2. Chapters:
1. Static force analysis: Introduction to static equilibrium. Conditions for static equilibrium of a
member under the action of two forces and three forces and its Free body diagram.Conditions for static
equilibrium of a member under the action of two forces and a couple, four force member and its Free
body diagram. Slider crank mechanisms, Analysis of a four-bar mechanism. Mechanisms involving plate
links. Analysis of Toggle mechanisms such as a stone crusher mechanism. Analysis of mechanisms
under the action of more than one force and more than one couple. Analysis of mechanisms involving
gears. Analysis of mechanisms involving cam. Analysis of mechanisms involving eight links
AdminAnalysis of drag link Mechanism, Analysis of Bell Crank Lever. Analysis of mechanisms
considering sliding friction when the forces are concurrent and non-concurrent.
2. Inertia forces: Determination of inertia forces and indicating them on various links of four bar
mechanism and slider crank mechanism. Determination of inertia forces and indicating them on various
links of slider crank mechanism. Crank effort diagram or Turning moment diagram. Types of Flywheel.
Determination of flywheel size for single cylinder I. C. engine. Determination of flywheel size for Multi-
cylinder I. C. engine and press work. Numericals on single and multi cylinder engines. Determination of
flywheel size for press work. Numericals on flywheel size for press work.
3. References:
1. Ballaney P. L., Theory of Machines and Mechanism, Khanna Publications, New-Delhi, 2009.
2. Rattan S. S, Theory of Machines, Tata Mc-Graw Hill Publishers Pvt. Ltd, New-Delhi,2009.
3. Singh V. P., Theory of Machines, Dhanpat Rai & Co. (P) Ltd, New-Delhi, 2010.
4. Robert L. Norton, Design of Machinery: An Introduction to the Synthesis and Analysis
ofMechanisms and Machines, McGraw-Hill, 2012.
5. Rao J.S., Rao V. Dukkipati, Mechanism and Machine Theory, New Age International
Publishers Year, 2012.
6. Sharma C. S. and Purohit Kamlesh, Theory of Mechanisms and Machines, PH Publishing
Pvt. Ltd., 2006.
4. Theory of Machines (TOM) is divided into two parts:-
Kinematics of Machinery: Study of motion of the components and basic geometry ofthe
mechanism and is not concerned with the forces which cause or affect motion. Study includes the
determination of velocity and acceleration of the machine members
Dynamics of Machinery: Analyses the forces and couples on the members of themachine due to
external forces (static force analysis) also analyses the forces and
Couples due to accelerations of machine members (Dynamic force analysis)
Deflections of the machine members are neglected in general by treating machine members as rigidbodies
(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 analyzed for its
role in transmitting force.
The forces associated with the principal function of the machine are usually known or assumed
5. Force analysis:
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.
Under estimation: 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
Graphical analysis of machine forces will be used here because of the simplification it offers to a problem,
especially in cases of complex machines. Moreover, the graphical analysis of forces is a direct application
of the equations of equilibrium
6. 0. A `particle’ is a small mass at some position in space.
1. When the sum of the forces acting on a particle is zero, its
velocity is constant;
2. The sum of forces acting on a particle of constant mass is
equal to the product of the mass of the particle and its acceleration;
3. The forces exerted by two particles on each other are equal
in magnitude and opposite in direction.
Fundamentals: Specifically, forces are defined through Newton’s laws of motion
Force Vector:
A force that acts on a point of a link carries the index of the point. For
example FP .
7. In planar systems, the moment of a forceabout an arbitrary
point is a moment vector along an axis perpendicular to the
plane (zaxis).
For example, the moment of the force FCabout O is a moment
in the positive z-direction (CCW) with a magnitude
MO = h×FC
where h is the distance from O to the axis of the force, also
called the moment arm. In the second example, the moment
of FB about O is a moment in the negative z-direction (CW)
with a magnitude
MO = h×FB
The direction of a moment can be determined using our
right-hand—the thumb would indicate the direction of the
moment when the other four fingers are curled about the
point in the direction of the force.
8. Force couples and torques:
Two parallel forces, equal in magnitude and opposite in direction,
acting on twodifferent points of a link form a couple. Themoment of
a couple, called a torque, is a vector in the z-direction and its
magnitude is
T = h×F
where h is the distance between the two axes and F = FA = FC is the
magnitude of either force.
The positive or negative direction of the torque can be determined
based on the right-hand method.
9. Common forces and torques:
Forces and torques (moments) that act on a link can be the result of gravity, springs, dampers,actuators,
friction, etc. These forces and torques can also be the result of reaction forces orreaction torques from
other links. These forces and torques can be categorized as applied,reaction, and friction.
Applied forces and torques
These are either known constants (gravity for example), or functions of positions (springs), or functions
of positions and velocities (dampers).
Reaction forces and torques
These are functions of applied forces/torques in static problems, and functions of the
appliedforces/torques and accelerations in dynamics problems.
Friction forces and torques
These may appear in machines as viscous (wet) or Coulomb (dry). Viscous friction dependson velocities;
therefore it can be categorized as an applied force/torque. Coulomb frictiondepends on reaction forces and
possibly velocities; therefore it can be categorized as a reaction force/torque.
10. Reaction forces and Torques:
Two links connected by a kinematic joint apply reaction forces
(and/or torques) on one another.
Pin joint
Two links connected by a pin joint apply reaction forces on each
other. The reaction forces are equal in magnitude and opposite in
direction. The magnitude anddirections that are shown on the free-
body-diagrams arearbitrary—they must be determined through an
analysis.
The reaction force on each link can be represented inters of its x and y
components, such as Fji( x) and Fji( y) .
For notational simplicity, we will use a single index to show each
component; for example, F1 and F2 . If the assigned direction to a
component is determined to be correct through an analysis, the
solution for that componentwill come out with a positive sign.
Otherwise the solutionwill end up with a negative sign indicating
that the assumeddirection for the component must be reversed.
11. Static Force Analysis:
A body is said to be in static equilibrium if under a set of applied forces
and torques its translational (linear) and rotational accelerations are
zeros (a body could be stationary or inmotion with a constant linear
velocity).
Planar static equilibrium equations for a single body that is acted
upon by forces and torques are expressed as
Equation (i) represents the sum of all the forces acting on the link, and
Eq. (ii) represents the sum of all the torques, Tj , and the moments, Mi ,
caused by all the forces acting on the bodywith respect to any reference
point.
equation)
ent
......(Mom
...(ii)...
..........
0.........
M
T
equation)
(Force
..(i).....
..........
0
F
0
F
0
F
i
j
i(y)
i(x)
i
12. Equilibrium Of Two Force Members
A member under the action of two forces will be in equilibrium if
– the forces are of the same magnitude,
– the forces act along the same line, and
– the forces are in opposite directions.
Figure shows such a member.
A member under the action of three forces will be in equilibrium if
– the resultant of the forces is zero, and
– the lines of action of the forces intersect at a point (known as point of concurrency).
13. Equilibrium Of Three Force Members
If only three forces act on a body that is in static equilibrium, their axes
intersect at a single point. This knowledge can help us simplify the solution
process in some problems. For example, if the axes of two of the forces are
known, the intersection of those two axes can assist us in determining the axis
of the third force. A special case of the three-force member is when three
forces meet at a pin joint that is connected between three links. When the
system is in static equilibrium, the sum of the three forces must be equal to
zero.
Figure (a) shows a member acted upon by three forces F1, F2 and
F3 and is in equilibrium as the lines of action of forces intersect at onepoint
O and the resultant is zero.
This is verified by adding the forces vectorially [Fig.(b)].
As the head of the last vector F3 meets the tail of the first vector F1,
the resultant is zero.
Figure (d) shows a case where the magnitudes and directions of
theforces are the same as before, but the lines of action of the forces do
not intersect at one point.
Thus, the member is not in equilibrium.
14. A member under the action of two forces and an applied
torque will be in equilibrium if
– the forces are equal in magnitude, parallel in direction and
opposite in sense and
– the forces form a couple which is equal and opposite to the
applied torque.
Figure shows a member acted upon by two equal forces
F1, and F2 and an applied torque T for equilibrium,
where T, F1 and F2 are the magnitudes of T, F1 and F2 respectively.
T is clockwise whereas the couple formed by F1, and F2
is counterclockwise.
Member with two forces and a torqque:
h
F
h
F
T
2
1
15. Equilibrium Of Four Force Members
A four-force member is completely
solvable if one force is known completely in
magnitude and direction along with lines of action
of the other three forces. The conditions of
equilibrium as stated above are sufficient.
Consider a system of four non-parallel forces
as shown in Fig.(a). Let O1, be the point of
intersection of the lines of action of F1 and F2.
Similarly, O2 is the point of intersection of the
lines of action of the forces F3 and F4. Join 0102.
The resultant of F1 and F2 and that of F3 and F4 is
parallel to 0102. The force polygon for the four
forces can be drawn as shown in Fig.(b) and the
forces F3 and F4 can be known completely.
16. Graphical force analysis:
Graphical force analysis employs scaled free-body diagrams and vector graphics in the determination of
unknown machine forces. The graphical approach is best suited for planar force systems. Since forces
are normally not constant during machine motion. analyses may be required for a number of mechanism
positions; however, in many cases, critical maximum-force positions can be identified and graphical
analyses performed for these positions only. An important advantage of the graphical approach is that it
provides useful insight as to the nature of the forces in the physical system.
This approach suffers from disadvantages related to accuracy and time. As is true of any graphical
procedure, the results are susceptible to drawing and measurement errors. Further, a great amount of
graphics time and effort can be expended in the iterative design of a machine mechanism for which
fairly thorough knowledge of force-time relationships is required. In recent years, the physical insight of
the graphics approach and the speed and accuracy inherent in the computer-based analytical approach
have been brought together through computer graphics systems, which have proven to be very effective
engineering design tools. There are a few special types of member loadings that are repeatedly
encountered in the force analysis of mechanisms, These include a member subjected to two forces, a
member subjected to three forces, and a member subjected to two forces and a couple. These special
cases will be considered in the following paragraphs, before proceeding to the graphical analysis of
complete mechanisms.
18. Free body Diagrams:
Engineering experience has demonstrated the importance and usefulness of free-body diagrams in
force analysis. A free-body diagram is a sketch or drawing of part or all of a system, isolated in
order to determine the nature of forces acting on that body. Sometimes a free-body diagram may
take the form of a mental picture; however, actual sketches are strongly recommended, especially
for complex mechanical systems. Generally, the first, and one of the most important, steps in a
successful force analysis is the identification of the free bodies to be used.
19. Analysis (Graphical):
• Member 4 is acted upon by three forces F, F34
and F14.
• Member 3 is acted upon by two forces F23 and
F43
• Member 2 is acted upon by two forces F32 and
F12 and a torque T.
• Initially, the direction and the sense of some of
the forces may not be
known.
• Link 3 is a two-force member and for its
equilibrium F23 and F43 must
act along BC.
• Thus, F34 being equal and opposite to F43 also
acts along BC.
20. • Assume that the force F on member 4 is known
completely.
• To know the other two forces acting on this
member completely, the
direction of one more force must be known.
• For member 4 to be in equilibrium, F14 passes
through the intersection
of F and F34 .
• By drawing a force triangle (F is completely
known), magnitudes of
F14 and F34 can be known [Fig.(e)].
Now F34 = F43 = F23 = F32
• Member 2 will be in equilibrium if F12 is equal,
parallel and opposite
to F32 and
