ce 102 statics

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Statics

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CE 102 Statics

Chapter 4

Equilibrium of Rigid Bodies



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Contents

Introduction

Free-Body Diagram

Reactions at Supports and Connections for a Two-Dimensional Structure

Equilibrium of a Rigid Body in Two Dimensions

Statically Indeterminate Reactions

Sample Problem 4.1

Sample Problem 4.2

Sample Problem 4.3

Equilibrium of a Two-Force Body

Equilibrium of a Three-Force Body

Sample Problem 4.4

Equilibrium of a Rigid Body in Three Dimensions

Reactions at Supports and Connections for a Three-Dimensional Structure

Sample Problem 4.5



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Introduction

• The necessary and sufficient condition for the static equilibrium of a body are that the resultant force and couple from all external forces form a system equivalent to zero,

00 FrMF O

000

000

zyx

zyx

MMM

FFF

• Resolving each force and moment into its rectangular components leads to 6 scalar equations which also express the conditions for static equilibrium,

• For a rigid body in static equilibrium, the external forces and moments are balanced and will impart no translational or rotational motion to the body.



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Free-Body DiagramFirst step in the static equilibrium analysis of a rigid body is identification of all forces acting on the body with a free-body diagram.

• Select the extent of the free-body and detach it from the ground and all other bodies.

• Include the dimensions necessary to compute the moments of the forces.

• Indicate point of application and assumed direction of unknown applied forces. These usually consist of reactions through which the ground and other bodies oppose the possible motion of the rigid body.

• Indicate point of application, magnitude, and direction of external forces, including the rigid body weight.



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• Reactions equivalent to a force with known line of action.



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• Reactions equivalent to a force of unknown direction and magnitude.

• Reactions equivalent to a force of unknown direction and magnitude and a couple.of unknown magnitude



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Equilibrium of a Rigid Body in Two Dimensions

• For all forces and moments acting on a two-dimensional structure,

Ozyxz MMMMF 00

• Equations of equilibrium become

000 Ayx MFF

where A is any point in the plane of the structure.

• The 3 equations can be solved for no more than 3 unknowns.

• The 3 equations can not be augmented with additional equations, but they can be replaced 000 BAx MMF



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Statically Indeterminate Reactions

• More unknowns than equations

• Fewer unknowns than equations, partially constrained

• Equal number unknowns and equations but improperly constrained



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Sample Problem 4.1

A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G.

Determine the components of the reactions at A and B.

SOLUTION:

• Create a free-body diagram for the crane.

• Determine B by solving the equation for the sum of the moments of all forces about A. Note there will be no contribution from the unknown reactions at A.

• Determine the reactions at A by solving the equations for the sum of all horizontal force components and all vertical force components.

• Check the values obtained for the reactions by verifying that the sum of the moments about B of all forces is zero.



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Sample Problem 4.1

• Create the free-body diagram.

• Check the values obtained.

• Determine B by solving the equation for the sum of the moments of all forces about A.

0m6kN5.23

m2kN81.9m5.1:0

BM A

kN1.107B

• Determine the reactions at A by solving the equations for the sum of all horizontal forces and all vertical forces.

0:0 BAF xx

kN1.107xA

0kN5.23kN81.9:0 yy AF

kN 3.33yA



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Sample Problem 4.2

A loading car is at rest on an inclined track. The gross weight of the car and its load is 5500 lb, and it is applied at at G. The cart is held in position by the cable.

Determine the tension in the cable and the reaction at each pair of wheels.

SOLUTION:

• Create a free-body diagram for the car with the coordinate system aligned with the track.

• Determine the reactions at the wheels by solving equations for the sum of moments about points above each axle.

• Determine the cable tension by solving the equation for the sum of force components parallel to the track.

• Check the values obtained by verifying that the sum of force components perpendicular to the track are zero.



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Sample Problem 4.2

• Create a free-body diagram

lb 2320

25sinlb 5500

lb 4980

25coslb 5500

y

x

W

W

• Determine the reactions at the wheels.

00in.5

in.6lb 9804in.25lb 2320:0

2

R

M A

lb 17582 R

00in.5

in.6lb 9804in.25lb 2320:0

1

R

M B

lb 5621 R

• Determine the cable tension.

0Tlb 4980:0 xF

lb 4980T



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Sample Problem 4.3

The frame supports part of the roof of a small building. The tension in the cable is 150 kN.

Determine the reaction at the fixed end E.

SOLUTION:

• Create a free-body diagram for the frame and cable.

• Solve 3 equilibrium equations for the reaction force components and couple at E.



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Sample Problem 4.3

• Create a free-body diagram for the frame and cable.

• Solve 3 equilibrium equations for the reaction force components and couple.

0kN1505.7

5.4:0 xx EF

kN 0.90xE

0kN1505.7

6kN204:0 yy EF

kN 200yE

:0EM

0m5.4kN1505.7

6

m8.1kN20m6.3kN20

m4.5kN20m7.2kN20

EM

mkN0.180 EM



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Equilibrium of a Two-Force Body

• Consider a plate subjected to two forces F1 and F2

• For static equilibrium, the sum of moments about A must be zero. The moment of F2 must be zero. It follows that the line of action of F2 must pass through A.

• Similarly, the line of action of F1 must pass through B for the sum of moments about B to be zero.

• Requiring that the sum of forces in any direction be zero leads to the conclusion that F1 and F2 must have equal magnitude but opposite sense.



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Equilibrium of a Three-Force Body

• Consider a rigid body subjected to forces acting at only 3 points.

• Assuming that their lines of action intersect, the moment of F1 and F2 about the point of intersection represented by D is zero.

• Since the rigid body is in equilibrium, the sum of the moments of F1, F2, and F3 about any axis must be zero. It follows that the moment of F3 about D must be zero as well and that the line of action of F3 must pass through D.

• The lines of action of the three forces must be concurrent or parallel.



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Sample Problem 4.4

A man raises a 10 kg joist, of length 4 m, by pulling on a rope.

Find the tension in the rope and the reaction at A.

SOLUTION:

• Create a free-body diagram of the joist. Note that the joist is a 3 force body acted upon by the rope, its weight, and the reaction at A.

• The three forces must be concurrent for static equilibrium. Therefore, the reaction R must pass through the intersection of the lines of action of the weight and rope forces. Determine the direction of the reaction force R.

• Utilize a force triangle to determine the magnitude of the reaction force R.



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Sample Problem 4.4

• Create a free-body diagram of the joist.

• Determine the direction of the reaction force R.

6.58

636.1414.1

313.2tan

m 2.313m 515.0828.2

m 515.020tanm 414.1)2545cot(

m 414.1

m828.245cosm445cos

21

AE

CE

BDBFCE

CDBD

AFAECD

ABAF



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Sample Problem 4.4

• Determine the magnitude of the reaction force R.

38.6sin

N 1.98

110sin4.31sin RT

N 8.147

N9.81

R

T



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Equilibrium of a Rigid Body in Three Dimensions

• Six scalar equations are required to express the conditions for the equilibrium of a rigid body in the general three dimensional case.

000

000

zyx

zyx

MMM

FFF

• These equations can be solved for no more than 6 unknowns which generally represent reactions at supports or connections.

• The scalar equations are conveniently obtained by applying the vector forms of the conditions for equilibrium,

00 FrMF O



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Sample Problem 4.5

A sign of uniform density weighs 270 lb and is supported by a ball-and-socket joint at A and by two cables.

Determine the tension in each cable and the reaction at A.

SOLUTION:

• Create a free-body diagram for the sign.

• Apply the conditions for static equilibrium to develop equations for the unknown reactions.



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Sample Problem 4.5

• Create a free-body diagram for the sign.

Since there are only 5 unknowns, the sign is partially constrain. It is free to rotate about the x axis. It is, however, in equilibrium for the given loading.

kjiT

kjiT

rr

rrTT

kjiT

kjiT

rr

rrTT

EC

EC

EC

ECECEC

BD

BD

BD

BDBDBD

72

73

76

32

31

32

7

236

12

848



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Sample Problem 4.5

• Apply the conditions for static equilibrium to develop equations for the unknown reactions.

0lb1080571.2667.2:

0714.1333.5:

0lb 270ft 4

0:

0lb 270:

0:

0lb 270

72

32

73

31

76

32

ECBD

ECBD

ECEBDBA

ECBDz

ECBDy

ECBDx

ECBD

TTk

TTj

jiTrTrM

TTAk

TTAj

TTAi

jTTAF

kjiA

TT ECBD

lb 22.5lb 101.2lb 338

lb 315lb 3.101

Solve the 5 equations for the 5 unknowns,

26

Problem 4.6

45o45o

OD

C

A

B

The semicircular rod ABCD ismaintained in equilibrium bythe small wheel at D and therollers at B and C. Knowingthat = 45o, determine thereactions at B , C , and D.

P

27

45o45o

OD

C

A

B

1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it.

P Solving Problems on Your Own


Problem 4.6

28

45o45o

OD

C

A

B

2. Write equilibrium equations and solve for the unknowns. For two-dimensional structure the three equations might be:

Fx = 0 Fy = 0 MO = 0where O is an arbitrary point in the plane of the structureor Fx = 0 MA = 0 MB = 0where point B is such that line AB is not parallel to the y axisor MA = 0 MB = 0 MC = 0where the points A, B , and C do not lie in a straight line.

P Solving Problems on Your Own


Problem 4.6

29

45o45o

OD

C

A

B

Problem 4.6 Solution

Draw a free-body diagramof the body.

45o45o

OD

C

A

B

P sin

P cos

D

C

R

B

C/ 2

C/ 2B/ 2

B/ 2

P

P

30


45o45o

OD

C

A

B

P sin

P cos

D

C

R

B

C/ 2

C/ 2B/ 2

B/ 2

Write three equilibriumequations and solvefor the unknowns.

+ MO = 0: (P sin) R _ D (R) = 0 D = P sin

+Fx = 0: P cos + B/ 2 _ C / 2 = 0 (2)

+Fy = 0: _P sin + B/ 2 + C / 2 _ P sin = 0

_2P sin + B/ 2 + C / 2 = 0 (3)

P

31


45o45o

OD

C

A

B

P sin

P cos

D

C

R

B

C/ 2

C/ 2B/ 2

B/ 2

(2) + (3) P(cos _ 2sin) + 2 B/ 2 = 0

B = (2sin _ cos) P (4)

(2) _ (3) P(cos + 2sin) _ 2 C/ 2 = 0

C = (2sin + cos) P (5)

22

22

P

32


45o45o

OD

C

A

B

P sin

P cos

D

C

R

B

C/ 2

C/ 2B/ 2

B/ 2For = 45o

sin = cos = 1/ 2

EQ. (4) : B = ( _ ) P = P ; B = P 45o

EQ. (5) : C = ( _ ) P = P ; C = P 45o

EQ. (1) : D = P/ 2 D = P/ 2

32

P

22 2

2 2

1

22 2

2 2

1 32

12

12

33

4 in4 in

2 in

3 in

3 in

A B40 lb20 lb

CD

E

Problem 4.7

The T-shaped bracket shown issupported by a small wheel at Eand pegs at C and D. Neglectingthe effect of friction, determinethe reactions at C , D , and Ewhen = 30o.

34


4 in4 in

2 in

3 in

3 in

A B40 lb20 lb

CD

E

Solving Problems on Your Own


Problem 4.7

35



4 in4 in

2 in

3 in

3 in

A B40 lb20 lb

CD

E



Problem 4.7

36

4 in4 in

2 in

3 in

3 in

A B40 lb20 lb

CD

E

4 in4 in

A40 lb20 lb

C2 in

3 in

3 in

B

D

EE

C

D

30o

Draw a free-body diagram of the body.


37

4 in4 in

A

C

40 lb20 lb

C2 in

3 in

3 in

B

D

EE

D

30o

Write equilibrium equations andsolve for the unknowns.

Fy = 0: E cos 30o _ 20 _ 40 = 0

E = = 69.28 lb

E = 69.3 lb 60o

60 lb

cos 30o


38

4 in4 in

A

C

40 lb20 lb

C2 in

3 in

3 in

B

D

EE

D

30oMD = 0:

( 20 lb)( 4 in) _ ( 40 lb)( 4 in) _ C ( 3 in) + E sin 30o ( 3 in) = 0

_ 80 _ 3C + 69.28 ( 0.5 )( 3 ) = 0

C = 7.974 lb C = 7.97 lb


39

Problem 4.7 Solution4 in4 in

A

C

40 lb20 lb

C2 in

3 in

3 in

B

D

EE

D

30o

+Fx= 0:

E sin 30o + C _ D = 0

( 69.28 lb )( 0.5 ) + 7.974 lb _ D = 0

D = 42.6 lb

40

Problem 4.8

1.2 m

1.2 m

y

1.5 m

z

x

5 kNA B

C

E

D

1 m2 m

A 3-m pole is supported by aball-and-socket joint at A andby the cables CD and CE.Knowing that the line of actionof the 5-kN force forms anangle =30o with the verticalxy plane, determine (a) thetension in cables CD and CE,(b) the reaction at A.

41

1.2 m

1.2 m

y

1.5 m

z

x

5 kNA B

C

E

D

1 m2 m



Problem 4.8

A 3-m pole is supported by aball-and-socket joint at A andby the cables CD and CE.Knowing that the line of actionof the 5-kN force forms anangle =30o with the verticalxy plane, determine (a) thetension in cables CD and CE,(b) the reaction at A.

42

2. Write equilibrium equations and solve for the unknowns. For three-dimensional body the six scalar equations

Fx = 0 Fy = 0 Fz = 0 Mx = 0 My = 0 Mz = 0

should be used and solved for six unknowns. These equationscan also be written as

F = 0 MO = (r x F ) = 0where F are the forces and r are position vectors.

1.2 m

1.2 m

y

1.5 m

z

x

5 kNA B

C

E

D

1 m2 m

A 3-m pole is supported by a ball-and-socket joint at A and by thecables CD and CE. Knowing thatthe line of actionof the 5-kN forceforms an angle =30o with thevertical xy plane, determine (a)the tension in cables CD and CE,(b) the reaction at A.

Solving Problems on Your OwnProblem 4.8

43



1.2 m

1.2 m

y

1.5 m

z

x

5 kNA B

C

E

D

1 m2 m

1.2 m

1.2 m

y

1.5 m

z

xA B

C

E

D

1 m2 m

30o

5 kN

TCETCD

Ay j

Ax i

Az k

44

Write equilibrium equationsand solve for the unknowns.

1.2 m

1.2 m

y

1.5 m

z

xA B

C

E

D

1 m2 m

30o

5 kN

TCETCD

Ay j

Ax i

Az k

5 unknowns and 6 equations ofequilibrium, but equilibrium ismaintained, MAC = 0 .

rB/A = 2 i rC/A = 3 i

Load at B, FB = _ ( 5 cos 30o ) j + ( 5 sin 30o ) k = _ 4.33 j + 2.5 k

CD = _ 3 i+ 1.5 j + 1.2 k CD = 3.562 m

TCD = TCD = (_ 3 i + 1.5 j + 1.2 k)

TCE = TCE = (_ 3 i + 1.5 j _ 1.2 k)

CDCD

TCD

3.562

TCD

3.562CECE


45

1.2 m

1.2 m

y

1.5 m

z

xA B

C

E

D

1 m2 m

30o

5 kN

TCETCD

Ay j

Ax i

Az k

MA = 0: rC/A x TCD + rC/A x TCE + rB/A x FB = 0

i j k 3 0 0_3 1.5 1.2

i j k 3 0 0_3 1.5 _1.2

i j k2 0 00 _4.33 2.5

TCD

3.562 +TCE

3.562+ = 0


46

1.2 m

1.2 m

y

1.5 m

z

xA B

C

E

D

1 m2 m

30o

5 kN

TCETCD

Ay j

Ax i

Az k

Equate coefficients of unit vectors to zero.

j: _3.6 + 3.6 _ 5 = 0

_3.6 TCD+3.6 TCE_17.81 = 0 (1)

TCD

3.562TCE

3.562

k: 4.5 + 4.5 _ 8.66 = 0

4.5 TCD+4.5 TCE = 30.85 (2)

TCD

3.562TCE

3.562

(2) + 1.25 (1): 9TCE _ 53.11 = 0 ; TCE = 5.90 kN

Eq. (1): _3.6TCD + 3.6 (5.902) _ 17.81 = 0 TCD = 0.954 kN


47

1.2 m

1.2 m

y

1.5 m

z

xA B

C

E

D

1 m2 m

30o

5 kN

TCETCD

Ay j

Ax i

Az k

F = 0: A + TCD + TCE + FB = 0

i: Ax + (_3) + (_3) = 0

Ax = 5.77 kN

3.5620.954

3.5625.902

j: Ay + (1.5) + (1.5) _ 4.33 = 0 Ay = 1.443 kN3.5620.954

3.5625.902

k: Az + (1.2) + (_1.2) + 2.5 = 0 Az = _ 0.833 kN3.5620.954

3.5625.902

A = ( 5.77 kN) i + ( 1.443 kN ) j - ( 0.833 kN ) k


48

A

B

C

a

20 in

10 in60 lb

Problem 4.9

Rod AC is supported by a pin andbracket at A and rests against apeg at B. Neglecting the effect offriction, determine (a) the reactionsat A and B when a = 8 in., (b) thedistance a for which the reactionat A is horizontal and thecorresponding magnitudes of thereactions at A and B.

49

A

B

C

a

20 in

10 in60 lb



Problem 4.9


50

A

B

C

a

20 in

10 in60 lb

2. For a three-force body, solution can be obtained byconstructing a force triangle. The resultants of the three forces must be concurrent or parallel. To solve a problem involving athree-force body with concurrent forces, draw the free-bodydiagram showing that the three forces pass through the samepoint. Complete the solution by using a force triangle.


Problem 4.9


51



A

B

C

a

20 in

10 in60 lb

C

10 in60 lb

A

8 in

12 inBB

A

21

GD

F E

10 in

(a) a = 8 in

tan =

= 26.57o

12

52

C

10 in60 lb

A

8 in

12 inBB

A

21

GD

F E

10 in

3 - FORCE BODY

Reaction at A passes through Dwhere B and 60-lb load intersect

AE = EB = (8) = 4 in.

EF = BG = 10 _ 4 = 6 in

DG = BG = (6) = 3 in.

FD = FG _ DG = 8 _ 3 = 5 in.

Tan = = ; = 26.57o

12

12

12

12

FDAF

510

Construct a force triangle.


53

C

10 in60 lb

A

8 in

12 inBB

A

21

GD

F E

10 in

FORCE TRIANGLE

60 lb

30 lb

30 lb

= 26.57o

= 26.57o

A

B

A = B = = 67.08 lb

A = 67.1 lb 26.6o

B = 67.1 lb 26.6o

sin 26.57o30 lb


54

A

B

C

a

20 in

10 in60 lb


(b) For A horizontal

C

60 lb

A

B

B

A

21

G

F

10 in

a

= 26.57o


55

C

60 lb

A

B

B

A

21

G

F

10 in

a

= 26.57o


Construct a force triangle.

ABF : BF = AF cos

BFG : FG = BF sin

a = FG = AF cos sin

a = (10 in.) cos 26.57o sin 26.57o

a = 4.00 in.

56

C

60 lb

A

B

B

A

21

G

F

10 in

a

= 26.57o


FORCE TRIANGLE

60 lb

A

B

= 26.57o

A = = 120 lb A = 120.0 lb

B = = 134.16 lb B = 134.2 lb 26.6o

tan 60 lb

sin 60 lb

57

DCBA

aaa

30o30o

P

Problem 4.10

Rod AD supports a vertical loadP and is attached to collars Band C, which may slide freely onthe rods shown. Knowing thatthe wire attached at D forms anangle = 30o with the vertical,determine (a) the tension in thewire, (b) the reactions at B and C.

58

DCBA

aaa

30o30o

P



Rod AD supports a vertical loadP and is attached to collars Band C, which may slide freely onthe rods shown. Knowing thatthe wire attached at D forms anangle = 30o with the vertical,determine (a) the tension in thewire, (b) the reactions at B and C.

Problem 4.10

59



DCBA

aaa

30o30o

P

Solving Problems on Your OwnRod AD supports a vertical loadP and is attached to collars Band C, which may slide freely onthe rods shown. Knowing thatthe wire attached at D forms anangle = 30o with the vertical,determine (a) the tension in thewire, (b) the reactions at B and C.

Problem 4.10

60



DCBA

aaa

30o30o

P

DCBA

30o

30o

P

aaa

C

B

30o30o30o30o

T

61

Write equilibrium equationsand solve for the unknowns.


DCBA

30o

30o

P

aaa

C

B

30o30o30o30o

T

F = 0: _ P cos 30o + T cos 60o = 0

T = P = P T = 3 P

30o

cos 30o

cos 60o

3 / 21 / 2

+ MB = 0: P a _ (C sin 30o) a + T cos 30o (2a) = 0

P a _ ( C ) a + 3 P ( ) 2a = 0

_ C + (1 + 3) P = 0; C = 8 P C = 8 P 30o

12

32

12

62

DCBA

30o

30o

P

aaa

C

B

30o30o30o30o

T

+F = 0: _ B cos 30o + C cos 30o _ T sin 30o = 0

_ B + 8 P _ 3 P ( ) = 0;

B = 7 P B = 7 P 30o

32

32

12


ce 102 statics

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