This document provides an overview of mathematical modeling of mechanical systems, including: - Translational systems with springs, masses, and dampers and examples of modeling simple systems. - Rotational systems with rotational springs, dampers, and moments of inertia along with examples. - Mechanical linkages including gears and gear trains. Properties of gears are discussed and gear ratios explained. Examples of modeling gear trains mathematically are provided. The document covers the basic elements and concepts for modeling both translational and rotational mechanical systems, along with examples, and also introduces mechanical linkages focused on gears and gear trains.

1. Lecture-5
Mathematical Modelling of Mechanical Systems
1

2. Outline of this Lecture
• Part-I: Translational Mechanical System
• Part-II: Rotational Mechanical System
• Part-III: Mechanical Linkages
2

3. Basic Types of Mechanical Systems
• Translational
– Linear Motion
• Rotational
– Rotational Motion
3

4. Basic Elements of Translational Mechanical Systems
Translational Spring
i)
Translational Mass
ii)
Translational Damper
iii)
Part-I
Translational Mechanical Systems

5. 5

6. Translational Spring
i)
Circuit Symbols
Translational Spring
• A translational spring is a mechanical element that
can be deformed by an external force such that the
deformation is directly proportional to the force
applied to it.
Translational Spring

7. Translational Spring
• If F is the applied force
• Then is the deformation if
• Or is the deformation.
• The equation of motion is given as
• Where is stiffness of spring expressed in N/m
2
x
1
x
0
2 
x
1
x
)
( 2
1 x
x 
)
( 2
1 x
x
k
F 

k
F
F

8. Translational Spring
• Given two springs with spring constant k1 and k2, obtain
the equivalent spring constant keq for the two springs
connected in:
8
(1) Parallel (2) Series

9. Translational Spring
9
(1) Parallel
F
x
k
x
k 
 2
1
F
x
k
k 
 )
( 2
1
F
x
keq 
• The two springs have same displacement therefore:
2
1 k
k
keq 

• If n springs are connected in parallel then:
n
eq k
k
k
k 


 
2
1

10. Translational Spring
10
(2) Series
F
x
k
x
k 
 2
2
1
1
• The forces on two springs are same, F, however
displacements are different therefore:
1
1
k
F
x 
2
2
k
F
x 
• Since the total displacement is , and we have
2
1 x
x
x 
 x
k
F eq

2
1
2
1
k
F
k
F
k
F
x
x
x
eq






11. Translational Spring
11
• Then we can obtain
2
1
2
1
2
1
1
1
1
k
k
k
k
k
k
keq




2
1 k
F
k
F
k
F
eq


• If n springs are connected in series then:
n
n
eq
k
k
k
k
k
k
k






2
1
2
1

12. Translational Spring
12
• Exercise: Obtain the equivalent stiffness for the following
spring networks.
3
k
i)
ii) 3
k

13. Translational Mass
Translational Mass
ii)
• Translational Mass is an inertia
element.
• A mechanical system without
mass does not exist.
• If a force F is applied to a mass
and it is displaced to x meters
then the relation b/w force and
displacements is given by
Newton’s law.
M
)
(t
F
)
(t
x
x
M
F 



14. Translational Damper
Translational Damper
iii)
• When the viscosity or drag is not
negligible in a system, we often
model them with the damping
force.
• All the materials exhibit the
property of damping to some
extent.
• If damping in the system is not
enough then extra elements (e.g.
Dashpot) are added to increase
damping.

15. Common Uses of Dashpots
Door Stoppers
Vehicle Suspension
Bridge Suspension
Flyover Suspension

16. Translational Damper
x
C
F 

• Where C is damping coefficient (N/ms-1).
)
( 2
1 x
x
C
F 
 


17. Translational Damper
• Translational Dampers in series and parallel.
2
1 C
C
Ceq 

2
1
2
1
C
C
C
C
Ceq



18. Modelling a simple Translational System
• Example-1: Consider a simple horizontal spring-mass system on a
frictionless surface, as shown in figure below.
or
18
kx
x
m 



0

 kx
x
m 


19. Example-2
• Consider the following system (friction is negligible)
19
• Free Body Diagram
M
F
k
f
M
f
k
F
x
M
• Where and are force applied by the spring and
inertial force respectively.
k
f M
f

20. Example-2
20
• Then the differential equation of the system is:
kx
x
M
F 
 

• Taking the Laplace Transform of both sides and ignoring
initial conditions we get
M
F
k
f
M
f
M
k f
f
F 

)
(
)
(
)
( s
kX
s
X
Ms
s
F 
 2

21. 21
)
(
)
(
)
( s
kX
s
X
Ms
s
F 
 2
• The transfer function of the system is
k
Ms
s
F
s
X

 2
1
)
(
)
(
• if
1
2000
1000



Nm
k
kg
M
2
001
0
2


s
s
F
s
X .
)
(
)
(
Example-2

22. Example-3
• Consider the following system
22
• Free Body Diagram
k
F
x
M
C
M
F
k
f
M
f
C
f
C
M
k f
f
f
F 



23. Example-3
23
Differential equation of the system is:
kx
x
C
x
M
F 

 


Taking the Laplace Transform of both sides and ignoring
Initial conditions we get
)
(
)
(
)
(
)
( s
kX
s
CsX
s
X
Ms
s
F 

 2
k
Cs
Ms
s
F
s
X


 2
1
)
(
)
(

24. Example-4
• Consider the following system
24
• Free Body Diagram (same as example-3)
M
F
k
f
M
f
B
f
B
M
k f
f
f
F 


k
Bs
Ms
s
F
s
X


 2
1
)
(
)
(

25. Example-5
• Consider the following system
25
• Mechanical Network
k
F
2
x
M
1
x B
↑ M
k
B
F
1
x 2
x

26. 26

28. 28

29. Example-6
• Find the transfer function X2(s)/F(s) of the following system.
1
M 2
M
k
B

30. 30

31. 31

32. Example-7
32

33. 33

34. 34

35. 35
Example-9

36. 36

37. 37

38. 38

39. Example-10
• Find the transfer function of the mechanical translational
system given in Figure-1.
39
Free Body Diagram
Figure-1
M
)
(t
f
k
f
M
f
B
f
B
M
k f
f
f
t
f 


)
(
k
Bs
Ms
s
F
s
X


 2
1
)
(
)
(

40. Example-11
40
• Restaurant plate dispenser

41. Example-12
41
• Find the transfer function X2(s)/F(s) of the following system.
Free Body Diagram
M1
1
k
f
1
M
f
B
f
M2
)
(t
F
1
k
f
2
M
f
B
f
2
k
f
2
k
B
M
k
k f
f
f
f
t
F 


 2
2
1
)
(
B
M
k f
f
f 

 1
1
0

42. Example-13
42
1
k
)
(t
u
3
x
1
M
4
B
3
B
2
M
2
x
2
B 5
B
2
k 3
k
1
x
1
B

43. Example-14: Automobile Suspension
43

44. Automobile Suspension
44

45. Automobile Suspension
45
)
.
(
)
(
)
( 1
0 eq




 i
o
i
o
o x
x
k
x
x
b
x
m 



2
eq.
i
i
o
o
o kx
x
b
kx
x
b
x
m 


 



Taking Laplace Transform of the equation (2)
)
(
)
(
)
(
)
(
)
( s
kX
s
bsX
s
kX
s
bsX
s
X
ms i
i
o
o
o 



2
k
bs
ms
k
bs
s
X
s
X
i
o



 2
)
(
)
(

46. Example-15: Train Suspension
46
Car Body
Bogie-2
Bogie
Frame
Bogie-1
Wheelsets
Primary
Suspension
Secondary
Suspension

47. Example: Train Suspension
47

48. ROTATIONAL MECHANICAL SYSTEMS
Part-I
48

49. Basic Elements of Rotational Mechanical Systems
Rotational Spring
)
( 2
1 
 
 k
T
2

1


50. Basic Elements of Rotational Mechanical Systems
Rotational Damper
2

1

)
( 2
1 
 
 
 C
T
T
C

51. Basic Elements of Rotational Mechanical Systems
Moment of Inertia


J
T 

T
J

52. Example-1
1

T 1
J
1
k
1
B
2
k
2
J
2

3

↑ J1
1
k
T
1
 3

1
B
J2
2

2
k

53. Example-2
↑ J1
1
k
1
B
T
1
 3

2
B
3
B J2
4
B
2

1

T 1
J
1
k
3
B
2
B
4
B
1
B
2
J
2

3


54. Example-3
1

T
1
J
1
k
2
B 2
J
2

2
k

55. Example-4

56. MECHANICAL LINKAGES
Part-III
56

57. Gear
• Gear is a toothed machine part, such
as a wheel or cylinder, that meshes
with another toothed part to
transmit motion or to change speed
or direction.
57

58. 58

59. 59

60. Fundamental Properties
• The two gears turn in opposite directions: one clockwise and
the other counterclockwise.
• Two gears revolve at different speeds when number of teeth
on each gear are different.

61. Gearing Up and Down
• Gearing up is able to convert torque to
velocity.
• The more velocity gained, the more torque
sacrifice.
• The ratio is exactly the same: if you get three
times your original angular velocity, you
reduce the resulting torque to one third.
• This conversion is symmetric: we can also
convert velocity to torque at the same ratio.
• The price of the conversion is power loss due
to friction.

62. Why Gearing is necessary?
62
• A typical DC motor operates at speeds that are far too
high to be useful, and at torques that are far too low.
• Gear reduction is the standard method by which a
motor is made useful.

63. Gear Trains
63

64. Gear Ratio
• You can calculate the gear ratio by using
the number of teeth of the driver
divided by the number of teeth of the
follower.
• We gear up when we increase velocity
and decrease torque.
Ratio: 3:1
• We gear down when we increase torque
and reduce velocity.
Ratio: 1:3
Gear Ratio = # teeth input gear / # teeth output gear
= torque in / torque out = speed out / speed in
Follower
Driver

65. 65

66. 66

67. Example of Gear Trains
• A most commonly used example of gear trains is the gears of
an automobile.
67

68. Mathematical Modelling of Gear Trains
• Gears increase or reduce angular velocity (while
simultaneously decreasing or increasing torque, such
that energy is conserved).
68
2
2
1
1 
 N
N 
1
N Number of Teeth of Driving Gear
1
 Angular Movement of Driving Gear
2
N Number of Teeth of Following Gear
2
 Angular Movement of Following Gear
Energy of Driving Gear = Energy of Following Gear

69. Mathematical Modelling of Gear Trains
• In the system below, a torque, τa, is applied to gear 1 (with
number of teeth N1, moment of inertia J1 and a rotational friction
B1).
• It, in turn, is connected to gear 2 (with number of teeth N2,
moment of inertia J2 and a rotational friction B2).
• The angle θ1 is defined positive clockwise, θ2 is defined positive
clockwise. The torque acts in the direction of θ1.
• Assume that TL is the load torque applied by the load connected
to Gear-2.
69
B1
B2
N1
N2

70. Mathematical Modelling of Gear Trains
• For Gear-1
• For Gear-2
• Since
• therefore
70
B1
B2
N1
N2
2
2
1
1 
 N
N 
1
1
1
1
1 T
B
J
a 

 

 

 Eq (1)
L
T
B
J
T 

 2
2
2
2
2 
 

 Eq (2)
1
2
1
2 

N
N
 Eq (3)

71. Mathematical Modelling of Gear Trains
• Gear Ratio is calculated as
• Put this value in eq (1)
• Put T2 from eq (2)
• Substitute θ2 from eq (3)
71
B1
B2
N1
N2
2
2
1
1
1
2
1
2
T
N
N
T
N
N
T
T



2
2
1
1
1
1
1 T
N
N
B
J
a 

 

 


)
( L
a T
B
J
N
N
B
J 



 2
2
2
2
2
1
1
1
1
1 



 





𝜏𝑎 = 𝐽1𝜃1 + 𝐵1𝜃1 +
𝑁1
𝑁2
(𝐽2
𝑁1
𝑁2
𝜃1 + 𝐵2
𝑁1
𝑁2
𝜃1 +
𝑁1
𝑁2
𝑇𝐿)
1
2
1
2 

N
N
 Eq (3)

72. Mathematical Modelling of Gear Trains
• After simplification
72
𝜏𝑎 = 𝐽1𝜃1 + 𝐵1𝜃1 +
𝑁1
𝑁2
(𝐽2
𝑁1
𝑁2
𝜃1 + 𝐵2
𝑁1
𝑁2
𝜃1 +
𝑁1
𝑁2
𝑇𝐿)
L
a T
N
N
B
N
N
B
J
N
N
J
2
1
1
2
2
2
1
1
1
1
2
2
2
1
1
1 



















 



 





L
a T
N
N
B
N
N
B
J
N
N
J
2
1
1
2
2
2
1
1
1
2
2
2
1
1 



































 

 


2
2
2
1
1 J
N
N
J
Jeq 








 2
2
2
1
1 B
N
N
B
Beq 









L
eq
eq
a T
N
N
B
J
2
1
1
1 

 

 



73. Mathematical Modelling of Gear Trains
• For three gears connected together
73
3
2
4
3
2
2
1
2
2
2
1
1 J
N
N
N
N
J
N
N
J
Jeq 


























3
2
4
3
2
2
1
2
2
2
1
1 B
N
N
N
N
B
N
N
B
Beq 



























74. Home Work
• Drive Jeq and Beq and relation between applied
torque τa and load torque TL for three gears
connected together.
74
J1 J2 J3
1

3

2

τa
1
N
2
N
3
N
1
B
2
B
3
B
L
T

75. END OF LECTURES-5-6-7
75