1. Control Systems
Lecture 5: Mathematical Modeling of Mechanical Systems and State Space Representation
Abdul Qadir Ansari, PhD

2. Lesson Objectives
Find the transfer function for linear, time-invariant translational mechanical systems.
Find the transfer function for linear, time-invariant rotational mechanical systems
Find the transfer functions for gear systems with no loss and for gear systems with loss.

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

4. 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 i)
Circuit Symbols
Translational Spring

5. Translational Spring
If F is the applied force
Then x1 is the deformation if x2=0
Or (x1-x2) is the deformation.
The Force Displacement equation is given as
Where k is stiffness of spring expressed in N/m

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

7. Translational Spring (1) Parallel
The two springs have same displacement therefore:
(1) Parallel
If n springs are connected in parallel then:

8. Translational Spring
The force on two springs is same, F, however displacements are different therefore:
(2) Series
Since the total displacement is , and we have

9. Translational Spring Then we can obtain
If n springs are connected in series then:

10. Translational Spring
Exercise: Obtain the equivalent stiffness for the following spring networks. i)
ii)

11. Translational Mass M Translational Mass is an inertia element.
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

12. Translational Damper
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.
Translational Damper
iii)

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

14. Translational Damper
Where C is damping coefficient (N/ms-1).

15. Translational Damper Translational Dampers in Parallel
Translational Dampers in Series

16. Force-velocity, force-displacement, and impedance relationships for springs, viscous dampers, and mass
where, K, f v, and M are called spring constant, coefficient of viscous friction, and mass, respectively.

17. Analogies Between Electrical and Mechanical Components
Mechanical systems, like electrical networks, have three passive, linear components.
Two of them, the spring and the mass, are energy-storage elements;
One of them, the viscous damper, dissipates energy.
The two energy-storage elements are analogous to the two electrical energy-storage elements, the inductor and capacitor.
The energy dissipater is analogous to electrical resistance.
The motion of translation is defined as a motion that takes place along a straight or curved path. The variables that are used to describe translational motion are acceleration, velocity, and displacement.

18. Newton’s Second Law
Newton's law of motion states that the algebraic sum of external forces acting on a rigid body in a given direction is equal to the product of the mass of the body and its acceleration in the same direction. The law can be expressed as
𝑭=𝑴𝒂

19. Steps to Obtain the Transfer Function of Mechanical System
The mechanical system requires just one differential equation, called the equation of motion, to describe it.
Assume a positive direction of motion, for example, to the right.
This assumed positive direction of motion is similar to assuming a current direction in an electrical loop.
First Step, draw a free-body diagram, placing on the body all forces that act on the body either in the direction of motion or opposite to it.
Second Step, use Newton’s law to form a differential equation of motion by summing the forces and setting the sum equal to zero.
Third Step, assuming zero initial conditions, we take the Laplace transform of the differential equation, separate the variables, and arrive at the transfer function.

21. Example-1(a): Find the transfer function of the mechanical translational system given in the Figure.
Free Body Diagram M
Figure

22. Example-1(b): Find the transfer function, X(s)/F(s), of the system.
Free Body Diagram (FBD)
First step is to draw the free-body diagram.
Place on the mass all forces felt by the mass.
We assume the mass is traveling toward the right. Thus, only the applied force points to the right; all other forces impede the motion and act to oppose it. Hence, the spring, viscous damper, and the force due to acceleration point to the left.
Second step is to write the differential equation of motion using Newton’s law to sum to zero all of the forces shown on the mass.

23. Example-1(b): Continue.
Third step is to take the Laplace transform, assuming zero initial conditions,
Finally, solving for the transfer function yields
Block Diagram

24. Impedance Approach to Obtain the Transfer Function of Mechanical System
Taking the Laplace transform of the force-displacement terms of mechanical components , we get
We can define impedance for mechanical components as
For the spring,
For the viscous damper,
and for the mass,

25. Example-2: Solve example-1 using the Impedance Approach.
Summing the forces in the Laplace Transformed FBD, we get
Which is in the form of
Laplace Transformed FBD

26. Example-3: Consider a simple horizontal spring-mass system on a frictionless surface, as shown in figure below.
The differential equation of the above system is or

27. Example-4: Find the transfer function, X(s)/F(s), of the system
Example-4: Find the transfer function, X(s)/F(s), of the system. Consider the system friction is negligible.
Free Body Diagram M
Where and are force applied by the spring and inertial force respectively.

28. M Example-4: continue Then the differential equation of the system is:
Taking the Laplace Transform of both sides and ignoring initial conditions we get

29. Example-4: continue.
The transfer function of the system is if

30. Example-4: continue.
The pole-zero map of the system is

31. Example-5: Find the transfer function, X(s)/F(s), of the following system, where the system friction is not negligible.
Free Body Diagram M

32. Example-5: continue. Differential equation of the system is:
Taking the Laplace Transform of both sides and ignoring
Initial conditions we get

33. Example-5: continue. if

34. Example-6: Find the transfer function, X(s)/F(s), of the following system.
Free Body Diagram M

35. Example-8: Find the transfer function Xo(s)/Xi(s) of the automobile suspension system.

36. Example-8: continue.

37. Example-13: continue. Taking Laplace Transform of the equation (2)
The transfer function of the system is

38. Example-15: Find the transfer function, X2(s)/F(s), of the system.
The system has two degrees of freedom, since each mass can be moved in the horizontal direction while the other is held still.
Thus, two simultaneous equations of motion will be required to describe the system.
The two equations come from free-body diagrams of each mass.
Superposition is used to draw the free body diagrams.
For example, the forces on M1 are due to (1) its own motion and (2) the motion of M2 transmitted toM1 through the system.
We will consider these two sources separately.

39. Case-I: Forces on M1 Example-15: Continue. Figure-1. Figure-1:
Combine (a) & (b)
All forces on M1
Figure-1.
Figure-1:
a. Forces on M1 due only to motion of M1;
b. Forces on M1 due only to motion of M2;
c. All forces on M1.

40. Case-I: Forces on M1 Example-15: Continue.
If we hold M2 still and move M1 to the right, we see the forces shown in Figure-1(a).
If we holdM1 still and moveM2 to the right, we see the forces shown in Figure 1(b).
The total force on M1 is the superposition, or sum of the forces, as shown in Figure-1(c).
Combine (a) & (b)
All forces on M1
Figure-1:
a. Forces on M1 due only to motion of M1;
b. Forces on M1 due only to motion of M2;
c. All forces on M1.
The Laplace transform of the equations of motion can be written from Figure-1 (c) as;
(1)

41. Case-II: Forces on M2 Example-15: Continue. Figure-2. Figure-2:
Combine (a) & (b)
All forces on M2
Figure-2.
Figure-2:
a. Forces on M2 due only to motion of M2;
b. Forces on M2 due only to motion of M1;
c. All forces on M2.

42. Case-II: Forces on M2 Example-15: Continue.
If we hold M1 still and move M2 to the right, we see the forces shown in Figure-2(a).
If we move M1 to the right and hold M2 still, we see the forces shown in Figure-2(b).
For each case we evaluate the forces on M2.
The total force on M2 is the superposition, or sum of the forces, as shown in Figure-2(c).
Combine (a) & (b)
All forces on M2
Figure-2:
a. Forces on M2 due only to motion of M2;
b. Forces on M2 due only to motion of M1;
c. All forces on M2.
The Laplace transform of the equations of motion can be written from Figure-2 (c) as;
(2)

43. From equation (1) and (2), the transfer function, X2(s)/F(s), is
Example-15: Continue.
(1)
(2)
From equation (1) and (2), the transfer function, X2(s)/F(s), is
Where,
Block Diagram

44. Example-15: Continue.

45. Example-16: Write, but do not solve, the equations of motion for the mechanical network shown below.
The system has three degrees of freedom, since each of the three masses can be moved independently while the others are held still.
M1 has two springs, two viscous dampers, and mass associated with its motion.
There is one spring between M1 and M2 and one viscous damper between M1 and M3.

47. Example-10: Find the transfer function Y(s)/F(s) of the restaurant plate dispenser system.

48. Example-14: Find the transfer function Y(s)/U(s) of the train suspension system.
Car Body
Bogie-2
Bogie
Frame
Bogie-1
Wheelsets
Primary
Suspension
Secondary

49. Example-14: continue.

50. Electric Circuit Analogs
An electric circuit that is analogous to a system from another discipline is called an electric circuit analog.
The mechanical systems with which we worked can be represented by equivalent electric circuits.
Analogs can be obtained by comparing the equations of motion of a mechanical system, with either electrical mesh or nodal equations.
When compared with mesh equations, the resulting electrical circuit is called a series analog.
When compared with nodal equations, the resulting electrical circuit is called a parallel analog.

51. Series Analog
For a direct analogy b/w Eq (1) & (2), convert displacement to velocity by divide and multiply the left-hand side of Eq (1) by s, yielding;
Equation of motion of the above translational mechanical system is;
Kirchhoff’s mesh equation for the above simple series RLC network is;
(1)
(2)
(3)
Comparing Eqs. (2) & (3), we recognize the sum of impedances & draw the circuit shown in Figure (c). The conversions are summarized in Figure (d).

52. Converting a Mechanical System to a Series Analog
Example-17: Draw a series analog for the mechanical system.
The equations of motion in the Laplace transform domain are;
(1)
(2)
Eqs (1) & (2) are analogous t0 electrical mesh equations after conversion to velocity. Thus,
(3)
(4)

53. Example-17: Continue. (3) (4)
Coefficients represent sums of electrical impedance.
Mechanical impedances associated withM1 form the first mesh,
whereas impedances between the two masses are common to the two loops.
Impedances associated with M2 form the second mesh.
The result is shown in Figure below, where v1(t) and v2(t) are the velocities of M1 and M2, respectively.

54. Skill-Assessment Exercise
PROBLEM: Find the transfer function, G)s) =X2(s)/F(s), for the translational
mechanical system shown in Figure

55. Answer Skill-Assessment Exercise

56. Parallel Analog
Equation of motion of the above translational mechanical system is;
Kirchhoff’s nodal equation for the simple parallel RLC network shown above is;
(1)
(2)
Comparing Eqs. (1) & (2), we identify the sum of admittances & draw the circuit shown in Figure (c).
The conversions are summarized in Figure 2.43(d).

57. Example-18: Draw a parallel analog for the mechanical system.
Converting a Mechanical System to a Parallel Analog
Example-18: Draw a parallel analog for the mechanical system.
Equations of motion after conversion to velocity are;
(1)
(2)

58. Example-18: Continue. (1) (2)
The Equation (1) and (2) are also analogous to electrical node equations.
Coefficients represent sums of electrical admittances.
Admittances associated with M1 form the elements connected to the first node,
whereas mechanical admittances b/w the two masses are common to the two nodes.
Mechanical admittances associated with M2 form the elements connected to the second node.
The result is shown in the Figure below, where v1(t) and v2(t) are the velocities of M1 and M2, respectively.

59. Rotational Mechanical Systems
Part-II

60. Basic Elements of Rotational Mechanical Systems
Rotational Spring

61. Basic Elements of Rotational Mechanical Systems
Rotational Damper

62. Basic Elements of Rotational Mechanical Systems
Moment of Inertia

63. Table: Torque-angular velocity, torque-angular displacement, and impedance rotational relationships for springs, viscous dampers, and inertia.

64. Example
Problem: Find the transfer function, G(s) = θ2(s)/T(s), for the rotational mechanical system shown in the Figure.

65. Answer: Skill-Assessment Exercise

66. Mechanical Linkages
Part-III

67. 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.

68. 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.

69. 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.

70. Why Gearing is necessary?
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.

71. Gear Trains

72. 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
Follower
Driver
Gear Ratio = # teeth input gear / # teeth output gear
= torque in / torque out = speed out / speed in

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

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

75. Mathematical Modelling of Gear Trains

76. Mathematical Modelling of Gear Trains

77. 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. B1 B2 N1 N2

78. Mathematical Modelling of Gear Trains
For Gear-1
For Gear-2
Since
therefore
Eq (1) B1 B2 N1 N2
Eq (2)
Eq (3)

79. 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) B1 B2 N1 N2

80. Mathematical Modelling of Gear Trains
After simplification

81. Mathematical Modelling of Gear Trains
For three gears connected together

82. Home Work
Drive Jeq and Beq and relation between applied torque τa and load torque TL for three gears connected together. J1 J2 J3 τa