1. CHAPTER 4
Equilibrium of Rigid Body

2. 　　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 solve rigid body equilibrium problems using the equations of equilibrium

3. Outline Application Definition Free Body Diagram
Free Body Diagram exercises
Reaction supports
Two-force & 3-force

4. 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?

5. APPLICATIONS (continued)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?

6. 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.

7. 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

8. How can we make use of an idealized model and a free body diagram to answer this question?

9. 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?

10. Idealized model
FBD

11. 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.

12. 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

13. Procedure for drawing an FBD
1. 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

14. 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.

15. 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.

16. 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.

17. 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.

18. 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.

19. 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.

20. Frictionless pin in slot
Rollers
Rocker
Support of Connection
Frictionless
surface
Short cable
Reaction
Short link
Collar on
frictionless rod
Frictionless pin in slot
Rough surface
Frictionless pin
or hinge
Fix support 1
Force and couple
Force of unknown
direction
Force with known
line of action
Number of
Unknowns 2 3
Figure 4.1
Reactions at support and connections

21. Reactions at Supports and Connections for a Two-Dimensional Structure
Reactions equivalent to a force with known line of action.

22. 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

23. 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.

24. For all forces and moments acting on a two-dimensional structure,
4.4 EQUILIBRIUM OF RIGID BODY IN TWO DIMENTIONS
For all forces and moments acting on a two-dimensional structure, P Q S Py Px Qy Qx A Ax B Sy Sx
Equations of equilibrium become
where A is any point in the plane of the structure.
The 3 equations can be solved for no more than 3 unknowns.
The 3 equations can not be augmented with additional equations, but they can be replaced

25. Equations of Equilibrium
(a) Equilibrium of forces
R = ΣF = 0
ΣFx = 0 ΣFy = 0

26. Equations of Equilibrium
(b) Equilibrium of moment
ΣMO = 0
ΣMA = 0 ΣMB = 0

27. More unknowns than equations
4.5 STATICALLY INDETERMINATE REACTION, PARTIAL CONSTRAINS
More unknowns than equations
Fewer unknowns than equations, partially constrained
Equal number unknowns and equations but improperly constrained

28. 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.

29. 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 = kN
 B = kN →
23.5kN
9.81kN Ay Ax
2400 x 9.8 = 23.5kN

30. +→ΣFX = 0; AX + B = 0, B = 107.1 AX + 107.1kN = 0 AX = -107.1 kN
Solution
To determine horizontal component of A,
we use horizontal equilibrium
+→ΣFX = 0;
AX + B = 0, B = 107.1
AX kN = 0
AX = kN
AX = kN ←
23.5kN Ay Ax
9.81kN
2400 x 9.8 = 23.5kN

31. To determine vertical component of A, we use vertical equilibrium
Solution
To determine vertical component of A,
we use vertical equilibrium
+↑ΣFY = 0
AY – 9.81k – 23.5k = 0
AY = kN
AY = 33.3 kN↑
23.5kN Ay Ax
9.81kN
2400 x 9.8 = 23.5kN

32. To find force A, A2 = AX2 + AY2 A = A = 112.2 kN tan θ = θ = 17.3
Solution
To find force A,
A2 = AX2 + AY2
A =
A = kN
tan θ =
θ = 17.3
 A = kN 
AX = 107.1
AY = 33.3
A = 112.2 θ

33. To check the answer, we calculate the moment about B
+ ΣMB = -(9.81kN)(2m)
-(23.5 kN)(6m)
+(107.1kN)(1.5m)
= 0
23.5kN
107.1kN
9.81kN
33.3kN

34. P 27kN Sample problems 4.2 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.

35. P Solution Free-Body Diagram Bx A By 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.
Each component is assumed to act
in the direction shown
27kN
70kN
1.8m
0.9m
0.6m A By Bx

36. Bx = 0 Solution Bx A By Equilibrium Equations 70kN 27kN 1.8m 0.9m
We write the following three equilibrium equations and solve for the reactions indicated :
Bx = 0

37. By = +95.33 kN By = 95.33 kN Solution Bx A By Equilibrium Equations
; -(70N)(0.9m) + By(2.7m) – (27 kN)(3.3m) – (27kN)(3.9m) = 0
By = kN
By = kN

38. = 0 ; A = +28.67 kN A = 28.67 kN Solution Bx A By
1.8m
0.9m
0.6m A By Bx
Equilibrium Equations
= 0 ;
- A(2.7m) + (70N)(1.8m) – (27 kN)(0.6m) – (27kN)(1.2m) = 0
A = kN
A = kN

39. = +28.67kN - 70kN +95.33kN - 27kN - 27kN = 0 Check Bx By A 70kN 27kN
1.8m
0.9m
0.6m Bx By
= kN A
= kN
The result are checked by adding the vertical components of all of the
external forces =
+28.67kN
- 70kN
+95.33kN
- 27kN
- 27kN
= 0

40. 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.
750mm
625mm
600mm

41. Solution Free-Body Diagram Wx = + (25 kN) cos 25˚ = +22.65 kN
600mm
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.
The 25 kN weight is then resolved into x and y components.
625mm
750mm
625mm
25 kN
Wx = + (25 kN) cos 25˚ = kN
Wy = - (25 kN) sin 25˚ = kN
25˚
22.65 kN
10.5 kN
25 kN
150mm
10.5 kN
625mm
22.65 kN
625mm

42. = 0 ; Equilibrium Equations 10.5 kN 22.65 kN
600mm
Equilibrium Equations
We take moments about A to eliminate T and R1
from the computation
625mm
750mm
625mm
- (10.5 kN)(625mm) – (22.65 kN)(150mm) + R2 (1250mm) = 0
R2 = + 8 kN
150mm
625mm
22.65 kN
10.5 kN
Now, taking moments about B to eliminate T and R2
from the computation, we write
(10.5 kN)(625mm) – (22.65 kN)(150mm) – R1 (1250mm) = 0
R1 = kN
R1 = 2.50 kN
= 0 ;

43. Equilibrium Equations
600mm
Equilibrium Equations
The value of T is found by writing
kN – T = 0
T =
T = kN
625mm
750mm
625mm
150mm
625mm
22.65 kN
10.5 kN

44. The computations are verified by writing
150mm
625mm
22.65 kN
10.5 kN
Check
22.65 kN
150mm
625mm
22.65 kN
10.5 kN
The computations are verified by writing
+2.50 kN + 8 kN kN = 0
2.50 kN
8 kN

45. PROBLEMS

46. The boom on a 4300kg truck is used to unload a pallet of singles
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
rear wheels B
(b) front wheel C

47. Solution A = 1600kg truck = 4300kg A 15˚ 6m 4.3m 0.5m 0.4m
free body diagram

48. free body diagram 6m
4.3m
0.5m
0.4m A
15˚ kN kN

49. free body diagram 6m
4.3m
0.5m
0.4m A
15˚ kN kN

50. Check kN kN kN kN

51. Problem 4.3
1.7 m
1.2 m
1.8 m
2.7 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
rear wheels A,
(b) front wheels B

52. Solution
Free body diagram
1.7 m
1.2 m
1.8 m
2.7 m
0.75 m

53. Free body diagram
1.7 m
1.2 m
1.8 m
2.7 m
0.75 m

54. Check
Free body diagram
1.7 m
1.2 m
1.8 m
2.7 m
0.75 m WC
2FA WD W
2FB

55. 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.

56. Free body diagram for MA (min)
4.5kg
45kg WA WG WB WD A
E=0 F E G D B
Free body diagram for MA (min)

57. Free body diagram for MA (max)
4.5kg
45kg WA WG WB WD A E
F=0 G D B F
Free body diagram for MA (max)

58. Consider a plate subjected to two forces F1 and F2
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.

59. 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.

60. 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).

61. Consider a rigid body subjected to forces acting at only 3 points.
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.

62. 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.

63. 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.

64. W = mg = (10kg)( 9.81 m/s ) = 98.1 N 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/s ) = 98.1 N 2

65. Solution
α= 58.6˚

66. T = 81.9 N
R = N
58.6˚

67. 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.

68. THE END