1

CE573 – Structural Dynamics Homework #12

1.

2

3

4

5

Note that one can use the Lagrange method to derive the equations of motion.

6

2. The fundamental modal properties, namely, the approximate fundamental modeshape and corresponding

natural frequency, of each structure are estimated using the Rayleigh’s Quotient approach. The initial guess vectors for the modeshapes are estimated using deflections that would be assumed by each structure under mass proportional lateral loading (as in, under unit lateral gravity field). One round of iteration to improve the guess modeshape is made for each structure. In the following pages, the Rayleigh’s Quotient results and the iteration process are given in tabular form. Below, the modeshapes are graphed to illustrate the differences.

Graphing the modeshapes, normalized to give unit displacement at roof, we see that in the case of the “soft ground story” structure, when the structure is responding in its fundamental mode, the largest story drift (i.e. story distortion) takes place at the ground story (65% of total displacement). When the “soft story” structure responds in its first mode, the displacements will concentrate at the soft-story. In the case of “soft third story”, 35% of the total displacement is taken by the soft third story.

Note that the “soft ground story” structure has the lowest fundamental frequency. The softening of the third story caused little change in the fundamental frequency.

0

1

2

3

4

5

0.00 0.20 0.40 0.60 0.80 1.00

Flo

or l

evel

Displacement

Modeshape (normalized at roof level to 1.0)

Identical k w/ k/4 ground story w/ k/2 3rd story

Iden

tical

k

Initi

al G

uess

Upd

atin

gN

orm

aliz

ed m

odes

hape

P.E

._m

axK

.E._

max

Upd

ated

roof

dis

p=1

Floo

r Lev

elFl

oor D

isp

Sto

ry D

isto

rtion

Sto

ry S

tiffn

ess

Stra

in E

nerg

yFl

oor M

ass

Floo

r Vel

Kin

etic

Ene

rgy

Floo

r Acc

elFl

oor F

orce

Sto

ry S

hear

Sto

ry D

isto

rtFl

oor D

isp

515

115

112.

515

1519

01.

001

10.

515

154

141

1498

1414

175

0.92

21

229

293

121

1272

1212

146

0.77

31

4.5

4141

29

19

40.5

99

105

0.55

41

850

501

51

512

.55

555

0.29

51

12.5

5555

00

00

00

0.00

27.5

335.

5

mul

tiplie

rk

km

wm

* w

^2w

^2w

^2 *

mw

^2 *

mw

^2 *

m /

kw

^2 *

m /

k

P.E

._m

ax =

K.E

._m

ax

w^2

=0.

0819

672

k/m

w =

0.28

6*

sqrt(

k/m

)

Itera

tion

1N

orm

aliz

ed m

odes

hape

P.E

._m

axK

.E._

max

Upd

ated

roof

dis

p=1

Floo

r Lev

elFl

oor D

isp

Sto

ry D

isto

rtion

Sto

ry S

tiffn

ess

Stra

in E

nerg

yFl

oor M

ass

Floo

r Vel

Kin

etic

Ene

rgy

Floo

r Acc

elFl

oor F

orce

Sto

ry S

hear

Sto

ry D

isto

rtFl

oor D

isp

519

01

190

1805

019

019

023

531.

0015

111

2.5

190

190

417

51

175

1531

2.5

175

175

2163

0.92

291

420.

536

536

53

146

114

610

658

146

146

1798

0.76

411

840.

551

151

12

105

110

555

12.5

105

105

1287

0.55

501

1250

616

616

155

155

1512

.555

5567

10.

2955

115

12.5

671

671

00

00

00

0.00

4136

5104

5.5

mul

tiplie

rk

km

wm

* w

^2w

^2w

^2 *

mw

^2 *

mw

^2 *

m /

kw

^2 *

m /

k

P.E

._m

ax =

K.E

._m

ax

w^2

=0.

0810

258

k/m

w =

0.28

5*

sqrt(

k/m

)

Gro

und

(firs

t) st

ory

k/4

Initi

al G

uess

Upd

atin

gN

orm

aliz

ed m

odes

hape

P.E

._m

axK

.E._

max

Upd

ated

roof

dis

p=1

Floo

r Lev

elFl

oor D

isp

Sto

ry D

isto

rtion

Sto

ry S

tiffn

ess

Stra

in E

nerg

yFl

oor M

ass

Floo

r Vel

Kin

etic

Ene

rgy

Floo

r Acc

elFl

oor F

orce

Sto

ry S

hear

Sto

ry D

isto

rtFl

oor D

isp

530

130

450

3030

805.

01.

001

10.

530

30.0

429

129

420.

529

2977

5.0

0.96

21

259

59.0

327

127

364.

527

2771

6.0

0.89

31

4.5

8686

.02

241

2428

824

2463

0.0

0.78

41

811

011

0.0

120

120

200

2020

520.

00.

6520

0.25

5013

052

0.0

00

00

00

0.00

6517

23

mul

tiplie

rk

km

wm

* w

^2w

^2w

^2 *

mw

^2 *

mw

^2 *

m /

kw

^2 *

m /

k

P.E

._m

ax =

K.E

._m

ax

w^2

=0.

0377

249

k/m

w =

0.19

4*

sqrt(

k/m

)

Itera

tion

1N

orm

aliz

ed m

odes

hape

P.E

._m

axK

.E._

max

Upd

ated

roof

dis

p=1

Floo

r Lev

elFl

oor D

isp

Sto

ry D

isto

rtion

Sto

ry S

tiffn

ess

Stra

in E

nerg

yFl

oor M

ass

Floo

r Vel

Kin

etic

Ene

rgy

Floo

r Acc

elFl

oor F

orce

Sto

ry S

hear

Sto

ry D

isto

rtFl

oor D

isp

580

5.0

180

5.0

3240

12.5

805

805

2139

11

30.0

145

0.00

805

805

477

5.0

177

5.0

3003

12.5

775

775

2058

60.

9623

6759

.01

1740

.50

1580

1580

371

6.0

171

6.0

2563

28.0

716

716

1900

60.

8885

0586

.01

3698

.00

2296

2296

263

0.0

163

0.0

1984

50.0

630

630

1671

00.

7811

711

0.0

160

50.0

029

2629

261

520.

01

520.

013

5200

.052

052

013

784

0.64

4383

520.

00.

2533

800.

0034

4613

784

00

00

00

045

738.

5012

1430

3

mul

tiplie

rk

km

wm

* w

^2w

^2w

^2 *

mw

^2 *

mw

^2 *

m /

kw

^2 *

m /

k

P.E

._m

ax =

K.E

._m

ax

w^2

=0.

0376

665

k/m

w =

0.19

4*

sqrt(

k/m

)

3rd

stor

y k/

2

Initi

al G

uess

Upd

atin

gN

orm

aliz

ed m

odes

hape

P.E

._m

axK

.E._

max

Upd

ated

roof

dis

p=1

Floo

r Lev

elFl

oor D

isp

Sto

ry D

isto

rtion

Sto

ry S

tiffn

ess

Stra

in E

nerg

yFl

oor M

ass

Floo

r Vel

Kin

etic

Ene

rgy

Floo

r Acc

elFl

oor F

orce

Sto

ry S

hear

Sto

ry D

isto

rtFl

oor D

isp

518

118

162

1818

276.

01.

001

10.

518

18.0

417

117

144.

517

1725

8.0

0.93

21

235

350

21

235

35.0

315

115

112.

515

1522

3.0

0.81

60.

59

5010

0.0

29

19

40.5

99

123.

00.

454

18

5959

.01

51

512

.55

564

.00.

235

112

.564

64.0

00

00

00

0.00

3247

2

mul

tiplie

rk

km

wm

* w

^2w

^2w

^2 *

mw

^2 *

mw

^2 *

m /

kw

^2 *

m /

k

P.E

._m

ax =

K.E

._m

ax

w^2

=0.

0677

966

k/m

w =

0.26

0*

sqrt(

k/m

)

Itera

tion

1N

orm

aliz

edm

odes

hape

Itera

tion

1N

orm

aliz

ed m

odes

hape

P.E

._m

axK

.E._

max

Upd

ated

roof

dis

p=1

Floo

r Lev

elFl

oor D

isp

Sto

ry D

isto

rtion

Sto

ry S

tiffn

ess

Stra

in E

nerg

yFl

oor M

ass

Floo

r Vel

Kin

etic

Ene

rgy

Floo

r Acc

elFl

oor F

orce

Sto

ry S

hear

Sto

ry D

isto

rtFl

oor D

isp

527

6.0

127

6.0

3808

8.0

276

276

4148

118

.01

162.

0027

627

64

258.

01

258.

033

282.

025

825

838

720.

9334

6235

.01

612.

5053

453

43

223.

01

223.

024

864.

522

322

333

380.

8047

2510

0.0

0.5

2500

.00

757

1514

212

3.0

112

3.0

7564

.512

312

318

240.

4397

359

.01

1740

.50

880

880

164

.01

64.0

2048

.064

6494

40.

2275

864

.01

2048

.00

944

944

00

00

00

070

63.0

010

5847

mul

tiplie

rk

km

wm

* w

^2w

^2w

^2 *

mw

^2 *

mw

^2 *

m /

kw

^2 *

m /

k

PE

max

=K

Em

axP

.E._

max

= K

.E._

max

w^2

=0.

0667

284

k/m

w =

0.25

8*

sqrt(

k/m

)

10

3. Using modal decomposition approach, we can convert the system of equations of motion for our 2-DOF structure into

( ) ( ) ( ) 1 ( ) 1,2T T T T

i i i i i i i i i i gM q t C q t K q t M x t i

or simply,

2( ) 2 ( ) ( ) ( ) 1, 2i i i i i i i gq t w q t w q t x t i

where

1

T

i

i T

i i

M

M

is modal participation factor for mode i.

Before proceeding, let’s find the modal properties of the structure. It has two natural modes with the following properties:

1 1 11

2 2 22

1.020.62 ; ;

1.62

1.021.62 ; ;

0.62

kw T

m w

kw T

m w

(Note that I have chosen the displacement of the first story as my first degree of freedom and set its value to 1.0 in the modeshapes.)

Using the above expressions we can see that if 1 0.3 secT , 2

0.620.3 0.11sec

1.62T .

These correspond to 1 20.9 rad / secw and 2 54.7 rad / secw . Also, we are assuming 2% of the critical

damping in each mode. Last but not least, 1 0.72 and 1 0.28 . (Aside: modal participation factors need

not add up to 1.0 even though in this solution they do.) Writing the equations of motion for each mode using the above information:

21 1 1

22 2 2

( ) 2 (0.02) (20.9) ( ) (20.9) ( ) 0.72 ( )

( ) 2 (0.02) (54.7) ( ) (54.7) ( ) 0.28 ( )

g

g

q t q t q t x t

q t q t q t x t

Given ( )gx t , we can compute the 1 2( ), ( )q t q t numerically, say using one of Newmark-β methods. We can re-

compose the displacements in our initial coordinate system using modeshapes and the principal coordinates

1 2( ), ( )q t q t as

1 1 11 1 2 2 1 2

2 1 2

( ) ( ) ( )1.0 1.0( ) ( ) ( ) ( )

( ) 1.62 ( ) 0.62 ( )1.62 0.62

x t q t q tq t q t q t q t

x t q t q t

11

Response to Sylmar record:

1( )q t

The inset shows the 4~7 sec range. Note the dominant frequency of vibration.

q1(t)

-6.0

-5.0

-4.0

-3.0

-2.0

-1.0

0.0

1.0

2.0

3.0

4.0

5.0

0 5 10 15 20 25 30

Time (sec)

Dis

pla

ce

me

nt

2 ( )q t

The inset shows the 3~7 sec range. Note the dominant higher frequency of vibration.

q2(t)

-0.15

-0.10

-0.05

0.00

0.05

0.10

0.15

0 5 10 15 20 25 30

Time (sec)

Dis

pla

ce

me

nt

The absolute maximum value of q1(t) is 4.92 cm and occurs at 4.24 sec. The absolute maximum value of q2(t) is 0.13 cm occurs at 4.20 sec. Typically, the peak values do not occur so close in time.

-6.0

-5.0

-4.0

-3.0

-2.0

-1.0

0.0

1.0

2.0

3.0

4.0

5.0

4.0 4.5 5.0 5.5 6.0 6.5 7.0

Time (sec)

Dis

pla

ce

me

nt

-0.15

-0.10

-0.05

0.00

0.05

0.10

0.15

4.0 4.5 5.0 5.5 6.0 6.5 7.0

Time (sec)

Dis

pla

cem

ent

12

Using modal combination as formulated earlier, we can find the first floor displacement as

First Floor Displacement, x1(t)

-6.0

-5.0

-4.0

-3.0

-2.0

-1.0

0.0

1.0

2.0

3.0

4.0

5.0

0 5 10 15 20 25 30

time (sec)

dis

pla

cem

ent

(cm

)

and the roof displacement as

Roof Displacement, x2(t)

-10.0

-8.0

-6.0

-4.0

-2.0

0.0

2.0

4.0

6.0

8.0

0 5 10 15 20 25 30

time (sec)

dis

pla

cem

ent

(cm

)

The floor displacements vary in the range of -4.9 to 4.1 cm at the first floor and -7.9 and 6.4 cm at the roof level. (So maximum absolute displacements are 4.9 cm and 7.9 cm, respectively.)

13

Using displacement response spectra for the ground motion, we can find the spectral displacements at 0.3 sec and 0.11 sec, the two natural periods, with 2% of the critical damping.

Displacement response spectra, Sylmar, 1994 Northridge Eq

0

10

20

30

40

50

60

70

80

90

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

Period (sec)

SD

(cm

)

2% damp.

Zooming into period range of interest:

Displacement response spectra, Sylmar, 1994 Northridge Eq

0

2

4

6

8

10

12

14

16

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40

Period (sec)

SD

(m

)

2% damp.

(0.3sec;2%) ~ 7 cmSD and (0.11sec;2%) ~ 0.4 cmSD which give

max1 1 1 1

max2 2 2 2

( ; ) 0.72 7 ~ 5 cm

( ; ) 0.28 0.4 ~ 0.1 cm

q SD T

q SD T

which, of course, agree with the absolute maxima of q1(t) and q2(t). Displacement values in each mode are

max max11 11 1

max max12 12 2

1.0 5 5 cm

1.0 0.1 0.1 cm

x q

x q

max max21 21 1

max max22 22 2

1.62 5 8 cm

0.62 0.1 0.06 cm

x q

x q

Using the SRSS method to combine modal contributions, we can estimate the maximum displacements as

2 2max max max1 11 12 5 cmx x x 2 2max max max

2 21 22 8 cmx x x

Aside: contribution of the second mode is minimal in both cases. Note that the maximum displacement estimates obtained through response spectrum approach agree very well with the maxima obtained using so-called time-history approach.