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1
Ishik University / Sulaimani
Civil Engineering Department
Engineering Mechanics I
Chapter -5-
Equilibrium of a Rigid Body 1
Ishik
Univ
ersity-S
ula
imani
Assista
nt L
ectu
rer - A
smaa A
bdulm
aje
ed
CHAPTER OBJECTIVES
To develop the equations of equilibrium for a rigid
body.
To introduce the concept of the free-body diagram for
a rigid body.
To show how to solve rigid-body equilibrium
problems using the
equations of equilibrium.
2
CHAPTER OUTLINE
Free-Body Diagrams
Equations of Equilibrium
Two and Three-Force Members
Constraints for a Rigid Body
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FBD is the best method to represent all the known and
unknown forces in a system.
FBD is a sketch of the outlined shape of the body, which
represents it being isolated from its surroundings.
Necessary to show all the forces and couple moments
that the surroundings exert on the body so that these
effects can be accounted for when equations of
equilibrium are applied.
Free-Body Diagrams
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Free-Body Diagrams
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Free-Body Diagrams
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Free-Body Diagrams
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Support Reactions
If the support prevents the translation of a body in a given
direction, then a force is developed on the body in that
direction.
If rotation is prevented, a couple moment is exerted on the
body.
Consider the three ways a horizontal member, beam is
supported at the end.
- roller, cylinder
- pin
- fixed support
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Support Reactions
Roller or cylinder
Prevent the beam from translating in the vertical direction.
Roller can only exerts a force on the beam in the vertical
direction.
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Support Reactions
Pin
The pin passes through a hold in the beam and two leaves that
are fixed to the ground.
Prevents translation of the beam in any direction Φ.
The pin exerts a force F on the beam in this direction.
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Support Reactions
Fixed Support
This support prevents both translation and rotation of the beam.
A couple and moment must be developed on the beam at its
point of connection.
Force is usually represented in x and y components.
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Cable exerts a force on the bracket
Type 1 connections
Rocker support for this bridge girder
allows horizontal movements so that
the bridge is free to expand and
contract due to temperature
Type 5 connections
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Concrete Girder rest on the ledge
that is assumed to act as a smooth
contacting surface
Type 6 connections
Utility building is pin supported
at the top of the column
Type 8 connections
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Floor beams of this building are
welded together and thus form fixed
connections
Type 10 connections
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External and Internal Forces
A rigid body is a composition of particles, both external
and internal forces may act on it.
For FBD, internal forces act between particles which are
contained within the boundary of the FBD, are not
represented.
Particles outside this boundary exert external forces on the
system and must be shown on FBD.
FBD for a system of connected bodies may be used for
analysis.
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Weight and Center of Gravity
When a body is subjected to gravity, each particle has a
specified weight.
For entire body, consider gravitational forces as a system
of parallel forces acting on all particles within the
boundary.
The system can be represented by a single resultant force,
known as weight W of the body.
Location of the force application is known as the center of
gravity.
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Idealized Models
Consider a steel beam used to support the roof joists of a building.
For force analysis, reasonable to assume rigid body since small
deflections occur when beam is loaded.
Bolted connection at A will allow for slight rotation when load is
applied => use Pin.
Support at B offers no resistance to horizontal movement => use Roller.
Building code requirements used to specify the roof loading (calculations of the joist forces).
Large roof loading forces account for extreme loading cases and for dynamic or vibration effects.
Weight is neglected when it is small compared to the load the beam supports.
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Example 5.1
Draw the free-body diagram of the uniform beam. The beam has a
mass of 100kg.
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Solution
Free-Body Diagram
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Solution
Support at A is a fixed wall.
Three forces acting on the beam at A denoted as Ax, Ay, Az,
drawn in an arbitrary direction.
Unknown magnitudes of these vectors.
Assume sense of these vectors.
For uniform beam,
Weight, W = 100(9.81) = 981N
acting through beam’s center of gravity, 3m from A.
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Draw the free-body diagram of member AB, which is supported
by a roller at A and a pin at B. Explain the significance of each
force on the diagram.
Example 5.2
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Solution
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Example 5.3
Draw the free-body diagram of the beam which supports the 80-
kg load and is supported by the pin at A and a cable which
wraps around the pulley at D. Explain the significance of each
force on the diagram.
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Solution
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Equations of Equilibrium
For equilibrium of a rigid body in 2D,
∑Fx = 0;
∑Fy = 0;
∑MO = 0
∑Fx and ∑Fy represent the algebraic sums of the x and y
components of all the forces acting on the body.
∑MO represents the algebraic sum of the couple moments and
moments of the force components about an axis perpendicular to
x-y plane and passing through arbitrary point O, which may lie
on or off the body.
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Alternative Sets of Equilibrium Equations
For coplanar equilibrium problems,
∑Fx = 0;
∑Fy = 0;
∑MO = 0 can be used
Two alternative sets of three independent equilibrium equations
may also be used.
∑Fa = 0;
∑MA = 0;
∑MB = 0
When applying these equations, it is required that a line passing
through points A and B is not perpendicular to the a axis.
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Alternative Sets of Equilibrium Equations
Consider FBD of an arbitrarily shaped body.
All the forces on FBD may be replaced by an
equivalent resultant force
FR = ∑F acting at point A and a
resultant moment MRA = ∑MA
If ∑MA = 0 is satisfied, MRA = 0
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Alternative Sets of Equilibrium Equations
A second set of alternative equations is
∑MA = 0;
∑MB = 0;
∑MC = 0
Points A, B and C do not lie on the same line.
Consider FBD, if ∑MA = 0, MRA = 0
∑MA = 0 is satisfied if line of action of FR passes through
point B.
∑MC = 0 where C does not lie on line AB.
FR = 0 and the body is in equilibrium.
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Example 5.4
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Solution
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Example 5.5
The cord supports a force of 500N and wraps over the
frictionless pulley. Determine the tension in the cord at C and
the horizontal and vertical components at pin A.
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Solution
Equations of Equilibrium
Tension remains constant as cord passes over the pulley (true for
any angle at which the cord is directed and for any radius of the pulley.
NT
mTmN
M A
500
0)2.0()2.0(500
;0
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NA
NNA
F
NAx
NA
F
y
y
y
x
x
933
030cos500500
;0
250
030sin500
;0
Solution
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Example 5.6
The link is pin-connected at a and rest a smooth support at B.
Compute the horizontal and vertical components of reactions
at pin A.
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Solution
FBD
Reaction NB is perpendicular to the link at B.
Horizontal and vertical components of reaction are represented
at A.
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Solution
Equations of Equilibrium;
NAx
NA
F
NN
mNmNmN
M
x
x
B
B
A
100
030sin200
;0
200
0)75.0()1(60.90
;0
NA
NNA
F
y
y
y
233
030cos20060
;0
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Example 5.7
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Solution
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Example 5.8
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Solution
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Example 5.9
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Solution
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Simplify some equilibrium problems by recognizing members
that are subjected top only 2 or 3 forces.
Two-Force Members
When a member is subject to no couple
moments and forces are applied at only two
points on a member, the member is called a
two-force member.
Two and Three-Force Members
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Two-Force Members
Hence, only the force magnitude must be determined or stated.
Other examples of the two-force members held in equilibrium
are shown in the figures to the right.
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Three-Force Members
If a member is subjected to only three forces, it
is necessary that the forces be either concurrent
or parallel for the member to be in equilibrium.
To show the concurrency requirement,
consider a body with any two of the three
forces acting on it, to have line of actions that
intersect at point O.
To satisfy moment equilibrium about O, the
third force must also pass through O, which
then makes the force concurrent.
If two of the three forces parallel, the point of
currency O, is considered at “infinity”.
Third force must parallel to the other two
forces to insect at this “point”.
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Example 5.10
Determine the normal reactions at A and B.
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Solution
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Example 5.11
Determine the tension in the cord and the horizontal and
vertical components of reaction at support A of the beam.
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Solution
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Determine the horizontal and vertical components of reaction at
C and the tension in the cable AB for the truss.
Example 5.12
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Solution
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Homework 5.1
Determine the horizontal and vertical components of reaction
at the pin A and the force in the cable BC. Neglect the
thickness of the members.
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Determine the horizontal and vertical components or reaction
at the pin A and the reaction on the beam fit C.
Homework 5.2
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Homework 5.3
Determine the normal reaction at the roller A and horizontal
and vertical components at pin B for equilibrium of the
member.
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