Chapter 15...Dynamic Force Analysis (Planar)

Theory of Machines and Machines

15-2

15.1 introduction

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 systemthat are not in equilibrium

Dynamic force analysis: dynamic forces are associated with accelerating mass.

15-3

Figure 15.1Particle of mass dmP at location P on a rigid body.

15.2 Centroid and center of mass

Gj

GjGjjQj

GjjGjjjjQij

GjPPQjp

PPQjPPQjjPQij

PQjPQjjQP

ppij

ppp

m

m

mmm

mdmmdm

dmdmdm

dm

dmd

A

RαRωωA

RαRωωAF

RR

RαRωωAF

RαRωωAA

AF

AF

=

×+××+=

×+××+=

==

×+××+=

×+××+=

=

=

∑∫∫

∫∑ ∫ ∫

∑ ∫

])([

)(

),(

)(

)(

15-4

Figure 15.2(a) Particles of mass distributed along a line. (b) Particles of mass distributed in a plane.

∫

∑∑

∑∑

=

=++++++

=

=++++

=

dmm

mm

mmmmmmmm

mxm

mmmxmxmxmx

jG

i

iiG

i

ii

RR

RRRRRR

14321

44332211

321

332211

The center of mass

15-5

Figure 15.3Center 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.

15-6

Figure 15.4Composite shape for Example 15.1 with dimensions in millimeters.

Example 15.1

mm

mm

mmmmmm

mmkgmmkgmmmmmmm

mmkgmmkgmmmmm

mmkgmmkgmmmmmmm

G

i

iiG

G

G

G

kjiR

RRRRR

kjiR

kjiR

kjiR

0.351.194.89

201030000,54)/)(20)(90)(60(5.0

4020160300,50)/)(40()20(

4020100000,640)/)(40)(200)(80(

321

332211

33

322

31

3

2

1

−+=

=++++

=

++===

−+=−=−=

−+===

∑∑

ρρ

ρπ

ρρ

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.

15-7

15.3 Mass Moments and Products of Inertia

∫= dmcedisI 2)tan(

dmRRI

dmRRI

dmRRI

inertiaofmonentmass

yxzz

zxyy

zyxx

])()[(

])()[(

])()[(

22

22

22

∫∫∫

+=

+=

+=

dmRRII

dmRRII

dmRRII

inertiaofproductsmass

xzxzzx

zyzyyz

yxyxxy

)(

)(

)(

∫∫∫

==

==

==

2

2

mdIImIkormkI

IIIIIIIII

I

G

GG

zzzyzx

yzyyyx

xzxyxx

+=

==

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−−

=

Inertia tensor

Principal mass moment of inertia

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

3

2

1

000000

II

II

Principal axes: all of the products of inertia becomes zero.

k: radius of gyration

Transfer or parallel-axis formulaIG: principal mass moments of inertiaI: mass moment of inertia about a parallel axis at distance d

15-8

Figure 15.5Connecting-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.

Example 15.2

2

22

2

222

22

2

222

22

2

2

3

2

2

223

220

33.1

)()(0931.0

12])1()13)[(/00657.0(

12)(

00396.02

])5.0()5.1)[(/00317.0(2

)(/00657.0

/386)75.0)(1)(13)(/26.0(

/00317.0/386

)75.0]()5.0()5.1[()/26.0(

)(

slbin

dmIdmIIIslbin

inininslb

hwmI

slbin

inininslb

rrmI

inslbsin

ininininlbwhlm

inslbsin

inininlb

lrrm

cylcylcylbarbarbarcylzz

bar

iocyl

bar

icyl

⋅⋅=

++++=

⋅⋅=

+⋅=

+=

⋅⋅=

+⋅=

+=

⋅=

==

⋅=

−=

−=

ρ

π

ρπ

1 .dai holes′′

15-9

Figure 15.6(a) An unbalanced set of forces on a rigid body. (b) The accelerations that result from the unbalanced forces.

15.4 Inertia Forces and D’Alembert’s Principle

0)(

0)(

321

=−+

=−+

=

=

++=

∑∑∑∑∑

jGG

Gjij

jGG

Gjij

I

m

I

m

ij

ij

α

α

M

AF

M

AF

FFFF

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.

∑∑

=

=

0

0

M

F

15-10

When a graphical solution by a force polygon is desired, 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.

G

G

AmIh

3

3α=

Equivalent Offset Inertia Force

1.The magnitude of the inertia force is .2.The direction of the inertia force is opposite to that of acceleration3.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

GmAGA

α

∑∑ == 00 MF and

15-11

Figure 15.8Solution 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.

Example 15.3

cwsradin

ftinsft

RA

BA

tBA

2

2

3

/85610

)/12)(/713(

=

=

=α

inftinsftslbin

sradslbinh 35.1)/12)(/444)(00570.0(

)/856)(0479.0(22

22

=⋅⋅

⋅⋅= lbA jF 27=

12.6 /AV ft s=21.2 /BAV ft s=

16.2 /BV ft s=

15-12

Example 15.4:

RAO2= 60 mm, RO4O2

= 100 mm, RBA = 220 mm, RBO4= 150 mm, RCO4

= RCB = 120 mm, RG3A = 90 mm, RG4O4= 90 mm, m3 = 1.5 kg,

m4 = 5 kg, IG2= 0.025 kg · m2, IG3

= 0.012 kg · m2, IG4= 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

ˆ ˆˆ

2

3

4

2

3

4

2

3

4

2

3

4

0

(1.5)(46.8 155 ) 70.2 233 ( )

(5.0)( 62.6 83.1 ) 313 415 ( )

0

(0.012)( 119 ) 1.43 ( )

(0.054)( 625 ) 33.8 ( )

G

G

G

G

G

G

m

m N

m N

I

I N m

I N m

− =

− = − − = − +

− = − − − = − +

− =

− = − − = ⋅

− = − − = ⋅

A

A i j i j

A i j i j

α

α k k

α k k

Calculate the inertia forces and inertia torques

15-13

Considering the free-body diagram of link 4 and 3 respectively,

4 4 4 4 4 4 4

3 3 3

4

4 4 34

3 3 43

34 34

34 34

34

( ) ( ) 0

( ) ( ) 0

25.2 33.8 96 ( 125 83 ) 0

18.6 1.43 (70.5 208 ) 0300 39

O G O G G CO c BO

A G A G G BA

x yO

x yA

m I

m I

F F

F F

= × − + − + × + × =

= × − + − + × =

= + − + − + =

= + + − =

= − −

∑∑∑∑

M R A α R F R F

M R A α R F

M k k k k

M k k kF i j

4

3

2

2 2 2

2

4 14 34 4

14

3 23 43 3

23

2 12 32 2

12 23 32

32 12 2

12 32

( ) 0

13 390 ( )( ) 0

230 238 ( )( ) 0

( ) 0

18.6 ( )

i C G

i G

i G

O AO G

AO

m

Nm

Nm

I

N m

= + + + − =

= − +

= + + − =

= − −

= + + − =

= = −

= × + + − =

= − × = ⋅

∑

∑

∑

∑

F F F F A

F i jF F F A

F i jF F F A

F F FM R F M α

M R F k

Summing forces on the link 2,3,4,

15-14

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 drivingfunctions 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.

15-15

Figure 15.10Example 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, IG2

= 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.

Make a kinematic analysis of the mechanism.In the acceleration polygon, the angular accelerations of link 3 and 4 are found to be

2 23 4148 / 604 /rad s ccw and rad s cwα α= =

15-16

Figure 15.11Free-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 lb; (c) F34 = 25 lb, F14 = 19.3 lb; (d) F34 = 94.3 lb, and F14 = 132 lb.

′ ′ ″ ″′″ ′″

4

4

4

4

4

4

44

4

(0.037)(604) 22.3

3.42 (349) 37.132.2

22.3 / 37.1 0.602

G

G

G

G

I in ib cw

m A lb

Ih lb

m A

α

α

= = ⋅

= =

= = =

15-17

Figure 15.12Free-body diagrams of link 3 of Example 15.5 showing superposition of forces: (a) F23 = F43 = 24.3 lb; (b) F23 = 145 lb, F43 = 94.8 lb; (c) F23 = F43 = 25 lb; (d) F23 = 145 lb, F43 = 94.3 lb.′″ ′″

′ ′ ″ ″

3

3

3

3

3

3

33

3

(0.625)(148) 92.25

7.13 (758) 16832.3

92.5 /168 0.550

G

G

G

G

I in ib ccw

m A lb

Ih lb

m A

α

α

= = ⋅

= =

= = =

15-18

Figure 15.13Free-body diagram of link 2 of Example 15.5: F32 = F12 = 145 lb, M12 = 226 in · lb.

12 2 32 (1.56)(145) 226M h F in lb cw= = = ⋅

15-19

Figure 15.14

15.6 Planar rotation about a fixed center

∑∑∑∑

=−

=−

=

+=+=

==

0

0

)()( 2

2

α

α

ααα

αω

OO

G

OO

GGGGGO

Gt

Gn

I

m

IM

mRImRRIM

mRFandmRF

M

AFG

G

GG

G

GGGG

RRkl

RmRIl

RmRIlmR

+=

+=

−+−=−

2

)()( ααα

Point P: the center of percussionThe inertial force that passes through P has zero moment about the center of percussion.

15-20

Figure 15.15Four-bar linkage.

15.7 Shaking forces and momentsShaking forces and shaking moments: transmitted to the frame or foundation of the machine owing to

the inertia of the moving part

∑∑∑

∑

−+−×=

−=

−+−+−=+=

=−+−+−++=

)()]([

)(

)()()(

0)()()(

432

432

432

4121

4321412

jGGjGs

Gjs

GGGs

s

GGG

jjj

j

Im

m

mmm

mmm

αARM

AF

AAAFFFF

AAAFFF

Fs : the resulting shaking force

Ms : the resulting shaking moment

• ASSIGNMENT : PROBLEM – 5, 7, 15 - 19