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 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
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 ∑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
5.7 Constraints for a Rigid Body
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
• 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,
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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