

c

 

  

d 1 θ  a  cos θ  b  cos θ θ 3.

Determine the angular velocity of link 3 using equation 6.22a. Open

 

ω θ 

a b



   ω cos θ θ  cos θ


Crossed

4.


 

ω θ 

a b



   ω cos θ θ  cos θ


 



 

 

VA θ  a  ω sin θ  j  cos θ

 

  

 

VAX θ  Re VA θ 5.

6.

  

VAY θ  Im VA θ


 

 

    

 

 

    

Open

VB1 θ  a  ω sin θ  b  ω θ  sin θ θ

Crossed

VB2 θ  a  ω sin θ  b  ω θ  sin θ θ

Using the Euler identity to expand equation 7.15b for AA, determine its x and y components.

 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ

 

  

 

AAX θ  Re AA θ 7.


Open

Crossed

8.

  


 

2

   2    b  cos θ θ 

 

2

 

a  α cos θ  a  ω  sin θ  b  ω θ  sin θ θ

 

a  α cos θ  a  ω  sin θ  b  ω θ  sin θ θ

α θ 

α θ 

   2    b  cos θ θ 


 

2     2  b  α θ  sin θ θ   b  ω θ  cos θ θ 

 

2     2  b  α θ  sin θ θ   b  ω θ  cos θ θ 

Open:

AB2 θ  a  α sin θ  a  ω  cos θ 

Crossed:

AB1 θ  a  α sin θ  a  ω  cos θ 

(See plots on the following pages.)


Plot the angular position of link 3 and the distance of pin B from the y axis for open (solid) and crossed (dashed) configurations.

ANGULAR POSITION OF LINK 3 250 200

Angle, deg

150 100 50 0  50

0

45

90

135

180

225

270

315

360

270

315

360

Crank Angle, deg Open Crossed

POSITION OF PIN B

Distance d, in

9.


6 5 4 3 2 1 0 1 2 3 4 5 6

0

45

90

135

180

Crank Angle, deg Open Crossed

225



10. Plot the angular velocity of link 3 and the velocities of pins A and B for open (solid) and crossed (dashed) configurations. ANGULAR VELOCITY OF LINK 3 4

Anglular Velocity, rad/sec

3 2 1 0 1 2 3 4

0

45

90

135

180

225

270

315

360

270

315

360

270

315

360

Crank Angle, deg

X & Y VELOCITY COMPONENTS OF PIN A

Velocity, in/sec

20

10

0  10  20

0

45

90

135

180

225

Crank Angle, deg

VELOCITY OF PIN B

Velocity, in/sec

20

10

0  10  20

0

45

90

135

180

Crank Angle, deg

225



11. Plot the angular acceleration of link 3 and the accelerations of pins A and B for open (solid) and crossed (dashed configurations. ANGULAR ACCELERATION OF LINK 3

Anglular Acceleration, rad/sec^2

60

30

0

 30

 60

0

45

90

135

180

225

270

315

360

270

315

360

270

315

360

Crank Angle, deg

X & Y ACCELERATION COMPONENTS OF PIN A


200

100

0  100  200

0

45

90

135

180

225

Crank Angle, deg

ACCELERATION OF PIN B


200 100 0  100  200

0

45

90

135

180

Crank Angle, deg

225




Write a program using an equation solver or any computer language to solve for the displacements, velocities, and accelerations in an inverted slider-crank linkage as shown in Figure P7-3. Plot the variation in all link's angular and pin's linear positions, velocities, and accelerations with a constant angular velocity input to the crank over one revolution for both open and crossed configurations of the linkage. To test the program, use data from row e of Table P7-3 except for the value of 2, which will be set to zero for this exercise.

Enter:


a  4  in


γ  30 deg

Link 2 velocity and accel.

ω  45

1.

rad

α  0 

sec

d  8  in

Link 1

rad sec

2

θ  0  deg 2  deg  360  deg

Crank position range: Solution:

c  2  in

Link 4


Draw the linkage and label it.

B y

 4

3 b A

a 2

c

2 d

02 2.

x

1

04


 

 



 



P θ  a  sin θ  sin γ  a  cos θ  d  cos γ

 

 



 



Q θ  a  sin θ  cos γ  a  cos θ  d  sin γ

 

R  c sin γ

Open:

 

 

S θ  R  Q θ



 

  



 

θ θ  2  atan2 2  S θ  T θ 

 

 

θ θ  θ θ  γ

 

b 1 θ 

 

 

T θ  2  P θ

   sin θ θ 

a  sin θ  c sin θ θ

 

 2  4 Sθ U θ

T θ

 

U θ  Q θ  R





 

  

 

θ θ  2  atan2 2  S θ T θ 

Crossed:

 

 2  4 Sθ U θ

T θ

 

θ θ  θ θ  γ

 

b 2 θ 

3.

 

   sin θ θ 

a  sin θ  c sin θ θ

Calculate the angular velocity of links 3 and 4 and the slip velocity using equations 6.30. Open:

 

ω θ 

 

bdot1 θ 

   b 1 θ  c cos γ

a  ω cos θ  θ θ

 

         c sinθθ cos θ θ 

a  ω sin θ  ω θ  b 1 θ  sin θ θ

 

 

ω θ  ω θ

 

Crossed: ω θ 

 

bdot2 θ 

   b 2 θ  c cos γ

a  ω cos θ  θ θ

 

         c sinθθ cos θ θ 

a  ω sin θ  ω θ  b 2 θ  sin θ θ

 

 

ω θ  ω θ 4.

Solve for the accelerations using equations (7.26) and (7.27). Open:

 

  

P1 θ  a  α cos θ θ  θ

 



 

 2   

 

 

R1 θ  c ω θ  sin θ θ  θ θ

 

 

  

2

Q1 θ  a  ω  sin θ θ  θ

  

 

 

T1 θ  b 1 θ  c cos θ θ  θ θ

 

α θ 

 

 

 

 

P1 θ  Q1 θ  R1 θ  S 1 θ

 

T1 θ

 

2

  

  

 

   



  



K1 θ  a  ω  b 1 θ  cos θ θ  θ  c cos θ θ  θ

   c sinθθ  θ

L1 θ  a  α b 1 θ  sin θ  θ θ M1 θ  ω θ   b 1 θ

 

 2

 

 

 2  c2  2 b1θ c cosθθ  θθ     

 

S 1 θ  2  bdot1 θ  ω θ

 

N1 θ  2  bdot1 θ  c ω θ  sin θ θ  θ θ




 

bddot1 θ  

 


 

 

 

 

K1 θ  L1 θ  M1 θ  N1 θ

 

T1 θ



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ

 

  

 

AAX θ  Re AA θ

 

  


       j  cosθθ  2  c ω θ   cos θ θ   j  sin θ θ  

AB1 θ  c α θ  sin θ θ

 

 

θAB1 θ  arg AB1 θ 

AB1 θ  AB1 θ Crossed:

 

  

P2 θ  a  α cos θ θ  θ

 



 

 2   

 

 

R2 θ  c ω θ  sin θ θ  θ θ

 

 

  

2

Q2 θ  a  ω  sin θ θ  θ

  

 

 

T2 θ  b 2 θ  c cos θ θ  θ θ

 

α θ 

 

 

 

 

P2 θ  Q2 θ  R2 θ  S 2 θ

 

T2 θ

 

2

  

  

 

   



  



K2 θ  a  ω  b 2 θ  cos θ θ  θ  c cos θ θ  θ

   c sinθθ  θ

L2 θ  a  α b 2 θ  sin θ  θ θ M2 θ  ω θ   b 2 θ

 2  c2  2 b2θ c cosθθ  θθ

 

 2

 

 

    

 

 

 

N2 θ  2  bdot2 θ  c ω θ  sin θ θ  θ θ

 

bddot2 θ  

 

 

 

K2 θ  L2 θ  M2 θ  N2 θ

 

T2 θ

       j  cosθθ  2  c ω θ   cos θ θ   j  sin θ θ  

AB2 θ  c α θ  sin θ θ

 

 

AB2 θ  AB2 θ

 

S 2 θ  2  bdot2 θ  ω θ

θAB2 θ  arg AB2 θ 




Plot the angular position of link 4 and the distance b for open (solid) and crossed (dashed) configurations.

ANGULAR POSITION OF LINK 4 400

Angle, deg

300

200

100

0

 100

0

45

90

135

180

225

270

315

360

225

270

315

360

Crank Angle, deg

LENGTH b 20

10 Distance b, in

5.


0

 10

 20

0

45

90

135

180

Crank Angle, deg


Plot the angular acceleration of link 4 and the accelerations of points A and B for open (solid) and crossed (dashed) configurations. ANGULAR ACCELERATION OF LINK 4

Anglular Acceleration, rad/sec^2

5000 3750 2500 1250 0  1250  2500  3750  5000

0

45

90

135

180

225

270

315

360

270

315

360

270

315

360

Crank Angle, deg

X & Y ACCELERATION COMPONENTS OF PIN A


10000

5000

0  5000  10000

0

45

90

135

180

225

Crank angle, deg

ACCELERATION OF PIN B 10000 Acceleration, in/sec^2

6.


7500

5000

2500

0

0

45

90

135

180

Crank Angle, deg

225




Write a program using an equation solver or any computer language to solve for the displacements, velocities, and accelerations in a geared fivebar linkage as shown in Figure P7-4. Plot the variation in all link's angular and pin's linear positions, velocities, and accelerations with a constant angular velocity input to the crank over one revolution for both open and crossed configurations of the linkage. To test the program, use data from row a of Table P7-4. Check your results with program FIVEBAR.

Enter:


f  6  in

Link 3

b  7  in

Link 2

a  1  in

Link 4

c  9  in

d  4  in

Link 5

Gear ratio, phase angle, and crank velocity and acceleration:

Solution: 1.

λ  2

ϕ  30 deg

ω  10 rad sec

Coupler data:

Rpa  6  in

δ  30 deg

1

α  0  rad sec

2



 

θ θ  λ θ  ϕ 2.


 







 

 







 

A θ  2  c d  cos λ θ  ϕ  a  cos θ  f



B θ  2  c d  sin λ θ  ϕ  a  sin θ

 

2



           D θ  C θ  A  θ E θ  2  B θ 2

2

2

2

C θ  a  b  c  d  f  2  a  f  cos θ   2  d  a  cos θ  f  cos λ θ  ϕ    2  a  d  sin θ  sin λ θ  ϕ



 





  a cosθ  f 

 





  a sinθ

 

 

 

F θ  A θ  C θ

G θ  2  b   d  cos λ θ  ϕ H θ  2  b   d  sin λ θ  ϕ

 

2

2

2

2



  

2

K θ  a  b  c  d  f  2  a  f  cos θ   2  d  a  cos θ  f  cos λ θ  ϕ    2  a  d  sin θ  sin λ θ  ϕ

        L θ  K θ  G θ M  θ  2  H  θ 3.



 

 

 

N θ  G θ  K θ


 





 

 

M θ

 





 

 

E θ

 





 

 

M θ

 





 

 

E θ

θ θ  2   atan2 2  L θ M θ  θ θ  2   atan2 2  D θ E θ 

CROSSED

θ θ  2   atan2 2  L θ M θ  θ θ  2   atan2 2  D θ E θ 

 2  4 Lθ N θ 

 2  4 Dθ F θ   2  4 Lθ N θ 

 2  4 Dθ F θ 



4.

Use equation 6.32c to find 5. ω  λ ω

5.

Calculate 3 and 4 and the pin velocities using equations 6.33. OPEN

 

ω θ 

        d ω sinθθ  θθ b   cos θ θ  2  θ θ   cos θ θ  

2  sin θ θ  a  ω sin θ  θ θ

      d ω sinθθ c sin θ θ  VA θ  a  ω  sin θ  j  cos θ   

ω θ 

 

 

a  ω sin θ  b  ω θ  sin θ θ

  

 


 

  


 

    j  cosθθ

VBA1 θ  b  ω θ  sin θ θ

 



 

 

 



    j  cosθθ

VC θ  d  ω sin θ θ

 

VB1 θ  VA θ  VBA1 θ

 

 

VB1x θ  Re VB1 θ



 

VB1y θ  Im VB1 θ

CROSSED

 

ω θ 

        d ω sinθθ  θθ b   cos θ θ  2  θ θ   cos θ θ  

2  sin θ θ  a  ω sin θ  θ θ

      d ω sinθθ c sin θ θ  VBA2 θ  b  ω θ   sin θ θ   j  cos θ θ    

ω θ 

 

a  ω sin θ  b  ω θ  sin θ θ

 

 

 



 

VB2 θ  VA θ  VBA2 θ

 

VB2x θ  Re VB2 θ

 



 

VB2y θ  Im VB2 θ

6.

Use equation 7.28c to find 5. α  λ α

7.

Calculate 3 and 4 and the pin accelerations using equations 7.29. OPEN

    a ω2 cosθ  θθ  2 2  b  ω θ  cos θ θ  θ θ   d  ω  cos θ θ  θ θ   2  d  α sin θ θ  θ θ   c ω θ b  sin θ θ  θ θ 

a  α sin θ  θ θ

 

α θ 

    a ω2 cosθ  θθ  2 2  c ω θ  cos θ θ  θ θ   d  ω  cos θ θ  θ θ   2  d  α sin θ θ  θ θ   b  ω θ c sin θ θ  θ θ  a  α sin θ  θ θ

 

α θ 


 




 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ

 

  

 

AAx θ  Re AA θ

  


 

       j  cosθθ  2  b  ω θ   cos θ θ   j  sin θ θ   AB1 θ  AA θ  ABA1 θ ABA1 θ  b  α θ  sin θ θ

 



 

AB1x θ  Re AB1 θ

 

 



 

AB1y θ  Im AB1 θ



    j  cosθθ  2  c ω   cos θ θ   j  sin θ θ   ACx  θ  Re AC θ  ACy  θ  Im AC θ  AC θ  c α sin θ θ

CROSSED

    a ω2 cosθ  θθ  2 2  b  ω θ  cos θ θ  θ θ   d  ω  cos θ θ  θ θ   2  d  α sin θ θ  θ θ   c ω θ b  sin θ θ  θ θ 

a  α sin θ  θ θ

 

α θ 

    a ω2 cosθ  θθ  2 2  c ω θ  cos θ θ  θ θ   d  ω  cos θ θ  θ θ   2  d  α sin θ θ  θ θ   b  ω θ c sin θ θ  θ θ  a  α sin θ  θ θ

 

α θ 

 

       j  cosθθ  2  b  ω θ   cos θ θ   j  sin θ θ   AB2 θ  AA θ  ABA2 θ ABA2 θ  b  α θ  sin θ θ

 



 

AB2x θ  Re AB2 θ

 



 

AB2y θ  Im AB2 θ




Find the acceleration of the slider in Figure 3-33 for 2 = 110 deg with respect to the global X axis assuming a constant 2 = 1 rad/sec CW. Use a graphical method.

Given:


a  2.170  in

Link 3 (A to B)

b  2.067  in

Link 4 (O4 to B)

c  2.310  in

Link 1 (O2 to O4)

d  1.000  in

Link 5 (B to C)


θ2XY  110  deg

  102  deg

Coordinate angle

Solution: 1.


  1  rad sec

Input crank angular acceleration

  0  rad sec

1

CW

2



Direction of AAt Y Direction of ABAt

148.950°

A 3

Direction of ABt

B

2 4 O2

5 X

57.635° y

158.818° 6

O4 C x

102.000° Direction of AC Direction of ACBt

  θ2XY   2.

  212.000 deg

In order to solve for the acceleration at point B, we will need 3, 3, and 4. From the graphical position solution above (in the local xy coordinate system), θ  148.950  deg  180  deg  

θ  70.950 deg

θ  57.635 deg  

θ  159.635 deg


a   sin θ    b sin θ  θ

 

 

ω  0.832 rad sec

ω 

a   sin   θ  c sin θ  θ

 

 

ω  0.591 rad sec

1

1




3.


4.


2


θABn  θ  180  deg  

5.

2

θABn  122.365 deg 2

AAn  a  

2

AAn  2.170 in sec

θAAn    180  deg  

θAAn  290.000 deg

AAt  a  


θAAt    90 deg  

θAAt  200.000 deg

2

2


2


θABAn  θ  180  deg  


Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn +  and AAn at an angle of AAn + . From the tip of AAn, draw AAt at an angle of AAt + . From the tip of AAt, draw ABAn at an angle of BAn + . Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABt, and ABAt. Y

0

1 IN/S/S

X

n

0.850 A B AB

Acceleration Scale

t B

103.340°

A

t BA

A

AA n

AA

n

A BA

6.


ka  1.000 

in sec

7.

2

AA  AAn

AA  2.170 in sec

AB  0.850  ka

AB  0.850 in sec

2 2

at an angle of 290.0 deg at an angle of -103.340 deg

From Problem 6-70a, the angular velocity of link 5 is   0.145  rad sec

1

CCW. From the graphical

layout,   158.818  deg in the global XY coordinate system. 8.

For point C, equation 7.4 becomes: AC = AB + (ACBt + ACBn) , where AB is known and 2

ACBn  e 


2


θACBn  



Y 0.005 AC X

0

0.5 IN/S/S t ACB

Acceleration Scale

AB n

A CB 9.


ka  0.50

in sec

AC  0.005  ka

2

AC  0.0025 in sec

2




PROBLEM 7-61b Find the acceleration of the slider in Figure 3-33 for 2 = 110 deg with respect to the global X

Statement:

axis assuming a constant 2 = 1 rad/sec CW. Use an analytical method. Given:


a  2.170  in

Link 3 (A to B)

b  2.067  in

Link 4 (O4 to B)

c  2.310  in

Link 1 (O2 to O4)

d  1.000  in

Link 5 (B to C)


θ2XY  110  deg

  102  deg

Coordinate angle

Solution: 1.


ω2  1  rad sec


α2  0  rad sec

1

CW

2



Y A 3 B

2 4

5 X

O2

y 6

O4 x

2.

C

102°

Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. Transform crank angle to the local xy coordinate system:

θ2  θ2XY   K1 

θ2  212.000 deg

d

K2 

a

K1  0.4608 2

K3 

d c

K2  0.4329 2

2

a b c d

2

2 a c

K3  0.6755

A  cos θ2  K1  K2 cos θ2  K3 C  K1   K2  1   cos θ2  K3 A  0.2662

B  1.0598

C  2.3515

B  2  sin θ2


3.







2

θ4xy  2  atan2 2  A B  4.

B  4 A  C



θ4xy  200.365 deg


2

d

K5 

b

2

2

c d a b

2

2 a b

K5  0.5178

D  cos θ2  K1  K4 cos θ2  K5

5.

7.

D  2.2370

E  2  sin θ2

E  1.0598

F  K1   K4  1   cos θ2  K5

F  0.3808






2

θ3xy  2  atan2 2  D E  6.

K4  0.4838

E  4  D F



θ3xy  289.050 deg


ω3 

a  ω2 sin θ4xy  θ2  b sin θ3xy  θ4xy

ω3  0.832

rad

ω4 

a  ω2 sin θ2  θ3xy  c sin θ4xy  θ3xy

ω4  0.591

rad

sec

sec

Use equations 7.12 to determine the angular accelerations of links 3 and 4. A  c sin θ4xy

B  b  sin θ3xy

D  c cos θ4xy

E  b  cos θ3xy

A  0.804 in

B  1.954 in

D  2.166 in

E  0.675 in

C  a  α2 sin θ2  a  ω2  cos θ2  b  ω3  cos θ3xy  c ω4  cos θ4xy 2

C  0.618

2

2

in sec

2

F  a  α2 cos θ2  a  ω2  sin θ2  b  ω3  sin θ3xy  c ω4  sin θ4xy 2

F  0.079

8.

2

C D  A  F A  E  B D

α3  0.267

rad sec

2


θ4  θ4xy   9.

Determine 5 using equation 4.17. Offset:

2

in sec

α3 

2

cc  0  in

θ4  302.365 deg

α4 

C  E  B F A  E  B D

α4  0.120

rad sec

2



 c sin θ4  cc  π e  

θ5  asin 

θ5  158.818 deg


ω5 

c cos θ4   ω4 e cos θ5

ω5  0.145

rad sec

11. Determine the angular acceleration of link 5 using equation 7.16d. c α4 cos θ4  c ω4  sin θ4  e ω5  sin θ5 2

α5 

2

α5  0.156

e cos θ5

sec

12. Use equation 7.16e for the acceleration of pin C. AC  c α4 sin θ4  c ω4  cos θ4  e α5 sin θ5  e ω5  cos θ5 2

AC  0.00161

in sec

2

rad

2

A positive sign means that AC is to the right.

2




Given:

Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular acceleration of link 4 and the linear acceleration of slider 6 in the sixbar crank-slider linkage of Figure 3-33 as a function of the angle of input link 2 for a constant 2 = 1 rad/sec CW. Plot Ac both as a function of 2 and separately as a function of slider position as shown in the figure. Link lengths: Link 2 (O2 to A)

a  2.170  in

Link 3 (A to B)

b  2.067  in

Link 4 (O4 to B)

c  2.310  in

Link 1 (O2 to O4)

d  1.000  in

Link 5 (B to C)

e  5.400  in   102  deg

Coordinate angle

Solution:

θ2XY  0  deg 1  deg  360  deg

Crank angle:


ω2  1  rad sec


α2  0  rad sec

1

CW

2


1.

Transform crank angle to the local xy coordinate system: θ2 θ2XY   θ2XY  

2.


d

K1  0.4608

a 2

K3 

2

2

a b c d

K2 

d

K2  0.4329

c

2

K3  0.6755

2 a c

A  θ2XY   cos θ2 θ2XY    K1  K2 cos θ2 θ2XY    K3 B θ2XY   2  sin θ2 θ2XY   C θ2XY   K1   K2  1   cos θ2 θ2XY    K3 3.






θ4xy θ2XY   2   atan2 2  A  θ2XY  B θ2XY   4.



B θ2XY   4  A  θ2XY   C θ2XY   2


2

d

K5 

b

2

2

c d a b

2

K4  0.4838

2 a b

K5  0.5178

D θ2XY   cos θ2 θ2XY    K1  K4 cos θ2 θ2XY    K5 E θ2XY   2  sin θ2 θ2XY   F  θ2XY   K1   K4  1   cos θ2 θ2XY    K5 5.






θ3xy θ2XY   2   atan2 2  D θ2XY  E θ2XY   6.



E θ2XY   4  D θ2XY   F  θ2XY   2



7.


ω3 θ2XY  

a  ω2 sin θ4xy θ2XY   θ2 θ2XY    b sin θ3xy θ2XY   θ4xy θ2XY  

ω4 θ2XY  

a  ω2 sin θ2 θ2XY   θ3xy θ2XY    c sin θ4xy θ2XY   θ3xy θ2XY  

Use equations 7.12 to determine the angular acceleration of link 4. A  θ2XY   c sin θ4xy θ2XY  

B θ2XY   b  sin θ3xy θ2XY  

D θ2XY   c cos θ4xy θ2XY  

E θ2XY   b  cos θ3xy θ2XY  

C θ2XY   a  α2 sin θ2 θ2XY    a  ω2  cos θ2 θ2XY    2

 b  ω3 θ2XY   cos θ3xy θ2XY    c ω4 θ2XY   cos θ4xy θ2XY   2

2

F  θ2XY   a  α2 cos θ2 θ2XY    a  ω2  sin θ2 θ2XY    2

 b  ω3 θ2XY   sin θ3xy θ2XY    c ω4 θ2XY   sin θ4xy θ2XY   2

α4 θ2XY  

2

C θ2XY   E θ2XY   B θ2XY   F  θ2XY  A  θ2XY   E θ2XY   B θ2XY   D θ2XY  ANGULAR ACCELERATION - LINK 4

Angular Acceleration, rad/s

2

1

0

1

0

45

90

135

180

225

Crank Angle, deg

8.


θ4 θ2XY   θ4xy θ2XY   

9.

Determine 5 using equation 4.17. Offset: cc  0  in

 c sin θ4 θ2XY    cc  π e  

θ5 θ2XY   asin 


ω5 θ2XY  

c cos θ4 θ2XY     ω4 θ2XY  e cos θ5 θ2XY  

270

315

360



11. Determine the angular acceleration of link 5 using equation 7.16d. c α4 θ2XY   cos θ4 θ2XY    c ω4 θ2XY   sin θ4 θ2XY    2

α5 θ2XY  

 e ω5 θ2XY   sin θ5 θ2XY   2

e cos θ5 θ2XY  

12. Use equation 7.16e for the acceleration of pin C. AC  θ2XY   c α4 θ2XY   sin θ4 θ2XY    c ω4 θ2XY   cos θ4 θ2XY    2

 e α5 θ2XY   sin θ5 θ2XY    e ω5 θ2XY   cos θ5 θ2XY   2

ACCELERATION - PIN C 6


4

2

0

2

4

0

45

90

135

180

Crank Angle, deg

225

270

315

360




Find the angular acceleration of link 6 in Figure 3-34b for 6 = 90 deg with respect to the global

Given:

Link lengths:

X axis assuming constant 2 = 10 rad/sec CW. Use a graphical method. Link 2 (O2 to A)

g  1.556  in

Link 3 (A to B)

f  4.248  in

Link 4 (O4 to C)

c  2.125  in

Link 5 (C to D)

b  2.158  in

Link 6 (O6 to D)

a  1.542  in

Link 5 (B to D)

p  3.274  in

Link 1 X-offset

d X  3.259  in

Link 1 Y-offset

d Y  2.905  in

Angle CDB

  36.0 deg


θ  90 deg Global XY system ω  10 rad sec


1

CW

See Figure 3-34b and Mathcad file P0763a.


Direction of ACDt Y

Direction of ADt

Direction of AAt X 2 A

D

O2

5

6

C

y

Direction of AABt

O6 3

5

4 x

Direction of ACt

O4

B Direction of ABDt

2.

Since this linkage is a Stephenson's II sixbar, we will have to start at link 6 and work back to link 2. We will assume a value for 6 and eventually find a value for 2. We will then multiply the magnitudes of all velocities by the ratio of the actual 2 to the found 2. Use Problem 6-73 for the link angular velocities. Assume:

  100  rad sec

2

CCW

From Problem 6-73 and the graphical layout (angles in the global XY coordinate system):   24.124 rad sec   14.64  rad sec   14.64  rad sec   3.016  rad sec

1

1 1 1

CW

  90.000 deg

CW

  217.003  deg

CW

  144.195  deg

CW

  144.628  deg

  253.003  deg



  10.000 rad sec

1

  221.690  deg

CW


3.


4.

For point C, this becomes: (ACt + ACn) = (ADt + ADn) + (ACDt + ACDn) , where 2

ACn  c 

ACn  455.450 in sec

θACn    180  deg

θACn  324.195 deg

ADn  a  

2

ADn  897.394 in sec

θADn    180  deg

θADn  270.000 deg

ADt  a  

ADt  154.200 in sec

θADt    90 deg

θADt  180.000 deg

2

5.

2

2

2

ACDn  b  

ACDn  462.523 in sec

θACDn    180  deg

θACDn  37.003 deg

2

Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ACn at an angle of ACn and ADn at an angle of ADn. From the tip of ADn, draw ADt at an angle of ADt + . From the tip of ADt, draw ACDn at an angle of CDn. Now that the vectors with known magnitudes are drawn, from the tips of ACn and ACDn, draw construction lines in the directions of ACt and ACDt, respectively. The intersection of these two lines are the tips of ACt, ACt, and ACDt.

n

AC

0

200 IN/S/S

Acceleration Scale

AC

t

AC 1.755

t

ACD AD n

ACD n

AD t

AD 0.425

6.


ka  200 

in sec

ACt  1.755  ka

2

ACt  351.00 in sec

2



ACt

 

  165.176 rad sec

c

ACDt  0.425  ka ACDt

  7.

ACDt  85.00 in sec

2

CW

For point B, equation 7.4 becomes: AB = AD + (ABDt + ABDn) , where 2

8.

CW

2

  39.388 rad sec

b

2

ABDn  p  

ABDn  701.715 in sec

θABDn    180  deg

θABDn  73.003 deg

ABDt  p  

ABDt  128.957 in sec

θABDt  θABDn  90 deg

θABDt  163.003 deg

2

1.010

2

AB

0

200 IN/S/S

Acceleration Scale

For point A, equation 7.4 becomes: (AAt + AAn) = AB + (AABt + AABn) , where 2

110.996°

t ABD

AAn  g  

AAn  155.600 in sec

θAAn    180  deg

θAAn  41.690 deg

2

AD

AABn  f  

2

AABn  38.641 in sec

θAABn    180  deg

2 n

ABD

θAABn  35.372 deg

0.153

t

AA

n

AA AA

2.021

AB

0 t AAB

100 IN/S/S

Acceleration Scale

n

AAB

9.

The requirement for this problem is that 2 = 0 (constant 2). For this to be so AAt must be zero. To do this the magnitude of AB must be smaller as shown above. Correction ratio:

r 

Corrected value of 6:

0.153

r  0.076

2.021   r 

  7.571

rad sec

2




Find the angular acceleration of link 6 in Figure 3-34b for 6 = 90 deg with respect to the global X axis assuming constant 2 = 10 rad/sec CW. Use an analytic method.

Given:

Solution: 1.


g  1.556  in

Link 3 (A to B)

f  4.248  in

Link 4 (O4 to C)

c  2.125  in

Link 5 (C to D)

b  2.158  in

Link 6 (O6 to D)

a  1.542  in

Link 5 (B to D)

p  3.274  in

Link 1 X-offset

d X  3.259  in

Link 1 Y-offset

d Y  2.905  in

Angle CDB

δ5  36.0 deg


θ6  90 deg

Global XY system


ω2  10 rad sec


α2  0.0 rad sec

1

CW

2


Since this linkage is a Stephenson's II sixbar an iterative solution must be used. This problem is suited to a longer-term project rather than a daily homework problem.




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular acceleration of link 6 in Figure 3-34b as a function of 2 for a constant 2 = 10 rad/sec CW.

Given:

Solution: 1.


g  1.556  in

Link 3 (A to B)

f  4.248  in

Link 4 (O4 to C)

c  2.125  in

Link 5 (C to D)

b  2.158  in

Link 6 (O6 to D)

a  1.542  in

Link 5 (B to D)

p  3.274  in

Link 1 X-offset

d X  3.259  in

Link 1 Y-offset

d Y  2.905  in

Angle CDB

δ5  36.0 deg


θ6  90 deg

Global XY system


ω2  10 rad sec


α2  0.0 rad sec

1

CW

2


Since this linkage is a Stephenson's II sixbar an iterative solution must be used. This problem is suited to a longer-term project rather than a daily homework problem.




Use a compass and straightedge to draw the linkage in Figure 3-35 with link 2 at 90 deg and find the angular acceleration of link 6 assuming a constant 2 = 10 rad/sec CCW. Use a graphical method.

Given:


a  1.000  in

Link 3 (A to B)

b  3.800  in

Link 4 (O4 to B)

c  1.286  in

Link 1 (O2 to O4)

d  3.857  in

Link 4 (O4 to D)

e  1.429  in

Link 5 (D to E)

f  1.286  in

Link 6 (O6 to E)

g  0.771  in

Link 1 (O4 to O6)

h  0.786  in

  90 deg

Angle BO4D

  157  deg

Crank angle:

Solution: 1.


  10 rad sec


  0  rad sec

1

CCW

2


Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest. Direction of AAt Direction of ADt

Y A

D

122.085° 33.359°

2

100.938° 5

3 4 O2

X

6 O4 O6 E

35.228°

B Direction of AEt Direction of ABAt

Direction of AEDt

Direction of ABt

2.

In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above, θ  33.359 deg

θ  33.359 deg

θ  122.085  deg

θ  122.085 deg

θ  122.085  deg  

θ  34.915 deg


 

 

ω  1.398 rad sec

 

 

ω  6.496 rad sec

ω 

a   sin θ    b sin θ  θ

ω 

a   sin   θ  c sin θ  θ

1

1




3.


4.


2


θABn  θ  180  deg

θABn  57.915 deg

2

AAn  a  

AAn  100.000 in sec

θAAn    180  deg

θAAn  270.000 deg

AAt  a  


θAAt    90 deg

θAAt  0.000 deg

2

5.

2

2

2

2



θABAn  θ  180  deg


Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn and AAn at an angle of AAn. From the tip of AAn, draw AAt at an angle of AAt + . From the tip of AAt, draw ABAn at an angle of BAn. Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt,

6.

respectively. The intersection of these two lines are the tips of AB, ABt, and ABAt. From the graphical solution at right, in

ka  25


sec AB  2.919  ka

AB  73.0 in sec

1.951

2 n

AB

2

ABt AB

at an angle of 15.969 deg ABt  1.951  ka   6.


ABt

2

  37.928 rad sec

c

15.969° t ABA

2

CCW 2.919

From the graphical layout,   100.938  deg  180  deg

  79.062 deg

  35.228 deg

  35.228 deg

0

AA n

AA

Acceleration Scale

Using equation (6.18), ω 

ω 

 

 

 

 

e ω sin   θ  f sin   

e ω sin θ    g sin   

25 IN/S/S

n

ABA

ω  9.804 rad sec

1

ω  15.885 rad sec

1


7.


For point E, this becomes: (AEt + AEn) = (ADt + ADn) + (AEDt + AEDn) , where 2

AEn  g  ω

AEn  194.560 in sec

θAEn    180  deg

θAEn  144.772 deg

2

ADn  e ω


θADn  θ  180  deg

θADn  214.915 deg

ADt  e 


θADt  θ  90 deg

θADt  124.915 deg

2

θAEDn    180  deg

θAEDn  100.938 deg

AE

AEt

n

AED

n

AE

0

50 IN/S/S

Acceleration Scale

AD ADt n

AD


in

ka  50

sec AEt  0.239  ka AEt g

2

AEDn  123.597 in sec

t AED

 

2

AEDn  f  ω

0.239

8.

2

AEt  11.950 in sec

2

2

  15.499 rad sec

2

CCW

2




Use a compass and straightedge to draw the linkage in Figure 3-35 with link 2 at 90 deg and find the angular acceleration of link 6 assuming a constant 2 = 10 rad/sec CCW. Use an analytical method.

Given:


a  1.000  in

Link 3 (A to B)

b  3.800  in

Link 4 (O4 to B)

c  1.286  in

Link 1 (O2 to O4)

d  3.857  in

Link 4 (O4 to D)

e  1.429  in

Link 5 (D to E)

f  1.286  in

Link 6 (O6 to E)

g  0.771  in

Link 1 (O4 to O6)

h  0.786  in

Angle BO4D

δ4  157  deg

θ2  90 deg

Crank angle:

Solution: 1.


ω2  10 rad sec


α2  0  rad sec

1

CCW

2



Y A

D

122.085° 2

3

33.359°

100.938° 5

4 O2

X

6 O4 O6 E

35.228°

B 2.


d

K2 

a

K1  3.8570 2

K3 

d c

K2  2.9992 2

2

a b c d

2

K3  1.2015

2 a c


B  2.0000






θ41  2  atan2 2  A B  4.

C  5.0585

2

B  4 A  C



θ41  237.915 deg



K4 

5.


2

d

K5 

b

7.

2

2

K4  1.0150

2 a b

D  cos θ2  K1  K4 cos θ2  K5

D  7.6284

E  2  sin θ2

E  2.0000

F  K1   K4  1   cos θ2  K5

F  0.0856





2

E  4  D F



θ3  326.641 deg


ω3 

a  ω2 sin θ41  θ2  b sin θ3  θ41

ω3  1.398

ω4 

a  ω2 sin θ2  θ3  c sin θ41  θ3

ω4  6.496

rad sec rad sec


θ42  θ41  157  deg  360  deg 8.

K5  3.7714


θ3  2  atan2 2  D E  6.

2

c d a b

θ42  34.915 deg


h

K2 

e

K1  0.5500 2

K3 

e f

h g

K2  1.0195 2

2

g h

2

K3  0.7263

2  e g

A  cos θ42  K1  K2 cos θ42  K3 B  2  sin θ42 C  K1   K2  1   cos θ42  K3 A  0.1603 9.

B  1.1447

C  0.3796






θ6  2  atan2 2  A B 

2

B  4 A  C



θ6  35.228 deg


h f

2

K5 

2

2  e f

D  cos θ42  K1  K4 cos θ42  K5 E  2  sin θ42

2

g h e f

2

K4  0.6112 K5  1.0119 D  0.2408 E  1.1447



F  K1   K4  1   cos θ42  K5

F  0.7807






2

θ5  2  atan2 2  D E 

E  4  D F



θ5  280.938 deg

12. Determine the angular velocity of links 5 and 6 for the open circuit using equations 6.18.

 e ω4  sin θ6  θ42   f  sin θ5  θ6

ω5  9.804

 e ω4  sin θ42  θ5   g  sin θ6  θ5

ω6  15.885

ω5   ω6  

rad sec rad sec

13. Use equations 7.12 to determine the angular accelerations of links 3 and 4 for the crossed circuit. A  c sin θ41

B  b  sin θ3

D  c cos θ41

E  b  cos θ3

A  1.090 in

B  2.090 in

D  0.683 in

E  3.174 in

C  a  α2 sin θ2  a  ω2  cos θ2  b  ω3  cos θ3  c ω4  cos θ41 2

2

2

in

C  35.034

sec

2

F  a  α2 cos θ2  a  ω2  sin θ2  b  ω3  sin θ3  c ω4  sin θ41 2

2

in

F  141.900

sec

α3 

2

2

C D  A  F

α3  36.545

A  E  B D

rad sec

2

α4 

C  E  B F A  E  B D

α4  37.931

rad sec

14. Use equations 7.12 to determine the angular accelerations of links 5 and 6 for the open circuit. Use link 4 as the input to the second fourbar stage. A  g  sin θ6

B  f  sin θ5

D  g  cos θ6

E  f  cos θ5

A  0.037 ft

B  0.105 ft

D  0.052 ft

E  0.020 ft

C  e α4 sin θ42  e ω4  cos θ42  f  ω5  cos θ5  g  ω6  cos θ6 2

2

2

ft

C  4.583

2.000

s

F  e α4 cos θ42  e ω4  sin θ42  f  ω5  sin θ5  g  ω6  sin θ6 2

F  1.588

2

2

ft 2.000

s

α5 

C D  A  F A  E  B D

α5  38.104

rad sec

2

α6 

C  E  B F A  E  B D

α6  15.486

rad sec

2

2




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular acceleration of link 6 in the sixbar linkage of Figure 3-35 as a function of 2 for a constant 2 = 1 rad/sec CCW.

Given:


a  1.000  in

Link 3 (A to B)

b  3.800  in

Link 4 (O4 to B)

c  1.286  in

Link 1 (O2 to O4)

d  3.857  in

Link 4 (O4 to D)

e  1.429  in

Link 5 (D to E)

f  1.286  in

Link 6 (O6 to E)

g  0.771  in

Link 1 (O4 to O6)

h  0.786  in

  1  rad sec


1

CCW

  0  rad sec

2

See Figure 3-35 and Mathcad file P0766.   0  deg 1  deg  360  deg

1.

Define the range of the input angle:

2.


d

K2 

a

K1  3.8570 2

2

a b c d

2

K3  1.2015

2 a c

 

c

K2  2.9992

2

K3 

d

 

 

 

A   cos   K1  K2 cos   K3

 

 

B   2  sin 

 

C   K1   K2  1   cos   K3 3.




 



 

 

   2   atan2 2  A  B   4.

 2  4 A  C 

B 


2

d

K5 

b

 

2

2

c d a b

2

2 a b

 

 

 

D   cos   K1  K4 cos   K5

 

K4  1.0150

K5  3.7714

 

E   2  sin 

 

F   K1   K4  1   cos   K5 5.


 





 

 

   2   atan2 2  D  E   6.

 2  4 D F  

E 


 

a  

 

a  

  

  

b

c



    sin      



  sin      

sin    



sin    


7.



 

 

      157  deg  360  deg 8.


h

K2 

e

K1  0.5500 2

e f

2

g h

2

K3  0.7263

2  e g

 

g

K2  1.0195

2

K3 

h

    K1  K2 cos  K3

A'   cos  

 

  

B'   2  sin  

 

     K3

C'   K1   K2  1   cos   9.




 



 

 2  4 A'  C' 

 

   2   atan2 2  A'  B'  

B' 


2

h

K5 

f

 

2

2

g h e f

2

K4  0.6112

2  e f

    K1  K4 cos  K5

D'   cos  

 

  

E'   2  sin  

 

     K5

F'   K1   K4  1   cos  


 





 

 

   2   atan2 2  D'  E'  

 2  4 D' F' 

E' 

12. Determine the angular velocity of link 6 for the open circuit using equations 6.18.   

 

 e    sin          f  sin      

 

 e    sin          g  sin      

  

13. Use equations 7.12 to determine the angular acceleration of link 4.

 

  

B   b  sin  

 

  

E   b  cos  

A   c sin  

D   c cos  

 

 

  

 

  

2     2 2  b     cos     c    cos   

C   a   sin   a    cos  

K5  1.0119


 


2     2 2  b     sin     c    sin   

F   a   cos   a    sin  

α4  

        A    E   B   D  C    E    B    F 

14. Use equations 7.12 to determine the angular accelerations of links 5 and 6 for the open circuit. Use link 4 as the input to the second fourbar stage.

 

  

B'   f  sin  

 

  

E'   f  cos  

A'   g  sin  

D'   g  cos  

 

  

 

  

 

      e 2 cos  2 2  f     cos     g     cos   

 

      e 2 sin  2 2  f     sin     g     sin   

C'   e α4   sin  

F'   e α4   cos  

α6  

        A'    E'   B'   D'  C'   E'   B'   F' 

ANGULAR ACCELERATION - LINK 6

Angular Acceleration, rad/sec^2

4

2

0

2

4

0

45

90

135

180

Crank Angle, deg

225

270

315

360




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular acceleration of link 8 in the linkage of Figure 3-36 as a function of 2 for a constant 2 = 1 rad/sec CCW.

Given:


a  0.450

First coupler (L3)

b  0.990

Common rocker (O4B)

c  0.590


d  1.000

Common rocker (O4C)

a'  0.590

Second coupler (CD)

b'  0.325

Output rocker (L6)

c'  0.325


Link 7 (L7)

e  0.938

Link 8 (L8)

f  0.572


p  0.823

Angle DCE

δ  7.0 deg

Angle BO4C

α  128.6  deg


  1  rad sec

1

CCW


See problem 4-43 for the position solution. The velocity and acceleration solutions will use the same vector loop equations for links 5, 6, 7, and 8, differentiated with respect to time. This problem is suitable for a project assignment and is probably too long for an overnight assignment.




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the magnitude and direction of the acceleration of point P in Figure 3-37a as a function of 2 for a constant 2 = 1 rad/sec CCW. Also calculate and plot the acceleration of point P versus point A.

Given:

Link lengths:

Solution: 1.

Link 2 (O2 to A)

a  0.136

Link 3 (A to B)

b  1.000

Link 4 (B to O4)

c  1.000

Link 1 (O2 to O4)

d  1.414

Coupler point:

Rpa  2.000

  0  deg

Crank speed:

  1  rad sec

1

  0  rad sec

2



2.


d

K2 

a

K1  10.3971 2

K3 

c

K2  1.4140

2

2

a b c d

2

K3  7.4187

2 a c

 

d

 

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

C θ  K1   K2  1   cos θ  K3 3.




 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  4.

B θ


 

2

d b

K5 

2

2

c d a b

2

2 a b

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

K4  1.4140

K5  7.4187

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5 5.


 





 

 

θ θ  2   atan2 2  D θ E θ  6.

 2  4 Dθ F θ 

E θ


 

a  

 

a  

ω θ 

ω θ 

b

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 






7.



 



 

   a 2 cosθ  j  sinθ

AA θ  a   sin θ  j  cos θ 8.


 

  

B' θ  b  sin θ θ

 

  

E' θ  b  cos θ θ

 

 

A' θ  c sin θ θ

D' θ  c cos θ θ

2

 

  

 

  

 

 2     c ωθ2 cosθθ

C' θ  a   sin θ  a    cos θ  b  ω θ  cos θ θ

 

2     C'  θ  D' θ  A'  θ  F' θ A'  θ  E' θ  B' θ  D' θ

 2     c ωθ2 sinθθ C'  θ  E' θ  B' θ  F' θ α θ  A'  θ  E' θ  B' θ  D' θ

F' θ  a   cos θ  a    sin θ  b  ω θ  sin θ θ

 

α θ 


 

           2  Rpa ω θ   cos θ θ    j  sin θ θ    AP θ  AA θ  APA θ APA θ  Rpa α θ  sin θ θ    j  cos θ θ  

 

 

θAp θ  arg AP θ 


 

ACCELERATION MAGNITUDE - POINTS P and A 2.2 2.0 1.8 1.6 1.4 1.2 1.0 0.0

45.0

90.0

135.0

180.0

225.0

Crank Angle, deg Point P Point A

θAA θ  arg AA θ 

270.0

 

AA θ  AA θ

θAp θ  if  θ  91 deg θAp θ  2  π θAp θ 


9.

315.0

360.0



ACCELERATION VECTOR DIRECTION - POINTS P and A 360

Acceleration Vector Angle, deg

315 270 225 180 135 90 45 0  45  90  135  180

0

45

90

135

180

Crank Angle, deg Point P Point A

225

270

315

360




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the magnitude and the direction of the acceleration of point P in Figure 3-37b as a function of 2. Also calculate and plot the velocity of point P versus point A.

Given:


a  0.50

First coupler (AB)

b  1.00

Rocker 4 (O4B)

c  1.00

Rocker 5 (L5)

c'  1.00

Ground link (O2O4)

d  0.75


b'  1.00

Coupler point (DP)

p  1.00


d'  1.50

Crank speed: Solution: 1.

  1  rad sec

1

  0  rad sec

2

See Figure 3-37b and Mathcad file P07-69.


 

 

 


 


2.

Define one revolution of the input crank: θ  0  deg 0.5 deg  360  deg

3.


d a

K1  1.5000

 

2

d

K2 

K3 

c

K2  0.7500

 

2

2

a b c d

2

2 a c

K3  0.8125

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 



 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ

 

 2  4 A θ Cθ   2 π

 

θ θ  2   atan2 2  A θ B θ  4.

B θ


 

2

d

K5 

b

2

2

c d a b

2

2 a b

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

K4  0.7500

K5  0.8125

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5 5.


 





 

 

θ θ  2   atan2 2  D θ E θ  6.

 2  4 Dθ F θ   2 π

E θ




 

    c cosθθ  d'

xP θ  d  b  cos θ θ

 

    c sinθθ

yP θ  b  sin θ θ 7.

Use equations 6.18 to calculate 3 and 4.

 

a  

 

a  

 θ 

 θ  8.

b

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 






 

  

B' θ  b  sin θ θ

 

  

E' θ  b  cos θ θ

 

 

2

 

 2     c θ2 cosθθ

 

 

2

 

 2     c θ2 sinθθ

A' θ  c sin θ θ

D' θ  c cos θ θ

 

  

 

  

C' θ  a   sin θ  a    cos θ  b   θ  cos θ θ F' θ  a   cos θ  a    sin θ  b   θ  sin θ θ

 

α θ 

 

α θ 

        A'  θ  E' θ  B' θ  D' θ C' θ  E' θ  B' θ  F' θ

Differentiate the position equations twice with respect to time to get the acceleration components. APx θ  b   α θ  sin θ θ

      θ2 cosθθ  2  c  α θ  sin θ θ    θ  cos θ θ  2 APy  θ  b   α θ  cos θ θ    θ  sin θ θ    2  c  α θ  cos θ θ    θ  sin θ θ    

 

AP θ 

 2  APyθ2

APx θ

VELOCITY MAGNITUDE of POINT P 0.6 0.4 Velocity, 1/sec

9.


0.2 0  0.2  0.4  0.6

0

45

90

135

180


225

270

315

360




Find the angular accelerations of links 3 and 4 and the linear accelerations of points A, B and P1 in the XY coordinate system for the linkage in Figure P7-27 in the position shown. Assume that 2 = 45 deg in the XY coordinate system and 2 = 10 rad/sec, constant. The coordinates of the point P1 on link 4 are (114.68, 33.19) with respect to the xy coordinate system. Use a graphical method.

Given:

Link lengths:

Solution: 1.

Link 2 (O2 to A)

a  14.00  in

Link 3 (A to B)

b  80.00  in

Link 4 (O4 to B)

c  51.26  in

Link 1 X-offset

d X  47.5 in

Link 1 Y-offset

d Y  76.00  in  12.00  in


p x  114.68 in


Crank angle:

  45 deg


α  126.582  deg


ω  10 rad sec


1

α  0  rad sec

CCW

2


Draw the linkage to a convenient scale. Indicate the directions of the velocity acceleration of interest. Direction of ABt Direction of ABAt B

x

4

P1 O4

29.063° Y

3

126.582°

99.055° Direction of AP1t A

45.000°

2

X O2

  45 deg  α   81.582 deg 2.

y

In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above (in the local xy coordinate system), θ  99.055 deg  α

θ  27.527 deg

θ  29.063 deg  α

θ  97.519 deg




ω 

a  ω sin θ    b sin θ  θ

 

 

ω  0.511 rad sec

ω 

a  ω sin   θ  c sin θ  θ

 

 

ω  2.353 rad sec


4.

For point B, this becomes: (ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where 2

ABn  283.838 in sec

θABn  θ  180  deg

2

θABn  277.519 deg

2

AAn  a  ω


θAAn    180  deg


AAt  a  α


θAAt    90 deg

θAAt  8.418 deg

2

5.

1


3.


1

2

2

2



θABAn  θ  180  deg


Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn +  and AAn at an angle of AAn + . From the tip of AAn, draw AAt at an angle of AAt + . From the tip of AAt, draw ABAn at an angle of BAn + . Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABt, and ABAt. x

0

Y

n

X

AB

20 in/s/s y

80.620°

Acceleration Scale 39.675 ABt

42.136

AB

AA t ABA n

ABA

6.


ka  20.0

in sec

AA  AAn

2

AA  1400 in sec

2



4.

2

AB  42.136 ka

AB  842.7 in sec

ABt  39.675 ka

ABt  793.500 in sec

  7.


ABt

  15.480 rad sec

c

at an angle of -80.62 deg 2


2

Determine the distance from O4 to P1 and the angle O4P1makes with the x axis. 2

2

Distance O2O4:

d 

dX  dY

Distance P1O4:

u 

px  d2  py2

Angle O4P1:

  atan

d  79.701 in u  48.219 in

 py    px  d 

  43.497 deg

Determine the acceleration at point P1. AP1t  u  

AP1t  746.431 in sec

AP1n  u  ω

2

AP1 

2

AP1t  AP1n

AP1n  267.001 in sec 2

AP1  792.748 in sec

 AP1t    9.921  deg  60.397 deg  AP1n 

atan

2 2

2

at an angle of 80.079 deg at an angle of -9.921 deg





Find the angular accelerations of links 3 and 4 and the linear accelerations of points A, B and P1 in the XY coordinate system for the linkage in Figure P7-27 in the position shown. Assume that 2 = 45 deg in the XY coordinate system and 2 = 10 rad/sec, constant. The coordinates of the point P1 on link 4 are (114.68, 33.19) with respect to the xy coordinate system. Use an analytical method.

Given:

Link lengths:

Solution: 1.

Link 2 (O2 to A)

a  14.00  in

Link 3 (A to B)

b  80.00  in

Link 4 (O4 to B)

c  51.26  in

Link 1 X-offset

d X  47.5 in

Link 1 Y-offset

d Y  76.00  in  12.00  in


p x  114.68 in


Crank angle:



α  126.582  deg


ω  10 rad sec


1

CCW

α  0  rad sec

2


Draw the linkage to scale and label it. B

x

4

P1 O4

29.063° Y

3

126.582°

99.055°

A

  θ2XY  α   81.582 deg d 

2

dX  dY

X O2

2 y

d  79.701 in 2.

45.000°

2


d

K2 

a

K1  5.6929 2

K3 

d c

K2  1.5548 2

2

a b c d

2

2 a c

 

K3  1.9340

 

A  cos   K1  K2 cos   K3

 

C  K1   K2  1   cos   K3

 

B  2  sin 


A  3.8401 3.


B  1.9785






2

θ  2  atan2 2  A B  4.

C  7.2529

B  4 A  C



θ  262.482 deg


2

d b

2

2

c d a b

K5 

2

K4  0.9963

2 a b

 

 

D  cos   K1  K4 cos   K5

D  10.0081

 

E  2  sin 

E  1.9785

 

F  K1   K4  1   cos   K5 5.



7.

F  1.0849




2

θ  2  atan2 2  D E  6.

K5  4.6074

E  4  D F



θ  332.475 deg

Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18. ω 

a  ω sin θ    b sin θ  θ

 

 

ω  0.511

ω 

a  ω sin   θ  c sin θ  θ

 

 

ω  2.353

rad sec

rad sec




 

   a ω2 cos  j  sin

AA  a  α sin   j  cos  AA  ( 205  1385i)

in


AA  1400

in sec

8.


AA  AA

2

sec

at 2

θAA  225 deg


 

 


 

B  b  sin θ 3

A  1.291  10 mm B  939.044 mm

 

 



D  170.343 mm

E  1.802  10 mm

2

 

2

 

2

2

 

2

 

2

 

C  a  α sin   a  ω  cos   b  ω  cos θ  c ω  cos θ C  260.638 in sec

2

 

 

F  a  α cos   a  ω  sin   b  ω  sin θ  c ω  sin θ 3

F  1.113  10 in sec

2

3


α 

9.


C D  A  F

α  14.227

A  E  B D

rad sec

α 

2

C  E  B F A  E  B D

α  15.479

sec




 

   c ω2 cosθ  j  sinθ

AB  c α sin θ  j  cos θ AB  ( 749  385i)

in sec

θAB  arg AB  α

AB  AB

2


in

AB  842.7

sec

at 2

θAB  279.4 deg

(Global)

10. Use equation 7.31 to determine the acceleration of the point P1 on link 4. u 

px  d2  py2

u  48.219 in

 py    px  d 

  141.014 deg



  u ω2 cosθ    j  sinθ  

  360  deg  θ  atan







AP1  u  α sin θ    j  cos θ   AP1  ( 320  725i)

in sec

2

θAP1  arg AP1  α

AP1  AP1

The acceleration of point P1 is AP1  792.7

in sec

at 2

θAP1  60.39 deg

rad

(Global)

2




Find the angular accelerations of links 3 and 4 and the linear accelerations of points A, B and P1 in the XY coordinate system for the linkage in Figure P7-27 in the position shown. Assume that 2 = 45 deg in the XY coordinate system and 2 = 10 rad/sec, constant. The coordinates of the point P1 on link 4 are (114.68, 33.19) with respect to the xy coordinate system. Use an analytical method.

Given:

Link lengths:

Solution: 1.

Link 2 (O2 to A)

a  14.00  in

Link 4 (O2 to B)

c  51.26  in

Link 1 X-offset

d X  47.5 in

Link 1 (O2 to O2)

d 

2

dX  dY

2

Link 3 (A to B)

b  80.00  in

Link 1 Y-offset

d Y  76.00  in  12.00  in

d  79.701 in


p x  114.68 in


Crank angle:

  0  deg 1  deg  360  deg


α  126.582  deg


ω  10 rad sec

1

Global XY system to local xy system α  0  rad sec

CCW



B

x

4

P1 O4

Y

3

A 2

X O2

y

2.


d

K2 

a

K1  5.6929 2

K3 

 

d c

K2  1.5548 2

2

a b c d 2 a c

 

2

K3  1.9340

 

A   cos   K1  K2 cos   K3

 

 

B   2  sin 

2



 

 

C   K1   K2  1   cos   K3 3.




 



 

 

θ   2   atan2 2  A  B   4.

 2  4 A  C 

B 


2

d

K5 

b

 

2

2

c d a b

2

K4  0.9963

2 a b

 

K5  4.6074

 

D   cos   K1  K4 cos   K5

 

 

E   2  sin 

 

 

F   K1   K4  1   cos   K5 5.




 



 

 

θ   2   atan2 2  D  E   6.


 

a  ω

 

a  ω

ω  

ω  

7.

 2  4 D F  

E 

b

c



    sin θ   θ  



  sin θ   θ  

sin θ   



sin   θ 


     D'   c cos θ  

     E'   b  cos θ  

A'   c sin θ 

 

 

B'   b  sin θ 

2

 

 2     c ω2 cosθ

C'   a  α sin   a  ω  cos   b  ω   cos θ 

 

2     C'    E'   B'   F'  A'    E'   B'   D' 

 2     c ω2 sinθ

F'   a  α cos   a  ω  sin   b  ω   sin θ 

 

α   8.

Use equation 7.31 to determine and plot the acceleration of the point P1 on link 4. u 

px  d2  py2

u  48.219 in

  141.067  deg

 

           2  u  ω    cos θ     j  sin θ    

AP1   u  α   sin θ     j  cos θ   


 




 



θAP1   arg AP1   α

AP1   AP1 

θAP1   if  θAP1   100  deg θAP1   2  π θAP1   MAGNITUDE OF Ap 2000


1500

1000

500

0

0

45

90

135

180

225

270

315

360

270

315

360

Crank Angle, deg

DIRECTION OF Ap (Global)


100

0

 100

 200

0

45

90

135

180

225

Crank Angle, deg

AP1 ( 81.582 deg)  792.706

in sec

2

θAP1( 81.582 deg)  60.447 deg




Find the angular accelerations of links 3 and 4 and the linear acceleration of point P in the XY coordinate system for the linkage in Figure P7-28 in the position shown. Assume that 2 = -94.121 deg in the XY coordinate system, 2 = 1 rad/sec, and 2 = 10 rad/sec2. The position of the coupler point P on link 3 with respect to point A is: p = 15.00, 3 = 0 deg. Use a graphical method.

Given:


a  9.17 in

Link 3 (A to B)

b  12.97  in

Link 4 (O4 to B)

c  9.57 in

Link 1 X-offset

d X  2.79 in

Link 1 Y-offset

d Y  6.95 in

Coupler point data:

p  15.00  in

δ  0  deg

Crank angle:

θ  26.000 deg α  68.121 deg


ω  1  rad sec


Global XY system to local xy system 1

  10 rad sec

CW

2


Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest. Y y

O2

X Direction ABAt 2 1

68.121°

Direction APAt

26.000° O4 B

4

P

A

Direction AAt

72.224° 3 60.472°

Direction ABt

x

2.


θ  72.224 deg

θ  60.472 deg

θ  60.472 deg


a  ω sin θ  θ  b sin θ  θ

 

 

ω  3.465 rad sec

ω 

a  ω sin θ  θ  c sin θ  θ

 

 

ω  4.656 rad sec

1

1




3.


4.


2

ABn  207.476 in sec

θABn  θ  180  deg  α

θABn  172.351 deg

AAn  a  ω

2


θAAn  θ  180  deg  α


AAt  a  


θAAt  θ  90 deg  α

θAAt  4.121 deg

2

5.

2

2

2



θABAn  θ  180  deg  α


2

Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn +  and AAn at an angle of AAn + . From the tip of AAn, draw AAt at an angle of AAt + . From the tip of AAt, draw ABAn at an angle of BAn + . Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABt, and ABAt.

6.

From the graphical solution at right, Acceleration scale factor

ka  10.0

in sec

AA  0.922  ka

2

AA  9.220 in sec

2

at an angle of 1.590 deg AB  7.156  ka

AB  71.560 in sec

2

at an angle of 99.204 deg t

ABAt  7.169  ka   7.

ABAt

ABAt  71.690 in sec   5.527 rad sec

b

2

AB

7.156

t

ABA

7.169

2 AB

For point P, equation 7.4 becomes: AP = AA + (APAt + APAn) , where APAn  p  ω

2

APAn  180.062 in sec

99.204°

Y

2

y

n

AA

1.590° t

AA

n

AB

X

n

A BA

θAPAn  θ  δ  180  deg  α

AA 0.922

θAPAn  184.103 deg APAt  p  


θAPAt  θAPAn  90 deg

2


x

0

20 IN/S/S

Acceleration Scale


8.


Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. n

A PA

t

APA AP

157.174°

Y y

9.065

0

20 IN/S/S

X AA

Acceleration Scale x

9.


ka  10.0

in sec

AP  9.065  ka

2

AP  90.650 in sec

2





Find the angular accelerations of links 3 and 4 and the linear acceleration of point P in the XY coordinate system for the linkage in Figure P7-28 in the position shown. Assume that 2 = -94.121 deg in the XY coordinate system, 2 = 1 rad/sec, and 2 = 10 rad/sec2. The position

Given:

of the coupler point P on link 3 with respect to point A is: p = 15.00, 3 = 0 deg. Use an analytical method. Link lengths: Link 2 (O2 to A)

a  9.174  in

Link 4 (O4 to B)

c  9.573  in

Link 1 X-offset

d X  2.790  in

Link 1(O2 to O4)

d 

2

dX  dY

Rpa  15.00  in

Crank angle:

θ  26.000 deg

Link 1 Y-offset

d Y  6.948  in

d  7.487 in δ  0  deg

α  68.121 deg


ω  1  rad sec


1.

b  12.971 in

2

Coupler point data:


Solution:

Link 3 (A to B)

1

  10 rad sec

CW


Draw the linkage to scale and label it. Y y

O2

X 2 68.121°

1 26.000°

O4 B

4

P

A 72.224° 3 60.472° x

2.


d

K2 

a

K1  0.8161 2

K3 

c

K2  0.7821 2

2

a b c d

 

d

2

2 a c

K3  0.3622

 

A  cos θ  K1  K2 cos θ  K3

 

C  K1   K2  1   cos θ  K3

 

B  2  sin θ

2



A  0.2581 3.

B  0.8767

C  0.4234






2

θ  2  atan2 2  A B 

B  4 A  C



θ  299.512 deg

θ  θ  360  deg 4.

θ  60.488 deg


2

d

K5 

b

2

2

c d a b

 

2

K4  0.5772

2 a b

 

D  cos θ  K1  K4 cos θ  K5

D  0.3096

 

E  2  sin θ

E  0.8767

 

F  K1   K4  1   cos θ  K5 5.

F  0.4749






2

θ  2  atan2 2  D E 

E  4  D F



θ  287.764 deg

θ  θ  360  deg 6.

7.

K5  0.9111

θ  72.236 deg


a  ω sin θ  θ  b sin θ  θ

 

 

ω  3.467

rad

ω 

a  ω sin θ  θ  c sin θ  θ

 

 

ω  4.658

rad

sec

sec




 

   a ω2 cosθ  j  sinθ

AA  a   sin θ  j  cos θ AA  ( 32.0  86.5i )

in sec


AA  92.198

in sec

8.


AA  AA

2

at 2

θAA  1.590 deg

(Global)


 

 

 

 





A  8.331 in

B  12.353 in

D  4.716 in

E  3.957 in

 

2

 

2

 

2

2

 

2

 

2

 

C  a   sin θ  a  ω  cos θ  b  ω  cos θ  c ω  cos θ C  86.722 in sec

2

 

 

F  a   cos θ  a  ω  sin θ  b  ω  sin θ  c ω  sin θ F  118.753 in sec

2


α  9.


C D  A  F

α  55.307

A  E  B D

rad sec

α 

2

C  E  B F A  E  B D

α  71.596




 

   c ω2 cosθ  j  sinθ

AB  c α sin θ  j  cos θ AB  ( 698.8  156.9i)

in sec


θAB  arg AB  α

AB  AB

2

AB  716.18

in sec

at 2

θAB  99.228 deg




      Rpa ω   cos θ  δ  j  sin θ  δ 

APA  Rpa α sin θ  δ  j  cos θ  δ 2

AP  AA  APA AP  AP

AP  830

in sec

2

 

arg AP  α  100.214 deg

rad sec

2




For the linkage in Figure P7-28, write a computer program or use an equation solver to calculate and plot the angular velocity and acceleration of links 3 and 4, and the magnitude and direction of the velocity and acceleration of point P as a function of 2 through its possible range of motion starting at the position shown. The position of the coupler point P on link 3 with respect to point A is: p = 15.00, 3 = 0 deg. Assume that, at t = 0, 2 = -94.121 deg in the XY coordinate system, 2 = 0, and 2 = 10 rad/sec2, constant.

Given:


a  9.174  in

Link 4 (O4 to B)

c  9.573  in

Link 1 X-offset

d X  2.790  in

Link 1(O2 to O4)

d 

2

dX  dY

Link 1 Y-offset

d Y  6.948  in

d  7.487 in

Rpa  15.00  in

Crank angle:

θ  26.000 deg 25.9 deg  15 deg

δ  0  deg

α  68.121 deg

θ  26 deg


ω  0  rad sec


1.

b  12.971 in

Coupler point data:


Solution:

2

Link 3 (A to B)

1

  10 rad sec



Y y

O2

X 2 1 26.000° O4 B

4

P

A

3

x

2.


d

K2 

a

K1  0.8161 2

K3 

 

d c

K2  0.7821 2

2

a b c d 2 a c

2

K3  0.3622

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

C θ  K1   K2  1   cos θ  K3

 

 

B θ  2  sin θ

2


3.





 



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  4.

B θ


2

d

K5 

b

 

2

2

c d a b

2

K4  0.5772

2 a b

 

K5  0.9111

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 5.


 





 

 

θ θ  2   atan2 2  D θ E θ  6.

 2  4 Dθ F θ 

E θ


 

ω θ 





2  θ  θ  

 

 

a  ω θ

 

a  ω θ

ω θ 

ω θ 

b

 

c



    sin θ θ  θ θ 



  sin θ θ  θ θ 





2

50 40

1.5

 

ω  θ 

 

sec rad

ω  θ 

1

0.5

rad

20 10

0  30

7.

sec 30

 25

 20

 15

0  26

 10

 22

θ

θ

deg

deg


 



 

   a ωθ2 cosθ  j  sinθ

AA θ  a   sin θ  j  cos θ 8.

 24


 

  

B' θ  b  sin θ θ

 

  

E' θ  b  cos θ θ

A' θ  c sin θ θ

D' θ  c cos θ θ

 

  

 

  

 20


 


 

 2  

 2     c ωθ2 cosθθ

   2   C'  θ  D' θ  A'  θ  F' θ A'  θ  E' θ  B' θ  D' θ

 2     c ωθ2 sinθθ C'  θ  E' θ  B' θ  F' θ α θ  A'  θ  E' θ  B' θ  D' θ

C' θ  a   sin θ  a  ω θ  cos θ  b  ω θ  cos θ θ

 

F' θ  a   cos θ  a  ω θ  sin θ  b  ω θ  sin θ θ

 

α θ 

14 150

12 2

 

sec

 

α θ 

10

rad

2

sec

100

rad

8

50

6  30

9.

 25

 20

 15

0  26

 10

 24

 22

θ

θ

deg

deg

 20


 


 

 

θAp θ  arg AP θ  α






θAp θ  if  θAp θ  0 θAp θ  2  π θAp θ 

4

MAGNITUDE

DIRECTION

1 10

3

8 10

 

AP θ 

2

sec in

3

θAp θ

3

deg

6 10 4 10

150

100 3

2 10

0  26

 24

 22

 20

50  26

 24

 22

θ

θ

deg

deg

 20




Derive analytical expressions for the accelerations of points A and B in Figure P7-29 as a function of 3, 3, 3 and the length AB of link 3. Use a vector loop equation.

Solution:


1.

Establish the global XY system such that the coordinates of the intersection of the slot centerlines is at (d X,d Y ). Then, define position vectors R1X, R1Y , R2, R3, and R4 as shown below.

Y 4 3 R3 R2

B

3

C 2

R4

A 1

R 1Y

X R 1X 2.


 π  π j   j θ   2 j ( 0) 3 j ( 0) 2 dY  e    a e  c e  dX  e  b e   = 0 j

3.

Differentiate this equation with respect to time.

d a  j  c  e dt 3.

 

j 

 j  π   2 d    b e    = 0 dt

Substituting the Euler identity into this equation gives: Va  jc    cos θ3  j  sin θ3   Vb j = 0

4.

Separate this equation into its real (x component) and imaginary (y component) parts, setting each equal to zero. Va  c  sin θ3 = 0

5.

Solve for the two unknowns Va and Vb in terms of the independent variables 3 and 3 Va = c  sin θ3

6.

c  cos θ3  Vb = 0 Vb = c  cos θ3

Differentiate again with respect to time to get the accelerations. Aa = c   cos θ3 2

Ab = c   sin θ3 2




The linkage in Figure P7-30a has link 2 at 120 deg in the global XY coordinate system. Find 6 and AD in the global XY coordinate system for the position shown if 2 = 10 rad/sec CCW and 2 = 50 rad/sec2 CW. Use the acceleration difference graphical method.

Given:


a  6.20 in

Link 3 (A to B)

b  4.50 in

Link 4 (O4 to B)

c  3.00 in

Link 5 (C to D)

e  5.60 in

Link 3 (A to C)

p  2.25 in

Link 4 (O4 to D)

f  3.382  in

Link 1 X-offset

d X  7.80 in

Link 1 Y-offset

d Y  0.62 in

Angle ACB

  0.0 deg

Angle BO4D

  110.0  deg

Input rocker angle:

θ  120  deg

Global XY system

ω  10 rad sec


1

  50 rad sec

CCW

2


Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest. Direction of AAt Direction of ACAt Axis of slip

A Y

87.972° 113.057° 5

D

C

110°

2

3


120° 175.455°

411.709° B

x Direction of ADCt

O4

  113.057  deg

X

O2 Direction of ABt

Direction of ABAt

  11.709 deg y

Coordinate rotation angle:   175.455  deg 2.

In order to solve for the accelerations at points A and B, we will need 3, 3, and 4. From the graphical position solution above (in the local xy coordinate system),

θ2xy  θ  

θ2xy  295.455 deg

θ3xy    

θ3xy  62.398 deg

θ4xy    

θ4xy  187.164 deg


a  ω sin θ4xy  θ2xy  b sin θ3xy  θ4xy

ω  15.924 rad sec

1


ω 


a  ω sin θ2xy  θ3xy  c sin θ4xy  θ3xy

ω  20.107 rad sec


3.


4.


2

ABn  1212.852 in sec

θABn    180  deg

1

2

θABn  191.709 deg

2

AAn  a  ω

AAn  620.000 in sec

θAAn  θ  180  deg

θAAn  300.000 deg

AAt  a  

AAt  310.000 in sec

θAAt  θ  90 deg

θAAt  30.000 deg

2

2

2


ABAn  1141.131 in sec

θABAn    180  deg


t ABA

2

5.727

t B

A

AB

4.158

Y n

A BA

131.968°

4.814 X

n

AB 0

AA

500 IN/S/S

A Acceleration Scale

1.386

n A

t A

A

33.435°


6.



in

ka  500 

sec

2

AA  1.386  ka

AA  693.0 in sec

AB  4.814  ka

AB  2407.0 in sec

ABAt  5.727  ka


 

ABAt b

 

ABt



2 2

CW

2

  693.000 rad sec

c


2

  636.333 rad sec

ABt  4.158  ka

7.

2

2

CCW

For point C, equation 7.4 becomes: AC = AA + (ACAt + ACAn) , where 2



θACAn    180  deg

θACAn  66.943 deg

ACAt  p  


θACAt  θACAn  90 deg

θACAt  156.943 deg t

ACA

2

2

Y

AC 126.220° 0

500 IN/S/S

n

A CA

1.745

X

Acceleration Scale

AA 8.


ka  500 

in sec

AC  1.745  ka 9.

2

AC  872.50 in sec

2


Determining the acceleration of point D requires the (graphical) solution of two equations simultaneously. They are: AD = AC + (ADCt + ADCn) and AD = ADt + ADn + ADcor + ADslip , where AC is known from the above analysis and From the velocity analysis (see Problem 6-91) ADCn  e 

2

  35.13  rad sec

1

ADCn  6911 in sec

Vslip  136.20 in sec 2

1



θADCn  177.972  deg  180  deg

θADCn  357.972 deg

ADt  f  


θADt      90 deg

θADt  211.709 deg

ADn  f  ω

2


θADn  θADt  90 deg

θADn  301.709 deg

ADcor  2  Vslip  ω

ADcor  5477.1 in sec

θADcor  θADt  180  deg

θADcor  31.709 deg

2

2

2

Y

n

A DC AC

X

ADt 33.485° A A

0

cor D

ADslip

n D

2000 IN/S/S

AD

3.700

t ADC

Acceleration Scale

8.


ka  2000

in sec

AD  3.700  ka

2

AD  7400 in sec   

2

at an angle of -33.485 deg   693.0 rad sec

2




The linkage in Figure P7-30a has link 2 at 120 deg in the global XY coordinate system. Find 6 and AD in the global XY coordinate system for the position shown if 2 = 10 rad/sec CCW and 2 = 50 rad/sec2 CW. Use an analytical method.

Given:

Solution: 1.


a  6.20 in

Link 3 (A to B)

b  4.50 in

Link 4 (O4 to B)

c  3.00 in

Link 5 (C to D)

h  5.60 in

Link 3 (A to C)

p  2.25 in

Link 4 (O4 to D)

f  3.382  in

Link 1 X-offset

d X  7.80 in

Link 1 Y-offset

d Y  0.62 in

Angle ACB

δ3  0.0 deg

Angle BO4D

δ4  110.0  deg

Input rocker angle:

θ2XY  120  deg

Global XY system


δ  175.455  deg


ω2  10 rad sec


α2  50 rad sec

1

(CCW)

2

(CW)



A

6

Y

71.487° D

5

C

110°

3

120°

55.455° B 16.254°

4

x

2

O4

175.455° X

O2 y d 

Calculate the distance O2O4:

Transform 2XY into the local coordinate system: 2.

2

dX  dY

2

d  7.825 in

θ2  θ2XY  δ

θ2  55.455 deg


d

K2 

a

K1  1.2620 2

K3 

d c

K2  2.6082 2

2

a b c d 2 a c

2

K3  2.3767



A  cos θ2  K1  K2 cos θ2  K3 B  2  sin θ2 C  K1   K2  1   cos θ2  K3 A  0.2028 3.

B  1.6474






2

θ4  2  atan2 2  A B  4.

B  4 A  C

2

d

K5 

b

7.

θ4  163.746 deg

2

2

c d a b

2

K4  1.7388

2 a b

D  cos θ2  K1  K4 cos θ2  K5

D  1.6967

E  2  sin θ2

E  1.6474

F  K1   K4  1   cos θ2  K5

F  0.3067

K5  1.9877






2

θ3  2  atan2 2  D E  6.




5.

C  1.5927

E  4  D F

  2 π

θ3  648.512 deg


ω3 

a  ω2 sin θ4  θ2  b sin θ3  θ4

ω3  15.924

ω4 

a  ω2 sin θ2  θ3  c sin θ4  θ3

ω4  20.107

rad sec rad sec

Use equations 7.12 to determine the angular accelerations of links 3 and 4 for the open circuit. A  c sin θ4

B  b  sin θ3

D  c cos θ4

E  b  cos θ3

A  21.328 mm

B  108.386 mm

D  73.154 mm

E  36.291 mm


3

C  2.134  10 in sec

2

2

2

F  a  α2 cos θ2  a  ω2  sin θ2  b  ω3  sin θ3  c ω4  sin θ4 2

3

F  1.087  10 in sec

α3  8.

C D  A  F A  E  B D

2

2

2

α3  636.368

rad sec

2

α4 

C  E  B F A  E  B D

α4  692.987

rad sec

Establish a vector loop that defines the position of point D with respect to O4 and write the loop equation.

2



A 2.517°

Y

71.487° R5

D

C 5

R BC

3 110° R4

RD

2 120° 175.455°

55.455° B

x O4

4

X

O2

16.254° y

RD = R4  RBC  R5

f e



 = c ejθ4  p  ej  θ3π  h ejθ5

j θ4 δ4

where f and the angles are functions of time and c, p, h, and 4 are constants. 9.

Solve the vector loop equation for 5 by substituting the Euler identity for the unit vectors and then separating into real and imaginary parts. f   cos θ4  δ4  j  sin θ4  δ4  = c  cos θ4  j  sin θ4   p   cos θ3  π  j  sin θ3  π    h   cos θ5  j  sin θ5  f  cos θ4  δ4 = c cos θ4  p  cos θ3  π  h  cos θ5 f  sin θ4  δ4 = c sin θ4  p  sin θ3  π  h  sin θ5 tan  θ5 =

f  sin θ4  δ4  c sin θ4  p  sin θ3  π

f  cos θ4  δ4  c cos θ4  p  cos θ3  π

θ5  atan2 f  cos θ4  δ4  c cos θ4  p  cos θ3  π f  sin θ4  δ4  c sin θ4  p  sin θ3  π  f  3.382 in

θ5  2.518 deg

10. Differentiate the position equaton and solve for 5 and fdot by substituting the Euler identity for the unit vectors and then separating into real and imaginary parts. f  ω4  sin θ4  δ4  j  cos θ4  δ4   fdot   cos θ4  δ4  j  sin θ4  δ4 

=

c ω4 sin θ4  j  cos θ4  p  ω3 sin θ3  π  j  cos θ3  π  h  ω5  sin θ5  j  cos θ5 f  ω4 sin θ4  δ4  fdot  cos θ4  δ4 = c ω4 sin θ4  p  ω3 sin θ3  π  h  ω5 sin θ5 f  ω4 cos θ4  δ4  fdot  sin θ4  δ4 = c ω4 cos θ4  p  ω3 cos θ3  π  hω5 cos θ5 Let

A  f  ω4 sin θ4  δ4  c ω4 sin θ4  p  ω3 sin θ3  π

A  71.929

in

B  f  ω4 cos θ4  δ4  c ω4 cos θ4  p  ω3 cos θ3  π

B  86.747

in

s s



then fdot  cos θ4  δ4 = A  h  ω5 sin θ5

fdot  sin θ4  δ4 = B  h  ω5 cos θ5

Eliminating fdot and solving for 5,

ω5 

A  tan  θ4  δ4  B

h   cos θ5  sin θ5  tan θ4  δ4 

and

A  h  ω5 sin θ5

fdot 

ω5  35.145

rad s in

fdot  136.252

cos θ4  δ4

s

11. Differentiate the velocity equaton and solve for 5 and fddot by substituting the Euler identity for the unit vectors and then separating into real and imaginary parts. f  α4 j  e





j  θ 4 δ 4

  f  ω 2 ej   θ4δ4  2 fdot  ω  j  ej   θ4δ4  fddot  ej   θ4δ4 4

c  j  α4 e

j  θ4

4

=

 j   θ3π  ω 2 ej   θ3π  h  j  α  ej  θ5  ω  ej  θ5 3 5   p  j  e   5 

2 j  θ4

 ω4  e

f  α4  sin θ4  δ4  j  cos θ4  δ4   f  ω4   cos θ4  δ4  j  sin θ4  δ4    2  fdot  ω4  sin θ4  δ4  j  cos θ4  δ4   fddot   cos θ4  δ4  j  sin θ4  δ4  2

=

c α4  sin θ4  j  cos θ4   c ω4   cos θ4  j  sin θ4   2

 p  α3  sin θ3  π  j  cos θ3  π   p  ω3   cos θ3  π  j  sin θ3  π   2

 h  α5  sin θ5  j  cos θ5   h  ω5   cos θ5  j  sin θ5  2

Let

C  f  α4 sin θ4  δ4  f  ω4  cos θ4  δ4  2  fdot  ω4 sin θ4  δ4  2

 c  α4 sin θ4  ω4  cos θ4   p   α3 sin θ3  π  ω3  cos θ3  π  2

2

 h  ω5  cos θ5 2

C  3003.51

in 2.00

s

D  f  α4 cos θ4  δ4  f  ω4  sin θ4  δ4  2  fdot  ω4 cos θ4  δ4  2

 c  α4 cos θ4  ω4  sin θ4  p   α3 cos θ3  π  ω3  sin θ3  π   2

2

 h  ω5  sin θ5 2

D  1295.03

in 2.00

s

then, similar to the velocity solution,

α5 

C tan  θ4  δ4  D

h   cos θ5  sin θ5  tan  θ4  δ4 

fddot 

C  h  α5 sin θ5 cos θ4  δ4

α5  1025.02

rad 2

s

fddot  5505.35

in 2.00

s



12. Using the terms in the acceleration equation in step 10, define the components of AD and then combine them to determine the total acceleration of point D.

ADtangential  f  α4  sin θ4  δ4  j  cos θ4  δ4  ADnormal  f  ω42  cos θ4  δ4  j  sin θ4  δ4  ADcoriolis  2  fdot  ω4  sin θ4  δ4  j  cos θ4  δ4  ADslip  fddot   cos θ4  δ4  j  sin θ4  δ4  AD  ADtangential  ADnormal  ADcoriolis  ADslip AD  ( 6592.71  3687.93i)

in

in the local coordinate frame

2.00

s Magnitude:

AD 

AD  7554.11

AD

in 2.00

s Angle in local coordinate frame:

 

θAD  arg AD

θAD  150.778 deg

Angle in global coordinate frame:

θADXY  θAD  δ

θADXY  326.233 deg

The angular accelaration of block 6 is the same as link 4, that is,

α6  α4

α6  692.987

rad 2

s




The linkage in Figure P7-30b has link 3 perpendicular to the X-axis and links 2 and 4 are parallel to each other. Find 4, AA, AB, and AP if 2 = 15 rad/sec CW and 2 = 100 rad/sec2 CW. Use the acceleration difference graphical method.

Given:


a  2.75 in

Link 3 (A to B)

b  3.26 in

Link 4 (O4 to B)

c  2.75 in

Link 1 (O4 to B)

d  4.43 in

Coupler point data:

p  1.63 in

δ  0  deg

ω  15 rad sec


1

  100  rad sec

CW

2



Direction of ABt Direction of ABA B

P

O2

O4 36.351°

A

θ  36.352 deg

Direction of AAt 2.


θ  90.000 deg

θ  180  deg  θ

θ  143.648 deg


 

 

ω  1.924  10

 

 

ω  15.000 rad sec

ω 

a  ω sin θ  θ  b sin θ  θ

ω 

a  ω sin θ  θ  c sin θ  θ

 15


4.

For point B, this becomes: (ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where 2

1

1


3.


rad sec

ABn  618.750 in sec

2



θABn  θ  180  deg

θABn  323.648 deg

2

AAn  a  ω

AAn  618.750 in sec

θAAn  θ  180  deg

θAAn  143.648 deg

AAt  a  

AAt  275.000 in sec

θAAt  θ  90 deg

θAAt  126.352 deg

2

5.

2

2

2



θABAn  θ  180  deg


Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn +  and AAn at an angle of AAn + . From the tip of AAn, draw AAt at an angle of AAt + . From the tip of AAt, draw ABAn at an angle of BAn + . Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of ABt, ABt, and ABAt. 3.841 167.610° t A

A

0

AB

Y

n

AA

t ABA

400 IN/S/S

9.425°

AA X

t

AB

celeration Scale 1.693

n

AB 2.218

6.


ka  400 

in sec

AA  677.2 in sec

AB  2.218  ka

AB  887.2 in sec

ABAt  3.841  ka


ABAt


2


2

  471.288 rad sec

b

2

CW

For point P, equation 7.4 becomes: AP = AA + (APAt + APAn) , where 2

2

APAn  0.000 in sec

θAPAn  θ  δ  180  deg


APAt  p  


θAPAt  θAPAn  90 deg


APAn  p  ω

8.

2

AA  1.693  ka

  7.

2

2

Since 3 = 0 the acceleration at P is one-half that at B and is in the same direction as AB.




The linkage in Figure P6-33b has link 3 perpendicular to the X-axis and links 2 and 4 are parallel to each other. Find 4, AA, AB, and AP if 2 = 15 rad/sec CW and 2 = 100 rad/sec2 CW. Use an analytical method.

Given:


a  2.75 in

Link 3 (A to B)

b  3.26 in

Link 4 (O4 to B)

c  2.75 in

Link 1 (O4 to B)

d  4.43 in

RPA  1.63 in

δ3  0  deg

Coupler point data:

Solution: 1.

1


ω2  15 rad sec


α2  100  rad sec

CW

2

CW



Y

B

O4

O2

X

36.351° A From the figure, 2.

θ2  36.351 deg

From the problem statement we have:

θ3  90 deg 3.

4.

θ4  θ2  180  deg

θ4  216.351 deg


ω3 

a  ω2 sin θ4  θ2  b sin θ3  θ4

ω3  1.924  10

ω4 

a  ω2 sin θ2  θ3  c sin θ4  θ3

ω4  15.000

 15 rad

sec

rad

Using the Euler identity to expand equation 7.13a for AA. Determine the magnitude, and direction. AA  a  α2  sin θ2  j  cos θ2   a  ω2   cos θ2  j  sin θ2  2

AA  ( 335  588i)

in sec

2



AA  AA AA  677.1

in sec

5.

CCW

sec

at 2

θAA  119.7 deg




A  c sin θ4

B  b  sin θ3

D  c cos θ4

E  b  cos θ3

A  1.630 in

B  3.260 in

D  2.215 in

E  0.0 in


C  833.683 in sec

2

2

2

F  a  α2 cos θ2  a  ω2  sin θ2  b  ω3  sin θ3  c ω4  sin θ4 F  954.989 in sec

α3  6.

2

2

2

α3  471.320

rad

2

C D  A  F A  E  B D

sec

α4 

2

C  E  B F A  E  B D

α4  431.175

Use equation 7.13c to determine the acceleration of point B. AB  c α4  sin θ4  j  cos θ4   c ω4   cos θ4  j  sin θ4  2

AB  ( 30509  14941i)

mm sec


AB  1337.5

in sec

7.


AB  AB

2

at 2

θAB  26.092 deg

Use equations 7.32 to find the acceleration of the point P. APA  RPA α3  sin θ3  δ3  j  cos θ3  δ3    RPA ω3   cos θ3  δ3  j  sin θ3  δ3  2

AP  AA  APA AP  AP

AP  730.37

in sec

2

 

arg AP  53.649 deg

rad sec

2




The crosshead linkage shown in Figure P7-30c has 2 DOF with inputs at cross heads 2 and 5. Find AB, AP3, and AP4 if the crossheads are each moving toward the origin of the XY coordinate system with a speed of 20 in/sec and are decelerating at 75 in/sec2. Use a graphical method.

Given:


b  34.32  in

Link 4 (B to C)

c  50.4 in

Link 2 Y-offset

a  59.5 in

Link 5 X-offset

d  57 in

AP3  31.5 in

BP3  22.2 in

BP4  41.52  in

CP 4  27.0 in

Coupler point data:

Solution:

Input velocities:

V2Y  20 in sec

Input accelerations

AA  75 in sec

1

2

V5X  20 in sec AC  75 in sec

1

2


1.


2.

In order to solve for the accelerations at point B, we will need 3, 3, 4, and 3. From the velocity solution (Problem 6-98),

Y

θ  320.123  deg   1.749  rad sec

1

P3

A

CCW Direction of AA

3

θ  122.718  deg   1.177  rad sec 3.

Direction of AP3At Direction of ABAt Direction of ABCt

2

1

Direction of AP4Ct B

5

Use equation 7.4 to (graphically) determine the magnitude of the acceleration at point B, the magnitude of the relative accelerations ABA, ABC and the angular accelerations of links 3 and 4. The equations to be solved (simultaneously) graphically are

X C Direction of AC

AB = AA + ABA

AB = AC + ABC

Eliminating AB by combing equations and expanding, AA + (ABAt + ABAn) = AC + (ABCt + ABCn) AA  75.000 in sec

P4

4

CW

2

AC  75.000 in sec

2

2

θAA  270  deg θAC  180  deg

ABAn  b  


θABAn  θ  180  deg


2

ABCn  c 

ABCn  69.821 in sec

θABCn  θ  180  deg

θABCn  57.282 deg

2

2



a.

Choose a convenient velocity scale and layout the known vectors AA and AC

b.

From the tip of AA, draw known vector ABAn.

c.

From the tip of AC, draw known vector ABCn.

d. From the tips of ABAn and ABCn draw construction lines in the known directions of these vectors. e. From the origin, draw a construction line to the intersection of the two construction lines in step d. f. Complete the vector triangle by drawing ABA from the tip of AA to the intersection of the AB construction line and drawing AB from the tail of AA to the intersection of the ABA construction line. And then do the same with the ACB vector

Y

AC X

5.701

A

n BC

141.390° A

n BA

AA

4.436

t ABC

AB 0

t ABA

50 IN/S/S

4.409 Acceleration Scale

6.


ka  50

in sec

2

AB  5.701  ka

AB  285.1 in sec

ABAt  4.409  ka


 

ABAt b

ABCt  4.436  ka   7.

2

ABCt c

  6.423 rad sec

2

2

ABCt  221.8 in sec   4.401 rad sec


CW

2

2

From the graphical solution, the angles from A to P3 and C to P4 are

CCW



θP3  9.250  deg 8.

The accelerations of points P3 and P4 with respect to A are: 2

2

AP3An  AP3  

AP3An  96.359 in sec

AP3At  AP3  

AP3At  202.336 in sec

AP4Cn  CP4 

2

AP4Cn  37.404 in sec

AP4Ct  CP4  3.

θP4  67.313 deg

2

2

AP4Ct  118.821 in sec

2

θAP3An  θP3  180  deg

θAP3An  170.750 deg

θAP3At  θP3  90 deg

θAP3At  80.750 deg

θAP4Cn  θP4  180  deg

θAP4Cn  112.687 deg

θAP4Ct  θP4  90 deg

θAP4Ct  157.313 deg

Use equation 7.4 to (graphically) determine the magnitude of the acceleration at points P3 and P4.

2.832

AP3 = AA + (AP3At + AP3An)

Y

AP4 = AC + (AP4Ct + AP4Cn)

6.

A P4

164.000°

t AP4C

From the graphical solution at right,

n

A P4C

Acceleration scale factor ka  200 

X

AC n

A P3A

AA 87.517°

in sec

AP3  4.577  ka

2

AP3  915.4 in sec

2

AP4  566.4 in sec

200 IN/S/S

Acceleration Scale

at an angle of -87.517 deg AP4  2.832  ka

0

2

4.577


t AP3A

A P3




The crosshead linkage shown in Figure P7-30c has 2 DOF with inputs at cross heads 2 and 5. Find AB, AP3, and AP4 if the crossheads are each moving toward the origin of the XY coordinate system with a speed of 20 in/sec and are decelerating at 75 in/sec2. Use an analytical method.

Given:


b  34.32  in

Link 4 (B to C)

c  50.4 in

Link 2 Y-offset

a  59.5 in

Link 5 X-offset

d  57 in

AP3  31.5 in

BP3  22.2 in

BP4  41.52  in

CP 4  27.0 in

Coupler point data:

Solution:

Input velocities:

V2Y  20 in sec

Input accelerations

AA  75 in sec

1

2

V5X  20 in sec AC  75 in sec

1

2


1.

Draw the linkage to a convenient scale and lable it.

2.

In order to solve for the accelerations at point B, we will need 3, 3, 4, and 3. From the velocity solution (Problem 6-98),

Y

θ  320.123  deg   1.749  rad sec 3.

2

θ  122.718  deg 1

  1.177  rad sec

CCW

1

CW

P3

A 3

Use equation 7.4 to (analytically) determine the magnitude of the acceleration at point B, the magnitude of the relative accelerations ABA, ABC and the angular accelerations of links 3 and 4. The equations to be solved simultaneously are AB = AA + ABA

B

C

AA + (ABAt + ABAn) = AC + (ABCt + ABCn) 2

AC  75.000 in sec

2

2

Let

θAA  270  deg θAC  180  deg

ABAn  b  


θABAn  θ  180  deg


2

5 X

AB = AC + ABC

Eliminating AB by combing equations and expanding,

AA  75.000 in sec

P4

4

2

2

ABCn  c 

ABCn  69.821 in sec

θABCn  θ  180  deg

θABCn  57.282 deg

AAy   AA

AAy  75.000 in sec

ABAnx  ABAn  cos θABAn

ABAnx  80.568 in sec

ABAny  ABAn  sin θABAn

ABAny  67.310 in sec

2 2

2


SOLUTION MANUAL 7-80-2 2

ACx   AC

ACx  75.000 in sec

ABCnx  ABCn  cos θABCn

ABCnx  37.738 in sec

ABCny  ABCn  sin θABCn

ABCny  58.743 in sec

θABAt  θABAn  90 deg

θABAt  50.123 deg

θABCt  θABCn  90 deg

θABCt  32.718 deg

mBCt  tan  θABCt

mBCt  0.642

mBAt  tan θABAt

2 2

mBAt  1.197

Write the equations for the two lines that represent the vectors ABAt and ABCt and solve them simultaneously to find the coordinates of AB: ABx 

mBCt   ACx  ABCnx   mBAt ABAnx  ABCny   AAy  ABAny  mBCt  mBAt

ABy  mBCt  ABx   ACx  ABCnx   ABCny ABx  222.804 in sec AB 

2

ABx  ABy

2

2

θAB  atan2 ABx ABy 

ABy  177.941 in sec AB  285.140 in sec

θAB  141.388 deg

A similar approach can be taken to find AP3 and AP4.

2

2




The crosshead linkage shown in Figure P7-30c has 2 DOF with inputs at cross heads 2 and 5. At t = 0, crosshead 2 is at rest at the origin of the global XY coordinate system and crosshead 5 is at rest at (70,0). Write a computer program to find and plot AP3 and AP4 for the first 5 sec of motion if A2 = 0.5 in/sec2 upward and A5 = 0.5 in/sec2 to the left.

Given:


b  34.32  in

Link 4 (B to C)

c  50.4 in

Link 2 Y-offset

a  59.5 in

Link 5 X-offset

d  57 in

AP3  31.5 in

BP3  22.2 in

  39.128 deg

BP4  41.52  in

CP 4  27.0 in

  55.405 deg

Coupler point data:

A2  0.5 in sec

Input accelerations Solution: 1.

2

A5  0.5 in sec

2


Draw the linkage to a convenient scale and lable it.

Y

2 P3

A 3

R3

b B a 1

c

R2

P4

4

 3

5

R4 X

4

R5

C d t  0  sec 0.1 sec  5  sec a ( t)  a 0  0.5 A2  t

a 0  0  in

2

d ( t)  d 0  0.5A5  t

V2( t)  A2  t 2.

d 0  70 in 2

V5( t)  A5  t

Define a vector loop as shown above. Summing the vectors around the loop, R2  R3  R4  R5 = 0

    a  b  sin   c sin  = 0

b  cos   c cos   d = 0

3.

Substitute the polar notation and Euler equivalents and solve for 3 and 4 using the method of Section 4.5 and the identities in equations 4.9. 2

K1( t) 

2

2

a( t)  b  c  d( t) 2 c

2



A ( t)  K1( t)  d ( t)

B( t)  2  a ( t)



2

( t)  2  atan2 2  A ( t) B( t)  2

K2( t)  

2

B( t )  4  A ( t )  C ( t )

2

a ( t)  b  c  d ( t)



2

2 b

D( t)  K2( t)  d ( t)

E( t)  2  a ( t)



F ( t)  K2( t)  d ( t)

2

( t)  2  atan2 2  D( t) E( t)  4.

C( t)  K1( t)  d ( t)

E( t)  4  D( t)  F ( t)



Differentiate the position equations with respect to time and solve for 3 and 4.

 

 

b   sin   c  sin   V5 = 0

 

 

V2  b   cos   c  cos  = 0 ( t)  

( t) 

5.







V2( t)  sin ( t)  V5( t)  cos ( t) b  sin ( t)  ( t)











V2( t)  sin ( t)  V5( t)  cos ( t)



c sin ( t)  ( t)







Differentiate the velocity equations with respect to time and solve for 3 and 4. b    sin     cos 

 

   c   sin  2 cos  A5 = 0

2

A2  b    cos     sin 

 

   c   cos  2 sin = 0

2



A' ( t)  b  sin ( t)





B'( t)  c sin ( t)



2





2





C' ( t)  b  ( t)  cos ( t)  c ( t)  cos ( t)  A5



D'( t)  b  cos ( t) 2





E'( t)  c cos ( t)





2







F'( t)  b  ( t)  sin ( t)  c ( t)  sin ( t)  A2

( t) 

6.

B'( t)  F'( t)  C' ( t)  E'( t)

( t) 

A' ( t)  E'( t)  B'( t)  D'( t)

C' ( t)  D'( t)  A' ( t)  F'( t) A' ( t)  E'( t)  B'( t)  D'( t)

Using equations 6.32, solve for the accelerations at P3.

         2  ( t)   cos ( t)    j  sin ( t)    

AP3A( t)  AP3  ( t)  sin ( t)    j  cos ( t)  

AP3( t)  j  A2  AP3A( t)

AP3 ( t)  AP3( t)

θAP3( t)  if  θAP3( t)  0 θAP3( t)  2  π θAP3( t) 

θAP3( t)  arg AP3( t) 


7.


Plot the magnitude and direction of the acceleration at P3. MAGNITUDE - P3 80

60

2

AP3 ( t ) 

sec

40

in

20

0

0

1

2

3

4

5

t sec

DIRECTION - P3 200

190

θAP3( t )

180

deg

170

160

0

1

2

3

4

5

t

8.

sec

Using equations 6.32, solve for the accelerations at P4.

         2  ( t)   cos ( t)    j  sin ( t)    

AP4A( t)  CP4 ( t)  sin ( t)    j  cos ( t)  

AP4( t)  j  A5  AP4A( t)

AP4 ( t)  AP4( t)

θAP4( t)  if  θAP4( t)  0 θAP4( t)  2  π θAP4( t) 

θAP4( t)  arg AP4( t) 


9.


Plot the magnitude and direction of the acceleration at P4. MAGNITUDE - P4 40

30

2

AP4 ( t ) 

sec

20

in

10

0

0

1

2

3

4

5

t sec

DIRECTION - P3 350

348

346

θAP4( t ) deg 344

342

340

0

1

2

3 t sec

4

5




The linkage in Figure P7-30d has the path of slider 6 perpendicular to the global X-axis and link 2 aligned with the global X-axis. Find 2 and AA in the position shown if the velocity of the slider is constant at 20 in/sec downward. Use the acceleration difference graphical method.

Given:


a  12 in

Link 3 (A to B)

b  24 in

Link 4 (O4 to B)

c  18 in

Link 5 (C to D)

f  24 in

Link 4 (O4 to C)

e  18 in

Link 1 X-offset

d X  19 in

Link 1 Y-offset

d Y  28 in

  0.0 deg

Angle BO4C


Input slider velocity Solution: 1.

1

θVD  90.0 deg


Draw the linkage to a convenient scale. Indicate the directions of the acceleration vectors of interest. Direction of ACt

Direction of ACDt C

Direction of ABt O4 19.963°

4

116.161°

B 5

Y

D 3

O2

2

114.410°

6 X

A

Direction of AD

Direction of AABt Direction of AAt

2.

In order to solve for the accelerations at points A, B and C, we will need 2, 2, 3, 3, and 4 4. From the graphical position solution above and the velocity solution (see Problem 6-99),   0.000  deg

  1.648  rad sec

θ  114.410  deg

  0.282  rad sec

θ  199.963  deg

  1.003  rad sec

1 1 1

CCW CCW CW

θ  19.963 deg   116.161  deg

  0.286  rad sec

1

CW


3.


4.

For point C, this becomes: (ACt + ACn) = AD + (ACDt + ACDn) , where



ACn  e 

ACn  18.108 in sec

θACn  θ  180  deg

θACn  199.963 deg

AD  0  in sec ACDn  f  

2

2

ACDn  1.963 in sec

θACDn    180  deg 5.

2

2

θACDn  296.161 deg

Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ACn at an angle of ACn +  and ADn at an angle of ADn + . From the tip of ADn, draw ADt at an angle of ADt + . From the tip of ADt, draw ACDn at an angle of CDn + . Now that the vectors with known magnitudes are drawn, from the tips of ACn and ACDn, draw construction lines in the directions of ACt and ACDt, respectively. The intersection of these two lines are the tips of ACt, AC, and ACDt. 0

5 IN/S/S Y

Acceleration Scale 3.706

X

n

A CD n

AC 147.759°

AC 0.788

6.

t CD

A

AtC

From the graphical solution above, ka  5.0


in sec

2

AC  3.706  ka

AC  18.530 in sec

ACt  0.788  ka

ACt  3.940 in sec

 

ACt


2

  0.219 rad sec

e

2

2

CW

Since link 4 is straight and the distance from O4 to B is the same as to C, AB  AC 7.

AB  18.530 in sec

2


For point A, equation 7.4 becomes: (AAt + AAn) = AB + (AABt + AABn) , where 2

AAn  a  


θAAn    180  deg

θAAn  180.000 deg

2

2

2

AABn  b  

AABn  1.909 in sec

θAABn  θ

θAABn  114.410 deg



0

10 IN/S/S

n

A AB Y

Acceleration Scale

AB

n

AA X 0.993

AAt

t AAB

AA

163.062°

3.407

8.


ka  10

in sec

2

AA  3.407  ka

AA  34.070 in sec

AAt  0.993  ka


 

AAt a

2


2

  0.828 rad sec

2

CW




The linkage in Figure P7-30d has the path of slider 6 perpendicular to the global X-axis and link 2 aligned with the global X-axis. Find 2 and AA in the position shown if the velocity of the slider is constant at 20 in/sec downward. Use an analytical method.

Given:


c  12 in

Link 3 (A to B)

b  24 in

Link 4 (O4 to B)

a  18 in

Link 5 (C to D)

f  24 in

Link 4 (O4 to C)

e  18 in

Link 1 X-offset

d X  19 in

Link 1 Y-offset

d Y  28 in

Link 1 (O2 to O4)

d 

2

dX  dY

Axis rotation angle:

  124.160  deg



Input slider acceleration AD  0  in sec Solution: 1.

2

1

d  33.838 in

Downward (positive x' direction)

2


Draw the linkage to a convenient scale and label it. C O4 y'

19.963°

4

116.161°

B 5

Y x'

y" D

3

O2 x"

2.

2

114.410°

6 X

A

124.160°

From the graphical position solution above and the velocity solution (see Problem 6-99), Local x"y"coordinate system: Output

Input

1

  124.160  deg

  1.648  rad sec

θ  58.570 deg

  0.282  rad sec

θ  324.123  deg

  1.003  rad sec

1 1

CCW CCW CW

Local x'y' coordinate system:

3.

θ  109.963  deg

  1.003  rad sec

  206.161  deg

  0.286  rad sec

Solve equations 7.16b and c for 2 (which is 4 in this case):

1 1

CW CW


  4.


 



2



AD  cos   e   cos   θ  f  



e sin   θ



2

  0.219  rad sec

2

Use equations 7.12 to determine the angular accelerations of link 2. A  c sin 

 


 

D  c cos 

 


A  9.930  in

B  20.479 in

D  6.738  in

E  12.515 in

 

2

 

2

 

2

2

 

2

 

2

 

 

C  a   sin θ  a    cos θ  b    cos θ  c   cos  C  36.278 in sec

2

 

 

F  a   cos θ  a    sin θ  b    sin θ  c   sin  F  32.758 in sec α 

9.

2

C  E  B F A  E  B D

α  0.827 

rad sec

2

Use equation 7.13c to determine the acceleration of point A.



 

   c 2 cos  j  sin

AA  c α sin   j  cos  AA  ( 26.510  21.397i ) 

in sec


2

θAA  arg AA  

AA  AA AA  34.068

in sec

at 2

θAA  163.068  deg

(Global)




The linkage in Figure P7-30a has the path of slider 6 perpendicular to the global X-axis and link 2 aligned with the global X-axis at t = 0. Write a computer program or use an equation solver to find and plot AD as a function of 2, over the possible range of motion of link 2, in the global XY coordinate system.

Given:

Link lengths:

Solution: 1.

Link 2 (O2 to A)

a  12 in

Link 3 (A to B)

b  24 in

Link 4 (O4 to B)

c  18 in

Link 5 (C to D)

f  24 in

Link 4 (O4 to C)

e  18 in

Link 1 X-offset

d X  19 in

Link 1 Y-offset

d Y  28 in

Link 1 (O2 to O4)

d 

2

dX  dY

2

d  33.838 in

Axis rotation angle:

  55.840 deg

  90.000 deg

Input velocity & accel:

ω2  1  rad sec

Crank angle:

θ2XY  70 deg 69 deg  180  deg

1

α2  0  rad sec

2

θ2 θ2XY   θ2XY  


Draw the linkage to a convenient scale and label it. C

x' O4 y" 4 B

5

Y x"

D y' 3 2

O2

2.

55.840°

6 X

A


d

K1  2.8198

a 2

K3 

2

2

a b c d

K2 

d c

K2  1.8799

2

2 a c

K3  2.4005

A  θ2XY   cos θ2 θ2XY    K1  K2 cos θ2 θ2XY    K3 B θ2XY   2  sin θ2 θ2XY   C θ2XY   K1   K2  1   cos θ2 θ2XY    K3 3.






θ4xy θ2XY   2   atan2 2  A  θ2XY  B θ2XY  



B θ2XY   4  A  θ2XY   C θ2XY   2


4.



2

d b

2

2

c d a b

K5 

2

K4  1.4099

2 a b

K5  2.6753

D θ2XY   cos θ2 θ2XY    K1  K4 cos θ2 θ2XY    K5 E θ2XY   2  sin θ2 θ2XY   F  θ2XY   K1   K4  1   cos θ2 θ2XY    K5 5.






θ3xy θ2XY   2   atan2 2  D θ2XY  E θ2XY   6.

7.



E θ2XY   4  D θ2XY   F  θ2XY   2


ω3 θ2XY  

a  ω2 sin θ4xy θ2XY   θ2 θ2XY    b sin θ3xy θ2XY   θ4xy θ2XY  

ω4 θ2XY  

a  ω2 sin θ2 θ2XY   θ3xy θ2XY    c sin θ4xy θ2XY   θ3xy θ2XY  

Use equations 7.12 to determine the angular acceleration of link 4. A  θ2XY   c sin θ4xy θ2XY  

B θ2XY   b  sin θ3xy θ2XY  

D θ2XY   c cos θ4xy θ2XY  

E θ2XY   b  cos θ3xy θ2XY  

C θ2XY   a  α2 sin θ2 θ2XY    a  ω2  cos θ2 θ2XY    2

 b  ω3 θ2XY   cos θ3xy θ2XY    c ω4 θ2XY   cos θ4xy θ2XY   2

2

F  θ2XY   a  α2 cos θ2 θ2XY    a  ω2  sin θ2 θ2XY    2

 b  ω3 θ2XY   sin θ3xy θ2XY    c ω4 θ2XY   sin θ4xy θ2XY   2

α4 θ2XY   8.

2

C θ2XY   E θ2XY   B θ2XY   F  θ2XY  A  θ2XY   E θ2XY   B θ2XY   D θ2XY 

Transform 4 to the local x"y"coordinate system and add 180 deg to get the link from O4 to C, which is the input angle to the crank-slider portion of the linkage.

θ4 θ2XY   θ4xy θ2XY   180  deg     9.

Determine 5 in the x"y" coordinate system using equation 4.17. Offset: cc  27.5 in

 e sin θ4 θ2XY    cc  π f  

θ5 θ2XY   asin 


ω5 θ2XY  

e cos θ4 θ2XY     ω4 θ2XY  f cos θ5 θ2XY  



11. Determine the angular acceleration of link 5 using equation 7.16d. e α4 θ2XY   cos θ4 θ2XY    e ω4 θ2XY   sin θ4 θ2XY    2

α5 θ2XY  

 f  ω5 θ2XY   sin θ5 θ2XY   2

f  cos θ5 θ2XY  

12. Use equation 7.16e for the acceleration of pin D. AD  θ2XY   e α4 θ2XY   sin θ4 θ2XY    e ω4 θ2XY   cos θ4 θ2XY    2

 f  α5 θ2XY   sin θ5 θ2XY    f  ω5 θ2XY   cos θ5 θ2XY   2

ACCELERATION - PIN D

20





 AD θ2XY 

2

sec

10

in

0

 100

 50

0

50

100

θ2XY deg

Positive is up (positive Y direction in the global coordinate system).

150

200




For the linkage of Figure P7-30e, write a computer program or use an equation solver to find and plot AD in the global coordinate system for one revolution of link 2 if 2 is constant at 10 rad/sec CW.

Given:


a  5.0 in

Link 3 (A to B)

b  5.0 in

Link 4 (O4 to B)

c  6.0 in

Link 1 (O2 to O4)

d  2.5 in

Link 5 (B to C)

e  15 in

Angle BO4C

  83.621 deg

θ2  0  deg 1  deg  360  deg

Crank angle:

ω2  10 rad sec


1

α2  0  rad sec

2


1.

This sixbar drag-link mechanism can be analyzed as a fourbar Grashof double crank in series with a crankslider mechanism using the output of the fourbar, link 4, as the input to the slider-crank.

2.


d

K1  0.5000

a 2

K3 

2

2

a b c d

K2 

d

K2  0.4167

c

2

K3  0.7042

2 a c

A  θ2  cos θ2  K1  K2 cos θ2  K3 B θ2  2  sin θ2 C θ2  K1   K2  1   cos θ2  K3 3.






θ41 θ2  2   atan2 2  A  θ2 B θ2  4.



B θ2  4  A  θ2  C θ2  2


2

d

K5 

b

2

2

c d a b

2

K4  0.5000

2 a b

D θ2  cos θ2  K1  K4 cos θ2  K5 E θ2  2  sin θ2 F  θ2  K1   K4  1   cos θ2  K5 5.






θ3 θ2  2   atan2 2  D θ2 E θ2  6.



E θ2  4  D θ2  F  θ2   2


ω3 θ2 

a  ω2 b



sin θ41 θ2  θ2

sin θ3 θ2  θ41 θ2 

K5  0.4050


a  ω2

ω4 θ2  7.

c




sin θ2  θ3 θ2 

sin θ41 θ2  θ3 θ2 

Use equations 7.12 to determine the angular acceleration of link 4. A  θ2  c sin θ41 θ2 

B θ2  b  sin θ3 θ2 

D θ2  c cos θ41 θ2 

E θ2  b  cos θ3 θ2 

C θ2  a  α2 sin θ2  a  ω2  cos θ2  2

 b  ω3 θ2  cos θ3 θ2   c ω4 θ2  cos θ41 θ2  2

2

F  θ2  a  α2 cos θ2  a  ω2  sin θ2  2

 b  ω3 θ2  sin θ3 θ2   c ω4 θ2  sin θ41 θ2  2

α4 θ2 

2

C θ2  E θ2  B θ2  F  θ2

A  θ2  E θ2  B θ2  D θ2

ANGULAR ACCEL - LINK 4 200

100

α4 θ2 

2

sec

0

rad  100

 200

0

45

90

135

180

225

270

θ2 deg

8.

Transform 4 to the slider-crank coordinate system (origin at O4) system.

θ42 θ2  θ41 θ2   9.

Determine 5 using equation 4.17. Offset: cc  0  in

 c sin θ42 θ2   cc  π e  

θ5 θ2  asin 


ω5 θ2 

c cos θ42 θ2    ω4 θ2 e cos θ5 θ2 

315

360



11. Determine the angular acceleration of link 5 using equation 7.16d. c α4 θ2  cos θ42 θ2   c ω4 θ2  sin θ42 θ2   2

α5 θ2 

 e ω5 θ2  sin θ5 θ2  2

e cos θ5 θ2 

12. Use equation 7.16e for the acceleration of pin D. AD  θ2  c α4 θ2  sin θ42 θ2   c ω4 θ2  cos θ42 θ2   2

 e α5 θ2  sin θ5 θ2   e ω5 θ2  cos θ5 θ2  2

ACCELERATION - PIN D

3

2 10

3

1 10

 

AD θ2 

2

sec in

0

3

 1 10

0

45

90

135

180

θ2 deg

225

270

315

360




The linkage of Figure P6-33f has link 2 at 130 deg in the global XY coordinate system. Find AD in the global coordinate system for the position shown if 2 = 15 rad/sec CW and 2 = 50 rad/sec2 CW. Use the acceleration difference graphical method.

Given:


a  5.0 in

Link 3 (A to B)

b  8.4 in

Link 4 (O4 to B)

c  2.5 in

Link 1 (O2 to O4)

d 1  12.5 in

Link 5 (C to E)

e  8.9 in

Link 5 (C to D)

h  5.9 in

Link 6 (O6 to E)

f  3.2 in

Link 7 (D to F)

k  6.4 in

Link 3 (A to C)

g  2.4 in

d 2  10.5 in

Link 1 (O2 to O6)

θ2  130  deg

Crank angle:

Slider axis offset and angle:

Solution: 1.

s  11.7 in

  150  deg


ω2  15 rad sec


α2  50 rad sec

1

2

CW CW



Direction of VF 8

Y

F

Direction of VFD

Direction of VE

7 E

Direction of VEC

6 Direction of VDC

5

Direction of VCA Direction VBA

C

Direction of VB

O6

D

Direction of VA

A

3

2

B 4 O4

O2

X

Angles measured from layout:

2.

θVB  24.351 deg

θVBA  79.348 deg

θVCA  θVBA

θVE  64.594 deg

θVEC  162.461  deg

θVDC  θVEC

θVF  150.00 deg

θVFD  198.690  deg

θ3  169.348  deg

θ4  65.649 deg

θ5  72.461 deg

θ6  154.594  deg

Use equation 6.7 to calculate the magnitude of the velocity at point A. VA  a  ω2

VA  75.000

in sec

θ7  108.690  deg



θVA  θ2  90 deg 3.

θVA  40.000 deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relative velocity VBA, and the angular velocity of link 3. The equations to be solved graphically are VB = VA + VBA

and

VC = VA + VCA

a. Choose a convenient velocity scale and layout the known vector VA. b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB construction line and drawing VB from the tail of VA to the intersection of the VBA construction line.

2.668

VA 0.943 V CA

0

VC

25 in/sec

3.302

Y

22.053° X

V BA VB 3.192

4.

From the velocity polygon we have: Velocity scale factor: VB  3.192  in kv

ω4 

VB c

VBA  3.302  in kv

ω3  5.

VBA b

kv 

25 in sec

1

in

VB  79.800

ω4  31.920

in rad

CW

sec

VBA  82.550

ω3  9.827

θVB  24.351 deg

sec

in

θVBA  79.348 deg

sec

rad

CCW

sec

Calculate the magnitude and direction of VCA and determine the magnitude and velocity of VC from the velocity polygon above. in VCA  g  ω3 VCA  23.586 θVCA  79.348 deg sec VCA Length of VCA on velocity polygon: vCA  vCA  0.943 in kv VC  2.668  in kv

VC  66.700

in sec

θVC  22.053deg


6.


Use equation 6.5 to (graphically) determine the magnitude of the velocity at point E, the magnitude of the relative velocity VED, and the angular velocity of link 5. The equations to be solved graphically are VE = VC + VEC

and

VD = VC + VDC

a. Choose a convenient velocity scale and layout the known vector VC. b. From the tip of VC, draw a construction line with the direction of VEC, magnitude unknown. c. From the tail of VC, draw a construction line with the direction of VE, magnitude unknown. d. Complete the vector triangle by drawing VEC from the tip of VC to the intersection of the VE construction line and drawing VE from the tail of VC to the intersection of the VEC construction line. 1.821

1.207 V EC VE

1.716 0

V DC

VD

25 in/sec VC Y

45.934°

X 1.900

7.

From the velocity polygon we have: Velocity scale factor: VE  1.716  in kv

ω6 

VE f

VEC  1.821  in kv

ω5  8.

VEC e

kv 

25 in sec

1

in

VE  42.900

ω6  13.406

in rad in

θVEC  162.461 deg

sec

rad

CCW

sec

Calculate the magnitude and direction of VDE and determine the magnitude and velocity of VD from the velocity polygon above. VDC  h  ω5

VDC  30.179

in

VD  1.900  in kv

VD  47.500

θVDC  162.461 deg

sec vDC 

Length of VDC on velocity polygon:

9.

CW

sec

VEC  45.525

ω5  5.115

θVE  64.594 deg

sec

in sec

VDC kv

vDC  1.207 in

θVD  45.934deg

Use equation 6.5 to (graphically) determine the magnitude of the velocity at point F, the magnitude of the relative velocity VFD, and the angular velocity of link 5. The equation to be solved graphically is VF = VD + VFD



a. Choose a convenient velocity scale and layout the known vector VD. b. From the tip of VD, draw a construction line with the direction of VFD, magnitude unknown. c. From the tail of VD, draw a construction line with the direction of VF, magnitude unknown. d. Complete the vector triangle by drawing VFD from the tip of VD to the intersection of the VF construction line and drawing VF from the tail of VD to the intersection of the VFD construction line.

2.454 Y 0

VD

25 in/sec

V FD 150.000° VF

45.934°

X 1.158

10. From the velocity polygon we have: Velocity scale factor:

kv 

25 in sec in

in

VF  1.158  in kv

VF  28.950

VFD  2.454  in kv

VFD  61.350

ω7 

VFD

ω7  9.586

k

1

sec

θVF  150.000 deg

in sec

rad

CCW

sec

11. The graphical solution for accelerations uses equation 7.4:


12. For point B, this becomes: (ABt + ABn) = (AAt + AAn) + (ABAt + ABAn) , where ABn  c ω4

2


θABn  θ4  180  deg 2

θABn  114.351 deg

AAn  a  ω2


θAAn  θ2  180  deg

θAAn  50.000 deg

AAt  a  α2


θAAt  θ2  90 deg

θAAt  40.000 deg

2

2

2

2

2

ABAn  b  ω3


θABAn  θ3  180  deg


13. Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw ABn at an angle of ABn and AAn at an angle of AAn. From the tip of AAn, draw AAt at an angle of AAt. From the tip of AAt, draw ABAn at an angle of BAn. Now that the vectors with known magnitudes are drawn, from the tips of ABn and ABAn, draw construction lines in the directions of ABt and ABAt, respectively. The intersection of these two lines are the tips of AB, ABt, and ABAt.



Y 0

1000 in/s/s

1.152 X

cceleration Scale

AA A

n

ABA 37.471°

AAt

n A

n

AB

3.358 t ABA

AB t B

A

14. From the graphical solution above, Acceleration scale factor

ka  1000

in sec

2 2

AA  1.152  ka

AA  1152 in sec

ABAt  3.358  ka

ABAt  3358 in sec

α3 

ABAt

α3  399.762

b

θAA  37.471 deg

2

rad sec

CW

2

15. For point C, : AC = AA + (ACAt + ACAn) , where Y

X 3

AA  1.152  10 in sec

2

θAA  37.471 deg

AA 50.502°

θAAn  θ2  180  deg

θAAn  50.000 deg

ACAt  g  α3


θACAt  θ3  90 deg

θACAt  79.348 deg

ACAn  g  ω3

2

θACAn  θ3  180  deg

n

ACA

2


4.150

2

θACAn  349.348 deg

0

500 in/s/s

Acceleration Scale

16. From the graphical solution above,

AC

t ACA



in

ka  500 


sec AC  4.150  ka

2

AC  2075 in sec

2

θAC  50.502 deg

17. For point E: (AEt + AEn) = AC + (AECt + AECn) , where AEn  f  ω6

2

AEn  575.1 in sec

θAEn  θ6  180  deg 2

Acceleration Scale

θAEn  25.406 deg

n

2

θAECn  θ5  180  deg

θAECn  107.539 deg ka  500 

in sec

α5 

2

AECt  1349 in sec

AECt

sec

AEt

AE

2

AC

rad

α5  151.517

e

X

AE

AECn  232.9 in sec

AECt  2.697  ka

500 in/s/s

2

AECn  e ω5


0

Y

CCW

t AEC

2

n

AEC 2.697

18. For point D, : AD = AC + (ADCt + ADCn) , where AC  2075.0 in sec

2

θAC  50.502 deg 2

ADCt  h  α5

ADCt  893.9 in sec

θADCt  θ5  90 deg

θADCt  162.461 deg

2

θADCn  θ5  180  deg

θADCn  107.539 deg

AD  3.194  ka

3.194

in sec

125.841°

AD 2

AC  2075 in sec

AE 2

θAD  125.841  deg 19. For point F, : AF = AD + (AFDt + AFDn) , where 2

Acceleration Scale

2

ADCn  154.4 in sec

ka  500 

AFDn  k ω7

AFDn  588.10 in sec

θAFDn  θ7  180  deg

θAFDn  71.310 deg

500 in/s/s

X

ADCn  h  ω5


0

Y

2

t

ADC

n

ADC



Y 0

500 in/s/s

4.033 X

cceleration Scale

30.000°

AF AD

t AFD

n

AFD


ka  500 

in sec

AF  4.033  ka

2

AF  2017 in sec

2

θAD  30.00  deg




Figure 3-14 shows a crank-shaper quick-return mechanism with the following dimensions: L2 = 4.80 in, L4 = 24.00 in, L5 = 19.50 in, the distance from link 4's fixed pivot (O4) to link 2's fixed pivot (O2) is 16.50 in, and the vertical distance from O2 to the pivot point on link 6 is 6.465 in. Use a graphical method to calculate the acceleration of link 6 when 2 = 45 deg. Assume that link 2 has a constant angular velocity of 2 rad/sec CW.

Given:

Link lengths and angles: Link 2

L2  4.80 in

Link 4

L4  24.00  in

Link 5

L5  19.50  in ω  2  rad sec



1

CW

α  0  rad sec

2



C 6

5

B

6.465" 4

O2

14.519° y

2 A 3

Axis of slip

13.538"


4 Direction of VA3 45.000°

O4 x

2.

Use equation 6.7 to calculate the magnitude of the velocity at point A on links 2 and 3. VA3  L2 ω

3.

VA3  9.600

in sec

θVA3  θ  90 deg

θVA3  45.0 deg




a. Choose a convenient velocity scale and layout the known vector VA3. b. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the Vslip construction line. 4.



5  in sec

1

0.975"

Vtrans

in

y

in

Vslip  1.654  in kv

Vslip  8.270


Vtrans  4.875

1.654"

sec in

x

Vslip

sec

0

5 in/sec

VA3 5.

The true velocity of point A on link 4 is Vtrans, VA4  Vtrans

6.

VA4  4.875

sec


VA4 c

VB  L4 ω 7.

in

c  13.538 in and

ω  0.360

rad

VB  8.642

in

θ  180  deg  14.519 deg θ  165.481 deg

CCW

sec sec

Draw links 1, 4, 5, and 6 to a convenient scale. Indicate the directions of the velocity vectors of interest. Direction of VCB C 6

5

B Direction of VB

Direction of VC

0.806° 4

O2

y

2 A 3 4 45.000°

O4 x


8.



9.



kv 

5  in sec in

VC  8.345

1.730"

0.434"

1

VB

VCB

in

y

VC

sec

0

θVC  180  deg

5 in/sec

x

VCB  0.434  in kv

VCB  2.170

in

1.669"

sec

10. Determine the angular velocity of link 5 using equation 6.7. θ  270.806  deg

From the linkage layout above: ω 

VCB

ω  0.111

L5

rad

CCW

sec

11. For the acceleration of point A, use equation 7.19: (AA2t + AA2n) = AA4t + AA4n + AAcor + AAslip , where 2

AA2n  L2 ω

AA2n  19.200 in sec

θAA  θ  180  deg

θAA  225.000 deg

2

2

AA4n  c ω

AA4n  1.755 in sec

θAA4n  θ  180  deg

θAA4n  345.481 deg



θAAcor  θ  90 deg

θAAcor  75.481 deg

(Vslip is negative)

2

2

AA2t  L2 α


AAcor AAslip

AA2

2.259" AA4 t AA4

n AA4

13. From the acceleration polygon above,

y

2.254" x

12. Repeat procedure of steps 3 and 7 for the equation in step 11. (See next page).

2

0

10 in/sec



ka  10


in sec

2 2

AA4  2.259  ka

AA4  22.6 in sec

AA4t  2.254  ka



2


AA4t

α  1.665

c

rad sec

CCW

2

14. The graphical solution for accelerations in pin-jointed fourbars uses equation 7.4: (APt + APn) = (AAt + AAn) + (APAt + APAn) 15. For point C, this becomes: AC = (ABt + ABn) + (ACBt + ACBn) , where ABn  L4 ω

2


2

θABn  θ  180  deg

θABn  345.481 deg

ABt  L4 α


θABt  θ  90 deg

θABt  75.481 deg

2

2

2

ACBn  L5 ω


θACBn  θ


16. Repeat procedure of step 12 for the equation in step 15.

3.748" AC

ABn

y t ACB

x 0

10 in/sec

ABt n ACB


ka  10

in sec

AC  3.748  ka

2

AC  37.5 in sec

2





Repeat Problem 7-87 using an analytical method to calculate the acceleration of point C for one revolution of the input crank (link 2).

Given:


d  16.50  in

Link 2 (L2)

a  4.80 in

Link 4 (L4)

a'  24.00  in

Link 5 (L5)

b'  19.50  in ω  2  rad sec


d'  22.965 in

Vertical offset from O4 to C

1

α  0  rad sec

2


Draw the mechanism to scale and define a vector loop for the input portion (links 1, 2, 3, and 4) using the fourbar crank-slider derivation in Section 7.3 as a model. θ  0  deg 0.5 deg  360  deg

c'

C 6

5

B

4

O2

y

2

A 3

d' 4

2 y R2

O4 R1

x R3

x


2.



j  θ

 d  b  e

j  θ


  

   d  bcosθ  j  sinθ



 

   a  sin θ sin θ θ 

 

θ θ  atan2 a  cos θ  d a  sin θ

 

b θ  3.

Differentiate the position equation, expand it and solve for 3 and bdot. a  j  ω e

j  θ

 

ω θ 

a  ω

 

b θ

j  θ



 bdot e

j  θ

 

 cos θ  θ θ

 

    sin θ θ 

  

a  ω cos θ  b θ  ω θ  cos θ θ

 

bdot θ  4.

 b  j  ω e


j  θ

2

2 j  θ

 a  j  ω  e

 bdot j  ω e 2

j  θ

2 j  θ

 b  j  ω  e

 

α θ  5.

1

 

b θ

j  θ

 bddot e



j  θ


j  θ

  a  α cos θ  θ θ

   a ω2 sinθθ  θ  2 bdotθ ωθ



The position, angular velocity, and angular acceleration are the same for links 3 and 4.

 

 

 

θ θ  θ θ 6.

 b  j   e

 

ω θ  ω θ

 

 

α θ  α θ

Write a vector loop equation for links 1, 4, 5, and 6, expand the result and separate into real and imaginary parts to solve for 5 and c' (horizontal distance from O4 to C).

 3 π   j  θ j  θ j π 2  c'  e   d' e = a' e  b' e j

R6  R7 = R4  R5 Substituting the Euler equivalents,

  

   b' cosθ  j  sinθ

j  c'  d' = a' cos θ  j  sin θ

Separating into real and imaginary components and solving for 5 and c'.

 

 a' cos θ θ   d'   b'  

θ θ  acos

 



    b' sinθθ

c' θ   a' sin θ θ


7.


Differentiate the position equation, expand it and solve for 5 and VC.

 3 π   j  θ j  θ 2  VC  e  = a' ω j  e  b' ω j  e j 




 

   b' ω sinθ  j  cosθ

VC = a' ω sin θ  j  cos θ

Separating into real and imaginary components and solving for 5 and VC.

        VC θ  a' ω θ  cos θ θ   b' ω θ  cos θ θ  a' sin θ θ ω θ     ω θ b' sin θ θ

 

Differentiate the position equation, expand it and solve for 5 and AC.

 3 π    2 j  θ  α  ej  θ  b' j   j  ω 2 ej  θ  α  ej  θ 2  AC  e  = a' j   j  ω  e       j 


      a'  sinθ  j  cosθ  2  b' ω   cos θ  j  sin θ   b'   sin θ  j  cos θ  2

AC = a' ω  cos θ  j  sin θ

Separating into real and imaginary components and solving for 5 and AC.

 

    a' ωθ2 sinθθ  b' ωθ2 cosθθ b' sin θ θ 

 

a' α θ  cos θ θ

 

      a' ωθ2 sinθθ  2  b' α θ  cos θ θ   b' ω θ  sin θ θ 

α θ 

AC θ  a' α θ  cos θ θ

ACCELERATION OF POINT C 100


8.

50

 

AC θ 

2

sec

0

in

 50

 100

0

45

90

135

180

225

θ deg Crank Angle, deg

270

315

360




Figure P7-22 shows a mechanism with dimensions. Use a graphical method to determine the accelerations of points A and B for the position shown. Ignore links 5 and 6. ω 2 = 24 rad/s clock-wise.

Given:


d  1.22 in


θ  56.5 deg

Link 2 (O2A)

a  1.35 in

Angle O2A makes with X axis

θ  14 deg

Link 4 (O4B)

e  1.36 in ω  24 rad sec


1

α  0  rad sec

CW

2



B

Axis of slip Y


4

0.939

132.661° A 3 2 X O2

Direction of VA3

2.


3.

VA3  32.400

in sec

θVA3  θ  90 deg

θVA3  76.0 deg




Y 0

12 in/sec

1.295" X Vtrans

VA3 2.369"

4.


5.

kv 

12 in sec in

in

Vslip  28.428


Vtrans  15.540

sec in sec


in sec


7.

1

Vslip  2.369  in kv


Vslip

VA4

c  0.939  in and

ω  16.550

c

rad

CW

sec


VB  22.507

in sec

θVA4  θ  90 deg 8.

θ  132.661  deg

θVA4  42.661 deg

For the acceleration of point A, use equation 7.19: (AA2t + AA2n) = AA4t + AA4n + AAcor + AAslip , where AA2n  a  ω

2

AA2n  777.600  in sec

θAA  θ  180  deg

θAA  194.000  deg

AA2t  a  α

AA2t  0.000  in sec 2

AA4n  c ω

2

2

AA4n  257.2  in sec

2


9.


θAA4n  θ

θAA4n  132.661  deg


AAcor  940.940  in sec


θAAcor  222.661  deg

2

Repeat procedure of steps 3 and 7 for the equation in step 8.

Y cor

AA

2.572" t

A A4 0

A A4

100 IN/S/S

3.647"

n A A4

87.504° X

cor AA

A A2 10. From the acceleration polygon above, Acceleration scale factor

ka  100 

in sec

2 2

AA4  3.647  ka

AA4  364.7  in sec

AA4t  2.572  ka

AA4t  257.2  in sec


2


AA4t

α  273.9 

c

rad sec

CCW

2

11. For point B: ABn  e ω

2

ABn  372.5  in sec

2

θABn  θ  180  deg

θABn  312.661  deg


ABt  372.5  in sec


θABt  222.661  deg

2




Figure P7-22 shows a mechanism with dimensions. Use an analytical method to calculate the accelerations of points A and B for the position shown. Ignore links 5 and 6. ω 2 = 24 rad/s clock-wise.

Given:


d  1.22 in

Angle link 2 makes with X axis θ  42.5 deg

Link 2 (O2A)

a  1.35 in

Angle link 3 makes with X axis θ  103.839  deg

Link 4 (O4B)

c  1.36 in ω  24 rad sec


1

α  0  rad sec

2


Draw the mechanism to scale and define a vector loop for links 1, 2, 3, and 4 using the fourbar crank-slider derivation in Section 7.3 as a model.

B Axis of slip X b = 0.939"

O4

4

103.839°

R3

d = 1.220"

Y


R1 42.500°

A 3

2

R2 O2 a = 1.350"

2.

Write the vector loop equation, expand the result and separate into real and imaginary parts to solve for 3 and b (distance from O4 to A). R2 = R1  R3 a e

j  θ

= d  b e

j  θ


  

  = d  bcosθ  j  sinθ


Separating into real and imaginary components and solving for b.


b  3.


  sin θ

a  sin θ

b  0.939 in


j  θ

a  ω b

bdot 

= b  j  ω e



j  θ

 bdot e

j  θ



 cos θ  θ

 

 

a  ω cos θ  b  ω cos θ

 

ω  16.544

rad

bdot  28.430

in

sin θ 4.

s


j  θ

2

2 j  θ

 a  j  ω  e

= bdot j  ω e

j  θ

α 

1 b

 b  j   e

2 j  θ

2

 b  j  ω  e

5.

s

j  θ

 bddot e



j  θ







2





j  θ

α  275.059 

rad 2

s

Determine the acceleration of point A on link 2 (in the local XY coordinate system) using equations 7.13a.



 

   a ω2 cosθ  j  sinθ

AA2  a  α sin θ  j  cos θ AA2  AA2

AA2  777.600 


in 2

θAA2  137.500  deg

s 6.

Determine the acceleration of points A and B on link 4 (in the local XY coordinate system) using equations 7.13a. Rename 3 to θ4:

θ  θ


α  α ω  ω



 

   b ω2 cosθ  j  sinθ

AA4  b  α sin θ  j  cos θ AA4  AA4

AA4  364.484 

in 2


θAA4  31.019 deg

s



 

   a ω2 cosθ  j  sinθ

AB  a  α sin θ  j  cos θ AB  AB

AB  523.844 

in 2

s


θAB  31.019 deg




Figure P7-23 shows a quick-return mechanism with dimensions. Use a graphical method to calculate the accelerations of points A, B, and C for the position shown. 2 = 16 rad/s. Ignore links 5 and 6.

Given:


d  1.69 in


Link 2 (L2)

a  1.00 in


Link 4 (L4)

e  4.76 in


ω  16 rad sec

1

CCW

θ  195.5  deg

α  0  rad sec

2



B


Direction of VA3

4


2 44.228° O2

X O4

2.


3.

VA3  16.000

in sec

θVA3  θ  90 deg

θVA3  189.0  deg




a. Choose a convenient velocity scale and layout the known vector VA3. b. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown. c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown. d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans from the tail of VA3 to the intersection of the Vslip construction line.

Y 0

8 in/sec

X

VA3

Vtrans

4.


5.

kv 

in


Vtrans  9.232 

in sec in sec

The true velocity of point A on link 4 is Vtrans, VA4  9.23

in sec


ω 

VA4

c  2.068  in and

ω  4.464 

c

rad

θ  44.228 deg

CCW

sec


VB  21.250

θVA4  θ  90 deg 8.

1

Vslip  13.072


7.

8  in sec

Vslip  1.634  in kv


1.634"

Vslip

1.154"

in sec

θVA4  45.772 deg

For the acceleration of point A, use equation 7.19: (AA2t + AA2n) = AA4t + AA4n + AAcor + AAslip , where AA2n  a  ω

2

AA2n  256.000  in sec

2



θAA  θ  180  deg

θAA  279.000  deg

AA2t  a  α

AA2t  0.000  in sec 2

9.

2 2

AA4n  c ω

AA4n  41.214 in sec

θAA4n  θ  180  deg

θAA4n  224.228  deg


AAcor  116.712  in sec


θAAcor  134.228  deg

2

Repeat procedure of steps 3 and 7 for the equation in step 8. 0

Y

25 IN/S/S

Acceleration Scale

n

X

AA4

4.042" 69.810°

AA4 t A4

A 3.711"

slip

AA

AA2 cor

AA


ka  25

in sec

2

AA4  4.042  ka

AA4  101.0  in sec

AA4t  3.711  ka

AA4t  92.8 in sec

2


2


AA4t c

α  44.9

rad sec

2

CW



11. For point B: ABn  e ω

2

ABn  94.9 in sec

2

θABn  θ  180  deg

θABn  224.228  deg


ABt  213.5  in sec


θABt  134.228  deg

2




Figure P7-23 shows a quick-return mechanism with dimensions. Use an analytical method to calculate the accelerations of points A and B for the position shown. Ignore links 5 and 6. ω 2 = 16 rad/s.

Given:


d  1.69 in

Angle link 2 makes with X axis θ  96.517 deg

Link 2 (O2A)

a  1.00 in

Angle link 3 makes with X axis θ  151.289  deg

Link 4 (O4B)

e  4.76 in


ω  16 rad sec

1

α  0  rad sec

2


Draw the mechanism to scale and define a vector loop for links 1, 2, 3, and 4 using the fourbar crank-slider derivation in Section 7.3 as a model.

B

Axis of transmission 4

Axis of slip A 3 2.068" 1.000"

R2 151.289°

96.517° R3

2

O2 R1

X O4

1.690

Y

2.

Write the vector loop equation, expand the result and separate into real and imaginary parts to solve for 3 and b (distance from O4 to A). R2 = R1  R3


a e

j  θ

= d  b e

SOLUTION MANUAL 7-92-2 j  θ


  

  = d  bcosθ  j  sinθ


Separating into real and imaginary components and solving for b. b  3.

  sin θ

a  sin θ

b  2.068 in


j  θ

a  ω b

bdot 

= b  j  ω e



j  θ

 bdot e

j  θ



 cos θ  θ

 

ω  4.463 

 

a  ω cos θ  b  ω cos θ

 

rad s

bdot  13.070

in

sin θ 4.


j  θ

2

2 j  θ

 a  j  ω  e

= bdot j  ω e

j  θ

α 

1 b

 b  j   e

2 j  θ

2

 b  j  ω  e

5.

s

j  θ

 bddot e



j  θ







2





j  θ

α  44.710

rad 2

s

Determine the acceleration of point A on link 2 (in the local XY coordinate system) using equations 7.13a.



 

   a ω2 cosθ  j  sinθ

AA2  a  α sin θ  j  cos θ AA2  AA2

AA2  256.000 


in 2

θAA2  83.483 deg

s 6.

Determine the acceleration of points A and B on link 4 (in the local XY coordinate system) using equations 7.13a. Rename 3 to θ4:

θ  θ


α  α



 

ω  ω

   b ω2 cosθ  j  sinθ

AA4  b  α sin θ  j  cos θ AA4  AA4

AA4  101.226 

in 2


θAA4  94.703 deg

s



 

   a ω  cosθ  j  sinθ

AB  a  α sin θ  j  cos θ AB  AB

AB  48.944

2

in 2

s


θAB  94.703 deg




Given:

The general linkage configuration and terminology for an offset fourbar slider-crank linkage are shown in Fig P7-2. The link lengths and the values of d, d-dot, and d-double-dot are defined in Table P7-5. For row a, find the acceleration of the pin joint A and the angular acceleration of the crank using a graphical method. Link lengths: a  1.4 in

Link 2 Offset Solution: 1.

c  1  in

Link 3

b  4  in

θ2  176.041  deg

ddot  10 in sec

1

dddot  0  in sec

2



Direction of VBA 13.052° B

. d = VB

176.041° A

000 b = 4. a = 1.400

c = 1.000"

O2

d = 2.500" 0

10 inches/sec

Direction of VA 2.

Use equation 6.6.24b to (graphically) determine the magnitude of the velocity at point A. The equation to be solved graphically is

VA

VA = VB + VBA a. Choose a convenient velocity scale and layout the known vector VB. b. From the tip of VB, draw a construction line with the direction of VBA, magnitude unknown. c. From the tail of VB, draw a construction line with the direction of VA, magnitude unknown. d. Complete the vector triangle by drawing VBA from the tip of VB to the intersection of the VA construction line and drawing VA from the tail of VB to the intersection of the VBA construction line. 3.

VBA

3.414" 3.330"

Y

86.041°


103.052°

VB

1.000"

X




kv 

10 in sec

1

in in

VA  3.330  in kv

VA  33.300

VBA  3.414  in kv

VBA  34.140

θVB  86.041 deg

sec in

θVBA  103.052  deg  180deg

sec

θVBA  76.948 deg 4.

Use equation 6.7 to find the angular velocity of the crank (link 2).

ω2  ω3 

VA

ω2  23.786

a VBA

ω3  8.535 

b

rad

CW

sec

rad

CW

sec

5.


6.

For point A, this becomes:

AAt + AAn = ABt - (ABAt + ABAn) , where

2

AAn  a  ω2

AAn  792.064  in sec

θAAn  θ2  180  deg

θAAn  356.041  deg

ABt  dddot

ABt  0.000  in sec

θABt  0  deg

θABt  0.000  deg

2

4.


2

2

ABAn  b  ω3

ABAn  291.385  in sec

θABAn  13.052 deg  180  deg

θABAn  193.052  deg

2

Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw AAn at an angle of AAn. From the origin, draw -ABAt at an angle of ABt. Now that the vectors with known magnitudes are drawn, from the tip AAn and the tip of -ABAt, draw construction lines in the directions of AAt and ABAn, respectively. The intersection of these two lines are the tips of AAt and ABAn.

5.

From the graphical solution below, in

Acceleration scale factor ka  200 

sec

2 2

AA  8.895  ka

AA  1779 in sec

AAt  7.965  ka

AAt  1593 in sec

α2 

AAt a

α2  1138

rad sec

2

2



Y


0

200 IN/S/S

Acceleration Scale

n -A BA

X

n

AA

67.521°

t

-A A4

AA t

AA

7.965" 8.895"



PROBLEM 7-94a The general linkage configuration and terminology for an offset fourbar slider-crank linkage are shown in Fig P7-2. The link lengths and the values of d, d-dot and d-double-dot are defined in Table P7-5. For row a, find the acceleration of pin joint A and the angular acceleration of the crank using the analytic method. Draw the linkage to scale and label it before setting up the equations. Link lengths:

Statement:

Given:

Link 2 (O2 A)

a  1.4 in

Slider position

d  2.5 in dddot  0 

Slider acceleration:

b  4  in

Link 3 (AB) Slider velocity in

ddot  10

Offset (yB) in s

2

s


Solution: 1.

Draw the linkage to scale and label it. 13.052° B

. .. d d

176.041° 000 b = 4.

A

a = 1.400

c = 1.000"

O2

d = 2.500"

2.

Determine the open value of 2 using equations 4.20 and 4.21. 2

2

2

K1  a  b  c  d

2

2

K2  2  a  c

K2  2.8 in

K3  2  a  d

K3  7 in

A  K1  K3

A  0.21 in

B  2  K2

B  5.6 in

C  K1  K3

C  13.79 in

2 2

2 2



2

θ2  2  atan2 2  A B  3.

2

K1  6.79 in

B  4 A  C



θ2  176.041  deg

Determine the value of 3 using equation 4.16a or 4.17.

β ( a b c d α) 

θ  asin



a  sin( α)  c 

 

b




a  sin( α)  c  b


c  1  in



θ3  β  a b c d θ2 4.

Determine the value of ω 2 and ω 3 using equations 6.24.

ω2 

ω3  5.

6.

θ3  193.052  deg

ddot cos θ3

ω2  23.785

a   cos θ2  sin θ3  sin θ2  cos θ3  a  ω2 cos θ2

ω3  8.525 

b  cos θ3

s

rad s

Determine the value of VA using equation 6.7. in

VA  a  ω2

VA  33.299

θVA  θ2  sign ω2  90 deg

θVA  86.041 deg

s

Determine the value of α2 using equation 7.17b. a  ω2   cos θ2  cos θ3  sin θ2  sin θ3   b  ω3  dddot cos θ3 2

α2 

α2  1139.37 

2

a   cos θ2  sin θ3  sin θ2  cos θ3 

rad 2

s 7.

rad

Determine the value of AA using equation 7.3. AA  a  α2  sin θ2  j  cos θ2   a  ω2   cos θ2  j  sin θ2  2

AA  AA

AA  1780.93

in sec


2

θAA  67.553 deg




Figure P8-1 shows the cam and follower from Problem 6-65. Using graphical methods, find and sketch the equivalent fourbar linkage for this position of the cam and follower.

Solution:


1.

Draw the cam and follower to scale. Axis of slip


4

2

O2

2.

O4

As described in Section 8.1 and Figure 8-1, find the centers of curvature of the cams, which are located on the common normal (axis of transmission). They will be the locations of the moving pivots of the effective pin-jointed fourbar linkage.

A 3 B

2

O2

4

O4




Figure P8-1 shows the cam and follower from Problem 6-65. Using graphical methods, find the pressure angle at the position shown.

Solution:


1.

Draw the cam and follower to scale. Axis of slip


2

4

O2

2.

O4

As defined in Section 8.6 and Figure 8-41, the pressure angle is the angle between the direction of motion of the follower and the direction of the axis of transmission. Establish those two directions and measure the angle between them. Direction of motion, link 4

55.3° Axis of transmission

2

O2

4

O4




Figure P8-2 shows the cam and follower. Using graphical methods, find and sketch the equivalent fourbar linkage for this position of the cam and follower.

Solution:


1.

Draw the cam and follower to scale.

4 Common Normal

3

2

2.


4

O4

B

3

A 2 O2




Figure P8-2 shows the cam and follower. Using graphical methods, find the pressure angle at the position shown.

Solution:


1.

Draw the cam and follower to scale.

4 Common Normal

3

2

2.

As defined in Section 8.6 and Figure 8-41, the pressure angle is the angle between the direction of motion of the follower and the direction of the axis of transmission. Establish those two directions and measure the angle between them. Direction of motion, link 4 Common Normal 4.9° 4

3

2




Figure P8-3 shows the cam and follower. Using graphical methods, find and sketch the equivalent fourbar linkage for this position of the cam and follower.

Solution:


1.

Draw the cam and follower to scale. Common Normal

4

3

2

2.


Slider 1

4

3 2




Figure P8-3 shows the cam and follower. Using graphical methods, find the pressure angle at the position shown.

Solution:


1.

Draw the cam and follower to scale. Common Normal

4

3

2

2.

As defined in Section 8.6 and Figure 8-41, the pressure angle is the angle between the direction of motion of the follower and the direction of the axis of transmission. Establish those two directions and measure the angle between them.

Direction of motion, link 4 13.8° Common Normal 4

3

2




Given:

Design a double-dwell cam to move a follower from 0 to 2.5 in in 60 deg, dwell for 120 deg, fall 2.5 in in 30 deg and dwell for the remainder. The total cycle must take 4 sec. Choose suitable programs for rise and fall to minimize accelerations. Plot the s v a j diagrams. RISE

DWELL

β1  60 deg

β2  120  deg

h 1  2.5 in Cycle time: Solution: 1.

3.

DWELL

β3  30 deg

h 2  0  in

β4  150  deg

h 3  2.5 in

h 4  0  in

τ  4  sec


The camshaft turns 2 rad during the time for one cycle. Thus, its speed is ω 

2.

FALL

2  π rad

ω  1.571

rad

sec τ From Table 8-3, the motion program with lowest acceleration that does not have infinite jerk is the modified trapezoidal. The modified trapezoidal motion is defined in local coordinates by equations 8.15 through 8.19. The numerical constants in these SCCA equations are given in Table 8-2. b  0.25

c  0.50

d  0.25

Cv  2.0000

Ca  4.8881

Cj  61.426

The SCCA equations for the rise or fall interval () are divided into 5 subintervals. These are: for 0 <= x <= b/2 where, for these equations, x is a local coordinate that ranges from 0 to 1

 b  b  2  π      sin  x  b   π  π

y1 ( x)  Ca x

y' 1 ( x)  Ca

 π  x  b 

y''1 ( x)  Ca sin

 π  b

y'''1 ( x)  Ca

 π  x    b 

  1  cos

π b

 π  x  b 

 cos

for b/2 <= x <= (1 - d)/2

 x2

y2 ( x)  Ca 

2 

 b  

1

π



1  2 1   x  b    2  2 8 π   1

y''2 ( x)  Ca

y' 2 ( x)  Ca x  b  



1

π



1 



2 

y'''2 ( x)  0

for (1 - d)/2 <= x <= (1 + d)/2

 b c  x   π 2  

y3 ( x)  Ca 

2 2 2  d   b2  1  1   ( 1  d )   d   cos π   x  1  d        8 2 8 2  d   π  π π   

 b  c  d  sin π   x  1    2 π 2 π  d 

y' 3 ( x)  Ca 

1  d  π y''3 ( x)  Ca cos   x   2  d 

d 

  y'''3 ( x)  Ca

1  d  π  sin   x   d 2  d 

π



for (1 + d)/2 <= x <= 1 - b/2

 x2  b  1  2  π 

1  1 2 2  1   x  2 d  b   2     2 4 8 π   b

y4 ( x)  Ca 

y' 4 ( x)  Ca  x 



b

1

π

b



2





y''4 ( x)  Ca

y'''4 ( x)  0

for 1 - b/2 <= x <= 1

2

b 2 d  b y5 ( x)  Ca   x  π 2 π  y' 5 ( x)  Ca

  (1  b)2  d2   b  2sin π (x  1)

2

   π

 b

y'''5 ( x)  Ca

π

  

 π   1  cos  ( x  1 ) b π    b

π  y''5 ( x)  Ca sin  ( x  1 ) b   4.

4

π   cos  ( x  1 ) b b  

The above equations can be used for a rise or fall by using the proper values of , , and h. To plot the SVAJ curves, first define a range function that has a value of one between the values of x1 and x2 and zero elsewhere. R( x x1 x2)  if [ ( x  x1)  ( x  x2) 1 0 ]

5.

The global SVAJ equations are composed of four intervals (rise, dwell, fall, and dwell). The local equations above must be assembled into a single equation each for S, V, A, and J that applies over the range 0 <=  <= 360 deg.

6.

Write the local svaj equations for the first interval, 0 <=  <= 1. Note that each subinterval function is multiplied by the range function so that it will have nonzero values only over its subinterval. For 0 <=  <= (Rise)

 b 1  d   y ( x)  R x 1  d 1  d   y ( x)     y1 ( x)  R x    2   3  2 2 2   2 2      b b 1d 1    y4 ( x)  R x 1  1  y5 ( x)   R x   2 2  2    

s1( x)  h 1  R x 0 

v1( x) 

a 1( x) 

  R x 0 

b 1  d 1d 1  y' 1 ( x)  R x    y' 2 ( x)  R x     β1 2 2 2  2 2     b b  1d   1    y' 4 ( x)  R x 1  1  y' 5 ( x)   R x  2 2  2    h1

h1

β1

j1 ( x) 

b

2

h1

β1

3

b

d

   y' 3 ( x)      

 b 1  d   y'' ( x)  R x 1  d 1  d   y'' ( x)     y''1 ( x)  R x    2   3  2 2 2   2 2      b b 1d 1    y''4 ( x)  R x 1  1  y''5 ( x)   R x   2 2  2    

  R x 0 

  R x 0 

b

 b 1  d   y''' ( x)  R x  1  d  1    y'''1 ( x)  R x    2  2 2 2  2 2     b b  1d   1    y'''4 ( x)  R x 1  1  y'''5 ( x)   R x  2 2  2    b

d

   y'''3 ( x)      


7.


Write the local svaj equations for the second interval, 1 <=  <= 1+ 2. For this interval, the value of S is the value of S at the end of the previous interval and the values of V, A, and J are zero because of the dwell. For 1 <=  <= 1+ 2 s2( x)  h 1

8.

v2( x)  0

a 2( x)  0

j2 ( x)  0

Write the local svaj equations for the third interval, 1 + 2 <=  <= 1.+ 2 + 3. For 1 + 2 <=  <= 1.+ 2 + 3

 b 1  d   y ( x)  R x 1  d 1    y1 ( x)  R x    2  2 2 2  2 2     1d b b    1    y4 ( x)  R x 1  1  y5 ( x)   R x  2 2 2    

s3( x)  h 31   R x 0 

  

v3( x)  

β3

h3

β3

j3 ( x)  

9.

2

h3

β3

d

   y3 ( x)      

 b 1  d   y' ( x)  R x 1  d 1  d   y' ( x)     y' 1 ( x)  R x    2   3  2 2 2    2 2      1d b b 1    y' 4 ( x)  R x 1  1  y' 5 ( x)   R x   2 2 2     

  R x 0 

h3

a 3( x)  

b

3

b

 b 1  d   y'' ( x)  R x  1  d  1    y''1 ( x)  R x    2  2 2 2  2 2     1d b b    1    y''4 ( x)  R x 1  1  y''5 ( x)   R x  2 2 2    

  R x 0 

b

d

 b 1  d   y''' ( x)  R x 1  d 1    y'''1 ( x)  R x    2  2 2 2  2 2     1d b b    1    y'''4 ( x)  R x 1  1  y'''5 ( x)   R x  2 2 2    

  R x 0 

b

   y''3 ( x)      

d

   y'''3 ( x)      

Write the local svaj equations for the fourth interval, 1 + 2 + 3 <=  <= 1.+ 2 + 3 + 4. For this interval, the values of S, V, A, and J are zero because of the dwell. For 1 + 2 + 3 <=  <= 1.+ 2 + 3 + 4 s4( x)  0

v4( x)  0

a 4( x)  0

j4 ( x)  0

10. Write the complete global equation for the displacement and plot it over one rotation of the cam, which is the sum of the four intervals defined above. Let

θ1  β1

θ2  θ1  β2

θ3  θ2  β3

θ  θ  θ1   R θ θ1 θ2  s2     θ1   θ2  θ1   θ  θ2   θ  θ3   R θ θ2 θ3  s3  R θ θ3 θ4  s4    θ3  θ2   θ4  θ3 

S ( θ )  s1

θ  0  deg 0.5 deg  360  deg

θ4  θ3  β4



DISPLACEMENT, S 3

Displacement, in

2.5 2 1.5 1 0.5 0

0

30

60

90

120

150

180

210

240

270

300

330

360

Cam Rotation Angle, deg

11. Write the complete global equation for the velocity and plot it over one rotation of the cam, which is the sum of the four intervals defined above.

θ  θ  θ1   R θ θ1 θ2  v2     θ1   θ2  θ1   θ  θ2   θ  θ3   R θ θ2 θ3  v3  R θ θ3 θ4  v4    θ3  θ2   θ4  θ3 

V ( θ )  v1

VELOCITY, V

Velocity, in

5

0

5

 10

0

30

60

90

120

150

180

210

240

270

300

330

360


12. Write the complete global equation for the acceleration and plot it over one rotation of the cam, which is the sum of the four intervals defined above.

θ  θ  θ1   R θ θ1 θ2  a 2     θ1   θ2  θ1   θ  θ2   θ  θ3   R θ θ2 θ3  a 3  R θ θ3 θ4  a 4    θ3  θ2   θ4  θ3 

A ( θ )  a 1



ACCELERATION, A 60

Acceleration, in

30

0

 30

 60

0

30

60

90

120

150

180

210

240

270

300

330

360


13. Write the complete global equation for the jerk and plot it over one rotation of the cam, which is the sum of the four intervals defined above.

θ  θ  θ1   R θ θ1 θ2  j2      θ1   θ2  θ1   θ  θ2   θ  θ3   R θ θ2 θ3  j3   R θ θ3 θ4  j4     θ3  θ2   θ4  θ3 

J ( θ )  j1 

JERK, J 1500

Jerk, in

1000 500 0  500  1000

0

60

120

180


240

300

360




Given:

Design a double-dwell cam to move a follower from 0 to 1.5 in in 45 deg, dwell for 150 deg, fall 1.5 in in 90 deg and dwell for the remainder. The total cycle must take 6 sec. Choose suitable programs for rise and fall to minimize velocities. Plot the SVAJ diagrams. RISE

DWELL

1.

β2  150  deg

β3  90 deg

β4  75 deg

h 1  1.5 in

h 2  0  in

h 3  1.5 in

h 4  0  in

τ  6  sec



2.

DWELL

β1  45 deg Cycle time: Solution:

FALL

2  π rad

ω  1.047

rad

sec τ The modified sinusoidal motion is defined in local coordinates by equations 8.15 through 8.19. The numerical constants in these SCCA equations are given in Table 8-2. b  0.25

c  0.00

d  0.75

Ca  5.5280 3.


 b  b  2  π      sin  x  b   π  π

y1 ( x)  Ca x

y' 1 ( x)  Ca

 π  x  b 

y''1 ( x)  Ca sin

 π  b

y'''1 ( x)  Ca

 π  x    b 

  1  cos

π b

 π  x  b 

 cos

for b/2 <= x <= (1 - d)/2

 x2 1 1 1  2 1 y2 ( x)  Ca   b      x  b    8 2 2  π 2 π    y''2 ( x)  Ca

y' 2 ( x)  Ca x  b  



1

π



1 



2 

y'''2 ( x)  0

for (1 - d)/2 <= x <= (1 + d)/2

 b c  x   π 2  

y3 ( x)  Ca 

2 2 2  d   b2  1  1   ( 1  d )   d   cos π   x  1       8 2 8 2 π d   π  π  

 b  c  d  sin π   x  1  d     2  π 2 π  d 

y' 3 ( x)  Ca 

1  d  π y''3 ( x)  Ca cos   x   2  d 

y'''3 ( x)  Ca

1  d  π  sin   x   d 2  d 

π

d 

  



for (1 + d)/2 <= x <= 1 - b/2

 x2  b  1  2  π 

1  1 2 2  1   x  2 d  b   2     2 4 8 π   b

y4 ( x)  Ca 

y' 4 ( x)  Ca  x 



b

1

π

b



2





y''4 ( x)  Ca

y'''4 ( x)  0

for 1 - b/2 <= x <= 1

2

b 2 d  b y5 ( x)  Ca   x  π 2 π  y' 5 ( x)  Ca

  (1  b)2  d2   b  2sin π (x  1)

2

4

 b

y'''5 ( x)  Ca

π

  

 π   1  cos  ( x  1 ) π  b  b

π  y''5 ( x)  Ca sin  ( x  1 ) b  4.

   π

π   cos  ( x  1 ) b b 


5.


6.



s1( x)  h 1  R x 0 

v1( x) 

a 1( x) 

  R x 0 


h1

β1

j1 ( x) 

b

2

h1

β1

3

b

d

   y' 3 ( x)      


  R x 0 

  R x 0 

b

 b 1  d   y''' ( x)  R x  1  d  1  d   y''' ( x)     y'''1 ( x)  R x    2   3  2 2 2   2 2      b b 1d 1    y'''4 ( x)  R x 1  1  y'''5 ( x)   R x   2 2  2     b


7.


Write the local svaj equations for the second interval, 1 <=  <= 1+ 2. For this interval, the value of S is the value of S at the end of the previous interval and the values of V, A, and J are zero because of the dwell. For 1 <=  <= 1+ 2 s2( x)  h 1

8.

v2( x)  0

a 2( x)  0

j2 ( x)  0


 b 1  d   y ( x)  R x 1  d 1  d   y ( x)     y1 ( x)  R x    2   3  2 2 2   2 2      1d b b 1    y4 ( x)  R x 1  1  y5 ( x)   R x   2 2 2     

s3( x)  h 31   R x 0 

  

v3( x)  

a 3( x)  

b 1  d 1d 1  y' 1 ( x)  R x    y' 2 ( x)  R x     β3 2 2 2  2 2     1d b b    1    y' 4 ( x)  R x 1  1  y' 5 ( x)   R x  2 2 2     h3 2

h3

β3

9.

  R x 0 

h3

β3

j3 ( x)  

b

3

b

d

   y'3 ( x)      

 b 1  d   y'' ( x)  R x  1  d  1  d   y'' ( x)     y''1 ( x)  R x    2   3  2 2 2   2 2      1d b b 1    y''4 ( x)  R x 1  1  y''5 ( x)   R x   2 2 2     

  R x 0 

b

 b 1  d   y''' ( x)  R x 1  d 1  d   y''' ( x)     y'''1 ( x)  R x    2   3  2 2 2   2 2      1d b b 1    y'''4 ( x)  R x 1  1  y'''5 ( x)   R x   2 2 2     

  R x 0 

b


v4( x)  0

a 4( x)  0

j4 ( x)  0


θ1  β1

θ2  θ1  β2

θ3  θ2  β3


S ( θ )  s1

θ  0  deg 1  deg  360  deg

θ4  θ3  β4



DISPLACEMENT, S

Displacement, in

1.5

1

0.5

0

0

60

120

180

240

300

360




V ( θ )  v1

VELOCITY, V

Velocity, in

4

2

0

2

0

60

120

180

240

300

360




A ( θ )  a 1



ACCELERATION, A 20

Acceleration, in

10

0

 10

 20

0

60

120

180

240

300

360




J ( θ )  j1 

JERK, J 300

Jerk, in

200

100

0

 100

0

60

120

180 Cam Rotation Angle, deg

240

300

360




Given:

Design a single-dwell cam to move a follower from 0 to 2.0 in in 60 deg, fall 2.0 in in 90 deg and dwell for the remainder. The total cycle must take 2 sec. Choose suitable programs for rise and fall to minimize velocities. Plot the SVAJ diagrams. RISE/FALL

DWELL

β  150  deg

β  210  deg

h  2.0 in

h 3  0.0 in τ  2  sec

Cycle time: Solution: 1.



2  π rad

ω  3.142

τ 2.

rad sec

Use a two-segment polynomial. Let the rise and fall, together, be one segment and the dwell be the second segment. Then, the boundary conditions are: at

 = 0:  = 1:

s = 0, s = h,

v = 0, v=0

a =0

 = 

s = 0,

v = 0,

a =0

This is a minimum set of 8 BCs. The v = 0 condition at  = 1 is required to keep the displacement from overshooting the lift, h. Define the total lift, the rise interval, the fall interval, and the ratio of rise to the total interval. Total lift:

h  2.000 in

Rise interval:

β  60 deg

β

Use the 8 BCs and equation 8.23 to write 8 equations in s, v, and a similar to those in example 8-9 but with 8 terms in the equation for s (the highest term will be seventh degree). For  = 0: s = v = a = 0 0  c0

0  c1

0  c2

For  = 1: s = h, v = 0 3

4

5

6

h  c3 A  c4 A  c5 A  c6 A  c7 A 2

3

4

7

5

0  3  c3 A  4  c4 A  5  c5 A  6  c6 A  7  c7 A

6

For  = : s = v = a = 0 0  c3  c4  c5  c6  c7 0  3  c3  4  c4  5  c5  6  c6  7  c7 0  6  c3  12 c4  20 c5  30 c6  42 c7 4.

A  0.400

β

β  90 deg

Fall interval: 3.

A 

Solve for the unknown polynomial coefficients. Note that C0 through C2 are zero



 A3 A4 A5 A6 A7     3 A 2 4 A 3 5 A 4 6 A 5 7 A 6   C   1 1 1 1   1  3 4 5 6 7    12 20 30 42   6

5.

 c3     c4   c5   C 1 H    c6  c   7

h   0 H   0  0   0

c3  289.352 in

c4  1229.745 in

c6  1374.421 in

c7  361.690 in

c5  1953.125 in

Write the SVAJ equations for the rise/fall segment.

θ S  θ  c3  

3

4

5

6

7

θ θ θ θ  c4    c5    c6    c7    β  β  β  β  β 2 3 4 5 6 θ θ θ θ θ  1         V θ   3  c3    4  c4    5  c5    6  c6    7  c7    β   β  β  β  β  β  A  θ 

1 2

β J  θ 

2 3 4 5  θ   12 c   θ   20 c   θ   30 c   θ   42 c   θ    4  5  6  7   β  β  β  β  β 



2 3 4  θ θ θ θ       6  c3  24 c4    60 c5    120  c6    210  c7    3  β  β  β  β  β 

1

Plot the displacement, velocity, acceleration, and jerk over the interval 0 <=  <= . θ  0  deg 1  deg  β DISPLACEMENT, S 2

Displacement, in

6.



 6  c3 

1

0

1

0

30

60

90


120

150



VELOCITY, V 4

Velocity, in

2

0 2 4

0

25

50

75

100

125

150

100

125

150

100

125

150


ACCELERATION, A

Acceleration, in

10

5

0 5  10

0

25

50


JERK, J 100

Jerk, in

50

0

 50

0

25

50





Design a three-dwell cam to move a follower from 0 to 2.5 in in 40 deg, dwell for 100 deg, fall 1.5 in in 90 deg, dwell for 20 deg, fall 1.0 in in 30 deg, and dwell for the remainder. The total cycle must take 10 sec. Choose suitable programs for rise and fall to minimize velocities. Plot the SVAJ diagrams.

Given:

RISE/FALL β1  40 deg

DWELL β2  100  deg

FALL β3  90 deg

DWELL β4  20 deg

h 1  2.5 in

h 2  0.0 in

h 3  1.5 in

h 4  0.0 in

β5  30 deg

β6  80 deg

h 5  1.0 in

h 6  0.0 in




2.

3.

τ  15 sec

2  π rad

ω  0.419

rad

sec τ From Table 8-3, the motion program with lowest velocity that does not have infinite jerk is the modified sinusoidal. The modified sinusoidal motion is defined in local coordinates by equations 8.15 through 8.19. The numerical constants in these SCCA equations are given in Table 8-2. b  0.25

c  0.00

d  0.75

Cv  1.7596

Ca  5.5280

Cj  69.466


 b  b  2  π      sin  x  b   π  π

y1 ( x)  Ca x

 π  x  b 

y''1 ( x)  Ca sin

y' 1 ( x)  Ca

 π  b

y'''1 ( x)  Ca

 π  x    b 

  1  cos

π b

 π  x  b 

 cos

for b/2 <= x <= (1 - d)/2

 x2

y2 ( x)  Ca 

2 

 b  

1

π



1  2 1   x  b    2  2 8 π   1

y''2 ( x)  Ca

y' 2 ( x)  Ca x  b  



1

π



1 



2 

y'''2 ( x)  0

for (1 - d)/2 <= x <= (1 + d)/2

 b c  x   π 2  

y3 ( x)  Ca 

2 2 2  d   b2  1  1   ( 1  d )   d   cos π   x  1       8 2 8 2 π d   π  π  

 b  c  d  sin π   x  1  d     2  π 2 π  d 

y' 3 ( x)  Ca 

d 

  



1  d  π y''3 ( x)  Ca cos   x   2  d 

1  d  π  sin   x   d 2  d 

π

y'''3 ( x)  Ca

for (1 + d)/2 <= x <= 1 - b/2

 x2  b  1  2  π 

1  1 2 2  1   x  2 d  b   2     2 4 8 π   b

y4 ( x)  Ca 

y' 4 ( x)  Ca  x 



b

1

π

b



2





y''4 ( x)  Ca

y'''4 ( x)  0

for 1 - b/2 <= x <= 1

2

b 2 d  b y5 ( x)  Ca   x  π 2 π  y' 5 ( x)  Ca

  (1  b)2  d2   b  2sin π (x  1)

2

4

   π

 b

y'''5 ( x)  Ca

π

  

 π   1  cos  ( x  1 ) π  b  b

π  y''5 ( x)  Ca sin  ( x  1 ) b 

π   cos  ( x  1 ) b b 

4.


5.

The global SVAJ equations are composed of six intervals (rise, dwell, fall, dwell, fall, and dwell). The local equations above must be assembled into a single equation each for S, V, A, and J that applies over the range 0 <=  <= 360 deg.

6.



s1( x)  h 1  R x 0 

v1( x) 

a 1( x) 

  R x 0 


h1

β1

j1 ( x) 

b

2

h1

β1

3

b

d

   y' 3 ( x)      


  R x 0 

  R x 0 

b



7.


Write the local svaj equations for the second interval, 1 <=  <= 1+ 2. For this interval, the value of S is the value of S at the end of the previous interval and the values of V, A, and J are zero because of the dwell. For 1 <=  <= 1+ 2 (dwell) s2( x)  h 1

8.

v2( x)  0

a 2( x)  0

j2 ( x)  0

Write the local svaj equations for the third interval, 1 + 2 <=  <= 1.+ 2 + 3. For 1 + 2 <=  <= 1.+ 2 + 3 (fall)

 b 1  d   y ( x)  R x 1  d 1  d   y ( x)    h  h   y1 ( x)  R x    2   3 1 3  2 2 2   2 2       1d b b 1    y4 ( x)  R x 1  1  y5 ( x)   R x   2 2 2     

s3( x)  h 31   R x 0 

  

v3( x)  

a 3( x)  

b 1  d 1d 1  y' 1 ( x)  R x    y' 2 ( x)  R x     β3 2 2 2  2 2     1d b b    1    y' 4 ( x)  R x 1  1  y' 5 ( x)   R x  2 2 2    

3

d

   y'3 ( x)      


2

h3

b

  R x 0 

h3

β3

9.

  R x 0 

h3

β3

j3 ( x)  

b

b


  R x 0 

b

Write the local svaj equations for the fourth interval, 1 + 2 + 3 <=  <= 1.+ 2 + 3 + 4. For this interval, the values of S, V, A, and J are zero because of the dwell. For 1 + 2 + 3 <=  <= 1.+ 2 + 3 + 4 (dwell) s4( x)  h 1  h 3

v4( x)  0

a 4( x)  0

j4 ( x)  0

10. Write the local svaj equations for the fifth interval, For 1 + 2 +3 + 4 <=  <= 1.+ 2 + 3 + 4 + 5 (fall) s5( x)  h 51   R x 0 

  

v5( x)  

 b 1  d   y ( x)  R x 1  d 1  d   y ( x)     y1 ( x)  R x    2   3  2 2 2   2 2      b b 1d 1    y4 ( x)  R x 1  1  y5 ( x)   R x   2 2  2    


  R x 0 

b

b

d

   y'3 ( x)      


a 5( x)  

h5

β5

j5 ( x)  

2

h5

β5

3


  R x 0 

 b 1  d   y'' ( x)  R x  1  d  1  d   y'' ( x)     y''1 ( x)  R x    2   3  2 2 2   2 2      b b 1d 1    y''4 ( x)  R x 1  1  y''5 ( x)   R x   2 2  2    

  R x 0 

b

b

 b 1  d   y''' ( x)  R x 1  d 1  d   y''' ( x)     y'''1 ( x)  R x    2   3  2 2 2   2 2      b b 1d 1    y'''4 ( x)  R x 1  1  y'''5 ( x)   R x   2 2  2    

11. Write the local svaj equations for the sixth interval. For this interval, the values of S, V, A, and J are zero because of the dwell. For 1 + 2 + 3+ 4+ 5 <=  <= 1.+ 2 + 3 + 4+ + 6 (dwell) s6( x)  0

v6( x)  0

a 6( x)  0

j6 ( x)  0

12. Write the complete global equation for the displacement and plot it over one rotation of the cam, which is the sum of the eight intervals defined above. Let

θ1  β1

θ5  θ4  β5

θ2  θ1  β2

θ3  θ2  β3

θ4  θ3  β4

θ6  θ5  β6

θ  θ  θ1   R θ θ1 θ2  s2     θ1   θ2  θ1   θ  θ2   θ  θ3   R θ θ2 θ3  s3  R θ θ3 θ4  s4     θ3  θ2   θ4  θ3   θ  θ4   θ  θ5   R θ θ4 θ5  s5  R θ θ5 θ6  s6    θ5  θ4   θ6  θ5 

S ( θ )  s1

θ  0  deg 0.5 deg  360  deg DISPLACEMENT, S 3

Displacement, in

2.5 2 1.5 1 0.5 0

0

60

120

180


240

300

360



15. Write the complete global equation for the velocity and plot it over one rotation of the cam, which is the sum of the eight intervals defined above.

θ  θ  θ1   R θ θ1 θ2  v2     θ1   θ2  θ1   θ  θ2   θ  θ3   R θ θ2 θ3  v3  R θ θ3 θ4  v4     θ3  θ2   θ4  θ3   θ  θ4   θ  θ5   R θ θ4 θ5  v5  R θ θ5 θ6  v6    θ5  θ4   θ6  θ5 

V ( θ )  v1

VELOCITY, V 8

Velocity, in

6 4 2 0 2 4

0

60

120

180

240

300

360


16. Write the complete global equation for the acceleration and plot it over one rotation of the cam, which is the sum of the eight intervals defined above.

θ  θ  θ1   R θ θ1 θ2  a 2     θ1   θ2  θ1   θ  θ2   θ  θ3   R θ θ2 θ3  a 3  R θ θ3 θ4  a 4     θ3  θ2   θ4  θ3   θ  θ4   θ  θ5   R θ θ4 θ5  a 5  R θ θ5 θ6  a 6    θ5  θ4   θ6  θ5 

A ( θ )  a 1

ACCELERATION, A

Acceleration, in

40

20

0  20  40

0

60

120

180

240

300

360



17. Write the complete global equation for the jerk and plot it over one rotation of the cam, which is the sum of the eight intervals defined above.

θ  θ  θ1   R θ θ1 θ2  j2      θ1   θ2  θ1   θ  θ2   θ  θ3   R θ θ2 θ3  j3   R θ θ3 θ4  j4      θ3  θ2   θ4  θ3   θ  θ4   θ  θ5   R θ θ4 θ5  j5   R θ θ5 θ6  j6     θ5  θ4   θ6  θ5 

J ( θ )  j1 

JERK, J 600

Jerk, in

360 120  120  360  600

0

60

120


240

300

360




Design a four-dwell cam to move a follower from 0 to 2.5 in in 40 deg, dwell for 100 deg, fall 1.5 in in 90 deg, dwell for 20 deg, fall 0.5 in in 30 deg, dwell for 40 deg, fall 0.5 in in 30 deg, and dwe for the remainder. The total cycle must take 15 sec. Choose suitable programs for rise and fall to minimize accelerations. Plot the SVAJ diagrams.

Given:

RISE/FALL β1  40 deg

DWELL β2  100  deg

FALL β3  90 deg

DWELL β4  20 deg

h 1  2.5 in

h 2  0.0 in

h 3  1.5 in

h 4  0.0 in

β5  30 deg

β6  40 deg

β7  30 deg

β8  10 deg

h 5  0.5 in

h 6  0.0 in

h 7  0.5 in

h 8  0.0 in




2.

3.

τ  15 sec

2  π rad

ω  0.419

rad

sec τ From Table 8-3, the motion program with lowest acceleration that does not have infinite jerk is the modified trapezoidal. The modified trapezoidal motion is defined in local coordinates by equations 8.15 through 8.19. The numerical constants in these SCCA equations are given in Table 8-2. b  0.25

c  0.50

d  0.25

Cv  2.0000

Ca  4.8881

Cj  61.426


 b  b  2  π      sin  x  b   π  π

y1 ( x)  Ca x

 π  x  b 

y''1 ( x)  Ca sin

y' 1 ( x)  Ca

 π  b

y'''1 ( x)  Ca

 π  x    b 

  1  cos

π b

 π  x  b 

 cos

for b/2 <= x <= (1 - d)/2

 x2

y2 ( x)  Ca 

2 

 b  

1

π



1  2 1   x  b    2  2 8 π   1

y''2 ( x)  Ca

y' 2 ( x)  Ca x  b  



1

π



1 



2 

y'''2 ( x)  0

for (1 - d)/2 <= x <= (1 + d)/2

 b c  x   π 2  

y3 ( x)  Ca 

2 2 2  d   b2  1  1   ( 1  d )   d   cos π   x  1       8 2 8 2 π d   π  π  

 b  c  d  sin π   x  1  d     2  π 2 π  d 

y' 3 ( x)  Ca 

d 

  



1  d  π y''3 ( x)  Ca cos   x   2  d 

1  d  π  sin   x   d 2  d 

π

y'''3 ( x)  Ca

for (1 + d)/2 <= x <= 1 - b/2

 x2  b  1  2  π 

1  1 2 2  1   x  2 d  b   2     2 4 8 π   b

y4 ( x)  Ca 

y' 4 ( x)  Ca  x 



b

1

π

b



2





y''4 ( x)  Ca

y'''4 ( x)  0

for 1 - b/2 <= x <= 1

2

b 2 d  b y5 ( x)  Ca   x  π 2 π  y' 5 ( x)  Ca

  (1  b)2  d2   b  2sin π (x  1)

2

4

   π

 b

y'''5 ( x)  Ca

π

  

 π   1  cos  ( x  1 ) b π    b

π  y''5 ( x)  Ca sin  ( x  1 ) b  

π   cos  ( x  1 ) b b  

4.


5.

The global SVAJ equations are composed of eight intervals (rise, dwell, fall, dwell, fall, dwell, fall, and dwell). The local equations above must be assembled into a single equation each for S, V, A, and J that applies over the range 0 <=  <= 360 deg.

6.

Write the local svaj equations for the first interval, 0 <=  <= 1. Note that each subinterval function is multiplied by the range function so that it will have nonzero values only over its subinterval. For 0 <=  <= (Rise) s1( x)  h 1  R x 0 

 b 1  d   y ( x)  R x 1  d 1    y1 ( x)  R x    2  2 2 2  2 2     b b  1d   1    y4 ( x)  R x 1  1  y5 ( x)   R x  2 2  2   

v1( x) 

a 1( x) 

  R x 0 

d

   y3 ( x)      

b 1  d 1  d 1  d  y' 1 ( x)  R x    y' 2 ( x)  R x    y' 3 ( x)      β1 2 2 2   2 2      b b  1d   1    y' 4 ( x)  R x 1  1  y' 5 ( x)   R x   2 2  2     h1

h1

β1

j1 ( x) 

b

2

h1

β1

3

b

  R x 0 

 b 1  d   y'' ( x)  R x 1  d 1    y''1 ( x)  R x    2  2 2 2  2 2     b b  1d   1    y''4 ( x)  R x 1  1  y''5 ( x)   R x  2 2  2   

  R x 0 

b

   y''3 ( x)      

d



7.


Write the local svaj equations for the second interval, 1 <=  <= 1+ 2. For this interval, the value of S is the value of S at the end of the previous interval and the values of V, A, and J are zero because of the dwell. For 1 <=  <= 1+ 2 (dwell) s2( x)  h 1

8.

v2( x)  0

a 2( x)  0

j2 ( x)  0

Write the local svaj equations for the third interval, 1 + 2 <=  <= 1.+ 2 + 3. For 1 + 2 <=  <= 1.+ 2 + 3 (fall)

 b 1  d   y ( x)  R x 1  d 1  d   y ( x)    h  h   y1 ( x)  R x    2   3 1 3  2 2 2   2 2       1d b b 1    y4 ( x)  R x 1  1  y5 ( x)   R x   2 2 2     

s3( x)  h 31   R x 0 

  

v3( x)  

a 3( x)  

b 1  d 1d 1  y' 1 ( x)  R x    y' 2 ( x)  R x     β3 2 2 2  2 2     1d b b    1    y' 4 ( x)  R x 1  1  y' 5 ( x)   R x  2 2 2     h3 2

h3

β3

9.

  R x 0 

h3

β3

j3 ( x)  

b

3

b

d

   y'3 ( x)      


  R x 0 

b


  R x 0 

b

Write the local svaj equations for the fourth interval, 1 + 2 + 3 <=  <= 1.+ 2 + 3 + 4. For this interval, the values of S, V, A, and J are zero because of the dwell. For 1 + 2 + 3 <=  <= 1.+ 2 + 3 + 4 (dwell) s4( x)  h 1  h 3

v4( x)  0

a 4( x)  0

j4 ( x)  0

10. Write the local svaj equations for the fifth interval, For 1 + 2 +3 + 4 <=  <= 1.+ 2 + 3 + 4 + 5 (fall)

 b 1  d   y ( x)  R x 1  d 1  d   y ( x)    h   y1 ( x)  R x    2   3 7  2 2 2   2 2       1d b b 1    y4 ( x)  R x 1  1  y5 ( x)   R x   2 2 2     

s5( x)  h 51   R x 0 

  

v5( x)  

b

b 1  d 1  d 1  d   y' 1 ( x)  R x    y' 2 ( x)  R x       y'3 ( x)   β5 2 2 2 2 2         1d b b    1    y' 4 ( x)  R x 1  1  y' 5 ( x)   R x   2 2 2      h5

  R x 0 

b


h5

a 5( x)  

β5

2

h5

j5 ( x)  

β5

3


 b 1  d   y'' ( x)  R x  1  d  1    y''1 ( x)  R x    2  2 2 2  2 2     1d b b    1    y''4 ( x)  R x 1  1  y''5 ( x)   R x  2 2 2    

  R x 0 

b

d

 b 1  d   y''' ( x)  R x 1  d 1    y'''1 ( x)  R x    2  2 2 2  2 2     1d b b    1    y'''4 ( x)  R x 1  1  y'''5 ( x)   R x  2 2 2    

  R x 0 

b

   y''3 ( x)      

d

   y'''3 ( x)      

11. Write the local svaj equations for the sixth interval. For this interval, the values of S, V, A, and J are zero because of the dwell. For 1 + 2 + 3+ 4+ 5 <=  <= 1.+ 2 + 3 + 4+ + 6 (dwell) s6( x)  h 5

v6( x)  0

a 6( x)  0

j6 ( x)  0

12. Write the local svaj equations for the seventh interval, For 1 + 2 +3 + 4 + 5 + 6 <=  <= 1.+ 2 + 3 + 4 + 5 + 6 + 7 (fall)

 b 1  d   y ( x)  R x 1  d 1    y1 ( x)  R x    2  2 2 2  2 2     1d b b    1    y4 ( x)  R x 1  1  y5 ( x)   R x  2 2 2    

s7( x)  h 71   R x 0 

  

v7( x)  

a 7( x)  

d

   y3 ( x)      

b 1  d 1  d 1  d   y' 1 ( x)  R x    y' 2 ( x)  R x       y'3 ( x)   β7 2 2 2 2 2         1d b b    1    y' 4 ( x)  R x 1  1  y' 5 ( x)   R x   2 2 2        R x 0 

h7

h7

β7

j7 ( x)  

b

2

h7

β7

3

b


  R x 0 

b


  R x 0 

b

13. Write the local svaj equations for the eighth interval. For this interval, the values of S, V, A, and J are zero because of the dwell. For 1 + 2 + 3+ 4+ 5 + + 7 <=  <= 1.+ 2 + 3 + 4+ + 6 + + 8 (dwell) s8( x)  0

v8( x)  0

a 8( x)  0

j8 ( x)  0

14. Write the complete global equation for the displacement and plot it over one rotation of the cam, which is the sum of the eight intervals defined above.


Let

θ1  β1

θ5  θ4  β5


θ2  θ1  β2

θ3  θ2  β3

θ4  θ3  β4

θ6  θ5  β6

θ7  θ6  β7

θ8  θ7  β8

θ  θ  θ1   R θ θ1 θ2  s2     θ1   θ2  θ1   θ  θ2   θ  θ3   R θ θ2 θ3  s3  R θ θ3 θ4  s4     θ3  θ2   θ4  θ3   θ  θ4   θ  θ5   R θ θ4 θ5  s5  R θ θ5 θ6  s6     θ5  θ4   θ6  θ5   θ  θ6   θ  θ7   R θ θ6 θ7  s7  R θ θ7 θ8  s8    θ7  θ6   θ8  θ7 

S ( θ )  s1

θ  0  deg 0.5 deg  360  deg DISPLACEMENT, S 3

Displacement, in

2.5 2 1.5 1 0.5 0

0

60

120

180

240

300

360


15. Write the complete global equation for the velocity and plot it over one rotation of the cam, which is the sum of the eight intervals defined above.

θ  θ  θ1   R θ θ1 θ2  v2     θ1   θ2  θ1   θ  θ2   θ  θ3   R θ θ2 θ3  v3  R θ θ3 θ4  v4     θ3  θ2   θ4  θ3   θ  θ4   θ  θ5   R θ θ4 θ5  v5  R θ θ5 θ6  v6     θ5  θ4   θ6  θ5   θ  θ6   θ  θ7   R θ θ6 θ7  v7  R θ θ7 θ8  v8    θ7  θ6   θ8  θ7 

V ( θ )  v1



VELOCITY, V 8

Velocity, in

6 4 2 0 2

0

60

120

180

240

300

360

16. Write the complete global equation for the acceleration and plot it over one rotation of the cam, which is the sum of the eight intervals defined above.

θ  θ  θ1   R θ θ1 θ2  a 2     θ1   θ2  θ1   θ  θ2   θ  θ3   R θ θ2 θ3  a 3  R θ θ3 θ4  a 4     θ3  θ2   θ4  θ3   θ  θ4   θ  θ5   R θ θ4 θ5  a 5  R θ θ5 θ6  a 6     θ5  θ4   θ6  θ5   θ  θ6   θ  θ7   R θ θ6 θ7  a 7  R θ θ7 θ8  a 8    θ7  θ6   θ8  θ7 

A ( θ )  a 1

ACCELERATION, A 40

Acceleration, in

20

0

 20

 40

0

60

120

180

240

300

360


17. Write the complete global equation for the jerk and plot it over one rotation of the cam, which is the sum of the eight intervals defined above.



θ  θ  θ1   R θ θ1 θ2  j2      θ1   θ2  θ1   θ  θ2   θ  θ3   R θ θ2 θ3  j3   R θ θ3 θ4  j4      θ3  θ2   θ4  θ3   θ  θ4   θ  θ5   R θ θ4 θ5  j5   R θ θ5 θ6  j6      θ5  θ4   θ6  θ5   θ  θ6   θ  θ7   R θ θ6 θ7  j7   R θ θ7 θ8  j8     θ7  θ6   θ8  θ7 

J ( θ )  j1 

JERK, J 600

Jerk, in

360 120  120  360  600

0

60

120


240

300

360




Size the cam from Problem 8-7 for a 1-in-radius roller follower considering pressure angle and radius of curvature. Use eccentricity only if necessary to balance those functions. Draw the cam profile. Repeat for a flat faced follower. Which would you use?

Units:


Given:

RISE

1.

3.

FALL

DWELL

β  120  deg

β  30 deg

β  150  deg

h 1  2.5 in

h 2  0  in

h 3  2.5 in

h 4  0  in

τ  4  sec



2.

DWELL

β  60 deg

Cycle time: Solution:

1

2  π rad

ω  1.571

rad

ω  15.000 rpm sec τ Problem 8-7 used a four-segment cam with modified trapezoidal acceleration for the rise and fall. Enter the above data into program DYNACAM. The input screen is shown below.

The cam was sized iteratively to balance the positive and negative extreme values of the pressure angle by varying the eccentricity and to reduce the maximum absolute pressure angle to 30 deg or less by increasing the prime circle radius. The resulting cam is shown below.


4.


The minimum and maximum pressure angles and radius of curvature are shown in the figure above. The design has the following dimensions: Prime circle radius

Rp  11.5 in

Roller follower radius

Rf  1.00 in

Follower eccentricity

ε  2.375  in

5.

Graphs of  and  for the roller follower are shown on the following page.

6.

The cam was sized for a flat-faced follower and the cam drawing is shown on the next page. In order to avoid undercutting, a base circle radius of 43 in is required. Obviously, the roller follower is to be preferred over the flat face follower in this case.








Size the cam from Problem 8-8 for a 1.5-in-radius roller follower considering pressure angle and radius of curvature. Use eccentricity only if necessary to balance those functions. Draw the cam profile. Repeat for a flat faced follower. Which would you use?

Units:


Given:

RISE

1.

3.

FALL

DWELL

β  150  deg

β  90 deg

β  75 deg

h 1  1.5 in

h 2  0  in

h 3  1.5 in

h 4  0  in

τ  6  sec



2.

DWELL

β  45 deg


1

2  π rad

ω  1.047

rad

ω  10.000 rpm sec τ Problem 8-8 used a four-segment cam with modified sinusoidal acceleration for the rise and fall. Enter the above data into program DYNACAM. The input screen is shown below.



4.


The minimum and maximum pressure angles and radius of curvature are shown in the figure above. The prime radius had to be increased above that which was necessary for the pressure angle constraint in order to provide a minimum radius of curvature that was twice the roller follower radius. The design has the following dimensions: Prime circle radius

Rp  6.5 in


Rf  1.50 in


ε  0.82 in

5.


6.

The cam was sized for a flat-faced follower and the cam drawing is shown on the next page. In order to avoid undercutting, a base circle radius of 12 in is required. Obviously, the roller follower is to be preferred over the flat face follower in this case.







Units:


Given:

RISE/FALL

1.

β  150  deg

h 1  2.0 in

h 2  0.0 in

3.

τ  2  sec



2.

DWELL

β  150  deg


1

2  π rad

ω  3.142

rad

ω  30.000 rpm sec τ Problem 8-9 used a two-segment cam with one polynomial segment for the rise and fall. Enter the above data in program DYNACAM. The input screen is shown below.



4.



Rp  4.40 in


Rf  0.50 in


ε  0.22 in

5.


6.

The cam was sized for a flat-faced follower and the cam drawing is shown on the next page. In order to avoid undercutting, a base circle radius of 7.81 in is required. Obviously, the roller follower is to be preferred over the flat face follower in this case.







Units:


Given:

RISE/FALL

1.

FALL

DWELL

β  100  deg

β  90 deg

β  20 deg

h 1  2.5 in

h 2  0.0 in

h 3  1.5 in

h 4  0.0 in

β  30 deg

β  80 deg

h 5  1.0 in

h 6  0.0 in τ  10 sec



2.

DWELL

β  40 deg


1

2  π rad

ω  0.628

rad

ω  6.000 rpm sec τ Problem 8-10 used a six-segment cam with modified sinusoidal acceleration for the rise and fall. Enter the above data into program DYNACAM. The input screen is shown below.



3.


4.

The minimum and maximum pressure angles and radius of curvature are shown in the figure above. The prime radius had to be increased above that which was necessary for the pressure angle constraint in order to provide a minimum radius of curvature that was twice the roller follower radius. The design has the following dimensions: Prime circle radius

Rp  10.50  in


Rf  2.00 in


ε  1.27 in

5.


6.








Units:


Given:

RISE/FALL

1.

FALL

DWELL

β  100  deg

β  90 deg

β  20 deg

h 1  2.5 in

h 2  0.0 in

h 3  1.5 in

h 4  0.0 in

β  30 deg

β  40 deg

β  30 deg

β  10 deg

h 5  0.5 in

h 6  0.0 in

h 7  0.5 in

h 8  0.0 in

τ  15 sec



2.

DWELL

β  40 deg


1

2  π rad

ω  0.419

rad

ω  4.000 rpm sec τ Problem 8-11 used an eight-segment cam with modified trapezoidal acceleration for the rise and fall. Enter the above data into program DYNACAM. The input screen is shown below.



3.


4.


Rp  7.50 in


Rf  0.50 in


ε  2.35 in

5.


6.







Given:

A high friction, high inertia load is to be driven. We wish to keep peak velocity low. Combine segments of polynomial displacements with a constant velocity segment on both rise and fall to reduce the maximum velocity below that obtainable with a full period modified sine acceleration alone (i.e., with no constant velocity portion). Compare the two designs and comment. Use an  of one for comparison. RISE β  90 deg h 1  1.0 in

Solution:

DWELL β  60 deg h 2  0  in

FALL β  50 deg h 3  1.0 in

DWELL β  160  deg h 4  0  in


1.

Use DYNACAM to get the data for comparison of these two cam profiles. Start with the CV-poly profile, which is shown below. In this design eight segments were used: three each for the rise and fall and one each for the two dwells. The (nearly) constant velocity segments were each 54 deg. The maximum acceleration and jerk are 12.2 and 446, respectively.

1.

The DYNACAM output screen for the modified sine profile is shown below. The maximum acceleration and jerk are 74.4 and 105, respectively. Obviously, for the CV-poly design choices that were made, the modified sine profile results in lower inertia forces and a smoother follower motion.






A constant velocity of 0.4 in/sec must be matched for 1.5 sec. Then the follower must return to your choice of starting point and dwell for 2 sec. The total cycle time is 6 sec. Design a cam for a follower radius of 0.75 in and a maximum pressure angle of 30 deg, absolute value.

Given:

Required constant velocity

Vcv  0.4 in sec

Dwell time

tdwell  2  sec

Cycle time

tcycle  6  sec

Follower radius

Rf  0.75 in

Solution: 1.

The camshaft turns 2 rad during the time for one cycle. Thus, its speed is 2  π rad

ω  1.047

tcycle

rad sec

Use a four-segment polynomial cam as described below. Segment 1 2 3 4

3.

Function Constant Velocity Polynomial (5 deg) None Polynomial (5 deg)

tcv tcycle tdwell tcycle

 2 π

β  90.000 deg

 2 π

β  120.000 deg

This leaves 150 deg to allocate to segments 2 and 4. Tentatively, let β  100  deg

β  50 deg

Calculate the lift for the constant velocity segment. L1  Vcv tcv

6.

β 

β 

Dwell segment

5.

Motion Rise Fall Dwell Rise

Calculate the two known cam rotation interval widths. Constant velocity segment

4.

L1  0.600 in

Make other initial design choices. Lift of segment 4 (and starting position for segment 1) Prime circle radius

7.

tcv  1.5 sec


ω  2.

1

L4  0.1 in

Rp  2.50 in

Enter the above data into program DYNACAM. The input screen and resulting SVAJ diagrams are shown on the next page.





8.

From SVAJ diagrams we see that all design goals are met but the peak acceleration in segment 2 is higher than the peak in segment 4. These can be more evenly balanced by increasing the segment 2 interval and decreasing the segment 4 interval. Iterate on these intervals until the peak accelerations are more nearly the same.

9.

The second interval was changed to 120 deg and the fourth to 30 deg, and the prime circle radius was increased to 2.625 in (to maintain min/Rf > 3). The resulting peak accelerations are more in balance and are shown in the SVAJ diagram below. The maximum pressure angle is well within the required limit.




A constant velocity of 0.25 in/sec must be matched for 3 sec. Then the follower must return to your choice of starting point and dwell for 3 sec. The total cycle time is 12 sec. Design a cam for a follower radius of 1.25 in and a maximum pressure angle of 35 deg, absolute value.

Given:



Dwell time

tdwell  3  sec

Cycle time

tcycle  12 sec

Follower radius

Rf  1.25 in

Solution: 1.


ω  0.524

tcycle

β 

β 

tcv tcycle tdwell tcycle

 2 π

β  90.000 deg

 2 π

β  90.000 deg

This leaves 180 deg to allocate to segments 2 and 4. Tentatively, let β  120  deg

β  60 deg

Calculate the lift for the constant velocity segment. L1  Vcv tcv

6.

Motion Rise Fall Dwell Rise

Calculate the two known cam rotation interval widths.

Dwell segment

5.

sec

Function Constant Velocity Polynomial (5 deg) None Polynomial (5 deg)

Constant velocity segment

4.

rad

Use a four-segment polynomial cam as described below. Segment 1 2 3 4

3.

L1  0.750 in

Make other initial design choices. Lift of segment 4 (and starting position for segment 1) Prime circle radius

7.

tcv  3  sec


ω  2.

1

L4  0.15 in

Rp  4.00 in






8.

From SVAJ diagrams we see that all design goals are met but the peak acceleration in segment 2 is higher than the peak in segment 4. These can be more evenly balanced by increasing the segment 2 interval and decreasing the segment 4 interval. Iterate on these intervals until the peak accelerations are more nearly the same. Note also that min is not greater than 3 times Rf.

9.

The second interval was changed to 140 deg and the fourth to 40 deg, and the prime circle radius was increased to 4.125 in (to maintain min/Rf > 3). The resulting peak accelerations are more in balance and are shown in the SVAJ diagram below. The maximum pressure angle is well within the required limit.




A constant velocity of 2 in/sec must be matched for 1 sec. Then the follower must return to your choice of starting point. The total cycle time is 2.75 sec. Design a cam for a follower radius of 0.5 in and a maximum pressure angle of 25 deg, absolute value.

Given:



Cycle time

tcycle  2.75 sec

Follower radius

Rf  0.5 in

Solution: 1.


ω  2.285

tcycle

rad sec

Use a two-segment polynomial cam similar to that used in Example 8-10 and described below. Segment 1 2

3.

Function Constant Velocity Polynomial (5 deg)

tcycle

 2 π

β  130.909 deg

Calculate the lift for the constant velocity segment. L1  2.000 in

Make other initial design choices. Prime circle radius

7.

tcv

β  229.091 deg

L1  Vcv tcv 6.

β 

This leaves 229.091 deg for segments 2. β  360  deg  β

5.

Motion Rise Fall

Calculate the known cam rotation interval width. Constant velocity segment

4.

tcv  1  sec


ω  2.

1

Rp  4.00 in






8.

From SVAJ diagrams we see that all design goals are met but the maximum pressure angle is considerably lower (in absolute value) than the minimum pressure angle. These can be balanced by making the follower eccentric with respect to the cam centerline. This will also allow the prime radius to be reduced. Iterate on these parameters until the maximum and minimum pressure angles are nearly the same (numerically) and reduce the prime circle radius until the pressure angle is at, or below, the required limit.

9.

The eccentricity was iterated to -0.23 in and the prime circle radius was reduced to 2.45 in. This resulted in a balanced maximum and minimum pressure angle that is just below the required limit. The minimum radius of curvature is almost 5 times the follower radius, which is ample. The SVAJ diagram for the final iteration is shown below.




Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot the s v a j diagrams for a modified trapezoidal acceleration cam function for any specified values of lift and duration. Test it using a lift of 20 mm over an interval of 60 deg at 1 rad/sec.

Enter:

Lift:

h 1  20 mm

Duration:

β1  60 deg

Solution:


1.

Enter values for lift and duration above.

2.

The numerical constants in these SCCA for the modified trapezoidal equations are given in Table 8-2.

3.

b  0.25

c  0.50

d  0.25

Cv  2.0000

Ca  4.8881

Cj  61.426


 b  b  2  π      sin  x  b   π  π

y1 ( x)  Ca x

y' 1 ( x)  Ca

 π  x  b 

y''1 ( x)  Ca sin

 π  b

y'''1 ( x)  Ca

 π  x    b 

  1  cos

π b

 π  x  b 

 cos

for b/2 <= x <= (1 - d)/2

 x2 1 1  1 1 1 2 1 y2 ( x)  Ca   b      x  b    y' 2 ( x)  Ca x  b       2 2  8 π 2   π 2  π    y''2 ( x)  Ca

y'''2 ( x)  0

for (1 - d)/2 <= x <= (1 + d)/2 2 2 2  b c 1  ( 1  d) d d 1  d  2 1 π    x     b         cos   x   8  π 2  2 8 2  π π d    π    

y3 ( x)  Ca 

 b  c  d  sin π   x  1    2 π 2 π  d 

y' 3 ( x)  Ca 

d 

 

1  d  π y''3 ( x)  Ca cos   x   2  d 

y'''3 ( x)  Ca

1  d  π  sin   x   d 2  d 

π

for (1 + d)/2 <= x <= 1 - b/2

 x2  b 1 1 b 1 2 2 y4 ( x)  Ca     1    x  2  d  b       2 8  4  2 π 2  π  



y' 4 ( x)  Ca  x 



for 1 - b/2 <= x <= 1

b π

1

b



2



y''4 ( x)  Ca

y'''4 ( x)  0







2 2 2 2 2 b 2 d  b ( 1  b)  d b π    y5 ( x)  Ca  x       sin  ( x  1 ) π 2 4 b   π π  

y' 5 ( x)  Ca

 π   1  cos  ( x  1 ) π  b  b

π  y''5 ( x)  Ca sin  ( x  1 ) b  4.

y'''5 ( x)  Ca

π   cos  ( x  1 ) b b 

π


5.


6.


 b 1  d   y ( x)  R x 1  d 1  d   y ( x)     y1 ( x)  R x    2   3  2 2 2   2 2      1d b b 1    y4 ( x)  R x 1  1  y5 ( x)   R x   2 2 2     

s1( x)  h 1  R x 0 

v1( x) 

a 1( x) 

b 1  d 1d 1  y' 1 ( x)  R x    y' 2 ( x)  R x     β1 2 2 2  2 2     1d b b    1    y' 4 ( x)  R x 1  1  y' 5 ( x)   R x  2 2 2       R x 0 

h1

h1

β1

j1 ( x) 

2

h1

β1

5.

b

3

b

d

   y' 3 ( x)      

 b 1  d   y'' ( x)  R x 1  d 1  d   y'' ( x)     y''1 ( x)  R x    2   3  2 2 2   2 2      1d b b 1    y''4 ( x)  R x 1  1  y''5 ( x)   R x   2 2 2     

  R x 0 

b

 b 1  d   y''' ( x)  R x  1  d  1  d   y''' ( x)     y'''1 ( x)  R x    2   3  2 2 2   2 2      1d b b 1    y'''4 ( x)  R x 1  1  y'''5 ( x)   R x   2 2 2     

  R x 0 

b

Plot the displacement, velocity, acceleration, and jerk functions over the lift interval: θ  0  deg 0.5 deg  β1



DISPLACEMENT, s

Displacement, mm

20

15

10

5

0

0

20


VELOCITY, v 40

Velocity, mm

30

20

10

0

0

20


60



ACCELERATION, a

Acceleration, mm

100

50

0

 50

 100

0

20

40

60

40

60


JERK, j

Acceleration, mm

2000

1000

0

 1000

 2000

0





Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot the s v a j diagrams for a modified sinusoidal acceleration cam function for any specified values of lift and duration. Test it using a lift of 20 mm over an interval of 60 deg at 1 rad/sec.

Enter:

Lift:

h 1  20 mm

Duration:

β1  60 deg

Solution:


1.


2.


3.

b  0.25

c  0.00

d  0.75

Cv  1.7596

Ca  5.5280

Cj  69.466


 b  b  2  π  y1 ( x)  Ca x     sin  x  b   π  π

y' 1 ( x)  Ca

 π  x  b 

y'''1 ( x)  Ca

y''1 ( x)  Ca sin

 π  b

 π  x    b 

  1  cos

π b

 π  x  b 

 cos

for b/2 <= x <= (1 - d)/2

 x2

y2 ( x)  Ca 

2 

 b  

1

π



1

  x  b  2

2

1

8 



 y' ( x)  C  x  2 a 2   π  1

y''2 ( x)  Ca

b  

1

π



1 



2 

y'''2 ( x)  0

for (1 - d)/2 <= x <= (1 + d)/2

 b c  x   π 2  

y3 ( x)  Ca 

2

2

2

 d   b2  1  1   ( 1  d )   d   cos π   x  1       8 2 8 2 d   π  π π  

d 

 b  c  d  sin π   x  1  d     2  π 2 π  d 

y' 3 ( x)  Ca 

1  d  π y''3 ( x)  Ca cos   x   2  d 

y'''3 ( x)  Ca

1  d  π  sin   x   d 2  d 

π

for (1 + d)/2 <= x <= 1 - b/2

 x2  b  1  2  π 

y4 ( x)  Ca 

y' 4 ( x)  Ca  x 



for 1 - b/2 <= x <= 1

b π

1

b



 1

   x  2 d  b 

2

b



2

2

2

π

2

y''4 ( x)  Ca



1 8





1 4



y'''4 ( x)  0

  



2

b 2 d  b y5 ( x)  Ca   x  π 2 π 

  (1  b)2  d2   b  2sin π (x  1)

2

   π

 

b

 π   1  cos  ( x  1 ) π  b  b

y' 5 ( x)  Ca

π  y''5 ( x)  Ca sin  ( x  1 ) b  4.

4

y'''5 ( x)  Ca

π   cos  ( x  1 ) b b 

π


5.


6.



s1( x)  h 1  R x 0 

v1( x) 

a 1( x) 

β1

h1

5.

2

h1

β1

d

   y3 ( x)      

 b 1  d   y' ( x)  R x  1  d  1  d   y' ( x)     y'1 ( x)  R x    2   3  2 2 2    2 2      1d b b 1    y' 4 ( x)  R x 1  1  y' 5 ( x)   R x   2 2 2     

  R x 0 

h1

β1

j1 ( x) 

b

3

b

 b 1  d   y'' ( x)  R x 1  d 1    y''1 ( x)  R x    2  2 2 2  2 2     1d b b    1    y''4 ( x)  R x 1  1  y''5 ( x)   R x  2 2 2    

  R x 0 

b

d

 b 1  d   y''' ( x)  R x  1  d  1    y'''1 ( x)  R x    2  2 2 2  2 2     1d b b    1    y'''4 ( x)  R x 1  1  y'''5 ( x)   R x  2 2 2    

  R x 0 

b

Plot the displacement, velocity, acceleration, and jerk functions over the lift interval: θ  0  deg 0.5 deg  β1

   y''3 ( x)      

d

   y'''3 ( x)      



DISPLACEMENT, s

Displacement, mm

30

20

10

0

0

20


VELOCITY, v 40

Velocity, mm

30

20

10

0

0

20

40

60

40

60


ACCELERATION, a

Acceleration, mm

200

100

0

 100

 200

0




JERK, j

Acceleration, mm

1500

1000

500

0

 500

0

20


60




Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot the s v a j diagrams for a cycloidal displacement cam function for any specified values of lift and duration. Test it using a lift of 20 mm over an interval of 60 deg at 1 rad/sec.

Enter:

Solution: 1.

Lift:

h  20 mm

Duration:

β  60 deg


Cycloidal motion is defined in local coordinates by equations 8.12. They are:

 θ  1  sin 2 π θ     β   β 2 π 

s θ  h  

a  θ  2  π

h 2

β

θ

 β

 β  h

 

  1  cos 2  π

j  θ  4  π 

2 h 3

β

 

θ 



β 

 cos 2  π

θ



β

Plot the displacement, velocity, acceleration, and jerk functions over the lift interval: θ  0  deg 0.5 deg  β DISPLACEMENT, s

Displacement, mm

20

15

10

5

0

0

20

40

60

40

60


VELOCITY, v 40 30 Velocity, mm

2.

 

 sin 2  π

v θ 

20 10 0

0




ACCELERATION, a

Acceleration, mm

200

100

0

 100

 200

0

20

40

60

40

60


JERK, J 1000

Jerk, mm

500

0

 500

 1000

0





Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot the s v a j diagrams for a 3-4-5 polynomial displacement cam function for any specified values of lift and duration. Test it using a lift of 20 mm over an interval of 60 deg at 1 rad/sec.

Enter:

Lift:

h  20 mm

Duration:

β  60 deg

Solution: 1.


The 3-4-5 polynomial is defined in local coordinates by equations 8.24. They are: 4   θ 3 θ   15       β  β

s θ  h  10  a  θ 

v θ 

3 4   θ 2  θ   30  θ    60       β   β  β  β 

h

 30 

2 3 2  θ θ θ  θ θ  h         60    180     120     j θ   60  360     360     2 3 β  β  β   β  β  β    β 

h

Plot the displacement, velocity, acceleration, and jerk functions over the lift interval: θ  0  deg 0.5 deg  β DISPLACEMENT, s

Displacement, mm

20

15

10

5

0

0

20

40

60

40

60


VELOCITY, v 40

30 Velocity, mm

2.

5  θ     β 

6 

20

10

0

0




ACCELERATION, a

Acceleration, mm

200

100

0

 100

 200

0

20

40

60

40

60


JERK, J 1500

Jerk, mm

1000

500

0

 500

 1000

0





Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot the s v a j diagrams for a 4-5-6-7 polynomial displacement cam function for any specified values of lift and duration. Test it using a lift of 20 mm over an interval of 60 deg at 1 rad/sec.

Enter:

Lift:

h  20 mm

Duration:

β  60 deg

Solution: 1.


The 4-5-6-7 polynomial is defined in local coordinates by equation 8.25. Differentiate it to get v, a, and j. 5 6 7   θ 4  θ   70  θ   20  θ    84           β  β  β  β 

s θ  h  35  v θ 

a  θ 

4 5 6   θ 3 θ θ θ       140     420     420     140     β   β  β  β  β 

h

h 2

β j  θ 

2 3 4 5  θ   1680  θ   2100  θ   840   θ           β  β  β  β 



2 3 4  θ θ θ θ       840     5040    8400    4200    3   β  β  β  β  β

h

Plot the displacement, velocity, acceleration, and jerk functions over the lift interval: θ  0  deg 0.5 deg  β

DISPLACEMENT, s 20

Displacement, mm

2.



 420  

15

10

5

0

0

20


60



VELOCITY, v 50

Velocity, mm

40 30 20 10 0

0

20

40

60

40

60

40

60


ACCELERATION, a

Acceleration, mm

200

100

0

 100

 200

0


JERK, J 1000

Jerk, mm

500

0

 500

 1000

0





Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot the s v a j diagrams for a simple harmonic displacement cam function for any specified values of lift and duration. Test it using a lift of 20 mm over an interval of 60 deg at 1 rad/sec. Lift: h  20 mm

Enter:

β  60 deg

Duration: Solution:

 2 

s θ 

h

a  θ 

π

 θ    β 

  1  cos π 2

h  θ   cos π  2 2  β β

v θ 

π h  θ   sin π  β 2  β

j  θ 

π

3

h  θ   sin π  3 2  β β

Plot the displacement, velocity, acceleration, and jerk functions over the lift interval: θ  0  deg 0.5 deg  β DISPLACEMENT, s 20

Displacement, mm

2.

The simple harmonic motion (SHM) is defined in local coordinates by equations 8.6. They are:

15

10

5

0

0

20

40

60

40

60


VELOCITY, v 30

Velocity, mm

1.


20

10

0

0




ACCELERATION, a 100

Acceleration, mm

50

0

 50

 100

0

20

40

60

40

60


JERK, J 300

Jerk, mm

200

100

0

0





Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot the pressure angle and radius of curvature for a modified trapezoidal acceleration cam function for any specified values of lift and duration. Test it using a lift of 20 mm over an interval of 60 deg at 1 rad/sec, and determine the prime circle radius needed to obtain a maximum pressure angle of 20 deg. What is the minimum diameter roller follower needed to avoid undercutting with these data?

Enter:

Lift:

h 1  20 mm

Eccentricity

ε  4  mm

Duration:

β1  60 deg

Prime circle radius

Rp  50 mm

Solution:


1.


2.


3.

b  0.25

c  0.50

d  0.25

Cv  2.0000

Ca  4.8881

Cj  61.426


 b  b  2  π      sin  x  b   π  π

y1 ( x)  Ca x

y' 1 ( x)  Ca

 π  x  b 

y''1 ( x)  Ca sin

 π  b

y'''1 ( x)  Ca

 π  x    b 

  1  cos

π b

 π  x  b 

 cos

for b/2 <= x <= (1 - d)/2


y'''2 ( x)  0

for (1 - d)/2 <= x <= (1 + d)/2 2 2 2  b c 1  ( 1  d) d d 1  d  2 1 π    x     b         cos   x   8  π 2  2 8 2  π π d    π    

y3 ( x)  Ca 

 b  c  d  sin π   x  1    2 π 2 π  d 

y' 3 ( x)  Ca 

d 

 

1  d  π y''3 ( x)  Ca cos   x   2  d 

y'''3 ( x)  Ca

1  d  π  sin   x   d 2  d 

π

for (1 + d)/2 <= x <= 1 - b/2

 x2  b 1 b 2 2    1    x  2  d  b      2 π 2 2  π b b y' 4 ( x)  Ca  x   1   y''4 ( x)  Ca 2 π  y4 ( x)  Ca 

for 1 - b/2 <= x <= 1





1 8





1 4



y'''4 ( x)  0








y' 5 ( x)  Ca

 π   1  cos  ( x  1 ) π  b  b

π  y''5 ( x)  Ca sin  ( x  1 ) b  4.

y'''5 ( x)  Ca

π   cos  ( x  1 ) b b 

π


5.


6.



s1( x)  h 1  R x 0 

v1( x) 

a 1( x) 

β1

h1

5.

2

h1

β1

d

   y3 ( x)      

 b 1  d   y' ( x)  R x  1  d  1  d   y' ( x)     y'1 ( x)  R x    2   3  2 2 2    2 2      1d b b 1    y' 4 ( x)  R x 1  1  y' 5 ( x)   R x   2 2 2     

  R x 0 

h1

β1

j1 ( x) 

b

3

b

 b 1  d   y'' ( x)  R x 1  d 1    y''1 ( x)  R x    2  2 2 2  2 2     1d b b    1    y''4 ( x)  R x 1  1  y''5 ( x)   R x  2 2 2    

  R x 0 

b

d

 b 1  d   y''' ( x)  R x  1  d  1    y'''1 ( x)  R x    2  2 2 2  2 2     1d b b    1    y'''4 ( x)  R x 1  1  y'''5 ( x)   R x  2 2 2    

  R x 0 

b

Using equations 8.31d and 8.33, write the pressure angle and radius of curvature functions.

θ   v1   ε   β1     ϕ θ Rp ε   atan  θ 2 2  s1 β   Rp  ε    1 

   y''3 ( x)      

d

   y'''3 ( x)      



3

  Rp  

ρ θ Rp 

2

2 2 θ θ  s1    v1    β1    β1  

2

2

 R  s  θ    2 v  θ   a  θ    R  s  θ   1 1  p 1 β      p 1 β     1   β1   β1    1  6.

Plot the pressure angle and radius of curvature functions over the lift interval: θ  0  deg 0.5 deg  β1 PRESSURE ANGLE

Pressure Angle, deg

30

20

10

0

 10

0

20

40

60


RADIUS OF CURVATURE 100

Radius of Curvature, mm

75 50 25 0  25  50  75  100

0

20

40

60


7.

The graphs above show the pressure angle and radius of curvature for the values of Rp and  entered on the first page. These values will be iterated below to obtain a balanced pressure angle whose absolute value is not greater than 20 deg. Rp  52 mm

ε  17.5 mm



PRESSURE ANGLE

Pressure Angle, deg

20

10

0

 10

 20

0

20

40

60


8.

From the graph of radius of curvature above, we see that the minimum value occurs at a cam angle of about 45 deg.

ρmin  ρ 45 deg Rp Using a multiple of 3, the maximum roller follower radius is

ρmin  32.123 mm Rf 

ρmin 3

Rf  10.7 mm




Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot the pressure angle and radius of curvature for a modified sinusoidal acceleration cam function for any specified values of lift and duration. Test it using a lift of 20 mm over an interval of 60 deg at 1 rad/sec, and determine the prime circle radius needed to obtain a maximum pressure angle of 20 deg. What is the minimum diameter roller follower needed to avoid undercutting with these data?

Enter:

Lift:

h 1  20 mm

Eccentricity

ε  4  mm

Duration:

β1  60 deg

Prime circle radius

Rp  50 mm

Solution:


1.


2.

The numerical constants in these SCCA for the modified sinusoidal equations are given in Table 8-2.

3.

b  0.25

c  0.00

d  0.75

Cv  1.7596

Ca  5.5280

Cj  69.466


 b  b  2  π      sin  x  b   π  π

y1 ( x)  Ca x

y' 1 ( x)  Ca

 π  x  b 

y''1 ( x)  Ca sin

 π  b

y'''1 ( x)  Ca

 π  x    b 

  1  cos

π b

 π  x  b 

 cos

for b/2 <= x <= (1 - d)/2


y'''2 ( x)  0

for (1 - d)/2 <= x <= (1 + d)/2 2 2 2  b c 1  ( 1  d) d d 1  d  2 1 π    x     b         cos   x   8  π 2  2 8 2  π π d    π    

y3 ( x)  Ca 

 b  c  d  sin π   x  1    2 π 2 π  d 

y' 3 ( x)  Ca 

d 

 

1  d  π y''3 ( x)  Ca cos   x   2  d 

y'''3 ( x)  Ca

1  d  π  sin   x   d 2  d 

π

for (1 + d)/2 <= x <= 1 - b/2

 x2  b 1 1 b 1 2 2 y4 ( x)  Ca     1    x  2  d  b       2 8  4  2 π 2  π   b b  y' 4 ( x)  Ca  x  1  y''4 ( x)  Ca y'''4 ( x)  0 2 π 



for 1 - b/2 <= x <= 1










y' 5 ( x)  Ca

 π   1  cos  ( x  1 ) π  b  b

π  y''5 ( x)  Ca sin  ( x  1 ) b  4.

y'''5 ( x)  Ca

π   cos  ( x  1 ) b b 

π


5.


6.


 b 1  d   y ( x)  R x 1  d 1  d   y ( x)     y1 ( x)  R x    2   3  2 2 2   2 2      1d b b 1    y4 ( x)  R x 1  1  y5 ( x)   R x   2 2 2     

s1( x)  h 1  R x 0 

v1( x) 

a 1( x) 

b 1  d 1  d 1  d  y' 1 ( x)  R x    y' 2 ( x)  R x    y' 3 ( x)      β1 2 2 2   2 2      1d b b    1    y' 4 ( x)  R x 1  1  y' 5 ( x)   R x   2 2 2        R x 0 

h1

h1

β1

j1 ( x) 

2

h1

β1

5.

b

3

b

 b 1  d   y'' ( x)  R x 1  d 1  d   y'' ( x)     y''1 ( x)  R x    2   3  2 2 2   2 2      1d b b 1    y''4 ( x)  R x 1  1  y''5 ( x)   R x   2 2 2     

  R x 0 

b

 b 1  d   y''' ( x)  R x  1  d  1  d   y''' ( x)     y'''1 ( x)  R x    2   3  2 2 2   2 2      1d b b 1    y'''4 ( x)  R x 1  1  y'''5 ( x)   R x   2 2 2     

  R x 0 

b


θ   v1   ε   β1     ϕ θ Rp ε   atan  θ  2 2  s1 β1   Rp  ε     



  Rp  

ρ θ Rp 

R   p  6.

s1

2

2 2 θ θ  s1    v1    β1    β1  

2

2

θ  θ θ θ  2  v1   a 1    Rp  s1      β1    β1   β1    β1  

Plot the pressure angle and radius of curvature functions over the lift interval: θ  0  deg 0.5 deg  β1 PRESSURE ANGLE

Pressure Angle, deg

30

20

10

0

 10

0

20

40

60




75 50 25 0  25  50  75  100

0

20

40

60


7.


ε  15.2 mm



PRESSURE ANGLE

Pressure Angle, deg

20

10

0

 10

 20

0

20

40

60


8.


ρmin  ρ 52 deg Rp

ρmin  25.562 mm

Using a multiple of 3, the maximum roller follower radius is

Rf 

ρmin 3

Rf  8.5 mm




Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot the pressure angle and radius of curvature for a cycloidal displacement cam function for any specified values of lift, duration, eccentricity, and prime circle radius. Test it using a lift of 20 mm over an interval of 60 deg at 1 rad/sec, and determine the prime circle radius needed to obtain a maximum pressure angle of 20 deg. What is the minimum diameter roller follower needed to avoid undercutting with these data?

Enter:

Solution: 1.

Lift:

h  20 mm

Eccentricity

ε  4  mm

Duration:

β  60 deg

Prime circle radius

Rp  50 mm


Cycloidal motion is defined in local coordinates by equations 8.12. They are:

 θ  1  sin 2 π θ     β   β 2 π 

s θ  h  

a  θ  2  π

2.

h

 

 sin 2  π

 β 

v θ 

θ

h

j  θ  4  π 

2 h

 β

 

  1  cos 2  π

 

θ 



β 

 cos 2  π

θ



2 3 β β β Using equations 8.31d and 8.33, write the pressure angle and radius of curvature functions.





v θ  ε    s θ  Rp2  ε2   

ϕ θ Rp ε  atan

3





ρ θ Rp 

2

Rp  sθ2  2 vθ2  aθ Rp  sθ

Plot the pressure angle and radius of curvature functions over the lift interval: θ  0  deg 0.5 deg  β PRESSURE ANGLE 30

Pressure Angle, deg

3.

 R  s θ  2  v θ 2  p 

20

10

0

 10

0

20


60





75 50 25 0  25  50  75  100

0

20

40

60


4.


ε  17.5 mm

PRESSURE ANGLE

Pressure Angle, deg

20

10

0

 10

 20

0

20

40

60


5.



ρmin  28.093 mm


Rf 

ρmin 3

Rf  9.4 mm




Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot the pressure angle and radius of curvature for a 3-4-5 polynomial displacement cam function for any specified values of lift, duration, eccentricity, and prime circle radius. Test it using a lift of 20 mm over an interval of 60 deg at 1 rad/sec, and determine the prime circle radius needed to obtain a maximum pressure angle of 20 deg. What is the minimum diameter roller follower needed to avoid undercutting with these data?

Enter:

Solution: 1.

Lift:

h  20 mm

Eccentricity

ε  4  mm

Duration:

β  60 deg

Prime circle radius

Rp  50 mm


The 3-4-5 polynomial is defined in local coordinates by equations 8.24. They are:



4 5 θ θ     15    6       β  β  β 

θ s θ  h  10  

v θ 

3 4   θ 2 θ θ     30    60    30    β   β  β  β 

h

2 3 2  θ    180   θ   120   θ   j  θ  h  60  360   θ   360   θ            2 3 β  β  β   β  β  β    β 

a  θ 

2.

3

h

 60 






v θ  ε    s θ  Rp2  ε2   


3





ρ θ Rp 

2

Rp  sθ2  2 vθ2  aθ Rp  sθ

Plot the pressure angle and radius of curvature functions over the lift interval: θ  0  deg 0.5 deg  β

PRESSURE ANGLE 30

Pressure Angle, deg

3.

 R  s θ  2  v θ 2  p 

20

10

0

 10

0

20


60





75 50 25 0  25  50  75  100

0

20

40

60


4.


ε  16.3 mm

PRESSURE ANGLE

Pressure Angle, deg

20

10

0

 10

 20

0

20

40

60


5.



ρmin  26.885 mm


Rf 

ρmin 3

Rf  9.0 mm




Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot the pressure angle and radius of curvature for a 4-5-6-7 polynomial displacement cam function for any specified values of lift, duration, eccentricity, and prime circle radius. Test it using a lift of 20 mm over an interval of 60 deg at 1 rad/sec, and determine the prime circle radius needed to obtain a maximum pressure angle of 20 deg. What is the minimum diameter roller follower needed to avoid undercutting with these data?

Enter:

Solution: 1.

Lift:

h  20 mm

Eccentricity

ε  4  mm

Duration:

β  60 deg

Prime circle radius

Rp  50 mm


The 4-5-6-7 polynomial is defined in local coordinates by equation 8.25. Differentiate it to get v, a, and j. 5 6 7   θ 4  θ   70  θ   20  θ    84           β  β  β  β 

s θ  h  35 

4 5 6  θ 3 θ θ θ       140     420     420     140     β   β  β  β  β 

v θ 

h

a  θ 

h 2

β 2.



2 3 4 5  θ   1680  θ   2100  θ   840   θ           β  β  β  β 

 420  








v θ  ε    s θ  Rp2  ε2   


3





ρ θ Rp 

2

Rp  sθ2  2 vθ2  aθ Rp  sθ


Pressure Angle, deg

3.

 R  s θ  2  v θ 2  p 

30 20 10 0  10

0

20


60





75 50 25 0  25  50  75  100

0

20

40

60


4.


ε  19.2 mm

PRESSURE ANGLE

Pressure Angle, deg

20

10

0

 10

 20

0

20

40

60


5.



ρmin  28.317 mm


Rf 

ρmin 3

Rf  9.4 mm




Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot the pressure angle and radius of curvature for a simple harmonic displacement cam function for any specified values of lift, duration, eccentricity, and prime circle radius. Test it using a lift of 20 mm over an interval of 60 deg at 1 rad/sec, and determine the prime circle radius needed to obtain a maximum pressure angle of 20 deg. What is the minimum diameter roller follower needed to avoid undercutting with these data?

Enter: Lift:

h  20 mm

Eccentricity

ε  4  mm

Duration:

β  60 deg

Prime circle radius

Rp  50 mm

Solution: 1.

2.


The simple harmonic motion (SHM) is defined in local coordinates by equations 8.6. They are:

 2 

s θ 

h

a  θ 

π

 θ    β 

  1  cos π 2

h  θ   cos π  2 2  β β

v θ 

π h  θ   sin π  β 2  β

j  θ 

π

3

h  θ   sin π  3 2  β β






v θ  ε    s θ  Rp2  ε2   


3





ρ θ Rp 

2

Rp  sθ2  2 vθ2  aθ Rp  sθ


Pressure Angle, deg

3.

 R  s θ  2  v θ 2  p 

20

10

0

 10

0

20


60





75 50 25 0  25  50  75  100

0

20

40

60


4.


ε  13.3 mm

PRESSURE ANGLE

Pressure Angle, deg

20

10

0

 10

 20

0

20

40

60


5.



ρmin  23.362 mm


Rf 

ρmin 3

Rf  7.8 mm




Derive equation 8.25 for the 4-5-6-7 polynomial.

Solution:


1.

2.

There are eight boundary conditions. They are: At  = 0: s = 0, v = 0, At  = : s = h, v = 0,

a = 0, a = 0,

j=0 j=0

Write equation 8.23 in the form of equation c in Example 8-5, but with eight terms. Differentiate it repeatedly to get equations for v, a, and j that are similar to equations d and e. Write eight equations in the unknown coefficients C0 through C7, applying the boundary conditions above. These eight equations are. For  = 0: s = 0, v = 0, a = 0, j = 0 0  C0

C0  0

0  C1

C1  0

0  2  C2

C2  0

0  6  C3

C3  0

For  = : s = h, v = 0, a = 0, j = 0 h  C4  C5  C6  C7 0  4  C4  5  C5  6  C6  7  C7 0  12 C4  20 C5  30 C6  42 C7 0  24 C4  60 C5  120  C6  210  C7 3.

Solve the last four equations for the four unknown coefficients C4, C5, C6, and C7. Set h = 1 then multiply all coefficients by h.

 1 4 C    12  24 

C4  35

 5 6 7  20 30 42   60 120 210  1

1

1

 1  0 H    0 0  

C5  84

 C4  C   5   C 1 H  C6     C7  C6  70

C7  20




Derive an expression for the pressure angle of a barrel cam with zero eccentricity.

Solution:


1.

A barrel cam is shown in Figure 8-4. The motion of the follower is parallel to the axis of rotation of the cam. The transmission axis is the common normal between the follower and the groove in which it travels. The common normal will go through and be perpendicular to the center of rotation of the follower and will be normal to the wall of the groove at the point of contact. Let s = axial displacement of the follower from some reference point, length units v = velocity of the follower, length/rad units r = mean radius of the groove, length units then, for an infinitesimal rotation of the cam, the follower will advance a distance ds while the distance along the groove will have advanced an amount rd. These two quantities form the legs of a right triangle whose hypotenuse is perpendicular to the common normal. Because of this relationship, we have

tan  ϕ =

d s = v  r  dθ  r

1



ϕ  atan

v

 r




Design a radial plate cam to move a translating roller follower through 30 mm in 30 deg, dwell for 100 deg, fall 10 mm in 10 deg, dwell for 20 deg, fall 20 mm in 20 deg, and dwell for the remainder Camshaft  = 200 rpm. Minimize the follower's peak velocity and determine the minimum prime circle radius that will give a maximum 25-deg pressure angle. Determine the minimum radii of curvature on the pitch curve.

Units:


Given:

RISE/FALL

Solution: 1.

1

DWELL

FALL

DWELL

β  30 deg

β  100  deg

β  10 deg

β  20 deg

h 1  30 mm

h 2  0.0 in

h 3  10 mm

h 4  0.0 in

β  20 deg

β  180  deg

h 5  20 mm

h 6  0.0 in

Shaft speed

ω  200  rpm


From Table 8-3, the motion program with lowest velocity that does not have infinite jerk is the modified sinusoidal. Use the modified sine for segments 1, 3, and 5. Segments 2, 4, and 6 are dwells. Enter the above data into program DYNACAM. The input screen is shown below.



2.


3.


Rp  205  mm


Rf  15 mm


ε  1.00 mm




Design a radial plate cam to move a translating roller follower through 30 mm in 30 deg, dwell for 100 deg, fall 10 mm in 10 deg, dwell for 20 deg, fall 20 mm in 20 deg, and dwell for the remainder Camshaft  = 200 rpm. Minimize the follower's peak acceleration and determine the minimum prime circle radius that will give a maximum 25-deg pressure angle. Determine the minimum radii of curvature on the pitch curve.

Units:


Given:

RISE/FALL

Solution: 1.

1

DWELL

FALL

DWELL

β  30 deg

β  100  deg

β  10 deg

β  20 deg

h 1  30 mm

h 2  0.0 in

h 3  10 mm

h 4  0.0 in

β  20 deg

β  180  deg

h 5  20 mm

h 6  0.0 in

Shaft speed

ω  200  rpm


From Table 8-3, the motion program with lowest acceleration that does not have infinite jerk is the modified trapezoidal. Use the modified trapezoidal for segments 1, 3, and 5. Segments 2, 4, and 6 are dwells. Enter the above data into program DYNACAM. The input screen is shown below.



2.


3.


Rp  235  mm


Rf  20 mm


ε  2.00 mm




Design a radial plate cam to move a translating roller follower through 30 mm in 30 deg, dwell for 100 deg, fall 10 mm in 10 deg, dwell for 20 deg, fall 20 mm in 20 deg, and dwell for the remainder. Camshaft  = 200 rpm. Minimize the follower's peak jerk and determine the minimum prime circle radius that will give a maximum 25-deg pressure angle. Determine the minimum radii of curvature on the pitch curve.

Units:


Given:

RISE/FALL

Solution: 1.

1

DWELL

FALL

DWELL

β  30 deg

β  100  deg

β  10 deg

β  20 deg

h 1  30 mm

h 2  0.0 in

h 3  10 mm

h 4  0.0 in

β  20 deg

β  180  deg

h 5  20 mm

h 6  0.0 in

Shaft speed

ω  200  rpm


From Table 8-3, the motion program with lowest jerk that does not have infinite jerk is the cycloidal displacement. Use the cycloidal for segments 1, 3, and 5. Segments 2, 4, and 6 are dwells. Enter the above data into program DYNACAM. The input screen is shown below.



2.


3.


Rp  234  mm


Rf  15 mm


ε  1.00 mm




Design a radial plate cam to lift a translating roller follower through 10 mm in 65 deg, return to 0 in 65 deg, and dwell for the remainder. Camshaft  = 3500 rpm. Minimize the cam size while not exceeding a 25-deg pressure angle. What size roller follower is needed?

Units:


Given:

RISE/FALL

Solution:

1

DWELL

β  130  deg

β  230  deg

h 1  10 mm

h 2  0  mm

Shaft speed

ω  3500 rpm


1.

Use a two-segment polynomial as described in Example 8-8. This will result in a 6-deg polynomial requiring 7 boundary conditions. Enter the above data into program DYNACAM. The input boundary conditions screen for the polynomial segment is shown below.

2.



3.



Rp  28 mm


Rf  8.0 mm


ε  0.0 mm




Design a cam-driven quick-return mechanism for a 3:1 time ratio. The translating roller follower should move forward and back 50 mm and dwell in the back position for 80 deg. It should take one-third the time to return as to move forward. Camshaft  = 100 rpm. Minimize the package size while maintaining a 25-deg maximum pressure angle. Draw a sketch of your design and provide svaj, , and  diagrams.

Units:


1

Given:

Solution: 1.

2.

Time ratio

tr  3

Dwell interval

β  80 deg

Lift

L  50 mm

Shaft speed

ω  100  rpm


Calculate the rise/fall interval widths. First segment interval

β  360  deg  β

Rise subinterval

βr  β

tr tr  1

β  280.000 deg

βr  210.000 deg

Use a three-segment polynomial. Enter the above data into program DYNACAM. The input screen is shown below.



3.


4.

The minimum and maximum pressure angles and radius of curvature are shown in the figure above. The design has the following dimensions:

5.

Prime circle radius

Rp  92 mm


Rf  20.0 mm


ε  25.0 mm

Graphs of s, v, a, j, , and  are shown on the following pages.








Design a cam-follower system to drive a linear translating piston at a constant velocity for 200 deg through a stroke of 100 mm at 60 rpm. Minimize the package size while maintaining a 25-deg maximum pressure angle. Draw a sketch of your design and provide svaj, , and  diagrams.

Units:


Given:

Constant velocity interval width and stroke β  200  deg

1

Cam rotation speed

h 1  100  mm

ω  60 rpm

Assumptions: A roller follower can be attached to the translating piston to interface with the cam. Solution: 1.


Calculate the cycle time and time for the cam to rotate through the CV segment. tcycle 

t1 

2.

tcycle  1.000 s

ω β

360  deg

 tcycle

t1  0.556 s

Calculate the required constant velocity. Vcv 

3.

2  π rad

h1 t1

Vcv  180.000

mm sec

Use a two-segment cam with both segments polynomials. Let the first segment be a 1-deg polynomial and the second be a 7-deg (8 boundary conditions). There are two design choices to be made concerning the cam motion. They are: Lift at end of second segment (and beginning of first)

L2  10 mm

Subinterval of second segment at which s = 0

βf  120  deg

4.

Enter the above data into program DYNACAM. The segment segment boundary condition input screen is shown on the next page followed by a cam drawing for the design choices made.

5.


6.

The minimum and maximum pressure angles and radius of curvature are shown in the cam drawing. The design has the following dimensions: Prime circle radius

Rp  122  mm


Rf  20 mm


ε  30.6 mm








Given:

Design a cam-follower system to rise 20 mm in 80 deg, fall 10 mm in 100 deg, dwell at 10 mm for 100 deg, fall 10 mm in 50 deg, and dwell at 0 for 30 deg. The total cycle must take 4 sec. Avoid unnecessary returns to zero acceleration. Minimize the package size and maximize the roller follower diameter while maintaining a 25 deg maximum pressure angle. Draw a sketch of your design and provide s v a j, , and  diagrams. RISE

1.

DWELL

FALL

β1  80 deg

β2  100  deg

β3  100  deg

β4  50 deg

h 1  20 mm

h 2  10 mm

h 3  10 mm

h 4  0  mm


FALL

τ  4  sec



2  π rad τ

ω  1.571

rad sec

2.

Use a four-segment for the for the rise, fall, dwell, fall, and dwell. Enter the above data into program DYNACAM. The input screen is shown below.

3.

The cam was sized iteratively to reduce package size while keeping the follower radius large, the pressure angle low, and the follower radius at least two times the radius of curvature . The resulting cam is shown below.


4.

5.



Rp  90 mm


Rf  30 mm


ε  0  mm

Graphs of svaj,  and  for the roller follower are shown on the following page.






Design a single-dwell cam to move a follower from 0 to 35 mm in 75 deg, fall 35 mm in 120 deg and dwell for the remainder. The total cycle must take 3 sec. Choose suitable programs for rise and fall to minimize velocities. Plot the SVAJ diagrams.

Given:

RISE/FALL

DWELL

β  195  deg

β  165  deg

h  35 mm

h 3  0.0 mm τ  3  sec




2  π rad

ω  2.094

τ 2.

rad sec


 = 0:  = 1:

s = 0, s = h,

v = 0, v=0

a =0

 = 

s = 0,

v = 0,

a =0

This is a minimum set of 8 BCs. The v = 0 condition at  = 1 is required to keep the displacement from overshooting the lift, h. Define the total lift, the rise interval, the fall interval, and the ratio of rise to the total interval. Total lift: h  35.000 mm β  75 deg

Rise interval:

β


0  c1

0  c2

For  = 1: s = h, v = 0 3

4

5

6

h  c3 A  c4 A  c5 A  c6 A  c7 A 2

3

4

7

5

0  3  c3 A  4  c4 A  5  c5 A  6  c6 A  7  c7 A

6


A  0.385

β

β  120  deg

Fall interval: 3.

A 




 A3 A4 A5 A6 A7     3 A 2 4 A 3 5 A 4 6 A 5 7 A 6   C   1 1 1 1   1  3 4 5 6 7    12 20 30 42   6

5.

 c3     c4   c5   C 1 H    c6  c   7

h   0 H   0  0   0

c3  5609.280 mm

c4  24548.849 mm

c6  28772.307 mm

c7  7721.009 mm

c5  39990.867 mm

Write the SVAJ equations for the rise/fall segment. 3

4

5

6

7

 θ   c  θ   c  θ   c  θ   c  θ   4   5  6   7   β  β  β  β  β 2 3 4 5 6 1  θ θ θ θ θ  V  θ   3  c3    4  c4    5  c5    6  c6    7  c7    β   β  β  β  β  β  S  θ  c3 

A  θ 

2 3 4 5  θ θ θ θ θ        6  c3    12 c4    20 c5    30 c6    42 c7    2  β  β  β  β  β  β 

J  θ 

1

1

3

β

2 3 4  θ   60 c   θ   120  c   θ   210  c   θ    5  6   7   β  β  β  β 



Plot the displacement, velocity, acceleration, and jerk over the interval 0 <=  <= . θ  0  deg 1  deg  β

DISPLACEMENT, S 40

30 Displacement, mm

6.



 6  c3  24 c4 

20

10

0

 10

0

50


150

200



VELOCITY, V 60

Velocity, mm

40 20 0  20  40  60

0

50

100

150

200


ACCELERATION, A

Acceleration, mm

200

100

0

 100

 200

0

50

100

150

200

150

200


JERK, J

Jerk, mm

1000

500

0

 500

0

50





Design a cam to move a follower at a constant velocity of 100 mm/sec for 2 sec then return to its starting position with a total cycle time of 3 sec.

Given:

Constant velocity: vc  100  mm sec

1

Time duration of cv segment: tcv  2  sec Solution: 1.



2.

Cycle time: τ  3  sec

2  π rad

τ

ω  2.094

rad sec

Use a two-segment polynomial as demonstrated in Example 8-12. The lift during the first segment and the svaj equations for the first segment are: vc Normalized velocity: vcv  vcv  47.746 mm ω tcv h cv  vc tcv h cv  200.000 mm β1   360  deg β1  240 deg

τ

s1( θ )  h cv 2.

θ β1

v1( θ )  vcv

a 1( θ )  0  mm

j1 ( θ )  0  mm

The boundary conditions for the second segment are: at

 = 1:

s = h cv, v = vcv

 = 360 deg

s = 0,

a =0

v = vcv, a = 0

This is a minimum set of 6 BCs. Define the total interval and the constant velocity interval, and the ratio of constant velocity interval to the total interval.

3.

Total interval:

β  360  deg

CV interval:

β1  240 deg

A 

β1 β

A  0.667

Use the 6 BCs and equation 8.23 to write 6 equations in s, v, and a similar to those in example 8-9 but with 6 terms in the equation for s (the highest term will be fifth degree). For  = 1:

s = h cv, v = vcv

a =0 2

3

4

h cv = c0  c1 A  c2 A  c3 A  c4 A  c5 A vcv =

1  2 3 4  c1  2  c2 A  3  c3 A  4  c4 A  5  c5 A  β  2

0 = 2  c2  6  c3 A  12 c4 A  20 c5 A For  = :

s = 0,

v = vcv, a = 0

0 = c0  c1  c2  c3  c4  c5 0=

1

β

5

  c 1  2  c 2  3  c 3  4  c 4  5  c 5

0 = 2  c2  6  c3  12 c4  20 c5

3


4.


Solve for the unknown polynomial coefficients.

1   0 C   0  1 0  0

2

A A

A

3

1 2 A 3 A

A 2

4

4 A

2

6 A

12 A

1

1

1

1

1

2

3

4

0

2

6

12

2

5

 hcv     β  vcv   0  H     0   β  vcv     0  5

c0  1.536  10 mm 6

 c0   c1  c   2   C 1 H  c3     c4   c5  6

c1  9.717  10 mm

c2  2.430  10 mm

6

c3  2.997  10 mm 5.

3

0

5

  4  5 A  3 20 A  1  5   20  A

5

c4  1.822  10 mm

c5  4.374  10 mm

Write the svaj equations for the second segment. 2

3

4

θ θ θ θ θ   c2    c3    c4    c5   β β β β         β

s2( θ )  c0  c1 

v2( θ ) 

a 2( θ ) 

β  1 2

1

β 4.

2 3 4 θ θ θ θ   3  c3    4  c4    5  c5    β β β β 



 c1  2  c2 

1

β j2 ( θ ) 

3



2 3 θ  θ   20 c   θ    12  c   4  5  β β β 

 2  c2  6  c3 





2 θ θ   60 c5    β β 

 6  c3  24 c4 



To plot the SVAJ curves, first define a range function that has a value of one between the values of a and b and zero elsewhere. R( θ a b )  if [ ( θ  a )  ( θ  b ) 1 0 ] S ( θ )  R θ 0 β1   s1( θ )  R θ β1 β   s2( θ ) V ( θ )  R θ 0 β1   v1( θ )  R θ β1 β   v2( θ ) A ( θ )  R θ 0 β1   a 1( θ )  R θ β1 β   a 2( θ ) J ( θ )  R θ 0 β1   j1 ( θ )  R θ β1 β   j2 ( θ )

6.

5

Plot the displacement, velocity, acceleration, and jerk over the interval 0 <=  <= . θ  0  deg 0.5 deg  β



DISPLACEMENT, S

Displacement, mm

300

200

100

0

 100

0

60

120

180

240

300

360

240

300

360

240

300

360


VELOCITY, V 100

Velocity, mm

0

 100

 200

 300

0

60

120


ACCELERATION, A

Acceleration, mm

400

200

0

 200

 400

0

60

120




JERK, J 1000

Jerk, in

0

 1000

 2000

0

60

120


240

300

360




Given:

Design a double-dwell cam to move a follower from 0 to 50 mm in 75 deg, dwell for 75 deg, fall 50 mm in 75 deg and dwell for the remainder. The total cycle must take 5 sec. Use a modified trapezoidal function for rise and fall and plot the s v a j diagrams. RISE

DWELL

1.

β2  75 deg

β3  75 deg

β4  135  deg

h 1  50 mm

h 2  0  mm

h 3  50 mm

h 4  0  mm

3.

τ  5  sec



2.

DWELL


FALL

2  π rad

ω  1.257

rad

sec τ The modified trapezoidal motion is defined in local coordinates by equations 8.15 through 8.19. The numerical constants in these SCCA equations are given in Table 8-2. b  0.25

c  0.50

d  0.25

Cv  2.0000

Ca  4.8881

Cj  61.426


 b  b  2  π      sin  x  b   π  π

y1 ( x)  Ca x

 π  x  b 

y''1 ( x)  Ca sin

y' 1 ( x)  Ca

 π  b

y'''1 ( x)  Ca

 π  x    b 

  1  cos

π b

 π  x  b 

 cos

for b/2 <= x <= (1 - d)/2

 x2

y2 ( x)  Ca 

2 

 b  

1

π



1  2 1   x  b    2  2 8 π   1

y''2 ( x)  Ca

y' 2 ( x)  Ca x  b  



1

π



1 



2 

y'''2 ( x)  0

for (1 - d)/2 <= x <= (1 + d)/2

 b c  x   π 2  

y3 ( x)  Ca 

2 2 2  d   b2  1  1   ( 1  d )   d   cos π   x  1  d        8 2 8 2  d   π  π π   

 b  c  d  sin π   x  1    2 π 2 π  d 

y' 3 ( x)  Ca 

1  d  π y''3 ( x)  Ca cos   x   2  d  for (1 + d)/2 <= x <= 1 - b/2

d 

 

y'''3 ( x)  Ca

1  d  π  sin   x   d 2  d 

π



 x2  b  1  2  π 

1  1 2 2  1   x  2 d  b   2     2 4 8 π   b

y4 ( x)  Ca 

y' 4 ( x)  Ca  x 



b

1

π





b



y''4 ( x)  Ca

2

y'''4 ( x)  0

for 1 - b/2 <= x <= 1

2

b 2 d  b y5 ( x)  Ca   x  π 2 π  y' 5 ( x)  Ca

  (1  b)2  d2   b  2sin π (x  1)

2

 b

   π

  

 π   1  cos  ( x  1 ) b π    b

 π y''5 ( x)  Ca sin  ( x  1 ) b   4.

4

y'''5 ( x)  Ca

 π  cos  ( x  1 ) b b  

π


5.


6.

Write the local svaj equations for the first interval, 0 <=  <= 1. Note that each subinterval function is multiplied by the range function so that it will have nonzero values only over its subinterval. For 0 <=  <= (Rise) s1( x)  h 1  R x 0 

 b 1  d   y ( x)  R x 1  d 1    y1 ( x)  R x    2  2 2 2  2 2     b b  1d   1    y4 ( x)  R x 1  1  y5 ( x)   R x  2 2  2   

v1( x) 

a 1( x) 

  R x 0 

h1 2

h1

β1

7.

   y3 ( x)      

d

b 1  d 1  d 1  d  y' 1 ( x)  R x    y' 2 ( x)  R x    y' 3 ( x)      β1 2 2 2   2 2      b b  1d   1    y' 4 ( x)  R x 1  1  y' 5 ( x)   R x   2 2  2     h1

β1

j1 ( x) 

b

3

b

  R x 0 

 b 1  d   y'' ( x)  R x 1  d 1    y''1 ( x)  R x    2  2 2 2  2 2     b b  1d   1    y''4 ( x)  R x 1  1  y''5 ( x)   R x  2 2  2   

  R x 0 

b

d

   y''3 ( x)      

b

 b 1  d   y''' ( x)  R x  1  d  1  d   y''' ( x)     y'''1 ( x)  R x    2   3  2 2 2   2 2      b b 1d 1    y'''4 ( x)  R x 1  1  y'''5 ( x)   R x   2 2  2    

Write the local svaj equations for the second interval, 1 <=  <= 1+ 2. For this interval, the value of S is the value of S at the end of the previous interval and the values of V, A, and J are zero because of the dwell.



For 1 <=  <= 1+ 2 s2( x)  h 1 8.

v2( x)  0

a 2( x)  0

j2 ( x)  0

Write the local svaj equations for the third interval, 1 + 2 <=  <= 1.+ 2 + 3. For 1 + 2 <=  <= 1.+ 2 + 3 s3( x)  h 31   R x 0 

 b 1  d   y ( x)  R x 1  d 1  d   y ( x)     y1 ( x)  R x    2   3  2 2 2    2 2      b b 1d 1    y4 ( x)  R x 1  1  y5 ( x)   R x   2 2  2    

  

v3( x)  

a 3( x)  


h3

β3

j3 ( x)  

2

h3

β3

9.

  R x 0 

b

3

b

  R x 0 

d

   y'3 ( x)      

 b 1  d   y'' ( x)  R x  1  d  1  d   y'' ( x)     y''1 ( x)  R x    2   3  2 2 2    2 2      b b 1d 1    y''4 ( x)  R x 1  1  y''5 ( x)   R x   2 2  2    

  R x 0 

b

b

 b 1  d   y''' ( x)  R x 1  d 1    y'''1 ( x)  R x    2  2 2 2  2 2     b b  1d   1    y'''4 ( x)  R x 1  1  y'''5 ( x)   R x  2 2  2   

d

   y'''3 ( x)      


v4( x)  0

a 4( x)  0

j4 ( x)  0


θ1  β1

θ2  θ1  β2

θ3  θ2  β3


S ( θ )  s1

θ  0  deg 0.5 deg  360  deg

θ4  θ3  β4



DISPLACEMENT, S 50

Displacement, mm

40 30 20 10 0

0

30

60

90

120

150

180

210

240

270

300

330

360




V ( θ )  v1

VELOCITY, V 100

Velocity, mm

50

0

 50

 100

0

30

60

90

120

150

180

210

240

270

300

330

360






A ( θ )  a 1

ACCELERATION, A

Acceleration, mm

200

100

0

 100

 200

0

30

60

90

120

150

180

210

240

270

300

330

360




J ( θ )  j1 

JERK, J 2000

Jerk, mm

1000

0

 1000

 2000

0

30

60

90

120

150

180

210


240

270

300

330

360




Given:

Design a double-dwell cam to move a follower from 0 to 50 mm in 75 deg, dwell for 75 deg, fall 50 mm in 75 deg and dwell for the remainder. The total cycle must take 5 sec. Use a modified sinusoidal function for rise and fall and plot the s v a j diagrams. RISE

DWELL

1.

β2  75 deg

β3  75 deg

β4  135  deg

h 1  50 mm

h 2  0  mm

h 3  50 mm

h 4  0  mm

τ  5  sec



2.

DWELL


FALL

2  π rad

ω  1.257

rad

sec τ The modified sinusoidal motion is defined in local coordinates by equations 8.15 through 8.19. The numerical constants in these SCCA equations are given in Table 8-2. b  0.25

c  0.00

d  0.75

Ca  5.5280 3.


 b  b  2  π      sin  x  b   π  π

y1 ( x)  Ca x

 π  x  b 

y''1 ( x)  Ca sin

y' 1 ( x)  Ca

 π  b

y'''1 ( x)  Ca

 π  x    b 

  1  cos

π b

 π  x  b 

 cos

for b/2 <= x <= (1 - d)/2

 x2

y2 ( x)  Ca 

2 

 b  

1

π



1  2 1   x  b    2  2 8 π   1

y''2 ( x)  Ca

y' 2 ( x)  Ca x  b  



1

π



1 



2 

y'''2 ( x)  0

for (1 - d)/2 <= x <= (1 + d)/2

 b c  x   π 2  

y3 ( x)  Ca 

2 2 2  d   b2  1  1   ( 1  d )   d   cos π   x  1  d        8 2 8 2  d   π  π π   

 b  c  d  sin π   x  1    2 π 2 π  d 

y' 3 ( x)  Ca 


d 

 

y'''3 ( x)  Ca

1  d  π  sin   x   d 2  d 

π



 x2  b  1  2  π 

1  1 2 2  1   x  2 d  b   2     2 4 8 π   b

y4 ( x)  Ca 

y' 4 ( x)  Ca  x 



b

1

π





b



y''4 ( x)  Ca

2

y'''4 ( x)  0

for 1 - b/2 <= x <= 1

2

b 2 d  b y5 ( x)  Ca   x  π 2 π  y' 5 ( x)  Ca

  (1  b)2  d2   b  2sin π (x  1)

2

4

 b

   π

  

 π   1  cos  ( x  1 ) b π    b

 π y''5 ( x)  Ca sin  ( x  1 ) b  

y'''5 ( x)  Ca

 π  cos  ( x  1 ) b b  

π

4.


5.


6.



s1( x)  h 1  R x 0 

v1( x) 

a 1( x) 

  R x 0 


h1

β1

j1 ( x) 

2

h1

β1

7.

b

3

b

  R x 0 

d

   y' 3 ( x)      


  R x 0 

b

b

 b 1  d   y''' ( x)  R x  1  d  1    y'''1 ( x)  R x    2  2 2 2  2 2     b b  1d   1    y'''4 ( x)  R x 1  1  y'''5 ( x)   R x  2 2  2   

d

   y'''3 ( x)      




For 1 <=  <= 1+ 2 s2( x)  h 1 8.

v2( x)  0

a 2( x)  0

j2 ( x)  0

Write the local svaj equations for the third interval, 1 + 2 <=  <= 1.+ 2 + 3. For 1 + 2 <=  <= 1.+ 2 + 3 s3( x)  h 31   R x 0 

 b 1  d   y ( x)  R x 1  d 1    y1 ( x)  R x    2  2 2 2  2 2     b b  1d   1    y4 ( x)  R x 1  1  y5 ( x)   R x  2 2  2   

  

v3( x)  

a 3( x)  

β3

h3

9.

2

h3

β3

d

   y3 ( x)      

 b 1  d   y' ( x)  R x 1  d 1  d   y' ( x)     y' 1 ( x)  R x    2   3  2 2 2    2 2      b b 1d 1    y' 4 ( x)  R x 1  1  y' 5 ( x)   R x   2 2  2    

h3

β3

j3 ( x)  

  R x 0 

b

3

b

  R x 0 

 b 1  d   y'' ( x)  R x  1  d  1    y''1 ( x)  R x    2  2 2 2  2 2     b b  1d   1    y''4 ( x)  R x 1  1  y''5 ( x)   R x  2 2  2   

  R x 0 

b

b

d

 b 1  d   y''' ( x)  R x 1  d 1    y'''1 ( x)  R x    2  2 2 2  2 2     b b  1d   1    y'''4 ( x)  R x 1  1  y'''5 ( x)   R x  2 2  2   

   y''3 ( x)      

d

   y'''3 ( x)      


v4( x)  0

a 4( x)  0

j4 ( x)  0


θ1  β1

θ2  θ1  β2

θ3  θ2  β3


S ( θ )  s1

θ  0  deg 0.5 deg  360  deg

θ4  θ3  β4



DISPLACEMENT, S

Displacement, mm

60

40

20

0

 20

0

30

60

90

120

150

180

210

240

270

300

330

360




V ( θ )  v1

VELOCITY, V 100

Velocity, mm

50

0

 50

 100

0

30

60

90

120

150

180

210

240

270

300

330

360




A ( θ )  a 1



ACCELERATION, A

Acceleration, mm

200

100

0

 100

 200

0

30

60

90

120

150

180

210

240

270

300

330

360




J ( θ )  j1 

JERK, J 2000

Jerk, mm

1000

0

 1000

 2000

0

30

60

90

120

150

180

210


240

270

300

330

360




Design a double-dwell cam to move a follower from 0 to 50 mm in 75 deg, dwell for 75 deg, fall 50 mm in 75 deg and dwell for the remainder. The total cycle must take 5 sec. Use a 4-5-6-7 polynomial function for rise and fall and plot the s v a j diagrams.

Given:

RISE

DWELL

FALL

DWELL

  75 deg

  75 deg

  75 deg

  135  deg

h 1  50 mm

h 2  0  mm

h 3  50 mm

h 4  0  mm

τ  5  sec




2.

2  π rad

ω  1.257

rad

sec τ The 4-5-6-7 polynomial is defined in local coordinates by equation 8.25. Differentiate it to get v, a, and j. 5 6 7  θ 4   84  θ   70  θ   20  θ            β  β  β  β 

s θ β h   h  35  v θ β h  

a  θ β h  

4 5 6  θ 3 θ θ θ       140     420     420     140     β   β  β  β  β 

h

h 2

β j  θ β h  



2 3 4 5  θ   1680  θ   2100  θ   840   θ           β  β  β  β 

 420  



2 3 4  θ θ θ θ       840     5040    8400    4200    3  β  β  β  β  β 

h

3.

The above equations can be used for a rise or fall by inserting the proper values of , , and h. To plot the SVAJ curves, first define a range function that has a value of one between the values of x and y and zero elsewhere. R θ x y   if  θ  x   θ  y  1 0

4.

Write the global SVAJ equations for the first interval, 0 <=  <= 1. For this interval, the local and global frames are coincident so the local equations can be used as written, substituting only for h 1 for h and 1 for . Note that each subinterval function is multiplied by the range function so that it will have nonzero values only over its subinterval. For 0 <=  <= 



 





 



S 1   R  0   s   h 1

A1    R  0   a   h 1 5.



 





 



V1   R  0   v   h 1 J1   R  0   j   h 1

Write the global SVAJ equations for the second interval, 1 <=  <= 1+ 2. For this interval, the value of S is the value of S at the end of the previous interval and the values of V, A, and J are zero because of the dwell. For 1 <=  <= 1+ 2 S 2   R       h 1





V2   R       0  mm







A2    R       0  mm



6.

J2   R       0  mm





Write the global SVAJ equations for the third interval, 1 + 2 <=  <= 1.+ 2 + 3. For 1 + 2 <=  <= 1.+ 2 + 3

    

Let

S 3   R       h 1  s     h 3







A3    R       a     h 3



7.



 



V3   R       v     h 3



 

J3   R       j     h 3





 





Write the global SVAJ equations for the fourth interval, 1 + 2 + 3 <=  <= 1.+ 2 + 3 + 4. For this interval, the values of S, V, A, and J are zero because of the dwell. For 1 + 2 + 3 <=  <= 1.+ 2 + 3 + 4 Let        S 4   R       0  mm

V4   R       0  mm

A4    R       0  mm

J4   R       0  mm







8.











Write the complete global equation for the displacement and plot it over one rotation of the cam, which is the sum of the four intervals defined above. S    S 1   S 2   S 3   S 4 

θ  0  deg 0.5 deg  360  deg

DISPLACEMENT, S

Displacement, mm

60

40

20

0

 20

0

30

60

90

120

150

180

210

240

270

300

330

360


9.

Write the complete global equation for the velocity and plot it over one rotation of the cam, which is the sum of the four intervals defined above. V    V1   V2   V3   V4 



VELOCITY, V 100

Velocity, mm

50 0  50  100

0

30

60

90

120

150

180

210

240

270

300

330

360


10. Write the complete global equation for the acceleration and plot it over one rotation of the cam, which is the sum of the four intervals defined above. A    A1    A2    A3    A4   ACCELERATION, A

Acceleration, mm

300 200 100 0  100  200  300

0

30

60

90

120

150

180

210

240

270

300

330

360


13. Write the complete global equation for the jerk and plot it over one rotation of the cam, which is the sum of the four intervals defined above. J    J1   J2   J3   J4  JERK, J 2000

Jerk, mm

1000 0  1000  2000

0

30

60

90

120

150

180

210


240

270

300

330

360





Given:

RISE/FALL

DWELL

β  270  deg

β  90 deg

h  65 mm

h 3  0.0 mm τ  2  sec




2  π rad

ω  3.142

τ 2.

rad sec


 = 0:  = 1:

s = 0, s = h,

v = 0, v=0

a =0

 = 

s = 0,

v = 0,

a =0

This is a minimum set of 8 BCs. The v = 0 condition at  = 1 is required to keep the displacement from overshooting the lift, h. Define the total lift, the rise interval, the fall interval, and the ratio of rise to the total interval. Total lift: h  65.000 mm β  90 deg

Rise interval:

β

A  0.333

β

β  180  deg

Fall interval: 3.

A 


0  c1

0  c2

For  = 1: s = h, v = 0 3

4

5

6

h  c3 A  c4 A  c5 A  c6 A  c7 A 2

3

4

7

5

0  3  c3 A  4  c4 A  5  c5 A  6  c6 A  7  c7 A

6





 A3 A4 A5 A6 A7     3 A 2 4 A 3 5 A 4 6 A 5 7 A 6   C   1 1 1 1   1  3 4 5 6 7    12 20 30 42   6

5.

 c3     c4   c5   C 1 H    c6  c   7

h   0 H   0  0   0

c3  582.985 in

c4  2798.327 in

c6  3731.102 in

c7  1.049  10 in

c5  4897.072 in

3

Write the SVAJ equations for the rise/fall segment.

θ S  θ  c3  

3

4

5

6

7

θ θ θ θ  c4    c5    c6    c7    β  β  β  β  β 2 3 4 5 6 θ θ θ θ θ  1         V θ   3  c3    4  c4    5  c5    6  c6    7  c7    β   β  β  β  β  β  A  θ 

1 2

β J  θ 

2 3 4 5  θ   12 c   θ   20 c   θ   30 c   θ   42 c   θ    4  5  6  7   β  β  β  β  β 



2 3 4  θ θ θ θ       6  c3  24 c4    60 c5    120  c6    210  c7    3  β  β  β  β  β 

1


Displacement, mm

6.



 6  c3 

50

0

 50

0


200



VELOCITY, V 100

Velocity, mm

50

0

 50

 100

0

45

90

135

180

225

270

180

225

270

180

225

270


ACCELERATION, A

Acceleration, mm

200

100

0

 100

 200

0

45

90


JERK, J

Jerk, mm

1000

500

0

 500

0

45

90






Given:


1




ω  2.


τ

ω  1.047

rad sec


τ

s1( θ )  h cv 3.

θ β1

v1( θ )  vcv

a 1( θ )  0  mm

j1 ( θ )  0  mm


 = 1:

s = h cv, v = vcv

 = 360 deg

s = 0,

a =0

v = vcv, a = 0


4.

Total interval:

β  360  deg

CV interval:

β1  180 deg

A 

β1 β

A  0.500


s = h cv, v = vcv

a =0 2

3

4


1  2 3 4  c1  2  c2 A  3  c3 A  4  c4 A  5  c5 A  β  2

0 = 2  c2  6  c3 A  12 c4 A  20 c5 A For  = :

s = 0,

v = vcv, a = 0

0 = c0  c1  c2  c3  c4  c5 0=

5

1

β

  c 1  2  c 2  3  c 3  4  c 4  5  c 5

0 = 2  c2  6  c3  12 c4  20 c5

3


5.



1   0 C   0  1 0  0

A A

2

A

3

1 2 A 3 A

A 2

4

4 A

2

6 A

12 A

1

1

1

1

1

2

3

4

0

2

6

12

2

5

5

c1  2.868  10 mm

6

c2  8.640  10 mm

5

c3  1.248  10 mm

5

c4  8.64  10 mm

c5  2.304  10 mm


3

4

θ θ θ θ θ   c2    c3    c4    c5   β β β β         β

s2( θ )  c0  c1 

v2( θ ) 

a 2( θ ) 

j2 ( θ ) 

2 3 4 θ θ θ θ   3  c3    4  c4    5  c5    β β β β 



β  1 2

1

β

5

 c1  2  c2 

1

β

7.

 c0   c1  c   2   C 1 H  c3     c4   c5 

 hcv     β  vcv   0  H     0   β  vcv     0  5

c0  3.720  10 mm

6.

3

0

4

  4  5 A  3 20 A  1  5   20  A

3



2 3 θ  θ   20 c   θ    12  c   4  5  β β β 

 2  c2  6  c3 





2 θ θ   60 c5    β β 

 6  c3  24 c4 




8.




DISPLACEMENT, S 800

Displacement, mm

600 400 200 0  200

0

60

120

180

240

300

360

240

300

360

240

300

360


VELOCITY, V 200

Velocity, mm

0

 200

 400

 600

0

60

120


ACCELERATION, A

Acceleration, mm

1000

500

0

 500

 1000

0

60

120




JERK, J 2000

Jerk, mm

1000

0

 1000

 2000

 3000

0

60

120


240

300

360




Size the cam from Problem 8-42 for a flat-faced follower considering follower face width and radius of curvature. Plot the radius of curvature and draw the cam profile.

Units:


Given:

RISE/FALL

DWELL

β  195  deg

β  165  deg

h  35 mm

h 3  0.0 mm


3.

τ  3  sec



2.

1

2  π rad

ω  2.094

rad

ω  20.000 rpm sec τ Problem 8-42 used a two-segment polynomial with the rise and fall together in one segment and the dwell in the second segment. Enter the above data into program DYNACAM. The input screen is shown below.

The cam was sized iteratively to have the smallest base circle diameter while having no negative curvature and at least a 10 mm positive curvature. The resulting cam is shown below.


4.

5.


The minimum and maximum radius of curvature are shown in the figure above. The design has the following dimensions: Base circle radius

Rb  87 mm

Follower width

facewidth  89 mm

Graphs of s, and v for the flat-faced follower are shown on the following page.






Given:

Size the cam from Problem 8-44 for a translating flat-faced follower considering follower face width and radius of curvature. Plot the radius of curvature and draw the cam profile. RISE

1.

3.

DWELL

β2  75 deg

β3  75 deg

β4  135  deg

h 1  50 mm

h 2  0  mm

h 3  50 mm

h 4  0  mm

τ  5  sec



2.

FALL

β1  75 deg Cycle time:

Solution:

DWELL

2  π rad

ω  1.257

rad

sec τ Problem 8-44 used modified trapezoidal rise and fall segments with two dwell segments. Enter the above data into program DYNACAM. The input screen is shown below.



4.

5.



Rb  116  mm

Follower width

facewidth  153  mm







Given:


DWELL

FALL

DWELL

β1  75 deg

β2  75 deg

β3  75 deg

β4  135  deg

h 1  50 mm

h 2  0  mm

h 3  50 mm

h 4  0  mm

Cycle time:

Solution: 1.



2.

3.

τ  5  sec

2  π rad

ω  1.257

rad

sec τ Problem 8-45 used modified sinusoidal rise and fall segments with two dwell segments. Enter the above data into program DYNACAM. The input screen is shown below.



4.

5.



Rb  123  mm

Follower width








Given:


DWELL

FALL

DWELL

  75 deg

  75 deg

  75 deg

  135  deg

h 1  50 mm

h 2  0  mm

h 3  50 mm

h 4  0  mm

Cycle time:

Solution: 1.



2.

3.

τ  5  sec

2  π rad

ω  1.257

rad

sec τ Problem 8-46 used 4-5-6-7 polynomials for the rise and fall segments with two dwell segments. Enter the above data into program DYNACAM. The input screen is shown below.



4.

5.



Rb  185  mm

Follower width









Given:

RISE/FALL

DWELL

β  220  deg

β  140  deg

h  50 mm

h 3  0.0 mm τ  1  sec




2  π rad

ω  6.283

τ 2.

rad sec


 = 0:  = 1:

s = 0, s = h,

v = 0, v=0

a =0

 = 

s = 0,

v = 0,

a =0

This is a minimum set of 8 BCs. The v = 0 condition at  = 1 is required to keep the displacement from overshooting the lift, h. Define the total lift, the rise interval, the fall interval, and the ratio of rise to the total interval. Total lift: h  50.000 mm β  100  deg

Rise interval:

β


0  c1

0  c2

For  = 1: s = h, v = 0 3

4

5

6

h  c3 A  c4 A  c5 A  c6 A  c7 A 2

3

4

7

5

0  3  c3 A  4  c4 A  5  c5 A  6  c6 A  7  c7 A

6


A  0.455

β

β  120  deg

Fall interval: 3.

A 




 A3 A4 A5 A6 A7     3 A 2 4 A 3 5 A 4 6 A 5 7 A 6   C   1 1 1 1   1  3 4 5 6 7    12 20 30 42   6

5.

 c3     c4   c5   C 1 H    c6  c   7

h   0 H   0  0   0

c3  193.740 in

c4  723.297 in

c6  619.969 in

c7  142.076 in

c5  1007.449 in

Write the SVAJ equations for the rise/fall segment. 3

4

5

6

7

 θ   c  θ   c  θ   c  θ   c  θ   4   5  6   7   β  β  β  β  β 2 3 4 5 6 1  θ θ θ θ θ  V  θ   3  c3    4  c4    5  c5    6  c6    7  c7    β   β  β  β  β  β  S  θ  c3 

A  θ 

2 3 4 5  θ θ θ θ θ        6  c3    12 c4    20 c5    30 c6    42 c7    2  β  β  β  β  β  β 

J  θ 

1

1

3

β

2 3 4  θ   60 c   θ   120  c   θ   210  c   θ    5  6   7   β  β  β  β 




40 Displacement, mm

6.



 6  c3  24 c4 

30

20

10

0

0

45

90


180

225

270



VELOCITY, V 60

Velocity, mm

40 20 0  20  40  60

0

45

90

135

180

225

270


ACCELERATION, A

Acceleration, mm

100

50

0

 50

 100

0

45

90

135

180

225

270

180

225

270


JERK, J 600

Jerk, mm

400 200 0  200  400

0

45

90






Given:


1




2.


2  π rad

τ

ω  1.571

rad sec


τ

s1( θ )  h cv 2.

θ β1

v1( θ )  vcv

a 1( θ )  0  mm

j1 ( θ )  0  mm


 = 1:

s = h cv, v = vcv

 = 360 deg

s = 0,

a =0

v = vcv, a = 0


3.

Total interval:

β  360  deg

CV interval:

β1  180 deg

A 

β1 β

A  0.500


s = h cv, v = vcv

a =0 2

3

4


1  2 3 4  c1  2  c2 A  3  c3 A  4  c4 A  5  c5 A  β  2

0 = 2  c2  6  c3 A  12 c4 A  20 c5 A For  = :

s = 0,

v = vcv, a = 0

0 = c0  c1  c2  c3  c4  c5 0=

1

β

5

  c 1  2  c 2  3  c 3  4  c 4  5  c 5

0 = 2  c2  6  c3  12 c4  20 c5

3


4.



1   0 C   0  1 0  0

2

A A

A

3

1 2 A 3 A

A 2

4

4 A

2

6 A

12 A

1

1

1

1

1

2

3

4

0

2

6

12

2

5

 hcv     β  vcv   0  H     0   β  vcv     0  5

c0  3.720  10 mm 6

 c0   c1  c   2   C 1 H  c3     c4   c5  5

c1  2.868  10 mm

c2  8.640  10 mm

5

c3  1.248  10 mm 5.

3

0

4

  4  5 A  3 20 A  1  5   20  A

5

c4  8.64  10 mm

c5  2.304  10 mm


3

4

θ θ θ θ θ   c2    c3    c4    c5   β β β β         β

s2( θ )  c0  c1 

v2( θ ) 

a 2( θ ) 

β  1 2

1

β 4.

2 3 4 θ θ θ θ   3  c3    4  c4    5  c5    β β β β 



 c1  2  c2 

1

β j2 ( θ ) 

3



2 3 θ  θ   20 c   θ    12  c   4  5  β β β 

 2  c2  6  c3 





2 θ θ   60 c5    β β 

 6  c3  24 c4 




6.

5




DISPLACEMENT, S 800

Displacement, mm

600 400 200 0  200

0

60

120

180

240

300

360

240

300

360

240

300

360


VELOCITY, V 200

Velocity, mm

0

 200

 400

 600

0

60

120


ACCELERATION, A

Acceleration, mm

1000

500

0

 500

 1000

0

60

120

180




JERK, J 2000

Jerk, mm

1000

0

 1000

 2000

 3000

0

60

120

180


240

300

360




Input:

Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot svaj diagrams for the family of SCCA cam functions for any specified values of lift and duration. It should allow the user to change the values of the SCCA parameters b, c, d, and Ca to generate and plot any member of the family. Test it using cycloidal motion with a lift of 100 mm in 100 deg, a dwell of 80 deg, return to zero in 120 deg, and dwell for the remainder of the cycle at 1 rad/sec. RISE

DWELL

β1  100  deg

FALL

β2  80 deg

DWELL

β3  120  deg

Total rise: h  100  mm

β4  60 deg ω  1  rad sec

Cam rotational velocity:

1

The numerical constants for the SCCA equations are given in Table 8-2 and listed below. Enter values of b, c, d, and Ca for the motion type that you desire. b

c

d

0.25 0.25 0.50

0.50 0.00 0.00

0.25 0.75 0.50

Function Modified trapezoidal Modified sine Cycloidal

b  0.50

Enter values: Program: 1.

Ca 4.8881 5.5280 6.2832

c  0.00

d  0.50

Ca  6.2832



 b  b  2  π      sin  x  b   π  π

y1 ( x)  Ca x

y' 1 ( x)  Ca

 π  x  b 

y''1 ( x)  Ca sin

 π  b

y'''1 ( x)  Ca

 π  x    b 

  1  cos

π b

 π  x  b 

 cos

for b/2 <= x <= (1 - d)/2

 x2 1 1 1  2 1 y2 ( x)  Ca   b      x  b    8 2 2  π 2 π    y''2 ( x)  Ca

y' 2 ( x)  Ca x  b  



1

π



1 



2 

y'''2 ( x)  0

for (1 - d)/2 <= x <= (1 + d)/2

 b c  x   π 2  

y3 ( x)  Ca 

2 2 2  d   b2  1  1   ( 1  d )   d   cos π   x  1       8 2 8 2 π d   π  π  

 b  c  d  sin π   x  1  d     2  π 2 π  d 

y' 3 ( x)  Ca 


y'''3 ( x)  Ca

1  d  π  sin   x   d 2  d 

π

d 

  



 x2  b  1  2  π 

1  1 2 2  1   x  2 d  b   2     2 4 8 π   b

y4 ( x)  Ca 

y' 4 ( x)  Ca  x 



b

1

π

b



2





y''4 ( x)  Ca

y'''4 ( x)  0

for 1 - b/2 <= x <= 1

2

b 2 d  b y5 ( x)  Ca   x  π 2 π  y' 5 ( x)  Ca

  (1  b)2  d2   b  2sin π (x  1)

2

   π

 b

y'''5 ( x)  Ca

π

  

 π   1  cos  ( x  1 ) b π    b

 π y''5 ( x)  Ca sin  ( x  1 ) b   2.

4

 π  cos  ( x  1 ) b b  


3.


4.

Write the local svaj equations for the first interval, 0 <=  <= 1. Note that each subinterval function is multiplied by the range function so that it will have nonzero values only over its subinterval. For 0 <=  <= (Rise) s1( x)  h   R x 0 

 b 1  d   y ( x)  R x  1  d  1    y1 ( x)  R x    2  2 2 2  2 2     b b  1d   1    y4 ( x)  R x 1  1  y5 ( x)   R x  2 2  2   

v1( x) 

a 1( x) 

  R x 0 

h 2

h

β1

5.

   y3 ( x)      

d

b 1  d 1d 1  y' 1 ( x)  R x    y' 2 ( x)  R x     β1 2 2 2  2 2     b b  1d   1    y' 4 ( x)  R x 1  1  y' 5 ( x)   R x  2 2  2    h

β1

j1 ( x) 

b

3

b

  R x 0 

d

   y' 3 ( x)      


  R x 0 

b

b

 b 1  d   y''' ( x)  R x  1  d  1    y'''1 ( x)  R x    2  2 2 2  2 2     b b  1d   1    y'''4 ( x)  R x 1  1  y'''5 ( x)   R x  2 2  2   

d

   y'''3 ( x)      




For 1 <=  <= 1+ 2 s2( x)  h 6.

v2( x)  0

a 2( x)  0

j2 ( x)  0


 b 1  d   y ( x)  R x  1  d  1  d   y ( x)     y1 ( x)  R x    2   3  2 2 2   2 2      1d b b 1    y4 ( x)  R x 1  1  y5 ( x)   R x   2 2 2     

s3( x)  h 1   R x 0 

  

v3( x)  

b 1  d 1  d 1  d  y' 1 ( x)  R x    y' 2 ( x)  R x    y' 3 ( x)      β3 2 2 2   2 2      1d b b    1    y' 4 ( x)  R x 1  1  y' 5 ( x)   R x   2 2 2      h

  R x 0 

h

a 3( x)  

β3

2

h

j3 ( x)  

β3

7.

b

3

b


  R x 0 

b


  R x 0 

b


8.

v4( x)  0

a 4( x)  0

j4 ( x)  0

Write the complete global equation for the displacement and plot it over one rotation of the cam, which is the sum of the four intervals defined above. Let

θ1  β1

θ2  θ1  β2

θ3  θ2  β3


S ( θ )  s1

θ  0  deg 0.5 deg  360  deg

θ4  θ3  β4



DISPLACEMENT, S

Displacement, mm

150

100

50

0

 50

0

30

60

90

120

150

180

210

240

270

300

330

360




V ( θ )  v1

VELOCITY, V

Velocity, mm

200

100

0

 100

0

30

60

90

120

150

180

210

240

270

300

330

360




A ( θ )  a 1



ACCELERATION, A 300

Acceleration, mm

150

0

 150

 300

0

30

60

90

120

150

180

210

240

270

300

330

360




J ( θ )  j1 

JERK, J 1000

Jerk, mm

500

0

 500

 1000

0

30

60

90

120

150

180

210


240

270

300

330

360




Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot the pressure angle for the cam of Problem 8-42 for any given prime circle radius and follower eccentricity. Test it using Rp = 45 mm and  = 10 mm.

Given:

Rp  45 mm

Solution:


1.

2.

ε  10 mm

The cam of Problem 8-42 used a two-segment polynomial with the following constants: c3  5609.280 mm

c4  24548.849  mm

c5  39990.867  mm

c6  28772.307  mm

c7  7721.009 mm

β  195  deg

From Problem 8-42 the s and v equations for the rise/fall segment are: 3

4

5

6

 θ   c  θ   c  θ   c  θ   c  θ   4   5  6   7   β  β  β  β  β

s θ  c3  v θ  3.

7

2 3 4 5 6  θ θ θ θ θ        3  c3    4  c4    5  c5    6  c6    7  c7    β   β  β  β  β  β 

1

Use equation 8-31d to calculate the pressure angle over the rise and fall segment. v θ  ε    s θ  Rp2  ε2   

ϕ θ  atan

Plot the pressure angle over the interval 0 <=  <= . θ  0  deg 1  deg  β

PRESSURE ANGLE 40

20 Pressure Angle, deg

4.

0

 20

 40

 60

0

50


150

200




Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot the pressure angle for the cam of Problem 8-43 for any given prime circle radius and follower eccentricity. Test it using Rp = 100 mm and  = -15 mm.

Given:

Rp  120  mm

Solution:


1.

ε  15 mm

The cam of Problem 8-43 used a two-segment polynomial with the following constants: c0  153600 mm

c1  971700 mm

c2  2430000  mm

c3  2997000  mm

c4  1822500  mm

c5  437400 mm

h cv  95.492 mm

vcv  47.746 mm

β  240  deg

β  360  deg 2.

From Problem 8-43 the s and v equations for the constant velocity rise segment are: s1 θ  h cv

3.

θ v1 θ  vcv β From Problem 8-42 the s and v equations for the second segment are: 2

3

4

 θ   c  θ   c  θ   c  θ   c  θ   2  3  4   5   β  β  β  β  β

s2 θ  c0  c1  v2 θ  4.

5

2 3 4  θ θ θ θ       c1  2  c2    3  c3    4  c4    5  c5    β   β  β  β  β 

1

To combine the s and v equations for both segments, first define a range function that has a value of one between the values of a and b and zero elsewhere. R θ a b   if  θ  a    θ  b  1 0

















s θ  R θ 0 β  s1 θ  R θ β β  s2 θ v θ  R θ 0 β  v1 θ  R θ β β  v2 θ 5.

Use equation 8-31d to calculate the pressure angle over the rise and fall segment. v θ  ε    s θ  Rp2  ε2   


4.

Plot the pressure angle over the interval 0 <=  <= . θ  0  deg 1  deg  β See next page



PRESSURE ANGLE 40

Pressure Angle, deg

20

0

 20

 40

 60

0

40

80

120

160

200


240

280

320

360




Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot the pressure angle for the rise segment of the cam of Problem 8-46 for any given prime circle radius and follower eccentricity. Test it using Rp = 75 mm and  = 20 mm.

Given:

Rp  75 mm

Solution:


1.

ε  20 mm

The cam of Problem 8-46 used a 4-5-6-7 polynomial for the rise with the following constants: h  50 mm

2.

β  75 deg

From Problem 8-46 the s and v equations for the rise segment are:



5 6 7 θ θ θ      84    70    20      β  β  β  β 

θ s θ  h  35   v θ 



h β

3.

4

3 4 5 6  θ   420   θ   420   θ   140   θ           β  β  β  β 

 140  



Use equation 8-31d to calculate the pressure angle for the rise segment. v θ  ε    s θ  Rp2  ε2   


Plot the pressure angle over the interval 0 <=  <= . θ  0  deg 1  deg  β

PRESSURE ANGLE 40

30

Pressure Angle, deg

4.

20

10

0

 10

 20

0

5

10

15

20

25

30

35

40

45


50

55

60

65

70

75




Design a cam to move a follower from 20.5 to 15 mm in 60 deg, fall an additional 15 mm in 90 deg, rise 20.5 mm in 110 deg, and dwell for the remainder. Use polynomial functions for the rise and falls. Some of the boundary conditions are given in Table P8-1, however, in order to make the polynomials piecewise continuous, other boundary conditions will have to be determined. The shaft speed is 250 rpm. Plot the s v a j diagrams.

Given:

FALL   60 deg

FALL   90 deg

RISE   110  deg

DWELL   100  deg

h 1  5.5 mm

h 2  15.0 mm

h 3  20.5 mm

h 4  20.5 mm

Shaft speed Solution: 1.

ω  250  rpm


Calculate the shaft speed in rad/sec for input into DYNACAM.

ω  26.180

rad sec

2.

Input the number of segments, which in this case is 4.

3.

Input 60 deg for the first segment and choose "poly" for the motion. Input 20.5 and 15 for the positions. Click on "Calculate". There are seven ICs to enter, all zeroes.

4.

Input 90 deg for the second segment and choose "poly" for the motion. Input 15 and 0 for the positions. Click on "Calculate". There 6 ICs to enter, again all zeroes. Click on "Plot". From the plot screen get the maximum value of the acceleration, which is 27,777.90.

5.

Input 110 deg for the third segment and choose "poly" for the motion. Input 0 and 20.5 for the positions. Click on "Calculate". There 8 ICs to enter, all zeroes except for the starting acceleration, which is 27,777.90.

6.

Input 100 deg for the last segment and choose "dwell" for the motion. Input 20.5 and 20.5 for the positions. Click on "Calculate". Click on "Next" in the command line. Plotting the results yields the diagram below.

Selected Cam-Follower Parameters

Accel (mm/s^2)

Velocity (mm/s)

Displcmnt (mm)

Displcmnt 20

Velocity

1

Accel

Jerk*E-3

2

3

4

10 0 500

0

20,000 0

Jerk*E-3 (mm/s^3)

-20,000 0 -5,000

0

30

60

90

120

150 180 210 Cam Angle (Deg)

240

270

300

330

360



PROBLEM 8-60 Design a cam to move a follower from 32 to 12 mm in 60 deg, fall an additional 12 mm in 50 deg, dwell 35 deg, rise 12 mm in 45 deg, rise an additional 20mm in 65 deg, and dwell for the remainder. Use polynomial functions for the rises and falls. Velocity and acceleration are zero at the beginning and end of each event and jerk is zero at θ = 0, 110, 145, and 255 deg. The shaft speed is 37.5 rpm. Plot the s v a j diagrams.

Statement:

Given:

FALL   60 deg

FALL   50 deg

DWELL   35 deg

RISE   45 deg

RISE   65 deg

DWELL   105  deg

h 1  20 mm

h 2  12 mm

h 3  0  mm

h 4  12 mm

h 3  20 mm

h 4  32 mm

Shaft speed

ω  37.5 rpm


Solution:

rad

1.

Calculate the shaft speed in rad/sec for input into DYNACAM. ω  3.927

2.

Input the number of segments, which in this case is 6.

3.

Input 60 deg for the first segment and choose "poly" for the motion. Input 32 and 12 for the positions. Click on "Calculate". There are 7 ICs to enter, all zeroes. Input 50 deg for the second segment and choose "poly" for the motion. Input 12 and 0 for the positions. Click on "Calculate". There 7 ICs to enter, again all zeroes. Input 35 deg for the third segment and choose "dwell" for the motion. Input 0 and 0 for the positions. Click on "Calculate". Input 45 deg for the fourth segment and choose "poly" for the motion. Input 0 and 12 for the positions. Click on "Calculate". There are 7 ICs to enter, all zeroes. Input 65 deg for the fifth segment and choose "poly" for the motion. Input 12 and 32 for the positions. Click on "Calculate". There 7 ICs to enter, again all zeroes. Input 105 deg for the last segment and choose "dwell" for the motion. Input 32 and 32 for the positions. Click on "Calculate". Click on "Next" in the command line. Plotting the results yields the diagram below.

6.

Selected Cam-Follower Parameters Displcmnt Displcmnt (mm)

7.

Velocity (mm/s)

6.

Accel (mm/s^2)

5.

Jerk*E-3 (mm/s^3)

4.

sec

30

Velocity

1

2

Accel 3

Jerk*E-3 4

5

6

20 10 0 100 0 -100 2,000 0 -2,000 100 0 -100 0

30

60

90

120

150 180 210 Cam Angle (Deg)

240

270

300

330

360




A 24-tooth gear has AGMA standard full-depth involute teeth with diametral pitch of 5. Calculate the pitch diameter, circular pitch, addendum, dedendum, tooth thickness, and clearance.

Given:

Tooth number N  24

Solution:

See Table 9-1 and Mathcad file P0901.

1.

Calculate the pitch diameter using equation 9.4c and the circular pitch using equation 9.4d. Pitch diameter

Circular pitch

2.

1

Diametral pitch p d  5  in

d 

p c 

N pd π pd

d  4.8000 in

p c  0.6283 in

Use the equations in Table 9-1 to calculate the addendum, dedendum, tooth thickness and clearance. Addendum

Dedendum

a 

1.0000

b 

1.2500

pd

pd

Tooth thickness

t  0.5 p c

Clearance

c 

0.2500 pd

a  0.2000 in

b  0.2500 in t  0.3142 in c  0.0500 in

Note: The circular tooth thickness is exactly half of the circular pitch, so the equation used above is more accurate than the one in Table 9-1. Also, all gear dimensions should be displayed to four decimal places since manufacturing tolerances for gear teeth profiles are usually expressed in ten-thousandths of an inch..





Given:


Solution:


1.


Circular pitch

2.

1

Diametral pitch p d  10 in

d 

p c 

N pd π pd

d  4.0000 in

p c  0.3142 in


Dedendum

a 

1.0000

b 

1.2500

pd

pd

Tooth thickness

t  0.5 p c

Clearance

c 

0.2500 pd

a  0.1000 in

b  0.1250 in t  0.1571 in c  0.0250 in






Given:


Solution:


1.


Circular pitch

2.

1


d 

p c 

N pd π pd

d  2.5000 in

p c  0.2618 in


Dedendum

a 

1.0000

b 

1.2500

pd

pd

Tooth thickness

t  0.5 p c

Clearance

c 

0.2500 pd

a  0.0833 in

b  0.1042 in t  0.1309 in c  0.0208 in





Using any available string, some tape, a pencil, and a drinking glass or tin can, generate and draw an involute curve on a piece of paper. With your protractor, show that all normals to the curve are tangent to the base circle.

Solution:

This is a "hands-on" student demonstration project. The result should look like Figure 9-5.




A spur gearset must have pitch diameters of 2.5 and 8 in. What is the largest standard tooth size, in terms of diametral pitch, that can be used without having any interference or undercutting and what are the number of teeth on each gear that result from using this diametral pitch? Assume that both gears are cut with a hob. a. For a 20-deg pressure angle b. For a 25-deg pressure angle

Given:

Pitch diameters: d 1  2.5 in

Solution:


d 2  8  in

1.

To avoid undercutting, use the minimum tooth numbers given in Table 9-4b.

a.

Pressure angle of 20 deg. Nmin  21

p dmin 

Nmin d1

1

p dmin  8.400  in

From Table 9-2, the smallest standard diametral pitch (largest tooth size) that can be used is 9. But, since the pinion pitch diameter is not an integer, using 9 would result in a noninteger number of teeth. 1

Therefore, we must go to the next larger (even) pitch (smaller tooth size) of p d  10 in

. The resulting

tooth numbers are: N1  p d  d 1 b.

N1  25

N2  p d  d 2

N2  80

Pressure angle of 25 deg Nmin  14

p dmin 

Nmin d1

1

p dmin  5.600  in

From Table 9-2, the smallest standard diametral pitch (largest tooth size) that can be used is 6. 1

Let p d  6  in

. The resulting tooth numbers are:

N1  p d  d 1

N1  15

N2  p d  d 2

N2  48




Design a simple, spur gear train for a ratio of -7:1 and a diametral pitch of 10. Specify pitch diameters and numbers of teeth. Calculate the contact ratio.

Given:

Gear ratio

mG  7

1

p d  10 in

Diametral pitch

Assumptions: The pinion is not cut by a hob and can, therefore, have fewer than 21 teeth (see Table 9-4b) for a 20-deg pressure angle. Design Choice: Pressure angle ϕ  20 deg Solution: 1.


From inspection of Table 9-5a, we see that 16 teeth is the least number that the pinion can have for a gear ratio of 7. therefore, let the number of teeth on the pinion be Np  16

2.

Ng  mG Np

Ng  112

Using equation 9.4c, calculate the pitch diameters of the pinion and gear. d p 

3.

and

Np pd

d p  1.6000 in

d g 

Ng pd

d g  11.2000  in

Calculate the contact ratio using equations 9.2 and 9.6b and those from Table 9-1. rp  0.5 d p a p 

1 pd

Center distance

Z 

rp  0.8000 in

rg  0.5 d g

a p  0.1000 in

a g 

C  rp  rg

C  6.4000 in

1 pd

rg  5.6000 in a g  0.1000 in

rp  ap2   rp cos ϕ 2  rg  ag 2  rg cosϕ 2  C sinϕ

Contact ratio

mp 

pd  Z

π cos ϕ

mp  1.682

Z  0.4964 in




Design a simple, spur gear train for a ratio of +6:1 and a diametral pitch of 5. Specify pitch diameters and numbers of teeth. Calculate the contact ratio.

Given:

Gear ratio

mG  6

Diametral pitch

1

p d  5  in

Assumptions: The pinion is not cut by a hob and can, therefore, have fewer than 21 teeth for a 20-deg pressure angle (see Table 9-4b). Design Choice: Pressure angle ϕ  20 deg Solution:

1.



2.

Np pd

Ng  96

d p  3.2000 in

d g 

Ng pd

d g  19.2000  in


1 pd

Center distance

Z 

rp  1.6000 in

rg  0.5 d g

a p  0.2000 in

a g 

C  rp  rg

C  11.2000  in

1 pd

rg  9.6000 in a g  0.2000 in


Contact ratio

4.

Ng  mG Np


3.

and

mp 

pd  Z

π cos ϕ

Z  0.9880 in

mp  1.673

An idler gear of any diameter is needed to get the positive ratio. If the idler is does not have the same number of teeth as the gear, the calculation of contact ratio (step 3) will not be correct.




Design a simple, spur gear train for a ratio of -7:1 and a diametral pitch of 8. Specify pitch diameters and numbers of teeth. Calculate the contact ratio.

Given:

Gear ratio

mG  7

Diametral pitch

1

p d  8  in

Assumptions: The pinion is not cut by a hob and can, therefore, have fewer than 21 teeth for a 20-deg pressure angle (see Table 9-4b). Design Choice: Pressure angle ϕ  20 deg Solution: 1.



2.

Ng  mG Np

Ng  119


3.

and

Np pd

d p  2.1250 in

d g 

Ng pd

d g  14.8750 in


1 pd

Center distance

Z 

rp  1.0625 in

rg  0.5 d g

a p  0.1250 in

a g 

C  rp  rg

C  8.5000 in

1 pd

rg  7.4375 in a g  0.1250 in


Contact ratio

mp 

pd  Z

π cos ϕ

mp  1.693

Z  0.6246 in




Design a simple, spur gear train for a ratio of +6.5:1 and a diametral pitch of 5. Specify pitch diameters and numbers of teeth. Calculate the contact ratio.

Given:

Gear ratio

mG  6.5

Diametral pitch

1

p d  5  in

Assumptions: The pinion is not cut by a hob and can, therefore, have fewer than 21 teeth for a 20-deg pressure angle (see Table 9-4b). Design Choice: Pressure angle ϕ  20 deg Solution: 1.


From inspection of Table 9-5a, we see that 17 teeth is the least number that the pinion can have for a gear ratio of 6.5. therefore, let the number of teeth on the pinion be (an even number so the gear tooth number will be an integer). Np  18

2.

Np pd

Ng  117

d p  3.6000 in

d g 

Ng pd

d g  23.4000 in


1 pd

Center distance

Z 

rp  1.8000 in

rg  0.5 d g

a p  0.2000 in

a g 

C  rp  rg

C  13.5000 in

1 pd

rg  11.7000 in a g  0.2000 in


Contact ratio

4.

Ng  mG Np


3.

and

mp 

pd  Z

π cos ϕ

Z  1.0033 in

mp  1.699

An idler gear of any diameter is needed to get the positive ratio. If the idler is does not have the same number of teeth as the gear, the calculation of contact ratio (step 3) will not be correct.




Design a compound, spur gear train for a ratio of -80:1 and diametral pitch of 12. Specify pitch diameters and numbers of teeth. Sketch the train to scale.

Given:

Gear ratio: mG  80

Solution:


1.

1

Diametral pitch: p d  12 in

Since the ratio is negative, we want to have an odd number of stages or an even number with an idler. Let the number of stages be 1

j  2 3  4

Possible number of stages j 

then

2.

Ideal, theoretical stage ratios

r( j )  mG

j

r( j )  2 3

8.944 4.309

4

2.991

Two stages would result in a stage ratio less than 10 but will require an idler, so we will use three stages. The average ratio for three stages is about 21:5. Using a pressure angle of 20 deg, let the stage ratios be Stage 1 ratio

r1 

20

Stage 2 ratio

5

r2 

20

Stage 3 ratio

5

and let the driver gears have tooth numbers of i  2 3  7

Tooth number index N  20

N  20

2

N  18

4

6

then the driven gears will have tooth numbers of N  r1 N 3

N  r2 N

2

5

N  80 d  i

d

i

in

7

i

pd



i 

Tooth numbers: N  i

2 3

1.6667 6.6667

2 3

20 80

4

1.6667

4

20

5

6.6667

5

80

6

1.5000

6

18

7

7.5000

7

90

N N N Checking the overall gear ratio:

6

N  90

5

N

i 

7

N  80

3

The pitch diameters are:

N  r3 N

4

3

5

N N N 2

4

7 6

 80.000

r3 

30 6




Design a compound, spur gear train for a ratio of 50:1 and diametral pitch of 8. Specify pitch diameters and numbers of teeth. Sketch the train to scale.

Given:

Gear ratio

Solution:


1.

mG  50

Diametral pitch

1

p d  8  in

Since the ratio is positive, we want to have an even number of stages or an odd number with an idler. Let the number of stages be 1

j  2 3  4


then

2.


r( j )  mG

j

r( j )  2 3

7.071 3.684

4

2.659

Two stages would result in a stage ratio less than 10 and about 7, so we will use two stages. The average ratio for two stages is about 50:7. Using a pressure angle of 20 deg, let the stage ratios be Stage 1 ratio

r1 

50

Stage 2 ratio

7

r2  7

and let the driver gears have tooth numbers of (note that N2 must be a multiple of 7) i  2 3  5


N  18

2

4


N  r2 N

2

5

N  150

N  126

3

5

N The pitch diameters are:

d  i

d i 

4

i

in 2 3

i

pd Tooth numbers: N  i i 

 2.6250 18.7500

2 3

4

2.2500

4

18

5

15.7500

5

126

N N Checking the overall gear ratio:

3

N N 2

5 4

 50.000

21 150




Design a compound, spur gear train for a ratio of 120:1 and diametral pitch of 5. Specify pitch diameters and numbers of teeth. Sketch the train to scale.

Given:

Gear ratio

Solution:


1.

mG  120

Diametral pitch

1

p d  5  in

Since the ratio is positive, we want to have an even number of stages or an odd number with an idler. Let the number of stages be 1

j  2 3  4


then

2.


r( j )  mG

j

r( j )  2 3

10.954 4.932

4

3.310

Two stages would result in a stage ratio greater than 10, so we will use three stages with an idler to get the required output direction. The average ratio for three stages is about 15:3. Using a pressure angle of 20 deg, let the stage ratios be Stage 1 ratio

r1  5

Stage 2 ratio

r2  6

Stage 3 ratio

and let the driver gears have tooth numbers of Tooth number index

i  2 3  8

N  18

N  18

2

N  18

4

6


N  r2 N

2

5

N  90

4

N  108

3

5

N  r3 N 8

6

N  72 8

N  18

The idler gear will have

7

N d 


i

i

pd d

i 

Tooth numbers: i

in 2 3



N 

i 

i

3.6000 18.0000

2 3

4

3.6000

4

18

5

21.6000

5

108

6

3.6000

6

18

7

3.6000

7

18

8

14.4000

8

72


3

5

N N N 2

4

8 6

 120.000

18 90

r3  4




Design a compound, spur gear train for a ratio of -250:1 and diametral pitch of 9. Specify pitch diameters and numbers of teeth. Sketch the train to scale.

Given:

Gear ratio

Solution:


1.

mG  250

Diametral pitch

1

p d  9  in

Since the ratio is negative, we want to have an odd number of stages or an even number with an idler. Let the number of stages be 1

j  2 3  4


then

2.


r( j )  mG

j

r( j )  2 3

15.811 6.300

4

3.976

Two stages would result in a stage ratio greater than 10 and will require an idler, so we will use three stages. The average ratio for three stages is about 20:3. Using a pressure angle of 20 deg, let the stage ratios be Stage 1 ratio

r1 

20

Stage 2 ratio

3

r2 

18

Stage 3 ratio

3

and let the driver gears have tooth numbers of i  2 3  7


N  18

2

N  20

4

6


N  r2 N

2

5

N  120

N  125

5

N d  i

d i 

7

N  108

3


N  r3 N

4

7

i

pd Tooth numbers:

i

in 2 3



N 

i 

i

2.0000 13.3333

2 3

4

2.0000

4

18

5

12.0000

5

108

6

2.2222

6

20

7

13.8889

7

125


3

5

N N N 2

4

7 6

 250.000

18 120

6

r3 

25 4




Design a compound, reverted, spur gear train for a ratio of 28:1 and diametral pitch of 8. Specify pitch diameters and numbers of teeth. Sketch the train to scale.

Given:

Gear ratio

Solution:


1.

mG  28

1

p d  8  in

Diametral pitch

Since the ratio is positive, we want to have an even number of stages. Let the number of stages be 1

j  2 3  4


then

2.

j

r( j )  2 3

5.292 3.037

4

2.300

Two stages would result in a stage ratio less than 10, so we will use two stages. The average ratio for two stages is about 21:4. Using a pressure angle of 25 deg, let the stage ratios be Stage 1 ratio

3.

r( j )  mG


r1  4

r2  7

Stage 2 ratio

Following the procedure of Example 9-3, N i  2 3  5 N  N = N  N = K

Tooth number index

2

Solving independently for N2 and N4,

3

N  2

4

5

K

N 

r1  1

Kmin   r1  1    r2  1 

where

and, r1 

4

N

3 2

N r2 

N

5 4

K r2  1

Kmin  40.000

By iteration, find a multiple of Kmin that will result in a minimum number of teeth on N2 and N4. K

K  3  Kmin

N 

K  120.000

N  24

2

r1  1

2

N  4

K r2  1

N  15 4

These are acceptable tooth numbers for gears with a 25-deg pressure angle that are cut by a hob. 4.

The driven gears will have tooth numbers of N  r1 N 3

N  96

2

3


d  i

N  r2 N 5

i 

i

in 2 3

Checking the overall gear ratio:

5

d

i

pd

N  105

4

 3.0000 12.0000

N  i

24 96

4

1.8750

15

5

13.1250

105

 N 3   N 5      28.000  N2   N4    





Given:

Gear ratio

Solution:


1.

mG  40

1

p d  8  in

Diametral pitch


j  2 3  4


then

2.

j

r( j )  2 3

6.325 3.420

4

2.515

Two stages would result in a stage ratio less than 10, so we will use two stages. The average ratio for two stages is about 20:3. Using a pressure angle of 25 deg, let the stage ratios be Stage 1 ratio

3.

r( j )  mG


r1  5

r2  8

Stage 2 ratio

Following the procedure of Example 9-3, N i  2 3  5

Tooth number index


N  N = N  N = Kand, 2

3

N  2

5

K

N 

r1  1

Kmin   r1  1    r2  1 

where

4

r1 

4

N

3 2

N r2 

N

5 4

K r2  1

Kmin  54.000

By iteration, find a multiple of Kmin that will result in a minimum number of teeth on N2 and N4. K

K  3  Kmin

N 

K  162.000

N  27

2

r1  1

2

N  4

K r2  1

N  18 4



N  135

2

3


d  i

N  r2 N 5

i 

i

in 2 3


5

d

i

pd

N  144

4

 3.3750 16.8750

N  i

27 135

4

2.2500

18

5

18.0000

144

 N 3   N 5      40.000  N2   N4    





Given:

Gear ratio

Solution:


1.

mG  65

1

p d  8  in

Diametral pitch


j  2 3  4


then

2.

j

r( j )  2 3

8.062 4.021

4

2.839

Two stages would result in a stage ratio less than 10, so we will use two stages. Using a pressure angle of 25 deg, let the stage ratios be Stage 1 ratio

3.

r( j )  mG


r1  6.5

r2  10

Stage 2 ratio

Following the procedure of Example 9-3, N i  2 3  5 N  N = N  N = K

Tooth number index

2


3

N  2

Kmin   r1  1    r2  1 

where

4

and, r1 

5

K

N 

r1  1

4

N

3 2

N r2 

N

5 4

K r2  1

Kmin  82.500

By iteration, find a multiple of Kmin that will result in a minimum, integer number of teeth on N2 and N4. K K K  2  Kmin N  N  2 4 r1  1 r2  1 K  165.000

N  22.0 2

N  15.0 4



N  143

2

3


d  i

N  r2 N 5

i 

i

in 2 3


5

d

i

pd

N  150

4

 2.7500 17.8750

N  i

22 143

4

1.8750

15

5

18.7500

150

 N 3   N 5      65.000  N2   N4    





Given:

Gear ratio

Solution:


mG  7

1

p d  4  in

Diametral pitch

1.

Since the ratio is positive, we want to have an even number of stages. Let the number of stages be 2.

2.

Using a pressure angle of 25 deg, let the stage ratios be Stage 1 ratio

3.

r1  3.5

r2  2

Stage 2 ratio


Tooth number index


N  N = N  N = Kand, 2

3

N  2

Kmin   r1  1    r2  1 

where

4

r1 

5

K

N 

r1  1

4

N

3 2

N r2 

N

5 4

K r2  1

Kmin  13.500


N  18 2

N  27 4



N  63

2

3


d  i

N  r2 N 5

i 

i

in 2 3


5

d

i

pd

N  54

4



N  i

4.5000 15.7500

18 63

4

6.7500

27

5

13.5000

54

 N 3   N 5      7.000  N2   N4    





Given:

Gear ratio

Solution:


mG  12

1

p d  6  in

Diametral pitch

1.

Since the ratio is positive, we want to have an even number of stages. Let the number of stages be 2.

2.


3.

r1  4

r2  3

Stage 2 ratio


Tooth number index


N  N = N  N = Kand, 2

3

N  2

Kmin   r1  1    r2  1 

where

4

r1 

5

K

N 

r1  1

4

N

3 2

N r2 

N

5 4

K r2  1

Kmin  20.000


N  16 2

N  20 4



N  64

2

3


d  i

N  r2 N 5

i 

i

in 2 3


5

d

i

pd

N  60

4



N  i

2.6667 10.6667

16 64

4

3.3333

20

5

10.0000

60

 N 3   N 5      12.000  N2   N4    




Design a compound, reverted, spur gear transmission that will give two shiftable ratios of +3:1 forward and -4.5:1 reverse with diametral pitch of 6. Specify pitch diameters and numbers of teeth. Sketch the train to scale.

Given:

Gear ratio

Solution:


mG  3

1

p d  6  in

Diametral pitch

1.

Since the forward ratio is positive, we want to have an even number of stages. Let the number of stages be 2.

2.


3.

r1  2

r2  1.5

Stage 2 ratio


Tooth number index


N  N = N  N = Kand, 2

3

N  2

Kmin   r1  1    r2  1 

where

4

r1 

5

K

N 

r1  1

4

N

3 2

N r2 

N

5 4

K r2  1

Kmin  7.500


N  15

N  18

2

4


The driven gears for the forward train will have tooth numbers of N  r1 N 3

N  30

2


N  r2 N

3

d  i

5

i 


 N 3   N 5      3.000  N2   N4     5.

i

in



N  i

2 3

2.5000 5.0000

15 30

4

3.0000

18

5

4.5000

27

The reverse train will also have two stages and use the first forward stage and an idler gear to get the change in direction. Let the stage ratios be Stage 1 ratio

6.

5

d

i

pd

N  27

4

r1  2

Stage 2 ratio

r2  2.25

Let the number of teeth on the reverse stage driver gear be N6  12 then the number of teeth on the driven gear will be Driven reverse gear

N7  r2 N6

N7  27




Design a compound, reverted, spur gear transmission that will give two shiftable ratios of +5:1 forward and -3.5:1 reverse with diametral pitch of 6. Specify pitch diameters and numbers of teeth. Sketch the train to scale.

Given:

Gear ratio

Solution:


mG  5

1

p d  6  in

Diametral pitch

1.


2.


3.

r1  2

r2  2.5

Stage 2 ratio


Tooth number index


N  N = N  N = Kand, 2

3

N  2

Kmin   r1  1    r2  1 

where

4

r1 

5

K

N 

r1  1

4

N

3 2

N r2 

N

5 4

K r2  1

Kmin  10.500


N  14

N  12

2

4



N  28

2


N  r2 N

3

d  i

5

i 


 N 3   N 5      5.000  N2   N4     5.

i

in



N  i

2 3

2.3333 4.6667

14 28

4

2.0000

12

5

5.0000

30

The reverse train will also have two stages and use the first forward stage and an idler gear to get the change in direction. Let the stage ratios be Stage 1 ratio

6.

5

d

i

pd

N  30

4

r1  2

Stage 2 ratio

r2  1.75


N7  r2 N6

N7  21




Design a compound, reverted, spur gear transmission that will give three shiftable ratios of +6:1, +3.5:1 forward and -4:1 reverse with diametral pitch of 8. Specify pitch diameters and numbers of teeth. Sketch the train to scale.

Given:

Gear ratio

Solution:


mG  6

1

p d  8  in

Diametral pitch

1.


2.

Using a pressure angle of 25 deg, let the stage ratios be 7 Stage 2 ratio Stage 1 ratio r1  3

3.

r2 

18 7


Tooth number index

N  N = N  N = Kand, 2


3

N  2

Kmin   r1  1    r2  1 

where

4

r1 

5

K

N 

r1  1

4

N

3 2

N r2 

N

5 4

K r2  1

Kmin  11.905

By iteration, find a multiple of Kmin that will result in a minimum, integer number of teeth on N2 and N4. K K K  4.2 Kmin N  N  2 4 r1  1 r2  1 K  50.000

N  15 2

N  14 4



N  35

2

3


d  i

N  r2 N 5


 N 3   N 5      6.000  N2   N4    

5

d

i

pd

N  36

4

i 

i

in



N  i

2 3

1.8750 4.3750

15 35

4

1.7500

14

5

4.5000

36

5.

The second forward train will also have two stages and use the first forward stage to get the overall ratio. Let the stage ratios be 7 Stage 1 ratio Stage 2 ratio r1  r2  1.5 3

6.

Let the number of teeth on the second forward stage driver gear be N6  20 then the number of teeth on the driven gear will be Second forward driven gear

N7  r2 N6

N7  30



7.

The reverse train will also have two stages and use the first forward stage and an idler gear to get the change in direction. Let the stage ratios be 7 12 Stage 1 ratio Stage 2 ratio r1  r2  3 7

8.


N9  r2 N8

N9  24




Design a compound, reverted, spur gear transmission that will give three shiftable ratios of +4.5:1, +2.5:1 forward and -3.5:1 reverse with diametral pitch of 5. Specify pitch diameters and numbers of teeth. Sketch the train to scale.

Given:

Gear ratio

Solution:


mG  4.5

1

p d  5  in

Diametral pitch

1.


2.

Using a pressure angle of 25 deg, let the stage ratios be 9 Stage 2 ratio Stage 1 ratio r1  6

3.

r2 

mG

r2  3.000

r1


Tooth number index

N  N = N  N = Kand, 2


3

N  2

Kmin   r1  1    r2  1 

where

4

r1 

5

K

N 

r1  1

4

N

3 2

N r2 

N

5 4

K r2  1

Kmin  10.000

By iteration, find a multiple of Kmin that will result in a minimum, integer number of teeth on N2 and N4. K K K  12 Kmin N  N  2 4 r1  1 r2  1 K  120.000

N  48 2

N  30 4



N  72

2

3


d  i

N  r2 N 5


 N 3   N 5      4.500  N2   N4    

5

d

i

pd

N  90

4

i 

i

in 2 3



N  i

9.6000 14.4000

48 72

4

6.0000

30

5

18.0000

90

5.

The second forward train will also have two stages and use the first forward stage to get the overall ratio. Let the stage ratios be 2.5 Stage 1 ratio Stage 2 ratio r1  1.500 r2  r2  1.667 r1

6.

Let the number of teeth on the second forward stage driver gear be N6  45 then the number of teeth on the driven gear will be Second forward driven gear

N7  r2 N6

N7  75



7.

The reverse train will also have two stages and use the first forward stage and an idler gear to get the change in direction. Let the stage ratios be 3.5 Stage 1 ratio Stage 2 ratio r1  1.500 r2  r2  2.333 r1

8.

Let the number of teeth on the reverse stage driver gear be N8  12 then the number of teeth on the driven gear will be

Driven reverse gear

N9  r2 N8

N9  28




Design the rolling cones for a -3:1 ratio and a 60-deg included angle between shafts. Sketch the train to scale.

Given:

Train ratio

Solution:


1.

Choose a back-cone pitch diameter for the input cone of 1.000 in. Then, from equation 9.1, d in  1.000  in

2.

mT  3

d out  mT  d in

d out  3.000 in

Using these back-cone pitch diameters and the given shaft included angle, draw the cones to scale.

Output

3.000 60.000°

1.000

Input




Design the rolling cones for a -4.5:1 ratio and a 40-deg included angle between shafts. Sketch the train to scale.

Given:

Train ratio

Solution:


1.

Choose a back-cone pitch diameter for the input cone of 1.000 in. Then, from equation 9.1, d in  1.000  in

2.

mT  4.5

d out  mT  d in

d out  4.500 in

Using these back-cone pitch diameters and the given shaft included angle, draw the cones to scale.

Output

4.500

40.000°

1.000




Figure P9-1 shows a compound planetary gear train (not to scale). Table P9-1 gives data for gear numbers of teeth and input velocities. For the data in row a, find 2.

Given:

Tooth numbers: N2  30 Input velocities:

Solution: 1.

N4  45

ω  20

ωarm  50

N5  50

N6  200


The formula method will be used in this solution. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be 2 and last be 6. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm ωFarm 2.

N3  25

=

ω  ωarm ω  ωarm

=R

Calculate R using equation 9.14 and inspection of Figure P9-1.

R 

N 2 N 3 N 5

R  0.08333

N 4 N 5 N 6

Note that N5 appears both in the numerator (a driver) and the denominator (driven). It could be left out completely because it is an idler. R is positive in this case because gears 2 and 6 rotate in the same direction. 3.

Solve the right-hand equation in step 1 for 2. ω 

ω  ωarm R

 ωarm

2 will be in the same direction as 6.

ω  790.000




Figure P9-2 shows a compound planetary gear train (not to scale). Table P9-2 gives data for gear numbers of teeth and input velocities. For the data in row a, find 2.

Given:

Tooth numbers: N2  50 Input velocities:

Solution: 1.

N4  45

ω  20

ωarm  50

N5  30

The formula method will be used in this solution. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be 3 and last be 6. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωFarm

=

ω  ωarm ω  ωarm

=R

Calculate R using equation 9.14 and inspection of Figure P9-2.

R 

N 3 N 5

R  0.4167

N 4 N 6

R is positive in this case because gears 3 and 6 rotate in the same direction. 3.

Solve the right-hand equation in step 1 for 3. ω 

4.

N6  40


ωLarm

2.

N3  25

ω  ωarm R

 ωarm

ω  118.000

Solve for 2 using equation 9.7. ω  

N3 N2

 ω

2 will be in the opposite direction as 6.

ω  59.000




Figure P9-3 shows a planetary gear train used in an automotive rear-end differential (not to scale). The car has wheels with a 16-in rolling radius and is moving forward in a straight line at 55 mph. The engine is turning 2500 rpm. The transmission is in direct drive (1:1) with the driveshaft. a. What is the rear wheel's rpm and the gear ratio between ring and pinion? b. As the car hits a patch of ice, the right wheel speeds up to 800 rpm. What is the speed of the left wheel? Hint: The average of both wheel's rpm is a constant. c. Calculate the fundamental train value of the epicyclic stage.

Given:

r  16 in

Solution:


1.

ωeng  2500 rpm

Calculate the rear wheel rpm and the gear ratio between ring and pinion.

ωwheel 

mG 

2.

V  55 mph

V r

ωeng ωwheel

ωwheel  577.732  rpm

mG  4.327

Using the hint given, calculate the speed of the left wheel when the right wheel hits the ice.

ωavg  ωwheel ωleft  2  ωavg  ωright

ωright  800  rpm ωleft  355.465  rpm




Design a speed-reducing planetary gearbox to be used to lift a 5-ton load 50 ft with a motor that develops 20 lb-ft of torque at its operating speed of 1750 rpm. The available winch drum has no more than a 16-in diameter when full of its steel cable. The speed reducer should be no larger in diameter than the winch drum. Gears of no more than 75 teeth are desired, and diametral pitch needs to be no larger than 6 to stand the stresses. Make multiview sketches of your design and show all calculations. How long will it take to raise the load with your design?

Units:


1

Given: Load

W  10000  lbf

Motor torque

Tm  20 ft  lbf

Maximum drum diameter

Motor speed

ωm  1750 rpm

d max  16 in

Assumptions: Minimum drum diameter (with no cable) is d min  8  in Solution: 1.


Determine the minimum gear ratio required. Worst load torque:

TL  W 

d max

TL  6667 ft  lbf

2

Minimum gear ratio required to lift load: TL

mG 

mG  333.333

Tm

2.

Use a combination of basic planetary gearsets, such as that shown in Figure 9-33, compounded in the following way. The input to each stage will be the sun gear. The output will be the arm, and the ring gear will be stationary. All stages will mesh with the same ring gear, which will be elongated to reach the planets of all stages. Except for the last stage, the arm of one stage will be coupled directly to the sun gear of the next stage. All stages will have the same number of teeth on the sun gears and the same (different from the sun gears) for the planets. Since they all mesh with the same ring gear, they must all have the same diametral pitch and pressure angle. Let the pressure angle be 25 deg.

3.

Calculate the number of stages required to achieve the required gear ratio. 1

Let the diametral pitch be

p d  4  in

Ring gear pitch diameter

d r  16 in

Nr  p d  d r

Number of teeth on each sun gear

Ns  12

Number of teeth on each planet gear

Np 

Nr  Ns 2

Nr  64

Np  26

For a single stage, the gear ratio for the planetary configuration of Figure 9-33 is -(Nr - Ns)/Ns. For multiple stages,

 Nr  Ns  mG     Ns 

n



where n is the number of stages. Solving for n,

 log mG    Nr  Ns    log Ns     

n  ceil

4.

stages

Calculate the actual gear ratio achieved.

 Nr  Ns  mG      Ns  5.

n4

n

mG  352.605

Calculate the time required to raise the load L  50 ft . Average drum diameter

Average cable velocity

Average time to raise load

d avg 

Vavg 

tavg 

1 2

  d max  d min

d avg ωm  2 mG L Vavg

d avg  12.000 in

Vavg  15.592

tavg  3.2 min

ft min




Determine all possible two-stage compound gear combinations that will give an approximation to the Naperian base 2.71828. Limit tooth numbers to between 18 and 80. Determine the arrangement that gives the smallest error.

Given: ratio  2.71828 Solution: 1.

Nmax  80

err  0.001%


Input the given data into the TK-Solver file compound.tkw. The n  8 tooth number combinations given below are the result. Let i  1 2  n N2 

N3 

N4 

N5 

25 29 30 30 31 31 31 35

67 57 32 64 48 64 79 67

70 47 31 62 45 60 75 50

71 65 79 79 79 79 80 71

i

2.

Nmin  18

i

i

i

least ( r n ) 

low  r

1

index  1 for i  2  n index  i if r  low i

low  r if r  low i

index

Determine the stage ratios and the overall ratio for each set. N3 Stage 1 ratio

ratio1  i

N5

i

ratio2 

Stage 2 ratio

i

N2

i

3.

Ratio  ratio1  ratio2

Error in ratio:

Error  ratio  Ratio

i

i

i

i i

Find the set with the least error.

N2  30 R

R3

N3  32 R

N4  31 R

N5  79 R

ratio1  1.06667

ratio2  2.54839

Ratio  2.718280

Error  4.30108  10

R

R

N4

i

Actual ratio for given set:

R  least ( Error n )

i

R

7

R

i




Determine all possible two-stage compound gear combinations that will give an approximation to 2. Limit tooth numbers to between 18 and 80. Determine the arrangement that gives the smallest error.


Nmin  15

Nmax  90

err  0.001%



N3 

N4 

N5 

16 23 25 28

63 75 51 85

47 41 25 43

75 79 77 89

i

i

i

least ( r n ) 

i

low  r

1



index 2.


ratio1  i

N5

i

ratio2 

Stage 2 ratio

i

N2

i

3.

i



Error in ratio:


i

i

i

i i

Find the set with the least error. R  least ( Error n ) N2  25 R

R3

N3  51 R

N4  25 R

N5  77 R

ratio1  2.04000

ratio2  3.08000

Ratio  6.283200

Error  1.50000  10

R

R

i

N4

R

5

R

i




Determine all possible two-stage compound gear combinations that will give an approximation to 3/2. Limit tooth numbers to between 18 and 80. Determine the arrangement that gives the smallest error.


Nmax  100

err  0.001%



N3 

N4 

N5 

22 22 22 25 29 32 33 33 36 37 38 41 41 43 43 43 44 44 45 46 47 47 50 55 55 56 57 58 61

31 34 49 51 44 63 34 51 55 59 49 57 75 50 55 58 62 67 64 75 63 65 51 62 67 85 77 88 68

61 61 95 100 85 94 61 61 71 67 55 77 92 57 57 79 61 95 67 82 64 81 50 61 76 86 86 85 66

68 62 67 77 88 75 93 62 73 66 67 87 79 77 70 92 68 98 74 79 75 92 77 85 98 89 100 88 93

i

2.

Nmin  20

i

i

least ( r n ) 

i

low  r

1



index


ratio1  i

N5

i

Stage 2 ratio

N2

i


ratio2  i

i

N4

i

Ratio  ratio1  ratio2 i

i

i

i




Error in ratio: 3.

i

i


R  12

N3  57 R

N4  77 R

N5  87 R

ratio1  1.39024

ratio2  1.12987

Ratio  1.570795

Error  9.41400  10

R

R

R

7

R




Determine all possible two-stage compound gear combinations that will give an approximation to 3. Limit tooth numbers to between 18 and 80. Determine the arrangement that gives the smallest error.


Nmin  20

Nmax  100

err  0.001%



N3 

N4 

N5 

22 27 37 37 38

68 65 68 80 77

61 47 39 39 43

93 92 100 85 100

i

i

i

least ( r n ) 

i

low  r

1



index 2.


ratio1  i

N5

i

ratio2 

Stage 2 ratio

i

N2

i

3.

i



Error in ratio:


i

i

i

i i


R4

N3  80 R

N4  39 R

N5  85 R

ratio1  2.16216

ratio2  2.17949

Ratio  4.712405

Error  1.47124  10

R

R

i

N4

R

5

R

i




Figure P-9-4a shows a reverted clock train. Design it using 25-deg nominal pressure angle gears of 24 p d having between 12 and 150 teeth. Determine the tooth numbers and nominal center distance. If the center distance has a manufacturing tolerance of plus or minus 0.006 in, what will the pressure angle and the backlash at the minute hand be at each extreme of tolerance?

Given:


Solution:


1.

1

Tolerance (one side) t  0.006  in

Write the speeds of the minute and hour hands and their ratio. Minute hand

ω 

2  π rad

ratio 

ω 

Hour hand

hr ω

2  π rad 12 hr

ratio  12.000

ω Assume some input speed in to the shaft with gears B and C on it. Then, ω  NB NA 2.

=

NB NA N3 N2

 ωin

NE

= r1

NC

NC NE

=

N5 N4

ω

 ωin

=

ω

NB NE  = ratio NA NC

ratio  r1 r2

= r2

Let r1 and r2 be factors of ratio. Say, Minute ratio

3.

ω 

r1  3

r2  4

Hour ratio

Following the procedure of Example 9-3,

N

i  2 3  5

Tooth number index


N  N = N  N = Kand, 2

3

N  2

K r1  1

Kmin   r1  1    r2  1 

where

4

r1 

5

N  4

N

3 2

N r2 

N

5 4

K r2  1

Kmin  20.000

By iteration, find a multiple of Kmin that will result in a minimum number of teeth on N2 and N4.

4.

K

K  3  Kmin

N 

K  60.000

N  15

2

r1  1

2

N  4

K r2  1

N  12 4

The other gears will have tooth numbers of N  r1 N 3

2

N  45 3

N  r2 N 5

4

N  48 5



N d 


i

d

i

i 

pd



C  0.5 d  d



i



in

N  i

2 3

0.6250 1.8750

15 45

4

0.5000

12

5

2.0000

48

A B C E

C  1.250 in

5.

The nominal center distance is

6.

Calculate the extreme values of center distance and their ratios with respect to the nominal center distance.

7.

3

Cmax  C  t

Cmax  1.2560 in

rmax 

Cmin  C  t

Cmin  1.2440 in

rmin 

Cmax C Cmin C

rmax  1.0048 rmin  0.9952

Using the equation in step 11 of Example 9-1, calculate the pressure angle that would result if the center distance were at its extreme values.

ϕmax  acos

cos( 25 deg) 

ϕmin  acos

cos( 25 deg) 



 

rmax



8.

2

 

rmin

ϕmax  25.581 deg

ϕmin  24.401 deg

The backlash in the minute hand is found using equation 9.3,

θBmax  43200  t

tan  ϕmax π d

θBmax  63

Minutes of arc

2

Depending on the nominal clearance in the gearset, decreasing the center distance will eventually cause the teeth to jamb and the gears may not be capable of assembly.




Figure P9-4b shows a three-speed shiftable transmission. Shaft F, with the cluster of gears E, G, and H, is capable of sliding left and right to engage and disengage the three gearsets in turn. Design the three reverted stages to give output speeds at shaft F of 150, 350, and 550 rpm for an input speed of 450 rpm on shaft D.

Units:


Given:

Output speeds:

1

ωF1  150  rpm

1.


Determine the gearset ratios r1 

ωD

r2 

ωF1

r1  3.0000 2.

ωF3  550  rpm

ωD  450  rpm

Input speed: Solution:

ωF2  350  rpm

ωD

r3 

ωF2

r2  1.2857

ωD ωF3

r3  0.8182

Following the procedure of Example 9-3, N N

A

N

E

=N N B

G

=N

C

N

H

= and, K

r1 

Solving independently for NA , NC and NB, N  A

N

E A

K r1  1

Kmin   r1  1    r2  1    r3  1 

where

N r2 

N

N

H

r3 

C

K

N 

r2  1

C

Kmin  1.000

G

N

B

N  B

K r3  1

1280 77

By iteration, find a multiple of Kmin that will result in a minimum, integer number of teeth on the drivers: K 

77 16

 Kmin

K  80.000

NA 

K r1  1

NA  20

NC 

K r2  1

NC  35

NB 

K r3  1

NB  44

These are acceptable tooth numbers for gears with a 25- or 20-deg pressure angle that are cut by a hob. 3.

The driven gears on shaft F will have tooth numbers of NE  r1 NA

NE  60

NG  r3 NB

NG  36

NH  r2 NC

NH  45




Figure P9-5a shows an epicyclic train used to drive a winch drum. Gear A, on shaft 1, is driven at 18 rpm CCW and gear D, on shaft 2, is fixed to ground. The tooth numbers are indicated in the figure. Determine the speed and direction of the drum. What is the efficiency of this train if the basic gearsets have E0 = 0.97?

Given:

Tooth numbers: NA  72

NB  16

ωA  18 rpm

Input speeds:

Basic gearset efficiency: Solution: 1.

NC  48

ND  40

ωD  0  rpm

E0  0.97


Determine the speed of the arm using the formula method for analyzing an epicyclic train. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be A and last be D. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm

=

ωFarm

ωD  ωarm ωA  ωarm

=R

Calculate R using equation 9.14 and inspection of Figure P9-5a.

 NA   NC      NB   ND 

R   

R  5.40000

Solve the right-hand equation above for arm with D = 0.

ωarm 

R R1

 ωA

ωarm  22.09  rpm

The arm drives the drum, so

ωdrum  ωarm 2.

Find the basic ratio  for the train using equation 9.15. ρ  R

3.

ωdrum  22.1 rpm

ρ  5.400

The combination of  > 1, shaft 2 fixed, and input to shaft 1 corresponds to Case 1 in Table 9-12 giving an efficiency of η 

ρ E0  1 ρ 1

η  0.963




Figure P9-5b shows an epicyclic train. The arm is driven CCW at 20 rpm. Gear A is driven at 40 rpm CW. The tooth numbers are indicated in the figure. Find the speed of the ring gear D.

Units:


Given:

Tooth numbers:

1

NA  17 Input speeds: Solution: 1.

NB  28

NC  15

ωA  40 rpm

ND  60

ωarm  20 rpm


Determine the speed of the ring gear using the formula method for analyzing an epicyclic train. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be A and last be D. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm ωFarm

=


=R

Calculate R using equation 9.14 and inspection of Figure P9-5b.

 NA   NC      NB   ND 

R   

R  0.15179

Solve the right-hand equation above for D .

ωD  R  ωA  ωarm  ωarm

ωD  29.11 rpm




Figure P9-6a shows an epicyclic train. The arm is driven at 50 rpm CCW and gear A, on shaft 1, is fixed to ground. The tooth numbers are indicated in the figure. Find the speed of gear D, on shaft 2. What is the efficiency of this train if the basic gearsets have E0 = 0.96?

Given:

Tooth numbers: NA  108 Input speeds:

NB  27

ωarm  50 rpm


NC  100

ND  35

ωA  0  rpm

E0  0.96


Determine the speed of the sun gear using the formula method for analyzing an epicyclic train. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be A and last be D. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm ωFarm

=


=R


 NA   NC      NB   ND 

R   

R  11.42857

Solve the right-hand equation above for D with A = 0.

ωD  ( 1  R)  ωarm 2.


3.

ωD  521.43 rpm

ρ  11.429

The combination of  > 1, shaft 1 fixed, and input to the arm corresponds to Case 4 in Table 9-12 giving an efficiency of

η 

ρ1 ρ  E0

η  0.996




Figure P9-6b shows a differential. Gear A is driven at 10 rpm CCW and gear B is driven CW at 24 rpm. The tooth numbers are indicated in the figure. Find the speed of gear D.

Units:


Given:

Tooth numbers:

1

NA'  18 Input speeds: Solution: 1.

NB'  18

NC  18

ωA  10 rpm

ND  30

ωB  24 rpm


Determine the speed of the sun gear using the formula method for analyzing an epicyclic train. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be A' and last be B'. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm ωFarm

=

ωB'  ωarm ωA'  ωarm

=R


 NA'    NB' 

R   

R  1.00000


ωA'  ωA ωarm 

ωB'  ωB

R ωA'  ωB' R1

ωarm  7.000 rpm

Gear D is attached to the arm shaft so,

ωD  ωarm

ωD  7.00 rpm




Figure P9-7a shows a gear train containing both compound-reverted and epicyclic stages. The tooth numbers are indicated in the figure. The motor is driven CW at 1500 rpm. Find the speeds of shafts 1 and 2.

Given:


NB  40

NC  20

ND  35

NE  48

NF  12

NG  72

NH  18

NJ  42

NK  15

Motor speed: Solution:

ωm  1500 rpm

CW


1.

The motor furnishes two inputs to the epicyclic train, which is composed of gears E (sun), F (planets), and G (ring). One input is to the sun gear through the reverted train ABCD. The other is directly from the motor to the arm.

2.

Determine the speed of the shaft containing gears D and E using equations 9.8.

 NA   NC        ωm  NB   ND 

3.

ωD   

ωD  321.429  rpm

ωE  ωD

ωE  321.429  rpm

Determine the speed of the ring gear G using the formula method for analyzing an epicyclic train. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be E and last be G. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm ωFarm

=

ωG  ωarm ωE  ωarm

ωarm  ωm

=R

Calculate R using equation 9.14 and inspection of Figure P9-7a. R  

NE

R  0.66667

NG

Solve the right-hand equation above for G.

ωG  R  ωE  ωarm  ωarm

4.

ωG  2285.71  rpm

Gears G and H rotate at the same speed and gear H drives the two output shafts. Calculate their speed using equation 9.7. ω  

ω  

NH NJ NH NK

 ωG

ω  979.6  rpm

 ωG

ω  2742.9 rpm




Figure P9-35b shows (schematically) a compound epicyclic train. The tooth numbers are 50, 25, 35, and 90 for gears 2, 3, 4, and 5, respectively. The arm is driven at 180 rpm CW and gear 5 is fixed to ground. Determine the speed and direction of gear 2. What is the efficiency of this train if the basic gearsets have E0 = 0.98?

Units:


Given:

Tooth numbers:

1

N2  50 Input speeds:

N3  25

ωarm  180  rpm


N4  35

N5  90

ω5  0  rpm

E0  0.98


Determine the speed of the sun gear 2 using the formula method for analyzing an epicyclic train. To start, choose a first and last gears that mesh with gears that have planetary motion. Let the first gear be 2 and last be 5. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm ωFarm

=

ω5  ωarm ω2  ωarm

=R


 N2   N4      N3   N5 

R   

R  0.77778

Solve the right-hand equation above for  with  = 0.

ω2 

R1 R

 ωarm

ω2  411.43 rpm

2.

Find the basic ratio  for the train using equation 9.15. 1 ρ  ρ  1.286 R

3.

The combination of  < -1, shaft 1 fixed, and input to the arm corresponds to Case 8 in Table 9-12 giving an efficiency of

η 

E0  ρ  1  ρ E0  1

η  0.991




Figure P9-8 shows a compound epicyclic train. Gear 2 is driven at 600 rpm CW and gear D is fixed to ground. The tooth numbers are indicated in the figure. Determine the speed and direction of gears 1 and 3.

Given:


NB  30

ω  600  rpm

Gear 2 speed: Solution: 1.

NC  72

ND  60

NE  120

NF  100

CW


Determine the speed of the arm using the formula method for analyzing an epicyclic train. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be A and last be D. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm

ωD  ωarm

=

ωFarm

ωA  ωarm

ωA  ω

=R

Calculate R using equation 9.14 and inspection of Figure P9-8. R  

 NC   NB  ND 

NA



R  0.96000

Solve the right-hand equation above for arm with D = 0.

ωarm 

R R1

 ωA

ωarm  293.88 rpm

The arm drives gear 1, so ω  ωarm 2.

ω  293.9  rpm

Determine the speed of gear F using the formula method for analyzing an epicyclic train. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be A and last be F. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm ωFarm

=

ωF  ωarm ωA  ωarm

=R

Calculate R using equation 9.14 and inspection of Figure P9-1. R 

 NA   NE   N   N   B  F 

R  0.96000

Solve the right-hand equation above for F .

ωF  R  ωA  ωarm  ωarm

ωF  587.755  rpm

Gear F drives gear 3, so ω  ωF

ω  587.8  rpm




Figure P9-9a shows a compound epicyclic train. Shaft 1 is driven at 300 rpm CCW and gear A is fixed to ground. The tooth numbers are indicated in the figure. Determine the speed and direction of shaft 2.

Units:


Given:


NB  18

Shaft 1 speed: Solution: 1.

1

NC  48

ω  300  rpm

ND  26

NE  60

NF  18

NG  68

CCW


Shaft 1 drives arm-1, the first stage arm, and arm-2, the second stage arm. The first stage is composed of gears A, B, C, and D, with gear A fixed. The second stage is composed of gears D, E, F, and G. Second stage inputs are from gear D and arm-2.

ωarm  ω 2.

Determine the speed of gear D using the formula method for analyzing an epicyclic train. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be A and last be D. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm ωFarm

ωD  ωarm

=

ωA  ωarm

ωA  0  rpm

=R


 NA   NC      NB   ND 

R   

R  5.74359


ωD  ( 1  R)  ωarm 3.

ωD  1423.08 rpm

Determine the speed of gear G using the formula method for analyzing an epicyclic train. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be D and last be G. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm ωFarm

=

ωG  ωarm ωD  ωarm

=R


 ND   NF      NE   NG 

R   

R  0.11471

Solve the right-hand equation above for G .

ωG  R  ωD  ωarm  ωarm

ωG  102.4 rpm

Gear G drives shaft 2, so ω  ωG

ω  102.4 rpm




Figure P9-10 shows a compound epicyclic train. Shaft 1 is driven at 60 rpm. The tooth numbers are indicated in the figure. Determine the speed and direction of gears G and M.

Given:


NB  60

NC  40

NDout  64

NDin  58

NE  18

NG  66

NH  24

NJ  54

NK  40

NL  38

NM  96

Shaft 1 speed: Solution:

ω  60 rpm

NF  26

CCW


1.

The motor furnishes inputs to both epicyclic trains. To the first stage, it drives the ring gear D through gear A, and it drives arm-1 through gears B and C. The second stage inputs are from shaft 1 through gears B and C to sun J, and from the output of the first stage through gear G to arm-2

2.

Determine the speed of gear D using equations 9.7.



   ω  NDout 

ωD    3.

NA

ωD  33.750 rpm

Determine the speed of arm-1 and gear J using equations 9.7.

 NB    ω  NC 

ωarm1    4.

ωarm1  90.000 rpm

ωJ  ωarm1

Determine the speed of the ring gear G using the formula method for analyzing an epicyclic train. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be D and last be G. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm ωFarm

=

ωG  ωarm1 ωD  ωarm1

=R

Calculate R using equation 9.14 and inspection of Figure P9-9b. R 

NDin NF  NE NG

R  1.26936

Solve the right-hand equation above for G.

ωG  R  ωD  ωarm1  ωarm1

5.

ωG  18.60  rpm

Determine the speed of gear L using the formula method for analyzing an epicyclic train. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be J and last be L. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm ωFarm

=

ωL  ωarm2 ωJ  ωarm2

ωarm2  ωG

=R


 NJ   NK      NH   NL 

R   

R  2.36842

Solve the right-hand equation above for L.

ωL  R  ωJ  ωarm2  ωarm2

ωL  187.71 rpm

ωM  ωL

ωM  187.71 rpm




Calculate the ratios in the Model T transmission shown in Figure 9-46 (p. 476).

Given:

Tooth numbers: N3  27

Solution: 1.

N4  33

N5  24

N6  27

N7  21

N8  30


For low gear, the gearset consists of gears 3, 4, 6, and 7. Determine the ratio of input (arm) to output (gear 6) using the formula method for analyzing an epicyclic train. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be gear 7 and last be gear 6. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm

=

ωFarm


=R

Calculate R using equation 9.14 and inspection of Figure 9-46.

 N3   N7      N6   N4 

R   

R  0.63636

Solve the right-hand equation above for the ratio arm /6 = mG for 7 = 0. mGL  2.

1

mGL  2.750

1R

For reverse gear, the gearset consists of gears 3, 5, 6, and 8. Determine the ratio of input (arm) to output (gear 6) using the formula method for analyzing an epicyclic train. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be gear 8 and last be gear 6. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm ωFarm

=


=R

Calculate R using equation 9.14 and inspection of Figure 9-46.

 N3   N8      N6   N5 

R   

R  1.25000

Solve the right-hand equation above for the ratio arm /6 = mG for 8 = 0. mGR 

1 1R

mGR  4.000




Figure P7-26 shows a V-belt drive. The sheaves (pulleys) have pitch diameters of 150 and 300 mm, respectively. The smaller sheave is driven at a constant 1750 rpm. For a cross-sectional, differential element of the belt, write the equations of its acceleration for one complete trip around both pulleys including its travel between the pulleys. Compute and plot the acceleration of the differential element versus time for one circuit around the belt path. What does your analysis tell you about the dynamic behavior of the belt?

Given:

Sheave radii and speed: r2  75 mm

r4  150  mm

ω  1750 rpm

Assumptions: Center distance: C  450  mm Solution: 1.


Draw a schematic representation of the V-belt drive to scale and label it.

443.706 B 3 9.594°

A 4

2

R150.000 R75.000

D

C

From the layout,

2.

Angle to point of tangency

β  9.594  deg

Distance from A to B

LAB  443.706  mm

Distance from B to C

LBC   π  2  β  r4

Distance from C to D

LCD  443.706  mm

Distance from D to A

LDA   π  2  β  r2

LDA  210.502 mm

Total path length

L  LAB  LBC  LCD  LDA

L  1619.387 mm

Calculate the angular velocity of each sheave. Sheave 2 Sheave 4

3.

LBC  521.473 mm

ω  183.260 ω 

r2 r4

 ω

rad sec ω  91.630

rad sec

Calculate the acceleration of an element on the belt for each portion of the path, starting at A. From A to B the acceleration is zero since the belt velocity is constant

AAB  0 

mm sec

2



From B to C the tangential acceleration is zero since the belt velocity is constant The normal component is ABC  r4 ω

6 mm

2

ABC  1.259  10

sec

2

ACD  0 

From C to D the acceleration is zero since the belt velocity is constant

mm sec

2

From D to A the tangential acceleration is zero since the belt velocity is constant The normal component is ADA  r2 ω

6 mm

2

ADA  2.519  10

sec 4.

2

Plot the acceleration over the entire path using a path variable of s. s  0  mm 5  mm  L Define a range function that will plot a variable only between two limits using the Heaviside step function . R( s a b )  Φ ( s  a )  Φ ( s  b ) The acceleration function for the entire path is. Let

s1  LAB

s2  LAB  LBC

s3  LAB  LBC  LCD

A ( s)  AAB  R s 0  mm s1  ABC  R s s1 s2  ACD  R s s2 s3  ADA  R s s3 L BELT ACCELERATION ALONG PATH


3000

2000

1000

0

0

500

1000

1500

2000

Distance Along Path, mm

5.

The graph shows the sudden change in acceleration as the belt enters and leaves a sheave. This results in infinite jerk at these points, which results in belt "hop" at these points, a condition that will cause eventual fatigue failure and wear in the belt. Because of the belt hop caused by infinite jerk at the initial belt/sheave tangency points the span of the belt is continuously vibrating.




Figure P9-11 shows an involute that has been generated from a base circle of radius rb. Point A is simultaneously on the base circle and on the involute. Point B is any point on the involute curve and point C is on the base circle where a line drawn from point B is tangent to the base circle. The angle B (angle BOC) is known as the involute pressure angle corresponding to point B (not to be confused with the pressure angle of two gears in mesh, which is defined in the text). The angle AOB is known as the involute of B and is often designated as inv B. Using the definitions of the involute tooth form and Figure 9-5, derive an equation for inv B as a function of B alone.

Solution:

See Figures P9-11 and 9-5 and Mathcad file P0946.

From Figure 9-5 and the definition of the involute curve, arc AC is equal to length BC. Therefore Angle AOC = (arc AC) / OC = BC / OC and thus, Angle AOC = tan B also, Angle AOB = angle AOC - B = tan B - B thus, inv B = tan B - B




Using the data and definitions from Problem 9-46, show that when the point B is at the pitch circle the involute pressure angle is equal to the pressure angle of two gears in mesh.

Solution:

See Figures P9-11, 9-6 and 9-7; and Mathcad file P0947.

When the point B in Figure P9-11 is at the pitch point, points B and P (of Figures 9-6 and 9-7) are coincident. The angle between a line through the tangency point of the line of action with the base circle and point O (in Figure 9-6) is the pressure angle of two gears in mesh, Looking at Figure P9-11, we see that this is also the angle designated as B, which is the involute pressure angle of point B.




Using the data and definitions from Problem 9-46 and with the point B at the pitch circle where the involute pressure angle B is equal to the pressure angle  of two gears in mesh, derive equation 9.4b.

Solution:

See Figures P9-11, 9-6 and 9-9; and Mathcad file P0948.

From equation 9.4a and Figure 9-9 p c =

Taking the ratio of p to p , b

pb pc

=

c

pb pc

rp cos ϕ rp

=

rb rc

π d N

=

π rp N

. Similarly, p b =

π d b N

=

π rb N

. But, from Figure 9-6, we see that rb = rp cos ϕ. Thus,

and

p b = rp cos ϕ




Using Figures 9-6 and 9-7, derive equation 9.2, which is used to calculate the length of action of a pair of meshing gears.

Solution:


1.

Define points E1 and E2 as shown in the figure below. Note that the distance O1E1 = rg cos  and O2E2 = rp cos  . Z = AB = E1B + E2A - E1E2 E1E2 = rg sin  + rp sin  = (rg + rp) sin  = C sin  O1B = rg + a g and O2A = rp + a p From triangles O1E1B and O2E2A we have E1B = Thus, Z=

 rg  ag2   rg cos ϕ 2

 rp  ap2   rp cos ϕ 2

E2A =

rp  ap 2  rp cosϕ 2   rg  ag2   rg cos ϕ 2  C sin ϕ 

Point of tangency of line of action and gear base circle, E2

Intersection of line of action with OD of gear 2. Beginning of contact, A B

 O2

A



O1

P Intersection of line of action with OD of gear 1. Leaving contact, B

Point of tangency of line of action and pinion base circle, E1




For lubrication purposes it is desired to have a backlash of 0.03 mm measured on the pitch circle of a 40-mm-diameter pinion in mesh with a 100-mm-diameter gear. if the gears are standard, full-depth, with 25 deg pressure angles, by how much should the center distance be increased?

Units:

Arc minute: arcmin 

Given:

Pitch diameterss: d p  40 mm d g  100  mm

1 60

 deg

Pressure angle: ϕ  25 deg

Desired backlash: b  0.03 mm Solution: 1.


Convert the desired backlash from a linear to an angular measurement on the 40-mm-diameter pinion.

θB  2.

2 b dp

θB  5.1566 arcmin

Use equation 9.3 to calculate the change in center distance required.

ΔC 

π d p θB

43200  tan ϕ  arcmin

ΔC  0.032 mm




For lubrication purposes it is desired to have a backlash of 0.0012 in measured on the pitch circle of a 2.000-in-diameter pinion in mesh with a 5.000-in-diameter gear. if the gears are standard, full-depth, with 25 deg pressure angles, by how much should the center distance be increased?

Units:

Arc minute: arcmin 

Given:

Pitch diameterss: d p  2.000  in d g  5.000  in

1 60

 deg Pressure angle: ϕ  25 deg

Desired backlash: b  0.0012 in Solution: 1.


Convert the desired backlash from a linear to an angular measurement on the 40-mm-diameter pinion.

θB  2.

2 b dp

θB  4.1253 arcmin

Use equation 9.3 to calculate the change in center distance required.

ΔC 

π d p θB

43200  tan ϕ  arcmin

ΔC  1.287  10

3

in




A 22-tooth gear has standard full-depth involute teeth with module of 6. Calculate the pitch diameter, circular pitch, addendum, dedendum, tooth thickness, and clearance using the AGMA specifications in Table 9-1, substituting m for 1/p d.

Given:


Solution:


1.

2.

Module m  6  mm


d  N  m

d  132.00 mm

Circular pitch

p c  π m

p c  18.85 mm


a  1.0000 m

a  6.00 mm

Dedendum

b  1.2500 m

b  7.50 mm

Tooth thickness

t  0.5 p c

t  9.42 mm

Clearance

c  0.2500 m

c  1.50 mm

Note: The circular tooth thickness is exactly half of the circular pitch, so the equation used above is more accurate than the one in Table 9-1. Also, all gear dimensions in mm should be displayed to two decimal places since manufacturing tolerances for gear teeth profiles are usually expressed in hundredths of a millimeter.





Given:


Solution:


1.

2.



d  N  m

d  120.00 mm

Circular pitch

p c  π m

p c  9.42 mm


a  1.0000 m

a  3.00 mm

Dedendum

b  1.2500 m

b  3.75 mm

Tooth thickness

t  0.5 p c

t  4.71 mm

Clearance

c  0.2500 m

c  0.75 mm






Given:


Solution:


1.

2.



d  N  m

d  60.00 mm

Circular pitch

p c  π m

p c  6.28 mm


a  1.0000 m

a  2.00 mm

Dedendum

b  1.2500 m

b  2.50 mm

Tooth thickness

t  0.5 p c

t  3.14 mm

Clearance

c  0.2500 m

c  0.50 mm





Determine the minimum number of teeth on a pinion with a 20 deg pressure angle for the following gear-to-pinion ratios such that there will be no tooth-to-tooh interference: 1:1, 2:1, 3:1, 4:1, and 5:1.

Given:

Pressure angle   20 deg

Solution:

See Tables 9-4 and 9-5 and Mathcad file P0955.

1.

From Table 9-5a, the minimum number of teeth on a 20-deg pressure angle pinion that can mesh with another gear is 13 if there are no more than 16 teeth on the other gear. So, for a 1:1 ratio, both gears can have 13 teeth. Looking further at Table 9-5a, we see that there must be more teeth on the pinion as the ratio increases. For instance, for ratios of 2:1 and 3:1, the pinion must have at least 15 teeth since a 15-tooth pinion can mesh with a gear with up to 45 teeth. If the raio is 4:1 or 5:1, the minimum number of teeth on the pinion is 16 since these ratios would require more than 45 teeth on the gear. SUMMARY Ratio 1:1 2:1 3:1 4:1 5:1

Min teeth on pinion 13 15 15 16 16

Number teeth on gear 13 30 45 64 80

However, from Table 9-4b, we see that none of these pinions could be cut with a hob without undercutting. They would have to be produced by other methods to avoid weakening the tooth by undercutting.




Determine the minimum number of teeth on a pinion with a 25 deg pressure angle for the following gear-to-pinion ratios such that there will be no tooth-to-tooh interference: 1:1, 2:1, 3:1, 4:1, and 5:1.

Given:

Pressure angle   25 deg

Solution:

See Tables 9-4 and 9-5 and Mathcad file P0956.

1.

From Table 9-5b, the minimum number of teeth on a 25-deg pressure angle pinion that can mesh with another gear is 9 if there are no more than 13 teeth on the other gear. So, for a 1:1 ratio, both gears can have 9 teeth. Looking further at Table 9-5b, we see that there must be more teeth on the pinion as the ratio increases. For instance, for ratios of 2:1 and 3:1, the pinion must have at least 10 teeth since a 10-tooth pinion can mesh with a gear with up to 32 teeth. If the raio is 4:1 or 5:1, the minimum number of teeth on the pinion is 11 since these ratios would require more than 32 teeth on the gear. SUMMARY Ratio 1:1 2:1 3:1 4:1 5:1

Min teeth on pinion 9 10 10 11 11

Number teeth on gear 9 20 30 44 55

However, from Table 9-4b, we see that none of these pinions could be cut with a hob without undercutting. They would have to be produced by other methods to avoid weakening the tooth by undercutting.




A pinion with a 3.000-in pitch diameter is to mesh with a rack. What is the largest standard tooth size, in terms of diametral pitch, that can be used without having any interference? a. For a 20-deg pressure angle b. For a 25-deg pressure angle

Given:

Pitch diameter:

Solution:


d  3.000  in

1.

Assume thatthe pinion is generated by means other than being cut by a hob.

a.


p dmin 

Nmin d

1

p dmin  6.000 in

1

From Table 9-2, the smallest standard diametral pitch (largest tooth size) that can be used is 6. p d  6  in

.

The resulting number of teeth on the pinion is: N  p d  d

b.

N  18


p dmin 

Nmin d

1

p dmin  4.000 in

1

From Table 9-2, the smallest standard diametral pitch (largest tooth size) that can be used is 4. p d  4  in The resulting number of teeth on the pinion is: N  p d  d

N  12

.




A pinion with a 75-mm pitch diameter is to mesh with a rack. What is the largest standard tooth size, in terms of diametral pitch, that can be used without having any interference? a. For a 20-deg pressure angle b. For a 25-deg pressure angle

Given:

Pitch diameter:

Solution:


d  75 mm

1.

Assume thatthe pinion is generated by means other than being cut by a hob.

a.


m 

d Nmin

m  4.167 mm

From Table 9-3, the largest standard module (largest tooth size) that can be used is 4 but, in order to have a whole number of teeth, the module must be 3 or 5. Since a module of 5 would result in fewer than 18 teeth, let m  3  mm. The resulting number of teeth on the pinion is: N  b.

d m

N  25


m 

d Nmin

m  6.250 mm

From Table 9-3, the largest standard module (largest tooth size) that can be used is 6 but, in order to have a whole number of teeth, the module must be 3 or 5. Since a module of 3 would result in more teeth than is necessary, let m  5  mm. The resulting number of teeth on the pinion is: N 

d m

N  15




In order to have a smooth-running gearset it is desired to have a contact ratio of at least 1.5. If the gears have a pressure angle of 25 deg and gear ratio of 4, what is the minimum number of teeth on the pinion that will yield the required minimum contact ratio?

Given:

Contact ratio:

Solution:


1.

mp  1.5

Gear ratio: mG  4

Pressure angle:   25 deg

Write equations 9.6b and 9.2 for the contact ratio and length of action, respectively. mp =

pd  Z

Z=

π cos 

rp  ap 2  rp cos 2  rg  ag 2  rg cos 2  C sin 

Note that pd =

N d

ap = ag =

1 pd

rg = mG rp

rp =

C = rp  rg = rp  1  mG

d 2

Substituting these identities into the equation for Z and collecting terms gives:

 2 N Z=     1  pd   2  1

2 2  2  N  cos    mG N  1   mG N  cos   N   1  m   sin    G  2 2   2   2  

Substituting this into the equation for mp and we have: 2

mp( N ) 

2.

2

2 2  N  1   N  cos    mG N  1   mG N  cos   N   1  m   sin      G 2 2  2   2   2 

π cos 

Notice that mp is independent of p d and depends only on the pressure angle, the number of teeth on the pinion, and the gear ratio. To determine the minimum number of teeth on the pinion, plot the function on the right side of the equation above over the range N  12 13  50.

1.6

From the graph we see that we need about 21 teeth. Try various values of N in the function until the calculated value of mp is greater than or equal to the required value.

1.55

m p( N ) 1.5

mp( 21)  1.500 Let

1.45

1.4 10

20

30 N

40

50

N  21




In order to have a smooth-running gearset it is desired to have a contact ratio of at least 1.5. If the gears have a pressure angle of 25 deg and 20-tooth pinion, what is the minimum gear ratio that will yield the required minimum contact ratio?

Given:

Contact ratio:

Solution:


1.

mp  1.5

Number of teeth: N  20

Pressure angle:   25 deg


pd  Z

π cos 

Z=


Note that pd =

N

ap = ag =

d

1

rg = mG rp

pd

rp =

C = rp  rg = rp  1  mG

d 2


 2 N Z=     1  pd   2 


1


mp mG  2.

2


π cos 

Notice that mp is independent of p d and depends only on the pressure angle, the number of teeth on the pinion, and the gear ratio. To determine the minimum number of teeth on the pinion, plot the function on the right side of the equation above over the range mG  1 1.5  9.

1.6

From the graph we see that we need a gear ratio of about 5. Try various values of mG in the function until the calculated value of mp is greater than or equal to the required value.

1.55

 

m p mG 1.5

mp( 5 )  1.502 Let

1.45

1.4

0

2

4

6 mG

8

10

mG  5




Calculate and plot the train ratio of a noncircular gearset, as a function of input angle, that is based on the centrodes of Figure 6-15b. The link length ratios ar L1/L2 = 1.60, L3/L2 = 1.60, and L4/L2 = 1.00.

Given:

Link lengths (assume L2 = 1.00):

Solution:

Link 2 (L2)

L2  1.00

Link 3 (L3)

L3  1.60

Link 4 (L4)

L4  1.00

Link 1 (L1)

L1  1.60


1.

In Figure 6-15b links 2 and 4 are the short links and 1 and 3 are the long links with link 1 fixed. To find the fixed and moving centrodes of I24 we will fix link 2 and find the intersection of links 1 and 3. The locus of these intersections is an ellipse that will be attached to link 2 when the linkage is reinverted such that link 1 is again grounded.

2.

Invert the linkage, fixing link 2 to ground and determine the range of motion for this Grashof double crank. a  L3

b  L4

c  L1

d  L2

a  1.600

b  1.000

c  1.600

d  1.000

θ  0.5 deg 1  deg  359.5  deg 3.

Determine the values of the constants needed for finding 4 on the inverted linkage (actually 1 on the original linkage) from equations 4.8a, 4.10a, 4.11b and 4.12. K1 

d

K2 

a

K1  0.6250 2

K3 

c

K2  0.6250 2

2

a b c d

2

K3  1.0000

2 a c

 

d

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

C θ  K1   K2  1   cos θ  K3 4.


 





 

 

B θ

 





 

 

B θ

θ θ  2   atan2 2  A θ B θ  θ θ  2   atan2 2  A θ B θ 

 2  4 A θ Cθ   2  4 A θ Cθ 

Use the crossed branch equation for the first 180 deg of crank motion and the open branch equation for the last 180 deg.

 



 

 

θ θ  if θ  π θ θ θ θ

If the calculated angle is negative, make it positive.

 



 

 

θ θ  if θ  π θ θ θ θ  2  π





Plot 1 as a function of 3 (link 3 is the driving crank and link 1 is the driven crank in the inverted linkage). 360 315 270

 

θ  θ

225 180

deg 135 90 45 0

0

45

90

135

180

225

270

315

360

θ deg

5.

Define the coordinates of the intersection of links 1 and 3 for the inverted linkage.

 

 

Equation of link 3: y = x m3 , where m3 θ  tan θ

 

  

Equation of link 1: y = ( x  d )  m1 , where m1 θ  tan θ θ The coordinates of the intersection (I24) of these two lines are

 

x24 θ 

 

m1 θ

 

 

m1 θ  m3 θ

 

d

   

y24 θ  x24 θ  m3 θ

LINK 2 CENTRODE

x24( 0.0001 deg)  1.300

1

y24 ( 0.0001 deg)  0.000 0.5

x24( 51.318 deg)  0.500 y24 ( 51.318 deg)  0.624

 

y24 θ

0

x24( 179.0001 deg)  0.300 y24 ( 179.0001 deg)  0.005

 0.5

θ( 51.318 deg)  128.682 deg 1  0.5

0

0.5

 

1

1.5

x24 θ

The ellipse has a major axis of 1.600 and a minor axis of 1.248 and the origin on the graph above is the point O2 The centrode for link 4 is a mirror image of this centrode.


6.


The linkage is shown below, reinverted such that link 1 is fixed, with the centrodes attached to links 2 and 4. They are tangent to each other at the instant center I24. The axis of slip is found by extending a line from the instant center I13 to I24. The axis of transmission is perpendicular to this line as shown.

Y I Axis of transmission

13

A 2 O4

O2 3

4 B

Axis of slip

X




Calculate and plot the train ratio of a noncircular gearset, as a function of input angle, that is based on the centrodes of Figure 6-15b. The link length ratios ar L1/L2 = 1.80, L3/L2 = 1.80, and L4/L2 = 1.00.

Given:

Link lengths (assume L2 = 1.00):

Solution:

Link 2 (L2)

L2  1.00

Link 3 (L3)

L3  1.80

Link 4 (L4)

L4  1.00

Link 1 (L1)

L1  1.80


1.

In Figure 6-15b links 2 and 4 are the short links and 1 and 3 are the long links with link 1 fixed. To find the fixed and moving centrodes of I24 we will fix link 2 and find the intersection of links 1 and 3. The locus of these intersections is an ellipse that will be attached to link 2 when the linkage is reinverted such that link 1 is again grounded.

2.

Invert the linkage, fixing link 2 to ground and determine the range of motion for this Grashof double crank. a  L3

b  L4

c  L1

d  L2

a  1.800

b  1.000

c  1.800

d  1.000

θ  0.5 deg 1  deg  359.5  deg 3.

Determine the values of the constants needed for finding 4 on the inverted linkage (actually 1 on the original linkage) from equations 4.8a, 4.10a, 4.11b and 4.12. K1 

d

K2 

a

K1  0.5556 2

K3 

c

K2  0.5556 2

2

a b c d

2

K3  1.0000

2 a c

 

d

 

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

C θ  K1   K2  1   cos θ  K3 4.


 





 

 

B θ

 





 

 

B θ

θ θ  2   atan2 2  A θ B θ  θ θ  2   atan2 2  A θ B θ 

 2  4 A θ Cθ   2  4 A θ Cθ 

Use the crossed branch equation for the first 180 deg of crank motion and the open branch equation for the last 180 deg.

 



 

 

θ θ  if θ  π θ θ θ θ

If the calculated angle is negative, make it positive.

 



 

 

θ θ  if θ  π θ θ θ θ  2  π





Plot 1 as a function of 3 (link 3 is the driving crank and link 1 is the driven crank in the inverted linkage). 360 315 270

 

θ  θ

225 180

deg 135 90 45 0

0

45

90

135

180

225

270

315

360

θ deg

5.

Define the coordinates of the intersection of links 1 and 3 for the inverted linkage.

 

 

Equation of link 3: y = x m3 , where m3 θ  tan θ

 

  

Equation of link 1: y = ( x  d )  m1 , where m1 θ  tan θ θ The coordinates of the intersection (I24) of these two lines are

 

x24 θ 

 

m1 θ

 

 

m1 θ  m3 θ

 

d

   

y24 θ  x24 θ  m3 θ

x24( 0.0001 deg)  1.400

LINK 2 CENTRODE 1

y24 ( 0.0001 deg)  0.000 x24( 56.251 deg)  0.500

0.5

y24 ( 56.251 deg)  0.748

 

y24 θ

x24( 179.0001 deg)  0.400

0

y24 ( 179.0001 deg)  0.007

 0.5

1  0.5

θ( 56.251 deg)  123.749 deg

0

0.5

 

1

1.5

x24 θ

The ellipse has a major axis of 1.800 and a minor axis of 1.496 and the origin on the graph above is the point O2 The centrode for link 4 is a mirror image of this centrode.


6.


The linkage is shown below, reinverted such that link 1 is fixed, with the centrodes attached to links 2 and 4. They are tangent to each other at the instant center I24. The axis of slip is found by extending a line from the instant center I13 to I24. The axis of transmission is perpendicular to this line as shown.

Y I Axis of transmission

13

A 2 O4

O2 3

4 B

Axis of slip

X




Figure P9-35b shows (schematically) a compound epicyclic train. The tooth numbers are 50, 25, 35, and 90 for gears 2, 3, 4, and 5, respectively. The arm is driven at 180 rpm CW and gear 5 is fixed to ground. Determine the speed and direction of gear 2. What is the efficiency of this train if the basic gearsets have E0 = 0.98?

Units:


Given:

Tooth numbers:

1


N3  25

ωarm  180  rpm


N4  35

N5  90

ω5  0  rpm

E0  0.98


Determine the speed of the sun gear 2 using the formula method for analyzing an epicyclic train. To start, choose a first and last gears that mesh with gears that have planetary motion. Let the first gear be 2 and last be 5. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm ωFarm

=


=R


 N2   N4      N3   N5 

R   

R  0.77778

Solve the right-hand equation above for  with  = 0.

ω2 

R1 R

 ωarm

ω2  411.43 rpm

2.

Find the basic ratio  for the train using equation 9.15. 1 ρ  ρ  1.286 R

3.

The combination of  < -1, shaft 1 fixed, and input to the arm corresponds to Case 8 in Table 9-12 giving an efficiency of

η 

E0  ρ  1  ρ E0  1

η  0.991




Figure P9-35h shows (schematically) a compound epicyclic train. The tooth numbers are 80, 20, 25, and 85 for gears 2, 3, 4, and 5, respectively. Gear 2 is driven at 200 rpm CCW and gear 5 is fixed to ground. Determine the speed and direction of the arm. What is the efficiency of this train if the basic gearsets have E0 = 0.98?

Units:


Given:

Tooth numbers:

1

N2  80

N3  20

ω2  200  rpm

Input speeds:


N4  25

N5  85

ω5  0  rpm

E0  0.98

See Figure P9-35h and Mathcad file P0964.

Determine the speed of the arm using the formula method for analyzing an epicyclic train. To start, choose first and last gears that mesh with gears that have planetary motion. Let the first gear be 2 and last be 5. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm

=

ωFarm


=R

Calculate R using equation 9.14 and inspection of Figure P9-7b. R 

 N2   N4   N   N   3  5

R  1.17647

Solve the right-hand equation above for arm with  = 0.

ωarm  2.

R R1

 ω2


3.

ωarm  1333 rpm

ρ  1.176

The combination of  > 1, shaft 1 fixed, and input to gear 2 corresponds to Case 3 in Table 9-12 giving an efficiency of

η 

ρ E0  1

E0  ρ  1 

η  0.884




Figure P9-35i shows (schematically) a compound epicyclic train. The tooth numbers are 24, 18, 20, and 90 for gears 2, 3, 4, and 5, respectively. The arm is driven at 100 rpm CCW and gear 2 is fixed to ground. Determine the speed and direction of gear 5. What is the efficiency of this train if the basic gearsets have E0 = 0.98?

Units:


Given:

Tooth numbers:

1


N3  18

ωarm  100  rpm


N4  20

N5  90

ω2  0  rpm

E0  0.98

See Figure P9-35i and Mathcad file P0965.

Determine the speed of the arm using the formula method for analyzing an epicyclic train. To start, choose first and last gears that mesh with gears that have planetary motion. Let the first gear be 2 and last be 5. Then, using equation 9.13c, write the relationship among the first, last, and arm.

ωLarm ωFarm

=


=R


 N2   N3   N4        N3   N4   N5 

R   

R  0.26667

Solve the right-hand equation above for 5 with  = 0.

ω5  ( R  1 )  ωarm 2.

Find the basic ratio  for the train using equation 9.15. ρ 

3.

ω5  126.67 rpm

1

ρ  3.750

R

The combination of  > 1, shaft 2 fixed, and input to the arm corresponds to Case 2 in Table 9-12 giving an efficiency of

η 

E0  ρ  1  ρ  E0

η  0.973




Using Figure 9-8, derive an equation for the operating pressure angle of two gears in mesh as a function of their base circle radii, the standard center distance, and the change in center distance.

Solution:


1.

Let:

Base circle radius of the gear be rbg Base circle radius of the pinion be rbp Change in center distance be C

2.

Then from Figure 9-8, the standard center distance is C =

3.

As the center distance increases the line of action remains tangent to the base circles as shown in  rbg  rbp  Figure 9-8b. Thus, the operating pressure angle is ϕ = acos   C  Δ C 

Standard pressure angle be  Standard center distance be C Operating pressure angle be O rbg  rbg cos ϕ




A pinion and gear in mesh have base circle radii of 1.8126 and 3.6252 in, respectively. If they were cut with a standard pressure angle of 25 o, determine their operating pressure angle if the standard center distance is increased by 0.062 in.

Given:

Base circle radii: rbp  1.8126 in rbg  3.6252 in Pressure angle: Change in center distance:

Solution: 1.


From Figure 9-8, the standard center distance is C 

2.

ϕ  25 deg

Δ  0.062  in

rbp  rbg cos ϕ

C  6.000  in

As the center distance increases the line of action remains tangent to the base circles as shown in Figure 9-8b. Thus, the operating pressure angle is

 rbp  rbg    CΔ 

ϕ  acos

ϕ  26.229 deg




A pinion and gear in mesh have base circle radii of 1.35946 and 2.26577 in, respectively. If they have a standard center distance of 4.000 in, determine the standard pressure angle and the operating pressure angle if the standard center distance is increased by 0.050 in.

Given:

Base circle radii: rbp  1.35946  in rbg  2.26577  in Pressure angle: Change in center distance:

Solution: 1.

Δ  0.050  in


From Figure 9-8, the standard pressure angle is

 rbp  rbg    C 

ϕ  acos 2.

C  4.000  in

ϕ  25.000 deg

As the center distance increases the line of action remains tangent to the base circles as shown in Figure 9-8b. Thus, the operating pressure angle is

 rbp  rbg    CΔ 

ϕ  acos

ϕ  26.476 deg




A 25-tooth pinion meshes with a 60-tooth gear. They have a diametral pitch of 4, a pressure angle of 20 o, and AGMA full-depth involute profiles. Find the gear ratio, circular pitch, base pitch, pitch diameters, standard center distance, addendum, dedendum, whole depth, clearance, outside diameters and contact ratio of the gearset.

Given:

Number of teeth: Np  25

Ng  60

Pressure angle:

  20 deg

1

Diametral pitch: p d  4  in Solution: 1.


From equation 9.5b, the gear ratio is: mG 

2.

Np

π pd

d g 

Np pd Ng pd

Np  Ng 2 pd

d g  15.000 in

C  10.625 in

The tooth dimensions are found from the specifications in Table 9-1 for   20 deg , coarse pitch: Addendum

Dedendum

7.

d p  6.250 in

The standard center distance is one half the summ of the pitch diameters. Using equation 9.4c, this can be written as: C 

6.

p b  0.738 in

From equation 9.4c, the pitch diameters and radii are: d p 

5.

p c  0.785 in

From equation 9.4b, the base pitch is p b  p c  cos 

4.

mG  2.400

From equations 9.4a and 9.4c, the circular pitch is: p c 

3.

Ng

a 

1.000

b 

1.250

pd

pd

a  0.2500 in b  0.3125 in

Whole depth

h t  a  b

h t  0.5625 in

Clearance

c  b  a

c  0.0625 in

Outside diameter

DOp  d p  2  a

DOp  6.7500 in

DOg  d g  2  a

DOg  15.5000 in

rp  0.5 d p

rg  0.5 d g

rp  3.125 in

rg  7.500 in

From equations 9.2 and 9.6a, the contact ratio is: Pitch radii



Length of contact Z 

rp  a 2  rp cos 2  rg  a2   rg cos  2  C sin

Z  1.2533 in Contact ratio

mp 

Z pb

mp  1.698




A 15-tooth pinion meshes with a 45-tooth gear. They have a diametral pitch of 5, a pressure angle of 25 o, and AGMA full-depth involute profiles. Find the gear ratio, circular pitch, base pitch, pitch diameters, standard center distance, addendum, dedendum, whole depth, clearance, outside diameters and contact ratio of the gearset.

Given:

Number of teeth: Np  15

Ng  45

Pressure angle:

  25 deg

1

Diametral pitch: p d  5  in Solution: 1.


From equation 9.5b, the gear ratio is: mG 

2.

Np

π pd

d g 

Np pd Ng pd

Np  Ng 2 pd

d g  9.000 in

C  6.000 in

The tooth dimensions are found from the specifications in Table 9-1 for   25 deg , coarse pitch: Addendum

Dedendum

7.

d p  3.000 in

The standard center distance is one half the summ of the pitch diameters. Using equation 9.4c, this can be written as: C 

6.

p b  0.569 in

From equation 9.4c, the pitch diameters and radii are: d p 

5.

p c  0.628 in

From equation 9.4b, the base pitch is p b  p c  cos 

4.

mG  3.000

From equations 9.4a and 9.4c, the circular pitch is: p c 

3.

Ng

a 

1.000

b 

1.250

pd

pd

a  0.2000 in b  0.2500 in

Whole depth

h t  a  b

h t  0.4500 in

Clearance

c  b  a

c  0.0500 in

Outside diameter

DOp  d p  2  a

DOp  3.4000 in

DOg  d g  2  a

DOg  9.4000 in

rp  0.5 d p

rg  0.5 d g

rp  1.500 in

rg  4.500 in

From equations 9.2 and 9.6a, the contact ratio is: Pitch radii



Length of contact Z 

rp  a 2  rp cos 2  rg  a2   rg cos  2  C sin

Z  0.8210 in Contact ratio

mp 

Z pb

mp  1.442




Design a compound, spur gear train that will reduce the speed of a 900-rpm synchronous AC motor to exactly 72 revolutions per hour with the output rotating in the same direction as the motor. Use gears with a pressure angle of 25 deg and minimize the package size.

Given:

Motor speed:

Solution:


n m  900  rpm

Output speed:

mG 

nm

n out  72

rev hr

mG  750

1.

Calculate the required gear ratio.

2.

Since the ratio must be positive, we want to have an even number of stages or an odd number with an idler. Let the number of stages be

n out

1

3.

Possible number of stages

j  2 3  5 Ideal, theoretical stage ratios

then

j 

r( j )  mG

r( j )  2 3

27.386 9.086

4

5.233

5

3.758

Four stages would result in a stage ratio less than 10 and about 5.25, so we will use four stages. Using a pressure angle of 25 deg, let the stage ratios be (choosing integer ratios whose product is mG  750 ) Stage 1 ratio

r1  6

Stage 2 ratio

r2  5

Stage 3 ratio

r3  5

Stage 4 ratio

r4  5

and let the driver gears have tooth numbers of Tooth number index i  2 3  9 N  14

N  14

2

4

N  14 6


N  84 3

2

N  r2 N 5

4

N  70 5

N  r3 N 7

9

N  70

N  70

7

5

N N N N Checking the resulting gear ratio:

N  r4 N

6

3

5

7

N N N N 2

4

6

9 8

 750

8

N  14 8

j




A gear-train designer wants to have a contact ratio of at least 1.5 for a gearset. If the gears have a pressure angle of 25 deg and the pinion has 21 teeth, what is the minimum gear-ratio that will yield the required minimum contact ratio?

Given:

Contact ratio: mp  1.5

Solution:

  25 deg See Mathcad file P0972.

1.

Number of teeth: N  21

Pressure angle:


pd  Z

π cos 

Z=


Note that pd =

N

ap = ag =

d

1

rg = mG rp

pd

rp =

C = rp  rg = rp  1  mG

d 2


 2 N Z=     1  pd   2 


1


mp mG 

2.

2


π cos 

Notice that mp is independent of p d and depends only on the pressure angle, the number of teeth on the pinion, and the gear ratio. To determine the minimum gear ratio, plot the function on the right side of the equation above over the range mG  2 2.1  20.

1.54

From the graph we see that we need a gear-ratio of about 4. Try various values of mG in the function until the calculated value of mp is greater than or equal to the required value.

1.52

 

m p mG 1.5

mp( 4 )  1.500 1.48

1.46

so, the minimum is 4.

0

5

10

15 mG

20

25

DESIGN OF MACHINERY



Provide a preliminary design (pitch diameters and numbers of teeth) for a gear set that will have a gear ratio of mG = 4, a diametral pitch p d = 8, and a contact ratio of at least 1.5.

Given:

Gear ratio

mG  4

Diametral pitch

1

p d  8  in

Design Choice: Pressure angle ϕ  25 deg Assumptions: The pinion is cut by a hob and can, therefore, have no fewer than 14 teeth for a 25-deg pressure angle (see Table 9-4b). Solution: 1.


From inspection of Table 9-5b, we see that 11 teeth is the least number that the pinion can have for a gear ratio of mG  4 . Therefore, let the number of teeth on the pinion be (an even number > 11). Iterate on the mumber of teeth until a satisfactory contact raio is achieved. Np  22

2.

Ng  mG Np

Ng  88


3.

and

Np pd

d p  2.7500 in

d g 

Ng pd

d g  11.0000  in


1 pd

Center distance

Z 

rp  1.3750 in

rg  0.5 d g

a p  0.1250 in

a g 

C  rp  rg

C  6.8750 in

1 pd

rg  5.5000 in a g  0.1250 in


Contact ratio

mp 

pd  Z

π cos ϕ

mp  1.506

Z  0.5358 in




Given:

The mallet shown in Figure 10-2 has the specifications given below. Find the location of its composite CG, and its moment of inertia and radius of gyration about axis ZZ. Assume the wood has a density equal to 0.9 times that of water. 3 Brass hammer head has specific weight: γd  0.31 lbf  in 3

γh  0.9 0.036  lbf  in

Wood handle has specific weight:

Solution: 1.

Head dimensions:

rd  0.625  in

h d  3.5 in

Handle dimensions:

rh1  0.750  in

Lh1  10 in

Calculate the volume and weight of each component.

Handle:

2

2

3

Vd  π rd  h d  π rh2  Lh2

Vd  3.705  in

Wd  γd  Vd

Wd  1.149  lbf

2

Vh1  17.671 in

3

Vh2  π rh2  Lh2

2

Vh2  0.590  in

Vh  Vh1  Vh2

Vh  18.261 in

Wh1  γh Vh1

Wh1  0.573  lbf

Wh2  γh Vh2

Wh2  0.019  lbf

Wh  Wh1  Wh2

Wh  0.592  lbf

Vh1  π rh1  Lh1

3 3

The CG of each component will lie along the XX axis. Find the distance from the origin to the CGs (distance to axes HH and DD). Head:

Xcgd  Lh1  rd

Handle:

Xcgh 

Xcgd  10.625 in

0.5 Lh1 Vh1   Lh1  0.5 Lh2  Vh2 Vh

Xcgh  5.182  in 3.

Find the location of the composite CG using equation 10.3d. XCg 

4.

Lh2  2  rd


Head:

2.

rh2  0.3875 in

Xcgd  Wd  Xcgh  Wh W d  Wh

XCg  8.774  in

Calculate the moment of inertia of the head with respect to the ZZ axis. IDDd 

Wd 12 g

  3  rd  h d

IZZd  IDDd 

2

Wd g

 Xcgd

2

2



3

IDDd  3.328  10

2

 lbf  sec  in

2

IZZd  0.339  lbf  sec  in


5.


Calculate the moment of inertia of the handle with respect to the ZZ axis. Ih1 

Wh1 12 g

  3  rh1  Lh1

2

2



2

 Lh1  IZZh1  Ih1    g  2  Wh1

Ih2 

Wh2 12 g



6

Ih2  8.301  10

2


IZZh2  5.595  10

IZZh  IZZh1  IZZh2

IZZh  0.055  lbf  sec  in

3

2


2

Add the moments of the two components about the ZZ axis to get the moment of inertia of the mallet about the ZZ axis. IZZ  IZZd  IZZh

7.

2

2 Lh2   IZZh2  Ih2    Lh1   g  2 

Wh2

6.

2

IZZh1  0.050  lbf  sec  in

  3  rh2  Lh2 2

2

Ih1  0.013  lbf  sec  in

2

IZZ  0.394  lbf  sec  in

Use equation 10.10b to calculate the radius of gyration about the ZZ axis.

k 

IZZ g W d  Wh

k  9.354  in




The mallet shown in Figure 10-2 has the specifications given below. Find the location of its composite CG, and its moment of inertia and radius of gyration about axis ZZ. Assume the wood has a density equal to 0.9 times that of water.

Given:

Wood has specific weight:

Solution: 1.

Head dimensions:

rd  1.25 in

h d  3.5 in

Handle dimensions:

rh  0.750  in

Lh  10 in


Calculate the volume and weight of each component. Head:

Handle:

2.

3.

Vd  π rd  h d

2

Vd  17.181 in

Wd  γ Vd

Wd  0.526  lbf

Vh  π rh  Lh

2

Vh  17.671 in

Wh  γ Vh

Wh  0.541  lbf

3

Head:

Xcgd  Lh  rd

Handle:

Xcgh 

Xcgd  11.250 in

Lh

Xcgh  5.000  in

2

Find the location of the composite CG using equation 10.3d. Xcgd  Wd  Xcgh  Wh W d  Wh

XCg  8.081  in

Calculate the moment of inertia of the head with respect to the ZZ axis. IDDd 

Wd 12 g

  3  rd  h d 2

IZZd  IDDd  5.

3

The CG of each component will lie along the XX axis. Find the distance from the origin to the CGs (distance to axes HH and DD).

XCg  4.

3

γ  0.85 0.036  lbf  in

Wd g

 Xcgd

2



2

3

IDDd  1.922  10

2


2

IZZd  0.174  lbf  sec  in

Calculate the moment of inertia of the handle with respect to the ZZ axis. IHHh 

Wh 12 g

  3  rh  Lh

IZZh  IHHh 

2

2

Wh g

 Xcgh



2

2

IHHh  0.012  lbf  sec  in 2

IZZh  0.047  lbf  sec  in

6.

Add the moments of the two components about the ZZ axis to get the moment of inertia of the mallet about the ZZ axis. 2 IZZ  IZZd  IZZh IZZ  0.221  lbf  sec  in

7.

Use equation 10.10b to calculate the radius of gyration about the ZZ axis.

k 

IZZ g W d  Wh

k  8.948  in




Calculate the location of the composite center of gravity, the mass moment of inertia and the radius of gyration with respect to the specified axis, for whichever of the following commonly available items that are assigned. (Note these are not short problems). a. A good-quality writing pen, about the pivot point at which you grip it to write. (How does placing the cap on the upper end of the pen affect these parameters when you write?) b. Two table knives, one metal and one plastic, about the pivot axis when held for cutting. Compare the calculated results and comment on what they tell you about the dynamic usability of the two knives (ignore sharpness considerations). c. A ball-peen hammer (available for inspection in any university machine shop), about the center of rotation (after you calculate it for the proper center of percussion). d. A baseball bat (see the coach) about the center of rotation (after you calculate its location for the proper center of percussion). e. A cylindrical coffee mug, about the handle hole.

Solution:

No solution is provided here due to the wide variety of possibilities in this open-ended problem.




Set up these equations in matrix form. Use program MATRIX, Mathcad, or a calculator that has matrix math capability to solve them. a. -5x -2y +12z -w = -9 x +3y -2z +4w = 10 -x -y +z = -7 3x -3y +7z +9w = -6 b. 3x -5y +17z -5w = -5 -2x +9y -14z +6w = 22 -x -y -2w = 13 4x -7y +8z +4w = -9

Solution:


1.

Place the coefficients of the unknowns in a 4 x 4 matrix.

 5 1 Ca    1 3  2.

2 12 1 

 3 2 Cb    1 4 

 3 2 4  1 1 0   3 7 9 

9

5 

  2   4 

14 6

1

0

7

8

Place the constants on the right-hand side of the equal sign in a 4 x 1 array.

 9  10 Ba     7   6    3.

5 17

 5  22 Bb     13   9   

Get the solution array by premultiplying the constant array by the inverse of the coefficient matrix.

 xa   ya   Ca 1 Ba  za   wa   

xa  3.547

ya  4.884

za  1.431

wa  1.334

 xb   yb   Cb 1 Bb  zb   wb   

xb  62.029

yb  0.235

zb  17.897

wb  24.397




Figure P10-1 shows a bracket made of steel. a. Find the location of its centroid referred to point B. b. Find its mass moment of inertia Ixx about the x axis through point B. c. Find its mass moment of inertia Iyy about the y axis through point B.

Given:

Dimensions in the figures below. Density: ρ  7800 kg m

Solution:


1.

3

Divide the bracket into five volumes, find the location of the CG and the mass moments for each of them and then add the results to get the CG and mass moments for the entire bracket. i  1 2  5

2.

Volume 1 is the rectangular prism with two negative cylinders shown below. c

Dimensions:

a

Y

Y

a  64 mm r

b  38 mm c  19 mm d  20 mm

X

Z

b

e  19 mm

d

r  5  mm

e

e

Determine the location of the CG in global coordinates. Xcg  1

c

Xcg  9.500 mm

a b Ycg  1

Zcg  0  mm

1

2 b 2

1

2

 2  π r  d d

2

Ycg  1.069 mm 1

a  b  2  π r

Determine the volume and mass of this segment. 2

V  a  b  c  2  π r  c 1

4

V  4.322  10 mm

3

M  ρ V

1

1

M  0.337 kg

1

1

Determine the mass moments about the local axes through the CGs. Rectangular prism:

ma  a  b  c ρ Iax  Iay  Iaz 

ma 12 ma 12 ma 12

ma  0.360 kg

2

2

2

2

2

2

 a b  a c  b c



Iax  1.664  10



Iay  1.339  10



Iaz  5.421  10

4

4

5

2

kg m

2

kg m

2

kg m



mb  π r  c ρ

Cylinder (one):

mb  0.012 kg

2

Ibx  Iby  Ibz 

mb r

7

Ibx  1.455  10

2

 12

2

2

 12

2

2

mb

 3 r  c

mb

 3 r  c



Iby  4.229  10



Ibz  4.229  10

7

7

2

kg m

2

kg m

2

kg m

Determine the mass moments about the global axes. Ixx  Iax  ma  Ycg



1

2

Iyy  Iay  ma  Xcg



1

Izz  Iaz  ma  Xcg



1

3.

  Zcg1   2  I 2

bx

1

  2

1

 Zcg

2



1



 2  Iby  mb 



2

bz

1

2

4

Ixx  1.581  10



1

c

2

4

 

Iyy  1.634  10

2

  Ycg1   2 I 2

 mb e

1

5



Izz  8.631  10 1

2

kg m

2

kg m

2

kg m

Volume 2 is the bend just above segment 1. It is a quarter hollow cylinder with dimensions shown below. Dimensions:

c

a  13 mm b  32 mm c  64 mm

Y, y

b

d  18 mm

x

a

Y

e  32 mm

y z d X

Z

e

Determine the location of the CG in global coordinates. Xcg  e  2

Ycg  d  2

4 3 π 4 3 π



3

3

2

2

3

3

2

2

b a b a



b a b a

Xcg  16.825 mm

Zcg  0  mm

2

2

Ycg  33.175 mm 2


V 

2

π b  c  π a  c

2

4

V  4.298  10 mm 2

4

3

M  ρ V 2

M  0.335 kg

2

2

Determine the mass moments about the local axes noted on the drawing of the segment. M Ix 

2

12



2

2

 3 a  3 b  c



2

4

Ix  2.144  10

2

kg m


M Iy 

12 M

Iz 

2

2

2




2

2

 3 a  3 b  c

2

 a b



4

2

Iy  2.144  10



4

2

Iz  2.000  10

2

kg m

2

kg m

Determine the mass moments about the global axes. Ixx  Ix  M  d 2

4

2

Ixx  3.230  10

2

2

Iyy  Iay  M  e 1

2

4

2

Iyy  4.771  10

2

Izz  Iaz  M  d  e 1

4.

2

1



4

2

Izz  5.061  10 1

2

kg m

2

kg m

2

kg m

Volume 3 is a rectangular prism with dimensions shown below. c

Dimensions:

a

a  64 mm b

b  19 mm

Y

c  38 mm d  40.5 mm

d

Y

e  51 mm X

Z

e

Determine the location of the CG in global coordinates. Xcg  e

Xcg  51.000 mm

3

Ycg  d

3

Ycg  40.500 mm

3

3

Zcg  0  mm 3


V  a  b  c

V  4.621  10 mm

3

3

M  ρ V

3

3

M  0.360 kg

3

3

Determine the mass moments about the local axes through the CGs. M Ix 

M Iy 

3

12 M

Iz 

3

12

3

12

2

2

2

2

2

2

 a b  a c  b c



Ix  1.339  10



Iy  1.664  10



Iz  5.421  10

4

4

5

2

kg m

2

kg m

2

kg m

Determine the mass moments about the global axes. Ixx  Ix  M   Ycg 3

3



  Zcg3  2

3

2

4

Ixx  7.250  10 3

2

kg m



Iyy  Iy  M   Xcg 3

3



Izz  Iz  M   Xcg 3

5.



3

  Zcg3 

Iyy  1.104  10

  Ycg3 

Izz  5.061  10

2

3

2

3

2

3

4

2

3

1

2

kg m

2

kg m

Volume 4 is a half cylinder with dimensions shown below.

Dimensions: a  19 mm b  40.5 mm c  70 mm

X r

r  32 mm c

Z

a

b

Y X

Determine the location of the CG in global coordinates. 4 r

Xcg  c  4

Xcg  83.581 mm

Ycg  b

4

3 π

Ycg  40.500 mm

4

3

Zcg  0  mm 4


V  a  4

π r

4

V  3.056  10 mm

3

4

2

M  ρ V 4

M  0.238 kg

4

4

Determine the mass moments about the local axes of the segment. M Ix 

M Iy 



2

 3 r  a



2

r

4

12



5

Ix  6.820  10

4

4 2

2 M

Iz 

4

12

Iy  1.221  10 2

 3 r  a



2

Determine the mass moments about the global axes.

5

Iz  6.820  10

2

kg m

2

kg m

2

kg m



Ixx  Ix  M   Ycg 4

4



4



Izz  Iz  M   Xcg 4

6.

4



Ixx  4.592  10

  Zcg4 

Iyy  1.787  10

  Ycg4 

Izz  2.124  10

2

Iyy  Iy  M   Xcg 4

  Zcg4  2

4

2

2

4

2

2

4

4

4

3

4

3

4

2

kg m

2

kg m

2

kg m

Volume 5 is a negative cylinder with dimensions shown below. Dimensions:

a r

a  70 mm b  40.5 mm c  19 mm r  13 mm

X

Z

c

b

Y X

Determine the location of the CG in global coordinates. Xcg  a

Xcg  70.000 mm

5

Ycg  b

5

Ycg  40.500 mm

5

5

Zcg  0  mm 5


4

V  c π r

V  1.009  10 mm

5

3

5

M  ρ V 5

M  0.079 kg

5

5

Determine the mass moments about the local axes of the segment. M Ix 

12 M

Iy 



2

 3 r  c



2

5 2

r

2 M

Iz 

5

5

12



Ix  5.691  10 Iy  6.649  10

2

 3 r  c



2

Iz  5.691  10

6

6

6

2

kg m

2

kg m

2

kg m



Determine the mass moments about the global axes. Ixx  Ix  M   Ycg 5

5



5



Izz  Iz  M   Xcg 5

a.

5



Ixx  1.348  10

  Zcg5 

Iyy  3.922  10

  Ycg5 

Izz  5.203  10

2

Iyy  Iy  M   Xcg 5

  Zcg5  2

5

2

2

5

2

2

5

4

5

4

5

5

4

2

kg m

2

kg m

2

kg m

Find the location of its centroid referred to point B.

 Xcg iM i i

XCg 



M

XCg  34.919 mm i

i

 YcgiM i i

YCg 



M

YCg  26.688 mm i

i

 ZcgiM i i

ZCg 



M

ZCg  0.000 mm i

i

b.

Find its mass moment of inertia Ixx about the X axis through point B. IXX 

 Ixxi

IXX  1.531  10

3

2

kg m

i

c.

Find its mass moment of inertia Iyy about the Y axis through point B. IYY 

 Iyyi i

3

IYY  2.976  10

2

kg m




Two springs are connected in series. One has a k of 34 and the other a k of 3.4. Calculate their effective spring constant. Which spring dominates? Repeat with the two springs in parallel. Which spring dominates? (Use any unit system.)

Given:

Spring constants: k1  34

Solution: 1.

N mm

k2  3.4

N mm


For springs in series, use equation 10.19c to find the effective spring rate. keff 

1  1 k   1 k2 

1

keff  3.091

N mm

The spring with the smaller rate dominates. 2.

For springs in parallel, use equation 10.20b to find the effective spring rate. keff  k1  k2 The spring with the larger rate dominates.

keff  37.400

N mm




Two springs are connected in series. One has a k of 125 and the other a k of 25. Calculate their effective spring constant. Which spring dominates? Repeat with the two springs in parallel. Which spring dominates? (Use any unit system.)

Given:

Spring constants: k1  125 

Solution: 1.

N mm

k2  25

N mm



1  1 k   1 k2 

1

keff  20.833

N mm

The spring with the smaller rate dominates. 2.

For springs in parallel, use equation 10.20b to find the effective spring rate. keff  k1  k2 The spring with the larger rate dominates.

keff  150.000

N mm




Two springs are connected in series. One has a k of 125 and the other a k of 115. Calculate their effective spring constant. Which spring dominates? Repeat with the two springs in parallel. Which spring dominates? (Use any unit system.)

Given:

Spring constants: k1  125 

Solution: 1.

N mm

k2  115 

N mm



1  1 k   1 k2 

1

keff  59.896

N mm

Neither spring dominates since they are both about equally stiff. 2.

For springs in parallel, use equation 10.20b to find the effective spring rate. keff  k1  k2

keff  240.000

N mm

Neither spring dominates since they are both about equally stiff.




Two dampers are connected in series. One has a damping factor of c1 = 12.5 and the other, c2 = 1.2. Calculate their effective damping constant. Which damper dominates? Repeat with the two dampers in parallel. Which damper dominates? (Use any unit system.)

Given:

Damping factors: c1  12.5

Solution: 1.

N  sec mm

c2  1.2

N  sec mm


For dampers in series, use equation 10.18a to find the effective damping factor. 1 1 ceff      c1 c2 

1

ceff  1.095

N  sec mm

The softer damper dominates. 2.

For dampers in parallel, use equation 10.18b to find the effective damping factor. ceff  c1  c2 The stiffer damper dominates.

ceff  13.700

N  sec mm




Two dampers are connected in series. One has a damping factor of c1 = 12.5 and the other, c2 = 2.5. Calculate their effective damping constant. Which damper dominates? Repeat with the two dampers in parallel. Which damper dominates? (Use any unit system.)

Given:


Solution: 1.

N  sec mm

c2  2.5

N  sec mm


For dampers in series, use equation 10.18a to find the effective damping factor. ceff 

1  1 c   1 c2 

1

ceff  2.083

N  sec mm

The softer damper dominates. 2.

For dampers in parallel, use equation 10.18b to find the effective damping factor. ceff  c1  c2 The stiffer damper dominates.

ceff  15.000

N  sec mm




Two dampers are connected in series. One has a damping factor of c1 = 12.5 and the other, c2 = 10. Calculate their effective damping constant. Which damper dominates? Repeat with the two dampers in parallel. Which damper dominates? (Use any unit system.)

Given:


Solution: 1.

N  sec mm

c2  10

N  sec mm


For dampers in series, use equation 10.18a to find the effective damping factor. ceff 

1  1 c   1 c2 

1

ceff  5.556

N  sec mm

Neither damper dominates since they are both about equally stiff. 2.

For dampers in parallel, use equation 10.18b to find the effective damping factor. ceff  c1  c2

ceff  22.500

N  sec mm

Neither damper dominates since they are both about equally stiff.




Given:

A mass of m = 2.75 and a spring with k = 48 are attached to one end of a lever at a radius of 5. Calculate the effective mass and effective spring constant at a radius of 10 on the same lever. (Use any unit system.) Mass:

Spring:

M  2.75 kg Solution: 1.

k  48

N mm

Radius: r1  5  mm


The effective mass and spring must have the same energy as the original. Taking the mass first and using equation 10.22b, 2

 r1  meff     M  r2  2.

r2  10 mm

meff  0.688 kg

To find the effective spring rate, use equation 10.23b. 2

 r1  keff     k  r2 

keff  12.000

N mm




Given:


Spring:


k  24

N mm

Radius: r1  3  mm



 r1  meff     M  r2  2.

r2  10 mm

meff  0.135 kg


 r1  keff     k  r2 

keff  2.160

N mm




Given:


Spring:


k  25

N mm

Radius: r1  15 mm



 r1  meff     M  r2  2.

r2  5  mm

meff  58.500 kg


 r1  keff     k  r2 

keff  225.000 

N mm




Refer to Figure 10-9 and Example 10-1. The data for the valve train are given below. Calculate the effective spring constant and effective mass of a single-DOF equivalent system placed on the cam side of the rocker arm. (Use ips unit system.)

Given:

Tappet (solid cylinder): d tp  0.75 in Ltp  1.25 in Pushrod (hollow cylinder): Rocker arm:

w  1.00 in

Cam shaft (cam in center): Valve spring:

d pid  0.250  in

h  1.50 in

d cs  1.00 in

a  2.00 in

Lpr  12 in b  3.00 in

Lcs  3.00 in

1

kvs  200  lbf  in

All parts are steel: Solution:

d pod  0.375  in

Modulus of elasticity

6

E  30 10  psi

γ  0.3

Spec. weight

1.

Break the system into individual elements as shown in Figure 10-9b.

2.

Define the individual spring constants of each of the six elements. Cam shaft (simply-supported beam with central load): 4

Spring constant

π d cs

Ics 

64 48 E Ics

kcs 

6 lbf

kcs  2.618  10

3

Lcs

in

Tappet (solid cylinder): Area

Spring constant

Atp 

π d tp

2

4 Atp  E

ktp 

7 lbf

ktp  1.060  10

Ltp

in

Pushrod (hollow cylinder): π  d pod  d pid 2

Area

Spring constant

Apr  kpr 

2



4 Apr  E Lpr

kpr  1.534  10

5 lbf

kra  3.164  10

6 lbf

krb  9.375  10

5 lbf

in

Rocker arm (side A): Moment of inertia

Spring constant

Ir  kra 

w h

3

12 3  E Ir a

3

in

Rocker arm (side B): Spring constant

krb 

3

in


Moment of inertia

lbf

3  E Ir b

3

in



3.

Damping will be neglected.

4.

Determine the mass of each of the elements. Tappet (solid cylinder): Volume

Vtp  Atp  Ltp

Mass

mtp 

γ Vtp g

mtp  4.291  10

4

2

1

2

1

2

1

2

1

2

1

2

1

lbf  sec  in

Pushrod (hollow cylinder): Volume

Vpr  Apr  Lpr

Mass

mpr 

γ Vpr g

4

mpr  5.721  10

lbf  sec  in

Rocker arm (side A): Volume

Vra  w h  a

Mass

mra 

γ Vra g

3

mra  2.331  10

lbf  sec  in

Rocker arm (side B): Volume

Vrb  w h  b

Mass

mrb 

γ Vrb g

3

mrb  3.497  10

lbf  sec  in

Omit the valve and valve spring because no data are available. 5.

Determine the effective mass and spring constant on either side of the rocker arm.. Left side:

mL  mtp  mpr  mra kL 

Right side:

mR  mrb kR 

6.

1  1  1  1 k   cs ktp kpr kra 

mL  3.332  10 1

lbf  sec  in

5 lbf

kL  1.368  10

mR  3.497  10

 1  1 k   rb kvs 

3

1

kR  199.957

in 3

lbf  sec  in

lbf in

Reduce the system to a single DOF. 2

b meff  mL     mR a 2

b keff  kL     kR a

2

1

meff  0.011 lbf  sec  in

5 lbf

keff  1.372  10

in




Figure P10-2 shows a cam-follower system. The dimensions and other data are given below. Find the arm's mass, center of gravity location and mass moment of inertia about both its CG and arm pivot. Create a linear, one-DOF lumped mass model of the dynamic system referenced to the cam follower. Ignore damping.

Given:

Arm dimensions:

Cutout dimensions:

Width (Z-direction):

a  2  in


a'  1.5 in

Height (Y-direction):

b  2.5 in


b'  2.5 in

Length (X-direction):

c  28 in


c'  3  in

Distance from left end of arm to roller center:

Lr  22 in

Distance from left end of arm to arm pivot:

Lp  10 in

Specific weight of aluminum: Solution: 1.

2.

6

E  10.4 10  psi


Determine the volume and weight of the arm. 3

Vsolid  a  b  c

Vsolid  140.000 in

Vslot  a' b' c'

Vslot  11.250 in

Va  Vsolid  Vslot

Va  128.750 in

Wsolid  γ Vsolid

Wsolid  14.000 lbf

Wslot  γ Vslot

Wslot  1.125 lbf

Wa  Wsolid  Wslot

Wa  12.875 lbf

3

3

Calculate the location of the arm CG with respect to the left end. Wsolid  XCg 

3.

3

γ  0.1 lbf  in

c 2

 Wslot  Lr

Wa

XCg  13.301 in

Determine the moment of inertia of the arm about the its CG. Wsolid

2

Solid portion about its own CG

Izsolid 

Slot portion about its own CG

Izslot 

Solid portion about arm CG

IZsolid  Izsolid 

12 g

 12 g

Wslot

 b c

2

 b'  c'

2



2



Wsolid

2

Izsolid  2.388 lbf  sec  in

Izslot  3.703  10

  XCg 



g 2

IZsolid  2.406 lbf  sec  in Slot portion about arm CG

IZslot  Izslot 

Wslot g

  Lr  XCg 

2

c



2

2

3

2

lbf  sec  in



IZslot  0.224 lbf  sec  in

4.

Wa g

  XCg  Lp

2

2

Ipivot  2.545 lbf  sec  in

Determine the mass and spring constant of the arm on the left side of the pivot. Iarm 

Moment of inertia

kL 

Spring constant

a b

3  E Iarm

4 lbf

kL  8.125  10

3

Volume

VL  a  b  Lp

Mass

mL 

γ  VL

in

2

1

mL  0.013 lbf  sec  in

g

Determine the mass and spring constant of the arm on the right side of the pivot. kR 

Spring constant

6.

3

12

Lp

6.

IZZ  2.181 lbf  sec  in

Determine the moment of inertia of the arm about the arm pivot. Ipivot  IZZ 

5.

2

IZZ  IZsolid  IZslot

Arm about its CG

3  E Iarm

4 lbf

kR  1.393  10

c  Lp 3

Volume

VR  a  b   c  Lp

Mass

mR 

γ  VR g

in

2

Find the effective mass and spring constant referenced to the cam-follower.. 2

 Lr  Lp  meff  mL     mR  Lp  2

 Lr  Lp  keff  kL     kR  Lp 

1

mR  0.023 lbf  sec  in

2

1


5 lbf

keff  1.0131  10

in




Use the method of virtual work to find the torque required to rotate the cam in Figure P10-2 through one revolution. The cam is a pure eccentric.

Given:

Cam eccentricity and speed:

e  0.5 in

ω  500  rpm

Cam radius and thickness:

rc  3  in

wc  0.75 in

Spring rate and preload:

k  123  lbf  in

1

(measured from Figure)

F50  173  lbf

Arm dimensions:

Cutout dimensions:


a  2  in


a'  1.5 in


b  2.5 in


b'  2.5 in


c  28 in


c'  3  in

Roller follower dimensions:

rf  1  in

wf  1.5 in


Lr  22 in

Distance from left end of arm to spring:

Ls  29 in

Distance from left end of arm to pivot:

Lp  10 in 3

Specific weight of aluminum:

γa  0.1 lbf  in

Specific weight of steel:

γs  0.3 lbf  in

3

Assumptions: Small angle theory applies. Solution:

See Figure P10-2, Problem 10-16, and Mathcad file P1017.

1.

Start with a kinematic analysis of the mechanism. Let the cam be link 2, the roller follower link 3, the arm link 4, and the spring link 5. Let the point where the spring connects to the arm be point B, and the center of the roller follower be point A.

2.

Calculate the motion of the roller follower center with respect to the X axis, which goes through the center of rotation of the cam. The cam and follower are shown at right. The origin of the coordinate frame is at the center of rotation. The cam, link 2, is shown rotated an amount 2 from the position at which the follower, link 3, is at its lowest position. A triangle is formed with sides e, rf + rc, and yf. This triangle will be used to determine the displacement of the follower, yf, with respect to the cam rotation angle, 2.

3

yf

r f + rc

 

X 2

 

e

 rf  rc

 

 sin θ

e

2



From the relationship among the angles of a triangle,

rc



Using the law of sines, β θ  asin

rf

Y

 

 

γ θ  θ  β θ



And, using the law of cosines,

yf θ  e   rf  rc  2  e  rf  rc  cos γ θ  e   rf  rc

Plotting yf as a function of 2:

θ  0  deg 2  deg  360  deg

 

2

2

  

2



FOLLOWER DISPLACEMENT

Follower displacement, in

1.5

 

1

yf θ in

0.5

0

0

45

90

135

180

225

270

315

360

θ deg Cam Rotation Angle, deg

3.

Differentiate the displacement function twice to get the velocity and acceleration functions.

 

vf θ 

 

vf θ 

 d y θ    ω  θ f   d  

 

a f θ 

40

2000

20

1000

sec in

 

af θ 

0

 20

 40

4.

 d v θ    ω  θ f   d  

2

sec

0

in  1000

0

90

180

270

360

 2000

0

90

180

θ

θ

deg

deg

Determine the angular displacement, velocity, and acceleration of the arm, link 4.

270

360


 

θ θ 

 

α θ  5.

1 Lr  Lp

 

 

 yf θ

1

ω θ 

Lr  Lp

 

 vf θ

 

 a f θ

Determine the displacement and velocity of point A, where the spring attaches to the arm.

 

y5 θ 

6.

1 Lr  Lp


Ls  Lp Lr  Lp

 

 

 yf θ

v5 θ 

Ls  Lp Lr  Lp

 

 vf θ

Calculate the spring force as a function of 2.

 

 

F5 θ  k y5 θ  F50 7.

The spring force is the only external force on the linkage that is not applied at a joint and is, therefore, the only force to be considered in the first term of equation 10.28b. Determine the first term in the equation.

 

   

EF θ  F5 θ  v5 θ

8.

Determine the components that make up the third term in equation 10.28b. Links 2 and 4 are rotating about fixed pivots so their input can be taken care of in the fourth term. The roller follower has both rotation and translation. The translation motion will be accounted for in this term. Mass of follower:

 

2

mf  π rf  wf 

γs

mf  1.414 lb

g

   

IF θ  mf  a f θ  vf θ

9.

Determine the components that make up the fourth term in equation 10.28b. There will be input from the rotation of the arm. The cam and follower have no angular acceleration so they have no contribution. From Problem 10-26, we have the following data: Mass of roller and arm combined:

mra 

14.289 lbf g

mra  14.289 lb

Distance from left end of arm to composite CG:

XCg  14.162 in

Composite moment of inertia with respect to CG:

ICG  2.433  lbf  sec  in

2

Moment with respect to the arm pivot: I4  ICG  mra  XCg  Lp

 

2

2

I4  3.074 lbf  sec  in

   

IT θ  I4 α θ  ω θ

10. Using equation 10.28b, solve for and plot the input torque as a function of cam angle.


 

T2 θ 

1 ω


  

 

 

 IF θ  IT θ  EF θ

INPUT TORQUE ON CAM 300

200

Torque, lbf-in

100

 

T2 θ

lbf  in

0

 100

 200

 300

0

45

90

135

180 θ deg Cam Angle, deg

225

270

315

360




Use the method of virtual work to find the torque required to rotate the cam in Figure P10-3 through one revolution. The cam is a pure eccentric.

Given:


e  20 mm


rc  3  in


k  10 N  m

ω  200  rpm wc  0.75 in

1


F50  0.2 N

Arm dimensions:

Cutout dimensions:


a  2  in


a'  1.5 in


b  2.5 in


b'  2.5 in


c  28 in


c'  3  in


rf  1  in

wf  1.5 in


Lr  22 in


Ls  29 in


Lp  10 in 3


γa  0.1 lbf  in


γs  0.3 lbf  in

3

Assumptions: Small angle theory applies. Solution:


1.


2.


3

yf

r f + rc

 

X 2

 

e

 rf  rc

 

 sin θ

e

2




rc




rf

Y

 

 

γ θ  θ  β θ






θ  0  deg 2  deg  360  deg

 

2

  

2

2



FOLLOWER DISPLACEMENT

Follower displacement, mm

50 40

  30

yf θ mm

20 10 0

0

45

90

135

180

225

270

315

360


3.


 

vf θ 

 d y θ    ω  θ f   d  

 

a f θ 

600

 d v θ    ω  θ f   d  

10000

400

5000

200

 

vf θ 

sec mm

 

af θ 

0

 200

4.

0

mm

 5000  10000

 400  600

2

sec

0

90

180

270

360

 15000

0

90

180

θ

θ

deg

deg

Determine the angular displacement, velocity, and acceleration of the arm, link 4.

270

360


 

θ θ 

 

α θ  5.

1 Lr  Lp

 

 yf θ

 

1

ω θ 

Lr  Lp

 

 vf θ

 

 a f θ

Determine the displacement and velocity of point A, where the spring attaches to the arm.

 

y5 θ 

6.

1 Lr  Lp


Ls  Lp Lr  Lp

 

 

 yf θ

v5 θ 

Ls  Lp Lr  Lp

 

 vf θ


 

 

F5 θ  k y5 θ  F50 7.


 

   

EF θ  F5 θ  v5 θ

8.


 

mf  1  kg

   

IF θ  mf  a f θ  vf θ

9.

Determine the components that make up the fourth term in equation 10.28b. There will be input from the rotation of the arm. The cam and follower have no angular acceleration so they have no contribution. From Problem 10-26, we have the following data: Mass of roller and arm combined:

mra 

14.289 lbf g

mra  6.481 kg


XCg  14.162 in


ICG  2.433  lbf  sec  in

2


 

2

2

I4  0.347 kg m

   

IT θ  I4 α θ  ω θ



 

T2 θ 

1 ω


  

 

 

 IF θ  IT θ  EF θ

INPUT TORQUE ON CAM 1

Torque, N-m

0.5

 

T2 θ Nm

0

 0.5

1

0

45

90

135


225

270

315

360




Use the method of virtual work to find the torque required to rotate the cam in Figure P10-3 through one revolution. The cam motion is a double harmonic.

Given:

Cam lift and speed:

h  20 mm

ω  200  rpm


rc  3  in

wc  0.75 in


k  10 N  m

1

F50  0.2 N

Arm dimensions:

Cutout dimensions:


a  2  in


a'  1.5 in


b  2.5 in


b'  2.5 in


c  28 in


c'  3  in


Solution:


rf  1  in

wf  1.5 in


Lr  22 in


Ls  29 in


Lp  10 in 3


γa  0.1 lbf  in


γs  0.3 lbf  in

3


1.


2.

Calculate the motion of the roller follower using equations 8.25. Define  and a range function so the rise and fall can be combined in a single equation. R θ a b   if  θ  a    θ  b  1 0

β  180  deg

θ    θ   1   h  yf θ  R θ 0 β    1  cos π      1  cos 2  π    2  β    β  4   θ  β    θ  β   1   h   R θ β 2  π    1  cos π      1  cos 2 π   2  β  4  β    

 







 



θ     θ     ω   π  h2   sin π    12  sin 2 π     β  β     β  θ  β      π h   θ  β  1  R θ β 2 π  β  2   sin π β   2  sin 2  π β          

vf θ  R θ 0 β 

 

 2   ω     2  R θ β 2  π  π  h   cos π θ  β   cos 2  π θ  β     2 2        β  β      β   





a f θ  R θ 0 β 

2

θ    h   θ     cos π   cos 2  π    2 2   β β   β π




θ  0  deg 2  deg  360  deg FOLLOWER DISPLACEMENT


20

15

 

yf θ

10

mm 5

0

0

45

90

135

180

225

270

315

360


3.

Plot the velocity and acceleration functions. 400

5000

200 0

 

vf θ 

sec mm

 

af θ 

0

2

sec

mm  5000

 200

 400

4.

0

90

 

 

α θ 

 10000

360

0

90

 

180

θ

θ

deg

deg

1 Lr  Lp 1 Lr  Lp

 

 yf θ

 

ω θ 

1 Lr  Lp

 

 vf θ

 

 a f θ

Determine the displacement and velocity of point A, where the spring attaches to the arm. y5 θ 

6.

270

Determine the angular displacement, velocity, and acceleration of the arm, link 4. θ θ 

5.

180

Ls  Lp Lr  Lp

 

 yf θ


 

v5 θ 

Ls  Lp Lr  Lp

 

 vf θ

270

360


 


 

F5 θ  k y5 θ  F50 7.


 

   

EF θ  F5 θ  v5 θ 8.


 

mf  1  kg

   

IF θ  mf  a f θ  vf θ 9.

Determine the components that make up the fourth term in equation 10.28b. There will be input from the rotation of the arm. The cam and follower have no angular acceleration so they have no contribution. From Problem 10-26, we have the following data: mra 

Mass of roller and arm combined:

14.289 lbf g

mra  6.481 kg


XCg  14.162 in


ICG  2.433  lbf  sec  in

2


 

2

2

I4  0.347 kg m

   

IT θ  I4 α θ  ω θ


 

T2 θ 

1 ω

  

 

 

 IF θ  IT θ  EF θ

INPUT TORQUE ON CAM 0.4

Torque, N-m

0.2

 

T2 θ Nm

0

 0.2

 0.4

0

45

90

135

180 θ deg

225

270

315

360




A 3000-lb automobile has a final drive ratio of 1:3 and transmission gear ratios of 1:4, 1:3, 1:2, and 1:1 in first through fourth speeds, respectively. What is the effective mass of the vehicle as felt by the engine flywheel in each gear?

Units:

blob  lbf  sec  in

Given:

Gear ratios:

2

1

Final

mGf  3

First

mG1  4

Third

mG3  2

Fourth

mG4  1

Weight of automobile: Solution: 1.

Calculate the mass of the vehicle. W

The effective mass of the vehicle at the flywheel while in first gear is, M

meff1  0.0540 blob

2

M


The effective mass of the vehicle at the flywheel while in third gear is, meff3   mGf  mG3

5.

2

The effective mass of the vehicle at the flywheel while in second gear is, meff2   mGf  mG2

4.

M  7.770 blob

g

meff1   mGf  mG1 3.

W  3000 lbf


M  2.

Second

2

M


The effective mass of the vehicle at the flywheel while in fourth gear is, meff4   mGf  mG4

2

M


mG2  3




Determine the effective spring constant and effective preload of the spring in Figure P10-2 as reflected back to the cam follower. See Problem 10-17 for additional data.

Given:

Spring data:

Solution: 1.

1

k  123  lbf  in

Fo  173  lbf

Distance from arm pivot to spring:

a  19 in

Distance from arm pivot to follower:

b  12 in


The effective spring constant at the follower is, 2

keff 

2.

 a  k   b

keff  308.354

lbf in

The effective preload at the follower is, Foeff 

a b

 Fo

Foeff  273.917 lbf




What is the effective inertia of a load applied at the drum of Figure P9-5 as reflected back to gear A?

Given:

From Problem 9-35: Ratio of drum speed to speed of gear A

Solution: 1.

ratio 

5.4 5.4  1

ratio  1.227


The energy of the effective inertia at gear A must be the same as that at the drum, 1 2

1

2

 Idrum ωdrum 

Ieff Idrum

2

2

 Ieff  ωA

2

 ωdrum  2 =  = ratio ω  A 

Thus, the ratio of the effective inertia to the drum inertia will be the square the ratio of drum speed to the speed of gear A. 2

Ratioinertia  ratio

Ratioinertia  1.506




What is the effective inertia of a load applied at the drum of Figure P9-7 as reflected back to the arm?

Given:

From Problem 9-40: Ratio of drum speed to speed of the arm

Solution: 1.

ratio  16.5


The energy of the effective inertia at the arm must be the same as that at the drum, 1 2

1

2

 Idrum ωdrum 

Ieff Idrum

2

2

 Ieff  ωarm

2

 ωdrum  2 =  = ratio ω  arm 

Thus, the ratio of the effective inertia to the drum inertia will be the square the ratio of drum speed to the speed of the arm. 2

Ratioinertia  ratio

Ratioinertia  272.250




Refer to Figure 10-8a (p. 507). For the data given below, find the equivalent mass at point A and the equivalent spring at point B.

Given:

a  100  mm

Solution:


1.

b  150  mm

1

kA  2000 N  m

mB  2  kg

Use equation 10.22b to calculate the equivalent mass at point A. 2

meff 

2.

 b  m   B a

meff  4.5 kg

Use equation 10.23b to calculate the equivalent spring constant at point B. 2

keff 

 a  k   A b

keff  888.9

N m




Refer to Figure 10-8a (p. 507). For the data given below, find the equivalent mass at point A and the equivalent spring at point B.

Given:

a  50 mm

Solution:


1.

b  150  mm

1

kA  1000 N  m

mB  3  kg

Use equation 10.22b to calculate the equivalent mass at point A. 2

meff 

2.

 b  m   B a

meff  27 kg

Use equation 10.23b to calculate the equivalent spring constant at point B. 2

keff 

 a  k   A b

keff  111.1

N m




For the cam-follower arm in Figure P10-2, determine the location of its fixed pivot that will have zero reaction force when the cam applies its force to the follower.

Given:

Arm dimensions:

Cutout dimensions:


a  2  in


a'  1.5 in


b  2.5 in


b'  2.5 in


c  28 in


c'  3  in

r  1  in

Roller dimensions:

w  1.5 in


Lr  22 in 3


γa  0.1 lbf  in


γs  0.3 lbf  in

3

Assumptions: The center of the steel roller is located on the horizontal centerline (X axis) of the arm. Solution: 1.


Determine the volume and weight of the roller and the arm. Roller:

Arm:

2.

2

Vr  4.712 in

Wr  γs Vr

Wr  1.414 lbf

Vsolid  a  b  c

Vsolid  140.000 in

Vslot  a' b' c'

Vslot  11.250 in

Va  Vsolid  Vslot

Va  128.750 in

Wsolid  γa Vsolid

Wsolid  14.000 lbf

Wslot  γa Vslot

Wslot  1.125 lbf

Wa  Wsolid  Wslot

Wa  12.875 lbf

Xcga 

3

3

c 2

 Wslot  Lr

Wa

Xcga  13.301 in

Calculate the location of the composite CG with respect to the left end. XCg 

4.

3

Calculate the location of the arm CG with respect to the left end. Wsolid 

3.

3

Vr  π r  w

Wr Lr  Wa Xcga W r  Wa

XCg  14.162 in

Determine the moment of inertia of the arm about the composite CG

Solid portion about its own CG

Izsolid 

Wsolid 12 g

2

 b c



2

2

Izsolid  2.388 lbf  sec  in



Slot portion about its own CG

Izslot 

Solid portion about composite CG

Wslot 12 g



2

 b'  c'

2

IZsolid  Izsolid 



2

Izslot  0.0037 lbf  sec  in

Wsolid

  XCg 



g

c

2

 2

2

IZsolid  2.389 lbf  sec  in Slot portion about composite CG

IZslot  Izslot 

Wslot

  Lr  XCg 

g

2

2

IZslot  0.183 lbf  sec  in Arm about composite CG

IZZa  IZsolid  IZslot 2

IZZa  2.206 lbf  sec  in 5.

Determine the moment of inertia of the roller about the composite CG Wr

Roller about its own CG

Izr 

Roller about composite CG

IZZr  Izr 

2 g

3

2

r

Izr  1.831  10 Wr g

  Lr  XCg 

2

2

IZZr  0.227 lbf  sec  in 6.

Determine the moment of inertia of the assembly about the composite CG 2

IZZ  IZZa  IZZr 7.

Calculate the radius of gyration using equation 10.11b.

k 

8.

IZZ  2.433 lbf  sec  in

IZZ g

k  8.108 in

Wa  Wr

Use equation 10.15b to calculate the location of the pivot point with respect to the left end.

Xpivot  XCg 

k

2

Lr  XCg

Xpivot  5.775 in

2

lbf  sec  in




Figure P10-4 shows a fourbar mechanism. The crank is 1.00in wide by 0.5 in thick. The coupler and rocker are both 0.75 in wide by 0.5 in thick. All links are made from steel. The ends of the links have a full radius equal to one half of the link width. The pivot pins all have a diameter of 0.25 in. Find the moment of inertia of the crank and rocker about their fixed pivots and the moment of inertia of the coupler about its CG.

Given:

Dimensions in the figure below. Specific weight of steel:   0.28 lbf  in

3

Pivot pin diameter: Solution:

d  0.25 in


1.

Divide each link into 3 volumes: 1) the main, rectangular body; 2) the half-cylinrical ends; and 3) the pivot pin holes.

2.

All links have the same basic shape, which is shown below.

d, 2 places

CG

w

L/2 t L 2.

Volume 1 is the main rectangular body of length L, width w, and thickness t. Mass:

m1( L w t) 

L w t  g I1( L w t) 

Moment of inertia about CG: 3.

m1( L w t) 12

2

 L w



2

Volume 2 is the half-cylinder of diameter w, and length t. There are two of these. 2

Mass:

m2( w t) 

π w  t   8 g

 m2( w t)  w2 2 L I2( L w t)  2    m2( w t)     8  2 


Volume 3 is the cylinder of diameter d, and length t. There are two of these and they are negative. 2

Mass:

m3( t)  

π d  t  


8 g

 m3( t)  d2 2 L    I3( L t)  2   m3( t)     8 2 

For the crank: L  2.00 in , w  1.00 in , t  0.50 in Mass:

mcrank  m1( L w t)  m2( w t)  m3( t)

mcrank  8.587  10

4

2

1

lbf  sec  in



Moment of inertia about CG: IcrankCG  I1( L w t)  I2( L w t)  I3( L t)

IcrankCG  6.046  10

4

2

lbf  sec  in

Moment of inertia about pivot axis: Icrank  IcrankCG  mcrank  

L

2

3

 2

6.

Icrank  1.463  10

2

lbf  sec  in

For the rocker: L  7.187  in , w  0.75 in , t  0.50 in Mass:

mrocker  m1( L w t)  m2( w t)  m3( t)

mrocker  2.026  10

3

2

1

lbf  sec  in

Moment of inertia about CG: IrockerCG  I1( L w t)  I2( L w t)  I3( L t)

2

IrockerCG  0.010 lbf  sec  in

Moment of inertia about pivot axis: Irocker  IrockerCG  mrocker 

L

2

 2

7.

2

Irocker  0.037 lbf  sec  in

For the coupler L  8.375  in , w  0.75 in , t  0.50 in Mass:

mcoupler  m1( L w t)  m2( w t)  m3( t)

3

mcoupler  2.349  10

2

Moment of inertia about CG: IcouplerCG  I1( L w t)  I2( L w t)  I3( L t)

1

lbf  sec  in

2

IcouplerCG  0.016 lbf  sec  in




The rocker in Figure 10-11a has the dimensions given below. When supported on knife-edges at A and B, the weights at the supports were found to be 4.3 N and 5.8 N, respectively. The rocker was supported at its pivot point with a low-friction ball bearing and the period of oscillation was found. What is the approximate moment of inertia of the rocker with respect to its pivot axis?

Given:

Dimensions: a  50.8 mm b  76.2 mm Total weight:

W  10.1 N

Period of oscillation: Solution: 1.

WA  4.3 N

WB  5.8 N

  0.75 sec


Determine the location of the CG by summing moments on the rocker about point A. The distance, r, from the pivot axis to the CG is: W  ( a  r)  WB ( a  b ) = 0 r 

2.

WB ( a  b )  W  a W

r  22.131 mm

Use equation 10.10g to calculate the approximate moment of inertia with respect to the pivot axis.

   Ipivot  W  r    2 π 

2

Ipivot  3185 kg mm

2




The arm in Problem 10-16 and Figure P10-2 has been redesigned such that the cross-section is no longer uniform and the material changed from aluminum to steel. However, the dimensions shown in the figure remain unchanged. The new arm has a total weight of 15.3 lb and, when supported on knife-edges at points 9.5 in to the left of the pivot and 17.5 in to the right of the pivot, the weights at the supports were found to be 7.1 lb and 8.2 lb, respectively. The arm was supported at its pivot point with a low-friction ball bearing and the period of oscillation was found. What is the approximate moment of inertia of the rocker with respect to its pivot axis?

Given:

Let the weight measurement point on the left of the pivot be A and the distance from the pivot axis to point A be a. Similarly, let the point on the right be B and its distance b. Dimensions: a  9.5 in b  17.5 in Total weight:

W  15.3 lbf

Period of oscillation: Solution: 1.

WA  7.1 lbf

WB  8.2 lbf

  2.0 sec


Determine the location of the CG by summing moments on the rocker about point A. The distance, r, from the pivot axis to the CG is: W  ( a  r)  WB ( a  b ) = 0 r 

2.

WB ( a  b )  W  a W

r  4.971 in

Use equation 10.10g to calculate the approximate moment of inertia with respect to the pivot axis.

     2 π 

Ipivot  W  r 

2

2

Ipivot  7.7 lbf  sec  in




Figure P10-5 shows a cam-follower system that drives slider 6 through an external output arm 3. Arms 2 and 3 are both rigidly attached to the 0.75-in-dia shaft X-X, which rotates in bearings that are supported by the housing. The pin-to-pin dimensions of the links are shown. The cross-section of arms 2, 3, and 5 are solid, rectangular 1.5 x 0.75 in steel. The ends of these links have a full radius equal to one half the link width. Link 4 is 1-in-dia x 0.125 in-wall, round steel tubing. Link 6 is a 2-in-dia x 6-in long solid steel cylinder. Find the effective mass and effective spring constant of the follower train referenced to the cam-follower roller.

Given:

Link 6 (solid cylinder): d 6  2.00 in L6  6.00 in Rocker arm 5:

w  0.75 in

h  1.50 in

a  10.0 in

Pushrod 4 (hollow cylinder): d 4od  1.00 in Output arm 3:

h  1.50 in

Spring:

ks  150  lbf  in

Roller arm 2:

w  0.75 in

h  1.50 in

L2  8  in 6

E  30 10  psi


L4  22 in

L3  16 in

1


w  0.75 in

d 4id  0.75 in

b  8.0 in

γ  0.28

Spec. weight

Break the system into individual elements as shown in Figure 10-9b.

2.

Define the individual spring constants of each of the six elements. Roller arm (cantilever beam with end load): Moment of inertia

Spring constant

I2  k2 

w h

3

12 3  E I2 L2

Output arm: Moment of inertia

Spring constant

I3  k3 

w h

3

k2  3.708  10

4 lbf

k3  4.635  10

3 lbf

k4  4.686  10

5 lbf

in

3

12 3  E I3 L3

3

in

Pushrod (hollow cylinder): π  d 4od  d 4id 2

Area

Spring constant

A4  k4 

2



4 A4  E L4

in

Rocker arm 5 (side B): Moment of inertia

Spring constant

I5  k5a 

w h

3

12 3  E I5 a

3

3

in

See Figures P10-5 and 10-9, and Mathcad file P1030.

1.

lbf

4 lbf

k5a  1.898  10

in



Rocker arm 5 (side C): Spring constant

k5b 

3  E I5 b

3.

Damping will be neglected.

4.

Determine the mass of each of the elements.

3

4 lbf

k5b  3.708  10

in

Roller arm 2 (cantilever beam with end load): Volume

V2  w h  L2

Mass

m2 

γ  V2

3

m2  6.527  10

g

2

1

2

1

lbf  sec  in

Outout arm 3 (cantilever beam with end load): Volume

V3  w h  L3

Mass

m3 

γ  V3

1

2

m3  0.013 lbf  sec  in

g

Pushrod 4 (hollow cylinder): Volume

V4  A4  L4

Mass

m4 

γ  V4

3

m4  5.482  10

g

lbf  sec  in

Rocker arm 5 (side B): Volume

V5a  w h  a

Mass

m5a 

γ V5a g

m5a  8.159  10

3

2

1

2

1

lbf  sec  in

Rocker arm 5 (side C): Volume

V5b  w h  b

Mass

m5b 

Cylinder 6: V6 

Volume

m6 

Mass

γ V5b g π d 6

3

lbf  sec  in

2

4 γ  V6 g

m5b  6.527  10

 L6 2

1

m6  0.014 lbf  sec  in

Omit the spring because no data are available. 5.

Determine the effective mass and spring constant on either side of the rocker arm 5, reflected to point A. Upper side:

mU  m4  m5a

2

1

mU  0.014 lbf  sec  in


kU 

Lower side:


1  1  1  k   s k4 k5a 

1

kU  148.777

lbf in 1

2

mL  m6  m5b


kL  k5b

kL  3.708  10

4 lbf

in

2

mAeff  mU 

 b  m   L a 2

kAeff  kU  6.

 b  k   L a

1

2

mAeff  0.027 lbf  sec  in

4 lbf

kAeff  2.388  10

in

Determine the effective mass and spring constant on either side of the output and roller arms, reflected to the roller. Output side:

kO 

Roller side:

1

2

mO  mAeff  m3

mO  0.040 lbf  sec  in

 1  1 k   Aeff k3 

1

kO  3.881  10

3 lbf

mR  m2

mR  6.527  10

kR  k2

kR  3.708  10

in 3

2

4 lbf

in

2

 L3  meff  mR     mO  L2  2

 L3  keff  kR     kO  L2 

2

1


4 lbf

keff  5.260  10

1

lbf  sec  in

in




The spring in Figure P10-5 has a rate of 150 lb/in with a preload of 60 lb. Determine the effective spring constant and preload of the spring as reflected back to the cam follower. See Problem 10-30 for a description of the system.

Given:

Spring data:

Solution: 1.

1

k  150  lbf  in

Fo  60 lbf

Distance from arm pivot to spring (L3):

a  16 in

Distance from arm pivot to follower (L2):

b  8  in


The effective spring constant at the follower is, 2

keff 

2.

 a  k   b

keff  600

lbf in

The effective preload at the follower is, Foeff 

a b

 Fo

Foeff  120 lbf




A company wants to manufacture chimes that are made from hollow tubes of various lengths. Regardless of length they will be hung from a hole that is 25 mm from one end of the tube. Develop an equation that will give the distance from this hole to the point where the chime should be struck such that there will be zero reaction force at the hole where the chime is hung. The distance should be a function of the length (L), outside diameter (OD), and inside diameter (ID) of the chime as well as the distance from the end to the hanging hole (25 mm) only. Solve your equation for the following dimensions: L = 300 mm, OD = 35 mm, ID = 30 mm.

Given:

Distance to hanging hole from end of tube: d  25 mm Test dimensions: Lt  300  mm

Solution: 1.

IDt  30 mm


Refering to Figure 10-3, let the distance from the hanging hole to the point where the chime should be struck be Ls. Then, Ls = x  lp

2.

ODt  35 mm

where

x=

IG

from equation 10.15a

m lp

The CG of the chime will be located at L/2 from either end. From Appendix C, for a hollow tube IG m

2

=

2

2

0.75 ID  0.75 OD  L 12

and, the distance from the CG to the hanging hole is x = 3.

2

d

Combine these equations to derive the equation for the strike distance, Ls, which is measured from the hanging hole. 2

Ls( L ID OD)  4.

L

2

2

0.75 ID  0.75 OD  L 12 ( 0.5 L  d )



 L  d   2 

Test the equation with the given values. Lstest  Ls Lt IDt ODt

Lstest  186.1 mm

where

d  25 mm




What is the amount by which the roller arm of Problem 10-30 must be extended on the opposite side of the pivot axis O2 in order to make the pivot axis a center of rotation if the point where the cam-follower is mounted is a center of percussion?

Given:

Length of roller arm without extension: L2  8  in Cross-section dimensions: w  1.50 in t  0.125  in

Solution: 1.


Refering to Figure 10-3, let x be the distance from O2 to the CG of the link after extension and lp be the distance from the CG (after extension) to the roller axis. The amount by which the link is extended is L The extended length of the link is L2 + L. L2 = x  lp

2.

where

IG m lp

=

L2  ΔL

from equation 10.15a

2

Eliminating x and lp from these three equations gives (ignoring the rounded ends on the link) 2

ΔL = L2  4 

3.

x=

IG

where

m

IG m

=

L2  ΔL 2  w2

(See Appendix C)

12

Solve for L iteratively.

  L2  ΔL  2  w2  f ( ΔL )  L2  4   12   2

Try

ΔL  3.953  in

Let

ΔL  4.0 in

Then the new length of link 2 is

f ( ΔL )  3.953 in

Lnew  L2  ΔL

Lnew  12.0 in




Figure P10-6 shows a sixbar mechanism with link lengths given in centimeters. The crank (2) is 30 mm wide by 10 mm thick. The couplers (3 and 5) are both 24 mm wide by 8 mm thick. The rocker (4) is 40 mm wide by 12 mm thick. All links are made from steel. The ends of the links have a full radius equal to one half of the link width. The pivot pins all have a diameter of 8 mm. Find the moment of inertia of the crank and rocker about their fixed pivots and the moment of inertia of the couplers about their CGs.

Given:

Dimensions in the figure below. Mass density of steel: ρ  7800 kg m

3

Pivot pin diameter:

Solution:

d  8  mm


1.

Divide each link into 3 volumes: 1) the main, rectangular body of length L; 2) the half-cylinrical ends; and 3) the pivot pin holes.

2.


d, 2 places

CG

w

L/2 t L 2.


m1( L w t)  L w t ρ I1( L w t) 


m1( L w t) 12

2

 L w



2


Mass:

m2( w t) 

π w  t  ρ 8

 m2( w t)  w2 2 L I2( L w t)  2    m2( w t)     8  2 



Mass:

m3( t)  

π d  t  ρ


8

 m3( t)  d2 2 L    I3( L t)  2   m3( t)     8 2 

For the crank (2): L  120  mm , w  30 mm , t  10 mm


Mass:


mcrank  m1( L w t)  m2( w t)  m3( t)

mcrank  0.306 kg

Moment of inertia about CG: IcrankCG  I1( L w t)  I2( L w t)  I3( L t)

IcrankCG  548.563 kg mm

2

Moment of inertia about pivot axis: Icrank  IcrankCG  mcrank  

L

2

 2

6.

3

2

4

2

Icrank  1.652  10 kg mm

For the rocker (4): L  360  mm , w  40 mm , t  12 mm Mass:

mrocker  m1( L w t)  m2( w t)  m3( t)

mrocker  1.404 kg

Moment of inertia about CG (which is also the pivot axis): Irocker  I1( L w t)  I2( L w t)  I3( L t) 7.

Irocker  1.842  10 kg mm

For the couplers (3 and 5) each L  240  mm , w  24 mm , t  8  mm Mass:

mcoupler  m1( L w t)  m2( w t)  m3( t)

mcoupler  0.372 kg

Moment of inertia about CG: IcouplerCG  I1( L w t)  I2( L w t)  I3( L t)

3

IcouplerCG  2.106  10 kg mm

2




A certain bat has a mass of 1 kg and a mass moment of inertia about its CG of 0.08 kg-m2. It's CG is located 630 mm from the end closest to the grip. If the center of a batter's grip is located 75 mm from the same end of the bat, at what point on the bat (measured from the end closest to the grip) should the batter hit the ball to produce no reaction at the grip?

Given:

Mass of the bat: mbat  1.0 kg Mass moment of inertia: I  0.08 kg m Distance from end to: xgrip  75 mm xCG  630  mm

Solution:


2

I

1.

Using equation 10.11b, calculate the radius of gyration of the bat: k 

2.

The distance from the CG to the center of the grip is x  xgrip  xCG

3.

Calculate the distance from the CG to the center of percussion for the given grip point using equation 10.15b: lp 

4.

k

2

x

lp  144.144 mm

The distance from the grip-end of the bat to the center of percussion is xball  xCG  lp

xball  774 mm

mbat

k  282.843 mm x  555.000 mm




The cam of Example 8-8 drives an aligned translating roller follower. The effective mass of the follower and the mechanism that it actuates is 0.45 kg. The follower spring has a rate of 8 N/m with a preload at zero displacement of 0.3 N. Use the method of virtual work to find and plot the torque required to rotate the cam through one rise-fall segment.

Given:

Cam speed:

ω  15 rad sec


k  8  N  m

Roller follower mass:

mf  0.45 kg

Rise-fall angle:

β  180  deg

Solution: 1.

1

1

F0  0.3 N


From Example 8-8, the follower displacement is



4 5 6 θ θ θ      192     192     64     in   β  β  β  β 

θ s θ  1  64  

3

Plotting s as a function of 2:

θ  0  deg 1  deg  180  deg FOLLOWER DISPLACEMENT


30

 

20

s θ

mm 10

0

0

30

60

90

120

150

180

θ deg

2.


 

vf θ 

3.

 d s θ    ω  θ   d  

 

a f θ 

 d v θ    ω  θ f   d  


 

 

Ff θ  k s θ  F0 4.


 

   

EF θ  Ff θ  vf θ


5.


Determine the components that make up the third term in equation 10.28b. The roller follower translates but does not rotate. The translation motion will be accounted for in this term.

 

   

IF θ  mf  a f θ  vf θ 6.

The fourth term in equation 10.28b is zero. The only rotating component is the cam itself and it rotates at constant speed.

7.

Using equation 10.28b, solve for and plot the input torque as a function of cam angle over the complete rise and fall. During the dwell the input torque is zero..

 

T2 θ 

1 ω

  

 

 IF θ  EF θ

INPUT TORQUE ON CAM 0.1

Torque, N-m

0.05

 

T2 θ Nm

0

 0.05

 0.1

0

30

60


120

150

180




Figure P10-6a shows a typical binary link with full-radius ends. Figure P10-6b shows a fullradius end and gives moments of inertia about its CG and about an axis through the center of the radius, R. Table P10-1 gives data for the length, L, between holes of diameter d; the end radius, R; the thickness, t; and the material of the link. For the row(s) assigned, find the moment of inertia of the link about one fixed pivot and its CG.

Given:

Mass density of steel: Length: Width: Thickness: Pivot pin diameter:

Solution:

3

3

ρ  7.8 10  kg m L  225  mm R  13 mm t  12 mm d  8  mm

w  2  R

w  26 mm


1.


2.


d, 2 places

CG

w

L/2 t L 2.


m1  L w t ρ

m1  0.548 kg

Moment of inertia about link CG: I1  3.

m1 12

2

 L w



2

I1  2341 kg mm

2


Mass (each):

m2 

π R  t  ρ

m2  0.025 kg

2

 m2 R2 2 L Moment of inertia about link CG: I2  2    m2     2 2  I2  633.151 kg mm 4.

2


Mass (each):

m3  

π d  t  ρ

m3  0.0024 kg

8

 m3 d 2

Moment of inertia about link CG: I3  2  

 8

2 L    m3   2 



I3  59.583 kg mm 5.

The total moment of inertia about the link CG is: ICG  I1  I2  I3

6.

2

ICG  2914 kg mm

2

Use equation 10.8 to calculate the moment of inertia about one pivot hole: Total mass of link:

mlink  m1  2   m2  m3 mlink  0.593 kg

Moment of inertia about pivot hole:

Ipivot  ICG  mlink  

L

 2

Ipivot  10414 kg mm

2

2




Using the definition of the binary link in Problem 10-37, write a computer program or use an equation solver to calculate the moment of inertia of the link about its fixed pivots and about its CG. Use the data in row a of Table P10-1 to test your program.

Solution:

See Figure P10-6, Table P10-1, and Mathcad file P1038.

1.


2.


d, 2 places

CG

w

L/2 t L 2.


m1( L w t ρ )  L w t ρ

Moment of inertia about link CG: I1( L w t ρ )  3.

m1( L w t ρ ) 12

2

 L w



2


Mass (each):

m2( R t ρ ) 

π R  t  ρ 2

 m2( R t ρ )  R2

Moment of inertia about link CG: I2( L R t ρ )  2  



4.

2

2  2 

 m2( R t ρ )  

L


Mass (each):

m3( t d ρ )  

π d  t  ρ 8

 m3( t d ρ )  d2

Moment of inertia about link CG: I3( L t d ρ )  2  



5.

8

2  2 

 m3( t d ρ )  

L

The total moment of inertia about the link CG is: ICG( L R w t d ρ )  I1( L w t ρ )  I2( L R t ρ )  I3( L t d ρ )

6.

Use equation 10.8 to calculate the moment of inertia about one pivot hole: Total mass of link: mlink ( L R w t d ρ )  m1( L w t ρ )  2   m2( R t ρ )  m3( t d ρ ) 



Moment of inertia about pivot hole: Ipivot( L R w t d ρ )  ICG( L R w t d ρ )  mlink ( L R w t d ρ )  

L

 2

7. Use the data of Table 10-1, row a to test the program. Mass density of steel: Length: Width: Thickness: Pivot pin diameter:

3

3

ρ  7.8 10  kg m L  225  mm R  13 mm t  12 mm d  8  mm

mlink ( L R w t d ρ )  0.593 kg ICG( L R w t d ρ )  2914 kg mm

2

Ipivot( L R w t d ρ )  10414 kg mm

2

These results agree with the solution to Problem 10-37a.

w  2  R

w  26 mm

2




Figure P10-7a shows a simplified ternary link (the vertices would normally be rounded). Figure P10-7b shows a triangular plate and gives the moment of inertia about its CG. Table P10-1 gives data for the length L, of the base of the link; the angle, δ, between the base and one side; the distance p along that side; the diameter, d, of the three holes; the thickness, t; and the material of the link. For the row a, find the moment of inertia of the link about its CG. As a first approximation, ignore the holes.

Given:

Mass density of steel: Length: Distance to apex: Angle: Thickness: Pivot pin diameter:

3

3

ρ  7.8 10  kg m L  225  mm p  185  mm δ  20 deg t  12 mm d  8  mm


Solution: 1.

Divide the link into 2 volumes: 1) the left right triangle, and 2) the remaining right triangle.

2.


y b2

b1

p

h



x xbar1

yCG

xCG xbar2 L 2.

Locate the CG of the constant thickness solid using the triangular areas. Base 1:

b 1  p  cos( δ)

b 1  173.843 mm

Height:

h  b 1 tan ( δ)

h  63.274 mm

Distance to CG from origin of xy coordinate frame for triangle 1: xbar1  ybar1  Base 2:

2 3 1 3

 b1

xbar1  115.895 mm

h

ybar1  21.091 mm

b 2  L  b 1

b 2  51.157 mm

Distance to CG from origin of xy coordinate frame for triangle 2:



xbar2  b 1 

1 3

 b2

xbar2  190.895 mm

ybar2  ybar1

ybar2  21.091 mm

Area of triangles: A1  0.5 b 1 h

A1  5499.9 mm

2

A2  0.5 b 2 h

A2  1618.4 mm

2

A  A1  A2

A  7118.3 mm

2

Location of overall CG with respect to xy axes: xCG 

xbar1 A1  xbar2 A2

xCG  132.948 mm

A

yCG  ybar1 3.

yCG  21.091 mm

Determine the mass and moment of inertia of volume 1 using given moment of inertia equation and equation 10.8: Mass:

m1  0.5 b 1 h  t ρ

m1  0.515 kg

Distance from volume 1 CG to CG of entire volume d 1  xCG  xbar1

d 1  17.052 mm

Moment of inertia of volume 1 about overall CG: I1  4.

m1 18

  b 1  h 2

2

2   m1 d1

2

I1  1128.5 kg mm


m2  0.5 b 2 h  t ρ

m2  0.151 kg


d 2  57.948 mm


m2 18

  b 2  h 2

2

  m2 d2

2

I2  564.4 kg mm

2

Add the moments of inertia to obtain the overall moment of inertia about the CG of the ternary link. I  I1  I2

I  1693 kg mm

2




Using the definition of the ternary link in Problem 10-39, write a computer program or use an equation solver to calculate the moment of inertia of the link about its CG. Use the data in row a of Table P10-1 to test your program. Include the holes in your calculation.

Input:

Mass density of steel: Length: Distance to apex: Angle: Thickness: Pivot pin diameter:

Solution: 1.

3

3

ρ  7.8 10  kg m L  225  mm p  185  mm δ  20 deg t  12 mm d  8  mm


Divide the link into 2 volumes: 1) the left right triangle, and 2) the remaining right triangle.

y b2

b1

p

b

h

 a

c

xbar1

x yCG

xCG xbar2 L 2.

Locate the CG of the constant thickness solid using the triangular areas. Base 1:

b 1  p  cos( δ)

b 1  173.843 mm

Height:

h  b 1 tan ( δ)

h  63.274 mm

Distance to CG from origin of xy coordinate frame for triangle 1:

Base 2:

xbar1 

2

ybar1 

1

3

3

 b1

xbar1  115.895 mm

h

ybar1  21.091 mm

b 2  L  b 1

b 2  51.157 mm

Distance to CG from origin of xy coordinate frame for triangle 2: xbar2  b 1 

1 3

ybar2  ybar1

 b2

xbar2  190.895 mm ybar2  21.091 mm



Area of triangles: A1  0.5 b 1 h

A1  5499.9 mm

2

A2  0.5 b 2 h

A2  1618.4 mm

2

A  A1  A2

A  7118.3 mm

2

Location of overall CG with respect to xy axes: xCG 

xbar1 A1  xbar2 A2 A

yCG  ybar1 3.

xCG  132.948 mm yCG  21.091 mm


m1  0.5 b 1 h  t ρ

m1  0.515 kg


d 1  17.052 mm


m1 18

  b 1  h 2

2

2   m1 d1

2

I1  1128.5 kg mm


m2  0.5 b 2 h  t ρ

m2  0.151 kg


d 2  57.948 mm


m2 18

  b 2  h 2

2

2   m2 d2

I2  564.4 kg mm

2

Determine the locations and moments of inertia of the hole centers. The equations of the lines parallel to and offset by 1.5d from the link boundaries are: y = 1.5 d

y = x tan( δ)  e

where

e 

1.5 d

e  12.770 mm

cos( δ)

α  atan

y = x tan( β )  f

 

h

 b2 

α  51.044 deg

β  180  deg  α f 

L h b2



1.5 d cos( α)

β  128.956 deg f  259.206 mm

Solving for the intersections of these three lines: xa 

1.5 d  e tan ( δ)

ya  1.5 d

xa  68.055 mm ya  12.000 mm



xb 

ef tan ( δ)  tan( β )

yb  xb tan ( δ)  e xc 

1.5 d  f

xb  169.897 mm yb  49.067 mm xc  199.867 mm

tan ( β )

yc  1.5 d

yc  12.000 mm

The distances from the hole centers to the link CG are: d a 

 xCG  xa 2  yCG  ya 2

d a  65.526 mm

d b 

 xCG  xb 2  yCG  yb 2

d b  46.346 mm

d c 

xCG  xc 2  yCG  yc 2

d c  67.534 mm

The moments of inertia of the holes with respect to the link CG are:

 d2

Ia  



8

 d2

Ib  

8

 d2

Ic  



6.

8

 π d 2   t ρ  4

Ia  20.239 kg mm

2

 π d 2   t ρ  4

Ib  10.143 kg mm

2

 π d 2  t ρ  4

Ic  21.495 kg mm

 da

2

 db

2

 dc   2

2

Add the moments of inertia to obtain the overall moment of inertia about the CG of the ternary link. I  I1  I2  Ia  Ib  Ic

I  1641 kg mm

2




Draw free body diagrams of the links in the geared fivebar linkage shown in Figure 4-11 and write the dynamic equations to solve for all forces plus the driving torque. Assemble the symbolic equations in matrix form for solution.

Solution:

No solution is provided to this algebraic exercise.




Draw free body diagrams of the links in the sixbar linkage shown in Figure 4-12 and write the dynamic equations to solve for all forces plus the driving torque. Assemble the symbolic equations in matrix form for solution.

Solution:





Table P11-1 shows kinematic and geometric data for several slider-crank linkages of the type and orientation shown in Figure P11-1. The point locations are defined as described in the text. For row a in the table, solve for forces and torques at the position shown. Also, compute the shaking force and the shaking torque. Consider the coefficient of friction  between slider and ground to be zero.

Units:


Given:

Link lengths:

1

2

Link 2 (O2 to A)

a  4.00 in

Link 3 (A to B)

Offset

c  0.00 in

Friction:

Crank angle and motion: θ  45 deg RP3  0.0 in

Coupler point: Mass:

2

IG2  0.10 blob in

α  20 rad sec

δFP3  0  deg 2

α  2.40 rad sec

m4  0.060  blob 2

RCG3  5.00 in

a G2  203.96 in sec 2

1.

a G3  371.08 in sec

2 2 2

θAG2  213.69 deg θAG3  200.84 deg θAG4  180.0  deg


Calculate the x and y components of the position vectors.





R12x  1.414 in





R12y  1.414 in

R12x  RCG2 cos θ  180  deg R12y  RCG2 sin θ  180  deg R32x  RCG2 cos θ

 

R32x  1.414 in

 

R32y  1.414 in

 

R23x  4.860 in

 

R23y  1.176 in

R32y  RCG2 sin θ

R23x  RCG3 cos θ R23y  RCG3 sin θ





R43x  6.804 in





R43y  1.646 in





RP3x  0.000 in





RP3y  0.000 in

R43x   b  RCG3  cos θ  180  deg R43y   b  RCG3  sin θ  180  deg

RP3x  RP3 cos θ  180  deg  δRP3 RP3y  RP3 sin θ  180  deg  δRP3 2.

δ  0  deg

T3  20 lbf  in

a G4  357.17 in sec Solution:

θ  166.40 deg

IG3  0.20 blob in

δ  0  deg

FP3  0  lbf

Force and torque: Accelerations:

ω  10 rad sec

m3  0.020  blob

RCG2  2.00 in

Mass center:

μ  0 1

δRP3  0.0 deg

m2  0.002  blob

Moment of inertia:

b  12.00  in

Calculate the x and y components of the acceleration of the CGs of all moving links in the global coordinate system (GCS).


3.

4.


a G2x  a G2 cos θAG2

a G2x  169.705 in sec

a G2y  a G2 sin θAG2

a G2y  113.136 in sec


a G3x  346.803 in sec


a G3y  132.015 in sec


a G4x  357.170 in sec

2 2 2 2 2

Calculate the x and y components of the external force at P on link 3 in the CGS. FP3x  FP3 cos δFP3

FP3x  0.000 lbf

FP3y  FP3 sin δFP3

FP3y  0.000 lbf

Substitute these given and calculated values into the matrix equation 11.10g, modified for this problem. Note that Mathcad requires that all elements in a matrix have the same dimension. Thus, the matrix and array in equation 11.10g will be made dimensionless and the dimensions will be put back in after solving it. 0 0 0 0 1  1  0 1 0 0 1 0   R12y R12x R32y R32x 0 0  in in in in  1 0 0 0 1 0 C    0 0 1 0 0 1  R23y R23x R43y R43x  0 0 in in in in   0 0 0 1 0 0  0 0 0 1 0  0

0 0

   1  0 0  0  0  0

0 0 0 0 0 0 μ 1

1   m2 a G2x lbf     1 m2 a G2y lbf     1 1 IG2 α lbf  in     1 m3 a G3x  FP3x   lbf    F      m3 aG3y  FP3y lbf  1     IG3 α  RP3x FP3y  RP3y FP3x  T3  lbf  1 in 1   1   m4 a G4x lbf   0  

F12x  R  lbf 1

F12x  28.7 lbf

R  C

F12y  R  lbf 2

1

F

F12y  5.87 lbf


F32x  R  lbf

F32x  28.4 lbf

F32y  R  lbf

F32y  6.10 lbf


F43x  21.4 lbf


F43y  8.74 lbf


F14y  8.74 lbf

T12  R  lbf  in

T12  99.6 lbf  in

3 5 7

8

5.


4 6

Calculate the shaking force and shaking torque using equations 11.15. F21  F12x  j  F12y

F41  j  F14y

Fs  F21  F41

Fs  ( 28.706  2.867j) lbf

Magnitude:

Fs  Fs

Fs  28.848 lbf

Angle:

θFs  arg Fs

θFs  5.703 deg

Ts  T12

Ts  99.6 lbf  in




Table P11-5 gives kinematic and geometric data for for a slider-crank linkage of the type and orientation shown in Figure 11-4. For the row(s) assigned in the table, solve for the input torque on link 2 using the method of virtual work at the position shown.

Units:


Given:

Link lengths:

2

1

Link 2 (O2 to A)

a  4.00 in

Link 3 (A to B)

b  12.00  in

Offset

c  0.00 in

Friction:

μ  0

Crank angle and motion: θ  45 deg RP3  0.0 in

Coupler point: Mass:

m3  0.020  blob 2

IG2  0.10 blob in

Moment of inertia:

RCG2  2.00 in

δ  0  deg

FP3  0  lbf

Force and torque: Accelerations:

δRP3  0.0 deg

m2  0.002  blob

Mass center:

θ  166.40 deg

α  20 rad sec

m4  0.060  blob 2

IG3  0.20 blob in

RCG3  5.00 in

δFP3  0  deg 2

α  2.40 rad sec

T3  20 lbf  in

a G2  203.96 in sec 2

a G3  371.08 in sec a G4  357.17 in sec

Velocities:

ω  10 rad sec

1

ω  2.43 rad sec

vG2  20.0 in sec 1

vG3  35.24  in sec

vP3  35.24  in sec

1.

2 2 2

1

vG4  35.14  in sec

Solution:

δ  0  deg

θAG2  213.69 deg θAG3  200.84 deg θAG4  180.0  deg θVG2  135.0  deg

1 1

1

Calculate the x and y components of the velocity vectors. vG2x  14.142 in sec

vG2y  vG2 sin θVG2

vG2y  14.142 in sec

vG3x  vG3 cos θVG3

vG3x  31.141 in sec


vG3y  16.495 in sec


vG4x  35.140 in sec

vP3x  vP3 cos θVP3

vP3x  31.127 in sec

θVG4  180.0  deg θVP3  152.04 deg

See Figure P11-1, Table P11-1, Table P11-2, and Mathcad file P1103a.


θVG3  152.09 deg

1

1 1

1 1

1



vP3y  vP3 sin θVP3 2.

3.

4.

vP3y  16.522 in sec

1

Calculate the x and y components of the acceleration of the CGs of all moving links in the global coordinate system (GCS). a G2x  a G2 cos θAG2

a G2x  169.705  in sec


a G2y  113.136  in sec


a G3x  346.803  in sec


a G3y  132.015  in sec


a G4x  357.170  in sec

2 2 2 2 2

Calculate the x and y components of the external force at P in the CGS. FP3x  FP3 cos δFP3

FP3x  0.000  lbf


FP3y  0.000  lbf

Substitute these given and calculated values into equation 11.16c and solve for the input torque.

T12 

1 ω

 m2  a G2x vG2x  a G2y vG2y  m3  a G3x vG3x  a G3y vG3y   m4  a G4x vG4x  IG2 α ω  IG3 α ω     F  v  F  v  P3y P3y   T3 ω   P3x P3x 

T12  99.687 lbf  in








Table P11-3 shows kinematic and geometric data for several pin-jointed fourbar linkages of the type and orientation shown in Figure P11-2. All have 1 = 0. The point locations are defined as described in the text. For row a in the table, solve for forces and torques at the position shown. Also, compute the shaking force and the shaking torque. Work in any units system you prefer.

Units:


Given:

Link lengths:

1

2

Link 2 (O2 to A)

a  4.00 in

Link 3 (A to B)

b  12.00  in

Link 4 (B to O4)

c  8.00 in

Link 3 (O2 to O4)

d  15.00  in

Crank angle and motion: θ  45 deg

ω  20 rad sec

Other link angles: θ  24.97  deg ω  5.62 rad sec

θ  99.30  deg ω  3.56 rad sec

Coupler point:

RP3  0.0 in

δRP3  0.0 deg

Rocker point:

RP4  8.0 in

δRP4  0.0 deg

m2  0.002  blob

Mass:

m3  0.020  blob 2

Mass center:

δ  0  deg

RCG4  4.00 in

δ  30 deg

δFP3  0  deg

T3  15 lbf  in α  20 rad sec

Accelerations:

IG4  0.50 blob in

RCG3  5.00 in

δ  0  deg

FP4  40 lbf

δFP4  30 deg

T4  25 lbf  in

2

α  75.29  rad sec

a G2  801.00 in sec 2

α  244.43 rad sec Solution:

2

IG3  0.20 blob in

RCG2  2.00 in

Force and torque: FP3  0  lbf

m4  0.100  blob 2

Moment of inertia: IG2  0.10 blob in

1.

1

1

2

2

a G3  1691.49  in sec a G4  979.02 in sec

2

2

θAG2  222.14 deg θAG3  208.24 deg θAG4  222.27 deg

See Figure P11-2, Table P11-3 and Mathcad file P1105a.






R12x  1.414 in





R12y  1.414 in

R12x  RCG2 cos θ  δ  180  deg R12y  RCG2 sin θ  δ  180  deg R32 

RCG2 sinδ2  a  RCG2 cosδ2 

 

 

R32  2.000 in

δ  atan2 a  RCG2 cos δ RCG2 sin δ

δ  0.000 deg

R32x  R32 cos θ  δ

R32x  1.414 in

  R32y  R32 sin θ  δ R23x  RCG3 cos θ  δ  180  deg R23y  RCG3 sin θ  δ  180  deg

R32y  1.414 in R23x  4.533 in R23y  2.111 in

1


R43 


RCG3 sinδ2  b  RCG3 cosδ2 

 

 

δ  atan2 b  RCG3 cos δ RCG3 sin δ



R43x  6.346 in





R43y  2.955 in

R43y  R43 sin θ  δ





R14x  2.534 in





R14y  3.095 in

R14x  RCG4 cos θ  δ  180  deg R14y  RCG4 sin θ  δ  180  deg

RCG4 sinδ2  c  RCG4 cosδ2 

 

 

δ  atan2 c  RCG4 cos δ RCG4 sin δ

R34x  1.241 in





R34y  4.799 in





RP3x  0.000 in





RP3y  0.000 in





RP4x  1.293 in





RP4y  7.895 in

RP3x  RP3 cos θ  δRP3 RP3y  RP3 sin θ  δRP3

RP4x  RP4 cos θ  δRP4 RP4y  RP4 sin θ  δRP4

4.

δ  23.794 deg



R34y  R34 sin θ  δ

3.

R34  4.957 in



R34x  R34 cos θ  δ

2.

δ  0.000 deg



R43x  R43 cos θ  δ

R34 

R43  7.000 in

Calculate the x and y components of the acceleration of the CGs of all moving links in the global coordinate system (GCS). 2 a G2x  a G2 cos θAG2 a G2x  593.948 in sec 2


a G2y  537.427 in sec


a G3x  1.490  10 in sec


a G3y  800.355 in sec


a G4x  724.459 in sec


a G4y  658.513 in sec

3

2

2 2 2

Calculate the x and y components of the external force on links 3 and 4 in the CGS. FP3x  FP3 cos δFP3

FP3x  0.000 lbf


FP3y  0.000 lbf

FP4x  FP4 cos δFP4

FP4x  34.641 lbf


FP4y  20.000 lbf

Substitute these given and calculated values into the matrix equation 11.9, modified for additional terms. Note that Mathcad requires that all elements in a matrix have the same dimension. Thus, the matrix and array in equation 11.10g will be made dimensionless and the dimensions will be put back in after solving it.



0 1 0 0 0 0 0  1  1 0 0 0 0 1 0  0  R12y R12x R32y R32x 0 0 0 0  in in in in  0 1 0 0 0 0 1  0  0 0 0 1 1 0 0 0 C   R23y R23x R43y R43x  0 0 0 0 in in  in in  0 0 1 0 0 0 1 0  0 0 1 0 0 0 1  0  R34y R34x R14y R14x 0 0 0  0 in in in in 

0



0



1



0 0

 0  0  0  0 

1   m2 a G2x lbf    1 m2 a G2y lbf     1 1 IG2 α lbf  in        m3 aG3x  FP3x lbf  1   F     m3 aG3y  FP3y lbf  1    IG3 α  RP3x FP3y  RP3y FP3x  T3  lbf  1 in 1      m4 aG4x  FP4x lbf  1   1   m  a  F  lbf  4 G4y P4y    IG4 α  RP4x FP4y  RP4y FP4x  T4  lbf  1 in 1  

1

F


F12x  124.0 lbf


F12y  62.3 lbf


F32x  122.8 lbf


F32y  61.2 lbf


F43x  93.0 lbf


F43y  45.2 lbf


F14x  14.10 lbf


F14y  0.676 lbf

T12  R  lbf  in

T12  176.4 lbf  in

1 3 5 7

9

5.

R  C

2 4 6 8

Calculate the shaking force and shaking torque using equations 11.15. F21  F12x  j  F12y

F41  F14x  j  F14y

Fs  F21  F41

Fs  ( 109.868  61.581j ) lbf

Magnitude: Ts  T12

Fs  Fs

Fs  125.949 lbf

Angle:

θFs  arg Fs

Ts  176.4 lbf  in

θFs  29.271 deg




Tables P11-3 and P11-4 show kinematic and geometric data for several pin-jointed fourbar linkages of the type and orientation shown in Figure P11-2. All have 1 = 0. The point locations are defined as described in the text. For row a in the table, solve for input torque on link 2, using the method of virtual work, at the position shown. Work in any units system you prefer.

Units:


Given:

Link lengths:

2

1

Link 2 (O2 to A)

a  4.00 in

Link 3 (A to B)

b  12.00  in

Link 4 (B to O4)

c  8.00 in

Link 3 (O2 to O4)

d  15.00  in

Link angles:

θ  45 deg

Coupler point:

RP3  0.0 in

δRP3  0.0 deg

Rocker point:

RP4  8.0 in

δRP4  0.0 deg

Mass:

θ  24.97  deg

m2  0.002  blob

m3  0.020  blob 2

δ  0  deg

RCG4  4.00 in

δ  30 deg


T3  15 lbf  in Accelerations:

α  20 rad sec

2 2 2

1

ω  5.62 rad sec ω  3.56 rad sec

RCG3  5.00 in

δ  0  deg

FP4  40 lbf

δFP4  30 deg

a G2  801.00 in sec

α  244.43 rad sec ω  20 rad sec

a G4  979.02 in sec vG2  40.0 in sec

1

1

2

2

1

vG3  54.44  in sec vG4  14.23  in sec

vP4  28.45  in sec

1.

2

a G3  1691.49  in sec

vP3  80.00  in sec

Solution:

IG4  0.50 blob in

T4  25 lbf  in

α  75.29  rad sec

Velocities:

2

IG3  0.20 blob in

RCG2  2.00 in


m4  0.100  blob 2

Moment of inertia: IG2  0.10 blob in Mass center:

θ  99.30  deg

1

1 1

See Figure P11-2, Table P11-3, Table P11-4, and Mathcad file P1106a.

Calculate the x and y components of the velocity vectors. vG2x  28.284 in sec


vG2y  28.284 in sec


vG3x  44.698 in sec


vG3y  31.077 in sec


vG4x  11.012 in sec

θAG3  208.24 deg θAG4  222.27 deg θVG2  135.0  deg

1


θAG2  222.14 deg

1

1 1

1 1

θVG3  145.19 deg θVG4  219.30 deg θVP3  135.00 deg θVP4  189.30 deg


2.

3.

4.

SOLUTION MANUAL 11-6a-2 1


vG4y  9.013 in sec


vP3x  56.569 in sec

vP3y  vP3 sin θVP3

vP3y  56.569 in sec


vP4x  28.076 in sec


vP4y  4.598 in sec

1

1 1

1

Calculate the x and y components of the acceleration of the CGs of all moving links in the global coordinate system (GCS). a G2x  a G2 cos θAG2 2 a G2x  593.948 in sec 2


a G2y  537.427 in sec


a G3x  1.490  10 in sec


a G3y  800.355 in sec


a G4x  724.459 in sec


a G4y  658.513 in sec

3

2

2 2 2

Calculate the x and y components of the external force on links 3 and 4 in the CGS. FP3x  FP3 cos δFP3

FP3x  0.000 lbf


FP3y  0.000 lbf

FP4x  FP4 cos δFP4

FP4x  34.641 lbf


FP4y  20.000 lbf

Substitute these given and calculated values into equation 11.16c and solve for the input torque. T12 

1 ω

 m2  a G2x vG2x  a G2y vG2y  m3  a G3x vG3x  a G3y vG3y    m4  a G4x vG4x  aG4y vG4y  IG2 α ω  IG3 α ω  IG4 α ω    F  v  F  v    F  v  F  v    P3x P3x P3y P3y P4x P4x P4y P4y    T3 ω  T4 ω 



T12  166.3 lbf  in






For row a in Table P11-3, input the associated disk file to program FOURBAR, calculate the linkage parameters for crank angles from zero to 360 deg by 5 deg increments, and design a steel disk flywheel to smooth the input torque using a coefficient of fluctuation of 0.05. Minimize the flywheel weight.

Units:


Given:

Link lengths:

2

1

Link 2 (O2 to A)

a  4.00 in

Link 3 (A to B)

b  12.00  in

Link 4 (B to O4)

c  8.00 in

Link 3 (O2 to O4)

d  15.00  in


ω  20 rad sec

Other link angles: θ  24.97  deg ω  5.62 rad sec

1

1

θ  99.30  deg ω  3.56 rad sec

Coupler point:

RP3  0.0 in

δRP3  0.0 deg

Rocker point:

RP4  8.0 in

δRP4  0.0 deg

Mass:

m2  0.002  blob m3  0.020  blob m4  0.100  blob 2


2

IG3  0.20 blob in

RCG2  2.00 in

δ  0  deg

RCG4  4.00 in

δ  30 deg



1

2

IG4  0.50 blob in

RCG3  5.00 in

δ  0  deg

FP4  40 lbf

δFP4  30 deg

T3  15 lbf  in T4  25 lbf  in Accelerations:

α  20 rad sec

2

α  75.29  rad sec

a G2  801.00 in sec 2

α  244.43 rad sec

Solution:

2

Coefficient of fluctuation:

k  0.05

Desired average speed:

ωavg  ω

2

a G3  1691.49  in sec a G4  979.02 in sec

θAG2  222.14 deg

2

2

θAG3  208.24 deg θAG4  222.27 deg

See Figure P11-2, Table P11-3 and Mathcad file P1107a.

1.

Enter the above data into program FOURBAR and determine the energy change from minimum to maximum speed by calculating the dynamic values and then plotting the input torque vs. crank angle. See screens below.

2.

Integrate the torque function using the values in the "Areas of Crank Torque Pulses" shown in the Linkage Balancing screen below as related to the Torque vs. Crank Angle screen above (see Table 11-1 in text). A to B B to C C to D D to A

+489.1 -402.9 +405.5 -383.3

+489.1 +86.2 max at C +491.7 min at D +108.4

Emaxspeed  86.2 in lbf

Eminspeed  491.7  in lbf

E  Emaxspeed  Eminspeed

E  405.5 in lbf

Is 

E

2

2

k ωavg

Is  20.3 blob in

The moment of inertia of the input crank and the motor armature can be subtracted from this value to obtain the required flywheel moment. There are an infinity of possible size/shape solutions for this problem.






Figure P11-3 shows a fourbar linkage and its dimensions. The steel crank and rocker have uniform cross sections 1 in wide by 0.5 in thick. The aluminum coupler is 0.75 in thick. In the instantaneous position shown, the crank O2A has  = 40 rad/sec and  = -20 rad/sec2. There is a horizontal force at P of F = 50 lb. Find all pin forces and the torque needed to drive the crank at this instant. 1

2

Units:


Given:


a  5.00 in

Link 3 (A to B)

b  4.40 in

Link 4 (B to O4)

c  5.00 in

Link 1 (O2 to O4)

d  9.50 in

Rpa  8.90 in

δ  56 deg

F  50 lbf

Coupler point:

Crank angle and motion: θ  50 deg Link cross-section dims: w2  1.00 in

t2  0.50 in

Material specific weight: Solution: 1.

2.

ω  40 rad sec

t3  0.75 in

α  20 rad sec

w4  1.00 in 3

γs  0.3 lbf  in

steel

1

T4  0  lbf  in 2

t4  0.50 in 3

γa  0.1 lbf  in

aluminum


Use program FOURBAR to determine the position, velocity, and acceleration of links 3 and 4. θ  10.105 deg

ω  41.552 rad sec

θ  113.008  deg

ω  26.320 rad sec

1

α  335.762  rad sec

1

α  2963.667 rad sec

2

2

Determine the distance to the CG in the LRCS on each of the three moving links. Links 2 and 4:

RCG2  0.5 a

Link 3:

RCG3x' 

RCG3y'  RCG3 

RCG2  2.500 in

RCG4  0.5 c

 

Rpa cos δ  b

RCG4  2.500 in

RCG3x'  3.126 in

3

 

Rpa sin δ

RCG3y'  2.459 in

3 2

RCG3x'  RCG3y'

2

RCG3  3.977 in

At an angle with respect to the local x' axis of δ  atan2 RCG3x' RCG3y'  3.

δ  38.199 deg

Determine the mass and moment of inertia of each link. m2  w2 t2 a 

γs g 3

m2  1.943  10 IG2 

m2 12

m3  blob

  w2  a 2

2



1 2

 

 b  Rpa sin δ  t3 3

m3  3.153  10

γa

m4  w4 t4 c

g

γs g 3

m4  1.943  10

blob 3

IG2  4.209  10

2

blob in

blob


IG3  IG4  4.

m3 6 m4 12


 b  Rpa sin δ

 2

IG3  0.039 blob in

  w4  c

IG4  4.209  10



2

2

2

2

3



2

blob in

Set up an LNCS xy coordinate system at the CG of each link, and draw all applicable vectors acting on the syste as shown in Figure 11-3. Draw a free-body diagram of each moving link as shown in Figure 11-3. P

y

F 32y

RP

a G2y Y

2

F 12y

F 32x R 32 x

a G2x R 12

R CG3 B

3

A

A

F 12x

O2

4 2

(b) FBD of Link 2 R CG4

R CG2

1

P

X

O2

F

O4

y

(a) The complete linkage with GCS

RP a G3y

a G3x

y

a G4y

F 34x

R 23

R 34 F 34y

F 23x x

a G4x R 14

R 43

F 14y

F 23y F 14x

(c) FBD of Link 3 (d) FBD of Link 4 5.






R12x  1.607 in





R12y  1.915 in


 

R32x  1.607 in

 

R32y  1.915 in






R23x  2.646 in





R23y  2.970 in

R23x  RCG3 cos δ  θ  180  deg R23y  RCG3 sin δ  θ  180  deg

x

F 43y F 43x



 





R43x  1.686 in

 

R43y  2.198 in

R43x  b  cos θ  RCG3 cos θ  δ







R43y   RCG3 sin θ  δ  b  sin θ

 

R34x  0.977 in

 

R34y  2.301 in

R34x  RCG4 cos θ R34y  RCG4 sin θ





R14x  0.977 in





R14y  2.301 in

R14x  RCG4 cos θ  180  deg R14y  RCG4 sin θ  180  deg





RPx  0.959 in





RPy  5.167 in

RPx  Rpa cos θ  δ  R23x RPy  Rpa sin θ  δ  R23y 6.




 

   a ω2 cosθ  j  sinθ

a G2  RCG2 α sin θ  j  cos θ a G2x  Re a G2

a G2x  5.104  10

3 in

sec a G2y  Im a G2

a G2y  6.160  10

3 in

sec



 

2

2

   a ω2 cosθ  j  sinθ

a A  a  α sin θ  j  cos θ



     2  RCG3 ω   cos θ  δ  j  sin θ  δ 

a CG3A  RCG3 α sin θ  δ  j  cos θ  δ

a G3x  Re a G3

a G3  a A  a CG3A

a G3x  8.636  10

3 in


a G3y  1.221  10

4 in

sec



 

2

2

   c ω2 cosθ  j  sinθ

a G4  RCG4 α sin θ  j  cos θ a G4x  Re a G4

a G4x  5.466  10

3 in


a G4y  6.084  10

3 in

sec 7.

2

Calculate the x and y components of the external force at P in the CGS. FPx  F

8.

2

FPy  0  lbf

Substitute these given and calculated values into the matrix equation 11.9. Note that Mathcad requires that all elements in a matrix have the same dimension. Thus, the matrix and array in equation 11.9 will be made dimensionless and the dimensions will be put back in after solving it.



0 1 0 0 0 0 0  1  1 0 0 0 0 1 0  0  R12y R12x R32y R32x 0 0 0 0  in in in in  0 1 0 0 0 0 1  0  0 0 0 1 1 0 0 0 C   R23y R23x R43y R43x  0 0 0 0 in in  in in  0 0 1 0 0 0 1 0  0 0 1 0 0 0 1  0  R34y R34x R14y R14x 0 0 0  0 in in in in  1   m2 a G2x lbf    1 m2 a G2y lbf     1 1 IG2 α lbf  in       m3 aG3x  FPx  lbf  1   F    m3 aG3y  FPy  lbf  1    IG3 α  RPx FPy  RPy FPx  lbf  1 in 1   1   m4 a G4x lbf   1   m4 a G4y lbf   1 1    α  T  lbf  in IG4  4  

0



0



1



0 0

 0  0  0  0 

R  C

1

F


F12x  37.1 lbf


F12y  42.4 lbf


F32x  27.2 lbf


F32y  54.4 lbf


F43x  50.0 lbf


F43y  92.9 lbf


F14x  39.3 lbf


F14y  104.7 lbf

T12  R  lbf  in

T12  279 lbf  in

1 3 5 7

9

2 4 6 8




Given:

Figure P11-4a shows a fourbar linkage and its dimensions in meters. The steel crank and rocker have uniform cross sections 50 mm wide by 25 mm thick. The aluminum coupler is 25 mm thick. In the instantaneous position shown, the crank O2A has  = 10 rad/sec and  = 5 rad/sec2. There is a vertical force at P of F = 100 N. Find all pin forces and the torque needed to drive the crank at this instant. Link lengths: Link 2 (O2 to A)

a  1.00 m

Link 3 (A to B)

b  2.06 m

Link 4 (B to O4)

c  2.33 m

Link 1 (O2 to O4)

d  2.22 m

Rpa  3.06 m

δ  31 deg

F  100  N

Coupler point:


ω  10 rad sec

1

T4  0  N  m

α  5  rad sec

2

Link cross-section dims: w2  50 mm

t2  25 mm


2.

t3  25 mm

w4  50 mm 3

γs  0.3 lbf  in

steel

t4  25 mm 3

γa  0.1 lbf  in

aluminum



ω  3.669  rad sec

θ  96.322 deg

ω  1.442  rad sec

1

α  55.752 rad sec

1

α  67.103 rad sec

2 2

Determine the distance to the CG in the LRCS on each of the three moving links. Links 2 and 4: RCG2  0.5 a RCG3x' 

Link 3:

RCG3y'  RCG3 

RCG2  0.500 m

RCG4  0.5 c

 


RCG4  1.165 m

RCG3x'  1.561 m

3

 

Rpa sin δ

RCG3y'  0.525 m

3 2

RCG3x'  RCG3y'

2

RCG3  1.647 m


δ  18.600 deg


γs

m3 

g

m2  10.380 kg IG2  IG3 

m2 12 m3 6

2

 

 b  Rpa sin δ  t3

γa

m4  w4 t4 c

g

m3  112.332 kg

  w2  a 2

2



2

 2

γs g

m4  24.185 kg IG2  0.867 kg m



 b  Rpa sin δ 2

1

2

IG3  125.951 kg m


IG4  4.

m4 12


  w4  c 2

2

2

IG4  10.947 kg m



Set up an LNCS xy coordinate system at the CG of each link, and draw all applicable vectors acting on the syste as shown in Figure 11-3. Draw a free-body diagram of each moving link as shown in Figure 11-3. B

y

Y

F 32y 3

a G2y

RP

R CG3

A

F 32x R 32 a G2x

P

F 12y

x

R 12 A

F 12x

4

2

R CG4

(b) FBD of Link 2 R CG2

X

1

O2

O4

y

F 34x


F 34y

R 34

F 43y B

F 43x

R 43 3

a G4y a G4x

F

a G3y

y

4

RP

x

P

a G3x

R 14 F 14y

R 23 F 23x

F 14x

A

F 23y

(d) FBD of Link 4 (c) FBD of Link 3

5.






R12x  0.250 m





R12y  0.433 m


 

R32x  0.250 m

 

R32y  0.433 m






R23x  1.479 m





R23y  0.725 m

R23x  RCG3 cos δ  θ  180  deg R23y  RCG3 sin δ  θ  180  deg

 





R43x  0.015 m

 

R43y  0.724 m









x



 

R34x  0.128 m

 

R34y  1.158 m






R14x  0.128 m





R14y  1.158 m






RPx  1.494 m





RPy  9.865  10

RPx  Rpa cos θ  δ  R23x

4

RPy  Rpa sin θ  δ  R23y 6.

m




 

   a ω2 cosθ  j  sinθ


m

a G2x  52.165


m

a G2y  85.353

sec



 

2

2

   a ω2 cosθ  j  sinθ

a A  a  α sin θ  j  cos θ



      RCG3 ω   cos θ  δ  j  sin θ  δ 

a CG3A  RCG3 α sin θ  δ  j  cos θ  δ 2

a G3x  Re a G3


a G3x  114.678

m sec

a G3y  Im a G3

a G3y  11.429

m sec



 

2

2

   c ω2 cosθ  j  sinθ


a G4x  77.166

m sec

a G4y  Im a G4

a G4y  13.424

m sec

7.

2

Calculate the x and y components of the external force at P in the CGS. FPx  0  N

8.

2

FPy  F




0 1 0 0 0 0 0  1  1 0 0 0 0 1 0  0  R12y R12x R32y R32x 0 0 0 0  m m m m  0 1 0 0 0 0 1  0  0 0 0 1 1 0 0 0 C   R23y R23x R43y R43x  0 0 0 0 m m  m m  0 0 1 0 0 0 1 0  0 0 1 0 0 0 1  0  R34y R34x R14y R14x 0 0 0  0 m m m m  1   m2 a G2x N   1 m2 a G2y N     1 1 IG2 α N  m       m3 aG3x  FPx  N  1   F    m3 aG3y  FPy  N  1    IG3 α  RPx FPy  RPy FPx  N  1 m 1   1   m4 a G4x N   1   m4 a G4y N   1 1   IG4 α  T4 N  m  

0



0



1



0 0

 0  0  0  0 

R  C

1

F

F12x  R  N

F12x  13559 N

F12y  R  N

F12y  12294 N

F32x  R  N

F32x  13018 N

F32y  R  N

F32y  11408 N

F43x  R  N

F43x  136 N

F43y  R  N

F43y  10224 N

F14x  R  N

F14x  1731 N

F14y  R  N

F14y  9899 N

T12  R  N  m

T12  5587 N  m

1 3 5 7

9

2 4 6 8




Given:

Figure P11-4b shows a fourbar linkage and its dimensions in meters. The steel crank and rocker have uniform cross sections 50 mm wide by 25 mm thick. The aluminum coupler is 25 mm thick. In the instantaneous position shown, the crank O2A has  = 15 rad/sec and  = -10 rad/sec2. There is a horizontal force at P of F = 200 N. Find all pin forces and the torque needed to drive the crank at this instant. Link lengths: Link 2 (O2 to A)

a  0.72 m

Link 3 (A to B)

b  0.68 m

Link 4 (B to O4)

c  0.85 m

Link 1 (O2 to O4)

d  1.82 m

Rpa  0.97 m

δ  54 deg

F  200  N

Coupler point:


ω  15 rad sec

1

T4  0  N  m

α  10 rad sec

2


t2  25 mm


2.

t3  25 mm

w4  50 mm 3

γs  0.3 lbf  in

steel

t4  25 mm 3

γa  0.1 lbf  in

aluminum



ω  16.412 rad sec

θ  132.283  deg

ω  1.570  rad sec

1

α  138.628  rad sec

1

α  427.881  rad sec

2

2

Determine the distance to the CG in the LRCS on each of the three moving links. Links 2 and 4: RCG2  0.5 a RCG3x' 

Link 3:

RCG3y' 

RCG3 

RCG2  0.360 m

RCG4  0.5 c

 


RCG4  0.425 m

RCG3x'  0.417 m

3

 

Rpa sin δ

RCG3y'  0.262 m

3 2

RCG3x'  RCG3y'

2

RCG3  0.492 m


δ  32.117 deg

Determine the mass and moment of inertia of each link. m2  w2 t2 a  m2  7.474 kg

γs g

m3 

1 2

 

 b  Rpa sin δ  t3

m3  18.463 kg

γa g

m4  w4 t4 c m4  8.823 kg

γs g


m2

IG2 

12 m3

IG3 

6 m4

IG4  4.

12


  w2  a 2

2

2

IG2  0.324 kg m




 2

IG3  3.318 kg m

  w4  c

IG4  0.533 kg m



2

2

2

2

2



Set up an LNCS xy coordinate system at the CG of each link, and draw all applicable vectors acting on the syste as shown in Figure 11-3. Draw a free-body diagram of each moving link as shown in Figure 11-3. P

y

F

Y

F 32y RP

a G2y

R 12

a G2x

x

4

3

A

F 32x

F 12y

B R CG3

R 32

F 12x

2

(b) FBD of Link 2

R CG4

R CG2

X O2

O4

y P F


RP y

a G3y a G3x

F 34x

R 34

F 43y x

R 43 R 23

a G4y

F 34y

B

A F 23x

a G4x F 14y

F 23y

R 14

x

F 14x

(c) FBD of Link 3 (d) FBD of Link 4

5.






R12x  0.312 m





R12y  0.180 m


 

R32x  0.312 m

 

R32y  0.180 m




R23x  RCG3 cos δ  θ  180  deg



R23x  0.279 m

F 43x





R23y  RCG3 sin δ  θ  180  deg

 





R23y  0.405 m



R43x  0.345 m

 

R43y  0.136 m









 

R34x  0.286 m

 

R34y  0.314 m






R14x  0.286 m





R14y  0.314 m






RPx  0.066 m





RPy  0.541 m

RPx  Rpa cos θ  δ  R23x RPy  Rpa sin θ  δ  R23y 6.




 

   a ω2 cosθ  j  sinθ


a G2x  138.496

m sec

a G2y  Im a G2

m

a G2y  84.118

sec



 

2

2

   a ω2 cosθ  j  sinθ

a A  a  α sin θ  j  cos θ





a G3x  Re a G3


a G3x  155.788

m sec

a G3y  Im a G3

a G3y  235.056

m sec



 

2

2

   c ω2 cosθ  j  sinθ


a G4x  133.128

m sec

a G4y  Im a G4

a G4y  123.897

m sec

7.

2

Calculate the x and y components of the external force at P in the CGS. FPx  F

8.

2

FPy  0  N




0 1 0 0 0 0 0  1  1 0 0 0 0 1 0  0  R12y R12x R32y R32x 0 0 0 0  m m m m  0 1 0 0 0 0 1  0  0 0 0 1 1 0 0 0 C   R23y R23x R43y R43x  0 0 0 0 m m  m m  0 0 1 0 0 0 1 0  0 0 1 0 0 0 1  0  R34y R34x R14y R14x 0 0 0  0 m m m m  1   m2 a G2x N    1 m2 a G2y N     1 1 IG2 α N  m       m3 aG3x  FPx  N  1   F    m3 aG3y  FPy  N  1    IG3 α  RPx FPy  RPy FPx  N  1 m 1   1   m4 a G4x N   1   m4 a G4y N   1 1    α  T  N  m IG4  4  

0



0



1



0 0

 0  0  0  0 

R  C

1

F

F12x  R  N

F12x  5484 N

F12y  R  N

F12y  5050 N

F32x  R  N

F32x  4449 N

F32y  R  N

F32y  4422 N

F43x  R  N

F43x  1373 N

F43y  R  N

F43y  81.8 N

F14x  R  N

F14x  198 N

F14y  R  N

F14y  1011 N

T12  R  N  m

T12  1168 N  m

1 3 5 7

9

2 4 6 8




Given:

Figure P11-5a shows a fourbar linkage and its dimensions in meters. The steel crank, coupler, and rocker have uniform cross sections 50 mm wide by 25 mm thick. In the instantaneous position shown, the crank O2A has  = 15 rad/sec and  = -10 rad/sec2. There is a vertical force at P of F = 500 N. Find all pin forces and the torque needed to drive the crank at this instant. Link lengths: Link 2 (O2 to A)

a  0.785  m

Link 3 (A to B)

b  0.356  m

Link 4 (B to O4)

c  0.950  m

Link 1 (O2 to O4)

d  0.544  m

Rpa  1.09 m

δ  0  deg

F  500  N

Coupler point:

Crank angle and motion: θ  96 deg Link cross-section dims: w2  50 mm t2  25 mm Material specific weight: Solution: 1.

2.

α  10 rad sec

2

w4  50 mm t4  25 mm

3

γs  0.3 lbf  in



ω  6.830  rad sec

θ  107.906  deg

ω  12.023 rad sec

1

1

α  106.282  rad sec α  49.372 rad sec

2

2

Determine the distance to the CG in the LRCS on each of the three moving links.

RCG3  0.5 Rpa

Link 3:

RCG2  0.393 m

RCG4  0.5 c

RCG4  0.475 m

RCG3  0.545 m


γs

m3  w3 t3 Rpa

g

m2  8.148 kg IG2  IG3  IG4 

4.

1

w3  50 mm t3  25 mm

Links 2 and 4: RCG2  0.5 a

3.

ω  15 rad sec

T4  0  N  m

m2 12 m3 12 m4 12

γs

m4  w4 t4 c

g

m3  11.314 kg

  w2  a 2

2

  w4  c 2

2



m4  9.861 kg 2

2

2

g

IG2  0.420 kg m



  w3  Rpa

γs



2

IG3  1.123 kg m

2

IG4  0.744 kg m

Set up an LNCS xy coordinate system at the CG of each link, and draw all applicable vectors acting on the system as shown in Figure 11-3. Draw a free-body diagram of each moving link as shown in Figure 11-3.



Y

P 3 3

A

y F 32y

RP

B

F 32x

R CG3

R 32

4

2

F 12x

X O2

x

a G2x

R 12 F 12y

R CG4

R CG2

a G2y

O4

(b) FBD of Link 2 (a) The complete linkage with GCS y

y F 43y

a G3y

RP

F 43x

F 23x

R 34 F 34y

a G4y

x

a G3x

R 23

F 34x

P

R 43

F

a G4x R 14

F 23y

F 14y F 14x

(c) FBD of Link 3 (d) FBD of Link 4 5.


  R12y  RCG2 sin θ  180  deg R12x  RCG2 cos θ  180  deg

R12x  0.041 m R12y  0.390 m

 

R32x  0.041 m

 

R32y  0.390 m

R32x  RCG2 cos θ R32y  RCG2 sin θ





R23x  0.511 m





R23y  0.189 m

R23x  RCG3 cos θ  180  deg R23y  RCG3 sin θ  180  deg





R43x  0.177 m





R43y  0.065 m

R43x   RCG3  b   cos θ  180  deg R43y   RCG3  b   sin θ  180  deg

 

R34x  0.146 m

 

R34y  0.452 m




R14x  RCG4 cos θ  180  deg



R14x  0.146 m

x





R14y  RCG4 sin θ  180  deg



R14y  0.452 m

RPx   Rpa  RCG3  cos θ

 

RPx  0.511 m

 

RPy  0.189 m

RPy   Rpa  RCG3  sin θ 6.




 

   a ω2 cosθ  j  sinθ


a G2x  22.366

m sec

a G2y  Im a G2

2

m

a G2y  175.247

sec



 

2

   a ω2 cosθ  j  sinθ

a A  a  α sin θ  j  cos θ



      RCG3 ω   cos θ  j  sin θ 

a CG3A  RCG3 α sin θ  j  cos θ 2

a G3x  Re a G3


m

a G3x  17.640


a G3y  129.301

2

m sec



 

2

   c ω2 cosθ  j  sinθ


a G4x  19.906

m sec

a G4y  Im a G4

a G4y  137.884

2

m sec

7.


8.

2

FPy  F




0 0 0 0 0 0 1  1  0 0 0 1 0 1 0  0  R12y R12x R32y R32x 0 0 0 0  m m m m  0 0 0 1 0 1 0  0  0 1 0 0 0 0 0 1 C   R23y R23x R43y R43x  0 0 0 0  m m m m  0 1 0 0 0 1 0 0  0 1 0 0 0 1 0  0  R34y R34x R14y R14x 0 0 0  0 m m m m  1   m2 a G2x N    1 m2 a G2y N     1 1 IG2 α N  m     1   a  F  N  m  3 G3x Px   1 F    m3 aG3y  FPy  N    IG3 α  RPx FPy  RPy FPx  N  1 m 1   1   m4 a G4x N   1   m4 a G4y N   1 1   IG4 α  T4 N  m  

0



0



1



0 0

 0  0  0  0 

R  C

1

F

F12x  R  N

F12x  231 N

F12y  R  N

F12y  2231 N

F32x  R  N

F32x  413 N

F32y  R  N

F32y  803 N

F43x  R  N

F43x  214 N

F43y  R  N

F43y  160 N

F14x  R  N

F14x  410 N

F14y  R  N

F14y  1519 N

T12  R  N  m

T12  372 N  m

1 3 5 7

9

2 4 6 8




Given:

Figure P11-5b shows a fourbar linkage and its dimensions in meters. The steel crank, coupler, and rocker have uniform cross sections of 60 mm diameter. In the instantaneous position shown, the crank O2A has  = -10 rad/sec and  = 10 rad/sec2. There is a horizontal force at P of F = 500 N. Find all pin forces and the torque needed to drive the crank at this instant. Link lengths: Link 2 (O2 to A)

a  0.86 m

Link 3 (A to B)

b  1.85 m

Link 4 (B to O4)

c  0.86 m

Link 1 (O2 to O4)

d  2.22 m

Rpa  1.33 m

δ  0  deg

F  500  N

Coupler point:

Crank angle and motion: θ  36 deg

ω  10 rad sec

1

T4  0  N  m

α  10 rad sec

2

Link cross-section dims: d link  60 mm Material specific weight: Solution: 1.

2.

3

γs  0.3 lbf  in


Use program FOURBAR to determine the position, velocity, and acceleration of links 3 and 4. ω  3.285  rad sec

θ  106.189  deg

ω  11.417 rad sec

1

RCG2  0.430  m

RCG3  0.5 b

Link 3:

α  43.426 rad sec

2

2

RCG4  0.5 c

RCG4  0.430  m

RCG3  0.925  m

Determine the mass and moment of inertia of each link. m2 

π d link 4

2

 a

γs g

m2  20.192 kg

IG2  IG3  IG4  4.

α  109.287  rad sec

Determine the distance to the CG in the LRCS on each of the three moving links. Links 2 and 4: RCG2  0.5 a

3.

1

θ  46.028 deg

m2 12 m3 12 m4 12

m3 

π d link 4

2

 b

γs

m4 

g

m3  43.436 kg

π d link 4

2

 c

γs g

m4  20.192 kg

3 2 2    d link  a  4  

IG2  1.249  kg m

3 2 2    d link  b  4  

IG3  12.398 kg m

3 2 2    d link  c  4  

IG4  1.249  kg m

2

2

2




y

y F 34x B

F 34y a G4y R 34

3 P

R CG4

O4

R CG2

F 14x

x

(d) FBD of Link 4

3

2

x

F 14y R 14

RP

O2

a G4x

4

R CG3 A F 43y


F 43x

y y

P F

F 12y

a G3y RP

F 12x

x a G2y

R 12

a G3x

a G2x

R 23

x F 32y

R 32

F 32x

F 23x

(b) FBD of Link 2

F 23y

(c) FBD of Link 3

5.

R 43






R12x  0.348  m





R12y  0.253  m

R12x  RCG2 cos θ  180  deg R12y  RCG2 sin θ  180  deg

 

R32x  0.348  m

 

R32y  0.253  m

R32x  RCG2 cos θ R32y  RCG2 sin θ





R23x  0.642  m





R23y  0.666  m

R23x  RCG3 cos θ  180  deg R23y  RCG3 sin θ  180  deg





R43x  0.642  m





R43y  0.666  m

R43x   RCG3  b   cos θ  180  deg R43y   RCG3  b   sin θ  180  deg



 

R34x  0.120  m

 

R34y  0.413  m






R14x  0.120  m





R14y  0.413  m

R14x  RCG4 cos θ  180  deg R14y  RCG4 sin θ  180  deg RPx   Rpa  RCG3  cos θ

 

RPx  0.281  m

 

RPy  0.291  m

RPy   Rpa  RCG3  sin θ 6.




 

   a ω2 cosθ  j  sinθ


m

a G2x  67.048


a G2y  54.028

m sec



 

2

2

   a ω2 cosθ  j  sinθ

a A  a  α sin θ  j  cos θ



      RCG3 ω   cos θ  j  sin θ 


a G3x  Re a G3


a G3x  1.302 

m sec

a G3y  Im a G3

a G3y  19.864

2

m sec



 

2

   c ω2 cosθ  j  sinθ


a G4x  49.187

m sec

a G4y  Im a G4

a G4y  102.448 

2

m sec

7.


8.

2

FPy  0  N




0 1 0 0 0 0 0  1  1 0 0 0 0 1 0  0  R12y R12x R32y R32x 0 0 0 0  m m m m  0 1 0 0 0 0 1  0  0 0 0 1 1 0 0 0 C   R23y R23x R43y R43x  0 0 0 0 m m  m m  0 0 1 0 0 0 1 0  0 0 1 0 0 0 1  0  R34y R34x R14y R14x 0 0 0  0 m m m m  1   m2 a G2x N    1 m2 a G2y N     1 1 IG2 α N  m       m3 aG3x  FPx  N  1   F    m3 aG3y  FPy  N  1    IG3 α  RPx FPy  RPy FPx  N  1 m 1   1   m4 a G4x N   1   m4 a G4y N   1 1    α  T  N  m IG4  4  

0



0



1



0 0

 0  0  0  0 

R  C

1

F

F12x  R  N

F12x  1851 N

F12y  R  N

F12y  1315 N

F32x  R  N

F32x  497  N

F32y  R  N

F32y  225  N

F43x  R  N

F43x  53.7 N

F43y  R  N

F43y  1087 N

F14x  R  N

F14x  1047 N

F14y  R  N

F14y  3156 N

T12  R  N  m

T12  45.28  N  m

1 3 5 7

9

2 4 6 8




Figure P11-6 shows a water jet loom laybar drive mechanism driven by a pair of Grashof crank rocker fourbar linkages. The crank rotates at 500 rpm. The laybar is carried between the coupler-rocker joints of the two linkages at their respective instant centers I3,4. The combined weight of the reed and laybar is 29 lb. A 540-lb beat-up force from the cloth is applied to the reed as shown. The steel links have a 2 x 1 in uniform cross-section. Find the forces on the pins for one revolution of the crank. Find the torque-time function required to drive the system.

Units:


Given:

Link lengths:

1

2


Link 2 (A to B)

a  2.00 in

Link 3 (B to C)

b  8.375  in

Link 4 (C to D)

c  7.187  in

Link 1 (A to D)

d  9.625  in

Rpa  0.0 in

δ  0  deg

Coupler point:

Crank angle and motion: ω  500  rpm Link cross-section dims:

w  2.00 in


2.

1

α  0  rad sec

2

t  1.00 in 3

γ  0.3 lbf  in

steel


Determine the distance to the CG in the LRCS on each of the three moving links. All three are located on the x' axis in the LRCS and their angle is zero deg. RCG2  0.5 a

RCG2  1.000 in

RCG4  0.5 c

RCG4  3.594 in

RCG3  0.5 b

RCG3  4.188 in

Determine the mass and moment of inertia of each link. m2  w t a 

γ

m3  w t b 

g 3

m2  3.108  10 IG2  IG3 

m2 12 m2 12

2

2



2

2

 w b

m4  w t c

g

m3  0.013 blob

blob



 w a

γ

γ g

m4  0.011 blob



IG2  0.00207 blob in



IG3  0.01920 blob in

2

2

Include one half of the mass of the laybar as a lumped mass at the end of link 4. IG4  3.

m4 12



2

 w c

R

2 14.5 lbf

2

CG4



Define any external forces, their locations and directions. Beat-up force

F  590  lbf

The angle in the CGS is 180 deg. 4.

g

2

IG4  0.53677 blob in

acting on link 4 at a distance

R  c  3.75 in R  10.937 in

Enter the above data into program FOURBAR and solve for the pin forces and driving torque. The dynamic input screen is shown below followed by a plot of dynamic pin forces..




5.

The input torque is plotted below.





Figure P11-7 shows a crimping tool. Find the force Fhand needed to generate a 2000 lb Fcrimp. Find the pin forces. What is the linkage's joint force transmission index (JFI) in this position?

Given:


a  0.80 in

Link 3 (B to C)

b  1.23 in

Link 2 (C to D)

c  1.55 in

Link 2 (A to D)

d  2.40 in

θ  49 deg

Link 2 angle:

RP4  1.00 in

Distance to crimp force from pivot D: FP4  2000 lbf

Crimp force:

(perpendicular to link 4) RP2  4.26 in

Distance to hand force from pivot A: Solution: 1.

2.

δ  0  deg


Enter the above data into program FOURBAR to determine link 3 and 4 angles and calculate the angle that the crimping force makes with respect to the fourbar coordinate frame.. θ  34.039 deg

θ  123.518  deg

θFP4  θ  90 deg

θFP4  213.518  deg

Determine the distance to the CG in the LRCS on each of the three moving links. Links 2 and 4: RCG2  0.5 a Link 3:

3.

δ  0  deg

RCG2  0.400  in

RCG3  0.5 b

RCG4  0.5 c

RCG3  0.615  in






R12x  0.262  in





R12y  0.302  in


 

R32x  0.262  in

 

R32y  0.302  in






R23x  0.510  in





R23y  0.344  in

R23x  RCG3 cos θ  180  deg R23y  RCG3 sin θ  180  deg R43x   b  RCG3  cos θ

 

R43x  0.510  in

 

R43y  0.344  in

R43y   b  RCG3  sin θ

 

R34x  0.428  in

 

R34y  0.646  in






R14x  0.428  in





R14y  0.646  in




RP2y   RP2  RCG2  sin θ  180  deg



RP2y  3.517  in

RCG4  0.775  in





RP2x   RP2  RCG2  cos θ  180  deg



 

RP4x  0.124  in

 

RP4y  0.188  in

RP4x   RP4  RCG4  cos θ RP4y   RP4  RCG4  sin θ 4.

5.

RP2x  3.057  in

Calculate the x and y components of the external crimp force at P on link 4 in the CGS. FP4x  FP4 cos θFP4

FP4x  1667.4 lbf

FP4y  FP4 sin θFP4

FP4y  1104.4 lbf

Substitute these given and calculated values into the matrix equation 11.9 modified to omit all mass and inertia terms. Note that Mathcad requires that all elements in a matrix have the same dimension. Thus, the matrix and array in equation 11.9 will be made dimensionless and the dimensions will be put back in after solving it. 0 0 0 1 0 0 0 1 0   1  0 0 0 0 0 1 0 1 0 1    RP2x RP2y   R12y R12x R32y R32x 0 0 0 0  in in in in in in   0  0 0 0 1 0 1 0 0 0   1 0 0 0 0 0 1 0 0   0  R23y R23x R43y R43x C   0 0  0 0 0 0  in in in in    0 1 0 0 0 1 0 0 0 0   0  0 1 0 0 0 1 0 0 0   R34y R34x R14y R14x   0 0 0 0 0  0  in in in in   0 0 0 0 0 0 0 1 tan  θ   0 0     0   0     0   0     0 F    1 FP4x  lbf     1  FP4y  lbf     RP4x FP4y  RP4y FP4x  lbf  1 in 1   0  

R  C

1

F


F12x  1117 lbf


F12y  681  lbf


F32x  1069 lbf


F32y  722  lbf

1 3

2 4



F43x  1069 lbf


F43y  722  lbf


F14x  598  lbf


F14y  382  lbf

Fhandx  R  lbf

Fhandx  47.7 lbf

Fhandy  R  lbf

Fhandy  41.5 lbf

5 7

9

6.

7.

6 8

10

Calculate the pin forces. 2

2

F12  1308 lbf

2

2

F32  1290 lbf

2

2

F43  1290 lbf

2

2

F14  710  lbf

Pin at A:

F12 

F12x  F12y

Pin at B:

F32 

F32x  F32y

Pin at C:

F43 

F43x  F43y

Pin at D:

F14 

F14x  F14y

Calculate the hand force. Fhand 

8.


2

Fhandx  Fhandy

2

Fhand  63.2 lbf

Use equation 11.23a to calculate the joint force index. JFI 

F32 FP4

JFI  0.645




Figure P11-8 shows a walking beam conveyor mechanism that operates a slow speed (25 rpm). The boxes being pushed each weigh 50 lb. Determine the pin forces in the linkage and the torque to drive the mechanism through one revolution. Neglect the masses of the links.

Solution:

No solution is given for this problem, which is suited to solution using the Working Model program.




Figure P11-9 shows a surface grinder table drive that operates at 120 rpm. The crank radius is 22 mm, the coupler is 157 mm, and its offset is 40 mm. The mass of the table and workpiece combined is 50 kg. Find the pin forces, slider side loads, and driving torque over one revolution.

Solution:





Figure P11-10 shows a power hacksaw that operates at 50 rpm. The crank is 75 mm, the coupler is 170 mm, and its offset is 45 mm. Find the pin forces, slider side loads, and driving torque over one revolution for a cutting force of 250 N in the forward direction and 50 N during the return stroke.

Solution:





Figure P11-11 shows a paper roll off-loading station. The paper rolls have a 0.9-m OD, 0.22-m ID, are 3.23 m long, and have a density of 984 kg/m3. The forks that support the roll are 1.2 m long. The motion is slow so inertial loading can be neglected. Find the force required of the air cylinder to rotate the roll through 90 deg.

Solution:





Derive an expression for the relationship between flywheel mass and the dimensionless parameter radius/thickness (r/t) for a solid disk flywheel of moment of inertia I. Plot this function for an arbitrary value of I and determine the optimum r/t ratio to minimize flywheel weight for that I.

Solution:





Figure P11-5a shows an oil field pump mechanism. The head of the rocker arm is shaped such that the lower end of a flexible cable attached to it will always be directly over the well head regardless of the position of the rocker arm 4. The pump rod, which connects to the pump in the well casing, is connected to the lower end of the cable. The force in the pump rod on the up stroke is 2970 lb and the force on the down stroke is 2300 lb. Link 2 weighs 598.3 lb and has a moment of inertia of 11.8 blob-in 2; both including the counterweight. Its CG is on the link cenerline, 13.2 in from O2. Link 3 weighs 108 lb and its CG is on the link centerline, 40 in from A. It has a mass moment of inertia of 150 blob-in 2. Link 4 weighs 2706 lb and has a moment of inertia of 10700 blob-in 2: both include the counterweight. Its CG is on the link centerline where shown. The crank turns at a constant speed of 4 rpm CCW. At the instant shown in the figure the crank angle is at 45 deg with respect to the global coordinate system. Find all pin forces and the torque needed to drive the crank for the position shown. Include gravity forces. 2

1

Units:


Given:

Link lengths: Link 2 (O2 to A):

a  14.0 in

Link 3 (A to B):

Link 4 (B to O4):

c  51.3 in

Link 1 (O2 to O4): d  79.7 in

Link 1 offsets:

d X  47.5 in

d Y  64 in

External load data:

F  2300 lbf

T4  0  lbf  in

Crank angle and motion: θ2xy  81.582 deg

Solution: 1.

ω  4  rpm

α  0  rad sec

Link CG positions:

RCG2  13.2 in

RCG3  40.0 in

RCG4  79.22  in

Link weights:

W2  598.3  lbf

W3  108  lbf

W4  2706 lbf

Moments of inertia:

IG2  11.8 blob in

2

2

IG3  150  blob in

2

2

IG4  10700  blob in

Angle between O4B and CG4B:

  143.11 deg

R34  32.00  in

Angle between O4B and CG4O4:

  14.03  deg

R14  79.22  in

Angle between O4B and CG4P:

  156.62 deg

Rp  124.44 in


Use Problems 6.84c and 7.70b with 2 = 4 rpm to determine the position, velocity, and acceleration of links 3 and 4. The angles are calculated in the xy coordinate system and then rotated into the XY coordinate system after calculating accelerations. Coordinate rotation angle:

2.

b  80.0 in

  atan2 d X d Y 

θ3xy  332.475  deg

ω  0.0214 rad sec

θ4xy  262.482  deg

ω  0.0986 rad sec

  126.582 deg 1

1

α  0.0250 rad sec α  0.0272 rad sec

2 2

Calculate the x and y components of the acceleration of the CGs of all moving links in the global coordinate system (GCS). a G2  RCG2 α  sin θ2xy  j  cos θ2xy   RCG2 ω   cos θ2xy  j  sin θ2xy  2

a G2  a G2

a G2  2.316 in sec

a G2x  a G2 cos θaG2 a G2x  1.638 in sec

2

2

θaG2  arg a G2   a G2y  a G2 sin θaG2 a G2y  1.638 in sec

2

θaG2  225.000 deg



a A  a  α  sin θ2xy  j  cos θ2xy   a  ω   cos θ2xy  j  sin θ2xy  2

a CG3A  RCG3 α  sin θ3xy  j  cos θ3xy   RCG3 ω   cos θ3xy  j  sin θ3xy  2

a G3  a A  a CG3A a G3  a G3

a G3  1.763 in sec

2


θaG3  arg a G3  

θaG3  244.955 deg

a G3y  a G3 sin θaG3

2

a G3y  1.598 in sec

2



      RCG4 ω   cos θ4xy    j  sin θ4xy   

a G4  RCG4 α sin θ4xy    j  cos θ4xy   2

a G4  a G4

a G4  2.288 in sec

2


θaG4  arg a G4  

θaG4  265.366 deg

a G4y  a G4 sin θaG4

2

a G4y  2.281 in sec

2

Transform the link angles to the global XY system:   θ2xy  

  θ3xy    360  deg

  θ4xy    360  deg

  45.000 deg

  99.057 deg

  29.064 deg

3.


4.






R12x  9.334 in





R12y  9.334 in

R12x  RCG2 cos   180  deg R12y  RCG2 sin   180  deg

 

R32x   a  RCG2  cos 

P

x

R32x  0.566 in

B R CG4

RP

Y 4

R 32 a G2y F 32y F 32x F 12y a G2x

3

O4 Y

R CG3 2

O2

R 12

A R CG2

y


X

F 12x

(b) FBD of Link 2

X



F 43y y

y

B

F 43x R 43 a G3y P

RP

F 14y O4

R 23

a G4y R 34 a G4x

x

a G3x

B

F 34x

F 14x

x

F 34y R 14

F

A

F 23x

(d) FBD of Link 4 F 23y

(c) FBD of Link 3

 

R32y   a  RCG2  sin 

R32y  0.566 in





R23x  6.297 in





R23y  39.501 in

R23x  RCG3 cos   180  deg R23y  RCG3 sin   180  deg





R43x  6.297 in





R43y  39.501 in

R43x   RCG3  b   cos   180  deg R43y   RCG3  b   sin   180  deg





R34x  31.702 in





R34y  4.357 in

R34x  R34 cos    R34y  R34 sin   





R14x  76.508 in





R14y  20.550 in

R14x  R14 cos     180  deg R14y  R14 sin     180  deg





RPx  123.828 in





RPy  12.326 in

RPx   Rp  cos    RPy   Rp  sin    7.

8.

Calculate the x and y components of the external force at P and the weight forces at the CGs. in the CGS. FPx  0  lbf

FPy  F

FG2y  W2

FG3y  W3

FG4y  W4

Substitute these given and calculated values into the matrix equation 11.9. Note that Mathcad requires that all elements in a matrix have the same dimension. Thus, the matrix and array in equation 11.9 will be made dimensionless and the dimensions will be put back in after solving it. m2  m4 

W2 g W4 g

m2  1.55 blob m4  7.01 blob

m3 

W3 g

m3  0.28 blob



0 1 0 0 0 0 0  1  1 0 0 0 0 1 0  0  R12y R12x R32y R32x 0 0 0 0  in in in in  0 1 0 0 0 0 1  0  0 0 0 1 1 0 0 0 C   R23y R23x R43y R43x  0 0 0 0 in in  in in  0 0 1 0 0 0 1 0  0 0 1 0 0 0 1  0  R34y R34x R14y R14x 0 0 0  0 in in in in 

1   m2 a G2x lbf   m2 aG2y  FG2y lbf  1     1 1 IG2 α lbf  in     1 m3 a G3x lbf     1 F    m3 aG3y  FG3y lbf   1 1   IG3 α lbf  in   1    m4 aG4x  FPx lbf   1    m4 aG4y  FPy  FG4y  lbf   I  α   R  F  R  F   lbf  1 in 1 Px Py Py Px   G4  

0



0



1



0 0

 0  0  0  0 

R  C

1

F


F12x  327 lbf


F12y  2682 lbf


F32x  324 lbf


F32y  2086 lbf


F43x  324 lbf


F43y  1978 lbf


F14x  323 lbf


F14y  3012 lbf

T12  R  lbf  in

T12  29442 lbf  in

1 3 5 7

9

2 4 6 8




Figure P11-5a shows an oil field pump mechanism. The head of the rocker arm is shaped such that the lower end of a flexible cable attached to it will always be directly over the well head regardless of the position of the rocker arm 4. The pump rod, which connects to the pump in the well casing, is connected to the lower end of the cable. The force in the pump rod on the up stroke is 2970 lb and the force on the down stroke is 2300 lb. Link 2 weighs 598.3 lb and has a moment of inertia of 11.8 blob-in 2; both including the counterweight. Its CG is on the link cenerline, 13.2 in from O2. Link 3 weighs 108 lb and its CG is on the link centerline, 40 in from A. It has a mass moment of inertia of 150 blob-in 2. Link 4 weighs 2706 lb and has a moment of inertia of 10700 blob-in 2: both include the counterweight. Its CG is on the link centerline where shown. The crank turns at a constant speed of 4 rpm CCW. Find and plot all pin forces and the torque needed to drive the crank for one revolution of the crank. Include gravity forces. 2

1

Units:


Given:


a  14.0 in

Link 3 (A to B):

Link 4 (B to O4):

c  51.3 in

Link 1 (O2 to O4): d  79.7 in

Link 1 offsets:

d X  47.5 in

d Y  64 in

External load data:

F  2300 lbf

T4  0  lbf  in


Solution: 1.

b  80.0 in

ω  4  rpm

α  0  rad sec

2

Link CG positions:

RCG2  13.2 in

RCG3  40.0 in

RCG4  32.0 in

Link weights:

W2  598.3  lbf

W3  108  lbf

W4  2706 lbf

Moments of inertia:

IG2  11.8 blob in

2

2

IG3  150  blob in

2

IG4  10700  blob in


  143.11 deg

R34  32.00  in


  14.03  deg

R14  79.22  in


  156.62 deg

Rp  124.44 in






Figure P11-5a shows an oil field pump mechanism. The head of the rocker arm is shaped such that the lower end of a flexible cable attached to it will always be directly over the well head regardless of the position of the rocker arm 4. The pump rod, which connects to the pump in the well casing, is connected to the lower end of the cable. The force in the pump rod on the up stroke is 2970 lb and the force on the down stroke is 2300 lb. Link 2 weighs 598.3 lb and has a moment of inertia of 11.8 blob-in 2; both including the counterweight. Its CG is on the link cenerline, 13.2 in from O2. Link 3 weighs 108 lb and its CG is on the link centerline, 40 in from A. It has a mass moment of inertia of 150 blob-in 2. Link 4 weighs 2706 lb and has a moment of inertia of 10700 blob-in 2: both include the counterweight. Its CG is on the link centerline where shown. The crank turns at a constant speed of 4 rpm CCW. At the instant shown in the figure the crank angle is at 45 deg with respect to the global coordinate system. Find the torque needed to drive the crank for the position shown using the method of virtual work. Include gravity forces. 2

1

Units:


Given:


a  14.0 in

Link 3 (A to B):

Link 4 (B to O4):

c  51.3 in

Link 1 (O2 to O4): d  79.7 in

Link 1 offsets:

d X  47.5 in

d Y  64 in

External load data:

FP4y  2300 lbf


Solution: 1.

b  80.0 in

ω  4  rpm

α  0  rad sec

2

Link CG positions:

RCG2  13.2 in

RCG3  40.0 in

RCG4  32.0 in

Link weights:

W2  598.3  lbf

W3  108  lbf

W4  2706 lbf

Moments of inertia:

IG2  11.8 blob in

2

2

IG3  150  blob in

2

IG4  10700  blob in


  143.11 deg

R34  32.00  in


  14.03  deg

R14  79.22  in


  156.62 deg

Rp  124.44 in


Use Problems 6.84c and 7.70b with 2 = 4 rpm to determine the position, velocity, and acceleration of links 3 and 4. The angles are calculated in the xy coordinate system and then rotated into the XY coordinate system, which is used in the remainder of the problem. Coordinate rotation angle:

  atan2 d X d Y 

θ  99.057 deg

ω  0.0214 rad sec

θ  29.064 deg

ω  0.0986 rad sec

Accelerations:

a G2  2.316  in sec a G3  1.764  in sec a G4  2.288  in sec a P4  1.391  in sec

Velocities:

vG2  5.529  in sec

  126.582 deg 1

1

2 2 2

2 1

α  0.0250 rad sec α  0.0272 rad sec

θAG2  225.00 deg θAG3  244.915  deg θAG4  265.366  deg θAP4  60.394 deg θVG2  135.0  deg

2 2



vG3  5.406  in sec vG4  7.809  in sec vP4  4.753  in sec 2.

3.

4.

1

1

θVG3  127.628  deg θVG4  105.034  deg θVP4  245.646  deg

Calculate the x and y components of the velocity vectors. vG2x  vG2 cos θVG2

vG2x  3.910 in sec


vG2y  3.910 in sec


vG3x  3.301 in sec




vG4x  2.026 in sec




vP4x  1.960 in sec


vP4y  4.330 in sec

1

1 1

1 1

1 1 1

Calculate the x and y components of the acceleration of the CGs of all moving links in the global coordinate system (GCS). a G2x  a G2 cos θAG2

a G2x  1.638 in sec


a G2y  1.638 in sec


a G3x  0.748 in sec


a G3y  1.598 in sec


a G4x  0.185 in sec


a G4y  2.281 in sec

2 2 2 2 2 2

Calculate the mass of each link. m2 

5.

1

W2 g

m3 

W3

m4 

g

W4 g


1 ω

 m2  a G2x vG2x  a G2y vG2y  m3  a G3x vG3x  a G3y vG3y    m4  a G4x vG4x  aG4y vG4y  IG2 α ω  IG3 α ω  IG4 α ω    W  v  W  v  W  v    F  v   2 G2y 3 G3y 4 G3y P4y P4y  

T12  10219 lbf  in








Use the information in problem 11-20 to find and plot the torque needed to drive the crank for one revolution of the crank using the method of virtual work.

Solution:


No solution for this problem is provided. This problem is more suitable for a longer-term project than for a short-term homework problem.




In Figure P11-13, links 2 and 4 each weigh 2 lb and there are 2 of each (another set behind). Their CGs are at their midpoints. Link 3 weighs 10 lb. The moments of inertia of links 2, 3, and 4 are 0.071, 0.430, and 0.077 blob-in 2, respectively. Find the torque needed to begin a slow CCW rotation of link 2 from the position shown using the method of virtual work. Include gravity forces.

Units:


Given:

Link lengths:

2

1

Link 2 (O2 to A)

a  9.174  in

Link 3 (A to B)

b  12.971 in

Link 4 (B to O4)

c  9.573  in

Link 3 (O2 to O4)

d  7.487  in

Link angles (LCS):

θ  26.0 deg

θ  72.239 deg

θ  60.491 deg

Weight:

W2  4  lbf

W3  10 lbf

W4  4  lbf

Mass:

m2 

W2

m3 

g

m2  0.010 blob 2

δ  0  deg

RCG4  4.786  in

δ  0  deg

α  0  rad sec


2

ω  0.010  rad sec

2

2

1

ω  0.0347 rad sec ω  0.0466 rad sec

1.

2

IG4  0.077  blob in

RCG3  7.086  in

δ  23.758 deg

FP4  0  lbf


a G2  0.0 in sec

α  0.0025 rad sec

Solution:

m4  0.010 blob

T4  0  lbf  in

α  0.00206  rad sec

Velocities:

g

2

T3  0  lbf  in Accelerations:

g

W4

IG3  0.430  blob in

RCG2  4.587  in


m4 

m3  0.026 blob

Moment of inertia: IG2  0.071  blob in Mass center:

W3

1 1

2

θAG2  85.879 deg

a G3  0.0174 in sec a G4  0.0159 in sec vG2  0.0459 in sec vG3  0.284  in sec

2

θAG3  100.159  deg

2

θAG4  123.308  deg

1

θVG2  41.21  deg

1

vG4  0.0223 in sec

θVG3  52.251 deg

1

θVG4  82.367 deg


Calculate the x and y components of the velocity vectors (global coordinate system). 1


vG2x  0.035 in sec


vG2y  0.030 in sec


vG3x  0.174 in sec




vG4x  2.962  10

1

1 1

3

in sec

1



vG4y  vG4 sin θVG4 2.

3.


1

Calculate the x and y components of the acceleration of the CGs of all moving links in the global coordinate system (GCS). a G2x  a G2 cos θAG2 2 a G2x  0.000 in sec 2


a G2y  0.000 in sec


a G3x  3.069  10


a G3y  0.017 in sec


a G4x  8.731  10


a G4y  0.013 in sec

3

in sec

2

2 3

in sec

2

2


1 ω

 m2  a G2x vG2x  a G2y vG2y  m3  a G3x vG3x  a G3y vG3y    m4  a G4x vG4x  aG4y vG4y  IG2 α ω  IG3 α ω  IG4 α ω     ω  T  ω  W  v  W  v  W  v  3 G3y 4 G4y  T3  4   2 G2y 



T12  221.32 lbf  in






Figure P11-3 shows a fourbar linkage and its dimensions. The steel crank and rocker have uniform cross sections 1 in wide by 0.5 in thick. The aluminum coupler is 0.75 in thick. In the instantaneous position shown, the crank O2A has  = 40 rad/sec and  = -20 rad/sec2. There is a horizontal force at P of F = 50 lb. Find the torque needed to drive the crank at this instant using the method of virtual work. 1

2

Units:


Given:


a  5.0 in

Link 3 (A to B)

b  7.4 in

Link 4 (B to O4)

c  8.0 in

Link 1 (O2 to O4)

d  9.5 in

Rpa  8.90 in

δ  56 deg

F  50 lbf

Coupler point:

Crank angle and motion: θ  50 deg Link cross-section dims: w2  1.0 in

t2  0.50 in


2.

ω  40 rad sec

t3  0.75 in

α  20 rad sec

w4  1.0 in 3

γs  0.3 lbf  in

steel

1

T4  0  lbf  in

t4  0.50 in 3

γa  0.1 lbf  in

aluminum



ω  41.552 rad sec

θ  113.008  deg

ω  26.320 rad sec

1

α  335.762  rad sec

1

α  2963.667 rad sec

2

2


RCG2  0.5 a

Link 3:

RCG3x'  RCG3y'  RCG3 

RCG2  2.500  in

RCG4  0.5 c

 


RCG4  4.000  in

RCG3x'  4.126  in

3

 

Rpa sin δ

RCG3y'  2.459  in

3 2

RCG3x'  RCG3y'

2

RCG3  4.803  in


2

δ  30.801 deg


γs g 3

m2  1.943  10 IG2 

m2 12

m3 

2

 

 b  Rpa sin δ  t3 3

 blob m3  5.303  10

  w2  a 2

1

2



 blob

γa g

m4  w4 t4 c

γs g 3

m4  3.108  10 3

IG2  4.209  10

 blob

2

 blob in


IG3 

IG4  4.

m3 6 m4 12



 2

IG3  0.097  blob in

  w4  c

IG4  0.017  blob in



2

2

2

2

2



Calculate the x and y components of the velocity of the CGs of all moving links in the global coordinate system (GCS).



 

 

v G2  RCG2 ω sin θ  j  cos θ vG2x  Re v G2

vG2x  76.604

vG2y  Im v G2



 

vG2y  64.279

 



v A  a  ω sin θ  j  cos θ

in s

in s









v CG3A  RCG3 ω sin θ  δ  j  cos θ  δ

v G3  v A  v CG3A vG3x  Re v G3

vG3x  22.521

in


vG3y  22.280

in









v PA  Rpa ω sin θ  δ  j  cos θ  δ

s s



v P  v A  v PA vPx  Re v P vPy  Im v P



 

vPx  184.907

in

vPy  21.240

in

s

 

s

v G4  RCG4 ω sin θ  j  cos θ

5.

vG4x  Re v G4

vG4x  96.905

in


vG4y  41.150

in

s s




 

   a ω2 cosθ  j  sinθ


3

a G2x  5.104  10 

in sec

a G2y  Im a G2

3

a G2y  6.160  10 

in sec



 

   a ω2 cosθ  j  sinθ

a A  a  α sin θ  j  cos θ





2

2




a G3x  Re a G3

in

a G3x  10277.3 


in

a G3y  12841.8 

sec



 

2

2

   c ω2 cosθ  j  sinθ


a G4x  8745.5

in sec

a G4y  Im a G4

a G4y  9734.6

in sec

6.

2

Calculate the x and y components of the external force at P in the CGS. FPx  F

7.

2



1 ω

 m2  a G2x vG2x  a G2y vG2y  m3  a G3x vG3x  a G3y vG3y    m4  a G4x vG4x  aG4y vG4y  IG2 α ω  IG3 α ω  IG4 α ω   F  v  F  v   Px Px Py Py 

T12  463  lbf  in








Figure P11-3 shows a fourbar linkage and its dimensions. The steel crank and rocker have uniform cross sections 1 in wide by 0.5 in thick. The aluminum coupler is 0.75 in thick. In the instantaneous position shown, the crank O2A has  = 40 rad/sec. There is a horizontal force at P of F = 50 lb. Find and plot all pin forces and the torque needed to drive the crank at a constant speed of 40 rad/sec for one revolution of the crank using the program FOURBAR. Change the lengths of links 3 and 4 to 7.4 and 8.0 in, respectively.

Units:


Given:

Link lengths:

1

2

Link 2 (O2 to A)

a  5.00 in

Link 3 (A to B)

b  7.40 in

Link 4 (B to O4)

c  8.00 in

Link 1 (O2 to O4)

d  9.50 in

Rpa  8.90 in

δ  56 deg

FP  50 lbf

Coupler point:

1

2

Crank motion: ω  40 rad sec α  0  rad sec Link cross-section dims: w2  1.00 in t2  0.50 in t3  0.75 in w4  1.00 in Material specific weight:

3

γs  0.3 lbf  in

steel

T4  0  lbf  in

aluminum

t4  0.50 in 3

γa  0.1 lbf  in

See Figure P11-3 and Mathcad file P1127. Solution: 1. Determine the distance to the CG in the LRCS on each of the three moving links. Links 2 and 4:

RCG2  0.5 a

Link 3:

RCG3x' 

RCG2  2.500 in

RCG4  0.5 c

 


RCG3x'  4.126 in

3

 

Rpa sin δ

RCG3y' 

RCG3y'  2.459 in

3 2

RCG3 

RCG4  4.000 in

RCG3x'  RCG3y'

2

RCG3  4.803 in

At an angle with respect to the local x' axis of δ  atan2 RCG3x' RCG3y'  δ  30.801 deg 2.


γs

m3 

g 3

m2  1.943  10 IG2  IG3  IG4  3.

m2 12 m3 6 m4 12

blob

  w2  a 2

1 2

 

 b  Rpa sin δ  t3 3

m3  5.303  10

2

blob

γa g

m4  w4 t4 c

3

 2

IG3  0.097 in lbf  sec

2

  w4  c

IG4  0.017 in lbf  sec

2



2

2



blob

2

 b  Rpa sin δ 2

g

m4  3.108  10 IG2  0.0042 in lbf  sec



γs

Calculate the x and y components of the external force at P in the CGS and the magnitude and angle of the position vector defining point P in the LRCS. FPx  FP




 

Rpay'  Rpa sin δ

Rpx'  Rpax'  RCG3x'

Rpy'  Rpay'  RCG3y'

Rpax'  4.977 in

Rpay'  7.378 in

Rpx'  0.851 in

Rpy'  4.919 in

Rp  4.992 in

θp  atan

Rp  4.

 

Rpax'  Rpa cos δ

2

Rpx'  Rpy'

2

 Rpy'    Rpx' 

θp  80.182 deg


See next page for plots of pin forces and torque.






Figure P11-3 shows a fourbar linkage and its dimensions. The steel crank and rocker have uniform cross sections 1 in wide by 0.5 in thick. The aluminum coupler is 0.75 in thick. In the instantaneous position shown, the crank O2A has  = 40 rad/sec. There is a horizontal force at P of F = 50 lb. Find and plot the torque needed to drive the crank at a constant speed of 40 rad/sec for one revolution of the crank using the method of virtual work. Change the lengths of links 3 and 4 to 7.4 and 8.0 in, respectively.

Units:


Given:

Link lengths:

1

2

Link 2 (O2 to A)

a  5.00 in

Link 3 (A to B)

b  7.40 in

Link 4 (B to O4)

c  8.00 in

Link 1 (O2 to O4)

d  9.50 in

Rpa  8.90 in

δ  56 deg

FP  50 lbf

Coupler point:

Crank motion: ω  40 rad sec Link cross-section dims: w2  1.00 in

t2  0.50 in


1

α  0  rad sec

t3  0.75 in

2

w4  1.00 in 3

γs  0.3 lbf  in

steel

T4  0  lbf  in

t4  0.50 in 3

γa  0.1 lbf  in

aluminum



RCG2  0.5 a

Link 3:

RCG3x' 

RCG2  2.500 in

RCG4  0.5 c

 


RCG3x'  4.126 in

3

 

Rpa sin δ

RCG3y' 

RCG3y'  2.459 in

3 2

RCG3 

RCG4  4.000 in

RCG3x'  RCG3y'

2

RCG3  4.803 in



γs

m3 

g 3

m2  1.943  10 IG2  IG3  IG4  3.

δ  30.801 deg

m2 12 m3 6 m4 12

blob

  w2  a 2

1 2

 

 b  Rpa sin δ  t3 3

m3  5.303  10

2

blob

γa g

m4  w4 t4 c

3

3


 2


  w4  c




2

2

2



g

m4  3.108  10 IG2  4.209  10



γs

2

blob in

2

2

Calculate the position, angular velocity, and acceleration of links 3 and 4 as functions of 2.

blob


K1 


d

K2 

a

K1  1.9000

 

2

d

K3 

c

K2  1.1875

 

2

2

a b c d

2

2 a c

K3  1.5561

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  K4 

2

d

K5 

b

 

 



 

a  ω

2

2

c d a b

 



b



 

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

2 a b

 

 

 2  4 Dθ F θ 

 

θ θ  2   atan2 2  D θ E θ  ω θ 

2

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

 

B θ

E θ

    sin θ θ  θ θ  A  θ  c sin θ θ  D θ  c cos θ θ 

  c sin θ θ  θ θ  B θ  b  sin θ θ  E θ  b  cos θ θ 


 

2

 

 

ω θ 

 2

a  ω






    c ωθ2 cosθθ

C θ  a  α sin θ  a  ω  cos θ  b  ω θ  cos θ θ

 

2     C θ  E θ  B θ  F  θ A  θ  E θ  B θ  D θ A'  θ  c sin θ θ 

 2     c ωθ2 sinθθ

F θ  a  α cos θ  a  ω  sin θ  b  ω θ  sin θ θ

 

α θ 

 

  

D' θ  c cos θ θ

 

 

2

 

 

  

 

  

B' θ  b  sin θ θ

E' θ  b  cos θ θ

 2     c ωθ2 cosθθ


 


 2     c ωθ2 sinθθ


 

α θ 

 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ

 


 

 


θAP θ  arg AP θ 


4.



 



 

 

v G2 θ  RCG2 ω sin θ  j  cos θ

 

  

 

vG2x θ  Re v G2 θ

 



 

  

vG2y θ  Im v G2 θ

 

v A θ  a  ω sin θ  j  cos θ

 

 

  



  



v CG3A θ  RCG3 ω θ  sin θ θ  δ  j  cos θ θ  δ

 

 

 

v G3 θ  v A θ  v CG3A θ

 

  


 

 

  



 

  

v PA θ  Rpa ω θ  sin θ θ  δ  j  cos θ θ  δ

 

 

  




 

v P θ  v A θ  v PA θ

 

  

vPx θ  Re v P θ

 

 

 

  

vPy θ  Im v P θ

    j  cosθθ

v G4 θ  RCG4 ω θ  sin θ θ

 

  

vG4x θ  Re v G4 θ 5.

 

  



 



 

   a ω2 cosθ  j  sinθ

a G2 θ  RCG2 α sin θ  j  cos θ

 

  

a G2x θ  Re a G2 θ

 



 

 

  

a G2y θ  Im a G2 θ

   a ω2 cosθ  j  sinθ

a A θ  a  α sin θ  j  cos θ

 

           2  RCG3 ω θ   cos θ θ  δ  j  sin θ θ  δ  a G3 θ  a A θ  a CG3A θ a CG3A θ  RCG3 α θ  sin θ θ  δ  j  cos θ θ  δ

 

  


 

 

 

  



a G4 θ  RCG4 α θ  sin θ θ

 

  

a G4x θ  Re a G4 θ 6.

  

Calculate the x and y components of the external force at P in the CGS. FPx  FP

7.

 




 

T12 θ 

1 ω



                                          

 m2 a G2x θ  vG2x θ  a G2y θ  vG2y θ    m3 a G3x θ  vG3x θ  aG3y θ  vG3y θ   m4 a G4x θ  vG4x θ  aG4y θ  vG4y θ   IG2 α ω  IG3 α θ  ω θ  IG4 α θ  ω θ  FPx vPx θ  FPy vPy θ 



 


8.


Plot the input torque for one revolution of the crank. θ  0  deg 2  deg  360  deg

1500

1000

500

 

T 12 θ in lbf

0

 500

 1000

0

30

60

90

120

150

180 θ deg

210

240

270

300

330

360




Given:

Figure P11-4a shows a fourbar linkage and its dimensions in meters. The steel crank and rocker have uniform cross sections 50 mm wide by 25 mm thick. The aluminum coupler is 25 mm thick. In the instantaneous position shown, the crank O2A has  = 10 rad/sec and  = 5 rad/sec2. There is a vertical force at P of F = 100 N. Find the torque needed to drive the crank at this instant using the method of virtual work. Link lengths: Link 2 (O2 to A)

a  1.00 m

Link 3 (A to B)

b  2.06 m

Link 4 (B to O4)

c  2.33 m

Link 1 (O2 to O4)

d  2.22 m

Rpa  3.06 m

δ  31 deg

F  100  N

Coupler point:


ω  10 rad sec

1

T4  0  N  m

α  5  rad sec

2


t2  25 mm


2.

t3  25 mm

w4  50 mm 3

γs  0.3 lbf  in

steel

t4  25 mm 3

γa  0.1 lbf  in

aluminum



ω  3.669  rad sec

θ  96.322 deg

ω  1.442  rad sec

1

α  55.752 rad sec

1

α  67.103 rad sec

2


RCG2  0.5 a

Link 3:

RCG3x' 

RCG2  0.500 m

 


 

Rpa sin δ

2

RCG3x'  RCG3y'

RCG4  1.165 m

RCG3y'  0.525 m

3

RCG3 

RCG4  0.5 c RCG3x'  1.561 m

3

RCG3y' 

2

RCG3  1.647 m


2

δ  18.600 deg


γs

m3 

g

m2  10.380 kg IG2  IG3 

m2 12 m3 6

2

2

2

γa g

m4  w4 t4 c

2

 2

γs g

m4  24.185 kg IG2  0.867 kg m






 

 b  Rpa sin δ  t3

m3  112.332 kg

  w2  a 2

1

2

IG3  125.951 kg m


IG4  4.

m4 12


  w4  c 2

2

2

IG4  10.947 kg m






 

 


vG2x  4.330




 

vG2y  2.500

 



v A  a  ω sin θ  j  cos θ



m s

m s









vG3x  5.999

m


vG3y  0.425

m










s s






 

vPx  5.995

m

vPy  5.906

m

 

s s


5.


vG4x  1.670

m


vG4y  0.185

m

s s




 

   a ω2 cosθ  j  sinθ


m

a G2x  52.165


m

a G2y  85.353

sec



 

2

2

   a ω2 cosθ  j  sinθ

a A  a  α sin θ  j  cos θ



     2  RCG3 ω   cos θ  δ  j  sin θ  δ 

a CG3A  RCG3 α sin θ  δ  j  cos θ  δ


a G3x  Re a G3

a G3x  114.678

m sec

2



a G3y  Im a G3

a G3y  11.429

m sec



 

2

   c ω2 cosθ  j  sinθ


a G4x  77.166

m sec

a G4y  Im a G4

a G4y  13.424

m sec

6.

2


7.

2

FPy  F


1 ω


T12  5588 N  m








Figure P11-4a shows a fourbar linkage and its dimensions in meters. The steel crank and rocker have uniform cross sections 50 mm wide by 25 mm thick. The aluminum coupler is 25 mm thick. In the instantaneous position shown, the crank O2A has  = 10 rad/sec. There is a vertical force at P of F = 100 N. Find and plot all pin forces and the torque needed to drive the crank at a constant speed of 10 rad/sec for one revolution of the crank using the program FOURBAR.

Given:


a  1.00 m

Link 3 (A to B)

b  2.06 m

Link 4 (B to O4)

c  2.33 m

Link 1 (O2 to O4)

d  2.22 m

Coupler point:

Rpa  3.06 m

δ  31 deg

FP  100  N

Crank motion:

ω  10 rad sec

1

α  0  rad sec

T4  0  N  m

2


t2  25 mm


t3  25 mm

w4  50 mm 3

γs  0.3 lbf  in

steel

t4  25 mm 3

γa  0.1 lbf  in

aluminum



RCG2  0.5 a

Link 3:

RCG3x' 

RCG2  0.500 m

 


 

Rpa sin δ

RCG3y'  0.525 m

3 2

RCG3 

RCG3x'  RCG3y'

RCG4  1.165 m

RCG3x'  1.561 m

3

RCG3y' 

RCG4  0.5 c

2

RCG3  1.647 m



γs

m3 

g

m2  10.380 kg IG2  IG3  IG4  3.

δ  18.600 deg

m2 12 m3 6 m4 12

1 2

m3  112.332 kg

  w2  a 2

 

 b  Rpa sin δ  t3

2

γa g

m4  w4 t4 c

2

IG2  0.867 kg m



 2

IG3  125.951 kg m

  w4  c

IG4  10.947 kg m



2

g

m4  24.185 kg

 b  Rpa sin δ 2

γs

2

2

2



Calculate the x and y components of the external force at P in the CGS and the magnitude and angle of the position vector defining point P in the LRCS. FPx  0  N

FPy  FP



 




Rpax'  2.623 m

Rpay'  1.576 m

Rpx'  1.062 m

Rpy'  1.051 m

Rp  1.494 m

θp  atan

Rp 

4.

 


2

Rpx'  Rpy'

2

 Rpy'    Rpx' 

θp  44.694 deg








Figure P11-4a shows a fourbar linkage and its dimensions in meters. The steel crank and rocker have uniform cross sections 50 mm wide by 25 mm thick. The aluminum coupler is 25 mm thick. In the instantaneous position shown, the crank O2A has  = 10 rad/sec. There is a vertical force at P of F = 100 N. Find and plot the torque needed to drive the crank at a constant speed of 10 rad/sec for one revolution of the crank using the method of virtual work.

Given:


a  1.00 m

Link 3 (A to B)

b  2.06 m

Link 4 (B to O4)

c  2.33 m

Link 1 (O2 to O4)

d  2.22 m

Coupler point:

Rpa  3.06 m

δ  31 deg

FP  100  N

Crank motion:

ω  10 rad sec

1

α  0  rad sec

T4  0  N  m

2


t2  25 mm


t3  25 mm

w4  50 mm 3

γs  0.3 lbf  in

steel

t4  25 mm 3

γa  0.1 lbf  in

aluminum



RCG2  0.5 a

Link 3:

RCG3x' 

RCG2  0.500 m

 


 

Rpa sin δ

RCG3y'  0.525 m

3 2

RCG3 

RCG3x'  RCG3y'

RCG4  1.165 m

RCG3x'  1.561 m

3

RCG3y' 

RCG4  0.5 c

2

RCG3  1.647 m



γs

m3 

g

m2  10.380 kg IG2  IG3  IG4  3.

δ  18.600 deg

m2 12 m3 6 m4 12

1 2

 

 b  Rpa sin δ  t3

γa

m4  w4 t4 c

g

m3  112.332 kg

  w2  a 2

2

2

IG2  0.867 kg m



 2

IG3  125.951 kg m

  w4  c

IG4  10.947 kg m



2

g

m4  24.185 kg

 b  Rpa sin δ 2

γs

2

2

2



Calculate the position, angular velocity, and acceleration of links 3 and 4 as functions of 2. K1 

d a

K2 

d c

2

K3 

2

2

a b c d 2 a c

2


K1  2.2200

 


K2  0.9528

 

K3  1.5265

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  K4 

2

d

K5 

b

 

 



 

a  ω

2

2

c d a b

 



b



 

 

 

D θ  cos θ  K1  K4 cos θ  K5

2 a b

 

 

 2  4 Dθ F θ 

 

θ θ  2   atan2 2  D θ E θ  ω θ 

2

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

 

B θ

E θ




 

ω θ 

a  ω






 

 

2

 

 2

    c ωθ2 cosθθ

 

 

2

 

 2     c ωθ2 sinθθ


 

α θ 

        A  θ  E θ  B θ  D θ A'  θ  c sin θ θ  C θ  E θ  B θ  F θ

 

  

D' θ  c cos θ θ

 

 

2

 

 

  

 

  

B' θ  b  sin θ θ

E' θ  b  cos θ θ

 2     c ωθ2 cosθθ


 


 2     c ωθ2 sinθθ


 

α θ 

 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ

 


 

 

θAP θ  arg AP θ 

AP θ  AP θ 4.


 



 

 




 

  

 


 



 

  


 

v A θ  a  ω sin θ  j  cos θ

 

 

  



  




 

 

 

v G3 θ  v A θ  v CG3A θ

 

  


 

 

  



 

  


 

 

  




 

v P θ  v A θ  v PA θ

 

  


 

 

 

  


    j  cosθθ

v G4 θ  RCG4 ω θ  sin θ θ

 

  

vG4x θ  Re v G4 θ 5.

 

  



 



 

   a ω2 cosθ  j  sinθ


 

  


 



 

 

  


   a ω2 cosθ  j  sinθ

a A θ  a  α sin θ  j  cos θ

 


 

  


 

 

 

  



a G4 θ  RCG4 α θ  sin θ θ

 

  

a G4x θ  Re a G4 θ 6.

  


7.

 


FPy  FP


 

T12 θ 

1 ω



                                           

 m2 a G2x θ  vG2x θ  a G2y θ  vG2y θ    m3 a G3x θ  vG3x θ  aG3y θ  vG3y θ   m4 a G4x θ  vG4x θ  aG4y θ  vG4y θ   IG2 α ω  IG3 α θ  ω θ  IG4 α θ  ω θ  FPx vPx θ  FPy vPy θ


8.



40000

20000

 

T 12 θ Nm

0

 20000

 40000

0

30

60

90

120

150

180 θ deg

210

240

270

300

330

360




Given:

Figure P11-4b shows a fourbar linkage and its dimensions in meters. The steel crank and rocker have uniform cross sections 50 mm wide by 25 mm thick. The aluminum coupler is 25 mm thick. In the instantaneous position shown, the crank O2A has  = 15 rad/sec and  = -10 rad/sec2. There is a horizontal force at P of F = 200 N. Find the torque needed to drive the crank at this instant using the method of virtual work. Link lengths: Link 2 (O2 to A)

a  0.72 m

Link 3 (A to B)

b  0.68 m

Link 4 (B to O4)

c  0.85 m

Link 1 (O2 to O4)

d  1.82 m

Rpa  0.97 m

δ  54 deg

F  200  N

Coupler point:


ω  15 rad sec

1

T4  0  N  m

α  10 rad sec

2


t2  25 mm


2.

t3  25 mm

w4  50 mm 3

γs  0.3 lbf  in

steel

t4  25 mm 3

γa  0.1 lbf  in

aluminum



ω  16.412 rad sec

θ  132.283  deg

ω  1.570  rad sec

1

α  138.628  rad sec

1

α  427.881  rad sec

2


RCG2  0.5 a

Link 3:

RCG3x' 

RCG2  0.360 m

 


 

Rpa sin δ

2

RCG3x'  RCG3y'

RCG4  0.425 m

RCG3y'  0.262 m

3

RCG3 

RCG4  0.5 c RCG3x'  0.417 m

3

RCG3y' 

2

RCG3  0.492 m


2

δ  32.117 deg


γs

m3 

g

m2  7.474 kg IG2  IG3 

m2 12 m3 6

2

2

2

γa g

m4  w4 t4 c m4  8.823 kg 2

IG2  0.324 kg m






 

 b  Rpa sin δ  t3

m3  18.463 kg

  w2  a 2

1

 2

2

IG3  3.318 kg m

γs g


IG4  4.

m4 12


  w4  c 2

2

2

IG4  0.533 kg m






 

 

v G2  RCG2 ω sin θ  j  cos θ




vG2x  2.700


vG2y  4.677

 

 



v A  a  ω sin θ  j  cos θ

m s

m s











vG3x  1.247

m

vG3y  4.769

m

vPx  Re v P

vPx  10.130

m

vPy  Im v P

vPy  5.850











s s



v P  v A  v PA



 

 

s

m s

v G4  RCG4 ω sin θ  j  cos θ vG4x  Re v G4 vG4y  Im v G4 5.

vG4x  0.494

m

vG4y  0.449

m

s s




 

   a ω2 cosθ  j  sinθ


a G2x  138.496

m sec

a G2y  Im a G2

m

a G2y  84.118

sec



 

2

2

   a ω2 cosθ  j  sinθ

a A  a  α sin θ  j  cos θ



     2  RCG3 ω   cos θ  δ  j  sin θ  δ  a G3  a A  a CG3A a G3x  Re a G3 a CG3A  RCG3 α sin θ  δ  j  cos θ  δ

a G3x  155.788

m sec

2



a G3y  Im a G3

a G3y  235.056

m sec



 

2

   c ω2 cosθ  j  sinθ


a G4x  133.128

m sec

a G4y  Im a G4

a G4y  123.897

m sec

6.

2


7.

2

FPy  0  N


1 ω


T12  1168 N  m








Given:

Figure P11-4b shows a fourbar linkage and its dimensions in meters. The steel crank and rocker have uniform cross sections 50 mm wide by 25 mm thick. The aluminum coupler is 25 mm thick. In the instantaneous position shown, the crank O2A has  = 15 rad/sec. There is a horizontal force at P of F = 200 N. Find and plot all pin forces and the torque needed to drive the crank at a constant speed of 15 rad/sec for one revolution of the crank using the program FOURBAR. Change the lengths of links 3 and 4 to 1.43 and 1.60 m, respectively. Link lengths: Link 2 (O2 to A)

a  0.72 m

Link 3 (A to B)

b  1.43 m

Link 4 (B to O4)

c  1.60 m

Link 1 (O2 to O4)

d  1.82 m

Coupler point:

Rpa  0.97 m

δ  54 deg

FP  200  N

Crank motion:

ω  15 rad sec

1

α  0  rad sec

T4  0  N  m

2


t2  25 mm


t3  25 mm

w4  50 mm 3

γs  0.3 lbf  in

steel

t4  25 mm 3

γa  0.1 lbf  in

aluminum



RCG2  0.5 a

Link 3:

RCG3x' 

RCG2  0.360 m

 


 

Rpa sin δ

RCG3y'  0.262 m

3 2

RCG3 

RCG3x'  RCG3y'

RCG4  0.800 m

RCG3x'  0.667 m

3

RCG3y' 

RCG4  0.5 c

2

RCG3  0.716 m



γs

m3 

g

m2  7.474 kg IG2  IG3  IG4  3.

δ  21.422 deg

m2 12 m3 6 m4 12

1 2

m3  38.828 kg

  w2  a 2

2

 

 b  Rpa sin δ  t3

γa g

m4  w4 t4 c

2

IG2  0.324 kg m



 2

IG3  17.218 kg m

  w4  c

IG4  3.546 kg m



2

2



g

m4  16.608 kg

 b  Rpa sin δ 2

γs

2

2

Calculate the x and y components of the external force at P in the CGS and the magnitude and angle of the position vector defining point P in the LRCS. FPx  FP FPy  0.N



 




Rpax'  0.570 m

Rpay'  0.785 m

Rpx'  0.097 m

Rpy'  0.523 m

Rp  0.532 m

θp  atan2 Rpx' Rpy' 

θp  100.458 deg

Rp  4.

 


2

Rpx'  Rpy'

2








Given:

Figure P11-4b shows a fourbar linkage and its dimensions in meters. The steel crank and rocker have uniform cross sections 50 mm wide by 25 mm thick. The aluminum coupler is 25 mm thick. In the instantaneous position shown, the crank O2A has  = 15 rad/sec. There is a horizontal force at P of F = 200 N. Find and plot the torque needed to drive the crank at a constant speed of 15 rad/sec for one revolution of the crank using the method of virtual work. Change the lengths of links 3 and 4 to 1.43 and 1.60 m, respectively. Link lengths: Link 2 (O2 to A)

a  0.72 m

Link 3 (A to B)

b  1.43 m

Link 4 (B to O4)

c  1.60 m

Link 1 (O2 to O4)

d  1.82 m

Coupler point:

Rpa  0.97 m

δ  54 deg

FP  200  N

Crank motion:

ω  15 rad sec

1

α  0  rad sec

T4  0  N  m

2


t2  25 mm


t3  25 mm

w4  50 mm 3

γs  0.3 lbf  in

steel

t4  25 mm 3

γa  0.1 lbf  in

aluminum



RCG2  0.5 a

Link 3:

RCG3x' 

RCG2  0.360 m

 


 

Rpa sin δ

RCG3y'  0.262 m

3 2

RCG3 

RCG3x'  RCG3y'

RCG4  0.800 m

RCG3x'  0.667 m

3

RCG3y' 

RCG4  0.5 c

2

RCG3  0.716 m



γs

m3 

g

m2  7.474 kg IG2  IG3  IG4  3.

δ  21.422 deg

m2 12 m3 6 m4 12

1 2

m3  38.828 kg

  w2  a 2

2

 

 b  Rpa sin δ  t3

γa g

m4  w4 t4 c

2

IG2  0.324 kg m



 2

IG3  17.218 kg m

  w4  c

IG4  3.546 kg m



2

2



g

m4  16.608 kg

 b  Rpa sin δ 2

γs

2

2

Calculate the position, angular velocity, and acceleration of links 3 and 4 as functions of 2.


K1 


d

K2 

a

K1  2.5278

 

2

d

K3 

c

K2  1.1375

 

2

2

a b c d

2

2 a c

K3  1.8862

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  K4 

2

d

K5 

b

 

 



 

a  ω

2

2

c d a b

 



b



 

 

 

D θ  cos θ  K1  K4 cos θ  K5

2 a b

 

 

 2  4 Dθ F θ 

 

θ θ  2   atan2 2  D θ E θ  ω θ 

2

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

 

B θ

E θ




 

ω θ 

a  ω






 

 

2

 

 2

    c ωθ2 cosθθ

 

 

2

 

 2     c ωθ2 sinθθ


 

α θ 


 

  

D' θ  c cos θ θ

 

 

2

 

 

  

 

  

B' θ  b  sin θ θ

E' θ  b  cos θ θ

 2     c ωθ2 cosθθ


 


 2     c ωθ2 sinθθ


 

α θ 

 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ

 


 

 

AP θ  AP θ 4.

θAP θ  arg AP θ 




 



 

 


 

  

 


 



 

  


 

v A θ  a  ω sin θ  j  cos θ

 

 

  



  




 

 

 

v G3 θ  v A θ  v CG3A θ

 

  


 

 

  



 

  


 

 

  




 

v P θ  v A θ  v PA θ

 

  


 

 

 

  


    j  cosθθ

v G4 θ  RCG4 ω θ  sin θ θ

 

  

vG4x θ  Re v G4 θ 5.

 

  



 



 

   a ω2 cosθ  j  sinθ


 

  


 



 

 

  


   a ω2 cosθ  j  sinθ

a A θ  a  α sin θ  j  cos θ

 


 

  


 

 

 

  



a G4 θ  RCG4 α θ  sin θ θ

 

  

a G4x θ  Re a G4 θ 6.

  

Calculate the x and y components of the external force at P in the CGS. FPx  FP

7.

 


FPy  0.N


 

T12 θ 

1 ω



                                          

 m2 a G2x θ  vG2x θ  a G2y θ  vG2y θ    m3 a G3x θ  vG3x θ  aG3y θ  vG3y θ   m4 a G4x θ  vG4x θ  aG4y θ  vG4y θ   IG2 α ω  IG3 α θ  ω θ  IG4 α θ  ω θ  FPx vPx θ  FPy vPy θ 



 


8.



10000

5000

 

T 12 θ Nm

0

 5000

 10000

0

30

60

90

120

150

180 θ deg

210

240

270

300

330

360




Given:

Figure P11-5a shows a fourbar linkage and its dimensions in meters. The steel crank, coupler, and rocker have uniform cross sections 50 mm wide by 25 mm thick. In the instantaneous position shown, the crank O2A has  = 15 rad/sec and  = -10 rad/sec2. There is a vertical force at P of F = 500 N. Find the torque needed to drive the crank at this instant using the method of virtual work. Link lengths: Link 2 (O2 to A)

a  0.785  m

Link 3 (A to B)

b  0.356  m

Link 4 (B to O4)

c  0.950  m

Link 1 (O2 to O4)

d  0.544  m

Rpa  1.09 m

δ  0  deg

F  500  N

Coupler point:

Crank angle and motion: θ  96 deg Link cross-section dims: w2  50 mm t2  25 mm Material specific weight: Solution: 1.

2.

3.

1

w3  50 mm t3  25 mm

α  10 rad sec

2

w4  50 mm t4  25 mm

3

γs  0.3 lbf  in



ω  6.830  rad sec

θ  107.906  deg

ω  12.023 rad sec

1

α  106.282  rad sec

1

α  49.372 rad sec

2

2


RCG2  0.5 a

RCG2  0.393 m

Link 3:

RCG3  0.5 Rpa

RCG3  0.545 m

RCG4  0.5 c

RCG4  0.475 m


γs

m3  w3 t3 Rpa

g

m2  8.148 kg IG2  IG3  IG4  4.

ω  15 rad sec

T4  0  N  m

m2 12 m3 12 m4 12

m3  11.314 kg

  w2  a 2

2

  w4  c 2

γs g

m4  9.861 kg 2

2

2

g

m4  w4 t4 c

IG2  0.420 kg m



  w3  Rpa

γs

2

IG3  1.123 kg m



2

2

IG4  0.744 kg m






 

 


vG2x  5.855

m


vG2y  0.615

m

s s





 

 



v A  a  ω sin θ  j  cos θ

 

 

v CG3A  RCG3 ω sin θ  j  cos θ

v G3  v A  v CG3A vG3x  Re v G3 vG3y  Im v G3





m

vG3x  10.421 vG3y  4.723






s

m s



v P  v A  v PA vPx  Re v P

vPx  9.132

m

vPy  Im v P

vPy  8.215

m



 

 

s s

v G4  RCG4 ω sin θ  j  cos θ vG4x  Re v G4 vG4y  Im v G4 5.

vG4x  5.434

m

vG4y  1.756

m

s s




 

   a ω2 cosθ  j  sinθ


a G2x  22.366

m sec

a G2y  Im a G2

2

m

a G2y  175.247

sec



 

2

   a ω2 cosθ  j  sinθ

a A  a  α sin θ  j  cos θ



      RCG3 ω   cos θ  j  sin θ 


a G3x  Re a G3


m

a G3x  17.640


a G3y  129.301

2

m sec



 

   c ω2 cosθ  j  sinθ


a G4x  19.906

m sec

a G4y  Im a G4

a G4y  137.884

2

m sec

2

2


6.


7.


FPy  F


1 ω


T12  372 N  m








Given:

Figure P11-5a shows a fourbar linkage and its dimensions in meters. The steel crank, coupler, and rocker have uniform cross sections 50 mm wide by 25 mm thick. In the instantaneous position shown, the crank O2A has  = 15 rad/sec. There is a vertical force at P of F = 500 N. Find and plot all pin forces and the torque needed to drive the crank at a constant speed of 15 rad/sec for one revolution of the crank using the program FOURBAR. Change the lengths of links 1, 2, and 3 to 1.000, 0.356, and 0.785 m, respectively. Link lengths: Link 2 (O2 to A)

a  0.356  m

Link 3 (A to B)

b  0.785  m

Link 4 (B to O4)

c  0.950  m

Link 1 (O2 to O4)

d  1.000  m

Rpa  1.09 m

δ  0  deg

FP  500  N

Coupler point:

Crank motion: ω  15 rad sec Link cross-section dims: w2  50 mm t2  25 mm Material specific weight: Solution: 1.

2.

2

w3  50 mm t3  25 mm

w4  50 mm t4  25 mm

3

γs  0.3 lbf  in


RCG2  0.5 a

RCG2  0.178 m

RCG4  0.5 c

Link 3:

RCG3  0.5 Rpa

RCG3  0.545 m

δ  0  deg

RCG4  0.475 m

Determine the mass and moment of inertia of each link.

γs

m3  w3 t3 Rpa

g

m2  3.695 kg IG2  IG3  IG4 

m2 12 m3 12 m4 12

γs g

m3  11.314 kg

  w2  a 2

2

  w4  c 2

γs g

m4  9.861 kg 2

2

2

m4  w4 t4 c

IG2  0.040 kg m



  w3  Rpa

2

IG3  1.123 kg m



2

2

IG4  0.744 kg m




FPy  FP

 

Rpax'  Rpa cos δ 4.

α  0  rad sec


m2  w2 t2 a 

3.

1

T4  0  N  m

Rpx'  Rpax'  RCG3

Rpax'  1.090 m

Rpx'  0.545 m









Given:

Figure P11-5a shows a fourbar linkage and its dimensions in meters. The steel crank, coupler, and rocker have uniform cross sections 50 mm wide by 25 mm thick. In the instantaneous position shown, the crank O2A has  = 15 rad/sec. There is a vertical force at P of F = 500 N. Find and plot the torque needed to drive the crank at a constant speed of 15 rad/sec for one revolution of the crank using the method of virtual work. Change the lengths of links 1, 2, and 3 to 1.000, 0.356, and 0.785 m, respectively. Link lengths: Link 2 (O2 to A)

a  0.356  m

Link 3 (A to B)

b  0.785  m

Link 4 (B to O4)

c  0.950  m

Link 1 (O2 to O4)

d  1.000  m

Rpa  1.09 m

δ  0  deg

FP  500  N

Coupler point:

Crank motion: ω  15 rad sec Link cross-section dims: w2  50 mm t2  25 mm

1.

2.

2

w3  50 mm t3  25 mm

w4  50 mm t4  25 mm

3



RCG2  0.5 a

RCG2  0.178 m

RCG4  0.5 c

Link 3:

RCG3  0.5 Rpa

RCG3  0.545 m

δ  0  deg

RCG4  0.475 m


γs

m3  w3 t3 Rpa

g

m2  3.695 kg IG2  IG3  IG4  3.

α  0  rad sec

γs  0.3 lbf  in

Material specific weight: Solution:

1

T4  0  N  m

m2 12 m3 12 m4 12

γs

m4  w4 t4 c

g

m3  11.314 kg

  w2  a 2

2

IG2  0.040 kg m

 2

2

  w4  c 2

g

m4  9.861 kg

2

  w3  Rpa

γs

2

IG3  1.123 kg m



2

2

IG4  0.744 kg m




d

K2 

a

K1  2.8090

 

2

d

K3 

c

K2  1.0526

 

2

2 a c

K3  2.0890

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 





 

 

C θ  K1   K2  1   cos θ  K3

 

 

θ θ  2   atan2 2  A θ B θ 

 2  4 A θ Cθ 

B θ

2

a b c d

2



d

K4 

K5 

b

 

 



 

a  ω

2

 



b



 

 

 

D θ  cos θ  K1  K4 cos θ  K5

2 a b

 

 

 2  4 Dθ F θ 

 

θ θ  2   atan2 2  D θ E θ  ω θ 

2

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

 

2

c d a b

E θ




 

ω θ 

a  ω






 

 

2

 

 2

    c ωθ2 cosθθ

 

 

2

 

 2     c ωθ2 sinθθ


 

α θ 


 

  

D' θ  c cos θ θ

 

 

2

 

  

 

  

B' θ  b  sin θ θ

E' θ  b  cos θ θ

 

 2     c ωθ2 cosθθ


 


 2     c ωθ2 sinθθ


 

α θ 

 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ

 


 

 

θAP θ  arg AP θ 

AP θ  AP θ 4.


 



 

 


 

  

 


 



 

  


 

v A θ  a  ω sin θ  j  cos θ

 

 

  



  





 


 

 

v G3 θ  v A θ  v CG3A θ

 

  


 

 

  



 

  


 

 

  




 

v P θ  v A θ  v PA θ

 

  


 

 

 

  


    j  cosθθ

v G4 θ  RCG4 ω θ  sin θ θ

 

  


5.

 

  



 



 

   a ω2 cosθ  j  sinθ


 

  


 



 

 

  


   a ω2 cosθ  j  sinθ

a A θ  a  α sin θ  j  cos θ

 


 

  


 

 

 

  



a G4 θ  RCG4 α θ  sin θ θ

 

  

a G4x θ  Re a G4 θ 6.

FPy  FP


 

T12 θ 

1 ω



                                         




8.

  


7.

 


 




1000

500

 

T 12 θ Nm

0

 500

 1000

0

30

60

90

120

150

180 θ deg

210

240

270

300

330

360




Given:

Figure P11-5b shows a fourbar linkage and its dimensions in meters. The steel crank, coupler, and rocker have uniform cross sections of 50 mm diameter. In the instantaneous position shown, the crank O2A has  = -10 rad/sec and  = 10 rad/sec2. There is a horizontal force at P of F = 300 N. Find torque needed to drive the crank at this instant using the method of virtual work. Link lengths: Link 2 (O2 to A)

a  0.86 m

Link 3 (A to B)

b  1.85 m

Link 4 (B to O4)

c  0.86 m

Link 1 (O2 to O4)

d  2.22 m

Rpa  1.33 m

δ  0  deg

F  300  N

Coupler point:

Crank angle and motion: θ  36 deg

ω  10 rad sec

Link cross-section dims: d link  50 mm Material specific weight: Solution: 1.

2.

3.

α  10 rad sec

2

3

γs  0.3 lbf  in


Use program FOURBAR to determine the position, velocity, and acceleration of links 3 and 4. 1

θ  46.028 deg

ω  3.285  rad sec

θ  106.189  deg

ω  11.417 rad sec

α  109.287  rad sec

1

α  43.426 rad sec

2

2


RCG2  0.5 a

RCG2  0.430 m

Link 3:

RCG3  0.5 b

RCG3  0.925 m

RCG4  0.5 c

RCG4  0.430 m


π d link 4

2

 a

γs

m3 

g

m2  14.022 kg IG2  IG3  IG4  4.

1

T4  0  N  m

π d link 4

2

 b

m3  30.164 kg

γs g

m4 

π d link

2

4

g

m4  14.022 kg

3 2 2    d link  a  12  4 

IG2  0.866 kg m

3 2 2    d link  b  12  4 

IG3  8.608 kg m

3 2 2    d link  c  12  4 

IG4  0.866 kg m

m2

γs

 c

2

m3

2

m4

2




 

 

v G2  RCG2 ω sin θ  j  cos θ vG2x  Re v G2 vG2y  Im v G2

vG2x  2.527

m

vG2y  3.479

m

s s





 

 



v A  a  ω sin θ  j  cos θ

 

 

v CG3A  RCG3 ω sin θ  j  cos θ


vG3x  7.242

m


vG3y  4.848

m










s s






 

vPx  8.199

m

vPy  3.924

m

 

s s


5.


vG4x  4.715

m


vG4y  1.369

m

s s




 

   a ω2 cosθ  j  sinθ


m

a G2x  67.048


a G2y  54.028

m sec



 

2

2

   a ω2 cosθ  j  sinθ

a A  a  α sin θ  j  cos θ



      RCG3 ω   cos θ  j  sin θ 


a G3x  Re a G3


a G3x  1.302

m sec

a G3y  Im a G3

a G3y  19.864

2

m sec



 

   c ω2 cosθ  j  sinθ


a G4x  49.187

m sec

a G4y  Im a G4

a G4y  102.448

2

m sec

2

2


6.


7.


FPy  0  N


1 ω


T12  7.10 N  m








Figure P11-5b shows a fourbar linkage and its dimensions in meters. The steel crank, coupler, and rocker have uniform cross sections of 50 mm diameter. In the instantaneous position shown, the crank O2A has  = 10 rad/sec. There is a horizontal force at P of F = 300 N. Find and plot all pin forces and the torque needed to drive the crank at a constant speed of 15 rad/sec for one revolution of the crank using the program FOURBAR. Change the length of link 4 to 1.86 m.

Given:


a  0.86 m

Link 3 (A to B)

b  1.85 m

Link 4 (B to O4)

c  1.86 m

Link 1 (O2 to O4)

d  2.22 m

Coupler point:

Rpa  1.33 m

δ  0  deg

FP  300  N

Crank motion:

ω  10 rad sec

Link cross-section dims: Material specific weight: Solution: 1.

2.

2

d link  50 mm 3

γs  0.3 lbf  in


RCG2  0.5 a

RCG2  0.430 m

Link 3:

RCG3  0.5 b

RCG3  0.925 m

RCG4  0.5 c

RCG4  0.930 m

δ  0  deg

Determine the mass and moment of inertia of each link. π d link 4

2

 a

γs g

m2  14.022 kg IG2  IG3  IG4 

m3 

π d link 4

2

 b

γs g

m3  30.164 kg

m4 

π d link 4

2

γs

 c

g

m4  30.327 kg

3 2 2    d link  a  12  4 

IG2  0.866 kg m

3 2 2    d link  b  12  4 

IG3  8.608 kg m

3 2 2    d link  c  12  4 

IG4  8.748 kg m

m2

2

m3

2

m4

2

Calculate the x and y components of the external force at P in the CGS and the magnitude and angle of the position vector defining point P in the LRCS. FPx  FP

FPy  0  N

 


α  0  rad sec


m2 

3.

1

T4  0  N  m


Rpax'  1.330 m

Rpx'  0.405 m









Given:

Figure P11-5b shows a fourbar linkage and its dimensions in meters. The steel crank, coupler, and rocker have uniform cross sections of 50 mm diameter. In the instantaneous position shown, the crank O2A has  = 10 rad/sec. There is a horizontal force at P of F = 300 N. Find torque needed to drive the crank at this instant using the method of virtual work. Change the length of link 4 to 1.86 m. Link lengths: Link 2 (O2 to A)

a  0.86 m

Link 3 (A to B)

b  1.85 m

Link 4 (B to O4)

c  1.86 m

Link 1 (O2 to O4)

d  2.22 m

Coupler point:

Rpa  1.33 m

δ  0  deg

FP  300  N

Crank motion:

ω  10 rad sec

Link cross-section dims:

1.

2.

2

3

γs  0.3 lbf  in



RCG2  0.5 a

RCG2  0.430 m

Link 3:

RCG3  0.5 b

RCG3  0.925 m

RCG4  0.5 c

RCG4  0.930 m

δ  0  deg


π d link 4

2

 a

γs

m3 

g

m2  14.022 kg IG2  IG3  IG4  3.

α  0  rad sec

d link  50 mm

Material specific weight: Solution:

1

T4  0  N  m

π d link

2

 b

4

γs g

m3  30.164 kg

m4 

π d link

2

γs

 c

4

g

m4  30.327 kg

3 2 2    d link  a  12  4 

IG2  0.866 kg m

3 2 2    d link  b  12  4 

IG3  8.608 kg m

3 2 2    d link  c  12  4 

IG4  8.748 kg m

m2

2

m3

2

m4

2


d

K2 

a

K1  2.5814

 

2

d

K3 

c

K2  1.1935

 

2

2

a b c d

2

2 a c

K3  1.7833

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 



 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ

 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  K4 

d b

2

K5 

2

B θ 2

c d a b 2 a b

2

 

 

 

D θ  cos θ  K1  K4 cos θ  K5


 


 

 

 



 

a  ω

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ



 

 2  4 Dθ F θ 

 

θ θ  2   atan2 2  D θ E θ  ω θ 

b



E θ




 

ω θ 

a  ω






 

 

2

 

 2

    c ωθ2 cosθθ

 

 

2

 

 2     c ωθ2 sinθθ


 

α θ 


 

  

D' θ  c cos θ θ

 

 

2

 

  

 

  

B' θ  b  sin θ θ

E' θ  b  cos θ θ

 

 2     c ωθ2 cosθθ


 


 2     c ωθ2 sinθθ


 

α θ 

 



 

   a ω2 cosθ  j  sinθ

AA θ  a  α sin θ  j  cos θ

 


 

 

θAP θ  arg AP θ 

AP θ  AP θ 4.


 



 

 


 

  

 


 



 

  


 

v A θ  a  ω sin θ  j  cos θ

 

 

  



  




 

 

 

v G3 θ  v A θ  v CG3A θ

 

  


 

  



 


 

  



  


 

 



 

v P θ  v A θ  v PA θ

 

  


 

 

 

  


    j  cosθθ

v G4 θ  RCG4 ω θ  sin θ θ

 

  

vG4x θ  Re v G4 θ 5.

 

  



 



 

   a ω2 cosθ  j  sinθ


 

  


 



 

 

  


   a ω2 cosθ  j  sinθ

a A θ  a  α sin θ  j  cos θ

 


 

  


 

 

 

  



a G4 θ  RCG4 α θ  sin θ θ

 

  

a G4x θ  Re a G4 θ 6.

FPy  0  N


 

T12 θ 

1 ω



                                         




8.

  

Calculate the x and y components of the external force at P in the CGS. FPx  FP

7.

 


 


See next page



2000

0

 

T 12 θ Nm

 2000

 4000

0

30

60

90

120

150

180 θ deg

210

240

270

300

330

360




Figure P11-3 shows a fourbar linkage and its dimensions. The steel crank and rocker have uniform cross sections 1 in wide by 0.5 in thick. The aluminum coupler is 0.75 in thick. In the instantaneous position shown, the crank O2A has  = 40 rad/sec. There is a horizontal force at P of F = 50 lb. Design a steel disc flywheel to smooth the input torque for the crank of Problem 11-26 using a coefficient of fluctuation of 0.05 while minimizing the flywheel weight.

Units:


Given:

Link lengths:

1

2

Link 2 (O2 to A)

a  5.00 in

Link 3 (A to B)

b  7.40 in

Link 4 (B to O4)

c  8.00 in

Link 1 (O2 to O4)

d  9.50 in

Rpa  8.90 in

δ  56 deg

FP  50 lbf

Coupler point:

1

2

Crank motion: ω  40 rad sec α  0  rad sec Link cross-section dims: w2  1.00 in t2  0.50 in t3  0.75 in w4  1.00 in 3

γs  0.3 lbf  in

Material specific weight:

steel


k  0.05


ωavg  ω

T4  0  lbf  in

aluminum

t4  0.50 in

3

γa  0.1 lbf  in

See Figure P11-3 and Mathcad file P1140. Solution: 1. Determine the distance to the CG in the LRCS on each of the three moving links. Links 2 and 4:

RCG2  0.5 a

Link 3:

RCG3x' 

RCG2  2.500  in

RCG4  0.5 c

 


RCG3x'  4.126  in

3

 

Rpa sin δ

RCG3y' 

RCG3y'  2.459  in

3 2

RCG3 

RCG4  4.000  in

RCG3x'  RCG3y'

2

RCG3  4.803  in

At an angle with respect to the local x' axis of δ  atan2 RCG3x' RCG3y'  δ  30.801 deg 2.


γs

m3 

g 3

m2  1.943  10 IG2  IG3  IG4 

m2 12 m3 6 m4 12

1 2

 

 b  Rpa sin δ  t3 3

 blob m3  5.303  10

  w2  a 2

2

 blob

γa g

γs

m4  w4 t4 c

g 3

m4  3.108  10 2

IG2  0.0042 in lbf  sec




 2

IG3  0.097  in lbf  sec

2

  w4  c

IG4  0.017  in lbf  sec

2



2

2

2



 blob


3.


Calculate the x and y components of the external force at P in the CGS and the magnitude and angle of the position vector defining point P in the LRCS. FPx  FP


 

 





Rpax'  4.977 in

Rpay'  7.378 in

Rpx'  0.851 in

Rpy'  4.919 in

Rp  4.992 in

θp  atan

Rp 

2

Rpx'  Rpy'

2

 Rpy'    Rpx' 

θp  80.182 deg

4.


5.


-473.6 +76.7 -533.8 +944.2

-473.6 -396.9 -930.7 +13.5

max at D min at A

Emaxspeed  930.7  in lbf

Eminspeed  13.5 in lbf


E  944.2  in lbf


6.


Use equation 11.22 to determine the required system mass moment of inertia. Is 

E

2

2

Is  11.8 blob in

k ωavg

The moment of inertia of the input crank and the motor armature can be subtracted from this value to obtain the required flywheel moment. However, in this case they are small compared with the required moment of inertia. 7.

There is an infinity of solutions for the dimensions of the flywheel for the required moment of inertia. One possible solution follows: Let the flywheel radius be rf  1.8 a

rf  9.000 in

From Appendix C the moment of inertia of a cylinder about its central axis is I = mr2/2. Solving for m, The required mass is

mf 

2  Is rf

2

mf  0.291  blob

The volume of a cylindrical disk is V = r2t and the mass is m = V/g. Eliminating V and solving for the thickness, t, for a steel disk tf 

mf  g 2

π rf  γs

tf  1.474 in




Figure P11-4a shows a fourbar linkage and its dimensions in meters. The steel crank and rocker have uniform cross sections 50 mm wide by 25 mm thick. The aluminum coupler is 25 mm thick. In the instantaneous position shown, the crank O2A has  = 10 rad/sec. There is a vertical force at P of F = 100 N. Design a steel disc flywheel to smooth the input torque for the crank of Problem 11-29 using a coefficient of fluctuation of 0.05 while minimizing the flywheel weight.

Given:


a  1.00 m

Link 3 (A to B)

b  2.06 m

Link 4 (B to O4)

c  2.33 m

Link 1 (O2 to O4)

d  2.22 m

Coupler point:

Rpa  3.06 m

δ  31 deg

FP  100  N

Crank motion:

ω  10 rad sec

1

α  0  rad sec

T4  0  N  m

2


Solution: 1.

t2  25 mm

t3  25 mm

w4  50 mm 3

γs  0.3 lbf  in


steel


k  0.05

t4  25 mm 3

γa  0.1 lbf  in

aluminum

ωavg  ω




RCG2  0.5 a

Link 3:

RCG3x' 

RCG2  0.500 m

 


 

Rpa sin δ

RCG3y'  0.525 m

3 2

RCG3 

RCG3x'  RCG3y'

RCG4  1.165 m

RCG3x'  1.561 m

3

RCG3y' 

RCG4  0.5 c

2

RCG3  1.647 m



γs

m3 

g

m2  10.380 kg IG2  IG3  IG4  3.

δ  18.600 deg

m2 12 m3 6 m4 12

1 2

m3  112.332 kg

  w2  a 2

 

 b  Rpa sin δ  t3

2

γa g

m4  w4 t4 c

2

IG2  0.867 kg m



 2

IG3  125.951 kg m

  w4  c

IG4  10.947 kg m



2

g

m4  24.185 kg

 b  Rpa sin δ 2

γs

2

2

2




FPy  FP



 

 





Rpax'  2.623 m

Rpay'  1.576 m

Rpx'  1.062 m

Rpy'  1.051 m

Rp  1.494 m

θp  atan

Rp 

2

Rpx'  Rpy'

2

 Rpy'    Rpx' 

θp  44.694 deg

4.


5.


-32191.6 +8802.8 -9137.9 +32476.5

-32191.6 -23388.8 -32526.7 max at D -50.2 min at A

Emaxspeed  32526.7  N  m

Eminspeed  50.2 N  m


E  32476.5 N  m


6.



E

2

2

Is  6495.3 kg m

k ωavg





Given:

Figure P11-4b shows a fourbar linkage and its dimensions in meters. The steel crank and rocker have uniform cross sections 50 mm wide by 25 mm thick. The aluminum coupler is 25 mm thick. In the instantaneous position shown, the crank O2A has  = 15 rad/sec. There is a horizontal force at P of F = 200 N. Design a steel disc flywheel to smooth the input torque for the crank of Problem 11-32 using a coefficient of fluctuation of 0.07 while minimizing the flywheel weight. Change the lengths of links 3 and 4 to 1.43 and 1.60 m, respectively. Link lengths: Link 2 (O2 to A)

a  0.72 m

Link 3 (A to B)

b  1.43 m

Link 4 (B to O4)

c  1.60 m

Link 1 (O2 to O4)

d  1.82 m

Coupler point:

Rpa  0.97 m

δ  54 deg

FP  200  N

Crank motion:

ω  15 rad sec

1

α  0  rad sec

T4  0  N  m

2


Solution: 1.

t2  25 mm

t3  25 mm

w4  50 mm 3

γs  0.3 lbf  in


steel


k  0.07

t4  25 mm 3

γa  0.1 lbf  in

aluminum

ωavg  ω




RCG2  0.5 a

Link 3:

RCG3x' 

RCG2  0.360 m

 


 

Rpa sin δ

RCG3y'  0.262 m

3 2

RCG3 

RCG3x'  RCG3y'

RCG4  0.800 m

RCG3x'  0.667 m

3

RCG3y' 

RCG4  0.5 c

2

RCG3  0.716 m



γs

m3 

g

m2  7.474 kg IG2  IG3  IG4  3.

δ  21.422 deg

m2 12 m3 6 m4 12

1 2

m3  38.828 kg

  w2  a 2

2

 

 b  Rpa sin δ  t3

γa g

m4  w4 t4 c

2

IG2  0.324 kg m



 2

IG3  17.218 kg m

  w4  c

IG4  3.546 kg m



2

2



g

m4  16.608 kg

 b  Rpa sin δ 2

γs

2

2

Calculate the x and y components of the external force at P in the CGS and the magnitude and angle of the position vector defining point P in the LRCS. FPx  FP FPy  0.N



 

 





Rpax'  0.570 m

Rpay'  0.785 m

Rpx'  0.097 m

Rpy'  0.523 m

Rp  0.532 m

θp  atan2 Rpx' Rpy' 

θp  100.458 deg

Rp 

2

Rpx'  Rpy'

2

4.


5.


-7512.0 +3135.5 -2947.2 +7386.1

-7512.0 -4376.5 -7323.7 max at D +62.4 min at A

Emaxspeed  7323.7 N  m

Eminspeed  62.4 N  m


E  7386.1 N  m


6.



E

2

2

Is  469.0 kg m

k ωavg





Given:

Figure P11-5a shows a fourbar linkage and its dimensions in meters. The steel crank, coupler, and rocker have uniform cross sections 50 mm wide by 25 mm thick. In the instantaneous position shown, the crank O2A has  = 15 rad/sec. There is a vertical force at P of F = 500 N. Design a steel disc flywheel to smooth the input torque for the crank of Problem 11-35 using a coefficient of fluctuation of 0.05 while minimizing the flywheel weight. Change the lengths of links 1, 2, and 3 to 1.000, 0.356, and 0.785 m, respectively. Link lengths: Link 2 (O2 to A)

a  0.356  m

Link 3 (A to B)

b  0.785  m

Link 4 (B to O4)

c  0.950  m

Link 1 (O2 to O4)

d  1.000  m

Rpa  1.09 m

δ  0  deg

FP  500  N

Coupler point:

Crank motion: ω  15 rad sec Link cross-section dims: w2  50 mm t2  25 mm Material specific weight: Coefficient of fluctuation: Solution: 1.

2.

2

w3  50 mm t3  25 mm

w4  50 mm t4  25 mm

3

γs  0.3 lbf  in k  0.05

ωavg  ω



RCG2  0.5 a

RCG2  0.178 m

RCG4  0.5 c

Link 3:

RCG3  0.5 Rpa

RCG3  0.545 m

δ  0  deg

RCG4  0.475 m

Determine the mass and moment of inertia of each link.

γs

m3  w3 t3 Rpa

g

m2  3.695 kg IG2  IG3  IG4 

m2 12 m3 12 m4 12

γs g

m3  11.314 kg

  w2  a 2

2

  w4  c 2

γs g

m4  9.861 kg 2

2

2

m4  w4 t4 c

IG2  0.040 kg m



  w3  Rpa

2

IG3  1.123 kg m



2

2

IG4  0.744 kg m




FPy  FP

 


α  0  rad sec


m2  w2 t2 a 

3.

1

T4  0  N  m


Rpax'  1.090 m

Rpx'  0.545 m



5.



-662.0 +251.7 -472.4 +932.3

-662.0 -410.3 -882.7 +49.6

max at D min at A

Emaxspeed  882.7  N  m

Eminspeed  49.6 N  m


E  932.3 N  m


6.



E

2

2

Is  82.9 kg m

k ωavg





Figure P11-5b shows a fourbar linkage and its dimensions in meters. The steel crank, coupler, and rocker have uniform cross sections of 50 mm diameter. In the instantaneous position shown, the crank O2A has  = 10 rad/sec. There is a horizontal force at P of F = 300 N. Design a steel disc flywheel to smooth the input torque for the crank of Problem 11-38 using a coefficient of fluctuation of 0.06 while minimizing the flywheel weight. Change the length of link 4 to 1.86 m.

Given:


a  0.86 m

Link 3 (A to B)

b  1.85 m

Link 4 (B to O4)

c  1.86 m

Link 1 (O2 to O4)

d  2.22 m

Coupler point:

Rpa  1.33 m

δ  0  deg

FP  300  N

Crank motion:

ω  10 rad sec

Link cross-section dims: Material specific weight: Coefficient of fluctuation: Solution: 1.

2.

2

d link  50 mm 3

γs  0.3 lbf  in k  0.06


ωavg  ω


RCG2  0.5 a

RCG2  0.430 m

Link 3:

RCG3  0.5 b

RCG3  0.925 m

RCG4  0.5 c

RCG4  0.930 m

δ  0  deg

Determine the mass and moment of inertia of each link. π d link 4

2

 a

γs g

m2  14.022 kg IG2  IG3  IG4 

m3 

π d link 4

2

 b

γs g

m3  30.164 kg

m4 

π d link 4

2

γs

 c

g

m4  30.327 kg

3 2 2    d link  a  12  4 

IG2  0.866 kg m

3 2 2    d link  b  12  4 

IG3  8.608 kg m

3 2 2    d link  c  12  4 

IG4  8.748 kg m

m2

2

m3

2

m4

2

Calculate the x and y components of the external force at P in the CGS and the magnitude and angle of the position vector defining point P in the LRCS. FPx  FP

FPy  0  N

 


α  0  rad sec


m2 

3.

1

T4  0  N  m


Rpax'  1.330 m

Rpx'  0.405 m



5.



-4318.9 +2471.1 -2135.2 +3862.2

-4318.9 max at B -1847.8 -3983.0 -120.8 min at A

Emaxspeed  4318.9 N  m

Eminspeed  120.8  N  m


E  4198.1 N  m


6.



E

2

2

Is  699.7 kg m

k ωavg





Table P11-5 gives kinematic and geometric data for several slider-crank linkages of the type and orientation shown in Figure 11-4 (p. 597). For the row(s) in the table assigned, solve for all pin forces and the resisting torque on the crank for the position shown.

Units:


Given:

Link lengths:

1

2

Link 2 (O2 to A)

a  4.00 in

Link 3 (A to B)

b  12.00  in

Offset

c  0.00 in

Friction:

μ  0.15

d  14 in

Slider position and motion:

m2  0.002  blob

Mass:

1.

m3  0.020  blob 2

IG2  0.10 blob in

Moment of inertia:

Solution:

ddot  400  in sec

1

Mass center:

RCG2  1.30 in δ  180  deg

Force :

FP  60 lbf

m4  0.060  blob

RCG3  3.00 in δ  0  deg

δFP  180  deg

Calculate the position of links 2 and 3 using equations 4.21, and 4.16 or 4.17. 2

2

2

2

2

K1  68 in

K2  2  a  c

K2  0

K3  2  a  d

K3  112 in

A  K1  K3

A  180 in

B  2  K2

B0

C  K1  K3

C  44 in

2

2

2



2

θ  2  atan2 2  A B 

β ( a b c d α) 

B  4 A  C

θ  asin





θ  52.617 deg

a  sin( α)  c 

 

b






θ  β a b c d θ 2.

a  sin( α)  c  b




θ  164.641  deg






R12x  0.789  in





R12y  1.033  in

R12x  RCG2 cos θ  180  deg R12y  RCG2 sin θ  180  deg

 

R32x   a  RCG2  cos θ

2

IG3  0.20 blob in

See Figure 11-4, Table P11-5, and Mathcad file P1145a.

K1  a  b  c  d

dddot  22760  in sec

R32x  1.639  in

2



 

R32y   a  RCG2  sin θ

R32y  2.145  in

 

R23x  2.893  in

 

R23y  0.795  in

R23x  RCG3 cos θ R23y  RCG3 sin θ





R43x  8.679  in





R43y  2.384  in

R43x   b  RCG3  cos θ  180  deg R43y   b  RCG3  sin θ  180  deg 3.

Calculate the angular velocity of links 2 and 3 using equations 6.24. ω 

ω 

4.

  a   cos θ  sin θ  sin θ  cos θ  ddot cos θ

  b  cos θ

a  ω cos θ

ω  21.831

rad s

rad s

Calculate the angular acceleration of links 2 and 3 using equations 7.17. 2

α 

α  5.

ω  104.019 

  

       b ω2  dddot cosθ a   cos θ  sin θ  sin θ  cos θ 

a  ω  cos θ  cos θ  sin θ  sin θ

 

  b  cos θ 2

2

 


α  0.58

α  2841

rad 2

s rad 2

s

Calculate the value of a G2 and a G3 using equations 7.3 and 7.32.



 

   RCG2 ω2 cosθ  j  sinθ

a G2  RCG2 α sin θ  j  cos θ a G2  a G2

a G2  14066.05

2

sec

θAG2  arg a G2



in

θAG2  127.380  deg

 

   a ω2 cosθ  j  sinθ

a A  a  α sin θ  j  cos θ



 

   RCG3 ω2 cosθ  j  sinθ

a G3  RCG3 α sin θ  j  cos θ a G3  a G3

θAG3  arg a G3 6.

a G3  8642.67 

in sec

2

θAG3  95.836 deg

Calculate the x and y components of the acceleration of the CGs of all moving links in the global coordinate system (GCS). 2


a G2x  8540 in sec


a G2y  11177  in sec


a G3x  879  in sec

2

2


7.

8.

SOLUTION MANUAL 11-45a-3 2


a G3y  8598 in sec

a G4x  dddot

a G4x  22760  in sec

2

Calculate the x and y components of the external force on link 4 in the CGS. FPx  FP cos δFP

FPx  60.000 lbf

FPy  FP sin δFP

FPy  0.000  lbf

Substitute these given and calculated values into the matrix equation 11.10g, modified for this problem. Note that Mathcad requires that all elements in a matrix have the same dimension. Thus, the matrix and array in equation 11.10g will be made dimensionless and the dimensions will be put back in after solving it. 0 0 0 0 1  1  1 0 0 1 0  0  R12y R12x R32y R32x 0 0  in in in in  1 0 0 0 1 0 C    0 0 1 0 0 1  R23y R23x R43y R43x  0 0 in in in in   0 0 0 1 0 0  0 0 0 1 0  0

0

0

0

0

0 0 0 0 μ 1

   1  0 0  0  0  0

 m2 a G2x lbf  1    1  m2 a G2y lbf    1 1  IG2 α lbf  in    1   m3 a G3x  lbf  F      m3 a G3y  lbf  1      IG3 α  lbf  1 in 1     m4 aG4x  FPx  lbf  1   1   m  g  lbf 4  

R  C


F12x  1392 lbf


F12y  173.6  lbf


F32x  1375 lbf


F32y  195.9  lbf


F43x  1357 lbf


F43y  367.9  lbf

1 3 5

1

F

2 4 6


F14y  R  lbf 7


F14y  344.7  lbf

F12  F12x  j  F12y

F32  F32x  j  F32y

F43  F43x  j  F43y

T12  R  lbf  in T12  4846 lbf  in 8

F12  F12

F12  1403 lbf

θF12  arg F12

θF12  172.891  deg

F32  F32

F32  1389 lbf

θF32  arg F32

θF32  8.111  deg

F43  F43

F43  1406 lbf

θF43  arg F43

θF43  15.166 deg




Table P11-5 gives kinematic and geometric data for several slider-crank linkages of the type and orientation shown in Figure 11-4 (p. 597). For the row(s) in the table assigned, solve for the torque available at the crank using the method of virtual work for the position shown assuming no friction loss.

Units:


Given:

Link lengths:

2

1

Link 2 (O2 to A)

a  4.00 in

Offset

c  0.00 in d  14 in

Slider position and motion:

1.

1

dddot  22760  in sec

m3  0.020  blob 2

IG2  0.10 blob in

Moment of inertia:

Solution:

ddot  400  in sec

m2  0.002  blob

Mass:

b  12.00  in

Link 3 (A to B)

Mass center:

RCG2  1.30 in δ  0  deg

Force :

FP  60 lbf

m4  0.060  blob 2

IG3  0.20 blob in

RCG3  3.00 in δ  0  deg

δFP  180  deg


Calculate the position of links 2 and 3 using equations 4.21, and 4.16 or 4.17. 2

2

2

K1  a  b  c  d

2

2

K1  68 in

K2  2  a  c

K2  0

K3  2  a  d

K3  112 in

A  K1  K3

A  180 in

B  2  K2

B0

C  K1  K3

C  44 in

2

2

2



2

θ  2  atan2 2  A B 

β ( a b c d α) 

B  4 A  C

θ  asin





θ  52.617 deg

a  sin( α)  c 

 

b






θ  β a b c d θ 2.

a  sin( α)  c 



b


θ  164.641  deg

Calculate the angular velocity of links 2 and 3 using equations 6.24. ω 

  a   cos θ  sin θ  sin θ  cos θ  ddot cos θ

ω  104.019 

rad s

2


ω 

3.


  b  cos θ

a  ω cos θ

ω  21.831

rad s

Calculate the value of VG2 and VG3 using equations 6.23 and 6.36.



 

 

VG2  RCG2 ω sin θ  j  cos θ

in

vG2  VG2

vG2  135.225

θVG2  arg VG2

θVG2  37.383 deg



 

s

 

VA  a  ω sin θ  j  cos θ



 

   VA

VG3  RCG3 ω sin θ  j  cos θ

4.

vG3  396.197

θVG3  arg VG3  180  deg

θVG3  28.568 deg 1


vG2x  107.449  in sec


vG2y  82.101 in sec


vG3x  347.959  in sec


vG3y  189.464  in sec

vG4x  ddot

vG4x  400.000  in sec

1

1 1 1

1

Calculate the angular acceleration of links 2 and 3 using equations 7.17. 2

α 

α  6.

s

Calculate the x and y components of the velocity vectors.

vG4y  0.0 in sec 5.

in

vG3  VG3

  

       b ω2  dddot cosθ a   cos θ  sin θ  sin θ  cos θ 

a  ω  cos θ  cos θ  sin θ  sin θ

 

  b  cos θ 2

2

 


α  2841

 

   RCG2 ω2 cosθ  j  sinθ

a G2  RCG2 α sin θ  j  cos θ a G2  a G2

a G2  14066.05

sec

θAG2  arg a G2



in 2

θAG2  127.380  deg

 

   a ω2 cosθ  j  sinθ

a A  a  α sin θ  j  cos θ



 

   RCG3 ω2 cosθ  j  sinθ

a G3  RCG3 α sin θ  j  cos θ

rad 2

s rad 2

s

Calculate the value of a G2 and a G3 using equations 7.3 and 7.32.



α  0.58



a G3  a G3

a G3  8642.67 

sec

θAG3  arg a G3 7.

8.

9.

in 2

θAG3  95.836 deg

Calculate the x and y components of the acceleration of the CGs of all moving links in the global coordinate system (GCS). 2


a G2x  8540 in sec


a G2y  11177  in sec


a G3x  879  in sec


a G3y  8598 in sec

a G4x  dddot

a G4x  22760  in sec

2

2 2 2

Calculate the x and y components of the external force on link 4 in the CGS. FPx  FP cos δFP

FPx  60.000 lbf

FPy  FP sin δFP

FPy  0.000  lbf


1 ω

 m2  a G2x vG2x  a G2y vG2y  m3  a G3x vG3x  a G3y vG3y   m4  a G4x vG4x  IG2 α ω  IG3 α ω     F  v  F  v   Px G4x Py G4y  

T12  4647 lbf  in








A system of two coplanar arms on a common shaft, as shown in Figure 12-1, is to be designed. For row a in Table P12-1, find the shaking force of the linkage when run unbalanced and design a counterweight to statically balance the system. Work in a any consistent unit system you prefer.

Given:

Masses and radii:

Solution: 1.

m1  0.20 kg

r1  1.25 m

θ  30 deg

m2  0.40 kg

r2  2.25 m

θ  120  deg


Resolve the position vectors into xy components in the arbitrary coordinate system associated with the freeze- frame position of the linkage chosen for analysis.

 

R1x  1.083 m

R1y  r1 sin θ

 

R2x  1.125 m

R2y  r2 sin θ

R1x  r1 cos θ R2x  r2 cos θ 2.

3.

 

R1y  0.625 m

 

R2y  1.949 m

Solve equation 12.2c for the mass-radius product components. mRbx  m1 R1x  m2 R2x

mRbx  0.233 kg m

mRby  m1 R1y  m2 R2y

mRby  0.904 kg m

Solve equations 12.2d and 12.2e for the position angle and mass-radius product required.

θb  atan2 mRbx mRby mRb 

2

2

mRbx  mRby

θb  75.524 deg mRb  0.934 kg m




The minute hand on Big Ben weighs 40 lb and is 10 ft long. Its CG is 4 ft from the pivot. Calculate the mR product and angular location needed to statically balance this link and design a physical counterweight, positioned close to the center. Select material and design the detailed shape of the counterweight, which is of 2 in uniform thickness in the Z direction.

Given:

Weight and radius data: W1  40 lbf

r1  4  ft t  2  in

Counterweight thickness: Design choices:

3

γ  0.28 lbf  in

Counterweight material is steel: Counter weight shape is cylindrical Solution:


1.

Since there is only one mass to balance, the balancing mass will be on the same centerline as the minute hand, on the side opposite the center of rotation of the hand.

2.

Solve equation 12.2c for the required mass-radius for balancing. mRb 

W1 g

 r1

mRb  4.973 slug ft 2

γ

3.

The mass of a solid cylinder is mb  π rc  t

4.

Combine the equations in steps 2 and 3 and solve for the cylinder radius as a function of the balance radius. Plot the result over a suitable range of Rb to get an idea of the sizes involved. rc Rb 

5.

mRb g

Rbal  5  in 6  in  20 in

π t Rb γ

Choose a balance radius of Rb  15 in. Then the required mass is mb 

mRb

mRb g π t Rb γ

16

Wb  128.000 lbf

Calculate the required cylinder diameter.

d c  2 

CYLINDER RADIUS vs BALANCE RADIUS

mb  3.978 slug

Rb

Wb  mb g 6.

g

d c  17.059 in

14





r c Rbal 12 in

10 8 6

5

10

15 Rbal in

20




A "V for victory" advertising sign is being designed to be oscillated about the apex of the V, on the billboard, as the rocker of a fourbar linkage. The angle between the legs of the V is 20 deg. Each leg is 8 ft long and 1.5 ft wide. Material is 0.25-in thick aluminum. Design the V-link for static balance.

Given:

V dimensions: lv  8  ft

Length:

Width:

wv  1.5 ft

Thickness:

tv  0.25 in

3

γa  0.1 lbf  in

Specific weight of aluminum: Design choices:

3

γs  0.28 lbf  in

Counterweight material is steel:

Counter weight shape is cylindrical disk with thickness Solution:

tc  3  in


1.

Since the V is symmetrical in geometry and mass, the counterweight will be located on the centerline of the V opposite the pivot point from the V.

2.

Assume that the centerlines of each leg of the V intersect at the pivot point. Then, the mass, distance to the CG, and angle with respect to the V centerline of each leg is: Mass of one leg

3.

γa g

mleg  43.200 lb

m1  mleg

r1  4  ft

θ  10 deg

m2  mleg

r2  4  ft

θ  10 deg


 

R1x  3.939 ft


 

R2x  3.939 ft


R1x  r1 cos θ R2x  r2 cos θ 4.

mleg  lv wv tv

 

R1y  0.695 ft

 

R2y  0.695 ft

Solve equation 12.2c for the mass-radius product. 3

mRbx  m1 R1x  m2 R2x

mRbx  4.084  10 in lb

mRby  m1 R1y  m2 R2y

mRby  0.000 in lb

mRb  mRbx

mRb  4084 in lb 2

γ

5.

The mass of a solid cylinder is mb  π rc  t

6.

Combine the equations in steps 4 and 5 and solve for the cylinder radius as a function of the balance radius. Plot the result over a suitable range of Rb to get an idea of the sizes involved. rc Rb 

mRb g π tc Rb γs

g

Rbal  5  in 6  in  20 in



CYLINDER RADIUS vs BALANCE RADIUS 18 16





r c Rbal 14 in

12 10 8

5

10

15 Rbal in

7.

Choose a balance radius of Rb  18 in. Then the required mass is mb 

8.

mRb

mb  226.900 lb

Rb

Calculate the required cylinder diameter.

d c  2 

mRb g π tc Rb γs

d c  18.5 in

20




A three-bladed ceiling fan has 1.5 ft by 0.25 ft equispaced rectangular blades that normally weigh 2 lb each. Manufacturing tolerances will cause the blade weight to vary up to plus or minus 5%. The mounting accuracy of the blades will vary the location of the CG versus the spin axis by plus or minus 10% of the blades' diameters. Calculate the weight of the largest steel counterweight needed at a 2-in radius to statically balance the worst-case blade assembly.

Given:

Blade dimensions: Length:

lb  1.5 ft

Manufacturing tolerances:

Width:

wb  0.25 ft

Weight

tw  0.05

Nominal weight: CG offset

Wbnom  2  lbf

tCG  0.10

Assumptions: 1. The blades are held in place by a bracket such that their base is 6 in from the center of rotation making the tip 24 in from the center. Thus, the blade sweep diameter is 48 in. 2. There is one heavy (maximum weight) blade at 0 deg and two light (minimum weight) ones at 120 and 240 deg, respectively. Thus, W1   1  tw  Wbnom W2   1  tw  Wbnom W3   1  tw  Wbnom Solution:

W1  2.100 lbf

r1  15 in

θ  0  deg

W2  1.900 lbf

r2  15 in

θ  120  deg

W3  1.900 lbf

r3  15 in

θ  240  deg


1.

There are two factors to be taken into account, the variation in blade weight and the error or eccentricity in the location of the global CG. The variation in blade weight about its spin axis will be considered first.

2.


  R2x  r2 cos θ R3x  r3 cos θ R1x  r1 cos θ

3.

R2x  7.500 in R3x  7.500 in

mRby 

W1 R1x  W2 R2x  W3 R3x g W1 R1y  W2 R2y  W3 R3y g

R1y  0.000 in R2y  12.990 in R3y  12.990 in

mRbx  3.000 in lb


Solve equations 12.2d and 12.2e for the position angle and mass-radius product required.

θb  atan2 mRbx mRby mRbb  5.

  R2y  r2 sin θ R3y  r3 sin θ R1y  r1 sin θ

Solve equation 12.2c for the mass-radius product components. mRbx 

4.

R1x  15.000 in

2

θb  180.000 deg 2

mRbx  mRby

mRbb  3.000 in lb

Now, account for the fact that the blades' spin axis can be eccentric from their CG. re  48 in tCG

Maximum eccentricity:

re  4.800 in

mR product due to eccentricity: mRbe 

W1  W2  W3 g

 re

mRbe  28.320 in lb


6.


Add the two mR products and divide by the 2-in radius specified for the counterweight to find the maximum weight required. Rcw  2.0 in

mcw 

mRb Rcw

mRb  mRbb  mRbe

mRb  31.320 in lb

mcw  15.66 lb

Note that 90% of the counterweight is required to balance the eccentricity. The manufacturer would be well advised to try to control this variation more tightly.




A system of three non coplanar weights is arranged on a shaft generally as shown in Figure 12-3. For row a in Table P12-2, find the shaking forces and shaking moment when run unbalanced at 100 rpm and specify the mR product and angle of the counterweights in planes A and B needed to dynamically balance the system. The correction planes are 20 units apart. Work in a any consistent unit system you prefer.

Given:

Masses and radii: m1  0.20 kg

r1  1.25 m

θ  30 deg

l1  2  m

m2  0.40 kg

r2  2.25 m

θ  120  deg

l2  8  m

m3  1.24 kg

r3  5.50 m

θ  30 deg

l3  17 m

Distance between correction planes: Solution: 1.



 

R1x  1.083 m


 

R2x  1.125 m


 

R3x  4.763 m

R3y  r3 sin θ

R1x  r1 cos θ R2x  r2 cos θ R3x  r3 cos θ 2.

mRBy 

 m1 R1x  l1   m2 R2x  l2   m3 R3x  l3 lB  m1 R1y  l1   m2 R2y  l2   m3 R3y  l3 lB

mRB 

3.

R1y  0.625 m

 

R2y  1.949 m

 

R3y  2.750 m

mRBx  4.862 kg m

mRBy  2.574 kg m

Solve equations 12.2d and 12.2e for the position angle and mass-radius product required in plane B.

θB  atan2 mRBx mRBy

4.

 

Solve equations 12.4e for summation of moments about O.

mRBx 

3.

lB  20 m

2

2

mRBx  mRBy

θB  152.101 deg mRB  5.501 kg m

Solve equations 12.4c for forces in x and y directions in plane A. mRAx  m1 R1x  m2 R2x  m3 R3x  mRBx

mRAx  0.811 kg m

mRAy  m1 R1y  m2 R2y  m3 R3y  mRBy

mRAy  0.069 kg m

Solve equations 12.2d and 12.2e for the position angle and mass-radius product required in plane A.

θA  atan2 mRAx mRAy mRA 

2

2

mRAx  mRAy

θA  175.160 deg mRA  0.814 kg m




A wheel and tire assembly has been run at 100 rpm on a dynamic balancing machine as shown in Figure 12-12. The force measured at the left bearing had a peak of 5 lb at a phase angle of 45 deg with respect to the zero reference angle on the tire. The force measured at the right bearing had a peak of 2 lb at a phase angle of -120 deg with respect to the zero reference on the tire. The center distance between the two bearings on the machine is 10 in. The left edge of the wheel rim is 4 in from the centerline of the closest bearing. The wheel is 7-in wide at the rim. Calculate the size and location with respect to the tire's zero reference angle, of balance weights needed on each side of the rim to dynamically balance the tire assembly. The wheel rim diameter is 15 in.

Given:

Bearing forces and plane locations with respect to correction plane A: F1  5  lbf

Left:

Right: F2  2  lbf

θ  45 deg

l1  14 in

θ  120  deg

l2  4  in

Distance between correction planes:

lB  7  in

Correction weight radii: RA  7.5 in ω  100  rpm

Tire rotational speed: Solution: 1.


Resolve the force vectors into xy components with respect to the zero reference angle of the tire.

  F2x  F2 cos θ F1x  F1 cos θ

2.

RB  7.5 in

F1x  3.536  lbf F2x  1.000  lbf

  F2y  F2 sin θ F1y  F1 sin θ

F1y  3.536  lbf F2y  1.732  lbf

Solve equations 12.4e for summation of moments about O, modified for the bearing forces. mRBx  

F1x l1  F2x l2 2

mRBx  22.883 in lb

lB ω mRBy   3.

F1y l1  F2y l2 2

lB ω Solve equations 12.2d and 12.2e for the position angle and mass-radius product required in plane B . Also, solve for the weight required at the given radius.

θB  atan2 mRBx mRBy mRB  WB  4.

mRBy  21.411 in lb

2

2

mRBx  mRBy mRB RB

g

θB  136.904  deg mRB  31.338 in lb WB  4.18 lbf

Solve equations 12.4c for forces in x and y directions in plane A . mRAx  

F1x  F2x 2

 mRBx

mRAx  13.956 in lb

 mRBy

mRAy  15.061 in lb

ω mRAy  

F1y  F2y 2

ω 5.

Solve equations 12.2d and 12.2e for the position angle and mass-radius product required in plane A .



θA  atan2 mRAx mRAy mRA  WA 

2

2

mRAx  mRAy mRA g RA

θA  47.180 deg mRA  20.533 in lb WA  2.74 lbf




A wheel and tire assembly has been run at 100 rpm on a dynamic balancing machine as shown in Figure 12-12. The force measured at the left bearing had a peak of 6 lb at a phase angle of -60 deg with respect to the zero reference angle on the tire. The force measured at the right bearing had a peak of 4 lb at a phase angle of 150 deg with respect to the zero reference on the tire. The center distance between the two bearings on the machine is 10 in. The left edge of the wheel rim is 4 in from the centerline of the closest bearing. The wheel is 7-in wide at the rim. Calculate the size and location with respect to the tire's zero reference angle, of balance weights needed on each side of the rim to dynamically balance the tire assembly. The wheel rim diameter is 16 in.

Given:

Bearing forces and plane locations with respect to correction plane A: F1  6  lbf

θ  60 deg

l1  14 in

Right: F2  4  lbf

θ  150  deg

l2  4  in

Left:


lB  7  in

Correction weight radii: RA  8  in ω  100  rpm

Tire rotational speed: Solution: 1.



  F2x  F2 cos θ F1x  F1 cos θ

2.

RB  8  in

F1x  3.000  lbf F2x  3.464  lbf

  F2y  F2 sin θ F1y  F1 sin θ

F1y  5.196  lbf F2y  2.000  lbf

Solve equations 12.4e for summation of moments about O, modified for the bearing forces. mRBx  

F1x l1  F2x l2 2

mRBx  14.155 in lb

lB ω mRBy   3.

F1y l1  F2y l2 2


θB  atan2 mRBx mRBy mRB  WB  4.

mRBy  32.565 in lb

2

2


g

θB  113.493  deg mRB  35.508 in lb WB  4.44 lbf

Solve equations 12.4c for forces in x and y directions in plane A . mRAx  

F1x  F2x 2

 mRBx

mRAx  15.789 in lb

 mRBy

mRAy  21.312 in lb

ω mRAy  

F1y  F2y 2

ω 5.

Solve equations 12.2d and 12.2e for the position angle and mass-radius product required in plane A .




2

2


θA  53.467 deg mRA  26.523 in lb WA  3.32 lbf




Table P11-3 shows the geometry and kinematic data for several fourbar linkages. For row a in Table P11-3: a. Calculate the size and angular locations of the counterbalance mass-radius products needed on links 2 and 4 to completely force balance the linkage by the method of Berkof and Lowen. Check your manual calculation with program FOURBAR. b. Calculate the input torque for the linkage both with and without the added balance weights and compare the results. Use program FOURBAR.

Units:


Given:

Link lengths:

1

2

Link 2 (O2 to A)

l2  4.00 in

Link 3 (A to B)

l3  12.00  in

Link 4 (B to O4)

l4  8.00 in

Link 3 (O2 to O4)

l1  15.00  in

Link angles:

θ  45 deg

θ  24.97  deg

θ  99.30  deg

Mass:

m2  0.002  blob

m3  0.020  blob

m4  0.100  blob

2


Solution: 1.

b 2  2.00 in

ϕ  0  deg

b 4  4.00 in

ϕ  30 deg

b 3  5.00 in

Solve equations 12.8c and 12.8d for the total mR product components for links 2 and 4.



l2



l3



l2



l3

mb2y  m3  b 3

l4



l3



l4



l3

mb4y  m3  b 3



 

 sin ϕ



mb4x  m3  b 3



 

 cos ϕ  l2

mb2x  0.0467 in blob

mb2y  0.0000 in blob



 

mb4x  0.0667 in blob

 


 cos ϕ

 sin ϕ





Determine the additional mR product components for links 2 and 4.

 

mR2x  0.0507 in blob

 

mR2y  0.0000 in blob

mR4x  mb4x  m4 b 4 cos ϕ

 

mR4x  0.4131 in blob

 

mR4y  0.2000 in blob

mR2x  mb2x  m2 b 2 cos ϕ mR2y  mb2y  m2 b 2 sin ϕ

mR4y  mb4y  m4 b 4 sin ϕ

2

IG4  0.50 blob in

See Table P11-3 and Mathcad file P1208a.

mb2x  m3  b 3

2.

2

IG3  0.20 blob in

ϕ  0  deg


3.


Solve equations 12.2d and 12.2e for the position angle and additional mass-radius product required.

θb2  atan2 mR2x mR2y mRb2 

2

2

mR2x  mR2y


2

2

mR4x  mR4y

θb2  180.000 deg mRb2  0.0507 blob in

θb4  154.165 deg mRb4  0.459 blob in

4.

Check the result using program FOURBAR. The linkage balancing screen below confirms the calculations.

5.

The maximum positive input torque without force balancing is 245.7 lb-in. After force balancing it is 460.7 lb-in.




Link 2 in Figure P12-1 rotates at 500 rpm. The links are steel with cross-sections of 1 x 2 in. Half of the 29-lb weight of the laybar and reed are supported by the linkage at point B. Design counterweights to force balance the linkage and determine its change in peak torque versus unbalanced condition. See Problem 11-13 for more information on the overall mechanism.

Units:


Given:

Link lengths:

1

2

Link 2 (A to B)

l2  2.00 in

Link 3 (B to C)

l3  8.375  in

Link 4 (C to D)

l4  7.187  in

Link 1 (A to D)

l1  9.625  in

Rpa  0.0 in

δ  0  deg

Coupler point:

Crank angle and motion: ω  500  rpm w  2.00 in

Link cross-section dims: Material specific weight: Solution: 1.

2.

steel

t  1.00 in 3

γ  0.28 lbf  in

Determine the distance to the CG in the LRCS on each of the three moving links. All three are located on the x' axis in the LRCS and their angle is zero deg. b 2  0.5 l2

b 2  1.000 in

b 3  0.5 l3

ϕ  0  deg

ϕ  0  deg

ϕ  0  deg

b 3  4.188 in

Determine the mass and moment of inertia of each link. γ

m3  w t l3

g

γ

m41  w t l4

g

γ

m3  0.012 blob

m41  0.010 blob

m4  m41  m42

m4  0.048 blob

b 4 

IG3 

m2 12 m2 12

  w  l2

2



  w  l3



2

IG4   l4  b 4

 m41   m42 m4  2  l4



2 g

b 4  6.406 in

2 2

IG3  0.01792 blob in

2 l4   2 2   m42   w  l4   m41  b 4   12  2 

m41

2

29 lbf

IG2  0.00193 blob in

2

2

m42 

g

m2  0.0029 blob

IG2 

2


Define any external forces, their locations and directions. F  590  lbf

Beat-up force

The angle in the CGS is 180 deg. 4.

2


m2  w t l2

3.

α  0  rad sec

acting on link 4 at a distance

R  l4  3.75 in R  10.937 in




l2



l3



l2



l3

mb2x  m3  b 3

mb2y  m3  b 3

l4



l3



 

 sin ϕ



mb4x  m3  b 3



 

 cos ϕ  l2




 

 cos ϕ

mb2x  0.0121 in blob



mb4x  0.0437 in blob




l4



l3

mb4y  m3  b 3 5.


 

 sin ϕ




 

mR2x  0.0150 in blob

 


 

mR4x  0.3510 in blob

 



mR4x  mb4x  m4 b 4 cos ϕ mR4y  mb4y  m4 b 4 sin ϕ 6.




2

2

mR2x  mR2y


2

2

mR4x  mR4y

θb2  180.000 deg mRb2  0.0150 blob in

θb4  180.000 deg mRb4  0.351 blob in

7.

Check the result using program FOURBAR. The linkage balancing screen below confirms the calculations.

8.

The maximum peak input torque without force balancing is 17.3 lb-in. After force balancing it is 34.9 lb-in.




Figure P12-2a shows a fourbar linkage and its dimensions in meters. The steel crank and rocker have uniform cross sections 50 mm wide by 25 mm thick. The aluminum coupler is 25 mm thick. The crank O2A rotates at a constant speed of  = 40 rad/sec. Design counterweights to force balance the linkage and determine its change in peak torque versus the unbalanced condition.

Given:


a  1.00 m

Link 3 (A to B)

b  2.06 m

Link 4 (B to O4)

c  2.33 m

Link 1 (O2 to O4)

d  2.22 m

Rpa  3.06 m

δ  31 deg

Coupler point:

Crank angle and motion: θ  60 deg Link cross-section dims: w2  50 mm

t2  25 mm


ω  40 rad sec

t3  25 mm

α  0  rad sec

w4  50 mm 3

γs  0.3 lbf  in

steel

1

2

t4  25 mm 3

γa  0.1 lbf  in

aluminum



RCG2  0.5 a

Link 3:

RCG3x 

RCG2  0.500 m RCG4  0.5 c

 


RCG3x  1.561 m

3

 

Rpa sin δ

RCG3y 

RCG3y  0.525 m

3 2

RCG3 

RCG4  1.165 m

2

RCG3x  RCG3y

RCG3  1.647 m

At an angle with respect to the local x' axis of δ  atan2 RCG3x RCG3y 2.

Determine the mass of each link. m2  w2 t2 a 

γs

m3 

g

m2  10.380 kg 3.

4.

δ  18.600 deg

1 2

 

 b  Rpa sin δ  t3

γa g

m3  112.332 kg

m4  w4 t4 c

Link lengths:

l2  a

l3  b

l4  c

CG position vectors:

b 2  RCG2

b 3  RCG3

b 4  RCG4

ϕ  0  deg

ϕ  δ

ϕ  0  deg




l2



l3



l2



l3

mb2y  m3  b 3

 



 cos ϕ  l2

 

 sin ϕ





g

m4  24.185 kg

Convert the above data for use in equations 12.8c and 12.8d.

mb2x  m3  b 3

γs

mb2x  27.2117 kg m

mb2y  28.6467 kg m




l4



l3



l4



l3

mb4x  m3  b 3

mb4y  m3  b 3 5.


 

mb4x  198.3298 kg m

 

mb4y  66.7469 kg m

 cos ϕ

 sin ϕ





Determine the additional mR product components for links 2 and 4. mR2x  mb2x  m2 b 2 cos ϕ

 

mR2x  32.4017 kg m

 

mR2y  28.6467 kg m

 

mR4x  226.5057 kg m

 

mR4y  66.7469 kg m

mR2y  mb2y  m2 b 2 sin ϕ




2

2

mR2x  mR2y

θb4  atan2 mR4x mR4y mRb4  7.

2

2

mR4x  mR4y

θb2  138.520 deg mRb2  43.25 kg m

θb4  163.581 deg mRb4  236.14 kg m

The maximum peak input torque without force balancing is found from program FOURBAR to be 464.9 kN-m. After force balancing it is 719.4 kN-m.




Figure P12-2b shows a fourbar linkage and its dimensions in meters. The steel crank and rocker have uniform cross sections 50 mm wide by 25 mm thick. The aluminum coupler is 25 mm thick. The crank O2A rotates at a constant speed of  = 50 rad/sec. Design counterweights to force balance the linkage and determine its change in peak torque versus the unbalanced condition.

Given:


a  0.72 m

Link 3 (A to B)

b  0.68 m

Link 4 (B to O4)

c  0.85 m

Link 1 (O2 to O4)

d  1.82 m

Rpa  0.97 m

δ  54 deg

Coupler point:

Crank angle and motion: θ  30 deg Link cross-section dims: w2  50 mm

t2  25 mm


ω  50 rad sec

t3  25 mm

α  0  rad sec

w4  50 mm 3

γs  0.3 lbf  in

steel

1

2

t4  25 mm

aluminum

3

γa  0.1 lbf  in



RCG2  0.5 a

Link 3:

RCG3x 

RCG2  0.360 m

RCG4  0.5 c

 


RCG3x  0.417 m

3

 

Rpa sin δ

RCG3y 

RCG3y  0.262 m

3 2

RCG3 

RCG4  0.425 m

2

RCG3x  RCG3y

RCG3  0.492 m

At an angle with respect to the local x' axis of δ  atan2 RCG3x RCG3y 2.

Determine the mass of each link. m2  w2 t2 a 

γs

m3 

g

m2  7.474 kg 3.

4.

δ  32.117 deg

1 2

 

 b  Rpa sin δ  t3

γa g

m3  18.463 kg

m4  w4 t4 c

γs g

m4  8.823 kg

Convert the above data for use in equations 12.8c and 12.8d. Link lengths:

l2  a

l3  b

l4  c

CG position vectors:

b 2  RCG2

b 3  RCG3

b 4  RCG4

ϕ  0  deg

ϕ  δ

ϕ  0  deg




mb2x  m3  b 3

l2

 



 cos ϕ  l2

 l3  l2  mb2y  m3  b 3  sin ϕ   l3 



mb2x  5.1471 kg m mb2y  5.1138 kg m




l4



l3



l4



l3

mb4x  m3  b 3

mb4y  m3  b 3 5.


 

mb4x  9.6175 kg m

 

mb4y  6.0371 kg m

 cos ϕ

 sin ϕ






 

mR2x  7.8375 kg m

 

mR2y  5.1138 kg m

 

mR4x  13.3673 kg m

 

mR4y  6.0371 kg m





2

2

mR2x  mR2y

θb4  atan2 mR4x mR4y mRb4  7.

2

2

mR4x  mR4y

θb2  146.877 deg mRb2  9.36 kg m

θb4  155.694 deg mRb4  14.67 kg m

The maximum peak input torque without force balancing is found from program FOURBAR to be 5.34 MN-m. After force balancing it is 10.63 MN-m.




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to solve for the mass-radius products that will force balance any fourbar linkage for which the geometry and mass properties are known.

Units:

Use any set of compatible units.

Enter data:

Link lengths:

Solution: 1.

Link 2 (O2 to A)

l2  4.00

Link 3 (A to B)

l3  12.00

Link 4 (B to O4)

l4  8.00

Link 1 (O2 to O4)

l1  15.00

Mass:

m2  0.002

m3  0.020

m4  0.100

Mass center:

b 2  2.00

b 3  5.00

b 4  4.00

ϕ  0  deg

ϕ  0  deg

ϕ  30 deg





l2



l3



l2



l3

mb2x  m3  b 3

mb2y  m3  b 3

l4



l3



l4



l3

mb4y  m3  b 3 2.



 

 sin ϕ



mb4x  m3  b 3



 

 cos ϕ  l2

mb2y  0.000



 

mb4x  0.0667

 

mb4y  0.000

 cos ϕ

 sin ϕ






 

mR2x  0.0507

 

mR2y  0.000

mR4x  mb4x  m4 b 4 cos ϕ

 

mR4x  0.413

 

mR4y  0.200


mR4y  mb4y  m4 b 4 sin ϕ 3.

mb2x  0.0467

Solve equations 12.2d and 12.2e for the position angle and additional mass-radius product required. mRb2 

2

2

mR2x  mR2y


2

2

mR4x  mR4y

θb4  atan2 mR4x mR4y

mRb2  0.0507

θb2  180.000 deg mRb4  0.459

θb4  154.165 deg




Figure P12-3 shows a system with two weights on a rotating shaft. For the given data below, determine the magnitudes and angles of the balance weights needed to dynamically balance the system.

Given:

Weights and radii:

Solution: 1.

W1  15 lbf

r1  6  in

θ  0  deg

l1  7  in

W2  20 lbf

r2  5  in

θ  270  deg

l2  9  in


lB  9  in

Correction weight radii:

RA  8  in


  R2x  r2 cos θ

R1x  6.000 in R2x  0.000 in

mRBy 

 W1 R1x  l1   W2 R2x  l2 lB g  W1 R1y  l1   W2 R2y  l2 lB g

2

mRB  W3 

R2y  5.000 in

mRBx  70.000 in lb mRBy  100.000 in lb

θB  55.008 deg

2

mRBx  mRBy

mRB  122.066 in lb

mRB g

W3  24.4 lbf

RB

Solve equations 12.4c for forces in x and y directions in plane A (4). mRAx  mRAy 

5.

R1y  0.000 in

Solve equations 12.2d and 12.2e for the position angle and mass-radius product required in plane B (3). Also, solve for the weight required at the given radius.


4.

  R2y  r2 sin θ R1y  r1 sin θ

Solve equations 12.4e for summation of moments about O, which is at plane 4. mRBx 

3.

Plane 3 RB  5  in


R1x  r1 cos θ

2.

Plane 4

W1 R1x  W2 R2x g W1 R1y  W2 R2y g

 mRBx

mRAx  160.000 in lb

 mRBy

mRAy  0.000 in lb

Solve equations 12.2d and 12.2e for the position angle and mass-radius product required in plane A (4).

θA  atan2 mRAx mRAy mRA  W4 

2

2


θA  180.000 deg mRA  160.000 in lb W4  20.0 lbf




Figure P12-4 shows a system with two weights on a rotating shaft. For the given data below, determine the radii and angles of the balance weights needed to dynamically balance the system.

Given:

Weights and radii:

Solution: 1.

W1  20 lbf

r1  6  in

θ  45 deg

l1  0  in

W2  15 lbf

r2  4  in

θ  300  deg

l2  9  in


lB  12 in

Correction weights:

W3  20 lbf Plane 4

Plane 3



  R2x  r2 cos θ R1x  r1 cos θ

2.

R1x  4.243  in R2x  2.000  in

mRBy 


mRB  R4 

R2y  3.464  in

mRBx  22.500 in lb mRBy  38.971 in lb

2

θB  120.000  deg

2

mRBx  mRBy

mRB  45.000 in lb

mRB g

R4  1.13 in

W4


5.

R1y  4.243  in

Solve equations 12.2d and 12.2e for the position angle and mass-radius product required in plane B (4). Also, solve for the radius required for the given weight.


4.

  R2y  r2 sin θ R1y  r1 sin θ


3.

W4  40 lbf

W1 R1x  W2 R2x g W1 R1y  W2 R2y g

 mRBx

mRAx  92.353 in lb

 mRBy

mRAy  71.862 in lb


θA  atan2 mRAx mRAy mRA  R3 

2

2

mRAx  mRAy

mRA g W3

θA  142.112  deg mRA  117.018  in lb R3  5.85 in




Figure P12-5 shows a system with two weights on a rotating shaft. For the given data below, determine the magnitudes and angles of the balance weights needed to dynamically balance the system.

Given:

Weights and radii:

Solution: 1.

W1  10 lbf

r1  3  in

θ  90 deg

l1  3  in

W2  15 lbf

r2  3  in

θ  240  deg

l2  7  in


lB  12 in


RA  3  in

R1x  0.000 in R2x  1.500 in

mRBy 


R1y  3.000 in R2y  2.598 in



θB  atan2 mRBx mRBy 2

mRB  W3 

θB  49.252 deg

2

mRBx  mRBy

mRB  20.108 in lb

mRB g

W3  6.70 lbf

RB


5.

  R2y  r2 sin θ R1y  r1 sin θ


4.

RB  3  in


  R2x  r2 cos θ

3.

Plane 3



2.

Plane 4

W1 R1x  W2 R2x g W1 R1y  W2 R2y g

 mRBx

mRAx  9.375 in lb

 mRBy

mRAy  6.262 in lb



2

2


θA  33.741 deg mRA  11.274 in lb W4  3.76 lbf




Figure P12-6 shows a system with three weights on a rotating shaft. For the given data below, determine the magnitudes and angles of the balance weights needed to dynamically balance the system.

Given:

Weights and radii:

Solution: 1.

W1  6  lbf

r1  5  in

θ  120  deg

l1  14 in

W2  12 lbf

r2  4  in

θ  240  deg

l2  4  in

W3  9  lbf

r3  8  in

θ  300  deg

l3  4  in


lB  14 in


RA  4  in

R1x  2.500  in R2x  2.000  in R3x  4.000  in

mRBy 

 W1 R1x  l1   W2 R2x  l2   W3 R3x  l3 lB g  W1 R1y  l1   W2 R2y  l2   W3 R3y  l3 lB g

R1y  4.330  in R2y  3.464  in R3y  6.928  in

mRBx  2.143  in lb mRBy  20.042 in lb



mRB  W5 

2

mRBx  mRBy mRB g

θB  83.897 deg mRB  20.157 in lb W5  5.04 lbf

RB


5.

  R2y  r2 sin θ R3y  r3 sin θ R1y  r1 sin θ


4.

RB  4  in


  R2x  r2 cos θ R3x  r3 cos θ

3.

Plane 5



2.

Plane 4


 mRBx

mRAx  0.857  in lb

 mRBy

mRAy  57.900 in lb



2

2


θA  89.152 deg mRA  57.906 in lb W4  14.48  lbf




Figure P12-6 shows a system with three weights on a rotating shaft. For the given data below, determine the magnitudes and angles of the balance weights needed to dynamically balance the system.

Given:

Weights and radii: W2  10 lbf

r2  3  in

θ  90 deg

l2  6  in

W3  10 lbf

r3  4  in

θ  180  deg

l3  12 in

W4  8  lbf

r4  4  in

θ  315  deg

l4  8  in

Solution: 1.


lB  8  in


RA  4  in

R2x  0.000 in R3x  4.000 in R4x  2.828 in

mRBy 

 W2 R2x  l2   W3 R3x  l3   W4 R4x  l4 lB g  W2 R2y  l2   W3 R3y  l3   W4 R4y  l4 lB g

R2y  3.000 in R3y  0.000 in R4y  2.828 in




mRB  W5 

2

mRBx  mRBy mRB g

θB  0.195 deg mRB  37.373 in lb W5  12.46 lbf

RB


5.

  R3y  r3 sin θ R4y  r4 sin θ R2y  r2 sin θ


4.

RB  3  in


  R3x  r3 cos θ R4x  r4 cos θ

3.

Plane 5


R2x  r2 cos θ

2.

Plane 1


 mRBx

mRAx  20.000 in lb

 mRBy

mRAy  7.500 in lb



2

2


θA  159.444 deg mRA  21.360 in lb W1  5.34 lbf




The 400-mm-dia steel roller in Figure P12-8 has been tested on a dynamic balancing machine at 100 rpm and shows the unbalance forces given below. Determine the angular locations and required diameters of 25-mm deep holes drilled radially inward from the surface in planes 2 and 3 to dynamically balance the system.

Given:

Forces and locations: F1  0.291  N

θ  45 deg

l1  200  mm

F4  0.514  N

θ  210  deg

l4  550  mm


lB  300  mm


RA  187.5  mm

1.

RB  187.5  mm


Resolve the force vectors into xy components in the arbitrary coordinate system associated with the test apparatus.

  F4x  F4 cos θ F1x  F1 cos θ

2.

Plane 3

ω  100  rpm

Test speed: Solution:

Plane 2

F1x  0.206 N F4x  0.445 N

  F4y  F4 sin θ F1y  F1 sin θ

F1y  0.206 N F4y  0.257 N

Solve equations 12.3 for summation of moments about O, which is at plane 1. Negative mass is used. mRBx 

F1x l1  F4x l4

3

mRBx  8.693  10

2

kg m

lB ω mRBy  3.

F1y l1  F4y l4

3

mRBy  5.547  10

2


mRB  W3 

2

mRBx  mRBy mRB g 3

d 3  2 

θB  147.455 deg mRB  0.0103 kg m W3  0.539 N

RB

ρ  7800 kg m

4.

kg m

lB ω Solve equations 12.2d and 12.2e for the position angle and mass-radius product required in plane B (3). Also, solve for the weight and drill diameter required at the given radius.

(removed)

depth  25 mm

W3 π ρ depth g

d 3  18.9 mm

Solve equations 12.4c for forces in x and y directions in plane A (2). mRAx 

F1x  F4x 2

 mRBx

mRAx  0.0109 kg m

 mRBy

mRAy  6.015  10

ω mRAy 

F1y  F4y 2

3

kg m

ω 5.




θA  atan2 mRAx mRAy 2

mRA  W2 

2

mRAx  mRAy mRA g

d 2  2 

θA  28.944 deg mRA  0.0124 kg m W2  0.650 N

RA W2 π ρ depth g

d 2  20.8 mm

(removed)




The 500-mm-dia steel roller in Figure P12-8 has been tested on a dynamic balancing machine at 100 rpm and shows the unbalance forces given below. Determine the angular locations and required diameters of 25-mm deep holes drilled radially inward from the surface in planes 2 and 3 to dynamically balance the system.

Given:

Forces and locations: F1  0.230  N

θ  30 deg

l1  200  mm

F4  0.620  N

θ  135  deg

l4  550  mm


lB  300  mm


RA  237.5  mm Plane 3 RB  237.5  mm

ω  100  rpm

Test speed: Solution: 1.



  F4x  F4 cos θ F1x  F1 cos θ

2.

Plane 2

F1x  0.199 N F4x  0.438 N

  F4y  F4 sin θ F1y  F1 sin θ

F1y  0.115 N F4y  0.438 N

Solve equations 12.3 for summation of moments about O, which is at plane 1. Negative mass is used. mRBx 

F1x l1  F4x l4

3

mRBx  8.540  10

2

kg m


F1y l1  F4y l4

mRBy  6.630  10

2

kg m

lB ω Solve equations 12.2d and 12.2e for the position angle and mass-radius product required in plane B (3). Also, solve for the weight and drill diameter required at the given radius.


mRB  W3 

2

mRBx  mRBy mRB g 3

d 3  2 

θB  142.176 deg mRB  0.0108 kg m W3  0.446 N

RB

ρ  7800 kg m

4.

3

(removed)

depth  25 mm

W3 π ρ depth g

d 3  17.241 mm


F1x  F4x 2

 mRBx

mRAx  0.0107 kg m

 mRBy

mRAy  0.012 kg m

ω mRAy 

F1y  F4y 2

ω 5.




θA  atan2 mRAx mRAy 2

mRA  W4 

2

mRAx  mRAy mRA g

d 4  2 

θA  47.441 deg mRA  0.0159 kg m W4  0.655 N

RA W4 π ρ depth g

d 4  20.876 mm

(removed)




The linkage in Figure P12-9a has rectangular steel links of 20 x 10 mm cross-section similar to that shown in Figure 12-10a. Design the necessary balance weights and other features necessary to completely eliminate the shaking force and shaking moment. State all assumptions.

Given:


a 2  78 mm

Link 3 (A to B)

a 3  109  mm

Link 4 (B to O4)

a 4  121  mm

Link 1 (O2 to O4)

a 1  54 mm

Link cross-section dims: Material density: steel Solution: 1.

3.

4.

2 a3  a3  e   3     1  2 2 h

e  39.366 mm

l3  a 3  2  e

l3  187.731 mm

Determine the distance to the CG in the LRCS on each of the three moving links. RCG2  0.5 a 2

RCG2  39.000 mm

r3  0.5 a 3

r3  54.500 mm

RCG4  0.5 a 4

RCG4  60.500 mm

Determine the mass of each link. m2  h  t a 2 ρ

m3  h  t l3 ρ

m4  h  t a 4 ρ

m2  0.122 kg

m3  0.293 kg

m4  0.189 kg

Solve equations 12.17a and 12.17b for the total mR product components for links 2 and 4.



mr2  m3  b 3

b 3  54.500 mm a2 

  a3   a4  mr4  m3  r3   a3 

6.

ρ  7800 kg m

Determine the amount by which link 3 must be elongated to make it physically like a pendulum by using equation 12.12a.

b 3  a 3  r3

5.

t  10 mm 3


h

2.

h  20 mm

mr2  0.0114 kg m mr4  0.0177 kg m

Determine the additional mR product components for links 2 and 4 to accomplish the force balance. 3

mR2  mr2  m2 RCG2

mR2  6.6760  10

mR4  mr4  m4 RCG4

mR4  6.2981  10

3

kg m kg m

Let the distance to the force-balance mass from the pivot point on links 2 and 4 be R2  75 mm

R4  70.8 mm

then the masses of the force balance weights will be (R4 was chosen to make the weights the same) m2b 

mR2 R2

m2b  0.0890 kg


m4b  7.

8.

mR4

m4b  0.0890 kg

R4

Calculate the new, total mass of links 2 and 4 and the distance from the pivot point to the composite CG. m'2  m2  m2b

m'2  0.211 kg

r2 

m'4  m4  m4b

m'4  0.278 kg

r4 

mr2

r2  54.2 mm

m'2 mr4

r4  63.8 mm

m'4

Calculate the new radius of gyration of links 2 and 4. m2  a 2  h 2

I'2  k' 2 

12

I'4  k' 4 

2

  m r  R 2  m R  r 2 2 2 CG2 2b 2 2

I'2

3

I'2  1.161  10

2

kg m

k' 2  74.244 mm

m'2 m4  a 4  h 2

9.


12

2

  m r  R 2  m R  r 2 4 4 CG4 4b 4 4

I'4

3

I'4  3.157  10

k' 4  106.625 mm

m'4

Solve equations 12.17d and 12.17e to determine the mass moment of inertia required for the two inertia counterweights that are geared to links 2 and 4.. Icw2  m'2  k' 2  r2  a 2 r2

Icw2  2.671  10

Icw4  m'4  k' 4  r4  a 4 r4

Icw4  6.432  10

2 2

2 2

3 3

2

kg m

2

kg m

2

kg m




The linkage in Figure P12-9a has rectangular steel links of 20 x 10 mm cross-section with "dog bone" ends of 50 mm dia, similar to that shown in Figure 12-10b. Design the necessary balance weights and other features necessary to completely eliminate the shaking force and shaking moment. State all assumptions.

Given:


a 2  78 mm

Link 3 (A to B)

Link 4 (B to O4)

a 4  121  mm

Link 1 (O2 to O4) a 1  54 mm

Link cross-section dims: Material density: steel

h  20 mm

t  10 mm

c  25 mm

3

ρ  7800 kg m

Assumption:

Only link 3 has the "dog bone" configuration.

Solution:


1.

a 3  109  mm

Determine the amount by which link 3 must be elongated to make it physically like a pendulum by using equations 12.13.

 a3    24 c

B  12 

B  76.320

 a3    26 c

C  24 

C  130.640

3

 a3   a3  D  2     13    12 π  10 c c

D  81.385

 D  C v    B  8  

 6.999  r   3.022     0.481 

r  polyroots ( v)

e  h  r

e  9.620 mm

3

2.

3.

4.


RCG2  39.000 mm

r3  0.5 a 3

r3  54.500 mm

RCG4  0.5 a 4 Determine the mass of each link.

RCG4  60.500 mm

m2  h  t a 2 ρ

m3  t h   a 3  2  c  2  e  2  π c   ρ

m4  h  t a 4 ρ

m2  0.122 kg

m3  0.428 kg

m4  0.189 kg

2

Solve equations 12.17a and 12.17b for the total mR product components for links 2 and 4. b 3  a 3  r3



mr2  m3  b 3

b 3  54.500 mm a2 

  a3   a4  mr4  m3  r3   a3  5.

Choosing the positive root,

mr2  0.0167 kg m mr4  0.0259 kg m

Determine the additional mR product components for links 2 and 4 to accomplish the force balance. mR2  mr2  m2 RCG2

mR2  0.0120 kg m

mR4  mr4  m4 RCG4

mR4  0.0145 kg m


6.



R4  91 mm

then the masses of the force balance weights will be (R4 was chosen to make the weights the same) m2b  m4b  7.

8.

mR2

m2b  0.159 kg

R2 mR4

m4b  0.159 kg

R4


m'2  0.281 kg

r2 

m'4  m4  m4b

m'4  0.348 kg

r4 

r2  59.4 mm

m'2 mr4

r4  74.5 mm

m'4


I'2 

k' 2 

I'4 

k' 4 

2

12

  m r  R 2  m R  r 2 2 2 CG2 2b 2 2

I'2

3

I'2  1.283  10

2

kg m

k' 2  67.555 mm

m'2

m4  a 4  h 2

9.

mr2

12

2

  m r  R 2  m R  r 2 4 4 CG4 4b 4 4

I'4

3

I'4  3.718  10

k' 4  103.358 mm

m'4


Icw2  3.579  10

Icw4  m'4  k' 4  r4  a 4 r4

Icw4  8.784  10

2 2

2 2

3 3

2

kg m

2

kg m

2

kg m




The linkage in Figure P12-9b has rectangular steel links of 20 x 10 mm cross-section similar to that shown in Figure 12-10a. Design the necessary balance weights and other features necessary to completely eliminate the shaking force and shaking moment. State all assumptions.

Given:


a 2  50 mm

Link 3 (A to B)

a 3  185  mm

Link 4 (B to O4)

a 4  90 mm

Link 1 (O2 to O4)

a 1  172  mm


3.

4.

2 a3  a3  e   3     1  2 2 h

e  67.402 mm

l3  a 3  2  e

l3  319.805 mm


RCG2  25.000 mm

r3  0.5 a 3

r3  92.500 mm

RCG4  0.5 a 4

RCG4  45.000 mm


m3  h  t l3 ρ

m4  h  t a 4 ρ

m2  0.078 kg

m3  0.499 kg

m4  0.140 kg

Solve equations 12.17a and 12.17b for the total mR product components for links 2 and 4.



mr2  m3  b 3

b 3  92.500 mm a2 

  a3   a4  mr4  m3  r3   a3 

6.

ρ  7800 kg m


b 3  a 3  r3

5.

t  10 mm 3


h

2.

h  20 mm

mr2  0.0125 kg m mr4  0.0225 kg m


mR2  0.0105 kg m

mR4  mr4  m4 RCG4

mR4  0.0161 kg m


R4  115  mm


mR2 R2

m2b  0.140 kg


m4b  7.

8.

mR4

m4b  0.140 kg

R4


m'2  0.218 kg

r2 

m'4  m4  m4b

m'4  0.281 kg

r4 

mr2

r2  57.1 mm

m'2 mr4

r4  80.0 mm

m'4


I'2 

k' 2 

12

I'4 

k' 4 

2

  m r  R 2  m R  r 2 2 2 CG2 2b 2 2

I'2

4

I'2  5.898  10

2

kg m

k' 2  51.980 mm

m'2 m4  a 4  h 2

9.


12

2

  m r  R 2  m R  r 2 4 4 CG4 4b 4 4

I'4

3

I'4  2.465  10

k' 4  93.707 mm

m'4


Icw2  1.926  10

Icw4  m'4  k' 4  r4  a 4 r4

Icw4  6.281  10

2 2

2 2

3 3

2

kg m

2

kg m

2

kg m




The linkage in Figure P12-9b has rectangular steel links of 20 x 10 mm cross-section with "dog bone" ends of 50 mm dia, similar to that shown in Figure 12-10b. Design the necessary balance weights and other features necessary to completely eliminate the shaking force and shaking moment. State all assumptions.

Given:


a 2  50 mm

Link 3 (A to B)

Link 4 (B to O4)

a 4  90 mm

Link 1 (O2 to O4) a 1  172  mm

Link cross-section dims: Material density: steel Assumption:

h  20 mm

t  10 mm

a 3  185  mm c  25 mm

3

ρ  7800 kg m

Only link 3 has the "dog bone" configuration.

See Figure P12-9b and Mathcad file P1223. Solution: 1. Determine the amount by which link 3 must be elongated to make it physically like a pendulum by using equations 12.13.

 a3    24 c

B  12 

B  112.800

 a3    26 c

C  24 

C  203.600

3

 a3   a3  D  2     13    12 π  10 c c

D  686.549

 D  C v    B  8  

 11.116  r   4.646  Choosing the positive root,    1.662 

r  polyroots ( v)

e  h  r

e  33.236 mm

3

2.

3.

4.


RCG2  25.000 mm

r3  0.5 a 3

r3  92.500 mm

RCG4  0.5 a 4

RCG4  45.000 mm


m3  t h   a 3  2  c  2  e  2  π c   ρ

m4  h  t a 4 ρ

m2  0.078 kg

m3  0.621 kg

m4  0.140 kg

2




mr2  m3  b 3

b 3  92.500 mm a2 

  a3   a4  mr4  m3  r3   a3  5.

mr2  0.0155 kg m mr4  0.0279 kg m


mR2  0.0136 kg m

mR4  mr4  m4 RCG4

mR4  0.0216 kg m


6.



R4  119.5  mm

then the masses of the force balance weights will be (R4 was chosen to make the weights the same) m2b  m4b  7.

8.

mR2

m2b  0.181 kg

R2 mR4

m4b  0.181 kg

R4


m'2  0.259 kg

r2 

m'4  m4  m4b

m'4  0.321 kg

r4 

r2  59.9 mm

m'2 mr4

r4  86.9 mm

m'4


I'2  k' 2 

I'4 

  m r  R 2  m R  r 2 2 2 CG2 2b 2 2

I'2

4

I'2  6.226  10

2

kg m

k' 2  49.041 mm

m'2

m4  a 4  h

k' 4 

2

12

2

9.

mr2

12

2

  m r  R 2  m R  r 2 4 4 CG4 4b 4 4

I'4

3

I'4  2.735  10

k' 4  92.276 mm

m'4


Icw2  2.328  10

Icw4  m'4  k' 4  r4  a 4 r4

Icw4  7.677  10

2 2

2 2

3 3

2

kg m

2

kg m

2

kg m




The device in Figure P12-10 is used to balance fan blade/hub assemblies. The center distance between the two bearings on the machine is 250 mm. The left edge of the fan hub (plane A) is 100 mm from the centerline of the closest bearing (at F2). The hub is 75 mm wide and has a diameter of 200 mm along the surfaces where balancing weights are fastened. The peak magnitude of force F1 is 0.5 N at a phase angle of 30 deg with respect to the rotating x' axis. Force F2 had a peak of 0.2 N at a phase angle of 130 deg. Calculate the magnitudes and locations, with respect to the x' axis, of balance weights placed in planes A and B of the hub to dynamically balance the fan assembly.

Given:

Bearing forces and plane locations with respect to correction plane A: F1  0.5 N

Left:

Right: F2  0.2 N

θ  30 deg

l1  350  mm

θ  130  deg

l2  100  mm

Distance between correction planes: lB  75 mm Correction weight radii: RA  100  mm RB  100  mm Fan rotational speed: ω  600  rpm Solution: 1.



  F2x  F2 cos θ F1x  F1 cos θ

2.

F1x  0.433 N F2x  0.129 N

  F2y  F2 sin θ F1y  F1 sin θ

F1y  0.250 N F2y  0.153 N

Solve equations 12.4e for summation of moments about O, modified for the bearing forces. mRBx 

F1x l1  F2x l2

mRBx  0.468 mm kg

2


F1y l1  F2y l2

mRBy  0.347 mm kg

2



mRB  WB  4.

mRB RB

g

mRB  0.583 mm kg WB  0.057 N

Solve equations 12.4c for forces in x and y directions in plane A . mRAx  mRAy 

5.

2

mRBx  mRBy

θB  36.550 deg

F1x  F2x 2

ω F1y  F2y 2

 mRBx

mRAx  0.391 mm kg

 mRBy

mRAy  0.245 mm kg

ω Solve equations 12.2d and 12.2e for the position angle and mass-radius product required in plane A .


2

2


θA  147.936 deg mRA  0.462 mm kg WA  0.045 N




The device in Figure P12-10 is used to balance fan blade/hub assemblies. The center distance between the two bearings on the machine is 250 mm. The left edge of the fan hub (plane A) is 100 mm from the centerline of the closest bearing (at F2). The hub is 55 mm wide and has a diameter of 150 mm along the surfaces where balancing weights are fastened. The peak magnitude of force F1 is 1.5 N at a phase angle of 60 deg with respect to the rotating x' axis. Force F2 had a peak of 2.0 N at a phase angle of -180 deg. Calculate the magnitudes and locations, with respect to the x' axis, of balance weights placed in planes A and B of the hub to dynamically balance the fan assembly.

Given:


Left:

Right: F2  2.0 N

θ  60 deg

l1  350  mm

θ  180  deg

l2  100  mm


Solution: 1.


RA  75 mm

Fan rotational speed:

ω  600  rpm

RB  75 mm



  F2x  F2 cos θ F1x  F1 cos θ

2.

lB  55 mm

F1x  0.750 N F2x  2.000 N

  F2y  F2 sin θ F1y  F1 sin θ

F1y  1.299 N F2y  0.000 N


F1x l1  F2x l2

mRBx  0.288 mm kg

2


F1y l1  F2y l2


2



mRB  WB  4.

2


g

θB  82.173 deg mRB  2.114 mm kg WB  0.28 N

Solve equations 12.4c for forces in x and y directions in plane A . mRAx 

F1x  F2x 2

 mRBx

mRAx  0.604 mm kg

 mRBy


ω mRAy  5.

F1y  F2y 2


θA  atan2 mRAx mRAy

θA  108.906 deg


mRA  WA 

2



mRA  1.866 mm kg WA  0.24 N




The device in Figure P12-10 is used to balance fan blade/hub assemblies. The center distance between the two bearings on the machine is 250 mm. The left edge of the fan hub (plane A) is 100 mm from the centerline of the closest bearing (at F2). The hub is 125 mm wide and has a diameter of 250 mm along the surfaces where balancing weights are fastened. The peak magnitude of force F1 is 1.1 N at a phase angle of 120 deg with respect to the rotating x' axis. Force F2 had a peak of 1.8 N at a phase angle of -93 deg. Calculate the magnitudes and locations, with respect to the x' axis, of balance weights placed in planes A and B of the hub to dynamically balance the fan assembly.

Given:


θ  120  deg

l1  350  mm

Right: F2  1.8 N

θ  93 deg

l2  100  mm

Left:


lB  125  mm

Correction weight radii: RA  125  mm

RB  125  mm

ω  600  rpm

Fan rotational speed: Solution: 1.



  F2x  F2 cos θ F1x  F1 cos θ

2.

F1x  0.550 N F2x  0.094 N

  F2y  F2 sin θ F1y  F1 sin θ

F1y  0.953 N F2y  1.798 N


F1x l1  F2x l2

mRBx  0.409 mm kg

2


F1y l1  F2y l2


2



mRB  WB  4.

2


g

θB  142.728 deg mRB  0.514 mm kg WB  0.040 N

Solve equations 12.4c for forces in x and y directions in plane A . mRAx 

F1x  F2x 2

 mRBx

mRAx  0.246 mm kg

 mRBy


ω mRAy  5.

F1y  F2y 2


θA  atan2 mRAx mRAy

θA  64.911 deg


mRA  WA 

2


2


mRA  0.580 mm kg WA  0.046 N




Figure P12-11a shows a fourbar linkage and its dimensions in mm. All links are 4-mm-thick steel. Link 3 has a uniform cross section and is 20 mm wide with ends that extend 10 mm beyond the pivot hole. Links 2 and 4 have a 10-mm radius at each end. Design counterweights to force balance the linkage using the method of Berkof and Lowen.

Given:


L2  58 mm

Link 3 (A to B)

L3  108  mm

Link 4 (B to O4)

L4  110  mm

Link 1 (O2 to O4)

L1  160  mm

Link cross section: h  20 mm Mass center:

t  4  mm

r  10 mm

b 2  0.5 L2

b 3  0.5 L3

b 4  0.5 L4

ϕ  0  deg

ϕ  0  deg

ϕ  0  deg

3

Solution: 1.

See Figure P12-11and Mathcad file P1227.

Calculate the mass of each link. m2   L2 h  π r   t  2

m3   L3  2  r  h  t 

m2  0.046 kg m3  0.080 kg

m4   L4 h  π r   t  2

2.

m4  0.078 kg




L2

mb2x  m3  b 3

 



 cos ϕ  L2

 L3  L  2  mb2y  m3  b 3  sin ϕ  L3  L  4  mb4x  m3  b 3  cos ϕ   L3  L  4  mb4y  m3  b 3  sin ϕ   L3  3.

mb2x  2.316 kg mm mb2y  0.000 kg mm mb4x  4.393 kg mm mb4y  0.000 kg mm


  mR2y  mb2y  m2 b 2 sin ϕ mR4x  mb4x  m4 b 4 cos ϕ mR4y  mb4y  m4 b 4 sin ϕ mR2x  mb2x  m2 b 2 cos ϕ

4.

3

  7.8 10  kg m

Mass density of steel:

mR2x  3.650 kg mm mR2y  0.000 kg mm mR4x  8.707 kg mm mR4y  0.000 kg mm


2

2

mR2x  mR2y


2

2

mR4x  mR4y


mRb2  3.6501 kg mm

θb2  180.000 deg mRb4  8.707 kg mm

θb4  180.000 deg




Use the data of Problem 12-27 to design the necessary balance weights and other features necessary to completely eliminate the shaking force and shaking moment the linkage exerts on the ground link.

Given:


a 2  58 mm

Link 3 (A to B)

a 3  108  mm

Link 4 (B to O4)

a 4  110  mm

Link 1 (O2 to O4)

a 1  160  mm


h  20 mm ρ  7800 kg m


Determine the amount by which link 3 must be elongated to make it physically like a pendulum by using equation 12.12a. 2 a3  a3  e   3     1  2 2 h

e  38.995 mm

l3  a 3  2  e

l3  185.989 mm

h

2.

3.

4.


RCG2  29.000 mm

r3  0.5 a 3

r3  54.000 mm

RCG4  0.5 a 4

RCG4  55.000 mm


m3  h  t l3 ρ

m4  h  t a 4 ρ

m2  0.036 kg

m3  0.116 kg

m4  0.069 kg

Solve equations 12.17a and 12.17b for the total mR product components for links 2 and 4. b 3  a 3  r3 a2 



a3 



a4 



a3 

mr4  m3  r3

6.

b 3  54.000 mm



mr2  m3  b 3

5.

t  4  mm 3





mr2  3.366 kg mm

mr4  6.383 kg mm


mR2  2.316 kg mm

mR4  mr4  m4 RCG4

mR4  2.608 kg mm


R4  84.5 mm




m4b  7.

8.

mR2

m2b  0.031 kg

R2 mR4

m4b  0.031 kg

R4


m'2  0.067 kg

r2 

m'4  m4  m4b

m'4  0.100 kg

r4 

r2  50.2 mm

m'2 mr4

r4  64.2 mm

m'4


I'2 

k' 2 

12

I'4 

k' 4 

2

  m r  R 2  m R  r 2 2 2 CG2 2b 2 2

I'2

4

I'2  2.573  10

2

kg m

k' 2  61.933 mm

m'2 m4  a 4  h 2

9.

mr2

12

2

  m r  R 2  m R  r 2 4 4 CG4 4b 4 4

I'4

3

I'4  1.059  10

k' 4  103.152 mm

m'4


Icw2  6.214  10

Icw4  m'4  k' 4  r4  a 4 r4

Icw4  2.170  10

2 2

2 2

4 3

2

kg m

2

kg m

2

kg m




Figure P12-11b shows a fourbar linkage and its dimensions in inches. All links have a uniform 0.5-in wide x 0.2-in thick cross-section and are made from steel. Link 3 has squared ends that extend 0.25 in from the pivot point centers. Links 2 and 4 have rounded ends that have a radius of 0.25 in. Design counterweights to force balance the linkage using the method of Berkof and Lowen.

Units:


Given:

Link lengths:

1

2

Link 2 (O2 to A)

L2  2.75 in

Link 3 (A to B)

L3  3.26 in

Link 4 (B to O4)

L4  2.95 in

Link 1 (O2 to O4)

L1  3.26 in

Link cross section: h  0.50 in

t  0.20 in

r  0.25 in

b 2  0.5 L2

b 3  0.5 L3

b 4  0.5 L4

ϕ  0  deg

ϕ  0  deg

ϕ  0  deg

Mass center:

3

Solution: 1.



4

m2  2.294  10 2.

blob

m3   L3  2  r  h  t  4

m3  2.744  10



L2

 



 cos ϕ  L2

 L3  L  2  mb2y  m3  b 3  sin ϕ  L3  L   4 mb4x  m3  b 3  cos ϕ   L3  L  4  mb4y  m3  b 3  sin ϕ   L3 

2

4

m4  2.440  10

blob

4

mb2x  3.773  10

blob in

mb2y  0.000 blob in 4

mb4x  4.048  10

blob in

mb4y  0.000 blob in



4.

blob

m4   L4 h  π r   t 

Solve equations 12.8c and 12.8d for the total mR product components for links 2 and 4. mb2x  m3  b 3

3.

3

  7.8 10  kg m


mR2x  6.927  10

4

blob in

mR2y  0.000 blob in mR4x  7.646  10

4

blob in

mR4y  0.000 blob in


2

2

mR2x  mR2y


2

2

mR4x  mR4y


mRb2  6.927  10

4

blob in

θb2  180.000 deg mRb4  7.646  10

θb4  180.000 deg

4

blob in





Units:


Given:

Link lengths:

1

2

Link 2 (O2 to A)

a 2  2.75 in

Link 3 (A to B)

a 3  3.26 in

Link 4 (B to O4)

a 4  2.95 in

Link 1 (O2 to O4)

a 1  3.26 in


3.

e  1.182 in

l3  a 3  2  e

l3  5.624 in


RCG2  1.375 in

r3  0.5 a 3

r3  1.630 in

RCG4  0.5 a 4

RCG4  1.475 in

Determine the mass of each link.

4

m2  2.294  10

blob

m4   a 4 h  π r   t  2

m3  h  t l3  4

m3  4.105  10

blob

4

m4  2.440  10

blob


b 3  1.630 in



a2 



a3 



a4 

mr2  m3  b 3

mr4  m3  r3



6.

  7800 kg m

2 a3  a3  e   3     1  2 2 h

2

5.

r  0.25 in


m2   a 2 h  π r   t 

4.

t  0.20 in 3


h

2.

h  0.50 in



 a3 

mr2  5.644  10

mr4  6.055  10

4

4

blob in

blob in


mR2  mr2  m2 RCG2

mR2  2.490  10

mR4  mr4  m4 RCG4

mR4  2.456  10

4

Let the distance to the force-balance mass from the pivot point on links 2 and 4 be

blob in blob in



R2  2  in

R4  1.973  in


m4b  7.

8.

mR2

m2b  1.245  10

R2 mR4

m4b  1.245  10

R4

4

blob

blob


m'2  3.539  10

m'4  m4  m4b

m'4  3.685  10

4

4

blob

r2 

blob

r4 

mr2

r2  1.595 in

m'2 mr4

r4  1.6 in

m'4


I'2 

k' 2 

12

I'4 

k' 4 

2

  m r  R 2  m R  r 2 2 2 CG2 2b 2 2

I'2

3

I'2  2.193  10

2

blob in

k' 2  2.489 in

m'2 m4  a 4  h 2

9.

4

12

2

  m r  R 2  m R  r 2 4 4 CG4 4b 4 4

I'4

3

I'4  2.568  10

k' 4  2.640 in

m'4


Icw2  4.645  10

Icw4  m'4  k' 4  r4  a 4 r4

Icw4  5.349  10

2 2

2 2

3 3

2

blob in

2

blob in

2

blob in




Figure P12-12 shows a fourbar linkage and its dimensions in inches. All links have a uniform 0.5-in wide x 0.2-in thick cross-section and are made from aluminum. Link 3 has squared ends that extend 0.25 in from the pivot point centers. Links 2 and 4 have rounded ends that have a radius of 0.25 in. Design counterweights to force balance the linkage using the method of Berkof and Lowen.

Units:


Given:

Link lengths:

1

2

Link 2 (O2 to A)

L2  3.44 in

Link 3 (A to B)

L3  7.40 in

Link 4 (B to O4)

L4  5.44 in

Link 1 (O2 to O4)

L1  8.88 in

Link cross section:

h  0.50 in

t  0.20 in

r  0.25 in

Mass center:

b 2  0.5 L2

b 3  0.5 L3

b 4  0.5 L4

ϕ  0  deg

ϕ  0  deg

ϕ  0  deg

3

3

Mass density of aluminum:   2.8 10  kg m Solution: 1.



4

m2  1.004  10 2.

blob

4

m3  2.070  10



L2

 



 cos ϕ  L2

 L3  L   2 mb2y  m3  b 3  sin ϕ  L3  L   4 mb4x  m3  b 3  cos ϕ   L3    L4 mb4y  m3  b 3  sin ϕ   L3 

4

m4  1.528  10

blob

blob

4

mb2x  3.560  10

blob in

mb2y  0.000 blob in 4

mb4x  5.630  10

blob in




4.

2


3.

m4   L4 h  π r   t 

m3   L3  2  r  h  t 

mR2x  5.287  10

4

blob in

mR2y  0.000 blob in mR4x  9.787  10

4

blob in



2

2

mR2x  mR2y


2

2

mR4x  mR4y


mRb2  5.287  10

4

blob in

θb2  180.000 deg mRb4  9.787  10

θb4  180.000 deg

4

blob in





Units:


Given:

Link lengths:

1

2

Link 2 (O2 to A)

a 2  3.44 in

Link 3 (A to B)

a 3  7.40 in

Link 4 (B to O4)

a 4  5.44 in

Link 1 (O2 to O4)

a 1  8.88 in

Link cross-section dims: Material density: alum Solution: 1.

3.

e  2.704 in

l3  a 3  2  e

l3  12.807 in


RCG2  1.720 in

r3  0.5 a 3

r3  3.700 in

RCG4  0.5 a 4

RCG4  2.720 in


4

m2  1.004  10

blob

m4   a 4 h  π r   t  2

m3  h  t l3  4

m3  3.356  10

blob

4

m4  1.528  10

blob


b 3  3.700 in



a2 



a3 



a4 

mr2  m3  b 3

mr4  m3  r3



6.

  2800 kg m

2 a3  a3  e   3     1  2 2 h

2

5.

r  0.25 in


m2   a 2 h  π r   t 

4.

t  0.20 in 3


h

2.

h  0.50 in



 a3 

mr2  5.772  10

mr4  9.127  10

4

4

blob in

blob in


mR2  mr2  m2 RCG2

mR2  4.044  10

mR4  mr4  m4 RCG4

mR4  4.971  10

4


Let the distance to the force-balance mass from the pivot point on links 2 and 4 be R2  2  in

R4  2.458  in




m4b  7.

8.

mR2

m2b  2.022  10

R2 mR4

m4b  2.022  10

R4

blob

4

blob


m'2  3.026  10

m'4  m4  m4b

m'4  3.550  10

4

4

blob r2  blob r4 

mr2

r2  1.907 in

m'2 mr4

r4  2.571 in

m'4


I'2  k' 2 

12

I'4  k' 4 

2

  m r  R 2  m R  r 2 2 2 CG2 2b 2 2

I'2

3

I'2  1.424  10

2

blob in

k' 2  2.169 in

m'2 m4  a 4  h 2

9.

4

12

2

  m r  R 2  m R  r 2 4 4 CG4 4b 4 4

I'4

3

I'4  4.660  10

2

blob in

k' 4  3.623 in

m'4


Icw2  4.510  10

Icw4  m'4  k' 4  r4  a 4 r4

Icw4  0.0120 blob in

2 2

2 2

3

2

blob in 2




Figure P12-13 shows a fourbar linkage and its dimensions in inches. Links 2 and 4 are rectangular steel with a 1-in wide x 0.12-in thick cross-section and 0.5-in radius ends. The coupler is 0.25-in- thick aluminum with 0.5-in radii at points A, B, and P. Design counterweights to force balance the linkage using the method of Berkof and Lowen.

Units:


Given:

Link lengths:

1

2

Link 2 (O2 to A)

L2  5.00 in

Link 3 (A to B)

L3  7.40 in

Link 4 (B to O4)

L4  8.00 in

Link 1 (O2 to O4)

L1  9.50 in

Link cross section: h  0.50 in Mass center:

t  0.20 in

b 2  0.5 L2

b 4  0.5 L4

Coupler point data:

ϕ  0  deg

ϕ  0  deg

p  8.9 in

Mass density of steel and aluminum: Solution: 1.

xbar  b 3 

h  p  sin 

L3  APx

h  7.378 in

3

2

xbar  ybar

2

b 3  4.803 in

L3 h

ybar 

APx  4.977 in

h

ybar  2.459 in

3

ϕ  atan

ybar 

  xbar 

3

m3  1.788  10

2

t3  0.25 in

APx  p  cos 

xbar  4.126 in

3

m3   t3

ϕ  30.801 deg

blob

Calculate the mass of links 2 and 4. m2   L2 h  π r   t 

m4   L4 h  π r   t 

2

3

m2  5.414  10

2

3

m4  8.645  10

blob

blob




mb2x  m3  b 3

L2

 



 cos ϕ  L2

 L3  L  2  mb2y  m3  b 3  sin ϕ  L3  L  4  mb4x  m3  b 3  cos ϕ   L3  L  4  mb4y  m3  b 3  sin ϕ   L3  3.

3

  2.8 10  kg m

Determine the location of the mass center and the mass of link 3. Take AB as the base of the triangle, Then,

The horizontal distance from A to P is:

2.

3

3

  7.8 10  kg m

  56 deg


The height is:

2.

r  0.25 in

3

mb2x  3.956  10 mb2y  2.972  10

3

blob in

3

mb4x  7.976  10

3

mb4y  4.755  10


 

mR2x  mb2x  m2 b 2 cos ϕ

blob in

mR2x  0.017 blob in




  mR4x  mb4x  m4 b 4 cos ϕ mR4y  mb4y  m4 b 4 sin ϕ mR2y  mb2y  m2 b 2 sin ϕ

4.

3

mR2y  2.972  10

blob in

mR4x  0.043 blob in mR4y  4.755  10

3

blob in


2

2

mR2x  mR2y


2

2

mR4x  mR4y


mRb2  0.0177 blob in

θb2  170.358 deg mRb4  0.0428 blob in

θb4  173.625 deg




Use the data of Problem 12-33, changing link 3 to be steel with the same cross-section dimensions as links 2 and 4, to design the necessary balance weights and other features necessary to completely eliminate the shaking force and shaking moment the linkage exerts on the ground link.

Units:


Given:

Link lengths:

1

2

Link 2 (O2 to A)

a 2  5.00 in

Link 3 (A to B)

a 3  7.40 in

Link 4 (B to O4)

a 4  8.00 in

Link 1 (O2 to O4)

a 1  9.50 in


3.

e  2.704 in

l3  a 3  2  e

l3  12.807 in


RCG2  2.500 in

r3  0.5 a 3

r3  3.700 in

RCG4  0.5 a 4

RCG4  4.000 in


4

m2  3.936  10

blob

m4   a 4 h  π r   t  2

m3  h  t l3  4

m3  9.348  10

blob

4

m4  6.126  10

blob




mr2  m3  b 3

b 3  3.700 in a2 



 a3 



a4 



a3 

mr4  m3  r3

6.

  7800 kg m

2 a3  a3  e   3     1  2 2 h

2

5.

r  0.25 in


m2   a 2 h  π r   t 

4.

t  0.20 in 3


h

2.

h  0.50 in



mr2  2.337  10

mr4  3.739  10

3

3

blob in

blob in


mR2  mr2  m2 RCG2

mR2  1.353  10

mR4  mr4  m4 RCG4

mR4  1.289  10

3

Let the distance to the force-balance mass from the pivot point on links 2 and 4 be




R2  2  in

R4  1.905  in


m4b  7.

8.

mR2

m2b  6.765  10

R2 mR4

m4b  6.766  10

R4

4

blob

blob


m'2  1.070  10

m'4  m4  m4b

m'4  1.289  10

3

3

blob

r2 

blob

r4 

mr2

r2  2.184 in

m'2 mr4

r4  2.900 in

m'4


I'2 

k' 2 

12

I'4 

k' 4 

2

  m r  R 2  m R  r 2 2 2 CG2 2b 2 2

I'2

3

I'2  9.4862  10

m4  a 4  h 12

2

  m r  R 2  m R  r 2 4 4 CG4 4b 4 4

I'4

2

I'4  0.0331 blob in

k' 4  5.069 in

m'4



Icw4  m'4  k' 4  r4  a 4 r4


2 2

2 2

2 2

2

blob in

k' 2  2.977 in

m'2 2

9.

4




Figure P12-14 shows a fourbar linkage and its dimensions in inches. All links are 0.08-in-thick steel and have a uniform cross-section of 0.26-in wide x 0.12-in thick. Links 2 and 4 have rounded ends with a 0.13-in radius. Link 3 has a squared ends that extend 0.13-in from the pivot point centers. Design counterweights to force balance the linkage using the method of Berkof and Lowen.

Units:


Given:

Link lengths:

1

2

Link 2 (O2 to A)

L2  5.52 in

Link 3 (A to B)

L3  4.88 in

Link 4 (B to O4)

L4  6.48 in

Link 1 (O2 to O4)

L1  2.72 in

Link cross section:

h  0.26 in

t  0.08 in

r  0.13 in

Mass center:

b 2  0.5 L2

b 3  0.5 L3

b 4  0.5 L4

ϕ  0  deg

ϕ  0  deg

ϕ  0  deg

3

Solution: 1.



5

m2  8.690  10 2.

blob

m3   L3  2  r  h  t  5

m3  7.803  10



L2

 



 cos ϕ  L2

 L3  L   2 mb2y  m3  b 3  sin ϕ  L3  L   4 mb4x  m3  b 3  cos ϕ   L3   L4  mb4y  m3  b 3  sin ϕ   L3 

2

4

m4  1.015  10

blob

4

mb2x  2.154  10

blob in

mb2y  0.000 blob in 4

mb4x  2.528  10

blob in




4.

blob

m4   L4 h  π r   t 


3.

3

  7.8 10  kg m


mR2x  4.552  10

4

blob in

mR2y  0.000 blob in mR4x  5.816  10

4

blob in



2

2

mR2x  mR2y


2

2

mR4x  mR4y


mRb2  4.552  10

4

blob in

θb2  180.000 deg mRb4  5.816  10

θb4  180.000 deg

4

blob in





Units:


Given:

Link lengths:

1

2

Link 2 (O2 to A)

a 2  5.52 in

Link 3 (A to B)

a 3  4.88 in

Link 4 (B to O4)

a 4  6.48 in

Link 1 (O2 to O4)

a 1  2.72 in

t  0.12 in

r  0.13 in

Link cross-section dims: h  0.26 in Material density: steel

3

  7800 kg m

See Figure P12-14 and Mathcad file P1236. Solution: 1. Determine the amount by which link 3 must be elongated to make it physically like a pendulum by using equation 12.12a. 2 a3  a3  e   3     1  2 2 h

e  1.784 in

l3  a 3  2  e

l3  8.448 in

h

2.

3.


RCG2  2.760 in

r3  0.5 a 3

r3  2.440 in

RCG4  0.5 a 4

RCG4  3.240 in

Determine the mass of each link. m2   a 2 h  π r   t  2

4

m2  1.304  10 4.

blob



mr2  m3  b 3

4

m3  1.924  10

blob

4

m4  1.522  10

blob

b 3  2.440 in a2 

  a3   a4  mr4  m3  r3   a3 

6.

2


5.

m4   a 4 h  π r   t 

m3  h  t l3 

mr2  5.310  10 mr4  6.233  10

4

4



mR2  mr2  m2 RCG2

mR2  1.712  10

mR4  mr4  m4 RCG4

mR4  1.302  10

4


Let the distance to the force-balance mass from the pivot point on links 2 and 4 be R2  2  in

R4  1.52 in


mR2 R2

m2b  8.561  10

5

blob


m4b  7.

8.

mR4

m4b  8.563  10

R4

5

blob


m'2  2.160  10

m'4  m4  m4b

m'4  2.378  10

4

4

blob

r2 

blob

r4 

mr2 m'2 mr4 m'4

r2  2.459 in r4  2.621 in


I'2  k' 2 

12

I'4  k' 4 

2

  m r  R 2  m R  r 2 2 2 CG2 2b 2 2

I'2

3

I'2  3.900  10

2

blob in

k' 2  4.250 in

m'2 m4  a 4  h 2

9.


12

2

  m r  R 2  m R  r 2 4 4 CG4 4b 4 4

I'4

3

I'4  5.865  10

2

blob in

k' 4  4.966 in

m'4


Icw2  0.00814 blob in

Icw4  m'4  k' 4  r4  a 4 r4


2 2

2 2

2

2




A manufacturing company makes 5-blade ceiling fans. Before assembling the fan blades onto the hub the blades are weighed and the location of the CG is determined as a distance from the center of rotation and an angular offset from the geometric center of the blade. At final assembly a technician is provided with the weight and CG data for the 5 blades. Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate the required weight and angular postion of a balance weight that is attached to the hub at a radius of 2.5 in. Use the geometric center of blade one as a reference axis. Test your program with the data given in Table P12-3.

Given:

Number of blades N  5 Blade counter i  1 2  N Blade test data: Blade

Weight W 

3 4 5

1.

δ 

i

1.50 lbf 1.48 lbf 1.54 lbf 1.55 lbf 1.49 lbf

1 2

CG Ang Offset

r 

i

i 

Solution:

Rad to CG

Hub radius rhub  2.5 in

i

12.01  in 11.97  in 11.95  in 12.03  in 12.04  in

0.25 deg 0.75 deg 0.25 deg 1.00 deg 0.50 deg


Resolve the position vectors into xy components with respect to an x-y frame whose positive x axis is coincident with the geometric center of blade one. Rx  r  cos( i  1 )  i

360  deg



i

N

 δ  i

Ry  r  sin( i  1 )  i

Rx 

-0.052 11.432

in

-9.698

i

in

6.982

-9.854

-6.900

3.620

-11.483

Solve equation 12.2c for the mass-radius product components. N



mRbx  

i 1

 Wi    Rxi  g 

mRbx  1.547 in lb

N



mRby  

i 1

 Wi    Ryi  g 


Solve equations 12.2d and 12.2e for the position angle (with respect to the centerline of blade one) and mass-radius product required. θb  atan2 mRbx mRby θb  7.818 deg mRb 

4.

 δ 

i

12.010 3.550

3.

N

Ry 

i

2.

360  deg



i

2

2

mRbx  mRby

mRb  1.561 in lb

Determine the balance weight required at the hub radius. Wb 

mRb g rhub

Wb  10.0 ozf




The motor rotor shown in Figure P12-14 has been tested on a dynamic balance machine at 1800 rpm and shows unbalanced forces of F1 = 2.43 lb @ 1 = 34.5 deg in the x-y plane at 1 and F4 = 5.67 lb @ 4 = 198 deg in the x-y plane at 4. Balance weights consisting of cylindrical disks whose center of rotation is a drilled hole located at a distance e from the center of the disk. The net weight of each disk is 0.50 lb and the disks are located on planes 2 and 3. Determine the angular locations (of the line through the drilled hole and the center of the disk with respect to the reference axis) and the eccentric distances e to dynamically balance the system. 1

2

Units:


Given:

Forces and locations: F1  2.43 lbf

θ  34.5 deg

l1  1.75 in

F4  5.67 lbf

θ  198  deg

l4  8.25 in


Wc  0.50 lbf

Correction weight :

ω  1800 rpm




 

F1x  2.003 lbf

F1y  F1 sin θ

 

F4x  5.392 lbf

F4y  F4 sin θ

F1x  F1 cos θ F4x  F4 cos θ 2.

F1y  1.376 lbf

 

F4y  1.752 lbf

F1x l1  F4x l4 mRBx  2.251  10

2

lB ω mRBy  

F1y l1  F4y l4 mRBy  7.910  10

2

lB ω

4

5

blob in

blob in

Solve equations 12.2d and 12.2e for the position angle and mass-radius product required in plane B (3). Also, solve for the eccentricty of the drilled hole in the balance weight.

θB  atan2 mRBx mRBy mRB 

2

2

mRBx  mRBy

eB  mRB 4.

 

Solve equations 12.3 for summation of moments about O, which is at plane 2. mRBx  

3.

lB  6.00 in

g

θB  19.360 deg mRB  2.3862  10

4

blob in

eB  0.184 in

Wc


F1x  F4x 2

ω

 mRBx

4

mRAx  1.2972  10

blob in


mRAy 

3.


F1y  F4y

 mRBy 5 mRAy  6.853  10 blob in 2 ω Solve equations 12.2d and 12.2e for the position angle and mass-radius product required in plane A (2). Also, solve for the eccentricty of the drilled hole in the balance weight.


2

2

mRAx  mRAy

eA  mRA

g Wc

θA  152.153 deg mRA  1.4671  10 eA  0.113 in

4

blob in




The motor rotor shown in Figure P12-14 has been tested on a dynamic balance machine at 1450 rpm and shows unbalanced forces of F1 = 4.82 lb @ 1 = 163 deg in the x-y plane at 1 and F4 = 7.86 lb @ 4 = 67.8 deg in the x-y plane at 4. Balance weights consisting of cylindrical disks whose center of rotation is a drilled hole located at a distance e from the center of the disk. The net weight of each disk is 0.375 lb and the disks are located on planes 2 and 3. Determine the angular locations (of the line through the drilled hole and the center of the disk with respect to the reference axis) and the eccentric distances e to dynamically balance the system. 1

2

Units:


Given:

Forces and locations: F1  4.82 lbf

θ  163  deg

l1  1.75 in

F4  7.86 lbf

θ  67.8 deg

l4  8.25 in


Wc  0.375  lbf

Correction weight :

ω  1450 rpm




 

F1x  4.609 lbf

F1y  F1 sin θ

 

F4x  2.970 lbf

F4y  F4 sin θ

F1x  F1 cos θ F4x  F4 cos θ 2.

F1y  1.409 lbf

 

F4y  7.277 lbf

F1x l1  F4x l4

mRBy  

4

mRBx  2.354  10

2

lB ω

F1y l1  F4y l4

4

mRBy  4.162  10

2

lB ω

blob in

blob in

Solve equations 12.2d and 12.2e for the position angle and mass-radius product required in plane B (3). Also, solve for the eccentricty of the drilled hole in the balance weight.

θB  atan2 mRBx mRBy mRB 

2

2

mRBx  mRBy

eB  mRB 4.

 

Solve equations 12.3 for summation of moments about O, which is at plane 2. mRBx  

3.

lB  6.00 in

g

θB  119.496 deg mRB  4.7814  10

4

blob in

eB  0.492 in

Wc


F1x  F4x 2

ω

 mRBx

mRAx  3.0653  10

4

blob in


mRAy 

5.


F1y  F4y

 mRBy 5 mRAy  3.941  10 blob in 2 ω Solve equations 12.2d and 12.2e for the position angle and mass-radius product required in plane A (2). Also, solve for the eccentricty of the drilled hole in the balance weight.


2

2

mRAx  mRAy

eA  mRA

g Wc

θA  7.327 deg mRA  3.0905  10 eA  0.318 in

4

blob in




Table P12-4 gives the geometry and kinematic data for several fourbar linkages similar to that shown in Figure P12-11. For row a in Table P12-4, design counterweights of the type shown in Figure P12-15 for links 2 and 4 to completely force balance the linkage by the method of Berkof and Lowen. The square ends of link 3 extend a distance, e, from the hole center. The other links' ends are full round with a radius, r, about the hole center. All pin holes have the same diameter, d, and all links have the same width, 2r, and thickness, t. The hole-to-hole link lengths are L1, L2, L3, and L4.

Given:

Link lengths: Link 2 (O2 to A) L2  100  mm

Link 3 (A to B)

Link 4 (B to O4) L4  200  mm

Link 1 (O2 to O4) L1  375  mm

Link features:

r  13 mm

e  13 mm 3

1.

3

Calculate the mass of each link. 2

2

  t ρ 2 m3   L3  2  e  2  r  0.5 π d   t ρ 2 2 m4   L4 2  r  π r  0.5 π d   t ρ

m2  0.096 kg m3  0.263 kg m4  0.177 kg

Solve equations 12.8c and 12.8d for the total mR product components for links 2 and 4. For these links, b = 0.5L and ϕ = 0. b 3  0.5 L3 ϕ  0



mb2x  m3  b 3

L2

 



 cos ϕ  L2

mb2x  13.134 kg mm

 L3  L   2 mb2y  m3  b 3  sin ϕ  L3  L  4  mb4x  m3  b 3  cos ϕ   L3  L  4  mb4y  m3  b 3  sin ϕ   L3  3.

t  4  mm

See Table P12-4, Figure P12-11, Figure P12-15, and Mathcad file P1240a.

m2   L2 2  r  π r  0.5 π d

2.

d  6  mm

ρ  7.8 10  kg m

Mass density of steel: Solution:

L3  300  mm

mb2y  0.000  kg mm mb4x  26.269 kg mm mb4y  0.000  kg mm

Determine the additional mR product components for links 2 and 4. b 2  0.5 L2

ϕ  0

 

mR2x  mb2x  m2 b 2 cos ϕ

b 4  0.5 L4

ϕ  0 mR2x  17.930 kg mm

 

mR2y  0.000  kg mm

 

mR4x  43.973 kg mm

 

mR4y  0.000  kg mm

mR2y  mb2y  m2 b 2 sin ϕ

mR4x  mb4x  m4 b 4 cos ϕ mR4y  mb4y  m4 b 4 sin ϕ


4.



2

2

mR2x  mR2y

mRb2  17.9304  kg mm


2

θb2  180.000  deg

2

mR4x  mR4y

mRb4  43.973 kg mm

θb4  atan2 mR4x mR4y 5.

θb4  180.000  deg

Determine the required dimensions for the counterweights on links 2 and 4. There are two independent variables that define the mR product of the counterweight, α and R. For each link, make a design choice for R and then solve for α. Repeat the process until you have satisfactory values for each. Link 2:

R2  8  r

α2 

2  mRb2 2 2 R2  R2  r   t ρ

mcw2  Link 4:

R2  104 mm

α2 2

  R2  r   t ρ 2

R4  10 r

α4 

2

2 2 R4  R4  r   t ρ

α4 2

mcw2  0.172 kg

R4  130 mm 2  mRb4

mcw4 

α2  59 deg

  R4  r   t ρ 2

2

α4  74 deg

mcw4  0.338 kg




A slider-crank linkage has the dimensions and speed at t = 0 given below. Its initial crank angle is zero. Calculate the piston acceleration at the time specified below. Use two methods, the exact solution, and the approximate Fourier series solution and compare the results.

Given:

Link lengths:

r  3  in l  12 in

Initial angular velocity: ω  200  rad sec Solution: 1.

t  1.0 sec

Calculate the exact acceleration using equation 13.1f.

2

    



r l  1  2  cos ω t 2

  r2sinωt4 

2

   

3

l2  r2sinωt2

2

in

a exact  42679.3

sec

2

Calculate the approximate acceleration using equation 13.3e. a approx  r ω   cos ω t  2



3.

Time span:


a exact  r ω  cos ω t 

2.

1

r l

 cos 2  ω t



in

a approx  42703.6

sec

Compare the results by calculating the error in the approximation as a percent of the exact. error 

a approx  a exact a exact

error  0.057 %

The approximate solution is a little less than 0.06% high.

2





Given:

Link lengths:

r  4  in l  15 in


t  0.9 sec


2

    



r l  1  2  cos ω t 2

  r2sinωt4 

2

   

3

l2  r2sinωt2

2

in

a exact  107395.7

sec

2




3.

Time span:



2.

1

r l

 cos 2  ω t



a approx  107857.8

in sec



error  0.430 %

The approximate solution is a little less than 0.5% high.

2




A slider-crank linkage has the dimensions given below. The peak gas pressure and crank angle are also given. Calculate the gas force and gas torque at this position.

Given:

Link lengths:

r  3  in l  12 in B  2  in

Piston bore:

Peak pressure:

p g  1000 psi

Assumptions: An approximate solution is acceptable. Solution: 1.


Calculate the gas force on the piston using equation 13.4. Fg 

2.

π 4

 pg  B

2

Fg  3142 lbf

Calculate the approximate gas torque on the crank using equation 13.8b. Tg21  Fg r sin θ   1 



r l

 cos θ



Tg21  2040 in lbf

Crank angle:

θ  10 deg





Given:

Link lengths:

r  4  in l  15 in B  3  in

Piston bore:

Peak pressure:

p g  600  psi




2.

π 4

 pg  B

2

Fg  4241 lbf




r l

 cos θ



Tg21  1871 in lbf

Crank angle:

θ  5  deg




A slider-crank linkage has the dimensions given below. The peak gas pressure and crank angle are also given. Calculate the exact gas torque at this position and compare it to that obtained by the approximate expression in equation 13.8b. What is the percent error?

Given:

Link lengths:

r  3  in l  12 in B  2  in

Piston bore: Solution: 1.

π 4

 pg  B

2

Fg  3142 lbf

Calculate the approximate gas torque on the crank using equation 13.8b.



r l

 cos θ 



Tg21a  2039.53 in lbf

Calculate the exact gas torque on the crank using equations 13.7b, 13.1d, and 13.6d. r sin θ   2   r  sin θ    l 1      l  

ϕ  atan

x  r cos θ  l 1  Tg21e  Fg tan ϕ  x 4.

Crank angle:

Calculate the gas force on the piston using equation 13.4.

Tg21a  Fg r sin θ   1  3.

p g  1000 psi


Fg  2.

Peak pressure:

 r  sin θ   l 

ϕ  2.488 deg

2

x  14.943 in Tg21e  2039.91 in lbf


Tg21a  Tg21e Tg21e

error  0.0186 %

θ  10 deg





Given:

Link lengths:

r  4  in l  15 in B  3  in


π 4

 pg  B

2

Fg  4241 lbf




r l

 cos θ 



Tg21a  1871.35 in lbf

Calculate the exact gas torque on the crank using equations 13.7b, 13.1d, and 13.6d. r sin θ   2    r  l 1    sin θ   l  

ϕ  atan

x  r cos θ  l 1  Tg21e  Fg tan ϕ  x 4.

Crank angle:


Tg21a  Fg r sin θ   1  3.

p g  600  psi


Fg  2.

Peak pressure:

 r  sin θ   l 

ϕ  1.332 deg

2

x  18.981 in Tg21e  1871.45 in lbf

Compare the results by calculating the error in the approximation as a percent of the exact.

error 


error  0.00567 %

θ  5  deg




The dimensions and mass properties of a connecting rod are given below. a. Calculate an exact dynamic model using two lumped masses, one at the wrist pin, point B, and one at whatever point is required. Define the lumped masses and their locations. b. Calculate an approximate dynamic model using two lumped masses, one at the wrist pin, point B, and one at the crank pin, point A. Define the lumped masses and their locations. c. Calculate the error in the mass moment of inertia of the approximate model as a percentage of the original mass moment of inertia.

Units:


Given:

Conrod length:

2

1

l  12 in

mass:

2

IG3  0.620  blob in

Mass moment of inertia:

Distance to CG (as fraction of l): Solution: 1.

Determine the exact model using equations 13.9d and 13.9e. Distance from point A to CG: la  ra l

la  4.800 in

Distance from point B to CG: lb  l  la

lb  7.200 in

Distance from CG to lumped mass at P:

lp 

mp  m3 mb  m3

lb lp  lb lp lp  lb

IG3 m3 lb

lp  4.306 in

mp  0.0125 blob mb  0.00748 blob

Determine the approximate model using equations 13.9d, letting lp = la. Let

lp  la

Masses:

m3a  m3

m3b  m3 3.

ra  0.4


Masses:

2.

m3  0.020  blob


m3a  0.0120 blob

m3b  0.00800 blob

Calculate the mass moment of inertia of the approximate model and compare it to the original. 2

2

IG3approx  m3a la  m3b lb error 

IG3approx  IG3 IG3

2

IG3approx  0.691 blob in error  11.48 %




The dimensions and mass properties of a connecting rod are given below. a. Calculate an exact dynamic model using two lumped masses, one at the wrist pin, point B, and one at whatever point is required. Define the lumped masses and their locations. b. Calculate an approximate dynamic model using two lumped masses, one at the wrist pin, point B, and one at the crank pin, point A. Define the lumped masses and their locations. c. Calculate the error in the mass moment of inertia of the approximate model as a percentage of the original mass moment of inertia.

Units:


Given:

Conrod length:

2

1

l  15 in

mass:

2

IG3  1.020  blob in


Distance to CG (as fraction of l): Solution: 1.

m3  0.025  blob ra  0.25


Determine the exact model using equations 13.9d and 13.9e. Distance from point A to CG:

la  ra l

la  3.750 in

Distance from point B to CG:

lb  l  la

lb  11.250 in


Masses:

mp  m3 mb  m3

2.


IG3 m3 lb

lp  3.627 in

mp  0.0189 blob mb  0.00609 blob

Determine the approximate model using equations 13.9d, letting lp = la. Let

lp  la

Masses:

m3a  m3 m3b  m3

3.

lp 


m3a  0.0188 blob m3b  0.00625 blob

Calculate the mass moment of inertia of the approximate model and compare it to the original. 2

2

IG3approx  m3a la  m3b lb error 

IG3approx  IG3 IG3

2

IG3approx  1.055 blob in error  3.40 %




The dimensions and mass properties of a crank are given below. Calculate a statically equivalent, two-lumped mass dynamic model with the lumps placed at the main pin and crankpin. What is the percent error in the model's moment of inertia about the crank pivot?

Units:


Given:

Crank length:

1

2

r  3.5 in

mass:

2

I2  0.300  blob in


Distance to CG (as fraction of r): Solution: 1.

ra  0.3


Determine the statically equivalent model using equations 13.11. rG2  ra r

Distance from O2 to CG: Mass: 2.

m2  0.060  blob

m2a  m2

rG2 r

rG2  1.050 in m2a  0.0180 blob

Calculate the mass moment of inertia of the statically equivalent model and compare it to the original. 2

IO2model  m2a r error 

IO2model  I2 I2

2

IO2model  0.2205 blob in error  26.50 %




The dimensions and mass properties of a crank are given below. Calculate a statically equivalent, two-lumped mass dynamic model with the lumps placed at the main pin and crankpin. What is the percent error in the model's moment of inertia about the crank pivot?

Units:


Given:

Crank length:

1

2

r  4  in

mass:

2

I2  0.400  blob in


Distance to CG (as fraction of r): Solution: 1.

ra  0.4


Determine the statically equivalent model using equations 13.11. rG2  ra r

Distance from O2 to CG: Mass: 2.

m2  0.050  blob

m2a  m2

rG2 r

rG2  1.600 in m2a  0.0200 blob

Calculate the mass moment of inertia of the statically equivalent model and compare it to the original. 2

IO2model  m2a r error 

IO2model  I2 I2

2

IO2model  0.3200 blob in error  20.00 %




The dimensions and mass properties of a crank, connecting rod, and piston are given below. Calculate the inertia force and inertia torque for the linkage.

Units:


Given:

Conrod length:

1

2

l  12 in

mass:

r3a  0.4

Distance to CG (as fraction of l): r  3.5 in

Crank length:

mass: m2  0.060  blob r2a  0.3

Distance to CG (as fraction of r):

Solution: 1.

m3  0.020  blob

Piston mass:

m4  0.012  blob

Crank speed:

ω  2000 rpm


Determine the approximate conrod model using equations 13.9d, letting lp = la. Distance from point A to CG:

la  r3a l

la  4.800 in


lb  l  la

lb  7.200 in

lp  la

Let

m3a  m3

m3b  m3

2.

m3a  0.0120 blob

lp  lb lp

m3b  0.00800 blob

lp  lb

rG2  r2a r

m2a  m2

Mass:

4.

lb

Determine the statically equivalent crank model using equations 13.11. Distance from O2 to CG:

3.

θ  45 deg

Crank angle:

rG2

rG2  1.050 in m2a  0.0180 blob

r

Calculate the lumped masses at points A and B using equations 13.12. mA  m2a  m3a

mA  0.0300 blob

mB  m3b  m4

mB  0.0200 blob

Calculate the inertia force and inertia torque using equations 13.14d and 13.15d.

 2  2  m   r ω  sin θ 

Fix  mA r ω  cos θ  mB r ω   cos θ  Fiy

Fi 

2





r l

 cos 2  θ 



Fiy  3257 lbf

A

2

2

Fix  Fiy

Fi  6330 lbf

Ti21  mB r  ω  sin θ   2

2

Fix  5428 lbf

r

 2 l

 cos θ 

at 3 r 2 l

atan2 Fix Fiy  30.964 deg

 cos 2  θ



Ti21  6482 in lbf





Units:


Given:

Conrod length:

1

2

l  12 in

mass:

r3a  0.4

Distance to CG (as fraction of l): r  4  in

Crank length:

mass: m2  0.050  blob r2a  0.4


Solution: 1.

m3  0.020  blob

Piston mass:

m4  0.019  blob

Crank speed:

ω  3000 rpm



la  r3a l

la  4.800 in


lb  l  la

lb  7.200 in

lp  la

Let

m3a  m3

m3b  m3

2.

m3a  0.0120 blob

lp  lb lp

m3b  0.00800 blob

lp  lb

rG2  r2a r

m2a  m2

Mass:

4.

lb


3.

θ  30 deg

Crank angle:

rG2

rG2  1.600 in m2a  0.0200 blob

r


mA  0.0320 blob

mB  m3b  m4

mB  0.0270 blob


 2  2  m   r ω  sin θ 


Fi 

2





r l

 cos 2  θ 



Fiy  6317 lbf

A

2

2

Fix  Fiy

Fi  22839 lbf

Ti21  mB r  ω  sin θ   2

2

Fix  21948 lbf

r

 2 l

 cos θ 

3 r 2 l

at


 cos 2  θ



Ti21  27345 in lbf





Units:


Given:

Conrod length:

1

2

l  15 in

mass:

r3a  0.25


Crank length:

mass: m2  0.060  blob r2a  0.3


Solution: 1.

m3  0.025  blob

Piston mass:

m4  0.023  blob

Crank speed:

ω  2500 rpm



la  r3a l

la  3.750 in


lb  l  la

lb  11.250 in

lp  la

Let

m3a  m3

m3b  m3

2.

m3a  0.0188 blob

lp  lb lp

m3b  0.00625 blob

lp  lb

rG2  r2a r

m2a  m2

Mass:

4.

lb


3.

θ  24 deg

Crank angle:

rG2

rG2  1.050 in m2a  0.0180 blob

r


mA  0.0368 blob

mB  m3b  m4

mB  0.0293 blob


 2  2  m   r ω  sin θ 


Fi 

2





r l

 cos 2  θ 



Fiy  3586 lbf

A

2

2

Fix  Fiy

Fi  15967 lbf

Ti21  mB r  ω  sin θ   2

2

Fix  15559 lbf

r

 2 l

 cos θ 

3 r 2 l

at


 cos 2  θ



Ti21  12630 in lbf





Units:


Given:

Conrod length:

1

2

l  15 in

mass:

r3a  0.25


Crank length:

mass: m2  0.050  blob r2a  0.4


Solution: 1.

m3  0.025  blob

Piston mass:

m4  0.015  blob

Crank speed:

ω  4000 rpm



la  r3a l

la  3.750 in


lb  l  la

lb  11.250 in

lp  la

Let

m3a  m3

m3b  m3

2.

m3a  0.0188 blob

lp  lb lp

m3b  0.00625 blob

lp  lb

rG2  r2a r

m2a  m2

Mass:

4.

lb


3.

θ  18 deg

Crank angle:

rG2

rG2  1.600 in m2a  0.0200 blob

r


mA  0.0388 blob

mB  m3b  m4

mB  0.0213 blob


 2  2  m   r ω  sin θ 


Fi 

2





r l

 cos 2  θ 



Fiy  8404 lbf

A

2

2

Fix  Fiy

Fi  44075 lbf

Ti21  mB r  ω  sin θ   2

2

Fix  43267 lbf

r

 2 l

 cos θ 

3 r 2 l

at


 cos 2  θ



Ti21  25956 in lbf




The dimensions and mass properties of a crank, connecting rod, and piston are given below. Calculate the pin forces on the linkage.

Units:


Given:

Conrod length:

1

2

l  12 in

r3a  0.4


Crank length:

mass: m2  0.060  blob r2a  0.3


Solution: 1.

m3  0.020  blob

mass:

Piston mass:

m4  0.022  blob

Gas force:

Crank speed:

ω  2000 rpm

Crank angle: θ  45 deg



la  r3a l

la  4.800 in


lb  l  la

lb  7.200 in

Let

lp  la

m3a  m3

m3b  m3

2.

rG2

lp

m3b  0.00800 blob

lp  lb

rG2  1.050 in m2a  0.0180 blob

r


mA  0.0300 blob

mB  m3b  m4

mB  0.0300 blob

Calculate the conrod angle and acceleration of the piston using equations 13.16e and 13.3e, respectively. r sin θ   2   r  sin θ    l 1      l  

ϕ  atan

a B  r ω   cos θ  2



5.

m3a  0.0120 blob

lp  lb

rG2  r2a r

m2a  m2

Mass:

4.

lb


3.

Fg  300  lbf

r l

ϕ  11.902 deg

 cos 2  θ



a B  108560.1

sec

Determine the sidewall force F41 using equation 13.20. F41   m4  m3b  a B  Fg  tan ϕ

in

F41  623.2 lbf

2


6.


Determine the wrist pin force F34 using equation 13.21. F34x  Fg  m4 a B

F34x  2088.3 lbf

F34y  Fg   m4  m3b  a B  tan ϕ

F34y  623.2 lbf

F34 

2

2

F34x  F34y

θ  atan2 F34x F34y 7.

F34  2179.3 lbf θ  163.384 deg

Determine the crankpin force F32 using equation 13.22. F32x  m3a r ω  cos θ   m3b  m4  a B  Fg

F32x  4259.5 lbf

F32y  m3a r ω  sin θ  Fg   m4  m3b  a B  tan ϕ

F32y  679.5 lbf

2 2

F32 

2

2

F32x  F32y

θ  atan2 F32x F32y 8.

F32  4313.4 lbf θ  9.064 deg

Determine the main pin force F21 using equations 13.19c and 13.22. F21x  m2a r ω  cos θ  F32x

F21x  6213.6 lbf

F21y  m2a r ω  sin θ  F32y

F21y  2633.6 lbf

2 2

F21 

2

2

F21x  F21y

θ  atan2 F21x F21y

F21  6748.7 lbf θ  22.969 deg





Units:


Given:

Conrod length:

1

2

l  12 in

r3a  0.4


Crank length:

mass: m2  0.050  blob r2a  0.4


Solution: 1.

m3  0.020  blob

mass:

Piston mass:

m4  0.019  blob

Gas force:

Crank speed:

ω  3000 rpm




la  r3a l

la  4.800 in


lb  l  la

lb  7.200 in

Let

lp  la

m3a  m3

m3b  m3

2.

rG2

lp

m3b  0.00800 blob

lp  lb

rG2  1.600 in m2a  0.0200 blob

r


mA  0.0320 blob

mB  m3b  m4

mB  0.0270 blob


ϕ  atan

a B  r ω   cos θ  2



5.

m3a  0.0120 blob

lp  lb

rG2  r2a r

m2a  m2

Mass:

4.

lb


3.

Fg  600  lbf

r l

ϕ  9.594 deg

 cos 2  θ



a B  407690.5

sec


in

F41  1759.2 lbf

2


6.



F34x  7146.1 lbf

F34y  Fg   m4  m3b  a B  tan ϕ

F34y  1759.2 lbf

F34 

2

2

F34x  F34y

θ  atan2 F34x F34y 7.

F34  7359.5 lbf θ  166.170 deg


F32x  14510.4 lbf


F32y  609.5 lbf

2 2

F32 

2

2

F32x  F32y

θ  atan2 F32x F32y 8.

F32  14523.2 lbf θ  2.405 deg


F21x  21348.2 lbf

F21y  m2a r ω  sin θ  F32y

F21y  4557.3 lbf

2 2

F21 

2

2

F21x  F21y

θ  atan2 F21x F21y

F21  21829.2 lbf θ  12.050 deg





Units:


Given:

Conrod length:

1

2

l  15 in

r3a  0.25


Crank length:

mass: m2  0.060  blob r2a  0.3


Solution: 1.

m3  0.025  blob

mass:

Piston mass:

m4  0.032  blob

Gas force:

Crank speed:

ω  2500 rpm




la  r3a l

la  3.750 in


lb  l  la

lb  11.250 in

Let

lp  la

m3a  m3

m3b  m3

2.

rG2

lp

m3b  0.00625 blob

lp  lb

rG2  1.050 in m2a  0.0180 blob

r


mA  0.0368 blob

mB  m3b  m4

mB  0.0383 blob


ϕ  atan

a B  r ω   cos θ  2



5.

m3a  0.0188 blob

lp  lb

rG2  r2a r

m2a  m2

Mass:

4.

lb


3.

Fg  900  lbf

r l

ϕ  5.446 deg

 cos 2  θ



a B  256600.5

sec


in

F41  849.9 lbf

2


6.



F34x  7311.2 lbf

F34y  Fg   m4  m3b  a B  tan ϕ

F34y  849.9 lbf

F34 

2

2

F34x  F34y

θ  atan2 F34x F34y 7.

F34  7360.5 lbf θ  173.369 deg


F32x  13024.0 lbf


F32y  979.5 lbf

2 2

F32 

2

2

F32x  F32y

θ  atan2 F32x F32y 8.

F32  13060.8 lbf θ  4.301 deg


F21x  16968.6 lbf

F21y  m2a r ω  sin θ  F32y

F21y  2735.8 lbf

2 2

F21 

2

2

F21x  F21y

θ  atan2 F21x F21y

F21  17187.7 lbf θ  9.159 deg





Units:


Given:

Conrod length:

1

2

l  15 in

r3a  0.25


Crank length:

mass: m2  0.050  blob r2a  0.4


Solution: 1.

m3  0.025  blob

mass:

Piston mass:

m4  0.014  blob

Gas force:

Crank speed:

ω  4000 rpm




la  r3a l

la  3.750 in


lb  l  la

lb  11.250 in

Let

lp  la

m3a  m3

m3b  m3

2.

rG2

lp

m3b  0.00625 blob

lp  lb

rG2  1.600 in m2a  0.0200 blob

r


mA  0.0388 blob

mB  m3b  m4

mB  0.0203 blob


ϕ  atan

a B  r ω   cos θ  2



5.

m3a  0.0188 blob

lp  lb

rG2  r2a r

m2a  m2

Mass:

4.

lb


3.

Fg  1200 lbf

r l

ϕ  4.727 deg

 cos 2  θ



a B  818901.3

sec


in

F41  1271.9 lbf

2


6.



F34x  10264.6 lbf

F34y  Fg   m4  m3b  a B  tan ϕ

F34y  1271.9 lbf

F34 

2

2

F34x  F34y

θ  atan2 F34x F34y 7.

F34  10343.1 lbf θ  172.936 deg


F32x  27898.2 lbf


F32y  2794.6 lbf

2 2

F32 

2

2

F32x  F32y

θ  atan2 F32x F32y 8.

F32  28037.8 lbf θ  5.720 deg


F21x  41247.9 lbf

F21y  m2a r ω  sin θ  F32y

F21y  7132.2 lbf

2 2

F21 

2

2

F21x  F21y

θ  atan2 F21x F21y

F21  41860.0 lbf θ  9.810 deg




The dimensions and mass properties of a crank, connecting rod, and piston are given below. a. Exactly balance the crank and recalculate the inertia force. b. Overbalance the crank with approximately two-thirds of the mass at the wrist pin placed at radius -r on the crank and recalculate the inertia force. c. Compare these results to those for the unbalanced crank.

Units:


Given:

Conrod length:

1

2

l  12 in

mass:


Crank length:

1.

r3a  0.4

mass: m2  0.060  blob


Solution:

m3  0.020  blob

Piston mass:

m4  0.012  blob

Crank speed:

ω  2000 rpm

r2a  0.3

Crank angle:



la  r3a l

la  4.800 in


lb  l  la

lb  7.200 in

lp  la

Let

m3a  m3

m3b  m3

2.

m2a  m2

Mass:

4.

lb

lp

m3b  0.00800 blob

lp  lb

rG2  r2a r rG2

rG2  1.050 in m2a  0.0180 blob

r


mA  0.0300 blob

mB  m3b  m4

mB  0.0200 blob

Calculate the inertia force for an exactly balanced crank using equations 13.14d and 13.15d. Fix  mB r ω   cos θ  2





r l

 cos 2  θ 

Fix  2171 lbf



Fiy  0  lbf Fib  5.

m3a  0.0120 blob

lp  lb


3.

θ  45 deg

Fiy  0 lbf 2

2

Fix  Fiy

Fib  2171 lbf

at


Calculate the inertia force for an overbalanced crank using equations 13.14d and 13.15d. mp 

mB 3

3

mp  6.6667  10

blob







Fix  mA r ω  cos θ  mB r ω   cos θ  2



2



  mA  mp  r ω  cos θ



2









r l

 cos 2  θ  

Fiy  mA r ω  sin θ   mA  mp  r ω  sin θ Fiob 

6.

2

2

2

Fix  Fiy

Fiob  1618 lbf

2

at





Fix  1447 lbf

Fiy  724 lbf atan2 Fix Fiy  26.565 deg

Compare the results to those for an unbalanced crank. From Problem 13-11, the inertia force for the unbalanced crank is Fi  6330 lbf . The percent differences for the balanced and overbalanced cranks are:

Exactly balanced:

Overbalanced:

Δ 

Δ 

Fib  Fi Fi Fiob  Fi Fi

Δ  65.7 %

Δ  74.4 %





Units:


Given:

Conrod length:

1

2

l  12 in

mass:


Crank length:

1.

r3a  0.4

mass: m2  0.050  blob


Solution:

m3  0.020  blob

Piston mass:

m4  0.019  blob

Crank speed:

ω  3000 rpm

r2a  0.4

Crank angle:



la  r3a l

la  4.800 in


lb  l  la

lb  7.200 in

Let

lp  la

m3a  m3

m3b  m3

2.

m2a  m2

Mass:

4.

lb

lp

m3b  0.00800 blob

lp  lb


rG2  1.600 in m2a  0.0200 blob

r


mA  0.0320 blob

mB  m3b  m4

mB  0.0270 blob






r l

 cos 2  θ 

Fix  11008 lbf



Fiy  0  lbf Fib  5.

m3a  0.0120 blob

lp  lb


3.

θ  30 deg

Fiy  0 lbf 2

2

Fix  Fiy

Fib  11008 lbf

at


Calculate the inertia force for an overbalanced crank using equations 13.14d and 13.15d.



mB

mp 

3

mp  9.0000  10

3





blob




2



  mA  mp  r ω  cos θ



2









r l

 cos 2  θ  


6.

2

2

2

Fix  Fiy

Fiob  8127 lbf

2

at





Fix  7931 lbf


Compare the results to those for an unbalanced crank. From Problem 13-12, the inertia force for the unbalanced crank is Fi  22839  lbf . The percent differences for the balanced and overbalanced cranks are:

Exactly balanced:

Overbalanced:

Δ 

Δ 


Δ  51.8 %

Δ  64.4 %





Units:


Given:

Conrod length:

1

2

l  15 in

mass:


Crank length:

1.

r3a  0.25

mass: m2  0.060  blob


Solution:

m3  0.025  blob

Piston mass:

m4  0.023  blob

Crank speed:

ω  2500 rpm

r2a  0.3

Crank angle:



la  r3a l

la  3.750 in


lb  l  la

lb  11.250 in

lp  la

Let

m3a  m3

m3b  m3

2.

m2a  m2

Mass:

4.

lb

lp

m3b  0.00625 blob

lp  lb


rG2  1.050 in m2a  0.0180 blob

r


mA  0.0368 blob

mB  m3b  m4

mB  0.0293 blob






r l

 cos 2  θ 

Fix  7506 lbf



Fiy  0  lbf Fib  5.

m3a  0.0188 blob

lp  lb


3.

θ  24 deg

Fiy  0 lbf 2

2

Fix  Fiy

Fib  7506 lbf

at



mB 3

3

mp  9.7500  10

blob










2



  mA  mp  r ω  cos θ



2









r l

 cos 2  θ  


6.

2

2

2

Fix  Fiy

Fiob  5453 lbf

2

at





Fix  5369 lbf



Exactly balanced:

Overbalanced:

Δ 

Δ 


Δ  53.0 %

Δ  65.9 %





Units:


Given:

Conrod length:

1

2

l  15 in

mass:


Crank length:

1.

r3a  0.25

mass: m2  0.050  blob


Solution:

m3  0.025  blob

Piston mass:

m4  0.015  blob

Crank speed:

ω  4000 rpm

r2a  0.4

Crank angle:



la  r3a l

la  3.750 in


lb  l  la

lb  11.250 in

lp  la

Let

m3a  m3

m3b  m3

2.

m2a  m2

Mass:

4.

lb

lp

m3b  0.00625 blob

lp  lb


rG2  1.600 in m2a  0.0200 blob

r


mA  0.0388 blob

mB  m3b  m4

mB  0.0213 blob






r l

 cos 2  θ 

Fix  17402 lbf



Fiy  0  lbf Fib  5.

m3a  0.0188 blob

lp  lb


3.

θ  18 deg

Fiy  0 lbf 2

2

Fix  Fiy

Fib  17402 lbf

at



mB 3

3

mp  7.0833  10

blob










2



  mA  mp  r ω  cos θ



2









r l

 cos 2  θ  


6.

2

2

2

Fix  Fiy

2

Fiob  12766 lbf

at





Fix  12674 lbf



Exactly balanced:

Overbalanced:

Δ 

Δ 


Δ  60.5 %

Δ  71.0 %




Combine the necessary equations to develop expressions that show how each of these dynamic parameters varies as a function of the crank/conrod ratio alone: a. Piston acceleration. b. Inertia force. c. Inertia torque. d. Pin forces. Plot the functions. Check your conclusions with program ENGINE.

Solution:


1.

Note that the stroke volume V of the cylinder must be held at some constant value while varying the r/l ratio in order to isolate the effects of that parameter. We must hold all other factors constant for this analysis as well, such as all masses, crank angle, and bore/stroke ratio. We will assume that the crank radius r is held constant as we vary the r/l ratio in order to keep the stroke volume constant. Note that we can vary the r/l ratio independently of the crank radius r since for any r we can choose a value of l to maintain r/l constant.

2.

(Part a.) The expression for the piston acceleration is shown in equation a. Note that the r/l ratio only appears in the second harmonic term. If we allow r/l to become very small (long conrod), then this term diminishes toward zero, leaving a pure harmonic primary component. Thus, as the conrod length approaches infinity, the acceleration approaches a pure cosine. The r/l term acts as a distortion factor that changes the shape of this harmonic function in a way that increases its peak negative value. The optimum value for r/l is zero. The longer the conrod, the lower the peak acceleration. a p  r ω   cos ω t  2



3.

r l

 cos 2  ω t 

(a)



(Part b.) The expression for inertia force is shown in equation b. Note that only the x-component contains an r/l term and it is within the piston acceleration term. Thus the effects of the r/l ratio on inertia force are the same as those described for piston acceleration above. The optimum value for r/l is then zero. The longer the conrod, the lower the peak inertia force in the x-direction.





Fix  mA r ω  cos ω t  mB r ω  cos θ 



2

Fiy  mA r ω  sin ω t 4.

2

2



l

 cos2   ω t





(b)

(Part c.) The expression for inertia torque is shown in equation c. Note that only the primary and tertiary components contain an r/l term. Thus, the effects of the r/l ratio on inertia torque are similar to those described for inertia force above. The dominant term in this equation is the second harmonic as it has the largest possible coefficient of 1. If the r/l ratio were zero (infinite conrod) then the inertia torque would be a pure harmonic of twice crank frequency. A nonzero r/l adds distortion to this curve in the form of the 1st and 3rd harmonics. The optimum value for r/l is zero. The longer the conrod, the lower the peak inertia torque. Ti21  mB r  ω  sin ω t   2

2

r

 2 l

5.



r

 cos ω t 

3 r 2 l

 cos2   ω t



(c)

(Part d.) The equation for the wrist pin force is shown in equation d. It contains both the acceleration of the piston (e) and tan (f). These are substituted into d to get g. This expression shows that the wrist pin force is directly proportional to both r/l and the sum of that ratio squared to form the fourth power as shown in h. The optimum value of r/l is then zero. The longer the conrod, the lower the wrist pin force. F34x  Fg  m4 a B F34y  Fg   m4  m3b  a B  tan ϕ

(d)



a B  r ω  cos ω t  2



r l

 cos2   ω t

(e)



r sin ω t   2  r    l 1    sin ω t   l  

ϕ  atan

(f)

F34x  Fg  m4 r ω  cos ω t  2





   

r l

 cos2   ω t



F34y  Fg   m4  m3b  r ω  cos ω t 

F34x

F34y

6.

2





(g)

r  cos2   ω t    l   l r

 

sin ω t

 2 r 1    sin ω t  l  

r

proportional to

l

(h)

 r2 r4      l2 l4   

proportional to

(Part d.) The equation for the crankpin force is shown in equation i. It contains both the acceleration of the piston (e) and tan (f). These are substituted into i to get j. This expression shows that the crankpin force is directly proportional to both r/l and the sum of that ratio squared and to the fourth power as shown in k. The optimum value of r/l is then zero. The longer the conrod, the lower the crankpin force. F32x  m3a r ω  cos ω t   m3b  m4  a B  Fg 2

(i) F32y  m3a r ω  sin ω t  Fg   m4  m3b  a B  tan ϕ 2

F32x  m3a r ω  cos ω t   m3b  m4  r ω  cos ω t  2

2





F32y  m3a r ω  sin ω t 

r l

 cos2   ω t  Fg



2

 Fg   m4  m3b  r ω  cos ω t  2







r l

r  cos2   ω t      l  

F32x

F32y

7.

proportional to

sin ω t

 2 r 1    sin ω t  l  

r l

proportional to

(j)

 r2 r4      l2 l4   

(Part d.) The equation for the mainpin force is shown in equation l. It is equal to the crankpin force just analyzed above plus another term. Note that the second term does not contain r/l, so the mainpin force has the same dependence on r/l as does the crankpin force.

(k)



F21x  m2a r ω  cos ω t  F32x 2

(l) F21y  m2a r ω  sin ω t  F32y 2

F21x

F21y

proportional to

r l

proportional to

 r2 r4      l2 l4   

(m)




Combine the necessary equations to develop expressions that show how each of these dynamic parameters varies as a function of the bore/stroke ratio alone: a. Gas force. b. Gas torque. c. Inertia force. d. Inertia torque. e. Pin forces. Plot the functions. Check your conclusions with program ENGINE.

Solution:


1.

The stroke volume V of the cylinder must be held at some constant value as shown in equation a while varying the B/S ratio in order to isolate the effects of that parameter. This relationship is rearranged to introduce the bore/stroke ratio B/S in equation b. We must hold all other factors constant for this analysis as well, such as all masses, crank angle, and crank/conrod ratio. Note that we can vary the B/S ratio independently of the crank/ conrod ratio r/l since for any r we can choose a value of l to maintain r/l constant. V 

but,

2.

2

B 

π 4

2

 B  S = constant

2

 B   S2   S

π

V 

so,

(a)

4

 

B

2

3  S S

(b)

(Part a.) The gas force equation is shown in equation c and the expression for (B/S)2 from b is substituted in equation c. Expressions b and c are each then solved for S and combined in d. Gas force Fg is proportional to B/S as shown in e. Any increase in the B/S ratio will increase the gas force and vice versa. Fg 

π 4

 pg  B

2

2

B 2 Fg   p g     S 4 S π

or

(c)

2

Fg  0.922635 p g  V  

B 

3

   S 

combine equations b and c to get:

so, for a given p g and V, Fg is proportional to (B/S)2/3. 3.

(d) (e)

(Part b.) The gas torque equation is shown in equation f. Expressions for its Fg and r terms are shown in equation g. These are substituted into f and shown as h. The first expression in parentheses is the stroke volume V, which is constant as shown in i. Thus, as shown in j, the gas torque is independent of B/S. Tg21  Fg r sin ω t   1 



r l

 cos ω t

π 2 Fg    p g  B 4



r 

S

π 4

(g)

2

1 r π 2  Tg21    p g    B  s  sin ω t   1   cos ω t  2 l 4    V 

(f)

2

 B  S = constant

1 r Tg21    p g  V  sin ω t   1   cos ω t 2 l  

(h)

(i) (j)


4.


(Part c.) The expression for inertia force is shown in equation k with the expression S = 2r substituted and the S factored out. These are simplified by grouping all other factors as parameters K1, K2, and K3 and the magnitude of the force found. The dimensionless parameter of B/S ratio is introduced in l. The stroke volume constraint is then introduced, combined with l and substituted in k to get the nonlinear inverse relationship of inertia force to the B/S ratio shown in m.

 ω2

Fix  S  

 mA cos ω t  mB  cos θ 

2 



 mA 2    ω  sin ω t   2 

=

Fiy  S  

2

B S  B   S

l

1



 cos 2  ω t 

B Fi  K3 B   S

2

π 4



2

K1  K2 = S  K3

(l)

1

2

B S =

π 4

 B   3

B

 S

1

(k)

1

introduce volume constraint, V=

S  K1

=



S  K2

S K12  S K2 2 =  S

2

Fix  Fiy =

Fi =

r

1

1

3

B=

 4  V   B  = 1.083852 V 3   B       S  π  S 

3

substitute this into the force equation, 

1

Fi  1.083852 K3 V   3

5.

B

2 3

 S

(m)

(Part d.) The inertia torque equation is shown simplified in equation n. Equation l is substituted into n to show the relationship of inertia torque to the B/S ratio in o. 2  2  mB ω

Ti21  S  

 8

 

r

 2 l

 sin ω t  sin 2  ω t 

3 r 2 l

 

 sin 3  ω t 

2

S  K4

=

(n)

2



Ti21 = K4 B 

 1

2  B   = K4 B     S  S

6.

B

2

2

3

4 B B = K4   V        π  S   S 

2

(o)

(Part e.) The equation for the wrist pin force is shown in equation p. It contains both the gas force (c) and the acceleration of the piston (q). These are substituted along with l into p to get r. This expression shows that the wrist pin force is directly proportional to both B2 and S as shown in s. This means that there may be an optimum value of B/S, which will minimize the wrist pin force. Note that the tan term contains only r/l terms so is constant in this analysis. F34x  Fg  m4 a B

(p)

F34y  Fg   m4  m3b  a B  tan ϕ a B  r ω  cos ω t  2



r l

 cos2   ω t



(q)



π S 2 r 2 F34x    p g  B  m4   ω  cos ω t   cos2   ω t 4 l  2  

(r)

S 2 r 2  π F34y    p g  B   m4  m3b    ω  cos ω t   cos2   ω t  tan  ϕ l  4  2   F34 7.

is proportional to

2

B S

(s)

(Part e.) The equation for the crankpin force is shown in equation t. It contains both the gas force Fg (c) and the acceleration of the piston (q). These are substituted along with l into t to get u. This expression shows that the crankpin force is directly proportional to both B2 and S as shown in v. This means that there may be an optimum value of B/S, which will minimize the crankpin force. Note that the tan term contains only r/l terms so is constant in this analysis. F32x  m3a r ω  cos ω t   m3b  m4  a B  Fg 2

 sin ω t  Fg   m4  m3b  a B  tan ϕ

2

F32y  m3a r ω

(t)

S 2 S 2 r π 2 F32x  m3a  ω  cos ω t   m3b  m4    ω  cos ω t   cos2   ω t   p g  B l  2  2   4 S 2 F32y  m3a  ω  sin ω t  2 π S 2 r 2      p g  B   m4  m3b    ω  cos ω t   cos2   ω t  tan  ϕ l  4 2  

8.

F32x

proportional to

F32y

proportional to

(u)

2

B S B

(v)

2

(Part e.) The equation for the mainpin force is shown in equation w. It is equal to the crankpin force just analyzed above plus another term. The result from u is substituted along with l into w to get x. This expression shows that the crankpin force is directly proportional to both B2 and S as shown in x. This means that there may be an optimum value of B/S, which will minimize the mainpin force. S 2 F21x  m2a  ω  cos ω t  F32x 2

(w)

S 2 F21y  m2a  ω  sin ω t  F32y 2 2

F21x

proportional to

B S

F21y

proportional to

B S

2

(x)




Develop an expression to determine the optimum bore/stroke ratio to minimize the wrist pin force. Plot the function.

Units:


1

2

Assumptions: Use the data from Problem 13-15 for the unvarying parameters. l  12 in

Conrod length:

r3a  0.4


Crank length:

mass: m2  0.060  blob r2a  0.3

Distance to CG (as fraction of r): Piston mass:

m4  0.022  blob

Gas pressure:

p g  300  psi

Crank speed:

ω  2000 rpm

Crank angle:

θ  45 deg

R/L ratio:

RoverL  0.2917

3

V  1  in

Stroke volume: Solution: 1.

m3  0.020  blob

mass:

See Problem 13-15 and Mathcad file P1325.


la  r3a l

la  4.800 in


lb  l  la

lb  7.200 in

Let

lp  la

m3a  m3

m3b  m3 2.

lb

m3a  0.0120 blob

lp  lb lp

m3b  0.00800 blob

lp  lb

Determine the stroke and crank radius as functions of B/S. 1

S ( BoverS ) 

Stroke as function of B/S:

3

   BoverS  π 

RoverL sin θ    1   RoverL sin θ  2

3

ϕ  11.903 deg

a B( BoverS )  r( BoverS )  ω   cos θ  RoverL cos 2  θ  2

Write an expression for the gas force as a function of the B/S ratio. Fg( BoverS ) 

6.

2

Calculate the conrod angle and acceleration of the piston using equations 13.16e and 13.3e, respectively. ϕ  atan

5.



r( BoverS )  0.5 S ( BoverS )

Crank length as a function of stroke: 4.

 4 V 

π 4

2

 p g  BoverS  S ( BoverS )

2

Determine the wrist pin force F34 using equation 13.21. F34x( BoverS )  Fg( BoverS )  m4 a B( BoverS )



F34y( BoverS )  Fg( BoverS )   m4  m3b  a B( BoverS )  tan ϕ F34( BoverS )  7.

2

F34x( BoverS )  F34y( BoverS )

2

Plot the wrist pin force F34 as a function of B/S. BoverS  0.1 0.2  10 WRIST PIN FORCE vs BORE/STROKE RATIO

3

2 10

1.3

3

Wrist Pin Force, lb

1.5 10

F34( BoverS)

3

1 10

lbf

500

0

0

2

4

6

8

10

BoverS B/S, dimensionless

8.

The graph shows that there is an optimum value of B/S that will minimize the wrist pin force, which in this case, is about B/S = 1.3. However, the optimum value is dependent on the other design choices (especially peak gas pressure) and can vary considerably from this value. The bore, stroke, and crank radius that would result from choosing this value are: S opt  S ( 1.3) Bopt 

lopt 

4 V π S opt ropt RoverL

S opt  0.910 in Bopt  1.183 in

lopt  1.560 in

This is probably not a practical design.

ropt  r( 1.3)

ropt  0.455 in




Use program ENGINE, your own computer program, or an equation solver to calculate the maximum value and the polar plot shape of the force on the main pin of a one-cubic-inch displacement, single-cylinder engine with bore = 1.12838 in for the following situations. a. Piston, conrod and crank masses = 0. b. Piston mass = 1 blob, conrod and crank masses = 0. c. Conrod mass = 1 blob, piston and crank masses = 0. d. Crank mass = 1 blob, conrod and piston masses = 0. Place the CG of the crank at 0.5r and the conrod at 0.33l. Compare and explain the differences in the main pin force under these different conditions with reference to the governing equations.

Assumptions: 1. 2. 3. 4. 5. Solution: 1.

Peak cylinder pressure is 600 psi. Link length ratio is L/R = 4. Main pin dia is 2.00 in. Crankpin dia is 1.500 in. Friction coefficient is zero.


Enter the above data and the masses for part a into program ENGINE to determine the maximum value of the force on the main pin and to get the polar plot of this force. a.

m2  0

m3  0

m4  0

F21max  604.3  lbf

Polar plot of main pin force for part a.

2.

Enter the above data and the masses for part b into program ENGINE to determine the maximum value of the force on the main pin and to get the polar plot of this force. b.

m2  0

m3  0

m4  1

F21max  79004  lbf



Polar plot of main pin force for part b.

3.

Enter the above data and the masses for part c into program ENGINE to determine the maximum value of the force on the main pin and to get the polar plot of this force. c.

4.

m2  0

m3  1

m4  0

F21max  68417  lbf

Enter the above data and the masses for part d into program ENGINE to determine the maximum value of the force on the main pin and to get the polar plot of this force. d.

m2  1

m3  0

m4  0

F21max  31696  lbf


Polar plot of main pin force for part c.

Polar plot of main pin force for part d.





Use program ENGINE, your own computer program, or an equation solver to calculate the maximum value and the polar plot shape of the force on the crankpin of a one-cubic-inch displacement, single-cylinder engine with bore = 1.12838 in for the following situations. a. Piston, conrod and crank masses = 0. b. Piston mass = 1 blob, conrod and crank masses = 0. c. Conrod mass = 1 blob, piston and crank masses = 0. d. Crank mass = 1 blob, conrod and piston masses = 0. Place the CG of the crank at 0.5r and the conrod at 0.33l. Compare and explain the differences in the crankpin force under these different conditions with reference to the governing equations.




Enter the above data and the masses for part a into program ENGINE to determine the maximum value of the force on the crankpin and to get the polar plot of this force. a.

m2  0

m3  0

m4  0

F32max  602  lbf

Polar plot of crankpin force for part a.

2.

Enter the above data and the masses for part b into program ENGINE to determine the maximum value of the force on the crankpin and to get the polar plot of this force. b.

m2  0

m3  0

m4  1

F32max  79004  lbf



Polar plot of crankpin force for part b.

3.

Enter the above data and the masses for part c into program ENGINE to determine the maximum value of the force on the crankpin and to get the polar plot of this force. c.

m2  0

m3  1

m4  0

F32max  68417  lbf

See polar plot on next page.

4.

Enter the above data and the masses for part d into program ENGINE to determine the maximum value of the force on the crankpin and to get the polar plot of this force. d.

m2  1


m3  0

m4  0

F32max  602  lbf


Polar plot of crankpin force for part c.

Polar plot of crankpin force for part d.





Use program ENGINE, your own computer program, or an equation solver to calculate the maximum value and the polar plot shape of the force on the wrist pin of a one-cubic-inch displacement, single-cylinder engine with bore = 1.12838 in for the following situations. a. Piston, conrod and crank masses = 0. b. Piston mass = 1 blob, conrod and crank masses = 0. c. Conrod mass = 1 blob, piston and crank masses = 0. d. Crank mass = 1 blob, conrod and piston masses = 0. Place the CG of the crank at 0.5r and the conrod at 0.33l. Compare and explain the differences in the wrist pin force under these different conditions with reference to the governing equations.




Enter the above data and the masses for part a into program ENGINE to determine the maximum value of the force on the wrist pin and to get the polar plot of this force. a.

m2  0

m3  0

m4  0

F34max  602  lbf

Polar plot of wrist pin force for part a.

2.

Enter the above data and the masses for part b into program ENGINE to determine the maximum value of the force on the wrist pin and to get the polar plot of this force. b.

m2  0

m3  0

m4  1

F34max  79004  lbf



Polar plot of wrist pin force for part b.

3.

Enter the above data and the masses for part c into program ENGINE to determine the maximum value of the force on the wrist pin and to get the polar plot of this force. c.

m2  0

m3  1

m4  0

F34max  2914 lbf


4.

Enter the above data and the masses for part d into program ENGINE to determine the maximum value of the force on the wrist pin and to get the polar plot of this force. d.

m2  1


m3  0

m4  0

F34max  602  lbf


Polar plot of wrist pin force for part c.

Polar plot of wrist pin force for part d.





Use program ENGINE, your own computer program, or an equation solver to calculate the maximum value and the polar plot shape of the force on the main pin of a one-cubic-inch displacement, single-cylinder engine with bore = 1.12838 in for the following situations. a. Engine unbalanced. b. Crank exactly balanced against mass at crankpin. c. Crank optimally overbalanced against masses at crankpin and wrist pin. Piston, conrod, and crank masses = 1. Place the CG of the crank at 0.5r and the conrod at 0.33l. Compare and explain the differences in the main pin force under these different conditions with reference to the governing equations.




Enter the above data and the masses for part a into program ENGINE to determine the maximum value of the force on the main pin and to get the polar plot of this force. a.

Engine unbalanced

F21max  179658 lbf

Polar plot of main pin force for part a.

2.

Enter the above data and the masses for part b into program ENGINE to determine the maximum value of the force on the main pin and to get the polar plot of this force.


b.


Crank exactly balanced against mass at crankpin

F21max  105075 lbf

Polar plot of main pin force for part b.

3.

Enter the above data and the masses for part c into program ENGINE to determine the maximum value of the force on the main pin and to get the polar plot of this force. c.

Crank optimally overbalanced.

F21max  52619  lbf




Use program ENGINE, your own computer program, or an equation solver to calculate the maximum value and the polar plot shape of the force on the crankpin of a one-cubic-inch displacement, single-cylinder engine with bore = 1.12838 in for the following situations. a. Engine unbalanced. b. Crank exactly balanced against mass at crankpin. c. Crank optimally overbalanced against masses at crankpin and wrist pin. Piston, conrod, and crank masses = 1. Place the CG of the crank at 0.5r and the conrod at 0.33l. Compare and explain the differences in the crankpin force under these different conditions with reference to the governing equations.




Enter the above data and the masses for part a into program ENGINE to determine the maximum value of the force on the crankpin and to get the polar plot of this force. a.

Engine unbalanced

F21max  147421 lbf

Polar plot of crankpin force for part a.

2.

Enter the above data and the masses for part b into program ENGINE to determine the maximum value of the force on the crankpin and to get the polar plot of this force.


b.



F21max  147421 lbf

Polar plot of crankpin force for part b.

3.

Enter the above data and the masses for part c into program ENGINE to determine the maximum value of the force on the crankpin and to get the polar plot of this force. c.


F21max  147421 lbf




Use program ENGINE, your own computer program, or an equation solver to calculate the maximum value and the polar plot shape of the force on the wrist pin of a one-cubic-inch displacement, single-cylinder engine with bore = 1.12838 in for the following situations. a. Engine unbalanced. b. Crank exactly balanced against mass at crankpin. c. Crank optimally overbalanced against masses at crankpin and wrist pin. Piston, conrod, and crank masses = 1. Place the CG of the crank at 0.5r and the conrod at 0.33l. Compare and explain the differences in the wrist pin force under these different conditions with reference to the governing equations.




Enter the above data and the masses for part a into program ENGINE to determine the maximum value of the force on the wrist pin and to get the polar plot of this force. a.

Engine unbalanced

F21max  79004  lbf

Polar plot of wrist pin force for part a.

2.

Enter the above data and the masses for part b into program ENGINE to determine the maximum value of the force on the wrist pin and to get the polar plot of this force. b.


Polar plot of wrist pin force for part b.

F21max  79004  lbf


3.


Enter the above data and the masses for part c into program ENGINE to determine the maximum value of the force on the wrist pin and to get the polar plot of this force. c.


F21max  79004  lbf




Use program ENGINE, your own computer program, or an equation solver to calculate the maximum value and the polar plot shape of the force on the shaking force of a one-cubic-inch displacement, single-cylinder engine with bore = 1.12838 in for the following situations. a. Engine unbalanced. b. Crank exactly balanced against mass at crankpin. c. Crank optimally overbalanced against masses at crankpin and wrist pin. Piston, conrod, and crank masses = 1. Place the CG of the crank at 0.5r and the conrod at 0.33l. Compare and explain the differences in the shaking force force under these different conditions with reference to the governing equations.




Enter the above data and the masses for part a into program ENGINE to determine the maximum value of the shaking force and to get the polar plot of this force. a.

Engine unbalanced

F21max  179023 lbf

Polar plot of shaking force for part a.

2.

Enter the above data and the masses for part b into program ENGINE to determine the maximum value of the shaking force and to get the polar plot of this force. b.


Polar plot of shaking force for part b.

F21max  105075 lbf


3.


Enter the above data and the masses for part c into program ENGINE to determine the maximum value of the shaking force and to get the polar plot of this force. c. Crank optimally overbalanced. F21max  52617  lbf




Figure P13-1 shows a single-cylinder air compressor stopped at top dead center (TDC). There is a static pressure P = 100 psi trapped in the 3-in-bore cylinder. The entire assembly weighs 30 lb. Draw the necessary free-body diagrams to determine the forces at points A, B, and C, and the supports R1 and R2, which are symmetrically located about the piston centerline. Assume that the piston remains stationary.

Given:

Pressure: p  100  psi Bore: d  3.00 in

Solution:


1.

Calculate the gas force:

 π d 2    4 

Fg  p   2.

Fg  706.9 lbf

Draw the FBDs. There will be four FBDs: the piston, the conrod, the crank, and the housing or body. The piston will have two downward forces; the gas force Fg and the piston weight; and one upward force F34 acting at point C, whose magnitude is the sum of the gas force and the piston weight. The conrod will have two downward forces; F43 and the conrod weight; and one upward force F23 acting at point B, whose magnitude is the sum of the gas force and the weights of the piston and the conrod. The crank will have two downward forces; F32 and the crank weight; and one upward force F12 acting at point A, whose magnitude is the sum of the gas force and the weights of the piston, conrod, and crank. The housing will have two downward forces; F21 acting at A and the weight of the housing acting at its CG; and three upward forces; Fg acting at the top of the cylinder, and the reaction forces R1 and R2, whose magnitudes will be one-half the total weight of the assembly, which is 15 lb each. The gas force causes compression in piston, conrod, and crank; and tension in the housing. The net force at the supports is the total assembly weight.




Calculate and plot the position, velocity, and acceleration of a slider-crank linkage with r = 3, l = 12, and  = 200 rad/sec over one cycle using the exact solution and the approximate Fourier solution. Also, calculate and plot the percent difference between the exact and approximate solutions for acceleration.

Given:

Link lengths:

r  3

l  12   200  rad sec

Angular velocity: Solution: 1.

1


Define the crank angle over one cycle.   0  deg 1  deg  360  deg

2.

Calculate and plot the exact and approximate position using equations 13.1d and 13.3c. Exact:

x   r cos   l 1 

Approximate:

x'    l 

2

r

4 l

 r  sin    l 

 r  cos  



r 4 l

2

 cos 2  



Exact & Approx Position 16

Position

14

x( ) x'( )

12

10

8

0

45

90

135

180

225

270

315

 deg Crank Angle

3.

Calculate and plot the exact and approximate velocity using equations 13.1e and 13.3d. Exact:

   

v   r  sin  

r 2 l



sin 2  

 2 r 1    sin    l  

360



v'    r   sin  

Approximate:



r 2 l

 sin 2   



Exact & Approx Velocity

3

1 10

Velocity

500

v( ) 0

v'( )

 500

 1 10

3

0

45

90

135

180

225

270

315

360

 deg Crank Angle

4.

Calculate and plot the exact and approximate acceleration using equations 13.1f and 13.3e. 2

a'   r    cos   2

Approximate:

5.

    

a    r   cos  

Exact:





r l  1  2  cos  2

  r2sin4 

2

3

l2  r2sin2 r l

2

 cos 2   



Compare the results by calculating the error in the approximation as a percent of the exact. error  

a'   a   a  

   



Exact & Approx Acceleration

5

1 10

Acceleration

0 a( ) a' ( ) 5

 1 10

 2 10

5

0

45

90

135

180

225

270

315

360

 deg Crank Angle

Acceleration Error - % 50

Error

0

error( ) %

 50

 100

 150

0

45

90

135

180  deg Crank Angle

225

270

315

360




Calculate and plot the position, velocity, and acceleration of a slider-crank linkage with r = 3, l = 15, and  = 100 rad/sec over one cycle using the exact solution and the approximate Fourier solution. Also, calculate and plot the percent difference between the exact and approximate solutions for acceleration.

Given:

Link lengths:

r  3

l  15   100  rad sec

Angular velocity: Solution: 1.

1


Define the crank angle over one cycle.   0  deg 1  deg  360  deg

2.

Calculate and plot the exact and approximate position using equations 13.1d and 13.3c. Exact:

x   r cos   l 1 

Approximate:

x'    l 

2

r

4 l

 r  sin    l 

 r  cos  



r 4 l

2

 cos 2  



Exact & Approx Position 18

Position

16 x( ) x'( ) 14

12

0

45

90

135

180

225

270

315

 deg Crank Angle

3.

Calculate and plot the exact and approximate velocity using equations 13.1e and 13.3d. Exact:

   

v   r  sin  

r 2 l



sin 2  

 2 r 1    sin    l  

360


Approximate:


v'    r   sin  



r 2 l

 sin 2   



Exact & Approx Velocity 400

Velocity

200

v( ) 0

v'( )

 200

 400

0

45

90

135

180

225

270

315

360

 deg Crank Angle

4.

Calculate and plot the exact and approximate acceleration using equations 13.1f and 13.3e. 2 r l  a    r   cos  

Exact:

   

a'   r    cos   2

Approximate:

5.





 1  2  cos 

2

  r2sin4 

2

3

l2  r2sin2 r l

2

 cos 2   



Compare the results by calculating the error in the approximation as a percent of the exact. error  

a'   a   a  

   



Exact & Approx Acceleration

4

4 10

4

a( ) 0

a' ( )

4

 2 10

 4 10

4

0

45

90

135

180

225

270

315

360

315

360

 deg Crank Angle

Acceleration Error - % 200

150

100 Error

Acceleration

2 10

error( ) % 50

0

 50

0

45

90

135

180  deg Crank Angle

225

270





Given:

Link lengths:

r  3  in l  9  in


t  0.01 sec


2

    



r l  1  2  cos ω t 2

  r2sinωt4 

2

   

3

l2  r2sinωt2

2

in

a exact  12133.3

sec

2




3.

Time span:



2.

1

r l

 cos 2  ω t



in

a approx  12047.6

sec


error 


error  0.706 %

2





Given:

Link lengths:

r  3  in l  15 in


t  0.02 sec


2

    



r l  1  2  cos ω t 2

  r2sinωt4 

2

   

3

l2  r2sinωt2

2

in

a exact  16436.6

sec

2




3.

Time span:



2.

1

r l

 cos 2  ω t



a approx  16406.3

in sec


error 


error  0.185 %

2




The equation given below is an approximation of the gas force over 180 deg of crank angle. Using this equation with  = 15 deg and Fgmax = 1200 lb, calculate and plot the approximate gas torque for r = 4 in and l = 12 in. What is the total energy delivered by the gas force over the 180 deg of motion? What is the average power delivered if the crank rotates at a constant speed of 1500 rpm?

Given:

Link lengths: r  4  in l  12 in Maximum gas force:

Speed: n  1500 rpm

Fgmax  1200 lbf

  15 deg

Gas force parameter:

Gas force equation:

  π Fg1   Fgmax sin    2 Fg   if    Fg1  Fg2 



Solution: 1.

Fg2  

Fgmax 2

 

       π   

  1  cos π




Define the crank angle over 180 deg.   0  deg 1  deg  180  deg

2.

Using equation 13.8b, calculate and plot the approximate gas torque as a function of crank angle. Tg21   Fg   r sin    1 





π

Tg21  d

0

E  485.7 lbf  ft

GAS TORQUE

3

Integrate the gas torque over the range of crank motion (180 deg) to determine the energy delivered.  E   

4.

l

 cos 

5 10

3

4 10 Gas Torque - in-lb

3.

r

Tg21( ) in lbf

3

3 10

3

2 10

3

1 10

Divide the energy delivered by the time for the crank to travel 180 deg to get the average power.

0

0

20

40

60

80

100 120 140 160 180 

deg

t 

180  deg

Pavg 

n E t

t  0.020 s Pavg  44.2 hp





Given:

Link lengths:

r  3.75 in l  11 in B  2.5 in

Piston bore:

Peak pressure:

p g  1150 psi




2.

π 4

 pg  B

2

Fg  5645 lbf




r l

 cos θ



Tg21  5869 in lbf

Crank angle:

θ  12 deg





Given:

Link lengths:

r  3.75 in l  11 in B  2.5 in


π 4

 pg  B

2

Fg  5645 lbf




r l

 cos θ 



Tg21a  5868.91 in lbf


ϕ  atan

x  r cos θ  l 1  Tg21e  Fg tan ϕ  x

4.

Crank angle:


Tg21a  Fg r sin θ   1  3.

p g  1150 psi


Fg  2.

Peak pressure:

 r  sin θ   l 

ϕ  4.064 deg

2

x  14.640 in Tg21e  5872.61 in lbf


error 


error  0.0630 %

θ  12 deg





Given:

Link lengths:

r  4.12 in l  14.5 in B  2.25 in

Piston bore:

Peak pressure:

p g  1325 psi




2.

π 4

 pg  B

2

Fg  5268 lbf




r l

 cos θ



Tg21  4348 in lbf

Crank angle:

θ  9  deg





Given:

Link lengths:

r  4.12 in l  14.5 in B  2.25 in


π 4

 pg  B

2

Fg  5268 lbf




r l

 cos θ 



Tg21a  4348.38 in lbf


ϕ  atan

x  r cos θ  l 1  Tg21e  Fg tan ϕ  x

4.

Crank angle:


Tg21a  Fg r sin θ   1  3.

p g  1325 psi


Fg  2.

Peak pressure:

 r  sin θ   l 

ϕ  2.548 deg

2

x  18.555 in Tg21e  4349.32 in lbf


error 


error  0.0217 %

θ  9  deg




The dimensions and mass properties of a slider-crank linkage are given below. Determine the approximately dynamically equivalent two-mass lumped parameter model for this linkage with the masses placed at the crank and wrist pins.

Units:


Given:

Masses:

2

1

m2  0.045  blob

m3  0.120  blob

Crank CG location as fraction of its length:

m4  0.15 blob r2  0.4

Assumption:

Two-thirds of the conrod mass is placed at the crank pin and one-third at the wrist pin.

Solution:


1.

Determine the lumped masses for link 3 using equations 13.10a and 13.10b and the assumption as to mass distribution. Masses:

2.

2

m3b  m3

1

3

3

m3a  0.0800 blob

m3b  0.04000 blob

Determine the lumped mass at point A due to the crank alone using equation 13.11. Masses:

3.

m3a  m3

m2a  m2 r2

m2a  0.0180 blob

Determine the two-mass model with masses placed at the crank and wrist pins using equations 13.12. mA  m2a  m3a

mA  0.098 blob

mB  m3b  m4

mB  0.190 blob




The dimensions and mass properties of a connecting rod are given below. Calculate the sizes of two dynamically equivalent masses and the location of one if the other is placed at point B (see Figure 13-10).

Units:


Given:

Conrod length:

Solution: 1.

2

1

l  12.5 in

mass:

m3  0.120  blob 2


IG3  0.15 blob in

Distance to CG :

la  4.5 in


Determine the exact model using equations 13.9d and 13.9e. Distance from point B to CG:

lb  l  la


Masses:

mp  m3

mb  m3


lp 

lb  8.000 in IG3 m3 lb

lp  0.156 in

mp  0.118 blob

mb  0.00230 blob




The dimensions and mass properties of a slider-crank linkage are given below. Determine the approximately dynamically equivalent two-mass lumped parameter model for this linkage with the masses placed at the crank and wrist pins.

Units:


Given:

Masses:

2

1

m2  0.060  blob

m3  0.180  blob

Crank CG location as fraction of its length:

m4  0.160  blob r2  0.38

Assumption:

Two-thirds of the conrod mass is placed at the crank pin and one-third at the wrist pin.

Solution:


1.

Determine the lumped masses for link 3 using equations 13.10a and 13.10b and the assumption as to mass distribution. Masses:

2.

2

m3b  m3

1

3

3

m3a  0.1200 blob

m3b  0.06000 blob

Determine the lumped mass at point A due to the crank alone using equation 13.11. Masses:

3.

m3a  m3

m2a  m2 r2

m2a  0.0228 blob

Determine the two-mass model with masses at the crank and wrist pins using equations 13.12. mA  m2a  m3a

mA  0.143 blob

mB  m3b  m4

mB  0.220 blob




The dimensions and mass properties of a connecting rod are given below. Calculate the sizes of two dynamically equivalent masses and the location of one if the other is placed at point B (see Figure 13-10).

Units:


Given:

Conrod length:

Solution: 1.

2

1

l  10.4 in

mass:

m3  0.180  blob 2


IG3  0.12 blob in

Distance to CG :

la  4.16 in


Determine the exact model using equations 13.9d and 13.9e. Distance from point B to CG:

lb  l  la


Masses:

mp  m3

mb  m3


lp 

lb  6.240 in IG3 m3 lb

lp  0.107 in

mp  0.177 blob

mb  0.00303 blob





Units:


Given:

Conrod length:

1

2

l  12.5 in

r3a  0.36

Distance to CG (as fraction of l):

r  3.13 in mass: m2  0.045  blob

Crank length:

r2a  0.4


Solution: 1.

m3  0.120  blob

mass:

Piston mass:

m4  0.015  blob

Crank speed:

ω  1800 rpm



la  r3a l

la  4.500 in


lb  l  la

lb  8.000 in

lp  la

Let

m3a  m3

m3b  m3

2.

m3a  0.0768 blob

lp  lb lp

m3b  0.04320 blob

lp  lb

rG2  r2a r

m2a  m2

Mass:

4.

lb


3.

θ  30 deg

Crank angle:

rG2

rG2  1.252 in m2a  0.0180 blob

r


mA  0.0948 blob

mB  m3b  m4

mB  0.0582 blob


 2  2  m   r ω  sin θ 


Fi 

2





r l

 cos 2  θ 



Fiy  5271 lbf

A

2

2

Fix  Fiy

Fi  16415 lbf

Ti21  mB r  ω  sin θ   2

2

Fix  15546 lbf

r

 2 l

 cos θ 

3 r 2 l

at


 cos 2  θ



Ti21  11943 in lbf





Units:


Given:

Conrod length:

1

2

l  10.4 in

r3a  0.40


r  2.60 in mass: m2  0.060  blob

Crank length:

r2a  0.38


Solution: 1.

m3  0.180  blob

mass:

Piston mass:

m4  0.016  blob

Crank speed:

ω  1850 rpm



la  r3a l

la  4.160 in


lb  l  la

lb  6.240 in

lp  la

Let

m3a  m3

m3b  m3

2.

m3a  0.1080 blob

lp  lb lp

m3b  0.07200 blob

lp  lb

rG2  r2a r

m2a  m2

Mass:

4.

lb


3.

θ  20 deg

Crank angle:

rG2

rG2  0.988 in m2a  0.0228 blob

r


mA  0.1308 blob

mB  m3b  m4

mB  0.0880 blob


 2  2  m   r ω  sin θ 


Fi 

2





r l

 cos 2  θ 



Fiy  4365 lbf

A

2

2

Fix  Fiy

Fi  22143 lbf

Ti21  mB r  ω  sin θ   2

2

Fix  21708 lbf

r

 2 l

 cos θ 

3 r 2 l

at


 cos 2  θ



Ti21  10324 in lbf





Units:


Given:

Conrod length:

1

2

l  10.4 in

r3a  0.36


r  2.60 in mass: m2  0.045  blob

Crank length:

r2a  0.40


Solution: 1.

m3  0.120  blob

mass:

Piston mass:

m4  0.015  blob

Crank speed:

ω  2000 rpm



la  r3a l

la  3.744 in


lb  l  la

lb  6.656 in

lp  la

Let

m3a  m3

m3b  m3

2.

m3a  0.0768 blob

lp  lb lp

m3b  0.04320 blob

lp  lb

rG2  r2a r

m2a  m2

Mass:

4.

lb


3.

θ  25 deg

Crank angle:

rG2

rG2  1.040 in m2a  0.0180 blob

r


mA  0.0948 blob

mB  m3b  m4

mB  0.0582 blob


 2  2  m   r ω  sin θ 


Fi 

2





r l

 cos 2  θ 



Fiy  4569 lbf

A

2

2

Fix  Fiy

Fi  17489 lbf

Ti21  mB r  ω  sin θ   2

2

Fix  16881 lbf

r

 2 l

 cos θ 

3 r 2 l

at


 cos 2  θ



Ti21  9280 in lbf





Units:


Given:

Conrod length:

1

2

l  12.5 in

r3a  0.40


r  3.13 in mass: m2  0.060  blob

Crank length:

r2a  0.38


Solution: 1.

m3  0.180  blob

mass:

Piston mass:

m4  0.015  blob

Crank speed:

ω  1500 rpm



la  r3a l

la  5.000 in


lb  l  la

lb  7.500 in

lp  la

Let

m3a  m3

m3b  m3

2.

m3a  0.1080 blob

lp  lb lp

m3b  0.07200 blob

lp  lb

rG2  r2a r

m2a  m2

Mass:

4.

lb


3.

θ  22 deg

Crank angle:

rG2

rG2  1.189 in m2a  0.0228 blob

r


mA  0.1308 blob

mB  m3b  m4

mB  0.0870 blob


 2  2  m   r ω  sin θ 


Fi 

2





r l

 cos 2  θ 



Fiy  3784 lbf

A

2

2

Fix  Fiy

Fi  17227 lbf

Ti21  mB r  ω  sin θ   2

2

Fix  16806 lbf

r

 2 l

 cos θ 

3 r 2 l

at


 cos 2  θ



Ti21  10419 in lbf




The dimensions and mass properties of a crank, connecting rod, and piston are given below. Calculate the pin forces torque on the linkage.

Units:


Given:

Conrod length:

1

2

l  12.5 in

mass:

r3a  0.36


r  3.13 in mass: m2  0.045  blob

Crank length:

r2a  0.40


Solution: 1.

m3  0.120  blob

Piston mass:

m4  0.150  blob

Gas force:

Crank speed:

ω  1800 rpm




la  r3a l

la  4.500 in


lb  l  la

lb  8.000 in

Let

lp  la

m3a  m3

m3b  m3

2.

rG2

lp

m3b  0.04320 blob

lp  lb

rG2  1.252 in m2a  0.0180 blob

r


mA  0.0948 blob

mB  m3b  m4

mB  0.1932 blob


ϕ  atan

a B  r ω   cos θ  2



5.

m3a  0.0768 blob

lp  lb

rG2  r2a r

m2a  m2

Mass:

4.

lb


3.

Fg  600  lbf

r l

ϕ  7.192 deg

 cos 2  θ



a B  110234.9

sec


in

F41  2612 lbf

2


6.



F34x  15935 lbf

F34y  Fg   m4  m3b  a B  tan ϕ

F34y  2611.9 lbf

F34 

2

2

F34x  F34y

θ  atan2 F34x F34y 7.

F34  16147.9 lbf θ  170.692 deg


F32x  28094 lbf


F32y  1658.6 lbf

2 2

F32 

2

2

F32x  F32y

θ  atan2 F32x F32y 8.

F32  28143 lbf θ  3.379 deg


F21x  29828 lbf

F21y  m2a r ω  sin θ  F32y

F21y  2659.5 lbf

2 2

F21 

2

2

F21x  F21y

θ  atan2 F21x F21y

F21  29946 lbf θ  5.095 deg





Units:


Given:

Conrod length:

1

2

l  10.4 in

mass:

r3a  0.40


r  2.60 in mass: m2  0.060  blob

Crank length:

r2a  0.38


Solution: 1.

m3  0.180  blob

Piston mass:

m4  0.160  blob

Gas force:

Crank speed:

ω  1850 rpm




la  r3a l

la  4.160 in


lb  l  la

lb  6.240 in

Let

lp  la

m3a  m3

m3b  m3

2.

rG2

lp

m3b  0.07200 blob

lp  lb

rG2  0.988 in m2a  0.0228 blob

r


mA  0.1308 blob

mB  m3b  m4

mB  0.2320 blob


ϕ  atan

a B  r ω   cos θ  2



5.

m3a  0.1080 blob

lp  lb

rG2  r2a r

m2a  m2

Mass:

4.

lb


3.

Fg  600  lbf

r l

ϕ  4.905 deg

 cos 2  θ



a B  110386.2

sec


in

F41  2146 lbf

2


6.



F34x  17062 lbf

F34y  Fg   m4  m3b  a B  tan ϕ

F34y  2146.3 lbf

F34 

2

2

F34x  F34y

θ  atan2 F34x F34y 7.

F34  17196.3 lbf θ  172.830 deg


F32x  34913 lbf


F32y  1458.2 lbf

2 2

F32 

2

2

F32x  F32y

θ  atan2 F32x F32y 8.

F32  34943 lbf θ  2.392 deg


F21x  37004 lbf

F21y  m2a r ω  sin θ  F32y

F21y  2219.2 lbf

2 2

F21 

2

2

F21x  F21y

θ  atan2 F21x F21y

F21  37070 lbf θ  3.432 deg





Units:


Given:

Conrod length:

1

2

l  10.4 in

mass:

r3a  0.36


r  2.60 in mass: m2  0.045  blob

Crank length:

r2a  0.40


Solution: 1.

m3  0.120  blob

Piston mass:

m4  0.150  blob

Gas force:

Crank speed:

ω  2000 rpm




la  r3a l

la  3.744 in


lb  l  la

lb  6.656 in

Let

lp  la

m3a  m3

m3b  m3

2.

rG2

lp

m3b  0.04320 blob

lp  lb

rG2  1.040 in m2a  0.0180 blob

r


mA  0.0948 blob

mB  m3b  m4

mB  0.1932 blob


ϕ  atan

a B  r ω   cos θ  2



5.

m3a  0.0768 blob

lp  lb

rG2  r2a r

m2a  m2

Mass:

4.

lb


3.

Fg  350  lbf

r l

ϕ  6.065 deg

 cos 2  θ



a B  121690.6

sec


in

F41  2461 lbf

2


6.



F34x  17904 lbf

F34y  Fg   m4  m3b  a B  tan ϕ

F34y  2460.8 lbf

F34 

2

2

F34x  F34y

θ  atan2 F34x F34y 7.

F34  18071.9 lbf θ  172.174 deg


F32x  31099 lbf


F32y  1240.9 lbf

2 2

F32 

2

2

F32x  F32y

θ  atan2 F32x F32y 8.

F32  31124 lbf θ  2.285 deg


F21x  32959 lbf

F21y  m2a r ω  sin θ  F32y

F21y  2108.5 lbf

2 2

F21 

2

2

F21x  F21y

θ  atan2 F21x F21y

F21  33027 lbf θ  3.660 deg





Units:


Given:

Conrod length:

1

2

l  12.5 in

mass:

r3a  0.40


r  3.13 in mass: m2  0.060  blob

Crank length:

r2a  0.38


Solution: 1.

m3  0.180  blob

Piston mass:

m4  0.150  blob

Gas force:

Crank speed:

ω  1500 rpm




la  r3a l

la  5.000 in


lb  l  la

lb  7.500 in

Let

lp  la

m3a  m3

m3b  m3

2.

rG2

lp

m3b  0.07200 blob

lp  lb

rG2  1.189 in m2a  0.0228 blob

r


mA  0.1308 blob

mB  m3b  m4

mB  0.2220 blob


ϕ  atan

a B  r ω   cos θ  2



5.

m3a  0.1080 blob

lp  lb

rG2  r2a r

m2a  m2

Mass:

4.

lb


3.

Fg  550  lbf

r l

ϕ  5.382 deg

 cos 2  θ



a B  85516.9

sec


in

F41  1737 lbf

2


6.



F34x  12278 lbf

F34y  Fg   m4  m3b  a B  tan ϕ

F34y  1736.9 lbf

F34 

2

2

F34x  F34y

θ  atan2 F34x F34y 7.

F34  12399.8 lbf θ  171.948 deg


F32x  26168 lbf


F32y  1387.7 lbf

2 2

F32 

2

2

F32x  F32y

θ  atan2 F32x F32y 8.

F32  26205 lbf θ  3.035 deg


F21x  27801 lbf

F21y  m2a r ω  sin θ  F32y

F21y  2047.3 lbf

2 2

F21 

2

2

F21x  F21y

θ  atan2 F21x F21y

F21  27876 lbf θ  4.212 deg





Units:


Given:

Conrod length:

1

2

l  12.5 in

mass:



1.

r3a  0.36

r  3.13 in mass: m2  0.045  blob

Crank length:

Solution:

m3  0.120  blob

Piston mass:

m4  0.015  blob

Crank speed:

ω  1800 rpm

r2a  0.4

Crank angle:



la  r3a l

la  4.500 in


lb  l  la

lb  8.000 in

Let

lp  la

m3a  m3

m3b  m3

2.

m2a  m2

Mass:

4.

lb

m3a  0.0768 blob

lp  lb lp

m3b  0.04320 blob

lp  lb

Determine the statically equivalent crank model using equations 13.11. rG2  r2a r

Distance from O2 to CG:

3.

θ  30 deg

rG2

rG2  1.252 in m2a  0.0180 blob

r


mA  0.0948 blob

mB  m3b  m4

mB  0.0582 blob






r l

 cos 2  θ 

Fix  6416 lbf



Fiy  0  lbf Fib 

Fiy  0 lbf 2

2

Fix  Fiy

Fib  6416 lbf

at



5.


Calculate the inertia force for an overbalanced crank using equations 13.14d and 13.15d. mB

mp 

mp  0.0194 blob

3








2



  mA  mp  r ω  cos θ



2









r l

 cos 2  θ  


6.

2

2

2

Fix  Fiy

Fiob  4673 lbf

2

at





Fix  4547 lbf


Compare the results to those for an unbalanced crank. From Problem 13-47, the inertia force for the unbalanced crank is Fi  16415  lbf . The percent differences for the balanced and overbalanced cranks are: Exactly balanced:

Overbalanced:

Δ 

Δ 


Δ  60.9 %

Δ  71.5 %





Units:


Given:

Conrod length:

1

2

l  10.4 in

mass:



1.

r3a  0.40

r  2.60 in mass: m2  0.060  blob

Crank length:

Solution:

m3  0.180  blob

Piston mass:

m4  0.016  blob

Crank speed:

ω  1850 rpm

r2a  0.38

Crank angle:



la  r3a l

la  4.160 in


lb  l  la

lb  6.240 in

Let

lp  la

m3a  m3

m3b  m3

2.

m2a  m2

Mass:

4.

lb

m3a  0.1080 blob

lp  lb lp

m3b  0.07200 blob

lp  lb



3.

θ  20 deg

rG2

rG2  0.988 in m2a  0.0228 blob

r


mA  0.1308 blob

mB  m3b  m4

mB  0.0880 blob






r l

 cos 2  θ 

Fix  9714 lbf




Fiy  0 lbf 2

2

Fix  Fiy

Fib  9714 lbf

at



5.



mp 

mp  0.0293 blob

3








2



  mA  mp  r ω  cos θ



2









r l

 cos 2  θ  


6.

2

2

2

Fix  Fiy

Fiob  7092 lbf

2

at





Fix  7024 lbf



Exactly balanced:

Overbalanced:

Δ 

Δ 


Δ  56.1 %

Δ  68.0 %





Units:


Given:

Conrod length:

1

2

l  10.4 in

mass:



1.

r3a  0.36

r  2.60 in mass: m2  0.045  blob

Crank length:

Solution:

m3  0.120  blob

Piston mass:

m4  0.015  blob

Crank speed:

ω  2000 rpm

r2a  0.40

Crank angle:



la  r3a l

la  3.744 in


lb  l  la

lb  6.656 in

Let

lp  la

m3a  m3

m3b  m3

2.

m2a  m2

Mass:

4.

lb

m3a  0.0768 blob

lp  lb lp

m3b  0.04320 blob

lp  lb



3.

θ  25 deg

rG2

rG2  1.040 in m2a  0.0180 blob

r


mA  0.0948 blob

mB  m3b  m4

mB  0.0582 blob






r l

 cos 2  θ 

Fix  7082 lbf




Fiy  0 lbf 2

2

Fix  Fiy

Fib  7082 lbf

at



5.



mp 

mp  0.0194 blob

3








2



  mA  mp  r ω  cos θ



2









r l

 cos 2  θ  


6.

2

2

2

Fix  Fiy

Fiob  5163 lbf

2

at





Fix  5077 lbf



Overbalanced:

Δ 

Δ 


Δ  59.5 %

Δ  70.5 %





Units:


Given:

Conrod length:

1

2

l  12.5 in

mass:



1.

r3a  0.36

r  2.60 in mass: m2  0.045  blob

Crank length:

Solution:

m3  0.180  blob

Piston mass:

m4  0.015  blob

Crank speed:

ω  2000 rpm

r2a  0.40

Crank angle:



la  r3a l

la  4.500 in


lb  l  la

lb  8.000 in

Let

lp  la

m3a  m3

m3b  m3

2.

m2a  m2

Mass:

4.

lb

m3a  0.1152 blob

lp  lb lp

m3b  0.06480 blob

lp  lb



3.

θ  25 deg

rG2

rG2  1.040 in m2a  0.0180 blob

r


mA  0.1332 blob

mB  m3b  m4

mB  0.0798 blob






r l

 cos 2  θ 

Fix  9465 lbf




Fiy  0 lbf 2

2

Fix  Fiy

Fib  9465 lbf

at



5.



mp 

mp  0.0266 blob

3








2



  mA  mp  r ω  cos θ



2









r l

 cos 2  θ  


6.

2

2

2

Fix  Fiy

Fiob  6837 lbf

2

at





Fix  6716 lbf



Overbalanced:

Δ 

Δ 


Δ  45.1 %

Δ  60.3 %




Draw a crank phase diagram for a three-cylinder inline engine with a 0, 120, 240-deg crankshaft and determine all possible firing orders for a. Four-stroke cycle. b. Two-stroke cycle. Select the best arrangement to give even firing for each stroke cycle.

Solution:


1.

Even firing is possible with both two and four stroke designs. A 1, 2, 3 firing order is required for the twostroke and a 1, 3, 2 firing order for the four-stroke.

a. Four-stroke cycle

TDC

Phase Angle

Cyl.

0

1

TDC Exhaust

Compress

Power

Intake TDC

Exhaust -120

2

TDC Compress

Intake

Power TDC

Compress -240

3

b. Two-stroke cycle

Exhaust Power

0

180

Cyl.

0

1

Intake

360

TDC

Phase Angle

TDC

540

TDC

Power

Power TDC

-120

2

TDC

Power

Power TDC

-240

3

TDC

Power 0

720 Crank Angle

180

360

Power 540

720 Crank Angle




Draw a crank phase diagram for an inline four-cylinder inline engine with a 0, 90, 270, 180-deg crankshaft and determine all possible firing orders for a. Four-stroke cycle. b. Two-stroke cycle. Select the best arrangement to give even firing for each stroke cycle.

Solution:


1.

The two-stroke has even firing but even firing is not possible with this four stroke design. The best firing order is 1, 2, 4, 3 for the two-stroke and 1, 4, 3, 2 for the four-stroke.


TDC

Phase Angle

Cyl.

0

1

TDC Exhaust

Compress

Power

Intake TDC Compress

-haust -90

TDC

2

Ex-

Intake

Power TDC

Compress -270

3

TDC Exhaust

-take

Power

In-

TDC

TDC

Compress -180

Power 0

b. Two-stroke cycle

Exhaust

4

Intake

180

360

540

TDC

Phase Angle

Cyl.

0

1

TDC

Power

Power TDC

-90

2

720 Crank Angle

TDC

Power

Power TDC

-270

3

TDC

Power

Power

TDC

-180

4

TDC

Power 0

180

Power 360

540

720 Crank Angle




Draw a crank phase diagram for a 45-deg vee, four-cylinder engine with a 0, 90, 270, 180-deg crankshaft and determine all possible firing orders for a. Four-stroke cycle. b. Two-stroke cycle. Select the best arrangement to give even firing for each stroke cycle.

Solution:


1.

Neither stroke has even firing with this design. The best firing order is 1, 2, 4, 3 for the two-stroke and 1, 4, 3, 2 for the four-stroke.


TDC

Phase Angle

Cyl.

0

1

TDC Exhaust

Compress

Power

Intake TDC

TDC Compress

-90

2

Intake

Power TDC

TDC

Compress -270

3

Exhaust

Intake

Power TDC

Compress -180

4

Exhaust Power

0

b. Two-stroke cycle

TDC

180

Intake

360

540

TDC

Phase Angle

Cyl.

0

1

TDC

Power

Power TDC

TDC

-90

2

720 Crank Angle

Power

Power TDC

TDC

-270

3

Power

Power TDC

TDC

-180

4

Power 0

180

360

Power 540

720 Crank Angle




Draw a crank phase diagram for a 45-deg vee, two-cylinder engine with a 0, 90-deg crankshaft and determine all possible firing orders for a. Four-stroke cycle. b. Two-stroke cycle. Select the best arrangement to give even firing for each stroke cycle.

Solution:


1.

Neither stroke has even firing with this design. The only possible firing order is 1, 2 for either stroke-cycle.


TDC

Phase Angle

Cyl.

0

1

TDC Exhaust

Compress

Power

Intake TDC Compress

Exhaust -90

2

Intake 0

b. Two-stroke cycle

TDC

180

Power 360

TDC

Phase Angle

Cyl.

0

1

540

TDC

Power

Power TDC

-90

2

TDC

Power 0

720 Crank Angle

180

Power 360

540

720 Crank Angle




Draw a crank phase diagram for a 90-deg vee, two-cylinder engine with a 0, 180-deg crankshaft and determine all possible firing orders for a. Four-stroke cycle. b. Two-stroke cycle. Select the best arrangement to give even firing for each stroke cycle.

Solution:


1.

Neither stroke has even firing with this design. The only possible firing order is 1, 2 for either stroke-cycle.


TDC

Phase Angle

Cyl.

0

1

TDC Exhaust

Compress

Power

Intake TDC

TDC Compress -180

2

Intake 0

b. Two-stroke cycle

Exhaust Power

180

Intake

360

540

TDC

Phase Angle

Cyl.

0

1

720 Crank Angle

TDC

Power

Power TDC

-180

2

Power 0

TDC

Power 180

360

Power 540

720 Crank Angle




Draw a crank phase diagram for a 180-deg opposed, two-cylinder engine with a 0, 180-deg crankshaft and determine all possible firing orders for a. Four-stroke cycle. b. Two-stroke cycle. Select the best arrangement to give even firing for each stroke cycle.

Solution:


1.

Both stroke-cycles are even firing but the two-stroke looks like a four-stroke in its firing pattern since the second set of firing pulses is on top of the first set. The only possible firing order is 1, 2 for either stroke-cycle.

TDC

Phase Angle

Cyl.

0

1

TDC


Exhaust

Compress Intake

Power

TDC

TDC Compress -180

2

Exhaust Power

Intake 0

180

360

TDC

Phase Angle

Cyl.

0

1

540

720 Crank Angle

TDC

b. Two-stroke cycle Power

Power TDC

-180

2

Power 0

TDC

Power 180

360

540

720 Crank Angle




Draw a crank phase diagram for a 180-deg opposed, four-cylinder engine with a 0, 180, 180, 0-deg crankshaft and determine all possible firing orders for a. Four-stroke cycle. b. Two-stroke cycle. Select the best arrangement to give even firing for each stroke cycle.

Solution:


1.

Both stroke-cycles are even firing but the two-stroke looks like a four-stroke in its firing pattern since the second set of firing pulses is on top of the first set. The best firing order is 1, 2, 4, 3 for the two-stroke-cycle and 1, 2, 3, 4 for the four-stroke.

TDC

Phase Angle

Cyl.

0

1

TDC


Exhaust

Compress

Power

Intake TDC

TDC

Compress -180

Exhaust

2

Power

Intake TDC

TDC

Compress -180

3

Exhaust

Intake

Power TDC

TDC

Exhaust 0

Intake 0

b. Two-stroke cycle

Compress

4 180

Power 360

TDC

Phase Angle

Cyl.

0

1

540

TDC

Power

Power TDC

-180

720 Crank Angle

2

TDC

Power

Power TDC

-180

3

Power

TDC

Power TDC

0

4

TDC

Power 0

180

Power 360

540

720 Crank Angle




Calculate the shaking force, torque and moment balance conditions through the second harmonic for an inline three-cylinder engine with a 0, 120, 240 deg crankshaft.

Assumptions: Use File F14-12.eng for the cylinder data required on the cylinder input screen. Solution: 1.

See program ENGINE files P14082.eng, P14084.eng, and Mathcad file P1408.mcd.

Open file P14082.eng in program ENGINE for the two-stroke cycle or file P14084.eng for the four-stroke cycle. The problem setup data is input from the disk file. Click the Calculate button and then the Done button, which will return you to the Home screen. Click the Balance button, accept the values shown and click Calculate and then Done, which will return you to the Home screen. Click the Assemble button, accept the values shown and click Calculate and then Done, which will return you to the Home screen. View the calculated results by clicking on the Charts drop-down menu, from the main option bar, or use the Plot or Print buttons on the Home screen.




Calculate the shaking force, torque and moment balance conditions through the second harmonic for an inline four-cylinder engine with a 0, 90, 270, 180 deg crankshaft.







Calculate the shaking force, torque and moment balance conditions through the second harmonic for an 45-deg vee, four-cylinder engine with a 0, 90, 270, 180 deg crankshaft.







Calculate the shaking force, torque and moment balance conditions through the second harmonic for an 45-deg vee, two-cylinder engine with a 0, 90 deg crankshaft.







Calculate the shaking force, torque and moment balance conditions through the second harmonic for an 90-deg vee, two-cylinder engine with a 0, 180 deg crankshaft.







Calculate the shaking force, torque and moment balance conditions through the second harmonic for an 180-deg opposed, two-cylinder engine with a 0, 180 deg crankshaft.







Calculate the shaking force, torque and moment balance conditions through the second harmonic for an 180-deg opposed, four-cylinder engine with a 0, 180, 180, 0 deg crankshaft.







Derive expressions, in general terms, for the magnitude and angular location with respect to the first crank throw, of the mass-radius products needed on the crankshaft to balance the shaking moment in a 90-deg vee-eight engine with a 0, 90, 270, 180-deg crankshaft.

Solution:





Derive expressions, in general terms, for the magnitude and angular location with respect to the first crank throw, of the mass-radius products needed on the crankshaft to balance the shaking moment in a 90-deg vee-six engine with a 0, 240, 120-deg crankshaft.

Solution:





Derive expressions, in general terms, for the magnitude and angular location with respect to the first crank throw, of the mass-radius products needed on the crankshaft to balance the shaking moment in a 90-deg vee-four engine with a 0, 180-deg crankshaft.

Solution:





Design a pair of Nakamura balance shafts to cancel the shaking force and reduce oscillations in the engine shown in Figure 14-18.

Units:


Given:

From Figure 14-18 and file F14-18.eng:

1

2

S  3.537  in

Stroke:

mB  0.0116 blob

Effective wrist pin mass: Solution: 1.

2.

See Figure 14-18, program ENGINE file F14-18.eng, and Mathcad file P1418.

Calculate the crank radius and conrod length. Crank radius:

r  0.5 S

r  1.768 in

Conrod length:

l  LoverR r

l  6.190 in

Use equation 14.18 to calculate the mr product needed for the balance shafts. mrbal 

r 2 l

 mB r

wrbal  mrbal g 3.

LoverR  3.50

L/R ratio:

mrbal  2.931  10

3

blob in

wrbal  1.131 lbf  in

The specific location of the two balance shafts will depend on the specific geometry of the engine being balanced. The y-dimensions must be symmetric with respect to the engine longitudinal center plane (y2 = -y1), and the vertical dimensions should be such that x1 - x2 = 0.7 l.




Using program ENGINE, data in Table P14-1 and the crank phase diagram from Problem 14-1, determine the maximum force magnitudes on main pin, crank pin, wrist pin, and piston for a 2-stroke engine with even firing. Over balance the crank, if necessary, to bring the balanced shaking force down to at least half of the unbalanced value.

Solution:

See program ENGINE file P1419.eng and Mathcad file P1419.mcd.

1.

Open file P1419.eng in program ENGINE. The problem setup data is input from the disk file. Click the Calculate button and then the Done button, which will return you to the Home screen. Click the Balance button, accept the values shown and click Calculate and then Done, which will return you to the Home screen. Click the Assemble button, accept the values shown and click Calculate and then Done, which will return you to the Home screen. Click the Flywheel button, accept the values shown and click Calculate and then Done, which will return you to the Home screen. View the calculated results by clicking on the Charts drop-down menu, from the main option bar, or use the Plot or Print buttons on the Home screen.





Solution:


1.






Solution:


1.






Solution:


1.





A four-cylinder inline engine with a 0, 180, 180, 0-deg crankshaft has a stroke of S = 3.500 in, a conrod length to crank radius ratio of L/R = 3.750, and an effective wrist pin mass of mB = 0.0215 blob. Design a pair of Nakamura balance shafts to cancel the shaking force and reduce the oscillations in the engine.

Units:


Given:

Stroke: S  3.500  in

2

1

L/R ratio:


2.

mB  0.0215 blob



r  0.5 S

r  1.750 in

Conrod length:


l  6.563 in


r 2 l

 mB r


LoverR  3.750

mrbal  5.017  10

3

blob in






A four-cylinder inline engine with a 0, 180, 180, 0-deg crankshaft has a stroke of S = 2.750 in, a conrod length to crank radius ratio of L/R = 3.00, and an effective wrist pin mass of mB = 0.0125 blob. Design a pair of Nakamura balance shafts to cancel the shaking force and reduce the oscillations in the engine.

Units:


Given:

Stroke: S  2.750  in

2

1

L/R ratio:


2.

mB  0.0125 blob



r  0.5 S

r  1.375 in

Conrod length:


l  4.125 in


r 2 l

 mB r


LoverR  3.00

mrbal  2.865  10

3

blob in





PEOBLEM 15-1 Statement:

Design a double-dwell cam to meet the specifications given below. Size a return spring and specify its preload to maintain contact between cam and follower. Calculate and plot the dynamic force and torque. Repeat for a form-closed cam. Compare the dynamic force, torque, and natural frequency for the form-closed design and the force-closed design.

Given:

RISE

DWELL

FALL

DWELL

β  60 deg

β  120  deg

β  30 deg

β  150  deg

h 1  2.5 in

h 2  0  in

h 3  2.5 in

h 4  0  in

Mod. sine Cycle time: Solution: 1.

Cycloidal τ  4  sec

Damping ratio:

ζ  0.2 Follower mass:

mf  2.2

See DYNACAM files P1501a and P1501b.


2  π rad τ

ω  1.571 

rad sec

ω  15.000 rpm

2.

Load the noted files into program DYNACAM. P1501a is the force-closed solution and P1501b is the formclosed solution. In each case, after opening the file, click on the Size Cam button and then click Calculate to see the pressure angle and minimum radius of curvature. Then click the Done button and you will see the cam drawing. After clicking the Done button on that screen, click the Dynamics button on the Home screen. Click its Calculate button to see the fully developed Dynamics screen. Once this sequence of commands has been entered, you may see further data by clicking on the Print or Plot buttons on the home screen. Accept all default values, which have been brought in from the file. It is not necessary to go to the SVAJ screen as that data has been brought in from the files.

3.

The cam is sized with an 11.125-in prime circle radius and -2.64 in eccentricity, which gives balanced pressure angles of 29.8 deg and a minimum radius of curvature of 2.94 in. The roller radius is 1.00 in.

4.

For the force-closed case, a spring constant of 100 lb/in with a preload of 170 lb is chosen to keep the follower force positive and prevent jump. The maximum and minimum follower force, maximum driving torque, undamped and damped natural frequencies, and damping coefficient are shown on the Cam Dynamics screen, which is shown below on the next page. The system natural frequency is about 4.3 times the driving frequency. This ratio could be increased at the expense of larger follower force by increasing the spring constant.

5.

For the form-closed case, it is only necessary to define spring constant and preload values of zero. The "b" diskfile does this. The maximum and minimum follower force, maximum driving torque, undamped and damped natural frequencies, and damping coefficient for this case are shown on the second Cam Dynamics screen, which is shown below on the next page.

6.

Comparing these two cases shows that in the force-closed case, the peak follower force is greater and the driving torque excursions are lower than in the form-closed case. This gives the form-closed cam the advantage in terms of lower stresses in cam and follower but a larger drive motor. However, this advantage comes at the expense of a more difficult and costly cam to make, since two cam surfaces must be machined instead of one.






Given:

Design a double-dwell cam to meet the specifications given below. Size a return spring and specify its preload to maintain contact between cam and follower. Calculate and plot the dynamic force and torque. Repeat for a form-closed cam. Compare the dynamic force, torque, and natural frequency for the form-closed design and the force-closed design. RISE

DWELL

FALL

β  45 deg

β  150  deg

β  90 deg

β  75 deg

h 1  1.5 in

h 2  0  in

h 3  1.5 in

h 4  0  in

3-4-5 poly Cycle time: Solution: 1.

DWELL

4-5-6-7 poly τ  6  sec

Damping ratio: ζ  0.1

Follower mass:

mf  1.4



2  π rad τ

ω  1.047

rad sec

ω  10.000 rpm

2.

Load the noted files into program DYNACAM. Diskfile P1502a contains the force-closed solution and P1502b the form-closed solution. In each case, after opening the file, click on the Size Cam button and then click Calculate to see the pressure angle and minimum radius of curvature. Then click the Done button and you will see the cam drawing. After clicking the Done button on that screen, click the Dynamics button on the Home screen. Click its Calculate button to see the fully developed Dynamics screen. Once this sequence of commands has been entered, you may see further data by clicking on the Print or Plot buttons on the home screen. Accept all default values, which have been brought in from the file. It is not necessary to go to the SVAJ screen as that data has been brought in from the files.

3.

The cam is sized with an 4.25-in prime circle radius and 0.75 in eccentricity, which gives balanced pressure angles of 30 deg and a minimum radius of curvature of 1.75 in. The roller radius is 1.00 in.

4.


5.


6.

Comparing these two cases shows that in the force-closed case both the peak follower force and driving torque excursions are greater than in the form-closed case. This gives the form-closed cam the advantage in terms of lower stresses in cam and follower and a smaller drive motor. However, this advantage comes at the expense of a more difficult and costly cam to make, since two cam surfaces must be machined instead of one.






Given:

Design a single-dwell cam to meet the specifications given below. Size a return spring and specify its preload to maintain contact between cam and follower. Calculate and plot the dynamic force and torque. Repeat for a form-closed cam. Compare the dynamic force, torque, and natural frequency for the form-closed design and the force-closed design. RISE

FALL

β  60 deg

DWELL

β  90 deg

h 1  2  in

β  210  deg h 2  0  in

Seventh degree polynomial Cycle time: Solution: 1.

τ  5  sec


Follower mass:

mf  3.2



2  π rad τ

ω  1.257

rad sec

ω  12.000 rpm

2.


3.

The cam is sized with an 4.40-in prime circle radius and 0.22 in eccentricity, which gives balanced pressure angles of 29.9 deg and a minimum radius of curvature of 2.53 in. The roller radius is 1.00 in.

4.


5.


6.







Given:

Design a three-dwell cam to meet the specifications given below. Size a return spring and specify its preload to maintain contact between cam and follower. Calculate and plot the dynamic force and torque. Repeat for a form-closed cam. Compare the dynamic force, torque, and natural frequency for the form-closed design and the force-closed design. RISE/FALL

1.

FALL

DWELL

β  40 deg

β  100  deg

β  90 deg

β  20 deg

h 1  2.5 in

h 2  0.0 in

h 3  1.5 in

h 4  0.0 in

β  30 deg

β  80 deg

h 5  1.0 in

h 6  0.0 in


DWELL

τ  10 sec


Follower mass:

mf  0.4



2  π rad τ

ω  0.628

rad sec

ω  6.000 rpm

2.


3.

Modified trapezoidal acceleration was used for the rise and both falls. The cam is sized with an 8.875-in prime circle radius and 1.45 in eccentricity, which gives balanced pressure angles of 29.7 deg and a minimum radius of curvature of 3.59 in. The roller radius is 1.00 in.

4.


5.


6.







Given:

Design a four-dwell cam to meet the specifications given below. Size a return spring and specify its preload to maintain contact between cam and follower. Calculate and plot the dynamic force and torque. Repeat for a form-closed cam. Compare the dynamic force, torque, and natural frequency for the form-closed design and the force-closed design. RISE/FALL

FALL

DWELL

β  40 deg

β  100  deg

β  90 deg

β  20 deg

h 1  2.5 in

h 2  0.0 in

h 3  1.5 in

h 4  0.0 in

β  30 deg

β  40 deg

β  30 deg

β  10 deg

h 5  0.5 in

h 6  0.0 in

h 7  0.5 in

h 8  0.0 in


DWELL

τ  15 sec


Follower mass:

mf  1.25



2  π rad τ

ω  0.419

rad sec

ω  4.000 rpm

2.


3.

Modified trapezoidal acceleration was used for the rise and all falls. The cam is sized with an 7.500-in prime circle radius and 2.35 in eccentricity, which gives balanced pressure angles of 30 deg and a minimum radius of curvature of 3.01 in. The roller radius is 1.00 in.

4.


5.


6.







A mass-spring-damper system as shown in Figure 15-1b has the values shown in Table P15-1. Find the undamped and damped natural frequencies and the value of critical damping for the system in row a of the table.

Given:

M  1.2 kg

Solution:

See Figure 15-1b, Table P15-1, and Mathcad file P1506a.

1.

k

ωn  3.416 

M

rad sec

Calculate the damped natural frequency using equation 15.3c.

ωd  3.

1

c  1.1 N  sec m

Calculate the undamped natural frequency using equation 15.1d.

ωn  2.

1

k  14 N  m

k M



 c     2 M 

2

ωd  3.385 

rad sec

Calculate the critical damping factor using equation 15.2i. cc  2  M  ωn

cc  8.198 

N  sec m




Figure P15-1 shows a cam-follower system. The dimensions and other data are given below. Find the arm's mass, center of gravity location and mass moment of inertia about both its CG and arm pivot. Create a linear, one-DOF lumped mass model of the dynamic system referenced to the cam follower and calculate the cam-follower force for one revolution. The cam is a pure eccentric with eccentricity and speed given below. Ignore damping.

Units:


Given:

Arm dimensions:

2

1

Cutout dimensions:


a  2  in


a'  1.5 in


b  2.5 in


b'  2.5 in


c  28 in


c'  3  in


Ls  19 in

Distance from arm pivot to roller center:

Lr  12 in


e  0.5 in

ω  500  rpm

rc  3  in


wc  0.75 in

Roller follower dimensions: rf  1  in

Solution: 1.

2.

1

F50  173  lbf


The effective mass and spring constant of the arm referenced to the cam follower were determined in Problem 10-16 as: 5 lbf meff  0.04652  blob kaeff  1.0131 10  in The spring rate and preload referenced to the cam follower are: 2  1  Lr  1  keff   kaeff   Ls   ks    

3.

wf  1.5 in

ks  123  lbf  in



1

keff  307.418

lbf

Fpl 

in


Ls Lr

 

e

 rf  rc

rf

3

yf

r f + rc

 

X 2

e

2




rc



 

 sin θ

Fpl  273.917 lbf

Y


 F50

 

 

γ θ  θ  β θ



1


 


4.

  

2

2




 

vf θ 

5.

2

 d y θ    ω  θ f   d  

 

a f θ 

 d v θ    ω  θ f   d  

Substitute the expressions for displacement, acceleration and spring preload into equation 15.9 and solve for the force on the follower as a function of cam angle.

 

 

 

Fc θ  meff  a f θ  keff  yf θ  Fpl Plot the force on the follower for one revolution of the cam.

θ  0  deg 2  deg  360  deg

FORCE ON FOLLOWER 600

500

400 Force, lb

6.

 

Fc θ

300

lbf 200

100

0

0

45

90

135

180

225


270

315

360




Repeat Problem 15-7 for a double-dwell cam to move the roller follower from 0 to 2.5 inches in 60 deg with modified sine acceleration, dwell for 120 deg, fall 2.5 inches in 30 deg with cycloidal motion, and dwell for the remainder. Cam speed is 100 rpm. Choose a suitable spring rate and preload to maintain follower contact. Select a spring from Appendix D. Assume a damping ratio of 0.10.

Units:


Given:

Arm dimensions:

2

1

Cutout dimensions:


a  2  in


a'  1.5 in


b  2.5 in


b'  2.5 in


c  28 in


c'  3  in


Ls  19 in


Lr  12 in

h  2.5 in

Cam lift and speed:

ω  100  rpm

Roller follower dimensions: rf  1  in Solution:

wf  1.5 in

See Figure P15-1, Problem 15-7, and DYNACAM file P1508.cam.

1.

The effective mass and spring constant of the arm referenced to the cam follower were determined in Problem 10-16 as: 5 lbf meff  0.04652  blob kaeff  1.0131 10  in

2.

There is no extension spring listed in Appendix D that is acceptable. A 2.5-in follower lift is nearly 4 in at the end of the arm where the spring attaches. There is no spring that has a safe maximum load that is 4 times the spring rate. We will, therefore, have to have a custom made spring with the following rate and preload when installed. Spring rate and preload:

3.

F50  37.9 lbf

The spring rate and preload referenced to the cam follower are: 2  1  Lr  1  keff   kaeff   Ls   ks    

4.

1

ks  75.8 lbf  in 1

keff  190 

lbf in

Fpl 

Ls Lr

 F50

Fpl  60 lbf

Enter the given cam data into the program DYNACAM and create a cam with acceptable pressure angles and radius of curvature. A cam drawing from DYNACAM that meets the requirements is shown on the next page. Find a suitable spring rate and preload to keep the minimum cam force positive and minimize the maximum cam force. One solution is shown in the Cam Dynamics screen shown below the cam drawing.






Repeat Problem 15-7 for a double-dwell cam to move the roller follower from 0 to 1.5 inches in 45 deg with 3-4-5 polynomial motion, dwell for 150 deg, fall 1.5 inches in 90 deg with 4-5-6 polynomial motion, and dwell for the remainder. Cam speed is 250 rpm. Choose a suitable spring rate and preload to maintain follower contact. Select a spring from Appendix D. Assume a damping ratio of 0.15.

Units:


Given:

Arm dimensions:

2

1

Cutout dimensions:


a  2  in


a'  1.5 in


b  2.5 in


b'  2.5 in


c  28 in


c'  3  in


Ls  19 in


Lr  12 in

h  2.5 in

Cam lift and speed:

ω  100  rpm


wf  1.5 in


1.


2.

There is no extension spring listed in Appendix D that is acceptable. A 1.5-in follower lift is nearly 2.5 in at the end of the arm where the spring attaches. There is no spring that has a safe maximum load that is 2.5 times the spring rate. We will, therefore, have to have a custom made spring with the following rate and preload when installed. Spring rate and preload:

3.

F50  94.7 lbf

The spring rate and preload referenced to the cam follower are: 2  1  Lr  1  keff   kaeff   Ls   ks    

4.

1

ks  120  lbf  in

1

keff  300

lbf in

Fpl 

Ls Lr

 F50

Fpl  150 lbf







Repeat Problem 15-7 for a single-dwell cam to move the roller follower from 0 to 2 inches in 60 deg, fall 2 inches in 90 deg, and dwell for the remainder. Use a seventh degree polynomial. Cam speed is 250 rpm. Choose a suitable spring rate and preload to maintain follower contact. Select a spring from Appendix D. Assume a damping ratio of 0.15.

Units:


Given:

2

1

Arm dimensions:

Cutout dimensions:


a  2  in


a'  1.5 in


b  2.5 in


b'  2.5 in


c  28 in


c'  3  in


Ls  19 in


Lr  12 in

h  2.5 in

Cam lift and speed:

ω  100  rpm


wf  1.5 in


1.


2.

There is no extension spring listed in Appendix D that is acceptable. A 2-in follower lift is over 3 in at the end of the arm where the spring attaches. There is no spring that has a safe maximum load that is 3 times the spring rate. We will, therefore, have to have a custom made spring with the following rate and preload when installed. Spring rate and preload:

3.

F50  31.6 lbf

The spring rate and preload referenced to the cam follower are: 2   1  Lr  1  keff      kaeff    Ls  ks

4.

1

ks  85.3 lbf  in

1

keff  213

lbf in

Fpl 

Ls Lr

 F50

Fpl  50 lbf







Repeat Problem 15-7 for a double-dwell cam to move the roller follower from 0 to 2 inches in 60 deg with cycloidal motion, dwell for 150 deg, fall 2 inches in 90 deg with modified sine acceleration, and dwell for the remainder. Cam speed is 200 rpm. Choose a suitable spring rate and preload to maintain follower contact. Select a spring from Appendix D. Assume a damping ratio of 0.15.

Units:


Given:

Arm dimensions:

2

1

Cutout dimensions:


a  2  in


a'  1.5 in


b  2.5 in


b'  2.5 in


c  28 in


c'  3  in


Ls  19 in


Lr  12 in

h  2.5 in

Cam lift and speed:

ω  100  rpm

Roller follower dimensions: rf  1  in Solution: 1.


The effective mass and spring constant of the arm referenced to the cam follower were determined in Problem 10-16 as: 5 lbf

meff  0.04652  blob 2.

in

1

ks  82.1 lbf  in

F50  94.7 lbf

The spring rate and preload referenced to the cam follower are: 2  1  Lr  1  keff      kaeff    Ls  ks

4.

kaeff  1.0131 10 

There is no extension spring listed in Appendix D that is acceptable. A 2-in follower lift is over 3 in at the end of the arm where the spring attaches. There is no spring that has a safe maximum load that is 3 times the spring rate. We will, therefore, have to have a custom made spring with the following rate and preload when installed. Spring rate and preload:

3.

wf  1.5 in

1

keff  205

lbf in

Fpl 

Ls Lr

 F50

Fpl  150 lbf







The cam in Figure P15-2 is a pure eccentric with eccentricity = 20 mm and turns at 200 rpm. The mass of the follower is 1 kg. The spring has a rate of 10 N/m and a preload of 0.2 N. Find the follower force over one revolution. Assume a damping ratio of 0.10. If there is follower jump, respecify the spring rate and preload to eliminate it.

Given:

Cam eccentricity and speed: Follower mass:

1.

2.

mf  1  kg 1

ζ  0.10

Fpl  0.2 N


Using the equation given in the figure, write functions for the displacement, velocity, and acceleration of the follower. Note that the displacement function is written such that at it is zero at t = 0. Displacement:

s θ  a   1  cos θ 

Velocity:

v θ  a  ω sin θ

Acceleration:

a  θ  a  ω  cos θ 2

Calculate the damping coefficient using equations 15.2i and 15.3a. c  2  ζ mf  k

3.

ω  200  rpm

Damping ratio:

k  10 N  m

Spring: Solution:

a  20 mm

c  0.632

N  sec m

Substitute the expressions for displacement, velocity, acceleration, and spring preload into equation 15.9 and solve for the force on the follower as a function of cam angle. Fc θ  mf  a  θ  c v θ  k s θ  Fpl Plot the force on the follower for one revolution of the cam.

θ  0  deg 2  deg  360  deg

FORCE ON FOLLOWER (ORIGINAL SPRING) 10

5 Force, N

4.

Fc( θ)

0

N 5

 10

0

60

120

180

240

θ deg Cam Rotation Angle, deg

300

360


The cam force becomes negative, which indicates that the follower will jump or lose contact with the cam. New values of spring rate and/or preload will be tried iteratively to make the force always positive. Let the new spring parameters be:

1

k  50 N  m

Fpl  7.5 N

c  2  ζ mf  k

c  1.414

N  sec m

Fcn θ  mf  a  θ  c v θ  k s θ  Fpl

FORCE ON FOLLOWER (NEW SPRING) 20

15 Force, N

5.


Fcn( θ)

10

N

5

0

0

60

120

180

240


300

360




The cam in Figure P15-2 has a symmetric double harmonic rise and fall of 20 mm and turns at 200 rpm. The mass of the follower is 1 kg. The spring has a rate of 10 N/m and a preload of 0.2 N. Find the follower force over one revolution. Assume a damping ratio of 0.10. If there is follower jump, respecify the spring rate and preload to eliminate it.

Given:

Cam rise/fall and speed: Follower mass: Spring:

Solution: 1.

h  20 mm

ω  200  rpm

mf  1  kg

Damping ratio:

1

k  10 N  m

ζ  0.10

Fpl  0.2 N


Using equations 8.25, write functions for the displacement, velocity, and acceleration of the follower. The value of  for both the rise and the fall is β  180  deg. To write the global equations, define a range function that has a value of unity between the limits of a and b and is zero elsewhere. R θ a b   if  θ  a    θ  b  1 0 Displacement:

2.

π h  θ    θ 1    R θ 0 β   sin π    sin 2  π    β 2 β    β 2   1 θ  β    sin 2 π θ  β    R θ β 2 π   sin π    β  2 β     

Velocity:

v θ  ω

Acceleration:

a  θ  ω 

2

h  θ    θ    R θ 0 β   cos π   cos 2  π    2 2 β    β  β  θ  β   cos 2 π θ  β    R θ β 2  π   cos π    β  β     

2 π

c  0.632

    

N  sec m

Substitute the expressions for displacement, velocity, acceleration, and spring preload into equation 15.9 and solve for the force on the cam as a function of cam angle. Fc θ  mf  a  θ  c v θ  k s θ  Fpl

4.

    


3.

θ   h   θ  1   s θ  R θ 0 β    1  cos π      1  cos 2  π    2  β    β  4   h  θ  β 1  θ  β       R θ β 2  π    1  cos π     1  cos 2  π   β  4  β   2   

Plot the force on the cam for one revolution of the cam.

θ  0  deg 2  deg  360  deg



FORCE ON CAM (ORIGINAL SPRING) 10

Force, N

5

Fc( θ)

0

N 5

 10

0

60

120

180

240

300

360



1

k  50 N  m

Fpl  8  N

c  2  ζ mf  k

c  1.414

N  sec m


FORCE ON CAM (NEW SPRING) 15

10 Force, N

5.

Fcn( θ) N 5

0

0

60

120

180

240


300

360




The cam in Figure P15-2 has a symmetric 3-4-5-6 polynomial rise and fall of 20 mm and turns at 200 rpm. The mass of the follower is 1 kg. The spring has a rate of 10 N/m and a preload of 0.2 N. Find the follower force over one revolution. Assume a damping ratio of 0.10. If there is follower jump, respecify the spring rate and preload to eliminate it.

Given:

Cam rise/fall and speed:

1

k  10 N  m

Spring:

1.

ω  200  rpm

mf  1  kg

Follower mass:

Solution:

h  20 mm

Damping ratio: Fpl  0.2 N


Using equations 8.26, write functions for the displacement, velocity, and acceleration of the follower. The value of  for the total rise and fall is β  360  deg.



4 5 6 θ θ θ      192     192     64      β  β  β  β  3

Displacement:

θ s θ  h  64  

Velocity:

h  θ θ θ θ v θ  ω  192     768     960     384    β   β  β  β  β

Acceleration:

2 a  θ  ω 

2

2.

3

4

 

2 3 4  θ θ θ θ       384     2304    3840    1920    2  β  β  β  β  β 

h

c  0.632

N  sec m

Substitute the expressions for displacement, velocity, acceleration, and spring preload into equation 15.9 and solve for the force on the cam as a function of cam angle. Fc θ  mf  a  θ  c v θ  k s θ  Fpl θ  0  deg 2  deg  360  deg

Plot the force on the cam for one revolution of the cam.

FORCE ON CAM (ORIGINAL SPRING) 6 4 Force, N

4.

5


3.

ζ  0.10

2 Fc( θ)

0

N 2 4 6

0

60

120

180 θ deg

240

300

360



1

k  30 N  m

Fpl  5  N

c  2  ζ mf  k

c  1.095

N  sec m


FORCE ON CAM (NEW SPRING) 10

8

Force, N

5.


Fcn( θ)

6

N 4

2

0

0

60

120

180

240


300

360




Given:

Design a double-dwell cam to meet the specifications given below. Size a return spring, selected from Appendix D, and specify its preload to maintain contact between cam and follower. Calculate and plot the dynamic force and torque. Repeat for a form-closed cam. Compare the dynamic force, torque, and natural frequency for the form-closed design and the force-closed design. RISE

DWELL

β  60 deg h 1  45 mm

FALL

β  120  deg h 2  0  mm

Mod. sine Cycle time: Solution: 1.

β  90 deg h 3  45 mm

DWELL β  90 deg h 4  0  mm

3-4-5 poly τ  1  sec


Follower mass:

mf  2  kg



2  π rad τ

ω  6.283

rad sec

ω  60.000 rpm

2.


3.

The cam is sized with an 92.5-mm prime circle radius and 10.8-mm eccentricity, which gives balanced pressure angles of 29.9 deg and a minimum radius of curvature of 52.4 mm. The roller radius is 25 mm.

4.

For the force-closed case, choose catalog number 447 from with a spring constant of 4.2 lb/in (735.8 N/m) with a preload of 10 N to keep the follower force positive and prevent jump. The maximum and minimum follower force, maximum driving torque, undamped and damped natural frequencies, and damping coefficient are shown on the Cam Dynamics screen, which is shown below on the next page. The system natural frequency is about 5.3 times the driving frequency. This ratio could be increased at the expense of larger follower force by increasing the spring constant.

5.


6.







Given:

Design a single-dwell cam to meet the specifications given below. Size a return spring, selected from Appendix D, and specify its preload to maintain contact between cam and follower. Calculate and plot the dynamic force and torque. Repeat for a form-closed cam. Compare the dynamic force, torque, and natural frequency for the form-closed design and the force-closed design. RISE

FALL

β  60 deg

DWELL

β  90 deg

h 1  25 mm

β  210  deg h 3  0  mm

Seventh-degree polynomial Cycle time: Solution: 1.

τ  2  sec


Follower mass:

mf  10 kg



2  π rad τ

ω  3.142

rad sec

ω  30.000 rpm

2.


3.

The cam is sized with an 85-mm prime circle radius and 2.8-mm eccentricity, which gives balanced pressure angles of 21.5 deg and a minimum radius of curvature of 50.0 mm. The roller radius is 25 mm.

4.

For the force-closed case, choose catalog number 447 from with a spring constant of 4.2 lb/in (735.8 N/m) with a preload of 10 N to keep the follower force positive and prevent jump. The maximum and minimum follower force, maximum driving torque, undamped and damped natural frequencies, and damping coefficient are shown on the Cam Dynamics screen, which is shown below on the next page. The system natural frequency is about 2.7 times the driving frequency. This ratio could be increased at the expense of larger follower force by increasing the spring constant.

5.


6.







Given:

Design a four-dwell cam to meet the specifications given below. Size a return spring and specify its preload to maintain contact between cam and follower. Calculate and plot the dynamic force and torque. Repeat for a form-closed cam. Compare the dynamic force, torque, and natural frequency for the form-closed design and the force-closed design. RISE/FALL

1.

FALL

DWELL

β  40 deg

β  100  deg

β  90 deg

β  20 deg

h 1  40 mm

h 2  0.0 in

h 3  20 mm

h 4  0.0 in

β  30 deg

β  40 deg

β  30 deg

β  10 deg

h 5  10 mm

h 6  0.0 in

h 7  10 mm

h 8  0.0 in


DWELL

τ  10 sec


Follower mass:

mf  3  kg



2  π rad τ

ω  0.628

rad sec

ω  6.000 rpm

2.


3.

Modified trapezoidal acceleration was used for the rise and all falls. The cam is sized with an 130-mm prime circle radius and 34-mm eccentricity, which gives balanced pressure angles of 29 deg and a minimum radius of curvature of 50 mm. The roller radius is 25 mm.

4.

For the force-closed case, a spring constant of 150 N/mm with a preload of 5 N is chosen to keep the follower force positive and prevent jump. The maximum and minimum follower force, maximum driving torque, undamped and damped natural frequencies, and damping coefficient are shown on the Cam Dynamics screen, which is shown below on the next page. The system natural frequency is about 11.3 times the driving frequency. This ratio could be increased at the expense of larger follower force by increasing the spring constant.

5.


6.







Units: Given:

Design a cam to drive an automotive valve train whose parameters are given below. Select a spring constant and preload to avoid jump to 3500 rpm. Fast opening and closing and maximum open time are desired. 3

kN  10  N h  12 mm

Cam rise/fall and speed: Valve train mass:

mv  0.2 kg

Roller follower:

Rf  5  mm

ω  3500 rpm ζ  0.30

Damping ratio:

Open-close event is 160 deg, low dwell is 200 deg. Solution: 1.


Use a 3-4-5 polynomial for both the rise and the fall. Although peak jerk is higher with this choice than with the 4-5-6-7 polynomial, peak acceleration is considerably lower. Using equation 8.24, write functions for the displacement, velocity, and acceleration of the follower. The value of  for the both the rise and fall is β  40 deg. To write the global equations, define a range function that has a value of unity between the limits of a and b and is zero elsewhere. R θ a b   if  θ  a    θ  b  1 0 Displacement: Rise:

β  β

β  40.000 deg 4   θ  3 θ   15        β   β 

s1 θ  h  10 

θ   β 

6 

High dwell:

β  160  deg  2  β

β  80.000 deg

Fall:

β  β

β  40.000 deg

5

   s2 θ  h

4   θ  β  β     15     β β       5   θ  β  β     6    β    





   s3 θ  h  1  10      



θ  β  β 



3





s θ  R θ 0 β  s1 θ  R θ β β  β  s2 θ   R θ β  β β  β  β  s3 θ

Global equation:

Velocity:





Rise:

v1 θ  ω

3 4   θ  2  θ   30  θ    60       β   β   β   β  

High dwell:

v2 θ  0 

m

Fall:

v3 θ  ω

h

 30 

sec

3 4   θ  β  β 2  θ  β  β   θ  β  β       30    60    30   β   β β β     

h











v θ  R θ 0 β  v1 θ  R θ β β  β  v2 θ   R θ β  β β  β  β  v3 θ

Global equation:





Acceleration: a 1 θ  ω 

2 3   θ   θ   120   θ    180          β   β   β  

h

2

Rise:

 60 

2

β High dwell:

a 2 θ  0 

m sec

2 a 3 θ  ω 

Fall:













Iteratively choose spring parameters that will prevent follower jump and then, calculate the damping coefficient using equations 15.2i and 15.3a. k  350  N  mm

Spring: c  2  ζ mv k 3.

2  θ  β  β   θ  β  β       60    180     2  β β     β   3    θ  β  β    120    β    

h

a  θ  R θ 0 β  a 1 θ  R θ β β  β  a 2 θ   R θ β  β β  β  β  a 3 θ

Global equation:

2.

2

1

Fpl  1200 N

c  158.745

N  sec m

Substitute the expressions for displacement, velocity, acceleration, and spring preload into equation 15.9 and solve for the force on the cam as a function of cam angle. Fc θ  mv a  θ  c v θ  k s θ  Fpl Plot the force on the cam for one revolution of the cam.

θ  0  deg 2  deg  360  deg

FORCE ON CAM 8

6 Force, kN

4.

Fc( θ)

4

kN

2

0

0

20

40

60

80

100


120

140

160




Figure P15-3 shows a cam-follower system that drives slider 6 through an external output arm 3. Arms 2 and 3 are both rigidly attached to the 0.75-in-dia shaft X-X, which rotates in bearings that are supported by the housing. The pin-to-pin dimensions of the links are shown. The cross-section of arms 2, 3, and 5 are solid, rectangular 1.5 x 0.75 in steel. The ends of these links have a full radius equal to one half the link width. Link 4 is 1-in-dia x 0.125 in-wall, round steel tubing. Link 6 is a 2-in-dia x 6-in long solid steel cylinder. Find the effective mass and effective spring constant of the follower train referenced to the cam-follower roller if the spring at A has a rate of 150 lb/in with a preload of 60 lb. Then determine and plot the kinetostatic follower force and camshaft torque over one cycle if the cam provides a 3-4-5 polynomial double dwell angular motion to roller arm 2 with a rise of 10 deg in 90 camshaft degrees, dwell for 90 deg, fall 10 deg in 90 deg and dwell for the remainder. The camshaft turns 100 rpm.

Units:


Given:

Follower rise:   10 deg

2

1

Cam speed: n  100  rpm Roller arm radius: r  8  in


w  0.75 in

h  1.50 in

a  10.0 in


h  1.50 in

Spring:

ks  150  lbf  in

Roller arm 2:

w  0.75 in

L4  22 in

L3  16 in

1


w  0.75 in

d 4id  0.75 in

b  8.0 in

Fso  60 lbf

h  1.50 in

L2  8  in 6

E  30 10  psi


Spec. weight

γ  0.28

lbf

1.

Use the method of Chapter 10 to find the equivalent lumped mass system. Break the system into individual elements as shown in Figure 10-9b.

2.

Define the individual spring constants of each of the six elements. Roller arm (cantilever beam with end load): Moment of inertia Spring constant

I2  k2 

w h

3

12 3  E I2 L2

Output arm: Moment of inertia Spring constant

I3  k3 

w h

3

k2  3.708  10

4 lbf

k3  4.635  10

3 lbf

k4  4.686  10

5 lbf

in

3

12 3  E I3 L3

3

in


Area Spring constant

A4  k4 

3

in


2



4 A4  E L4

in



Rocker arm 5 (side B): Moment of inertia Spring constant

I5 

w h

k5a 

3

12 3  E I5 a

3

4 lbf

k5a  1.898  10

in


k5b 

3  E I5 b

3.

3

4 lbf

k5b  3.708  10

in

Determine the mass of each of the elements. Roller arm 2 (cantilever beam with end load): Volume

V2  w h  L2

Mass

m2 

γ  V2

3

m2  6.527  10

g

2

1

2

1

lbf  sec  in


V3  w h  L3

Mass

m3 

γ  V3

1

2

m3  0.013 lbf  sec  in

g


V4  A4  L4

Mass

m4 

γ  V4

3

m4  5.482  10

g

lbf  sec  in


V5a  w h  a

Mass

m5a 

γ V5a g

m5a  8.159  10

3

2

1

2

1

lbf  sec  in


V5b  w h  b

Mass

m5b 

Cylinder 6: Volume

Mass

V6  m6 


3

lbf  sec  in

2

4 γ  V6 g

m5b  6.527  10

 L6 2

1

m6  0.014 lbf  sec  in


Determine the effective mass and spring constant on either side of the rocker arm 5, reflected to point A.



Upper side: mU  m4  m5a kU 

2

1

2

1


1  1  1  k   s k4 k5a 

1

kU  148.777

lbf in

Lower side: mL  m6  m5b


kL  k5b

kL  3.708  10

4 lbf

in

2

mAeff  mU 

 b  m   L a 2


 b  k   L a

1

2

mAeff  0.027 lbf  sec  in 4 lbf

kAeff  2.388  10

in


kO 

1

2



 1  1 k   Aeff k3 

1

kO  3.881  10

3 lbf

in

Roller side: mR  m2

mR  6.527  10

kR  k2

kR  3.708  10

3

2

4 lbf

in

2

meff  mR 

 L3   L   mO  2 2

 L3  keff  kR     kO  L2  6.

1

lbf  sec  in

2

1


4 lbf

keff  5.260  10

in

The program DYNACAM will be used for this solution. Calculate the speed in rad/sec and the effective lift in inches for input to the SVAJ screen of DYNACAM. rad   n   10.472 h  r  h  1.4 in sec Choose poly functions for the rise and fall segments and, when calculating these segments, choose 6 BCs, setting V and A to zero at the beginning and end of the segments. This will result in 3-4-5 polynomials for the rise and fall.

7.

Click Done and then, at the Home screen, click Size Cam. Choose a follower radius and then a prime circle radius that will make the radius of curvature, min, significantly larger than the follower radius.




8.


Click Done and then, at the Home screen, click Dynamics. Enter the effective mass, spring rate, and spring preload. The results are shown below.




Figure P15-3 shows a cam-follower system that drives slider 6 through an external output arm 3. Arms 2 and 3 are both rigidly attached to the 0.75-in-dia shaft X-X, which rotates in bearings that are supported by the housing. The pin-to-pin dimensions of the links are shown. The cross-section of arms 2, 3, and 5 are solid, rectangular 1.5 x 0.75 in steel. The ends of these links have a full radius equal to one half the link width. Link 4 is 1-in-dia x 0.125 in-wall, round steel tubing. Link 6 is a 2-in-dia x 6-in long solid steel cylinder. Find the effective mass and effective spring constant of the follower train referenced to the cam-follower roller if the spring at A has a rate of 150 lb/in with a preload of 60 lb. Then determine and plot the kinetostatic follower force and camshaft torque over one cycle if the cam provides a cycloidal displacement double dwell angular motion to roller arm 2 with a rise of 10 deg in 90 camshaft degrees, dwell for 90 deg, fall 10 deg in 90 deg and dwell for the remainder. The camshaft turns 100 rpm.

Units:


Given:

Follower rise:   10 deg

2

1

Cam speed: n  100  rpm Roller arm radius: r  8  in


w  0.75 in

h  1.50 in

a  10.0 in


h  1.50 in

Spring:

ks  150  lbf  in

Roller arm 2:

w  0.75 in

L4  22 in

L3  16 in

1


w  0.75 in

d 4id  0.75 in

b  8.0 in

Fso  60 lbf

h  1.50 in

L2  8  in 6

E  30 10  psi


Spec. weight

γ  0.28

lbf

1.

Use the method of Chapter 10 to find the equivalent lumped mass system. Break the system into individual elements as shown in Figure 10-9b.

2.

Define the individual spring constants of each of the six elements. Roller arm (cantilever beam with end load): Moment of inertia Spring constant

I2  k2 

w h

3

12 3  E I2 L2

Output arm: Moment of inertia Spring constant

I3  k3 

w h

3

k2  3.708  10

4 lbf

k3  4.635  10

3 lbf

k4  4.686  10

5 lbf

in

3

12 3  E I3 L3

3

in


Area Spring constant

A4  k4 

3

in


2



4 A4  E L4

in



Rocker arm 5 (side B): Moment of inertia Spring constant

I5 

w h

k5a 

3

12 3  E I5 a

3

4 lbf

k5a  1.898  10

in


k5b 

3  E I5 b

3.

3

4 lbf

k5b  3.708  10

in

Determine the mass of each of the elements. Roller arm 2 (cantilever beam with end load): Volume

V2  w h  L2

Mass

m2 

γ  V2

3

m2  6.527  10

g

2

1

2

1

lbf  sec  in


V3  w h  L3

Mass

m3 

γ  V3

1

2

m3  0.013 lbf  sec  in

g


V4  A4  L4

Mass

m4 

γ  V4

3

m4  5.482  10

g

lbf  sec  in


V5a  w h  a

Mass

m5a 

γ V5a g

m5a  8.159  10

3

2

1

2

1

lbf  sec  in


V5b  w h  b

Mass

m5b 

Cylinder 6: Volume

Mass

V6  m6 


3

lbf  sec  in

2

4 γ  V6 g

m5b  6.527  10

 L6 2

1

m6  0.014 lbf  sec  in


Determine the effective mass and spring constant on either side of the rocker arm 5, reflected to point A.



Upper side: mU  m4  m5a kU 

2

1

2

1


1  1  1  k   s k4 k5a 

1

kU  148.777

lbf in

Lower side: mL  m6  m5b


kL  k5b

kL  3.708  10

4 lbf

in

2

mAeff  mU 

 b  m   L a 2


 b  k   L a

1

2

mAeff  0.027 lbf  sec  in 4 lbf

kAeff  2.388  10

in


kO 

1

2



 1  1 k   Aeff k3 

1

kO  3.881  10

3 lbf

in

Roller side: mR  m2

mR  6.527  10

kR  k2

kR  3.708  10

3

2

4 lbf

in

2

meff  mR 

 L3   L   mO  2

2

1


2

 L3  keff  kR     kO  L2  6.

4 lbf

keff  5.260  10

in

The program DYNACAM will be used for this solution. Calculate the speed in rad/sec and the effective lift in inches for input to the SVAJ screen of DYNACAM. rad   n   10.472 h  r  h  1.4 in sec Choose full cycloidal for the rise and fall segments. Set the units to in-lb-s.

7.

1

lbf  sec  in

Click Done and then, at the Home screen, click Size Cam. Choose a follower radius and then a prime circle radius that will make the radius of curvature, min, significantly larger than the follower radius.




8.


Click Done and then, at the Home screen, click Dynamics. Enter the effective mass, spring rate, and spring preload. The results are shown below.




A single-dwell cam-follower system similar to that shown in Figure 15-1a provides a twosegment polynomial for a rise of 35 mm in 75 deg, a fall of 35 mm in 120 deg, and a dwell for the remainder of the cycle. Using equations 15.8 and 15.10, calculate and plot the dynamic force and torque for one cycle if the roller follower train weighs 2.34 N, the system has a damping ratio of  = 0.06, and the spring has a rate of 1.5 N/mm with a preload of 10 N. The cam turns at 20 rpm.

Given:

RISE/FALL

DWELL

β  195  deg

β  165  deg

h  35 mm

h 3  0.0 mm

Cycle time:

Roller-follower train weight W  2.34 N k  1.5 N  mm

Spring rate and preload Solution: 1.

τ  3  sec

ζ  0.06

Damping ratio 1

Fpl  10 N


The camshaft speed is ω  20 rpm

ω  2.094

rad sec

The damping coefficient (see equation 15.3) is c  2  k 2.

W g

 3 N s

ζ

c  2.270  10

mm

Using a two-segment polynomial, let the rise and fall, together, be one segment and the dwell be the second segment. Then, the boundary conditions are: at v = 0, a = 0  = 0: s = 0, v=0  = 1: s = h,  = 

s = 0,

v = 0,

a =0


h  35.000 mm

Rise interval:

β  75 deg

A 

β

A  0.385

β

β  120  deg

Fall interval:

Solving the polynomial equation for these boundary conditions yields the following coefficients:

3.

c3  5609.280 mm

c4  24548.849  mm

c6  28772.307  mm

c7  7721.009 mm

c5  39990.867  mm

Write the SVA equations for the rise/fall segment. 3

4

5

6

7

 θ   c  θ   c  θ   c  θ   c  θ   4  5   6  7    β  β  β  β  β 2 3 4 5 6  1 θ θ θ θ θ  V1 θ   3  c3    4  c4    5  c5    6  c6    7  c7    β   β  β  β  β  β  S 1 θ  c3 

A1  θ 

2 3 4 5  θ θ θ θ θ        6  c3    12 c4    20 c5    30 c6    42 c7    2   β  β  β  β  β  β

1


4.

Write the SVA equations for the dwell segment. S 2 θ  0  mm

5.


V2 θ  0  mm

A2  θ  0  mm

To define the SVA curves for the complete cycle define a range function that has a value of one between the values of a and b and zero elsewhere. R( x a b )  if [ ( x  a )  ( x  b ) 1 0 ]

6.

Write the SVA equations for the entire cycle in absolute units. S  θ  R θ 0  deg β  S 1 θ  R θ β 360  deg  S 2 θ



V  θ  ω R θ 0  deg β  V1 θ  R θ β 360  deg  V2 θ





A  θ  ω  R θ 0  deg β  A1  θ  R θ β 360  deg  A2  θ 2

7.



Plot the force on the cam using equation 15.8. Fc θ 

W g

 A  θ  c V  θ  k S  θ  Fpl

θ  0  deg 0.5 deg  360  deg

80

Force, N

60 Fc( θ)

40

N 20

0

0

30

60

90

120

150

180

210

240

270

300

330

360

θ deg

Plot the driving torque on the cam using equation 15.10. Tc θ 

Fc θ  V  θ ω

2000

1000 Torque, N-mm

8.

T c( θ) N  mm

0

 1000

 2000

0

30

60

90

120

150

180 θ deg

210

240

270

300

330

360




A single-dwell cam-follower system similar to that shown in Figure 15-1a provides a twosegment polynomial for a rise of 35 mm in 75 deg, a fall of 35 mm in 120 deg, and a dwell for the remainder of the cycle. Using equation 15.8, size a return spring and specify its preload to maintain contact between cam and follower. Then calculate and plot the dynamic force for one cycle if the roller follower train weighs 3.55 N, the system has a damping ratio of  = 0.06, and the cam turns at 100 rpm.

Given:

RISE/FALL

DWELL

β  195  deg

β  165  deg

h  35 mm

h 3  0.0 mm

Roller-follower train weight W  3.55 N Solution: 1.

ζ  0.06

Damping ratio


The camshaft speed is ω  100  rpm

ω  10.472

rad

sec Interactively chose a spring rate and preload, do the calculations and return to this point until satisfactory values are chosen. k  0.25 N  mm

1

Fpl  2.0 N


W g

 3 N s

ζ

c  1.142  10

mm

Using a two-segment polynomial, let the rise and fall, together, be one segment and the dwell be the second segment. Then, the boundary conditions are: at v = 0, a = 0  = 0: s = 0, = s = h, v =0  1:  = 

s = 0,

v = 0,

a =0


h  35.000 mm

Rise interval:

β  75 deg

A 

β

A  0.385

β

β  120  deg

Fall interval:

Solving the polynomial equation for these boundary conditions yields the following coefficients:

3.

c3  5609.280 mm

c4  24548.849  mm

c6  28772.307  mm

c7  7721.009 mm

c5  39990.867  mm

Write the SVA equations for the rise/fall segment. 3

4

5

6

7

 θ   c  θ   c  θ   c  θ   c  θ   4  5   6  7    β  β  β  β  β 2 3 4 5 6 1  θ θ θ θ θ  V1 θ   3  c3    4  c4    5  c5    6  c6    7  c7    β   β  β  β  β  β  S 1 θ  c3 


2 3 4 5  θ θ θ θ θ         6  c3    12 c4    20 c5    30 c6    42 c7    2   β  β  β  β  β  β

A1  θ 

4.


1

Write the SVA equations for the dwell segment. S 2 θ  0  mm

5.

V2 θ  0  mm

A2  θ  0  mm


6.

Write the SVA equations for the entire cycle in absolute units. S  θ  R θ 0  deg β  S 1 θ  R θ β 360  deg  S 2 θ

V  θ  ω R θ 0  deg β  V1 θ  R θ β 360  deg  V2 θ





A  θ  ω  R θ 0  deg β  A1  θ  R θ β 360  deg  A2  θ 2




W g

 A  θ  c V  θ  k S  θ  Fpl

θ  0  deg 0.5 deg  360  deg

8

6

Force, N

7.



Fc( θ)

4

N

2

0

0

30

60

90

120

150

180 θ deg

210

240

270

300

330

360




A single-dwell cam-follower system similar to that shown in Figure 15-1a provides a constant velocity to the follower of 100 mm/sec for 2 sec then returns to its starting position with a total cycle time of 3 sec. Using equations 15.8 and 15.10, calculate and plot the dynamic force and torque for one cycle if the roller follower train weighs 4.5 N, the system has a damping ratio of  = 0.06, and the spring has a rate of 2.5 N/mm with a preload of 50 N.

Given:


Time duration of cv segment: tcv  2  sec


Roller-follower train weight W  4.5 N

Damping ratio

k  2.5 N  mm

Spring rate and preload Solution: 1.

The camshaft speed is ω 

c  2  k

W g

Fpl  50 N

2  π rad

ω  2.094

rad sec  3 N s

ζ

c  4.064  10

mm

Use a two-segment polynomial as demonstrated in Example 8-12. The lift during the first segment and the sva equations for the first segment are: vc Normalized velocity: vcv  vcv  47.746 mm ω tcv h cv  vc tcv h cv  200.000 mm β1   360  deg β1  240 deg τ s1( θ )  h cv

3.

1

ζ  0.06


τ The damping coefficient (see equation 15.3) is

2.

1

θ β1

v1( θ )  vcv

a 1( θ )  0  mm


 = 1:

s = h cv, v = vcv

 = 360 deg

s = 0,

a =0

v = vcv, a = 0


4.

5.

Total interval:

β  360  deg

CV interval:

β1  240 deg

A 

β1

A  0.667

β

Use the 6 BCs and equation 8.23 to write 6 equations in s, v, and a similar to those in example 8-9 but with 6 terms in the equation for s (the highest term will be fifth degree). The resulting coefficients are: c0  153600 mm

c1  971700 mm

c2  2430000  mm

c3  2997000  mm

c4  1822500  mm

c5  437400 mm

Write the sva equations for the return segment. 2

3

4

θ θ θ θ θ s2( θ )  c0  c1    c2    c3    c4    c5   β β β β β v2( θ ) 

1



2 3 4 θ  θ   4  c   θ   5  c   θ    3  c   3  4  5  β β β β 

 c1  2  c2 

β 

5


a 2( θ )  6.



2 3 θ θ θ    12 c4    20 c5    β β β 

 2  c2  6  c3 

1

β


2




7.

Write the SVA equations for the entire cycle in absolute units.









S  θ  R θ 0  deg β1  s1 θ  R θ β1 360  deg  s2 θ

 







V  θ  ω R θ 0  deg β1  v1 θ  R θ β1 360  deg  v2 θ

 









A  θ  ω  R θ 0  deg β1  a 1 θ  R θ β1 360  deg  a 2 θ 2

8.




W g

 A  θ  c V  θ  k S  θ  Fpl

θ  0  deg 0.5 deg  360  deg

Force, N

600

400 Fc( θ) N 200

0

0

30

60

90

120

150

180

210

240

270

300

330

360

270

300

330

360

θ deg

Plot the driving torque on the cam using equation 15.10. Tc θ 

Fc θ  V  θ ω

50

Torque, N-m

9.

0 Tc( θ) Nm  50

 100

0

30

60

90

120

150

180 θ deg

210

240




A single-dwell cam-follower system similar to that shown in Figure 15-1a provides a constant velocity to the follower of 100 mm/sec for 2 sec then returns to its starting position with a total cycle time of 3 sec. Using equation 15.8, size a return spring and specify its preload to maintain contact between cam and follower. Then calculate and plot the dynamic force for one cycle if the roller follower train weighs 5.0 N and the system has a damping ratio of  = 0.06.

Given:


Solution: 1.

1

Time duration of cv segment: tcv  2  sec


Roller-follower train weight W  5.0 N

Damping ratio

ζ  0.06


The camshaft speed is ω 

2  π rad

ω  2.094

rad

sec τ Interactively chose a spring rate and preload, do the calculations and return to this point until satisfactory values are chosen. k  0.25 N  mm

1

Fpl  4.0 N


W g

c  1.355  10

mm

Use a two-segment polynomial as demonstrated in Example 8-12. The lift during the first segment and the sva equations for the first segment are: vc Normalized velocity: vcv  vcv  47.746 mm ω tcv h cv  vc tcv h cv  200.000 mm β1   360  deg β1  240 deg τ s1( θ )  h cv

3.

 3 N s

ζ

θ β1

v1( θ )  vcv

a 1( θ )  0  mm


 = 1:

s = h cv, v = vcv

 = 360 deg

s = 0,

a =0

v = vcv, a = 0


4.

5.

Total interval:

β  360  deg

CV interval:

β1  240 deg

A 

β1

A  0.667

β

Use the 6 BCs and equation 8.23 to write 6 equations in s, v, and a similar to those in example 8-9 but with 6 terms in the equation for s (the highest term will be fifth degree). The resulting coefficients are: c0  153600 mm

c1  971700 mm

c2  2430000  mm

c3  2997000  mm

c4  1822500  mm

c5  437400 mm

Write the sva equations for the return segment. s2( θ )  c0  c1 

2

3

4

θ θ θ θ θ   c2    c3    c4    c5   β β β β β

5


v2( θ ) 

a 2( θ ) 



2 3 4 θ θ θ θ    3  c3    4  c4    5  c5    β β β β 

 c1  2  c2 

1

β 



1

β 6.


2

2 3 θ  θ   20 c   θ    12  c   4  5  β β β 

 2  c2  6  c3 




7.

Write the SVA equations for the entire cycle in absolute units.









S  θ  R θ 0  deg β1  s1 θ  R θ β1 360  deg  s2 θ

 







V  θ  ω R θ 0  deg β1  v1 θ  R θ β1 360  deg  v2 θ

 









A  θ  ω  R θ 0  deg β1  a 1 θ  R θ β1 360  deg  a 2 θ 2


W g

 A  θ  c V  θ  k S  θ  Fpl

θ  0  deg 0.5 deg  360  deg

60

40 Force, N

8.



Fc( θ) N 20

0

0

30

60

90

120

150

180 θ deg

210

240

270

300

330

360




As stated in Section 15.2, "A common rule of thumb is to design the (cam-follower) system to have a fundamental frequency ω n at least ten times the highest forcing frequency expected in service..." Since this is not always possible, calculate and plot the amplitude ratio from 5 <= ω n/ωf <= 10 for damping ratios of 0, 0.02, 0.04, 0.06, 0.08, and 0.10.

Given:

Frequency ratio: fr  0.1 0.11  0.2 (frequency ratio is the reciprical of ω n/ωf)

Solution:


Calculate the amplitude ratio as a function of frequency ratio and damping ratio using equation 15.6e. AR ( fr dr ) 

1

1  fr 2

2

 ( 2  dr  fr )

2

AMPLITUDE RATIO 1.05

1.04

Amplitude Ratio

1.

1.03

1.02

1.01

1 0.1

0.12

0.14

0.16

Frequency Ratio dr = 0.0 dr = 0.02 dr = 0.04 dr = 0.06 dr = 0.08 dr = 0.1

0.18

0.2




A cam-follower system similar to that shown in Figure P15-2 has a lumped mass of 0.1 kg, lumped stiffness of 0.10 N/mm, and a damping coefficient of 0.40 kg/s. If the forcing frequency is 50 rpm, determine the resulting amplitude ratio.

Given:

M  0.1 kg

Solution:


1.

1

c  0.40 kg s

ωf  50 rpm

k

ωn  31.623

M

rad sec

ωf  5.236 

rad sec

Calculate the damping ratio using equation 15.3a.

ζ 

3.

1


ωn 

2.

k  0.10 N  mm

c

ζ  0.063

2  M  ωn

Calculate the amplitude ratio using equation 15.6e.

1

ampratio 

 1  

2

2 2 ωf   ωf     2  ζ  ω     ωn   n  

ampratio  1.028




A cam-follower system similar to that shown in Figure P15-1 has a lumped mass of 2.5 kg, lumped stiffness of 4 N/mm, and a damping coefficient of 12.0 kg/s. If the forcing frequency is 80 rpm, determine the resulting amplitude ratio.

Given:

M  2.5 kg

Solution:


1.

1

c  12.0 kg s

ωf  80 rpm

k

ωn  40.000

M

rad sec

ωf  8.378 

rad sec

Calculate the damping ratio using equation 15.3a.

ζ 

3.

1


ωn 

2.

k  4.0 N  mm

c

ζ  0.060

2  M  ωn

Calculate the amplitude ratio using equation 15.6e.

1

ampratio 

 1  

2

2 2 ωf   ωf     2  ζ  ω     ωn   n  

ampratio  1.046




Design a double-dwell motion to translate a slider from 0 to 75 mm in 60-deg (timing angle) with modified sine acceleration, dwell for 120-deg, fall 75 mm in 30-deg with cycloidal motion, and dwell for the remainder. Calculate and plot the displacement function of the slider using DYNACAM and modify that function to drive the crank of a non-offset crank-slider with the dimensions: crank = 100 mm, coupler = 300 mm such that at maximum stroke of the slider, the crank is at zero degrees. The slider must follow the specified motion. Plot the resulting crank input motion.

Given:

Crank-slider geometry: L2  100  mm L3  300  mm

Offset  0  mm

Motion data: RISE

DWELL

FALL

DWELL

β  60 deg

β  120  deg

β  30 deg

β  150  deg

h 1  75 mm

h 2  0  mm

h 3  75 mm

h 4  0  mm

modified sine

cycloidal

Table 8-2 Parameters and coefficients for the modified sine acceleration: b  0.25 Solution: 1.

c  0.00

d  0.75

Ca  5.5280


Using equations 8.12d and 8.14 through 8.19, write the displacement equations for the rise and fall of the slider Rise: modified sine

 b  b  2  π      sin  x  b   π  π

y1 ( x)  Ca x

 x2

y2 ( x)  Ca 

2 

 b  

1

π



 b c y3 ( x)  Ca     x   π 2  

1  2 1   x  b    2  2 8 π   1

2 2 2  d   b2  1  1   ( 1  d )   d   cos π   x  1       8 2 8 2 π d   π  π  

 x2  b  1  2  π 

2

  

1  1 2 2  1   x  2 d  b   2     2 4 8 π   b

y4 ( x)  Ca 

b 2 d  b y5 ( x)  Ca   x  π 2 π 

d 





  (1  b)2  d2   b  2sin π (x  1)

2

4

   π

 b

  

Fall: cycloidal y ( x)   x 



2.

1 2 π

 sin( 2  π x) 



The above equations can be used for a rise or fall by using the proper values of , , and h. To plot the displacement curve, first define a range function that has a value of one between the values of x1 and x2 and zero elsewhere. R( x x1 x2)  if [ ( x  x1)  ( x  x2) 1 0 ]



3.

The global displacement equation is composed of four intervals (rise, dwell, fall, and dwell). The local equation above must be assembled into a single equation for the displacement that applies over the range 0 <=  <= 360 deg of the timing angle.

4.

Write the local displacement equation for the first interval, 0 <=  <= 1. Note that each subinterval function is multiplied by the range function so that it will have nonzero values only over its subinterval. For 0 <=  <= (Rise)

 b 1  d   y ( x)  R x 1  d 1    y1 ( x)  R x    2  2 2 2  2 2     1d b b    1    y4 ( x)  R x 1  1  y5 ( x)   R x  2 2 2    

s1( x)  h 1  R x 0 

b

d

   y3 ( x)      

For β1 <= θ <= β1+ β2 (dwell) s2( x)  h 1 For β1+ β2 <= θ <= β1+ β2 + β 3 (fall) s3( x)  h 3 ( 1  y ( x) ) For β1+ β2 + β 3 <= θ <= 2π (dwell) s4( x)  0  mm Write the complete global equation for the displacement and plot it over one timing cycle, which is the sum of the four intervals defined above.

θ1  β

Let

θ2  θ1  β

θ3  θ2  β

θ4  θ3  β


S ( θ )  s1

θ  0  deg 0.5 deg  360  deg SLIDER DISPLACEMENT, S 80

Displacement, mm

5.

60

40

20

0

0

30

60

90

120

150

180

210

Timing Angle, deg

240

270

300

330

360


7.

Write the equation for the slider position as a function of the timing angle, θ Top dead center (θ 2 = 0-deg):

TDC  L2  L3

TDC  400.000 mm

Slider position:

d ( θ )  TDC  S ( θ )

Using equations 4.20 and 4.21, write the equation for the crank angle, θ2 as a function of the timing angle, θ K1 θ  L2  L3   d  θ  2

2

2

K2  0  mm

A  θ  K1 θ  K3 θ

K3 θ  2  L2 d  θ

2

C θ  K1 θ  K3 θ

B  2  K2



θ2 θ  2  atan2 2  A  θ B 

B  4  A  θ  C θ 2



CRANK DISPLACEMENT, DEG 80

Crank Angle, degrees

6.


60

40

20

0

0

30

60

90

120

150

180

210

240

270

300

330

360

Timing Angle, degrees

Maximun crank angle:

θ2max  θ2( 60 deg)

θ2max  66.782 deg




Design a double-dwell motion to translate a slider from 0 to 65 mm in 80-deg (timing angle) with 3-4-5 polynomial, dwell for 100-deg, fall 65 mm in 40-deg with 3-4-5 polynomial motion, and dw for the remainder. Calculate and plot the displacement function of the slider using DYNACAM and modify that function to drive the crank of a non-offset crank-slider with the dimensions: crank = 100 mm, coupler = 300 mm such that at maximum stroke of the slider, the crank is at zero degrees. The slider must follow the specified motion. Plot the resulting crank input motion.

Given:

Crank-slider geometry: L2  100  mm L3  300  mm


Motion data: RISE

DWELL

FALL

DWELL

β  80 deg

β  100  deg

β  40 deg

β  140  deg

h 1  65 mm

h 2  0  mm

h 3  65 mm

h 4  0  mm

3-4-5 poly Solution: 1.

3-4-5 poly


Using equation 8.24, write the displacement equation for the rise and fall of the slider. Rise and fall: 3-4-5 polynomial 3

4

y ( x)  10 x  15 x  6  x

5

2.

The above equations can be used for a rise or fall by using the proper values of , , and h. To plot the displacement curve, first define a range function that has a value of one between the values of x1 and x2 and zero elsewhere. R( x x1 x2)  if [ ( x  x1)  ( x  x2) 1 0 ]

3.


4.

Write the local displacement equation for each interval.

5.

For 0 <=  <= (Rise)

s1( x)  h 1 y ( x)

For β1 <= θ <= β1+ β2 (dwell)

s2( x)  h 1

For β1+ β2 <= θ <= β1+ β2 + β 3 (fall)

s3( x)  h 3 ( 1  y ( x) )

For β1+ β2 + β 3 <= θ <= 2π (dwell)

s4( x)  0  mm

Write the complete global equation for the displacement and plot it over one timing cycle, which is the sum of the four intervals defined above. Let

θ1  β

θ2  θ1  β

θ3  θ2  β


S ( θ )  R θ 0 θ1  s1

θ  0  deg 0.2 deg  360  deg

θ4  θ3  β



SLIDER DISPLACEMENT, S

Displacement, mm

80

60

40

20

0

0

30

60

90

120

150

180

210

240

270

300

330

360

Timing Angle, deg

7.

Write the equation for the slider position as a function of the timing angle, θ Top dead center (θ 2 = 0-deg):

TDC  L2  L3

TDC  400.000 mm

Slider position:

d ( θ )  TDC  S ( θ )


2

2

K2  0  mm

A  θ  K1 θ  K3 θ

K3 θ  2  L2 d  θ

2

C θ  K1 θ  K3 θ

B  2  K2



θ2 θ  2  atan2 2  A  θ B 

B  4  A  θ  C θ 2





6.

60

40

20

0

0

30

60

90

120

150

180

210

240

270

300

330

360


Maximun crank angle:

θ2max  θ2( 65 deg)

θ2max  59.468 deg




Design a single-dwell polynomial motion to translate a slider from 0 to 75 mm and return to 0 in 120-deg (timing angle) and dwell for the remainder. Calculate and plot the displacement function of the slider using DYNACAM and modify that function to drive the crank of a non-offset crank-slider with the dimensions: crank = 90 mm, coupler = 270 mm such that at maximum stroke of the slider, the crank is at 180 degrees. The slider must follow the specified motion. Plot the resulting crank input motion.

Given:

Crank-slider geometry: L2  90 mm L3  270  mm


Motion data: RISE-FALL

DWELL

β  120  deg

β  240  deg

h 1  75 mm

h 2  0  mm

polynomal Solution: 1.


Using equation (a) from Example 8-8, write the displacement equation for the rise and fall of the slider. Rise and fall: siingle dwell polynomial 3

4

5

y ( x)  64 x  192  x  192  x  64 x 2.

6

The above equation can be used for a rise or fall by using the proper values of , , and h. To plot the displacement curve, first define a range function that has a value of one between the values of x1 and x2 and zero elsewhere. R( x x1 x2)  if [ ( x  x1)  ( x  x2) 1 0 ]

3.

The global displacement equation is composed of two intervals (rise-fall, and dwell). The local equation above must be assembled into a single equation for the displacement that applies over the range 0 <=  <= 360 deg of the timing angle.

4.

Write the local displacement equation for each of the two intervals.

5.

For 0 <=  <= (Rise-Fall)

s1( x)  h 1 y ( x)

For β 1 <= θ <= 360 (dwell)

s2( x)  0  mm


θ1  β

θ2  360  deg θ  θ  θ1   R θ θ1 θ2  s2    θ1   θ2  θ1 

S ( θ )  R θ 0 θ1  s1

θ  0  deg 0.2 deg  360  deg



SLIDER DISPLACEMENT, S

Displacement, mm

80

60

40

20

0

0

30

60

90

120

150

180

210

240

270

300

330

360

Timing Angle, deg

6.

Write the equation for the slider position as a function of the timing angle, θ Bottom dead center (θ 2 = 180-deg): BDC  L3  L2 d ( θ )  BDC  S ( θ )

Slider position:


2

2

K2  0  mm

A  θ  K1 θ  K3 θ

K3 θ  2  L2 d  θ

2

C θ  K1 θ  K3 θ

B  2  K2



θ2 θ  2  atan2 2  A  θ B 

B  4  A  θ  C θ 2




160 Crank Angle, degrees

7.

BDC  180.000 mm

140

120

100

80

0

30

60

90

120

150

180

210

240

270

300

330

360


Minimun crank angle:

θ2max  θ2( 60 deg)

θ2max  89.719 deg




Using the constant velocity motion program of Example 16-1 applied to the slider of a nonoffset fourbar crank-slider with constant velocity stroke of 150 mm at 40 mm/s, calculate and plot the s, v, a, j functions for this motion applied directly to the slider using DYNACAM. Calculate the modified program needed to drive the crank of the linkage to obtain the specified motion at the slider. Crank = 200 mm, coupler = 600 mm, crank start angle = 110 deg. The cam rotates at a constant 10 rpm.

Given:

Crank-slider geometry: a  200  mm b  600  mm Cam data:

ω  10 rpm

Slider data:

VCV  40

1.

mm

ΔdCV  150  mm

s

θ2start  110  deg

Crank data: Solution:

c  0  mm


Determine the cam rotation angle for the constant velocity and return segments. tCV 

Constant velocity period:

2.

ΔdCV

tCV  3.750 s

VCV

CV cam rotation angle:

βCV  ω tCV

βCV  225.000  deg

Return cam rotation angle:

βret  2  π  βCV

βret  135.000  deg

Using equation 8.28a from Example 8-12, write the sva equation for the constant velocity and return segments o the slider. s1 ( θ )  ΔdCV 

Constant velocity segment:

θ βCV

v1  VCV

a1  0 

sec

Return segment: 2

s( θ ) = C0  C1

3.

3

4

θ θ  θ  θ  θ   C2   C3   C4   C5      βret  βret   βret   βret   βret 

5

v( θ ) =

2 3 4 ω  θ   θ   4 C   θ   5  C   θ    C1  2  C2   3  C  3  4 5      βret   βret   βret   βret   βret  

a(θ) =

2 2 3  ω   2  C  6 C   θ   12 C   θ   20 C   θ   2 3 4  5 β      ret    βret   βret   βret  

Applying the BCs given in Example 8-12: C0  ΔdCV

1 1 1  A   3 4 5     6 12 20 

C1 

VCV  βret

C2  0  mm

ω

 C0  C1  B   0  mm     0 mm 

C  A

1

B

mm

 2.400  103     3 C  3.600  10  mm    3  1.440  10 

2


C3  2400 mm


C4  3600 mm 2

s2 ( θ )  C0  C1

4.

C5  1440 mm 3

4


5

v2( θ ) 

2 3 4 ω  θ   θ   4  C   θ   5 C   θ    C1  2  C2   3  C  3 4  5     βret   βret   βret   βret   βret  

a2( θ ) 

2 2 3  ω   2 C  6  C   θ   12 C   θ   20 C   θ   2 3  4 5  β      ret    βret   βret   βret  

To plot the sva curves, first define a range function that has a value of one between the values of x1 and x2 and zero elsewhere. R( x x1 x2)  if [ ( x  x1)  ( x  x2) 1 0 ]

5.

The global displacement equation is composed of two intervals (constant velocity and return). The local equations above must be assembled into a single equation for the displacement that applies over the range 0 <=  <= 360 deg of the timing angle.

6.

Write the local sva equations for each of the two intervals. For 0 <=  <= CV(Constant velocity) s1( θ )  s1 ( θ ) v1( θ )  VCV a 1( θ )  0 

mm sec

2

s2( θ )  s2 ( θ )

For β CV <= θ <= 360 (Return)

v2( θ )  v2( θ ) a 2( θ )  a2( θ ) 7.


θ1  βCV

θ2  360  deg

S ( θ )  R θ 0 θ1  s1( θ )  R θ θ1 θ2  s2 θ  θ1 V ( θ )  R θ 0 θ1  v1( θ )  R θ θ1 θ2  v2 θ  θ1 A ( θ )  R θ 0 θ1  a 1( θ )  R θ θ1 θ2  a 2 θ  θ1 θ  0  deg 0.2 deg  360  deg



SLIDER DISPLACEMENT, S 200

Displacement, mm

150 100 50 0  50

0

30

60

90

120

150

180

210

240

270

300

330

360

Timing Angle, deg

SLIDER VELOCITY, V 50

Velocity, mm/sec

0  50  100  150  200

0

30

60

90

120

150

180

210

240

270

300

330

360

270

300

330

360

Timing Angle, deg

SLIDER ACCELERATION, A

Acceleration, mm/sec/sec

400

200

0

 200

 400

0

30

60

90

120

150

180

210

Timing Angle, deg

240


8.


Write the equation for the slider position as a function of the timing angle, θ Let

A  1

then

d 0 

B  2  a  cos θ2start 

B 

2

2

B  4 A  C

d 0  501.402 mm

2 A d ( θ )  d 0  S ( θ )

Slider position:

Using equations 4.20 and 4.21, write the equation for the crank angle, θ2 as a function of the timing angle, θ K1 θ  a  b   d  θ  2

2

2

K2  0  mm

A  θ  K1 θ  K3 θ

K3 θ  2  a  d  θ

2

C θ  K1 θ  K3 θ

B  2  K2



θ2 θ  2  atan2 2  A  θ B 

B  4  A  θ  C θ 2



CRANK DISPLACEMENT, DEG 120 110 Crank Angle, degrees

9.

2

C  a  b

100 90 80 70 60

0

30

60

90

120

150

180

210


240

270

300

330

360




Using the constant velocity motion program of Example 16-1 applied to the slider of a nonoffset fourbar crank-slider with constant velocity stroke of 5.5 in at 0.75 in/s, calculate and plot the s, v, a, j functions for this motion applied directly to the slider using DYNACAM. Calculate the modified program needed to drive the crank of the linkage to obtain the specified motion at the slider. Crank = 7.5 in, coupler = 18 in, crank start angle = 120 deg. The cam rotates at a constant 5 rpm.

Given:

Crank-slider geometry: a  7.5 in b  18.0 in Cam data:

ω  5  rpm

Slider data:

VCV  0.75

Crank data: Solution: 1.

c  0  in

in

ΔdCV  5.5 in

s

θ2start  120  deg


Determine the cam rotation angle for the constant velocity and return segments. tCV 

Constant velocity period:

2.

ΔdCV

tCV  7.333 s

VCV

CV cam rotation angle:

βCV  ω tCV

βCV  220.000  deg

Return cam rotation angle:

βret  2  π  βCV

βret  140.000  deg

Using equation 8.28a from Example 8-12, write the sva equation for the constant velocity and return segments of the slider. s1 ( θ )  ΔdCV 

Constant velocity segment:

θ βCV

v1  VCV

a1  0 

sec

Return segment: 2

3

4

θ θ  θ  θ  θ  s( θ ) = C0  C1  C2   C3   C4   C5      βret  βret   βret   βret   βret 

3.

5

v( θ ) =

2 3 4 ω  θ  θ  θ  θ   C1  2  C2   3  C3   4  C4   5  C5       βret   βret   βret   βret   βret  

a(θ) =

2 2 3  ω   2  C  6 C   θ   12 C   θ   20 C   θ   2 3 4  5 β      ret    βret   βret   βret  

Applying the BCs given in Example 8-12: C0  ΔdCV

1 1 1  A   3 4 5     6 12 20 

C1 

VCV  βret

C2  0  mm

ω

 C0  C1  B   0  mm     0 mm 

C  A

1

B

mm

 90.000  C   135.000  in    54.000 

2


C3  90 in


C4  135  in 2

s2 ( θ )  C0  C1

4.

C5  54 in 3

4


5

v2( θ ) 

2 3 4 ω  θ   θ   4  C   θ   5 C   θ    C1  2  C2   3  C  3 4  5     βret   βret   βret   βret   βret  

a2( θ ) 

2 2 3  ω   2 C  6  C   θ   12 C   θ   20 C   θ   2 3  4 5  β      ret    βret   βret   βret  

To plot the sva curves, first define a range function that has a value of one between the values of x1 and x2 and zero elsewhere. R( x x1 x2)  if [ ( x  x1)  ( x  x2) 1 0 ]

5.

The global displacement equation is composed of two intervals (constant velocity and return). The local equations above must be assembled into a single equation for the displacement that applies over the range 0 <=  <= 360 deg of the timing angle.

6.

Write the local sva equations for each of the two intervals. For 0 <=  <= CV(Constant velocity) s1( θ )  s1 ( θ ) v1( θ )  VCV a 1( θ )  0 

mm sec

2

s2( θ )  s2 ( θ )

For β CV <= θ <= 360 (Return)

v2( θ )  v2( θ ) a 2( θ )  a2( θ ) 7.


θ1  βCV

θ2  360  deg

S ( θ )  R θ 0 θ1  s1( θ )  R θ θ1 θ2  s2 θ  θ1 V ( θ )  R θ 0 θ1  v1( θ )  R θ θ1 θ2  v2 θ  θ1 A ( θ )  R θ 0 θ1  a 1( θ )  R θ θ1 θ2  a 2 θ  θ1 θ  0  deg 0.2 deg  360  deg



SLIDER DISPLACEMENT, S 6

Displacement, in

4

2

0

2

0

30

60

90

120

150

180

210

240

270

300

330

360

240

270

300

330

360

270

300

330

360

Timing Angle, deg

SLIDER VELOCITY, V 1

Velocity, in/sec

0

1

2

3

0

30

60

90

120

150

180

210

Timing Angle, deg

SLIDER ACCELERATION, A 3

Acceleration, in/sec/sec

2 1 0 1 2 3

0

30

60

90

120

150

180

210

Timing Angle, deg

240


8.


Write the equation for the slider position as a function of the timing angle, θ Let

A  1

then

d 0 

B  2  a  cos θ2start 

B 

2

2

B  4 A  C

d 0  13.037 in

2 A d ( θ )  d 0  S ( θ )

Slider position:

Using equations 4.20 and 4.21, write the equation for the crank angle, θ2 as a function of the timing angle, θ K1 θ  a  b   d  θ  2

2

2

K2  0  mm

A  θ  K1 θ  K3 θ

K3 θ  2  a  d  θ

2

C θ  K1 θ  K3 θ

B  2  K2



θ2 θ  2  atan2 2  A  θ B 

B  4  A  θ  C θ 2





9.

2

C  a  b

120

100

80

60

0

30

60

90

120

150

180

210


240

270

300

330

360




A non-offset crank-slider with crank and coupler lengths of 3.25 in and 10.875 in, respectively, is to be driven by a servo through a gear reducer. The required slider timing diagram is shown in Figure P16-1. Choose a servo from Table 16-2 to drive the crank with the lowest resulting vibration and calculate the resulting crank displacement for one cycle. Based on this crank input, calculate the resulting slider motion and compare the two by plotting their normalized displacements. The crank angle at t = 0 is 70-deg, and the crank turns clockwise to accomplish the first motion of the slider.

Given:

Crank-slider geometry: a  3.25 in b  10.875 in

c  0  in

Motion data:

Solution:

RISE

DWELL

FALL

DWELL

τ1  10 sec

τ2  10 sec

τ3  12 sec

τ4  8  sec

h 1  2.0 in

h 2  0  in

h 3  2.0 in

h 4  0  in

Crank angle at t = 0:

θ20  70 deg

Cycle time:

τ  40 sec

See Figure P16-1, Table 16-2 and Mathcad file P1606.

1.

From Table 16-2, choose the S-curve servo (cycloidal motion).

2.

Using equations 4.16a and b, calculate the position of the slider at t = 0 and the displacement of the crank equivalent to the stroke of the slider using the law of cosines.

 a  sin θ20  b 

θ30  asin 

c

π 

d 0  a  cos θ20  b  cos θ30

θ21  acos

d 0  11.549 in

  d 0  h 1  a  b   2  a   d 0  h 1    2

2

2

Δθ  θ20  θ21 3.

θ21  30.515 deg Δθ  39.485 deg

Using equation 8.12d, for the crank angle during the slider extention and return. y ( x)   x 



4.

θ30  163.690  deg

1 2 π

 sin( 2  π x) 



To plot the crank displacement curve, first define a range function that has a value of one between the values of x1 and x2 and zero elsewhere. R( x x1 x2)  if [ ( x  x1)  ( x  x2) 1 0 ]

3.


4.

Write the local displacement equation for the intervals τ1, τ2, τ3, and τ4 : For 0 <= t <= τ1(Rise) s1( x)  Δθ  y ( x)



For β1 <= θ <= β1+ β2 (dwell) s2( x)  Δθ For β1+ β2 <= θ <= β1+ β2 + β 3 (fall) s3( x)  Δθ  ( 1  y ( x) ) For β1+ β2 + β 3 <= θ <= 2π (dwell) s4( x)  0  deg 5.


t1  τ1

t2  t1  τ2

t3  t2  τ3

t4  t3  τ4

  R t t t   s  t  t1   1 2 2    t1   t2  t1   t  t2   t  t3   R t t2 t3  s3  R t t3 t4  s4    t3  t2   t4  t3 

S ( t)  R t 0  sec t1  s1

t

t  0  sec 0.05 sec  τ CRANK DISPLACEMENT - DEGREES

Displacement, deg

40

30

20

10

0

0

10

20

30

40

Time, sec

6.

Write the equation for the slider position as a function of the crank angle, θ2 (S(t)) using the law of cosines.

2

d  2  a  d  cos θ2  a  b 2

Let

Then

=0

2

B( t)  2  a  cos θ20  S ( t) 

A  1

d ( t) 

B( t) 

2

B( t )  4  A  C 2 A

 d0

2

C  a  b

2



SLIDER DISPLACEMENT - IN 2.5

Slider Displacement, in

2

1.5

1

0.5

0

0

10

20

30

40

Time, sec

Plot the normalized crank and slider displacements by dividing each by its total stroke.

NORMALIZED SLIDER AND CRANK DISPLACEMENTS

0.8

Displacement

7.

0.6

0.4

0.2

0

0

10

20 Time, sec

Crank Slider

30

40




Repeat Problem 16-6 using the timing diagram of Figure P16-1 but with a total stroke of 75 mm. The crank length is 110 mm, the coupler length is 275 mm, and the initial crank angle is 90 deg.

Given:

Crank-slider geometry: a  110  mm b  275  mm

c  0  mm

Motion data:

Solution:

RISE

DWELL

FALL

DWELL

τ1  10 sec

τ2  10 sec

τ3  12 sec

τ4  8  sec

h 1  75 mm

h 2  0  mm

h 3  75 mm

h 4  0  mm

Crank angle at t = 0:

θ20  90 deg

Cycle time:

τ  40 sec

See Figure P16-1, Table 16-2 and Mathcad file P1606.

1.

From Table 16-2, choose the S-curve servo (cycloidal motion).

2.

Using equations 4.16a and b, calculate the position of the slider at t = 0 and the displacement of the crank equivalent to the stroke of the slider using the law of cosines.

 a  sin θ20  b 

θ30  asin 

c

π 

d 0  a  cos θ20  b  cos θ30

3.

d 0  252.042 mm

  d 0  h 1 2  a2  b 2 θ21  acos  2  a   d 0  h 1   

θ21  52.869 deg

Δθ  θ20  θ21

Δθ  37.131 deg

Using equation 8.12d, for the crank angle during the slider extention and return. y ( x)   x 



4.

θ30  156.422  deg

1 2 π

 sin( 2  π x) 



To plot the crank displacement curve, first define a range function that has a value of one between the values of x1 and x2 and zero elsewhere. R( x x1 x2)  if [ ( x  x1)  ( x  x2) 1 0 ]

3.


4.

Write the local displacement equation for the intervals τ1, τ2, τ3, and τ4 : For 0 <= t <= τ1(Rise) s1( x)  Δθ  y ( x) For β1 <= θ <= β1+ β2 (dwell) s2( x)  Δθ For β1+ β2 <= θ <= β1+ β2 + β 3 (fall)



s3( x)  Δθ  ( 1  y ( x) ) For β1+ β2 + β 3 <= θ <= 2π (dwell) s4( x)  0  deg 5.


t1  τ1

t2  t1  τ2

t3  t2  τ3

t4  t3  τ4

  R t t t   s  t  t1   1 2 2    t1   t2  t1   t  t2   t  t3   R t t2 t3  s3  R t t3 t4  s4    t3  t2   t4  t3 

S ( t)  R t 0  sec t1  s1

t

t  0  sec 0.05 sec  τ CRANK DISPLACEMENT - DEGREES

Displacement, deg

40

30

20

10

0

0

10

20

30

40

Time, sec

6.

Write the equation for the slider position as a function of the crank angle, θ2 (S(t)) using the law of cosines.

2

d  2  a  d  cos θ2  a  b 2

Let

A  1

Then

d ( t) 

=0

2

B( t)  2  a  cos θ20  S ( t)  B( t) 

2

B( t )  4  A  C 2 A

 d0

2

C  a  b

2



SLIDER DISPLACEMENT - mm

Slider Displacement, mm

80

60

40

20

0

0

10

20

30

40

Time, sec

Plot the normalized crank and slider displacements by dividing each by its total stroke.

NORMALIZED SLIDER AND CRANK DISPLACEMENTS

0.8

Displacement

7.

0.6

0.4

0.2

0

0

10

20 Time, sec

Crank Slider

30

40

DESIGN OF MACHINERY -5th Ed SOLUTION MANUAL

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