Solved MDOF Example

7/28/2019 Solved MDOF Example

1/8

Earthquake Engineering Ahmed ElgamalMichael Fraser

Solved MDOF and Modal Analysis Example

1

3m

m

u2

u1

h1

h2

Ic

Ic

Elevation View

Use:

Esteel = 29,000 ksiIc = 164.8 in

4

h1 = 15 ft

h2 = 12 ft

in

skip

in

slb

sin

ftft

ft

lb

g

Wm

22

2209317.017.93

4.386

304030

=

=

==

Plan View

30 ft

40 ft

(a) For the two-story building shown above, define the two-degree-of-freedom free vibration matrix

equation in terms of k and m. Using this matrix equation, determine the natural frequencies 1 and2(in terms of k and m). Using these expressions for1 and2,plug in numerical values and determine and2 in radians. For each natural frequency, define and sketch the corresponding mode shape.

(b) Verify that the modes are orthogonal as expected.

(c) Normalize the first mode such that 1Tm1 = 1.0(d) Use the normalized first mode (from above) to verify that 1Tk1 = 12(e) Use the El Centro Response Spectrum and a damping ratio of 5% to estimate the maximum base shear

and moment.

(f) Find 0a and 1a in kmc 10 aa += for a viscous damping of 5% in modes 1 and 2.

3i

c

3i

c

floor3i

c

column h

48EI

h

EI

124kh

12EI

k =

=

= i


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2

Write the equation of motion in matrix form:

=

+

+

0

0

0

03

2

1

22

221

2

1

u

u

kk

kkk

u

u

m

m

&&

&&

substitute in u2=&& and rearrange to get

=

+0

03

2

1

2

22

2

2

21

u

u

mkk

kmkk

Take the determinant and set equal to zero

032

22

2

2

21 =+

mkkkmkk

04321

2

2

2

1

42 =+ kkmkmkm

solving for2 yields

m

kkkkkk

6

1644 22212

1212

1

++=

m

kkkkkk

6

1644 22212

1212

2

+++=

( )( )( )

inkipin

inksi

h

EIk c 335.39

180

8.1642900048483

4

3

1

1 ===

( )( )

( )inkip

in

inksi

h

EIk c 826.76

144

8.1642900048483

4

3

2

2 ===

in

skipm

2

09317.0

=

substituting in k1, k2, and m:


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3

2

22

1 95.101s

rad=

2

22

2 1138

s

rad=

s

rad097.101 =

s

rad734.332 =

Hzf 607.12

11 ==

Sec

fT 6223.0

1

1

1 ==

Hzf 369.5

2

22 ==

Sec

f

T 1863.01

2

2 ==

Determining 1st

Mode Shape:

Plug k1, k2, m, and1 = 10.097 rad/sec into

=

+

0

03

21

11

2

122

2

2

121

mkk

kmkk

( )( )( )( )

=

+0

0

09317.0097.10826.76826.76

826.7609317.03097.10826.76335.39

21

11

2

2

=

0

0

327.67826.76

826.76665.87

21

11

0826.76665.872111

=

let 121 = , then

( ) 01826.76665.87 11 =

876.0665.87

826.7611 ==

=

=000.1

876.0

21

11

1


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4

Determining 2nd

Mode Shape:

Plug k1, k2, m, and2 = 33.734 rad/sec into

=

+

0

03

22

12

2

222

2

2

221

mkk

kmkk

( )( )( )( )

=

+

0

0

09317.0734.33826.76826.76

826.7609317.03734.33826.76335.39

22

12

2

2

=

0

0

200.29826.76

826.76917.201

22

12

0826.76917.201 2212 =

let 122 = , then

( ) 01826.76917.201 12 =

380.0917.201

826.7612

=

=

=

=

000.1

380.0

22

12

2

0

5

10

15

20

25

30

0 0.5 1 1.5

h

eight(ft)

11 = 0.876

21 = 1.000

0

5

10

15

20

25

30

-0.5 0 0.5 1 1.5

height(ft)

22 = 1.000

12 = -0.380

1

stMode Shape 2

ndMode Shape


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5

(b) Verify that the modes are orthogonal as expected.

To verify that the modes are orthogonal, need to show that 0== jk TiTi , for ij

[ ]( ) [ ] 0

1

380.010317.9024485.0

1

380.0

09317.00

009317.031876.0 21 =

=

= xT 2

[ ] [ ] 01

380.05264.99310.24

1

380.0

826.76826.76

826.76826.76335.391876.01 =

=

+=T

Can also show 022 == 1k TT

(c) Normalize the first mode such that 1Tm1 = 1.0

[ ]( )

30766.01

876.0

09317.00

009317.031876.01 =

=T

Divide 1 by 30766.0 , therefore

=

=8029.1

5793.1

21

11

1

Check: [ ]( )

00.18029.1

5793.1

09317.00

009317.038029.15793.11 =

=T ok

(d) Use the normalized first mode (from above) to verify that 1Tk1 = 12

[ ] 212

2

2

1 097.10950.1018029.1

5793.1

826.76826.76

826.76826.76335.398029.15793.1 =

==

+=

s

rad

s

radT 2


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6

(e) Use the El Centro Response Spectrum and a damping ratio of 5% to estimate the maximum base shear and

moment.

T1

T2

From Response Spectrum:

For T1 = 0.622 sec and = 5%, D1 = 2.8 in. and A1 = 0.8g = 309.12 in/s2

For T2 = 0.0.186 sec and = 5%, D2 = 0.31 in. and A2 = 0.9g = 347.76 in/s2

jnjn

n

njn mA

M

Lf =


7/8



7

=

==2

1

2

2

1

j

jij

j

jij

i

i

m

m

M

L

( )( ) ( )( )( )( ) ( )( )

0987.1109317.0876.027951.0

109317.0876.027951.0222

212

2

111

212111

2

1

2

1

2

1

1

1

1 =+

+=

++

==

=

=

mm

mm

m

m

M

L

j

jj

j

jj

( )( ) ( )( )( )( ) ( )( )

0977.0109317.0380.027951.0

109317.0380.027951.0222

222

2

121

222121

2

1

2

2

2

1

2

2

2 =++

=++

==

=

=

mm

mm

m

m

M

L

j

jj

j

jj

( ) ( ) kipsin

skip

s

inmA

M

Lf 159.83876.027951.012.3090987.1

2

21111

1

111 =

==

( ) ( ) kipsin

skip

s

inmA

M

Lf 609.3380.027951.076.3470977.0

2

21212

2

212 =

==

( ) ( ) kipsin

skip

s

inmA

M

Lf 643.31109317.012.3090987.1

2

22121

1

121 =

==

( ) ( ) kipsin

skip

s

inmA

M

Lf 166.3109317.076.3470977.0

2

22222

2

222 =

==

Base Shear

=

=N

j

jnn fV1

0

kipskipskipsfffVj

j 80.114643.31159.832111

2

1

101 =+=+== =

kipskipskipsfffVj

j 443.0166.3609.32212

2

1

202 ==+== =

Calculate Maximum Base Shear, V0 max


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8

( ) ( ) ( ) ( ) kipskipskipsVVV 80.114443.080.114 222022

01max0 =+=+=

Base Moment

j

N

j

jnn dfM =

=1

0

( ) ( ) inkipsinkipsinkipsdfdfdfM jj

j =+=+== =

0.25221324643.31180159.83221111

2

1

101

( ) ( ) inkipsinkipsinkipsdfdfdfM jj

j ==+== =

16.376324166.3180609.3222112

2

1

202

Calculate Maximum Base Moment, M0 max

( ) ( ) ( ) ( ) ftkipsinkipsinkipsinkipsMMM ==+=+= 0.21028.2522316.3760.25221 222022

01max0

(f) Find 0a and 1a in kmc 10 aa += for a viscous damping of 5% in modes 1 and 2.

1 = 2 = = 0.05

( )( )( )

sradsradsrad

sradsrada

ji

ji777.0

734.33097.10

734.33097.10205.0

20 =+

=+

=

( )sradsradsrad

aji

100228.0

734.33097.10

205.0

21 =+

=+

=

srada 777.00 = & radsa 00228.01 =

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Solved MDOF Example