homework DDS Exo Corrigé

7/17/2019 homework DDS Exo Corrig

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California Polytechnic State University Department of Architectural Engineering

Spring Quarter 2009 Instructor: Ansgar Neuenhofer

3/30/2009 12:43 PM C:\calpol y\arce412\homework\spring_2009\hw1.doc 1

March 30, 2009

ARCE 412: STRUCTURAL DYNAMICS

Homework 1 (Due 04-01-09)

Problem 1

For each of the frame structures shown, calculate the stiffnessk of the frame in the direction of the force. Assume uniform

flexural stiffnessEIand neglect axial deformation.

Solution:

Structure (a) (b) (c)

3

1[ ]m

k

EI 0.0417 0.121 0.538

Problem 2

For the two structures above, calculate the lateral stiffnessk .

Solution:

Structure (a) (b)

[k/ft]k 31.46 60.46

6 ft

15

ft

220,000k-ft

EI =

5000 kEA =

moment connection!

(b)

6 ft

8ft

7ft

5000 kEA =

220, 000k-ftEI =

(a)

5 m

3m

5 m

3m

5 m

3m

(a) (b) (c)


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4/1/2009 7:00 AM C:\calpol y\arce412\homework\spring_2009\hw1_sol.doc 1

April 1, 2009


Homework 1-Solution

Problem 1

For each of the frame structures shown, calculate the stiffness k of the frame in the direction of the force. Assume uniform

flexural stiffnessEIand neglect axial deformation.

Moment Diagrams for Unit Force

Displacement Calculation

( )

( )

( )

2

2

2 2 2

1 1(a) 3 3.00 5.00 24 0.0417

31 1

(b) 1.5 2 3.00 2.50 8.25 0.1212

31 1 1(c) 0.913 0.587 0.587 0.913 2 3.00 0.587 2 2.50 1.86 0.538

3 3

EI k EI

EI k EI

EI k EI

= + = = =

= + = = =

= + + = = =

Deflected Shape (for illustration only)

5 m

3m

5 m

3m

5 m

3m

(a) (b) (c)

0.9130 0.9130

0.5870

0.58701.500

1.5003.00

(a) (b) (c)

(a) (b) (c)


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4/1/2009 7:00 AM C:\calpol y\arce412\homework\spring_2009\hw1_sol.doc 2

Problem 2

For the two structures above, calculate the lateral stiffnessk .

2 2

-1 2

2

3 31 2

(a)

1 1 17 (7 8) 3.125 10 0.01225 0.01953 0.03178 ft/k

20000 3 5000

1= 31.46k/ft( )

0.03178

(b)

15tan 68.2 cos 0.3714 cos 0.1379

63 3 20000 5000

cos 0.1379 17.78 415 16.155

k ans

EI EAk

L L

= + + = + =

=

= = = =

= + = + = +

2.69 60.46k/ft( )ans=

6 ft

15

ft

220,000k-ft

EI =

5000 kEA=

moment connection!

(b)

6 ft

8

ft

7

ft

5000kEA=

220, 000k-ftEI =

(a)

7

3.125N =

M


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4/1/2009 7:01 AM C:\calpoly\arce412 \homework\sp ring_2009\hw2.doc 1

April 1, 2009



Reading: Chopra Sections 2.1 and 2.2, Review mass moment of inertia

Problem 1

Find the natural circular frequency /k m = of the above systems/structures.Comments:

(d) For torsional vibration of the disk of mass m(circular shaft massless). The shear modulus of the shaft isG .

(h) For vibration in the x -ory direction. The platform of weightW is braced laterally in each side by two steel cables. The

cables have axial stiffness EA . Due to high prestressing, the compression cables contribute to the structural stiffness.

(i) Consider axial deformation only in cableBD.

(j) Use both a flexibility (apply unit force in direction ofu) and a stiffness approach. In the stiffness approach, start with a

3x3 stiffness matrix corresponding to the three dofs or , ,B B Cu , then eliminate the two rotations to get a scalar stiffnessrelation involvinguonly (this is the static condensation technique learned in ARCE 306).

Solution:

Structure a b c d e f

[rad/sec] 1 2k k

m

+

( )1 2

1 2

k k

k k m+

( )

( )1 2 3

1 2 3

k k k

k k k m

+

+ +

4

216

G d

mLR

3

48EI

L m 3192EI

L m 0.100 EI

m

(h) (i) (j)

[rad/sec] 1.189 EA

hm12.57 0.1171

EI

m

Note: /m W g=

(a) (b) (c)

(f)(e)

8 ft

6

ft

6

ft

5000 kEA =

220,000 k-ftEI =

10.35kW =

A

B

C

D

(i)

20 ft 4 ft 6 ft 6 ft 4 ft 20 ft

W

u

EI

(d)

(g)

(h)

1 0 ft 10 ft

EI W

u

(j)A B C


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4/1/2009 7:01 AM C:\calpoly\arce412 \homework\sp ring_2009\hw2.doc 2

Problem 2

(a) Plot three cycles of a free vibration response of a SDF system with a mass of 1.0 k-s2 /inand a stiffness of 5 k/in

subjected to an initial displacement of 2 inches and zero initial velocity. Label0

, , (0), (0) and .T f u u u

(b) Double the stiffness of the SDF system in (a) and replot the response. Label0

, (0), (0)and .T u u u

(c) Plot the response of the system in (b) for an initial displacement of 2inches and an initial velocity of 10 in/sec. Label

0, (0), (0)and .T u u u


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4/1/2009 7:21 AM C:\calpoly\arce412 \homework\sp ring_2009\hw2_sol .doc 1

April 6, 2009

ARCE 412: Structural Dynamics

Homework 2-Solution

Find the natural circular frequency of the above systems/structures.Comments:

(d) For torsional vibration of the disk of mass m(circular shaft massless). The shear modulus of the shaft isG .(h) For vibration in the x -ory direction. The platform of weightW is braced laterally in each side by two steel cables. The

cables have axial stiffness EA . Due to high prestressing, the compression cables contribute to the structural stiffness.

(i) Consider axial deformation only in cableDE .(j) Use both a flexibility (apply unit force in direction ofu) and a stiffness approach. In the stiffness approach, start with a

3x3 stiffness matrix corresponding to the three dofs or , ,B B Cu , then eliminate the two rotations to get a scalar stiffnessrelation involvinguonly.

(a) (b) (c)

(f)(e)

8 ft

6

ft

6

ft

5000kEA=

220, 000k-ftEI =

10.35kW =

A

B

C

D

(i)

20 ft 4 ft 6 ft 6 ft 4 ft 20 ft

W

u

EI

(d)

(g)

(h)

10 ft 10 ft

EI W

u

(j)A B C


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Problem 1

( )

( ) ( )

1 21 2

1 2 1 2

1 2 1 2

1 2

1 2 3 1 2 3

1 2 3 1 2 3 1 2 3

12 3 1 2 3 1 2 3

4 42

2

(a) ( )

1b ( )1 1 ( )

1 1 1c ( )

1 1 1 1 ( ) ( )

1(d) , (

32 2 16

k kk k k ans

m

k k k k k ans k k k k m

k k

k k k k k k k ans

k k k k k k k k k m

k k k k k k k k

GJ G d k G d k MMI mR

L L MMI LmR

+= + =

= = =+ ++

+ += = = = =

+ + + + + ++ ++ +

= = = = =

( )

( )

3 3

3 3 3

2 2 2

2

)

48 48(e) ( )

2 12 192 192(f) ( )

2

1 1 1(g) = 2 20 2 4 3 6 2 100 0.01 0.1 ( )

3 3 3 100

(h) cos (stiffness of each of the 4 cable

ans

EI EI k ans

L L m

EI EI EI k ans

L L mL

EI EI EI k EI ans

m

EAk

L

= =

= = =

+ + = = = =

=

( )

2

2 2

2 2

s)

4 cos 45 4 0.5 2 2 1.189 ( )2 2

1 1 1 50.76 32.2(i) 6 12 2.5 10 0.01970 ft/k 50.76 k/ft 12.57 rad/sec( )20,000 5,000 0.01970 10.35

1 1(j) 3.75 3.125 3.75 3.125 10 3.12

3 3

EA EA EA EA EAk ans

h h h hm hm

k ans

EI

= = = = =

= + = = = = =

= + +

25 10 72.917

0.013710.01371k/ft 0.1171 ( )

72.917

EI EI EI k ans

m m

=

= = = =

work for g,i,j shown on next page


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work for g i

work for j

stiffness approach for j

stiffness matrix

use 1 dofs are 1,2,3 , ,

[2 3](eliminate) 1(retain)

stiffness submatrices correspoding to 2 dofs to

A A BEI

e r

= = =

= =

K

0.024 0 0.060

0 0.800 0.200

0.060 0.200 0.400

[ ]

1

1con

be eliminated and single dof to be retained

= =

= =

=

ee rr

er re

ee

rr re ee er

=

K K

K K

K

K K K K K

0.80 0.20

0.02400.20 0.40

00 0.06

0.06

1.4286 -0.7143

-0.7143 2.8571

=0.024 0[ ]

0.0137k EI

=

=

1.4286 -0.7143 0

0.06 0.0137-0.7143 2.8571 0.06

Flexibility and stiffness approaches lead to the same scalar stiffnessk and hence to the same natural frequency.

8 ft

6

ft

6

ft

5000 kEA=

220, 000k-ftEI =

A

B

C

D

1P=

6

2.5BD

N =

M

2 2

3

+

1P=

M

[k-ft]

2 2

3

+

1P= 1P=

M

[k-ft]

3.125

3.75

+

M

[k-ft]

10 ft 10 ft

1P=A

B


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Problem 2

(a) Plot three cycles of a free vibration response of a SDF system with a mass of 1.0 k-s2 /inand a stiffness of 5 k/in

subjected to an initial displacement of 2 inches and zero initial velocity. Label0

, , (0), (0)and .T f u u u

(b) Double the stiffness of the SDF system in (a) and replot the response. Label0

, (0), (0)and .T u u u

(c) Plot the response of the system in (b) for an initial displacement of 2inches and an initial velocity of 10 in/sec. Label

0, (0), (0)and .T u u u

0 2 4 6 8 104

3

2

1

0

1

2

3

4

Time [sec]

Displacement[in]

(a)

(b)

(c)

(0)u

(0) slope of

( ) at 0

u

u t t

=

=

0u

0u 0u

T

TT


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4/3/2009 12:54 PM C:\calpoly\arce412\homework\ spring_2009\hw3.doc

April 6, 2009

ARCE 412: Structural Dynamics


Reading: Chopra Section 2.2

Problem 1:

The structure above with given lateral stiffnessk is set into free vibration with an initial displacement of 0.5 in. and an

initial velocity of 10 in/sec. Make the following assumptions regarding damping:

(a) undamped

(b) 1% damped ( 0.01=

)(c) 20% damped ( 0.20= )Find the solution for the displacement ( )u t and use MATLAB to plot three cycles of vibration.

Problem 2:

(a) What is the amplitude of motion of the system in Problem 1 for the undamped case?

(b) What is the maximum displacement of the system in Problem 1 for the damped cases?

(c) What is the required damping ratio to reduce the displacement at 1.5 sec to 1?

For (b) and (c) use MATLAB to calculate the response ( )u t for closely spaced time points. Then use the maxcommand for(b) and trial and error for (c).

Problem 3(Chopra 2.11):

For a system with damping ratio , determine the number of free vibration cycles required to reduce the displacement am-

plitude to 10% of the initial amplitude; the initial velocity is zero.

Solution: 10%ln(10) 0.366

2j

= =


The vertical suspension system of an automobile is idealized as a viscously damped SDF system. Under the 3000-lb weightof the car the suspension system deflects 2 in. The suspension is designed to be critically damped.

(a) Calculate the damping and stiffness coefficients of the suspension.

(b) With four l60lb passengers in the car, what is the effective damping ratio?(c) Calculate the natural vibration frequency for case (b).


The stiffness and damping properties of a mass-spring-damper system are to be determined by a free vibration test; the massis given as

20.1lb-sec /inm = . In this test the mass is displaced 1 in. by a hydraulic jack and then suddenly released. Atthe end of 20 complete cycles, the time is 3 sec and the amplitude is 0.2 in. Determine the stiffness and damping coeffi-

cients.

22 k sec /inm=

50 k/ink=

( )u t

Solution:

( ) 1500 lb/in

215.9lb-sec/in

(b) 0.908

(c) 5.28 rad/sec

cr

D

a k

c c

=

= =

=

=

Solution:

175.5 lb/in, 0.107 lb-sec/ink c= =


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4/10/2009 1:04 PM C:\calpol y\arce412\homework\spring_2009\hw3_sol.doc 1

April 6, 2009


Homework 3-Solution

Problem 1

Use MATLAB to plot three cycles of displacement response ( )u t of the structure above (rigid girder, column fixed at thebase). The structure is set into free vibration with an initial displacement of 0.5 in. and an initial velocity of 10 in/sec. Make

the following assumptions regarding damping:

(a) undamped

(b) 1% damped ( 0.01= )(c) 20% damped ( 0.20= )

Problem 2:

(a) What is the amplitude of motion of the system in Problem 1 for the undamped case?

(b) What is the maximum displacement of the system in Problem 1 for the damped cases?

(c) What is the required damping ratio to reduce the displacement at 1.5 sec to 1?

[ ]2 2

2 2

0

max

max

(0) 10(a) (0) 0.5 2.06 "

5

(b) 2.04 " at 0.27 sec ( 0.01)

1.71 " at 0.24 s ec ( 0.20)

(c) try 0.1 ( 1.5) 0.989 (good enough)

n

uu u

u t

u t

u t

= + = + =

= = =

= = =

= = =

Problem 3 (Chopra 2.11) MOVED TO HW #4

For a system with damping ratio , determine the number of free vibration cycles required to reduce the displacement am-

plitude to 10% of the initial amplitude; the initial velocity is zero.

1

10%

1 10%

1 1 1 ln(10) 0.366ln 2 ln 2

0.1 2j

uj

j u j

+

= = = =

0 0.5 1 1.5 2 2.5 3 3.5 42.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

Time [sec]

Displacement[in]

0 =0.01 =

0.20 =

22ksec /inm=

50k/ink=

( )u t


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4/10/2009 1:04 PM C:\calpol y\arce412\homework\spring_2009\hw3_sol.doc 2

Problem 4(Chopra 2.14)

The vertical suspension system of an automobile is idealized as a viscously damped SDF system. Under the 3000-lb weight

of the car the suspension system deflects 2 in. The suspension is designed to be critically damped.

(a) Calculate the damping and stiffness coefficients of the suspension.(b) With four l60lb passengers in the car, what is the effective damping ratio?

(c) Calculate the natural vibration frequency for case (b).

(a) The stiffness coefficient is

30001500 lb/in

2k = =

The damping coeffcient is

30002 2 1500 215.9 lb-sec/in

386crc c km = = = =

(b) With passengers, weight is 3640 lbW = . The damping ratio is

215.90.908

36402 2 1500386

cr

c c

c km

= = = =

(c) The natural vibration frequency for case (b) is

2 21500 3861 1 0.908 5.28 rad/sec3640

nD

= = =

Problem 5 (Chopra 2.15)MOVED TO HW #4

The stiffness and damping properties of a mass-spring-damper system are to be determined by a free vibration test; the mass

is given as m 20.1lb-sec /inm= . In this test the mass is displaced 1 in. by a hydraulic jack and then suddenly released. At

the end of 20 complete cycles, the time is 3 sec and the amplitude is 0.2 in. Determine the stiffness and damping coeffi-

cients.

(1) Determine and n

1

1

1 1 1ln ln 0.0128 1.28%

2 2 20 0.2j

u

j u

+

= = = =

Therefore, assumption of small damping in the above equation is valid.

3 20.15 sec 0.15 sec 41.89 rad/sec

20 0.15n nD D

T T T

= = = = =

(2) The stiffness coefficient is

2 241.89 0.1 175.5lb/innk m= = =

(3) The damping coefficient is

2 2 0.1 41.89 8.377 lb-sec/in 0.0128 8.377 0.107 lb-sec/incr n cr c m c c = = = = = =


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California Polytechnic State University Department of Architectural EngineeringSpring Quarter 2009 Instructor: Ansgar Neuenhofer

4/10/2009 7:29 AM C:\calpoly\arce412\homework\spring_2006\hw4.doc

April 10, 2009



Reading: Chopra Section 3.1, 3.2


For a system with damping ratio , determine the number of free vibration cycles required to reduce the displacement

amplitude to 10% of the initial amplitude; the initial velocity is zero.

Solution:

10%

ln(10) 0.366

2j

= =


The stiffness and damping properties of a mass-spring-damper system are to be determined by a free vibration test; the

mass is given as20.1lb-sec /inm = . In this test the mass is displaced 1 in. by a hydraulic jack and then suddenly re-

leased. At the end of 20 complete cycles, the time is 3 sec and the amplitude is 0.2 in. Determine the stiffness and dam-

ping coefficients.Solution:

175.5 lb/in, 0.107 lb-sec/ink c= =

Problem 3: An SDF structure is excited by a sinusoidal force. At resonance the amplitude of displacement was mea-

sured to be 2 in. At an exciting frequency of one-tenth the natural frequency of the system, the displacement amplitude

was measured to be 0.2 in. Estimate the damping ratio of the system.

Problem 4: In a forced vibration test under harmonic excitation it was noted that the amplitude of motion at resonance

was exactly four times the amplitude at an excitation frequency 20% higher than the resonant frequency. Determine the

damping ratio of the system.

Problem 5: The displacement response of a SDF structure to harmonic excitation and initial conditions (0)u and (0)u is

given by

steady-stae vibrationtransient vibration

( ) [ cos sin ] sin cosnt D Du t e A t B t C t D t = + + +

Find expressions for the constants of integration A and Bin terms of , , (0), (0), , , , andn DC D u u .

Solution:

[ ]

(0)

(0) (0)n

D

A u D

u u D C B

=

+ =


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California Polytechnic State University Department of Architectural EngineeringSpring Quarter 2009 Instructor: Ansgar Neuenhofer

4/10/2009 7:29 AM C:\calpoly\arce412\homework\spring_2006\hw4.doc

Problem 6:

Determine and plot the response of the frame above ( 0 5 sect ). Assume the girder is rigid and neglect the mass ofthe columns.

(a) Assume at rest initial conditions and zero damping.

(b) Assume 5% damping and initial conditions (0) 1inu = and (0) 50 in/secu = .Submit two figures, one for (a) and one for (b), containing three plots each (the transient, stead-state and total

responses). Write the numerical values for constants , , ,A B C D on the figures.

Problem 7:

(a) Calculate the vertical displacement of the cantilever tip due to gravity.

Assume 5% damping and consider steady state motion only:

(b) For0 10 kp = and 5,10,15 rad/sec = calculate the amplitude of motion. Which of the three forcing frequencies

causes the largest displacement? Explain.(c) For 15rad/sec = calculate the maximum allowable amplitude

0p of the forcing function such that the deflection of

the cantilever due to gravity plusdynamic action is downward at all times.

20k/ink =50 kW =

0( ) sin( )p t p t =

( ) 15sin(10 ) [k]p t t=

20ft

15ft

40 kW=

4

29000ksi

175in

E

I

=

=


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4/17/2009 10:37 AM C:\calpoly\arce412\homework\spring_2009\hw4_sol .doc

April 17, 2009


Homework 4-Solution

Problem 1For a system with damping ratio , determine the number of free vibration cycles required to reduce the displacement

amplitude to 10% of the initial amplitude; the initial velocity is zero.

110%

1 10%

ln(10)1 1 1 0.366ln 2 ln 2

0.1 2j

uj

j u j

+

= = = =

Problem 2 (Chopra 2.15)

The stiffness and damping properties of a mass-spring-damper system are to be determined by a free vibration test; the

mass is given as m 20.1lb-sec /inm= . In this test the mass is displaced 1 in. by a hydraulic jack and then suddenly

released. At the end of 20 complete cycles, the time is 3 sec and the amplitude is 0.2 in. Determine the stiffness and

damping coefficients.

(1) Determine and n

1

1

1 1 1ln ln 0.0128 1.28%

2 2 20 0.2j

u

j u

+

= = = =

Therefore, assumption of small damping in the above equation is valid.

3 20.15 sec 0.15 sec 41.89 rad/sec

20 0.15n nD D

T T T

= = = = =

(2) The stiffness coefficient is2 241.89 0.1 175.5 lb/innk m= = =

(3) The damping coefficient is

2 2 0.1 41.89 8.377 lb-sec/in 0.0128 8.377 0.107 lb-sec/incr n cr c m c c = = = = = =

Problem 3

A SDF system is excited by a sinusoidal force. At resonance, the amplitude of displacement was measured to be 2in. At

an exciting frequency of one-tenth the natural frequency of the system, the displacement amplitude was measured to be

0.2 in. estimate the damping ratio of the system.

At n =

(a) 0 01

( ) 22

stu u

= =

At 0.1 n =

0 0( ) 0.2 ( 0.1 excitation is so slow, that it can be considered static)stu u = =

Substituting0

( ) 0.2st

u = in (a) gives0.05=

If we dont want to make the assumption of static response for 0.1= , we can use the approach followed in Problem 4(see below).


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Problem 4

In a forced vibration test under harmonic excitation it was noted that the amplitude of motion at resonance was exactly

four times the amplitude at an excitation frequency 20% higher than the resonant frequency. Determine the damping

ratio of the system.

We assume that damping is small enough to justify the approximation that the resonant frequency is n and theresonance amplitude is1/2. The given data then implies:

(a) 0 01

( ) ( )2n

stu u

= =

(b)

( ) ( ) ( ) ( )0 1.2 0 0

2 22 22 2

1 1( ) ( ) ( )

1 2 1 1.2 2 1.2n st st

u u u

= = =

+ +

Combining Eqs. (a) and (b)

( ) ( )

2

20

2 2 2 20 1.2

( )1 1 1 1 0.0576( )( ) 4 ( 0.44) (2.4 )2 2

n

n

u ansu

=

=

= = = +

Assumption of small damping is reasonable.


17/60




Problem 5

The displacement response of a SDF structure to harmonic excitation and initial conditions (0)u and (0)u is given by

steady-stae vibrationtransient vibration

( ) [ cos sin ] sin cosnt D Du t e A t B t C t D t

= + + +

Find expressions for the constants of integration A and Bin terms of , , (0), (0), , , , andn DC D u u .

[ ]

(0)

(0) ( )

( sin cos ) ( cos sin )( )

cos sin

(0)

(0) (0)(0)( )

n nt tD D D D n D D

D n

nn

D D

u A D

A u D ans

e A t B t e A t B t u t

C t D t

u B A C

u u D C u A CB ans

= +

=

= + +

+

= +

+ + = =

Problem 6

Determine and plot the response of the frame above ( 0 5 sect ). Assume the girder is rigid and neglect the mass ofthe columns. Identify the steady state and the transient portion of the response (both in the equations and on the plot).

(a) Assume at rest initial conditions and zero damping.

(b) Assume 5% damping and initial conditions (0) 1inu = and (0) 50 in/secu = .Submit two figures, one for (a) and one for (b), containing three plots each (the transient, stead-state and total

responses). Indicate the numerical values for constants , , ,A B C Don the figures.(a)

( ) ( ) ( ) ( )

( ) ( )

3 3

0

20 0

2 22 22 2

02 2

24 24 29000 17520.88 k/in

(15 12)

20.88 386.414.20 rad/sec

40

15 k

10 rad/sec

10= 0.7042

14.20

1 2

1 2 1 2

0

151.425

1 20.88 1 0.7042

0

0

1.4

n

n

n

EIk

L

p

p pC D

k k

pC

k

D

A

B C C

= = =

= =

=

=

= =

= =

+ +

=

= = =

=

=

= = =

0 02 2

25 0.7042 1.0036

1( ) sin sin

1 1

transientresponse steady state response

n

p pu t t wt

k k

=

= +

( ) 15sin(10 ) [k]p t t=

20ft

15ft

40 kW=

4

29000 ksi

175in

E

I

=

=


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(b)

( ) ( ) ( ) ( )

20 0

2 22 22 2

( ) ( ) ( ) ( cos sin ) sin cossteady state responsetransientresponse

(0) (0)1 2(0)

1 2 1 2

n tc p D D

n

u t u t u t e A t B t C t D t

u up pC D A u D B

k k

= + = + + +

+ = = = =

+ +

[ ]

1.195 in 2.600 in 1.397 in 0.195 in

D

D C

A B C D

= = = =

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 52.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

Time [sec]

Displacement[in]

totaltransientsteady-state

(0) 0, (0) 0, 0

0, 1.0036 1.425 0

u u

A B C D

= = =

= = = =

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 54

3

2

1

0

1

2

3

4

Time [sec]

D

isplacement[in]

totaltransientsteady-state

(0) 1" (0) 50 "/ sec 5%

1.195 2.600 1.397 0.195

u u

A B C D

= = =

= = = =


19/60




Problem 7

(a) Calculate the vertical displacement of the cantilever tip due to gravity.

Assume 5% damping and consider steady state motion only:

(b) For0 10 kp = and 5,10,15 rad/sec = calculate the amplitude of motion. Which of the three forcing frequencies

causes the largest displacement? Explain.

(c) For 15rad/sec = calculate the maximum allowable amplitude0

p of the forcing function such that the deflection of

the cantilever due to gravity plusdynamic action is downward at all times.

1 1 0

2 2 0

3 3 0

50(a) 2.5 "( )

20

20 386.4(b) 12.43 rad/sec

5010k5

5 rad/sec 0.402 1.192 0.596 in( )12.43 20 k/in

10k1010 rad/sec 0.804 2.763 1.38 in( )

12.43 20 k/in

11515 rad/sec 1.206

12.43

st

n

Wu ans

k

u ans

u ans

u

= = =

= =

= = = = =

= = = = =

= = = =

2

0 00

0k2.122 1.06in( )

20k/in

Excitation frequency is closest to resonance largest amplitude

(c) 2.5 2.122 0 23.6k( )20

st d

ans

p pu u R p ans

k

=

= = >

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