eighth edition - welcome to apl100 -...

VECTOR MECHANICS FOR ENGINEERS:

STATICS

Eighth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

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

4 Equilibrium of

Rigid Bodies


Vector Mechanics for Engineers: Statics

Eig

hth

E

ditio

n

4 - 2

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



Eig

hth

E

ditio

n

4 - 3

Contents

Sample Problem 4.1

Sample Problem 4.3

Sample Problem 4.4

Equilibrium of a Two-Force Body

Equilibrium of a Three-Force Body

Sample Problem 4.6



Eig

hth

E

ditio

n

4 - 4

Contents

Equilibrium of a Rigid Body in Three

Dimensions

Reactions at Supports and Connections

for a Three-Dimensional Structure

Sample Problem 4.8



Eig

hth

E

ditio

n

Introduction

•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.


Se

ve

nth

E

ditio

n

Equilibrium Essentiality

The force and couple

moment system acting on a

body can be reduced to an

equivalent resultant force

and resultant couple moment

at any arbitrary point O on

or off the body.

../../WINDOWS/Temporary Internet Files/Content.IE5/0T2RWDYF/contents.ppt





Se

ve

nth

E

ditio

n

Equilibrium Essentiality

lf this resultant force and

couple moment are both

equal to zero, then the

body is said to be in

equilibrium

FR = 0 and ( MR)O = 0






Se

ve

nth

E

ditio

n

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,






Se

ve

nth

E

ditio

n

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

Free-Body Diagram






Se

ve

nth

E

ditio

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

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






Se

ve

nth

E

ditio

n •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.

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







Eig

hth

E

ditio

n

4 - 12

Reactions at Supports and Connections for a

Two-Dimensional Structure

•Reactions

equivalent

to a force

with

known

line of

action.



Eig

hth

E

ditio

n

4 - 13

Reactions at Supports and Connections for a

Two-Dimensional Structure

•Reactions equivalent to a force of

unknown direction and magnitude.



Eig

hth

E

ditio

n

4 - 14

Reactions at Supports and Connections for a Two-

Dimensional Structure

•Reactions equivalent to a force of

unknown direction and magnitude and

a couple of unknown magnitude



Eig

hth

E

ditio

n

4 - 15



Eig

hth

E

ditio

n



Eig

hth

E

ditio

n

4 - 19



Eig

hth

E

ditio

n

4 - 20



Eig

hth

E

ditio

n

4 - 21


Se

ve

nth

E

ditio

n

Rocker Bearing used to Support the

Roadway of a Bridge







Eig

hth

E

ditio

n

4 - 23

Equilibrium of a Rigid Body in Two Dimensions

•For all forces and

moments acting on a two-

dimensional structure,

Ozyxz MMMMF 00



Eig

hth

E

ditio

n

4 - 24

Equilibrium of a Rigid Body in Two Dimensions

•Equations of

equilibrium become 000 Ayx MFF

where A is any point

in the plane of the

structure.



Eig

hth

E

ditio

n

Alternative form of equilibrium equations

4 - 25



Eig

hth

E

ditio

n


4 - 26

If MA = 0, then R must pass through A

If Fx = 0 where x is arbitrary then R not

only passes through A but is also

perpendicular to x

If MB = 0 where B is any point such that

AB is not perpendicular to x direction

then R = 0



Eig

hth

E

ditio

n


4 - 27



Eig

hth

E

ditio

n

4 - 28

If MA = 0, then R must pass through A

If MB = 0, then R must pass through B

If MC = 0, then R must be zero where C

is not collinear with A and B



Eig

hth

E

ditio

n

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

0

0

0

B

A

x

M

M

F

0

0

0

C

B

A

M

M

M



Eig

hth

E

ditio

n

Statically Indeterminate Reactions

•More unknowns than equations

Four unknown reaction force unknown, three equations available



Eig

hth

E

ditio

n

•Fewer unknowns than equations, partially constrained

Two unknown reaction force unknown, three equations available

Partially Constrained, or referred as unstable



Eig

hth

E

ditio

n

•Equal number unknowns and equations but improperly constrained

Improperly Constrained, also indeterminate



Eig

hth

E

ditio

n

Free to move horizontally

Fx=0, then two equation left for three unknowns



Eig

hth

E

ditio

n

4 - 34

If the reactions are

concurrent or parallel,

the structure is

improperly constrained.

Improperly Constrained, also indeterminate

AM 0



Eig

hth

E

ditio

n

4 - 35



Eig

hth

E

ditio

n

4 - 36

Sample Problem 4.1

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.

A fixed crane has a

mass of 1000 kg

and is used to lift a

2400 kg crate.



Eig

hth

E

ditio

n

4 - 37

Sample Problem 4.1

Determine the

components of the

reactions at A and B.



Eig

hth

E

ditio

n

4 - 38

Sample Problem 4.1

SOLUTION:

•Create a free-body

diagram for the

crane.

•Determine reactions

at B and A.



Eig

hth

E

ditio

n

4 - 39

Sample Problem 4.1

•Free-body

diagram.



Eig

hth

E

ditio

n

4 - 40

Sample Problem 4.1

•Determine B by finding

the sum of the moments

of all forces about A.

0m6kN5.23

m2kN81.9m5.1:0 BM A

kN1.107B



Eig

hth

E

ditio

n

4 - 41

Sample Problem 4.1

•Determine the reactions at

A by sum of all horizontal

forces and all vertical

forces. 0:0 BAF xx

kN1.107xA

0kN5.23kN81.9:0 yy AF

kN 3.33yA



Eig

hth

E

ditio

n



Eig

hth

E

ditio

n

4 - 47



Eig

hth

E

ditio

n

4 - 48

RA = 238 N

RB = 62 N RA = 250 N

RB = 50 N



Eig

hth

E

ditio

n

4 - 49



Eig

hth

E

ditio

n

4 - 50



Eig

hth

E

ditio

n

4 - 51



Eig

hth

E

ditio

n

4 - 52

Sample Problem 4.3 Determine the tension

in the cable and the

reaction at each pair

of wheels.



Eig

hth

E

ditio

n

4 - 53

Sample Problem 4.3

and its load is 5500 N,

and it is applied at G.

The cart is held in

position by the cable.

A loading car is at

rest on an inclined

track. The gross

weight of the car



Eig

hth

E

ditio

n

4 - 54

Sample Problem 4.3

SOLUTION:

•Create a free-body diagram

for the car with the

coordinate system aligned

with the track.



Eig

hth

E

ditio

n

4 - 55

Sample Problem 4.3

•Determine the cable

tension by summing force

components parallel to the

track.

•Determine the reactions at

the wheels by summing

moments about points

above each axle.



Eig

hth

E

ditio

n

4 - 56

Sample Problem 4.3

• Create a free-body diagram

N 2320

25sinN 5500

N 4980

25cosN 5500

y

x

W

W



Eig

hth

E

ditio

n

4 - 57

Sample Problem 4.3

• Determine the reactions at the wheels.

0250mm1

mm150N 9804mm625N 2320:0

2R

M A

N 17582R

00mm125

mm150N 9804mm625N 2320:0

1R

M B

.4N 5621R

• Determine the cable tension.

0TN 4980:0xF

N 4980T



Eig

hth

E

ditio

n

4 - 58

Sample Problem 4.4

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.



Eig

hth

E

ditio

n

4 - 59

Sample Problem 4.4

SOLUTION:


diagram for the

frame and cable.

•Solve 3 equilibrium equations for

the reaction force components and

couple at E.



Eig

hth

E

ditio

n

4 - 60

Sample Problem 4.4

• Create a free-body diagram for the

frame and cable.



Eig

hth

E

ditio

n

4 - 61

Sample Problem 4.4

• 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

mkN0.180EM



Eig

hth

E

ditio

n

4 - 62

Draw the free body diagram of the foot

lever shown in Figure. The operator

applies a vertical force to the pedal so

that the spring is stretched 1.5 in. and

the force in the short link at B is 20 lb.



Eig

hth

E

ditio

n

4 - 63

Two smooth pipes, each having a mass

of 300 kg. are supported by the fork of

the tractor in Fig. Draw the free body

diagrams

for each pipe and both pipes together.



Eig

hth

E

ditio

n

4 - 64


•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.



Eig

hth

E

ditio

n

4 - 65


•Similarly, the line of action

of F1 must pass through B

for the sum of moments

about B to be zero.



Eig

hth

E

ditio

n

4 - 66


•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.



Eig

hth

E

ditio

n

Two and Three force members

4 - 67

Find reactions at

A, B, D.



Eig

hth

E

ditio

n

Two and Three force members

4 - 68



Eig

hth

E

ditio

n



Eig

hth

E

ditio

n

4 - 74



Eig

hth

E

ditio

n

Two force member

4 - 75

The bucket link AB on the back hoe is

a typical example of a two-force

member since it is pin connected at its

ends and provided its weight is

neglected no other force acts on this

member.



Eig

hth

E

ditio

n

4 - 76


•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.



Eig

hth

E

ditio

n

4 - 77


•Since the rigid body is in

equilibrium, the sum of the

moments of F1, F2, and F3

about any axis must be zero.

•The moment of F3 about D

must be zero as well and that

the line of action of F3 must

pass through D.



Eig

hth

E

ditio

n

4 - 78


•The lines of action of the

three forces must be

concurrent or parallel.



Eig

hth

E

ditio

n

4 - 79

Sample Problem 4.6

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.



Eig

hth

E

ditio

n

4 - 80

Sample Problem 4.6 •The three forces must

be concurrent for

static equilibrium.

•Reaction R passes

through the

intersection of the

lines of action of the

weight and rope

forces.



Eig

hth

E

ditio

n

4 - 81

•Determine the direction of

R.

636.1414.1

313.2tan

m 2.313m 515.0828.2

m 515.020tanm 414.1)2045cot(

m 414.1

m828.245cosm445cos

21

AE

CE

BDBFCE

CDBD

AFAECD

ABAF

6.58



Eig

hth

E

ditio

n

4 - 82

Sample Problem 4.6

•Determine the

magnitude of the

reaction force R.

38.6sin

N 1.98

110sin4.31sin

RT

N 8.147

N9.81

R

T



Eig

hth

E

ditio

n

4 - 83

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



Eig

hth

E

ditio

n

4 - 84

Equilibrium of a Rigid Body in Three Dimensions

•These equations can be solved for no

more than 6 unknowns which generally

represent reactions at supports or

connections.

00 FrMF O



Eig

hth

E

ditio

n

4 - 85

Reactions at Supports and Connections for a Three-




Eig

hth

E

ditio

n

4 - 86



Eig

hth

E

ditio

n

4 - 87



Eig

hth

E

ditio

n

4 - 88



Eig

hth

E

ditio

n

4 - 89





Eig

hth

E

ditio

n

Universal Joints

universal joint 3.flv



Eig

hth

E

ditio

n

4 - 91





Eig

hth

E

ditio

n

4 - 92



Eig

hth

E

ditio

n

4 - 93



Eig

hth

E

ditio

n

Sample problem 4.7

• A 20-kg ladder used to reach

high shelves in a storeroom is

supported by two flanged

wheels A and B mounted on a

rail and by an unflanged wheel

C resting against a rail fixed to

the wall. An 80-kg man stands

on the ladder and leans to the

right. The line of action of the

combined weight W of the man

and ladder intersects the floor at

point D. Determine the

reactions at A, B, and C.



Eig

hth

E

ditio

n

4 - 95



Eig

hth

E

ditio

n

(80 20)(9.81) 981

0

981 0

981 0

0

y z y z

y y

z z

W mgj j j

F

A j A k B j B k j Ck

A B

A B C



Eig

hth

E

ditio

n

0

1.2 0.9 0.6 981

0.6 3 1.2 0

A

y z

M

i B j B k i k j

i j k Ck



Eig

hth

E

ditio

n

02.136.0

9816.09.02.1

0

0

0981

981)81.9)(2080(

kCkji

jkikBjBi

M

CBA

BA

jjjmgW

zy

A

zz

yy

5 unknowns and 5 equations,

so the problem can be solved



Eig

hth

E

ditio

n

4 - 99

Sample Problem 4.8 A sign of uniform density weighs 540 N

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.



Eig

hth

E

ditio

n

4 - 100

Sample Problem 4.8



Eig

hth

E

ditio

n

4 - 101

Sample Problem 4.8

SOLUTION:


diagram.

•Apply static

equilibrium to

develop equations

for the unknown

reactions.



Eig

hth

E

ditio

n

4 - 102

Sample Problem 4.8



Eig

hth

E

ditio

n

4 - 103

Sample Problem 4.8

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.



Eig

hth

E

ditio

n

4 - 104

Sample Problem 4.8

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



Eig

hth

E

ditio

n

4 - 105

Sample Problem 4.

623 7

313 7

2 23 7

540N 0

: 0

: 270 N 0

: 0

1.2 m 540N 0

: 1.6 0.5143 0

: .8 .771 648 N 0

BD EC

x BD EC

y BD EC

z BD EC

A B BD E EC

BD EC

BD EC

F A T T j

i A T T

j A T T

k A T T

M r T r T i j

j T T

k T T



Eig

hth

E

ditio

n

4 - 106

Sample Problem 4.

623 7

313 7

2 23 7

: 0

: 270 N 0

: 0

: 1.6 0.5143 0

: .8 .771 648 N 0

x BD EC

y BD EC

z BD EC

BD EC

BD EC

i A T T

j A T T

k A T T

j T T

k T T



Eig

hth

E

ditio

n

4 - 107

Sample Problem 4.

kjiA

TT ECBD

N 45.02N 202.3N 675.3

N 3.630N 6.202

Solve the 5 equations

for the 5 unknowns,



Eig

hth

E

ditio

n

What if…?

4 - 108

Can the sign be in static

equilibrium if cable BD

is removed?



Eig

hth

E

ditio

n

4 - 109

The sign cannot be in static equilibrium

because TEC causes a moment about the y-

axis (due to the existence of TEC,Z) which

must be countered by an equal and

opposite moment.

This can only be provided by a cable

tension that has a z-component in the

negative-z direction, such as what TBD

has.


Se

ve

nth

E

ditio

n

Sample Problem 4.9

• A uniform pipe cover of radius r = 240 mm and mass 30 kg is held in a horizontal position by the cable CD. Assuming that the bearing at B does not exert any axial thrust, determine the tension in the cable and the reactions at A and B.






Se

ve

nth

E

ditio

n

The supports do

not exert an

appreciable

couple during

normal operation

and these are not

included in

analysis.






Se

ve

nth

E

ditio

n

0294

7

2

7

3

7

622

0

07

2

2947

3

7

6

0294

0

kkrir

kTjTiTkrirkAjAiAkr

M

kTA

jTBAiTBA

jTjBiBkAjAiA

F

zyx

B

z

yyxx

yxzyx

6 unknowns and 6 equations






Se

ve

nth

E

ditio

n

Sample Problem 4.10






Se

ve

nth

E

ditio

n • A 450-lb load hangs from the comer C of a rigid piece of pipe ABCD which has been bent as shown. The pipe is supported by the ball-and-socket joints A and D, which are fastened, respectively, to the floor and to a vertical wall, and by a cable attached at the midpoint E of the portion BC of the pipe and at a point G on the wall. Determine (a) where G should be located if the tension in the cable is to be minimum, (b) the corresponding minimum value of the tension.






Se

ve

nth

E

ditio

n






Se

ve

nth

E

ditio

n

18

61212

54004501212

0

0

kji

AD

AD

kjjiWAC

WACTAE

M AD






Se

ve

nth

E

ditio

n

12 12 65400 1800

18

4 2 4

4 2 4

6

4 2 4 12 12 6

6 18

i j kAE T k

AE T T AE

AE i j k

i j kunit vector along AE

T must be along this unit vector

i j k i j kT

min

6 12 1800

6 1800

200 100 200

i j

T

T i j k






Se

ve

nth

E

ditio

n

Must be

the same

direction x=0, y=15 ft

6 ( 12) 0 6

6 12 0 6

200 100 200

,

EG x i y j k

x y

x y are the coordinates of G





eighth edition - welcome to apl100 -...

Documents