Theory of Machines and Machines
Chapter 15...Dynamic Force Analysis (Planar)
15.1 introduction Dynamic force analysis: dynamic forces are associated with accelerating mass.
Figure (a) A mass that is moving in a circular path experiences centripetal acceleration, and there is a dynamic force, referred to as centrifugal force, associated with this acceleration. This force acts radially outward and will be transmitted to the support, producing a bearing reaction. (b) A rotating shaft with an eccentric mass The purpose of this chapter: (1) to learn how much acceleration will result from a system of unbalanced forces. (2) to learn how these dynamic forces can be assessed for system that are not in equilibrium
15-2
15.2 Centroid and center of mass Figure 15.1 Particle of mass dmP at location P on a rigid body.
dFp = A p dm p
∑F = ∫ A ij
p
dm p
A P = A Q + ω j × (ω j × R PQ ) + α j × R PQ
∑ F = A ∫ dm + ω × (ω × ∫ R dm ) + α × ∫ R ( ∫ dm = m , ∫ R dm = m R ) ∑ F = A m + ω × (ω × m R ) + α × m R ij
Q
p
ij
j
Q
j
P
j
j
PQ
P
j
j
j
PQ
P
j
PQ
dmP
G
j
G
j
j
G
= m j [ A Q + ω j × (ω j × R G ) + α j × R G ] = m j AG
15-3
The center of mass Figure 15.2 (a) Particles of mass distributed along a line. (b) Particles of mass distributed in a plane.
x=
m1 x1 + m2 x2 + m3 x3 = m1 + m2 + m3
RG = RG =
∑m x ∑m
i i i
m1R1 + m2 R 2 + m3 R 3 + m4 R 4 = m1 + m2 + m3 + m4
∑m R ∑m i
i
i
1 Rdm mj ∫
15-4
Figure 15.3 Center of mass location for (a) a right circular solid, (b) a rectangular solid, and (c) a triangular prism.
When the mass is evenly distributed over a plane area or a volume, The center of mass can often be found by symmetry.
15-5
Example 15.1 Figure 15.4 Composite shape for Example 15.1 with dimensions in millimeters.
When the body is of a more irregular shape, the center of mass can often still be found by considering it to be combination of simpler subshapes. m1 = (80mm)(200mm)(40mm)( ρ kg / mm 3 ) = 640,000 ρ kg R G1 = 100i + 20 j − 40k mm m2 = π (20mm) 2 (40mm)(− ρ kg / mm 3 ) = −50,300kg R G2 = 160i + 20 j − 40k mm m3 = 0.5(60mm)(90mm)(20mm)( ρ kg / mm 3 ) = 54,000 ρ kg R G3 = 30i + 10 j + 20k mm RG =
m1R1 + m2 R 2 + m3R 3 = m1 + m2 + m3
∑m R ∑m i
i
i
R G = 89.4i + 19.1j − 35.0k mm
15-6
15.3 Mass Moments and Products of Inertia I = ∫ (dis tan ce) 2 dm mass monent of inertia I xx = ∫ [( R y ) 2 + ( R z ) 2 ]dm I yy = ∫ [( R x ) 2 + ( R z ) 2 ]dm I = ∫ [( R ) +( R ) ]dm zz
x 2
y 2
Inertia tensor
⎡ I xx ⎢ I = ⎢− I yx ⎢ − I zx ⎣
− I xy I yy − I zy
I G = k 2 m or k = I = I G + md 2
I yz = I zy = ∫ ( R y R z )dm I zx = I xz = ∫ ( R z R x )dm
IG m
k: radius of gyration
Transfer or parallel-axis formula IG: principal mass moments of inertia I: mass moment of inertia about a parallel axis at distance d
mass products of inertia I xy = I yx = ∫ ( R x R y )dm
− I xz ⎤ ⎥ − I yz ⎥ I zz ⎥⎦
Principal mass moment of inertia
⎡ I1 I = ⎢⎢ 0 ⎢⎣ 0
0 I2 0
0⎤ 0 ⎥⎥ I 3 ⎥⎦
Principal axes: all of the products of inertia becomes zero.
15-7
Example 15.2 Figure 15.5 Connecting-rod shape for Example 15.2. The connecting rod is made of ductile iron with density of 0.260 lb/in3. Find the mass moment of inertia about the z axis. mcyl = ρπ (r02 − ri 2 )l (0.26lb / in 3 )π [(1.5in) 2 − (0.5in) 2 ](0.75) 386in / s 2 = 0.00317lb ⋅ s 2 / in =
1′′ dai. holes
(0.26lb / in 3 )(13in)(1in)(0.75in) 386in / s 2 = 0.00657lb ⋅ s 2 / in
mbar = ρwhl =
m(ro2 + ri 2 ) I cyl = 2 (0.00317lb ⋅ s 2 / in)[(1.5in) 2 + (0.5in) 2 ] = 2 2 = 0.00396in ⋅ lb ⋅ s m( w 2 + h 2 ) 12 (0.00657lb ⋅ s 2 / in)[(13in) 2 + (1in) 2 ] = 12 2 = 0.0931in ⋅ lb ⋅ s
I bar =
2 2 I zz = I cyl + ( I bar + mbar d bar ) + ( I cyl + mcyl d cyl )
= 1.33in ⋅ lb ⋅ s 2
15-8
15.4 Inertia Forces and D’Alembert’s Principle Figure 15.6 (a) An unbalanced set of forces on a rigid body. (b) The accelerations that result from the unbalanced forces.
∑F = F + F + F ∑F = m A ∑M = I α ∑ F + (−m A ) = 0 ∑M + ( − I α ) = 0 1
2
j
ij
Gij
3
G
G
j
j
ij
Gij
G
G
j
D’Alembert’s Principle: The vector sum of all external forces and inertia forces acting upon a system of rigid bodies is zero. The vector sum of all external moments and inertia torques acting upon a system of rigid bodies is zero.
∑F = 0 ∑M = 0 15-9
Equivalent Offset Inertia Force When a graphical solution by a force polygon is desired, ∑ F = 0 and
∑M = 0
can be combined.
Figure 15.7 (a) Unbalanced forces and resulting accelerations. (b) Inertia force and inertia couple. (c) Inertia force offset from center of mass.
h=
I Gα 3 m3 AG
1.The magnitude of the inertia force is mAG . AG 2.The direction of the inertia force is opposite to that of acceleration 3.The perpendicular offset distance from the center of mass to the line of action of the force is given by the above equation. 4.The force is offset from the center of mass so as to produce a moment about the center of mass that is opposite in sense to acceleration α
15-10
Example 15.3 Figure 15.8 Solution for Example 15.3: (a) Scale drawing with RBA = 10 in, RGA = 5 in, RAO = 8 in, and RBO = 6 in. (b) Acceleration polygon. (c) Free-body diagram and force polygon.
VBA = 21.2 ft / s
VA = 12.6 ft / s
VB = 16.2 ft / s
α3 =
t ABA RBA
(713 ft / s 2 )(12in / ft ) = 10in = 856rad / s 2 cw
h=
(0.0479in ⋅ lb ⋅ s 2 )(856rad / s 2 ) = 1.35in (0.00570in ⋅ lb ⋅ s 2 )(444 ft / s 2 )(12in / ft )
FA = 27 jlb
15-11
Example 15.4: ˆ R ˆ RAO2 = 60 mm, RO4O2 = 100 mm, RBA = 220 mm, RBO4 = 150 mm, RCO4 = RCB = 120 mm, RG3A = 90 mm, G4O4 = 90 mm, m3 = 1.5 kg, m4 = 5 kg, IG2 = 0.025 kg · m2, IG3 = 0.012 kg · m2, IG4ˆ= 0.054 kg · m2, α 2 = 0, α 3 = –119k rad/s2, α 4 = –625k rad/s2, AG3 = 162∠– 73.2° m/s2, AG4 = 104∠233° m/s2, FC = –0.8 j kN.
Figure 15.9
Calculate the inertia forces and inertia torques − m2 A G2 = 0 − m3 A G3 = −(1.5)(46.8i − 155 j) = −70.2i + 233 j( N ) − m4 A G4 = −(5.0)(−62.6i − 83.1j) = −313i + 415 j( N ) − I G2 α 2 = 0 − I G3 α 3 = −(0.012)(−119k ) = 1.43k ( N ⋅ m) − I G4 α 4 = −(0.054)(−625k ) = 33.8k ( N ⋅ m)
15-12
Considering the free-body diagram of link 4 and 3 respectively,
∑M ∑M ∑M ∑M
O4 A O4 A
= R G4O4 × (−m4 A G4 ) + (− I G4 α 4 ) + R CO4 × Fc + R BO4 × F34 = 0 = R G3 A × (−m3 A G3 ) + (− I G3 α 3 ) + R BA × F43 = 0 = 25.2k + 33.8k − 96k + (−125 F34x + 83F34y )k = 0 = 18.6k + 1.43k + (70.5 F34x − 208 F34y )k = 0
F34 = −300i − 39 j Summing forces on the link 2,3,4,
∑F
i4
= F14 + F34 + FC + (− m4 A G4 ) = 0
F14 = −13i + 390 j( N )
∑F
i3
= F23 + F43 + (− m3 A G3 ) = 0
F23 = −230i − 238 j( N )
∑F
i2
= F12 + F32 + (− m2 A G2 ) = 0
F12 = F23 = −F32
∑M
O2
= R AO2 × F32 + M12 + (− I G2 α 2 ) = 0
M12 = − R AO2 × F32 = 18.6k ( N ⋅ m)
15-13
15.5 The Principle of Superposition Linear System: the response or output of a system is directly proportional to the drive or input to the system. In the absence of Coulomb or dry friction, most mechanisms are linear for force analysis purpose. The principle of superposition: for linear systems the individual responses to several disturbances or driving functions can be superposed on each other to obtain the total response of the system. Example: nonlinear factor: static or Coulomb friction, systems with clearances or backlash systems with springs that change stiffness as they are deflected
Complete dynamic force analysis of a planar motion mechanism: (1) make a kinematic analysis of the mechanism. (2) make a complete static force analysis of the mechanism. (3) calculate the inertia forces and inertia torques for each link or element of the mechanism. make another complete force analysis of the mechanism. (4) add the results of steps 2 and 3 to obtain the resultant forces and torques on each link.
15-14
Figure 15.10 Example 15.5: RAO2 = 3 in, RO4O2 = 14 in, RBA = 20 in, RBO4 = 10 in, RCO4 = 8 in, RCB = 6 in, RG3A = 10 in, RG4O4 = 5.69 in, w3 = 7.13 lb, w4 = 3.42 lb, IG = 0.25 in · lb · s2, IG3 = 0.625 in · lb ·s2, IG4 = 0.037 in · lb · s2, ω 2 = 60 rad/s, and α 2 = 0. 2
Make a kinematic analysis of the mechanism. In the acceleration polygon, the angular accelerations of link 3 and 4 are found to be α 3 = 148rad / s 2ccw and α 4 = 604rad / s 2cw
15-15
Figure 15.11
Free-body diagrams of link 4 of Example 15.5 showing superposition of forces: (a) F34′ = 24.3 lb, F14′ = 44.3 lb; (b) F34″ = –F14″ = 94.8 ′″ = 25 lb, F14 ′″ = 19.3 lb; (d) F34 = 94.3 lb, and F14 = 132 lb. lb; (c) F34
I G4 α 4 = (0.037)(604) = 22.3in ⋅ ib cw m4 AG4 = h4 =
3.42 (349) = 37.1lb 32.2
I G4 α 4 m4 AG4
= 22.3 / 37.1 = 0.602lb
15-16
Figure 15.12
′ = F43′ = 24.3 lb; (b) F23 ″ = 145 lb, F43″ = 94.8 lb; Free-body diagrams of link 3 of Example 15.5 showing superposition of forces: (a) F23 (c) F23 ′″ = F43 ′″ = 25 lb; (d) F23 = 145 lb, F43 = 94.3 lb.
I G3 α 3 = (0.625)(148) = 92.25in ⋅ ib ccw m3 AG3 = h3 =
7.13 (758) = 168lb 32.3
I G3 α 3 m3 AG3
= 92.5 /168 = 0.550lb
15-17
Figure 15.13 Free-body diagram of link 2 of Example 15.5: F32 = F12 = 145 lb, M12 = 226 in · lb.
M 12 = h2 F32 = (1.56)(145) = 226in ⋅ lb cw
15-18
15.6 Planar rotation about a fixed center
Figure 15.14
F n = mRGω 2 and
F t = mRGα
∑ M O = I Gα + RG (mRGα ) = ( I G + mRG2 )α ∑M = I α ∑ F − mA = 0 ∑M − I α = 0 O
O
G
O
O
(− mRGα )l = − I Gα + (− mRGα ) RG l=
IG + RG mRG
k2 l= + RG RG
Point P: the center of percussion The inertial force that passes through P has zero moment about the center of percussion.
15-19
15.7 Shaking forces and moments Shaking forces and shaking moments: transmitted to the frame or foundation of the machine owing to the inertia of the moving part
Figure 15.15 Four-bar linkage.
∑F = F
12
+ F14 + (− m2 A G2 ) + (−m3 A G3 ) + (−m4 A G4 ) = 0
Fs = F21 + F41 Fs = (− m2 A G2 ) + (−m3 A G3 ) + (−m4 A G4 ) Fs = ∑ (− m j A G j ) M s = ∑ [R G j × ( − m j A G j )] + ∑ (− I G j α j )
• ASSIGNMENT : PROBLEM – 5,
Fs : the resulting shaking force
Ms : the resulting shaking moment
7, 15 - 19
15-20
