# Equilibrium of a Rigid Body

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## in SI Units, 12e

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### Chapter Objectives

• Develop the equations of equilibrium for a rigid body

• Concept of the free-body diagram for a rigid body

• Solve rigid-body equilibrium problems using the equations of equilibrium

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### Chapter Outline

• Conditions for Rigid Equilibrium

• Free-Body Diagrams

• Equations of Equilibrium

• Two and Three-Force Members

Free Body Diagrams

Equations of Equilibrium

• Constraints and Statical Determinacy

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### Two-Force Members

• When forces are applied at only two points on a member, the member is called a two-force member

• Only force magnitude must be determined

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### Three-Force Members

• When subjected to three forces, the forces are concurrent or parallel

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### Example 5.13

The lever ABC is pin-supported at A and connected to a short link BD. If the weight of the members are negligible, determine the force of the pin on the lever at A.

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### Solution

Free Body Diagrams

• BD is a two-force member

• Lever ABC is a three-force member Equations of Equilibrium

Solving,

kN F

kN FA

32 . 1

07 . 1

0 45

sin 3

. 60 sin

; 0

0 400

45 cos 3

. 60 cos

; 0

3 . 4 60

. 0

7 . tan 1 0

F F

F

N F

F F

A y

A x

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### Support Reactions

As in the two-dimensional case:

• A force is developed by a support

• A couple moment is developed when rotation of the attached member is prevented

• The force’s orientation is defined by the coordinate angles , and

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### Example 5.14

Several examples of objects along with their associated free- body diagrams are shown. In all cases, the x, y and z axes are established and the unknown reaction components are indicated in the positive sense. The weight of the objects is neglected.

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### 5.6 Equations of Equilibrium

Vector Equations of Equilibrium

• For two conditions for equilibrium of a rigid body in vector form,

∑F = 0 ∑MO = 0

Scalar Equations of Equilibrium

• If all external forces and couple moments are expressed in Cartesian vector form

∑F = ∑Fxi + ∑Fyj + ∑Fzk = 0

∑MO = ∑Mxi + ∑Myj + ∑Mzk = 0

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### Redundant Constraints

• More support than needed for equilibrium

• Statically indeterminate: more unknown loadings than equations of equilibrium

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### Improper Constraints

• Instability caused by the improper constraining by the supports

• When all reactive forces are concurrent at this point, the body is improperly constrained

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### Procedure for Analysis

Free Body Diagram

• Draw an outlined shape of the body

• Show all the forces and couple moments acting on the body

• Show all the unknown components having a positive sense

• Indicate the dimensions of the body necessary for computing the moments of forces

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### Procedure for Analysis

Equations of Equilibrium

• Apply the six scalar equations of equilibrium or vector equations

• Any set of non-orthogonal axes may be chosen for this purpose

Equations of Equilibrium

• Choose the direction of an axis for moment summation such that it insects the lines of action of as many unknown forces as possible

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### Example 5.15

The homogenous plate has a mass of 100kg and is subjected to a force and couple moment along its edges. If it is supported in the horizontal plane by means of a roller at A, a ball and socket joint at N, and a cord at C, determine the components of

reactions at the supports.

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### Solution

Free Body Diagrams

• Five unknown reactions acting on the plate

• Each reaction assumed to act in a positive coordinate direction

Equations of Equilibrium

0 981

300

; 0

0

; 0

0

; 0

N N

T B

A F

B F

B F

C z

z z

y y

x x

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### Solution

Equations of Equilibrium

• Components of force at B can be eliminated if x’, y’ and z’

axes are used

0 )

3 ( .

200 )

5 . 1 ( 981 )

5 . 1 ( 300

; 0

0 ) 2 ( )

2 ( 300 )

1 ( 981

; 0

0 .

200 )

3 ( )

3 ( )

5 . 1 ( 981 )

5 . 1 ( 300

; 0

0 ) 2 ( )

1 ( 981 )

2 (

; 0

' '

m T

m N m

N m

N M

m A

m N

m N

M

m N m

A m

B m

N m

N M

m B

m N

m T

M

C y

z x

z z

y

Z C

x

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### Solution

Solving,

Az = 790N Bz = -217N TC = 707N

• The negative sign indicates Bz acts downward

• The plate is partially constrained as the supports cannot prevent it from turning about the z axis if a force is applied in the x-y plane

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