C
OLLEGE OF
E
NGINEERING
M ECHANICAL E NGINEERING
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C
OLLEGE OF
E
NGINEERING
M ECHANICAL E NGINEERING
GENERAL INSTRUCTIONS:
This examination contains five problems. You are required to select and solve four of the five problems. Clearly indicate the problems you wish to be graded. If you attempt solving all of them without indicating which four of your choice, the four problems with the worst grades will be considered. Note that
.
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Mechanical Engineering Ph.D. Preliminary Qualifying Examination
Vibration – January 26, 2009
You are required to work four of the five problems, one of which is Problem No. 5 . Clearly indicate which problems you are choosing. Show all work on the exam sheets provided and write your student personal identification (PID) number on each sheet. Do not write your name on any sheet.
Your PID number:____________________________
Question #1
A uniform bar of length L and weight W is suspended symmetrically by two unstrechable strings as shown in the figure. If the bar is given small initial rotation about the vertical axis, a.
Draw the free body diagram of the bar during its free oscillation. b.
Write down the equation of motion for small angular oscillation about axis O-O. c.
Determine the period of the free oscillation of the bar. a
O h
O
L

You are required to work four of the five problems. Clearly indicate which problems you are choosing. Show all work on the exam sheets provided and write your student personal identification (PID) number on each sheet. Do not write your name on any sheet.
Your PID number:____________________________
Question #2
The system shown in the figure is in its static equilibrium position (SEP). It consists of a uniform rod of mass m and length L and is supported by spring of stiffness k and dashpot of coefficient c . a.
Draw the free body diagram of each system as it oscillates about the SEP. b.
Derive the equation of motion of each system using Newton’s second law. c.
Determine the undamped natural frequency. d.
Determine the damping ratio, the critical damping coefficient, and the damped natural frequency.

You are required to work four of the five problems. Clearly indicate which problems you are choosing. Show all work on the exam sheets provided and write your student personal identification (PID) number on each sheet. Do not write your name on any sheet.
Your PID number:____________________________
Question #3
The system shown consists of a cylinder of mass m with a piston, which imparts resistance proportional to the velocity of a linear viscous damping c , the cylinder is restrained by a spring of stiffness k
(a) draw the free body diagram of the cylinder,
(b) write down the equation of motion using Newton’s second law, and
(c) determine the response amplitude and phase angle using Complex Algebra . y t

Y sin
( )
 t

Y sin
 t t

You are required to work four of the five problems. Clearly indicate which problems you are choosing. Show all work on the exam sheets provided and write your student personal identification (PID) number on each sheet. Do not write your name on any sheet.
Your PID number:____________________________
Question #4
The system shown below consists of two rotors coupled by a discontinuous shaft of modulus of rigidity is G
  6 lb in
2
/ rad :

Draw the free-body diagram of each rotor,

Derive the equations of motion,

Determine the natural frequencies of free torsional oscillations and provide the physical meaning of each value,

Draw the normal mode shape and evaluate the value of the twist at the junction of the two shafts, i.e.
 
1
or
 
2

MANDATORY PROBLEM (EVERYONE IS REQUIRED TO SOLVE THIS PROBLEM)
Your PID number:____________________________
Question #5
Consider a rigid body of mass m and mass moment of inertia J c
with respect to center of gravity
C g
. Suppose that the body is supported by two springs of stiffness k that are attached at distances
2 l and l with respect to the center of gravity C g
as shown in Figure 5a. Let m = 10 kg, J c
= 5 kgm 2 k = 100 N/m, and l = 1 m.
Part I:
(a) Derive the equations of motion for this body using coordinates x and

.
(b) Determine the natural frequencies of the system.
(c) Draw the natural mode shapes of the system.
2 l m , l
J
2 l m , J l
(5a) (5b)
Next, consider the same rigid body as shown in Figure 5b.
Part II:
(d) Derive the equations of motion for this body using coordinates x
1
and x
2
.
(e) Determine the natural frequencies of the system.
(f) Draw the natural mode shapes of the system.
Part III:
(g) State if there are differences in the equations of motion, natural frequencies and mode shapes obtained in each case and explain why.
(h) What are the couplings in equations of motion, respectively, in these two cases?

Problem 1
A uniform bar of length L and weight W is suspended symmetrically by two strings as shown in the figure. If the bar is given small initial rotation about the vertical axis, d.
Draw the FBD of the bar during its free oscillation. e.
Write down the equation of motion for small angular oscillation about axis O-O. f.
Determine the period of the free oscillation of the bar.
Solution :
Figure 5.
FBD
From the static equilibrium position we write
2T
 mg
I I

T sin

 
O a
O h
L h
(1)
Under free vibration of the bar and in an arbitrary position

, the bar will be raised up slightly, and will be displaced by a distance ( / 2)

from the its suspended string. The string also be tilted by an angle

from the vertical such that, ( / 2) h . This geometric relation gives
 
( / 2h)

.
Now writing the equation of motion by taking moments about axis OO, gives
I
0
   
T T a a
2h

Using equation (1), equation (2) takes the form
I
0
  mg a
2
4h mg a
2
       
0
(2)
(3)

where mg a
2
 mg a
2

4 m

L
2
12


 h

3g a
2 hL
2
(4)
Problem 2
FBD
I I
(b) From the free-body diagram, we write the equation of motion from the static equilibrium position using Newton’s second law of moments about the hinge axis O
I
0
  
 
L
  ka
  a
 mL
2

3
 cL
2
  ka
2
 
0 (1)
(c) The undamped natural frequency is obtained by dividing both sides of the equation of motion (1) by the coefficient of

, i.e.,
 
3 c m
 ka
2

3 mL
2
  n

3 ka
2
 a k
3
2 mL L m
(2)
(d) The damping ratio is obtained by writing the equation of motion in the form
 
2
    n n

0 (3)
Thus we can write km 2
 n

3 c m
  
2
3 c
 n

3 c m a k
2 3
L m m
The critical damping coefficient is obtained from (4) as

3 cL
2 3 kma
(4)
  c
 c cr
3 cL
2 3 kma
 c cr

2 3 kma
3 L
The damped natural frequency, frequency,

, as n

, is written in terms of the undamped natural nd
 nd
  n
1
 
2  a k
3 1
L m



3 cL
2 3 kma
 2
 a k
3 1
L m

9
2 2 c L
12 kma
2

Problem # 3
Using Newton’s second law and with the help of the FBD, the equation of motion is mx
  
(
 x ) (1)
Rearranging mx
 cx
 kx
 cy (1a)
But ( )

Y sin
 
Y Im e (2)
Also ( ) Y cos t i Y Im e
Note that one can write i
 e i
 
/ 2

(3) because e i
 
/ 2

 cos
 
/ 2
  i sin
 
/ 2
  i
Thus one can write equation (1a) in the form mx
 cx
 kx
 icY

Im e
 cY

Im e i

 

2
(4)
The response must oscillate at the same frequency of the excitation in the steady state at amplitude and phase angle to be determined, thus one can write the response in the form
( )

X Ime
 
X Ime where X

X Ime
 i

(5)
We need the first and second time derivatives of ( ) , i.e.
( )
 
, ( )
  2
X Ime (6)
Substituting expressions (5) and (6) into equation (4), gives
X Ime
 i 2
 n

X Ime
  2 n
X Ime

2
 n

Y Ime i

 

2
(7)
Canceling out Ime from both sides of equation (7) and rearranging, gives
 2
X
 i 2
 n
X
 n
2
X

2
 n

Y Ime i

2

2
 n

Y , where Y

Y Ime i

2 (8)
Rearranging
Y
X


 n
2
2
 n

 i 2
 n

(9)
Multiplying and dividing by the conjugate of the denominator, gives
X
Y
 
 n
2
2
 n

 i 2
 n




 n
2
 n
2

 i 2
 n
 i 2
 n




 2 n
  2
2
 n

2
 

2
 n
  2

 n
2
 i 2
 n

(10)
Dividing the numerator and denominator by
 2 n
, and setting takes the form r
 
/
 n
, equation (10)
Y
X


1
 r
2

2

2
 r

2
 r

2

1
 r
2

 i 2
 r (11)
Multiplying and dividing each expression by

1
 r
2

2  
2
 r

2 gives
X
Y

2
 r

1


1 r

2 r

2
2


2


2


2
 r

2 r
 2





1

 r
2
1


2 r

2


2
 r

2
 i

1
 r
2

2

2 r
 
2
 r

2




(12)
With the help of the shown triangle equation (12)may be written in the form
X

Y
2
 r

1
 r
2

2  
2
 r

2
 cos
  i sin
 
(13) (1
 r )
2 r
Expressing X and Y in terms of their original
Definitions (5) and (8), gives
X Ime
 i

Y Ime i

/ 2


1
 r
2
2


2 r
 
2
 r
 2
 cos
  i sin
 
X Ime
Y i
 
/ 2

X

2
 r

1
 r
2

2
 
2
 r
 2

1
 r
2

2
 
2
 r
 2
2
 r
, and
 cos
  i sin
  
(1 r ) (2 r )
 
2
 r

1
 r
2

2
 
2
 r
 2
IM e
 i

, thus i
 
/ 2
  i

or
  
/ 2
   tan

1
2

1
 r r
2
 
/ 2

Problem 4
Insert a virtual disk at the shaft discontinuity of moment of inertia corresponding twist angle is
J

0 , the n n

. The equations of motion of the three degrees of freedom are
J
 
K
1
(
 n
 
1
) , J
 
K
1
(
 n
 
1
)

K
2
(

2
  n
)

0 , J from the second equation we have:
( K
1

K
Thus

 n
2
2
)


  n
2

1
K
1
0.021


1
2

K
2
  6
 n

2
  
 
K
2
(

2
  n
)
  6

Problem 5
Solution to Problem 5:
Part I:
The equations of motion can be derived as follows:


F
M
C
 m

J
 x 
C
 

 m  x 
J
C
  


 k
 x
2 kl
 x


2 l

2 l

  x
 l
 
  kl
 x
 l
  which leads to the following matrix equations

 m
0
0
J
C


 x 
  




2 k kl

5 kl kl
2

 x


0
0
(1)
Assume
Eq. (1). x

X sin
 t and
   sin
 t , and substitute the assumed form solutions into


2 k

 m kl

2
5 kl
2

 kl
J
C

2


X

 
0
0
, (2)
The natural frequencies can be determined by setting the determinant of Eq. (2) to zero det



2 k

 m kl

2
5 kl
2

 kl
J
C

2

2 k





0 , which yields m

2

5 kl
2 
J
C

2

 k
2 l
2 
0 (3)
Equation (3) is the characteristic equation of the system. Substitute J
C
= 0.5
ml 2 into (3) m
2

4 
12 mk

2 
18 k
2 
0 , (4)
Therefore, the natural frequencies are

1


6

3 2
 k m

4 .
2 (rad/sec) and

2


6

3 2
 k m

10 .
1 (rad/sec). (5)
Substituting the natural frequencies into the homogeneous part of Eq. (2) gives the natural modes of the system:
X l


2 k
 k m

1
2

1

4

3 2

4 .
1 and
X l


2 k
 k m

2
2

1

4

3 2
 
0 .
1 . (6) l l

(1) = 1 l l

(1) = 1
1 st mode shape 2 nd mode shape

Part II:
The equations of motion can be derived as follows:


F
M
C
 m

J
 x 
C
 

 m


 x 
1
3
J
C
 x 
2
3 l

2  x 
2
3

 x 
3 l
1
  kx
1

 kx
2
2 klx
1
 klx
2 which leads to the following matrix equations, after substituting J
C
= 0.5
ml 2


 m m
2 m m


 x 
1
 x 
2




3 k
12 k
3 k
6 k

 x
1 x
2

0
0
(7)
Assume x
1

Eq. (1).
X
1 sin
 t and x
2

X
2 sin
 t , and substitute the assumed form solutions into




3 k
12 k
 m

 m
2

2
3 k
6 k


2 m

2 m

2





X
1
X
2



0
0
, (8)
The natural frequencies can be determined by setting the determinant of Eq. (8) to zero det




3 k
12 k
 m

 m
2

2

3 k

3
6 k k


2 m m

2 m



6 k
2
2





0 m

2
, which yields
 
3 k

2 m

2

12 k
 m

2


0 (9)
Equation (9) can be simplified to m
2  4 
12 mk
 2 
18 k
2 
0 , (10)
Therefore the natural frequencies remain the same as before. However, the natural modes of the system:
X
X
1
 
2

3 k
3 k


2 m

1
2 m

1
2

3
3


2

6

6


3
3
2
2

 
0 .
4 and
X
X
1
 
2

3 k
3 k


2 m m


2
2
2
2

3
3


2

6

6


3
3
2
2

  
2 .
4 .
(11)
X
1
(1)
X
1
(2)
1 st mode shape 2 nd mode shape

Part III: Natural frequencies remain the same, but equations of motion and mode shapes are different. The first case is static coupling and the second is both dynamic and static coupling.