COLLEGE OF ENGINEERING

MECHANICAL ENGINEERING

Ph.D. Preliminary Qualifying Examination

Signature Page Vibration Examination (Modify)

January 26, 2009 (Monday)--Modify

9:00 am – 12:00 noon Room 2145 Engineering Building For identification purposes, please fill out the following information in ink. Be sure to print and sign your name. This cover page is for attendance purposes only, and will be separated from the rest of the exam before the exam is graded. Write your student number on all exam pages. Do NOT write your name on any of the other exam pages besides the cover page. Name (print in INK) Signature (in INK)

Student Number (in INK)

Do all your work on provided sheets of paper. If you need extra sheets, please request them from proctor. When you are finished with the test, return the exam plus any additional sheets to the proctor.

COLLEGE OF ENGINEERING

MECHANICAL ENGINEERING

Ph.D. Preliminary Qualifying Examination

Cover Page

9:00 am – 12:00 noon Room 2145 Engineering Building 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 Problem number 5 is mandatory.

Do all your work on the provided sheets of paper. If you need extra sheets, please request them from the proctor. When you are finished with the test, return the exam plus any additional sheets to the proctor.

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.

O

OL

a

h

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.

O

ck

a

L

SEPO

ck

a

L

SEP

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.

( ) siny t Y t ( ) siny t Y t ( ) siny t Y t

( ) siny t Y t ( ) siny t Y t ( ) siny t Y 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 6 211.5 10 / /G lb in 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/n or 2/n

1

2

n

1

2

n

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 Jc with respect to center of gravity Cg. Suppose that the body is supported by two springs of stiffness k that are attached at distances 2l and l with respect to the center of gravity Cg as shown in Figure 5a. Let m = 10 kg, Jc = 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. (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 x1 and x2.

(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?

gC

x

SEP

m, JC

k k2l l

gC

x

SEP

m, JC

k k2l l2l l

gC

x1

SEP

x2

m, JC

k k2l l

gC

x1

SEP

x2

m, JC

k k2l l2l l

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.

Figure 5. Solution:

FBD From the static equilibrium position we write

2T mg

(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)a from the its suspended string. The

string also be tilted by an angle from the vertical such that, ( / 2) ha . This

geometric relation gives ( / 2h)a .

Now writing the equation of motion by taking moments about axis OO, gives

0I 2(Tsin )( / 2) T T2h

aa a a

(2) Using equation (1), equation (2) takes the form

20 n

0

mg mgI 0 0 0

4h 4I h

2 2a a

(3)

O

OL

a

h

T T

mg

SEP

0I

TT

a

ha/2

h

Tsin

Tsin

T T

mg

SEP

0I

TT

a

ha/2

h

Tsin

Tsin

where n 220

mg mg 3g

4I h hLL4 m h

12

2 2 2a a a

(4) Problem 2

FBD (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

2

2 2

0 03

mLI cL L ka a cL ka (1)

(c) The undamped natural frequency is obtained by dividing both sides of the equation of

motion (1) by the coefficient of , i.e.,

2 2

2 23 3 0 3 3n

c ka ka a k

m mL mL L m (2)

(d) The damping ratio is obtained by writing the equation of motion in the form 22 0n n (3)

Thus we can write km3 3 3

2 32 2 3

2 3

n

n

c c c cL

m m a k kmam

L m

(4)

The critical damping coefficient is obtained from (4) as

3 2 3

32 3cr

cr

c cL kmac

c Lkma

The damped natural frequency, nd , is written in terms of the undamped natural

frequency, n , as

2 2 22

2

3 91 3 1 3 1

122 3nd n

a k cL a k c L

L m L m kmakma

L

a k

c

m

OSEP

0I

R

O

kacL

L

a k

c

m

O

L

a k

c

m

OSEP

0I

R

O

kacL

Problem # 3 Using Newton’s second law and with the

help of the FBD, the equation of motion is

( )mx kx c y x (1)

Rearranging

mx cx kx cy (1a)

But ( ) sin Im i ty t Y t Y e (2)

Also ( ) cos Im i ty t Y t i Y e (3)

Note that one can write /2 /2

because cos /2 sin /2i i

i e e i i

Thus one can write equation (1a) in the form

2Im Imi t

i tmx cx kx icY e cY e

(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

( ) where i t i i t ix t X Ime X Ime X X Ime (5)

We need the first and second time derivatives of ( )x t , i.e.

2( ) , ( )i t i tx t i X Ime x t X Ime (6)

Substituting expressions (5) and (6) into equation (4), gives

2 22 2i t

i t i t i tn n nX Ime i X Ime X Ime Y Ime

(7)

Canceling out i tIme from both sides of equation (7) and rearranging, gives

2 2 2 22 2 2 , where i i

n n n nX i X X Y Ime Y Y Y Ime

(8)

Rearranging

2 2

2

2

n

n n

X

Y i

(9)

Multiplying and dividing by the conjugate of the denominator, gives

2 2

2 2 2 2

22

2 2

n nn

n n n n

iX

Y i i

2 2

2 22 2

22

2

nn n

n n

i

(10)

Dividing the numerator and denominator by 2n , and setting / nr , equation (10)

takes the form

2

2 22

21 2

1 2

X rr i r

Y r r

(11)

Multiplying and dividing each expression by 2 221 2r r gives

2 22 2

2 2 2 22 2 22 2

2 1 2 1 2

1 2 1 2 1 2

r r r rX ri

Y r r r r r r

(12)

With the help of the shown triangle equation (12)may be written in the form

2 22

2cos sin

1 2

X ri

Yr r

(13)

Expressing X and Y in terms of their original

Definitions (5) and (8), gives

/ 22 22

2cos sin

1 2

i

i

X Ime ri

Y Imer r

/ 2

2 22 22 2

2 2cos sin

1 2 1 2

i iiX Ime r r

i IM eY

r r r r

, thus

1

22 22

2 2, and / 2 or / 2 tan / 2

11 2

r rX i i i

rr r

2 r

2(1 )r

2 2 2(1 ) (2 )r r

2

2tan

1

r

r

2 r

2(1 )r

2 2 2(1 ) (2 )r r

2

2tan

1

r

r

Problem 4

Insert a virtual disk at the shaft discontinuity of moment of inertia 0n

J , the

corresponding twist angle is n . The equations of motion of the three degrees of freedom

are

1 1 1 1( )nJ K , 1 1 2 2( ) ( ) 0n n n nJ K K , 2 2 2 2( )nJ K

from the second equation we have:

6 6 611 2 1 2

2 2 2

( ) 0.1536 10 0.0941 10 ( 0.5979) 0.0596 10n nK K K K

Thus 2

0.021

1

n

Problem 5 Solution to Problem 5: Part I: The equations of motion can be derived as follows:

xmF lxklxkxm 2

CC JM lxkllxklJC 22

which leads to the following matrix equations

0

0

5

2

0

02

x

klkl

klkx

J

m

C

(1)

Assume tXx sin and t sin , and substitute the assumed form solutions into Eq. (1).

0

0

5

222

2 X

Jklkl

klmk

C

, (2)

The natural frequencies can be determined by setting the determinant of Eq. (2) to zero

05

2det

22

2

CJklkl

klmk, which yields

052 22222 lkJklmk C (3)

Equation (3) is the characteristic equation of the system. Substitute JC = 0.5ml2 into (3)

01812 2242 kmkm , (4)

Therefore, the natural frequencies are

2.42361 mk (rad/sec) and 1.102362 mk (rad/sec). (5)

Substituting the natural frequencies into the homogeneous part of Eq. (2) gives the natural modes of the system:

1.4234

1

2 2

1

1

1

mk

k

l

X and

1.0234

1

2 2

2

2

2

mk

k

l

X. (6)

1st mode shape 2

nd mode shape

X(1) = 4.1

l(1) = 1

X(1) = 4.1

l(1) = 1

X(1) = -0.1

l(1) = 1

X(1) = -0.1

l(1) = 1

Part II:

The equations of motion can be derived as follows:

xmF 2121

3

2

3kxkx

xxm

CC JM 2112 2

33klxklx

l

x

l

xJC

which leads to the following matrix equations, after substituting JC = 0.5ml2

0

0

612

332

2

1

2

1

x

x

kk

kk

x

x

mm

mm

(7)

Assume tXx sin11 and tXx sin22 , and substitute the assumed form solutions into

Eq. (1).

0

0

612

233

2

1

22

22

X

X

mkmk

mkmk

, (8)

The natural frequencies can be determined by setting the determinant of Eq. (8) to zero

0612

233det

22

22

mkmk

mkmk, which yields

0122363 2222 mkmkmkmk (9)

Equation (9) can be simplified to

01812 2242 kmkm , (10)

Therefore the natural frequencies remain the same as before. However, the natural modes of the system:

4.02363

23623

3

232

1

2

1

1

2

1

1

mk

mk

X

X and

4.22363

23623

3

232

2

2

2

2

2

2

1

mk

mk

X

X.

(11)

1st mode shape 2

nd mode shape

X1(1) = 0.4

X1(1) = 1

X1(1) = 0.4

X1(1) = 1

X1(2) = 0.4

X1(2) = 1

X1(2) = 0.4

X1(2) = 1

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.