Theory of
Machine
INTRODCUTION
• Science of Mechanics is a branch of
engineering science which deals with study of
relative motion between the various parts of
machine and forces which act on them
• Statics – analysis of stationary system
• Dynamics- systems that change with time
• Kinetics-Which deals with inertia forces
which arise from combined effect of the
mass and motion of the machine parts
• Kinematics-Relative motion of a body
without consideration of forces causing the
motion
• It deals with geometry of the motion and
concepts like displacement velocity and
acceleration as a function of time.
Mechanism
• A mechanism is a combination of rigid or
restraining bodies so shaped and connected
that they move upon each other with
definite relative motion ex: slider crank
mechanism
Slider crank Mechanism
Machine
• A machine is a mechanism or collection of
mechanism which transmits force from
the source of power to the load to
overcome and thus performs a useful
work
• A system can be defined as a mechanism
or a machine on the basis of its primary
objective
Machine
• Transfer and transform the motion ( without
consideration of the forces involved ) the
systems said to be a mechanism if the system
is used to transferring mechanical energy then
it is called a machine
• Mechanical work is always associated with
movement
• Every machine is a mechanism but not vice
versa.
https://www.youtube.com/watch?v=O9tfIfwlmz8
Plane and space mechanism
(a) Plane Mechanism If
all points of
mechanism move in a
parallel planes
• Simplest type of
motion and is along a
straight line path
(b) Space mechanism
All the points of the
mechanisms do not
move in a parallel
Types of Constrained Motions
• 1. Completely constrained motion
• When the motion between a pair is
limited to a definite direction
irrespective of the direction of force
applied, then the motion is said to be a
completely constrained motion.
2. Incompletely constrained motion
When the motion between a pair can
take place in more than one direction,
then the motion is called an incompletely
constrained motion. The change in the
direction of impressed force may alter
the direction of relative motion between
the pair.
• 3. Successfully constrained motion.
• When the motion between two elements of a
pair is possible in more than one direction
but is made to have motion only in one
direction by using some external means, it is
a successfully constrained motion.
https://www.youtube.com/watch?v=RMBsBRWX7QE
https://www.youtube.com/watch?v=FnjQEXBqYdY
Kinematic link or element
• A resistant body or a group of resistant bodies with
rigid connection preventing their relative
movement is known as a link.
• A Slider crank mechanism consists of four links:
Frame and guides , crank , connecting rod and
slider
• https://www.youtube.com/watch?v=ZO8QEG4x0wY
Types of link
• Rigid link: which does not undergo any
deformation while transmitting motion.
• Flexible link: which is partially
deformed in a manner not to affect the
transmission of motion Ex: belts, rope ,
chains etc.
• Fluid link: which is formed by having a
fluid in a receptacle and the motion is
transmitted through the fluid by pressure
ex hydraulic jacks presses and brakes
Kinematic pair:
• The two links or elements of a machine
when in contact with each other are said
to form a pair.
• If the relative motion between them is
completely constrained, the pair is
known as kinematic pair.
Classification of Kinematic Pairs
a. turning pair
b. Prismatic pair ( sliding joint)
c. Screw pair:
d. Cylindrical pair:
e. spherical pair
f. planar pair
Kinematic Pairs according to Nature of Contact
• Lower Pair: A pair of links having surface
or area contact between the members is
known as a lower pair. The contact surfaces
of the two links are similar.
• Examples: Nut turning on a screw. shaft
rotating in a bearing. all pairs of a slider-
crank mechanism, universal joint, etc.
Kinematic Pairs according to Nature of Contact
• Higher Pair: When a pair has a point or line
contact between the links. it is known as a
higher pair. The contact surfaces of the two
links are dissimilar.
• Examples: Wheel rolling on a surface, cam
and follower pair, tooth gears, ball and roller
bearings, etc.
https://www.youtube.com/watch?v=TYq_0v66ucM
Types of Joints
• 1. Binary joint: When two links are joined at the
same connection, the joint is known as binary joint.
Example Joint B
• 2. Ternary joint: When three links are joined at
the same connection, the joint is known as ternary
joint. It is equivalent to two binary joints as one of
the three links joined carry the pin for the other two
links. Example : Joint T
Types of Joints
• 1. 3. Quaternary joint:
• When four links are joined at the same
connection, the joint is called a quaternary
joint. It is equivalent to three binary joints
• In general, when l number of links are joined
at the same connection, the joint is equivalent
to (l – 1) binary joints.
Degrees of Freedom
• An unconstrained rigid body moving in space can
describe the following independent motions
• 1. Transla1ional motions along any three mutually
perpendicular axes x, y and z
• 2. Rotational motions about these axes
Thus, a rigid body possesses six degrees
of freedom. The connection of a link
with another imposes certain
constraints on their relative motion.
The number of restraints can never
be zero
• Degrees of freedom =6 - Number of
restraints
Consider a four bar chain in a plane motion, as
shown in Fig. (a). one variable such as θ is
needed to define the relative positions of all the
links. Therefore the number of degrees of
freedom of a four bar chain is one.
consider a five bar chain, as shown in Fig. (b).
In this case two variables such as θ1 and θ2 are
needed to define completely the relative
positions of all the links. Thus, we say that the
number of degrees of freedom is two.
Kutzbach criterion for the movability
of a mechanism having plane motion.
• Now let us consider a plane mechanism
with l number of links. Since in a
mechanism, one of the links is to be fixed,
therefore the number of movable links will
be (l – 1) and thus the total number of
degrees of freedom will be 3 (l – 1) before
they are connected to any other link.
• In general, a mechanism with l number of
links connected by j number of binary
joints or lower pairs (i.e. single degree of
freedom pairs) and h number of higher
pairs (i.e. two degree of freedom pairs),
then the number of degrees of freedom of
a mechanism is given by
n = 3 (l – 1) – 2 j – h (i)
If there are no two degree of freedom pairs
(i.e. higher pairs), then h = 0. Then
n = 3 (l – 1) – 2 j
Application of Kutzbach Criterion
to Plane Mechanisms
The number of degrees of freedom or
movability (n) for some simple
mechanisms as shown in Fig. are
determined as follows
1. The mechanism, as shown in Fig. (a), has
three links and three binary joints, i.e. l = 3
and j = 3.
n = 3 (3 – 1) – 2 × 3 = 0
2. The mechanism, as shown in Fig. (b), has
four links and four binary joints, i.e. l = 4
and j = 4.
n = 3 (4 – 1) – 2 × 4 = 1
Simillary calculate the degrees of freedom
for the rest of the planar motion
(a)When n = 0, then the mechanism forms a
structure and no relative motion between
the links is possible,
(b) When n = 1, then the mechanism can be
driven by a single input motion,
(c) When n = 2, then two separate input
motions are necessary to produce constrained
motion for the mechanism.
(d) When n = – 1 or less, then there are
redundant constraints in the chain and it
forms a statically indeterminate structure.
Grubler’s Criterion for Plane Mechanisms
• The Grubler’s criterion applies to mechanisms with
only single degree of freedom joints.
• Substituting n = 1 and h = 0 in Kutzbach equation,
we have
1 = 3 (l – 1) – 2 j or 3l – 2j – 4 = 0
This equation is known as the Grubler's criterion for
plane mechanisms with constrained motion.
The simplest possible mechanisms of this type are a
four bar mechanism and a slider-crank mechanism in
which l = 4 and
j = 4.
https://www.youtube.com/watch?v=vOFM8eG8kVc
Types of Kinematic Chains
1. Four bar chain or quadric cyclic chain,
2. Single slider crank chain, and
3. Double slider crank chain.
Four Bar Chain or Quadric Cycle Chain
4- the crank
3- coupler
2- lever
1- frame
https://www.youtube.com/watch?v=KBFFwgCCP0U
Grashof ’s law for a four bar
mechanism,
The sum of the shortest and longest link
lengths should not be greater than the sum of
the remaining two link lengths.
Shortest link + largest link <= sum of the other two
links
A very important consideration in designing a
mechanism is to ensure that the input crank
makes a complete revolution relative to the
other links
In a four bar chain, one of the links, in particular the
shortest link, will make a complete revolution
relative to the other three links, if it satisfies the
Grashof’s law. Such a link is known as crank or
driver.
AD (link 4 ) is a crank. The link BC (link 2) which
makes a partial rotation or oscillates is known as
lever or rocker or follower and the link CD (link
3) which connects the crank and lever is called
connecting rod or coupler. The fixed link AB (link
1) is known as frame of the mechanism.
https://www.mekanizmalar.com/
http://www.mechanisms.co/four-bar-mechanism.html
L2+L1<L3+L4
Shortest link beside the fixed
https://www.youtube.com/watch?v=4tIo3AQQiU8
L1 shortest
L1+L2<L3+L4
https://www.youtube.com/watch?v=9F3RLX4Ps-I
Shortest link is opposite to
the fixed link (L1) is the
one making a full turn
https://www.youtube.com/watch?v=NvOwwRX7KXI
If each two links have the same
length
https://www.youtube.com/watch?v=zMa4cbnBQWY
https://www.youtube.com/watch?v=h8bz4ni6mdY
Single Slider Crank Chain
• A single slider crank chain is a modification of the
basic four bar chain. It consist of one sliding pair
and three turning pairs. It is usually, found in
reciprocating steam engine mechanism. This type of
mechanism converts rotary motion into
reciprocating motion and vice versa.
β
https://www.youtube.com/watch?v=s3G3au-EyAQ
Displacement
Velocity and Acceleration
Analysis
Velocity
Determine “how fast” parts of a machine are
moving.
Linear Velocity (v):
Important when concerned with the timing of
a mechanism. v B
B
Magnitude:
Straight line, instantaneous speed
of a point. A
Direction: v
A
The instantaneous direction of
movement.
Angular Velocity (w):
vB
Magnitude: B
Instantaneous speed of the
rotation of a link.
Direction: 2
vA
The instantaneous A
angular direction (cw or
w2
ccw) of movement.
To calculate angular velocity: d
w
dt
Linear and Angular Velocity:
v v = rw
B
B
rB The directions of v and
v
2
w are always
A
A rA consistent.
w2
w must be in radians.
Reiterating:
Points have linear velocity !
Links have rotational velocity !
Example:
If the speed of point B is 50 m/s, determine the
velocity of point A, B and link 2, if the link is moving
clockwise.
B
18 cm
A 2
8 cm
600
450
Relative-Motion Analysis: Velocity
• The general plane motion of a rigid body can be a
combination of translation and rotation
• We use a relative-motion analysis involving two sets
of coordinate axes
• X’, Y’ coordinate system has a known motion
• The axes of this coordinate
system will only be allowed
to translate with respect to
the fixed frame
Relative Velocity:
vA vB
The velocity of a point, as seen
from another point.
Velocity of B relative to A (vB/A)
vB/A = vB vA
Example: B
The speed of point B is 50 ft/s,
down and to the right. Determine 18 in
A
velocity of A relative to B.
8 in
600
450
Relative-Motion Analysis: Velocity
Position
• Position vector rA specifies the location of the “base
point” A, and the rB/A locates point B with respect to
point A
• By vector addition, the position of B is
Displacement rB rA rB / A
• A and B undergo
displacements drA and drB
Relative-Motion Analysis: Velocity
Displacement
• Due to the rotation about A, drB/A = rB/A dθ, and the
displacement of B is
drB drA drB / A
due to rotation about A
due to translation about A
due to translation and rotation
Relative-Motion Analysis: Velocity
Velocity
• To determine the relationship between the
velocities of points A and B,
drB drA drB / A
dt dt dt
Relative-Motion Analysis: Velocity
Velocity
• vB is determined by considering the entire body to
translate with a velocity of vA, and rotate about A
with an angular velocity ω
• Vector addition of these two effects, applied to B,
yields vB
• vB/A represents the effect of circular motion, about
A. It can be expressed by the cross product
v B v A w rB / A
Example:
The link is guided by two block A and B, which move in
the fixed slots. If the velocity of A is 2 m/s downward,
determine the velocity of B at the instant θ = 45°.
Solution
Vector Analysis
Kinematic Diagram
Since points A and B are restricted to move and vA is
directed downward, vB must be directed horizontally to
the right. This motion causes the link to rotate CCW
Apply the velocity equation v B v A v B / A
to points A and B in order to
solve for the two unknown
magnitudes vB and ω
Solution
Vector Analysis
Velocity Equation
We have
v B v A w rB / A
vB i 2 j [wk (0.2 sin 45 i 0.2 cos 45 j)]
vB i 2 j 0.2w sin 45 j 0.2w cos 45 i
Equating the i and j components gives
vB 0.2w cos 45
0 2 0.2w sin 45 w 14.1 rad/s , vB 2 m/s
Example
The link is guided by two block A and B, which move
in the fixed slots. If the velocity of C is 2 m/s
downward, determine the velocity of B at the instant
θ = 45°.
Solution
Vector Analysis
Kinematic Diagram
CB and AB rotate counterclockwise.
Velocity Equation
Link CB (general plane motion):
v B v C wCB rB / C
vB i 2 j wCBk (0.2i 0.2 j)
vB i 2 j 0.2wCB j 0.2wCBi
vB 0.2wCB
0 2 0.2wCB
wCB 10 rad/s and vB 2 m/s
Solution
Velocity Equation
Link AB (rotation about a fixed axis)
v B w AB rB
2i w AB k (0.2 j)
2 0.2w AB
w AB 10 rad/s
Same results can be obtained using Scalar Analysis.
Velocity Analysis
Relative Velocity Method
Two points on a rigid body can only
have a relative velocity:
Perpendicular to the line that connects them.
B
B
A “the motion of B,
as seen from A”
Graphical Velocity Analysis
The relative velocity equation permits the creation of a
vector triangle. vB = vA +> vB/A
Mechanism and vector triangle can be drawn on CAD to
determine unknown velocities
vA vB/A
vB
vB/A
B vA
vB = vA +> vB/A
A
vB
Problem for student :
The 2 inch crank is rotating at 300 rpm,
clockwise. Determine the velocity of the
piston.
650
2 in 8 in
1.5 in
Instantaneous Center of Zero Velocity
• Velocity of any point B located on a rigid body can be
obtained if base point A has zero velocity
• Since vA = 0, therefore vB = ω x rB/A.
• Point A is called the instantaneous center of zero velocity
(IC) and it lies on the instantaneous axis of zero velocity .
Magnitude of vB is ωrB/IC
• Due to circular motion,
direction of vB must be
perpendicular to rB/IC
Instantaneous Center of Zero Velocity
• For wheel rolling without slipping, the point of contact
with the ground has zero velocity
• Hence this point represents the IC for the wheel
Location of the IC
• Velocity of a point on the body is always perpendicular
to the relative-position vector extending from the IC to
the point
• IC is located along the line drawn perpendicular to vA,
distance from A to the IC
is rA/IC = vA/ω
Instantaneous Center of Zero Velocity
Location of the IC
• Construct A and B line segments that are
perpendicular to vA and vB.
• Extending these perpendicular to their point of
intersection as shown locates the IC
• IC is determined by proportional triangles
Instantaneous Center of Zero Velocity
Procedure for Analysis
• Motion can be determined with reference to
instantaneous center of zero velocity
• The body is imagined as “extended and pinned” at
the IC where it rotates about this pin with its ω
• Magnitude of velocity can be
determined by using v = ωr
• Velocity has a sense of direction
Example
Show how to determine the location of the
instantaneous center of zero velocity for (a) member BC
shown in Fig 1; and (b) the link AB shown in Fig 2.
Fig 1
Solution
Part (a)
vB is caused by the clockwise rotation of link AB.
vB is perpendicular to AB and acts at an angle θ from the
horizontal.
Motion of point B causes the
piston to move horizontally
with a velocity vC.
When the line are drawn
perpendicular to vB and vC,
the intersect at the IC.
Example 16.10
Solution
Part (b)
Points B and C follow circular paths of motion since rods
AB and DC are each subjected to rotation about a fixed
axis.
Since the velocity is tangent to the path, vC and vB are
directed vertically downward, along CB.
rC / IC rB / IC
wCB (vC / rC / IC ) 0
Fig 2.
Solution
Part (b)
As a result, rod CB momentarily translates.
CB will move and cause the instantaneous center to
move to some finite location.
Relative-Motion
Analysis:Acceleration
• The velocity v B v A w rB / A
• An equation that relates the accelerations of two
points on a rigid body subjected to general plane
motion,
d vB d v A d vB/ A
dt dt dt
• aB/A can be expressed in terms of its tangential
and normal components of motion
a B a A (a B / A )t (a B / A )n
Relative-Motion
Analysis:Acceleration
• The terms can be represented graphically
= +
a B a A rB / A w rB / A
2
Relative-Motion
Analysis:Acceleration
Velocity Analysis.
• Determine the angular velocity ω of the body by
using a velocity analysis
VECTOR ANALYSIS
Kinematics Diagram
• Establish the directions of the fixed x, y coordinates
and draw the kinematics diagram
• If points A and B move along curved paths, their
tangential and normal accelerations components
should be indicated
Relative-Motion
Analysis:Acceleration
Acceleration Equation.
• To apply aB = aA + α x rB/A – ω2rB/A express the
vectors in Cartesian vector form and substitute
them into the equation
• When solution yields a negative answer for an
unknown magnitude, it indicates that the sense
of direction of the vector is opposite
Relative-Motion
Analysis:Acceleration
SCALAR ANALYSIS
Kinematics Diagram
• If the equation a B a A (a B / A )t (a B / A )n is
applied, then the magnitudes and directions of the
relative-acceleration components (aB/A)t and (aB/A)n
must be established.
• To do this, draw a kinematic diagram.
Relative-Motion
Analysis:Acceleration
SCALAR ANALYSIS
Acceleration Equation.
• Represent the vectors in
a B a A (a B / A )t (a B / A )n
graphically by showing their magnitudes and
directions underneath each term. The scalar
equations are determined from the x and y
components of these vectors
Example
The crankshaft AB of an engine turns with a clockwise
angular acceleration of 20 rad/s2. Determine the
acceleration of the piston at this instant AB is in the
shown. At this instant ωAB = 10 rad/s and ωBC = 2.43
rad/s.
Solution
Kinematic Diagram
aC is vertical since C moves along a straight-line path.
Acceleration Equation
Expressing each of the position
vectors in Cartesian vector form
rB {0.25 sin 45 i 0.25 cos 45 j} m
{0.177i 0.177j} m
rC / B {0.75 sin13.6 i 0.75 cos13.6 j} m
{0.176i 0.729j} m
Solution
Crankshaft AB (rotational about a fixed axis):
a B AB rB w AB
2
rB
(20k ) (0.177i 0.177j) (10) 2 (0.177i 0.177j)
{21.21i 14.14j} m/s 2
Connecting Rod BC (general plane motion):
aC a B BC rC / B w BC
2
rC / B
aC j 21.21i 14.14j ( CBk ) (0.176i 0.729j) (2.43) 2 (0.176i 0.729j)
0 20.17 0.729 BC aC 0.176 BC 18.45
BC 27.7 rad/s 2 and aC 13.6 m/s2
• The hydraulic cylinder is extending with the velocity
and acceleration shown. Determine the angular
acceleration of crank AB and link BC at the instant
shown.
Relative-Motion
Analysis:Acceleration
SCALAR ANALYSIS
Kinematics Diagram
• If the equation a B a A (a B / A )t (a B / A )n is
applied, then the magnitudes and directions of the
relative-acceleration components (aB/A)t and (aB/A)n
must be established.
• To do this, draw a kinematic diagram.
Relative-Motion
Analysis:Acceleration
SCALAR ANALYSIS
Acceleration Equation.
• Represent the vectors in
a B a A (a B / A )t (a B / A )n
• graphically by showing their magnitudes and directions
underneath each term. The scalar equations are
determined from the x and y components of these vectors
The total acceleration can also be determined
through a graphical procedure using either CAD or
traditional drawing techniques,
