1. DNT 122 – APPLIED MECHANICS
CHAPTER 4 :
EQUILIBRIUM OF A RIGID BODIES
BY : MS FARAH HANAN BINTI MOHD FAUDZI

2. CHAPTER OBJECTIVES
To developed the equations of equilibrium for a rigid body
To introduce the concept of the free-body diagram for rigid body
To show how to solve rigid-body equilibrium problems using the equations of equilibrium

3. CHAPTER OUTLINE Conditions for Rigid Equilibrium Free-Body Diagrams
Equations of Equilibrium (2D)
Equilibrium in Three Dimensions
Constraints for a Rigid Body

4. 4.1 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

5. 4.2 FREE-BODY DIAGRAM
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

6. 4.2 FREE-BODY DIAGRAM

7. 4.2 FREE-BODY DIAGRAM

8. 4.2 FREE-BODY DIAGRAM

9. 4.2 FREE-BODY DIAGRAM 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

10. 4.2 FREE-BODY DIAGRAM 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

11. 4.2 FREE-BODY DIAGRAM 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

12. 4.2 FREE-BODY DIAGRAM External and Internal Forces
A rigid body is a composition of particles, both external and internal forces may act on it
Weight and Center of Gravity
When a body is subjected to gravity, each particle has a specified weight
Idealized Models
Results that approximate as closely as possible the actual situation.

13. THE PROCESS OF SOLVING RIGID BODY EQUILIBRIUM PROBLEMS
For analyzing an actual physical system, first we need to create an idealized model.
Then we need to draw a free-body diagram showing all the external (active and reactive) forces.
Finally, we need to apply the equations of equilibrium to solve for any unknowns.

14. PROCEDURE FOR DRAWING A FREE BODY DIAGRAM (Section 5.2)
Idealized model
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 the external forces and couple moments. These typically include: a) applied loads, b) support reactions, and, c) the weight of the body.

15. PROCEDURE FOR DRAWING A FREE BODY DIAGRAM (Section 5.2) (continued)
Idealized model
Free body diagram
3. Label loads and dimensions: 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.

16. 4.2 FREE-BODY DIAGRAM Example 4.2 Draw the free-body diagram of
the foot lever. The operator
applies a vertical force to the
pedal so that the spring is
stretched 40mm and the force
in the short link at B is 100N.

17. 4.2 FREE-BODY DIAGRAM Solution
Free-Body Diagram Idealized model of the lever
Pin support at A exerts components Ax and Ay on the lever, each force with a known line of action but unknown magnitude
Link at B exerts a force 100N acting in the direction of the link
Spring exerts a horizontal force on the lever
Fs = ks = 5N/mm(40mm) = 200N
Operator’s shoe apply a vertical force F on the pedal
Compute the moments using the dimensions on the FBD
Compute the sense by the equilibrium equations

18. CLASS EXERCISES
Draw a FBD of the dumpster D of the truck,which has a weight of 25kN and a center of gravity at G. It is supported by a pin at A and the hydraulic cylinder BC (treat as a short link).

19. PROBLEM SOLVING
1.5 m 1m 3m

20. 4.3 Equations of Equilibrium (2D)
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

21. 4.3 Equations of Equilibrium (2D)
Example 4.3.1
Determine the horizontal and vertical
components of reaction for the beam loaded.
Neglect the weight of the beam in the
calculations.

22. 4.3 Equations of Equilibrium (2D)
Solution
FBD
600N force is represented by its x and y components
200N force acts on the beam at B and is
independent of the
force components
Bx and By, which
represent the effect of
the pin on the beam

23. 4.3 Equations of Equilibrium (2D)
Solution
Equations of Equilibrium
Summing forces at x directions

24. 4.3 Equations of Equilibrium (2D)
Solution
A direct solution of Ay can be obtained by applying ∑MB = 0 about point B
Forces 200N, Bx and By all create zero moment about B

25. 4.3 Equations of Equilibrium (2D)
Solution
Checking,

26. 4.3 Equations of Equilibrium (2D)
Example 4.3.2
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.

27. 4.3 Equations of Equilibrium (2D)
Solution
FBD of the cord and pulley
Principle of action: equal but opposite reaction observed in the FBD
Cord exerts an unknown load
distribution p along part of
the pulley’s surface
Pulley exerts an equal but
opposite effect on the cord

28. 4.3 Equations of Equilibrium (2D)
Solution
FBD of the cord and pulley
Easier to combine the FBD of the pulley and
contracting portion of the cord so that the
distributed load becomes internal to the system
and is eliminated from the
analysis

29. 4.3 Equations of Equilibrium (2D)
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

30. 4.3 Equations of Equilibrium (2D)
Solution

31. 4.4 Equilibrium in Three Dimensions
Support Reactions

32. 4.4 Equilibrium in Three Dimensions
Support Reactions

33. 4.4 Equilibrium in Three Dimensions
Support Reactions

34. 4.4 Equilibrium in Three Dimensions
Support Reactions

35. 4.4 Equilibrium in Three Dimensions
Example

36. 4.4 Equilibrium in Three Dimensions
Example

37. 4.4 Equilibrium in Three Dimensions
CLASS EXERSICE
SOLUTION

38. 4.4 Equilibrium in Three Dimensions
Equations of Equilibrium
Vector Equations - For two conditions for equilibrium of a rigid body in vector form,
∑F = 0
∑MO = 0
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

39. 4.4 Equilibrium in Three Dimensions
Equations of Equilibrium
Scalar Equations - If all the applied external forces and couple moments are expressed in Cartesian vector form
∑F = ∑Fxi + ∑Fyj + ∑Fzk = 0
∑MO = ∑Mxi + ∑Myj + ∑Mzk = 0
i, j and k components are independent from one another

40. 4.4 Equations of Equilibrium (3D)
Six scalar equilibrium
∑Fx = 0
∑Fy = 0
∑Fz = 0
-shows that the sum of the external force
components acting in the x, y and z directions
must be zero
∑Mx = 0
∑My = 0
∑Mz = 0
-shows that the sum of the moment
components about the x, y and z axes to be
zero

41. 4.5 Constraints for a Rigid Body
To ensure the equilibrium of a rigid body, it is necessary to satisfy the equations equilibrium and have the body properly held or constrained by its supports
Redundant Constraints
More support than needed for equilibrium
Statically indeterminate: more unknown loadings on the body than equations of equilibrium available for their solution

42. 4.5 Constraints for a Rigid Body
Redundant Constraints
Example
For the 2D and 3D problems, both are statically indeterminate because of additional supports reactions
In 2D, there are 5 unknowns but 3 equilibrium equations can be drawn

43. 4.5 Constraints for a Rigid Body
Redundant Constraints
Example
In 3D, there are 8 unknowns but 6 equilibrium equations can be drawn
Additional equations
involving the physical
properties of the body
are needed to solve
indeterminate problems

44. 4.5 Constraints for a Rigid Body
Improper Constraints
Instability of the body caused by the improper constraining by the supports
In 3D, improper constraining occur when the support reactions all intersect a common axis
In 2D, this axis is perpendicular to the plane of the forces and appear as a point
When all reactive forces are concurrent at this point, the body is improperly constrained

45. 4.5 Constraints for a Rigid Body
Improper Constraints
Example
From FBD, summation of moments about the x axis will not be equal to zero, thus rotation occur
In both cases,impossible to solve completely for the unknowns

46. 4.5 Constraints for a Rigid Body
Example 4.5.1
The homogenous plate has a mass of 100kg
and is subjected to a force and couple
moment along its edges. If it is supported in
the horizontal plane by means of a roller at
A, a ball and socket joint
at B, and a cord at C,
determine the components
of reactions at the supports.

47. 4.5 Constraints for a Rigid Body
Solution
FBD
Five unknown reactions acting on the plate
Each reaction assumed to act in a positive coordinate direction

48. 4.5 Constraints for a Rigid Body
Solution
Equations of Equilibrium
Moment of a force about an axis is equal to the product of the force magnitude and the perpendicular distance from line of action of the force to the axis
Sense of moment determined from right-hand rule

49. 4.5 Constraints for a Rigid Body
Solution
Forces that parallel to an axis or pass through it create no moment about the axis

50. 4.5 Constraints for a Rigid Body
Solution
Solving,
Az = 790N Bz = -217N TC = 707N
The negative sign indicates Bz acts downward
The plate is partially constrained since the supports cannot prevent it from turning about the z axis if a force is applied in the x-y plane

51. 4.5 Constraints for a Rigid Body
Example 5.15
Determine the tension in cables BC and BD and the reactions at the ball-and-socket joint A for the mast shown in figure

52. 4.5 Constraints for a Rigid Body
Solution (Vector Analysis)
FBD
Equations of equilibrium

53. 4.5 Constraints for a Rigid Body
Solution (Vector Analysis)

54. 4.5 Constraints for a Rigid Body
Solution (Vector Analysis)
Summing moments at point A
Evaluating the cross product