Theory of Machines

1. Prof A B Karpe
By
Prof. A B Karpe
Karpe.ajit@gmail.com
Department of Mechanical Engineering
STES’s, Smt. Kashibai Navale College of Engineering

2. CONTENTS
• Introduction
• Analytical method for Displacement, Velocity and Acceleration analysis
of slider crank Mechanism.
• Position analysis of links with vector and complex algebra methods
• Loop closure equation
• Chase solution
• Velocity and acceleration analysis of four bar and slider crank
mechanisms using vector and
complex algebra methods.
• Hooke’s joint, Double Hooke’s joint.A B Karpe

3. APPROXIMATE ANALYTICAL METHOD
 
 OPStrokeofLinewithRodConnectingofAngle
OPStrokeofLinewithCrankofAngle
r
l
RatioObliquityn
OCCrankofRadiusr
CPRodConnectingofLengthl
,
,
,
,







A B Karpe

4. • Find: Crank Position (ϴ) using
Displacement of Piston
• Find Velocity of Piston using (Vp)
Velocity of Piston
• Find Acceleration of piston (Ap)
Acceleration of Piston
• Find angular velocity of connecting rod (ωp)
Angular velocity of C.R
• Find Angular acceleration of Connecting rod (αp)
Angular acceleration of C.R.







n
rx
2
sin
)cos1(
2







n
rvP
2
2sin
sin.







n
raP


2cos
cos.2
nn
CP




cos.
sin
cos.
22



2/322
22
)sin(
)1(sin.






n
n
CPA B Karpe










 22
sin2
2sin
sin.
n
rvP
Uniform speed of
crank
Non-Uniform speed of
crank

5. NUMERICAL
• IN an IC engine the rod of rotation is 2000 rpm. The
connecting rod is 270 mm long and crank radius is 60
mm. Determine at 30 % of outstroke.
i. Linear velocity of piston
ii. Angular position of crank
iii. Linear acceleration of piston
iv. Angular acceleration of connecting rod
v. Angular velocity of connecting rod
vi. Crank angle for maximum piston velocity
A B Karpe

6. COMPLEX ALGEBRA METHOD
»This method is preferred when high level
accuracy is desired or the analysis is to be
repeated for a large number of configurations.
• Complex algebra method is also called as complex variable
method.
• Complex numbers are not vectors but that can be used to
represent vector
• In Polar form of fig, vector OR of magnitude r and direction
θ is written as,
• In Complex form the same can be
represented as
 rROR
i
reR  A B Karpe

7. COMPLEX ALGEBRA METHOD
»Where, r = magnitude of vector R
θ = Direction of vector R w. r. t. Real axis
• The location of any point in fig (b) is specified by its
corresponding real and imaginary coordinates
1i
(a) (b)
yx
iRRR 
A B Karpe

8. COMPLEX ALGEBRA METHOD
• Employing complex rectangular notation for vector R
• Euler’s Equation from trigonometry
Hence,
and
 sincos iRRRR 

sincos iei

)sin(cos  iR 
i
R Re
)
2
(





i
i
eie
A B Karpe

9. DEFINITION OF COMPLEX
NUMBER:
• A complex number is a number of the form a + bi,
where a and b are real number and i is an imaginary unit
satisfying
Properties of Complex number:
• Equity: Two complex numbers are equal if and only if their
real parts and their imaginary parts are respectively equal.
• Addition: To add two complex numbers, add the real parts
to one another and the imaginary parts to one another.
1i
A B Karpe

10. • In complex form vector OC of
magnitude R is written as,
------ (i)
Where, R is magnitude and θ is position of
vector
On Differentiate Eq (i) w. r. t. Time, we
get velocity
Velocity,
VELOCITY ANALYSIS OF A LINK BY
COMPLEX ALGEBRA
i
OC Re






)(
)(Re


i
dt
d
eR
dt
d
V
i
i
C
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11.  



i
i
er
dt
d
ieR
.
.












 )
2
(
..


i
C erV•
---(ii)
• Where, r.ω = Magnitude of velocity
and = Direction of velocity of link OC which is
900 ahead from original position of OC
)
2
( i
e
A B Karpe

12. ACCELERATION ANALYSIS OF A
LINK BY COMPLEX ALGEBRA
• On Differentiating equation (ii) w. r. t. Time, we get
acceleration of link,
 )(.. )2/( 
 
 i
c er
dt
d
a




 
dt
d
e
dt
d
iera ii
c

  )2/()2/(
...
t
c
r
cc
ii
c
aaa
erera

  )2/()(2
.... 

A B Karpe

13. Velocity and acceleration of a link in complex variable
form
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14. LOOP CLOSURE EQUATION
• The sum of the relative position vectors for the links
forming a close loop in the mechanism take by order is always
zero.
• As the links in the mechanism are rigid, hence even after
changing the loop position the magnitude of position vector
do not change w. r. t. Time.
• The four general cases are discussed
 Loop closure equation for four bar chain
 Loop closure equation for slider crank chain
 Loop closure equation for offset slider crank chain
 Loop closure equation crank and slotted lever mechanism
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15. LOOP CLOSURE EQUATION FOR FOUR
BAR MECHANISM
DADA
CDCD
BCBC
ABAB




DOD
COC
BOB
AOA



From fig Absolute position
vectors A, B, C & D are given as
Relative position
vectors
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16. LOOP CLOSURE EQUATION FOR FOUR BAR CHAIN
0 DACDBCABDACDBCAB
0 DACDBCAB
• Adding all relative positions vectors as under
• The above equation represents Loop closure
equation
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17. LOOP CLOSURE EQUATION FOR
SLIDER CRANK CHAIN
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18. LOOP CLOSURE EQUATION FOR
SLIDER CRANK CHAIN
From fig Absolute position
vectors A, B, C & D are given as
COC
BOB
AOA



Relative position vectors
CACA
BCBC
ABAB



CABCABCABCAB 
0 CABCAB
Adding all relative positions vectors as under
The above equation represents Loop closure
equationA B Karpe

19. LOOP CLOSURE EQUATION FOR
OFFSET SLIDER CRANK
MECHANISM
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20. LOOP CLOSURE EQUATION FOR
SLIDER CRANK CHAIN
From fig Absolute position
vectors A, B, C & D are given as
DOD
COC
BOB
AOA




Relative position vectors
DADA
CACD
BCBC
ABAB




CABCABCABCAB 
0 CABCAB
Adding all relative positions vectors as under
The above equation represents Loop closure
equationA B Karpe

21. NUMERICAL
• In slider crank mechanism, the crank radius is 100
mm and length of connecting rod is 500mm. The
crank is rotating in counter-clockwise direction at an
angular velocity of 15 rad/sec and the angular
acceleration of piston and angular acceleration
connecting rod when the crank is at 600 from IDC.
A B Karpe

22. STEPS
• For Loop Closure Equation
• Find, θ3
• Angular velocity of connecting rod
• Find, ω3
• Acceleration of piston
324 RRR 
xeler ii
 32
.. 
)cos.cos.()sin.sin..( 33223322  lrilrVp 
A B Karpe

23. VECTOR ALGEBRA METHOD
• Vectors are used to represent the magnitude and
position of kinematic links
• Let the loop closure equation for a slider crank
mechanism is
• The above equation in vector form is written as
• Properties of Vector
 Dot product of two perpendicular vector is zero
 Cross product of two parallel vectors is zero
BAC 

 BBAACC
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24. • Chace soutions are closed form solutions of two and
three dimensional vector equations.
• Case 1: magnitude and direction of C are unknown
Chace Solution-I
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25. • Magnitude of A and B are unknown
eliminating B by taking dot product with
every term with (B cross K).
Chace Solution-II
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26. • Magnitude of A and direction B are unknown
(eliminating A)
Chace Solution-III
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27. VELOCITY ANALYSIS USING VECTOR METHOD
• Let R the magnitude of the vector of a link
• Differentiate w. r. t. Time to get velocity
• Where,
• --------------------(i)

 RRR

 R
dt
d
R
dt
dR
RR
dt
d
)(
)(



RkRR
dt
d

)()(

 RkRRRR
dt
d
V 
A B Karpe

28. ACCELERATION ANALYSIS OF
LINK USING VECTOR METHOD
• Differentiate Equation (i), to get acceleration
• For non uniform motion of link
• For non-uniform motion of link
• The first part of equation is tangential component of
acceleration whereas second part is centripetal
component of acceleration






)()( RkRRR
dt
d
V
dt
d
a 
)(2

 RRa 
)()( 2

 RRRkRa 
A B Karpe

29. NUMERICAL
• In IC engine mechanism, crank is 40 mm long
and length of connecting rod is 160 mm. The
crank angle is 400 with the TDC position. Find
the angle made by connecting rod with the
line of stroke and the distance between the
crank and piston using chace solution.
A B Karpe

30. STEPS
• To form Loop closure equation
• Angular position of connecting rod (θ3
• Distance between crank and piston

 332211 ... RRRRRR
A B Karpe

31. HOOK JOINT
• Used to connect two non parallel intersecting shaft
• It is used when angular misalignment between shaft
is up to 400
• The main disadvantage of single hook joint, velocity
ratio is not constant.
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32. UNIVERSAL JOINT
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33. Hook’s Joint Analysis
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34. MAXIMUM AND MINIMUM
SPEED OF DRIVEN SHAFT
• Angular velocity ratio
• or
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35. MAXIMUM AND MINIMUM SPEED OF DRIVEN
SHAFT
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36. ANGULAR ACCELERATION OF THE DRIVEN SHAFT
• Angular velocity of driving and driven shaft is
equal
CONDITION FOR EQUAL SPEED
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37. MAXIMUM FLUCTUATION OF
SPPED
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38. Polar diagram of angular
velocities
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39. DOUBLE HOOK JOINT
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40. NUMERICALS
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41. STEPS
• Torque required at driving shaft
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42. • Value of α for the total fluctuation speed to be
24 rpm
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43. THANK YOU
A B Karpe