Spring Quarter 2009 Instructor: Ansgar Neuenhofer …aneuenho/teaching/arce412homework.pdf · ARCE...

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

3/30/2009 12:43 PM C:\calpoly\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 stiffness EI and neglect axial deformation. Solution: Structure (a) (b) (c)

3

1[ ]m

kEI

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

8 f

t 7

ft

5000 kEA =

220, 000 k-ftEI =

(a)

5 m 3

m 5 m

3 m

5 m

3 m

(a) (b) (c)


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

April 1, 2009 ARCE 412: STRUCTURAL DYNAMICS

Homework 1-Solution

Problem 1 For each of the frame structures shown, calculate the stiffnessk of the frame in the direction of the force. Assume uniform flexural stiffness EI and 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 0.121231 1 1

(c) 0.913 0.587 0.587 0.913 2 3.00 0.587 2 2.50 1.86 0.5383 3

EI k EI

EI k EI

EI k EI

Δ = ⋅ ⋅ + = → = = ⋅Δ

Δ = ⋅ ⋅ ⋅ + = → = = ⋅Δ

Δ = ⋅ + − ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ = → = = ⋅Δ

Deflected Shape (for illustration only)

5 m 3

m 5 m

3 m

5 m

3 m

(a) (b) (c)

0.9130 0.9130

0.58700.58701.500

1.5003.00

(a) (b) (c)

(a) (b) (c)



Problem 2 For the two structures above, calculate the lateral stiffnessk .

2 2

-1 2

23 3

1 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

α α α

α

Δ = ⋅ ⋅ ⋅ + + ⋅ ⋅ = + =

=

⎛ ⎞⎟⎜= = = =⎟⎜ ⎟⎜⎝ ⎠⋅ ⋅

= + = + ⋅ = + 2.69 60.46k/ft( )ans=

6 ft

15

ft

220, 000k-ft

EI =

5000 kEA =

moment connection!

(b)

6 ft

8 f

t 7

ft

5000 kEA =

220, 000 k-ftEI =

(a)

7

3.125N =

M


4/1/2009 7:01 AM C:\calpoly\arce412\homework\spring_2009\hw2.doc 1



Reading: Chopra Sections 2.1 and 2.2, Review mass moment of inertia Problem 1 Find the natural circular frequency /k mω = of the above systems/structures. Comments: (d) For torsional vibration of the disk of massm (circular shaft massless). The shear modulus of the shaft isG . (h) For vibration in thex -ory direction. The platform of weightW is braced laterally in each side by two steel cables. The cables have axial stiffnessEA . 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 ϕ ϕΔ , then eliminate the two rotations to get a scalar stiffness relation involvingu only (this is the static condensation technique learned in ARCE 306). Solution:

Structure (a) (b) (c) (d) (e) (f) (g)

[rad/sec]ω 1 2k km+

( )

1 2

1 2

k kk k m+

( )

( )1 2 3

1 2 3

k k k

k k k m

+

+ +

4

216G dmLRπ 3

48EIL m

3192EIL m

0.100EIm

(h) (i) (j)

[rad/sec]ω 1.189EAhm

12.57 0.1171EIm

Note: /m W g=

(a) (b) (c)

(f)(e)

8 ft

6 f

t 6

ft

5000 kEA =

220, 000 k-ftEI =

10.35 kW =

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


4/1/2009 7:01 AM C:\calpoly\arce412\homework\spring_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 /in and a stiffness of 5 k/in subjected to an initial displacement of 2 inches and zero initial velocity. Label

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

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

0, (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



April 6, 2009 ARCE 412: Structural Dynamics

Homework 2-Solution

Find the natural circular frequencyω of the above systems/structures. Comments: (d) For torsional vibration of the disk of massm (circular shaft massless). The shear modulus of the shaft isG . (h) For vibration in thex -ory direction. The platform of weightW is braced laterally in each side by two steel cables. The cables have axial stiffnessEA . 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 ϕ ϕΔ , then eliminate the two rotations to get a scalar stiffness relation involvingu only.

(a) (b) (c)

(f)(e)

8 ft

6 f

t 6

ft

5000 kEA =

220, 000 k-ftEI =

10.35 kW =

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



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 kk ans

k k k k mk k

k k k k k kk ansk k k k k k k k k m

k k k k k k k k

GJ G d k G dk MMI mR

L L MMI LmR

ω

ω

ω

π πω

+= + → =

⋅= = → =

+ ++

+ += = = = → =+ + + + + ++ +

+ +

= = = → = =

( )

( )

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 EIk ans

L L m

EI EI EIk ans

L L mL

EI EIEI k EI ans

m

EAk

L

ω

ω

ω

α

= → =

⋅= = → =

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜⎝ ⎠

⎛ ⎞⎟⎜Δ ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ ⋅ = → = = → =⎟⎜ ⎟⎜⎝ ⎠

=

( )

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

ω

ω

= = ⋅ = → = =

⋅Δ = ⋅ ⋅ + ⋅ ⋅ = → = = → = =

Δ = ⋅ + − ⋅ ⋅ + ⋅ 25 10 72.917

0.013710.01371k/ft 0.1171 ( )

72.917EI EI EI

k ansm m

ω

⋅ =

= = → = =

work for g,i,j shown on next page



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

ϕ ϕ

⎡ ⎤⎢ ⎥⎢ ⎥= = = Δ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

= =

K0.024 0 0.0600 0.800 0.2000.060 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 0.200.02400.20 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 f

t 6

ft

5000 kEA =

220, 000 k-ftEI =

A

B

C

D

1P =

6

2.5BDN =

M2 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



Problem 2 (a) Plot three cycles of a free vibration response of a SDF system with a mass of 1.0 k-s2 /in and a stiffness of 5 k/in subjected to an initial displacement of 2 inches and zero initial velocity. Label

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

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

0, (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 10−4

−3

−2

−1

0

1

2

3

4

Time [sec]

Dis

plac

emen

t [in

]

(a)

(b)

(c)

(0)u

(0) slope of ( ) at 0u

u t t=

=0u

0u 0u

T

TT


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 ( 0.01ζ= ) (c) 20% damped ( 0.20ζ= )

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 ζ 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 max command for (b) and trial and error for (c). Problem 3 (Chopra 2.11): For a system with damping ratio ζ , 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) 0.3662

jπζ ζ

= =

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). 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 mass

is given as 20.1 lb-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.

22 k sec /inm=

50 k/ink=

( )u t

Solution:

( ) 1500 lb/in

215.9 lb-sec/in

(b) 0.908

(c) 5.28 rad/sec

cr

D

a k

c c

ζ

ω

=

= =

=

=

Solution:

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


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


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 the base). 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 ( 0.01ζ= ) (c) 20% damped ( 0.20ζ= )

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 ζ to reduce the displacement at 1.5 sec to 1”?

[ ]2 2

2 20

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 sec ( 0.20)

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

n

uu u

u tu t

u t

ω

ζζ

ζ

⎡ ⎤ ⎡ ⎤⎢ ⎥= + = + =⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

= = == = =

= → = =

Problem 3 (Chopra 2.11) MOVED TO HW #4 For a system with damping ratio ζ , 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.

110%

1 10%

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

0.1 2j

uj

j u jπζ πζ

πζ ζ+

⎡ ⎤ ⎡ ⎤⎢ ⎥ = → = = =⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

0 0.5 1 1.5 2 2.5 3 3.5 4−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

Time [sec]

Dis

plac

emen

t [in

]

0ζ =0.01ζ =

0.20ζ =

22 k sec /inm=

50 k/ink=

( )u t


4/10/2009 1:04 PM C:\calpoly\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

364022 1500

386cr

c cc km

ζ = = = =⋅

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

2 21500 3861 1 0.908 5.28 rad/sec

3640nDω ω ζ

⋅= − = − =

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 20.1 lb-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 ζ and nω

1

1

1 1 1ln ln 0.0128 1.28%

2 2 20 0.2j

u

j uζ

π π+

⎡ ⎤ ⎡ ⎤⎢ ⎥= = = =⎢ ⎥⎢ ⎥ ⎢ ⎥⋅ ⎣ ⎦⎢ ⎥⎣ ⎦

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 DT T T

πω= = ≈ = = =

(2) The stiffness coefficient is 2 241.89 0.1 175.5 lb/innk mω= ⋅ = ⋅ =

(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 crc m c cω ζ= = ⋅ ⋅ = = = ⋅ =


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



Reading: Chopra Section 3.1, 3.2 Problem 1 (Chopra 2.11): For a system with damping ratio ζ , 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) 0.3662

jπζ ζ

= =

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 20.1 lb-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 cosntD Du t e A t B t C t D tζω ω ω ω ω−= ⋅ + + +

Find expressions for the constants of integrationA andB in terms of , , (0), (0), , , , and n DC D u u ω ω ω ζ . Solution:

[ ](0)

(0) (0)n

D

A u D

u u D CB

ω ζ ωω

= −

+ − −=



Problem 6: Determine and plot the response of the frame above ( 0 5 sect≤ ≤ ). Assume the girder is rigid and neglect the mass of the columns. (a) Assume at rest initial conditions and zero damping. (b) Assume 5% damping and initial conditions (0) 1 inu = 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) For

010 kp = and 5,10,15 rad/secω = calculate the amplitude of motion. Which of the three forcing frequencies

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

0p of the forcing function such that the deflection of

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

20 k/ink =50 kW =

0( ) sin( )p t p tω=

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

20 ft

15 ft

40 kW=

4

29000 ksi

175 in

E

I

==


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


Homework 4-Solution

Problem 1 For a system with damping ratio ζ , 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 jπζ πζ

πζ ζ+

⎡ ⎤ ⎡ ⎤⎢ ⎥ = → = = =⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

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 20.1 lb-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 ζ and nω

1

1

1 1 1ln ln 0.0128 1.28%

2 2 20 0.2j

uj u

ζπ π+

⎡ ⎤ ⎡ ⎤⎢ ⎥= = = =⎢ ⎥⎢ ⎥ ⎢ ⎥⋅ ⎣ ⎦⎢ ⎥⎣ ⎦

Therefore, assumption of small damping in the above equation is valid. 3 2

0.15 sec 0.15 sec 41.89 rad/sec20 0.15n nD D

T T Tπ

ω= = ≈ = = =

(2) The stiffness coefficient is 2 241.89 0.1 175.5 lb/innk mω= ⋅ = ⋅ =

(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 crc m c cω ζ= = ⋅ ⋅ = = = ⋅ =

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 ζ of the system. At nω ω=

(a) 0 01

( ) 22stu uζ

= =

At 0.1 nω ω= 0 0( ) 0.2 ( 0.1 excitation is so slow, that it can be considered static)stu u β≈ = = →

Substituting0

( ) 0.2stu = in (a) gives 0.05ζ =

If we don’t want to make the assumption of static response for 0.1β = , we can use the approach followed in Problem 4 (see below).



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 nω and the resonance amplitude is1/2ζ . The given data then implies:

(a) 0 01

( ) ( )2n stu uω ω ζ= =

(b)( ) ( ) ( ) ( )

0 1.2 0 02 22 22 2

1 1( ) ( ) ( )

1 2 1 1.2 2 1.2n st stu u uω ω

β ζ β ζ= = =

− + − +

Combining Eqs. (a) and (b)

( ) ( )

2 20

2 2 2 20 1.2

( )1 1 1 10.0576( )

( ) 4 ( 0.44) (2.4 )2 2n

n

uans

uω ω

ω ωζ

ζζ ζ=

=

⎛ ⎞ ⎛ ⎞⎟⎜ ⎟ ⎟⎜ ⎜= = ⇔ =⎟ ⎟⎜ ⎜ ⎟⎜⎟ ⎝ ⎠⎜ ⎟ − +⎜⎝ ⎠

Assumption of small damping is reasonable.



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 cosntD Du t e A t B t C t D tζω ω ω ω ω−= ⋅ + + +

Find expressions for the constants of integrationA andB in terms of , , (0), (0), , , , and n DC D u u ω ω ω ζ .

[ ]

(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 tu tC t D t

u B A C

u u D Cu A CB ans

ω ζ ω ζω ω ω ω ω ζ ω ωω ω ω ω

ω ω ζ ω

ω ζ ωω ζ ωω ω

− −

= +→ = −

= − + − ++ −

= − +

+ − −+ −→ = =

Problem 6 Determine and plot the response of the frame above ( 0 5 sect≤ ≤ ). Assume the girder is rigid and neglect the mass of the 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) 1 inu = 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 D on 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

4015 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

ω

ω

ωβ

ω

β ζβ

β ζβ β ζβ

ζ

β

ωβ

ω

⋅ ⋅= = =

⋅

⋅= =

=

=

= =

− −= =

− + − +

=

= = =− −

=

=

= − = − = −

0 02 2

25 0.7042 1.0036

1( ) sin sin

1 1transient response steady state response

np p

u t t wtk k

βω

β β

⋅ = −

= − +− −

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

20 ft

15 ft

40 kW=

4

29000 ksi

175 in

E

I

==



(b)

( ) ( ) ( ) ( )

20 0

2 22 22 2

( ) ( ) ( ) ( cos sin ) sin cossteady state responsetransient response

(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

ω ζ ω ω ω ω

ω ζβ ζβ

β ζβ β ζβ

−= + = + + +

+− −= = = − =

− + − +

[ ]

1.195 in 2.600 in 1.397 in 0.195 in

D

D C

A B C D

ωω

− −

= = = = −

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

Time [sec]

Dis

plac

emen

t [in

]

totaltransientsteady-state

(0) 0, (0) 0, 00, 1.0036 1.425 0

u uA B C D

ζ= = == = − = =

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−4

−3

−2

−1

0

1

2

3

4

Time [sec]

Dis

plac

emen

t [in

]

totaltransientsteady-state

(0) 1" (0) 50 "/ sec 5%1.195 2.600 1.397 0.195

u uA B C D

ζ= = == = = = −



Problem 7

(a) Calculate the vertical displacement of the cantilever tip due to gravity. Assume 5% damping and consider steady state motion only: (b) For

010 kp = and 5,10,15 rad/secω = calculate the amplitude of motion. Which of the three forcing frequencies

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

0p of the forcing function such that the deflection of

the cantilever due to gravity plus dynamic 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 20k/in

10k1010 rad/sec 0.804 2.763 1.38 in( )

12.43 20k/in

11515 rad/sec 1.206

12.43

st

n

Wu ans

k

u ans

u ans

u

ω

ω β

ω β

ω β

= = =

⋅= =

= = = = ⋅ =

= = = = ⋅ =

= = = =

2

0 00

0 k2.122 1.06 in( )

20k/in

Excitation frequency is closest to resonance largest amplitude

(c) 2.5 2.122 0 23.6k( )20st d

ans

p pu u R p ans

k

ω

⋅ =

→

= − ⋅ = − ⋅ > → <

20 k/ink =50 kW =

0( ) sin( )p t p tω=


C:\calpoly\arce412\homework\spring_2009\hw5.doc 4/17/2009

April 17, 2009

ARCE 412: STRUCTURAL DYNAMICS

Homework 5 (Due 04-20-09) Problem The single bay/single story moment frame above with floor weight of 400 kips above is subject to ten cycles of harmonic ground motion

2( ) 10 in sin 0 10gu t t t T T

πω

ω= ⋅ < < =

whose frequencyω is twice that of the natural frequency nω . Determine the constants 1 1, , ,A B C D in the expressions below for the displacement of the floor relative to the moving ground

( )1 12

( ) cos sin sin cos 0 10ntD Du t e A t B t C t D t t T Tζω π

ω ω ω ωω

−= ⋅ + + + < < =

After 10 cycles the ground motions stops and the structure vibrates freely. Determine the constants 2 2,A B in the expression below for 2 seconds of free vibration after the ground has stopped moving.

( )11 2 1 2 1 1( 10 ) cos sin 0 2 secnt

D Du t T e A t B t tζω ω ω−+ = ⋅ + < <

Plot ( ) and ( )gu t u t as a function of time. Partial solution:

1 1

2 2

0.351sec

0.175 sec

1.747 in, 26.51in, 13.10 in, 1.747 in

1.852 in, 25.36 in

nT

T

A B C D

A B

=

=

= − = − = =

= =

12

ft

15 ft

W 36 x135

W36 x170 W 36 x170

( )u t

( )gu t

=10%ζ

400kW =

0 1 2 3 4

−30

−20

−10

0

10

20

30

Time [sec]

Dis

plac

emen

t [in

]

displacement ( ) of fground displacement

loor (mass) relative(

to ground)g

u tu t


C:\calpoly\arce412\homework\spring_2009\hw5_sol.doc 4/17/2009 7:19 AM - 1 -


Homework 5-Solution

Problem The single bay/single story moment frame above with floor weight of 400 kips above is subject to harmonic ground motion 2

( ) 10 in sin 0 10gu t t t T Tπ

ωω

= ⋅ < < = whose frequencyω is twice that of the natural frequency nω . Determine the constants 1 1, , ,A B C D in the expressions below for the displacement of the floor relative to the moving ground for 10 cycles of ground motion

( )1 12

( ) cos sin sin cos 0 10ntD Du t e A t B t C t D t t T Tζω π

ω ω ω ωω

−= ⋅ + + + < < =

After 10 cycles the ground motions stops and the structure vibrates freely. Determine the constants 2 2,A B in the expression below for 2 seconds of free vibration after the ground has stopped moving.

( )11 2 1 2 1 1( 10 ) cos sin 0 2 secnt

D Du t T e A t B t tζω ω ω−+ = ⋅ + < <

Plot ( ) and ( )gu t u t as a function of time.

12

ft

15 ft

W36x135

W 36 x170 W36 x170

( )u t

( )gu t

=10%ζ

400kW =

0 1 2 3 4

−30

−20

−10

0

10

20

30

Time [sec]

Dis

plac

emen

t [in

]

displacement ( ) of fground displacement

loor (mass) relative(

to ground)g

u tu t



Vibration properties

6 2

6 2

2 2 36 6

3

10,500 29, 000 304.5 10 k-in

7, 800 29, 000 226.2 10 k-in

2

1 1 1 172 144 2 72 180 3.0094 10 in/k

304.5 10 3 226.2 10 3

1 1332.29k/in

3.0094 10 in/k

400/386.42 2 0.351sec

332.29

C

G

n

n

EI

EI

f

kf

mT

k

ωβ

ω

π π

ω

−

−

= ⋅ = ⋅

= ⋅ = ⋅

= =

= ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ = ⋅⋅ ⋅

= = =⋅

= = =

( )

( ) ( )

n

2 2

2 20 0

2 217.92 rad/sec

0.351

10.5 0.351 0.1753 sec

2 2= 35.84 rad/sec

0.1753

10 in sin 35.84

35.83 10 sin 35.84 12,840 in/sec sin 35.83

400k/386.4 in/sec 12, 840 in/sec 13292k

n

n

g

g

g

T

T T

T

u t

u t t

p m u

π π

β

π πω

= = =

= ⋅ = ⋅ =

= =

= ⋅

= ⋅ ⋅ = ⋅

= − ⋅ = ⋅ = −

72k"

72k"

for unit lateral loadM



Constants in response equation (see Hw #4)

( ) ( )

( ) ( )

1 1

20

2 22

02 22

1

1

( ) ( ) ( ) ( cos sin ) sin cossteady state responsetransient response

1

1 2

2

1 2

(0)

(0)

n tc p D D

n

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

pC

k

pD

k

A u D

uB

ω ζ ω ω ω ω

β

β ζβ

ζβ

β ζβ

ω ζ

−= + = + + +

−=

− +

−=

− +

= −

+=

[ ]

( ) ( )

( ) ( )

2

2 22

2 22

2

(0)

13292 1 213.01in

332.29 1 2 2 0.1 2

13292 2 0.1 21.747 in

332.29 1 2 2 0.1 2

0 1.747 1.747 in

0 17.92 0.1 1.747 13.01 35.8426.51in

17.92 1 0.1

D

u D C

C

D

A

B

ωω

− −

−= − ⋅ =

− + ⋅ ⋅

− ⋅ ⋅= − ⋅ =

− + ⋅ ⋅

= − = −

+ ⋅ ⋅ − ⋅= = −

⋅ −

In order to describe the free vibration response after the ground motion has stopped, we need to find the displacement ( 10 )u t T= and the velocity ( 10 )u t T= at the end of the ground excitation. These values serve as initial conditions for the free vibration.

( )1 1

2

2

( ) cos sin sin cos 0 10 1.753 sec

( 10 ) 1.852 in

( sin cos ) ( cos sin ) cos sin( )10 448.8 in/sec( )

(0) 1.852 in

(0)

n

n n

tD D

t tD D D D n D D

n

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

u Tt

e A t B t e A t B t C t D tu tTu t

A u

uB

ζω

ω ζ ω ζ

ω ω ω ω

ω ω ω ω ω ζ ω ω ω ω ω ω

ω ζ

−

− −

= ⋅ + + + < < == =

= − + − + + −= =

= =

+=

( )

2

2 2

(0) 448.8 17.92 0.1 1.85225.36 in

17.92 1 0.1

( 10 ) cos sin 0 2 secn

D

tD D

u

u t T e A t B t tζω

ω

ω ω−

+ ⋅ ⋅= =

⋅ −

+ = ⋅ + < <

plot see page 1





Reading: Chopra Sections 5.1, 5.4 Problem 1 An SDF system has the following properties: 20.2533 k sec /in, =10 k/in, =0.05m k ζ= . Use the NEWMARK average acceleration method to calculate and plot the displacement response ( )u t of this system to a force ( ) 10 sin( )p t tπ= (at rest initial conditions). (a) Use hand calculation with 0.1sectΔ = for as many times it that you need to understand the procedure. (b) Implement the NEWMARK average acceleration method for a SDF system in MATLAB by writing a function function u = newmark(m,c,k,p,delt) where , ,m c k are the mass, viscous damping coefficient, and stiffness, respectively of the SDF structure, p is a row vector containing the loading function and tΔ is the time step. The output argument u is a vector (say row vector) containing the displacement response of the SDF system. (c) Test your algorithm by calculating the displacement response ( )u t for 0 10 sect≤ ≤ . Use time steps 0.5, 0.1, 0.02 sectΔ = . Also plot the exact solution. The exact solution is (see HW 4) ( ) exp( ) ( cos sin ) sin cosn D Du t t A t B t C t D tζω ω ω ω ω= − ⋅ + + + Submit: (a) Hand calculations. (b) A single plot containing the solution with 0.5, 0.1, 0.02 sectΔ = and the exact solution. Use different colors or different linestyles to distinguish solutions for different tΔ . Problem 2 Use the NEWMARK average acceleration method to calculate and plot the displacement response (the lateral deformation) of two SDF systems with natural periods 2secnT = and 0.5 secnT = , respectively to the 1989 Loma Prieta earthquake (recorded at Gilroy). Assume 5% damping. The record lomaprieta.dat (on blackboard) contains 1001ground acceleration data points in cm/sec2 at a spacing of 0.02 sec.

Loma Prieta Ground Acceleration Submit: Single page with two plots (ground acceleration in units of g and relative displacement response in [in] (the output of the NEWMARK function), single plot for the two periods).

0 2 4 6 8 10 12 14 16 18 20

−600

−400

−200

0

200

400

600

Time [s]

Acc

eler

atio

n [c

m/s

2 ]




Homework 6-Solution

Problem 1 An SDF system has the following properties: 20.2533 k sec /in, =10 k/in, =0.05m k ζ= . Use the NEWMARK average acceleration method to calculate and plot the displacement response ( )u t of this system to a force ( ) 10 sin( )p t tπ= (at rest initial conditions). (a) Use hand calculation with 0.1sectΔ = for as many times it that you need to understand the procedure. (b) Implement the NEWMARK average acceleration method for a SDF system in MATLAB by writing a function function u = newmark(m,c,k,p,delt) where , ,m c k are the mass, viscous damping coefficient, and stiffness, respectively of the SDF structure, p is a row vector containing the loading function and tΔ is the time step. The output argument u is a vector (say row vector) containing the displacement response of the SDF system. (c) Test your algorithm by calculating the displacement response ( )u t for 0 10 sect≤ ≤ . Use time steps 0.5, 0.1, 0.02 sectΔ = . Also plot the exact solution. The exact solution is (see HW 4) ( ) exp( ) ( cos sin ) sin cosn D Du t t A t B t C t D tζω ω ω ω ω= − ⋅ + + + Submit: (a) Hand calculations. (b) A single plot containing the solution with 0.5, 0.1, 0.02 sectΔ = and the exact solution. i t ( )p t ( )u t ( )u t ( )u t 1 0.00 0.0000 0.0000 0.0000 0.00002 0.10 3.0902 10.7951 0.5398 0.02703 0.20 5.8779 16.1621 1.8876 0.14844 0.30 8.0902 13.5749 3.3745 0.41155 0.40 9.5106 3.6228 4.2343 0.79196 0.50 10.0000 -10.2937 3.9008 1.19877 0.60 9.5106 -23.2601 2.2231 1.50498 0.70 8.0902 -30.6321 -0.4715 1.59249 0.80 5.8779 -29.6599 -3.4861 1.3946

10 0.90 3.0902 -20.3904 -5.9886 0.920811 1.00 0.0000 -5.5699 -7.2866 0.2571

Numerical (Newmark Average Acceleration Method using different time steps) and analytical response.

0 2 4 6 8 10−3

−2

−1

0

1

2

3

Time [sec]

Dis

plac

emen

t [in

]

exact(

0.5

vir

0.1

tua0. ll0 y identi

0

cal2, )

t

t

t

Δ =

Δ =

Δ =



Problem 2 Use the NEWMARK average acceleration method to calculate and plot the acceleration, velocity and displacement response of two SDF systems with natural periods 2secnT = and 0.5 secnT = , respectively to the 1989 Loma Prieta earthquake (recorded at Gilroy). Assume 5% damping. The record (\ansgar\lomaprieta.dat on the ARCE shared drive) contains 1001ground acceleration data points in cm/sec2 at a spacing of 0.02 sec.

Loma Prieta Ground Acceleration Submit: Single page with two plots (ground acceleration in units of g and relative displacement response in [in] (the output of the NEWMARK function), single plot for the two periods). Loma Prieta ground acceleration and displacement response of two SDF structures (displacement is relative to the ground) with 5% damping.

0 2 4 6 8 10 12 14 16 18 20

−600

−400

−200

0

200

400

600

Time [s]

Acc

eler

atio

n [c

m/s

2 ]

0 5 10 15 20

−20

−10

0

10

20

Time [s]

Dis

plac

emen

t [in

]

2 se0 ec

c.5 sn

nTT =

=

0.05ζ =

a

max

m x

20.6 iin

3.35 nuu

==

0 5 10 15 20−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

t [sec]

Gro

und

Acc

eler

atio

n [g

]

max 0.665gu g=

California Polytechnic State University Department of Architectural Engineering Spring Quarter 2009 nstructor: Ansgar Neuenhofer




Reading: Chopra, Chapter 6 Problem 1 Calculate the 5% damped displacement (in cm) and acceleration response spectra (in g) (also called the spectral displacement and spectral acceleration, respectively) of the Loma Prieta and Northridge earthquakes (on Blackboard). Con-sider 51 natural periods nT equally spaced between 0.2 secnT = and 3 secnT = . Submit: Two plots (spectral displacement and acceleration), single plot for the two EQ motions and table with spectral values (see below) Hint: Cut and paste spectral valuesA andD from MATLAB into EXCEL Solution: # [sec]nT [cm]D LP [g]A LP [cm]D NR [g]A NR

1 0.2 ... ... 51 3 Problem 2 A 10-ft-long vertical cantilever made of a steel pipe supports a 3000-lb weight attached at the tip as shown. The properties of the pipe are: outside diameter=6.625, inside diameter=6.065 in. Find (1) the peak deformation, (2) the equivalent static force and (3) the bending stress resulting from the Loma Prieta and Northridge earthquakes. Ignore the weight of the pipe and assume 5% damping. CAUTION: Earthquake records are in 2cm/sec

Partial Solution:

0 0.5 1 1.5 2 2.5 30

10

20

30

40

50

60

70

80

Tn [sec]

Spe

ctra

l Dis

plac

emen

t [cm

]

Northridge

0.05ζ =

0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.5

Tn [sec]

Spe

ctra

l Acc

eler

atio

n [g

]

0.05ζ =Loma Prieta

Northridge

Loma Prieta

Loma Prieta: 50.0ksi( )

Northridge: 36.2ksi( )

ans

ans

σ

σ

=

=




Homework 7-Solution

Problem 1 Calculate the 5% damped displacement (in cm) and acceleration response spectra (in g) for the Loma Prieta and North-ridge earthquakes (in directory ansgar on the ARCE shared drive). Consider 51 natural periods nT equally spaced be-tween 0.2 secnT = and 3 secnT = . Submit: Two plots (spectral displacement and acceleration), single plot for the two motions. Table (see below) with spectral values. Hint: Cut and paste spectral valuesA andD from MATLAB into EXCEL

0 0.5 1 1.5 2 2.5 30

10

20

30

40

50

60

70

80

Tn [sec]

Spe

ctra

l Dis

plac

emen

t [cm

]

Lo

No

ma

rthr

Pr

idge

ieta

0.05ζ =

0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.5

Tn [sec]

Spe

ctra

l Acc

eler

atio

n [g

]

Lo

No

ma

rthr

Pr

idge

ieta

0.05ζ =



MATLAB statements %HW 7, NEWMARK METHOD, ground acceleration, response spectrum clear all close('all') width = 1.5; width1 = 1.0 size1=16; size2=20; load c:\calpoly\arce412\eqrecords\lomaprieta.dat; load c:\calpoly\arce412\eqrecords\northridge.dat; input = [lomaprieta northridge]; t = [0:0.02:20]; zeta = 0.05; m=1;delt = 0.02; delT = (3 - 0.2)/50; T = [0.2:delT:3]; for i=1:2 for j=1:length(T); j TT = T(j); k = 4*pi^2 ./ TT^2; c = zeta * 2 * sqrt( k * m); [u,udot,uddot] = newmark(m,c,k,-input(:,i)'*m,delt); dis(i,j) = max(abs(u)); acc(i,j) = dis(i,j) .* k / 981; end end col = ['b';'r']; st=['- ';'--'] figure(1) hold on for i=1:2 plot(T,dis(i,:),col(i,:),'linewidth',width,'linestyle',st(i,:)); ty = [col(i,:) '*'] plot(T,dis(i,:),ty,'linewidth',width1); end xlabel('\itT_n \rm[sec]','fontsize',size2); ylabel('Spectral Displacement [cm]','fontsize',size2); set(gca,'plotboxaspectratio',[1.5 1 1]); set(gca,'fontsize',size1); print -depsc -r1200 -tiff c:\calpoly figure(2) hold on for i=1:2 plot(T,acc(i,:),col(i,:),'linewidth',width,'linestyle',st(i,:)); ty = [col(i,:) '*'] plot(T,acc(i,:),ty,'linewidth',width1); end xlabel('\itT_n \rm [sec]','fontsize',size2); ylabel('Spectral Acceleration [g]','fontsize',size2); set(gca,'plotboxaspectratio',[1.5 1 1]); set(gca,'fontsize',size1); stop %build table res(:,[1 2 4 3 5]) = [ T' dis' acc']; %print -depsc -r1200 -tiff c:\calpoly



Numerical results

[sec]nT [cm]D LP [g]A LP [cm]D NR [g]A NR

0.20 2.54 2.55 0.65 0.650.26 1.98 1.22 1.51 0.930.31 4.73 1.96 2.83 1.170.37 3.15 0.94 2.56 0.760.42 4.08 0.91 4.27 0.960.48 7.37 1.29 5.33 0.930.54 9.15 1.28 7.42 1.040.59 8.25 0.95 8.72 1.000.65 7.57 0.73 8.97 0.860.70 9.91 0.80 8.71 0.710.76 11.94 0.83 12.13 0.840.82 13.11 0.79 18.02 1.090.87 14.16 0.75 20.94 1.110.93 15.32 0.72 19.72 0.920.98 17.00 0.71 15.70 0.651.04 19.10 0.71 12.83 0.481.10 21.30 0.71 14.06 0.471.15 23.52 0.71 15.43 0.471.21 25.75 0.71 20.51 0.571.26 27.95 0.70 24.87 0.631.32 30.07 0.69 27.32 0.631.38 32.14 0.68 28.16 0.601.43 33.96 0.67 27.81 0.551.49 35.51 0.65 25.87 0.471.54 37.48 0.63 26.17 0.441.60 40.05 0.63 27.17 0.431.66 41.92 0.62 28.05 0.411.71 43.82 0.60 28.79 0.401.77 45.37 0.58 29.43 0.381.82 46.45 0.56 32.01 0.391.88 48.94 0.56 37.64 0.431.94 50.91 0.55 43.14 0.461.99 52.19 0.53 48.10 0.492.05 52.60 0.50 53.08 0.512.10 52.73 0.48 57.91 0.532.16 53.21 0.46 61.66 0.532.22 54.83 0.45 65.75 0.542.27 56.25 0.44 69.01 0.542.33 57.33 0.43 71.14 0.532.38 58.01 0.41 72.36 0.512.44 58.25 0.39 72.62 0.492.50 58.08 0.38 73.39 0.472.55 57.58 0.36 74.55 0.462.61 56.94 0.34 75.03 0.442.66 56.22 0.32 74.80 0.422.72 55.38 0.30 73.86 0.402.78 54.38 0.28 72.30 0.382.83 53.22 0.27 70.33 0.352.89 51.96 0.25 69.56 0.342.94 50.63 0.24 70.16 0.333.00 49.24 0.22 70.40 0.31



Problem 2 A 10-ft-long vertical cantilever made of a 6-in nominal-diameter standard steel pipe supports a 3000-lb weight attached at the tip as shown. The properties of the pipe are: outside diameter=6.625, inside diameter=6.6065 in. Find (1) the peak deformation, (2) the equivalent static force and (3) the bending stress resulting from the Northridge and Loma Prieta earthquakes. Ignore the weight of the pipe.

Note: Section properties are calculated based on nominal dimensions. Use of steel manual results in smaller section properties and correspondingly larger stresses. The results are

4 4 4

3 3

3

from Newmark

(6.625 6.065 ) 28.14 in643 3 29, 000 28.14

1.417 k/in120

3000 lbs

30002 0.465 sec

1.417 10 386.4

Spectral displacement,force,moment,stress

6.34cmLoma Prieta: ( 0.465, 0.05)

n

n

I

EIk

LW

T

D T

π

π

ζ

= − =

⋅ ⋅= = =

=

= =⋅ ⋅

= = = 2.50 in( )2.54 cm/in

2.50 1.417 3.54 k( )

3.54 120 424 k-in

4246.625/2 50.0 ksi( )

28.14

4.59cmNorthridge: ( 0.465, 0.05) 1.81 in ( )

2.54 cm/in

( )1.81 1.417 2.56 k

2.56 120 307 k-in

307

n

ans

F ans

M

ans

D T ans

ansF

M

σ

ζ

σ

=

= ⋅ =

= ⋅ =

= ⋅ =

= = = =

= ⋅ =

= ⋅ =

= 6.625/2 36.2 ksi( )28.14

ans⋅ =

58.1ksi(Loma Prieta)

42.0 ksi(Northridge)

σ

σ

=

=





Reading: Chopra Section 9.1, 9.2 Problem 1 A uniform rigid bar of total massm is supported on two springs

1k and

2k at the two ends and

subjected to dynamic forces as shown in the figure. Formulate the equations of motion with respect to the two DOFs defined at the left end of the bar. Partial Solution:

1 2 2

2 22 2

2

2 3

mLmk k k L

k L k L mL mL

⎡ ⎤⎢ ⎥⎡ ⎤+ ⎢ ⎥⎢ ⎥= = ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥⎢ ⎥⎣ ⎦

k m

Problem 2 Find the 2x2 (a) mass, (b) flexibility (using PVF hand calculation) (c) stiffness (use beam element in MATLAB, eliminate unwanted DOFs) matrices of the beam for the two DOFs shown. Check whether the stiffness and flexibility matrices are the inverse of each other. Partial Solution:

3 8 7

7 8486LEI

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

f

Problem 3 A rigid bar is supported by a weightless column as shown. Find the 2x2 (a) mass (b) flexibility (using PVF hand calculation) (c) stiffness (by inverting the flexibility matrix) matrices of the system defined for the two DOFs shown. Find the stiffness matrix by inverting the flexibility matrix.

3 2 5

5 146LEI

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

f



Problem 4 The figure shows a uniform slab supported on four columns rigidly attached to the slab and clamped at the base. The slab has a total mass m and is rigid in plane and out of plane. Each column is of circular cross section with moment of inertia I or 2I . With the DOFs selected as , ,x yu u u

θ

(1) at the center of mass of the slab (as shown) (2) at the center of stiffness of the slab (not shown) find the mass and stiffness matrices in terms ofm and the lateral stiffness 312 /k EI h= of the smaller column. Problem 5 Find the mass and stiffness matrices with respect to the three in plane degrees-of-freedom , ,x y zu u ϕ . In-plane stiffness is provided by the three diagonal braces.

16 '

16 '

16 '

12 '

12 '

12 '

EA

3C

0.5 ft

150#/ft

t

γ

=

=

xy z

16 '

16 '

16 '

12 '

12 '

12 '

EA

3C

0.5 ft

150#/ft

t

γ

=

=

xy z

slab

Solution

0.064 0 0

0 0.0295 0.7071 [ft,k]

0 0.7071 26.19

2.683 0 0

0 2.683 0 [ft, k, sec]

0 0 644.0

86.4 k

EA

W

⎡ ⎤⎢ ⎥⎢ ⎥

= −⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

⎡ ⎤⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

=

K

M

3

2

2

Solution for (4.1)

6 0 012

0 6 ,with

0 3

0 0

0 0

0 06

EIk b k

hb b

m

m

bm

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥= − =⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

K

M


4/28/2009 8:29 PM C:\calpoly\arce412\homework\spring_2009\hw8_sol.doc - 1 -

May 6, 2009 ARCE 412: STRUCTURAL DYNAMICS

Homework 8-Solution

Relative displacement, velocity and acceleration response time history for Loma Prieta Earthquake( 5%)ζ = .



Problem 2c Using the beam element with four degrees-of-freedom the stiffness matrix with respect to the 8 free degrees-of-freedom (4 rotations and 2 translations) is

free

K =

−− −

−− −

12 54 6 0 0 0

54 648 0 324 54 0

6 0 24 54 6 0

0 324 54 648 0 53

0 54 6 0 24 6

0 0

use 1, 1, 1E I L

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ = = =⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦0 53 6 12

Since zero mass is associated with the rotations, we eliminate the four rotations to generate a stiffness matrix with respect to the two translations only.

3 ( )finalEI

ansL

⎡ ⎤⎢ ⎥= ⎢ ⎥⎣ ⎦

K 259.2 -226.8

-226.8 259.2

Below are the corresponding MATLAB statements free = [2 3 4 5 6 8]; Kfree = K(free,free); r = [2 4]; e = [1 3 5 6]; Krr = Kfree(r,r); Kre = Kfree (r,e); Ker = Kfree (e,r); Kee = Kfree (e,e); %Final 2x2 st. matrix Kfinal = Krr - Kre*inv(Kee) * Ker Problem 4.2 When we apply the degrees-of-freedom at the center of rigidity all off-diagonal terms of the stiffness matrix a zero (that is the definition of the center of rigidity). Furthermore, the diagonal elements related to the displacements are the same as before. Hence we only need to calculate element 33K of the stiffness matrix. The mass matrix is no longer diagonal, coupling exists between the acceleration in they -direction and the moment (and also, by symmetry) between the angular acceleration and they -force). Element 33M of the mass matrix also changes.

( )

1 2

2 2 2 2 233 1 2

22 2 2 2

33

3

22

13 25,

36 3613 25 17

2 2 2 4 236 36 6

1 1 712 6 36

6 0 0 6 0 012

0 6 0 0 6 0 ( ), with

17 0 0 2.8330 06

1 0 0

10 1

61

06

d b d b

K kd kd k b k b b k

M m b b m b mb

EIk k ans k

hbb

b

b

= =

= ⋅ ⋅ + ⋅ = + =

⎛ ⎞⎟⎜= + + ⋅ =⎟⎜ ⎟⎜⎝ ⎠⎡ ⎤

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥= = =⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ ⎦⎢ ⎥⎣ ⎦

=

K

M22

1 0 0

0 1 0.1667 ( )

0 0.1667 0.1944736

m b m ans

b bb

⎡ ⎤⎢ ⎥ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦



Problem 5 Find the mass and stiffness matrices with respect to the three in plane degrees-of-freedom , ,x y zu u ϕ . In-plane stiffness is provided by the three diagonal braces.

2 211

21 12

31 13

2 222

32 23

2 2 2 233

2 cos 2 0.8 0.064020

0

0

cos 0.707 0.0295016.971

0.02950 24 0.7071

1 12 0.80 12 0.7071 24 26.19

20 16.971

0.0640 0 0

0 0.0295 0

EA EAk EA

Lk k

k k

EA EAk EA

Lk k EA EA

k EA EA

EA

α

α

= ⋅ = ⋅ =

= =

= =

= = =

= = − ⋅ = −

⎛ ⎞⎟⎜= ⋅ ⋅ ⋅ + ⋅ ⋅ =⎟⎜ ⎟⎜⎝ ⎠

= −K

3slab

2

2 2 2

.7071 k, ft( )

0 0.7071 26.19

48 ' 24 ' 0.5 ' 0.150 k/ft 86.4 k

86.4 k-sec2.683

32.2 ft1

(24 48 ) 2.683 643.9k-sec -ft12

2.683 0 0

0 2.683 0 [ft, k, sec]( )

0 0 644.0

ans

W

m

MMI

ans

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

= ⋅ ⋅ ⋅ =

= =

= ⋅ + ⋅ =

⎡ ⎤⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

M

16 '

16 '

16 '

12 '

12 '

12 '

EA

3C

0.5 ft

150#/ft

t

γ

=

=

xy z

16 '

16 '

16 '

12 '

12 '

12 '

EA

3C

0.5 ft

150#/ft

t

γ

=

=

xy z

16 ft 16 ft 16 ft

12

ft 1

2 ft

16 ft 16 ft 16 ft 1

2 ft

12

ft

16 ft 16 ft 16 ft

12

ft 1

2 ft

33k

23k

13k

20.8 1220

0.384

EA

EA

⋅ ⋅

=

0.384EA

20.707 2416.971

0.7071

EA

EA

⋅ ⋅

=

11k

21k

31k

1 2 31, 0, 0u u u= = =

12k32k

22k

1 2 30, 1, 0u u u= = =

1 2 30, 0, 1u u u= = =

0.032EA

0.032EA0.0295EA


5/8/2009 7:53 AM C:\calpoly\arce412\homework\spring_2009\hw_9.doc



Reading: Chopra Section 10.1, 10.2, 10.3 Problem 1 Determine the natural vibration frequencies and mode shapes of the structure for

1k k= and

22k k= . Sketch the mode shapes.

Use the mass and stiffness matrices determined in HW 8. Solution:

1 2 1 2

1 11.592 3.076

0.634 / 2.366/k k

L Lm mω ω

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= = = =⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

φ φ

Problem 2 Determine the natural vibration frequencies and mode shapes of the structure in terms of ,m EI andL . Sketch the mode shapes. Use the mass and stiffness matrices determined in HW 8. Solution:

1 2 1 24 4

1 19.859 38.184

1 1EI EI

mL mLω ω

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= = = =⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

φ φ

Problem 3: Determine the natural vibration frequencies and mode shapes of the structure (rigid beams) in terms of ,m EI andh . Sketch the mode shapes. Use MATLAB’s eig function to calculate the eigenvalues (natural frequencies) and eigenvectors (mode shapes). Normalize the mode shapes such that the value at the roof level has unit value. In Matlab, use 1, 1, 1, 1E I h m= = = = . Model the frame in RISA or ETABS, perform a frequency analysis to find the mode shapes and frequencies. Compare the results. A correct model should give you the exact same frequencies and mode shapes. Solution:

1 2 3 1 2 33 3 3

0.314 0.5 3.186

2.241 4.899 7.140 0.686 0.5 2.186

1 1 1

EI EI EImh mh mh

ω ω ω

⎡ ⎤ ⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥

= = = = = − = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

φ φ φ

(total mass)m



. May 14, 2009 ARCE 412: STRUCTURAL DYNAMICS

Homework 9-Solution



Problem 10.12





Reading: Chopra Sections 10.7-10.8 Problem 1 For the structure of HW 8, P 4 (rigid slab on four columns), find the three natural periods nT and corresponding mode shapes Φ for in-plane vibrations. Define the degrees-of-freedom (1) at the center of mass of the slab (as shown) (2) at the center of stiffness of the slab (not shown) and show that both frequencies and mode shapes are independent of where we apply the degrees-of-freedom. Order the periods in descending order and sketch the mode shapes. Use MATLAB. (3) Model the structure in ETABS or RISA 3D and calculate natural periods and mode shapes. Include plots of the mode shapes in your submittal. Model the columns as fixed-fixed. Use 332.2 k (total weight of slab), 30 ft, 12 / 100 k/ftW b k EI h= = = = . Solution: 1 2 3 0.267 sec, 0.256 sec 0.146 secT T T= = =

Problem 2 Shape 4[in ]I

W 14x48 485 W 14x90 999

girders assumed rigid

12

ft 1

2 ft

12

ft

25 k

1u

2u

3u

50 k

50 k

14 x 48W

14 x 90W

14 x120W



W 14x120 1380 (1) Determine the natural periods nT and modes nφ of the three-story frame above. Sketch the mode shapes and

identify the associated natural frequencies. Normalize each mode so that the modal mass Tn n nM φ φ= m has

unit value. Solution:

[ ]0.270 0.124 0.084 secT =

for in k, sec, ft

for in

Φ

Φ

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

m

m

0.2554 0.4103 0.64070.5313 0.3876 -0.46000.7701 -0.8068 0.2097

0.885 1.421 2.2191.840 1.343 -1.5932.668 -2.795 0.726

k, sec, in

(2) Determine the free vibration response of the frame for the following three vectors of initial displacements (the initial velocity is zero). Neglect damping. Consider 0 1.0 sect≤ ≤ .

I II III1 2 3 1 0.25 1 1 1 1 [in]T T T⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = − = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦u u u

Comment on the relative contribution of the three vibration modes to the response produced by the three initial displace-ments. Submit:

(1) Summary of analysis. Sketch of mode shapes. (2) One plot for each vector of initial displacements showing the roof response in modes 1, 2, and 3 and the total

roof response (i.e. 4 lines per plot). (3) Use hand calculation to calculate at 1sect = for the three initial conditions:

3.1 The total (all modes combined) first floor displacement. 3.2 The roof displacement in mode 2. 3.3 The second floor displacement in mode 1. Partial Solution Important arrays and their dimensions:

number of degrees of freedom(dof) =number of modes

number of time points

structure response ( x matrix)

modal response ( x matrix)

mode shapes ( x matrix)

( , ) value of mode shape at dof

N

NT

N NT

N NT

N N

i j j iϕ

Φ

=

u

q

u matrix product= row times column = modal supoerpositionΦ q

Roof displacement for IC

mode 2mode

mo

3

2

t

de 1

otal


mode 2mode

mo

3

2

t

de 1

otaldisplacement in [in]

3.1 3.2 3.3

IC1 0.26 0.026 0.57

IC2 0.90 0.75 0.061

IC 3 0.69 0.40 0.025

− −

− −

−


5/18/2009 10:00 AM C:\calpoly\arce412\homework\spring_2009\hw10_sol.doc - 1 -


Homework 10-Solution

Problem 1 For the structure of HW 8, P 4 (rigid slab on four columns) find the three natural periods nT and corresponding mode shapes Φ for in-plane vibrations. Define the degrees-of-freedom (1) at the center of mass of the slab (as shown) (2) at the center of stiffness of the slab (not shown) and show that both frequencies and mode shapes are independent of where we apply the degrees-of-freedom. Order the periods in descending order and sketch the mode shapes. Use 332.2 k (total weight of slab), 30 ft, 12 / 100 k/ftW b k EI h= = = = . Center of mass (compare HW 8 solution)

2 2 2 2

6 0 0 6 0 0 600 0 0 1 0 0 1 0 0 1 0 0

0 6 100 0 6 30 0 600 3, 000 0 1 0 1 0 1 0 0 1 0

0 3, 000 270, 000 0 0 1500 3 0 30 3 30 300 0 0 0

6 6

k b m

b b b

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= − = − = − = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − ⋅ ⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

K M

⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Center of stiffness (compare HW 8 solution)

2 22 2

1 0 0 1 0 06 0 0 6 0 0 600 0 0

1 10 6 0 100 0 6 0 0 600 0 0 1 1 0 1 30

6 617 17 0 0 255, 000 1 7 1 7 300 0 0 0 30 0 0 306 6 6 36 6 36

k m b

bb b

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= = = = = ⋅⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⋅⎣ ⎦⋅⎢ ⎥ ⎢ ⎥ ⋅⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦

K M

1 0 0

0 1 5

0 5 175

⎡ ⎤⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎥⎢ ⎥⎥ ⎣ ⎦⎥

⎥

MATLAB statemenets [phi lam] = eig(K,M); %frequencies omega = sqrt(diag(lam)) %sort [omega j] = sort(omega); phi = phi(:,j); T = 2*pi ./ omega

1 2 3 0.267 sec, 0.256 sec 0.146 secT T T= = =

Since 0.9813 0.9027 5 0.0157= + ⋅ and 0.1926 0.5932 5 0.0801− = − + ⋅ , the two mode shape describe identical motions.

CM

CR

0 1.000 0

0.9813 0 0.1926

0.0157 0 0.0801

0 1.0000 0

0.9027 0 0.5932

0.0157 0 0.0801

⎡ ⎤⎢ ⎥⎢ ⎥

= −⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

⎡ ⎤⎢ ⎥⎢ ⎥

= −⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Φ

Φ

Mode2 Mode 3

CMCR1

23 1

23

Mode2 Mode 3

CMCR1

23 1

23



Sketch of mode shapes Problem 2 Shape 4[in ]I

W 14x48 485 W 14x90 999 W 14x120 1380

(1) Determine the natural periods nT and modes nφ of the three-story frame above. Sketch the mode shapes and identify the associated natural frequencies. Normalize each mode so that the modal mass T

n n nM φ φ= m has unit value. Solution:

[ ]0.270 0.124 0.084 secT =⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

0.885 1.421 2.2191.840 1.343 -1.5932.668 -2.795 0.726

Φ

(2) Determine the free vibration response of the frame for the following three vectors of initial displacements (the initial velocity is zero). Neglect damping. Consider 0 1.0 sect≤ ≤ .

I II III1 2 3 1 0.25 1 1 1 1 [in]T T T⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = − = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦u u u

Comment on the relative contribution of the three vibration modes to the response produced by the three initial displace-ments.


25 k

1u

2u

3u

50 k

50 k

14 x 48W

14 x 90W

14 x120W


25 k

1u

2u

3u

50 k

50 k

14 x 48W

14 x 90W

14 x120W

Mode1 Mode2 Mode 3



Submit: (1) Summary of analysis. Sketch of mode shapes. (2) One plot for each vector of initial displacements showing the roof response in modes 1, 2, and 3 and the total

roof response (i.e. 4 lines per plot). (3) Use hand calculation to calculate at 1sect = for the three initial conditions:

3.1 The total (all modes combined) first floor displacement. 3.2 The roof displacement in mode 2. 3.3 The second floor displacement in mode 1. MATLAB code %3-story frame (rigid girders) clear all close('all') %story mass and stiffnesses m = 50/386.4; E = 29000; k1 = 24*E*1380 /144/12^3; k2 = 24*E* 999 /144/12^3; k3 = 24*E* 485 /144/12^3; K = [k1+k2 -k2 0; -k2 k2+k3 -k3; 0 -k3 k3]/12; M = [m 0 0; 0 m 0; 0 0 m/2]; %frequencies [phi lam] = eig(K,M); omega = sqrt(diag(lam)) %sort frequencies and modes [omega j] = sort(omega); phi= phi(:,j);T= 2*pi ./ omega N = length(T); %normalize phi for i=1:N Mn(i) = phi(:,i)' * M * phi(:,i); phi(:,i) = phi(:,i) ./ sqrt(Mn(i)); end %initial conditions %u0 = [1 2 3]'; u0 = [-1 0.25 1]'; %u0 = [1 -1 1]'; q0 = inv(phi) * u0; t = [0:0.001:1]; NT= length(t); umode=zeros(N,NT,N); %modal displacements for i=1:3 q(i,:) = q0(i) .* cos(omega(i) .* t); end %structure displacements in individual modes for i=1:N umode(:,:,i) = phi(:,i) * q(i,:); end %total structure displacements (all modes combined) utot = phi * q; col = ['r';'g';'b';'k']; figure hold on for i=1:3 plot(t,umode(3,:,i),col(i,:)); end plot(t,utot(3,:),col(4,:)); axis([0 1 -3 3]);



Summary of matrices and hand calculations

2k/in k sec /in

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Φ =

K M

554.5 -232.8 0 0.1294 0 0

-232.8 345.9 -113.0 0 0.1294 0

0 -113.0 113.0 0 0 0.0647

0.885 1.4

0 0

0,1 0,1

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥Φ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

= Φ

= Φ =

q u

q u

-1

-1

-1

21 2.219 0.1145 0.2381 0.1726

1.840 1.343 -1.593 = 0.1839 0.1737 -0.1808

2.668 -2.795 0.726 0.2872 -0.2062 0.0470

0.1

0,2 0,2

1

2

3

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

= Φ =q u-1

145 0.2381 0.1726 1.1086

0.1839 0.1737 -0.1808 -0.0111

0.2872 -0.2062 0.0470 0.0158

0.1145 0.2381 0.1726

0.1839

0,3 0,3

1

0.25

1

⎡ ⎤⎡ ⎤ ⎡ ⎤−⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

= Φ =q u-1

0.1177

0.1737 -0.1808 -0.3213

0.2872 -0.2062 0.0470 -0.2917

0.1145 0.2381 0.1726

0.1839 0.1737 -0.1808

0.2872

1

2

3

1

1

1

1

cos 1

cos 1

cos 1

For IC 1

3.1 0.885 1.1086u

ω

ω

ω

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− =⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

⎡ ⎤ ⎡ ⎤⋅⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⋅ =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⋅ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

= ⋅ ⋅

0.0489

-0.1706

-0.2062 0.0470 0.5404

-0.2812

0.8305

0.7550

( ) ( )

( )

( )

3,2

2,1

0.2812 1.421 0.0111 0.8305 2.219 0.0158 0.7550 0.263 in

3.2 2.795 0.0111 0.8305 0.026 in

3.3 1.840 1.1086 0.2812 0.574 in other IC analogous

u

u

− + ⋅ − ⋅ + ⋅ ⋅ = −

= − ⋅ − ⋅ =

= ⋅ ⋅ − = −

Summary displacement in [in]

3.1 3.2 3.3

IC1 0.26 0.026 0.57

IC2 0.90 0.75 0.061

IC 3 0.69 0.40 0.025

− −

− −

−



Plots Sketch of mode shapes

0 0.2 0.4 0.6 0.8 1−3

−2

−1

0

1

2

3

time [sec]

roof

dis

plac

emen

t [in

]


mode 2mode

mo

3

1

t

de 1

otal

0 0.2 0.4 0.6 0.8 1−3

−2

−1

0

1

2

3

time [sec]

roof

dis

plac

emen

t [in

]


mode 2mode

mo

3

2

t

de 1

otal

0 0.2 0.4 0.6 0.8 1−3

−2

−1

0

1

2

3

time [sec]

roof

dis

plac

emen

t [in

]


mode 2mode

mo

3

3

t

de 1

otal


5/22/2009 1:12 PM C:\calpoly\arce412\homework\spring_2009\hw_11.doc



Calculate the response of the three-story frame above to the Loma Prieta ground motion. Select 5% damping. Response quantities of interest are: (1-3) The inter-story drift ratio in % for the three stories. Story heights are 12 ft for stories 2 and 3 and 15 ft for story 1. (4) The overturning moment in k-ft. Plot the response in modes 1 and 2 and the total response for 0 12 sect< < . Don't include the third mode (its contribution is virtually zero and the plots would become too busy). Distinguish the three lines by color or line-style. On your plot, clearly mark the maximum absolute value of the response in each mode as well as the maximum of the total response. Why is the maximum value of the sum (sum of all modes) not equal to the sum of the maximum modal values? Develop a table summarizing the maximum value of each result quantity in the first mode and the total ma-ximum value. Comment on the effects of the second mode on the total maximum. Which of the four response quantities is most affected by the second mode? Some help on MATAB implementation %Total structure response u = phi * q %Structure response (floor displacements) in individual modes u1 = phi(:,1) * q(1,:) u2 = phi(:,2) * q(2,:) u2 = phi(:,3) * q(3,:) %1st story drift in three modes and total d1 = u(1,:); d11 = u1(1,:); d12 = u2(1,:); d13 = u3(1,:); %2nd story drift in three modes and total

18 f

girders assumed rigid = story stiffnessk

1u

2u

3u

96.6 k

96.6 k

48.3 k

1500 k/ftk =

2625 k/ftk =

2625 k/ftk =



d2 = u(1,:) - u(2,:); d21 = u1(1,:)- u1(2,:); d22 = u2(1,:)-u2(2,:); d23 = u3(1,:)-u3(2,:); %don't forget to convert drift into drift ratio %3rd story drift in three modes and total your turn... %forces acting at floor level (three modes and total) F = K * u; F1 = K * u1; F2 = K * u2; F3 = K * u3; %overturning moment(three modes and total) your turn... Some Solutions:

1

2

3

1

2

3

0.500 sec

0.162 sec

0.111sec

2.6980

0.4606

0.09207

T

T

T

=

=

=

Γ =

Γ = −

Γ =

Note: Γ 's are obtained with mass-orthonormal mode shape using units of feet.



Mode 1 Mode 2 Total

02

46

810

12−

0.4

−0.3

−0.2

−0.1 0

0.1

0.2

0.3

0.4

t [s]

Interstory 1 Drift [%]


6/1/2009 12:58 PM C:\calpoly\arce412\homework\spring_2009\hw_11_sol.doc - 1 -


Homework 11-Solution

Calculate the response of the three-story frame above to the Loma Prieta ground motion. Select 5% damping. Response quantities of interest are: (1-3) The inter-story drift ratio in % for the three stories. Story heights are 12 ft for stories 2 and 3 and 15 ft for story 1. (4) The overturning moment in k-ft. Plot the response in modes 1 and 2 and the total response for 0 12 sect< < . Don't include the third mode (its contribution is virtually zero and the plots would become too busy). Distinguish the three lines by color or line-style. On your plot, clearly mark the maximum absolute value of the response in each mode as well as the maximum of the total response. Why is the maximum value of the sum (sum of all modes) not equal to the sum of the maximum modal values? Develop a table summarizing the maximum value of each result quantity in the first mode and the total ma-ximum value. Comment on the effects of the second mode on the total maximum. Which of the four response quantities is most affected by the second mode? Some help on MATAB implementation %Total structure response u = phi * q %Structure response (floor displacements) in individual modes u1 = phi(:,1) * q(1,:) u2 = phi(:,2) * q(2,:) u3 = phi(:,3) * q(3,:) %1st story drift in three modes and total d1 = u(1,:); d11 = u1(1,:); d12 = u2(1,:); d13 = u3(1,:); %2nd story drift in three modes and total

18 f

girders assumed rigid = story stiffnessk

1u

2u

3u

96.6 k

96.6 k

48.3 k

1500 k/ftk =

2625 k/ftk =

2625 k/ftk =



d2 = u(1,:) - u(2,:); d21 = u1(1,:)- u1(2,:); d22 = u2(1,:)-u2(2,:); d23 = u3(1,:)-u3(2,:); %don't forget to convert drift into drift ratio %3rd story drift in three modes and total your turn... %forces acting at floor level (three modes and total) F = K * u; F1 = K * u1; F2 = K * u2; F3 = K * u3; %overturning moment(three modes and total) your turn... Some Solutions:

1

2

3

1

2

3

0.500 sec

0.162 sec

0.111sec

2.6980

0.4606

0.09207

T

T

T

=

=

=

Γ =

Γ = −

Γ =

Note: Γ 's are obtained with mass-orthonormal mode shape using units of feet. Solution: Table 1: Modal and combined maximum values of four response quantities considered. Mode 1 Mode 2 Mode 3 Total DR 1 [%] 0.3341 0.0389 0.0012 0.3143DR 2 [%] 0.1630 0.0558 0.0048 0.2072DR 3 [%] 0.0578 0.0435 0.0071 0.0961OTM [k-ft] 1962 244 13 2116 Note: Mode 2 and Mode 3 columns not required in HW. In time history analysis we first calculate the sum of the individual modal time histories and then take the maximum of this summation. We thus calculate the maximum of the sum. Since the modal maximum values usually occur at different time, the maximum of the sum is commonly smaller than the sum of the maximum values. The sum of the maximum values gives us an overly conservative estimate of the maximum response. The maximum of the sum over all modes can also be smaller than the maximum in mode 1, since higher modes may contribute with different signs. The combined drift ratio in story 1 (0.3143), for example, is smaller than that in mode 1 (0.3341%) since the effects of modes 2 and 3 is negative. Higher modes increase the drift ratio in story 2 by about 27% (0.2072/0.1630=1.27) and the drift ratio in story 3 by about 66%. As already mentioned higher modes reduce the drift ration in story 1. Higher modes only have a small effect on the overturning moment, which is clearly dominated by mode 1 (2116/1962=1.08).



Mode 1 Mode 2 Total

02

46

810

12−

0.4

−0.3

−0.2

−0.1 0

0.1

0.2

0.3

0.4

t [s]




Mode 1 Mode 2 Total

02

46

810

12−

2500

−2000

−1500

−1000

−500 0

500

1000

1500

2000

2500

t [s]

Overturning Moment [k−ft]



ETABS model: A frequency analysis in ETABS yields the following results: MATLAB 0.4991 sec 0.1620 sec 0.1108 sec A time history analysis in ETABS yields the following results for the inter-story drift ratio (MATLAB results are from Table 1): The results are reasonably close to those from MATLAB.`

MATLAB0.000961

0.00207

0.00314

Spring Quarter 2009 Instructor: Ansgar Neuenhofer …aneuenho/teaching/arce412homework.pdf · ARCE...

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