8/7/2019 20100205043742!530_343lecture05

http://slidepdf.com/reader/full/20100205043742530343lecture05 1/6

ME 530.343: Design and Analysis of Dynamic SystemsSpring 2009

Lecture 5 – Mass-Spring Systems andEquivalent Mechanical Systems

Wednesday, February 11, 2009

1 Today’s Objectives

Purpose of today’s lecture:

• Degrees of freedom

• Identify equivalent systems: meqx + beqx + keqx = f eq

• Springs and dampers in series and in parallel

Reading: Palm Chapter 4.1, 4.2, 4.4, 4.5 (ignore Laplace transform examples for now)

2 Degrees of Freedom

To begin modeling any dynamic system, you first know how many degrees of freedom  there are inthe system. This will tell you:

• How many equations of motion you will finding

• The size of the matrix equation describing the system (if you need to put it in that form, as wewill in the last half of this course)

• How many and which system variables you will select

The number of degrees of freedom is the number of parameters that can be independently

varied in a system.

Test for simple mass-spring-damper systems: Can you hold one mass in a mechanical system at aconstant position, but still move another mass in the system?

mm

k 

For complex or continuous systems (e.g., buildings), the concept of (finite) degrees of freedom is notas well defined. You typically want to create a model that has enough degrees of freedom to appro-priately characterize the system’s behavior, but not so many that the computational requirementsare too high.

1

8/7/2019 20100205043742!530_343lecture05

http://slidepdf.com/reader/full/20100205043742530343lecture05 2/6

3 Equivalent System Example

     x     x     x     x     x     x     x     x     x     x     x     x     x     x     x     x     x     x     x

k 

k m

1

m

 b

y

x

r J

q

r 

1 1

1

2

2

2

m

m

r 

f  f 

f 

f k1

 b2

1

1

1

2

2

2

k2

f 

f 

r J

q

f 

1 1

First question: How many degrees of freedom? 1Next question: In which variable(s) do we want to write the EOM? y

From last time, in terms of x:m1 + m2

r2r1

2+

J 1

r21

x + b

r2r1

2x + (k1 + k2)

r2r1

2x = 0

equivalent mass: meq = m1 + m2r22

r21

+ J 1r21

equivalent damping: beq = br22

r21

equivalent stiffness: keq = (k1 + k2)r22

r21

The equation of motion of a second order, one-degree-of-freedom system should always be of theform: meqx + beqx + keqx = f eq

Putting this in terms of y, we havem2 + m1

r1r2

2+

J 1

r22

y + by + (k1 + k2)y = 0 Left as an excercise to prove this!

So, for in terms of y, get a equivalent mass: meq = m2 + m1(r1

r2 )2

+J 1

r22equivalent damping: beq = b

equivalent stiffness: keq = k1 + k2

Note that this is quite different!

2

8/7/2019 20100205043742!530_343lecture05

http://slidepdf.com/reader/full/20100205043742530343lecture05 3/6

4 Elements in Parallel and in Series

What is meant by “parallel” and “series”?

Springs:

m

f 

k 

ym

m

y

x

k 

k 1

k1

k1

k2

k2

1

2

2

k 

mf 

f 

f 

f 

x

y

k2

Parallel: my = −f k1 − f k2my = −k1y − k2y

my + (k1 + k2)y = 0=⇒ keq = k1 + k2

Series: 0 = −f k1 + f k2f k1 = k1y

f k2 = k2(x− y)mx = −f k2mx = −k2(x− y)f k1 = f k2 = k1y = k2(x− y)

(k1 + k2)y = k2xy = k2x

k1+k2

mx + k2(−y + x) = 0

mx + k2(− k2xk1+k2

+ k1+k2k1+k2

x) = 0

mx + k1k2k1+k2

x = 0

keq = k1k2k1+k2

1keq

= 1k1

+ 1k2

+ ...

Questions: in parallel or in series?

m

xxxxxxxxxxxxxx

xxxxxxxxxxxxxx

 b b

 b1

22

f 

 b1

      x      x     x     x     x     x     x     x     x     x     x

     x     x     x     x     x     x     x     x     x     x     x     x     x    x    x    x   x   x   x   x   x  x

  x  x  x x x

 xxxx x x x  x  x  x  x  x   

x   x   x   x    x    x    x    x     x      x x  x  x  x  x

   x   x   x

   x    x

    x    x    x    x     x     x     x      x      x

     x     x     x     x     x     x     x     x     x     x     x    x    x   x   x  x

  x x xxx x  x  x  x   

x   x    x    x      x  x  x

   x   x

    x    x     x     x

3

8/7/2019 20100205043742!530_343lecture05

http://slidepdf.com/reader/full/20100205043742530343lecture05 4/6

mixed systems:

m

y

q k  k 2 2

11

J

q

k  k 

Laws for dampers → same thing!series: 1

beq= 1

b1+ 1

b2+ ...

parallel: beq = b1 + b2 + ...

4.1 Why are these simplifications useful?

Translational example

m

k 1 2

3

4

5

6

7

eq

k 

k 

k k 

k 

k 

m

k 

x

Want to write: meqx + keqx = 0. Exercise: compute meq and keq.

Rotational example:

m

J

q

J1

1

2

2

3

3

4

4

JJ

xxx

xxx

xxxxxx

 b b b

 b

x

meqx + beqx = 0 or J eqθ + beq θ = 0

Exercise: compute J eq and beq.

4

8/7/2019 20100205043742!530_343lecture05

http://slidepdf.com/reader/full/20100205043742530343lecture05 5/6

4.2 Example: Modeling the human fingertip

m

y

k k 

k 1 1

2 2

eq eq

3

 b

 b

 bk 

m

y

1beq

= 1b1

+ 1b2→ beq = b1b2

b1+b2keq2,3 = k2 + k31

keq= 1

k1+ 1

keq2,3→ keq =

k1keq2,3k1+keq2,3

keq = k1(k2+k3)k1+k2+k3

meqx + beqx + keqx = 0

5 Revisit Example From Lecture 4

m

xxxxxxxxxxxxxxxxxxx

 b

k 

1

1

1

2

2

2

3

 b

m

xxxxxxxxxxxxxx k 

 b

First question: How many degrees of freedom? 2

Next question: In which variable(s) do we want to write the EOM? x1 and x2

Note: x2 is defined with respect to movement of first block, x1Recall from last time:

mass 1: m1x1 = f k1 + f b2 − f b1 − f b3= −k1x1 + b2x2 − b1x1 − b3x1

m1x1 + (b1 + b3)      beq

x1 + k1x1 − b2x2 = 0

5

8/7/2019 20100205043742!530_343lecture05

http://slidepdf.com/reader/full/20100205043742530343lecture05 6/6