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CHAPTER 5
EQUILIBRIUM OF A RIGID BODY
5.1 Conditions of Rigid Body Diagram
In this section, you will learn how to develop equations required for equilibrium of a rigid
body.
Referring to figure below, a rigid body is fixed in the x, y, z reference and is either at rest
or moves with the reference at constant velocity.
Figure (b) shows a free-body diagram of the arbitrary ith particle of the body. There are
two types of forces which act on it.
The resultant internal force, fi, is caused by the interactions with adjacent particles.
The resultant external force, Fi, represent, for example, the effects of gravitational,
electrical or magnetic.
If the particle is in equilibrium;
All these equations are added together vectorially;
Fi + fi = 0 ;
∑Fi + ∑fi = 0 ;
The summation of the internal forces will equal zero since the internal forces between
particles within the body will occur in equal but opposite collinear pairs.
Only the sum of the external forces will remain;
∑Fi = ∑F ;
Then, the above equation can be written as
∑F = 0 ;
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School of Mechatronics, KUKUM
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STATICS
Referring to figure below, the moments of the forces acting on the ith particle about the
arbitrary point O.
Using the above particle equilibrium equation and the distributive law of the vector cross
product,
ri x (Fi + fi ) = ri x Fi + ri x fi = 0 ;
All these equations are added together vectorially;
∑ri x Fi + ∑ri x fi = 0 ;
The second term is zero because the internal forces occur in equal but opposite collinear
pairs;
∑Mo = ∑ ri x Fi ;
The resultant moment of each pair of forces about point O is zero ;
∑Mo = 0 ;
The two equations of equilibrium for a rigid body :
∑F = 0 ;
∑Mo = 0 ;
(Eq. 1)
5.1 Free Body Diagram (FBD) in Two-Dimension
A free-body diagram is a sketch of the outlined shape of the body, which represents it as
being isolated or “free” from its surroundings. On this sketch it is 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 the equations of equilibrium are applied.
Table 1 shows supports for rigid bodies subjected to two-dimensional force systems.
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School of Mechatronics, KUKUM
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STATICS
5.2 Support Reactions
There are various types of reactions that occur at supports and points of support between
bodies subjected to coplanar force systems.
As a general rule, if a support prevents the translation of a body in a given direction, then
a force is developed on the body in that direction. Likewise, if rotation is prevented, a
couple moment is exerted on the body.
There are three ways in which a horizontal member, such as a beam, is supported at its
end. Figure shows one method consists of a roller or cylinder.
Since this support only prevents the beam from translating in the vertical direction, the
roller can only exert a force on the beam in this direction.
The beam can be supported in a more restrictive manner by using a pin as shown in
figure.
The pin passes through a hole in the beam and two leaves which are fixed to the ground.
Here the pin can prevent translation of the beam in any direction Φ and the pin must exert
a force F on the beam in this direction.
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School of Mechatronics, KUKUM
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STATICS
For purpose of analysis, it is generally easier to represent this resultant force F by its two
components Fx and Fy. If Fx and Fy are known, then F and Φ can be calculated.
The most restrictive way to support the beam would be to use a fixed support. This
support will prevent both translation and rotation of the beam.
Figure shows a force and couple moment which are developed on the beam at its point of
connection to prevent both translation and rotation.
The force is usually represented by its components Fx and Fy. Types of support reactions,
and forces and couple moments involved :
EXAMPLES
Example 5-1 (pp 202)
Example 5-2 (pp 203)
Example 5-3 (pp 204)
Example 5-4 (pp 205)
Example 5-5 (pp 206)
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School of Mechatronics, KUKUM
ENT151/4
STATICS
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School of Mechatronics, KUKUM
ENT151/4
STATICS
5.3 Equations of Equilibrium
∑Fx = 0
∑Fy = 0
∑Mo = 0
When a body is subjected to a system of forces, which all lie in the x-y plane, then the
forces can be resolved into their x and y components.
The conditions for equilibrium for two dimensions are;
∑Fx = 0 ;
∑Fy = 0 ;
∑Mo = 0 ;
(Eq. 2)
∑Fx and ∑Fy represent, respectively, the algebraic sums of the x and y components of all
the forces acting on the body.
∑Mo represents the algebraic sums of the couple moments and the moments of all the
force components about an axis perpendicular to the x-y plane and passing through the
arbitrary point O, which may lie either on or off the body.
Procedure for Analysis :
Coplanar force equilibrium problems for a rigid body can be solved using the following
procedure:
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School of Mechatronics, KUKUM
ENT151/4
STATICS
Free-Body Diagram
1) Establish the x, y coordinate axes in any suitable orientation.
2) Draw an outlined shape of the body.
3) Show all the forces and couple moments acting on the body.
4) Label all the loading and specify their directions relative to the x, y axes. The sense of
a force or couple moment having an unknown magnitude but known line of action can
be assumed.
5) Indicate the dimensions of the body necessary for computing the moments of forces.
Equation of Equilibrium
6) Apply the moment equation of equilibrium, ∑Mo = 0, about a point (O) that lies at the
intersection of the lines of action of two unknown forces. In this way, the moments of
these unknowns are zero about O and a direct solution for the third unknown can be
determined.
7) When applying the force equilibrium equations, ∑Fx = 0 and ∑Fy = 0, orient the x
and y axes along lines that will provide the simplest resolution of the forces into their x
and y components.
8) If the solution of the equilibrium equations yields a negative scalar for a force or
couple moment magnitude, this indicates that the sense is opposite to that which was
assumed on the free-body diagram
EXAMPLES
Example 5-6 (pp 211)
Example 5-7 (pp 212)
Example 5-8 (pp 213)
Example 5-9 (pp 214)
Example 5-10 (pp 215)
Example 5-11 (pp 216)
Example 5-12 (pp 217)
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School of Mechatronics, KUKUM
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STATICS
5.4 Two and three-force members
Two force members :
Three-force members
EXAMPLE :
Example 5-13 (pp 220)
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School of Mechatronics, KUKUM
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STATICS
5.5 Equilibrium in Three Dimensions
The general procedure for establishing the free-body diagram of a rigid body for three
dimensions is stated as in the case of two dimensions. Essentially it required first
“isolating” the body by drawing its outlined shape. This is followed by a careful labeling
of all the forces and couple moments in reference to an established x, y, z coordinate
system. As a general rule, components of reaction having an unknown magnitude are
shown acting on the free-body diagram in the positive sense. In this way, if any negative
values are obtained, they will indicate that the components act in negative coordinate
directions.
Support reactions :
Table 5-2 shows supports for rigid bodies subjected to three-dimensional force systems.
It is important to recognize the symbols used to represent each of these supports and to
understand clearly how the forces and couple moments are developed by each support.
A force is developed by a support that restricts the translation of the attached members
whereas a couple moment is develop when rotation of the attached member is prevented.
For example, the ball-and-socket joint prevents any translation of the connecting member
(refer to Table 5-2). Therefore, a force must act on the member at the point of connection.
These force has three components having unknown magnitude, Fx, Fy, Fz. Provided these
components are known, one can obtained the magnitude of force and the force’s
orientation defined by the coordinate direction angles α, β, γ.
Magnitude of force : F = √ (Fx2 + Fy2 + Fz2)
The third direction angle : Cos2 α + Cos2 β + Cos2 γ = 1
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School of Mechatronics, KUKUM
ENT151/4
STATICS
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School of Mechatronics, KUKUM
ENT151/4
STATICS
EXAMPLES
Example 5-14 (pp 211)
5.6
Example 5-15 (pp 212)
Equation of Equilibrium
The conditions for equilibrium of a rigid body subjected to a three-dimensional force
system require that both the resultant force and resultant couple moment acting on the
body be equal to zero.
Vector equations of equilibrium
Two conditions for equilibrium of a rigid body;
∑F = 0 ; ∑Mo = 0
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(Eq. 1)
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School of Mechatronics, KUKUM
ENT151/4
STATICS
where ∑F is the vector sum of all the external forces acting on the body and ∑Mo is the
sum of the couple moments and the moments of all the forces about any point O located
either on or off the body.
Scalar equations of equilibrium
If all the applied external forces and couple moments are expressed in Cartesian vector
form and substituted into Eq. 1;
∑F = ∑Fx i + ∑Fy j + ∑Fz k
=0
∑Mo = ∑Mx i + ∑My j + ∑Mz k
=0
Where ;
∑Fx = 0 ; ∑Fy = 0 ; ∑Fz = 0
(Eq. 2a)
∑Mx = 0 ; ∑My = 0 ; ∑Mz = 0
(Eq. 2b)
These six scalar equilibrium equations (Eq. 2a and Eq. 2b) may be used to solve for at
most six unknowns shown on the free-body diagram.
Note :
If a support prevents translation of a body in a specific direction, then the support exerts a
force on the body in that direction. If rotation about an axis is prevented, then the support
exerts a couple moment on the body about the axis.
Procedure for Analysis
Three-dimensional equilibrium problems for a rigid body can be solved using the
following procedure:
Free-Body Diagram
1) Draw an outlined shape of the body.
2) Show all the forces and couple moments acting on the body.
3) Establish the origin of the x, y, z axes at a convenient point and orient the axes so that
they are parallel to as many of the external forces and moments as possible.
4) Label all the loadings and specify their directions relative to the x, y, z axes.
5) Indicate the dimensions of the body necessary for computing the moment of forces.
Equation of Equilibrium
6) If the x, y, z force and moment components seem easy to determine, then apply six
scalar equations of equilibrium; otherwise use the vector equations.
7) It is not necessary that the set of axes chosen for force summation coincide with the set
of axes chosen for moment summation.
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School of Mechatronics, KUKUM
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STATICS
8) Choose the direction of an axis for moment summation such that it intersects the lines
of action of as many unknown forces is possible. The moment of forces passing
through points on this axis and forces which are parallel to the axis will then be zero.
5.7
Constraints for a Rigid Body
Redundant constraints
When the body has a redundant support more supports then are necessary to hold it in
equilibrium, it becomes statically indeterminate. Statically indeterminate – there will be
more unknown loadings on the body then equations of equilibrium available for their
solution.
Eg :
Two dimensional case – five unknown but only three equation.
Three dimensional case – eight unknown but only five equation
Improper Constraints
Instability of the body because of improper constraining by the supports.
3-Dimensional – if the support reaction all intersec a common axis
2-Dimensional – the axis is perpendicular to the plane of the forces
When all the reactive forces are concurrent at the point, the body is improperly constraint.
Improper constraining leads to instability occurs when the reactive forces are all parallel.
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School of Mechatronics, KUKUM
ENT151/4
STATICS
Proper constraining requires
1. the line of action of the reactive forces does not intersect point on a common axis
2. the reactive force must not all be parallel on one another
EXAMPLES
Example 5-15 (pp 242)
Example 5-16 (pp 243)
Example 5-17 (pp 244)
Example 5-18 (pp 245)
Example 5-19 (pp 246)
EXERCISES
Exercise 5 - 5
Exercise 5 - 19
Exercise 5 - 22
Exercise 5 - 24
Exercise 5 - 89
Exercise 5 - 95
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School of Mechatronics, KUKUM