CHAPTER 4
Equilibrium of Rigid Body
1
Chapter Objectives
i.
ii.
iii.
To develop the equations of equilibrium for a
rigid body.
To introduce the concept of the free-body
diagram for a rigid body.
To solve rigid body equilibrium problems
using the equations of equilibrium
2
Outline






Application
Definition
Free Body Diagram
Free Body Diagram exercises
Reaction supports
Two-force & 3-force
3
APPLICATIONS
A 200 kg platform is suspended off an oil rig.
How do we determine the force reactions
at the joints and the forces in the cables?
How are the idealised model and
the free body diagram used to do this?
Which diagram above is the idealised
model?
4
APPLICATIONS (continued)
A
B
A steel beam is used to support roof joists.
How can we determine the support reactions at A & B?
Again, how can we make use of an idealized model
and a free body diagram to answer this question?
5
Definition
Static equilibrium for a rigid body:
A body (or any part of it) which is currently
stationary will remain stationary if the
resultant force and resultant moment are zero
for all the forces and couples applied on it.
6
CONDITIONS FOR RIGID-BODY EQUILIBRIUM
In contrast to the forces on a particle, the
forces on a rigid-body are not usually
concurrent and may cause rotation of the
body (due to the moments created by the
forces).
Forces on a particle
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 and  MO = 0
Forces on a rigid body
7
How can we make use of an idealized model and a free body
diagram to answer this question?
8
How are the idealized model and the free body diagram
used to do this?
Which diagram above is the idealized model and free body
diagram?
9
Idealized model
FBD
10
4.2 Free Body Diagram
In solving a problem concerning the equilibrium of a rigid body, it is essential
to consider all of the forces acting on the body.
It is equally important to exclude any forces which is not directly applied to the
body.
Omitting a force or adding an extraneous one would destroy the conditions
of equilibrium.
Therefore, the first step in the solution of the problem should be to draw
a free body diagram of the rigid body under consideration.
Free body diagram have already been used on many occasions in chapter 2.
11
Sample Photo
A free body diagram of the tractor shown would include
all of the external forces acting on the tractor:
the weight of the tractor
the weight of the load in the bucket
the forces exerted by the ground on the tires
In chapter 6, we will discuss how to determine the internal
forces in structures made of several connected pieces,
such the forces in the members that support the bucket
of the tractor
12
Procedure for drawing an FBD
1.
2.
-
Draw outlined shape.
Isolate the body from its constraints and connections.
Show all forces.
Identify all external forces and couple moments that
act on the body.
Place each force and couple at the point that it is
applied
13
Procedure for drawing an FBD
3. Identify each loading and give dimensions.
- The forces and couple moments that are
known should be labelled with their
magnitudes and directions.
14
SUPPORT REACTIONS IN 2-D
A few examples are shown above.
As a general rule, if a support prevents translation of a
body in a given direction, then a force is developed on
the body in the opposite direction. Similarly, if rotation
is prevented, a couple moment is exerted on the body.
15
Free Body Diagram


No equilibrium problem should be solved
without first drawing the free-body diagram.
Internal forces are never shown on the FBD
since they occur in equal but opposite collinear
pairs and therefore cancel out.
16
Free Body Diagram (cont)


The weight of a body is an external force,
and its effect is shown as a single resultant
force acting through the body’s centre of
gravity.
Couple moments can be placed anywhere on
the FBD since they are free vectors. Forces
can act at any point along their lines of action
since they are sliding vectors.
17
4.3 REACTIONS AT SUPPORT AND CONNECTIONS
FOR A TWO DIMENTIONAL STRUCTURE
1.Reactions Equivalent to a Force with known Line of Action
Support and connections causing reactions of this type include rollers, rockers, frictionless surface,
short links and cables, collars on frictionless rods, and frictionless pins in slots.
Each of these supports and connections can prevent motion in one direction only.
They are shown in figure 4.1 together with the reactions they produce.
Each of these reactions involves one unknown, namely the magnitude of the reaction;
this magnitude should be denoted by an appropriate letter.
The line of action of the reaction is known and should be indicated clearly in the free body diagram.
2.Reactions Equivalent to a Force of Unknown Direction and Magnitude
Support and connections causing reactions of this type include frictionless pins in fitted holes, hinges, and
rough surface.
The can prevent translation of the free body in all directions, but they cannot prevent the body from rotating
about the connection.
Reactions of this group involve two unknowns and are usually represented by their x and y components.
In the case of rough surface, the component normal to the surface must be directed away from the surface
, and thus is directed toward the free body diagram.
18
4.3 REACTIONS AT SUPPORT AND CONNECTIONS
FOR A TWO DIMENTIONAL STRUCTURE
3. Reactions Equivalent to a Force and a Couple
These reactions are caused by fixed supports, which oppose any motion of the free body
and thus constrain it completely.
Fixed supports actually produce forces over the entire surface of contact;
these forces, however, form a system which can be reduced to a force and couple.
Reactions of this group involve three unknowns, consisting usually of the two components
of the force and the moment of the couple.
19
Support of Connection
Reaction
Figure 4.1
Number of
Unknowns
1
Reactions at support and connections
Rocker
Rollers
Frictionless
surface
Force with known
line of action
1
Short link
Short cable
Force with known
line of action
1
Collar on
frictionless rod
Frictionless pin in slot
Force with known
line of action
2
Frictionless pin
or hinge
Rough surface
Force of unknown
direction
3
Fix support
Force and couple
20
Reactions at Supports and Connections for a Two-Dimensional Structure
• Reactions equivalent to a
force with known line of
action.
21
Reactions at Supports and Connections for a Two-Dimensional Structure
• Reactions equivalent to a
force of unknown direction
and magnitude.
• Reactions equivalent to a
force of unknown
direction and magnitude
and a couple.of unknown
magnitude
22
Picture shows the rocker expansion bearing of a plate
girder bridge.
The convex surface of the rocker allows the support
of the girder to move horizontally.
The abutment mounted rocker bearing shown is used
to support the roadway of a bridge.
23
4.4 EQUILIBRIUM OF RIGID BODY IN TWO DIMENTIONS
Q
P
• For all forces and moments acting on a twodimensional structure,
Fz  0 M x  M y  0 M z  M O
S
• Equations of equilibrium become
Py
Px
Qy
Qx
Sy
 Fx  0  Fy  0  M A  0
where A is any point in the plane of the
structure.
Sx
• The 3 equations can be solved for no more
than 3 unknowns.
Ax
A
B
• The 3 equations can not be augmented with
additional equations, but they can be replaced
 Fx  0  M A  0  M B  0
24
Equations of Equilibrium
(a) Equilibrium of forces
R = ΣF = 0
ΣFx = 0
ΣFy = 0
25
Equations of Equilibrium
(b) Equilibrium of moment
ΣMO = 0
ΣMA = 0
ΣMB = 0
26
4.5 STATICALLY INDETERMINATE REACTION, PARTIAL CONSTRAINS
• More unknowns than
equations
• Fewer unknowns than • Equal number unknowns
and equations but
equations, partially
improperly constrained
constrained
27
Sample problems 4.1:
A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate.
It is held in place by a pin at A and a rocker at B.
The centre of gravity of the crane is located at G.
Determine the components of the reactions at A and B.
28
Solution
To determine force B,
calculate moments about point A and the
summation of moment is equal to zero
+ ΣMA = 0
+B (1.5) – (9.81k)(2) – (23.5k)(6) = 0
B = + 107.1 kN
 B = 107.1 kN →
Ay
Ax
23.5kN
9.81kN
29
2400 x 9.8 = 23.5kN
Solution
To determine horizontal
component of A,
we use horizontal equilibrium
+→ΣFX = 0;
AX + B = 0, B = 107.1
Ay
Ax
23.5kN
9.81kN
AX + 107.1kN = 0
AX
= -107.1 kN
AX
= 107.1 kN ←
30
2400 x 9.8 = 23.5kN
Solution
To determine vertical
component of A,
we use vertical equilibrium
+↑ΣFY = 0
AY – 9.81k – 23.5k = 0
AY = + 33.3kN
Ay
Ax
23.5kN
AY = 33.3 kN↑
9.81kN
31
2400 x 9.8 = 23.5kN
Solution
To find force A,
A2 = AX2 + AY2
A=
A = 112.2
θ
AX = 107.1
AY = 33.3
AX  AY
A=
2
2
(107.1) 2  (33.3) 2
A = 112.2 kN
tan θ =
33.3
107.1
θ = 17.3
 A = 112.2 kN
17.3
32
Check
To check the answer, we
calculate the moment about
B
+ ΣMB = -(9.81kN)(2m)
-(23.5 kN)(6m)
+(107.1kN)(1.5m)
33.3kN
=0
107.1kN
107.1kN
23.5kN
9.81kN
33
Sample problems 4.2
P
27kN
27kN
1.8m
0.9m
0.6m
0.6m
Three loads are applied to a beam as shown.
The beam is supported by a roller at A and by a pin at B.
Neglecting the weight of the beam,
determine the reactions at A and B when P = 70kN.
34
Solution
Free-Body Diagram
P
27kN
27kN
A free body diagram of the beam
is drawn. The reaction at A is vertical
and is denoted by A.
The reaction at B is represented by
components Bx and By.
1.8m
0.9m
0.6m
70kN
0.6m
Each component is assumed to act
in the direction shown
27kN
27kN
Bx
A
By
1.8m
0.9m
0.6m
0.6m
35
Solution
70kN
27kN
27kN
Bx
A
By
1.8m
0.9m
0.6m
0.6m
Equilibrium Equations
We write the following three equilibrium equations and solve for the reactions indicated :
Bx = 0
36
Solution
70kN
27kN
27kN
Bx
A
By
1.8m
0.9m
0.6m
0.6m
Equilibrium Equations
; -(70N)(0.9m) + By(2.7m) – (27 kN)(3.3m) – (27kN)(3.9m) = 0
By = +95.33 kN
By = 95.33 kN
37
Solution
70kN
27kN
27kN
Bx
A
By
1.8m
0.9m
0.6m
0.6m
Equilibrium Equations
=0;
- A(2.7m) + (70N)(1.8m) – (27 kN)(0.6m) – (27kN)(1.2m) = 0
A = +28.67 kN
A = 28.67 kN
38
Check
70kN
27kN
27kN
Bx
A
By
= +28.67kN
0
= +95.33kN
1.8m
0.9m
0.6m
0.6m
The result are checked by adding the vertical components of all of the
external forces
= +28.67kN - 70kN
+95.33kN - 27kN - 27kN = 0
39
Sample problem 4.3
A loading car is at rest on a track forming an angle
of 25˚ with the vertical.
The gross weight of the car and its load is 25kN, and
it is applied at a point 750mm from the track, halfway
between the two axles.
The car is held by a cable attached 600mm from the track.
Determine the tension in the cable and the reaction at each
pair of wheels.
625mm
40
Solution
Free-Body Diagram
A free body diagram of the car
is drawn. The reaction at each wheel is perpendicular
to the track, and the tension force T is parallel to the
track. For convenience, we choose the x axis parallel
to the track and the y axis perpendicular to the track.
625mm
625mm
25 kN
The 25 kN weight is then resolved into x and y components.
Wx = + (25 kN) cos 25˚ = +22.65 kN
Wy = - (25 kN) sin 25˚ = -10.5 kN
10.5 kN
625mm
625mm
150mm
25˚
22.65 kN
25 kN
22.65 kN
10.5 kN
41
Equilibrium Equations
We take moments about A to eliminate T and R1
from the computation
625mm
625mm
- (10.5 kN)(625mm) – (22.65 kN)(150mm) + R2 (1250mm) = 0
R2 = + 8 kN
R2 = + 8 kN
Now, taking moments about B to eliminate T and R2
from the computation, we write
10.5 kN
625mm
625mm
150mm
=0;
22.65 kN
(10.5 kN)(625mm) – (22.65 kN)(150mm) – R1 (1250mm) = 0
R1 = + 2.5 kN
R1 = 2.50 kN
42
Equilibrium Equations
The value of T is found by writing
625mm
+ 22.65 kN – T = 0
T = + 22.65
625mm
T = 22.65 kN
10.5 kN
625mm
150mm
22.65 kN
625mm
43
10.5 kN
625mm
150mm
22.65 kN
Check
625mm
22.65 kN
The computations are verified by writing
2.50 kN
10.5 kN
625mm
150mm
22.65 kN
+2.50 kN + 8 kN -10.5 kN = 0
625mm
8 kN
44
PROBLEMS
45
Problem 4.1
6m
4.3m
0.5m
0.4m
The boom on a 4300kg truck is used to unload a pallet of singles
of mass 1600 kg. Determine the reaction at each of the two
(a) rear wheels B
(b) front wheel C
46
Solution
A = 1600kg
6m
A
truck = 4300kg
15˚
4.3m
0.5m
0.4m
free body diagram
47
free body diagram
6m
A
42.183 kN
15˚
15.696 kN
4.3m
0.5m
0.4m
48
free body diagram
6m
A
42.183 kN
15˚
15.696 kN
4.3m
0.5m
0.4m
49
Check
42.183 kN
15.696 kN
24.266 kN
33.616 kN
50
Problem 4.3
1.7 m
2.7 m
1.8 m
1.2 m
0.75 m
Two crates each having a mass of 110 kg, are placed as shown in the bed of a 13.5 kN
pick up truck. Determine the reactions at each of the two
(a) rear wheels A,
(b) front wheels B
51
Solution
Free body diagram
1.7 m
2.7 m
1.8 m
1.2 m
0.75 m
52
Free body diagram
1.7 m
2.7 m
1.8 m
1.2 m
0.75 m
53
Check
Free body diagram
1.7 m
2.7 m
1.8 m
1.2 m
0.75 m
WC
2FA
WD
W
2FB
54
Problem 4.9
Four boxes are placed on a uniform 14kg, wooden plank which rests on two sawhorses.
Knowing that the masses of boxes B and D are 4.5kg and 45kg, respectively,
Determine the range of values of the mass of box A so that the plank remains in equilibrium
when box C is removed.
55
45kg
4.5kg
WA
WB
A
WD
B
E
E=0
D
G
WG
F
F
Free body diagram for MA (min)
56
45kg
4.5kg
WA
WB
A
WD
B
E
D
G
E
WG
F
F=0
Free body diagram for MA (max)
57
4.6 EQUILIBRIUM OF A TWO FORCE BODY
• Consider a plate subjected to two forces F1 and F2
• For static equilibrium, the sum of moments about A
must be zero. The moment of F2 must be zero. It
follows that the line of action of F2 must pass
through A.
• Similarly, the line of action of F1 must pass
through B for the sum of moments about B to be
zero.
• Requiring that the sum of forces in any direction be
zero leads to the conclusion that F1 and F2 must
have equal magnitude but opposite sense.
58
EQUILIBRIUM OF A TWO FORCE
BODY
• When a body is subjected to forces at two
points on the body.
• No couple moment.
• These forces maintain the force
equilibrium provided they are equal in
magnitudes and opposite in direction.
59
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). 60
4.7 EQUILIBRIUM OF A THREE FORCE BODY
• Consider a rigid body subjected to forces acting at
only 3 points.
• Assuming that their lines of action intersect, the
moment of F1 and F2 about the point of intersection
represented by D is zero.
• Since the rigid body is in equilibrium, the sum of the
moments of F1, F2, and F3 about any axis must be
zero. It follows that the moment of F3 about D must
be zero as well and that the line of action of F3 must
pass through D.
• The lines of action of the three forces must be
concurrent or parallel.
61
EQUILIBRIUM OF A THREE
FORCE BODY
• If a member is subjected to only 3 forces,
then it is necessary that the forces be
either concurrent of parallel for equilibrium
condition.
• All 3 forces must have lines of action that
intersect at one point.
62
Sample problems 4.6
A man raises a 10kg joist, of length 4m, by pulling on a rope.
Find the tension T in the rope and the reaction at A.
63
Solution
Free-body diagram
The joist is a three-force body, since it is acted
upon by three forces;
it’s weight W,
the force T exerted by the rope,
and the reaction R of the ground at A.
We note that,
W = mg = (10kg)( 9.81 m/s2) = 98.1 N
64
Solution
α= 58.6˚
65
T = 81.9 N
R = 147.8 N
58.6˚
66
Newton’s Law
Newton’s First Law
If the resultant force acting on a particle is zero, the particle will remain at rest
or will move with constant speed in a straight line.
Newton’s Second Law
If the resultant force acting on a particle is not zero, the particle will have an
acceleration proportional to the magnitude of the resultant and in the direction
of this resultant force.
Newton’s Third Law
The forces of action and reaction between bodies in contact have the
same magnitude, same line of action, and opposite sense.
67
THE END
68