Machine Design Problems

MACHINE DESIGN - An Integrated Approach, 4th Ed.

1-1-1

PROBLEM 1-1 Statement:

It is often said, "Build a better mousetrap and the world will beat a path to your door." Consider this problem and write a goal statement and a set of at least 12 task specifications that you would apply to its solution. Then suggest 3 possible concepts to achieve the goal. Make annotated, freehand sketches of the concepts.

Solution: Goal Statement: Create a mouse-free environment. Task Specifications: 1. Cost less than $1.00 per use or application. 2. Allow disposal without human contact with mouse. 3. Be safe for other animals such as house pets. 4. Provide no threat to children or adults in normal use. 5. Be a humane method for the mouse. 6. Be environmentally friendly. 7. Have a shelf-life of at least 3 months. 8. Leave no residue. 9. Create minimum audible noise in use. 10. Create no detectable odors within 1 day of use. 11. Be biodegradable. 12. Be simple to use with minimal written instructions necessary. Concepts and sketches are left to the student. There are an infinity of possibilities.

© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0101.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


1-2-1


A bowling machine is desired to allow quadriplegic youths, who can only move a joystick, to engage in the sport of bowling at a conventional bowling alley. Consider the factors involved, write a goal statement, and develop a set of at least 12 task specifications that constrain this problem. Then suggest 3 possible concepts to achieve the goal. Make annotated, freehand sketches of the concepts.

Solution: Goal Statement: Create a means to allow a quadriplegic to bowl. Task Specifications: 1. Cost no more than $2 000. 2. Portable by no more than two able-bodied adults. 3. Fit through a standard doorway. 4. Provide no threat of injury to user in normal use. 5. Operate from a 110 V, 60 Hz, 20 amp circuit. 6. Be visually unthreatening. 7. Be easily positioned at bowling alley. 8. Have ball-aiming ability, controllable by user. 9. Automatically reload returned balls. 10. Require no more than 1 able-bodied adult for assistance in use. 11. Ball release requires no more than a mouth stick-switch closure. 12. Be simple to use with minimal written instructions necessary. Concepts and sketches are left to the student. There are an infinity of possibilities.



1-3-1


A quadriplegic needs an automated page turner to allow her to read books without assistance. Consider the factors involved, write a goal statement, and develop a set of at least 12 task specifications that constrain this problem. Then suggest 3 possible concepts to achieve the goal. Make annotated, freehand sketches of the concepts.

Solution: Goal Statement: Create a means to allow a quadriplegic to read standard books with minimum assistance. Task Specifications: 1. Cost no more than $1 000. 2. Useable in bed or from a seated position 3. Accept standard books from 8.5 x 11 in to 4 x 6 in in planform and up to 1.5 in thick. 4. Book may be placed, and device set up, by able-bodied person. 5. Operate from a 110 V, 60 Hz, 15 amp circuit or by battery power. 6. Be visually unthreatening and safe to use. 7. Require no more than 1 able-bodied adult for assistance in use. 8. Useable in absence of assistant once set up. 9. Not damage books. 10. Timing controlled by user. 11. Page turning requires no more than a mouth stick-switch closure. 12. Be simple to use with minimal written instructions necessary. Concepts and sketches are left to the student. There are an infinity of possibilities.



1-4-1

PROBLEM 1-4 Statement: Units:

Convert a mass of 1 000 lbm to (a) lbf, (b) slugs, (c) blobs, (d) kg. blob :=

lbf  sec

2

in Given:

Mass

Solution:

See Mathcad file P0104.

M := 1000  lb

1. To determine the weight of the given mass, multiply the mass value by the acceleration due to gravity, g. W := M  g

W = 1000 lbf

2. Convert mass units by assigning different units to the units place-holder when displaying the mass value. Slugs

M = 31.081 slug

Blobs

M = 2.59 blob

Kilograms

M = 453.592 kg



1-5-1


A 250-lbm mass is accelerated at 40 in/sec2. Find the force in lb needed for this acceleration.

Given:

Mass

M := 250 lb

Acceleration

in a := 40 sec

Solution: 1.

2


To determine the force required, multiply the mass value, in slugs, by the acceleration in feet per second squared. Convert mass to slugs:

M = 7.770 slug

Convert acceleration to feet per second squared: F := M  a

a = 3.333s

-2

 ft

F = 25.9 lbf



1-6-1


Express a 100-kg mass in units of slugs, blobs, and lbm. How much does this mass weigh?

Units:

blob 

Given:

M  100  kg

lbf  sec

2

in

Assumptions: The mass is at sea-level and the gravitational acceleration is g  32.174

ft sec

Solution: 1.

or 2

g  386.089 

in sec

or 2

g  9.807 

m sec

2


Convert mass units by assigning different units to the units place-holder when displaying the mass value. The mass, in slugs, is

M  6.85 slug

The mass, in blobs, is

M  0.571  blob

The mass, in lbm, is

M  220.5  lb

Note: Mathcad uses lbf for pound-force, and lb for pound-mass. 2.

To determine the weight of the given mass, multiply the mass value by the acceleration due to gravity, g. The weight, in lbf, is

W  M  g

W  220.5  lbf

The weight, in N, is

W  M  g

W  980.7  N

© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


1-7-1


Prepare an interactive computer program (using, for example, Excell, Mathcad, or TKSolver) from which the cross-sectional properties for the shapes shown in the inside front cover can be calculated. Arrange the program to deal with both ips and SI unit systems and convert the results between those systems.

Solution:

See the inside front cover and Mathcad file P0107.

1.

Rectangle, let: b  3  in

h  4  in

Area

A  b  h

2

A  12.000 in

2

A  7742 mm Moment about x-axis

Moment about y-axis

Ix 

Iy 

b h

3

12 h b

4

Ix  16.000 in

6

4

6

4

Ix  6.660  10  mm 3

4

Iy  9.000  in

12

Iy  3.746  10  mm Radius of gyration about x-axis

Radius of gyration about y-axis

Polar moment of inertia

kx 

ky 

Ix

kx  1.155  in

A

kx  29.329 mm

Iy

ky  0.866  in

A

ky  21.997 mm

Jz  Ix  Iy

4

Jz  25.000 in

7

4

6

4

6

4

Jz  1.041  10  mm 2.

Solid circle, let: D  3  in 2

Area

A 

π D 4

Ix 

π D 64

Iy 

π D 64

4

Ix  3.976  in

Ix  1.655  10  mm 4

Moment about y-axis

2

A  4560 mm 4

Moment about x-axis

2

A  7.069  in

4

Iy  3.976  in


kx 

Ix A

kx  0.750  in kx  19.05  mm




1-7-2

Iy

ky 

ky  0.750  in ky  19.05  mm

A 4

Jz 


3.

π D

4

Jz  7.952  in

32

6

4

6

4

6

4

6

4

5

4

5

4

Jz  3.310  10  mm

Hollow circle, let: D  3  in

d  1  in A 

Area

Moment about x-axis

Ix 

 4

2



4

π

 D d

π 64



2

2

A  6.283  in

2

A  4054 mm

 D d



4

4

Ix  3.927  in

Ix  1.635  10  mm Moment about y-axis

Iy 

 64 π

4

 D d



4

4

Iy  3.927  in




4.

kx 

ky 

Jz 

Ix

kx  0.791  in

A

kx  20.08  mm

Iy

ky  0.791  in

A

ky  20.08  mm

 32 π

4

 D d



4

4

Jz  7.854  in

Jz  3.269  10  mm

Solid semicircle, let: D  3  in

R  0.5 D

R  1.5 in

2

Area

A 

π D

2

A  3.534  in

8

2

A  2280 mm Moment about x-axis

Ix  0.1098 R

4

4

Ix  0.556  in

Ix  2.314  10  mm Moment about y-axis

Iy 

π R 8

4

4

Iy  1.988  in

Iy  8.275  10  mm



Radius of gyration about x-axis



1-7-3

kx 

ky 

Ix A Iy A

Jz  Ix  Iy

kx  0.397  in kx  10.073 mm ky  0.750  in ky  19.05  mm 4

Jz  2.544  in

6

4

4

4

4

4

Jz  1.059  10  mm Distances to centroid

5.

a  0.4244 R

a  0.637  in a  16.17  mm

b  0.5756 R

b  0.863  in b  21.93  mm

Right triangle, let: b  2  in Area

Moment about x-axis

Moment about y-axis

h  1  in A 

Ix 

Iy 

b h 2

b h

A  645  mm 3

2

4

Ix  0.056  in

36 h b

2

A  1.000  in

Ix  2.312  10  mm 3

36

4

Iy  0.222  in




kx 

ky 

Ix A

Iy A

Jz  Ix  Iy

kx  0.236  in kx  5.987  mm ky  0.471  in ky  11.974 mm 4

Jz  0.278  in

5

Jz  1.156  10  mm

4



1-8-1


Prepare an interactive computer program (using, for example, Excell, Mathcad, or TKSolver) from which the mass properties for the solids shown in the page opposite the inside front cover can be calculated. Arrange the program to deal with both ips and SI unit systems and convert the results between those systems.

Units:

blob 

Solution:

See the page opposite the inside front cover and Mathcad file P0108.

1.

lbf  sec

2

in

a  2  in

Rectangular prism, let:

b  3  in

c  4  in

3

V  a  b  c

Volume

3

γ  0.28 lbf  in V  24.000 in

V  393290 mm M 

Mass

Moment about x-axis

Moment about y-axis

Ix 

Iy 

V γ

3

M  0.017  blob

g

M  3.048  kg

2

M a  b



2

2

Ix  0.019  blob in

12

2

M a  c

Ix  2130.4 kg mm



2

2

Iy  0.029  blob in

12

Iy  3277.6 kg mm

Moment about z-axis

Iz 

2

M b  c



2


Radius of gyration about z-axis

2.Cylinder, let:

r  2  in Volume

kx 

ky 

kz 

2

Ix

2

kx  1.041  in

M

kx  26.437 mm

Iy

ky  1.291  in

M

ky  32.791 mm

Iz

kz  1.443  in

M

kz  36.662 mm 3

L  3  in

γ  0.30 lbf  in 2

V  π r  L

3

V  37.699 in

V  617778 mm Mass

2

Iz  0.036  blob in

12

Iz  4097.0 kg mm


2

M 

V γ g

3

M  0.029  blob M  5.13 kg



1-8-2 2

Moment about x-axis

Moment about y-axis

Moment about z-axis

Ix 

Iy 

Iz 

M r

2

Ix  0.059  blob in

2



2

Ix  6619.4 kg mm



2

M  3 r  L

2

Iy  0.051  blob in

12



2

Iy  5791.9 kg mm



2

M  3 r  L

2



3.

Ix

kx 

kx  35.921 mm

Iy

ky  1.323  in

M

ky  33.601 mm

Iz

kz 

2

kx  1.414  in

M

ky 

2

Iz  0.051  blob in

12

Iz  5791.9 kg mm Radius of gyration about x-axis

2

kz  1.323  in

M

kz  33.601 mm

Hollow cylinder, let: a  2  in

b  3  in

Volume

3

L  4  in

γ  0.28 lbf  in

2



2

3

V  π b  a  L

V  62.832 in

V  1029630  mm Mass

Moment about x-axis

Moment about y-axis

Moment about z-axis



M 

Ix 

Iy 

Iz 

kx 

ky 

V γ

M  0.046  blob

g

M  7.98 kg

2

M

 a b

2

3



2

2

Ix  0.296  blob in 4

Ix  3.3  10  kg mm

M 12 M 12



2

2



2

 3 a  3 b  L

2

Iy  0.209  blob in 4

Iy  2.4  10  kg mm



2

2

Ix M Iy M



2

 3 a  3 b  L

2

2

2

Iz  0.209  blob in 4

Iz  2.4  10  kg mm

2

kx  2.550  in kx  64.758 mm ky  2.141  in ky  54.378 mm


MACHINE DESIGN - An Integrated Approach, 4th Ed. Radius of gyration about z-axis

4.

1-8-3 Iz

kz 

kz  2.141  in

M

kz  54.378 mm

Right circular cone, let: r  2  in

3

h  5  in

γ  0.28 lbf  in 2

Volume

V 

π r  h

3

V  20.944 in

3

V  343210 mm Mass

Moment about x-axis

Moment about y-axis

M 

Ix 

V γ

M  0.015  blob

g

3

M  2.66 kg 2

10

Iy  M 

3

2

M r

Ix  0.018  blob in

12r2  3h2 80

Ix  2059.4 kg mm

2

Iy  0.023  blob in

Iy  2638.5 kg mm

Moment about z-axis

Iz  M 

12r2  3h2 80



5.

Ix

kx 

2

kx  27.824 mm

Iy

ky  1.240  in

M

ky  31.495 mm

Iz

kz 

2

kx  1.095  in

M

ky 

2

Iz  0.023  blob in

Iz  2638.5 kg mm Radius of gyration about x-axis

2

kz  1.240  in

M

kz  31.495 mm

Sphere, let: r  3  in Volume

Mass

Moment about x-axis

V 

M 

Ix 

4 3

3

 π r

V  1853333  mm

V γ

5

3

M  0.082  blob

g 2

3

V  113.097  in

M  14.364 kg 2

M r

2

Ix  0.295  blob in

Ix  33362  kg mm

2


MACHINE DESIGN - An Integrated Approach, 4th Ed. Moment about y-axis

Iy 

1-8-4 2 5

2

M r

2

Iy  0.295  blob in

Iy  33362  kg mm Moment about z-axis




Iz 

kx 

ky 

kz 

2 5

2

M r

2

Iz  0.295  blob in

Iz  33362  kg mm Ix M Iy M Iz M

2

2

kx  1.897  in kx  48.193 mm ky  1.897  in ky  48.193 mm kz  1.897  in kz  48.193 mm



1-9-1


Convert the template in Problem 1-7 to have and use a set of functions or subroutines that can be called from within any program in that language to solve for the cross-sectional properties of the shapes shown on the inside front cover.

Solution:

See inside front cover and Mathcad file P0109.

1. Rectangle:

Area

A ( b h )  b  h

Moment about x-axis

Ix( b h ) 

Moment about y-axis

Iy( b h ) 

3

b h 12

3

h b 12 2

2. Solid circle:

Area

A ( D) 

π D 4

4

Moment about x-axis

Ix( D) 

π D 64

4

Moment about y-axis

3. Hollow circle: Area

Moment about x-axis

Moment about y-axis

Iy( D) 

π D 64

A ( D d )  Ix( D d ) 

Iy( D d ) 

π



2

2



4

4



4

4

 D d

4

π 64

π 64

 D d

 D d

  

4. Solid semicircle: 2

Area

A ( D) 

π D 8

Moment about x-axis

Ix( R)  0.1098 R

Moment about y-axis

Iy( R) 

π R

4

4

8

5. Right triangle: Area

Moment about x-axis

Moment about y-axis

A ( b h ) 

Ix( b h )  Iy( b h ) 

b h 2 b h

3

36 h b

3

36



1-10-1


Convert the template in Problem 1-8 to have and use a set of functions or subroutines that can be called from within any program in that language to solve for the cross-sectional properties of the shapes shown on the page opposite the inside front cover.

Solution:

See the page opposite the inside front cover and Mathcad file P0110.

1

Rectangular prism: Volume

V ( a b c)  a  b  c

Mass

M ( a b c γ) 

Moment about x-axis

Moment about y-axis

Moment about z-axis 2.

Ix( a b c γ)  Iy( a b c γ) 

Iz( a b c γ) 

V ( a b c)  γ g

2

2

2

2

2

2

M ( a b c γ)  a  b 12

M ( a b c γ)  a  c





12 M ( a b c γ)  b  c



12

Cylinder: 2

Volume

V ( r L)  π r  L

Mass

M ( r L γ) 

V ( r L)  γ g 2

Moment about x-axis

Moment about y-axis

Moment about z-axis

3.

Ix( r L γ)  Iy( r L γ)  Iz( r L γ) 

M ( r L γ)  r 2



2

2

2

12





2

M ( r L γ)  3  r  L M ( r L γ)  3  r  L



12

Hollow cylinder:

2



2

Volume

V ( a b L)  π b  a  L

Mass

M ( a b L γ) 

Moment about x-axis

Moment about y-axis

Moment about z-axis

Ix( a b L γ)  Iy( a b L γ)  Iz( a b L γ) 

V ( a b L)  γ g M ( a b L γ) 2 M ( a b L γ) 12 M ( a b L γ) 12

2

 a b



2



2

2

2



2

2

2

 3 a  3 b  L  3 a  3 b  L

 


MACHINE DESIGN - An Integrated Approach, 4th Ed. 4. Right circular cone:

1-10-2 2

Volume

Mass

Moment about x-axis

Moment about y-axis

Moment about z-axis 5.

π r  h

V ( r h ) 

3

M ( r h γ) 

Ix( r h γ) 

V ( r h )  γ g 3 10

2

 M ( r h γ)  r

2 2  12 r  3  h  I ( r h γ)  M ( r h γ)  y

80

Iz( r h γ)  M ( r h γ) 

12r2  3h2 80

Sphere: Volume

Mass

Moment about x-axis

Moment about y-axis

Moment about z-axis

V ( r) 

4 3

3

 π r

M ( r γ) 

V ( r)  γ g

Ix( r γ) 

2

Iy( r γ) 

2

Iz( r γ) 

2

5

5

5

2

 M ( r γ)  r

2

 M ( r γ)  r

2

 M ( r γ)  r



2-1-1


Figure P2-1 shows stress-strain curves for three failed tensile-test specimens. All are plotted on the same scale. (a) Characterize each material as brittle or ductile. (b) Which is the stiffest? (c) Which has the highest ultimate strength? (d) Which has the largest modulus of resilience? (e) Which has the largest modulus of toughness?

Solution:

See Figure P2-1 and Mathcad file P0201.

1.

The material in Figure P2-1(a) has a moderate amount of strain beyond the yield point, P2-1(b) has very little, and P2-1(c) has considerably more than either of the other two. Based on this observation, the material in Figure P2-1(a) is mildly ductile, that in P2-1(b)is brittle, and that in P2-1(c) is ductile.

2.

The stiffest material is the one with the grearesr slope in the elastic range. Determine this by dividing the rise by the run of the straight-line portion of each curve. The material in Figure P2-1(c) has a slope of 5 stress units per strain unit, which is the greatest of the three. Therefore, P2-1(c) is the stiffest.

3.

Ultimate strength corresponds to the highest stress that is achieved by a material under test. The material in Figure P2-1(b) has a maximum stress of 10 units, which is considerably more than either of the other two. Therefore, P2-1(b) has the highest ultimate strength.

4.

The modulus of resilience is the area under the elastic portion of the stress-starin curve. From observation of the three graphs, the stress and strain values at the yield points are: P2-1(a)

σya := 5

εya := 5

P2-1(b)

σyb := 9

εyb := 2

P2-1(c)

σyc := 5

εyc := 1

Using equation (2.7), the modulus of resiliency for each material is, approximately, P21a :=

P21b :=

P21c :=

1 2 1 2 1 2

⋅ σya ⋅ ε ya

P21a = 12.5

⋅ σyb ⋅ ε yb

P21b = 9

⋅ σyc⋅ ε yc

P21c = 2.5

P2-1 (a) has the largest modulus of resilience 5.

The modulus of toughness is the area under the stress-starin curve up to the point of fracture. By inspection, P2-1 (c) has the largest area under the stress-strain curve therefore, it has the largest modulus of toughness.



2-2-1


Determine an approximate ratio between the yield strength and ultimate strength for each material shown in Figure P2-1.

Solution:


1.

2.

The yield strength is the value of stress at which the stress-strain curve begins to be nonlinear. The ultimate strength is the maximum value of stress attained during the test. From the figure, we have the following values of yield strength and tensile strength: Figure P2-1(a)

S ya := 5

S ua := 6

Figure P2-1(b)

S yb := 9

S ub := 10

Figure P2-1(c)

S yc := 5

S uc := 8

The ratio of yield strength to ultimate strength for each material is: Figure P2-1(a)

Figure P2-1(b)

Figure P2-1(c)

ratioa :=

ratiob :=

ratioc :=

S ya S ua S yb S ub S yc S uc

ratioa = 0.83

ratiob = 0.90

ratioc = 0.63



2-3-1


Which of the steel alloys shown in Figure 2-19 would you choose to obtain (a) Maximum strength (b) Maximum modulus of resilience (c) Maximum modulus of toughness (d) Maximum stiffness

Given:

Young's modulus for steel

Solution:

See Figure 2-19 and Mathcad file P0203.

1.

E  207  GPa

Determine from the graph: values for yield strength, ultimate strength and strain at fracture for each material. Steel

Yield Strength

Ultimate Strength

Fracture Strain

AISI 1020:

Sy1020  300  MPa

Sut 1020  400  MPa

εf 1020  0.365

AISI 1095:

Sy1095  550  MPa

Sut 1095  1050 MPa

εf 1095  0.11

AISI 4142:

Sy4142  1600 MPa

Sut 4142  2430 MPa

εf 4142  0.06

Note: The 0.2% offset method was used to define a yield strength for the AISI 1095 and the 4142 steels. 2.

From the values of Sut above it is clear that the AISI 4142 has maximum strength.

3.

Using equation (2-7) and the data above, determine the modulus of resilience. 2

UR1020 

1 Sy1020  2 E

UR1020  0.22

3

m 2

1 Sy1095 UR1095   2 E

UR1095  0.73

MN  m 3

m 2

UR4142 

MN  m

1 Sy4142  2 E

UR4142  6.18

MN  m 3

m Even though the data is approximate, the AISI 4142 clearly has the largest modulus of resilience. 4.

Using equation (2-8) and the data above, determine the modulus of toughness. UT1020  UT1095  UT4142 

1 2 1 2 1 2

  Sy1020  Sut 1020  εf 1020

UT1020  128 

MN  m 3

m   Sy1095  Sut 1095  εf 1095

UT1095  88

MN  m 3

m   Sy4142  Sut 4142  εf 4142

UT4142  121 

MN  m 3

m

Since the data is approximate, there is no significant difference between the 1020 and 4142 steels. Because of the wide difference in shape and character of the curves, one should also determine the area under the curves by graphical means. When this is done, the area under the curve is about 62 square units for 1020 and 66 for 4142. Thus, they seem to have about equal toughness, which is about 50% greater than that for the 1095 steel. 5.

All three materials are steel therefore, the stiffnesses are the same.



2-4-1


Which of the aluminum alloys shown in Figure 2-21 would you choose to obtain (a) Maximum strength (b) Maximum modulus of resilience (c) Maximum modulus of toughness (d) Maximum stiffness

Given:

Young's modulus for aluminum

Solution:


1.

E  71.7 GPa

Determine, from the graph, values for yield strength, ultimate strength and strain at fracture for each material. Alum

Yield Strength

Ultimate Strength

Fracture Strain

1100:

Sy1100  120  MPa

Sut 1100  130  MPa

εf 1100  0.170

2024-T351:

Sy2024  330  MPa

Sut 2024  480  MPa

εf 2024  0.195

7075-T6:

Sy7075  510  MPa

Sut 7075  560  MPa

εf 7075  0.165

Note: The 0.2% offset method was used to define a yield strength for all of the aluminums. 2.

From the values of Sut above it is clear that the 7075-T6 has maximum strength.

3.


UR1100 

1 Sy1100  2 E

UR1100  0.10

1 Sy2024  2 E

3

m

2

UR2024 

MN  m

UR2024  0.76

MN  m 3

m 2

1 Sy7075 UR7075   2 E

UR7075  1.81

MN  m 3

m

Even though the data is approximate, the 7075-T6 clearly has the largest modulus of resilience. 4.

Using equation (2-8) and the data above, determine the modulus of toughness. UT1100  UT2024  UT7075 

1 2 1 2 1 2

  Sy1100  Sut 1100  εf 1100

UT1100  21

MN  m 3

m   Sy2024  Sut 2024  εf 2024

UT2024  79

MN  m 3

m   Sy7075  Sut 7075  εf 7075

UT7075  88

MN  m 3

m Even though the data is approximate, the 7075-T6 has the largest modulus of toughness. 5.

All three materials are aluminum therefore, the stiffnesses are the same.



2-5-1


Which of the thermoplastic polymers shown in Figure 2-22 would you choose to obtain (a) Maximum strength (b) Maximum modulus of resilience (c) Maximum modulus of toughness (d) Maximum stiffness

Solution:


1.

Determine, from the graph, values for yield strength, ultimate strength, strain at fracture, and modulus of elasticity for each material. Plastic

Yield Strength

Ultimate Strength

Fracture Strain

Mod of Elasticity

Nylon 101:

SyNylon  63 MPa

Sut Nylon  80 MPa

εf Nylon  0.52

ENylon  1.1 GPa

HDPE:

SyHDPE  15 MPa

Sut HDPE  23 MPa

εf HDPE  3.0

EHDPE  0.7 GPa

PTFE:

SyPTFE  8.3 MPa

Sut PTFE  13 MPa

εf PTFE  0.51

EPTFE  0.8 GPa

2.

From the values of Sut above it is clear that the Nylon 101 has maximum strength.

3.


URNylon 

1 SyNylon  2 ENylon

URNylon  1.8

MN  m 3

m 2

URHDPE 

1 SyHDPE  2 EHDPE 1 SyPTFE  2 EPTFE

3

m

2

URPTFE 

MN  m

URHDPE  0.16

URPTFE  0.04

MN  m 3

m

Even though the data is approximate, the Nylon 101 clearly has the largest modulus of resilience. 4.

Using equation (2-8) and the data above, determine the modulus of toughness. UTNylon 

UTHDPE 

UTPTFE 

  SyNylon  Sut Nylon  εf Nylon

1 2

2

2

MN  m 3

m

1

1

UTNylon  37

  SyHDPE  SutHDPE  εf HDPE

  SyPTFE  SutPTFE  εf PTFE

UTHDPE  57

MN  m 3

m UTPTFE  5 

MN  m 3

m

Even though the data is approximate, the HDPE has the largest modulus of toughness. 5.

The Nylon 101 has the steepest slope in the (approximately) elastic range and is, therefore, the stiffest of the three materials..



2-6-1


A metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. What is the modulus of elasticity? What is the strain energy at the elastic limit? Assume that the test speimen is 12.8-mm dia and has a 50-mm gage length. Can you define the type of metal based on the given data?

Given:

Elastic limit: Strength

S el  414  MPa

Strain

Test specimen: Diameter d o  12.8 mm Solution: 1.

Length Lo  50 mm


The modulus of elasticity is the slope of the stress-strain curve, which is a straight line, in the elastic region. Since one end of this line is at the origin, the slope (modulus of elasticity) is E 

2.

ε el  0.002

S el

E  207  GPa

ε el

The strain energy per unit volume at the elastic limit is the area under the stress-strain curve up to the elastic limit. Since the curve is a straight line up to this limit, the area is one-half the base times the height, or U'el 

1 2

 S el ε el

U'el  414 

kN  m 3

m

The total strain energy in the specimen is the strain energy per unit volume times the volume,

Uel  U'el

3.

π d o 4

2

 Lo

Uel  2.7 N  m

Based on the modulus of elasticity and using Table C-1, the material is steel.



2-7-1


A metal has a strength of 41.2 kpsi (284 MPa) at its elastic limit and the strain at that point is 0.004. What is the modulus of elasticity? What is the strain energy at the elastic limit? Assume that the test speimen is 0.505-in dia and has a 2-in gage length. Can you define the type of metal based on the given data?

Given:


S el  41.2 ksi

Strain

Test specimen: Diameter d o  0.505  in Solution: 1.

S el  284  MPa

Length Lo  2.00 in



2.

ε el  0.004

S el

6

E  10.3 10  psi

ε el

E  71 GPa

The strain energy per unit volume at the elastic limit is the area under the stress-strain curve up to the elastic limit. Since the curve is a straight line up to this limit, the area is one-half the base times the height, or

U'el 

1 2

 S el ε el

U'el  82.4

lbf  in 3

U'el  568 

in

kN  m 3

m

The total strain energy in the specimen is the strain energy per unit volume times the volume, Uel  U'el 3.

π d o 4

2

 Lo

Uel  33.0 in lbf

Based on the modulus of elasticity and using Table C-1, the material is aluminum.



2-8-1


A metal has a strength of 134 MPa at its elastic limit and the strain at that point is 0.006. What is the modulus of elasticity? What is the strain energy at the elastic limit? Assume that the test speimen is 12.8-mm dia and has a 50-mm gage length. Can you define the type of metal based on the given data?

Given:


S el  134  MPa

Strain

Test specimen: Diameter d o  12.8 mm Solution: 1.

Length Lo  50 mm



2.

ε el  0.003

S el

E  45 GPa

ε el


1 2

 S el ε el

U'el  201 

kN  m 3

m


Uel  U'el

3.

π d o 4

2

 Lo

Uel  1.3 N  m

Based on the modulus of elasticity and using Table C-1, the material is magnesium.



2-9-1


A metal has a strength of 100 kpsi (689 MPa) at its elastic limit and the strain at that point is 0.006. What is the modulus of elasticity? What is the strain energy at the elastic limit? Assume that the test speimen is 0.505-in dia and has a 2-in gage length. Can you define the type of metal based on the given data?

Given:


S el  100  ksi

Strain

ε el  0.006

S el  689  MPa Test specimen: Diameter d o  0.505  in Solution: 1.



2.

Length Lo  2.00 in

S el

6

E  16.7 10  psi

ε el

E  115  GPa


1 2

 S el ε el

U'el  300 

lbf  in 3

3 kN  m

U'el  2  10 

in

3

m


Uel  U'el

3.

π d o 4

2

 Lo

Uel  120.18 in lbf

Based on the modulus of elasticity and using Table C-1, the material is titanium.



2-10-1


A material has a yield strength of 689 MPa at an offset of 0.6% strain. What is its modulus of resilience?

Units:

MJ  10  joule

Given:

Yield strength

S y  689  MPa

Yield strain

ε y  0.006

Solution: 1.

6


The modulus of resilience (strain energy per unit volume) is given by Equation (2.7) and is approximately

UR 

1 2

 S y ε y

UR  2.067 

MJ 3

UR  2.1 MPa

m



2-11-1


A material has a yield strength of 60 ksi (414 MPa) at an offset of 0.2% strain. What is its modulus of resilience?

Units:

MJ  10  joule

Given:

Yield strength

S y  60 ksi

Yield strain

ε y  0.002

Solution: 1.

6

S y  414  MPa


The modulus of resilience (strain energy per unit volume) is given by Equation (2.7) and is approximately

UR 

1 2

 S y ε y

UR  60

in lbf 3

in

UR  0.414 

MJ 3

UR  0.414  MPa

m



2-12-1


A steel has a yield strength of 414 MPa, an ultimate tensile strength of 689 MPa, and an elongation at fracture of 15%. What is its approximate modulus of toughness? What is the approximate modulus of resilience?

Given:

S y  414  MPa

Solution:


1.

ε f  0.15

Determine the modulus of toughness using Equation (2.8).

UT 

2.

S ut  689  MPa

 Sy  S ut     εf  2 

UT  82.7

MN  m 3

UT  82.7 MPa

m

Determine the modulus of resilience using Equation (2.7) and Young's modulus for steel: E  207  GPa 2

1 Sy UR   2 E

UR  414 

kN  m 3

UR  0.41 MPa

m



2-13-1


The Brinell hardness of a steel specimen was measured to be 250 HB. What is the material's approximate tensile strength? What is the hardness on the Vickers scale? The Rockwell scale?

Given:

Brinell hardness of specimen

Solution:


HB  250

1. Determine the approximate tensile strength of the material from equations (2.10), not Table 2-3. S ut  0.5 HB ksi

S ut  125  ksi

S ut  862  MPa

2. From Table 2-3 (using linear interpolation) the hardness on the Vickers scale is HV 

HB  241 277  241

 ( 292  253 )  253

HV  263

3. From Table 2-3 (using linear interpolation) the hardness on the Rockwell C scale is HRC 

HB  241 277  241

 ( 28.8  22.8)  22.8

HRC  24.3



2-14-1


The Brinell hardness of a steel specimen was measured to be 340 HB. What is the material's approximate tensile strength? What is the hardness on the Vickers scale? The Rockwell scale?

Given:

Brinell hardness of specimen

Solution:


HB  340

1. Determine the approximate tensile strength of the material from equations (2.10), not Table 2-3. S ut  0.5 HB ksi

S ut  170  ksi

S ut  1172 MPa

2. From Table 2-3 (using linear interpolation) the hardness on the Vickers scale is HV 

HB  311 341  311

 ( 360  328 )  328

HV  359

3. From Table 2-3 (using linear interpolation) the hardness on the Rockwell C scale is HRC 

HB  311 341  311

 ( 36.6  33.1)  33.1

HRC  36.5



2-15-1


What are the principal alloy elements of an AISI 4340 steel? How much carbon does it have? Is it hardenable? By what techniques?

Solution:


1. Determine the principal alloying elements from Table 2-5 for 43xx steel.. 1.82% Nickel 0.50 or 0.80% Chromium 0.25% Molybdenum 2. From "Steel Numbering Systems" in Section 2.6, the carbon content is From the last two digits, the carbon content is 0.40%. 3. Is it hardenable? Yes, all of the alloying elements increase the hardenability. By what techniques? It can be through hardened by heating, quenching and tempering; and it can also be case hardened (See Section 2.4).



2-16-1



Solution:


1. Determine the principal alloying elements from Table 2-5 for 10xx steel. Carbon only, no alloying elements 2. From "Steel Numbering Systems" in Section 2.6, the carbon content is From the last two digits, the carbon content is 0.95%. 3. Is it hardenable? Yes, as a high-carbon steel, it has sufficient carbon content for hardening. By what techniques? It can be through hardened by heating, quenching and tempering; and it can also be case hardened (See Section 2.4).



2-17-1



Solution:


1. Determine the principal alloying elements from Table 2-5 for 61xx steel.. 0.15% Vanadium 0.60 to 0.95% Chromium 2. From "Steel Numbering Systems" in Section 2.6, the carbon content is From the last two digits, the carbon content is 0.80%. 3. Is it hardenable? Yes, all of the alloying elements increase the hardenability. By what techniques? It can be through hardened by heating, quenching and tempering; and it can also be case hardened (See Section 2.4).



2-18-1


Which of the steels in Problems 2-15, 2-16, and 2-17 is the stiffest?

Solution:


1. None. All steel alloys have the same Young's modulus, which determines stiffness.



2-19-1


Calculate the specific strength and specific stiffness of the following materials and pick one for use in an aircraft wing spar.

Given:

Material

Code

Steel

st  0

Ultimate Strength Sut  80 ksi st

Young's Modulus 6

E  30 10  psi st

Weight Density lbf

γ  0.28 st

3

in Aluminum

al  1

Sut

al

 60 ksi

E

6

al

 10.4 10  psi

γ  0.10

lbf

al

3

in Titanium

ti  2

Sut  90 ksi ti

6

E  16.5 10  psi ti

γ  0.16 ti

lbf 3

in Index Solution: 1.

i  0 1  2


Specific strength is the ultimate tensile strength divided by the weight density and specific stiffness is the modulus of elasticity divided by the weight density. The text does not give a symbol to these quantities. Sut Specific strength

γ

i 1



i

in

E 

286·103 600·103 563·103

2.

Specific stiffness

i 1



γ in



i

Steel Aluminum Titanium

107·106 104·106 103·106

Based on the results above, all three materials have the same specific stiffness but the aluminum has the largest specific strength. Aluminum for the aircraft wing spar is recommended.



2-20-1


If maximum impact resistance were desired in a part, which material properties would you look for?

Solution:


1. Ductility and a large modulus of toughness (see "Impact Resistance" in Section 2.1).



2-21-1

PROBLEM 2-21

_____

Statement:

Refer to the tables of material data in Appendix A and determine the strength-to-weight ratios of the following material alloys based on their tensile yield strengths: heat-treated 2024 aluminum, SAE 1040 cold-rolled steel, Ti-75A titanium, type 302 cold-rolled stainless steel.

Given:

Material

Yield Strength

Mat  "2024 Aluminum, HT" Sy  290  MPa 1

1

Specific Weight 3

γ  0.10 lbf  in

γ  27.14 

1

1

kN 3

m Mat  "1040 CR Steel" 2

Sy  490  MPa 2

3

γ  0.28 lbf  in

γ  76.01 

2

2

kN 3

m Mat  "Ti-75A Titanium" 3

Sy  517  MPa 3

3

γ  0.16 lbf  in

γ  43.43 

3

3

kN 3

m Mat  "Type 302 CR SS" 4

Sy  1138 MPa 4

3

γ  0.28 lbf  in

γ  76.01 

4

4

kN 3

m i  1 2  4 Solution: 1.


Calculate the strength-to-weight ratio for each material as described in Section 2.1.

SWR  i

Sy

SWR

γ

10  m

i

i

4

i

 "2024 Aluminum, HT"  "1040 CR Steel"  Mat   i  "Ti-75A Titanium"   "Type 302 CR SS"   



1.068 0.645 1.190 1.497



2-22-1

PROBLEM 2-22

_____

Statement:

Refer to the tables of material data in Appendix A and determine the strength-to-weight ratios of the following material alloys based on their ultimate tensile strengths: heat-treated 2024 aluminum SAE 1040 cold-rolled steel, unfilled acetal plastic, Ti-75A titanium, type 302 cold-rolled stainless steel.

Given:

Material

Tensile Strength

Specific Weight 3

Mat  "2024 Aluminum, HT" Sut  441  MPa

γ  0.10 lbf  in

Mat  "1040 CR Steel"

Sut  586  MPa

γ  0.28 lbf  in

Mat  "Acetal, unfilled"

Sut  60.7 MPa

γ  0.051  lbf  in

Mat  "Ti-75A Titanium"

Sut  586  MPa

γ  0.16 lbf  in

Mat  "Type 302 CR SS"

Sut  1310 MPa

γ  0.28 lbf  in

1

1

2 3 4 5

2 3 4 5

3

γ  27.14  kN  m

1

1

3

3

γ  76.01  kN  m

2

2

3

3

3

3

3

γ  43.43  kN  m

4

4

3

5

3

γ  13.84  kN  m

3

γ  76.01  kN  m 5

i  1 2  5 Solution: 1.


Calculate the strength-to-weight ratio for each material as described in Section 2.1.

Sut SWR  i

γ

SWR

i

i

4



10  m

i

 "2024 Aluminum, HT" 

 "1040 CR Steel"    Mat   "Acetal, unfilled"  i  "Ti-75A Titanium"     "Type 302 CR SS" 

1.625 0.771 0.438 1.349 1.724



PROBLEM 2-23

2-23-1

_____

Statement:

Refer to the tables of material data in Appendix A and calculate the specific stiffness of aluminum, titanium, gray cast iron, ductile iron, bronze, carbon steel, and stainless steel. Rank them in increasing order of this property and discuss the engineering significance of these data.

Units:

Mg  10  kg

Given:

Material

3

Modulus of Elasticity

Density 3

Mat  "Aluminum"

E  71.7 GPa

ρ  2.8 Mg m

Mat  "Titanium"

E  113.8  GPa

ρ  4.4 Mg m

Mat  "Gray cast iron"

E  103.4  GPa

ρ  7.2 Mg m

Mat  "Ductile iron"

E  168.9  GPa

ρ  6.9 Mg m

Mat  "Bronze"

E  110.3  GPa

ρ  8.6 Mg m

Mat  "Carbon steel"

E  206.8  GPa

ρ  7.8 Mg m

Mat  "Stainless steel"

E  189.6  GPa

ρ  7.8 Mg m

1

1

2

2

3

3

4

4

5

5

6

6

7

7

1

3

2

3

3

3

4

3

5

3

6

3

7

i  1 2  7 Solution: 1.


Calculate the specific stiffness for each material as described in Section 2.1. E E'  i

2.

ρ

E'

i i

 "Aluminum"     "Titanium"   "Gray cast iron"  Mat   "Ductile iron"  i    "Bronze"   "Carbon steel"   "Stainless steel"   

10

i 6

2



s

2



m

25.6 25.9 14.4 24.5 12.8 26.5 24.3

Rank them in increasing order of specific stiffness. E' Mat  "Bronze"

2 5 s

5

10 E' Mat  "Gray cast iron"

10 E'



2

6



2

10

6

 14.4

m

2 7 s

7

 12.8

m

2 3 s

3

Mat  "Stainless steel"

6



2

 24.3

m



E' Mat  "Ductile iron"

2 4 s

4

10 E' Mat  "Aluminum"

10 E'

10 E'

6



2

6

10

6

 24.5

 25.6

m 

2

 25.9

m

2 6 s

6

3.

2

m

2 2 s

2

Mat  "Carbon steel"



2 1 s

1

Mat  "Titanium"

6

2-23-2



2

 26.5

m

Bending and axial deflection are inversely proportional to the modulus of elasticity. For the same shape and dimensions, the material with the highest specific stiffness will give the smallest deflection. Or, put another way, for a given deflection, using the material with the highest specific stiffness will result in the least weight.



2-24-1


Call your local steel and aluminum distributors (consult the Yellow Pages) and obtain current costs per pound for round stock of consistent size in low-carbon (SAE 1020) steel, SAE 4340 steel, 2024-T4 aluminum, and 6061-T6 aluminum. Calculate a strength/dollar ratio and a stiffness/dollar ratio for each alloy. Which would be your first choice on a cost-efficiency basis for an axial-tension-loaded round rod (a) If maximum strength were needed? (b) If maximum stiffness were needed?

Solution:

Left to the student as data will vary with time and location.



2-25-1


Call your local plastic stock-shapes distributors (consult the Yellow Pages) and obtain current costs per pound for round rod or tubing of consistent size in plexiglass, acetal, nylon 6/6, and PVC. Calculate a strength/dollar ratio and a stiffness/dollar ratio for each alloy. Which would be your first choice on a cost-efficiency basis for an axial-tension-loaded round rod or tube of particular diameters. (a) If maximum strength were needed? (b) If maximum stiffness were needed?

Solution:

Left to the student as data will vary with time and location.



2-26-1


A part has been designed and its dimensions cannot be changed. To minimize its deflections under the same loading in all directions irrespective of stress levels, which material woulod you choose among the following: aluminum, titanium, steel, or stainless steel?

Solution:


1.

Choose the material with the highest modulus of elasticity because deflection is inversely proportional to modulus of elasticity. Thus, choose steel unless there is a corrosive atmosphere, in which case, choose stainless steel.



2-27-1


Assuming that the mechanical properties data given in Appendix Table A-9 for some carbon steels represents mean values, what is the value of the tensile yield strength for 1050 steel quenched and tempered at 400F if a reliability of 99.9% is required?

Given:

Mean yield strength

Solution:


1.

S y  117  ksi

S y  807  MPa

From Table 2-2 the reliability factor for 99.9% is Re  0.753. Applying this to the mean tensile strength gives S y99.9  S y Re

S y99.9  88.1 ksi

S y99.9  607  MPa



2-28-1


Assuming that the mechanical properties data given in Appendix Table A-9 for some carbon steels represents mean values, what is the value of the ultimate tensile strength for 4340 steel quenched and tempered at 800F if a reliability of 99.99% is required?

Given:

Mean ultimate tensile strength

Solution:


1.

S ut  213  ksi

S ut  1469 MPa

From Table 2-2 the reliability factor for 99.99% is Re  0.702. Applying this to the mean ultimate tensile strength gives S ut99.99  S ut Re

S ut99.99  150  ksi

S ut99.99  1031 MPa



2-29-1


Assuming that the mechanical properties data given in Appendix Table A-9 for some carbon steels represents mean values, what is the value of the ultimate tensile strength for 4130 steel quenched and tempered at 400F if a reliability of 90% is required?

Given:

Mean ultimate tensile strength

Solution:


1.

S ut  236  ksi

S ut  1627 MPa

From Table 2-2 the reliability factor for 90% is Re  0.897. Applying this to the mean ultimate tensile strength gives S ut99.99  S ut Re

S ut99.99  212  ksi

S ut99.99  1460 MPa



2-30-1


Assuming that the mechanical properties data given in Appendix Table A-9 for some carbon steels represents mean values, what is the value of the tensile yield strength for 4140 steel quenched and tempered at 800F if a reliability of 99.999% is required?

Given:

Mean yield strength

Solution:


S y  165  ksi

S y  1138 MPa

1. From Table 2-2 the reliability factor for 99.999% is Re  0.659. Applying this to the mean tensile strength gives S y99.9  S y Re S y99.9  109  ksi S y99.9  750  MPa



2-31-1


A steel part is to be plated to give it better corrosion resistance. Two materials are being considered: cadmium and nickel. Considering only the problem of galvanic action, which would you chose? Why?

Solution:


1.

From Table 2-4 we see that cadmium is closer to steel than nickel. Therefore, from the standpoint of reduced galvanic action, cadmium is the better choice. Also, since cadmium is less noble than steel it will be the material that is consumed by the galvanic action. If nickel were used the steel would be consumed by galvanic action.



2-32-1


A steel part with many holes and sharp corners is to be plated with nickel. Two processes are being considered: electroplating and electroless plating. Which process would you chose? Why?

Solution:


1.

Electroless plating is the better choice since it will give a uniform coating thickness in the sharp corners and in the holes. It also provides a relatively hard surface of about 43 HRC.



2-33-1


What is the common treatment used on aluminum to prevent oxidation? What other metals can also be treated with this method? What options are available with this method?

Solution:


1.

Aluminum is commonly treated by anodizing, which creates a thin layer of aluminum oxide on the surface. Titanium, magnesium, and zinc can also be anodized. Common options include tinting to give various colors to the surface and the use of "hard anodizing" to create a thicker, harder surface.



2-34-1


Steel is often plated with a less nobel metal that acts as a sacrificial anode that will corrode instead of the steel. What metal is commonly used for this purpose (when the finished product will not be exposed to saltwater), what is the coating process called, and what are the common processes used to obtain the finished product?

Solution:


1.

The most commonly used metal is zinc. The process is called "galvanizing" and it is accomplished by electroplating or hot dipping.



2-35-1


A low-carbon steel part is to be heat-treated to increase its strength. If an ultimate tensile strength of approximately 550 MPa is required, what mean Brinell hardness should the part have after treatment? What is the equivalent hardness on the Rockwell scale?

Given:

Approximate tensile strength

Solution:


1.

Use equation (2.10), solving for the Brinell hardness, HB. S ut = 3.45 HB

2.

S ut  550  MPa

HB 

S ut 3.45 MPa

HB  159

From Table 2-3, the equivalent hardness on the Rocwell scale is 83.9HRB.



2-36-1


A low-carbon steel part has been tested for hardness using the Brinell method and is found to have a hardness of 220 HB. What are the approximate lower and upper limits of the ultimate tensile strength of this part in MPa?

Given:

Hardness

Solution:


1.

HB  220

Use equation (2.10), solving for ultimate tensile strength. Minimum:

S utmin  ( 3.45 HB  0.2 HB)  MPa

S utmin  715  MPa

Maximum:

S utmax  ( 3.45 HB  0.2 HB)  MPa

S utmax  803  MPa



2-37-1


Figure 2-24 shows "guide lines" for minimum weight design when failure is the criterion. The guide line, or index, for minimizing the weight of a beam in bending is f2/3/, where f is the yield strength of a material and  is its mass density. For a given cross-section shape the weight of a beam with given loading will be minimized when this index is maximized. The following materials are being considered for a beam application: 5052 aluminum, cold rolled; CA-170 beryllium copper, hard plus aged; and 4130 steel, Q & T @ 1200F. The use of which of these three materials will result in the least-weight beam? 3

Units:

Mg  kg

Given:

5052 Aluminum

3

S ya  255  MPa

ρa  2.8 Mg m

3

CA-170 beryllium copper S yb  1172 MPa 4130 steel Solution:

ρb  8.3 Mg m

3

S ys  703  MPa

ρs  7.8 Mg m


1.

The values for the mass density are taken from Appendix Table A-1 and the values of yield strength come from from Tables A-2, A-4, and A-9 for aluminum, beryllium copper, and steel, respectively.

2.

Calculate the index value for each material. Index S y ρ  

3

0.667

Sy

ρ



Mg m

0.667

MPa

Aluminum

Ia  Index S ya ρa 

Ia  14.4

Beryllium copper

Ib  Index S yb ρb 

Ib  13.4

Steel

Is  Index S ys ρs 

Is  10.2

The 5052 aluminum has the highest value of the index and would be the best choice to minimize weight.



2-38-1


Figure 2-24 shows "guide lines" for minimum weight design when failure is the criterion. The guide line, or index, for minimizing the weight of a member in tension is f/, where f is the yield strength of a material and  is its mass density. The weight of a member with given loading will be minimized when this index is maximized. For the three materials given in Problem 2-37, which will result in the least weight tension member? 3

Units:

Mg  kg

Given:

5052 Aluminum

3

S ya  255  MPa

ρa  2.8 Mg m

CA-170 beryllium copper S yb  1172 MPa 4130 steel Solution:

3

ρb  8.3 Mg m

3

S ys  703  MPa

ρs  7.8 Mg m


1.

The values for the mass density are taken from Appendix Table A-1 and the values of yield strength come from from Tables A-2, A-4, and A-9 for aluminum, beryllium copper, and steel, respectively.

2.

Calculate the index value for each material. Index S y ρ  

S y Mg m 3  MPa ρ

Aluminum

Ia  Index S ya ρa 

Ia  91.1

Beryllium copper

Ib  Index S yb ρb 

Ib  141.2

Steel

Is  Index S ys ρs 

Is  90.1

The beryllium copper has the highest value of the index and would be the best choice to minimize weight.



2-39-1


Figure 2-23 shows "guide lines" for minimum weight design when stiffness is the criterion. The guide line, or index, for minimizing the weight of a beam in bending is E1/2/, where E is the modulus of elasticity of a material and  is its mass density. For a given cross-section shape the weight of a beam with given stiffness will be minimized when this index is maximized. The following materials are being considered for a beam application: 5052 aluminum, cold rolled; CA-170 beryllium copper, hard plus aged; and 4130 steel, Q & T @ 1200F. The use of which of these three materials will result in the least-weight beam? 3

Units:

Mg  kg

Given:

5052 Aluminum

ρa  2.8 Mg m

3

CA-170 beryllium copper Eb  127.6  GPa

ρb  8.3 Mg m

Es  206.8  GPa

ρs  7.8 Mg m

4130 steel Solution:

3

Ea  71.7 GPa

3


1.

The values for the mass density and modulus are taken from Appendix Table A-1.

2.

Calculate the index value for each material.

Index( E ρ ) 

E

3

0.5

ρ



Mg m GPa

0.5

Aluminum

Ia  Index Ea ρa 

Ia  3.0

Beryllium copper

Ib  Index Eb ρb 

Ib  1.4

Steel

Is  Index Es ρs 

Is  1.8

The 5052 aluminum has the highest value of the index and would be the best choice to minimize weight.



2-40-1


Figure 2-24 shows "guide lines" for minimum weight design when stiffness is the criterion. The guide line, or index, for minimizing the weight of a member in tension is E/, where E is the modulus of elasticity of a material and  is its mass density. The weight of a member with given stiffness will be minimized when this index is maximized. For the three materials given in Problem 2-39, which will result in the least weight tension member?

Units:

Mg  kg

Given:

5052 Aluminum

3

ρa  2.8 Mg m

3

CA-170 beryllium copper Eb  127.6  GPa

ρb  8.3 Mg m

Es  206.8  GPa

ρs  7.8 Mg m

4130 steel Solution:

3

Ea  71.7 GPa

3


1.

The values for the mass density and modulus are taken from Appendix Table A-1.

2.

Calculate the index value for each material. 3

Index( E ρ ) 

E Mg m  ρ GPa

Aluminum

Ia  Index Ea ρa 

Ia  25.6

Beryllium copper

Ib  Index Eb ρb 

Ib  15.4

Steel

Is  Index Es ρs 

Is  26.5

The steel has the highest value of the index and would be the best choice to minimize weight.



3-1-1


Which load class from Table 3-1 best suits these systems? (a) Bicycle frame (b) Flag pole (c) Boat oar (d) Diving board (e) Pipe wrench (f) Golf club.

Solution:


1. Determine whether the system has stationary or moving elements, and whether the there are constant or time-varying loads. (a) Bicycle frame Class 4 (Moving element, time-varying loads) (b) Flag pole

Class 2 (Stationary element, time-varying loads)

(c) Boat oar

Class 2 (Low acceleration element, time-varying loads)

(d) Diving board Class 2 (Stationary element, time-varying loads) (e) Pipe wrench Class 2 (Low acceleration elements, time-varying loads) (f) Golf club

Class 4 (Moving element, time-varying loads)



3-2a-1

PROBLEM 3-2a Statement:

Draw free-body diagrams for the system of Problem 3-1a (bicycle frame).

Assumptions: 1. A two-dimensional model is adequate. 2. The lower front-fork bearing at C takes all of the thrust load from the front forks. 3. There are no significant forces on the handle bars. Solution: 1.

See Mathcad file P0302a.

A typical bicycle frame is shown in Figure 3-2a. There are five points on the frame where external forces and moments are present. The rider's seat is mounted through a tube at A. This is a rigid connection, capable of transmitting two force components and a moment. The handle bars and front-wheel forks are supported by the frame through two bearings, located at B and C. These bearings are capable of transmitting radial and axial loads. The pedal-arm assembly is supported by bearings at D. These bearings are capable of transmitting radial loads. The rear wheel-sprocket assembly is supported by bearings mounted on an axle fixed to the frame at E.

Ma

Rb

Ra

A

B Fbr

Fax

Fct Fcr

Rc

α C

Fay Rd Re Fey

Fex

E

Fdx

D

Fdy

FIGURE 3-2a Free Body Diagram for Problem 3-2a

2. The loads at B and C can be determined by analyzing a FBD of the front wheel-front forks assembly. The loads at D can be determined by analyzing a FBD of the pedal-arm and front sprocket (see Problem 3-3), and the loads at E can be determined by analyzing a FBD of the rear wheel-sprocket assembly. 3. With the loads at B, C, D, and E known, we can apply equations 3.3b to the FBD of the frame and solve for Fax , Fay , and Ma.

Σ Fx :

−Fax − Fbr⋅ cos( α) + Fcr⋅ cos( α) − Fct⋅ sin( α) − Fdx + Fex = 0

(1)

Σ Fy :

−Fay − Fbr⋅ sin( α) + Fcr⋅ sin( α) + Fct⋅ cos( α) − Fdy + Fey = 0

(2)

Σ Mz:

Ma + ( Rbx ⋅ Fby − Rby⋅ Fbx ) + ( Rcx⋅ Fcy − Rcy⋅ Fcx) ... = 0 + ( R ⋅ F − R ⋅ F ) + ( R ⋅ F − R ⋅ F )  ex ey ey ex  dx dy dy dx 

(3)

© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0302a.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


3-2e-1

PROBLEM 3-2e Statement:

Draw free-body diagrams for the system of Problem 3-1e (pipe wrench).

Assumptions: A two-dimensional model is adequate. Solution:

See Mathcad file P0302e.

1. A typical pipe wrench with a pipe clamped in its jaw is shown in Figure 3-2e(a). When a force Fhand is applied on the wrench, the piping system provides an equal and opposite force and a resisting torque, Tpipe.

Fhand

Tpipe Fhand a

(a) FBD of pipe wrench and pipe

Fbt Fbn Fax

A α Fay d

b

(b) FBD of pipe wrench only FIGURE 3-2e Free Body Diagrams for Problem 3-2e

2. The pipe reacts with the wrench at the points of contact A and B. The forces here will be directed along the common normals and tangents. The jaws are slightly tapered and, as a result, the action of Fhand tends to drive the wrench further into the taper, increasing the normal forces. This, in turn, allows for increasing tangential forces. It is the tangential forces that produce the turning torque. 3.

4.

Applying equations 3.3b to the FBD of the pipe wrench,

Σ Fx :

−Fax + Fbn⋅ cos( α) − Fbt⋅ sin( α) = 0

(1)

Σ Fy :

−Fay + Fbn⋅ sin( α) + Fbt⋅ cos( α) − Fhand = 0

(2)

Σ M A:

d ⋅ ( Fbt⋅ cos( α) + Fbn⋅ sin( α) ) − ( d + a ) ⋅ Fhand = 0

(3)

These equations can be solved for the vertical forces if we assume α is small so that sin(α) = 0 and cos (α) = 1.

© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0302e.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


3-3-1


Draw a free-body diagram of the pedal-arm assembly from a bicycle with the pedal arms in the horizontal position and dimensions as shown in Figure P3-1. (Consider the two arms, pedals and pivot as one piece). Assuming a rider-applied force of 1500 N at the pedal, determine the torque applied to the chain sprocket and the maximum bending moment and torque in the pedal arm.

Given:

a  170  mm

b  60 mm

Frider  1.5 kN

Assumptions: The pedal-arm assembly is supprted by bearings at A and at B. Solution:


1. The free-body diagram (FBD) of the pedal-arm assembly (including the sprocket) is shown in Figure 3-3a. The rider-applied force is Frider and the force applied by the chain (not shown) is Fchain. The radial bearing reactions are Fax, Faz, Fbx, and Fbz. Thus, there are five unknowns Fchain, Fax, Faz, Fbx, and Fbz. In general, we can write six equilibrium equations for a three-dimensional force system, but in this system there are no forces in the y-direction so five equations are available to solve for the unknowns. z Fchain Sprocket

Faz Fbz

a Frider

A B

Arm

b

Arm (sectioned)

Fax Fbx

y

Pedal x

(a) FBD of complete pedal-arm assembly z

a

Tc

Frider b

Mc Arm

Fc

y

Pedal x

(b) FBD of pedal and arm with section through the origin FIGURE 3-3 Dimensions and Free Body Diagram of the pedal-arm assembly for Problem 3-3

2.

The torque available to turn the sprocket is found by summing moments about the sprocket axis. From Figure 3-3a, it is



 Ty-axis:

3-3-2

a  Frider  r Fchain = a  Frider  Tsprocket = 0 Tsprocket  a  Frider

Tsprocket  255  N  m

where r is the sprocket pitch radius. 3.

In order to determine the bending moment and twisting torque in the pedal arm, we will cut the arm with a section plane that goes through the origin and is parallel to the y-z plane, removing everything beyond that plane and replacing it with the internal forces and moments in the pedal arm at the section. The resulting FBD is shown in Figure 3-3b. The internal force at section C is Fc the internal bending moment is Mc, and the internal twisting moment (torque) is Tc. We can write three equilibrium equations to solve for these three unknowns: Shear force in pedal arm at section C

 Fz :

Fc  Frider = 0

Fc  Frider

Fc  1.5 kN

Mc  a  Frider

Mc  255  N  m

Tc  b  Frider

Tc  90 N  m

Bending moment in pedal arm at section C

 My-axis:

a  Frider  Mc = 0

Twisting moment in pedal arm at section C

 Mx-axis:

b  Frider  Tc = 0


MACHINE DESIGN - An Integrated Approach,4th Ed.

3-4-1


The trailer hitch from Figure 1-1 has loads applied as shown in Figure P3-2. The tongue weight of 100 kg acts downward and the pull force of 4905 N acts horizontally. Using the dimensions of the ball bracket in Figure 1-5 (p. 15), draw a free-body diagram of the ball bracket and find the tensile and shear loads applied to the two bolts that attach the bracket to the channel in Figure 1-1.

Given:

a  40 mm Mtongue  100  kg

b  31 mm Fpull  4.905  kN

c  70 mm t  19 mm

d  20 mm

Assumptions: 1. The nuts are just snug-tight (no pre-load), which is the worst case. 2. All reactions will be concentrated loads rather than distributed loads or pressures. Solution: 1.


The weight on the tongue is Wtongue  Mtongue g

Wtongue  0.981  kN

2. The FBD of the hitch and bracket assembly is shown in Figure 3-4. The known external forces that act on the ball are Fpull and Wtongue . The reactions on the bracket are at points C and D. The bolts at C provide tensile (Fc2x) and shear (Fc2y) forces, and the bracket resists rotation about point D where the reaction force Fd2 is applied by the channel to which the bracket is bolted. 3. Solving for the reactions by summing the horizontal and vertical forces and the moments about D:

 Fx :

 Fpull  Fc2x  Fd2 = 0

(1)

 Fy :

Fc2y  Wtongue = 0

(2)

 MD :

Fc2x d  Fpull  ( a  t  b  d )  Wtongue c = 0

(3)

W tongue 70 = c

1

F pull

1

40 = a 2

B

A

2

19 = t 31 = b

C

Fc2x

20 = d D

B

C D

Fd2 F c2y

FIGURE 3-4 Dimensions and Free Body Diagram for Problem 3-4

4.

Solving equation (3) for Fc2x



Fc2x  5.

Fpull  ( a  t  b  d )  Wtongue c d

(4)

Fd2  25.505 kN

(5)

Fc2y  0.981  kN

(6)

Solving (2) for Fc2y Fc2y  Wtongue

7.

Fc2x  30.41  kN

Substituting into (1) and solving for Fd2 Fd2  Fc2x  Fpull

6.

3-4-2

The loads applied to the two bolts that attach the bracket to the channel are: Axial force on two bolts

Fc2x  30.4 kN

Shear force on two bolts

Fc2y  0.98 kN

We assume that each bolt would carry one half of these loads.



3-5-1


For the trailer hitch of Problem 3-4, determine the horizontal force that will result on the ball from accelerating a 2000-kg trailer to 60 m/sec in 20 sec.

Given:

Mass of trailer

Mtrailer  2000 kg

Final velocity

vf  60

Time to reach velocity

τ  20 sec

m sec

Assumptions: 1. Acceleration is constant. 2. The rolling resistance of the tires and the wheel bearings is negligible. Solution: 1.


From elementary kinematics, the acceleration required is a 

2.

vf

τ

a  3.00

m sec

(1) 2

Using Newton's second law to find the force required to accelerate the trailer, Fhitch  Mtrailer  a

Fhitch  6.00 kN

(2)



3-6-1


For the trailer hitch of Problem 3-4, determine the horizontal force that will result on the ball from an impact between the ball and the tongue of the 2000-kg trailer if the hitch deflects 2.8 mm dynamically on impact. The tractor weighs 1000 kg. The velocity at impact is 0.3 m/sec.

Given:

Mass of trailer

Mtrailer  2000 kg

Dynamic deflection

δi  2.8 mm

Mass of tractor

Mtractor  1000 kg

Impact velocity

vi  0.3

m sec

Assumptions: 1. The tractor is the "struck member" because the hitch is on the tractor and it is the hitch that deflects. 2. Equations (3.9) and (3.10) can be used to model the impact. Solution: 1.


The weight of the trailer (the "striking member") is Wtrailer  Mtrailer  g

2.

The correction factor, from equation (3-15), is 1

η  1

3.

Wtrailer  19.613 kN

Mtractor

η  0.857

3  Mtrailer

Eliminating E from equations (3.9a) and (3.10) and solving for the horizontal force on the ball Fi yields 1 2  Fi δi = η   Mtrailer  vi  2 2  1

Fi 

η Mtrailer  vi δi

2

Fi  55.1 kN



3-7-1


The piston of an internal-combustion engine is connected to its connecting rod with a "wrist pin." Find the force on the wrist pin if the 0.5-kg piston has an acceleration of 2 500 g.

Given:

Mass of piston

Mpiston  0.5 kg

Acceleration of piston

a piston  2500 g

Assumptions: The force on the wrist pin due to the weight of the piston is very small compared with the acceleration force. Solution: 1.


The acceleration in m/s is

4

a piston  2.452  10 

m sec

2.

2

Using Newton's Second Law expressed in equation (3.1a), the force on the wrist pin is Fwristpin  Mpiston  a piston

Fwristpin  12.258 kN



PROBLEM 3-8

3-8-1

_____

Statement:

A cam-follower system similar to that shown in Figure 3-15 has a mass m = 1 kg, a spring constant k = 1000 N/m, and a damping coefficient d = 19.4 N-s/m. Find the undamped and damped natural frequencies of this system.

Units:

cps := 2 ⋅ π⋅ rad⋅ sec

Given:

Mass

−1

M := 1 ⋅ kg,

Spring constant

−1

k := 1000⋅ N ⋅ m

−1

Damping coefficient d := 19.4⋅ N ⋅ s⋅ m Solution: 1.


Calculate the undamped natural frequency using equation 3.4.

ωn := 2.

k

ωn = 31.6

M

rad sec

ωn = 5.03 cps

Calculate the undamped natural frequency using equation 3.7.

ωd :=

d  −   M  2⋅ M  k

2

ωd = 30.1

rad sec

ωd = 4.79 cps



3-9-1


A ViseGrip plier-wrench is drawn to scale in Figure P3-3. Scale the drawing for dimensions. Find the forces acting on each pin and member of the assembly for an assumed clamping force of P = 4000 N in the position shown. What force F is required to keep it in the clamped position shown?

Given:

Clamping force Dimensions

P  4.00 kN a  50.0 mm b  55.0 mm c  39.5 mm d  22.0 mm

e  28.0 mm f  26.9 mm g  2.8 mm h  21.2 mm

α  21.0 deg β  129.2  deg

Assumptions: Links 3 and 4 are in a toggle position, i.e., the pin that joins links 3 and 4 is in line with the pins that join 1 with 4 and 2 with 3. Solution: 1.


The FBDs of the assembly and each individual link are shown in Figure 3-9. The dimensions, as scaled from Figure P3-3 in the text, are given above and are shown on the link FBDs. 4

F

P 1

2

3

P

F 55.0 = b

50.0 = a

39.5 = c

F

F14 22.0 = d

129.2°

1



4 F34

F41

F21

P



28.0 = e  

F43



F12

3 F23

F32

P

2.8 = g

21.2 = h

2

F 26.9 = f

FIGURE 3-9 Free Body Diagrams for Problem 3-9

2.

Looking first at Part 3, we see that it is a three-force body. Therefore, the lines of action of the three forces must intersect at a point. But, since Parts 3 and 4 are in a toggle position, F43 and F23 are colinear, which means that their x- and y-components must be equal and opposite, leading to the conclussion that F = 0.

© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0309.xmcd 3-9-1 mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


3.

3-9-2

Now, looking at Part 1, we see that (for F = 0) it is also a three-force body, as is Part 2. In fact, the forces on Part 1 are identical to those on Part 2. Solving for the unknown reactions on Parts 1 and 2,  Fx:

F41 cos( 180  deg  α)  F21 cos( β  180  deg) = 0

(a)

 Fy:

F41 sin( 180  deg  α)  F21 sin( β  180  deg)  P = 0

(b)

Solving equation (a) for F21 F21 = 

F41 cos( 180  deg  α)

(c)

cos( β  180  deg)

Substituting equation (c) into (b) F41 sin( 180  deg  α) 

F41 cos( 180  deg  α) cos( β  180  deg)

 sin( β  180  deg)  P = 0

(d)

Solving equation (d) for F41 P

F41   sin( 180  deg  α) 

F21  

cos( 180  deg  α) cos( β  180  deg)

 sin( β  180  deg)

F41 cos( 180  deg  α)

F41  5.1 kN

cos( β  180  deg) F21  7.5 kN

Checking moment balance on Part 1, F41 sin( α)  c  F21 sin( β  90 deg)  d  P a  0  kN  m The result is, within the accuracy of the scaled dimensions, zero as it must be. 4.

5.

6.

The x and y components of the pin forces on Part 1 are F41x  F41 cos( 180  deg  α)

F41x  4.749  kN

F41y  F41 sin( 180  deg  α)

F41y  1.823  kN

F21x  F21 cos( β  180  deg)

F21x  4.749  kN

F21y  F21 sin( β  180  deg)

F21y  5.823  kN

The forces on the pins at the ends of Part 4 are F14  F41

F14  5.1 kN

F34  F14

F34  5.1 kN


F43  5.1 kN

F23  F43

F23  5.1 kN



7.

3-9-3


F12  7.5 kN

F32  F23

F32  5.1 kN

Checking moment equilibrium on Part 2, F12 ( e cos( β  90 deg)  g  sin( β  90 deg) )   0  kN  m  F32 ( h  cos( α)  f  sin( α) ) which is zero, as it must be.



3-10-1


Given:

An overhung diving board is shown in Figure P3-4a. Find the reaction forces and construct the shear and moment diagrams for this board with a 100 kg person standing at the free end. Determine the maximum shear force, maximum moment and their locations. Beam length Distance to support Mass at free end

L  2000 mm a  700  mm M  100  kg

2000 = L R1

Assumptions: The weight of the beam is negligible compared to the applied load and so can be ignored.

P

R2 700 = a

Solution:

See Figure 3-10 and Mathcad file P0310. FIGURE 3-10A Free Body Diagram for Problem 3-10

1. From inspection of Figure 3-10, write the load function equation q(x) = -R1-1 + R2-1 - P-1 2. Integrate this equation from - to x to obtain shear, V(x) V(x) = -R10 + R20 - P0 3. Integrate this equation from - to x to obtain moment, M(x) M(x) = -R11 + R21 - P1 4. Determine the magnitude of the force, P

P  M  g

P  980.7  N

5. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0 R1  P

La a

R2  P  R1 6. Define the range for x

V = R1  R2  P = 0

M = R1 L  R2 ( L  a ) = 0

R1  1821 N R2  2802 N

x  0  in 0.005  L  L

7. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 8. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  R1 S ( x 0  in)  R2 S ( x a )  P S ( x L) M ( x)  R1 S ( x 0  in)  x  R2 S ( x a )  ( x  a )  P S ( x L)  ( x  L) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


3-10-2

9. Plot the shear and moment diagrams. 1000

Shear Diagram

0 V ( x) N  1000

 2000

0

500

1000

1500

2000

x mm 0

Moment Diagram

 375 M ( x) Nm

 750

 1125

 1500

0

500

1000

1500

2000

x mm

FIGURE 3-10B Shear and Moment Diagrams for Problem 3-10

10. The maximum value of the shear force ocuurs throughout the distance from x = 0 to x = a and is R1  1821 N 11. Find the maximum value of the bending moment by determining the value of x where the shear is zero. Inspection of the shear diagram shows that this occurs at x = a. Mmax  M ( a )

Mmax  1275 N  m



3-11-1


Given:

Determine the impact force and dynamic deflection that will result when the 100-kg person in Problem 3-10 jumps up 250 mm and lands back on the board. Assume that the board weighs 29 kg and deflects 131 mm statically when the person stands on it. Find the reaction forces and construct the shear and moment diagrams for this dynamic loading. Determine the maximum shear force, maximum moment and their locations along the length of the board. Beam length

L  2000 mm

Distance to support

a  700  mm

Mass of person

mpers  100  kg

Mass of board

mboard  29 kg

Static deflection

δst  131  mm

Height of jump

h  250  mm

2000 = L R1

R2 700 = a

Assumptions: Equation (3.15) can be used to approximate a mass correction factor. Solution:

Fi

FIGURE 3-11A Free Body Diagram for Problem 3-11


1. The person is the moving object and the board is the struck object. The mass ratio to be used in equation (3.15) for the correction factor is massratio 

mpers

massratio  3.448

mboard

2. From equation (3.15), the correction factor is 1

η  1

η  0.912

1 3  massratio

3. The weight of the moving mass is

Wpers  mpers g

Wpers  0.981  kN

4. The dynamic force is found by solving equation (3.14) for Fi.



Fi  Wpers  1 



1

2  η h 

Fi  3.056  kN



δst 

From this we see that the dynamic force ratio is

Fi Wpers

 3.12

5. From inspection of Figure 3-11, write the load function equation q(x) = -R1-1 + R2-1 - Fi-1 6. Integrate this equation from - to x to obtain shear, V(x) V(x) = -R10 + R20 - Fi0 7. Integrate this equation from - to x to obtain moment, M(x) M(x) = -R11 + R21 - Fi1 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


3-11-2

8. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. V = R1  R2  Fi = 0

At x = L+, V = M = 0 R1  Fi

L a

M = R1 L  R2 ( L  a ) = 0

R1  5676 N

a

R2  Fi  R1

R2  8733 N

x  0  in 0.005  L  L

9. Define the range for x

10. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 11. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  R1 S ( x 0  in)  R2 S ( x a )  Fi S ( x L) M ( x)  R1 S ( x 0  in)  x  R2 S ( x a )  ( x  a )  Fi S ( x L)  ( x  L) 12. Plot the shear and moment diagrams. Shear Diagram

V ( x) kN

Moment Diagram

4

0

2

1

0

M ( x)

2 kN  m

2

3

4 6

0

0.5

1

1.5

2

4

0

0.5

1

x

x

m

m

1.5

2


13. The maximum value of the shear force ocuurs throughout the distance from x = 0 to x = a and is R1  5676 N 14. Find the maximum value of the bending moment by determining the value of x where the shear is zero. Inspection of the shear diagram shows that this occurs at x = a. Mmax  M ( a )

Mmax  3973 N  m



3-12-1


Given:

An overhung diving board is shown in Figure P3-4b. Find the reaction forces and construct the shear and moment diagrams for this board with a 100 kg person standing at the free end. Determine the maximum shear force, maximum moment and their locations. Beam length

L  1300 mm

Mass at free end

M  100  kg

Assumptions: 1. The weight of the beam is negligible compared to the applied load and so can be ignored. Solution:

2000 1300 = L P

M1

R1

See Figure 3-12 and Mathcad file P0312. 700

1. From inspection of Figure 3-12, write the load function equation q(x) = -M1-2 + R1-1 - P-1


2. Integrate this equation from - to x to obtain shear, V(x) V(x) = -M1-1 + R10 - P0 3. Integrate this equation from - to x to obtain moment, M(x) M(x) = -M10 + R11 - P1 4. Determine the magnitude of the force, P

P  M  g

P  980.7  N

5. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0

V = R1  P = 0

R1  P

R1  981  N

M1  R1 L

M1  1275 m N


M = M1  R1 L = 0

x  0  in 0.005  L  L

7. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 8. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  R1 S ( x 0  mm)  P S ( x L) M ( x)  M1 S ( x 0  mm)  R1 S ( x 0  mm)  ( x  0  mm)  P S ( x L)  ( x  L) 9. Plot the shear and moment diagrams. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


3-12-2

1000

Shear Diagram

800

V ( x) N

600 400 200 0

0

0.5

1

1.5

2

1.5

2

x m 0

Moment Diagram

 300

M ( x) Nm

 600  900  1200  1500

0

0.5

1 x m


10. The maximum value of the shear force ocuurs throughout the distance from x = 0 to x = L and is R1  981  N 11. Find the maximum value of the bending moment by determining the value of x where the shear is zero. Inspection of the shear diagram shows that this occurs at x = 0. Mmax  M ( 0  mm)

Mmax  1275 N  m



3-13-1


Given:

Determine the impact force and dynamic deflection that will result when a 100-kg person jumps up 25 cm and lands back on the board. Assume the board weighs 19 kg and deflects 8.5 cm statically when the person stands on it. Find the reaction forces and construct the shear and moment diagrams for this dynamic loading. Determine the maximum shear force, maximum moment, and their locations along the length of the board. Total board length

b  2000 mm

Supported length

a  700  mm

Mass of board


Static board deflection

δstat  85 mm

Mass of person

mperson  100  kg

Height of jump

h  250  mm

2000 1300 = L Fi

M1

Assumptions: 1. The board can be modelled as a cantilever beam with maximum shear and moment at the edge of the support. Solution: 1.


The person impacts the board upon landing. Thus, the board is the struck object and the person is the striking object. To determine the force exerted by the person we will first need to know the impact correction factor from equation (3.15). 1 1

η  0.94

mboard

(1)

3  mperson

We can now use equation (3.14) to determine the impact force, Fi,

 

Fi  mperson g   1 

3.

700


η 

2.

R1

1

2  η h 



δstat 

Fi  3.487  kN

(2)

Write an equation for the load function in terms of equations 3.17 and integrate the resulting function twice using equations 3.18 to obtain the shear and moment functions. Note use of the unit doublet function to represent the moment at the wall. For the beam in Figure 3-13, q(x) = -M1-2 + R1-1 - Fi-1

(3)

V(x) = -M1-1 + R10 - Fi0 + C1

(4)

M(x) = -M10 + R11 - Fi1 + C1x+ C2

(5)

The reaction moment M1 at the wall is in the z direction and the forces R1 and Fi are in the y direction in equation (4). All moments in equation (5) are in the z direction. 4.

Because the reactions have been included in the loading function, the shear and moment diagrams both close to zero at each end of the beam, making C1 = C2 = 0.

5.

The reaction force R1 and the reaction moment M1 can be calculated from equations (4) and (5) respectively

by substituting the boundary conditions x = l+, V = 0, M = 0. Note that we can substitute l for l+ since their difference is vanishingly small. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


3-13-2

l  b  a

Unsupported beam length

l  1300 mm

V(l) = -M1-1 + R10 - Fi0 = 0 V = R1  Fi = 0

(6)

R1  Fi

R1  3.487  kN

M(l) = -M10 + R11 - Fi1 = 0 M = M1  R1 l  Fi ( l  l) = 0

(7)

M1  R1 l

M1  4533 N  m

6. To generate the shear and moment functions over the length of the beam, equations (4) and (5) must be evaluated for a range of values of x from 0 to l, after substituting the above values of C1, C2, R1, and M1 in them. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than the dummy variable z, and a value of one when it is greater than or equal to z. It will have the same effect as the singularity function. Range of x

x  0  in 0.005  l  l

Unit step function

S ( x z)  if ( x  z 1 0 )

Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. 0

V ( x)  R1 S ( x 0  in)  ( x  0 )  Fi S ( x l)  ( x  l) 0

0 1

M ( x)  M1 S ( x 0  in)  ( x  0 )  R1 S ( x 0  in)  ( x  0 )  Fi S ( x l )  ( x  l)

(8)

1

Plot the shear and moment diagrams (see below). Shear Diagram

Moment Diagram 0 1

3 V ( x) kN

M ( x) 2

kN  m 3

1

0

2

4

0

0.5

1

5

0

0.5

1

x

x

m

m

1.5

7. The graphs show that the shear force and the moment are both largest at x = 0. The function values of these points can be calculated from equations (4) and (5) respectively by substituting x = 0 and evaluating the singularity functions: Vmax = V(0) = R1<0 - 0>0 - Fi<0 - l>0 = R1

(9)


MACHINE DESIGN - An Integrated Approach, 4th Ed. Vmax  R1

3-13-3 Vmax  3.49 kN

M.max = M(0) = -M1<0 - 0>0 + R1<0 - 0>1 - Fi<0 - l>1 = -M1 Mmax  M1

(10)

Mmax  4533 N  m



3-14-1


Units: Given:

Figure P3-5 shows a child's toy called a pogo stick. The child stands on the pads, applying half her weight on each side. She jumps off the ground, holding the pads up against her feet, and bounces along with the spring cushioning the impact and storing energy to help each rebound. Find the natural frequency of the system, the static deflection of the spring with the child standing still, and the dynamic force and deflection when the child lands after jumping 2 in off the ground.

blob :=

lbf ⋅ sec

2

in

Child's weight

Wc := 60⋅ lbf

Spring constant

k := 100⋅ lbf ⋅ in

Pogo stick weight

Wp := 5 ⋅ lbf

Height of drop

h := 2 ⋅ in

−1

Assumptions: 1. An approximate energy method will be acceptable. 2. The correction factor for energy dissipation will be applied. Solution:


Fi /2

1. Find the natural frequency of the (child/spring) system. Mass of child (striker)

m :=

Mass of stick (struck)

mb :=

Natural frequency

ω :=

f :=

Wc

Fi /2

m = 0.155⋅ blob

g Wp

mb = 0.013⋅ blob

g k

ω = 25.367⋅

m

rad sec

P

ω 2⋅ π

f = 4.037⋅ Hz FIGURE 3-14 Free Body Diagram for Problem 3-14

2. The static deflection of the spring with the child standing still is Static deflection of spring

δst :=

Wc

δst = 0.6⋅ in

k

3. Determine the mass ratio correction factor from equation (3.15): Correction factor

1

η := 1+

mb

η = 0.973

3⋅ m

4. Using equation (3.14), determine the dynamic force.

 

Fi := Wc⋅  1 +

1+

2 ⋅ η⋅ h 



δst 

Fi = 224⋅ lbf



3-15-1


A pen plotter imparts a constant acceleration of 2.5 m/sec2 to the pen assembly, which travels in a straight line across the paper. The moving pen assembly weighs 0.5 kg. The plotter weighs 5 kg. What coefficient of friction is needed between the plotter feet and the table top on which it sits to prevent the plotter from moving when the pen accelerates?

Given:

Acceleration of pen ass'y a  2.5 m sec Mass of pen ass'y mpen  0.5 kg mplot  5  kg

Mass of plotter Solution: 1.

2


The force imparted to the pen assembly by the internal drive mechanism must be reacted at the table top by the plotter feet. The horizontal force at the feet will be equal to the force on the pen assembly and must be less than or equal to the maximum friction force, which is the product of the coefficient of friction and the normal force, which is the weight of the plotter. Horizontal driving force on pen ass'y

Fpen  mpen a

Fpen  1.25 N

Weight of plotter

Wplot  mplot  g

Wplot  49.033 N

Minimum coefficient of friction

μ 

Fpen Wplot

μ  0.025



3-16-1


A track to guide bowling balls is designed with two round rods as shown in Figure P3-6. The rods are not parallel to one another but have a small angle between them. The balls roll on the rods until they fall between them and drop onto another track. The angle between the rods is varied to cause the ball to drop at different locations. Each rod's unsupported length is 30 in and the angle between them is 3.2 deg. The balls are 4.5 in dia and weigh 2.5 lb. The center distance between the 1-in-dia rods is 4.2 in at the narrow end. Find the distance from the narrow end at which the ball drops through and determine the worst-case shear and moment maximum for the rods as the ball rolls a distance from the narrow end that is 98% of the distance to drop. Assume that the rods are simply supported at each end and have zero deflection under the applied loading. (Note that assuming zero deflection is unrealistic. This assumption will be relaxed in the next chapter after deflection has been discussed.)

Given:

Unsupported rod length Half-angle between rods Bowling ball diameter

Solution:


1.

L  30 in α  1.6 deg D  4.5 in

Bowling ball weight Rod diameter Half width of rod gap

Calculate the distance between the ball and rod centers. Distance between centers

h 

D d 2

A

W  2.5 lbf d  1.0 in c  2.1 in

A

h  2.75 in

c TOP VIEW

 F

u



x W/2 F

SECTION A-A

width(x) (a) Distance between the roll axis and the rod axis.

(b) Partial FBD of the bowling ball.

FIGURE 3-16 Dimensions and Free Body Diagrams for Problem 3-16


MACHINE DESIGN - An Integrated Approach, 4th Ed. 2.

3-16-2

Let x be the distance along the roll axis, and u be the corresponding distance to the point of contact between the ball and rods, measured along the rods. Then the distance from the center plane of the ball to the center of a rod as shown in Figure 3-16(a) is, width( x)  c cos( α)  x sin( α)

(1)

And the distance from the narrow end to the point at which the ball drops (assuming rigid rods) is xdrop 

h  c cos( α)

xdrop  23.31  in

sin( α)

The distance along the rod corresponding to xdrop is u drop 

3.

xdrop  h  sin( α)

The angle made by a line through the ball-rod centers and the horizontal plane (see Figure 3-16b) is

θ ( x)  acos



4.

width( x)  h

 

When x = 0, this is

θ0  θ ( 0  in)

θ0  40.241 deg

When x = 0.98xdrop, this is

θ98%  θ  0.98 xdrop

θ98%  5.577  deg

The loading on the ball is symmetric about its center plane along the x-axis. Figure 3-16(b) shows a FBD of one half of the ball with the internal forces along the plane of symmetry due to the reaction at the other rod omitted. With these forces omitted we may only sum forces in the vertical direction.

 Fy :

F  sin( θ )  μ  F  cos( θ ) 

F=

5.

u drop  23.24  in

cos( α)

W 2

(2)

=0

W

(3)

2  ( sin( θ )  μ  cos( θ ) )

The ball will drop through the rods when  is zero. If there were no friction force present ( = 0) then F would become very large as  approached zero. The presence of the friction term in the denominator of equation (3) limits F to finite values. However, with the assumption that the rods are rigid, there is no way for the rods to provide a normal force when  reaches zero. Thus, we will need to limit the range of  for this analysis. Let

μ  0

Then

xmax 

u max  Fmax 

and

θmin  θ98%

h  cos θmin  c cos( α) sin( α) xmax  h  sin( α) cos( α) W

2  sin θmin

xmax  22.84  in

u max  22.77  in Fmax  12.86  lbf

(4)



6.

3-16-3

Determine the worst-case shear and moment maximum for the rods as the ball rolls along their length from Figure B-2(a) in Appendix B where a in the figure is u max. Then,

 

Mmax  Fmax u max  1 

u max  L

 

Mmax  70.6 in lbf

(5)

For the shear, we must find the reactions, which are

 

R1  Fmax  1  R2  Fmax  R1

u max  L

 

R1  3.10 lbf R2  9.76 lbf

The maximum absolute value of shear is the larger of these two. Thus Vmax  R2

Vmax  9.8 lbf

(6)



3-17-1


A pair of ice tongs is shown in Figure P3-7. The ice weighs 50 lb and is 10 in wide across the tongs. The distance between the handles is 4 in, and the mean radius r of a tong is 6 in. Draw free-body diagrams of the two tongs and find all forces acting on them. Determine the bending moment at point A.

Given:

Weight of ice Distances

W  50 lbf a x  11.0 in

a y  6.0 in

b x  5.0 in

b y  12.0 in

cx  2.0 in

cy  3.5 in

Assumptions: Assume that the horizontal force at C (the handle) is zero, thus Fc  0  lbf Solution:

(1)


F

F

F C FC O 11.0 = ax

3.5 = cy

FO 2.0 = cx

A

12.0 = by 5.0 = bx

FB B W/2

W FIGURE 3-17A Free Body Diagrams for Problem 3-17

1.

Summing forces and moments on a single tong (see FBD above right).

 Fx

FO  FB  FC = 0

 Fy

W

 MC

FO cy  FB  b y  cy 

2

(2)

F=0

(3) W 2

  b x  cx = 0

2.

From equations (1) and (2), FO = FB

3.

Eliminating FO from equations (4) and (5) and solving for FB FB 

W   b x  cx 2 by

(4) (5)

FB  14.58  lbf



3-17-2

F 4. From equation (3), the vertical force on one handle is F 

W

FC

F  25 lbf

2

O

5. From Figure 3-17B we see that, at any section that we might take through the tong, there will be an internal moment, shear force, and axial force present. The bending moment will be a maximum at point A because it is the fartherest point from the centroid of the system. Summing forces and moments:

 Fx

C

-FDs cos  + FDn sin  (6) + FO = 0

11.0 = ax

A

3.5 = cy

FO 2.0 = cx

 D

FDs M D FDn FIGURE 3-17B Free Body Diagram with section at D for Problem 3-17

 Fy

-FDs sin  - FDn cos  +F=0 (7)

 MO

F cx - M D - (FDs cos + FDn sin )(ay + rc sin ) + (FDs sin + FDn cos )[ax - rc (1 - cos)] = 0 (8)

6.

Solving equations (6) and (7) for FDs and FDn FDn = F  cos( θ )  FO sin( θ )

7.

FDs =

FDn sin( θ )  FO cos( θ )

The maximum value of MD will occur at  = 0 deg. At  = 0 deg, FO  FB FDn  F

FDn  25 lbf

FDs  FO

FDs  14.58  lbf

MD  F  cx  FDs a y  FDn a x

MD  237.5  lbf  in



3-18-1


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 44 415 lb of paper rolls in two rows by two high as shown in Figure P3-8. 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 Nominal coefficient of friction

r := 50⋅ ft μnom := 0.43

Coefficient of friction uncertainty

u μ := 0.08

Trailer width Height of CG from pavement

w := 8 ⋅ ft 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 Figure 3-18A, Tilt angle

θ := 3 ⋅ deg

a := h ⋅ tan( θ )

a = 0.393⋅ ft

b :=

w 2

−a

xbar := b ⋅ cos( θ ) ybar := b⋅ sin( θ ) +

3°

b = 3.607⋅ ft

7.500'

h cos( θ )

ybar = 7.699⋅ ft

a

The coordinates of the CG of the loaded trailer with respect to the lower outside corner of the tires are: xbar = 3.602⋅ ft

ybar

xbar = 3.602⋅ ft

ybar = 7.699⋅ ft

b xbar 4.000'

FIGURE 3-18A Location of CG for Problem 3-18


MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-18-2 2. The trailer is on the verge of tipping over when the copule 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 3-18B),

ΣM

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)

Rx

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

Ry xbar (4)

mtot

FIGURE 3-18B FBD of Trailer on the Verge of Tipping

or, vtip =

3.

Fc⋅ r⋅ g

(5)

Fw

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 in Figure 3-18C, we see that Normal force between paper and floor

Fn = Wp⋅ cos( θ ) − Fcp ⋅ sin( θ )

(6)



3-18-3

Tangential force tending to slide the paper Ft = Wp⋅ sin( θ ) + Fcp ⋅ cos( θ )

(7)

Fcp

Centrifugal force on the paper

Fcp =

Wp g

Wp

2

⋅ as =

W p vs ⋅ g r

(8)

Ft Fn

But, the maximum friction force is (9)

Ff = μ ⋅ Fn = Ft

FIGURE 3-18C FBD of Paper on the Verge of Sliding

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 ) )   μ ⋅ sin( θ ) + cos( θ )  ⋅ r⋅ g  

(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. Maximium coefficient

μmax := μnom + u μ

μmax = 0.51

Minimium coefficient

μmin := μnom − uμ

μmin = 0.35

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.



3-19-1


Assume that the CG of the paper rolls in Problem 3-18 is 2.5 ft above the floor of the trailer. At what speed on the same curve will the pile of rolls tip over (not slide) with respect to the trailer?

Given:

Weight of paper

Wp := 44415⋅ lbf

Radius of curve Paper roll length Height of CG from floor

r := 50⋅ ft L := 38⋅ in h := 2.5⋅ ft

L = 3.167⋅ ft

Assumptions: The paper rolls act as a single, lumped mass and tip about one corner where they are braced against sliding. The brace provides no moment support. Solution:


1. First, calculate the location of the paper's CG with respect to the outside corner when it is on the reverse-banked curve. From Figure 3-19, Tilt angle

θ := 3⋅ deg

a := h⋅ tan ( θ)

a = 0.131⋅ ft

b := L − a

b = 3.036⋅ ft

xbar := b ⋅ cos ( θ)

xbar = 3.031⋅ ft

ybar := b⋅ sin ( θ) +

Fcp 2.500' Wp a

ybar

Rx b xbar R y 3.167'

h FIGURE 3-19

cos ( θ)

FBD of Paper on the Verge of Tipping

ybar = 2.662⋅ ft The coordinates of the CG of the paper with respect to the lower outside corner are: xbar = 3.031⋅ ft 2.

ybar = 2.662⋅ ft

The paper 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 paper acting through its CG and the vertical reaction at the outside edge of the rolls. At this instant, it is assumed that the entire weight of the paper is reacted at the outside corner. Any increase in tangential velocity of the tractor and trailer will result in tipping. Summing moments about the outside corner nearest the floor (see Figure 3-19),

ΣM

Wp⋅ xbar − Fcp ⋅ ybar = 0

(1)

where Fcp 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

2

(2)

r

and the force necessary to keep the tractor trailer following a circular path is Fcp = mp⋅ a tip

(3)

where mp is the mass of the paper. Combining equations (2) and (3) and solving for vtip, we have


MACHINE DESIGN - An Integrated Approach, 4th Ed. vtip =

3-19-2

Fcp ⋅ r

(4)

mp

or, vtip =

3.

Fcp ⋅ r⋅ g

(5)

Wp

Calculate the minimum paper tipping velocity of the tractor/trailer. From equations (1) and (5), Centrifugal force required to tip the paper

Fcp :=

Minimum tipping speed

vtip :=

xbar ybar

⋅ Wp

Fcp ⋅ r⋅ g Wp

Fcp = 50574⋅ lbf

vtip = 29.2⋅ mph

Thus, with the assumptions that we have made, the paper would not begin to tip over until the tractor/trailor reached a speed of vtip = 29.2⋅ mph



3-20-1


Assume that the load of paper rolls in Problem 3-18 will slide sideways at a truck speed of 20 mph on the curve in question. Estimate the impact force of the cargo against the trailer wall. The force-deflection characteristic of the trailer wall has been measured as approximately 400 lb/in.

Given:

Weight of paper

Wp := 44415⋅ lbf

Weight of trailer

Wt := 14000⋅ lbf

Speed of tractor/trailer

vt := 20⋅ mph

Radius of curve Trailer width Paper roll length

r := 50⋅ ft w := 8 ⋅ ft L := 38⋅ in lbf k := 400 in

Trailer wall stiffness

L = 3.167⋅ ft

Assumptions: 1. The paper rolls act as a monolith since they are tightly strapped together with steel bands. 2. The worst case will result if friction between the floor and the paper is neglected. Solution: 1.


Calculate the distance that the rolls will slide before impacting the wall. s :=

2.

1 2

vt

2

r

2

vi = 64.266⋅

in sec

With the paper as the moving mass and the trailer as the stationary or struck mass, calculate the correction factor using equation (3.15) 1 1+

η = 0.905

Wt 3 ⋅ Wp

Calculate the static deflection caused by the paper against the trailer wall.

δst := 6.

sec

2⋅ a p⋅ s

η :=

5.

in

a p = 206.507⋅

From elementary particle dynamics, estimate the velocity at impact due to the centripetal acceleration vi :=

4.

s = 10⋅ in

Determine the centripetal acceleration at 20 mph. a p :=

3.

⋅ ( w − 2⋅ L)

Wp

δst = 111.037⋅ in

k

Using equation (3.12), estimate the dynamic force of the paper rolls impacting the trailer wall.

Fi := Wp⋅ vi⋅

η g ⋅ δst

Fi = 13114⋅ lbf



3-21-1


Figure P3-9 shows an automobile wheel with two common styles of lug wrench being used to tighten the wheel nuts, a single-ended wrench in (a), and a double-ended wrench in (b). In each case two hands are required to provide forces respectively at A and B as shown. The distance between points A and B is 1 ft in both cases. The wheel nuts require a torque of 70 ft-lb. Draw free body diagrams for both wrenches and determine the magnitudes of all forces and moments on each wrench. Is there any difference between the way these two wrenches perform their assigned task? Is one design better than the other? If so, why? Explain.

Given:

Distance between A and B Tightening torque

d AB := 1 ⋅ ft T := 70⋅ ft ⋅ lbf

Assumptions: 1. The forces exerted by the user's hands lie in a plane through the wrench that is also parallel to the plane of the wheel. 2. The applied torque is perpendicular to the plane of the forces. 3. By virtue of 1 and 2 above, this is a planar problem that can be described in a 2D FBD. Solution:


1. Summing moments about the left end of the wrench (for either case)

12" = dAB F

T − F ⋅ d AB = 0 2. Solving for F T

F :=

T d AB

F = 70⋅ lbf

F (a) Single-ended Wrench

3. This result is the same for both wrenches. 12" = dAB

Is there any difference between the way these two wrenches perform their assigned task? No, they both require the same two-handed exertion of 70 lb from each hand. Is one design better than the other? If so, why? Explain. Design (b) has advantages over (a) because it is balanced about the wheel nut. This allows the user to spin the wrench once the nut is loosened. It is also slightly easier to apply the upward and downward forces (F) in a plane with design (b).

F

6"

T F (b) Double-ended Wrench




3-22-1


A roller-blade skate is shown in Figure P3-10. The polyurethane wheels are 72 mm dia. The skate-boot-foot combination weighs 2 kg. The effective "spring rate" of the person-skate subsystem is 6000 N/m. Find the forces on the wheels' axles for a 100-kg person landing a 0.5-m jump on one foot. (a) Assume all 4 wheels land simultaneously. (b) Assume that one wheel absorbs all the landing force.

Given:

Mass of struck member

Msys  2  kg

Stiffness of struck member

k  6000

N

Mass of striking member

m Mperson  100  kg

Height of drop

h  0.5 m

Assumptions: Equation (3.14) applies in this case. Solution: 1.


The weight of the striking mass is Wperson  Mperson g

2.

The static deflection of the subsystem is

δst  3.

Wperson

δst  163.444  mm

k

The correction factor is 1

η  1

4.

Wperson  980.7  N

η  0.993

Msys 3  Mperson

From equation (3.14), the force of impact is



Fi   1 



1

2  η h 

  Wperson

δst 

Fi  3.59 kN

(a) If this will be absorbed by 4 wheel axles, the force per axle is Fa 

Fi

Fa  897  N

4

(b) If one wheel absorbs all force

Fb  Fi

Fb  3.59 kN



3-23a-1


Given:

A beam is supported and loaded as shown in Figure P3-11a. Find the reactions, maximum shear, and maximum moment for the data given in row a from Table P3-1. Beam length

L  1  m

Distance to distributed load

a  0.4 m

L b a

Distance to concentrated load b  0.6 m 1

Distributed load magnitude

w  200  N  m

Concentrated load

F  500  N

F

w

R2

R1

Solution:

See Figures 3-23 and Mathcad file P0323a. FIGURE 3-23A Free Body Diagram for Problem 3-23

1. From inspection of Figure P3-11a, write the load function equation q(x) = R1-1 - w0 + w0 - F-1 + R2-1 2. Integrate this equation from - to x to obtain shear, V(x) V(x) = R10 - w1 + w1 - F0 + R20 3. Integrate this equation from - to x to obtain moment, M(x) M(x) = R11 - w2/2 + w2/2 - F1 + R21 4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0 V = R1  w ( L)  w ( L  a )  F  R2 = 0 M = R 1 L  R1 

w 2

L 

w 2

2

L 

F L

w 2

2

 ( L  a)  F  ( L  b) = 0

 ( L  b) 

w 2 L

 ( L  a)

2

R1  264  N

R2  w a  F  R1 5. Define the range for x

R2  316  N

x  0  m 0.005  L  L

6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  R1 S ( x 0  m)  w S ( x 0  m)  ( x)  w S ( x a )  ( x  a )  F  S ( x b )  R2 S ( x L) M ( x)  R1 S ( x 0  m)  x 

w 2

2

 S ( x 0  m)  x 

w 2

2

 S ( x a )  ( x  a )  F  S ( x b )  ( x  b )



3-23a-2


Shear Diagram

200 V ( x) 0

N

 200  400

0

0.2

0.4

0.6

0.8

x m

Moment Diagram

150

100 M ( x) Nm 50

0

0

0.2

0.4

0.6

0.8

x m

FIGURE 3-23aB Shear and Moment Diagrams for Problem 3-23a

9. Determine the maximum shear and maximum moment from inspection of the diagrams. Maximum shear:

Vmax  V ( b )

Vmax  316  N

Maximum moment occurs where V is zero, which is x = b: Mmax  M ( b )

Mmax  126.4  N  m



3-24a-1

PROBLEM 3-24a Statement: Given:

A beam is supported and loaded as shown in Figure P3-11b. Find the reactions, maximum shear, and maximum moment for the data given in row a from Table P3-1. Beam length

L  1  m


a  0.4 m

a

1

w  200  N  m


F w

F  500  N

Concentrated load Solution:

L

See Figures 3-24 and Mathcad file P0324a.

1. From inspection of Figure P3-11b, write the load function equation

M1 R1


q(x) = -M1-2 + R1-1 - w0 - F-1 2. Integrate this equation from - to x to obtain shear, V(x) V(x) = -M1-1 + R10 - w1 - F0 3. Integrate this equation from - to x to obtain moment, M(x) M(x) = -M10 + R11 - w2/2 - F1 4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0 V = R1  [ w  ( L  a )  F ] = 0 M = M1  R1 L 

w 2

2

 ( L  a) = 0

R1  w ( L  a )  F M1 

w 2


R1  620  N

2

 ( L  a )  R1 L

M1  584  N  m

x  0  m 0.005  L  L

6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  R1 S ( x 0  m)  w S ( x a )  ( x  a )  F  S ( x L) M ( x)  M1  R1 S ( x 0  m)  x 

w 2

2

 S ( x a )  ( x  a )  F  S ( x L)  ( x  L)



3-24a-2

8. Plot the shear and moment diagrams. Shear Diagram 600 V ( x) N

400 200 0

0

0.2

0.4

0.6

0.8

0.6

0.8

x m

Moment Diagram

0

 150 M ( x) Nm

 300

 450

 600

0

0.2

0.4 x m



Vmax  V ( 0  m)

Vmax  620  N

Maximum moment occurs where V is zero, which is x = 0: Mmax  M ( 0  m)

Mmax  584  N  m



3-25a-1


Given:

A beam is supported and loaded as shown in Figure P3-11d. Find the reactions, maximum shear, and maximum moment for the data given in row a from Table P3-1. Beam length

L  1  m


a  0.4 m

L b a

Distance to concentrated load b  0.6 m

F

1


w  200  N  m

Concentrated load

F  500  N

w

R2

R1

Solution:

See Figures 3-25 and Mathcad file P0325a. FIGURE 3-25A Free Body Diagram for Problem 3-25

1. From inspection of Figure P3-11c, write the load function equation q(x) = R1-1 - w0 + R2-1 - F-1 2. Integrate this equation from - to x to obtain shear, V(x) V(x) = R10 - w1 + R20 - F0 3. Integrate this equation from - to x to obtain moment, M(x) M(x) = R11 - w2/2 + R21 - F1 4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0 V = R1  w  ( L  a )  R2  F = 0 M = R 1 L  R1 

w 2

2

 ( L  a )  R2 ( L  b ) = 0

w 2    ( L  a )  F  ( L  b )  w ( L  a )  ( L  b ) b 2  1

R2  w ( L  a )  F  R1 5. Define the range for x

R1  353  N R2  973  N

x  0  m 0.005  L  L

6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  R1 S ( x 0  m)  w S ( x a )  ( x  a )  R2 S ( x b )  F  S ( x L) M ( x)  R1 S ( x 0  m)  x 

w 2

2

 S ( x a )  ( x  a )  R2 S ( x b )  ( x  b )



3-25a-2


Shear Diagram

0.5 V ( x) 0

kN

 0.5 1

0

0.2

0.4

0.6

0.8

x m

Moment Diagram

0

 75 M ( x) Nm

 150

 225

 300

0

0.2

0.4

0.6

0.8

x m



Vmax  V ( b )

Vmax  580.0  N

Maximum moment occurs where V is zero, which is x = a: Mmax  M ( b )

Mmax  216  N  m



3-26a-1


Solution:

A beam is supported and loaded as shown in Figure P3-11d. Find the reactions, maximum shear, and maximum moment for the data given in row a from Table P3-1. Beam length

L  1  m


a  0.4 m

Distance to reaction load

b  0.6 m

L b a

1


w  200  N  m

Concentrated load

F  500  N

F w

R2

R1



1. From inspection of Figure 3-26aA, write the load function equation q(x) = R1-1 - w0 + R2-1 - F-1 2. Integrate this equation from - to x to obtain shear, V(x) V(x) = R10 - w1 + R20 - F0 3. Integrate this equation from - to x to obtain moment, M(x) M(x) = R11 - w2/2 + R21 - F1 4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0 V = R1  w  ( L  a )  R2  F = 0 M = R 1 L  R1 

w 2

2

 ( L  a )  R2 ( L  b )  F ( L  a ) = 0

w 2    ( L  a )  F  ( b  a )  w ( L  a )  ( L  b ) b 2  1

R2  w ( L  a )  F  R1 5. Define the range for x

R1  147  N R2  473  N

x  0  m 0.005  L  L

6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  R1 S ( x 0  m)  w S ( x a )  ( x  a )  R2 S ( x b )  F  S ( x a ) M ( x)  R1 S ( x 0  m)  x 

w 2

2

 S ( x a )  ( x  a )  R2 S ( x b )  ( x  b )  F  S ( x a )  ( x  a )



3-26a-2


Shear Diagram

250 V ( x) 0

N

 250  500

0

0.2

0.4

0.6

0.8

x m

Moment Diagram

60

40 M ( x) Nm

20

0

 20

0

0.2

0.4

0.6

0.8

x m



Vmax  V ( b  0.001  mm)

Vmax  393  N

Maximum moment occurs where V is zero, which is x = a: Mmax  M ( a )

Mmax  58.7 N  m



3-27-1


A storage rack is to be designed to hold the paper roll of Problem 3-8 as shown in Figure P3-12. Determine the reactions and draw the shear and moment diagrams for the mandrel that extends 50% into the roll.

Given:

Paper roll dimensions

OD  1.50 m ID  0.22 m Lroll  3.23 m

Roll density

ρ  984  kg m

3

Assumptions: 1. The paper roll's weight creates a concentrated load acting at the tip of the mandrel. 2. The mandrel's root in the stanchion experiences a distributed load over its length of engagemen Solution:


W 1. Determine the weight of the roll and the length of the mandrel. W 

 4

π

2

2



 OD  ID  Lroll  ρ  g

W  53.9 kN

M1

Lm R1

Lm  0.5 Lroll

FIGURE 3-27 Lm  1.615  m

Free Body Diagram for Problem 3-27

2. From inspection of Figure 3-27, write the load function equation q(x) = -M1-2 + R1-1 - W-1 3. Integrate this equation from - to x to obtain shear, V(x) V(x) = -M1-1 + R10 - W0 4. Integrate this equation from - to x to obtain moment, M(x) M(x) = -M10 + R11 - W1 5. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0 V = R1  W = 0

M = M1  R1 L = 0

R1  W

R1  53.895 kN

M1  R1 Lm

M1  87.040 kN  m


x  0  m 0.005  Lm  Lm



3-27-2

7. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 8. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  R1 S ( x 0  m)  W  S  x Lm M ( x)  M1  R1 S ( x 0  m)  x  W  S  x Lm   x  Lm 9. Plot the shear and moment diagrams. Shear Diagram 40 V ( x) kN 20

0

0

0.5

1

1.5

2

x m

Moment Diagram

20

1.615

 10 M ( x) kN  m

 40

 70

 100

0

0.5

1

1.5

2

x m




3-28-1


Figure P3-13 shows a forklift truck negotiating a 15 deg ramp to to drive onto a 4-ft-high loading platform. The truck weighs 5 000 lb and has a 42-in wheelbase. Determine the reactions and draw the shear and moment diagrams for the worst case of loading as the truck travels up the ramp.

Given:

Ramp angle Platform height Truck weight Truck wheelbase

θ  15 deg h  4  ft W  5000 lbf Lt  42 in

h  48 in

Assumptions: 1. The worst case is when the truck CG is located at the center of the beam's span. 2. Use a coordinate frame that has the x-axis along the long axis of the beam. 3. Ignore traction forces and the weight components along the x-axis of the beam. 4. There are two ramps, one for each side of the forklift. 5. The location of the CG in Figure P3-13 is 32 in from the front wheel and 10 in from the rear wheel. CGa  32 in Solution:

CGb  10 in


L b a

CG a

y

CG b

R1  Fa

Wa

Fb

x Wb

R2

FIGURE 3-28A Dimensions and Free Body Diagram for Problem 3-28

1. Determine the length of the beam between supports and the distances a and b. Length of beam

With the CG at midspan, we have

and

L 

h

L  15.455 ft

sin( θ )

a  CGa =

L 2

a 

L

b 

L

2

2

 CGa

a  5.061  ft

 CGb

b  8.561  ft



3-28-2

2. The weight distribution on the wheels is determined from the distance from the front wheel to the CG. Each wheel weight is divided by 2 to get the weight on a single ramp. Weight on front wheel

Wa 

CGb W  Lt 2

Wa  595  lbf

Weight on rear wheel

Wb 

W

Wb  1905 lbf

2

 Wa

3. The normal force on the ramp at each wheel is adjusted for the ramp angle. Load at front wheel

Fa  Wa cos( θ )

Fa  575  lbf

Load at rear wheel

Fb  Wb cos( θ )

Fb  1840 lbf

4. From inspection of Figure 3-28A, write the load function equation q(x) = R1-1 - Fa-1 - Fb-1 + R2-1 5. Integrate this equation from - to x to obtain shear, V(x) V(x) = R10 - Fa0 - Fb0 + R20 6. Integrate this equation from - to x to obtain moment, M(x) M(x) = R11 - Fa1 - Fb1 + R21 7. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0 V = R1  Fa  Fb  R2 = 0 M = R1 L  Fa ( L  a )  Fb ( L  b ) = 0 R1 

1 L

 Fa ( L  a )  Fb ( L  b )

R2  Fa  Fb  R1 8. Define the range for x

R1  1207 lbf R2  1207 lbf

x  0  m 0.005  L  L

9. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 10. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  R1 S ( x 0  m)  Fa S ( x a )  Fb S ( x b )  R2 S ( x L) M ( x)  R1 S ( x 0  m)  x  Fa S ( x a )  ( x  a )  Fb S ( x b )  ( x  b )  R2 S ( x L)  ( x  L)



3-28-3


Shear Diagram

1000 V ( x) 0

lbf

 1000  2000

0

2

4

6

8

10

12

14

16

x ft

Moment Diagram

10000

15.455

8000

M ( x)

6000

ft  lbf 4000 2000 0

0

2

4

6

8

10

12

14

16

x ft




3-29-1

PROBLEM 3-29

_____

Statement:

Run the TKSolver or Mathcad model for Case Study 1A and move the point of application of the hand force along the lever by changing the values of Rb2, recalculate and observe the changes to the forces and moments.

Problem:

Determine the forces on the elements of the bicycle brake lever assembly shown in Figure 3-1 during braking.

Given:

The geometry of each element is known. The average human's hand can develop a grip force of about 267 N (60 lb) in the lever position shown. Magnitude of handle force Fb2

Fb2 := 267⋅ N

Direction of handle force Fb2

θb2 := 270⋅ deg

Direction of cable force Fc2

θc2 := 184⋅ deg

Direction of cable force Fcable

θcable := 180⋅ deg

Position vector components (Change the value of Rb2x and note the results) Rb2x := 19⋅ mm

Rc2x := −25⋅ mm

R12x := −12⋅ mm

Rb2y := −4⋅ mm

Rc2y := 0⋅ mm

R12y := −7⋅ mm

R21x := 7⋅ mm

Rb1x := 47.5⋅ mm

R31x := −27⋅ mm

R21y := 19⋅ mm

Rb1y := −14⋅ mm

R31y := 30⋅ mm

Assumptions: The accelerations are negligible. All forces are coplanar and two-dimensional. A class 1 load model is appropriate and a static analysis is acceptable. The higher applied load will be used as a worst case, assuming that it can be reached before bottoming the tip of the handle on the handgrip. If that occurs, it will change the beam's boundary conditions and the analysis. Solution: 1.

See Figures 3-1, 3-2, and Mathcad file P0329.

Figure 3-1 shows the hand brake lever assembly, which consists of three subassemblies: the handlebar (1), the lever (2), and the cable (3). The lever is pivoted to the handlebar and the cable is connected to the lever. The cable runs within a plastic-lined sheath (for low friction) down to the brake caliper assembly at the bicycle's wheel rim. The user's hand applies equal and opposite forces at some point on the lever and handgrip. These forces are transformed to a larger force in the cable by reason of the lever ratio of part 2. Figure 3-1 is a free-body diagram of the entire assembly since it shows all the forces and moments acting on it except for its weight, which is small compared to the applied forces and is thus neglected for this analysis. The "broken away" portion of the handlebar provides internal x and y force components and a moment. These are arbitrarily shown as positive in sign. Their actual signs will "come out in the wash" in the calculations. The known applied forces are shown in their actual directions and senses.

2.

Figure 3-2 shows the three subassembly elements separated and drawn as free-body diagrams with all relevant forces and moments applied to each element, again neglecting the weights of the parts. The lever (part 2) has three forces on it, Fb2, Fc2, and F12. The two-character subscript notation used here should be read as, force of element 1 on 2 (F12) or force at B on 2 (Fb2), etc. This defines the source of the forces (first subscript) and the element on which it acts (second subscript). This notation will be used consistently throughout this text for both forces and position vectors such as Rb2, Rc2, and R12 in Figure 3-2, which serve to locate the above three forces in a local, non rotating coordinate system whose origin is at the center of gravity (CG) of the element or subassembly being analyzed. (See foot note on page 83 of the text). On this brake lever, Fb2 is an applied force whose magnitude and direction are known. Fc2 is the force in the



3-29-2

cable. Its direction is known but not its magnitude. Force F12 is provided by part 1 on part 2 at the pivot pin. Its magnitude and direction are both unknown. We can write equations 3.3b for this element to sum forces in the x and y directions and sum moments about the CG. Note that all unknown reactive forces and moments are initially assumed positive in the equations. Their true signs will come out in the calculation. (See foot note on page 84 of the text).

ΣFx = Fb2x + Fc2x + F12x = 0 (a)

ΣFy = Fb2y + Fc2y + F12y = 0 ΣMz = ( R12 × F12) + ( Rb2 × Fb2) + ( Rc2 × Fc2 ) = 0 The cross products in the moment equation represent the "turning forces" or moments created by the application of these forces at points remote from the CG of the element. Recall that these cross products can be expanded to

ΣMz = ( R12x⋅ F12y − R12y⋅ F12x) ...  = 0

+ ( Rb2x⋅ Fb2y − Rb2y⋅ Fb2x) ... + ( R ⋅ F − R ⋅ F )   c2x c2y c2y c2x 

(b)

We have three equations and four unknowns (F12x, F12y, Fc2x, Fc2y) at this point, so we need another equation. It is available from the fact that the direction of Fc2 is known. (The cable can pull only along its axis). We can express one component of the cable force Fc2 in terms of its other component and the known angle θc2 of the cable. (c) Fc2y = Fc2x⋅ tan( θc2 ) We will now use a Mathcad solve block to solve equations a through c. Calculate components of Fb2 Fb2x := Fb2⋅ cos( θb2)

Fb2x = −0 ⋅ N

Fb2y := Fb2⋅ sin( θb2)

Fb2y = −267⋅ N

Guess

F12x := 1000⋅ N

Given

Fb2x + Fc2x + F12x = 0

Fc2x := −1000⋅ N

F12y := 1000⋅ N

Fc2y := −1000⋅ N

Fb2y + Fc2y + F12y = 0

( R12x⋅ F12y − R12y⋅ F12x) ...  = 0 + ( Rb2x⋅ Fb2y − Rb2y⋅ Fb2x) ... + ( R ⋅ F − R ⋅ F )   c2x c2y c2y c2x  Fc2y = Fc2x⋅ tan( θc2 )



3-29-3

 F12x  F   12y  := Find ( F , F , F , F ) 12x 12y c2x c2y  Fc2x     Fc2y  Components of the unknown forces F12, and Fc2 F12x = 1047⋅ N 3.

Fc2x = −1047⋅ N

F12y = 340⋅ N

Fc2y = −73.2⋅ N

Part 3 in Figure 3-2 is the cable that passes through a hole in part 1. This hole is lined with a low friction material, which allows us to assume no friction at the joint between parts 1 and 3. We will further assume that the three forces F13, Fc3, and Fcable form a concurrent system of forces acting through the CG and thus create no moment. With this assumption, only a summation of forces is necessary for this element.

ΣFx = Fcablex + F13x + Fc3x = 0 (d)

ΣFy = Fcabley + F13y + Fc3y = 0 Using Newton's third law, we have Fc3x := −Fc2x and Fc3y := −Fc2y. We also assume that the cable entering from the left is horizontal and that the reaction F13 is vertical, thus Fcabley := 0 ⋅ N

and

(e)

F13x := 0 ⋅ N

We can now solve for the forces on part 3 directly, Fcablex := −F13x − Fc3x

Fcablex = −1047⋅ N

F13y := −Fcabley − Fc3y

F13y = −73.2⋅ N

The assembly of elements labeled part 1 in Figure 3-2 has both force and moments on it (i.e., it is not a concurrent system), so the three equations 3.3b are needed.

ΣFx = F21x + Fb1x + F31x + Px + Fsheathx = 0 (f)

ΣFy = F21y + Fb1y + F31y + Py = 0 ΣMz = Mh + ( R21 × F21) + ( Rb1 × Fb1) + ( R31 × F31) ... = 0 + ( Rp × Fp) + ( Rd × Fsheath ) Expanding cross products in the moment equation gives the moment magnitude as

ΣMz = Mh + ( R21x⋅ F21y − R21y⋅ F21x) ... = 0

+ ( Rb1x⋅ Fb1y − Rb1y⋅ Fb1x) ... + ( R31x⋅ F31y − R31y⋅ F31x) ... + ( R ⋅ F − R ⋅ F ) ...  Px Py Py Px + ( 0 − Rdy ⋅ Fsheathx)

    

(g)

Using Newton's third law, we have F31x := −F13x

F21x := −F12x

Fb1x := −Fb2x (h)

F31y := −F13y

F21y := −F12y

Fb1y := −Fb2y



3-29-4

Fsheathx := −Fcablex

Given

RPx := −27⋅ mm

RPy := 0 ⋅ mm

Rdx := −41⋅ mm

Rdy := 27⋅ mm

We will now use a Mathcad solve block to solve equations (f) through (h). Guess

Px := 1000⋅ N

Given

F21x + Fb1x + F31x + Px + Fsheathx = 0

Mh := −100⋅ N ⋅ m

Py := 0 ⋅ N

F21y + Fb1y + F31y + Py = 0

Mh + ( R21x⋅ F21y − R21y⋅ F21x) ... = 0 + ( Rb1x⋅ Fb1y − Rb1y⋅ Fb1x) ...  + ( R31x⋅ F31y − R31y⋅ F31x) ...  + ( R ⋅ P − R ⋅ P ) ...   Px y Py x  + ( 0⋅ N ⋅ m − Rdy⋅ Fsheathx) 

 Px     Py  := Find( Px , Py , Mh) M   h Summarizing, the results obtained for a grip force Fb2 = 267⋅ N are: Handlebar (1)

Fb1x = 0 ⋅ N

Fb1y = 267⋅ N

F21x = −1047⋅ N

F21y = −340⋅ N

F31x = 0 ⋅ N

F31y = 73.2⋅ N −6

Px = 1 × 10

⋅N

Py = 0 ⋅ N

Mh = 0.0⋅ N ⋅ m Lever (2)

Cable (3)

Fc2x = −1047⋅ N

Fc2y = −73.2⋅ N

F12x = 1047⋅ N

F12y = 340⋅ N

Fc3x = 1047⋅ N

Fc3y = 73.2⋅ N

F13x = 0 ⋅ N

F13y = −73.2⋅ N

Fcablex = −1047⋅ N

Fcabley = 0 ⋅ N



3-30-1

PROBLEM 3-30

_____

Statement:

Run the TKSolver or Mathcad model for Case Study 2A and move the point of application of the crimp force along the jaw by changing the values of Rhand, recalculate and observe the changes to the forces and moments.

Problem:

Determine the forces on the elements of the crimping tool shown in Figure 3-3 during a crimp operation.

Given:

The geometry is known and the tool develops a crimp force of 2000 lb (8896 N) at closure in the position shown. Applied crimp force

Fc4x := −1956.30⋅ lbf

Fc4y := 415.82⋅ lbf

Position vector components (Change the value of Rhand and note the results) Rc4x := 0.454⋅ in

R12x := 1.399⋅ in

R32x := 2.199⋅ in

Rc4y := 0.337⋅ in

R12y := 0.049⋅ in

R32y := 0.077⋅ in

R23x := −0.602⋅ in

R43x := 0.602⋅ in

R14x := −0.161⋅ in

R23y := 0.127⋅ in

R43y := −0.127⋅ in

R14y := −0.758⋅ in

R34x := 0.161⋅ in

R34y := 0.758⋅ in

Rhand := −4.40⋅ in

Assumptions: The accelerations are negligible. All forces are coplanar and two-dimensional. A class 1 load model is appropriate and a static analysis is acceptable. Solution: 1.

See Figures 3-3, 3-4, and Mathcad file P0330.

Figure 3-3 shows the tool in the closed position, in the process of crimping a metal connector onto a wire. The user's hand provides the input forces between links 1 and 2, shown as the reaction pair Fhand. The user can grip the handle anywhere along its length but we are assuming a nominal moment arm of Rhand for the application of the resultant of the user's grip force (see Figure 3-4). The high mechanical advantage of the tool transforms the grip force to a large force at the crimp. Figure 3-3 is a free-body diagram of the entire assembly, neglecting the weight of the tool, which is small compared to the crimp force. There are four elements, or links, in the assembly, all pinned together. Link 1 can be considered to be the "ground" link, with the other links moving with respect to it as the jaw is closed. The desired magnitude of the crimp force Fc is defined and its direction will be normal to the surfaces at the crimp.

2.

Figure 3-4 shows the elements of the crimping tool assembly separated and drawn as free-body diagrams with all forces applied to each element, again neglecting their weights as being insignificant compared to the applied forces. The centers of gravity of the respective elements are used as the origins of the local, non rotating coordinate systems in which the points of application of all forces on the element are located. (See footnote on page 116 of the text).

3.

We will consider link 1 to be the ground plane and analyze the remaining moving links. Note that all unknown forces and moments are initially assumed positive. Link 4 has three forces acting on it: Fc4 is the known (desired) force at the crimp, and F14 and F34 are the reaction forces from links 1 and 3, respectively. The magnitudes of these two forces are unknown as is the direction of F14. The direction of F34 will be the same as link 3, since it is a two-force member. Writing equations 3.3b for this element:

ΣFx = F14x + F34x + Fc4x = 0 ΣFy = F14y + F34y + Fc4y = 0

(a)

ΣMz = ( R14x⋅ F14y − R14y⋅ F14x) ...  = 0

+ ( R34x⋅ F34y − R34y⋅ F34x) ... + ( R ⋅ F − R ⋅ F )   c4x c4y c4y c4x 



3-30-2

We have three equations and four unknowns (F14x, F14y, F34x, F34y) at this point, so we need another equation. It is available from the fact that the direction of F34 is known. We can express one component of the force F34 in terms of its other component and the known angle θ3 of link 3. (b) F34y = F34x⋅ tan( θ3) (c)

where

θ3 := 168⋅ deg

Guess

F14x := 500⋅ lbf

Given

F14x + F34x + Fc4x = 0

F34x := 1000⋅ lbf

F14y := −100⋅ lbf

F34y := −100⋅ lbf

F14y + F34y + Fc4y = 0

( Rc4x⋅ Fc4y − Rc4y⋅ Fc4x) ...  = 0 + ( R14x⋅ F14y − R14y⋅ F14x) ... + ( R ⋅ F − R ⋅ F )   34x 34y 34y 34x  F34y = F34x⋅ tan( θ3)

 F14x  F   14y  := Find ( F , F , F , F ) 14x 14y 34x 34y  F34x     F34y  Components of the unknown forces F14, and F34 F14x = 442.9⋅ lbf 4.

5.

F14y = −94.1⋅ lbf

F34x = 1513.4⋅ lbf

F34y = −321.7⋅ lbf

Link 3 has two forces on it, F23 and F43. Because this is a two-force link, these two forces are equal in magnitude and opposite in direction. Also, from Newton's third law, F43 = - F34. Thus, F43x := −F34x

F43y := −F34y

F23x := −F43x

F23y := −F43y

F43x = −1513.4⋅ lbf

F43y = 321.7⋅ lbf

F23x = 1513.4⋅ lbf

F23y = −321.7⋅ lbf

(d)

Link 2 has three forces acting on it: Fhand is the unknown force from the hand, and F12 and F32 are the reaction forces from links 1 and 3, respectively. Force F12 is provided by part 1 on part 2 at the pivot pin and force F32 is provided by part 3 acting on part 2 at their pivot pin. The magnitude and direction of F32 is known and the direction of Fhand is known. Using equations 3.3b, we can solve for the magnitude of Fhand and the two components of F12. From the third law, F32x := −F23x

F32y := −F23y

F32x = −1513.4⋅ lbf

F32y = 321.7⋅ lbf

ΣFx = F12x + F32x = 0 (e)

ΣFy = Fhand + F12y + F32y = 0 ΣMz = ( R12 × F12) + ( R32 × F32) ... = 0 + ( Rhand × Fhand ) Guess

F12x := 1500⋅ lbf

F12y := −100⋅ lbf

Fhand := 100⋅ lbf


MACHINE DESIGN - An Integrated Approach, 4th Ed. Given

3-30-3

F12x + F32x = 0 F12y + F32y + Fhand = 0

( R12x⋅ F12y − R12y⋅ F12x) ...  = 0 + ( R32x⋅ F32y − R32y⋅ F32x) ... + R ⋅ F   hand hand   F12x     F12y  := Find( F12x , F12y , Fhand ) F   hand  F12x = 1513.4⋅ lbf 6.

7.

F12y = −373.4⋅ lbf

Fhand = 51.7⋅ lbf

The four forces on link 1 can now be determined using the third law. F21x := −F12x

F21y := −F12y

F41x := −F14x

F41y := −F14y

F21x = −1513.4⋅ lbf

F21y = 373.4⋅ lbf

F41x = −442.9⋅ lbf

F41y = 94.1⋅ lbf

Fc1x := −Fc4x

Fc1y := −Fc4y

Fc1x = 1956.3⋅ lbf

Fc1y = −415.8⋅ lbf

The solution to this problem for the scaled dimensions in Figure 3-3 assuming a 2000-lb (8896-N) force applied at the crimp, normal to the crimp surface, is given above. The total forces at the pivot points are: F12 :=  F12x + F12y

2

0.5

Pivot A

F32 :=  F32x + F32y

2

0.5

Pivot B

F43 :=  F43x + F43y

2

0.5

Pivot C

F14 :=  F14x + F14y

2

0.5

Pivot D

2 2 2 2



F12 = 1559⋅ lbf



F32 = 1547⋅ lbf



F43 = 1547⋅ lbf



F14 = 453⋅ lbf

The moment that must be applied to the handles to generate the crimp force of Crimp force

Fc4 :=  Fc4x + Fc4y

Moment

Mh := Rhand ⋅ Fhand

2

2

0.5



Fc4 = 2000⋅ lbf Mh = 227⋅ lbf ⋅ in

This moment can be obtained with a force of Fhand = 52⋅ lbf applied at mid-handle. This force is within the physiological grip-force capacity of the average human.



3-31-1

PROBLEM 3-31

_____

Statement:

Run the TKSolver or Mathcad model for Case Study 2A and move the point of application of P along the x direction by changing the values of Rpx, recalculate and observe the changes to the forces and moments. What happens when the vertical force P is centered on link 3? Also, change the angle of the applied force P to create an x component and observe the effects on the forces and moments on the elements.

Problem:

Determine the forces on the elements of the scissors-jack in the position shown in Figure 3-5.

Given:

The geometry is known and the jack supports a force of 1000 lb (4448 N) in the position shown. Support force

Px := 0.0⋅ lbf

Py := −1000⋅ lbf

Position vector components (Change the value of Rpx and note the results) Rpx := −0.50⋅ in

R12x := −3.12⋅ in

R32x := 2.08⋅ in

Rpy := 0.87⋅ in

R12y := −1.80⋅ in

R32y := 1.20⋅ in

R42x := 2.71⋅ in

R23x := −0.78⋅ in

R43x := 0.78⋅ in

R42y := 1.00⋅ in

R23y := −0.78⋅ in

R43y := −0.78⋅ in

R14x := 3.12⋅ in

R24x := −2.58⋅ in

R34x := −2.08⋅ in

R14y := −1.80⋅ in

R24y := 1.04⋅ in

R34y := 1.20⋅ in

Angle of gear teeth common normal

θ := −45.0⋅ deg

Assumptions: The accelerations are negligible. The jack is on level ground. The angle of the elevated car chassis does not impart an overturning moment to the jack. All forces are coplanar and two-dimensional. A class 1 load model is appropriate and a static analysis is acceptable. Solution:

See Figures 3-5 through 3-8, and Mathcad file P0331.

1.

Figure 3-5 shows a schematic of a simple scissors jack used to raise a car. It consists of six links that are pivoted and/or geared together and a seventh link in the form of a lead screw that is turned to raise the jack. While this is clearly a three-dimensional device, it can be analyzed as a two-dimensional one if we assume that the applied load (from the car) and the jack are exactly vertical (in the z direction). If so, all forces will be in the xy plane. This assumption is valid if the car is jacked from a level surface. If not, then there will be some forces in the yz and xz planes as well. The jack designer needs to consider the more general case, but for our simple example we will initially assume two-dimensional loading. For the overall assembly as shown in Figure 3-5, we can solve for the reaction force Fg, given force P, by summing forces: Fg = -P.

2.

Figure 3-6 shows a set of free-body diagrams for the entire jack. Each element or subassembly of interest has been separated from the others and the forces and moments shown acting on it (except for its weight, which is small compared to the applied forces and is thus neglected for this analysis). The forces and moments can be either internal reactions at interconnections with other elements or external loads from the "outside world." The centers of gravity of the respective elements are used as the origins of the local, non rotating coordinate systems in which the points of application of all forces on the element are located. In this design, stability is achieved by the mating of two pairs of crude (non involute) gear segments acting between links 2 and 4 and between links 5 and 7. These interactions are modeled as forces acting along a common normal shared by the two teeth. This common normal is perpendicular to the common tangent at the contact point. There are 3 second-law equations available for each of the seven elements allowing 21 unknowns. An additional 10 third-law equations will be needed for a total of 31. This is a cumbersome system to solve for such a simple device, but we can use its symmetry to advantage in order to simplify the problem.

3.

Figure 3-7 shows the upper half of the jack assembly. Because of the mirror symmetry between the upper and lower portions, the lower half can be removed to simplify the analysis. The forces calculated for this half will be duplicated in the other. If we wished, we could solve for the reaction forces at A and B using equations 3.3b from this free-body diagram of the half-jack assembly.



3-31-2

4.

Figure 3-8a shows the free-body diagrams for the upper half of the jack assembly, which are essentially the same as those of Figure 3-6. We now have four elements but can consider the subassembly labeled 1 to be the "ground," leaving three elements on which to apply equations 3.3. Note that all forces and moments are initially assumed positive in the equations.

5.

Link 2 has three forces acting on it: F42 is the unknown force at the gear tooth contact with link 4; F12 and F32 are the unknown reaction forces from links 1 and 3, respectively. Force F12 is provided by part 1 on part 2 at the pivot pin and force F32 is provided by part 3 acting on part 2 at their pivot pin. The magnitudes and the directions of these pin forces and the magnitude of F42 are unknown. The direction of F42 is along the common normal shown in Figure 3-8b. Write equations 3.3b for this element to sum the forces in the x and y directions and sum moments about the CG (with the cross products expanded). ΣFx = F12x + F32x + F42x = 0 (a)

ΣFy = F12y + F32y + F42y = 0 ΣMz = ( R12x⋅ F12y − R12y⋅ F12x) ...  = 0

+ ( R32x⋅ F32y − R32y⋅ F32x) ... + ( R ⋅ F − R ⋅ F )   42x 42y 42y 42x 

6.

Link 3 has three forces acting on it: P, F23 and F43. Only P is known. Writing equations 3.3b for this element gives

ΣFx = F23x + F43x + Px = 0 (b)

ΣFy = F23y + F43y + Py = 0 ΣMz = ( R23x⋅ F23y − R23y⋅ F23x) ...  = 0

+ ( R43x⋅ F43y − R43y⋅ F43x) ... + ( R ⋅ P − R ⋅ P )   px y py x 

7.

Link 4 has three forces acting on it: F24 is the unknown force from link 2; F14 and F34 are the unknown reaction forces from links 1 and 3, respectively.

ΣFx = F14x + F24x + F34x = 0 (c)

ΣFy = F14y + F24y + F34y = 0 ΣMz = ( R14x⋅ F14y − R14y⋅ F14x) ...  = 0

+ ( R24x⋅ F24y − R24y⋅ F24x) ... + ( R ⋅ F − R ⋅ F )   34x 34y 34y 34x 

8.

9.

The nine equations in sets a through c have 16 unknowns in them, F12x, F12y, F32x, F32y, F23x, F23y, F43x, F43y, F14x , F14y, F34x, F34y, F24x, F24y, F42x, F42y. We can write the third-law relationships between action-reaction pairs at each of the joints to obtain six of the seven additional equations needed: F32x = −F23x

F32y = −F23y

F34x = −F43x

F34y = −F43y

F42x = −F24x

F42y = −F24y

(d)

The last equation needed comes from the relationship between the x and y components of the force F24 (or F42) at the tooth/tooth contact point. Such a contact (or half) joint can transmit force (excepting friction force) only along the common normal , which is perpendicular to the joint's common tangent as shown in Figure 3-8b. The common normal is also called the axis of transmission.



3-31-3

The tangent of the angle of this common normal relates the two components of the force at the joint: (e)

F24y = F24x⋅ tan( θ ) 10. Equations (d) and (e) will be substituted into equations (a) through (c) to create a set of nine simultaneous equations for solution. Guess

Given

F12x := 500⋅ lbf

F12y := 500⋅ lbf

F14x := −500⋅ lbf

F23x := 500⋅ lbf

F23y := 500⋅ lbf

F24x := 500⋅ lbf

F43x := −500⋅ lbf

F43y := 500⋅ lbf

( R12x⋅ F12y − R12y⋅ F12x) ... + ( −R32x⋅ F23y + R32y⋅ F23x) ... + ( −R ⋅ F ⋅ tan( θ ) + R ⋅ F ) 42x 24x 42y 24x 

 =0   

F14y := 500⋅ lbf

F12x − F23x − F24x = 0 F12y − F23y − F24x⋅ tan( θ ) = 0

( R23x⋅ F23y − R23y⋅ F23x) ...  = 0 + ( R43x⋅ F43y − R43y⋅ F43x) ... + ( R ⋅ P − R ⋅ P )   px y py x 

F23x + F43x + Px = 0

( R14x⋅ F14y − R14y⋅ F14x) ...  =0 + ( R24x⋅ F24x⋅ tan( θ ) − R24y⋅ F24x) ... + ( −R ⋅ F + R ⋅ F )  34x 43y 34y 43x  

F14x + F24x − F43x = 0

F23y + F43y + Py = 0

F14y + F24x⋅ tan( θ ) − F43y = 0

 F12x     F12y   F14x     F14y   F23x  := Find ( F , F , F , F , F , F , F , F , F ) 12x 12y 14x 14y 23x 23y 24x 43x 43y    F23y  F   24x   F43x     F43y  Results:

F14x = −877.8⋅ lbf

F14y = 469.6⋅ lbf

F24x = 290.1⋅ lbf

F24y := F24x⋅ tan( θ )

F34x := −F43x

F34y := −F43y

F23x = 587.7⋅ lbf

F23y = 820.5⋅ lbf

F43x = −587.7⋅ lbf

F43y = 179.5⋅ lbf

F12x = 877.8⋅ lbf

F12y = 530.4⋅ lbf

F32x := −F23x

F32y := −F23y

F42x := −F24x

F42y := −F24y

F24y = −290.1⋅ lbf



3-32-1

PROBLEM 3-32

_____

Statement:

Figure P3-14 shows a cam-follower arm. If the load P = 200 lb, what spring force is needed at the right end to maintain a minimum load between cam and follower of 25 lb? Find the maximum shear force and bending moment in the follower arm. Plot the shear and moment diagrams.

Given:

Load at left end of beam Load at cam follower

P := 200⋅ lbf Pcam := 25⋅ lbf

Distance from left end to: Pivot point Cam follower Spring Solution: 1.

a := 10⋅ in b := 22⋅ in c := 29⋅ in


Draw a FBD of the cam-follower arm (beam). c b

P

C

R

2.

Pcam

Fspring

From inspection of the FBD, write the load function equation q(x) = -P-1 + R-1 + Pcam-1 - Fspring-1

3.

Integrate this equation from -∞ to x to obtain shear, V(x) V(x) = -P0 + R0 + Pcam0 - Fspring0

4.

Integrate this equation from -∞ to x to obtain moment, M(x) M(x) = -P1 + R1 + Pcam1 - Fspring1

5.

Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = c, where both are zero. At x = c+, V = M = 0

V = −P + R + Pcam − Fspring = 0 M = −P⋅ c + R⋅ ( c − a ) + Pcam⋅ ( c − b ) = 0

Fspring :=

P⋅ a + Pcam⋅ ( b − a ) c−a

R := Fspring + P − Pcam 6.

Define the range for x

x := 0 ⋅ in , 0.002⋅ c .. c

Fspring = 121.05⋅ lbf R = 296.05⋅ lbf



3-32-2

For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x , z) := if ( x ≥ z , 1 , 0 )

8.

Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x) := −P⋅ S ( x , 0 ⋅ in) + R⋅ S ( x , a ) + Pcam⋅ S ( x , b ) − Fspring⋅ S ( x , c) M ( x) := −P⋅ S ( x , 0 ⋅ in) ⋅ x + R⋅ S ( x , a ) ⋅ ( x − a) + Pcam⋅ S ( x , b ) ⋅ ( x − b ) − Fspring⋅ S ( x , c) ⋅ ( x − c)

9.

Plot the shear and moment diagrams and find the maximum shear force and bending moment.

SHEAR DIAGRAM 200 100

V ( x) lbf

0 − 100 − 200 − 300

0

10

20

30

x in

Vmax := V ( 0 ⋅ in)

Vmax = 200⋅ lbf

MOMENT DIAGRAM 0

− 500 M ( x) in⋅ lbf

− 1000

− 1500

− 2000

0

10

20

30

x in

Mmax := M ( a )

Mmax = 2000⋅ in⋅ lbf



PROBLEM 3-33

3-33-1

_____

Statement:

Write a computer program or equation solver model to calculate all the singularity functions listed in equations 3.17. Set them up as functions that can be called from within any other program or model.

Solution:


1.

No solution is provided for this programming problem.



3-34a-1


Given:

A beam is supported and loaded as shown in Figure P3-15. Find the reactions, maximum shear, and maximum moment for the data given in row a from Table P3-2. Beam length

L := 20⋅ in

Distance to RH bearing

a := 16⋅ in

Distance to concentrated load

b := 18⋅ in

Concentrated load

P := 1000⋅ lbf

b

P

R2

R1 a

FIGURE 3-34aA Solution: 1.

See Figure 3-34 and Mathcad file P0334a.


From inspection of Figure 3-34, write the load function equation q(x) = R1-1 + R2-1 - P-1

2.

Integrate this equation from -∞ to x to obtain shear, V(x) V(x) = R10 + R20 - P0

3.

Integrate this equation from -∞ to x to obtain moment, M(x) M(x) = R11 + R21 - P1

4.

Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b, where both are zero. At x = b +, V = M = 0 V = R1 + R2 − P = 0 M = R1 ⋅ b + R2 ⋅ ( b − a ) = 0 R1 :=

P a

⋅ (a − b)

R2 := P − R1

R1 = −125⋅ lbf R2 = 1125⋅ lbf

5.


6.

For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z.

x := 0 ⋅ m , 0.002⋅ L .. L

S ( x , z) := if ( x ≥ z , 1 , 0 ) 7.

Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x) := R1⋅ S ( x , 0 ⋅ in) + R2⋅ S ( x , a ) − P⋅ S ( x , b ) M ( x) := R1⋅ S ( x , 0 ⋅ in) ⋅ x + R2⋅ S ( x , a ) ⋅ ( x − a ) − P⋅ S ( x , b ) ⋅ ( x − b)

8.

Plot the shear and moment diagrams.



3-34a-2

SHEAR DIAGRAM

MOMENT DIAGRAM

1000

1000

0 500 V ( x)

M ( x)

lbf

in⋅ lbf

− 1000

0 − 2000

− 500

0

5

10

15

− 3000

20

0

5

10

x

x

in

in

15

20


9.

Determine the maximum shear and maximum moment from inspection of the diagrams. Maximum shear:

Vmax := V ( a )

Vmax = 1000⋅ lbf

Maximum moment occurs where V is zero, which is x = a: Mmax := M ( a )




3-35a-1


Input data:

A beam is supported and loaded as shown in Figure P3-15. Write a computer program or equation solver model to find the reactions and calculate and plot the loading, shear, and moment functions. Test the program with the data given in row a from Table P3-2. b

Enter data in highlighted areas Beam length

L := 20⋅ in


a := 16⋅ in


b := 18⋅ in

Concentrated load

F := 1000⋅ lbf

F

R2

R1 a




From inspection of Figure 3-35, write the load function equation q(x) = R1-1 + R2-1 - F-1

2.

Integrate this equation from -∞ to x to obtain shear, V(x) V(x) = R10 + R20 - F0

3.

Integrate this equation from -∞ to x to obtain moment, M(x) M(x) = R11 + R21 - F1

4.

Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b, where both are zero. At x = b +, V = M = 0 V = R1 + R2 − F = 0 M = R1 ⋅ b + R2 ⋅ ( b − a ) = 0 R1 :=

F a

⋅ ( a − b)

R2 := F − R1

R1 = −125⋅ lbf R2 = 1125⋅ lbf

5.


6.


x := 0 ⋅ m , 0.002⋅ L .. L

S ( x , z) := if ( x ≥ z , 1 , 0 ) 7.

Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x) := R1⋅ S ( x , 0 ⋅ in) + R2⋅ S ( x , a ) − F ⋅ S ( x , b ) M ( x) := R1⋅ S ( x , 0 ⋅ in) ⋅ x + R2⋅ S ( x , a ) ⋅ ( x − a ) − F ⋅ S ( x , b ) ⋅ ( x − b )

8.




3-35a-2

SHEAR DIAGRAM

MOMENT DIAGRAM

1000

1000

0 500 V ( x)

M ( x)

lbf

in⋅ lbf

− 1000

0 − 2000

− 500

0

5

10

15

− 3000

20

0

5

10

x

x

in

in

15

20


9.


Vmax := V ( a )

Vmax = 1000⋅ lbf

Maximum moment occurs where V is zero, which is x = a: Mmax := M ( a )




3-36a-1


Solution: 1.


L := 20⋅ in


L := 20⋅ in

Distance to start of load

a := 16⋅ in

Distance to end of load

b := 18⋅ in

Distributed load

p := 1000⋅

b a

L

R1

lbf in

p

R2

FIGURE 3-36aA



From inspection of Figure 3-36, write the load function equation q(x) = R1-1 - p0 + p0 + R2-1

2.

Integrate this equation from -∞ to x to obtain shear, V(x) V(x) = R10 - p1 + p1 + R20

3.

Integrate this equation from -∞ to x to obtain moment, M(x) M(x) = R11 - p2/2 + p2/2 + R21

4.

Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0 V = R1 − p ⋅ ( L − a ) + p ⋅ ( L − b ) + R2 = 0 M = R1 ⋅ L −

R1 :=

p 2⋅ L

p 2

2

⋅ (L − a) +

p 2

2

⋅ ( L − b ) + R2⋅ ( L − b) = 0

⋅ 2 ⋅ ( b − a ) ⋅ L + a − b 2

2



R1 = 300⋅ lbf

R2 := p ⋅ ( b − a ) − R1

R2 = 1700⋅ lbf

5.


6.


x := 0 ⋅ in , 0.002⋅ L .. L

S ( x , z) := if ( x ≥ z , 1 , 0 ) 7.

Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x) := R1⋅ S ( x , 0 ⋅ in) − p ⋅ S ( x , a ) ⋅ ( x − a) + p ⋅ S ( x , b ) ⋅ ( x − b ) + R2⋅ S ( x , L) M ( x) := R1⋅ S ( x , 0 ⋅ in) ⋅ x −

p 2

2

⋅ S(x , a)⋅ ( x − a) +

p 2

2

⋅ S ( x , b ) ⋅ ( x − b ) + R2⋅ S ( x , L) ⋅ ( x − L)



3-36a-2


SHEAR DIAGRAM

MOMENT DIAGRAM

1000

5000

4000 0 V ( x)

M ( x)

lbf

in⋅ lbf

3000

2000 − 1000 1000

− 2000

0

5

10

15

20

0

0

5

10

x

x

in

in

15

20


9.


Vmax := V ( b )

Vmax = 1700⋅ lbf

Maximum moment occurs where V is zero, which is x = c, where:

c :=

R1 ⋅ b + R2 ⋅ a R1 + R2

Mmax := M ( c)

c = 16.3⋅ in




3-37a-1


Input data:

A beam is supported and loaded as shown in Figure P3-16. Write a computer program or equation solver model to find the reactions and calculate and plot the loading, shear, and moment functions. Test the program with the data given in row a from Table P3-2. Enter data in highlighted areas Beam length

Solution: 1.

b

L := 20⋅ in


L := 20⋅ in

Distance to start of load

a := 16⋅ in

Distance to end of load

b := 18⋅ in

Distributed load

p := 1000⋅

a

p

L

R1

R2

lbf

FIGURE 3-37aA

in



From inspection of Figure 3-37, write the load function equation q(x) = R1-1 - p0 + p0 + R2-1

2.

Integrate this equation from -∞ to x to obtain shear, V(x) V(x) = R10 - p1 + p1 + R20

3.

Integrate this equation from -∞ to x to obtain moment, M(x) M(x) = R11 - p2/2 + p2/2 + R21

4.

Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0 V = R1 − p ⋅ ( L − a ) + p ⋅ ( L − b ) + R2 = 0 M = R1 ⋅ L −

R1 :=

p 2⋅ L

p 2

2

⋅ (L − a) +

p 2

2

⋅ ( L − b ) + R2⋅ ( L − b) = 0

⋅ 2 ⋅ ( b − a ) ⋅ L + a − b 2

2



R1 = 300⋅ lbf

R2 := p ⋅ ( b − a ) − R1

R2 = 1700⋅ lbf

5.


6.


x := 0 ⋅ in , 0.002⋅ L .. L

S ( x , z) := if ( x ≥ z , 1 , 0 ) 7.

Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x) := R1⋅ S ( x , 0 ⋅ in) − p ⋅ S ( x , a ) ⋅ ( x − a) + p ⋅ S ( x , b ) ⋅ ( x − b ) + R2⋅ S ( x , L) M ( x) := R1⋅ S ( x , 0 ⋅ in) ⋅ x −

p 2

2

⋅ S(x , a)⋅ ( x − a) +

p 2

2

⋅ S ( x , b ) ⋅ ( x − b ) + R2⋅ S ( x , L) ⋅ ( x − L)



3-37a-2


SHEAR DIAGRAM

MOMENT DIAGRAM

1000

5000

4000 0 V ( x)

M ( x)

lbf

in⋅ lbf

3000

2000 − 1000 1000

− 2000

0

5

10

15

20

0

0

5

10

x

x

in

in

15

20


9.


Vmax := V ( b )

Vmax = 1700⋅ lbf

Maximum moment occurs where V is zero, which is x = c, where:

c :=

R1 ⋅ b + R2 ⋅ a R1 + R2

Mmax := M ( c)

c = 16.3⋅ in




3-38a-1


Given:


b

L := 20⋅ in


a := 16⋅ in


b := 18⋅ in

Concentrated load

P := 1000⋅ lbf

Distributed load

p := 1000⋅ lbf ⋅ in

P

p

R2

R1 a

−1


2.



Determine the distance from the origin to the left and right ends of the roller. Distance to left end

e := 0.1⋅ a

e = 1.600⋅ in

Distance to right end

f := 0.9⋅ a

f = 14.400in ⋅

From inspection of Figure 3-38, write the load function equation q(x) = R1-1 - p0 + p0 + R2-1 - P-1

3.

Integrate this equation from -∞ to x to obtain shear, V(x) V(x) = R10 - p1 + p1 + R20 - P0

4.

Integrate this equation from -∞ to x to obtain moment, M(x) M(x) = R11 - p2/2 + p2/2 + R21 - P1

5.

Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b, where both are zero. At x = b +, V = M = 0 V = R1 − p ⋅ ( b − e ) + p ⋅ ( b − f ) + R2 − P = 0 M = R1 ⋅ b −

R1 :=

p 2

2

⋅ ( b − e) +

p 2

2

⋅ ( b − f ) + R2⋅ ( b − a) = 0

 e2 − f 2  b − a  + f − e ⋅ p −  ⋅P  2⋅ a   a 

R2 := p ⋅ ( f − e) − R1 + P

R1 = 6275⋅ lbf R2 = 7525⋅ lbf

6.


7.


x := 0 ⋅ m , 0.002⋅ L .. L

S ( x , z) := if ( x ≥ z , 1 , 0 ) 8.

Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x) := R1⋅ S ( x , 0 ⋅ m) − p ⋅ S ( x , e) ⋅ ( x − e) + p ⋅ S ( x , f ) ⋅ ( x − f ) + R2⋅ S ( x , a ) − P⋅ S ( x , b ) p 2 2 ⋅ S ( x , e) ⋅ ( x − e) + ⋅ S ( x , f ) ⋅ ( x − f ) ... 2 2 + R2⋅ S ( x , a ) ⋅ ( x − a) − P⋅ S ( x , b ) ⋅ ( x − b )

M ( x) := R1⋅ S ( x , 0 ⋅ m) ⋅ x −

p



9.

3-38a-2


SHEAR DIAGRAM

MOMENT DIAGRAM

10000

30000

5000

20000

V ( x) lbf

M ( x) 0

in⋅ lbf

10000

− 5000

− 10000

0

0

5

10

15

− 10000

20

0

5

10

x

x

in

in

15

20


9.


Vmax := V ( f )

Vmax = 6525⋅ lbf

Maximum moment occurs where V is zero, which is x = c: c−e R1

=

f −c R2 − P

Mmax := M ( c)

c :=

f ⋅ R1 + e ⋅ R2 − e ⋅ P R1 + R2 − P

c = 7.875⋅ in




3-39a-1


Input data:

A beam is supported and loaded as shown in Figure P3-17. Write a computer program or equation solver model to find the reactions and calculate and plot the loading, shear, and moment functions. Test the program with the data given in row a from Table P3-2. Enter data in highlighted areas Beam length

L := 20⋅ in


a := 16⋅ in


b := 18⋅ in

Concentrated load

P := 1000⋅ lbf

Distributed load Solution: 1.

2.

b

p := 1000⋅ lbf ⋅ in

P

p

R2

R1 a

−1

FIGURE 3-39aA Free Body Diagram for Problem 3-39


Determine the distance from the origin to the left and right ends of the roller. Distance to left end

e := 0.1⋅ a

e = 40.64⋅ mm


f := 0.9⋅ a

f = 365.76mm ⋅

From inspection of Figure 3-39, write the load function equation q(x) = R1-1 - p0 + p0 + R2-1 - P-1

3.

Integrate this equation from -∞ to x to obtain shear, V(x) V(x) = R10 - p1 + p1 + R20 - P0

4.

Integrate this equation from -∞ to x to obtain moment, M(x) M(x) = R11 - p2/2 + p2/2 + R21 - P1

5.

Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b, where both are zero. At x = b +, V = M = 0 V = R1 − p ⋅ ( b − e ) + p ⋅ ( b − f ) + R2 − P = 0 M = R1 ⋅ b −

R1 :=

p 2

2

⋅ ( b − e) +

p 2

2

⋅ ( b − f ) + R2⋅ ( b − a) = 0

  e2 − f 2 b − a  + f − e ⋅ p −  ⋅P 2 ⋅ a    a 

R2 := p ⋅ ( f − e) − R1 + P

R1 = 6275⋅ lbf R2 = 7525⋅ lbf

6.


7.


x := 0 ⋅ m , 0.002⋅ L .. L

S ( x , z) := if ( x ≥ z , 1 , 0 ) 8.

Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x) := R1⋅ S ( x , 0 ⋅ m) − p ⋅ S ( x , e) ⋅ ( x − e) + p ⋅ S ( x , f ) ⋅ ( x − f ) + R2⋅ S ( x , a ) − P⋅ S ( x , b )



3-39a-2

p 2 2 ⋅ S ( x , e) ⋅ ( x − e) + ⋅ S ( x , f ) ⋅ ( x − f ) ... 2 2 + R2⋅ S ( x , a ) ⋅ ( x − a) − P⋅ S ( x , b ) ⋅ ( x − b )

M ( x) := R1⋅ S ( x , 0 ⋅ m) ⋅ x −

9.

p


SHEAR DIAGRAM

MOMENT DIAGRAM

10000

30000

5000

20000

V ( x) lbf

M ( x) 0

in⋅ lbf

10000

− 5000

− 10000

0

0

5

10

15

− 10000

20

0

5

10

x

x

in

in

15

20


9.


Vmax := V ( f )

Vmax = 6525⋅ lbf

Maximum moment occurs where V is zero, which is x = c: c−e R1

=

f −c R2 − P

Mmax := M ( c)

c :=

f ⋅ R1 + e ⋅ R2 − e ⋅ P R1 + R2 − P

c = 7.875⋅ in




3-40a-1


A beam is supported and loaded as shown in Figure P3-18. Find the reactions, maximum shear, and maximum moment for the data given in row a from Table P3-2. a

Given:

Distance to gear 2

L := 20⋅ in

Distance to gear 1

a := 16⋅ in


b := 18⋅ in

Concentrated load at gear 2

P2 := 1000⋅ lbf


P1 := 0.4⋅ P2

P1

P2

R2

R1 b L

FIGURE 3-40a Solution: 1.



From inspection of Figure 3-40, write the load function equation q(x) = R1-1 - P1-1 + R2-1 - P2-1

2.

Integrate this equation from -∞ to x to obtain shear, V(x) V(x) = R10 - P10 + R20 - P20

3.

Integrate this equation from -∞ to x to obtain moment, M(x) M(x) = R11 - P11 + R21 - P1

4.

Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0 V = R1 − P1 + R2 − P2 = 0 M = R1⋅ L − P1⋅ ( L − a ) + R2⋅ ( b − a ) = 0 R1 := P1⋅  1 −



  + P2⋅  1 − b  a

L



b

R2 := P1 + P2 − R1

R1 = −67⋅ lbf R2 = 1467⋅ lbf

5.


6.


x := 0 ⋅ m , 0.002⋅ L .. L

S ( x , z) := if ( x ≥ z , 1 , 0 ) 7.

Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x) := R1⋅ S ( x , 0 ⋅ in) − P1⋅ S ( x , a ) + R2⋅ S ( x , b ) − P2⋅ S ( x , L) M ( x) := R1⋅ S ( x , 0 ⋅ mm) ⋅ ( x − 0 ⋅ mm) − P1⋅ S ( x , a ) ⋅ ( x − a ) ... + R2⋅ S ( x , b ) ⋅ ( x − b) − P2⋅ S ( x , L) ⋅ ( x − L)

8.




3-40a-2

SHEAR DIAGRAM

MOMENT DIAGRAM

1000

0

500

− 1000

V ( x)

M ( x)

lbf

in⋅ lbf 0

− 500

− 2000

0

5

10

15

20

− 3000

0

5

10

x

x

in

in

15

20


9.


Vmax := V ( b )

Vmax = 1000⋅ lbf

Maximum moment occurs where V is zero, which is x = b: Mmax := M ( b )




3-41a-1


Input data:

A beam is supported and loaded as shown in Figure P3-18. Write a computer program or equation solver model to find the reactions and calculate and plot the loading, shear, and moment functions. Test the program with the data given in row a from Table P3-2. a

Enter data in highlighted areas Distance to gear 2

L := 20⋅ in

Distance to gear 1

a := 16⋅ in


b := 18⋅ in


P2 := 1000⋅ lbf


P1 := 0.4⋅ P2

P1

P2

R2

R1 b L




From inspection of Figure 3-41, write the load function equation q(x) = R1-1 - P1-1 + R2-1 - P2-1

2.

Integrate this equation from -∞ to x to obtain shear, V(x) V(x) = R10 - P10 + R20 - P20

3.

Integrate this equation from -∞ to x to obtain moment, M(x) M(x) = R11 - P11 + R21 - P1

4.

Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0 V = R1 − P1 + R2 − P2 = 0 M = R1⋅ L − P1⋅ ( L − a ) + R2⋅ ( b − a ) = 0 R1 := P1⋅  1 −



  + P2⋅  1 − b  a

L



b

R2 := P1 + P2 − R1

R1 = −67⋅ lbf R2 = 1467⋅ lbf

5.


6.


x := 0 ⋅ m , 0.002⋅ L .. L

S ( x , z) := if ( x ≥ z , 1 , 0 ) 7.

Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( z) := R1⋅ S ( z , 0 ⋅ in) − P1⋅ S ( z , a) + R2⋅ S ( z , b) − P2⋅ S ( z , L) M ( z) := R1⋅ S ( z , 0 ⋅ mm) ⋅ ( z − 0 ⋅ mm ) − P1⋅ S ( z , a) ⋅ ( z − a ) ... + R2⋅ S ( z , b) ⋅ ( z − b ) − P2⋅ S ( z , L) ⋅ ( z − L)



3-41a-2


SHEAR DIAGRAM

MOMENT DIAGRAM

1000

0

500

− 1000

V ( x)

M ( x)

lbf

in⋅ lbf 0

− 500

− 2000

0

5

10

15

20

− 3000

0

5

10

x

x

in

in

15

20


9.


Vmax := V ( b )

Vmax = 1000⋅ lbf

Maximum moment occurs where V is zero, which is x = b: Mmax := M ( b )




PROBLEM 3-42

3-42-1

_____

Statement:

A 1000 kg speedboat reaches a speed of 16 kph at the instant it takes up the slack in a 100 m-long tow rope attached to a surfboard carrying a 100 kg passenger. If the rope has k = 5 N/m, what is the dynamic force exerted on the surfboard?

Given:

Mass of speedboat

ms := 1000⋅ kg

Speed of boat

vi := 16⋅ kph

Mass of passenger

mp := 100⋅ kg

Rope stiffness

k := 5 ⋅ N ⋅ m

−1

Assumptions: 1. The water does not influence the dynamic force. 2. An impact model can be used to estimate the dynamic force. Solution: 1.


For the impact model, the passenger is the "struck" mass and the speedboat is the "striking mass". Thus, from equation 3.15, the energy correction factor is: 1

η := 1+

2.

mp

η = 0.97

3 ⋅ ms

Use equation 3.11 to estimate the dynamic force on the surfboard/passenger.

Fi := vi⋅ η⋅ ms⋅ k

Fi = 309⋅ N



3-43-1


Figure P3-19 shows an oil-field pump jack. For the position shown, draw free-body diagrams of the crank (2), connecting rod (3) and walking beam (4) using variable names similar to those used in Case Studies 1A and 2A. Assume that the crank turns slowly enough that accelerations can be ignored. Include the weight acting at the CG of the walking beam and the crank but not the connecting rod.

Assumptions: 1. A two-dimensional model is adequate. 2. Inertia forces may be ignored. Solution: 1.


Isolate each of the elements to be analyzed, starting with the walking beam, since the external forces on it are known. Place the known force, Fcable, at the point P and the known weight at the CG. Assume the forces at the interfaces O4 and B to be positive. The position vectors R14, R34, and Rp will be known as will the angle, θ3,that the connecting rod makes with the horizontal axis.

F34 R34 y

R 14 RP

head end P

F

B x

14y

4

counterweight

O4

F cable 2.

θ3

F

14x

W4

F43

The connecting rod is a two-force member with the forces acting at the interfaces A and B along the line joining points A and B. The assumption made in step 1 is that these are compressive forces on link 3.

y B R 43

3

θ3 x

R23

A

F 23



3-43-2

The crank is acted on by forces at A and O2, its weight at its CG, and a torque which we will assume to be positive (CCW). As in step 1, assume that the unknown reaction force at O2 is positive.

F32 y

F

A

12y

θ3 x

2 O2

F

12x

T2 W2



3-44-1


For the pump jack of Problem 3-43 and the data of Table P3-3, determine the pin forces on the walking beam, connecting rod, and crank and the reaction torque on the crank.

Given:

R12 := 13.2⋅ in

θ12 := 135⋅ deg

R14 := 79.22⋅ in

θ14 := 196⋅ deg

R32 := 0.80⋅ in

θ32 := 45⋅ deg

R34 := 32.00⋅ in

θ34 := 169⋅ deg

Fcable := 2970⋅ lbf

W2 := 598⋅ lbf

θ3 := 98.5⋅ deg Solution: 1.

W4 := 2706⋅ lbf

RP := 124.44⋅ in θP := 185⋅ deg

See Mathcad files P0343 and P0344.

Draw free-body diagrams of each element (see Problem 3-43).

F34 R34 y

R 14

θ3

RP

head end

F

P

B x

14y

4

counterweight

O4

F

14x

W4

F cable

F43 y B

F32 y

F

R 43

A

12y

3

θ3

θ3 x

x 2 O2

F

12x

R23

T2 W2 A

F 23 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0344.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

MACHINE DESIGN - An Integrated Approach, 4th Ed. 2. Calculate the x- and y-components of the position vectors.

3.

4.

3-44-2

R12x := R12⋅ cos( θ12)

R12x = −9.334⋅ in

R12y := R12⋅ sin( θ12)

R12y = 9.334⋅ in

R14x := R14⋅ cos( θ14)

R14x = −76.151⋅ in

R14y := R14⋅ sin( θ14)

R14y = −21.836⋅ in

R32x := R32⋅ cos( θ32)

R32x = 0.566⋅ in

R32y := R32⋅ sin( θ32)

R32y = 0.566⋅ in

R34x := R34⋅ cos( θ34)

R34x = −31.412⋅ in

R34y := R34⋅ sin( θ34)

R34y = 6.106⋅ in

RPx := RP⋅ cos( θP)

RPx = −123.966 ⋅ in

RPy := RP⋅ sin( θP)

RPy = −10.846⋅ in

Write equations 3(b) for link 4, the walking beam.

Σ Fx:

F14x + F34x = 0

(1)

Σ Fy:

−Fcable + F14y + F34y − W4 = 0

(2)

Σ Mz:

Rpx⋅ Fcable + ( R14x⋅ F14y − R14y⋅ F14x) + ( R34x⋅ F34y − R34y⋅ F34x) = 0

(3)

The direction (but not the sense) of F34 is known so write the equation that relates the x- and y-components of this force. F34y − F34x⋅ tan( θ3) = 0

5.

(4)

There are four unknowns in the four equations above. Solving for F14x, F14y, F34x, and F34y, 0 1 0   1  0 1 0 1    A :=  −R14y R14x −R34y R34x   in in in  in   0 −tan( θ3) 1   0 F14x = 2446⋅ lbf

6.

7.

8.

F14y = −10687⋅ lbf

0   F + W 4  cable   lbf B :=    −RPx ⋅ Fcable   in⋅ lbf    0 F34x = −2446⋅ lbf

 F14x  F   14y  := A − 1⋅ B⋅ lbf  F34x     F34y  F34y = 16363⋅ lbf

From Newton's thrid law and, since the connecting rod (3) is a two-force member F43x := −F34x

F43x = 2446⋅ lbf

F43y := −F34y

F43y = −16363⋅ lbf

F23x := −F43x

F23x = −2446⋅ lbf

F23y := −F43y

F23y = 16363⋅ lbf

Write equations 3(b) for link 2, the crank.

Σ Fx:

F12x + F32x = 0

(5)

Σ Fy:

F12y + F32y − W2 = 0

(6)

Σ Mz:

T2 + ( R12x⋅ F12y − R12y⋅ F12x) + ( R32x⋅ F32y − R32y⋅ F32x) = 0

(7)

There are three unknowns in the three equations above. Solving for F12x, F12y, and T2, since F32x := −F23x

F32x = 2446⋅ lbf

F32y := −F23y

F32y = −16363⋅ lbf



3-44-3

−F32x   lbf     W2 − F32y B :=   lbf    −( R32x⋅ F32y − R32y⋅ F32x)    in⋅ lbf  

0 0  1   0 1 0  A :=  −R12y R12x  1  in in  

F12x = −2446

lbf

T2 = 146128

in-lbf

F12y = 16961

 F12x    −1  F12y  := A ⋅ B  T   2 

lbf



3-45-1


Figure P3-20 shows an aircraft overhead bin mechanism in end view. For the position shown, draw free-body diagrams of links 2 and 4 and the door (3) using variable names similar to those used in Case Studies 1A and 2A. There are stops that prevent further clockwise motion of link 2 (and the identical link behind it at the other end of the door) resulting in horizontal forces being applied to the door at points A. Assume that the mechanism is symmetrical so that each set of links 2 and 4 carry one half of the door weight. Ignore the weight of links 2 and 4 as they are negligible.

Assumptions: 1. A two-dimensional model is adequate. 2. Inertia forces may be ignored as the mechanism is at rest against stops. Solution: 1.


Isolate each of the elements to be analyzed, starting with the door. Place the force, Fstop, at the point A and the known weight at the CG. Assume the forces in links 2 and 4 to be positive (tensile). The position vectors R43 and R23 will be known as will the angles θ2 and θ4 that links 2 and 4 make with the horizontal axis.

F 23 θ4 F 43

θ2

y

F stop

R43

R23

A

3

B x

W3 2

2.

F 12

Links 2 and 4 are two-force members with the forces acting at the pinned ends along the line joining the pin centers. The assumption made in step 1 is that these are tensile forces on links 2 and 4.

θ2

y O2 R12 2

x θ4

y

R32

F 14

A O4

R14

x 4

R34

B

F 34

F 32



3-46-1


For the overhead bin mechanism of Problem 3-45 and the data of Table P3-4, determine the pin forces on the door (3), and links 2 & 4 and the reaction force on each of the two stops.

Given:

R23 := 180.0⋅ mm θ23 := 160.345⋅ deg W3 := 45⋅ N

Solution: 1.

θ43 := 27.862⋅ deg

R43 := 180.0⋅ mm

θ2 := 85.879⋅ deg

θ4 := 172.352 ⋅ deg



F 12

θ2

y O2

θ4

y

F 14

R12 2

O4

x

x

R14

4

R34

F 34

B

R32 A F 23 θ4 F 43

θ2

F 32

y

F stop

A

R23

R43 3

B x

W3 2 2.

3.

Calculate the x- and y-components of the position vectors on the door (3). R23x := R23⋅ cos( θ23)

R23x = −169.512 ⋅ mm

R23y := R23⋅ sin( θ23)

R23y = 60.544⋅ mm

R43x := R43⋅ cos( θ43)

R43x = 159.134 ⋅ mm

R43y := R43⋅ sin( θ43)

R43y = 84.122⋅ mm

Write equations 3(b) for link 3, the door.

Σ Fx:

Fstop + F23x + F43x = 0

(1)

Σ © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0346.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

MACHINE DESIGN - An Integrated Approach, 4th Ed. F23y + F43y − 0.5⋅ W3 = 0 Σ Fy:

Σ Mz: 4.

5.

3-46-2 (2)

−R23x⋅ Fstop + ( R23x⋅ F23y − R23y⋅ F23x) + ( R43x⋅ F43y − R43y⋅ F43x) = 0

(3)

The directions (but not the sense) of F 23 and F43 are known so write the equations that relates the x- and y-components of these forces. F23y − F23x⋅ tan( θ2) = 0

(4)

F43y − F43x⋅ tan( θ4) = 0

(5)

There are five unknowns in the five equations above. Solving for F23x, F23y, F43x, F43y, and Fstop:

0 1 0 1   1   1 0 1 0   0  −R23y R23x −R43y R43x −R23x  A :=   mm mm mm mm mm   0 0 0   −tan( θ2) 1  0 0 −tan( θ4) 1 0  

F23x = 1.49⋅ N

F23y = 20.63⋅ N

 0   0.5⋅ W3    N   B :=  0   0     0 

F43x = −13.96⋅ N

 F23x     F23y   F43x  := A − 1⋅ B⋅ N    F43y  F   stop 

F43y = 1.87⋅ N

The pin forces at A and B are: F23 :=

2

F23x + F23y

The force on each stop is: 6.

2

F23 = 20.68⋅ N

F43 :=

2

F43x + F43y

2

F43 = 14.08⋅ N

Fstop = 12.47⋅ N

From Newton's thrid law and, since links 2 and 4 are two-force members F34x := −F43x

F34x = 13.96⋅ N

F34y := −F43y

F34y = −1.87⋅ N

F32x := −F23x

F32x = −1.49⋅ N

F32y := −F23y

F32y = −20.63⋅ N

The pin forces at O2 and O4 are numerically equal to those at A and B, respectively.



3-47-1


A particular automobile front suspension consists of two A-arms , the wheel (with tire), a coil spring, and a shock absorber (damper). The effective stiffness of the suspension (called the "ride rate") is a function of the coil spring stiffness and the tire stiffness. The A-arms are designed to give the wheel a nearly vertical displacement as the tire rides over bumps in the road. The entire assembly can be modeled as a spring-mass-damper system as shown in Figure 3-15(b). If the sprung mass (mass of the portion of the vehicle supported by the suspension system) weighs 675 lb, determine the ride rate that is required to achieve an undamped natural frequency of 1.4 Hz. What is the static deflection of the suspension for the calculated ride rate?

Units:

Hz := 2 ⋅ π⋅ rad⋅ sec

Given:

Sprung mass

Solution:

See Figure 3-15(b) and Mathcad file P0347.

−1

Ws := 675⋅ lbf

Ws

Natural frequency ωn := 1.4⋅ Hz

2

1.

Calculate the sprung mass Ms :=

2.

Using equation 3.4, calculate the required ride rate Ride rate

3.

−1

Ms = 1.748 lbf ⋅ sec ⋅ in

g

2

k := ωn ⋅ Ms

k = 135.28

lbf in

Calculate the static deflection using equation 3.5 Static deflection

δ :=

Ws k

δ = 4.99in



3-48-1


The independent suspension system of Problem 3-47 has an unsprung weight (the weight of the axle, wheel, A-arms, etc.) of 106 lb. Calculate the natural frequency (hop resonance) of the unsprung mass if the sum of the tire and coil spring stiffnesses is 1100 lb/in.

Units:


Given:

Unsprung mass

Solution:

−1

Wu := 106⋅ lbf

Stiffness

k := 1100⋅

lbf in

See Figure 3-15(b) and Mathcad file P0348. Wu

2

1.

Calculate the unsprung mass Mu :=

2.

Using equation 3.4, calculate the natural frequency Natural frequency

−1

Mu = 0.275 lbf ⋅ sec ⋅ in

g

ωn :=

k Mu

ωn = 10.1 Hz



3-49-1


The independent suspension system of Problem 3-47 has a sprung weight of 675 lb and a ride rate of 135 lb/in. Calculate the damped natural frequency of the sprung mass if the damping coefficient of the shock absorber is a constant 12 lb-sec/in.

Units:


Given:

Sprung mass

−1

Ws := 675⋅ lbf

Damping coefficient

Solution:

d := 12⋅

Ride rate

k := 135⋅

lbf in

lbf ⋅ sec in

See Figure 3-15(b) and Mathcad file P0349. Ws

2

1.

Calculate the sprung mass Ms :=

2.

Using equation 3.7, calculate the damped natural frequency

g

Ms = 1.748lbf ⋅ sec ⋅ in

Damped natural frequency

ωd :=

k Ms

−1

− 

   2⋅ Ms  d

2

ωd = 1.29Hz



3-50-1

PROBLEM 3-50_______________________________________________________ Statement:

Figure P3-22 shows a powder compaction mechanism. For the position shown, draw free-body diagrams of the input arm (2), connecting rod (3) and compacting ram (4) using variable names similar to those used in Case Studies 1A and 2A. Assume that the input arm turns slowly enough that accelerations can be ignored. Ignore the weights of the arm, connecting rod, and compacting ram. Neglect friction.

Assumptions: 1. A two-dimensional model is adequate. 2. Inertia forces may be ignored. 3. The reactions at slider bearings E and F can be modeled as concentrated forces acting horizontally at the center of each bearing. See Mathcad file P0350.

Solution: 1.

y

Isolate each of the elements to be analyzed, starting with the compacting rod, since the external force on it is known. Place the known force, Fcom , at the point P. The position vectors R14E,

F14E

R14F, and R p will be known as will the angle, q3,that the compacting ram makes with the vertical axis. 2.

3.

R14E

E

R34

D

The connecting rod is a two-force member with the forces acting at the interfaces B and D along the line joining points B and D. The assumption made in step 1 is that these are tensile forces on link 3. The input arm is acted on by forces at A, B, and C. Assume that the unknown reaction force at A is positive.

F14F

x

F34 3

F43

F

R14F RP

P Fcom

D

Compacting Ram (4)

y R43 x

C

R23 B

F23

Connecting Rod (3)

y Fin

F32 x

Rin B R32

F12y A

R12

F12x

Input Arm (2) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0350.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


3-51-1

PROBLEM 3-51______________________________________________________ Statement:

For the compaction mechanism of Problem 3-50 and the data of Table P3-5, determine the pin forces on the compacting ram, connecting rod, and input arm.

Given:

R12 := 148.4⋅ mm θ12 := −45⋅ deg

R14E := 57.0⋅ mm θ14E := 90⋅ deg

R14F := 62.9⋅ mm θ14F := 270⋅ deg

R34 := 32.00⋅ in

R23 := 87.6⋅ mm

θ23 := 254.36⋅ deg

R32 := 42.9⋅ mm

θ32 := 74.36⋅ deg

Solution: 1.

θ43 := 74.36⋅ deg

R43 := 87.6⋅ mm

θ34 := 90⋅ deg

R34 := 15.0⋅ mm

Rin := 152.6⋅ mm θin := 225⋅ deg Fcom := 100⋅ N

θ34 := −105.64⋅ deg

RP := 105.0⋅ mm θP := 270⋅ deg

θ3 := 254.36⋅ deg



y

F43

R14E

E

F14E

R34

D

D

x

F34

y

F14F

R43

R14F

F θ3

RP

P

x

Fcom

R23

Compacting Ram (4)

B C

F23 y Connecting Rod (3)

Fin

F32 x

Rin B R32 Input Arm (2)

F12y A

R12

F12x



3.

4.

3-51-2

Calculate the x- and y-components of the position vectors. R12x := R12⋅ cos( θ12)

R12x = 104.935 ⋅ mm

R12y := R12⋅ sin( θ12)

R12y = −104.935 ⋅ mm

R14Ex := R14E⋅ cos( θ14E)

R14Ex = 0 ⋅ mm

R14Ey := R14E⋅ sin( θ14E)

R14Ey = 57.000⋅ mm

R14Fx := R14F ⋅ cos( θ14F )

R14Fx = −0.000⋅ mm

R14Fy := R14F ⋅ sin( θ14F )

R14Fy = −62.900⋅ mm

R23x := R23⋅ cos( θ23)

R23x = −23.616⋅ mm

R23y := R23⋅ sin( θ23)

R23y = −84.357⋅ mm

R32x := R32⋅ cos( θ32)

R32x = 11.566⋅ mm

R32y := R32⋅ sin( θ32)

R32y = 41.312⋅ mm

R34x := R34⋅ cos( θ34)

R34x = 0.000⋅ mm

R34y := R34⋅ sin( θ34)

R34y = 15.000⋅ mm

R43x := R43⋅ cos( θ43)

R43x = 23.616⋅ mm

R43y := R43⋅ sin( θ43)

R43y = 84.357⋅ mm

RPx := RP⋅ cos( θP)

RPx = −0.000⋅ mm

RPy := RP⋅ sin( θP)

RPy = −105.000 ⋅ mm

Rinx := Rin⋅ cos( θin)

Rinx = −107.904 ⋅ mm

Riny := Rin⋅ sin( θin)

Riny = −107.904 ⋅ mm

Write equations 3(b) for link 4, the compacting ram.

Σ Fx:

F14E + F14F + F34x = 0

(1)

Σ Fy:

Fcom + F34y = 0

(2)

Σ Mz:

(−R14Ey⋅ F14E) + (−R14Fy⋅ F14F ) + (R34x⋅ F34y − R34y⋅ F34x) = 0

(3)


5.

(4)

There are four unknowns in the four equations above. Solving for F14x, F14y, F34x, and F34y, 1 1 0   1  0 0 0 1    A :=  −R14Ey −R14Fy −R34y R34x   mm mm mm mm    0 −tan( θ3) 1   0 F14E = 18.2⋅ N

6.

7.

F14F = 9.8⋅ N

 0   F   − com  B :=  N   0     0  F34x = −28.0⋅ N

 F14E  F   14F  := A − 1⋅ B⋅ N  F34x     F34y  F34y = −100.0⋅ N


F43x = 28.0⋅ N

F43y := −F34y

F43y = 100.0⋅ N

F23x := −F43x

F23x = −28.0⋅ N

F23y := −F43y

F23y = −100.0⋅ N

Write equations 3(b) for link 2, the input arm.

Σ Fx:

F12x + F32x + Finx = 0

(5)

Σ Fy:

F12y + F32y + Finy = 0

(6)

Σ Mz:

(R12x⋅ F12y − R12y⋅ F12x) + (R32x⋅ F32y − R32y⋅ F32x) + (Rinx⋅ Finy − Riny⋅ Finx) = 0

(7)



3-51-3

The direction (but not the sense) of Fin is known so write the equation that relates the x- and y-components of this force. Finy − Finx⋅ tan( θin) = 0

9.

(8)

There are four unknowns in the four equations above. Solving for F12x, F12y, Finx, and Finy, since F32x := −F23x

F32x = 28⋅ N

0 1 0   1  0 1 0 1    −Riny Rinx  A :=  −R12y R12x  mm mm mm mm    0 −tan( θin) 1   0

F12x = 36.0⋅ N F12 :=

F12y = −36.0⋅ N

 F12x2 + F12y2  

F12 = 51⋅ N

F32y = 100⋅ N

F32y := −F23y −F32x     N    −F32y    B := N    −( R32x⋅ F32y − R32y⋅ F32x)    N ⋅ mm   0   Finx = −64.0⋅ N Fin :=

 F12x  F   12y  := A − 1⋅ B⋅ N  Finx     Finy 

Finx = −64.0⋅ N

 Finx2 + Finy2  

Fin = 91⋅ N



3-52-1


Figure P3-23 shows a drag link slider crank mechanism. For the position shown, draw free-body diagrams of links 2 through 6 using variable names similar to those used in Case Studies 1A and 2A. Assume that the crank turns slowly enough that accelerations can be ignored. Ignore the weights of the links and any friction forces or torques.

Assumptions: 1. A two-dimensional model is adequate. 2. Inertia forces may be ignored. 3. Links 4 and 6 are three-force bodies. Solution: 1.


Isolate each of the elements to be analyzed, starting with the slider, link 6, since the external forces on it are known. Place the known force, FP, at the point P. This is a three-force member

y F56

θ5

so the forces are coincident at point D and there is no turning moiment on the link. The angle, θ5,that link 5 makes with the horizontal axis is known. 2.

D

FP

F16

Link 5 is a two-force member with the forces acting at the interfaces C and D along the line joining points C and D. The assumption made in step 1 is that these are compressive forces on link 5.

F45

P

x

Slider block 6

C

y

x

θ5

R45

D

R65

F65

Link 5

3.

Link 4 is a three-force body with the three forces meeting at a point. The position vectors R 14, R34, and R54 will be known as will the angles,θ 3 and θ5,that links 3 and 5, respectively, make with the horizontal axis.



3-52-2

C

y R34

F54

E

R54

x F14y

B

O4

F14x R14

F34

F43

Link 4

y B 4.

Link 3 is a two-force member with the forces acting at the interfaces A and B along the line joining points A and B.

x

R23 5.

R43

The crank is acted on by forces at A and O2, and a torque which we will assume to be positive (CCW). As in step 1, assume that the unknown reaction force at O2 is positive.

A

F23 Link 3

F y

12y

T F12x

R32 O2

x R12

A

F32 Link 2



3-53-1


For the drag link slider crank mechanism of Problem 3-52 and the data of Table P3-6, determine the pin forces on the slider, connecting rods, and crank and the reaction torque on the crank.

Given:

R12 := 63.5⋅ mm

θ12 := 45.38⋅ deg

R14 := 93.6⋅ mm

θ14 := −55.89⋅ deg

R23 := 63.5⋅ mm

θ23 := 267.8⋅ deg

R32 := 63.5⋅ mm

θ32 := 225.38⋅ deg

R34 := 103.5⋅ mm θ34 := 202.68⋅ deg

θ43 := 87.80⋅ deg

R45 := 190.5⋅ mm θ45 := 156.65⋅ deg

R54 := 103.5⋅ mm θ54 := 45.34⋅ deg

R65 := 190.5⋅ mm θ65 := −23.35⋅ deg

R43 := 63.5⋅ mm

θ5 := 156.65deg

FP := 85⋅ N Solution: 1.

θ3 := 87.80⋅ deg



y F56

θ5

Slider block 6

P

FP

x

D

F16

F45

C

y

x

θ5

R45

D

Link 5

2.

R65

F65

Calculate the x- and y-components of the position vectors. R12x := R12⋅ cos( θ12)

R12x = 44.602⋅ mm

R12y := R12⋅ sin( θ12)

R12y = 45.198⋅ mm

R14x := R14⋅ cos( θ14)

R14x = 52.489⋅ mm

R14y := R14⋅ sin( θ14)

R14y = −77.497⋅ mm

R23x := R23⋅ cos( θ23)

R23x = −2.438⋅ mm

R23y := R23⋅ sin( θ23)

R23y = −63.453⋅ mm

R32x := R32⋅ cos( θ32)

R32x = −44.602⋅ mm

R32y := R32⋅ sin( θ32)

R32y = −45.198⋅ mm

R34x := R34⋅ cos( θ34)

R34x = −95.497⋅ mm

R34y := R34⋅ sin( θ34)

R34y = −39.908⋅ mm

R43x := R43⋅ cos( θ43)

R43x = 2.438⋅ mm

R43y := R43⋅ sin( θ43)

R43y = 63.453⋅ mm



3-53-2

R45x := R45⋅ cos( θ45)

R45x = −174.898 ⋅ mm

R45y := R45⋅ sin( θ45)

R45y = 75.504⋅ mm

R54x := R54⋅ cos( θ54)

R54x = 72.75⋅ mm

R54y := R54⋅ sin( θ54)

R54y = 73.619⋅ mm

R65x := R65⋅ cos( θ65)

R65x = 174.898 ⋅ mm

R65y := R65⋅ sin( θ65)

R65y = −75.504⋅ mm

C

F43

y

y

R34 B

F54

E

R43 x

R54

x F14y

B

O4

F14x R23

A

R14

F34

Link 4

F23 Link 3

F y

12y

T F12x

R32 O2

x R12

A

F32 Link 2 3.

4.

Write equations 3(b) for link 5, the slider.

Σ Fx:

F56x − FP = 0

(1)

Σ Fy:

F16 + F56y = 0

(2)


5.

There are three unknowns in the three equations above. Solving for F56x, F56y, and F16,

(3)



 FP    N  B :=   0   0   

0 0  1 0 1 1 A :=     −tan( θ5) 1 0 

F56x = 85.0⋅ N 6.

3-53-3

F56y = −36.7⋅ N

 F56x    −1  F56y  := A ⋅ B⋅ N F   16  F16 = 36.7⋅ N


F65x = −85⋅ N

F65y := −F56y

F65y = 36.7⋅ N

F45x := −F65x

F45x = 85⋅ N

F45y := −F65y

F45y = −36.7⋅ N

F54x = −85⋅ N

F54y := −F45y

F54y = 36.7⋅ N

and, for link 4 F54x := −F45x 7.

8.

Write equations 3(b) for link 4, the rocker.

Σ Fx:

F34x + F54x + F14x = 0

(4)

Σ Fy:

F34y + F54y + F14y = 0

(5)

Σ Mz:

(R14x⋅ F14y − R14y⋅ F14x) + (R34x⋅ F34y − R34y⋅ F34x) + (R54x⋅ F54y − R54y⋅ F54x) = 0


9.

(7)

There are four unknowns in the four equations above. Solving for F34x, F34y, F14x, and F14y,

0 1 0   1  0 1 0 1    A :=  −R34y R34x −R14y R14x   mm mm mm mm    0 0   −tan( θ3) 1

F34x = 3.5⋅ N

F34y = 90.9⋅ N

−F54x     N    −F54y    B := N    −( R54x⋅ F54y − R54y⋅ F54x)    N ⋅ mm   0   F14x = 81.5⋅ N

 F34x  F   34y  := A − 1⋅ B⋅ N  F14x     F14y 

F14y = −127.6⋅ N

10. From Newton's thrid law and, since the connecting rod (3) is a two-force member F43x := −F34x

F43x = −3.5⋅ N

F43y := −F34y

F43y = −90.9⋅ N

F23x := −F43x

F23x = 3 ⋅ N

F23y := −F43y

F23y = 90.9⋅ N

F32x = −3.5⋅ N

F32y := −F23y

F32y = −90.9⋅ N

and, for link 2 F32x := −F23x



3-53-4

11. Write equations 3(b) for link 2, the crank.

Σ Fx:

F12x + F32x = 0

(8)

Σ Fy:

F12y + F32y = 0

(9)

Σ Mz:

T2 + ( R12x⋅ F12y − R12y⋅ F12x) + ( R32x⋅ F32y − R32y⋅ F32x) = 0

(10)

12. There are three unknowns in the three equations above. Solving for F12x, F12y, and T2

0 0  1   0 1 0  A :=  −R12y R12x   mm mm 1   

F12x = 3.5 N

−F32x   N     −F32y B :=   N    −( R32x⋅ F32y − R32y⋅ F32x)    N ⋅ mm  

F12y = 90.9

N

 F12x    −1  F12y  := A ⋅ B  T   2 

T2 = −7796

N*mm



4-1a-1


A differential stress element has a set of applied stresses on it as indicated in each row of Table P4-1. For row a, draw the stress element showing the applied stresses, find the principal stresses and maximum shear stress using Mohr's circle diagram, and draw the rotated stress element showing the principal stresses.

Given:

σx := 1000

σy := 0

σz := 0

τxy := 500

τyz := 0

τzx := 0

Solution:

See Figure 4-1a and Mathcad file P0401a.

500 y

1. Draw the stress element, indicating the x and y axes.

1000

x

2. Draw the Mohr's circle axes, indicating the τ and σ axes with CW up and CCW down. 3. Plot the positive x-face point, which is (+1000, -500), and label it with an "x."

FIGURE 4-1aA

4. Plot the positive y-face point, which is (0, +500), and label it with a "y."

Stress Element for Problem 4-1a

5. Draw a straight line from point x to point y. Using the point where this line intersects the σ-axis as the center of the Mohr circle, draw a circle that goes through points x and y. 6. The center of the circle will be at

σc :=

σx + σy

σc = 500

2

7. The circle will intersect the σ-axis at two of the principal stresses. In this case, we see that one is positive and the other is negative so they will be σ1 and σ3. The third principal stress is σ2 = 0. 2

8. Calculate the radius of the circle

 σx − σy  2   + τxy  2 

R :=

R = 707.1

τ CW

τ CW

τ1-3 τ 1-2 500

500

y

-500

500 σ3

1000

1500 σ

0

σ1

τ2-3

-500 σ3

500

1000

1500 σ

0

σ2

σ1

2φ

500

x

τ CCW

500

τ CCW

FIGURE 4-1aB 2D and 3D Mohr's Circle Diagrams for Problem 4-1a © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0401a.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

MACHINE DESIGN - An Integrated Approach, 4th Ed. 9. Calculate the principal stresses

4-1a-2

σ1 := σc + R

σ1 = 1207

σ3 := σc − R

σ3 = −207

σ2 := 0

10. Draw the three Mohr's circles to represent the complete 3D stress state. y

11. Calculate the principal shear stresses

τ12 := 0.5⋅ ( σ1 − σ2)

τ12 = 603.6

τ23 := 0.5⋅ ( σ2 − σ3)

τ23 = 103.6

τ13 := 0.5⋅ ( σ1 − σ3)

τ13 = 707.1

207 1207 22.5° x

The maximum principal stress is always τ13. 12. Determine the orientation of the principal normal stress (σ1) with respect to the x-axis. From the 2D Mohr's circle diagram, we see that the angle 2φ from x to σ1 is CCW and is given by

 σx − σc  ϕ := ⋅ acos  2  R  1

ϕ = 22.5 deg

FIGURE 4-1aC Rotated Stress Element for Problem 4-1a

13. Draw the rotated 2D stress element showing the two nonzero principal stresses.



4-1h-1

PROBLEM 4-1h Statement:

A differential stress element has a set of applied stresses on it as indicated in each row of Table P4-1. For row h, draw the stress element showing the applied stresses, find the principal stresses and maximum shear stress and draw the Mohr's circle diagram.

Given:

σx := 750

σy := 500

σz := 250

τxy := 500

τyz := 0

τzx := 0

Solution:

See Figures 4-1h and Mathcad file P0401h.

z 250

1. Calculate the coefficients (stress invariants) of equation (4.4c). 3

C2 := σx + σy + σz

C2 = 1.500 × 10

 σx τxy   σx τzx   σy τyz    +   +    τxy σy   τzx σz   τyz σz 

C1 :=

750

5

C1 = 4.375 × 10

500

500

x

 σx τxy τzx    C0 :=  τxy σy τyz  τ τ σ   zx yz z 

7

y

FIGURE 4-1hA

C0 = 3.125 × 10

Stress Element for Problem 4-1h 3

2. Find the roots of the triaxial stress equation:

 −C0    C1   v :=  −C2     1 

500

r := polyroots ( v)

2

σ − C2⋅ σ + C 1⋅ σ − C0 = 0

 110   r = 250     1140  τ CW

3. Extract the principal stresses from the vector r by inspection.

σ1 := r

σ1 = 1140

3

σ2 := r

σ2 = 250

σ3 := r

σ3 = 110

2 1

τ12 := τ23 :=

τ2-3

-500

4. Using equations (4.5), evaluate the principal shear stresses.

τ13 :=

τ1-3 τ 1-2

500

σ1 − σ3 2

σ1 − σ2 2

σ2 − σ3 2

τ13 = 515

500

1000

1500

σ

0

σ3

σ2

σ1

500

τ12 = 445 τ23 = 70

τ CCW

FIGURE 4-1hB 5. Draw the three-circle Mohr diagram.

The Three Mohr's Circles for Problem 4-1h



4-2-1


A 400-lb chandelier is to be hung from two 10-ft-long solid steel cables in tension. Choose a suitable diameter for the cable such that the stress will not exceed 5000 psi. What will be the deflection of the cables? State all assumptions.

Given:

Weight of chandelier Length of cable Allowable stress

W := 400⋅ lbf L := 10⋅ ft L = 120 in σallow := 5000⋅ psi

Number of cables

N := 2

Young's modulus

E := 30⋅ 10 ⋅ psi

6

Assumptions: The cables share the load equally. Solution:

See Mathcad file P0402. W

1.

Determine the load on each cable

2.

The stress in each cable will be equal to the load on the cable divided by its cross-sectional area. Using equation (4.7), and setting the stress equal to the allowable stress, we have

P :=

P = 200 lbf

N

4⋅ P

σallow =

π⋅ d 3.

2

Solve this equation for the unknown cable diameter. d :=

4⋅ P

d = 0.226 in

π⋅ σallow

4.

Round this up to the next higher decimal equivalent of a common fractional size:

5.

Using equation (4.8), determine the deflection in each cable. Cross-section area

Cable deflection

A :=

π⋅ d

∆s :=

2

4 P⋅ L A ⋅E

d := 0.250⋅ in

2

A = 0.049 in

∆s = 0.016 in



4-3-1


For the bicycle pedal-arm assembly in Figure 4-1 with rider-applied force of 1500 N at the pedal, determine the maximum principal stress in the pedal arm if its cross-section is 15 mm in dia. The pedal attaches to the pedal arm with a 12-mm screw thread. What is the stress in the pedal screw?

Given:

Distances (see figure) Rider-applied force

a  170  mm Frider  1.5 kN

Pedal arm diameter

d pa  15 mm

Screw thread diameter

d sc  12 mm

b  60 mm

z

Solution:

See Figure 4-3 and Mathcad file P0403. a

1. From the FBD in Figure 4-3A (and on the solution for Problem 3-3), we see that the force from the rider is reacted in the pedal arm internally by a moment, a torque, and a vertical shear force. There are two points at section C (Figure 4-3B) that we should investigate, one at z = 0.5 d pa (point A), and one at y = 0.5 d pa (point B).

Frider a  Mc = 0

 M x:

Frider b  Tc = 0

Mc

b

Arm y

Fc Pedal x

2. Refering to the FBD resulting from taking a section through the arm at C, the maximum bending moment Mc is found by summing moments about the y-axis, and the maximum torque Tc is found by summing moments about the x-axis.  M y:

Tc

C

Frider


z

Section C

A Maximum bending moment: Mc  Frider a

Mc  255  N  m

Maximum torque: Tc  Frider b

Fc  Frider

3.

x

Tc  90 N  m

Vertical shear:

B

Arm

y FIGURE 4-3B

Fc  1.500  kN

Points A and B at Section C

Determine the stress components at point A where we have the effects of bending and torsion, but where the transverse shear due to bending is zero because A is at the outer fiber. Looking down the z-axis at a stress element on the surface at A, Distance to neutral axis

cpa  0.5 d pa

cpa  7.5 mm


MACHINE DESIGN - An Integrated Approach, 4th Ed. 4

π d pa

Moment of inertia of pedal-arm

Ipa 

Bending stress (x-direction)

σx 

Stress in y-direction

σy  0  MPa

Torsional stress due to Tc

τxy 

Principal stresses at A, equation (4.6a)

Mc cpa

Tc cpa

τxy  135.8  MPa

2  Ipa

σx  σy 2

σx  σy 2

CW

2  σx  σy  2     τxy 2   2  σx  σy  2     τxy  2 

σ2A  0  MPa

σ3A  23 MPa

Determine the stress components at point B where we have the effects of transverse shear and torsion, but where the bending stress is zero because B is on the neutral plane. Looking down the y-axis at a stress element at B, 2

π d pa

Cross-section area of pedal-arm

Apa 

Torsional stress due to Tc and shear stress due to Fc

τzx 

Normal stresses

σx  0  MPa

Principal stresses at B

σ1B  124  MPa

Apa  176.7  mm

4 4 Fc   τxy 3 Apa

σ1B 

σ3B 

5.

4

σx  769.6  MPa

Ipa

σ1A 

σ1A  793  MPa

3

Ipa  2.485  10  mm

64

σ3A 

4.

4-3-2

τzx  124.5  MPa

CW

σz  0  MPa σx  σz 2

σx  σz 2

2  σx  σz  2     τzx 2   2  σx  σz  2     τzx  2 

σ2B  0  MPa

The maximum principal stress is at point A and is

2

σ3B  124  MPa

σ1A  793  MPa



4-3-3

Determine the stress in the pedal screw. Bending moment

Msc  Frider b

Msc  90 N  m

Distance to neutral axis

csc  0.5 d sc

csc  6  mm

Moment of inertia of pedal screw

Isc 

Bending stress (y-direction)

σy 

Stress in z-direction

σz  0  MPa

Torsional stress

τxy  0  MPa

4

π d sc

3

Isc  1.018  10  mm

64 Msc csc Isc

4

σy  530.5  MPa

Since there is no shear stress present at the top of the screw where the bending stress is a maximum, the maximum principal stress in the pedal screw is

σ1  σy

σ1  530.5  MPa



4-4-1


The trailer hitch shown in Figure P4-2 and Figure 1-1 (p. 12) has loads applied as defined in Problem 3-4. The tongue weight of 100 kg acts downward and the pull force of 4905 N acts horizontally. Using the dimensions of the ball bracket shown in Figure 1-5 (p. 15), determine: (a) The principal stresses in the shank of the ball where it joins the ball bracket. (b) The bearing stress in the ball bracket hole. (c) The tearout stress in the ball bracket. (d) The normal and shear stresses in the 19-mm diameter attachment holes. (e) The principal stresses in the ball bracket as a cantilever.

Given:

a  40 mm b  31 mm Mtongue  100  kg Fpull  4.905  kN

c  70 mm d sh  26 mm

d  20 mm t  19 mm

Assumptions: 1. The nuts are just snug-tight (no pre-load), which is the worst case. 2. All reactions will be concentrated loads rather than distributed loads or pressures. Solution:

See Figure 4-4 and Mathcad file P0404. Wtongue  Mtongue g

1. The weight on the tongue is


2. Solving first for the reactions on the ball by summing the horizontal and vertical forces and the moments about A. W tongue 70 = c

1

F pull

1

40 = a 2

A B

A

19 = t B

F b1

31 = b

F a1x

C

F a1y 20 = d

F a2y

D

Fa2x 2

Fc2x

F b2 C D

Fd2 F c2y


 Fx :

Fpull  Fa1x  Fb1 = 0

(1)

 Fy :

Fa1y  Wtongue = 0

(2)


MACHINE DESIGN - An Integrated Approach, 4th Ed.  MA:

4-4-2

Fb1 t  Fpull  a = 0

(3) Fpull  a

3. Solving equation (3) for Fb1

Fb1 

Fb1  10.326 kN

4. Substituting into (1) and solving for Fa1x

Fa1x  Fpull  Fb1

Fa1x  15.231 kN

5. Solving (2) for Fa1y

Fa1y  Wtongue

Fa1y  0.981  kN

t

6. Now, refering to the FBD of the bracket, we can apply the equations of equilibrium to determine the reactions at C and D on the bracket.  Fx :

Fa2x  Fb2  Fc2x  Fd2 = 0

(4)

 Fy :

Fc2y  Fa2y = 0

(5)

 MC:

Fd2 d  Fb2 b  Fa2x ( b  t)  Fa2y c = 0

(6)

7. Note also that the interface forces between part 1 (ball) and part 2 (bracket) have been drawn on their respective FBDs in opposite senses. Therefore, Fa2x  Fa1x

Fa2y  Fa1y

8. Solving equation (6) for Fd2

Fd2 

Fb2  Fb1

Fa2x ( b  t)  Fa2y c  Fb2 b

Fd2  25.505 kN

d

9. Substituting into (4) and solving for Fc2x

Fc2x  Fa2x  Fb2  Fd2

Fc2x  30.41  kN


Fc2y  Fa2y

Fc2y  0.981  kN

11. Determine the principal stresses in the shank of the ball where it joins the ball bracket. The internal bending moment at A on the FBD of the ball is


Moment of inertia of shank


M  Fpull  a

M  196.2  N  m

csh  0.5 d sh

csh  13 mm

Ish 

σx 

π  d sh 64 M  csh Ish


σy  0  MPa

Shear stress at A


4 4

Ish  2.243  10  mm

4

σx  113.7  MPa

Since the shear stress is zero, x is the maximum principal stress, thus

σ1  σx

σ1  114  MPa

σ2  0  MPa

σ3  0  MPa



4-4-3

12. Determine the bearing stress in the ball bracket hole. Bearing area

Abearing  d sh t

Bearing stress

σbearing 

Abearing  494  mm

Fpull

2

σbearing  9.93 MPa

Abearing

13. Determine the tearout stress in the ball bracket.

Tearout length

Shear area (see Figure 4-4B) 2

Atear = 2  t R  ( 0.5 d )

2

Atear  2  t ( 32 mm)   0.5 d sh 2

Atear  1111 mm

2

2

Stress

R

d

τtear 

Fpull Atear

FIGURE 4-4B Tearout Diagram for Problem 4-4

τtear  4.41 MPa d bolt  19 mm

14. Determine the normal and shear stresses in the attachment bolts if they are 19-mm dia. Bolt cross-section area (2 bolts)

Abolt  2 

Normal stress (tension)

σbolt 

π d bolt

Fc2x Abolt

4

2

Abolt  567.1  mm

2

σbolt  53.6 MPa


MACHINE DESIGN - An Integrated Approach, 4th Ed. Shear stress

τbolt 

4-4-4 W tongue

Fc2y

τbolt  1.7 MPa

Abolt

1

F pull

15. Determine the principal stresses in the ball bracket as a cantilever (see Figure 4-4C).

a

Bending moment

2

M  Fpull  a  Wtongue c Width of bracket Moment of inertia Total tensile stress

M  264.8  N  m M

w  64 mm I 

w t

R

3

I  36581  mm

12

σ 

c

M t 2 I



Fpull w t

4

FIGURE 4-4C Cantilever FBD for Problem 4-4

σ  72.8 MPa

Since there are no shear stress at the top and bottom of the bracket where the bending stresses are maximum, they are also the principal stresses, thus

σ1  σ τmax 

σ1  72.8 MPa σ 2

σ2  0  MPa

σ3  0  MPa

τmax  36.4 MPa



4-5-1


Repeat Problem 4-4 for the loading conditions of Problem 3-5, i.e., determine the stresses due to a horizontal force that will result on the ball from accelerating a 2000-kg trailer to 60 m/sec in 20 se Assume a constant acceleration. From Problem 3-5, the pull force is 6000 N. Determine: (a) The principal stresses in the shank of the ball where it joins the ball bracket. (b) The bearing stress in the ball bracket hole. (c) The tearout stress in the ball bracket. (d) The normal and shear stresses in the 19-mm diameter attachment holes. (e) The principal stresses in the ball bracket as a cantilever.

Given:

a  40 mm b  31 mm Mtongue  100  kg Fpull  6  kN

c  70 mm d sh  26 mm

d  20 mm t  19 mm




Wtongue  Mtongue g



1

F pull

1

40 = a 2

A B

A

19 = t B

F b1

31 = b

F a1x

C

F a1y 20 = d

F a2y

D

Fa2x 2

Fc2x

F b2 C D

Fd2 F c2y


 Fx :


(1)

 Fy :


(2)


MACHINE DESIGN - An Integrated Approach, 4th Ed.  MA:

4-5-2


(3) Fpull  a


Fb1 

Fb1  12.632 kN



Fa1x  18.632 kN


Fa1y  Wtongue

Fa1y  0.981  kN

t


Fa2x  Fb2  Fc2x  Fd2 = 0

(4)

 Fy :

Fc2y  Fa2y = 0

(5)

 MC:


(6)


Fa2y  Fa1y


Fd2 

Fb2  Fb1

Fa2x ( b  t)  Fa2y c  Fb2 b

Fd2  30.432 kN

d



Fc2x  36.432 kN


Fc2y  Fa2y

Fc2y  0.981  kN






M  240  N  m

csh  0.5 d sh

csh  13 mm

Ish 

σx 

π  d sh

4

64 M  csh Ish


σy  0  MPa

Shear stress at A


4

Ish  2.243  10  mm

4

σx  139.1  MPa


σ1  σx

σ1  139  MPa

12. Determine the bearing stress in the ball bracket hole.

σ2  0  MPa

σ3  0  MPa


MACHINE DESIGN - An Integrated Approach, 4th Ed. Bearing area


Bearing stress

σbearing 

4-5-3 Abearing  494  mm

Fpull

2

σbearing  12.15  MPa

Abearing


Tearout length


Atear = 2  t R  ( 0.5 d )

2

Atear  2  t ( 32 mm)   0.5 d sh 2


2

2

Stress

R

d

τtear 

Fpull Atear

FIGURE 4-5B

τtear  5.4 MPa

Tearout Diagram for Problem 4-5

d bolt  19 mm


Abolt  2 

σbolt 


π d bolt

2

Abolt  567.1  mm

4

Fc2x

σbolt  64.2 MPa

Abolt

Shear stress

τbolt 

2

W tongue

Fc2y


Abolt

1

F pull


a

Bending moment

2

M  Fpull  a  Wtongue c Width of bracket Moment of inertia

M  308.6  N  m M

w  64 mm I 

w t

c R

3

I  36581  mm

12

4


Total tensile stress

σ 

M t 2 I



Fpull w t

σ  85.1 MPa



4-5-4



σ1  85.1 MPa σ 2

σ2  0  MPa

σ3  0  MPa




4-6-1


Repeat Problem 4-4 for the loading conditions of Problem 3-6, i.e., determine the stresses due to a horizontal force that will results from an impact between the ball and the tongue of the 2000-kg trailer if the hitch deflects 2.8 mm dynamically on impact. The tractor weighs 1000 kg and the velocity at impact is 0.3 m/sec. Determine: (a) The principal stresses in the shank of the ball where it joins the ball bracket. (b) The bearing stress in the ball bracket hole. (c) The tearout stress in the ball bracket. (d) The normal and shear stresses in the 19-mm diameter attachment holes. (e) The principal stresses in the ball bracket as a cantilever.

Given:

a  40 mm b  31 mm Mtongue  100  kg Fpull  55.1 kN

c  70 mm d sh  26 mm

d  20 mm t  19 mm




Wtongue  Mtongue g



1

F pull

1

40 = a 2

A B

A

19 = t B

F b1

31 = b

F a1x

C

F a1y 20 = d

F a2y

D

Fa2x 2

Fc2x

F b2 C D

Fd2 F c2y


 Fx :


(1)



4-6-2

 Fy :


(2)

 MA:


(3) Fpull  a


Fb1 

Fb1  116  kN



Fa1x  171.1  kN


Fa1y  Wtongue

Fa1y  0.981  kN

t


Fa2x  Fb2  Fc2x  Fd2 = 0

(4)

 Fy :

Fc2y  Fa2y = 0

(5)

 MC:


(6)


Fa2y  Fa1y


Fd2 

Fb2  Fb1

Fa2x ( b  t)  Fa2y c  Fb2 b

Fd2  251.382  kN

d



Fc2x  306.482  kN


Fc2y  Fa2y

Fc2y  0.981  kN





3


M  2.204  10  N  m

csh  0.5 d sh

csh  13 mm

Ish 

σx 

π  d sh 64 M  csh Ish


σy  0  MPa

Shear stress at A


4 4

Ish  2.243  10  mm

4

σx  1277 MPa


σ1  σx

σ1  1277 MPa

σ2  0  MPa

σ3  0  MPa



4-6-3

12. Determine the bearing stress in the ball bracket hole. Bearing area


Bearing stress

σbearing 

Abearing  494  mm

Fpull

2

σbearing  111.54 MPa

Abearing


Tearout length


Atear = 2  t R  ( 0.5 d )

2

Atear  2  t ( 32 mm)   0.5 d sh 2


2

2

Stress

R

d

τtear 

Fpull Atear

FIGURE 4-6B Tearout Diagram for Problem 4-6

τtear  49.59  MPa d bolt  19 mm


Abolt  2 

σbolt 


π d bolt

2

Abolt  567.1  mm

4

Fc2x

σbolt  540  MPa

Abolt

Shear stress

τbolt 

2

W tongue

Fc2y


Abolt

1

F pull


a

Bending moment

Width of bracket Moment of inertia

M  2.3  10  N  m M

w  64 mm I 

2

3

M  Fpull  a  Wtongue c

w t

c R

3

I  36581  mm

12

4


Total tensile stress

σ 

M t 2 I



Fpull w t

σ  635.5  MPa



4-6-4



σ1  636  MPa σ 2

σ2  0  MPa

σ3  0  MPa

τmax  318  MPa



4-7-1


Design the wrist pin of Problem 3-7 for a maximum allowable principal stress of 20 ksi if the pin is hollow and loaded in double shear.

Given:

Force on wrist pin

Fwristpin  12.258 kN

Allowable stress

σallow  20 ksi od  0.375  in

Assumptions: Choose a suitable outside diameter, say Solution:

Fwristpin  2756 lbf

See Figure 4-12 in the text and Mathcad file P0407.

1. The force at each shear plane is

F 

Fwristpin

F  1378 lbf

2

2. With only the direct shear acting on the plane, the Mohr diagram will be a circle with center at the origin and radius equal to the shear stress. Thus, the principal normal stress is numerically equal to the shear stress, which in this case is also the principal shear stress, so we have  = 1 = allow.

3. The shear stress at each shear plane is

4. Solving for the inside diameter,

τ=

id 

F A

2

=

od 

4 F

2 2 π  od  id 

4 F

π σallow

= σallow

id  0.230  in



4-8-1


A paper mill processes rolls of paper having a density of 984 kg/m3. The paper roll is 1.50 m outside dia (OD) by 0.22 m inside dia (ID) by 3.23 m long and is on a simply supported, hollow, steel shaft. Find the shaft ID needed to obtain a maximum deflection at the center of 3 mm if the shaft OD is 22 cm.

Given:

Paper density

ρ  984 

kg 3

m Roll dimensions Outside diameter Inside diameter Lemgth

od  220  mm E  207  GPa δ  3  mm

Shaft outside dia Young's modulus Allowable deflection

OD  1.50 m ID  0.22 m L  3.23 m

Assumptions: The shaft (beam) supporting the paper roll is simply-supported at the ends and is the same length as the paper roll. The paper acts as a distributed load over the length of the shaft. Solution: 1.


The weight of the paper roll is equal to its volume times the paper density times g. Wroll 

 4

π

2



2

 OD  ID  L ρ  g

Wroll  53.89  kN Wroll

The intensity of the distributed load is w 

3.

Using Figure B-2(b) in Appendix B with a = 0, the maximum deflection is at the midspan and is y =

w x 24 E I



2

3

w  16.686

N

2.

L



3

 2  L x  x  L

mm

4

For x = L/2, this reduces to

y =

5  w L

384  E I 4

Letting  = -y and solving for I, we have

4.

I 

5  w L

7

I  3.808  10  mm

384  E δ

The area moment of inertia for a hollow circular cross-section is I =

π 64



4

4



4

 od  id

1

Solving this for the id yields

id   od  4



Round this down (for slightly less deflection) to

64 I 

4



π 

id  198.954  mm

id  198  mm



4-9-1


A ViseGrip plier-wrench is drawn to scale in Figure P4-3, and for which the forces were analyzed in Problem 3-9, find the stresses in each pin for an assumed clamping force of P = 4000 N in the position shown. The pins are 8-mm dia and are all in double shear.

Given:

Pin forces as calculated in Problem 3-9: Member 1 F21  7.5 kN

F41  5.1 kN

Member 2

F12  7.5 kN

F32  5.1 kN

Member 3

F23  5.1 kN

F43  5.1 kN

Member 4

F14  5.1 kN

F34  5.1 kN

d  8  mm

Pin diameter

Assumptions: Links 3 and 4 are in a toggle position, i.e., the pin that joins links 3 and 4 is in line with the pins that join 1 with 4 and 2 with 3. Solution: 1.


The FBDs of the assembly and each individual link are shown in Figure 4-9. The dimensions, as scaled from Figure P4-3 in the text, are shown on the link FBDs. 4

F

P 1

2

3

P

F 55.0 = b

50.0 = a

39.5 = c

F

F14 22.0 = d

129.2°

1



4 F34

F41

F21

P



28.0 = e  

F43



F12

3 F23

P

F32

2.8 = g

21.2 = h

2

F 26.9 = f


2.

The cross-sectional area for all pins is the same and is

A 

π d 4

2

A  50.265 mm

2



4-9-2

The pin that joins members 1 and 2 is the most highly stressed while the stress on each of the remaining pins is the same. Since the pins are in double shear, we will divide the pin load by 2 in each case.

Pin joining 1 and 2

τ12 

All other pins

τ14 

F12 2 A F14 2 A

τ12  74.6 MPa

τ14  50.7 MPa



4-10-1


Given:

The over-hung diving board of problem 3-10 is shown in Figure P4-4a. Assume cross-section dimensions of 305 mm x 32 mm. The material has E = 10.3 GPa. Find the largest principal stress at any location in the board when a 100-kg person is standing at the free end. W  100  kgf

Weight of person

2000 = L

Board dimensions

R1

Distance to support

a  0.7 m

Length of board

L  2  m

Cross-section

w  305  mm

P

R2

t  32 mm

700 = a

FIGURE 4-10 Assumptions: The weight of the beam is negligible compared to the applied load and so can be ignored. Solution: 1.


From the FBD of the diving board and Figure B-3 (Appendix B), the reactions at the supports are R1  W   1 



R2  W  

L



R1  1821 N

a

L

 a

2.

R2  2802 N

Also from Figure D-3, the maximum bending moment occurs at the right-hand support where, in the FBD above, x = a. Mmax  R1 a

3.


Mmax  1275 N  m

The maximum bending stress will occur on the top and bottom surfaces of the board at the section where the maximum bending moment occurs which, in this case, is at x = a. The only stress present on the top or bottom surface of the board is the bending stress x. Therefore, on the top surface where the stress is tensile, x. is the principal stress 1 . Thus, Distance to extreme fiber

Moment of inertia

Bending stress

Maximum principal stress

c  I 

σx 

t

c  16 mm

2 w t

3

12 Mmax c

σ1  σx

5

I  8.329  10  mm

I

4

σx  24.492 MPa σ1  24.5 MPa



4-11-1


Given:

Repeat Problem 4-10 using the loading conditions of Problem 3-11. Assume the board weighs 29 kg and deflects 13.1 cm statically when the person stands on it. Find the largest principal stress at any location in the board when the 100-kg person in Problem 4-10 jumps up 25 cm and lands back on the board. Find the maximum deflection. Beam length

L  2000 mm

Distance to support

a  700  mm

Mass of person

mpers  100  kg

Mass of board


Static deflection

δst  131  mm

Height of jump

h  250  mm

Cross-section

w  305  mm

2000 = L R1

Fi

R2 700 = a

FIGURE 4-11

t  32 mm


Assumptions: The apparent Young's modulus for fiberglas is 4

E  1.03 10  MPa Solution:


1. From Problem 3-11, the dynamic load resulting from the impact of the person with the board isFi  3.056  kN 2. From the FBD of the diving board and Figure B-3(a) (Appendix B), the reactions at the supports are R1  Fi  1 



R2  Fi 

L



R1  5.675  kN

a

L

 a

R2  8.731  kN

3. Also from Figure D-3(a), the maximum bending moment occurs at the right-hand support where, in the FBD above, x = a. Mmax  R1 a

Mmax  3.973  kN  m

4. The maximum bending stress will occur on the top and bottom surfaces of the board at the section where the maximum bending moment occurs which, in this case, is at x = a. The only stress present on the top or bottom surface of the board is the bending stress x. Therefore, on the top surface where the stress is tensile, x is the principal stress 1 . Thus, c 

Distance to extreme fiber

I 

Moment of inertia

σx 

Bending stress

t

c  16 mm

2 w t

3

12 Mmax c I

σ1  σx


5

I  8.329  10  mm

4

σx  76.322 MPa σ1  76.3 MPa

5. Calculate the maximum deflection from the equation given in Figure D-3(a) at x = L. Let b in the figure be a in our problem and let a in the figure be equal to L, then ymax 

Fi 6  a  E I

 ( a  L)  L  L ( L  a )  a  ( L  a )  L 3

3

2

ymax  401.4  mm



4-12-1


Repeat Problem 4-10 using the cantilevered diving board design in Figure P4-4b.

Given:

Beam length

L  1300 mm

Weight at free end

P  100  kgf

Cross-section

w  305  mm

2000 1300 = L P

t  32 mm Assumptions:

The apparent Young's modulus for fiberglas is

M1

R1

4


700

FIGURE 4-12



1.

2.

From the FBD of the diving board and Figure B-1(a) (Appendix B), the reactions at the supports are R1  P

R1  981  N

M1  P L

M1  1275 N  m

Also from Figure D-1, the maximum bending moment occurs at the support where, in the FBD above, x = 0. Mmax  M1

Mmax  1275 N  m

3. The maximum bending stress will occur on the top and bottom surfaces of the board at the section where the maximum bending moment occurs which, in this case, is at x = 0. The only stress present on the top or bottom surface of the board is the bending stress x. Therefore, on the top surface where the stress is tensile, x is the principal stress 1 . Thus, Distance to extreme fiber

Moment of inertia

Bending stress


c  I 

σx 

t

c  16 mm

2 w t

3

12 Mmax c

σ1  σx

5

I  8.329  10  mm

I

4

σx  24.492 MPa σ1  24.5 MPa



4-13-1


Given:

Repeat Problem 4-11 using the diving board design shown in Figure P4-4b. Assume the board weighs 19 kg and deflects 8.5 cm statically when the person stands on it. Unsupported length

L  1300 mm

Mass of board


Static board deflection

δstat  85 mm

Mass of person

mperson  100  kg

Height of jump

h  250  mm

Cross-section

w  305  mm

2000 1300 = L Fi

M1

t  32 mm

R1

700

Assumptions: The apparent Young's modulus for fiberglas is FIGURE 4-13

4




1.

From Problem 3-13, the dynamic load resulting from the impact of the person with the board isFi  3.487  kN

2.

From the FBD of the diving board and Figure B-1(a) (Appendix B), the reactions at the supports are

3.

R1  Fi

R1  3487 N

M1  Fi L

M1  4533 N  m

Also from Figure D-1, the maximum bending moment occurs at the support where, in the FBD above, x = 0. Mmax  M1 Mmax  4533 N  m

4. The maximum bending stress will occur on the top and bottom surfaces of the board at the section where the maximum bending moment occurs which, in this case, is at x = 0. The only stress present on the top or bottom surface of the board is the bending stress x. Therefore, on the top surface where the stress is tensile, x is the principal stress 1 . Thus, Distance to extreme fiber

Moment of inertia

Bending stress

Maximum principal stress 5.

c  I 

σx 

t

c  16 mm

2 w t

3

5

I  8.329  10  mm

12 Mmax c

4

σx  87.086 MPa

I

σ1  σx

σ1  87.1 MPa

Calculate the maximum deflection from the equation given in Figure D-3(a) at x = L. Let b in the figure be a in our problem and let a in the figure be equal to L, then 3

ymax 

Fi L

3  E I

ymax  297.7  mm



4-14-1


Figure P4-5 shows a child's toy called a pogo stick. The child stands on the pads, applying half her weight on each side. She jumps off the ground, holding the pads up against her feet, and bounces along with the spring cushioning the impact and storing energy to help each rebound. Design the aluminum cantilever beam sections on which she stands to survive jumping 2 in off the ground. Assume an allowable stress of 20 ksi. Define and size the beam shape.

Given:

Allowable stress

σallow  20 ksi

Young's modulus

E  10.3 10  psi

6

Assumptions: The beam will have a rectangular cross-section with the load applied at a distance of 5 in from the central support. L  5  in Solution:


1. From Problem 3-14, the total dynamic force on both foot supports is Fi  224  lbf

Fi /2

Therefore, the load on each support is P 

Fi

Fi /2

P  112  lbf

2

2. To give adequate support to the childs foot, let the width of the support beam be w  1.5 in 3. From Figure B-1(a) in Appendix B, the maximum bending moment at x = 0 is M  P L 4.

We can now calculate the minimum required section modulus, Z = I/c.

Bending stress Solving for Z,

5.

P

M  560  in lbf

σ= Z 

For a rectangular cross-section, I =

Solving for t,

M

M

t 

Z  458.8  mm

σallow 3

12


= σallow

Z

w t

FIGURE 4-14

and c =

6 Z w

t 2

so Z =

w t

3

2

6

t  0.335  in

Round this up to the next higher decimal equivalent of a common fraction,

t  0.375  in



4-15-1


Design a shear pin for the propeller shaft of an outboard motor if the shaft through which the pin is placed is 25-mm diameter, the propeller is 20-cm diameter, and the pin must fail when a force > 400 N is applied to the propeller tip. Assume an ultimate shear strength for the pin material of 100 MPa.

Given:

Propeller shaft dia Propeller dia Max propeller tip force

d  25 mm D  200  mm Fmax  400  N

Ultimate shear strength

S us  100  MPa

Fpin

T

Propeller Hub

Shear Pin

Assumptions: A shear pin is in direct, double shear. Solution:

Fpin


d

FIGURE 4-15

1. Calculate the torque on the propeller shaft that will result from a tip force on the propeller of Fmax. T  Fmax 2.

2


T  40000  N  mm

This will be reacted by the shear pin's couple on the shaft. Determine the magnitude of the direct shear force. Fpin 

3.

D

Propeller Shaft

T d

Fpin  1600 N

Determine the maximum pin diameter that will shear at this force.

Direct shear stress

τ=

Fpin A

=

4  Fpin

π d pin 4  Fpin

Solving for the pin diameter

d pin 

Round this to

d pin  4.5 mm

π S us

2

= S us

d pin  4.514  mm



4-16-1


Given:

A track to guide bowling balls is designed with two round rods as shown in Figure P4-6. The rods are not parallel to one another but have a small angle between them. The balls roll on the rods until they fall between them and drop onto another track. The angle between the rods is varied to cause the ball to drop at different locations. Find the maximum stress and deflection in the rods assuming that they are (a) simply supported at each end, and (b) fixed at each end. Rod length

L  30 in

Rod diameter

d  1.00 in

Distance to load

a  23.15  in

Young's modulus

E  30 10  psi

Fball

a

6

R1

Assumptions: The analysis of Problem 3-16 yielded the following for a simply supported beam:

R2

L

FIGURE 4-16A Free Body Diagram for Problem 4-16(a), taken on a plane through the rod axis and ball center

Max ball load

Fball  13.89  lbf

Max moment

Mmax  73.4 in lbf

Reactions

R1  3.17 lbf R2  10.72  lbf

Solution:


1. The maximum bending stress will occur at the outer fibers of the rod at the section where the maximum bending moment occurs which, in this case, is at x = a. The only stress present on the top or bottom surface of the rod is the bending stress x. Therefore, on the bottom surface where the stress is tensile, x is the principal stress 1 . Thus, for a simply supported rod, c 


I 

Moment of inertia

σx 

Bending stress

c  0.5 in

2

π d

4

4

I  0.0491 in

64 Mmax c

σx  748  psi

I

σ1  σx


d

σ1  748  psi

Calculate the maximum deflection for the simply supported case from the equation given in Figure D-2(a), ymax 

Fball 6  E I



  2 a 



3

a

4

L

 L a

2

 

3. For the case where the rod is built in at each end, the beam is statically indeterminate. As seen in Figure 4-16B, there are four unknown reactions and only two equilibrium equations can be written using statics. We will find the reactions using Example 4-7 as a model.

ymax  0.0013 in Fball

a

M1

R1

L

R 2 M2

FIGURE 4-16B Free Body Diagram for Problem 4-16(b), taken on a plane through the rod axis and ball center



4-16-2

4. Write an equation for the load function in terms of equations 3.17 and integrate the resulting function four times using equations 3.18 to obtain the shear and moment functions. Note use of the unit doublet function to represent the moment at the wall. For the beam in Figure 4-16B, q(x) = -M1-2 + R1-1 - F-1 + R2-1 + M2-2 V(x) = -M1-1 + R10 - F0

+ R2
- L>0 + M2-1 + C1

M(x) = -M10 + R11 - F1 + R21 + M20 + C1x+ C2 (x) = ( -M11 + R12/2 - F2/2

+ R2
- L>2/2 + M21 + C1x2/2 + C2x + C3) / EI

y(x) = ( -M12/2 + R13/6 - F3/6 + R23/6 + M22 /2+ C1x3/6 + C2x2/2 + C3x + C4) / EI 5. Because the reactions have been included in the loading function, the shear and moment diagrams both close to zero at each end of the beam, making C1 = C2 = 0. This leaves six unknowns; the four reactions and the constants of integration, C3 and C4. There are four boundary conditions that we can use and two equilibrium equations. The boundary conditions are: at x = 0,  = 0 and y = 0; and at x = L,  = 0 and y = 0. Applying the boundary conditions at x = 0 results in C3 = C4 = 0. Applying the BCs at x = L results in the following two equations, which are solved for R1 and M1. At x = L,

θ=0

0=

y =0

0=

R1 2 R1 6

2

 L  M 1 L  3

L 

M1 2

2

F 2

L 

 ( L  a) F 6

2

 ( L  a)

3

Solving these two equations simultaneously for R1 and M1, M1 

Fball

R1  2 

6.

L M1 L



( L  a)

 ( L  a )  2



 Fball 

L

( L  a)

3

 

M1  16.765 in lbf

2

2

R1  1.842  lbf

L

The remaing two reactions can be found by using the equations of equilibrium.

 Fy = 0:

R1  Fball  R2 = 0

 M = 0:

M1  Fball  a  R2 L  M2 = 0

Solving these two equations simultaneously for R2 and M2,

7.

R2  Fball  R1

R2  12.048 lbf

M2  M1  Fball  a  R2 L

M2  56.657 in lbf

Define the range for x, x  0  in 0.005  L  L

8. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 9. Write the shear, moment, slope, and deflection equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


4-16-3

V ( x)  R1 S ( x 0  mm)  Fball  S ( x a )  R2 S ( x L) M ( x)  M1 S ( x 0  mm)  R1 S ( x 0  mm)  ( x  0  mm)  Fball  S ( x a )  ( x  a )   M2 S ( x L)  R2 S ( x L)  ( x  L)

θ ( x) 

y ( x) 



1 E I

1 E I

 M1 S ( x 0  mm)  x 

R1 2

2

 S ( x 0  mm)  ( x  0  mm) 

Fball 2

 R  M2 S( x L)  ( x  L)  2  S( x L)  ( x  L) 2 2 

 M1



2

 S ( x 0  mm)  x 

2 M2

R1 6

3

 S ( x 0  mm)  ( x  0  mm) 

  

Fball

 R   S ( x L)  ( x  L) 2  2  S ( x L)  ( x  L) 3 6  2



2

 S ( x a )  ( x  a ) 

6

3



 S ( x a )  ( x  a ) 

  

10. Plot the shear, moment, slope, and deflection diagrams. (a) Shear Diagram

(b) Moment Diagram

5

40 20 Moment, M - lb in

Shear, V - lb

0

5

 10

 15

0  20  40

0

10

20

 60

30

0

Distance along beam, x - in

30

(d) Deflection Diagram 0 Deflection - thousandths of in

0.1 Slope - Thousands of Rad

20


(c) Slope Diagram

0

 0.1

10

0

10

20

30

 0.2

 0.4

 0.6

 0.8


0

10

20

30


FIGURE 4-16C Shear, Moment, Slope, and Deflection Diagrams for Problem 4-16(b) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


4-16-4

11 The maximum moment occurs at x = L and is Mmax  M2

Mmax  56.7 in lbf

12 Calculate the maximum bending and principal stresses.

σx 

Bending stress

Mmax c I

σ1  σx


σx  577  psi σ1  577  psi

13. To find the maximum deflection, first determine at what point on the beam the slope is zero. Let this be at x = e. From the slope diagram, we see that e < a. Using the slope equation and setting it equal to zero, we have For  = 0

0 = M1 e 

Solving for e

e 

Maximum deflection

2  M1 R1

ymax  y ( e)

R1 2 e 2 e  18.204 in ymax  0.00063  in



4-17-1


A pair of ice tongs is shown in Figure P4-7. The ice weighs 50 lb and is 10 in wide across the tongs. The distance between the handles is 4 in, and the mean radius r of the tong is 6 in. The rectangular cross-sectional dimensions are 0.75 x 0.312 in. Find the stress in the tongs.

Given:

Assumptions: beam.

Mean radius of tong

rc  6.00 in

Tong width

w  0.312  in

Tong depth

h  0.75 in

F C FC O

The tong can be analyzed as a curved 11.0 = ax

See Problem 3-17, Figure 4-17, Solution: and Mathcad file P0417.

3.5 = cy

FO 2.0 = cx

A

12.0 = by 5.0 = bx

1. The maximum bending moment and axial force in the tong were found in Problem 3-17 at point A. They are

FB B

Maximum moment

MA  237.5  in lbf

Axial force at D

FAn  25 lbf

W/2 FIGURE 4-17 Free Body Diagram for Problem 4-17

2.

3.

Calculate the section area, inside radius and outside radus. A  h  w

A  0.234  in

Inside and outside radii of section

ri  rc  0.5 h

ri  5.625  in

ro  rc  0.5 h

ro  6.375  in

Use the equation in the footnote on page 195 of the text to calculate the radius of the neutral axis. Radius of neutral axis

4.

rn 

ro  ri

 ro  ln   ri 

rn  5.992  in

Calculate the eccentricty and the distances from the neutral axis to the extreme fibers. Eccentricity

e  rc  rn

e  0.007821 in

Distances from neutral axis to extreme fibers

ci  rn  ri

ci  0.3672 in

co  ro  rn

co  0.3828 in

Stresses at inner and outer radii

σi 

MA c i FAn   e A ri A

 MA co  FAn   A  e A ro 

σo   5.

2

Area of section

σi  8.58 ksi

σo  7.69 ksi

The shear stress is zero at the outer fibers. Therefore, these are the principal stresses. At the inner surface

σ1  σi

σ1  8.58 ksi

σ2  0  ksi

σ3  0  ksi



4-18-1


A set of steel reinforcing rods is to be stretched axially in tension to create a tensile stress of 30 ksi prior to being cast in concrete to form a beam. Determine how much force will be required to stretch them the required amount and how much deflection is required. There are 10 rods; each is 0.75-in diameter and 30 ft long.

Given:

Desired stress

σ  30 ksi

Rod diameter

d  0.75 in

Number of rods Rod length

Nrods  10 L  30 ft

Young's modulus

E  30 10  psi

6

Assumptions: The rods share the load equally. Solution:


π d

2

2

1.

Calculate the cross-sectional area of one rod. A 

2.

Determine the force required to achieve the desired stress level in one rod.

σ= 3.

F A

4

F  13.254 kip

Determine the total force required to achieve the desired stress level in all rods. Ftotal  Nrods F

4.

F  σ  A

A  0.442  in

Ftotal  132.5  kip

Determine the amount the rods will deflect under the applied load.

δ 

F L A E

δ  0.360  in



4-19-1


The clamping fixture used to pull the rods in Problem 4-18 is conected to the hydraulic ram by a clevis like that shown in Figure P4-8. Determine the size of the clevis pin needed to withstand the applied force. Assume an allowable shear stress of 20 000 psi and an allowable normal stress of 40 000 psi. Determine the required outside radius of the clevis end to not exceed the above allowable stresses in either tear out or bearing if the clevis flanges are each 0.8 in thick.

Given:

Desired rod stress σrod  30 ksi Nrods  10 L  30 ft S sallow  20 ksi

Number of rods Rod length Clevis strength

d  0.75 in

Rod diameter

Young's modulus Clevis flange thickness

6

E  30 10  psi t  0.8 in

S ballow  40 ksi Assumptions: The rods share the load equally, and there is one clevis for all ten rods. Solution:

See Figures 4-12 and 4-13 in the text, Figure 4-19, and Mathcad file P0419. A 

π d

2

2

A  0.442  in

1.

Calculate the cross-sectional area of one rod.

2.

Determine the force required to achieve the desired stress level in one rod. F σrod = F  σrod  A F  13.254 kip A

3.


4

Ftotal  132.5  kip

This force is transmitted through the clevis pin, which is in double shear. 4.

Calculate the minimum required clevis pin diameter for the allowable shear stress. Divide the load by 2 because of the double shear loading.

τpin =

Ftotal 2  Apin


=

2  Ftotal

π d

2

= S sallow

d 

2  Ftotal

π S sallow

d  2.054  in

Round this up to the next higher decimal equivalent of a common fraction ( 2 1/8) 5.

d  2.125  in

Check the bearing stress in the clevis due to the pin on one side of the clevis. Bearing stress area

Ab  d  t

Bearing force

Fb 

Bearing stress

σb 

Ftotal 2 Fb Ab

2

Ab  1.700  in

Fb  66.268 kip

σb  39.0 ksi

Since this is less than S ballow, this pin diameter is acceptable. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


4-19-2

Determine the tearout stress in the clevis. Shear area (see Figure 4-19)

2

Tearout length

2

Atear = 2  t R  ( 0.5 d )

Shear force Ftear 

Ftotal

Ftear  66.268 kip

2

Shear stress and strength R

d

τ=

Ftear Atear

=

Ftear 2

2  t R  ( 0.5 d )

2

FIGURE 4-19

= S sallow


2

Solving for the clevis radius, R

 Ftear  2 R     ( 0.5 d) 2  t  S sallow  

R  2.328  in

Round this up to the next higher decimal equivalent of a common fraction ( 2 3/8)

R  2.375  in



4-20-1


Repeat Problem 4-19 for 12 rods, each 1 cm in diameter and 10 m long. The desired rod stress is 20 MPa. The allowable normal stress in the clevis and pin is 280 MPa and their allowable shear stress is 140 MPa. Each clevis flange is 2 cm wide.

Units:

kN  10  newton

Given:

Desired rod stress σrod  200  MPa

3

6

9

MPa  10  Pa

GPa  10  Pa d  10 mm

Rod diameter

Number of rods

Nrods  12

Young's modulus

E  207  GPa

Rod length Clevis strength

L  10 m S sallow  140  MPa

Clevis flange thickness

t  20 mm

S ballow  280  MPa Assumptions: The rods share the load equally, and there is one clevis for all twelve rods. Solution:

See Figures 4-12 and 4-13 in the text, Figure 4-20, and Mathcad file P0420. A 

π d

2

A  78.54 mm

1.

Calculate the cross-sectional area of one rod.

2.

Determine the force required to achieve the desired stress level in one rod.

σrod = 3.

F

F  σrod  A

A

4

2

F  15.708 kN


Ftotal  188.5 kN

This force is transmitted through the clevis pin, which is in double shear. 4.

Calculate the minimum required clevis pin diameter for the allowable shear stress. Divide the load by 2 because of the double shear loading.

τpin =

Ftotal 2  Apin


=

2  Ftotal

π d

2

= S sallow

2  Ftotal

d 

π S sallow d  30 mm

Round this up to the next higher even mm 5.

d  29.277 mm

Check the bearing stress in the clevis due to the pin on one side of the clevis. Bearing stress area

Ab  d  t

Bearing force

Fb 

Bearing stress

σb 

Ftotal 2 Fb Ab

Ab  600 mm

2

Fb  94.248 kN

σb  157.1 MPa

Since this is less than S ballow, this pin diameter is acceptable. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0420.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


4-20-2

Determine the tearout stress in the clevis. Shear area (see Figure 4-19)

2

Tearout length

2

Atear = 2  t R  ( 0.5 d )

Shear force Ftear 

Ftotal

Ftear  94.248 kN

2

Shear stress and strength R

d

τ=

Ftear Atear

=

Ftear 2

2  t R  ( 0.5 d )

2

= S sallow

FIGURE 4-20 Tearout Diagram for Problem 4-20

2


 Ftear  2 R     ( 0.5 d) 2  t  S sallow  

Round this up to the next higher even mm

R  22.544 mm

R  24 mm



4-21-1


Figure P4-9 shows an automobile wheel with two common styles of lug wrench being used to tighten the wheel nuts, a single-ended wrench in (a), and a double-ended wrench in (b). In each case two hands are required to provide forces respectively at A and B as shown. The distance between points A and B is 1 ft in both cases and the handle diameter is 0.625 in. The wheel nuts require a torque of 70 ft-lb. Find the maximum principle stress and maximum deflection in each wrench design.

Given:

Distance between A and B

d AB  1  ft

Tightening torque Wrench diameter

T  70 ft  lbf d  0.625  in


See Figure 4-21 and Mathcad file P0421. 12" = dAB

1. In Problem 3-21 we found that for both cases

F

F  70 lbf 2. From examination of the FBDs, we see that, in both cases, the arms are in bending and the stub that holds the socket wrench is in pure torsion. The maximum bending stress in the arm will occur near the point where the arm transitions to the stub. The stress state at this transition is very complicated, but we can find the nominal bending stress there by treating the arm as a cantilever beam, fixed at the transition point. For both cases the torque in the stub is the same.

T F (a) Single-ended Wrench

12" = dAB F

6"

Case (a) T

2. The bending moment at the transition is F

Ma  F  d AB

(b) Double-ended Wrench

Ma  840  lbf  in FIGURE 4-21

3. The tensile stress at this point is found from

Moment of inertia

4.

I 

Free Body Diagrams for Problem 4-21

π d

4

64

Dist to extreme fibre

c  0.5 d

Stress

σx 

M a c I

4

I  0.00749  in c  0.313  in

σx  35.05  ksi

There are no other stress components present at this point, so x is the maximum principle stress here and

σ1  σx

σ1  35.0 ksi

σ2  0  psi

σ3  0  psi

T  840  in lbf

5.

The torque in the stub is

6.

The shear stress at any point on the outside surface of the stub is found from



7.


J  2  I

Shear stress

τxy 

4

J  0.0150 in

Tc

τxy  17.52  ksi

J

There are no other stress components present along the outside surface of the stub, so

σ1  τxy 8.

4-21-2

σ1  17.5 ksi

σ2  0  psi

σ3  σ1

Thus, the maximum principle stress for case (a) is on the upper surface of the handle (arm) near the point where it transitions to the stub. There will be two deflection components that we can calculate separately and then add (superposition). One will come from the bending of the arm and one will come from the twisting of the stub, projected out to the end of the arm.

9.

Deflection of the arm due to bending only for a stub length of stub  3  in: 6

E  30 10  psi

Assuming that the wrenches are made from steel

6

G  11.7 10  psi

3

F  d AB

From Figure B-1(a), Appendix B,

yarm 

yarm  0.179  in

From equation (4.24), the angular twist of the stub is

θstub 

The deflection at the end of the arm due to the stub twist is

ystub  d AB θstub

ystub  0.173  in

So, the total deflection is

ya  yarm  ystub

ya  0.352  in

3  E I T  stub

θstub  0.014  rad

J G

Case (b) Mb 

10. The bending moment at the transition is

F  d AB 2

Mb  420  lbf  in


σx 

Stress

M b c I

σx  17.52  ksi

12. There are no other stress components present at this point, so x is the maximum principle stress here and

σ1  σx 13. The torque in the stub is 14.

σ1  17.5 ksi

σ2  0  psi

σ3  0  psi

T  840  in lbf

The shear stress at any point on the outside surface of the stub is found from Shear stress

τxy 

Tc J

τxy  17.52  ksi

15. There are no other stress components present along the outside surface of the stub, so

σ1  τxy

σ1  17.5 ksi

σ2  0  psi

σ3  σ1



4-21-3

16. Thus, the maximum principle stress for case (b) is the same on the upper surface of the handle (arm) near the point where it transitions to the stub, and on the outside surface of the stub. There will be two deflection components that we can calculate separately and then add (superposition). One will come from the bending of the arm and one will come from the twisting of the stub, projected out to the end of the arm. F   0.5 d AB

Deflection of the arm due to bending only: From Figure B-1(a), Appendix B,

yarm 

From equation (4.24), the angular twist of the stub is

θstub 

The deflection at the end of the arm due to the stub twist is

ystub 

So, the total deflection is

yb  yarm  ystub

3  E I T  stub J G d AB 2

 θstub

3

yarm  0.022  in

θstub  0.014  rad

ystub  0.086  in

yb  0.109  in



4-22-1


A roller-blade skate is shown in Figure P4-10. The polyurethane wheels are 72 mm dia. The skate-boot-foot combination weighs 2 kg. The effective "spring rate" of the person-skate subsystem is 6000 N/m. The axles are 10-mm-dia steel pins in double shear. Find the stress in the pins for a 100-kg person landing a 0.5-m jump on one foot. (a) Assume all 4 wheels land simultaneously. (b) Assume that one wheel absorbs all the landing force.

Given:

Axle pin diameter

Solution:


d  10 mm

Fa  897  N

Fb  3.59 kN

1.

From Problem 3-22, we have the forces for cases (a) and (b):

2.

In both cases, this is the force on one axle. The shear force will be one half of these forces because the pins are in double shear. Shear area

As 

π d

2

As  78.54  mm

4

2

Shear stress Case (a) all wheels landing

τa 

Case (b) one wheel landing

τb 

Fa 2  As Fb 2  As

τa  5.71 MPa

τb  22.9 MPa



4-23a-1


A beam is supported and loaded as shown in Figure P4-11a. Find the reactions, maximum shear, maximum moment, maximum slope, maximum bending stress, and maximum deflection for the data given in row a from Table P4-2.

Given:

Beam length

L  1  m


a  0.4 m

L b

Distance to concentrated load b  0.6 m w  200  N  m

Concentrated load

F  500  N 8

I  2.85 10

Moment of inertia

a

1


Distance to extreme fiber c  2.00 10

4

m

F

w

R2

R1

2

m


Solution: 1.

2.


The reactions, maximum shear and maximum moment were all found in Problem 3-23a. Those results are summarized here. Load function

q(x) = R1-1 - w0 + w0 - F-1 + R2-1

Shear function

V(x) = R10 - w1 + w1 - F0 + R20

Moment function

M(x) = R11 - w2/2 + w2/2 - F1 + R21

Modulus of elasticity

E  207  GPa

Reactions

R1  264.0  N

Maximum shear

Vmax  316  N

Maximum moment

Mmax  126.4  N  m

R2  316.0  N (negative, from x = b to x =L) (at x = b)

Integrate the moment function, multiplying by 1/EI, to get the slope. (x) = [R12/2 - w3/6 + w3/6 - F2/2 + R22/2 + C3]/EI

3.

Integrate again to get the deflection. y(x) = [R13/6 - w4/24 + w4/24 - F3/6 + R23/6 + C3x +C4]/EI

4.

Evaluate C3 and C4 At x = 0 and x = L, y = 0, therefore, C4 = 0. 0=

R1

C3 

6

3

L 

w 24

4

L 

w 24

4

 ( L  a) 

F 6

3

 ( L  b )  C 3 L

 R1 3 w 4 w F 4 3 L  L   ( L  a)   ( L  b)  L  6 24 24 6  1

 

2

C3  31.413 N  m

x  0  m 0.005  L  L

5.


6.

For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 )

7.

Write the slope and deflection equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. See Figure 4-23aB where these functions are plotted.



θ ( x) 

y ( x) 

 R1

1 E I



2

E I

w 6

w

3

 S ( x 0  in)  x 

6



3

 S ( x a )  ( x  a ) 

 R  2  S ( x L)  ( x  L) 2   F  S( x b)  ( x  b ) 2  C3 2  2

 R1

1

2

 S ( x 0  in)  x 

4-23a-2



6

3

 S ( x 0  in)  x 

w 24

w

4

 S ( x 0  in)  x 

  

4

24

 R  2  S( x L)  ( x  L) 3   F  S ( x b )  ( x  b) 3  C3 x 6  6

  

θmax  θ ( L)

8. Maximum slope occurs at x = L



 S ( x a )  ( x  a ) 

θmax  0.335  deg

9. Maximum deflection occurs at x = c, where  = 0 and c < b.

θ0 =

 R1 2 w 3 w  3  c   c   ( c  a )  C3 = 0 E I  2 6 6  1



w

a

Solving for c, R1

A 

2

 3

6

B  3 

A  92.000 N B 

c 

w 6

a

2

C  C3 

3

C  33.547 N  m

2

B  4 A  C

c  0.523  m

2 A

ymax  y ( c)

ymax  1.82 mm

SLOPE, radians

DEFECTION, mm

0.01

0

0.005

 0.5 y ( x)

0

mm

 0.005  0.01

6

a

2

B  16.000 N  m

Substituting c into the deflection equation,

θ( x)

w

1  1.5

0

0.2

0.4

0.6

0.8

1

2

0

0.2

0.4

0.6

x

x

m

m

0.8

1

FIGURE 4-23aB Slope and Deflection Diagrams for Problem 4-23a

10. The maximum bending stress occurs at x = b, where the moment is a maximum. For c  2.00 10

σmax 

2

m

Mmax c I

c  20 mm

σmax  88.7 MPa



4-24a-1


Given:

A beam is supported and loaded as shown in Figure P4-11b. Find the reactions, maximum shear, maximum moment, maximum slope, maximum bending stress, and maximum deflection for the data given in row a from Table P4-2. Beam length

L  1  m


a  0.4 m

L a F

Distance to concentrated load b  0.6 m 1


w  200  N  m

Concentrated load

F  500  N 8

I  2.85 10

Moment of inertia

Distance to extreme fiber c  2.00 10 Solution: 1.

2.

w

M1

4

m

R1

2

m




q(x) = -M1-2 + R1-1 - w0 - F-1

Shear function

V(x) = -M1-1 + R10 - w1 - F0

Moment function

M(x) = -M10 + R11 - w2/2 - F1


E  207  GPa

Reactions

R1  620.0  N

Maximum shear

Vmax  620  N

Maximum moment

Mmax  584  N  m

M1  584.0  N  m (positive, at x = 0) (negative, at x = 0)

Integrate the moment function, multiplying by 1/EI, to get the slope. (x) = [-M11 + R12/2 - w3/6 - F2/2 + C3]/EI

3.

Integrate again to get the deflection. y(x) = [-M12/2 + R13/6 - w4/24 - F3/6 + C3x +C4]/EI

4.

Evaluate C3 and C4. At x = 0,  = 0 and y = 0, therefore, C3 = 0 and C4 = 0.

5.


6.


x  0  m 0.005  L  L

S ( x z)  if ( x  z 1 0 ) 7.


θ ( x) 

1 E I



 M1 S ( x 0  in)  x 

R1 2

 F    S( x L)  ( x  L) 2  2

2

 S ( x 0  in)  x 

w 6

3



 S ( x a )  ( x  a ) 

  


MACHINE DESIGN - An Integrated Approach, 4th Ed. y ( x) 

 M1

1 E I

 

2

 S ( x 0  in)  x 

R1

6  2F 3    S ( x L)  ( x  L)  6

4-24a-2 3

 S ( x 0  in)  x 

w 24

4



 S ( x a )  ( x  a ) 

  



θmax  2.73 deg

9. Maximum deflection occurs at x = L

ymax  y ( L)

ymax  32.2 mm

10. The maximum bending stress occurs at x = 0, where the moment is a maximum. For

σmax 

M1 c

σmax  410  MPa

I

SLOPE, radians

DEFLECTION, mm

0

0

 0.01

 10

 0.02

y ( x)

θ( x)

mm

 0.03

 20

 30

 0.04  0.05

c  20 mm

0

0.2

0.4

0.6

0.8

1

 40

0

0.2

0.4

0.6

x

x

m

m

0.8

1




4-25a-1


A beam is supported and loaded as shown in Figure P4-11c. Find the reactions, maximum shear, maximum moment, maximum slope, maximum bending stress, and maximum deflection for the data given in row a from Table P4-2.

Given:

Beam length

L  1  m


a  0.4 m

Distance to reaction load

b  0.6 m

L b a


w  200  N  m

Concentrated load

F  500  N 8

I  2.85 10

Moment of inertia

F

1


4

m

2

m

w

R2

R1


Solution: 1.

2.



q(x) = R1-1 - w0 + R2-1 - F-1

Shear function

V(x) = R10 - w1 + R20 - F0

Moment function

M(x) = R11 - w2/2 + R21 - F1


E  207  GPa

Reactions

R1  353.3  N

Maximum shear

Vmax  580  N

Maximum moment

Mmax  216  N  m

R2  973.3  N (positive, at x = b) (negative, at x = b)

Integrate the moment function, multiplying by 1/EI, to get the slope. (x) = [R12/2 - w3/6 + R22/2 - F2/2 + C3]/EI

3.

Integrate again to get the deflection. y(x) = [R13/6 - w4/24 + R23/6 - F3/6 + C3x +C4]/EI

4.

Evaluate C3 and C4 At x = 0 and x = b, y = 0, therefore, C4 = 0. 0=

R1 6

3

b 

w 24

4

 ( b  a )  C3 b

1  R1 3 w 4 C3      b   ( b  a)  b 6 24 

2

C3  21.22  N  m

x  0  m 0.005  L  L

5.


6.


7.




θ ( x) 

y ( x) 

 R1

4-25a-2

  E I 2 6  R   2  S ( x b )  ( x  b) 2   F  S( x L)  ( x  L) 2  C3  2  2  1



1



 R1

E I

2

 S ( x 0  in)  x 

w

3

 S ( x a )  ( x  a ) 

   R   2  S( x b)  ( x  b ) 3   F  S ( x L)  ( x  L) 3  C3 x  6  6  6

3

 S ( x 0  in)  x 

w

24

4

 S ( x a )  ( x  a ) 



θmax  0.823  deg

9. Maximum deflection occurs at x = L.

ymax  y ( L)

ymax  4.81 mm

10. The maximum bending stress occurs at x = b, where the moment is a maximum. For

σmax 

Mmax c


I

SLOPE, radians

DEFLECTION, mm

0.005

2

0

0 y ( x)

θ( x)  0.005

mm

 0.01

 0.015

c  20 mm

2

4

0

0.2

0.4

0.6

0.8

1

6

0

0.2

0.4

0.6

x

x

m

m

0.8

1




4-26a-1


A beam is supported and loaded as shown in Figure P4-11d. Find the reactions, maximum shear, maximum moment, maximum slope, maximum bending stress, and maximum deflection for the data given in row a from Table P4-2.

Given:

Beam length

L  1  m a  0.4 m

Distance to R2

b  0.6 m


w  200  N  m

Concentrated load

F  500  N

b

8


4

m

2

m

F w

R2

R1

R3


E  207  GPa


a

1

I  2.85 10

Moment of inertia

Solution:

L



1. From inspection of Figure P4-11d, write the load function equation q(x) = R1-1 - F-1 - w0 + R2-1 - R3-1 2. Integrate this equation from - to x to obtain shear, V(x) V(x) = R10 - F0 - w1 + R20 - R30 3. Integrate this equation from - to x to obtain moment, M(x) M(x) = R11 - F1 - w2/2 + R21 - R31 4. Integrate the moment function, multiplying by 1/EI, to get the slope. (x) = [R12/2 - F2/2 - w3/6 + R22/2 + R32/2 + C3]/EI 5. Integrate again to get the deflection. y(x) = [R13/6 - F3/6 - w4/24 + R23/6 + R33/6 + C3x + C4]/EI 6. Evaluate R1, R2, R3, C3 and C4 At x = 0, x = b, and x = L; y = 0, therefore, C4 = 0. At x = L+, V = M = 0 R1  100  N

Guess

R2  100  N

2

R3  100  N

C3  5  N  m

Given R1 6 R1 6

b 

F

3

F

3

L 

6

6

3

 ( b  a) 

3

 ( L  a) 

w 24 w 24

4

3

 ( b  a )  C3 b = 0  N  m

4

 ( L  a) 

R2 6

3

3

 ( L  b )  C 3 L = 0  N  m

R1  F  w  ( L  a )  R2  R3 = 0  N R 1 L  F  ( L  a ) 

w 2

2

 ( L  a )  R 2 ( L  b ) = 0  N  m



4-26a-2

 R1  R   2   Find  R R R C  1 2 3 3  R3     C3  R1  112.33 N

R2  559.17 N

2

R3  51.50  N

C3  5.607  N  m

x  0  in 0.002  L  L


8. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 9. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  R1 S ( x 0  in)  F  S ( x a )  w S ( x a )  ( x  a )  R2 S ( x b )  R3 S ( x L) M ( x)  R1 S ( x 0  in)  x  F  S ( x a )  ( x  a )   R2 S ( x b )  ( x  b )

w 2

2

 S ( x a )  ( x  a ) 

10. Plot the shear and moment diagrams.

SHEAR, N

V ( x) N

MOMENT, N-m

200

60

0

35 M ( x)

 200

10

Nm

 400

 15

 600

 40 0

200

400

600

800

3

1 10

0

200

400

600

x

x

mm

mm

800

3

1 10


11. From the diagram, we see that maximum shear occurs at x = b -, Vmax  V ( b  0.001  mm)

Vmax  428  N

12. The maximum moment occurs at x = a, Mmax  M ( a )

Mmax  44.9 N  m

13. Write the slope and deflection equations in Mathcad form, using the function S as a multiplying factor to get t effect of the singularity functions. See Figure 4-26aB where these functions are plotted. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


θ ( x) 

y ( x) 

1 E I

1 E I

 R1



2

2

 S ( x 0  in)  x 

F 2

4-26a-3 2

 S ( x a )  ( x  a ) 

w 6

3

 R R  2  S ( x b )  ( x  b) 2  3  S ( x L)  ( x  L) 2  C3 2  2

 R1



6

3

 S ( x 0  in)  x 

F 6

3

 S ( x a )  ( x  a ) 

w 24

  

4

 R R  2  S( x b)  ( x  b ) 3  3  S( x L)  ( x  L) 3  C3 x 6  6

θmax  0.0576 deg

15. Maximum deflection occurs between x = 0 and x = a

ymax  0.200  mm

  

16. The maximum bending stress occurs at x = a, where the moment is a maximum. For Mmax c

DEFLECTION, mm

0.1

0.1

0.05

0

θ( x)

y ( x) 0

mm

 0.05

 0.1

c  20 mm


I

SLOPE, deg.

deg



 S ( x a )  ( x  a ) 

14. Maximum slope occurs between x = a and x = b

σmax 



 S ( x a )  ( x  a ) 

 0.1

 0.2

0

0.2

0.4

0.6

0.8

1

 0.3

0

0.2

0.4

0.6

x

x

m

m

0.8

1

FIGURE 4-26aC Slope and Deflection Diagrams for Problem 4-26a



4-27-1


A storage rack is to be designed to hold the paper roll of Problem 4-8 as shown in Figure P4-12. Determine suitable values for dimensions a and b in the figure. Consider bending, shear, and bearing stresses. Assume an allowable tensile/compressive stress of 100 MPa and an allowable shear stress of 50 MPa for both stanchion and mandrel, which are steel. The mandrel is solid and inserts halfway into the paper roll. Balance the design to use all of the material strength. Calculate the deflection at the end of the roll.

Given:


Roll density

OD  1.50 m

Material properties

S y  100  MPa

ID  0.22 m

S ys  50 MPa

Lroll  3.23 m

E  207  GPa

3

ρ  984  kg m

Assumptions: 1. The paper roll's weight creates a concentrated load acting at the tip of the mandrel. 2. The mandrel's root in the stanchion experiences a distributed load over its length of engagement Solution:

See Figures 4-27 and Mathcad file P0427.

1. In Problem 3-27, we were concerned only with the portion of the mandrel outside of the stanchion. Therefore, we modeled it as a cantilever beam with a shear and moment reaction at the stanchion. Unfortunately, this tells us nothing about the stress or force distributions in the portion of the mandrel that is inside the stanchion. To do this we need to modify the model by replacing the concentrated moment (and possibly the concentrated shear force) with a force system that will yield information about the stress distribution in the mandrel on that portion that is inside the stanchion. Figure 4-27A shows the FBD used in Problem 3-27. Figure 4-27B is a simple model, but is not representative of a built-in condition. It would be appropriate if the hole in the stanchion did not fit tightly around the mandrel. Figure 4-27C is an improvement that will do for our analysis.

y

x

M1

 4

π

2

2





W  53.9 kN

Lm R1

FIGURE 4-27A Free Body Diagram for Problem 3-27 y W

R1

x

2. Determine the weight of the roll and the length of the mandrel. W 

W

Lm R2

FIGURE 4-27B Simplified Free Body Diagram, not used

Lm  1.615  m

3. From inspection of Figure 4-27C, write the load function equation

W

y w

q(x) = -w0 + w0 + R-1 - W-1

a x

4. Integrate this equation from - to x to obtain shear, V(x) V(x) = -w1 + w1 + R0 - W0 5. Integrate this equation from - to x to obtain moment, M(x)

Lm

b R

FIGURE 4-27C Free Body Diagram used in Problem 4-27



4-27-2

M(x) = -(w/2)2 + (w/2)2 + R1 - W1 6. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b + Lm, where both are zero. At x = (b + Lm)+ , V = M = 0 0 = w  b  Lm  w  Lm  R  W

R = W  w b

w w 2 w 2 2 w 2 0 =    b  Lm   Lm  R Lm =    b  Lm   Lm  ( W  w b )  Lm 2 2 2 2

w=

2  W  Lm b

2

Note that R is inversely proportional to b and w is inversly proportional to b 2. 7. To see the value of x at which the shear and moment are maximum, let b  400  mm

w 

then

2  W  Lm b

R  W  w b

and

2

L  b  Lm

x  0  mm 0.002  L  L


9. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 10. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  w S ( x 0  mm)  x  w S ( x b )  ( x  b )  R S ( x b )  W  S ( x L) M ( x) 

w 2

2

 S ( x 0  mm)  x 

w 2

2

 S ( x b )  ( x  b )  R S ( x b )  ( x  b )  W  S ( x L)  ( x  L)

11. Plot the shear and moment diagrams. Shear Diagram

Moment Diagram

200

50

0

V ( x) kN

0

 200

M ( x) kN  m

 400

 50

 600  800

0

400

800

1200

1600

2000

 100

0

400

800

1200

x

x

mm

mm

1600

2000

FIGURE 4-27D Shear and Moment Diagram Shapes for Problem 4-27

12. From Figure 4-27D, the maximum internal shear and moment occur at x = b and are Vmax =

2  W  Lm b

Mmax  W  Lm

Mmax  87.04  kN  m



4-27-3

13. The bending stress will be a maximum at the top or bottom of the mandrel at a section through x = b.

σmax =

Mmax a

where

2 I

I=

π a

4

so,

σmax =

64

32 Mmax

π a

1

 32 W  Lm   π S  y  

Solving for a,

a 

Round this to

a  210  mm

3

= Sy

3

a  206.97 mm

14. Using this value of a and equation 4.15c, solve for the shear stress on the neutral axis at x = b.

τmax =

4  Vmax 3 A

=

8  W  Lm

 π a 2   b 3   4 

= S ys

8  W  Lm

Solving for b

b 

Round this to

b  134  mm

b  134.026  mm

 π a 2    Sys 3   4 

15. These are minimum values for a and b. Using them, check the bearing stress. Magnitude of distributed load

w 

2  W  Lm b

Bearing stress

σbear 

w  9695

2

w b

N mm

σbear  46.2 MPa

a b

Since this is less than S y, the design is acceptable for a  210  mm and b  134  mm 16. Assume a cantilever beam loaded at the tip with load W and a mandrel diameter equal to a calculated above. Moment of inertia

I 

π a

4

7

I  9.547  10  mm

64

4

3

Deflection at tip (Appendix B)

ymax  

W  Lm

3  E I

ymax  3.83 mm

This can be accomodated by the 220-mm inside diameter of the paper roll.



4-28-1


Figure P4-13 shows a forklift truck negotiating a 15 deg ramp to to drive onto a 4-ft-high loading platform. The truck weighs 5 000 lb and has a 42-in wheelbase. Design two (one for each side) 1-ft-wide ramps of steel to have no more than 1-in deflection in the worst case of loading as the truck travels up them. Minimize the weight of the ramps by using a sensible cross-sectional geometry.

Given:

Ramp angle Platform height

θ  15 deg

Truck weight Truck wheelbase

W  5000 lbf Lt  42 in

h  4  ft

Ramp width Allowable deflection

w  12 in δmax  1.0 in

Young's modulus

E  30 10  psi

6

Assumptions: 1. The worst case is when the truck CG is located at the center of the beam's span. 2. Use a coordinate frame that has the x-axis along the long axis of the beam. 3. Ignore traction forces and the weight components along the x-axis of the beam. 4. There are two ramps, one for each side of the forklift. Solution:


L b a

CG a

y

CG b

R1  Fa

Wa

Fb

x Wb

R2


1. Determine the length of the beam between supports and the distances a and b for the worst-case loading. h

Length of beam

L 

From Problem 3-28,

a  5.061  ft

sin( θ )

L  15.455 ft b  8.561  ft

2. The load distribution of the wheels on a single ramp is given in Problem 3-28 as Fa  575.0  lbf

Fb  1839.9 lbf

3. From inspection of Figure 4-28A, write the load function equation q(x) = R1-1 - Fa-1 - Fb-1 + R2-1



4-28-2

4. Integrate this equation from - to x to obtain shear, V(x) V(x) = R10 - Fa0 - Fb0 + R20 5. Integrate this equation from - to x to obtain moment, M(x) M(x) = R11 - Fa1 - Fb1 + R21 R1  1207.4 lbf

6. The reactions are given in Problem 3-28 as

R2  1207.4 lbf

7. Integrate the moment function, multiplying by 1/EI, to get the slope. (x) = [R12/2 - Fa2/2 - Fb2/2 + R22/2 + C3]/EI 8. Integrate again to get the deflection. y(x) = [R13/6 - Fa3/6 - Fb3/6 + R23/6 + C3x +C4]/EI 9. Evaluate C3 and C4 At x = 0 and x = L, y = 0, therefore, C4 = 0. 3

3

3

0 = R1 L  Fa ( L  a )  Fb ( L  b )  6  C3 L 1

C3 

 R1 L  Fa ( L  a )  Fb ( L  b ) 3

6 L

3

3

6

2

C3  4.983  10  lbf  in



x  0  m 0.005  L  L


9. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 10. Write the slope and deflection equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. Use an assumed value of I so that the value of x that corresponds to ymax 4

can be found. Let I  10 in

θ ( x) 

y ( x) 

1 E I

1 E I

 R1



2

2

 S ( x 0  m)  x 

Fa 2

2

 S ( x a )  ( x  a ) 

 R  2  S ( x L)  ( x  L) 2  C3  2

 R1



6

3

 S ( x 0  m)  x 

Fa 6

3

 S ( x a )  ( x  a ) 

 R  2  S( x L)  ( x  L) 3  C3 x  6

Fb 2

Fb 6

2



 S ( x b )  ( x  b ) 

  

3



 S ( x b )  ( x  b ) 

  

11. Plot the shear and moment diagrams using the assumed value of I.



4-28-3

SLOPE, radians

DEFLECTION, in

0.02

0

0.01  0.5

θ( x)

y ( x) 0

in 1

 0.01

 0.02

0

4

8

12

 1.5

16

0

4

8

x

x

ft

ft

12

16

FIGURE 4-28B Slope and Deflection Diagrams for Problem 4-28, Using an Assumed Value for I

12. Maximum deflection occurs at x = c, where  = 0 and c < b.

θ0 =

 R1 2 Fa  2 c   ( c  a )  C3 = 0 E I  2 2  1



Solving for c, A 

R1 2



2

Fa

B  a  Fa

2

4

A  316.200  lbf c 

C  C3 

a  Fa 2 6

B  3.492  10  lbf  in

2

C  6.043  10  in  lbf

2

B 

B  4 A  C

c  7.804  ft

2 A

13. The maximum deflection occurs at x = c and is

ymax =

1 E I

 R 1 c 3



 6



Fa 6



 ( c  a )  C3 c = δmax 3



Solving for I I 

1 E δmax

 R 1 c 3





6



Fa 6



 ( c  a )  C3 c 3

4

I  10.159 in



This is the minimum allowable value of the moment of inertia. 14. Assume a channel section such as that shown in Figure 4-28C. To keep it simple, let the thickness of the flanges and web be the same. Choose 3/8-in thick plate, which is readily available. Then, t  0.375  in 15. The cross-sectional area of the ramp is 16. The distance to the CG is

cg( h ) 

A ( h )  w t  2  t ( h  t) 1 A (h)

 w t 2



 2

2

 t h  t

2

 



4-28-4

17. The moments of inertia of the web and a flange are w t

3

Iweb( h )   w t  cg( h )  12  Ifl ( h ) 

t ( h  t) 12

3

  2 t

 h  t  cg( h ) 



Flange

2

Web h  t

2

t

 2 

18. Using the known moment of inertia, solve for the unknown flange height, h. Guess h  1  in Given I = Iweb( h )  2  Ifl ( h ) h  Find ( h ) Round this up to

h

h  3.988  in

h  4.00 in

w FIGURE 4-28C Channel Section for Problem 4-28

19. Summarizing, the ramp design dimensions are: Length

L  185.5  in

Flange height

h  4.00 in

Shape

channel

Width

w  12.00  in

Thickness

t  0.375  in

Material

steel



4-29a-1


Find the spring rate of the beam in Problem 4-23 at the applied concentrated load for row a in Table P4-2.

Given:

Beam length

L  1  m


a  0.4 m

L b a


Solution:

1


w  200  N  m

Concentrated load

Fb  500  N 8

Moment of inertia

I  2.85 10


E  207  GPa

F

w

R2

R1

4

m

FIGURE 4-29 Free Body Diagram for Problem 4-23


1. The deflection equation was found in Problem 4-23. Those results are summarized here. Load function

q(x) = R1-1 - w0 + w0 - F-1 + R2-1

Shear function

V(x) = R10 - w1 + w1 - F0 + R20

Moment function

M(x) = R11 - w2/2 + w2/2 - F1 + R21

Slope function

(x) = [R12/2 - w3/6 + w3/6 - F2/2 + R22/2 + C3]/EI

Deflection function

y(x) = [R13/6 - w4/24 + w4/24 - F3/6 + R23/6 + C3x +C4]/EI

2. To determine the stiffness under the load F we will need to find the incremental beam deflection due to F alone. The procedure will be to find the deflection at x = b when F = 0, and then find it when Fb  500  N . The stiffness will then be the force divided by the incremental deflection. 3. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 4. Write the reactions (from Problem 3-23), integration constant, and deflection (from problem 4-23) equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. R1( F ) 

w 2

L 

F L

 ( L  b) 

w 2 L

 ( L  a)

2

R2( F )  w a  F  R1( F ) C3( F )  y ( x F ) 

 R1( F ) 3 w 4 w F 4 3 L  L   ( L  a)   ( L  b)  6 L  24 24 6  1

  1

E I

 R1( F )



6

3

 S ( x 0  in)  x 

w 24

w

4

 S ( x 0  in)  x 

24

4



 S ( x a )  ( x  a ) 

 R (F )  2  S( x L)  ( x  L) 3   F  S ( x b )  ( x  b) 3  C3( F )  x 6  6

  

5. The deflection at x = b for F  0  N is

y0  y ( b F )

y0  0.137  mm

6. The deflection at x = b for F  Fb is

yF  y ( b F )

yF  1.765  mm

Δy  yF  y0

Δy  1.627  mm

7. The deflection due to F alone is

8. The stiffness of the beam under the load F at x = b is

k 

F

Δy

k  307 

N mm



4-30a-1



Given:

Beam length

Solution:

L  1  m

L


a  0.4 m


w  200  N  m

Concentrated load

FL  500  N

a

1

8

Moment of inertia

I  2.85 10


E  207  GPa

F w

4

m

M1 R1


FIGURE 4-30

1. The deflection equation was found in Problem 4-24. Those results are summarized here.


Load function

q(x) = -M1-2 + R1-1 - w0 - F-1

Shear function

V(x) = -M1-1 + R10 - w1 - F0

Moment function

M(x) = -M10 + R11 - w2/2 - F1

Slope function

(x) = [-M11 + R12/2 - w3/6 - F2/2 + C3]/EI

Deflection function

y(x) = [-M12/2 + R13/6 - w4/24 - F3/6 + C3x +C4]/EI

2. To determine the stiffness under the load F we will need to find the incremental beam deflection due to F alone. The procedure will be to find the deflection at x = L when F = 0, and then find it when FL  500  N . The stiffness will then be the force divided by the incremental deflection. 3. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 4. Write the reaction (from Problem 3-24) and deflection (from problem 4-24) equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. R1( F )  w ( L  a )  F M1( F ) 

y ( x F ) 

w 2

2

 ( L  a )  R 1( F )  L

1 E I

 M1( F )

 

2 F

2

 S ( x 0  in)  x 

    S( x L)  ( x  L) 3  6

R1( F ) 6

3

 S ( x 0  in)  x 

w 24

4

  

5. The deflection at x = L for F  0  N is

y0  y ( L F )

y0  3.912  mm

6. The deflection at x = L for F  FL is

yF  y ( L F )

yF  32.163 mm

Δy  yF  y0

Δy  28.251 mm


8. The stiffness of the beam under the load F at x = L is

k 

F

Δy



 S ( x a )  ( x  a ) 

k  17.7

N mm



4-31a-1



Given:

Beam length

L  1  m


a  0.4 m

L b a


Solution:

1


w  200  N  m

Concentrated load

FL  500  N 8

Moment of inertia

I  2.85 10


E  207  GPa

F w

R2

R1

4

m




q(x) = R1-1 - w0 + R2-1 - F-1

Shear function

V(x) = R10 - w1 + R20 - F0

Moment function

M(x) = R11 - w2/2 + R21 - F1

Slope function

(x) = [R12/2 - w3/6 + R22/2 - F2/2 + C3]/EI

Deflection function

y(x) = [R13/6 - w4/24 + R23/6 - F3/6 + C3x +C4]/EI

2. To determine the stiffness under the load F we will need to find the incremental beam deflection due to F alone. The procedure will be to find the deflection at x = L when F = 0, and then find it when FL  500  N . The stiffness will then be the force divided by the incremental deflection. 3. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 4. Write the reactions (from Problem 3-25), integration constant, and deflection (from problem 4-25) equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. R1( F ) 

w 2    ( L  a )  F  ( L  b )  w ( L  a )  ( L  b ) b 2  1

R2( F )  w ( L  a )  F  R1( F ) 1  R1( F ) 3 w 4 C3( F )     b   ( b  a)  b  6 24  y ( x F ) 

  6 E I 24  R (F )   2  S( x b)  ( x  b ) 3   F  S ( x L)  ( x  L) 3  C3( F )  x  6  6  1

 R1( F )



3

 S ( x 0  in)  x 

w

4

 S ( x a )  ( x  a ) 

5. The deflection at x = L for F  0  N is

y0  y ( L F )

y0  0.288  mm

6. The deflection at x = L for F  FL is

yF  y ( L F )

yF  4.808  mm


MACHINE DESIGN - An Integrated Approach, 4th Ed. 7. The deflection due to F alone is

4-31a-2

Δy  yF  y0

8. The stiffness of the beam under the load F at x = L is

Δy  4.52 mm k 

F

Δy

k  111 

N mm



4-32a-1



Given:

Beam length

L  1  m


a  0.4 m

L b a


Solution:

F w

1


w  200  N  m

Concentrated load

Fa  500  N 8

Moment of inertia

I  2.85 10


E  207  GPa

R2

R1

R3

4

m




q(x) = R1-1 - F-1 - w0 + R2-1 - R3-1

Shear function

V(x) = R10 - F0 - w1 + R20 - R30

Moment function

M(x) = R11 - F1 - w2/2 + R21 - R31

Slope function

(x) = [R12/2 - F2/2 - w3/6 + R22/2 + R32/2 + C3]/EI

Deflection function y(x) = [R13/6 - F3/6 - w4/24 + R23/6 + R33/6 + C3x + C4]/EI 2. To determine the stiffness under the load F we will need to find the incremental beam deflection due to F alone. The procedure will be to find the deflection at x = a when F = 0, and then find it when Fa  500  N . The stiffness will then be the force divided by the incremental deflection. 3. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 4. Write the reactions, integration constant, and deflection (from problem 4-26) equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. Let

f1 ( F ) 

F 6

3

 ( b  a)  w

f3 ( F )  F  ( L  a ) 

2

w 24

 ( b  a)

 ( L  a)

4

f2 ( F ) 

F 6

3

 ( L  a) 

w 24

 ( L  a)

4

2

then R1( F )  R2( F ) 

 L

3 L b  ( L  b ) 1 ( L  b)



b

 f1 ( F )  f2 ( F ) 

  f3 ( F )  L R1( F ) 

R3( F )  F  w ( L  a )  R1( F )  R2( F ) C3( F ) 

1 b

 f1 ( F ) 

b

( L  b) 6

2



 f3 ( F )



R1 Fa  112.333  N R2 Fa  559.167  N R3 Fa  51.500 N

2

6

 R 1( F )



y ( x F ) 

1 E I

 R1( F )



6

3

 S ( x 0  in)  x 

F 6

4-32a-2

3

 S ( x a )  ( x  a ) 

w 24

4

 R (F ) R (F )  2  S( x b)  ( x  b ) 3  3  S ( x L)  ( x  L) 3  C3( F )  x 6  6

5. The deflection at x = a for F  0  N is

y0  y ( a F )

y0  0.00126  mm

6. The deflection at x = a for F  Fa is

yF  y ( a F )

yF  0.177  mm

Δy  yF  y0

Δy  0.176  mm


8. The stiffness of the beam under the load F at x = a is

k 

F

Δy



 S ( x a )  ( x  a ) 

  

k  2844

N mm



4-33a-1


For the bracket shown in Figure P4-14 and the data in row a of Table P4-3, determine the bending stress at point A and the shear stress due to transverse loading at point B. Also the torsional shear stress at both points. Then determine the principal stresses at points A and B.

Given:

Tube length

L  100  mm

F

y

Arm length

a  400  mm

Arm thickness

t  10 mm

Arm depth

h  20 mm

Applied force

F  50 N

Tube OD

OD  20 mm

A B

T

T

x

M L R

Tube ID

ID  14 mm


E  207  GPa

FIGURE 4-33 Free Body Diagram of Tube for Problem 4-33

Solution:


1. Determine the bending stress at point A. From the FBD of the tube in Figure 4-33 we see that Reaction force

R  F

R  50.0 N

Reaction moment

M  F  L

M  5.00 N  m

Distance from NA to outside of tube

ct  0.5 OD

ct  10.0 mm

Moment of inertia

It 

Bending stress at point A

σxA 

 64 π

4

 OD  ID

4



M  ct

It  5968 mm

4

σxA  8.38 MPa

It

2. Determine the shear stress due to transverse loading at B. Cross-section area

A 

π 4



2

 OD  ID

Maximum shear

V  R

Maximum shear stress (Equation 4.15d)

τVmax  2 

2



V

A  160.2  mm

2

τVmax  0.624  MPa

A

3. Determine the torsional shear stress at both points. Using equation 4.23b and the FBD above Torque on tube

T  F  a


J 

Maximum torsional stress at surface

τTmax 

T  20.0 N  m

 32 π

4

 OD  ID T  ct J

4



J  11936  mm

4

τTmax  16.76  MPa

4. Determine the principal stress at point A. Stress components

σxA  8.378  MPa

σzA  0  MPa

τxz  τTmax

τxz  16.76  MPa



4-33a-2

Principal stresses

σ1 

σxA  σzA 2

2  σxA  σzA  2     τxz 2  

σ1  21.46  MPa

2  σxA  σzA  2     τxz 2  

σ3  13.08  MPa

σ2  0  MPa

σ3 

τ13 

σxA  σzA 2

σ1  σ3

τ13  17.27  MPa

2

5. Determine the principal stress at point B. Stress components

σxB  0  MPa

σyB  0  MPa

τxy  τTmax  τVmax

τxy  16.13  MPa

Principal stresses

σ1 

σxB  σyB 2

2  σxB  σyB  2     τxy 2  

σ1  16.13  MPa

2  σxB  σyB  2     τxy 2  

σ3  16.13  MPa

σ2  0  MPa

σ3 

τ13 

σxB  σyB 2

σ1  σ3 2

τ13  16.13  MPa



4-34a-1


For the bracket shown in Figure P4-14 and the data in row a of Table P4-3, determine the deflection at load F.

Given:

Tube length Arm length Arm thickness Arm depth

Solution:


L  100  mm a  400  mm t  10 mm h  20 mm

Applied force Tube OD Tube ID Modulus of elasticity Modulus of rigidity

F  50 N OD  20 mm ID  14 mm E  207  GPa G  80.8 GPa

1.

The deflection at load F can be determined by superimposing the rigid-body deflection of the arm due to the twisting of the tube with the beam deflection of the tube and the arm alone.

2.

Determine the rigid-body deflection due to twisting of the tube. Refering to Figure 4-34, the torque in the tube is Torque on tube

T  F  a


Jt 

Tube angle of twist

θ 

T  20.0 N  m

 32 π

4

 OD  ID

4



Jt  11936  mm

TL

4

θ  2.07368  10

J t G

3

 rad

θ  0.119  deg Deflection at F due to  3.

δθ  a  θ

δθ  0.829  mm

Determine the rigid-body deflection due to bending of the tube. It 

Moment of inertia Deflection of tube end and arm end (see Appendix B)

Jt

It  5968 mm

2

4

3

δtb 

F L

δtb  0.013  mm

3  E It

F

F

y

a

y A B

T

T

x

z

M

h T

L F

R

FIGURE 4-34 Free Body Diagrams of Tube and Arm for Problem 4-34

4.

Determine the beam bending of arm alone. Moment of inertia

Deflection at F 5.

Ia 

δa 

t h

3

Ia  6667 mm

12 F a

4

3

δa  0.773  mm

3  E Ia

Determine the total deflection by superposition.

δtot  δθ  δtb  δa

δtot  1.616  mm

downward



4-35a-1


For the bracket shown in Figure P4-14 and the data in row a of Table P4-3, determine the spring rate of the tube in bending, the spring rate of the arm in bending, and the spring rate of the tube in torsion. Combine these into an overall spring rate in terms of the force F and the linear deflection at F.

Given:

Tube length Arm length Arm thickness Arm depth

Solution:


1.

2.

L  100  mm a  400  mm t  10 mm h  20 mm

Applied force Tube OD Tube ID Modulus of elasticity Modulus of rigidity

F  50 N OD  20 mm ID  14 mm E  207  GPa G  80.8 GPa

Determine the spring rate due to bending of the tube.

 64 π

Moment of inertia

It 

Deflection of tube end and arm end (see Appendix B)

δtb 

Spring rate due to bending in tube

ktb 

4

4

 OD  ID



It  5968 mm

4

3

F L

δtb  0.013  mm

3  E It F

ktb  3706

δtb

N mm

Determine the spring rate due to beam bending of arm alone. t h

Ia 

Moment of inertia

3

Ia  6667 mm

12

Deflection at F

δa 

Spring rate due to bending in arm

ka 

F a

4

3

δa  0.773  mm

3  E Ia

F

ka  64.7

δa

N mm F

F

y

a

y A B

T

T

x

z

M

h T

L F

R

FIGURE 4-35 Free Body Diagrams of Tube and Arm for Problem 4-35

3.

Determine the spring rate of the tube in torsion. Refering to Figure 4-35, the torque in the tube is Torque on tube Polar moment of inertia

T  F  a Jt 

 32 π

4

 OD  ID

4



T  20.0 N  m Jt  11936  mm

4


MACHINE DESIGN - An Integrated Approach, 4th Ed. Tube angle of twist

θ 

4-35a-2

TL

θ  2.07368  10

J t G

3

 rad

θ  0.119  deg

4.

Deflection at F due to q

δθ  a  θ

Spring rate due to torsion in tube

kθ 

F

kθ ktb ka

1 koa

=

1 kθ



1 ktb



δtot 

F koa

N mm

1 ka

koa  30.9

ktb ka  kθ ka  kθ ktb

Checking,

kθ  60.28 

δθ

Determine the overall spring rate. The springs are in series, thus

koa 

δθ  0.829  mm

N mm

δtot  1.616  mm

which is the same total deflection gotten in Problem 4-34.



4-36a-1


For the bracket shown in Figure P4-14 and the data in row a of Table P4-3, redo Problem 4-33 considering the stress concentration at points A and B. Assume a stress concentration factor of 2.5 in both bending and torsion.

Given:

Tube length

L  100  mm

Arm length

a  400  mm

Arm thickness

t  10 mm

Arm depth

h  20 mm

Applied force

F  50 N

Tube OD

OD  20 mm

Tube ID

ID  14 mm


E  207  GPa

Stress-concentration factors

Ktb  2.5

Solution: 1.

2.

3.

4.

F

y A B

T

T

x

M L R


Kts  2.5


Determine the bending stress at point A. From the FBD of the tube in Figure 4-36 we see that Reaction force

R  F

R  50.0 N

Reaction moment

M  F  L

M  5.00 N  m

Distance from NA to outside of tube

ct  0.5 OD

ct  10.0 mm

Moment of inertia

It 

Bending stress at point A

σxA  Ktb

 64 π

4

 OD  ID

4



M  ct

It  5968 mm

4

σxA  20.94  MPa

It

Determine the shear stress due to transverse loading at B.

 4

π

Cross-section area

A 

Maximum shear

V  R

Maximum shear stress (Equation 4.15d)

τVmax  2 

2

 OD  ID

2



V

A  160.2  mm

2

τVmax  0.624  MPa

A

Determine the torsional shear stress at both points. Using equation 4.23b and the FBD above Torque on tube

T  F  a


J 

Maximum torsional stress at surface

τTmax  Kts

 32 π

T  20.0 N  m 4

 OD  ID T  ct J

4



J  11936  mm

4

τTmax  41.89  MPa

Determine the principal stress at point A. Stress components

σxA  20.944 MPa

σzA  0  MPa



τxz  τTmax

4-36a-2

τxz  41.89  MPa

Principal stresses

σ1 

σxA  σzA 2

2  σxA  σzA  2     τxz 2  

σ1  53.6 MPa

2  σxA  σzA  2     τxz 2  

σ3  32.71  MPa

σ2  0  MPa

σ3 

τ13  5.

σxA  σzA 2

σ1  σ3

τ13  43.18  MPa

2

Determine the principal stress at point B. Stress components

σxB  0  MPa

σyB  0  MPa

τxy  τTmax  τVmax

τxy  41.26  MPa

Principal stresses

σ1 

σxB  σyB 2

2  σxB  σyB  2     τxy 2  

σ1  41.26  MPa

2  σxB  σyB  2     τxy 2  

σ3  41.26  MPa

σ2  0  MPa

σ3 

τ13 

σxB  σyB 2

σ1  σ3 2

τ13  41.26  MPa



4-37-1


Given:

A semicircular, curved beam as shown in Figure 4-37 has the dimensions given below. For the load pair applied along the diameter and given below, find the eccentricity of the neutral axis and the stress at the inner and outer fibers. Outside diameter

od  150  mm

Inside diameter

id  100  mm

Width of beam

w  25 mm

Load

F  14 kN

w

F od

Solution:

id


F

1. Calculate the section depth, area, inside radius and outside radus. Section depth

h 

od  id 2

Area of section

A  h  w

Centroid radius

rc 

Inside and outside radii of section

(a) Entire Beam

h  25 mm A  625  mm

od  id

2

rc  62.5 mm

4

ri  rc  0.5 h

ri  50 mm

ro  rc  0.5 h

ro  75 mm

F M F rc (b) Critical Section

2. The critical section is the one that is along the horizontal centerline. There, the bending moment is Bending moment 3.


M  0.875  kN  m

Use the equation in the footnote of the text to calculate the radius of the neutral axis. Radius of neutral axis

4.

M  F  rc

FIGURE 4-37

rn 

ro  ri

rn  61.658 mm

 ro  ln   ri 

Calculate the eccentricty and the distances from the neutral axis to the extreme fibers. Eccentricity

e  rc  rn

e  0.8424 mm

Distances from neutral axis to extreme fibers

ci  rn  ri

ci  11.66  mm

co  ro  rn

co  13.34  mm

Stresses at inner and outer radii

σi 

M



ci

e A ri



F A

 M co  F    e A ro  A

σo  

σi  409.9  MPa

σo  273.2  MPa



4-38-1


Design a solid, straight, steel torsion bar to have a spring rate of 10 000 in-lb per radian per foot of length. Compare designs of solid round and solid square cross-sections. Which is more efficient in terms of material use?

Given:

Length of rod

L  12 in

Spring rate

k  10000 

Solution: 1.

Modulus of rigidity

6

G  11.7 10  psi

in lbf rad


Determine the rod diameter and volume for a round rod. Spring rate

k=

J G

J =

L

π d

4

32

1

2.

Rod diameter

d 

Volume of rod

V 

 32 L k   π G    π d 4

4

d  0.569  in

2

3

L

V  3.046  in

Determine the rod width and volume for a square rod using equation 4.26b and Table 4-6. Spring rate

k=

K G

K = 2.25 a

L

4

1

3.

 L k   2.25 G   

Rod half-width

a 

Volume of rod

V  ( 2  a )  L

2

4

a  0.260  in

2  a  0.520  in 3

V  3.241  in

Even though the square rod width is less than the round rod diameter, it takes slightly more material when a square rod is used than when a round rod is used. Thus, the round rod is more efficient.



4-39-1


Design a 1-ft-long steel, end-loaded cantilever spring for a spring rate of 10 000 lb/in. Compare designs of solid round and solid square cross-sections. Which is more efficient in terms of material use?

Given:

Length of rod

L  12 in

Spring rate

k  10000 

Solution: 1.

lbf in

Determine the rod diameter and volume for a round rod. k=

3  E I

I=

3

L

π d

4

64

1

Rod diameter

 64 L3 k   d    3  π E 

Volume of rod

V 

π d

4

d  1.406  in

2

4

3

L

V  18.64  in

Determine the rod width and volume for a square rod using equation 4.26b and Table 4-6. Spring rate

k=

3  E I 3

L

3.

6

E  30 10  psi

See Figure B-1(a) in Appendix B and Mathcad file P0439.

Spring rate

2.

Modulus of rigidity

Rod width

 4 L3 k   a    E 

Volume of rod

V  a  L

2

I= 1

a

4

12

4

a  1.232  in 3

V  18.215 in

It takes slightly more material when a round rod is used than when a square rod is used. Thus, the square rod is more efficient.



4-40-1


Redesign the roll support of Problem 4-8 to be like that shown in Figure P4-16. The stub mandrels insert to 10% of the roll length at each end. Choose appropriate dimensions a and b to fully utilize the mandrel's strength, which is the same as in Problem 4-27. See Problem 4-8 for additional data.

Given:


Roll density

OD  1.50 m

S y  100  MPa

Material properties

ID  0.22 m

S ys  50 MPa

Lroll  3.23 m

E  207  GPa

3

ρ  984  kg m

Assumptions: 1. The paper roll's weight creates a concentrated load acting at the tip of the mandrel. 2. The mandrel's root in the stanchion experiences a distributed load over its length of engagement Solution:


1. Model the support in such a way that stresses in the portion of the mandrel that is inside the stanchion can be determined. There are several assumptions that can be made about the loads on this portion of the mandrel. Figure 4-40A shows the one that will be used for this design.

w a x

2. Determine the weight of the roll, the load on each support, and the length of the mandrel. W 

π 4



2

2




F

y

Lm

b R

W  53.9 kN

FIGURE 4-40A Free Body Diagram used in Problem 4-40

F  0.5 W

F  26.95  kN


Lm  323  mm

3. From inspection of Figure 4-40A, write the load function equation q(x) = -w0 + w0 + R-1 - F-1 4. Integrate this equation from - to x to obtain shear, V(x) V(x) = -w1 + w1 + R0 - F0 5. Integrate this equation from - to x to obtain moment, M(x) M(x) = -(w/2)2 + (w/2)2 + R1 - F1 6. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b + Lm, where both are zero. At x = (b + Lm)+ , V = M = 0 0 = w  b  Lm  w  Lm  R  F

R = F  w b

w w 2 w 2 2 w 2 0 =    b  Lm   Lm  R Lm =    b  Lm   Lm  ( F  w b )  Lm 2 2 2 2

w=

2  F  Lm b

2

Note that R is inversely proportional to b and w is inversly proportional to b 2. 7. To see the value of x at which the shear and moment are maximum, let © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

MACHINE DESIGN - An Integrated Approach, 4th Ed. b  200  mm

w 

then

2  F  Lm b

4-40-2 R  F  w b

and

2

L  b  Lm

x  0  mm 0.002  L  L


9. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 10. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  w S ( x 0  mm)  x  w S ( x b )  ( x  b )  R S ( x b )  F  S ( x L) M ( x) 

w 2

w

2

 S ( x 0  mm)  x 

2

2

 S ( x b )  ( x  b )  R S ( x b )  ( x  b )  F  S ( x L)  ( x  L)


Moment Diagram

200

2

100

1

V ( x) kN

M ( x) 0

kN  m

 100

 200

4

7

0

100

200

300

400

500

 10

600

0

100

200

300

x

x

mm

mm

400

500

600

FIGURE 4-40B Shear and Moment Diagram Shapes for Problem 4-40

12. From Figure 4-40B, the maximum internal shear and moment occur at x = b and are Vmax =

2  F  Lm

Mmax  F  Lm

b

Mmax  8.704  kN  m

13. The bending stress will be a maximum at the top or bottom of the mandrel at a section through x = b.

σmax =

Mmax a 2 I

where

I=

π a 64

4

so,

σmax =

π a

1

 32 W  Lm   π S  y  

Solving for a,

a 

Round this to

a  125  mm

32 Mmax 3

= Sy

3

a  121.037  mm



4-40-3

14. Using this value of a and equation 4.15c, solve for the shear stress on the neutral axis at x = b.

τmax =

4  Vmax 3 A

=

8  F  Lm

 π a 2   b 3   4 

8  F  Lm

Solving for b

b 

Round this to

b  38 mm

= S ys

b  37.828 mm

 π a 2    Sys 3   4 

15. These are minimum values for a and b. Using them, check the bearing stress. Magnitude of distributed load

w 

2  F  Lm b

Bearing stress

σbear 

2

w b a b

w  12055 

N mm


Since this is less than S y, the design is acceptable for a  125  mm and b  38 mm



4-41-1


A 10-mm ID steel tube carries liquid at 7 MPa. Determine the principal stresses in the wall if its thickness is: a) 1 mm, b) 5 mm.

Given:

Tubing ID

Assumption:

The tubing is long therefore the axial stress is zero.

Solution:


(a) Wall thickness is

ID  10 mm

Inside pressure

p i  7  MPa

t  1  mm

1. Check wall thickness to radius ratio to see if this is a thick or thin wall problem. ratio 

t 0.5 ID

ratio  0.2

Since the ratio is greater than 0.1, this is a thick wall problem. 2. Using equations 4.48a and 4.48b, the stresses are maximum at the inside wall where Inside radius

ri  0.5 ID

ri  5  mm

Outside radius

ro  ri  t

ro  6  mm

Tangential stress

2  ro    σt   1 2 2  2 ri  ro  ri 

σt  38.82  MPa

Radial stress

σr 

2

ri  p i



2

ri  p i

1 

 ro  ri  2

2

ro ri

2



2

σr  7.00 MPa



3. Determine the principal stresses (since, for this choice of coordinates, the shear stress is zero),

σ1  σt

σ1  38.82  MPa

σ2  0  MPa σ3  σr

σ3  7.00 MPa

The maximum shear stress is

τmax 

σ1  σ3 2

τmax  22.91  MPa

t  5  mm

(b) Wall thickness is

1. Check wall thickness to radius ratio to see if this is a thick or thin wall problem. ratio 

t 0.5 ID

ratio  1

Since the ratio is greater than 0.1, this is a thick wall problem. 2. Using equations 4.48a and 4.48b, the stresses are maximum at the inside wall where Inside radius

ri  0.5 ID

ri  5  mm

Outside radius

ro  ri  t

ro  10 mm



4-41-2

Tangential stress

2  ro    σt   1 2 2  2 ri  ro  ri 

σt  11.67  MPa

Radial stress

2  ro    σr   1 2 2  2 ri  ro  ri 

σr  7.00 MPa

2

ri  p i

2

ri  p i

3. Determine the principal stresses (since, for this choice of coordinates, the shear stress is zero),

σ1  σt

σ1  11.67  MPa

σ2  0  MPa σ3  σr

σ3  7.00 MPa


τmax 

σ1  σ3 2




4-42-1


A cylindrical tank with hemispherical ends is required to hold 150 psi of pressurized air at room temperature. Find the principal stresses in the 1-mm-thick wall if the tank diameter is 0.5 m and its length is 1 m.

Given:

Tank ID Wall thickness Inside pressure

Solution: 1.

ID  500  mm t  1  mm p i  150  psi

p i  1034 kPa


Check wall thickness to radius ratio to see if this is a thick or thin wall problem. ratio 

3

t

ratio  4  10

0.5 ID

Since the ratio is less than 0.1, this is a thin wall problem. 2.

3.

Using equations 4.49a, 4.49b and 4.49c, the stresses are Radius

r  0.5 ID

Tangential stress

σt 

pi  r t

Radial stress

σr  0  MPa

Axial stress

σa 

pi  r 2 t

r  250  mm

σt  258.55 MPa σr  0.00 MPa σa  129.28 MPa

Determine the principal stresses (since, for this choice of coordinates, the shear stress is zero),

σ1  σt

σ1  259  MPa

σ1  37.5 ksi

σ2  σa

σ2  129  MPa

σ2  18.75  ksi

σ3  0  MPa

σ3  0.00 MPa

σ3  0.00 MPa


τmax 

σ1  σ3 2

τmax  129  MPa



4-43-1


Figure P4-17 shows an off-loading station at the end of a paper rolling machine. The finished paper rolls are 0.9-m OD by 0.22-m ID by 3.23-m long and have a density of 984 kg/m3. The rolls are transfered from the machine conveyor (not shown) to the forklift truck by the V-linkage of the off-load station, which is rotated through 90 deg by an air cylinder. The paper then rolls onto the waiting forks of the truck. The forks are 38-mm thick by 100-mm wide by 1.2-m long and are tipped at a 3-deg angle from the horizontal. Find the stresses in the two forks on the truck when the paper rolls onto it under two different conditions (state all assumptions): (a) The two forks are unsupported at their free end. (b) The two forks are contacting the table at point A.

Given:


OD  0.90 m ID  0.22 m Lroll  3.23 m

Roll density

ρ  984  kg m

Fork dimensions t  38 mm w  100  mm Lfork  1200 mm

3

θfork  3  deg

Assumptions: 1. The greatest bending moment will occur when the paper roll is at the tip of the fork for case (a) and when it is midway between supports for case (b). 2. Each fork carries 1/2 the weight of a paper roll. 3. For case (a), each fork acts as a cantilever beam (see Appendix B-1(a)). 4. For case (b), each fork acts as a beam that is built-in at one end and simply-supported at the other. Solution:

See Figure 4-43 and Mathcad file P0443. F

1. Determine the weight of the roll and the load on each fork.

π

W 

4



2

2




F  0.5 W

L fork

t

W  18.64  kN R1

F  9.32 kN

M1

Case (a), Cantilever Beam

2. The moment of inertia and the distance to the extreme fiber for a fork are I  c 

w t

3

12 t 2

5

I  4.573  10  mm

4

0.5 L fork t

c  19 mm

L fork R1

Case (a)

R2

M2

Case (b), Fixed-Simply Supported Beam

3. From Figure D-1(a), the moment is a maximum at the support and is Mmax  F  Lfork

F

Mmax  11.186 kN  m

4. The bending stress is maximum at the support and is

FIGURE 4-43A Free Body Diagrams used in Problem 4-43

σa 

Mmax c I

σa  464.8  MPa

Case (b) 5. This beam is statically indeterminate. However, using singularity functions and the method shown in Example 4-7, we can determine the reactions and find the maximum moment. 6. Calculate the distance from the left support to the load and the distance between supports.



a  0.5 Lfork

a  600  mm

L  Lfork

L  1200 mm

4-43-2

7. From inspection of Figure 4-43A, write the load function equation q(x) = R1-1 - F-1 + R2-1 + M2-2 8. Integrate this equation from - to x to obtain shear, V(x) V(x) = R10 - F0 + R20 + M2-1 9. Integrate this equation from - to x to obtain moment, M(x) M(x) = R11 - F1 + R21 + M20 10. Integrate the moment function, multiplying by 1/EI, to get the slope. (x) = [R12/2 - F2/2 + R22/2 + M21 + C3]/EI 11. Integrate again to get the deflection. y(x) = [R13/6 - F3/6 + R23/6 + M22/2 + C3x + C4]/EI 12. Evaluate R1, R2, M2, C3 and C4 At x = 0 and x = L; y = 0, therefore, C4 = 0. At x = L,  = 0 At x = L+, V = M = 0 Guess

R1  1  kN

R2  1  kN

M2  1  kN  m

2

C3  1  kN  m

Given 3

R 1 L 6



2

R 1 L 2



F  ( L  a)

3

6 F  ( L  a)

3

 C3 L = 0  kN  m

2

2

2

 C3 = 0  kN  m

R1  R2  F = 0  kN R1 L  F  ( L  a )  M2 = 0  kN  m

 R1  R   2   Find  R R M C  1 2 2 3  M2     C3  R1  2.913  kN 13. Define the range for x

R2  6.409  kN

M2  2.097  kN  m

2

C3  0.419  kN  m

x  0  in 0.002  L  L

14. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


4-43-3

15. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  R1 S ( x 0  in)  F  S ( x a )  R2 S ( x L) M ( x)  R1 S ( x 0  in)  x  F  S ( x a )  ( x  a )  R2 S ( x L)  ( x  L) 16. Plot the shear and moment diagrams. Shear Diagram

Moment Diagram

10

2 1

5 V ( x) kN

M ( x) 0

kN  m 1

5

 10

0

2

0

200

400

600

800

1000 1200

3

0

200

400

600

x

x

mm

mm

800

1000 1200


17. The maximum moment occurs at x = L,

Mmax  M ( L)

18. The bending stress is maximum at the support and is

Mmax  2.097  kN  m

σa 

Mmax c I

σa  87.2 MPa



4-44-1


Determine a suitable thickness for the V-links of the off-loading station of Figure P4-17 to limit their deflections at the tips to 10-mm in any position during their rotation. Two V-links support the roll, at the 1/4 and 3/4 points along the roll's length, and each of the V arms is 10-cm wide by 1-m long. The V arms are welded to a steel tube that is rotated by the air cylinder. See Problem 4-43 for more information.

Given:

Roll OD

OD  0.90 m

Arm width

wa  100  mm

Roll ID

ID  0.22 m

Arm length

La  1000 mm

Roll length

Lroll  3.23 m

Max tip deflection

δtip  10 mm

Roll density

ρ  984  kg m

Mod of elasticity

E  207  GPa

3

Assumptions: 1. The maximum deflection on an arm will occur just after it begins the transfer and just before it completes it, i.e., when the angle is either zero or 90 deg., but after the tip is no longer supported by the base unit. 2. At that time the roll is in contact with both arms ("seated" in the V) and will remain in that state throughout the motion. When the roll is in any other position on an arm the tip will be supported. 3. The arm can be treated as a cantilever beam with nonend load. 4. A single arm will never carry more than half the weight of a roll. 5. The pipe to which the arms are attached has OD = 160 mm. Solution:


1. Determine the weight of the roll and the load on each V-arm. W 

 4

π

2

2




450

W  18.64  kN

F  0.5 W

F  9.32 kN

2. From Appendix B, Figure B-1, the tip deflection of a cantilever beam with a concentrated load located at a distance a from the support is ymax =

F a

2

6  E I

 ( a  3  L)

1000 = L 370 = a

where L is the beam length and I is the cross-section moment of inertia. In this case M

3

I= 3. Setting

w a t a

F

12

ymax = δtip

F

FIGURE 4-44 a  370  mm

and

Free Body Diagram used in Problem 4-44

substituting for I and solving for ta 1

 2 F  a2  3 La  a  ta    E δtip  wa   Let the arm thickness be

3

ta  31.889 mm ta  32 mm



4-45-1


Determine the critical load on the air cylinder rod in Figure P4-17 if the crank arm that rotates it is 0.3 m long and the rod has a maximum extension of 0.5 m. The 25-mm-dia rod is solid steel with a yield strength of 400 MPa. State all assumptions.

Given:

Rod length Rod diameter

L  500  mm d  25 mm

E  207  GPa S y  400  MPa

Young's modulus Yield strength

Assumptions: 1. The rod is a fixed-pinned column. 2. Use a conservative value of 1 for the end factor (see Table 4-7 in text). Solution: 1.


Calculate the slenderness ratio that divides the unit load vs slenderness ratio graph into Johnson and Euler regions. S rD  π

2.

2 E

Calculate the cross-section area and the moment of inertia. Area Moment of inertia

3.

S rD  101.07

Sy

A 

π 2

I 

π

d

4 64

d

4

4

2

I  1.92  10  mm

4

Using Table 4-7, calculate the effective column length. Leff  1  L

4.

A  490.87 mm

Leff  500  mm

Calculate the slenderness ratio for the column. Radius of gyration

Slenderness ratio

k 

S r 

I A Leff k

k  6.25 mm

S r  80.00

Since the Sr for this column is less than SrD, it is a Johnson column. 5.

Calculate the critical load using the Johnson equation. 2  1  S y S r    Pcr  A  S y     E  2 π   

Pcr  134.8  kN



4-46-1


The V-links of Figure P4-17 are rotated by the crank arm through a shaft that is 60 mm dia by 3.23 m long. Determine the maximum torque applied to this shaft during motion of the V-linkage and find the maximum stress and deflection for the shaft. See Problem 4-43 for more information.

Given:


OD  900  mm ID  220  mm

Shaft dims

d  60 mm Lshaft  3230 mm

Lroll  3230 mm 3

ρ  984  kg m

Roll density

G  79 GPa

Modulus of rigidity

Assumptions: The greatest torque will occur when the link is horizontal and the paper roll is located as shown in Figure P4-17 or Figure 4-46. Solution:

See Figure 4-46 and Mathcad file P0446. y

1. Determine the weight of the roll on the V-arms.

 4

π

W 

2

2




W  18.64  kN 2. Summing moments about the shaft center, T 

OD 2

W

T  8.390  kN  m

3. Calculate the polar moment of inertia. J 

π d

4

W

6

J  1.272  10  mm

32

4

T

Ry

4. The maximum torsional stress will be at the outside diameter of the shaft. The radius of the OD is, r 

d

r  30 mm

2

60-mm-dia shaft

450.0

FIGURE 4-46 Free Body Diagram used in Problem 4-46

5. Determine the maximum torsional stress using equation (4.23b).

τmax 

Tr J

τmax  197.8  MPa

6. Use equation (4.24) to determine the angular shaft deflection.

θ 

T  Lshaft J G

θ  15.447 deg



4-47-1


Determine the maximum forces on the pins at each end of the air cylinder of Figure P4-17. Determine the stress in these pins if they are 30-mm dia and in single shear.

Given:


OD  0.90 m ID  0.22 m Lroll  3.23 m

Roll density

ρ  984  kg m

Pin diameter

d  30 mm

3

Assumptions: 1. The maximum force in the cylinder rod occurs when the transfer starts. 2. The cylinder and rod make an angle of 8 deg to the horizontal at the start of transfer. 3. The crank arm is 300 mm long and is 45 deg from vertical at the start of transfer. 4. The cylinder rod is fully retracted at the start of the transfer. At the end of the transfer it will have extended 500 mm from its initial position. Solution:


1. Determine the weight of the roll on the forks. W 

 4

π

2

2

y




W  18.64  kN 2. From the assumptions and Figure 4-47, the x and y distances from the origin to point A are, Rax  300  cos( 45 deg)  mm W

Ray  300  sin( 45 deg)  mm

Rx

Rax  212.132  mm

x 212.1

Ry

A

Ray  212.132  mm

F

8°

212.1 450.0

3. From Figure 4-47, the x distance from the origin to point where W is applied is,

FIGURE 4-47 Free Body Diagram at Start of Transfer for V-link of Problem 4-47

Rwx  4.

OD 2

Rwx  450  mm

Sum moments about the pivot point and solve for the compressive force in the cylinder rod. W  Rwx  F  Rax sin( 8  deg)  F  Ray cos( 8  deg) = 0 F 

W  Rwx Ray cos( 8  deg)  Rax sin( 8  deg)

F  46.469 kN

This is the shear force in the pins



4-47-2

Determine the cross-sectional area of the pins and the direct shear stress. Shear area

Shear stress

A 

τ 

π d 4 F A

2

A  706.858  mm

2

τ  65.7 MPa



4-48-1


A 100-kg wheelchair marathon racer wants an exerciser that will allow indoor practicing in any weather. The design shown in Figure P4-18 is proposed. Two free-turning rollers on bearings support the rear wheels. A platform supports the front wheels. Design the 1-m-long rollers as hollow tubes of aluminum to minimize the height of the platform and also limit the roller deflections to 1 mm in the worst case. The wheelchair has 650-mm-dia drive wheels separated by a 700-mm track width. The flanges shown on the rollers limit the lateral movement of the chair while exercising and thus the wheels can be anywhere between those flanges. Specify suitable sized steel axles to support the tubes on bearings. Calculate all significant stresses.

Given:

Mass of chair M  100  kg Wheel diameter d w  650  mm Track width

T  700  mm

Aluminum

Ea  71.7 GPa

Roller length

Lr  1000 mm

Steel

Es  207  GPa

Assumptions: 1. The CG of the chair with rider is sufficiently close to the rear wheel that all of the weight is taken by the two rear wheels. 2. The small camber angle of the rear wheels does not significantly affect the magnitude of the forces on the rollers. 3. Both the aluminum roller and the steel axle are simply supported. The steel axles that support the aluminum tube are fixed in the mounting block and do not rotate. The aluminum tube is attached to them by two bearings (one on each end of the tubes, one for each axle). The bearings' inner race is fixed, and the outer race rotates with the aluminum tube. Each steel axle is considered to be loaded as a simply supported beam. Their diameter must be less than the inner diameter of the tubes to fit the roller bearings between them. Solution:

δ  1  mm

Maximum deflection Modulus elasticity

F 



Free Body Diagram of One Wheel used in Problem 4-48

1. Calculate the weight of the chair with rider. W  M  g

F

FIGURE 4-48A


Weight of chair

W/2

W  980.7  N

2. Calculate the forces exerted by the wheels on the rollers (see Figure 4-48A). From the FBD of a wheel, summing vertical forces 2  F  cos( θ )  Let

3.

θ  20 deg

W 2

=0 then

F 

W 4  cos( θ )

F  260.9  N

The worst condition (highest moment at site of a stress concentration) will occur when the chair is all the way to the left or right. Looking at the plane through the roller that includes the forces exerted by the wheels (the FBD is shown in Figure 4-48B) the reactions R1 and R2 come from the bearings, which are inside the hollow roller and are, themselves, supported by the steel axle.

4. Solving for the reactions. Let the distance from R1 to F be a  15 mm


MACHINE DESIGN - An Integrated Approach, 4th Ed.  M1

R2 Lr  F  ( a  T )  F  a = 0

 Fy

R1  2  F  R2 = 0

R2 

F  (2 a  T )

4-48-2 700

F

F

R2  190.5  N

Lr

15

R1  2  F  R2

R2

R1

R1  331.3  N

1000

FIGURE 4-48B Free Body Diagram of One Tube used in Problem 4-48

5. The maximum bending moment will be at the right-hand load and will be Mrmax  R2 Lr  ( a  T )

Mrmax  54.3 N  m

Note, if the chair were centered on the roller the maximum moment would be Mc  F 

Lr  T

Mc  39.1 N  m

2

and this would be constant along the axle between the two loads, F. 6. Note that the bearing positions are fixed regardless of the position of the chair on the roller. Because of symmetry, Ra1  R1

Ra1  331.3  N

Ra2  R2

Ra2  190.5  N

1000

65 R1

7. The maximum bending moment occurs at R1 and is for b  65 mm Mamax  Ra1 b

R2

R a1

R a2 1130

FIGURE 4-48C Free Body Diagram of One Axle used in Problem 4-48

Mamax  21.5 N  m

8. Determine a suitable axle diameter. Let the factor of safety against yielding in the axle be Nsa  3 9. Tentatively choose a low-carbon steel for the axle, say AISI 1020, cold rolled with S y  393  MPa 10. At the top of the axle under the load R1 there is only a bending stress. Set this stress equal to the yield strength divided by the factor of safety.

σx =

32 Mamax 3

π d a

=

Sy Nsa 1

 32 Nsa Mamax    π S y  

Solving for the axle diameter, d a

d a 

Let the axle diameter be

d a  15 mm

3

d a  11.875 mm

made from cold-rolled AISI 1020 steel.


MACHINE DESIGN - An Integrated Approach, 4th Ed. 11. Suppose that bearing 6302 from Chapter 10, Figure 10-23. It has a bore of 15 mm and an OD of 42 mm. Thus, the inside diameter of the roller away from the bearings where the moment is a maximum will be d i  40 mm. This will provide a 1-mm shoulder for axial location of the bearings.

4-48-3

150

700 F

F

F 15

12. The maximum deflection of the roller will occur when the chair is in the center of the roller. For this case the reactions are both equal to the loads, F (see Figure 4-48D). The maximum deflection is at the center of the roller.

F 1000

FIGURE 4-48D Free Body Diagram of Roller with Chair in the Center.

13. Write the load function and then integrate four times to get the deflection function. q(x) = F-1 - F-1 - F-1 + F-1 y(x) = F[3 - 3 - 3 + 3 + C3x]/(6EI) where

C3 =

1 L

 ( L  a )  a  L  3

3

3

14. Write the deflection function at x = L/2 for a  150  mm ymax

 L  3 =     6  Ea I  2  F

3  L  a  1  ( L  a) 3  a3  L3 2  2   

15. Set this equation equal to the allowed deflection  and solve for the required moment of inertia, I.

 Lr  3 I      6  Ea δ  2  F

3   Lr  1  3 3 3   a    Lr  a   a  Lr  2 2  

4

I  6.618  10  mm

4

16. Knowing the inside diameter of the tube, solve for the outside diameter. 1

π  4 4 I=   d o  d i  64 Round this up to

64 I 4 d o    d i   π 

4

d o  44.463 mm

d o  46 mm

DESIGN SUMMARY Axles

Rollers

Material

AISI 1020 steel, cold-rolled

Material

2024-T4 aluminum

Diameter

d a  15 mm

Outside diameter

d o  46 mm

Length

1220 mm

Inside diameter

d i  40 mm

Length

1040 mm

Spacing

c   d w  d o  sin( θ ) c  238  mm



4-49a-1


A hollow, square column has the dimensions and properties below. Determine if it is a Johnson or an Euler column and find the critical load: (a) If its boundary conditions are pinned-pinned. (b) If its boundary conditions are fixed-pinned. (c) If its boundary conditions are fixed-fixed. (d) If its boundary conditions are fixed-free.

Given:

Length of column Outside dimension

Material L  100  mm Yield strength so  4  mm

Steel S y  300  MPa

Inside dimension

si  3  mm

E  207  GPa

Solution: 1.



2.


2 E

S rD  116.7

Sy

Calculate the cross-section area and the moment of inertia. 2

Area

A  so  si

Moment of inertia

I 

1 12



2

4

 so  si

A  7.00 mm



4

2

I  14.58  mm

4

(a) pinned-pinned ends 3.


4.

Leff  100  mm


k 

S r 

Slenderness ratio

I

k  1.443  mm

A Leff k

S r  69.28

Since the S r for this column is less than S rD, it is a Johnson column. 5.

Calculate the critical load using the Johnson equation.



Pcr  A  S y 



2  Sy S r    E  2 π  

1



Pcr  1.73 kN

(b) fixed-pinned ends 6.

Using Table 4-7, calculate the effective column length. Leff  0.8 L

7.

Leff  80 mm


k 

I A

k  1.443  mm


MACHINE DESIGN - An Integrated Approach, 4th Ed. Slenderness ratio

S r 

Leff k

4-49a-2 S r  55.43

Since the S r for this column is less than S rD, it is a Johnson column. 8.

Calculate the critical load using the Johnson equation. 2  1  S y S r    Pcr  A  S y     E  2 π   

Pcr  1.86 kN

(c) fixed-fixed ends 9.


Leff  65 mm

10. Calculate the slenderness ratio for the column. Radius of gyration

Slenderness ratio

k  S r 

I A Leff k

k  1.443  mm S r  45.03

Since the S r for this column is less than S rD, it is a Johnson column. 11. Calculate the critical load using the Johnson equation. 2  S y S r   1  Pcr  A  S y     E  2 π   

Pcr  1.94 kN

(d) fixed-free ends 12. Using Table 4-7, calculate the effective column length. Leff  2.1 L

Leff  210  mm

13. Calculate the slenderness ratio for the column. Radius of gyration

Slenderness ratio

k  S r 

I A Leff k

k  1.443  mm S r  145.49

Since the S r for this column is greater than S rD, it is an Euler column. 14. Calculate the critical load using the Euler equation. 2

Pcr  A 

π E Sr

2

Pcr  676  N



4-50a-1


A hollow, round column has the dimensions and properties below. Determine if it is a Johnson or an Euler column and find the critical load: (a) If its boundary conditions are pinned-pinned. (b) If its boundary conditions are fixed-pinned. (c) If its boundary conditions are fixed-fixed. (d) If its boundary conditions are fixed-free.

Given:

Length of column Outside diameter

L  1500 mm Material Yield strength od  20 mm


Inside diameter

id  14 mm

E  207  GPa

Solution: 1.



2.

3.


2 E

S rD  116.7

Sy

Calculate the cross-section area, moment of inertia, and the radius of gyration. Area

A 

Moment of inertia

I 

Radius of gyration

k 

 4

2

2

 64

4

4

π

π

 od  id  od  id



A  160.22 mm



I  5968 mm

I

2

4

k  6.103  mm

A

Define functions to determine column type and critical load. Type

type S r 

"Euler" if S r  S rD "Johnson" otherwise

Critical load

Pcr S r 

2

return A 

π E Sr

2

if type S r = "Euler"

2  1  S y S r    A  Sy     otherwise E  2 π   



5.

Calculate the slenderness ratio for the column. Slenderness ratio

6.

Leff  1500 mm

S r 

Leff k

S r  245.77

Determine the type and critical load using the functions defined above. type S r  "Euler"

Pcr S r  5.42 kN



4-50a-2



8.


9.

Leff  1200 mm

S r 

Leff k

S r  196.62

Determine the type and critical load using the functions defined above. type S r  "Euler"

Pcr S r  8.47 kN

(c) fixed-fixed ends 10. Using Table 4-7, calculate the effective column length. Leff  0.65 L

Leff  975  mm

11. Calculate the slenderness ratio for the column. Slenderness ratio

S r 

Leff k

S r  159.75

12. Determine the type and critical load using the functions defined above. type S r  "Euler"

Pcr S r  12.8 kN


Leff  3150 mm


S r 

Leff k

S r  516.12

15. Determine the type and critical load using the functions defined above. type S r  "Euler"

Pcr S r  1.23 kN



4-51a-1


A solid, rectangular column has the dimensions and properties below. Determine if it is a Johnson or an Euler column and find the critical load: (a) If its boundary conditions are pinned-pinned. (b) If its boundary conditions are fixed-pinned. (c) If its boundary conditions are fixed-fixed. (d) If its boundary conditions are fixed-free.

Given:

Length of col. Thickness

L  100  mm t  10 mm

Material Yield strength


Height

h  20 mm


E  207  GPa

Solution: 1.



2.

2 E


A  h  t

Moment of inertia

I 

Radius of gyration

3.

S rD  116.7

Sy

k 

h t

A  200.00 mm

3

I  1667 mm

12 I

2

4

k  2.887  mm

A




Critical load


2

return A 

π E Sr



A  S y 



2


2  S y Sr    otherwise E  2 π  

1





5.


6.

Leff  100  mm

S r 

Leff k

S r  34.64

Determine the type and critical load using the functions defined above. type S r  "Johnson"

Pcr S r  57.36  kN



4-51a-2



8.


9.

Leff  80 mm

S r 

Leff k

S r  27.71

Determine the type and critical load using the functions defined above. type S r  "Johnson"

Pcr S r  58.31  kN

(c) fixed-fixed ends 10. Using Table 4-7, calculate the effective column length. Leff  0.65 L

Leff  65 mm


S r 

Leff k

S r  22.52

12. Determine the type and critical load using the functions defined above. type S r  "Johnson"

Pcr S r  58.9 kN


Leff  210  mm


S r 

Leff k

S r  72.75

14. Determine the type and critical load using the functions defined above. type S r  "Johnson"

Pcr S r  48.34  kN



4-52a-1


A solid, circular column, loaded eccentrically, has the dimensions and properties below. Find the critical load: (a) If its boundary conditions are pinned-pinned. (b) If its boundary conditions are fixed-pinned. (c) If its boundary conditions are fixed-fixed. (d) If its boundary conditions are fixed-free.

Given:

Length of column Outside diameter

L  100  mm od  20 mm

Material Yield strength


Eccentricity (t)

e  10 mm


E  207  GPa

Solution: 1.


Calculate the cross-section area, distance to extreme fiber, and the moment of inertia. A 

Area

4.

π 4

2

 od


c  0.5 od

Moment of inertia

I 

π

4

64

 od

A  314.16 mm

2

c  10 mm I  7854 mm

4

Calculate the radius of gyration and eccentricity ratio for the column. k 

Radius of gyration

Er 

Eccentricity ratio

I A e c k

2

k  5.00 mm Er  4.0



4.

Leff  100  mm

Calculate the slenderness ratio for the column. S r 

Slenderness ratio 5.

Leff k

S r  20.00

Calculate the critical load using the Secant equation. Guess

P  1  kN

Given

S y A

P=

 

  4  E A  P

1  Er sec S r Pcr  Find ( P)

Pcr  18.63  kN



7.

Leff  80 mm

Calculate the slenderness ratio for the column.


MACHINE DESIGN - An Integrated Approach, 4th Ed. S r 

Slenderness ratio 8.

Leff k

4-52a-2 S r  16.00

Calculate the critical load using the Secant equation. Guess

P  1  kN

Given

S y A

P=

 

 

P

1  Er sec S r

4  E A 

Pcr  Find ( P)

Pcr  18.71  kN



Leff  65 mm

10. Calculate the slenderness ratio for the column. S r 

Slenderness ratio

Leff k

S r  13.00

11. Calculate the critical load using the Secant equation. Guess

P  1  kN

Given

S y A

P=

  4  E A 

 

P


Pcr  18.76  kN


Leff  210  mm

13. Calculate the slenderness ratio for the column. S r 

Slenderness ratio

Leff k

S r  42

14. Calculate the critical load using the Secant equation. Guess

P  1  kN

Given

S y A

P=

 


  4  E A  P

Pcr  17.93  kN



4-53-1


Design an aluminum, hollow, circular column for the conditions given below for (a) pinned-pinned ends and (b) fixed-free ends.

Given:

Length of column Wall thickness

Factor of safety L  3  m Yield strength t  5  mm

FS  3 S yc  150  MPa

Load supported

F  900  N

E  71.7 GPa

Solution: 1.


Start by calculating the slenderness ratio that divides the unit load vs slenderness ratio graph into Johnson and Euler regions. 2 E

S rD  π

2.


S rD  97.136

S yc


3.

Leff  3000 mm

To start the iterative process, assume that the final design will be an Euler column with the critical load equal to FS*F. From equation 4.38b, 2

Pcr =

π  E A  k

2

I

2

and

k =

2

A

L

2

Substituting for k2

Pcr =

π  E I 2

= FS  F

L 2

I 

Solving for I

Leff  FS  F 2

π E The required moment of inertia, assuming an Euler column is I  34339  mm 4.

Using the relationships given on the inside cover, solve for the outside diameter of the tube. Guess

D  20 mm

Given

I=

π  4 4  D  ( D  2  t) 

64

D  Find ( D) 5.

4

D  30.64  mm

Using this diameter, calculate the slenderness ratio and compare to S rD. If it is greater than S rD the assumption o an Euler column is correct, if not, recalculate using the Johnson equation. Inside diameter

d  D  2  t

Area

Ar 

Radius of gyration

kr 

Slenderness ratio

S r 

 4

π

2

 D d

I Ar Leff kr

d  20.64  mm



2

Ar  402.7  mm

2

kr  9.234  mm

S r  324.9



4-53-2

Since this is greater than S rD, the assumption of an Euler column is correct and the minimum outside diameter is D  30.64  mm (b) fixed-free ends 6.


7.

Leff  6300 mm

To start the iterative process, assume that the final design will be an Euler column with the critical load equal to FS*F. From equation 4.38b, 2

Pcr =

π  E A  k

2

I

2

and

k =

2

A

L

2

Substituting for k2

Pcr =

π  E I 2

= FS  F

L 2

I 

Solving for I

Leff  FS  F 2

π E 5

The required moment of inertia, assuming an Euler column is I  2  10  mm 8.

Using the relationships given on the inside cover, solve for the outside diameter of the tube. Guess

D  20 mm

Given

I=

π  4 4  D  ( D  2  t) 

64

D  Find ( D) 9.

4

D  47.37  mm

Using this diameter, calculate the slenderness ratio and compare to S rD. If it is greater than S rD the assumption of an Euler column is correct, if not, recalculate using the Johnson equation. Inside diameter

d  D  2  t

Area

Ar 

Radius of gyration

kr 

Slenderness ratio

S r 

 4

π

2

 D d

I Ar Leff kr

d  37.37  mm



2

Ar  665.6  mm

2

kr  15.084 mm

S r  417.7

Since this is greater than S rD, the assumption of an Euler column is correct and the minimum outside diameter is D  47.37  mm © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


4-54-1


Three round, 1.25-in-dia bars are made of SAE 1030 hot-rolled steel but are of different lengths, 5 in, 30 in, and 60 in, respectively. They are loaded axially in compression. Compare the load supporting capability of the three bars if the ends are assumed to be: (a) Pinned-pinned. (b) Fixed-pinned. (c) Fixed-fixed. (d) Fixed-free.

Given:

Outside diameter Lengths Material

Solution: 1.

2

SAE 1030 Steel

L  60 in

i  1 2  3

3

Yield strength

S y  38 ksi


E  30 10  psi

6

Calculate the slenderness ratio that divides the unit load vs. slenderness ratio graph into Johnson and Euler regions. 2 E

S rD  124.8

Sy


Moment of inertia

Radius of gyration 3.

L  30 in

1


S rD  π 2.

d  1.25 in L  5  in

A 

π 2

I 

π

k 

d

4

64

d

A  791.73 mm

4

I  49882  mm

I

2

4

k  7.938  mm

A




Critical load


2

return A 

π E Sr

2


2  S y S r   1  A  S y      otherwise E  2 π   


Using Table 4-7, calculate the effective column length.

Leff  1  L 5.

5 Leff   30   in    60 




S r 

Slenderness ratio

6.

4-54-2

 16  S r   96     192 

Leff k

Determine the type and critical load using the functions defined above.

 "Johnson"  Type   "Johnson"  i    "Euler" 

 

Type  type S r i

i

 

Pcr S r

i

lbf

46250 32844 9857



 4.0  Leff   24.0   in    48.0 

Leff  0.8 L

8.


Slenderness ratio

9.

S r 

 12.8  S r   76.8     153.6 

Leff k


 


i

 "Johnson"  Type   "Johnson"  i    "Euler" 

 

Pcr S r

i

lbf

46388 37808 15401



Leff  0.65 L

11.

 3.3  Leff   19.5   in    39.0 


Slenderness ratio

S r 

Leff k

 10.4  S r   62.4     124.8 



12.

4-54-3


 


i

 "Johnson"  Type   "Johnson"  i    "Johnson" 

 

Pcr S r

i

lbf

46471 40807 23329

(d) fixed-free ends 13.


 10.5  Leff   63.0   in    126.0 

Leff  2.1 L

14.


Slenderness ratio

15.

S r 

Leff k

 33.6  S r   201.6     403.2 


 


i

 "Johnson"  Type   "Euler"  i    "Euler" 

 

Pcr S r

i

lbf

44944 8940 2235



4-55-1

PROBLEM 4-55

_____

Statement:

Figure P4-19 shows a 1.5-in-dia, 30-in-long steel rod subjected to tensile loads P = 10000 lb applied at each end of the rod, acting along its longitudinal Y axis and through the centroid of its circular cross section. Point A is 12 in below the upper end and point B is 8 in below A. For this bar with its loading, find: (a) All components of the stress tensor matrix (equation 4.1a) for a point midway between A and B. (b) The displacement of point B relative to point A. (c) The elastic strain in the section between A and B. (d) The total strain in the section between A and B.

Given:

Tensile load

P  10000  lbf

Diameter Lengths

Modulus of elasticity d  1.50 in L  30 in LA  12 in LAB  8  in

Solution: 1.


Calculate the cross-section area of the rod. A 

2.

6

E  30 10  psi

π d

2

2

A  1.767  in

4

(a) The loading is simple axial tension so all components of the stress tensor are zero except yy, which is found using equation 4.7.

σyy 

P

σyy  5659 psi

A

This stress is uniform across the rod and has the same value at any cross section along the longitudinal axis except close to the ends where the load is applied. 3.

(b) The displacement of point B relative to A can be found using equation 4.8.

ΔsBA  4.

A E

ΔsBA  1.509  10

3

 in

(c) The elastic strain in the rod can be found using Hooke's law (equation 2.2)

ε 

5.

P LAB

σyy E

ε  1.886  10

4

(d) Assuming that the yield strength of this steel is greater than yy, the strain calculated in step 4 is the total strain.



4-56-1

PROBLEM 4-56

_____

Statement:

The rod in Figure P4-19, with the loading of Problem 4-55, is subjected to a reduction of temperature from 80F to 20F after the load is applied. The coefficient of thermal expansion for steel is approximately 6 in/in/degF. Find: (a) All components of the stress tensor matrix (equation 4.1a) for a point midway between A and B. (b) The displacement of point B relative to point A. (c) The elastic strain in the section between A and B. (d) The total strain in the section between A and B.

Units:

Temperature scale F  1

Given:

Tensile load

P  10000  lbf

Diameter Lengths

Modulus of elasticity d  1.50 in L  30 in LA  12 in LAB  8  in

Temperatures

T1  80 F

Solution: 1.

Coefficient of thermal expansion See Mathcad file P0456.

T2  20 F

α  6  10

6

F

1

Calculate the cross-section area of the rod. A 

2.

6

E  30 10  psi

π d

2

2

A  1.767  in

4

(a) The loading is simple axial tension so all components of the stress tensor are zero except yy, which is found using equation 4.7.

σyy 

P

σyy  5659 psi

A

This stress is uniform across the rod and has the same value at any cross section along the longitudinal axis except close to the ends where the load is applied. The change in temperature does not affect the stress since the ends are free. 3.

(b) The displacement of point B relative to A can be found by summing equation 4.8 for the elastic portion and the thermal expansion equation from elementary mechanics of materials for the thermal portion.

ΔsBA  4.

A E

 α  T2  T1  LAB

3

ΔsBA  1.371  10

 in

(c) The elastic strain in the rod can be found using Hooke's law (equation 2.2)

ε  5.

P LAB

σyy E

ε  1.886  10

4

(d) Assuming that there is no plastic strain in the rod, the total strain is the sum of the elastic strain found in step 4 plus the thermal strain.

ε tot  ε  α  T2  T1

ε tot  1.714  10

4



4-57-1

PROBLEM 4-57

_____

Statement:

Figure P4-20 shows a steel bar fastened to a rigid ground plane with two 0.25-in-dia hardened steel dowel pins. For P = 1500 lb, find: (a) The shear stress in each pin. (b) The direct bearing stress in each pin and hole. (c) The minimum value of dimension h to prevent tearout failure if the steel bar has a shear strength of 32500 psi.

Given:

Pin diameter Distance between pins

d  0.250  in a  2.0 in

Thickness of bar

t  0.25 in

Solution: 1.

Applied load Shear strength of bar Distance from right pin to load

P  1500 lbf S s  32.5 ksi b  4.0 in


Draw a free-body diagram and find the shear forces (reactions) on each pin.

a

b

RL

h

RR P Write equations 3.3b for the bar and solve for the reactions.

 F: RL  2.

b a

P

π d

2

RR  P  RL

RR  4500 lbf

4

2

A  0.0491 in

(a) Use equation 4.9 to determine the shear stress in each pin. Left pin

Right pin 4.

RL  3000 lbf

RL  a  P b  0

Calculate the cross-section area of a pin. A 

3.

 M:

RL  RR  P  0

τL  τR 

RL

τL  61.1 ksi

A RR

τR  91.7 ksi

A

(b) Calculate the bearing area from equation 4.10 and use it to determine the bearing stress in each pin. Bearing area

Abear  d  t

σL 

RL Abear

2

Abear  0.0625 in

σL  48.0 ksi



σR 

5.

RR Abear

4-57-2

σR  72.0 ksi

h  d

   t , where (h - d)/2 is the distance from the edge of the hole to 2    the outside of the bar. Substitute this area in equation 4.9 for the shear area and substitute the shear strength for xy, solving then for the unknown distance h. (c) The tearout area is

Atear  2  

Left pin

h L 

Right pin

h R 

Minimum value of h

RL S s t RR S s t

d

h L  0.619  in

d

h R  0.804  in

h min  h R

h min  0.804  in



4-58-1

PROBLEM 4-58

_____

Statement:

Figure P4-20 shows a steel bar fastened to a rigid ground plane with two 0.25-in-dia hardened steel dowel pins. For P = 2200 lb, find: (a) The shear stress in each pin. (b) The direct bearing stress in each pin and hole. (c) The minimum value of dimension h to prevent tearout failure if the steel bar has a shear strength of 32500 psi.

Given:

Pin diameter Distance between pins

d  0.250  in a  2.0 in

Thickness of bar

t  0.25 in

Solution: 1.

Applied load Shear strength of bar Distance from right pin to load

P  2200 lbf S s  32.5 ksi b  4.0 in



a

b

RL

h


 F: RL  2.

b a

P

π d

2

4

RR  P  RL

RR  6600 lbf

2

A  0.0491 in

(a) Use equation 4.9 to determine the shear stress in each pin. Left pin

Right pin 4.

RL  4400 lbf

RL  a  P b  0


3.

 M:

RL  RR  P  0

τL  τR 

RL A RR A

τL  89.6 ksi τR  134.5  ksi

(b) Calculate the bearing area from equation 4.10 and use it to determine the bearing stress in each pin. Bearing area

Abear  d  t

2

Abear  0.0625 in



σL  σR 

5.

(c) The tearout area is

Atear  2  

RL Abear RR Abear

4-58-2

σL  70.4 ksi σR  105.6  ksi

h  d

   t , where (h - d)/2 is the distance from the edge of the hole to  2  

the outside of the bar. Substitute this area in equation 4.9 for the shear area and substitute the shear strength for xy, solving then for the unknown distance h. Left pin

h L 

Right pin

h R 

Minimum value of h

RL S s t RR S s t

d

h L  0.792  in

d

h R  1.062  in

h min  h R

h min  1.062  in



4-59-1

PROBLEM 4-59

_____

Statement:

Figure P4-21 shows a rectangular section aluminum bar subjected to off-center forces P = 4000 N applied as shown. (a) Solve for the maximum normal stress in the mid-region of the bar well away from the eyes where the loads are applied. (b) Plot the normal stress distribution across the cross section at this mid-region. (c) Sketch a "reasonable" plot of the normal stress distribution across the cross section at the ends, close to the applied loads.

Given:

Depth of bar h  40 mm Thickness of bar t  10 mm

Solution:


1.

Applied loads Location of eye

P  4000 N d  35 mm (from bottom edge)

Draw a free-body diagram of the bar, cut at any section along the length of the bar. FACE OF CUT SURFACE P d

P

M

0.5h

h

SECTION CENTROIDAL AXIS

Equilibrium requires that there be a force directed along the centroidal axis of the cross section that is equal and opposite to the applied force and a bending moment to react the couple formed by the applied force and the reaction force. Thus, since the reaction moment is clockwise, M  ( d  0.5 h )  P 2.

Calculate the cross-section area, moment of inertia, and distance from the centroid to the outer surface. A  h  t I 

t h

A  400.0  mm

3

4

12

4

c  20.000 mm

(a) The normal stress on a section well away from the ends is a combination of uniform tension, as given by equation 4.7, and bending, as given by equation 4.11a.

σ ( y )  

M y I



P A

This will be a maximum at y = c. σmax  σ ( c) 3.

2

I  5.333  10  mm

c  0.5 h 2.

M  60.000 N  m


(b) Plot the normal stress distribution across the cross section at the mid-region of the bar for y  c c  1  mm  c



4-59-2

NORMAL STRESS ON SECTION 40

Stress, MPa

30 20

σ( y ) MPa

10 0

 10  20  20

 10

0

10

20

30

y mm Distance from neutral axis, mm

4.

(c) Sketch a "reasonable" plot of the normal stress distribution across the cross section at the ends, close to the applied loads. Use the "force flow" analogy show in Figures 4-37 and 4-38 as a guide to the stress distribution. Near the applied load the stress will be highly concentrated. As the distance from the point of load application increases the stress will become more evenly distributed.



4-60-1

PROBLEM 4-60

_____

Statement:

Figure P4-22 shows a bracket machined from 0.5-in-thick steel flat stock. It is rigidly attached to a support and loaded with P = 5000 lb at point D. Find: (a) The magnitude, location, and the plane orientation of the maximum normal stress at section A-A. (b) The magnitude, location, and the plane orientation of the maximum shear stress at section A-A. (c) The magnitude, location, and the plane orientation of the maximum normal stress at section B-B. (d) The magnitude, location, and the plane orientation of the maximum shear stress at section B-B.

Given:

Distance from support to: Section A-A Point D d  8  in Depth of section h  3  in Applied load P  5000 lbf

Centroid of B-B b  18.5 in a  10 in Thickness of section t  0.5 in

Assumptions: The bracket remains flat and does not buckle (out-of-plane) under the applied load. Solution: 1.


Calculate the cross-section area and moment of inertia at sections A-A and B-B, which are the same. 2

A  h  t 2.

A  1.500  in

I 

t h

3

12

4

I  1.1250 in

For parts (a) and (b), draw a free-body diagram of the portion of the bracket that is to the right of section A-A.

V a

A M

y

h

A

d

x

D

P 3.

Use the equilibrium equations 3.3a to calculate the shear force and bending moment on section A-A.

 F:

V  P 4.

 M:

V  P  0 V  5000 lbf

P ( a  d )  M  0 M  P ( a  d )

M  10000  in lbf

(a) The maximum normal stress in the bracket at section A-A is determined using equation 4.11b. It is located a the bottom of the section and is oriented in the positive x direction, i.e., it is tensile. Distance from neutral axis to extreme fiber

c  0.5 h

c  1.500  in



σmax 

Maximum normal stress 5.

4-60-2

M c

σmax  13.33  ksi

I

(b) The maximum shear stress in the bracket at section A-A is either at the neutral axis (due to the transverse shear, which is a maximum at the NA) or it is at the top or bottom of the section (due to the bending stress at those points, which is numerically the same).

τmax 

At the neutral axis, using equation 4.14b

3 V  2 A

τmax  5.000  ksi

At the bottom edge the stress state is: σx  σmax, σy  0  ksi, τxy  0  ksi. Using equation 4.6a, the principal stresses are

σ1 

σ2 

σx  σy 2

σx  σy 2

2  σx  σy  2     τxy  2 

σ1  13.333 ksi

2  σx  σy  2     τxy  2 

σ2  0.000  ksi

σ3  0  ksi And, from equation 4.6b, the maximum shear stress is

τmax 

σ1  σ3

τmax  6.667  ksi

2

As seen from the Mohr's Circle in Figure 4-8, this stress is oriented 45 degrees from the positive x axis. 6.

For parts (c) and (d), draw a free-body diagram of the portion of the bracket that is below section B-B.

b

y

F M

B

d

B

x

h

D

P

7.

Use the equilibrium equations 3.3a to calculate the normal force and bending moment on section B-B.

 F:

F  P  0

 M:

P ( b  d )  M  0


MACHINE DESIGN - An Integrated Approach, 4th Ed. F  P 8.

4-60-3

F  5000 lbf

M  P ( b  d )

(c) The maximum normal stress in the bracket at section B-B is a combination of uniform tension and bending and is determined by summing equations 4.7 and 4.11b. It is located at the left edge of the section and is oriented in the positive y direction, i.e., it is tensile. c  0.5 h

Distance from neutral axis to extreme fiber Maximum normal stress 9.

M  52500  in lbf

σmax 

M c I



F A

c  1.500  in

σmax  73.33  ksi

(d) The maximum shear stress in the bracket at section B-B is at the left edge of the section (due to the combined tensile and bending stresses). Since there is no transverse shear on this section, the shear stress at the neutral axis is zero. At the left edge the stress state is: σx  0  ksi, σy  σmax, τxy  0  ksi. Using equation 4.6a, the principal stresses are

σ1 

σ2 

σx  σy 2

σx  σy 2

2  σx  σy  2     τxy 2  

σ1  73.333 ksi

2  σx  σy  2     τxy 2  

σ2  0.000  ksi

σ3  0  ksi And, from equation 4.6b, the maximum shear stress is

τmax 

σ1  σ3 2

τmax  36.667 ksi

As seen from the Mohr's Circle in Figure 4-8, this stress is oriented 45 degrees from the positive x axis.



4-61-1

PROBLEM 4-61

_____

Statement:

For the bracket of Problem 4-60, solve for the deflection of point C.

Given:

Distance from support to: Point D d  8  in Depth of section h  3  in Applied load

Point C a  18.5 in Thickness of section t  0.5 in

P  5000 lbf

6

E  30 10  psi


Assumptions: 1. The bracket remains flat and does not buckle (out-of-plane) under the applied load. 2. The bracket can be modeled using its centroidal axis length dimensions. Solution: 1.


Calculate the moment of inertia along the segment AC. I 

2.

t h

3

4

I  1.1250 in

12

Draw idealized free-body diagrams of the portions of the bracket from the support to point C and from point C to point D.

y

P a

MA

MC

C

x

A P d P

MC

C

D P 3.

Calculate the magnitude of the moments on segment AC using equilibrium equation 3.3a. MC  P ( a  d )

MC  52500  in lbf

MA  P d

MA  40000  in lbf



4-61-2

From inspection of the FBD, write the load function equation q(x) = -MA-2 + P-1 - P-1 - MC-2

5.

Integrate this equation from - to x to obtain shear, V(x) V(x) = -MA-1 + P0 - P0 - MC-1

6.

Integrate this equation from - to x to obtain moment, M(x) M(x) = -MA0 + P1 - P1 - MC0

7.

Integrate the moment function, multiplying by 1/EI, to get the slope. (x) = [-MA1 + P2/2 - P2/2 - MC1 + C3]/EI

8.

Integrate again to get the deflection. y(x) = [-MA2/2 + P3/6 - P3/6 - MC2/2 + C3x +C4]/EI

9.

Evaluate C3 and C4. At x = 0,  = 0 and y = 0, therefore, C3 = 0 and C4 = 0.

10. Evaluate  and y at x = a using the equations in steps 7 and 8, respectively.

θC 

yC 

1 E I

  MA a 



P 2

a

2

 

 MA 2 P 3 a  a  E I  2 6  1



θC  0.196  deg

yC  0.0465 in



4-62-1

PROBLEM 4-62

_____

Statement:

Figure P4-23 shows a 1-in-dia steel bar supported and subjected to the applied load P = 500 lb. Solve for the deflection at the load and the slope at the roller support.

Given:

Diameter Applied load

Solution:


1.

6

Modulus of elasticity d  1.00 in Dimensions: P  500  lbf a  20 in

E  30 10  psi L  40 in

Draw a free-body diagram. L a R2

M1 R1

2.

This is a statically indeterminate beam because there are three unknown reactions, R1, M1, and R2. To solve for these unknowns, follow the method presented in Example 4-7. First, calculate the moment of inertia for the round section. I 

3.

P

π d

4

4

I  0.0491 in

64

From inspection of the FBD, write the load function equation q(x) = -M1-2 + R1-1 - R2-1 + P-1

4.

Integrate this equation from - to x to obtain shear, V(x) V(x) = -M1-1 + R10 - R20 + P0

5.

Integrate this equation from - to x to obtain moment, M(x) M(x) = -M10 + R11 - R21 + P1

6.

Integrate the moment function, multiplying by 1/EI, to get the slope. (x) = [ -M11 + R12/2 - R22/2 + P2/2 + C3]/EI

7.

Integrate again to get the deflection. y(x) = [-M12/2 + R13/6 - R23/6 + P3/6 + C3x + C4]/EI

8.

Evaluate R1, M1, R2, C3 and C4 At x = 0, y = 0 and  = 0, therefore, C3 = 0 and C4 = 0. At x = a, y = 0 At x = L+, V = M = 0 Guess

M1  1000 in lbf

Given

y(a) = 0:

V(L) = 0:

R1  500  lbf 

M1 2

2

a 

R1 6

3

R2  1000 lbf 3

 a = 0  lbf  in

R1  R2  P = 0  lbf



M1  R1 L  R2 ( L  a ) = 0  lbf  in

M(L) = 0:

 M1     R1   Find  M1 R1 R2 R   2 9.

4-62-2

M1  5000 in lbf

R1  750  lbf R2  1250 lbf

Evaluate y at x = L to get the deflection at the load.

yL 

 M1 2 R1 3 R2 3 L  L   ( L  a)  E I  2 6 6  1



yL  1.584  in

10. Evaluate  at x = a to get the slope at the roller support.

θA 

 E I  1

  M1 a 

R1 2

a

2

 

θA  0.0340 rad



4-63-1

PROBLEM 4-63

_____

Statement:

Figure P4-24 shows a 1.25-in-dia solid steel shaft with several twisting couples applied in the directions shown. For TA = 10000 lb-in, TB = 20000 lb-in, TC = 30000 lb-in, find: (a) The magnitude and location of the maximum shear stress in the shaft. (b) The corresponding principal stresses for the location determined in part (a). (c) The magnitude and location of the maximum shear strain in the shaft.

Given:

Modulus of rigidity Shaft diameter d  1.25 in Torque magnitudes: TA  10 kip in TB  20 kip in LAB  18 in

Segment lengths: Solution: 1.

LBC  12 in

6

G  11.7 10  psi TC  30 kip in LCD  10 in


Looking at the shaft from the left end (A), TA and TC are clockwise (negative) and TB is counterclockwise (positive). For the shaft to be in equilibrium, the applied torques must sum to zero. Write the equilibrium equation and solve for the unknown reaction TD. TA  TB  TC  TD  0 TD  TA  TB  TC

2.

3.

The net torque on each shaft segment is now TAB  TA

TAB  10 kip in

TBC  TAB  TB

TBC  10 kip in

TCD  TBC  TC

TCD  20 kip in

Calculate the outside radius and the polar moment of inertia of the shaft. r 

4.

d

r  0.625  in

2

J 

π d

4

4

J  0.240  in

32

(a) Since the shaft is uniform in cross-section, the maximum shear stress will occur in that segment that has the largest absolute value of torque applied to it. In this case, that is segment CD. Use equation 4.23b to calculate the maximum shear stress in segment CD.

τmax  5.

TD  20 kip in

TCD  r

τmax  52.2 ksi

J

(b) Mohr's circle for pure shear is centered at 0,0 and has a radius equal to the shear stress on the stress element. Thus, for this case, the two nonzero principal stresses are

σ1  τmax

σ1  52.2 ksi

σ3  τmax

σ3  52.2 ksi

The third principal stress is zero, σ2  0  ksi 6.

(c) The shear strain in any given segment is proportional to the shear stress so the maximum shear strain will occur in segment CD, where the shear stress is a maximum. Hooke's law for shear is similar to that given in equation 2.2.

γmax 

τmax G

3

γmax  4.46  10

 rad



4-64-1

PROBLEM 4-64

_____

Statement:

If the shaft of Problem 4-63 were rigidly attached to fixed supports at each end (A and D) and loaded only by the couples TB and TC, then find: (a) The reactions TA and TD at each end of the shaft. (b) The rotation of section B with respect to section C. (c) The magnitude and location of the maximum shear strain.

Given:

Shaft diameter Torque magnitudes:

d  1.25 in TB  20 kip in

Modulus of rigidity TC  30 kip in

Segment lengths:

LAB  18 in

LBC  12 in

Solution: 1.

LCD  10 in


Calculate the outside radius and the polar moment of inertia of the shaft. r 

2.

6

G  11.7 10  psi

d

r  0.625  in

2

J 

π d

4

4

J  0.240  in

32

(a) Looking at the shaft from the left end (A), TC is clockwise (negative) and TB is counterclockwise (positive). For the shaft to be in equilibrium, the applied torques must sum to zero. Since there are two unknown reactions in the equilibrium equation, we cannot solve for them without another equation. An equation that expresses the fact that the total rotational deflection from A to D is zero is called the compatibility equation. Write the equilibrium and compatibility equations and solve for the unknown reactions TA and TD. TA  TB  TC  TD  0

θAB  θBC  θCD  0 TA  10 kip in

Guess

Given



TA  TB LBC J G



TD LCD J G

 0

TD  30 kip in

TA LAB J G



TA  TB LBC J G

 TA     Find  TA TD  TD 

4.

J G



TA  TB  TC  TD = 0  kip in 

3.

TA LAB



TD LCD J G

= 0  rad

TA  3.50 kip in

clockwise

TD  13.50  kip in

counterclockwise

The net torque on each shaft segment is now TAB  TA

TAB  3.5 kip in

TBC  TAB  TB

TBC  16.5 kip in

TCD  TBC  TC

TCD  13.5 kip in

(b) Use equation 4.24 to calculate the rotation of section B with respect to C.



θBC   5.

TBC LBC J G

4-64-2

θBC  0.0706 rad

θBC  4.045  deg

(c) Since the shaft is uniform in cross-section, the maximum shear stress will occur in that segment that has the largest absolute value of torque applied to it. In this case, that is segment BC. Use equation 4.23b to calculate the maximum shear stress in segment BC.

τmax 

TBC  r J


The shear strain in any given segment is proportional to the shear stress so the maximum shear strain will occur in segment BC, where the shear stress is a maximum. Hooke's law for shear is similar to that given in equation 2.2.

γmax 

τmax G

3

γmax  3.68  10

 rad



4-65-1

PROBLEM 4-65

_____

Statement:

Figure P4-25 shows a pivot pin that is press-fit into part A and is slip fit in part B. If F = 100 lb and l = 1.50 in, what pin diameter is needed to limit the maximum stress in the pin to 50 kpsi?

Given:

Applied force Total length, l

F  100  lbf l  1.50 in

Maximum stress σ  50 ksi Beam length L  0.5 l

Assumptions: 1. Since there is a slip fit between the pin and part B, part B offers no resistance to bending of the pin and, since the pin is press-fit into part A, it can be modeled as a cantilever beam of length l/2. 2. Part B distributes the concentrated force F so that, at the pin, it is uniformly distributed over the exposed length of the pin. Solution: 1.


Calculate the intensity of the uniformly distributed load acting over the length of the pin. w 

2.

F

w  133.3 

L

lbf in

A cantilever beam with uniform loading is shown in Figure B-1(b) in Appendix B. In this case, the dimension a in the figure is zero. As shown in the figure, when a = 0, the maximum bending moment occurs at the support and is 2

Mmax  3.

w L

Mmax  37.50  lbf  in

2

The bending stress in a beam is given in equation 4.11c, which can be solved for the required section modulus, Z. Z 

Mmax

4

Z  7.500  10

σ

where, for a round cross-section 1

d min 

 32 Z   π   

z=

I c

=

π d

3

 in

3

Solving for d,

32

3

d min  0.197  in



PROBLEM 4-66

4-66-1

_____

Statement:

Figure P4-25 shows a pivot pin that is press-fit into part A and is slip fit in part B. If F = 100 N and l = 64 mm, what pin diameter is needed to limit the maximum stress in the pin to 250 MPa?

Given:

Applied force Total length, l

F  100  N l  64 mm

Maximum stress σ  250  MPa Beam length L  0.5 l




2.

F

w  3.125 

L

N mm


Mmax  3.

w L

Mmax  1600.0 N  mm

2

The bending stress in a beam is given in equation 4.11c, which can be solved for the required section modulus, Z. Z 

Mmax

Z  6.400  mm

σ

where, for a round cross-section

z=

I c

=

π d

3

3

Solving for d,

32

1

d min 

 32 Z   π   

3

d min  4.0 mm



4-67-1


Figure P4-25 shows a pivot pin that is press-fit into part A and is slip fit in part B. Determine the l/d ratio that will make the pin equally strong in shear and bending if the shear strength is equal to one-half the bending strength.



The intensity of the uniformly distributed load acting over the exposed length of the pin is w 

2.

2 F l

A cantilever beam with uniform loading is shown in Figure B-1(b) in Appendix B. In this case, the dimension a in the figure is zero. As shown in the figure, when a = 0, the maximum bending moment for a beam of length L occurs at the support and is 2

Mmax = 3.

=

2

M Z

 F  l    32  = 8  F  l  4   3    π d  π d 3

=

Figure B-1(b) in Appendix B shows that the maximum shear occurs at the support and, for a = 0, is Vmax = w L =

5.

 2 F    l  = F  l  2   

From equation 4.15c, the maximum shear stress due to the transverse loading is

τmax =

6.

1 2 F  l  F l    = 2 l 2 4

From equation 4.11c, the bending stress is

σmax =

4.

2

w L

4 V 4 4 16 F  = F  = 3 A 3 2 2 π d 3  π d

For equal shear and bending strength, let the shear stress equal one half the bending stress. 16 F 3  π d Solving for l/d,

= 2

1 8 F  l  2 3 π d l d

=

4 3



4-69-1


Figure P4-26a shows a C-clamp with an elliptical body dimensioned as shown. The clamp has a T-section with a uniform thickness of 3.2 mm at the throat as shown in Figure P4-26b. Find the bending stress at the inner and outer fibers of the throat if the clamp force is 2.7 kN.

Given:

Clamping force F  2.7 kN Distance from center of screw to throat Section dimensions:

Solution: 1.

ri  63.5 mm

Flange b  28.4 mm

Determine the location of the CG of the T-section and the distance from the centerline of the screw to the centroid of the section at the throat. 0.5 t ( b  t)  0.5 ( h  t)  ( h  t) t

yCG  9.58 mm

b t  ( h  t)  t

rc  ri  yCG

rc  73.08  mm

Using equation 4.12a and Figure 4-16, calculate the distance to the neutral axis, rn, and the distance from the centroidal axis to the neutral axis, e. Distance from the screw centerline to the outside fiber Cross section area


A  b  t  ( h  t)  t rn 

ri t

i

Distance from centroidal to neutral axis

A  182.4  mm

2

rn  71.86  mm

ro

 t dr   dr r r  r  t

b

i

e  rc  rn

e  1.21 mm

M  rc F

5

M  1.97  10  N  mm

Calculate the distances from the neutral axis to the inner and outer fibers. ci  rn  ri

5.

ro  95.30  mm

Take a section through the throat area and draw a FBD. There will be a vertical axial force through the section CG (at a distance rc from the screw centerline) which will form a couple of magnitude rc x F. This couple will be balanced by an internal moment of equal magnitude. Internal moment

4.

ro  ri  h

A    r

3.

t  3.2 mm


yCG 

2.

Web h  31.8 mm

ci  8.364  mm

co  ro  rn

co  23.436 mm

Using equations 4.12d and 4.12e, calculate the stresses at the inner and outer fibers of the throat section.

σi 

 ci  F  e A  ri  A M

σo  



 co  F  e A  ro  A M



σi  132.2  MPa

σo  204.3  MPa



4-70-1


A C-clamp as shown in Figure P4-26a has a rectangular cross section as in Figure P4-26c. Find the bending stress at the inner and outer fibers of the throat if the clamping force is 1.6 kN.

Given:

Clamping force F  1.6 kN Distance from center of screw to throat Section dimensions:

Solution: 1.

Width b  6.2 mm

Depth h  31.8 mm


Determine the distance from the centerline of the screw to the centroid of the section at the throat. rc  ri 

2.

ri  63.5 mm

h

rc  79.40  mm

2



A  b  h rn 

A ro

2

rn  78.33  mm dr

i

e  rc  rn

e  1.07 mm


M  rc F

5

M  1.27  10  N  mm


5.

b r


4.

ro  95.30  mm

A  197.160  mm

   r

3.

ro  ri  h

ci  14.827 mm

co  ro  rn

co  16.973 mm


σi 

 ci  F  e A  ri  A M

σo  



 co  F  e A  ro  A M



σi  148.3  MPa

σo  98.8 MPa



4-71-1


Given:

A C-clamp as shown in Figure P4-26a has an elliptical cross section as in Figure P4-26d. Dimensions of the major and minor axes of the ellipse are given. Determine the bending stress at the inner and outer fibers of the throat if the clamping force is 1.6 kN. Clamping force F  1.6 kN Distance from center of screw to throat ri  63.5 mm Section dimensions:

Solution: 1.




2.

Width b  9.6 mm

h

rc  79.40  mm

2

Using equation 4.12a and Figure 4-16, calculate the distance to the neutral axis, rn, and the distance from the centroidal axis to the neutral axis, e. ro  ri  h

Distance from the screw centerline to the outside fiber Cross section area


b h A  π  2 2

A  239.766  mm A

rn 

ro

      r

2   1   r  rc  2 b 4 2  h  

dr

i

e  rc  rn

e  0.805  mm

M  rc F

5

M  1.27  10  N  mm


5.

0.5


4.

2

rn  78.595 mm

r

Distance from centroidal to neutral axis 3.

ro  95.30  mm

ci  15.095 mm

co  ro  rn

co  16.705 mm


σi 

 ci  F  e A  ri  A M

σo  



 co  F  e A  ro  A M



σi  163.2  MPa

σo  108.7  MPa



4-72-1


Given:

A C-clamp as shown in Figure P4-26a has a trapezoidal cross section as in Figure P4-26e. Determine the bending stress at the inner and outer fibers of the throat if the clamping force is 1.6 kN. Clamping force F  1.6 kN Distance from center of screw to throat ri  63.5 mm Section dimensions:

Solution: 1.

b o  3.2 mm

Determine the distance from the centerline of the screw to the centroid of the section at the throat. h bi  2 bo  3 bi  bo

rc  76.75  mm



A 

bi  bo 2

h

ro

bi 

bi  bo h

2

rn  75.771 mm   r  ri dr

i

e  rc  rn

e  0.979  mm


M  rc F

5

M  1.228  10  N  mm


5.

ro  95.30  mm

A  203.520  mm

r


4.

ro  ri  h

A

rn 

     r

3.



rc  ri  2.

Width b i  9.6 mm

ci  12.271 mm

co  ro  rn

co  19.529 mm


σi 

 ci  F  e A  ri  A M

σo  



 co  F  e A  ro  A M



σi  126.9  MPa

σo  118.4  MPa



4-73-1


We want to design a C-clamp with a T-section similar to the one shown in Figure P4-26. The depth of the section will be 31.8 mm as shown but the width of the flange (shown as 28.4 mm) is to be determined. Assuming a uniform thickness of 3.2 mm and a factor of safety against static yielding of 2, determine a suitable value for the width of the flange if the C-clamp is to be made from 60-40-18 ductile iron and the maximum design load is 1.6 kN.

Given:

Maximum clamping force F  1.6 kN Distance from center of screw to throat

ri  63.5 mm

Section dimensions: Web h  31.8 mm Factor of safety N  2 Yield strength S y  324  MPa Solution: 1.

t  3.2 mm


Determine the location of the CG of the T-section and the distance from the centerline of the screw to the centroid of the section at the throat as functions of the unknown flange width, b. yCG ( b ) 

0.5 t ( b  t)  0.5 ( h  t)  ( h  t) t b  t  ( h  t)  t

rc( b )  ri  yCG ( b ) 2.

Using equation 4.12a and Figure 4-16, calculate the distance to the neutral axis, rn, and the distance from the centroidal axis to the neutral axis, e,as functions of b. ro  ri  h

Distance from the screw centerline to the outside fiber A ( b )  b  t  ( h  t)  t

Cross section area


A (b)

rn( b ) 

r t

 i   r i


e( b )  rc( b )  rn( b )

Calculate the distances from the neutral axis to the inner and outer fibers. co( b )  ro  rn( b )


σi( b )  6.

i

M ( b )  rc( b )  F

ci( b )  rn( b )  ri 5.

r

 o t dr   dr r r  r t

b


4.

ro  95.3 mm

 ci( b )  F   e( b )  A ( b )  ri  A ( b ) M (b)



Set the tensile stress on the inner fiber equal to the yield strength divided by the factor of safety and solve for the flange width, b.



7.

Guess

b  12 mm

Given

σi( b ) =

Sy N

b  Find ( b )

4-73-2

b  10.13  mm

Using the calculated value of b, check the stresses at the inner and outer fibers..

σi( b ) 

 ci( b )  F   e( b )  A ( b )  ri  A ( b )

σo( b )  

M (b)



 c o( b )  F  e( b )  A ( b )  ro  A ( b ) M ( b)



σi( b )  162  MPa

σo( b )  149.4  MPa

A suitable minimum value for the flange width is b  10.1 mm



4-74-1


A round steel bar is 10 in long and has a diameter of 1 in. (a) Calculate the stress in the bar when it is subjected to a 1000-lb force in tension. (b) Calculate the bending stress in the bar if it is fixed at one end (as a cantilever beam) and has a 1000-lb transverse load at the other end. (c) Calculate the transverse shear stress in the bar of part (b). (d) Calculate the torsional shear stress when the 1000-lb force is displaced 10 inches radially from the centerline (axis) of the cantilever beam. (e) Calculate the maximum bending stress in the bar if it is formed into a semicircle with a centroidal radius of 10/ in and 1000-lb opposing forces are applied at the ends in the plane of the of the ends. Assume that there is no distortion of the cross section during bending. (f) Calculate the direct bearing stress that would result on the bar of (a) if it were the pin in a pin-and-clevis connection that is subjected to a 1000-lb pull if the center part (the eye or tongue) is 1-in wide. (g) Determine how short the bar must be when loaded as a cantilever beam for its maximum flexural bending stress and its maximum transverse shear stress to provide equal tendency to failure. Find the length as a fraction of the diameter if the failure stress in shear is half the failure stress in bending. (h) If the force on the cantilever beam in (f) is eccentric, inducing torsional as well as bending stress, what fraction of the diameter would the eccentricity need to be in order to give a torsional stress equal to the transverse shear stress?

Given:

Length of bar L  10 in Force F  1000 lbf

Solution:


Diameter Load Radius

d  1.00 in R  10 in

(a) Use equation 4.7 to calculate the axial stress. Cross sectional area

Axial stress

A  π

σ 

d

2

2

A  0.785  in

4

F

σ  1.27 ksi

A

(b) The beam loading diagram is shown in Appendix Figure B-1a with the concentrated load at a = L. The maximum bending stress occurs at x = 0 and is given by Equation 4.11b. Bending moment

M  L F

Radius of bar

c 

Moment of inertia

Maximum bending stress

M  10000  in lbf

d

c  0.5 in

2

I  π

d

4

4

I  0.049  in

64

σ 

M c I

σ  101.9  ksi

(c) The maximum transverse shear stress occurs at y = 0 and is given by Equation 4.15c and in Figure 4-20b. Maximum transverse shear stress in a solid, round bar

τmax 

4 F  3 A


(d) The maximum torsional shear shear stress occurs at y = d/2 and is given by Equation 4.23b. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0474.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


4-74-2

Twisting torque

T  F  R


J  π

τ 

Max torsional shear stress

d

T  10000  in lbf 4

4

J  0.098  in

32

Td

τ  50.93  ksi

2 J

(e) The maximum bending stress for a curved beam occurs at r = ri and is given by Equation 4.12d. 10

Centroidal radius

rc 

Inside radius

ri  rc  0.5 d

ri  2.683  in

Outside radius

ro  rc  0.5 d

ro  3.683  in

Cross section area

A  π

rn 


 in

π

d

rc  3.183  in

2

2

A  0.785  in

4 A

rn  3.163  in

r

 o   2   r

2

 d   r  r 2   c 2 dr r

i

Distance from centroid to neutral axis

e  rc  rn

M  rc F

Internal moment

e  0.020  in

M  3183 in lbf

Distances from the neutral axis to the inner and outer fibers ci  rn  ri

ci  0.480  in

co  ro  rn

co  0.520  in

Stress at the inner fibers of the throat section

σi 

 ci  F  e A  ri  A M



σi  37.9 ksi

(f) The direct bearing stress is given in Equations 4.7 and 4.10. Given length of bearing contact

l  1  in

Projected area of contact

Abearing  l d

Bearing stress

σbearing 

2

Abearing  1  in F

Abearing

σbearing  1.0 ksi

(g) Determine how short the bar must be when loaded as a cantilever beam for its maximum flexural bending stress and its maximum transverse shear stress to provide equal tendency to failure. Find the length as a fraction of the diameter if the failure stress in shear is half the failure stress in bending. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0474.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


Bending stress

M c

σ=

I

=

4-74-3

32 F  L

π d

3

4 F 16 F  = 3 A 2 3  π d

Transverse shear

τ=

Equating

σ = 2 τ

32 F  L

Solving for L

L=

π d

=

3

32 3



F

π d

2

d 3

(h) If the force on the cantilever beam in (f) is eccentric, inducing torsional as well as bending stress, what fraction of the diameter would the eccentricity need to be in order to give a torsional stress equal to the transverse shear stress? Torsional shear stress

Transverse shear

Equating

τtor =

Tc

τtrans =

J

=

16 F  e

π d

3

4 F 16 F  = 3 A 2 3  π d 16 F  e

τtor = τtrans

π d

Solving for the eccentricity, e

e=

3

=

16 F 3  π d

2

d 3



4-75a-1


For a filleted flat bar in tension similar to that shown in Appendix Figure C-9 and the data from row a from Table P4-4, determine the nominal stress, the geometric stress concentration factor, and the maximum axial stress in the bar.

Given:

Widths Thickness Force

Solution:

See Appendix Figure C-9 and Mathcad file P0475a.

1.

P

σnom  40.0 MPa

h d

Determine the geometric stress concentration factor using Appendix Figure C-9. Width ratio

D d

3.

d  20 mm Radius r  4  mm

Determine the nominal stress in the bar using equation 4.7.

σnom  2.

D  40 mm h  10 mm P  8000 N

 2.00

From Figure E-9

A  1.0966

SCF

Kt  A  

r

b  0.32077 b

 d

Kt  1.838

Determine the maximum stress in the bar using equation 4.31.

σmax  Kt σnom




4-76a-1


For a filleted flat bar in bending similar to that shown in Appendix Figure C-10 and the data from row a from Table P4-4, determine the nominal stress, the geometric stress concentration factor, and the maximum bending stress in the bar.

Given:

Widths Thickness Moment

Solution:


1.

d

c  10 mm

2

σnom 

M c

I 

h d

3

3

I  6.667  10  mm

12

4

σnom  120.0  MPa

I


D d

3.

d  20 mm Radius r  4  mm

Determine the nominal stress in the bar using equation 4.11b. c 

2.

D  40 mm h  10 mm M  80 N  m

 2.00

From Figure E-9

A  0.93232

SCF

Kt  A  

r

b  0.30304

b

 d

Kt  1.518



σmax  182.2  MPa



4-77a-1


For a shaft, with a shoulder fillet, in tension similar to that shown in Appendix Figure C-1 and the data from row a from Table P4-4, determine the nominal stress, the geometric stress concentration factor, and the maximum axial stress in the shaft.

Given:

Widths Radius Force

Solution:


1.

D  40 mm r  4  mm P  8000 N

Determine the nominal stress in the bar using equation 4.7.

σnom 

4 P

π d 2.

σnom  25.5 MPa

2


D d

3.

d  20 mm

 2.00

From Figure E-1

A  1.01470

SCF

Kt  A  

r

b  0.30035

b

 d

Kt  1.645






4-78a-1


For a shaft, with a shoulder fillet, in bending similar to that shown in Appendix Figure C-2 and the data from row a from Table P4-4, determine the nominal stress, the geometric stress concentration factor, and the maximum bending stress in the shaft.

Given:

Widths Radius Moment

Solution:


1.

d

c  10 mm

2

σnom 

M c

I 

π d

4

3

I  7.854  10  mm

64

4

σnom  101.9  MPa

I


D d

3.

d  20 mm

Determine the nominal stress in the bar using equation 4.11b. c 

2.

D  40 mm r  4  mm M  80 N  m

 2.00

From Figure E-2

A  0.90879

SCF

Kt  A  

r

b  0.28598

b

 d

Kt  1.44



σmax  146.7  MPa



4-79-1


A differential stress element has a set of applied stresses on it as shown in Figure 4-1. For σx = 850, σy = -200, σz = 300, τxy = 450, τyz = -300, and τzx = 0; find the principal stresses and maximum shear stress and draw the Mohr's circle diagram for this three-dimensional stress state.

Given:

σx  850

σy  200

σz  300

τxy  450

τyz  300

τzx  0


Solution:

1. Calculate the coefficients (stress invariants) of equation (4.4c). C2  σx  σy  σz

C1 

C2  950.000

 σx τxy   σx τzx   σy τyz           τxy σy   τzx σz   τyz σz 

 σx τxy τzx    C0   τxy σy τyz  τ τ σ   zx yz z  3

σ2  r

σ2  388

σ3  r

σ3  470

3 2 1

 CW

 1-3

500  1-2

 2-3

-500

4. Using equations (4.5), evaluate the principal shear stresses.

τ13  τ12  τ23 

σ1  σ3 2

σ1  σ2 2

σ2  σ3 2

8

 470  r   388     1032 

3. Extract the principal stresses from the vector r by inspection.

σ1  1032

C0  1.882  10 2

r  polyroots ( v)

σ1  r

5

σ  C2 σ  C1 σ  C0 = 0

2. Find the roots of the triaxial stress equation:

 C0    C1   v   C2     1 

C1  2.675  10

3

500

1000

2

0

1

1500 

τ13  751 500

τ12  322 τ23  429

5. Draw the three-circle Mohr diagram.

 CCW

FIGURE 4-79 The Three Mohr's Circles for Problem 4-79



4-80-1

PROBLEM 4 - 80 Statement:

Write expressions for the normalized (stress/pressure) tangential stress as a function of the normalized wall thickness (wall thickness/outside radius) at the inside wall of a thick-wall cylinder and for a thin-wall cylinder, both with internal pressure only. Plot the ratio of these two expressions and determine the range of the wall thickness to outside radius-ratio for which the stress predicted by the thin-wall expression is at least 5% greater than that predicted by the thick-wall expression.

Solution:


1.

Let the σt/p ratio be S' and the t/ro ratio be t', then For the thick-wall cylinder at the inside wall, using equation 4.48a 2

S'thick ( t') 

2  2  t'  t' 2

2  t'  t'

and, for the thin-wall cylinder, using equation 4.49a S'thin ( t') 

1 t'

2.

Choose a range for the normalized thickness ratio, t'  0.01 0.02  0.99

3.

Plot the difference between the two functions. Δ ( t') 

S'thin( t')  S'thick( t')  S'thick ( t')

25

20

15

Δ ( t') % 10

5

0

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t' © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0480.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


4.

Determine the values of t' for which the difference is 5%.

Δ ( 0.10)  5.0 % 5.

4-80-2

Δ ( 0.946 )  5.1 %

The range of the normalized thickness for which the thin-wall stress is at least 5% greater than the thick-wall stress is from 0.10 to 0.946.



4-81-1


A hollow square torsion bar such as that shown in Table 4-3 has dimensions a = 25 mm, t = 3 mm, and l = 300 mm. If it is made of steel with a modulus of rigidity of G = 80.8 GPa, determine the maximum shear stress in the bar and the angular deflection under a torsional load of 500 N-m.

Given:

Dimensions Modulus

Solution:

See Table 4-3 Mathcad file P0481.

1.

a  25 mm G  80.8 GPa

2

2  t  ( a  t) 2 a t  2 t

Q  2  t ( a  t)

4

2

2

K  31944  mm

Q  2904 mm

4

3

Using equation 4.26a, calculate the maximum shear stress.

τmax  3.

l  300  mm T  500  N  m

Calculate the factors K and Q for a hollow square from Table 4-3.

K 

2.

t  3  mm Load

T Q

τmax  172.2  MPa

Using equation 4.26b, calculate the angular deflection.

θ 

Tl K G

θ  0.058 radians θ  3.33 deg



4-82-1


Design a hollow rectangular torsion bar such as that shown in Table 4-3 that has dimensions a = 45 mm, b = 20 mm, and l = 500 mm. It is made of steel with a shear yield strength of 90 MPa and has an applied torsional load of 135 N-m. Use a factor of safety against yielding of 2.

Given:

Dimensions a  45 mm b  20 mm l  500  mm Modulus G  80.8 GPa Load T  135  N  m Shear yield strength S sy  90 MPa Factor of safety N  2

Solution:

See Table 4-3 Mathcad file P0482.

1.

Calculate the Q-factor for a hollow rectangle from Table 4-3. Q( t)  2  t ( a  t)  ( b  t)

2.

Calculate the maximum shear stress as a function of thickness, t, using equation 4.26a.

τ( t)  3.

T Q( t)

Define a function that relates the maximum shear stress to the shear strength divided by the factor of safety and solve for the thickness, t. Guess a value of t

t  3  mm

Define the design function

f ( t)  τ( t) 

t  root( f ( t) t)

S sy N

t  1.927  mm

Let t = 2 mm (note that this solution does not check for buckling under the applied load)



4-83-1


A pressure vessel with closed ends has the following dimensions: outside diameter, OD = 450 mm, and wall thickness, t = 6 mm. If the internal pressure is 690 kPa, find the principal stresses on the inside surface away from the ends. What is the maximum shear stress at the point analyzed?

Given:

Dimensions Pressure

Solution:


1.

2.

OD  450  mm p  690  kPa

t  6  mm

Convert the given dimensions to inside and outside radii. ro  0.5 OD

ro  225  mm

ri  ro  t

ri  219  mm

Determine whether to use thick-wall or thin-wall theory. ro 10

 22.5 mm

Since the wall thickness, t  6  mm, is much less than one tenth the outside radius, use thin wall theory. 3.

Calculate the principal stresses using equations 4.49. Tangential (y-direction)

σt 

p  ro t

Radial (x-direction)

σr  0  MPa

Axial (z-direction)

σa 

p  ro 2 t

σt  25.9 MPa σr  0.0 MPa σa  12.9 MPa

The principal stresses are:

4.

σ1  σt

σ1  25.9 MPa

σ2  σa

σ2  12.9 MPa

σ3  σr

σ3  0.0 MPa

Using equation 4.6b, calculate the maximum shear stress.

τmax 

σ1  σ3 2




4-84-1


A simply supported steel beam of length, l, with a concentrated load, F, acting at midspan has a rectangular cross-section with width, b, and depth, h. If the strain energy due to transverse shear loading is Us and that due to bending loading is Ub, derive an expression for the ratio Us/Ub and plot it as a function of h/l over the range 0 to 0.10.

Solution:

See Mathcad file P0484. l

1.

From equation 4.22e, the strain energy in transverse loading is:

 2 3  V Us =   dx 5  G A  0

l

2.

From equation 4.22d, the strain energy in bending loading is:

 2 1  M Ub =   dx 2  E I  0

l

3.

Let

U' =

Us Ub

 2  V dx 

U' =

, then:

0 6 E I   5 G A l  2  M dx  0

4.

For a rectangular cross-section:

A = b h

I=

and

b h

3

12 l

5.

E

And, for steel:

G

=

5 2

 2  V dx 2  0 h U' =  l 4  2  M dx 

therefore

0

6.

7.

For the given loading: F

For x between 0 and l/2,

V=

For x between l/2 and l,

V=

and

2 F 2

and

M=

M= F x 2



F x 2 F l 2

Substituting these expressions into the equation for U' and integrating gives: l    0.5 l   2 2   F F   dx    dx   12.0   2 12.0 2  2 h          2   2 2  2   h  0 0.5 l l l  6.0 h       4 4  0.5 l l 2    l 2 2    F  x F  x F  l   dx      2   2  2  dx          0.5 l  0 


MACHINE DESIGN - An Integrated Approach, 4th Ed. h' =

h

4-84-2 2

U'( h')  6  h'

then

8.

Let

9.

Plotting the strain energy ratio over the range:

l

h'  0 0.001  0.10

STRAIN ENERGY RATIO vs DEPTH TO LENGTH RATIO 6

Strain Energy Ratio - Percent

5

4

U'( h' ) %

3

2

1

0

0

0.02

0.04

0.06

0.08

0.1

h' Depth to Length Ratio



4-85a-1


A beam is supported and loaded as shown in Figure P4-27(a). Find the reactions for the data given in row a from Table P4-2.

Given:

Beam length Distance to R2

L  1  m a  0.4 m


w  200  N  m

1

L

a w

R1

R2

R3


Solution:

See Figure P4-27(a) and Mathcad file P0485a.

1.

Consider the reaction force R1 to be redundant and remove it temporarily. The beam will then be statically determinant and will deflect at x = 0. Now consider the reaction force R1 to be an unkown applied load that will force the deflection to be zero. Write an equation for the deflection at x = 0 in terms of the force R1 with the deflection set to zero.

2.

Write equation 4.21 for the deflection y1 at the unknown applied load R1 in terms of the strain energy in y1 =

the beam at that point:

3.

Substitute equation 4.22d and differentiate:

  y1 =   

 U R1

L

M E I





 R1



M  dx

(a)



0

4.

5.

Write an expression for the bending moment and its partial derivative with respect toR1 as a function of x. For x between 0 and a,

M = R 1 x 

For x between a and l,

M = R 1 x 

w x

2

2 w x 2

2

 R 2 ( x  a )

 M =x R1

(b)

 M =x R1

(c)

Substitute equations (b) and (c) into (a), set equal to zero and integrate.



    

a

2   R1 x  w x   x dx  2  

0

6.

    

L

4-85a-2

2   R1 x  w x  R2 ( x  a)  x dx = 0 2  

a

Solving for R1 and R2 and summing forces and moments about x = 0:

From strain energy

R1 3

3

m 

4  L3 a  L2 a 3 a3      R2  w  L = 0   2 8 3 2  3

Summing forces

R1  R2  R3  w  L = 0

Summing moments

R 2 a  R 3 L 

2

7.

w L

=0

2

R1  65 N

Use these three equations to solve for R1, R2, and R3. Guess Given R1 3

3

m 

R2  70 N

R3  65 N

4  L3 a  L2 a 3 a3      R2  w  L = 0   2 8 3 2  3

R1  R2  R3  w  L = 0 2

R 2 a  R 3 L 

w L

=0

2

 10.714  R   148.81   N    40.476 

R  Find  R1 R2 R3

R  10.7 N 1

R  148.8  N 2

R  40.5 N 3



4-86a-1


A beam is supported and loaded as shown in Figure P4-27(b). Find the reactions for the data given in row a from Table P4-2.

Given:

Beam length Distance to R2

L  1.0 m a  0.4 m

Distributed load magnitude Distance to concentrated load Concentrated load

w  200  N  m b  0.6 m F  500  N

1

L b a F w

R1

R2

R3


Solution:

See Figure P4-27(b) and Mathcad file P0485a.

1.

Consider the reaction force R1 to be redundant and remove it temporarily. The beam will then be statically determinant and will deflect at x = 0. Now consider the reaction force R1 to be an unkown applied load that will force the deflection to be zero. Write an equation for the deflection at x = 0 in terms of the force R1 with the deflection set to zero.

2.

Write equation 4.21 for the deflection y1 at the unknown applied load R1 in terms of the strain energy in y1 =

the beam at that point:

3.

Substitute equation 4.22d and differentiate:

  y1 =   

 U R1

L

M E I







M  dx

(a)

 R1 

0

4.

Write an expression for the bending moment and its partial derivative with respect toR1 as a function of x. w x

2

For x between 0 and a,

M = R 1 x 

For x between a and b,

M = R1 x  w a   x 

2



a

  R 2 ( x  a )

2

 M =x R1

(b)

 M =x R1

(c)



4-86a-2

M = R1 x  w a   x 

For x between b and L,



a

  R 2 ( x  a )  F  ( x  b )

2

(d)

 M =x R1 5.

Substitute equations (b), (c) and (d) into (a), set equal to zero and integrate.     

a

2   R1 x  w x   x dx  2  

0

   

b

R  x  w a   x   1   

a

   R2 ( x  a )  x dx  = 0 

2

a

   

L

R  x  w a   x   1   

a

   R2 ( x  a )  F  ( x  b )  x dx 

2

b

6.

Solving for R1 and R2 and summing forces and moments about x = 0:

From strain energy

7.

2 3  L3   3    R1   L  a  L  a   R2  =0 2 6  3 3 2 3  a2 L2 a L3 a4   3   w   L  b L  b   F    3 2 24  6   6 3

Summing forces

R1  R2  R3  w  a  F = 0

Summing moments

R 2 a  R 3 L 

w a

2

 F b = 0

2

Use these three equations to solve for R1, R2, and R3. Guess R1  100  N Given

R2  400  N

R3  200  N

2 3  L3   3    R1   L  a  L  a   R2  =0 2 6  3 3  a2 L2 a L3 a4   L3 b  L2 b 3    F    w     3 2 24  6   6 3

R1  R2  R3  w  a  F = 0 R 2 a  R 3 L 

w a

2

 F b = 0

2

 81.143  R   575.238   N    85.905 

R  Find  R1 R2 R3

R  81.1 N 1

R  575.2  N 2

R  85.9 N 3



5-1a-1


A differential stress element has a set of applied stresses on it as indicated in row a of Table P5-1. For row a, draw the stress element showing the applied stresses. Find the principal stresses and the von Mises stress.

Given:

σx  1000

σy  0

σz  0

τxy  500

τyz  0

τzx  0

Solution:

See Figure 5-1a and Mathcad file P0501a.

1. Draw the stress element, indicating the x and y axes.

500

2. From Problem 4-1a, the principal stresses are

σ1  1207

σ2  0

y

σ3  207

x

1000

3. Using equatoion 5.7c, the von Mises stress is

σ' 

2

σ1  σ1 σ3  σ3

2

σ'  1323

FIGURE 5-1aA Stress Element for Problem 5-1a



5-1h-1


A differential stress element has a set of applied stresses on it as indicated in each row of Table P5-1. For row h, draw the stress element showing the applied stresses, find the principal stresses and the von Mises stress.

Given:

σx  750

σy  500

σz  250

τxy  500

τyz  0

τzx  0

Solution:

See Figure 5-1h and Mathcad file P0501h. z

1. Draw the stress element (see Figure 5-1h).

250

2. From Problem 4-1h, the principal stresses are

σ1  1140

σ2  250

σ3  110

3. Using equation 5.7, the von Mises stress is

σ' 

1 2

σ'  968

  σ1  σ2   σ2  σ3   σ1  σ3 2

2

750

2



500

500

x

500 y

FIGURE 5-1h Stress Element for Problem 5-1h

© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0501h.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


5-2-1


A 400-lb chandelier is to be hung from two 10-ft-long solid, low- carbon steel cables in tension. Size the cables for a safety factor of 4. State all assumptions.

Given:

Weight of chandelier Length of cable Design Safety factor

W  400  lbf L  10 ft Nd  4

Number of cables

N  2

Young's modulus

E  30 10  psi

L  120 in

6

Assumptions: The material is AISI 1010 hot-rolled steel with S y  26 ksi Solution:

See Mathcad file P0502. P 

W

P  200 lbf

1.

Determine the load on each cable

2.

Using the distortion-energy failure theory,

3.

In this case, the only stress in the axial direction is the tensile stress. Therefore, this is the principal stress and also the von Mises stress. 4 P σ' = σ1 = σ = 2 π d

4.

Substitute the equation in step 3 into the design equation in step 2 and solve for the minimum diameter, d.

N Nd =

Sy

σ'

1

 4  P N d  d     π S y  5.

2

d  0.198 in

Round up to an available size (see Table 13-2) and check the actual factor of safety against static failure. 2

d  0.207  in

Ns 

π d  S y 4 P

Ns  4.4



5-3-1


For the bicycle pedal-arm assembly in Figure P5-1 with rider-applied force of 1500 N at the pedal, determine the von Mises stress in the 15-mm-dia pedal arm. The pedal attaches to the arm with a 12-mm thread. Find the von Mises stress in the screw. Find the safety factor against static failure if the material has S y = 350 MPa.

Given:

Distances (see figure) Rider-applied force

a  170  mm Frider  1.5 kN

b  60 mm Screw thread diameter

d sc  12 mm

Pedal arm diameter

d pa  15 mm

Material yield strength

S y  350  MPa


z

Solution:

1. From problem 4-3, the maximum principal stresses in the pedal arm are at point A and are

σ1  793  MPa

a

σ2  0  MPa

σ' 

2

Mc

b

σ3  23 MPa 2. Using equation 5.7c, the von Mises stress is

σ1  σ1 σ3  σ3

Tc

C

Frider Arm y

Fc Pedal

2

x

FIGURE 5-3A

σ'  805 MPa


3. The factor of safety for the pedal arm is N 

Sy

σ'

z

N  0.43

4. From Problem 4-3 solution, the stresses at the top of the screw where it joins the pedal arm are

σx  530.5  MPa

Section C

A

σz  0  MPa

B

Arm

τzx  0  MPa

x

5. From this, we see that the principal stresses are

σ1  σx

y

σ2  0  MPa FIGURE 5-3B

σ3  0  MPa


6. The von Mises stress is

σ'  σ1

7. The factor of safety for the screw is

N 

Sy

σ'

σ'  530.5 MPa N  0.66



5-4-1


The trailer hitch shown in Figure P4-2 and Figure 1-1 (p. 12) has loads applied as defined in Problem 3-4. The tongue weight of 100 kg acts downward and the pull force of 4905 N acts horizontally. Using the dimensions of the ball bracket shown in Figure 1-5 (p. 15) and S y = 300 MPa ductile steel, determine static safety factors for: (a) The shank of the ball where it joins the ball bracket. (b) Bearing failure in the ball bracket hole. (c) Tearout failure in the ball bracket. (d) Tensile failure in the 19-mm diameter attachment holes. (e) Bending failure in the ball bracket as a cantilever.

Given:

a  40 mm b  31 mm c  70 mm d  20 mm Mtongue  100  kg Fpull  4.905  kN d sh  26 mm t  19 mm S y  300  MPa


See Figure 5-4 and Mathcad file P0504. W tongue 70 = c

1

F pull

1

40 = a 2

A B

A

19 = t B

F b1

31 = b

F a1x

C

F a1y 20 = d

F a2y

D

Fa2x 2

Fc2x

F b2 C D

Fd2 F c2y


1.

From Problem 4-4, the principal stresses in the shank of the ball where it joins the ball bracket are:

σ1  114  MPa

σ2  0  MPa

σ3  0  MPa

2. Since 1 is the only nonzero principal stress, it is also the von Mises stress.The factor of safety against a static failure at the shank of the ball is © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0504.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


σ'  σ1 3.

Na 

5-4-2

Sy

Na  2.6

σ'

From Problem 4-4, the principal stresses at the bearing area in the ball bracket hole are:

σ1  9.93 MPa

σ2  0  MPa

σ3  0  MPa

4. Since 1 is the only nonzero principal stress, it is also the von Mises stress.The factor of safety against a static bearing failure in the ball bracket hole is

σ'  σ1

Nb 

Sy

Nb  30.2

σ'

Tearout length

5. From Problem 4-4, the shear stress in the tearout area in the ball bracket is:

τ  4.41 MPa 2

6. For pure shear, the von Mises stress is σ'  3  τ and the factor of safety against a static tearout failure is Nc 

Sy

Nc  39.3

σ'

7. From Problem 4-4, the principal stresses in the attachment bolts if they are 19-mm diameter are:

σx  53.6 MPa

σy  0  MPa

R

d FIGURE 5-4B


τxy  1.7 MPa 8.

The von Mises stress and the factor of safety against a static failure in the attachment bolts are:

σ' 

2

σ'  53.7 MPa

9.

2

σx  σy  σx σy  3  τxy

2

Nd 

Sy

σ'

Nd  5.6

From Problem 4-4, the principal stresses in the bracket due to bending in the ball bracket as a cantilever are:

σ1  72.8 MPa

σ2  0  MPa

σ3  0  MPa


σ'  σ1

Ne 

Sy

σ'

Ne  4.1



5-5-1


Repeat Problem 5-4 for the loading conditions of Problem 3-5, i.e., determine the horizontal force that will result on the ball from accelerating a 2000-kg trailer to 60 m/sec in 20 sec. Assume a constant acceleration. From Problem 3-5, the pull force is 6000 N. Determine static safety factors for: (a) The shank of the ball where it joins the ball bracket. (b) Bearing failure in the ball bracket hole. (c) Tearout failure in the ball bracket. (d) Tensile failure in the 19-mm diameter attachment holes. (e) Bending failure in the ball bracket as a cantilever.

Given:

a  40 mm b  31 mm Mtongue  100  kg Fpull  6  kN

c  70 mm d  20 mm d sh  26 mm t  19 mm

S y  300  MPa Assumptions: 1. The nuts are just snug-tight (no pre-load), which is the worst case. 2. All reactions will be concentrated loads rather than distributed loads or pressures. Solution:

See Figures 5-5 and Mathcad file P0505. W tongue 70 = c

1

F pull

1

40 = a 2

A B

A

19 = t B

F b1

31 = b

F a1x

C

F a1y 20 = d

F a2y

D

Fa2x 2

Fc2x

F b2 C D

Fd2 F c2y


1. From Problem 4-5, the principal stresses in the shank of the ball where it joins the ball bracket are:

σ1  139  MPa

σ2  0  MPa

σ3  0  MPa



σ'  σ1

Na 

5-5-2

Sy

Na  2.2

σ'

3. From Problem 4-5, the principal stresses at the bearing area in the ball bracket hole are:

σ1  12.15  MPa

σ2  0  MPa

σ3  0  MPa


σ'  σ1

Nb 

Sy

Nb  24.7

σ'

Tearout length


τ  5.4 MPa 2


Sy

Nc  32.1

σ'


σx  64.2 MPa

σy  0  MPa

R

d FIGURE 5-5B


τxy  1.7 MPa 8. The von Mises stress and the factor of safety against a static failure in the attachment bolts are:

σ' 

2

2


σ'  64.3 MPa

2

Nd 

Sy

σ'

Nd  4.7

9. From Problem 4-5, the principal stresses in the bracket due to bending in the ball bracket as a cantilever are:

σ1  85.1 MPa

σ2  0  MPa

σ3  0  MPa


σ'  σ1

Ne 

Sy

σ'

Ne  3.5



5-6-1


Repeat Problem 5-4 for the loading conditions of Problem 3-6, i.e., determine the horizontal force that will results from an impact between the ball and the tongue of the 2000-kg trailer if the hitch deflects 2.8 mm dynamically on impact. The tractor weighs 1000 kg and the velocity at impact is m/sec. Determine static safety factors for: (a) The shank of the ball where it joins the ball bracket. (b) Bearing failure in the ball bracket hole. (c) Tearout failure in the ball bracket. (d) Tensile failure in the 19-mm diameter attachment holes. (e) Bending failure in the ball bracket as a cantilever.

Given:

a  40 mm b  31 mm c  70 mm d  20 mm Mtongue  100  kg Fpull  55.1 kN d sh  26 mm t  19 mm S y  300  MPa



W tongue 70 = c

1

F pull

1

40 = a 2

A B

A

19 = t B

F b1

31 = b

F a1x

C

F a1y 20 = d

F a2y

D

Fa2x 2

Fc2x

F b2 C D

Fd2 F c2y


1. From Problem 4-6, the principal stresses in the shank of the ball where it joins the ball bracket are:

σ1  1277 MPa

σ2  0  MPa

σ3  0  MPa



σ'  σ1

Na 

5-6-2

Sy

Na  0.23

σ'

3. From Problem 4-6, the principal stresses at the bearing area in the ball bracket hole are:

σ1  111.5  MPa

σ2  0  MPa

σ3  0  MPa


σ'  σ1

Nb 

Sy

Nb  2.7

σ'

Tearout length


τ  49.6 MPa 2


Sy

Nc  3.5

σ'


σx  540.5  MPa

σy  0  MPa

R

d FIGURE 5-6B


τxy  1.7 MPa 8. The von Mises stress and the factor of safety against a static failure in the attachment bolts are:

σ' 

2

2


σ'  540.5 MPa

2

Nd 

Sy

σ'

Nd  0.56

9. From Problem 4-6, the principal stresses in the bracket due to bending in the ball bracket as a cantilever are:

σ1  635.5  MPa

σ2  0  MPa

σ3  0  MPa


σ'  σ1

Ne 

Sy

σ'

Ne  0.47



5-7-1


Design the wrist pin of Problem 3-7 for a safety factor of 3 and S y = 100 ksi if the pin is hollow and loaded in double shear.

Given:

Force on wrist pin


Yield strength

S y  100  ksi

Design safety factor

Nd  3 od  0.375  in


Fwristpin  2756 lbf

See Figure 4-12 in the text and Mathcad file P0507.

1. The force at each shear plane is

F 

Fwristpin

F  1378 lbf

2

2. With only the direct shear acting on the plane, the Mohr diagram will be a circle with center at the origin and radius equal to the shear stress. Thus, the principal normal stress is numerically equal to the shear stress, which in this case is also the principal shear stress, so we have  = 1 = '.

3. The shear stress at each shear plane is

4. Using the distortion-energy failure theory,

5. Solving for the inside diameter,

F

τ=

id 

=

A

4 F



2

π od  id Sy

Nd =

σ'

2

od 



= σ'



2

2



2

π od  id  S y

=

4 F

4  F  Nd

π S y

6. Round this down to the decimal equivalent of a common fraction (9/32),

7. The realized factor of safety is,

N 



2



id  0.281  in

2

π od  id  S y 4 F

id  0.297 in

N  3.5



5-8-1


Given:

A paper mill processes rolls of paper having a density of 984 kg/m3. The paper roll is 1.50-m outside diameter (OD) x 0.22-m inside diameter (ID) x 3.23-m long and is on a simple supported, hollow, steel shaft with S y = 300 MPa. Find the shaft ID needed to obtain a static safety factor of 5 if the shaft OD is 22 cm. Paper roll: Density

ρ  984 

kg

y

3

Outside dia.

m OD  1500 mm

Inside dia.

ID  220  mm

Length

L  3230 mm

w x L

R

R

Shaft: Strength

V

S y  300  MPa

Outside dia.

od  220  mm

Factor of safety

Ns  5

R L/2

x -R

Assumptions: 1. The shaft is stiffer than the paper roll so the weight of the roll on the shaft can be modelled as a uniformly distributed load. 2. The bearings that support the shaft are close to the ends of the paper roll and are thin with respect to the length of the roll so we can consider the distance between the shaft supports to be the same as the length of the roll. Solution:

L

0

M

2

wL /8

x

0 L/2


L

FIGURE 5-8 Load, Shear, and Moment Diagrams for Problem 5-8

1. The weight of the paper roll is,

 4

π

2

Volume

V 

Weight

W  ρ  g  V

2



3

 OD  ID  L

V  5.585  m

(1)

W  53.895 kN

(2)

2. From Figure 5-8, we see that the bending moment in the shaft is a maximum at the center of the span. First, determine the magnitude of the distributed load, then find the maximum bending moment using Figure D-2(b) in Appendix B with a = 0 and x = L/2. Distributed load

w 

W

w  16.686

L 2

Maximum moment

Mmax 

w L

newton 7

Mmax  2.176  10  newton  mm

8

(3)

mm (4)

3. Using equation 4.11b, find the maximum bending stress as a function of the unkown shaft inside diameter, id. Bending stress at midspan

σmax =

M c I

=

32 Mmax od



4



(5)

4

π od  id

4. This is the only stress element present at this point on the shaft and there is no shear stress at this point so max = 1 and 2 = 3 = 0. Furthermore, since 2 and 3 are zero, max = '. Equation 5.8a can be used to find the unknown id, © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


Factor of safety

Ns =

5-8-2

Sy

(6)

σ'

Substituting equation 5 into 6 and solving for id, we have 1

Shaft id

 π Sy od4  32 Ns Mmax od  id    π S y  

4

id  198  mm

(7)



5-9-1


A ViseGrip plier-wrench is drawn to scale in Figure P5-3, and for which the forces were analyzed in Problem 3-9 and the stresses in Problem 4-9, find the safety factors for each pin for an assumed clamping force of P = 4000 N in the position shown. The pins are 8-mm dia, S y = 400 MPa, and are all in double shear.

Given:

Pin stresses as calculated in Problem 4-9: Pin 1-2 τ12  74.6 MPa Pin 1-4

τ14  50.7 MPa

Pin 2-3

τ23  50.7 MPa

Pin 3-4

τ34  50.7 MPa S y  400  MPa

Yield strength

Assumptions: Links 3 and 4 are in a toggle position, i.e., the pin that joins links 3 and 4 is in line with the pins th join 1 with 4 and 2 with 3. Solution: 1.



F

P 1

2

3

P

F 55.0 = b

50.0 = a

39.5 = c

F

F14 22.0 = d

129.2°

1



4 F34

F41

F21

P



28.0 = e  

F43



F12

3 F23

F32

P

2.8 = g

21.2 = h

2

F 26.9 = f


2.

The pins are in pure shear, so the principal stresses are



3.

5-9-2

Pin joining 1 and 2

σ'12 

3  τ12

σ'12  129.211 MPa

All other pins

σ'14 

3  τ14

σ'14  87.815 MPa

Using the distortion-energy failure theory, the factors of safety are Pin joining 1 and 2

All other pins

N12 

N14 

Sy

σ'12 Sy

σ'14

N12  3.1

N14  4.6



5-10-1


Given:

An over-hung diving board is shown in Figure P5-4a. Assume cross-section dimensions of 305 mm x 32 mm. Find the largest principal stress in the board when a 100-kg person is standing at the free end. What is the static safety factor if the material is brittle fiberglass with S ut = 130 MPa in the longitudinal direction? Maximum principal stresses due to bending at R2 from Problem 4-10

2000 = L R1

P

σ1  24.5 MPa σ2  0  MPa

R2

σ3  0  MPa Ultimate strength

S ut  130  MPa

700 = a


Solution: 1.


The diving board will be in tension at the top of the board and compression along the bottom. At the top, over the right-hand support, the nonzero principal stress is positive and the load line on the 1-3 diagram is along the 1 axis. Using the Modified-Mohr failure theory, the static safety factor is Ns 

S ut

σ1

Ns  5.3



5-11-1


Given:

Repeat Problem 5-10 assuming the 100-kg person in Problem 5-10 jumps up 25 cm and lands back on the board. Assume the board weighs 29 kg and deflects 13.1 cm statically when the person stands on it. What is the static safety factor if the material is brittle fiberglass with S ut = 130 MPa i the longitudinal direction? Maximum principal stresses due to bending at R2 from Problem 4-11

2000 = L R1

P

σ1  76.3 MPa σ2  0  MPa

R2

σ3  0  MPa Ultimate strength

700 = a

S ut  130  MPa FIGURE 5-11 Free Body Diagram for Problem 5-11

Solution: 1.


The diving board will be in tension at the top of the board and compression along the bottom. At the top, ove the right-hand support, the nonzero principal stress is positive and the load line on the 1-3 diagram is along the 1 axis. Using the Modified-Mohr failure theory, the static safety factor is Ns 

S ut

σ1

Ns  1.7



5-12-1

PROBLEM 5-12 Statement: Given:

Repeat Problem 5-10 using the cantilevered diving board design in Figure P5-4b. 2000

Maximum principal stresses due to bending at support from Problem 4-12

1300 = L

σ1  24.5 MPa

P

σ2  0  MPa M1

σ3  0  MPa Ultimate strength Solution:

S ut  130  MPa


R1

700


1.

The diving board will be in tension at the top of the board and compression along the bottom. At the top, at the built-in support, the nonzero principal stress is positive and the load line on the 1-3 diagram is along the 1 axis. Using the Modified-Mohr failure theory, the static safety factor is Ns 

S ut

σ1

Ns  5.3



5-13-1


Given:

Repeat Problem 5-11 using the cantilevered diving board design in Figure P5-4b. Assume the board weighs 19 kg and deflects 8.5 cm statically when the person stands on it. 2000

Maximum principal stresses due to bending at support from Problem 4-13

1300 = L

σ1  87.1 MPa

P

σ2  0  MPa M1

σ3  0  MPa Ultimate strength Solution:

S ut  130  MPa


R1

700


1.

The diving board will be in tension at the top of the board and compression along the bottom. At the top, at the built-in support, the nonzero principal stress is positive and the load line on the 1-3 diagram is along the 1 axis. Using the Modified-Mohr failure theory, the static safety factor is Ns 

S ut

σ1

Ns  1.5



5-14-1


Figure P4-5 shows a child's toy called a pogo stick. The child stands on the pads, applying half he weight on each side. She jumps off the ground, holding the pads up against her feet, and bounces along with the spring cushioning the impact and storing energy to help each rebound. Design the aluminum cantilever beam sections on which she stands to survive jumping 2 in off the ground with a safety factor of 2. Use 1100 series aluminum. Define and size the beam shape.

Given:

Cold rolled 1100 aluminum: Yield strength

S y  22 ksi

Safety factor

Ns  2

Assumptions: The beam will have a rectangular cross-section with the load applied at a distance of 5 in from the central support. L  5  in Solution:


1. From Problem 3-14, the total dynamic force on both foot supports is

Fi /2

Fi /2

Fi  224  lbf Therefore, the load on each support is P 

Fi

P  112  lbf

2

2. To give adequate support to the childs foot, let the width of the support beam be w  1.5 in 3. From Figure B-1(a) in Appendix B, the maximum bending moment at x = 0 is M  P L

P FIGURE 5-14

M  560  in lbf


4. We can now calculate the minimum required section modulus, Z = I/c. Using the distortion-energy failure theor the bending stress will also be the only nonzero principal stress, which will also be the von Mises stress. Design equation

Ns =

Bending stress

σ=

Solving for Z,

5. For a rectangular cross-section,

Solving for t,

Z 

Sy

σ' M Z

Sy Ns

N s M

I= t 

= σ' =

Z  834.3  mm

Sy w t

3

12 6 Z w

and

c=

t 2

so

Z=

3

w t

2

6

t  0.451  in


t  0.500  in



5-15-1


What is the safety factor for the shear pin as defined in Problem 4-15?

Solution: Any part whose stress equals its strength has a safety factor of 1 by definition.



5-16-1


A track to guide bowling balls is designed with two round rods as shown in Figure P5-6. The rods are not parallel to one another but have a small angle between them. The balls roll on the rods unt they fall between them and drop onto another track. The angle between the rods is varied to cause the ball to drop at different locations. Find the static safety factor for the 1-in dia SAE 1045 normalized steel rods. (a) Assume rods are simply supported at each end. (b) Assume rods are fixed at each end. S y  58 ksi

Given:

Yield strength

Solution:


Fball

a

R1

1. The maximum bending stress will occur at the outer fibers of the rod at the section where the maximum bending moment occurs which, in this case, is at x = a. The only stress present on the top or bottom surface of the rod is the bending stress x. Therefore, on the

R2

L


bottom surface where the stress is tensile, x is the principal stress 1 . Thus, from Problem 4-16, for a simply supported rod, Maximum principal stress

σ1  748  psi

σ'a  σ1

2. Using the distortion-energy failure theory, the safety factor against a static failure is Nsa 

Sy

σ'a

Nsa  78

3. For the built-in case, the maximum bending stress will occur at the outer fibers of the rod at the section where the maximum bending moment occurs which, in this case, is at x = L. The only stress present on the top or bottom surface of the rod is the bending stress x. Therefore, on the bottom surface where the stress is tensile, x is the principal stress 1 . Thus, from Problem 4-16, for a simply supported rod, Maximum principal stress

Fball

a

M1

R1

L

R 2 M2


σ1  577  psi

σ'b  σ1

4. Using the distortion-energy failure theory, the safety factor against a static failure is Nsb 

Sy

σ'b

Nsb  101



5-17-1


Given:

A pair of ice tongs is shown in Figure P5-7. The ice weighs 50 lb and is 10 in wide across the tongs. The distance between the handles is 4 in, and the mean radius r of the tong is 6 in. The rectangular cross-sectional dimensions are 0.75 x 0.312 in. Find the safety factor for the tongs if their S y = 30 ksi. F Yield strength

C

S y  30 ksi

FC O

See Problem 4-17, Figure 5-17, and Solution: Mathcad file P0517. 11.0 = ax

1. The maximum bending stress in the tong was found in Problem 4-17 at point A. Vertical direction

3.5 = cy

FO 2.0 = cx

A

12.0 = by

σi  8.58 ksi

5.0 = bx

FB

All other components are zero

B 2. There are no other stress components present so

σ1  σi

σ2  0  ksi

σ3  0  ksi

and

σ'  σ1

σ'  8.58 ksi

W/2 FIGURE 5-17

3. The factor of safety is (using the distortion energy theory)


N 

Sy

σ'

N  3.5



5-18-1


A pair of ice tongs is shown in Figure P5-7. The ice weighs 50 lb and is 10 in wide across the tongs. The distance between the handles is 4 in, and the mean radius r of the tong is 6 in. The rectangular cross-sectional dimensions are 0.75 x 0.312 in. Find the safety factor for the tongs if they are made from Class 20 gray cast iron.

F Given:

Tensile strength

S ut  22 ksi

Compressive strength

S uc  83 ksi

C FC O

See Problem 4-18, Figure 5-18, and Solution: Mathcad file P0518. 1. The maximum bending stress in the tong was found in Problem 4-17 at point A.

All other components are zero

12.0 = by 5.0 = bx

FB B

2. Therefore, the principal stresses are

σ2  0  ksi

2.0 = cx

A

σi  8.58 ksi

Vertical direction

σ1  σi

11.0 = ax

3.5 = cy

FO

W/2

σ3  0  ksi

3. The load line on the 1-3 diagram is along the 1 axis. Using the Modified-Mohr failure theory, the static safety factor is S ut N  N  2.6


σ1



5-19-1


Determine the size of the clevis pin, shown in Figure P5-8, needed to withstand an applied force of 130 000 lb. Also determine the required outside radius of the clevis end to not fail in either tear out or bearing if the clevis flanges are each 2.5 in thick. Use a safety factor of 3 for all modes of failure. Assume S y = 89.3 ksi for the pin and S y = 35.5 ksi for the clevis.

Given:

Applied force Clevis strength

P  130  kip S yclevis  35.5 ksi

Safety factor

Ns  3

Solution: 1.

Clevis flange thickness t  2.50 in Pin strength S ypin  89.3 ksi

See Figures P5-8 in the text and Mathcad file P0519.

Determine the force carried by each of the two flanges of the clevis. F  0.5 P

F  65 kip

This force is transmitted through each end of the clevis pin, which is in double shear. 2.

σ'pin = 3  τpin =

The pin is in direct (pure) shear. Therefore, the von Mises stress is

4 3 F

π d 3.

Calculate the minimum required clevis pin diameter using the distortion-energy failure theory. Ns =

S ypin

σ'pin

2

=


π d  S ypin 4 3 F 4  3  F  Ns

d 

d  2.194 in

π S ypin

Round this up to the next higher decimal equivalent of a common fraction ( 2 1/4) 4.

2

d  2.250  in

Check the bearing stress in the clevis due to the pin on one side of the clevis. 2

Bearing stress area

Ab  d  t

Ab  5.625 in

Bearing force

Fb  F

Fb  65 kip

Bearing stress

σb 

Fb

σb  11.6 ksi

Ab

Tearout length

5. Determine the safety factor against a static bearing failure. Nbear 

S yclevis

σb

Nbear  3.1

Since this is greater than 3, the pin diameter is acceptable. 6. Determine the tearout stress in the clevis. Shear area (see Figure 5-19)

2

Atear = 2  t R  ( 0.5 d )

2

d

R

FIGURE 5-19 Shear force Ftear  F


Ftear  65 kip



Shear stress and distortion-energy equation

Ns =

S yclevis

σ'tear

=

S yclevis 3  τtear

τtear =

2

=

5-19-2

2  t S yclevis R  ( 0.5 d )

Ftear Atear

Ftear

=

2

2  t R  ( 0.5 d )

2

2

3  Ftear

2


 3  Ftear Ns  2 R     ( 0.5 d ) 2  t S  yclevis  

R  2.211 in

Round this up to the next higher decimal equivalent of a common fraction ( 2 1/4)

The tearout area for each flange is

2

Atear  2  t R  ( 0.5 d )

2

R  2.250  in 2

Atear  9.743 in

7. Design summary: Pin diameter

d  2.250 in

Clevis flange radius

R  2.25 in



5-20-1


A 100 N-m torque is applied to a 1-m-long, solid, round shaft. Design it to limit its angular deflection to 2 deg and select a steel alloy to have a yielding safety factor of 2.

Given:

Applied torque Maximum deflection

T  100  N  m θmax  2  deg

Safety factor

Ns  2

L  1000 mm G  79 GPa

Shaft length Modulus of rigidity

Assumptions: A ductile steel will be chosen. Solution: 1.


Using the angular deflection requirement and equation (4.24), determine the required polar moment of inertia an the minimum diameter.

θ=

TL

J 

J G

TL

4

J  3.626  10  mm

θmax G

4

1

J =

π d

d 

32

d  24.653 mm

d

 2

τmax 

τmax  34.47  MPa

J

For this case of pure shear, use the distortion-energy theory and equations (5.8) and (5.9) to solve for the minimum required yield strength.

Ns =

4.

4

Determine the shear stress at the outside diameter of the shaft using equation (4.23b). T  

3.

 32 J   π   

d  25 mm

Round this up to 2.

4

Sy

σ'

=

Sy 3  τmax

S y 

3  τmax Ns

S y  119.4  MPa

Using this value of S y, choose a steel from Table A-9 in Appendix A. Any of the steels listed in Table A-9 will be adequate. The least expensive is AISI 1010, hot rolled.



5-21-1


Figure P5-9 shows an automobile wheel with two common styles of lug wrench being used to tighten the wheel nuts, a single-ended wrench in (a), and a double-ended wrench in (b). The distance between points A and B is 1 ft in both cases and the handle diameter is 0.625 in. What is the maximum force possible before yielding the handle if the material S y = 45 ksi?

Given:


d AB  1  ft

Wrench diameter Yield strength

d  0.625  in S y  45 ksi



1. From examination of the FBDs, we see that, in both cases, the arms are in bending and the stub that holds the socket wrench is in pure torsion. The maximum bending stress in the arm will occur near the point where the arm transitions to the stub. The stress state at this transition is very complicated, but we can find the nominal bending stress there by treating the arm as a cantilever beam, fixed at the transition point. For both cases the torque in the stub is the same.

F


12" = dAB

Case (a)

F

6"

2. The bending moment at the transition is Ma = Fa d AB T

3. The tensile stress at this point is found from F (b) Double-ended Wrench

Moment of inertia I 

π d

4

FIGURE 5-21

4

I  0.00749 in

64



c  0.5 d

Stress

σx =

c  0.313 in

Ma c I


σ2  0  psi

σ1 = σx

σ3  0  psi

5. Since there is only one nonzero principal stress, the von Mises stress is

σ' = σ1 = σx =

Ma c I

=

Fa d AB c I

6. Using the distortion-energy theory, solve for the maximum applied force. Ns =

Sy

σ'

=

I Sy Fa d AB c

=1

Fa 

I  Sy d AB c

Fa  89.882 lbf



5-21-2

σ' 

7. The von Mises stress in the handle at the transition point is

T  Fa d AB

8. Determine the torque in the stub.

Fa d AB c

σ'  45 ksi

I T  1079 in lbf

9. The shear stress at any point on the outside surface of the stub is found from Polar moment of inertia

J  2  I

Shear stress

τxy 

4

J  0.0150 in

Tc

τxy  22.5 ksi

J


σ1  τxy and

σ' 

σ1  22.5 ksi 2

σ1  σ1 σ3  σ3

σ2  0  psi

2

σ3  σ1

σ'  39.0 ksi

11. Thus, the maximum von Mises stress for case (a) is on the upper surface of the handle (arm) near the point where it transitions to the stub, and the maximum force that can be applied to the handle without yielding is Fa  89.9 lbf Case (b) 12. The bending moment at the transition is

Mb =


σx =

Fb d AB 2 Mb c I


σ2  0  psi

σ1 = σx

σ3  0  psi

13. Since there is only one nonzero principal stress, the von Mises stress is σ' = σ1 = σx =

Mb c I

=

Fb d AB c 2 I

14. Using the distortion-energy theory, solve for the maximum applied force. Ns =

Sy

σ'

=

2 I  Sy Fb d AB c

Fb 

=1

2 I Sy

Fb  179.763 lbf

d AB c

15. The von Mises stress in the handle at the transition point is T  Fb d AB

16. The torque in the stub is

σ' 

Fb d AB c 2 I

σ'  45 ksi

T  2157 in lbf

14. The shear stress at any point on the outside surface of the stub is found from Shear stress

τxy 

Tc J

τxy  45 ksi


σ1  τxy

σ1  45.0 ksi

σ2  0  psi

σ3  σ1



and

σ' 

2

σ1  σ1 σ3  σ3

5-21-3

2

σ'  77.9 ksi

16. Since the von Mises stress in the stub due to torsion is greater than the yield strength, the force in the handle will be limited by the shear stress in the stib and by the bending stress in the handle. Ns =

Fb 

Sy

σ'

=

Sy 3  τxy

J  Sy 3  d AB c

=

J  Sy 3 T  c

=

J  Sy 3  Fb d AB c

=1

Fb  103.8 lbf

17. Thus, the maximum von Mises stress for case (b) is on the stub, and the maximum force that can be applied to the handles without yielding is Fb  103.8 lbf



5-22-1


A roller-blade skate is shown in Figure P5-10. The polyurethane wheels are 72 mm dia and spaced on 104-mm centers. The skate-boot-foot combination weighs 2 kg. The effective "spring rate" of the person-skate subsystem is 6000 N/m. The axles are 10-mm-dia steel pins in double shear with S y = 400 MPa. Find the safety factor for the pins when a 100-kg person lands a 0.5-m jump on one foot. (a) Assume all 4 wheels land simultaneously. (b) Assume that one wheel absorbs all the landing force.

Given:

Axle pin diameter

Solution:


d  10 mm

1.

From Problem 4-22, we have the stresses for cases (a) and (b):

2.

Using the distortion-energy failure theory, Case (a) all wheels landing

Nsa 

Case (b) one wheel landing

Nsb 

Sy 3  τa Sy 3  τb

Yield strength

S y  400  MPa

τa  5.71 MPa

τb  22.9 MPa

Nsa  40.4

Nsb  10.1



5-23a-1


Given:

A beam is supported and loaded as shown in Figure P5-11a. For the data given in row a from Table P5-2, find the static safety factor: (a) If the beam is a ductile material with S y = 300 MPa, (b) If the beam is a cast-brittle material with S ut = 150 MPa, S uc = 570 MPa. Ductile yield strength

L

S y  300  MPa

b

Brittle ultimate tensile strength S ut  150  MPa

Solution:

a

F

w


R2

R1


1.

The maximum bending stress occurs under the concentrated load F at x = b. It was determined in Problem 4-23a as

σx  88.7 MPa 2.

Since this is the only stress component present in the given coordinate frame, x is equal to 1 and the other two principal stresses are zero.

σ1  σx 3.

4.

σ2  0  MPa

σ3  0  MPa

For case (a), use the distortion-energy failure theory. With only one nonzero principal stress, the von Mises stress is the same as 1. von Mises stress

σ'  σ1

Safety factor, case (a)

Nsa 

Sy

σ'

σ'  88.7 MPa Nsa  3.4

For case (b), use the Modified Mohr failure theory. The nonzero principal stress is positive and the load line on the s1-s3 diagram is along the 1 axis. Safety factor, case (b)

Nsb 

S ut

σ1

Nsb  1.7



5-24a-1


A beam is supported and loaded as shown in Figure P5-11b. For the data given in row a from Tabl P5-2, find the static safety factor: (a) If the beam is a ductile material with S y = 300 MPa, (b) If the beam is a cast-brittle material with S ut = 150 MPa, S uc = 570 MPa. L

Given:

Ductile yield strength

a

S y  300  MPa

F

Brittle ultimate strength S ut  150  MPa

w

M1

Solution:


R1


1.

The maximum bending stress occurs at the support where x = 0. It was determined in Problem 4-24a as

σx  410  MPa 2.


σ1  σx 3.

4.

σ2  0  MPa

σ3  0  MPa


σ'  σ1


Nsa 

Sy

σ'

σ'  410 MPa Nsa  0.73

For case (b), use the Modified Mohr failure theory. The nonzero principal stress is positive and the load line on the 1-3 diagram is along the 1 axis. Safety factor, case (b)

Nsb 

S ut

σ1

Nsb  0.37



5-25a-1


A beam is supported and loaded as shown in Figure P5-11c. For the data given in row a from Table P5-2, find the static safety factor: (a) If the beam is a ductile material with S y = 300 MPa, (b) If the beam is a cast-brittle material with S ut = 150 MPa, S uc = 570 MPa. L

Given:


S y  300  MPa

b

Brittle ultimate strength S ut  150  MPa

Solution:

a

F w


R2

R1


1.

The maximum bending stress occurs at the right-hand support where x = b. It was determined in Problem 4-25a as σx  151.6  MPa

2.


σ1  σx 3.

4.

σ2  0  MPa

σ3  0  MPa


σ'  σ1


Nsa 

Sy

σ'

σ'  151.6 MPa Nsa  2.0


Nsb 

S ut

σ1

Nsb  0.99



5-26a-1


A beam is supported and loaded as shown in Figure P5-11d. For the data given in row a from Tabl P5-2, find the static safety factor: (a) If the beam is a ductile material with S y = 300 MPa, (b) If the beam is a cast-brittle material with S ut = 150 MPa, S uc = 570 MPa. L

Given:


S y  300  MPa

b

Brittle ultimate strength S ut  150  MPa Solution: 1.

a

F w


R2

R1

R3

FIGURE 5-26

The maximum bending stress occurs under the concentrated load F, where x = a. It was determined in Problem 4-26a as


σx  31.5 MPa 2.


σ1  σx 3.

4.

σ2  0  MPa

σ3  0  MPa


σ'  σ1


Nsa 

Sy

σ'

σ'  31.5 MPa Nsa  9.5


Nsb 

S ut

σ1

Nsb  4.8



5-27-1


A storage rack is to be designed to hold the paper roll of Problem 5-8 as shown in Figure P5-12. Determine suitable values for dimensions a and b in the figure. Make the static factor of safety at least 1.5. The mandrel is solid and inserts halfway into the paper roll. (a) The beam is a ductile material with Sy = 300 MPa (b) The beam is a cast-brittle material with Sut = 150 MPa, S uc = 570 MPa.

Given:



3

OD  1.50 m ID  0.22 m Lroll  3.23 m

Factor of safety

Ns  1.5

S y  300  MPa

Brittle ultimate strength

S ut  150  MPa

Roll density

Assumptions: The paper roll's weight creates a concentrated load acting at the tip of the y mandrel. The mandrel's root in the w stanchion experiences a distributed load a over the length of engagement (see the solution to Problem 3-27 for further discussion of this point). The required diameter a of the mandrel root section b (over the length b) will be sized to use the allowable tensile strength in bending. R The length b will be sized to use the FIGURE 5-27 allowable transverse shear strength.

ρ  984  kg m

W

x

Lm


Solution: 1.

2.


Determine the weight of the roll and the length of the mandrel.

Length


2  W  Lm b



W  53.9 kN Lm  1.615 m

Mmax  W  Lm

Mmax  87.04 kN  m

Part (a) - The bending stress will be a maximum at the top or bottom of the mandrel at a section where the mandrel root leaves the stanchion. Mmax a

where

2 I

I=

π a

4

so,

64

σmax =

32 Mmax

π a

3

At this point the only nonzero stress component is max therefore

σ2  0  MPa

σ1 = σmax 5.

2


The maximum internal shear and moment occur at a section where the mandrel root leaves the stanchion. and are

σmax = 4.

2

W 

Vmax =

3.

 4

π

Weight

σ3  0  MPa

All three of the ductile failure theories have the same fail/safe boundary for this condition (slope of load line is zero) Ns =

Sy

σ1

or

Ns σ1 = S y



5-27-2

1

6.

7.

Solving for a,

 32 Ns W  Lm  a     π S y 

Round this to

a  166  mm

4  Vmax 3 A

8  W  Lm

=

 π a 2   b  4 

3 

At this point, this is the only nonzero stress component therefore, the principal stresses are

σ2  0  MPa

σ1 = τmax 9

a  164.272 mm

Using this value of a and equation (4.15c), solve for the shear stress on the neutral axis at the same section.

τmax =

8.

3

σ3 = τmax

Using the distortion energy theory, the von Mises stress is

σ' = 3  τmax

and

b 

Solving for b

Ns =

Sy

σ'

8  Ns W  Lm

Sy 3

b  92.9 mm

 π a 2  S y  3   4  3 a  166 mm

Rounding to higher even values, let

Ns τmax =

or

b  94 mm

for case (a).

10. Part (b) - The bending stress will be a maximum at the top or bottom of the mandrel at a section where the mandrel root leaves the stanchion.

σmax =

Mmax a

where

2 I

I=

π a

4

so,

64

σmax =

32 Mmax

π a

3

11. At this point the only nonzero stress component is max therefore

σ2  0  MPa

σ1 = σmax

σ3  0  MPa

12. All three of the brittle failure theories have the same fail/safe boundary for this condition (slope of load line is zero) Ns =

S ut

σ1

=

S ut

σmax

=

2  I  S ut Mmax a

3

=

π a  S ut 32 Mmax 1

13. Solving for a, Round this to

 32 Ns Mmax  a     π Sut 

3

a  206.97 mm

a  208  mm

14. Using this value of a and equation (4.15c), solve for the shear stress on the neutral axis at the same section.



τmax =

4  Vmax 3 A

5-27-3

8  W  Lm

=

 π a 2   b 3   4 

15. At this point, this is the only nonzero stress component therefore, the principal stresses are

σ2  0  MPa

σ1 = τmax

σ3 = τmax

16. Using the Modified Mohr theory,

Ns =

Solving for b

S ut

σ1

=

 π a 2    b S ut  4 

3 

S ut

τmax

=

b 

8  W  Lm 8  Ns W  Lm

 π a 2    Sut 3   4 

Rounding to higher even values, let

a  208 mm

b  68.3 mm

b  70 mm

for case (b).



5-28-1


Figure P5-13 shows a forklift truck negotiating a 15 deg ramp to to drive onto a 4-ft-high loading platform. The truck weighs 5 000 lb and has a 42-in wheelbase. Design two (one for each side) 1-ft-wide ramps of steel to have a safety factor of 3 in the worst case of loading as the truck travels up them. Minimize the weight of the ramps by using a sensible cross-sectional geometry. Choose an appropriate steel or aluminum alloy.

Given:

Ramp angle θ  15 deg Platform height h  4  ft Truck wheelbase Lt  42 in

Ramp width Truck weight

w  12 in W  5000 lbf

Assumptions: 1. The worst case is when the truck CG is located at the center of the beam's span. 2. Use a coordinate frame that has the x-axis along the long axis of the beam. 3. Ignore traction forces and the weight components along the x-axis of the beam. 4. There are two ramps, one for each side of the forklift. See Figure 5-28 and Mathcad file P0528. Solution:

L b a

CG a

y

CG b

R1  Fa

Wa

Fb

x Wb

R2


1. From Problem 3-28 the maximum bending moment in the ramp occurs at the rear wheel of the truck and is Mmax  8324 ft  lbf

Mmax  99888 in lbf

2. The bending stress is the only stress component present and is, therefore, also the only nonzero principal stress and is also the von Mises stress. The governing design equations then are

σ' =

Mmax Z

and

Ns =

Sy

σ'

3. The approach will be to 1) choose a suitable factor of safety, 2) choose a suitable material and determine its yiel strength, 3) from the equations above determine the required value of the section modulus, 4) choose an appropriate cross-section for the ramp, and 5) determine the dimensions of the cross-section. 4. The following design choices have been made for this problem: © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0528.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


5-28-2

Design factor of safety

Nsd  3

Material

7075 Aluminum, heat treated

Yield strength

S y  73 ksi

5. Solve the design equations for the minimum section modulus, Z. Nsd  Mmax

Z 

3

Z  4.105 in

Sy

This is the minimum allowable value of the section modulus. 6. Assume a channel section such as that shown in Figure 5-28B. To keep it simple, let the thickness of the flanges and web be the same. Choose 1/2-in thick plate, which is readily available. Then, t  0.50 in A ( h )  w t  2  t ( h  t)

7. The cross-sectional area of the ramp is cg( h ) 

8. The distance to the CG is

 w t 2

1 A (h)



 2

2

 t h  t

2

 

9. The moments of inertia of the web and a flange are Iweb( h ) 

Ifl ( h ) 

w t

3

12

 w t  cg( h ) 



t ( h  t)

3

12

t

 

Flange

2

Web

2

h  t  h  t  cg( h )   2  

t

2

I ( h )  Iweb( h )  2  Ifl ( h )

h

11. The maximum stress will occur in the flange at the top and is compressive. The distance from the centroid up to the top of the flange is

w

c( h )  h  cg( h ) 12. Using the known section modulus, solve for the unknown flange height, h. Guess h  1  in

FIGURE 5-28B Channel Section for Problem 5-28

Given Z=

I (h) c( h )

h  Find ( h )

h  3.843 in

Round this up to

h  4.00 in

13. Summarizing, the ramp design dimensions are: Width

w  12.00 in

Flange height

h  4.00 in

Shape

channel

Thickness

t  0.5 in

Material

7075 aluminum



5-29-1


A differential element is subected to the stresses given below and a ductile material has the strengths given below. Calculate the safety factor and draw 1-3 diagrams of each theory showing the stress state using: (a) Maximum shear-stress theory, and (b) Distortion-energy theory.

Given:

Principal stresses

σ1  10 ksi

σ2  0  ksi

σ3  20 ksi

Material properties


S y  40 ksi


Solution:


1. Calculate the slope of load line. (The load line is the line from the origin through the stress point.) m 

σ3

m  2

σ1

2. The safety factor equation for the distortion-enrgy theory is the same regardless of which quadrant the load line falls in. However, the equation for the maximum shear-stress factor of safety is different for each of the three quadrants that the load line (1st, 3rd, or 4th) can fall in. In this case, the load line falls in the 4th quadrant. The factors of safety are: (a) Maximum shear-stress theory

Sy

Na 

Na  1.3

σ1  σ3

3

(b) Distortion energy theory 40

σ1  σ1 σ3  σ3

σ'  26.5 ksi Nb 

Sy

σ'

(a) Maximum shear stress boundary

2

Nb  1.5

3. Plot the 1-3 diagram showing the safe-fail boundaries, the stress state point (10 ksi, -20 ksi) and the load line. Note that if 1 > 3 , then only that area on the graph that is to the right of and below the diagonal line can contain valid stress points. The factor of safety is the distance along the load line from the origin to the intersection of the load line with the failure boundary, divided by the distance from the origin to the stress point. Since the distance from the origin to the distortion-energy boundary is greater than the distance to the maximum shear-stress baoundary, its factor of safety is greater.

30 (b) Distortion energy boundary MINIMUM NONZERO PRINCIPAL STRESS, KSI

σ' 

2

20 10 0

1

sy

-10 (10,-20) -20 -30

Stress states at which failure will occur

-sy

-40

Load Line

-50 -60 -40

-30

-20

-10

0

10

20

30

40

50

MAXIMUM PRINCIPAL STRESS, KSI

FIGURE 5-29 1 -  3 Diagram for Problem 5-29



5-30-1


A differential element is subected to the stresses and strengths given below. Calculate the safety factor and draw 1-3 diagrams of each theory showing the stress state using: (a) Coulomb-Mohr theory, and (b) Modified Mohr theory.

Given:

Principal stresses

σ1  10 ksi

σ2  0  ksi

σ3  20 ksi

Material properties


S y  40 ksi


Solution:



σ3 σ1

m  2

2. The safety factor equation for both theories is different for each quadrant the load line falls in. The equation for the modified Mohr factor of safety is different for each of the two regions in the 4th quadrant that the load line can fall in. In this case, the load line falls in the 4th quadrant, below the -1 slope line.. The factors of safety are: (a)Coulomb-Mohr theory

50 40

S uc

Nb  3.2 3. Plot the 1-3 diagram showing the safe-fail boundaries, the stress state point (10 ksi, -20 ksi) and the load line. Note that if 1 > 3 , then only that area on the graph that is to the right of and below the diagonal line can contain valid stress points. The factor of safety is the distance along the load line from the origin to the intersection of the load line with the failure boundary, divided by the distance from the origin to the stress point. Since the distance from the origin to the modified Mohr boundary is greater than the distance to the Coulomb-Mohr boundary, its factor of safety is greater.

30 MINIMUM NONZERO PRINCIPAL STRESS, KSI

 S uc  S ut   S   σ1  σ3 ut  

Na  2.4

S uc σ1  S ut σ3

3

(b) Modified Mohr theory Nb 

S uc S ut

Na 

20

(a) Coulomb-Mohr boundary

10

1

0 -10 (10,-20) -20 -30 -40 Stress states at which failure will occur

-50 -60

-S

(b) Modified Mohr boundary

-70

Load Line

-80 -90

-S

ut

uc

-100 -100 -90 -80 -70 -60 -50 -40 -30 -20 -10

0

10

20

30

40

50





5-31-1


Design a jack-stand in a tripod configuration that will support 2 tons of load with a safety factor of 3. Use SAE 1020 steel and minimize its weight.

Solution:

This open-ended design problem has many valid solutions that are left to the student.



5-32-1


A part has the combined stress state and strengths given below. Choose an appropriate failure theory based on the given data, find the effective stress and factor of safety against static failure.

Given:

Stresses: σx  10 ksi Strengths:

Solution:

σy  5  ksi

S y  18 ksi

τxy  4.5 ksi




1.

Because S uc is greater than S ut, this is an uneven material, which is characteristic of a brittle material. Therefore, use the modified Mohr theory.

2.

Find the maximum shear stress and principal stresses that result from this combination of applied stresses using equations 4.6.

Maximum shear stress

τmax 

Principal stresses

σ1  σ2 

2  σx  σy  2    τxy 2  

σx  σy 2

σx  σy 2

τmax  5.148 ksi

 τmax

σ1  12.648 ksi

 τmax

σ2  2.352 ksi

σ3  0  psi 3.

4.

Find the Dowling factors C1, C2, C3 using equations 5.12b: C1 

1

C2 

1

C3 

1

2

2

2



S uc  2  S ut



S uc



S uc  2  S ut



S uc



S uc  2  S ut



S uc

  σ1  σ2 

  σ2  σ3 

  σ3  σ1 



  σ1  σ2

C1  8.898 ksi

 

  σ2  σ3

C2  1.764 ksi

 

  σ3  σ1

C3  9.486 ksi



Then find the largest of the six stresses C1, C2, C3 , 1, 2, 3: C   1     C2    C  3 σeff  max      σ1       σ2     σ3  

σeff  12.6 ksi

which is the modified-Mohr effective stress. 5.

The safety factor can now be found using equation 5.12d.

N 

S ut

σeff

N  1.6



5-33a-1


For the bracket shown in Figure P5-14 and the data in row a of Table P5-3, determine the von Mises stresses at points A and B.

Solution:


1.

From Problem 4-33a the principal stresses at point A are

σ1  21.46  MPa 2.

2

σ1  σ1 σ3  σ3

2

σ'A  30.2 MPa

From Problem 4-33a the principal stresses at point B are

σ1  16.13  MPa 4.

σ3  13.08  MPa

Use equation (5.7c) to find the von Mises stress at point A.

σ'A  3.

σ2  0  MPa

σ2  0  MPa

σ3  16.13  MPa

Use equation (5.7c) to find the von Mises stress at point B.

σ'B 

2

σ1  σ1 σ3  σ3

2

σ'B  27.9 MPa

F

y A B

T

T

x

M L R




5-34a-1


Calculate the safety factor for the bracket in Problem 5-33 using the distortion energy, the maximum shear stress, and the maximum normal-stress theories. Comment on their appropriateness. Assume a ductile material strength as given below.

Given:

Yield strength

Solution:


1.

S y  400  MPa


σ1A  21.46  MPa 2.

σ2A  0  MPa

σ3A  13.08  MPa

Using the two nonzero stresses, the slope of the load line on a 1-3 graph is m 

σ3A

m  0.61

σ1A

This intersects the failure boundaries in the fourth quadrant. 3.

Calculate the von Mises effective stress at point A using equation (5.7c).

σ'A  4.

5.

2

2

σ1A  σ1A σ3A  σ3A

Determine the factor of safety at point A Distortion energy

NADE 


NAMS 

Maximum normal stress

NANS 

Sy

NADE  13.2

σ'A Sy

σ1A  σ3A Sy

σ1A

NAMS  11.6

NANS  18.6

From Problem 4-33a, the principal stresses at Point B are

σ1B  16.13  MPa 6.

σ'A  30.205 MPa

σ2B  0  MPa

σ3B  16.13  MPa


σ3B

m  1

σ1B

This intersects the failure boundaries in the fourth quadrant. 7.

Calculate the von Mises effective stress at point A using equation (5.7c).

σ'B  8.

2

2

σ1B  σ1B σ3B  σ3B

σ'B  27.938 MPa

Determine the factor of safety at point B



9.

Distortion energy

NBDE 


NBMS 

Maximum normal stress

NBNS 

5-34a-2

Sy

NBDE  14.3

σ'B Sy

σ1B  σ3B Sy

σ1B

NBMS  12.4

NBNS  24.8

Whichever theory is used, the critical point (lowest factor of safety) is point A. The distortion energy theory should be used because experimental data follows its failure boundary more nearly than the maximum shear stress in all quadrants. Using the maximum normal stress theory would give an overestimate of the actual safety factor.



5-35a-1


Calculate the safety factor for the bracket in Problem 5-33 using the Coulomb-Mohr and the modified Mohr effective stress theories. Comment on their appropriateness. Assume a brittle material strength as given below.

Given:

Tensile strength

Solution:


1.

S ut  350  MPa

S uc  1000 MPa


σ1A  21.46  MPa 2.

Compressive strength

σ2A  0  MPa

σ3A  13.08  MPa


σ3A

m  0.61

σ1A

This intersects the failure boundaries in the fourth quadrant. For the Coulomb-Mohr diagram (see Figure 5-9 in the text) there is a single, straight line in this quadrant. For the modified-Mohr theory, the load line will intersect the boundary at a point similar to B' in Figure 5-11 in the text. 3.

4.

Determine the factor of safety at point A Coulomb-Mohr

NACM 

Modified-Mohr

NAMM 

S uc σ1A  S ut σ3A S ut

σ1A

NACM  13.4

NAMM  16.3

From Problem 4-33a, the principal stresses at Point B are

σ1B  16.13  MPa 5.

S ut S uc

σ2B  0  MPa

σ3B  16.13  MPa


σ3B

m  1

σ1B

This intersects the failure boundaries in the fourth quadrant. For the Coulomb-Mohr diagram (see Figure 5-9 in the text) there is a single, straight line in this quadrant. For the modified-Mohr theory, the load line will intersect the boundary at the point (S ut -S ut) Figure 5-11 in the text. 6.

7.

Determine the factor of safety at point B Coulomb-Mohr

NBCM 

Modified-Mohr

NBMM 

S ut S uc S uc σ1B  S ut σ3B S ut

σ1B

NBCM  16.1

NBMM  21.7

Whichever theory is used, the critical point (lowest factor of safety) is point A. The modified-Mohr theory should be used because experimental data follows its failure boundary more nearly than the Coulomb-Mohr when the slope of the load line is in the fourth quadrant. Using the Coulomb-Mohr would give an underestimat of the actual safety factor.



5-35a-2

Calculating factor of safety using Modified Mohr and equations (5.12c, d, and e) Point A C1 

1

C2 

1

C3 

1

2

2

2



S uc  2  S ut



S uc



S uc  2  S ut



S uc



S uc  2  S ut



S uc

  σ1A  σ2A 

  σ2A  σ3A 

  σ3A  σ1A 

1 is maximum so



  σ1A  σ2A

 

  σ2A  σ3A

 

  σ3A  σ1A



N 

S ut

C1  13.9 MPa

C2  4.6 MPa

C3  18.5 MPa

N  16.3

σ1A

Point B

C1 

1

C2 

1

C3 

1

2

2

2



S uc  2  S ut



S uc



S uc  2  S ut



S uc



S uc  2  S ut



S uc

  σ1B  σ2B 

  σ2B  σ3B 

  σ3B  σ1B 

1 is maximum so



  σ1B  σ2B

 

  σ2B  σ3B

 

  σ3B  σ1B



N 

S ut

σ1B

C1  10.5 MPa

C2  5.6 MPa

C3  16.1 MPa

N  21.7



5-36a-1


For the bracket shown in Figure P5-14 and the data in row a of Table P5-3, redo Problem 5-33 considering the stress concentration at points A and B. Assume a stress concentration factor of 2.5 in both bending and torsion.

Given:

Factors of safety: Bending

Solution: 1. 2.

4.

Kfs  2.5

Torsion


From Problem 4-36a the principal stresses at point A are σ1  53.6 MPa σ2  0  MPa

σ3  32.7 MPa

Use equation (5.7c) to find the von Mises stress at point A.

σ'A  3.

Kf  2.5

2

σ1  σ1 σ3  σ3

2

σ'A  75.5 MPa

From Problem 4-36a the principal stresses at point B are σ1  41.3 MPa σ2  0  MPa

σ3  41.3 MPa

Use equation (5.7c) to find the von Mises stress at point B.

σ'B 

2

σ1  σ1 σ3  σ3

2

σ'B  71.5 MPa F

y A B

T

T

x

M L R




5-37-1


Given: S y  700  MPa

A semicircular, curved beam as shown in Figure 5-37 has the dimensions given below. For a load pair F = 14 kN applied along the diameter, find the safety factor at the inner and outer fibers: (a) If the beam is a ductile material with Sy = 700 MPa, (b) If the beam is a cast-brittle material with Sut = 420 MPa, Suc = 1200 MPa.

(b) Tensile strength Solution: 1.

w

(a) Yield strength S ut  420  MPa

Compressive strength S uc  1200 MPa See Figure 5-37 and Mathcad file P0537.

From Problem 4-37, the stresses at the inside radius and outside radius are: Inside

σi  409.9  MPa

Outside

σo  273.2  MPa

F od

id F

(a) Entire Beam

These are the only stress components present on their respective surfaces so they are also principal stresses. Thus,

σ1i  409.9  MPa

σ2i  0  MPa

σ3i  0  MPa

σ1o  0  MPa

σ2o  0  MPa

σ3o  273.2  MPa

F M F

Part (a) 2.

rc

Use the distortion energy theory for the ductile material.

3. Since 1 is the only nonzero principal stress, it is also the von Mises effective stress,

(b) Critical Section


4.

σ'i  σ1i

σ'i  409.9 MPa

σ'o  σ3o

σ'o  273.2 MPa

The factor of safety against a static failure for this ductile material is Inside surface

Nai 

Outside surface

Nao 

Sy

σ'i Sy

σ'o

Nai  1.7 Nao  2.6

Part (b) 5.

Use the modified-Mohr theory for the brittle material.

6.

The load line on the 1-3 graph for the inside surface is along the positive 1 axis. In this case, the factor of safety equation simplifies to Inside surface

7.

Nbi 

S ut

σ1i

Nbi  1.0

The load line on the 1-3 graph for the outside surface is along the negative 3 axis. In this case, the factor of safety equation simplifies to Outside surface

Nbo 

S uc

σ3o

Nbo  4.4



5-38-1


Assume that the curved beam of Problem 5-37 has a crack on its inside surface of half-width a = 2 mm and a fracture toughness of 50 MPa-m0.5. What is its safety factor against sudden fracture?

Given:

Outside diameter Width of section

od  150  mm t  25 mm

Inside diameter Load

id  100  mm F  14 kN

Half crack length

a  2  mm

Fracture toughness

Kc  50 MPa m

Solution:


1.

From Problem 4-37, the nominal stress at the inside radius is: Nominal inside stress σi  409.9  MPa

2.

Calculate the half-width of the beam.

3.

Calculate the geometry and stress intensity factors. π a  β  sec  β  1.016   2 b  K  β  σi π a

4.

b  0.5 t

b  12.5 mm

K  33.01 MPa m

Determine the factor of safety against sudden fracture failure

NFM 

Kc K

NFM  1.5

w

F od

id F

(a) Entire Beam

F M F rc (b) Critical Section




5-39-1


Consider the failed 260-in dia by 0.73-in wall rocket case of Figure 5-14. The steel had S y = 240 k and a fracture toughness Kc = 79.6 ksi-in 0.5. It was designed for an internal pressure of 960 psi but failed at 542 psi. Failure was attributed to a small crack that precipitated a sudden, brittle, fracture-mechanics failure. Find the nominal stress in the wall and the yielding safety factor at the failure conditions and estimate the size of the crack that caused it to explode. Assume b = 1.0.

Given:

Solution: 1.

2.

3.

Case diameter

d  260  in

Fracture toughness

Kc  79.6 ksi in

Wall thickness

t  0.73 in

Design pressure

p d  960  psi

Yield strength

S y  240  ksi

Failure pressure

p f  542  psi


Find the nominal stress in the wall. The ratio of the wall thickness to the radius of the case is such that we can use thin-wall theory. Thus Case radius

r  0.5 d

Tangential stress

σt 

r  130 in

pd  r

σt  171.0 ksi

t pd  r

Axial stress

σa 

Radial stress

σr  0  psi

σa  85.5 ksi

2 t

Find the yielding safety factor at the failure conditions. Since, for these directions, there are no shear stresses present, these are the principal stresses. The von Mises stress is Von Mises stress

σ' 

Factor of safety against yielding

Ns 

2

σt  σt σa  σa Sy

σ'

2

σ'  148.1 ksi Ns  1.6

Estimate the size of the crack that caused it to explode. Tangential stress

Axial stress

σt  σa 

pf  r t pf  r 2 t

σt  96.5 ksi σa  48.3 ksi

(a) Assume that the crack was longitudinal (growing in the axial direction) Nominal stress

σnom  σt

Stress intensity factor

K = σnom a  π

σnom  96.5 ksi

Setting the stress intensity factor equal to the fracture toughness of the material and solving for the crack length

 Kc2     σ 2 π   nom 

Half-length

a 

Crack length

2  a  0.433 in

a  0.216 in



5-39-2

(b) Assume that the crack was tangential (growing in the tangential direction) Nominal stress

σnom  σa

Stress intensity factor

K = σnom a  π

σnom  48.3 ksi

Setting the stress intensity factor equal to the fracture toughness of the material and solving for the crack length

Half-length

 Kc2   a    σ 2 π   nom 

Crack length

2  a  1.732 in

a  0.866 in



5-40-1


Redesign the roll support of Problem 5-8 to be like that shown in Figure P5-16. The stub mandrels insert to 10% of the roll length at each end. Design dimension a for a factor of safety of 2. See Problem 5-8 for additional data. (a) The beam is a ductile material with S y = 300 MPa (b) The beam is a cast-brittle material with S ut = 150 MPa, S uc = 570 MPa.

Given:

Paper roll dimensions:

OD  1.50 m ID  0.22 m

Material properties: Yield strength

Lroll  3.23 m

Tensile strength S ut  150  MPa Comp strength

Roll density

S y  300  MPa

3

ρ  984  kg m

S uc  570  MPa Ns  2

Factor of safety

Assumptions: 1. The paper roll's weight creates a concentrated load acting at the tip of the mandrel. 2. The base of the mandrel (the portion that inserts into the stanchion) is solid and fits tightly into the stanchion. Therefore, the mandrel can be treated as a cantilever beam. 3. The length of ther mandrel base is b  100  mm. Solution: 1.

2.

For the assumptions made, it is not necessary to determine the stress distribution on the mandrel base inside the stanchion. From Figure 5-40, we see that we can determine the diameter a by applying the beam stress equation at the section where the mandrel transitions from the base to the full diameter.

π 4



2

2




y

x

a M1

Lm

b

W  53.9 kN

R

F  0.5 W

F  26.95 kN

FIGURE 5-40


Lm  323 mm


From Figure 5-40, the maximum internal bending moment occurs at x = 0 and is Mmax  F  Lm

4.

F

Determine the weight of the roll, the load on each support, and the length of the mandrel. W 

3.


Mmax  8.704 kN  m

The bending stress will be a maximum at the top or bottom of the mandrel at a section through x = 0.

σmax =

Mmax a 2 I

where

I=

π a

4

64

There are no other stress components at this point so σmax = σ1 and

σ2  0  MPa 5.

σ3  0  MPa

For the ductile material of part (a), the maximum principal stress is also the von Mises stress so

σmax = σ' =

32 Mmax

π a

3

=

Sy Ns



5-40-2

1

5.

Solving for a,

 32 Ns F  Lm  a     π S y 

Round this to

a  84 mm

3

a  83.922 mm

for the ductile material of part (a)

For the brittle material of part (b), the load line on the 1-3 diagram is along the positive 1 axis where both brittle material failure theories have the same boundary, which is 1 = S ut. Thus, for the brittle case of part (b),

σmax = σ1 =

32 Mmax

π a

3

=

S ut Ns

1

Solving for a,

 32 Ns F  Lm  a     π S ut 

Round this to

a  106  mm

3

a  105.735 mm

for the brittle material of part (b)



5-41-1


A 10-mm ID steel tube carries liquid at 7 MPa. The steel has S y = 400 MPa Determine the safety factor for the wall if its thickness is: a) 1 mm, b) 5 mm.

Given:

Yield strength

Assumption:

The tubing is long therefore the axial stress is zero.

Solution:

See Mathcad file P0541. t  1  mm

(a) Wall thickness is 1.

S y  400  MPa

From Problem 4-41, this is a thick wall cylinder and the principal stresses are:

σ1a  38.82  MPa 2.

2

2

σ1a  σ1a σ3a  σ3a

Sy

Na  9.4

σ'a t  5  mm

(b) Wall thickness is


σ1b  11.67  MPa 5.

σ2b  0  MPa

σ3b  7.00 MPa

Calculate the von Mises effective stress using equation (5.7c).

σ'b  6.

σ'a  42.752 MPa

Using the distortion energy theory, the factor of safety is

Na 

4.

σ3a  7.00 MPa

Calculate the von Mises effective stress using equation (5.7c).

σ'a  3.

σ2a  0  MPa

2

2

σ1b  σ1b σ3a  σ3b

σ'b  16.336 MPa

Using the distortion energy theory, the factor of safety is

Nb 

Sy

σ'b

Nb  24.5



5-42-1


A cylindrical tank with hemispherical ends is required to hold 150 psi of pressurized air at room temperature. The steel has S y = 400 MPa. Determine the safety factor if the tank diameter is 0.5 m with 1 mm wall thickness, and its length is 1 m.

Given:

Yield strength

Solution:


1.

S y  400  MPa

From Problem 4-42, the maximum principal stresses in the wall are

σ1  259  MPa

σ2  129  MPa σ' 

2

σ1  σ1 σ2  σ2

σ3  0  MPa 2

σ'  224.301 MPa

2.

The von Mises stress is

3.

Using the distortion-energy theory, the factor of safety against a static failure is Ns 

Sy

σ'

Ns  1.8



5-43-1


The paper rolls in Figure P5-17 are 0.9-m OD by 0.22-m ID by 3.23-m long and have a density of 984 kg/m3. The rolls are transfered from the machine conveyor (not shown) to the forklift truck by the V-linkage of the off-load station, which is rotated through 90 deg by an air cylinder. The paper then rolls onto the waiting forks of the truck. The forks are 38-mm thick by 100-mm wide by 1.2-m long and are tipped at a 3-deg angle from the horizontal and have Sy = 600MPa. Find the safety factor for the two forks on the truck when the paper rolls onto it under two different conditions (state all assumptions): (a) The two forks are unsupported at their free end. (b) The two forks are contacting the table at point A. F

Given:

S y  600  MPa

Yield strength

Assumptions: 1. The greatest bending moment will occur when the paper roll is at the tip of the fork for case (a) and when it is midway between supports for case (b). 2. Each fork carries 1/2 the weight of a paper roll. 3. For case (a), each fork acts as a cantilever beam (see Appendix B-1(a)). 4. For case (b), each fork acts as a beam that is built-in at one end and simply-supported at the other. Solution:

L fork

t

R1 Case (a), Cantilever Beam

0.5 L fork

F

t

L fork

See Figure 5-43 and Mathcad file P0543. R1

1.

From Problem 4-43, the maximum stresses in the forks are: Case (a)

M1

R2

M2


FIGURE 5-43

σa  464.8  MPa

Free Body Diagrams used in Problem 5-43

at the base of the fork. Case (b)

σb  87.2 MPa

also at the base of the fork.

Since there are no other stress components present, these are also the maximum principal stresses and the von Mises stresses. Thus, σ'a  σa and σ'b  σb. Case (a) 2.

The factor of safety against a static failure is

Nsa 

Sy

σ'a

Nsa  1.3

Case (b) 3.

The factor of safety against a static failure is

Nsb 

Sy

σ'b

Nsb  6.9



5-44-1


Determine a suitable thickness for the V-links of the off-loading station of Figure P5-17 to limit their deflections at the tips to 10-mm in any position during their rotation. Two V-links support the roll, at the 1/4 and 3/4 points along the roll's length, and each of the V arms is 10-cm wide by 1-m long. What is their safety factor against yielding when designed to limit deflection as above?

Given:

Roll OD

OD  0.90 m

Arm width

wa  100  mm

Roll ID

ID  0.22 m

Arm length

La  1000 mm

Roll length

Lroll  3.23 m

Max tip deflection


Roll density Yield strength

ρ  984  kg m

Mod of elasticity

E  207  GPa

3

S y  400  MPa

Assumptions: 1. The maximum deflection on an arm will occur just after it begins the transfer and just before it completes it, i.e., when the angle is either zero or 90 deg., but after the tip is no longer supported b the base unit. 2. At that time the roll is in contact with both arms ("seated" in the V) and will remain in that state throughout the motion. When the roll is in any other position on an arm the tip will be supported. 3. The arm can be treated as a cantilever beam with nonend load. 4. A single arm will never carry more than half the weight of a roll. 5. The pipe to which the arms are attached has OD = 160 mm. Solution:


1. Determine the weight of the roll and the load on each V-arm. W 

 4

π

2

2




450

W  18.64  kN

F  0.5 W

F  9.32 kN

2. From Appendix B, Figure B-1, the tip deflection of a cantilever beam with a concentrated load located at a distance a from the support is ymax =

F a

2

6  E I

 ( a  3  L)

1000 = L 370 = a

where L is the beam length and I is the cross-section moment of inertia. In this case

F

3

I= 3. Setting

w a t a

M

12

ymax = δtip

F

a  370  mm

and




3

ta  31.889 mm ta  32 mm

4. The maximum bending stress in the arm will be at its base where it joins the 160-mm-dia pipe. The bending moment, moment of inertia, and distance to the outside fiber at that point are:



Bending moment Moment of inertia Distance to outer fiber

M  a  F I 

wa ta

5-44-2

M  3449 N  m 3

12

c  0.5 t a

5

I  2.731  10  mm

4

c  16 mm

5. The bending stress, which is also the von Mises stress, is

σ' 

M c I

σ'  202.1  MPa

6. Using the distortion-energy theory, the factor of safety against a static failure is Ns 

Sy

σ'

Ns  2.0



5-45-1


Determine the safety factor based on critical load on the air cylinder rod in Figure P5-17 if the crank arm that rotates it is 0.3 m long and the rod has a maximum extension of 0.5 m. The 25-mm-dia rod is solid steel with a yield strength of 400 MPa. State all assumptions.

Given:

Rod length Rod diameter

L  500  mm d  25 mm

Young's modulus Yield strength

E  207  GPa S y  400  MPa

Assumptions: 1. The rod is a fixed-pinned column. 2. Use a conservative value of 1 for the end factor (see Table 4-7 in text). Solution:

See Problems 4-45, 4-47, and Mathcad file P0545.

1.

From Problem 4-45, the critical load on the air cylinder rod is

Pcr  134.8  kN

2.

From Problem 4-47, the maximum load on the air cylinder rod is

F  46.47  kN

3.

The factor of safety against a buckling failure is Nbuck 

Pcr F

Nbuck  2.9



5-46-1


The V-links of Figure P5-17 are rotated by the crank arm through a shaft that is 60 mm dia by 3.23 m long. Determine the maximum torque applied to this shaft during motion of the V-linkage and find the static safety factor against yielding for the shaft if its S y = 400 MPa. See Problem 5-43 for more information. y

Given:

Yield strength

S y  400  MPa

Assumptions: The greatest torque will occur when the link is horizontal and the paper roll is located as shown in Figure P5-17 or Figure 5-46. Solution: P0546.

See Figure 5-46 and Mathcad file

1. From Problem 4-46, the maximum torsional stress in the shaft is

W

τmax  197.88 MPa

T

2. Using the distortion-energy theory, the factor of safety against static yielding is

Ry 60-mm-dia shaft

Ns 

Sy 3  τmax

Ns  1.2

450.0




5-47-1


Determine the maximum forces on the pins at each end of the air cylinder of Figure P4-17. Determine the safety factor for these pins if they are 30-mm dia and in single shear. S y = 400 MPa.

Given:


OD  0.90 m ID  0.22 m

Pin diameter Yield strength

d  30 mm S y  400  MPa

Lroll  3.23 m 3

ρ  984  kg m

Roll density

Assumptions: 1. The maximum force in the cylinder rod occurs when the transfer starts. 2. The cylinder and rod make an angle of 8 deg to the horizontal at the start of transfer. 3. The crank arm is 300 mm long and is 45 deg from vertical at the start of transfer. 4. The cylinder rod is fully retracted at the start of the transfer. At the end of the transfer it will have extended 500 mm from its initial position. Solution:


1. Determine the weight of the roll on the forks. W 

 4

π

2

2

y




W  18.64 kN 2. From the assumptions and Figure 4-47, the x and y distances from the origin to point A are, Rax  300  cos( 45 deg)  mm W

Ray  300  sin( 45 deg)  mm

Rx

Rax  212.132 mm

x 212.1

Ry

A

Ray  212.132 mm

F

8°

212.1 450.0

3. From Figure 4-47, the x distance from the origin to point where W is applied is,


Rwx 

OD 2

Rwx  450 mm

4. Sum moments about the pivot point and solve for the compressive force in the cylinder rod. W  Rwx  F  Rax sin( 8  deg)  F  Ray cos( 8  deg) = 0 F 

W  Rwx Ray cos( 8  deg)  Rax sin( 8  deg)

F  46.469 kN

This is the shear force in the pins 5. Determine the cross-sectional area of the pins and the direct shear stress. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0547.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


Shear area

Shear stress

A 

τ 

π d

2

4 F

A  706.858 mm

5-47-2

2

τ  65.7 MPa

A

6. Using the distortion-energy theory, the factor of safety against a static yielding failure is Ns 

Sy 3 τ

Ns  3.5



5-48-1


Figure P5-18 shows an exerciser for a 100-kg wheelchair racer. The wheel chair has 65 cm dia drive wheels separated by a 70-cm track width. Two free-turning rollers on bearings support the rear wheels. The lateral movement of the chair is limited by the flanges. Design the 1-m-long rollers as hollow tubes of aluminum (select alloy) to minimize the height of the platform and also limit the roller deflections to 1 mm in the worst case. Specify suitable sized steel axles to support the tubes on bearings. Calculate all significant stresses.

Given:


T  700  mm

Aluminum

Ea  71.7 GPa

Roller length

Lr  1000 mm

Steel

Es  207  GPa

Assumptions: 1. The CG of the chair with rider is sufficiently close to the rear wheel that all of the weight is taken by the two rear wheels. 2. The small camber angle of the rear wheels does not significantly affect the magnitude of the forces on the rollers. 3. Both the aluminum roller and the steel axle are simply supported. The steel axles that support the aluminum tube are fixed in the mounting block and do not rotate. The aluminum tube is attached to them by two bearings (one on each end of the tubes, one for each axle). The bearings' inner race is fixed, and the outer race rotates with the aluminum tube. Each steel axle is considered to be loaded as a simply supported beam. Their diameter must be less than the inner diameter of the tubes to fit the roller bearings between them. Solution:

δ  1  mm

Maximum deflection Modulus elasticity:

W/2

F

F 



FIGURE 5-48A Free Body Diagram of One Wheel used in Problem 5-48


1. Calculate the weight of the chair with rider. Weight of chair

W  M  g

W  980.7 N

2. Calculate the forces exerted by the wheels on the rollers (see Figure 5-48A). From the FBD of a wheel, summing vertical forces 2  F  cos( θ )  Let

θ  20 deg

W 2

=0 then

F 

W 4  cos( θ )

F  260.9 N

3. The worst condition (highest moment at site of a stress concentration) will occur when the chair is all the way to the left or right. Looking at the plane through the roller that includes the forces exerted by the wheels (the FBD is shown in Figure 5-48B) the reactions R1 and R2 come from the bearings, which are inside the hollow roller and are, themselves, supported by the steel axle. 4. Solving for the reactions. Let the distance from R1 to F be a  15 mm



 M1

R2 Lr  F  ( a  T )  F  a = 0

 Fy

R1  2  F  R2 = 0

R2 

F  (2 a  T )

5-48-2 700

F

R2  190.5 N

Lr

15

R2

R1

R1  2  F  R2

F

1000

R1  331.3 N FIGURE 5-48B

5. The maximum bending moment will be at the right-hand load and will be

Free Body Diagram of One Tube used in Problem 5-48

Mrmax  R2 Lr  ( a  T )

Mrmax  54.3 N  m


Lr  T

Mc  39.1 N  m

2

and this would be constant along the axle between the two loads, F. 6. Note that the bearing positions are fixed regardless of the position of the chair on the roller. Because of symmetry, Ra1  R1

Ra1  331.3 N

Ra2  R2

Ra2  190.5 N

1000

65 R1

7. The maximum bending moment occurs at R1 and is for b  65 mm

R2

R a1

R a2 1130

Mamax  Ra1 b

FIGURE 5-48C Free Body Diagram of One Axle used in Problem 5-48

Mamax  21.5 N  m

8. Determine a suitable axle diameter. Let the factor of safety against yielding in the axle be Nsa  3 9. Tentatively choose a low-carbon steel for the axle, say AISI 1020, cold rolled with S y  393  MPa 10. At the top of the axle under the load R1 there is only a bending stress, which is also the von Mises stress. Set th stress equal to the yield strength divided by the factor of safety.

σ' =

32 Mamax

π d a

3

=

Sy Nsa 1


 32 Nsa Mamax  d a    π S y  


d a  15 mm

3

d a  11.875 mm




5-48-3

11. Suppose that bearing 6302 from Chapter 10, Figure 10-23, page 684 is used. It has a bore of 15 mm and an OD of 42 mm. Thus, the inside diameter of the roller away from the bearings where the moment is a maximum will be d i  40 mm. This will provide a 1-mm shoulder for axial location of the bearings. 12. The maximum deflection of the roller will occur when the chair is in the center of the roller. For this case the reactions are both equal to the loads, F (see Figure 4-48D). The maximum deflection is at the center of the roller.

150

700 F

F

F 15

F 1000

13. Write the load function and then integrate four times to get the deflection function.


q(x) = F-1 - F-1 - F-1 + F-1 y(x) = F[3 - 3 - 3 + 3 + C3x]/(6EI) where

C3 =

1 L

 ( L  a )  a  L  3

3

3

14. Write the deflection function at x = L/2 for a  150  mm

ymax =

3  L 3  L   1  ( L  a) 3  a3  L3   a    2  6  Ea I  2  2   

 

F

15. Set this equation equal to the allowed deflection  and solve for the required moment of inertia, I. 3  Lr  3  Lr   1 3 3 3  I       a    Lr  a   a  Lr  6  Ea δ  2  2 2  

F

4

I  6.618  10 mm

4

16. Knowing the inside diameter of the tube, solve for the outside diameter. 1

I=

π  4 4   d o  d i  64

Round this up to

d o 

 64 I  d 4  π i   

4

d o  44.463 mm

d o  46 mm

DESIGN SUMMARY Axles

Rollers

Material


Material

2024-T4 aluminum

Diameter

d a  15 mm

Outside diameter

d o  46 mm

Length

1220 mm

Inside diameter

d i  40 mm

Length

1040 mm

Spacing

c   d w  d o  sin( θ ) c  238 mm




5-49-1

_____

A part made of ductile steel with Sy = 40 ksi is subjected to a three-dimensional stress state of 1 = -80 ksi, 2 = -80 ksi, 3 = -80 ksi. What is the maximum shear stress? Will the part fail?

Solution: 1.


This is a case of hydrostatic stress. As explained in Section 5.1, the maximum shear stress is zero. Parts loaded hydrostatically can withstand stresses well in excess of their yield strength. One example of this is that parts on the ocean floor such as those retrieved from the Titanic are intact and undistorted even though they are surrounded by water at great pressure.



5-50-1

PROBLEM 5-50

_____

Statement:

A component in the shape of a large sheet is to be fabricated from 7075-T651 aluminum, which has a fracture toughness Kc = 24.2 MPa-m0.5 and a tensile yield strength of 495 MPa. Determine the largest edge crack that could be tolerated in the sheet if the nominal stress does not exceed one half the yield strength.

Given:

Fracture toughness

Kc  24.2 MPa m

Yield strength

S y  495  MPa

Solution: 1.

Mathcad file P0550.

Calculate the nominal stress based on the yield strength and the stress level given in the problem statement.

σnom  2.

0.5

Sy

σnom  247.5 MPa

2

Determine the value of the geometry factor  from the discussion in Section 5.3 for a plate with an edge crack.

β  1.12 3.

Using equation 5.14b, calculate the critical crack length for this material under the given stress condition.

 Kc  a     π  β  σnom  1

2

a  2.4 mm



5-51-1

PROBLEM 5-51

_____

Statement:

A component in the shape of a large sheet is to be fabricated from 4340 steel, which has a fracture toughness Kc = 98.9 MPa-m0.5 and a tensile yield strength of 860 MPa. The sheets are inspected for crack flaws after fabrication, but the inspection device cannot detect flaws smaller than 5 mm. The part is too heavy as designed. An engineer has suggested that the thickness be reduced and the material be heat-treated to increase its tensile strength to 1515 MPa, which would result in decreasing the fracture toughness to 60.4 MPa-m0.5. Assuming that the stress level does not exceed one half the yield strength, is the suggestion feasible? If not, why not.

Given:

Fracture toughness

Kc1  98.9 MPa m

Kc2  60.4 MPa m

Yield strength

S y1  860  MPa

S y2  1515 MPa

Solution: 1.

0.5


Calculate the nominal stress for the two material conditions based on the yield strength and the stress level given in the problem statement.

σnom1 

σnom2 

2.

0.5

S y1

σnom1  430 MPa

2 S y2

σnom2  757.5 MPa

2

Determine the value of the geometry factor  from the discussion in Section 5.3 for a large plate.

β  1 3.

Using equation 5.14b, calculate the critical crack length for each material condition under the given stress condition.

a 1 

2  Kc1   π  β  σnom1 

1



 Kc2  a 2     π  β  σnom2  1

4.

a 1  16.8 mm

2  a 1  33.7 mm

a 2  2.0 mm

2  a 2  4.0 mm

2

The suggestion to increase the strength of the material so that its thickness can be decreased to save weight is not feasible because the critical crack size of the material in the second condition is less than that which can be detected by the inspection equipment.



5-52-1

PROBLEM 5-52

_____

Statement:

A large plate is subjected to a nominal tensile stress of 350 MPa. The plate has a central crack that is 15.9 mm long. Calculate the stress intensity factor at the tip of the crack.

Given:

Nominal stress

σnom  350  MPa

Crack length

lcrack  15.9 mm

Solution: 1.


Calculate the half-width of the crack a  0.5 l crack

2.

a  7.95 mm

Determine the value of the geometry factor  from the discussion in Section 5.3 for a plate with an edge crack.

β  1 3.

Using equation 5.14b, calculate the stress intensity factor. K  β  σnom π a

0.5

K  55.3 MPa m



5-53-1

PROBLEM 5-53

_____

Statement:

A movie scene calls for a stuntman to hang from a rope that is suspended 3 m above a pit of poisonous spiders. The rope is attached to a glass sheet that is 3000 mm long by 100 mm wide and 1.27 mm thick. The stuntman knows that the glass sheet contains a central crack with total length of 16.2 mm that is oriented parallel to the ground. The fracture toughness of the glass is 0.83 MPa-m0.5. Should he do the stunt? Show all assumptions and calculations in support of your answer.

Given:

Fracture toughness

Kc  0.83 MPa m

Glass dimensions Total crack length

L  3000 mm W  100  mm lcrack  16.2 mm

Assumptions: Weight of stuntman Desired safety factor Solution: 1.

2.

0.5

Weight  900  N NFMd  3


Calculate the nominal stress based on the assumed weight of the stuntman and the glass dimensions. Cross-section area

A  W  t

Nominal stress

σnom 

A

2

σnom  7.087 MPa

a  0.5 l crack

a  8.1 mm

Glass half-width

b  0.5 W

b  50 mm

sec 

π a 

  2 b 

β  1.017

Using equation 5.14b, calculate the stress intensity factor for the given assumptions. 0.5

K  1.149 MPa m

Using equation 5.15, calculate the safety factor against sudden failure for the given assumptions. NFM 

5.

Weight

Crack half-width

K  β  σnom π a 4.

A  127 mm

Determine the value of the geometry factor  from equation 5.14c for a plate with a central crack.

β 

3.

t  1.27 mm

Kc K

NFM  0.72

The stuntman should definitely not do the stunt since the factor of safety is not only less than the desired value, but is less than one.



5-54-1

PROBLEM 5-54

_____

Statement:

A material has a fracture toughness of 50 MPa-m0.5 and a yield strength of 1000 MPa and is to be made into a large panel. If the panel is stressed to one-half the yield stress, what is the maximum central crack size that can be tolerated without catastrophic failure?

Given:

Fracture toughness

Kc  50 MPa m

Yield strength

S y  1000 MPa

Solution: 1.


Calculate the nominal stress based on the yield strength and the stress level given in the problem statement.

σnom  2.

0.5

Sy

σnom  500 MPa

2

Determine the value of the geometry factor  from the discussion in Section 5.3 for a large plate with a central crack.

β  1 3.


 Kc  a     π  β  σnom  1

lcritical  2  a

2

a  3.18 mm

lcritical  6.4 mm



5-55-1

PROBLEM 5-55

_____

Statement:

A material that has a fracture toughness of 33 MPa-m0.5 is to be made into a large panel that is 2000 mm long by 250 mm wide and 4 mm thick. If the minimum allowable total crack length is 4 mm, what is the maximum tensile load in the long direction that can be applied without catastrophic failure with a safety factor of 2.5?

Given:

Fracture toughness

0.5

Kc  33 MPa m

Panel dimensions L  2000 mm Total allow. crack length lcrack  4  mm Safety factor Solution: 1.

NFM  2.5

Calculate the allowable stress intensity factor using equation 5.15. Kc NFM

0.5

Kallow  13.2 MPa m

Determine the value of the geometry factor  from equation 5.14c for a plate with a central crack. Crack half-width

a  0.5 l crack

a  2 mm

Panel half-width

b  0.5 W

b  125 mm

β 

3.

sec 

π a 

  2 b 

β  1.00

Using equation 5.14b, calculate the allowable nominal stress in the panel.

σallow 

4.

t  4  mm


Kallow 

2.

W  250  mm

Kallow

β  π a

σallow  166.5 MPa

Calculate the allowable load for the given conditions. Fallow  σallow W  t

Fallow  167 kN



5-56-1

PROBLEM 5-56

_____

Statement:

Figure P5-19 shows an SAE 1020 cold-rolled steel bar fastened to a rigid ground plane with two 0.25-in-dia A8 steel dowel pins, hardened to HRC52. For P = 1500 lb and t = 0.25 in, find: (a) The safety factor for each pin. (b) The safety factor for direct bearing stress in each hole. (c) The safety factor for tearout failure if h = 1 in.

Given:

Pin diameter Applied load Distance between pins

d  0.250  in P  1500 lbf a  2.0 in

Depth of section Distance from right pin to load Yield strength of bar

h  1.0 in b  4.0 in S yb  57 ksi

Thickness of bar

t  0.25 in

Yield strength of pin

S yp  225  ksi

Solution: 1.



a

b

RL

h


 F: RL  2.

b a

P

π d 4

RR  P  RL

RR  4500 lbf

2

2

A  0.0491 in

Use equation 4.9 to determine the shear stress in each pin. Left pin

Right pin 4.

RL  3000 lbf

RL  a  P b  0


3.

 M:

RL  RR  P  0

τL  τR 

RL

τL  61.1 ksi

A RR

τR  91.7 ksi

A

(a) From equations 5.8c and 5.9b, the safety factor against failure in the pins is

Left pin

NL 

0.577  S yp

τL

NL  2.1



NR 

Right pin

5.

0.577  S yp

τR

5-56-2

NR  1.4

Calculate the bearing area from equation 4.10 and use it to determine the bearing stress in each hole. Abear  d  t

Bearing area

σL  σR 

RL Abear RR Abear

2

Abear  0.0625 in

σL  48.0 ksi σR  72.0 ksi

These are principal stresses 1. 6.

7.

(b) Calculate the safety factor for direct bearing from equation 5.8c where 2 and 3 are both zero. Left hole

NL 

Right hole

NR 

The tearout area is

Atear  2  

S yb

σL S yb

σR

NL  1.2

NR  0.8

h  d

   t , where (h - d)/2 is the distance from the edge of the hole to the  2  

outside of the bar. Substitute this area in equation 4.9 for the shear area and solve for the shear strength xy. Atear  2  



Left hole

Right hole

8.

h  d 2

   t  

τL 

τR 

2

Atear  0.187 in RL Atear RR Atear

τL  16.00 ksi

τR  24.00 ksi

(c) From equations 5.8c and 5.9b, the safety factor against tearout failure in the holes is Left hole

NL 

Right hole

NR 

0.577  S yb

τL 0.577  S yb

τR

NL  2.1

NR  1.4



5-57-1

PROBLEM 5-57

_____

Statement:

Figure P5-19 shows a class 50 cast iron bar fastened to a rigid ground plane with two 0.25-in-dia A8 steel dowel pins, hardened to HRC52. For P = 1500 lb and t = 0.25 in, find: (a) The safety factor for each pin. (b) The safety factor for direct bearing stress in each hole. (c) The safety factor for tearout failure if h = 1 in.

Given:

Pin diameter Applied load Distance between pins

d  0.250  in P  1500 lbf a  2.0 in

Depth of section Distance from right pin to load Tensile strength of bar

h  1.0 in b  4.0 in S utb  52 ksi

Thickness of bar

t  0.25 in

Yield strength of pin

S yp  225  ksi

Solution: 1.



a

b

RL

h


 F: RL  2.

b a

P

π d 4

RR  P  RL

RR  4500 lbf

2

2

A  0.0491 in

Use equation 4.9 to determine the shear stress in each pin. Left pin

Right pin 4.

RL  3000 lbf

RL  a  P b  0


3.

 M:

RL  RR  P  0

τL  τR 

RL

τL  61.1 ksi

A RR

τR  91.7 ksi

A

(a) From equations 5.8c and 5.9b, the safety factor against failure in the pins is Left pin

NL 

0.577  S yp

τL

NL  2.1



NR 

Right pin

5.

0.577  S yp

τR

5-57-2

NR  1.4

Calculate the bearing area from equation 4.10 and use it to determine the bearing stress in each hole. Abear  d  t

Bearing area

σL  σR 

RL Abear RR Abear

2

Abear  0.0625 in

σL  48.0 ksi σR  72.0 ksi

These are principal stresses 1. 6.

7.

(b) Calculate the safety factor for direct bearing from equation 5.12a where 2 and 3 are both zero. Left hole

NL 

Right hole

NR 

The tearout area is

Atear  2  

S utb

σL S utb

σR

NL  1.1

NR  0.7

h  d

   t , where (h - d)/2 is the distance from the edge of the hole to the  2  

outside of the bar. Substitute this area in equation 4.9 for the shear area and solve for the shear strength xy. Atear  2  



Left hole

Right hole

8.

h  d 2

   t  

τL 

τR 

2

Atear  0.187 in RL Atear RR Atear

τL  16.00 ksi

τR  24.00 ksi

(c) For pure shear the Mohr circle is centered at 0,0 and has a radius equal to the shear stress. This results in 1 = . Using the Modified-Mohr failure theory and Figure 5-11, we see that we can use equation 5.12a for the safety factor against tearout. NL 

S utb

τL

NL  3.3

NR 

S utb

τR

NR  2.2



5-58-1

PROBLEM 5-58

_____

Statement:

Figure P5-20 shows a bracket machined from 0.5-in-thick SAE 1045 cold-rolled steel flat stock. It is rigidly attached to a support and loaded with P = 5000 lb at point D. Find: (a) The safety factor against static failure at point A. (b) The safety factor against static failure at point B.

Given:

Distance from support to: Point D d  8  in Depth of section h  3  in Applied load P  5000 lbf

Points B and C b  17 in Thickness of section t  0.5 in Tensile yield strength S y  77 ksi



Calculate the cross-section area and moment of inertia at A, B, and C, which are the same. 2

A  h  t 2.

A  1.500 in

t h

I 

3

4

I  1.1250 in

12

For part (a), draw a free-body diagram of the entire bracket.

V A y

h

M

B x

h C d

h

D

P 3.

Use the equilibrium equations 3.3a to calculate the shear force and bending moment at the support.

 F:

V  P 4.

 M:

V  P  0 V  5000 lbf

MA  P ( d )

MA  40000 in lbf

The normal stress in the bracket at point A is determined using equation 4.11b. c  0.5 h

Distance from neutral axis to extreme fiber Normal stress at point A 5.

P ( d )  M  0

σA  

MA c I

c  1.500 in

σA  53.33 ksi

(a) The transverse shear stress in the bracket at point A is zero, therefore A is a principal stress. There are no stress components in the y or z directions so this is a case of uniaxial stress. Thus, equations 5.7 reduce to

σ' 

2

σA

σ'  53.3 ksi

Use equation 5.8a to calculate the factor of safety against a static failure at point A.


MACHINE DESIGN - An Integrated Approach, 4th Ed. NA 

6.

Sy

5-58-2

NA  1.4

σ'

For part (b), draw a free-body diagram of the portion of the bracket that is below point B.

b F y M

B

x

d

h

D

P 7.

Use the equilibrium equations 3.3a to calculate the normal force and bending moment on the section shown.

 F:

F  P 8.

 M:

F  P  0 F  5000 lbf

MB1  P ( b  0.5 h  d )

MB1  52500 in lbf

The normal stress in the bracket at point B in the y direction is a combination of uniform tension and bending and is determined by summing equations 4.7 and 4.11b. Normal stress at B in y direction

9.

P ( b  0.5 h  d )  M  0

σBy 

MB1 c I



F A

σBy  73.33 ksi

The normal stress in the bracket at point B in the x direction is bending and is determined from equation 4.11b, using the FBD from part (a). MB2  V  b  MA

MB2  45000 in lbf

Normal stress at B in x direction

σBx 

MB2 c I

σBx  60.00 ksi

10. (b) The transverse shear at B due to the shear force V is zero so Bx and By are the only stress components at B. Use equations 5.7d and 5.8a to determine the factor of safety against a static failure at B (ignoring the stress concentration there).

σ'  NB 

2

2

σBx  σBy  σBx σBy Sy

σ'

σ'  67.66 ksi NB  1.1



5-59-1

PROBLEM 5-59

_____

Statement:

Figure P5-20 shows a bracket machined from 1-in-thick class 60 cast iron flat stock. It is rigidly attached to a support and loaded with P = 5000 lb at point D. Find: (a) The safety factor against static failure at point A. (b) The safety factor against static failure at point B.

Given:

Distance from support to: Point D d  8  in Depth of section h  3  in Applied load P  5000 lbf

Points B and C b  17 in Thickness of section t  0.5 in Ultimate tensile strength S ut  62 ksi S uc  187  ksi

Ultimate comp. strength



Calculate the cross-section area and moment of inertia at A, B, and C, which are the same. 2

A  h  t 2.

A  1.500 in

t h

I 

3

4

I  1.1250 in

12

For part (a), draw a free-body diagram of the entire bracket.

V A y

h

M

B x

h C d

h

D

P 3.

Use the equilibrium equations 3.3a to calculate the shear force and bending moment at the support.

 F:

V  P 4.

 M:

V  P  0 V  5000 lbf

MA  P ( d )

MA  40000 in lbf

The normal stress in the bracket at point A is determined using equation 4.11b. c  0.5 h

Distance from neutral axis to extreme fiber Normal stress at point A 5.

P ( d )  M  0

σA  

MA c I

c  1.500 in

σA  53.33 ksi

(a) The transverse shear stress in the bracket at point A is zero, therefore A is a principal stress. There are no stress components in the y or z directions so this is a case of uniaxial stress. Thus, use equation 5.12a (adapted to a compressive stress state) to calculate the factor of safety against a static failure at point A. NA 

S uc

σA

NA  3.5



6.

5-59-2

For part (b), draw a free-body diagram of the portion of the bracket that is below point B.

b F y M

B

x

d

h

D

P 7.

Use the equilibrium equations 3.3a to calculate the normal force and bending moment on the section shown.

 F:

F  P 8.

 M:

F  P  0 F  5000 lbf

MB1  P ( b  0.5 h  d )

MB1  52500 in lbf

The normal stress in the bracket at point B in the y direction is a combination of uniform tension and bending and is determined by summing equations 4.7 and 4.11b.

σBy 

Normal stress at B in y direction 9.

P ( b  0.5 h  d )  M  0

MB1 c I



F A

σBy  73.33 ksi

The normal stress in the bracket at point B in the x direction is bending and is determined from equation 4.11b, using the FBD from part (a). MB2  V  b  MA

MB2  45000 in lbf

Normal stress at B in x direction

σBx 

MB2 c I

σBx  60.00 ksi

10. (b) The transverse shear at B due to the shear force V is zero so Bx and By are the only stress components at B. Use equations 4.6 to determine the principal stresses and 5.12a to determine the factor of safety against a static failure at B (ignoring the stress concentration there).

σ1 

σ2 

σBx  σBy 2

σBx  σBy 2

σ3  0  ksi

2  σBx  σBy     2  

σ1  73.333 ksi

2  σBx  σBy     2  

σ2  60.000 ksi

NB 

S ut

σ1

NB  0.85



5-60-1

PROBLEM 5-60

_____

Statement:

Figure P5-21 shows a 1-in-dia SAE 1040 hot-rolled, normalized steel bar supported and subjected to the applied load P = 500 lb. Find the safety factor against static failure.

Given:


d  1.00 in P  500  lbf

Dimensions:

a  20 in

Solution: 1.

6

Modulus of elasticity E  30 10  psi Yield strength S y  54 ksi

L  40 in



M1 R1

2.

This is a statically indeterminate beam because there are three unknown reactions, R1, M1, and R2. To solve for these unknowns, follow the method presented in Example 4-7. First, calculate the moment of inertia and distance to the extreme fiber for the round section. I 

3.

P

π d

4

4

I  0.0491 in

64

c  0.5 d

c  0.500 in


4.


5.


6.


5.


7.


M1  1000 in lbf

Given

y(a) = 0: V(L) = 0:

R1  500  lbf 

M1 2

2

a 

R1 6

3

R2  1000 lbf 3

 a = 0  lbf  in

R1  R2  P = 0  lbf



5-60-2

M1  R1 L  R2 ( L  a ) = 0  lbf  in

M(L) = 0:

 M1     R1   Find  M1 R1 R2 R   2

M1  5000 in lbf

R1  750 lbf R2  1250 lbf

x  0  in 0.02 L  L

8.


9.


10. Write the moment equation in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. M ( x)  M1  R1 S ( x 0  in)  x  R2 S ( x a )  ( x  a )  P S ( x L)  ( x  L) 11. Plot the moment equation and determine the maximum bending moment.

MOMENT DIAGRAM

As expected, the maximum bending moment occurs under the support at x = a. Mmax  M ( a )

10

Mmax  10.0 kip in

5 M ( x) kip  in 0

5

0

10

20

30

40

x in

12. Use equation 4.11b to calculate the maximum bending stress in the bar.

σmax 

Mmax c I

σmax  101.9 ksi

13. There are no other stress components present (the transverse shear is zero at the extreme fiber) so this is a principal stress and the other two principal stresses are zero. Thus, this is a case of uniaxial stress. Determine the safety factor against static failure using equations 5.7c and 5.8a, which reduce to N 

Sy

σmax

N  0.53



5-61-1

PROBLEM 5-61

_____

Statement:

Figure P5-21 shows a 1.5-in-dia class 60 cast iron bar supported and subjected to the applied load = 500 lb. Find the safety factor against static failure.

Given:


d  1.50 in P  500  lbf

Dimensions:

a  20 in

Solution: 1.

6

Modulus of elasticity E  30 10  psi Tensile strength S ut  54 ksi

L  40 in



M1 R1

2.

This is a statically indeterminate beam because there are three unknown reactions, R1, M1, and R2. To solve for these unknowns, follow the method presented in Example 4-7. First, calculate the moment of inertia and distance to the extreme fiber for the round section. I 

3.

P

π d

4

4

I  0.2485 in

64

c  0.5 d

c  0.750 in


4.


5.


6.


5.


7.


M1  1000 in lbf

Given

y(a) = 0: V(L) = 0:

R1  500  lbf 

M1 2

2

a 

R1 6

3

R2  1000 lbf 3

 a = 0  lbf  in

R1  R2  P = 0  lbf



5-61-2

M1  R1 L  R2 ( L  a ) = 0  lbf  in

M(L) = 0:

 M1     R1   Find  M1 R1 R2 R   2

M1  5000 in lbf

R1  750 lbf R2  1250 lbf

x  0  in 0.02 L  L

8.


9.


10. Write the moment equation in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. M ( x)  M1  R1 S ( x 0  in)  x  R2 S ( x a )  ( x  a )  P S ( x L)  ( x  L) 11. Plot the moment equation and determine the maximum bending moment.

MOMENT DIAGRAM

As expected, the maximum bending moment occurs under the support at x = a. Mmax  M ( a )

10

Mmax  10.0 kip in

5 M ( x) kip  in 0

5

0

10

20

30

40

x in

12. Use equation 4.11b to calculate the maximum bending stress in the bar.

σmax 

Mmax c I

σmax  30.2 ksi

13. There are no other stress components present (the transverse shear is zero at the extreme fiber) so this is a principal stress and the other two principal stresses are zero. Thus, this is a case of uniaxial stress. Determine the safety factor against static failure using equation 5.12a. N 

S ut

σmax

N  1.8



5-62-1

PROBLEM 5-62

_____

Statement:

Figure P5-22 shows a pivot pin that is press-fit into part A and is slip fit in part B. If F = 100 lb, l = 2 in, and d = 0.5 in, what is the pin's safety factor against yielding when made of SAE 1020 cold-rolled steel?

Given:

Applied force

F  100  lbf

Yield strength

S y  57 ksi

Total length, l Pin dia

l  2.00 in d  0.5 in

Beam length

L  0.5 l

L  1.000  in




2.

F

w  100.0 

L

lbf in

A cantilever beam with uniform loading is shown in Figure B-1(b) in Appendix B. In this case, the dimension a the figure is zero. As shown in the figure, when a = 0, the maximum bending moment occurs at the support and is 2

w L

Mmax  3.

2

Calculate the moment of inertia and distance to the extreme fiber of the pin. The bending stress in the beam is then found using equation 4.11b. I 

π d

4

I  3.068  10

64

c  0.5 d

σ 

4.

Mmax  50.00  lbf  in

Mmax c I

3

4

 in

c  0.250  in

σ  4074 psi

There are no other stress components present (the transverse shear is zero at the extreme fiber) so this is a principal stress and the other two principal stresses are zero. Thus, this is a case of uniaxial stress. Determine the safety factor against static failure using equations 5.7c and 5.8a, which reduce to N 

Sy

σ

N  14



5-63-1

PROBLEM 5-63

_____

Statement:

Figure P5-22 shows a pivot pin that is press-fit into part A and is slip fit in part B. If F = 100 N, l = 50 mm, and d = 16 mm, what is the pin's safety factor against yielding when made of class 50 cast iron?

Given:

Applied force

F  100  N

Tensile strength

S ut  359  MPa


l  50 mm d  16 mm

Beam length

L  0.5 l

L  25 mm




2.

F

w  4.0

L

N mm

A cantilever beam with uniform loading is shown in Figure B-1(b) in Appendix B. In this case, the dimension a the figure is zero. As shown in the figure, when a = 0, the maximum bending moment occurs at the support and is 2

w L

Mmax  3.

2

Calculate the moment of inertia and distance to the extreme fiber of the pin. The bending stress in the beam is then found using equation 4.11b. I 

π d

4

σ 

3

I  3.217  10  mm

64

c  0.5 d

4.

Mmax  1250 N  mm

Mmax c I

4

c  8.000  mm

σ  3.108  MPa

There are no other stress components present (the transverse shear is zero at the extreme fiber) so this is a principal stress and the other two principal stresses are zero. Thus, this is a case of uniaxial stress. Determine the safety factor against static failure using equation 5.12a. N 

S ut

σ

N  115



5-64-1


A differential element is subjected to the stresses (in ksi): x = 10, y = -20, xy = -20. The material is uneven and has strengths (in ksi) of S ut = 50, S y = 40, and S uc = 90. Calculate the safety factor and draw a a-b diagram showing the boundary for each theory with the stress state and load line using: (a) Coulomb-Mohr theory, and (b) Modified Mohr theory.

Given:

Stress components

σx  10 ksi

σy  20 ksi

τxy  20 ksi

Material properties


S y  40 ksi


Solution: 1.


Calculate the nonzero principal stresses using equation 4.6a.

σa 

σb  2.

4.

2

σx  σy 2

2  σx  σy  2     τxy 2  

σa  20 ksi

2  σx  σy  2     τxy 2  

σb  30 ksi

Calculate the slope of load line. (The load line is the line from the origin through the stress point.) m 

3.

σx  σy

σb σa

m  1.5

The safety factor equation for both theories is different for each quadrant the load line falls in. The equation for the modified Mohr factor of safety is different for each of the two regions in the 4th quadrant that the load line can fall in. In this case, the load line falls in the 4th quadrant, below the -1 slope line.. The factors of safety are: (a) Coulomb-Mohr theory

Na 

(b) Modified Mohr theory

Nb 

S uc S ut S uc σa  S ut σb

S uc

 S uc  S ut   S   σa  σb ut  

Na  1.4

Nb  2

Plot the a-b diagram showing the safe-fail boundaries, the stress state point (20 ksi, -30 ksi) and the load line. Note that if a > b , then only that area on the graph that is to the right of and below the diagonal line can contain valid stress points. The factor of safety is the distance along the load line from the origin to the intersection of the load line with the failure boundary, divided by the distance from the origin to the stress poin Since the distance from the origin to the modified Mohr boundary is greater than the distance to the Coulomb-Mohr boundary, its factor of safety is greater. See Figure 5-62 on the following page.



5-64-2

b 50 40

MINIMUM NONZERO PRINCIPAL STRESS, KSI

30 20


10

a

0 -10 -20 (20,-30) -30 -40 Stress states at which failure will occur

-50 -60

-S


-70

Load Line

-80 -90

-S

ut

uc

-100 -100 -90 -80 -70 -60 -50 -40 -30 -20 -10

0

10

20

30

40

50


FIGURE 5-64 a -  b Diagram for Problem 5-64



5-65-1


A differential element is subjected to the stresses (in ksi): x = 10, y = -5, xy = 15. The material is uneven and has strengths (in ksi) of S ut = 50, S y = 40, and S uc = 90. Calculate the safety factor and draw a a-b diagram showing the boundary for each theory with the stress state and load line using: (a) Coulomb-Mohr theory, and (b) Modified Mohr theory.

Given:

Stress components

σx  10 ksi

σy  5  ksi

τxy  15 ksi

Material properties


S y  40 ksi


Solution: 1.



σa 

σb  2.

4.

2

σx  σy 2

2  σx  σy  2     τxy  2 

σa  19.3 ksi

2  σx  σy  2     τxy  2 

σb  14.3 ksi


3.

σx  σy

σb σa

m  0.741

The safety factor equation for both theories is different for each quadrant the load line falls in. The equation for the modified Mohr factor of safety is different for each of the two regions in the 4th quadrant that the load line can fall in. In this case, the load line falls in the 4th quadrant, above the -1 slope line.. The factors of safety are: (a) Coulomb-Mohr theory

Na 


Nb 

S uc S ut S uc σa  S ut σb

S ut

σa

Na  1.8

Nb  2.6

Plot the a-b diagram showing the safe-fail boundaries, the stress state point (19.3 ksi, -14.3 ksi) and the load line. Note that if a > b , then only that area on the graph that is to the right of and below the diagonal line can contain valid stress points. The factor of safety is the distance along the load line from the origin to the intersection of the load line with the failure boundary, divided by the distance from the origin to the stress poin Since the distance from the origin to the modified Mohr boundary is greater than the distance to the Coulomb-Mohr boundary, its factor of safety is greater. See Figure 5-63 on the following page.



5-65-2

b 50 40


30 20


10

a

0 (19.3,-14.3)

-10 -20 -30 -40

Load Line Stress states at which failure will occur

-50 -60

-S

ut


-70 -80 -90

-S

uc

-100 -100 -90 -80 -70 -60 -50 -40 -30 -20 -10

0

10

20

30

40

50





5-66-1


A differential element is subjected to the stresses (in ksi): x = -20, y = -15, xy = 15. The material is uneven and has strengths (in ksi) of S ut = 50, S y = 40, and S uc = 90. Calculate the safety factor and draw a a-b diagram showing the boundary for each theory with the stress state and load line using: (a) Coulomb-Mohr theory, and (b) Modified Mohr theory.

Given:

Stress components

σx  20 ksi

σy  15 ksi

τxy  15 ksi

Material properties


S y  40 ksi


Solution: 1.



σa 

σb  2.

4.

2

σx  σy 2

2  σx  σy  2     τxy  2 

σa  2.29 ksi

2  σx  σy  2     τxy  2 

σb  32.7 ksi


3.

σx  σy

σb σa

m  14.263

(third quadrant since both principal stresses are negative)

The safety factor equation for both theories is the same when the load line falls in the third quadrant. The factors of safety are:

(a) Coulomb-Mohr theory

Na  


Nb  

S uc

σb

S uc

σb

Na  2.8

Nb  2.8

Plot the a-b diagram showing the safe-fail boundaries, the stress state point (-2.29 ksi, -32.7 ksi) and the load line. Note that if a > b , then only that area on the graph that is to the right of and below the diagonal line can contain valid stress points. The factor of safety is the distance along the load line from the origin to the intersection of the load line with the failure boundary, divided by the distance from the origin to the stress point. Since the distance from the origin to the modified Mohr boundary is the same as the distance to the Coulomb-Mohr boundary, its factor of safety is the same. See Figure 5-63 on the following page.



5-66-2

b 50 40


30 20


10

a

0 -10 -20 -30 (-2.29,-32.7) -40 Stress state at which failure will occur for both theories

-50 -60

-S

ut

-70 -80 -90

-S


uc

Load Line -100 -100 -90 -80 -70 -60 -50 -40 -30 -20 -10

0

10

20

30

40

50





5-67-1

PROBLEM 5-67

_____

Statement:

Derive the von Mises effective stress equation 5.7d for the two-dimensional case.

Solution:


1.

Start with equation 5.7c, which gives the von Mises stress in terms of the two nonzero principal stresses.

σ'  2.

σ3 

(a)

σx  σy 2

σx  σy 2

2  σx  σy  2     τxy  2  2

 σx  σy  2     τxy 2  

σx  σy

and

2

6.

(c)

2  σx  σy  2 R     τxy  2 

(d)

σ3  σc  R

(e)

Substitute equations d into b and c.

σ1  σc  R 5.

(b)

To make the manipulations easier, define:

σc  4.

2

Define 1 and 3 in terms of x, y, and xy using equations 4.6a.

σ1 

3.

2

σ1  σ1 σ3  σ3

Substitute equations e into a, expand, collect terms and simplify.

σ' 

σc  R 2  σc  R  σc  R  σc  R 2

σ' 

σc  2  R σc  R  σc  R  σc  2  R σc  R

σ' 

σc  3  R

2

2

2

2

2

2

2

(f)

2

Substitute equations d into f, expand, collect terms and simplify to obtain the derived equation.

2 2  σx  σy   σx  σy  2 σ'     3    3  τxy 2 2    

σ' 

2

2


2

(5.7d)



5-68-1


Figure P5-23 shows an oil-field pump jack. The crank drive shaft at O2 is loaded in torsion and bending with maximum values of 6500 in-lb and 9800 in-lb, respectively. The point on the shaft with maximum stress is located away from the key that connects the shaft to the crank. Using a factor of safety of 2 against static yielding, determine a suitable diameter for the shaft if it is to be made of SAE 1040 cold-rolled steel.

Given:

Yield strength

S y  71 ksi

Torque T  6500 in lbf Solution: 1.

Bending moment M  9800 in lbf


Express the torsional and bending stresses as a functions of the unknown shaft diameter, d

Bending stress

σx( d ) 

32 M

π d Torsional stress

3

16 T

τxy( d ) 

π d

2.

3

Use these two stresses in an expression for the von Mises effective stress, equation 5.7d withy = 0.

σ'( d ) 

von Mises effective stress 3.

Ns  2

Factor of safty

2

σx( d )  3  τxy( d )

2

Use equation 5.8a as a design relationship to solve for the diameter, d. Design equation

2

2

σx( d )  3  τxy( d ) = 2

Sy Ns 2

 32 M   3  16 T  =  S y   3  3   Ns   π d   π d   

2

Solving for d 1

 ( 32 M ) 2  3  ( 16 T ) 2 d    2   2  Sy  π       Ns  

6

A suitable diameter for the given design requirements is

d  1.480 in

d  1.500  in



5-69-1


Figure P5-24a shows a C-clamp with an elliptical body dimensioned as shown. The clamp has a T-section with a uniform thickness of 3.2 mm at the throat as shown in Figure P5-24b. Find the static factor of safety if the clamping force is 2.7 kN and the material is class 40 gray cast iron.

Given:

Clamping force F  2.7 kN Distance from center of screw to throat Section dimensions: t  3.2 mm Material properties

Solution: 1.

ri  63.5 mm

Flange b  28.4 mm

Web h  31.8 mm

S ut  290  MPa

S uc  965  MPa


Determine the location of the CG of the T-section and the distance from the centerline of the screw to the centroid of the section at the throat. yCG 

0.5 t ( b  t)  0.5 ( h  t)  ( h  t) t

yCG  9.578  mm

b t  ( h  t)  t

rc  ri  yCG 2.

rc  73.078 mm


A  b  t  ( h  t)  t

ri t

   r

i

6.

rn  71.864 mm

 t dr   dr r r  r  t i

e  rc  rn

e  1.214  mm

M  rc F

M  197  N  m


5.

2


4.

ro

b


ro  95.3 mm

A  182.4  mm A

rn 


ro  ri  h

ci  8.364  mm

co  ro  rn

co  23.436 mm


 ci  F  e A  ri  A M

Inner radius

σi 

Outer radius

σo  



 co  F  e A  ro  A M



σi  132.2  MPa

σo  204.3  MPa

These are the principal stresses so,



7.

5-69-2

Inner radius

σ1i  σi

σ2i  0  MPa

σ3i  0  MPa

Outer radius

σ1o  0  MPa

σ2o  0  MPa

σ3o  σo

Calculate the factor of safety using equations 5.12c, 5.12d, and 5.12e. Inner radius

C1i 

1

C2i 

1

C3i 

1

2

2

2



S uc  2  S ut



S uc



S uc  2  S ut



S uc



S uc  2  S ut



S uc

  σ1i  σ2i 

  σ2i  σ3i 

  σ3i  σ1i 



  σ1i  σ2i

C1i  92.46  MPa





  σ2i  σ3i

C2i  0.00 MPa





  σ3i  σ1i

C3i  92.46  MPa



σeff  max C1i C2i C3i σ1i σ2i σ3i Ni 

S ut

σeff  132.182  MPa Ni  2.2

σeff

Outer radius

C1o 

1

C2o 

1

C3o 

1

2

2

2



S uc  2  S ut



S uc



S uc  2  S ut



S uc



S uc  2  S ut



S uc

  σ1o  σ2o 

  σ2o  σ3o 

  σ3o  σ1o 



  σ1o  σ2o



  σ2o  σ3o

S ut

σeff

C2o  61.41  MPa





  σ3o  σ1o

C3o  61.41  MPa



σeff  max C1o C2o C3o σ1o σ2o σ3o No 

C1o  0.00 MPa



σeff  61.41  MPa No  4.7



5-70-1


A C-clamp as shown in Figure P5-24a has a rectangular cross section as in Figure P5-24c. Find the static factor of safety if the clamping force is 1.6 kN and the material is class 50 gray cast iron.

Given:

Clamping force F  1.6 kN Distance from center of screw to throat Section dimensions: Material properties

Solution: 1.

Width b  6.4 mm S ut  359  MPa

Depth h  31.8 mm S uc  1131 MPa



2.

ri  63.5 mm

h

rc  79.4 mm

2



A  b  h rn 

A ro

rn  78.327 mm dr

i

e  rc  rn

e  1.073 mm

M  rc F

M  127 N  m


ci  14.827 mm

co  ro  rn

co  16.973 mm


σi 

 ci  F  e A  ri  A M

σo   6.

2


5.

b r


4.

ro  95.300 mm

A  203.520 mm

   r

3.

ro  ri  h



σi  143.7 MPa

 co  F  e A  ro  A M



σo  95.8 MPa

These are the principal stresses so, Inner radius

σ1i  σi

σ2i  0  MPa

σ3i  0  MPa

Outer radius

σ1o  0  MPa

σ2o  0  MPa

σ3o  σo



7.

5-70-2


C1i 

1

C2i 

1

C3i 

1

2

2

2



S uc  2  S ut



S uc



S uc  2  S ut



S uc



S uc  2  S ut



S uc

  σ1i  σ2i 

  σ2i  σ3i 

  σ3i  σ1i 



  σ1i  σ2i

C1i  98.09 MPa





  σ2i  σ3i

C2i  0.00 MPa





  σ3i  σ1i

C3i  98.09 MPa




S ut

σeff  143.707 MPa Ni  2.5

σeff

Outer radius

C1o 

1

C2o 

1

C3o 

1

2

2

2



S uc  2  S ut



S uc



S uc  2  S ut



S uc



S uc  2  S ut



S uc

  σ1o  σ2o 

  σ2o  σ3o 

  σ3o  σ1o 



  σ1o  σ2o





  σ2o  σ3o





  σ3o  σ1o




S ut

σeff

C1o  0.00 MPa

C2o  30.39 MPa

C3o  30.39 MPa

σeff  30.394 MPa No  11.8



5-71-1


A C-clamp as shown in Figure P5-24a has an elliptical cross section as in Figure P5-24d. Dimensions of the major and minor axes of the ellipse are given. Find the static factor of safety if the clamping force is 1.6 kN and the material is class 60 gray cast iron.

Given:

Clamping force F  1.6 kN Distance from center of screw to throat Section dimensions: Material properties

Solution: 1.

Width b  9.6 mm S ut  427  MPa

Depth h  31.8 mm S uc  1289 MPa



2.

ri  63.5 mm

h

rc  79.4 mm

2



b h A  π  2 2

ro 2   1   r  rc  2 b 4 2  h  

dr

i

e  rc  rn

e  0.805 mm

M  rc F

M  127 N  m


ci  15.095 mm

co  ro  rn

co  16.705 mm


σi 

 ci  F  e A  ri  A M

σo   6.

0.5


5.

2

rn  78.595 mm

r


4.

ro  95.300 mm

A  239.766 mm A

rn 

      r

3.

ro  ri  h



 co  F  e A  ro  A M



σi  163.2 MPa

σo  108.7 MPa

These are the principal stresses so,



7.

5-71-2

Inner radius

σ1i  σi

σ2i  0  MPa

σ3i  0  MPa

Outer radius

σ1o  0  MPa

σ2o  0  MPa

σ3o  σo


C1i 

1

C2i 

1

C3i 

1

2

2

2



S uc  2  S ut



S uc



S uc  2  S ut



S uc



S uc  2  S ut



S uc

  σ1i  σ2i 

  σ2i  σ3i 

  σ3i  σ1i 



  σ1i  σ2i



  σ2i  σ3i

C2i  0.00 MPa





  σ3i  σ1i

C3i  109.12 MPa




C1i  109.12 MPa



S ut

σeff  163.169 MPa Ni  2.6

σeff

Outer radius

C1o 

1

C2o 

1

C3o 

1

2

2

2



S uc  2  S ut



S uc



S uc  2  S ut



S uc



S uc  2  S ut



S uc

  σ1o  σ2o 

  σ2o  σ3o 

  σ3o  σ1o 



  σ1o  σ2o





  σ2o  σ3o





  σ3o  σ1o




S ut

σeff

C1o  0.00 MPa

C2o  36.02 MPa

C3o  36.02 MPa

σeff  36.016 MPa No  11.9



5-72-1


A C-clamp as shown in Figure P5-24a has a trapezoidal cross section as in Figure P5-24e. Find the static factor of safety if the clamping force is 350 lb and the material is class 40 gray cast iron.

Given:

Clamping force F  1.6 kN Distance from center of screw to throat

Solution: 1.

Section dimensions:

Width b i  9.6 mm

b o  3.2 mm

Material properties

S ut  290  MPa

S uc  965  MPa

Determine the distance from the centerline of the screw to the centroid of the section at the throat. h bi  2 bo  3 bi  bo

rc  76.75 mm



A 

bi  bo 2

h

ro

bi 

h

rn  75.771 mm   r  ri dr

i

e  rc  rn

e  0.979 mm

M  rc F

M  123 N  m


ci  12.271 mm

co  ro  rn

co  19.529 mm


σi 

 ci  F  e A  ri  A M

σo   6.

bi  bo

2


5.

ro  95.300 mm

A  203.520 mm

r


4.

ro  ri  h

A

rn 

     r

3.



rc  ri  2.

ri  63.5 mm



σi  126.9 MPa

 co  F  e A  ro  A M



σo  118.4 MPa

These are the principal stresses so, Inner radius

σ1i  σi

σ2i  0  MPa

σ3i  0  MPa

Outer radius

σ1o  0  MPa

σ2o  0  MPa

σ3o  σo



5-72-2


C1i 

1

C2i 

1

C3i 

1

2

2

2



S uc  2  S ut



S uc



S uc  2  S ut



S uc



S uc  2  S ut



S uc

  σ1i  σ2i 

  σ2i  σ3i 

  σ3i  σ1i 



  σ1i  σ2i



  σ2i  σ3i

C2i  0.00 MPa





  σ3i  σ1i

C3i  88.77 MPa




C1i  88.77 MPa



S ut

σeff  126.907 MPa Ni  2.3

σeff

Outer radius

C1o 

1

C2o 

1

C3o 

1

2

2

2



S uc  2  S ut



S uc



S uc  2  S ut



S uc



S uc  2  S ut



S uc

  σ1o  σ2o 

  σ2o  σ3o 

  σ3o  σ1o 



  σ1o  σ2o





  σ2o  σ3o





  σ3o  σ1o




S ut

σeff

C1o  0.00 MPa

C2o  35.58 MPa

C3o  35.58 MPa

σeff  35.577 MPa No  8.2



5-73-1


The connecting rod (3) on the oil-field pump jack shown in Figure P5-23 is, in fact, made up of two rods, one connecting on each side of the walking beam (4). Determine a suitable width of 1/2-inch-thick SAE 1020 cold-rolled bar stock to use if the maximum tensile load on the bars is 3500 lb each. Use a factor of safety of 4 against static yielding.

Given:

Yield strength

S y  57 ksi

Factor of safty

Ns  4

Thickness

t  0.50 in

Tensile force

F  3500 lbf

Solution: 1.


Express the tensile stress as a functions of the unknown width, w. Tensile stress

2.

w t

The tensile stress is the only stress present so it is also the von Mises effective stress.

σ'( w) 

von Mises effective stress 3.

F

σx( w) 

F w t

Use equation 5.8a as a design relationship to solve for the diameter, d. Design equation

Solving for w

F w t

w 

=

Sy Ns

N s F t Sy

w  0.491 in

A suitable size for the given design requirements is

w  0.500  in



5-74-1


A work platform is elevated on the end of a boom that has the ability to extend its length and vary its angle with respect to ground. The platform width is large compared to the boom diameter so that it is possible to load the boom eccentrically resulting in a combination of bending, torsion and direct compression in the base of the boom. At the base the boom is a hollow tube with an outside diameter of 8 in and a wall thickness of 0.75 in. It is made from SAE 1030 CR steel. Determine the factor of safety against static failure if the loading at a point at the base of the boom is: M = 600 kip-in, T = 76 kip-in, and an axial compression of 4800 lb.

Given:

Yield strength SAE 1030 CR steel S y  64 ksi D  8.00 in

Boom dimensions Loading Solution: 1.

2.

3.

5.

F  4800 lbf

Calculate the bending stress at the point of interest. Inside diameter

d  D  2  twall

Moment of inertia

I 

Distance to outer fiber

c  0.5 D

Bending stress

σbend 

 64 π

4

 D d

d  6.500 in



4

4

I  113.438 in c  4.000 in

M c

σbend  21.157 ksi

I

Calculate the axial stress due to the compressive load at the point of interest.

 4

π

Cross-section area

A 

Axial stress

σaxial 

2

 D d



2

2

A  17.082 in

F

σaxial  0.281 ksi

A

Combine the bending and axial stresses to get the maximum normal stress on the compressive side of the boom.

σx  σbend  σaxial

σx  21.438 ksi

Calculate the torsional stress at the point of interest. Polar moment

J  2  I

Torsional stress

τxy 

4

J  226.876 in

Tc

τxy  1.34 ksi

J

Calculate the von Mises effective stress using equation 5.7d. von Mises stress

6.

T  76 kip in


Max. normal stress 4.

M  600  kip in

twall  0.75 in

σ' 

2

σx  3  τxy

2

σ'  21.563 ksi

Calculate the factor of safety using equation 5.8a. Factor of safety

N 

Sy

σ'

N  2.97



5-75-1


Repeat Problem 5-74 for a boom that is made from class 20 gray cast iron. At the base the boom is hollow tube with an outside diameter of 10 in and a wall thickness of 1.00 in.

Given:

Strength Class 20 gray cast iron



Boom dimensions

D  10.00  in

twall  1.00 in

Loading Solution: 1.

2.

3.

4.

5.

M  600  kip in

T  76 kip in

F  4800 lbf


Calculate the bending stress at the point of interest. Inside diameter

d  D  2  twall

Moment of inertia

I 

Distance to outer fiber

c  0.5 D

Bending stress

σbend 

 64 π

4

 D d

d  8.000 in



4

4

I  289.812 in c  5.000 in

M c

σbend  10.352 ksi

I

Calculate the axial stress due to the compressive load at the point of interest.

 4

π

Cross-section area

A 

Axial stress

σaxial 

2

 D d



2

F

2

A  28.274 in

σaxial  0.17 ksi

A

Combine the bending and axial stresses to get the maximum normal stress on the tensile and compressive sides of the boom. Max compressive

σxc  σbend  σaxial

σxc  10.521 ksi

Max tensile

σxt  σbend  σaxial

σxt  10.182 ksi

Calculate the torsional stress at the point of interest. Polar moment

J  2  I

Torsional stress

τxy 

4

J  579.624 in

Tc

τxy  0.656 ksi

J

Calculate the principal stresses on the tensile and compressive sides of the boom. 2

Compressive side

 σxc  2 τmaxc     τxy  2  σ1c 

σxc 2

 τmaxc

τmaxc  5.301 ksi

σ1c  0.041 ksi

σ2c  0  ksi σ3c 

σxc 2

 τmaxc

σ3c  10.562 ksi



5-75-2

2

 σxt  2 τmaxt     τxy 2  

Tensile side

σ1t 

σxt 2

τmaxt  5.133 ksi

 τmaxc

σ1t  10.392 ksi

σ2t  0  ksi σ3t  6.

σxt 2

 τmaxc

σ3t  0.210 ksi

Calculate the factor of safety using equations 5.12c, 5.12d, and 5.12e. Compressive side

C1c 

1

C2c 

1

C3c 

1

2

2

2



S uc  2  S ut



S uc



S uc  2  S ut



S uc



S uc  2  S ut



S uc

  σ1c  σ2c 

  σ2c  σ3c 

  σ3c  σ1c 



  σ1c  σ2c



  σ2c  σ3c

C2c  2.80 ksi

 

  σ3c  σ1c

C3c  2.83 ksi



σeff  max C1c C2c C3c σ1c σ2c σ3c Nc 

C1c  0.03 ksi



S ut

σeff  2.829 ksi Nc  7.8

σeff

Tensile side C1t 

1

C2t 

1

C3t 

1

2

2

2



S uc  2  S ut



S uc



S uc  2  S ut



S uc



S uc  2  S ut



S uc

  σ1t  σ2t 

  σ2t  σ3t 

  σ3t  σ1t 



  σ1t  σ2t



  σ2t  σ3t

S ut

σeff

C2t  0.06 ksi





  σ3t  σ1t

C3t  7.69 ksi



σeff  max C1t C2t C3t σ1t σ2t σ3t Nt 

C1t  7.64 ksi



σeff  10.392 ksi Nt  2.1



5-76-1


Assume that the curved beam of Problem 5-70 has a crack on its inside surface of half-width a = 1.5 mm and a fracture toughness of 35 MPa-m0.5. What is its safety factor against sudden fracture?

Given:

Width of section

t  31.8 mm

Half crack length

a  1.5 mm

Solution:

Fracture toughness


1.

From Problem 5-70, the nominal stress at the inside radius is: Nominal inside stress σi  143.7  MPa

2.

Calculate the half-width of the beam.

3.

Calculate the geometry and stress intensity factors. π a  β  sec  β  1.006   2 b  K  β  σi π a

4.

Kc  35 MPa m

b  0.5 t

b  15.9 mm

K  9.92 MPa m

Determine the factor of safety against sudden fracture failure

NFM 

Kc K

NFM  3.5



5-77-1

PROBLEM 5-77

_____

Statement:

A large aircraft panel is to be made from 7075-T651 aluminum bar. From test data it is found that the nominal tensile stress in the panel is 200 MPa. What is the average maximum central crack size that can be tolerated without catastrophic failure?

Given:

Fracture toughness

Kc  22 ksi in

Nominal stress

σnom  200  MPa

Solution: 1.

0.5

0.5

Kc  24.2 MPa m


Determine the value of the geometry factor  from the discussion in Section 5.3 for a large plate with a central crack.

β  1 2.


 Kc  a     π  β  σnom  1

lcritical  2  a

2

a  4.65 mm

lcritical  9.3 mm



5-78-1


Design the connecting rod (link 3) of Problem 3-50 for a safety factor of 4 if the link is made from SAE 1010 hot-rolled steel sheet, the pin hole diameter at each end is 6 mm, and the maximum applied tensile load is 2000 N. There are two links carrying the load.

Given:

Force on links

Ftotal  2000 N

Yield strength

S y  179  MPa


Nd  4

Pin hole diameter

d  6  mm w  3  d

Assumptions: Choose a suitable width, say Solution:

w  18 mm

See Figure P3-22 and Mathcad file P0577. F 

1. The force on each link is

Ftotal

F  1000 N

2

2. With only a tensile force acting on the link, the tensile stress will be the principal stress and it will also be the von Mises effective stress, so we have σx = 1 = '. 3. The tensile stress on each link is

F

σx =

4. Using the distortion-energy failure theory,

A

Nd =

t 

5. Solving for the thickness,t,

=

Sy

σ'

F t w

=

= σ'

t w Sy F

F  Nd

t  1.241  mm

w S y

t  2  mm

6. Round this up to the next higher integer value, N 

7. The realized factor of safety against tensile failure is, 8.

t  w S y

N  6.4

F

Check the factor of safety against bearing failure in the pin holes. Bearing area

Abear  w t

σbear 

Abear  36.0 mm

F Abear

2


This is the principal stresses 1. 9.

Calculate the safety factor for direct bearing from equation 5.8c where 2 and 3 are both zero. Pin hole

10. The tearout area is

Nbear 

2

Sy

σbear

Nbear  6.4

2

Atear = 2  t R  ( 0.5 d ) , where R  0.5 w (see figure below). Substitute this area in

equation 4.9 for the shear area and solve for the shear strength xy. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


5-78-2 Tearout length

Shear area

2

Atear  2  t R  ( 0.5 d ) Atear  33.941 mm

Shear stress

τxy 

2

2

F Atear

τxy  29.46  MPa d

8.

R

From equations 5.8c and 5.9b, the safety factor against tearout failure in the holes is

Ntear 

0.577  S y

τxy

Ntear  3.5

This is slightly less than the design FS of Nd  4 so, choose t = 2.5 mm or increase w to 4*d.



5-79-1


Design the compacting ram (link 4) of Problem 3-50 for a safety factor of 4 if the ram is made from SAE 1010 hot-rolled steel bar, the pin hole diameter at the joint where link 3 attaches is 6 mm, and the applied load Fcom = 2000 N. The piston has a diameter of 35 mm.

Given:

Force at point P

Fcom  2000 N

Yield strength

S y  179  MPa


Nd  4

Pin hole diameter

d  6  mm

Assumptions: The points of maximum stress on a plane through point D are sufficiently removed from point D that there is no stress concentration at those points. Solution:


1. From Problem 3-51 the forces and reactions on the ram are: F34x  553  N

F34y  2000 N

F14E  357  N

F14F  196  N

42.5

E

F14E

D

F34x

120.0

2. The maximum bending moment is at point D and is: M  42.5 mm F14E

F34y

M  15172.5 N  mm

F14F

The section modulus and area for the ram are 3

Z ( D) 

π D

π D

P

4

3. Between points D and P there is a compressive force of Fcom  2000 N. Thus, there is a compressive stress due

Fcom

to this force in addition to the bending stress at point D.

Compacting Ram (4)

On the left side of the ram at the section through point D

σbL ( D) 

M Z ( D)

F

2

A ( D) 

32

σa( D) 

77.5

Fcom

σL ( D)  σbL ( D)  σa( D)

A ( D)

On the right side of the ram at the section through point D

σbR( D)  

M Z ( D)

σa( D) 

4  Fcom

σR( D)  σbR( D)  σa( D)

2

π D

The compressive stress on the right side will be numerically greater than that on the left side. 4. Since the shear stress due to bending is zero at these points, the axial stress will be the principal stress and it will also be the von Mises effective stress, so we have σx = 1 = '. 5. Using the distortion-energy failure theory,

Nd =

Sy

σ'

=

Sy

σR( D)

6. Solving for the diameter, D, © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

MACHINE DESIGN - An Integrated Approach, 4th Ed. Guess

5-79-2

D  10 mm

f ( D)  Nd  σR( D)  S y

D  root( f ( D) D)

7. Round this up to the next higher even integer value, say 8. The realized factor of safety against axial yeilding is,

N 

Sy

σR( D)

D  16.368 mm D  18 mm N  5.2

9. The axial stress on each side of the ram on a section through D is:

σL ( D)  18.6 MPa

σR( D)  34.4 MPa



5-80-1


A differential element is subected to the stresses given below and a ductile material has the strengths given below. Calculate the safety factor and draw 1-3 diagrams of each theory showing the stress state using: (a) Maximum shear-stress theory, and (b) Distortion-energy theory.

Given:

Principal stresses

σ1  70 MPa

Material properties

S ut  350  MPa S y  280  MPa

Solution:

σ2  0  ksi

σ3  140  MPa S uc  350  MPa



σ3

m  2

σ1

2. The safety factor equation for the distortion-enrgy theory is the same regardless of which quadrant the load line falls in. However, the equation for the maximum shear-stress factor of safety is different for each of the three quadrants that the load line (1st, 3rd, or 4th) can fall in. In this case, the load line falls in the 4th quadrant. The factors of safety are: Na 

(a) Maximum shear-stress theory

Sy

Na  1.3

σ1  σ3

(b) Distortion energy theory

3

280

σ' 

2

σ1  σ1 σ3  σ3

(a) Maximum shear stress boundary

2 210

(b) Distortion energy boundary

σ'  26.9 ksi Sy

σ'

Nb  1.5

3. Plot the 1-3 diagram showing the safe-fail boundaries, the stress state point (70MPa, -140 MPa) and the load line. Note that if 1 > 3 , then only that area on the graph that is to the right of and below the diagonal line can contain valid stress points. The factor of safety is the distance along the load line from the origin to the intersection of the load line with the failure boundary, divided by the distance from the origin to the stress point. Since the distance from the origin to the distortion-energy boundary is greater than the distance to the maximum shear-stress baoundary, its factor of safety is greater.

MINIMUM NONZERO PRINCIPAL STRESS, MPa

Nb 

140

70

0 sy

1

-70 (70,-140) -140

-210

Stress states at which failure will occur

-s y

-280

Load Line

-350

-420 -280

-210

-140

-70

0

70

140

210

280

350

MAXIMUM PRINCIPAL STRESS, MPa




5-81-1


A part has the combined stress state and strengths given below. Using the Distortion-Energy failure theory, find the von Mises effective stress and factor of safety against static failure.

Given:

Stresses: σx  70 MPa

S y  126  MPa

Strengths: Solution: 1.

σy  35 MPa

τxy  31.5 MPa

S ut  140  MPa S uc  140  MPa




2  σx  σy  2 τmax     τxy  2 

Principal stresses

σ1  σ2 

σx  σy 2

σx  σy 2


 τmax

σ1  88.5 MPa

 τmax

σ2  16.5 MPa

σ3  0  psi 2.

Find the von Mises effective stress using equation 5.7d:

σ' 

3.

2

2

σx  σx σy  σy  3  τxy

2

The safety factor can now be found using equation 5.8a.

σ'  81.6 MPa

N 

Sy

σ'

N  1.5



5-82-1


Repeat Problem 5-78 for the connecting rod made from class 20 cast iron.

Given:

Force on links

Ftotal  2000 N

Strength

S ut  152  MPa


Nd  4

Pin hole diameter

d  6  mm w  4  d

Assumptions: Choose a suitable width, say Solution:

S uc  572  MPa

w  24 mm

See Figure P3-22 and Mathcad file P0582. F 

1. The force on each link is

Ftotal

F  1000 N

2

2. With only a tensile force acting on the link, the tensile stress will be the principal stress so we have σx = 1.

3. The tensile stress on each link is

F

σx =

4. Using the modified-Mohr failure theory,

A

Nd =

t 

5. Solving for the thickness,t,

=

S ut

σ1

F t w

= σ1

t w S ut

=

F

F  Nd

t  1.096  mm

w S ut

t  2  mm

6. Round this up to the next higher integer value, N 

7. The realized factor of safety against tensile failure is, 8.

t w S ut

N  7.3

F

Check the factor of safety against bearing failure in the pin holes. Bearing area

Abear  w t

σbear 

Abear  48.0 mm

F Abear

2


This is the principal stresses 1. 9.

Calculate the safety factor for direct bearing from equation 5.8c where 2 and 3 are both zero. Pin hole

10. The tearout area is

Nbear  2

S uc

σbear

Nbear  27.5

2

Atear = 2  t R  ( 0.5 d ) , where R  0.5 w (see figure below). Substitute this area in

equation 4.9 for the shear area and solve for the shear strength xy.



5-82-2

Tearout length

Shear area

2

Atear  2  t R  ( 0.5 d ) Atear  46.476 mm

Shear stress

τxy 

2

2

F Atear

τxy  21.52  MPa Principal stress 8.

σ1  τxy

d

R

For pure shear the Mohr circle is centered at 0,0 and has a radius equal to the shear stress. This results in 1 = . Using the Modified-Mohr failure theory and Figure 5-11, we see that we can use equation 5.12a for the safety factor against tearout. Ntear 

S ut

σ1

Ntear  7.1



5-83-1


Repeat Problem 5-79 for the part made from class 20 cast iron.

Given:

Force at point P

Fcom  2000 N

Tensile strength

S ut  152  MPa


Nd  4

Pin hole diameter

d  6  mm

Assumptions: The points of maximum stress on a plane through point D are sufficiently removed from point D that there is no stress concentration at those points. Solution:

See Figure P3-22 and Mathcad files P0579 and P0583.

1. From Problem 3-51 the forces and reactions on the ram are: F34x  553  N

F34y  2000 N

F14E  357  N

F14F  196  N

42.5

E

F14E

D

F34x

120.0

2. The maximum bending moment is at point D and is: M  42.5 mm F14E

F34y 77.5

M  15172.5 N  mm

F14F

The section modulus and area for the ram are 3

Z ( D) 

F

2

π D

A ( D) 

32

π D

P

4

3. Between points D and P there is a compressive force of Fcom  2000 N. Thus, there is a compressive stress due

Fcom

to this force in addition to the bending stress at point D.

Compacting Ram (4)

On the left side of the ram at the section through point D

σbL ( D) 

M Z ( D)

σa( D) 

Fcom

σL ( D)  σbL ( D)  σa( D)

A ( D)

On the right side of the ram at the section through point D

σbR( D)  

M Z ( D)

σa( D) 

4  Fcom

σR( D)  σbR( D)  σa( D)

2

π D

The tensile stress on the left side will be critical for an uneven, brittle material. 4. With only a tensile stress acting on the ram at this point, it will be the principal stress so we have σL = 1. 5. Using the modified-Mohr failure theory,

Nd =

S ut

σ1

=

S ut

σL

6. Solving for the diameter, D, Guess

D  10 mm

f ( D)  Nd  σL ( D)  S ut

D  root( f ( D) D)

D  14.567 mm



7. Round this up so that

D  3  d

8. The realized factor of safety against axial failure is,

N 

S ut

σL ( D)

5-83-2

D  18 mm N  8.2

9. The axial stress on each side of the ram on a section through D is:

σL ( D)  18.6 MPa

σR( D)  34.4 MPa



5-84-1


A differential element is subected to the stresses and strengths given below. Calculate the safety factor and draw 1-3 diagrams of each theory showing the stress state using: (a) Coulomb-Mohr theory, and (b) Modified Mohr theory.

Given:

Principal stresses

σ1  70 MPa

Material properties

S ut  350  MPa S uc  630  MPa

Solution:

σ2  0  MPa

σ3  140  MPa



σ3 σ1

m  2

2. The safety factor equation for both theories is different for each quadrant the load line falls in. The equation for the modified Mohr factor of safety is different for each of the two regions in the 4th quadrant that the load line can fall in. In this case, the load line falls in the 4th quadrant, below the -1 slope line.. The factors of safety are: Na 

(a)Coulomb-Mohr theory

(b) Modified Mohr theory S uc

280

 S uc  S ut   S   σ1  σ3 ut  

210

Nb  3.2 3. Plot the 1-3 diagram showing the safe-fail boundaries, the stress state point (70 MPa,-140 MPa) and the load line. Note that if 1 > 3 , then only that area on the graph that is to the right of and below the diagonal line can contain valid stress points. The factor of safety is the distance along the load line from the origin to the intersection of the load line with the failure boundary, divided by the distance from the origin to the stress point. Since the distance from the origin to the modified Mohr boundary is greater than the distance to the Coulomb-Mohr boundary, its factor of safety is greater.

Na  2.4

S uc σ1  S ut σ3

3

350

140 MINIMUM NONZERO PRINCIPAL STRESS, MPa

Nb 

S uc S ut


70

1

0 -70 (70,-140) -140 -210 -280 Stress states at which failure will occur

-350 -420

-S

ut


-490

Load Line

-560 -630 -S

uc

-700 -700 -630 -560 -490 -420 -350 -280 -210 -140 -70

0

70

140 210 280 350

MAXIMUM PRINCIPAL STRESS, MPa




5-85-1


A part has the combined stress state and strengths given below. Using the Modified-Mohr failure theory, find the effective stress and factor of safety against static failure.

Given:

Stresses: σx  70 MPa Strengths:

Solution:

σy  35 MPa

S y  126  MPa

τxy  31.5 MPa

S ut  140  MPa S uc  560  MPa


1.

Because S uc is greater than S ut, this is an uneven material, which is characteristic of a brittle material.

2.



2  σx  σy  2 τmax     τxy  2 

Principal stresses

σ1  σ2 

σx  σy 2

σx  σy 2


 τmax

σ1  88.5 MPa

 τmax

σ2  16.5 MPa

σ3  0  psi 3.

4.

Find the Dowling factors C1, C2, C3 using equations 5.12b: C1 

1

C2 

1

C3 

1

2

2

2



S uc  2  S ut



S uc



S uc  2  S ut



S uc



S uc  2  S ut



S uc

  σ1  σ2 

  σ2  σ3 

  σ3  σ1 



  σ1  σ2

C1  62.3 MPa

 

  σ2  σ3

C2  12.3 MPa

 

  σ3  σ1

C3  66.4 MPa



Then find the largest of the six stresses C1, C2, C3 , 1, 2, 3: C   1     C2    C  3 σeff  max      σ1       σ2     σ3  

σeff  88.5 MPa

which is the modified-Mohr effective stress. 5.

The safety factor can now be found using equation 5.12d.

N 

S ut

σeff

N  1.6



6-1a-1


For the data in row a in Table P6-1, find the stress range, alternating stress component, mean stress component, stress ratio, and amplitude ratio.

Given:

σmax  1000

Solution:


1.

σmin  0

Use equations (6.1) to calculate the required quantities. Stress range

Δσ  σmax  σmin

Alternating stress

σa 

Mean stress

σm 

Stress ratio

R 

Amplitude ratio

A 

σmax  σmin 2

σmax  σmin 2

σmin σmax σa σm

Δσ  1000 σa  500

σm  500

R0

A1



6-2a-1


For the strength data in row a in Table P6-2, calculate the uncorrected endurance limit and draw th strength-life (S-N) diagram for the material, assuming it to be steel.

Given:

Tensile strength

Solution:


1.


Using equation (6-5a), calculate the uncorrected endurance limit. S'e 

return 0.5 S ut if S ut  200  ksi

S'e  45.0 ksi

100  ksi otherwise S m  0.9 S ut

S m  81.0 ksi

2.

Using equation (6.9), calculate the fatigue strength at N = 10 3 cycles.

3.

The equation for the S-N curve in the HCF region is given by equation (6.10a):

4.

Determine the constants a and b from equations (6.10c) and (6.10a). From Table 6-5, for N = 10 6 , z  3.000 b 

a 

5.

1 z

 Sm    S'e 

 log

b

b  0.0851

Sm

103

S'f = a  N

a  145.8 ksi

b

To draw the S-N graph over the range 10 3 <= N <= 10 8, define a piecewise continuous function. S'f ( N ) 

return a  N

b

if N  10

6

S'e otherwise 6. Plot the S-N curve over the range

3

5

N  10 10  10

8

100

S' f ( N ) ksi

10 3 1 10

4

1 10

5

6

1 10

1 10

7

1 10

8

1 10

N

FIGURE 6-2a S-N Diagram for Steel for Problem 6-2a



6-3-1


For the bicycle pedal-arm assembly in Figure P6-1 assume a rider-applied force that ranges from 0 1500 N at the pedal each cycle. Determine the fluctuating stresses in the 15-mm-dia pedal arm. Fi the fatigue safety factor if S ut = 500 MPa.

Given:

Material yield strength

S y  350  MPa

S ut  500  MPa

Applied load

Fmax  1500 N

Fmin  0  N

Pedal arm diameter

d  15 mm

Solution: 1.


From problem 4-3, the maximum principal stresses in the pedal arm due to Fmax are at point A and are

σ1max  793  MPa 2.

σ2max  0  MPa

Using equation 5.7c, the maximum von Mises stress is 2

σ'max 

σ1max  σ1max σ3max  σ3max

2

σ'max  804.7 MPa

σ'min  0  MPa

3.

The minimum von Mises stress is zero.

4.

The alternating and mean components of the von Mises stress are:

σ'a  σ'm  5.

σ3max  23 MPa

σ'max  σ'min

σ'a  402.4 MPa

2

σ'max  σ'min

σ'm  402.4 MPa

2 S'e  0.5 S ut

Calculate the unmodified endurance limit.

S'e  250 MPa

z

z

a

A

Tc

C

Section C

Frider Mc

b

Arm

Arm Fc

Pedal

y

x y

x

FIGURE 6-3A

FIGURE 6-3B



6.

B

Calculate the endurance limit modification factors for a nonrotating round beam. Load

Cload  1



A95  0.010462 d

Size

d equiv 

6-3-2

2

A95  2.354 mm

A95

d equiv  5.544 mm

0.0766

 d equiv  Csize  1.189     mm  A  4.51

Surface

Csurf

7.

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability

Creliab  0.753

 0.097

Csize  1.007

b  0.265

Csize  1

(machined)

b

Csurf  0.869

(R = 99.9%)

Calculate the modified endurance limit. S e  Cload  Csize Csurf  Ctemp Creliab S'e

8.

2

S e  163.56 MPa

Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety. Nf 

S e S ut

σ'a S ut  σ'm S e

Nf  0.31


MACHINE DESIGN - An Integrated Approach

6-4a-1


For the strength data in row a in Table P6-2, calculate the uncorrected fatigue strength at 5E8 cycles and draw the strength-life (S-N) diagram for the material, assuming it to be an aluminum alloy.

Given:

Tensile strength

Solution:


1.


Using equation (6-5c), calculate the uncorrected fatigue strength at 5E8 cycles. S'f5E8 

return 0.4 S ut if S ut  48 ksi

S'f5E8  19.0 ksi

19 ksi otherwise S m  0.9 S ut

S m  81.0 ksi

2.


3.


4.

Determine the constants a and b from equations (6.10c) and (6.10a). From Table 6-5, for N = 5E8 , z  5.699 b 

a 

1 z

 Sm    S'f5E8 

 log

b

b  0.1105

Sm

103

S'f = a  N

a  173.772 ksi

b

5.

To draw the S-N graph over the range 10 3 <= N <= 10 8,

6.

Plot the S-N curve over the range

S'f ( N )  a  N

3

5

b

8

N  10 1.01 10  10

100

S' f ( N ) ksi

10 3 1 10

4

1 10

5

6

1 10

1 10

7

1 10

8

1 10

N

FIGURE 6-4a S-N Diagram for Aluminum for Problem 6-4a



6-5a-1


For the data in row a in Table P6-3, find the corrected endurance strength (or limit), create equations for the S-N line, and draw the S-N diagram.

Given:

Material Tensile strength

steel S ut  110  ksi

Surface finish Loading

surface  "ground" load  "torsion"

Shape Size (diameter)

round d  2  in

Temperature Reliability

T  72

Solution: 1.

R  0.999


Using equation (6-5a), calculate the uncorrected endurance limit. S'e 

return 0.5 S ut if S ut  200  ksi

S'e  55.0 ksi

100  ksi otherwise 2.

Calculate the endurance limit modification factors for a nonrotating round rod. Load

Cload 

return 1 if load = "bending"

Cload  1

return 1 if load = "torsion" return 0.7 if load = "axial" Size

d equiv  d

 d equiv    in 

Surface

 0.097

Csize  0.869  

Csize  0.812

A 

A  1.34

return 1.34 if surface = "ground" return 2.70 if surface = "machined" return 2.70 if surface = "cold_rolled" return 14.4 if surface = "hot_rolled" return 39.9 if surface = "forged"

b 

return 0.085 if surface = "ground"

b  0.085

return 0.265 if surface = "machined" return 0.265 if surface = "cold_rolled" return 0.718 if surface = "hot_rolled" return 0.995 if surface = "forged"

 S ut    ksi 

Temperature

b

Csurf  A  

Csurf  0.899

Ctemp 

Ctemp  1

return 1 if T  840 1  0.0032 ( T  840 ) otherwise


MACHINE DESIGN - An Integrated Approach, 4th Ed. Creliab 

Reliability

6-5a-2 Creliab  0.753

return 1.000 if R = 0.50 return 0.897 if R = 0.90 return 0.814 if R = 0.99 return 0.753 if R = 0.999 return 0.702 if R = 0.9999 return 0.659 if R = 0.99999

3.


4.

S e  30.2 ksi

Using equation (6.9), calculate the fatigue strength at N = 10 3 cycles. S m 

return 0.75 S ut if load = "axial"

S m  99.0 ksi

0.9 S ut otherwise Sf = a N

b

5.


6.


a 

1 z

 Sm    Se 

 log Sm

10  3

7.

b  0.1717

a  324.120 ksi

b

To draw the S-N graph over the range 10 3 <= N <= 10 8, define a piecewise continuous function. S f ( N ) 

return a  N

b

if N  10

6

S e otherwise

8.

Plot the S-N curve over the range

3

5

N  10 10  10

8

100

Sf ( N ) ksi

10 3 1 10

4

1 10

5

6

1 10

1 10

7

1 10

8

1 10

N

FIGURE 6-5a S-N Diagram for Problem 6-5a © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0605a.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


6-6-1


For the trailer hitch from Problem 3-6 on p. 169 (also see Figures P6-2 and 1-5), find the infinite-lif fatigue safety factors for all modes of failure assuming that the horizontal impact force of the traile on the ball is fully reversed. Use steel with S ut = 600 MPa and S y = 450 MPa. Determine safety factors for: (a) The shank of the ball where it joins the ball bracket. (b) Bearing failure in the ball bracket hole. (c) Tearout failure in the ball bracket. (d) Tensile failure in the 19-mm diameter attachment holes. (e) Bending failure in the ball bracket as a cantilever.

Given:

a  40 mm b  31 mm Mtongue  100  kg Fpull  55.1 kN S y  300  MPa

c  70 mm d sh  26 mm

d  20 mm t  19 mm

w  64 mm

R  32 mm

S ut  600  MPa


See Figures 6-6 and Mathcad file P0606. W tongue 70 = c

1

F pull

1

40 = a 2

A B

A

19 = t B

F b1

31 = b

F a1x

C

F a1y 20 = d

F a2y

D

Fa2x 2

Fc2x

F b2 C D

Fd2 F c2y


1.

The dynamic loading in this problem is fully reversed so the mean stresses are zero and the alternating stresses are the same as those calculated in Problem 4-6. From Problem 4-6, the alternating components of the principal stresses in the shank of the ball where it joins the ball bracket are:

σa1  1277 MPa 2.

σa2  0  MPa

Since 1 is the only nonzero principal stress, it is also the von Mises stress.

σa3  0  MPa



σ'a  σa1

6-6-2

σ'a  1277 MPa S'e  0.5 S ut

S'e  300 MPa

3.


4.


Cload  1

Size

A95  0.010462 d sh d equiv 

(bending load) 2

A95

A  4.51

Csurf

5.

 0.097

Csize  0.955

b  0.265

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


2


0.0766

 d equiv  Csize  1.189     mm  Surface

A95  7.072 mm

(machined)

b

Csurf  0.828

(R = 99.9%)


S e  178.54 MPa Na 

Se

Na  0.14

6.

Calculate the factor of safety for the ball shank.

7.

From Problem 4-6, the alternating components of the principal stresses at the bearing area in the ball bracket ho are:

σa1  111.5  MPa 8.

σa2  0  MPa

σa3  0  MPa

Since 1 is the only nonzero principal stress, it is also the von Mises stress.

σ'a  σa1 9.

σ'a

σ'a  111.5 MPa

Calculate the endurance limit modification factors that are different from those in step 4. Load

Cload  0.7

(axial load)

Size

Csize  1

(axial load)

10. Calculate the modified endurance limit. S e  Cload  Csize Csurf  Ctemp Creliab S'e

11. Calculate the factor of safety for the bearing.

S e  130.91 MPa Nb 

Se

σ'a

Nb  1.17



6-6-3

12. From Problem 4-6, the alternating components of the von Mises stress at the tearout shear area in the ball bracket hole is:

σ'a  85.91  MPa 13. Calculate the endurance limit modification factors that are different from those in step 9. Load

Cload  1

Size

A95  2  t ( 32 mm)   0.5 d sh

(simulated bending load) 2

d equiv 

2

A95  1111 mm

A95

2


0.0766

 d equiv    mm 

 0.097

Csize  1.189  

Csize  0.747


S e  139.71 MPa Nc 

15. Calculate the factor of safety against tearout.

Se

Nc  1.6

σ'a

16. From Problem 4-6, the alternating components of the von Mises stress in the attachment bolts is:

σ'a  540.5  MPa 17. Calculate the endurance limit modification factors that are different from those in step 9. Load

Cload  0.7

(axial load)

Size

Csize  1

(axial load)


19. Calculate the factor of safety against bolt tensile failure.

S e  130.91 MPa Nd 

Se

Nd  0.24

σ'a

20. From Problem 4-6, the alternating components of the principal stresses in the cantilever beam are:

σa1  635.5  MPa

σa2  0  MPa

σa3  0  MPa

21. Since 1 is the only nonzero principal stress, it is also the von Mises stress.

σ'a  σa1

σ'a  635.5 MPa

22. Calculate the endurance limit modification factors that are different from those in step 17. Load

Cload  1

Size

A95  0.05 w t

(bending load) A95  60.8 mm

2



d equiv 

6-6-4

A95


0.0766

 d equiv  Csize  1.189     mm 

 0.097

Csize  0.86


24. Calculate the factor of safety for the cantilever beam.

S e  160.85 MPa Ne 

Se

σ'a

Ne  0.25



6-7-1


Design the wrist pin of Problem 3-7 for infinite life with a safety factor of 1.5 if the 2500-g acceleration is fully reveresed and S ut = 130 ksi.

Given:

Force on wrist pin


Tensile strength

S ut  130  ksi


Nd  1.5 od  0.375  in


Fwristpin  2756 lbf

See Figure 4-12 in the text and Mathcad file P0607. Fwristpin

F 

F  1378 lbf

1.

The force at each shear plane is

2.

With only the direct shear acting on the plane, the Mohr diagram will be a circle with center at the origin and radius equal to the shear stress. Thus, the principal normal stress is numerically equal to the shear stress, which in this case is also the principal shear stress, so we have  = 1 = '.

3.

The shear stress at each shear plane is

τ=

2

F A

=



4 F 2

= σ'

 2 2 π  od  id   S 2

π od  id Se

For fully reversed loading the factor of safety is, Nd =

5.


6.

Calculate the endurance limit modification factors for a nonrotating round pin (uniformly stressed). Load

Cload  1

Size

A95 ( id) 

=

e

4.

σ'

S'e  0.5 S ut



2



7.

d equiv( id) 

4

A  1.34

Csurf

S'e  65 ksi

2

π od  id

 d equiv( id)  Csize( id)  0.869     in  Surface

4 F

Temperature

Ctemp  1

Reliability


0.0766

 0.097

b  0.085

 S ut   A     ksi 

A95 ( id)

(ground)

b

Csurf  0.886

(R = 50%)

Calculate the modified endurance limit. S e( id)  Cload  Csize( id)  Csurf  Ctemp Creliab S'e

8.

Solving for the inside diameter, guess id  0.2 in Given Nd =





2

2

π od  id  S e( id) 4 F



id  Find ( id) 9.

6-7-2

id  0.299 in

Round this down to the decimal equivalent of a common fraction (9/32),

id  0.281  in

10. The realized factor of safety is,

Nf 





2

2

π od  id  S e( id) 4 F

Nf  1.8



6-8-1


Given:

A paper machine processes rolls of paper having a density of 984 kg/m3. The paper roll is 1.50-m OD x 0.22-m ID x 3.23 m long and is on a simply supported, hollow, steel shaft with S ut = 400 MPa Find the shaft ID needed to obtain a dynamic safety factor of 2 for a 10-year life if the shaft OD is 22 cm and the roll turns at 50 rpm. y

Paper roll:

ρ  984 

Density Outside dia. Inside dia. Length Shaft: Strength

kg

w

3

m OD  1500 mm ID  220  mm L  3230 mm

x L

R

R

V

S ut  400  MPa

R

Outside dia. Design safety factor

od  220  mm Nfd  2

0

Design life Shaft speed

Life  10 yr ω  50 rpm

L/2

L x -R

M

2

wL /8

Assumptions: 1. The shaft is stiffer than the paper roll so the weight of the roll on the shaft can be modelled as a uniformly distributed load. 2. The bearings that support the shaft are close to x 0 the ends of the paper roll and are thin with respect L/2 L to the length of the roll so we can consider the distance between the shaft supports to be the FIGURE 6-8 same as the length of the roll. Load, Shear, and Moment Diagrams 3. There are no stress concentrations near the for Problem 6-8 point of maximum moment on the shaft. 4. The paper mill operates 3 shifts/day, 365 days/year. 5. The shaft is machined and the material reliabilty is 99.9%. See Figure 6-8 and Mathcad file P0608. Solution: 1.

This is a case of fully reversed bending. The FBD for this loading case is shown in Appendix B, Figure B-2b, with the dimension a equal to 0. That is, the distributed load starts at the left support and ends at the right support.

2.

Calculate the number of stress cycles to see if we will design for finite or infinite life. Nlife  Life ω

9

Nlife  1.652  10

cycles

This is well beyond 10 6 cycles, so we will design for infinite life. 3.

4.

Determine the weight of the paper roll and the magnitude of the distributed load on the shaft.

 4

π

Roll volume

V 

Roll weight

W  V  ρ  g

Distributed load on shaft

w 

2

2



 OD  ID  L

9

V  5.585  10  mm

3

W  53.895 kN

W

w  16.686

L

N

(a

mm

Figure D-2b shows that the maximum bending moment occurs at the center of the shaft and is 2

Mmax 

w L 8

7

Mmax  2.176  10  N  mm


MACHINE DESIGN - An Integrated Approach, 4th Ed. Ma  Mmax

This is fully reversed bending so 5.

Mm  0  N  mm

and

The stress in the shaft at the point of maximum bending moment will depend upon the, as yet, unknown id. Tha is, I ( id) 

Area moment of inertia

π 64

σa( id) 

Alternating stress 6.

6-8-2



4



4

 od  id

(b)

Ma od (c)

2  I ( id)

Calculate the modified endurance strength of the shaft. S'e  0.5 S ut

Unmodified endurance limit

S'e  200  MPa

Modification factors: Load

Cload  1

Size

Csize  1.189  

   mm  od

 0.097

 Sut   4.51    MPa 

Surface

Csurf

Temperature

Ctemp  1

Reliability


Csize  0.705  0.265

Csurf  0.922

Modified endurance limit S e  Cload  Csize Csurf  Ctemp Creliab S'e 7.

S e  97.8 MPa

(d)

Use the factor of safety equation as a design equation to solve for the unknown id. For fully reversed bending, the factor of safety is Nf =

Se

(e)

σa

Substituting equations b and c into e and solving for id, 1 4  4 32 Nfd  Ma od  id   od   π S e  

Rounding this, let the shaft ID be

id  190  mm

This gives a wall thickness of

t 

1 2

 ( od  id)

id  191.526  mm

t  15 mm



6-9-1


For the Vise Grip plier-wrench is drawn to scale in Figure P6-3, and for which the forces were analyzed in Problem 3-9 and the stresses in Problem 4-9, find the safety factors for each pin for an assumed clamping force of P = 4000 N in the position shown. The pins are 8-mm dia, S y = 400 MP S ut = 520 MPa, and are all in double shear. Assume a desired finite life of 5E4 cycles.

Given:

Pin stresses as calculated in Problem 4-9: Pin 1-2 τ12  74.6 MPa

S y  400  MPa

Yield strength

Pin 1-4

τ14  50.7 MPa

Tensile strength S ut  520  MPa

Pin 2-3

τ23  50.7 MPa

Pin diameter

d  8  mm

Pin 3-4

τ34  50.7 MPa

Desired life

Nlife  5  10

4

Assumptions: 1. Links 3 and 4 are in a toggle position, i.e., the pin that joins links 3 and 4 is in line with the pins that join 1 with 4 and 2 with 3. Solution: 1.



F

P 1

2

3

P

F 55.0 = b

50.0 = a

39.5 = c

F

F14 22.0 = d

129.2°

1



4 F34

F41

F21

P



28.0 = e  

F43



F12

3 F23

F32

P

2.8 = g

21.2 = h

2

F 26.9 = f

FIGURE 6-9 Free Body Diagrams for Problem 6-9 2.

The pins are in pure shear, so the principal stresses are



3.

6-9-2

Pin joining 1 and 2

σ'12 

3  τ12

σ'12  129.211 MPa

All other pins

σ'14 

3  τ14

σ'14  87.815 MPa

This is a case of repeated fatigue loading. The alternating and mean von Mises stress components are: Pin joining 1 and 2

σ'12a  0.5 σ'12

σ'12m  σ'12a

All other pins

σ'14a  0.5 σ'14

σ'14m  σ'14a

S'e  0.5 S ut

S'e  260 MPa

4.


5.

Calculate the endurance limit modification factors for a non rotating round pin (uniformly stressed). Load

Cload  1

Size

A95 

π d

2

A95  50.265 mm

4

d equiv 

A95


0.0766

 d equiv  Csize  1.189     mm  A  4.51

Surface

 Sut   A     MPa 

Csurf

6.

Temperature

Ctemp  1

Reliability


2

 0.097

Csize  0.868

b  0.265

(machined)

b

Csurf  0.86

(R = 50%)


S e  194.07 MPa S m  0.9 S ut

S m  468 MPa

7.


8.


9.


a 

1 z

 Sm    S'e 

 log Sm

10  3

b

Sf = a N

b

b  0.0851

a  842.4 MPa



6-9-3

4

10. Calculate the corrected fatigue strength at Nlife  5  10 cycles. S f  a  Nlife

b

S f  335.49 MPa

11. Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety. Pin joining 1 and 2

Nf 

All other pins

Nf 

S f  S ut

σ'12a S ut  σ'12m S f S f  S ut

σ'14a S ut  σ'14m S f

Nf  3.2

Nf  4.6



6-10-1


An overhung diving board is shown in Figure P6-4a. A 100-kg person is standing on the free end. Assume cross-sectional dimensions of 305 mm x 32 mm. What is the fatigue safety factor for infinite life if the material is brittle fiberglass with S f = 39 MPa @ N = 5E8 cycles and S ut = 130 MP in the longitudinal direction? 2000 = L

Given:

Mass of person Board dimensions Load dimensions Material properties

M  100  kg w  305  mm t  32 mm b  700  mm L  2000 mm S ut  130  MPa

R1

P

R2

S f5E8  39 MPa

700 = b


Assumptions: 1. The given fatigue strength is fully corrected. 2. There are no stress-concentrations near the point of maximum moment on the diving board. Solution:


1.

This is a case of repeated bending. The FBD for this loading case is shown in Appendix B, Figure B-3a, with th dimension a equal to L. That is, the concentrated force F is at the end of the overhung beam.

2.

Determine the weight of the person on the end of the board. W  M  g

Weight 3.

W  980.7  N

(a)

Figure B-3a in Appendix B shows that the maximum bending moment occurs at the right-hand support and is 6

Mmax  W  ( L  b )

Mmax  1.275  10  N  mm

This is repeated bending so Ma  4.

Mmax

The stress in the board at the point of maximum bending moment is Area moment of inertia

5.

Mm  Ma

and

2

I 

w t

Alternating stress

σa 

Mean stress

σm 

3

12 Ma t 2 I Mm t 2 I

5

I  8.329  10  mm

4

(b)

σa  12.2 MPa

(c)

σm  12.2 MPa

(d)

For repeated (fluctuating) bending, the factor of safety for Case 3 loading is Nf 

S f5E8 S ut

σa S ut  σm S f5E8

Nf  2.4

(e)



6-11-1


Repeat Problem 6-10 assuming the 100-kg person in Problem 6-10 jumps up 25 cm and lands back on the board. Assume the board weighs 29 kg and deflects 13.1 cm statically when the person stands on it. What is the fatigue safety factor for finite life if the material is brittle fiberglass with S = 39 MPa @ N = 5E8 cycles and S ut = 130 MPa in the longitudinal direction?

Given:

Maximum principal stresses due to bending at R2 from Problem 4-11 σ1max  76.3 MPa

2000 = L R1

P

σ2max  0  MPa σ3max  0  MPa Ultimate strength

S ut  130  MPa

Fatigue strength

S f  39 MPa

Fatigue life

Ncycle  5  10

R2

8

700 = b

FIGURE 6-11 Solution: 1.



The dynamic loading in this case is repeated, i.e., the stresses go from zero to the maximum values given above. Thus, the minimum and maximum von Mises stresses are:

σ'max 

2


2

σ'max  76.3 MPa 2.


σ'a  σ'm  3.


σ'max  σ'min 2

σ'max  σ'min 2

σ'a  38.1 MPa σ'm  38.1 MPa

Assuming a Case 3 load line, use equation (6.18e) to calculate the factor of safety.

Nf 

S f  S ut

σ'a S ut  σ'm S f

Nf  0.79



6-12-1


Repeat Problem 6-10 using the cantilevered diving board design in Figure P6-4b.

Given:

Mass of person Board dimensions

2000

Load dimensions Material properties

M  100  kg w  305  mm t  32 mm L  1300 mm S ut  130  MPa

1300 = L P

M1

S f5E8  39 MPa

R1

Assumptions: 1. The given fatigue strength is fully corrected. 700 2. There are no stress-concentrations near the point of maximum moment on the diving board. FIGURE 6-12 Free Body Diagram for Problem 6-12

Solution:


1.

This is a case of repeated bending. The FBD for this loading case is shown in Appendix B, Figure B-1a, with the dimension a equal to L. That is, the concentrated force F is at the end of the cantilever beam.

2.

Determine the weight of the person on the end of the board. W  M  g

Weight 3.

W  980.7  N

(a)

Figure B-1a in Appendix B shows that the maximum bending moment occurs at the support and is 6

Mmax  W  L

Mmax  1.275  10  N  mm

This is repeated bending so Ma  4.

Mmax

The stress in the board at the point of maximum bending moment is Area moment of inertia

5.

Mm  Ma

and

2

I 

w t

Alternating stress

σa 

Mean stress

σm 

3

12 Ma t 2 I Mm t 2 I

5

I  8.329  10  mm

4

(b)

σa  12.2 MPa

(c)

σm  12.2 MPa

(d)

For repeated (fluctuating) bending, the factor of safety for Case 3 loading is Nf 

S f5E8 S ut

σa S ut  σm S f5E8

Nf  2.4

(e)



6-13-1


Repeat Problem 6-11 using the cantilevered diving board design in Figure P6-4b. Assume the board weighs 19 kg and deflects 8.5 cm statically when the person stands on it.

Given:

Maximum principal stresses due to bending at support from Problem 4-13 σ1max  87.1 MPa

2000 1300 = L P

σ2max  0  MPa σ3max  0  MPa Ultimate strength

S ut  130  MPa

Fatigue strength

S f  39 MPa

Fatigue life

Ncycle  5  10

M1

8

R1

700

FIGURE 6-13 Solution: 1.



The dynamic loading in this case is repeated, i.e., the stresses go from zero to the maximum values given above. Thus, the minimum and maximum von Mises stresses are:

σ'max 

2


2

σ'max  87.1 MPa 2.


σ'a  σ'm  3.


σ'max  σ'min 2

σ'max  σ'min 2

σ'a  43.5 MPa σ'm  43.5 MPa

Assuming a Case 3 load line, use equation (6.18e) to calculate the factor of safety. Nf 

S f  S ut


Nf  0.69



6-14-1


Figure P6-5 shows a child's toy called a pogo stick. The child stands on the pads, applying half her weight on each side. She jumps off the ground, holding the pads up against her feet, and bounces along with the spring cushioning the impact and storing energy to help each rebound. Design the aluminum cantilever beam sections on which she stands to survive jumping 2 in off the ground with a dynamic safety factor of 2 for a finite life of 5E4 cycles. Use 2000 series aluminum. Define and size the beam shape.

Given:

Heat treated 2024 aluminum: Tensile strength S ut  64 ksi Design safety factor

Nfd  2

Design life

Nlife  5  10

4

Assumptions: The beam will have a rectangular crosssection with the load applied at a distance of 5 in from the central support. L  5  in Solution: 1.


Fi /2

From Problem 3-14, the total dynamic force on both foot supports is Fimax  224  lbf

Fi /2

Fimin  0  lbf

Therefore, the load on each support is Pmax  Pmin 

Fimax

Pmax  112  lbf

2 Fimin

Pmin  0  lbf

2

2.

To give adequate support to the childs foot, let the width of the support beam be w  1.5 in

3.

From Figure B-1(a) in Appendix B, the maximum bending moment at x = 0 is Mmax  Pmax L

4.

P FIGURE 5-14 Free Body Diagram for Problem 5-14

Mmax  560  in lbf

Mmin  0  in lbf

Calculate the alternating and mean components of the bending moment. Ma  Mm 

Mmax  Mmin

Ma  280  in lbf

2 Mmax  Mmin

Mm  280  in lbf

2

S'e  19 ksi @ 5E8 cycles

5.

Determine the unmodified endurance limit.

6.

Calculate the endurance limit modification factors for a nonrotating rectangular beam. Load

Cload  1

Size

A95 ( t)  0.05 w t

d equiv( t) 

A95 ( t) 0.0766



6-14-2

 d equiv( t)  Csize( t)  0.869     in  A  2.7

Surface

Csurf

7.

 0.097

b  0.265

 S ut   A     ksi 

Temperature

Ctemp  1

Reliability


(machined)

b

Csurf  0.897

(R = 50%)

Calculate the modified endurance limit. S e( t)  Cload  Csize( t)  Csurf  Ctemp Creliab S'e S m  0.9 S ut

8.


9.


S m  57.6 ksi

Sf = a N

b

10. Determine the constants a and b from equations (6.10c) and (6.10a). From Table 6-5, for N = 5E8 , z  5.699 1

b ( t) 

z

 Sm    Se( t) 

 log

a ( t) 

Sm

103

b( t )

4

11. Determine the corrected fatigue strength at Nlife  5  10 cycles.

S f ( t)  a ( t)  Nlife

b( t )

12. We can now determine the minimum required section depth, t. Using the distortion-energy failure theory with the modified Goodman diagram, the bending stress will also be the only nonzero principal stress, which will also be the von Mises stress. Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety. Guess t  10 mm. Bending stress

σ=

M c I

Given Nfd =

t 12 6 M = M  = 2 3 2 w t w t

w t 6

2



S f ( t)  S ut Ma S ut  Mm S f ( t)

t  Find ( t)

t  0.304  in


t  0.375  in

Using this value of t, the values of the functions of t are: Csize( t)  0.912

S e( t)  15.545 ksi

S f ( t)  38.981 ksi

The realized safety factor is Nf 

w t 6

2



S f ( t)  S ut Ma S ut  Mm S f ( t)

Nf  3.0



6-15a-1


For a notched part having a notch dimension r, geometric stress concentration factor Kt, and material strength S ut as shown in row a of Table P6-4, find the Neuber factor a, the material's notch sensitivity q, and the fatigue stress-concentration factor Kf.

Given:

Ultimate tensile strength

S ut  100  ksi

Geometric stress-concentration factor

Kt  3.3

Notch radius

r  0.25 in

Material is steel

Loading  "Bending" Solution:

See Mathcad file P0615a. 1 2

1.

From Table 6-6, the Neuber constant for S ut  100 ksi is

a  0.062  in

2.

Using equation 6-13, the notch sensitivity is

q 

1 1

3.

a  0.062 in

2

q  0.890 a r

The fatigue stress-concentration factor, from equation 6.11b, is Kf  1  q   Kt  1 

Kf  3.05



6-16-1


A track to guide bowling balls is designed with two round rods as shown in Figure P6-6. The rods are not parallel to one another but have a small angle between them. The balls roll on the rods until they fall between them and drop onto another track. The angle between the rods is varied to cause the ball to drop at different locations. Find the infinite-life safety factor for the 1-in dia SAE 1010 cold-rolled steel rods. (a) Assume rods are simply supported at each end. (b) Assume rods are fixed at each end.

Given:

Tensile strength


Rod diameter

d  1.00 in

Solution:

Fball

a

See Figure 6-16 and Mathcad file P0616. R1

1.

The maximum bending stress will occur at the outer fibers of the rod at the section where the maximum bending moment occurs which, in this case, is at x = a. The only stress present on the top or bottom surface of the rod is the bending stress x. Therefore, on the bottom surface where the stress is tensile, sx is the principal stress 1 . Thus, from Problem 4-16, for a simply supported rod, Maximum principal stress

2.

R2

L


σ1  748  psi

The dynamic loading is repeated from 0 to 1 for each ball that rolls down the track. The alternating and mean components of the von Mises stress are: Alternating von Mises stress

σ'a  0.5 σ1

σ'a  374 psi

Mean von Mises stress

σ'm  0.5 σ1

σ'm  374 psi

S'e  0.5 S ut

S'e  26.5 ksi

3.


4.


Cload  1

Size

A95  0.010462 d

2

d equiv 

 d equiv  Csize  0.869     in  Surface

A  2.70

Csurf

 S ut   A     ksi 

Temperature

Ctemp  1

Reliability


A95 0.0766

 0.097

b  0.265

Csize  0.957

(machined)

b

Csurf  0.943

(R = 99.999%)



6-16-2


6.

Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety. Case a

7.

Nfa 

S e S ut

9.

Nfa  32


For the built-in case, the maximum bending stress will occur at the outer fibers of the rod at the section where the maximum bending moment occurs which, in this case, is at x = L. The only stress present on the top or bottom surface of the rod is the bending stress x. Therefore, on the bottom surface where the stress is tensile, sx is the principal stress 1 . Thus, from Problem 4-16, for a simply supported rod,

Fball

a

M1

R1

L

R 2 M2


σ1  577  psi


S e  15.759 ksi

The dynamic loading is repeated from 0 to 1 for each ball that rolls down the track. The alternating and mean components of the von Mises stress are: Alternating von Mises stress

σ'a  0.5 σ1

σ'a  288.5 psi


σ'm  0.5 σ1

σ'm  288.5 psi

Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety. Case b

Nfb 

S e S ut


Nfb  42



6-17-1


A pair of ice tongs is shown in Figure P6-7. The ice weighs 50 lb and is 10 in wide across the tongs. The distance between the handles is 4 in, and the mean radius r of the tong is 6 in. The rectangular cross-sectional dimensions are 0.75 x 0.312 in. Find the safety factor for the tongs for 5E5 cycles if their S ut = 50 ksi. F

Given:

Tensile strength


Cross-section: Width Depth

w  0.312  in h  0.75 in

Life

Nf  5  10

C FC O

5

11.0 = ax

3.5 = cy

FO 2.0 = cx

A

12.0 = by

Assumptions: The tongs are forged. Use 99.99% reliability. Operating temperature is between 32F and 70F.

5.0 = bx

FB Solution:

See Problem 4-17, Figure 6-17, and Mathcad file P0617.

B

1. The maximum bending stress in the tong was found in Problem 4-17 at point A. Vertical direction

W/2 FIGURE 6-17 Free Body Diagram for Problem 6-17

σi  8.58 ksi

All other components are zero. 2.

There are no other stress components present so

σ1max  σi 3.

σ1max  8.58 ksi

σ3max  0  ksi

The dynamic loading in this case is repeated, thus

σ1min  0  ksi 4.


σ2min  0  ksi


Even though this is a brittle material, for HCF analysis, determine the von Mises effective stresses. Since there i only one nonzero stress,

σ'max  σ1max

σ'max  8.58 ksi

σ'min  σ1min

σ'min  0 ksi

σ'a  σ'm 

σ'max  σ'min

σ'a  4.29 ksi

2

σ'max  σ'min

σ'm  4.29 ksi

2

S'e  0.5 S ut

S'e  25 ksi

5.


6.


Cload  1

Size

A95  0.05 w h d equiv 

A95 0.0766

A95  7.548 mm

2




6-17-2

 d equiv  Csize  1.189     mm  A  39.9

Surface

 S ut   A     ksi 

Csurf

7.

Temperature

Ctemp  1

Reliability


 0.097

Csize  0.952

b  0.995

(forged)

b

Csurf  0.814

(R = 99.99%)


S e  13.59 ksi = 10 3

S m  0.9 S ut

8.

Using equation (6.9), calculate the fatigue strength at N

cycles.

9.


Sf = a N

S m  45 ksi b

10. Determine the constants a and b from equations (6.10c) and (6.10a). From Table 6-5, for N = 10 6 , z  3.000 b 

a 

1 z

 Sm    Se 

 log Sm

10  3

b

11. Using equation (6.10a), determine the fatigue strength.

b  0.1733

a  148.991 ksi

S f5E5  a  Nf

b

S f5E5  15.326 ksi

12. Assuming a Case 3 load line, use equation (6.18e) to calculate the factor of safety.

Nf5E5 

S f5E5 S ut

σ'a S ut  σ'm S f5E5

Nf5E5  2.7



6-18-1


A pair of ice tongs is shown in Figure P6-7. The ice weighs 50 lb and is 10 in wide across the tongs. The distance between the handles is 4 in, and the mean radius r of the tong is 6 in. The rectangular cross-sectional dimensions are 0.75 x 0.312 in. Find the safety factor for the tongs for 5E5 cycles if they are made of Class 40 gray cast iron.

Given:

Tensile strength


Cross-section: Width Depth

w  0.312  in h  0.75 in Nf  5  10

Life

F C FC 3.5 = cy

O

5

FO

11.0 = ax

Assumptions: The tongs are as-cast. Use 99.99% reliability. Operating temperature is between 32F and 70F. Set Csurf to 1 for a cast finish, which does not need a surface factor. See Problem 4-18, Figure 6-18, and Solution: Mathcad file P0618.

2.0 = cx

A

12.0 = by 5.0 = bx

FB B W/2

1.

The maximum bending stress in the tong was found in Problem 4-17 at point A.

σi  8.58 ksi

Vertical direction


All other components are zero. 2.

There are no other stress components present so

σ1max  σi 3.

σ1max  8.58 ksi


The dynamic loading in this case is repeated, thus

σ1min  0  ksi 4.




Even though this is a brittle material, for HCF analysis, determine the von Mises effective stresses. Since there is only one nonzero stress,

σ'max  σ1max

σ'max  8.58 ksi

σ'min  σ1min

σ'min  0 ksi

σ'a  σ'm 

σ'max  σ'min

σ'a  4.29 ksi

2

σ'max  σ'min

σ'm  4.29 ksi

2

S'e  0.4 S ut

S'e  16.8 ksi

5.


6.


Cload  1

Size

A95  0.05 w h

A95  7.548 mm

2



d equiv 

A95


0.0766

 d equiv  Csize  1.189     mm 

7.

6-18-2

Surface

Csurf  1

Temperature

Ctemp  1

Reliability


 0.097

Csize  0.952

(R = 99.99%)


S e  11.22 ksi S m  0.9 S ut

8.


9.


Sf = a N

S m  37.8 ksi b


a 

1 z

 Sm    Se 

 log Sm

10  3

b

11. Using equation (6.10a), determine the fatigue strength.

b  0.1758

a  127.305 ksi

S f5E5  a  Nf

b

S f5E5  12.678 ksi

12. Assuming a Case 3 load line, use equation (6.18e) to calculate the factor of safety.

Nf5E5 

S f5E5 S ut

σ'a S ut  σ'm S f5E5

Nf5E5  2.3



6-19-1


Determine the size of the clevis pin shown in Figure P6-8 needed to withstand an applied repeated force of 0 to 130000 lb for infinite life. Also determine the required outside radius of the clevis end to not fail in either tearout or bearing if the clevis flanges are each 2.5 in thick. Use a safety factor of 3. Assume S ut = 140 ksi for the pin and S ut = 80 ksi for the clevis.

Given:

Minimum force

Pmin  0  lbf

Maximum force

Pmax  130  kip

Pin

S utp  140  ksi

Flange thickness

t  2.5 in

Clevis

S utc  80 ksi

Material strength:

Nf  3

Factor of safety against fatigue failure

Assumptions: The parts are machined. Use 90% reliability and room temperature. Solution: 1.


Calculate the alternating and mean components of the forces on the clevis and link. Pa  Pm 

Pmax  Pmin

Pa  65 kip

2 Pmax  Pmin

Pm  65 kip

2

Stress in Pin 2.

The pin is in double shear and there is no stress-concentration. The alternating and mean loads at one section on the pin are

τa =

1 Pa  2 Apin

τm =

Apin ( d ) 

1 Pm  2 Apin π d

2

(2)

3.

The cross-section area of the pin is

4.

The alternating and mean shear stresses and von Mises stresses are

τa( d ) 

σ'a( d ) 

Pa

4

Pm

τm( d ) 

2  Apin ( d ) 3  τa( d )

(1)

(3)

2  Apin ( d )

σ'm( d ) 

3  τm( d )

(4)

Pin Strength S'ep  0.5 S utp

S'ep  70 ksi

5.


6.

Calculate the endurance limit modification factors for a nonrotating round pin in bending. Load

Cload  1

Size

A95 ( d )  0.010462 d

2

 dequiv( d)    in 

d equiv( d ) 

A95 ( d ) 0.0766

 0.097

Csize( d )  0.869   Surface

A  2.70

b  0.265

(machined)



 S utp   A     ksi 

Csurf

7.

Temperature

Ctemp  1

Reliability


6-19-2

b

Csurf  0.729

(R = 90%)

Calculate the modified endurance limit. S ep( d )  Cload  Csize( d )  Csurf  Ctemp Creliab S'ep

(5)

Design Equation 8.

Using the modified-Goodman failure criterion and a case 3 load line, the factor of safety is given by equation 6-18e as Nf =

9.

S e S ut

(6)


Substituting equations 4 and 5 into 6 and solving for d yields Given Nf =

Guess

d  2.0 in

S ep( d )  S utp

σ'a( d )  S utp  σ'm( d )  S ep( d )

d  Find ( d )

d  2.632 in d  2.750  in

Rounding to the next higher eighth of an inch, let With this value of d, we have

σ'a( d )  9.5 ksi

σ'm( d )  9.5 ksi

S ep( d )  39.71 ksi

and the realized factor of safety against fatigue failure in the pin is Nf 

S ep( d )  S utp

Nf  3.3

σ'a( d )  S utp  σ'm( d )  S ep( d )

Tearout length

Clevis Tearout (See Figure 6-19) 10. Let the outside radius of the clevis be R. Then the tearout area is 2

Atear ( R)  2  t  R  ( 0.5 d )

2

11. The alternating and mean shear stresses and von Mises stresses are

τa( R) 

Pa 2  Atear ( R)

τm( R) 

Pm 2  Atear ( R)

d

(7)

R

FIGURE 6-19 Tearout Diagram for Problem 6-19



σ'a( R) 

3  τa( R)

σ'm( R) 

6-19-3

3  τm( R)

(8)

Clevis Strength 12. Calculate the unmodified endurance limit.

S'ec  0.5 S utc

S'ec  40 ksi

13. Calculate the endurance limit modification factors for a nonrotating rectangular shear area (uniformly stressed). Load

Cload  1

Size

A95 ( R)  Atear ( R)

d equiv( R) 

 d equiv( R)  Csize( R)  0.869     in  A  2.70

Surface

Csurf Temperature

Ctemp  1

Reliability


0.0766

 0.097

b  0.265

 S utc   A     ksi 

A95 ( R)

(machined)

b

Csurf  0.845

(R = 90%)

14. Calculate the modified endurance limit. S ec( R)  Cload  Csize( R)  Csurf  Ctemp Creliab S'ec

(9)

Design Equation 15. Using the modified-Goodman failure criterion and a case 3 load line, the factor of safety is given by equation 6-18e as S e S ut (10) Nf = σ'a S ut  σ'm S e 16. Substituting equations 8 and 9 into 10 and solving for R yields Given Nf =

Guess

R  2  in

S ec( R)  S utc

σ'a( R)  S utc  σ'm( R)  S ec( R)

R  Find ( R)

R  2.624 in

R  2.625  in

Bearing Stress The maximum bearing stress in the hole in each flange is

σmaxbear 

 Pa  Pm 2 d t

σmaxbear  9.5 ksi

This is small compared to the ultimate strength of the clevis. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0619.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


6-20-1


A ±100 N-m torque is applied to a 1-m-long, solid round steel shaft. Design it to limit its angular deflection to 2 deg and select a steel alloy to have a fatigue safety factor of 2 for infinite life.

Given:

Applied torque

Ta  100  N  m

Shaft length Max deflection

L  1000 mm θmax  2  deg


Nfd  2

Modulus of rigidity

G  80.8 GPa

Tm  0  N  m

Assumptions: There are no stress-concentrations anywhere on the shaft. The shaft is machined, reliability is 99.9%, and the it is at room temperature. Solution: 1.


This is a case of fully reversed torsion. We will use the von Mises effective stress so the load factor will be 1. Tmax  Ta  Tm

The maximum torque is 2.

Tmax  100 N  m

The diameter of the shaft can be found from equations 4.24 and 4.25 with  = max.

θmax =

Tmax L J G

=

32 Tmax L 4

π d  G 1

3.

Solving for d,

 32 Tmax L  d     π θmax G 

Rounding, let

d  24.5 mm

4

d  24.514 mm

Now, we can solve for the stress in the shaft. Polar moment of inertia

J 

Torsional stress

τa 

π 32

d

4

Ta d 2 J

4

J  3.537  10 mm

4

τa  34.632 MPa

The corresponding von Mises normal stress is von Mises stress 4.

3  τa

σ'a  59.984 MPa

Using the factor of safety equation for reversed loading, calculate the required endurance limit Nf =

5.

σ'a 

Se

σ'a

S e  Nfd  σ'a

S e  119.967 MPa

This endurance limit is a function of the unknown ultimate tensile strength. Use the endurance limit modificati equation to determine the required S ut. S e = Cload  Csize Csurf  Ctemp Creliab S'e

6.

Calculate the endurance limit modification factors for a solid, round steel shaft.


MACHINE DESIGN - An Integrated Approach, 4th Ed. Load

Cload  1

Size

Csize  1.189  

Surface

A  4.51

   mm 

Csurf

7.

6-20-2

d

 0.097

Csize  0.872

b  0.265

 S ut  = A    MPa 

Temperature

Ctemp  1

Reliability


Uncorrected endurance strength

S'e = 0.5 S ut

(machined)

b

(R = 99.9%)

Substituting these into the equation above and solving for S ut, 1

Se   S ut     0.5 A  Csize Creliab MPa 

b 1

 MPa

S ut  395 MPa

Based on this requirement, choose AISI 1020 cold-rolled steel that will be machined to size. 8.

Check the actual factor of safety based on the material chosen. For this material, S ut  469  MPa

 Sut   A     MPa 

b

Surface factor

Csurf


S'e  0.5 S ut

Corrected endurance strength

S e  Cload  Csize Csurf  Ctemp Creliab S'e

Csurf  0.884 S'e  234.5 MPa

S e  136.0 MPa

Factor of safety

Nf 

Se

σ'a

Nf  2.3



6-21-1


Figure P6-9 shows an automobile wheel with two common styles of lug wrench being used to tighten the wheel nuts, a single-ended wrench in (a), and a double-ended wrench in (b). The distance between points A and B is 1 ft in both cases and the handle diameter is 0.625 in. How many cycles of tightening can be expected before a fatigue failure if the average tightening torque is 100 ft-lb and the material S ut = 60 ksi?

Given:


d AB  1  ft

Minimum torque Tmin  0  ft  lbf

Wrench diameter

d  0.625  in

Maximum torque Tmax  100  ft  lbf

Tensile strength


Assumptions: 1. The forces exerted by the user's hands lie in a plane through the wrench that is also parallel to the plane of the wheel. 2. The applied torque is perpendicular to the plane of the forces. 3. By virtue of 1 and 2 above, this is a planar problem that can be described in a 2D FBD. 4. The surface is as-forged, the reliability is 50%, and the wrench will not be used in extremely ho or cold environments. Solution: 1.


From examination of the FBDs, we see that, in both cases, the arms are in bending and the stub that holds the socket wrench is in pure torsion. The maximum bending stress in the arm will occur near the point where the arm transitions to the stub. The stress state at this transition is very complicated, but we can find the nominal bending stress there by treating the arm as a cantilever beam, fixed at the transition point. For both cases the torque in the stub is the same.

F


12" = dAB

Case (a) 2.

F

6"

The bending moment at the transition is M = F  d AB = T Ma 

T

Tmax  Tmin F

2

(b) Double-ended Wrench

Ma  600 in lbf FIGURE 6-21 Mm  Ma 3.


The alternating and mean components of the bending stress at this point are found from Moment of inertia I 

π d


64 c  0.5 d

Alternating stress

σxa 

Mean stress

σxm 

4

I  in

4

c  0.313 in

Ma c I Mm c I

σxa  25.033 ksi σxm  25.033 ksi



There are no other stress components present at this point, so x is the maximum principle stress here and

σ2  0  psi

σ1 = σx 5.

σ3  0  psi

Since there is only one nonzero principal stress, the von Mises stress is

σ'a  σxa 6.

6-21-2

σ' = σ1 = σx

σ'm  σxm

Assuming a Case 3 load line, use equation (6.18e) to solve for the fatigue strength at which the wrench will fail (safety factor of 1). Nf =

S f  S ut


=1

S f 

σ'a S ut

S f  42.954 ksi

S ut  σ'm

7.

Using equation (6-5a), calculate the uncorrected endurance limit. S'e  0.5 S ut

8.


Cload  1

Size

A95  0.010462 d

2

 d equiv  Csize  0.869     in  A  39.9

Surface

Csurf

9.

and

d equiv 

Temperature

Ctemp  1

Reliability


A95 0.0766

 0.097

Csize  1.002

b  0.995

 S ut   A     ksi 

S'e  30 ksi

(as forged)

b

Csurf  0.679

(R = 50%)


10. Using equation (6.9), calculate the fatigue strength at N = 10 3 cycles.

S e  20.398 ksi S m  0.9 S ut

11. The equation for the S-N curve in the HCF region is given by equation (6.10a):

Sf = a N

S m  54 ksi b


a 

1 z

 Sm    Se 

 log Sm

10  3

b

b  0.1409

a  142.955 ksi



6-21-3

1

13. Calculate the number of cycles to failure using equation (6.10a)

 Sf  Na    a

b 3

Na  5.1  10

Case (b)


F  d AB

M=

14. The bending moment at the transition is

σx =

M c I

2

=

T 2

Tc

=

2 I

16. The bending stress in the handle for case (b) is one half that of case (a). However, the torque in the stub is the same in both cases. The shear stress at any point on the outside surface of the stub is found from Polar moment of inertia

J  2  I


τxymax  τxymin 

Minimum shear stress

τa 

Alternating shear stress

τm 

Mean shear stress

4

J  0.0150 in Tmax c

τxymax  25.03 ksi

J Tmin c

τxymin  0 ksi

J

τxymax  τxymin

τa  12.52 ksi

2

τxymax  τxymin

τm  12.52 ksi

2


σ1a  τa

and

σ'a 

σ1a  12.5 ksi

2

σ2a  0  psi

2


σ3a  σ1a

σ'a  21.7 ksi

σ'm  σ'a

σ'm  21.7 ksi

18. Assuming a Case 3 load line, use equation (6.18e) to solve for the fatigue strength at which the wrench will fail (safety factor of 1). Nf =

S f  S ut


=1

S f 

σ'a S ut

S f  33.944 ksi

S ut  σ'm 1

19. Calculate the number of cycles to failure using equation (6.10a)

Nb 

 Sf    a

b 4

Nb  2.7  10



6-22-1


A roller-blade skate is shown in Figure P6-10. The polyurethane wheels are 72 mm dia and spaced on 104-mm centers. The skate-boot-foot combination weighs 2 kg. The effective "spring rate" of the person-skate subsystem is 6000 N/m. The axles are 10-mm-dia steel pins in double shear with S ut = 550 MPa. Find the fatigue safety factor for the pins when a 100-kg person lands a 0.5-m jump on one foot assuming infinite life. (a) Assume all 4 wheels land simultaneously. (b) Assume that one wheel absorbs all the landing force.

Given:

Axle pin diameter

d  10 mm


Assumptions: Pins are machined and reliability is 99.999%. Solution:


τa  5.71 MPa

τb  22.9 MPa

1.

From Problem 4-22, we have the stresses for cases (a) and (b):

2.

The dynamic loading in this case is repeated so the stresses given in step 1 are the maximum and the minimum stresses are zero. Determine the minimum, maximum, alternating, and mean von Mises stresses. Part (a)

σ'maxa  σ'aa  σ'ma 

Part (b)

σ'mb 

σ'maxa  9.89 MPa

σ'maxa  σ'mina

σ'mina  0  MPa

σ'aa  4.945 MPa

2

σ'maxa  σ'mina

σ'ma  4.945 MPa

2

σ'maxb  σ'ab 

3  τa

3  τb

σ'maxb  39.664 MPa

σ'maxb  σ'minb

σ'minb  0  MPa

σ'ab  19.832 MPa

2

σ'maxb  σ'minb

σ'mb  19.832 MPa

2 S'e  0.5 S ut

S'e  275 MPa

3.


4.

Calculate the endurance limit modification factors for a nonrotating round pin. Load

Cload  1

Size

A95 

π d

2

A95  78.54 mm

4

d equiv 

A95


0.0766


A  4.51

Csurf

 Sut   A     MPa 

2

 0.097

b  0.265

Csize  0.849

(machined)

b

Csurf  0.847



5.

Temperature

Ctemp  1

Reliability


6-22-2

(R = 99.999%)


6.

S e  130.42 MPa

Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety. Part (a)

Nfa 

Part (b)

Nfb 

S e S ut

σ'aa S ut  σ'ma S e S e S ut

σ'ab S ut  σ'mb S e

Nfa  21.3

Nfb  5.3



6-23a-1


The beam in Figure P6-11a is subjected to a sinusoidal force-time function with Fmax = F and Fmin = -F/2, where F and the beam's other data are given in row a of Table P6-5. Find the stress state in the beam due to this loading and choose a material specification that will give a safety factor of 3 for N = 5E8 cycles.

Given:

Beam length Distance to concen. load Concentrated load Moment of inertia 8

I  2.85 10

L

L  1  m b  0.6 m F  500  N

b F

4

m

Distance to extreme fiber c  2.00 10 Design safety factor Nd  3

2

m

8

Solution:

Cycle life Nf  5  10 See Figure 6-23 and Mathcad file P0623a.

R2

R1


1. The minimum, maximum, alternating, and mean components of the loads are: Fmax  F Fa  2.

Fmax  Fmin 2



Mm  Fm b   1 



Fa  375  N

Fm 

σ'm 

b



Fmin  250  N

2

Fmax  Fmin 2

Fm  125  N

Ma  90 N  m

L b



Mm  30 N  m

L

M a c

σ'a  63.158 MPa

I Mm c

σ'm  21.053 MPa

I

Calculate the beam cross-section dimensions from I and c. Beam depth

h  2  c

Beam width

w 

12 I h

5.

F

Calculate the alternating and mean components of the maximum bending stress in the beam using equation (4.11b). These are principal stresses and also von Mises stresses.

σ'a 

4.

Fmin 

Calculate the alternating and mean components of the maximum bending moment on the beam using the equati in Figure B-2(a) in Appendix B. Ma  Fa b   1 

3.

Fmax  500  N

3

h  40 mm w  5.344  mm


Cload  1



A95  0.05 w h

Size

d equiv 

Surface

Csurf  1

Temperature

Ctemp  1

Reliability

Creliab  1

A95  10.688 mm

A95

2

d equiv  11.812 mm

0.0766

 d equiv  Csize  1.189     mm 

6.

6-23a-2

 0.097

Csize  0.936

(R = 50%)

Determine the modified endurance limit as a function of the unknown endurance limit. S e S ut  Cload  Csize Csurf  Ctemp Creliab 0.5 S ut

7.

Assuming a Case 3 load line, use equation (6.18e) as the design equation.

Nd =

8.

S e S ut  S ut

σ'a S ut  σ'm S e S ut

Solve the equations in steps 6 and 7 simultaneously for the desired S ut. S ut 

Nd   2  σ'a  Cload  Csize Csurf  Ctemp Creliab σ'm Cload  Csize Csurf  Ctemp Creliab

S ut  468  MPa 9.

Choose AISI 1020, cold-rolled steel (see Appendix A, Table A-9).



6-24a-1


The beam in Figure P6-11b is subjected to a sinusoidal force-time function with Fmax = F and Fmin = F/2, where F and the beam's other data are given in row a of Table P6-5. Find the stress state in the beam due to this loading and choose a material specification that will give a safety factor of 1.5 for N = 5E8 cycles.

Given:

Beam length Concentrated load Moment of inertia

L  1  m F  500  N 8

I  2.85 10

4

m

Distance to extreme fiber c  2.00 10 Design safety factor Nd  1.5 Nf  5  10

Cycle life Solution:

L

F

2

m M1

8

R1


FIGURE 6-24 1. The minimum, maximum, alternating, and mean components of the loads are: Fmax  F Fa 

2.

3.

2

Fmax  500  N

Fmin 

Fa  125  N

Fm 

Ma  Fa L

Ma  125  N  m

Mm  Fm L

Mm  375  N  m

Fmin  250  N

2

Fmax  Fmin 2

Fm  375  N


σ'm 

M a c

σ'a  87.719 MPa

I Mm c

σ'm  263.158  MPa

I


h  2  c

Beam width

w 

12 I h

5.

F

Calculate the alternating and mean components of the maximum bending moment on the beam using the equati in Figure B-1(a) in Appendix B.

σ'a 

4.

Fmax  Fmin


3

h  40 mm w  5.344  mm


Cload  1

Size

A95  0.05 w h

A95  10.688 mm

2



d equiv 

A95

6.

Csurf  1

Temperature

Ctemp  1

Reliability

Creliab  1


0.0766


6-24a-2

 0.097

Csize  0.936

(R = 50%)

Determine the modified endurance limit as a function of the unknown endurance limit assuming the material is steel. S e S ut  Cload  Csize Csurf  Ctemp Creliab 0.5 S ut

7.


Nd =

8.





S ut  676  MPa 9.

Choose AISI 1060 hot-rolled steel (see Appendix A, Table C-9).



6-25a-1


The beam in Figure P6-11c is subjected to a sinusoidal force-time function with Fmax = F and Fmin = 0, where F and the beam's other data are given in row a of Table P6-5. Find the stress state in the beam due to this loading and choose a material specification that will give a safety factor of 2.5 for N = 5E8 cycles.

Given:

Beam length L  1  m Distance between supports b  0.6 m Concentrated load F  500  N Moment of inertia 8

I  2.85 10


Fmax  Fmin 2

Fmax  500  N

Fmin  0  N

Fa  250  N

Fm 

Ma  Fa ( b  L)

Ma  100  N  m

Mm  Fm ( b  L)

Mm  100  N  m

Fmin  0  N

Fmax  Fmin 2

Fm  250  N


σ'm 

M a c

σ'a  70.175 MPa

I Mm c

σ'm  70.175 MPa

I


h  2  c

Beam width

w 

12 I h

5.

8

R2

R1

Calculate the alternating and mean components of the maximum bending moment on the beam using the equati in Figure B-1(a) in Appendix B.

σ'a 

4.

m

The minimum, maximum, alternating, and mean components of the loads are:

Fa 

3.

2


Fmax  F

2.

F

4

Nf  5  10

Cycle life

1.

b

m

Distance to extreme fiber c  2.00 10 Design safety factor Nd  2.5

Solution:

L

3

h  40 mm w  5.344  mm


Cload  1



A95  0.05 w h

Size

d equiv 

Surface

Csurf  1

Temperature

Ctemp  1

Reliability

Creliab  1

A95  10.688 mm

A95

2


0.0766

 d equiv  Csize  1.189     mm 

6.

6-25a-2

 0.097

Csize  0.936

(R = 50%)

Determine the modified endurance limit as a function of the unknown endurance limit assuming the material is steel. S e S ut  Cload  Csize Csurf  Ctemp Creliab 0.5 S ut

7.


Nd =

8.





S ut  550  MPa 9.

Choose AISI 1035 cold-rolled steel (see Appendix A, Table A-9).



6-26a-1


Given:

The beam in Figure P6-11d is subjected to a sinusoidal force-time function with Fmax = F and Fmin = -F, where F and the beam's other data are given in row a of Table P6-5. Find the stress state in the beam due to this loading and choose a material specification that will give a safety factor of 6 for N = 5E8 cycles. Beam length L  1  m Distance to concentrated load a  0.4 m Distance to 2nd support b  0.6 m Concentrated load F  500  N Moment of inertia 8

I  2.85 10

Distance to extreme fiber c  2.00 10 Design safety factor Nfd  6

Solution: 1.

b a

F

4

m

Nlife  5  10

Cycle life

L

2

R2

R1

R3

m FIGURE 6-26A

8



To determine the stresses, we must first get the maximum bending moment. From inspection of Figure P6-26, write the load function equation q(x) = R1-1 - F-1 + R2-1 - R3-1

2.

Integrate this equation from - to x to obtain shear, V(x) V(x) = R10 - F0 + R20 - R30

3.

Integrate this equation from - to x to obtain moment, M(x) M(x) = R11 - F1 + R21 - R31

4.

Integrate the moment function, multiplying by 1/EI, to get the slope. (x) = [R12/2 - F2/2 + R22/2 + R32/2 + C3]/EI

5.

Integrate again to get the deflection. y(x) = [R13/6 - F3/6 + R23/6 + R33/6 + C3x + C4]/EI

6.

Evaluate R1, R2, R3, C3 and C4 At x = 0, x = b, and x = L; y = 0, therefore, C4 = 0. At x = L+, V = M = 0 R1  100  N

Guess

R2  100  N

2

R3  100  N

C3  5  N  m

Given R1 6 R1 6

b 

F

3

F

3

L 

6

6

3

3

 ( b  a )  C 3 b = 0  N  m

3

 ( L  a) 

R2 6

3

3

 ( L  b )  C3 L = 0  N  m

R1  F  R2  R3 = 0  N R1 L  F  ( L  a )  R2 ( L  b ) = 0  N  m © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


6-26a-2

 R1  R   2   Find  R R R C  1 2 3 3  R3     C3  R1  111.11 N

R2  472.22 N

2

R3  83.33  N

C3  5.556  N  m

x  0  in 0.002  L  L

7.


8.


9.

Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)   R1 S ( x 0  in)  F  S ( x a )   R2 S ( x b )  R3 S ( x L) M ( x)  R1 S ( x 0  in)  x  F  S ( x a )  ( x  a )  R2 S ( x b )  ( x  b )


V ( x) N

Moment Diagram

200

60

0

35 M ( x)

 200

10

Nm

 400

 15

 600

 40 0

200

400

FIGURE 6-26aB

600

800

3

1 10

0

200

400

600

x

x

mm

mm

800

3

1 10

Shear and Moment Diagrams for Problem 6-26a

11. From Figure 6-26aB, the maximum moment occurs at x = a. The maximum, minimum, alternating and mean bending moments at x = a are: Mmax  M ( a ) Ma  Mm 

Mmax  Mmin 2 Mmax  Mmin 2

Mmax  44.4 N  m

Mmin  Mmax

Ma  44.444 N  m Mm  0  N  m



6-26a-3

12. Calculate the alternating and mean components of the maximum bending stress in the beam using equation (4.11b). These are principal stresses and also von Mises stresses.

σ'a 

M a c

σ'a  31.189 MPa

I Mm c

σ'm 

σ'm  0  MPa

I

13. Calculate the beam cross-section dimensions from I and c. Beam depth

h  2  c

Beam width

w 

h  40 mm

12 I h

w  5.344  mm

3

14. Calculate the endurance limit modification factors for a nonrotating rectangular beam. Load

Cload  1

Size

A95  0.05 w h d equiv 

A95  10.688 mm

A95


0.0766


 0.097

Csize  1.189  

Surface

A  4.51

Temperature

Ctemp  1

Reliability

Creliab  1

2

Csize  0.936

b  0.265

 S ut    MPa 

Csurf  S ut  A  

b

(R = 50%)

15. Determine the modified endurance limit as a function of the unknown endurance limit. S e S ut  Cload  Csize Csurf  S ut  Ctemp Creliab 0.5 S ut 16. Assuming a Case 3 load line, use equation (6.18e) as the design equation and solve for S ut. Guess

S ut  100  MPa

Given

Nfd =



S ut  Find  S ut

S ut  447  MPa

Csurf  S ut  0.895 18. Choose AISI 1020, cold-rolled steel (see Appendix A, Table A-9). © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


6-27-1


A storage rack is to be designed to hold the paper roll of Problem 6-8 as shown in Figure P6-12. Determine a suitable value for dimension a in the figure for an infinite-life fatigue safety factor of 2. Assume dimension b = 100 mm and that the mandrel is solid and inserts halfway into the paper roll. (a) The beam is a ductile material with S ut = 600 MPa. (b) The beam is a cast-brittle material with S ut = 300 MPa.

Given:


Ductile tensile strength

3

OD  1.50 m ID  0.22 m

Roll density Design safety factor

ρ  984  kg m

Lroll  3.23 m

Brittle tensile strength

S utb  300  MPa

Nfd  2

S uta  600  MPa W

Assumptions: The paper roll's weight creates a concentrated load acting at the tip of the mandrel. The mandrel's root fits tightly in the stanchion so it can be modeled as a cantilever beam. The mandrel is machined, reliability is 90%, and it operates at room temperature. Solution: 1.

M1

Lm R1

FIGURE 6-27



Determine the weight of the roll and the length of the mandrel. W 

Weight

π 4



2

2




W  53.9 kN


Length 2.

a

Lm  1.615 m

The maximum moment occurs at a section where the mandrel root leaves the stanchion and is Mmax  W  Lm

Mmax  87.04 kN  m

3.

The dynamic loading is repeated from 0 to Mmax on each stress cycle, thus Mmin  0  kN  m

4.

Part (a) - Calculate the alternating and mean components of the bending moment. Ma  Mm 

Mmax  Mmin

Ma  43520 N  m

2 Mmax  Mmin

Mm  43520 N  m

2

S'e  0.5 S uta

S'e  300 MPa

5.


6.


Cload  1

Size

A95 ( a )  0.010462 a

2

 dequiv( a)  Csize( a )  1.189     mm 

d equiv( a ) 

A95 ( a ) 0.0766

 0.097


MACHINE DESIGN - An Integrated Approach, 4th Ed. A  4.51

Surface

Csurf

7.

6-27-2 b  0.265

b

 S uta   A     MPa 

Temperature

Ctemp  1

Reliability


(machined)

Csurf  0.828

(R = 90%)

Calculate the modified endurance limit. S e( a )  Cload  Csize( a )  Csurf  Ctemp Creliab S'e

8.

We can now determine the minimum required diameter, a. Using the distortion-energy failure theory with the modified Goodman diagram, the bending stress will also be the only nonzero principal stress, which will also be the von Mises stress. Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety. Guess a  100  mm. Bending stress

σ=

M c I

Given Nfd =

a 64 32 M = M  = 2 4 3 π a π a

π a

3

32



S e( a )  S uta Ma S uta  Mm S e( a )

a  Find ( a )

a  186.864 mm a  190  mm

Round this up to the next higher even value Using this value of a, the values of the functions of a are: Csize( a )  0.787

S e( a )  175.371 MPa

The realized safety factor is Nfa 

9.

π a 32

3



S e( a )  S uta

Nfa  2.1

Ma S uta  Mm S e( a ) S'e  0.4 S utb

Part (b) - Determine the unmodified endurance limit.

S'e  120 MPa

10. Calculate the endurance limit size modification factor for a nonrotating rectangular beam. Size

A95 ( a )  0.010462 a

2


d equiv( a ) 

A95 ( a ) 0.0766

 0.097

11. Calculate the modified endurance limit. S e( a )  Cload  Csize( a )  Csurf  Ctemp Creliab S'e



6-27-3

12. We can now determine the minimum required diameter, a. Using the distortion-energy failure theory with the modified Goodman diagram, the bending stress will also be the only nonzero principal stress, which will also be the von Mises stress. Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety. Guess a  100  mm. Bending stress

σ=

M c I

Given Nfd =

a 64 32 M = M  = 2 4 3 π a π a

π a 32

3



S e( a )  S utb Ma S utb  Mm S e( a )

a  Find ( a ) Round this up to the next higher even value

a  251.687 mm a  252  mm

Using this value of a, the values of the functions of a are: Csize( a )  0.766

S e( a )  68.253 MPa

The realized safety factor is Nfb 

π a 32

3




Nfb  2.0



6-28-1


Figure P6-13 shows a forklift truck negotiating a 15 deg ramp to to drive onto a 4-ft-high loading platform. The truck weighs 5 000 lb and has a 42-in wheelbase. Design two (one for each side) 1-ft-wide ramps of steel to have a safety factor of 2 for infinite life in the worst case of loading as the truck travels up them. Minimize the weight of the ramps by using a sensible cross-sectional geometry. Choose an appropriate steel or aluminum alloy.

Given:

Ramp angle θ  15 deg Platform height h  4  ft Truck wheelbase Lt  42 in

Ramp width Truck weight

w  12 in W  5000 lbf

Assumptions: 1. The worst case is when the truck CG is located at the center of the beam's span. 2. Use a coordinate frame that has the x-axis along the long axis of the beam. 3. Ignore traction forces and the weight components along the x-axis of the beam. 4. There are two ramps, one for each side of the forklift. See Figures 6-28 and Mathcad file P0628. Solution:

L b a

CG a

y

CG b

R1  Fa

Wa

Fb

x Wb

R2


1.

From Problem 3-28 the maximum bending moment in the ramp occurs at the rear wheel of the truck and is Mmax  8324 ft  lbf

Mmax  99888 in lbf

Mmin  0  in lbf

The alternating and mean components of the bending moment are: Ma 

Mm  2.

Mmax  Mmin 2 Mmax  Mmin 2

Ma  49944 in lbf

Mm  49944 in lbf

The bending stress is the only stress component present and is, therefore, also the only nonzero principal stress and is also the von Mises stress. The governing design equations then are



σ'a =

Ma

σ'm =

Z

6-28-2

Mm

Nfd =

Z

Combining these into a single equation

S e S ut


Nfd =

Z  S e S ut Ma S ut  Mm S e

3.

The approach will be to 1) choose a suitable factor of safety, 2) choose a suitable material and determine its tensile strength and endurance limit, 3) from the equation above determine the required value of the section modulus, 4) choose an appropriate cross-section for the ramp, and 5) determine the dimensions of the cross-section.

4.

The following design choices have been made for this problem: Design factor of safety

Nfd  2

Material

AISI 1095 steel, hot-rolled

Tensile strength

S ut  120  ksi S'e  0.5 S ut

S'e  60 ksi

5.


6.


Cload  1

Size

Csize  0.732

Surface

A  14.4

Csurf

7.

(initially guessed, and then found by iteration) b  0.718

 S ut   A     ksi 

Temperature

Ctemp  1

Reliability


(hot-rolled)

b

Csurf  0.463

(R = 90%)


8.

S e  18.237 ksi

Solve the design equation for the minimum section modulus, Z. Z  Nfd 

Ma Sut  Mm Se

3

Z  6.309 in

S e S ut

This is the minimum allowable value of the section modulus. 9.

Assume a channel section such as that shown in Figure 6-28B. To keep it simple, let the thickness of the flanges and web be the same. Choose 5/8-in thick plate, which is readily available. Then, t  0.625  in

10. The cross-sectional area of the ramp is A ( h )  w t  2  t ( h  t) 11. The distance to the CG is

cg( h ) 

1 A (h)

 w t 2



 2

2

 t h  t

2

 



6-28-3

12. The moments of inertia of the web and a flange are Iweb( h ) 

Ifl ( h ) 

w t

3

12

 w t  cg( h ) 



3

t ( h  t) 12

t

 

Flange

2

Web

2

h  h  t  cg( h )  2 

t

t

2

 

I ( h )  Iweb( h )  2  Ifl ( h )

h

13. The maximum stress will occur in the flange at the top and is compressive. The distance from the centroid up to the top of the flange is

w

c( h )  h  cg( h ) 14. Using the known section modulus, solve for the unknown flange height, h. Guess h  1  in Given

Z=

I (h) c( h )

h  Find ( h )

FIGURE 6-28B Channel Section for Problem 6-28

h  4.304 in

Round this to

h  4.25 in

15. Check the size modification factor. A95  0.05 w cg( h )  t ( h  cg( h ) ) d equiv 

A95

d equiv  5.858 in

0.0766

 d equiv  Csize  0.869     in 

2

A95  2.628 in

 0.097

Csize  0.732

16. Summarizing, the ramp design dimensions are: Width

w  12.00 in

Flange height

h  4.25 in

Shape

channel

Thickness

t  0.625 in

Material

1095 steel



6-29-1


A bar, 22 mm x 30 mm in cross-section, is loaded axially in tension with Fmin = -8 kN and Fmax = +8 kN. A 10-mm hole passes through the center of the 30-mm side. Find the safety factor for infinite life if the material has S ut = 500 MPa.

Given:

Bar width

w  30 mm

Maximum load

Fmax  8  kN

Bar thickness

h  22 mm

Minimum load

Fmin  8  kN

Hole diameter Tensile strength

d  10 mm S ut  500  MPa

Infinite life

F

Assumptions: Machined surfaces, temperature of 37C, and reliability of 99.999%. Solution: 1. N =


For completely reversed loading, the factor of safety is Se Kf  σ'athe uniform axial stress is the only stress component present, Since and

σ'a = σa 2.

N =

ø10

Calculate the endurance limit modification factors for an axial bar. Load

Cload  0.7

(axial loading)

Size

Csize  1

(axial loading)

Surface

A  4.51

22

b  0.265

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


b

Csurf  0.869

F FIGURE 6-29 (R = 99.999%)



5.

Kf  σa

S'e  250  MPa

Csurf

4.

Se

Calculate the unmodified endurance limit. S'e  0.5 S ut

3.

30

S e  100.2  MPa

Determine the nominal (not increased by a stress concentration factor) alternating component of stress at the hole. Area

A  ( w  d )  h

Alternating load

Fa 

Alternating stress

σa 

Fmax  Fmin 2 Fa A

A  440  mm

2

Fa  8  kN

σa  18.182 MPa



6-29-2

Determine the geometric stress concentration factor from Appendix C, Figure C-13.

7.

d

 7.9735 

d

2

3

d  9.2659     w  w w 4 5 d d  1.8145    2.9684   w  w

Kt  3.0039  3.753 

Determine the notch sensitivity of the material. Note from Figure 6-35 that the Neuber constant for steel in tension is slightly lower that for torsional loading. However, comparison of values of a 1/2 obtained from the dashed red curve with those in Table 6-6 indicates that, for tension as well as torsion, a value of 20 ksi should be added to S ut to obtain a 1/2 from Table 6-6. Lookup value of S ut

S'ut  S ut  20 ksi

Neuber constant

a  0.068  in

Notch radius

r  0.5 d

Notch sensitivity

q 

2

0.5

a  0.068  in

q  0.867 a r

Determine the fatigue stress concentration factor from equation (6.11b). Kf  1  q   Kt  1 

9.

S'ut  93 ksi

r  5  mm

1 1

8.

Kt  2.33

Kf  2.153

Determine the factor of safety against fatigue failure for the assumptions made.

Nf 

Se Kf  σa

Nf  2.6



6-30-1


A bar, 22 mm x 30 mm in cross-section, is loaded axially in tension with Fmin = 0 kN and Fmax = 16 kN. A 10-mm hole passes through the center of the 30-mm side. Find the safety factor for infinite life if the material has S ut = 500 MPa.

Given:

Bar width

w  30 mm

Maximum load

Fmax  16 kN

Bar thickness

h  22 mm

Minimum load

Fmin  0  kN


d  10 mm S ut  500  MPa

Infinite life

Assumptions: Machined surfaces, temperature of 37C, and reliability of 99.999%. Solution:


F 1.

For fluctuating loading, the factor of safety is

N =

S e S ut Kf  σ'a S ut  Kfm σ'm S e

30 Since the uniform axial stress is the only stress component present,

σ'a = σa 2.

and

σ'm = σm


3.

S'e  250  MPa


Cload  0.7

(axial loading)

Size

Csize  1

(axial loading)

Surface

A  4.51

 Sut    MPa 

b

Temperature

Ctemp  1

Reliability


Csurf  0.869

F FIGURE 6-30 (R = 99.999%)



5.

22

b  0.265


4.

ø10

S e  100.2  MPa

Determine the nominal (not increased by a stress concentration factor) alternating and mean components of stress at the hole. Area

A  ( w  d )  h

A  440  mm

2


MACHINE DESIGN - An Integrated Approach, 4th Ed. Alternating load

Fm 

Mean load

Alternating stress

σa 

σm 

Mean stress 6.

Fa 

6-30-2

Fmax  Fmin

Fa  8  kN

2 Fmax  Fmin

Fm  8  kN

2 Fa

σa  18.182 MPa

A Fm

σm  18.182 MPa

A


7.

d

 7.9735 

d

2

3

d  9.2659     w  w w 4 5 d d  1.8145    2.9684   w  w

Kt  3.0039  3.753 



Neuber constant

a  0.068  in

Notch radius

r  0.5 d

Notch sensitivity

q 

S'ut  93 ksi

2

0.5

a  0.068  in r  5  mm

1 1

8.

Kt  2.33

q  0.867 a r

Determine the fatigue stress concentration factors from equations (6.11b) and (6.17). Kf  1  q   Kt  1 

Kf  2.153

Kf   σm  σa  78 MPa

Assuming the yield strength for this material is about 400 MPa, we can use the first of equations (6.17) and Kfm  Kf 9.

Kfm  2.153


Nf 

S e S ut Kf  σa S ut  Kfm σm S e

Nf  2.1



6-31-1


A bar, 22 mm x 30 mm in cross-section, is loaded axially in tension with Fmin = 8 kN and Fmax = 24 kN. A 10-mm hole passes through the center of the 30-mm side. Find the safety factor for infinite life if the material has S ut = 500 MPa.

Given:

Bar width

w  30 mm

Maximum load


Bar thickness

h  22 mm

Minimum load



d  10 mm S ut  500  MPa

Infinite life

Assumptions: Machined surfaces, temperature of 37C, and reliability of 99.999%. Solution: 1.

F



N =


30

Since the uniform axial stress is the only stress component present, and

σ'a = σa 2.


3.

ø10

σ'm = σm

S'e  250  MPa


Cload  0.7

(axial loading)

Size

Csize  1

(axial loading)

Surface

A  4.51

22

b  0.265

 Sut    MPa 

b


Csurf  0.869

F FIGURE 6-31

Temperature

Ctemp  1

Reliability



(R = 99.999%)

4. Calculate the modified endurance limit. S e  Cload  Csize Csurf  Ctemp Creliab S'e 5.

S e  100.2  MPa


A  ( w  d )  h

Alternating load

Fa 

Fmax  Fmin 2

A  440  mm

2

Fa  8  kN



Fm 

Mean load

Alternating stress

σm 

Mean stress 6.

σa 

6-31-2

Fmax  Fmin

Fa

σa  18.182 MPa

A Fm

σm  36.364 MPa

A

Determine the geometric stress concentration factor from Appendix C, Figure C-13. d

 7.9735 

d

2

  w 4 5 d d  1.8145    2.9684   w  w

Kt  3.0039  3.753 

7.

Fm  16 kN

2

w

 9.2659 

3

  w

Kt  2.33



Neuber constant

a  0.068  in

Notch radius

r  0.5 d

Notch sensitivity

q 

S'ut  93 ksi

2

0.5

a  0.068  in r  5  mm

1 1

8.

d

q  0.867 a r


Kf  2.153

Kf   σm  σa  117  MPa


Kfm  2.153


Nf 


Nf  1.8



6-32-1


A bar, 22 mm x 30 mm in cross-section, is loaded axially in tension with Fmin = -4 kN and Fmax = 12 kN. A 10-mm hole passes through the center of the 30-mm side. Find the safety factor for infinite life if the material has S ut = 500 MPa.

Given:

Bar width

w  30 mm

Maximum load


Bar thickness

h  22 mm

Minimum load



d  10 mm S ut  500  MPa

Infinite life

Assumptions: Machined surfaces, temperature of 37C, and reliability of 99.999%. Solution: 1.


F


N =


30


σ'a = σa 2.


3.

S'e  250  MPa

Calculate the endurance limit modification factors for an axial bar. Load Size

Cload  0.7 Csize  1

Surface

A  4.51

b  0.265

 Sut    MPa 

b

Temperature

Ctemp  1

Reliability


Csurf  0.869

F FIGURE 6-32 (R = 99.999%)



5.

22

(axial loading) (axial loading)


4.

ø10

σ'm = σm

S e  100.2  MPa


A  ( w  d )  h

Alternating load

Fa 

Fmax  Fmin 2

A  440  mm

2

Fa  8  kN



Fm 

Mean load

Alternating stress

σm 

Mean stress 6.

σa 

6-32-2

Fmax  Fmin

Fm  4  kN

2 Fa

σa  18.182 MPa

A Fm

σm  9.091  MPa

A


7.

d

 7.9735 

d

2

3

d  9.2659     w  w w 4 5 d d  1.8145    2.9684   w  w

Kt  3.0039  3.753 



Neuber constant

a  0.068  in

Notch radius

r  0.5 d

Notch sensitivity

q 

S'ut  93 ksi

2

0.5

a  0.068  in r  5  mm

1 1

8.

Kt  2.33

q  0.867 a r


Kf  2.153

Kf   σm  σa  59 MPa


Kfm  2.153


Nf 


Nf  2.3



6-33a-1


For the bracket shown in Figure P6-14 subjected to a sinusoidal force-time function with Fmax = F and Fmin = -F, where F and the beam's other data are given in row a of Table P6-6. Find the stress states at points A and B due to this fully reversed loading and choose a ductile steel material specification that will give a safety factor of 2 for infinite life. Assume a geometric stress-concentration factor of 2.5 in bending and 2.8 in torsion.

Given:

Outside diameter Geometric stress

od  20 mm Kt  2.5

concentration factors

Kts  2.8


Nd  2

A B

T

Assumptions: The finish is machined, reliability is 50%, and the bracket operates at room temperature. The notch sensitivity q = 1 so that Kf = Kt. Solution:

F

y

T

x

M L R

See Figure 6-33 and Mathcad file P0633a. FIGURE 6-33 Free Body Diagram of Tube for Problem 6-33

1.

From Problem 4-33a the stress components at point A are σx  8.38 MPa

2.

Calculate fatigue stresses and principal stresses. Fatigue stresses

Principal stresses

τzx  16.76  MPa

σb  Kt σx

σb  20.95 MPa

τs  Kts τzx

τs  46.93 MPa 2

 σb  2 σ1      τs 2 2 σb

σ1  58.56 MPa σ2  0  MPa

σ3 

3.

σb 2

2



 σb  2    τs 2  

σ3  37.61 MPa

Calculate the alternating von Mises effective stress (the mean component is zero).

σ'a 

2

2

σ1  σ1 σ3  σ3

σ'a  83.94 MPa

S'e S ut  0.5 S ut

4.

Calculate the unmodified endurance limit

5.

Determine the endurance limit modification factors Load

Cload  1

Size

A95  0.010462 od

2

d eq 

A95 0.0766

(nonrotating round section)

d eq  7.391 mm



6-33a-2

 deq  Csize  1.189     mm  Surface

 0.097

Csize  0.979

A  4.51

b  0.265

 S ut  Cs S ut  A     MPa 

b

(cold-drawn tubing)

Csurf  S ut  if  Cs S ut  1 1 Cs S ut 

6.

Temperature

Ctemp  1

Reliability

Creliab  1.0

(R = 50%)

Calculate the modified endurance limit S e S ut  Cload  Csize Csurf  S ut  Ctemp Creliab S'e S ut

7.

Use the equation for the factor of safety for fully reversed loading to solve for S ut. Guess S ut  100  MPa

Given

8.

The varables that depend on S ut are:

Nd =

S e S ut

σ'a

Csurf  S ut  0.946


S ut  362 MPa

S e S ut  167.88 MPa



6-34a-1


For the bracket shown in Figure P6-14 subjected to a sinusoidal force-time function with Fmax = F and Fmin = 0, where F and the beam's other data are given in row a of Table P6-6. Find the stress states at points A and B due to this repeated loading and choose a ductile steel material specification that will give a safety factor of 2 for infinite life. Assume a geometric stress-concentration factor of 2.8 in bending and 3.2 in torsion.

Given:


od  20 mm Kt  2.8


Kts  3.2


Nd  2

A B

T

Assumptions: The finish is machined, reliability is 50%, and the bracket operates at room temperature. The notch sensitivity q = 1 so that Kf = Kt. Solution:

F

y

T

x

M L R

See Figure 6-33 and Mathcad file P0633a. FIGURE 6-34 Free Body Diagram of Tube for Problem 6-34

1.

2.

From Problem 4-33a the stress components at point A are

σxmax  8.38 MPa

τzxmax  16.76  MPa

σxmin  0  MPa

τzxmin  0  MPa

Calculate the alternating and mean stress components.

σxa  σxm  τzxa  τzxm  3.

σxmax  σxmin

σxa  4.19 MPa

2

σxmax  σxmin

σxm  4.19 MPa

2

τzxmax  τzxmin

τzxa  8.38 MPa

2

τzxmax  τzxmin

τzxm  8.38 MPa

2

Calculate fatigue stresses and principal stresses. Fatigue stresses

σba  Kt σxa

σba  11.73 MPa

τsa  Kts τzxa

τsa  26.82 MPa

σbm  Kt σxa

σbm  11.73 MPa

τsm  Kts τzxa

τsm  26.82 MPa

Principal stresses 2

 σba  2 σ1a      τsa 2  2  σba

σ1a  33.32 MPa σ2a  0  MPa



6-34a-2

2

 σba  2 σ3a      τsa 2 2   σba

σ3a  21.58 MPa

2

 σbm  2 σ1m      τsm 2  2  σbm

σ1m  33.32 MPa σ2m  0  MPa

2

 σbm  2 σ3m      τsm 2  2  σbm

4.

σ3m  21.58 MPa

Calculate the alternating and mean von Mises effective stress components. 2

2

σ'a 


σ'm 

σ1m  σ1m σ3m  σ3m

2

σ'a  47.91 MPa 2

σ'm  47.91 MPa

S'e S ut  0.5 S ut

5.


6.


Cload  1

Size

A95  0.010462 od

2

d eq 


A95

d eq  7.391 mm

0.0766

 deq  Csize  1.189     mm  Surface

 0.097

Csize  0.979

A  4.51

b  0.265

 S ut  Cs S ut  A     MPa 

b

(machined tubing)

Csurf  S ut  if  Cs S ut  1 1 Cs S ut 

7.

Temperature

Ctemp  1

Reliability

Creliab  1.0

(R = 50%)

Calculate the modified endurance limit S e S ut  Cload  Csize Csurf  S ut  Ctemp Creliab S'e S ut

8.

Use the equation for the factor of safety for repeated loading assuming a Case 3 load line and using equation (6.18e). Guess S ut  100  MPa



Given

Nd =



S ut  Find  S ut 9.


6-34a-3

Csurf  S ut  1

S ut  291 MPa S e S ut  142.72 MPa



6-35a-1


For the bracket shown in Figure P6-14 subjected to a sinusoidal force-time function with Fmax = F and Fmin = -F, where F and the beam's other data are given in row a of Table P6-6. Find the stress states at points A and B due to this fully reversed loading and choose a cast iron material specification that will give a safety factor of 2 for infinite life. Assume a geometric stress-concentration factor of 2.5 in bending and 2.8 in torsion.

Given:


od  20 mm Kt  2.5


Kts  2.8


Nd  2

A B

T

Assumptions: The finish is as-cast, reliability is 50%, and the bracket operates at room temperature. There ia a fillet at the wall with radius r  2  mm. However, set the stress concentration factors and Csurf to 1 since cast iron's internal flaws mask these effects. Solution:

F

y

T

x

M L R



1

From Problem 4-33a the stress components at point A are σx  8.38 MPa

2.

Calculate fatigue stress concentration factors and principal stresses. Let Fatigue stresses

τzx  16.76  MPa Kf  1

Kfs  1

σb  Kf  σx

σb  8.38 MPa

τs  Kfs τzx

τs  16.76 MPa

Principal stresses

σ1 

σb 2

2



 σb  2    τs 2

σ1  21.47 MPa

2

 σb  2 σ3      τs 2 2 σb

σ3  13.09 MPa

σ2  0  MPa 3.

Calculate the alternating von Mises effective stress (the mean component is zero).

σ'a 

2

2

σ1  σ1 σ3  σ3

σ'a  30.21 MPa

S'e S ut  0.4 S ut

4.


5.


Cload  1

Size

A95  0.010462 od

2




d eq 

6-35a-2

A95

d eq  7.391 mm

0.0766

 deq  Csize  1.189     mm 

6.

Surface

Csurf  1

Temperature

Ctemp  1

Reliability

Creliab  1.0

 0.097

Csize  0.979 ( cast iron)

(R = 50%)

Calculate the modified endurance limit S e S ut  Cload  Csize Csurf  Ctemp Creliab S'e S ut

7.

Use the equation for the factor of safety for fully reversed loading to solve for S ut. Guess S ut  100  MPa Given

8.


Nd =

S e S ut

σ'a


S ut  154 MPa

S e S ut  60.43 MPa



6-36a-1


For the bracket shown in Figure P6-14 subjected to a sinusoidal force-time function with Fmax = F and Fmin = 0, where F and the beam's other data are given in row a of Table P6-6. Find the stress states at points A and B due to this repeated loading and choose a cast iron material specification that will give a safety factor of 2 for infinite life. Assume a geometric stress-concentration factor o 2.8 in bending and 3.2 in torsion.

Given:


od  20 mm Kt  2.8


Kts  3.2


Nd  2

Assumptions: The finish is as-cast, reliability is 50%, and the bracket operates at room temperature. There ia a fillet at the wall with radius r  2  mm. However, set the stress concentration factors and Csurf to 1 since cast iron's internal flaws mask these effects. Solution: 1.

2.

Free Body Diagram of Tube for Problem 6-36


From Problem 4-33a the stress components at point A are

σxmax  8.38 MPa

τzxmax  16.76  MPa

σxmin  0  MPa

τzxmin  0  MPa

Calculate the alternating and mean stress components.

σxa  σxm  τzxa  τzxm  3.

FIGURE 6-36

σxmax  σxmin

σxa  4.19 MPa

2

σxmax  σxmin

σxm  4.19 MPa

2

τzxmax  τzxmin

τzxa  8.38 MPa

2

τzxmax  τzxmin

τzxm  8.38 MPa

2

Calculate fatigue stress concentration factors (set 1 for cast iron) and principal stresses Fatigue stress concentration factors Fatigue stresses

Kf  1

Kfs  1

Kfm  1

Kfsm  1

σba  Kf  σxa

σba  4.19 MPa

τsa  Kfs τzxa

τsa  8.38 MPa

σbm  Kfm σxa

σbm  4.19 MPa

τsm  Kfsm τzxa

τsm  8.38 MPa



6-36a-2

Principal stresses 2

 σba  2 σ1a      τsa 2 2   σba

σ1a  10.73 MPa σ2a  0  MPa

2

 σba  2 σ3a      τsa 2 2   σba

σ3a  6.54 MPa

2

 σbm  2 σ1m      τsm 2  2  σbm

σ1m  10.73 MPa σ2m  0  MPa

2

 σbm  2 σ3m      τsm 2  2  σbm

4.

σ3m  6.54 MPa

Calculate the alternating von Mises effective stress (the mean component is zero). 2

2

σ'a 


σ'm 

σ1m  σ1m σ3m  σ3m

2

σ'a  15.11 MPa 2

σ'm  15.11 MPa

S'e S ut  0.4 S ut

5.


6.


Cload  1

Size

A95  0.010462 od

2

d eq 

A95

d eq  7.391 mm

0.0766

 deq  Csize  1.189     mm 

7.


Surface

Csurf  1

Temperature

Ctemp  1

Reliability

Creliab  1.0

 0.097

Csize  0.979 (cast iron)

(R = 50%)

Calculate the modified endurance limit S e S ut  Cload  Csize Csurf  Ctemp Creliab S'e S ut

8.

Use the equation for the factor of safety for repeated loading assuming a Case 3 load line and using equation (6.18e). Guess S ut  100  MPa Given

Nd =




S ut  107 MPa



6-37-1


A semicircular, curved beam as shown in Figure 5-37 has the dimensions given below. For a load pair F = ±3 kN applied along the diameter, find the safety factor at the inner and outer fibers: (a) If the beam is steel with S ut = 700 MPa, (b) If the beam is cast-iron with S ut = 420 MPa.

Given:

(a) Tensile strength

S uta  700  MPa

(b) Tensile strength

S utb  420  MPa

Maximum load

Fmax  3  kN

Minimum load


Solution: 1.


From Problem 4-37, the stresses at the inside radius are:

σi  409.9  MPa

Inside

These are based on a load of 14 kN. Since the stress in a curved beam is directly proportional to the applied load, we can determine the stresses at the inside surface for this problem by applying the ratio 3/14 to this stress. Thus,

σmax  σmin 

3 14 3 14

 σi

σmax  87.836 MPa

 σi

σmin  87.836 MPa FIGURE 6-37

These are the only stress components present on their respective surfaces so they are also von Mises stresses. 2.


The dynamic loading in this problem is fully reversed. Determine the alternating stress component.

σ'a 

σmax  σmin

σ'a  87.836 MPa

2

Part (a) S'ea  0.5 S uta

S'ea  350 MPa

3.


4.


Cload  0.7

(combined axial and bending loads)

Size

Section dims

w  25 mm

A95  0.05 w h d equiv 

A95  31.25 mm

A95


A  4.51

2


0.0766  0.097

Csize  1.189   Surface

h  25 mm

b  0.265

Csize  0.888 (machined)



Csurf

5.

 S uta   A     MPa 

Temperature

Ctemp  1

Reliability


6-37-2

b

Csurf  0.795

(R = 90%)

Calculate the modified endurance limit. S ea  Cload  Csize Csurf  Ctemp Creliab S'ea

6.

S ea  155.15 MPa

Assuming no stress concentration, the fatigue factor of safety at the inner fiber for fully reversed loading is Nfa 

S ea

Nfa  1.8

σ'a

Part (b) S'eb  160  MPa

7.

Calculate the unmodified endurance limit using equation (6.5b).

8.


Cload  0.7

(combined axial and bending loads)

Size

Section dims

w  25 mm

A95  0.05 w h d equiv 

A95  31.25 mm

A95

A  4.51

Csurf

9.

 0.097

Csize  0.888

b  0.265

 S utb   A     MPa 

Temperature

Ctemp  1

Reliability


2


0.0766


h  25 mm

(machined)

b

Csurf  0.910

(R = 90%)

Calculate the modified endurance limit. S eb  Cload  Csize Csurf  Ctemp Creliab S'eb

S eb  81.21 MPa

10. Assuming no stress concentration, the fatigue factor of safety at the inner fiber for fully reversed loading is Nfb 

S eb

σ'a

Nfb  0.92



6-38-1


A 42-mm-dia steel shaft with a 19-mm transverse hole is subjected to a sinusoidal combined loading of  = ±100 MPa bending stress and steady torsion of 110 MPa. Find the safety factor for infinite life if S ut = 1000 MPa.

Given:

Shaft diameter

D  42 mm

Max bending stress

σmax  100  MPa

Hole diameter

d  19 mm

Min bending stress

σmin  100  MPa

Steady torsion

τm  110  MPa

Tensile strength S ut  1000 MPa Infinite life

Assumptions: Stresses given include stress concentration effects. Ground surfaces, temperature of 37C, and reliability of 50%. Solution: 1.


Calculate the alternating and mean von Mises stress components.

σ'a 

σmax  σmin

σ'm  2.

σ'a  100 MPa

2 3  τm

σ'm  190.526 MPa


3.

S'e  500 MPa

Calculate the endurance limit modification factors for a rotating, round shaft. Cload  1

Load Size

Csize  1.189  

Surface

A  1.58

D

 

 mm 

Csurf

4.

(combined bending and torsion)  0.097

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability

Creliab  1

Csize  0.827 b  0.085

b

Csurf  0.878

(R = 50%)


5.

(ground)

S e  363.37 MPa

Assuming a Case 3 load line, determine the factor of safety against fatigue failure.

Nf 

S e S ut


Nf  2.1



6-39-1


A 42-mm-dia steel shaft with a 19-mm transverse hole is subjected to a combined loading of  = ±100 MPa bending stress and an alternating torsion of ±110 MPa, which are 90 deg out of phase. Find the safety factor for infinite life if S ut = 1000 MPa.

Given:

Shaft diameter

D  42 mm

Max bending stress

σmax  100  MPa

Hole diameter

d  19 mm

Min bending stress

σmin  100  MPa

Max torsional stress

τmax  110  MPa

Min torsional stress

τmin  110  MPa

Tensile strength S ut  1000 MPa

ϕ  90 deg

Phase angle

Assumptions: Stresses given include stress concentration effects. Ground surfaces, temperature of 37C, and reliability of 50%. Solution: 1.


The dynamic loading is fully reversed so both mean stresses are zero. Calculate the alternating SEQA stress component using equation (6.23). Q  2 

Stress ratio

τmax

Q  2.2

σmax

1

SEQAa 

2.

σmax 

3



4

2

1 

2

Q 

1

3 2

2

 Q  cos( 2  ϕ) 

 

SEQAa  190.526 MPa

S'e  500 MPa

Cload  1

(combined bending and torsion)

   mm 

Size

Csize  1.189  

Surface

A  1.58

D

 0.097

 Sut    MPa 

Temperature

Ctemp  1

Reliability

Creliab  1

Csize  0.827 b  0.085

(ground)

b


Csurf  0.878

(R = 50%)


5.

Q

2

Calculate the endurance limit modification factors for a rotating, round shaft. Load

4.

16

4


3.

9

S e  363.37 MPa

Assuming a Case 3 load line, determine the factor of safety against fatigue failure. Nf 

Se SEQAa

Nf  1.9



6-40-1


Redesign the roll support of Problem 6-8 to be like that shown in Figure P6-16. The stub mandrels insert to 10% of the roll length at each end. Design dimension a for an infinite-life factor of safety of 2. See Problem 6-8 for additional data. (a) The beam is a ductile material with S y = 450 MPa, S ut = 600 MPa (b) The beam is a cast-brittle material with S ut = 300 MPa

Given:


OD  1.50 m ID  0.22 m

3

ρ  984  kg m

Roll density Design safety factor

Nfd  2

Lroll  3.23 m Ductile tensile strength

S uta  600  MPa Brittle tensile strength

S utb  300  MPa

Assumptions: The paper roll's weight creates a concentrated load acting at the tip of the mandrel. The mandrel's root fits tightly in the stanchion so it can be modeled as a cantilever beam. The mandrel is machined, reliability is 99.999%, and it operates at room temperature. Solution: 1.


Determine the weight of the roll, the load on each support, and the length of the mandrel. Weight of paper W 

 4

π

2

2

FIGURE 6-40





W  53.9 kN

2.

Load on one mandrel

F  0.5 W

F  26.95 kN

Length of mandrel


Lm  0.323 m

The maximum moment occurs at a section where the mandrel root leaves the stanchion and is Mmax  F  Lm

Mmax  8.704 kN  m

3.

The dynamic loading is repeated from 0 to Mmax on each stress cycle, thus Mmin  0  kN  m

4.

Part (a) - Calculate the alternating and mean components of the bending moment. Ma  Mm 

Mmax  Mmin

Ma  4352 N  m

2 Mmax  Mmin

Mm  4352 N  m

2

S'e  0.5 S uta

S'e  300 MPa

5.


6.


Cload  1

Size

A95 ( a )  0.010462 a

2

d equiv( a ) 

A95 ( a ) 0.0766



6-40-2

 dequiv( a)  Csize( a )  1.189     mm  A  4.51

Surface

Csurf

7.

 0.097

b  0.265 b

 S uta   A     MPa 

Temperature

Ctemp  1

Reliability


(machined)

Csurf  0.828

(R = 99.999%)

Calculate the modified endurance limit. S e( a )  Cload  Csize( a )  Csurf  Ctemp Creliab S'e

8.

We can now determine the minimum required diameter, a. Using the distortion-energy failure theory with the modified Goodman diagram, the bending stress will also be the only nonzero principal stress, which will also be the von Mises stress. Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety. Guess a  100  mm. Bending stress

σ=

M c I

Given Nfd =

a 64 32 M = M  = 2 4 3 π a π a

π a

3

32



S e( a )  S uta Ma S uta  Mm S e( a )

a  Find ( a )

a  92.421 mm a  94 mm

Round this up to the next higher even value Using this value of a, the values of the functions of a are: Csize( a )  0.843

S e( a )  137.942 MPa

The realized safety factor is Nfa 

9.

π a 32

3



S e( a )  S uta

Nfa  2.1

Ma S uta  Mm S e( a ) S'e  0.4 S utb

Part (b) - Determine the unmodified endurance limit.

S'e  120 MPa

10. Calculate the endurance limit size modification factor for a nonrotating rectangular beam. Size

A95 ( a )  0.010462 a

2


d equiv( a ) 

A95 ( a ) 0.0766

 0.097

11. Calculate the modified endurance limit. S e( a )  Cload  Csize( a )  Csurf  Ctemp Creliab S'e © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0640.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


6-40-3

12. We can now determine the minimum required diameter, a. Using the distortion-energy failure theory with the modified Goodman diagram, the bending stress will also be the only nonzero principal stress, which will also be the von Mises stress. Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety. Guess a  100  mm. Bending stress

σ=

M c I

Given Nfd =

a 64 32 M = M  = 2 4 3 π a π a

π a 32

3




a  Find ( a ) Round this up to the next higher even value

a  124.874 mm a  125  mm

Using this value of a, the values of the functions of a are: Csize( a )  0.82

S e( a )  53.672 MPa

The realized safety factor is Nfb 

π a 32

3




Nfb  2.0



6-41-1


A 10-mm ID steel tube carries liquid at 7 MPa. The pressure varies periodically from zero to maximum. The steel has S ut = 400 MPa Determine the infinite-life fatigue safety factor for the wall if its thickness is: a) 1 mm, b) 5 mm.

Given:

Tensile strength

S ut  400  MPa

Assumption: The tubing is long therefore the axial stress is zero. The finish is machined, reliability is 99.999% and the tubing is at room temperature. Solution:

See Mathcad file P0641. t  1  mm

(a) Wall thickness is 1.

From Problem 4-41, this is a thick wall cylinder and the maximum principal stresses are:

σ1maxa  38.82  MPa 2.

σ2maxa  0  MPa

σ3maxa  7.00 MPa

Calculate the minimum, maximum, alternating, and mean von Mises effective stress using equation (5.7c).

σ'min  0  MPa σ'maxa 

2

σ1maxa  σ1maxa σ3maxa  σ3maxa

σ'maxa  σ'min

σ'aa 

σ'maxa  σ'min

σ'ma  21.376 MPa

2

S'e  0.5 S ut

3.


4.

Calculate the endurance limit modification factors for axial loading. Load Size

Cload  0.7 Csize  1

Surface

A  4.51

Csurf

5.

σ'maxa  42.752 MPa

σ'aa  21.376 MPa

2

σ'ma 

2

S'e  200 MPa

(axial loading) (axial loading) b  0.265

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


( machined )

b

Csurf  0.922

(R = 99.999%)


6.

S e  85.04 MPa

Assuming a Case 3 load line, the factor of safety from equation (6.18e) is

Na 

S e S ut

σ'aa S ut  σ'ma S e

Na  3.3


MACHINE DESIGN - An Integrated Approach, 4th Ed. t  5  mm

(b) Wall thickness is 7.


σ1maxb  11.67  MPa 8.

σ2maxb  0  MPa

σ3maxb  7.00 MPa


σ'maxb 

2

σ1maxb  σ1maxb σ3maxb  σ3maxb

σ'maxb  σ'min

σ'ab 

2

σ'mb  9.

6-41-2

σ'maxb  σ'min 2

2

σ'maxb  16.336 MPa

σ'ab  8.168 MPa

σ'mb  8.168 MPa

The endurance limit does not change from part a to b. Assuming a Case 3 load line, the factor of safety from equation (6.18e) is

Nb 

S e S ut

σ'ab S ut  σ'mb S e

Nb  8.6



6-42-1


A cylindrical tank with hemispherical ends is required to hold 150 psi of pressurized air at room temperature. The pressure cycles from zero to maximum. The steel has S ut = 500 MPa. Determine the infinite-life fatigue safety factor if the tank diameter is 0.5 m with 1 mm wall thickness, and its length is 1 m.

Given:

Tensile strength

Assumption:

The finish is machined, reliability is 99.999% and the tank is at room temperature.

Solution:


1.

S ut  500  MPa

From Problem 4-42, this is a thin wall cylinder and the maximum principal stresses are:

σ1max  259  MPa 2.

σ2max  129  MPa

σ3max  0  MPa


σ'min  0  MPa σ'max 

σ'a 

2


σ'max  σ'min

σ'max  224.301 MPa

σ'a  112.151 MPa

2

σ'max  σ'min

σ'm 

2

σ'm  112.151 MPa

2

S'e  0.5 S ut

3.


4.

Calculate the endurance limit modification factors for axial loading. Load

Cload  0.7

(axial loading)

Size

Csize  1

(axial loading)

Surface

A  4.51

b  0.265

 Sut    MPa 

5.

Ctemp  1

Reliability


( machined )

b

Csurf  A   Temperature

S'e  250 MPa

Csurf  0.869

(R = 99.999%)


6.

S e  100.2 MPa

Assuming a Case 3 load line, the factor of safety from equation (6.18e) is Nf 

S e S ut


Nf  0.74



6-43-1


The paper rolls in Figure P6-17 are 0.9-m OD by 0.22-m ID by 3.23-m long and have a density of 984 kg/m3. The rolls are transfered from the machine conveyor (not shown) to the forklift truck by the V-linkage of the off-load station, which is rotated through 90 deg by an air cylinder. The paper then rolls onto the waiting forks of the truck. The forks are 38-mm thick by 100-mm wide by 1.2-m long and are tipped at a 3-deg angle from the horizontal and have S ut = 600MPa. Find the infinite-life safety factor for the two forks on the truck when the paper rolls onto it under two different conditions (state all assumptions): (a) The two forks are unsupported at their free end. (b) The two forks are contacting the table at point A.

Given:

Tensile strength S ut  600  MPa Fork width Fork thickness

F

L fork

t

w  100  mm t  38 mm

Assumptions: 1. The greatest bending moment will occur when the paper roll is at the tip of the fork for case (a) and when it is midway between supports for case (b). 2. Each fork carries 1/2 the weight of a paper roll. 3. For case (a), each fork acts as a cantilever beam (see Appendix B-1(a)). 4. For case (b), each fork acts as a beam that is built-in at one end and simply-supported at the other. 5. The forks are machined, the reliability is 90%, and they operate at room temperature.

R1

M1

Case (a), Cantilever Beam

0.5 L fork

F

t

L fork R1

R2

M2


Solution:

See Figure 6-43 and Mathcad file P0643. FIGURE 6-43

1.

From Problem 4-43, the maximum stresses in the forks are:

Free Body Diagrams used in Problem 6-43

Case (a)

σmaxa  464.8  MPa

at the base of the fork.

Case (b)

σmaxb  87.2 MPa

also at the base of the fork.

Both cases

σmin  0  MPa

Since there are no other stress components present, these are also the maximum principal stresses and the von Mises stresses. This is a repeated load problem.

Case (a) 2.

The dynamic loading is repeated from 0 to 1 for each paper roll that is transfered. The alternating and mean components of the von Mises stress are: Alternating von Mises stress

σ'a  0.5  σmaxa  σmin

σ'a  232.4  MPa


σ'm  0.5  σmaxa  σmin

σ'm  232.4  MPa

S'e  0.5 S ut

S'e  300  MPa

3.


4.


Cload  1



Size

6-43-2

A95  0.05 w t

d equiv 

 d equiv  Csize  0.869     in  Surface

A  4.51

Csurf

5.

Temperature

Ctemp  1

Reliability


Csize  0.814

(machined)

b

Csurf  0.828

(R = 90%)


6.

0.0766

 0.097

b  0.265

 Sut   A     MPa 

A95

S e  181.357  MPa


Nfa 

S e S ut


Nfa  0.60

Case (b) 7.

8.

The dynamic loading is repeated from 0 to 1 for each paper roll that is transfered. The alternating and mean components of the von Mises stress are: Alternating von Mises stress

σ'a  0.5  σmaxb  σmin

σ'a  43.6 MPa


σ'm  0.5  σmaxb  σmin

σ'm  43.6 MPa


Nfb 

S e S ut


Nfb  3.2



6-44-1


Determine a suitable thickness for the V-links of the off-loading station of Figure P6-17 to limit th deflections at the tips to 10 mm in any position during their rotation. Two V-links support the roll at the 1/4 and 3/4 points along the roll's length and that each of the V-links is 10 cm wide by 1 m long. What is the infinite-life safety factor when designed to limit deflection as above? S ut = 600 MPa. See Problem 4-43 for more information.

Given:

Roll OD

OD  0.90 m

Arm width

wa  100  mm

Roll ID

ID  0.22 m

Arm length

La  1000 mm

Roll length

Lroll  3.23 m

Max tip deflection


Roll density

ρ  984  kg m

Mod of elasticity Tensile strength

E  207  GPa S ut  600  MPa

3

Assumptions: 1. The maximum deflection on an arm will occur just after it begins the transfer and just before it completes it, i.e., when the angle is either zero or 90 deg., but after the tip is no longer supported b the base unit. 2. At that time the roll is in contact with both arms ("seated" in the V) and will remain in that state throughout the motion. When the roll is in any other position on an arm the tip will be supported. 3. The arm can be treated as a cantilever beam with nonend load. 4. A single arm will never carry more than half the weight of a roll. 5. The pipe to which the arms are attached has OD = 160 mm. 6. The V-links are machined, reliability is 90%, and they operate at room temperature. Solution: 1.


Determine the weight of the roll and the load on each V-arm. W 

 4

π

2

2




F  0.5 W 2.

W  18.64  kN F  9.32 kN

From Appendix B, Figure B-1, the tip deflection of a cantilever beam with a concentrated load located at a distance a from the support is ymax =

F a

2

6  E I

 ( a  3  L)

where L is the beam length and I is the cross-section moment of inertia. In this case 3

I= 3.

w a t a 12

Setting ymax = δtip and a  370  mm,




3

ta  31.889 mm ta  32 mm



6-44-2

The maximum bending stress in the arm will be at its base where it joins the 160-mm-dia pipe. The bending moment, moment of inertia, and distance to the outside fiber at that point are: Mmax  F  a

Bending moment

Mmax  3.449  kN  m

Mmin  0  kN  m Ma 

1

Mm 

1

2

  Mmax  Mmin

Ma  1.725  kN  m

  Mmax  Mmin

Mm  1.725  kN  m

2

c  0.5 t a

Distance to n.a.

I 

Moment of inertia

wa ta

5

I  2.731  10  mm

M a c I Mm c

σmnom  5.

3

12

σanom 

Nom tensile stress

c  16 mm

I

4

σanom  101  MPa σmnom  101  MPa

Determine the stress concentration factors. Figure E-10 comes the closest to our situation. Assuming that the effective D/d-ratio is 2 and r/d is about 0.25, Kt  1.4. For a material with 3

ksi  10  psi

2

S ut  87 ksi

a  0.073  in

and, for the assumed value of r/d,

r  0.25 ta

r  8  mm

The notch sensitivity factor is

q 

1 1

q  0.885 a r

and the fatigue stress concentration factor is Kf  1  q   Kt  1  6.

Kf  1.35

Assuming that Kfm  Kf , the actual alternating and mean components of stress at the point where the V-link meets the central hub are

σa  Kf  σanom

σa  136.8  MPa

σm  Kfm σmnom

σm  136.8  MPa

7.

Since there are no other nonzero stress components at this point on the top of the arm, the von Mises stresses are and σ'a  σa σ'm  σm

8.

Determine the modified material strength. Unmodified endurance limit

S'e  0.5 S ut

Load

Cload  1

Size

A95  0.05 wa ta

S'e  300  MPa

A95  160  mm

2



d eq 

A95

9.

d eq  45.703 mm

0.0766

 deq  Csize  1.189     mm 

 0.097

 Sut    MPa 

Surface

Csurf  4.51 

Temperature

Ctemp  1

Reliability


Endurance limit

6-44-3


Csize  0.821  0.265

Csurf  0.828

(R = 90%) S e  182.8  MPa

Calculate the factor of safety. Using the distortion energy theory and the modified Goodman theory, the fatigue factor of safety for a V-link thickness of ta  32 mm is Nf 

S ut S e


Nf  1.0



6-45-1


Determine the infinite-life fatigue safety factor based on the tension load on the air-cylinder rod in Figure P6-17. The tension load cycles from zero to maximum (compression loads below the critical buckling load will not affect the fatigue life). The crank arm that it rotates is 0.3 m long and the rod has a maximum extension of 0.5 m. The 25-mm-dia rod is solid steel with S ut = 600 MPa. State all assumptions.

Given:


OD  0.90 m ID  0.22 m

Rod diameter Tensile strength

d  25 mm S ut  600  MPa

Lroll  3.23 m 3

ρ  984  kg m

Roll density

Assumptions: 1. The maximum force in the cylinder rod occurs when the transfer starts. 2. The cylinder and rod make an angle of 5.5 deg to the horizontal at the end of transfer. 3. The crank arm is 300 mm long and is 45 deg from vertical at the end of transfer. 4. The finish is machined, reliability is 90%, and the cylinder operates at room temperature. 5. The cylinder rod is fully retracted at the start of the transfer. At the end of the transfer it will have extended 500 mm from its initial position. Solution: 1.


Determine the weight of the roll on the forks. W 

 4

π

2

2

y




W  18.64 kN 2.

From the assumptions and Figure 6-45, the x and y distances from the origin to point A are, Rax  300  cos( 45 deg)  mm

450.0

Ray  300  sin( 45 deg)  mm

W Rx

Rax  212.132 mm

x 212.1

Ray  212.132 mm

Ry

A F

3.

From Figure 6-45, the x distance from the origin to point where W is applied is, Rwx 

4.

OD 2

5.5°

212.1

Rwx  450 mm

Sum moments about the pivot point and solve for the tensile force in the cylinder rod.

FIGURE 6-45 Free Body Diagram at End of Transfer for V-link of Problem 6-45

W  Rwx  Fmax Rax sin( 5.5 deg)  Fmax Ray cos( 5.5 deg) = 0 Fmax 

5.

W  Rwx Ray cos( 8  deg)  Rax sin( 8  deg)

Assume that the dynamic load is repeated so

Fmax  35.017 kN

tension




6-45-2

Determine the alternating and mean components of axial stress in the rod.

π d

A 

Area

Fm 

Mean load

σa 

Alternating stress

σm 

Mean stress

A  490.874 mm

4

Fa 

Alternating load

2

Fmax  Fmin

Fa  17.508 kN

2 Fmax  Fmin

Fm  17.508 kN

2 Fa

σa  35.668 MPa

A Fm

σm  35.668 MPa

A Nf =

S e S ut

7.


8.


σ'a = σa 9.

2


σ'm = σm


S'e  0.5 S ut

S'e  300 MPa

10. Calculate the endurance limit modification factors for an axial bar. Load

Cload  0.7

(axial loading)

Size

Csize  1

(axial loading)

Surface

A  4.51

Csurf

b  0.265

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


(machined)

b

Csurf  0.828

(R = 90%)


S e  155.95 MPa

12. Determine the factor of safety against fatigue failure for the assumptions made. Nf 

S e S ut

σa S ut  σm S e

Nf  3.5



6-46-1


The V-links of Figure P6-17 are rotated by the crank arm through a shaft that is 60 mm dia by 3.23 m long. Determine the maximum torque applied to this shaft during motion of the V-linkage and find the infinite-life fatigue safety factor for the shaft if its S ut = 600 MPa. See Problem 6-43 for more information. y

Given:

Tensile strength S ut  600  MPa Shaft diameter

d  60 mm

Assumptions: 1. The greatest torque will occur when the link is horizontal and the paper roll is located as shown in Figure P6-17 or Figure 6-46. 2. The V-links are machined, use a reliability of 90%, and operate at room temperature. Solution: 1.


W

From Problem 4-46, the maximum torsional stress in the shaft is

T

τmax  197.88 MPa

Ry 60-mm-dia shaft

2.

3.

Although not exactly true, assume that the load is fully reversed, then the minimum torque is τmin  197.88 MPa


Calculate the alternating component of the torsional stress in the shaft. Alternating stress

4.

450.0

τa 

τmax  τmin

τa  197.88 MPa

2

Convert this to the corresponding component of the von Mises stress. Alternating stress

σ'a 

3  τa

σ'a  342.738 MPa S'e  0.5 S ut

5.

Calculate the unmodified endurance limit using equation (6.5a).

6.

Calculate the endurance limit modification factors for a solid, round steel shaft. Load

Cload  1

Size

Csize  1.189  

Surface

A  4.51

   mm 

Csurf

d

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


S'e  300 MPa

 0.097

b  0.265

Csize  0.799

(machined)

b

Csurf  0.828

(R = 90%)




8.

6-46-2

S e  178.07 MPa

Calculate the factor of safety for the shaft.

Nf 

Se

σ'a

Nf  0.52



6-47-1


Determine the maximum forces on the pins at each end of the air cylinder of Figure P6-17. Determine the infinite-life fatigue safety factor in these pins if they are 30-mm dia and in single shear. S ut = 600 MPa. See Problem 6-43 for more information.

Given:

Pin diameter

d  30 mm

S ut  600  MPa

Tensile strength

Assumptions: 1. The maximum force in the cylinder rod occurs when the transfer starts. 2. The dynamic loading is fully reversed. 3. The finish is machined, reliability is 90%, and the pins are at room temperature. Solution:

See Figure 6-47 and Mathcad file P0647. y

W Rx

x 212.1

Ry

A F

8°

212.1 450.0


τmax  65.7 MPa

1.

From Problem 4-47 the maximum shear stress on the pins is

2.

This is the only stress component so the alternating von Mises stress is σ'a 

3.


4.

Calculate the endurance limit modification factors for a nonrotating direct shear. Size Load

A95 

S'e  0.5 S ut

π d

S'e  300 MPa

2

A95  706.858 mm

4

2

Cload  1 d equiv 

A95


0.0766


3  τmax σ'a  113.796 MPa

A  4.51

 0.097

b  0.265




Csurf

5.

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


6-47-2

b

Csurf  0.828

(R = 90%)


S e  170.12 MPa

6. Assuming a Case 3 load line, use equation (6.14) to determine the factor of safety. Nf 

Se

σ'a

Nf  1.5



6-48-1


Figure P6-18 shows an exerciser for a 100-kg wheelchair racer. The wheelchair has 65-cm-dia drive wheels separated by a 70-cm track width. Two free-turning rollers on bearings support the rear wheels. The lateral movement of the chair is limited by the flanges. Design the 1-m-lomg rollers as hollow tubes of aluminum (select alloy) to minimize the height of the platform and also limit the roller deflections to 1 mm in the worst case. Specify suitable sized steel axles to support the tubes on bearings. Calculate the fatigue safety factors at a life of 5E8 cycles.

Given:


T  700  mm

Aluminum

Ea  71.7 GPa

Roller length

Lr  1000 mm

Steel

Es  207  GPa

Assumptions: 1. The CG of the chair with rider is sufficiently close to the rear wheel that all of the weight is taken by the two rear wheels. 2. The small camber angle of the rear wheels does not significantly affect the magnitude of the forces on the rollers. 3. Both the aluminum roller and the steel axle are simply supported. The steel axles that support the aluminum tube are fixed in the mounting block and do not rotate. The aluminum tube is attached to them by two bearings (one on each end of the tubes, one for each axle). The bearings' inner race is fixed, and the outer race rotates with the aluminum tube. Each steel axle is considered to be loaded as a simply supported beam. Their diameter must be less than the inner diameter of the tubes to fit the roller bearings between them. 4. All surfaces are machined, reliability is 90%, and parts are at room temperature. Solution:

δ  1  mm

Maximum deflection Modulus elasticity

W/2

F

F 

 FIGURE 6-48A

Free Body Diagram of One Wheel used in Problem 6-48

See Figures 6-48 and Mathcad file P0648. W  M  g

W  980.7  N

1.

Calculate the weight of the chair with rider. Weight of chair

2.

Calculate the forces exerted by the wheels on the rollers (see Figure 5-48A). From the FBD of a wheel, summing vertical forces 2  F  cos( θ )  Let

θ  20 deg

W 2

=0 then

F 

W 4  cos( θ )

F  260.9  N

3.

The worst condition (highest moment at site of a stress concentration) will occur when the chair is all the way to the left or right. Looking at the plane through the roller that includes the forces exerted by the wheels (the FBD is shown in Figure 6-48B) the reactions R1 and R2 come from the bearings, which are inside the hollow roller and are, themselves, supported by the steel axle.

4.

Solving for the reactions. Let the distance from R1 to F be a  15 mm



 M1

R2 Lr  F  ( a  T )  F  a = 0

 Fy

R1  2  F  R2 = 0

R2 

F  (2 a  T )

700

F

F

R2  190.5  N

Lr

15

R1  2  F  R2 5.

6-48-2

R2

R1

R1  331.3  N

The maximum bending moment will be at the right-hand load and will be

1000

FIGURE 6-48B Free Body Diagram of One Tube used in Problem 6-48

Mrmax  R2 Lr  ( a  T )

Mrmax  54.3 N  m


Lr  T

Mc  39.1 N  m

2

and this would be constant along the axle between the two loads, F. 6.

Note that the bearing positions are fixed regardless of the position of the chair on the roller. Because of symmetry,

7.

Ra1  R1

Ra1  331.3  N

Ra2  R2

Ra2  190.5  N

1000

65 R1

R2

R a1

The maximum bending moment occurs at R1 and is for b  65 mm

R a2 1130

FIGURE 6-48C

Mamax  Ra1 b

Free Body Diagram of One Axle used in Problem 6-48

Mamax  21.5 N  m 8.

Determine a suitable axle diameter. Let the factor of safety against yielding in the axle be Nsa  3

9.

Tentatively choose a low-carbon steel for the axle, say AISI 1020, cold rolled with S y  393  MPa

10. At the top of the axle under the load R1 there is only a bending stress, which is also the von Mises stress. Set this stress equal to the yield strength divided by the factor of safety.

σ' =

32 Mamax

π d a

3

=

Sy Nsa 1


 32 Nsa Mamax  d a    π S y  


d a  15 mm

3

d a  11.875 mm


11. Suppose that bearing 6302 from Chapter 10, Figure 10-23, page 684 is used. It has a bore of 15 mm and an OD of 42 mm. Thus, the inside diameter of the roller away from the bearings where the moment is a maximum will be d i  40 mm. This will provide a 1-mm shoulder for axial location of the bearings. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

MACHINE DESIGN - An Integrated Approach, 4th Ed. 12. Design for a factor of safety of Nra  3. Tentatively choose

6-48-3

150

700 F

2024-T4 aluminum with

F

S ut  440  MPa and S'e5E8  138  MPa

F 15

F 1000

13. A point on the outside diameter of the roller will see completely reversed bending, which will also be the only nonzero principal stress. Thus,

σx =

Kf  Mrmax Z

= σ' =


Se Nfr

where Kf is the fatigue stress concentration at the shoulder and S e is the modified endurance limit. 14. Tentatively choose (these values arrived at by iteration): Outside diameter

d o  45 mm

Shoulder diameter

D  54 mm

Fillet radius

r  5  mm

15. Determine the fatigue stress concentration factor. From Figure E-2 and Table 6-6 for r do

D

 0.111

do

Kt  0.97098  

  d  o r

 1.2

 0.21796

Kt  1.57

4

S ut  6.38  10  psi 1

q  1

0.102

q  0.813

r in Kf  1  q   Kt  1 

Kf  1.46

16. Calculate the alternating von Mises stress component. I 

π  4 4   d o  d i  64

c  0.5 d o

σ'a 

4

I  7.563  10  mm

4

c  22.5 mm

Kf  Mrmax  c I

σ'a  23.6 MPa



6-48-4

17. Destermine the endurance limit at 5E8 cycles Load factor

Cload  1

Size factor

 do  Csize  1.189     mm 

 0.097

 Sut   4.51    MPa 

Surface factor as machined

Csurf

Reliability at 90%


Csize  0.822  0.265

Csurf  0.899

Modified endurance limit S e5E8  Cload  Csize Csurf  Creliab S'e5E8

S e5E8  91.44  MPa

18. Determine the factor of safety for repeated loading. Nfr 

S e5E8

Nfr  3.87

σ'a

19. The maximum deflection of the roller will occur when the chair is in the center of the roller. For this case the reactions are both equal to the loads, F. Using Figure B-2(a) in Appendix B, the maximum deflection is at the center of the roller and is for a  150  mm

x  0.5 Lr

x  500  mm

E  71.7 GPa ymax  2 

F 6  E I

  1 

a 3 3   x  ( x  a )   L r  a 2 2     a  3 a  Lr  2  Lr   x L  r

    

ymax  0.875  mm

This design meets the deflection requirement and has a reasonable factor of safety against fatigue failure while allowing sufficient space for the bearings. DESIGN SUMMARY Axle

Roller

Material


Material

2024-T4 aluminum

Diameter

d a  15 mm

Outside diameter

d o  45 mm

Length

1220 mm

Inside diameter

d i  40 mm

Shoulder dia

D  54 mm

Fillet radius

r  5  mm

Length

1040 mm

Center line spacing

c   d w  d o  sin( θ ) c  238  mm



6-49-1

PROBLEM 6-49

_____

Statement:

Figure P6-19 shows a machined pivot pin that is press-fit into part A and is slip fit in part B. If F = 100 lb, l = 2 in, and d = 0.5 in, what is the pin's safety factor against fatigue when made of SAE 1020 cold-rolled steel? The loading is fully reversed and a reliability of 90% is desired. There is a bending stress concentration factor Kt = 1.8 at the section where the pin leaves part A on the right-hand side.

Given:

Applied force

F  100  lbf

Material strength S y  57 ksi


l  2.00 in d  0.5 in

Beam length


L  0.5 l




2.

F

w  100.0 

L

lbf in


w L

Mmax  3.

2

Calculate the moment of inertia and distance to the extreme fiber of the pin. The nominal alternating bending stress in the beam is then found using equation 4.11b. I 

π d

4

I  3.068  10

64

c  0.5 d

3

4

 in

c  0.250  in

Mmax c

σanom  4.

Mmax  50.00  lbf  in

I

σanom  4074 psi

From Table 6-6, the Neuber constant for S ut  68 ksi is 1 2

a  0.096  in 5.

2

Using equation 6-13, the notch sensitivity for r  0.5 d is q 

1 1

6.

a  0.096  in

q  0.839 a r

The fatigue stress-concentration factor for Kt  1.8, from equation 6.11b, is Kf  1  q   Kt  1 

Kf  1.67



7.

6-49-2

Because the stress state in the pin is simple, uniaxial stress, the alternating principal stress is equal to the alternating tensile stress and is also equal to the alternating von Mises stress. Thus,


σa  6.81 ksi

σ'  σa S'e  0.5 S ut

S'e  34.0 ksi

8.


9.

Calculate the endurance limit modification factors for a non rotating round pin in bending. Load

Cload  1

Size

A95  0.010462 d

d equiv 

2

A95 0.0766

A95  2.615  10

3

2

 in

d equiv  0.185  in

This is less than the lower limit in equation 6.7b, so use Csize  1 (machined) A  2.7 b  0.265

Surface

Csurf

 S ut   A     ksi 

Temperature

Ctemp  1

Reliability


b

Csurf  0.883

(R = 90%)


S e  26.9 ksi

11. Using equation 6.14, calculate the factor of safety against a fatigue failure for this case of fully reversed bending.

Nf 

Se

σ'

Nf  4.0



6-50-1

PROBLEM 6-50

_____

Statement:

Figure P6-19 shows a machined pivot pin that is press-fit into part A and is slip fit in part B. If F = 100 N, l = 50 mm, and d = 16 mm, what is the pin's safety factor against fatigue when made of class 50 cast iron? The loading is fully reversed and a reliability of 90% is desired. There is a bending stress concentration factor Kt = 1.8 at the section where the pin leaves part A on the right-hand side.

Given:

Applied force

F  100  N


Total length, l

l  50 mm

Beam length

L  0.5 l

Pin dia d  16 mm




2.

F

w  4.0

L

N mm


Mmax  3.

w L 2

Calculate the moment of inertia and distance to the extreme fiber of the pin. The nominal alternating bending stress in the beam is then found using equation 4.11b. I 

π d

4

3

I  3.217  10  mm

64

c  0.5 d

σanom  4.

Mmax  1250 N  mm

4

c  8.000  mm

Mmax c I

σanom  3.108  MPa

Since this is a brittle material, so the full value of the geometric stress concentration factor Kt  1.8 will be applied to the nominal stress using equation 4.31.

5.

Because the stress state in the pin is simple, uniaxial stress, the alternating principal stress is equal to the alternating tensile stress and is also equal to the alternating von Mises stress. Thus,

σa  Kt σanom

σa  5.60 MPa

σ'  σa S'e  0.5 S ut

S'e  179.5  MPa

6.


7.

Calculate the endurance limit modification factors for a non rotating round pin in bending. Load

Cload  1



A95  0.010462 d

Size

d equiv 

A95

Csurf

2

d equiv  5.913  mm

0.0766

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


b

Csurf  0.949

(R = 90%)


9.

A95  2.678  mm

This is less than the lower limit in equation 6.7b, so use Csize  1 (machined) A  4.51 b  0.265

Surface

8.

2

6-50-2

S e  152.7  MPa

Using equation 6.14, calculate the factor of safety against a fatigue failure for this case of fully reversed bending Nf 

Se

σ'

Nf  27



6-51-1

PROBLEM 6-51

_____

Statement:

A component in the shape of a large sheet is to be fabricated from 7075-T651 aluminum, which has a fracture toughness Kc = 24.2 MPa-m0.5 and a tensile yield strength of 495 MPa. Determine the number of loading cycles that can be endured if the nominal stress varies from 0 to one half the yield strength and the initial crack had a total length of 1.2 mm. The values of the coefficient and exponent in equation 6.4 for this material are A = 5 x 10 -11 (mm/cyc) and n = 4.

Units:

cycle  1

Given:

Fracture toughness

Kc  24.2 MPa m

Yield strength

S y  495  MPa

Initial crack length

lo  1.2 mm

Coeff. and exponent

A  5  10

Solution: 1.

0.5

 11

 mm cycle

1

n  4


Calculate the minimum and maximum nominal stresses based on the yield strength and the stress level given in the problem statement.

σmin  0  MPa σmax  2.

Sy

σmax  247.5 MPa

2

Determine the value of the geometry factor  from the discussion in Section 5.3 for a plate with a central crack.

β  1 3.

Using equation 5.14b, calculate the critical crack length for this material at the maximum stress condition.

 Kc  a c     π  β  σmax  1

4.

2

a c  3.0 mm

Calculate the initial crack half-length. a o  0.5 l o

5.

a o  0.60 mm

Using equations 6.3, write the stress intensity factor range as a function of crack half-length.

ΔK ( a )  β  π a   σmax  σmin 6.

Integrate equation 6.4 to find the number of cycles to failure. 1  Nc    A     

ac

1

 ΔK ( a )   0.5   MPa m 

4

da

5

Nc  7.2  10 cycle

ao



6-52-1

PROBLEM 6-52

_____

Statement:

A component in the shape of a large sheet is to be fabricated from 4340 steel, which has a fracture toughness Kc = 98.9 MPa-m0.5. The sheets are inspected for crack flaws after fabrication, but the inspection device cannot detect flaws smaller than 5 mm. Determine the minimum thickness required for the sheet to have a minimum cycle life of 10 6 cycles (using fracture-mechanics criteria if its width is 400 mm and the load normal to the crack varies from 20 to 170 kN. The values of th coefficient and exponent in equation 6.4 for this material are A = 4 x 10 -9 (mm/cyc) and n = 3.

Units:

cycle  1

Given:

Fracture toughness

Kc  98.9 MPa m

Width of sheet Load

W  400  mm Fmin  20 kN

Coeff. and exponent

A  4  10

Cycles to failure

Nf  10  cycle

0.5

Fmax  170  kN

9

 mm cycle

The initial total crack length is lo  5  mm

Solution:


Write equations for the minimum and maximum nominal stresses as a function of the unknown thickness.

σmin( t)  2.

n  3

6

Assumption:

1.

1

Fmin

σmax( t) 

W t

Fmax W t

Determine the value of the geometry factor  from the discussion in Section 5.3 for a plate with a central crack.

β  1 3.

Using equation 5.14b, write an equation for the critical crack length as a function of t for this material at the maximum stress condition.

 Kc  a c( t)     π  β  σmax( t)  1

4.

2

Calculate the initial crack half-length. a o  0.5 l o

5.

a o  2.50 mm

Using equations 6.3, write the stress intensity factor range as a function of crack half-length and sheet thickness

ΔK ( a t)  β  π a   σmax( t)  σmin( t)  6.

Use equation 6.4 to find the minimum sheet thickness. First, guess a value: t  4  mm

Given

 Nf  A =      

ac( t )

1

 ΔK ( a t) 

n

da

t  Find ( t)

t  3.2 mm

 0.5   MPa m 

ao



6-53-1

PROBLEM 6-53

_____

Statement:

A closed, thin-wall cylinder is made from an aluminum alloy that has a fracture toughness of 38 MPa-m0.5 and has the following dimensions: length = 200 mm, OD = 84 mm, and ID = 70 mm. A 2.8-mm-deep semicircular crack is discovered on the inner diameter away from the ends, oriented along a line parallel to the cylinder axis. If the cylinder is repeatedly pressurized from 0 to 75 MPa, how many pressure cycles can it withstand? The values of the coefficient and exponent in equation 6.4 for this material are A = 5 x 10 -12 (mm/cyc) and n = 4. (Hint: the value of the geometry factor for a semicircular surface flaw is  = 2/and the crack grows in the radial direction).

Units:

cycle  1

Given:

Fracture toughness

0.5

Kc  38 MPa m

a o  2.8 mm

Initial crack depth

Solution: 1.

Cylinder dimensions Internal pressure

L  200  mm p min  0  MPa

Coeff. and exponent Geometry factor

A  5  10  mm cycle β  0.6367

 12

1

n  4


Calculate the nominal hoop stress (tangential direction, normal to cylinder axis) using equation 4.49a based on the pressure levels given in the problem statement. r  0.5 ID

r  35.0 mm

σmin  p min 

r

t  0.5 ( OD  ID)

t  7.0 mm

σmin  0.0 MPa

t

σmax  p max  2.

OD  84 mm ID  70 mm p max  75 MPa

r

σmax  375.0 MPa

t

Using equation 5.14b, calculate the critical crack length for this material at the maximum stress condition.

 Kc  a c     π  β  σmax  1

2

a c  8.1 mm

However, since this is larger than the wall thickness, failure will occur when the crack reaches the OD so a c  t 3.

a c  7.0 mm

Using equations 6.3, write the stress intensity factor range as a function of crack depth.

ΔK ( a )  β  π a   σmax  σmin 4.

Integrate equation 6.4 to find the number of cycles to failure. 1  Nc    A     

ac

1

 ΔK ( a )   0.5   MPa m 

4

da

6

Nc  1.34  10 cycle

ao



6-54-1

PROBLEM 6-54

_____

Statement:

A non rotating, hot-rolled, steel beam has a channel section with h = 64 mm and b = 127 mm. It is loaded in repeated bending with the neutral axis through the web. Determine its corrected fatigue strength with 90% reliability if it is used in an environment that has a temperature that is below 450C and has an ultimate tensile strength of 320 MPa.

Given:


S ut  320  MPa

Reliability Dimensions

R  0.90 h  64 mm

Solution: 1.


Calculate the uncorrected endurance limit using equation 6.5a. S'e  0.5 S ut

2.

b  127  mm

S'e  160.0 MPa

Determine the loading factor from equation 6.7a. Cload  1

3.

Determine the size factor from equations 6.7b and 6.7d, and Figure 6-25. Area stressed above 95% of max

A95  0.05 b  h

Equivalent diameter

d equiv 

Size factor

4.

 d equiv  Csize  1.189     mm 

A95  406.4 mm

A95

2

d equiv  72.8 mm

0.0766

 0.097

Csize  0.784

Determine the surface factor from equation 6.7e and Table 6-3. From Table 6-3

Surface factor

A  57.7

Csurf

 Sut   A     MPa 

b  0.718 b

Csurf  0.917

5.

Determine the temperature factor from equation 6.7f. Since T < 450C, Ctemp  1.

6.

Determine the reliability factor from Table 6-4, Creliab  0.897.

7.

Using equation 6.6, calculate the corrected endurance limit. S e  Cload  Csize Csurf  Ctemp Creliab S'e

S e  103.3 MPa



6-55-1

PROBLEM 6-55

_____

Statement:

A non rotating, machined, steel rod has a round section with d = 50 mm. It is loaded with a fluctuating axial force. Determine its corrected fatigue strength with 99% reliability if it is used in an environment that has a temperature below 450C and has an ultimate tensile strength of 480 MPa.

Given:


S ut  480  MPa


R  0.99 d  50 mm

Solution: 1.



2.

S'e  240.0 MPa

Determine the loading factor from equation 6.7a. Cload  0.70

3.

The size factor for an axially loaded member is Csize  1

4.


Surface factor

A  4.51

Csurf

 Sut   A     MPa 

b  0.265 b

Csurf  0.878

5.


6.


7.


S e  120.1 MPa



6-56-1

PROBLEM 6-56

_____

Statement:

A non rotating, cold-drawn, steel rod has a round section with d = 76 mm. It is loaded in repeated torsion. Determine its corrected fatigue strength with 99% reliability if it is used in an environment that has a temperature of 500C and has an ultimate tensile strength of 855 MPa.

Given:


S ut  855  MPa

Reliability Dimensions Temperature

R  0.99 d  76 mm T  500  C

Solution: 1.



2.

S'e  427.5  MPa


3.

Determine the size factor from equation 6.7b. Size factor

4.

   mm 

Csize  1.189  

 0.097

Csize  0.781

Determine the surface factor from equation 6.7e and Table 6-3. A  4.51

From Table 6-3

Surface factor

5.

d

Csurf

b  0.265

 Sut   A     MPa 

b

Csurf  0.754

Determine the temperature factor from equation 6.7f. Ctemp  1  0.0058 



T  450  C  C

 

6.


7.


Ctemp  0.710

S e  145.5  MPa



6-57-1

PROBLEM 6-57

_____

Statement:

A non rotating, ground, steel rod has a rectangular section with h = 60 mm and b = 40 mm. It is loaded in repeated bending. Determine its corrected fatigue strength with 99.9% reliability if it is used in an environment that has a temperature that is below 450C and has an ultimate tensile strength of 1550 MPa.

Given:


S ut  1550 MPa


R  0.999 h  60 mm

Solution: 1.

b  40 mm


Calculate the uncorrected endurance limit using equation 6.5a (S ut exceeds 1400 MPa). S'e  700  MPa

2.


3.

Determine the size factor from equations 6.7b and 6.7d, and Figure 6-25. Area stressed above 95% of max

A95  0.05 b  h

Equivalent diameter

d equiv 

Size factor

4.

 d equiv  Csize  1.189     mm 

A95  120.0 mm

A95

2

d equiv  39.6 mm

0.0766

 0.097

Csize  0.832


Surface factor

A  1.58

Csurf

 Sut   A     MPa 

b  0.085 b

Csurf  0.846

5.


6.


7.


S e  371.2 MPa



6-58-1

PROBLEM 6-58

_____

Statement:

A steel, grooved shaft similar to that shown in Figure C-5 (Appendix C) is to be loaded in bending Its dimensions are: D = 57 mm, d = 38 mm, r = 3 mm. Determine the fatigue stress- concentration factor if the material S ut = 1130 MPa.

Given:

Dimensions: D  57 mm Tensile strength S ut  1130 MPa

Solution:

See Figure C-5 and Mathcad file P0658.

1.

r  3  mm

The geometric stress-concentration factor is found from the equation in Figure C-5. For

D d

Kt  A  

 1.500 r

A  0.93894

b  0.32380

and

b

 d

2.

d  38 mm

Kt  2.14

The Neuber constant is found by linear interpolation of the values in Table 6-6. S ut  163.9  ksi 2

a 1  0.031  in a  3.

S ut  S 2 S1  S2

S 1  160  ksi

  a 1  a 2  a 2

2

a 2  0.024  in

S 2  180  ksi

0.5

a  0.030  in

Calculate the notch sensitivity using equation 6.13. q 

1 1

q  0.920 a r

4.

The fatigue stress-concentration factor can now be found from equation 6.11b. Kf  1  q   Kt  1 

Kf  2.05



6-59-1

PROBLEM 6-59

_____

Statement:

A steel shaft with a transverse hole similar to that shown in Figure C-8 (Appendix C) is to be loaded in torsion. Its dimensions are: D = 32 mm, d = 3 mm. Determine the fatigue stressconcentration factor if the material S ut = 808 MPa.

Given:

Dimensions: Tensile strength

Solution:


1.

D  32 mm S ut  808  MPa

d  3  mm

The geometric stress-concentration factor is found from an equation in Figure C-8. Although the maximum torsional stress is on the surface, it will be almost a maximum just below the surface so it will be conservative to use curve B.

2.

d

3

Kt  3.34

The Neuber constant is found by linear interpolation of the values in Table 6-6. However, since the loading is torsional, 20 ksi must be added to the value of S ut that is used in the table (see the text in Figure 6-36, Part 1). S utt  S ut  20 ksi 2

a 1  0.044  in a 

3.

d

 27.159 

2

d  30.231     D  D  D 4 5 6 d d d  393.19    650.39    15.451    D  D  D

Kt  3.9702  9.292 

S utt  S 2 S1  S2

S utt  137.2  ksi S 1  130  ksi

  a 1  a 2  a 2

2

a 2  0.039  in

S 2  140  ksi

0.5

a  0.040  in

Calculate the notch sensitivity using equation 6.13. Let r  0.5 d q 

1 1

q  0.857 a r

4.


Kf  3.00



6-60-1

PROBLEM 6-60

_____

Statement:

A hardened aluminum filleted flat bar similar to that shown in Figure C-9 (Appendix C) is to be loaded axially. Its dimensions are: D = 1.20 in, d = 1.00 in, r = 0.10 in. Determine the fatigue stress-concentration factor if the material S ut = 76 ksi.

Given:

Dimensions: Tensile strength

Solution:


1.

D d

Kt  A  

 1.200 r

r  0.100  in

A  1.03510

b  0.25084

and

b

 d

Kt  1.84

The Neuber constant is found by linear interpolation of the values in Table 6-8. 2

a 1  0.144  in a 

3.

d  1.00 in

The geometric stress-concentration factor is found from the equation in Figure C-9. For

2.

D  1.20 in S ut  76 ksi

S ut  S 2 S1  S2

S 1  70 ksi

  a 1  a 2  a 2

2

a 2  0.131  in

S 2  80 ksi

0.5

a  0.136  in

Calculate the notch sensitivity using equation 6.13. q 

1 1

q  0.699 a r

4.


Kf  1.59



6-61-1


A rotating shaft with a shoulder fillet seated in the inner race of a rolling contact bearing with the shoulder against the edge of the bearing is shown in Figure P6-20. The bearing has a slight eccentricity that induces a fully reversed bending moment in the shaft as it rotates. Measurements indicate that the resulting alternating stress amplitude due to bending is a = 57 MPa. The torque on the shaft fluctuates from a high of 90 N-m to a low of 12 N-m and is in phase with the bending stress. The shaft is ground and its dimensions are: D = 23 mm, d = 19 mm, and r = 1.6 mm. The shaft material is SAE 1040 cold-rolled steel. Determine the infinite-life fatigue safety factor for the shaft for a reliability of 99%.

Given:

Strength SAE 1040 CR

S ut  586  MPa Alternating bending stress σxa  57 MPa

Fluctuating torque

Tmax  90 N  m Tmin  12 N  m

Shaft dimensions

D  23 mm

Solution: 1.

Determine the mean and alternating components of the fluctuating torsional stress. c  J 


d

c  9.5 mm

2

π d

4

4

J  1.279  10  mm

32

τxymax 

Torsional stress

τxymin 

Tmax c J Tmin c J

τxymax  τxymin

τxym 

2

τxymax  τxymin

τxya 

2

4

τxymax  66.827 MPa τxymin  8.91 MPa τxym  37.869 MPa τxya  28.958 MPa

Using Appendix C, determine the geometric stress concentration factors for the bending and torsional stresses. Bending (Fig. C-2): For

D d

r

 1.211

Kt  A  

r

d

 0.084

Torsion (Fig. C-3): For

D d

 1.211

Kts  A  

A  0.97098

b  0.21796

b

 d

r

 d

3.

r  1.6 mm


Distance to outside fiber

2.

d  19 mm

Kt  1.665 r d

 0.084

A  0.83425

b  0.21649

b

Kts  1.425

Calculate the notch sensitivity of the material for bending and torsion using Table 6-6.



6-61-2

Bending: 2

Neuber constant (for S ut  586.0  MPa) 1

q b 

Notch sensitivity

a  0.075  in q b  0.77

a

1

r

Torsion: Neuber constant (for S ut  20 MPa  606.0  MPa) 1

q s 

Notch sensitivity

1 4.

5.

2

a  0.0585  in q s  0.811

a r

Calculate the fatigue stress concentration factors for bending and torsion using equation 6.11b. Bending

Kf  1  q b  Kt  1 

Kf  1.512

Torsion

Kfs  1  q s  Kts  1 

Kfs  1.345

Determine what, if any, fatigue stress concentration factor should be applied to the mean torsional stress. S y  490  MPa

Yield strength SAE 1040 CR Evaluate

Kfs  2  τxymax  179.8  MPa

which is less than S y so

Kfsm  Kfs

6.

7.

Calculate the mean and alternating components of the stresses increased by the appropriate fatigue stress concentration factors. Bending

σm  0  MPa

σa  Kf  σxa

σa  86.188 MPa

Torsion

τm  Kfsm τxym

τm  50.934 MPa

τa  Kfs τxya

τa  38.949 MPa

Find the mean and alternating von Mises stresses using equations 6.22b (with y = 0). 2

Mean

σ'm 

3  τm

Alternating

σ'a 

σa  3  τa

2

σ'm  88.22  MPa 2

σ'a  109.451  MPa

8. Calculate the unmodified endurance limit. S'e  0.5 S ut

S'e  293  MPa

9. Calculate the endurance limit modification factors for a rotating, round shaft. Load

Cload  1

(combined bending and torsion)



Size

Csize  1.189  

  mm  

Surface

A  1.58

d

6-61-3

 0.097

Csize  0.894 b  0.085

 Sut    MPa 

b


Ctemp  1

Reliability


(ground)

Csurf  0.919

(R = 99%)


S e  196  MPa

11. Assuming a Case 2 load line, determine the factor of safety against fatigue failure.

Nf 

Se



1 

σ'a 

σ'm  S ut

 

Nf  1.5



6-62-1


A tension member in a machine is filleted as shown in Figure P6-21. The member has a manufacturing defect that causes the fluctuating tension load to be applied eccentrically resulting in a fluctuating bending load as well. Measurements indicate that the maximum bending stress is 16.4 MPa and the minimum is 4.1 MPa. The tensile load fluctuates from a high of 3.6 kN to a low of 0.90 kN and is in phase with the bending stress. The member is machined and its dimensions are: D = 33 mm, d = 25 mm, h = 3 mm and r = 3 mm. The material is SAE 1020 cold-rolled steel. Determine the infinite-life fatigue safety factor for the member for a reliability of 90%.

Given:


S ut  380  MPa σxmax  16.4 MPa

Fluctuating tension

Fmax  3.6 kN

Dimensions Solution: 1.

D  33 mm

Fmin  0.90 kN

d  25 mm

r  3  mm

Determine the mean and alternating components of the fluctuating stresses.

σmb  σab 

Tension

σxmax  σxmin σxmax  σxmin

σat 

σab  6.150  MPa

2

σxtmax 

σmt 

σmb  10.250 MPa

2

σxtmin 

Fmax

σxtmax  48.000 MPa

h d Fmin

σxtmin  12.000 MPa

h d

σxtmax  σxtmin

σmt  30.000 MPa

2

σxtmax  σxtmin

σat  18.000 MPa

2

Using Appendix C, determine the geometric stress concentration factors for the bending and tensile stresses. Bending (Fig. C-10): For

D d

r

 1.32

Ktb  A  

d r

 0.12

Tension (Fig. C-9): For

D d

Ktt  A  

d r

 d

b  0.27269

Ktb  1.709 r

 1.32

A  0.95880

b

 d

3.

h  3  mm


Bending

2.

σxmin  4.1 MPa

 0.12

A  1.05440

b  0.27021

b

Ktt  1.87

Calculate the notch sensitivity of the material for bending and tension using Table 6-6.



6-62-2

Bending and tension: 2

Neuber constant (for S ut  55.1 ksi) Notch sensitivity

a  0.118  in 1

q 

1

4.

5.

Bending

Kfb  1  q   Ktb  1 

Kfb  1.528

Tension

Kft  1  q   Ktt  1 

Kft  1.648

Determine what, if any, fatigue stress concentration factor should be applied to the mean stresses.

Evaluate

S y  207  MPa

Kft   σxmax  σxtmax  106.1  MPa Kfbm  Kfb

which is less than S y so

Kftm  Kft

Calculate the mean and alternating components of the stresses increased by the appropriate fatigue stress concentration factors. Bending

Tension

7.

r

Calculate the fatigue stress concentration factors for bending and tension using equation 6.11b.

Yield strength SAE 1040 CR

6.

q  0.744 a

σmb  Kfb σmb

σmb  15.662 MPa

σab  Kfb σab

σab  9.397  MPa

σmt  Kftm σmt

σmt  49.427 MPa

σat  Kft  σat

σat  29.656 MPa

Find the mean and alternating von Mises stresses using equations 6.22b (with y = 0). Mean

σ'm  σmb  σmt

σ'm  65.089 MPa

Alternating

σ'a  σab  σmb

σ'a  25.06  MPa


S'e  190  MPa


Cload  0.70

Size

Csize  1

Surface

A  4.51

Csurf

 Sut   A     MPa 

b  0.265

(machined)

b

Csurf  0.934



Temperature

Ctemp  1

Reliability


6-62-3

(R = 90%)


S e  111.48 MPa

11. Assuming a Case 3 load line, determine the factor of safety against fatigue failure.

Nf 

S e S ut


Nf  2.5



6-63a-1


For a filleted flat bar in tension similar to that shown in Appendix Figure C-9 and the data from row a from Table P6-7, determine the alternating and mean axial stresses as modified by the appropriate stress concentration factors in the bar.

Given:


D  40 mm h  10 mm Pmin  8000 N

Widths Thickness Force Solution: 1.

Determine the nominal mean and alternating components of the fluctuating stresses. Pmax

σxm  σxa 

σxmax  160.0  MPa

h d Pmin

σxmin 

σxmin  40.0 MPa

h d

σxmax  σxmin

σxm  100.0  MPa

2

σxmax  σxmin

σxa  60.0 MPa

2

Using Appendix C-9, determine the geometric stress concentration factor. For

D d

r

2

d

Kt  A  

r

 0.2

A  1.09960

b  0.32077

b

 d

3.

d  20 mm Radius r  4  mm Pmax  32000  N

See Table P6-7 and Mathcad file P0663a.

σxmax 

2.

S ut  469  MPa

Kt  1.843

Calculate the notch sensitivity of the material using Table 6-6. 2

Neuber constant (for S ut  68 ksi) Notch sensitivity

q 

a  0.096  in 1

1 4.

r

Calculate the fatigue stress concentration factor using equation 6.11b. Kf  1  q   Kt  1 

5.

q  0.805 a

Kf  1.679

Determine what, if any, fatigue stress concentration factor should be applied to the mean stress. Yield strength SAE 1020 CR

S y  393  MPa

Evaluate



Kfm 

6-63a-2

S 1  Kf  σxmax S 2  Kf  σxmax  σxmin return Kf if S 1  S y return

S y  Kf  σxa

σxm

if  S 1  S y  S 2  2  S y

0 otherwise Kfm  1.679

6.

Calculate the mean and alternating components of the stresses increased by the appropriate fatigue stress concentration factors.

σm  Kfm σxm

σm  167.9  MPa


σa  100.7  MPa



6-64a-1


For a filleted flat bar in bending similar to that shown in Appendix Figure C-10 and the data from row a from Table P6-7, determine the alternating and mean bending stresses as modified by the appropriate stress concentration factors in the bar.

Given:


D  40 mm d  20 mm Radius r  4  mm h  10 mm Mmin  80 N  m Mmax  320  N  m

Widths Thickness Moment Solution: 1.


Determine the nominal mean and alternating components of the fluctuating stresses. c 

d

c  10 mm

2 Mmin c

σxmin 


h d

3

3

I  6.667  10  mm

12

Mmax c

σxmax  480  MPa

I

σxmax  σxmin

σxm  300.0  MPa

2

σxmax  σxmin

σxa  180.0  MPa

2

D d

r

2

d

Kt  A  

r

 0.2

A  0.93232

b  0.30304

b

 d

Kt  1.518



q 

a  0.096  in 1

1 4.

q  0.805 a r


5.

4


3.

I 

σxmin  120  MPa

I

σxmax 

2.

S ut  469  MPa

Kf  1.417

Determine what, if any, fatigue stress concentration factor should be applied to the mean stresses. Yield strength SAE 1020 CR

S y  393  MPa



6-64a-2

Evaluate Kfm 


S y  Kf  σxa

σxm

if  S 1  S y  S 2  2  S y

0 otherwise Kfm  0.460 6.



σm  137.9  MPa


σa  255.1  MPa



6-65a-1


For a shaft, with a shoulder fillet, in tension similar to that shown in Appendix Figure C-1 and the data from row a from Table P6-7, determine the alternating and mean axial stresses as modified by the appropriate stress concentration factors in the shaft.

Given:


D  40 mm h  10 mm Pmin  8000 N

Widths Thickness Force Solution: 1.

Determine the nominal mean and alternating components of the fluctuating stresses. 4  Pmax

π d σxmin 


σxmax  101.9  MPa

2

4  Pmin

π d


2

σxmax  σxmin

σxm  63.7 MPa

2

σxmax  σxmin

σxa  38.2 MPa

2


D d

r

2

d

Kt  A  

r

 0.2

A  1.01470

b  0.30035

b

 d

3.

d  20 mm Radius r  4  mm Pmax  32000  N


σxmax 

2.

S ut  469  MPa

Kt  1.645



q 

a  0.096  in 1

1 4.

r


5.

q  0.805 a

Kf  1.52


S y  393  MPa



Evaluate

Kfm 

6-65a-2


S y  Kf  σxa

σxm

if  S 1  S y  S 2  2  S y




σm  96.7 MPa


σa  58.0 MPa



6-66a-1


For a shaft, with a shoulder fillet, in bending similar to that shown in Appendix Figure C-2 and the data from row a from Table P6-7, determine the alternating and mean bending stresses as modified by the appropriate stress concentration factors in the shaft.

Given:


D  40 mm d  20 mm Radius r  4  mm h  10 mm Mmin  80 N  m Mmax  320  N  m

Widths Thickness Moment Solution: 1.


Determine the nominal mean and alternating components of the fluctuating stresses. c 

d

c  10 mm

2 Mmin c

σxmin 


π d

4

3

I  7.854  10  mm

64

Mmax c

σxmax  407.437  MPa

I

σxmax  σxmin

σxm  254.6  MPa

2

σxmax  σxmin

σxa  152.8  MPa

2

D d

r

2

d

Kt  A  

r

 0.2

A  0.90879

b  0.28598

b

 d

Kt  1.44



q 

a  0.096  in 1

1 4.

q  0.805 a r


5.

4


3.

I 

σxmin  101.859  MPa

I

σxmax 

2.

S ut  469  MPa

Kf  1.354


S y  393  MPa



Evaluate

Kfm 

6-66a-2


S y  Kf  σxa

σxm

if  S 1  S y  S 2  2  S y




σm  186.1  MPa


σa  206.9  MPa



6-67-1


A machine part is subjected to fluctuating, simple, multiaxial stresses. The fully corrected nonzero stress ranges are: σxmin = 50 MPa, σxmax = 200 MPa, σymin = 80 MPa, σymax = 320 MPa, τxymin = 120 MPa, τxymax = 480 MPa. The material properties are: S e = 525 MPa and S ut = 1200 MPa. Using a Case 3 load line, calculate and compare the infinite-life safety factors given by the Sines and von Mises Methods.

Given:

Material properties:

σxmin  50 MPa

Stresses:

Solution: 1.

S ut  1200 MPa

S e  525  MPa

σxmax  200  MPa

σymin  80 MPa

σymax  320  MPa

τxymin  120  MPa

τxymax  480  MPa


Calculate the alternating, and mean components of the given stresses.

σxa  σxm 

σya  σym  τxya  τxym 

σxmax  σxmin

σxa  75 MPa

2

σxmax  σxmin

σxm  125 MPa

2

σymax  σymin

σya  120 MPa

2

σymax  σymin

σym  200 MPa

2

τxymax  τxymin 2

τxymax  τxymin 2

τxya  180 MPa τxym  300 MPa

(a) Sines Method 2.

Calculate the equivalent alternating and mean stresses using equations 6.21b. 2

σ'a 

2

σxa  σya  σxa σya  3  τxya

2

σ'a  329.0 MPa

σ'm  σxm  σym 3.

σ'm  325.0 MPa

The factor of safety for the Sines Method, using equation (6.18e) is Nfs 

S e S ut

Nfs  1.11


(b) von Mises Method 4.

Calculate the equivalent alternating and mean stresses using equations 6.22b.

σ'a 

2

2

σxa  σya  σxa σya  3  τxya

2

σ'a  329.0 MPa



σ'm  5.

2

2

σxm  σym  σxm σym  3  τxym

σ'm  548.3 MPa

The factor of safety for the von Mises Method, using equation (6.18e) is Nfvm 

6.

2

6-67-2

S e S ut


Nfvm  0.92

This example shows that the von Mises method is more conservative when the endurance limit is modified by such factors as surface finish and when there is a high mean shear stress.



6-68-1


A cylindrical tank with hemispherical ends has been built. It was made from hot rolled steel that has S ut = 380 MPa. The tank outside diameter is 300 mm with 20 mm wall thickness. The pressure may fluctuate from 0 to an unknown maximum. For an infinite-life fatigue safety factor of 4 with 99.99% reliability, what is the maximum pressure to which the tank may be subjected?

Given:

Ultimate strength

S ut  380  MPa

Tank dimensions Reliability & FS

OD  300  mm R  0.9999

Solution: 1.

t  20 mm Nf  4


Determine the maximum principal stresses as functions of the unkown pressure, which occur at the inside wall, for this thick-wall cylinder: Outside radius

ro  0.5 OD

ro  150 mm

Inside radius

ri  ro  t

ri  130 mm

Tangential

σt( p ) 

Radial

2  ro    σr( p )  1 2 2 2 ri  ro  ri 

2   1  ro  2 2 2 ri  ro  ri  2

ri  p

2

ri  p

2

ri  p

σa( p ) 

Axial

2

ro  ri

2

These are principal stresses so, using equation 5.7a:

σ'( p ) 

2

2

2

σt( p )  σr( p )  σa( p )  σt( p )  σa( p )  σa( p )  σr( p )  σt( p )  σr( p )

σ'min  0  MPa 2.

Determine the alternating, and mean von Mises effective stress using equations 6.1.

σ'a( p ) 

σ'( p )  σ'min

σ'm( p ) 

2

σ'( p )  σ'min

S'e  0.5 S ut

3.


4.

Calculate the endurance limit modification factors for axial loading. Load

Cload  0.7

(axial loading)

Size

Csize  1

(axial loading)

Surface

A  57.7

Csurf Temperature

2

b  0.718

 Sut   A     MPa 

S'e  190  MPa

(hot rolled)

b

Csurf  0.811

Ctemp  1


MACHINE DESIGN - An Integrated Approach, 4th Ed. Reliability 5.


(R = 99.99%)


6.

6-68-2

S e  75.7 MPa

Assuming a Case 3 load line, solve equation (6.18e) for p max Guess

p  1  MPa

Given

Nf =

S e S ut

σ'a( p )  S ut  σ'm( p )  S e

p max  Find ( p ) 7.

Nf  4.00

p max  4.54 MPa

The principal stresses at p max are, respectively:

σ1  σt p max 

σ2  σa p max 

σ3  σr p max 

σ1  31.9 MPa

σ2  13.7 MPa

σ3  4.5 MPa



6-69-1


A rotating shaft has been designed and fabricated from SAE 1040 HR steel. It is made from tubing that has an outside diameter of 60 mm and a wall thickness of 5 mm. Strain gage measurements indicate that there is a fully reversed axial stress of 68 MPa and a torsional stress that fluctuates from 12 MPa to 52 MPa in phase with the axial stress at the critical point on the shaft. Determine the infinite-life fatigue safety factor for the shaft for a reliability of 99%.

Given:

Strength SAE 1040 HR

S ut  524  MPa Alternating bending stress σxa  68 MPa

Torsional stress

τxymax  52 MPa

τxymin  12 MPa

Shaft dimensions

OD  60 mm

t  5  mm

See Mathcad file P0669. Solution: 1. Calculate the alternating, and mean components of the given stresses.

2.

σxm  0  MPa τxymax  τxymin τxya 

σxa  68 MPa

τxym 

τxym  32 MPa

τxya  20 MPa

2 τxymax  τxymin 2

Find the mean and alternating von Mises stresses using equations 6.22b (with y = 0). 2

Mean

σ'm 

3  τxym

σ'm  55.426 MPa

Alternating

σ'a 

σxa  3  τxya

2

2

σ'a  76.315 MPa


S'e  262  MPa


Cload  1

Size

OD  Csize  1.189     mm 

Surface

A  57.7

Csurf

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


(combined bending and torsion)  0.097

Csize  0.799 b  0.718

(hot rolled)

b

Csurf  0.644

(R = 99%)


S e  110  MPa

6. Assuming a Case 3 load line, determine the infinite-life fatigue safety factor. Nf 

S e S ut


Nf  1.25



6-70a-1


For the data in row a in Table P6-8, find the safety factor for each of the four loading cases based on the Modified-Goodman diagram if S e = 100, S y = 150, and S ut = 200.

Given:

S e  100

S y  150

σ'm  50

σ'a  30

Solution: 1.

S ut  200

See Table 6-8 and Mathcad file P0670a.

Calculate the coordinates of point D in Figure 6-46 using equations (6.16). Dm 

S ut  S y  S e

Dm  100

S ut  S e

Da  S y  Dm 1.

Da  50

Use equations (6.18) and (6.16) to calculate the required quantities. Case 1: Nf1 



Sy

σ'a 

1 

 if σ'a  Da  S ut  σ'a  1   otherwise σ'm  Se  σ'm 

Nf1  2.40

Sy

Case 2: Nf2 

Se



1 

σ'm 

 if σ'm  Dm

Nf2  2.50

σ'a  S ut  S y  σ'm

otherwise

σ'a Case 3:

Nf3 

S e S ut


if

Se

σ'a σ'm

S y  σ'a

 Dm

Nf3  1.82

S ut

otherwise

σ'm Case 4:



Se

S ut  S e  S e σ'a  S ut σ'm 2

σ'mSut 

2

S e  S ut

σ'aSut  S e  OZ 

2

Se S ut

2

 σ'mSut 2

σ'a  σ'm



6-70a-2

 σ'm  σ'mSut 2  σ'a  σ'aSut2

ZSut 

Se

if σ'm 

σ'a σ'm



σ'mSy 

σ'm  σ'mSut2   σ'a  σ'aSut 2



Se S ut

otherwise

S y  σ'a  σ'm 2

σ'aSy  σ'mSy  σ'a  σ'm

 σ'm  σ'mSy2   σ'a  σ'aSy 2

ZSy 

Sy

if σ'm 

1 

Nf4 

σ'm  σ'mSy 2  σ'a  σ'aSy2

OZ  ZSut OZ OZ  ZSy OZ

if σ'mSut  Dm

σ'a σ'm

otherwise

Nf4  1.69

otherwise



6-70h-1


For the data in row h in Table P6-8, find the safety factor for each of the four loading cases based on the Modified-Goodman diagram if S e = 100, S y = 150, and S ut = 200.

Given:

S e  100

S y  150

σ'm  80

σ'a  80

Solution: 1.

S ut  200

See Table 6-8 and Mathcad file P0670h.

Calculate the coordinates of point D in Figure 6-46 using equations (6.16). Dm 

S ut  S y  S e

Dm  100

S ut  S e

Da  S y  Dm 1.

Da  50

Use equations (6.18) and (6.16) to calculate the required quantities. Case 1: Nf1 



Sy

σ'a 

1 

 if σ'a  Da  S ut  σ'a  1   otherwise σ'm  Se  σ'm 

Nf1  0.50

Sy

Case 2: Nf2 

Se



1 

σ'm 

 if σ'm  Dm

Nf2  0.75

σ'a  S ut  S y  σ'm

otherwise

σ'a Case 3:

Nf3 

S e S ut


if

Se

σ'a σ'm

S y  σ'a

 Dm

Nf3  0.83

S ut

otherwise

σ'm Case 4:



Se

S ut  S e  S e σ'a  S ut σ'm 2

σ'mSut 

2

S e  S ut

σ'aSut  S e  OZ 

2

Se S ut

2

 σ'mSut 2

σ'a  σ'm



6-70h-2

 σ'm  σ'mSut 2  σ'a  σ'aSut2

ZSut 

Se

if σ'm 

σ'a σ'm



σ'mSy 

σ'm  σ'mSut2   σ'a  σ'aSut 2



Se S ut

otherwise

S y  σ'a  σ'm 2

σ'aSy  σ'mSy  σ'a  σ'm

 σ'm  σ'mSy2   σ'a  σ'aSy 2

ZSy 

Sy

if σ'm 

1 

Nf4 

σ'm  σ'mSy 2  σ'a  σ'aSy2

OZ  ZSut OZ OZ  ZSy OZ

if σ'mSut  Dm

σ'a σ'm

otherwise

Nf4  0.84

otherwise



6-71-1


A rotating shaft with a shoulder fillet, in torsion similar to that shown in Appendix Figure C-3 is made from SAE 1020 CR steel and has dimensions D = 40 mm, d = 20 mm, and r = 4 mm. The shaft is ground and is subjected to a fully reversed torque of +/- 80 N-m. Determine the infinite life safety factor for the shaft for a reliability of 99.9%.

Given:


D  40 mm Ta  80 N  m

Dimensions Torque Solution: 1.

S ut  469  MPa d  20 mm Tm  0  N  m

r  4  mm


Determine the nominal mean and alternating components of the stresses.

τxym  0  MPa τxya 

16 Ta

π d 2.

τxya  50.9 MPa

3


D

Kt  A  

r

2

d

r

d

 0.2

A  0.86331

b

 d

3.

Kt  1.268

Calculate the notch sensitivity of the material using Table 6-6 adding 20 ksi to S ut because of the torsional load Neuber constant (for S ut  20 ksi  88 ksi) q 

Notch sensitivity

1 1

4.

6.

q  0.846 a r

Kf  1.226


τm  τxym

τm  0  MPa

τa  Kf  τxya

τa  62.46  MPa

Calculate the von Mises normal stress. von Mises stress

7.

2

a  0.072  in


5.

b  0.23865

σ'a 

3  τa

σ'a  108.19 MPa

Calculate the endurance limit modification factors for a rotating, solid, round steel shaft. Load

Cload  1

(pure torsion)



Size

Csize  1.189  

   mm 

Surface

A  1.58

Csurf

d

6-71-2

 0.097

Csize  0.889

b  0.085

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


(ground)

b

Csurf  0.937

(R = 99.9%) S'e  0.5 S ut


S'e  234.5  MPa 8.

Calculate the endurance limit. S e  Cload  Csize Csurf  Ctemp Creliab S'e

9.

S e  147.1  MPa

Using equation 6.14, calculate the infinite-life factor of safety.

Nf 

Se

σ'a

Nf  1.4



7-1-1


Two 3 x 5 cm blocks of steel with machined finish Ra = 0.6 mm are rubbed together with a normal force of 400 N. Estimate the true area of contact between them if their S y = 400 MPa.

Given:

Length of block Width of block

L  5  cm w  3  cm

Normal force Yield strength

F  400  N S y  400  MPa

Assumptions: The compressive yield strength is the same as the tensile yield strength. Then, S yc  S y. Solution: 1.


Using equation 7.1, the true area of contact is estimated as Ar 

3

F

Ar  3.33333  10

3  S yc

2.

The apparent area of contact is

3.

The ratio of apparent area to true area is

Aa  w L

2

cm

2

Aa  15 cm

Aa Ar

 4500



7-2-1


Given:

Estimate the dry coefficient of friction between the two pieces in Problem 7-1 if their S ut = 600 MPa. Length of block

L  5  cm

Yield strength

S y  400  MPa

Width of block

w  3  cm

Ultimate strength

S ut  600  MPa

Normal force

F  400  N

Assumptions: The compressive yield strength is the same as the tensile yield strength. Then, S yc  S y. Solution:


Using equations 7.3 and 7.4, the coefficient of friction is estimated as S us  0.80 S ut

μ 

S us 3  S yc

S us  480 MPa

μ  0.40



7-3-1


For the bicycle pedal-arm assembly in Figure P7-1 assume a rider-applied force that ranges from 0 to 400 N at the pedal each cycle. Determine the maximum contact stresses at one sprocket tooth-chain roller interface. Assume that the one tooth takes all the applied torque, that the chain roller is 8-mm dia, the sprocket has a nominal (pitch) dia of 100 mm, and that the sprocket tooth is essentially flat at the point of contact. The roller and sprocket are made of SAE X1340 steel, induction hardened to HRC 45-58. The roller and sprocket contact over a length of 8 mm. Assuming rolling plus 9% sliding, estimate the number of cycles to failure for this particular tooth-roller combination.

Given:

Roller radius

R1  4  mm

Sprocket radius

R2  ∞ mm

Sprocket width Pedal force

w  8  mm Frider  400  N

The parts are steel. Therefore: E  207  GPa

ν  0.28 Sprocket pitch dia

d p  100  mm

Pedal arm length len  170  mm Assumptions: The coefficient of friction is μ  0.33 Solution: 1.

2.


Determine the maximum contact force. Torque on sprocket

Tmax  Frider len

Contact force

Fcmax 

2  Tmax

Tmax  68 N  m Fcmax  1.36 kN

dp

Find the material constants from equation 7.9a. 2

Material constants

m1 

1ν

6

m1  4.452  10

E

1 MPa

m2  m1 Geometry constant

B 

1 2

 

1

 R1



1

 

1

B  125 m

R2  1

Contact patch half-width 3.

4.

a 

 2 m1  m2 Fcmax      B w  π

2

a  0.0878 mm

The average and maximum contact pressure can now be found from equations 7.14b and c. Fcmax

Average pressure

p avg 

Maximum pressure

p max 

Tangential pressure

fmax  μ  p max

2 a w 2  Fcmax

π a  w

p avg  968.1 MPa p max  1233 MPa fmax  406.8 MPa

With m = 0.33, the principal stresses in the contact zone will be maximal on the surface (z = 0) at x = 0.3a from th centerline as shown in Figures 7-20 and 7-22. The applied stress components are found from equation 7.23a for the normal force and equation 7.23b for the tangential force. For

x  0.3 a



σxn  p max  1 

x

2

a

σxt  2  fmax 

σxn  1176 MPa

2

x

σxt  244.05 MPa

a

σzn  p max  1 

x

2

a

σzn  1176 MPa

2

σzt  0  MPa τxzt  fmax  1 

τxzn  0  MPa x

2

a 5.

6.

τxzt  388.0 MPa

2

Equations 7.24a and 7.24b can now be solved for the total applied stresses along the x, y, and z axes.

σx  σxn  σxt

σx  1420 MPa

σz  σzn  σzt

σz  1176 MPa

τxz  τxzn  τxzt

τxz  388.019 MPa

Assuming the rollers are short, we expect a plane stress condition to exist. The stress in the third dimension is then:

σy  0  MPa 7.

7-3-2

also,


τyz  0  MPa

Unlike the pure-rolling case, these stresses are not principal because of the applied shear stress. The principal stresses are found from equation 4.4 using a cubic root finding solution. C2 

σx  σy  σz

C2  2.596  10

MPa

τxz  τyz   σx τxy   σx  σy       MPa MPa  MPa MPa  MPa MPa     C1     τxy  τxz σz   τyz σz  σy   MPa MPa   MPa MPa   MPa MPa       

C1  1.519  10

 σx τxy τxz   MPa MPa MPa   τxy σy τyz  C0     MPa MPa MPa   τxz τyz σz   MPa MPa MPa   

C0  0

3

3

6

2

f ( σ )  σ  C2 σ  C1 σ  C0



 C0    C1   v   C2     1 

r  polyroots ( v)  MPa

 1704.6  r   891.1  MPa    0.0 

Principal stresses:

σ1  r

σ1  0 MPa

σ2  r

σ2  891.1 MPa

σ3  r

σ3  1704.6 MPa

3 2 1

8.

The maximum normal stress calculated in step 7 is 1705 MPa, compressive. Its K-factor can be calculated from equation 7.25d. K  π  m1  m2  σ3

2

K-factor 9.

7-3-3

K  81.3 MPa

From Table 7-7, Part 1, Line 4 the slope and intercept factors of this cast iron for rolling with 9% sliding are

λ  8.51

ζ  41.31

10. These are used in equation 7.26 along with the value of K from above to find the number of cycles that can be expected at this load before pitting begins. log ( K) =

ζ  log  Nlife λ

K   psi 

ζ λ log

Nlife  10

6

Nlife  4.6  10

cycles



7-4-1


For the trailer hitch from Problem 3-4 on p. 169, determine the contact stresses in the ball and ball cup. Assume that the ball is 2-in dia and the ill-fitting cup that surrounds the it is an internal spherical surface 10% larger in diameter than the ball.

F Given:

Solution:

Ball diameter Cup diameter Pull force

d  2  in D  2.2 in Fpull  4.905  kN

Tongue weight

Wtong  0.981  kN

Poisson's ratio

ν  0.28


E  30.0 10  psi

6


Total force

2

F 

Fpull  Wtong

2

FIGURE 7-4 Diagram Showing Contact Force for Problem 7-4

F  1125 lbf Ball radius

R1  0.5 d

R1  1.000 in

Cup radius

R2  0.5 D

R2  1.100 in

Geometry constant

B 

1 2

 

1



 R1

  R2 

1

1

B  0.045 in

2

Material constants

m1 

1ν

8 1

m1  3.072  10

E

psi

m2  m1 1

Contact patch radius

a 

Contact area

A  π a

Average pressure

p avg 

F

p max 

3

Maximum pressure

 3 m1  m2  F   B 8 

3

2

2

A  0.022 in

p avg  52.1 ksi

A

2

a  0.0829 in

 p avg

p max  78.1 ksi

Stresses Axial

σzmax  p max

In-plane

σxmax  

1  2 ν 2

σzmax  78.1 ksi  p max

σxmax  60.9 ksi

σymax  σxmax



Max shear stress

τyzmax 

p max 2

 



7-4-2 1  2 ν 2



2 9

 ( 1  ν)  2  ( 1  ν)



τyzmax  26.4 ksi Depth at max shear stress

zτmax  a 

2  2 ν 7  2 ν

zτmax  0.05228 in



7-5-1


Given:

Solution:

For the trailer hitch from Problem 3-5 on p. 169, determine the contact stresses in the ball and ball cup. Assume that the ball is 2-in dia and the ill-fitting cup that surrounds the it is an internal spherical surface 10% larger in diameter than the ball. Ball diameter Cup diameter Trailer mass

d  2  in D  2.2 in mtrail  2000 kg

Terminal velocity Time to reach vel. Tongue weight

v  60 m sec t  20 sec Wtong  0.981  kN

Poisson's ratio

ν  0.28


E  30.0 10  psi

1

F

6

See Figure 7-5 and Mathcad file P0705. Acceleration Pull force

v

a 

m

a  3.00

t

sec

Fpull  mtrail a

2

FIGURE 7-5 Diagram Showing Contact Force for Problem 7-5

Fpull  6.00 kN 2

Fpull  Wtong

2

Total force

F 

Ball radius

R1  0.5 d

R1  1.000 in

Cup radius

R2  0.5 D

R2  1.100 in

Geometry constant

B 

1 2

 

1



 R1

F  1367 lbf

  R2 

1

1

B  0.045 in

2

Material constants

m1 

1ν

8 1

m1  3.072  10

E

psi

m2  m1 1


a 

Contact area

A  π a

Average pressure

p avg 

F

p max 

3

Maximum pressure

 3 m1  m2  F   B 8  2

a  0.0885 in 2

A  0.025 in

p avg  55.6 ksi

A

2

3

 p avg

p max  83.3 ksi

Stresses Axial


σzmax  83.3 ksi


MACHINE DESIGN - An Integrated Approach, 4th Ed. In-plane

σxmax  

7-5-2

1  2 ν 2

 p max


σymax  σxmax Max shear stress

τyzmax 

p max 2

 



1  2 ν 2



2 9

 ( 1  ν)  2  ( 1  ν)




zτmax  a 

2  2 ν 7  2 ν

zτmax  0.05579 in



7-6-1


For the trailer hitch from Problem 3-6 on p. 169, determine the contact stresses in the ball and ball cup. Assume that the ball is 2-in dia and the ill-fitting cup that surrounds the it is an internal spherical surface 10% larger in diameter than the ball.

Given:

Ball diameter Cup diameter Trailer mass

d  2  in D  2.2 in mtrail  2000 kg

Tongue weight

Wtong  0.981  kN

Poisson's ratio

ν  0.28


E  30.0 10  psi

Solution:

F

6


From Problem 3-6, the impact (pull) force is Pull force

Fpull  55.1 kN

Total force

F 

2

Fpull  Wtong

FIGURE 7-6

2

Diagram Showing Contact Force for Problem 7-6

F  12389 lbf Ball radius

R1  0.5 d

R1  1.000 in

Cup radius

R2  0.5 D

R2  1.100 in

Geometry constant

B 

1 2

 

1



 R1

  R2 

1

1

B  0.045 in

2

Material constants

m1 

1ν

8 1

m1  3.072  10

E

psi

m2  m1 1


a 

Contact area

A  π a

Average pressure

p avg 

F

p max 

3

Maximum pressure

 3 m1  m2  F   B 8 

3

2

2

A  0.107 in

p avg  115.9 ksi

A

2

a  0.1845 in

 p avg

p max  173.8 ksi

Stresses Axial


In-plane

σxmax  

1  2 ν 2

σzmax  173.8 ksi

 p max


σymax  σxmax © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P0706.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


Max shear stress

τyzmax 

p max 2

 



7-6-2

1  2 ν 2



2 9

 ( 1  ν)  2  ( 1  ν)




zτmax  a 

2  2 ν 7  2 ν

zτmax  0.11632 in



7-7-1


For the 12-mm dia steel wrist pin of Problem 3-7, find the maximum contact stress if the 2500 g acceleration is fully reversed. The aluminum piston has a hole for the wrist pin that is 2% larger than the pin and an engagement length of 20 mm.

Given:

Wrist pin dia Piston hole multiplier

d  12 mm k  1.02

Acceleration

a  2500 g

E1  206.8  GPa

M  0.5 kg

Aluminum piston ν2  0.34

Piston mass

L  2  cm

Length of contact Solution:

Material properties: Steel wrist pin ν1  0.28

E2  71.7 GPa

See Mathcad file P0707. Contact force

F  M  a

F  12.26 kN

Pin radius

R1  0.5 d

R1  6 mm

Hole radius

R2  k R1

R2  6.12 mm

Geometry constant

B 

1 2

 

1

 R1



  R2  1

B  1.63399  10

2

Material constants

m1 

1  ν1

6

m1  4.456  10

E1 2

m2 

3

1  ν2 E2

1

1 MPa

5

m2  1.233  10

mm

1 MPa

1

Contact patch half-width

 2 m1  m2 F  a      B L π

Contact area

A  2  L a

Average pressure

p avg 

Maximum pressure

p max 

F

2

a  2.002 mm A  80.10 mm

2

p avg  153.0 MPa

A 2 F

π a  L

p max  194.9 MPa

Wrist pin 1.

2.

The maximum normal stresses in the center of the contact patch at the surface of the steel wrist pin are found using equations 7.17a. Axial


σzmax  194.9 MPa

In-plane

σy1max  2  ν1 p max

σy1max  109.1 MPa

The maximum shear stress and its location under the surface are found from equations 7.17b. Max shear stress

τ13max  0.304  p max

τ13max  59.2 MPa

Depth at max shear stress

zτmax  0.786  a

zτmax  1.57 mm



7-7-2

Piston hole 3.

4.

The maximum normal stresses in the center of the contact patch at the surface of the aluminum piston hole are found using equations 7.17a. Axial



In-plane



The maximum shear stress and its location under the surface are found from equations 7.17b. Max shear stress

τ13max  0.304  p max


zτmax  0.786  a

τ13max  59.2 MPa zτmax  1.57 mm



7-8-1


A paper mill processes rolls of paper having a density of 984 kg/m3. The paper roll is 1.50 m outside dia (OD) by 0.22 m inside dia (ID) by 3.23 m long and has an effective modulus of elasticity in compression of 14 MPa and n = 0.3. Determine the width of its contact patch when it sits on a flat steel surface, loaded by its own weight.

Given:

Paper properties

3

ρ1  984  kg m

Roll dimensions:

E1  14 MPa

Outside diameter OD  1.50 m

ν1  0.3 E2  207  GPa

Steel properties

ID  0.22 m

Inside diameter

L  3.23 m

Lemgth

ν2  0.28 Solution: 1.


The weight of the paper roll is equal to its volume times the paper density times g. Wroll 

2.

3.

 4

π

2

2



 OD  ID  L ρ1  g

Wroll  53.895 kN

Determine the radii of the contacting bodies. Roll

R1  0.5 OD

Plate

R2  ∞ mm

R1  750 mm

Determine the geometry and material constants. Geometry constant

B 

1 2

 

1

 R1



1

 

1

B  0.667 m

R2 

2

Material constants

m1 

1  ν1

m1  0.065

E1 2

m2 

4.

1  ν2

1 MPa 6

m2  4.452  10

E2

1 MPa

Calculate the contact patch hakf-width and width. 1


 2 m1  m2 Wroll  a      B L  π

Contact patch width

width  2  a

2

a  32.1832 mm width  64.4 mm



7-9-1


For the ViseGrip plier-wrench for which the forces were analyzed in Problem 3-9, find the force in link 4 needed to create a 0.25-mm-wide flat on each side of a 2-mm dia aluminum pin squeezed in its 5-mm-wide jaws.

Given:

Pin radius

R1  1  mm

Anvil curvature

R2  ∞ mm

Steel anvil

Contact patch

a  0.125  mm

E2  206.8  GPa

Contact length

L  5  mm

Aluminum pin


ν2  0.28 ν1  0.34

E1  71.7 GPa Solution: 1.


Determine the geometry and material constants. Geometry constant

B 

1 2

 

1

 R1



1

 

R2 

1

B  500 m

2

Material constants

m1 

1  ν1 E1

5

m1  1.233  10

2

m2 

1  ν2 E2

MPa

6

m2  4.456  10

1

1 MPa

2

F 

π  a  B L

2.

Calculate the contact force

3.

Get the geometry from Problem 3-9 and calculate the pin force.

2   m1  m2

α  21.0 deg

F  3.654 kN

β  129.2  deg F

F4  sin( 180  deg  α) 

cos( 180  deg  α) cos( β  180  deg)

 sin( β  180  deg)

F4  4.65 kN



7-10-1


An overhung diving board is shown in Figure P7-4a. A 100-kg person is standing on the free end. The board sits on a fulcrum that has a cylindrical contact surface of 5-mm radius. What is the size of the contact patch between the board and the fulcrum if the board material is fiberglass with E = 10.3 GPa and n = 0.3?

Given:

Fulcrum radius

R1  5  mm

Board curvature

R2  ∞ mm

Mass of person Material properties: Aluminum fulcrum

M  100  kg

2000 = L R1

ν1  0.34 E1  71.7 GPa

Fiberglass board

R2

ν2  0.30 E2  10.3 GPa

700 = a

Board dimensions: Width (Prob 4-10) w  305  mm Distance to right support a'  0.7 m Contact length L  2  m Solution:

P


See Figure 7-10 and Mathcad file P0710. Weight of person

P  M  g

P  0.981 kN

Summing moments about the support on the left end of the board, Fulcrum reaction

Geometry constant

F  P B 

1 2

L

F  2.802 kN

a'

 

P L  F  b = 0

1

 R1



1

 

B  0.100 mm

R2 

1

2

Material constants

m1 

1  ν1

5

m1  1.233  10

E1 2

m2 

1  ν2

5

m2  8.835  10

E2

1 MPa 1 MPa

1

Contact patch half-width Contact patch width

 2 m1  m2 F  a      B w π a2  2  a

2

a  0.0767 mm a2  0.153 mm



7-11-1


Repeat Problem 7-10 assuming the 100-kg person in Problem 7-10 jumps up 25 cm and lands back on the board. Assume the board weighs 29 kg and deflects 13.1 cm statically when the person stands on it. What is the size of the contact patch between the board and the 5-mm-radius aluminum fulcrum if the board material is fiberglass with E = 10.3 GPa and n = 0.3?

Given:

Fulcrum radius

R1  5  mm

Board curvature

R2  ∞ mm

Mass of person Material properties: Aluminum fulcrum

M  100  kg

2000 = L R1

ν1  0.34 E1  71.7 GPa

Fiberglass board

R2

ν2  0.30 E2  10.3 GPa

700 = a

Board dimensions: Width (Prob 4-10) w  305  mm Distance to right support a'  0.7 m Contact length L  2  m Solution: 1.

P



From Problem 3-11, the dynamic load resulting from the impact of the person with the board is P  3.056  kN Summing moments about the support on the left end of the board, Fulcrum reaction

Geometry constant

F  P 1

B 

2

L

F  8.731 kN

a'

 

P L  F  b = 0

1

 R1



  R2  1

B  0.100 mm

1

2

Material constants

m1 

1  ν1

5

m1  1.233  10

E1 2

m2 

1  ν2

5

m2  8.835  10

E2

1 MPa 1 MPa

1


a 

Contact patch width

a2  2  a

 2 m1  m2 F      B w π

2

a  0.1355 mm a2  0.271 mm



7-12-1


Estimate the volume of adhesive wear to expect from an HB270 steel shaft of 40-mm diameter rotating at 250 rpm for 10 years in a plain bronze bushing if the transverse load is 1000 N. (a) For conditions of poor lubrication. (b) For conditions of good lubrication.

Units:

rpm  2  π rad min

Given:

Journal diameter Journal speed

1

Journal hardness

kilo  kg g d  40 mm n  250  rpm kilo HB  270  2 mm

yr  260  day Load Life

rev  2  π rad F  1000 N Life  10 yr

Assumptions: The bushing is softer than the journal (shaft) so the wear will take place predominately on the bushing. Solution: 1.


From Figure 7-6 in the text we see that iron (the principal ingredient of steel) is metalurgically compatible with copper and tin, the principal ingredients of bronze. From Figure 7-7 we have the following values of adhesive wear coefficient for metalurgically compatible and poor and good lubrication, respectively. Adhesive wear coefficients

4

Ka  10

5

Kb  10 The length of sliding is 2.

π d rev

 n  Life

good lubrication 8

L  1.176  10 m

Using equation 7.7a we can estimate the volume of wear for each lubrication condition (a) poor lubrication

(b) good lubrication

3.

L 

poor lubrication

Va  Ka

Vb  Kb

F L HB F L HB

6

3

5

3

Va  4.4  10 mm

Vb  4.4  10 mm

These numbers are entirely too large. The bushing will fail long before the ten years have gone by.



7-13-1


Estimate how long it will take to file 1 mm off a 2-cm cube of HB150 steel if the machinist applies 100 N over a 10-cm stroke at 60 strokes per minute. (a) If done dry. (b) If done lubricated.

Units:

kilo  kg g

Given:

Cube dimension Depth of wear

a  2  cm d  1  mm

Stroke rate Stroke

Force on file

F  100  N

Steel hardness

1

n  60 min s  10 cm kilo HB  150  2 mm

Assumptions: Only one face will be filed. Solution: 1.


This is a two-body abrasion problem. From Table 7-2, the wear coefficients for dry and lubricated abrasion using a file are Wear coefficients

Ka  5  10 Kb  1  10

2

dry

1

lubricated

2.

The length of sliding is L = s n  t where t is the time required.

3.

The area of a face is Aa = a , and the depth is d = K

4.

Combining these three equations and solving for the time, t

F L

2

(a) dry

ta 

d  HB a

HB Aa

2

Ka F  s n

strokesa  ta n

(b) lubricated

tb 

d  HB a

ta  20 min strokesa  1177

2

Kb F  s n

strokesb  tb n

tb  10 min strokesb  588



7-14-1


Figure P7-5 shows a child's toy called a pogo stick. The child stands on the pads, applying half her weight on each side. She jumps off the ground, holding the pads up against her feet, and bounces along with the spring cushioning the impact and storing energy to help each rebound. Estimate the abrasivbe wear rate for the tip, which impacts the ground assuming a condition of dry, loose abrasive grains (sand). Express the wear rate in number of jumps to remove 0.02 in from the 1-in-dia aluminum tip if its S ut = 50 ksi.

Units:

kilo  kg g

Given:

Tip diameter

d  1.00 in

Wear depth

a  0.02 in

Tensile strength


Assumptions: The tip slides a distance of one tip diameter per jump. Solution:


1. From Problem 3-14, the impact force for each jump is P  224  lbf 2. Use equation (2.10) to estimate the Brinell hardness of the steel tip. HB 

 Sut  kilo  530  psi  2 mm 1



HB  94

Fi /2

Fi /2

kilo mm

2

3. This is a two-body abrasion problem. From Table 7-2, the wear coefficient for dry abrasion using loose abrasive grains is 3

K  1  10

4. The length of sliding (see assumption) is

L = d  Njumps

where d is the tip diameter and Njumps is the number of jumps.

5. The area of the tip is

and the wear depth is

Aa 

π d

a = K

2

4

2

Aa  0.785 in

P FIGURE 7-14 Free Body Diagram for Problem 7-14

P L HB Aa

6. Substituting for L and solving for the number of jumps,

Njumps 

a  HB Aa K d P

Njumps  9410



7-15-1


Create a table of acceptable materials to run against a steel shaft based on their metallurical compatibility. Rank them as to suitability.

Solution:

See Figure 7-6 in text.



7-16-1


Determine the size of the contact patch and the maximum contact stresses for a 20-mm-dia steel ball rolled against a flat aluminum plate with 1 kN.

Given:

Ball radius

R1  10 mm

Plate curvature

R2  ∞ mm

Load

F  1  kN

Material properties Steel ball

ν1  0.28 E1  206.8  GPa

ν2  0.34

Aluminum plate

E2  71.7 GPa Solution:


1. Calculate geometry and material constants, contact patch dimension, and pressures. Geometry constant

B 

1 2

 

1

 R1



  R2  1

B  0.05 mm

1

2

Material constants

m1 

1  ν1

6

m1  4.456  10

E1 2

m2 

1  ν2

1 MPa

5

m2  1.233  10

E2

1 MPa

1


 3 m1  m2  a    F B 8 

Contact area

A  π a

Average pressure

p avg 

F

p max 

3

Maximum pressure

3

2

A  0.789 mm

2

p avg  1267 MPa

A

2

a  0.501 mm

 p avg

p max  1900 MPa

2. Determine the stresses in the ball at the surface Axial


In-plane

σxmax1  

σzmax  1900 MPa

1  2  ν1 2

 p max

σxmax1  1482 MPa

σymax1  σxmax1 3. Determine the stresses in the ball below the surface Max shear stress

τyzmax1 

 1  2 ν1 2     1  ν1  2   1  ν1 9 2 2  

p max



τyzmax1  641.5 MPa




zτmax1  a 

2  2  ν1 7  2  ν1

7-16-2

zτmax1  0.316 mm

4. Determine the stresses in the plate at the surface Axial


In-plane

σxmax2  


1  2  ν2 2

 p max


σymax2  σxmax2 5. Determine the stresses in the plate below the surface

Max shear stress

τyzmax2 

 1  2 ν2 2     1  ν2  2   1  ν2 9 2 2  

p max



τyzmax2  615.2 MPa Depth at max shear stress


2  2  ν2 7  2  ν2




7-17-1


Determine the size of the contact patch and the maximum contact stresses for a 20-mm-dia steel ball rolled against a 30-mm diameter aluminum ball with 800 N.

Given:

Steel ball radius

R1  10 mm


Aluminum ball radius

R2  15 mm

Steel ball

Load

F  0.8 kN

ν1  0.28 E1  206.8  GPa

Aluminum ball

ν2  0.34 E2  71.7 GPa

Solution: 1.


Calculate geometry and material constants, contact patch dimension, and pressures. Geometry constant

1

B 

2

 

1

 R1



1

 

B  0.083 mm

R2 

1

2

Material constants

m1 

1  ν1

6

m1  4.456  10

E1 2

m2 

1  ν2

1 MPa

5

m2  1.233  10

E2

1 MPa

1


a 

Contact area

A  π a

Average pressure

p avg 

F

p max 

3

Maximum pressure

2.

 3 m1  m2  F   B 8 

3

2

A  0.484 mm

2

p avg  1653 MPa

A

2

a  0.392 mm

 p avg

p max  2480 MPa

Determine the stresses in the steel ball at the surface Axial


In-plane

σxmax1  


1  2  ν1 2

 p max


σymax1  σxmax1 3.

Determine the stresses in the steel ball below the surface

Max shear stress

τyzmax1 

 1  2 ν1 2     1  ν1  2   1  ν1 9 2 2  

p max



τyzmax1  837.1 MPa




4.


7-17-2

2  2  ν1 7  2  ν1


Determine the stresses in the aluminum ball at the surface Axial


In-plane

σxmax2  


1  2  ν2 2

 p max


σymax2  σxmax2 5.

Determine the stresses in the aluminum ball below the surface

Max shear stress

τyzmax2 

 1  2 ν2 2     1  ν2  2   1  ν2 9 2 2  

p max





2  2  ν2 7  2  ν2




7-18-1


Determine the size of the contact patch and the maximum contact stresses for a 40-mm-dia steel cylinder, 25-cm long, rolled against a flat aluminum plate with 4 kN.

Given:

Cylinder radius

R1  20 mm


Plate curvature

R2  ∞ mm

Steel cylinder

Load

F  4  kN

Contact length

L  250  mm

ν1  0.28 E1  206.8  GPa

Aluminum plate ν2  0.34 E2  71.7 GPa

Solution: 1.



B 

1 2

 

1

 R1



  R2  1

B  0.025 mm

1

2

Material constants

m1 

1  ν1

6

m1  4.456  10

E1 2

m2 

1  ν2

1 MPa

5

m2  1.233  10

E2

1 MPa

1


 2 m1  m2 F  a      B L π

Contact area

A  2  L a

Average pressure

p avg 

Maximum pressure 2.

p max 

F

2

a  0.0827 mm A  41.36 mm

2

p avg  96.7 MPa

A 2 F

π a  L

p max  123.1 MPa

Determine the stresses in the cylinder at the surface. The maximum normal stresses in the center of the contact patch at the surface of the steel cylinder are found using equations 7.17a. Axial



In-plane



3. Determine the stresses in the cylinder below the surface. The maximum shear stress and its location under the surface are found from equations 7.17b.

4.

Max shear stress

τ13max  0.304  p max


zτmax  0.786  a


Determine the stresses in the plate at the surface. The maximum normal stresses in the center of the contact patch at the surface of the aluminum plate are found using equations 7.17a. Axial





In-plane 5.


7-18-2


Determine the stresses in the plate below the surface. The maximum shear stress and its location under the surface are found from equations 7.17b. Max shear stress

τ13max  0.304  p max


zτmax  0.786  a




7-19-1


Determine the size of the contact patch and the maximum contact stresses for a 40-mm-dia steel cylinder, 25-cm long, rolled against a parallel 50-mm-dia steel cylinder with 10 kN.

Given:

Cylinder radius

R1  20 mm


Cylinder radius

R2  25 mm

Steel cylinder

Load

F  10 kN

Contact length

L  250  mm

ν1  0.28 E1  206.8  GPa

ν2  0.28

Steel cylinder

E2  206.8  GPa Solution: 1.



B 

1 2

 

1

 R1



  R2  1

B  0.045 mm

1

2

Material constants

m1 

1  ν1

6

m1  4.456  10

E1 2

m2 

1  ν2

1 MPa

6

m2  4.456  10

E2

1 MPa

1


 2 m1  m2 F  a      B L π

Contact area

A  2  L a

Average pressure

p avg 

Maximum pressure 2.

3.

p max 

F

2

a  0.0710 mm A  35.51 mm

2

p avg  281.6 MPa

A 2 F

π a  L

p max  358.6 MPa

Determine the stresses in either cylinder at the surface. The maximum normal stresses in the center of the contact patch at the surface of the steel cylinder are found using equations 7.17a. Axial



In-plane



Determine the stresses in either cylinder below the surface. The maximum shear stress and its location under the surface are found from equations 7.17b. Max shear stress

τ13max  0.304  p max


zτmax  0.786  a

τ13max  109.0 MPa zτmax  0.0558 mm



7-20-1


Determine the size of the contact patch and the maximum contact stresses for a 20-mm-dia steel ball rolled against a 40-mm-dia steel cylinder, 25-mm long with 10 kN force.

Given:

Ball radii

R1  10.00  mm

The parts are steel. Therefore:

R'1  10.00  mm

E  206.8  GPa ν  0.28

R2  20.00  mm (radial)

Cylinder radii

R'2  ∞ mm

(axial)

(normal to contact plane) Radial load F  10 kN Angle between planes of R1 and R2 θ  0  deg Assumptions: The relative motion is rolling with < 1% sliding. Solution: 1.


Find the material constants from equation 7.9b. 2

1ν

m1 

Material constants

6

m1  4.456  10

E



1 MPa

m2  m1 2.

Two geometry constants are needed from equations 7.19a. Geometry constants

A 

1 2

 

1

 R1



1 R'1

1



R2



1

 

A  0.1250 mm

R'2 

1

1 2 2  1 1   1  1   B       R  R'1  2  R1  2 R'2   1 1  1 1   2  R  R'    R  R'   cos( 2  θ ) 1  2 2   1

     

1

Angle  Factors from equations 7.19e

B 180 ϕ  acos   A π

2

B  0.0250 mm

1

ϕ  78.46

 0.86215

ka  50.192 ϕ

ka  1.167 2

kb  0.0045333  0.043581 ϕ  0.0017292  ϕ   3.7374 10  1.4207 10 3.

5

3

9

5

 ϕ  3.7418 10 ϕ

7

4

 ϕ  kb  0.875

Determine the contact patch dimensions using the material and geometry constants in equations 7.19d. 1

Major axis half-width

3  3 m1  m2  a  ka   F A 4 

a  0.947  mm

1

Minor axis half-width

3  3 m1  m2  b  kb   F A 4 

b  0.710  mm


MACHINE DESIGN - An Integrated Approach, 4th Ed. A  π a  b

Contact area 4.

A  2.11354  mm

2

The average and maximum contact pressure can now be found from equations 7.18b and c. Average pressure Maximum pressure

5.

7-20-2

p avg 

F

p max 

3

p avg  4731 MPa

A 2

 p avg

p max  7097 MPa

The maximum normal stresses in the center of the contact patch at the surface are then found using equations 7.21a. In-plane

σx  2  ν  ( 1  2  ν) 



σy  2  ν  ( 1  2  ν) 



Axial

b

 p  max

σx  5312 MPa

 p  max

σy  5759 MPa

a  b a

a  b

σz  p max

σz  7097 MPa

These stresses are principal:

σ1  σx

σ2  σy

σ3  σz

The maximum shear stress associated with them at the surface is

τ13  6.

8.

τ13  892  MPa

2

The maximum shear stress under the surface on the z-axis is approximately Max shear stress

7.

σ1  σ3

τ13max  0.34 p max

τ13max  2413 MPa

All of the stresses found so far exist on the centerline of the patch. At the edge of the patch, at the surface, there will also be a shear stress. Two constants are found from equation 7.21b for this calculation. k3 

b

k4 

1

k3  0.75

a

a

2

 a b

2

k4  0.662

These constants are used in equations 7.21c and d to find the shear stresses on the surface at the ends of the major and minor axes.

Major axis

τxy  ( 1  2  ν) 

k3 k4

Minor axis

τxy  ( 1  2  ν) 

2

k3 k4

2

 

1

 k4

 atanh k4  1  p max





k3



k4

1 

 k4      pmax  k3  

 atan

τxy  1084 MPa

τxy  966  MPa



7-21-1


A cam-follower system has a dynamic load of 0 to 2 kN. The cam is cylindrical with a minimum radius of curvature of 20 mm. The roller follower is crowned with radii of 15 mm in one direction and 150 mm in the other. Find the contact stresses if the follower is steel and the cam is nodular iron.

Given:

Roller radius

R1  15.00  mm

Crown radius

R'1  150.00 mm(90 deg to roller rad)

Cam curvature

R2  20.00  mm (radial)

Cam curvature

R'2  ∞ mm

(axial)

(normal to contact plane) Radial load F  2  kN Angle between planes of R1 and R2 θ  0  deg Material properties

E1  206.8  GPa ν1  0.28

steel

E2  172.4  GPa ν2  0.30

cast iron

Assumptions: The relative motion is rolling with < 1% sliding. Solution: 1.



Material constants

m1 

1  ν1

3

m1  4.456  10

E1 2

m2  2.

1  ν2

3



m2  5.278  10

E2



1 GPa 1 GPa


A 

1 2

 

1

 R1



1 R'1

1



R2



  R'2  1

A  0.0617 mm

1

1

     

2 2  1 1   1  1   B       R  R'1  2  R1  2 R'2   1 1  1 1   2  R  R'    R  R'   cos( 2  θ ) 1  2 2   1

1

Angle 

Factors from equations 7.19e

B 180 ϕ  acos   A π

2

B  0.0550 mm

1

ϕ  26.89

 0.86215

ka  50.192 ϕ

ka  2.939 2

kb  0.0045333  0.043581 ϕ  0.0017292  ϕ   3.7374 10  1.4207 10

5

3

9

5

 ϕ  3.7418 10

7

4

 ϕ 

ϕ

kb  0.477



7-21-2

Determine the contact patch dimensions using the material and geometry constants in equations 7.19d. 1 3  3 m1  m2  a  ka   F A 4 


a  1.818  mm

1

4.


3  3 m1  m2  b  kb   F A 4 

b  0.295  mm

Contact area

A  π a  b

A  1.68582  mm

The average and maximum contact pressure can now be found from equations 7.18b and c. Average pressure

Maximum pressure

5.

2

p avg 

F

p max 

3


A

2

 p avg


The maximum normal stresses in the steel follower at the center of the contact patch at the surface are then found using equations 7.21a. In-plane

σx  2  ν1   1  2  ν1 



σy  2  ν1   1  2  ν1 



Axial

b

 p  max

σx  1106 MPa

 p  max

σy  1670 MPa

a  b a

a  b

σz  p max

σz  1780 MPa


σ1  σx

σ2  σy

σ3  σz


τ13  6.

τ13  337  MPa

2


7.

σ1  σ3

τ13max  0.34 p max

τ13max  605  MPa


b

k4 

1

k3  0.162

a

a

2

 a b

2

k4  0.987



8.


Major axis

τxy   1  2  ν1 

k3 2

 

Minor axis

τxy   1  2  ν1 

k3 2

1

 k4

k4

 atanh k4  1  p max


 k4      p max  k3  






k3



k4

1 

k4 9.

7-21-3

 atan

The maximum normal stresses in the nodular iron cam at the center of the contact patch at the surface are then found using equations 7.21a. In-plane

σx  2  ν2   1  2  ν2 



σy  2  ν2   1  2  ν2 



Axial

b

 p  max

σx  1167 MPa

 p  max

σy  1680 MPa

a  b a

a  b

σz  p max

σz  1780 MPa


σ1  σx

σ2  σy

σ3  σz


τ13 

σ1  σ3

τ13  306  MPa

2

10. The maximum shear stress under the surface on the z-axis is approximately Max shear stress

τ13max  0.34 p max

τ13max  605  MPa

11. All of the stresses found so far exist on the centerline of the patch. At the edge of the patch, at the surface, there will also be a shear stress. Two constants are found from equation 7.21b for this calculation. k3 

b

k4 

1

k3  0.162

a

a

2

 a b

2

k4  0.987

12. These constants are used in equations 7.21c and d to find the shear stresses on the surface at the ends of the major and minor axes. Major axis

τxy   1  2  ν2 

k3 2

k4 Minor axis

τxy   1  2  ν2 

k3 2

k4

 

1

 k4

 atanh k4  1  p max





k3



k4

1 

 k4      p max  k3  

 atan


τxy  91 MPa



7-22-1


An "inline" skate is shown in Figure P7-10. The polyurethane wheels are 72 mm dia. by 12-mm thick with a 6-mm crown radius and are spaced on 104- mm centers. The skate-boot-foot combination weighs 2 kg. The effective "spring rate" of the person-skate subsystem is 6000 N/m. Find the contact stresses in the wheels when a 100-kg person lands a 0.5-m jump on one foot on concrete. Assume the urethane wheels and concrete have the properties below. (a) Assume that all 4 wheels land simultaneously. (b) Assume that one wheel absorbs all the landing force.

Given:

Force per wheel (from Problem 3-22): Case (a) Fa  897  N Material properties: Urethane E1  600  MPa

ν1  0.4

E2  21 GPa

Concrete

Fb  3.59 kN

Case (b)

ν2  0.2

Wheel dimensions

R1  36 mm

R'1  6  mm

Concrete dimensions

R2  ∞ mm

R'2  ∞ mm

Contact angle

θ  0  deg




m1 

Material constants

1  ν1

m1  1.400 

E1

1 GPa

2

m2  2.

1  ν2

m2  0.0457

E2

1 GPa


A 

1 2

 

1

 R1



1 R'1



1



R2

  R'2  1

A  0.0972 mm

1

1

Angle 


2

2 2  1 1   1  1   B       R  R'1  2  R1  2 R'2   1 1  1 1   2  R  R'    R  R'   cos( 2  θ ) 1  2 2   1

     

B 180 ϕ  acos   A π

ϕ  44.42

1

 0.86215

ka  50.192 ϕ

B  0.0694 mm

1

ka  1.906 2

kb  0.0045333  0.043581 ϕ  0.0017292  ϕ   3.7374 10  1.4207 10

5

3

9

5

 ϕ  3.7418 10

7

4

 ϕ 

ϕ

kb  0.593 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


7-22-2

Part (a) 3.

Determine the contact patch dimensions using the material and geometry constants in equations 7.19d. 1 3  3 m1  m2  a  ka    Fa A 4 


a  4.108  mm

1

4.


3  3 m1  m2  b  kb    Fa A 4 

b  1.278  mm

Contact area

Ac  π a  b

Ac  16.49024 mm

The average and maximum contact pressure can now be found from equations 7.18b and c. p avg 

Average pressure

p max 

Maximum pressure

5.

2

Fa

p avg  54.4 MPa

Ac 3 2

 p avg

p max  81.6 MPa


σx  2  ν1   1  2  ν1 



σy  2  ν1   1  2  ν1 



Axial

b

 p  max

σx  69.1 MPa

 p  max

σy  77.7 MPa

a  b a

a  b

σz  p max

σz  81.6 MPa


σ1  σx

σ2  σy

σ3  σz


τ13  6.

τ13  6.22 MPa

2


7.

σ1  σ3

τ13max  0.34 p max

τ13max  27.7 MPa


b a

k3  0.311



k4  8.

1 a

2

 a b

7-22-3

2

k4  0.95

These constants are used in equations 7.21c and d to find the shear stresses on the surface at the ends of the major and minor axes. k3

τxy   1  2  ν1 

Major axis

2

 

 k4

k4

k3

τxy   1  2  ν1 

Minor axis

2

1

 atanh k4  1  p max

τxy  5.24 MPa

 k4      p max  k3  




k3



k4

1 

k4



 atan

Part (b) 9.

Determine the contact patch dimensions using the material and geometry constants in equations 7.19d. 1 3  3 m1  m2  a  ka    Fb A 4 


a  6.522  mm

1


 3 m1  m2  b  kb    Fb A 4 

Contact area

Ac  π a  b

3

b  2.029  mm

Ac  41.56824 mm

2

10. The average and maximum contact pressure can now be found from equations 7.18b and c. p avg 

Average pressure

p max 

Maximum pressure

Fb

p avg  86.4 MPa

Ac 3 2

 p avg

p max  129.5  MPa

11. The maximum normal stresses in the center of the contact patch at the surface are then found using equations 7.21a. In-plane

σx  2  ν1   1  2  ν1 



σy  2  ν1   1  2  ν1 



Axial

b

 p  max

σx  109.8  MPa

 p  max

σy  123.4  MPa

a  b a

a  b

σz  p max

σz  129.5  MPa


σ1  σx

σ2  σy

σ3  σz

The maximum shear stress associated with them at the surface is © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


τ13 

7-22-4

σ1  σ3

τ13  9.88 MPa

2

12. The maximum shear stress under the surface on the z-axis is approximately Max shear stress

τ13max  0.34 p max

τ13max  44.0 MPa

13. All of the stresses found so far exist on the centerline of the patch. At the edge of the patch, at the surface, there will also be a shear stress. Two constants are found from equation 7.21b for this calculation. k3 

b

k4 

1

k3  0.311

a

a

2

 a b

2

k4  0.95

14. These constants are used in equations 7.21c and d to find the shear stresses on the surface at the ends of the major and minor axes. Major axis

τxy   1  2  ν1 

k3 2

k4 Minor axis

τxy   1  2  ν1 

k3 2

k4

 

1

 k4

 atanh k4  1  p max


 k4      p max  k3  






k3



k4

1 

 atan



7-23-1


A pair of 12-in dia cylindrical steel rolls run together with 9% slip. Find their contact stresses for a radial contact forces of 1000 lb/in of length.

Given:

Roller radii

R1  6.00 in

The parts are hardened steel. Therefore: 6

R2  6.00 in Radial load/in

E  30 10  psi

ν  0.28

1

F'  1000 lbf  in

Assumptions: The coefficient of friction is μ  0.33. Solution: 1.



m1 

Material constants

1ν

8 1

m1  3.072  10

E

psi

m2  m1 B 

Geometry constant

1 2

 

1

 R1



  R2 

1

1

B  0.167 in 1


3.

 2 m1  m2  a     F' B π 

2

a  0.0153 in

The average and maximum contact pressure can now be found from equations 7.14b and c. F'

Average pressure

p avg 

Maximum pressure

p max 

Tangential pressure


p avg  32.6 ksi

2 a 2  F'

p max  41.6 ksi

π a

fmax  13.7 ksi

With m = 0.33, the principal stresses in the contact zone will be maximal on the surface (z = 0) at x = 0.3a from the centerline as shown in Figures 7-20 and 7-22. The applied stress components are found from equation 7.23a for the normal force and equation 7.23b for the tangential force. For

x  0.3 a

σxn  p max  1 

x

2

a

σxt  2  fmax 

x

σxt  8.23 ksi

a

σzn  p max  1 

x

2

a

σzt  0  ksi

σxn  39.6 ksi

2

σzn  39.6 ksi

2

τxzn  0  ksi


MACHINE DESIGN - An Integrated Approach, 4th Ed. x

τxzt  fmax  1 

2

a 4.

5.

τxzt  13.1 ksi

2



σx  47.871 ksi


σz  39.642 ksi


τxz  13.082 ksi

Assuming the rollers are long, we expect a plane strain condition to exist. The stress in the third dimension is found from equation 7.23b:

σy  ν  σx  σz 6.

7-23-2

σy  24.504 ksi

Unlike the pure-rolling case, these stresses are not principal because of the applied shear stress. The principal stresses are found from equation 4.4 using a cubic root finding solution.

σx 

σx

σy 

ksi

τxy  0

σy ksi

τyz  0

C2  σx  σy  σz C1 

τxz 

σz ksi

τxz ksi

C2  112.017

 σx τxy   σx τxz   σy τyz           τxy σy   τxz σz   τyz σz 

 σx τxy τxz    C0   τxy σy τyz  τ τ σ   xz yz z  3

σz 

3

C1  3.871  10

C0  4.231  10

4

2

f ( σ )  σ  C2 σ  C1 σ  C0

 C0    C1   v   C2     1 

s  polyroots ( v)  ksi

 57.5  s   30.0  ksi    24.5 

Principal stresses:

σ1  s

σ1  24.5 ksi

σ2  s

σ2  30.0 ksi

σ3  s

σ3  57.5 ksi

3 2 1


τ13  7.

σ1  σ3 2

τ13  16.5 ksi

The principal stresses are maximum at the surface as seen in Figures 7-20 and 7-22 in the text.



7-24-1


Estimate the cycle life of the rolls in Problem 7-23 if they are made of class 30 gray cast iron, austempered to HB270.

Given:

Roller radii

R1  6.00 in

The parts are gray cast iron. Therefore: 6

R2  6.00 in Radial load/in

E  15 10  psi

ν  0.28

1

F'  1000 lbf  in

Assumptions: The coefficient of friction is μ  0.33. Solution: 1.



m1 

Material constants

1ν

8 1

m1  6.144  10

E

psi

m2  m1 B 

Geometry constant

1 2

 

1

 R1



  R2 

1

1

B  0.167 in 1


3.

 2 m1  m2  a     F' B π 

2

a  0.0217 in


Average pressure

p avg 

Maximum pressure

p max 

Tangential pressure


p avg  23.1 ksi

2 a 2  F'

p max  29.38 ksi

π a

fmax  9.70 ksi

With m = 0.33, the principal stresses in the contact zone will be maximal on the surface (z = 0) at x = 0.3a from the centerline as shown in Figures 7-20 and 7-22. The applied stress components are found from equation 7.23a for the normal force and equation 7.23b for the tangential force. For

x  0.3 a

σxn  p max  1 

x

2

a

σxt  2  fmax 

x

σxt  5.82 ksi

a

σzn  p max  1 

x

2

a



2


2





x

2

a 4.

5.


2



σx  33.850 ksi


σz  28.031 ksi


τxz  9.250 ksi

Assuming the rollers are long, we expect a plane strain condition to exist. The stress in the third dimension is found from equation 7.23b:

σy  ν  σx  σz 6.

7-24-2

σy  17.327 ksi


σx 

σx

σy 

ksi

τxy  0

σy

σz 

ksi

τyz  0

τxz 

C2  σx  σy  σz C1 

ksi

τxz ksi

C2  79.208



σz

3

C1  1.935  10

C0  1.496  10

4

2

f ( σ )  σ  C2 σ  C1 σ  C0

 C0    C1   v   C2     1 

s  polyroots ( v)  ksi

 40.6  s   21.2  ksi    17.3 

Principal stresses:

σ1  s

σ1  17.3 ksi

σ2  s

σ2  21.2 ksi

σ3  s

σ3  40.6 ksi

3 2 1

7.

The maximum normal stress calculated in step 6 is 40.9 ksi, compressive. Its K-factor can be calculated from equation 7.25d. K-factor

K  π  m1  m2  σ3

2

K  637.5 psi



8.

From Table 7-7, Part 2, Line 33 the slope and intercept factors of this cast iron for rolling with 9% sliding are

λ  7.87 9.

7-24-3

ζ  35.90

These are used in equation 7.26 along with the value of K from above to find the number of cycles that can be expected at this load before pitting begins.

log ( K) =


K   psi 

ζ λ log

Nlife  10

13

Nlife  6.7  10

cycles



7-25-1


A 12-mm-dia, class 30 gray cast iron pad is supported by a steel bar made from SAE 4340 steel, quenched and tempered at 800F. The force on the pad is 3.8 kN. Estimate the real area of contact and the ratio of the real area to the apparent area of contact.

Given:

Pad diameter d  12 mm Compressive yield strengths: Contact force

Steel

S ycs  1469 MPa

Cast iron

S ycci  752  MPa

F  3.8 kN

Assumptions: The compressive yield strength of the cast iron is approximately the same as the compressive strength. Solution: 1.


Calculate the apparent area of contact. Aa 

2.

2

4

Aa  113.097 mm

2

Use equation 7.1 and the cast iron strength, which is the weaker of the two, to estimate the real area of contact. Ar 

3.

π d

F 3  S ycci

Ar  1.6844 mm

2

The ratio of real to apparent area of contact is: ratio 

Ar Aa

ratio  1.5 %


MACHINE DESIGN - An Integrated Approach 4th Ed.

7-26-1


Estimate the dry coefficient of friction between the two materials in Problem 7-35 if the shear strength of the cast iron is S us = 310 MPa. How does this compare to the value given in Table 7-1?

Given:

Pad diameter Shear strength:

d  12 mm Cast iron

Compressive yield strength

S us  310  MPa S yc  752  MPa


See Table 7-1 and Mathcad file P0726.

Using equation 7.3, the coefficient of friction is approximately:

μ 

S us 3  S yc

μ  0.14

From Table 7-1, the coefficient of friction for mild steel on cast iron (lubricated) is 0.183.



PROBLEM 7-27

7-27-1

_____

Statement:

Two 0.5-in x 1-in 1040 hot-rolled steel pads are in contact with a force of 900 lb. Estimate the real area of contact and the ratio of the real area to the apparent area of contact.

Given:

Pad dimensions Compressive yield strength:

L  1.00 in S yc  42 ksi

Contact force

F  900  lbf

w  0.5 in

Assumptions: The compressive yield strength of the steel is approximately the same as the tensile yield strength. Solution: 1.


Calculate the apparent area of contact. Aa  L w

2.

Use equation 7.1 and the compressive yield strength to estimate the real area of contact. Ar 

3.

2

Aa  0.500 in

F 3  S yc

2

Ar  0.0071 in

The ratio of real to apparent area of contact is:

ratio 

Ar Aa

ratio  1.43 %



PROBLEM 7-28

7-28-1

_____

Statement:

Estimate the dry coefficient of friction between the two materials in Problem 7-27. How does this compare to the value given in Table 7-1?

Given:

Pad dimensions Ultimate tensile strength:

L  1.0 in S ut  76 ksi

Compressive yield strength:

S yc  42 ksi

w  0.50 in



Estimate the ultimate shear strength using equation 7.4 for steel. S us  0.8 S ut

2.

S us  60.8 ksi


μ 

S us 3  S yc

μ  0.48



7-29-1

PROBLEM 7-29

_____

Statement:

Two materials have been tested to determine the amount of adhesive wear that takes place when they are run together. The average depth of wear results are given in Table P7-1 along with the test parameters for a total of 350 tests. What is the average wear coefficient for the materials tested if the Brinell hardness of the softer of the two is HB 277.

Given:

Hardness

HB  277  kgf  mm

i  1  4

Data from Table P7-1: F 

Solution: 1.

A  10 mm

2

i

l 

i

100  N 200  N 200  N 400  N

2

NT 

i

d 

i

5000 m 5000 m 7500 m 10000  m

100 75 75 100

i

0.180  mm 0.372  mm 0.550  mm 1.47 mm

See Table P7-1 and Mathcad file P0729.

The total number of tests is: Total 

 NT i

Total  350

i

2.

The average force, apparent contact area, length of sliding, and depth of wear are: Favg 

 F iNT i

1



Total

Favg  228.571 N

i

Aavg 

A  NT  i Total   i 1



Aavg  10.000 mm

2

i

lavg 

l  NT  i Total   i 1



3

lavg  6.964  10 m

i

d avg 

1 Total

 diNT i



d avg  0.669 mm

i

3.

Calculate the average adhesive wear coefficient using equation 7.7b and the values calculated above.

Kavg 

HB Aavg  d avg Favg lavg

5

Kavg  1.14  10



7-30-1

PROBLEM 7-30

_____

Statement:

A piece of mild steel with HB = 280 has its thickness reduced in an abrasive grinder. Both the grinding wheel and the steel part have the same width, which is 20 mm. On each pass through the grinder 0.1 mm is removed. If the abrasive wear coefficient for this operation is 5E-1, what is the approximate normal force on the grinding wheel?

Given:

Hardness Depth of wear per stroke Wear coefficient

Solution: 1.

K  5  10

From equation 7.7a, the volume of wear per stroke is F l H

Solving for the normal force F, F 

3.

w  20 mm

1


V  K 2.

2

H  280  kgf  mm Width of wear d  0.1 mm

V H K l

where

V  d  w l

Substituting for V and canceling the l-terms, F 

d  w H K

F  11.0 kN



7-31-1

PROBLEM 7-31

_____

Statement:

Two steel gears with involute tooth profiles are in mesh. At the line of contact between the gears they can be modeled as two cylinders in contact. When the contact is away from the pitch point there is a combination of rolling and sliding. Determine and plot the dynamic contact principal stresses on the surface of the teeth for the following gears if the contact force is 500 lb and the coefficient of friction is 0.15: R1 = 2.000 in, R2 = 6.000 in. The thickness (face width) of both gears is 0.5 in. Also find the value of x/a for which the principal stresses have an extreme value.

Given:

The parts are hardened steel. Therefore: E  30 10  psi ν  0.28 Radii of curvature R1  2.00 in R2  6.00 in

6

Contact force Gear tooth thickness Coefficient of friction Solution: 1.


Calculate the force per unit of length along the line of contact. F' 

2.

W  500  lbf F  0.5 in (face width) μ  0.15

1

W

F'  1000 lbf  in

F


Material constants

1ν

m1 

8 1

m1  3.072  10

E

psi

m2  m1

Geometry constant

1

B 

2

 

1

 R1



  R2 

1

1

B  0.333 in 1


3.

4.

 2 m1  m2  a     F' B π 

2

a  0.0108 in


Average pressure

p avg 

p avg  46.2 ksi

Maximum pressure

p max 

Tangential pressure


2 a 2  F'

p max  58.8 ksi

π a

fmax  8.8 ksi

The applied stress components are found from equation 7.23a for the normal force and equation 7.23b for the tangential force.



σxn( x)  if  x  a p max  1 

 

x

2

a

2



0  psi 

 

σzn( x)  σxn( x)



7-31-2


σxt( x) 

x

return 2  fmax  



2



 1 if x  a

a  2 a   2 x  x return 2  fmax     1 if x  a a  2 a   2  fmax 

x a


otherwise



x

 

a

τxzt( x)  if  x  a fmax  1 

5.

x

2 2



0  ksi

 

Equations 7.24a can now be written for the total applied stresses along the x and z axes.

σx( x)  σxn( x)  σxt( x) σz( x)  σzn( x)  σzt τxz( x)  τxzn  τxzt( x) 6.

The face width of the gears is short with respect to the other dimensions, therefore we expect a plane stress condition to exist. The stress in the third dimension is

σy  0  ksi 7.

Unlike the pure-rolling case, these stresses are not principal because of the applied shear stress. The principal stresses are found from equation 4.6a for the plane stress case.

σ1( x) 

σ2( x) 

σx( x)  σz( x) 2

σx( x)  σz( x) 2

2  σx( x)  σz( x)  2     τxz( x) 2  

2  σx( x)  σz( x)  2     τxz( x) 2  

σ3  0  ksi

8.

Plot the two nonzero principal stresses over the range x  3  a 2.999  a  3  a



7-31-3

PRINCIPAL CONTACT STRESSES 0.5

0.18

Normalized stress

0

σ1( x) pmax

σ2( x)

 0.5

pmax 1

 1.5 2

1

0

1

2

x a Normalized width, x/a

9.

Note that for  = 0.15, the extreme values of the principal stresses occur at approximately x/a = 0.18.



7-32-1

PROBLEM 7-32

_____

Statement:

Two steel gears with involute tooth profiles are in mesh. At the line of contact between the gears they can be modeled as two cylinders in contact. When the contact is away from the pitch point there is a combination of rolling and sliding. Determine the dynamic contact stresses on the surface of the teeth for the following gears if the contact force is 1500 lb and the coefficient of friction is 0.33: R1 = 2.500 in, R2 = 5.000 in. The thickness (face width) of both gears is 0.625 in.

Given:

The parts are hardened steel. Therefore E  30 10  psi ν  0.28 Radii of curvature R1  2.500  in R2  5.000  in

6

Contact force Gear tooth thickness Coefficient of friction Solution: 1.



2.

W  1500 lbf F  0.625  in (face width) μ  0.33

1

W

F'  2400 lbf  in

F


1ν

m1 

Material constants

8 1

m1  3.072  10

E

psi

m2  m1 B 

Geometry constant

1 2

 

1

 R1



  R2 

1

1

B  0.300 in 1


3.

4.

 2 m1  m2  a     F' B π 

2

a  0.0177 in


Average pressure

p avg 

Maximum pressure

p max 

Tangential pressure


2 a 2  F'

π a

p avg  67.8 ksi p max  86.4 ksi fmax  28.5 ksi

With  = 0.33, the principal stresses in the contact zone will be maximal on the surface (z = 0) at x = 0.3a from th centerline as shown in Figures 7-20 and 7-22. The applied stress components are found from equation 7.23a for the normal force and equation 7.23b for the tangential force. For

x  0.3 a

σxn  p max  1 

x

2

a

2




σxt  2  fmax 

x

σxt  17.10 ksi

a

σzn  p max  1 

x

2

a


2

σzt  0  ksi τxzt  fmax  1 

τxzn  0  ksi x

2

a 5.

6.

7-32-2


2



σx  99.497 ksi


σz  82.395 ksi


τxz  27.190 ksi

The face width of the gears is short with respect to the other dimensions, therefore we expect a plane stress condition to exist. The stress in the third dimension is

σy  0  ksi 7.


σ1  0  ksi

σ2 

σ3 

σx  σz 2

σx  σz 2

2  σx  σz  2     τxz  2 



2  σx  σz  2    τxz 2  

σ2  62.443 ksi

σ3  119.449 ksi


τ13 

σ1  σ3 2

τ13  59.7 ksi



7-33-1

PROBLEM 7-33

_____

Statement:

Two contacting rollers are needed for a machine application. They run together with a combination of rolling and 9% sliding. Both are to be made from 1144 CD steel. The radial contact force is 1200 N and the coefficient of friction is 0.33. The rollers are to have the same radii and are both 10 mm long. If the design life is 8E08 cycles, determine a suitable radius for the rollers.

Given:

The parts are hardened steel. Therefore Contact force W  1200 N Roller length w  10 mm Coefficient of friction

μ  0.33

1.

8

Nd  8  10

ζ  21.79



2.

ν  0.28

Design life

λ  4.10

Material data from Table 7-7 Solution:

6

E  30 10  psi

W

F'  120.00 N  mm

w

1


Material constants

m1 

1ν

6

m1  4.456  10

E



1 MPa

m2  m1

Geometry constant

1

B( R) 

2

 

1

R



1



R 1


3.

4.

 2 m1  m2  a ( R)     F'  π B( R) 

2


Average pressure

p avg ( R) 

Maximum pressure

p max ( R) 

Tangential pressure

fmax ( R)  μ  p max ( R)

2  a ( R) 2  F'

π  a ( R)

With  = 0.33, the principal stresses in the contact zone will be maximal on the surface (z = 0) at x = 0.3a from th centerline as shown in Figures 7-20 and 7-22. The applied stress components are found from equation 7.23a for the normal force and equation 7.23b for the tangential force. For

x( R)  0.3 a ( R) 2

σxn( R)  p max ( R)  1  0.3

2

σzn( R)  p max ( R)  1  0.3

τxzn  0  MPa © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


σxt( R)  2  fmax ( R)  0.3

7-33-2

σzt  0  MPa 2

τxzt( R)  fmax ( R)  1  0.3 5.


σx( R)  σxn( R)  σxt( R) σz( R)  σzn( R)  σzt τxz( R)  τxzn  τxzt( R) 6.

The face width of the rollers is short with respect to the other dimensions, therefore we expect a plane stress condition to exist. The stress in the third dimension is

σy  0  MPa 7.


σ1  0  MPa

σ2( R) 

σ3( R) 

8.

σx( R)  σz( R) 2

σx( R)  σz( R) 2

2  σx( R)  σz( R)  2     τxz( R) 2  

2  σx( R)  σz( R)  2     τxz( R) 2  

Calculate the K-factor for this material and the number of cycles before the onset of pitting for this design using equation 7.26. ζ  log Nd 

K  10 9.

λ

 psi

K  1390 psi

Use equation 7.25e to solve for the roller radius. First, guess R  40 mm Given

K = π  m1  m2  σ3( R) R  Find ( R)

2

R  47.9 mm



7-34-1

PROBLEM 7-34

_____

Statement:

Two contacting rollers are needed for a machine application. They run together with a combination of rolling and 9% sliding. Both are to be made from Meehanite. The rollers have the same radii (30 mm) and are both 45 mm long. If the design life is 1E08 cycles, determine the allowable load that may be applied to these rollers.

Given:

Material data from Table 7-7 Roller radii R1  30 mm Roller length

Solution: 1.

L  45 mm


Find the geometry constant from equation 7.9a. Geometry constant

2.

K  1450 psi R2  30 mm

B 

1 2

 

1

 R1



1

 

R2 

B  0.033

1 mm

Calculate the allowable force on the rollers using equation 7.25d.

Allowable force

F 

K L 2 B

F  6.75 kN



7-35-1


A 25-mm-dia, Class 20 gray cast iron pad is supported by a steel bar made from SAE 4130 steel, quenched and tempered at 800F. The force on the pad is 2800 N. Estimate the real area of contact and the ratio of the real area to the apparent area of contact.

Given:

Pad diameter d  25 mm Compressive yield strengths: Contact force

Pad

S ycp  572  MPa

Bar

S ycb  1193 MPa

F  2800 N

Assumptions: The compressive yield strength of the cast iron is approximately the same as its compressive strength. Solution: 1.


Calculate the apparent area of contact. Aa 

2.

2

4

Aa  490.9 mm

2

Use equation 7.1 and the pad strength, which is the weaker of the two, to estimate the real area of contact. Ar 

3.

π d

F 3  S ycp

Ar  1.632 mm

2


Ar Aa

ratio  0.3 %



7-36-1


Estimate the dry coefficient of friction between the two materials in Problem 7-35 if the shear strength of the cast iron is S us = 310 Mpa. How does this compare to the value given in Table 7-1?

Given:

Pad diameter Shear strength:

d  25 mm Cast iron

Compressive yield strength

S us  310  MPa S yc  572  MPa




μ 

S us 3  S yc

μ  0.18

From Table 7-1, the coefficient of friction for mild steel on cast iron (lubricated) is 0.183.



7-37-1


Two 25-mm x 40-mm SAE 1020 hot-rolled steel pads are in contact with a force of 9 kN. Estimate the real area of contact and the ratio of the real area to the apparent area of contact.

Given:

Pad dimensions w  25 mm Compressive yield strength: Contact force

L  40 mm S yc  379  MPa

F  9000 N

Assumptions: The compressive yield strength of the steel is approximately the same as its tensile yield strength. Solution: 1.


Calculate the apparent area of contact. Aa  L w

2.

2

Use equation 7.1 and the pad strength, which is the weaker of the two, to estimate the real area of contact. Ar 

3.

Aa  1000.0 mm

F 3  S yc

Ar  7.916 mm

2


Ar Aa

ratio  0.79 %



7-38-1


Estimate the dry coefficient of friction between the two materials in Problem 7-37. How does this compare to the value given in Table 7-1?

Given:

Pad dimensions w  25 mm L  40 mm Ultimate tensile strength: S ut  379  MPa Compressive yield strength:

S yc  379  MPa



Estimate the ultimate shear strength equation 7.4 for steel. S us  0.8 S ut

2.

S us  303.2 MPa


μ 

S us 3  S yc

μ  0.27

From Table 7-1, the coefficient of friction for mild steel on mild steel is 0.74.



7-39-1


A 25-mm-diameter steel shaft of hardness HB420 rotates at 700 rpm in a 40-mm-long plain bronze bushing with an average radial load of 500 N. Estimate the time it would take to remove 0.05 mm of bushing material by adhesive wear if the lubrication were suddenly lost assuming a uniform wear rate around the bushing.

Units:

kilo  kg g

Given:

Journal diameter d  25 mm Journal speed n  700  rpm

rev  2  π rad

kilo

Journal hardness HB  420 

mm Solution: 1.

1

Load Bushing length

F  500  N lbush  40 mm

Depth of wear

d  0.05 mm

2


From Figure 7-6 in the text we see that iron (the principal ingredient of steel) is metalurgically compatible with copper and tin, the principal ingredients of bronze. From Figure 7-7 we have the following value of the adhesive wear coefficient for metalurgically compatible materials with no lubrication. Wear coefficient

2.

rpm  rev  min

4

K  8  10

Using equation 7.7b, estimate the time for the bushing to wear d  0.05 mm Apparent area of contact

Aa  π d  lbush

Length of contact in time t

lcont( t) 

Solving equation 7.7b for l

lcont =

Substituting and solving for t

t 

In a time of t  29.4 min the shaft will turn

π d rev

 n t

d  HB Aa K F

d  HB lbush rev n K F

θ  n  t

t  29.4 min

4

θ  2  10 rev



7-40-1


A machine has a tripod base that utilizes 15-mm-dia Nylon 11 balls as support pads at its feet. The tripod rests on a flat steel plate. The 360N weight of the machine is distributed equally to the three legs of the tripod. Determine the size of the contact patch and the contact stresses in the nylon balls. Assume that Poisson's ratio for Nylon is 0.25.

Given:

Ball radius

R1  7.5 mm

Plate curvature

R2  ∞ mm

Load

F  120  N

Material properties Nylon ball

ν1  0.25 E1  1.3 GPa

Steel plate

ν2  0.28 E2  207  GPa

Solution:


1. Calculate geometry and material constants, contact patch dimension, and pressures. Geometry constant

1

B 

2

 

1

 R1



  R2  1

B  0.067 mm

1

2

Material constants

m1 

1  ν1

4

m1  7.212  10

E1 2

m2 

1  ν2

6

m2  4.452  10

E2

1 MPa 1 MPa

1


a 

Contact area

A  π a

Average pressure

Maximum pressure

 3 m1  m2  F   B 8 

3

a  0.788 mm

2

A  1.952 mm

p avg 

F

p avg  61 MPa

p max 

3

A

2

 p avg

2

p max  92 MPa

2. Determine the stresses in the ball at the surface Axial


In-plane

σxmax1  


1  2  ν1 2

 p max


σymax1  σxmax1 3. Determine the stresses in the ball below the surface Max shear stress

τyzmax1 

  1  2 ν1 2    1  ν1  2   1  ν1 9 2 2  

p max





2  2  ν1 7  2  ν1




7-41-1


A ball bearing consists of a number of balls (separated by a ball cage) and two rings with raceways as shown in Figure P7-7. The raceways have compound curvature. In a plane containing the axis of the bearing the curvature is concave and conforms closely to the ball radius. In a plane perpendicular to the axis the curvature is convex for the inner raceway and is related to the bore size of the bearing. Determine the size of the contact patch and the maximum contact stresses between a ball and the inner raceway with a radial load of 5200 N in a steel bearing with the following dimensions: ball dia = 8 mm, raceway radius for concave surface = 4.05 mm, raceway radius for convex surface = 13 mm.

Given:

Ball radii

R1  4.00 mm

The parts are steel. Therefore

R'1  4.00 mm

E  206.8  GPa

R2  13.0 mm

Raceway radii

ν  0.28

(radial)

R'2  4.05 mm (axial) Radial load F  5200 N Angle between planes of R1 & R2 Solution: 1.

(normal to contact plane) θ  0  deg



m1 

Material constants

1ν

6

m1  4.456  10

E



1 MPa

m2  m1 2.


A 

1 2

 

1

 R1



1 R'1

1



R2



1

 

A  0.1650 mm

R'2 

1

1 2 2  1 1   1  1   B       R  R'1  2  R1  2 R'2   1 1  1 1   2  R  R'    R  R'   cos( 2  θ ) 1  2 2   1

     

1

Angle 


B 180 ϕ  acos   A π

2

B  0.1619 mm

1

ϕ  11.10

 0.86215

ka  50.192 ϕ

ka  6.301 2

kb  0.0045333  0.043581 ϕ  0.0017292  ϕ   3.7374 10  1.4207 10

5

3

9

5

 ϕ  3.7418 10

7

4

 ϕ 

ϕ

kb  0.321



7-41-2



a  3.749  mm

1

4.


3  3 m1  m2  b  kb   F A 4 

b  0.191  mm

Contact area

A  π a  b

A  2.24904  mm


Maximum pressure 5.

2

p avg 

F

p max 

3


A

2

 p avg



σx  2  ν  ( 1  2  ν) 



σy  2  ν  ( 1  2  ν) 



Axial

b

 p  max

σx  2016 MPa

 p  max

σy  3394 MPa

a  b a

a  b

σz  p max

σz  3468 MPa


σ1  σx

σ2  σy

σ3  σz


τ13  6.

σ1  σ3

The maximum shear stress under the surface on the z-axis is approximately

τ13max  0.34 p max

Max shear stress 7.

8.

τ13  726  MPa

2

τ13max  1179 MPa


b

k4 

1

k3  0.051

a

a

2

 a b

2

k4  0.999




Major axis

τxy  ( 1  2  ν) 

k3 k4

Minor axis

τxy  ( 1  2  ν) 

2

k3 k4

2

 

1

 k4

7-41-3

 atanh k4  1  p max





k3



k4

1 

 k4      pmax  k3  

 atan


τxy  72 MPa



7-42-1


A pair of steel rollers used in a manufacturing process roll together with a combination of rolling and sliding. One roller has a diameter of 75 mm and the other has a diameter of 50 mm. They are both 200-mm long. The contact force, which is normal to the contact plane, is 18500 N. Assuming that the coefficient of friction between the rollers is 0.33, determine the maximum tensile, compressive, and shear stresses in the rollers.

Given:

Roller radii

R1  25 mm

The parts are hardened steel. Therefore

R2  37.5 mm L  200  mm F  18500  N

Roller length Radial load Solution: 1.

E  206.8  GPa

ν  0.28

Coefficient of friction

μ  0.33



m1 

Material constants

1ν

6

m1  4.456  10

E

1 MPa

m2  m1 B 

Geometry constant

1 2

 

1

 R1



  R2  1

B  0.033 mm

1

1


3.

 2 m1  m2 F  a      B L π

2

a  0.1255 mm

The average and maximum contact pressure can now be found from equations 7.14b and c. F

Average pressure

p avg 

Maximum pressure

p max 

Tangential unit force


2 a L 2 F

π a  L

p avg  368.6 MPa p max  469.3 MPa fmax  154.9 MPa

With = 0.33, the principal stresses in the contact zone will be maximal on the surface (z = 0) at x = 0.3a from th centerline as shown in Figures 7-20 and 7-22. The applied stress components are found from equation 7.23a for the normal force and equation 7.23b for the tangential force. For

x  0.3 a

σxn  p max  1 

x

2

a

σxt  2  fmax 

2

x

σxt  92.92 MPa

a

σzn  p max  1 

σxn  447.7 MPa

x

2

a

2

σzn  447.7 MPa



σzt  0  MPa

τxzn  0  MPa


x

2

a 4.

5.

τxzt  147.7 MPa

2



σx  540.591 MPa


σz  447.672 MPa


τxz  147.732 MPa

Since the rollers are long, we expect a plane strain condition to exist. The stress in the third dimension is found from equation 7.23b:

σy  ν  σx  σz 6.

7-42-2

σy  276.714 MPa


σx 

σx

σy 

MPa

τxy  0

σy MPa

τyz  0

C2  σx  σy  σz C1 

τxz 

σz MPa

τxz MPa

C2  1.265  10



σz 

3

5

C1  4.936  10

C0  6.093  10

7

2

f ( σ )  σ  C2 σ  C1 σ  C0

 C0    C1   v   C2     1 

s  polyroots ( v)  MPa

 649.0  s   339.3  MPa    276.7 

Principal stresses:

σ1  s

σ1  276.7 MPa

σ2  s

σ2  339.3 MPa

σ3  s

σ3  649.0 MPa

3 2 1



7-42-3


τ13  7.

σ1  σ3 2

τ13  186.1 MPa




7-43-1


Repeat Problem 7-41 for contact beween a ball and outer raceway. The outer raceway radius for concave surface = 4.05 mm and the outer raceway radius for concave surface = 17.02 mm.

Given:

Ball radii

R1  4.00 mm

The parts are steel. Therefore

R'1  4.00 mm

E  206.8  GPa

ν  0.28

R2  17.02  mm (radial)

Raceway radii

R'2  4.05 mm (axial) Radial load F  5200 N Angle between planes of R1 & R2 Solution: 1.

(normal to contact plane) θ  0  deg



m1 

Material constants

1ν

6

m1  4.456  10

E



1 MPa

m2  m1 2.


A 

1 2

 

1

 R1



1 R'1

1



R2



1

 

A  0.0972 mm

R'2 

1

1 2 2  1 1   1  1   B       R  R'1  2  R1  2 R'2   1 1  1 1   2  R  R'    R  R'   cos( 2  θ ) 1  2 2   1

     

1

Angle 


B 180 ϕ  acos   A π

2

B  0.0941 mm

1

ϕ  14.48

 0.86215

ka  50.192 ϕ

ka  5.011 2

kb  0.0045333  0.043581 ϕ  0.0017292  ϕ   3.7374 10  1.4207 10 3.

5

3

9

5

 ϕ  3.7418 10 ϕ

7

4

 ϕ  kb  0.371

Determine the contact patch dimensions using the material and geometry constants in equations 7.19d. 1


 3 m1  m2  a  ka   F A 4 

3

a  3.557  mm 1


 3 m1  m2  b  kb   F A 4 

3

b  0.263  mm


MACHINE DESIGN - An Integrated Approach, 4th Ed. A  π a  b

Contact area 4.

A  2.94254  mm

2


Maximum pressure 5.

7-43-2

p avg 

F

p max 

3


A

2

 p avg



σx  2  ν  ( 1  2  ν) 



σy  2  ν  ( 1  2  ν) 



Axial

b

 p  max

σx  1565 MPa

 p  max

σy  2570 MPa

a  b a

a  b

σz  p max

σz  2651 MPa


σ1  σx

σ2  σy

σ3  σz


τ13  6.

σ1  σ3

The maximum shear stress under the surface on the z-axis is approximately

τ13max  0.34 p max

Max shear stress 7.

8.

τ13  543  MPa

2

τ13max  901  MPa


b

k4 

1

k3  0.074

a

a

2

 a b

2

k4  0.997

These constants are used in equations 7.21c and d to find the shear stresses on the surface at the ends of the major and minor axes. Major axis

τxy  ( 1  2  ν) 

k3 k4

Minor axis

τxy  ( 1  2  ν) 

2

k3 k4

2

 

1

 k4

 atanh k4  1  p max





k3



k4

1 

 k4      pmax  k3  

 atan


τxy  77 MPa



7-44-1


A machine has two crowned, cylindrical rollers rolling against each other with a dynamic load of 0 to 3.5 kN. The first roller has a major radius of 14 mm with a crown radius of 80 mm. The second roller has a major radius of 75 mm and a crown radius of 100 mm. The two axes of rotation have a 30 degree angle between them. Find the contact stresses if both rollers are steel.

Given:

Roller 1 radius

R1  14.00  mm

Crown radius

R'1  80.0 mm

Roller 2 radius

R2  75.00  mm (radial)

Crown radius

R'2  100.0  mm (axial)

(90 deg to roller rad)

(normal to contact plane) Radial load F  3.5 kN Angle between planes of R1 and R2 θ  30 deg E1  206.8  GPa ν1  0.28

Material properties

steel




Material constants

m1 

1  ν1

3

m1  4.456  10

E1



1 GPa

m2  m1 2.


A 

1 2

 

1

 R1



1 R'1

1



R2



1

 

A  0.0536 mm

R'2 

1

1

     

2 2  1 1   1  1   B       R  R'1  2  R1  2 R'2   1 1  1 1   2  R  R'    R  R'   cos( 2  θ ) 1  2 2   1

1

Angle  Factors from equations 7.19e

B 180 ϕ  acos   A π

2

B  0.0303 mm

1

ϕ  55.56

 0.86215

ka  50.192 ϕ

ka  1.572 2

kb  0.0045333  0.043581 ϕ  0.0017292  ϕ   3.7374 10  1.4207 10

5

3

9

5

 ϕ  3.7418 10

7

4

 ϕ 

ϕ

kb  0.685



7-44-2



a  1.192  mm

1

4.


3  3 m1  m2  b  kb   F A 4 

b  0.519  mm

Contact area

A  π a  b

A  1.94438  mm


Maximum pressure

5.

2

p avg 

F

p max 

3


A

2

 p avg


The maximum normal stresses in the steel follower at the center of the contact patch at the surface are then found using equations 7.21a. In-plane

σx  2  ν1   1  2  ν1 



σy  2  ν1   1  2  ν1 



Axial

b

 p  max

σx  1872 MPa

 p  max

σy  2340 MPa

a  b a

a  b

σz  p max

σz  2700 MPa


σ1  σx

σ2  σy

σ3  σz


τ13  6.

τ13  414  MPa

2


7.

σ1  σ3

τ13max  0.34 p max

τ13max  918  MPa


b

k4 

1

k3  0.436

a

a

2

 a b

2

k4  0.9



7-44-3


Major axis

τxy   1  2  ν1 

k3

 

1

2  k4

 atanh k4  1  p max


 k4      p max  k3  




k4 Minor axis

τxy   1  2  ν1 

k3 2

k4



k3



k4

1 

 atan



7-45-1


A cam-follower system has a motion with a combination of rolling and sliding. The cam is cylindrical with a minimum radius of curvature of 80 mm. The roller follower is also cylindrical with a radius of 14 mm. They are both 18-mm long. The maximum contact force, which is normal to the contact plane, is 3200 N. Both the cam and roller are made from hardened steel. Assuming that the coefficient of friction between the cam and roller follower is 0.33, determine the maximum tensile, compressive, and shear stresses in the cam.

Given:

Roller radius

R1  14 mm

Cam radius

R2  80 mm

Roller length Radial load

L  18 mm F  3200 N

Solution: 1.

The parts are hardened steel. Therefore E  206.8  GPa

ν  0.28

Coefficient of friction

μ  0.33



m1 

Material constants

1ν

6

m1  4.456  10

E



1 MPa

m2  m1 B 

Geometry constant

1 2

 

1

 R1



1

 

B  0.042  mm

R2 

1

1


3.

a 

 2 m1  m2 F      B L π

2

a  0.1550 mm

The average and maximum contact pressure can now be found from equations 7.14b and c. F

Average pressure

p avg 

Maximum pressure

p max 

Tangential unit force


2 a L 2 F

π a  L

p avg  573.3  MPa p max  730.0  MPa fmax  240.9  MPa

With = 0.33, the principal stresses in the contact zone will be maximal on the surface (z = 0) at x = 0.3a from th centerline as shown in Figures 7-20 and 7-22. The applied stress components are found from equation 7.23a for the normal force and equation 7.23b for the tangential force. For

x  0.3 a

σxn  p max  1 

x

2

a

σxt  2  fmax 

2

x

σxt  144.53 MPa

a

σzn  p max  1 

σxn  696.4  MPa

x

2

a

2

σzn  696.4  MPa



σzt  0  MPa

τxzn  0  MPa


x

2

a 4.

5.

τxzt  229.8  MPa

2



σx  840.886  MPa


σz  696.351  MPa


τxz  229.796  MPa

Since the contact patch is short, we expect a plane stress condition to exist. The stress in the third dimension is:

σy  0  MPa 6.

7-45-2

σy  0.000  MPa


σx 

σx

σy 

MPa

τxy  0

σy MPa

τyz  0

C2  σx  σy  σz C1 

τxz 

σz MPa

τxz MPa

C2  1.537  10



σz 

3

5

C1  5.327  10

C0  0

2

f ( σ )  σ  C2 σ  C1 σ  C0

 C0    C1   v   C2     1 

s  polyroots ( v)  MPa

 1009.5  s   527.7   MPa    0.0 

Principal stresses:

σ1  s

σ1  0  MPa

σ2  s

σ2  527.7  MPa

σ3  s

σ3  1009.5 MPa

3 2 1



7-45-3


τ13  7.

σ1  σ3 2

τ13  504.8  MPa




7-46-1


Estimate how long it will take to remove 2 μm of material from the 5000 mm2 surface of a block of HB110 steel if a coarse polishing machine applies 80 N over a 400-mm stroke at 120 strokes per minute. (a) If done dry. (b) If done lubricated.

Units:

kilo  kg g

Given:

Block surface

Aa  5000 mm Stroke rate

n  120  min

Depth of wear

d  2  μm

Stroke

Force on file

F  80 N

Steel hardness

s  400  mm kilo HB  110  2 mm

2

1

Assumptions: Only one face will be polished. Solution: 1.


This is a two-body abrasion problem. From Table 7-2, the wear coefficients for dry and lubricated abrasion in a coarse polishing operation are Wear coefficients

Ka  1  10 Kb  2  10

4 4

2.

The length of sliding is L = s n  t where t is the time required.

3.

The depth of material removed is d = K

4.

Combining these two equations and solving for the time, t

(a) dry

lubricated

F L HB Aa

ta 

d  HB Aa Ka F  s n

strokesa  ta n

(b) lubricated

dry

tb 

d  HB Aa Kb F  s n

strokesb  tb n

ta  28 min strokesa  3371

tb  14 min strokesb  1686



7-47-1

PROBLEM 7-47

_____

Statement:

Loose abrasive grains are introduced in error into the lubricating system of a flat bronze thrust bearing that has a hardness of 60HB and a surface area of 500 mm2. If a hardened steel part exerts a force of 50 N on the bearing while oscillating across it at 200 strokes/min with a stroke of 30 mm, what depth of wear will occur in an 8-hour shift?

Units:

stroke  1

Given:

Hardness

H  60 kgf  mm

Shift time

t  8  hr

Wear coefficient

K  2  10

1.

Stroke

s  30 mm

2

1

From equation 7.7b, the depth of wear is F l H  Aa

The length of wear is, l = s n  t

3.

Aa  500  mm


d = K 2.

Area of wear

n  200  stroke min F  50 N

Stroke rate Contact force Solution:

3

2

where t is the elapsed time

Substituting for l ,

d  K

F  s n  t H  Aa

d  1.0 mm



7-48-1


Two rollers are in contact with a 9% sliding combined with rolling and the resulting maximum compressive principal stress in the contact zone is 15500 psi. Both rollers are made from 6061-T6 hard anodized aluminum. The design life of the rollers is 4 years of 2-shift operation at 260 days/year and they each turn at 200 rpm. What is the expected safety factor against pitting for the roller pair?

Given:

Roller material properties E  10.4 10  psi

Solution: 1.

6

Roller speed

n  200  rpm

Max stress Design life

σmax  15500  psi

ν  0.34

shift  2  day

life d  4  yr

Calculate the required cycle life. 9

1ν

8 1

m1  8.504  10

E

K  π  m1  m2  σmax

2

K  128.4  psi

From Table 7-7, Part 2, Line 36 the slope and intercept factors of this aluminum for rolling with 9% sliding are

λ  5.02

ζ  20.12

These are used in equation 7.26 along with the value of K from above to find the number of cycles that can be expected at this load before pitting begins. log ( K) =




K



ζ λ log   psi  Nlife  10

6.

m2  m1

psi

The maximum normal stress is σmax  15.5 ksi, compressive. Its K-factor can be calculated from equation 7.25d. K-factor

5.

cycles  1.255  10

Calculate the material constants. 2

4.

1


m1  3.

hrspershift  8  hr daysperyr  260  day yr

cycles  n  shift hrspershift daysperyr life d 2.

1

9

Nlife  3.4  10

cycles

Calculate the factor of safety against pitting. Nf 

Nlife cycles

Nf  2.7



7-49-1

PROBLEM 7-49

_____

Statement:

Two contacting rollers run together in pure rolling. Both are made from Class 20 Gray iron, HB 130-180. One roller has a diameter of 2.75 in and the other has a diameter of 3.25 in. Both are 10 in long. The applied load is 5500 lbf. If the design life is 1E08 cycles, determine the factor of safety against pitting failure.

Given:

Material data from Table 7-7 Roller radii R1  2.75 in Roller length Applied force

Solution: 1.

L  10.0 in F  5500lbf


Find the geometry constant from equation 7.9a. Geometry constant

2.

B 

1 2

 

1

 R1



  R2  1

B  0.336 

1 in

Calculate the allowable force on the rollers using equation 7.25d.

Allowable force

3.

K  960  psi R2  3.25 in

Fallow 

K L 2 B

Fallow  14300  lbf

Calculate the factor of safety using equation f in Example 7-5.

Safety factor

Nf 

Fallow F

Nf  2.6



10-1a-1


A simply supported shaft is shown in Figure P10-1. A constant magnitude transverse load P is applied as the shaft rotates subject to a time-varying torque that varies from Tmin to Tmax. For the data in row a of Table P10-1, find the diameter of shaft required to obtain a safety factor of 2 in fatigue loading if the shaft is steel of S ut = 108 ksi and S y = 62 ksi. The dimensions are in inches, the force in pounds, and the torque is in lb-in. Assume no stress concentrations are present.

Given:

Distance between bearings a  16 in Applied load P  1000 lbf

Distance to P Tensile strength

b  18 in S ut  108  ksi

Minimum torque

Tmin  0  lbf  in

Yield strength

S y  62 ksi

Maximum torque

Tmax  2000 lbf  in


Nd  2

Assumptions: The finish is machined, reliability is 99%, and the shaft is at room temperature. Solution: 1.

See Figure P10-1 and Mathcad file P1001a.

The maximum moment in the shaft occurs at the right bearing as seen in the moment diagram in Figure B-3(a) in Appendix B (note that in the figure a is the distance to the load and b is the distance between bearings). Using the equation given in the figure, calculate the alternate bending moment (the mean is zero). Ma  P ( a  b )

2.

Ma  2000 in lbf

Calculate the mean and alternating components of torque. Tm  Ta 

Tmax  Tmin

Tm  1000 in lbf

2 Tmax  Tmin

Ta  1000 in lbf

2

S'e  0.5 S ut

S'e  54 ksi

3.


4.

Determine the endurance limit modification factors for a rotating round shaft. Load

Cload  1

Size

Csize( d )  0.869  

Surface

A  2.70

   in 

Csurf

5.

 S ut   A     ksi 

Temperature

Ctemp  1

Reliability


d

 0.097

b  0.265

(machined)

b

Csurf  0.781

(R = 99%)

Determine the modified endurance limit as a function of the unknown diameter, d. S e( d )  Cload  Csize( d )  Csurf  Ctemp Creliab S'e

6.

Use equation (10.8) with unity for all stress concentration factors as a design equation to find d.



Guess

10-1a-2

d  1  in

Given 1

3 2   2   32 Nd  Ma  4  Ta T 3 m  d=     4 S ut  S e( d )  π  d  Find ( d )

3

d  1.188  in

Using this value for d, the size modification factor and endurance limit are: Size modification factor

Csize( d )  0.855

Endurance limit

S e( d )  29.3 ksi



10-2a-1


A simply supported shaft is shown in Figure P10-2. A constant magnitude distributed load p is applied as the shaft rotates subject to a time-varying torque that varies from Tmin to Tmax. For the data in row a of Table P10-1, find the diameter of shaft required to obtain a safety factor of 2 in fatigue loading if the shaft is steel of S ut = 745 MPa and Sy = 427 MPa. The dimensions are in cm, the distributed force in N/cm, and the torque is in N-m. Assume no stress concentrations are present.

Given:

Dist. between bearings Distance to end of p

L  20 cm b  18 cm

Applied distributed load Minimum torque

p  1000 N  cm Tmin  0  N  m

Maximum torque

Tmax  2000 N  m

Distance to p a  16 cm Tensile strength S ut  745  MPa 1

Yield strength S y  427  MPa Design safety factor Nfd  2



The maximum moment in the shaft occurs between a and b. See the appendix to this problem below for the determination of Ma. Ma  48.45  N  m

2.


Tmax  Tmin

Tm  1000 N  m

2 Tmax  Tmin

Ta  1000 N  m

2

S'e  0.5 S ut

S'e  372.5  MPa

3.


4.


Cload  1

Size

Csize( d )  1.189  

d

Surface

A  4.51

b'  0.265

   mm 

Csurf

5.

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


 0.097

(machined)

b'

Csurf  0.782

(R = 99%)


6.

Use equation (10.8) with unity for all stress concentration factors as a design equation to find d.



10-2a-2

d  1  in

Guess Given

1

3 2   2   32 Nfd  Ma  4  Ta 3 Tm  d=     4 S ut  S e( d )  π  d  Find ( d )

3

d  48.6 mm


Csize( d )  0.816

Endurance limit

S e( d )  193.4  MPa

APPENDIX - Maximum bending moment 1.

From inspection of Figure P10-2, write the load function equation q(x) = R1-1 - p0 + p0 + R2-1

2.

Integrate this equation from - to x to obtain shear, V(x) V(x) = R10 - p1 + p1 + R20

3.

Integrate this equation from - to x to obtain moment, M(x) M(x) = R11 - p2/2 + p2/2 + R21

4.

Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0 V = R1  p  ( L  a )  p  ( L  b )  R2 = 0 M = R 1 L  R1 

p 2 L

p 2

2

 ( L  a) 

p 2

 ( L  a )  ( L  b ) 2

2

 ( L  b) = 0 2



R2  p  ( b  a )  R1

R1  300  N R2  1700 N

x  0  m 0.005  L  L

5.


6.


7.

Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions.



10-2a-3

V ( x)  R1 S ( x 0  m)  p  S ( x a )  ( x  a )  p  S ( x b )  ( x  b )  R2 S ( x L) M ( x)  R1 S ( x 0  m)  x  8.

p 2

2

 S ( x a )  ( x  a ) 

p 2

 S ( x b )  ( x  b )

2

Plot the shear and moment diagrams. Shear Diagram 500 0 V ( x)  500 N

 1000  1500  2000

0

5

10

15

20

x cm

Moment Diagram 50

35 M ( x) 20

Nm

5

 10

0

5

10

15

20

x cm

FIGURE 10-2a Shear and Moment Diagrams for Problem 10-2a

9.

Determine the maximum maximum moment from inspection of the diagrams. Maximum moment occurs where zero, which is x = c. From the shear diagram, ca R1

=

bc R2

Mmax  M ( c)

c 

a  R2  b  R1 R1  R2

c  16.300 cm Mmax  48.45  N  m



10-3-1


For the bicycle pedal-arm assembly in Figure P6-1 assume a rider-applied force that ranges from 0 to 1500 N at the pedal each cycle. Design a suitable shaft to connect the two pedal arms. Use a fatigue safety factor of 2 and a material with S ut = 500 MPa. The shaft has a square extension on each end where it inserts into the pedal arms.

Given:

Material tensile strength

S ut  500  MPa


Applied load by rider

Fridermax  1500 N

Fridermin  0  N

Nfd  2

Assumptions: 1. The configuration is similar to that shown in Figure 10-3 with the dimensions a, b, and c below. 2. The moments in the shaft are greater when the right pedal is loaded. 3. The torque is transmitted from the sprocket to the shaft through the square shaft extension. 4. The shaft is machined, reliability is 50%, and the bicycle is not operated in extreme temperatures. Shaft dimensions: a  10 mm b  22 mm c  80 mm See Figure 10-3 and Mathcad file P1003. Solution: 1.

From problems 3-3 and 7-3, the moment, torque, and chain force Fchain are:

Sprocket

Fchain

Frame Bearing (2 Places)

R 1z

Pedal Arm (Both Ends)

Mmax  90 N  m

T

Mmin  90 N  m

x

T

Tmax  255  N  m Tmin  0  N  m

Horizontal Section (Top View)

d sprocket  100  mm Fchainmax 

y

2  Tmax

c

d sprocket

a R 2y

Frider

Fchainmax  5100 N

2.

R 2z

z

Note that in Problem 3-3 the pedal-arm torque becomes a concentrated bending moment on this shaft and the pedal-arm bending moment becomes the sprocket torque, which is carried along the shaft from the y-axis to the centerplane of the sprocket, the distance a (see Figure 10-3A).

d

x

M

R 1y b Vertical Section (Front View) FIGURE 10-3A Free Body Diagram for Problem 10-3

3.

Find the moment distribution along the shaft from x = 0 to x = c. This will be done by finding the moment paral to the z-axis by looking at the forces and moments in the x-y plane, then finding the moment parrallel to the y-ax by looking at the forces in the x-z plane. These two moment distributions will be added vectorially to get the total moment distribution.

4.

Determine the bearing reactions R1y and R2y in the x-y plane.

 Fy:

Frider  R1y  R2y = 0



 Mz:

10-3-2

M  b  Frider  ( c  b )  R2y = 0 R2ymax 

Mmax  b  Fridermax

R2ymax  2121 N

cb

R1ymax  Fridermax  R2ymax 5.

R1ymax  3621 N

Use singularity functions to determine the moment distribution of the forces in the x-y plane. q xy = -M-2 - Frider-1 + R1y-1 - R2y-1 Vxy = -M-1 - Frider0 + R1y0 - R2y0 Mxy = -M0 - Frider1 + R1y1 - R2y1 x  0  mm 0.005  c  c

6.


7.

For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x u )  if ( x  u 1 0 )

8.

Write the moment equation in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. Mxy( x)  Mmax  Fridermax S ( x 0  mm)  x  R1ymax S ( x b )  ( x  b )  R2ymax S ( x c)  ( x  c)

9.

Plot the moment diagram (see Figure 10-3B). 0

10. Determine the bearing reactions R1z and R2z in the x-z plane.

 Fz :

Fchain  R1z  R2z = 0

 Mz:

b  Fchain  ( c  b )  R2z = 0

R2zmax 

b  Fchainmax

 50 Mxy ( x) Nm  100

cb

R2zmax  1934 N

 150

0

20

60

80

x

R1zmax  Fchainmax  R2zmax R1zmax  7034 N

40 mm

FIGURE 10-3B Moment Diagram for x-y Plane for Problem 10-3

11. Use singularity functions to determine the moment distribution of the forces in the x-z plane. q xz = - Fchain-1 + R1z-1 - R2z-1 Vxz = - Fchain0 + R1z0 - R2z0 Mxz = - Fchain1 + R1z1 - R2z1 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


10-3-3

12. Write the moment equation in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. Mxz( x)  Fchainmax S ( x 0  mm)  x  R1zmax  S ( x b )  ( x  b )  R2zmax  S ( x c)  ( x  c) 13. Plot the moment diagram (see Figure 10-3C). 14. Determine the total moment distribution. Mtot ( x) 

2

Mxy( x)  Mxz( x)

0

2

This is plotted in Figure 10-3D.

200

150  50 Mxz( x)

Mtot ( x)

Nm

Nm

100

 100 50

 150

0

20

40

60

0

80

0

20

40

x mm

FIGURE 10-3C

60

80

x mm

FIGURE 10-3D

Moment Diagram for x-z Plane for Problem 10-3

Total Moment Diagram for Problem 10-3

15. From Figure 10-3D, we see that the moment is a maximum at x = b. However, the torque is carried only from x = to x = a so we will investigate the section just to the right of x = a where there is a stress concentration due to the shoulder. Moment at x = a

Mamax  Mtot ( a )

Mamax  116.73 N  m

Moment at x = b

Mbmax  Mtot ( b )

Mbmax  166.487  N  m

16. The bending moment is fully reversed while the torque is repeated from zero to Tmax. Therefore, we have the following alternating and mean components of moment and torque at x = a. Alternating moment

Ma  Mamax

Ma  116.73 N  m

Mean moment

Mm  0  N  m

Mm  0  N  m

Alternating torque

Ta  0.5 Tmax

Ta  127.5  N  m

Mean torque

Tm  0.5 Tmax

Tm  127.5  N  m

16. There will be a stress concentration due to the step just to the right of the sprocket and also because of the transition from square to round at the shoulder. Estimate this to be Kf  2. Also, let Kfm  Kf

Kfs  Kf

17. Calculate the unmodified endurance limit.

Kfsm  Kfs S'e  0.5 S ut

S'e  250  MPa



10-3-4

18. Calculate the endurance limit modification factors for a rotating round beam. Load

Cload  1

Size

Csize( d )  1.189  

d

Surface

A  4.51

b  0.265

   mm 

Csurf

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability

Creliab  1.0

 0.097

(machined)

b

Csurf  0.869

(R = 50%)

19. Calculate the modified endurance limit. S e( d )  Cload  Csize( d )  Csurf  Ctemp Creliab S'e 20. Use equation (10.8) to solve for the diameter at x = a. Guess d  12 mm Given 1

3   2 2  32 Nfd   Kf  Ma  4   Kfs Ta d=   S e( d )  π  d  Find ( d )

Kfm Mm

2

 S ut

3 4

  Kfsm Tm

2

3

  

d  35.526 mm

This is too big for a practical design. A stronger material should be chosen to bring this down to about 16 mm.



10-4a-1


Determine the maximum deflections in torsion and in bending of the shaft shown in Figure P10-1 for the data in row a in Table P10-1 if the steel shaft diameter is 1.75 in.

Given:

Shaft length Distance to P

L  20 in b  18 in

Distance between bearings a  16 in Shaft diameter d  1.750  in

Applied load

P  1000 lbf

Young's modulus

E  30 10  psi

Modulus of rigidity

G  11.5 10  psi

Maximum torque Tmax  2000 lbf  in Solution: 1.

Calculate the area moment of inertia and the polar moment of inertia. I  J 


π d

4

4

I  0.460  in

64

π d

4

4

J  0.921  in

32

Use the equation in Figure B-3(a) in Appendix B for this overhung beam with concentrated load to determine the maximum deflection in bending. Note that in the figure a is the distance to the load and b is the distance between bearings. This deflection occurs at the right end of the shaft, under the load P. ymax 

3.

6



2.

6

P 6  E I

 

ab

 a

3

L 

b a

 ( L  a )  ( L  b )  a  ( b  a )  L 3

3



ymax  0.00357  in

Calculate the torsional deflection using equation (10.9a).

θmax 

Tmax L G J

θmax  0.216  deg



10-5a-1


Determine the maximum deflections in torsion and in bending of the shaft shown in Figure P10-2 for the data in row a in Table P10-1 if the steel shaft diameter is 40 mm.

Given:

Distance between bearings Distance to end of p

Solution: 1.

L  20 cm b  18 cm 1


p  1000 N  cm Tmin  0  N  m

Maximum torque

Tmax  2000 N  m

Young's modulus E  206.8  GPa Mod of rigidity G  80.8 GPa

Calculate the area moment of inertia and the polar moment of inertia. I  J 


π d

4

I  125664 mm

64

π d

4

4

J  251327 mm

32

4

Calculate the torsional deflection using equation (10.9a).

θmax  3.

a  16 cm d  40 mm



2.

Distance to p Shaft diameter

Tmax L

θmax  1.129  deg

G J

From inspection of Figure P10-2, write the load function equation and integrate it twice to get the shear and moment equations. q(x) = R1-1 - p0 + p0 + R2-1 V(x) = R10 - p1 + p1 + R20 M(x) = R11 - p2/2 + p2/2 + R21

4.


p 2 L

p 2

2

 ( L  a) 

p 2

 ( L  a )  ( L  b ) 2

2

 ( L  b) = 0 2



R2  p  ( b  a )  R1 5.

R1  300  N R2  1700 N

Integrate twice more to get the slope and deflection equations.

(x) = [R12/2 - p3/6 + p3/6 + R22/2 + C3]/EI y(x) = [R13/6 - p4/24 + p4/24 + R23/6 + C3x + C4]/EI 6.

Solve for the constants of integration with the boundary conditions y = 0 at x = 0 and x = L. At x = 0

C4 = 0


MACHINE DESIGN - An Integrated Approach, 4th Ed. At x = L

0=

R1 6

C3 

3

L 

p 24

10-5a-2 4

 ( L  a) 

p 24

4

 ( L  b )  C 3 L

R1 3 p p 4 4  ( L  a)   ( L  b)  L  L  24 24 6  1



2

C3  1.950  N  m

x  0  m 0.005  L  L

7.


8.


9.

Write the slope and deflection equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions.

θ ( x) 

 R1  p 2 p 3 3  S ( x 0  mm)  x   S ( x a )  ( x  a )   S ( x b )  ( x  b )  C3 E I  2 6 6 

y ( x) 

 R1  p p 3 4 4  S ( x 0  mm)  x   S ( x a )  ( x  a )   S ( x b )  ( x  b )  C3 x E I  6 24 24 

1

1





10. Plot the slope and deflection diagrams (see Figure 10-5). 11. The maximum deflection occurs at the value of x for which  is zero. Let this be x = c, then guess c  10 cm Given

θ ( c) = 0

c  Find ( c)

12. The maximum deflection occurs at x = c and is

c  11.402 cm

ymax  y ( c)

ymax  0.00570  mm

SLOPE

DEFLECTION

0.15

0

0.1 3

 2 10

θ( x) 10

0.05

y ( x)

3

mm 0

3

 4 10

 0.05  0.1

0

FIGURE 10-5

5

10

15

20

 6 10

3

0

5

10

x

x

cm

cm

15

20

Slope and Deflection Diagrams for Problem 10-5a



10-6a-1


Determine the size of key necessary to give a safety factor of at least 2 against both shear and bearing failure for the design shown in Figure P10-3 using the data from row a in Table P10-1 and the data given below. Assume a shaft diameter of 1.75 in.

Given:

Shaft diameter

d  1.750  in

Shaft properties: S ut1  108  ksi

Maximum torque Tmax  2000 lbf  in

S y1  62 ksi

Minimum torque Tmin  0  lbf  in

Key properties:

Nd  2


S ut2  88 ksi S y2  52 ksi

Assumptions: 1. The finish is machined, reliability is 90%, and the shaft is at room temperature. 2. Note that the given magnitudes of the radial forces shown acting on gear(s) in this problem are not necessarily consistent with a load associated with the given torque for any real gear of reasonable pressure angle. Since gears are taken up in a later chapter, these shaft design problems ignore the real gear loadings and use an arbitrary value to provide a shaft design exercise. Solution: 1.


As recommended in Table 10-2, for a shaft diameter of d  1.75 in, use a square key of width w  0.375  in

Key width 2.

Determine the alternating and mean key shear force components. Mean and alternating force components

3.

4.

1 Tmax  Tmin  2 0.5 d

Fa  1143 lbf

Fm 

1 Tmax  Tmin  2 0.5 d

Fm  1143 lbf

Write the equations for the mean and alternating components of the shear stress and the effective von Mises stress. Key shear area as Ashear = w L function of length Fa

Mean and alternating shear stresses

τa =

Mean and alternating von Mises stresses

σ'a = 3  τa = 3 

τm =

w L Fa w L

Fm w L

σ'm = 3  τm = 3 

Fm w L

Using the modified Goodman failure criterion, the design equation is

Nd =

S e S ut S ut σ'a  S e σ'm

Solving for L 5.

Fa 

=

S e S ut

 3   S F  S F    ut a e m  w L 

L=

3  Nd   S ut Fa  S e Fm w S ut S e

Determine the key material endurance limit.




10-6a-2

S'e  0.5 S ut2

S'e  44 ksi

Correction factors: Load

Cload  1

Size

A95 = w L

d eq =

w L 0.0766

 0.097

Csize = 0.869  d eq Cs  0.869  

  0.0766  in   w

 0.5 0.097

Cs  0.805

 0.0485

Csize = 0.805  L

 Sut2   2.7    ksi 

Surface (machined)

Csurf

Temperature

Ctemp  1

Reliability


 0.265

(R = 90%)

 0.0485

S e = Cload  Csize Csurf  Ctemp Creliab S'e = 1  Cs L

6.

 0.0485

 Csurf  1  Crelaib S'e = Ce L

Ce  Cs Csurf  Creliab S'e

where

Csurf  0.824

Ce  26.175 ksi

Substitute S e into the design equation. 3  Nd   S ut Fa  Ce L

 Fm

 0.0485

L=

 0.0485

w S ut Ce L Solving by iteration, let

RHS( L) 

 0.0485   L    Fm 3  Nd  S ut2 Fa  Ce     in  

L w S ut2 Ce    in  Guess

L  0.5 in

RHS( L)  0.510  in

L  RHS( L)

RHS( L)  0.510  in

Tentatively, let 7.

 0.0485

L  0.625  in

(5/8 in)

Check the realised factor of safety against fatigue failure. Endurance limit

S e  Ce 

L

  in 

 0.0485

S e  26.778 ksi



Realised factor of safety

8.

S e S ut2

 3   S F  S F    ut2 a e m  w L 

Nf  2.4

Check worst-case static bearing stress. Bearing area

9.

Nf 

10-6a-3

Abear 

1 2

2

 w L

Abear  0.117  in

Maximum force

Fmax  Fa  Fm

Maximum bearing stress on key

σmax 

Factor of safety against static bearing yield

Ns 

Fmax

Fmax  2286 lbf

σmax  19.50  ksi

Abear

S y2

Ns  2.7

σmax

Design Summary Key width

w  0.375  in

(3/8 x 3/8)

Key length

L  0.625  in

(5/8 in)



10-7a-1


Determine the size of key necessary to give a safety factor of at least 2 against both shear and bearing failure for the design shown in Figure P10-4 using the data from row a in Table P10-1 and the data given below. Assume a shaft diameter of 40 mm.

Given:

Shaft diameter

d  40 mm

Maximum torque

Tmax  2000 N  m

Minimum torque

Tmin  0  N  m


Nd  2

Shaft properties: S ut1  745  MPa S y1  427  MPa Key properties:

S ut2  600  MPa S y2  360  MPa

Assumptions: 1. The finish is machined, reliability is 90%, and the shaft is at room temperature. 2. Note that the given magnitudes of the radial forces shown acting on gear(s) in this problem are not necessarily consistent with a load associated with the given torque for any real gear of reasonable pressure angle. Since gears are taken up in a later chapter, these shaft design problems ignore the real gear loadings and use an arbitrary value to provide a shaft design exercise. Solution: 1.


As recommended in Table 10-2, for a shaft diameter of d  40 mm, use a square key of width w  10 mm

Key width 2.


3.

4.

1 Tmax  Tmin  2 0.5 d

Fa  50.0 kN

Fm 

1 Tmax  Tmin  2 0.5 d

Fm  50.0 kN



τa =


σ'a = 3  τa = 3 

τm =

w L Fa w L

Fm w L

σ'm = 3  τm = 3 

Fm w L


Nd =


Solving for L 5.

Fa 

=

S e S ut

 3   S F  S F    ut a e m  w L  L=


Determine the key material endurance limit. Uncorrected endurance strength

S'e  0.5 S ut2

S'e  300  MPa



10-7a-2


Cload  1

Size

A95 = w L

d eq =

w L 0.0766

 0.097

Csize = 1.189  d eq Cs  1.189  

  0.0766  mm   w

 0.5 0.097

Cs  0.939

 0.0485

Csize = Cs L

 S ut2   4.51    MPa 

Surface (machined)

Csurf

Temperature

Ctemp  1

Reliability


 0.265

(R = 90%)

 0.0485


6.

 0.0485

 Csurf  1  Creliab S'e = Ce L


where

Csurf  0.828

Ce  209.145  MPa


 Fm

 0.0485

L=

 0.0485


RHS( L) 

 0.0485   L     Fm 3  Nd  S ut2 Fa  Ce     mm  

L  w S ut2 Ce    mm  Guess

L  10 mm

RHS( L)  121.468  mm

L  RHS( L)

RHS( L)  133.39 mm

L  RHS( L)

RHS( L)  133.866  mm

Tentatively, let 7.

 0.0485

L  134  mm

(both keys, divide by 2 for one)

Check the realised factor of safety against fatigue failure.

Endurance limit

S e  Ce 

   mm  L

 0.0485

S e  164.923  MPa




8.

S e S ut2

 3   S F  S F    ut2 a e m  w L 

Nf  2.0


9.

Nf 

10-7a-3

Abear 

1 2

 w L

Maximum force



σmax 


Ns 

Abear  670  mm

2

Fmax  100  kN

Fmax

σmax  149.25 MPa

Abear

S y2

Ns  2.4

σmax


w  10 mm

Key length

L1  0.5 L

L1  67 mm



10-8-1


A paper mill processes rolls of paper having a density of 984 kg/m3. The paper roll is 1.50-m outside diameter (OD) x 0.22-m inside diameter (ID) x 3.23-m long and is on a simple supported, hollow, steel shaft with S ut = 400 MPa. Find the shaft ID needed to obtain a dynamic safety factor of 2 for a 10-yr life if the shaft OD is 22 cm and the roll turns at 50 rpm with 1.2 hp absorbed.

Given:

Paper roll: Density Outside dia. Inside dia. Length Shaft: Strength

ρ  984 

y

kg 3

w

m OD  1500 mm ID  220  mm L  3230 mm

x

S ut  400  MPa

Outside dia. Factor of safety

od  220  mm Nf  2

Power Speed

HP  1.2 hp ω  50 rpm

1.

V L/2

L x

0

-R M

2

wL /8

x

0 L/2

L

FIGURE 10-8 Load, Shear, and Moment Diagrams for Problem 10-8


The weight of the paper roll is, Volume Weight

2.

R

R

Assumptions: 1. The shaft is stiffer than the paper roll so the weight of the roll on the shaft can be modelled as a uniformly distributed load. 2. The bearings that support the shaft are close to the ends of the paper roll and are thin with respect to the length of the roll so we can consider the distance between the shaft supports to be the same as the length of the roll. 3. The shaft is machined, reliability is 99.9%, and it is at room temperature. 4. The paper machine operates 3 shifts/day. Solution:

L

R

V 

π 4



2

2



3

 OD  ID  L

V  5.585  m

W  ρ  g  V

W  53.895 kN

From Figure 10-8, we see that the bending moment in the shaft is a maximum at the center of the span. First, determine the magnitude of the distributed load, then find the maximum bending moment using Figure B-2(b) in Appendix B with a = 0 and x = L/2. Distributed load

w 

W

w  16.686

L 2

Maximum moment

w L

Mmax 

newton mm 7

Mmax  2.176  10  newton  mm

8

3. Using equation 4.11b, find the maximum bending stress as a function of the unkown shaft inside diameter, id. This is the only alternating stress element present at this point on the shaft and there is no alternating shear stress at this point so max = 1 and 2 = 3 = 0. Furthermore, since 2 and 3 are zero, max = 'a. Bending stress at midspan

σmax =

M c I

=

32 Mmax od



4



4

π od  id

= σ'a



4.

5.

10-8-2

Calculate the steady shaft torque using equation (10.1a) and the mean von Mises stress using equation (5.7d). HP

Mean torque

Tm 

Tm  171  N  m

Mean torsional stress

τm =


σ'm = 3  τm

ω 16 Tm od



4



4

π od  id

Determine the number of stress cycles. Let Nshifts  3

shift_hours 

2080 hr yr

Lcycle  10 yr

Ncycles  ω shift_hours  Nshifts Lcycle

Stress cycles

Ncycles  1.176  10

9

Since this is greater than 10 6, design for infinite life. S'e  0.5 S ut

S'e  200  MPa

6.


7.


Cload  1

Size

Csize  1.189  

Surface

A  4.51

  mm  

Csurf

8.

od

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


 0.097

b  0.265

(machined)

b

Csurf  0.922

(R = 99.9%)

Determine the modified endurance limit as a function of the unknown diameter, d. S e  Cload  Csize Csurf  Ctemp Creliab S'e

9.

Csize  0.705

S e  97.819 MPa

Use equation (10.7b) with unity for all stress concentration factors as a design equation to find id. Nf =

S e S ut


Substituting the expressions for 'a and 'm found above,



Nf =

10-8-3

S e S ut 32 Mmax od



4



4

π od  id

 S ut 

16 3  Tm od



4



4

π od  id

 Se

Solving for the unknown inside diameter, 1

 4 16 Nf  od  2  Mmax Sut  3 Tm S e id  od   π S e S ut   Round this down (for a thicker wall) to

id  190  mm

The wall thickness will be

t  0.5 ( od  id)

4

id  191.467  mm

t  15 mm



10-9a-1


Repeat Problem 10-1 taking the stress concentration at the keyway shown in Figure P10-3 into account.

Given:

Distance between bearings Applied load

a  16 in P  1000 lbf

Distance to P b  18 in Tensile strength S ut  108  ksi

Minimum torque


Yield strength

S y  62 ksi

Design safety factor Nd  2 Maximum torque Tmax  2000 lbf  in Assumptions: 1. The finish is machined, reliability is 99%, and the shaft is at room temperature. 2. The notch radius in the keyway is r  0.015  in. 3. Note that the given magnitudes of the radial forces shown acting on gear(s) in this problem are not necessarily consistent with a load associated with the given torque for any real gear of reasonable pressure angle. Since gears are taken up in a later chapter, these shaft design problems ignore the real gear loadings and use an arbitrary value to provide a shaft design exercise. Solution:

See Figures P10-1 and P10-3, and Mathcad file P1009a.

1.

There is no stress concentration at the point on the shaft where the bending moment is a maximum. On the other hand, at the gear where there is a stress concentration, the bending moment is zero. The question is: which point requires the larger diameter in order to meet the safety factor requirement? To answer this question, find the diameter required at each point and choose the larger. Start with the point where the bending moment is maximum.

2.

The maximum moment in the shaft occurs at the right bearing as seen in the moment diagram in Figure B-3(a) in Appendix B (note that in the figure a is the distance to the load and b is the distance between bearings). Using the equation given in the figure, calculate the alternating bending moment (the mean is zero). Ma  P ( a  b )

3.

Ma  2000 in lbf


Tmax  Tmin

Tm  1000 in lbf

2 Tmax  Tmin

Ta  1000 in lbf

2

S'e  0.5 S ut

S'e  54 ksi

4.


5.


Cload  1

Size

Csize( d )  0.869  

Surface

A  2.70

  in  

Csurf

 S ut   A     ksi 

Temperature

Ctemp  1

Reliability


d

 0.097

b  0.265

(machined)

b

Csurf  0.781

(R = 99%)



6.

10-9a-2


7.

Use equation (10.8) with unity for all stress concentration factors as a design equation to find d. d  1  in

Guess Given

1

3 2   2   32 Nd  Ma  4  Ta T 3 m  d=     4 S ut  S e( d )  π 

3

Diameter required at point of maximum moment: d 1  Find ( d )

d 1  1.188  in

Using this value for d, the size modification factor and endurance limit are:

8.

Size modification factor

Csize d 1  0.855

Endurance limit

S e d 1  29.3 ksi

At the gear the bending moment is zero but the alternating and mean torques are the same as above. Also, there is a stress concentration due to the keyway that must be applied to the alternating and mean shear stresses present at the gear. First, set the moment equal to zero: Ma  0  lbf  in, then determine the values of the alternating and mean fatigue stress concentration factors.

9.

Determine the geometric stress concentration factor from the upper curve in Figure 10-16 using a curve-fit equation. Kts( d )  1.251  

r

 0.230

 d

10. Determine the notch sensitivity of the material. Note from Figure 6-36, part 1, that a value of 20 ksi should be added to S ut to obtain a 1/2 from Table 6-6. Using a curve-fit to Table 6-6,



2 3   3.10030  10 7  S    in    ksi   ksi   ksi   4 5 6   9 S  12 S  15 S  1.38322  10     3.28018  10     3.21209  10       ksi   ksi   ksi  

a ( S )  8.69657  10

2

3 S

 2.75956  10



 

5

 3.94116  10

Neuber constant

a  a  S ut  20 ksi

Notch radius

r  0.015  in

Notch sensitivity

q 

1 1

S

0.5

a  0.045  in

q  0.732 a r



10-9a-3

11. Determine the fatigue stress concentration factor from equation (6.11b). Alternating factor

Kfs( d )  1  q   Kts( d )  1 

Mean factor

Kfsm( d )  Kfs( d )

12. Repeat step 7 using these stress concentration factors with the torsional stresses. Given 1

3   2 2   32 Nd  Ma  4  Kfs( d )  Ta  T 3 m d=    Kfsm( d )   4 S e( d ) S ut   π 

Diameter required at point of gear attachment:

d 2  Find ( d )

Putting the two calculated diameters into a vector

trial_dia 

3

d 2  1.129  in

 d1     d2 

13. The larger of the two must be used as the minimum shaft diameter. Thus, d  max( trial_dia ). d  1.188  in 14. The factor of safety at the point of maximum bending moment is Nd  2 . The factor of safety at the gear is equal to the ratio of the two diameters cubed times Nd. Thus, 3

 d1  Nfgear     Nd  d2 

Nfgear  2.3



10-10a-1


Repeat Problem 10-2 taking the stress concentration at the keyway shown in Figure P10-4 into account.

Given:

Distance between bearings Distance to end of p

L  20 cm b  18 cm 1


p  1000 N  cm Tmin  0  N  m

Maximum torque

Tmax  2000 N  m

Distance to p Tensile strength

a  16 cm S ut  745  MPa

Yield strength Design safety factor

S y  427  MPa Nfd  2

Assumptions: 1. The finish is machined, reliability is 99%, and the shaft is at room temperature. 2. The ratio of notch radius to shaft diameter in the keyway is roverd  0.021. 3. The stress concentration factors for the mean stresses are Kfm  1 and Kfsm  1. Solution:

See Figures P10-2 and P10-4, and Mathcad file P1010a.

1.

There is no stress concentration at the point on the shaft where the bending moment is a maximum, which is at x = 16.3 cm (see the appendix to this problem, below). The keys are at a  16 cm and b  18 cm. On the other hand, the key at a  16 cm looks as if it extends into the section where the moment is a maximum so, use the maximum moment as the alternating bending moment.

2.


3.


4.

Tmax  Tmin 2 Tmax  Tmin 2

Tm  1000 N  m Ta  1000 N  m

Determine the geometric stress concentration factors from Figure 10-16 using curve-fit equations. Bending

Kt  0.4521 ( roverd)

Torsion

Kts  1.251  ( roverd)

 0.4115

 0.230

Kt  2.2 Kts  3.0

5. Determine the notch sensitivity of the material for bending and torsion. Note from Figure 6-36, part 1, that a value of 20 ksi should be added to S ut to obtain a 1/2 from Table 6-6 for the torsional Neuber constant. Bending

Lookup value of S ut

S ut  108  ksi

Neuber constant

a N  0.056  in

Notch sensitivity

q ( d ) 

2

1 aN

1

Torsion

0.5

a N  0.056  in

roverd d



Neuber constant

a N  0.045  in

2

S'ut  128  ksi 0.5

a N  0.045  in



10-10a-2

1

q s( d ) 

Notch sensitivity

aN

1 6.

roverd d

Determine the fatigue stress concentration factors from equation (6.11b). Bending factor

Kf ( d )  1  q ( d )   Kt  1 

Torsion factor

Kfs( d )  1  q s( d )   Kts  1  S'e  0.5 S ut

S'e  372.5  MPa

7.


8.


Cload  1

Size

Csize( d )  1.189  

d

Surface

A  4.51

b'  0.265

  mm  

 Sut    MPa 

 0.097

(machined)

b'


9.

Temperature

Ctemp  1

Reliability


Csurf  0.782

(R = 99%)


10. Use equation (10.8) as a design equation to find d. Guess

d  25 mm

Given 1

3   2 2   32 Nfd   Kf ( d)  Ma  4   Kfs( d)  Ta K  T 3 fsm m d=     4 S ut  S e( d )  π  d  Find ( d )

3

d  65.3 mm

Using this value for d, the size modification factor, stress concentration factors, and endurance limit are: Size modification factor

Csize( d )  0.793

Endurance limit

S e( d )  187.9  MPa

Stress concentration factors

Kf ( d )  2.0

Kfs( d )  2.7



10-10a-3

APPENDIX - Maximum bending moment 1. From inspection of Figure P10-2, write the load function equation q(x) = R1-1 - p0 + p0 + R2-1 2. Integrate this equation from - to x to obtain shear, V(x) V(x) = R10 - p1 + p1 + R20 3. Integrate this equation from - to x to obtain moment, M(x) M(x) = R11 - p2/2 + p2/2 + R21 4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = L+, V = M = 0 V = R1  p  ( L  a )  p  ( L  b )  R2 = 0 M = R 1 L  R1 

p

p 2

2

 ( L  a) 

2

2

 ( L  b) = 0

 ( L  a )  ( L  b ) 2

2 L

p

2



R1  300  N

R2  p  ( b  a )  R1

R2  1700 N

x  0  m 0.005  L  L


6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  R1 S ( x 0  m)  p  S ( x a )  ( x  a )  p  S ( x b )  ( x  b )  R2 S ( x L) M ( x)  R1 S ( x 0  m)  x 

p 2

2

 S ( x a )  ( x  a ) 

p 2

 S ( x b )  ( x  b )

2


Shear Diagram

0 V ( x)  500 N

 1000  1500  2000

0

5

10

15

20

x cm © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


Moment Diagram

10-10a-4

50 40 30

M ( x) 20

Nm

10 0  10

0

5

10

15

20

x cm

FIGURE 10-10a Shear and Moment Diagrams for Problem 10-10a

9. Determine the maximum maximum moment from inspection of the diagrams. Maximum moment occurs where zero, which is x = c. From the shear diagram, ca R1

=

bc R2

c 

Mmax  M ( c) Moment at x = b

a  R2  b  R1 R1  R2

c  16.300 cm Mmax  48.45  N  m

M ( b )  34 N  m



10-11a-1


Determine the amount of diametral interference needed to provide a suitable interference fit for the 6-in diameter by 1-in thick gear of Figure P10-3 using a shaft diameter of 1.75 in, such that the stresses in the hub and shaft will be safe and the torque from row a in Table P10-1 can be transmitted through the interference fit. Assume that both parts are steel with the properties given below.

Given:

Gear hub diameter

d hub  6.00 in

Shaft diameter:

Gear hub length Peak shaft torque

L  1.00 in Tp  2000 lbf  in

Young's modulus E  30 10  psi Poisson's ratio ν  0.28

d shaft  1.75 in 6

Assumptions: 1. Material properties are: S ut  108  ksi, and S y  62 ksi. 2. Note that the given magnitudes of the radial forces shown acting on gear(s) in this problem are not necessarily consistent with a load associated with the given torque for any real gear of reasonable pressure angle. Since gears are taken up in a later chapter, these shaft design problems ignore the real gear loadings and use an arbitrary value to provide a shaft design exercise. 3. The coefficient of friction between the hub and shaft is μ  0.15. Solution:


1.

As a design choice, let the design factor of safety on torque capacity and hub failure be

2.

Calculate hub and shaft radii.

3.

Nd  2

Hub radius

ro  0.5 d hub

ro  3  in

Nominal interface radius

r  0.5 d shaft

r  0.875  in

Shaft inside diameter

ri  0  in

The minimum interference is determine by the desired torque capacity. Use equation 10.14c to solve for the minimum diametral interference.

π L r μ  δmin E  ro  r  2

Torque capacity

T=



Nd

2  ro

2

2

2

Solving for min

δmin 

Let the minimum diametral interference be

2  ro  Nd  Tp

π L r μ  E  ro  r  2

2

δmin  0.00071  in

δmin  0.0007 in

Note that it is not practical to specify a dimension in inches that has more than 4 decimal places. 4.

Find the von Mises stress in the hub as a function of the unknown maximum diametral interference .

δ E  ro  r  2

Interference pressure Stress in shaft

p ( δ) 

2

2

4  r ro

Tangential

σti ( δ)  p ( δ)

Radial

σri( δ)  p ( δ)



10-11a-2

Stress in hub

σto( δ)  p ( δ) 

Tangential

2

2

2

2

ro  r ro  r

σro( δ)  p ( δ)

Radial

These are principal stresses. The tangential stress is 1 and the radial is 3. Use equation (5.7c) to find the von Mises stress.

δ E

σ'o( δ) 

von Mises

4  r ro

  3  ro  r 4

2

4

0.5



Kt  1

5.

There is no bending stress in the shaft at the gear, therefore

6.

The safety factors against failure (yielding in the shaft and hub) during press fit can now be used to find the maximum diametral interference: Guess

δ  δmin

Shaft

Given

Nd =

S y Kt σti ( δ)

δsmax  Find ( δ) Hub

δsmax  0.0040 in Nd =

Given

Sy Kt σ'o( δ)

δhmax  Find( δ)

δhmax  0.0021 in δmax  δhmax

Using the smaller of the two, 7.

8.

δmax  0.0021 in

The torque capacity of the joint (with an assumed coefficient of friction of μ  0.15 ) is about 2 times the peak transmitted torque and the safety factors against failure of the shaft and hub exceed 2 therefore, the minimum and maximum diametral interference below is acceptable. Minimum diametral interference

δmin  0.0007 in

Minimum diametral interference

δmax  0.0021 in

If we divide the tolerance on the shaft and hub equally and use the basic hole system, the shaft and hub bore specifications are: Tolerance on shaft or hub

t  0.5  δmax  δmin

t  0.0007 in

Hub bore diameter:

Dmin  d shaft

Dmin  1.7500 in

Dmax  Dmin  t

Dmax  1.7507 in

d max  d shaft  δmax

d max  1.7521 in

d min  d max  t

d min  1.7514 in

δmax  d max  Dmin

δmax  0.0021 in

Shaft diameter:

Maximum interference



Minimum interference 9.

10-11a-3

δmin  d min  Dmax

δmin  0.0007 in

Check the safety factors against torque capacity and static failure of the hub.

π L r μ  δmin E  ro  r  2

Torque

Hub strength

Ntorque 

Nhub 

Tp Sy

Kt σ'o δmax



2  ro

2

2

Ntorque  2.0

Nhub  2.0



10-12a-1


Assume that the device shown keyed to the shaft of Figure P10-3 is a Class 50, cast iron flywheel of 20-in outside diameter and 1-in thickness. The hub is 4-in dia and 3-in thick. Determine the maximum speed at which it can safely be run using a safety factor of 2. Use dimensions and other appropriate data from Problem 10-6 and row a in Table P10-1. Consider the transverse force P to be zero in this case.

Given:

Flywheel :

odfw  20 in

Safety factor

Nos  2

thickfw  1  in

Tensile strength

S ut  52.5 ksi

Hub:

odhub  4  in thickhub  3  in

Specific weight Poisson's ratio

γ  0.26 lbf  in ν  0.26

Shaft diameter

d shaft  1.75 in

3

Assumptions: The hub is thicker (in the axial direction) than the flywheel disk. This has the effect of strengthening the flywheel from the shaft out to the od of the hub. This effect will be ignored and the flywheel analyzed as if the hub had the same thickness as the disk portion. Solution: 1.


The stress is maximum at the inside radius of a flywheel, which in this case is at the shaft diameter. At this radiu the radial stress is zero so that the tangential stress is the only nonzero stress and is 1. With the other principal stresses equal to zero, the load line is along the s1 axis on the 1-3 diagram. For that case, the factor of safety equation is the same for all brittle material failure theories and is Ns =

2.

S ut

σ1

Use equation (10.23a) with t replaced by S ut (Ns = 1) to determine the speed at which the flywheel will fail. Inside radius

ri  0.5 d shaft

ri  0.875  in

Outside radius

ro  0.5 odfw

ro  10 in

Critical radius

r  ri

r  0.875  in 1

Failure speed



g  S ut

ωfail  

2 2 ri  ro   3  ν   2 2  γ  8    ri  ro  2 r  

   1  3  ν 2   r   3ν 

2

ωfail  9332 rpm 3.

The factor of safety on the operating speed (see Example 10-7) is defined as

Maximum safe operating speed

ωos 

ωfail Nos

Nos =

ωfail ωos

ωos  4666 rpm

Using this definition results in the static strength factor of safety, Ns, being equal to the square of Nos.



10-13a-1


Determine the critical frequency of shaft whirl for the assembly shown in Figure P10-3 using dimensions from row a of Table P10-1 and a steel shaft diameter of 2 in. Use the flywheel dimensions of Problem 10-12.

Units:

blob 

Given:

Flywheel :

lbf  sec

2

in

Hub:

1.

Shaft:

thickfw  1  in

Distance between bearings

a  16 in

odhub  4  in

Distance to flywheel

b  18 in

thickhub  3  in

Shaft diameter

d shaft  2.00 in

Young's modulus

E  30 10  psi

3

γ  0.26 lbf  in

Specific weight Solution:

odfw  20 in

6


Determine the shaft stiffness at the flywheel using Figure B-3(a) in Appendix B. Note that in Figure B-3(a) the dimensions a and b are the dimensions b and a, respectively in Figure P10-3. The deflection under the load (x=a in Figure B-3(a)) is y =

F 6  E I

 

ba

 b

3

a 

a b

 ( a  b )  b  ( a  b )  a 3



To use the symbols of Figure P10-3, change a to b and b to a. y =

F 6  E I

 

ab

 a

3

b 

b a

 ( b  a )  a  ( b  a )  b 3



The stiffness of the beam at the point of the applied load is k =

F y

. Solve the above equation for F/y = k

while substituting d 4/64 for I. 4

k 

3  π d shaft  E 32 

ab

 a

2.

3

b 

a

 ( b  a )  a  ( b  a )  b 3

in



Calculate the mass of the flywheel and hub. Volume of hub

Volume of flywheel Total mass

3.

b

5 lbf

k  9.817  10 

Vhub  Vfw  mtot 

π  2 2   odhub  d shaft   thickhub 4

π  2 2   odfw  odhub   thickfw 4

γ g

3

Vhub  28.274 in

3

Vfw  301.593  in

  Vhub  Vfw

mtot  0.222  blob

k

ωn  2102

Calculate the critical frequency.

ωn 

mtot

rad sec

ωn  20075  rpm © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

MACHINE DESIGN - An Integrated Approach, 4th Ed. fn 

4.

10-13a-2

ωn

fn  335  Hz

2 π

Deflection at the flywheel due to its own weight. δ 

mtot  g k

5

δ  8.736  10

 in



10-14a-1


Determine the critical frequency of shaft whirl for the assembly shown in Figure P10-4 using dimensions from row a of Table P10-1 and a steel shaft diameter of 40 mm. The cast-iron roller diameter is 3 times the shaft diameter.

Given:

Distance between bearings L  20 cm

Shaft Young's modulus

E  206.8  GPa

Distance to start of roller a  16 cm Distance to end of roller b  18 cm

Roller specific weight Roller diameter

γ  70580  N  m

d shaft  40 mm

Shaft diameter Solution: 1.

d roller  120  mm

Calculate the area moment of inertia of the shaft. I 

π d shaft

4 5

I  1.257  10  mm

64

4

Calculate the weight of the roller. Vroller 

Volume of roller

π   d 4 

2

 d shaft

roller

5

Vroller  2.011  10  mm Weight of roller

Wroller  γ Vroller

Mass of roller

mroller 

2

  ( b  a)

3

Wroller  14.191 N

Wroller

mroller  1.447 kg

g p 

Distributed load on shaft due to weight of roller 3.

d roller  3  d shaft


Area moment of inertia 2.

3

Wroller ba

p  709.548 

N m

From inspection of Figure P10-4, write the load function equation and integrate it twice to get the shear and moment equations. q(x) = R1-1 - p0 + p0 + R2-1 V(x) = R10 - p1 + p1 + R20 M(x) = R11 - p2/2 + p2/2 + R21

4.


p 2 L

p 2

2

 ( L  a) 

p 2

 ( L  a )  ( L  b ) 2

2

 ( L  b) = 0 2



R2  p  ( b  a )  R1 5.

R1  2.129  N R2  12.06  N




10-14a-2

(x) = [R12/2 - p3/6 + p3/6 + R22/2 + C3]/EI y(x) = [R13/6 - p4/24 + p4/24 + R23/6 + C3x + C4]/EI 6.

Solve for the constants of integration with the boundary conditions y = 0 at x = 0 and x = L. At x = 0

C4 = 0

At x = L

0=

R1

C3 

6

p

3

L 

4

24

 ( L  a) 

p

4

24

 ( L  b )  C 3 L

R1 3 p p 4 4  ( L  a)   ( L  b)  L  L  24 24 6  1

2



C3  0.014  N  m

7. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 8.

Write the deflection equation in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. y ( x) 

9.

 R1  p p 3 4 4  S ( x 0  mm)  x   S ( x a )  ( x  a )   S ( x b )  ( x  b )  C3 x E I  6 24 24  1



Determine the approximate stiffness of the shaft by dividing the weight of the roller by the deflection of the sha due to the roller weight at the mid-point of the roller. Roller midpoint

c 

ab

c  170  mm

2

Deflection at x = c

δc  y ( c)

Stiffness

k 

5

δc  2.345  10

Wroller

 mm

8 N

k  6.051  10 

δc

m

10. Calculate the critical frequency.

ωn 

k mroller

ωn  20449 

rad sec

ωn  195275 rpm fn 

ωn 2 π

fn  3255 Hz



10-15a-1


What are the maximum, minimum, and average power values for the shaft shown in Figure P10-1 for the data in row a of Table P10-1 if the shaft speed is 750 rpm?

Given:

Shaft speed

ω  750  rpm

Minimum torque

Tmin  0  in lbf

Maximum torque

Tmax  2000 in lbf

Solution: 1.

rad sec


Use equations (10.1) to calculate the power values.

Average torque

2.

ω  78.54 

Tavg 

Tmax  Tmin 2

Tavg  1000 in lbf

Maximum power

Pmax  Tmax ω

Pmax  23.8 hp

Minimum power

Pmin  Tmin ω

Pmin  0  hp

Average power

Pavg  Tavg ω

Pavg  11.9 hp

Note, if doing these calculations by hand, that suitable unit conversion factors must be used.



10-16a-1


What are the maximum, minimum, and average power values for the shaft shown in Figure P10-2 for the data in row a of Table P10-1 if the shaft speed is 50 rpm?

Given:

Shaft speed

ω  50 rpm

Minimum torque

Tmin  0  N  m

Maximum torque

Tmax  2000 N  m

Solution: 1.

rad sec


Use equations (10.1) to calculate the power values. Average torque

2.

ω  5.236 

Tavg 

Tmax  Tmin 2

Tavg  1000 N  m

Maximum power


Pmax  10.5 kW

Minimum power


Pmin  0  kW

Average power


Pavg  5.24 kW




10-17a-1


Figure P10-5 shows a roller assembly driven by a gear. The roller extends over 80% of the length a and is centered in that dimension. The roller occupies 95% of the exposed shaft length between the bearing faces. The shaft is steel of S ut = 745 MPa and S y = 427 MPa. For the data in row a of Table P10-1, find: (a) The safety factor against fatigue failure for a shaft diameter of 40 mm. (b) The maximum torsional deflection between gear and roller. (c) The torsional natural frequency of the shaft.

Given:

Distance to end of shaft Distance to gear

L  20 cm b  18 cm


p  1000 N  cm Tmin  0  N  m

Yield strength Shaft diameter

S y  427  MPa d  40 mm

Maximum torque

Tmax  2000 N  m

Concentrated load

P  1000 N

Distance between bearings Tensile strength 1

a  16 cm S ut  745  MPa

Assumptions: 1. The finish is machined, reliability is 90%, and the shaft is at room temperature. 2. Use SI units with Table P10-1 consistent with Problem 10-2. 3. The ratio of notch radius to shaft diameter in the keyway is roverd  0.021. 4. The roller stiffens the shaft torsionally so that the torsional deflection will occur largely between the right end of the roller (at the second or middle key) and the gear (right key). 5. Note that the given magnitudes of the radial forces shown acting on gear(s) in this problem are not necessarily consistent with a load associated with the given torque for any real gear of reasonable pressure angle. Since gears are taken up in a later chapter, these shaft design problems ignore the real gear loadings and use an arbitrary value to provide a shaft design exercise. Solution:


Part (a) 1.

The maximum moment in the shaft occurs at approximately the middle of the roller. See the appendix to this problem below for the determination of Mmax. The maximum moment is also the alternating moment. Thus, Mmax  297.3  N  m

2.

Ma  Mmax

Ma  297.3  N  m

Calculate the mean and alternating components of torque, assuming that one half is taken out at the right-hand key and the other half at the left-hand key. At the middle of the roller only one half of the torque is present.

 Tmax  Tmin   2  2 

Tm  500  N  m

 Tmax  Tmin   2  2 

Ta  500  N  m

Tm 

1

Ta 

1





S'e  0.5 S ut

S'e  372.5  MPa

3.


4.


Cload  1

Size

Csize  1.189  

Surface

A  4.51

  mm   d

 0.097

b'  0.265




 Sut    MPa 

10-17a-2

b'


5.

Temperature

Ctemp  1

Reliability


Csurf  0.782

(R = 90%)

Determine the modified endurance limit as a function of the unknown diameter, d. S e  Cload  Csize Csurf  Ctemp Creliab S'e

6.

Use equation (10.8) with unity for all stress concentration factors to find the factor of safety at the point in the shaft where the bending moment is maximum. 3 2  2 3  Ma   Ta 4 π d Nfc    32  Se

7.

1

Nfc  2.1

Ma  Mf

Ma  84.4 N  m

Tm 

 Tmax  Tmin    2  

Tm  1000 N  m

Ta 

 Tmax  Tmin    2  

Ta  1000 N  m

Determine the geometric stress concentration factors from Figure 10-16 using the assumed r/d. For an r/d ratio of

9.

  Tm  4  S ut  3

Although the bending moment is lower at the ends of the roller where the keys are, there is a stress concentration there that is not present at the point where the bending moment is a maximum. Of the two key locations, the one on the right is critical (even though the moment is lower there) because the full torque is present in the shaft at this point. The bending moment and torque at the right keyway (x = f) are: Mf  84.4 N  m

8.

S e  217.148  MPa

roverd  0.021

Kt  2.2

Kts  3.0

Determine the notch sensitivity of the material for bending and torsion. Note from Figure 6-36, part 1, that a value of 20 ksi should be added to S ut to obtain a 1/2 from Table 6-6 for the torsional Neuber constant.

Bending

Notch radius

r  roverd d


S ut  108  ksi

Neuber constant

a N  0.056  in

Notch sensitivity

q 

2

1 1

Torsion

r  0.84 mm

0.5

a N  0.056  in q  0.765

aN r



Neuber constant

a N  0.045  in

2

S'ut  128  ksi 0.5

a N  0.045  in



10-17a-3

1

q s 

Notch sensitivity

1

q s  0.802

aN r

10. Determine the fatigue stress concentration factors from equation (6.11b). Bending factor

Kf  1  q   Kt  1 

Kf  1.917

Torsion factor

Kfs  1  q s  Kts  1 

Kfs  2.603

Maximum and minimum nominal stress

τanom 

16 Ta

π d

3

τanom  79.6 MPa

τmnom 

16 Tm

π d

τmnom  79.6 MPa

3

τmaxnom  τmnom  τanom

τmaxnom  159.155  MPa

τminnom  τmnom  τanom

τminnom  0  MPa

Using equations 6.17, the mean-stress fatigue-concentration factors are Kfm τmax τmin K S  

return K if K τmax  S return

S  K τanom

τmnom

if K τmax  S

return 0 if K τmax  τmin  2  S Kfsm  Kfm τmaxnom τminnom Kfs 0.577  S y

Kfsm  0.49

11. Use equation (10.8) with the stress concentration factors above to find the factor of safety at the point in the shaft where the right-hand roller key is located. 3  2 2 3   Kf  Ma    K fs Ta 4 π d Nff    32  Se

  Kfsm Tm 4  S ut  3

1

Nff  0.57

So, in this case, the critical section on the shaft is not at the point of maximum bending moment. This is for two reasons, 1) there is no stress concentration at this point but there is elsewhere, and 2) the torque is maximum at the point of a stress concentration. Part (b) 12. For the assumptions made, the length of the shaft that will deflect torsionally is Lbf  b  f

13. Calculate the polar moment of inertia of the shaft.

J 

π d

Lbf  36 mm

4

32

5 4 14. J  Using 2.513 equation 10  mm(10.9a), calculate the angular deflection between the gear and the right-hand key for the maximu applied torque. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


10-17a-4

Modulus of rigidity

G  80.8 GPa

Angular deflection

θ 

Tmax Lbf

θ  0.203  deg

J G

Part (c) 15. Use equations (10.27) to determine the shaft spring constant, mass moment of inertia, and natural frequency.

ρ  76500 

Density of steel

kg 3

m

π d

mshaft 

Mass of shaft

Mass moment of inertia

2

d

mshaft  19.227 kg

2

Im  3.845  10



2

kt  564090

Lbf kt

ωn 

Natural frequency

 

 L ρ

G J

kt 

Spring constant

4

mshaft

Im 

2

ωn  12112 

Im

ωn

fn 

3 2

m  kg

N m rad rad sec

fn  1928 Hz

2 π

APPENDIX - Maximum bending moment 1. Determine the distance from the origin to the left and right ends of the roller. Distance to left end

e  0.1 a

e  16 mm


f  0.9 a

f  144  mm

2. From inspection of Figure P10-5, write the load function equation q(x) = R1-1 - p0 + p0 + R2-1 - P-1 2. Integrate this equation from - to x to obtain shear, V(x) V(x) = R10 - p1 + p1 + R20 - P0 3. Integrate this equation from - to x to obtain moment, M(x) M(x) = R11 - p2/2 + p2/2 + R21 - P1 4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = b +, V = M = 0 V = R1  p  ( b  e )  p  ( b  f )  R 2  P = 0 M = R 1 b  R1 

p 2

2

 ( b  e) 

p 2

2

 ( b  f )  R 2 ( b  a ) = 0

 e2  f 2  b   f  e  p    2 a   a

a

P 

R1  6275 N



10-17a-5

R2  p  ( f  e)  R1  P

R2  7525 N

x  0  m 0.005  b  b


6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z, and a value of one when it is greater than or equal to z. S ( x z)  if ( x  z 1 0 ) 7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  R1 S ( x 0  m)  p  S ( x e)  ( x  e)  p  S ( x f )  ( x  f )  R2 S ( x a )  P S ( x b ) M ( x)  R1 S ( x 0  m)  x 

p 2

2

 S ( x e)  ( x  e) 

p 2

2

 S ( x f )  ( x  f )  R2 S ( x a )  ( x  a )

8. Plot the shear and moment diagrams.

SHEAR

MOMENT

10

300

5

200

V ( x) kN

M ( x) 0

Nm

5

 10

100

0

0

5

10

FIGURE 10-17a

15

 100

20

0

5

10

x

x

cm

cm

15

20


9. Determine the maximum maximum moment from inspection of the diagrams. Maximum moment occurs where V zero, which is x = c. From the shear diagram, ce R1

=

f c R2  P

c 

f  R1  e  R2  e  P R1  R2  P

Mmax  M ( c)

c  7.875  cm Mmax  297.3  N  m

The moment at the left edge of the roller (at key) is

Me  M ( e)

Me  100.4  N  m

The moment at the right edge of the roller (at key) is

Mf  M ( f )

Mf  84.4 N  m

The moment at the right-hand bearing is

Ma  M ( a )

Ma  20 N  m



10-18a-1


Figure P10-5 shows a roller assembly driven by a gear. The roller extends over 80% of the length a and is centered in that dimension. The roller occupies 95% of the exposed shaft length between the bearing faces. For the data in row a of Table P10-1, find the maximum bending deflection of th 40-mm-diameter shaft.

Given:

Distance to end of shaft Distance to gear

L  20 cm b  18 cm

Applied distributed load Maximum torque

p  1000 N  cm Tmax  2000 N  m

Minimum torque

Tmin  0  N  m

Distance between bearings Shaft diameter 1

Concentrated load Young's modulus

a  16 cm d  40 mm P  1000 N E  206.8  GPa

Assumptions: 1. The roller does not stiffen the shaft in bending but it does transmit the distributed load directly to the shaft. 2. Note that the given magnitudes of the radial forces shown acting on gear(s) in this problem are not necessarily consistent with a load associated with the given torque for any real gear of reasonable pressure angle. Since gears are taken up in a later chapter, these shaft design problems ignore the real gear loadings and use an arbitrary value to provide a shaft design exercise. Solution: 1.


Determine the distance from the origin to the left and right ends of the roller and the area moment of inertia. Distance to left end

e  0.1 a

e  16 mm


f  0.9 a

f  144  mm

I 

Moment of inertia 2.

π d

4

5

I  1.257  10  mm

64

4

From inspection of Figure P10-5, write the load function equation and integrate it twice to get the shear and moment equations. q(x) = R1-1 - p0 + p0 + R2-1 - P-1 V(x) = R10 - p1 + p1 + R20 - P0 M(x) = R11 - p2/2 + p2/2 + R21 - P1

3.

Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where both are zero. At x = b +, V = M = 0 V = R1  p  ( b  e )  p  ( b  f )  R 2  P = 0 M = R 1 b  R1 

p 2

2

 ( b  e) 

p 2

2

 ( b  f )  R 2 ( b  a ) = 0

 e2  f 2  b  a   f  e  p   P 2  a    a 

R2  p  ( f  e)  R1  P 4.

R1  6275 N R2  7525 N


(x) = [R12/2 - p3/6 + p3/6 + R22/2 - P2/2 + C3]/EI y(x) = [R13/6 - p4/24 + p4/24 + R23/6 - P3/6 + C3x + C4]/EI © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


10-18a-2

Solve for the constants of integration with the boundary conditions y = 0 at x = 0 and x = a. At x = 0

C4 = 0

At x = L

0=

R1

C3 

6

3

a 

p 24

4

 ( a  e) 

p 24

4

 ( a  f )  C 3 a

R1 3 p p 4 4  ( a  e)  (a  f )  a  a  24 24 6  1

2



C3  15.578 N  m

x  0  m 0.005  L  L

6.


7.


8.

Write the slope and deflection equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions.

θ ( x) 

y ( x) 

9.

 R1

1 E I



2

E I

p 6

3

 S ( x e)  ( x  e) 

p 6



3

 S ( x f )  ( x  f ) 

 R  2  S ( x a )  ( x  a) 2  P  S( x b)  ( x  b ) 2  C3 2  2

 R1

1

2

 S ( x 0  mm)  x 



6

p

3

 S ( x 0  mm)  x 

24

4

 S ( x e)  ( x  e) 

p 24

  

4



 S ( x f )  ( x  f ) 

 R  2  S( x a)  ( x  a ) 3  P  S ( x b )  ( x  b) 3  C3 x 6  6

  

Plot the slope and deflection diagrams (see Figure 10-18).

SLOPE

DEFLECTION

1

0.04

0.5

0.02

θ( x) 10

y ( x)

3

0

mm

 0.5

1

0

 0.02

0

FIGURE 10-18

5

10

15

20

 0.04

0

5

10

x

x

cm

cm

15

20

Slope and Deflection Diagrams for Problem 10-18a © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


10-18a-3

10. The maximum deflection occurs at the value of x for which  is zero. Let this be x = c, then guess c  8  cm Given

θ ( c) = 0

c  Find ( c)

c  7.955  cm

11. The maximum deflection occurs at x = c and is ymax  y ( c)

ymax  0.0300 mm

yL  y ( L)

yL  0.0229 mm

12. The deflection at the end of the shaft is



10-19a-1


Figure P10-6 shows two gears on a common shaft. Assume that the constant radial force P1 is 40% of P2. For the data in row a from Table P10-1, find the diameter of the shaft required to obtain a safety factor of 2 in fatigue loading if the shaft is steel with properties given below.

Given:

Maximum torque


Minimum torque


Radial load

P2  1000 lbf

Load factor

n  0.4

a  16 in

Shaft dimensions:

Tensile strength

S ut  108  ksi b  18 in

Yield strength

S y  62 ksi

Design safety factor L  20 in Nd  2 Assumptions: There are no applied axial loads. Steel will be used for infinite life. The shaft is machined, reliability is 90%, and the shaft is at room temperature. The ratio of the fillet radius in the keyway to shaft diameter is roverd  0.02. Solution: 1.


Calculate the radial load on gear 1: P1  n  P2

2.

P1  400  lbf

Using Figures D-2(a) and D-3(a), the reaction and moment equations are R1  P1  1 



a

   P2  1  

b

L



R1  67 lbf

b

R2  P1  P2  R1

R2  1467 lbf

M(z) = R11 - P11 + R21 - P21 z  0  in 0.005  L  L

3.

Define the range for z

4.

For a Mathcad solution, define a step function S. This function will have a value of zero when z is less than u, and a value of one when it is greater than or equal to u. S ( z u )  if ( z  u 1 0 )

5.

Write the shear and moment in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( z)  R1 S ( z 0  in)  P1 S ( z a )  R2 S ( z b )  P2 S ( z L) M ( z)  R1 S ( z 0  mm)  ( z  0  mm)  P1 S ( z a )  ( z  a )   R2 S ( z b )  ( z  b )  P2 S ( z L)  ( z  L)

6.

Plot the shear and moment diagrams (see Figure 10-19a). The maximum moment occurs at the right-hand bearing and is MB  M ( b ) . The moment at the left-hand gear is MA  M ( a ) . The moment at the right-hand gear is zero so the two points of interest are A and B. At A the bending moment is lower than at B but there is a stress concentration due to the key seat. At B the moment is a maximum but there is no stress concentration. We will investigate both points. MA  1067 in lbf

MB  2000 in lbf



10-19a-2

Shear Diagram

Moment Diagram

1000

0

 500 500 V ( z)

M ( z)  1000 lbf  in

lbf 0

 1500

 500

0

5

10

FIGURE 10-19a

15

 2000

20

0

5

10

z

z

in

in

15

20


7.


Tmax  Tmin

Tm  1000 in lbf

2 Tmax  Tmin

Ta  1000 in lbf

2

Right-hand bearing (point B) S'e  0.5 S ut

S'e  54 ksi

8.


9.


Cload  1

Size

Csize( d )  0.869  

Surface

A  2.70

   in  d

 0.097

b  0.265

 S ut    ksi 

b


Ctemp  1

Reliability


(machined)

Csurf  0.781

(R = 90%)

10. Determine the modified endurance limit as a function of the unknown diameter, d. S e( d )  Cload  Csize( d )  Csurf  Ctemp Creliab S'e 11. Use equation (10.8) with unity for all stress concentration factors as a design equation to find d. Guess

d  1  in



10-19a-3

Given 1

3 2   2   32 Nd  MB  4  Ta T 3 m  d=     4 S ut  S e( d )  π  d B  Find ( d )

3

d B  1.153  in


Csize d B  0.857

Endurance limit

S e d B  32.4 ksi

Left-hand gear (point A) 12. Determine the geometric stress concentration factors from Figure 10-16 using a curve-fit equation.  0.4115

Bending

Kt  0.452  ( roverd)

Torsion

Kts  1.251  ( roverd)

 0.230

Kt  2.3 Kts  3.1

13. Determine the notch sensitivity of the material in bending and torsion. Note from Figure 6-36, part 1, that a value of 20 ksi should be added to S ut to obtain a 1/2 from Table 6-6 when the loading is torsional. Bending: Lookup value of S ut

S'ut  S ut

S'ut  108  ksi

Neuber constant

a  0.057  in

Notch radius

r( d )  roverd d

Notch sensitivity

q ( d ) 

2

0.5

a  0.057  in

1 1

a r( d )

Torsion: Lookup value of S ut


Neuber constant

a s  0.045  in

Notch radius

r( d )  roverd d

Notch sensitivity

q s( d ) 

2

S'ut  128  ksi 0.5

a s  0.045  in

1 1

as r( d )

14. Determine the fatigue stress concentration factors from equation 6.11b and 6.17. Bending

Kf ( d )  1  q ( d )   Kt  1 

Torsion

Kfs( d )  1  q s( d )   Kts  1 



10-19a-4


Mean factor

15. Repeat step 11 using these stress concentration factors with the mean and alternating stresses. Given

1

3   2 2   32 Nd   Kf ( d )  MA  4   Kfs( d)  Ta Tm  3 d=    Kfsm( d )   4 S e( d ) S ut  π  d A  Find ( d )

Diameter required at point A:

3

d A  1.337  in

16. Check the assumption that Kfsm = Kfs using equation 6.17.

τmaxnom 

16

π d

3

  Tm  Ta

Kfs( d )  τmaxnom  26.2 ksi

τmaxnom  10.2 ksi 0.577  S y  35.774 ksi

Since Kfsmax is less than S sy, the assumption was valid. 17. Since d A is larger than d B, it must be used as the minimum shaft diameter. Thus, d  d A.

d  1.337  in

18. The factor of safety at A is Nd  2 . The safety factor at B is 3 2  2  3  MB   Ta T 4 π d 3 m NB      32  4 S ut  S e( d )

1

NB  3.1



10-20-1


A 300-mm-long, solid, straight shaft is supported in self-aligning bearings at each end. A gear is attached at the middle of the shaft with a 10-mm square, steel key in a slot. The geometric stress-concentration factor in the keyslot is 2.5 and its corner radius is 0.5 mm. The gear drives a fluctuating load that creates a bending moment that varies from +10 N-m to +100 N-m and a torque that varies from -35 N-m to +170 N-m each cycle. The material chosen is cold-drawn 4140 steel, hardened and tempered to RC 45 (S ut = 1250 MPa). Design the shaft for infinite life and determine the diameter needed for a safety factor of 1.5.

Given:

Minimum moment

Mmin  10 N  m

Maximum moment

Mmax  100  N  m

Minimum torque

Tmin  35 N  m

Maximum torque

Tmax  170  N  m

Stress conc. factor

r  0.5 mm

Notch radius Tensile strength

Kt  2.5

S ut  1250 MPa

Design safety factor Nfd  1.5

Assumptions: 1. The finish is machined, reliability is 90%, and the shaft is at room temperature. 2. The geometric stress concentration factors are the same for bending and torsion. 3. Note that the given magnitudes of the radial forces shown acting on gear(s) in this problem are not necessarily consistent with a load associated with the given torque for any real gear of reasonable pressure angle. Since gears are taken up in a later chapter, these shaft design problems ignore the real gear loadings and use an arbitrary value to provide a shaft design exercise. Solution: 1.


Calculate the mean and alternating components of bending moment and torque. Mm  Ma  Tm  Ta 

Mmax  Mmin 2 Mmax  Mmin 2 Tmax  Tmin 2 Tmax  Tmin 2

Mm  55 N  m Ma  45 N  m Tm  68 N  m Ta  103  N  m

2. Determine the notch sensitivity of the material for bending and torsion. Note from Figure 6-36, part 1, that a value of 20 ksi should be added to S ut to obtain a 1/2 from Table 6-6 for the torsional Neuber constant. Bending


S ut  181  ksi

Neuber constant

a N  0.024  in

Notch sensitivity

q 

2

1

q  0.854 aN

1 Torsion

r



Neuber constant

a N  0.018  in

Notch sensitivity

q s 

2

1 1

0.5

a N  0.024  in

aN

S'ut  201  ksi 0.5

a N  0.018  in q s  0.886

r



10-20-2

Determine the fatigue stress concentration factors from equation (6.11b). Bending factor

Kf  1  q   Kt  1 

Kf  2.281

Torsion factor

Kfs  1  q s  Kt  1 

Kfs  2.329

Assume

Kfm  Kf

Kfsm  Kfs

S'e  0.5 S ut

S'e  625  MPa

4.


5.


Cload  1

Size

Csize( d )  1.189  

d

Surface

A  4.51

b'  0.265

   mm 

Csurf

6.

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


 0.097

(machined)

b'

Csurf  0.682

(R = 90%)


7.

Use equation (10.8) as a design equation to find d. Guess

d  20 mm

Given 1

  2 3 2  32 Nfd   Kf  Ma  4   Kfs Ta d=   S e( d )  π  d  Find ( d )

d  23.395 mm

Kfm Mm

2



3 4

S ut Round this to


2

3

  

d  22 mm


Csize( d )  0.881

Endurance limit

S e( d )  337  MPa



10-21-1

PROBLEM 10-21

_____

Statement:

Figure P10-7 shows a dial assembly. The dial is 500-mm-dia by 16-mm-thick, solid steel and is bolted to the top flange of the shaft. The shaft is 70 mm dia over length L1, 40 mm dia over length L2, and is carried on tapered roller bearings. All shaft fillet radii are 0.5 mm. A 5 kW motor drives the shaft through a 20:1 reduction gearbox and a commercial bellows coupling. The maximum rated torque from the motor is 17.75 N-m and its stall torque is 3x its rated torque. Find the maximum stress and angular deflection in the shaft under conditions of dial locked and motor at stall.

Given:

Rated motor torque

TMR  17.75  N  m

Stall torque factor

fstall  3

Gear reduction ratio

mG  20

Modulus of rigidity

G  80.8 GPa

Shaft dimensions

d 1  70 mm

Solution: 1.

L2  24 mm

For a locked dial the motor will be stalled. The torque on the shaft at the gearbox output will be Tss  1065 N  m

Calculate the polar moment of inertia and outside radius of each section of the shaft. J1 

J2  3.

d 2  40 mm


Tss  TMR fstall  mG 2.

L1  144  mm

π d 1

4

32

π d 2

6

4

r1  0.5 d 1

r1  35 mm

5

4

r2  0.5 d 2

r2  20 mm

J1  2.357  10  mm 4

J2  2.513  10  mm

32

Calculate the stress in each section of the shaft when the dial is locked.

τ1 

Tss r1

τ1  15.8 MPa

J1

τ2 

Tss r2 J2

τ2  84.8 MPa

The maximum stress for a locked dial is in the smaller diameter shaft and is

τmax  τ2 4.


Calculate the deflection of the shaft under the stall torque.

θstall 

Tss L1 J 1 G



Tss L2 J2 G

θstall  0.118  deg



10-22-1

PROBLEM 10-22

_____

Statement:

For the dial assembly of Problem 10-21 and Figure P10-7, find the first torsional natural frequency the shaft-dial assembly as felt at the motor shaft. The gearbox has a torsional stiffness of 1.56E5 N-m/rad and the coupling has a torsional stiffness of 5.1E5 N-m/rad. Assume the dial is empty of tooling and that ground support is infinitely stiff.

Given:

Dial dimensions:

D  500  mm

t  16 mm

Gear reduction ratio Shaft dimensions

mG  20 d 1  70 mm

Density of steel L1  144  mm

5

Gearbox stiffness kG  1.56 10  N  m Solution: 1.

 π D2     t ρ  4 

mD  24.504 kg

1 2

 mD 

D

2

2.000



ID  0.766 m

2

 kg

Calculate the stiffness of the shaft using equations 10.27. For steel, G  80.8 GPa

k1 

π d 1

4

32 J 1 G L1

kshaft 

1  1 k   1 k2 

6

J1  2.357  10  mm

4

J2 

6 N m

k1  1.323  10 

k2 

rad

1

π d 2

4 5

J2  2.513  10  mm

32 J 2 G

4

5 N m

k2  8.461  10 

L2

rad

5 N m

kshaft  5.160  10 

rad

Calculate the stiffness of the shaft-coupling-gearbox assembly. ktotal 

5.

5

kC  5.1 10  N  m

Coupling stiffness

Calculate the mass moment of inertia of the dial.

J1 

4.

L2  24 mm

Calculate the mass of the dial (the other masses will be neglected).

ID 

3.

d 2  40 mm


mD  2.

3

ρ  7800 kg m

 1  1  1 k   shaft kC kG 

1

4 N m

ktotal  9.700  10 

rad

Using equation 10.27a, calculate the torsional natural frequency.

ωn 

ktotal ID

ωn  356 

rad sec

fn 

ωn 2 π

fn  56.6 Hz



10-23-1

PROBLEM 10-23

_____

Statement:

For the dial assembly of Problem 10-21 and Figure P10-7, find the first torsional natural frequency the shaft-dial-tool assembly as felt at the motor shaft. The gearbox has a torsional stiffness of 1.56E5 N-m/rad and the coupling has a torsional stiffness of 5.1E5 N-m/rad. The dial is fitted with 20 sets of tooling bolted to its top surface, equispaced on a 416-mm bolt circle. Each tooling assembly weighs 75.62 N. Assume all ground support is infinitely stiff.

Given:

Dial dimensions:

D  500  mm

t  16 mm


mG  20 d 1  70 mm


5

kG  1.56 10  N  m

Gearbox stiffness Tooling: Solution: 1.

 π D2     t ρ  4 

mD  24.504 kg

2

mt  Nt

2.000

ID  0.766 m

Nt  20

Wt

mt  154.222 kg

g

 Dt  It  mt   2

 kg

2 2.000

It  6.672 m

 kg

2.000

Itotal  ID  It

Itotal  7.438 m

 kg

Calculate the stiffness of the shaft using equations 10.27. For steel, G  80.8 GPa J1 

k1 

π d 1

4 6

J1  2.357  10  mm

32 J 1 G

4

J2 

6 N m

k1  1.323  10 

L1

1 1 kshaft      k1 k2 

k2 

rad

1

π d 2

4 5

J2  2.513  10  mm

32 J 2 G

4

5 N m

k2  8.461  10 

L2

rad

5 N m

kshaft  5.160  10 

rad


5.

Wt  75.62  N

Weight

Calculate the mass moment of inertia of the dial and the tools. D ID   mD   2 2

4.

5

kC  5.1 10  N  m

Calculate the mass of the dial and the tools.

1

3.

L2  24 mm


mD  2.

d 2  40 mm

Coupling stiffness

Dt  416  mm

Bolt circle dia

3

ρ  7800 kg m

 1  1  1 k   shaft kC kG 

1

4 N m

ktotal  9.700  10 

rad


ωn 

ktotal Itotal

ωn  114 

rad sec

fn 

ωn 2 π

fn  18.2 Hz



10-24-1

PROBLEM 10-24

_____

Statement:

For the dial assembly of Problem 10-21 and Figure P10-7, find the first torsional natural frequency the shaft-dial assembly as felt at the motor shaft. The gearbox has a torsional stiffness of 1.56E5 N-m/rad, the coupling has a torsional stiffness of 5.1E5 N-m/rad, and the motor mount has a torsional stiffness of 2.44E5 N-m. Assume the dial is empty of tooling.

Given:

Dial dimensions:

D  500  mm

t  16 mm


mG  20 d 1  70 mm


5

Gearbox stiffness kG  1.56 10  N  m

1.

5

 π D2     t ρ  4 

1 2

 mD 

D

2

2.000

 2

ID  0.766 m

 kg

Calculate the stiffness of the shaft using equations 10.27. For steel, G  80.8 GPa

k1 

π d 1

4

32 J 1 G L1

kshaft 

1  1 k   1 k2 

6

J1  2.357  10  mm

4

J2 

6 N m

k1  1.323  10  1

k2 

rad

π d 2

4 5

J2  2.513  10  mm

32 J 2 G

4

5 N m

k2  8.461  10 

L2

rad

5 N m

kshaft  5.160  10 

rad

Calculate the stiffness of the shaft-coupling-gearbox-motor-mount assembly. ktotal 

5.

mD  24.504 kg

Calculate the mass moment of inertia of the dial.

J1 

4.

5

kC  5.1 10  N  m

Calculate the mass of the dial (the other masses will be neglected).

ID 

3.

Coupling stiffness

L2  24 mm


mD  2.

d 2  40 mm

km  2.44 10  N  m

Motor mount stiffness Solution:

3

ρ  7800 kg m

 1  1  1  1 k   shaft kC kG km 

1

4 N m

ktotal  6.941  10 

rad


ωn 

ktotal ID

ωn  301 

rad sec

fn 

ωn 2 π

fn  47.9 Hz



10-25-1

PROBLEM 10-25

_____

Statement:

The tooling on the dial assembly of Problem 10-21 and Figure P10-7 imparts a time varying torque to the dial that ranges from a peak of 30% of the motor's rated torque (at the motor) to zero, 20 time per dial revolution. If the motor speed is 600 rpm, find the worst-case stress-time and deflection-time functions for the shaft. Choose a shaft material to give a safety factor of at least 3 against failure. Disregard the response of the system to the forced vibration.

Given:

Rated motor torque

TMR  17.75  N  m

Stall torque factor

Gear reduction ratio

mG  20

Modulus of rigidity

Shaft dimensions

d 1  70 mm

Motor speed

n m  600  rpm

Tool torque factor

ftool  0.3


Nd  3

Number of tools

Nt  20

Solution: 1.

Maximum torque

Tmax  ftool  TMR mG

Minimum torque

Tmin  0  N  m

Mean torque

Tm 

π d 1

Tmax  Tmin

Tm  53.25  N  m

2 Tmax  Tmin

Ta  53.25  N  m

2

4

π d 2

6

4

r1  0.5 d 1

r1  35 mm

5

4

r2  0.5 d 2

r2  20 mm

J1  2.357  10  mm

32 4

J2  2.513  10  mm

32

Calculate the mean and alternating stress components in the smaller diameter shaft.

τ2m  σ'm 

Tm r2 J2 3  τ2m

τ2m  4.24 MPa

τ2a 

σ'm  7.34 MPa

σ'a 

Ta r2 J2 3  τ2a

τ2a  4.24 MPa σ'a  7.34 MPa

Calculate the mean and alternating deflection of the shaft.

θm 

θa  5.

Ta 

Tmax  106.5  N  m

Calculate the polar moment of inertia and outside radius of each section of the shaft.

J2 

4.

L2  24 mm

Calculate the mean and alternating torque components on the dial shaft.

J1 

3.

d 2  40 mm


Alternating torque 2.

L1  144  mm

fstall  3 G  80.8 GPa

Tm L1 J 1 G Ta L1 J1 G





Tm L2 J 2 G Ta L2 J2 G

3

θm  5.913  10

3

θa  5.913  10

 deg

 deg

Calculate the endurance limit correction factors using equations 6.7 and the modified endurance limit, as a function of S ut, using equations 6.5a and 6.6.



10-25-2

 d2  Csize  1.189     mm 

Cload  1

 S ut  Csurf  S ut  4.51    MPa 

 0.097

Csize  0.831

 0.265

Ctemp  1


(R = 0.90)

S e S ut  Cload  Csize Csurf  S ut  Ctemp Creliab 0.5 S ut 6.

Calculate the fatigue stress-concentration factors at the step where shafts 1 and 2 join for a fillet radius of r  0.5 mm. Use Figure C-3 in appendix C to calculate the geometric stress-concentration factor and equations 6.11 and 6.17 to calculate the fatigue factors. Doverd  A  b 

d1

Doverd  1.750

d2

2  Doverd 2  1.33 2  Doverd 2  1.33

Kts  A  

  d 2   r

 ( 0.84897  0.86331 )  0.86331

A  0.8580

 ( 0.23161  0.23865 )  0.23865

b  0.2360

b

Kts  2.41

Neuber constant function:



2 3   3.10030  10 7  S    in   ksi  ksi   ksi     4 5 6   9 S  12 S  15 S  1.38322  10     3.28018  10     3.21209  10       ksi   ksi   ksi  

a ( S )  8.69657  10

2

3 S

 2.75956  10

Notch sensitivity



q  S ut 

S

1 1

7.

 

5

 3.94116  10

a  S ut  20 ksi r

Alternating factor

Kfs S ut  1  q  S ut   Kts  1 

Mean factor

Kfsm S ut  Kfs S ut

Using equation 10.8, solve for the minimum required ultimate tensile strength. Guess S ut  200  MPa 1

Given

  3  32 Nd  4  Kfs Sut  Ta d2 =    S e S ut  π  S ut  Find  S ut

  Kfsm S ut  Tm  4  S ut  3

3

S ut  112  MPa



10-26-1

PROBLEM 10-26

_____

Statement:

Combine the data from Problems 10-23 and 10-25 and (a) Find the ratio between the torsional forcing frequency and the first torsional natural frequency of the dial assembly. (b) Using the frequency ratio calculated in part (a), repeat Problem 10-25 taking into account the forced vibration response of the system if the damping ratio  = 0.20.

Given:

Dial dimensions:

D  500  mm

t  16 mm


mG  20 d 1  70 mm


5

Gearbox stiffness kG  1.56 10  N  m Tooling:

Solution: 1.

Motor speed

n m  600  rpm

Rated motor torque

TMR  17.75  N  m

 π D2     t ρ  4 

Nd  3


mt  Nt

Wt

mt  154.222 kg

g

2

2.000

ID  0.766 m

 Dt  It  mt   2

 kg

2 2.000

It  6.672 m

 kg

2.000

Itotal  7.438 m

 kg

Calculate the stiffness of the shaft using equations 10.27. For steel, G  80.8 GPa J1 

k1 

π d 1

4 6

J1  2.357  10  mm

32 J 1 G

1  1 k   1 k2 

4

J2 

6 N m

k1  1.323  10 

L1

kshaft 

k2 

rad

1

π d 2

4 5

J2  2.513  10  mm

32 J 2 G

4

5 N m

k2  8.461  10 

L2

rad

5 N m

kshaft  5.160  10 

rad


 1  1  1 k   shaft kC kG 

1

4 N m

ktotal  9.700  10 

rad


ωn  6.

Nt  20

ftool  0.3

Tool torque factor

mD  24.504 kg

Itotal  ID  It

5.

Wt  75.62  N

Weight

Calculate the mass moment of inertia of the dial and the tools. D ID   mD   2 2

4.

5

kC  5.1 10  N  m

Calculate the mass of the dial and the tools.

1

3.

L2  24 mm


mD  2.

d 2  40 mm

Coupling stiffness

Dt  416  mm

Bolt circle dia

3

ρ  7800 kg m

ktotal Itotal

ωn  114 

rad sec

fn 

ωn 2 π

fn  18.2 Hz

Determine the forcing frequency.



n D 

Dial speed

ωf  Nt n D 7.

nm

10-26-2

n D  30.0 rpm

mG

ωf  62.8

rad

ff 

sec

ωf 2 π

ff  10.0 Hz

(a) Calculate the ratio of forcing frequency to natural frequency. Fratio 

ωf

Fratio  0.55

ωn

8.

Determine the amplitude ratio of the dial-tooling-shaft-coupling-gearbox system, when subjected to the externally forced vibration, using the graph in Figure 10-26(a). For the frequency ratio Fratio  0.55 , and damping ratio  = 0.20, the amplitude (and torque) magnification is Y  1.46 .

9.

Calculate the mean and alternating torque response components on the dial shaft. Maximum torque

Tmax  Y  ftool  TMR mG

Minimum torque

Tmin  0  N  m

Mean torque

Tm 

Alternating torque

Ta 

Tmax  155.5  N  m

Tmax  Tmin

Tm  77.75  N  m

2 Tmax  Tmin

Ta  77.75  N  m

2

10. Calculate the polar moment of inertia and outside radius of each section of the shaft. J1  J2 

π d 1

4

32

π d 2

6

4

r1  0.5 d 1

r1  35 mm

5

4

r2  0.5 d 2

r2  20 mm

J1  2.357  10  mm 4

J2  2.513  10  mm

32

11. Calculate the mean and alternating stress components in the smaller diameter shaft.

τ2m  σ'm 

Tm r2 J2 3  τ2m

τ2m  6.19 MPa

τ2a 

σ'm  10.72  MPa

σ'a 

Ta r2 J2 3  τ2a

τ2a  6.19 MPa σ'a  10.72  MPa

12. Calculate the mean and alternating deflection of the shaft.

θm  θa 

Tm L1 J 1 G Ta L1 J1 G

 

Tm L2 J 2 G Ta L2 J2 G

3

θm  8.632  10

3

θa  8.632  10

 deg

 deg

13. Calculate the endurance limit correction factors using equations 6.7 and the modified endurance limit, as a function of S ut, using equations 6.5a and 6.6. Cload  1

 d2  Csize  1.189     mm 

 0.097

Csize  0.831



 S ut  Csurf  S ut  4.51    MPa 

10-26-3

 0.265

Ctemp  1


(R = 0.90)

S e S ut  Cload  Csize Csurf  S ut  Ctemp Creliab 0.5 S ut 14. Calculate the fatigue stress-concentration factors at the step where shafts 1 and 2 join for a fillet radius of r  0.5 mm. Use Figure C-3 in appendix C to calculate the geometric stress-concentration factor and equations 6.11 and 6.17 to calculate the fatigue factors. Doverd  A  b 

d1

Doverd  1.750

d2

2  Doverd 2  1.33 2  Doverd 2  1.33

Kts  A  

  d  2 r

 ( 0.84897  0.86331 )  0.86331

A  0.8580

 ( 0.23161  0.23865 )  0.23865

b  0.2360

b

Kts  2.41





a ( S )  8.69657  10

2

3 S

 2.75956  10

Notch sensitivity



q  S ut 

 

5

 3.94116  10

S

1 1

a  S ut  20 ksi r

Alternating factor

Kfs S ut  1  q  S ut   Kts  1 

Mean factor


15. (b) Using equation 10.8, solve for the minimum required ultimate tensile strength. Guess S ut  200  MPa 1

Given


3 4



 Kfsm S ut  Tm  S ut

3

 

S ut  187  MPa



10-27-1

PROBLEM 10-27

_____

Statement:

An impulsive impact load of 500 N is applied tangentially to the dial rim 20 times per dial revolution. If the motor speed is 30 rpm, find the worst-case stress-time and deflection-time functions for the shaft. Choose a shaft material to give a safety factor of at least 1.5 against failure.

Given:

Dial diameter Gear reduction ratio

D  500  mm mG  20

Modulus of rigidity

Shaft dimensions

d 1  70 mm

L1  144  mm

Motor speed

n m  30 rpm

Impact load

F  500  N


Nd  1.5

Number of tools

Nt  20

Solution: 1.

D

Mean torque

Tm  Ta 

Tmax  375.0  N  m

2

Tmin  0  N  m Tmax  Tmin

Tm  187.50 N  m

2 Tmax  Tmin

Ta  187.50 N  m

2

Calculate the polar moment of inertia and outside radius of each section of the shaft.

J2 

π d 1

4

π d 2

6

4

r1  0.5 d 1

r1  35 mm

5

4

r2  0.5 d 2

r2  20 mm

J1  2.357  10  mm

32 4

J2  2.513  10  mm

32

Calculate the mean and alternating stress components in the smaller diameter shaft.

τ2m  σ'm 

Tm r2 J2 3  τ2m

τ2m  14.92  MPa

τ2a 

σ'm  25.84  MPa

σ'a 

Ta r2 J2 3  τ2a

τ2a  14.92  MPa σ'a  25.84  MPa

Calculate the mean and alternating deflection of the shaft.

θm 

θa  5.

Tmax  3  F 

Minimum torque

J1 

4.

L2  24 mm

Calculate the mean and alternating torque components on the dial shaft. Assume that the impact load causes a reaction in the dial shaft that is 3 times the applied load. (See discussion of force impact loads in Section 3.8).

Alternating torque

3.

d 2  40 mm


Maximum torque

2.

G  80.8 GPa

Tm L1 J 1 G Ta L1 J1 G





Tm L2 J 2 G Ta L2 J2 G

θm  0.021  deg

θa  0.021  deg

Calculate the endurance limit correction factors using equations 6.7 and the modified endurance limit, as a function of S ut, using equations 6.5a and 6.6.



10-27-2

 d2  Csize  1.189     mm 

Cload  1

 S ut  Csurf  S ut  4.51    MPa 

 0.097

Csize  0.831

 0.265

Ctemp  1


(R = 0.90)

S e S ut  Cload  Csize Csurf  S ut  Ctemp Creliab 0.5 S ut 6.

Calculate the fatigue stress-concentration factors at the step where shafts 1 and 2 join for a fillet radius of r  0.5 mm. Use Figure C-3 in appendix C to calculate the geometric stress-concentration factor and equations 6.11 and 6.17 to calculate the fatigue factors. Doverd  A  b 

d1

Doverd  1.750

d2

2  Doverd 2  1.33 2  Doverd 2  1.33

Kts  A  

  d 2   r

 ( 0.84897  0.86331 )  0.86331

A  0.8580

 ( 0.23161  0.23865 )  0.23865

b  0.2360

b

Kts  2.41




2 3   3.10030  10 7  S    in    ksi   ksi   ksi   4 5 6   9 S  12 S  15 S  1.38322  10     3.28018  10     3.21209  10       ksi   ksi   ksi  

a ( S )  8.69657  10

2

3 S

 2.75956  10

Notch sensitivity



q  S ut 

S

1 1

7.

 

5

 3.94116  10

a  S ut  20 ksi r

Alternating factor

Kfs S ut  1  q  S ut   Kts  1 

Mean factor


Using equation 10.8, solve for the minimum required ultimate tensile strength. Guess S ut  200  MPa 1

Given


3 4



 Kfsm S ut  Tm  S ut

3

 

S ut  244  MPa



10-28-1

PROBLEM 10-28

_____

Statement:

For the dial assembly of Problem 10-21 and Figure P10-7, size a square key to couple the 40-mm-d dial shaft to the coupling.

Given:

Shaft stall torque Tss  1065 N  m

Shaft diameter

d  40 mm

Design choices: Shaft material: SAE 1050, normalized steel S y1  427  MPa

S ut1  745  MPa

S y2  393  MPa

S ut2  469  MPa

Key material: SAE 1020, cold-rolled steel

Design safety factor Nd  1, let the key shear at stall torque to protect the motor. Key is machined, reliability is 90%, and the dial operates at room temperature.

Solution: 1.

Maximum torque

Tmax  Tss

Minimum torque

Tmin  0  N  m


As recommended in Table 10-2, for a shaft diameter of d  40.0 mm, use a square key of width w  12 mm

Key width 2.


3.

4.

1 Tmax  Tmin  2 0.5 d

Fa  26625  N

Fm 

1 Tmax  Tmin  2 0.5 d

Fm  26625  N

Write the equations for the mean and alternating components of the shear stress and the effective von Mises stress. Key shear area as function of length

Ashear = w L


τa =


σ'a = 3  τa = 3 

Fa

τm =

w L Fa w L

Fm w L

σ'm = 3  τm = 3 

Fm w L


Nd =


Solving for L 5.

Fa 

=

S e S ut

 3   S F  S F    ut a e m  w L 

L=


Determine the key material endurance limit.


MACHINE DESIGN - An Integrated Approach, 4th Ed. Uncorrected endurance strength

10-28-2

S'e  0.5 S ut2

S'e  235  MPa


Cload  1

Size

A95 = w L

d eq =

w L 0.0766

 0.097

Csize = 1.189  d eq Cs  1.189  

  0.0766  mm   w

 0.5 0.097

Cs  0.931

 0.0485

Csize = 0.931  L

 S ut2   4.51    MPa 

Surface (machined)

Csurf

Temperature

Ctemp  1

Reliability


 0.265

(R = 90%)

 0.0485


6.

 0.0485



where

Csurf  0.884

Ce  172.973  MPa


 Fm

 0.0485

L=

 0.0485


RHS( L) 

 0.0485   L     Fm 3  Nd  S ut2 Fa  Ce     mm  


L  20 mm

RHS( L)  33.886 mm

L  RHS( L)

RHS( L)  34.551 mm

L  RHS( L)

RHS( L)  34.576 mm

Tentatively, let 7.

 0.0485

L  35 mm

Check the realized factor of safety against fatigue failure. Endurance limit

S e  Ce 

   mm  L

 0.0485

S e  145.6  MPa



Realized factor of safety

8.

S e S ut2

 3   S F  S F    ut2 a e m  w L 

Nf  1.0


9.

Nf 

10-28-3

Abear 

1 2

 w L


Maximum force



σmax 


Ns 

Fmax

2

Fmax  53.3 kN


Abear

S y2

Ns  1.5

σmax


w  12 mm

Key length

L  35 mm

(12 x 12 mm)



10-29-1

PROBLEM 10-29

_____

Statement:


Given:

Shaft torque

Tdial  106.5  N  m

Shaft diameter

d  40 mm


S ut1  745  MPa

S y2  393  MPa

S ut2  469  MPa

Key material: SAE 1020, cold-rolled steel Design safety factor Nd  3

Key is machined, reliability is 90%, and the dial operates at room temperature.

Solution: 1.

Maximum torque

Tmax  Tdial

Minimum torque

Tmin  0  N  m



Key width 2.


3.

4.

1 Tmax  Tmin  2 0.5 d

Fa  2663 N

Fm 

1 Tmax  Tmin  2 0.5 d

Fm  2663 N


Ashear = w L


τa =


σ'a = 3  τa = 3 

Fa

τm =

w L Fa w L

Fm w L

σ'm = 3  τm = 3 

Fm w L


Nd =


Solving for L 5.

Fa 

=

S e S ut

 3   S F  S F    ut a e m  w L  L=



S'e  0.5 S ut2

S'e  235  MPa



10-29-2


Cload  1

Size

A95 = w L

d eq =

w L 0.0766

 0.097

Csize = 1.189  d eq Cs  1.189  

  0.0766  mm   w

 0.5 0.097

Cs  0.931

 0.0485

Csize = 0.931  L

 S ut2   4.51    MPa 

Surface (machined)

Csurf

Temperature

Ctemp  1

Reliability


 0.265

(R = 90%)

 0.0485


6.

 0.0485



where

Csurf  0.884

Ce  172.973  MPa


 Fm

 0.0485

L=

 0.0485


RHS( L) 

 0.0485   L     Fm 3  Nd  S ut2 Fa  Ce     mm  


L  20 mm

RHS( L)  10.166 mm

L  RHS( L)

RHS( L)  9.917  mm

L  RHS( L)

RHS( L)  9.908  mm

Tentatively, let 7.

 0.0485

L  10 mm

Check the realized factor of safety against fatigue failure.

Endurance limit

S e  Ce 

   mm  L

 0.0485

S e  154.7  MPa




8.

S e S ut2

 3   S F  S F    ut2 a e m  w L 

Nf  3.0


9.

Nf 

10-29-3

Abear 

1 2

 w L

Abear  60 mm

Maximum force



σmax 


Ns 

Fmax

2

Fmax  5.3 kN

σmax  89 MPa

Abear

S y2

Ns  4.4

σmax


w  12 mm

Key length

L  10 mm

(12 x 12 mm)



10-30-1

PROBLEM 10-30

_____

Statement:


Given:

Shaft torque

Tdial  375  N  m

Shaft diameter

d  40 mm


S ut1  745  MPa

S y2  393  MPa

S ut2  469  MPa

Key material: SAE 1020, cold-rolled steel Design safety factor Nd  3

Key is machined, reliability is 90%, and the dial operates at room temperature.

Solution: 1.

Maximum torque

Tmax  Tdial

Minimum torque

Tmin  0  N  m



Key width 2.


3.

4.

1 Tmax  Tmin  2 0.5 d

Fa  9375 N

Fm 

1 Tmax  Tmin  2 0.5 d

Fm  9375 N


Ashear = w L


τa =


σ'a = 3  τa = 3 

Fa

τm =

w L Fa w L

Fm w L

σ'm = 3  τm = 3 

Fm w L


Nd =


Solving for L 5.

Fa 

=

S e S ut

 3   S F  S F    ut a e m  w L  L=



S'e  0.5 S ut2

S'e  235  MPa



10-30-2


Cload  1

Size

A95 = w L

d eq =

w L 0.0766

 0.097

Csize = 1.189  d eq Cs  1.189  

   0.0766 mm  w

 0.5 0.097

Cs  0.931

 0.0485

Csize = 0.931  L

 S ut2    MPa 

Surface (machined)

Csurf  4.51 

Temperature

Ctemp  1

Reliability


 0.265

Csurf  0.884

(R = 90%)

 0.0485



where 6.

 0.0485


Ce  172.973  MPa


 Fm

 0.0485

L=

 0.0485




3  Nd  S ut2 Fa  Ce 

  mm  



RHS( L) 

w S ut2 Ce 

L

L

 

 0.0485



 Fm



 0.0485

 mm 

Guess

L  20 mm

RHS( L)  35.795 mm

L  RHS( L)

RHS( L)  36.572 mm

L  RHS( L)

RHS( L)  36.601 mm

Tentatively, let 7.

L  38 mm

Check the realized factor of safety against fatigue failure.

Endurance limit

S e  Ce 

  mm   L

 0.0485

S e  145.0  MPa




8.

S e S ut2

 3   S F  S F    ut2 a e m  w L 

Nf  3.1


9.

Nf 

10-30-3

Abear 

1 2

 w L


Maximum force



σmax 


Ns 

Fmax

2

Fmax  18.8 kN

σmax  82 MPa

Abear

S y2

Ns  4.8

σmax


w  12 mm

Key length

L  38 mm

(12 x 12 mm)



10-31a-1

PROBLEM 10-31a

_____

Statement:

A simply supported shaft with overhanging load is shown in Figure P10-1. A constant magnitude transverse force P is applied as the shaft rotates. The shaft is also subject to a steady torque of Tmax. For the data in the row(s) assigned from Table P10-1 (ignoring Tmin), find the diameter of shaft required to obtain a safety factor of 2.5 in fatigue loading if the shaft is steel of S ut = 118 kpsi and S y = 102 kpsi. The dimensions are in inches, the force in pounds, and the torque is in lb-in. Assume no stress concentrations are present, the shaft is machined, the required reliability is 90%, and the shaft operates at room temperature.

Given:

Distance between bearings a  16 in Applied load P  1000 lbf

Solution: 1.

Maximum torque



Nd  2.5

S y  102  ksi

Ma  2000 in lbf

Calculate the unmodified endurance limit. S'e  59.0 ksi


Cload  1

Size

Csize( d )  0.869  

Surface

A  2.70

   in 

Csurf

4.

Yield strength

The maximum moment in the shaft occurs at the right bearing as seen in the moment diagram in Figure B-3(a) in Appendix B (note that in the figure a is the distance to the load and b is the distance between bearings). Using the equation given in the figure, calculate the alternating bending moment (the mean is zero).

S'e  0.5 S ut 3.

b  18 in S ut  118  ksi

See Figure P10-1, Table P10-1, and Mathcad file P1031a.

Ma  P ( a  b ) 2.

Distance to P Tensile strength

 S ut   A     ksi 

Temperature

Ctemp  1

Reliability


d

 0.097

b  0.265

(machined)

b

Csurf  0.763

(R = 90%)


5.

Use equation 10.6a with unity for all stress concentration factors as a design equation to find d. Guess

Given

d  1.00 in 1   2  32 Nd  Ma  2 3  Tmax  2  d=     4  S      π  S e( d )   y  

1 3

d  Find ( d )

d  1.153  in © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


10-32a-1


_____

Repeat Problem 10-31 taking the stress concentration at the keyway shown in Figure P10-3 into account. Distance to P Distance between bearings a  16 in b  18 in Tensile strength Applied load P  1000 lbf S ut  118  ksi Tmax  2000 lbf  in

Maximum torque

Yield strength

S y  102  ksi

Design safety factor Nd  2.5 Assumptions: 1. The finish is machined, reliability is 90%, and the shaft is at room temperature. 2. The notch radius in the keyway is r  0.015  in 3. Note that the given magnitudes of the radial forces shown acting on gear(s) in this problem are not necessarily consistent with a load associated with the given torque for any real gear of reasonable pressure angle. Since gears are taken up in a later chapter, these shaft design problems ignore the real gear loadings and use an arbitrary value to provide a shaft design exercise. Solution:


1.

There is no stress concentration at the point on the shaft where the bending moment is a maximum. On the other hand, at the gear where there is a stress concentration, the bending moment is zero. The question is: which point requires the larger diameter in order to meet the safety factor requirement? To answer this question, find the diameter required at each point and choose the larger. Start with the point where the bending moment is maximum.

2.

The maximum moment in the shaft occurs at the right bearing as seen in the moment diagram in Figure B-3(a) in Appendix B (note that in the figure a is the distance to the load and b is the distance between bearings). Using the equation given in the figure, calculate the alternating bending moment (the mean is zero). Ma  P ( a  b )

3.

Ma  2000 in lbf


4.

S'e  59.0 ksi


Cload  1

Size

Csize( d )  0.869  

Surface

A  2.70

   in 

Csurf

5.

 S ut   A     ksi 

Temperature

Ctemp  1

Reliability


d

 0.097

b  0.265

(machined)

b

Csurf  0.763

(R = 90%)


6.


d  1.00 in



10-32a-2

1 1   2  32 Nd  Ma  2 3  Tmax  2  d=     4  S      π  S e( d )   y  

Given

d 1  Find ( d ) 7.

3

d 1  1.153  in

At the gear the bending moment is zero but the steady torque is the same as above. Also, there is a stress concentration due to the keyway that must be applied to the mean shear stress present at the gear. First, set the moment equal to zero: Ma  0  lbf  in, then determine the values of the alternating and mean fatigue stress concentration factors.

8.

Determine the geometric stress concentration factor from the upper curve in Figure 10-16 using a curve-fit equation. Kts( d )  1.251  

r

 0.230

 d

9.

Determine the notch sensitivity of the material. Note from Figure 6-36, part 1, that a value of 20 ksi should be added to S ut to obtain a 1/2 from Table 6-6. Using a curve-fit to Table 6-6,




a ( S )  8.69657  10

2

3 S

 2.75956  10



 

5

 3.94116  10

Neuber constant

a  a  S ut  20 ksi

Notch radius

r  0.015  in

Notch sensitivity

q 

S

0.5

a  0.040  in

1 1

q  0.754 a r

10. Determine the fatigue stress concentration factor from equation (6.11b). Alternating factor

Kfs( d )  1  q   Kts( d )  1 

Mean factor


11. Repeat step 6 using these stress concentration factors with the torsional stresses. 1

Given 1   2 2   32 Nd  Ma  2 3  Tmax   d=      Kfsm( d )      4  S y     π  S e( d ) 

3



Diameter required at point of gear attachment:

10-32a-3

d 2  Find ( d )

Putting the two calculated diameters into a vector

trial_dia 

d 2  1.061  in

 d1     d2 

12. The larger of the two must be used as the minimum shaft diameter. Thus, d  max( trial_dia ). d  1.153  in 13. The factor of safety at the point of maximum bending moment is Nd  2.500 . The factor of safety at the gear is equal to the ratio of the two diameters cubed times Nd. Thus, 3

 d1  Nfgear     Nd  d2 

Nfgear  3.2



10-33a-1

PROBLEM 10-33a

_____

Statement:

A simply supported shaft is shown in Figure P10-2. A constant magnitude distributed unit load p is applied as the shaft rotates. The shaft is also subject to a steady torque of Tmax. For the data in the row(s) assigned from Table P10-1 (ignoring Tmin), find the diameter of shaft required to obtain a safety factor of 2.5 in fatigue loading if the shaft is steel of S ut = 814 MPa and S y = 703 MPa. The dimensions are in cm, the distributed force in N/cm, and the torque is in N-m. Assume no stress concentrations are present, the shaft is machined, the required reliability is 90%, and the shaft operates at room temperature.

Given:

Distance between bearings L  20 cm Distance to end of p b  18 cm 1

p  1000 N  cm Tmax  2000 N  m

Applied distributed load Maximum torque

Distance to start of p Tensile strength

a  16 cm S ut  814  MPa


S y  703  MPa Nfd  2.5

See Figure P10-2, Table P10-1, and Mathcad file P1033a. Solution: 1. The maximum moment in the shaft occurs between a and b. See the appendix to this problem below for the determination of Ma. Ma  48.45  N  m 2.


3.

S'e  407.0  MPa


Cload  1

Size

Csize( d )  1.189  

d

Surface

A  4.51

b'  0.265

  mm  

Csurf

4.

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


 0.097

(machined)

b'

Csurf  0.764

(R = 90%)


5.


d  20 mm 1

Given

1   2  32 Nfd  Ma  2 3  Tmax  2  d=     4  S      π  Se( d)   y  

3

d  Find ( d )

d  39.8 mm





10-33a-2


3.


4.


p 2 L

p 2

p

2

 ( L  a) 

2

2

 ( L  b) = 0

 ( L  a )  ( L  b ) 2

2



R1  300.000  N

R2  p  ( b  a )  R1

R2  1700 N

5.

Define the range for x x  0  m 0.005  L  L

6.


7.

Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  R1 S ( x 0  m)  p  S ( x a )  ( x  a )  p  S ( x b )  ( x  b )  R2 S ( x L) M ( x)  R1 S ( x 0  m)  x 

8.


p 2

2

 S ( x a )  ( x  a ) 

p 2

 S ( x b )  ( x  b )

Shear Diagram

2

Moment Diagram

500

50

0 35 V ( x)  500 N

M ( x) Nm

 1000  1500  2000

20

5

0

5

10

15

20

x cm FIGURE 10-33a

 10

0

5

10

15

20

x cm




9.

10-33a-3

Determine the maximum maximum moment from inspection of the diagrams. Maximum moment occurs where V is zero, which is x = c. From the shear diagram, ca R1

=

bc R2

Mmax  M ( c)

c 

a  R2  b  R1 R1  R2

c  16.300 cm Mmax  48.450 N  m



10-34a-1


_____

Repeat Problem 10-33 taking the stress concentration at the keyways shown in Figure P10-4 into account. Distance to start of p Distance between bearings L  20 cm a  16 cm Tensile strength Distance to end of p b  18 cm S ut  814  MPa Applied distributed load Maximum torque

1

p  1000 N  cm Tmax  2000 N  m


S y  703  MPa Nfd  2.5

Assumptions: 1. The finish is machined, reliability is 90%, and the shaft is at room temperature. 2. The ratio of notch radius to shaft diameter in the keyway is roverd  0.021 3. Note that the given magnitudes of the radial forces shown acting on gear(s) in this problem are not necessarily consistent with a load associated with the given torque for any real gear of reasonable pressure angle. Since gears are taken up in a later chapter, these shaft design problems ignore the real gear loadings and use an arbitrary value to provide a shaft design exercise. Solution:


1.

There is no stress concentration at the point on the shaft where the bending moment is a maximum, which is at x = 16.3 cm (see the appendix to this problem, below). The keys are at a  16 cm and b  18 cm. On the other hand, the key at a  16 cm looks as if it extends into the section where the moment is a maximum so, use the maximum moment as the alternating bending moment.

2.


3.


4.

S'e  407.0  MPa


Cload  1

Size

Csize( d )  1.189  

d

Surface

A  4.51

b'  0.265

   mm 

Csurf

5.

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


 0.097

(machined)

b'

Csurf  0.764

(R = 90%)


6.

Determine the geometric stress concentration factors from Figure 10-16 using curve-fit equations. Bending

Kt  0.4521 ( roverd)

Torsion

Kts  1.251  ( roverd)

 0.4115

 0.230

Kt  2.2 Kts  3.0



7.

10-34a-2

Determine the notch sensitivity of the material for bending and torsion. Note from Figure 6-36, part 1, that a value of 20 ksi should be added to S ut to obtain a 1/2 from Table 6-6 for the torsional Neuber constant. Bending



S ut  118  ksi



aN ( S )  8.69657  10

2

 2.75956  10

3 S



 3.94116  10

Neuber constant

a N  aN  S ut

Notch sensitivity

q ( d ) 

9.

0.5

a N  0.050  in

aN roverd d S'ut  S ut  20 ksi


Neuber constant

a N  aN  S ut  20 ksi

Notch sensitivity

q s( d ) 

S'ut  138  ksi 0.5

a N  0.040  in

1 1

8.

S

1 1

Torsion

 

5

aN roverd d

Determine the fatigue stress concentration factors from equation 6.11b. Kf ( d )  1  q ( d )   Kt  1 

Kfm( d )  Kf ( d )

Kfs( d )  1  q ( d )   Kts  1 


Use equation 10.6a as a design equation to find d. Guess

d  1.00 in 1

Given

1   2  32 Nfd  Kf ( d )  Ma  2 3  Kfsm( d)  Tmax  2  d=     4     Sy  π  S e( d )    

d 1  Find ( d )

3

d 1  55.0 mm





2.

10-34a-3


3.


4.


p 2 L

p 2

p

2

 ( L  a) 

2

2

 ( L  b) = 0

 ( L  a )  ( L  b ) 2

2



R1  300.000  N

R2  p  ( b  a )  R1

R2  1700 N

5.

Define the range for x x  0  m 0.005  L  L

6.


7.

Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions. V ( x)  R1 S ( x 0  m)  p  S ( x a )  ( x  a )  p  S ( x b )  ( x  b )  R2 S ( x L) M ( x)  R1 S ( x 0  m)  x 

8.

p 2

2

 S ( x a )  ( x  a ) 

p 2

 S ( x b )  ( x  b )

2

Plot the shear and moment diagrams. Shear Diagram

Moment Diagram

500

50

0

40 30

V ( x)  500 N

M ( x) Nm

 1000

20 10

 1500 0  2000

0

5

10

15

20

x cm

FIGURE 10-34a

 10

0

5

10

15

20

x cm




9.

10-34a-4

Determine the maximum maximum moment from inspection of the diagrams. Maximum moment occurs where zero, which is x = c. From the shear diagram, ca R1

=

bc R2

Mmax  M ( c)

c 

a  R2  b  R1 R1  R2

c  16.300 cm Mmax  48.450 N  m



10-35-1


Figure P10-8 shows the last stage in a gearbox with dual output. The gear is manufactured integrally with the shaft. The shaft is supported by two self-aligning ball bearings. Crank arms are connected to each end of the shaft. The load on the cranks produces equal fluctuating transverse forces on the shaft ends as well as equal fluctuating torques. The torque is transmitted through end-milled keyways in the crank and shaft and a parallel key that fits snugly in each keyway. The crank is located axially by a shoulder that is 50-mm from the plane in which the transverse load acts. The fillet radius to shaft diameter is r/d = 0.05 and the shoulder to shaft diameter ratio is D/d = 1.2. The shaft/gear material is SAE 4130 steel Q&T @ 1200F. The transverse force fluctuates from 8 kN to 16.5 kN and the torque fluctuates from 1.1 kN-m to 2.2 kN-m. For a factor of safety of 2.5 against an infinite-life fatigue failure, determine a suitable shaft diameter, d.

Given:

Tensile strength S ut  814  MPa Ratios: Doverd  1.2 Loads:

Pmin  8  kN

Pmax  16.5 kN

Tmin  1.1 kN  m l  50 mm

Distance from center of crank to shoulder Solution: 1.

Alternating

Pm 

Pa  Ta 

Pmin  Pmax

Nf  2.5

Pm  12.3 kN

2 Tmin  Tmax

Tm  1.65 kN  m

2 Pmax  Pmin

Pa  4.3 kN

2 Tmax  Tmin

Ta  0.55 kN  m

2

Calculate the mean and alternating bending moment at the shoulder. Mean

Mm  Pm l

Mm  0.61 kN  m

Alternating

Ma  Pa l

Ma  0.21 kN  m

Using Appendix C, determine the geometric stress concentration factors for the bending and torsional stresses at the shoulder. Bending (Fig. C-2): For

Doverd  1.2

roverd  0.05

Kt  A  ( roverd) Torsion (Fig. C-3):

b

A  0.83425 Kts  A  ( roverd)

4.

Safety factor

Calculate the mean and alternating components of the loads.

Tm 

3.

Tmax  2.2 kN  m


Mean

2.

roverd  0.05

A  0.97098

b  0.21796

Kt  1.865 b  0.21649

b

Kts  1.596

Calculate the notch sensitivity of the material for bending and torsion using Table 6-6. Assume a value for the notch radius and iterate the solution if necessary. Assume d  50 mm r  roverd d r  2.500  mm



10-35-2

Bending: 2

Neuber constant (for S ut  814  MPa) 1

q b 

Notch sensitivity

a  0.050  in q b  0.863

a

1

r

Torsion: Neuber constant (for S ut  20 MPa  834  MPa) 1

q s 

Notch sensitivity

5.

6.

q s  0.887

a

1

2

a  0.040  in

r

Calculate the fatigue stress concentration factors for bending and torsion using equation 6.11b. Bending

Kf  1  q b  Kt  1 

Kf  1.746

Torsion

Kfs  1  q s  Kts  1 

Kfs  1.528

Assume that the stresses are such that the conditions of equation 6.17 will result in Kfm  Kf

Kfsm  Kfs

and

S'e  0.5 S ut

7.

Calculate the unmodified endurance limit (equation 6.5a).

8.

Calculate the endurance limit modification factors for a rotating round beam. Load

Cload  1

Size

Csize( d )  1.189  

d

Surface

A  4.51

b  0.265

9.

Combined bending and torsion

  mm  

Csurf

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability

Creliab  1.0

S'e  407  MPa

 0.097

(machined)

b

Csurf  0.764

(R = 50%)

Calculate the modified endurance limit. S e( d )  Cload  Csize( d )  Csurf  Ctemp Creliab S'e

10. Use equation (10.8) to solve for the diameter at the shoulder. Given 1

3   2 2  32 Nf   Kf  Ma  4   Kfs Ta d=   S e( d )  π 

Kfm Mm

2

 S ut

3 4


2

3

  


MACHINE DESIGN - An Integrated Approach, 4th Ed. d  Find ( d )

10-35-3

d  54.177 mm

d  56 mm

Say

10. Calculate the minimum, maximum, mean, and alternating bending stresses to check the assumption regarding th mean stress concentration factors. 32 Pmin l

σxmin 

π d

32 Pmax l

σxmax 

π d σxm 

σxmax  47.851 MPa

3

32 Pm l

π d σxa 


3

σxm  35.526 MPa

3

32 Pa l

π d

σxa  12.325 MPa

3

S y  703  MPa

Yield strength SAE 4130 Q&T @ 1200F Evaluate

Kfm 


S y  Kf  σxa

σxm

if  S 1  S y  S 2  2  S y

0 otherwise Kfm  1.746

Kf  1.746

Which checks with the assumption made in step 6 above. 11. DESIGN SUMMARY Shaft diameter

d  56 mm

Shoulder dia

D  Doverd d

D  67.2 mm

Fillet radius

r  roverd d

r  2.8 mm



10-36-1


Determine the size of key necessary to give a safety factor of at least 2 against both shear and bearing failure for the crank/shaft connection of Problem 10-35. Assume a shaft diameter of 58 mm and a key made from SAE 1040 CR steel.

Given:

Shaft diameter

d  58 mm

Shaft properties: S ut1  814  MPa

Maximum torque Tmax  2.2 kN  m

S y1  703  MPa

Minimum torque Tmin  1.1 kN  m

Key properties:

Nd  2


S ut2  586  MPa S y2  490  MPa



As recommended in Table 10-2, for a shaft diameter of d  58 mm w  16 mm

Key width 2.


3.

4.

1 Tmax  Tmin  2 0.5 d

Fa  18.97  kN

Fm 

1 Tmax  Tmin  2 0.5 d

Fm  56.90  kN



τa =


σ'a = 3  τa = 3 

τm =

w L Fa w L

Fm w L

σ'm = 3  τm = 3 

Fm w L


Nd =


Solving for L

5.

Fa 

=

S e S ut

 3   S F  S F    ut a e m  w L 

L=



S'e  0.5 S ut2

S'e  293  MPa



10-36-2


Cload  1

Size

A95 = w L

d eq =

w L 0.0766

 0.097

Csize = 1.189  d eq Cs  1.189  

   0.0766 mm  w

 0.5 0.097

Cs  0.918

 0.0485

Csize = 0.805  L

 S ut2    MPa 

Surface (machined)

Csurf  4.51 

Temperature

Ctemp  1

Reliability


 0.265

(R = 50%)

 0.0485


6.

 0.0485



where

Csurf  0.833

Ce  224  MPa


 Fm

 0.0485

L=

 0.0485




3  Nd  S ut2 Fa  Ce 

 0.0485

  in 



RHS( L) 

L

w S ut2 Ce 

L



 Fm



 0.0485

  in 

Guess

L  25 mm

RHS( L)  39.339 mm

L  RHS( L)

RHS( L)  39.747 mm

Tentatively, let 7.

L  40 mm

Check the realised factor of safety against fatigue failure.

Endurance limit

S e  Ce 

L

  in 

 0.0485

S e  219.1  MPa




8.

S e S ut2

 3   S F  S F    ut2 a e m  w L 

Nf  2.0


9.

Nf 

10-36-3

Abear 

1 2

 w L


Maximum force



σmax 


Ns 

Fmax

2

Fmax  75.9 kN

σmax  237.07 MPa

Abear

S y2

Ns  2.1

σmax


w  16 mm

Key length

L  40 mm

(16 x 10)



10-37-1


As an alternative to the keyed connection in Problem 10-35 determine the amount of diametral interference needed to provide a suitable interference fit for the crank of Figure P10-8 using a shaft diameter of 58 mm, such that the stresses in the hub and shaft will be safe and the maximum torque can be transmitted through the interference fit. The crank material is the same as the shaft and its length along the shaft is 64 mm. The effective outside diameter of the crank is 150 mm.

Given:

Gear hub diameter

d hub  150  mm

Shaft diameter:


L  64 mm Tp  2.2 kN  m

Young's modulus E  209  GPa Poisson's ratio ν  0.28

Material properties are:

S ut  814  MPa, and S y  703  MPa.

d shaft  58 mm

Assumptions: The coefficient of friction between the hub and shaft is μ  0.15. Solution:


1.


2.


3.

Nd  2

Hub radius

ro  0.5 d hub

ro  75 mm



r  29 mm


ri  0  mm


π L r μ  δmin E  ro  r  2

Torque capacity

T=



Nd

2  ro

2

2

2

Solving for min

δmin 

Let the minimum diametral interference be 4.

2  ro  Nd  Tp 2 2 π L r μ  E  ro  r 

δmin  0.057  mm

δmin  0.06 mm


δ E  ro  r  2


p ( δ) 

2

2

4  r ro

Tangential

σti ( δ)  p ( δ)

Radial

σri( δ)  p ( δ)

Stress in hub Tangential

σto( δ)  p ( δ) 

2

2

2

2

ro  r ro  r

Radial

σro( δ)  p ( δ)

These are principal stresses. The tangential stress is σ1 and the radial is σ3. Use equation (5.7c) to find the von Mises stress. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


10-37-2

δ E

σ'o( δ) 

von Mises

4  r ro

  3  ro  r

4

4

2

0.5



Kt  1

5.

There is no bending stress in the shaft at the crank, therefore

6.


δ  δmin

Shaft

Given

Nd =



δsmax  0.23 mm Nd =

Given

Sy Kt σ'o( δ)


δhmax  0.11 mm δmax  δhmax


8.


δmin  0.06 mm


δmax  0.12 mm


t  0.5  δmax  δmin

t  0.03 mm

Hub bore diameter:

Dmin  d shaft

Dmin  58.00  mm


Dmax  58.03  mm


d max  58.12  mm


d min  58.09  mm



δmax  0.12 mm

Minimum interference


δmin  0.06 mm

Shaft diameter:

9.

δmax  0.11 mm


π L r μ  δmin E  ro  r  2

Torque

Hub strength

Ntorque  Nhub 

Tp Sy




2  ro

2

2

Ntorque  2.1 Nhub  1.9



10-38-1


An electric motor-driven shaft that has a fluctuating load torque turns at an average speed of 1600 rpm. The load torque varies sinusoidally once per revolution of the shaft with a peak torque of 29500 in-lb. The shaft diameter is 2.0 in. Design a suitable flywheel for this system that will provide a coefficient of fluctuation of 0.05 and an overspeed safety factor of at least 4. The flywheel is to be a hollow circular disc of constant thickness made from SAE 1020 CR steel.

Given:

Yield strength of SAE 1020 CR steel S y  57 ksi Average shaft speed ωavg  1600 rpm d i  2.000  in


3

ν  0.28

γ  0.28 lbf  in

Coeff. of fluct.

Cf  0.05

Peak torque

Tpeak  29500  in lbf


Plot the load torque over one revolution of the shaft. Tl( θ )  Tpeak  sin( θ )

θ  0 0.1  2  π

4

Load Torque

4 10

4

2 10 T l( θ)

0

in lbf

4

 2 10

4

 4 10

0

60

120

180

240

300

360

θ deg Shaft Position

2.

The average load torque is

Tavg  0  in lbf

The minimum shaft speed occurs at

θmin  180  deg

The maximum shaft speed occurs at

θmax  360  deg

Calculate the required change in kinetic energy by integrating the left-hand side of equation 10.18d.

Energy variation 3.

4.

θ  max Ek   Tl( θ ) dθ θ min

Ek  59000  in lbf

Calculate the required system moment of inertia using equation 10.22. Ek

System moment of inertia

Is 

Let

Im  Is

2

Is  42.032 lbf  in sec

2

Cf  ωavg

As a design choice, let the disc thickness be t  0.4 d i , t  0.800  in (this value of t was repeatedly changed to arrive at a value of 4.0 for Nos). Calculate the outside radius required for this thickness using equation 10.17d.



Outside radius

5.

10-38-2

 2  Im g  d i  4 ro      π γ  t  2  

0.25

ro  14.655 in

Let the maximum tangential stress (at r = ri) be equal to the yield stress and solve equation 10.23a for the yield speed.

Yield speed

0.5 S y g   ωyield    2    3  ν   di  2 d 1  3 ν   i   2     2  ro    γ         8   2   3  ν   2  

ωyield  6376 rpm Factor of safety against overspeeding

6.

Nos 

ωyield ωavg

Nos  4.0

DESIGN SUMMARY Inside radius

ri  0.5 d i

Outside radius

ro  14.655 in

Tickness

t  0.800  in

ri  1  in



10-39-1


Repeat Problem 10-38 for a peak load torque of 40500 in-lb and a torque vs shaft position function of Tl = Tpeak (sin  + sin 2).

Given:

Yield strength of SAE 1020 CR steel S y  57 ksi Average shaft speed ωavg  1600 rpm d i  2.000  in


3

ν  0.28

γ  0.28 lbf  in

Coeff. of fluct.

Cf  0.05

Peak torque

Tpeak  40500  in lbf


Plot the load torque over one revolution of the shaft. Tl( θ )  Tpeak  ( sin( θ )  sin( 2  θ ) )

θ  0 0.1  2  π

5

Load Torque

1 10

4

5 10 T l( θ)

0

in lbf

4

 5 10

5

 1 10

0

60

120

180

240

300

360

θ deg Shaft Position

2.

The average load torque is

Tavg  0  in lbf

The minimum shaft speed occurs at




Calculate the required change in kinetic energy by integrating the left-hand side of equation 10.18d.

Energy variation

3.

4.

(and again at 240 deg)

θ  max Ek   Tl( θ ) dθ θ min

Ek  91125  in lbf



Is 

Let

Im  Is

Is  64.919 lbf  in sec

2

2

Cf  ωavg

As a design choice, let the disc thickness be t  0.6 d i , t  1.200  in (this value of t was repeatedly changed to arrive at a value of 4.0 for Nos). Calculate the outside radius required for this thickness using equation 10.17d.

Outside radius

 2  Im g  d i 4  ro      π γ  t  2  

0.25

ro  14.762 in



10-39-2


Yield speed

 ωyield    3   γ    8

  2 2   ν   d i  2  1  3 ν   di      2  r    2 o           3  ν   2   S y g

0.5

ωyield  6330 rpm Factor of safety against overspeeding

6.

Nos 

ωyield ωavg

Nos  4.0


ri  0.5 d i

Outside radius

ro  14.762 in

Tickness

t  1.200  in

ri  1.000  in



10-40a-1


Given:

As an alternative to the keyed connection in Problem 10-6 determine the amount of diametral interference needed to provide a suitable interference fit for the gear of Figure P10-3 using the data of Problem 10-6 and row a in Table P10-1. The gear hub diameter is 3.50 in, its length along the shaft is 2.00 in, and it has the same properties as the shaft material. Shaft diameter: d shaft  1.75 in Gear hub diameter d hub  3.50 in 6


L  2.00 in Tp  2000 lbf  in



S ut  108  ksi, and S y  62 ksi.



1.


2.


3.

Nd  2

Hub radius

ro  0.5 d hub

ro  1.75 in



r  0.875  in


ri  0  in


π L r μ  δmin E  ro  r  2

Torque capacity

T=



Nd

2  ro

2

2

2

Solving for min

δmin 


2  ro  Nd  Tp 2 2 π L r μ  E  ro  r 

δmin  0.00043  in

δmin  0.0005 in



δ E  ro  r  2


p ( δ) 

2

2

4  r ro

Tangential

σti ( δ)  p ( δ)

Radial

σri( δ)  p ( δ)


σto( δ)  p ( δ) 

2

2

2

2

ro  r ro  r

Radial

σro( δ)  p ( δ)

These are principal stresses. The tangential stress is σ1 and the radial is σ3. Use equation (5.7c) to find the von Mises stress. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


10-40a-2

δ E

σ'o( δ) 

von Mises

4  r ro

  3  ro  r

4

4

2

0.5



Kt  1

5.

There is no bending stress in the shaft at the gear, therefore

6.


δ  δmin

Shaft

Given

Nd =



δsmax  0.0048 in Nd =

Given

Sy Kt σ'o( δ)




8.


δmin  0.0005 in


δmax  0.0021 in


t  0.5  δmax  δmin

t  0.0008 in

Hub bore diameter:

Dmin  d shaft

Dmin  1.7500 in


Dmax  1.7508 in


d max  1.7521 in


d min  1.7513 in



δmax  0.0021 in



δmin  0.0005 in

Shaft diameter:

9.

δmax  0.0021 in


π L r μ  δmin E  ro  r  2

Torque

Hub strength


Tp Sy




2  ro

2

2




10-41-1


Figure P10-6 shows a shaft, running at 400 rpm, with two attached gears. The right-hand gear (2) furnishes the load torque and the left-hand gear (1) furnishes an equal but opposite direction input torque. The torque varies between 2.2 kN-m and 6.2 kN-m. Determine the minimum, maximum, and average power transmitted by the shaft.

Given:

Shaft speed

ω  400  rpm

Minimum torque

Tmin  2.2 kN  m

Maximum torque

Tmax  6.2 kN  m

Solution: 1.

rad sec



Average torque

2.

ω  41.888

Tavg 

Tmax  Tmin 2

Tavg  4  kN  m

Maximum power


Pmax  259.7  kW

Minimum power


Pmin  92 kW

Average power


Pavg  175.9  kW




10-42-1


Figure P10-6 shows a shaft, running at 750 rpm, with two attached gears. The right-hand gear (2) furnishes the load torque and the left-hand gear (1) furnishes an equal but opposite direction input torque. The forces on a gear act at the pitch diameter and have a radial (P in the figure) and a tangential component (not shown). If the pitch diameter of gear 1 is 250 mm and the tangential force varies between 15 kN and 60 kN, determine the minimum, maximum, and average power transmitted by the shaft.

Given:

Shaft speed

ω  750  rpm

1.


Determine the minimum and maximum torque. Minimum torque

Tmin  0.5 d  Fmin

Tmin  1.875  kN  m

Maximum torque

Tmax  0.5 d  Fmax

Tmax  7.5 kN  m


Average torque

2.

sec


Maximum tangential force

1.

rad

d  250  mm Fmin  15 kN

Pitch diameter Minimum tangential force

Solution:

ω  78.54 

Tavg 

Tmax  Tmin 2

Tavg  5  kN  m

Maximum power


Pmax  589.0  kW

Minimum power


Pmin  147  kW

Average power


Pavg  368.2  kW




10-43-1


Determine the amount of diametral interference needed to provide a suitable interference fit (instead of the keyed attachment shown) for the 5-in diameter by 1.5-in thick gear (1) of Figure P10-6 using a shaft diameter of 1.5 in, such that the stresses in the hub and shaft will be safe and the input torque of 1500 lbf-in can be transmitted through the interference fit. Both parts are SAE 4130 steel normalized @ 1650F.

Given:

Gear hub diameter

d hub  5.00 in

Shaft diameter:


L  1.50 in Tp  1500 lbf  in




S y  63 ksi

d shaft  1.50 in 6



1.


2.


3.

Nd  2

Hub radius

ro  0.5 d hub

ro  2.5 in



r  0.75 in


ri  0  in


π L r μ  δmin E  ro  r  2

Torque capacity

T=



Nd

2  ro

2

2

2

Solving for min

δmin 


2  ro  Nd  Tp 2 2 π L r μ  E  ro  r 

δmin  0.00041  in

δmin  0.0004 in



δ E  ro  r  2


p ( δ) 

2

2

4  r ro

Tangential

σti ( δ)  p ( δ)

Radial

σri( δ)  p ( δ)


σto( δ)  p ( δ) 

2

2

2

2

ro  r ro  r

Radial

σro( δ)  p ( δ)



10-43-2

These are principal stresses. The tangential stress is σ1 and the radial is σ3. Use equation (5.7c) to find the von Mises stress.

δ E

σ'o( δ) 

von Mises

4  r ro

  3  ro  r

4

4

2

0.5



5.

There is very little bending stress in the shaft at the gear, therefore Kt  1

6.


δ  δmin

Shaft

Given

Nd =



δsmax  0.0035 in Nd =

Given

Sy Kt σ'o( δ)




8.


δmin  0.0004 in

Maximum diametral interference

δmax  0.0018 in


t  0.5  δmax  δmin

t  0.0007 in

Hub bore diameter:

Dmin  d shaft

Dmin  1.5000 in


Dmax  1.5007 in


d max  1.5018 in


d min  1.5011 in



δmax  0.0018 in



δmin  0.0004 in

Shaft diameter:

9.

δmax  0.0018 in


π L r μ  δmin E  ro  r  2

Torque

Hub strength


Tp Sy




2  ro

2

2




10-44-1


Determine the amount of diametral interference needed to provide a suitable interference fit (instead of the keyed attachment shown) for the 125-mm diameter by 75-mm thick gear (1) of Figure P10-6 using a shaft diameter of 80 mm, such that the stresses in the hub and shaft will be safe and the input torque of 170 N-m can be transmitted through the interference fit. Both parts are SAE 4140 steel normalized @ 1650F.

Given:

Gear hub diameter

d hub  125  mm

Shaft diameter:


L  75 mm Tp  170  N  m

Young's modulus E  209  GPa Poisson's ratio ν  0.28


S ut  1020 MPa

S y  655  MPa

d shaft  80 mm



1.


2.


3.

Nd  2

Hub radius

ro  0.5 d hub

ro  62.5 mm



r  40 mm


ri  0  mm


π L r μ  δmin E  ro  r  2

Torque capacity

T=



Nd

2  ro

2

2

2

Solving for min

δmin 


2  ro  Nd  Tp 2 2 π L r μ  E  ro  r 

δmin  0.0039 mm

δmin  0.004  mm

Note that it is not practical to specify a dimension in milimeters that has more than 3 decimal places. 4.


δ E  ro  r  2


p ( δ) 

2

2

4  r ro

Tangential

σti ( δ)  p ( δ)

Radial

σri( δ)  p ( δ)


σto( δ)  p ( δ) 

2

2

2

2

ro  r ro  r

Radial

σro( δ)  p ( δ)



10-44-2

These are principal stresses. The tangential stress is σ1 and the radial is σ3. Use equation (5.7c) to find the von Mises stress.

δ E

σ'o( δ) 

von Mises

4  r ro

  3  ro  r

4

4

2

0.5



5.

There is very little bending stress in the shaft at the gear, therefore Kt  1

6.


δ  δmin

Shaft

Given

Nd =



δsmax  0.425  mm Nd =

Given

Sy Kt σ'o( δ)


δhmax  0.141  mm δmax  δhmax


8.


δmin  0.004  mm

Maximum diametral interference

δmax  0.140  mm


t  0.5  δmax  δmin

t  0.0680 mm

Hub bore diameter:

Dmin  d shaft

Dmin  80.000 mm


Dmax  80.068 mm


d max  80.140 mm


d min  80.072 mm



δmax  0.140  mm



δmin  0.004  mm

Shaft diameter:

9.

δmax  0.140  mm


π L r μ  δmin E  ro  r  2

Torque

Hub strength


Tp Sy




2  ro

2

2




10-45-1


Table P10-2 shows the energy pulses delivered to (positive) and from (negative) a rotating system along with the shaft angles at which the torque function crosses the average torque line in a torque-time function. Using this data, determine the shaft angles at which minimum and maximum shaft speed occur and the total change in energy from the shaft position at which maximum speed occurs to the position at which minimum speed occurs. Range

ΔE

θ0  0  deg to θ1  75 deg

ΔE1  1040 N  m

θ1  75 deg to θ2  195  deg

ΔE2  2260 N  m

θ2  195  deg to θ3  330  deg

ΔE3  2950 N  m

θ3  330  deg to θ4  360  deg

ΔE4  1740 N  m

Given:

Solution:


1.

The minimum shaft speed occurs at the shaft angle at the end of the largest positive energy pulse. In this case, it is at θ2  195  deg.

2.

The maximum shaft speed occurs at the shaft angle at the end of the largest negative energy pulse. In this case, it is at θ3  330  deg.

3.

The total energy change is the algebraic difference between the accumulated energy between the maximum and minimum shaft speeds. E1  ΔE1

E1  1040 N  m

E2  E1  ΔE2

E2  1220 N  m

E3  E2  ΔE3

E3  1730 N  m

E4  E3  ΔE4

E4  10 N  m

ΔEnergy  E3  E2

ΔEnergy  2950 N  m



10-46-1


An electric motor-driven shaft that has a fluctuating load torque turns at an average speed of 1950 rpm. The energy pulses to (positive) and from (negative) the driven system are given in Table P10-2. The shaft diameter is 50 mm. Design a suitable flywheel for this system that will provide a coefficient of fluctuation of 0.05 and an overspeed safety factor of at least 5. The flywheel is to be a hollow circular disc of constant thickness made from SAE 1040 CR steel.

Given:

Yield strength of SAE 1040 CR steel S y  490  MPa Coeff. of fluct. Average shaft speed ωavg  1950 rpm Shaft diameter

Solution: 1.

d i  50 mm

3

ρ  7800 kg m Cf  0.05

ν  0.28


Calculate the change in kinetic energy from the data in Table P10-2. The minimum shaft speed occurs at




The change in kinetic energy between the maximum and minimum speeds is Ek  2950 N  m 2.

3.



Is 

Let

Im  Is

2

Is  1.415  kg m

2

Cf  ωavg

As a design choice, let the disc thickness be t  0.4 d i , t  20.0 mm (this value of t was repeatedly changed to arrive at a value of 5.0 for Nos). Calculate the outside radius required for this thickness using equation 10.17d.

Outside radius

4.

 2 Im  d i  4 ro      π ρ  t  2  

0.25

ro  276  mm


Yield speed

 ωyield    3  ρ   8

  2 2   ν   d i  2  1  3 ν   di     2   2  ro   3  ν    2            Sy

0.5

ωyield  9580 rpm Factor of safety against overspeeding 5.

Nos 

ωyield ωavg

Nos  5


ri  0.5 d i

ri  25 mm

Outside radius

ro  276  mm

Tickness

t  20 mm



12-1-1


A 20-deg pressure angle, 27-tooth spur gear has a diametral pitch of 5. Find the pitch diameter, addendum, dedendum, outside diameter, and circular pitch.

Given:

Diametral pitch

p d  5  in

Pressure angle

ϕ  20 deg

Solution:

1

Number of teeth N  27

See Mathcad file P1201. Pitch diameter

Addendum

Dedendum

d 

N pd

a 

1.0

b 

1.25

pd

pd

d  5.400  in

a  0.200  in b  0.250  in

Outside diameter

Do  d  2  a

Do  5.800  in

Circular pitch

p c 

π

p c  0.628  in

pd



12-2-1



Given:

Diametral pitch

p d  8  in

Pressure angle

ϕ  25 deg

Solution:

1

Number of teeth N  43


Addendum

Dedendum

d 

N pd

a 

1.0

b 

1.25

pd

pd

d  5.375  in

a  0.125  in b  0.156  in

Outside diameter

Do  d  2  a

Do  5.625  in

Circular pitch

p c 

π

p c  0.393  in

pd



12-3-1


A 57-tooth spur gear is in mesh with a 23-tooth pinion. For the diametral pitch and pressure angle given below, find the contact ratio.

Given:

Tooth numbers:

Solution:

1

pinion

Np  23

Diametral pitch

p d  6  in

gear

Ng  57

Pressure angle

ϕ  25 deg

See Mathcad file P1203. Circular pitch

p c 

π pd

p b  p c  cos( ϕ)

Base pitch

p c  0.524  in p b  0.475  in

Pinion: Pitch dia Pitch rad

d p 

Np pd

rp  0.5 d p

d p  3.833  in rp  1.917  in

Gear: Pitch dia Pitch rad

d g 

Center distance

pd

rg  0.5 d g a 

Addendum

Ng

C 

1.0 pd Np  Ng 2 pd

d g  9.500  in rg  4.750  in a  0.167  in

C  6.667  in

Length of action

Z 

rp  a 2  rp cos( ϕ)  2   rg  a 2  rg cos( ϕ)  2  C sin( ϕ)

Z  0.708  in Contact ratio

mp 

Z pb

mp  1.491



12-4-1


A 78-tooth spur gear is in mesh with a 27-tooth pinion. For the diametral pitch and pressure angle given below, find the contact ratio.

Given:

Tooth numbers:

Solution:

1

pinion

Np  27

Diametral pitch

p d  6  in

gear

Ng  78

Pressure angle

ϕ  20 deg


p c 

π pd


Base pitch

p c  0.524  in p b  0.492  in


d p 

Np pd

rp  0.5 d p

d p  4.500  in rp  2.250  in


d g 

Center distance

pd

rg  0.5 d g a 

Addendum

Ng

C 


d g  13.000 in rg  6.500  in a  0.167  in

C  8.750  in

Length of action

Z 



mp 

Z pb

mp  1.726


MACHINE DESIGN - An

12-5-1


What will the pressure angle be if the center distance of the spur gearset in Problem 12-3 is increased by 5%?

Given:

Tooth numbers: pinion Np  23

1.

Base Circle Old Pitch Circle New Pitch Circle

Ng  57

gear

Solution:

new

1

Diametral pitch

p d  6  in

Pressure angle

ϕ  25 deg


rbp

fc rp

C + C fc r g

rbg

Define the factor by which the center distance changes. Let

C  ΔC C

= fc

Dividing numerator and denominator of the left side by C, fc = 1 

FIGURE 12-5

ΔC

Diagram Showing Center Change for Problem 12-5

C

2.

In this case, fc  1  0.05

fc  1.05

3.

Determine the pitch diameter and pitch radius of the pinion. d p 

Pitch diameter

pd

rp  0.5 d p

Pitch radius 4.

Np

d p  3.833  in rp  1.917  in

When the centers are moved apart the base circle diameters don't change but new pitch circle diameters are defined by the intersection of the new line of action, which is always tangent to the two base circles, with the line of centers. The new pitch radii are proportional to fc. Thus, from Figure 12-5, the new pressure angle is cos ϕnew =

rbp fc  rp

Substituting rp cos( ϕ) for rbp and solving for new,

 rp cos( ϕ)    fc  rp 

ϕnew  acos

ϕnew  30.33  deg


MACHINE DESIGN - An

12-6-1



Given:

Tooth numbers: pinion Np  27

1.


Ng  78

gear

Solution:

new

1

Diametral pitch

p d  6  in

Pressure angle

ϕ  20 deg


rbp

fc rp

C + C fc r g

rbg


C  ΔC C

= fc

Dividing numerator and denominator of the left side by C, fc = 1 

FIGURE 12-6

ΔC


C

2.


fc  1.07

3.

Determine the pitch diameter and pitch radius of the pinion. d p 

Pitch diameter

pd

rp  0.5 d p

Pitch radius 4.

Np

d p  4.500  in rp  2.250  in

When the centers are moved apart the base circle diameters don't change but new pitch circle diameters are defined by the intersection of the new line of action, which is always tangent to the two base circles, with the line of centers. The new pitch radii are proportional to fc. Thus, from Figure 12-5, the new pressure angle is cos ϕnew =

rbp fc  rp


 rp cos( ϕ)    fc  rp 


ϕnew  28.57  deg



12-7-1


If the spur gearsets in Problems 12-3 and 12-4 are compounded as shown in Figure 12-14, what wil the overall train ratio be?

Given:

Tooth numbers

Solution:

N2  57

N3  23

N4  78

N5  27


1. Using equation 12.9b and assuming that gears 2 and 4 are the driver gears, the train (velocity) ratio is mv 

N2 N4 N3 N5

mv  7.159

2. On the other hand, if gears 3 and 5 are the driver gears then the train ratio is mv 

N3 N5 N2 N4

mv  0.1397



12-8-1


A paper mill processes rolls of paper having a density of 984 kg/m3. The paper roll is 1.50-m outside diameter (OD) x 0.22-m inside diameter (ID) x 3.23-m long and is on a simple supported, hollow, steel shaft with S ut = 400 MPa. Design a 2.5:1 reduction spur gearset to drive this roll shaft to obtain a minimum dynamic safety factor of 2 for 10-year life if the roll turns at 50 rpm with 1.2 hp absorbed.

Units:

yr  2080 hr

Given:

Bending factor of safety

Nfb  2

Life (years)

Life  10 yr

Surface factor of safety

Nfs  2

Gear speed

n g  50 rpm

Power to be transmitted

H  1.2 hp

Gear ratio

mg  2.5

Assumptions: 1. If both pinion and gear are the same material, it will only be necessary to determine the pinion size as it will be governing for the set. 2. If the gears are not surface hardened, it will only be necessary to design to the surface requirement as it will be governing for both bending and surface stresses. Solution:


Design Choices: Pressure angle

ϕ  25 deg

AGMA Quality level

Np  26

Reliability

Number of pinion teeth

Qv  7 R  0.99

Material: AGMA Grade 2 steel for both pinion and gear, through hardened to HB  350 S ac  ( 27000  364  HB)  psi S ac  154400 psi 1.

2.

Determine the number of gear teeth, the pinion speed, and the cycle life. Number of gear teeth

Ng  Np mg

Ng  65

Rotational speed of pinion (rpm)

n p  n g mg

n p  125  rpm

Cycle life:

N  Life

N  1.56  10

2 π

8

Determine the surface geometry factor, I. I 

3.

np

 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

I  0.1368

Write the equations for pitch diameter, pitch-line velocity, and transmitted load in terms of the unknown diametral pitch, pd. Note that, in Mathcad, unit conversion factors are not included. d p p d  

Pitch diameter of pinion (in)

Vt p d  

Pitch-line velocity (fpm)

Wt p d  

Transmitted load (lbf) 4.

Set the application factor, Ca

5.

Write the equation for the dynamic load factor, C

Np

d g p d  

pd

Ng pd

d p p d   n p 2 H

Vt  p d 

Ca  1

v

B  0.25  12  Qv

0.6667

A  50  56 ( 1  B)

B  0.731 A  65.062



12-8-2

Cv p d  

A     min  A  Vt  p d    ft  

B

6.

Tentatively choose the mounting factor, Cm (Assume 0 < F < 2 in

Cm  1.6

7.

Choose an elastic coefficient from Table 12-18 (steel on steel).

Cp  2300 psi

8.

The surface stress equation for the pinion is

9.

Determine the endurance strength of the pinion.

0.5

C a W t  p d   C m

σcp p d F   Cp

Cv p d   F  d p p d   I

 0.056

Life factor

CL  2.466  N

CL  0.857

Reliability

CR  0.7  0.15 log( 1  R)

CR  1

Temperature factor

CT  1

Material surface strength (psi) Grade 2, 350 HB

S ac  154400 psi S fcp 

Endurance strength

S ac CL

S fcp  132381 psi

CT  CR

10. Write the design equations using the range of face-width to diametral-pitch ratio given in the text, and the bending stress equation, solved for the unknown face width. Limits and safety factor FU  p d  

d

 Sfc  Nfc =    σc 

8 pd

2

2

 Cp  F  p d      Nfs Cv p d   d p p d   I  S fcp  p d 

6 6.5 16   in in in

12. From the graph, choose a standard value of p d from Table 12-2. 1

3

 

F pd in

p d  10 in

 

2

FL pd

13. The calculated value of F is

in

F  p d   1.393  in

 

FU pd 1

14. Round this up to the decimal equivalent of a common fractional value. F  1.375  in

in

0

15. Then, the parameters that depend on p d and F are: d p p d   2.600  in

pd C a W t  p d   C m

Face width

11. Plot F(P) vs p over the range

FL  p d  

16

d g p d   6.500  in

6

8

10

12

14

16

pd  in

FIGURE 12-8 Graph of Face Width and Limits for the Pinion (surface) in Problem 12-8



Vt p d   85

ft min

Wt p d   465  lbf

12-8-3

σcp p d F   94207  psi

16. There is a wide range of choice for the number of teeth on the pinion. The total weight of the gears goes down very slightly with increasing pinion tooth number. The choice of Np  26 was somewhat arbitrary but was arrived at after trying values from 18 to 26.



12-9-1


Design a two-stage compound spur gear train for an overall ratio of approximately 47:1. Specify tooth numbers for each gear in the train.

Given:

Approximate ratio

mG  47

n  2

Number of stages

Assumptions: The gears will be cut with a hob and the pressure angle will be ϕ  25 deg Solution:

See Mathcad file P1209. 1

mGs  mG

n

mGs  6.856

1.

For equal stages, the stage ratio is

2.

The minimum number of teeth that we can have without interference on a 25-deg gear cut with a hob is 14. Try pinions with 14, 15, 16, etc. teeth to see if the mating gear will have close to an integral number of teeth. 14 mGs  95.979 15 mGs  102.835 16 mGs  109.69 mG 

 96   14   

n

3.

The first trial is very close to an integer so try 96 teeth:

4.

We can get slightly closer by using two 14-tooth pinions with 95 and 97-tooth gears mG 

5.

95 97 14 14

mG  47.02

mG  47.015

However, it may be harder to find the 97-tooth gear and, it may be less expensive to have two gears with the same number of teeth than to have different numbers of teeth.



12-10-1


Design a three-stage compound spur gear train for an overall ratio of approximately 656:1. Specify tooth numbers for each gear in the train.

Given:

Gear ratio

Solution:


mG  656

s  3

Number of stages

1

ravg  mG

s

ravg  8.689

1.

The average stage ratio is

2.

If the minimum number of teeth is (20 deg pa)

Nmin  18

then the number of teeth on the driven gear for each stage is

or

NG1  floor  ravg Nmin

NG1  156

NG2  ceil ravg Nmin

NG2  157

3. But, the prime factors of 656 are 2 4 and 41. This suggests that one of the gears have a number of teeth that is a multiple of 41, say 164 164 ravg

 18.875

Then, the remaing 2-stage ratio is

Thus,

N5 N7 N4 N6

N2  18

Let

mG2  mG

if

= 72

N2 N3

N3  164

mG2  72

N4  18

and

N6  18

5 6

then

N5 N7 = 72 18 18 = 2  3

Try

N5  2  3

4 2

Summarizing,

1 4

N7  2  3

N2  18

N3  164

N4  18

N5  144

N6  18

N7  162

then

4.

Since the driven gears all have less than 180 teeth, no stage ratio is greater than 10.

5.

Checking the overall gear ratio mG 

N 3 N 5 N 7 N 2 N 4 N 6

N5 N7  23328

mG  656.00



12-11-1


An epicycle spur gear train as shown in Figure 12-16 has a sun gear of 33 teeth and a planet gear of 21 teeth. Find the required number of teeth in the ring gear and determine the ratio between the arm and the sun gear if the ring gear is held stationary.

Given:

Sun tooth number

Ns  33

Arm speed

ωa  1  rpm

Planet tooth number

Np  21

Ring speed

ωr  0  rpm

Solution: 1.


From Figure 12-16 we see that the diameter of the ring gear is equal to the sum of the diameter of the sun gear and two times the diameter of the planet gear. Since sun, planets, and ring gear must have the same diametral pitch, Nr  Ns  2  Np

Number of teeth on ring: 2.

Choose the sun gear as the first gear and the ring gear as the last. Then, write equation (12.11c).

ωra ωsa

=

ωr  ωa ωs  ωa

Solving for the sun speed,

3.

Nr  75

=

Ns Nr

ωs 

Nr ωa  Ns ωa  Nr ωr

The ratio between the arm and sun gear is

Ns

ωa ωs

ωs  3.273  rpm

 0.306



12-12-1


An epicycle spur gear train as shown in Figure 12-16 has a sun gear of 23 teeth and a planet gear of 31 teeth. Find the required number of teeth in the ring gear and determine the ratio between the arm and the ring gear if the sun gear is held stationary.

Given:

Sun tooth number

Ns  23

Arm speed

ωa  1  rpm

Planet tooth number

Np  31

Sun speed

ωs  0  rpm

Solution: 1.





ωra ωsa

=


Solving for the ring speed,

3.

Nr  85

=

Ns Nr

ωr 

Nr ωa  Ns ωa  Ns ωs

The ratio between the arm and ring gear is

Nr

ωa ωr

ωr  1.271  rpm

 0.787



12-13-1


An epicycle spur gear train as shown in Figure 12-16 has a sun gear of 23 teeth and a planet gear of 31 teeth. Find the required number of teeth in the ring gear and determine the ratio between the sun and the ring gear if the arm is held stationary.

Given:

Sun tooth number

Ns  23

Arm speed

ωa  0  rpm

Planet tooth number

Np  31

Sun speed

ωs  1  rpm

Solution: 1.





ωra ωsa

=


Solving for the ring speed,

3.

Nr  85

=

Ns

ωr 

Nr Nr ωa  Ns ωa  Ns ωs

The ratio between the sun and ring gear is

Nr

ωs ωr

ωr  0.271  rpm

 3.696



12-14-1


If the gearset in Problem 12-3 transmits 125 HP at 1000 pinion rpm, find the torque on each shaft.

Given:

Tooth numbers: pinion Np  23 gear

Ng  57

Pinion speed

ωp  1000 rpm

Transmitted power

P  125  hp

Assumptions: There is no loss of power in the gear mesh (100% efficiency). Solution: 1.


For the pinion shaft

Tp 

P

Tp  7878 in lbf

ωp

Tp  656.5  ft  lbf 2.

The gear shaft will rotate at a lower speed, which is determined by the gear ratio. (The speed will be decreased in proportion to the ratio and the torque will be increased by the reciprocal of the ratio).

3.

For the gear shaft

ωg  Tg 

Np Ng P

ωg

 ωp

ωg  403.509  rpm Tg  19524  in lbf Tg  1627 ft  lbf

4.

We could have calculated the torque on the gear shaft directly without finding the gear shaft speed, Ng Tg   Tp Tg  19524  in lbf Np Tg  1627 ft  lbf



12-15-1


If the gearset in Problem 12-4 transmits 33 kW at 1600 pinion rpm, find the torque on each shaft.

Given:

Tooth numbers: pinion Np  27 gear

Ng  78

Pinion speed

ωp  1600 rpm

Transmitted power

P  33 kW

Assumptions: There is no loss of power in the gear mesh (100% efficiency). Solution:

See Mathcad file P1215. Tp 

P

Tp  197  N  m

1.


2.

The gear shaft will rotate at a lower speed, which is determined by the gear ratio. (The speed will be decreased i proportion to the ratio and the torque will be increased by the reciprocal of the ratio).

3.

For the gear shaft

ωg  Tg 

4.

ωp

Np Ng P

ωg

 ωp

ωg  553.846  rpm Tg  569  N  m

We could have calculated the torque on the gear shaft directly without finding the gear shaft speed, Ng Tg   Tp Tg  569  N  m Np



12-16-1


Size the spur gears in problem 12-14 for a bending factor of safety of at least 2 assuming a steady torque, 25 deg pressure angle, full depth teeth, quality index of 9, an AISI 4140 steel pinion, and a class 40 cast iron gear.

Given:

Factor of safety

Nfb  2

Power to be transmitted (hp) Rotational speed of pinion

Reliability H  125  hp R  0.99 n  1000 rpm AGMA Quality level Qv  9


Np  23

Solution:

Ng  57

Number of gear teeth

N  10

Life (cycles)

7


Pinion 1.

Determine the bending geometry factor, J (Table 12-13) Jp  0.42

2.

Write the equations for pitch diameter, pitch-line velocity, and transmitted load in terms of the unknown diametral pitch, pd. Note that, in Mathcad, unit conversion factors are not included. d  p d  


Vt p d  


Wt p d  

Transmitted load (lbf)

3.

Set the application factor, Ka

4.

Write the equation for the dynamic load factor, K

Np pd dpd  n 2 H

Vt  p d 

Ka  1 v

B  0.25  12  Qv

0.6667

B  0.52

A  50  56 ( 1  B) A  Kv p d     min  A  Vt  p d    ft   5.

Tentatively choose the mounting factor, Km (Assume 2 < F < 6 in) Km  1.7

6.

The bending stress equation for the pinion is

7.


σbp p d F  

A  76.878 B

Ka Wt p d   p d  Km Kv p d   F  Jp

 0.0323

Life factor

KL  1.6831 N

Reliability

KR  0.7  0.15 log( 1  R)

KL  1 KR  1

Temperature factor

KT  1

Material bending strength (psi) AISI 4140 Nitrided steel

S atp  40000  psi

Endurance strength

S fbp 

S atp KL KT  KR

S fbp  40001  psi



Write the design equations using the range of face-width to diametral-pitch ratio given in the text, and the bending stress equation, solved for the unknown face width. FU  p d  

Upper limit

FL  p d  

Lower limit

9.

12-16-2

Nfb =

Face width

F  p d  

d

p d 

pd 8 pd

S fb

Safety factor

Plot F(P) vs. p over the range

16

σb Ka Wt p d   p d  Km Nfb Kv p d   Jp S fbp

1 1.5 6   in in in


4

 

F pd

p d  4  in

in


FL pd

3

 

F  p d   2.743  in

in

12. Round this up to the decimal equivalent of a common fractional value.

2

 

FU pd in

1

F  2.750  in 13. Then, the parameters that depend on p d and F are: d  p d   5.750  in

Kv p d   0.809

Vt p d   1505

Wt p d   2740 lbf

ft min

σbp p d F   19952  psi

0

2

2.5

3

3.5

4

4.5

5

5.5

6

pd  in

FIGURE 12-16A Graph of Face Width and Limits for the Pinion in Problem 12-16

The assumption made in step 5 is correct so no further iteration is required. Gear 14. Determine the bending geometry factor, J (Table 12-13) Jg  0.49 15. The bending stress equation for the gear is

σbg p d F  

Ka Wt p d   p d  Km Kv p d   F  Jg

16. Determine the endurance strength of the pinion. Material bending strength (psi) Class 40 cast iron Endurance strength

S atg  13000  psi S fbg 

S atg KL KT  KR

S fbg  13000  psi



12-16-3

17. Write the design equations using the range of face-width to diametral-pitch ratio given in the text, and the bending stress equation, solved for the unknown face width. Upper limit

Lower limit

FU  p d  

16

FL  p d  

8

Nfb =

Face width

F  p d  

d

pd

S fb

Safety factor

18. Plot F(P) vs. p over the range p d 

pd

σb Ka Wt p d   p d  Km Nfb Kv p d   Jg S fbg

1 1.5 6   in in in

19. From the graph, choose a standard value of p d from Table 12-2.

6

1

p d  3  in

 

F pd


in

F  p d   4.178  in

4.5

 

FL pd in


3

 

FU pd

F  4.250  in

in

1.5

22. Then, the parameters that depend on P and F are: d  p d   7.667  in

Kv p d   0.788

Vt p d   2007

Wt p d   2055 lbf

ft min

0

2

2.5

3

3.5

4

4.5

5

5.5

6

pd  in

FIGURE 12-16B Graph of Face Width and Limits for the Gear in Problem 12-16

σbg p d F   6391 psi

23. The gear dimensions are larger (smaller diametral pitch means bigger teeth) than for the pinion. This means that we will accept the gear requirements for the pinion, thus, for the set 1

Diametral pitch

p d  3  in

Face width

F  4.250  in

24. Determine the realized factor of safety for the gear using the above values for F and p d. Gear factor of safety

Nfbg 

S fbg

σbg p d F 

Nfbg  2.0

25. Check the factor of safety on the pinion: Pinion factor of safety

Nfbp 

S fbp

σbp p d F 

Nfbp  5.4



12-17-1


Size the spur gears in problem 12-15 for a bending factor of safety of at least 2.5 assuming a steady torque, 20 deg pressure angle, full depth teeth, quality index of 11, an AISI 4340 steel pinion, and an A-7-d nodular iron gear.

Given:

Factor of safety

Nfb  2.5


Ng  78


H  33 kW n  1600 rpm

Reliability AGMA Quality level

R  0.99 Qv  11


Np  27

Life (cycles)

N  10

Solution: Pinion

7


1.


2.

Write the equations for pitch diameter, pitch-line velocity, and transmitted load in terms of the unknown diametral pitch, p d. Note that, in Mathcad, unit conversion factors are not included. d  p d  


Vt p d  


Wt p d  



4.



Vt  p d 

Ka  1 v

B  0.25  12  Qv

0.6667

B  0.25



6.


7.


B


 0.0323

Life factor

KL  1.6831 N

Reliability

KR  0.7  0.15 log( 1  R)

KL  1 KR  1

Temperature factor

KT  1


S atp  42000  psi

Endurance strength 8.

σbp p d F  

A  92

S fbp 


S fbp  42001  psi

Write the design equations using the range of face-width to diametral-pitch ratio given in the text, and the bending stress equation, solved for the unknown face width.



FU  p d  

Upper limit

FL  p d  

Lower limit

9.

Nfb =

Face width

F  p d  

d

p d 

16 pd 8 pd

S fb

Safety factor


12-17-2


1 1.5 10   in in in


4

 

F pd

p d  6  in

in


FL pd

3

 

F  p d   1.318  in

in


2

 

FU pd in

1

F  1.375  in 13. Then, the parameters that depend on p d and F are: d  p d   4.500  in

Kv p d   0.908

Vt p d   1885

Wt p d   775  lbf

ft min

σbp p d F   16103  psi

0

4

5

6

7

8

9

10

pd  in



σbg p d F  


16. Determine the endurance strength of the pinion. Material bending strength (psi) A-7-d nodular iron Endurance strength

S atg  34000  psi S fbg 


S fbg  34001  psi

17. Write the design equations using the range of face-width to diametral-pitch ratio given in the text, and the bending stress equation, solved for the unknown face width. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


Upper limit

Lower limit

FU  p d  

16

FL  p d  

8

Nfb =

Face width

F  p d  

d

pd

pd

S fb

Safety factor


12-17-3


1 1.5 10   in in in


6

1

p d  6  in

 

F pd


in

F  p d   1.434  in

4.5

 

FL pd in


3

 

FU pd

F  1.500  in

in

1.5


Kv p d   0.908

Vt p d   1885

Wt p d   775  lbf

ft min

0

4

5

6

7

8

9

10

pd  in


σbg p d F   13004  psi

23. The gear dimensions are larger than those for the pinion. This means that we will accept the gear requirements for the pinion, thus, for the set 1

Diametral pitch

p d  6  in

Face width

F  1.500  in


Nfbg 

S fbg


Nfbg  2.6


Nfbp 

S fbp


Nfbp  2.8



12-18-1


Size the spur gears in problem 12-14 for a surface factor of safety of at least 2 assuming a steady torque, 25 deg pressure angle, full depth teeth, quality index of 9, an AISI 4140 steel pinion, and a class 40 cast iron gear.

Given:

Factor of safety

Nfs  2


Reliability H  125  hp R  0.99 n  1000 rpm AGMA Quality level Qv  9

Number of pinion teeth Pressure angle

Np  23 ϕ  25 deg

Solution:

Ng  57


N  10

Life (cycles)

7


Pinion 1.


2.

 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

I  0.1365

Write the equations for pitch diameter, pitch-line velocity, and transmitted load in terms of the unknown diametral pitch, p d. Note that, in Mathcad, unit conversion factors are not included. d p p d  


Vt p d  


Wt p d  


3.


4.


Np pd d p p d   n 2 H

Vt  p d 

Ca  1 v

B  0.25  12  Qv

0.6667

B  0.52

A  50  56 ( 1  B)

A  76.878

A  Cv p d     min  A  Vt  p d    ft   5.

Tentatively choose the mounting factor, Cm (Assume 2 < F < 6 in Cm  1.7

6.

Choose an elastic coefficient from Table 12-18. (Steel on cast-iron)

6.


7.

Determine the endurance strength of the pinion. Life factor

Cp  2100 psi

σcp p d F   Cp

CL  2.466  N

B

 0.056

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I

CL  1



12-18-2 CR  0.7  0.15 log( 1  R)

Reliability Temperature factor

CT  1

Material surface strength (psi) AISI 4140 Nitrided steel

S acp  165000 psi S fcp 


S acp CL

S fcp  164997 psi

CT  CR

Write the design equations using the range of face-width to diametral-pitch ratio given in the text, and the surface stress equation, solved for the unknown face width. Upper limit

Lower limit

9.

CR  1

FU  p d  

16

FL  p d  

8

Nfc =

Face width

F  p d  

d

pd

 Sfc  σ   c

Safety factor

Plot F(P) vs. p over the range p d 

pd

2

C a W t  p d   C m

2  Cp    Nfs Cv p d   d p p d   I  S fcp 



1 1.5 6   in in in


4

 

F pd

1

p d  4  in

in

3

 


FL pd

F  p d   2.379  in

2

in


 

FU pd in

1

F  2.375  in 13. Then, the parameters that depend on p d and F are: d p p d   5.750  in

Cv p d   0.809

Vt p d   1505

Wt p d   2740 lbf

ft min

σcp p d F   116767 psi

0

2

2.5

3

3.5

4

4.5

5

5.5

6

pd  in


The assumption made in step 5 is correct so no further iteration is required. Gear 14. Write the equation for the gear pitch diameter

d g p d  

Ng pd



12-18-3 C a W t  p d   C m

σcg p d F   Cp

15. The surface stress equation for the gear is

Cv p d   F  d g p d   I

16. Determine the endurance strength of the pinion. S acg  80000  psi

Material bending strength (psi) Class 40 cast iron Hardness factor

Hp  230

Hp

Hg  200

From equation 12.26,

A  0

Gear ratio

mg 

Hg

Ng

 1.2

mg  2.478

Np

Hardness ratio factor

CH  1  A   mg  1 

Endurance strength

S fcg 

CH  1

S acg CL  CH

S fcg  79999  psi

CT  CR

17. Write the design equations using the range of face-width to diametral-pitch ratio given in the text, and the surface stress equation, solved for the unknown face width. Upper limit Lower limit

Safety factor

16

FL  p d  

8

pd pd

 Sfc  Nfc =    σc 

2

C a W t  p d   C m

2

 Cp  F  p d      Nfs Cv p d   d g p d   I  S fcg 

Face width 18. Plot F(P) vs. p over the range p d  d

FU  p d  

1 1.5 6   in in in 6


 

F pd

p d  4  in

in


FL pd

 

F  p d   4.083  in

22. Then, the parameters that depend on p d and F are:

Vt p d   1505

ft min

3

in


d g p d   14.250 in

4.5

Cv p d   0.809 Wt p d   2740 lbf

 

FU pd in

1.5

0

2

2.5

3

3.5

4

4.5

5

5.5

6

pd  in




12-18-4

σcg p d F   56282  psi 23. The gear face width is larger than for the pinion. This means that we will accept the gear requirements for the pinion thus, for the set 1

Diametral pitch

p d  4  in

Face width

F  4.125  in

24. Determine the realized factor of safety for the gear using the above values for F and p d.

Gear factor of safety

 Sfcg   σ p F   cg d  

2

 Sfcp  Nfsp     σcp pd F  

2

Nfsg 

Nfsg  2.0


Nfsp  3.5



12-19-1


Size the spur gears in problem 12-15 for a surface factor of safety of at least 1.2 assuming a steady torque, 20 deg pressure angle, full depth teeth, quality index of 11, an AISI 4340 steel pinion, and an A-7-d nodular iron gear.

Given:

Factor of safety

Nfs  1.2


Reliability H  33 kW R  0.99 n  1600 rpm AGMA Quality level Qv  11


Np  27 ϕ  20 deg

Solution:

Ng  78


N  10

Life (cycles)

7


Pinion 1.


2.

 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

I  0.1194



Vt p d  


Wt p d  


3.


4.


Np pd d p p d   n 2 H

Vt  p d 

Ca  1 v

B  0.25  12  Qv

0.6667

B  0.25

A  50  56 ( 1  B) Cv p d  

A  92

A     min  A  Vt  p d    ft  

5.


6.

Choose an elastic coefficient from Table 12-18. (Steel on nodular-iron)

6.


7.


B

Cp  2160 psi

σcp p d F   Cp

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I

 0.056

Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1



12-19-2

Temperature factor

CT  1


S acp  162000 psi S fcp 


S fcp  161997 psi

CT  CR

Write the design equations using the range of face-width to diametral-pitch ratio given in the text, and the surface stress equation, solved for the unknown face width. Upper limit

Lower limit

9.

S acp CL

FU  p d  

16

FL  p d  

8

Nfc =

Face width

F  p d  

d

pd

 Sfc  σ   c

Safety factor

Plot F(P) vs. p over the range p d 

pd

2

C a W t  p d   C m




6 6.5 16   in in in


2

 

F pd

1

p d  10 in

in

1.5

 


FL pd

F  p d   1.478  in

1

in


 

FU pd in

0.5


Cv p d   0.925

Vt p d   1131

Wt p d   1291 lbf

ft min

σcp p d F   146808 psi

0

6

8

10

12

14

16

pd  in


The assumption made in step 5 is correct so no further iteration is required. Gear Ng

14. Write the equation for the gear pitch diameter

d g p d  


σcg p d F   Cp


Cv p d   F  d g p d   I



12-19-3

16. Determine the endurance strength of the pinion. S acg  97000  psi

Material bending strength (psi) A-7-d nodular iron Hardness factor

Hp  235

Hp

Hg  230


A  0

Gear ratio

mg 

Hg

Ng

 1.0

mg  2.889

Np


CH  1  A   mg  1 

Endurance strength

S fcg 

CH  1

S acg CL  CH

S fcg  96998  psi

CT  CR

17. Write the design equations using the range of face-width to diametral-pitch ratio given in the text, and the surface stress equation, solved for the unknown face width. Upper limit Lower limit

Safety factor

16

FL  p d  

8

pd pd

 Sfc  Nfc =    σc 

2

C a W t  p d   C m

2


Face width 18. Plot F(P) vs. p over the range p d  d

FU  p d  

6 6.5 16   in in in 3


 

F pd

p d  10 in

in


FL pd

 

F  p d   1.427  in

in

21. Round this up to the decimal equivalent of a common fractional value. F  1.500  in 22. Then, the parameters that depend on p d and F are: d g p d   7.800  in

Cv p d   0.925

Vt p d   1131

Wt p d   1291 lbf

ft min

2

 

FU pd 1 in

0

6

8

10

12

14

16

pd  in


σcg p d F   86374  psi © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


12-19-4

23. The gear face width is the same as that for the pinion. This means that we will accept the gear requirements for the set 1

Diametral pitch

p d  10 in

Face width

F  1.500  in



 Sfcg   σ p F   cg d  

2

 Sfcp  Nfsp     σcp pd F  

2

Nfsg 

Nfsg  1.3


Nfsp  1.2



12-20-1


If the gearset in Problem 12-11 transmits 83 kW at 1200 arm rpm, find the torque on each shaft.

Given:

Sun tooth number

Ns  33

Transmitted power

P  83 kW

Planet tooth number

Np  21

Arm speed

ωa  1200 rpm

Ring gear speed

ωr  0  rpm

Nr  Ns  2  Np

Nr  75

Solution:


1.

Determine the number of teeth on the ring:

2.


ωra ωsa

=


Solving for the planet speed,

3.

=

Ns Nr

ωs 

Nr ωa  Ns ωa  Nr ωr Ns

ωs  3927 rpm

Calculate the torque on the sun and arm shafts. Arm shaft

Ta 

Sun shaft

Ts 

P

ωa P

ωs

Ta  660  N  m

Ts  202  N  m



12-21-1


If the gearset in Problem 12-12 transmits 39 HP at 2600 arm rpm, find the torque on each shaft.

Given:

Tooth numbers: sun

Ns  23

ωa  2600 rpm

Arm speed

Transmitted power

planet

Np  31

Sun speed

ωs  0  rpm

P  39 hp


See Mathcad file P1221. Ta 

P

Ta  945  in lbf

1.

For the arm shaft

2.

The ring gear will rotate at a higher speed, which is determined by the gear ratio.

3.

For the ring gear (see Problem 12-12)

ωa

Number of teeth on ring:

ωr 

Tr 

Nr  Ns  2  Np


ωr  3303.529 rpm

Nr P

ωr

Nr  85

Tr  744  in lbf



12-22-1


If the gearset in Problem 12-13 transmits 23 kW at 4800 sun rpm, find the torque on each shaft.

Given:

Tooth numbers: sun planet Transmitted power

Ns  23

Arm speed

ωa  0  rpm

Np  31

Sun speed

ωs  4800 rpm

P  23 kW


See Mathcad file P1222. Ts 

P

Ts  45.8 N  m

1.

For the sun shaft

2.

The ring gear will rotate at a lower speed, which is determined by the gear ratio.

3.

For the ring gear (see Problem 12-13) Number of teeth on ring:

ωr 

Tr 

ωs

Nr  Ns  2  Np


ωr  1298.8 rpm

Nr P

ωr

Nr  85

Tr  169  N  m



12-23-1


Size the spur gears in problem 12-20 for a bending factor of safety of at least 2.8 assuming a steady torque, 25 deg pressure angle, full depth teeth, quality index of 9, an AISI 4140 steel pinion, and a class 40 cast iron gear.

Given:

Factor of safety

Nfb  2.8

Power to be transmitted Rotational speed of pinion

Reliability H  83 kW R  0.99 n  1200 rpm AGMA Quality level Qv  9


Np  21

Solution:

Ng  33


N  10

Life (cycles)

7


Pinion 1.


2.

Write the equations for pitch diameter, pitch-line velocity, and transmitted load in terms of the unknown diametral pitch, p d. Note that, in Mathcad, unit conversion factors are not included. Pitch diameter of pinion and gear (in)

Vt p d  


Wt p d  



4.


d p p d  

Np

d g p d  

pd

Ng pd

d g p d   n 2 H

Vt  p d 

Ka  1

v

B  0.25  12  Qv

0.6667

B  0.52

A  50  56 ( 1  B) Kv p d  

A     min  A  Vt  p d    ft  

5.


6.


7.


σbp p d F  

A  76.878 B


 0.0323

Life factor

KL  1.6831 N

Reliability

KR  0.7  0.15 log( 1  R)

KL  1 KR  1

Temperature factor

KT  1


S atp  40000  psi

Endurance strength

S fbp 


S fbp  40001  psi




Upper limit

FL  p d  

Lower limit

9.

12-23-2

Nfb =

Face width

F  p d  

d

p d 

pd 8 pd

S fb

Safety factor


16


1 1.5 6   in in in


6

 

F pd

p d  4  in

in


FL pd

 

F  p d   2.196  in

in


5 4 3

 

FU pd 2 in 1


Kv p d   0.768

Vt p d   2592

Wt p d   1417 lbf

ft min

σbp p d F   13212  psi

0

2

3

4

5

6

pd  in



σbg p d F  


16. Determine the endurance strength of the pinion. Material bending strength (psi) Class 40 cast iron Endurance strength

S atg  13000  psi S fbg 


S fbg  13000  psi



12-23-3


Lower limit

FU  p d  

16

FL  p d  

8

Nfb =

Face width

F  p d  

d

pd

S fb

Safety factor


pd


1 1.25 6   in in

in


6

1

p d  3  in

 

F pd


in

F  p d   3.486  in

5

 

FL pd in


4

 

FU pd

F  3.500  in

in

3

22. Then, the parameters that depend on P and F are: d g p d   11.000 in

Kv p d   0.744

Vt p d   3456

Wt p d   1063 lbf

ft min

2

2

3

4

pd  in


σbg p d F   4624 psi

23. The gear dimensions are larger (smaller diametral pitch means bigger teeth) than for the pinion. This means that we will accept the gear requirements for the pinion, thus, for the set 1

Diametral pitch

p d  3  in

Face width

F  3.500  in


Nfbg 

S fbg


Nfbg  2.8


Nfbp 

S fbp


Nfbp  7.7



12-24-1


Size the spur gears in Problem 12-21 for a bending factor of safety of at least 2.4 assuming a steady torque, 20 deg pressure angle, full depth teeth, quality index of 11, an AISI 4340 steel pinion, and an A-7-d nodular iron gear.

Given:

Factor of safety

Nfb  2.4

Power to be transmitted (hp) Number of pinion (sun) teeth

H  39 hp Np  23


Number of gear (planet) teeth

Ng  31

Life (cycles)

Solution:

R  0.99 Qv  11 N  10

7


Take the sun as the pinion and the planet as the gear, which is also an idler. With a stationary sun, the planet gear will have one point (where it meshes with the sun) that has an instantaneous zero velocity. All other points on the planet will have a linear velocity that is proportional to its distance from this instant center. The point that meshes with the ring gear has maximum velocity. The pitchline velocity at the ring gear is a function of the tooth numbers and the diametral pitch of the set. But the absolute planet velocity depends only on the tooth numbers and the arm velocity. For the data of Problem 12-21, the absolute planet (gear) rotational velocity is For

ωa  2600 rpm ωg 

Np  2  Ng Ng

ωr  3303.53  rpm  ωr

ωg  9058.1 rpm

The sun (pinion) is stationary and, therefore, has a zero pitchline velocity. However, the planet (gear) mesh point rotates about the pitch circle of the sun at the rotational velocity of the arm. Therefore, take the pinion pitchline velocity to be Vtp = rp ωa =

Np 2 pd

 ωa

The maximum pitchline velocity of the planet (gear) is at the point where it meshes with the ring gear, which is also the pitchline velocity of the ring gear. Thus, for the gear, the pitchline velocity is Vtg = rr ωr =

Np  2  Ng 2 pd

 ωr

The torques on the arm and ring were determined in Problem 12-21 to be Ta  945.38 in lbf and Tr  744.05 in lbf . Load analysis of the gearset reveals that the transmitted load on the sun, planet, and ring gear teeth is Wt =

2  p d  Tr Np  2  Ng

Pinion 1.


2.

Write the equations for pitch diameter, pitch-line velocity, and transmitted load in terms of the unknown diametral pitch, pd. Note that, in Mathcad, unit conversion factors are not included. Pitch diameter of pinion (in)

d  p d  

Np pd



12-24-2


Vtp p d  


Wt p d  

Np 2 pd

 ωa

2  p d  Tr Np  2  Ng

Ka  1

3.


4.


B  0.25  12  Qv

v

0.6667

B  0.25

A  50  56 ( 1  B) Kvp p d  

A     min  A  Vtp pd    ft  

5.


6.


7.


σbp p d F  

Kvp p d   F  Jp

 0.0323

KL  1.6831 N

Reliability

KR  0.7  0.15 log( 1  R)

KL  1 KR  1

Temperature factor

KT  1


S atp  42000  psi S fbp 


S fbp  42001  psi

Write the design equations using the range of face-width to diametral-pitch ratio given in the text, and the bending stress equation, solved for the unknown face width. Upper limit

Lower limit

9.

B

Ka Wt p d   p d  Km

Life factor


A  92

16

FL  p d  

8

Nfb =

Face width

F  p d  

p d 

pd

pd

S fb

Safety factor

Plot F(P) vs. p over the range d

FU  p d  

σb Ka Wt p d   p d  Km Nfb Kvp p d   Jp S fbp

10 11.5 20   in in in

10. From the graph, choose a standard value of p d from Table 12-2. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


12-24-3

1

p d  14 in

2


 

F pd

F  p d   0.997  in

in

 


FL pd

1

in

F  1.000  in 13. Then, the parameters that depend on p d and F are: d  p d   1.643  in Vtp p d   1118

1.5

Kvp p d   0.925 ft

min

 

FU pd in

0.5

Wt p d   245  lbf

0 10

12

14

16

18

20

pd  in

σbp p d F   17449  psi FIGURE 12-24A The assumption made in step 5 is correct so no further iteration is required.

Graph of Face Width and Limits for the Pinion in Problem 12-24

Gear 14. Determine the bending geometry factor, J (Table 12-9)

Jg  0.37

15. Write the equations for pitch diameter and pitchline velocity for the gear in terms of the unknown diametral pitc p d. Note that, in Mathcad, unit conversion factors are not included. Pitch diameter of pinion (in)

Pitchline velocity (fpm)

dg p d  

Ng

Vtg p d  

Np  Ng

pd

Kvg p d  

16. Write the equation for the dynamic load factor, K v

2 pd

A     min  A  Vtg pd    ft  

17. Account for the fact that the gear is an idler by setting the idler factor

σbg p d F  

17. The bending stress equation for the gear is

 ωr

B

KI  1.42 Ka Wt p d   p d  Km KI Kvg p d   F  Jg

18. Determine the endurance strength of the pinion. Material bending strength (psi) A-7-d nodular iron Endurance strength

S atg  34000  psi S fbg 


S fbg  34001  psi

19. Write the design equations using the range of face-width to diametral-pitch ratio given in the text, and the bending stress equation, solved for the unknown face width. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


FU  p d  

Upper limit

16 pd

FL  p d  

Lower limit

Nfb =

Face width

F  p d  

p d 

8 pd

S fb

Safety factor

17. Plot F(P) vs. p d over the range

12-24-4

σb Ka Wt p d   p d  Km KI  Nfb Kvg p d   Jg S fbg

6 6.5 16   in in in 2


 

F pd

1

in

p d  12 in

 

FL pd


in

F  p d   1.244  in

 

in

F  1.250  in

0.5

0

21. Then, the parameters that depend on P and F are: dg p d   2.583  in

Kvg p d   0.879

Vtg p d   3892

Wt p d   210  lbf

min

1

FU pd


ft

1.5

6

8

FIGURE 12-24B

10

12

14

16

pd  in

Graph of Face Width and Limits for the Gear in Problem 12-24

σbg p d F   14095  psi 22. The gear dimensions are larger (smaller diametral pitch means bigger teeth) than for the pinion. This means that we will accept the gear requirements for the pinion, thus, for the set 1

Diametral pitch

p d  12 in

Face width

F  1.250  in


Nfbg 

S fbg


Nfbg  2.4


Nfbp 

S fbp


Nfbp  4.1



12-25-1


Size the spur gears in problem 12-20 for a surface factor of safety of at least 1.7 assuming a steady torque, 25 deg pressure angle, full depth teeth, quality index of 9, an AISI 4140 steel pinion, and a class 40 cast iron gear.

Given:

Factor of safety

Nfs  1.7


Reliability H  83 kW R  0.99 n g  1200 rpm AGMA Quality level Qv  9


Np  21 ϕ  25 deg

Solution:

Ng  33


N  10

Life (cycles)

7


Pinion 1.


2.

 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

I  0.1170

Write the equations for pitch diameter, pitchline velocity, and transmitted load in terms of the unknown diametral pitch, p d. Note that, in Mathcad, unit conversion factors are not included. Pitch diameters of pinion and gear (in)

Vt p d  

pitchline velocity (fpm)

Wt p d  


3.


4.


d p p d  

Np

d g p d  

pd

Ng pd

d g p d   n g 2 H

Vt  p d 

Ca  1 v

B  0.25  12  Qv

0.6667

B  0.52

A  50  56 ( 1  B) Cv p d  

A  76.878

A     min  A  Vt  p d    ft  

5.


6.

Choose an elastic coefficient from Table 12-18. (Steel on cast iron). Cp  2100 psi

7.


8.


σcp p d F   Cp

B

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I

 0.056

Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1


MACHINE DESIGN - An Integrated Approach, 4th Ed. Temperature factor

CT  1


S acp  165000 psi S fcp 


12-25-2

S acp CL

S fcp  164997 psi

CT  CR


Upper limit

FL  p d  

Lower limit

pd 8 pd

 Sfc  Nfc =    σc 

Safety factor

2

C a W t  p d   C m

2

 Cp  F  p d      Nfs Cv p d   d p p d   I  S fcp 

Face width


16

p d 

1 1.5 6   in in in


4

 

F pd in

1

p d  5  in

3

 

FL pd


in

F  p d   2.149  in

2

 

FU pd


in

1

0


Cv p d   0.785

Vt p d   2073

Wt p d   1771 lbf

ft min

2

3

FIGURE 12-25A

4

5

6

pd  in


σcp p d F   123671 psi The assumption made in step 5 is correct so no further iteration is required. Gear Ng

15. Write the equation for the gear pitch diameter

d g p d  


σcg p d F   Cp


Cv p d   F  d g p d   I



12-25-3

17. Determine the endurance strength of the pinion. Material bending strength (psi) Class 40 cast iron

S acg  80000  psi

Hp  230

Hardness factor

Hp

Hg  200


A  0

Gear ratio

mg 

Hg

Ng

 1.2

mg  1.571

Np


CH  1  A   mg  1 

Endurance strength

S fcg 

CH  1

S acg CL  CH

S fcg  79999  psi

CT  CR


Lower limit

FU  p d  

16

FL  p d  

8 pd

 Sfc  Nfc =    σc 

Safety factor

2

C a W t  p d   C m

2


Face width

p d 


1 1.5 6   in in in 6


 

F pd in

p d  4  in

21. The calculated value of F is F  p d   3.807  in

4.5

 

FL pd in

3

 

FU pd


in

1.5

0

23. Then, the parameters that depend on P and F are: d g p d   8.250  in

pd

Cv p d   0.768

2

2.5

FIGURE 12-25B

3

3.5

4

4.5

5

5.5

6

pd  in



MACHINE DESIGN - An Integrated Approach, 4th Ed. Vt p d   2592

ft min

12-25-4

Wt p d   1417 lbf

σcg p d F   59858  psi

24. The gear dimensions are larger (smaller diametral pitch means bigger teeth) than for the pinion. This means that we will accept the gear requirements for the pinion, thus for the set 1

Diametral pitch

p d  4  in

Face width

F  4.000  in


 Sfcg   σ p F   cg d  

2

 Sfcp  Nfsp     σcp pd F  

2

Nfsg 

Nfsg  1.8


Nfsp  4.8



12-26-1


Size the spur gears in Problem 12-21 for a surface factor of safety of at least 1.3 assuming a steady torque, 20 deg pressure angle, full depth teeth, quality index of 11, an AISI 4340 steel pinion, and an A-7-d nodular iron gear.

Given:

Factor of safety

Nfs  1.3

Pressure angle

ϕ  20 deg

Power to be transmitted (hp) Number of pinion (sun) teeth

H  39 hp Np  23


R  0.99 Qv  11

Number of gear (planet) teeth

Ng  31

Life (cycles)

Solution:

N  10

7


Take the sun as the pinion and the planet as the gear, which is also an idler. With a stationary sun, the planet gear will have one point (where it meshes with the sun) that has an instantaneous zero velocity. All other points on the planet will have a linear velocity that is proportional to its distance from this instant center. The point that meshes with the ring gear has maximum velocity. The pitchline velocity at the ring gear is a function of the tooth numbers and the diametral pitch of the set. But the absolute planet velocity depends only on the tooth numbers and the arm velocity. For the data of Problem 12-21, the absolute planet (gear) rotational velocity is For

ωa  2600 rpm ωg 

Np  2  Ng Ng

ωr  3303.53  rpm  ωr

ωg  9058.1 rpm

The sun (pinion) is stationary and, therefore, has a zero pitchline velocity. However, the planet (gear) mesh point rotates about the pitch circle of the sun at the rotational velocity of the arm. Therefore, take the pinion pitchline velocity to be Vtp = rp ωa =

Np 2 pd

 ωa

The maximum pitchline velocity of the planet (gear) is at the point where it meshes with the ring gear, which is also the pitchline velocity of the ring gear. Thus, for the gear, the pitchline velocity is Vtg = rr ωr =

Np  2  Ng 2 pd

 ωr

The torques on the arm and ring were determined in Problem 12-21 to be Ta  945.38 in lbf and Tr  744.05 in lbf . Load analysis of the gearset reveals that the transmitted load on the sun, planet, and ring gear teeth is Wt =

2  p d  Tr Np  2  Ng

Pinion 1.


2.

 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

I  0.0923

Write the equations for pitch diameter, pitchline velocity, and transmitted load in terms of the unknown diametr pitch, p d. Note that, in Mathcad, unit conversion factors are not included. Pitch diameter of pinion (in)

d p p d  

Np pd



12-26-2


Vtp p d  


Wt p d  

Np 2 pd

 ωa

2  p d  Tr Np  2  Ng

Ca  1

3.


4.


B  0.25  12  Qv

v

0.6667

B  0.25

A  50  56 ( 1  B) Cvp p d  

A  92

A     min  A  Vtp pd    ft  

5.

Tentatively choose the mounting factor, Cm (Assume 0 < F < 2 in)

Cm  1.6

6.

Choose an elastic coefficient from Table 12-18.

Cp  2100 psi

7.


8.


σcp p d F   Cp

0.5

Ca Wt p d   Cm

Cvp p d   F  d p p d   I

 0.056

Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1

Temperature factor

CT  1


S acp  162000 psi S fcp 


B

S acp CL

S fcp  161997 psi

CT  CR


Lower limit

Safety factor

Face width

FU  p d  

16

FL  p d  

8

Nfc =

pd

pd

 Sfc  σ   c

2

C a W t  p d   C m

2

 Cp  F  p d      Nfs Cvp p d   d p p d   I  S fcp 



p d 

10. Plot F(P) vs. p over the range d

12-26-3

6 6.5 16   in in in


2

 

1

F pd

p d  14 in

in


1.5

 

FL pd

F  p d   0.611  in

1

in

 


FU pd in

F  0.625  in

0.5


0

6

8

10

12

14

16

d p p d   1.643  in

Cvp p d   0.925

FIGURE 12-26A

Vtp p d   1118

Wt p d   245  lbf


ft min

pd  in

σcp p d F   140460 psi The assumption made in step 5 is correct so no further iteration is required. Gear 15. Write the equations for pitch diameter and pitchline velocity for the gear in terms of the unknown diametral pitc p d. Note that, in Mathcad, unit conversion factors are not included. d g p d  


Vtg p d  


16. Write the equation for the dynamic load factor, C v

Ng pd Np  Ng 2 pd

A  Cvg p d     min  A  Vtg pd    ft  

σcg p d F   Cp


 ωr

B

Ca Wt p d   Cm

Cvg p d   F  d g p d   I

18. Determine the endurance strength of the pinion. Material bending strength (psi) A-7-d nodular iron Hardness factor From equation 12.26,

S acg  97000  psi

Hp  235

Hg  230

Hp Hg

 1.0

A  0



12-26-4

Ng

mg 

Gear ratio

mg  1.348

Np


CH  1  A   mg  1 

Endurance strength

S fcg 

CH  1

S acg CL  CH

S fcg  96998  psi

CT  CR

19. Write the design equations using the range of face-width to diametral-pitch ratio given in the text, and the bending stress equation, solved for the unknown face width. Upper limit Lower limit

FU  p d  

16

FL  p d  

8

pd pd

 Sfc  Nfc =    σc 

Safety factor

2

C a W t  p d   C m

p d 


6 6.5 16   in in in 3


 

F pd

1

p d  12 in

in

 


2 1.5

in

 

FU pd


1

in 0.5

F  1.000  in

0

24. Then, the parameters that depend on P and F are: d g p d   2.583  in

Cvg p d   0.879

Vtg p d   3892

Wt p d   210  lbf

min

2.5

FL pd

F  p d   0.978  in

ft

2

 Cp  F  p d      Nfs Cvg p d   d g p d   I  S fcg 

Face width

6

8

10

12

14

16

pd  in


σcg p d F   84140  psi

25. The gear dimensions are larger than that for the pinion. Thus, for the set 1

Diametral pitch

p d  12 in

Face width

F  1.000  in

26. Determine the realized factor of safety for the gear using the above values for F and p d. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.



12-26-5

 Sfcg   σ p F   cg d  

2

 Sfcp  Nfsp     σcp pd F  

2

Nfsg 

Nfsg  1.3


Nfsp  2.9



12-27-1


If the gearset in Problem 12-10 transmits 190 kW at 1800 input pinion rpm, find the torque on each of the four shafts.

Given:

Transmitted power

P  190  kW

Input speed

ωp1  1800 rpm

Assumptions: Tooth numbers:

Solution: 1.

Stage 1

Np1  18

Ng1  164

Stage 2

Np2  18

Ng2  144

Stage 3

Np3  18

Ng3  162


Confirm that the gear ratio is correct. mG 

2.

Np1 Np2 Np3

mG  656.000

Calculate the gear speeds. Stage 1

Stage 2

Stage 3

3.

Ng1 Ng2 Ng3

ωp1  1800 rpm

ωg1 

ωp2  ωg1

ωg2 

ωp3  ωg2

ωg3 

Np1 Ng1 Np2 Ng2 Np3 Ng3

 ωp1

ωg1  197.561  rpm

 ωp2

ωg2  24.695 rpm

 ωp3

ωg3  2.744  rpm

Calculate the shaft torque. Shaft 1

T1 

Shaft 2

T2 

Shaft 3

T3 

Shaft 4

T4 

P

ωp1 P

ωg1 P

ωg2 P

ωg3

T1  1008 N  m

Input

T2  9184 N  m

T3  73471  N  m

T4  661236 N  m

Output



12-28-1


Size the first-stage spur gears in problem 12-27 for a bending factor of safety of at least 3.2 and a surface factor of safety of at least 1.7 assuming a steady torque, 25 deg pressure angle, full depth teeth, Qv = 8, and AISI 4140 steel for all gears.

Given:

Pressure angle

ϕ  25 deg


Nfb  3.2

Reliability

R  0.99


Nfs  1.7

AGMA Quality level

Qv  8

Power to be transmitted (hp)

H  190  kW

Life (cycles)

N  10


Np  18


Ng  164

7

Rotational speed of pinion (rpm) n  1800 rpm Assumptions: Since both pinion and gear from each stage are the same material, it will only be necessary to determine the pinion size as it will be governing for the set. Solution:


Stage 1 Pinion - Bending Jp  0.38

1.

Determine the bending geometry factor, J (Table 12-13)

2.

Write the equations for pitch diameter, pitchline velocity, and transmitted load in terms of the unknown diametral pitch, pd. Note that, in Mathcad, unit conversion factors are not included.

Vt p d  


Ng pd

d p p d   n 2 H

Vt  p d 

Ka  1

3.


4.

Write the equation for the dynamic load factor, Kv 0.6667

B  0.63

A  50  56 ( 1  B) Kv p d  

A     min  A  Vt  p d    ft  

5.

Tentatively choose the mounting factor, Km (Assume 2 < F < 6 in)

6.


A  70.721 B

Km  1.7

σbp p d F   7.

d g p d  

pd

Wt p d  


B  0.25  12  Qv

Np

d p p d  

Pitch diameter of pinion and gear (in)



KL  1.6831 N

 0.0323

KL  1


MACHINE DESIGN - An Integrated Approach, 4th Ed. KR  0.7  0.15 log( 1  R)

Reliability

KT  1


S atp  40000  psi S atp KL

S fbp 

S fbp  40001  psi

KT  KR


Lower limit

9.

KR  1

Temperature factor


12-28-2

FU  p d  

16

FL  p d  

8

Nfb =

Face width

F  p d  

p d 

d

pd

S fb

Safety factor


pd


1 1.1 5   in in in


8

 

1

F pd

p d  3  in

in


 

FL pd

F  p d   4.546  in

4

in


 

FU pd in


6

2

0

2

2.5

3

3.5

4

pd  in

d g p d   54.667 in FIGURE 12-28A

Vt p d   2827

ft min

σbp p d F   12627  psi

Wt p d   2974 lbf

Graph of Face Width and Limits for the Stage 1 Pinion (bending) in Problem 12-28

Kv p d   0.702

The assumption made in step 5 is correct so no further iteration is required. Stage 1 Pinion - Surface Fatigue 1.

Determine the surface geometry factor, I.



I  2.

 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

12-28-3

I  0.1726


Vt p d  


pd d p p d   n

Wt p d  


2 H

Vt  p d 

Ca  1

3.


4.

Write the equation for the dynamic load factor, Cv B  0.25  12  Qv

Np

d p p d  


0.6667

A  50  56 ( 1  B) A  Cv p d     min  A  Vt  p d    ft  

B  0.63

5.


6.

Choose an elastic coefficient from Table 12-18. (Steel on steel).

7.


Cp  2300 psi

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I

Determine the endurance strength of the pinion.  0.056

Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1

Temperature factor

CT  1


S acp  165000 psi S fcp 


B

Cm  1.7

σcp p d F   Cp 8.

A  70.721

S acp CL CT  CR

S fcp  164997 psi


FU  p d  

16 pd


MACHINE DESIGN - An Integrated Approach, 4th Ed. FL  p d  

Lower limit

Safety factor

Nfc =

12-28-4

8 pd

 Sfc  σ   c

2

C a W t  p d   C m

2


Face width


p d 

1 1.1 6   in in in 6


p d  4  in

 

F pd in


 

4

FL pd in


 

FU pd 2 in

F  4.000  in 0

14. The bending requirement is governing in this case. The diametral pitch and face width for the stage 2 gearset are: 1

p d  3  in

2

3

4

5

6

pd  in

FIGURE 12-28B F  4.500  in

Graph of Face Width and Limits for the Stage 1 Pinion (surface) in Problem 12-28



12-29-1


Size the second-stage spur gears in problem 12-27 for a bending factor of safety of at least 3.2 and a surface factor of safety of at least 1.7 assuming a steady torque, 25 deg pressure angle, full depth teeth, quality index of 8, and AISI 4140 steel for all gears.

Given:

Pressure angle Bending factor of safety

ϕ  25 deg Nfb  3.2

Reliability

R  0.99


Nfs  1.7

AGMA Quality level

Qv  8

Power to be transmitted Number of pinion teeth

H  190  kW Np  18

Life (cycles) Number of gear teeth

N  10 Ng  144


n  197.561  rpm

7

Assumptions: Since both pinion and gear from each stage are the same material, it will only be necessary to determine the pinion size as it will be governing for the set. See Mathcad file P1229. Solution: Stage 2 Pinion - Bending Jp  0.38

1.


2.

Write the equations for pitch diameter, pitchline velocity, and transmitted load in terms of the unknown diametral pitch, p d. Note that, in Mathcad, unit conversion factors are not included.

Vt p d  


d g p d  

pd

pd

2 H

Vt  p d 

Ka  1

3.


4.


B  0.63

A  50  56 ( 1  B) Kv p d  

A     min  A  Vt  p d    ft  

5.


6.


σbp p d F   7.

Ng

d p p d   n

Wt p d  


B  0.25  12  Qv

Np

d p p d  


A  70.721 B

Km  1.8



Life factor

KL  1.6831 N

 0.0323

KL  1



Reliability

KT  1



S fbp 

S fbp  40001  psi

KT  KR


Lower limit

9.

KR  1

Temperature factor


12-29-2

FU  p d  

16 pd

FL  p d   S fb

Safety factor

Nfb =

Face width

F  p d  

p d 


8 pd


1 1.1 3   in in in


20

1

p d  1.5 in

 

F pd


in

F  p d   9.313  in

15

 

FL pd


10

in

 

FU pd

F  9.375  in

in

5

13. Then, the parameters that depend on p d and F are: d p p d   12.000 in

d g p d   96.000 in

Vt p d   621 

Wt p d   13547  lbf

ft min

σbp p d F   12417  psi

Kv p d   0.827

0

1

1.5

2

2.5

pd  in

FIGURE 12-29A Graph of Face Width and Limits for the Stage 2 Pinion (bending) in Problem 12-29




MACHINE DESIGN - An Integrated Approach, 4th Ed. I  2.

 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

12-29-3 I  0.1702

Write the equations for pitch diameter, pitchline velocity, and transmitted load in terms of the unknown diametral pitch, p d. Note that, in Mathcad, unit conversion factors are not included.

Vt p d  



Wt p d  


2 H

Vt  p d 

Ca  1

3.


4.


Np

d p p d  


0.6667


B  0.63

5.


6.


7.


Cp  2300 psi

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I


Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1

Temperature factor

CT  1


S acp  165000 psi S fcp 


B

Cm  1.8

σcp p d F   Cp 8.

A  70.721


S fcp  164997 psi

Write the design equations using the range of face-width to diametral-pitch ratio given on page 740 of the text, and the bending stress equation, solved for the unknown face width. Upper limit

FU  p d  

16 pd



FL  p d  

Lower limit

8 pd

Safety factor

 Sfc  Nfc =    σc 

Face width

F  p d  


p d 

12-29-4

2

C a W t  p d   C m




1 1.1 3   in in in 20


p d  2  in

 

F pd in


 

FL pd in


15

10

 

FU pd in

5

F  8.250  in 0

14. The bending requirement is governing in this case. The diametral pitch and face width for the stage 2 gearset are: 1

p d  1.5 in

F  9.375  in

1

1.5

2

2.5

3

pd  in

FIGURE 12-29B Graph of Face Width and Limits for the Stage 2 Pinion (surface) in Problem 12-29



12-30-1


Size the third-stage spur gears in problem 12-27 for a bending factor of safety of at least 3.2 and a surface factor of safety of at least 1.7 assuming a steady torque, 25 deg pressure angle, full depth teeth, quality index of 8, and AISI 4140 steel for all gears.

Given:

Pressure angle Bending factor of safety

ϕ  25 deg Nfb  3.2

Reliability

R  0.99


Nfs  1.7

AGMA Quality level

Qv  8

Power to be transmitted Number of pinion teeth

H  190  kW Np  18

Life (cycles) Number of gear teeth

N  10 Ng  162

7

n  24.695 rpm

Rotational speed of pinion

Assumptions: Since both pinion and gear from each stage are the same material, it will only be necessary to determine the pinion size as it will be governing for the set. Solution:



1.


2.

Write the equations for pitch diameter, pitchline velocity, and transmitted load in terms of the unknown diametral pitch, p d. Note that, in Mathcad, unit conversion factors are not included. d p p d  


Vt p d  


Wt p d  

Transmitted load (lbf) Set the application factor, Ka

4.


B  0.63

pd

Ng pd

d p p d   n 2 H

Vt  p d 

A  50  56 ( 1  B) Kv p d  

A     min  A  Vt  p d    ft  

5.


6.


A  70.721 B

Km  2.0

σbp p d F   7.

d g p d  

Ka  1

3.

B  0.25  12  Qv

Np



KL  1.6831 N

 0.0323

KL  1



Reliability

KT  1



S fbp 

S fbp  40001  psi

KT  KR


Lower limit

9.

KR  1

Temperature factor


12-30-2

FU  p d  

16

FL  p d  

8

Nfb =

Face width

F  p d  

p d 

d

pd

S fb

Safety factor


pd


0.5 0.6 2   in in in


40

 

1

F pd

p d  0.75 in

in


 

FL pd

F  p d   18.954 in

in



20

 

FU pd in

F  19.000 in

d p p d   24.000 in

30

10

0 0.5

0.75

1

1.25

1.5

1.75

2

pd  in

d g p d   216.000  in FIGURE 12-30A

Vt p d   155 

ft min

σbp p d F   12470  psi

Wt p d   54189  lbf

Graph of Face Width and Limits for the Stage 3 Pinion (bending) in Problem 12-30

Kv p d   0.903




MACHINE DESIGN - An Integrated Approach, 4th Ed. I  2.

 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

12-30-3 I  0.1724



Vt p d  


Wt p d  


pd d p p d   n 2 H

Vt  p d 

Ca  1

3.


4.


Np

0.6667

A  50  56 ( 1  B)

B  0.63

Cv p d  

A     min  A  Vt  p d    ft  

5.


Cm  2.0

6.


Cp  2300 psi

7.


σcp p d F   Cp 8.

B

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I


Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1

Temperature factor

CT  1


S acp  165000 psi S fcp 

Endurance strength

9.

A  70.721


S fcp  164997 psi


FU  p d  

16 pd



Lower limit

Safety factor

Nfc =

12-30-4

8 pd

 Sfc  σ   c

2

C a W t  p d   C m

2


Face width


p d 

0.5 0.6 2   in in in 40


p d  1  in

 

F pd in

12. The calculated value of F is F  p d   16.825 in

 

FL pd

20

in


30

 

FU pd in

10

F  16.875 in 14. The bending requirement is governing in this case. The diametral pitch and face width for the stage 2 gearset are: 1

p d  0.75 in

F  19.000 in

0 0.5

0.75

1

1.25

1.5

1.75

2

pd  in

FIGURE 12-30B Graph of Face Width and Limits for the Stage 3 Pinion (surface) in Problem 12-30



12-31-1



Given:

Approximate ratio

mG  78

Number of stages

n  2

Assumptions: The gears will be cut with a hob and the pressure angle will be ϕ  25 deg Solution: 1.


For equal stages, the stage ratio is 1

mGs  mG

2.

3.

mGs  8.832

The minimum number of teeth that we can have without interference on a 25-deg gear cut with a hob is 12. Try pinions with 12, 13, 14, etc. teeth to see if the mating gear will have close to an integral number of teeth. 12 mGs  105.981

15 mGs  132.476

13 mGs  114.813

16 mGs  141.308

14 mGs  123.645

17 mGs  150.14

The first trial is very close to an integer so try 106 teeth: mG 

4.

n

 106   12   

n

mG  78.028

We can get the exact ratio by using 12-tooth and 16-tooth pinions with 104- and 144-tooth gears Let

Stage 1

Np1  12

Ng1  104

Stage 2

Np2  16

Ng2  144

Then, the gear ratio is

mG 

Ng1 Ng2 Np1 Np2

mG  78.000



12-32-1


Figure P12-1 shows the same paper machine that was analyzed in Problem 6-46 and in other problems in previous chapters. Using the data of Problem 4-46, design a suitable spur-gear set for this application for a 10-yr life against surface failure. State all assumptions.

Units:

yr  2080 hr

Given:

Gear ratio

mg  2

Torque on gear

Tg  74240  in lbf


Life (years)

Assumptions: 1. If both pinion and gear are the same material, it will only be necessary to determine the pinion size as it will be governing for the set. 2. If the gears are not surface hardened, it will only be necessary to design to the surface requirement as it will be governing for both bending and surface stresses. 3. Pinion speed is n p  3  rpm (5 seconds for a 90-deg rotation). Design Choices: Pressure angle Number of pinion teeth

ϕ  25 deg

AGMA Quality level

Qv  8

Np  26

Reliability

R  0.99

Material: AGMA Grade 2 steel for both pinion and gear, through hardened to HB  300 S ac  ( 27000  364  HB)  psi S ac  136200 psi Surface factor of safety Solution: 1.

2.

Nfs  2

shifts  3

Number of shifts


Determine the number of gear teeth and the cycle life. Number of gear teeth

Ng  Np mg

Cycle life:

N  Life shifts

Ng  52 np

7

N  1.123  10

2 π


 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

I  0.1277

3. Write the equations for pitch diameter, pitchline velocity, and transmitted load in terms of the unknown diametral pitch, p d. Note that, in Mathcad, unit conversion factors are not included. d p p d  


Vt p d  


Wt p d  



5.


Np

d g p d  

pd

Ng pd

d p p d   n p 2 2  Tg

d g p d 

Ca  1 v

B  0.25  12  Qv

0.6667

A  50  56 ( 1  B)

B  0.63 A  70.721



12-32-2 A  Cv p d     min  A  Vt  p d    ft  

B

6.


Cm  1.7

7.


Cp  2300 psi

8.


9.


0.5

C a W t  p d   C m

σcp p d F   Cp

Cv p d   F  d p p d   I

 0.056

Life factor

CL  2.466  N

CL  0.993

Reliability

CR  0.7  0.15 log( 1  R)

CR  1

Temperature factor

CT  1


S ac  136200 psi S fcp 

Endurance strength

S ac CL

S fcp  135315 psi

CT  CR

10. Write the design equations using the range of face-width to diametral-pitch ratio given in the text, and the bending stress equation, solved for the unknown face width. Limits and safety factor FU  p d  

8 pd

Nfc =

 Sfc  σ   c

3.5

4

2

2


1 1.5 10   in in in


8

 

F pd in

1

p d  2.5 in

6

 

FL pd


4

in

F  p d   5.414  in

 

FU pd


in

2

0



Face width


FL  p d  

16

d g p d   20.800 in

1

1.5

2

2.5

3

pd  in



MACHINE DESIGN - An Integrated Approach, 4th Ed. Vt  p d   8 

ft min

Wt p d   7138 lbf

12-32-3

σcp p d F   94935  psi

14. There is a wide range of choice for the number of teeth on the pinion. The total weight of the gears goes down very slightly with increasing pinion tooth number. The choice of Np  26 was somewhat arbitrary but was arrived at after trying values from 18 to 26.



12-33-1


Design a non reverted compound transmission based on the arrangement shown in Figure 12-14a for an overall train ratio of approximately 90:1. It should be capable of transmitting 50 hp at 1000 rpm input shaft speed. State all assumptions.

Given:


H  50 hp

Approximate gear ratio

mv  90

Input speed (rpm)

n in  1000 rpm

Number of stages

n  2

Assumptions: 1. If both pinion and gear from each stage are the same material, it will only be necessary to determine the pinion size as it will be governing for the set. 2. If the gears are not surface hardened, it will only be necessary to design to the surface requirement as it will be governing for both bending and surface stresses. Design Choices: Pressure angle Surface factor of safety

ϕ  25 deg

AGMA Quality level

Qv  8

Nfs  1.7

Reliability

R  0.99 7

Life (cycles) N  10 Material: AGMA Grade 2 steel for both pinion and gear, both stages, through hardened to HB  300 S ac  ( 27000  364  HB)  psi S ac  136200 psi Solution: 1.


For equal stage ratios, the stage ratio is 1

mvs  mv 2.

3.

mvs  9.487

The minimum number of teeth that we can have without interference on a 25-deg gear cut with a hob is 14. Try pinions with 14, 15, 16, etc. teeth to see if the mating gear will have close to an integral number of teeth. 14 mvs  132.816

20 mvs  189.737

26 mvs  246.658

32 mvs  303.579

15 mvs  142.302

21 mvs  199.223

27 mvs  256.144

33 mvs  313.065

16 mvs  151.789

22 mvs  208.71

28 mvs  265.631

34 mvs  322.552

17 mvs  161.276

23 mvs  218.197

29 mvs  275.118

35 mvs  332.039

18 mvs  170.763

24 mvs  227.684

30 mvs  284.605

36 mvs  341.526

19 mvs  180.25

25 mvs  237.171

31 mvs  294.092

37 mvs  351.013

The last trial is very close to an integer so try 351 teeth: mv 

4.

n

For each stage let

 351   37   

n

Np  37

mv  89.993 Ng  351

Stage 1 Pinion Given: 1.


n 1  n in

n 1  1000 rpm


 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

I  0.1732



12-33-2



Vt p d  


Wt p d  


d g p d  

pd


4.

Write the equation for the dynamic load factor, Cv 0.6667

2 H

Vt  p d 

Cv p d  


6.


7.


B

Cm  1.6 Cp  2300 psi

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I


Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1

Temperature factor

CT  1


S ac  136200 psi S fcp 


A  70.721

A     min  A  Vt  p d    ft  

5.

8.

pd

d p p d   n 1

A  50  56 ( 1  B)

B  0.63

σcp p d F   Cp

Ng

Ca  1

3.

B  0.25  12  Qv

Np

S ac CL CT  CR

S fcp  136198 psi


Lower limit

FU  p d   FL  p d  

16 pd 8 pd



 Sfc  Nfc =    σc 

Safety factor

d

6 6.5 16   in in in

p d 

From the graph, choose a standard value of p d from Table 12-2.

3

1

p d  8  in 12.

 

F pd

The calculated value of F is

in

F  p d   1.697  in 13.

 

2

FL pd

Round this up to the decimal equivalent of a common fractional value.

in

 

FU pd 1

F  1.750  in 14.

2



11.

2

C a W t  p d   C m

Face width

10.

12-33-3

in

Then, the parameters that depend on p d and F are: d p p d   4.625  in Vt p d   1211

ft min

d g p d   43.875 in Wt p d   1363 lbf

0

6

8

10

12

14

16

pd  in

FIGURE 12-33A Graph of Face Width and Limits for the Stage 1 Pinion (surface) in Problem 12-33

σcp p d F   102877 psi Stage 2 Pinion Given: 1.


n 2 

Ng

 n in

n 2  105.413  rpm



Vt p d  


Wt p d  


Np

Np pd d p p d   n 2 2 H

Vt  p d 

Write the equation for the dynamic load factor, Cv A  Cv p d     min  A  Vt  p d    ft  

B



12-33-4

3.


4.


σcp p d F   Cp 5.

C a W t  p d   C m

Cv p d   F  d p p d   I


Lower limit

Safety factor

16

FL  p d  

8

pd

pd

 Sfc  σ   c

2

C a W t  p d   C m

2



FU  p d  

Nfc =

Face width

6.

Cm  1.7

p d 

2 2.5 12   in in in 8

7.

From the graph, choose a standard value of p d from Table 12-2. 1

in

p d  4  in 8.

 

5.333

FL pd

The calculated value of F is

in

F  p d   3.779  in 9.

 

F pd

 

FU pd 2.667

Round this up to the decimal equivalent of a common fractional value. F  3.750  in

in

0


2

3

4

5

6

7

8

pd  in

d p p d   9.250  in

d g p d   87.750 in

FIGURE 12-33B

Vt p d   255 

Wt p d   6464 lbf


ft min

σcp p d F   104866 psi



12-34-1


Design a reverted compound transmission based on the arrangement shown in Figure 12-14b for an overall train ratio of approximately 80:1. It should be capable of transmitting 30 hp at 1500 rpm input shaft speed. State all assumptions.

Given:


H  30 hp

Approximate gear ratio

mv  80

Input speed (rpm)

n in  1500 rpm

Number of stages

n  2

Assumptions: 1. If both pinion and gear from each stage are the same material, it will only be necessary to determine the pinion size as it will be governing for the set. 2. If the gears are not surface hardened, it will only be necessary to design to the surface requirement as it will be governing for both bending and surface stresses. 3. If both stages have the same tooth numbers, then both must have the same diametral pitch in order that the center distances be the same. In this case, the dimensions of the second stage will govern the design. Design Choices: Pressure angle Surface factor of safety

ϕ  25 deg

AGMA Quality level

Qv  8

Nfs  1.7

Reliability

R  0.99 7

Life (cycles) N  10 Material: AGMA Grade 2 steel for both pinion and gear, both stages, through hardened to HB  300 S ac  ( 27000  364  HB)  psi S ac  136200 psi Solution: 1.


For equal stage ratios, the stage ratio is 1

mvs  mv 2.

3.

mvs  8.944

The minimum number of teeth that we can have without interference on a 25-deg gear cut with a hob is 14. Try pinions with 14, 15, 16, etc. teeth to see if the mating gear will have close to an integral number of teeth. 14 mvs  125.22

20 mvs  178.885

26 mvs  232.551

32 mvs  286.217

15 mvs  134.164

21 mvs  187.83

27 mvs  241.495

33 mvs  295.161

16 mvs  143.108

22 mvs  196.774

28 mvs  250.44

34 mvs  304.105

17 mvs  152.053

23 mvs  205.718

29 mvs  259.384

35 mvs  313.05

18 mvs  160.997

24 mvs  214.663

30 mvs  268.328

36 mvs  321.994

19 mvs  169.941

25 mvs  223.607

31 mvs  277.272

37 mvs  330.938

The next to last trial is very close to an integer so try 322 teeth: mv 

4.

n

For each stage let

 322   36   

n

Np  36

mv  80.003 Ng  322

Stage 2 Pinion Given:


n 2 

Np Ng

 n in

n 2  167.702  rpm




2.

12-34-2

 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

I  0.1723



Vt p d  


Wt p d  


d g p d  

pd


4.

Write the equation for the dynamic load factor, Cv 0.6667

2 H

Vt  p d 

Cv p d  


6.


7.


B

Cm  1.7 Cp  2300 psi

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I


Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1

Temperature factor

CT  1


S ac  136200 psi S fcp 


A  70.721

A     min  A  Vt  p d    ft  

5.

8.

pd

d p p d   n 2

A  50  56 ( 1  B)

B  0.63

σcp p d F   Cp

Ng

Ca  1

3.

B  0.25  12  Qv

Np

S ac CL CT  CR

S fcp  136198 psi


FU  p d  

16 pd



Lower limit

Safety factor

Face width


FL  p d  

12-34-3

8 pd

 Sfc  Nfc =    σc 

2

C a W t  p d   C m

2


2 2.1 12   in in in


4

 

F pd

1

p d  5  in

in

3

 


FL pd

F  p d   2.397  in

in


2

 

FU pd in

1


0

2

3

4

5

6

7

8

9 10 11 12

pd  in

d g p d   64.400 in FIGURE 12-34A

Vt p d   316 

ft min

Wt p d   3132 lbf


σcp p d F   104942 psi 15. This diametral pitch and face width should be used for both stages in the gearset.



PROBLEM 12-35

12-35-1

_____

Statement:

If the 23-tooth pinion in the gearset in Problem 12-3 is on the input shaft, find the velocity ratio, torque ratio, and gear ratio for the gearset.

Given:

Input tooth number

Solution:


1.

Nout  57

Nin Nout

mV  0.404

Calculate the torque ratio using equation 12.1b. mA 

3.

Output tooth number

Calculate the velocity ratio using equation 12.1a. Note that, for a given diametral pitch, the pitch radius of a gear is directly proportional to the number of teeth on the gear. mV  

2.

Nin  23

1 mV

mA  2.478

Calculate the gear ratio using equation 12.1c. mG  mA

mG  2.478



PROBLEM 12-36

12-36-1

_____

Statement:

If the 78-tooth gear in the gearset in Problem 12-4 is on the input shaft, find the velocity ratio, torque ratio, and gear ratio for the gearset.

Given:

Input tooth number

Solution:


1.

Nout  27

Nin Nout

mV  2.889

Calculate the torque ratio using equation 12.1b. mA 

3.

Output tooth number

Calculate the velocity ratio using equation 12.1a. Note that, for a given diametral pitch, the pitch radius of a gear is directly proportional to the number of teeth on the gear. mV  

2.

Nin  78

1 mV

mA  0.346

Calculate the gear ratio using equation 12.1c. mG  mV

mG  2.889



12-37-1

PROBLEM 12-37

_____

Statement:

Figure P12-2 shows an involute of a circle that starts at point A(0, rb) and continues to point P(x,y)

Given:

The angle  is known as the roll angle and  is the involute pressure angle. Derive expressions for the x,y coordinates of P in terms of the base circle radius rb and the involute pressure angle only. Plot y vs. x over the range 0 <  < 40 deg for rb = 2 in. Base circle radius rb  2.00 in

Solution:


1.

From the definition of the involute, the distance from P to Q is the arc-length along the base circle subtended by the angle . Thus, PQ  rb θ

2.

The distance OP = rp is then rp 

3.

2

2

rp  rb 1  θ

2

Looking at triangle OPQ, we see that

tan ( ϕ) 

4.

2

rb  rb  θ

rb θ

θ  tan( ϕ)

rb

Writing the x and y coordinates of P in terms of the above relationships, we have 2

x( ϕ)  rb 1  tan ( ϕ)  sin( tan ( ϕ)  ϕ)

2

y ( ϕ)  rb 1  tan( ϕ)  cos( tan( ϕ)  ϕ) 5.

Plot y vs. x over the range ϕ  0  deg 1  deg  40 deg 2.6

2.4 y ( ϕ) in 2.2

2

0

0.1

0.2

0.3

0.4

x( ϕ) in © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


12-38-1

PROBLEM 12-38

_____

Statement:

Derive equation 12.2 using Figure 12-5.

Solution:


1.

Label points on Figure 12-5 as follows: Center of pinion, Op Center of gear, Og Beginning of contact, E Pitch point, P Leaving contact, F Extend line of action to tangency with pinion base circle, A (not shown on figure) Extend line of action to tangency with gear base circle, B (not shown on figure).

2.

The length of action, Z, is equal to the distance EF = EP + PF.

3.

The distance PF = AF - AP and EP = BE - BP. Thus, Z = AF + BE - (AP + BP).

4.

The distance OpF = rp + a p, and OgE = rg + a g, where a p is the pinion addendum and a g is the gear addendum .

5.

Since the line of action is tangent to both base circles, angle AOpP = angle BOgP = , the gear pressure angle. Also, the distance OpA = rbp = rpcos() and OgB = rbg = rgcos().

6.

From the relationships above,

AF 

rp  ap2   rp cos( ϕ) 2

AP  rp sin( ϕ) 7.

BE 

 rg  ag2   rg cos( ϕ) 2

BP  rg sin( ϕ)

Substituting the equations in step 6 into the equation for Z in step 3,

Z 

rp  ap2   rp cos( ϕ) 2  rg  ag 2  rg cos( ϕ)  2  rp  rg  sin( ϕ)

But, rp + rg = C, the center distance. So,

Z 

rp  ap2   rp cos( ϕ) 2  rg  ag 2  rg cos( ϕ)  2  C sin( ϕ)

which is equations 12-2.



12-39-1

PROBLEM 12-39

_____

Statement:

A 39-tooth spur gear is in mesh with an 18-tooth pinion. The p d = 8 and  = 25 deg. Find the contact ratio.

Given:

Tooth numbers: Pinion Diametral pitch

Solution:

Np  18 1

p d  8  in

Gear

Ng  39

Pressure angle

ϕ  25 deg


p c 

π pd


Base pitch

p c  0.393  in p b  0.356  in


d p 

Np pd

rp  0.5 d p

d p  2.250  in rp  1.125  in


d g 

Center distance

pd

rg  0.5 d g a 

Addendum

Ng

C 


d g  4.875  in rg  2.438  in a  0.125  in

C  3.563  in

Length of action

Z 



mp 

Z pb

mp  1.450



12-40-1

PROBLEM 12-40

_____

Statement:

A 79-tooth spur gear is in mesh with a 20-tooth pinion. The p d = 8 and  = 20 deg. Find the contact ratio.

Given:

Tooth numbers: Pinion

Np  20

Diametral pitch

p d  8  in

Solution:

1

Gear

Ng  79

Pressure angle

ϕ  20 deg


p c 

π pd


Base pitch

p c  0.393  in p b  0.369  in


d p 

Np pd

rp  0.5 d p

d p  2.500  in rp  1.250  in


d g 

Center distance

pd

rg  0.5 d g a 

Addendum

Ng

C 


d g  9.875  in rg  4.938  in a  0.125  in

C  6.188  in

Length of action

Z 



mp 

Z pb

mp  1.690



12-41-1

PROBLEM 12-41

_____

Statement:


Given:


Np  18

Diametral pitch

p d  8  in

Solution: 1.

1

Gear

Ng  39

Pressure angle

ϕ  25 deg



C  ΔC C

new


= fc

Dividing numerator and denominator of the left side by C, ΔC fc = 1  C

fc rp

2.


fc r g

3.

Determine the pitch diameter and pitch radius of the pinion.

fc  1.060

d p 

Pitch diameter

rbp C + C rbg

Np pd

d p  2.250  in FIGURE 12-41 rp  0.5 d p

Pitch radius


rp  1.125  in 4.

When the centers are moved apart the base circle diameters don't change but new pitch circle diameters are defined by the intersection of the new line of action, which is always tangent to the two base circles, with the line of centers. The new pitch radii are proportional to f c. Thus, from Figure 12-41, the new pressure angle is cos ϕnew =

rbp fc  rp


 rp cos( ϕ)    fc  rp 


ϕnew  31.24  deg



12-42-1

PROBLEM 12-42

_____

Statement:


Given:


Np  20

Diametral pitch

p d  8  in

Solution: 1.

1

Gear

Ng  79

Pressure angle

ϕ  20 deg



C  ΔC C

new


= fc

Dividing numerator and denominator of the left side by C, ΔC fc = 1  C 2.


3.

Determine the pitch diameter and pitch radius of the pinion.

C + C fc r g

rbg

fc  1.050

d p 

Pitch diameter

rbp

fc rp

Np pd

d p  2.500  in

FIGURE 12-42 Diagram Showing Center Change for Problem 12-42

rp  0.5 d p

Pitch radius

rp  1.250  in 4.

When the centers are moved apart the base circle diameters don't change but new pitch circle diameters are defined by the intersection of the new line of action, which is always tangent to the two base circles, with the line of centers. The new pitch radii are proportional to f c. Thus, from Figure 12-41, the new pressure angle is cos ϕnew =

rbp fc  rp


 rp cos( ϕ)    fc  rp 


ϕnew  26.50  deg



12-43-1


If the spur gearsets in Problems 12-39 and 12-40 are compounded as shown in Figure 12-14, what will the overall train ratio be?

Given:

Tooth numbers

Solution: 1.

N3  39

N4  20

N5  79


Using equation 12.9b and assuming that gears 2 and 4 are the driver gears, the train (velocity) ratio is mv 

2.

N2  18

N2 N4 N3 N5

mv  0.117

On the other hand, if gears 3 and 5 are the driver gears then the train ratio is mv 

N3 N5 N2 N4

mv  8.5583



12-44-1

PROBLEM 12-44

_____

Statement:


Given:

Number of teeth

N  23

ϕ  20 deg

1

p d  6  in

Diametral pitch Solution:

Pressure angle


Addendum

Dedendum

d 

N pd

a 

1.0

b 

1.25

pd

pd

d  3.833  in

a  0.167  in b  0.208  in

Outside diameter

Do  d  2  a

Do  4.167  in

Circular pitch

p c 

π

p c  0.524  in

pd



12-45-1

PROBLEM 12-45

_____

Statement:


Given:

Number of teeth

N  32

ϕ  20 deg

1

p d  4  in

Diametral pitch Solution:

Pressure angle


Addendum

Dedendum

d 

N pd

a 

1.0

b 

1.25

pd

pd

d  8.000  in

a  0.250  in b  0.313  in

Outside diameter

Do  d  2  a

Do  8.500  in

Circular pitch

p c 

π

p c  0.785  in

pd



PROBLEM 12-46

12-46-1

_____

Statement:


Given:

Approximate ratio

mG  53

Number of stages

n  2



For equal stage ratios, the stage ratio is approximately 1

mGs  mG 2.

n

mGs  7.280

The minimum number of teeth that we can have without interference on a 25-deg gear cut with a hob is 14. Try pinions with 14, 15, 16, etc. teeth to see if the mating gear will have close to an integral number of teeth. 14 mGs  101.922 15 mGs  109.202 16 mGs  116.482

3.

The first trial is very close to an integer so try 102 teeth: mG 

4.

 102   14   

mG  53.082

We can get slightly closer by using two 14-tooth pinions with 101 and 103-tooth gears mG 

5.

n

101  103 14 14

mG  53.077

However, it may be harder to find the 101 and 103-tooth gears and, it may be less expensive to have two gears with the same number of teeth than to have different numbers of teeth.



12-47-1

PROBLEM 12-47

_____

Statement:

Design a three-stage compound spur gear train for an overall ratio of approximately 592:1. Specify tooth numbers for each gear in the train.

Given:

Approximate ratio

Solution:


mG  592

s  3

Number of stages

1

ravg  mG

s

ravg  8.397

1.

The average stage ratio is

2.

If the minimum number of teeth is (for a 20 deg pressure angle) Nmin  18 then the number of teeth on the driven gear for each stage is

or 3.

NG1  floor  ravg Nmin

NG1  151

NG2  ceil ravg Nmin

NG2  152

But, the prime factors of 592 are 2 4 and 37. This suggests that one of the gears have a number of teeth that is a multiple of 37, say 148. 148 ravg

 17.626

N5 N7

Thus,

N4 N6

N2  18

Let

if

= 72

N4  18

N3  148

N6  18

and

5 6

then

N5 N7 = 72 18 18 = 2  3

Try

N5  2  3

4 2

Summarizing,

1 4

N7  2  3 N2  18

N3  148

N4  18

N5  144

N6  18

N7  162

then

N5 N7  23328

4.

Since the driven gears all have less than 180 teeth, no stage ratio is greater than 10.

5.

Checking the overall gear ratio mG 

N 3 N 5 N 7 N 2 N 4 N 6

mG  592.00



PROBLEM 12-48

12-48-1

_____

Statement:

Design a planetary gear train similar to that shown in Figure 12-16 for an overall velocity ratio of exactly 0.2 if the sun gear is the input, the arm is the output, and the ring gear is stationary. Specify tooth numbers for each gear in the train.

Given:

Exact ratio

mv  0.2



Let the first gear be the sun gear (gear 2) and the last be the ring gear (gear 4). Then, equation 12.11c can be written as

ω4  ωarm ω2  ωarm 2.

ωin

=

ωarm ω2

=

N2 N2  N4

1  mv mv

 N2

= mv

N2  20 , then the number of teeth required for the ring gear is N4  80

In order to mesh properly, the number of teeth on the ring gear must be equal to the sum of the number of teeth on the sun gear plus two times the number of teeth on the planet gear(s). Thus, N3 

5.

N4

Let the sun gear have tooth number

N4 

4.

N2

Taking 4 = 0 and solving for out/in, we have

ωout

3.

 

N4  N2

N3  30

2

Check the overall velocity ratio and summarize the tooth numbers.

Overall velocity ratio

Sun gear

N2  20

Planet gear(s)

N3  30

Ring gear

N4  80

mv 

N2 N2  N4

mv  0.200



PROBLEM 12-49

12-49-1

_____

Statement:

Design a planetary gear train similar to that shown in Figure 12-16 for an overall velocity ratio of exactly 4/3 if the sun gear is stationary, the arm is the input, and the ring gear is the output. Specify tooth numbers for each gear in the train.

Given:

Exact ratio

4

mv 

3





ωin

=

ω4 ωarm

=

N4  N2 N4

N2

= mv

N2  20 , then the number of teeth required for the ring gear is N4  60

mv  1


5.

N4


N4 

4.

N2


ωout

3.

 

N4  N2

N3  20

2



Sun gear

N2  20

Planet gear(s)

N3  20

Ring gear

N4  60

mv 

N4  N2 N4

mv  1.333



12-50-1

PROBLEM 12-50

_____

Statement:

A 21-tooth pinion rotating at 1800 rpm meshes with a 33-tooth gear in a spur gear reducer. Both pinion and gear are manufactured to a quality level of 9. A reliability of 0.9 has been specified, and the transmitted tangential load is 2800 lb. Conditions are such that Km = 1.7. It is proposed that standard 25-degree, full-depth teeth be used, with both pinion and gear hobbed from an AISI 4140 nitrided steel. The diametral pitch is 6, and the face width 2.000 in. Estimate the number of cycles of bending stress (using the AGMA equations) that the gearset can withstand.

Given:

Tooth numbers

Np  21

Ng  33

Pinion speed

n  1800 rpm

Mounting factor

Km  1.7

Diametral pitch

p d  6  in

Quality index

Qv  9

Face width Tangential load

F  2.000  in Wt  2800 lbf

Pressure angle Reliability

ϕ  25 deg

1

R  0.90

Assumptions: The life of the pinion will be less than that of the gear. Solution: 1.


Calculate the pitch diameter of the pinion using equation 12.4a and the pitch-line velocity of the gearset. d 

2.

Np

d  3.500  in

pd

d n

V 

2

V  1649

ft min

Calculate the dynamic factor Kv using equations 12.16 and 12.17. B  0.25  12  Qv Kv 

0.6667

A     V  min   A      ft  

B  0.520

A  76.878

B

Kv  0.802

3.


4.

Assume an application factor, Ka

5.

Calculate the bending stress using equation 12.15.

σb 

A  50  56 ( 1  B)

J  0.40

Ka  1

Ka Wt p d  Km

σb  44.51  ksi

Kv F  J

S'fb  39.5 ksi

6.

Determine the material bending fatigue strength from Table 12-20 using the average value.

7.

Determine the reliability factor for R  0.90

8.

Assume a temperature factor, KT

9.

Set the bending stress equal to the corrected bending-fatigue strength in equation 12.24 and solve for the life factor, KL . KL 

KT  KR σb S'fb

from Table 12-19.

KR  0.85

KT  1

KL  0.958

10. Assume that the fatigue life is over 10 6 cycles and use the lower curve in Figure 12-24 to solve for the cycle life 1

KL = 1.6831 N

 0.0323

N 

 1.6831   K   L 

0.0323

N  3.80  10

7



12-51-1

PROBLEM 12-51

_____

Statement:

A 21-tooth pinion rotating at 1800 rpm meshes with a 33-tooth gear in a spur gear reducer. Both pinion and gear are manufactured to a quality level of 9. A reliability of 0.9 has been specified, and the transmitted tangential load is 2800 lb. Conditions are such that Km = 1.7. It is proposed that standard 25-degree, full-depth teeth be used, with both pinion and gear hobbed from an AISI 4140 nitrided steel. The diametral pitch is 6, and the face width 2.000 in. Estimate the number of cycles of contact (surface) stress (using the AGMA equations) that the gearset can withstand.

Given:

Tooth numbers

Np  21

Ng  33

Pinion speed

n  1800 rpm

Mounting factor

Cm  1.7

Diametral pitch

p d  6  in

Quality index

Qv  9


F  2.000  in Wt  2800 lbf


ϕ  25 deg

1

R  0.90



Calculate the pitch diameter of the pinion using equation 12.4a and the pitch-line velocity of the gearset. Np

d  2.

pd

d  3.500  in

d n

V  1649

2

ft min

Calculate the dynamic factor Kv using equations 12.16 and 12.17. B  0.25  12  Qv Cv 

3.

V 

0.6667

A     V  min   A      ft  

B  0.520

A  50  56 ( 1  B)

A  76.878

B

Cv  0.802

Calculate the geometry factor, I I 

 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

I  0.117

4.

Determine the elastic coefficient from Table 12-18.

5.

Assume an application factor, Ca

6.

Calculate the contact stress using equation 12.21.

Cp  2300 psi

0.5

Ca  1

1

 Ca Wt 1 Cm  σc  Cp      Cv F  d I 

2

σc  195.76 ksi

7.

Determine the material surface-fatigue strength from Table 12-21 using the average value.

8.


9.

Assume a temperature factor, CT

from Table 12-19.

S'fc  167.5  ksi

CR  0.85

CT  1

10. Set the contact stress equal to the corrected surface-fatigue strength in equation 12.25 and solve for the life factor, CL . © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


CL 

CT  CR σc S'fc

12-51-2

CL  0.993

11. Use the lower curve in Figure 12-26 to solve for the cycle life. 1

CL = 2.466  N

 0.056

N 

 2.466   C   L 

0.056

N  1.13  10

7



12-52-1


If the gearset in Problem 12-46 transmits 7.5 kW at 1750 input pinion rpm, find the torque on each of the three shafts.

Given:

Transmitted power

P  7.5 kW

Input speed

ωp1  1750 rpm


Solution: 1.

Stage 1

Np1  14

Ng1  101

Stage 2

Np2  14

Ng2  103



2.

mG  53.077

Np1 Np2


Stage 2 3.

Ng1 Ng2

ωp1  1750 rpm

ωg1 

ωp2  ωg1

ωg2 

Np1 Ng1 Np2 Ng2

 ωp1

ωg1  242.574  rpm

 ωp2

ωg2  32.971 rpm

Calculate the shaft torque Shaft 1

T1 

Shaft 2

T2 

Shaft 3

T3 

P

ωp1 P

ωg1 P

ωg2

T1  40.9 N  m

Input

T2  295  N  m

T3  2172 N  m

Output



12-53-1


Size the first-stage spur gears in problem 12-52 for a bending factor of safety of at least 2.8 and a surface factor of safety of at least 1.8 assuming a steady torque, 25-deg pressure angle, full depth teeth, Qv = 9, and AISI 4340 steel for all gears.

Given:

Pressure angle

ϕ  25 deg


Nfb  2.8

Reliability

R  0.99


Nfs  1.8

AGMA Quality level

Qv  9


H  10 hp

Life (cycles)

N  10


Np  14


Ng  101

7




1.


2.


Vt p d  


Ng pd

d p p d   n 2 H

Vt  p d 

Ka  1

3.


4.


B  0.52

A  50  56 ( 1  B) Kv p d  

A     min  A  Vt  p d    ft  

5.


6.


A  76.878 B

Km  1.6

σbp p d F   7.

d g p d  

pd

Wt p d  


B  0.25  12  Qv

Np

d p p d  




Life factor

KL  1.6831 N

Reliability

KR  0.7  0.15 log( 1  R)

KL  1 KR  1


MACHINE DESIGN - An Integrated Approach, 4th Ed. Temperature factor

KT  1


S atp  41500  psi S fbp 


S atp KL

S fbp  41501  psi

KT  KR


Lower limit

9.

12-53-2

FU  p d  

16

FL  p d  

8

Nfb =

Face width

F  p d  

p d 

d

pd

S fb

Safety factor


pd


5 5.1 10   in in in


4

 

1

F pd

p d  8  in

in


 

FL pd

F  p d   1.195  in

in



d g p d   12.625 in

Vt p d   802 

Wt p d   412  lbf

min

σbp p d F   14175  psi

Kv p d   0.850

2

 

FU pd in

F  1.250  in

ft

3

1

0

5

6

7

8

9

10

pd  in




 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

I  0.1682



2.

12-53-3


Vt p d  



Wt p d  


2 H

Vt  p d 

Ca  1

3.


4.


Np

d p p d  


0.6667


B  0.52

5.


6.


7.


Cp  2300 psi

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I


Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1

Temperature factor

CT  1


S acp  162500 psi S fcp 


B

Cm  1.6

σcp p d F   Cp 8.

A  76.878


S fcp  162497 psi


Lower limit

FU  p d  

16

FL  p d  

8

pd

pd



Safety factor

Nfc =

 Sfc  σ   c

12-53-4

2

C a W t  p d   C m

2


Face width


p d 


5 5.1 10   in in in 4

 

F pd

1

p d  8  in

in


 

FL pd

2

in


3

 

FU pd in

1

F  1.000  in 14. The bending requirement is governing in this case. The diametral pitch and face width for the stage 1 gearset are: 1

p d  8  in

0

5

6

7

8

9

10

pd  in

FIGURE 12-53B F  1.250  in




12-54-1


Size the second-stage spur gears in problem 12-52 for a bending factor of safety of at least 2.8 and a surface factor of safety of at least 1.8 assuming a steady torque, 25-deg pressure angle, full depth teeth, Qv = 9, and AISI 4340 steel for all gears.

Given:

Pressure angle

ϕ  25 deg


Nfb  2.8

Reliability

R  0.99


Nfs  1.8

AGMA Quality level

Qv  9


H  10 hp

Life (cycles)

N  10


Np  14


Ng  103

7

Rotational speed of pinion (rpm) n  242.574  rpm Assumptions: Since both pinion and gear from each stage are the same material, it will only be necessary to determine the pinion size as it will be governing for the set. Solution:



1.


2.

Write the equations for pitch diameter, pitchline velocity, and transmitted load in terms of the unknown diametral pitch, pd. Note that, in Mathcad, unit conversion factors are not included. d p p d  


Vt p d  


Wt p d  


d g p d  

pd


4.


B  0.52

pd

d p p d   n 2 H

Vt  p d 

A  50  56 ( 1  B) Kv p d  

A     min  A  Vt  p d    ft  

5.

Tentatively choose the mounting factor, Km (Assume 1.99 < F < 6 in)

6.


σbp p d F   7.

Ng

Ka  1

3.

B  0.25  12  Qv

Np

A  76.878 B

Km  1.7



KL  1.6831 N

 0.0323

KL  1



Reliability

KT  1


S atp  41500  psi S fbp 

S atp KL

S fbp  41501  psi

KT  KR


Lower limit

9.

KR  1

Temperature factor


12-54-2

FU  p d  

16 pd

FL  p d   S fb

Safety factor

Nfb =

Face width

F  p d  

p d 


8 pd


2 2.1 6   in in in


6

 

1

F pd

p d  4  in

in


 

4

FL pd

F  p d   2.134  in

in


 

FU pd 2 in


d g p d   25.750 in

Vt p d   222 

Wt p d   1485 lbf

ft min

σbp p d F   14058  psi

Kv p d   0.912

0

1

2

3

4

5

6

pd  in






I  2.

 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

12-54-3

I  0.1686



Vt p d  


Wt p d  


pd d p p d   n 2 H

Vt  p d 

Ca  1

3.


4.


Np

0.6667


B  0.52

5.

Tentatively choose the mounting factor, Cm (Assume 1.99 < F < 6 in)

6.


7.


Cp  2300 psi

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I


Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1

Temperature factor

CT  1


S acp  162500 psi S fcp 


B

Cm  1.7

σcp p d F   Cp 8.

A  76.878


S fcp  162497 psi

Write the design equations using the range of face-width to diametral-pitch ratio given in of the text, and the bending stress equation, solved for the unknown face width. Upper limit

FU  p d  

16 pd



FL  p d  

Lower limit

Safety factor

Nfc =

12-54-4

8 pd

 Sfc  σ   c

2

C a W t  p d   C m

2


Face width


p d 


2 2.1 6   in in in 4

 

F pd

1

p d  4  in

in


 

FL pd

2

in


3

 

FU pd in

1


p d  4  in

0

1

2

3

4

5

6

pd  in

FIGURE 12-54B F  2.250  in




12-55-1


If the gearset in Problem 12-47 transmits 18.5 kW at 1184 input pinion rpm, find the torque on each of the four shafts.

Given:

Transmitted power

P  18.5 kW

Input speed

ωp1  1184 rpm


Solution: 1.

Stage 1

Np1  18

Ng1  148

Stage 2

Np2  18

Ng2  144

Stage 3

Np3  18

Ng3  162



2.

Np1 Np2 Np3

mG  592.000


Stage 2

Stage 3

3.

Ng1 Ng2 Ng3

ωp1  1184 rpm

ωg1 

ωp2  ωg1

ωg2 

ωp3  ωg2

ωg3 

Np1 Ng1 Np2 Ng2 Np3 Ng3

 ωp1

ωg1  144  rpm

 ωp2

ωg2  18 rpm

 ωp3

ωg3  2  rpm

Calculate the shaft torque. Shaft 1

T1 

Shaft 2

T2 

Shaft 3

T3 

Shaft 4

T4 

P

ωp1 P

ωg1 P

ωg2 P

ωg3

T1  149.2  N  m

Input

T2  1227 N  m

T3  9815 N  m

T4  88331  N  m

Output



12-56-1


Size the first-stage spur gears in problem 12-55 for a bending factor of safety of at least 2.4 and a surface factor of safety of at least 2.0 assuming a steady torque, 25-deg pressure angle, full depth teeth, Qv = 10, and AISI 4140 steel for all gears.

Given:

Pressure angle

ϕ  25 deg


Nfb  2.4

Reliability

R  0.99


Nfs  2.0

AGMA Quality level

Qv  10


H  25 hp

Life (cycles)

N  10


Np  18


Ng  148

7




1.


2.



Vt p d  


Wt p d  


d g p d  

pd


4.


B  0.397

pd

d p p d   n 2 H

Vt  p d 

A  50  56 ( 1  B) Kv p d  

A     min  A  Vt  p d    ft  

5.


6.


σbp p d F   7.

Ng

Ka  1

3.

B  0.25  12  Qv

Np

A  83.776 B

Km  1.7



KL  1.6831 N

 0.0323

KL  1



Reliability

KT  1


S atp  40000  psi S fbp 

S atp KL

S fbp  40001  psi

KT  KR


Lower limit

9.

KR  1

Temperature factor


12-56-2

FU  p d  

16

FL  p d  

8

Nfb =

Face width

F  p d  

p d 

d

pd

S fb

Safety factor


pd


5 5.1 10   in in in


4

 

1

F pd

p d  6  in

in


 

FL pd

F  p d   1.575  in

in



d g p d   24.667 in

Vt p d   930 

Wt p d   887  lbf

min

σbp p d F   14997  psi

Kv p d   0.884

2

 

FU pd in

F  1.750  in

ft

3

1

0

5

6

7

8

9

10

pd  in






I  2.

 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

12-56-3

I  0.1707



Vt p d  


Wt p d  


pd d p p d   n 2 H

Vt  p d 

Ca  1

3.


4.


Np

0.6667


B  0.397

5.


6.


7.


Cp  2300 psi

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I


Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1

Temperature factor

CT  1


S acp  165000 psi S fcp 


B

Cm  1.7

σcp p d F   Cp 8.

A  83.776


S fcp  164997 psi


FU  p d  

16 pd



FL  p d  

Lower limit

Safety factor

Nfc =

12-56-4

8 pd

 Sfc  σ   c

2

C a W t  p d   C m

2


Face width


p d 


5 5.1 10   in in in 4

 

F pd

1

p d  6  in

in


 

FL pd

2

in


3

 

FU pd in

1


p d  6  in

0

5

6

7

8

9

10

pd  in

FIGURE 12-56B F  1.750  in




12-57-1


Size the second-stage spur gears in problem 12-55 for a bending factor of safety of at least 2.4 and a surface factor of safety of at least 2.0 assuming a steady torque, 25-deg pressure angle, full depth teeth, Qv = 10, and AISI 4140 steel for all gears.

Given:

Pressure angle

ϕ  25 deg


Nfb  2.4

Reliability

R  0.99


Nfs  2.0

AGMA Quality level

Qv  10


H  25 hp

Life (cycles)

N  10


Np  18


Ng  144

7

Rotational speed of pinion (rpm) n  144  rpm Assumptions: Since both pinion and gear from each stage are the same material, it will only be necessary to determine the pinion size as it will be governing for the set. Solution:



1.


2.



Vt p d  


Wt p d  


d g p d  

pd


4.


B  0.397

pd

d p p d   n 2 H

Vt  p d 

A  50  56 ( 1  B) Kv p d  

A     min  A  Vt  p d    ft  

5.


6.


σbp p d F   7.

Ng

Ka  1

3.

B  0.25  12  Qv

Np

A  83.776 B

Km  1.7



KL  1.6831 N

 0.0323

KL  1



Reliability

KT  1


S atp  40000  psi S fbp 

S atp KL

S fbp  40001  psi

KT  KR


Lower limit

9.

KR  1

Temperature factor


12-57-2

FU  p d  

16 pd

FL  p d   S fb

Safety factor

Nfb =

Face width

F  p d  

p d 


8 pd


1 1.1 6   in in in


6

 

1

F pd

p d  3  in

in


 

4

FL pd

F  p d   3.055  in

in


 

FU pd 2 in


d g p d   48.000 in

Vt p d   226 

Wt p d   3647 lbf

ft min

σbp p d F   15669  psi

Kv p d   0.937

0

1

2

3

4

5

6

pd  in





MACHINE DESIGN - An Integrated Approach, 4th Ed. I 

2.

 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

12-57-3 I  0.1702



Vt p d  


Wt p d  


pd d p p d   n 2 H

Vt  p d 

Ca  1

3.


4.


Np

0.6667

A  50  56 ( 1  B)

B  0.397

Cv p d  

A     min  A  Vt  p d    ft  

5.


6.


7.


Cp  2300 psi

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I


Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1

Temperature factor

CT  1


S acp  165000 psi S fcp 


B

Cm  1.7

σcp p d F   Cp 8.

A  83.776


S fcp  164997 psi


FU  p d  

16 pd



Lower limit

8 pd

 Sfc  Nfc =    σc 

Safety factor

12-57-4

2

C a W t  p d   C m

2


Face width


p d 


1 1.1 6   in in in 6

 

F pd

1

p d  3  in

in


 

4

FL pd in


 

FU pd 2 in


p d  3  in

0

1

2

3

4

5

6

pd  in

FIGURE 12-57B F  3.250  in




12-58-1


Size the third-stage spur gears in problem 12-55 for a bending factor of safety of at least 2.4 and a surface factor of safety of at least 2.0 assuming a steady torque, 25-deg pressure angle, full depth teeth, Qv = 10, and AISI 4140 steel for all gears.

Given:

Pressure angle

ϕ  25 deg


Nfb  2.4

Reliability

R  0.99


Nfs  2.0

AGMA Quality level

Qv  10


H  25 hp

Life (cycles)

N  10


Np  18


Ng  162

7




1.


2.



Vt p d  


Wt p d  


d g p d  

pd


4.


B  0.397

pd

d p p d   n 2 H

Vt  p d 

A  50  56 ( 1  B) Kv p d  

A     min  A  Vt  p d    ft  

5.


6.


σbp p d F   7.

Ng

Ka  1

3.

B  0.25  12  Qv

Np

A  83.776 B

Km  1.8



KL  1.6831 N

 0.0323

KL  1



Reliability

KT  1


S atp  40000  psi S fbp 

S atp KL

S fbp  40001  psi

KT  KR


Lower limit

9.

KR  1

Temperature factor


12-58-2

FU  p d  

16

FL  p d  

8

Nfb =

Face width

F  p d  

p d 

d

pd

S fb

Safety factor


pd


1 1.1 6   in in in


10

 

1

F pd

p d  1.5 in

8

in


 

FL pd

F  p d   6.27 in

in



d g p d   108.000  in

Vt p d   57

Wt p d   14589  lbf

min

σbp p d F   16078  psi

 

FU pd in

F  6.500  in

ft

6

Kv p d   0.966

4

2

0

1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 pd  in






I  2.

 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

12-58-3

I  0.1724



Vt p d  


Wt p d  


pd d p p d   n 2 H

Vt  p d 

Ca  1

3.


4.


Np

0.6667


B  0.397

5.


6.


7.


Cp  2300 psi

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I


Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1

Temperature factor

CT  1


S acp  165000 psi S fcp 


B

Cm  1.8

σcp p d F   Cp 8.

A  83.776


S fcp  164997 psi


FU  p d  

16 pd



FL  p d  

Lower limit

8 pd

Safety factor

 Sfc  Nfc =    σc 

Face width

F  p d  


p d 


C a W t  p d   C m




1 1.1 6   in in in 10

 

8

in


 

FL pd

6

in

13. Round this up to the decimal equivalent of a common fractional value. F  5.250  in 14. The bending requirement is governing in this case. The diametral pitch and face width for the stage 3 gearset are: 1

2

F pd

1

p d  1.5 in

p d  1.5 in

12-58-4

 

FU pd in

4

2

0

1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 pd  in

FIGURE 12-58B F  6.500  in




12-59-1


The pinion of an external gearset has pitch radius rp = 40 mm and the gear has pitch radius rg = 160 mm. If the pinion is the input member of the set, determine the velocity ratio, the torque ratio, and the gear ratio of the set.

Given:

Pitch radii

Solution:


1.

rout  rg

rout  160 mm

 rin    rout 

mV  

mV  0.25

Using equation 12.1b, calculate the torque ratio using the minus sign because this is an external set. Torque ratio

4.

rin  40 mm

Using equation 12.1a, calculate the velocity ratio using the minus sign because this is an external set. Velocity ratio

3.

rg  160  mm

Equate the pinion pitch radius to rin and the gear pitch radius to rout (in this case). rin  rp

2.

rp  40 mm

mA 

1 mV

mA  4

Using equation 12.1c, calculate the gear ratio using the torque ratio since it is greater than 1. Gear ratio

mG  mA

mG  4

This is usually stated as being 4:1.



12-60-1


A pinion having 20 teeth and a diametral pitch of 8 (in -1) is in mesh with a rack. If the pinion rotates one revolution, how far will the rack move?

Given:

Number of teeth

Solution:


1.

Diametral pitch

1

p d  8  in

Calculate the pitch diameter using equation 12-4a. Pitch diameter

2.

Np  20

d 

Np pd

d  2.500  in

The rack and pinion can be modeled as a cylinder, with diameter equal to the pitch diameter of the gear, in contact with a plane along the pitch line of the rack. When the gear is rotated one revolution, the rack will move a distance equal to the circumference of the cylinder. Distance moved

s  π d

s  7.854 in



12-61-1


A gearset with full-depth teeth is designed to have a pinion with 24 teeth, a gear with 54 teeth, and a diametral pitch of 6. Compare the contact ratio for this set for pressure angles of 14.5, 20, and 25 degrees.

Given:

Number of teeth Np  24 Diametral pitch

Solution: 1.


Calculate the pitch radus and addendum for each gear, and the nominal center distamce.

Addendum:

Center distance

1 Np  2 pd

rp  2.000 in

rg 

1 Ng  2 pd

rg  4.500 in

a p 

1 pd

a p  0.167 in

a g  a p

a g  0.167 in

C  rp  rg

C  6.500 in

rp  ap 2  rp cos( ϕ)  2   rg  ag2   rg cos( ϕ) 2  C sin( ϕ)

Write the equation for the contact ratio as a function of pressure angle using equation 12.7b. mp( ϕ) 

4.

rp 

Write the equation for the length of action as a function of pressure angle using equation 12.2. Z ( ϕ) 

3.

1

p d  6  in

Pitch radius:

2.

Ng  54

p d  Z ( ϕ)

π cos( ϕ)

Evaluate the contact ratio for the three pressure angles. 14.5 degrees

mp( 14.5 deg)  2.007

20 degrees

mp( 20 deg)  1.685

25 degrees

mp( 25 deg)  1.492



PROBLEM 12-62

12-62-1

_____

Statement:

Design a planetary gear train similar to that shown in Figure 12-16 for an overall velocity ratio of exactly 5 if the ring gear is stationary, the arm is the input, and the sun gear is the output. Specify tooth numbers for each gear in the train.

Given:

Exact ratio

mV  5





N2 N4


ωout ωin 3.

=

=

ω2 ωa

=

N4 N2

 1 = mv

N4  N2  mV  1  4.

N4  80


5.

N2  20 , then the number of teeth required for the ring gear is


N4  N2

N3  30

2



Sun gear

N2  20

Planet gear(s)

N3  30

Ring gear

N4  80

mV 

N4 N2

1

mV  5.000



12-63-1

PROBLEM 12-63

_____

Statement:

A 22-tooth pinion rotating at 1650 rpm meshes with a 66-tooth gear in a spur gear reducer. Both pinion and gear are manufactured to a quality level of 10. A reliability of 0.9 has been specified, and the transmitted tangential load is 5000 lb. Conditions are such that Km = 1.7. It is proposed that standard 25-degree, full-depth teeth be used, with both pinion and gear hobbed from an AISI 4340 nitrided steel. The diametral pitch is 5, and the face width 2.500 in. Estimate the number of cycles of bending stress (using the AGMA equations) that the gearset can withstand.

Given:

Tooth numbers

Np  22

Ng  66

Pinion speed

n  1650 rpm

Mounting factor

Km  1.7

Diametral pitch

p d  5  in

Quality index

Qv  10


F  2.500  in Wt  5000 lbf


ϕ  25 deg

1

R  0.90



Calculate the pitch diameter of the pinion using equation 12.4a and the pitch-line velocity of the gearset. d 

2.

Np

d  4.400  in

pd

d n

V 

2

V  1901

ft min

Calculate the dynamic factor Kv using equations 12.16 and 12.17. B  0.25  12  Qv Kv 

0.6667

A     V  min   A      ft  

B  0.397

A  83.776

B

Kv  0.847

3.


4.

Assume an application factor, Ka

5.

Calculate the bending stress using equation 12.15.

σb 

A  50  56 ( 1  B)

J  0.42

Ka  1

Ka Wt p d  Km

σb  47.80  ksi

Kv F  J

S'fb  41.5 ksi

6.

Determine the material bending fatigue strength from Table 12-20 using the average value.

7.


8.

Assume a temperature factor, KT

9.

Set the bending stress equal to the corrected bending-fatigue strength in equation 12.24 and solve for the life factor, KL . KL 

KT  KR σb S'fb

from Table 12-19.

KR  0.85

KT  1

KL  0.979

10. Assume that the fatigue life is over 10 6 cycles and use the lower curve in Figure 12-24 to solve for the cycle life 1

KL = 1.6831 N

 0.0323

N 

 1.6831   K   L 

0.0323

N  1.93  10

7



12-64-1

PROBLEM 12-64

_____

Statement:

A 22-tooth pinion rotating at 1650 rpm meshes with a 66-tooth gear in a spur gear reducer. Both pinion and gear are manufactured to a quality level of 10. A reliability of 0.9 has been specified, and the transmitted tangential load is 5000 lb. Conditions are such that Km = 1.7. It is proposed that standard 25-degree, full-depth teeth be used, with both pinion and gear hobbed from an AISI 4340 nitrided steel. The diametral pitch is 5, and the face width 2.500 in. Estimate the number of cycles of contact (surface) stress (using the AGMA equations) that the gearset can withstand.

Given:

Tooth numbers

Np  22

Ng  66

Pinion speed

n  1650 rpm

Mounting factor

Cm  1.7

Diametral pitch

p d  5  in

Quality index

Qv  10


F  2.500  in Wt  5000 lbf


ϕ  25 deg

1

R  0.90



Calculate the pitch diameter of the pinion using equation 12.4a and the pitch-line velocity of the gearset. Np

d  2.

pd

d  4.400  in

d n

V  1901

2

ft min

Calculate the dynamic factor Kv using equations 12.16 and 12.17. B  0.25  12  Qv Cv 

3.

V 

0.6667

A     V  min   A      ft  

B  0.397

A  50  56 ( 1  B)

A  83.776

B

Cv  0.847

Calculate the geometry factor, I I 

 sin( ϕ)  cos( ϕ)   Ng   N N 2   p g

I  0.144

4.

Determine the elastic coefficient from Table 12-18.

5.

Assume an application factor, Ca

6.

Calculate the contact stress using equation 12.21.

Cp  2300 psi

0.5

Ca  1

1

 Ca Wt 1 Cm  σc  Cp      Cv F  d I 

2

σc  183.32 ksi

7.

Determine the material surface-fatigue strength from Table 12-21 using the average value.

8.


9.

Assume a temperature factor, CT

from Table 12-19.

S'fc  162.5  ksi

CR  0.85

CT  1

10. Set the contact stress equal to the corrected surface-fatigue strength in equation 12.25 and solve for the life factor, CL . © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


CL 

CT  CR σc S'fc

12-64-2

CL  0.959

11. Use the lower curve in Figure 12-26 to solve for the cycle life. 1

CL = 2.466  N

 0.056

N 

 2.466   C   L 

0.056

N  2.11  10

7



12-65-1


The pinion of an internal gearset has pitch radius rp = 30 mm and the gear has pitch radius rg = 150 mm. If the pinion is the input member of the set, determine the velocity ratio, the torque ratio, and the gear ratio of the set.

Given:

Pitch radii

Solution:


1.

rout  rg

rout  150 mm

mV 

 rin  r   out 

mV  0.2

Using equation 12.1b, calculate the torque ratio using the minus sign because this is an external set. Torque ratio

4.

rin  30 mm

Using equation 12.1a, calculate the velocity ratio using the plus sign because this is an internal set. Velocity ratio

3.

rg  150  mm

Equate the pinion pitch radius to rin and the gear pitch radius to rout (in this case). rin  rp

2.

rp  30 mm

mA 

1 mV

mA  5

Using equation 12.1c, calculate the gear ratio using the torque ratio since it is greater than 1. Gear ratio

mG  mA

mG  5

This is usually stated as being 5:1.



12-66-1


A pinion having 18 teeth and a diametral pitch of 10 (in -1) is in mesh with a rack. If the rack moves 1 in, how many degrees will the pinion rotate?

Given:

Number of teeth

Solution:


1.

Diametral pitch

1

p d  10 in

Calculate the pitch diameter using equation 12-4a. Pitch diameter

2.

Np  18

d 

Np pd

d  1.800  in

The rack and pinion can be modeled as a cylinder, with diameter equal to the pitch diameter of the gear, in contact with a plane along the pitch line of the rack. When the rack is moved 1 in, the pinion will rotate through an angle θ = d/2s, where d is the diameter of the cylinder and s is the distance that the rack moves. Distance moved

s  1  in

Pinion rotation

θ 

d 2 s

θ  51.566 deg



12-67-1


A gearset with 25 deg, full-depth teeth is designed to have a pinion with 24 teeth, a gear with 54 teeth, and a diametral pitch of 6. Compare the contact ratio for this set for a range of center distances from 0.90C to 1.10C.

Given:

Number of teeth Np  24 Diametral pitch

Solution: 1.

rp 

1 Np  2 pd

rp  2.000 in

rg 

1 Ng  2 pd

rg  4.500 in

Base circle radius:

rbp  rp cos( ϕ)

rbp  1.813 in

Addendum:

a p 

1

a p  0.167 in

pd

a g  a p

a g  0.167 in

C  rp  rg

C  6.500 in

Write the equation for the pressure angle as a function of variation in center distance. Let f be a factor which, when multiplied by C gives the actual center distance C'. Then, the actual pitch radius of the pinion will be rp' frp and the actual pressure angle will be ϕ = acos(rbp/frp).

 rbp    f  rp 

Note that ϕ'( 1 )  25 deg

Write the equation for the length of action as a function of actual pressure angle and center distance using equation 12.2. Z ( f ) 

rp  ap2   rp cos( ϕ'( f ) )  2   rg  ag2   rg cos( ϕ'( f ) )  2  f  C sin( ϕ'( f ) )

Write the equation for the contact ratio as a function of pressure angle using equation 12.7b. mp( f ) 

5.

ϕ  25 deg

Calculate the pitch radus, base radius, and addendum for each gear, and the nominal center distamce.

ϕ'( f )  acos

4.

Pressure angle


Center distance

3.

1

p d  6  in

Pitch radius:

2.

Ng  54

pd  Z ( f )

π cos( ϕ'( f ) )

Evaluate the contact ratio for the range of center distance. Let f  0.90 0.905  1.10 See the graph on the next page.



12-67-2

CONTACT RATIO vs CHANGE IN CENTER DISANCE 4

Contact Ratio

3

m p( f ) 2

1

0 0.9

0.95

1

1.05

1.1

f Change in Center Distance Factor



PROBLEM 12-68

12-68-1

_____

Statement:

Design a planetary gear train similar to that shown in Figure 12-16 for an overall velocity ratio of exactly 1.25 if the sun gear is stationary, the arm is the input, and the ring gear is the output. Specify tooth numbers for each gear in the train.

Given:

Exact ratio

mV  1.25





ωin

=

ω4 ωa

=

N2 N4

 1 = mv

N2  20 , then the number of teeth required for the ring gear is

N2

N4  80

mV  1


5.

N4


N4 

4.

N2


ωout

3.

=

N4  N2

N3  30

2



Sun gear

N2  20

Planet gear(s)

N3  30

Ring gear

N4  80

mV 

N2 N4

1

mV  1.250



13-1-1


A 20-deg pressure angle, 30-deg helix angle, 27-tooth helical gear has a diametral pitch p d = 5. Find the pitch diameter, addendum, dedendum, outside diameter, normal, transverse, and axial pitch.

Given:

Pressure angle

ϕ  20 deg

Helix angle

ψ  30 deg

Number of teeth

N  27

Diametral pitch

p d  5  in

Solution: 1.


Use equations 13.1 to find the pitch diameter, transverse pitch, normal pitch, and axial pitch. Pitch diameter

Transverse pitch

2.

1

d  p t 

N

d  5.400  in

pd

π pd

Normal pitch

p n  p t  cos( ψ)

Axial pitch

p x 

pn sin( ψ)

p t  0.628  in p n  0.544  in p x  1.088  in

Use the equations in Table 12-1 to calculate the addendum, dedendum, and outside diameter. Addendum

Dedendum Outside diameter

a  b 

1 pd 1.25 pd

D  d  2  a

a  0.200  in b  0.250  in D  5.800  in



13-2-1


A 25-deg pressure angle, 20-deg helix angle, 43-tooth helical gear has a diametral pitch p d = 8. Find the pitch diameter, addendum, dedendum, outside diameter, normal, transverse, and axial pitch.

Given:

Pressure angle

ϕ  25 deg

Helix angle

ψ  20 deg

Number of teeth

N  43

Diametral pitch

p d  8  in

Solution: 1.


Use equations 13.1 to find the pitch diameter, transverse pitch, normal pitch, and axial pitch. Pitch diameter

Transverse pitch

2.

1

d  p t 

N

d  5.375  in

pd

π pd

Normal pitch


Axial pitch

p x 

pn sin( ψ)

p t  0.393  in p n  0.369  in p x  1.079  in

Use the equations in Table 12-1 to calculate the addendum, dedendum, and outside diameter. Addendum

Dedendum Outside diameter

a  b 

1 pd 1.25 pd

D  d  2  a

a  0.125  in b  0.156  in D  5.625  in



13-3-1


A 57-tooth, 10-deg helix angle, helical gear is in mesh with a 23-tooth pinion. The p d = 6 and  = 25 deg. Find the transverse and axial contact ratios.

Given:

Pressure angle

ϕ  25 deg

Helix angle

ψ  10 deg


Np  23

Diametral pitch

p d  6  in


Ng  57

1

Assumptions: The face width factor is fwf  10. Solution: 1.


Use equation 12.7 to find the transverse contact ratio. Circular pitch

π

p c 

p c  0.524  in

pd


Base pitch

p b  0.475  in


Np

d p 

d p  3.833  in

pd

rp  0.5 d p

rp  1.917  in


Center distance

d g  9.500  in

pd

rg  0.5 d g a 

Addendum

Ng

d g 

rg  4.750  in

1.0

a  0.167  in

pd Np  Ng

C 

2 pd

C  6.667  in

Length of action

Z 


Z  0.708  in mp 

Transverse contact ratio 2.

Z pb

mp  1.491

Use equation 13.5 to find the axial contact ratio. Face width

Axial contact ratio

F 

mF 

fwf

F  1.667  in

pd F  p d  tan ( ψ)

π

mF  0.561



13-4-1


A 78-tooth, 30-deg helix angle, helical gear is in mesh with a 27-tooth pinion. The p d = 6 and  = 20 deg. Find the transverse and axial contact ratios.

Given:

Pressure angle

ϕ  20 deg

Helix angle

ψ  30 deg


Np  27

Diametral pitch

p d  6  in


Ng  78

1

Assumptions: The face width factor is fwf  12. Solution: 1.



π

p c 

p c  0.524  in

pd


Base pitch

p b  0.492  in


Np

d p 

d p  4.500  in

pd

rp  0.5 d p

rp  2.250  in


Center distance

d g  13.000 in

pd

rg  0.5 d g a 

Addendum

Ng

d g 

rg  6.500  in

1.0

a  0.167  in

pd Np  Ng

C 

2 pd

C  8.750  in

Length of action

Z 


Z  0.849  in mp 

Transverse contact ratio 2.

Z pb

mp  1.726


Axial contact ratio

F 

mF 

fwf

F  2.000  in


π

mF  2.205



13-5-1


A 90-deg straight bevel gearset is needed to give a 9:1 reduction. Determine the pitch cone angles, pitch diameters, and gear forces if the 25-deg pressure angle pinion has 14 teeth of p d = 6, and the transmitted power is 746 W at 1000 pinion rpm.

Given:

Power transmitted

H  746  W

Gear ratio

mG  9

Pinion speed

ωp  1000 rpm

Diametral pitch

p d  6  in

Teeth on pinion

Np  14

Pressure angle

ϕ  25 deg

Solution:


1.

Use equation 13.7b to calculate number of teeth on the gear.

2.

Use equation 12.4a to calculate the pitch diameters. Pinion

Gear

3.

d g 

Np

Ng  126

d p  2.333  in

pd Ng

d g  21.000 in

pd

Pinion

αp  atan

Gear

αg  atan mG

1

 

αp  6.340  deg

 mG 

αg  83.660 deg

αp  αg  90 deg

Determine the torque on the pinion shaft and the transmitted force. Pinion

Transmitted force

5.

d p 

Ng  mG Np

Use equation 13.7b to calculate the pitch cone angles.

Note that 4.

1

Tp 

H

Tp  63.051 in lbf

ωp

Wt 

2  Tp

Wt  54.043 lbf

dp

Use equations 13.8a to calculate the pinion and gear forces. Pinion

Gear

Total force

Wap  Wt tan ( ϕ)  sin αp

Wap  2.78 lbf

Wrp  Wt tan( ϕ)  cos αp

Wrp  25.05  lbf

Wag  Wt tan ( ϕ)  sin αg

Wag  25.05  lbf

Wrg  Wt tan( ϕ)  cos αg

Wrg  2.78 lbf

W  Wt ( cos( ϕ) )

1

W  59.63  lbf



13-6-1


A 90-deg straight bevel gearset is needed to give a 4.5:1 reduction. Determine the pitch cone angles, pitch diameters, and gear forces if the 20-deg pressure angle pinion has 18 teeth of p d = 5, and the transmitted power is 7460 W at 800 pinion rpm.

Given:

Power transmitted

H  7460 W

Gear ratio

mG  4.5

Pinion speed

ωp  800  rpm

Diametral pitch

p d  5  in

Teeth on pinion

Np  18

Pressure angle

ϕ  20 deg

Solution:


1.


2.


Gear

3.

d g 

Np

Ng  81

d p  3.600  in

pd Ng

d g  16.200 in

pd

Pinion


Gear


1

 

αp  12.529 deg

 mG 

αg  77.471 deg



Transmitted force

5.

d p 

Ng  mG Np


Note that 4.

1

Tp 

H

Tp  788.134  in lbf

ωp

Wt 

2  Tp

Wt  437.852  lbf

dp


Gear

Total force


Wap  34.57  lbf


Wrp  155.57 lbf


Wag  155.57 lbf


Wrg  34.57  lbf

W  Wt ( cos( ϕ) )

1

W  465.953  lbf



13-7-1


A 90-deg straight bevel gearset is needed to give a 5:1 reduction. Determine the pitch cone angles, pitch diameters, and gear forces if the 20-deg pressure angle pinion has 16 teeth of p d = 7, and the transmitted power is 3 hp at 600 pinion rpm.

Given:

Power transmitted

H  3  hp

Gear ratio

mG  5

Pinion speed

ωp  600  rpm

Diametral pitch

p d  7  in

Teeth on pinion

Np  16

Pressure angle

ϕ  20 deg

1

Assumptions: The spiral angle is ψ  35 deg Solution:


1.


2.

Use equation 12.4a to calculate the pitch diameters. d p 

Pinion

d g 

Gear

3.

5.

d p  2.286  in

pd Ng

d g  11.429 in

pd

Pinion


Gear


1

 

 mG 

αp  11.310 deg

αg  78.690 deg



Tp 

Transmitted force

Wt 

H

ωp 2  Tp dp

Tp  315.127  in lbf

Wt  275.736  lbf

Use equations 13.2 to calculate the pressure angle in the normal plane.

ϕn  atan( tan( ϕ)  cos( ψ) ) 6.

Ng  80


Note that 4.

Np

Ng  mG Np

ϕn  16.602 deg

Use equations 13.8b to calculate the pinion and gear forces. Pinion Wap  Wrp 

Wt cos( ψ) Wt cos( ψ)

  tan ϕn  sin αp  sin( ψ)  cos αp 

Wap  169.64 lbf

  tan  ϕn  cos αp  sin( ψ)  sin αp 

Wrp  136.28 lbf



13-7-2

Gear Wag  Wrg 


Total force

  tan ϕn  sin αg  sin( ψ)  cos αg 

Wag  60.55  lbf

  tan  ϕn  cos αg  sin( ψ)  sin αg 

Wrg  209.01 lbf

W  Wt ( cos( ϕ) )

1

W  293.432  lbf



13-8-1


A paper mill processes rolls of paper having a density of 984 kg/m3. The paper roll is 1.50-m outside diameter (OD) x 0.22-m inside diameter (ID) x 3.23-m long and is on a simple supported, hollow, steel shaft with S ut = 400 MPa. Design a 2.5:1 reduction helical gearset to drive this roll shaft to obtain a minimum dynamic safety factor of 2 for 10-year life if the roll turns at 50 rpm with 1.2 hp absorbed.

Given:


Nfc  2

Life (years)



H  1.2 hp

Gear speed

n g  50 rpm

Gear ratio

mg  2.5

Assumptions: 1. If both pinion and gear are the same material, it will only be necessary to determine the pinion size as it will be governing for the set. 2. If the gears are not surface hardened, it will only be necessary to design to the surface requirement as it will be governing for both bending and surface stresses. Design Choices: Pressure angle

ϕ  25 deg Np  14


Qv  11

AGMA Quality level

R  0.99

Relibility

Helix angle ψ  20 deg Material: AGMA Grade 1 steel for both pinion and gear, through hardened to HB  230 S ac  ( 26000  327  HB)  psi S ac  101210 psi Solution: 1.

2.


Determine the number of gear teeth, the pinion speed, and the cycle life. Number of gear teeth

Ng  Np mg

Ng  35


n p  n g mg

n p  125  rpm

Cycle life:

N  Life

np

8

N  6.574  10

2 π

Determine the surface geometry factor, I. a  p d  

Addendum

1 pd Np

Pitch radii

rp p d  

rg p d  

Center distance

C p d   rp p d   rg p d 

2 pd

Ng 2 pd

Radii of curvature

ρp  p d  

2 2 0.5  rp p d   C pd   rg pd     rp p d   cos( ϕ) 

ρg  p d   C p d   sin( ϕ)  ρp  p d  Normal plane pressure angle

ϕn  atan( tan( ϕ)  cos( ψ) )

Base helix angle

ψb  acos cos( ψ) 

 

ϕn  23.662 deg

cos ϕn  cos( ϕ)

 

ψb  18.256 deg



13-8-2

Length of action Z  p d  

  r  p   a  p   2   r  p   cos( ϕ)  2  d p d  p d  r p  a p 2  r p  cos( ϕ) 2  C p  sin( ϕ)   d   g d   d    g d  Z pd 

mp p d  

Transverse contact ratio

   

π pd

 cos( ϕ) F  p d  tan( ψ)

Axial contact ratio

mF  p d F  

Fractional parts of mp and mF

n r p d   mp p d   floor  mp p d  

π

n a p d F   mF  p d F   floor  mF  p d F   p t  p d  

Axial pitch

p x  p d  

p n  p d   p t  p d   cos( ψ)

π pd pn  pd  sin( ψ)

Minimum length of lines of contact Lmin p d F  

return

return

3.

mp pd  F  napd F  nrpd   pxpd   if n a p d F    1  n r p d   cos ψb mp pd   F   1  na p d F     1  nr pd    px  p d  if n a p d F    1  n r p d   cos ψb

Load sharing ratio

mN  p d F  

Geometry factor

I  p d F  

F

Lmin p d F  cos( ϕ)

 1  1   2 r  p   m  p F   ρ p  ρ p   p d N d g d   p d

Write the equations for pitch diameter, pitchline velocity, and transmitted load in terms of the unknown diametr pitch, p d. Note that, in Mathcad, unit conversion factors are not included. Pitch diameter of pinion (in)

pitchline velocity (fpm) Transmitted load (lbf)

d p p d   Vt p d   Wt p d  

Np

d g p d  

pd

Ng pd

d p p d   n p 2 H

Vt  p d 



13-8-3

Ca  1

4.


5.


B  0.25  12  Qv

v

0.6667

B  0.250

A  50  56 ( 1  B) Cv p d  

A  92.000

A     min  A  Vt  p d    ft  

B

6.


Cm  1.7

7.


Cp  2300 psi

8.


9.


σcp p d F   Cp

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I  p d F 

 0.056

Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

CL  0.791 CR  1.000

Temperature factor

CT  1


S ac  101210 psi S fcp 

Endurance strength

S ac CL

S fcp  80060  psi

CT  CR


Lower limit

Safety factor

FU  p d  

16

FL  p d  

8

pd

pd

 Sfc  Nfc =    σc 

2

C a W t  p d   C m

2

Face width

 Cp  F' p d F      Nfc Cv p d   d p p d   I  p d F   S fcp 

Guess value for F

F  2.35 in

Found by iteration with value in step 11.



p d 


13-8-4

2 2.5 12   in in in


4



F' pd F in

1

p d  5  in



3

 

FL pd


2

in

F' p d F   2.355  in

 

FU pd in


1

0

13. Then, the parameters that depend on p d and F are: d p p d   2.800  in Vt p d   92

4

5

6

7

8

9 10 11 12


Wt p d   432  lbf

min

3

pd  in

d g p d   7.000  in

ft

2

σcp p d F   56448  psi

mp p d   1.420

mF  p d F   1.376

n r p d   0.420

n a p d F   0.376

mN  p d F   0.728

Lmin p d F   3.264  in

I  p d F   0.188

14. The realized factor of safety against surface failure in the pinion is Nf  p d F  

 Sfcp   σ p F   cp d  

2

Nf  p d F   2.0



13-9-1


A 2-start wormset has the dimensions given below. Find the lead, lead angle, worm gear diameter, and center distance. Will it self-lock? The input speed is 2200 rpm.

Given:

Worm pitch diameter

d  50 mm

Gear ratio

mG  22

Axial pitch

p x  10 mm

Input (worm) speed

n w  2200 rpm

Threads on worm

Nw  2

L  p x  Nw

L  20 mm

Solution:


1.

Use equation 13.13 to calculate the lead.

2.

Use equation 13.12 to calculate the lead angle.

λ  atan

3.

Calculate the number of teeth on the gear

Ng  mG Nw

4.

Use equation 13.13 to calculate gear diameter.

p c  p x

   π d 

d g  C 

L

λ  7.256  deg Ng  44

p c  Ng

d g  140.06 mm

π d  dg

C  95.03  mm

5.

Use equation 13.17 to calculate the center distance.

6.

If the lead angle per worm tooth is less than 6 deg, the set is self-locking. The lead angle per tooth is

λt 

λ Nw

λt  3.628  deg

2

Self-locking



13-10-1



Given:

Worm pitch diameter

d  1.750  in

Gear ratio

mG  17

Axial pitch

p x  0.200  in

Input (worm) speed

n w  1400 rpm

Threads on worm

Nw  3

L  p x  Nw

L  0.600  in

Solution:


1.


2.



3.


Ng  mG Nw

4.


p c  p x

L

 

 π d 

d g  C 

λ  6.228  deg Ng  51

p c  Ng

d g  3.247  in

π d  dg

C  2.498  in

5.


6.


λt 

λ Nw

λt  2.076  deg

2

Self-locking



13-11-1



Given:

Worm pitch diameter

d  40 mm

Gear ratio

mG  82

Axial pitch

p x  5  mm

Input (worm) speed

n w  4500 rpm

Threads on worm

Nw  1

L  p x  Nw

L  5  mm

Solution:


1.


2.



3.


Ng  mG Nw

4.


p c  p x

L

 

 π d 

d g  C 

λ  2.279  deg Ng  82

p c  Ng

d g  130.51 mm

π d  dg

C  85.25  mm

5.


6.


λt 

λ Nw

λt  2.279  deg

2

Self-locking



13-12-1


Determine the power transmitted and the torques and forces in the mesh for the wormset in Problem 13-9 if it turns at 1000 worm rpm.

Units:

fpm  ft  min

Given:

Worm pitch diameter

d w  50 mm

Gear ratio

mG  22

Axial pitch

p x  10 mm

Input (worm) speed

n w  1000 rpm

Threads on worm

Nw  2

Pressure angle

ϕ  20 deg

L  p x  Nw

L  20 mm

Solution:

1


1.


2.



3.


Ng  mG Nw

4.


p c  p x

5.


d g  C 

6.

L

 

 π d w 

p c  Ng

λ  7.256  deg Ng  44

d g  140.06 mm

π dw  dg 2

C  95.03  mm

Find the maximum recommended face width from equation 13.19. Fmax  0.67 d w

Fmax  33.5 mm

7.

Find the materials factor Cs from equation 13.24. Since C < 203 mm, Cs  1000.

8.

Find the ratio correction factor Cm from equations 13.25. Based on mG  22 , the second of the expressions in that equation set will be used. 2

Cm  0.0107 mG  56 mG  5145 9.

Cm  0.821

Find the tangential velocity Vt from equation 13.27. Vt 

nw dw

Vt  519.5  fpm

2  cos( λ)

10. Use this velocity to find the velocity factor Cv from equations 13.26. For this value of Vt, the first of these equations is appropriate.  0.0011

Vt fpm

Cv  0.659  e

Cv  0.372

11. Find the tangential load Wt from equation 13.23.

 dg    mm 

Wtg  Cs Cm Cv 

0.8

 Fmax

N 75.948 mm

Wtg  7028 N

12. Find the coefficient of friction from the third expression in equation 13.29.

 Vt   0.110    fpm  μ  0.103  e

0.450

 0.012

μ  0.028



13-12-2

13. Find the friction force Wf from equation 13.28. Wf 

μ  Wtg cos( λ)  cos( ϕ)

Wf  214.5  N

14. Find the rated output power from equation 13.21.

Φ o 

nw dg   Wtg mG 2

Φ o  2.34 kW

15. Find the power lost in the mesh from equation 13.22.

Φ l  Vt Wf

Φ l  0.57 kW

16. Find the rated input power from equation 13.20.

Φ  Φ o  Φ l

Φ  2.91 kW

17. The efficiency of the gearset is e 

Φo

e  80.5 %

Φ

18. Find the rated output torque from equation 13.31. Tg  Wtg

dg 2

Tg  492  N  m



13-13-1


Determine the power transmitted and the torques and forces in the mesh for the wormset in Problem 13-10 if it turns at 500 worm rpm.

Units:


Given:

Worm pitch diameter

d w  1.75 in

Gear ratio

mG  17

Axial pitch

p x  0.200  in

Input (worm) speed

n w  500  rpm

Threads on worm

Nw  3

Pressure angle

ϕ  20 deg

L  p x  Nw

L  0.600  in

Solution:

1


1.


2.



3.


Ng  mG Nw

4.


p c  p x

5.


d g 

 

 π d w 

C  6.

L

p c  Ng

λ  6.228  deg Ng  51

d g  3.247  in

π dw  dg 2

C  2.498  in


Fmax  1.173  in

7.

Find the materials factor Cs from equation 13.24. Since C < 8 in, Cs  1000.

8.

Find the ratio correction factor Cm from equations 13.25. Based on mG  17 , the first of the expressions in that equation set will be used. 2

Cm  0.0200 mG  40 mG  76  0.46 9.

Cm  0.815


nw dw

Vt  230.4  fpm

2  cos( λ)


Vt fpm

Cv  0.659  e

Cv  0.511


 dg  Wtg  Cs Cm Cv    in 

0.8

 Fmax

lbf in

Wtg  1254 lbf




  fpm  

 0.110 

μ  0.103  e

Vt

0.450

 0.012

μ  0.041



13-13-2



Wf  54.8 lbf


Φ o 


Φ o  0.95 hp



Φ l  0.38 hp


Φ  Φ o  Φ l

Φ  1.33 hp


Φo

e  71.3 %

Φ


dg 2

Tg  2035 in lbf



13-14-1


If the gearset in Problem 13-3 transmits 125 HP at 1000 pinion rpm, find the torque on each shaft.

Given:

Tooth numbers: Number of pinion teeth

Np  23 Ng  57


Pinion speed Transmitted power

ωp  1000 rpm P  125  hp



See Mathcad file P1214. P Solution: 1. For the pinion shaft Tp 

Tp  7878 in lbf

ωp

Tp  656.5  ft  lbf

2.


3.

For the gear shaft

ωg  Tg 

4.

Np Ng

 ωp

P

ωg  403.509  rpm Tg  19524  in lbf

ωg

Tg  1627 ft  lbf

We could have calculated the torque on the gear shaft directly without finding the gear shaft speed, Tg 

Ng Np

 Tp

Tg  19524  in lbf

Tg  1627 ft  lbf



13-15-1



Given:


Np  27


Ng  78

Pinion speed Transmitted power

ωp  1600 rpm P  33 kW



P

Tp  197  N  m

1.


2.


3.

For the gear shaft

ωg  Tg 

4.

ωp

Np Ng

 ωp

P

ωg  553.846  rpm Tg  569  N  m

ωg


Ng Np

 Tp

Tg  569  N  m



13-16-1


Size the helical gears in problem 13-14 for a bending factor of safety of at least 2 assuming a steady torque, 25 deg pressure angle, full depth teeth, a face width factor of 10, quality index of 9, an AISI 4140 steel pinion, and a class 40 cast iron gear.

Given:

Factor of safety

Nfb  2


Ng  57


H  125  hp n  1000 rpm


R  0.99 Qv  9


Np  23 ϕ  25 deg

Life (cycles) Helix angle

N  10 ψ  10 deg

Solution:

7


Pinion Jp  0.57

1.


2.

Write the equations for pitch diameter, pitchline velocity, and transmitted load in terms of the unknown diametral pitch, p d. Note that, in Mathcad, unit conversion factors are not included. d  p d  


Vt p d  


Wt p d  


3.


4.



Vt  p d 

Ka  1 v

B  0.25  12  Qv

0.6667

B  0.52



6.


7.


σbp p d F  

A  76.878 B

Km  1.7 Ka Wt p d   p d  Km Kv p d   F  Jp

 0.0323

Life factor

KL  1.6831 N

Reliability

KR  0.7  0.15 log( 1  R)

KL  1 KR  1

Temperature factor

KT  1


S atp  40000  psi



13-16-2 S fbp 


S atp KL

S fbp  40001  psi

KT  KR


Lower limit

FU  p d  

16

FL  p d  

8

Nfb =

Face width

F  p d  

p d 

pd

S fb

Safety factor


pd


1 1.5 6   in in in 4


p d  5  in

 

F pd in

3

 

FL pd


in


2

 

FU pd in

1

F  3.125  in 0


2

2.5

3

3.5

4

4.5

5

5.5

6

pd  in

d  p d   4.600  in

Kv p d   0.824

FIGURE 13-16A

Vt p d   1204

Wt p d   3425 lbf


ft min

σbp p d F   19839  psi The assumption made in step 5 is correct so no further iteration is required. Gear 14. Determine the bending geometry factor, J (Table 13-4)

Jg  0.63


σbg p d F  


16. Determine the endurance strength of the pinion.



13-16-3

Material bending strength (psi) Class 40 cast iron

S atg  13000  psi S fbg 

Endurance strength

S atg KL

S fbg  13000  psi

KT  KR


Lower limit

FU  p d  

16

FL  p d  

8

Nfb =

Face width

F  p d   p d 

pd

S fb

Safety factor


pd


1 1.5 6   in in in


6

 

1

F pd

p d  3  in

in


 

FL pd

F  p d   3.250  in

3

in

 


FU pd in

F  3.250  in 22. Then, the parameters that depend on P and F are: d  p d   7.667  in

4.5

1.5

0

Kv p d   0.788

2

2.5

3

3.5

4

4.5

5

5.5

6

pd  in

FIGURE 13-16B Vt p d   2007

ft min


Wt p d   2055 lbf

σbg p d F   6500 psi 23. The gear dimensions are larger (smaller diametral pitch means bigger teeth) than for the pinion. This means that we will accept the gear requirements for the pinion, thus, for the set 1

Diametral pitch

p d  3  in

Face width

F  3.250  in




Nfbg 

S fbg


13-16-4

Nfbg  2.0


Nfbp 

S fbp


Nfbp  5.6



13-17-1


Size the helical gears in problem 13-15 for a bending factor of safety of at least 2.5 assuming a steady torque, 20 deg pressure angle, full depth teeth, a face width factor of 12, quality index of 11, an AISI 4340 steel pinion, and an A-7-d nodular 40 iron gear.

Given:

Factor of safety

Nfb  2.5


Ng  78


H  33 kW n  1600 rpm


R  0.99 Qv  11


Np  27 ϕ  20 deg


N  10 ψ  30 deg

Solution:

7



1.


2.



Vt p d  


Wt p d  


3.


4.



Vt  p d 

Ka  1 v

B  0.25  12  Qv

0.6667

B  0.25



6.


7.


σbp p d F  

A  92 B


 0.0323

Life factor

KL  1.6831 N

Reliability

KR  0.7  0.15 log( 1  R)

KL  1 KR  1

Temperature factor

KT  1


S atp  42000  psi



13-17-2

S fbp 


S atp KL

S fbp  42001  psi

KT  KR


Upper limit

pd

FL  p d  

Lower limit

Nfb =

Face width

F  p d  

p d 

8 pd

S fb

Safety factor


16


1 1.5 10   in in in 4


p d  8  in

 

F pd in

3

 

FL pd


in


2

 

FU pd in

1

F  2.000  in 0


4

5

6

7

8

9

10

pd  in

d  p d   3.375  in

Kv p d   0.918

FIGURE 13-17A

Vt p d   1414

Wt p d   1033 lbf


ft min


Jg  0.51


σbg p d F  


16. Determine the endurance strength of the pinion.



13-17-3

Material bending strength (psi) A-7-d nodular iron

S atg  34000  psi S fbg 

Endurance strength

S atg KL

S fbg  34001  psi

KT  KR


Lower limit

FU  p d  

16

FL  p d  

8

Nfb =

Face width

F  p d   p d 

pd

S fb

Safety factor


pd


1 1.5 10   in in in 6


 

F pd

1

p d  8  in

in

4.5

 


FL pd

F  p d   2.207  in

3

in

 

FU pd


in

1.5

F  2.250  in 0


4

5

6

7

8

9

10

pd  in

Kv p d   0.918 FIGURE 13-17B

Vt p d   1414

ft min


Wt p d   1033 lbf


Diametral pitch

p d  8  in

Face width

F  2.250  in




Nfbg 

S fbg


13-17-4

Nfbg  2.5


Nfbp 

S fbp


Nfbp  3.0



13-18-1


Size the helical gears in problem 13-14 for a surface factor of safety of at least 1.6 assuming a steady torque, 25 deg pressure angle, full depth teeth, a face width factor of 10, quality index of 9, an AISI 4140 steel pinion, and a class 40 cast iron gear.

Given:

Factor of safety

Nfc  1.6


Ng  57


H  125  hp n  1000 rpm


R  0.99 Qv  9


Np  23 ϕ  25 deg


N  10 ψ  10 deg

Solution:

7


Pinion 1.


Addendum

1 pd Np

Pitch radii

rp p d  

rg p d  

Center distance

C p d   rp p d   rg p d 

2 pd

Ng 2 pd

Radii of curvature

ρp  p d  




Base helix angle


 

ϕn  24.666 deg


 

ψb  9.079  deg


  r  p   a  p   2   r  p   cos( ϕ)  2  d p d  p d  r p  a p 2  r p  cos( ϕ) 2  C p  sin( ϕ)   d   g d   d    g d 


mp p d  

   

Z pd 

π pd


Axial contact ratio

mF  p d F  



π



13-18-2


Axial pitch

p x  p d  

p n  p d   p t  p d   cos( ψ)



return

return

2.


Load sharing ratio

mN  p d F  

Geometry factor

I  p d F  

F





Vt p d  



4.


d g p d  

pd

Ng pd

d p p d   n

Wt p d  


Np

2 H

Vt  p d 

Ca  1 v

B  0.25  12  Qv

0.6667

B  0.52

A  50  56 ( 1  B) Cv p d  

A  76.878

A     min  A  Vt  p d    ft  

B

5.


Cm  1.7

6.

Choose an elastic coefficient from Table 12-18 (steel on CI).

Cp  2100 psi

7.


σcp p d F   Cp

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I  p d F 



13-18-3

Determine the endurance strength of the pinion. CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1

Temperature factor

CT  1

Material surface strength (psi) AISI 4140 steel

S ac  165000 psi S fcp 


 0.056

Life factor

S ac CL

S fcp  164997 psi

CT  CR

Write the design equations using the range of face-width to diametral-pitch ratio given in the text, and the bending stress equation, solved for the unknown face width. Limits, safety factor

FU  p d  

FL  p d  

16 pd

 Sfc  Nfc =    σc 

8 pd

C a W t  p d   C m

2

Face width


Guess value for F

F  2.347  in


p d 

2


2 2.5 12   in in in


4



F' pd F

1

p d  5  in

in



3

 


FL pd

F' p d F   2.347  in

2

in


 

FU pd in

1


d g p d   11.400 in

0

2

3

4

5

6

7

8

9 10 11 12

pd  in

FIGURE 13-18A Vt p d   1204

ft min

Wt p d   3425 lbf

Graph of Face Width and Limits for the Pinion (surface) in Problem 13-18

σcp p d F   129197 psi

mp p d   1.491

mF  p d F   0.667

n r p d   0.491

n a p d F   0.667

mN  p d F   0.798

Lmin p d F   2.975  in

I  p d F   0.171



13-18-4

Gear C a W t  p d   C m

σcg p d F   Cp


Cv p d   F  d g p d   I  p d F 

16. Determine the endurance strength of the gear.  0.056

Life factor

CL  2.466  N

CL  1

Reliability

CR  0.7  0.15 log( 1  R)

CR  1

Temperature factor

CT  1

Material surface strength (psi) Class 40 CI

S ac  80000  psi S fcg 

Endurance strength

S ac CL

S fcg  79999  psi

CT  CR

17. Write the design equations using the range of face-width to diametral-pitch ratio given in the text, and the bending stress equation, solved for the unknown face width. Limits, safety factor

FU  p d  

FL  p d  

16 pd

8

Nfc =

pd

C a W t  p d   C m

Face width

Guess value for F

F  2.745  in p d 

2

2

 Cp  F' p d F      Nfc Cv p d   d g p d   I  p d F   S fcg 


 Sfc  σ   c


2 2.5 12   in in in 8




F' pd F in

p d  4  in



6

 

FL pd


4

in

F' p d F   2.746  in

 

FU pd


in

2

0


2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 pd  in

d p p d   5.750  in

d g p d   14.250 in

FIGURE 13-18B

Vt p d   1505

Wt p d   2740 lbf

Graph of Face Width and Limits for the Gear (surface) in Problem 13-18

ft min

σcp p d F   99427  psi

mp p d   1.491

mF  p d F   0.617

n r p d   0.491


MACHINE DESIGN - An Integrated Approach, 4th Ed. n a p d F   0.617

mN  p d F   0.84

13-18-5 Lmin p d F   3.276  in

I  p d F   0.163

23. The gear teeth are larger than for the pinion. This means that we will accept the gear requirements for the pinion thus, for the set 1

Diametral pitch

p d  4  in

Face width

F  2.750  in



 Sfcg  Nfsg     σcg pd F  

2

 Sfcp   σ p F   cp d  

2

Nfsg  1.6


Nfsp 

Nfsp  2.8



13-19-1


Size the helical gears in problem 13-15 for a surface factor of safety of at least 1.2 assuming a steady torque, 20 deg pressure angle, full depth teeth, a face width factor of 12, quality index of 11, an AISI 4340 steel pinion, and an A-7-d nodular iron gear.

Given:

Factor of safety

Nfc  1.2


Ng  78


H  33 kW n  1600 rpm


R  0.99 Qv  11


Np  27 ϕ  20 deg


N  10 ψ  30 deg

Solution:

7


Pinion 1.


Addendum

1 pd Np

Pitch radii

rp p d  

rg p d  

Center distance

C p d   rp p d   rg p d 

2 pd

Ng 2 pd

Radii of curvature

ρp  p d  




Base helix angle


 

ϕn  17.495 deg


 

ψb  28.481 deg




mp p d  

   

Z pd 

π pd


Axial contact ratio

mF  p d F  



π



13-19-2


Axial pitch

p x  p d  

p n  p d   p t  p d   cos( ψ)



return

return

2.


Load sharing ratio

mN  p d F  

Geometry factor

I  p d F  

F


   ρ  p   ρ  p    2 rp pd   mN  p d F  g d   p d 1

1



Vt p d  



4.


d g p d  

pd

Ng pd

d p p d   n

Wt p d  


Np

2 H

Vt  p d 

Ca  1 v

B  0.25  12  Qv

0.6667

B  0.25

A  50  56 ( 1  B) Cv p d  

A  92

A     min  A  Vt  p d    ft  

B

5.


Cm  1.6

6.

Choose an elastic coefficient from Table 11-18 (steel on NI).

Cp  2160 psi

7.


σcp p d F   Cp

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I  p d F 



13-19-3


Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1

Temperature factor

CT  1


S ac  160000 psi S fcp 


 0.056

Life factor

S ac CL

S fcp  159997 psi

CT  CR


FU  p d  

FL  p d  

16 pd

 Sfc  Nfc =    σc 

8 pd

C a W t  p d   C m

2

Face width


Guess value for F

F  1.140  in


p d 

2


6 6.1 16   in in in


2



F' pd F

1

p d  12 in

in



1.5

 


FL pd

F' p d F   1.140  in

1

in


 

FU pd in

0.5


d g p d   6.500  in

0

6

7

8

9 10 11 12 13 14 15 16 pd  in

FIGURE 13-19A Vt p d   942 

ft min

Wt p d   1550 lbf


σcp p d F   138282 psi

mp p d   1.726

mF  p d F   2.757

n r p d   0.726

n a p d F   0.757

mN  p d F   0.516

Lmin p d F   2.42 in

I  p d F   0.231



13-19-4


σcg p d F   Cp


Cv p d   F  d g p d   I  p d F 


Life factor

CL  2.466  N

CL  1

Reliability

CR  0.7  0.15 log( 1  R)

CR  1

Temperature factor

CT  1

Material surface strength (psi) A-7-d nodular iron

S ac  100000 psi S fcg 

Endurance strength

S ac CL

S fcg  99998  psi

CT  CR


FU  p d  

FL  p d  

16 pd

8

Nfc =

pd

C a W t  p d   C m

Face width

Guess value for F

F  1.028  in p d 

2

2



 Sfc  σ   c


6 6.1 16   in in in 2




F' pd F in

p d  12 in



1.5

 

FL pd


1

in

F' p d F   1.030  in

 

FU pd


in

0.5

0


6

7

8

9 10 11 12 13 14 15 16 pd  in

d p p d   2.250  in

d g p d   6.500  in

FIGURE 13-19B

Vt p d   942 

Wt p d   1550 lbf


ft min

σcp p d F   147201 psi

mp p d   1.726

mF  p d F   2.481

n r p d   0.726



mN  p d F   0.527

13-19-5 Lmin p d F   2.136  in

I  p d F   0.227

23. The gear teeth are the same size as the pinion. This means that we will accept the gear requirements for the pinion thus, for the set 1

Diametral pitch

p d  12 in

Face width

F  1.125  in



 Sfcg  Nfsg     σcg pd F  

2

 Sfcp   σ p F   cp d  

2

Nfsg  1.3


Nfsp 

Nfsp  1.2



13-20-1


Size the bevel gears in problem 13-5 for a bending factor of safety of at least 2 assuming a 5-year, 1-shift life, steady torque, quality index of 9, an AISI 4140 steel pinion and gear.

Given:

Factor of safety

Nfb  2


Ng  126

Power to be transmitted (hp) Rotational speed of pinion (rpm)

Reliability H  746  W n p  1000 rpm AGMA Quality level

R  0.99 Qv  9


Np  14

Life  5  yr

Solution:

Life (years)



1.

Determine the bending geometry factor, J (Figure 13-5)

2.

Write the equations for pitch diameter, pitchline velocity, and transmitted torque in terms of the unknown diametral pitch, p d. Note that, in Mathcad, unit conversion factors are not included. Pitch diameter of pinion (in)


Transmitted torque (in-lbf)

Np

d p p d   Vt p d   Tp 

d g p d  

pd

pd

d p p d   n p 2

H

Tp  63.051 in lbf

np

3. Set the application, size, and type factors.

Ka  1

4. Write the equation for the dynamic load factor, Kv

B  0.25  12  Qv

Ks  1

Kx  1

0.6667

B  0.52

A  50  56 ( 1  B) Kv p d  

A  76.878

A     min  A  Vt  p d    ft  

B

Km  1.6

5. Tentatively choose the mounting factor, Km (Assume 0 < F < 2 in) 6. The bending stress equation for the pinion is

Ng

σbp p d F  

2  Tp



pd



Ka Km Ks

d p p d  F  Jp Kv p d   Kx

7. Determine the endurance strength of the pinion. np

Cycle life

N  Life

N  2.63  10

Life factor

KL  1.6831 N

Reliability

KR  0.7  0.15 log( 1  R)

2 π  0.0323

9

KL  0.835 KR  1

Temperature factor

KT  1


S atp  40000  psi

Endurance strength

S fbp 


S fbp  33412  psi



13-20-2

8. Write the design equations using the face-width to pitch-cone length ratio given in the text, and the bending stress equation, solved for the unknown face width. Upper limit, F = L/3

Np

FLover3 p d  



 Np     Ng  

6  p d  sin atan



S fb

Safety factor

Nfb =

Face width

F  p d  


p d 

σb 2  Tp p d  Ka Km Ks



Nfb

d p p d   Jp Kv p d   Kx S fbp

16 16.5 26   in in in


2

 

1.5

F pd

1

p d  18 in

in

 

FLover3 pd


1

in 0.5

12. Round this to the decimal equivalent of a common fractional value. 0 16

F  1.125  in

18

20

22

24

26

pd  in

13. Then, the parameters that depend on p d and F are: d p p d   0.778  in Vt p d   204 

Kv p d   0.915


ft min

σbp p d F   15912  psi The assumption made in step 5 is correct so no further iteration is required. GEAR 14. Determine the bending geometry factor, J (Figure 13-5)

Jg  0.220


σbg p d F  

2  Tp



pd



Ka Km Ks

d g p d  F  Jg Kv p d   Kx

16. Write the design equations using the face-width to pitch-cone length ratio given in the text, and the bending stress equation, solved for the unknown face width. F = L/3


Np



 Np     Ng  






Safety factor

Nfb =

S fb


F  p d  

Face width

17. Plot F(P) vs. p d over the range p d 

13-20-3



Nfb

d g p d   Jg Kv p d   Kx S fbp

30 30.5 40   in in in

2


 

1.5

F pd

1

p d  36 in

in

 


FLover3 pd

F  p d   0.602  in

in

1

0.5

20. Round this to the decimal equivalent of a common fractional value. 0 30

F  0.625  in

32

34

36

38

40

pd  in

21. Then, the parameters that depend on p d and F are: d g p d   3.500  in Vt p d   102 

ft min

FIGURE 13-20B

Kv p d   0.938


σbg p d F   16093  psi

22. The gear dimensions are smaller (larger diametral pitch means smaller teeth) than for the pinion. This means that we will accept the pinion requirements for the gear, thus, for the set 1

Diametral pitch

p d  18 in

Face width

F  1.125  in

Pitch cone angle

αp  atan

Pitch-cone length

L 

 Np    Ng 

d p p d 

2  sin αp

αp  6.34 deg

L  3.522  in

L 3

 1.174  in

23. Determine the realized factor of safety for the pinion using the above values for F and p d. Pinion factor of safety

Nfbp 

S fbp


Nfbp  2.1

25. Check the factor of safety on the gear: Gear factor of safety

Nfbg 

S fbp


Nfbg  14.6



13-21-1


Size the bevel gears in problem 13-6 for a bending factor of safety of at least 2.5 assuming a 15-year, 3-shift life, steady torque, quality index of 11, an AISI 4340 steel pinion and gear.

Units:

yr  2080 hr

Given:

Factor of safety

Nfb  2.5


Ng  81


H  7460 W n p  800  rpm


R  0.99 Qv  11


Np  18

Life (years)


Solution:



1.


2.

Write the equations for pitch diameter, pitchline velocity, and transmitted torque in terms of the unknown diametral pitch, p d. Note that, in Mathcad, unit conversion factors are not included.




Np

d p p d   Vt p d   Tp 

d g p d  

pd

pd

d p p d   n p 2

H

Tp  788.134  in lbf

np


Ka  1


B  0.25  12  Qv

Ks  1

Kx  1

0.6667

B  0.25

A  50  56 ( 1  B) Kv p d  

A  92

A     min  A  Vt  p d    ft  

σbp p d F  

B

Km  1.6

5. Tentatively choose the mounting factor, Km (Assume 0 < F < 2 in)

6. The bending stress equation for the pinion is

Ng

2  Tp



pd



Ka Km Ks


7. Determine the endurance strength of the pinion. Cycle life

shifts  3


np 2 π  0.0323

Life factor

KL  1.6831 N

Reliability

KR  0.7  0.15 log( 1  R)

9

N  1.498  10 KL  0.851 KR  1

Temperature factor

KT  1


S atp  42000  psi



13-21-2

S fbp 

Endurance strength

S atp KL

S fbp  35727  psi

KT  KR


Np




 Np     Ng  




S fb

Safety factor

Nfb =

Face width

F  p d  


p d 


Nfb




2 2.5 12   in in in 4


 

p d  6  in

3

F pd in


 2

F  p d   1.404  in

FLover3 pd in

12. Round this to the decimal equivalent of a common fractional value. F  1.500  in

0


1

Kv p d   0.942

ft

2

4

6

8

10

12

pd  in


min


Jg  0.221


σbg p d F  

2  Tp



pd



Ka Km Ks


16. Write the design equations using the face-width to pitch-cone length ratio given in the text, and the bending stress equation, solved for the unknown face width. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


13-21-3 Np


F = L/3



 Np     Ng  




Safety factor

Nfb =

S fb


F  p d  

Face width




Nfb


6 6.5 16   in in in 2


 

p d  10 in

1.5

F pd in


 

F  p d   1.034  in

FLover3 pd

1

in

20. Round this to the decimal equivalent of a common fractional value.

0.5

F  1.125  in

0


ft min

6

8

10

12

14

16

pd  in

Kv p d   0.953


σbg p d F   13137  psi

22. The gear dimensions are smaller (larger diametral pitch means smaller teeth) than for the pinion. This means th we will accept the pinion requirements for the gear, thus, for the set 1

Diametral pitch

p d  6  in

Face width

F  1.500  in

Pitch cone angle

αp  atan

Pitch-cone length

L 

 Np    Ng 

d p p d 

2  sin αp

αp  12.529 deg

L  6.915  in

L 3

 2.305  in


Nfbp 

S fbp


Nfbp  2.7


Nfbg 

S fbp


Nfbg  9.9



13-22-1


Size the spiral gears in problem 13-7 for a bending factor of safety of at least 2.2 assuming a 10-year, 3-shift life, steady torque, quality index of 8, an AISI 4340 steel pinion and gear.

Units:

yr  2080 hr

Given:

Factor of safety

Nfb  2.2


Reliability H  3  hp R  0.99 n p  600  rpm AGMA Quality level Qv  8


Np  16

Solution:

Ng  80



Life (years)



1.


2.

Write the equations for pitch diameter, pitchline velocity, and transmitted torque in terms of the unknown diametral pitch, p d. Note that, in Mathcad, unit conversion factors are not included. d p p d  


Vt p d  


Tp 


3.

Set the application, size, and type factors.

4.


Np

d g p d  

pd

2

H

Tp  315.127  in lbf

np Ks  1

B  0.25  12  Qv

Kx  1.15

0.6667

B  0.63

A  50  56 ( 1  B) Kv p d  

5.


6.


7.

Determine the endurance strength of the pinion. Cycle life

shifts  3

pd

d p p d   n p

Ka  1 v

Ng

A  70.721

A     min  A  Vt  p d    ft  

σbp p d F  


B

Km  1.6 2  Tp



pd



Ka Km Ks

d p p d  F  Jp Kv p d   Kx np 2 π  0.0323

Life factor

KL  1.6831 N

Reliability

KR  0.7  0.15 log( 1  R)

9

N  2.246  10 KL  0.84 KR  1

Temperature factor

KT  1


S atp  42000  psi



13-22-2

S fbp 


S atp KL

S fbp  35262  psi

KT  KR

Write the design equations using the face-width to pitch-cone length ratio given in the text, and the bending stress equation, solved for the unknown face width. Upper limit, F = L/3

Np




 Np     Ng  




S fb

Safety factor

Nfb =

Face width

F  p d  


p d 


Nfb




2 2.5 12   in in in


p d  10 in

4

 

3

F pd


in

F  p d   1.273  in

 2

FLover3 pd


in 1

F  1.375  in 0


Kv p d   0.880

ft

2

4

6

8

10

12

pd  in


min

σbp p d F   14844  psi The assumption made in step 5 is correct so no further iteration is required. Gear 14. Determine the bending geometry factor, J (Figure 13-8)

Jg  0.302


σbg p d F  

2  Tp



pd



Ka Km Ks




13-22-3

Np


F = L/3



 Np     Ng  




Safety factor

Nfb =

S fb


F  p d  

Face width




Nfb


10 11.5 20   in in in 2


 

p d  16 in

1.5

F pd in


 

F  p d   0.642  in

FLover3 pd

1

in


0.5

F  0.750  in

0 10


ft min

12

14

16

18

20

pd  in

Kv p d   0.902

FIGURE 13-21B

σbg p d F   13730  psi


22. The gear dimensions are smaller (larger diametral pitch means smaller teeth) than for the pinion. This means tha we will accept the pinion requirements for the gear, thus, for the set 1

Diametral pitch

p d  10 in

Face width

F  1.375  in

Pitch cone angle

αp  atan

Pitch-cone length

L 

 Np    Ng 

d p p d 

2  sin αp

αp  11.31  deg

L  4.079  in

L 3

 1.360  in


Nfbp 

S fbp


Nfbp  2.4


Nfbg 

S fbp


Nfbg  11.8



13-23-1


Size the bevel gears in problem 13-5 for a minimum safety factor of 1.4 for any mode of failure of pinion or gear assuming a 5-year, 1-shift life, steady torque, quality index of 9, an AISI 4140 steel pinion and gear.

Units:

yr  2080 hr

Given:

Factor of safety

Nfc  1.4


Ng  126


Reliability H  746  W n p  1000 rpm AGMA Quality level

R  0.99 Qv  9


Np  14


Life (years)


See Mathcad file P1323. I  0.092

1.

Determine the surface geometry factor, I, from Figure 13-6.

2.




Vtp p d  


Tp 

Np

d g p d  

pd Np 2 pd


4.


v

H

Tp  63.051 in lbf

np Cs  1

B  0.25  12  Qv

Cf  1

0.6667

B  0.52

A  50  56 ( 1  B) Cvp p d  

5. 6. 7.

Tentatively choose the mounting factor, Cmd Choose an elastic coefficient from Table 11-18 (steel on steel) and a stress adjustment constant.

pd

 np

Ca  1

3.

Ng

A  76.878

A     min  A  Vtp pd    ft  

Cmd  3.0

B

Cxc  1

Cp  2100 psi

0.5

Cb  0.634


shifts  1


np 2 π  0.056

Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

N  6.24  10

8

CL  0.793 CR  1



13-23-2

Hardness

CH  1

Temperature factor

CT  1


S acp  167500 psi S fcp 


S fcp  132885 psi

CT  CR

Write the equation for the design pinion torque. TD p d F  

9.

S acp CL

2  Sacp d p pd  0.774  CH    2 Cs Cmd  Cf  Ca Cxc  Cp Cb CT  CR 

F

I  Cvp p d 





z p d F  

Define the exponent z.

return 0.667 if Tp  TD p d F  1 otherwise

10. The surface stress equation for the pinion is

 2 TD pd F    Tp  σcp p d F   Cp Cb     F  d p pd  2 I   TD p d F    



z pd F

 

Ca Cmd

Cvp p d 

 Cs Cf  Cxc

11. Write the design equations using the face-width to pitch-cone length ratio given in the text, and the bending stress equation, solved for the unknown face width. FLover3 p d  

Upper limit, F = L/3

 Sfc  Nfc =    σc 

Np



 Np     Ng  




Face width

 Cp Cb  F' p d F      S fcp  12. Plot F(P) vs. p d over the range

 2  TD p d F    Tp      d p pd  2 I   TD pd F    

2

p d 

12 12.5 22   in in in



z pd F

 

Guess value for F

Ca Cmd

Cvp p d 

2

 Cs Cf  Cxc Nfc

F  0.955  in

2




1

F' pd F

p d  16 in



1.5

in


 

FLover3 pd

F' p d F   0.956  in

in

15. Round this to the decimal equivalent of a common fractional value. F  1.000  in 16. Then, the parameters that depend on p d and F are: d p p d   0.875  in

1

d g p d   7.875  in

0.5

0 12

14

16

18

20

22

pd  in

FIGURE 13-23 Graph of Face Width and Limits for for Problem 13-23


MACHINE DESIGN - An Integrated Approach, 4th Ed. Vtp p d   229 

ft min

Cvp p d   0.911

13-23-3

σcp p d F   1  10  psi 5

TD p d F   101  in lbf

17. When both pinion and gear are of the same material, the stress in the gear will always be lower than in the pinion Therefore, the face width and diametral pitch found for the pinion should also be used for the gear. 1

Diametral pitch

p d  16 in

Face width

F  1.000  in

Pitch cone angle

αp  atan

Pitch-cone length

L 

 Np    Ng 

αp  6.34 deg

d p p d 

L  3.962  in

2  sin αp

L 3

 1.321  in

18. Determine the realized factor of safety for the pinion using the above values for F and p d.

Pinion factor of safety

 Sfcp  Nfcp     σcp pd F  

2

Nfcp  1.4



13-24-1


Size the bevel gears in problem 13-6 for a surface factor of safety of at least 1.3 assuming a 15-year, 3-shift life, steady torque, quality index of 11 an AISI 4340 steel pinion and gear.

Units:

yr  2080 hr

Given:

Factor of safety

Nfc  1.3


Ng  81


H  7460 W n p  800  rpm


R  0.99 Qv  11


Np  18

Life (years)


Solution:


1.


2.




Vtp p d  


Tp 

Np

d g p d  

pd Np 2 pd


4.


v

H

Tp  788.134  in lbf

np Cs  1

B  0.25  12  Qv

Cf  1

0.6667

B  0.25

A  50  56 ( 1  B) Cvp p d  

5. 6. 7.

Tentatively choose the mounting factor, Cmd Choose an elastic coefficient from Table 12-18 (steel on steel) and a stress adjustment constant.

pd

 np

Ca  1

3.

Ng

A  92

A     min  A  Vtp pd    ft  

Cmd  3.0

B

Cxc  1

Cp  2100 psi

0.5

Cb  0.634


shifts  3


np 2 π  0.056

Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

9

N  4.493  10 CL  0.71 CR  1

Hardness

CH  1

Temperature factor

CT  1


S acp  162000 psi



13-24-2

S fcp 


S fcp  115071 psi

CT  CR


9.

S acp CL


F






z p d F  







z pd F

 

Ca Cmd

Cvp p d 


11. Write the design equations using the face-width to pitch-cone length ratio given in the text, and the bending stress equation, solved for the unknown face width. Np





Nfc =

 Np     Ng  




Face width



2

p d 

2 2.5 12   in in in



z pd F

 

Guess value for F

Ca Cmd

Cvp p d 

 Sfc  σ   c

2


F  1.325  in

4




1

F' pd F

p d  6  in



3

in


 2

FLover3 pd

F' p d F   1.325  in

in

15. Round this to the decimal equivalent of a common fractional value. F  1.375  in

1

0


d g p d   13.500 in

2

4

6

8

10

12

pd  in

FIGURE 13-24 Graph of Face Width and Limits for for Problem 13-24

Vtp p d   628 

ft min

Cvp p d   0.942

σcp p d F   99697  psi

TD p d F   1567 in lbf



13-24-3


Diametral pitch

p d  6  in

Face width

F  1.375  in

Pitch cone angle

αp  atan

Pitch-cone length

L 

 Np    Ng 

αp  12.529 deg

d p p d 

L  6.915  in

2  sin αp

L 3

 2.305  in



 Sfcp  Nfcp     σcp pd F  

2

Nfcp  1.3



13-25-1


Size the spiral gears in problem 13-7 for a surface factor of safety of at least 1.4 assuming a 10-year, 3-shift life, steady torque, quality index of 8, an AISI 4340 steel pinion and gear.

Units:

yr  2080 hr

Given:

Factor of safety

Nfc  1.4


Reliability H  3  hp R  0.99 n p  600  rpm AGMA Quality level Qv  8


Np  16

Solution:

Ng  80



Life (years)


Pinion I  0.158

1.


2.




Vtp p d  


Tp 

Np

d g p d  

pd Np 2 pd


4.


v

H

Tp  315.127  in lbf

np Cs  1

B  0.25  12  Qv

Cf  1

0.6667

B  0.63

A  50  56 ( 1  B)

A  70.721

A  Cvp p d     min  A  Vtp pd    ft   5.

Tentatively choose the mounting factor, Cmd

Cmd  3.0

6.

Choose an elastic coefficient from Table 11-18 (steel on steel) and a stress adjustment constant.

Cp  2100 psi

7.

pd

 np

Ca  1

3.

Ng

B

Cxc  1 0.5

Cb  0.634


shifts  3


np 2 π  0.056

Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

9

N  2.246  10 CL  0.738 CR  1

Hardness

CH  1

Temperature factor

CT  1


S acp  162000 psi



13-25-2

S fcp 


S fcp  119626 psi

CT  CR


9.

S acp CL


F






z p d F  







z pd F

 

Ca Cmd

Cvp p d 




 Sfc  Nfc =    σc 

Np



 Np     Ng  




Face width

 Cp Cb  F' p d F      S fcp 

F  1.141  in

Guess value for F 12. Plot F(P) vs. p d over the range


2

p d 

1

p d  10 in

 

Ca Cmd

Cvp p d 



2 2.5 12   in in in




z pd F

2

4



F' pd F



3

in

 2

FLover3 pd

14. The calculated value of F is F' p d F   1.141  in

in 1

15. Round this to the decimal equivalent of a common fractional value. 0

F  1.250  in

2

3

4

5

6

7

8

9 10 11 12

pd  in


d g p d   8.000  in

FIGURE 13-25 Graph of Face Width and Limits for the Pinion in Problem 13-25


MACHINE DESIGN - An Integrated Approach, 4th Ed. Vtp p d   251 

ft min

Cvp p d   0.880

13-25-3

σcp p d F   98091  psi

TD p d F   658  in lbf


Diametral pitch

p d  10 in

Face width

F  1.250  in

Pitch cone angle

αp  atan

Pitch-cone length

L 

 Np    Ng 

αp  11.31  deg

d p p d 

L  4.079  in

2  sin αp

L 3

 1.360  in



 Sfcp  Nfcp     σcp pd F  

2

Nfcp  1.5



13-26-1


Find the rated power and rated output torque of the wormset in Problem 13-9 with an input speed o 2200 rpm.

Units:


Given:

Worm pitch diameter

d w  50 mm

Gear ratio

mG  22

Axial pitch

p x  10 mm

Input (worm) speed

n w  2200 rpm

Threads on worm

Nw  2

Pressure angle

ϕ  20 deg

L  p x  Nw

L  20 mm

Solution:

1


1.


2.



3.


Ng  mG Nw

4.


p c  p x

L

 

 π d w 

d g  C 

p c  Ng

λ  7.256  deg Ng  44

d g  140.056  mm

π dw  dg

5.


6.

Find the maximum recommended face width from equation 13.19.

2

Fmax  0.67 d w

Fmax  33.5 mm

7.


8.


Cm  0.0107 mG  56 mG  5145 9.

Cm  0.821


nw dw

Vt  1143 fpm

2  cos( λ)

10. Use this velocity to find the velocity factor Cv from equations 13.26. For this value of Vt, the second of these equations is appropriate.

 Vt  Cv  13.31     fpm 

 0.571

Cv  0.239


 dg  Wtg  Cs Cm Cv    mm 

0.8

 Fmax

N 75.948 mm

Wtg  4510 N




  fpm  

 0.110 

μ  0.103  e

Vt

0.450

 0.012

μ  0.020



13-26-2



Wf  94.5 N


Φ o 


Φ o  3.31 kW



Φ l  0.55 kW


Φ  Φ o  Φ l

Φ  3.86 kW


Φo

e  85.8 %

Φ


dg 2

Tg  316  N  m



13-27-1


Find the rated power and rated output torque of the wormset in Problem 13-10 with an input speed of 1400 rpm.

Units:


Given:

Worm pitch diameter

d w  1.75 in

Gear ratio

mG  17

Axial pitch

p x  0.200  in

Input (worm) speed

n w  1400 rpm

Threads on worm

Nw  3

Pressure angle

ϕ  20 deg

L  p x  Nw

L  0.600  in

Solution:

1


1.


2.



3.


Ng  mG Nw

4.


p c  p x

L

 

 π d w 

d g  C 

p c  Ng dw  dg


6.


Ng  51

d g  3.247  in

π

5.

λ  6.228  deg

2

C  2.498  in

Fmax  0.67 d w

Fmax  1.173  in

7.


8.

Find the ratio correction factor Cm from equations 13.25. Based on mG  17 , the first of the expressions in that equation set will be used. 2

Cm  0.0200 mG  40 mG  76  0.46 9.

Cm  0.815


nw dw

Vt  645  fpm

2  cos( λ)


Vt fpm

Cv  0.659  e

Cv  0.324


 dg    mm 


0.8

 Fmax

N 75.948 mm

Wtg  794  lbf


 Vt   0.110    fpm  μ  0.103  e

0.450

 0.012

μ  0.026



13-27-2



Wf  21.8 lbf


Φ o 


Φ o  1.69 hp



Φ l  0.43 hp


Φ  Φ o  Φ l

Φ  2.11 hp


Φo

e  79.8 %

Φ


dg 2

Tg  1290 in lbf



13-28-1


Find the rated power and rated output torque of the wormset in Problem 13-11 with an input speed of 4500 rpm.

Units:


Given:

Worm pitch diameter

d w  40 mm

Gear ratio

mG  82

Axial pitch

p x  5  mm

Input (worm) speed

n w  4500 rpm

Threads on worm

Nw  1

Pressure angle

ϕ  20 deg

L  p x  Nw

L  5  mm

Solution:

1


1.


2.



3.


Ng  mG Nw

4.


p c  p x

L

 

 π d w 

d g  C 

p c  Ng dw  dg


6.


Ng  82

d g  130.51 mm

π

5.

λ  2.279  deg

2

C  85.25  mm

Fmax  0.67 d w

Fmax  26.8 mm

7.


8.

Find the ratio correction factor Cm from equations 13.25. Based on mG  82 , the third of the expressions in that equation set will be used. Cm  1.1483  0.00658  mG

9.

Cm  0.609


nw dw

Vt  1857 fpm

2  cos( λ)


 Vt  Cv  13.31     fpm 

 0.571

Cv  0.181


 dg  Wtg  Cs Cm Cv    mm 

0.8

 Fmax

N 75.948 mm

Wtg  1915 N




  fpm  

 0.110 

μ  0.103  e

Vt

0.450

 0.012

μ  0.016



13-28-2



Wf  32.6 N


Φ o 


Φ o  718  W



Φ l  307  W


Φ  Φ o  Φ l

Φ  1026 W


Φo

e  70.0 %

Φ


dg 2

Tg  125  N  m



13-29-1

PROBLEM 13-29

_____

Statement:

A 23-tooth helical gear is cut with a 20-degree-pressure-angle hob at a helix angle of 25 deg. The hob has a standard diametral pitch of 5. The resulting teeth have standard spur-gear dimensions in the normal plane. Find the pitch diameter, addendum, dedendum, outside diameter, normal, transverse, and axial pitch, and the transverse pressure angle.

Given:

Normal plane data:

ϕn  20 deg

Pressure angle

Other data: Number of teeth N  23 Solution: 1.

ψ  25 deg

1

p d  4.532  in

Use equations 13.1 to find the pitch diameter, transverse pitch, normal pitch, and axial pitch.

Transverse pitch

d  p t 

N

d  5.076  in

pd

π

p t  0.693  in

pd

Normal pitch


Axial pitch

p x 

pn sin( ψ)

p n  0.628  in p x  1.487  in

Use the equations in Table 12-1 with the normal diametral pitch to calculate the addendum, dedendum, and outside diameter. Addendum

Dedendum

Outside diameter 4.

Helix angle

Use equation 13.1d to calculate the transverse diametral pitch.

Pitch diameter

3.

p nd  5  in


p d  p nd  cos( ψ) 2.

1

Diametral pitch

a 

b 

1

a  0.200  in

p nd 1.25

b  0.250  in

p nd

D  d  2  a

D  5.476  in

Use equation 13.2 to find the pressure angle in the transverse plane.

 tan ϕn    cos( ψ) 

ϕt  atan

ϕt  21.880 deg



13-30-1

PROBLEM 13-30

_____

Statement:

A 38-tooth helical gear is cut with a 25-degree-pressure-angle hob at a helix angle of 30 deg. The hob has a standard diametral pitch of 4. The resulting teeth have standard spur-gear dimensions in the normal plane. Find the pitch diameter, addendum, dedendum, outside diameter, normal, transverse, and axial pitch, and the transverse pressure angle.

Given:

Normal plane data:

ϕn  25 deg

Pressure angle

Other data: Number of teeth N  38 Solution: 1.

ψ  30 deg

1

p d  3.464  in

Use equations 13.1 to find the pitch diameter, transverse pitch, normal pitch, and axial pitch.

Transverse pitch

d  p t 

N

d  10.970 in

pd

π

p t  0.907  in

pd

Normal pitch


Axial pitch

p x 

pn sin( ψ)

p n  0.785  in p x  1.571  in

Use the equations in Table 12-1 with the normal diametral pitch to calculate the addendum, dedendum, and outside diameter. Addendum

Dedendum

Outside diameter 4.

Helix angle

Use equation 13.1d to calculate the transverse diametral pitch.

Pitch diameter

3.

p nd  4  in


p d  p nd  cos( ψ) 2.

1

Diametral pitch

a 

b 

1

a  0.250  in

p nd 1.25

b  0.313  in

p nd

D  d  2  a

D  11.470 in

Use equation 13.2 to find the pressure angle in the transverse plane.

 tan ϕn    cos( ψ) 

ϕt  atan

ϕt  28.300 deg



13-31-1


_____

A 39-tooth, 20-deg helix angle, helical gear is in mesh with an 18-tooth pinion. The p d = 8 and  = 25 deg. Find the transverse and axial contact ratios.

Given:

Tooth numbers: Ng  39 Pressure angle Diametral pitch

Np  18

ϕ  25 deg

Helix angle

ψ  20 deg

1

p d  8  in

Assumptions: The face width factor is fwf  12 Solution: 1.



π

p c 

p c  0.393  in

pd


Base pitch

p b  0.356  in


Np

d p 

d p  2.250  in

pd

rp  0.5 d p

rp  1.125  in


Center distance

d g  4.875  in

pd

rg  0.5 d g a 

Addendum

Ng

d g 

C 

rg  2.438  in

1.0

a  0.125  in

pd Np  Ng 2 pd

C  3.563  in

Length of action Z 


Z  0.516  in Transverse contact ratio 2.

mp 

Z

mp  1.450

pb


Axial contact ratio

F 

mF 

fwf

F  1.500  in


π

mF  1.390



13-32-1


_____

A 79-tooth, 30-deg helix angle, helical gear is in mesh with an 20-tooth pinion. The p d = 6 and  = 20 deg. Find the transverse and axial contact ratios.

Given:

Tooth numbers: Ng  79 Pressure angle Diametral pitch

Np  20

ϕ  20 deg

Helix angle

ψ  30 deg

1

p d  6  in

Assumptions: The face width factor is fwf  12 Solution: 1.



π

p c 

p c  0.524  in

pd


Base pitch

p b  0.492  in


Np

d p 

d p  3.333  in

pd

rp  0.5 d p

rp  1.667  in


Center distance

d g  13.167 in

pd

rg  0.5 d g a 

Addendum

Ng

d g 

C 

rg  6.583  in

1.0

a  0.167  in

pd Np  Ng 2 pd

C  8.250  in

Length of action Z 


Z  0.832  in Transverse contact ratio 2.

mp 

Z

mp  1.690

pb


Axial contact ratio

F 

mF 

fwf

F  2.000  in


π

mF  2.205



13-33-1

PROBLEM 13-33

_____

Statement:

If the gearset in Problem 13-31 transmits 135 HP at 1200 pinion rpm, find the torque on each shaft

Given:

Tooth numbers: Ng  39 Pinion speed

Np  18

ωp  1200 rpm

Transmitted power

P  135  hp

Assumptions: There is no loss of power in the gear mesh (100% efficiency). Solution: 1.


For the pinion shaft Tp 

P

ωp

Tp  7090 in lbf

Tp  590.9  ft  lbf

2.


3.

For the gear shaft

ωg  Tg 

4.

Np Ng

 ωp

P

ωg  553.846  rpm Tg  15362  in lbf

ωg

Tg  1280 ft  lbf


Ng Np

 Tp

Tg  15362  in lbf

Tg  1280 ft  lbf



13-34-1



Given:


Np  20


Ng  79Transmitted power

ωp  1200 rpm

Pinion speed

P  30 kW



P

Tp  239  N  m

1.


2.


3.

For the gear shaft

ωg  Tg 

4.

ωp

Np Ng

 ωp

P

ωg  303.797  rpm Tg  943  N  m

ωg


Ng Np

 Tp

Tg  943  N  m



13-35-1


Given:

Solution:

Size the helical gears in problem 13-33 for a bending factor of safety of at least 2 assuming a steady torque, 25 deg pressure angle, full depth teeth, quality index of 9, an AISI 4140 steel pinion, and a class 40 cast iron gear. Number of gear teeth Factor of safety Nfb  2 Ng  39 Power to be transmitted Rotational speed of pinion

H  135  hp n  1200 rpm


R  0.99 Qv  9


Np  18 ϕ  25 deg


N  10 ψ  10 deg

7



1.


2.



Vt p d  


Wt p d  


3.


4.



Vt  p d 

Ka  1 v

B  0.25  12  Qv

0.6667

B  0.52

A  50  56 ( 1  B) Kv p d  

5.


6.


7.


A     min  A  Vt  p d    ft  

σbp p d F  

A  76.878 B


 0.0323

Life factor

KL  1.6831 N

Reliability

KR  0.7  0.15 log( 1  R)

KL  1 KR  1

Temperature factor

KT  1


S atp  40000  psi



13-35-2 S fbp 


S atp KL

S fbp  40001  psi

KT  KR


Lower limit

FU  p d  

16

FL  p d  

8

Nfb =

Face width

F  p d  

p d 

pd

S fb

Safety factor


pd


1 1.5 6   in in in 4


p d  3  in

 

F pd in

3

 

FL pd


in


2

 

FU pd in

1

F  1.375  in 0


2

2.5

3

3.5

4

4.5

5

5.5

6

pd  in

d  p d   6.000  in

Kv p d   0.792

FIGURE 13-35A

Vt p d   1885

Wt p d   2363 lbf


ft min


Jg  0.64


σbg p d F  


16. Determine the endurance strength of the pinion. Material bending strength (psi) Class 40 cast iron

S atg  13000  psi



13-35-3

S fbg 

Endurance strength

S atg KL

S fbg  13000  psi

KT  KR


Lower limit

FU  p d  

16

FL  p d  

8

Nfb =

Face width

F  p d   p d 

pd

S fb

Safety factor


pd


1 1.5 6   in in in


6

 

1

F pd

p d  3  in

in


 

FL pd

F  p d   3.657  in

3

in

 


FU pd

1.5

in

F  3.750  in 22. Then, the parameters that depend on P and F are: d  p d   6.000  in

4.5

0

Kv p d   0.792

2

2.5

3

3.5

4

4.5

5

5.5

6

pd  in

FIGURE 13-35B Vt p d   1885

ft min


Wt p d   2363 lbf

σbg p d F   6339 psi 23. The gear dimensions are larger (smaller diametral pitch means bigger teeth) than for the pinion. This means that we will accept the gear requirements for the pinion, thus, for the set 1

Diametral pitch

p d  3  in

Face width

F  3.750  in


Nfbg 

S fbg


Nfbg  2.1



13-35-4


Nfbp 

S fbp


Nfbp  5.6



13-36-1


Size the helical gears in problem 13-34 for a bending factor of safety of at least 2.5 assuming a steady torque, 20 deg pressure angle, full depth teeth, quality index of 11, an AISI 4340 steel pinio and an A-7-d nodular 40 iron gear.

Given:

Factor of safety

Nfb  2.5


Ng  79


H  30 kW n  1200 rpm


R  0.99 Qv  11


Np  20 ϕ  20 deg


N  10 ψ  30 deg

Solution:

7



1.


2.



Vt p d  


Wt p d  


3.


4.



Vt  p d 

Ka  1 v

B  0.25  12  Qv

0.6667

B  0.25



6.


7.


σbp p d F  

A  92 B


 0.0323

Life factor

KL  1.6831 N

Reliability

KR  0.7  0.15 log( 1  R)

KL  1 KR  1

Temperature factor

KT  1


S atp  42000  psi



13-36-2 S fbp 


S atp KL

S fbp  42001  psi

KT  KR


Lower limit

FU  p d  

16

FL  p d  

8

Nfb =

Face width

F  p d  

p d 

pd

S fb

Safety factor


pd


1 1.5 10   in in in 4


p d  6  in

 

F pd in

3

 

FL pd


in


2

 

FU pd in

1

F  1.875  in 0


4

5

6

7

8

9

10

pd  in

d  p d   3.333  in

Kv p d   0.927

FIGURE 13-36A

Vt p d   1047

Wt p d   1268 lbf


ft min


Jg  0.50


σbg p d F  


16. Determine the endurance strength of the pinion. Material bending strength (psi) A-7-d nodular iron

S atg  34000  psi



13-36-3

S fbg 

Endurance strength

S atg KL

S fbg  34001  psi

KT  KR


Lower limit

FU  p d  

16

FL  p d  

8

Nfb =

Face width

F  p d   p d 

pd

S fb

Safety factor


pd


1 1.5 10   in in in 6


 

F pd

1

p d  6  in

in

4.5

 


FL pd

F  p d   2.050  in

3

in

 

FU pd


1.5

in

F  2.125  in 0


4

5

6

7

8

9

10

pd  in

Kv p d   0.927 FIGURE 13-36B

Vt p d   1047

ft min


Wt p d   1268 lbf


Diametral pitch

p d  6  in

Face width

F  2.125  in


Nfbg 

S fbg


Nfbg  2.6



13-36-4


Nfbp 

S fbp


Nfbp  2.9



13-37-1


Size the helical gears in problem 13-33 for a surface factor of safety of at least 1.6 assuming a steady torque, 25 deg pressure angle, full depth teeth, quality index of 9, an AISI 4140 steel pinion, and a class 40 cast iron gear.

Given:

Factor of safety

Nfc  1.6


Ng  39


H  135  hp n  1200 rpm


R  0.99 Qv  9


Np  18 ϕ  25 deg


N  10 ψ  20 deg

Solution:

7


Pinion 1.


Addendum

1 pd Np

Pitch radii

rp p d  

rg p d  

Center distance

C p d   rp p d   rg p d 

2 pd

Ng 2 pd

Radii of curvature

ρp  p d  




Base helix angle


 

ϕn  23.662 deg


 

ψb  18.256 deg




mp p d  

   

Z pd 

π pd


Axial contact ratio

mF  p d F  



π



13-37-2


Axial pitch

p x  p d  

p n  p d   p t  p d   cos( ψ)



return

return

2.


Load sharing ratio

mN  p d F  

Geometry factor

I  p d F  

F


   ρ  p   ρ  p    2 rp pd   mN  p d F  g d   p d 1

1



Vt p d  



4.


d g p d  

pd

Ng pd

d p p d   n

Wt p d  


3.

Np

2 H

Vt  p d 

Ca  1 v

B  0.25  12  Qv

0.6667

B  0.52

A  50  56 ( 1  B) Cv p d  

A  76.878

A     min  A  Vt  p d    ft  

B

5.


Cm  1.7

6.

Choose an elastic coefficient from Table 12-18 (steel on CI).

Cp  2100 psi

7.


σcp p d F   Cp

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I  p d F 



13-37-3


Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1

Temperature factor

CT  1


S ac  165000 psi S fcp 


 0.056

Life factor

S ac CL

S fcp  164997 psi

CT  CR


FU  p d  

FL  p d  

16 pd

 Sfc  Nfc =    σc 

8 pd

C a W t  p d   C m

2

Face width


Guess value for F

F  3.059  in


p d 

2


2 2.5 12   in in in


4



F' pd F

1

p d  5  in

in



3

 


FL pd

F' p d F   3.059  in

2

in


 

FU pd in

1


d g p d   7.800  in

0

2

3

4

5

6

7

8

9 10 11 12

pd  in

FIGURE 13-37A Vt p d   1131

ft min

Wt p d   3939 lbf


σcp p d F   128455 psi

mp p d   1.45

mF  p d F   1.81

n r p d   0.45

n a p d F   0.81

mN  p d F   0.682

Lmin p d F   4.582  in

I  p d F   0.192



13-37-4


σcg p d F   Cp


Cv p d   F  d g p d   I  p d F 


Life factor

CL  2.466  N

CL  1

Reliability

CR  0.7  0.15 log( 1  R)

CR  1

Temperature factor

CT  1

Material surface strength (psi) Class 40 CI

S ac  80000  psi S fcg 

Endurance strength

S ac CL

S fcg  79999  psi

CT  CR


FU  p d  

FL  p d  

16 pd

8

Nfc =

pd

C a W t  p d   C m

Face width

Guess value for F

F  3.887  in p d 

2

2



 Sfc  σ   c


2 2.5 12   in in in 6




F' pd F in

p d  4  in



4.5

 

FL pd


3

in

F' p d F   3.887  in

 

FU pd


in

1.5

0


2

2.5

3

3.5

4

4.5

5

5.5

6

pd  in

d p p d   4.500  in

d g p d   9.750  in

FIGURE 13-37B

Vt p d   1414

Wt p d   3151 lbf


ft min

σcp p d F   93300  psi

mp p d   1.45

mF  p d F   1.796

n r p d   0.45



mN  p d F   0.684

13-37-5 Lmin p d F   5.662  in

I  p d F   0.191

23. The gear teeth are larger than for the pinion. This means that we will accept the gear requirements for the pinion thus, for the set 1

Diametral pitch

p d  4  in

Face width

F  3.875  in



 Sfcg  Nfsg     σcg pd F  

2

 Sfcp   σ p F   cp d  

2

Nfsg  1.6


Nfsp 

Nfsp  3.1



13-38-1


Size the helical gears in problem 13-34 for a surface factor of safety of at least 1.2 assuming a stead torque, 20 deg pressure angle, full depth teeth, quality index of 11, an AISI 4340 steel pinion, and A-7-d nodular iron gear.

Given:

Factor of safety

Nfc  1.2


Ng  79


H  30 kW n  1200 rpm


R  0.99 Qv  11


Np  20 ϕ  20 deg


N  10 ψ  30 deg

Solution:

7


Pinion 1.


Addendum

1 pd Np

Pitch radii

rp p d  

rg p d  

Center distance

C p d   rp p d   rg p d 

2 pd

Ng 2 pd

Radii of curvature

ρp  p d  




Base helix angle


 

ϕn  17.495 deg


 

ψb  28.481 deg




mp p d  

   

Z pd 

π pd


Axial contact ratio

mF  p d F  



π



13-38-2


Axial pitch

p x  p d  

p n  p d   p t  p d   cos( ψ)



return

return

2.


Load sharing ratio

mN  p d F  

Geometry factor

I  p d F  

F





Vt p d  



4.


d g p d  

pd

Ng pd

d p p d   n

Wt p d  


3.

Np

2 H

Vt  p d 

Ca  1 v

B  0.25  12  Qv

0.6667

B  0.25

A  50  56 ( 1  B) Cv p d  

A  92

A     min  A  Vt  p d    ft  

B

5.


Cm  1.6

6.

Choose an elastic coefficient from Table 12-18 (steel on NI).

Cp  2160 psi

7.


σcp p d F   Cp

0.5

C a W t  p d   C m

Cv p d   F  d p p d   I  p d F 



13-38-3


Reliability

CR  0.7  0.15 log( 1  R)

CL  1 CR  1

Temperature factor

CT  1


S ac  160000 psi S fcp 


 0.056

Life factor

S ac CL

S fcp  159997 psi

CT  CR


FU  p d  

FL  p d  

16 pd

 Sfc  Nfc =    σc 

8 pd

C a W t  p d   C m

2

Face width


Guess value for F

F  1.599  in


p d 

2


6 6.5 16   in in in


2



F' pd F

1

p d  10 in

in



1.5

 


FL pd

F' p d F   1.598  in

1

in


 

FU pd in

0.5


d g p d   7.900  in

0

6

7

8

9 10 11 12 13 14 15 16 pd  in

FIGURE 13-38A Vt p d   628 

ft min

Wt p d   2113 lbf


σcp p d F   144632 psi

mp p d   1.69

mF  p d F   2.986

n r p d   0.69

n a p d F   0.986

mN  p d F   0.52

Lmin p d F   3.123  in

I  p d F   0.246



13-38-4


σcg p d F   Cp


Cv p d   F  d g p d   I  p d F 


Life factor

CL  2.466  N

CL  1

Reliability

CR  0.7  0.15 log( 1  R)

CR  1

Temperature factor

CT  1

Material surface strength (psi) A-7-d nodular iron

S ac  100000 psi S fcg 

Endurance strength

S ac CL

S fcg  99998  psi

CT  CR


FU  p d  

FL  p d  

16 pd

8

Nfc =

pd

C a W t  p d   C m

Face width

Guess value for F

F  1.040  in p d 

2

2



 Sfc  σ   c


6 6.5 16   in in in 2




F' pd F in

p d  10 in



1.5

 

FL pd


1

in

F' p d F   1.041  in

 

FU pd


in

0.5

0


6

7

8

9 10 11 12 13 14 15 16 pd  in

d p p d   2.000  in

d g p d   7.900  in

FIGURE 13-38B

Vt p d   628 

Wt p d   2113 lbf


ft min

σcg p d F   88013  psi

mp p d   1.69

mF  p d F   2.067

n r p d   0.69



mN  p d F   0.527

13-38-5 Lmin p d F   2.135  in

I  p d F   0.243

23. The gear teeth are smaller than for the pinion. This means that we will accept the pinnion requirements for the gear thus, for the set 1

Diametral pitch

p d  10 in

Face width

F  1.625  in



 Sfcg  Nfsg     σcg pd F  

2

 Sfcp   σ p F   cp d  

2

Nfsg  1.9


Nfsp 

Nfsp  1.2



13-39-1

PROBLEM 13-39

_____

Statement:

A 90-deg straight bevel gearset is needed to give a 3:1 reduction. Determine the pitch cone angles, pitch diameters, and gear forces if the 25-deg pressure angle pinion has 15 teeth of p d = 4, and the transmitted power is 8 hp at 550 pinion rpm.

Given:

Power transmitted

H  8  hp

Gear ratio

mG  3

Pinion speed

ωp  550  rpm

Diametral pitch

p d  4  in

Teeth on pinion

Np  15

Pressure angle

ϕ  25 deg

Solution:


1.


2.


Gear

3.

d p 

d g 

Ng  45

d p  3.750  in

pd Ng

d g  11.250 in

pd

Pinion


Gear


1

 

αp  18.435 deg

 mG 

αg  71.565 deg

αp  αg  90.000 deg


Transmitted force 5.

Np

Ng  mG Np


Note that 4.

1

Tp 

H

Tp  916.7  in lbf

ωp

Wt 

2  Tp

Wt  488.9  lbf

dp


Gear

Total force


Wap  72.10  lbf


Wrp  216.29 lbf


Wag  216.29 lbf


Wrg  72.10  lbf

W  Wt ( cos( ϕ) )

1

W  539.468  lbf



13-40-1

PROBLEM 13-40

_____

Statement:

A 90-deg straight bevel gearset is needed to give a 6:1 reduction. Determine the pitch cone angles, pitch diameters, and gear forces if the 20-deg pressure angle pinion has 20 teeth of p d = 8, and the transmitted power is 3 kW at 900 pinion rpm.

Given:

Power transmitted

H  3  kW

Gear ratio

mG  6

Pinion speed

ωp  900  rpm

Diametral pitch

p d  8  in

Teeth on pinion

Np  20

Pressure angle

ϕ  20 deg

Solution:


1.


2.


Gear

3.

d p 

d g 

Ng  120

d p  2.500  in

pd Ng

d g  15.000 in

pd

Pinion


Gear


1

 

αp  9.462  deg

 mG 

αg  80.538 deg

αp  αg  90.000 deg


Transmitted force 5.

Np

Ng  mG Np


Note that 4.

1

Tp 

H

Tp  281.7  in lbf

ωp

Wt 

2  Tp

Wt  225.4  lbf

dp


Gear

Total force


Wap  13.49  lbf


Wrp  80.92  lbf


Wag  80.92  lbf


Wrg  13.49  lbf

W  Wt ( cos( ϕ) )

1

W  239.847  lbf



13-41-1

PROBLEM 13-41

_____

Statement:

A 90-deg spiral bevel gearset is needed to give an 8:1 reduction. Determine the pitch cone angles, pitch diameters, and gear forces if the 20-deg pressure angle pinion has 21 teeth of p d = 10, and the transmitted power is 2.5 kW at 1100 pinion rpm.

Given:

Power transmitted

H  2.5 kW

Gear ratio

mG  8

Pinion speed

ωp  1100 rpm

Diametral pitch

p d  10 in

Teeth on pinion

Np  21

Pressure angle

ϕ  20 deg

1

Assumptions: The spiral angle is ψ  35 deg. Solution:


1.


2.

Use equation 12.4a to calculate the pitch diameters. d p 

Pinion

d g 

Gear

3.

5.

d p  2.100  in

pd Ng

d g  16.800 in

pd

Pinion


Gear


1

 

 mG 

αp  7.125  deg αg  82.875 deg

αp  αg  90.000 deg


Tp 

Transmitted force

Wt 

H

ωp 2  Tp dp

Tp  192.1  in lbf

Wt  182.9  lbf

Use equations 13.2 to calculate the pressure angle in the normal plane.

ϕn  atan( tan( ϕ)  cos( ψ) ) 6.

Ng  168


Note that 4.

Np

Ng  mG Np

ϕn  16.602 deg

Use equations 13.8b to calculate the pinion and gear forces. Pinion Wap 

Wt cos( ψ)

  tan ϕn  sin αp  sin( ψ)  cos αp 

Wap  118.85 lbf



Wrp 

Wt cos( ψ)

13-41-2

  tan  ϕn  cos αp  sin( ψ)  sin αp 

Wrp  81.96  lbf

  tan ϕn  sin αg  sin( ψ)  cos αg 

Wag  50.18  lbf

  tan  ϕn  cos αg  sin( ψ)  sin αg 

Wrg  135.37 lbf

Gear Wag  Wrg 


Total force

W  Wt ( cos( ϕ) )

1

W  194.681  lbf



PROBLEM 13-42

13-42-1

_____

Statement:

A 1-start wormset has d = 2.00 in, p x = 0.25 in, mG = 40. Find the lead, lead angle, worm gear diameter, and center distance. Will it self-lock? The input speed is 1100 rpm.

Given:

Threads on worm

Nw  1

Worm pitch diameter

d  2.00 in

Gear ratio

mG  40

Axial pitch

p x  0.25 in

Input (worm) speed

n w  1100 rpm

Solution: 1.


Use equation 13.13 to calculate the lead. L  p x  Nw

2.

L  0.250  in



L

 

 π d 

3.

Calculate the number of teeth on the gear Ng  mG Nw

4.

d g 

p c  Ng

π

d g  3.183  in

Use equation 13.17 to calculate the center distance. C 

6.

Ng  40

Use equation 13.13 to calculate gear diameter. p c  p x

5.

λ  2.279  deg

d  dg 2

C  2.592  in


λt 

λ Nw

λt  2.279  deg

Self-locking



PROBLEM 13-43

13-43-1

_____

Statement:

A 2-start wormset has d = 2.50 in, p x = 0.30 in, mG = 50. Find the lead, lead angle, worm gear diameter, and center distance. Will it self-lock? The input speed is 1800 rpm.

Given:

Threads on worm

Nw  2

Worm pitch diameter

d  2.50 in

Gear ratio

mG  50

Axial pitch

p x  0.30 in

Input (worm) speed

n w  1800 rpm

Solution: 1.



2.

L  0.600  in



L

 

 π d 

3.


4.

d g 

p c  Ng

π

d g  9.549  in


6.

Ng  100


5.

λ  4.369  deg

d  dg 2

C  6.025  in


λt 

λ Nw

λt  2.184  deg

Self-locking



PROBLEM 13-44

13-44-1

_____

Statement:

A 3-start wormset has d = 60 mm, p x = 12 mm, mG = 60. Find the lead, lead angle, worm gear diameter, and center distance. Will it self-lock? The input speed is 2500 rpm.

Given:

Threads on worm

Nw  3

Worm pitch diameter

d  60 mm

Gear ratio

mG  60

Axial pitch

p x  12 mm

Input (worm) speed

n w  1800 rpm

Solution: 1.



2.

L  36.000 mm



L

 

 π d 

3.


4.

d g 

p c  Ng

π

d g  687.55 mm


6.

Ng  180


5.

λ  10.812 deg

d  dg 2

C  373.77 mm


λt 

λ Nw

λt  3.604  deg

Self-locking



PROBLEM 13-45

13-45-1

_____

Statement:

Determine the power transmitted and the torques and forces in the mesh for the wormset in Problem 13-42 if it runs at 800 worm rpm.

Units:


Given:

Threads on worm

Nw  1

Pressure angle

ϕ  20 deg

Worm pitch diameter

d w  2.00 in

Gear ratio

mG  40

Axial pitch

p x  0.25 in

Input (worm) speed

n w  800  rpm

Solution: 1.

1



2.

L  0.250  in



L

 

λ  2.279  deg

 π d w 

3.


4.

Ng  40


5.

p c  Ng

π

d g  3.183  in


6.

d g 

dw  dg

C  2.592  in

2


Fmax  1.340  in

7.


8.


Cm  0.0107 mG  56 mG  5145 9.

Cm  0.814


nw dw

Vt  419.2  fpm

2  cos( λ)


Cv  0.659  e

Vt fpm

Cv  0.416



13-45-2


 dg    in 


0.8

 Fmax

lbf in

Wtg  1144 lbf


 Vt   0.110    fpm  μ  0.103  e

0.450

 0.012

μ  0.031



Wf  38.4 lbf


Φ o 


Φ o  0.58 hp



Φ l  0.49 hp


Φ  Φ o  Φ l

Φ  1.07 hp


Φo

e  54.2 %

Φ


dg 2

Tg  1821 in lbf



PROBLEM 13-46

13-46-1

_____

Statement:

Determine the power transmitted and the torques and forces in the mesh for the wormset in Problem 13-43 if it runs at 1200 worm rpm.

Units:


Given:

Threads on worm

Nw  2

Pressure angle

ϕ  20 deg

Worm pitch diameter

d w  2.50 in

Gear ratio

mG  50

Axial pitch

p x  0.30 in

Input (worm) speed

n w  1200 rpm

Solution: 1.

1



2.

L  0.600  in



L

 

λ  4.369  deg

 π d w 

3.


4.

Ng  100


5.

p c  Ng

π

d g  9.549  in


6.

d g 

dw  dg

C  6.025  in

2


Fmax  1.675  in

7.


8.


Cm  0.0107 mG  56 mG  5145 9.

Cm  0.790


nw dw

Vt  787.7  fpm

2  cos( λ)


 Vt  Cv  13.31     fpm 

 0.571

Cv  0.295

11. Find the tangential load Wt from equation 13.23. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


 dg    in 

0.8


 Fmax

lbf in

13-46-2

Wtg  2375 lbf


   fpm  

 0.110 

μ  0.103  e

Vt

0.450

 0.012

μ  0.023



Wf  59.0 lbf


Φ o 


Φ o  4.32 hp



Φ l  1.41 hp


Φ  Φ o  Φ l

Φ  5.73 hp


Φo

e  75.4 %

Φ


dg 2

Tg  11342  in lbf



13-47-1

PROBLEM 13-47

_____

Statement:

A 2-start wormset has L = 2.00 in, C = 9.00 in, mG = 20, and the angle between the shafts is 90 deg. Find the pitch diameters of the worm and worm gear, the lead angle, and the axial pitch.

Given:

Threads on worm

Nw  2

Center distance

C  9.00 in

Worm lead

L  2.00 in

Gear ratio

mG  20

Solution: 1.


Determine the pitch diameter of the worm gear using equation 13.13. L Nw

=

π d g

L = π d g

Ng

Nw Ng

=

π d g mG

Solving for d g, d g 

2.

mG L

d g  12.732 in

π

Determine the pitch diameter of the worm using equation 13.17 solved for d w. d w  2  C  d g

3.

d w  5.268  in

Calculate the lead angle using equation 13.12.


L

 

 π d w 

4.

λ  6.891  deg

Calculate the axial pitch using equation 13.13. p x 

L Nw

p x  1.000  in



13-48-1

PROBLEM 13-48

_____

Statement:

A 5-start wormset has  = 20 deg, C = 2.75 in, Ng = 33, and the angle between the shafts is 90 deg. Find the pitch diameters of the worm and worm gear, the lead, and the axial pitch.

Given:

Threads on worm

Nw  5

Center distance

C  2.75 in

Worm lead angle

λ  20 deg

Teeth on worm gear

Ng  33

Solution: 1.


Determine the pitch diameter of the worm gear using equations 13.12, 13.13, and 13.17. tan ( λ) =

L Nw

=

L

L = π d w tan ( λ)

π d w

π d g

L = π d g

Ng

Nw Ng

Eliminating L and solving for d w, dw =

Nw Ng tan ( λ)

 dg

Substitute into equation 13.17 and solve for d g, dg = 2 C 

2.

Nw Ng tan ( λ)

1

Nw

d g  3.883  in

Ng tan ( λ)

d w  1.617  in

Calculate the lead using equation 13.12. L  π d w tan( λ)

4.

2 C

d g 

Determine the pitch diameter of the worm using equation 13.17 solved for d w. d w  2  C  d g

3.

 dg

L  1.848  in

Calculate the axial pitch using equation 13.13. p x 

L Nw

p x  0.370  in



13-49-1


Size the bevel gears in problem 13-40 for a bending factor of safety of at least 2.5 assuming a 5-year, 2-shift life, steady torque, quality index of 8, and an AISI 4140 steel pinion and gear.

Units:

yr  2080 hr

Given:

Factor of safety

Nfb  2.5


Ng  120


H  3  kW n p  900  rpm


R  0.99 Qv  8


Np  20

Life (years)


Solution:



1.


2.

Write the equations for pitch diameter, pitchline velocity, and transmitted torque in terms of the unknown diametral pitch, p d. Note that, in Mathcad, unit conversion factors are not included. Pitch diameter of pinion (in)



Np

d p p d   Vt p d   Tp 

d g p d  

pd

pd

d p p d   n p 2

H

Tp  281.728  in lbf

np


Ka  1


B  0.25  12  Qv

Ks  1

Kx  1

0.6667

B  0.63

A  50  56 ( 1  B) Kv p d  

A  70.721

A     min  A  Vt  p d    ft  

σbp p d F  

B

Km  1.6

5. Tentatively choose the mounting factor, Km (Assume 0 < F < 2 in)

6. The bending stress equation for the pinion is

Ng

2  Tp



pd



Ka Km Ks



shifts  2


np 2 π  0.0323

Life factor

KL  1.6831 N

Reliability

KR  0.7  0.15 log( 1  R)

9

N  1.123  10 KL  0.859 KR  1

Temperature factor

KT  1


S atp  40000  psi



13-49-2

S fbp 

Endurance strength

S atp KL

S fbp  34343  psi

KT  KR


Np




 Np     Ng  




S fb

Safety factor

Nfb =

Face width

F  p d  


p d 




Nfb


8 8.5 18   in in in


4

1

p d  10 in

 

3

F pd


in

F  p d   1.295  in

 2

FLover3 pd


in 1

F  1.375  in 13. Then, the parameters that depend on p d and F are: d p p d   2.000  in Vt p d   471 

0

8

10

12

ft

σbp p d F   12935  psi

16

18

pd  in

Kv p d   0.845

min

14


The assumption made in step 5 is correct so no further iteration is required.

GEAR 14. Determine the bending geometry factor, J (Figure 13-5)

Jg  0.245


σbg p d F  

2  Tp



pd



Ka Km Ks


16. Write the design equations using the face-width to pitch-cone length ratio given in the text, and the bending stress equation, solved for the unknown face width.



13-49-3 Np


F = L/3



 Np     Ng  




Safety factor

Nfb =

S fb


F  p d  

Face width




Nfb


14 14.5 24   in in in


2

1

p d  20 in

 

1.5

F pd


in

F  p d   1.01 in

 

FLover3 pd

1

in


0.5

F  1.000  in 0 14

21. Then, the parameters that depend on p d and F are: d g p d   6.000  in

16

18

20

22

24

pd  in

Kv p d   0.884 FIGURE 13-49B

Vt p d   236 

ft min

σbg p d F   13881  psi



Diametral pitch

p d  10 in

Face width

F  1.375  in

Pitch cone angle

αp  atan

Pitch-cone length

L 

 Np    Ng 

d p p d 

2  sin αp

αp  9.462  deg

L  6.083  in

L 3

 2.028  in


Nfbp 

S fbp


Nfbp  2.7


Nfbg 

S fbp


Nfbg  13.0



13-50-1


Size the bevel gears in problem 13-40 for a minimum safety factor of 1.8 for any mode of failure of pinion or gear assuming a 5-year, 2-shift life, steady torque, quality index of 8, and an AISI 4140 steel pinion and gear.

Units:

yr  2080 hr

Given:

Factor of safety

Nfc  1.8


Ng  120


H  3  kW n p  900  rpm


R  0.99 Qv  8


Np  20

Life (years)




1. Determine the surface geometry factor, I, from Figure 13-6.

2. Write the equations for pitch diameter, pitchline velocity, and transmitted load in terms of the unknown diametral pitch, p d. Note that, in Mathcad, unit conversion factors are not included. Pitch diameter of pinion (in)

d p p d  


Vtp p d  


Tp 

Np

d g p d  

pd Np 2 pd

Ng pd

 np

H

Tp  281.728  in lbf

np

3. Set the application factor, Ca

Ca  1

Cs  1

4. Write the equation for the dynamic load factor, Cv

B  0.25  12  Qv

Cf  1

0.6667

B  0.63

A  50  56 ( 1  B)

A  70.721

A  Cvp p d     min  A  Vtp pd    ft   5. Tentatively choose the mounting factor, Cmd

Cmd  3.0

6. Choose an elastic coefficient from Table 12-18 (steel on steel) and a stress adjustment constant.

Cp  2100 psi

B

Cxc  1 0.5

Cb  0.634


shifts  2


np 2 π  0.056

Life factor

CL  2.466  N

Reliability

CR  0.7  0.15 log( 1  R)

9

N  1.123  10 CL  0.768 CR  1



13-50-2

Hardness

CH  1

Temperature factor

CT  1


S acp  167500 psi S fcp 

Endurance strength

S acp CL

S fcp  128583 psi

CT  CR

8. Write the equation for the design pinion torque. I  Cvp p d 

 Sacp d p pd  0.774  CH  TD p d F       2 Cs Cmd  Cf  Ca Cxc  Cp Cb CT  CR  F

9. Define the exponent z.

z p d F  

2






z pd F

 

Ca Cmd

Cvp p d 




 Sfc  Nfc =    σc 

Np



 Np     Ng  




Face width



2

p d 

8 8.5 18   in in in



z pd F

 

Ca Cmd

Cvp p d 

2


F  1.21 in

Guess value for F 4


p d  10 in



F' pd F



3

in

14. The calculated value of F is F' p d F   1.217  in 15. Round this to the decimal equivalent of a common fractional value. F  1.250  in 16. Then, the parameters that depend on p d and F are:

 2

FLover3 pd in

1

0

8

10

12

14

16

18

pd  in


MACHINE DESIGN - An Integrated Approach, 4th Ed. d p p d   2.000  in

d g p d   12.000 in

13-50-3 FIGURE 13-50 Graph of Face Width and Limits for for Problem 13-50

Vtp p d   471 

ft min

Cvp p d   0.845

σcp p d F   95063  psi

TD p d F   714  in lbf

17. When both pinion and gear are of the same material, the stress in the gear will always be lower than in the pinion. Therefore, the face width and diametral pitch found for the pinion should also be used for the gear. 1

Diametral pitch

p d  10 in

Face width

F  1.250  in

Pitch cone angle

αp  atan

Pitch-cone length

L 

 Np    Ng 

αp  9.462  deg

d p p d 

L  6.083  in

2  sin αp

L 3

 2.028  in



Nfcp 

 Sfcp   σ p F   cp d  

2

Nfcp  1.8



13-51-1


Size the spiral gears in problem 13-41 for a bending factor of safety of at least 2.0 assuming a 7-year, 3-shift life, steady torque, quality index of 8, an AISI 4340 steel pinion and gear.

Units:

yr  2080 hr

Given:

Factor of safety

Nfb  2.0


Ng  168

Power to be transmitted (kW) Rotational speed of pinion (rpm)

Reliability H  3  kW n p  1100 rpm AGMA Quality level

R  0.99 Qv  8


Np  21


Solution:

Life (years)


Pinion 1. Determine the bending geometry factor, J (Figure 13-8)

Jp  0.360

(assumed)

2. Write the equations for pitch diameter, pitchline velocity, and transmitted torque in terms of the unknown diametral pitch, p d. Note that, in Mathcad, unit conversion factors are not included. Pitch diameter of pinion (in)



d p p d   Vt p d   Tp 

Np

d g p d  

pd

pd

d p p d   n p 2

H

Tp  230.505  in lbf

np


Ka  1


B  0.25  12  Qv

Ks  1

Kx  1.15

0.6667

B  0.63

A  50  56 ( 1  B)

A  70.721

A  Kv p d     min  A  Vt  p d    ft  

σbp p d F  

B

Km  1.6

5. Tentatively choose the mounting factor, Km (Assume 0 < F < 2 in) 6. The bending stress equation for the pinion is

Ng

2  Tp



pd



Ka Km Ks



shifts  3


np 2 π  0.0323

Life factor

KL  1.6831 N

Reliability

KR  0.7  0.15 log( 1  R)

9

N  2.883  10 KL  0.833 KR  1

Temperature factor

KT  1


S atp  42000  psi



13-51-2

S fbp 

Endurance strength

S atp KL

S fbp  34979  psi

KT  KR


Np




 Np     Ng  




S fb

Safety factor

Nfb =

Face width

F  p d  


p d 


4

 

3

F pd

1

p d  16 in

Nfb


2 2.5 20   in in in




in

 2

FLover3 pd


in 1

12. Round this to the decimal equivalent of a common fractional value. 0

F  1.500  in

2

4

6

8

10 12 14 16 18 20 pd  in


Kv p d   0.858


ft min


Jg  0.360


σbg p d F  

2  Tp



pd



Ka Km Ks




13-51-3 Np


F = L/3



 Np     Ng  




Safety factor

Nfb =

S fb


F  p d  

Face width




Nfb


10 11.5 50   in in in


2

 

1.5

F pd

1

p d  32 in

in

 


FLover3 pd

F  p d   0.694  in

in

1

0.5

20. Round this to the decimal equivalent of a common fractional value. 0 10 15

F  0.750  in

20 25

30

35 40

45 50

pd  in


ft min

FIGURE 13-51B

Kv p d   0.894


σbg p d F   16194  psi


Diametral pitch

p d  16 in

Face width

F  1.500  in

Pitch cone angle

αp  atan

Pitch-cone length

L 

 Np    Ng 

d p p d 

2  sin αp

αp  7.125  deg

L  5.291  in

L 3

 1.764  in


Nfbp 

S fbp


Nfbp  2.1


Nfbg 

S fbp


Nfbg  16.6



14-1-1


A linear spring is to give 200 N at its maximum deflection of 150 mm and 40 N at its minimum deflection of 50 mm. What is the spring rate?

Given:

Working force

Fwork  200  N

Initial force

Finit  40 N

Working deflection

ywork  150  mm

Initial deflection

yinit  50 mm

Solution: 1.


Spring rate is the slope of the load vs. deflection function. Thus, for a linear spring Spring rate

k 

Fwork  Finit ywork  yinit

k  1.60

N mm



14-2-1


Find the ultimate tensile strength, the ultimate shear strength, and the torsional yield strength of a 1.8-mm-dia, A229 oil-tempered steel wire.

Given:

Wire diameter

Solution:


1.

d  1.8 mm

From Table 14-4, for A229 A  1831.2 MPa

2.

Using equation 14.3, Tensile strength

3.

S ut  A  

  mm   d

b

S ut  1644 MPa

From Table 14-6, with a factor of 0.50, Maximum torsional yield strength

4.

b  0.1833

S ys  0.50 S ut

S ys  822  MPa

S us  0.67 S ut

S us  1102 MPa

Using equation 14.4, Ultimate shear strength



14-3-1


Find the torsional yield and ultimate shear strength of an 0.105-in-dia, unset A230 wire to be used in a helical compression spring.

Given:

Wire diameter

Solution:


1.

d  0.105  in

ASTM A230 is not listed in Table 14-4, but it is shown in Figure 14-3. At the wire diameter given, it has a tensile strength that is approximately the same as that for A229. From Table 14-4, for A229 A  146.78 ksi

2.

S ut  A  

   in  d

b

S ut  222  ksi


4.

b  0.1833

Using equation 14.3, Tensile strength

3.

d  2.667  mm

S ys  0.50 S ut

S ys  111  ksi

S us  0.67 S ut

S us  149  ksi




14-4-1


What is the torsional fatigue strength of the wire in Problem 14-3 at N = 5E6 cycles?

Given:

Wire diameter

d  0.105  in

Cycle life

N  5  10

Solution: 1.

d  2.667  mm

6


ASTM A230 is not listed in Table 14-4, but it is shown in Figure 14-3. At the wire diameter given, it has a tensile strength that is approximately the same as that for A229. From Table 14-4, for A229 A  146.78 ksi

b  0.1833

2. Using equation 14.3, Tensile strength 3.

S ut  A  

   in  d

b

S ut  222  ksi

The required life falls between the values given in Table 14-7 for 1E6 and 1E7 cycles. We will use an exponential fit to the data for interpolation. Using equation 6.10

 6 b 7 = a   10  b

4.

At 1E6 cycles

0.40 S ut = a  10

At 1E7 cycles

0.38 S ut

Solving both equations for a, equating the results and solving for b 0.40 S ut

106

0.38 S ut

=

107

b

b

b

 107    = 0.38 0.40  106    b  log 

0.38 

b  log( 10) = log

  0.40 

1

  0.40  log( 10)

a 

5.

0.40 S ut

106

b

0.38 

b  0.022

a  120.726  ksi

Interpolating, Fatigue strength at 5E6 cycles

S fs  a  N

b

S fs  85.6 ksi



14-5-1


Draw the Modified Goodman diagram for the wire of Problem 14-3.

Given:

Wire diameter

d  0.105  in

Cycle life

N  5  10

Solution: 1.

ASTM A230 is not listed in Table 14-4, but it is shown in Figure 14-3. At the wire diameter given, it has a tensile strength that is approximately the same as that for A229. From Table 14-4, for A229 b  0.1833

Using equation 14.3, S ut  A  

   in 

Tensile strength 3.

S ut  221.9  ksi

S ys  0.50 S ut

S ys  110.9  ksi

S us  0.67 S ut

S us  148.6  ksi

Using Table 14-7 and a cycle life of 1E6, for unpeened wire Torsional fatigue strength (R=0)

6.

b


5.

d


4.

6


A  146.78 ksi 2.

d  2.667  mm

S fw  .40 S ut

S fw  88.7 ksi

From equation (d) in Example 14-2, S fs  0.5

S fw S us

S fs  63.3 ksi

S us  0.5 S fw S fs

7.

Now, the equation for the Goodman line is

τa τm  

8.

Plotting this over the range

τm  0  ksi 10 ksi  150  ksi

S us

 τm  S fs


100 80 60 40 20 0

0

25

50

75

100

125

150

Mean shear stress



14-6-1


What are the spring rate and spring index of a squared and ground compression spring with the data given below?

Given:

Wire diameter

d  1  mm

Mean coil dia

D  10 mm

Total coils

Nt  12

Assumptions: The spring wire is steel so that G  80.8 GPa. Solution:


1.

From Figure 14-9, the number of active coils is

2.

Using equation 14.7,

Na  Nt  2

Na  10

4

Spring rate

k 

d G 3

k  1.01

8  D  Na 3.

N mm

Using equation 14.5, Spring Index

C 

D d

C  10



14-7-1


Find the natural frequency of the spring in Problem 14-6.

Given:

Wire diameter

d  1  mm

Mean coil dia

D  10 mm

Nt  12

Total coils 3

Assumptions: The spring wire is steel so that G  80.8 GPa and γ  0.28 lbf  in Solution:

3

, γ  76005  N  m


1.


2.

Using equation 14.7,

Na  Nt  2

Na  10

4

Spring rate

d G

k 

k  1.01

3

8  D  Na 3.

mm


4.

N

C 

D

C  10

d

Calculate the natural frequency using equation 14.11c. Natural frequency

fn 

2

π N a



d 2

D



G g 32 γ

fn  363.4  Hz



14-8-1


A paper mill processes rolls of paper having a density of 984 kg/m3. The paper roll is 1.50 m outside dia (OD) by 0.22 m inside dia (ID) by 3.23 m long and is on a simply supported, hollow, steel shaft with 22-cm OD x 20-cm ID and as long as the paper roll. Find the spring rate of the shaft and the fundamental natural frequency of the shaft-roll assembly.

Given:

Paper density

ρ  984 

kg

Roll dimensions:

3

m od  220  mm id  200  mm E  207  GPa

Shaft outside dia Shaft inside dia Young's modulus

Outside diameter Inside diameter Length

OD  1.50 m ID  0.22 m L  3.23 m

Assumptions: The shaft (beam) supporting the paper roll is simply-supported at the ends and is the same length as the paper roll. The paper acts as a distributed load over the length of the shaft. Solution: 1.


The mass of the paper roll is equal to its volume times the paper density. mroll 

 4

π

2

2



 OD  ID  L ρ

mroll  5495.74  kg I 

π



4



4

 od  id

7

I  3.645  10  mm

4

2.

The area moment of inertia of the shaft is

3.

The spring rate (stiffness) of the shaft can be found from the deflection equation in Figure B-2(b) in Appendix B When the distributed load covers the entire span, the maximum deflection is 4

ymax =

5  w L

384  E I k 

Solving for k,

3

5 W  L

=

384  E I

384  E I 3

k=

64

W y

=

384  E I 3

5 L

4 N

k  1.720  10 

5 L 4.

mm

Use equations 3.4 to find the natural frequency of the system.

ωn 

fn 

k mroll

ωn  55.9

rad sec

ωn 2 π

fn  8.90 Hz



14-9-1


Determine the minimum allowable bending radius for an 50 HRC strip steel spring of 1-mm thickness.

Given:

Thickness

Solution:


t  1  mm

1.

From Table 14-5, the bend factor for spring steel with a 50 HRC is bf  5

2.

Use the bend factor definition given in the text to calculate the minimum bend radius. rmin 

bf  t 2

rmin  2.5 mm



14-10-1


An over-hung diving board is shown in Figure P14-1a. A 100-kg person is standing at the free end Assume cross-section dimensions of 305 mm x 32 mm and a material E = 10.3 GPa. What is the spring rate and fundamental natural frequency of the diver-board combination? 2000 = L

Given:

W  100  kgf

Weight of person Board dimensions: Distance to support Length of board Cross-section

R1

a  0.7 m L  2  m w  305  mm t  32 mm E  10.3 GPa

Young's modulus

R2 700 = a

Assumptions: The weight of the board is negligible compared to the applied load and so can be ignored. Solution:

P


See Figure 14-10 and Mathcad file P1410. I 

w t

3

5

I  8.329  10  mm

4

1.

The area moment of inertia of the board is

2.

The spring rate (stiffness) of the board can be found from the deflection equation in Figure B-3(a) in Appendix B. When the load is at the end of the beam, the maximum deflection is ymax =

F L 3  E I k 

Solving for k,

 ( L  a)

2

3.

k=

3  E I L ( L  a )

12

2

F y

3  E I

=

L ( L  a )

k  7.614 

2

N mm


ωn 

fn 

k g W

ωn  8.73

rad sec

ωn 2 π

fn  1.39 Hz



14-11-1


Design a helical compression spring for a static load of 45 lb at a deflection of 1.25 in with a safety factor of 2.5. Use C = 7.5. Specify all parameters necessary to manufacture the spring.

Given:

Working force Working deflection

Fwork  45 lbf ywork  1.25 in

Spring index

C  7.5

Design choices: Clash allowance

α  0.15

G  11.5 10  psi Ns  2.5

Set removed

Km  0.65

A  184.649  ksi b  0.1625

ASTM A228 wire Solution:

6

Shear modulus Safety factor

See Mathcad file P1411. k 

Fwork

k  36

lbf

1.

Determine the desired spring rate.

2.

Use the design equation from Example 14-3A (Mathcad Supplement) to determine the wire diameter.

ywork

in

1

d 

2 b  8 Ns ( C  0.5)  Fwork  ( 1  α)  in 2   π Km A  in  

Wire diameter 3.

d  0.125  in

Let

d  0.125  in

Calculate the mean coil diameter and number of active coils. Mean coil dia

D  C d

Number of active coils

Na 

D  0.938  in

4

d G 3

8 D  k

Na  11.831 Na  11.75

Note that we round it to the nearest 1/4 coil as the manufacturing tolerance cannot achieve better than that accuracy. We must now calculate the actual (corrected) spring rate: Corrected spring rate 4.

7.

3

k  36.249

8  D  Na

lbf in

Nt  Na  2

Nt  13.75

The shut height can now be determined. Shut height

6.

d G

Assume squared and ground ends making the total number of coils, from Figure 14-9: Total coils

5.

4

k 

Ls  d  Nt

Ls  1.719  in

The free length (see Figure 14-8) can now be found from Deflection to shut height

yshut  ywork  α ywork

yshut  1.438  in

Free length

Lf  Ls  yshut

Lf  3.156  in

To check for buckling, two ratios need to be calculated, Lf/D and y max/Lf.



Deflection ratio

sr  y' 

14-11-2

Lf

sr  3.367

D ywork

y'  0.396

Lf

Take these two values to Figure 14-14 and find that their coordinates are safely within the zones that are stable against buckling for either end-condition case. 8.

9.

The inside and outside coil diameters are Inside coil dia

Di  D  d

Di  0.813  in

Outside coil dia

Do  D  d

Do  1.063  in

The smallest hole and largest pin that should be used with this spring are Smallest hole

holemin  Do  0.05 D

holemin  1.11 in

Largest pin

pin max  Di  0.05 D

pin max  0.77 in

10. The total weight of the spring is Weight density

3

ρ  0.28 lbf  in 2 2

Weight

Wt 

π  d  D Nt ρ

Wt  0.14 lbf

4

11. We now have a complete design specification for this A228 wire spring: Wire diameter

d  0.125  in

Outside diameter

Do  1.063  in

Total coils

Nt  13.75

Free length

Lf  3.156  in

ends squared and ground



14-12-1


Repeat Problem 14-10 using the cantilevered diving board design in Figure P14-1b. 2000 1300 = L

Given:

Weight at free end Board dimensions: Length of board Cross-section Young's modulus

P  100  kgf P

L  1.3 m w  305  mm t  32 mm E  10.3 GPa

M1

R1

700

Assumptions: The weight of the board is negligible compared to the applied load and so can be ignored. Solution:


See Figure 14-12 and Mathcad file P1412. I 

w t

3

5

I  8.329  10  mm

4

1.

The area moment of inertia of the board is

2.

The spring rate (stiffness) of the board can be found from the deflection equation in Figure B-1(a) in Appendix B. When the load is at the end of the beam, the maximum deflection is

12

3

F L

ymax =

Solving for k,

k 

3  E I

3  E I 3

k=

F y

=

k  11.71 

L 3.

3  E I 3

L

N mm


ωn 

fn 

k g P

ωn 2 π

ωn  10.82 

rad sec

fn  1.72 Hz



14-13-1


Given:

Solution: 1.

For the data given below, find Na, D, Lf, Lshut, yinitial, and the minimum hole diameter for the spring. Infinite life is desired with a safety factor of 1.4. Choose an acceptable spring index. Setting will be done. Minimum force

Fmin  50 lbf

Maximum force

Fmax  250  lbf

Diameter

Working deflection Clash allowance

Δy  0.75 in α  0.15

Unpeened Set after winding

Shear modulus

G  11.5 10  psi

Squared ends

Find the the mean and alternating forces from equation 14.15a:

Mean force

3.

6

d  0.312  in


Alternating force

2.

Chrome-vanadium wire properties:

Fa  Fm 

Fmax  Fmin

Fa  100.0  lbf

2 Fmax  Fmin

Fm  150.0  lbf

2

Assume a spring index (found by trial and error to get the correct safety factor) and calculate the mean coil diameter D from equation 14.5. Spring index

C  4.4

Mean coil diameter

D  C d

D  1.373  in

Find the direct shear factor Ks and use it to calculate the shear stress i at the initial deflection (lowest defined force), and the mean stress m. Direct shear factor

Stress at Fmin

0.5

Ks  1 

τi  Ks

Ks  1.114

C

8  Fmin D

π d Stress at Fm

τm  Ks

8  Fm D

π d 4.

τm  19.2 ksi

3

Find the Wahl factor Kw and use it to calculate the alternating shear stress a in the coil. Wahl factor

Alternating stress

Kw 

4 C  1 4 C  4

τa  Kw



0.615 C

8  F a D

π d 5.

τi  6.4 ksi

3

Kw  1.36

τa  15.7 ksi

3

Find the ultimate tensile strength of this wire material from equation 14.3 and Table 14-4 and use it to find the ultimate shear strength from equation 14.4 and the torsional yield strength from Table 14-6, assuming that the set has been removed and using the low end of the recommended range. From Table 14-4, for A228 music wire Ultimate tensile strength

A  173.128  ksi S ut  A  

  in   d

b  0.1453

b

S ut  205.1  ksi



6.

7.

Shear yield strength

S ys  0.65 S ut

S ys  133.3  ksi


S us  0.67 S ut

S us  137.4  ksi

Find the wire endurance limit for unpeened springs in repeated loading from equation 14.12 and convert it to fu reversed endurance strength with equation 14.16c. Wire endurance limit

S ew  45 ksi

Fully reversed endurance limit

S es  0.5

S es  26.91  ksi

S us  0.5 S ew

Nfs 

S es  S us  τi

Nfs  1.41

S es  τm  τi  S us τa

The spring rate is defined from the two specified forces at their relative deflection. Spring rate

9.

S ew S us

The safety factor is calculated from equation 14.16b. Fatigue factor of safety

8.

14-13-2

k 

Fmax  Fmin

k  266.7 

Δy

lbf in

To get the defined spring rate, the number of active coils must satisfy equation 14.7, solving for Na yields: Number of active coils

4

Na 

d G 3

Na  19.744

Na  19.75

8 D  k

Note that we round it to the nearest 1/4 coil as the manufacturing tolerance cannot achieve better than that accuracy. Having rounded the number of active coils, we must now calculate the spring rate using equation 14. Corrected spring rate

4

d G

k 

3

k  266.59

8  D  Na

lbf in

10. For squared ends the total number of coils, from Figure 14-9: Total coils

Nt  Na  2

Nt  21.75

11. The shut height can now be determined. Shut height

Lshut  d  Nt

Lshut  6.786  in

12. The initial deflection to reach the smaller of the two loads is Initial deflection

yinit 

Fmin k

yinit  0.188  in

13. For the given clash allowance of 15% of the working deflection: Clash allowance

Δyclash  α Δy

Δyclash  0.112  in

14. The free length (see Figure 14-8) can now be found from Lf  Lshut  Δyclash  Δy  yinit

Lf  7.836  in

15. The deflection to the shut height is © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


14-13-3

yshut  Lf  Lshut

yshut  1.05 in

16. The force at this shut height is Fshut  k yshut

Fshut  279.9  lbf

17. The shut-height stress and safety factor are Stress at Fshut

τshut  Ks

8  Fshut D

π d Safety factor at working deflection

Ns 

3

S ys

τshut

τshut  35.9 ksi

Ns  3.71

which is acceptable. 18. To check for buckling, two ratios need to be calculated, Lf/D and ymax /Lf. Slenderness ratio

Deflection ratio

sr  y' 

Lf

sr  5.708

D yinit  Δy Lf

y'  0.12

Take these two values to Figure 14-14 and find that their coordinates are safely within the zones that are stable against buckling for either end-condition case. 19. The inside and outside coil diameters and minimum hole dia are Inside coil dia

Di  D  d

Di  1.061  in

Outside coil dia

Do  D  d

Do  1.685  in

Minimum hole dia

d hole  Do  0.05 D

d hole  1.753  in

20. Summarizing the results: Number of active coils

Nt  21.75

Mean coil diameter

D  1.373  in

Free length

Lf  7.836  in

Shut length

Lshut  6.786  in

Deflection to Fmin

yinit  0.188  in

Minimum hole dia

d hole  1.753  in



14-14-1


Figure P14-2 shows a child's toy called a pogo stick. The child stands on the pads, applying half her weight on each side. She jumps off the ground, holding the pads up against her feet, and bounces along with the spring cushioning the impact and storing energy to help each rebound. Assume a 60-lb child and a spring constant of 100 lb/in. Design the helical compression spring to survive jumping 2 in off the ground with a dynamic safety factor of 2 for a finite life of 5E4 cycles Determine the fundamental natural frequency of the system.

Given:

Weight of child

W  60 lbf

Spring life

Ncycles  5  10

4

Design Choices: Clash allowance α  0.15 Spring index C  7 Squared and ground ends Design safety factor Nfsd  1.5 Solution:

1

Spring constant

k  100  lbf  in

Shear modulus

G  11.5 10  psi

6

Music wire properties: Strength A  184.65 ksi Set after winding, unpeened

b  0.1625


1.

From Problem 3-14,

Maximum force

Fmax  224.2  lbf

2.

Let the force of holding the pads against the feet be

Minimum force


3.

Then, the working deflection is

4.

Find the the mean and alternating forces from equation 14.15a: Alternating force Mean force

5.

6.

Δy 

Fa  Fm 

Fmax  Fmin

Fmax  Fmin k

Δy  2.042  in

Fa  102.1  lbf

2 Fmax  Fmin

Fm  122.1  lbf

2

Calculate the factors necessary to find the wire diameter. Direct shear factor

Ks  1 

Wahl factor

Kw 

0.5

Ks  1.071

C

4 C  1 4 C  4



0.615

Kw  1.213

C

Yield strength factor

Kys  0.60

(Table 14-6, set removed)

Fatigue strength factor

Kfw  0.37

(Table 14-7, unpeened)

Ultimate shear strength factor

KU  0.67

(equation 14.4)

Solving for the wire diameter using an equation similar to equation g in Example 14-4A (Mathcad Supplement) where S fw is used in equation 14.16b instead of S es and S fw = Kfw S ut. 1

Nfsd  1  8 C Nfsd   d    K s  Fm   Ks Fmin  Nfsd  π KU  A  in2     2 K    U Kfw    K w  Fa      Kfw     

2 b

 in


MACHINE DESIGN - An Integrated Approach, 4th Ed. d  0.281  in 7.

8.

d  0.281  in

Let the wire diameter be

Calculate the ultimate tensile strength, ultimate shear strength, and the fatigue strength at 5E4 cycles.

  in  

b


S ut  A  


S us  KU  S ut

S us  152  ksi

Wire fatigue strength

S fw  Kfw S ut

S fw  84.0 ksi


S ys  Kys S ut

S ys  136  ksi

Fatigue strength

S fs  0.5

d

S ut  227  ksi

S fw S us S us  0.5 S fw

S fs  58.0 ksi

Calculate the mean coil diameter D from equation 14.5. Mean coil diameter

9.

14-14-2

D  C d

D  1.967  in

Calculate the shear stress i at the initial deflection (lowest defined force), and the mean stress m. Stress at Fmin

τi  Ks

8  Fmin D


τm  Ks

τi  4.8 ksi

3

8  Fm D

π d

τm  29.5 ksi

3

10. Calculate the alternating shear stress a in the coil. Alternating stress

8  F a D

τa  Kw

π d

τa  28.0 ksi

3

11. The safety factor is calculated from equation 14.16b. Fatigue factor of safety

Nfs 

S fs  S us  τi

Nfs  1.50

S fs  τm  τi  S us τa

12. To get the defined spring rate, the number of active coils must satisfy equation 14.7, solving for Na yields: Number of active coils

4

Na 

d G 3

Na  11.777

Na  11.75

8 D  k

Note that we round it to the nearest 1/4 coil as the manufacturing tolerance cannot achieve better than that accuracy. Having rounded the number of active coils, we must now calculate the spring rate using equation 14.

Corrected spring rate

4

k 

d G 3

k  100.23

8  D  Na

lbf in

13. For squared and ground ends the total number of coils, from Figure 14-9: Total coils

Nt  Na  2

Nt  13.75



14-14-3



Lshut  3.864  in


Fmin

yinit 

yinit  0.200  in

k

16. For the given clash allowance factor: Clash allowance




Lf  6.412  in

18. The deflection to the shut height is yshut  Lf  Lshut

yshut  2.548  in




τshut  Ks

8  Fshut D

π d Safety factor at shut height

Ns 


3

S ys

Ns  2.20

τshut


Deflection ratio

sr  y' 

Lf

sr  3.26

D yinit  Δy

y'  0.35

Lf

Take these two values to Figure 14-14 and find that their coordinates are safely within the zones that are stable against buckling for either end-condition case. 22. The inside and outside coil diameters and minimum hole dia are Inside coil dia

Di  D  d

Di  1.686  in

Outside coil dia

Do  D  d

Do  2.248  in

Maximum pin dia

d pin  Di  0.05 D

d pin  1.588  in


Nt  13.75

Deflection to Fmin

yinit  0.2 in

Mean coil diameter

D  1.967  in

Maximum pin dia

d pin  1.588  in

Free length

Lf  6.412  in

Shut length

Lshut  3.864  in



14-15-1


Draw the Modified Goodman diagram for the spring data below and find its safety factor.

Given:

Fatigue strength (R=0)

S fw  40 ksi



S us  200  ksi

Mean shear stress τm  95 ksi Initial shear stress

Solution: 1.

3.

τi  75 ksi


From equation (d) in Example 14-2, S fw S us

S fs  0.5

2.

τa  12 ksi

S fs  22.2 ksi

S us  0.5 S fw

S fs

Now, the equation for the Goodman line is

τ'a τ'm  

Equation for load line

τaL τ'm 


τ'm  0  ksi 10 ksi  200  ksi


25

S us

τa τm

 τ'm  S fs

 τ'm

τm ksi

20 15

τa ksi

10 5 0

0

25

50

75

100

125

150

175

200

Mean shear stress

4.

Use equation 14.16b to calculate the safety factor.

Nfs 

S fs  S us  τi

S fs  τm  τi  S us τa

Nfs  0.98



14-16-1


Problem 6-16 describes a track for bowling balls that are 4.5-in dia and 2.5-lb weight. Design a spring-loaded launcher that will allow quadriplegic bowlers to launch the balls down the bowling alley from the point where the track of Problem 6-16 drops them with only a switch closure that releases the launcher. The launcher's plunger will be cocked by an assistant and the energy stored in the helical compression spring, which you will design, will drive the plunger into the ball and roll it down the bowling alley.

Given:

Ball weight

Wball  2.5 lbf

Design Choices: Clash allowance Spring index Design safety factor

Solution: 1.

α  0.10

vi  150  in sec

Deflection to Fmax

Δy  5.000  in

Squared and ground ends

Determine the required spring rate by equating the spring potential energy with the ball's initial velocity when the spring force reaches zero. 1 1 2 2 PE = KE  k Δy =  mball  vi 2 2 Wball

mball  6.475  10

g 2

k  5.828 

 3 lbf  sec



2

in

lbf in

Determine the maximum spring force and set the minimum force to zero. Fmin  0  lbf

Fmax  k Δy

Fmax  29.138 lbf

Find the the mean and alternating forces from equation 14.15a: Alternating force

Mean force 4.

1


 vi  k  mball     Δy 

3.

Music wire properties: Strength A  184.65 ksi b  0.1625 Set after winding, unpeened

C  12 Nfsd  2

Ball initial velocity

mball 

2.

6

G  11.5 10  psi

Shear modulus

Fa  Fm 

Fmax  Fmin

Fa  14.6 lbf

2 Fmax  Fmin

Fm  14.6 lbf

2


Ks  1 

Wahl factor

Kw 

0.5

Ks  1.042

C

4 C  1 4 C  4



0.615 C

Kw  1.119


Kys  0.60



KU  0.67

(equation 14.4)

Torsional endurance limit


(unpeened)



14-16-2

Solve for the wire diameter using equation g in Example 14-4A (Mathcad Supplement). Guess

d  0.2 in

Given

1

Nfsd  1  8 C Nfsd  d=  Ks Fm   Ks Fmin  Nfsd 2  0.67 π A  in  b     d A     0.67  in     1  Kw Fa       0.50 S ew  d  Find ( d ) 6.

7.

Let the wire diameter be

 in

d  0.207  in

   in 

b


S ut  A  



S us  160  ksi



S ys  143  ksi

Fatigue strength

S es  0.5

d

S ut  239  ksi

S ew S us S us  0.5 S ew

S es  26.2 ksi

Calculate the mean coil diameter D from equation 14.5. D  C d

D  2.484  in


τi  Ks

8  Fmin D


τm  Ks

τi  0.0 ksi

3

8  Fm D

π d 9.

2 b

Calculate the ultimate tensile strength, ultimate shear strength, and the shear endurance limit.

Mean coil diameter 8.

d  0.209  in

      

τm  10.8 ksi

3

Calculate the alternating shear stress a in the coil. Alternating stress

8  F a D

τa  Kw

π d

τa  11.6 ksi

3

10. The safety factor is calculated from equation 14.16b. Fatigue factor of safety

Nfs 



Nfs  2.0


4

Na 

d G 3

Na  29.549

Na  29.5

8 D  k



14-16-3


4

d G

k 

k  5.84

3

lbf

8  D  Na

in


Nt  Na  2

Nt  31.50



Lshut  6.52 in


Fmin

yinit 

yinit  0.000  in

k





Lf  12.021 in


yshut  5.500  in


Fshut  32.1 lbf


τshut  Ks

8  Fshut D


Ns 

3

S ys

τshut


Ns  6.00

which is acceptable. 20. To check for buckling, two ratios need to be calculated, Lf/D and ymax /Lf. Slenderness ratio Deflection ratio

sr  y' 

Lf D yinit  Δy Lf

sr  4.839 y'  0.416

Take these two values to Figure 14-14 and find that their coordinates are safely within the zone that is stable against buckling for parallel ends. 21. The inside and outside coil diameters and minimum hole dia are Inside coil dia

Di  D  d

Di  2.277  in

Outside coil dia

Do  D  d

Do  2.691  in

Maximum pin dia

d pin  Di  0.05 D

d pin  2.153  in



14-16-4


Nt  31.50

Mean coil diameter

D  2.484  in

Free length

Lf  12.021 in

Shut length

Lshut  6.520  in

Deflection to Fmin

yinit  0  in

Maximum pin dia

d pin  2.153  in



14-17-1


Design a helical extension spring to handle a dynamic load that varies from 175 N to 225 N over 8.5 mm working deflection. Use music wire and standard hooks. The forcing frequency is 1500 rpm. Infinite life is desired. Minimize the package size. Choose appropriate safety factors against fatigue, yielding, and surging.

Given:

Minimum force

Fmin  175  N

Life

L  ∞

Maximum force

Fmax  225  N

Shear modulus

G  80.8 GPa

Working deflection

Δy  8.5 mm

Forcing freq

ωf  1500 rpm

Nfb  1.5

Spring index

C  8

S ew  310  MPa

Wire strength

A  2153.5 MPa

Design Choices: Fatigue safety factor Wire endurance limit

b  0.1625 Solution: 1.


Find the mean and alternating loads. Fa  Fm 

2.

Fmax  Fmin

Fa  25.0 N

2 Fmax  Fmin

Fm  200.0  N

2

Calculate the hook bending factor. 2

Kb  3.

4 C  C  1

Kb  1.103

4 C ( C  1)

Solve for d by iteration using an equation derived from the safety factor equation for bending in the hooks. 1

d=

 4   4  Kb C  1   Nfb Fm   Nfb  1   Fmin  π A    0.577   0.67 A  db  0.5 S ew       Nfb Fa    0.5 0.67 S ew   

     

2 b

1

 4  4 Kb C  1 RHS( d )    Nfb Fm   Nfb  1   Fmin   π A  mm2  b    d       0.5 S ew  0.577  0.67 A         mm    Nfb Fa     0.5 0.67 S ew    

4.

d  1.000  mm

RHS( d )  3.561  mm

d  RHS( d )

RHS( d )  3.329  mm

d  RHS( d )

RHS( d )  3.340  mm

d  RHS( d )

RHS( d )  3.339  mm

d  3.5 mm

D  C d

D  28 mm

      

2 b

 mm

Calculate the mean coil diameter. Mean coil diameter



Use the assumed value of C to find an appropiate value of initial coil stress i from equations 14.21: 3 2   3 2   2.987  C  139.7  C  3427 C  38404   psi

τi1  4.231  C  181.5  C  3387 C  28640  psi

τi1  75.8 MPa

τi2

τi2  126.9  MPa

τi  6.

τi1  τi2

τi  101.3  MPa

2

Find the direct shear factor Ks : Direct shear factor

7.

14-17-2

0.5

Ks  1 

Ks  1.063

C

Use the value of i from (c) in equation 14.8 to find the corresponding initial coil-tension force Fi: 3

Fi 

π d  τi 8  K s D

Fi  57.3 N

Check that this force is less than the required minimum applied force Fmin, which in this case, it is. Any applied force smaller than Fi will not deflect the spring. 8.

Use the direct shear factor Ks and previously assumed values to find the mean stress m: Stress at Fm

τm  Ks

8  Fm D

π d 9.

τm  353.4  MPa

3


Alternating stress

Kw 

4 C  1 4 C  4



τa  Kw

0.615 C

8  F a D

π d

3

Kw  1.184

τa  49.2 MPa

10. Find the ultimate tensile strength of this wire material from equation 14.3 and Table 14-4 and use it to find the ultimate shear strength from equation 14.4 and the torsional yield strength for the coil body from Table 14-10, assuming no set removal.

  mm  

b


S ut  A  


S ys  0.45 S ut

S ys  790.6  MPa


S us  0.67 S ut

S us  1177.1 MPa

d

S ut  1756.8 MPa

11. Find the wire endurance limit for unpeened springs from equation 14.13 and convert it to fully reversed endurance strength with equation 14.16c. Fully reversed endurance limit

S es  0.5


S es  178.51 MPa

12. The fatigue safety factor for the coils in torsion is calculated from equation 14.16b. Minimum stress

Fatigue safety factor

τmin  τm  τa Nfs 

S es  S us  τmin

S es  τm  τmin  S us τa

τmin  304.2  MPa Nfs  2.34



14-17-3

Note that the minimum stress due to force Fmin is used in this calculation, not the coil-winding stress from step 5 13. The stresses in the end hooks also need to be determined. The bending stresses in the hook are found from equation 14.23: C1 =

2  R1 d

2 D

=

C1  C

=C

2 d

C1  8.00

2

Kb 

4 C1  C1  1

Kb  1.103

4  C1  C1  1  16 D Fa

σa  Kb

π d σm  Kb

3

π d

16 D Fm

π d σmin  Kb

4  Fa





3

4  Fm

π d

16 D Fmin

π d

3

σa  94.29  MPa

2



σm  754.3  MPa

2

4  Fmin

π d

2

σmin  660.01 MPa

14. Convert the torsional endurance strength to a tensile endurance strength with the von Mises relationship and use it and the ultimate tensile strength in equation 14.16 to find a fatigue safety factor for the hook in bending: S e 

S es

S e  309.37 MPa

0.577 S e  S ut  σmin

Nfb 

S e  σm  σmin  S ut σa

Nfb  1.74

15. The torsional stresses in the hook are found from equation 14.24 using an assumed value of C2  5. R2 

C2 d

Kw2 

R2  8.75 mm

2 4  C2  1

Kw2  1.188

4  C2  4

τBa  Kw2

8  Fa D

π d τBm  Kw2

τBa  49.4 MPa

3

8  Fm D

π d τBmin  Kw2

3

8  Fmin D

π d

3

τBm  395.0  MPa

τBmin  345.6  MPa

16. The fatigue safety factor for the hook in torsion is calculated from equation 14.16b. Nfs 

S es  S us  τBmin

S es  τBm  τBmin  S us τBa

Nfs  2.22



14-17-4

17. The spring rate is defined from the two specified forces at their relative deflection. k 

Spring rate

Fmax  Fmin

k  5882.4

Δy

N m

18. To get the defined spring rate, the number of active coils must satisfy equation 14.7, solving for Na yields: 4


Na 

d G

Na  11.737

3

Na  11.75

8 D  k Note that we round it to the nearest 1/4 coil as the manufacturing tolerance cannot achieve better than that accuracy. Having rounded the number of active coils, we must now calculate the spring rate using equation 14. Corrected spring rate

4

d G

k 

k  5876.0

3

8  D  Na

N m

19. The total number of coils in the body and the body length are Total coils

Nt  Na  1

Nt  12.75

Body length

Lb  Nt d

Lb  44.6 mm

20. The free length can now be determined. The length of a standard hook is equal to the coil inside diameter: Hook length

Lhook  D  d

Lhook  24.5 mm

Free length

Lf  Lb  2  Lhook

Lf  93.6 mm

21. The initial coil tension force must be found again in order to obtain the deflection to reach the larger of the two loads. 3 2   3 2   2.987  C  139.7  C  3427 C  38404   psi

τi1  4.231  C  181.5  C  3387 C  28640  psi

τi1  75.8 MPa

τi2

τi2  126.9  MPa

τi 

τi1  τi2

τi  101.3  MPa

2 3

Fi 

π d  τi

Fi  57.3 N

8  K s D

ymax 

Fmax  Fi

ymax  28.53  mm

k

22. The inside and outside coil diameters are Inside coil dia

Di  D  d

Di  24.50  mm

Outside coil dia

Do  D  d

Do  31.50  mm

23. The weight of the spring's active coils is found from equation 14.11b and is Weight density

3

ρ  0.28 lbf  in 2 2

Weight

Wa 

π  d  D Na ρ 4

Wa  0.756  N

24. The natural frequency of this spring is found from equation 14.11a and is: © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

MACHINE DESIGN - An Integrated Approach, 4th Ed. Natural frequency

fn 

1 2



14-17-5

k g

fn  138.1  Hz

Wa

25. The ratio between the natural frequency and the forcing frequency is Forcing frequency

Frequency ratio

ff  fn ff

ωf

ff  25 Hz

2 π

 5.5

which could be higher.

26. We now have a complete design specification for this A228-wire spring: Wire diameter

d  3.50 mm

Total coils

Nt  12.75

Outside diameter

Do  31.50  mm

Free length

Lf  93.63  mm

Standard hooks



14-18-1


Design a helical extension spring with standard hooks to handle a dynamic load that varies from 300 lb to 500 lb over 2 in working deflection. Use chrome-vanadium wire. The forcing frequency is 1000 rpm. Infinite life is desired. Minimize the package size. Choose appropriate safety factors against fatigue, yielding, and surging.

Given:

Minimum force

Fmin  300  lbf

Life

L  ∞

Maximum force Working deflection

Fmax  500  lbf Δy  2.00 in

Shear modulus Forcing freq.

G  11.7 10  psi ωf  1000 rpm

Nfb  1.28

Spring index

C  5.5


Wire strength

A  173.13 ksi


6

b  0.1453 Solution: 1.



2.

Fmax  Fmin

Fa  100.0  lbf

2 Fmax  Fmin

Fm  400.0  lbf

2


Kb  3.

4 C  C  1

Kb  1.157

4 C ( C  1)


d=


     

2 b

1

 4  4 Kb C  1 RHS( d )    Nfb Fm   Nfb  1   Fmin    π A  in2 b    d     0.577  0.67 A     0.5 S ew       in    Nfb Fa     0.5 0.67 S ew    

4.

      

d  0.1 in

RHS( d )  0.438  in

d  RHS( d )

RHS( d )  0.404  in

d  RHS( d )

RHS( d )  0.406  in

d  RHS( d )

RHS( d )  0.406  in

d  0.406  in

D  C d

D  2.233  in

2 b

 in





τi1  4.231  C  181.5  C  3387 C  28640  psi

τi1  14.8 ksi

τi2

τi2  23.3 ksi

τi  6.

τi1  τi2

τi  19.0 ksi

2


7.

14-18-2

Ks  1 

0.5

Ks  1.091

C


Fi 

π d  τi 8  K s D

Fi  205.4  lbf



τm  Ks

8  Fm D

π d 9.

τm  37.1 ksi

3


Alternating stress

Kw 

4 C  1 4 C  4

τa  Kw



0.615 C

8  F a D

π d

3

Kw  1.278

τa  10.9 ksi

10. Find the ultimate tensile strength of this wire material from equation 14.3 and Table 14-4 and use it to find the ultimate shear strength from equation 14.4 and the torsional yield strength for the coil body from Table 14-10, assuming no set removal. b


d S ut  A     in 


S ys  0.50 S ut

S ys  98.7 ksi


S us  0.67 S ut

S us  132.2  ksi

S ut  197.4  ksi


S es  0.5


S es  27.11  ksi






τmin  26.214 ksi Nfs  1.66



14-18-3

Note that the minimum stress due to force Fmin is used in this calculation, not the coil-winding stress from step 5. 13. The stresses in the end hooks also need to be determined. The bending stresses in the hook are found from equation 14.23: C1 =

2  R1 d

2 D

=

C1  C

=C

2 d

C1  5.50

2

Kb 

4 C1  C1  1

Kb  1.157

4  C1  C1  1  16 D Fa

σa  Kb


3

π d

16 D Fm


4  Fa





3

4  Fm

π d

16 D Fmin

π d

3

σa  20.43  ksi

2



σm  81.7 ksi

2

4  Fmin

π d

2

σmin  61.28  ksi


S es

S e  46.99  ksi

0.577 S e  S ut  σmin

Nfb 


Nfb  1.28


C2 d

Kw2 

R2  1.015  in

2 4  C2  1

Kw2  1.188

4  C2  4

τBa  Kw2

8  Fa D


τBa  10.1 ksi

3

8  Fm D


3

8  Fmin D

π d

3

τBm  40.4 ksi

τBmin  30.3 ksi




Nfs  1.72


MACHINE DESIGN - An Integrated Approach, 4th Ed. 17. The spring rate is defined from the two specified forces at their relative deflection. k 

Spring rate

Fmax  Fmin

14-18-4

lbf

k  100.0 

Δy

in



d G

Na 

Na  35.689

3

Na  35.75

8 D  k Note that we round it to the nearest 1/4 coil as the manufacturing tolerance cannot achieve better than that accuracy. Having rounded the number of active coils, we must now calculate the spring rate using equation 14. 4


d G

k 

k  99.8

3

8  D  Na

lbf in


Nt  Na  1

Nt  36.75

Body length

Lb  Nt d

Lb  14.92  in



Lhook  1.827  in

Free length


Lf  18.57  in


τi1  4.231  C  181.5  C  3387 C  28640  psi

τi1  14.8 ksi

τi2

τi2  23.3 ksi

τi 

τi1  τi2

τi  19.0 ksi

2 3

Fi 

π d  τi

Fi  205.4  lbf

8  K s D

ymax 

Fmax  Fi

ymax  2.95 in

k


Di  D  d

Di  1.83 in

Outside coil dia

Do  D  d

Do  2.64 in


3

ρ  0.28 lbf  in 2 2

Weight

Wa 

π  d  D Na ρ 4

Wa  9.091  lbf

24. The natural frequency of this spring is found from equation 14.11a and is: Natural frequency

fn 

1 2



k g Wa

fn  32.6 Hz



14-18-5

25. The ratio between the natural frequency and the forcing frequency is

Forcing frequency Frequency ratio

ff  fn ff

ωf

ff  16.667 Hz

2 π

 2.0



d  0.41 in

Total coils

Nt  36.75

Outside diameter

Do  2.64 in

Free length

Lf  18.57  in

Standard hooks



14-19-1


Design a helical compression spring to handle a dynamic load that varies from 780 N to 1000 N over a 22-mm working deflection. Use squared and ground, unpeened music wire and a 10% clash allowance. The forcing frequency is 500 rpm. Infinite life is desired. Minimize the package size. Choose appropriate safety factors against fatigue, yielding, and surging.

Given:

Minimum force

Fmin  780  N

Forcing freq.

ωf  500  rpm

Maximum force

Fmax  1000 N

Shear modulus

G  79.6 GPa

Clash factor

α  0.10

Working deflection Δy  22 mm Music wire properties: Strength A  2153.5 MPa Set after winding, unpeened Design Choices: Design safety factor Nfsd  1.1 Solution: 1.

Find the the mean and alternating forces from equation 14.15a: Fa  Fm 

Mean force

3.

C  9

Spring index


Alternating force

2.

b  0.1625

Fmax  Fmin

Fa  110.0  N

2 Fmax  Fmin

Fm  890.0  N

2


Ks  1 

Wahl factor

Kw 

0.5

Ks  1.056

C

4 C  1 4 C  4



0.615 C

Kw  1.162


Kys  0.60



KU  0.67

(equation 14.4)


S ew  310  MPa

(unpeened)


d  5  mm

Given

1

Nfsd  1  8  C Nfsd  d=  Ks Fm   Ks Fmin  Nfsd 2  π  A  mm 0.67   b      d   A       0.67  mm     1  Kw Fa    S ew    0.50  d  Find ( d ) 4.

d  6.172  mm Let the wire diameter be

      

2 b

 mm

d  6.5 mm




5.

  mm  

S ut  A  



S us  1064 MPa



S ys  953  MPa

Fatigue strength

S es  0.5


S es  181.4  MPa

D  58.5 mm

Calculate the shear stress i at the initial deflection (lowest defined force), and the mean stress m.

Stress at Fm

τi  Ks

8  Fmin D

τi  446.6  MPa

3

8  Fm D

τm  Ks

π d

τm  509.6  MPa

3


8  F a D

τa  Kw

π d

τa  69.3 MPa

3


9.

S ut  1589 MPa

D  C d

π d

8.

d

Calculate the mean coil diameter D from equation 14.5.

Stress at Fmin

7.

b


Mean coil diameter 6.

14-19-2

Nfs 


Nfs  1.3


The spring rate is defined in this problem because of the two specified forces at a particular relative deflection. Spring rate

k 

Fmax  Fmin

k  10.00 

Δy

N mm


4

Na 

d G 3

Na  8.872

Na  8.75

8 D  k


4

k 

d G 3

k  10.14 

8  D  Na

N mm


Nt  Na  2

Nt  10.75



14-19-3



Lshut  69.9 mm


Fmin

yinit 

yinit  76.9 mm

k



Δyclash  2.2 mm


Lf  171.0  mm


yshut  101.1  mm


Fshut  1025 N


τshut  Ks

8  Fshut D


Ns 

3

S ys

τshut  587.1  MPa

Ns  1.62

τshut


Deflection ratio

sr  y' 

Lf

sr  2.923

D yinit  Δy

y'  0.579

Lf

Take these two values to Figure 14-14 and find that their coordinates are safely within the zones that are stable against buckling for both cases. 21. The inside and outside coil diameters are Inside coil dia

Di  D  d

Di  52 mm

Outside coil dia

Do  D  d

Do  65 mm

22. The smallest hole and largest pin that should be used with this spring are Smallest hole


holemin  67.9 mm

Largest pin

pin max  Di  0.05 D

pin max  49.1 mm

23. The weight of the springs active coils is found from equation 14.11b Weight density

3

ρ  0.285  lbf  in



14-19-4

2 2

Weight

Wa 

π  d  D Na ρ

Wa  4.13 N

4


fn 

1 2



k g

fn  77.6 Hz

Wa


Frequency ratio

ff  fn ff

ωf

ff  8.333  Hz

2 π

 9.3

which is sufficiently high. 26. We now have a complete design specification for this A228 wire spring: Wire diameter

d  6.5 mm

Outside diameter

Do  65.0 mm

Total coils

Nt  10.75

Free length

Lf  171  mm


25. The initial and working forces are Initial force

Finit  k yinit

Finit  780  N

Working force

Fwork  k  yinit  Δy 

Fwork  1003 N

26. Safety factors: S ys

Yielding

Nys 

Fatigue

Nfs 

Surging

Nsurge 

τshut S es  S us  τi

S es  τm  τi  S us τa fn ff

Nys  1.6

Nfs  1.32

Nsurge  9.3



14-20-1


Design a helical compression spring to handle a dynamic load that varies from 135 N to 220 N over a 32-mm working deflection. Use squared, peened chrome-vanadium wire and a 15% clash allowance. The forcing frequency is 250 rpm. Infinite life is desired. Minimize the package size. Choose appropriate safety factors against fatigue, yielding, and surging.

Given:

Minimum force

Fmin  135  N

Forcing freq.

ωf  250  rpm

Maximum force

Fmax  220  N

Shear modulus

G  79.6 GPa

Clash factor

α  0.15

Working deflection Δy  32 mm Chrome-vanadium wire properties: Strength A  1909.9 MPa Set after winding, unpeened Design Choices: Design safety factor Nfsd  1.1 Solution: 1.

Find the the mean and alternating forces from equation 14.15a: Fa  Fm 

Mean force

3.

C  10.5

Spring index


Alternating force

2.

b  0.1453

Fmax  Fmin

Fa  42.5 N

2 Fmax  Fmin

Fm  177.5  N

2


Ks  1 

Wahl factor

Kw 

0.5

Ks  1.048

C

4 C  1 4 C  4



0.615 C

Kw  1.138


Kys  0.65



KU  0.67

(equation 14.4)


S ew  465  MPa

(peened)


d  5  mm

Given

1

d=

d  Find ( d ) 4.

Nfsd  1  8  C Nfsd   Ks Fm   Ks Fmin   Nfsd 2  0.67 π A  mm  b      d   A     0.67  mm     1  Kw Fa    S ew    0.50  d  3.081  mm Let the wire diameter be

      

2 b

 mm

d  3.5 mm




5.


S ut  A  



S us  1067 MPa



S ys  1035 MPa

Fatigue strength

S es  0.5

Stress at Fm


S es  297.3  MPa

D  36.8 mm

τi  Ks

8  Fmin D

τi  308.7  MPa

3

8  Fm D

τm  Ks

π d

τm  405.9  MPa

3


8  F a D

τa  Kw

π d

τa  105.5  MPa

3


9.

S ut  1592 MPa

D  C d

π d

8.

 


7.

d

 mm 


6.

14-20-2 b

Nfs 


Nfs  1.6



k 

Fmax  Fmin

k  2.66

Δy

N mm


4

Na 

d G 3

Na  11.325

Na  9.75

8 D  k


4

k 

d G 3

8  D  Na

k  3.09

N mm


Nt  Na  2

Nt  11.75



Lshut  1.619  in



14-20-3


Fmin

yinit 


k





Lf  121.7  mm


yshut  80.6 mm


Fshut  248.5  N


τshut  Ks

8  Fshut D


Ns 

3

S ys


Ns  1.82

τshut


Deflection ratio

sr  y' 

Lf

sr  3.311

D yinit  Δy

y'  0.623

Lf


Di  D  d

Di  33.25  mm

Outside coil dia

Do  D  d

Do  40.25  mm




Largest pin

pin max  Di  0.05 D



3

ρ  0.285  lbf  in 2 2

Weight

Wa 

π  d  D Na ρ 4

Wa  0.188  lbf

24. The natural frequency of this spring is found from equation 14.11a and is: © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


Natural frequency

fn 

1 2

14-20-4

k g



fn  95.02  Hz

Wa


Forcing frequency

Frequency ratio

ff  fn ff

ωf

ff  4.167  Hz

2 π

 22.8

which is sufficiently high. 26. We now have a complete design specification for this A228 wire spring: Wire diameter

d  3.5 mm

Outside diameter

Do  40.3 mm

Total coils

Nt  11.75

Free length

Lf  122  mm

ends squared



Finit  135  N

Working force


Fwork  234  N


Yielding

Nys 

Fatigue

Nfs 

Surging

Nsurge 



Nys  1.8

Nfs  1.6

Nsurge  23



14-21-1


Design a helical compression spring for a static load of 400 N at a deflection of 45 mm with a safety factor of 2.5. Use C = 8. Specify all parameters necessary to manufacture the spring.

Given:

Working force

Fwork  400  N

Shear modulus

G  80.8 GPa

Working deflection

ywork  45 mm

Safety factor

Ns  2.5

Set removed

Km  0.65

C  8

Spring index Design choices: Clash allowance

α  0.15 A  2153.5 MPa b  0.1625

ASTM A228 wire Solution:

See Mathcad file P1421. Fwork

k 

k  8.889 

N

1.


2.


ywork

mm

1

d 

2 b  8 Ns ( C  0.5)  Fwork  ( 1  α)  mm 2   π Km A  mm  

Wire diameter 3.

d  4.789  mm

Let

d  5  mm

Calculate the mean coil diameter and number of active coils. Mean coil dia

D  C d


Na 

D  40 mm

4

d G 3

8 D  k

Na  11.096 Na  11


7.

3

k  8.967 

8  D  Na

N mm

Nt  Na  2

Nt  13.00


6.

d G


5.

4

k 

Ls  d  Nt

Ls  65 mm



yshut  51.75  mm

Free length


Lf  116.75 mm

To check for buckling, two ratios need to be calculated, Lf/D and y max/Lf. Slenderness ratio

sr 

Lf D

sr  2.919


MACHINE DESIGN - An Integrated Approach, 4th Ed. Deflection ratio

y' 

14-21-2

ywork

y'  0.385

Lf


9.


Di  D  d

Di  35 mm

Outside coil dia

Do  D  d

Do  45 mm



holemin  47.00  mm

Largest pin

pin max  Di  0.05 D

pin max  33.00  mm


3

ρ  0.28 lbf  in 2 2

Weight

Wt 

π  d  D Nt ρ

Wt  2.44 N

4


d  5  mm

Outside diameter

Do  45 mm

Total coils

Nt  13.00

Free length

Lf  116.75 mm




14-22-1


Given:

Design a straight-ended helical torsion spring for a static load 200 N-m at a deflection of 45 deg with a safety factor of 1.8. Specify all parameters necessary to manufacture the spring. State all assumptions. Applied moment M  200  N  m Bending modulus E  206.8  GPa Deflection at load θ  45 deg θ  0.125  rev

Design Choices: Use unpeened oil tempered wire with 40-mm-long, straight ends. The coil is loaded to close it. Spring index Design safety factor Nyd  1.8 C  11

Solution: 1.


Ks  0.85 (from Table 14-13)

Length of ends

L1  40 mm

Material strength factors

A  1831.2 MPa b  0.1833 (ASTM A229 wire)

L2  40 mm


Calculate the Wahl bending factors for inside surface. 2

Kbi  2.

4 C  C  1

Kbi  1.073

4  C ( C  1 )

Solve for d using the static yield criterion. 1

 32 Kbi Nyd  M  d    π Ks A  mm3    3.

4.

 mm

d  16.144 mm

Use a wire diameter from the available sizes in Table 14-2. Calculate the mean coil diameter D from equation 14.5. Wire diameter

d  16 mm

Spring index

C  11

Mean coil diameter

D  C d

D  176  mm

Calculate the maximum compressive stress in the coil at the inner surface.

σimax  Kbi

32 M

π d 5.

3 b

σimax  533.5  MPa

3

Find the ultimate tensile strength of this oil tempered material from equation 14.3 and Table 14-4 and use it to find the bending yield strength from Table 14-13, assuming stress relieving. Ultimate tensile strength

S ut  A  

  mm  

Bending yield strength

S y  Ks S ut

d

b

S ut  1102 MPa S y  936  MPa Nyb 

Sy

Nyb  1.8

6.

The realized static safety factor against yielding is

7.

The spring rate is defined from the two specified moments at their relative deflection. Spring rate

8.

k 

M

θ

σimax

k  1600

N m rev

To get the defined spring rate, the number of active coils must satisfy equation 14.28, solving for Na yields:



14-22-2

4


Na 

d E 10.8 D k rev

Na  4.46

Note that to force k to be in units of N-m per rev we must multiply k by rev. 9.

The ends contribute to the active coils from equation 14.26a as Ne 

L1  L2 3  π D

Ne  0.05

and, from equation 14.26b, the number of body coils in the spring are Nb  Na  Ne

Nb  4.4

10. Check the angular deflection at the specified load from equation 14.27c.

θ  10.8

M  D Na 4

 rev

θ  45.0 deg

d E



14-23-1


Design a straight-ended helical torsion spring for a static load 300 in-lb at a deflection of 75 deg with a safety factor of 2. Specify all parameters necessary to manufacture the spring. State all assumptions.

Given:

Applied moment Deflection at load

M  300  in lbf θ  75 deg

6

Bending modulus E  30 10  psi θ  0.208  rev

Design Choices: Use unpeened music wire with 2-in-long, straight ends. The coil is loaded to close it. Spring index Design safety factor Nyd  2 C  7.5

Solution: 1.


Ks  1 (from Table 14-13)

Length of ends

L1  2  in

L2  2  in


A  184.65 ksi

b  0.1625 (ASTM A228 wire)



Kbi  2.

4 C  C  1

Kbi  1.110

4  C ( C  1 )


 32 Kbi Nyd  M  d    π Ks A  in3    3.

4.

 in

d  0.312  in


d  0.312  in

Spring index

C  7.5

Mean coil diameter

D  C d

D  2.34 in



32 M

π d 5.

3 b

σimax  111.7  ksi

3


S ut  A  

  in  



d

b

S ut  223  ksi S y  223  ksi Nyb 

Sy

Nyb  2.0

6.


7.


k 

M

θ

σimax

k  1440

in lbf rev



14-23-2

To get the defined spring rate, the number of active coils must satisfy equation 14.28, solving for Na yields: 4


Na 

d E 10.8 D k rev

Na  7.81



L1  L2 3  π D

Ne  0.18


Nb  7.6


θ  10.8

M  D Na 4

 rev

θ  75.0 deg

d E



14-24-1


Given:

Design a straight-ended helical torsion spring for a dynamic load 50-105 N-m at a deflection of 80 deg with a safety factor of 2. Specify all parameters necessary to manufacture the spring. State all assumptions. Bending modulus Minimum moment Mmin  50 N  m E  206.8  GPa Mmax  105  N  m

Maximum moment

Deflection at load

Δθ  80 deg

Design Choices: Use unpeened oil tempered wire with 40-mm-long, straight ends. The coil is loaded to close it. Design safety factor Nd  2.5 Yield strength factor


Length of ends

L1  40 mm

Material strength factors Wire endurance limit

A  1831.2 MPa b  0.1833 (ASTM A229 wire) Spring index S ewb  537  MPa

L2  40 mm C  8.3

See Mathcad file P1424. Solution: 1. Find the mean and alternating loads. Ma  Mm  2.

Mmax  Mmin

Ma  27.5 N  m

2 Mmax  Mmin

Mm  77.5 N  m

2

Calculate the Wahl bending factors for inside and outside surfaces. 2

Kbi 

4 C  C  1

Kbi  1.099

4  C ( C  1 ) 2

Kbo  3.

4 C  C  1

Kbo  0.916

4  C ( C  1 )

Solve for d using the static yield criterion. Use this value of d for starting the iteration in the next step. 1

 32 Kbi Nd  Mmax  d    π Ks A  mm3    4.

3 b

 mm

d  13.739 mm

Solve for d by iteration using the fatigue criterion. 1 b     d    A       N  M   mm   M   min   32 Kbo  fb a 0.5 S ewb  d= π A  

3 b

1 b     d    A     32 Kbo Nd  Ma  mm   Mmin    0.5 S ewb    RHS( d )    3 π A  mm  

3 b

 mm



14-24-2

d  RHS( d )

d  14.120 mm

d  RHS( d )

d  14.099 mm

d  RHS( d )

d  14.100 mm

Since this results in a slightly larger wire diameter, the fatigue criterion is governing and the spring must be designed for fatigue. 3.

Use a wire diameter from the available sizes in Table 14-2. Calculate the mean coil diameter D from equation 14 d  15 mm C  8.3 D  C d

Wire diameter Spring index Mean coil diameter 4.



32 Mmax

Calculate the maximum, minimum, alternating, and mean tensile stresses in the coil at the outer surface.

σomin  Kbo

32 Mmin

π d σomax  Kbo

σm 

σa 

7.

8.

32 Mmax

σomax  290.3  MPa

3

σomax  σomin

σm  214.3  MPa

2

σomax  σomin

σa  76.0 MPa

2


S ut  A  



  in   d

b

S ut  2017 MPa S y  2017 MPa

Convert the wire bending endurance limit for unpeened springs from equation 14.33 to fully reversed endurance strength with equation 14.34b. S ewb S ut Fully reversed S e  0.5 S e  309.74 MPa S ut  0.5 S ewb endurance limit The fatigue safety factor for the coils in bending is calculated from equation 14.34a. Fatigue factor of safety

9.

σomin  138.2  MPa

3

π d

6.


3

π d 5.

D  124.5  mm

Nfb 

S e  S ut  σomin

Nfb  3.3

S e  σm  σomin  S ut σa

The static safety factor against yielding is Static factor of safety

Nyb 

Sy

σimax

Nyb  5.8



14-24-3

These are both acceptable safety factors. 10. The spring rate is defined from the two specified moments at their relative deflection. k 

Spring rate

Mmax  Mmin

Δθ

k  247.5 

N m rev


4

Na 

d E 10.8 D k rev

Na  31.46

Note that to force k to be in units of N-m per rev we must multiply k by rev. 12. The ends contribute to the active coils from equation 14.26a as Ne 

L1  L2

Ne  0.07

3  π D


Nb  31

13. The angular deflections at the specified loads from equation 14.27c are

θmin  10.8

Mmin D Na 4

 rev

θmin  73 deg

d E

θmax  10.8

Mmax D Na 4

 rev

θmax  153  deg

d E



14-25-1


Design a straight-ended helical torsion spring for a dynamic load 150-350 in-lb over a deflection of 50 deg with a safety factor of 1.4. Specify all parameters necessary to manufacture the spring. State all assumptions.

Given:

Minimum moment Maximum moment

Mmin  150  in lbf Mmax  350  in lbf

Bending modulus Deflection at load

6

E  30 10 psi Δθ  50 deg

Design Choices: Use unpeened music wire with 2-in-long, straight ends. The coil is loaded to close it. Design safety factor Nd  1.4

Solution: 1.


Ks  0.80 (from Table 14-13)

Length of ends

L1  2  in


A  184.65 ksi b  0.1625 (ASTM A228 wire) Spring index S ewb  77.99  ksi C  4.5


Find the mean and alternating loads. Ma  Mm 

2.

L2  2  in

Mmax  Mmin

Ma  100.0  in lbf

2 Mmax  Mmin

Mm  250.0  in lbf

2


Kbi 

4 C  C  1

Kbi  1.198

4  C ( C  1 ) 2

Kbo  3.

4 C  C  1

Kbo  0.854

4  C ( C  1 )


 32 Kbi Nd  Mmax  d    π Ks A  in3    4.

3 b

 in

d  0.323  in

Solve for d by iteration using the fatigue criterion. 1 b     d    A       N  M   mm   M   min   32 Kbo  d a 0.5 S ewb  d= π A  

3 b

1 b     d   A     32 Kbo Nd  Ma  in   Mmin    0.5 S ewb    RHS( d )    3 π A  in  

3 b

 in


MACHINE DESIGN - An Integrated Approach, 4th Ed. d  RHS( d )

d  0.334  in

d  RHS( d )

d  0.334  in

14-25-2

Since this results in a slightly larger wire diameter, the fatigue criterion is governing and the spring must be designed for fatigue. 3.

4.

Use a wire diameter from the available sizes in Table 14-2. Calculate the mean coil diameter D from equation 14 Wire diameter

d  0.343  in

Spring index

C  4.5

Mean coil diameter

D  C d



32 Mmax



32 Mmin


σm 

σa 

7.

32 Mmax

σm  53.9 ksi

2

σomax  σomin

σa  21.5 ksi

2

Find the ultimate tensile strength of this oil tempered material from equation 14.3 and Table 14-4 and use it to find the bending yield strength from Table 14-13, assuming stress relieving.

  in  


S ut  A  



d

b

S ut  220  ksi S y  176  ksi

Convert the wire bending endurance limit for unpeened springs from equation 14.33 to fully reversed endurance strength with equation 14.34b. S e  0.5

S ewb S ut S ut  0.5 S ewb

S e  47.41  ksi

The realized fatigue safety factor for the coils in bending is calculated from equation 14.34a. Fatigue factor of safety

9.

σomax  75.4 ksi

3

σomax  σomin

Fully reversed endurance limit 8.

σomin  32.3 ksi

3

π d

6.


3

π d 5.

D  1.544  in

Nfb 



Nfb  1.5

The static safety factor against yielding is


MACHINE DESIGN - An Integrated Approach, 4th Ed. Static factor of safety

Nyb 

Sy

14-25-3 Nyb  1.7

σimax

These are both acceptable safety factors. 10. The spring rate is defined from the two specified moments at their relative deflection. k 

Spring rate

Mmax  Mmin

Δθ

k  1440

in lbf rev


4

d E

Na 

10.8 D k rev

Na  17.30

Note that to force k to be in units of in-lb per rev we must multiply k by rev. 12. The ends contribute to the active coils from equation 14.26a as Ne 

L1  L2

Ne  0.27

3  π D


Nb  17

13. The angular deflections at the specified loads from equation 14.27c are

θmin  10.8

Mmin D Na 4

 rev

θmin  38 deg

d E

θmax  10.8

Mmax D Na 4

 rev

θmax  88 deg

d E



14-26-1


Design a Belleville spring to give a constant 400 N ±10% force over a 1-mm deflection.

Given:

Nominal force

Fflat  400  N

Working deflection

ywork  1  mm

Design Choices: 6

The diameter ratio is Rd  2. Use unset carbon spring steel 50HRC. Properties, E  30 10  psi ,

ν  0.28 . The outside diameter of the spring is Do  39.55  mm Solution: 1.


Since a constant force spring is needed, the h/t ratio (see Figure 14-30) is hovert  1.414

h/t ratio 2.

The required force variation of not more than ±10% can be met by choosing an appropriate deflection range to operate in from Figure 14-31. If the deflection is kept between about 53% and 146% of the flat deflection, this tolerance will be achieved. The nominal force will then occur at the flat position and the spring must operate on both sides, so must be mounted in similar fashion to that shown in Figure 14-33.

3.

Use the above assumptions and the specified nominal force in equation 14.37a to find an appropriate spring thickness t: 1

2  Fflat Do  1   t    10  132.4  MPa hovert 

4

t  0.760  mm

h  hovert t

h  1.075  mm

4.

The height h can now be found:

5.

Based on the choices in step 2, find the minimum and maximum deflections: ymin  0.53 h ymax  1.46 h

t  0.760  mm

Let

ymin  0.57 mm ymax  1.57 mm

The difference between these distances is equal to the required deflection range, so the force tolerance can be met over that range. 6.

Figure 14-34 shows that the worst stress state will occur at the largest deflection ymax, so solve equations 14.36 for stresses at that deflection:

  Rd  1  2   R2  d  

K1 

6

π ln Rd 



K2 

6

π ln Rd 



K3 

6



K4  K5 

π ln Rd 

 Rd  1  ln Rd 

K1  0.689



 1



 Rd  1    2 

 Rd  ln Rd    Rd  1  Rd    ln Rd      Rd  1  2 Rd

2   Rd  1 

K2  1.22

K3  1.378

K4  1.115 K5  1


MACHINE DESIGN - An Integrated Approach, 4th Ed. 4  E ymax

σc  

2





2

K1 Do  1  ν

 

ymax 

   K3 t 2  

σc  1831 MPa

 4  E ymax  K1 Do2 1  ν2 

ymax       K2  h    K3 t 2     

σti  906.1  MPa

 4  E ymax  K1 Do2 1  ν2 

ymax       K4  h    K5 t 2     

σto  1416 MPa

σti  



σto  

7.

 

 K2  h 

14-26-2







Table 14-5 gives S ut  246  ksi for this material. Table 14-15 indicates that 120% of this value can be used for an unset spring. The safety factor for static loading is then Ns 

1.20 S ut

σc

Ns  1.1

which is acceptable. 8.

A summary of the spring design is Outside diameter

Do  39.55  mm

Inside diameter

Di 

Do Rd

Thickness

t  0.76 mm

Height

h  1.07 mm

Di  19.77  mm



14-27-1


Design a Belleville spring for bimodal operation between ±50 N.

Given:

Nominal force

Fflat  50 N

Design Choices: The diameter ratio is Rd  2. Use unset carbon spring steel 50HRC. Properties: 6

E  30 10  psi, ν  0.28 . The outside diameter of the spring is Do  24.50  mm Solution: 1.


Since a bimodal force spring is needed, the h/t ratio (see Figure 14-30) is hovert  2.828

h/t ratio 2.

The required force variation of not more than ±100% can be met by choosing an appropriate deflection range to operate in from Figure 14-30. If the deflection is kept between about 0% and 200% of the flat deflection, this will be achieved. A zero force will then occur at the flat position and the spring must operate on both sides, so must be mounted in similar fashion to that shown in Figure 14-32.

3.

Use the above assumptions and the specified nominal force in equation 14.37a to find an appropriate spring thickness t: 1 2  Do  Fflat  t    7 hovert   19.2 10  psi 

4

t  0.299  mm h  hovert t

t  0.30 mm

Let

h  0.848  mm

4.


5.

Based on the choices in step 2, find the minimum and maximum deflections: ymin  0  h

ymin  0  mm

ymax  2.00 h

ymax  1.70 mm

The difference between these distances is equal to the required deflection range, so Working deflection 6.

ywork  ymax

ywork  1.70 mm


  R  1  2 d   R2  d  

K1 

6

π ln Rd 



K2 

6

π ln Rd 



K3 

6



K4 

K5 

π ln Rd 

 Rd  1  ln Rd 

K1  0.689



 1



 Rd  1    2 

 Rd  ln Rd    Rd  1  Rd    ln Rd      Rd  1  2   Rd

2   Rd  1 

K2  1.22

K3  1.378

K4  1.115

K5  1


MACHINE DESIGN - An Integrated Approach, 4th Ed. 4  E ymax

σc  

2





2

K1 Do  1  ν

 

ymax 

   K3 t 2  

σc  1523 MPa

 4  E ymax  K1 Do2 1  ν2 

ymax       K2  h    K3 t 2     

σti  1523 MPa

 4  E ymax  K1 Do2 1  ν2 

ymax       K4  h    K5 t 2     

σto  1105 MPa

σti  



σto  

7.

 

 K2  h 

14-27-2







Table 14-5 gives S ut  246  ksi for this material. Table 14-15 indicates that 120% of this value can be used

for an unset spring. The safety factor for static loading is then 1.20 S ut Ns  Ns  1.3

σc



Do  24.5 mm

Inside diameter

Di 

Do Rd

Thickness

t  0.3 mm

Height

h  0.85 mm

Di  12.25  mm



14-28-1


Given the data below for a helical compression spring loaded in fatigue, design the spring for infinite life. State all assumptions and empirical data used.

Given:

Minimum force

Fmin  225  N

Forcing freq.

ωf  625  rpm

Maximum force

Fmax  450  N

Shear modulus

G  79.3 GPa

Working deflection Δy  20 mm Music wire properties: Strength A  2153.5 MPa b  0.1625 Wire diameter d  8  mm Solution: 1.


Find the the mean and alternating forces from equation 14.15a: Alternating force

Mean force 2.

3.

4.

Fa  Fm 

Fmax  Fmin

Fa  112.5  N

2 Fmax  Fmin

Fm  337.5  N

2

Calculate the factors necessary to find the fatigue factor of safety. Direct shear factor

Ks  1 

Wahl factor

Kw 

0.5

Ks  1.059

C

4 C  1 4 C  4



0.615 C

Kw  1.172


Kys  0.60



KU  0.67

(equation 14.4)


S ew  310  MPa

(unpeened)


  mm  

b


S ut  A  



S us  1029 MPa



S ys  922  MPa

Fatigue strength

S es  0.5

d


S ut  1536 MPa

S es  182.5  MPa


5.

Clash factor α  0.15 Spring index C  8.5 Set after winding, unpeened

D  C d

D  68.0 mm


τi  Ks

8  Fmin D

π d

3

τi  80.6 MPa



Stress at Fm

14-28-2

8  Fm D

τm  Ks

π d 6.


8  F a D

τa  Kw

π d 7.

Nfs 


Nfs  3.3



9.

τa  44.6 MPa

3


8.

τm  120.9  MPa

3

k 

Fmax  Fmin

k  11.25 

Δy

N mm


4

Na 

d G 3

Na  11.478

Na  11.5

8 D  k


4

d G

k 

3

k  11.23 

8  D  Na

N mm


Nt  Na  2

Nt  13.50



Lshut  108  mm


yinit 

Fmin k






Lf  151  mm




Fshut  483.3  N



14-28-3


τshut  Ks

8  Fshut D


3

S ys

Ns 


Ns  5.3

τshut


Deflection ratio

Lf

sr 

sr  2.221

D yinit  Δy

y' 

y'  0.265

Lf


Di  D  d

Di  60 mm

Outside coil dia

Do  D  d

Do  76 mm




Largest pin

pin max  Di  0.05 D



3

ρ  0.285  lbf  in 2 2

Weight

Wa 

π  d  D Na ρ 4

Wa  9.6 N


fn 

1 2



k g Wa

fn  53.68  Hz


Forcing frequency Frequency ratio

ff  fn ff

ωf 2 π

ff  10.42  Hz

 5.2

which is a little low.



14-28-4


d  8  mm

Outside diameter

Do  76.0 mm

Total coils

Nt  13.50

Free length

Lf  151.038  mm

ends squared



Finit  225  N

Working force


Fwork  450  N


Yielding

Nys 

Fatigue

Nfs 

Surging

Nsurge 



Nys  5.3

Nfs  3.3

Nsurge  5.2



14-29-1


A helical extension spring, loaded in fatigue, has been designed for infinite life with the data given below. Find the safety factors for failure in the standard hooks. State all assumptions and sources of empirical data.

Given:

Minimum force

Fmin  665  N

Life

L  ∞

Maximum force

Fmax  935  N

Shear modulus

G  79.3 GPa

Working deflection Wire endurance limit

Δy  2.00 in

Spring index Wire strength

C  9 A  2059.2 MPa

Wire diameter Number active coils

d  8  mm Na  13.75

Solution: 1.

(Chrome-silicon) b  0.0934



2.

S ew  310  MPa

Fmax  Fmin

Fa  135.0  N

2 Fmax  Fmin

Fm  800.0  N

2


Kb  3.

4 C  C  1


4.

D  72 mm

3 2   3 2   2.987  C  139.7  C  3427 C  38404   psi

τi1  4.231  C  181.5  C  3387 C  28640  psi

τi1  67.4 MPa

τi2

τi2  115.1  MPa

τi1  τi2

τi  91.3 MPa

2


6.

D  C d

Use the value of C to find an appropiate value of initial coil stress i from equations 14.21:

τi  5.

Kb  1.090

4 C ( C  1)

Ks  1 

0.5


τm  Ks

8  Fm D

π d 7.

Ks  1.056

C

τm  302.4  MPa

3


Alternating stress

Kw 

4 C  1 4 C  4

τa  Kw



8  F a D

π d

3

0.615 C

Kw  1.162

τa  56.2 MPa



9.

14-29-2

Find the ultimate tensile strength of this wire material from equation 14.3 and Table 14-4 and use it to find the ultimate shear strength from equation 14.4 and the torsional yield strength for the coil body from Table 14-10, assuming no set removal.

   mm 

b


S ut  A  


S ys  0.50 S ut

S ys  847.9  MPa


S us  0.67 S ut

S us  1136.1 MPa

d

S ut  1695.7 MPa

Find the wire endurance limit for unpeened springs from equation 14.13 and convert it to fully reversed endurance strength with equation 14.16c. S es  0.5

Fully reversed endurance limit


S es  179.49 MPa

10. The stresses in the end hooks also need to be determined. The bending stresses in the hook are found from equation 14.23: C1 =

2  R1 d

=

2 D

C1  C

=C

2 d

C1  9.00

2

Kb 

4 C1  C1  1

Kb  1.09

4  C1  C1  1  16 D Fa

σa  Kb




3

π d

16 D Fm


3



3

σa  108.1  MPa

2

4  Fm

π d

16 D Fmin

π d

4  Fa



σm  640.6  MPa

2

4  Fmin

π d

2

σmin  532.5  MPa


S es

S e  311.1  MPa

0.577 S e  S ut  σmin

Nfb 


Nfb  1.67


C2 d

Kw2 

2 4  C2  1 4  C2  4

R2  20 mm Kw2  1.188



τBa  Kw2

8  Fa D



3

8  Fm D


3

8  Fmin D

π d

14-29-3

3

τBm  340.2  MPa





Nfs  2.0



14-30-1


Given the following data for a helical torsion spring, loaded in fatigue, find the spring index, unloaded coil diameter, minimum loaded coil diameter, and safety factor in fatigue. State all assumptions and sources of empirical data used.

Given:

Minimum deflection

θmin  0.25 rev

Maximum deflection Number active coils

θmax  0.75 rev Na  20

Bending modulus

6

E  30 10  psi

Spring rate k  60 in lbf  rev Wire dia. d  0.192  in

1

Design Choices: Use unpeened music wire with 2-in-long, straight ends. The coil is loaded to close it. Yield strength factor Ks  0.80 (from Table 14-13)

Solution: 1.

Length of ends

L1  2  in

L2  2  in


A  184.65 ksi b  0.1625 (ASTM A228 wire) S ewb  77.99  ksi


Find the minimum, maximum, mean, and alternating loads. Mmin  k θmin

Mmin  15 in lbf

Mmax  k θmax

Mmax  45 in lbf

Mmax  Mmin

Ma 

Ma  15.0 in lbf

2

Mm 

Mmax  Mmin

Mm  30.0 in lbf

2

3

2.

k=

Use equation 14.28 to calculate the spring index.

d E 10.8 C Na

3

C  3.

d E

C  16.38

10.8 k rev Na


Kbi 

4 C  C  1

Kbi  1.048

4  C ( C  1 ) 2

Kbo  4.

4 C  C  1


5.

D  C d

D  3.146  in



32 Mmax 3

π d 6.

Kbo  0.956

4  C ( C  1 )

σimax  67.9 ksi



32 Mmin

π d

3

σomin  20.6 ksi



σomax  Kbo

32 Mmax

π d σm 

σa  7.

8.

σomax  61.9 ksi

3

σomax  σomin

σm  41.3 ksi

2

σomax  σomin

σa  20.6 ksi

2


S ut  A  



  in   d

b

S ut  241  ksi S y  193  ksi

Convert the wire bending endurance limit for unpeened springs from equation 14.33 to fully reversed endurance strength with equation 14.34b. Fully reversed endurance limit

9.

14-30-2

S e  0.5

S ewb S ut

S e  46.51  ksi

S ut  0.5 S ewb

The fatigue safety factor for the coils in bending is calculated from equation 14.34a. Fatigue factor of safety

Nfb 


Nfb  1.7


10. The ends contribute to the active coils from equation 14.26a as

Ne 

L1  L2 3  π D

Ne  0.13


Nb  20

11. The unloaded coil diameter and minimum loaded coil diameters are: Unloaded coil diameter

Di  D  d

Loaded coil diameter

Dimin 

Di  2.954  in D Nb

Nb 

θmax

d

Dimin  2.839  in

rev



14-31-1


A helical compression spring is required to provide a minimum force of 150 lb at installation and have a working deflection of 1 in. The spring rate is 75 lb/in. The coil must fit in a 2.1-in-dia hole with 0.1-in clearance. Use 0.250-in diameter, unpeened, music wire and squared/ground ends. Using a 15% clash allowance, find: (a) The stresses and safety factor for infinte life in fatigue. (b) The shut height. (c) The stress and safety factor at shut height. (d) The total number of coils. (e) The free length. (f) The natural frequency in Hz. (g) Draw a Goodman diagram and show the safety factor from (a) on it.

Given:

Minimum force


Spring rate

6

Spring OD OD  2.000  in Working deflection Δy  1.000  in Music wire properties: Strength A  184.65 ksi b  0.1625 Solution: 1.

2.

From the given information, find the maximum force and the spring index. Maximum force

Fmax  Fmin  k Δy

Spring index

C 

Fmax  225  lbf

OD  d

C7

d


Mean force

4.

Shear modulus G  11.5 10  psi Clash factor α  0.15 Wire diameter d  0.250  in Set after winding, unpeened


Alternating force

3.

1

k  75 lbf  in

Fa  Fm 

Fmax  Fmin

Fa  37.5 lbf

2 Fmax  Fmin

Fm  187.5  lbf

2


Ks  1 

Wahl factor

Kw 

0.5

Ks  1.071

C

4 C  1 4 C  4



0.615 C

Kw  1.213


Kys  0.60



KU  0.67

(equation 14.4)



(unpeened)

Calculate the ultimate tensile strength, ultimate shear strength, and the shear endurance limit. Ultimate tensile strength

S ut  A  

  in  



d

b

S ut  231  ksi S us  155  ksi



5.



Fatigue strength

S es  0.5

Stress at Fm

S es  26.3 ksi

D  1.750  in

τi  Ks

8  Fmin D

τm  Ks

τi  45.8 ksi

3

8  Fm D

π d

τm  57.3 ksi

3


8  F a D

τa  Kw

π d

τa  13.0 ksi

3


9.

S us  0.5 S ew

D  C d

π d

8.

S ew S us


7.

S ys  139  ksi


6.

14-31-2

Nfs 


Nfs  1.2



4

Na 

d G 3

Na  13.97

Na  14

8 D  k


4

d G

k 

3

k  74.84 

8  D  Na

lbf in


Nt  Na  2

Nt  16.00



Lshut  4  in


yinit 

Fmin k

yinit  2.004  in




14. The free length (see Figure 14-8) can now be found from © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


14-31-3

Lf  Lshut  Δyclash  Δy  yinit

Lf  7.154  in


yshut  3.154  in




τshut  Ks

8  Fshut D


Ns 


3

S ys

Ns  1.92

τshut

which is acceptable. 18. The inside and outside coil diameters are Inside coil dia

Di  D  d

Di  1.5 in

Outside coil dia

Do  D  d

Do  2  in




Largest pin

pin max  Di  0.05 D


20. The weight of the springs active coils is found from equation 14.11b 3

Weight density

ρ  0.285  lbf  in

Weight

Wa 

2 2

π  d  D Na ρ

Wa  1.077  lbf

4


fn 

1 2



k g Wa

fn  81.91  Hz

22. We now have results for questions (a) through (f): (a) Fatigue stresses and safety factor

τi  45.8 ksi

τm  57.3 ksi

Nfs  1.2

(b) Shut height

Lshut  4.000  in

(c) Stress at shut height


and safety factor

τa  13 ksi

Ns  1.9

(d) Total coils

Nt  16.00

(e) Free length

Lf  7.154  in

(f) Natural frequency

fn  81.91  Hz




14-31-4

23. (g) Plot the Goodman diagram.

τ'a τ'm  

The equation for the Goodman line is

S es S us

τa

 τ'm  S es

Equation for load line

τaL τ'm 


τ'm  0  ksi 10 ksi  200  ksi


30

 τ'm

τm

τm

25

ksi

20 15

τa

10

ksi

5 0

0

25

50

75

100

125

150

175

200

Mean shear stress

The Goodman boundary is the solid line with negative slope. The stress state is at the intersection of the vertical and horizontal dashed lines. The safety margin is the distance along the load line from the stress state point to the intersection of the load line with the Goodman boundary.



14-32-1


A helical compression spring is required to provide a time-varying force that ranges from a mimimum of 100 lb to a maximum of 300 lb over a deflection of 1 in. It needs to work freely over shaft of 1.25-in dia. Use a cold-drawn carbon steel wire having an S ut = 250 ksi. A spring index of 6, a clash allowance of 15%, and squared and ground ends are desired.

Given:

Minimum force


Spring rate

k  75 lbf  in

Maximum force Working deflection Wire strength

Fmax  300  lbf Δy  1.000  in S ut  250  ksi

Shear modulus Clash factor Spring index

G  11.5 10  psi α  0.15 C  6

Shaft OD

ODshaft  1.250  in

Unset after winding, unpeened Solution: 1.

2.

From the given information, find a trial wire diameter. Maximum shaft dia

ODshaft = IDspring  0.05 D = ( D  d )  0.05 D = C d  d  0.05 C d

Trial wire diameter

d 

Standard wire size

d  0.343  in

ODshaft

d  0.266  in

0.95 C  1

(Found by iteration, see note in step 8)


Mean force

4.

6


Alternating force

3.

1

Fa  Fm 

Fmax  Fmin

Fa  100.0  lbf

2 Fmax  Fmin

Fm  200.0  lbf

2


Ks  1 

Wahl factor

Kw 

0.5

Ks  1.083

C

4 C  1 4 C  4



0.615 C

Kw  1.252


Kys  0.45



KU  0.67

(equation 14.4)



(unpeened)

Calculate the ultimate tensile strength, ultimate shear strength, and the shear endurance limit. S ut  250  ksi

Ultimate tensile strength Ultimate shear strength


S us  168  ksi



S ys  113  ksi

Fatigue strength

S es  0.5


S es  26.0 ksi




6.

D  C d

D  2.058  in


τi  Ks

8  Fmin D


τm  28.1 ksi

3


8  F a D

τa  Kw

π d 8.

τi  14.1 ksi

3

8  Fm D

τm  Ks

π d 7.

14-32-2

τa  16.3 ksi

3


Nfs 


Nfs  1.3


This was originally too low. The wire diameter in step 1 was repeatedly changed to get a satisfactory safety factor. 9.


k 

Fmax  Fmin

k  200.00

Δy

lbf in


4

Na 

d G 3

Na  11.413

Na  11.5

8 D  k


4

d G

k 

3

k  198.50

8  D  Na

lbf in


Nt  Na  2

Nt  13.50



Lshut  4.631  in


yinit 

Fmin k

yinit  0.504  in






14-32-3


Lf  6.284  in


yshut  1.654  in




τshut  Ks

8  Fshut D


Ns 


3

S ys

Ns  2.4

τshut

which is acceptable. 19. The inside and outside coil diameters are Inside coil dia

Di  D  d

Di  1.715  in

Outside coil dia

Do  D  d

Do  2.401  in




Largest pin

pin max  Di  0.05 D



3

ρ  0.285  lbf  in 2 2

Weight

Wa 

π  d  D Na ρ

Wa  1.958  lbf

4


fn 

1 2



k g Wa

fn  98.92  Hz

23. We now have a complete design specification for this cold-drawn wire spring: Wire diameter

d  0.343  in

Outside diameter

Do  2.401  in

Total coils

Nt  13.50

Free length

Lf  6.284  in




14-33-1

PROBLEM 14-33

_____

Statement:

Find the ultimate tensile strength and the ultimate shear strength of ASTM A228 music wire in the following preferred diameters: 0.5 mm, 1.0 mm, 2.0 mm, 4.0 mm, and 6.0 mm.

Given:

Wire diameters: i  1 2  5 d  0.5 mm d  1.0 mm 1

Solution: 1.

d  2.0 mm

2

4

d  6.0 mm 5


Calculate the ultimate tensile strength using equation 14.3 and the coefficient and exponent from Table 14-4. A  2153.5 MPa

For ASTM A228 music wire,

 di   S ut  A   i  mm 

2.

d  4.0 mm

3

b

d

i

mm



S ut

i

GPa

b  0.1625



0.5

2.41

1.0

2.15

2.0

1.92

4.0

1.72

6.0

1.61

Calculate the ultimate shear strength using equation 14.4. d S us  0.67 S ut i

i

i

mm



S us

i

GPa



0.5

1.61

1.0

1.44

2.0

1.29

4.0

1.15

6.0

1.08



14-34-1

PROBLEM 14-34

_____

Statement:

Find the ultimate tensile strength and the ultimate shear strength of ASTM A228 music wire in the following preferred diameters: 0.020 in, 0.038 in, 0.081 in, 0.162 in, and 0.250 in.

Given:

Wire diameters: i  1 2  5 d  0.020  in d  0.038  in 1

Solution: 1.

3

d  0.162  in 4

d  0.250  in 5


Calculate the ultimate tensile strength using equation 14.3 and the coefficient and exponent from Table 14-4. A  184.649  ksi

For ASTM A228 music wire,

 di  S ut  A    i  in 

2.

d  0.081  in

2

b

d

i

in



S ut

i

ksi

b  0.1625



0.020

348.7

0.038

314.1

0.081

277.8

0.162

248.2

0.250

231.3

Calculate the ultimate shear strength using equation 14.4. d S us  0.67 S ut i

i

i

in



S us

i

ksi



0.020

234

0.038

210

0.081

186

0.162

166

0.250

155



PROBLEM 14-35

14-35-1

_____

Statement:

Select the preferred diameter for an ASTM A227 cold-drawn wire that will have an ultimate tensile strength as close to, but not less than, 180 kpsi.

Given:

Minimum tensile strength: S ut  180  ksi

Solution:


1.

Calculate the maximum wire diameter using equation 14.3 and the coefficient and exponent from Table 14-4. For ASTM A227 cold-drawn wire,

A  141.04 ksi

b  0.1822

1 b

 S ut  d trial     in A 2.

d trial  0.262  in

Choose the next lower diameter from Table 14-2 as it will have a higher ultimate strength than the next higher diameter. d  0.250  in

3.

Verify that the strength of a wire of this diameter is greater than the minimum required value.

  in  

S ut  A  

d

b

S ut  182  ksi



PROBLEM 14-36

14-36-1

_____

Statement:

Select the preferred diameter for an ASTM A229 oil-tempered wire that will have an ultimate tensile strength as close to, but not less than, 1430 MPa.

Given:

Minimum tensile strength: S ut  1430 MPa

Solution:


1.

Calculate the maximum wire diameter using equation 14.3 and the coefficient and exponent from Table 14-4. For ASTM A229 oil-tempered wire, A  1831.2 MPa

b  0.1833

1 b

 S ut  d trial     mm A 2.

d trial  3.85 mm

Choose the next lower diameter from Table 14-2 as it will have a higher ultimate strength than the next higher diameter. d  3.5 mm

3.

Verify that the strength of a wire of this diameter is greater than the minimum required value.

  mm  

S ut  A  

d

b

S ut  1455 MPa



14-37-1


Given:

Design a helical extension spring to handle a dynamic load that varies from 275 N to 325 N over 10 mm working deflection. Use chrome silicon wire and standard hooks. The forcing frequency is 800 rpm. Infinite life is desired. Minimize the package size. Choose appropriate safety factors against fatigue, yielding, and surging. Life Minimum force Fmin  275  N L  ∞ Maximum force

Fmax  325  N

Shear modulus

G  80.8 GPa

Working deflection

Δy  10.0 mm

Forcing freq.

ωf  800  rpm

Nfb  1.5

Spring index

C  9

S ew  310  MPa

Wire strength

A  2059.2 MPa


b  0.0934 Solution: 1.



2.

Fmax  Fmin

Fa  25.0 N

2 Fmax  Fmin

Fm  300.0  N

2


Kb  3.

4 C  C  1

Kb  1.090

4 C ( C  1)


 4   4  K b C  1  d=  Nfb Fm   Nfb  1   Fmin  π A    0.577   0.67 A  db  0.5 S ew       Nfb Fa    0.5  0.67 S  ew   

     

2 b

1

RHS( d ) 

4.

 4  4 Kb C  1  Nfb Fm   Nfb  1   Fmin   π A  mm2  b     d     0.5  S 0.577  0.67 A    ew    mm      Nfb Fa    0.5 0.67 S ew    

d  1.000  mm

RHS( d )  4.011  mm

d  RHS( d )

RHS( d )  3.879  mm

d  RHS( d )

RHS( d )  3.882  mm

d  RHS( d )

RHS( d )  3.882  mm

d  4.0 mm

D  C d

D  36 mm

      

2 b

 mm





τi1  4.231  C  181.5  C  3387 C  28640  psi

τi1  67.4 MPa

τi2

τi2  115.1  MPa

τi  6.

τi1  τi2

τi  91.3 MPa

2


7.

14-37-2

Ks  1 

0.5

Ks  1.056

C


Fi 

π d  τi 8  K s D

Fi  60.4 N



τm  Ks

8  Fm D

π d 9.

τm  453.6  MPa

3


Alternating stress

Kw 

4 C  1 4 C  4



τa  Kw

0.615 C

8  F a D

π d

3

Kw  1.162

τa  41.6 MPa

10. Find the ultimate tensile strength of this wire material from equation 14.3 and Table 14-4 and use it to find the ultimate shear strength from equation 14.4 and the torsional yield strength for the coil body from Table 14-10, assuming no set removal. b


d  S ut  A     mm 


S ys  0.45 S ut

S ys  814.1  MPa


S us  0.67 S ut

S us  1212.1 MPa

S ut  1809.1 MPa


S es  0.5


S es  177.73 MPa


τmin  τm  τa

τmin  412.0  MPa


MACHINE DESIGN - An Integrated Approach, 4th Ed. Nfs 


14-37-3



Nfs  2.5


2  R1 d

2 D

=

C1  C

=C

2 d

C1  9.00

2

Kb 

4 C1  C1  1

Kb  1.09

4  C1  C1  1  16 D Fa

σa  Kb


3

π d

16 D Fm


4  Fa





3

4  Fm

π d

16 D Fmin

π d

3

σa  80.07  MPa

2



σm  960.9  MPa

2

4  Fmin

π d

2

σmin  880.82 MPa


S es

S e  308.02 MPa

0.577 S e  S ut  σmin

Nfb 


Nfb  1.7


C2 d

Kw2 

R2  10 mm

2 4  C2  1

Kw2  1.188

4  C2  4

τBa  Kw2

8  Fa D



3

8  Fm D


3

8  Fmin D

π d

3

τBm  510.3  MPa




14-37-4



Nfs  2.2



Spring rate

Fmax  Fmin

k  5000.0

Δy

N m



Na 

d G

Na  11.084

3

Na  11.0

8 D  k Note that we round it to the nearest 1/4 coil as the manufacturing tolerance cannot achieve better than that accuracy. Having rounded the number of active coils, we must now calculate the spring rate using equation 14. Corrected spring rate

4

k 

d G

k  5038.0

3

8  D  Na

N m


Nt  Na  1

Nt  12.00

Body length

Lb  Nt d

Lb  48.0 mm



Lhook  32.0 mm

Free length


Lf  112.0  mm


τi1  4.231  C  181.5  C  3387 C  28640  psi

τi1  67.4 MPa

τi2

τi2  115.1  MPa

τi 

τi1  τi2

τi  91.3 MPa

2 3

Fi 

π d  τi

Fi  60.4 N

8  K s D

ymax 

Fmax  Fi

ymax  52.53  mm

k


Di  D  d

Di  32.00  mm

Outside coil dia

Do  D  d

Do  40.00  mm


3

ρ  0.28 lbf  in



14-37-5

2 2

Weight

Wa 

π  d  D Na ρ

Wa  1.188  N

4


fn 

1 2



k g

fn  102  Hz

Wa


Frequency ratio

ff  fn ff

ωf

ff  13.333 Hz

2 π

 7.6



d  4.00 mm

Total coils

Nt  12.00

Outside diameter

Do  40.00  mm

Free length

Lf  112.00 mm

Standard hooks



14-38-1


Design a helical extension spring with standard hooks to handle a dynamic load that varies from 60 lb to 75 lb over 0.5 in working deflection. Use music wire. The forcing frequency is 1200 rpm. Infinite life is desired. Minimize the package size. Choose appropriate safety factors against fatigue, yielding, and surging.

Given:

Minimum force


Life

L  ∞

Maximum force Working deflection

Fmax  75 lbf Δy  0.50 in

Shear modulus Forcing freq.

G  11.7 10  psi ωf  1200 rpm

Nfb  1.5

Spring index

C  9


Wire strength

A  184.649  ksi


6

b  0.1625 Solution: 1.



2.

Fmax  Fmin

Fa  7.5 lbf

2 Fmax  Fmin

Fm  67.5 lbf

2


Kb  3.

4 C  C  1

Kb  1.090

4 C ( C  1)


d=


     

2 b

1

 4  4 Kb C  1 RHS( d )    Nfb Fm   Nfb  1   Fmin    π A  in2 b    d     0.577  0.67 A     0.5 S ew       in    Nfb Fa     0.5 0.67 S ew    

4.

      

d  1.00 in

RHS( d )  0.153  in

d  RHS( d )

RHS( d )  0.167  in

d  RHS( d )

RHS( d )  0.166  in

d  RHS( d )

RHS( d )  0.166  in

d  0.177  in

D  C d

D  1.593  in

2 b

 in





τi1  4.231  C  181.5  C  3387 C  28640  psi

τi1  9.77 ksi

τi2

τi2  16.70  ksi

τi  6.

τi1  τi2

τi  13.2 ksi

2


7.

14-38-2

0.5

Ks  1 

Ks  1.056

C


Fi 

π d  τi 8  K s D

Fi  17.1 lbf



τm  Ks

8  Fm D

π d 9.

τm  52.1 ksi

3


Alternating stress

Kw 

4 C  1 4 C  4

τa  Kw



0.615 C

8  F a D

π d

3

Kw  1.162

τa  6.4 ksi

10. Find the ultimate tensile strength of this wire material from equation 14.3 and Table 14-4 and use it to find the ultimate shear strength from equation 14.4 and the torsional yield strength for the coil body from Table 14-10, assuming no set removal.

  in  

b


S ut  A  


S ys  0.50 S ut

S ys  122.3  ksi


S us  0.67 S ut

S us  163.9  ksi

d

S ut  244.7  ksi

11. Find the wire endurance limit for unpeened springs from equation 14.13 and convert it to fully reversed endurance strength with equation 14.16c. S ew S us Fully reversed S es  0.5 S es  26.08  ksi S us  0.5 S ew endurance limit 12. The fatigue safety factor for the coils in torsion is calculated from equation 14.16b. Minimum stress Fatigue safety factor




τmin  45.75  ksi Nfs  2.5



14-38-3


2  R1 d

2 D

=

C1  C

=C

2 d

C1  9.00

2

Kb 

4 C1  C1  1

Kb  1.09

4  C1  C1  1  16 D Fa

σa  Kb


3

π d

16 D Fm


4  Fa





3

4  Fm

π d

16 D Fmin

π d

3

σa  12.27  ksi

2



σm  110.4  ksi

2

4  Fmin

π d

2

σmin  98.15  ksi


S es

S e  45.20  ksi

0.577 S e  S ut  σmin

Nfb 


Nfb  1.9


C2 d

Kw2 

R2  0.442  in

2 4  C2  1

Kw2  1.188

4  C2  4

τBa  Kw2

8  Fa D


τBa  6.5 ksi

3

8  Fm D


3

8  Fmin D

π d

3

τBm  58.6 ksi

τBmin  52.1 ksi




Nfs  2.4



14-38-4


Spring rate

Fmax  Fmin

k  30.0

Δy

lbf in



d G

Na 

Na  11.836

3

Na  11.75

8 D  k Note that we round it to the nearest 1/4 coil as the manufacturing tolerance cannot achieve better than that accuracy. Having rounded the number of active coils, we must now calculate the spring rate using equation 14. 4


d G

k 

k  30.2

3

8  D  Na

lbf in


Nt  Na  1

Nt  12.75

Body length

Lb  Nt d

Lb  2.3 in



Lhook  1.42 in

Free length


Lf  5.09 in


τi1  4.231  C  181.5  C  3387 C  28640  psi

τi1  9.8 ksi

τi2

τi2  16.7 ksi

τi 

τi1  τi2

τi  13.2 ksi

2 3

Fi 

π d  τi

Fi  17.1 lbf

8  K s D

ymax 

Fmax  Fi

ymax  1.91 in

k


Di  D  d

Di  1.416  in

Outside coil dia

Do  D  d

Do  1.770  in


3

ρ  0.28 lbf  in 2 2

Weight

Wa 

π  d  D Na ρ 4

Wa  0.405  lbf


fn 

1 2



k g Wa

fn  84.9 Hz



14-38-5


Frequency ratio

ff  fn ff

ωf

ff  20 Hz

2 π

 4.2



d  0.177  in

Total coils

Nt  12.75

Outside diameter

Do  1.770  in

Free length

Lf  5.089  in

Standard hooks



14-39-1


Given:

Design a straight-ended helical torsion spring for a static load 50 N-m at a deflection of 60 deg with a safety factor of 2. Specify all parameters necessary to manufacture the spring. State all assumptions. Bending modulus Applied moment M  50 N  m E  206.8  GPa Deflection at load θ  60 deg θ  0.167  rev

Design Choices: Use unpeened oil tempered wire with 40-mm-long, straight ends. The coil is loaded to close it. Design safety factor Nyd  2 Spring index C  10

Solution: 1.


Ks  0.85 (from Table 14-13)

Length of ends

L1  40 mm


A  1831.2 MPa b  0.1833 (ASTM A229 wire)

L2  40 mm



Kbi  2.

4 C  C  1

Kbi  1.081

4  C ( C  1 )


 32 Kbi Nyd  M  d    π Ks A  mm3    3.

4.

 mm

d  10.272 mm


d  11 mm

Spring index

C  10

Mean coil diameter

D  C d

D  110  mm



32 M

π d 5.

3 b


3


S ut  A  

  mm  



d

b

S ut  1180 MPa S y  1003 MPa Nyb 

Sy

Nyb  2.4

6.


7.


8.

k 

M

θ

σimax

k  300 

N m rev




14-39-2

4


Na 

d E 10.8 D k rev

Na  8.50



L1  L2 3  π D

Ne  0.08


Nb  8.4


θ  10.8

M  D Na 4

 rev

θ  60.0 deg

d E



14-40-1


Design a straight-ended helical torsion spring for a static load 430 in-lb at a deflection of 55 deg with a safety factor of 2. Specify all parameters necessary to manufacture the spring. State all assumptions.

Given:

Applied moment Deflection at load

6

M  430  in lbf θ  55 deg

Bending modulus E  30 10  psi θ  0.153  rev

Design Choices: Use unpeened music wire with 2-in-long, straight ends. The coil is loaded to close it. Spring index Design safety factor Nyd  2 C  9

Solution: 1.



Length of ends

L1  2  in

L2  2  in


A  184.65 ksi

b  0.1625 (ASTM A228 wire)



Kbi  2.

4 C  C  1

Kbi  1.090

4  C ( C  1 )


 32 Kbi Nyd  M  d    π Ks A  in3    3.

4.

 in

d  0.352  in


d  0.362  in

Spring index

C9

Mean coil diameter

D  C d

D  3.258  in



32 M

π d 5.

3 b


3


S ut  A  

   in 



d

b

S ut  218  ksi S y  218  ksi Nyb 

Sy

Nyb  2.2

6.


7.


8.

k 

M

θ

σimax

k  2815

in lbf rev




14-40-2

4


Na 

d E 10.8 D k rev

Na  5.20



L1  L2 3  π D

Ne  0.13


Nb  5.1


θ  10.8

M  D Na 4

 rev

θ  55.0 deg

d E



14-41-1


Given the following data for a helical torsion spring, loaded in fatigue, find the spring index, unloaded coil diameter, minimum loaded coil diameter, and safety factor in fatigue. State all assumptions and sources of empirical data used. Deflection at assembly = 0.15 rev, working deflection = 0.35 rev, k = 10 N-m/rev, Na = 25, 4.50 mm oil-tempered wire, unpeened.

Given:

Minimum deflection

θmin  0.15 rev

Bending modulus

E  206.8  GPa

Maximum deflection Number active coils

θmax  0.50 rev

Spring rate Wire diameter

k  10 N  m rev d  4.50 mm

Na  25

1

Design Choices: Use unpeened oil-tempered wire with 40-mm-long, straight ends. The coil is loaded to close it. Yield strength factor Ks  0.85 (from Table 14-13)

Solution: 1.

Length of ends

L1  40 mm

L2  40 mm


A  1831.2 MPa b  0.1833 (ASTM A229 wire) S ewb  537  MPa


Find the minimum, maximum, mean, and alternating loads. Mmin  k θmin

Mmin  1.50 N  m

Mmax  k θmax

Mmax  5.00 N  m

Mmax  Mmin

Ma 

Ma  1.75 N  m

2

Mm 

Mmax  Mmin

Mm  3.25 N  m

2

3

2.

k=

Use equation 14.28 to calculate the spring index.

d E 10.8 C Na

3

C  3.

d E

C  6.98

10.8 k rev Na


Kbi 

4 C  C  1

Kbi  1.119

4  C ( C  1 ) 2

Kbo  4.

4 C  C  1


5.

D  C d

D  31.41  mm



32 Mmax

π d 6.

Kbo  0.902

4  C ( C  1 )

3






32 Mmin


σa  7.

8.

32 Mmax

σomax  503.9  MPa

3

σomax  σomin

σm  327.5  MPa

2

σomax  σomin

σa  176.4  MPa

2


S ut  A  



  mm   d

b

S ut  1390 MPa S y  1181 MPa

Convert the wire bending endurance limit for unpeened springs from equation 14.33 to fully reversed endurance strength with equation 14.34b. Fully reversed endurance limit

9.

σomin  151.2  MPa

3

π d σm 

14-41-2

S e  0.5

S ewb S ut

S e  332.78 MPa

S ut  0.5 S ewb

The fatigue safety factor for the coils in bending is calculated from equation 14.34a. Fatigue factor of safety

Nfb 


Nfb  1.4


10. The ends contribute to the active coils from equation 14.26a as

Ne 

L1  L2 3  π D


Ne  0.27 Nb  25

11. The unloaded coil diameter and minimum loaded coil diameters are: Unloaded coil diameter

Di  D  d

Loaded coil diameter

Dimin 

Di  26.9 mm D Nb

Nb 

θmax

d

Dimin  26.3 mm

rev



14-42-1


Design a Belleville spring to give a static force of approximately 2000 lb at a maximum deflection of 0.05 in with a nearly constant spring rate.

Given:

Maximum force Fflat  2000 lbf

Maximum deflection

ymax  0.05 in 6

Assumptions: The diameter ratio is Rd  2. Use unset carbon spring steel 50HRC. Properties, E  30 10  psi ,

ν  0.28 . Solution: 1.


Since a constant rate spring is needed, the h/t ratio (see Figure 14-30) is hovert  0.40

h/t ratio 2.

Use the above choice and the specified maximum deflection to find the spring thickness t: h  ymax

3.

h

t 

t  0.125  in

hovert

Use equation 14.37 to find the force at maximum deflection. This must be done by iterating on the outside diameter, Do. Do  3.0 in

Guess:

7

19.2 10  psi  hovert t

Fflat 

4

Fflat  2083.33  lbf

2

Do 4.


  Rd  1  2  K1   π ln Rd   2   Rd  6

K2 

K3 

K4 

K5 

 Rd  1

6



π ln Rd   ln Rd  6

π ln Rd 

 Rd   2

K1  0.689



 1

K2  1.220



1



 

K3  1.378


K5  1

2   Rd  1 

σc  

4  E ymax 2







2

K1 Do  1  ν

 

 

 K2  h 

ymax  2

   K3 t  

ymax       K2  h    K3 t 2    K1 Do2 1  ν2     

σti  

4  E ymax



K4  1.115



σc  213  ksi

σti  148.8  ksi



 4  E ymax  K1 Do2 1  ν2 

σto  

5.





14-42-2

ymax       K4  h    K5 t 2     

σto  161  ksi


1.20 S ut

σc

Ns  1.4



Do  3.000  in

Inside diameter

Di 

Do Rd

Thickness

t  0.125  in

Height

h  0.050  in

Di  1.500  in



14-43-1


Design a Belleville spring to give a static force of 400 lb at 50% deflection to flat and 200 lb at fla

Given:

Force at flat

Fflat  200  lbf 6

Assumptions: The diameter ratio is Rd  2. Use unset carbon spring steel 50HRC. Properties, E  30 10  psi ,

ν  0.28 . The outside diameter of the spring is Do  4.00 in Solution:


1.

Since a spring with 200% force to flat at 50% deflection to flat is needed, the h/t ratio (see Figure 14-30) is

2.

h/t ratio hovert  2.828 Use the above assumptions and the specified nominal force in equation 14.37a to find an appropriate spring thickness t: 1 4

2  Do  Fflat  t    7 hovert   19.2 10  psi 

t  0.049  in

h  hovert t

h  0.139  in

3.


4.

Figure 14-34 shows that the worst stress state will occur at the largest deflection ymax, so solve equations 14.36 for stresses at that deflection: ymax  h

  Rd  1  2   R2  d  

K1 

6

π ln Rd 



K2 

6

π ln Rd 



K3 

6



K4 

π ln Rd 

 Rd  1  ln Rd 

K1  0.689



 1

K2  1.220



 Rd  1    2 

K3  1.378


K5 

K5  1

2   Rd  1  4  E ymax

σc  

2





 4  E ymax  K1 Do2 1  ν2  



 



 

 K2  h 

2

K1 Do  1  ν

σti  

ymax 

   K3 t 2  

ymax       K2  h    K3 t 2     

ymax       K4  h    K5 t 2    K1 Do2 1  ν2     

σto  

4  E ymax



K4  1.115



σc  252  ksi

σti  28.2 ksi

σto  209  ksi



14-43-2


1.20 S ut

σc

Ns  1.2



Do  4.000  in

Inside diameter

Di 

Do Rd

Thickness

t  0.049  in

Height

h  0.139  in

Di  2.000  in



14-44-1


Design a helical compression spring for a static load of 60 lb at a deflection of 1.50 in with a safety factor of 2.0 to work in a 1.06-in hole. Specify all parameters necessary to manufacture the spring.

Given:

Working force Working deflection

6

Fwork  60 lbf ywork  1.50 in

Shear modulus Safety factor

G  11.5 10  psi Ns  2.0

Clash allowance

α  0.15

Set removed

Km  0.65

ASTM A228 wire

A  184.649  ksi b  0.1625

Spring index C  7.0 (found by trial-and-error)

Design choices:

Solution:


Fwork

k  40

lbf

1.


2.


ywork

in

1

d 

2 b  8 Ns ( C  0.5)  Fwork  ( 1  α)  in 2   π Km A  in  

Wire diameter 3.

d  0.125  in

Let

d  0.125  in

Calculate the mean and outside coil diameters and number of active coils. Mean coil dia

D  C d

D  0.875  in

Outside coil dia

Do  D  d

Do  1.000  in


Na 

4

d G 3

8 D  k

Na  13.097 Na  13.0


3

k  40.298

8  D  Na

lbf in

Nt  Na  2

Nt  15.00


6.

d G


5.

4

k 

Ls  d  Nt

Ls  1.875  in



yshut  1.725  in

Free length


Lf  3.600  in



14-44-2

To check for buckling, two ratios need to be calculated, Lf/D and y max/Lf. Slenderness ratio

Deflection ratio

sr  y' 

Lf

sr  4.114

D ywork

y'  0.417

Lf


9.


Di  D  d

Di  0.750  in

Outside coil dia

Do  D  d

Do  1.000  in




Largest pin

pin max  Di  0.05 D



3

ρ  0.28 lbf  in 2 2

Weight

Wt 

π  d  D Nt ρ

Wt  0.14 lbf

4


d  0.125  in

Outside diameter

Do  1.000  in

Total coils

Nt  15.00

Free length

Lf  3.600  in




14-45-1


Design a helical compression spring for a static load of 200 N at a deflection of 40 mm with a safety factor of 1.8 to work in a 25-mm hole. Specify all parameters necessary to manufacture the spring.

Given:

Working force

Fwork  200  N

Shear modulus

G  80.8 GPa


Safety factor

Ns  1.8

Clash allowance

α  0.15

Set removed

Km  0.65

ASTM A228 wire

A  2153.5 MPa b  0.1625

Spring index C  7.5 (found by trial-and-error)

Working deflection Design choices:

Solution:


Fwork

k  5.000 

N

1.


2.


ywork

mm

1 2 b  8 Ns ( C  0.5)  Fwork  ( 1  α) d   mm 2   π Km A  mm  

Wire diameter 3.

d  2.658  mm

Let

d  2.8 mm

Calculate the mean and outside coil diameters and number of active coils. Mean coil dia

D  C d

D  21.000 mm

Outside coil dia

Do  D  d

Do  23.800 mm


Na 

4

d G 3

8 D  k

Na  13.407 Na  13.5


7.

3

k  4.965 

8  D  Na

N mm

Nt  Na  2

Nt  15.50


6.

d G


5.

4

k 

Ls  d  Nt

Ls  43.4 mm




Free length


Lf  89.4 mm

To check for buckling, two ratios need to be calculated, Lf/D and y max/Lf.



Deflection ratio

sr  y' 

14-45-2

Lf

sr  4.257

D ywork

y'  0.447

Lf


9.


Di  D  d

Di  18.2 mm

Outside coil dia

Do  D  d

Do  23.8 mm



holemin  24.85  mm

Largest pin

pin max  Di  0.05 D

pin max  17.15  mm


3

ρ  0.28 lbf  in 2 2

Weight

Wt 

π  d  D Nt ρ

Wt  0.48 N

4


d  2.8 mm

Outside diameter

Do  23.8 mm

Total coils

Nt  15.50

Free length

Lf  89.4 mm




14-46-1


Three springs are arranged in series similar to the configuration shown in Figure 3-1(a). They have spring rates k1 = 50 N/mm, k2 = 150 N/mm, and k3 = 500 N/mm, respectively. Determine the total spring rate, the deflection of each spring, and the overall deflection if a load of F = 600 N is applied.

Given:

Spring rates: Load:

Solution: 1.

N

mm F  600  N

N

k3  500 

mm

N mm

Using equation 14.1, calculate the deflection of each spring.

y2 

y3 

F

y1  12 mm

k1 F

y2  4  mm

k2 F

y3  1.2 mm

k3

The sum of the three is the overall deflection. ytotal  y1  y2  y3

3.

k2  150 


y1 

2.

k1  50

ytotal  17.2 mm

Alternatively, use equation 14.2b to calculate the total spring rate and use it to determine the total deflection. ktotal  ytotal 

k1 k2 k3 k2 k3  k1 k3  k1 k2 F ktotal

ktotal  34.884

N mm

ytotal  17.2 mm



14-47-1


Three springs are arranged in parallel similar to the configuration shown in Figure 3-1(b). They have spring rates k1 = 50 N/mm, k2 = 150 N/mm, and k3 = 500 N/mm, respectively. Determine the total spring rate, the force carried by each spring, and the overall deflection if a load of F = 600 N is applied.

Given:

Spring rates: Load:

Solution: 1.

k1  50

4.

N mm

k3  500 

N mm

Using equation 14.2a, calculate the total spring rate. ktotal  700 

N mm

Calculate the deflection, which is the same for each spring (assuming that the plate that carries the load does not rotate). y 

3.

k2  150 


ktotal  k1  k2  k3 2.

N

mm F  600  N

F ktotal

y  0.857  mm

Calculate the force carried by each of the three springs using equation 14.1 F1  y  k1

F1  42.857 N

F2  y  k2

F2  128.571  N

F3  y  k3

F3  428.571  N

The sum of the three is the total load carried by the plate. Ftotal  F1  F2  F3

Ftotal  600  N



14-48-1


A spring made from ASTM A228 wire with ends squared and ground, wire diameter d = 3 mm, outside diameter Do = 27 mm, 14 total coils, and free length Lf = 80 mm has been chosen for an application. Determine the static safety factor if the spring is subjected to a static load of 175 N.

Given:

Working force

Fwork  175  N

Wire diameter Outside diameter Free length

Shear modulus

G  80 GPa

d  3  mm Do  27 mm

Total coils

Nt  14

Lf  80 mm

ASTM A228 wire A  2153.5 MPa

Ends squared and ground Ne  2 Assumption: Solution: 1.

b  0.1625

Set removed See Mathcad file P1448.

Determine the mean diameter, spring index, shear factor, number of active coils, and spring rate. Mean diameter

D  Do  d

Spring index

C 

Shear factor

D  24 mm

D

C8

d

Ks  1 

0.5

Na  Ks  1.063

C


Na  Nt  Ne

Spring rate

k 

Na  12

4

d G

k  4.883 

3

8  D  Na 2.

3.

Calculate the ultimate tensile strength of the wire material from equation 14.3 and Table 14-4 and use it to find the torsional yield strength from Table 14-6.

   mm 


S ut  A  


S ys  0.60 S ut

d

b

S ut  1801 MPa S ys  1081 MPa

Calculate the shear stress in the coil at the working load and the factor of safety against yielding under the working load. Working shear stress

τwork  Ks

8  Fwork  D

π d Factor of safety 4.

N mm

S ys

Nswork 

τwork

3

τwork  420.9  MPa Nswork  2.6

Calculate the shear stress in the coil at shut height and the factor of safety against yielding at shut height. Shut height

Ls  d  Nt

Ls  42 mm

Deflection to shut height yshut  Lf  Ls Force at shut height

Fs  k yshut

Shut height stress

τs  Ks

π d Factor of safety

Nss 

Fs  185.547  N

8  Fs  D

S ys

τs

yshut  38 mm

3

τs  446.2  MPa Nss  2.4



14-49-1


A spring with ends squared and ground, wire diameter d = 4 mm, outside diameter Do = 40 mm, 18 total coils, and free length Lf = 140 mm has been chosen for an application where the initial deflection is 15 mm and the working deflection is 50 mm. Determine minimum working length, shut height, clash allowance, spring index, and spring rate for this spring.

Given:

Working deflection


Initial deflection yinit  15 mm

Wire diameter Outside diameter

d  4  mm Do  40 mm

Shear modulus Total coils

Free length

Lf  140  mm

G  80 GPa Nt  18

Ends squared and ground Ne  2 Solution: 1.


Determine the mean diameter, spring index, number of active coils, and spring rate. Mean diameter

D  Do  d

Spring index

C 

D

D  36 mm C9

d


Na  Nt  Ne

Spring rate

k 

Na  16

4

d G 3

k  3.429 

8  D  Na 2.

N mm

Calculate the minimum working length, shut height, and clash allowance. Minimum work length

Lmin  Lf  yinit  ywork Lmin  75 mm

Shut height

Ls  d  Nt

Clash allowance

α 

Lmin  Ls ywork

Ls  72 mm

α  6 %



14-50-1


An extension spring has wire diameter d = 3 mm and outside diameter Do = 27 mm. Determine the preferred preload for this spring.

Given:

Wire diameter

Solution:


1.

Mean diameter

D  Do  d

Spring index

C 

D

C8

d

Ks  1 

D  24 mm

0.5 C

Na  Ks  1.063

Use the value of C to find an appropiate value of initial coil stress i from equations 14.21: 3 2   3 2   2.987  C  139.7  C  3427 C  38404   psi

τi1  4.231  C  181.5  C  3387 C  28640  psi

τi1  75.8 MPa

τi2

τi2  126.9  MPa

τi  3.

Outside diameter Do  27 mm

Determine the mean diameter, spring index, and shear factor.

Shear factor 2.

d  3  mm

τi1  τi2

τi  101.3  MPa

2

Use the value of i from step 2 in equation 14.8b to find the corresponding initial coil-tension force Fi: 3

Fi 

π d  τi 8  K s D

Fi  42.1 N



13-51-1


A series-stack of Belleville springs is required for more deflection in a design. The stack will be guided by an internal pin as shown in Figure 14-35(b). The minimum inside diameter of the individual springs in the stack is Di = 25 mm. Determine the recommended surface conditions for the pin and the maximum pin diamter. (Hint: go to Spirol.com on the internet and look up disc springs, stacking).

Given:

Inside diameter

Solution:

See Figure 14-35(b) and Mathcad file P1451.

Di  25 mm

1. The Spirol recommended surface conditions are "case hardened to a depth not less than 0.6 mm and a hardness of 58 HRC. A surface finish of less than or equal to 4 microns is also recommended." 2.

From the table of recommended clearances c  0.4 mm. Maximum pin dia.

d pin  Di  c

d pin  24.6 mm



14-52-1


Equation 14.9a defines a combined direct shear and stress concentration factor, Kw, to be used with helical springs that are made with round wire. Equation 14.11 defines a similar factor, Krw, that is to be used to calculate shear stress when the spring is made with rectangular wire. Determine and plot the ratio Krw/Kw for square wire with values of spring index, C, ranging from 1.2 to 10. Constant values from Table 14-6 for b/h = 1/1: S 0  1.6844 S 1  2.8219 S 2  2.4577 S 3  1.0591 S 4  0.1721

Given:

Solution: 1.


The Wahl factor for round wire is, from equation 14.9a: Kw( C) 

2.

4 C  1 4 C  4



0.615 C

The stress concentration factor for rectangular wire is, from equation 14.11: s( C)  ln( C)

 S S  s( C) S  s( C) S  s( C) S  s( C)  2

Krw ( C)  e 3.

0

1

2

3

4

3

4

C  1.2 1.21  10

Plot the ratio of these two factors over the range:

STRESS CONCENTRATION FACTOR RATIO FOR SQUARE vs. ROUND WIRE

Stress Concentration Factor Ratio, Square/Round

1

0.9

Krw ( C) Kw( C)

0.8

0.7

0.6

0

2

4

6

8

C Spring Index, C



14-53-1


Repeat Problem 14-6 with 1-mm square wire instead of round wire.

Given:

Wire dimension Mean coil dia

b  1  mm D  10 mm

h  1  mm Total coils

Nt  12

Na  Nt  2

Na  10

Assumptions: The spring wire is steel so that G  80.8 GPa. Solution:


1.


2.

Using equation 14.11d and Table 14-7, Shape factor

K1  0.180

Spring rate

k  K1

3

b h  G 3

k  1.454 

D  Na 3.

N mm


C 

D b

C  10



14-54-1


A helical compression spring is required to provide a minimum force of 650 N at installation and have a working deflection of 25 mm. The spring rate is 13 N/mm. The coil must fit in a 53-mm-dia hole with 3-mm clearance. Use 6-mm square, music wire with squared/ground ends. Using a 15% clash allowance, find: (a) The stress at the working deflection. (b) The shut height. (c) The stress at the shut height. (d) The total number of coils. (e) The free length. (f) The natural frequency in Hz.

Given:

Minimum force Fmin  650  N Spring OD OD  50 mm Working deflection Δy  25 mm Music wire dims: b  6  mm Constant values from Table 14-6 for b/h = 1/1: S 0  1.6844 S 1  2.8219 S 2  2.4577

Solution: 1.

2.

1

Spring rate Shear modulus Clash factor h  6  mm

k  13 N  mm G  80.8 GPa α  0.15

S 3  1.0591

S 4  0.1721




Spring index

C 

OD  b

Fmax  975  N C  7.333

b

Calculate the stress concentration factor from equation 14.11a. s  ln( C)

 S S  s S  s  S  s  S  s  2

Krw  e

0

1

2

3

3

4

4

Krw  1.166

3.


4.

Mean coil diameter D  C b D  44.0 mm Calculate the shear stress w at the working deflection using the shape factor from Table 14-7 and equation 14.11b. Shape factors

K1  0.180

Stress at Fmax

τw  Krw 

K2  2.41

K2 Fmax D b h

5.

2

τw  558.2  MPa

To get the defined spring rate, the number of active coils must satisfy equation 14.11d, solving for Na yields: 3


Na  K1

b h  G 3

Na  17.021

Na  17

D k

Note that we round it to the nearest 1/4 coil as the manufacturing tolerance cannot achieve better than that accuracy. Having rounded the number of active coils, we must now calculate the spring rate using equation 14.11d: Corrected spring rate

3

k  K1

b h  G 3

D  Na

k  13.02 

N mm


MACHINE DESIGN - An Integrated Approach, 4th Ed. 6. For squared ends the total number of coils, from Figure 14-9: Total coils 7.

Nt  19.00

Lshut  h  Nt


The initial deflection to reach the smaller of the two loads is Initial deflection

9.

Nt  Na  2


8.

14-54-2

yinit 

Fmin

yinit  49.94  mm

k

For the given clash allowance factor: Clash allowance




Lf  193  mm


yshut  79 mm


Fshut  1024 N

13. The shut-height stress is Stress at Fshut

τshut  Krw 

K2 Fshut D b h

2



Di  D  b

Di  38 mm

Outside coil dia

Do  D  b

Do  50 mm




Largest pin

pin max  Di  0.05 D


16. The weight of the spring's active coils is found from equation 14.11b 3

Weight density

ρ  0.285  lbf  in

Weight

Wa  b  h  π D Na ρ

Wa  6.545  N


fn 

1 2



k g Wa

fn  69.83  Hz



14-54-1


A helical compression spring is required to provide a minimum force of 650 N at installation and have a working deflection of 25 mm. The spring rate is 13 N/mm. The coil must fit in a 53-mm-dia hole with 3-mm clearance. Use 6-mm square, music wire with squared/ground ends. Using a 15% clash allowance, find: (a) The stress at the working deflection. (b) The shut height. (c) The stress at the shut height. (d) The total number of coils. (e) The free length. (f) The natural frequency in Hz.

Given:

Minimum force Fmin  650  N Spring OD OD  50 mm Working deflection Δy  25 mm Music wire dims: b  6  mm Constant values from Table 14-6 for b/h = 1/1: S 0  1.6844 S 1  2.8219 S 2  2.4577

Solution: 1.

2.

1

Spring rate Shear modulus Clash factor h  6  mm

k  13 N  mm G  80.8 GPa α  0.15

S 3  1.0591

S 4  0.1721




Spring index

C 

OD  b

Fmax  975  N C  7.333

b

Calculate the stress concentration factor from equation 14.11a. s  ln( C)

 S S  s S  s  S  s  S  s  2

Krw  e

0

1

2

3

3

4

4

Krw  1.166

3.


4.

Mean coil diameter D  C b D  44.0 mm Calculate the shear stress w at the working deflection using the shape factor from Table 14-7 and equation 14.11b. Shape factors

K1  0.180

Stress at Fmax

τw  Krw 

K2  2.41

K2 Fmax D b h

5.

2

τw  558.2  MPa

To get the defined spring rate, the number of active coils must satisfy equation 14.11d, solving for Na yields: 3


Na  K1

b h  G 3

Na  17.021

Na  17

D k

Note that we round it to the nearest 1/4 coil as the manufacturing tolerance cannot achieve better than that accuracy. Having rounded the number of active coils, we must now calculate the spring rate using equation 14.11d: Corrected spring rate

3

k  K1

b h  G 3

D  Na

k  13.02 

N mm


MACHINE DESIGN - An Integrated Approach, 4th Ed. 6. For squared ends the total number of coils, from Figure 14-9: Total coils 7.

Nt  19.00

Lshut  h  Nt


The initial deflection to reach the smaller of the two loads is Initial deflection

9.

Nt  Na  2


8.

14-54-2

yinit 

Fmin

yinit  49.94  mm

k

For the given clash allowance factor: Clash allowance




Lf  193  mm


yshut  79 mm


Fshut  1024 N

13. The shut-height stress is Stress at Fshut

τshut  Krw 

K2 Fshut D b h

2



Di  D  b

Di  38 mm

Outside coil dia

Do  D  b

Do  50 mm




Largest pin

pin max  Di  0.05 D


16. The weight of the spring's active coils is found from equation 14.11b 3

Weight density

ρ  0.285  lbf  in

Weight

Wa  b  h  π D Na ρ

Wa  6.545  N


fn 

1 2



k g Wa

fn  69.83  Hz



15-1-1


Compare the tensile load capacity of a 5/16-18 UNC thread and a 5/16-24 UNF thread made of the same material. Which is stronger? Make the same comparison for M8 x 1.25 and M8 x 1 ISO threads. Compare them all to the strength of a 5/16-14 Acme thread.

Assumptions: The material strength for all threads is S  100  ksi. Solution: 1.


Get the tensile stress area for each thread specification from Tables 15-1 (for Unified National Standard sizes), 15-2 (for metric sizes), and 15-3 (for Acme threads). Thread specification

2.

Tensile Stress Area 2

5/16-18 UNC

a tUNC  0.0524 in

5/16-24 UNF

a tUNF  0.0581 in

M8 x 1.25

a tMC  36.61  mm

2

M8 x 1

a tMF  39.17  mm

2

5/16-14 Acme

a tAcme  0.053  in

2

2

Using the assumed strength, determine the allowable load for each thread specification. Thread specification

Equation

Allowable Load

5/16-18 UNC

FUNC  a tUNC S

FUNC  5240 lbf

5/16-24 UNF

FUNF  a tUNF  S

FUNF  5810 lbf

M8 x 1.25

FMC  a tMC S

FMC  25.24  kN

M8 x 1

FMF  a tMF  S

FMF  27.01  kN

5/16-14 Acme

FAcme  a tAcme S

FAcme  5300 lbf FAcme  23.58  kN

3.

State the conclusions. 1. The fine thread has a higher capacity than the coarse thread for both the UNS and ISO threads. 2. The Acme thread has a higher capacity than the UNS coarse thread, but has a lower cpacity than the fine UNS thread. 3. The Acme thread has a lower capacity than either of the metric threads.



15-2-1


A 3/4-6 Acme thraed screw is used to lift a 2-kN load. The mean collar diameter is 40 mm. Find the torque to lift and to lower the load using a ball-bearing thrust washer. What are the efficiencies? Is it sel-locking?

Given:

Screw diameter Collar diameter

d  0.750  in d c  40 mm

Threads per inch

Nt  6  in

Applied load Radial thread angle

P  2  kN α  14.5 deg

1

Assumptions: 1. The thread coefficient of friction is μ  0.15. 2. The collar coefficient of friction is μc  0.02. Solution:

See and Mathcad file P1502.

1.

Get the thread pitch diameter from Table 15-3.

2.

Determine the thread pitch and lead.

3.

Td 

L  p

L  0.167  in

P d p 2



μ  π dp  L cos( α)  π dp cos( α)  μ  L

 μc  P



 μ  π dp  L cos( α)   π dp cos( α)  μ  L

 μc  P

P d p 2

dc

Tu  42.68  in lbf

2 dc

Td  18.25  in lbf

2

Use equation 15.7c to determine the lifting (up) and lowering (down) efficiencies. eu 

ed 

5.

p  0.167  in

Use equations 15.5 to determine the lifting (up) and lowering (down) torques. Tu 

4.

d p  0.667  in 1 p  Nt

P L

eu  27.9 %

2  π Tu P L

ed  65.4 %

2  π Td

Use equation 15.6a to determine if the screw is self-locking. self_locking 

return "yes" if μ 

L

π d p

 cos( α)

"no" otherwise self_locking  "yes"



15-3-1


A 1 3/8-4 Acme thraed screw is used to lift a 1-ton load. The mean collar diameter is 2 in. Find the torque to lift and to lower the load using a ball-bearing thrust washer. What are the efficiencies? Is it sel-locking?

Given:

Screw diameter Collar diameter

d  1.375  in d c  2.00 in

Threads per inch

Nt  4  in

Applied load Radial thread angle

P  2000 lbf α  14.5 deg

1

Assumptions: 1. The thread coefficient of friction is μ  0.15. 2. The collar coefficient of friction is μc  0.02. Solution:

See and Mathcad file P1503.

1.

Get the thread pitch diameter from Table 15-3.

2.


3.

Td 

L  p

L  0.250  in

P d p 2




 μc  P




 μc  P

P d p 2

dc

Tu  316.0  in lbf

2 dc

Td  153.0  in lbf

2

Use equation 15.7c to determine the lifting (up) and lowering (down) efficiencies. eu 

ed 

5.

p  0.250  in


4.

d p  1.250  in 1 p  Nt

P L

eu  25.2 %

2  π Tu P L

ed  52.0 %

2  π Td

Use equation 15.6a to determine if the screw is self-locking. self_locking 


L

π d p

 cos( α)

"no" otherwise self_locking  "yes"



15-4-1


The trailer hitch from Figure 1-1 (p. 12) has loads applied as shown in Figure P3-2. The tongue weight of 100 kg acts downward and the pull force of 4905 N acts horizontally. Using the dimensions of the ball bracket in Figure 1-5 (p. 15), draw a free-body diagram of the ball bracket and find the tensile and shear loads applied to the two bolts that attach the bracket to the channel in Figure 1-1. Size and specify the bolts and their preload for a safety factor of at least 1.7.

Given:

Hitch dimensions:

a  40 mm

Tongue weight

Mtongue  100  kg

b  31 mm

Pull force

Fpull  4.905  kN

c  70 mm

Number of bolts

Nbolts  2

d  20 mm t  19 mm

Young's modulus Design safety factor

E  206.8  GPa Nd  1.7

Bolt modulus

Ebolt  E

Member modulus

Ememb  E

Assumptions: The shear load will be taken by friction between the hitch and the support. Design Choices: Use M12 x 1.75 , class 8.8 bolts. Material properties for class 8.8:

Bolt diameter

d b  12 mm

Proof strength

S p  600  MPa

Clamp length

l  30 mm

Yield strength

S y  660  MPa

Preload fraction

fp  0.59 W tongue

70 = c

1

F pull

1

40 = a 2

B

A

2

19 = t 31 = b

C

Fc2x

20 = d

B

C D

Fd2

D

F c2y

FIGURE S15-4 Dimensions and Free Body Diagram for Problem 15-4

Solution: 1.

See Figure S15-4 and Mathcad file P1504.


2.


The FBD of the hitch and bracket assembly is shown in Figure 3-4. The known external forces that act on the ball are Fpull and Wtongue . The reactions on the bracket are at points C and D. The bolts at C provide tensile (Fc2x) and shear (Fc2y) forces, and the bracket resists rotation about point D where the reaction force Fd2 is applied by the channel to which the bracket is bolted.



4.

Solving for the reactions by summing the horizontal and vertical forces and the moments about D:

 Fx :


(1)

 Fy :


(2)

 MD :


(3)

Solving equation (3) for Fc2x Fc2x 

5.

15-4-2


Fd2  25.505 kN

(5)

Fc2y  0.981  kN

(6)


7.

(4)


6.

Fc2x  30.41  kN


Fc2x  30.4 kN

Shear force taken by friction

Fc2y  0.98 kN P 

8.

Determine the load per bolt.

9.

Get the tensile stress area from Table 15-2.

10. Calculate the preload.

Ptot  Fc2x

Ptot

P  15.20  kN

Nbolts

At  84.27  mm

Fi  fp  S p At

2

Fi  29.83  kN

11. Determine the relevant ratios for this joint from equations 15-18a and b. Joint aspect ratio:

Plate to bolt modulus:

j  r 

db

j  0.400

l Ememb

r1

Ebolt

12. Calculate Cr = C using equation 15.19 and the coefficients p i from Table 15-8 for j  0.4 . Coefficients from Table 15-8:

p 0  0.7351 p 1  1.2612 p 2  1.1111 p 3  0.3779

Joint stiffness constant:

3

2

C  p 3  r  p 2  r  p 1  r  p 0

C  0.207



15-4-3

13. The portions of the applied load P felt by the bolt and the material can now be found from equations 15.13. Pb  C P

Pb  3.15 kN

Pm  ( 1  C)  P

Pm  12.1 kN

14. Find the resulting loads in bolt and material after the load P is applied. Fb  Fi  Pb

Fb  32.98  kN

Fm  Fi  Pm

Fm  17.78  kN

15. The maximum tensile stress in the bolt is

σb 

Fb At

σb  391.4  MPa

16. This is a uniaxial stress situation, so the principal stress and von Mises stress are identical to the applied tensile stress. The safety factor against yielding for class 8.8 with S y  660  MPa is then Ny 

Sy

σb

Ny  1.7

17. The load required to separate the joint and the safety factor against joint separation are found from equations 15.14c and 15.14d. P0 

Fi 1C

Nsep 

P0 P

P0  37.6 kN Nsep  2.5



15-5-1


Given:

For the hitch of Problem 3-4, determine the horizontal force that will result on the ball from accelerating a 2000-kg trailer to 60 m/sec in 20 sec. Size and specify the bolts and their preload for a safety factor of at least 1.7. Tongue weight Hitch dimensions: a  40 mm Mtongue  100  kg b  31 mm

Pull force

Fpull  6  kN

c  70 mm

Number of bolts

Nbolts  2

d  20 mm t  19 mm


E  206.8  GPa Nd  1.7

Bolt modulus

Ebolt  E

Member modulus

Ememb  E

Assumptions: The shear load will be taken by friction between the hitch and the support. Design Choices: Use M12 x 1.75 , class 8.8 bolts. Material properties for class 8.8:

Bolt diameter

d b  12 mm

Proof strength

S p  600  MPa

Clamp length

l  30 mm

Yield strength

S y  660  MPa

Preload fraction


70 = c

1

F pull

1

40 = a 2

B

A

2

19 = t 31 = b

C

Fc2x

20 = d

B

C D

Fd2

D

F c2y


Solution: 1.



2.





4.


 Fx :


(1)

 Fy :


(2)

 MD :


(3)


5.

15-5-2


Fd2  30.43  kN

(5)

Fc2y  0.981  kN

(6)


7.

(4)


6.

Fc2x  36.43  kN


Fc2x  36.4 kN


Fc2y  0.98 kN P 

8.


9.



Ptot

Ptot  Fc2x

P  18.22  kN

Nbolts

At  84.27  mm


2

Fi  29.33  kN

11. Determine the relevent ratios for this joint from equations 15-18a and b. Joint aspect ratio:


j  r 

db

j  0.400

l Ememb

r1

Ebolt

12. Calculate Cr = C using equation 15.19 and the coefficients p i from Table 15-8 for j  0.4 . Coeficients from Table 15-8:

p 0  0.7351 p 1  1.2612 p 2  1.1111 p 3  0.3779


3

2

C  p 3  r  p 2  r  p 1  r  p 0

C  0.207



15-5-3


Pb  3.77 kN

Pm  ( 1  C)  P

Pm  14.4 kN


Fb  33.10  kN

Fm  Fi  Pm

Fm  14.88  kN


σb 

Fb At

σb  392.8  MPa


Sy

σb

Ny  1.7


Fi 1C

Nsep 

P0 P

P0  37.0 kN Nsep  2.0



15-6-1


For the trailer hitch of Problem 3-4, determine the horizontal force that will result on the ball from an impact between the ball and the tongue of the 2000-kg trailer if the hitch deflects 2.8 mm dynamically on impact. Size and specify the bolts and their preload for a safety factor of at least 1.7.

Given:

Hitch dimensions:

a  40 mm

Tongue weight

Mtongue  100  kg

b  31 mm

Pull force

Fpull  55.1 kN

c  70 mm

Number of bolts

Nbolts  2

d  20 mm t  19 mm


E  206.8  GPa Nd  1.7

Bolt modulus

Ebolt  E

Member modulus

Ememb  E

Assumptions: The shear load will be taken by friction between the hitch and the support. Design Choices: Use M24 x 3 , class 12.9 bolts. Material properties for class 12.9:

Bolt diameter

d b  24 mm

Clamp length

l  30 mm

Proof strength

S p  970  MPa

Yield strength

S y  1100 MPa Preload fraction


70 = c

1

F pull

1

40 = a 2

B

A

2

19 = t 31 = b

C

Fc2x

20 = d

B

C D

Fd2

D

F c2y


Solution: 1.



2.





4.


 Fx :


(1)

 Fy :


(2)

 MD :


(3)


5.

15-6-2


Fd2  251.38 kN

(5)

Fc2y  0.981  kN

(6)


7.

(4)


6.

Fc2x  306.48 kN


Fc2x  306.5  kN


Fc2y  0.98 kN P 

8.


9.



Ptot

Ptot  Fc2x

P  153.24 kN

Nbolts

At  352.5  mm


2

Fi  188.06 kN

11. Determine the relevant ratios for this joint from equations 15-18a and b. Joint aspect ratio:


j  r 

db

j  0.800

l Ememb

r1

Ebolt

12. Calculate Cr = C using equation 15.19 and the coefficients p i from Table 15-8 for j  0.8 . Coefficients from Table 15-8:

p 0  0.7800 p 1  1.2503 p 2  1.0672 p 3  0.3571


3

2

C  p 3  r  p 2  r  p 1  r  p 0

C  0.240



15-6-3


Pb  36.75  kN

Pm  ( 1  C)  P

Pm  116.5  kN


Fb  224.81 kN

Fm  Fi  Pm

Fm  71.56  kN


σb 

Fb At

σb  637.7  MPa

16. This is a uniaxial stress situation, so the principal stress and von Mises stress are identical to the 3

applied tensile stress. The safety factor against yielding for class 8.8 with S y  1.1  10  MPa is then Ny 

Sy

σb

Ny  1.7


Fi 1C

Nsep 

P0 P

P0  247.4  kN Nsep  1.6



15-7-1


A 1/2-in dia UNC, class 7 bolt with rolled threads is preloaded to 80% of its proof strength when clamping a 3-in-thick sandwich of solid steel. Find the safety factors against static yielding and joint separation when a static 1000-lb external load is applied. Use 99% reliability.

Given:

Bolt diameter d  0.500  in Preload fraction fp  0.80

Material properties for class 7: Proof strength S p  105  ksi

l  3  in

Clamp length

Yield strength

Number of bolts Nbolts  1

Solution:

6

Ptot  1000 lbf Ebolt  E

Applied load Bolt modulus

Young's modulus E  30 10  psi Member modulus Ememb  E


Ptot

P  1000 lbf

1.


2.


3.

Calculate the preload.

4.

Determine the relevant ratios for this joint from equations 15-17a and b.

Nbolts

2

At  0.1419 in


Joint aspect ratio:


5.

S y  115  ksi

Ultimate strength S ut  133  ksi

Fi  11920  lbf

j  r 

d

j  0.167

l Ememb

r1

Ebolt

Calculate Cr = C using equation 15.18 and the coefficients p i from Table 15-8 for j  0.167 . Use linear interpolation between ja  0.1 and jb  0.2 pb  pa

Interpolation equation:

p  p a p b  

Coefficients from Table 15-8:

p 0  p ( 0.4389 0.6118)

jb  ja

  j  ja   p a

p 1  p ( 0.9197 1.1715) p 2  p ( 0.8901 1.0875) p 3  p ( 0.3187 0.3806)

Joint stiffness constant: 6.

7

3

2

C  p 3  r  p 2  r  p 1  r  p 0

p 0  0.5542 p 1  1.0876 p 2  1.0217 p 3  0.3600 C  0.128

The portions of the applied load P felt by the bolt and the material can now be found from equations 15.13. Pb  C P

Pb  128.3  lbf

Pm  ( 1  C)  P

Pm  871.7  lbf

Find the resulting loads in bolt and material after the load P is applied. Fb  Fi  Pb

Fb  12048  lbf


MACHINE DESIGN - An Integrated Approach, 4th Ed. Fm  Fi  Pm 8

Fm  11048  lbf

The maximum tensile stress in the bolt is

σb  9.

15-7-2

Fb At

σb  84.9 ksi

This is a uniaxial stress situation, so the principal stress and von Mises stress are identical to the applied tensile stress. The safety factor against yielding for Grade 7 with S y  115  ksi is then Ny 

Sy

σb

Ny  1.4


Fi 1C

Nsep 

P0 P

P0  13674  lbf Nsep  13.7



15-8-1


Given:

An M14 x 2, class 8.8 bolt with rolled threads is preloaded to 75% of its proof strength when clamping a 30-mm-thick sandwich of solid aluminum. Find the safety factors against static yielding and joint separation when a static 5-kN external load is applied. Material properties for class 8.8: Bolt diameter d  14 mm Proof strength Preload fraction fp  0.75 S p  600  MPa l  30 mm

Clamp length

Yield strength

Number of bolts Nbolts  1 Ptot  5  kN

Applied load

S y  660  MPa

Ultimate strength S ut  830  MPa Young's modulus Ememb  71.7 GPa Young's modulus Ebolt  206.8  GPa

Solution:

See Table 15-8 and Mathcad file P1508. P 

Ptot

P  5  kN

1.


2.


3.


4.


Nbolts


Joint aspect ratio:


5.

At  115.44 mm

2

Fi  52 kN

j  r 

d

j  0.467

l Ememb

r  0.347

Ebolt

Calculate Cr = C using equation 15.19 and the coefficients p i from Table 15-8 for j  0.467 . Use linear interpolation between ja  0.4 and jb  0.5 p  p a p b  


p 0  p ( 0.7351 0.7580)

p 0  0.7504

p 1  p ( 1.2612 1.2632)

p 1  1.2625

p 2  p ( 1.1111 1.0979)

p 2  1.1023

p 3  p ( 0.3779 0.3708)

p 3  0.3732


7

pb  pa


3

jb  ja

  j  ja   p a

2

C  p 3  r  p 2  r  p 1  r  p 0

C  0.430


Pb  2.1 kN

Pm  ( 1  C)  P

Pm  2.9 kN


Fb  54.1 kN

Fm  Fi  Pm

Fm  49.1 kN




σb  9.

15-8-2

Fb At

σb  468.6  MPa

This is a uniaxial stress situation, so the principal stress and von Mises stress are identical to the applied tensile stress. The safety factor against yielding for Grade 8.8 with S y  660  MPa is then Ny 

Sy

σb

Ny  1.4


Fi 1C

Nsep 

P0 P

P0  91.1 kN Nsep  18.2



15-9-1


A 7/16-in dia UNC, Grade 7 bolt with rolled threads is preloaded to 70% of its proof strength when clamping a 2.75-in-thick sandwich of solid steel. Find the safety factors against fatigue failure, yielding, and joint separation when a 1000-lb (peak) fluctuating external load is applied. Use 99% reliability.

Given:

Bolt diameter d  0.4375 in Preload fraction fp  0.70

Material properties for Grade 7: Proof strength S p  105  ksi

l  2.75 in

Clamp length

Yield strength


Solution:

6

Ptot  1000 lbf Ebolt  E

Applied load Bolt modulus

Young's modulus E  30 10  psi Member modulus Ememb  E

See Mathcad file P1509. Pmax 

Ptot

Pmax  1000 lbf

1.


2.


3.


4.


Nbolts



2

At  0.1063 in


Joint aspect ratio:

5.

S y  115  ksi

Ultimate strength S ut  133  ksi

Fi  7813 lbf

j  r 

d

j  0.159

l Ememb

r1

Ebolt



p 0  p ( 0.4389 0.6118)

p 0  0.5411

p 1  p ( 0.9197 1.1715)

p 1  1.0685

p 2  p ( 0.8901 1.0875)

p 2  1.0067

p 3  p ( 0.3187 0.3806)

p 3  0.3553


7.

pb  pa


3

jb  ja

  j  ja   p a

2

C  p 3  r  p 2  r  p 1  r  p 0

C  0.124

The portions of the applied load P felt by the bolt and the material can now be found from equations 15.13. Pb  C Pmax

Pb  124.0  lbf

Pm  ( 1  C)  Pmax

Pm  876.0  lbf


Fb  7937 lbf



15-9-2

Fm  Fi  Pm 8.

Calculate the alternating and mean components of the fluctuating bolt load. Falt 

Fb  Fi

Fmean  9.

Fm  6937 lbf

Falt  62.0 lbf

2 Fb  Fi

Fmean  7875 lbf

2

The nominal mean and alternating stresses in the bolt are:

σmnom 

Nominal mean stress

σanom 

Nominal alternating stress

Fmean

σmnom  74.083 ksi

At Falt

σanom  583  psi

At

10. The fatigue stress-concentration factor for this thread is found from equation 15.15c and the mean stress-concentration factor factor Kfm is found from equation 6.17. Kf  5.7 

Fatigue factor

0.6812 d in

Kf  6.00

Nominal maximum stress

σmaxnom  σmnom  σanom

σmaxnom  74.667 ksi

Nominal minimum stress

σminnom  σmnom  σanom

σminnom  73.5 ksi

Kfm 

return Kf if Kf  σmaxnom  S y return

S y  Kf  σanom

σmnom

if Kf  σmaxnom  S y

Kfm  1.505

return 0 if Kf  σmaxnom  σminnom  2  S y

σalt  Kf 

Falt

σalt  3.50 ksi

At

σmean  Kfm

Fmean At

σmean  111.50 ksi

11. The stress at the initial preload is

σinit  Kfm

Fi At

σinit  110.62 ksi

12. An endurance strength must be found for this material. Using the methods of Section 6.6 we find for S ut  133  ksi S'e  0.5 S ut

S'e  66.5 ksi

13. From the tables and formulas in Section 6.6 we have: Load

Cload  0.70

Size

Csize  1


MACHINE DESIGN - An Integrated Approach, 4th Ed. A  2.70

Surface

15-9-3

b  0.265

 S ut    ksi 

b


Ctemp  1

Reliability


Csurf  0.739

and the endurance limit is S e  Cload  Csize Csurf  Ctemp Creliab S'e

S e  28.00  ksi

14. The corrected endurance strength and the ultimate tensile strength are used in equation 15.16 to find the safety factor from the Goodman line. Nf 

S e  S ut  σinit

S e  σmean  σinit  S ut σalt

Nf  1.3

15. Calculate the maximum bolt stress and the safety factor against yielding for S y  115  ksi.

σb  Ny 

Fb

σb  74.667 ksi

At Sy

Ny  1.5

σb

16. The load required to separate the joint and the safety factor against joint separation are found from equations 15.14c and 15.14d. Nsep 

Fi Pmax ( 1  C)

Nsep  8.9



15-10-1


An M12 x 1.25, class 9.8 bolt with rolled threads is preloaded to 85% of its proof strength when clamping a 5-cm-thick sandwich of aluminum. Find the safety factors against fatigue failure, yielding, and joint separation when a 2.5-kN (peak) fluctuating external load is applied. Use 99% reliability.

Given:

Bolt diameter d  12 mm Preload fraction fp  0.85

Material properties for class 9.8 bolt: Proof strength S p  650  MPa

l  50 mm

Clamp length

Yield strength

Number of bolts Nbolts  1 Ptot  2.5 kN

Applied load

Solution:

Young's modulus Ebolt  206.8  GPa Member modulus

Ememb  71.7 GPa

Pmax  3  kN

Pmin  0  kN

See Mathcad file P1510. Pmax 

Ptot

1.


2.


3.


4.


Nbolts

At  92.07  mm


Joint aspect ratio:


5.

S y  720  MPa

Ultimate strength S ut  900  MPa

2

Fi  50.87  kN

j  r 

d

j  0.240

l Ememb

r  0.347

Ebolt



p 0  p ( 0.6118 0.6932)

p 0  0.6444

p 1  p ( 1.1715 1.2426)

p 1  1.1999

p 2  p ( 1.0875 1.1177)

p 2  1.0996

p 3  p ( 0.3806 0.3845)

p 3  0.3822


7.

pb  pa


3

jb  ja

  j  ja   p a

2

C  p 3  r  p 2  r  p 1  r  p 0

C  0.345

The portions of the applied load P felt by the bolt and the material can now be found from equations 15.13. Pb  C Pmax

Pb  0.9 kN

Pm  ( 1  C)  Pmax

Pm  1.6 kN


Fb  51.73  kN



15-10-2

Fm  Fi  Pm 8.


Fb  Fi

Fmean  9.

Fm  49.23  kN

Falt  0.43 kN

2 Fb  Fi

Fmean  51.3 kN

2


σmnom 

Nominal mean stress

σanom 


Fmean

σmnom  557.2  MPa

At Falt

σanom  4.68 MPa

At


Fatigue factor

0.02682  d mm

Kf  6



σmaxnom  561.9  MPa



σminnom  552.5  MPa

Kfm 



σmnom


Kfm  1.242

return 0 if Kf  σmaxnom  σminnom  2  S y 11. The local mean and alternating stresses in the bolt are then: Local mean stress


σm  691.8  MPa

Local alternating stress


σa  28.2 MPa



Fi At

σinit  686.02 MPa

13. An endurance strength must be found for this material. Using the methods of Section 6.6 we find for S ut  130.534  ksi S'e  0.5 S ut

S'e  450.0  MPa


Cload  0.70

Size

Csize  1


MACHINE DESIGN - An Integrated Approach, 4th Ed. Surface

A  4.51 Csurf

15-10-3

b  0.265

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


b

Csurf  0.744


S e  190.65 MPa



S e  σm  σinit  S ut σa

Nf  1.5

16. Calculate the maximum bolt stress and the safety factor against yielding for S y  720  MPa.

σb  Ny 

Fb

σb  561.9  MPa

At Sy

Ny  1.3

σb



Nsep  31.0



15-11-1


Find the tightening torque required for the bolt in Problem 15-7.

Given:

Bolt diameter d  0.500  in Preload fraction fp  0.80 Applied load

Solution:

Ptot  1000 lbf

Material properties for Grade 7: Proof strength S p  105  ksi Number of bolts

Nbolts  1


Ptot

P  1000 lbf

1.


2.


3.


4.

Calculate the tightening torque required for each bolt using equation 15.23.

Nbolts


Ti  0.21 Fi d

2

At  0.1419 in Fi  11920  lbf

Ti  1252 in lbf



15-12-1



Given:

Bolt diameter d  14 mm Preload fraction fp  0.75 Applied load

Solution:

Ptot  5  kN

Material properties for class 8.8: Proof strength S p  600  MPa Nbolts  1

Number of bolts


Ptot

P  5  kN

1.


2.


3.


4.


Nbolts


Ti  0.21 Fi d

At  115.44 mm

2

Fi  51.9 kN

Ti  153  N  m



15-13-1



Given:

Bolt diameter d  0.4375 in Preload fraction fp  0.70 Applied load

Solution:


Material properties for Grade 7: Proof strength S p  105  ksi Number of bolts

Nbolts  1


Ptot

P  5000 lbf

1.


2.


3.


4.


Nbolts


Ti  0.21 Fi d

2

At  0.1063 in Fi  7813 lbf

Ti  718  in lbf



15-14-1



Given:

Bolt diameter d  12 mm Preload fraction fp  0.85 Applied load

Solution:

Ptot  20 kN

Material properties for class 9.8: Proof strength S p  650  MPa Number of bolts

Nbolts  1


Ptot

P  20 kN

1.


2.


3.


4.


Nbolts


Ti  0.21 Fi d

At  92.07  mm

2

Fi  50.9 kN

Ti  128  N  m



15-15-1


An automobile manufacturer would like a feasibility study of the concept of building-in electric, motor-powered, screw jacks at each end of the car to automatically jack the car wheels off the ground for service. Assuming a 2-ton vehicle with a 60/40 front/rear weight distribution, design a self locking screw jack capable of lifting either end of the car. The jack body will be attached to the car frame and the screw will extend downward to engage the ground. Assume a minimum installed clearance of 8 in under the retracted screw in the up position. It must lift the car frame at least an additional 8 in. Use rolling element thrust bearings. Determine a minimu screw size, safe against column buckling. Determine its required lifting torque and efficiency, and the power required to lift it to full height in 45 sec. What is your recommendation as to the feasibility of this idea?

Given:

Weight of car Noload extension

Wcar  2000 lbf

Front weight factor

xnoload  8  in

Loaded extension

ffront  0.6 xload  8  in

tup  45 sec

Time to rertract

Design Choices: 1. The thread coefficient of friction is μ  0.15. 2. The collar coefficient of friction is μc  0.02. 6

3. Use AISI 1050 steel, Q&T @ 400F, with S y  117  ksi, and E  30 10  psi. 4. The column is fixed-free with end condition constant C  2. 5. Use the pitch diameter of the screw to compute the column radius of gyration. 6. Use a buckling design safety factor of Nbd  4. 7. Use a mean collar diameter of d c  2.00 in. 8. The length of engagement of the nut on the screw is Lnut  0.75 in. Solution: 1.

2.


Using the buckling critera (see Problem 4-53), find the minimum pitch diameter for the screw. Column load

P  ffront  Wcar

P  1200 lbf

Effective column length

Leff  C  xload  xnoload 

Leff  32 in

Start by calculating the slenderness ratio that divides the unit load vs slenderness ratio graph into Johnson and Euler regions. S rD  π

3.

2 E

S rD  71.143

Sy

To start the iterative process, assume that the final design will be an Euler column with the critical load equal to Nbd*P. From equation 4.38b, 2

Pcr =

π  E A  k

2

2

and

k =

2

L

I A

2

Substituting for k2

Pcr =

π  E I 2

= Nbd  P

L 2

Solving for I

I 

Leff  Nbd  P 2

π E 4

The required moment of inertia, assuming an Euler column is I  0.0166 in



15-15-2

Using the equation for the moment of inertia of a solid, round rod, solve for the pich diameter of the screw. 1

I= 5.

π d

4

d p 

64

 64 I   π   

4

d p  0.763  in

Using this diameter, calculate the slenderness ratio and compare to S rD. If it is greater than S rD the assumption of an Euler column is correct, if not, recalculate using the Johnson equation.

π

Ar 

Area

4

S r 

Slenderness ratio

2

2

Ar  0.457  in

I

kr 

Radius of gyration

 dp

kr  0.191  in

Ar Leff

S r  167.9

kr

Since this is greater than S rD, the assumption of an Euler column is correct. 6.

7.

Enter Table 15-3 with this minimum pitch diameter and chose a tentative screw size of 7/8-6. d  0.875  in

Threads per inch

Nt  6  in

Radial thread angle

α  14.5 deg

Pich daimeter

d p  0.792  in


p 

1

p  0.167  in

Nt

L  p 8.

L  0.167  in

Use equations 15.5 to determine the extending and retracting torques. Text 

Tret 

9.

1

Screw diameter

P d p 2 P d p 2



 μ  π dp  L cos( α)   π dp cos( α)  μ  L

 μc  P



μ  π dp  L cos( α)  π dp cos( α)  μ  L

 μc  P

dc 2 dc 2

Text  130.6  in lbf

Tret  65.4 in lbf

Use equation 15.7c to determine the lifting (up) and lowering (down) efficiencies. eext  eret 

P L

eext  24.4 %

2  π Text P L

eret  48.7 %

2  π Tret

10. Use equation 15.6a to determine if the screw is self-locking. self_locking 


L

π d p

 cos( α)

"no" otherwise self_locking  "yes" © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


15-15-3

11. Check the safety factor against thread shear by stripping using equations 15.8. The area factor for thread stripping is given in Table 15-5 as wi  0.77 for the minor diameter of an Acme thread. Minor diameter

d r  0.708  in

Total shear area

As  π d r L Lnut Nt

Shear stress

τs 

Factor of safety

Ns 

P

2

As  1.668  in

τs  719.3  psi

As 0.577  S y

τs

Ns  94

12. Determine the power required to lift the car. Rotational speed

Power required

ω 

xload  xnoload tup L

H  Text ω

ω  2.133 

rad sec

H  0.042  hp



15-16-1


Design a manual screw jack similar to that shown in Figure 15-4 for a 20-ton lift capacity and a 100-mm lift stroke. Assume that the operator can apply a 400-N force at the tip of its bar handle to turn either the screw or the nut, depending on your design. Design the cylindrical bar handle to fail in bending at the design load before the jackscrew fails so that one cannot lift an overload and fail the screw. Use rolling element thrust bearings. Seek a safety factor of 3 for thread or column failure. State all assumptions.

Given:

Design jack load

P  40000  lbf

Design handle load

Fhandle  400  N

Lift stroke

stroke  100  mm


Nd  3

Young's modulus

E  30 10  psi

6

Design Choices: 1. The thread coefficient of friction is μ  0.15. 2. The collar coefficient of friction is μc  0.02. 3. For the screw, use AISI 1050 steel, Q&T @ 400F, with S yscrew  117  ksi. 4. For the handle, use AISI 1020 cold rolled steel, with S yhandle  57 ksi. 5. The column is fixed-free with end condition constant C  2. 6. Use the pitch diameter of the screw to compute the column radius of gyration. 7. Use a mean collar diameter of d c  2.00 in. 8. The length of engagement of the nut on the screw is Lnut  2.00 in. 9. The length of the handle from its base to midpoint of grip is Lhandle  10 in Solution: 1.


Using the buckling critera (see Problem 4-53), find the minimum pitch diameter for the screw. Leff  C stroke

Effective column length 2.

Start by calculating the slenderness ratio that divides the unit load vs slenderness ratio graph into Johnson and Euler regions. S rD  π

3.

Leff  7.874  in

2 E

S rD  71.143

S yscrew

To start the iterative process, assume that the final design will be a Johnson column with the critical load equal to Nbd*P. From equation 4.38b, Pcr = S y A 

Substituting for k

2

Solving for d p

 Sy   E  2 π  A



2

 

L

2

2

 k

k =

I A

I=

64

4

A=

π d

2

4

2

 Sy  A 2 L2 Pcr = S y A    = Nd  P   I E  2 π  1 1

2  4 Nd  P 16 S yscrew  Leff   d p     π Syscrew  E  2 π   

The minimum required pitch diameter, assuming a Johnson column is 5.

π d

2

d p  1.185  in

Using this diameter, calculate the slenderness ratio and compare to S rD. If it is less than S rD the assumption of a Johnson column is correct, if not, recalculate using the Euler equation. Area

Ar 

π 4

 dp

2

2

Ar  1.103  in


MACHINE DESIGN - An Integrated Approach, 4th Ed. I 

Moment of inertia

π 64

4

I  0.097  in

kr  0.296  in

Ar Leff

S r 

Slenderness ratio

4

I

kr 

Radius of gyration

 dp

15-16-2

S r  26.6

kr

Since this is less than S rD, the assumption of a Johnson column is correct. 6.

7.

Enter Table 15-3 with this minimum pitch diameter and chose a tentative screw size of 1 3/8-5. d  1.375  in

Threads per inch

Nt  4  in

Radial thread angle

α  14.5 deg

Pich daimeter

d p  1.250  in


p 

1

p  0.250  in

Nt

L  p 8.

L  0.250  in


Td 

9.

1

Screw diameter

P d p 2




 μc  P




 μc  P

P d p 2

dc

Tu  6319 in lbf

2 dc

Td  3060 in lbf

2

Use equation 15.7c to determine the lifting (up) and lowering (down) efficiencies. eu  ed 

P L

eu  25.2 %

2  π Tu P L

ed  52.0 %

2  π Td

10. Use equation 15.6a to determine if the screw is self-locking. self_locking 


L

π d p

 cos( α)

"no" otherwise self_locking  "yes" 11. Check the safety factor against thread shear by stripping using equations 15.8. The area factor for thread stripping is given in Table 15-5 as wi  0.77 for the minor diameter of an Acme thread. Minor diameter

d r  0.708  in

Total shear area

As  π d r L Lnut Nt

Shear stress

τs 

P As

2

As  4.448  in

τs  8.99 ksi



Factor of safety

Ns 

0.577  S yscrew

τs

15-16-3 Ns  7.5

12. Find a suitable length and diameter for the handle using a safety factor against yielding of one at the handle design load. Bending moment

M  Tu

Handle length

Lhandle 

Section modulus

Z 

M  6319 in lbf M Fhandle

M

Lhandle  70.3 in

3

Z  0.1109 in

S yhandle 1

Handle diameter

d 

Use a handle with diameter

 32 Z   π   

3

d  1.041  in

d  1.00 in

However, the length is unrealistic (nearly 6 feet).



15-17a-1


Determine the effective spring constant of an aluminum, copper-asbestos, steel sandwich under compressive load. It is uniformly loaded over its 10-cm2 area. The first and third members are each 10 mm thick and the middle member is 1 mm thick, making a total thickness of 21 mm. Which material dominates the calculation?

Given:

Thicknesses

t1  10 mm

1.

t2  1  mm

E2  93 GPa

t3  10 mm

E3  206.8  GPa

2


Member stiffnesses k1 

Aluminum (1)

Copper-asbestos (2)

A  E1 t1 A  E2 t2 A  E3 t3

9 N

k1  7.170  10 

m

10 N

k2  9.300  10 

m

10 N

k3  2.068  10 

m

Sandwich stiffness: 1

ktot  k1 3.

k2 

k3 

Steel (3)

2.

E1  71.7 GPa

A  10 cm

Area Solution:

Modulus

1

1

 k2

9 N

 k3

1

ktot  5.04  10 

m

The aluminum dominates the calculation.



15-18a-1


Determine the effective spring constant of an aluminum, copper-asbestos, steel sandwich under compressive load. It is uniformly loaded over its 1.5-in 2 area. The first and third members are each 0.4 in thick and the middle member is 0.04 in thick, making a total thickness of 0.84 in. Which material dominates the calculation?

Given:

Thicknesses

t1  0.40 in

1.

E2  13.5 10  psi

t3  0.40 in

E3  30 10  psi

6

2


Member stiffnesses k1 

Copper-asbestos (2)

k2 

k3 

Steel (3)

A  E1 t1 A  E2 t2 A  E3 t3

7 lbf

k1  3.900  10 

in

8 lbf

k2  5.063  10 

in

8 lbf

k3  1.125  10 

in

Sandwich stiffness: 1

ktot  k1 3.

6

t2  0.04 in

Aluminum (1)

2.

6

E1  10.4 10  psi

A  1.5 in

Area Solution:

Modulus

1

1

 k2

7 lbf

 k3

1

ktot  2.74  10 

in

The aluminum dominates the calculation.



15-19a-1

PROBLEM 15-19a

_____

Statement:

A preloaded steel bolt similar to that shown in Figure 15-31(a) clamps two flanges of total thickness l. Using the data given in the row(s) assigned in Table P15-1, find the joint stiffness constant.

Given:

Bolt specification: M8 x 1 Bolt diameter d  8  mm Member material matl  "steel"

Clamped length l  30 mm Bolt modulus Ebolt  206.8  GPa Ememb 

Member modulus of elasticity function:

return 206.8  GPa if matl = "steel" return 71.0 GPa if matl = "alum"

Solution: 1.

See Figure 15-31(a), Table P15-1, Table P15-8, and Mathcad file P1519a.

Determine the relevant ratios for this joint from equations 15-18a and b. Joint aspect ratio:


2.

j  r 

d

j  0.267

l Ememb

r  1.000

Ebolt



p  p a p b  


p 0  p ( 0.6118 0.6932)

p 0  0.6661

p 1  p ( 1.1715 1.2426)

p 1  1.2189

p 2  p ( 1.0875 1.1177)

p 2  1.1076

p 3  p ( 0.3806 0.3845)

p 3  0.3832


3

jb  ja

2

  j  ja   p a

C  p 3  r  p 2  r  p 1  r  p 0

C  0.172



15-20-1


A single-cylinder air compressor head sees forces that range from 0 to 18.5 kN each cycle. The head is 80-mm-thick aluminum, the unconfined gasket is 1-mm-thick Teflon, and the block is aluminum. The effective clamp length of the cap screw is 120 mm. The piston is 75 mm dia and the cylinder is 140 mm outside dia. Specify a suitable number, class, preload, and bolt circle for the cylinder head cap screws to give a minimum safety factor of 1.2 for any possible failure mode.

Given:

Total load

Ptot  18.5 kN

Gasket thickness

Clamp length

l  120  mm

Member mod.


Piston diameter

id  75 mm

Bolt modulus

Ebolt  206.8  GPa

Teflon mod.

Eg  240  MPa

Cylinder diameter Design safety factor

od  140  mm Nd  1.2

Design Choices: Use 10 M12 x 1.75 , class 8.8 cap screws. Material properties for class 8.8: Proof strength S p  600  MPa

Solution: 1.

tg  1  mm

Bolt diameter Number of bolts

d  12 mm Nbolts  8

Yield strength

S y  660  MPa

Preload fraction

fp  0.90

Ultimate strength

S ut  830  MPa

Bolt circle diameter

d bc  107.5  mm



Ptot

P 

P  2313 N

Nbolts

Pmin  0  N

Pmax  P 2

2.


At  84.27  mm

3.



4.

Determine the relevant ratios for the joint without the gasket from equations 15-18a and b. Joint aspect ratio:

Plate to bolt modulus: 5.

j  r 

Fi  45.51  kN

d

j  0.1

l Ememb

r  0.347

Ebolt

Calculate Cr = C using equation 15.19 and the coefficients p i from Table 15-8 for j  0.1 . Coefficients from Table 15-8:

p 0  0.4389 p 1  0.9197 p 2  0.8901 p 3  0.3187


3

2

C  p 3  r  p 2  r  p 1  r  p 0

C  0.214

We can estimate the stiffness of the bolt kb' from equation 15.17 using its tensile stress area from Table 15-1 and then estimate the material stiffness km for the no gasket case by using the expression for C in equation 15.13c, given kb' and C. Estimated bolt stiffness

kb' 

At  Ebolt l

  1  

d

 l

1

kb'  132.0 

kN mm



Estimated member stiffness 7.

( 1  C)

km  486 

C

kN mm

Ag 

π

  od  id  Nbolts  d 2

4  Nbolts

2

2

Ag  1259 mm



2

The stiffness of this piece of gasket material is found from equation 15.11c, Gasket stiffness per bolt

9.

km  kb' 

Now we will address the unconfined gasket. The bolt stiffness is not affected by the gasket but the material stiffness is. We now have two springs in series, the metal, whose stiffness is calculated above, and the gasket. These combine according to equation 14.2b. The portion of the unconfined gasket subjected to the clamp force can be assumed to be from the outside diameter of the flange shown in Figure 15-33 to the inside diameter of the vessel. The bolt hole should be subtracted from the gasket area. The area of the clamped gasket around one bolt is: Gasket area per bolt

8.

15-20-2

kg 

Ag  Eg

kg  302.1 

tg

kN mm

The combined stiffness of the gasketed joint (from equation 14.2b) is Let the aluminum and gasket stiffnesses be km1  km and km2  kg Combined gasket and cover stiffness

km 

1 1 km1



1

km  186.3 

kN mm

km2

10. The joint constant with the unconfined gasket is now kb'

Joint constant with unconfined gasket

C 

and

( 1  C)  0.585

km  kb'

C  0.415


Pb  958.97 N

Pm  ( 1  C)  P

Pm  1353.5 N

12. Find the resulting loads in bolt and material after the load P is applied. Fb  Fi  Pb Fb  46.46  kN Fm  Fi  Pm

Fm  44.15  kN

13. Calculate the alternating and mean components of the fluctuating bolt load. Falt 

Fb  Fi

Fmean 

2 Fb  Fi 2

Falt  0.479  kN Fmean  45.99  kN

14. Because these loads are fluctuating, we need to calculate the mean and alternating components of the force felt in the bolt. The mean and alternating forces are Mean force

Fmean 

Fb  Fi 2




15-20-3

Falt 

Alternating force

Fb  Fi

Falt  479  N

2

15. The nominal mean and alternating stresses in the bolt are:

σmnom 

Nominal mean stress

σanom 


Fmean


At Falt


At


Fatigue factor

0.02682  d

Kf  6

mm







Kfm 



σmnom


Kfm  1.147



σm  625.7  MPa



σa  34.3 MPa



Fi At


19. An endurance strength must be found for this material. Using the methods of Section 6.6 we find for S ut  830  MPa S'e  0.5 S ut

S'e  415.0  MPa


Cload  0.70

Size

Csize  1

Surface

A  4.51

b  0.265

 Sut    MPa 


b

Csurf  0.76

Ctemp  1



15-20-4


Reliability and the endurance limit is


S e  179.64 MPa




Nf  1.3


σb  Ny 

Fb

σb  551.38 MPa

At Sy

Ny  1.2

σb



Nsep  34

24. Check the screw spacing, comparing it to the rules given in the text. Bolt spacing

Space/dia ratio

space bolt  ratio 

π d bc Nbolts

space bolt d

space bolt  42.2 mm ratio  3.5

which is OK



15-21-1


A single-cylinder engine head sees forces that range from 0 to 4000 lb each cycle. The head is 2.5-in-thick cast iron, the unconfined gasket is 0.125-in-thick copper-asbestos, and the block is cast iron. The effective clamp length of the cap screw is 3.125 in. The piston is 3 in dia and the cylinder is 5.5 in outside dia. Specify a suitable number, class, preload, and bolt circle for the cylinder head cap screws to give a minimum safety factor of 1.2 for any possible failure mode.

Given:

Total load


Gasket thickness tg  0.125  in

Clamp length

l  3.125  in

Member mod.

Ememb  15 10  psi

Piston diameter

id  3.00 in

Bolt modulus

Ebolt  30 10  psi

Cylinder dia.

od  5.50 in

Gasket modulus Eg  13.5 10  psi


Nd  1.2

1.

Bolt diameter d  0.3125 in Number of bolts Nbolts  8

Yield strength

S y  92 ksi

Preload fraction fp  0.90

Ultimate strength

S ut  120  ksi

Bolt circle diameter d bc  4.250  in



P 

Ptot

P  500  lbf

Nbolts


Pmax  P 2

2.


At  0.0524 in

4.



4.



6

6

Design Choices: Use 8 5/16-18 UNC , grade 5 cap screws. Material properties for grade 5: Proof strength S p  85 ksi

Solution:

6

j  r 

Fi  4009 lbf

d

j  0.1

l Ememb

r  0.5

Ebolt


p 0  0.4389 p 1  0.9197 p 2  0.8901 p 3  0.3187


3

2

C  p 3  r  p 2  r  p 1  r  p 0

C  0.162

We can estimate the stiffness of the bolt kb' from equation 15.17 using its tensile stress area from Table 15-1 and then estimate the material stiffness km for the no gasket case by using the expression for C in equation 15.13c, given kb' and C.



Estimated bolt stiffness

Estimated member stiffness 7.



l

km  kb' 

  1 

d

1

kb'  457.3 

 l

( 1  C)

km  2370

C

kip in

kip in

Ag 

π

  od  id  Nbolts  d 2

4  Nbolts

2

2

2

Ag  2.01 in



The stiffness of this piece of gasket material is found from equation 15.11c, Gasket stiffness per bolt

9.

At  Ebolt

Now we will address the unconfined gasket. The bolt stiffness is not affected by the gasket but the material stiffness is. We now have two springs in series, the metal, whose stiffness is calculated above, and the gasket. These combine according to equation 14.2b. The portion of the unconfined gasket subjected to the clamp force can be assumed to be from the outside diameter of the flange shown in Figure 15-33 to the inside diameter of the vessel. The bolt hole should be subtracted from the gasket area. The area of the clamped gasket around one bolt is: Gasket area per bolt

8.

kb' 

15-21-2

kg 

Ag  Eg

kg  217028

tg

kip in

The combined stiffness of the gasketed joint (from equation 14.2b) is Let the cast iron and gasket stiffnesses be km1  km and km2  kg Combined gasket and head stiffness

km 

1 1 km1



1

km  2345

kip in

km2

10. The joint constant with the unconfined gasket is now Joint constant with unconfined gasket

C 

kb' km  kb'

C  0.163


Pb  81.61  lbf

Pm  ( 1  C)  P

Pm  418.4  lbf


Fb  4090 lbf

Fm  Fi  Pm

Fm  3590 lbf

13. Calculate the alternating and mean components of the fluctuating bolt load. Falt 

Fb  Fi

Fmean 

2 Fb  Fi 2

Falt  40.80  lbf Fmean  4.05 kip

14. The nominal mean and alternating stresses in the bolt are: © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


15-21-3

σmnom 

Nominal mean stress

σanom 


Fmean

σmnom  77.3 ksi

At Falt

σanom  779  psi

At


Fatigue factor

0.02682  d

Kf  5.9

mm







Kfm 



σmnom


Kfm  1.131



σm  602.6  MPa



σa  31.7 MPa



Fi At

σinit  86.51  ksi

18. An endurance strength must be found for this material. Using the methods of Section 6.6 we find for S ut  120  ksi S'e  0.5 S ut

S'e  60.0 ksi


Cload  0.70

Size

Csize  1

Surface

A  2.7

b  0.265

 S ut    ksi 


Ctemp  1

Reliability


b

Csurf  0.759


S e  25.957 ksi



15-21-4




Nf  1.5

21. Calculate the maximum bolt stress and the safety factor against yielding for S y  92 ksi.

σb  Ny 

Fb

σb  78.06  ksi

At Sy

Ny  1.2

σb



Nsep  10

23. Check the screw spacing, comparing it to the rules given in the text. Bolt spacing

Space/dia ratio

space bolt  ratio 

π d bc Nbolts

space bolt d

space bolt  1.67 in ratio  5.3

which is OK



15-22-1


The forged steel connecting rod for the engine of problem 15-21 is split around the 38-mm-dia crankpin and fastened with two bolts and nuts that hold its two halves together. The total load on the two bolts varies from 0 to 8.5 kN each cycle. Design these bolts for infinite life. Specify their size, class, and preload.

Given:

Total force

Ptot  8.5 kN

Member mod.

Ememb  206.8  GPa

Number of bolts

Nbolts  2

Bolt modulus


Design Choices: Use M12 x 1.25 , class 9.8 bolts. Material properties for class 9.8: Proof strength S p  650  MPa

Solution: 1.

Bolt diameter Preload fraction

d  12 mm fp  0.70

Yield strength

S y  720  MPa


Nd  1.5

Ultimate strength

S ut  900  MPa

Clamp length

l  2.5 d


Determine the load per bolt. P 

Ptot

P  4.25 kN

Nbolts

Pmin  0  N

Pmax  P At  92.07  mm

2.


3.


4.



Joint aspect ratio:


2

Fi  41.9 kN

j  r 

d

j  0.400

l Ememb

r1

Ebolt


p 0  0.7351 p 1  1.2612 p 2  1.1111 p 3  0.3779


7.

3

2

C  p 3  r  p 2  r  p 1  r  p 0

C  0.207


Pb  0.88 kN

Pm  ( 1  C)  P

Pm  3.4 kN


Fb  42.77  kN



15-22-2

Fm  Fi  Pm 8.


Fb  Fi

Fmean  9.

Fm  38.52  kN

Falt  0.440  kN

2 Fb  Fi

Fmean  42.33  kN

2


σmnom 

Nominal mean stress

σanom 


Fmean


At Falt


At


Fatigue factor

0.02682  d mm

Kf  6.0



σmaxnom  67.379 ksi



σminnom  65.992 ksi

Kfm 



σmnom


Kfm  1.503



σm  691.2  MPa



σa  28.78  MPa



Fi

σinit  684.0  MPa

At


S'e  450.0  MPa


Cload  0.70

Size

Csize  1

Surface

A  4.51

b  0.265



Csurf

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


15-22-3

b

Csurf  0.744


S e  190.65 MPa




Nf  1.5


σb  Ny 

Fb

σb  464.6  MPa

At Sy

Ny  1.5

σb



Nsep  12.4



15-23a-1


(See also Problem 4-33.) The bracket in Figure P15-2 is fastened to the wall by 4 cap screws equispaced on a 10-cm-dia bolt circle and arranged as shown. The wall is the same material as the bracket. The bracket is subjected to a static force F, where F and the beam's other data are given in row a in Table P15-1. Find the forces acting on each of the 4 cap screws due to this loading and choose a suitable cap screw diameter, length, and preload that will give a minimum safety factor of 2 for any possible mode of failure.

Given:

Tube length Arm length Wall plate thickness

L  100  mm a  400  mm t  10 mm

Applied force Modulus of elasticity Number of bolts


d bc  100  mm

(top or bottom)


Nd  2

Bolt modulus

Ebolt  E

Clamp length

l  2  t

F  50 N E  207  GPa Nbolts  2

Ememb  E

Member modulus

Assumptions: 1. The cap screws and wall provide the couple shown in the FBD of Figure 15-23. This is an idealization, but is conservative. 2. The torsion reaction and the vertical reaction at the wall are provided by frictional forces between the wall and the wall plate on the bracket and the screws are subjected to the horizontal forces only. Design Choices: Use M5 X 0.8 , class 4.6 cap screws. Material properties for class 4.6: Proof strength S p  225  MPa S y  240  MPa

Yield strength

Cap screw diameter

d  5  mm

Preload fraction

fp  0.54

y F

dbc Ptot T

d

T Ptot

x

L R

FIGURE 15-23 Free Body Diagram of Tube and Wall Plate for Problem 15-23

Solution: 1.

2.


Calculate the magnitude of the couple in Figure 15-23 by summing moments. Distance d'

d'  d bc sin( 45 deg)

d'  70.711 mm

Total screw force

Ptot  F 

L

Ptot  70.71  N

Determine the load per screw.

d'

P 

Ptot Nbolts

Ptot  d'  F  L = 0

P  35.36  N



15-23a-2

At  14.18  mm

3.


4.


5.



j 

Joint aspect ratio:


r 

2

Fi  1.72 kN

d

j  0.25

l Ememb

r1

Ebolt



p 0  p ( 0.6118 0.6932)

p 0  0.6525

p 1  p ( 1.1715 1.2426)

p 1  1.2070

p 2  p ( 1.0875 1.1177)

p 2  1.1026

p 3  p ( 0.3806 0.3845)

p 3  0.3825


8.

9.

pb  pa


3

jb  ja

  j  ja   p a

2

C  p 3  r  p 2  r  p 1  r  p 0

C  0.165

The portions of the applied load P felt by the screw and the material can now be found from equations 15.13. Pb  C P

Pb  5.85 N

Pm  ( 1  C)  P

Pm  29.5 N

Find the resulting loads in screw and material after the load P is applied. Fb  Fi  Pb

Fb  1.73 kN

Fm  Fi  Pm

Fm  1.69 kN

The maximum tensile stress in the screw is

σb 

Fb At

σb  121.9  MPa


Sy

σb

Ny  2.0


Fi 1C

Nsep 

P0 P

P0  2.1 kN Nsep  58



15-24a-1


(See also Problem 6-33.) The bracket in Figure P15-2 is fastened to the wall by 4 cap screws equispaced on a 10-cm-dia bolt circle and arranged as shown. The wall is the same material as the bracket. The bracket is subjected to a sinusoidal force time function with Fmax = F, and Fmin = -F, where F and the beam's other data are given in row a in Table P15-1. Find the forces acting on each of the 4 cap screws due to this loading and choose a suitable cap screw diameter, length, and preload that will give a minimum safety factor of 1.5 for any possible mode of failure for N = 5E8 cycles.

Given:

Tube length Arm length

L  100  mm a  400  mm

Applied force Modulus

Wall plate thickness

t  10 mm



d bc  100  mm

(top or bottom)


Nd  1.5

F  50 N Ememb  207  GPa

Assumptions: 1. The cap screws and wall provide the couple shown in the FBD of Figure 15-23. This is an idealization, but is conservative. 2. The torsion reaction and the vertical reaction at the wall are provided by frictional forces between the wall and the wall plate on the bracket and the screws are subjected to the horizontal forces only. Design Choices: Use M4 x 0.7 , class 4.8 cap screws. Material properties for class 4.8: Cap screw diameter d  4  mm Clamp length Proof strength S p  310  MPa l  16 mm Yield strength

S y  340  MPa

Preload fraction

Ultimate strength

S ut  420  MPa

Modulus

fp  0.75 Ebolt  207  GPa

y F

dbc Ptot T

d

T Ptot

x

L R


Solution: 1.

2.




d'  70.711 mm

Total screw force

Ptot  F 

L

Ptot  70.71  N

Determine the load per screw. P 

Ptot Nbolts

d'

Ptot  d'  F  L = 0

P  35.36  N



15-24a-2

Pmin  P

Pmax  P At  8.78 mm

3.


4.


5.



r 


6.

Fi  2.04 kN

j 

Joint aspect ratio:

2

d

j  0.25

l Ememb

r1

Ebolt



p 0  p ( 0.6118 0.6932)

p 0  0.6525

p 1  p ( 1.1715 1.2426)

p 1  1.2070

p 2  p ( 1.0875 1.1177)

p 2  1.1026

p 3  p ( 0.3806 0.3845)

p 3  0.3825

8.

9.

jb  ja

3

  j  ja   p a

2

C  p 3  r  p 2  r  p 1  r  p 0


pb  pa


C  0.165


Pb  5.85 N

Pm  ( 1  C)  P

Pm  29.5 N


Fb  2.05 kN

Fm  Fi  Pm

Fm  2.01 kN

Calculate the alternating and mean components of the fluctuating screw load. Falt 

Fb  Fi

Fmean 

Falt  2.926  N

2 Fb  Fi


2

10. The nominal mean and alternating stresses in the bolt are: Nominal mean stress

σmnom 

Fmean At




15-24a-3

σanom 


Falt


At


Fatigue factor

0.6812 d

Kf  5.8

in







Kfm 



σmnom


Kfm  1.452



σm  338.1  MPa



σa  1.935  MPa



Fi


At


S'e  210.0  MPa


Cload  0.70

Size

Csize  1

Surface

A  4.51 Csurf

b  0.265

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


b

Csurf  0.91


S e  108.88 MPa

16. The corrected endurance strength and the ultimate tensile strength are used in equation 15.16 to find the safety factor from the Goodman line. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


Nf 



15-24a-4

Nf  10


σb  Ny 

Fb

σb  233.166  MPa

At Sy

Ny  1.5

σb

18. The load required to separate the joint and the safety factor against joint separation are found from equations 15.14c and 15.14d.

Nsep 


Nsep  69



15-25a-1


Given:

(See also Problem 6-34.) The bracket in Figure P15-2 is fastened to the wall by 4 cap screws equispaced on a 10-cm-dia bolt circle and arranged as shown. The wall is the same material as the bracket. The bracket is subjected to a sinusoidal force time function with Fmax = F, and Fmin = 0, where F and the beam's other data are given in row a in Table P15-1. Find the forces acting on each of the 4 cap screws due to this loading and choose a suitable cap screw diameter, length, and preload that will give a minimum safety factor of 1.5 for any possible mode of failure for N = 5E8 cycles. Applied force Tube length L  100  mm F  50 N Modulus Arm length a  400  mm Ememb  207  GPa Wall plate thickness

t  10 mm



d bc  100  mm

(top or bottom)


Nd  1.5

Assumptions: 1. The cap screws and wall provide the couple shown in the FBD of Figure 15-23. This is an idealization, but is conservative. 2. The torsion reaction and the vertical reaction at the wall are provided by frictional forces between the wall and the wall plate on the bracket and the screws are subjected to the horizontal forces only. Design Choices: Use M4 x 0.7 , class 4.8 cap screws. Material properties for class 4.8: Cap screw diameter d  4  mm Clamp length Proof strength S p  310  MPa l  16 mm Yield strength

S y  340  MPa

Preload fraction

Ultimate strength

S ut  420  MPa

Modulus

fp  0.75 Ebolt  207  GPa

y F

dbc Ptot T

d

T Ptot

x

L R


Solution: 1.

2.




d'  70.711 mm

Total screw force

Ptot  F 

L

Ptot  70.71  N


d'

P 

Ptot Nbolts

Ptot  d  F  L = 0

P  35.36  N



15-25a-2

Pmin  0  N

Pmax  P At  8.78 mm

3.


4.


5.



j 

Joint aspect ratio:

r 


6.

2

Fi  2.04 kN

d

j  0.25

l Ememb

r1

Ebolt



p 0  p ( 0.6118 0.6932)

p 0  0.6525

p 1  p ( 1.1715 1.2426)

p 1  1.2070

p 2  p ( 1.0875 1.1177)

p 2  1.1026

p 3  p ( 0.3806 0.3845)

p 3  0.3825

8.

9.

jb  ja

3

  j  ja   p a

2

C  p 3  r  p 2  r  p 1  r  p 0


pb  pa


C  0.165


Pb  5.85 N

Pm  ( 1  C)  P

Pm  29.5 N


Fb  2.05 kN

Fm  Fi  Pm

Fm  2.01 kN

Calculate the alternating and mean components of the fluctuating screw load. Falt 

Fb  Fi

Fmean 

Falt  2.926  N

2 Fb  Fi


2

10. The nominal mean and alternating stresses in the bolt are: Nominal mean stress

σmnom 

Fmean At




15-25a-3

σanom 


Falt


At


Fatigue factor

0.6812 d in

Kf  5.8







Kfm 



σmnom


Kfm  1.452



σm  338.1  MPa



σa  1.935  MPa



Fi


At


S'e  210.0  MPa


Cload  0.70

Size

Csize  1

Surface

A  4.51 Csurf

b  0.265

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


b

Csurf  0.91


S e  108.88 MPa



Nf 



15-25a-4

Nf  10


σb  Ny 

Fb

σb  233.2  MPa

At Sy

Ny  1.5

σb

18. The load required to separate the joint and the safety factor against joint separation are found from equations 15.14c and 15.14d.

Nsep 


Nsep  69



15-26-1


(See also Problem 6-42.) A cylindrical tank with hemispherical ends is required to hold 425 kPa of pressurized air at room temperature. The pressure cycles from zero to maximum. The tank diameter is 0.5 m and its length is 1 m. The hemispherical ends are attached by some number of bolts through mating flanges on each part of the tank. A 0.5-mm-thick, compressed asbestos, unconfined gasket is used between the 10-mm-thick steel flanges. Determine a suitable number, class, preload for, and size of bolts to fasten the ends of the tank. Specify the bolt circle and outside diameter of the flange needed to prevent leakage. A minimum safety factor of 2 is desired against leakage and a safety factor of 1.5 against bolt failure for infinite life.

Given:

Maximum pressure

p max  425  kPa

Gasket thickness

Flange thickness

tf  10 mm

Member mod.


Tank diameter

id  500  mm

Bolt modulus


Leakage safety factor

Nsep  2

Asbestos mod.

Eg  480  MPa

Bolt safety factor

Nb  1.5

Design Choices: Use 40 M16 x 1.5 , class 12.9 bolts. Material properties for class 10.9: Proof strength S p  970  MPa

Solution:

Yield strength

S y  1100 MPa

Ultimate strength

S ut  1220 MPa

tg  0.5 mm


d  16 mm Nbolts  40

Preload fraction

fp  0.75


1.


Ptot  P 

π id 4

 p max

Ptot

P  2086 N

Nbolts

Pmin  0  N

2.


3.


4.

Calculate the clamp length. Clamp length

5.

Pmax  P At  167.25 mm


Fi  121.67 kN

l  2  t f  tg

l  20.5 mm

2



Ptot  83.4 kN

j  r 

d

j  0.78

l Ememb

r1

Ebolt



p  p a p b  


p 0  p ( 0.7773 0.7800)

jb  ja

  j  ja   p a p 0  0.7795




Estimated member stiffness

9.

p 1  p ( 1.2543 1.2503)

p 1  1.2511

p 2  p ( 1.0735 1.0672)

p 2  1.0684

p 3  p ( 0.3595 0.3571)

p 3  0.3576

3

2

C  p 3  r  p 2  r  p 1  r  p 0

C  0.239

We can estimate the stiffness of the bolt kb' from equation 15.17 using its tensile stress area from Table 15-1 and then estimate the material stiffness km for the no gasket case by using the expression for C in equation 15.13c, given kb' and C. Estimated bolt stiffness

8.

15-26-2

kb' 

At  Ebolt



l

km  kb' 

  1 

d l

1

 

kb'  947.6 

( 1  C)

km  3013

C

kN mm

kN mm

Now we will address the unconfined gasket case. The bolt stiffness is not affected by the gasket but the material stiffness is. We now have two springs in series, the metal, whose stiffness is calculated above, and the gasket. These combine according to equation 14.2b. The portion of the unconfined gasket subjected to the clamp force can be assumed to be from the outside diameter of the flange shown in Figure 15-33 to the inside diameter of the vessel. The bolt hole should be subtracted from the gasket area. The area of the clamped gasket around one bolt is: Flange OD

od  id  6  d

Gasket area per bolt

Ag 

π

od  596  mm   od  id  Nbolts  d 2

4  Nbolts

2

2



Ag  1865 mm

2

The stiffness of this piece of gasket material is found from equation 15.11c, where, from Table 15-10. Gasket stiffness per bolt

kg 

Ag  Eg

kg  1790.3

tg

kN mm

10. The combined stiffness of the gasketed joint (from equation 14.2b) is Let the steel and gasket stiffnesses be km1  km and km2  kg Combined gasket and cover stiffness

km 

1 1 km1



1

km  1123.0

kN mm

km2

11. The joint constant with the unconfined gasket is now kb'

Joint constant with unconfined gasket

C 

and

( 1  C)  0.542

km  kb'

C  0.458


Pb  0.95 kN

Pm  ( 1  C)  P

Pm  1131.5 N



15-26-3


Fb  122.63 kN

Fm  Fi  Pm

Fm  120.54 kN

14. Because these loads are fluctuating, we need to calculate the mean and alternating components of the force felt in the bolt. The mean and alternating forces are Fmean 

Mean force

Falt 

Alternating force

Fb  Fi 2

Fb  Fi

Fmean  122.2  kN Falt  477  N

2


σmnom 

Nominal mean stress

σanom 


Fmean


At Falt


At


Fatigue factor

0.02682  d mm

Kf  6.1







Kfm 



σmnom


Kfm  1.482



σm  1082.5 MPa



σa  17.5 MPa



Fi At

3

σinit  1.08  10  MPa

19. An endurance strength must be found for this material. Using the methods of Section 6.6 we find for S ut  1220 MPa S'e  0.5 S ut

S'e  610.0  MPa


Cload  0.70

Size

Csize  1



15-26-4

A  4.51

Surface

b  0.265

 Sut    MPa 

b


Ctemp  1

Reliability


Csurf  0.686


S e  238.42 MPa



Nf  1.5


22. Calculate the maximum bolt stress and the safety factor against yielding for S y  1100 MPa.

σb  Ny 

Fb

σb  733.2  MPa

At Sy

Ny  1.5

σb


Fi

Nsep  108

Pmax ( 1  C)

24. Check the bolt spacing, comparing it to the rule given in the text. Bolt circle dia

d bc  0.5 ( od  id)

Bolt spacing

space bolt 

Space/dia ratio

ratio 

π d bc Nbolts

space bolt d

d bc  548  mm space bolt  43 mm ratio  2.7

which is OK



15-27-1


(See also Problem 6-42.) A cylindrical tank with hemispherical ends is required to hold 250 kPa of pressurized air at room temperature. The pressure cycles from zero to maximum. The tank diameter is 0.5 m and its length is 1 m. The hemispherical ends are attached by some number of bolts through mating flanges on each part of the tank. A confined o-ring gasket is used between the 10-mm-thick steel flanges. Determine a suitable number, class, preload for, and size of bolts to fasten the ends of the tank. Specify the bolt circle and outside diameter of the flange needed to prevent leakage. A minimum safety factor of 2 is desired against leakage and a safety factor of 1.5 against bolt failure for infinite life.

Given:

Maximum pressure

p max  250  kPa

Bolt safety factor

Flange thickness

tf  10 mm

Member mod.


Tank diameter

id  500  mm

Bolt modulus


Leakage safety factor

Nsep  2

Nb  1.5

Design Choices: Use 24 M12 x 1.25 , class 10.9 cap screws. Material properties for class 10.9: Proof strength S p  830  MPa


d  12 mm Nbolts  24

Yield strength

S y  940  MPa

Preload fraction

fp  0.75

Ultimate strength

S ut  1040 MPa

Solution:


1.


Ptot  P 

π id 4

 p max

Ptot

Ptot  49.1 kN P  2045.31  N

Nbolts

Pmin  0  N

Pmax  P 2

2.


At  92.07  mm

3.



Fi  57.31  kN

4.

Calculate the clamp length. l  2  t f

l  20 mm

Clamp length 5.



j  r 

d l Ememb Ebolt

j  0.6 r1


p 0  0.7709 p 1  1.2600 p 2  1.0851 p 3  0.3647



8.

9.

2

C  p 3  r  p 2  r  p 1  r  p 0


15-27-2 C  0.231


Pb  0.47 kN

Pm  ( 1  C)  P

Pm  1.6 kN


Fb  57.79  kN

Fm  Fi  Pm

Fm  55.74  kN


Fb  Fi

Fmean 

Falt  0.237  kN

2 Fb  Fi

Fmean  57.55  kN

2


σmnom 

Nominal mean stress

σanom 


Fmean


At Falt


At


Fatigue factor

0.02682  d mm

Kf  6







Kfm 



σmnom


Kfm  1.479



σm  924.5  MPa



σa  15.5 MPa



Fi At


14. An endurance strength must be found for this material. Using the methods of Section 6.6 we find for S ut  1040 MPa © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

MACHINE DESIGN - An Integrated Approach, 4th Ed. S'e  0.5 S ut

15-27-3

S'e  520.0  MPa


Cload  0.70

Size

Csize  1

Surface

A  4.51

Csurf

b  0.265

 Sut   A     MPa 

Temperature

Ctemp  1

Reliability


b

Csurf  0.716


S e  212.03 MPa




Nf  1.5


σb  Ny 

Fb

σb  627.6  MPa

At Sy

Ny  1.5

σb



Nsep  36

19. Check the bolt spacing, comparing it to the rules given in the text. Flange OD

od  id  6  d

od  572  mm

Bolt circle dia

d bc  0.5 ( od  id)

d bc  536  mm

Bolt spacing

space bolt 

Space/dia ratio

ratio 

π d bc Nbolts

space bolt d

space bolt  70.2 mm ratio  5.8

which is OK



15-28-1


Calculate the proof loads (load that causes a tensile stress equal to the proof strength) for 1/2-13 UNC bolts in each SAE grade listed in Table 15-6.

Given:

Bolt specification: 1/2-13 UNC Bolt diameter

Solution: 1.

d  0.500  in


There are eight grade numbers in Table 15-6 that are applicable to the bolt size specified. The grade numbers an proof strength values are: Grade  i

i  1 2  8

"1" "2" "4" "5" "5.2" "7" "8" "8.2" 2.

2

At  0.1419 in

Tensile stress area

S p  i

33 ksi 55 ksi 65 ksi 85 ksi 85 ksi 105  ksi 120  ksi 120  ksi

Use equation 15.2 to calculate the proof load.

Fp  At  S p i

Fp i

 "1"   "2"     "4"   "5"  Grade    i  "5.2"   "7"   "8"     "8.2" 

i

lbf

 4683 7805 9224

12062 12062 14900 17028 17028



15-29-1


Calculate the proof loads (load that causes a tensile stress equal to the proof strength) for M20 x 2.50 bolts in each class listed in Table 15-7.

Given:

Bolt specification: M20 x 2.50 Bolt diameter

Solution: 1.

Tensile stress area

At  244.79 mm

2


There are seven grade numbers in Table 15-7. The grade numbers and proof strength values are:

i  1 2  7

2.

d  20 mm

i

S p 

"4.6" "4.8" "5.8" "8.8" "9.8" "10.9" "12.9"

225  MPa 310  MPa 380  MPa 600  MPa 650  MPa 830  MPa 970  MPa

Class 

i


Fp  At  S p i

Fp i

 "4.6"     "4.8"   "5.8"  Class   "8.8"  i    "9.8"   "10.9"   "12.9"   

i

kN

 55 76 93

147 159 203 237



15-30-1

PROBLEM 15-30

_____

Statement:

Determine the joint stiffness constant for the bolt and members in Problem 15-7.

Given:


d  0.500  in

Member material matl  "steel"

2

Tensile area

At  0.1419 in

Bolt modulus

Ebolt  30 10  psi

1.

6

See Problem 15-9, Table 15-8, and Mathcad file P1530.



2.

6

return 30 10  psi if matl = "steel" return 10.4 10  psi if matl = "alum"

Member stiffness parameters: Solution:

l  3  in

6

Ememb 


Grip

j  r 

d

j  0.167

l Ememb

r  1.000

Ebolt



p  p a p b  


p 0  p ( 0.4389 0.6118)

p 0  0.5542

p 1  p ( 0.9197 1.1715)

p 1  1.0876

p 2  p ( 0.8901 1.0875)

p 2  1.0217

p 3  p ( 0.3187 0.3806)

p 3  0.3600


3

jb  ja

2

  j  ja   p a

C  p 3  r  p 2  r  p 1  r  p 0

C  0.128



15-31-1

PROBLEM 15-31

_____

Statement:


Given:

Bolt specification: M14 x 2 Bolt diameter d  14 mm Member material matl  "alum"




Solution: 1.

See problem 15-8, Table P15-8, and Mathcad file P1531.



2.

j  r 

d

j  0.467

l Ememb

r  0.343

Ebolt



p  p a p b  


p 0  p ( 0.7351 0.7580)

p 0  0.7504

p 1  p ( 1.2612 1.2632)

p 1  1.2625

p 2  p ( 1.1111 1.0979)

p 2  1.1023

p 3  p ( 0.3779 0.3708)

p 3  0.3732


3

jb  ja

2

  j  ja   p a

C  p 3  r  p 2  r  p 1  r  p 0

C  0.432



15-32-1

PROBLEM 15-32

_____

Statement:


Given:

Bolt specification: 7/16-14 UNC Bolt diameter d  0.4375 in

Clamped length l  2.75 in

Member material matl  "steel"

Bolt modulus

6

Ebolt  30 10  psi 6

Ememb 


return 30 10  psi if matl = "steel" 6

return 10.4 10  psi if matl = "alum" Solution: 1.

See Problem 15-9, Table 15-8, and Mathcad file P1532.



2.

j  r 

d

j  0.159

l Ememb

r  1.000

Ebolt



p  p a p b  


p 0  p ( 0.4389 0.6118)

p 0  0.5411

p 1  p ( 0.9197 1.1715)

p 1  1.0685

p 2  p ( 0.8901 1.0875)

p 2  1.0067

p 3  p ( 0.3187 0.3806)

p 3  0.3553


3

jb  ja

2

  j  ja   p a

C  p 3  r  p 2  r  p 1  r  p 0

C  0.124



15-33-1

PROBLEM 15-33

_____

Statement:


Given:

Bolt specification: M12 x 1.25 Bolt diameter d  12 mm Member material matl  "alum"




Solution: 1.

See problem 15-10, Table 15-8, and Mathcad file P1533.



2.

j  r 

d

j  0.240

l Ememb

r  0.343

Ebolt



p  p a p b  


p 0  p ( 0.6118 0.6932)

p 0  0.6444

p 1  p ( 1.1715 1.2426)

p 1  1.1999

p 2  p ( 1.0875 1.1177)

p 2  1.0996

p 3  p ( 0.3806 0.3845)

p 3  0.3822


3

jb  ja

2

  j  ja   p a

C  p 3  r  p 2  r  p 1  r  p 0

C  0.347



15-34a-1

PROBLEM 15-34a

_____

Statement:

Figure P15-3 shows a bolted joint eccentrically loaded in shear. The shear loads are taken by the dowel pins, the number and size of which are given in Table P15-3. Although the figure shows 5 dowel pins, that is not the case for every row in the table. For a = 4 in, b = 4 in, l = 10 in, P = 2500 lb, and the data in the row(s) assigned from Table P15-3, find the magnitude and direction of the total shear force on each dowel.

Given:

Eccentric load Dimensions: a  4  in b  4  in l  10 in P  2500 lbf Dowel diameters and position coordinates with respect to position C: i  1 2  5 A: d  0.250  in x  0.5 a y  0.5 b x  2.000  in y  2.000  in 1

1

1

1.

x  0.5 a

y  0.5 b

x  2.000  in

y  2.000  in

C:

d  0.250  in

x  0  in

y  0  in

x  0.000  in

y  0.000  in

D:

d  0.250  in

x  0.5 a

y  0.5 b

x  2.000  in

y  2.000  in

E:

d  0.250  in

x  0.5 a

y  0.5 b

x  2.000  in

y  2.000  in

2

2

3

2

3

4

3

4

5

4

5

π

4 5

n  5

4

 i

 d

2

 A iy i

i



xbar  0.000  in A

ybar 

i

i

i



ybar  0.000  in A

i

i

Determine the radial distance from the group centroid to each dowel.

 

i

r  if d  0  in  x  xbar i

i

  y i  ybar  2

2

 

0  in

Calculate the direct shear at each dowel due to the force P acting through the centroid using equation 15.25a. P F1  if  d  0  in  0  lbf  i i n  

F1  500.0  lbf 1

Calculate the magnitude of the moment acting on the group. M  P ( l  xbar)

6.

5

3

Use equations 15.24 to find the centroid of the dowel group with respect to position C.

xbar 

5.

4

5

 A ixi

4.

3

2

Calculate the cross-sectional (shear) area of each dowel. i

3.

2


A  2.

1

d  0.250  in

Number of dowels in group: Solution:

1

B:

M  25000  in lbf

Calculate the magnitude of the indirect shear at each dowel due to the moment acting on the group using equation 15.25b. M r j  1 2  5

F2  i

i

 rj 

2

j

7.

Calculate the angle that the vector F2i makes with the positive x axis.



15-34a-2

s  if ( M  0  in lbf 1 1 )







i



α  if  x  xbar  0  in  y  ybar  0  in atan2 x  xbar y  ybar 0  deg  s 90 deg i



i

θ  if d  0  in α 0  deg i

8.

i

i

i





Calculate the vector components of F2i.

 i

F2x  F2  cos θ i

9.

i

i

 i

F2y  F2  sin θ i

i

Determine the total shear force at each dowel. F  i

F   F 2

2x i

1 i

 F2y

i



2

10. Calculate the angle that Fi makes with the positive x axis.

i



 

θF  if d  0  in atan2 F2x F1  F2y 0 i

i

i

i

11. Summarize results (magnitude and direction in the last two columns): A

i 2



in

θ

r

i

in



i

deg



F2x

i

lbf

F2y

i



lbf

F 

i

lbf



θF

i

deg



0.0491

2.828

45.0

1563

1563

1890

34.2

0.0491

2.828

-45.0

1563

-1563

2588

-52.9

0.0491

0.000

-90.0

0

0

500

-90.0

0.0491

2.828

-225.0

-1563

1563

1890

145.8

C D

0.0491

2.828

-135.0

-1563

-1563

2588

-127.1

E

A B



15-35a-1

PROBLEM 15-35a

_____

Statement:

Figure P15-3 shows a bolted and doweled joint eccentrically loaded in shear. The shear loads are taken by dowel pins, the number and size of which are given in Table P15-3. Although the figure shows 5 dowels, that is not the case for every row in the table. For a = 4 in, b = 4 in, l = 10 in, P = 2500 lb, and the data in the row(s) assigned from Table P15-3, find the total shear stress in each dowel.

Given:


1

1

1.

x  0.5 a

y  0.5 b

x  2.000  in

y  2.000  in

C:

d  0.250  in

x  0  in

y  0  in

x  0.000  in

y  0.000  in

D:

d  0.250  in

x  0.5 a

y  0.5 b

x  2.000  in

y  2.000  in

E:

d  0.250  in

x  0.5 a

y  0.5 b

x  2.000  in

y  2.000  in

2

2

3

2

3

4

4

5

4 5

n  5

π 4

 i

 d

2


 A iy i

i



xbar  0.000  in A

ybar 

i

i

i



ybar  0.000  in A

i

i


 

i

r  if d  0  in  x  xbar i

i

  y i  ybar  2

2

 

0  in


F1  500.0  lbf 1


6.

5

3

Calculate the cross-sectional (shear) area of each dowel.

xbar 

5.

4

5

 A ixi

4.

3

2


i

3.

2

3

4

5

A  2.

1

d  0.250  in


1

B:

M  25000  in lbf


F2  i

i

 rj 

2

j

7.




15-35a-2

s  if ( M  0  in lbf 1 1 )







i



α  if  x  xbar  0  in  y  ybar  0  in atan2 x  xbar y  ybar 0  deg  s 90 deg i



i

θ  if d  0  in α 0  deg i

8.

i

i

i






 i


9.

i

i

 i


i

Determine the total shear force at each dowel.

F   F 2

F 

2x i

i

1 i

 F2y

i



2

10. Calculate the total shear stress at each dowel.

  

τ  if  d  0  in  i

i

F A

i



0  ksi

i

 

11. Summarize results (the shear stress is in the last column): A

i 2



in

θ

r

i

in



i

deg



F2x

i

lbf

F2y

i



lbf

F 

i

lbf

τ 

i

ksi



0.0491

2.828

45.0

1563

1563

1890

38.5

0.0491

2.828

-45.0

1563

-1563

2588

52.7

0.0491

0.000

-90.0

0

0

500

10.2

0.0491

2.828

-225.0

-1563

1563

1890

38.5

C D

0.0491

2.828

-135.0

-1563

-1563

2588

52.7

E

A B



15-36a-1

PROBLEM 15-36a

_____

Statement:

Figure P15-3 shows a bolted and doweled joint eccentrically loaded in shear. The shear loads are taken by alloy steel dowel pins, the number and size of which are given in Table P15-3. Although the figure shows 5 dowel pins, that is not the case for every row in the table. For a = 4 in, b = 4 in, l = 10 in, P = 2500 lb, SAE grade 8 bolts, and the data in the row(s) assigned from Table P15-3, find the factor of safety against yielding in shear for each dowel pin. See Table 15-12 for strength data.

Given:


1

1

1.

x  0.5 a

y  0.5 b

x  2.000  in

y  2.000  in

C:

d  0.250  in

x  0  in

y  0  in

x  0.000  in

y  0.000  in

D:

d  0.250  in

x  0.5 a

y  0.5 b

x  2.000  in

y  2.000  in

E:

d  0.250  in

x  0.5 a

y  0.5 b

x  2.000  in

y  2.000  in

Shear strength

S ys  117  ksi

2

2

3

2

3

4

4

5

n  5

π 4

 i

 d

4 5

2


 A iy i

i



xbar  0.000  in A

ybar 

i

i

i



ybar  0.000  in A

i

i


 

i

r  if d  0  in  x  xbar i

i

  y i  ybar  2

2

 

0  in


F1  500.0  lbf 1


6.

5

3

Calculate the cross-sectional (shear) area of each dowel.

xbar 

5.

4

5

 A ixi

4.

3

2


i

3.

2

3

4

5

A  2.

1

d  0.250  in


1

B:

M  25000  in lbf


F2  i

i

 rj 

2

j

7.




15-36a-2

s  if ( M  0  in lbf 1 1 )











α  if  x  xbar  0  in  y  ybar  0  in atan2 x  xbar y  ybar 0  deg  s 90 deg i i i i  i 



θ  if d  0  in α 0  deg i

8.

i

i




 i


9.

i

 i


i

Determine the total shear force at each dowel.

F   F 2

F 

2x i

i

 F2y

1 i

i



2

10. Calculate the total shear stress at each dowel.

  

τ  if  d  0  in  i

i

F A

i



0  ksi

 

i

11. Using equation 5.9a, calculate the factor of safety against yielding in shear.



Ny  if  d  0  in  i



i

S ys

τ



0

i



12. Summarize results (the safety factor is in the last column): A

i 2



in

θ

r

i

in



i

deg



F2x

i

lbf

F2y

i



lbf

F 

i

lbf

τ 

i

ksi



Ny  i

0.0491

2.828

45.0

1563

1563

1890

38.5

3.0

0.0491

2.828

-45.0

1563

-1563

2588

52.7

2.2

0.0491

0.000

-90.0

0

0

500

10.2

11.5

0.0491

2.828

-225.0

-1563

1563

1890

38.5

3.0

C D

0.0491

2.828

-135.0

-1563

-1563

2588

52.7

2.2

E

A B



15-37-1


The coefficient of friction for an oil-lubricated, single-start, power screw thread-nut combination is 0.10. Which of the American Standard Acme Threads in Table 15-3 will be self-locking for this thread-nut combination? Which will be the least likely to back down in the presence of dynamic loading? Which will be the most likely to back down in the presence of a dynamic load?

Given:

Screw-nut coefficient of friction

Solution:


μ  0.10

Thread angle

α  14.5 deg

i  1 2  23

1.

There are 23 entries in Table 15-3. Establish an index variable with range 1 .. 23:

2.

Input the values of the major diameter, d; thread pitch, p; and pitch diameter, d p. Note that for a single-start thread L = p. Major Diameter

Thread Pitch

d 

L 

i

i

0.250 0.313 0.375 0.438 0.500 0.625 0.750 0.875 1.000 1.125 1.250 1.375 1.500 1.750 2.000 2.250 2.500 2.750 3.000 3.500 4.000 4.500 5.000

3.

0.063 0.071 0.083 0.083 0.100 0.125 0.167 0.167 0.200 0.200 0.200 0.250 0.250 0.250 0.250 0.333 0.333 0.333 0.500 0.500 0.500 0.500 0.500

Pitch Diameter d p  i

0.219 0.277 0.333 0.396 0.450 0.563 0.667 0.792 0.900 1.025 1.150 1.250 1.375 1.625 1.875 2.083 2.333 2.583 2.750 3.250 3.750 4.250 4.750

Calculate the right-hand side of inequality 15.6a. L selflock  i

i

π d p

 cos( α) i



4.

Display the selflock calculations. If the result is less than  the thread will be self locking if not, it will back dow under a static load. d  i

5.

15-37-2

selflock  i

0.250

0.089

0.313

0.079

0.375

0.077

0.438

0.065

0.500

0.068

0.625

0.068

0.750

0.077

0.875

0.065

1.000

0.068

1.125

0.060

1.250

0.054

1.375

0.062

1.500

0.056

1.750

0.047

2.000

0.041

2.250

0.049

2.500

0.044

2.750

0.040

3.000

0.056

3.500

0.047

4.000

0.041

4.500

0.036

5.000

0.032

All of the selflock values are less than  so all threads will be selflocking under a static load. The 5.000-inch major diameter thread has the lowest selflock value and will, therefore, be the least likely to back down in the presence of a dynamic load. The 0.250-inch major diameter thread has the highest selflock value and is, therefore, the most likely to back down in the presence of a dynamic load.



15-38-1


The coefficient of friction for an oil-lubricated, single-start, power screw thread-nut combination is 0.20. Which of the American Standard Acme Threads in Table 15-3 will have the greatest efficiency (neglecting thrust-collar friction)?

Given:

Screw-nut coefficient of friction

Solution:


μ  0.20

Thread angle

α  14.5 deg

i  1 2  23

1.

There are 23 entries in Table 15-3. Establish an index variable with range 1 .. 23:

2.

Input the values of the major diameter, d; thread pitch, p; and pitch diameter, d p. Note that for a single-start thread L = p. Major Diameter

Thread Pitch

d 

d p 

L 

i

i

0.250 0.313 0.375 0.438 0.500 0.625 0.750 0.875 1.000 1.125 1.250 1.375 1.500 1.750 2.000 2.250 2.500 2.750 3.000 3.500 4.000 4.500 5.000

3.

Pitch Diameter i

0.063 0.071 0.083 0.083 0.100 0.125 0.167 0.167 0.200 0.200 0.200 0.250 0.250 0.250 0.250 0.333 0.333 0.333 0.500 0.500 0.500 0.500 0.500

0.219 0.277 0.333 0.396 0.450 0.563 0.667 0.792 0.900 1.025 1.150 1.250 1.375 1.625 1.875 2.083 2.333 2.583 2.750 3.250 3.750 4.250 4.750

Calculate the efficiency using equation 15.7d.

L e  i

i



π d p  cos( α)  μ  L

i

i

π d p π μ  d p  L  cos( α) i

i

i



Display the efficiency calculations. d  i

5.

15-38-2

e  i

0.250

0.301

0.313

0.278

0.375

0.273

0.438

0.241

0.500

0.251

0.625

0.251

0.750

0.274

0.875

0.242

1.000

0.251

1.125

0.228

1.250

0.209

1.375

0.232

1.500

0.216

1.750

0.190

2.000

0.169

2.250

0.196

2.500

0.179

2.750

0.164

3.000

0.216

3.500

0.190

4.000

0.169

4.500

0.152

5.000

0.139

Determine the maximum efficiency and its corresponding major diameter. emax  max( e)

emax  0.301

The major diameter of the Acme thread with the greatest efficiency is 0.250 in. As a general rule, the greater the major diameter the less efficient the screw thread is. However, as can be seen from the table above, this is not always the case.



15-39-1


Determine the number of engaged screw threads required to make the total stripping-shear area of the engaged threads equal to twice the tensile stress area for each of the fine pitch screw threads in Table 15-2.

Solution:

See Table 15-2, Table 15-5, and Mathcad file P1539.

1.

There are 14 entries in the fine-thread columns Table 15-2. Establish an index variable with range 1 .. 14: i  1 2  14

2.

Input the values of the major diameter, d; the pitch, p; the minor diameter, d r; and tensile stress area, At. Major Diameter, mm

Pitch, mm

d 

p 

i

i

8 10 12 14 16 18 20 22 24 27 30 33 36 39

3.

1.00 1.25 1.25 1.50 1.50 1.50 1.50 1.50 2.00 2.00 2.00 2.00 3.00 3.00

Minor Diameter, mm d r  i

6.77 8.47 10.47 12.16 14.16 16.16 18.16 20.16 21.55 24.55 27.55 30.55 32.32 35.32

Tensile Stress Area, mm2 At  i

39.17 61.20 92.07 124.55 167.25 216.23 271.50 333.06 384.42 495.74 621.2 760.80 864.94 1028.39

Determine the number of threads required to make the total stripping-shear area equal to twice the tensile stress area. n  As = 2  At where n is the number of threads and As is given by equation 15.8a. Solving for n,

n  i

2  At

i

0.8 π d r  p i

i

where the factor, 0.8, is taken from Table 15-5.



4.

15-39-2

Display the values of n:

d  i

n  i

8

4.6

10

4.6

12

5.6

14

5.4

16

6.3

18

7.1

20

7.9

22

8.8

24

7.1

27

8.0

30

9.0

33

9.9

36

7.1

39

7.7



15-40-1


Calculate the ultimate tensile loads (load that causes a tensile stress equal to the tensile strength) for 1/2-13 UNC bolts in each SAE grade listed in Table 15-6.

Given:


Solution: 1.

d  0.500  in

Tensile stress area


There are eight grade numbers in Table 15-6 that are applicable to the bolt size specified. The grade numbers an tensile strength values are: Grade  i

i  1 2  8

"1" "2" "4" "5" "5.2" "7" "8" "8.2" 2.

2

At  0.1419 in

S ut  i

60 ksi 74 ksi 115  ksi 120  ksi 120  ksi 133  ksi 150  ksi 150  ksi


Fut  At  S ut i

Fut i

 "1"   "2"     "4"   "5"  Grade    i  "5.2"   "7"   "8"     "8.2" 

lbf

i

 8514

10501 16319 17028 17028 18873 21285 21285



15-41-1


Calculate the ultimate tensile loads (load that causes a tensile stress equal to the tensile strength) for M20 x 2.50 bolts in each class listed in Table 15-7.

Given:


Solution: 1.

d  20 mm

Tensile stress area

2


There are seven grade numbers in Table 15-7. The grade numbers and tensile strength values are:

i  1 2  7

S ut 

Class  i

"4.6" "4.8" "5.8" "8.8" "9.8" "10.9" "12.9"

2.

At  244.79 mm

i

400  MPa 420  MPa 520  MPa 830  MPa 900  MPa 1040 MPa 1220 MPa

Use equation 15.2 to calculate the ultimate tensile load.

Fut  At  S ut i

Fut i

 "4.6"     "4.8"   "5.8"  Class   "8.8"  i    "9.8"   "10.9"   "12.9"   

i

kN

 98

103 127 203 220 255 299



15-42-1


3/8-in dia UNC bolts and nuts clamp a 0.75-in-thick aluminum cover plate to a 0.50-in-thick steel flange. Determine the stiffness factor of the joints.

Given:

Bolt diameter

d  0.375  in

Bolt modulus

Ebolt  30 10  psi

Cover plate

lc  0.75 in

Cover plate

Ec  10.4 10  psi

Flange

lf  0.50 in

Flange

Ef  30 10  psi

Solution: 1.

6

6


Determine the relevant ratios for this joint from equations 15-18. Clamped length:

l  lc  lf

Joint aspect ratio:

j 


Low and high modulus plate thicknesses:

l  1.25 in

d

j  0.3

l

rH  rL 

2.

6

Ef

rH  1

Ebolt Ec

rL  0.347

Ebolt

TL  lc

TL  0.75 in

TH  lf

TH  0.5 in

Plate thickness ratio:

t 

Material modulus ratio:

m 

TL

t  0.6

TL  TH Ec

m  0.347

Ef

Calculate the Cr terms CH and CL for rH and rL , respectively, using equation 15.19 and the coefficients p i from Table 15-8 for j  0.3 . Coefficients from Table 15-8:

p 0  0.6932

p 1  1.2426

p 2  1.1177

p 3  0.3845 3

2

CL  p 3  rL  p 2  rL  p 1  rL  p 0 3

CL  0.381

2

CH  p 3  rH  p 2  rH  p 1  rH  p 0

3.

CH  0.184

Use equation 15.20 to calculate Ct. taking the coefficients from Table 15-9 for j  0.3 . Coeficients from Table 15-9:

q 0  0.0861

q 1  8.2344

q 2  22.274

q 3  13.963 3

2

Ct  q 3 t  q 2 t  q 1 t  q 0

Ct  0.024



15-42-2

Calculate a correction factor to linearize the result from the previous step using equation 15.21. Correction factor:

a  e

 0.0598 ln( j ) 30.1385 ln( j ) 20.4350 ln( j ) 2.3516

a  0.177 5.

Using equation 15.22, calculate the joint stiffness factor C. Joint stiffness factor:

C  CH   t  a  Ct   CL  CH 

C  0.303



15-43-1


M14 x 2 bolts and nuts clamp a 16-mm-thick aluminum cover plate to a 12-mm-thick steel flange. Determine the stiffness factor of the joints.

Given:

Bolt diameter

d  14 mm

Bolt modulus


Cover plate

lc  16 mm

Cover plate

Ec  71.7 GPa

Flange

lf  12 mm

Flange

Ef  206.8  GPa

Solution: 1.



l  lc  lf

Joint aspect ratio:

j 


Low and high modulus plate thicknesses:

2.

d

j  0.5

l

rH  rL 

l  28 mm

Ef

rH  1

Ebolt Ec

rL  0.347

Ebolt

TL  lc

TL  16 mm

TH  lf

TH  12 mm

Plate thickness ratio:

t 

Material modulus ratio:

m 

TL

t  0.571

TL  TH Ec

m  0.347

Ef


p 0  0.7580

p 1  1.2632

p 2  1.0979

p 3  0.3708 3

2

CL  p 3  rL  p 2  rL  p 1  rL  p 0 3

CL  0.437

2

CH  p 3  rH  p 2  rH  p 1  rH  p 0

3.

CH  0.222


q 0  0.0533

q 1  7.8676

q 2  19.357

q 3  11.457 3

2

Ct  q 3 t  q 2 t  q 1 t  q 0 4.

Ct  0.366

Calculate a correction factor to linearize the result from the previous step using equation 15.21.



Correction factor:

a  e

15-43-2

 0.0598 ln( j ) 30.1385 ln( j ) 20.4350 ln( j ) 2.3516

a  0.135 5.


C  CH   t  a  Ct   CL  CH 

C  0.355



15-44-1


M16 x 2.0 bolts with nuts clamp a 50-mm-thick aluminum cover plate to a 30-mm-thick steel flange. Determine the stiffness factor of the joints.

Given:

Bolt diameter

d  16 mm

Bolt modulus


Cover plate

lc  50 mm

Cover plate

Ec  71.7 GPa

Flange

lf  30 mm

Flange

Ef  206.8  GPa

Solution: 1.



l  lc  lf

Joint aspect ratio:

j 


d

Plate thickness ratio: 2.

j  0.2

l

rH 

Ef

rH  1

Ebolt Ec

rL  Low and high modulus plate thicknesses:

l  80 mm

rL  0.347

Ebolt

TL  lc

TL  50 mm

TH  lf

TH  30 mm

t 

TL

t  0.625

TL  TH


p 0  0.6118

p 1  1.1715

p 2  1.0875

p 3  0.3806 3

2

CL  p 3  rL  p 2  rL  p 1  rL  p 0 3

CL  0.320

2

CH  p 3  rH  p 2  rH  p 1  rH  p 0 3.

CH  0.147


q 0  0.1010

q 1  8.5465

q 2  24.166

q 3  15.497 3

2

Ct  q 3 t  q 2 t  q 1 t  q 0 4.

Ct  0.214


a  e

 0.0598 ln( j ) 30.1385 ln( j ) 20.4350 ln( j ) 2.3516

a  0.214 5.


C  CH   t  a  Ct   CL  CH 

C  0.248



15-45-1


5/16-18 UNC bolts with nuts clamp a 1.625-in-thick gray cast iron cover plate to a 1.5-in-thick steel flange. Determine the stiffness factor of the joints.

Given:

Bolt diameter

d  0.3125 in

Bolt modulus

Ebolt  30000  ksi

Cover plate

lc  1.625  in

Cover plate

Ec  15000  ksi

Flange

lf  1.500  in

Flange

Ef  30000  ksi

Solution: 1.



l  lc  lf

Joint aspect ratio:

j 


d


j  0.1

l

rH 

Ef

rH  1

Ebolt Ec


l  3.125  in

rL  0.5

Ebolt

TL  lc

TL  1.625  in

TH  lf

TH  1.5 in

t 

TL

t  0.52

TL  TH


p 0  0.4389

p 1  0.9197

p 2  0.8901

p 3  0.3187 3

2

CL  p 3  rL  p 2  rL  p 1  rL  p 0 3

CL  0.162

2

CH  p 3  rH  p 2  rH  p 1  rH  p 0 3.

CH  0.091


q 0  0.0079

q 1  17.040

q 2  92.832

q 3  202.44

q 4  209.38

q 5  82.726

5

4

3

2

Ct  q 5 t  q 4 t  q 3 t  q 2 t  q 1 t  q 0 4.

Ct  0.068


a  e

 0.0598 ln( j ) 30.1385 ln( j ) 20.4350 ln( j ) 2.3516

a  0.260 5.


C  CH   t  a  Ct   CL  CH 

C  0.129



15-46-1


M16 x 1.5 bolts with nuts clamp a 8-mm-thick titanium cover plate to a 8-mm-thick stainless steel flange. Determine the stiffness factor of the joints.

Given:

Bolt diameter

d  16 mm

Bolt modulus


Cover plate

lc  8  mm

Cover plate

Ec  113.8  GPa

Flange

lf  8  mm

Flange

Ef  189.6  GPa

Solution: 1.



l  lc  lf

Joint aspect ratio:

j 


d


j 1

l

rH 

Ef

rH  0.917

Ebolt Ec


l  16 mm

rL  0.55

Ebolt

TL  lc

TL  8  mm

TH  lf

TH  8  mm

t 

TL

t  0.5

TL  TH

Calculate the Cr terms CH and CL for rH and rL , respectively, using equation 15.19 and the coefficients p i from Table 15-8 for j  1 . Coefficients from Table 15-8:

p 0  0.4389

p 1  0.9197

p 2  0.8901

p 3  0.3187 3

2

CL  p 3  rL  p 2  rL  p 1  rL  p 0 3

CL  0.149

2

CH  p 3  rH  p 2  rH  p 1  rH  p 0 3.

CH  0.098

Use equation 15.20 to calculate Ct. taking the coefficients from Table 15-9 for j  1 . Coeficients from Table 15-9:

q 0  0.0079

q 1  17.040

q 2  92.832

q 3  202.44

q 4  209.38

q 5  82.726

5

4

3

2

Ct  q 5 t  q 4 t  q 3 t  q 2 t  q 1 t  q 0 4.

Ct  0.124


a  e

 0.0598 ln( j ) 30.1385 ln( j ) 20.4350 ln( j ) 2.3516

a  0.095 5.


C  CH   t  a  Ct   CL  CH 

C  0.124



15-47-1


An M12 X 1.25 soft steel nut is assembled with a hardened steel bolt. The nut is 11 mm thick and has a shear yield-strength of 120 MPa. Determine the axial force that will cause stripping failure of the nut if the nut threads fail before the bolt fails.

Given:

Pitch Major dia. of bolt d  12 mm Nut thickness h  11 mm Shear strength S ys  120  MPa

p  1.25 mm

Assumptions: The nut major diameter is equal to the bolt major diameter. Solution: 1.

See Table 15-2, Table 15-5, and Mathcad file P1547.

Calculate the approximate number of threads in the nut. Number of threads

2.

p

n  8.8

wo  0.88

Calculate the shear area in the nut using equation 15.8b and the number of threads from step 1. Shear area

4.

h

Determine the area factor for nut thread stripping from Table 15-5. Area factor

3.

n 

As  n  π d  wo p

As  364.9 mm

2

Using equation 15.8c, calculate the tensile force that will cause the nut threads to fail in shear. Failure load

Ffail  As  S ys

Ffail  43.8 kN



15-48-1


Compare the yield loads (load that causes a tensile stress equal to the yield strength) to the proof loads (load that causes a tensile stress equal to the proof strength) for M12 x 1.25 bolts in each class listed in Table 15-7.

Given:


Solution: 1.

d  12 mm

2

See Tables 15-2 and 15-7 and Mathcad file P1548.

There are seven grade numbers in Table 15-7. The grade numbers, proof strength, and yield strength values are:

i

S p 

"4.6" "4.8" "5.8" "8.8" "9.8" "10.9" "12.9"

225  MPa 310  MPa 380  MPa 600  MPa 650  MPa 830  MPa 970  MPa

Class 

i  1 2  7

2.

At  92.07  mm

Tensile stress area

S y 

i

i

240  MPa 340  MPa 420  MPa 660  MPa 720  MPa 940  MPa 1100 MPa

Use equation 15.2 to calculate the proof load and the yield load. The percent difference is in the last column. Fp  At  S p i

i

Fy  At  S y i

i

Fp

 "4.6"     "4.8"   "5.8"  Class   "8.8"  i    "9.8"   "10.9"   "12.9"   

i

kN



Fy

i

kN

Fy  Fp i



i

Fp

i

21

22

6.7

29

31

9.7

35

39

10.5

55

61

10.0

60

66

10.8

76

87

13.3

89

101

13.4



1 %





15-49-1


Given:

An M16 x 1.50, class 4.8 bolt with cut threads is preloaded to 85% of its proof strength when clamping a 20-mm-thick sandwich of solid steel. Find the safety factors against static yielding and joint separation when a static 3-kN external load is applied. Material properties for class 4.8: Bolt diameter d  16 mm Proof strength Preload fraction fp  0.85 S p  310  MPa l  20 mm

Clamp length

Yield strength


Solution:

Applied load

Ptot  3  kN

Bolt modulus

Ebolt  E

Young's modulus E  206.8  GPa Ememb  E

Member modulus

See Mathcad file P1549. Ptot

P 

P  3  kN

1.


2.


3.


4.


Nbolts

At  167.25 mm


r 


2

Fi  44 kN

j 

Joint aspect ratio:

5.

S y  340  MPa

Ultimate strength S ut  420  MPa

d

j  0.8

l Ememb

r1

Ebolt


p 0  0.7800 p 1  1.2503 p 2  1.0672 p 3  0.3571


7

8

3

2

C  p 3  r  p 2  r  p 1  r  p 0

C  0.240


Pb  0.7 kN

Pm  ( 1  C)  P

Pm  2.3 kN


Fb  44.8 kN

Fm  Fi  Pm

Fm  41.8 kN


σb 

Fb At

σb  38.8 ksi



15-49-2


Sy

σb

Ny  1.3


Fi 1C

Nsep 

P0 P

P0  58.0 kN Nsep  19.3



15-50-1


Given:

A 15-mm-thick steel cap is to be fastened to a 15-mm-thick steel flange with 6 bolts and nuts. The external load on the cap is 30 kN. Size and specify the bolts for a safety factor of at least 1.5 and specify the torque required on each bolt to obtain the preload if the threads are lubricated. Number of bolts Nbolts  6 Clamp length l  30 mm Bolt modulus


Applied load


Design Choices: Use M10 x 1.5 , class 5.8 bolts. Material properties for class 5.8: Proof strength S p  380  MPa S y  420  MPa

Yield strength Solution:

Member mod.


Bolt diameter

d  10 mm

Preload fraction

fp  0.70


Ptot

P  5.00 kN

1.


2.


3.


4.


Nbolts


Joint aspect ratio:


At  57.99  mm

j  r 

2

Fi  15.43  kN

d

j  0.333

l Ememb

r1

Ebolt



p 0  p ( 0.6932 0.7351)

p 0  0.7072

p 1  p ( 1.2426 1.2612)

p 1  1.2488

p 2  p ( 1.1177 1.1111)

p 2  1.1155

p 3  p ( 0.3845 0.3779)

p 3  0.3823


7

pb  pa


3

jb  ja

  j  ja   p a

2

C  p 3  r  p 2  r  p 1  r  p 0

C  0.192


Pb  1.0 kN

Pm  ( 1  C)  P

Pm  4.0 kN

Find the resulting loads in bolt and material after the load P is applied.



8

Fb  Fi  Pb

Fb  16.4 kN

Fm  Fi  Pm

Fm  11.4 kN


σb  9.

15-50-2

Fb At

σb  282.5  MPa


Sy

σb

Ny  1.5


Fi 1C

Nsep 

P0 P

P0  19.1 kN Nsep  3.8

11. Use equation 15.23d to calculate the tightening torque. Ti  0.21 Fi d

Ti  32.4 N  m



15-51-1


Repeat Problem 15-50 with a total external load on the six bolts that varies from 0 to 30 kN per cycle. Design these bolts for infinite life with a factor of safety of at least 1.5. Specify their size, class, preload, and tightening torque.

Given:

Total force


Member mod.


Number of bolts

Nbolts  6

Bolt modulus


Design Choices: Use M10 x 1.5 , class 5.8 bolts with rolled threads. Material properties for class 5.8: Proof strength S p  380  MPa

Solution: 1.


d  10 mm fp  0.70

Yield strength

S y  420  MPa


Nd  1.5

Ultimate strength

S ut  520  MPa

Clamp length

l  30 mm



Ptot

P  5  kN

Nbolts

Pmin  0  N

Pmax  P At  57.99  mm

2.


3.


4.



Joint aspect ratio:


2

Fi  15.4 kN

j  r 

d

j  0.333

l Ememb

r1

Ebolt



p 0  p ( 0.6932 0.7351)

p 0  0.7072

p 1  p ( 1.2426 1.2612)

p 1  1.2488

p 2  p ( 1.1177 1.1111)

p 2  1.1155

p 3  p ( 0.3845 0.3779)

p 3  0.3823


pb  pa


3

jb  ja

  j  ja   p a

2

C  p 3  r  p 2  r  p 1  r  p 0

C  0.192


Pb  0.96 kN


MACHINE DESIGN - An Integrated Approach, 4th Ed. Pm  ( 1  C)  P 7.

8.

Pm  4.0 kN


Fb  16.38  kN

Fm  Fi  Pm

Fm  11.38  kN


Fb  Fi

Fmean  9.

15-51-2

Falt  0.479  kN

2 Fb  Fi

Fmean  15.90  kN

2

The fatigue stress-concentration factor Kf  2.2 for rolled threads is taken from Table 15-8. The mean stress-concentration factor is taken as Kfm  1 in this case. The mean and alternating stresses in the bolt are:

σalt  Kf 

Falt

σalt  18.17  MPa

At


Fmean At

σmean  274.26 MPa



Fi


At


S'e  260.0  MPa


Cload  0.70

Size

Csize  1

Surface

A  4.51

b  0.265

 Sut    MPa 


Ctemp  1

Reliability


b

Csurf  0.86


S e  127.39 MPa



Nf 

15-51-3


Nf  3.1



σb  Ny 

Fb

σb  282.5  MPa

At Sy

Ny  1.5

σb



Nsep  3.8


Ti  32.4 N  m



15-52-1


A 20-mm-thick aluminum cap is to be fastened to a 20-mm-thick aluminum flange with 8 bolts and nuts. The external load on the cap varies from 0 to 40 kN per cycle. Size and specify the bolts for infinite life and a safety factor of at least 1.5 and specify the torque required on each bolt to obtain the preload if the threads are lubricated.

Given:

Total force


Member mod.


Number of bolts

Nbolts  8

Bolt modulus


Design Choices: Use M10 x 1.5 , class 8.8 bolts with rolled threads. Material properties for class 8.8: Proof strength S p  600  MPa

Solution: 1.


d  10 mm fp  0.70

Yield strength

S y  660  MPa


Nd  1.5

Ultimate strength

S ut  830  MPa

Clamp length

l  40 mm



Ptot

P  5  kN

Nbolts

Pmin  0  N

Pmax  P At  57.99  mm

2.


3.


4.



Joint aspect ratio:


2

Fi  24.4 kN

j  r 

d

j  0.250

l Ememb

r  0.347

Ebolt



p 0  p ( 0.6118 0.6932)

p 0  0.6525

p 1  p ( 1.1715 1.2426)

p 1  1.2070

p 2  p ( 1.0875 1.1177)

p 2  1.1026

p 3  p ( 0.3806 0.3845)

p 3  0.3825


pb  pa


3

jb  ja

  j  ja   p a

2

C  p 3  r  p 2  r  p 1  r  p 0

C  0.351


Pb  1.75 kN



Pm  ( 1  C)  P 7.

8.

Pm  3.2 kN


Fb  26.11  kN

Fm  Fi  Pm

Fm  21.11  kN


Fb  Fi

Fmean  9.

15-52-2

Falt  0.877  kN

2 Fb  Fi

Fmean  25.23  kN

2

The fatigue stress-concentration factor Kf  3.0 for rolled threads is taken from Table 15-8. The mean stress-concentration factor is taken as Kfm  1 in this case. The mean and alternating stresses in the bolt are:

σalt  Kf 

Falt

σalt  45.34  MPa

At


Fmean At

σmean  435.11 MPa



Fi


At


S'e  415.0  MPa


Cload  0.70

Size

Csize  1

Surface

A  4.51

b  0.265

 Sut    MPa 


Ctemp  1

Reliability


b

Csurf  0.76


S e  179.64 MPa



Nf 



15-52-3

Nf  1.8


σb  Ny 

Fb

σb  450.2  MPa

At Sy

Ny  1.5

σb



Nsep  7.5


Ti  51.1 N  m



16-1-1


A submerged-arc complete joint penetration (CJP) butt weld was made between two sections of A36 hot-rolled steel plate. The plate is 10-in wide by 1/2-in-thick. E70 electrode was used. How much tensile load can the assembly withstand across the weld without yielding in either base metal or weld?

Given:

Plate:

L  10 in

Electrode:


T  0.5 in

S y  36 ksi

Assumptions: There are no stress concentrations or residual stresses. The yield strength of the electrode material is S ey  0.75 S ut S ey  52.5 ksi Solution: 1.


Calculate the cross-sectional area of both the plate and the weld. A  L T

2.

2

A  5  in

Both the plate and weld are in pure tension so the tensile stress is the only nonzero principle stress and is also the von Mises stress. Calculate the load that will cause failure in the weld and in the plate. The lesser of the two is the maximum load that can be applied to the joint. Weld

Pweld  A  S ey

Pweld  262.5  kip

Plate

Pplate  A  S y

Pplate  180  kip

The maximum load that can be applied to the joint without yielding is Pplate  180  kip



16-2-1


The plate of Problem 16-1 has partial joint penetration (PJP) welds applied from each side. If each weld has a throat dimension of 1/4 in, determine the maximum allowable tensile load across the weld.

Given:

Plate:

L  10 in

Electrode:


Weld:

t  0.25 in two places

T  0.5 in

S y  36 ksi

Assumptions: The weld will fail in shear along the weld throat and, from equation 16-1, the maximum allowable shear stress is τmax  0.3 S ut , τmax  21 ksi Solution: 1.

2.


Calculate the cross-sectional areas of the weld and the plate. 2

Aweld  L 2  t

Aweld  5  in

Aplate  L T

Aplate  5  in

2

Calculate the load that will cause failure in the weld and in the plate. Weld:

Pweld  Aweld  τmax

Pweld  105  kip

Plate:

Pplate  Aplate  S y

Pplate  180  kip

The limiting load is Pweld  105  kip



16-3-1


A tee bracket similar to that shown in Figure 16-11 is to have 1/2-in-thick A572 Grade 42 steel welded with 3/16-in fillet welds along both inside corners using an E70 electrode. If it will be subjected to a 20-kip tensile load on the leg of the tee, determine the minimum required length, L, of the bracket based on full-length welds.

Given:

Plates:

T  0.5 in

S put  70.3 ksi

Electrode: Exx  70 ksi Weld: w  0.1875 in two places Load: P  20 kip Assumptions: The weld will fail in shear along the weld throat and, from equation 16-1, the maximum allowable shear stress is τmax  0.3 Exx , τmax  21 ksi Solution: 1.

See Figure 16-11, Table 16-3, and Mathcad file P1603.

Determine the throat dimension of the welds. t  0.707  w

2.

3.

t  0.133 in

Determine the shear area in the throats of the welds as a function of L and then solve for L using equation 4.9. Ashear = L 2  t

L=

Solving for L,

L 

Ashear

and

2 t

τxy = τmax =

P

P Ashear

L  3.592 in

2  t τmax

Check whether this value for L gives an acceptable safety factor against yielding in the fused base metal. The shear area on the leg along the width of the welds is Afusion  2  L w , and, from Table 16-3 the yield strength of the base metal is S y  42 ksi. The shear yield strength of the base metal is S sy  0.577  S y . The factor of safety against yielding in the fused base metal is:

Nyield =

where

S sy

τxy

=

Afusion P

 S sy

Afusion

Nyield 

P 2

Afusion  1.347 in

 S sy

and

Nyield  1.63 S sy  24.2 ksi



16-4-1


Figure P16-1 shows a bar welded to a base on three sides with 3/16-in fillet welds using an E70 electrode. Material is A572 Grafde 50 hot-rolled steel. What is your recommended maximum static load P that can safely be applied?

Given:

Bar dims: Electrode: Weld: Material:

a  6  in Exx  70 ksi w  0.1875 in S y  50 ksi

b  3  in

d  2  in

T  0.25 in

three places S ut  65 ksi

Assumptions: The weld will fail in shear along the weld throat and, from equation 16-1, the maximum allowable shear stress is τmax  0.3 Exx , τmax  21 ksi. Consider the weld as a line. The weld pattern matches that of part 5 of Figure 16-15. The weld will limit the load as Category F. Solution: 1.

See Figure P16-1, Figure 16-15, and Mathcad file P1604.


2.

t  0.133 in

Determine the shear area in the throats of the welds as a line function. 2

Aw  d  2  b 3.

b

xbar  1.125 in

2 b  d

Find the moment arm from the applied load to the centroid.

P

T ( P)  P r

Aw

From part 5 of Figure 16-15 the torsional geometry factor is: Jw 

7.

r  7.875 in

Find the unit load due to direct shear and the applied torque as functions of P. fs ( P) 

6.

in

2

r  a  b  xbar 5.

in

Find the centroid of the weld pattern with the equation given in part 5 of Figure 16-5. xbar 

4.

Aw  8 

( 2 b  d)

3

12

2



b  ( b  d)

2

4

Jw  14.542

2 b  d

in

in

The unit load due to torsional shear is a vector that is perpendicular to a line from the centroid to any point on the weld. It is a maximum at a point furthest from the centroid. In this case, it is the furtherest point to the right on either the top or bottom of the bar. In either case, it will have horizontal and vertical components. The vertical component will add to the direct shear and that total will add vectorially to the horizontal component. The moment arm for the horizontal component is d/2, and for the vertical component it is b xbar. Horizontal unit load:

fth ( P) 

Vertical unit load:

ftv ( P) 

Total unit load:

FR( P) 

T ( P)  d 2  Jw T ( P)  ( b  xbar) Jw fth ( P)   fs ( P)  f tv ( P)  2

2



8.

16-4-2

Solve for P setting the shear stress equal to the maximum allowable stress, which is (0.3)E70. Guess

P  3  kip

Given t=

FR( P)

P  Find ( P)

τmax

Recommended maximum static load: P  2.2 kip Primary shear:

fs ( P)  0.276 

kip in

Shear due to moment: Horizontal component:

Vertical component: Total unit load:

fth ( P)  1.194 

kip

ftv ( P)  2.239 

kip

FR( P)  2.784 

kip

in

in in



16-5-1


Figure P16-1 shows a bar welded to a base on three sides with 3/16-in fillet welds using an E70 electrode. Material is A572 Grade 50 hot-rolled steel. What is your recommended maximum dynamic repeated load, zero to Pmax that can be applied for 10E8 cycles with a safety factor of 1.6?

Given:

Bar dims: Electrode: Weld:

a  6  in Exx  70 ksi w  0.1875 in

b  3  in

d  2  in

three places

Nfr  1.6

T  0.25 in

Assumptions: The weld will fail in shear along the weld throat. Consider the weld as a line. The weld pattern matches that of part 5 of Figure 16-15. The weld will limit the load as Category F. Solution: 1.



2.

t  0.133 in

Determine the shear area in the throats of the welds as a line function. 2

Aw  d  2  b 3.

b

xbar  1.125 in

2 b  d

Find the moment arm from the applied load to the centroid.

P

( 2 b  d)

3

12

2



b  ( b  d)

2

4

Jw  14.542

2 b  d

in

in

The unit load due to torsional shear is a vector that is perpendicular to a line from the centroid to any point on the weld. It is a maximum at a point furthest from the centroid. In this case, it is the furtherest point to the right on either the top or bottom of the bar. In either case, it will have horizontal and vertical components. The vertical component will add to the direct shear and that total will add vectorially to the horizontal component. The moment arm for the horizontal component is d/2, and for the vertical component it is b xbar. Horizontal unit load:

fth ( P) 

Vertical unit load:

ftv ( P) 

Total unit load: 8.

T ( P)  P r

Aw

From part 5 of Figure 16-15 the torsional geometry factor is: Jw 

7.

r  7.875 in

Find the unit load due to direct shear and the applied torque as functions of P. fs ( P) 

6.

in

2

r  a  b  xbar 5.

in

Find the centroid of the weld pattern with the equation given in part 5 of Figure 16-5. xbar 

4.

Aw  8 

FR( P) 

T ( P)  d 2  Jw T ( P)  ( b  xbar) Jw fth ( P)   fs ( P)  f tv ( P)  2

2

From Table 16-5a, a Category F joint has a shear stress-range endurance strength of S ers  8  ksi. Using the given safety factor, the allowable shear stress is



Allowable shear stress: 9.

τallow 

16-5-2

S ers

τallow  5  ksi

Nfr

Solve for P setting the shear stress equal to the maximum allowable stress. Guess

P  1  kip

Given t=

FR( P)

Pmax  Find ( P)

τallow Pmax  525  lbf

Recommended maximum dynamic load:

Primary shear:

fs  Pmax  65.6

lbf in

Shear due to moment: fth  Pmax  284.3 

lbf

Vertical component:

ftv  Pmax  533.1 

lbf

Total unit load:

FR Pmax  662.8 

lbf

Horizontal component:

in

in in



16-6a-1


Figure P16-2 shows a bracket welded to a wall with a fillet weld using an E70 electrode. For the row a in Table P16-1, determine the fillet weld size needed between the tube and the wall for a static load F. The pipe and wall material is A36 steel.

Bracket dims: OD  3.5 in ID  3.068  in a  2  OD l  2.5 OD Electrode: Exx  70 ksi Load: F  2.5 kip Assumptions: The weld will fail in shear along the weld throat and, from equation 16-1, the maximum allowable shear stress is τmax  0.3 Exx , τmax  21 ksi. Consider the weld as a line. The weld pattern Given:

matches that of part 9 of Figure 16-5. The weld will limit the load as Category F. Solution: 1.

See Figure P16-2, Figure 16-15, and Mathcad file P1606a.

Determine the shear area, bending factor, and torsion factor in the throats of the welds as line functions from Figure 16-15, part 9. 2

Aw  π OD S w  π

OD

Jw  π

OD

Aw  10.996 2

4

in 3

S w  9.621 

in

3

4

in

in 4

Jw  33.674

in

in

2.

The weld has a combined bending and torsion load. The maximum stress will occur at the top, center of the pipe at the wall. There will be a direct shear component, a bending component, and a torsional moment component.

3.

Find the unit load due to direct shear. fs 

4.

F Aw

fb 

M Sw

fb  2.274 

kip

ft  0.909 

kip

in

ft 

T  OD 2  Jw

in

Find the magnitude of the resultant unit load at the critical point ( fs is in the negative y-direction, f b is in the positive x-direction, and ft is in the negative z-direction). Total unit load:

7.

in

Find the unit load due to the torsional moment. T  F  a

6.

lbf

Find the unit load due to bending at the critical point. M  F  l

5.

fs  227.364 

FR 

2

2

fs  fb  ft

2

FR  2.459 

kip in

Using equation 16.1, solve for the minimum required weld size. t 

FR

τmax

t  0.117 in

w  1.414  t w  0.166 in

Say w = 3/16 in



16-7a-1


Figure P16-2 shows a bracket welded to a wall with a fillet weld using an E70 electrode. For the row a in Table P16-1, determine the fillet weld size needed between the tube and the wall for a dynamic load that varies from - 0.1F to +0.2F using a safety factor of 1.5. The pipe and wall material is A36 steel.

Given:

Bracket dims: Electrode: Load:

OD  3.5 in Exx  70 ksi F  2.5 kip

ID  3.068  in

a  2  OD

l  2.5 OD

Nfr  1.5


See Figure P16-2, Figure 16-15, and Mathcad file P1607a.

Determine the dynamic load range.

ΔF  0.2 F  ( 0.1F ) 2.

ΔF  750  lbf

Determine the shear area, bending factor, and torsion factor in the throats of the welds as line functions from Figure 16-15, part 9. 2

Aw  π OD S w  π

OD

Jw  π

OD

Aw  10.996 2

4

in 3

S w  9.621 

in

in

3

4

in

4

Jw  33.674

in

in

3.

The weld has a combined bending and torsion load. The maximum stress will occur at the top, center of the pipe at the wall. There will be a direct shear component, a bending component, and a torsional moment component.

4.


5.

ΔF Aw

fb 

M Sw

fb  0.682 

kip

ft  0.273 

kip

in

ft 

T  OD 2  Jw

in

Find the magnitude of the resultant unit load at the critical point ( fs is in the negative y-direction, f b is in the positive x-direction, and ft is in the negative z-direction). Total unit load:

8.

in

Find the unit load due to the torsional moment. T  ΔF  a

7.

lbf

Find the unit load due to bending at the critical point. M  ΔF  l

6.

fs  68.209

FR 

2

2

fs  fb  ft

2

FR  0.738 

kip in

From Table 16-5a, a Category F joint has a shear stress-range endurance strength of S ers  8  ksi. Using the given safety factor, the allowable shear stress is


MACHINE DESIGN - An Integrated Approach, 4th Ed. Allowable shear stress: 9.

τallow 

S ers Nfr

16-7a-2

τallow  5.333  ksi

Using equation 16.1, solve for the minimum required weld size.

t 

FR

τallow

t  0.138 in

w  1.414  t w  0.196 in

Say w = 1/4 in



16-8-1


Figure P16-3 shows a bracket machined from 12-mm-thick A572 Grade 50 hot-rolled steel flat stock. It is welded to a support with a fillet weld all around using an E80 electrode. Determine the fillet weld size needed between the bracket and the support for a static load of P = 12 kN.

Given:

Distance from support to point D Bracket dims: d  75 mm Electrode: Exx  80 ksi Load: P  12 kN

a  200  mm b  12 mm Exx  552  MPa

Assumptions: The weld will fail in shear along the weld throat and, from equation 16-1, the maximum allowable shear stress is τmax  0.3 Exx , τmax  165  MPa. Consider the weld as a line. The weld pattern matches that of part 7 of Figure 16-5. The weld will limit the load as Category F. Solution: 1.


Determine the shear area and bending factor, in the throats of the welds as line functions from Figure 16-15, part 7. Aw  2  b  2  d S w  b  d 

d

mm

Aw  174 

2

3

S w  2775

2

mm mm

3

mm

2.

The weld has a combined bending and shear load. The maximum stress will occur at the top, center of the bracket at the support. There will be a direct shear component and a bending component.

3.


4.

P Aw

mm

fb 

M

fb  0.865 

Sw

kN mm

Find the magnitude of the resultant unit load at the critical point. Total unit load:

6.

kN

Find the unit load due to bending at the critical point. M  P a

5.

fs  0.069 

FR 

2

fs  fb

2

FR  0.868 

kN mm


FR

τmax

t  5.24 mm

w  1.414  t w  7.41 mm

Say w = 8 mm



16-9-1


Figure P16-3 shows a bracket machined from 12-mm-thick A572 Grade 50 hot-rolled steel flat stock. It is welded to a support with a fillet weld all around using an E90 electrode. Determine the fillet weld size needed between the bracket and the support for a dynamic load that varies from 0 to +3 kN using a safety factor of 1.8.

Given:

Distance from support to point D Bracket dims: d  75 mm Electrode: Exx  90 ksi Load: F  3  kN

a  200  mm b  12 mm Exx  621  MPa Nfr  1.8



Determine the dynamic load range.

ΔF  F  0  kN 2.

ΔF  3  kN

Determine the shear area and bending factor, in the throats of the welds as line functions from Figure 16-15, part 7. Aw  2  b  2  d S w  b  d 

d

mm

Aw  174 

2

3

S w  2775

2

mm mm

3

mm

3.

The weld has a combined bending and shear load. The maximum stress will occur at the top, center of the bracket at the support. There will be a direct shear component and a bending component.

4.

Find the unit load due to direct shear. F fs  Aw

5.

fb 

M

fb  0.216 

Sw

kN mm

Find the magnitude of the resultant unit load at the critical point. Total unit load:

8.

kN mm

Find the unit load due to bending at the critical point. M  F  a

6.

fs  0.017 

FR 

2

fs  fb

2

FR  0.217 

kN mm

From Table 16-5b, a Category F joint has a shear stress-range endurance strength of S ers  55 MPa. Using the given safety factor, the allowable shear stress is Allowable shear stress:

9.

τallow 

S ers Nfr

τallow  30.6 MPa


FR

τallow

t  7.099 mm

w  1.414  t w  10.037 mm

Say w = 10 mm



16-10-1


Two 8-mm-thick by 50-mm-wide, A572 Grade 42 steel straps are welded together with fillet welds in a lap joint using an E70 electrode. The tensile load on the straps is 45 kN. What is your recommended weld size for the two full-length welds?

Given:

Straps:

T  8  mm

W  50 mm

Electrode: Load:

Exx  70 ksi P  45 kN

Exx  483  MPa

S y  345  MPa

Assumptions: Each weld will carry half the load and will fail in shear along the weld throat and, from equation 16-1, the maximum allowable shear stress is τmax  0.3 Exx , τmax  145  MPa Solution: 1.

See Figures 16-2 and 16-3, Table 16-1, and Mathcad file P1610.

Determine the transverse load carried by each weld. Transverse load on weld

2.

3.

Pt 

P

Pt  22.5 kN

2

Determine the shear area in the throats of the welds as a function of t and then solve for t using equation 16.1. Ashear = W  t

t=

Solving for t,

t 

Ashear

and

τxy = τmax =

W Pt

Pt Ashear

t  3.1 mm

W  τmax

Convert this throat dimension to a leg width w assuming an equal-leg fillet. w 

Leg width

t

w  4.4 mm

cos( 45 deg)

4.

Check this against the recommended minimum weld size for this thickness strap. Table 16-2 indicates that a 8-mm-thick part needs at least a 5-mm weld leg width, so increase the weld leg width to 5 mm. Set w  5  mm

5.

Check whether this value for w gives an acceptable safety factor against yielding in the fused base metal. The shear area on the leg along the width of the welds is Afusion  W  w , and, from Table 16-3 the yield strength of the base metal is S y  290  MPa. The shear yield strength of the base metal is S sy  0.577  S y . The factor of safety against yielding in the fused base metal is:

Nyield =

where

S sy

τxy

=

Afusion Pt

 S sy

Nyield  Afusion  250  mm

Afusion Pt 2

 S sy

and

Nyield  1.9 S sy  167  MPa

This is slightly less than the safety factor against static failure in the weld (see Table 16-1) but is acceptable.



16-11-1


Given:

Two 8-mm-thick by 50-mm-wide, A572 Grade 42 steel straps are welded together with fillet welds in a lap joint using an E70 electrode. Determine the fillet weld size needed for a dynamic load that varies from 0 to +12 kN using a safety factor of 1.5 for an infinite life of the two full-length welds. Straps:

T  8  mm

W  50 mm

Electrode: Load:

S ut  485  MPa

Exx  70 ksi Fmin  0  kN

Exx  483  MPa Fmax  12 kN Nfrs  1.5

Assumptions: Each weld will carry half the load and will fail in shear along the weld throat. Since the weld takes the load directly this is a Category F weldment. Solution: 1.


Determine the dynamic load range per weld.

ΔF  2.

Fmax  Fmin 2

ΔF  6  kN

From Table 16-5b, a Category F joint has a shear stress-range endurance strength of S ers  55 MPa. Using the given safety factor, the allowable shear stress is Allowable shear stress:

3.

4.

τallow 

S ers Nfrs


Using equation 4.9, solve for the allowable shear area and the resulting minimum value of the throat dimension t. Allowable shear area on each throat

Ashear 

Minimum throat dimension

t 

ΔF τallow

Ashear

Ashear  163.6 mm

2

t  3.273 mm

W

Convert this throat dimension to a leg width w assuming an equal-leg fillet. Leg width

w 

t cos( 45 deg)

w  4.6 mm

5.

Check this against the recommended minimum weld size for this thickness strap. Table 16-2 indicates that a 8-mm-thick part needs at least a 5-mm weld leg width, so increase the weld leg width to 5 mm. Set w  5  mm

6.

Check whether this value for w gives an acceptable safety factor against fatigue failure in the fused base metal by assuming that this is a Category C weldment. The most likely point of failure is at the weld toe. The weld leg area is Afusion  W  w , and, from Table 16-5b the endurance strength is S er  69 MPa Max stress at toe of each weld

σtoe 

Factor of safety in base metal

Nf 

Fmax 2  Afusion S er

σtoe

σtoe  24 MPa

Nf  2.9

Since Nf is larger than Nfrs, the initial assumption that this was a Category F weldment was correct.



16-12-1


Two 12-mm-thick by 50-mm-wide, weldable aluminum straps are welded together with fillet welds in a lap joint using an aluminum electrode. Determine the fillet weld size needed for a dynamic load that varies from 0 to +5 kN using a safety factor of 2.0 for an infinite life of the two full-length welds.

Given:

Straps: Load:

T  12 mm Fmin  0  kN

W  50 mm Fmax  5  kN

Nfrs  2.0

Assumptions: Each weld will carry half the load and will fail in shear along the weld throat. Since the weld takes the load directly this is a Category F weldment. Solution: 1.


Determine the dynamic load range per weld.

ΔF  2.

Fmax  Fmin 2

ΔF  2.5 kN

From Table 16-5b and the notation in the text that the values in Table 16-5 can be reduced by a factor of three for aluminum, a Category F joint has a shear stress-range endurance strength of S ers  0.333  55 MPa S ers  18.3 MPa. Using the given safety factor, the allowable shear stress is Allowable shear stress:

3.

4.

τallow 

S ers Nfrs


Using equation 4.9, solve for the allowable shear area and the resulting minimum value of the throat dimension t. Allowable shear area on each throat

Ashear 

Minimum throat dimension

t 

ΔF τallow

Ashear

Ashear  273 mm

2

t  5.46 mm

W

Convert this throat dimension to a leg width w assuming an equal-leg fillet. Leg width

w 

t cos( 45 deg)

w  7.7 mm

5.

Check this against the recommended minimum weld size for this thickness strap. Table 16-2 indicates that a 12-mm-thick part needs at least a 5-mm weld leg width. The calculated leg width is adequate but increase it to the next higher mm integer. Set w  8  mm

6.

Check whether this value for w gives an acceptable safety factor against fatigue failure in the fused base metal by assuming that this is a Category C weldment. The most likely point of failure is at the weld toe. The weld leg area is Afusion  W  w , and, from Table 16-5b the endurance strength is S er  0.333  69 MPa S er  23.0 MPa Max stress at toe of each weld

σtoe 

Factor of safety in base metal

Nf 

Fmax 2  Afusion S er

σtoe

σtoe  6.25 MPa

Nf  3.7

Since Nf is larger than Nfrs, the initial assumption that this was a Category F weldment was correct. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, P1612.xmcd mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


17-1-1


Find the torque that a 2-surface, dry disk clutch can transmit if the outside and inside lining diameters are 120 mm and 70 mm, respectively, and the applied axial force is 10 kn. Assume uniform wear and m = 0.4. Is the pressure on the lining acceptable? What lining materials would be suitable?

Given:

Number of surfaces

Ns  2

Clutch dimensions:

Axial force Friction coefficient

F  10 kN μ  0.4

Outside diameter Inside diameter

od  120  mm id  70 mm

Assumptions: Uniform wear model. Solution: 1.


Calculate the outside and inside radii of the disk. Outside radius ro  0.5 od Inside radius

ro  60 mm

ri  0.5 id

ri  35 mm T  Ns μ  F 

2.

Using equation 17.6, calculate the torque capacity.

3.

Calculate the maximum lining pressure using equation 17.5a. Maximum pressure

p max 

ro  ri 2

T  380  N  m

F

2  π ri  ro  ri

p max  1819 kPa 4.

From Table 17-1, we see that either a molded or sintered metal lining are suitable.



17-2-1


Find the torque that a 2-surface, dry disk clutch can transmit if the outside and inside lining diameters are 120 mm and 70 mm, respectively, and the applied axial force is 10 kn. Assume uniform pressure and  = 0.4. Is the pressure on the lining acceptable? What lining materials would be suitable?

Given:

Number of surfaces

Ns  2


F  10 kN μ  0.4

Clutch dimensions: Outside diameter Inside diameter

od  120  mm id  70 mm

Assumptions: Uniform pressure model. Solution: 1.


Calculate the outside and inside radii of the disk. Outside radius

ro  0.5 od

ro  60 mm

Inside radius

ri  0.5 id

ri  35 mm

  2  ro  ri  T  Ns μ  F   3  2 2  ro  ri  3

2.


3.

Calculate the lining pressure using equation 17.2c. Maximum pressure

4.

p 

3 2



T 3 3 π μ  Ns  ro  ri 

3

T  389  N  m

p  1340 kPa

From Table 17-1, we see that either a molded or sintered metal lining are suitable.



17-3-1


Design a single-surface disk clutch to transmit 100 N-m of torque at 750 rpm using a molded lining with a maximum pressure of 1000 kPa and  = 0.25. Assume uniform wear. Find the outside and inside diameters required if ri = 0.577 ro. What is the power transmitted?

Given:

Number of surfaces

Ns  1

Torque capacity

T  100  N  m

Friction coefficient

μ  0.25

Maximum pressure

p max  1000 kPa

Disk radius factor

fr  0.577

Rotational speed

ω  750  rpm



Use equation 17.5b and the disk radius factor to solve for the required outside and inside radii. T = π μ  ri p max   ro  ri 2

2

ri = fr  ro

 1

ro 

T    2    π μ  fr  p max   1  fr  

Round this to

3

ro  70 mm ri  fr  ro

Then, the inside radius is Round this to 2.

3.

ro  69.16  mm

ri  40.39  mm

ri  40 mm

Convert the radii to diameters. Outside diameter

od  2  ro

od  140  mm

Inside diameter

id  2  ri

id  80 mm

H  T  ω

H  7.85 kW

Calculate the maximum transmitted power. Transmitted power



17-4-1


Design a single-surface disk clutch to transmit 100 N-m of torque at 750 rpm using a molded lining with a maximum pressure of 1000 kPa and  = 0.25. Assume uniform pressure. Find the outside and inside diameters required if ri = 0.577 ro. What is the power transmitted?

Given:

Number of surfaces

Ns  1

Torque capacity

T  100  N  m

Friction coefficient Disk radius factor

μ  0.25

Maximum pressure Rotational speed

p  1000 kPa ω  750  rpm

fr  0.577



Use equation 17.2c and the disk radius factor to solve for the required outside and inside radii. T=

2 3

 π p  μ   ro  ri 3

3

  Ns

ri = fr  ro 1

ro 

3 T    3    2  π μ  p  N s   1  f r  

Round this to

3

ro  62 mm ri  fr  ro

Then, the inside radius is Round this to 2.

3.

ro  61.832 mm

ri  35.774 mm

ri  36 mm


od  2  ro

od  124  mm

Inside diameter

id  2  ri

id  72 mm

H  T  ω

H  7.85 kW




17-5-1


How many surfaces are needed in a wet disk clutch to transmit 120 N-m of torque at 1000 rpm using a sintered lining with a maximum pressure of 1800 kPa and  = 0.06. Assume uniform wear. Find the outside and inside diameters required if ri = 0.577 ro. How many disks are needed? What is the power transmitted?

Given:


μ  0.06

Rotational speed

ω  1000 rpm

Torque capacity Maximum pressure

fr  0.577

T  120  N  m p max  1800 kPa

Assumptions: Uniform wear model. Design Choices: The number of surfaces depends on the disk dimensions. Let ri  30 mm Solution:


1.

Calculate the outside radius.

2.

Convert the radii to diameters.

3.

ro  51.993 mm

fr

od  2  ro

od  104  mm

Inside diameter

id  2  ri

id  60 mm

ro  52 mm

Calculate the required axial force using equation 17.5a. F  2  π ri p max   ro  ri

F  7.464  kN

Using equation 17.6, calculate the number of friction surfaces required. Number surfaces

5.

ri

Outside diameter

Axial force 4.

ro 

Ns  ceil

2 T

 

 μ  F   ro  ri

Ns  7

H  T  ω

H  12.6 kW




17-6-1


How many surfaces are needed in a wet disk clutch to transmit 120 N-m of torque at 1000 rpm using a sintered lining with a maximum pressure of 1800 kPa and  = 0.06. Assume uniform pressure. Find the outside and inside diameters required if ri = 0.577 ro. How many disks are needed? What is the power transmitted?

Given:


μ  0.06

Rotational speed

ω  1000 rpm

Torque capacity Pressure

fr  0.577

T  120  N  m p  1800 kPa

Assumptions: Uniform pressure model. Design Choices: The number of surfaces depends on the disk dimensions. Let ri  30 mm Solution:


1.

Calculate the outside radius.

2.

Convert the radii to diameters.

3.

ro  51.993 mm

fr

od  2  ro

od  104  mm

Inside diameter

id  2  ri

id  60 mm

ro  52 mm

Calculate the required axial force using equation 17.1b. F  π p   ro  ri 2

2



F  10.201 kN

Using equation 17.3, calculate the number of friction surfaces required.

Number surfaces

5.

ri

Outside diameter

Axial force 4.

ro 

 3 T   ro2  ri2       μ  F   r 3  r 3  i  2 o

Ns  5

H  T  ω

H  12.6 kW

Ns  ceil




17-7-1


Figure P17-1 shows a single short-shoe drum brake. Find its torque capacity and required actuating force for the dimensions given below. What value of c will make it self-locking?

Given:

Pivot to load Pivot to y-axis Pivot to x-axis Maximum pressure

a  100mm b  70 mm e  20 mm p max  1300 kPa

Drum radius Drum width Shoe angle Friction coeff.

r  30 mm w  50 mm θ  35 deg μ  0.3

Assumptions: Short-shoe theory is appropriate. The drum rotates CCW. Solution: 1.


Determine the normal force on the drum from equation 17.8. Fn  p max  r θ  w

Normal force 2.

Use equation 17.10 to calculate the torque capacity. T  μ  Fn r

Torque capacity 3.

4.

Fn  1.191  kN

T  10.7 N  m

Determine the required actuating force from equation 17.11b and the brake geometry. Distance c

c  r  e

Actuation force

Fa  Fn

c  10 mm

b  μ c a

Fa  798  N

Check to see if the brake is self-locking using the relationship given in the text. self_locking 

return "yes" if μ  c  b "no" otherwise

self_locking  "no" 5.

Calculate the value of c that would make the brake self-locking use the above relationship. Value of c to self-lock

clock 

b

μ

clock  233.3  mm



17-8-1



Given:



Drum radius Drum width Shoe angle Friction coefficient

r  30 mm w  50 mm θ  35 deg μ  0.3

Assumptions: Short-shoe theory is appropriate. The drum rotates CW. Solution: 1.


Determine the normal force on the drum from equation 17.8. Normal force

2.

4.

Fn  1.191  kN

Use equation 17.10 to calculate the torque capacity. Torque capacity

3.

Fn  p max  r θ  w

T  μ  Fn r

T  10.7 N  m

Determine the required actuating force from equation 17.11b and the brake geometry modified for CW rotation. Distance c

c  r  e

Actuation force

Fa  Fn

c  10 mm

b  μ c a

Fa  870  N

Since the brake is not self-energizing with CW rotation, it cannot be self-locking for any value of c.



17-9-1



Given:


a  8.00 in b  6.00 in e  4.00 in p max  250  psi


r  5.00 in w  1.50 in θ  30 deg μ  0.35

Assumptions: Short-shoe theory is appropriate. The drum rotates CCW. Solution: 1.



Normal force 2.

Use equation 17.10 to calculate the torque capacity. T  μ  Fn r

Torque capacity 3.

4.

Fn  981.7  lbf

T  1718 in lbf


c  r  e

Actuation force

Fa  Fn

c  1.000  in

b  μ c a

Fa  693  lbf





clock 

b

μ

clock  17.1 in



17-10-1



Given:


a  8.00 in b  6.00 in e  4.00 in p max  250  psi


r  5.00 in w  1.50 in θ  30 deg μ  0.35

Assumptions: Short-shoe theory is appropriate. The drum rotates CW. Solution: 1.



2.

4.

Fn  981.7  lbf


3.

Fn  p max  r θ  w

T  μ  Fn r

T  1718 in lbf

Determine the required actuating force from equation 17.11b and the brake geometry modified for CW rotation. Distance c

c  r  e

Actuation force

Fa  Fn

c  1.000  in

b  μ c a

Fa  779  lbf

Since the brake is not self-energizing with CW rotation, it cannot be self-locking for any value of c.



17-11-1


Figure P17-2 shows a double short-shoe drum brake. Find its torque capacity and required actuating force for the dimensions given below. What value of c will make it self-locking? Hint: Calculate the effects of each shoe separately and superpose them.

Given:




r  40 mm w  60 mm θ  25 deg μ  0.25

Assumptions: Short-shoe theory is appropriate. The drum rotates CCW. Solution:


Top shoe - self energizing 1.

Determine the normal force on the drum from equation 17.8. Fn1  p max  r θ  w

Normal force 2.

Use equation 17.10 to calculate the torque capacity. T1  μ  Fn1 r

Torque capacity 3.

4.

Fn1  1.571  kN

T1  15.71  N  m


c  r  e

Actuation force

Fa  Fn1

c  10 mm b  μc a

Fa  1353 N





clock 

b

clock  320  mm

μ

Bottom shoe - non self energizing 6.

The actuation force is the same on both shoes and is equal to the actuation force on the top shoe. Use equation 17.11b to solve for the normal force on the bottom shoe. Normal force

7.

Fn2  Fa

a b  μc

Fn2  1476 N


T2  μ  Fn2 r

T2  14.76  N  m

Both shoes 8.

The total torque capacity is

Ttot  T1  T2

Ttot  30.5 N  m



17-12-1


Figure P17-2 shows a double short-shoe drum brake. Find its torque capacity and required actuating force for the dimensions given below. What value of c will make it self-locking? Hint: Calculate the effects of each shoe separately and superpose them.

Given:


a  12.00  in b  8.00 in e  3.00 in p max  200  psi


r  6.00 in w  2.00 in θ  25 deg μ  0.28




Determine the normal force on the drum from equation 17.8. Fn1  p max  r θ  w

Normal force 2.

Use equation 17.10 to calculate the torque capacity. T1  μ  Fn1 r

Torque capacity 3.

4.

Fn1  1047 lbf

T1  1759 in lbf


c  r  e

Actuation force

Fa  Fn1

c  3.000  in b  μc a

Fa  625  lbf





clock 

b

clock  28.6 in

μ

Bottom shoe - non self energizing 6.


7.

Fn2  Fa

a b  μc

Fn2  848  lbf


T2  μ  Fn2 r

T2  1425 in lbf

Both shoes 8.



Ttot  3184 in lbf



17-13-1


Figure P17-3 shows a single long-shoe drum brake. Find its torque capacity and required actuating force for the dimensions given below.

Given:

Pivot to load

a x  100mm

Drum radius

r  30 mm

Pivot to Y-axis

b x  70 mm

Drum width

w  50 mm

Pivot to X-axis

b y  20 mm

Shoe start angle

θ1  25 deg

Shoe end angle

θ2  125  deg

p max  1300 kPa

Maximum pressure

μ  0.3

Friction coeff.

Assumptions: Long-shoe theory is appropriate. The drum rotates CCW and the brake is self-energizing. Solution: 1.


θmax 

Determine the maximum shoe angle.

return 90 deg if θ2  90 deg

θ2 otherwise θmax  90 deg 2.

Use equation 17.15 to calculate the torque capacity. 2

T  μ  w r 

Torque capacity

p max

sin θmax

  cos θ1  cos θ2 

T  26.0 N  m 3.

Determine the required actuating force from equations 17.14 and the brake geometry. b 

Dimension b

2

2

bx  by

b  72.801 mm

Normal moment MFn  w r b 

1 1     θ2  θ1    sin 2  θ2  sin 2  θ1  2 4 sin θmax   p max

MFn  184.423  N  m Friction moment MFf  μ  w r

p max

sin θmax

 r  cos θ2  cos θ1  



b 2

  sin θ2  sin θ1 2

2



MFf  15.487 N  m

Actuating force

Fa 

MFn  MFf ax

Fa  1689 N



17-14-1


Figure P17-3 shows a single long-shoe drum brake. Find its torque capacity and required actuating force for the dimensions given below.

Given:

Pivot to load

a x  8.00 in

Drum radius

r  5.00 in

Pivot to Y-axis

b x  6.00 in

Drum width

w  1.50 in

Pivot to X-axis

b y  4.00 in

Shoe start angle

θ1  35 deg

Maximum pressure

p max  250  psi

Shoe end angle

θ2  155  deg


μ  0.35



θmax 





T  μ  w r 

Torque capacity

p max

sin θmax

  cos θ1  cos θ2 

T  5662 in lbf 3.


Dimension b

2

2

bx  by

b  7.211  in



MFn  19925  in lbf Friction moment MFf  μ  w r

p max

sin θmax

 r  cos θ2  cos θ1  



b 2

  sin θ2  sin θ1 2

2

 

MFf  6017 in lbf

Actuating force

Fa 

MFn  MFf ax

Fa  1738 lbf



17-15-1


Figure P17-4 shows a double long-shoe drum brake. Find its torque capacity and required actuating force for the dimensions given below.

Given:

Pivot to load

a x  90mm

Drum radius

r  40 mm

Pivot to Y-axis

b x  80 mm

Drum width

w  30 mm

Pivot to X-axis

b y  30 mm

Shoe start angle

θ1  30 deg

Maximum pressure

p max  1500 kPa

Shoe end angle

θ2  160  deg

Friction coeff.

μ  0.25

Assumptions: Long-shoe theory is appropriate. The drum rotates CCW. Solution:


Top shoe - self-energizing 1.


θmax 




T1  μ  w r 

Torque capacity

p max

sin θmax

  cos θ1  cos θ2 

T1  32.50  N  m 3.


Dimension b

2

2

bx  by

b  85.44  mm


1 1     θ2  θ1    sin 2  θ2  sin 2  θ1  4 sin θmax  2  p max


p max

sin θmax

 r  cos θ2  cos θ1  



b 2

  sin θ2  sin θ1 2

2



MFf  35.06  N  m

Actuating force

Fa 

MFn  MFf ax

Fa  2194 N

Bottom shoe - nonself-energizing 4.

The magnitude of the actuation force is the same on both shoes and is equal to the actuation force on the top shoe. This will result in a lower maximum pressure on the lower, nonself-energizing, shoe. Determine the maximum pressure on the lower shoe by solving equation 17.14c (using the plus sign) for p max2.



p max2 

a x Fa w r



17-15-2

sin θmax θ θ  2  2 2  b  sin( θ ) dθ   μ  sin( θ )  ( r  b  cos( θ ) ) dθ θ θ 1 1

p max2  1107 kPa 5.


2

T2  μ  w r 

p max2

sin θmax

  cos θ1  cos θ2 

T2  23.98  N  m 6.

The torque capacity of the double brake is the sum of the torque capacity of the two. Total torque capacity


Ttot  56.5 N  m



17-16-1


Figure P17-4 shows a double long-shoe drum brake. Find its torque capacity and required actuating force for the dimensions given below.

Given:

Pivot to load

a x  12.00  in

Drum radius

r  6.00 in

Pivot to Y-axis

b x  8.00 in

Drum width

w  2.00 in

Pivot to X-axis

b y  3.00 in

Shoe start angle

θ1  25 deg

Maximum pressure

p max  200  psi

Shoe end angle

θ2  145  deg

Friction coeff.

μ  0.28

Assumptions: Long-shoe theory is appropriate. The drum rotates CCW. Solution:




θmax 




T1  μ  w r 

Torque capacity

p max

sin θmax

  cos θ1  cos θ2 

T1  6957 in lbf 3.


Dimension b

2

2

bx  by

b  8.544  in




p max

sin θmax

 r  cos θ2  cos θ1  



b 2

  sin θ2  sin θ1 2

2




Actuating force

Fa 

MFn  MFf ax

Fa  1974 lbf




MACHINE DESIGN - An Integrated Approach, 4th Ed. p max2 

a x Fa w r



17-16-2


p max2  129.0  psi 5.


2

T2  μ  w r 

p max2

sin θmax

  cos θ1  cos θ2 

T2  4486 in lbf 6.

The torque capacity of the double brake is the sum of the torque capacity of the two. Total torque capacity


Ttot  11443  in lbf



17-17a-1


The short-shoe approximation is considered to be valid for brake shoes with an included angle of up to about 45 deg. For the brake shown in Figure P17-3, calculate its torque capacity and required actuating force by both the short-shoe method and the long-shoe method and compare the results for the dimensions given below.

Given:

Pivot to load

a x  90mm

Drum radius

r  40 mm

Pivot to Y-axis

b x  80 mm

Drum width

w  30 mm

Pivot to X-axis

b y  30 mm

Shoe start angle

θ1  75 deg

Shoe end angle

θ2  105  deg

p max  1500 kPa

Maximum pressure

μ  0.3


Assumptions: The drum rotates CCW and the brake is self-energizing. Solution:


Short-shoe method

θ  θ2  θ1

θ  30 deg

1.

Calculate the total shoe angle.

2.


Normal force 3.

Use equation 17.10 to calculate the torque capacity. Tshort  μ  Fn r

Torque capacity 4.

Fn  0.942  kN

Tshort  11.3 N  m


c  r  b y

c  10 mm

Actuation force

Fashort  Fn

bx  μ  c ax

Fashort  806  N

Long-shoe method 5.

θmax 





Tlong  μ  w r 

Torque capacity

p max

sin θmax

  cos θ1  cos θ2 

Tlong  11.2 N  m 7.


Dimension b

2

2

bx  by

b  85.44  mm


p max

sin θmax

1 1     θ2  θ1    sin 2  θ2  sin 2  θ1  2 4  

MFn  78.711 N  m



17-17a-2

Friction moment MFf  μ  w r

p max

sin θmax

 r  cos θ2  cos θ1  



b 2

  sin θ2  sin θ1 2

2



MFf  11.181 N  m

Actuating force

8.

Falong 

MFn  MFf ax

Falong  750  N

Compare the results.

Method

Torque Capacity

Actuation Force

Short shoe



Long shoe

Tlong  11.2 N  m




17-18a-1



Given:

Pivot to load

a x  90mm

Drum radius

r  40 mm

Pivot to Y-axis

b x  80 mm

Drum width

w  30 mm

Pivot to X-axis

b y  30 mm

Shoe start angle

θ1  75 deg

Shoe end angle

θ2  105  deg

Maximum pressure Friction coeff.

p max  1500 kPa

μ  0.3

Assumptions: The drum rotates CCW. Solution:


Short-shoe method Top shoe - self energizing

θ  θ2  θ1

θ  30 deg

1.


2.


3.

Fn  0.942  kN


4.

Fn  p max  r θ  w Tshort1  μ  Fn r

Tshort1  11.3 N  m


c  r  b y

Actuation force


c  10 mm bx  μ  c ax


Bottom shoe - nonself energizing 5.


6.

Fn2  Fashort 

ax bx  μ  c

Fn2  874  N


Tshort2  μ  Fn2 r

Tshort2  10.49  N  m

Both shoes 7.


Tshort  Tshort1  Tshort2


Long-shoe method Top shoe - self energizing 8.


θmax 




2

Tlong1  μ  w r 

p max

sin θmax

  cos θ1  cos θ2 



17-18a-2

Tlong1  11.2 N  m 10. Determine the required actuating force from equations 17.14 and the brake geometry. b 

Dimension b

2

2

bx  by

b  85.44  mm



MFn  78.711 N  m Friction moment MFf  μ  w r

p max

sin θmax

 r  cos θ2  cos θ1  



b 2

  sin θ2  sin θ1 2

2



MFf  11.181 N  m Falong 

Actuating force

MFn  MFf ax


Bottom shoe - nonself-energizing 11. The magnitude of the actuation force is the same on both shoes and is equal to the actuation force on the top shoe. This will result in a lower maximum pressure on the lower, nonself-energizing, shoe. Determine the maximum pressure on the lower shoe by solving equation 17.14c (using the plus sign) for p max2. p max2 

a x Falong w r




p max2  1127 kPa 12. Use equation 17.15 to calculate the torque capacity. Torque capacity

2


p max2

sin θmax

  cos θ1  cos θ2 

Tlong2  8.40 N  m 13. The torque capacity of the double brake is the sum of the torque capacity of the two. Total torque capacity

Tlong  Tlong1  Tlong2

Tlong  19.6 N  m

14. Compare the results. Method

Torque Capacity

Actuation Force

Short shoe



Long shoe

Tlong  19.6 N  m




17-19a-1



Given:

Pivot to load

a x  8.00 in

Drum radius

r  5.00 in

Pivot to Y-axis

b x  6.00 in

Drum width

w  1.50 in

Pivot to X-axis

b y  4.00 in

Shoe start angle

θ1  75 deg

Maximum pressure

p max  250  psi

Shoe end angle

θ2  105  deg

Friction coeff.

μ  0.35

Assumptions: The drum rotates CCW and the brake is self-energizing. Solution:


Short-shoe method

θ  θ2  θ1

θ  30 deg

1.


2.


Normal force 3.

Use equation 17.10 to calculate the torque capacity. Tshort  μ  Fn r

Torque capacity 4.

Fn  982  lbf

Tshort  1718 in lbf


c  r  b y

c  1.000  in

Actuation force


bx  μ  c ax

Fashort  693  lbf

Long-shoe method 5.


θmax 




Tlong  μ  w r 

Torque capacity

p max

sin θmax

  cos θ1  cos θ2 

Tlong  1698 in lbf 7.


Dimension b

2

2

bx  by

b  183.162  mm



MFn  6920 in lbf



17-19a-2

Friction moment MFf  μ  w r

p max

sin θmax

 r  cos θ2  cos θ1  



b 2

  sin θ2  sin θ1 2

2




Actuating force

8.

Falong 

MFn  MFf ax

Falong  653  lbf

Compare the results.

Method

Torque Capacity

Actuation Force

Short shoe



Long shoe

Tlong  1698 in lbf




17-20a-1



Given:

Pivot to load

a x  8.00 in

Drum radius

r  5.00 in

Pivot to Y-axis

b x  6.00 in

Drum width

w  2.00 in

Pivot to X-axis

b y  4.00 in

Shoe start angle

θ1  75 deg

Maximum pressure

p max  250  psi

Shoe end angle

θ2  105  deg

Friction coeff.

μ  0.35

Assumptions: The drum rotates CCW. Solution:


Short-shoe method Top shoe - self energizing

θ  θ2  θ1

θ  30 deg

1.


2.


3.

Fn  1309 lbf


4.

Fn  p max  r θ  w Tshort1  μ  Fn r

Tshort1  2291 in lbf


c  r  b y

Actuation force


c  1.000  in bx  μ  c ax




6.

Fn2  Fashort 

ax bx  μ  c

Fn2  1165 lbf


Tshort2  μ  Fn2 r

Tshort2  2038 in lbf

Both shoes 7.


Tshort  Tshort1  Tshort2


Long-shoe method Top shoe - self energizing 8.


θmax 



Use equation 17.15 to calculate the torque capacity.




Torque capacity

17-20a-2 p max

sin θmax

  cos θ1  cos θ2 

Tlong1  2265 in lbf 10. Determine the required actuating force from equations 17.14 and the brake geometry. b 

Dimension b

2

2

bx  by

b  7.211  in



MFn  9227 in lbf Friction moment MFf  μ  w r

p max

sin θmax

 r  cos θ2  cos θ1  



b 2

  sin θ2  sin θ1 2

2



MFf  2265 in lbf Falong 

Actuating force

MFn  MFf ax


Bottom shoe - nonself-energizing 11. The magnitude of the actuation force is the same on both shoes and is equal to the actuation force on the top shoe. This will result in a lower maximum pressure on the lower, nonself-energizing, shoe. Determine the maximum pressure on the lower shoe by solving equation 17.14c (using the plus sign) for p max2.

p max2 

a x Falong w r




p max2  151.5  psi 12. Use equation 17.15 to calculate the torque capacity. Torque capacity

2


p max2

sin θmax

  cos θ1  cos θ2 

Tlong2  1372 in lbf 13. The torque capacity of the double brake is the sum of the torque capacity of the two. Total torque capacity

Tlong  Tlong1  Tlong2


14. Compare the results. Method

Torque Capacity

Actuation Force

Short shoe



Long shoe





17-21-1

PROBLEM 17-21 Statement: Given:

Figure P17-2 shows a double short-shoe drum brake. Find the reaction forces in the global XY system for the dimensions given below. Pivot to load

a  90 mm

Drum radius

r  40 mm

Pivot to y-axis

b  80 mm

Drum width

w  60 mm

Pivot to x-axis

e  30 mm

Shoe angle

θ  25 deg

Maximum pressure p max  1500 kPa

Friction coefficient μ  0.25



Top shoe - self energizing 1. Determine the normal force on the drum from equation 17.8. Normal force 2. Calculate the friction force.

Fn1  p max  r θ  w

Fn1  1571 N

Ff1  μ  Fn1

Ff1  392.7  N

3. Determine the required actuating force from equation 17.11b and the brake geometry modified for CW rotation. Distance c

c  r  e

Actuation force

Fa1  Fn1

c  10.000 mm b  μc a

Fa1  1353 N

4. The reaction forces on the top shoe pivot are: Rx1  Ff1

Rx1  392.7  N

Ry1  Fa1  Fn1

Ry1  218.2  N

Bottom shoe - nonself energizing 5. The magnitude of the actuation force is the same on both shoes and is equal to the actuation force on the top shoe. Use equation 17.11b to solve for the normal force on the bottom shoe. Actuation force

Fa2  Fa1

Normal force

Fn2  Fa2

6. Calculate the friction force.

Fa2  1353 N a b  μc

Ff2  μ  Fn2

Fn2  1476 N Ff2  368.9  N

7. The reaction forces on the top shoe pivot are: Rx2  Ff2

Rx2  368.9  N

Ry2  Fa2  Fn2

Ry2  123.0  N



17-22-1


Figure P17-2 shows a double short-shoe drum brake. Find the reaction forces in the global XY system for the dimensions given below.

Given:


a  12.00  in b  8.00 in e  3.00 in p max  200  psi


r  6.00 in w  2.00 in θ  25 deg μ  0.28




2.

Determine the normal force on the drum from equation 17.8. Normal force Fn1  p max  r θ  w Calculate the friction force.

Ff1  μ  Fn1

Fn1  1047.2 lbf Ff1  293.2  lbf

3. Determine the required actuating force from equation 17.11b and the brake geometry modified for CW rotation. Distance c c  r  e c  3.000  in b  μc Actuation force Fa1  Fn1 Fa1  624.828  lbf a 4. The reaction forces on the top shoe pivot are: Rx1  Ff1

Rx1  293.2  lbf

Ry1  Fa1  Fn1

Ry1  422.4  lbf


The magnitude of the actuation force is the same on both shoes and is equal to the actuation force on the top shoe. Use equation 17.11b to solve for the normal force on the bottom shoe. Actuation force

Fa2  Fa1

Normal force

Fn2  Fa2

Fa2  625  lbf a b  μc

Ff2  μ  Fn2

6.

Calculate the friction force.

7.

The reaction forces on the top shoe pivot are: Rx2  Ff2 Ry2  Fa2  Fn2

Fn2  848.183  lbf Ff2  237.5  lbf Rx2  237.5  lbf Ry2  223.4  lbf



17-23-1


Find the reaction forces at the arm pivot in the global XY system for the brake in Figure P17-3 for the dimensions given below.

Given:

Pivot to load

a x  100mm

Drum radius

r  30 mm

Pivot to Y-axis

b x  70 mm

Drum width

w  50 mm

Pivot to X-axis

b y  20 mm

Shoe start angle

θ1  25 deg

Maximum pressure

p max  1300 kPa

Shoe end angle

θ2  125  deg


μ  0.3



θmax 





Dimension b

2

2

bx  by

b  72.801 mm




p max

sin θmax

 r  cos θ2  cos θ1  



b 2

  sin θ2  sin θ1 2

2



MFf  15.487 N  m

Actuating force 3.

MFn  MFf ax

Fa  1689 N

Let the angle between the global X axis and the x1 axis be . Rotation angle

4.

Fa 

 by    bx 

α  atan

α  15.945 deg

First, find the reaction forces in the x1-y1 system by summing forces in the x1 and y1 directions

 Fx1

  Fa sin( α)   cos( θ ) dFn   sin( θ ) dFf  Rx1 = 0  

 Fy1

  Fa cos( α)   sin( θ ) dFn   cos( θ ) dFf  Ry1 = 0  

where Rx1 and Ry1 are both assumed to act in their respective positive directions. 5.

Expand the integrals and solve these two equations for Rx1 and Ry1, respectively.



17-23-2

θ   θ2   2 2   Rx1  Fa sin( α)  w r   sin( θ )  cos( θ ) dθ  μ   sin( θ ) dθ  sin θmax  θ θ 1  1 

p max

Rx1  744.0  N θ   θ2   2 2   Ry1  Fa cos( α)  w r   sin( θ ) dθ  μ   sin( θ )  cos( θ ) dθ  sin θmax  θ θ 1  1 

p max

Ry1  1052.9 N 6.

Transform these forces to the global XY system. RX  Rx1 cos( α)  Ry1 sin( α)

RX  1005 N

RY  Rx1 sin( α)  Ry1 cos( α)

RY  808  N



17-24-1



Given:

Pivot to load

a x  8.00 in

Drum radius

r  5.00 in

Pivot to Y-axis

b x  6.00 in

Drum width

w  1.50 in

Pivot to X-axis

b y  4.00 in

Shoe start angle

θ1  35 deg

Maximum pressure

p max  250  psi

Shoe end angle

θ2  155  deg


μ  0.35



θmax 





Dimension b

2

2

bx  by

b  7.211  in




p max

sin θmax

 r  cos θ2  cos θ1  



b 2

  sin θ2  sin θ1 2

2




Actuating force 3.

MFn  MFf ax

Fa  1738 lbf


4.

Fa 

 by    bx 

α  atan

α  33.69  deg


 Fx1


 Fy1



Expand the integrals and solve these two equations for Rx1 and Ry1, respectively.



17-24-2

θ   θ2   2 2   Rx1  Fa sin( α)  w r   sin( θ )  cos( θ ) dθ  μ   sin( θ ) dθ  sin θmax  θ θ 1  1 

p max

Rx1  2072.3 lbf θ   θ2   2 2   Ry1  Fa cos( α)  w r   sin( θ ) dθ  μ   sin( θ )  cos( θ ) dθ   sin θmax  θ 1  θ1 

p max

Ry1  1267.3 lbf 6.

Transform these forces to the global XY system. RX  Rx1 cos( α)  Ry1 sin( α)

RX  2427 lbf

RY  Rx1 sin( α)  Ry1 cos( α)

RY  95.1 lbf



17-25-1



Given:

Pivot to load

a x  90mm

Drum radius

r  40 mm

Pivot to Y-axis

b x  80 mm

Drum width

w  30 mm

Pivot to X-axis

b y  30 mm

Shoe start angle

θ1  30 deg

Maximum pressure

p max  1500 kPa

Shoe end angle

θ2  160  deg


μ  0.25

Assumptions: Long-shoe theory is appropriate. The drum rotates CCW. Solution: 1.



θmax 



Determine the required actuating force from equations 17.14 and the brake geometry. 2

b 

Dimension b

2

bx  by

b  85.44  mm




p max

sin θmax

 r  cos θ2  cos θ1  



b 2

  sin θ2  sin θ1 2

2



MFf  35.06  N  m

Actuating force 3.

Fa 

MFn  MFf ax

Fa  2194 N


 by    bx 

α  atan

α  20.556 deg



 Fx1


 Fy1


where Rx1 and Ry1 are both assumed to act in their respective positive directions. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


17-25-2

Expand the integrals and solve these two equations for Rx1 and Ry1, respectively. θ   θ2   2 2   Rx1  Fa sin( α)  w r   sin( θ )  cos( θ ) dθ  μ   sin( θ ) dθ  sin θmax  θ θ 1  1 

p max

Rx1  1570.2 N θ   θ2   2 2   Ry1  Fa cos( α)  w r   sin( θ ) dθ  μ   sin( θ )  cos( θ ) dθ  sin θmax  θ θ 1  1 

p max

Ry1  637.2  N 6.

Transform these forces to the global XY system. RXtop  Rx1 cos( α)  Ry1 sin( α)

RXtop  1694 N

RYtop  Rx1 sin( α)  Ry1 cos( α)

RYtop  45.3 N



p max2 

a x Fa w r




p max2  1107 kPa 8. First, find the reaction forces in the x2-y2 system by summing forces in the x2 and y2 directions. However, for a nonself-energizing shoe, the signs of the Ff terms are opposite to those for the self-energizing case. Note also that the y2 axis points toward the shoe (opposite from what is shown in Figure P17-4).

 Fx2

  Fa sin( α)   cos( θ ) dFn   sin( θ ) dFf  Rx2 = 0  

 Fy2

  Fa cos( α)   sin( θ ) dFn   cos( θ ) dFf  Ry2 = 0  


Expand the integrals and solve these two equations for Rx2 and Ry2, respectively. θ   θ2   2 2   Rx2  Fa sin( α)  w r   sin( θ )  cos( θ ) dθ  μ   sin( θ ) dθ  sin θmax  θ θ 1  1 

p max2


MACHINE DESIGN - An Integrated Approach, 4th Ed. Rx2  356.6  N

17-25-3

θ   θ2   2 2   Ry2  Fa cos( α)  w r   sin( θ ) dθ  μ   sin( θ )  cos( θ ) dθ   sin θmax  θ 1  θ1 

p max2

Ry2  24.0 N 10. Transform these forces to the global XY system. RXbot  Rx2 cos( α)  Ry2 sin( α)

RXbot  325  N

RYbot  Rx2 sin( α)  Ry2 cos( α)

RYbot  147.7  N



17-26-1



Given:

Pivot to load

a x  12.00  in

Drum radius

r  6.00 in

Pivot to Y-axis

b x  8.00 in

Drum width

w  2.00 in

Pivot to X-axis

b y  3.00 in

Shoe start angle

θ1  25 deg

Maximum pressure

p max  200  psi

Shoe end angle

θ2  145  deg


μ  0.28

Assumptions: Long-shoe theory is appropriate. The drum rotates CCW. Solution: 1.



θmax 



Determine the required actuating force from equations 17.14 and the brake geometry. 2

b 

Dimension b

2

bx  by

b  8.544  in




p max

sin θmax

 r  cos θ2  cos θ1  



b 2

  sin θ2  sin θ1 2

2




Actuating force 3.

Fa 

MFn  MFf ax

Fa  1974 lbf


 by    bx 

α  atan

α  20.556 deg



 Fx1


 Fy1


where Rx1 and Ry1 are both assumed to act in their respective positive directions. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


17-26-2

Expand the integrals and solve these two equations for Rx1 and Ry1, respectively. θ   θ2   2 2   Rx1  Fa sin( α)  w r   sin( θ )  cos( θ ) dθ  μ   sin( θ ) dθ  sin θmax  θ θ 1  1 

p max

Rx1  1503.1 lbf θ   θ2   2 2   Ry1  Fa cos( α)  w r   sin( θ ) dθ  μ   sin( θ )  cos( θ ) dθ  sin θmax  θ θ 1  1 

p max

Ry1  1738.6 lbf 6.

Transform these forces to the global XY system. RXtop  Rx1 cos( α)  Ry1 sin( α)

RXtop  2018 lbf

RYtop  Rx1 sin( α)  Ry1 cos( α)

RYtop  1100 lbf



p max2 

a x Fa w r




p max2  129.0  psi 8. First, find the reaction forces in the x2-y2 system by summing forces in the x2 and y2 directions. However, for a nonself-energizing shoe, the signs of the Ff terms are opposite to those for the self-energizing case. Note also that the y2 axis points toward the shoe (opposite from what is shown in Figure P17-4).

 Fx2

  Fa sin( α)   cos( θ ) dFn   sin( θ ) dFf  Rx2 = 0  

 Fy2

  Fa cos( α)   sin( θ ) dFn   cos( θ ) dFf  Ry2 = 0  


Expand the integrals and solve these two equations for Rx2 and Ry2, respectively. θ   θ2   2 2   Rx2  Fa sin( α)  w r   sin( θ )  cos( θ ) dθ  μ   sin( θ ) dθ  sin θmax  θ θ 1  1 

p max2

Rx2  61.7 lbf © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


Ry2  Fa cos( α)  w r

17-26-3

θ   θ2   2 2     sin( θ ) dθ  μ   sin( θ )  cos( θ ) dθ   sin θmax  θ 1  θ1 

p max2

Ry2  399.3  lbf 10. Transform these forces to the global XY system. RXbot  Rx2 cos( α)  Ry2 sin( α)

RXbot  82.5 lbf

RYbot  Rx2 sin( α)  Ry2 cos( α)

RYbot  396  lbf



17-27-1


A clutch is needed for an electric motor that transmits 20 kW at 1100 rpm. The clutch will attach directly to the motor housing face plate and is to have the same housing diameter as the motor, which is 125 mm. The minimum radial clearance between the housing OD and the clutch disk OD is 5 mm. The clutch output shaft will have the same diameter as the motor shaft, which is 15 mm. Design a multiple disk clutch for this application. State all assumptions and design choices. Specify the clutch material, outside disk radius, inside disk radius, and the required actuation force.

Given:

Shaft diameter

d s  15 mm

Transmitted power Housing dia.

H  20 kW d h  125  mm

Rotational speed Disk radial clearance

ω  1100 rpm cr  5  mm

Assumptions: Uniform pressure model. Design choices: Use a molded clutch material running against steel with: Maximum pressure Friction coefficient μ  0.25 Disk outside radius ro  0.5 d h  cr ro  57.5 mm Solution: 1.


Calculate the transmitted torque. T 

2.

H

T  173.6  N  m

ω

Tentatively choose an inside radius for the clutch discs and use equation 17.2c to solve for the required number of friction faces. Let Ns 

3.

p  1000 kPa

ri  0.5 d s  5  mm 3 T 2  π p  μ   ro  ri 3

2

T=

3

 π p  μ   ro  ri 3

3

  Ns

Ns  1.762

3



Increase this to the next higher integer and solve for the required inside radius. 1

Let

Round this to 4.

5.

ri   ro  3

Ns  2



3 T

  2  π p  μ  N s 

3

ri  28.969 mm

ri  29 mm


od  2  ro

od  115  mm

Inside diameter

id  2  ri

id  58 mm

Calculate the actuation force required using equation 17.3. 3  T   ro  ri 2

Actuation force

F 

2



2  μ  Ns  ro  ri 3

3

F  7748 N





17-28-1


A clutch is needed for an electric motor that transmits 25 hp at 800 rpm. The clutch will attach directly to the motor housing face plate and is to have the same housing diameter as the motor, which is 5.5 in. The minimum radial clearance between the housing OD and the clutch disk OD is 0.25 in. The clutch output shaft will have the same diameter as the motor shaft, which is 0.625 in. Design a multiple disk clutch for this application. State all assumptions and design choices. Specify the clutch material, outside disk radius, inside disk radius, and the required actuation force.

Given:

Shaft diameter

d s  0.625  in

Transmitted power Housing dia.

H  25 hp d h  5.5 in

Rotational speed Disk radial clearance

ω  800  rpm cr  0.25 in

Assumptions: Uniform pressure model. Design choices: Use a molded clutch material running against steel with: Maximum pressure Friction coefficient μ  0.25 Disk outside radius ro  0.5 d h  cr ro  2.50 in Solution: 1.


Calculate the transmitted torque. T 

2.

H

T  1970 in lbf

ω

Tentatively choose an inside radius for the clutch discs and use equation 17.2c to solve for the required number of friction faces. 2 3 3 Let ri  0.5 d s  0.25 in T =  π p  μ   ro  ri   Ns 3 Ns 

3.

p  150  psi

3 T 2  π p  μ   ro  ri 3

Ns  1.623

3



Increase this to the next higher integer and solve for the required inside radius. 1

Let

Round this to 4.

5.

ri   ro  3

Ns  2



3 T

  2  π p  μ  N s 

3

ri  1.456  in

ri  1.50 in


od  2  ro

od  5.00 in

Inside diameter

id  2  ri

id  3.00 in

Calculate the actuation force required using equation 17.3. 3  T   ro  ri 2

Actuation force

F 

2



3 2  μ  Ns  ro  ri  3

F  1929 lbf



17-29-1


Find the torque that a dual-pad, caliper disc brake with pad angle of 60 deg can transmit if the outside and inside lining diameters are 160 mm and 90 mm, respectively, and the applied axial force is 3 kN. Assume uniform wear and = 0.35. Is the pressure on the lining acceptable? What lining materials would be suitable?

Given:

Number of pads

Np  2


F  3  kN μ  0.35

Clutch dimensions: Outside diameter Inside diameter Pad angle

od  160  mm id  90 mm θ  0.333  π




ro  80 mm

ri  0.5 id

ri  45 mm T  Np μ  F 

2.


3.

Calculate the maximum lining pressure using equation 17.5a. Maximum pressure

p max 

ro  ri 2

T  131.25 N  m

F

θ  ri  ro  ri

p max  1821 kPa 4.

From Table 17-1, we see that either a molded or sintered metal lining is suitable.



17-30-1


Find the torque that a 2-pad, caliper disc brake with pad angle of 60 deg can transmit if the outside and inside lining diameters are 160 mm and 90 mm, respectively, and the applied axial force is 3 kN. Assume uniform pressure and = 0.35. Is the pressure on the lining acceptable? What lining materials would be suitable?

Given:

Number of pads

Np  2


F  3  kN μ  0.35

Clutch dimensions: Outside diameter Inside diameter Pad angle

od  160  mm id  90 mm θ  0.333  π




ro  80 mm

ri  0.5 id

ri  45 mm 3

2.


3.

Calculate the lining pressure using equation 17.2c. Pressure

p 

3

2 ro  ri T  Np μ  F   3 2 2 ro  ri

T  134.7  N  m

3 T

θ  μ  Np  ro  ri  3

3

p  1311 kPa

4.

From Table 17-1, we see that either a molded or sintered metal lining is suitable.



17-31-1


Design a dual-pad caliper disc brake to provide a braking force of 240 N at the periphery of a 750-mm-dia wheel that is rotating at 670 rpm. Use an inside radius to outside radius ratio of 0.577. Assume uniform wear. State all assumptions and design choices. Specify the brake material, outside pad radius, inside pad radius, pad angle, and the required actuation force.

Given:

Number of pads

Np  2

Wheel speed

Peripheral force

Fw  240  N

Wheel diameter d w  750  mm

Disk radius factor

fr  0.577

ω  670  rpm

Assumptions: Uniform wear model. Design Choices: Friction coefficient Pad angle (90 deg) Solution:

μ  0.30 θ  0.50 π

Pad material Max. pressure

T  0.5 d w Fw

T  90 N  m

Sintered metal p max  1500 kPa


1.

Calculate the torque required

2.

Using equation 17.5b and the disk radius factor, solve for the required outside and inside radii. T=

θ  μ  ri p max 2

  ro  ri 2

2

  Np

ri = fr  ro 1

ro 

2 T    2     θ  μ  pmax  fr   1  fr   Np

ro  69.16  mm

3.

3

round this to

Then the inside radius is

ri  fr  ro

Round this to

ri  40 mm

ro  70 mm ri  40.39  mm

Solve equation 17.5a for the actuating force, F. Actuating force

F  θ  ri p max   ro  ri F  2.827  kN

Rated torque

T  Np μ  F 

ro  ri 2

T  93.3 N  m 4.

DESIGN SUMMARY Brake material Coefficient of friction Pad inside radius

Sintered metal μ  0.30 ri  40 mm

Pad outside radius

ro  70 mm

Pad angle Actuating force Maximum pressure

θ  90 deg

Rated torque

T  93.3 N  m

F  2.83 kN p max  1500 kPa



17-32-1


Design a dual-pad caliper disc brake to provide a braking force of 240 N at the periphery of a 750-mm-dia wheel that is rotating at 670 rpm. Use an inside radius to outside radius ratio of 0.577. Assume uniform pressure. State all assumptions and design choices. Specify the brake material, outside pad radius, inside pad radius, pad angle, and the required actuation force.

Given:

Number of pads

Np  2

Wheel speed

Peripheral force

Fw  240  N

Wheel diameter d w  750  mm

Disk radius factor

fr  0.577

ω  670  rpm

Assumptions: Uniform wear model. Design Choices: Friction coefficient Pad angle (90 deg) Solution:

μ  0.30 θ  0.50 π

Pad material Sintered metal Uniform pressure p  1500 kPa

T  0.5 d w Fw

T  90 N  m


1.

Calculate the torque required

2.

Using equation 17.2c and the disk radius factor, solve for the required outside and inside radii. T=

θμ p  3 3   ro  ri   Np

ri = fr  ro

3

1

3 T  ro     3   θ  μ  p   1  fr   Np

3

ro  61.832 mm

3.

ro  62 mm

round this to


ri  fr  ro

Round this to

ri  36 mm

ri  35.774 mm

Solve equation 17.1b for the actuating force, F.

θp  2 2   ro  ri 

Actuating force

F 

Rated torque

2 ro  ri T  Np μ  F   3 2 2 ro  ri

2 F  3.002  kN

3

3

T  90.3 N  m 4.


Sintered metal μ  0.30 ri  36 mm

Pad outside radius

ro  62 mm

Pad angle Actuating force Uniform pressure Rated torque

θ  90 deg F  3.00 kN p  1500 kPa T  90.3 N  m



17-33-1


An ultra-light solar racecar weighs 500 lb with driver. It has two 20-in-dia bicycle wheels in front that are to have dual-pad caliper disk brakes on each wheel. The brakes must be capable of bringing the car to a stop in a distance of 150 feet from a speed of 45 mph. Neglecting aerodynamic and rolling resistance forces, design dual-pad caliper disc brakes for the car. Use an inside radius to outside radius ratio of 0.577. Assume uniform wear. State all assumptions and design choices. Specify the brake material, outside pad radius, inside pad radius, pad angle, and the required actuation force.

Given:

Number of pads

Np  4

Car weight

Stopping distance

Ds  150  ft

Wheel diameter d w  20 in

Disk radius factor

fr  0.577

Initial speed

Assumptions: Uniform wear model

Constant deceleration

Level surface

Design Choices: Friction coefficient Pad angle

μ  0.30 θ  60 deg

Pad material Max. pressure

Solution: 1.

Vi  45 mph

Molded p max  225  psi


Calculate the torque required. For uniform deceleration, the stopping distance, initial velocity, and deceleration are related by

Ds =

Vi

2

d 

solving for the deceleration,

2 d

Vi

2

2

2  Ds

Fw 

W

d g T  0.5 d w Fw

The force required between the tires and ground is Thus, the torque required by the brakes is 2.

W  500  lbf

d  14.52 ft  s

Fw  225.6  lbf T  188.04 lbf  ft

Using equation 17.5b and the disk radius factor, solve for the required outside and inside radii. T=

θ  μ  ri p max 2

  ro  ri 2

2

  Np

ri = fr  ro 1

ro 

2 T     2   θ  μ  pmax  fr   1  fr   Np

ro  3.461  in

3.

3

round this to


ri  fr  ro

Round this to

ri  2.00 in

ro  3.50 in ri  2.019  in

Solve equation 17.5a for the actuating force, F. Actuating force

F  θ  ri p max   ro  ri F  706.9  lbf

Rated torque

T  Np μ  F 

ro  ri 2

T  194.4  lbf  ft © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


4.


17-33-2

Molded μ  0.30 ri  2.00 in

Pad outside radius

ro  3.50 in

Pad angle Actuating force Maximum pressure

θ  60 deg

Rated torque

T  194  lbf  ft

F  706.9  lbf p max  225  psi



17-34-1


An ultra-light solar racecar weighs 500 lb with driver. It has two 20-in-dia bicycle wheels in front that are to have dual-pad caliper disk brakes on each wheel. The brakes must be capable of bringing the car to a stop in a distance of 150 feet from a speed of 45 mph. Neglecting aerodynamic and rolling resistance forces, design dual-pad caliper disc brakes for the car. Use an inside radius to outside radius ratio of 0.577. Assume uniform pressure. State all assumptions and design choices. Specify the brake material, outside pad radius, inside pad radius, pad angle, and the required actuation force.

Given:

Number of pads

Np  4

Car weight

Stopping distance

Ds  150  ft

Wheel diameter d w  20 in

Disk radius factor

fr  0.577

Initial speed

Assumptions: Uniform pressure model

Constant deceleration

Level surface

Design Choices: Friction coefficient Pad angle

μ  0.30 θ  60 deg

Pad material Pressure

Solution: 1.

Vi  45 mph

Molded p  225  psi


Calculate the torque required. For uniform deceleration, the stopping distance, initial velocity, and deceleration are related by

Ds =

Vi

2

2 d

solving for the deceleration,

The force required between the tires and ground is Thus, the torque required by the brakes is 2.

W  500  lbf

d 

Vi

2

2

2  Ds

Fw 

W

d g T  0.5 d w Fw

d  14.52 ft  s

Fw  225.6  lbf T  188.04 lbf  ft

Using equation 17.2c and the disk radius factor, solve for the required outside and inside radii. T=

θμ p  3 3   ro  ri   Np

ri = fr  ro

3

1

3 T  ro     3   θ  μ  p   1  fr   Np

3

ro  3.095  in

3.

ro  3.125  in

round this to


ri  fr  ro

Round this to

ri  1.750  in

ri  1.803  in

Solve equation 17.1b for the actuating force, F.

θp  2 2   ro  ri 

Actuating force

F 

Rated torque

2 ro  ri T  Np μ  F   3 2 2 ro  ri

2 F  789.7  lbf

3

3

T  197.6  lbf  ft © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


4.


17-34-2

Molded μ  0.30 ri  1.750  in

Pad outside radius

ro  3.125  in

Pad angle Actuating force Uniform pressure Rated torque

θ  60 deg F  789.7  lbf p  225  psi T  198  lbf  ft


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