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ME 302 DYNAMICS OF ME 302 DYNAMICS OF MACHINERYMACHINERY

Dynamic Force Analysis II

Dr. Sadettin KAPUCU

© 2007 Sadettin Kapucu

2Gaziantep University

ExampleExample

The slender bar of mass m is released form rest in the horizontal position as shown. At that instant, determine the force exerted on the bar by the support A.

Freebody Diagram

Equations Of Motion

GamF

IM

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ExampleExample

The slender bar of mass m is released form rest in the horizontal position as shown. At that instant, determine the force exerted on the bar by the support A.

Kinematics of the slender rod

?? yxG aaa

?

Freebody DiagramEquations Of Motion

GamF

IM

xx maF yy maF AA IM

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ExampleExample

The slender bar of mass m is released form rest in the horizontal position as shown. At that instant, determine the force exerted on the bar by the support A.

Kinematics of the slender rod

?? yxG aaa

?

)x(x rax

ray

x

r

At the instant bar is released, its angular velocity 0 0xa

jlilkay

2

1

2

1x

0xa

lay 2

1

5Gaziantep University

ExampleExampleThe slender bar of mass m is released form rest in the horizontal position as shown. At that instant, determine the force exerted on the bar by the support A.

Freebody DiagramEquations Of Motion

AA IM

00

xx

xx

AmA

maF

lmmgAmamgA

maF

yyy

yy

2

14

mgAy

3212

222 mll

mml

mdII GA l

gmllmg

2

3

32

1 2

0xa

lay 2

1

6Gaziantep University

D’Alembert’s PrincipleD’Alembert’s Principle

D’Alembert’s principle permits the reduction of a problem in dynamics to one in statics. This is accomplished by introducing a fictitious force equal in magnitude to the product of the mass of the body and its acceleration, and directed opposite to the acceleration. The result is a condition of kinetic equilibrium.

fictitious force and torque

0 aaaF

mmm

0 ααατIII

The meaning of the equation; i.e. indication of a dynamic case still holds true, but equation, having zero on right hand side becomes very easy to solve, like that in a “static force analysis” problem.

CGF

m, Ia

CGF

m, Ia

-ma

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Solution of a Solution of a DDynamic ynamic PProblem roblem UUsing sing D’Alembert’s D’Alembert’s PPrinciplerinciple

1. Do an acceleration analysis and calculate the linear acceleration of the mass centers of each moving link. Also calculate the angular acceleration of each moving link.

2. Masses and centroidal inertias of each moving link must be known beforehand.

3. Add one fictitious force on each moving body equal to the mass of that body times the acceleration of its mass center, direction opposite to its acceleration, applied directly onto the center of gravity, apart from the already existing real forces.

4. Add fictitious torque on each moving body equal to the centroidal inertia of that body times its angular acceleration, direction or sense opposite to that of acceleration apart from the already existing real torques.

5. Solve statically.

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Example 1Example 1

AB=10 cm, AG3=BG3=5 cm, =60o m2=m4=0.5 kg, m3=0.8 kg, I3=0.01 kg.m2

In the figure, a double- slider mechanism working in horizontal plane is shown. The slider at B is moving rightward with a constant velocity of 1 m/sec. Calculate the amount of force on this mechanism in the given kinematic state.

4

B

2A

3

G3

x

BABA VVV

?AV

smVB /1

ABtoVB

A ?

BV

AV

BAV

smVA /5774.0

smVB

A /1547.1

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Example 1 contExample 1 contIn the figure, a double- slider mechanism working in horizontal plane is shown. The slider at B is moving rightward with a constant velocity of 1 m/sec. Calculate the amount of force on this mechanism in the given kinematic state.

4

B

2A

3

G3

x

BABA VVV

BtoAfromsmAB

Va B

An

BA

22

2

/33.131.0

1547.1

?Aa

BABA aaa

0

BAA aa

t

BA

n

BA aa

ABtoa t

BA

?

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Example 1 contExample 1 cont

4

B

2A

3

G3

x

BtoAfromsmAB

Va B

An

BA

22

2

/33.131.0

1547.1

?Aa

BAA aa

t

BA

n

BA aa

ABtoa t

BA

?

2/698,7 sma t

BA

t

BA

a

Aa

2/396.15 smaA

G3

3Ga

n

BA

a

2/698.72/3

smaa AG

3ABa t

BA

CCWsradAB

a t

BA

23 /98.78

1.0

698.7

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Example 1 contExample 1 cont

4

B

2A

3

G3

x

t

BA

a

Aa

2/396.15 smaA

G3

3Ga

n

BA

a

2/698.72/3

smaa AG

CCWsradAB

a t

BA

23 /98.78

1.0

698.7

3Gma

Ama

3I

D’Alembert forces and moments

090698.7396.15*5.0 NmaA

0901584.6698.7*8.03

NmaG

CWNmI 7689.098.76*01.03

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Example 1 contExample 1 cont

4

B

2A

3

G3

x

N698.7

N1584.6

Nm7689.0

BF4

B

2A

3

G3

N698.7

N1584.6

Nm7689.0

BF

12F

14F

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Example 1 contExample 1 cont

4

B

2A

3

G3

N698.7

N1584.6

Nm7689.0

BF

12F

14F

1212 0;0 FFFFF BBx

NF

FFy

86.13

01584.6698.7;0

14

14

NF

F

M B

11.150867.0

3087.1

060sin*1.0*60cos*05.0*1584.6

60cos*1.0*698.77698.0

;0

12

12

NFB 11.15

+

x

y

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Example 1 contExample 1 cont

4

B

2A

3

G3

x

3Gma

Ama

3I

090698.7396.15*5.0 NmaA

0901584.6698.7*8.03

NmaG

CWNmI 7689.098.76*01.03

3Gma

3I

3Gma

3Gma

h

hmaI G *33

3

3

Gma

Ih

mh 125.01589.6

7689.0

h3G

ma

3Gma

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Example 1 contExample 1 cont

4

B

2A

3

G3

x

h

N1584.6

N698.7

BF4

B

2A

3G3

N698.7

BF

12F

14F

mh 125.01589.6

7689.0

h

N1584.6

16Gaziantep University

Example 1 contExample 1 cont

4

B

2A

3G3

N698.7

BF

12F

14F

h

N1584.6

1212 0;0 FFFFF BBx

NF

FFy

86.13

01584.6698.7;0

14

14

NF

F

M B

11.150867.0

3087.1

060sin*1.0*60cos*1.0*698.7

60cos*)125.005.0(*1584.6;0

12

12

NFB 11.15

+

x

y

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Example 2Example 2Crank AB of the mechanism shown is balanced such that the mass center is at A. Mass center of the link CD is at its mid point. At the given instant, link 4 is translating rightward with constant velocity of 5 m/sec. Calculate the amount of motor torque required on crank AB to keep at the given kinematics state.

14

2

3

A

C

B

D

=45

F

AC=10 cm, CB=BD=4 cm, AF=2 cm CG3=4 cm, m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2

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Example 2 cont.Example 2 cont.Crank AB of the mechanism shown is balanced such that the mass center is at A. Mass center of the link CD is at its mid point. At the given instant, link 4 is translating rightward with constant velocity of 5 m/sec. Calculate the amount of motor torque required on crank AB to keep at the given kinematics state.

14

2

3

A

C

B

D

=45

F

AC=10 cm, CB=BD=4 cm, AF=2 cm CG3=4 cm, m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2

CBCB VVV

ABtoVB ?

sec/5 mVC

BCtoVC

B ?

smVC /5

5 m/s

smVC

B /05.5smVB /85.3

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Example 2 cont.Example 2 cont.Crank AB of the mechanism shown is balanced such that the mass center is at A. Mass center of the link CD is at its mid point. At the given instant, link 4 is translating rightward with constant velocity of 5 m/sec. Calculate the amount of motor torque required on crank AB to keep at the given kinematics state.

14

2

3

A

C

B

D

=45

F

AC=10 cm, CB=BD=4 cm, AF=2 cm CG3=5 cm, m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2

CBCB aaa

0Can

CB

n

CB

tB

nB aaaa

AtoBfromsmAB

Va BnB

222

/5.192077.0

85.3

BCtoa t

CB ?

CtoBfromsmBC

Va C

Bn

CB

22

2

/6.63704.0

05.5

ABtoa tB ?

2/5.192 smanB

2/6.637 sman

CB

2/483 sma t

CB

2/776 sma tB

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Example 2 cont.Example 2 cont.Crank AB of the mechanism shown is balanced such that the mass center is at A. Mass center of the link CD is at its mid point. At the given instant, link 4 is translating rightward with constant velocity of 5 m/sec. Calculate the amount of motor torque required on crank AB to keep at the given kinematics state.

14

2

3

A

C

B

D

=45

F

AC=10 cm, CB=BD=4 cm, AF=2 cm CG3=5 cm, m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2

??,?, 32 Ba

2/5.192 smanB

2/6.637 sman

CB

2/483 sma t

CB

2/776 sma tB

CCWsradABa tB

222 /10078

077.0

776*

CWsradBCa t

CB

233 /20000

04.0

800*

Ba

262/800 2 smaB

B

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Example 2 contExample 2 contD’Alembert forces and moments

14

2

3

A

C

B

D

=45

F

2/5.192 smanB

2/6.637 sman

CB

2/483 sma t

CB

2/776 sma tB

CCWsrad 22 /10078

CWsrad 23 /20000

Ba

262/800 2 smaB

B

82400080053 N*am B

CWNm*.I 5041007805022

CCWNm*.I 10002000005033

m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2

-m3aB

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Example 2 contExample 2 cont

14

2

3

A

C

B

D

=45

F

-m3aB

4

23

A

C

B

D

=45

F

C

B

A22°

504 Nm

4000 N

1000 Nm

CyF

CxF

BxFByF

ByF

BxF

CyFCxF

AxFAyF

4AF

A4

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Example 2 contExample 2 cont

4

23

A

C

B

D

=45

F

C

B

A22°

504 Nm

4000 N

1000 Nm

CyF

CxF

BxFByF

ByF

BxF

CyFCxF

AxFAyF

4AF

A4

BxAxBxAxx FFFF;F 00

ByAyByAyy FFFF;F 00

007140028280504

022225040

ByBx

ByBxA

F*.F*.

cos*AB*Fsin*ABFT;M

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Example 2 contExample 2 cont

4

23

A

C

B

D

=45

F

C

B

A22°

504 Nm

4000 N

1000 Nm

CyF

CxF

BxFByF

ByF

BxF

CyFCxF

AxFAyF

4AF

A4

069.556;082cos*4000;0 BxCxBxCxx FFFFF

007.3961;082sin*4000;0 ByCyByCyy FFFFF

0*02828.0*02828.01000;045cos**45sin*1000;0 CyCxCyCxB FFBCFBCFM

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Example 2 contExample 2 cont

4

23

A

C

B

D

=45

F

C

B

A22°

504 Nm

4000 N

1000 Nm

CyF

CxF

BxFByF

ByF

BxF

CyFCxF

AxFAyF

4AF

A4

00 Cxx F;F

00 4AyCyy FF;F

00 4 CA*FT;M CyAA

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Example 2 contExample 2 cont

4

23

A

C

B

D

=45

F

C

B

A504 Nm

4000 N

1000 Nm

CyF

CxF

BxFByF

ByF

BxF

CyFCxF

AxFAyF

4AF

A4

BxAxBxAxx FFFF;F 00

ByAyByAyy FFFF;F 00

0*0714.0*02828.0504

022cos**22sin*504;0

ByBx

ByBxA

FFT

ABFABFTM

069.556

;082cos*4000;0

BxCx

BxCxx

FF

FFF

007.3961

;082sin*4000;0

ByCy

ByCyy

FF

FFF

0*02828.0*02828.01000

;045cos**45sin*1000;0

CyCx

CyCxB

FF

BCFBCFM

0;0 Cxx FF

00 4AyCyy FF;F

00 4 CA*FT;M CyAA

N.FBx 69556 N..

FCy 3435355028280

1000

N...FBy 41393160739613435355

CCWNm.T

.*.).(*.

933326

041393160714069556028280504