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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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California Polytechnic State University Department of Architectural Engineering
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April 1, 2009
ARCE 412: STRUCTURAL DYNAMICS
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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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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California Polytechnic State University Department of Architectural Engineering
Spring Quarter 2009 Instructor: Ansgar Neuenhofer
4/1/2009 7:01 AM C:\calpoly\arce412 \homework\sp ring_2009\hw2.doc 1
April 1, 2009
ARCE 412: STRUCTURAL DYNAMICS
Homework 2 (Due 04-06-09)
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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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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Spring Quarter 2009 Instructor: Ansgar Neuenhofer
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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Spring Quarter 2009 Instructor: Ansgar Neuenhofer
4/1/2009 7:21 AM C:\calpoly\arce412 \homework\sp ring_2009\hw2_sol .doc 2
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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Spring Quarter 2009 Instructor: Ansgar Neuenhofer
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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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California Polytechnic State University Department of Architectural Engineering
Spring Quarter 2009 Instructor: Ansgar Neuenhofer
4/3/2009 12:54 PM C:\calpoly\arce412\homework\ spring_2009\hw3.doc
April 6, 2009
ARCE 412: Structural Dynamics
Homework 3 (Due 04-10-09)
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
= =
Problem 4(Chopra 2.14):
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).
Problem 5(Chopra 2.15):
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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April 6, 2009
ARCE 412: STRUCTURAL DYNAMICS
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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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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April 10, 2009
ARCE 412: STRUCTURAL DYNAMICS
Homework 4 (Due 04-17-09)
Reading: Chopra Section 3.1, 3.2
Problem 1(Chopra 2.11):
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
= =
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 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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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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April 17, 2009
ARCE 412: STRUCTURAL DYNAMICS
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.
7/17/2019 homework DDS Exo Corrig
17/60
California Polytechnic State University Department of Architectural Engineering
Spring Quarter 2009 Instructor: Ansgar Neuenhofer
4/17/2009 10:37 AM C:\calpoly\arce412\homework\spring_2009\hw4_sol .doc
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
=
=
7/17/2019 homework DDS Exo Corrig
18/60
California Polytechnic State University Department of Architectural Engineering
Spring Quarter 2009 Instructor: Ansgar Neuenhofer
4/17/2009 10:37 AM C:\calpoly\arce412\homework\spring_2009\hw4_sol .doc
(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
= = =
= = = =
7/17/2019 homework DDS Exo Corrig
19/60
California Polytechnic State University Department of Architectural Engineering
Spring Quarter 2009 Instructor: Ansgar Neuenhofer
4/17/2009 10:37 AM C:\calpoly\arce412\homework\spring_2009\hw4_sol .doc
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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