Page : 1 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
UCSI University
Faculty of Engineering
Kuala Lumpur, Malaysia
Department of Mechatronics
Lecture 3
Modeling Mechanical Systems I
Mohd Sulhi bin Azman
Lecturer
Department of Mechatronics
UCSI University
1 August 2011
Page : 3 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Contents
• Mechanical Systems
• Basic Law
– Time derivative motion
• displacement, velocity & acceleration
– Newton’s Law of Motion
• Mass, Spring and Damper
• Equivalent System
• Modeling Examples
Mechanical System
• A mechanical system is basically a system of elements
that interact based on fundamental mechanical
principles.
• There are two types of mechanical system:
– Translational system
• The motion is generally linear
– Rotational system
• The motion is rotational (revolution)
• We need to consider torque and moments of inertia
• In this lecture, we shall be discussing modeling
Page : 5 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Time Derivative Motion
• Time derivative is the rate of change of a function with
respect to time.
• Displacement
– A vector quantity representing the length of separation
between two objects or bodies,
• Velocity
– The rate of change of displacement
• Acceleration
– The rate of change of velocity
Time Derivative Motion
Displacement
Velocity
Acceleration
Differentiate
Page : 7 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Newton’s Law of Motion
• Law I : Mass (inertia)
– An object will stay at rest until there is an external
force applied to it.
• Law II : Acceleration/gravity
– F=ma
• Law III : Action/Reaction
– For every action, there will always be an equal and
opposite reaction, that is : F
action
=F
reaction
=-F
action
.
Mass
• A mass is simply the numerical value of its inertia.
• Put it in another way, a mass is a property of an object that resists
acceleration. It has the property of an inertia.
• Mathematically:
• This equation illustrates how mass relates to the inertia of a body.
• Consider two objects with different masses. If we apply an
identical force to each, the object with a bigger mass will
experience a smaller acceleration, and the object with a smaller
mass will experience a bigger acceleration. We might say that the
larger mass exerts a greater "resistance" to changing its state of
motion in response to the force.
Page : 9 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Spring
• A mechanical device that
stores energy.
• Most spring obeys Hooke’s
Law, that is, the force with
which the spring pushes
back is linearly proportional
to the distance from its
equilibrium length.
• Mathematically:
Hooke's law accurately models
the physical properties of common
mechanical springs for small
changes in length.
(Source: Wikipedia, 2010).
F
= −
k x
Spring
• Given the following spring.
• Then:
(
1
2
)
0
(
1
2
)
t
Page : 11 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Equivalent System : Series Spring
• Since the spring force is the same:
1 2
1 2
x
x
x
F
F
x
k
k
= +
= +
1 2 eq eqF
k
x
F
k
F
F
k
k
=
=
+
1 2 1 2 1 2 1 2 11
1
eq eq i n i eq iF
k
k
k
F
k k
k k
k
k
k
k
k
= ==
+
=
+
=
∑
Also:
1x
2x
x
Equivalent System : Series Spring
• Spring constant:
• Spring force:
1
2
1
1
1
eq
k
= +
k
k
1
2
eq
F
=
F
=
F
1
1
i n
1
Page : 13 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Equivalent System : Parallel Spring
• For springs in parallel, the elongation is
the same, hence:
x
1
2
1
2
x
x
x
F
F
x
k
k
= =
=
=
1
2
1
2
1
2
eq
eq
eq
F
F
F
k x
k x
k x
k
k
k
= +
=
+
= +
Equivalent System : Parallel Spring
• Spring constant:
• Spring force:
(
)
1
2
1
2
1
2
eq
eq
eq
eq
eq
F
F
F
F
k x
k x
F
k
k
x
F
k x
= +
=
+
=
+
=
1
i n
eq
i
i
k
k
=
=
=
∑
In parallel spring, the
total force is the sum of
the individual force
acting on a particular
spring. Note that the
displacement (or
Page : 15 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Damper (Dashpot)
• Also known as dashpot.
• It is a mechanical device that resist the motion
via viscous friction.
Schematic diagram
Modeling Dampers
• Damping constant or dashpot impedance.
• Symbol :
µ
,
f
v
or
b
.
• For the purpose of this class, we shall use the
symbol
b
.
( )
d
dx
F
b v t
b x
b
dt
Page : 17 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Damper vs Viscous Friction
•
Damper or dashpot is a mechanical device, and
apparently, it has the same property to that of
viscous friction.
•
Viscous friction however, is generated when two
surfaces separated by a liquid slide against each
other.
•
The damping force due to friction opposes the
motion, and depends on the nature of fluid flow
between the surfaces.
•
The relationship between the relative velocity of
sliding mass and the generated damping is given by
the relationship:
damping
( )
v
dx
F
b v t
b x
b
F
dt
=
=
ɺ
=
=
v
f
x
Modeling Dampers
• Given the following damper:
• The modeling equation is:
(
1
2
)
(
1
2
)
Page : 19 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Equivalent System : Series Damper
• Damping constant:
• Damping force:
1
b
b
21
2
eq
F
=
F
=
F
1
2
1
1
1
eq
b
= +
b
b
1
1
i n
1
i
eq
i
b
b
=
=
=
∑
Equivalent System : Parallel Damper
• Damping constant:
• Damping force:
1
b
2
b
(
)
1
2
1
2
1
2
eq
eq
eq
eq
eq
F
F
F
F
b x b x
F
b
b
x
F
b x
= +
=
+
=
+
=
ɺ
ɺ
ɺ
ɺ
1
2
eq
b
= +
b
b
1
i n
eq
i
i
b
b
=
=
Page : 21 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Modeling Methods & Steps
•
Step I : Sketch Free Body Diagram & Define Important Parameters
– Draw (sketch) a suitable free body diagram and define important parameters.
– Be careful and consistent with signs : if you define your motion to the right to
be positive, then use that convention throughout your calculation.
– It doesn’t really matter what convention you use for as long as your are
consistent, you will get the right equation.
•
Step II : Obtain a Set of Differential Equation
– Start your modeling by using Newton’s Law and conserve all forces i.e. Fin=Fout.
– From there, obtain a set of differential equation describing your system.
•
Step III : Represent Your DE in Transfer Function or State Space
– Next, you may represent your differential equation in state space format or in
transfer function, depending on the requirements i.e. the question.
– If you wish to represent your equation in transfer function, then take the
forward Laplace transform and assume zero initial conditions.
– If you wish to represent your equation in state space, then generate the state
space matrix.
Modeling Example 1
Page : 23 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Solution to Modeling Example 1
• Step I : Draw a FBD.
• Step II : Obtain a set of differential equations of
motion.
0
0
0
mass
spring
damper
F
F
F
F
m x
k x b x
=
+
+
=
+
+
=
∑
ɺɺ
ɺ
Mass
F
springF
massF
damperx
Modeling Example 2
• Find the transfer function and state space
representation for the following system :
Page : 25 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Solution to Modeling Example 2
• Step I : Draw a FBD
F
due to spring+ F
due to damper+ F
due to mass= F
applied forceSolution to Modeling Example 2
• Step II : Obtain a set of differential equation of motion.
• Step III : Determine the transfer function. Take the forward
Laplace transform and assume zero initial condition throughout.
Remember that the transfer function is the ratio of the output to
the input.
mass damper spring applied
2 2
( )
( )
( )
( )
( )
vv
F
F
F
F
d x t
dx t
M
f
k x t
f t
dt
dt
M x
f x
k x
f t
+
+
=
+
+
=
+
+
=
ɺɺ
ɺ
{
}
(
)
2
2
( )
( )
( )
( )
( )
( )
( )
vv
M x
f x
k x
f t
Ms X s
f sX s
k X s
F s
Ms
f s
k X s
F s
+
+
=
+
+
=
+
+
=
ɺɺ
ɺ
Page : 27 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Solution to Modeling Example 2
• Next, define the input and the output:
• Then, compute the transfer function:
input
s
F
output
s
X
֏
֏
)
(
)
(
K
s
f
Ms
s
F
s
X
s
G
v
+
+
=
=
21
)
(
)
(
)
(
Modeling Example 3
• Find the transfer function for the following
system:
Page : 29 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Solution to Modeling Example 3
• Step I : Draw a FBD for each block of mass.
Mass I
(M
1)
x
1f(t)
F
k1F
v3F
k2Mass II
(M
2)
x
2F
v3F
k2F
k3f
v3(x
1-x
2)
k
2(x
1-x
2)
Mass I
(M
1)
f(t)
k
1x
1.
.
Mass II
(M
2)
k
3x
2k
2(x
2-x
1)
f
v3(x
2-x
1)
.
.
F
v1F
M1F
v2F
M2f
v2x
2.
M
2x
2..
f
v1x
.
1M
1x
1..
Solution to Modeling Example 3
• Step II : Write the differential equation of motion.
– For mass 1 (M
1)
– For mass 2 (M
2)
(
)
(
)
(
)
(
)
(
)
1 3 1 2
1 3
1 3 3
in out 1
1 1 1 1 2 1 1 2 1 2 1 1 1 1 2 1 2 2 2
( )
( )
( )
M v v k k v v
v v v
F
F
f t
F
F
F
F
F
f t
M x
f x
f
x
x
k x
k
x
x
f t
M x
f
f
x
k
k
x
f x
k x
=
=
+
+
+
+
=
+
+
−
+
+
−
=
+
+
+
+
−
+
ɺɺ
ɺ
ɺ
ɺ
ɺɺ
ɺ
ɺ
(
)
(
)
(
)
(
)
(
)
2 2 3 2 3
2 3
in out
2 2 2 2 1 2 2 1 3 2
0
0
0
M v v k k v v
F
F
F
F
F
F
F
M x
f x
f
x
x
k
x
x
k x
M x
f
f
x
k
k
x
f x
k x
=
=
+
+
+
+
=
+
+
−
+
−
+
=
+
+
+
+
−
+
ɺɺ
ɺ
ɺ
ɺ
Page : 31 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Solution to Modeling Example 3
• Step III : To find the transfer function, take the
forward Laplace transform and assume zero initial
condition. Remember to also define your input and your
output.
• Simplify:
(
)
(
)
(
)
(
)
(
)
(
)
1 3 3
2 3 3
1 1 1 1 2 1 2 2 2 2 2 2 2 3 2 1 2 1
( )
0
v v v
v v v
f t
M x
f
f
x
k
k
x
f x
k x
M x
f
f
x
k
k
x
f x
k x
=
+
+
+
+
−
+
=
+
+
+
+
−
+
ɺɺ
ɺ
ɺ
ɺɺ
ɺ
ɺ
L
(
)
(
)
(
)
(
)
(
)
(
)
1 3 3
2 3 3
2
1 1 2 1 2 2 2
2 2 3 2 2 1
( )
( )
( )
( )
( )
0
v v v
v v v
M s
f
f
s
k
k
X s
f s
k
X s
F s
M s
f
f
s
k
k
X s
f s
k
X s
+
+
+
+
−
+
=
+
+
+
+
−
+
=
Solution to Modeling Example 3
• Form a matrix:
• Use Cramer’s Rule to solve for X
1
(s)/F(s):
(
)
(
)
(
)
(
)
(
)
(
)
1 3 3
3 2 3
2
1 1 2 2 1
2
2
2 2 2 3
( )
( )
( )
0
v v v
v v v
M s
f
f
s
k
k
f s
k
X s
F s
X s
f s
k
M s
f
f
s
k
k
+
+
+
+
−
+
=
−
+
+
+
+
+
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
1 3 2 31 3 3
3 2 3
( ) 1
2 2
2 2 3
1 2
1 1 2 2
2
2 2 2 3
( )
( )
0
( )
X s v v vv v v
v v v
D
X s
D
F s
f s
k
M s
f
f
s
k
k
X s
M s
f
f
s
k
k
f s
k
f s
k
M s
f
f
s
k
k
Page : 33 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Solution to Modeling Example 3
• Simplifying:
• Finally:
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2 3
1 3 2 3 3
2
2 2 3
1 2
2 2
1 1 2 2 2 3 2
( )
( )
v vv v v v v
F s
M s
f
f
s
k
k
X s
M s
f
f
s
k
k
M s
f
f
s
k
k
f s
k
+
+
+
+
=
+
+
+
+
+
+
+
+
−
+
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2 3
1 3 2 3 3
2
2 2 3
1
2
2 2
1 1 2 2 2 3 2
( )
( )
v v
v v v v v
M s
f
f
s
k
k
X s
F s
M s
f
f
s
k
k
M s
f
f
s
k
k
f s
k
+
+
+
+
=
+
+
+
+
+
+
+
+
−
+
Solution to Modeling Example 3
• Note:
– You can think of X
1
(s)/F(s) to mean the displacement
of M
1
after being forced by F(s).
– The same case goes for X
2
(s)/F(s).
Page : 35 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
Knock Yourself Out!
• Derive the modeling equations for the following
system and represent them in state space.
Next Step
• Textbook reference : Chapter 2.
• Homework 3 has been posted on the course
website. Attempt them. You do not have to
submit Homework 3 as it will not be graded.
Page : 37 EE406 Control Systems Lecture 3 : Modeling Mechanical Systems I
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