1. Chapter 4: Equilibrium of Rigid Body
Applied Mechanics

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 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
Two and Three-Force Members
Constraints for a Rigid Body

4. 4.1 Conditions for Rigid-Body Equilibrium
Consider rigid body fixed in the x, y and z reference and is either at rest or moves with reference at constant velocity
Two types of forces that act on it, the resultant internal force and the resultant external force
Resultant internal force fi is caused by interactions with adjacent particles

5. 4.1 Conditions for Rigid-Body Equilibrium
When equation of equilibrium is applied to each of the other particles of the body, similar equations will result
Adding all these equations vectorially,
∑Fi + ∑fi = 0
Summation of internal forces = 0 since internal forces between particles in the body occur in equal but opposite collinear pairs (Newton’s third law)

6. 4.1 Conditions for Rigid-Body Equilibrium
Only sum of external forces will remain
Let ∑Fi = ∑F, ∑F = 0
Consider moment of the forces acting on the ith particle about the arbitrary point O
By the equilibrium equation and distributive law of vector cross product,
ri X (Fi + fi) = ri X Fi + ri X fi = 0
Resultant moment of each pair of forces about point O is zero
Using notation ∑MO = ∑ri X Fi,
∑MO = 0

7. 4.1 Conditions for Rigid-Body Equilibrium
Equations of Equilibrium for Rigid Body
∑F = 0
∑MO = 0

8. 4.2 Free-Body Diagrams
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

9. 4.2 Free-Body Diagrams

10. 4.2 Free-Body Diagrams

11. 4.2 Free-Body Diagrams

12. 4.2 Free-Body Diagrams Support Reactions
If the support prevents the translation of a body in a given direction, then a force is developed on the body in that direction
If rotation is prevented, a couple moment is exerted on the body
Consider the three ways a horizontal member, beam is supported at the end
- roller, cylinder
- pin
- fixed support

13. 4.2 Free-Body Diagrams 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

14. 4.2 Free-Body Diagrams 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

15. 4.2 Free-Body Diagrams 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

16. 4.2 Free-Body Diagrams External and Internal Forces
A rigid body is a composition of particles, both external and internal forces may act on it
For FBD, internal forces act between particles which are contained within the boundary of the FBD, are not represented
Particles outside this boundary exert external forces on the system and must be shown on FBD
FBD for a system of connected bodies may be used for analysis

17. 4.2 Free-Body Diagrams Weight and Center of Gravity
When a body is subjected to gravity, each particle has a specified weight
For entire body, consider gravitational forces as a system of parallel forces acting on all particles within the boundary
The system can be represented by a single resultant force, known as weight W of the body
Location of the force application is known as the center of gravity

18. 4.2 Free-Body Diagrams Weight and Center of Gravity
Center of gravity occurs at the geometric center or centroid for uniform body of homogenous material
For non-homogenous bodies and usual shapes, the center of gravity will be given

19. 4.2 Free-Body Diagrams Idealized Models
Consider a steel beam used to support the roof joists of a building
For force analysis, reasonable to assume rigid body since small deflections occur when beam is loaded
Bolted connection at A will allow for slight rotation when load is applied => use Pin

20. 4.2 Free-Body Diagrams
Support at B offers no resistance to horizontal movement => use Roller
Building code requirements used to specify the roof loading (calculations of the joist forces)
Large roof loading forces account for extreme loading cases and for dynamic or vibration effects
Weight is neglected when it is small compared to the load the beam supports

21. 4.2 Free-Body Diagrams Procedure for Drawing a FBD
1. Draw Outlined Shape
Imagine body to be isolated or cut free from its constraints
Draw outline shape
2. Show All Forces and Couple Moments
Identify all external forces and couple moments that act on the body

22. 4.2 Free-Body Diagrams Example 5.1
Draw the free-body diagram of the uniform
beam. The beam has a mass of 100kg.

23. 4.2 Free-Body Diagrams
Solution
Free-Body Diagram

24. 4.2 Free-Body Diagrams Example 5.2 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.

25. 4.2 Free-Body Diagrams 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 exert vertical force F on the pedal
Compute the moments using the dimensions on the FBD
Compute the sense by the equilibrium equations

26. 4.3 Equations of Equilibrium
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

27. 4.3 Equations of Equilibrium
Example 5.6
Determine the horizontal and vertical
components of reaction for the beam loaded.
Neglect the weight of the beam in the
calculations.

28. 4.3 Equations of Equilibrium
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

29. 4.3 Equations of Equilibrium
Solution
Equations of Equilibrium
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

30. 4.3 Equations of Equilibrium
Solution

31. 4.3 Equations of Equilibrium
Solution
Checking,

32. 4.3 Equations of Equilibrium
Example 5.7
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.

33. 4.3 Equations of Equilibrium
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

34. 4.3 Equations of Equilibrium
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

35. 4.3 Equations of Equilibrium
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

36. 4.3 Equations of Equilibrium
Solution

37. 4.4 Two- and Three-Force Members
Simplify some equilibrium problems by recognizing embers that are subjected top only 2 or 3 forces
Two-Force Members
When a member is subject to
no couple moments and forces
are applied at only two points
on a member, the member is
called a two-force member

38. 4.4 Two- and Three-Force Members
Two-Force Members
Example
Forces at A and B are summed to obtain their respective resultants FA and FB
These two forces will maintain translational and force equilibrium provided FA is of equal magnitude and opposite direction to FB
Line of action of both forces is known and passes through A and B

39. 4.4 Two- and Three-Force Members
Two-Force Members
Hence, only the force magnitude must be determined or stated
Other examples of the two-force members held in equilibrium are shown in the figures to the right

40. 4.4 Two- and Three-Force Members
Bucket link AB on the back hoe is a typical example of a two-force member since it is pin connected at its end provided its weight is neglected, no other force acts on this member
The hydraulic cylinder is pin connected at its ends, being a two-force member

41. 4.4 Two- and Three-Force Members
Example 5.13
The lever ABC is pin-supported
at A and connected to a short
link BD. If the weight of the
members are negligible,
determine the force of the pin
on the lever at A.

42. 4.4 Two- and Three-Force Members
View Free Body Diagram
Solution
FBD
Short link BD is a two-force member, so the resultant forces at pins D and B must be equal, opposite and collinear
Magnitude of the force is unknown but line of action known as it passes through B and D
Lever ABC is a three-force member

43. 4.4 Two- and Three-Force Members
Solution
FBD
For moment equilibrium, three non-parallel forces acting on it must be concurrent at O
Force F on the lever at B is equal but opposite to the force F acting at B on the link
Distance CO must be 0.5m since lines of action of F and the 400N force are known

44. 4.4 Two- and Three-Force Members
Solution
Equations of Equilibrium
Solving,

45. 4.5 Equations of Equilibrium
Vector Equations of Equilibrium
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

46. 4.5 Equations of Equilibrium
Scalar Equations of Equilibrium
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

47. 4.5 Equations of Equilibrium
Scalar Equations of 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

48. 4.6 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

49. 4.6 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

50. 4.6 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

51. 4.6 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

52. 4.6 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

53. 4.6 Constraints for a Rigid Body
Improper Constraints
Proper constraining requires
- lines of action of the reactive forces do not insect points on a common axis
- the reactive forces must not be all parallel to one another
When the minimum number of reactive forces is needed to properly constrain the body, the problem is statically determinate and equations of equilibrium can be used for solving

54. 4.6 Constraints for a Rigid Body
Procedure for Analysis
Free Body Diagram
Draw an outlined shape of the body
Show all the forces and couple moments acting on the body
Establish the x, y, z axes at a convenient point and orient the axes so that they are parallel to as many external forces and moments as possible
Label all the loadings and specify their directions relative to the x, y and z axes

55. 4.6 Constraints for a Rigid Body
Procedure for Analysis
Free Body Diagram
In general, show all the unknown components having a positive sense along the x, y and z axes if the sense cannot be determined
Indicate the dimensions of the body necessary for computing the moments of forces

56. 4.6 Constraints for a Rigid Body
Procedure for Analysis
Equations of Equilibrium
If the x, y, z force and moment components seem easy to determine, then apply the six scalar equations of equilibrium,; otherwise, use the vector equations
It is not necessary that the set of axes chosen for force summation coincide with the set of axes chosen for moment summation
Any set of nonorthogonal axes may be chosen for this purpose

57. 4.6 Constraints for a Rigid Body
Procedure for Analysis
Equations of Equilibrium
Choose the direction of an axis for moment summation such that it insects the lines of action of as many unknown forces as possible
In this way, the moments of forces passing through points on this axis and forces which are parallel to the axis will then be zero
If the solution yields a negative scalar, the sense is opposite to that was assumed

58. 4.6 Constraints for a Rigid Body
Example 5.15
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 N, and a cord at C,
determine the components
of reactions at the supports.

59. 4.6 Constraints for a Rigid Body
Solution
FBD
Five unknown reactions acting on the plate
Each reaction assumed to act in a positive coordinate direction

60. 4.6 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

61. 4.6 Constraints for a Rigid Body
Solution
Components of force at B can be eliminated if x’, y’ and z’ axes are used

62. 4.6 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

63. Chapter Summary Free-Body Diagram
Draw a FBD, an outline of the body which shows all the forces and couple moments that act on the body
A support will exert a force on the body in a particular direction if it prevents translation of the body in that direction
A support will exert a couple moment on a body is it prevents rotation

64. Chapter Summary Free-Body Diagram
Angles are used to resolved forces and dimensions used to take moments of the forces
Both must be shown on the FBD

65. Chapter Summary Two Dimensions
∑Fx = 0; ∑Fy = 0; ∑MO = 0 can be applied when solving 2D problems
For most direct solution, try summing the forces along an axis that will eliminate as many unknown forces as possible

66. Chapter Review

67. Chapter Review

68. Chapter Review