EQUILIBRIUM OF A RIGID BODY
&
FREE-BODY DIAGRAMS
Chapter 5 Objectives:
Students will be able to:
a) Identify support reactions
(2D and 3D)
b) Draw a free-body diagram.
c) Identify 2-force and multiforce members
d) Apply equations of
equilibrium (2D and 3D)
APPLICATIONS
Real world example
Idealized model
FBD
The truck ramp has a weight of 400 lb.
The ramp is pinned to the body of the truck and held in the
position by the cable. How can we determine the cable tension
and support reactions?
Key: Properly representing the reactions at the pinned joints and
cables, as well as the weight, in a Free Body Diagram (FBD)
APPLICATIONS (continued)
FBD
Two smooth pipes, each having a mass of 300 kg, are
supported by the tines of the tractor fork attachment.
How can we determine all the reactive forces ?
Key: Properly representing the reactions from the tines, as
well as the weight, in a Free Body Diagram (FBD)
CONDITIONS FOR RIGID-BODY EQUILIBRIUM (Section 5.1)
In Chapters 1-3 we only considered
forces acting on a particle (concurrent
forces). In this case rotation is not a
concern, so equilibrium could be
satisfied by:
 F = 0 (no translation)
Forces on a particle
We will now consider cases where forces are not
concurrent so we are also concerned that the rigid
body does not rotate. In order for a rigid body to
be in equilibrium, the net force as well as the net
moment about any arbitrary point O must be
equal to zero.
 F = 0 (no translation)
Forces on a rigid body
and  MO = 0 (no rotation)
THE PROCESS OF SOLVING RIGID BODY EQUILIBRIUM PROBLEMS
Procedure:
1) Draw a free-body diagram (FBD) showing all the external
(active and reactive) forces.
2) Apply the equations of equilibrium to solve for any unknowns.
FREE-BODY DIAGRAMS
(Section 5.2)
Free-body diagram (FBD)
Procedure for drawing a Free-Body Diagram (FBD)
1. Draw an outlined shape. Imagine the body to be isolated or cut “free”
from its constraints and draw its outlined shape.
2. Show all of the external forces and couple moments. These typically include:
a) applied loads
b) support reactions (see table on following pages for reactions)
c) weight of the body
3. Label loads and dimensions on the FBD: All known forces and couple moments
should be labeled with their magnitudes and directions. For the unknown forces and
couple moments, use letters like Ax, Ay, MA, etc.. Indicate any necessary dimensions.
Support Reactions in 2D (Table 5-1)
memorize this table!
The 2D reactions shown below are the ones shown in Table 5-1 in the
text. As a general rule:
1) if a support prevents translation of a body in a given direction, then
a force is developed on the body in the opposite direction.
2) if rotation is prevented, a couple moment is exerted on the body in
the opposite direction
EXAMPLE – Drawing a FBD
Given: The operator applies a vertical
force to the pedal so that the
spring is stretched 1.5 in. and the
force in the short link at B is
20 lb.
Draw: A an idealized model and freebody diagram of the foot pedal.
“Weightless link” (see Table 5-1)
“Smooth pin”
Example
Draw a FBD of member ABC, which is supported
by a smooth collar at A, rocker at B, and link CD.
EQUATIONS OF EQUILIBRIUM (Section 5.3)
As noted in Section 5.1
If a rigid body is in equilibrium, then:
• The sum of the forces acting on the
body equals zero
• The sum of the moments (about any
point) acting on the body equals
zero
• For 2D equilibrium, these equations
are most commonly expressed as:
 Fx  0
 Fy  0
 M O  0 (for any point O)
(Most common equations for 2D equilibrium)
Alternate Equations for 2D Equilibrium
Although we will typically use the most common equations for 2D
equilibrium shown below, the alternate equations are sometimes useful
(especially in Chapter 6 when we analyze truss, frames, and machines)
Most common equations for 2D equilibrium
 Fx  0
 Fy  0
 M O  0 (for any point O)
Alternate equations of 2D equilibrium
 Fx  0
 MA  0
 MB  0 (A and B not on a vertical line)
 Fy  0
 MA  0
 M B  0 (A and B not on a horizontal line)
 MA  0
 MB  0
 M C  0 (A, B, and C not on a line)
Note that with any of these sets of three equations, we will typically have
three unknowns on our FBD.
TWO-FORCE MEMBERS & THREE FORCE-MEMBERS (Section 5.4)
When a member is subjected to only two forces:
- We refer to the member as a “2-force member”
- The forces must be axial (along the axis of A and B below)
- The forces at A and B must be equal and opposite
When a member is subjected three or more forces:
- it is sometimes referred to as a “multi-force member”
- represent the reactions as indicated in Table 5-1
TWO-FORCE MEMBERS & THREE FORCE-MEMBERS (Section 5.4)
Sketch a FBD for member AB below in each case.
100 lb
EXAMPLE OF TWO-FORCE MEMBERS
In the cases above, members AB can be considered as two-force
members, provided that their weight is neglected.
This fact simplifies the equilibrium
analysis of some rigid bodies since the
directions of the resultant forces at A and B
are thus known (along the line joining
points A and B).
THE PROCESS OF SOLVING RIGID BODY EQUILIBRIUM PROBLEMS
Procedure for 2D problems:
1) Draw a free-body diagram (FBD) showing all the external (active
and reactive) forces.
a) refer to Table 5-1 for reactions for common supports
b) identify any 2-force members and represent the reactions as axial
forces
2) Apply the equations of equilibrium to solve for any unknowns.
a) The most common set of equations are:
 Fx  0
 Fy  0
 M O  0 (for any point O)
b) See earlier slide for other possible sets of equations
c) We are generally solving for 3 unknowns with these 3 equations
IMPORTANT NOTES
1. If there are more unknowns than the number of independent
equations, then we have a statically indeterminate situation.
We cannot solve these problems using just statics.
2. The order in which we apply equations may affect the
simplicity of the solution. For example, if we have two
unknown vertical forces and one unknown horizontal force,
then solving  FX = 0 first allows us to find the horizontal
unknown quickly.
3. If the answer for an unknown comes out as a negative
number, then the sense (direction) of the unknown force is
opposite to that assumed when starting the problem.
EXAMPLE
Given: The 4kN load at B of
the beam is supported
by pins at A and C .
Find: The support reactions
at A and C.
Plan:
1. Put the x and y axes in the horizontal and vertical directions,
respectively.
2. Determine if there are any two-force members.
3. Draw a complete FBD of the boom.
4. Apply the E-of-E to solve for the unknowns.
EXAMPLE (Continued)
FBD of the beam:
AY
4 kN
1.5 m
1.5 m
AX
A
C
45°
B
FC
Note: Upon recognizing CD as a two-force member, the number of
unknowns at C are reduced from two to one. Now, using the three
equations of equilibrium:
MA = FC sin 45  1.5 – 4  3 = 0
Fc = 11.31 kN or 11.3 kN
FX = AX + 11.31 cos 45 = 0;
AX = – 8.00 kN
FY = AY + 11.31 sin 45 – 4 = 0;
AY = – 4.00 kN
Note that the negative signs means that the reactions have the
opposite direction to that shown on FBD.
Example – 2D Equilibrium
Given: The jib crane is supported by
a pin at C and rod AB. The
load has a mass of 2000 kg
with its center of mass located at G.
Assume x = 5 m.
Find: Support reactions at B and C.
Example – 2D Equilibrium
For the frame and loading shown,
determine the reactions at C and D.
Ex: 2D Equilibrium (Fundamental Problem F5-4)
The 25-kg bar has a center of mass at G. If it
is supported by a smooth peg at C, a roller at
A, and cord AB, determine the reactions at
these supports.
3D Equilibrium
Applying the equations of equilibrium in 3D is similar to what we have
just covered in 2D except:
- We need to become familiar with common 3D reactions (Table 5-2)
- We will have 6 equations of equilibrium instead of 3!
What sort of reactions do you think that we will have for each of the 3D
supports below?
Ball-and-socket joint
Journal bearing
Pinned connection
APPLICATIONS – 3D Equilibrium
The tie rod from point A is used to support the overhang at the entrance
of a building. It is pin connected to the wall at A and to the center of the
overhang B.
If A is moved to a lower position D, will the force in the rod change
or remain the same? By making such a change without understanding
if there is a change in forces, failure might occur.
APPLICATIONS – 3D Equilibrium
The floor crane,
which weighs 300 lb,
is supporting a oil
drum.
How do you
determine the largest
oil drum weight that
the crane can support
without overturning ?
Support Reactions in 3D (Table 5-2)
memorize this table!
The 3D reactions shown below are the ones shown in Table 5-2 in the text. Recall that
in general:
1) if a support prevents translation of a body in a given direction, then a force is
developed on the body in the opposite direction.
2) if rotation is prevented, a couple moment is exerted on the body in the opposite
direction
IMPORTANT NOTE
A single bearing or hinge can prevent rotation by providing a resistive
couple moment. However, it is usually preferred to use two or more
properly aligned bearings or hinges. Thus, in these cases, only force
reactions are generated and there are no moment reactions created.
EQUATIONS OF EQUILIBRIUM (Section 5.6)
As stated earlier, when a body is in equilibrium, the net force
and the net moment equal zero, i.e.,
 F = 0 and  MO = 0
In 2D the equations above resulted in three scalar equations.
However, in 3D the result is six scalar equations.
FX = 0
FY = 0
FZ = 0
MX = 0
MY = 0
MZ = 0
6 equations for
3D equilibrium
Note: The moment equations can be determined about any point.
Usually, choosing the point where the maximum number of
unknown forces are present simplifies the solution. Any forces
occurring at the point where moments are taken do not appear in the
moment equation since they pass through the point.
CONSTRAINTS AND STATICAL DETERMINACY (Section 5.7)
This FBD has
8 unknowns!
Redundant Constraints: When a body has more supports than
necessary to hold it in equilibrium, it becomes statically indeterminate.
A problem that is statically indeterminate has more unknowns than
equations of equilibrium. Additional techniques will be introduced in
Mechanics of Materials for dealing with such problems.
IMPROPER CONSTRAINTS
We have 6 unknowns in the FBD above, but there is nothing restricting
rotation about the AB axis.
In some cases, there may be as many
unknown reactions as there are
equations of equilibrium.
M
A
0
However, if the supports are not
properly constrained, the body may
become unstable for some loading cases.
EXAMPLE – 3D Equilibrium
Given:The rod, supported by thrust
bearing at A and cable BC, is
subjected to an 80 lb force.
Find: Reactions at the thrust
bearing A and cable BC.
Plan:
a)
b)
c)
d)
Establish the x, y and z axes.
Draw a FBD of the rod.
Write the forces using scalar equations.
Apply scalar equations of equilibrium to solve for the
unknown forces.
EXAMPLE (continued)
FBD of the rod:
Applying scalar equations of equilibrium in appropriate order, we get
 F X = AX = 0 ;
So
AX = 0
 F Z = AZ + FBC – 80 = 0 ;
 M Y = – 80(1.5) + FBC (3.0) = 0 ;
So FBC = 40 lb, AZ = 40 lb
Note: We used the vector equation  MA = 0. Since the x, y, and z axes go through
point A, we reduced this to three equations summing moments about the x, y, and z
axes. If we had picked point B, the axes would be different (x’, y’, and z’). We picked
point A since the largest number of forces went though this point.
EXAMPLE (continued)
FBD of the rod:
= 40 lb
Continuing to write scalar equations of equilibrium:
M X = MAX + 40(6) – 80(6) = 0 ;
MZ = MAZ = 0 ;
MAZ= 0
MAX= 240 lb ft
Example – 3D Equilibrium
A rod is supported by smooth
journal bearings at A, B, and
C. Assume the rod is properly
aligned.
Find the reactions at all the
supports for the loading
shown.
Example – 3D Equilibrium
The 200 x 200-mm square plate shown
has a mass of 25 kg and is supported
by three vertical wires. Determine the
tension in each wire.
Ex: 3D Equilibrium (Problem 5-84) Determine the largest weight of the
oil drum that the 300 lb floor crane
can support without overturning.
Also find the vertical reactions at
wheels A, B, and C.
Example – 3D Equilibrium
A small door weighing 16 lb is attached by hinges A and
B to a wall and is held in the horizontal position shown
by rope EFH. The rope passes around a small,
frictionless pulley at F and is tied to a fixed cleat at H.
Assuming that the hinge at A does not exert any axial
thrust, determine the tension in the rope and the
reactions at A and B.
z