1. DEFINITIONS
Dynamics of Machinery
Analyses the forces and couples on the members of the machine due to external forces (static force analysis), also analyses the forces and couples due to accelerations of machine members ( Dynamic force analysis)
Rigid Body
Deflections of the machine members are neglected in general by treating machine members as rigid bodies (also called rigid body dynamics). In other words the link must be properly designed to withstand the forces without undue deformation to facilitate proper functioning of the system.
In order to design the parts of a machine or mechanism for strength, it is necessary to determine the forces and torques acting on individual links. Each component however small, should be carefully analysed for its role in transmitting force.
The forces associated with the principal function of the machine are usually known or assumed.

3. FORCES IN MACHINE SYSTEMS
A machine system is considered to be a system of an arbitrary group of bodies (links), which will be considered rigid. We are involved with different types of forces in such systems. Note that when the word "force" is discussed it will refer to a "generalised force" which will also include moments.
a) Joint (Reaction) Forces: are commonly called the joint forces in machine systems since the action and reaction between the bodies involved will be through the contacting kinematic elements of the links that form a joint. The joint forces are along the direction for which the degree-of-freedom is restricted. e.g. in constrained motion direction. For example, consider a revolute joint in a planar mechanism. In such a joint there is a rotational freedom and any moment along the axis of the revolute joint will not be transmitted from one link to the other, but there will be a force transmission in any general direction which will be determined by the forces acting on the links. If the revolute joint is in a spatial mechanism, there will be moment reaction components perpendicular to the revolute joint axis and a force reaction in a general direction (e.g. there will be three force components and two moment components). In case of a prismatic joint in a planar mechanism there will be no reaction force component along the axis of the slide but a force perpendicular to the slider axis and a couple along the z-axis will be transmitted between the links joined.

4. b) Physical Forces : As the physical forces acting on a rigid body we shall include external forces applied on the rigid body, the weight of the rigid body, driving force, or forces that are transmitted by bodies that are not rigid such as springs or strings attached to the rigid body. In case of springs, the magnitude and the direction of the force acting will depend on the geometry of the mechanism at the instant considered. In case of a string, when the string is tight, the force will be in the direction of the string and its sense must be such that it keeps the string in tension; otherwise the string force will be zero.
In machine systems, if the force distribution within the rigid body considered is not our concern, the weight of the rigid body can be considered to be equivalent to a force applied at the center of gravity of the rigid body, in the sense and direction of the gravity field.
c) Friction or Resisting Force: In general the resisting forces are those that result due to motion and which resist the motion. Since the rigid body assumption is made, one can neglect the internal friction forces that will exist within the body. In such a case friction forces are at the joints in the direction of the relative motion but in opposite sense or in the members that are specially designed to create the friction force (dampers). Friction forces will be discussed in more detail in coming section .
d) Inertial Forces. Are the forces due to the inertia of the rigid bodies involved. These forces will be discussed in the coming sections.

5. FORCE ANALYSIS
Apart from static forces, mechanism also experiences inertia forces when subjected to
acceleration, called dynamic forces.
Static forces are predominant at lower speeds and
dynamic forces are predominant at higher speeds.
Here, the analysis is aimed at determining the forces transmitted from one point to another,
essentially from input point to output point. This would be the starting point for strength
design of a component/ system, basically to decide the dimensions of the components
Force analysis is essential to avoid either overestimation or under estimation of forces on
machine member.
Underestimation: leads to design of insufficient strength and to early failure.
Overestimation: machine component would have more strength than required.
Over design leads to heavier machines, costlier and becomes not competitive

6. General Principle of force analysis:
A machine / mechanism is a three dimensional object, with forces acting in three dimensions.
For a complete force analysis, all the forces are projected on to three mutually perpendicular planes. Then, for each reference plane, it is necessary that, the vector sum of the applied forces in zero and that, the moment of the forces about any axis perpendicular to the reference plane or about any point in the plane is zero for equilibrium.

8. TWO FORCE MEMBER

9. 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.

11. 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.

12. THREE FORCE MEMBER

14. TWO FORCE and ONE MOMENT (TORQUE) MEMBER
F1 = F2 =F and T= F x h

16. FORCES BETWEEN MEMBERS

18. The all Gravity forces (mg) were neglected compared to the Joint forces.

19. -

21. One can reduce the number of equations to be solved if the free-body diagrams are analysed to some detail. One need not write the forces in terms of its x and y components if the direction is known and one can identify the forces that are of equal magnitude before attempting for a solution. The free-body diagrams of the links in the four-bar mechanism are redrawn below.

22. In this case to simplify the calculations we note that Fij = -Fji for the joint forces. Furthermore, since link 3 is a two-force member, F23 and F43 are equal, opposite and their line of action is along AB. Hence F23 =F23 <q13, and q13 is known from the kinematic analysis. Also link 2 is a two-force plus a moment member. Therefore: F13 = - G12
Hence:
F 43= -F32 =-F34= -G12= -F23
Now, one can solve for the unknown forces if we write the 3 equilibrium equations for link 4 and one moment equilibrium equation for link 2, which are
There are four equations with four unknowns (F34, T12, G14,F14 or F34, T12, G14x, G14y). If the magnitudes come out negative, the direction of the force or torque is opposite to that indicated on the free-body diagram.

23. VECTORAL CALCULATION OF MOMENTy x
Where;
F is the magnitude of F and v is a unit vector in the direction of the force F.
r is the distance from point C to a point on the line of action of F and u is a unit vector in the direction of r.
r must be directed from the point that moment will be taken according to this point,toward a point at which force is applied.

26. Example
Figure shows a slider crank mechanism in which the resultant gas pressure 8 x 104 N/m2 acts on the piston of cross sectional area 0.1 m2 . The system is kept in equilibrium as a result of the couple applied to the crank 2, through the shaft at O2. Determine forces acting on all the links (including the pins) and the couple on 2.

27. Example

28. Example
Since link 3 is acted upon by only two forces, F43 and F23 are collinear, equal in magnitude and opposite in direction

29. Example

30. Example
Determine the torque T2 required to keep the given mechanism in equilibrium.

31. THE FREE BODY DIAGRAMS F34x F23x F34y F23y 3 FC=1000N A B F23y F12y
390
FC=1000N A B
F23y
F12y
F12x T2 O2 A
450 2 B
F14y
F41y
F34y 4

32. For the Connecting Rod (The member 3)
F23y
F12y
F12x T2 O2 A
450
F23x 2
F34x
F23x
F34y
F23y 3
390
FC=1000N A B B
F14y
F41y
F34y 4
F34x
For the Connecting Rod (The member 3)
For the piston (The member 4)
From the eqn.(2)
For the Crank (The member 2)

33. F23y FC=1000N F23x A 3 F34y B A A FC=1000N F34y C 3 B B
The another way (Vectoral method) to calculate the moment for the Connecting Rod (The member 3)
F23x
F34y
F23y 3
390
FC=1000N A B
To show the best, how to calculate the moment by vectorial method; the forces, position vectors ,and their unit vectors are drawn separately as shown below 3
390
FC=1000N A B C
F34y A B

34. The Second way for solution
Three force member
Two force and one moment member

35. Example

37. Example
Free Body Diagrams

38. For the member 4
We have 4 unknowns and 3 equations, we can not find them still.

39. F14x
F34x
F14y
F34y 4
P=200N B O4 C B O4
The Calculation of the moment by vector method
It gave the same result

40. The Calculation of the moment by vector method
For the member 3
The Calculation of the moment by vector method
It gave the same result
The number of the unknowns are 6:
The number of the equations are 6
Since the member 2 is two force and one Moment member

41. The Calculation of the moment by vector method
For the member 3
F34y
S=500N
F34x B 3
F23x A
F23y
The Calculation of the moment by vector method
It gave the same result
The number of the unknowns are 6:
The number of the equations are 6
Since the member 2 is two force and one Moment member

42. From the eqn (8)
From the eqn (2)
From the eqn (3)
From the eqn (1)
From the eqn (4)

43. For the Crank (The member 2)
F23y
F12y
F12x T2 O2 A
450
F23x 2
For the Crank (The member 2)

44. Example
A four link mechanism is acted upon by forces as shown in the figure. Determine the torque T2 to be applied on link 2 to keep the mechanism in equilibrium.
AD=50mm, AB=40mm, BC=100mm, Dc=75mm, DE= 35mm,

45. Example
Free Body Diagrams

46. Example
Free Body Diagrams

47. Example
Free Body Diagrams

48. Example
Free Body Diagrams

49. Example
Free Body Diagrams

50. Example: Figure shows a two-dof articulated robot having the same link dimensions. The robot is interacting with the environment surface in a horizontal plane. Obtain the joint torques 1,2 for pushing the surface with an endpoint force of F=(Fx,Fy). Assume no friction.
It had been taken from

51. Link 0
Link 1
Link 2 x y θ1 L1 L2 θ2 Fx Fy

52. 2 1 2 1 FREE BODY DIAGRAM Fy Fx Link 2 L2 A B A Link 1 L1 B A' B'
The sign of the torques is reverse since they are reaction torques.