**Equilibrium of a Rigid Body**

**5**

**Engineering Mechanics: Statics **

**in SI Units, 12e**

**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

**5.4 Two- and Three-Force Members**

**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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**5.4 Two- and Three-Force Members**

### Three-Force Members

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

**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*
*F*_{A}

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*

### 5.5 Free-Body Diagrams

### 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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### 5.5 Free-Body Diagrams

### 5.5 Free-Body Diagrams

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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.

### Solution

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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 ∑M**_{O} = 0

Scalar Equations of Equilibrium

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

**∑F = ∑F**_{x}**i + ∑F**_{y}**j + ∑F**_{z}**k = 0**

**∑M**_{O} = ∑M_{x}**i + ∑M**_{y}**j + ∑M**_{z}**k = 0**

**5.7 Constraints for a Rigid Body**

**Redundant Constraints**

**Redundant Constraints**

• More support than needed for equilibrium

• Statically indeterminate: more unknown loadings than equations of equilibrium

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**5.7 Constraints for a Rigid Body**

**Improper Constraints**

**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

**5.7 Constraints for a Rigid Body**

### 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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**5.7 Constraints for a Rigid Body**

### 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

**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*

**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,

**A**_{z} **= 790N B**_{z} = -217N **T**_{C} = 707N

**• The negative sign indicates B**_{z} 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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