Chapter 4

Namas ChandraIntroduction to Mechanical engineering

HibbelerChapter 4-1

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Chapter 4: EquilibriumChapter 4: Equilibrium

Equilibrium means balance of forces to prevent body from translating, and balance of moments to prevent body from rotating.

Vector analysis in 3-D is the preferred method of solution.


HibbelerChapter 4-2

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4.1 Conditions for Equilibrium

Equilibrium means that the object is at rest (if originally at rest), or in constant velocity (if originally moving).

F 0 (Static)

F ma (dynamic) but a 0 for static

i i

i i

i

Let F be the external force and f be the

internal force. Then F + f = 0.

Since for equilibrium f = 0, we have F 0


HibbelerChapter 4-3

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4.1 Conditions of Equilibrium Con’t

Moment about any point O,

M = 0


HibbelerChapter 4-4

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4.2 Free Body Diagrams

Need to know how to represent support and contact conditions.

If a support prevents translation in any direction, we have a reaction force in that direction.

If a support prevents rotation in any orientation, then we have a couple moment exerted.


HibbelerChapter 4-5

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4.2 Free Body Diagrams Con’t

Weight always acts at the center of gravity.

W = m g

Consider the case of springs.


HibbelerChapter 4-6

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4.2 Free Body Diagrams Con’t

Consider the cantilever beam supported by a fixed support at A.

Free body diagram


HibbelerChapter 4-7

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Free Body Diagrams of a Platform

Consider the platform

Exclude all other effects except the platform now!


HibbelerChapter 4-8

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4.3 Equations of Equilibrium

In two dimensions (x-y plane)

0

Rarely though

0

0

x y z

a A B

A B C

F F F

F M M

M M M


HibbelerChapter 4-9

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4.3 Supports and Reactions

Roller allows motion along the plane. Reaction force is perpendicular to the surface.

Supports are idealized first. Reaction forces (magnitude and direction) and moments then depends on the type of support.

Rocker allows rotation at that point. Reaction force is perpendicular to the surface.


HibbelerChapter 4-10

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4.3 Supports and Reactions-2

Pin connected to a collar. Reaction force is perpendicular to the rod.

Hinge allows motion both in x and y direction but no rotation. Reaction force in x and y only. F not along member

Fixed allows no rotation and no translation. Reaction force vector and moment will result.



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4.3 Analysis Precedure

First draw the free body diagram for the loading shown to the right.

Apply equations of equilibrium through force and moment balance.



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Problem 4-3 (page 138, Section 4.1-4.3)Problem 4-3 (page 138, Section 4.1-4.3)

4.3 Draw the free-body diagram of the automobile, which has a mass of 5 Mg and center of mass at G. The tires are free to roll, so rolling resistance can be neglected. Explain the significance of each force on the diagram.

Solution:

effect of gravity

(weight) on the car.

= effect of the cable

on the car.

and = effect of the

road surface on the car. A B

W

T

N N



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4.7 Draw the free body diagram of the beam. The incline at B is smooth.

Solution:



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4.17 Determine the stretch of each spring for equilibrium of 20-Kg block. The springs are shown in their equilibrium position.

Solution:

y

y

x

Equilibrium:

Spring

+ F = 0; F 20(9.81) 0

Spring and

4 + F = 0; F F sin 45 =0

53

+ F = 0; F F cos 45 196.2=04

AD

AB AC

AB AC

AD

AB AC



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Solution-Con’t (slide 2)

Solving Eq. 1 and 2 yields:

F 158.55 N F 140.14N

Spring elongation:

140.14 0.467m

300158.55

0.793m200

196.2 0.490m

400

AC AB

AB

AC

AD

x

x

x

4.17 Determine the stretch of each spring for equilibrium of 20-Kg block. The springs are shown in their equilibrium position.



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4.4 Two-Force Members

Two force members are trusses that have forces (tension or compression) but not couple moments.

1 2 3 4 5 6

Resolve all the forces at A and B so that

and

If and are collinear then and .

Then the body is a two-force member.

It c

an be either in compression < 0

or ten

0

sion

A B

A B A B

A

A

F F F F F F F F

F F F F M

F

F

> 0



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4.4 Two and Three-Force Members

If a member is subjected to three coplanar forces, then the forces should either be concurrent or coplanar for equilibrium.

Hydraulic cylinder is a two-force member



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4.26 Determine the horizontal and vertical components of reaction at the pin A and the force in the short link BD.

Solution:

CCW + M 0;

8 (1.5 cos 30 ) - F (0.5 sin 30 ) = 0

F 41.57 kN = 41.6 kN

+ F 0;

41.57 - 0 41.6 kN

F 0;

A 8 0 A 8 kN

A

BD

BD

x

x x

y

y y

A A

��



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4.5 Equilibrium in 3-D

The concept of equilibrium in 3-D is similar. Here we need to solve all the known and unknown in 3-D. Once again we need to know the reaction forces and moments for each type of support, see Table 4-2

Ball- Only 3 forces

Bearing- 2 forces+2 moments Fixed- All 6 forces & moments

Pin- All except 1 moment



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4.6 Equations of Equilibrium

In 3-D we apply all the equations of mo

0

tion

0

F

M

x y z

x y z

The same equations can be written as a set of s

F

ix scalar equat

F F

ions.

0

0M M M

Procedure for Analysis:

1. Draw Free body diagram for the body under analysis

2. Mark all the reaction and external forces/moments.

3. Use the above equations to solve.



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4.68 Member AB is supported by a cable BC and at A by a smooth fixed square rod which fits loosely through the square hole of the collar. If the force lb, determine the tension in cable BC and the x, y, z components of reaction at A.

F = - 45k

2 2 2

Force Vector:

12i + 4j + 6k F

( 12) 4 6

6 2 3 = - i + j + k

7 7 7Equilibrium:

3 0; - 45 = 0

7 105 lb

BC BC

BC BC BC

z BC

BC

F

F F F

F F

F

Solution:



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F = - 45k

0;

6 (105) 0 90 lb

7

0;

2 (105) 0 30 lb

7

0;

3 (105)(4) 0 180 lb ft

7

x

x x

y

y y

x

Ax Ax

F

A A

F

A A

M

M M





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0;

3 45 (12) (105) (12) 0

7 0

0;

2 6 (105) (12) + (105) (4) = 0

7 7 720 lb ft

y

Ay

Ay

z

Az

Az

M

M

M

M

M

M

F = - 45k





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4.7 Friction

Friction is the force of resistance offered by a body that prevents or retards a body from motion relative to the first.

Friction always acts tangent to the surface and opposing any possible motion.

Friction is caused by small asperities as shown here.

We should consider all the minor surface asperities to get a distributed load .



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4.7 Friction-2

Frictional coefficient changes from static to kinetic when the value reduces..

Consider the motion of the following structure.

A

Seven unknowns:

N , , , , , ,A x y c cF B B P N F

Two sets of 3 equations

and

Friction F = Ns



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4.7 Friction-3But we use simple law suggested by Coloumb for two possible cases.

Impending (or possible) motion-statics

Actual motion-kinetic

First we still need to establish equilibirum to find

s s

k k

F N

F N

N

For static conditions, we use friction sF N

For dynamics conditions, we use friction kF N



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4.7 Tipping or impending motion

Tipping during motion or sliding depends if the clockwise moment at the bottom corner is CW or CCW.

Evaluate the location of N with respect to W.



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Example of pipes stackedExample of pipes stacked

Solution:

The concrete pipes are stacked. Determine the minimum coefficient of static friction so that the pile does not collapse.

Draw the Free body diagrams first.

Top pipe

0 0; 0

0; sin30 cos30 sin30 cos30 0

0;2 cos30 2 sin30 0

A B A B

x A B

y

M F r F r F F

F N F N F

F N F W



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Example of pipes stackedExample of pipes stacked

Solution:

Bottom pipe

0 0; . 0

0; sin30 cos30 0 (2)

0; cos30 sin30 0 (3)

C C

x

y C

M F r F r F F

F N F F

F N W N F

Use and A AF F N N

min

(2),

0.268s

From

F

N



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Example of man on a plankExample of man on a plank

Solution:

s

Determine how far the man can walk up the plank without causing

the plank to slip. between and is 0.3. Man weighs 200 lb.

d

A B

0; cos30 cos 60 0 (1)

0;2 sin 30 sin 60 200 0 (2)

0; cos10(15) sin10(15)

200sin 20(3) 200cos 20( ) 0 (3)

x A B B

y A B B

A B B

F F F N

F N F N

M N F

d



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Example of man on a plank-2Example of man on a plank-2

If the plank is on the verge of moving, slipping would occur at A.

Hence and 0.3

Substituting in eqns (1

), (2) and (3),

a

10.2 ft

110.09lb nd 88.14lb

A A B B B

A B

d

N N

F N F N N

Solution continued



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4.100 Two boys, each weighing 60 lb, sit at the ends of a uniform board, which has a weight of 30 lb. If the board rests at its center on a post having a coefficient of static friction of with the board, determine the greatest angle of tilt before slipping occurs. Neglect the size of post and the thickness of the board in the calculations.

( ) 0.6s

0;

0.6 cos - sin =0

tan = 0.6

= 31.0

xF

N N

��Solution:



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Problem 4-112 (page 195, Review Problems)Problem 4-112 (page 195, Review Problems)

4.112 The horizontal beam is supported by springs at its ends. If the stiffness of the spring at A is , determine the required stiffness of the spring at B so that if the beam is loaded with the 800-N force it remains in the horizontal position both before and after loading.

5 kN/mAk

Equilibrium:

CCW + 0;

(3) - 800 (1) = 0

266.67 N

CCW+ 0;

800(2) - (3)=0

533.33 N

A

B

B

B

A

A

M

F

F

M

F

F

Solution:



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Spring Force Formula:

533.33 266.67

5000

2500 N/m=2.50 kN/m

A B

B

B

Fx

kx x

k

k

4.112 The horizontal beam is supported by springs at its ends. If the stiffness of the spring at A is , determine the required stiffness of the spring at B so that if the beam is loaded with the 800-N force it remains in the horizontal position both before and after loading.


5 kN/mAk



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Chapter 4: Equilibrium.. concludesChapter 4: Equilibrium.. concludes

Chapter 4

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