Mechanics of Machines
advertisement
DEPARTMENT OF AUTOMOBILE ENGINEERING
SCHOOL OF MECHANICAL ENGINEERING
VI Semester
U4AEA09 MECHANICS OF MACHINES
LESSON NOTES
U4AEA09 MECHANICS OF MACHINES
LTPC
3104
Objective: - To make the students to understand the Structural analysis of various
Mechanical components at different load conditions, torque, & Power transmission,
velocity, acceleration, friction, Balancing & vibration.
UNIT I Mechanisms
9
Machine Structure – Kinematic link, pair and chain – Grueblers criteria – Constrained
motion – Degrees of freedom - Slider crank and crank rocker mechanisms –
Inversions – Applications – Kinematic analysis of simple mechanisms –
Determination of velocity and acceleration.(Graphical method)
UNIT II Friction
9
Surface contacts – sliding and rolling friction - Friction in screw and nut –Plate and
disc clutches – Belt (flat and V) and rope drives. Ratio of tensions –Condition for
maximum power transmission – Open and crossed belt drive.
UNIT III Gearing and cams
9
Gear profile and geometry – Nomenclature of spur and helical gears – Gear trains:
Simple, compound gear trains and epicylic gear trains - Determination of speed and
torque - Cams – Types of cams – Design of profiles – Knife edged and roller ended
followers with and without offsets for various types of follower motions
UNIT IV Balancing
9
Static and dynamic balancing – Single and several masses in different planes –
Balancing of reciprocating masses- primary balancing and concepts of secondary
balancing – Single and multi cylinder engines (Inline) – Balancing of radial V engine
– direct and reverse crank method
UNIT V Vibration
9
Free, forced and damped vibrations of single degree of freedom systems –
Force transmitted to supports – Vibration isolation – Vibration absorption – Torsional
vibration of shaft – Single and multi rotor systems – Geared shafts – Critical speed of
shaft.
TOTAL: 45+15(Tutorial) = 60 periods
TEXT BOOKS
1. Rattan.S.S, “Theory of Machines”, Tata McGraw–Hill Publishing Co, New
Delhi,2004.
2. Ballaney.P.L, “Theory of Machines”, Khanna Publishers, New Delhi, 2002.
3. R.S.Khurmi “ Theory of Machines”.
REFERENCE BOOKS
1.
Rao, J.S and Dukkipati, R.V, “Mechanism and Machine Theory”, Second
Edition, Wiley Eastern Ltd., 1992.
2.
Malhotra, D.R and Gupta, H.C., “The Theory of Machines”, Satya Prakasam,
Tech. India Publications, 1989.
3.
Gosh, A. and Mallick, A.K., “Theory of Machines and Mechanisms”, Affiliated
East West Press, 1989.
4.
Shigley, J.E. and Uicker, J.J., “Theory of Machines and Mechanisms”, McGrawHill, 1980.
UNIT – I
MECHANISMS
Mechanism:
It is a set of machine elements or components or parts, arranged in a specific
order to produce a specified motion.
Machine:
When a mechanism is used to transmit forces and couples it becomes a
machine.
Kinematics:
It is defined as the study of motion of the components and the basic geometry
of the mechanism.
Link:
All the rigid bodies forming the network (Kinematic chain) of a machine are
called links.
Element:
The part of a link which is connected to a neighbouring link is called an
element.
Kinematic pair:
When two elements belonging to two links, are connected to give relative
motion, it is called as kinematic pair.
Classification of Kinematic pairs:
(i)
According to the type of relative motion between the elements
a)
Sliding pair
b)
Turning pair
c)
Rolling pair
d)
Screw pair
e)
Spherical.
(ii)
According to the type of contact between the elements
a)
Lower pair
b)
Higher pair
According to the type of closure
a) Self closed pair
b) Force closed pair.
(iii)
Kinematic chain:
When the Kinematic pairs are coupled in such a way that the last time link is
joined to the first link to transmit definite motion, it is called a Kinematic chain.
Give the relation between the number of pairs(p) and the number of links(l) of a
Kinematic chain.
= 2p – 4
Give the relation between the number of links(l) and the number of joints(i) of a
Kinematic chain.
j=
3
2
2
'Degree of freedom' or Mobility.
It is defined as the number of input parameters which must be controlled
independently in order to bring the device into a particular position.
The Kutchzbach criterion for the movability of a mechanism having plane motion.
n= 3( – 1) – 2j – h where :
n = No of degrees of freedom
= No of links
j = No of binary joints
h = No of higher pair.
Inversion of mechanism?
It is a method of obtaining different mechanisms by fixing different links in a
Kinematic chain.
The important inversions of four bar chain.
1.
2.
3.
Crank and lever mechanism – Beam engine
Double crank mechanism – Coupling rod of locomotive
Double lever mechanism – Watt's indicator mechanism.
The inversions of single slider crank chain.
1.
2.
3.
4.
5.
Bull engine (or) Pendulum pump
Oscillating cylinder engine
Rotary internal combustion engine (or) Gnome engine
Crank and slotted lever quick return motion mech.
Whitworth quick return motion mechanism.
Double Slider Crank Chain
A Kinematic chain which consists of two turning pairs and two sliding pairs
is known as double slider crank chain.
The Inversions Of Double Slider Crank Chain.
1.
Elliptical trammel
2.
Scotch Yoke mechanism.
3.
Oldham's coupling
Pantograph
It is a device used to reproduce a displacement exactly in an enlarged or
reduced scale. It works on the principle of four bar chain mechanism.
Differentiate Kinematic pair and Kinematic chain.
Kinematic pair
1. A pair is a joint of two links or
elements of a machine.
Ex: 1.Crank and connecting road.
2.Piston and cylinder.
Kinematic chain
1. When the Kinematic pairs are coupled in such
a way that the last link is joined to the first link
to transmit definite motion, it is called a
Kinematic chain.
Ex: 1. Four bar chain.
2. Double slider crank chain.
Structure
An assembly of resistant bodies, known as members having no relative
motion between them and meant for carrying loads is called structure.
Compare Machine and Structure:
Machine
Structure
1. The different parts of a machine has
relative motion.
2. It transforms energy into useful work.
3. Links are used to transmit motion.
4. Example: Shaping m/c
1. There is no relative motion between the
members.
2. It does not convert the energy into work.
3. Members are used for carrying loads.
4. Example: bridges, roof trusses m/c
frames.
Mechanical advantage of a mechanism.
It is defined as the ratio of output torque to the input torque. It is also defined
as the ration of load to the effort.
Differents in Kinematic chain and Mechanism.
Kinematic chain
1. It is a combination
Kinematic pairs.
2. Types:
a) Four bar chain,
b) Slider crank chain
Mechanism
of
1. If any one of the link in the Kinematic chain
is fixed then the Kinematic chain is called a
mechanism.
2. Types:
a) Simple mechanism
b)Compound mechanism.
Transmission angle?
In a four bar chain mechanism, the angle between the coupler and the
follower link is called as the transmission angle.
Diffrents in mechanism and machine:
Mechanism
1.
Mechanism
transmits
modifies motion.
and
2. It is the outline of the m/c to
produce definite motion between
various links.
Machine
1.M/C transmits forces and couples.
2. Machine may have many mechanisms for
transmitting power or work.
The Grashof's law for a four bar mechanism.
Grashof's law states that the sum of the shortest and longest links cannot be
greater than the sum of the remaining two link lengths, if there is to be continuous
relative motion between the two members.
The significance of Grashof's law:
(i)
Grashof's law specifies the order in which the links are connected in a
Kinematic chain.
(ii)
It specifies, which link of the four bar chain is fixed.
(iii) A four bar chain should satisfy Grashaf's law, otherwise no link will make a
complete revolution relative to one another.
i.e. ( S + l) * (P + Q)
where
S = length of the shortest link
L = Length of the longest link
P & Q = Length of the other two links.
The working mechanism of bicycle bells:
Bicycle bells are working on the principle of snap action mechanism
(otherwise called as toggle mechanism (or) flip – flop mechanism).
The working Mechanism of screw jack.
Screw jack works on linear actuator mechanism.
Movability and Mobility
(i)
(ii)
Movability includes six degrees of freedom of the devices as a whole, as if the
ground link were not fixed and thus applies to a Kinematic chain.
Mobility neglects these and considers only the internal relative motions, thus
applying to a mechanism.
Quick return motion mechanism is the inversion of
Single slider crank chain.
Classify the 'Constrained motion', and example for each.
1. Completely constrained motion [Ex: Square bar moving in
square hole]
2. Incompletely constrained motion [Ex:circular shaft in a hole]
3. Successfully constrained motion [Ex: Motion of IC engine value]
The Grubler's criterion for plane mechanisms.
n = 3( – 1) – 2j
for single degree of freedom joints => 3
where:
= No of links
j = No of binary joints.
– 2j = 4
In a four bar chain, which link is called crank.
In a four bar chain, the smallest link, which makes complete revolution is
called crank.
In a crank and slotted lever quick return mechanism, the distance between the
fixed centres is 150mm and the driving crank is 75mm long. Determine the ratio
of the time taken on the cutting and return strokes.
Solution :
Time of cutting stroke 360
Time of return storke
length of crank
sin 90
2 length of fixed link
= 75/150 = ½
= sin –1(0.5) = 30º
2
= 90 – 30 = 60
2
= 120º
360 120
Stroke ratio =
120
= 2.
90 -
Elliptical Trammel
It is an inversion of double slider crank chain. It is used as an instrument for
drawing ellipses. It has two sliding pairs and one turning pair.
Scotch yoke mechanism and its use.
It is an inversion of double slider crank chain. It is used for converting rotary
motion into reciprocating motion.
The distance between two parallel shafts is 18mm and they are connected by an
oldham's coupling. The driving shaft revolves at 160 rpm. What will be the
maximum speed of the tongue of the intermediate piece along its groove?
Solution :
r = Distance between the area of shafts = 18 mm = 0.018m
N = 160 rpm
ω = 16.75 rad/ rad/s
Maximum sliding velocity V = r.ω
= (0.018)(16.75)
= 0.302 m/s
Toggle Position
It is the position of a mechanism at which the mechanical advantage is infinite
and the sine of angle between the coupler and driving link is zero.
Motion Adjustment Mechanism
The mechanism used to adjust or modify the motion of the link are known as
motion adjustment mechanisms. Motion adjustment is obtained by wedges, levers
and rack and pinion.
Define instantaneous centre of rotation and write the equation to determine the
number of instantaneous centres of a mechanism.
The combined motion of rotation and translation of the link AB may be
assumed to be a motion of pure rotation about some centre I, called the
instantaneous centre, which changes from instant to instant No of instantaneous
centres.
N=
n n 1
2
n = No of links
What is the magnitude of linear velocity of a point B on a link AB relative to A?
The magnitude of linear velocity of a point B on a link AB, which rotates with
W angular velocity with respect to A is;
VBA = ωBA × AB.
Instantaneous axis.
It is a line drawn through an instantaneous centre and perpendicular to the plane of
motion.
Space centrode:
The locus of the the instantaneous centre in space during a definite motion of the
body is called the space centrode.
Body centrode:
The locus of the instantaneous centre relative to the body itself is called the body
centrode.
The properties of instantaneous centre.
1. The two rigid links have no linear velocity relative to another link at the
instantaneous centre.
2. A rigid link rotates instantaneous relative to another link at the instantaneous
centre for the configuration of the mechanism considered.
The Different Types Of Instantaneous Centre
1.
2.
3.
Fixed Instantaneous centre
Permanent Instantaneous centre
Neither fixed – nor permanent Instantaneous centre.
The location of instantaneous centre for the following.
(i)
(ii)
When the two links are connected by pin joint
When the two links have a pure rolling contact.
Solution :
(i)
(ii)
When the two links are connected by pin joint, the instantaneous centre lies on
the centre of the pin.
When the two links have a pure rolling contact, the instantaneous centre lies on
their point of contact.
Kennedy's theorem.
It states that if three bodies move relative to each other, they have three
instantaneous centres and they lie on a straight line.
A pin joins two links A and B, which rotates with angular velocities W A and WB
respectively in opposite direction. What is the rubbing velocity of that pin?
Rubbin velocity = (ωA + ωB) x r
r = radius of the pin.
Coincident point.
When a point on one link is sliding along another rotating link, then the point
is known as coincident point.
The Expression For Coriolis Component Of Acceleration.
aCAB = 2VωOA
where
ωOA = Angular velocity of 'OA' and
V = Linear velocity of 'B'
The Two Components Of Acceleration
1.
Radial component of acceleration.
ArOB = 2OB × OB
2.
Tangential component of acceleration
alOB = OB × OB
where:
OB = Angular velocity of link OB
OB = Angular acceleration of link OB and
OB = length of link OB.
In a revolving stage with a speed of 3 rpm, a person is walking with a speed of 0.5
m/s along a radial path. Determine the magnitude of the coriolis component of
acceleration in this motion. – NOV 2003
Solution :
Coriolis acceleration
=2Vω
2 3
= 2 x (0.5)
60
= 0.3141 rad/sec.
In a four bar chain mechanism crank AB = 0.4m rotates with 150 rpm and lever CD
oscillates with a linear velocity of 7 m/s and what is the mechanical advantage of
the system.
Solution :
Mechanical advantage =
VAB
VDC
VAB = ωAB × AB
2N AB
=
× AB
60
2150
=
× 0.4
60
= 6.2 m/s
VDC = 7 m/s
6 .2
Mech. Advantage =
= 0.898.
7
Fixed Instantaneous Centre
In a mechanism, the instantaneous centre which remain in the same place for
all configurations of the mechanism is called as fixed instantaneous centre.
Location Of The Instantaneous Centre
The instantaneous centre lies at the centre of curvature of the curved surface.
Solved Problems
1. For the Kinematic linkages shown in fig, calculate the number of binary links,
ternary links, links and degree of freedom.
Solution :
(a) For fig (i):
binary links = 4 higher pairs (h) = 0
fig.
ternary links = 4
total links = 8
Degree of freedom,
n = 3( – 1) – 2 j – h
=3(8 – 1) – 2 x 11 – 0
n = –1.
The link is a unconstrained chain.
(b)
For fig(ii)
binary links = 4
Ternary links = 4
Total links = 8
Degree of freedom , n = 3( – 1)–2j – h
n = 3(8 – 1) – 2 x 10
n=1
The linkage is a constrained chain.
(c)
For fig (iii)
binary links = 7
Ternary links = 2
Total links = 11
Degrees of freedom, n = 3( – 1)–2j – h
=3(11 – 1) – 2 x 15
n=0
The linkage is a locked chain.
2. Determine the mobility of all the linkages shown in fig.
Solution :
For Fig(i)
Mobility (or) No of degree of freedom
n = 3( – 1)–2j – h
From the Fig (a), = 5; h = 0; It has two binary joints and two ternary joints,
so
j=2+2x2=6
n = 3( – 1)–2j – h
n = 3(5 – 1)–2 x 6 – 0
n=0
Ans : Therefore linkage forms structure.
(2)
For Fig.(b): From the Fig.(b), = 6; h = 0;
It has seven binary joints, so j = 7
n = 3( – 1) – 2j – h
n = 3(6 – 1) – 2 x 7 – 0 = 1
Ans: Therefore it has single degree of freedom.
(3)
For Fig.1(C):From the Fig.(C),
= 7;h = 0; It has eight binary joints, so j = 8.
n = 3( – 1) – 2j – h
n = 3 (7 – 1) – 2 x 8 – 0 = 2
Ans: Therefore, the linkage has two degrees of freedom.
(4)
For Fig.(d): = 4; h = 1 (i.e., roller constitutes a higher pair); and
j = 3 (i.e., it has three binary joints)
n = 3( – 1) – 2j – h
=3(4 – 1) – 2 x 3 – 1 = 2
Ans: Therefore the linkage has two degrees of freedom.
(5)
For Fig.(e): From the Fig(e), = 3; j = 2 and h = 1(i.e., roller constitutes a higher
pair)
n = 3( – 1) – 2j – h
=3(3 – 1) – 2 x 2 – 1
Ans: Therefore the linkage has single degree of freedom.
3.
Det
ermine
the degree
of
freedom
of
linkages
shown in
Fig.
Solution :
(i)
For Fig.(i): = 6; j = 7; h = 0
Degree of freedom,
n = 3( – 1) – 2j – h
n = 3(6 – 1) – 2 x 7 – 0 = 1.
(ii)
For Fig (ii): 1 = 8; j = 10; h = 0
– 1) – 2 j – h
=3( 8 – 1 ) - 2 x 10 – 10 = 1 Ans
(iii)
For Fig.(iii):
(iv)
For Fig.(iv):
= 4; h = 0
– 1) – 2j – h
= 3(4 – 1) – 2 x 4 – 0 = 0
= 4; j = 4 (three turning pair and one sliding pair); h = 0
n = 3( – 1) – 2j – h
= 3 (4 – 1) – 2 x 4 – 0 = 1
4.
(v)
For Fig.(v): = 3; j = 3; h = 0
– 1) – 2j – h
=3(3 – 1) – 2 x 3 – 0 = 0
(vi)
For Fig.(vi): = 4; j = 4; h – 0
– 1) – 2j – h
=3 (4 – 1) – 2 x 4 – 0 = 1
(vii)
For Fig (vii): = 6; j = 7; h – 0
n = 3( – 1) – 2j – h
=3(6 – 1) – 2 x 7 – 0 = 1
For the kinematic linkages shown in Fig. Find the degrees of freedom.
Solution :
(1)
1 = 6; j = 7; h = 0
Number of degree of freedom,
n = 3( – 1) – 2j – h
n = 3(6 – 1) – 2 x 7 – 0 = 1
Thus the motion is constrained.
(2)
For Fig.(ii):
= 7; j = 8; h = 0
n =
– 1) – 2j – h
= 3(7 – 1) – 2 x 8 = 2
Thus the motion is constrained.
(3)
For Fig.(iii): = 5; j = 6; h = 0
– 1) – 2j – h
= 3 (5 – 10 – 2 x 6 – 0 = 0
Thus it is a structure. Ans.
(4)
For Fig.(iv): In this Fig (iv), the link 3 can slide without causing any
movement in the rest of the mechanism. Thus this link is said to have redundant
degree of freedom.
L = 4;j = 4;h = 0; r = 1
– 1) – 2j – h – r
=3 (4 – 1) – 2 x 4 – 0 – 1 = 0
Thus it is a structure.
5. Find the degrees of freedom for the linkages shown in Fig below.
Solution:
(1)
(2)
(3)
For Fig.(i):
= 14;j = 18;h = 1
– 1) – 2j – h
= 3(14 – 1) – 2 x 18 – 1 = 2 Ans
For Fig.(ii): = 7; j = 8; h = 0
– 1) – 2j – h
= 3(7 – 1) – 2 x 8 – 0 = 2 Ans
For Fig.(iii): = 8;
j = 10 [ Therefore two binary joints and four ternary joints, so j = 2 + 4 x
2 = 10]; h = 0
– 1) – 2j – h
= 3(8 – 1) – 2 x 10 – 0 = 1
(4)
For Fig.(iv): = 7; j = 7; h = 1
– 1) – 2j – h
= 3(7 – 1) – 2 x 7 – 1 = 3 Ans.
6.
Examine the mechanism shown in fig and indicate the cases where unique
relation between the motions of the input and output links exists.
Solution:
(i)
For Fig.(i):
= 6; j = 7; h = 0
n = 3( – 1) – 2j – h
=3(6 – 1) – 2 x 7 – 0 = 1
Thus it is a constrained motion. This mechanism is driven by single input motion.
(2)
For Fig.(ii): In the Fig.(ii), the link 3 can slide without causing any
movement in the rest of the mechanism. Thus this link is said to have redundant
degree of freedom.
L = 4; j = 4; h = 0; r = 1
– 1) – 2j – h – r
=3(4 – 1) – 2 x 4 – 0 – 1 = 0
Thus it is a structure.
(3)
For Fig.(iii): = 10; h = 0; h = 0; j = 12
[because ten binary joints and one ternary joint,
so 10 + (1 x 2) = 12]
– 1) – 2j – h
=3(10 – 1) – 2 x 12 – 0 = 3
So it does not have unique relation between input and output link's motions.
7.
Prove that the mechanism shown in Fig. Below is a constrained kinematic
chain.
Solution:
From the Fig., = 6; j = 7 and h = 0; p = 5
Therefore Degree of freedom,
– 1) – 2j – h
=3(6 – 1) – 2 x 7 – 0 = 1
Thus it is a single degree of freedom system.
Two conditions to be satisfied are:
= 2p – 4
3
and j =
l–2
2
where
.... (i)
....(ii)
= Number of links = 6
p = Number of pairs = 5
j = Number of joints = 7
From equation (i), 6 = 2 x 5 – 4 = 6
So equation (i) is satisfied.
3
From equation (ii) 7 = x 6 – 2 = 7
2
L.H.S. = R.H.S.
So equation (ii) is also satisfied.
Hence, the given mechanism is a constrained kinematic chain. Ans.
8. Explain the four bar chain and its inversions stating its applications with neat
sketches.
Solution :
Four bar chain:
A four link mechanism or linkage is the most fundamental and the simplest
kinematic chain. It consists of four turning pairs.
Four links of a four bar chain are:
1. Crank (or) driver: A link that makes complete revolutions is called as crank.
2. Coupler: The link opposite to the fixed link is known as coupler.
3. Lever or Rocker or follower: The link which makes a partial rotation or
oscillation is known as lever.
4. Frame: The fixed link is known as frame.
Inversions of four bar chain:
I. First Inversion (1. Crank and lever Mechanism):
As shown in Fig (a), link 1 is the crank, link 4 is fixed and link 3 oscillates
whereas in Fig. (b),lnk 2 is fixed and link 3 oscillates. The mechanism is also known
as crank-rocker mechanism or a crank-lever mechanism or a rotary-oscillating
converter.
Application: Beam engine
2. Beam Engine:
This is an example of crank-lever mechanism, where one link oscillates, while
the other rotates about the fixed link, as shown in Fig:
This mechanism is used to convert rotary motion into reciprocating motion.
In this mechanism, when the crank rotates about a fixed centre A, the lever
oscillates about a fixed centre D. The end E of the lever CDE is connected to a piston
rod which reciprocates due to rotation of the crank.
II. Second inversion: (Double crank mechanism)
If the shortest link, i.e., link 1 (crank) is fixed, the adjacent links 2 and 4 would
make complete revolutions, as shown in Fig. The mechanism thus obtained is
known as crank-crank or double crank mechanism or rotary-rotary converter.
Application: Coupling rod of a locomotive.
4. Coupling Rod of a Locomotive:
This is an example of a double crank mechanism where both cranks rotate
about the points in the fixed link. It consists of four links. The opposite links are
equal in length, since links 1 and 3 work as two cranks, the mechanism is also known
as rotary-rotary converter. Refer Fig.
III. Third inversion:
If the link opposite to shortest link is fixed. i.e., link 3 is fixed, then the
shortest link (link 1) is made coupler and the other two links 2 and 4 would oscillate
as shown in Fig. The mechanism thus obtained is known as rocker:- rocker or
double-rocker or double-lever mechanism or an oscillating-oscillating converter.
Applications:
1.
1.Watt's indicator mechanism
2.Pantograph
3.Ackermann steering
Watt's Indicator Mechanism:
This mechanism was invented by Watt for his steam engine to guide the
piston rod. It is also known as simplex indicator. It consists of four links: fixed link
at A, link BC and link DEF, connected to the piston of the indicator cylinder. Links
BC and DEF work as levers and due to this, the mechanism is also known as double
lever mechanism. The displacement of the lever DEF is directly proportional to the
steam or gas pressure in the indicator cylinder.
In Fig. full lines depict the initial position of the mechanism, whereas the
dotted line shows the position of the mechanism when gas or steam pressure acts on
the indicator plunger.
2.
Pantograph:
Pantograph is a device which is used to reproduce a displacement exactly in
an enlarged or reduced scale. It is used in drawing offices, for duplicating the
drawings, maps, plans, etc. As shown in Fig. it is a four bar mechanism in the form
of a parallelogram ABCD with link BC extended to P.
From the similar triangles ApD and CpP, it is clear that p and P will follow
Ap BC
or
similar path. Ratio of the motion of p to the motion of P will
.
AP BP
3.
Ackermann Steering:
The Ackermann steering mechanism for a motion car is shown in Fig. In
Ackermann steering, the mechanism ABCD is a four bar chain. The shorter links AB
and DC equal in length, and the longer links AD and BC are of unequal length.
When the vehicle moves along a straight path, the longer links AD and BC
remain parallel as shown in Fig.(a). When the vehicle is steering to the left ( or
right), the position of the mechanism is shown in Fig.(b). The length of the links are
so proportioned that the lines drawn from the axes of all the four wheels intersect at
a common point p1. This fact ensures that the relative motion between the tyres and
the road surface is pure rolling.
9.
(i)Derive the equation to determine the degree of freedom of planar
mechanism. Prove that a com-roller follower mechanism is an exception for the
above equation.
(ii) Explain with neat sketches and their kinematic differences, two
different inversions of a single slider crank chain that can be used for the same
application in machine tools.
(AU -Nov 2003)
Solution :
(i)
Degree of freedom of a planar mechanism,
F = 3( – 1) – 2 j – h
where
= no of links
j = no of binary joints
h = no of higher pair.
Max degree of freedom of one link = 3
Max degree of freedom of '
' links = 3 x
Degree of freedom lost by fixed link = 3
No. of binary joints = 2
No of higher pair = 1
F = 3 ( - 1) – 2 x j – h
Cam-Roller follower:
= 4
j=3
h=1
F = 3(4 – 1) – 2 x 3 – 1
= 9–6–1
F=2
But actual F = 1, i.e. One input and one output.
Therefore this is an exception.
(ii)
The two inversions of single slider crank chain which are used for the same
application (i.e. Quick return motion) are:
(a)
(b)
Whitworth quick – return mechanism
Crank and slotted lever mechanism
Kinematic difference:
(i)
(ii)
In crank and slotted lever mechanism the output line oscillates.
In whit worth mechanism the output link rotates.
Description:
(i)
Whitworth quick return mechanism:
It is a mechanism used in workshop i.e. In shaping and slotting machines. In this
mechanism, link 2 (i.e. Crank) is fixed, link 3 rotates, link 4 reciprocates and link 1
oscillates.
Initially, let the slider 4 be at B1 point C2, then the tool will be in its extreme left
position. When the crank further rotates and the slider 4 reaches to point C 1 , the
tool will be in its extreme right position. The distance between extreme left and right
positions is the stroke length.
When the link AE rotates from the position AB2 to AB2 , then the ram moves
from left to right and corresponding movement of the crank link will be from BC 2 to
B C1. When the line AE further rotates, link AE move from AB2 to AB1, then the ram
moves from right to left and the crank BC correspondingly moves from BC2 to BC1 .
When crank moves from BC2 to BC1, it is the backward stroke. The time taken
during the left to right motion of the ram will be equal to the time taken by the crank
from BC1 portion to BC2 . The time taken during the right to left motion of the ram
will be equal to the time taken by the crank from BC2 position to BC1. When the tool
is in its left extreme position, it is the return or backward stroke.
Let
Then,
(b)
= obtuse angle C1BC1 at B
= acute angle at C1 B2 at B.
Time of cutting stroke
360º
(or )
Time of return stroke
360º
Crank and slotted Lever Quick return mechanism:-
This mechanism is mostly used in shaping and slotting machines. This
mechanism is also obtained by fixing the link 3, as shown in Fig. From the figure,
OA is the driving crank, BP is the slotted lever, PT is the connecting rod and T is the
cutting tool.
As the link OA revolves clockwise, the link BP oscillates between the two
extreme positions BP1 and BP2. In the extreme pistons, BP1 and BP2 are tangential to
the circle with A as center and AO radius. Hence the two strokes of the tool are
made while the crank makes over the arcs FGE and EHF respectively. The cutting
stroke occurs when the crank moves over the arc FGE.
Let,
Then,
= Obtuse angle FAE at A
= Acute angle FAE atA
Time of cutting stroke
360º
(or )
Time of return stroke
360º
The travel of the tool or length of stroke P1P2:
P1P 2 = 2 BP sin P1BA
= 2BP sin (90º - )
2
= 2BP cos
2
AO
P1P2 = 2BP ×
AB
The ratio of the times given by the above formula can be varied by varying the
distance AB.
10.
Sketch the single slider crank chain and its inversions. Explain any one of
its applications.
Solution :
Single slider crank chain is a modification of four bar chain. It consists of one
sliding pair and three turning pair. This is used to convert reciprocating motion to
rotary motion and vice versa.
Inversions of single slider crank chain
(i)
First inversion:
First inversion is obtained when link 1 (i.e. Frank is frame is fixed and link 2
and 4 are made the crank and slider respectively.
(ii)
Second inversion:
Second inversion is obtained by fixing the link 2 (i.e. crank) of a slider crank
chain. As shown in Fig., when the link 2 is fixed then the link 3 along with the slider
at its end E becomes a crank. This makes link 1 to rotate about O along with the
slider which also reciprocates on it.
(iii) Third inversion:
(iv)
This is obtained by fixing the link 3 of the slider crank mechanism. Link2 acts as a
crank and link 4 oscillates.
(iv)
Fourth inversion:
Fourth Inversion:
Fourth inversion is obtained by fixing the link 4 of the slider crank
mechanism. As shown in Fig. In the inversion, link 3 can oscillate about the fixed
pivot B on link 4. This makes end A of link 2 to oscillate about B and end O to
reciprocate along the axis of the fixed link 4.
Application:
Whitworth quick return mechanism:
(Refer previous problem).
11.
Sketch the double slider crank chain and its inversions.
working of old ham's coupling.
Explain the
Solution :
DOUBLE SLIDER CRANK CHAIN
As its name implies, in double slider crank chain, there are two sliders and
one crank. It is also a four bar pneumatic chain consists of two sliding pairs and two
turning pairs as shown in Fig. The link 1 and link 2 form one sliding pair and link 4
and link 1 from the second sliding pair. The link 2 and link 3 form one turning pair
and link 3 and link 4 form second turning pair.
Inversions:
(i)
First inversion:
First inversion is obtained by fixing the link 1. In this, the two adjacent pairs 2
– 3 and 3 – 4 are turning pairs and the other two pairs 1 – 2 and 1 – 4 are sliding
pairs. Refer Fig (a) Application: Elliptical trammel
(ii)
Second inversion:
Second inversion is obtained by fixing any one of the slider blocks of the first
inversion. When link 4 is fixed end B of crank 3 rotates about A and link 1
reciprocates in the horizontal direction.
Application: Scotch yoke mechanism
(iii)
Third inversion;
Third inversion is obtained by fixing the link 3 of the first inversion.
inversion link 1 is free to move.
Application: Oldham's coupling.
In this
Oldham's Coupling:
This mechanism is used for transmitting motion between two shafts which
are parallel but not coaxial. This inversion is obtained by fixing the link 3 as shown
in Fig. It consists of a driving shaft, fitted with a flange(link 2) having a diametrical
slot on its face; a driven shaft fitted with flange C (link 4) also has diametrical slot on
its face. This whole makes link 4. The slots on the two flanges are at right to each
other. An intermediate piece circular shape and having tongues X and Y at right
angles on opposite sides, is fitted in between the flanges of the two shafts in such a
way that
the tongues of the intermediate piece get fitted closely in the slots of
flanges.
The intermediate circular piece E forms link 1 which slides between the
flanges C and D. When driving shaft rotates through certain angle the driven shaft
also rotates through the same angle. Motion is transmitted through intermediate
link1. If the distance between the axis of the shafts is x, it will be the diameter of a
circle traced by the center of intermediate piece.
12.
In a crank and slotted level quick return motion mechanism, the length of
the fixed link is 300m and that of the driving crank is 150 mm. Determine the
maximum angle the slotted lever will make with the fixed link. Also determine
the ration of time of cutting and return strokes.
If the length of the slotted lever is 700mm, what would be the length of the
stroke assuming that the line of the stroke passes through the extreme positions of
the free end of the slotted lever?
Solution :
Given data :
AB = 300mm = 0.3
AE = 150mm = 1.15m
BP1 = 7==mm = 0.7m
Solution: (i) Inclination of the slotted bar with the fixed link:
Let ABE = inclination of the slotted bar with the vertical
Fig.1.67 shows the extreme positions of the crank.
We know that, sin ABE = sin 90º
2
AE
=
AB
0.15
0.5
=
0.3
ABE = 90º –
2
= sin–1(0.5)
= 30º Ans.
(ii)
Time ratio of cutting stroke to the return stroke:
We know that 90º –
= 30º
2
or = 120º
Time of cutting stroke
360º
=
=2
Time of return stroke
(iii)
Length of the stroke:
Length of the stroke
= P1 P2 = 2(P1 P)
= 2(BP) sin 90º
2
º
= 2 x 0.7 sin (90 – 60º)
L = 450 mm
:
Time of cutting stroke
Time of return stroke
=
360º
360º 96.4º
2.735
=
96.4º
13.In a four bar chain ABCD, AD is fixed and is 120 mm long and rotates at 100
rpm. Clockwise while the link CD = 60mm, Oscillates about D, BC and AD are of
equal length. Find the angular velocity of link CD when angle BAD = 600.
Solution:
(Relative velocity method)
Given Data: AD = 120mm; AB = 30mm;
CD = 60mm; NBA = 100 r.p.m.;
2N BA 2 100
ω BA =
= 10.47 rad/sec
60
60
(a)
Configuration Diagram:
(b) Velocity Diagram
Scale : 1 cm = 20mm Scale: 0.0628 m/s = 1 cm
Vba = WBA x AB = 1.047 x 0.03
Vba = 0.3141
Procedure:
(a)
Configuration diagram: Refer Fig(a)
1.
2.
3.
4.
With suitable scale draw AD and AB with an angle 600.
From point B draw an arc of 120 mm radius.
From point D draw another arc of radius 60mm.
Take the intersecting point as C, and join BC and CD , as shown in Fig
(b)
Velocity diagram: Refer Fig (b)
1. With suitable scale draw VBA (i.e., ab ) perpendicular to AB at any point.
2. Since the link AD is fixed whose relative velocity is zero hence a and d are lie
on the same point.
3. Draw lines perpendicular to CD and BC from points 'd' and 'b' respectively
so as to get the intersection point.
4. Locate the intersecting point as C, as shown in Fig.(b)
From Fig.(b),
VBA = ab on vector velocity diagram,
VCB = bc on vector velocity diagram and
VCD = cd (or) ca on vector velocity diagram
Result: By measurement, we find that
VCD = 0.238 m/s
0.2387 0.2387
WCD =
CD
0.06
= 4 rad/sec (clockwise about D) Ans.
14.
In a four, link mechanism, the crank AB rotates at 36 rod/sec. The length of
the links are AB = 200 mm, BC = 400mm, CD = 450 mm and AD = 600 mm. AD is
the fixed link. At the instant when AB is at right angle to AD. Determine the
velocity of :
(i)
(ii)
and AD.
The mid-point of link BC.
A point on link CD, 100mm from the points connecting the links CD
Solution :
Given Data:
WBA = 36 rad/sec and <BAD = 900
AB = 200mm BC = 400 mm
CD = 450mm
AD = 600mm
AD = Fixed.
(a)
Configuration diagram
Scale: 100 mm = 1 cm
(b) Velocity diagram
Scale: 1.44m/s = 1 cm
VBA = WBA x AB
= 36 x 0.2 = 7.2m/s
Procedure:
(a)
Configuration diagram: Ref Fig.(a). Draw the space diagram as to the
dimensions.
(b)
Velocity diagram: Refer Fig(b). Draw the velocity diagram as discussed in the
previous
Example.
1.
Locate the point 'e' at the centre of vector bc to find velocity of mid point of
link BC.
2.
Join 'e' and 'a'. Now the vector 'ea' will give the velocity of the mid point E.
i.e., VEA = ea = (4.55 x 1.44) = 6.552 m/s
To find out the velocity of point 'F' on CD
DF Vdf
DC Vcd
Vdf Vcd
DF
DC
By measurement,
VCD = 4.6 x 1.44 = 6.624 m/sec [From Fig.(b)]
ωVdf = 6.624 x
DF
= 1.472 m/sec Ans.
DC
Results:
1.
Velocity of mid point of link BC with respect to A is
VEA = 6.552 m/s Ans.
2.
Velocity of a point on a link CD 100 mm from D is
VAF = VDF = 1.472 m/s Ans.
15.
In a slider crank mechanism, the length of crank OB and connecting rod AB
are 125mm and 150mm respectively. The centre of gravity G of the connecting rod
is 275mm from the slider A. The crank speed is 600 rpm clockwise. When the
crank has turned 450 from the inner dead centre position, determine:
1.
2.
3.
Velocity of slider A,
Velocity of the point G, and
Angular velocity of the connecting rod AB.
Solution :
Given Data:
OB = 125mm
AB = 500mm
AG = 275mm.
NBO = 600 rpm(C ω )
2 600
ωBO =
= 62.83 r/s
60
= 45º from inner dead centre
VBO = ω BO x OB = 62.83 x 0.125 = 7.854 m/s
(a)
Configuration diagram: Scale : 62.5 mm = 1 cm
(b)
Velocity diagram: Scale: 1.96 m/s = 1 cm
VaG AG
Vab
AB
VaG 2.85 1.963
275
500
VaG = 3.07 m/s
Procedure:
(a)
Configuration diagram: Ref Fig (a).
1. First draw line of stroke.
2. At any point on the line of stroke draw a crank OB of length 125 mm and =
45º
3. From point B draw an arc of radius 500 to cut the stroke, then the cutting point
be A.
4. Locate G on Ab at a distance AG = 275 mm.
(b)
Velocity diagram: Ref Fig.(b).
1. Draw ob vector perpendicular to OB with suitable scale
2. From O draw an horizontal vector parallel to the line of stroke.
3. From point 'b' draw a line perpendicular to AB, it will intersect the line of
stroke OA at 'a'.
4. From velocity diagram it is understood that
VOA = oa = 3.3 x 1.96 = 6.468m/s
VBA = ab = 2.85 x 1.96 = 5.586 m/s
16.
To locate the point 'g' on vector velocity diagram, take Vag = 3.07 m/s
(already calculated) and join the point 'o' and 'g' which will give the velocity with
respect to 'o'.
VGO = 3.45 x 1.96 = 6.77 m/s
Results: Velocity of the slider A = VOA = 6.468 m/s Ans.
Velocity of the gravity G = VOG = VGO = 6.77 m/s Ans.
Angular velocity of the connecting rod AB is given by,
ωAB =
VAB 5.586
AB
0.5
ωAB = 11.1 rad/sec
17.
In the mechanism as shown in Fig., the crank rotates at 20 rpm
anticlockwise and gives motion to the sliding blocks B and D. The dimensions of
various links are oA = 300 mm; AB = 1200 mm; BC = 450 mm and CD = 450 mm.
For the given configuration, determine.
(i)
(ii)
Velocity of sliding at B and D
Angular velocity of CD.
Solution
NOA = 20 r.p.m. Anticlockwise;
OA = 300 mm;
AB = 1200 MM;
BC = 450 mm;
CD = 450 mm;
2 20
ω OA =
= 2.094 rad/sec;
60
VOA = 0.6283 m/s.
Procedure:
(a)
Configuration diagram: Refer Fig.(a). Draw the configuration diagram as per
the given data.
(b)
Velocity diagram: Refer Fig:(b);
1.
2.
Draw oa vector from point 'o' in the direction perpendicular to the crank OA.
From 'o' draw a line parallel to the line of action of the slider 'B'
3.
oab.
4.
5.
6.
From 'a' draw a line perpendicular to the link AB to complete vector triangle
From point 'o' draw a line parallel to line of action of slider 'D'.
To locate point 'C' on ab'
AB AC
AB AB
AC
750
AC AB
0.553
AB
1200
[ ωAC = ab x scale = 4.4 x 0.1256 = 0.553 m/s]
ωAC = 0.345 m/s
From the point 'c' draw a line towards 'd' perpendicular to the link CD .
Results:
(i)
(ii)
Velocity if slider at B:
VB = Vector ob = 3.2 x 0.1256 = 0.401 m/s Ans.
And Velocity of slider at D:
VD = 1.9 x 0.1256 = 0.238 m/s Ans.
2.9 0.1256
Angular velocity of CD: CD CD
CD
0.45
0.364
0.809rad / sec .
ωCD =
0.45
18.
In a mechanism shown in Fig. The crank OA is 100 mm long and rotates
clockwise about O at 120 rpm. The connecting rod AB is 400 mm long. At a point
C on AB, 150 mm from A, the rod CE 350 mm long is attached. This rod CE slides
in a slot in a trunion at D. The end E is connected by a link EF, 300 mm long to the
horizontally moving slider F.
Configuration diagram : Scale : 50 mm = 1 cm
Velocity diagram : Scale : 0.252 m/s = 1 cm
For the mechanism in the position shown, Find
1.
2.
3.
Velocity of F
Velocity of sliding of CE in the trunion and
Angular velocity of CE.
Solution:
Given Data:
OA = 100
NOA = 120 r.p.m.
AB = 400 mm;
AC = 150 mm;
CE = 350 mm;
EF = 300 mm;
2N OA
ωOA =
60
2 120
ωOA =
= 12.5664 rad/sec
60
ωOA = ωOA × OA = 12.566 × 0.1 = 1.2566 m/s
Procedure:
(a)
(b)
Configuration diagram: Refer Fig a.
Velocity diagram: Refer Fig b.
1. Draw VOA = 1.256 m/s, from 0 to a in the direction perpendicular to the crank
OA.
2. From point 'o' draw a line parallel to line of stroke of the slider B.
3. From point 'a' draw a line perpendicular to link AB. Name the point of
intersection as 'b'.
4. To locate 'C' on 'ab'.
BC bc BC
BA ba BA
VBC = bc = 1.6875 × 0.252
VBC = 0.425 m/s.
5.
6.
7.
From 'c' draw a line perpendicular to the link CE.
From point 'f' draw a line parallel to the link CE.
To locate point 'e':
CE ce CE
CD cd CD
VCE = ce = cd ×
VCE = 0.441 m/s.
350
350
= 0.315 ×
250
250
[ = cd × scale]
19.
From point 'e', draw a line perpendicular to the link EF and draw a line
parallel to the line of stroke 'F' at 'O' and locate the cutting point as 'f'.
Result:
1.
Velocity of F:
VF = Vector 'of' = 2.1 x 0.252
= 0.529 m/s Ans.
2.
3.
[ of = 2.1 cm;Scale = 0.252]
Velocity of sliding of CE in the trunnion:
VCE = Vector 'ec' = 0.441 m/s Ans
Angular velocity of CE:
0.441
ωCE = CE
= 1.26 rad/sec
CE
0.35
20. An engine mechanism shown in Fig. The crank CB =100 mm and the
connecting rod BA=300mm with centre of gravity G, 100mm from B. In the
position shown the crankshaft has a speed of 75 rad/s and an angular acceleration
of 1200 rad/s2. find: 1: velocity of G and angular velocity of AB, and 2 acceleration
of G and angular acceleration of AB.
G
120
Solution: Given : BC=75 rad/s; BC=1200 rad/s2, CB =100 mm =0.1mm=0.1 m, B.A =300
mm=0.3m
We know that velocity of B with respect to C or velocity of B,
vB=vB=BC x CB = 75 x 0.1=7.5m/s
(perpendicular to BC)
Since the angular acceleration of the crankshaft, BC =1200 rad/s2, therefore
tangential component of the acceleration of B with respect to C,
'BC BC xCB 1200 x 0.1= 120 m/s2
Note: When the angular acceleration is not given, ten there will be no tangential
component of the acceleration.
1. Velocity of g and angular velocity of AB
First of all, draw the space diagram, to some suitable scale as shown in fig (a). Now
the velocity diagram, as shown in fig (b0 , is draw as discussed below:
1. Draw vector cb perpendicular to CB, to some suitable scale, to represent the
velocity of B with respect to C or velocity of B (i.e. vBC or VB), such that
Vector cb =vBC =vB=75 m/s
2. From point b, draw vector ba perpendicular to BA to represent the velocity of
A with respect to B i.e vAB, and from point c, draw vector ca parallel to the
path of motion of A (which is along AC) to represent the velocity of A i.e. vA.
The vectors ba and ca intersect at a.
3. Since the point G lies on AB, therefore divide vector ab at g in the same ratio
as G divides AB in the space diagram. In other words,
Ag/ab = AG /AB
The vector cg represents the velocity of G.
By measurement, we find that velocity of G,
VG= vector cg = 6.8m/s Ans.
From velocity diagram, we find that velocity of A, with respect to B,
VAB= vector ba = 4 m/s
We know the angular velocity of AB,
AB
VAB
4
13.3rad / s(Clockwise ) Ans.
BA 0.3
(a) Space diagram.
(b) Velocity diagram.
2. Acceleration of G and angular acceleration of AB
We know that radial component of the acceleration of B with respect to C,
ar BC
v 2BC (7.5)2
562.5m / s 2
CB
0.1
and radial component of the acceleration of A with respect to B,
ar BC
v 2BC (7.5)2
53.3m / s 2
BA
0.3
Now the acceleration diagram, as shown in Fig © is drawn.
1. Draw vector c’ b” parallel to CB, to some suitable scale, to represent the radial
component of the acceleration of B with respect to C. i.e a’BC, such that Vector c’ b”
= a rBC=562.5m/s2
2. From point b” ,draw vector b”b’ perpendicular to vector c’ b” or CB to represent
the tangential component of the acceleration of B with respect to C i.e a1BC, such that
Vector b” b’ = a’BC=120m/s2
…(Given)
3. Join c’b’. the vector c’b’ represents the total acceleration of B with respect to C i.e
aBC acceleration of A with respect to C i.e. a1BC.
4. From point b’, draw vector ‘x parallel to BA to represent radial component of the
acceleration of A with respect to B i.e arAB such that
Vector b” x’ = a rBC=53.3m/s2
5. From point x, draw vector x’ perpendicular to vector b’x or BA to represent
tangential component of the acceleration of A with respect to B i.e a rAB. Whose
magnitude is yet known.
6. Now draw vector c’ a’ parallel to the path of motion of A (which is along AC) to
represent the acceleration of A i.e aA. The vectors xa’ and c’a’ intersect at a’ Join b’ a’.
The vector b’ a’ represents the acceleration of A with respect to B i.e aAB.
7. In order to find the acceleration of G, divide vector a’ b’ in g’ in the same ratio as G
divides BA in Fig .8.6 (a) . Join c’g’. The vector c’g’ represents the acceleration of G.
By measurement, we find that acceleration of G,
aG vector c' g'=414 m/s2 Ans.
From acceleration diagram, we find that tangential component of the
acceleration of A with respect toB.
A’AB=vector xa’ =546 m/s.. (By measurement )
Angular acceleration of AB,
AB
a ' AB 546
1820rad / s 2 (Clockwise )Ans.
BA
0.3
21. In the mechanism shown in fig, the slider C is moving to the right with a
velocity o 1 m/s and acceleration of 2.5 m/s2.
The dimensions of various links are AB= 3m inclined at 45 with the
horizontal Determine: 1. the magnitude of vertical and horizontal component of the
acceleration of the point B, and 2. the angular acceleration of the links AB and BC.
Solution: Given: vC=1m/s. ac=2.5m/s2; AB=3m; BC=1.5m
First of all, draw the space diagram, as shown in fig. (a),to some suitable scale.
Now the velocity diagram, as shown in fig. (b) , is drawn as discussed below:
1. Since the points A and D are fixed point, therefore they lie at one place in the
velocity diagram. Draw vector dc parallel to Dc, to some suitable scale, which
represents the velocity of slider C with respect to D or simply velocity of C,
such that
Vector dc=vCD=vC=1 m/s.
2. Since point B has two motion, one with respect to A and the other with
respect to C, therefore from point a, draw vector ab perpendicular to AB to
represent the velocity of B with respect to A, i.e.v BA and from point c draw
vector cb perpendicular to CB to represent the velocity of B with respect to C
i.evBC. The vectors ab and cb intersect at b.
By measurement we find that velocity of B with respect to A.
VBA= vector ab =0.72 m/s and velocity of B with respect to C,
v BC vector cb=0.72m/s
We know that radial component of acceleration of B with respect to C,
ar BC
v 2BC (0.72)2
0.346m / s 2
CB
1.5
and radial component of acceleration of B with respect to A.
ar BA
v 2BA (0.72)2
0.173m / s 2
AB
3
Now the acceleration diagram, as shown in Fig 8.8 ©, is drawn as discussed below:
1.* Since the points A and D are fixed points, therefore they lie at one place in the
acceleration diagram. Draw vector d’c’ parallel to DC, to some suitable scale to
represent the acceleration of C with respect to D or simply acceleration of C i.e a CD or
aC such that
vector d’c’=aCD=ac=2.5 m/s2
2. The acceleration of B with respect to C will have components, i.e. one radial
component of B with respect to C (arBC) and the other tangential component of B
with respect to C(a’BC). Therefore from point ‘c’, draw vector c’x parallel to CB to
represent arBC= 0.346m/s2
3. Now from point x, draw vector xb’ perpendicular to vector c’x or CB to represent
atBA). Whose magnitude is yet unknown.
4.The acceleration of B with respect to A will also have two components of B with
respect to A (arBA) and other tangential component of B with respect to A (atBA).
Therefore from point a’ draw vector a’y parallel to AB to represent atBA, such that
vector a’y =arBA=0.173 m/s2
From point y, draw vector yb’ perpendicular to vector a’y or AB to represent a tBA.
The vector yb’ interest the vector xb’at b’.Join a’ b’ and c’ b’. The vector a’b’
represents the acceleration of point B (aB) and the vector c’ b’ represents the
acceleration of B with respect to C.
1. Magnitude of vertical and horizontal component of the acceleration of the point B
Draw b’ b” is the horizontal component of the acceleration of the point B. By
measurement,
Vector b’ b” = 1.13 m/s2 and vector a’ b” = 0.9 m/s2 Ans.
2. Angular acceleration of AB and BC
By measurement from acceleration diagram, we find that tangential
component of acceleration of the point B. with respect to A.
atBA = vector yb’ = 1.41 m/s2
and tangential component of acceleration of the point B with respect to C,
atBC= vector xb’ = 1.94 m/s2
We know that angular acceleration of AB,
AB
at BA 1.41
0.47rad / s 2 Ans.
AB
3
and angular acceleration of BC,
BC
at BC 1.94
1.3rad / s 2 Ans.
BC
1.5
22. PQRS is a four bar chain with link PS fixed. The lengths of the link are PQ
=62.5 mm; QR=175mm; RS=112.5 mm; and PS =20mm. The crank PQ rotates at 10
rad/s clockwise. Draw the velocity and acceleration diagram when angle QPS=60
abd Q abd R lie on the same side of PS. find the angular velocity and angular
acceleration of links QR and RS.
Solution: Given QP= 10 rad/s; PQ= 625 mm = 0.0625m; QR=175mm=0.175m;
RS=112.5mm= 0.1125m; PS=200=0.2m
We know that velocity of Q with respect to P or velocity of Q.
VQP=vQ=QP x PQ = 10 x 0.0625 = 0.625 m/s.
(Perpendicular to PQ)
Angular velocity of links (QR and Rs)
First of all, draw the space diagram of a four bar chain, to some suitable scale,
as shown in Fig (a).Now the velocity diagram as shown in fig (b), is drawn as
discussed below:
1. Since P and S are fixed points, therefore these points lie at one place in velocity
diagram. Draw vector pq perpendicular to PQ, to some suitable scale, to represent
the velocity of Q with respect to P or velocity of Q i.e.vQP or VQ such that
vector pq= vQP =vQ =0.625 m/s.
2.From point q, draw vector qr perpendicular to QR to represent the velocity of R
with respect to Q (i.e.vRQ) and from point s, draw vector sr perpendicular to SR to
represent that velocity of R with respect to S or velocity of R (i.e v RS or vR). The
vectors qr and sr intersect at r. By measurement ,we find that.
VRQ=vector qr = 0.333m/s, and vRS=vR = vector sr=0.426 m/s.
We know that angular velocity of link QR.
QR=
v RQ
QR
0.333
1.9rad / s (Anticlockwise) Ans.
0.175
and angular velocity of links RS,
RS=
v RS
0.426
3.78rad / s (clockwise) Ans.
SR 0.1125
Angular acceleration of links QR and Rs.
Since the angular acceleration of the crank PQ is not given, therefore there
will be no tangential component of the acceleration of Q with respect to P.
We know that radial component of the acceleration of Q with respect to P(or
the acceleration of Q)
a r QP aQP aQ
v 2QP
PQ
(0.625)2
6.25m / s 2
0.0625
Radial component of the acceleration of R with respect to Q.
a r RQ
2
v RQ
QR
(0.333)2
0.634m / s 2
0.175
And radial component of the acceleration of R with respect to S (or the acceleration
of R).
ar RS aRS aR
v 2RS (0.426)2
1.613m / s 2
SR
0.1125
The acceleration diagram , as shown in Fig © is drawn as follows:
1. Since P and S are fixed points, therefore these point lie at one place in the
acceleration diagram. Draw vector p’q’ parallel to PQ, to some suitable scale, to
represent the radial component of acceleration of Q with respect to P or
acceleration of Q i.e arQP or aQ such that
Vector p’q’=arQP = aQ =6.25 m/s2
2. From point q’ draw vector q’ x parallel to QR to represent the radial component of
acceleration of R with respect to Q i.e. arRQ such that
Vector q’ x = arRQ = 0.634 m/s2
3. From point x, draw vector xr’ perpendicular to QR to represent the tangential
component of acceleration of R with respect to Q i.e. atRC whose magnitude is not yet
known.
4. Now from points’ draw vectors ’y parallel to SR to represent the radial component
of the acceleration of R with respect to S i.e. arRS. Such that
Vectors ’y =arRS = 1.613 m/s2
5. From point y, draw vector yr’ perpendicular to SR to represent the tangential
component of acceleration of R with respect to S i.e atRS.
6. The vector xr’ and yr’ intersect at r’.Join p’r and q’ r’. By measurement we find that
t
t
aRQ
vectorxr ' 4.1m / s2and aRS
vectoryr ' 5.3m / s 2
We know that angular acceleration of link QR,
QR
at RQ
QR
4.1
23.43 rad/s2 ( Anticlockwise )Ans.
0.175
and angular acceleration of link RS.
RS
t
aRS
5.3
47.1 rad/s2 ( Anticlockwise)Ans.
SR 0.1125
23. The dimension and configuration of the four bar mechanism, shown in Fig, are
as follows:
P1A = 300mm; P2B=360 mm; AB = 360 mm, and P1P1=600 mm.0
The angle AP1P2= 60. The crank P1A has an angular velocity o10 rad/ and an
angular acceleration of 30 rad/s2, both clockwise. Determine the angular velocities
and angular accelerations of P2B, and AB and the velocity and acceleration of the P2b,
and the velocity and acceleration of the joint B.
Solution: Given:AP1=10 rad/s; AP1=30 rad/s, P1A= 30mm= 0.3 m; P2B=AB=360 mm=
0.36m
We know that the velocity of A with respect to P1 or velocity of A,
VAP1=vA=AP1 x p1A= 10 x 0.3 = 3 m/s.
Velocity of B and angular velocities of P2B and AB
First of all, draw the space diagram, to some suitable scale, as shown in
fig.(a).Now the velocity diagram, as shown in fig (b), is drawn as discussed below.
1. Since P1 and P2 are fixed points, therefore these points lie at one place in
velocity diagram. Draw vector p1 perpendicular to P1A, to some suitable
scale to represent the velocity of A with respect to P1 or velocity of Ai.e.
vAP1=vA=3m/s
2. From point a, draw vector ab perpendicular to AB to represent velocity of B
with respect to A (i.e.vBA) and from point p2 draw vector p2b perpendicular to
P2B to represent the velocity of B with respect to P2or velocity of B i.e vBP2 or
vB. The vectors ab and p2b intersect at b.
By measurement ,we find that
VBP2=vB= vector p2b=2.2 m/s Ans.
And
vBA=vectorab= 2.05m/s
We know that angular velocity of P2B,
p 2B
v BP 2
2.2
6.1 rad/s (clockwise) Ans.
P2B 0.36
and angular velocity of AB,
AB
v BA 2.05
5.7rad/s (clockwise) Ans.
BA 0.36
Acceleration of B and angular acceleration of P2B and AB
We know that tangential component of the acceleration of A with respect to P1,
at AP 1 AP 1 P1A 30 0.3 9m / s 2
Radial component of the acceleration of A with respect to P1,
a r AP 1
v 2 AP 1
2 AP 1 P1A 102 x 0.3 30m / s 2
P1A
Radial component of the acceleration of B with respect to A
ar BA
v 2BA (2.05)2
11.67m / s 2
AB
0.36
and radial component of the acceleration of B with respect to P2’
a r BP 2
v 2BP 2 (2.2)2
13.4m / s 2
P2B
0.36
The acceleration diagram as shown in fig ©, is drawn as follows.
1. Since P1 and P2 are fixed points, therefore thee points will lie at one place in the
acceleration diagram. Draw vector p1’x parallel to P1A, to some suitable scale, to
represent the radial component of the acceleration of A with respect to P1, such that.
Vector p’1x=arAP1 = 30 m/s2
2. From point x, draw vector xa’ perpendicular to p1Ato represent the tangential
component of the acceleration of A with respect to P1, such that vector xa’=atAP1=
9m/s2
3. Join P1’ a’. the vector p1 ‘a’ represents the acceleration of A. By measurement, we
find that the acceleration of A.
aA=aAP1=31.6 m/s2
4. From point a’, draw vector a’ y parallel to AB to represent the radial component of
the acceleration of B with respect to A, such that
vector a’y=a’BA=11.67 m/s2
5. From Point y, draw vector yb’ perpendicular to AB to represent the tangential
component of the acceleration of B with respect to A(i.e. atBA) whose magnitude is yet
unknown.
6. Now from point P2’ draw vector p2z parallel to p2B to represent
component of the acceleration B with respect to P2, such that
the radial
Vector p2’z =arBP2 =13.44m/s2
7. From point z, draw vector zb’ perpendicular to P2B to represent the tangential
component of the acceleration of B with respect to P2i.e atBP2.
8. The vectors yb’ and zb’ intersect at b’. Now the vector p’ 2b’ represents the
acceleration of B with respect to P2 or the acceleration of B i.e aBP2 or aB. By
measurement, we find that
ABP2=aB=vector p2’b’= 29.6 m/s2 Ans.
Also
vector yb’ =atBA=13.6m/s2, and vector zb’=atBP2= 26.6m/s2
We know that angular acceleration of P2B,
P 2B
t
aBP
26.6
2
73.8rad / s 2 ( Anticlockwise )Ans.
P2B 0.36
and angular acceleration of AB= AB
t BA
AB
13.6
37.8rad / s 2 ( Anticlockwise)Ans.
0.36
24. In the mechanism as shown in fig , the crank OA rotates at 20 r.p.m.
anticlockwise and gives motion to the sliding blocks B and D. The dimensions of
the various links are oA=300mm; AB= 1200mm; BC= 450 mm and CD = 450m.
For the given configuration , determine :1. velocities of sliding at B and D, 2 angular
velocity of CD, 3.linear acceleration of D, and 4, angular acceleration of CD.
Solution: Given NAO=20 r.p.m. or AO=2x 20/60=2.1 rad/s; OA=300 mm = 0.3m;
AB=1200 mm = 1.2 m; BC=CD=450 mm =0.45m
We know that linear velocity of a with respect to O or velocity of A,
VAO =vA=AO x OA = 2.1 x 0.3 = 0.63 m/s
(Perpendicular to OA)
1.Velocites of sliding at B and D.
First of all, draw the space diagram ,to some suitable scale, as showing fig (a)
new the velocity diagram, as sown in Fig (b), is drawn as discussed below.
1. Draw vector on perpendicular to OA, to some suitable scale, to represent the
velocity of A with respect to O (or simply velocity of A), such that. Vector
oa =vAO=vA= 0.63 m/s
2. From point a, draw vector ab perpendicular to AB to represent the velocity of B
with respect to A (i.e vBA) and from point 0 draw vector ab parallel to path of motion
b (which is along BO) to represent the velocity of B with respect to O (or simply
velocity of B). The vectors ab and ab intersect at b.
3. Divide vector ab at c in the same ratio as C divides AB in the space diagram .In
other words.
BC/CA = bc/ca
4. Now from point c, draw vector CD to represent the velocity of D with respect to C
(i.e. vDC) and from point a draw vector ad parallel to the path of motion of d (which
along the vertical direction ) ro represent the velocity of sliding at B,
By measurement, we find that velocity of sliding at B,
VB=vector ob = 0.4m/s Ans.
And velocity of sliding at D. vD= vector od = 0.24 m/s Ans.
2. angular velocity of CD
By measurement from velocity diagram, we find that velocity of D with
respect to C,
VDC= vector cd= 0.37 m/s
Angular velocity of CD.
CD
v DC 0.37
0.82rad / s( Anticlockwise ) Ans
CD 0.45
3. Linear acceleration of D
We know that the radial component of the acceleration of A with respect to 0 or
acceleration of A.
ar AO aA
v 2 AO
2 AO OA (2.1)2 0.3 1.323m / s 2
OA
Radial component of the acceleration of B with respect to A.
a r BA
v 2BA (0.54)2
0.243m / s 2
AB
1.2
… (BY measurement , vBA= 0.54 m/s.
Radial component of the acceleration of D with respect to C,
ar DC
v 2DC (0.37)2
0.304m / s
CD
0.45
Now the acceleration diagram, as show in fig (c), is drawn as discussed below :
1.
Draw vector o’ a’ parallel to OA to some suitable scale ,to represent the radial
component of the acceleration of a with respect to o or simply the acceleration of
,such that
Vector o’a’ =a’AO=aA = 1.323 m/s2
2. From point a’ draw vector a ‘ x parallel to AB to represent the radial component
often acceleration of B with respect to A, such that
Vector a’x=arBA=0.243 m/s2
3. From point x, draw vector xb’ perpendicular to AB to represent the tangential
component of the acceleration of B with respect to A (i.e atBA) whose magnitude is
not yet known.
4. From point x draw vector o’b’ parallel to the path of motion of B (which is along
BO) to represent the acceleration of B(aB). The vectors xb’ and o’b’ intersect at b’. Join
a’b’. the vector a’b’ represents the acceleration of b with respect to A.
5. Divide vector a’ b’ at c’ in the same ratio as C divides AB in the space diagram. In
other words,
BC/BA=b’ c’/b’a’
6. From point x’, draw vector c’ y parallel to CD to represent the radial component of
the acceleration of D with respect to C, such that
vector c’y =arDC= 0.304 m/s2
7. From point y, draw yd’ perpendicular to CD to represent the tangential
component of acceleration of d with respect to C (i.e.atDC) whose magnitude is not yet
known.
8. From point 0’ draw vector of o’ d’ parallel to the path of motion of D (which is
along the vertical direction) to represent the acceleration of D (aD). The vectors yd’
and o’d’ intersect at d’.
By measurement, we find that linear acceleration of D.
AD=vector 0’d’=0.16m/s2 Ans.
4. Angular acceleration of CD
From the acceleration diagram ,we find that the tangential component of the
acceleration of D with respect to C,
AtDC=vector yd’ =1.28m/s2
..(By measurement)
Angular acceleration of CD,
CD=
a 'DC 1.28
2.84rad / s 2 (clockwise ) Ans.
Cd
0.45
25. Find out the acceleration of the slider D and the angular acceleration of link
CD for the engine mechanism shown in fig .
The crank OA rotates uniformly at 180 r.p.m. in clockwise direction. The
various length are: OA=150mm; AB=450 mm; PB=240mm ; BC=210mm; CD=660 mm.
Solution: Given :NAO= 180 r.p.m. or AO= 2 x 180/60= 18.85 rad/s; OA= 150mm=0.15
m ; Ab=450mm=0.45m; PB=240 mm=0.24m : CD=660mm =0.66m
We know that velocity of A with respect to or velocity of A.
VAO=vA=AO x OA
=18.85 x 0.15=2.83 m/s.
(perpendicular to OA)
First of all draw the space diagram, to some suitable scale, are shown in fig
(a). Now the velocity diagram, as shown in fig (b) is drawn as discussed.
1.
Since O and P are fixed points, therefore these points lie at one place in the
velocity diagram. Draw vector oa perpendicular to OA, to some suitable scale, to
represent the velocity of A with respect to O or velocity o A (i.e. v AO or vA) such
that
vector oa = vaO=vA=2.83 m/s
2.
Since the point B moves with respect to A and also with respect to P, therefore
draw vector ab perpendicular to AB to represent the velocity of B with respect of
A i.e vBA, and from point p draw vector pb perpendicular to PB to represent the
velocity of B with respect to Por velocity of B (i.evBP or vB). the vector ab and pb
intersect at b.
3.
Since the point C lies on PB produced, therefore divide 3 vector pb at c in the
same ratio as cdivide 3s PB in the space diagram. In other words, pb/pc=PB/PC.
4. From point c, draw vector cd perpendicular to CD to represent the velocity of D
with respect to C and from point o draw vector od parallel to the path of motion of
the slider D (which is vertical), to represent the velocity of D, i.e. v D.
By measurement, we find that velocity of the slider D,
vD = vector od = 2.36m/s
Velocity of D with respect to C,
VDC = vector cd = 1.2m/s
Velocity of B with respect to A,
VBA = vector ab = 1.8m/s
and velocity of B with respect to P, vBP = vector pd = 1.5m/s
Acceleration of the slider D
We know that radial component of the acceleration of A with respect to O or
acceleration of A,
arAO a A 2AO AO 18.85 0.15 53.3m / s2
2
Radial component of the acceleration of B with respect to A,
r
aBA
2
vBA
(1.8)2
7.2m / s2
AB 0.45
Radial component of the acceleration of B with respect to P,
r
BP
a
2
vBP
(1.5)2
9.4m / s2
PB 0.24
Radial component of the acceleration of D with respect to C,
2
1.2 2.2m / s2
VDC
CD
0.66
2
r
DC
a
Now the acceleration diagram, as shown in fig. (c) is drawn as discussed below:
1.
Since O and P are fixed points, therefore these points lie at one place in the
acceleration diagram. Draw vector o’a’ parallel to OA, to some suitable scale, to
represent the radial component of the acceleration of A with respect to O or the
acceleration of A (i.e arAO or aA ) such that.
Vector o’a’ = arAO = aA = 53.3 m/s2
2.
From point a’, draw vector a’x parallel to AB to represent the radial component
r
of the acceleration of B with respect to A (i.e aBA
) such that
r
Vector a’x = aBA
= 7.2m/s2
3.
From point x, draw vector xb’ perpendicular to the vector a’x or AB to represent
the tangential component of the acceleration of B with respect to A i.e.
t
whose magnitude is yet unknown.
aBA
4.
Now from point p’, draw vector p’ y parallel to PB to represent the radial
r
component of the acceleration of B with respect to P(i.e aBP
), such that
r
Vector p’y = aBP
= 9.4 m/s2.
5.
6.
7.
From pointy, draw vector yb’ perpendicular to vector b’y or PB to represent the
r
tangential component of the acceleration of B, i.e aBP
. The vectors xb’ and yb’
intersect at b’. Join p’b’. The vector p’b’ represents the acceleration of B, i.e.aB.
Since the point C lies on PB produced, therefore divide vector p’b’ at c’ in the
same ratio as C divides PB in the space diagram. In other words, p’b’/p’c’ =
PB/PC
From point c’, draw vector c’z parallel to CD to represent the radial component
r
of the acceleration of D with respect to C i.e. aDC
such that
r
vector c'z aDC
2.2m/ s2
8.
9.
From point z, draw vector zd’ perpendicular to vector c’z or CD to represent the
t
tangential component of the acceleration of D with respect to C i.e. aDC
, whose
magnitude is yet unknown.
From point o’, draw vector o’d’ parallel to the path of the motion of D (which is
vertical) to represent the acceleration of D, i.e.aD. The vectors zd’ and o’d’
intersect at d’. Join c’d’.
By measurement, we find that acceleration of D,
aD= vector o’d’ = 69.6m/s2 Ans.
Angular acceleration of CD
From acceleration diagram, we find that tangential component of the
acceleration of D with respect to C,
t
aDc
vector zd'=17.4m/s2
...(By measurement)
We know that angular acceleration of CD,
CD
t
aDC
17.4
26.3rad / s2 (Anticlockwise) Ans.
CD 0.66
26. In the toggle mechanism shown in fig, the slider D is constrained to move on a
horizontal path. The crank OA is rotating in the counter-clockwise direction at a
speed
Fig.
Of 180 r.p.m increasing at the rate of 50 rad/s2. The dimensions of the various links
are as follows:
OA = 180mm; CB = 240 mm; AB = 360mm; and BD = 540 mm.
For the given configuration, find 1. Velocity of slider D and angular velocity of BD,
and Acceleration of slider D and angular acceleration of BD.
Solution: Given: NAO = 180 r.p.m or AO = 2 x 180/60 = 18.85 rad/s; OA = 180 mm =
0.18 m; CB = 240 mm =0.24 m; AB = 360mm =0.36m; BD = 540 mm = 0.54m
We know that velocity of A with respect to O or velocity of A,
vAO = vA = AO x OA = 18.85 x 0.18 = 3.4 m/s
1. Velocity of slider D and angular velocity of BD
First of all, draw the space diagram to some suitable scale, as shown in fig 8.17
(a). Now the velocity diagram, as shown in fig(b) is drawn as discussed below:
1.
Since O and Care fixed points, therefore these points lie at one place in the
velocity diagram. Draw vector oa perpendicular to OA, to some suitable scale, to
represent the velocity of A with respect to O or velocity of A i.e. vAO or vA, such
that
Vector oa = vAO = vA = 3.4m/s
2.
Since B moves with respect to A and also with respect to C, therefore draw
vector ab perpendicular to AB to represent the velocity of B with respect to A i.e.
vBA, and draw vector cb perpendicular to CB to represent the velocity of B with
respect to C ie., vBC. The vectors ab and cd intersect at b.
3. From point b, draw vector bd perpendicular to BD to represent the velocity of D
with respect to B i.e. vDB, and from point c draw vector cd parallel to CD (i.e., in
the direction of motion of the slider D) to represent the velocity of D i.e. v D.
By measurement, we find that velocity of B with respect to A,
vBA = vector ab = 0.9m/s
velocity of B with respect to C,
vBC = vector cb, = 2.8m/s
Velocity of D with respect to B,
VDB = vector bd = 2.4m/s
and velocity of slider D,
vD = vector cd = 2.05 m/s Ans.
Angular velocity of BD
We know that the angular velocity of BD,
BD
v DB
2.4
4.5rad / s Ans.
BD 0.54
2. Acceleration of slider D and angular acceleration of BD.
Since the angular acceleration of OA increases at the rate of 50 rad/s2, i.e. AO =
50rad/s2 therefore
Tangential component of the acceleration of A with respect to O,
tAO AO OA 50 0.18 9m/ s2
Radial component of the acceleration of A with respect to O,
3.4 63.9m / s2
v2
AO
OA
0.18
2
r
AO
a
Radial component of the acceleration of B with respect to A,
0.9 2.25m / s2
v2
BA
AB
0.36
2
r
BA
a
Radial component of the acceleration of B with respect to C,
2
2.8 32.5m / s2
vBC
CB
0.24
2
r
BC
a
and radial component of the acceleration of D with respect to B,
2
2.4 10.8m / s2
vDB
BD
0.54
2
r
DB
a
Now the acceleration diagram, as shown in Fig (c), is drawn as discussed below:
1.Since O and C are fixed points, therefore these points lie at one place in the
acceleration diagram. Draw vector o’x parallel to OA, to some suitable scale, to
represents the radial component of the acceleration of A with respect to O
i.e. arAo such that
Vector o’x = arAo = 63.9 m/s2
2. From point x, draw vector xa’ perpendicular to vector o’x or OA to represent the
tangential component of the acceleration of A with respect to O i.e. atAo , such that
vector xa’ = atAo = 9m/s2
3. Join o’a’. The vector o’a’ represents the total acceleration of A with respect to O or
acceleration of A with respect to A i.e. aAO or aA.
4. Now from point a’, draw vector a’y parallel to AB to represent the radial
r
component of the acceleration of B with respect to A i.e. aBA
, such that
r
vector a’y = aBA
=2.25m/s2
5. From point y, draw vector yb’ perpendicular to vector a’yor AB to represent the
t
tangential component of the acceleration of B with respect to A i.e aBC
whose
magnitude is yet unknown.
6. Now from point c’, draw vector c’z parallel to CB to represent the radial
r
component of the acceleration of B with respect to C i.e, aBC
such that
r
vector c’z = aBC
=32.5 m/s2
7. From point z, draw vector zb’ perpendicular to vector c’z or CB to represent the
t
tangential component of the acceleration of B with respect to C i.e, aBC
. The vectors
yb’ and zb’ intersect at b’. Join c’b’. The vector c’b’ represents the acceleration of B
with respect to C i.e. aBC
8. Now from point b’, draw vector b’s parallel to BD to represent the radial
r
component of the acceleration of D with respect to B i.e. aDB
such that
r
vector b’s = aDB
=10.8 m/s2
9. From point s, draw vector sd’ perpendicular to vector b’s or BD to represent the
'
tangential component of the acceleration of D with respect to B i.e. a DB
whose
magnitude is yet unknown.
10. From point C’, draw vector c’d’ parallel to the path of motion of D (which is
along CD) to represent the acceleration of D i.e. aD. The vectors sd’ and c’d’ interest
at d’.
By measurement we find that acceleration of slider D,
AD = vector c’d’ = 13.3 m/s2 Ans.
Angular acceleration of BD
By measurement we find that tangential component of the acceleration of D with
respect to B,
AtDB = vector sd’=38.5m/s2
We know that angular acceleration of BD,
BD
t
aDB
38.5
71.3rad/ s2 (Clockwise)
BD 0.54
UNIT –II
FRICTION
Comparision Dry, Greasy and Fluid friction.
Object
Definition
Dry Friction
The friction that
exists
between
two unlubricated
Surfaces
Other
names
Solid friction
Greasy Friction
When the two surfaces
in contact have a
minute thin layer of
lubricant
between
them, then it is called
as greasy friction
Skin friction
Fluid Friction
When the two surfaces
in
contact
are
completely separated by
a lubricant then it is
called as fluid friction.
Film friction or viscous
friction
Comparesion of the laws of solid friction and fluid friction.
S.No
Laws of solid or Dry friction
1
The frictional force is directly
proportional to the normal reaction
between the surfaces.
2
The frictional force opposes the
motion or its tendency to the motion
and depends upon the nature of the
surface in contact
3
The frictional force is independent of
the area and the shape of the
contacting surfaces.
Laws of Fluid Friction
The frictional force is almost
independent of load
The frictional force is independent of
the substances of the bearing surfaces
and opposing tendency is less.
The frictional force reduces with
increase in temperature of the
lubricant
Slope of a thread
It is the inclination of the thread with horizontal.
Lead screw
Slope of thread = tan 1
circumference of screw
Differentiate between angle of repose and limiting angle of friction
S.No
Angle of Repose
1
The angle of repose is defined as
the maximum inclination of a
plane at which a body remains in
equilibrium over the inclined plane
by the assistance of friction only.
2
The angle of repose is equal to the
limiting angle of friction
Limiting Angle of Friction
The limiting angle of friction is defined
as the angle at which the resultant
reaction R makes with the normal
reaction RN.
The limiting angle is not equal to angle
of reposes
The effects of limiting angle of friction?
1.
If limiting angle of friction is equal tan-1 µ then the body will move over
the plane irrespective of the magnitude of the force (F) limiting force of friction.
2.
If < tan-1 then no motion of body on plane is possible irrespective of how
large the magnitude of F may be
Co-efficient of friction (µ).
It is defined as the ratio of the limiting friction F to the normal reaction (RN) between
the two bodies.
Limiting force of friction F
Normal reaction
RN
Differentiate coefficient of friction in square thread and V thread.
(a ) In square thread =
(b) In V thread 1
F
RN
cos
Where
F = Limiting force of friction
RN = Normal reaction and
2β = Angle of ‘V’ in a ‘V’ thread
A block of 50 N rests on a horizontal plane and whose co efficient of friction is
0.25 what is the force required to pull the block at an angle 30 to the horizontal.
Given data:
W = 50 N
µ = 0.25
= 30
tan 1 14.04
Solution :
Required force pulling
w sin
p
cos -
p
50 sin 14.04
12.62 N
cos(30-14.04)
In the above problem what is the minimum force required to just pull the body.
Given Data: W = 50 kg
µ = 0.25
tan 1 14.04
Solution : Pmin = W sin
At minimum force position
pmin W sin =50 sin 14.04
p min 12.129 N
The efficiency of inclined plane?
The efficiency of an inclined plane is defined as the ratio between effort without
friction p0 and the effort with friction p.
Compare laws of static friction and dynamic friction.
S.No
Laws of static friction
1
The frictional force always acts in a
direction opposite to the direction of
motion of the body.
2
The magnitude of frictional force is
exactly equal to the force
3
The magnitude of the limiting
friction bears a constant ratio to the
normal reaction between the two
surface
Laws of Dynamic Friction
The frictional force always acts in a
direction, opposite to the direction of
motion of the body
Frictional force remains constant for
moderate speeds.
The magnitude of the kinematic
friction bears a constant ratio to the
normal reaction between the two
surfaces.
A body of 50N is placed on a 20 inclined plane whose coefficient of friction µ=0.3
what is the force required to hold the body at its position in the horizontal
direction.
Given data :
W
µ
= 50 N
= 0.3 tan 1 16.69
α
= 20
Solution : since = 90 Indirectly given- Horizontal force
Holding Force
w sin ( + ) 50sin 20 16.69
sin( - - ) sin 90 20 16.69
p 37.255 N
p
What is the mechanical efficiency of an inclined plane which needs 50N of force
to raise a block on its surface when it is purely lubricated and 65N when the
surface is dry?
Given Data
P0
P
= 50 N
= 65 N
without friction
with considering friction
Solution: Efficiency of the inclined surface is given by
p 0 50
0.769
p 65
up 76.92%
up
A inclined plane α inclination to the horizontal having limiting angle of friction
is used to lower a load down, what is the efficiency of the inclined plane if the
effort is applied in the horizontal direction.
Solution:
down
cot
cot
down
tan( )
tan
What is the effort required to lift a 50 tonn of lorry using screw jack ? µ=0.3 and
α=20.
Solution :
tan 1 ( ) 16.69
w 50 tonne =50 10 9.81
=490.5kN
W tan( )
Effort required to lift lorry p= 490.5 tan(16.69 20)
365.59kN
The preferred ‘α’ value for a screw to have maximum efficiency?
4
2
The maximum efficiency of a screw jack?
max
1 sin
1 sin
where =tan -1
Self locking screws have lesser efficiency
Self locking needs some friction on the thread surface of the screw and nut
hence it needs higher effort to lift a body and hence automatically the efficiency
decreases.
The functions of clutches
1.
2.
3.
4.
It supplies power to the transmission system.
It stops the vehicle by disconnecting the engine from transmission system
It is used to change the gear and idling the engine.
It gives gradual increment of speed to the wheels.
In a five plate clutch drive, the inner and outer radii of friction plate are 100 mm
and 200 mm respectively. The coefficient of friction is 0.25 and the axial force is 15
kN. What is the frictional torque acting on the plate using uniform wear principle.
Given Data
np
r1
r2
µ
w
T
=5 and n=5 -1=4
=100mm
= 200mm
= 0.25
= 15kN
=?
Solution:
T n WR
r r
n W 1 2
2
100 200
3
4 0.25 15 103
10
2
T 2250 N m
The axial force required at the engagement and disengagement of cone clutch?
Axial force required
= W Wn (1 cot )
Wn = Normal load
µ = Co efficient of friction. And
α =Cone angle of the clutch.
The difference between cone clutch and centrifugal clutch?
Cone clutch works on the principle of friction alone. But centrifugal clutch
uses principle of centrifugal force in addition with it.
Compare plate clutches and cone clutches.
S.no
Plate clutches
1
Works on the principle of friction.
2
Friction lined flat plates are used.
3
Single plate and multiplate clutch is
possible depending on the load
condition
Cone clutches
Works on the principle friction
Friction lined frustum of cone is used
It is not possible
4
It uses principle of uniform wear and
uniform pressure.
It also uses the principle of uniform
wear and uniform pressure
Friction
Friction is the important factor in engineering and physical applications such as belt
and ropes jibs, clutches and brakes, nut and bolts, so it is the necessary one.
If the friction exceeds certain value it will cause heat damage and wear when applied
so it is called necessary evil.
The belt materials
1. Leather 2. Cotton or fabric 3. Rubber 4. Balata and 5. Nylon.
The types of belt device
1. Parallel belt devices
a. Open belt
b. Cross belt
i. One idler pulley
ii. Many idler pulley
iii. Intermediate pulley
2. Angular belt drives (with or without idler pulley)
a. Right angle
b. Intersecting drives
c. Non intersecting drives
Velocity ratio.
It is defined as the ratio between velocity of the driver and the follower or driven.
The thickness of belt addicts the velocity ratio?
Yes but it is negligible
N 2 d1 t
N1 d 2 t
Where
N1 and N2 = Speed of the driver and the driven respectively.
D1 and d2 = Diameters of the driver and the driven respectively and
T = Belt thickness.
The effect slip on velocity ratio of a belt drive?
N 2 d1
s
The effect of slip is to decrease the speed of belt on the
1
N1 d 2 100
driving shaft and
where
to decrease the speed of the driven shaft..
S= Slip of the belt drive.
The law of belting.
Law of belting states that the centre line of the belt as it approaches the pulley
must lie in a plane perpendicular to the axis of the pulley or must lie in the plane of
the pulley. Otherwise the belt will run off the pulley.
In a open belt drive distance between pulleys is x and their diameters are D1 and
D2 what is the length of belt required by this system?
( D D2 )
L ( D1 D2 ) 2 x 1
2
4x
Angle of contact
It is angle made by a common normal drawn to the tangent line at the point of
engagement and at the point of disengagement of the belt on a pulley at its centre.
Lab angle for open belt drive = = 180 2
Lap angle for cross belt drive = = 180 2
180
180
rad
rad
Compare slip and creep in belt drive.
S.No
1
Slip
The relative motion between belt and pulley due
to insufficient is called as slip
2
The effect of slip is to reduce the speed of driven
3
It can be avoided by proper friction lining on the
surface of the pulley or on belt
Creep
The phenomenon of sudden contraction and
expansion of belt when it passes from slack
side to tight side is called as creep.
The effect of creep is to reduce the speed of the
driven pulley.
It can be improved avoided by arranging the
driving system such that the tight side should
be the lower one and slack side be upper one.
4
Its effect on velocity ratio
N 2 d1
N1 d 2
s
1 100
N 2 d1 E 2
where
N1 d 2 E 1
E=
Young’s
modulus of the belt material and
σ1, σ2 stress on tight side and slack sides of the
belt
The centrifugal effect on belts
During operation as the belt passes over a pulley the centrifugal effect due to
its self weight tends to lift the belt from the pulley surface. This reduces the normal
reaction and hence the friction resistance.
The centrifugal force produce additional tension in the belt.
Condition for maximum power transmission in belts.
p TV mv3 C
1
C 1
e
The disadvantage of V belt drive over flat belt
1.
2.
3.
4.
V belt cannot be used over large distance
It is not as durable as flat belt
Since the V belt subjected to certain amount of creep therefore it is not suitable
for constant speed applications such as synchronous machines, and timing
devices.
It is a costlier system.
The ratio of driving tension in flat belt V belt and in rope drives
T1
e For flat belt drive
T2
T1
e coses For V belt and rope drive
T2
where T1 Tension in tight side
T2 = Tension in loose side
= Angle of contact and
2 =Angle of V groove
The condition for transmission of optimum or maximum power in belt drive
Power transmitted shall be a maximum when the centrifugal tension TC is
one third of the belt strength T or when the belt runs at the velocity of
T
3m
Where
T = Total belt tension and
M = Mass of belt per unit length.
Cross belt used instead of open belt
1.
2.
Cross belt is used where the direction of rotation of driven pulley is opposite to
driving pulley.
Where we need more power transmission there we can use cross belt drive.
wipping
IF the centre distance between two pulleys are too long then the belt begin to
vibrate in a direction perpendicular to the direction of motion of belt. This
phenomenon is called as wipping.
Wipping can be avoided by using idler pulleys
Lubrication reduces friction
In practical all the manting surfaces are having roughness with it. It causes
friction. If the surfaces are smooth then friction is very less. Lubrication smoothens
the manting surface by introducing oil film between it. The fluids are having high
smoothness than solids and thus lubrication reduces friction.
The S.I unit for frictional moment
Frictional moment = N-m
The maximum angle of wedge for self locking
2
Angle of wedge
=Limitiing friction angle
Crowing in pulley
The process of increasing the frictional resistance on the pulley surface is
known as crowning. It is done in order to avoid slipping of the belt.
Initial tension in belts
In order to increase the frictional grip between the belt and pulleys, the belt is
tightened up. Due to this the belt gets subjected to some tension even when the
pulleys are stationary. This tension in the belt is called initial tension T0.
List out the commonly used breaks.
1.
2.
3.
Hydraulic brakes : e.g. Pumps or hydrodynamic brake and fluid agitator
Electric brakes: e.g. Eddy current brakes
Mechanical brakes: e.g. Radial brakes and Axial brakes
Expression for the ratio of tension between the tight and slack sides of a band and
block brake.
n
T1 1 tan
T2 1 tan
T1 = Tension in that side
T2 = Tension in slack side
µ =Coefficient of friction between block and drum
2 = Subtending angle and
n = Number of blocks
Brake
Brake is a device by means of which motion of a body is retarded for showing
down or to bring it to rest which works on the principle of frictional force, it acts
against the driving force.
Self energizing.
When moments of efforts applied on the brake drum and frictional force are
in the same direction the breaking torque becomes maximum frictional force aids the
braking action. In such a case the brake is said to be partially self actuating or self
energizing.
The intensity of pressure acting brake shoe is assumed to be uniform
The intensity of pressure is assumed to be content when the break shoe has small
angle of contact.
For large angle of contact it is assumed that the rate of wear of the shoe remains
constant.
The P.I.V drive system used?
P.I.V. Positive infinitely Variable drive is used in an infinitely varying speed
systems.
The various Types of Friction.
(i) Static friction:
It is the friction, experienced by a body, when at rest.
(ii)
Dynamic friction:
It is the friction, experienced by a body, when in motion. The dynamic
friction is also called Kinetic friction and is less than the static friction. It is of the
following three types:
(a)
Sliding friction:
It is the friction, experienced by a body, when it slides over another body.
(b)
Rolling friction:
It is the friction, experienced between the surface which has balls or rollers
interposed between them.
Pivot friction:
It is the friction, experienced by a body, due to the motion of rotation as in ease of
foot step bearings.
Coefficient of Friction:
It is defined as the ratio of the limiting friction(F) to the normal reaction (Rn)
between the two bodies. It is generally denoted by w. mathematically coefficient of
friction.
W = F/RN
Limiting Angle of Friction:
Consider that a body A of weight (W) is resting on a horizontal plane (B) as shown
in Fig:
Limiting angle of friction
In the limiting case, when the motion just begins, the body will be in equilibrium
order the action of the following three cases.
1.
2.
3.
Weight of the body (W)
Applied horizontal force (P) and
Reaction (R) between the body A and the plane B
The reaction R must, therefore be equal and opposite to the resultant of W
and P and will be inclined at an angle to the normal reaction RN. This angle is
known as to limiting angle of friction. It may be
defined as the angle which the
resultant reaction R makes with the normal reaction RN.
tan = F/RN
= RN/RN
=
Screw Friction:
The screw bolts, studs, nuts etc are widely used in various machine and
structure for temporary fastening. These fastenings have screw threads, which are
made by cutting a continuous helical groove on a cylindrical surface. If the threads
are at on the other surface of a solid rod, these are known as external threads. But if
the threads are cut on the interval surface of a hollow rod, there are known as
internal threads. The screw threads are mainly of two types i.e., V-threads and
square threads.
In general the V – threads are used for the purpose of tightening pieces
together as bolts and nuts etc,
The various terms used in screw threads.
The following terms are important for the study of screw:
1.
Helix:
It is the curve traced by a particle while describing a circular path at a
uniform speed and advancing in the axial direction at a uniform rate. In other words
it is the curve traced by a particle while moving along a screw thread.
2.
Pitch:
It is the distance from a point of a screw to a corresponding point on the next
thread measured parallel to the axis of the screw.
3.
Lead:
It is the distance, a screw thread advances axially in one turn.
4.
Depth of thread:
It is the distance between the top and bottom surfaces of a thread
as erest and root of a thread)
( also known
5.
Single – threaded screw :
If the lead of a screw is equal to its pitch, it is known as single threaded screw.
6.
Multi – threaded screw:
If more than one thread is cut in one lead distance of a screw, it is known as
multi = thread screw e.g. In a double threaded screw two threads are cut in one lead
length. In such cases all the threads run independently along the length of the rod
mathematically.
Lead = Pitch x Number of threads.
7.
Slope of thread:
It is the inclination of the thread with the horizontal Mathematically,
tan =
lead of screw
circumfere nce of screw
= p/d
= n.p/d
where
p = Pitch of the screw
d = mean diameter of the screw
n = Number of threads in one lead.
Define Screw Jack and derive an expression for torque required to lift a body.
Screw Jack:
The screw jack is a device, for lifting huge loads, by applying a comparatively
smaller effort at its handle.
The above Fig shows a common form of a screw jack, which consists of a
square threaded rod (also called screw rod or simply screw) which fits into the inner
threads of the nut. The load to be raised or lowered is placed on the head of the
square threaded rod which is rotated by the application of an effort at the end of the
lever for lifting or lowering the load.
Torque required to lift the load by a screw Jack.
If one complete form of a screw thread by imagined to be unwound from the
body of the screw and developed it will form an inclined plane as shown in Fig:
Let
p
d
P
= Pitch of the screw
= Mean diameter of the screw
= Helix angle
= Effort applied at the circumference of the
screw to lift the load
W = Load to be lifted ad
= Coefficient of friction, between the screw
and nut = tan , where is the friction
angle.
From the geometry of the Fig:
tan = p/ d
T1 = P × d/2
= W tan (+) d/2
T1 – W tan ( + )d/2
1.
When the nominal diameter (do) and the core diameter (dc) of the screw
thread is given, then the mean diameter of the screw
do dc
2
p
= do
2
p
= dc
2
d=
2.
Since the mechanical advantage is the ratio of load lifted(W) to the effort
applied (P1) at the end of the level, therefore mechanical advantage:
M.A. =
W W 2l
P1
P.d
W 2l
=
w tan d
=
2l
d. tan
P.d
P1 2l
Torque Required to lower the load by a Screw Jack
Torque required to over come friction between the screw and nut
T = P × d/2
= W tan( –)d/2
Note: When, then P = W tan ( – )
Expression for efficiency of a Screw Jack.
Efficiency of a Screw Jack
Efficiency
Ideal effort
Actual effort
Po
P
W tan
=
W tan
=
Which shows that the efficiency of a screw jack is independent of the load raised.
Maximum efficiency of a Screw Jack:
sin 90º sin
sin 90º sin
sin 90º sin
=
sin 90º sin
1 sin
=
1 sin
The Friction of V – threads.
1.
When coefficient of friction 1
is considered, then the V – thread is
cos
equivalent to a square thread.
2.
All the equation of square threaded screw also hold good for V – thread. In
case of V – threads, 1 (ie tan ) may be substituted in place of (i.e tan ). Thus for
V – threads
P = w tan 1
1= Virtual friction angle, such
that tan1 = 1
Friction of a V – thread
We know that the normal reaction in case of a square threaded screw is RN = W cos
where = Helix angle.
But in case of V – thread (or acme or trapezoid threads) the normal reaction, between
the screw and nut is increased because the axial components of this normal reaction
must be equal to the axial load W,
Let
2
= Angle of the V – thread
= Semi angle of the V – thread
W
RN =
cos
and frictional force
F
= WRN.
=
where
W
cos
1 known as virtual coefficient of friction.
cos
The various Friction clutches with neat sketches.
A friction clutch has its principal application in the transmission of power of
shafts and machines which must be started and stopped frequently. Its application
is also found in cases in which power is to be delivered to machines partially or fully
loaded.
The friction clutches of the following types are important from the subject point of
view:
1.
2.
3.
Disc or plate clutches
(single disc or multiple disc clutch)
Cone clutches and
Centrifugal clutch.
Single disc or plate clutch
A single disc or plate clutch as in Fig consists of a clutch plate whose both
sides are faced with a friction material (usually of femora).
Now consider two friction surfaces, maintained in contact by an axial through W as
shown in above figure.
T = Torque transmitted by the clutch
p = Intensity of axial pressure with which the contact surfaces are held together.
r1 and r2 = External and Internal radii of friction faces and
= Coefficient of friction
Consider an elementary ring of radius r and thickness dr as on Fig (b)
We shall now consider the following two cases:
1.
2.
1.
When there is a uniform pressure and
When there is a uniform wear.
Considering uniform pressure:
When the pressure is uniformly distributed over the entire are of the friction face
that the intensity of pressure
W
p=
2
2
r1 r2
W = Axial thrust with which the contract or friction surfaces are held together.
r1 2 r2 3
2
T = .W
.W.R.
2
2
3
r1 r2
2.
Considering uniform wear:
Let P be the normal intensity of pressure at a distance r from the axis of the clutch.
Since the intensity of pressure varies inversely with the distance, therefore
p.r = C ( a constant)
or p = C/r
and the normal force on to ring
Total frictional torque on the friction surface.
= ½ x *(r1 + r2)
= .W.R.
7. Explain the Multiple Disc clutch and derive an expression for torque
transmitted.
A multiple disc clutch as shown in Fig may be used when a large torque is to
be transmitted. The inside discs (usually of steel) are fastened to the driver shaft to
permit axial motion (except for the last disc). The outside discs (usually of bronze)
are held by bolts and the fastened to be housing which is keyed to the driving shaft.
The multiple disc clutches are extensively used in motor cars, machine tools etc:
n1 = number of discs on the driving shaft and
shaft
Therefore the Number of constant surfaces
; n2 = number of discs on the driver
n=n1 + n2 –1
and total frictional torque acting on the friction surfaces or on the clutches T
.W.R.
Where
R = Mean radius of the friction surfaces
3
3
2 r1 r2
R= 2
3 r`1 r2 2
R=
r1 r2
2
(for uniform pressure)
(for uniform wear)
=
n.
8. A shaft runs at 80 rpm and drives another shaft at 150 rpm through belt drive.
The diameter of the driving pulley is 600 mm. Determine the diameter of the
driven pulley in the following cases:
(i)
(ii)
(iii)
(iv)
Neglecting belt thickness
Taking belt thickness as 5 mm,
Assuming for case (ii) a total slip of 4%, and
Assuming for case (ii) a slip of 2% on each pulley.
Solution :
N1 80 rpm, D1 600mm, N2 150 rpm
i
N2 D1
150 600
or
or D2 320 mm
N1 D2
80
D2
ii
N2 D1 1
150 600 5
or
or D2 317.7mm
N1 D2 1
80
D2 5
iii
N2 D1 t 100 S
150 600 5 100 4
or
N1 D2 1 100
80 D2 5 100
D2 304.8mm
iv
N2 D1 t 100 S
N1 D2 1 100
where S=S1 S2 0.01S1S2 2 2 0.01 2 2 3.96
150 600 5 100 3.96
,D2 304.9mm.
80 D2 5
100
9. Two parallel shafts, connected by a crossed-belt, are provided with pulleys 480
mm and 640 mm in diameters. The distance between the centre lines of the shafts
in 3 m. Find by how much the length of the belt should be changed if it is desired
to alter the direction of rotation of the driven shaft.
Solution:
R = 320 mm, C=3m, R=240 mm
For Cross belt
R r
0.32 0.24
sin
sin
sin 0.1867
C
3
10o 45 ' or
0.1878 rad
cos =0.9825
Lc 2 R r 2C cos
= +2 0.1878 0.32 0.24 2 3 0.9825 7.865 m
For open belt
R r
0.32 0.24
sin
Sin
sin 0.0267
3
C
o
1 32'
As the angle is very small, the approximate relation can be used.
Lo R 1
R r
C
0.32 0.24
2
2C
0.32 0.24
3
2
2 3 7.761 m
The length of the belt should be reduced by
Lc Lo 7.865 7.761 0.104m or 104 mm.
10. Design a set of stepped pulleys to drive a machine from a counter shaft that
runs at 220 rpm. The distance between centres of the two sets of pulleys is 2 m the
diameter of the smallest step on the countershaft is 160 mm. The machine is to
run at 80 100 and 130 rpm and should be able to rotate in either direction.
Solution:
As the driven shaft is to rotate in either direction, both the cases of a crossedbelt and an open-belt are to be considered.
(i) For Crossed-belt system
The smallest step on the countershaft will correspond to the biggest step on
the machine shaft (or the minimum speed of the machine shaft).
n1 n2 n3 220 rpm, r1 80 mm
N1,N2 ,N3 80, 100, 130 rpm respectively
(a) For First Step
R1 n1
R
220
or 1
or R1 220 mm
r1 N1
80
80
(b) For Second Step
R 2 n2 220
or R 2 2.2r2
r2
N2 100
Also R 2 r2 R1 r1.
2.2r2 r2 220 80 or 3.2r2 300
r2 93.75 mm, R2 93.7 2.2 206.3 mm
(c) For Third Step
R3 220
or R3 1.69r3
r3
130
Also R3+r3 = R1 + r1
1.69 r3 + r3 = 220 + 80 = 300
r3 = 111.5 mm, R3 = 111.5 x 1.69 = 188.5 mm
(ii) For Open-belt System
(a) For First Step
r1 = 80 mm R1 = 220 mm as before
(b) For Second Step
R2 r2
R2 r2
C
2.2r2 r2
10.05r2
2
R1 r1
2.2r2 r2
2
2
R1 r1
2
C
0.22 0.08
0.22 0.08
2
1.44 2
r2 0.9523
2
r22 13.958r2 1.323
r2 6.979
2
1.323 6.979 50.029 7.073
2
r2 7.073 6.979 0.094 m or 94 mm
R 2 2.2 94 206.8 mm
(c) For Third Step
2
2
1.69r3 r3
8.451r3
r
2
3
1.69r3 r3
2
2
0.9523
0.476 2
r3 09523 or r32 35.508r3 4.001
2
17.754
2
4.001 17.754 319.206 17.866
2
2
r3 17.866 17.754 0.112 m or 112 mm
R3 112 1.69 189.3 mm
11. A 100 mm wide and 10 mm and thick belt transmits 5 kW between two parallel
shafts. The distance between the shaft centers is 1.5 m and the diameter of the
smaller pulley is 440 mm. The driving and the driven shafts rotate at 60 rpm and
150 rpm respectively. Find the stress in the belt if the two pulleys are connected
by: (i) an open belt, and (ii) a cross belt. The coefficient of friction is 0.22.
Solution:
P 5kW, b=100 mm, C=1.5m, t=10mm, r=220mm, N1 60rpm, 0.22,N2 150rpm.
t 2 N2
t 2 150
10
2 r
r
220 3535 mm/s or 3.535 m/s
60 2
60
2
2
P T2 T2 or 5000= T1 T2 3.535 or T1 T2 1414.5N
T1
e , where = angle of lap or contact
T2
Remember that in designing the belt drive, the lesser of the angle of contact
on the smaller and the larger pulleys has to be considered. Also note that the smaller
pulley has lesser angle of contact.
150
220
220
R r
60
2 2sin1
2sin1
1500
C
o
25.4 0.443
2.698 rad.
T1
e0.222.698 1.81 or T1 1.81T2
T2
(i) Open-belt Drive
But
T1 T2 1414.5
1.81 T2 T1 1414.5 or 0.81 T2 1414.5
T2 17463.N and T1 1746.3 1.81 3160.8N
Stress in the belt, f1
T1
3160.8
3.16N/ mm2
b t 100 10
(i) Cross-belt Driven Drive
R r
C
2 2 sin1
or
150
220
220
o
60
= +2sin-1
61.8
1500
or
= +1.08=4.22 rad
T1
e0.224.22 2.53
T2
But
T1 T2 1414.5
2.53 T2 T1 1414.5
or
T2 924.5N and T1 924.5 2.53 2339N
ft
2339
2.339 N/mm2 .
100 10
12. A belt drive is required to transmit 10 kW from a motor running at 600 rpm.
The belt is 12 mm thick and has a mass density of 0.001 g/mm3. Safe stress in the
belt is not to exceed 2.5 N/mm2. Diameter of the driving pulley is 250 mm,
whereas the speed of the driven pulley is 220 rpm. The two shafts are 1.25 m apart.
The coefficient of friction is 0.25.
Determine the width of the belt.
Solution:
P 10kW,t 12mm, 0.001gm / mm3 1000kg / m3 ,R1 125mm,N1 600mm,
C 1.25m,N2 220rpm, 0.25, ft 2.5 N/mm2 2.5 10 6 N / m2
To calculate the width of the belt, we need to know the maximum tension in
the belt which is the sum of the tight side tension and the centrifugal tension, i.e.,
T=T1+Tc. Calculation of T1
P T1 T2
t 2 N t
= R1
R1
2 60 2
where
=
2 600
12
125 8230 mm / s or 8.23 m/s
60
2
10 000= T1 T2 8.23
or
T1 T2 1215
Also
T1
R R1
e , where = - 2 = 2sin1 2
T2
C
or
600
125
125
220
= -sin
1250
i
= -19.9o 0.347 2.79
or
T1
e0.252.79 2.01 or T1 2.01T2
T2
ii
From i and ii ,2.01 T2 T1 1215 or T2 1203 N and T1 2418 N
Calculation of Tc
Tc m 2
=mass per unit length 2
volume per unit length density 2
x sec tional area length density 2
width thickness length density 2
812.8b N
T T1 Tc ft b t
b in m
2418 812.8b 2.5 10 6 b 0.012 or 29 187 b = 2418
b=0.0828 m or 82.8 mm.
13. The groove on the pulleys of a multiple rope drive have an angle of 50o and
accommodate ropes of 22 mm diameter having a mass of 0.8 kg per meter length
for which a safe operating tension 1200 N has been laid down. The two pulleys
are of equal size. The drive is designed for maximum power conditions. Speed of
both the pulleys is 180 rpm. Assuming coefficient of friction as 0.25, determine
the diameters of the pulleys and the number of ropes when the power transmitted
is 150 kW.
Solution :
T 1200 N,P=150 000 W,m=0.8kg/m length, =180o... two pulley are of equal size
50
25o , 0.25, =0.25, N1 N2 180 rpm
2
Under maximum power conditions.
2
2
T 1200 800 N
3
3
Tc T T1 1200 800 400 N
T1
Also Tc m 2 or 400=0.8 2 or = 22.36 m/s
Or
D1N1
60
22.36 or
D1 180
60
22.36
D1 2.37 m, Also D2 2.37 m
T1
180
1
e / sin e0.25
6.41
T2
180 sin 25o
T2
T1
800
124.8N
6.41 6.41
P T1 T2 .n
150 000 = 800-124.8 22.36 n
n 9.94 say 10 ropes.
14. The following data relate to a rope drive:
Power transmitted = 20 kW, Diameter of pulley = 480 mm, Speed = 80 rpm, Angle
of lap on smaller pulley = 160o, Number of ropes = 8, Mass of rope/m length = 48
G2 kg, Limiting working tension = 132 G2 kN, Coefficient of friction = 0.3, Angle of
groove = 44o
If G is the girth of rope in m, determine the initial tension and the diameter of
each rope.
Solution:
Power transmitted/rope = 20 000/8 = 2500 W
Velocity of rope
DN
60
0.48 80
60
2.01 m/s
Now,
P T1 T2
or
2500= T1 T2 2.01
or
T1 T2 1244 N
i
Also
T1
160
1
e / sin e0.3
9.359
T2
180 sin22o
ii
From (i) and (ii),
9.359 T2 T2 1244
T2 148.8 N, T1 148.8 9.359 1392.8N
T1 T2 1392.8 148.8
770.8N
2
2
Bow, Working tension = Tension on tight side + Centrifugal tension
Initial tension
T1 m 2
132 000 G2 1392.8 48 G2. 2.01
2
131 806 G2 1392.8
G2 0.01056 or G=0.1028
Now girth (circumference) of rope = d = 0.1028 or d = 0.327 mm.
15. 2.5 kW of power is transmitted by an open-belt drive. The linear velocity of
the belt is 2.5 m/s. The angle of lap on the smaller pulley is 165 o. The coefficient
of friction is 0.3.
Determine the effect on power transmission in the following cases:
(i)
(ii)
(iii)
Initial tension in the belt increase by 8%
Initial tension in the belt is decreased by 8%
Angle of lap is increased by 8% by the use of an idler pulley, for the
same speed and the tension on the tight side, and
Coefficient of friction is increased by 8% by suitable dressing to the
friction surface of the belt.
(iv)
Solution:
P 2.5kW, 0.3, =165o , =2.5m/s
P= T1 T2
2500 T1 T2 2.5 or T1 T2 1000 N,
T1
e e0.3165 / 180 2.37
T2
or
T1 2.37T2
2.37T2 T2 1000 or T2 729.9N
T1 729.9 2.37 1729.9N
Initial tension, T0
(i)
T1 T2 1729.3 729.9
1229.9N
2
2
When initial tension is increased by 8%
T '0 1229.9 1.08 1328.3N
or
T1 T2
1328.3 or T1 T2 2656.6
2
As remain unchanged, e or
2.37T2 T2 2556.6
T1
is same.
T2
T2 788.3 N, T1 1868.3N
P T1 T2 1868.3 788.3 2.5 2700 W or 2.7 kW
Increase in power
(ii)
2.7 2.5
0.08 or 8%
2.5
When initial tension is decreased by 8%
T'0 1229.9 1 0.08 1131.5
or
T1 T2
1131.5 or T1 T2 2263
2
3.37 T2 = 2263
T2 671.5N, T1 1591.5 N
P= 1591.5-67.5 2.5 2300 W or 2.3 kW
Decrease in power
2.5 2.3
0.08 or 8%
2.5
(iii)
T1
e
T2
T1 is the same as before, whereas increases by 8%
1651.08
0.3
1729.9
180
e
2.54
T2
T2 680.5N
P 1729.9 680.5 2.5 2624W or 2.624 kW
Increase in power
iv
2.624 2.5
0.0496 ot 4.96%
2.5
165
0.31.08
T1
180
e e
2.54
T2
or T1 2.54 T2
T1 T2 1229.9 2 2459.8
T2 694.9N
T1 694.9 2.54 1764.9N
P 1764.9 694.9 2.5 2675Wor2.675kW
Increase in power
2.675 2.5
0.07 or 7%
2.5
16. In a belt drive, the mass of the belt is 1 kg/m length and its speed is 12 m/s.
The drive transmits 19.2 kW of power. Determine the initial tension in the belt
and the strength of the belt. Coefficient of frictions 0.25 and the angle of lap on
smaller pulley in 220o.
Solution:
P T1 T2
or
19 200= T1 T2 12
or
T1 T2 1600
i
220
0.25
T1
180
e / sin e
e0.96 2.61
T2
or
T1 2.61T2
From (i) and (ii),
2.61T2 T2 1600
T2 994 N, T1 2594 N
ii
Centrifugal tension m2 1 122 144 N
Initial tension To
T1 T2
2594 994
Tc
144 1938N
2
2
Strength of the belt = Total tension on he tight side
T1 Tc 2594 144 2738 N.
17. The initial tension in a belt drive is found to be 600 N and the ratio of frictions
1.8. The mass of the belt is 0.8 kg/m length. Determine (i) the velocity of the belt
for maximum power transmission (ii) tension on the tight side of the belt when it
is started (iii) tension on the tight side of the belt when running at maximum
speed.
Solution:
From equation, for maximum power transmission with consideration of
initial tension,
To
3m
600
15.8 m/s
3 0.8
i
=
ii
Tension on the tight side of the belt when it is started
T1
iii
2kTo 2 1.8 600
771.4 N
k 1
1.8 1
Tension on the tight side of the belt when running at
maximum speed Centrifugal tension = m 2 0.8 15.8 2 199.7N
Initial tension
T0
600
T1 T2
Tc
2
T1 T1 /1.8
199.7
2
0.778 T1 400.3 or T1 514.6 N
It can also be found by applying the relation
T1
2k To Tc
k 1
2 1.8 600 199.7
1.8 1
514.6 N.
UNIT - III
GEARINGS & CAMS
The advantages of Gear drive.
1.
2.
3.
4.
Exact velocity ratio is obtained, since there is no slip.
It is capable of transmitting larger power than that of the belt and chain drives.
It is more efficient and effective means of power transmission.
It requires less space compared to belt and rope drives.
Prove or disprove that in a spur gear pair, pure rolling occurs only at one point
along the path of contact.
(AU –NOV 2003).
Solution :
Velocity of sliding in spur gear pair = (W1 + W2) x AP
where:
W1, W2 = angular velocities of the gears.
AP = The distance of point A in the path of contact to the pitch point. Where point A
coincides with P.
AP – 0, then pure rolling will take place at P only.
epicyclic gear train
When the gears are arranged in such a manner that one or more gears move
upon and around another gear, then the gear train is known as epicyclical gear train.
Examples:
– Differential gears of automobile
– Back gear of lathe
– Wrist watches
The limitations of gear drive.
1.
2.
3.
The manufacture of gear requires special tools and equipments.
The manufacturing cost is comparatively high.
The error in cutting teeth may cause vibrations and noise during operation.
Classify gears?
a. Based on the position of teeth on the wheel:
(i)
Straight gears
(ii)
Helical gears
(iii) Heving bone gears
(iv) Curved teeth gears.
b. Based on peripheral speed: (v)
(i)
Low velocity – V < 3 m/s
(ii)
Medium velocity gears – V = 3 to 15 m/s
(iii) High velocity gears – V > 15 m/s.
c. Based on the relative of the shafts
(i)
Row gears
(ii)
Planetary gears.
Pitch circle in gears.
It is an imaginary circle which by pure rolling action, would give the same
motion as the actual gear.
Pitch point.
It is a common point of contact between two pitch circles.
Pitch surface:
It is the surface of the rolling discs which the meshing gears have replaced at the
pitch circle.
circular pitch with reference to gears.
It is the distance measured along the circumference of the pitch circle from a
point on one tooth to the corresponding point on the adjacent tooth.
PC =
D
D = Diameter of pitch circle,
T
T = No of teeth on the wheel.
diametral pitch in reference to gears.
It is the ratio of number of teeth to the pitch circle diameter.
T
Pd =
T = No of teeth
D
D = Pitch circle diameter
the relationship between Diametral pitch and circular pitch.
Pd =
PC
Pd = Diametral pitch
PC = Circular pitch.
module
It is the ratio of the pitch circle diameter to the number of teeth.
D
m=
T
addendum circle.
It is the circle drawn through the top of the teeth and is concentric with the
pitch circle.
addendum.
It is the radial distance of a tooth from the pitch circle to the top of the tooth.
addendum circle.
It is the circle drawn through the bottom of the tooth and is concentric with
the pitch circle.
clearance with reference to the terminology of gears.
It is the radial distance from the top of the tooth to the bottom of the tooth, in a
meshing gear.
Backlash in gears.
It is the difference between the tooth space and the tooth thickness along the
pitch circle.
Backlash = Tooth – Tooth thickness.
path of contact.
It is the path traced by the point of contact of two teeth from the beginning to
the end of engagement.
arc of contact
It is the path traced by a point on the pitch circle from the beginning to the
end of engagement of a given pair of teeth.
arc of approach
It is the portion of the path of contact from the beginning of the engagement to the
pitch point.
arc of recess
It is the portion of the path of contact from the pitch point to the end of the
engagement of a pair of teeth.
Define velocity ratio.
It is the ratio of speed of driving gear to the speed of the driven gear.
N A TB
N B TA
NA, NB = Speed of the driver and driven.
TA, TB = No of teeth on driver and driven.
Define contact ratio.
It is the ratio of the length of arc of contact to the circular pitch is known as
contact ratio. The value gives the number of pairs of teeth in contact.
Prob: A toothed wheel has 120 teeth, its module is 1.5mm. Find the circular pitch,
pitch diameter and the diametral pitch:
Solution :
T = 120
m = 1.5mm
(i)
Pitch diameter:
module m = D/T
D = m.T = 1.5 x 120 = 180 mm.
D = 180mm.
(ii)
Circular pitch (PC):
D 180
PC =
= 4.712mm.
T
120
(iii)
Diametral pitch(Pd) =
T 120
= 0.667 tooth/mm.
D 180
State the law of gearing.
It states that for obtaining constant velocity ratio, at any instant of teeth the
common normal at each point of contact should always pass through a pitch point,
situated on the line joining the centre of rotation of the pair of making gears.
normal and axial pitch in helical gears.
Normal pitch:
It is the distance between similar faces of adjacent teeth, along a helix on the
pitch cylinder normal to the teeth.
Axial pitch:
It is the distance measured parallel to the axis between similar faces of adjacent teeth.
the differences between involute and cycloidal tooth profile
S.No
Involute tooth profile
Cycloidal tooth profile
The centre distance should not
1. Variation in centre distance does not
affect the velocity ration.
vary.
2. Pressure angle remains constant Pressure angle varies. It is zero at
throughout the teeth.
the pitch point and maximum at the
start and end of engagement.
3. Interference occurs
No interference occurs
4. Weaker teeth
Stronger teeth.
cam.
Cam is a rotating machine element which gives reciprocating (or) oscillating
motion to another known as follower.
Classify followers.
(a)
According to the surface in contact
1.
Knife edge follower
2.
Roller follower
3.
Flat faced follower
4.
Spherical faced follower
(b)
According to the motion of the follower
Tangent cam.
When the flanks of the cam are straight and tangential to the base circle and
nose circle the cam is known as tangent cam.
Different motions of the follower
(i)
(ii)
(iii)
(iv)
Uniform motion
Simple harmonic
Uniform acceleration and retardation
Cycloidal motion.
Dwell period:
The period during which the follower remains at rest is called dwell period.
Offset follower.
When the motion of the follower is along an axis away from the axis of the
cam centre, it is called offset follower.
Base circle:
It is the smallest circle that can be drawn to the cam profile.
cam profile.
The surface of the cam which comes into contact of with the follower is
known as cam profile.
trace point:
It is a reference point on the follower to trace the cam profile
pressure angle of cam:
It is the angle between the direction of the follower motion and a normal to
the pitch curve.
pitch point in cams
It is a point on the pitch curve at which the pressure angle is maximum.
pitch circle with reference to cams.
It is the circle passing through the pitch point and concentric with the base
circle.
Define cam angle.
It is the angle of rotation of the cam for a definite displacement of the
follower.
Define lift of cam.
It is the maximum displacement of the follower from the base circle of the
cam. It is also called as the throw of the cam.
High surface stress in flat faced follower be minimised
High surface stress in the follower is minimised by matching the flat end of
the follower to a spherical shape.
Distinguish radial cam and cylindrical cam.
Radial cam
Cylindrical cam
1. In this cam, the follower reciprocates 1. In this the follower reciprocates (or)
(or)
oscillates
in
a
direction oscillates in a direction parallel to the cam
perpendicular to the cam axis.
axis.
The equation for maximum acceleration of a follower moving with cycloidal
motion.
(i)
Maximum acceleration during out stroke:
22S
aO =
2
O
where:
ω = Angular velocity of cam
S = Lift of follower
Q = Angular displacement
(ii)
During return stroke:
22S
QR =
2
R
The equation for maximum acceleration of a follower moving with simple
harmonic motion.
(i)
During outstroke:
aO =
(ii)
22S
2
2 0
During return stroke:
aR =
22S
2
2R
the equation for maximum velocity of a follower moving with SHM.
s
2
Where :
V=
ω = Angular Velocity of cam
S = Lift of follower
A = Angular displacement of cam.
The equation for maximum velocity and max-acceleration for a follower moving
with uniform acceleration and retardation motion.
(i)
Maximum Velocity:
Vmax =
(ii)
2s
Maximum acceleration:
42S
amax =
2
where:
ω = Angular velocity of cam
S = Lift or Stroke of follower
Q = Angular displacement of cam.
The equation for maximum and minimum acceleration of a Roller follower with
tangent cam, when the roller has contact with straight flanks.
Minimum acceleration of the follower:
amin = ω2(r1 + r3)
Maximum acceleration of the follower:
2 cos 2
2
amax = ω (r1 + r3)
3
cos
where
ωs = Angular velocity of the cam
r1 = Least radius of the cam
r3 = Radius of the roller.
The equation for acceleration of the Roller follower with tangent cam when the
roller has contact with the nose.
3
3
r sin 2
r cos 2
a = ω2r cos
3
2
2
2
2 2 2
l
r
sin
4 l r sin 2
Where
ω = Angular velocity of the cam
r = Distance between the cam and nose centre
= Angle turned by the cam from the
beginning of the roller displacement
= Semi-angle of action of cam or angle of ascent.
Type of motion gives the best possible motion for high speed operations and low
speed operations?
Cycloidal motion provides the best possible cam follower motion for high
speed operations.
Simple harmonic motion is suitable for low and medium speed operations.
Define undercutting in cams.
If the curvature of the pitch curve is too sharp, then the part of the cam shape
would be lost and thereafter the intended cam motion would not be achieved. Such
a cam is said to be undercut.
Undercutting in cam profile be avoided
(i)
(ii)
(iii)
By decreasing the desired follower lift, L
By increasing the cam rotation angle, and
By increasing the cam size.
The pressure angle can be reduced for a given motion requirement?
(i)
(ii)
By increasing the cam size
By adjusting the offset of the follower.
Monogram
In monogram, by knowing the values of total lift of the follower © and the
cam rotation angle (B) for each segment of the displacement diagram, we can read
directly the maximum pressure angle occuring in the segment for a particular choice
of prime circle radius(Ro).
Compare Roller and Mushroom follower.
Roller follower
1. Roller followers are extensively used
where more space is available.
2. It is used in stationary, gas engines,
oil engine and aircraft valves in engines.
Mushroom follower
1. The mushroom followers are
generally used where space is limited.
2. It is used in cams which operate the
valves in automobile engines.
The roller follower extensively used
Roller followers are extensively used where more space is available such as in
stationary gas oil engines and air craft engines.
Define cam and explain the classification of Cam and followers.
A cam is a rotating member which gives reciprocating or oscillating motion to
another element known as follower.
Classification of Followers
According to the surface in contact.
contact are as Follows:
(a)
The followers, according to the surface in
Knife edge follower :
When the contacting end of the follower has a sharp knife edge, it is called a knife
edge follower.
(b)
Roller follower:
When the contacting end of the follower is a roller, it is called a roller follower.
(c)
Flat faced or mushroom follower:
When the contacting end of the follower in a perfectly flat face, it is called faced
follower.
(d)
Spherical faced follower:
When to contacting end of the follower is of spherical shape, it is called a spherical
faced follower.
Classification of Cams:
1.
Radial or disc cam:
In radial cams, to follower reciprocates or oscillates in a direction
perpendicular to cam axis. The Cams as shown in Fig are all radial cams.
2.
Cylindrical cam:
In cylindrical cams, to follower reciprocates or oscillates in a direction parallel
to the cam axis. The follower rides in a groove at its cylindrical surface.
The various terms used in radial cam
Terms Used in Radial Cams:
1.
Base Circle :
It is the smallest circle that can be drawn to the Cam profile.
2.
Trace point:
It is the reference point on to follower and is used to generate the pitch curve.
In case of knife edge follower, the knife edge represents the trace point and the pitch
curve corresponds to the Cam profile. In a roller follower the centre of the roller
represents to trace point
3.
Pressure angle:
It is the angle between the direction of to follower motion end a
normal to
the pitch curve. The angle is very important in designing a Cam profile. It to
pressure angle is too large a reciprocating follower will jam in its bearing.
4.
Pitch point:
It is a point on the pitch curve having to maximum pressure angle.
5.
Pitch Circle:
It is a circle drawn from the centre of the Cam through to pitch point.
6.
Pitch curve:
It is the curve generated by the trace point as the follower moves relative to
the Cam. For Knife edge follower, the pitch curve and the Cam profile are same
where as for a roller follower they are separated by to radius of the roller.
7.
Prime Circle:
It is the smallest circle that can be drawn from the centre of the Cam and
tangent to the pitch curve. For a Knife edge and a flat face follower, the prime circle
and the base circle are identical. For a roller, follower, to prime circle is larger than
to base circle by the radius of the roller.
8.
Lift or stroke:
It is the maximum travel of the follower from its lowest position to the top
most position.
The Uniform velocity motion of the follower with velocity and acceleration
diagrams.
Motion of the follower:
Displacement, velocity and Acceleration diagrams when the follower moves with
uniform velocity
The displacement, velocity and acceleration diagrams when a Knife edge follower
moves with uniform velocity in Fig.
The Simple harmonic motion of the follower with velocity and acceleration
diagrams.
Simple Harmonic motion.
The displacement, Velocity and acceleration diagrams when the follower
moves with simple harmonic motion are shown in Fig: (a) (b) (c) respectively. The
displacement diagram is drawn as follows:
1.
2.
3.
4.
Draw a semi-circle on to follower stroke as diameter.
Divide to semi-circle into any number of even equal parts (Say eight)
Divide to angular displacement of the Cam during out stroke and return stroke
into same number of equal parts.
The displacement diagram is obtained by projecting to points as it in Fig.
The velocity and acceleration diagram are shown in Fig(a) & (b) respectively
Let
S = Stroke of the follower
ωo and ωR = Angular displacement of the Cam during out stroke and return stroke of
the follower respectively in radians and
ω = Angular velocity of the Cam in rad/s
ω
Time required for to out stroke of the follower in seconds
to = ω o / ω
S 1 s
ωp=
2 to
2 o
and maximum velocity of the follower on to stroke
S s
VO = Vp =
2 o
2 o
We know that the centripetal acceleration of the point P,
VP 2
2
..s 2 22s
ap =
2
OP
2
o s 2o
Maximum acceleration of the follower on to outstroke,
2 2s
aO = ap =
2
20
Similarly, maximum velocity of the follower on to return stroke.
s
VR =
2R
and maximum acceleration of the follower on to return stroke
22s
aR =
2
2R
The Uniform Acceleration and Retardation motions of the follower with velocity
and acceleration diagrams.
Displacement, Velocity and Acceleration Diagram when the follower moves
with uniform Acceleration and Retardation.
The displacement, velocity and acceleration diagrams when the follower
moves with uniform acceleration and retardation are shown in Fig (a), (b) and (c)
respectively we see that the displacement diagram consist of a parabolic curve and
may be drawn as discussed below:
1.
2.
3.
4.
Divide the angular displacement of the cam during outstroke (*0) in to any even
number of equal parts (say eight) and draw vertical lines through these points as
shown in Fig: (a)
Divide the stroke of the follower (S) in to the same number of equal even parts.
Join Aa to intersect to vertical line through point 1 at B. Similarly, obtain the
other points C; D etc as shown in Fig (a). Now join these points to obtain the
parabolic curve for the out stroke of the follower.
In the similar way as discussed above the displacement diagram for the follower
during return stroke may be drawn
Since the acceleration and retardation are uniform, therefore the velocity
varies directly with the time. The velocity diagram is shown in Fig.
Let S = Stroke of the follower
& R = Angular displacement of the cam during out stroke and return
stroke of the follower respectively and
= Angular velocity of the cam.
o
We know that time required for the follower during out stroke
tO = o /
and time required for the follower during return stroke.
TR = R
Mean velocity of the follower during out stroke:
= S/tO
and mean velocity of the follower during return stroke
= S/tR
Since the maximum velocity of the follower is equal to twice the mean
velocity, therefore maximum velocity of the follower during out stroke
Displacement, Velocity and Acceleration Diagram when the follower moves with
cycloidal motion.
VO =
2S 2s
to
o
Similarly, maximum velocity of the follower during return stroke.
2.s
VR =
R
Maximum acceleration of the follower during out stroke,
o
2 2.S 4 2 .S
aO =
to / 2
t o o
o 2
Similarly, maximum acceleration of the follower during return stroke
4.S
aR =
R 2
the Cycloidal motion of the follower with velocity and acceleration diagrams.
Cycloidal Motion
The displacement, Velocity and acceleration diagrams when the follower moves
with cycloidal motion are shown in Fig (a), (b) and (c) respectively. We know that
cycloid is a curve traced by a point on a circle when the circle rolls without slipping
on a straight line.
Let
= Angle through which the cam rotates in time + seconds and
= Angular velocity of the cam
We know that displacement of the follower after time t seconds
2
1
sin
x = S
o 2 o
Velocity of the follower after time + seconds
d
2 d
dx
2
S
cos
dt
o t 2o
o dt
(Differentiate equation (1))
2
S d
1 cos
=
o t
o
2
.S
=
1 cos
o
o
. . . (ii)
The velocity is maximum, when
2
2
1 (or)
cos
o
o
(or)
=
o
/2
Substituting
=
O/2 in equation (ii) we gave maximum velocity of the
follower during outstroke
.S
2.S
(1 1)
VO
o
R
Similarly maximum velocity of the follower during return stroke
..S
VR = 2
R
Now acceleration of the follower after time + Sec,
d 2 x .S 2 2 d
sin
dt 2
o o
o dt
[Differentiating equation (ii)]
2
2 .S 2
sin
o 2 o
d
dt
The acceleration is maximum, when
2
2
1
sin
o
o 2
(or)
O/4
Substituting = O/4 in equation (iii), we have maximum acceleration of the
follower during out stroke.
22 .S
aO =
o 2
Similarly, maximum acceleration of the follower during return stroke
22 .S
aR =
R 2
The velocity and acceleration diagram as shown in Fig (b) and (c) respectively.
Tangent cam Derive an expression for displacement, velocity and acceleration for
a tangent cam
Tangent cam
When the flanks of the cam are straight and tangential to the base circle and nose
circle, then the cam is known as tangent cam.
Let
r1 = Radius of the base circle (or) minimum radius of the cam
r2 = Radius of the roller
r3 = radius of nose
= Semi – angle of action of cam
angle of ascent
= Angle turned by the cam from the beginning of the roller with the straight
flank.
= Angular velocity of the cam.
1. When the roller has contact with straight flanks. A roller having contact with
straight flanks.
V
=
dx
dt
d
d
d
dt
d
sin d
r 1 r 2
dt
cos 2 dt
=
sin
(r1 + r2)
2
cos
Maximum velocity of the follower
sin
Vmax = (r1 + r2)
2
cos
The acceleration is maximum when
when the roller just leaves contact
with the straight flank at G or when the straight flank merges into a circular nose.
Maximum acceleration of the follower
amax =
2 cos 2
(r1 + r2)
3
cos
2
ii. When the roller has contact with the nose
A roller having contact with the circular nose at G is shown in Fig
r sin 21
= .r sin 1
1
2 L2 r 2 sin 2 1 2
L2 .r cos 21 r 3 sin 4 1
= .r cos 1
3
L2 r 2 sin 2 1 2
2
Circular Arc cam and Derive an expression for displacement, velocity and
acceleration for a,
Circular Arc cam
Circular Arc cam with flat-faced follower when the flank of the cam
connecting the base circle and nose arc of converse circular arcs then the cam is
known as circular arc cam.
Let
r1 = Minimum radius of the cam of radius of the base circle = OE
r2 = Radius of nose
R = Radius of circular flank = PE
2
1.
2.
= Semi – angle of action of cam or angle of ascent = angle EOK and
= Angle of action of cam on the circular flank.
When the flat face of the follower has contact on the circular flank and
When the flat face of the follower has contact on the nose
(i)
When the flat face of the follower has contact on the circular flank.
First of all. Let us consider that the flat face of the follower has contact at EC
i.e at the junction of the circular flank and base circle)
Maximum Velocity of the follower
Vmax = (R – r1) sin
Maximum acceleration of the follower
amax =
(R – r1)
2
and minimum acceleration of the follower
amin = 2(R – r1)cos
2.
When the flat face of the follower has contact on the nose. The flat face at the
follower having contact on the nose at C
dx dx d dx
V
=
dt d dt d
= OQ sin ( – B)
=
× OQ sin
d d d d
a=
dt d dt d
=
× OQ Cos
2 × OQ Cos
=
A cam is to give the following motion to a knife – edged follower:
1. Outstroke during 600of cam rotation: 2. Dwell for the next 300 of cam rotation: 3.
Return stroke during next 600 of cam rotation, and 4. Dwell for the remaining 2100 of
cam rotation.
The stroke of the follower is 40mm and the minimum radius of the cam is 50 mm.
The follower moves with uniform velocity during the both the outstroke and return
strokes. Draw the profile of the cam when (a) the axis of the follower passes through
the axis of the cam shaft, and
Construction
First of all, the displacement diagram, as shown in Fig. is drawn as discussed in the
following steps:
1.
2.
Draw a horizontal line AX=360 to some suitable scale. On this line, mark AS=60
to represent outstroke of the follower, ST=30 to represent dwell, TP=60 to
represent return stroke and PX=210 to represent dwell.
Draw vertical line AY equal to the stroke of the follower (i.e. 40mm) and
complete the rectangle as shown in Fig.
3.
Divide the angular displacement during outstroke and return stroke into any
equal number of even parts (say six) and draw vertical lines through each point.
4.
Since the follower moves with uniform velocity during outstroke and return and
return stroke, therefore the displacement diagram consists of straight lines. join
AG and HP.
5.
The complete displacement diagram is shown by AGHPX in Fig.
Profit of the cam when the axis of follower passes through the axis of cam shaft
The profile of the cam when the axis of the cam shaft, as shown in Fig. is drawn as
discussed in the following steps:
1.
2.
3.
4.
5.
6.
7.
Draw a base circle with radius equal to the minimum radius of the cam (i.e. 50
mm) with O as centre.
Since the axis of the follower passes through the axis of the cam shaft, therefore
mark trace point A, as shown in Fig. 20.11.
From OA, Mark angle AOS=60 to represent outstroke, angle SOT=30 to
represent dwell and angle TOP=60 to represent return stroke.
Divide the angular displacement during outstroke and return stroke (i.e. angle
AOS and angle TOP) into the same number of equal even parts as in
displacement diagram.
Join the points 1,2,3…… etc. and 0’, 1’, 2’, 3’,…. etc. from the displacement
diagram.
Now set off 1B, 2C, 3D….. etc. from the displacement diagram.
Join the points A, B, C,…..M, N, P with a smooth curve. The curve AGHPA is the
complete profile of the cam.
Notes: The points B, C, D ……L, M, N may also be obtained as follows:
1.
2.
Mark AY=40 mm on the axis of the follower, and set of Ab, Ac, Ad….. etc. equal
to the distances IB, 2C, 3D, … etc, as in displacement diagram.
From the centre of the cam O draw arcs with radii Ob, Oc, Od, etc. The arcs
intersect the produced lines O1, O2, etc. at B, C, D…L, M, N.
Profit of the cam when the axis of the follower is offset by 20 mm from the axis of
the cam shaft.
The profile of the cam when the axis of the follower is offset from the axis of the cam
shaft, as shown in Fig. is drawn as discussed in the following steps:
1. Draw a base circle with radius equal to the minimum radius of the cam (i.e.
50mm) with O as centre.
2. Draw the axis of the follower at the distance of 20mm from the axis of the cam,
which intersect the base circle at A.
3. Joint AO and draw an offset circle of radius 20mm with centre O.
4. From OA, mark angle AOS=60 to represent outstroke angle SOT=30 represent
dwell and angle TOP=60 to represent return stroke.
5. Divide the angular displacement during outstroke and return stroke (i.e. angle
AOS and angle TOP) into the same number of equal even parts as in
displacement diagram.
6. Now from the points 1,2, 3… etc and 0’, 1’, 2’, 3’, …..etc. on the base circle, draw
tangents to the offset circle and produce these tangents beyond the base circle as
shown in Fig. 20.12.
7. Now set off IB, 2C, 3D… etc and o’H,1’J….etc from the displacement diagram.
8. Joint the points A, B, C….M, N, P with a smooth curve. The curve AGHPA is the
complete profile of the cam.
A cam is to be designed for a knife edge follower with the following data:
1.
2.
3.
4.
(a)
(b)
Cam lift = 40 mm during 90 of cam rotation, the follower returns to its original
position with simple harmonic motion.
Dwell during the next 30
During the next 60 of cam rotation, the follower returns to its original position
with simple harmonic motion.
Dwell during the remaining 180
The line of stroke of the follower passes through the axis of the shaft and
The line of the stroke is offset 20 mm from the axis of the shaft.
The radius of the base circle of the cam is 40 mm. Determine the maximum velocity
and acceleration of the follower during its ascent, if the com rotates at 240 r.p.m.
Solution: Given
S = 40 mm = 0.04m; o 90 / 2 rad = 1.571 rad:
R 60 / 3 rad = 1.047 rad : N=240 r.p.m.
First of all the displacement diagram, as shown in Fig is drawn as discussed in the
following steps:
1.
2.
3.
4.
5.
6.
Draw horizontal line AX=360 to some suitable scale. On this line, make AS=90
to represent outstroke ; SR=30 to represent dwell; RP=60 to represent return
stroke and PX=180 to represent dwell.
Draw vertical line AY=40 mm to represent the cam lift or stroke of the follower
and complete the rectangle as shown in Fig.
Divide the angular displacement during outstroke and return into any equal
number of even parts (say six) and draw vertical lines through each point.
Since the follower moves with simple harmonic motion, therefore draw a
semicircle with AY as diameter and divided into six equal parts.
From points a, b, c….etc. draw horizontal lines intersecting the vertical lines
drawn through 1, 2, 3,…etc and 0’, 1’, 2’,…..etc. at B, C, D,…..M, N, P.
Join the points A,B, C,…etc with a smooth curve as shown in Fig. This is the
required displacement diagram.
a) Profile of the cam when the line of stroke of the follower passes through the axis
of the ca shift.
The profile of the cam when the line of stroke of the follower passes through the axis
of the cam shaft, a shown in the similar way as is discussed in Example.
b) Profit of the com when the line of stroke of the follower is offset 20mm from the
axis of the cam shaft.
The profit of the cam when the line of stroke of the follower is offset 20 mm
from the axis of the cam shaft, as shown in Fig.
Maximum velocity of the follower during its ascent and descent
We know that angular velocity of the cam,
2 N 2 240
25.14 rad/s
60
60
We also know that the maximum velocity of the follower during its ascent,
S 25.14 0.04
vO
1 m/s Ans.
2 O
2 1.571
and maximum velocity of the follower during its descent,
vR
S 25.14 0.04
1.51 m/s Ans.
2 R
2 1.047
Maximum acceleration of the follower during its ascent and descent
We know that the maximum acceleration of the follower during its ascent
2 2 S 25.14 0.04
aO
50.6 m/s 2 Ans.
2
2
2 O
2 1.571
2
2
and maximum acceleration of the following during its descent,
2 2 S 25.14 0.04
aR
113.8 m/s 2 Ans.
2
2
2 R
2 1.047
2
2
A cam, with a minimum radius of 25 mm, rotating clockwise at a uniform speed is
to be designed to give a roller follower, at the end of a valve rod, motion described
below:
1. To raise the valve through 50 mm during
120 rotation of the cam;
2. To keep the valve fully raised through next
30:
3. To lower the valve during next 60; and
4. To keep the valve closed during rest of the revolution i.e. 150 .
Draw the profil of the cam when (a) the line of stroke of the valve of the rod
passes through the units of the cam shaft, and (b) the line of the stroke is offset
15mm from the axis of the cam shaft.
The displacement of the valve, while being raised and lowered, is to take
place with simple harmonic motion. Determine the maximum acceleration of the
valve rod when the cam shaft rotates 100 r.p.m.
Draw the displacement, the velocity and the acceleration diagram for one complete
revolution of the cam.
Solution:
Given : S =50mm = 0.05m; O 120 2 / 3 rad;
R 60 / 3 1.047 rad ; N=100r.p.m.
Since the valve is being raised and lowered with simple harmonic motion,
therefore the displacement diagram, as shown in Fig. (a) is drawn in the similar
manner as discussed in the serious example.
a) Profile of the cam when the line of the valve rod passes through the axis of the
cam shaft.
The profile of the cam, as shown in Fig , is drawn as discussed in the following steps.
1.
Draw a base circle with centre O and radius equal to the minimum radius of the
cam (i.e. 25 mm)
2.
3.
Draw a prime circle with centre O and radius,
Draw angle AOS = 120 represent raising or out stroke of the valve, angle
SOT=30 to represent dwell and angle TCP=60 to represent lowering or return
stroke of the valve.
Divide the angular displacement of the cam during raising and lowering of the
valve (i.e. angle AOS and TOP) into the same number of equal even parts as in
displacement diagrams.
4.
5.
6.
7.
Joint the points 1,2, 3, etc. with the centre O and produce the lines beyond prime
circle shown in Fig 20.17.
Set off IB, 2C, 3D etc. equal to the displacement diagram.
Joint the points A, B, C, ….N, P, A. The curve drawn through these points is
known as period curve.
From the points A, B, C…..N, P, draw circles of radius equal to the radius of the
roller.
Joint the bottoms of the circles with a smooth curve as shown in Fig. This is the
required profile of the cam.
8.
9.
b) Profit of the cam when the line of stroke is offset 15 mm from the axis of the cam
shaft
The profile of the cam when the line of stroke is offset from the axis of the cam shaft,
as shown in Fig. may be drawn as discussed in the following steps:
1.
2.
3.
4.
Draw a base circle with center O and radius equal to 25 mm.
Draw a prime circle with centre O radius OA = 35 mm.
Draw an off-set circle with centre O and radius equal to 15 mm.
Join OA from OA draw the angular displacement of cam i.e. draw angle AOS =
120 angle SOT = 30 and angle TOP = 60.
5. Divide the angular displacement of the cam during raising and lowering of the
valve into the same number of equal even parts (i.e. six parts) as in displacement
diagram.
6. From points 1,2,3…. etc. and 0’, 1’, 3’, … etc on the prime circle, draw tangents to
the offset circle.
7. Set off 1B, 2C, 3D… etc. equal to displacement as measured from displacement
diagram.
8. By joining the points A, B, C,….M, N, P with a smooth curve, we get a pitch
curve.
9. Now A, B, C…etc, as centre, draw circles with radius equal to the radius of
roller.
10. Join the bottoms of the circles with a smooth curve as shown in Fig. This is the
required profile of the cam.
Maximum acceleration of the valve rod
We know that angular velocity of the cam shaft,
2 N 2 100
10.47 rad/s
60
60
We also known that maximum velocity of the valve rod to raise valve,
vO
S 10.47 0.05
0.39 m/s
2 O
2 2.1
and maximum velocity of the valve rod to lower the valve,
S 10.74 0.05
vR
0.785 m/s
2 R
2 1.047
The velocity diagram for one complete revolution of the cam is shown in
Fig.(b). we know that the maximum acceleration of the valve rod to raise the valve,
2
2 2 S 10.47 0.05
aO
6.13 m/s 2 Ans.
2
2
2 O
2 2.1
2
and maximum acceleration of the valve rod to lower the valve,
2
2 2 S 10.47 0.05
OR
24.67 m/s 2 Ans
2
2
2 O
2 1.047
2
The acceleration diagram for one complete revolution of the cam is shown in Fig.
A cam drives a flat reciprocating of the cam, flower in the following manner:
During first 120 rotation of the cam, follower moves outwards through a
distance of 20 mm with simple harmonic motion. The follower dwells during next
30 of cam rotation. During and 120 of cam rotation, the follower moves inwards
with simple harmonic motion. The following dwells for the next 90 of cam rotation.
The minimum radius of the cam is 25 mm. Draw the profile of the cam.
Construction
Since the follower moves outwards and inwards with simple harmonic
motion, therefore the displacement diagram, as shown in Fig is drawn in the similar
manner as discussed earlier.
Now the profile of the cam driving a flat reciprocating follower, as shown in Fig. is
drawn as discussed in the following steps:
1.
2.
Draw a base circle with centre O and radius OA equal to the minimum radius of
the cam (i.e. 25 mm)
Draw angle AOS=120 to represent the outward stroke, angle SOT = 30 to
represent dwell and angle TOP = 120 to represent in ward stroke.
3.
4.
5.
6.
7.
Divide the angular displacement during outward stroke and inward stroke (i.e.
angles AOS and TOP) into the same number of equal even parts as in the
displacement diagram.
Join the points 1, 2, 3…. etc. with centre O and produce beyond the base circle.
From points 1, 2, 3…. etc., set off 1B, 2C, 3D …. etc. equal to the distances
measured from the displacement diagram.
Now at points B, C, D…. M, N, P, draw the position of the flat-faced follower.
The axis of the follower at all these positions passes through the cam centre.
The curve drawn tangentially to the flat side of the follower is the required
profile of the cam, as shown in Fig.
Draw a cam profile to drive an oscillating roller follower to the specifications
given below:
(a)
Follower to move outwards through an angular displacement of 20 during
the first 120
rotation of the cam;
(b)
Follower to dwell during the next 120 of cam rotation.
(c)
Follower to the dwell during the next 120 of cam rotation.
The distance between pivot centre and roller centre =120 mm: distance
between pivot centre and cam axis = 130 mm: minimum radius of cam = 40mm:
radius of roller =10mm: inward and outward strokes take place with simple
harmonic motion.
Construction
We know that the angular displacement of the roller follower
20 20 /180 / 9 rad
Since the distance between the pivot centre and the roller centre (i.e. the radius A 1 A)
is 120 mm, therefore length of the are AA2 as shown in Fig. along which the
displacement of the roller actually takes place.
120 / 9 41.88 mm
…( Length of are = Radius of are x Angle subtended by the are at the centre in
radians)
Since the angle is very small, therefore length of chord AA 2, is taken equal to the
length of are AA2. Thus in order to draw the displacement diagram, we shall take lift
of the follower equal to length of chord AA2 .i.e. 41.88 mm.
The outward and inward strokes take place with simple harmonic motion, therefore
the displacement diagram, as shown in Fig. is drawn in the similar way as discussed
in Example.
The profile of the cam to drive an oscillating roller follower, as shown in Fig. is
drawn as discussed in the following steps;
1. First of all, draw a base circle with centre O and radius equal to the minimum
radius of the cam (i.e. 40 mm)
2. Draw a prime circle with centre O and radius OA
a. = min. radius of cam + radius of roller
b. = 40 +10 = 50 mm
3. Now locate the pivot centre A1 such that OA1=130mm and AA1 =120 mm. Draw a
pivot circle with centre O and radius OA1=130 mm.
1.
2.
3.
4.
5.
6.
7.
Join OA, Draw angle A1OS = 120 to represent the outward stroke of the
follower, angle SOT = 120 to represent the inward stroke of the follower and
angle TOA1=120 to represent the dwell.
Divide angles A1 OS and SOT into the same number of equal even parts as in the
displacement diagram and mark points 1, 2, 3, …..4’, 5’, 6’, on the pivot circle.
Now with points 1,2,3……4’, 5’, 6’, (on the pivot circle) as centre radius equal to
A1A (i.e.120mm) draw circle arcs to intersect the prime circle at points 1, 2,
3,…..4’, 5’, 6’,.
Set off the distances 1B, 2C, 3D…..4’L, 5’M along the arcs drawn equal to the
distances as measured from the displacement diagram.
The curve passing through the points A, B, C, D….L, M, N is known as pitch
curve.
Now draw circles with A,B, C, D…L, M, N as centre and radius equal to the
radius of roller.
Join the bottoms of the circles with a smooth curve as shown in Fig. This is the
required profile of the cam.
A cam, with a minimum radius of 50mm, rotating clockwise at a uniform speed, is
required to give a knife edge follower the motion as described below:
1. To move outwards through 40 mm during 100 rotation of the cam; 2. To dwell
from for next 80 ; 3. To return to its starting position during next 90; and 4. To
dwell for the rest period of a reduction i.e. 90.
Draw the profile of the cam
(i)
When the line of stroke of the follower passes through the centre of the
cam shaft, and
(ii)
When the line of stroke of the follower is off-set by 15 mm.
The displacement of the follower is to take place with uniform acceleration and
uniform retardation. Determine the maximum velocity and acceleration of the
follower when the cam shaft rotates at 900 r.p.m.
Draw the displacement, velocity and acceleration diagram for one complete
revolution of the cam.
Solution. Given.
S 40mm 0.04m; o 100 100 /180 1.745 rad;
R 90 / 2 1.571 rad; N=900 r.p.m.
First of all, the displacement diagram, as shown in Fig. (a), is drawn as discussed is
the following steps:
1. Draw a horizontal line ASTPQ such that AS represents the angular displacement
of the cam during period of 80 after outward stroke. The line TP represents the
angular displacement of the cam during return stroke (i.e.90) and the line PQ
represent the dwell period of 90 after return stroke.
2. Divide AS and TP into any number of equal even parts (say six).
3. Draw vertical lines through points 0, 1, 2, 3, etc, and equal to the lift of the valve
i.e. 40 mm.
4. Divide the vertical lines 3-f and 3’-f’ into six equal parts as shown by points a, b, c,
….and a’, b’, c’, …. in Fig. 20.24(a).
5. Since the follower moves with equal uniform acceleration and uniform
retardation, there fore the displacement diagram of the outward and return stroke
consists of a double parabola.
6. Join Aa, Ab and Ac intersecting the vertical lines through 1,2 and 3 at B, C and D
respectively.
7. Join the points B, C and D with a smooth curve. This is the required parabola for
the half outstroke of the valve. Similarly the other curves may be drawn as shown in
Fig.
8. The curve ABC…..NI’Q is the required displacement diagram.
(i) Profile of the cam when the line of stroke of the follower passes through the centre
of the cam shaft
The profile of the cam when the line of stroke of the follower passes through
the centre of cam shaft, as shown in Fig. 20.25, may be drawn as discussed in the
following steps:
1.
2.
3.
4.
5.
6.
Draw a base circle with centre O and radius 50 mm (equal to minimum radius of
the cam).
Divide the base circle such that angle AOS = 1000; angle SOT = 800 and angle TOP
= 900.
Divide angles AOS and TOP into the same number of equal even parts as in
displacement diagram (i.e. six parts).
join the points 1,2,3…and 1’,2’,3’,…..with centre O and produce these lines
beyond the base circle.
From points 1,2,3,….and 1’,2’,3’,….. mark the displacements 1B, 2C, 3D…. etc, as
measured from the displacement diagram.
Join the points A,B,C……M,N,P with a smooth curve as shown in Fig. This is the
required profile of the cam.
ii) Profile of the cam when the line of stroke the follower is offset by 15 mm.
The profile of the cam when the line of stroke of the follower is offset may be
drawn as discussed in Example. The profile of cam is shown in Fig.
Maximum velocity of the follower during out stroke and return stroke
We know that angular velocity of the cam shaft,
2V 2 900
94.26 rad/s
60
60
We also know that the maximum velocity of the following during out stroke,
vO
2S 2 94.26 0.04
4.32 m / s
O
1.745
and maximum velocity of the follower during return stroke,
vR
2S 2 94.26 0.04
4.8 m/s
R
1.571
The velocity diagram is shown in Fig.
Maximum acceleration of the follower during out stroke and return stroke
We know that the maximum acceleration of the follower during out stroke,
aO
42S
O
2
4(94.26)
1.571
2
576m / s2
The acceleration diagram is shown in Fig. (c)
Design a cam for operating the exhaust valve of an oil engine. It is required to
give equal uniform acceleration and retardation during opening and closing of the
valve each of which corresponds to 600 of cam rotation. The valve must remain in
the fully open position for 200 of cam rotation.
The lift of the valve is 37.5 mm and the least radius of the cam is 40mm. The
follower is provided with a roller of radius 20 mm and its line of stroke passes
through the axis of the cam.
Construction:
First of all, the displacement diagram, as shown in Fig. is drawn as discussed
in the following steps:
1. Draw a horizontal line ASTP such that AS represents the angular displacement of
the cam during opening (i.e. out stroke) of the valve (equal to 600), to some suitable
scale. The line ST represents the dwell period of 200 i.e. the period during which the
valve remains fully open and TP represents the angular displacement during closing
(i.e return stroke) of the valve which is equal to 600.
2. Divide AS and TP into any number of equal even parts (say six).
3. Draw vertical lines through points 0,1,2,3 etc. and equal to lift of the valve i.e. 37.5
mm.
4. Divide the vertical lines 3f and 3’f’ into six equal parts as shown by the points
a,b,c,….. and a’, b’, c’…in Fig.
5. Since the valve moves with equal uniform acceleration and retardation. Therefore
the displacement diagram for opening and closing of a valve consists of double
parabola.
6. Complete the displacement diagram as shown in Fig.
Now the profile of the cam, with a roller follower when its line of stroke
passes through the axis of cam, as shown in Fig. is drawn in the similar way as
discussed in Example.
A canrotating clockwise at a uniform speed of 1000 r.p.m. is required to give a
roller follower the motion defined below:
1.
2.
3.
4.
Follower to move outwards though 50 mm during 1200 of cam rotation.
Follower to dwell for next 600 of cam rotation.
Follower to return to its starting position during next 900 of cam rotation.
Follower to dwell for the rest of the cam rotation.
The minimum radius of the cam is 50 mm and the diameter of roller is 10 mm.
The line of stroke of the follower is off-set by 20 mm from the axis of the cam shaft.
If the displacement of the follower takes place with uniform and equal acceleration
and retardation on both the outward and return strokes, draw profile of the cam and
find the maximum velocity and acceleration during and stroke and return stroke.
Solution:
Given : N =1000 r.p.m, ; S = 50 mm ; O = 1200= 2 /3 rad = 2.1 rad ; OR = 900 =
/2 rad = 1.571 rad
Since the displacement of the follower takes place with uniform and equal
acceleration and retardation on both outward and return strokes, therefore the
displacement diagram, as shown in Fig. is drawn in the similar manner as discussed
in the previous example. But in this case, the angular displacement and stroke of the
follower is divided into eight equal parts.
Now, the profile of the cam, as shown in Fig, is drawn as discussed in the following
steps:
1.
2.
3.
4.
5.
6.
7.
8.
9.
Draw a base circle with centre O and radius equal to the minimum radius of the
cam (i.e. 50 mm).
Draw a prime circle with centre O and radius
OA = Minimum radius of the cam + radius of roller = 50+5 = 55 mm.
Draw and off set circle with centre O and radius equal to 20 mm.
Divide the angular displacements of the cam during out stroke and return stroke
into eight equal parts as shown by points 0,1,2…….and 0’,1’,2’……etc. on the
prime circle in Fig.
From thee points draw tangents to the off-set circle.
Set off 1B, 2C,3D…. etc. equal to the displacements as measured from the
displacement diagram.
By joining the points A,B,C……T,U,A with a smooth curve, we get a pitch curve.
Now from points A,B,C……T,U, draw circles with radius equal to the radius of
the roller.
Join the bottoms of these circles with a smooth curve to obtain the profile of the
cam as shown in Fig.
Maximum velocity of the follower during out stroke and return stroke
We know that angular velocity of the cam.
2N 2 100
104.7rad/s
60
60
We also know that the maximum velocity of the follower during outstroke,
O
2S 2 104.7 0.05
5 m/s
O
2.1
and maximum velocity of the follower during return stroke,
R
2S 2 104.7 0.05
6.66 m / s
R
1.571
Maximum acceleration of the follower during out stroke and return stroke
We know the maximum acceleration of the follower during out stroke,
aO
42 .S
O
2
4 104.7 0.05
2
2.1
2
497.2 m/s2
and maximum acceleration of the follower during return stroke,
aR
42 .S
R
2
4(104.7)2 0.05
1.571
2
888 m/s2
Construct the profile of a cam to suit the following specifications:
Cam shaft diameter = 40 mm; Least radius of cam = 25 mm ; Diameter of roller
= 25 mm Angle of lift = 1200 ; Angle of fall = 1500 ; Lift of the follower = 40 mm ;
Number of pauses are two of equal interval between motions.
During the lift, the motion is S.H.M. During the fall the motion is uniform
acceleration and deceleration. The speed of the cam shaft is uniform. The line of
stroke of the follower is off-set 12.5 mm from the centre of the cam.
Construction:
First of all the displacement diagram, as shown in Fig. is drawn as discussed
in the following steps :
1.
Since the follower moves with simple harmonic motion during lift (i.e. for
0
120 of cam rotation), therefore draw the displacement curve ADG in the similar
manner as discussed in Example
2.
Since the follower moves with uniform acceleration and deceleration during
fall (i.e. for 1500 of cam rotation”), therefore draw the displacement curve HLP
consisting of double parabola as discussed in Example.
Now the profile of the cam, when the line of stroke of the follower is off-set 12.5 mm
from the centre of the cam, as shown in Fig. is drawn as discussed in the following
steps:
1.
Draw a base circle with centre O and radius equal to the least radius of cam
(i.e. 25 mm).
2.
3.
4.
5.
Draw a prime circle with centre O and radius,
OA = Least radius of cam + radius of roller = 25 +25/2 = 37.5 mm
Draw a circle with centre O and radius equal to 20 mm to represent the cam
shaft.
Draw an offset circle with centre O and radius equal to 12.5 mm.
Join OA. From OA draw angular displacements of the cam, i.e. draw angle
AOS = 1200 to represent lift of the follower, angle SOT = 450 to represent
pause, angle TOP = 1500 to represent fall of the follower and angle POA = 450
to represent pause.
6.
7.
8.
9.
10.
11.
3600 1200 1500
2
45
0
Divide the angular displacements during lift and fall (i.e. angle AOS and
TOP) into the same number of equal even parts (i.e. six parts) as in the
displacement diagram.
From points 1,2,3……etc. and 0’,1’,2’,3’….etc, on the prime circle, draw
tangents to the off-set circle.
Set off 1B, 2C, 3D….etc, equal to the displacements as measured form the
displacement diagram.
By joining the points A,B,C…. M,N,P with a smooth curve, we get a pith
curve.
Now with A,B,C….. etc. as centre, draw circles with radius equal to the
radius of roller.
Join the bottom of the circles with a smooth curve as shown in Fig. this is the
required profile of the cam.
It is required to set out the profile of a cam to give the following motion to the
reciprocating follower with a flat mushroom contact face:
i) Follower to have a stroke of 20 mm during 1200 of cam rotation.
ii) Follower to dwell for 300 of cam rotation:
iii) Follower to return to its initial position during 1200 of cam rotation ; and
iv) Follower to dwell for remaining 900 of cam rotation.
The minimum radius of the cam in 25 mm. the out stoke of the follower is
performed with simple harmonic motion and the return stroke with equal uniform
acceleration and retardation.
Construction:
Since the out stroke of the follower is performed with simple harmonic
motion and the return stroke with uniform acceleration and retardation, therefore
the displacement diagram, as shown in Fig, is drawn in the similar manner as
discussed in the previous example.
The profile of the cam with a flat mushroom contact face reciprocating
follower, as shown in Fig. is drawn in the similar way as discussed in Example.
It is required to set out the profile of a cam with oscillating follower for the
following motion:
a) Follower to move outward through an angular displacement of 20 0 during 900 of
cam rotation ; (b) Follower to dwell for 450 of cam rotation ; (c) Follower to return its
original position of Zero displacement in 750 of cam rotation ; and (d) Follower to
dwell for the remaining period of the revolution of the cam.
The distance between the pivot centre and the follower roller centre is 70 mm
and the roller diameter is 20 mm. The minimum radius of the cam corresponds to
the starting position of the follower as given in (a). The location of the pivot point is
70 mm to the left and 60 mm above the axis of rotation of the cam. The motion of
the follower is to take place with S.H.M. during out stroke and with uniform
acceleration and retardation during return stroke.
Construction:
We know that the angular displacement of the roller follower,
200 20 /180 / 9 rad
Since the distance between the pivot centre and the roller centre (i.e. radius
A1A2) is 70 mm, therefore length of arc AA2, as shown in Fig. along which the
displacement of the roller actually takes place
= 70x /9=24.5 mm
Since the angle is very small, therefore length of chord AA2 is taken equal to
the length of are AA2. Thus in order to draw the displacement diagram, we shall take
lift of the follower equal to the length of chord AA2 i.e. mm.
The follower moves it simple harmonic motion during out stroke and with
uniform acceleration and retardation during return stroke.
Therefore, the
displacement diagram, as shown in Fig. is drawn in the similar way as discussed in
the previous example.
The profile of the cum. A shown in fig, is drawn as discussed in the following steps:
1.
2.
3.
First of all, locate the pivot point A1 which is 70 mm to the left and 60 mm
above the axis of the cam.
Since the distance between the pivot centre A1 and the follower roller centre
A is 70 mm and the roller diameter is 20 mm, therefore draw a circle with
centre A and radius equal to the radius of roller i.e. 10 mm.
we find that the minimum radius of the cam
= 60-10 =50 mm
Radius of the prime circle,
OA = Min. radius of cam +Radius of roller = 50+10=60 mm
4. Now complete the profile of the cam in the similar way as discussed in
Example.
Draw the profile of the cam when the roller follower moves with cycloidal motion
during out stroke and return stroke, as given below:
1.
Out stroke with maximum displacement of 31.4 mm during 1800 of cam
rotation.
2.
Return stroke for the next 1500 of cam rotation.
3.
Dwell for the remaining 300 of cam rotation.
The minimum radius of the cam is 15 mm and the roller diameter of the follower
is 10 mm. the axis of the roller follower is offset by 10 mm towards right from the
axis of cam shaft.
Construction:
First of all, the displacement diagram, as shown in Fig. is drawn as discussed in the
following steps:
1.
4.
Draw horizontal line ASP such that AS = 1800 to represent the out stroke, SN =
300 to represent the dwell period.
Divide AS and SN into any number of even equal parts (say six).
From the points 1,2,3….. etc. draw vertical lines and set-off equal to the stroke of
the follower.
From a point G draw a generating circle of radius,
5.
r
6.
Divide the generating circle into six equal parts and from these points draw
horizontal lines to meet the vertical diameter at a’ G and b’.
Join AG and GN. From point a’ draw, lines parallel to AG and GN to interest the
vertical lines drawn through 1, 2, 4’ and 5’ at B, C, L and M respectively.
Similarly draw parallel lines from b’ interesting the vertical lines through 4,5,1’
and 2’ at E, F, H and J respectively.
2.
3.
7.
Stroke 31.4
5mm
2
2
8. Join the points A,B,C…L,M,N with a smooth curve.
9. The curve A B C……LMN is the required displacement diagram.
Now the profile of the cam, as shown in Fig.
Prob: In a symmetrical tangent cam operating a roller follower, the least of the cam
is 30 mm and roller radius is 17.5mm. The angle of ascent is 750 and the total lift is
17.5mm. the speed of the cam shaft is 600 r.p.m. Calculate: 1. the principal
dimensions of the cam; 2. the accelerations of the follower at the beginning of the
lift, where straight flank merges into the circular nose and at the apex of the
circular nose. Assume that there is no dwell between ascent and descent .
Solution: Given: r1=30mm; r2=17.5mm; =750; Total lift = 17.5 mm; N=600 r.p.m or
=2 600/60=62.84 rad/s
1.
Principal dimensions of the cam
Let
r=OK=Distance between cam centre and nose centre,
r3= Nose radius, and
= Angle of contact of cam with straight flanks.
From the geometry of Fig. 20.42,
R+r3=r1+Total lift
= 30 +17.5=47.5 mm
r=47.5-r3
…(i)
Also, OE=OPPE or
r1=OP+r3
OP=r1 –r3=30-r3
…(ii)
Now from right angled triangle OKP,
Or
OP=OK x cos
…( cos =OP/OK)
0
30-r3 = (47.5 r3)cos75 =(47.5-r3) 0.2588=12.3-0.2588r3
…( OK=r)
and
r3=23.88mm
r=OK=47.5-23.88=23.62mm
Again, from right angled triangle ODB,
tan
DB KP OK sin 23.62sin750
0.4803
OB OB
r2
30 17.5
=25.60
2. Acceleration of the follower at the beginning of the lift
We know that acceleration of the follower at the beginning of the lift, i.e.
when the roller has contact at E on the straight flank,
Amin=2(+r2)=(62.84)2(30+17.5)2=187 600 mm/s2
=187.6m/s2
Acceleration of the follower where straight flank merges into a circular nose.
We know that acceleration of the follower where straight flank merges into a
circular nose i.e. when the roller just leaves contact at G.
2 cos2
2 cos2 25.60
2
amax 2 ( r2 )
62.84 30 17.5
2
3
0
cos
cos 25.6
2-0.813
2
2
=187600
303 800mm / s 303.8 m/s
0.733
Acceleration of the follower at the apex of the circular nose
We know that acceleration of the follower for contact with the circular nose,
L2 .r cos 21 r 3 sin4 1
a 2 .r cos 1
3/2
L2r 2 sin2 1
Since 1 is measured from the top position of the follower, therefore for the follower
to have contact at the apex of the circular nose (i.e. at point H). 1=0.
Acceleration of the follower at the apex of the circular nose,
L2 .r
r
r
a 2 .r 1 3 2 .r 1 2 .r 1 2 3
L
L
r r
23.62
2
= 62.84 23.62 1
146 530mm / s2 ...( L r 2 r 3 )
17.5 23.88
=146.53m/s2
A cam has straight working faces which are tangential to a base circle of diameter
90mm. The follower is a roller of diameter 40 mm and the centre of roller moves
along a straight line passing through the centre line of the cam shaft. The angle
between the tangential faces of the cam is 900 and the faces are joined by a nose
circle of 10mm radius. The speed of rotation of the cam is 120 revolutions per min.
Find the acceleration of the roller centre 1. When during the lift, the roller is
just about to leave the straight flank; and 2. When the roller is at the outer end of its
lift.
Solution: Given: d1=90mm or r1=45mm;d2=40mm or r2=20mm; 2 =900 or =45;
r3=10mm; N=120r.p.m. or =2 x 120/60 = 12.57 rad/s
The tangent cam operating a roller follower is shown in Fig.
First of all. Let us find the *angle turned by the cam () when the roller is just about
to leave the straight flank at G. The centre of the roller at this position lies at D.
Since the cam is assumed to be stationary. is the angle turned by the roller when it
is just about to leave the straight flank at G.
From the geometry of the figure,
BD=PK=OP=OE-PE
=OE-KG
=r1=r3 = 45-10=35mm
Now from triangle OBD.
BD
BD
OB OE EB
BD
35
=
0.5385
r1 r2 45 20
tan
=28.30
1. Acceleration of the roller centre when roller is just about to leave the straight
flank.
We know that acceleration of the roller centre when the roller is just about to
leave the straight flank,
2 cos2
2 cos2 28.30
2
a 2 ( r2 )
12.57
45
20
2
3
0
cos
cos 28.3
=18 500mm / s2 18.5 m/s2
2. Acceleration of the roller centre when the roller is at the outer end of the lift
First of all, let us find the values of OK and KD, From the geometry of the figure,
OK r (OP)2 (PK)2 2 OP
...(
OP=PK)
= 2 OE EP 2 45 10 49.5 mm
KD=L=KG+GD=r3 r2 10 20 30mm
We know that acceleration of the roller centre when the roller is at the outer
end of the lift. i.e. when the roller has contact at the top of the nose,
L2 .r cos 21 r 3 sin4 1
r
a 2 .r cos 1
2 .r 1
3/2
2
2
2
L
L r sin 1
...( At the outer end of the lift, 1 0)
49.5
=12.572 49.5 1
20 730mm/s 2 20.73m / s2
30
A symmetrical circular cam operating a flat – faced follower has the following
particulars: Minimum radius of the cam =30mm; Total lift =20mm; Angle of lift
=750; Nose radius =5mm; speed =600 r.p.m. Find: 1. the principal dimensions of the
cam, and 2. the acceleration of the follower at the beginning of the lift, at the end
of contact with the circular flank, at the beginning of contact with the circular
flank, at the beginning of contract with nose and at the apex of the nose.
Solution: Given: r1 =OE=30mm;x=JK=20mm;=750; r2=QF=QK=5mm; N=600 r.p.m. or
=2x600/60=62.84 rad/s
1. Principal dimensions of the cam
A symmetrical circular cam operating a flat faced follower is shown in Fig.
OQ=Distance between cam centre and nose centre,
R=PE=Radius of circular flank, and
= Angle of contact on the circular flank.
Let
We know that lift of the follower (x),
20=OQ+r2-r1=OQ+5-30=OQ-2f5
OQ=20+25=45 mm
We know that PQ=PF-FQ=PE-FQ=OP+OE-FQ
=OP+30-5=(OP+25) mm
Now from a triangle OPQ,
2xOPx0Q cos
(PQ)2=(OP)2+(OQ)2-2XOPX OQ cos
(OP+25)2 =(OP)2+452-2 X OP X 45cos(1800-750)
(OP)2 +50OP+625=(OP)2 +45cos (1800-750)
(OP)2 +50Op+625=(OP)2+2025+23.3OP
50OP-23.3OP=2025-625
or
and
26.7 OP = 1400
OP=1400/26.7=52.4 mm
Radius of circular flanks,
R=PE=OP+OE=52.4+30
=82.4mm
and
PQ = OP+25=52.4+25
=77.4mm
In order to find angle , consider a triangle OPQ. We know that
OQ
PQ
sin sin
0
0
OQ sin 45 sin 180 75
sin=
0.5616
PQ
77.4
=34.20
or
2. Acceleration of the follower
We know that acceleration of the follower at the beginning of the lift,
a= 2(R-r1)cos=2(R-r1) …..( At the beginning of lift. =00)
=(62.84)2 (82.4 – 30)=206 930 mm/s = 206.93 m/s2
Acceleration of the follower at the end of contact with the circular flank,
a= 2(R-r1)cos=2(R-r1)cos
…….( At the end of contact with the circular flank, =)
=(62.84)2 (82.4 – 30)cos 34.20=171 130 mm/s2 = 171.13 m/s2
Acceleration of the follower at the beginning of contact with nose,
a 2 OQcos 2 OQcos
...
At the begining of contact with nose,=
=- 62.84 45cos 750 34.20 134 520mm/s2 134.52m / s2
2
2
=134.52m/s (Re tardation)
and acceleration of the follower at the apex of nose,
a 2 OQcos 2 OQ...
At the apex of nose,-=0
=- 62.84 45 177 700mm/s2 177.7m / s2
2
=177.7m/s2 (Re tardation)
A symmetrical cam with convex flanks operates a flat-footed follower. The lift is
8mm, base circle radius 25mm and the nose radius 12mm. The total angle of the
cam action is 1200.
1. Find the radius of convex flanks, 2. Draw the profile of the cam, and 3.
Determine the maximum velocity and the maximum acceleration when the cam
shaft rotates at 500 rpm.
Solution: Given: x=JK=8mm; r1=OE=OJ=25mm; r2=QF=QK=12mm; 2=<EOG=1200 or
=<EOK=600;N=500 r.p.m. or =2500/60=52.37 rad/s
1. Radius of convex flanks
Let
R =Radius of convex flanks = PE=P’G
A symmetrical cam with convex flanks operating a flat footed follower is
shown in fig. 20.47. From the geometry of the figure,
and
OQ=OJ+JK-QK=r1+X-r2
=25+8-12=21mm
PQ=PF-QF=PE-QF=(R-12)mm
OP=PE-OE=(R-25)mm
Now consider the triangle OPQ. We know that
(PQ)2 = (OP)2 + (OQ)2 – 2OP x OQ x cos
(R-12)2 =(R-25)2 +(21)2 – 2(R-25)21cos(1800 – 60o)
R2- 24R+144= R2 – 50R + 625 + + R – 525
-24R + 144 =-29R+ or 5R = 397
R = 397/5 = 79.4 mm Ans.
2. Profile of the cam
The profile of the cam, as shown in fig. is drawn as discussed in the following steps:
(a)
First of all, draw a base circle with centre O and radius OE = r1 = 25 mm.
(b)
(c)
(d)
(e)
Draw angle EOK = 600 such that the total angle of can action is 1200.
On line OK mark OQ = 21mm (as calculated above).Now Q as centre, draw a
circle of radius equal to the nose radius r 2 = QK = OF = 12mm. This circle cuts
the line OK at J. Now JK represents the lift of the follower (i.e. 8mm).
Produce EO and GO as shown in fig.20.47.Now with Q as centre and radius
equal to PQ = R – r2 = 79.4 – 12 = 67.4 mm, draw arcs intersecting the lines EO
and GO produced at P and P’ respectively. The centre P’ may also be obtained
by drawing arcs with centres O and Q and radii OP and PQ respectively.
Now with P and P’as centres and radius equal to R = 79.4 mm, draw arcs EF
and GH which represent the convex flanks. EFKHGAE is the profile of the cam.
3. Maximum velocity and maximum acceleration.
First of all, let us find the angle . From triangle OPQ,
OQ
PQ
sin sin
OQ
21
sin
sin
sin 1800 600
PQ
79.4 12
0.2698
15.650
We know that maximum velocity,
Vmax= (R-)sin=52.37 (79.4 – 25) sin 15.650 = 770 mm/s
= 0.77 m/s Ans.
and maximum acceleration,
amax = 2(R-) = (52.37)2 (79.4 – 25) = 149200 mm/s2=149.2 m/s2 Ans.
The following particulars relate to a symmetrical circular can operating a flat faced
follower:
Least radius = 16 mm, nose radius = 3.2 mm, distance between cam shaft centre and
nose centre = 25 mm, angle of action of cam= 1500, and cam shaft speed = 600 r.p.m.
Assuming that there is no dwell between ascent or descent, determine the lift
of the valve, the flank radius and the acceleration and retardation of the follower at a
point where circular nose merges into circular flank.
Solution: Given : r1 = OE = OJ = 16 mm; r2 = QK= QF = 3.2 mm; OQ = 25 mm; 2 =
1500 or = 750; N = 600 r.p.m or = 2 x 600/60 = 62.84 rad/s
Lift of the value
A symmetrical circular cam operating a flat faced follower is shown in fig.
We know that lift of the valve,
x = JK = OK – OJ
= OQ + QK – OJ = OQ + r2 – r1
=25 + 3.2 – 16 = 12.2 mm Ans.
Flank radius
Let
R = PE = Flank radius.
First of all, let us find out the values of OP and PQ. From the geometry of fig.
and
OP = PE – OE = R-16
PQ = PF – FQ = R-3.2
Now consider the triangle OPQ. We know that
(PQ)2 = (OP)2 + (OQ)2 – 2OP x OQ x cos
Substituting the values of OP and PQ in the above expression,
(R-3.2)2 = (R-16)2 + (25)2 – 2(R-16) x 25cos(1800-750)
R2 – 6.4R + 10.24 = R2 – 32R + 256 + 625 – (50R-800)
(-0.2588)
-6.4R+10.24=-19.06R+673.96 or 12.66 R = 663.72
R = 52.43 mm Ans.
Acceleration and retardation of the follower at a point where circular nose merges
into a circular flank
From Fig. we see that a point F, the circular nose merges into a circular flank. Let
be the angle of action of cam at point F. From triangle OPQ,
OQ
PQ
sin sin
OQ
OQ
sin
sin 180o 75o
sin105o
PQ
PF FQ
25
0.966 0.4907
52.43 3.2
29.4o
We know that acceleration of the follower,
a 2 OP cos 2 R cos
...(
=)
= (62.84)2 (52.43 – 16) cos 29.40 = 125.330 mm/s2
= 125.33 m/s2 Ans.
We also know that retardation of the follower,
a 2 OQ cos 2 OQ cos( ) ...(
=)
=(62.84)2 25 cos(750 29.40 ) 69.110mm / s2
69.11m / s2 Ans.
A flat ended valve tappet is operated by a symmetrical cam with circular arc for
flank and nose. The straight line path of the tappet passes through the cam axis.
Total angle of action = 1500, Lift = 6mm. Base circle diameter = 30 mm. Period of a
acceleration is half the period of retardation during the lift. The cam rotates at
1250 r.p.m. Find \: 1. flan and nose radii; 2. maximum acceleration and retardation
during the lift.
Solution:
Given : 2 1500 or = 750; x = JK = 6mm;
d1 =30mm or r1 = OE = OJ = 15mm; N = 1250 r.p.m or
= 2 x 1250/60 = 131 rad/s
1. Flank and nose radii
The circular are cam operating a flat ended valve tappet is shown in fig.
Let R = PE = Flank radius, and
r2 = OF = OK = Nose radius.
First of all, let us find the values of OP, OQ and PQ. The acceleration takes place
while the follower is on the flank and retardation while the follower is on nose. Since
the period of acceleration is half the period of retardation during the lift, therefore
1
2
...(i)
We know that
1800 1800 750 1050
750 1800 1800 1050 750 ... ii
From equations (i) and (ii),
250 and =500
Now from the geometry,
OQ = OJ+JK-QK= r1 + x- r2 = 15 + 6 – r2 = 21 – r2 ….(iii)
and PQ = PF – FQ = PE – FQ = (OP+OE) – FQ = OP+15-r2 …..(iv)
Now from triangle OPQ,
OP
OQ
PQ
sin sin sin
21 r2 OP 15 r2
OP
or
o
sin50
sin 250
sin1050
21-r2
21 r2
sin500
0.766 38 1.8r2 ...(v)
o
sin25
0.4226
OP+15-r2
OP 15 r2
Also OP=
sin500
0.766
0
sin105
0.966
OP=
= 0.793 x OP + 11.9 – 0.793 r2
0.207 OP = 11.9 – 0.7993 r2 or OP=57.5 - 3.83 r2 ….(vi)
From equations (v) and (vi),
38 – 1.8 r2 = 57.5 – 3.83 r2 or 2.03 r2 = 19.5
r2 = 9.6 mm Ans.
We know that
OP = 38 – 1.8 r2 = 38-1.8 x 9.6 = 20.7 mm
….[ From equation (v) ]
R = PE = OP + OE = 20.7+15 = 35.7 mm Ans.
2. Maximum acceleration and retardation during the lift
We know that maximum acceleration
2 R r1 2 OP 131 20.7 355230mm / s2
2
=355.23m/s2 Ans.
and maximum retardation,
2 OQ 2 OQ 2 (21 r2 )
...[from equation(iii)]
= (131)2 (21-9.6) =195 640 mm/s2 = 195.64 m/s2 Ans.
A cam consists of a circular disc of diameter 75 mm with its centre displaced
25mm from the camshaft axis. The follower has a flat surface (horizontal) in
contact with the cam and the line of action of the follower is vertical and passes
through the shaft axis as shown in fig. The mass of the follower is 2.3kg and is
pressed downwards by a spring which has a stiffness of 3.5 N/mm. In the lowest
position the spring force is 45N.
1.
Derive an expression for the acceleration of the follower in terms of the angle
of rotation from the beginning of the lift.
2.
As the cam shaft speed is gradually increased, a value is reached at which the
follower begins to lift from the cam surface. Determine the camshaft speed for this
condition.
Solution:
Given: d =75 mm or r = OA = 37.5 mm; OQ = 25 mm; m = 2.3 kg; s = 3.5 N/mm; S =
45 N
1.
Expression for the acceleration of the follower
The cam in its lowest position is shown by full lines in fig. and by dotted lines where
it has rotated through an angle .
From the geometry of the figure, the displacement of the follower,
x = AB = OS = OQ – QS
= OQ – PQcos
= OQ – OQ cos
….(PQ = OQ)
= OQ(1-cos) = 25(1 - cos) ….(i)
Differentiating equation (i) with respect to t, we get velocity of the follower,
v
dx dx d dx
dt d dt d
…(Substituting d /dt = )
= 25sin x = 25sin
…(ii)
Now differentiating equation (ii) with respect to t, we get acceleration of the
follower,
dv dv d
25 cos
dt d dt
252 cos mm/s2 0.025 cos m / s2 Ans.
a
2. Cam shaft speed
Let
N = Cam shaft speed in r.p.m
We know that accelerating force
=m.a = 2.3x 0.0252cos = 0.0572cos N
Now for any value , the algebraic sum of the spring force, weight of the
follower and the accelerating force is equal to the vertical reaction between the cam
and follower. When this reaction is zero, then the follower will just begin to leave the
cam.
S + s.x + m.g + m.a = 0
45 + 3.5 x 25(1-cos) + 2.3 x 9.81 + 0.05752cos = 0
45 + 87.5 – 87.5 cos + 22.56 + 0.05752cos = 0
155.06 – 87.5cos + 0.0572cos = 0
2697 – 1522cos + 2cos = 0
..(Dividing by 0.0575)
2cos =1522cos - 2697 or 2 = 1522 – 2697sec
Since sec +1 or, - 1, therefore the minimum value of 2 occurs when = 1800
therefore
2 =1522 – (-2697) = 4219
…[ Substituting sec = -1]
= 65 rad/s
and maximum allowable cam shaft speed,
N
60 65 60
621 r.p.m Ans.
2
2
UNIT - IV
BALANCING
The importance of balancing.
If the moving part of a machine are not balanced completely then the inertia
forces are set up which may cause excessive noise, vibration, wear and tear of the
system. So balancing of machine is necessary.
Why balancing of dynamic forces is necessary?
If dynamic force are not balanced, they will cause effects as wear and tear on
bearings and excessive vibrations on machines. It is very common in cam shafts,
steam turbine rotors, engine crank shafts, and centrifugal pumps etc.
Unbalanced effects of shafts in high speed machines are to be closely looked into
why?
The dynamic forces centrifugal forces or a result of unbalanced masses are a
function the angular velocity of rotation.
i.e
Fc = m2r
2 N
60
In high speed engines the angular velocity is more hence unbalanced effect on shaft
may cause serious effects.
different types of balancing.
1. Balancing of rotating masses
Static balancing
Dynamic balancing
2. Balancing of reciprocating masses
Define static balancing.
A system of rotating masses is said to be in static balance if the combined
mass centre of the system lies on the axis of rotation.
State the condition for static balancing.
The net dynamic force acting on the shaft is equal to zero. This requires that
the line of action of their centrifugal forces must be same.
Why balancing is necessary for rotating masses?
The unbalanced forces caused by the rotating masses causes forces exerted on
the frame which are time varying, which will impart vibratory motion to the frame
and produce noise.
Dynamic balancing implies static balancing – Justify.
Condition for dynamic balancing are,
1. The net dynamic force acting on the shaft is zero. This is called as condition
for static balancing.
2. The net couple due to dynamic forces acting on the shaft is zero.
. Write the condition for complete balancing.
1. The resultant centrifugal force must be zero and The resultant couple must be
zero
The product of rotating mass and perpendicular distance between the rotating
mass and reference plane is called as Mass moment.
Write the equation for balancing a single rotating mass by a single mass.
For balancing single rotating mass by a single rotating mass, the equation is m 1r1 =
m2r2
Write the equations used to solve balancing of several masses rotating in a single
plane.
Horizontal component of force : ∑H = m1 r1 cos 1 + m2 r2 cos 2 + ….
Vertical component of force : ∑V = m1 r1 sin 1 + m2 r2 sin 2 + ….
Magnitude of balancing force (F0) : Fc =
(H ) 2 (V ) 2
1
V
H
Angle with respect to horizontal : = tan
Why complete balancing is not possible in reciprocating engine?
Balancing of reciprocating masses is done by introducing the balancing mass
opposite to the crank. The vertical component of the dynamic force of this balancing
mass gives rise to “Hammer blow”. In order to reduce the Hammer blow, a part of
the reciprocating mass is balanced. Hence complete balancing is not possible in
reciprocating engines.
Differentiate between the unbalanced force due to a reciprocating mass and that
due to a revolving masses.
1. Complete balancing of revolving mass can be possible. But fraction of
reciprocating mass only balanced.
2. The unbalanced fore due to reciprocating mass varies in magnitude but
constant in direction. But in the case of revolving masses, the unbalanced force
is constant in magnitude but varies in direction.
The various cases of balancing of revolving masses?
i)
Balancing of single rotating mass by a single mass rotating in the same
plane.
ii)
Balancing of a single rotating mass by two masses rotating in different
planes.
iii)
Balancing of several rotating masses in a single plane.
iv)
Balancing of several rotating masses in different planes.
Primary balancing
Primary balance is the balance achieved by compensating for the eccentricities of the
masses in the rotating system, including the connecting rods. Primary balance is
controlled by adding or removing mass to or from the crankshaft, typically at each
end, at the required radius and angle, which varies both due to design and
manufacturing tolerances. In theory any conventional engine design can be balanced
perfectly for primary balance.
Secondary balancing
It was states earlier that the secondary acceleration force is defined as
Secondary force = mr2
cos 2
n
Its frequency is twice that of the primary force and the magnitude 1/n times the
magnitude of the primary force.
The expression can also be written as mr(2)2
cos 2
4n
Now, consider the cranks of an engine (fig) one actual one and the other imaginary,
with the following specifications.
Angular velocity
Actual
Imaginary
2
Length of crank
r
Mass at the crank pin
m
r
4n
m
Thus, when the actual crank has turned through an angle =t, the imaginary crank
would have turned an angle of 2=2t.
mr(2)2
4n
2
mr(2)
Component of this force along line of stroke
cos2
4n
Centrifugal force induced in the imaginary crank
Thus, the effect of the secondary force is equivalent to an imaginary crank of length
r/4n rotating at double the angular velocity, i.e. twice of the engine speed.
The imaginary crank coincides with the actual at inner top-dead centre. At other
times, it makes an angle with the line of stroke equal to twice that of the engine
crank.
The secondary couple about a reference plane is given by the multiplication of the
secondary force with the distance l of the plane from the reference plane.
Why the cranks of a locomotive are generally at right angles to one another?
In order to facilitate the starting of locomotive in any position (i.e in order to
have uniformly in turning moment) the cranks of a locomotive are generally at 90 to
one another.
If two cylinders are placed in between the planes of two driving wheel in a
locomotive engine then it is known as Inside Cylinder Locomotive. What are the
effects of an unbalanced primary force along the line of stroke of two cylinder
locomotive
i)
ii)
Variation in tractive force along the line of stroke, and
Swaying couple.
tractive force.
The resultant unbalanced force due to the two cylinders along the line of stroke, is
known as tractive force.
The tractive force is maximum or minimum when the angle of inclination of the
crank to the line of stroke () is equal to 135 and 315.
Swaying couple
The unbalanced force acting at a distance between the line of stroke of two
cylinders, constitute s couple in the horizontal direction. This couple is known as
swaying couple.
The swaying couple is maximum when the angle of inclination of the crank to the
line of stroke () is equal to 45 and 25.
In-line engines
Multi cylinder engines with the cylinder centre lines in the same plane and on
the same side of the centre line of the crank shaft, are known as in-line engine.
The condition to be satisfied for complete balance of in-line engine
1. The algebraic sum of the primary and secondary forces must be zero, and
2. The algebraic sum of the couples due to primary and secondary forces must
be zero.
Radial engines are preferred
In radial engines the connecting rods are connected to a common crank and
hence the plane of rotation of the various cranks is same, therefore there are no
unbalanced primary or secondary couples. Hence radial engines are preferred.
A rotor rotating around a vertical spindle, has the following masses placed on it.
(i)
(ii)
Determine the magnitude and angular position of a mass, that should be
placed at 262.5 mm, to give balance when rotating.
Also determine the unbalanced force on the spindle when the rotor is
rotating at 250 rpm.
Given Data: mA = 2.5 kg; A = 0; rA = 260 mm; mB = 3.5 kg; B = 60; rB = 300 mm; mC =
5.0 kg; C = 150; rC = 225mm; b = 262.5 mm
Solution
1. By Analytical Method:
According to the space diagram Fig (a) resolves the forces vertically and
horizontally.
Forces are: mArA = 2.5 x 0.26 = 0.65 kg-m; mBrB = 3.5 x 0.3 = 1.05 kg-m;
mCrC = 5.0 x 0.225 = 1.125 kg-m
sum of horizontal forces :
H = mArA cosθA + mBrB cosθB + mCrC cosθC
H = 0.65 cos 0° + 1.05 cos 60° + 1.125 cos 150° = 0.2007 kg-m
sum of vertical forces :
V = mArA sinθA + mBrB sinθB + mCrC sinθC
V = 0.65 sin 0° + 1.05 sin 60° + 1.125 sin 150° = 1.4718 kg-m.
Resultant,
R ( H ) 2 ( V ) 2
= 0.2007 2 1.47182 1.4854kg m.
We know that R = B’.b = 1.4854
B’ = 1.4854 / b = 1.4854/0.2625 = 5.6587 kg
Angle of resultant force B, θ = 180 + θ
V
1.4718
tan 1
82.23
H
0.2007
=180 + 82.23 262.23
' tan 1
But
II By Graphical Method :
Procedure :
Step 1:
Draw the space diagram with the positions of the several masses as
shown in fig.
Step 2:
Find out the centrifugal forces (i.e) mass
moments.
Step 3:
Now draw the force polygon with the obtained centrifugal forces, such
that oa represents the centrifugal force exerted by the mass mA in
magnitude and direction to given scale. Similarly, draw ab and bc to
represent centrifugal forces of other masses mB and mC.
Step4 :
The closing side of the force polygon oc represents the resultant force
in magnitude and direction as shown in fig.
oc
Step 5 :
= B’.b = 1.485 kg-m.
The balancing force is equal to the resultant force but in opposite
direction.
=262
Step 6 :
Now find out the magnitude of the balancing mass B’ at a given radius
b = 0.2625 m, such that
B’b = 1.485 kg –m
B’ = 1.485 / 0.2625 = 5.657 kb
Unbalanced force calculation (FC) :
2 250
FC B. .b 5.657
0.2625
60
1017.78N.
2
2
2 N
= 60
Three masses are attached to a shaft as follows : 10 kg at 90 mm radius; 15 kg at 120
mm radius and 9 kg at 150 mm radius. The masses are to be arranged so that the
shaft is in complete balance. Determine the angular position of masses relative to
10 kg mass. All the masses are in the same plane.
Given Data : mA = 10kg; rA = 90 mm; mB = 15 kg; rB = 120 mm; mC = 9 kg; rC = 150 mm.
Solution :
Let us solve the problem by graphical method. The procedure is as below.
Procedure :
Step 1 :
Calculate the centrifugal forces (i.e mass
,moments)
mA.rA = 10 x 0.09 = 0.9 kg –m
mB.rB = 15 x 0.12 = 1.8 kg-m
mC.rC = 9 x 0.15 = 1.35 kg –m
Step 2 :
Draw oa to represent 0.9 kg-m in any
convenient direction, to the given scale as shown in fig.
Step 3 :
With ‘a’ as center and radius equal to 1.8 kgm draw an arc.
Step 4 :
With ‘o’ as center and radius equal to 1.35
kg-m draw another arc intersecting the first arc at ‘b’.
Step 5 :
Since the masses are in complete balance, there will not be any out of
balance mass moment, diagram (i.e couple polygon) will be a closed
triangle whose sides are proportional to the values in the Step 1. so join
ob and ab .
Step 6: Directly measure the directions from the triangle oab.
Angle between 10 kg and 15 kg = 135°.
Angle between 10 kg and 9 kg = 75°.
A rotating shaft carries four unbalanced masses 18kg, 14kg, 16kg and 12 kg at radii
5 cm, 6 cm, 7 cm and 6 cm respectively. The 2nd, 3rd and 4th masses revolve in planes
8 cm, 16 cm and 28 cm respectively measured from the plane of the first mass and
are angularly located at 60°, 135° and 270° respectively measured clockwise from
the first mass looking from this mass end of the shaft. The shaft is dynamically
balanced by two masses, both located at 5 cm radii and revolving in planes midway between those of 1st and 2nd masses and mid-way between those 3rd and 4th
masses. Determine graphically or otherwise, the magnitudes of the masses and
their respective angular positions.
Given data :
m1 = 18 kg; r1 = 5cm; θ1 = 0; m2 = 14 kg; r2 = 6 cm; θ2 = 60º;
m3 = 16 kg; r3 = 7cm; θ3 = 135º; m4 = 12 kg; r4 = 6 cm; θ4 = 270º;
Solution :
The position of the planes and angular positions of masses is assumed as shown in
fig. (a) and (b). position of mass m1 is assumed to be in horizontal direction.
Procedure :
1. First of all, draw the couple polygon from the data given in column(6) of Table, as
shown in fig. © to the given scale. The vector d’ o’ represents the balancing couple.
Since the balancing couple is proportional to 0.009 mB, therefore by measurement.
0.009mB = Vector d’ o’ = 0.120
mB = 0.120 / 0.009
1. By measurement angular position of balancing mass B is = B = 25 (CW).
2. Now draw the force polygon from the data given in column (4) at table as shown
in fig, the vector eo represents the balancing force. Since the balancing force is
proportional to 0.05mA, therefore by measurement.
0.05mA = Vector eo = 0.5405 or mA = 10.81 kg.
3. Angular position of balancing mass mA, by measurement is = A = 275 CW.
A rotating shaft carries four masses A, B, C and D which are redially attached
to it. The mass centers are 30 cm, 38 cm, 40 cm and 35 cm respectively from the
axis of rotation. The masses A, C and D are 7.5 kg, 5 kg and 4 kg respectively.
The axial distances between the planes of rotation of A and B is 400 mm and
between B and C is 500 mm. The masses A and C are at right angles to each
other. Find for a complete balance.
1. The angles between the masses B and D from mass A,
2. The axial distance between the planes of rotation of C and D and
3. The magnitude of mass B.
Solution :
Let us take B as a reference, the data can be tabulated as given in Table.
The position of the planes and angular position of the masses is as shown in fig. A)
and b). mass A is assumed to be horizontal.
Procedure :
1. First of all draw the couple polygon using the data in the last column (6) of table
5.3 as shown in fig (c) to the given scale. The vector o’c represents the couple due
to mass ‘D’. by measurement to the chosen scale.
0.14y = 0.142 or y = 1.014m
The axial distance between plane C and D = y-0.5 = 0.514 m
2. Directly measured the angle between mass A and mass D from fig. By
measurement,
We get D = 475
3. Draw the force polygon (as shown fig) using the data in the column (4) of table.
Such that od represents the magnitude and direction of the balancing mass ‘B’.
by measurement to the chosen scale.
Vector ‘od’ = 0.038 x = 0.351 or x = 9.24
4. Measure the angular position of mass B with respect to A from fig.
By measurement, we get B = 198
The following data relate to a single – cylinder reciprocating engine :
mass of reciprocating parts
= 60 kg
mass of revolving parts
Speed
Stroke
= 40 kg at 160 mm radius
= 160 r.p.m
= 30 mm.
If two-thirds of the reciprocating parts and the whole of revolving parts are to be
balanced. Determine.
i)
ii)
the balancing mass required at a radius of 350 mm, and
the resultant unbalanced force when the crank has turned 50 from
TDC.
Given Data :
M=60 kg; m1=40kg;
r=160 mm=0.16 m; N=160 r.p.m;
Stroke =320 mm=0.32m ; c=2/3; b=350 mm =0.35m; =50
To find : (i) Balance mass required (B), and
(ii)
Resultant unbalanced force , when =50
Solution : Angular velocity of crank,
2 N 2 (160)
16.755 rad/s
60
60
crank radius,
(i)
r
Balance mass required (B):
We know that B.b
B(0.35)
Or
stroke 0.320
0.16m
2
2
=(m1+c.m)r
=(40+2/3 x 60)0.16=12.8
B=36.57 kg . Ans.
(ii)
Resultant unbalanced force when =50
We know that the resultant unbalanced force at any instant,
m 2 r (1 c)2 cos 2 c 2 sin 2
60(16.755) 2 0.16 (1 2 / 3) 2 cos 2 50 (2 / 3) 2 sin 2 50
1492.55 N.
Ans.
Prove that the resultant unbalanced force is minimum when half of the
reciprocating masses are balanced by rotating masses, i.e, when c=1/2.
Solution: We know that the resultant unbalanced force
F m 2 r (1 c) 2 cos 2 c 2 sin 2
....(i)
This force will be maximum or minimum when dF=0
Dc
Squaring the equation (i) on both sides.
F 2 m 2 4 r 2 [1 c cos 2 c 2 sin 2 ]
2
...(ii)
Differentiating equation (ii) with respect to c, we get
2F.
dF
(m 2 r ) 2 {[2(1 c)( 1) cos 2 ] 2c sin 2 }
dc
dF (m 2 r ) 2
=
[2c sin 2 -2(1-c)cos 2 ]
dc
2F
(m 2 r ) 2
=
[2c sin 2 -2 cos 2 2c cos 2 ]
2F
(m 2 r ) 2
=
[2c (sin 2 +cos 2 )-2cos 2 ]
2F
(m 2 r ) 2
=
[2c 2 cos 2 ]
2F
For maximum and minimum value, dF=0 Dc
(m 2 r ) 2
[2c 2 cos 2 ] 0 or c=cos 2
...(i)
2F
Substituting the value of c=cos 2 in equation(i) , we get
F=m 2 r (1 cos 2 ) 2 cos 2 cos 4 sin 2
=m 2 r sin 4 cos 2 cos 4 sin 2
...[ 1-cos 2 sin 2 ]
=m 2 r sin cos sin 2 cos 2
=1/2 m 2 r 2sin cos
...[ sin 2 cos 2 1]
=1/2 m 2 r sin 2
The value of F will be minimum when sin 2=-1 or 2=90 or =-45
From equation (iii), c= cos2 =cos2(-45)=1/2.
Thus at c=1/2 , the resultant unbalanced force is minimum. Ans.
A V-twin engine has the cylinder axes at right angles and the connecting rods
operate a common crank. The reciprocating mass per cylinder is 10 kg and the
crank radius is 70mm. The length of the connecting rod is 350mm. Show that the
engine may be balanced for primary forces by means of a revolving balance mass.
If the engine speed is 500r.p.m. , what is the value of maximum resultant
secondary force
Given Data: 2=90 or =45; m=10kg ; r=70mm =0.07m;
L=350 mm =0.35m;N=500r.p.m
Solution:
1.
2 N 2 (500)
52.37 rad / s
60
60
Proof:
We know that resultant primary force.
FP 2m 2 r (sin 2 sin )2 (cos 2 cos )2
When =45; FP 2m 2 r (sin 2 45 sin )2 (cos 2 45 cos )2
sin cos
2m 2 r
m 2 r Centrifugal force
2 2
As FP is equal to centrifugal force produced by a mass ‘m’ at crank radius ‘r’ when
rotating at ‘’ rad/s., Primary forces can be balanced by revolving masses. Ans.
2
2.
2
Maximum resultant secondary force:
We know that resultant secondary force,
2m
2 r sin 2
(When 2 =90)
n
This is maximum , when sin 2=1 or=45or 135
Maximum resultant secondary force,
2m
( FS ) max
2r
....(substituting =45)
n
10
= 2
(52.37)2 0.07 543 N. Ans.
0.35 / 0.07)
FS
The pistons of a 60twin V-engine has strokes of 120mm. The connecting rod
driving a common crank has a length of 200mm. The mass of the reciprocating
parts per cylinder is 1 kg and the speed of the crank shaft is 2500 r.p.m. Determine
the magnitude of the primary and secondary forces.
Given Data: : 2=60 or =30; Stroke=120 mm=0.12m; r=stroke =0.06m
2
L=200 mm =0.2m; m-1kg;N=2500r.p.m
To find : Magnitude of the primary and secondary forces.
Solution : Angular velocity ,
2 N 2 (2500)
261.8rad / s
60
60
0.2
n l/r
3.33
0.06
1. Magnitude of the primary force,
We know that resultant primary force.
FP 2m 2 r (sin 2 sin ) 2 (cos 2 cos ) 2
=2(1) (261.8) 2 0.06 (sin 2 30 sin ) 2 (cos 2 30 cos ) 2
FP 8224.71 0.0625sin 2 0.5625cos 2 Ans.
2. Magnitude of the secondary force,
We know that resultant secondary force,
Fs
2m
2 r (sin sin 2 .sin 2 ) 2 (cos cos 2 cos ) 2
n
=
2(1) (261.8) 2 0.06
(sin 30 sin 60.sin 2 ) 2 (cos 30 cos 60 cos ) 2
3.33
2469.88 (0.433sin 2 ) 2 (0.433cos 2 ) 2
=1069.1 sin 2 2 cos 2 2 1069.1 Ans.
The reciprocating mass per cylinder in a 60 V-twin engine is 1.5 kg. The stroke is
100 mm for each cylinder. If the engine runs at 1800 r.p.m Determine the
maximum and minimum values of the primary forces and find out the
corresponding crank position.
Given Data: : 2=60 or =30; Stroke=100 mm=0.12m; r=stroke =0.05m
2
N=2500r.p.m; m-1.5kg
Solution : Angular velocity ,
2 N 2 (1800)
188.49rad / s
60
60
We know that for a V- engine , the resultant primary force is given by
FP 2m 2 r (sin 2 sin ) 2 (cos 2 cos ) 2
Where symbols have their usual meanings
For =30; FP 2m 2 r (sin 2 30 sin ) 2 (cos 2 30 cos ) 2
=2m 2 r
FP
1
9
.sin 2 .cos2
16
16
m 2 r
sin 2 9 cos 2
2
For maximum and minimum values of FP,
dFp
=0
d
dFp m 2 r 1 [2sin .cos 9 2.cos .sin ]
=
d
2
2
sin 2 9 cos 2
m 2 r 2sin .cos 18.sin .cos
0
4
sin 2 9 cos 2
m 2 r 16sin .cos
2
2
4
sin 9 cos
0
m 2 r
8.sin 2 .
0
2
2
4
sin 9 cos
Here
or
m 2 r
8sin 2
0;
0
4
sin 2 9 cos 2
sin2 =0 or 2 =0 and
=0 and /2 Ans.
Maximum primary force:
Substituting =0 in equation (i) we get,
m 2 r
3
( FP ) max
sin 2 0 9 cos 2 0 m 2 r
2
2
3
= 1.5(188.49)2 0.05 3996.95 N . Ans
2
Minimum Primary force:
Substituting =/2 in equation (i) , we get
m 2 r
sin 2 / 2 9 cos 2 / 2
2
m 2 r 1.5(188.49) 2 0.05
=
1332.3 N . Ans
2
2
( FP ) min
Balancing of inline engines.
If a reciprocating mass is transferred to the crankpin, the axial component
parallel to the cylinder axis of the resulting centrifugal force represents the primary
unbalanced force.
Consider a shaft consisting of three equal cranks asymmetrically spaced. The
crankpins carry equivalents of three unequal reciprocating masses. Then
mr
Primary force =
Pr imary couple =
2
1
cos
mr lcos
Secondary force =
2
2
mr
Secondary couple =
2
mr
4n
2
cos 2 mr
2
4n
2
2
2
n
l cos 2 mr
2
lcos 2
n
3
4
In order to solve the above equations graphically, first draw the mr cos
polygon (2 is common to all forces). Then the axial component of the resultant force
(Fr cos ) multiplied by 2 provides the primary unbalanced force on
The system at that moment. This unbalanced force is zero when =90o and a
maximum when =0o.
In case the force polygon encloses, the resultant as well as the axial
component will always be zero and thus, the system will be in primary balance.
Then
F
ph
0 and
F
pv
0.
To find the secondary unbalance force, firs find the positions of the imaginary
secondary cranks. Then transfer the reciprocating masses at the crankpins. Draw
the mr polygon. Measure the resultant axial component and multiply the same by
(2)2/4n or 2/4 to get the secondary force.
In the same way primary and secondary couple (mrl) polygons can be drawn
for primary and secondary couples.
In the following paragraphs, some multi-crank arrangements have been
examined.
In-line Four-cylinder Four-stroke Engine
Such an engine has two outer as well as inner cranks (throws) in line. The
inner throws are at 180o to the outer throws. Thus, the angular positions for the
cranks are: for the first, (180o+) for the second, (180o+ ) for the third and for the
fourth Fig.
For convenience, choose a plane passing through the middle bearing about
which the arrangement is symmetrical as the reference plane.
Primary force
mr2 cos cos 180o cos 180o cos 0
Primary couple
3l
1
1
mr2 cos cos 180o
2
2
2
31
cos 180o cos 0
2
Secondary force
mr2
cos 2 cos 360o 2 cos 360o 2 cos 2
n
4me2
cos 2
n
Maximum value =
4mr2
at 2=0o ,180o ,360o and 540o or =0o ,90o ,180o and 270o
n
Secondary couple
3l
1
1
31
mr2 cos 2 cos 360o 2 cos 360o 2 cos 2 0
2
2
2
2
Graphical solution has been shown in Fig. Thus, this engine is not balanced in
secondary forces.
Six Cylinder Four-stroke Engine
Only a graphical solution is being given for simplicity. In a four stroke
engine, the cycle is completed in two revolution of the crank and the cranks are 120o
apart.
Crank positions for different cylinders for the firing order 142635 for
clockwise rotation of the crankshaft are
For first, =0o
For second, =240o
For third, = 120o
For fourth, = 120o
For fifth, =240o
For sixth, = 0o
Assuming m and r equal for all cylinders and taking a vertical plane passing
through the middle of shaft as the reference plane, the force and the couple polygons
are drawn as shown in Fig.
Since all the force and couple polygons close, it is and inherently balanced
engine for primary and secondary forces and couples.
Exmples :
1. A four-cylinder oil engine is in complete primary balance. The arrangement of
the reciprocating masses in different planes is as shown in Fig. (a). The stroke of
each piston is 2r mm. Determine the reciprocating mass of the cylinder 2 and the
relative crank positions.
Solution:
Crank length = 2r/2 = r
Tank 2 as the reference plane and 3 = 0o
m1r1l1 380 r -1.3 494r
m1r1 380r
m3r3l3 590r 2.8 1652r
m3r3 590r
m4r4l4 480r 2.8 1.3 1968r m 4r4 480r
494 r cos 1 1652 r cos 0 o 1968r cos 4 0
or 494 cos 1 1652 1968 cos 4
and -494 r sin 1 1652r sin0o 1968 r sin 4 0
or
494 sin 1 1968 sin 4
Squaring and adding (i) and (ii),
494
2
1652 1968cos 4 1968 sin 4
2
2
= 1652 1968 cos2 4 2 1652 1968 cos 4 1968 sin 4
2
2
2
= 1652 1968 2 1652 1968 cos 4
2
2
cos 4 0.978 or 4 167.9 o or 192.1o
Choo sing one value, say 4 16 7.9o
Dividing ii by i , tan1
=
1968 sin167.9o
1652 1968cos167.9 o
+412.53
1.515
272.28
1 123.4o
Writing the force equation, (r is common),
380 cos 123.4o m2 cos 2 590cos0o 480cos167.9o 0
or
iii
m2 cos 2 88.5
and 380 sin 123.4o m2 sin 2 590 sin0o 480 sin167.9o 0
or
iv
m2 sin 2 417.9
Squaring and adding iii and iv ,m2 427.1kg
Dividing iii by iv , tan 2
or
417.9
4.72
88.5
2 282o
Figure: (b) shows the relative crank positions.
Had we chosen 4 = 192.1o, a different set of values of m2, 1 and 2 would
have come.
To solve the problem graphically, draw the couple polygon (triangle) as
shown in Fig. (c) from the three known values. This provides the relative direction
of the masses m1, m3 and m4. Now, complete the force polygon [Fig.(d)] and obtain
the magnitude and direction of m2. The results obtained are 4 = 168o, 1=123o,
2=282o.
Also m2r = 427 r or m2 = 427 kg
Note that the couple triangle can be drawn in more than one way. However,
only two sets of answers are obtained. Also m1r1l1 is negative and, therefore, its
direction is reversed in the diagram.
2. The arrangement of the cranks of a 4-crank symmetrical engine is shown in Fig.
The reciprocating masses at cranks 1 and 4 are each equal to m 1 and of the crank 2
and 3 are each equal to m2. Show that the arrangement is balanced for primary
forces and couples and for secondary forces if
m1 cos l1 tan
1
,
,cos cos =
m2 cos l2 tan
2
Determine also the magnitude of the out of balance secondary couple if the
system rotates at rad/s.
Solution:
As particular positions of the cranks are being considered, horizontal and
vertical components of primary and secondary forces and couples must be taken.
(i)
Primary Forces
m1 cos m2 cos 180o
2
o
m2 cos 180
Fph r
o
+m
cos
360
1
2r2 m1 cos m2 cos
m1 sin m2 sin 180o m2 sin 180o
Fpv r
m2 sin 360o a
2
0
For primary balance of forces, Fph must be zero,
i.e., m1cos - m2 cos = 0
m1 cos
m2 cos
or
(ii) Primary Couples Take reference plane at the middle of shaft about which the
system is symmetrical.
Cph
m1 li cos m2 l2 cos 180o
r
o
o
m2 l2 cos 180 m1 l1 cos 360
2
r2 m1l1 cos m2l2 cos m2l2 cos m1l1 cos
0
C
pv
2r2 m2l2 sin m1l1 sin
Thus, for balancing of primary couples,
M2l2 sin - m1l1 sin = 0
or
m1 li sin m2 l2 sin 180 o
r2
o
o
m2 l2 sin 180 m1 l1 sin 360
l1 m2 sin cos sin tan
l2 m2 sin cos sin tan
(iii) Secondary Forces
Fsh
r2
m1 cos 2 m2 cos 2 180o m2 cos 2 180 o m1 cos 2 360 o
n
r2
m1 cos 2 m2 cos 2 m2 cos 2 m1 cos 2
n
2rw 2
m1 cos 2 m2 cos 2
n
Fsv
r2
m1 sin2 m2 sin2 180o m2 sin2 180o m1 sin2 360o
n
r2
m1 sin2 m2 sin2 m2 sin2 m1 sin2
n
0
For the balancing o secondary forces,
m1 cos 2 m2 cos 2 0
or
m1
cos cos 2 0
m2
or
cos
2cos2 1 2cos2 1 0
cos
or
2coscos2 cos 2cos2 cos cos 0
or
2coscos cos+cos cos cos 0
or
2cos+cos 2cos cos 1 0
As cos +cos#0, 2coscos-1=0
1
or cos cos =
2
(iv) Secondary Couples
C
sh
r2
m1 li cos 2 m2 l2 cos 2 180o m2 l2 cos 2 180 o m1 l1 cos 2 360 o
n
r2
m1l1 cos 2 m2l2 cos 2 m2l2 cos 2 m1l1 cos 2
n
0
C
sv
r2
m1 li sin2 m2 l2 sin2 180o m2 l2 sin2 180o m1 l1 sin2 360 o
n
2r2
m1l1 sin2 m2l2 sin2
n
Out of balance secondary couple
2r2
m1l1 sin2 m2l2 sin2
n
Graphical Solution
From the polygon of primary forces
m1 cos m2r cos or
m1 cos
m2 cos
From the polygon of primary couples,
m2l2 sin m1l1 sin
l1 m2 sin cos sin tan
l2 m1 sin cos sin tan
From the polygon of secondary forces,
m1 cos 2 m2 cos 2
or
m1 cos 2 m2 cos 2 0
Simplifying as in (iii) of analytical solution above, cos cos
1
2
From the polygon of secondary couples,
Resultant mrl m1rl1 sin2 m2rl1 sin 180o 2 m2rl1 sin 180o 2 m1rl1 sin2
or out of balance secondary couple
2r2
m1l1 sin2 m2l2 sin2
n
Each crank and the connecting rod of a four-crank in-line engine are 200 mm and
800 mm respectively. The outer cranks are set at 120o to each other and each has a
reciprocating mass of 200 kg. The spacing between adjacent planes of cranks are
400 mm, 60 mm and 500 mm. If the engine is in complete primary balance,
determine the reciprocating masses of the inner cranks and their relative angular
positions. Also find the secondary unbalanced force if the engine speed is 210
rpm.
Solution:
2 210
22rad / s,n 800 / 200 4
60
Figure represents the relative position of the cylinders and the cranks,
Taking 2 as the reference plane,
Primary couples about the RF,
m1r1l1 200 0.2 0.4 16
m2r2l2 0
m3r3l3 m2 0.2 0.6 0.12m3
m4r4l4 200 0.2 1.1 44
The couple polygon is drawn in fig.
m3r3l3 of crank 3 from the diagram = 53.7 at 135o
m3r3l3=m3 x 0.12 = 53.7 or m3=448 kg
As its direction is to be negative, its direction is (135o +180o) or 315o.
Primary force (mr) along each of outer cranks = 200 x 0.2 = 40
Primary force (mr) along crank 3 = 448 x 0.2 = 89.6
The force polygon is drawn in Fig.
m2r2 or crank 2 from the diagram = 87.6 At 161.4o
m2r2=m2x0.2 = 87.6 or m2 = 438 kg
Its angular position is 161.4o.
Figure (b) represents the relative position of the cylinders and the cranks,
From secondary unbalanced force polygon, mr = 198
Maximum unbalanced force 198
2
222
198
23 958N
n
n
The successive cranks of a five-cylinder in-line engine are at 144o apart. The
spacing between cylinder centre lines is 400 mm. The lengths of the crank and the
connecting rod are 100 mm and 450 mm respectively and the reciprocating mass
for each cylinder is 20 kg. The engine speed is 630 rpm. Determine the maximum
values of the primary and secondary forces and couples and the position of the
central crank at which these occur.
Solution:
2 630
60 rad/s
60
Figure (a) represents the relative position of the cylinders and the cranks.
Primary force (mr) along each crank = 20 x 0.1 = 2
The primary force polygon is a closed polygon Fig, therefore, no unbalanced
primary force.
Primary couples about the mid-pane, m1r1l1 = 2 x 0.8 = 1.6
m2r2l2 2 0.4 0.8, m3r3l3 0, m4r4l4 0.8, m5r5l5 1.6
The couple polygon is drawn in Fig.
Unbalanced mrl on measurement = 2.1
The unbalanced primary couple = 2.1 2 2.1 662 9148N
The maximum value of the secondary couple will occur when it coincides
with the line of stroke, i.e. when the crankshaft rotates through 18 o and 198o
clockwise. As initial position of mid-crank 3 is 288o, its positions for maximum
primary couple will be (288o-15o) and (288o-198o) or 270o and 90o.
The positions of the cranks for secondary forces and couples will as shown in
Fig.(d)
Secondary force (mr) along each crank = 20 x 0.1 =2
The force polygon is a closed polygon Fig(e), therefore, no unbalanced
secondary force.
Secondary couples about the mid-pane, m1r1l1,m2r2l2 are the same as above for
primary couples.
The couple polygon is shown in Fig. (f). It does not close.
Unbalanced mrl on measurement = 4.24
The unbalanced couple 3.41
2
1
3.41
662 3301N.m
n
450 /100
The maximum value of the secondary couple will occur when it coincide with
the line of stroke, i.e. when the crankshaft rotates through 126o and 306o clockwise.
As initial position of mid-crank 3 is 216o, its positions for maximum secondary
couple will be (216o-126o), (216o-306o) or 90o and-90o or 90o and 27o. However, since
the secondary crank position are taken at double the angles, the original crank will
rotten through 45o and 135o. As the crank rotates through a full revolution, the
maximum secondary couple will also occur at 225o and 315o.
Each crank and the connecting rod of a six-cylinder four stroke in-line engine are
60 mm and 240 mm respectively. The pitch distances between the cylinder centre
lines are 80mm, 80mm, 100mm, 80mm respectively. The reciprocating mass of
each cylinder is 1.4 kg. The engine speed is 1000 rpm. Determine the out of
balance primary and secondary forces and couples on the engine if the firing
order be 142635. Take a plane midway between the cylinders 3 and 4 as the
reference plane.
Solution
Figure (a) represents the relative position of the cylinders and the cranks for the
firing order 142635 for clockwise rotation of the crankshaft. As the engine is a fourstroke engine, firing takes place once in two revolutions of the crank and the angle
between the cranks is 120o.
Primary force (mr) along each crank = 1.4 x 60 = 84
The force polygon can exactly be drawn in the same manner as shown in Fig.
It is closed polygon, therefore, no unbalanced primary force.
Primary couples about he mid-pane,
m1r1l1 84 210 17 640
m2r2l2 84 130 10 920
m3r3l3 84 50 4200
m4r4l4 84 50 4200
m5r5l5 84 130 10 920
m6r6l6 84 210 17 640
The couple polygon is again exactly similar to as shown in Fig. (b). It is a
closed polygon, therefore, no unbalanced primary couple.
The positions of the cranks for secondary forces and couples will be as shown
in Fig.(b).
Secondary force (mr) along each crank = 1.4 x 60 = 84
The force polygon can exactly be drawn in the same manner as shown in Fig.
(d). It is a closed polygon, therefore, no unbalanced secondary force.
Secondary couples about the mid-pane,
m1r1l1, m2r2l2… are the same as above for primary couples.
The couple polygon is again exactly similar to as shown in Fig. (d). It is a
closed polygon, therefore, no unbalanced secondary couple.
The stroke of each piston of a six-cylinder two-stroke in-line engine is 320 mm
and the connecting rod is 800 mm long. The cylinder centre lines are spaced at 500
mm. The cranks are at 60o apart and the firing order is 145236. The reciprocating
mass per cylinder is 100 kg and the rotating parts 50 kg per crank. Determine the
out of balance forces and couples about the mid plane if the engine rotates at 200
rpm.
Solution
Figure (a) represents the relative position of the cylinders and the cranks for
the Firing order 145236 for clockwise rotation of the crankshaft.
Total mass at the crank pin = 100 + 50 = 150 kg
Primary force (mr) along each crank = 150 x 0.16 = 24
The force polygon [Fig.(b)] is a closed polygon, therefore, no
unbalanced primary force.
Primary couples about the mid-pane,
m1r1l1 24 1.25 30
m2r2l2 24 0.75 18
m3r3l3 24 0.25 6
m4r4l4 6
m5r5l5 18
m6r6l6 30
The couple polygon 14.27 (c) is again a closed polygon, therefore, no
unbalanced primary couple. As it is not necessary to add the vectors it order, the
couple polygon can also be drawn as in Fig.(d).
The positions of the cranks for secondary forces and couples will as shown in
fig.(e).
Secondary force (mr) along each crank = 100 x 0.16 = 16 (The rotating masses
do not affect the secondary forces as they are only due to second harmonics of the
piston acceleration).
The force polygon is a closed polygon Fig. (f), therefore, no unbalanced
secondary force.
Secondary couples about the mid-pane,
m1r1l1 16 1.25 20
m2r2l2 16 0.75 12
m3r3l3 16 0.25 4
m4r4l4 4
m5r5l5 12
m6r5l6 20
The couple polygon is shown in Fig. (g). It does not close.
Unbalanced mrl on measurement = 55.43.
2
The unbalanced couple 55.43
2
1 2 200
55.43
4863 N.m
n
5 60
UNIT - V
VIBRATION
Vibration isolation
In order to diminish the transmission of force to the foundation machines are often
mounted on springs and dampers.
Ccauses of vibration
The causes of vibration are unbalanced forces, elastic nature of the system,
self excitations, winds and earthquakes.
Period and cycle of vibration.
Period :
It is the time interval after which the motion is repeated itself.
Cycle :
It is defined as the motion completed during one time period.
Frequency of vibration.
It is the number of cycles described ib one second, unit Hz.
Classify vibration
1.
2.
3.
Free vibrations
(a) Longitudinal vibration,
(b) Transverse vibration
(c) Torsional vibration
Forced vibrations, and
Damped vibration
Free vibration and forced vibrations
Free or natural vibrations : When no external force acts on the body, after giving it an
initial displacement, then the body is said to be free or natural vibrations.
Forced vibrations : When the body vibrates under the influence of external force,
then the body is said to be under forces\d vibration.
Damping and damped vibration
Damping : The resistance against the vibration is called damping.
Damping vibration : When there us a reduction in amplitude over cycle of vibration,
then the motion is said to be damped vibration.
Resonance.
When the frequency of external force is equal to the natural frequency of a
vibrating body, the amplitude of vibration becomes excessively large. This
phenomenon is known as resonance.
Degree of freedom or movability
The number of independent coordinates required to completely define the
motion of a system is known as degree of freedom of the system.
A cantilever beam has infinite number of degrees of freedom.
steady state and transient vibrations.
In ideal systems, the free vibration continue indefinitely as there is no
damping. Such vibration is termed as steady state vibrations.
In real systems, the amplitude of vibration decays continuously because of
natural damping and vanishes finally. Such vibration in real system is called
transient vibration.
Damping.
Resistance to the oscillation is known as damping. Vibrating systems are all to an
extent, subjected to damping due toe energy dissipated by friction and other
resistances.
Undamped free vibrations.
When an elastic system free from external impressed forces is disturbed from
its equilibrium positions, it vibrates under the influence of inherent forces, and is
said to be in a state of free vibration. It will vibrate at its natural frequency, and its
amplitude will gradually become smaller with time due to the energy being
dissipated by motion.
Forced vibrations.
When a system vibrates under the influence of external force, the vibrations
are said to be forced vibrations. It follows that the forced vibrations take place at the
frequency of the external alternating force and it is independent of the natural
frequency of vibration.
Equivalent spring stiffness
Equivalent spring stiffness (Seqv) is the measure of overall spring stiffness of any
system having more than one spring connected in series or parallel.
(a) For spring in series :
(b) For springs in parallel :
1 1 1 1
1
.....
Seqv S1 S2 S3
Sn
Seqv S1 S2 S3 ..... Sn
List out the various methods of finding the natural frequency of free longitudinal
vibrations.
1. Energy method
2. Equilibrium method
3. Rayleigh’s method
Principle of rayleigh’s method of finding natural frequency of vibrations
The principle of rayleigh’s method is the maximum kinetic energy at the
mean position is equal to the maximum potential energy (or strain energy) at the
extreme position.
The natural frequency of free longitudinal vibration is
fn
1
2
s
1
m 2
g
0.4985
where m=mass of body; s = spring stiffness and = Static deflection.
If mass of the constraint (i.e, effect of inertia of constraint) is taken into account,
the natural frequency of a longitudinal vibration is given by,
fn
1
2
S
m
m c
3
where mc = mass of the constraint
A shaft supported in long bearing is assumed to have both ends fixed for solving
transverse vibration problems.
A car having a mass of 100 kg deflects its spring 4 cm under its load. Determine
the natural frequency of the car in the vertical direction.
Solution :
m = 1000 kg; = 4 cm = 0.04m.
fn
1
2
g
1
2
9.81
2.492 Hz
0.04
The damping force per unit velocity is known as damping coefficient.
Write the differential equation and characteristic equation for spring mass damper
system.
d 2 x c dx s
Differential equation :
x0
dt 2 m dt m
c
s
2
Characteristic equation : k k
0 with usual notations.
m
m
Distinguish between critical damping and large damping.
If system is critically damped, the mass moves back very quickly to its
equilibrium position within no time. Whereas in large (over) damping, the mass
moves slowly to the equilibrium position.
When do you say a vibrating system in under damped?
The equation of motion of a free damped vibration is given by
d 2 x c dx s
x0
dt 2 m dt m
2
s c
If
, then radical becomes negative. The two roots k1 and k2 are known as
m 2m
complex conjugate. Then the vibrating system is known as under damping.
Explain the Dunkerley’s method used in natural transverse vibration for a shaft
carrying a number of point loads and uniformly distributed load is obtained by
Dunkerley’s formula.
Dunkerley’s Formula :
1
1
1
1
1
.....
2
2
2
2
f n ( f n1 ) ( f n 2 ) ( f n3 )
( f ns )2
fn1, fn2, fn3 etc : Natural frequency of transverse vibration at each point loads, and
fns = Natural frequency of transverse vibration of the UDL.
Define critical or whirling or whipping speed of a shaft.
The speed at which resonance occurs is called critical speed of the shaft. In
other words, the speed at which the shaft runs so that the additional deflection of the
shaft from the axis of rotation becomes infinite, is known as critical speed.
What are the factors that affects that the critical speed of a shaft?
The critical speed essentially depends on
a) The eccentricity of the C.G of the rotating masses from the axis of
rotation of the shaft.
b) Diameter of the disc
c) Span (length) of the shaft, and
d) Type of supports connections at its ends.
In a shaft whirling situation, let y=additional displacement of C.G from axis of
rotation due to centrifugal force. Discuss as to how ‘y’ varies with respect to the
operating speed and critical speed.
Solution : y
e
2
Nc
1
N
If the operating sped is equal to critical speed, then
2
Nc
- 1 = 0 ; hence y = 0 ; deflection is zero.
N
0.4985
and N c
rps
Critical speed of shaft is the same as the natural frequency of transverse vibration
– Justify.
We know that critical or whirling speed c = n
s
g
Hz
m
c n
If Nc is the critical speed in rps, then
2 Nc
g
Nc
1
2
g
0.4985
rps Hence proved.
the causes of critical speed?
i)
ii)
iii)
Eccentric mountings
Bending due to self weight, and
Non-uniform distribution of rotor material.
Damping ratio ().
It is defined as the ratio of actual damping coefficient (c) to the critical damping
coefficient (CC).
Mathematically, Damping ratio
logarithmic decrement.
c
c
cc 2m n
Logarithmic decrement is defines as the natural logarithm of the amplitude
reduction factor. The amplitude reduction factor is the ratio of any two successive
amplitudes on the same side of the mean position.
x
x
=log e 1 log e n
x2
xn1
It is a convenient way to measure the amount of damping is to measure the
rate of decay of oscillation. This is best expressed by the term L.D. it is defined as
the natural logarithm of the ratio of any two successive amplitudes.
A vibrating system consist of a mass of 7 kg and a spring stiffness 50 N/cm and
damper of damping coefficient 0.36 Ncm-1 sec. find damping factor.
Solution :
m= 7 kg; s=50 N/cm = 5000 N/m ; c = 0.36 N/cm/sec = 36 N/m/sec
n
s
5000
26.72 rad/sec
m
7
cc = 2 m n = 2 X 7 X 26.72 = 374.16 N/m/s
Damping factor =
c
0.0962
cc
What is the relationship between frequencies of undamped and damped vibration
Solution :
d
f d 2
f n n
2
2
2
d wn a
n
n
fd
c2
1
fn
2ms
Dynamic magnifier or magnification factor
It is the ratio of maximum displacement of the forced vibration (x max) to the
deflection due to the static force F(x0).
D
xmax
x0
1
c 2
1 2
s2
n
2
2
2
Sketch the vector relationship between the dynamic existing force, spring force,
damping force, inertia force of a single degree of freedom system subjected to
forced vibration.
Give the equation for damping factor and damped frequency d
(i) Damping factor, =
c
c
c c 2m n
fd
I 2 n
2
c = Damping coefficient;
Cc = Critical damping coefficient; and
n= Natural or undamped frequency.
(ii) Damped frequency, d
Where
transmissibility
When a machine is supported by a spring, the spring transmits the force
applied on the machine to the fixed support or foundation. This is called as
transmissibility.
transmissibility ratio or isolation factor.
The ratio of force transmitted (FT) to the force applied (F) is known as
transmissibility ratio.
FT xmax s 2 c2 2
F
F
It is a measure of the effectiveness of the vibration isolating material.
Derive the relationship for transmissibility.
Vibrations are produced in machines having unbalanced masses. These
vibrations will be transmitted to the foundation upon which the machines are
installed. This is usually undesirable. To diminish the transmitted forces, machines
are usually mounted on springs or dampers, or on some other vibration isolating
material. Then the vibratory forces can be reach the foundation only through these
springs, dampers, or the isolating material used.
elastic suspension.
When machine components are suspended from elastic members, the
vibrational force produced by the machine components will not be transmitted to the
foundation. This is called as elastic suspension.
Specify any two industrial applications where the transmissibility effects of
vibration are important.
1. All machine tools
2. All turbo machine
With the help of transmissibility Vs frequency curve explain the effect of
damping on transmissibility
If /n = 1 and no damping, then magnification factor is infinity (∞). That is, the
system vibrates with maximum amplitude depending upon damping ratio and f/f n3
the amplitude of vibration varies,
if c/cc = 0.5, f/fn = 1,
magnification factor = 2.
Importance of vibration isolation.
When an unbalanced machine is installed on the foundation, it produces vibration
in the foundation. So in order to prevent these vibrations or to minimize the
transmission of forces to the foundation, vibration isolation is important.
Methods of isolating the vibration
1. High speed engines / machines mounted on foundation and supports cause
vibrations of excessive amplitude because of the unbalanced forces. It can be
minimized by providing “spring – damper” etc.
2. The materials used for vibration isolation are rubber, felt cork, etc. these are
placed between the foundations and vibrating body.
Torsional vibrations.
When the particles of a shaft or disc move in a circle about the axis of the shaft,
then the vibrations are known as torsional vibrations.
Differentiate between transverse and torsional vibrations.
1. In transverse vibrations, the particles of the shaft move approximately
perpendicular to the axis of the shaft. But in torsional vibrations, the particles
of the shaft move in a circle about the axis of the shaft.
2. Due to transverse vibrations, tensile and compressive stresses are induced.
Due to torsional shear stresses are induced in the shaft.
Write down the expression for natural frequency of free torsional vibration (a)
without considering the effect of inertia of the constraint, and (b) considering the
effects of inertia of the constraints.
Solution :
a) without considering the effect of the inertia of constraint
natural frequency of torsional vibration f n
1
2
b) with the effect of the inertia of constraint
fx
1
2
where
q
I
1 c
3
q = Torsional stiffness of shaft in N-m
I = Mass moment of inertia of disc in kg-m2 = m k2, and
q
I
Ic = Mass moment of inertia of constraint (i.e) shaft of spring etc.
Define node in torsional vibration.
Node is the point or the section of the shaft at which amplitude of the
torsional vibration is zero. At nodes, the shaft remains unaffected by the vibration.
torsional equivalent shaft.
A shaft having variable diameter for different lengths can be theoretically
replaced by an equivalent shaft of uniform diameter such that they have the same
total angle of twist when equal opposing torques are applied at their ends. Such a
theoretically replaced shaft is known as torsionally equivalent shaft.
Expression for the frequency of vibration of a two rotor system.
Natural frequency of torsional vibration, f x
Where
1
2
CJ
Il
C = rigidity modulus of shaft,
I = Mass M.I of rotor
J = Polar M.I of shaft, and
I = Length of node from rotor.
The node in the case of torsional vibration of two rotor system will be situated
closer to the rotor having larger mass moment of inertia – Justify.
Two equations for two rotor system are :
1) lA IA = lB IB and (2) l = lA + lB
2) From equation (1), if IB value is large, then lB =
lA -I A
; then lB value will
IB
be lesser that the value of lA. It means that the rotor having larger mass
moment of inertia will have node closer to it.
Derive the equation used in determine the position of node in two rotor system.
Solution :
We know that,
f na
1
2
CJ
Ia Ia
f nb
1
2
CJ
Ib Ib
f na f nb
1
2
CJ
1
I a I a 2
CJ
Ib Ib
I a I a Ib Ib
(1)
from fig. l=la lb (2)
Derive the equation used to obtain torsionally equivalent shaft of a stepped shaft.
The total angle of twist of the shaft is equal to the sum of the angle of twists of
different lengths.
= 1+2+3
So
Or
Tl
Tl
Tl
Tl
1 2 3
CJ CJ1 CJ 2 CJ 3
l l1 l2 l3
J J1 J 2 J 3
l
32
d4
l1
32
d14
l2
32
d 22
l3
32
d32
Let d = d1, then
4
d
d
l l1 l2 1 l3 1
d2
d3
where
4
= Length of a torsionally equivalent shaft.
A mass of 50 kg suspended form a spring produces a statical deflection of 17 mm
and when in motion it experiences viscous damping force of value 250 N at a
velocity of 0.3m/s. Calculate the periodic time of damped vibration. If the mass is
then subjected to a periodic disturbing force having a maximum value of 260 N
and making 2 cycles, find the amplitude of ultimate motion.
Given data: m= 50 kg; =17mm; F= 250N; v= 0.3m/s
To find:
(i)
Time period of damped vibration (td), and
(ii)
Amplitude of forced vibration (xmax).
Solution:
(i)
Periodic time of damped vibrations:
Since the motion experiences a viscous damping force of value 250 N at a velocity of
0.3m/s. therefore damping coefficient , c= 250/0.3=833.333 N/m/s.
Stiffness of the spring, s=
We know that, n
m.g 50 9.81
28852.94 N / m
0.017
s
28852.94
24rad / s
m
50
We know that the critical damping coefficient,
cc 2mn 2 50 24 2402.2 N / m / s
Damping factor, =
c 833.333
0.347
c v 2402.2
.... m.g=s.
The circular frequency of damped vibrations is given by
d 1 2 n 1 0.347 24 22.51rad / s.
2
Therefore, the time period of damped vibrations,
td
2
d
2
0.279 s.
22.51
ii) Amplitude of forced vibrations:
Given that, periodic disturbing force, Fo=200N.
And frequency of disturbing force =2 cycles/s =2Hz.
2 2
rad / s
f
2
We know that the amplitude of forced vibrations,
Therefore, angular velocity of disturbing force, =
xmax
FO / s
2
2
1 2
n n
200 / 28852.94
2
2
1 2 0.347
24
24
7.023 103 mor 7.023mm.
A mass of 50kg is supported by an elastic structure of total stiffness 20KN/m. The
damping ratio of the system is 0.2. A simple harmonic disturbing force acts on the
mass and at any time t seconds, the force is 60 sin 10t newtons. Find the amplitude
of the vibrations and the phase angle caused by the damping:
Given data:
M= 50kg; s=20 kN/m=20 x103 N/m; =0.2; F=60 sin 10t.
To find:
(i)
Amplitude of forced vibrations (xmax), and
(ii)
Phase angle caused by the damping ().
Solution:
Since the period force, F= Fo sin t 60 sin 10t, therefore static force, Fo = 60N and
angular velocity of the periodic disturbing force, =10 rad/s
We know that natural circular frequency of free vibrations,
n
s
20 103
20rad / s
m
50
(i)
Amplitude of forced vibrations (xmax):
FO / s
We know that, xmax
2
2
2
1 2
n n
60 / 20 10
3
2
2
2
10
1 2 0.2
20
20
3.865 10 3 mor3.865mm.
(ii)
Phase angle caused by the damping ():
2 2 0.2 10
20 0.2666
n
We know that, tan
2
2
10
1
1
20
n
or the phase angle, tan 1 0.2666 14.930.
A mass of 500 kg is mounted on supports having a total stiffness of 100 KN / m
and which provides viscous damping, damping ratio being 0.4. The mass is
constrained to move vertically and is subjected to a vertical disturbing force of
the types Fo sin t. Determine the frequency at which resonance will occur and
the maximum allowable value of Fo if the amplitude at resonance is to be
restricted to 5 mm.
Given data: m=500kg; s= 100KN/m=100x103 N/m; =0.4; xmax = 5mm=5x10-3m.
To find:
(i)
Frequency at which resonance will occur (f) and
(ii)
Maximum allowable value of Fo.
Solution:
(i)
Frequency at which resonance will occur (f):
We know that the resonance will occur at =n.
s
100 103
Therefore, = n
14.14rad / s
m
500
The frequency at which resonance will occur, is given by
f fn
14.14
2.25 Hz.
2
2
(ii)
Maximum allowable value of F0:
We know that the maximum displacement of forced vibration at resonance,
X max
FO
2 .s
5 103
FO
orFO 400 N .
2 0.4 100 103
A body having a mass of 15kg is suspended from a spring which deflects 12 mm
under the weight of the mass. Determine the frequency of the free vibrations.
What is the viscous damping force needed to make the motion a periodic at a
speed of 1mm/s? If, when damped to this extent, a disturbing force having a
maximum value of 100 N and vibrating at 6 Hz is made to act on the body,
determine the amplitude of the ultimate motion.
Given data:
M=15kg; =12mm=0.012m; FO=100N; f=6Hz.
To find:
(i)
(ii)
(iii)
Frequency of the free vibrations(fn).
Damping force needed to make the motion aperiodic at a speed of 1mm/s.
and
Amplitude of forced vibration (xmax).
Solution:
(i)
Frequency of the free vibrations (fn):
We know that, f n
1
2
g
1
s 2
9.81
4.55Hz.
0.012
(ii) Damping force needed to make the motion a periodic at a speed of 1 mm/s:
We know that the motion becomes aperiodic when the damped frequency is zero
or when it is critically damped (i.e. =1).
=n
g
9.81
28.59rad / s
s
0.012
Damping coefficient , C=Cc =2m n=2x15x28.59.
=857N/m/s or 0.857 N/mm/s.
Thus, the force needed is 0.857 N at a speed of 1 mm/s.
(iii) Amplitude of forced vibrations (xmax)
We know that angular velocity of forced vibration,
2 f 2 6 37.7rad / s
and stiffness of the spring, s=
m.g
15 9.81
12262.5 N / m.
0.0012
Amplitude of forced vibration,
X max
=
FO
s m c
2 2
2
100
12262.5-15 37.7 2 857 37.7 2
2
X max 0.00298m(or )2.98mm.
A spring mass system has a natural frequency of 1.25Hz. When the mass is at
rest, the support is made to move up with displacement y= 18 sin 2t (where t
is in seconds and y in millimeters) measured from the beginning of the
motion. Determine the distance through which the mass moves in the first 0.4
seconds.
Given data:
Fn=1.25Hz; y=18 sin 2t
To find: Distance through which the mass moves in the first 0.4 seconds.
Solution:
The given system is shown in Fig. Since the displacement of the support
Y= y sin t = 18 sin 2t,
Initial amplitude, y= 18mm = 0.018m.
and angular velocity, =2=6.28 rad/s
Natural circular frequency, n=2xfn=2x1.25=7.85 rad/s
The differential equation for the system shown in Fig is given by.
d 2x
m. 2 s y x
dt
d 2x
or m. 2 s.x s. y
dt
d 2x
or m. 2 s.x s. y sin t
dt
y=Y sin t
...(i)
The solution of this differential equation is,
x A cos nt B sin nt
y
2
1 2
sin t
....(ii)
where A and B are constants.
Substituting Y, and n values in equation (ii), we get
x A cos 7.85t B sin 7.85t
0.018
6.28 2
1
7.85
x A cos 7.85t B sin 7.85t 0.05sin 6.28t
sin 6.28t
....(iii)
The values of constants A and B are obtained by applying its initial conditions.
(i)
(ii)
when t=0, then x=0 therefore, from equation (iii), A=0
When t=0, then dx/dt =0
Therefore differentiating equation (iii) and equating to zero, we have
or
dx
7.85 B cos 7.85t 0.05(6.28) cos 6.28 0....( A 0)
dt
7.85Bcos7.85t=-0.3145cos6.28
B=
-0.3145
0.04.....( t 0)
7.85
Now the equation (iii) becomes, x=-0.04 sin 7.85t+0.05 sin6.28t
Vertical distance through which the mass is moved in the first 0.1 seconds (i.e.
t=0.1s).
180
180
x=-0.04.sin 7.85 0.4
+0.05sin 6.28 0.4
=0.0293m or 29.3 mm.
The support of a spring – mass system is vibrating with an amplitude of 6 mm and
a frequency of 20Hz. If the mass is 1.1kg and the spring has a stiffness of 2000
N/m. determine the amplitude of vibration of the mass. What amplitude will
result if a damping factor of 0.25 is included in the system.
Given data:
Y=6mm=0.006m; f=20Hz; m=1.1kg; s=2000N/m.
To find:
(i)
(ii)
Amplitude of vibration when =0, and
Amplitude of vibration when =0.25
Solution:
s
2000
42.64rad / s
m
1.1
Frequency of the support, =2xf = 2x20=125.66 rad/s
Natural frequency, n
Amplitude of vibration when =0:
(i)
We know that amplitude of forced vibrations due to excitation of the support.
X max
Y
1 2
n
2
2
2
2
1- 2
n n
...[From Equation (8.27)]...(i)
or the phase angle, tan 1 0.2666 14.930.
But given is a spring mass system without damper. Therefore, substituting =0,
we get
X max
Y
(ii)
1
2
1-
n
X max
1
2
0.006
125.66
1
42.64
Amplitude of Vibration when =0.25:
Substituting =0.25 in equation (i), we get
X max
Y
125.66
1 2 0.25
42.64
2
2
2
125.66 2
125.66
1
2
0.25
42.64
42.64
or X max 1.365 10 3 m or 1.365mm.
Determine the critical speed of a 1000kg automobile traveling on a concrete road
with expansion joints spaced 12 meters apart, if the static deflection of the spring
system is 50mm.
Given data:
M=1000kg; space =12m; =50mm=0.05m.
To find: Critical speed of the automobile.
Solution:
Natural frequency of the system,
fn
1
2
s
1
m 2
g
1
2
9.81
2.23Hz.
0.05
Let v= Critical speed of the automobile in m/s.
Then the corresponding forcing frequency is v/12 cycles / sec. This must be equal to
the natural frequency fn.
v
2.23Hz
12
22.76 60 60
or Critical speed, v=26.76m/s=
km / hr 96.34km / hr.
1000
A compressor supported symmetrically on four springs has a mass of 100kg. the
mass of the reciprocating parts is 2kg which move through a vertical stroke of
80mm with SHM. Neglecting damping, determine the combined stiffness of the
springs so that the force transmitted to the foundation is 1/25th of the impressed
force. The machine crankshaft rotates at 1000r.p.m.
When the compressor is actually supported on the springs, it is found that the
damping reduces the amplitude of successive free vibrations by 25% find
(i)
the force transmitted to the foundation at 1000 r.pm.
(ii)
(iii)
the force transmitted to the foundation at resonance, and
the amplitude of the vibrations at resonance.
Given data:
M=100kg; mu=2kg; stroke = 80mm=0.08m; =1/25=0.04; N=1000 r.p.m.
To find:
(a) Combined stiffness of the springs(s),
(b) Force transmitted to the foundation at 1000 r.p.m. (FT,
(c) Force transmitted to the foundation at the resonance, and
(d) Amplitude of the forced vibrations at resonance.
Solution:
2 N 2 (1000)
104.7rad / s
60
60
storke 0.08
Eccentricity, e=
0.04m
2
2
Angular velocity, =
(a) Combined stiffness of the springs(s):
In the absence of damping (i.e =0), the transmissibility is given by
1
0.04
2
1
n
1
n 20.53rad / s
We know that, n
s
s
20.53 rad/s
m
100
combined stiffness, s=42161.88N/m.
2
104.7
1
n
(b) Force transmitted to the foundation at 1000 r.pm. (FT):
Since the damping reduces the amplitude of successive free vibrations by 25%
therefore final amplitude of vibration, x2=0.75x1.
x
2
We know that ln 1
1 2
x2
x
2
ln 1
1 2
0.75 x1
on solving, we get damping factor, =0.0458.
or
1 2
n
We know that,
2
2 2 2
1
n n
2
104.7
1 2 0.0458
20.53
2
2
104.7 2
104.7
1
2
0.0458
2053
2053
Transmissibility, =0.044
The maximum unbalanced force on the compressor due to the reciprocating parts,
FO =mu2.e=2(104.7)20.04=876.967N
Force transmitted to the foundation,
FT=FO=0.044 x 876.967=38.586N
(d)
Force transmitted to the foundation at resonance:
At resonance, /n =1, therefore transmissibility ratio is given by
1 2
2
2
1 2 0.0458
2
2 0.0458
10.963
Maximum unbalanced force on the machine due to reciprocating parts at resonance,
FO mu.(n)2e=2(20.5320.04=33.72N
Force transmitted to the foundation at resonance,
FT =.FO=10.963 x 33.72=369.65N.
e) Amplitude of the forced vibrations at resonance:
we know that amplitude of the forced vibration at resonance,.
Force transmitted at resonance
369
8.77 103 m
Combined stiffness
42161.88
Amplitude=8.77mm.
A shaft of 100mm diameter and 1 metre long is fixed at one end and the other end
carries a flywheel of mass 1 tonne. The radius of gyration of the fly wheel is 0.5m.
Find the frequency of torsional vibrations. If the modulus of rigidity for the shaft
material is 80 GN/m2
Given data:
D=100mm=0.1m;l=1m;m=1tonne=1000kg;
K=0.5m;C=80GN/m2 =80x109N/m2.
To find:
Frequency of torsional vibrations (fn).
Solution:
We know that polar moment of inertia of the shaft.
j
32
d4
32
(0.1) 4 9.82 106 m 4
Torsional stiffness of the shaft is given by
C.J 80 109 (9.82 106 )
q=
785.6 103 N m.
l
1
We also know that mass moment of inertia of the shaft.
1=m.k2=1000(0.5)2=250kg-m2.
Frequency of torsional vibrations.
fn
1
2
q
1
l 2
785.6 103
8.922 Hz.
250
In the above problem, if the mass moment of inertia of the constraint is 50kg – m3,
find the natural frequency of torsional vibration.
Given data:
Ic=50kg-m2.
To find: Natural frequency of torsional vibrations (fn).
Solution: We know that the natural frequency of torsional vibration when
considering the inertia of the constraint,
fn
1
2
q
1
I
l c 2
3
785.6 103
8.638Hz.
50
250
3
A shaft of 75mm diameter and 2m long is fixed at both the ends vertically, as
shown in Fig.9.2. A fly wheel of mass one tonne is provided on the shaft at a
distance of 1.5m from its upper end as shown. The flywheel’s radius of gyration is
800mm. Find the natural torsional frequency of vibration of the system. Take
C=8x104 N/mm2.
Given data:
D=75mm=0.075m;l=2m;m=1tonne=1000kg;
K=800mm=0.8m; C=8x104 N/mm2=8x1010N/m2
To find:
Natural frequency of torsional vibration (fn).
Solution:
Mass moment of inertia. 1=mk2=1000(0.8)2=640kg-m4
Torsional stiffness of the shaft for length l1 is given by.
10
6
CJ 8 10 3.110
q1
165.67 103 N m
l2
1.5
similarly the torsional stiffness of the shaft for length l2 is given by
10
6
CJ 8 10 3.110
q2
497 103 N m
l2
0.5
Total torsional stiffness of the shaft,
q=q1+q2=165.67 x 103 +497 x 103 = 662.68 x103 N-m,
Therefore the normal frequency of torsional vibration,
fn
1
2
q
1
l 2
662.68 103
5.12 Hz.
640
A vibrating system consists of a mass of 8 kg, spring of stiffness 5.6 N/mm and a
dashpot of damping coefficient of 40 N/m/s. Find:
(a) the critical damping coefficient,
(b) the damping factor,
(c) the natural frequency of damped vibration,
(d) the logarithmic decrement,
(e) the ratio of two consecutive amplitudes, and
(f) the number of cycles after which the original amplitude is reduced to 20
percent
Given data:
m = 8 kg; s = 5.6 N/mm = 5.6 x 103 N/m;
C = 40 N/ m/ s
Solution:
(a) Critical damping coefficient (cc):
We know that
c c 2mn 2m
s
2 s.m
m
c c 2 5.6 103 8 422.32N/m/s
(b) Damping factor ():
We know that,
Damping factor,
c
40
0.0945
c c 422.32
(c) Natural frequency of damped vibration (fd):
We know that the circular frequency of damped vibrations,
d 1 2 .n
s
5.6 103
where n
26.34rad / s
m
8
d 1 0.0945 26.34 26.22rad / s
2
Natural frequency of damped vibration of the system
d 26.22
4.173Hz
2
2
(d) Logarithmic decrement ():
fd
We know that logarithmic decrement,
2
1 2
2 0.0945
1 0.0945
2
0.596
x
(e) Ratio of two consecutive amplitudes n :
xn + 1
Let xn and xn+1 = Magnitudes of two consecutive amplitudes
The logarithmic decrement can also be given by
x
x
ln n or n e
xn 1 xn 1
xn
e0.596 1.8156
xn 1
(f) Number of cycles after which the amplitude is reduced to 20% (n):
Let
x1 = Amplitude at the starting position
xn = Amplitude after n cycle = 20% x1 = 0.2x1
The logarithmic decrement in terms of number of cycles (n) is given by
or
or
1 x
.ln 1
n xn
1 x
0.596= .ln 1
n 0.2x1
n = 2.7cycles
A vibrating system consists of a mass of 20 kg, a spring of stiffness 20 kN/m and a
damper. The damping provided is only 30% of the critical value. Determine:
(i)
the damping factor,
(ii)
the critical damping coefficient,
(iii) the natural frequency of damped vibrations,
(iv) the logarithmic decrement, and
(v)
the ratio of the consecutive amplitudes
Given Data:
m = 20kg; s = 20kN/m = 20 x 103N/m;
c = 30%; c = 0.3cc
Solution:
(i) Damping factor ():
We know that,
Damping factor,
c 0.3c c
0.3
cc
cc
(ii) Critical damping coefficient (cc):
The critical damping coefficient is given by
c c 2 s.m 2 20 103 20
1264.91N/m/s
(iii) Natural frequency of damped vibrations (fd):
We know that the circular frequency of damped vibrations,
d 1 2 .n
where n
s
20 103
31.622rad / s
m
20
d 1 0.3 .31.622 30.165rad / s
2
Therefore the natural frequency of damped vibrations is given by
fd
d 30.165
4.8Hz
2
2
(iv) Logarithmic decrement ():
We know that logarithmic decrement,
2
1 2
2 0.3
1 0.3
2
1.976
(v) Ratio of two consecutive amplitudes:
Let xn and xn+1 = Magnitudes of two consecutive amplitudes
The logarithmic decrement can also be given by
x
x
ln n or n e
xn 1 xn 1
xn
e1.976 7.213
xn 1
The machine has a mass of 200 kg. It is placed on two different isolators and the
corresponding free vibration record is shown in figure Determine:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Damping factor,
Natural frequency of damped vibration, and
Natural frequency of undamped vibration,
Spring stiffness,
Critical damping coefficient, and
Damping coefficient ,
1.52
x (mm)
0.95
0.594
0.37
0.23
t
0.38sec
Figure Free vibration record
Given data:
m = 20kg;
x1 = 1.52 mm;
x2 = 0.95 mm; x3 = 0.594 mm;
x4 = 0.37 mm; x5 = 0.23 mm
Solution:
(i) Damping factor ():
From the figure, we can write
x1 x 2 x3 x 4
1.6
x 2 x3 x 4 x5
We know that the logarithmic decrement,
x
1.52
ln 1 ln
ln 1.6 0.47
0.95
x2
The logarithmic decrement () in terms of damping factor () is given by
or
2
1 2
0.47=
2
1- 2
or 0.47
2
4 2 2
1 2
or Damping factor, ζ = 0.0744
(ii) Natural frequency of damped vibrations (fd):
From the figure it is known that the time taken for four complete cycles is 0.38 sec.
Frequency of damped vibration,
fd = Number of cycles per =
4
10.526Hz
0.38
and the circular frequency of damped vibrations,
d = fd .2 = 10.256 x 2 = 66.14 rad/s
(iii) Natural frequency of undamped vibrations (fn):
The relationship between d and n is given by
d 1 2 n
or
66.14= 1- 0.0744
4
n
Circular frequency of undamped vibrations, n = 66.32 rad/s
Then the natural frequency of undamped vibration,
n 66.32
2
2
10.56Hz
fn
(iv) Spring stiffness (s):
We know that,
n
or
s
m
66.32=
s
200
Spring stiffness, s= 879668 N/m = 879.668 kN/m
(v) Critical damping coefficient (cc):
We know that,
cc = 2 m.n = 2 x 200 x 66.32 = 26528 N/m/s
(vi) Damping coefficient (c):
We know that,
c
or c=.c c
cc
0.0744 26528 1973.68N/m/s
The successive amplitudes of vibrations of vibratory system as obtained under
free vibration are 12, 9.3, 6.4, 3.8 and 1.2 mm respectively. Determine the damping
ratio of the system.
Given Data:
x1 = 12mm;
x2 = 9.3 mm;
x3 = 6.4mm; x4 = 3.8mm;
x5 = 1.2mm
To find: Damping ratio ()
Solution:
In this problem, the individual successive ratios are not same. So we have to
take the first and last amplitudes of vibrations to calculate the logarithmic
decrement. In this case, there are four cycles. i.e., n = 4.
We know that the logarithmic decrement,
1 x
= ln 1
n xn
1 x
= ln 1
n x5
1 12
= ln
0.575
4 1.2
We also know that,
2
1 2
or
0.575
2
1 2
or 0.575
2
4 2 2
1 2
Damping factor, = 0.0912
An instrument vibrates with a frequency of 1.24 Hz when there is no damping.
When the damping is provided, the frequency of damped vibrations was observed
to be 1.03 Hz. Find:
(i)
(ii)
the damping factor, and
the logarithmic decrement
Given data: fn = 1.24 Hz; fd = 1.03 Hz
Solution:
(i) Damping factor ():
We know that the natural circular frequency of undamped vibrations,
n 2 fn
= 2 1.24 = 7.791 rad/s
and circular frequency of damped vibrations,
d 2 fd
= 2 1.03 = 6.472 rad/s
We also know that circular frequency of damped vibrations (d);
d 1 2 n
or
6.472 1 2 7.791
Damping factor, = 0.556
(ii) Logarithmic decrement ():
We know that the logarithmic decrement () in terms of damping factor (),
2
1 2
2 0.556
1 0.556
2
4.21
A shaft 30mm diameter and 1.5 long has a mass of 16 kg per metre length. It is
simply supported at the ends and carries three isolated loads 1 kN, 1.5 kN and 2
kN at 0.4 m, 0.6 m and 0.8 m respectively from the left support. Find the frequency
of the transverse vibrations: 1. Neglecting the mass of the shaft, and
2.
considering the mass of the shaft. Take E = 200 GPa.
Given Data:
d = 30 mm = 0.03m; l = 1.5m;
m = 16 kg/m;
E=200 GPa = 200 x 109N/m2
To Find: Frequency of the transverse vibrations by (a) neglecting the mass of the
shaft, and (b) considering the mass of the shaft.
1 kN
1.5 kN
2 kN
16 kg/m
B
A
0.4m
0.6m
0.8m
1.5m
Solution:
The shaft carrying the loads is shown in figure
Moment of inertia of the shaft,
4
d
64
4
0.03 3.976 10 8 m4
64
I
Static deflection due to a load of 1 kN,
1000 0.4 1.1
Wa2b2
1
3EI.l
3 200 109 3.976 10 8 1.5
2
2
=5.41 10-3m
....Here a=0.4m, and b=1.1m
Similarly, static deflection due to a load of 1.5 kN,
1500 0.6 0.9
Wa2b2
2
3EI.l
3 200 109 3.976 10 8 1.5
=0.01222m
2
2
....Here a=0.6m, and b=0.9m
Static deflection due to a load of 2 kN,
2000 0.8 0.7
Wa2b2
3
3EI.l
3 200 109 3.976 10 8 1.5
=0.01752m
2
2
....Here a=0.8m, and b=0.7m
and static deflection due to the mass of the shaft (i.e., a udl)
16 9.811.5
5
wl4
5
s
384 EI 384 200 109 3.976 10 8
1.301 103 m
4
(a) Neglecting the mass of the shaft:
The natural frequency of transverse vibrations, according to the Dunkerley’s
equation is given by
fn
0.4985
1 2 3
0.4985
3
5.41 10 0.01222 0.01752
= 2.659Hz
(b) Considering the mass of the shaft:
The natural frequency of transverse vibrations, according to the Dunkerley’s
equation is given by
0.4985
fn
1 2 3
s
1.27
0.4985
1.301 10 3
5.41 10 3 0.01222 0.01752
1.27
2.62Hz
A shaft 1.5 m long is supported in flexible bearings at the end and carries two
wheels each of 50 kg mass. One wheel is situated at the centre of the shaft and the
other at a distance of 0.4m from the centre towards right. The shaft is hollow of
external diameter 75 mm and inner diameter 37.5 mm. The density of the shaft
material is 8000 kg/m3. Find the frequency of transverse vibration. Take E = 200
GPa.
Given Data: l = 1.5m;
Do = 75mm = 0.075m;
Di = 37.5 mm = 0.0375 m;
= 8000 kg/m3;
E = 200 GPa = 200 x 109 N/m2
To find: Frequency of transverse vibration (fn)
50 kg
50 kg
26.51 kg/m
A
B
C
0.75m
D
0.4m
1.5m
Solution: A shaft supported in flexible bearings is assumed to be a simply supported
beam. The given shaft is shown in figure.
Since the density of shaft material is given as 8000 kg/m3, therefore mass of
the shaft per metre length,
ms = Area x Length x Density
0.0752 0.03752 1 8000
4
26.51kg / m
We know that moment of inertia of the shaft,
D04 Di4
64
4
4
0.075 0.0375
64
6
4
1.456 10 m
1
We know that static deflection due to a mass of 50 kg at C,
1
W1a2b2
3EI.l
50 9.81 0.75 0.75
2
2
3 200 109 1.456 10 6 1.5
1.184 10 4 m
....Here a=0.75m, and b=0.75m
Similarly, static deflection due to a mass of 50 kg at D,
2
W2a2b2
3EI.l
50 9.811.15 0.35
2
2
3 200 109 1.456 10 6 1.5
6.064 10 5 m
....Here a=1.15m, and b=0.35m
We know that static deflection due to mass of the shaft,
5
wl4
s
384 EI
26.51 9.811.5
5
384 200 109 1.456 10 6
4
5.887 10 5 m
....Here w=ms g
Therefore the frequency of transverse vibration is given by
fn
0.4985
1 2
s
1.27
0.4985
5.887 10 5
1.184 10 4 6.064 10 5
1.27
33.2Hz
Note:
WHIRLING OR CRITICAL SPEED OF SHAFTS
Critical or whirling or whipping speed is the speed at which the shaft tends to
vibrate violently in the transverse direction. In other words, the speed at which
resonance occurs are known as the critical speed.
At critical speeds, the amplitude of vibration of rotors is excessively large and
a large amount of force is transmitted to the bearings of foundations. The system
may even fail because of violent nature of vibrations in the transverse direction.
Therefore, it is important to find the natural frequency of the shaft to avoid the
occurrence of critical speeds.
Causes of the Critical Speeds:
The critical speed may occur due to one or more of the following reasons:
(i)
(ii)
(iii)
Eccentric mountings like gears, flywheels, pulleys, etc.
Bending of the shaft due to its own weight,
Non – uniform distribution of rotor material, etc.
Determine the critical speed of the shaft of Example (8) loaded in the same way.
Given data: Refer Example (8)
To find: Critical speed of the shaft (Ncr)
Solution:
We know that the critical speed of the shaft in r.p.s is equal to the natural frequency
of transverse vibrations in Hz.
From Example (8)
Natural frequency, fn = 33.2 Hz
Critical speed of the shaft, Ncr = 33.2 r.p.s.
or
Ncr = 33.2 x 60 = 1992 r.p.m
A shaft of diameter 10 mm carries at its centre a mass of 12 kg. It is supported by
two short bearings, the centre distance of which is 400 mm. Find the whirling
speed: (i) Neglecting the mass of the shaft, and (ii) considering the mass of the
shaft. The density of shaft material is 7500 kg/m3. Take E = 200 GN/m2.
Given Data: d = 10mm = 0.01m;
m = 12 kg;
= 7500 kg/m3;
E = 200 GN/m2
= 200 x 109 N/m2
50 kg
6.51 kg/m
B
C
0.2m
0.4m
To find Whirling speed by (i) neglecting the mass of the shaft, and (ii) considering
the mass of the shaft
Solution:
A shaft supported in short bearings is assumed to be a simply supported beam. The
given shaft is shown in figure.
Since the density of shaft material is given as 7500 kg/m3, therefore mass of
the shaft per metre length,
ms Area Length Density
2
0.01 1 7500 0.589kg / m
4
4
Moment of inertia, I= d4
0.01 4.91 10 10 m4
64
64
=
We known the static deflection due to a mass of 12 kg at C,
1
Wa2b2
3EI.l
12 9.81 0.2 0.2
2
2
3 200 109 4.91 10 10 0.4
1.598 10 3 m
......Here a = 0.2m, and b=0.2m
and the static deflection due to mass of the shaft,
0.589 9.81 0.4
5
wl4
5
s
384 EI 384 200 109 4.91 1010
1.961 105 m
4
(i) Neglecting the mass of the shaft:
The natural frequency of transverse vibrations is given by
fn
0.4985
0.4985
12.47Hz
1.598 103
We know that whirling speed (Ncr) of the shaft in r.p.s is equal to the frequency of
transverse vibration in Hz.
1
Ncr = 12.47 r.p.s. = 12.47 x 60 = 748.22 r.p.m
(ii) Considering the mass of the shaft:
The natural frequency of transverse vibrations is given by
fn
0.4985
1
s
1.27
0.4985
1.961 10 5
1.598 10 3
1.27
12.41Hz
Therefore, whirling speed,
Ncr = 12.41 r.p.s. = 12.41 x 60 = 744.63 r.p.m
The following data relate to a shaft held in long bearings.
Length of shaft = 1.2m; Diameter of shaft =14mm; Mass of a rotor at midpoint =
16kg; Eccentricity of centre of mass of rotor from centre of rotor =0.4mm; modulus
of elasticity of shaft material =200 GN/m2; permissible stress in shaft material =70
x 106 N/m2.
Determine the critical speed of the shaft and the rang of speed over which it
is unsafe to run the shaft. Assume the shaft to be mass less.
Given data: l=1.2m; d=14mm; m=16kg; e=0.4mm; E=200 GN/m2;=70 x 106 N/m2.
To find:
(i)
(ii)
Critical speed of the shaft (Ncr); and
Range of speed.
Solution:
I
d4
14 10
64
3
4
64
=1.886 10-9 m 4
(i)
Critical speed of he shaft:
As the shaft is held in long bearings, it may be assumed to be fixed at both ends.
We know that the static deflection at the centre of shaft.
16 9.81 1.2
Wl 3
192EI 192 200 109 1.886 10 9
3
=3.745 10 -3 m
Natural frequency, fn
0.4985
0.4985
8.145Hz.
3.745 10 3
Critical speed, Ncr =8.145 r.p.s.
=8.145 x 60
=488.72 rpm
(ii)
Range of speed:
=
When the shaft rotates, the additional dynamic load (W1=m1.g) on the shaft can be
obtained from the relation,
M
y
I
WI.l
d
and y =
2
8
WI.l
8
I
d
2
M=
Where
WI 1.2
6
8
70 10
1.886 10 9 14 10 3
2
or
WI 125.7N
or
Additional deflection due to this load,
y=
WI
125.7
3.745 10 3
W
16.81
=0.003m
We also know that the additional deflection,
Thus the range of unsafe speed is from 459rpm to 525rpm.
ADDITIONAL PROBLEMS
Example 1 : A machine part having a mass of 2.5 kg vibrates in a various medium.
A harmonic exciting force of 30 N acts on the part and causes a resonant amplitude
of 14 mm with a period of 0.22 second. Find the damping coefficient.
If the frequency of the exciting force is changed to 4 Hz, determine the increase in
the amplitude of the forced vibrations upon the removal of the damper.
Solution :
m = 2.5 kg, F0 = 30 N, A = 14 mm, T = 0.22 s,
2 2
28.56 rad/s
T 0.22
(i) at resonance, = n
or n
or
s
28.56 rad/s
m
s
28.56,s 2039N / m or 2.039 N/mm
m
Now, A=
or
Or
Or
A=
F0 / s
2
2 2 2
1
n n
F0 / s
2
30 / 2039
2
= 0.526
0.014
c = 2m n = 2 2.5 28.56 0.526
= 75.04 N/m/s
= 0.075 04 N/mm/s
ii) =fn 2 = 4 2 = 25.13 rad/s
with damper :
1
n
A=
30/2039
2
2
25.13 2
25.13
1
2
0.526
28.56
28.56
30 / 2039
(0.2258) 2 (0.9248) 2
0.0155 m
Without damper : = 0
A
30 / 2039
0.00652 m
0.2258
Increase in magnitude = 0.0652 – 0.0155 = 0.0497 m or 49.7 mm
Example : A single cylinder vertical diesel engine has a mass of 400 kg and is
mounted on a steel chassis frame. The static deflection owing to the weight of the
chassis is 2.4 mm. the reciprocating masses of the engine amounts to 18 kg and the
stroke of the engine is 160 mm. A dashpot with a damping coefficient of 2 N/mm/s
is also used to dampen the vibrations. In the steady-state of the vibrations,
determine,
i) The amplitude of the vibrations if the driving shaft rotates at 500 rpm
ii) The speed of the driving shaft when the resonance occurs.
Solution :
m = 400 kg
c = 200 N/m/s
N = 500 rpm
r = 80 mm
= 2.4 mm = 0.0024 m
2 500
52.36 rad/s
60
Now, s = mg
s 0.0024 = 400 9.81
s = 1.635 106 N/m
centrifugal force due to reciprocating parts (or the static force),
F0 = mr2 = 18 0.08 (52.36)2 = 3948 N
i) A=
=
F0
(s-m2 ) 2 (c) 2
3948
[1.635 10 400(52.36)2 ]2 (2000 52.36)2
6
=0.0072 m or 7.2mm
ii) Resonant speed :
n
or
s
1.635 106
63.93 rad/s
m
400
2N
63.93, N=610.5 rpm
60
Example:
A refrigerator unit having a mass of 35 kg is to be supported on three
springs, each having a spring stiffness s. the unit operates at 480 rpm. Find the
value of stiffness s if only 10% of the shaking force is allowed to be transmitted to
the supporting structure.
Solution :
As no damper is used.
1
2
1
n
=
2 480
16 and =0.1
60
0.1=
or
1
16 2
1
n
2
16
0.1 0.1
1
n
If the positive sign is taken,
16
9 which is possible.
n
Therefore, taking the negative sign,
16
11
n
or n 15.15 rad/s
or
s
s
15.15
m
35
Equivalent stiffness,
s = 8037 N/m = 8.037 N/mm
Stiffness of each spring =
8.037
2.679 N/mm
3
Example : A machine supported symmetrically on four springs has a mass of 80
kg. The mas of the reciprocating parts is 2.2 kg which move through a vertical
stroke of 100 mm with simple harmonic motion. Neglecting damping, determine
the combined stiffness of the springs so that the force transmitted to the
foundation is 1/20th of the impressed force. The machine crankshaft rotates at 800
rpm.
If, under actual working conditions, the damping reduces the amplitudes of
successive vibrations by 30%, find,
i) the force transmitted to the foundation at 800 rpm,
ii) the force transmitted to the foundation at resonance, and
iii) the amplitude of the vibrations at resonance.
Solution :
M 80 kg, =
1
0.05
20
m 2.2kg, N=800 rpm
r=
100
50mm
2
2N 2 800
83.78 rad/s
60
60
In the absence of damping,
or
1
2
1
n
0.05=
1
2
83.78
1
n
n 18.28 rad/s
or
s
s
18.28
M
80
Combined stiffness, s = 26.739 N/m = 26.739 N/mm
i)
x
1
ln 1 ln
1 0.3
1- 2
x2
2
2
0.003 23,
1 2
=0.0567
=
=
1+(2/n ) 2
[1 ( / n ) 2 ]2 (2 / n ) 2
83.78
1+ 2 0.0567
18.28
2
2
2
83.78 2
83.78
1-
2 0.0567
18.28
18.28
0.0563
The maximum unbalanced force on the machine due to the reciprocating parts,
F = mr2 = 2.2 0.05 (83.78)2 = 772.1 N
F
t
F
or 0.0563 =
ii) At resonance,
F
or Ft 43.47N
772.1
(1+(2))2
1 (2 0.0567) 2
1, =
8.875
n
2
2 0.0567
maximum unbalanced force on the machine due to reciprocating parts at resonance,
i.e when =n,
F = 2.2 0.05 (18.28)2 = 36.76 N
Ft = F = 8.875 36.76 = 326.25 N
iii) Aplitude =
=
Force transmitted at resonance
stiffness
326.25
12.2mm
26.739
Example: The shaft show in fig. carries two masses. The mass A is 300 kg with
radius of gyration of 0.75 m and the mass B is 500 kg with radius of gyration of 0.9
m. Determine the frequency of the torsional vibrations. It is desired to have the
node at the mid-section of the shaft of 120 mm diameter by changing the diameter
of the section having a 90 mm diameter. What will be the new diameter?
Solution :
a) Reducing the shaft to torsionally equivalent shaft of 100 mm diameter.
Fig.
4
4
4
100
100
100
100
300
160
125
400
100
150
120
90
= 300 + 31.6 + 60.2 + 609.7
= 1001.5 mm or
To locate the node point,
1.0015 m
a
a
=
b
b
4
mbk b2
or
mak a2
or
300 (0.75)2
or
a
a
b
500 (0.9)2 (1.0015
a
)
0.707m
a
1
2
fn
(0.1)4
GJ
1
32
13.2 Hz
2 300 (0.75)2 0.707
a a
84 109
b) If the node point is to be at the mid-section of the shaft with diameter 120 mm.
a
300 31.6
mak a2
a
60.2
361.7mm or 0.3617 m
2
mbk b2
b
300 (0.75)2 0.3617 500(0.9)2
or
b
b
0.1507m
Let d be the new diameter of the last section of the shaft.
Then,
b
4
4
1
100
100
125
400 d
2
200
60.2
100
or 150.7=
400
2
d
4
d=135 mm
Example : A torsional system is shown in fig. find the frequencies of torsional
vibrations and the positions of the nodes. Also find the amplitudes of vibrations
G= 84 109 N/m2.
Solution :
Let the engine, propeller and the flywheel are represented by the rotors a, B and C
respectively. Reducing the system to torsionally equivalent shaft of 40 mm diameter,
4
40
1 2
1.247m
45
a a
30
a
b b
50
or
b
b
0.6
a
Fig.
Also,
1
1
2
1
1
1
30 a 90 1.247
a a
On solving,
1
c
1
1
a
a
a
1
4 0.6
a
= 4.821 m and
b = 2.893 m and
0.913 m
0.548 m
a
for two-node frequency [above fig],
a
= 0.913 m and
=0.548 m
b
4
However, actual,
=
a
1
fn
2
a
GJ
a a
40
0.913
1.462 m
45
1
2
(0.04)4
32
4.41 Hz
30 0.913
84 109
Amplitudes
Ac
Aa
c1
or A c 1
a
Ab
Ac
b
1.247 0.913
0.366 rad (assuming A a 1 rad)
0.913
or A b 0.366
c2
0.548
0.058 rad
4.0 0.548
For single-node vibration or for fundamental frequency (fig.d)
a
= 4.821 m and
fn
1
2
b
= 2.893 m
(0.04)4
32
1.92 Hz
30 0.821
84 109
The node occurs at a distance of 2.893 m from B.
A c 1
4.821 1.247
0.741 rad
4.821
A b 0.741
2.893
1.937 rad
4 2.893
Example: The following data refer to the transmission gear of a motor ship (below
fig).
Moment of inertia of flywheel = 4800 kg.m2
Moment of inertia of propeller = 3200 kg.m2
Modulus of rigidity of shaft material = 80 109 N/m2
Equivalent MOI per cylinder = 400 kg.m2
Assuming the diameter of the torsionally equivalent crankshaft to be 320
mm and treating the arrangement as a three-rotor system, determine the frequency
of free torsional vibrations.
Solution :
Replace the four cylinders by a single rotor at the centre of their combined mass
(refer fig.b).
The length of the torsionally equivalent shaft between the flywheel nd the propeller,
4
2
4
320
320 320
4
6
300
280 300
4
5.178 10.236 6.473 21.89m
a a
b b
(400 4)
1
1
c
b b
1
3200
a
b
3200
1
1
a
b
, i.e
1
2
1 1
4800 4 2
b
b
a
b
1
21.89
On solving,
= 14.76 m and 1.48 m,
a = 29.52 m and 2.96 m
b
2
b
fig.
fn1
1 GJ
1
2 Iblb 2
(0.32)4
32
6.65 Hz
3200 14.76
84 109
(single-node)
14.76
21 Hz
(Two-node)
1.48
Example : A reciprocating IC engine is coupled to a centrifugal pump through a
pair of gears. The shaft from the flywheel of the engine to the gear wheel has a 48
mm diameter and is 800 mm long. The shaft from the pinion to the pump has 32
mm diameter and is 280 mm long. Pump speed is four times the engine speed.
Moments of inertia of flywheel, gear-wheel, pinion and pump impeller are 1000
kg.m2, 5kg.m2 respectively. Find the natural frequency of the torsional oscillations
of the system G = 80 GN/m2.
fn2 6.65
Solution :
Let rotors A and B in fig. represent the engine flywheel and the centrifugal
pump respectively. As the pump speed is four times the engine speed (the pinion is
supported on the pump shaft and the gear on the engine shaft), therefore,
Gr
Also,
1
0.25
4
da 48 mm
1
800 mm
a
1000kg.m2
db 32 mm
2
280 mm
b
18 kg.m2
ca
14 kg.m2
'
b
'
c
b
2
r
G
ca
cb
5 kg.m2
G=80 GN/m2
18
5
14
94 kg.m2
2
2
(0.25)
(0.25)
cb
2
r
G
14
5
94 94 kg.m2
2
(0.25)
Thus, the system becomes a three-rotor one.
4
48
(0.25) 280
88.6 mm
32
'
2
2
= 0.0886 m
Now,
or
a a
' '
b b
1000
1
Also,
288
a
1
'
c
a a
1
1000
a
a
'
b
1
1
'
b
a
1 1
94 0.8
3.47la
'
2
1
a
'
b
1
0.886 3.47
a
0.2354m or 0.0059m
Single node frequency
1
fn1
2
GJ
a a
fn2 2.12
1
2
(0.048)4
32
2.12 Hz
1000 0.2354
80 109
0.2354
13.4 Hz
0.0059
*******
