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ME 302 DYNAMICS OF ME 302 DYNAMICS OF MACHINERYMACHINERY

STATIC FORCE ANALYSIS

Dr. Sadettin KAPUCU

© 2007 Sadettin Kapucu

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StaticEquilibriumStaticEquilibrium

00

00

00

zz

yy

xx

M;F

M;F

M;F

From Newton’s first law, a body is in static equilibrium if the resultant of all the forces acting on a rigid body is zero. This condition can be expressed mathematicaly as:

00 MF

and

In space, these two vector equation yields six scalar equations:

In planar, there are three scalar equations:

0

00

z

yx

M

;F;F

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Force SystemsForce Systems

In coplanar force system under which the body acted on is in equilibrium, we can have following force systems:

1. Two force member : If there are only two forces acting on a body, it is called a “ two force member”.

To satisfy sum of torque is equal to zero distance between the forces must

be zero. This means that the forces are collinear.

F1

F2

To satisfy sum of forces equal to zero, the two forces should be equal and opposite.

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Force SystemsForce Systems

2. Three force member : If there are only three forces acting on a body, it is called a “ three force member”.

To satisfy sum of torque is equal to zero, the lines of application of all the three forces intersects at one single point. This point is called the point of concurrency.

F1

F2

To satisfy sum of forces equal to zero, the vectors must form a closed polygon and coplanar.

F3

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Solving static force problemsSolving static force problems

• Separate the mechanism into its links, considering each a free body with all the acting external and constraint forces on it.

• Apply the rules of statics each free body which are

00 MF

and

Solution of vector equations can be by arithmetical and or graphical.

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Solving static force problemsSolving static force problems

Graphical Approach : We draw straight lines to represent vectors which are in proper directions and lengths proportional to the magnitudes of the vectors and in an articulated manner as depicted in Figure. Vectors form a closed polygon called a “vector loop”

F1

F2

F3

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Solving static force problemsSolving static force problems

Arithmetical Approach : The simplest arithmetical approach is to separate vector equation into components.

F1

F2

F3

1

3

2

0coscoscos 332211 FFF

0332211 sinFsinFsinFThese two component equations are not no longer vector equations. They are

scalar and can be simultanously solved to find max two of the following;

1F 2F 3F1 2 3

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ExampleExampleAn external force of 10 N is acting horizontally on the rocker

link, 30 mm from the point D. Find the amount of torque to be applied to the crank AB to keep the mechanism in static equilibrium.

A

1

B

2

D

43

C

a1

a2

a3 a4

30

10 N

a1 = 80 mm

a2 = 30 mm

a3 = 70 mm

a4 = 50 mm

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ExampleExampleFirst step is a position analysis to find the angles of the crank CD and the

coupler links. The simplest way is to draw the mechanism to scale and measure the

required angles by a protractor directly from the figure. Or we can take a purely arithmetical approach.

A

1

B

2

D

43

C

a1

a2

a3 a430

10 N

02

tan2

tan2

CBA

132 )1(cos KKKA sin2B

132 )1(cos KKKC

2

11 a

aK

4

12 a

aK

42

24

23

22

21

3 2 aa

aaaaK

67.230

801 K

60.150

802 K

66.150*30*2

2500490090064003

K

30.167.266.1)60.11(60cos A73.160sin2 B

02.367.266.1)60.11(60cos C

002.32

tan*73.12

tan*30.1 2

86.891

and

52.1332

A

1

B

D

C

a1

a2

a3 a4

Ø1

Ø2

06,2075,65

92,23tan 1

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ExampleExampleArithmetical method: Separate the mechanism into free bodies of links, Put all the acting and interacting forces, Then, apply the law of statics for each free body.

D

C

10 N

89,86°

B

3C

20,06°

A

B

2

60°

T

4

FDx

FDy

FCy

FCy

FCx

FCx

FBx

FBy

FBx

FBy

FAx

FAy

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D

C

10 N

89,86°

4

FDx

FDy

FCy

FCx

Static equations for link 4;

10010;0 CxDxCxDxx FFFFF

CyDyCyDyy FFFFF 0;0

08568950868950868930100 ,cos**F,sin**F,sin**;M CyCxD

99,299*12,0*99,49 CyCx FF

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B

3C

20,06°

FCy

FCx

FBx

FBy

Static equations for link 3;

CxBxCxBxx FFFFF 0;0

CyByCyByY FFFFF 0;0

006,20cos*70*06,20sin*70*;0 CyCxB FFM

0*75,60*01,24 CyCx FF

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A

B

2

60°

T

FBx

FBy

FAx

FAy

Static equations for link 2;

BxAxBxAxx FFFFF 0;0

ByAyByAyy FFFFF 0;0

TFFM BxByB 00,60sin*30*00,60cos*30*;0

TFF BxBy *98,25*00,15

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D

C

10 N

B

3C

20,06°

A

B

2

60°

T

4

FDx

FDy

FCy

FCy

FCx

FCx

FBx

FBy

FBx

FBy

FAx

FAy

We have 9 equations to solve simultaneously; FAx, FAy, FBx, FBy, FCx, FCy, FDx, FDy, and T.

N,FCy 372

N,FCx 995

N,FFFF AyByDyCy 372

NFFFF DxAxCxBx 99,510

CWmm.N,T 07120 ANS.

10010;0 CxDxCxDxx FFFFF

CyDyCyDyy FFFFF 0;0

0856,89cos*50*86,89sin*50*86,89sin*30*10;0 CyCxD FFM

99,299*12,0*99,49 CyCx FF

CxBxCxBxx FFFFF 0;0

CyByCyByY FFFFF 0;0

006,20cos*70*06,20sin*70*;0 CyCxB FFM

0*75,60*01,24 CyCx FF

BxAxBxAxx FFFFF 0;0

ByAyByAyy FFFFF 0;0

TFFM BxByB 00,60sin*30*00,60cos*30*;0

TFF BxBy *98,25*00,15

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ExampleExampleGraphical method: Draw the mechanism in scale, Measure the unknown quantities directly from

the scaled drawing, Separate the mechanism into free bodies of links

(scaled drawing), State whether the link is two force - three force

member and then put all the acting and interacting forces,

Apply the law of statics for each free body.

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D

C

10 N

89,86°

B

3C

20,06°

A

B

2

60°

T

4

Link 3 is two force member,Link 4 is three force member

F23

F43

F34

F32

F14

F12F14 F34

10 N

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D

C

10 N

B

3C

A

B

2

4

F43

F34

F23

F14

F32

d1

T12

OF

10 N

F14 F34

F12

F14 & F34 are measured directly from the scaled

force polygon.10 N stands for 50 mmF14 stands for 22.5 mm

F34 stands for 32.5 mm

NF 5.450

5.22*1014

NF 5.650

5.32*1034

N..*.F*dT 751265651932112