MA3005 Topic 3b - Mechanical Systems

MA3005: Control TheoryTopic 3: Mathematical Modeling

of Mechanical Systems

Tegoeh Tjahjowidodo

+-

Controller Plant++

D(s)

U(s) Y(s)

1

Constituents of Mechanical Systems

• Inertia elements

– Mass (kg) or Moment of Inertia (kg-m2)

• Spring elements

– Deformed by an external force (torque) where the deformation (angular disp.) is directly proportional to the force (torque) applied to it

• Damper elements

– Force (torque) acting on the damper is proportional to the velocity (angular velocity) difference of both ends

2

Translational Spring

F = kx = k(x1 – x2)

where k is called spring constant

3

F = kx = k(x1 – x2)

where k is called spring constant

Translational Spring

4

Rotational Spring

T = kq = k(q1 – q2)

where k is called torsional spring constant

q

1

q

2

kT T

5

Translational Damper

where b is the viscous friction coefficient

b

6

)( 21 xxbxbF

Rotational Damper

b is the torsional viscous friction coefficient

b

7

)( 21 bbT

Modeling of Mechanical Systems

• Obtain Equation of Motion using Newton’s second law

• For translational motion

F = ma

• For rotational motion

T = Jα

where, F is sum of all external forces acting in a given direction on mass, m and a is absolute acceleration in that direction

where, T is sum of all torques acting about a given axis, J is moment of inertia about that axis, α is absolute acceleration in that direction

8

Spring-mass-damper system

• Used to model flexible structures, suspension systems

• For input force (f), the output is the response (displacement, x) of the mass

• How does the output change for different force inputs?

bk

mxf(t)

9

Assumptions for this model

• Mass, spring, and damper constants are fixed and known

• Springs satisfy Hooke’s law

• Damper is (linear) viscous force, proportional to velocity

10

Free Body Diagram

• Applying Newton’s 2nd law, the eqn. of motion is

• Free Vibration Response:

If f(t) = 0 and response of

spring-mass-damper system is

only due to initial conditions

f(t) x

kx bx•

( )F f t kx bx mx

mx bx kx f t

Note: x is measured from the equilibrium location

11

TF of Mass-Spring-Damper System

Taking Laplace Transform of both sides and assuming zero initial conditions

or,

[m s2 + b s + k] X(s) = F(s)

mx bx kx f t

)]0()0()([

)]0()([)()(2 xsxsXsm

xssXbskXsF

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The transfer function is

where X(s) is the displacement (output) and F(s)the external force (input)

kbsmssFsX

2

1)()(

13

TF for Rotational System

Eqn. of Motion

Taking LT and zero i.c.

2

( ) 1

( )a

s

s Js Bs K

aKBJ

Note: Torque a is the input Angular displacement is the output

14

Example 1: B-3-14 (pp. 148)

Determine the eqns. of motion and TF for the system

15

y

Define the displacement of a point of spring k1

y

k1

k1y

16

Assuming y is moving to the right side (assumption#1):

The spring is stretched and essentially it pull all parts in its right to the left side.

k2(x-y) u(t)m

y x

k2

k2(x-y)k1y2 1( )k x y k y

17

)()( 2 yxktuxm

Define the displacement of x and y:

Assuming x moves to right and x > y (assumption#2):

Consider the mass m:

• from the 1st equation:

• The equation of motion:

xkk

ky

xkykyk

21

2

221

uxkk

kkxm

21

21

18

k2

k1y

y xk2(y-x)

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How if we take different assumption?

For assumption#2, let’s now take and x < y

xkxyk 12 )( xkyxk 12 )(

resulting in the same equation

Assuming x moves to right and x > y (assumption#2):y x

k2

k2(x-y)k1y2 1( )k x y k y

System Transfer Function

The equation of motion is

20

uxkk

kkxm

21

21

Taking Laplace transform of the equation:

uxkk

kkxm LLL

21

21

)()()0()0()(21

212 sUsXkk

kkxxssXsm

For given initial conditions 0)0()0( xx

)()()(21

212 sUsXkk

kksXms

System Transfer Function

The transfer function is

21

212

1)()(

kkkk

mssUsX

21

What is equivalent spring constant for two springs attached in series?

What is equivalent spring constant for two springs attached in parallel?

Example 2:

• Derive equations of motion for two degrees of freedom(2-DOF) system

22

Free body diagram (FBD)

Consider for the spring k0, k1, and k2; inertia J1 and J2

23

T1

11 2

2

T2T0 T0T1

T0 T0 T2

k1k0 k2

T1 = k11

24

1

T1

k1


Assuming 1 is moving clockwise:

The spring is stretched in clockwise direction

T0 = k0(2 – )

25

2k0


Assuming 2 is moving clockwise and 2 > 1 :

1

T0 T0

The spring is also stretched in clockwise direction

T2 = k22

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2

T2

k2


1 has been assumed moving clockwise:

The spring is stretched in clockwise direction

27

Define the equation of motion on J1

Applying Newton’s 2nd law:

The total torque is equivalent to the inertial torque

1

T1T0

1101 JTT

11J

0)( 2101111 kkJ

cw = positive

28

Define the equation of motion on J1

Applying Newton’s 2nd law:

The total torque is equivalent to the inertial torque 22J

22

T0 T2

2220 JTT

0)( 1202222 kkJ

Eqns. of motion (2 D.O.F. system)

Hence, the Equations of Motion are

29

What is the input to this system?Is it possible to find transfer function for this system?

0)( 2101111 kkJ

0)( 1202222 kkJ

Initial Condition Response

• Taking Laplace transform of Eqn. (1) and (2)

For given initial conditions θ1 (0) = θ10; θ2 (0) = θ20 ;

Eqn 3 and 4 can be solved to get initial condition response Θ1(s) and Θ2(s)

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1 10 2 200 ; 0

21 1 1 1 1 1 0 1 2

2 22 2 2 2 2 2 0 2 1

0 0 0 (3)

0 0 0 (4)

J s s s k s k s s

J s s s s k s k s s

General Guidelines to Model Mass-Spring-Damper Systems

1) Define a (+)ve direction

2) Define displacements from rest or equilibrium position for each inertia element in the (+)ve direction

3) Assume relative displacement across every spring and damper elements to visualise if they are in compression/ elongation

4) Determine all the forces in the spring and damper elements according to step 3. Forces on inertial element will have opposite directions

5) For each inertia element, use Newton’s 2nd Law to determine resultant force producing accel. in (+)ve direction

6) Laplace transform makes the ODE into algebraic equation for ease of manipulation of the position variable 31

Reading Assignment

32

• Modern Control Engineering, 5th Ed.

Ch. 3: Sec. 3.2

Other Mechanical systems

• Many systems have multiple components

• Power can be transmitted through pulleys, gears, levers, etc

• Can be interconnected with electrical systems such as in motors.

Now, we explore few examples of such multi-body systems or and later understand systems that involve modeling of both

mechanical as well as electrical systems

33

Gear Trains

If radii of gear 1 and gear 2 are r1 and r2

Numbers of teeth on gear 1 and gear 2 are n1 and n2

Then,

2

1

2

1

nn

rr

34

The surface speeds at the point of contact of the two gears must be identical

r1 1 = r2 2

Therefore,

2

1

2

1

1

2

nn

rr

35

• If frictional losses are neglected,

• No power loss, the torque applied to the input shaft (T1) is related to the torque transmitted to the output shaft (T2) by

T1 1 = T2 2

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Example 1. Gear Trains

Viscous rotational damping coeff., b1

Viscous rotational damping coeff., b2

37

• Load is driven by motor through gear train

• Shafts of the gear train are stiff

• No backlash or elastic deformation

• Number of teeth on each gear to radius of the gear

• Subscript 1 refers to motor shaft, gear 1 etc.

• Subscript 2 refers to load shaft, gear 2

38

Eqn. of Motion for Motor Shaft

Newton’s 2nd law:

where, • Tm is the torque developed by

the motor • T1 is the load torque on gear 1

due to the rest of the gear train

• J1 is the moment of inertia for the gear 1 along with its shaft

39

Tm

T1

b1

mTTbJ 11111 (1)

Eqn. of Motion of Load Shaft

where,

• T2 is the torque transmitted to gear 2

• TL is the load torque

• J1 is the moment of inertia for the gear 1 along with its shaft

2 11 2 2

1 2

3n

T T Tn

Since the transmitted power remains unchanged

Note: If n1/n2 < 1, then, 2/1 < 1 40

T2

2

TLb2

22222 TTbJ L (2)

The gear ratio reduces the speed, and increases the torque T2, elimination of T1 and T2 yields

This equation can be written in terms of 1 or 2

41

mL TTbJn

nbJ )( 2222

2

11111

Eqn. of motion in terms of 1

Since 2 = (n1/n2)1

Defining equivalent inertia and viscous damping coefficient (for shaft 1) for the gear train as

42

2 2

1 11 1 2 1 1 2

2 2

, eq eq

n nJ J J B b b

n n

mL TTn

nb

n

nbJ

n

nJ

2

112

2

2

1112

2

2

11

• Equation of Motion can be simplified as

where N = n1/n2

• Note: If n1/n2 << 1, the effect of J2 (b2) on the equivalent inertia J1eq (friction B1eq) is negligible

Tm

ω1

J1eqb1eq

NTL

Eqn. of Motion in terms of 1(Simplified)

43

Equivalent system for the gear train

mLeqeq TNTBJ 1111 (4)

TF of the gear coupling

44

In (4)

Assuming that the load torque TL is zero Taking the Laplace transform Dropping all subscripts – all quantities referred to

input

1 1 mJs B T s

1 1

m

s

T s Js B

BJs 1

n1/n2

Tm Motor Speed

2

Load Speed

Eqn. of motion in terms of 2

• Since 1 = (n2/n1)2

45

mL TTbJn

nbJ 2222

2

11111

mL Tn

nTbb

n

nJJ

n

n

1

2221

2

1

2221

2

1

2

• Equivalent inertia and viscous friction coeff. with reference to shaft 2 are

• Equation of Motion can be simplified as

46

2 2

2 22eq 2 1 2eq 2 1

1 1

, n n

J J J B b bn n

where n = n1/n2

mLeqeq Tn

TBJ1

2222

Example 2. Rack and pinion gear

Used to convert rotational motion to translational motion

Torsionalspring constant

Input

Output

47

Governing Equations

J

K(-A)

a

1B

K(-A)

fc

M

fc

2B x

To develop an input-output transfer function the redundant variable fc,, A must be eliminated

48

aAKBJ )(1 (1)

)( Ac KRf (2)

cfxBxM 2 (3)

• In addition, the geometric relationship

must hold because of the contact between the rack and the pinion gear

• Taking LT of (1) with zero ICs and substituting for A

from (4)

4AR x

21

5a

KX s T s

RsJs B s K

49

• Taking LT of (2) and substituting for fc from (3)

• Eliminating A using (4) and using (5)

22 A

KMs X s B sX s s s

R

KsRBB

sRMKBBJK

sRMBJBJsMR

KRsΤsX

a

221

2221

3212

42)()(

50

• If shaft is assumed to be stiff, governing eqns are

• Resulting transfer function from input torque to output linear displacement x becomes

sRBBsMRJ

RsΤsX

a2

2122)(

)(

51

acRfBJ 1

xR

cfxBxM 2

Modeling of an automobile

52

Half-Car Model

A simplified model

53

Quarter Car

Model

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MA3005 Topic 3b - Mechanical Systems