Response of MDOF structures to ground motion 1. If damping is well-behaving, or can be approximated...

Response of MDOF structures to ground motion

1

( ) ( ) ( ) 1 ( )gM x t C x t K x t M x t && & &&%% % %

If damping is well-behaving, or can be approximated using equivalent viscous damping, we can decouple the equations of motion using modal decomposition:

1

21 2

( )

( )( ) ( )

( )

N

N

q t

q tx t q t

q t

LM% % % % %

and separate the system into its natural modes.

2

( ) ( ) ( ) 1 ( )gM x t C x t K x t M x t && & &&%% % %

becomes

( ) ( ) ( ) 1 ( )

1,2, ,

T 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 N

&& & &&%% % % % % % %

L

or when normalized with respect to modal mass T

i iM % %

2( ) 2 ( ) ( ) ( ) 1,2, ,i i i i i i i gq t w q t w q t x t i N && & && L

3

2( ) 2 ( ) ( ) ( ) 1,2, ,i i i i i i i gq t w q t w q t x t i N && & && L

where , called modal participation factor for mode i.

1

T

i

i T

i i

M

M

%%

% %

1

2

1

1

N

Tj ji

i ji T N

i i j jij

mM

M m

%%

% %

4

2( ) 2 ( ) ( ) ( ) 1,2, ,i i i i i i i gq t w q t w q t x t i N && & && L

For a lightly damped (underdamped) system that is initially at rest, solution can be found using the convolution/Duhamel’s integral:

( ),

, 0

( ) ( ) sin ( )i i

tw ti

i g d id i

q t x e w t dw

&&

Once you have you can find the contribution of the i-th mode to the response of the structure.

( )iq t

5

( ) ( ) 1, 2, ,ji ji ix t q t j N K

Contribution of the i-th mode to the displacement at the j-th floor:( )jx t

Interstory drift, i.e. story distortion, in story j is given by the difference of displacements of the floor above and floor below:

( 1)( ) ( )ji ji j it x t x

Using the modal response, we can find various response values in each mode.

6

To find internal forces (story shears, moments, etc.) associated with deformations convenient, we can introduce the concept of equivalent static lateral forces.

Equivalent static lateral forces are external forces F which, if applied as static forces, would cause structural displacements x at given time instant.

At any instant of time, the equivalent lateral forces associated with displacements due to contribution by mode i :

1

2

( )

( )( )

( )

( ) ( )

i

ii

Ni

i i

F t

F tF t

F t

F t K x t

% M

% %7

2 21 1 1 11

2 22 2 2 2 2

2 2

( ) ( )( )

( ) ( ) ( )( )

( ) ( ) ( )

i i i i ii

i i i i i ii

Ni i N Ni i i N Ni

w m q t w m x tF t

F t w m q t w m x tF t

F t w m q t w m x t

% M M M

Similarly, we can use inertial forces to find the equivalent lateral forces,

the velocity term is at least an order of magnitude smaller than the displacement term, and as such, neglected.

2

2

( ) ( ) ( ) 1 ( )

( ) ( )

( 2 ( ) ( ))

( )

absi i i i i g

i i i g

i i i i i i

i i i

F t M x t M q t x t

M q t x t

M wq t w q t

w M q t

&& && &&% %% %

&& &&%

&%

%

8

for underdamped structures, the equivalent static lateral force at the j-th floor can be found from

Internal forces can be determined by static analysis of the structure loaded by the equivalent static lateral forces.

( )2,

, 0

( ) ( ) sin ( )i i

tw ti

ji i j ji g d id i

F t w m x e w t dw

&&

( ),

, 0

( ) ( ) sin ( )i i

tw ti

i g d id i

q t x e w t dw

&&As

9

Story shear at j-th story due to response in i-th mode may be calculated by summing the modal inertial forces above and at story j :

( ) ( )N

ji kik j

V t F t

Total shear force at the foundation level (“base shear”) due to response in i-th mode:

01

( ) ( )N

i jij

V t F t

Total overturning moment at the foundation level (“base overturning moment”) due to response in i-th mode:

01

( ) ( )N

i ji jj

M t F t h

jh : elev. of story j above the base

10

We can write the base shear for i-th mode as

20

1 1 1

( ) ( ) ( )N N N

i ji j ji i j jiej j j

V t F t m x t w m x

&&

( )20 ,

1 , 0

2( )

0 ,, 0

2( )

0 ,, 0

( ) ( ) sin ( )

( ) 1 ( ) sin ( )

1( ) 1 ( ) sin ( )

i i

i i

i i

tNw ti

i i j ji g d ij d i

tT w ti

i i i g d id i

Tt

iT w tii i g d iT

d ii i

V t w m x e w t dw

wV t M x e w t d

w

M wV t M x e w t d

wM

&&

&&% %

% &&%% %

% %

1 1 1T TT T

i i iM M M % % %% % %

But

11

2

2( )

0 ,, 0

2

21 ( )

0 ,2

, 0

1

1( ) ( ) sin ( )

( ) ( ) sin ( )

i i

i i

Tti

w tii g d iT

d ii i

N

ji j tj w ti

i g d iNd i

j jij

M wV t x e w t d

wM

mw

V t x e w t dwm

%% &&

% %

&&

The term

2

1

2

1

N

ji jj

N

j jij

m

m

is called the “effective modal mass” of mode i.

12

The overturning base moment for i-th mode could be written as

2

( )0 ,

1 , 0

1( ) ( ) sin ( )i i

TtN

i w tii j j ji g d iT

j d ii i

M wM t h m x e w t d

wM

% &&%

% %

01

( ) ( )N

i ji jj

M t F t h

13

1 10,

0

1 1

( )( )

( ) ( )

N N

ji j j j jij ji

equiv i N Ni

ji j jij j

F t h h mM t

hV t F t m

ii

i

mode mode

mode

Base Overturning MomentEquivalent Height =

Base Shear

,equiv ih

14

2

1

2

1

N

ji jj

N

j jij

m

m

“Effective modal mass” of mode i =

10,

0

1

( )

( )

N

j j jiji

equiv i Ni

j jij

h mM t

hV t m

15

Building Mode 1 Mode 2 Mode 3 Mode 4 Mode 5 Mode 6 Mode 7 Mode 8 Mode 9 Mode 10

Effective Modal Mass 1.00Equivalent Height 1.00

Effective Modal Mass 0.95 0.05Equivalent Height 0.81 -0.31

Effective Modal Mass 0.91 0.07 0.01Equivalent Height 0.75 -0.27 0.19

Effective Modal Mass 0.89 0.08 0.02 0.00Equivalent Height 0.72 -0.25 0.16 -0.13

Effective Modal Mass 0.88 0.09 0.02 0.01 0.00Equivalent Height 0.70 -0.24 0.15 -0.12 0.10

Effective Modal Mass 0.87 0.09 0.03 0.01 0.00 0.00Equivalent Height 0.69 -0.24 0.15 -0.11 0.09 -0.09

Effective Modal Mass 0.86 0.09 0.03 0.01 0.01 0.00 0.00Equivalent Height 0.68 -0.23 0.14 -0.11 0.09 -0.08 0.07

Effective Modal Mass 0.86 0.09 0.03 0.01 0.01 0.00 0.00 0.00Equivalent Height 0.68 -0.23 0.14 -0.10 0.08 -0.07 0.07 -0.06

Effective Modal Mass 0.85 0.09 0.03 0.01 0.01 0.00 0.00 0.00 0.00Equivalent Height 0.67 -0.23 0.14 -0.10 0.08 -0.07 0.06 -0.06 0.06

Effective Modal Mass 0.85 0.09 0.03 0.01 0.01 0.00 0.00 0.00 0.00 0.00Equivalent Height 0.67 -0.22 0.14 -0.10 0.08 -0.07 0.06 -0.06 0.05 -0.05

9-Story

10-Story

5-Story

6-Story

7-Story

8-Story

1-Story

2-Story

3-Story

4-Story

Identical mass, identical story stiffness building

16

The total response of the structure is obtained by combining the modal responses in all the modes of vibration.

The displacement at the j-th floor, the lateral force at the j-th floor, the base shear, and the base moment are given by

1

1

0 01

0 01

( ) ( )

( ) ( )

( ) ( )

( ) ( )

N

j jii

N

j jii

N

ii

N

ii

x t x t

F t F t

V t V t

M t M t

17

MODAL DECOMPOSITION APPROACH TO ANALYSE BASE-EXCITED STRUCTURES

The response of an idealized multistory building to earthquake ground motion can be computed by the following procedure:

1. Find the ground acceleration. 2. Define the structural properties:

a. Compute mass and stiffness distribution (i.e. [M] and [K]) b. Estimate modal damping ratios.

3. Find natural frequencies 1 2, , , Nw w w and modeshapes 1 2, , , N

of

vibration (i.e. solve the eigenvalue problem). 4. Compute the response of individual modes of vibration by repeating the following

steps for each mode: a. Compute the modal response ( )iq t by numerical evaluation of the

Duhamel integral. b. Compute the floor displacements. c. Compute story drifts from the floor displacements. d. Compute the equivalent lateral forces. e. Compute internal forces –story shears and moments– by static analysis of

the structure subject to the equivalent forces. 5. Determine the total value of any response quantity by combining the modal

contributions that response quantity.

18

Examples

19

T1=1.2 sec

T2=0.7 sec

T3=0.4 sec

Total wt=900 kip

20

21

22

23

El Centro 5/18/1940 EQ ground motion – NS component. The length of the record is 30 sec.

24

25

26

Use the response records to compute inertia forces developed in the structure.

Ex: Inertial forces that develop in the structure during 1st mode response .

27

Distribution of the modal inertial forces follow the respective modeshape

28

Distribution of shear forces in the structure for the first three modes:

29

Modal base shear demand

30

31

Question: Is there an easier way to estimate maximum response?

YES!use response spectra

32

( ),

, ,0

( ) ( ) sin ( ) ( , , , )i i

tw ti i

i g d i i g i id i d i

q t x e w t d W t x ww w

&& &&

( ),

0

( , , , ) ( ) sin ( )i i

tw t

i g i i g d iW t x w x e w t d && &&

For lightly damped structures , so we can approximate ,d i iw w

( ) ( , , , )ii i g i i

i

q t W t x ww

&&

where

1 1

( ) ( ) ( ) ( , , , ) ( )N N

ii i i g i i i

i i i

x t q t t W t x w tw

&&% % %

For example, displacements are

33

Interested in the “maxima” – the absolute maximum quantities, such as peak displacement, peak interstory drift (story distortion), and such.

Ex: Maximum displacement of floor j.

First, we find the maximum story displacement for each story and in each mode.

Say, we want to find, the displacement of j-th story due to response in i-th mode.( )jix t

( ) ( )ji ji ix t q t

max maxmax ( ) max ( ) ( ) max ( , , , )iji ji ji i ji i ji i g i i

t t ti

x x t q t q t W t x ww

&&

1max ( , , , ) ( , | )i g i i i i gt

i

W t x w SD T xw

&& &&

The maximum displacement (relative to ground) of a single-degree-of-freedom system with period

iT and damping ratio i when excited with the given ground motion . gx&&

34

max ( , | )ji ji i i i gx SD T x &&

max maxmax ( ) max ( ) ( ) max ( , , , )iji ji ji i ji i ji i g i i

t t ti

x x t q t q t W t x ww

&&

How do we find total response?

35

1 2( ) ( ) ( ) ( )j j j jNx t x t x t x t L

max1 2max ( ) ( ) ( )j j j jN

tx x t x t x t L

1 2 1 2max ( ) ( ) ( ) max ( ) max ( ) max ( )j j jN j j jNt t t tx t x t x t x t x t x t L L

max max1 2

1 1

max ( ) max ( ) max ( ) max ( )N N

j j j jN ji jit t t t

i i

x x t x t x t x t x

L

2max max

1

N

j jii

x x

Square-root of Sum of Squares (SRSS) combination

Absolute Sum approach (Absolute combination)

36

( 1)( ) ( ) ( )ji ji j it x t x t

1

( ) ( )N

j jii

t t

2max max

1

( )N

j jii

t

CAUTION: When you want to combine the effects of the modes to estimate a reasonable value for the maximum of a response parameter (story displacement, interstory drift, story force, etc.), you need to find value of the response parameter for each mode and then combine using any of the combination rules.

Do not use already-combined response parameters (say, story displacement estimates that considered contributions from all modes) to estimate other response parameters (say, story forces); such an approach will result in erroneous estimates.

37

2

2( )

0 ,, 0

2

( ),

0

1( ) ( ) sin ( )

1( ) sin ( )

i i

i i

Tti

w tii g d iT

d ii i

Tti

w ti g d iT

i i

M wV t x e w t d

wM

Mw x e w t d

M

%% &&

% %

%% &&

% %

Base shear

2

max0 0

1max ( ) max ( , , , )

T

i

i i i i g i iTt ti i

MV V t w W t x w

M

%% &&

% %

max ( , , , ) ( , | )i i g i i i i gt

w W t x w PSA T x && &&

38

2

2max max0 0

1 1

effective modal mass ( , | )N N

i i i gii i

V V PSA T x

&&g

2

max0

max0

1( , | )

Effective modal mass ( , | )

T

i

i i i gT

i i

i i i i g

MV PSA T x

M

V PSA T x

%% &&

% %

&&

How do we find total base shear?

39

Example: 7-story building

40

The spectral displacement values at the first four periods of our 7-story structure are

1 1

2 2

3 3

4 4

0.68sec 4.43 in

0.23sec 0.44 in

0.14sec 0.15 in

0.11sec 0.07 in

d

d

d

d

T S

T S

T S

T S

giving

max1 1 1

max2

max3

max4

4.33 x 4.43 in=19.2 in

0.62 in

0.12 in

0.04 in

q SD

q

q

q

41

1 1

2 2

3 3

4 4

0.68sec 4.43 in

0.23sec 0.44 in

0.14sec 0.15 in

0.11sec 0.07 in

d

d

d

d

T S

T S

T S

T S

giving

max1 1 1

max2

max3

max4

4.3292 x 4.43 in=19.2 in

0.62 in

0.12 in

0.04 in

q SD

q

q

q

max max71 71 1

max max72 72 2

max max73 73 3

(3.1% of the first-mode response.)

(0.6% of the first-mode response.)

0.2914 x 19.2 in 5.6 in

0.17in

0.03 in

u q

u q

u q

Roof displacement

Note that these maxima match the maxima in the corresponding response histories.

42

43

max max max max max max max max7 71 72 73 74 75 76 77

5.59 0.17 0.03 0.008 0.0027 0.00092 0.000015 5.80 in

u u u u u u u u

2 2 2 2 2 2 2max max max max max max max max7 71 72 73 74 75 76 77

5.59 in

u u u u u u u u

Absolute Sum approach (Absolute combination)

Square-root of Sum of Squares (SRSS) combination

Roof total displacement estimate

44

45

Rule of thumb:

Maximum displacement at the roof is 1.2~1.5 times the spectral displacement of the fundamental mode.

More like 1.2 for frame and 1.5 for shearwall buildings.

46

Building Mode 1 Mode 2 Mode 3 Mode 4 Mode 5 Mode 6 Mode 7 Mode 8 Mode 9 Mode 10

Effective Modal Mass 1.00Equivalent Height 1.00Roof Disp to SD ratio 1.00

Effective Modal Mass 0.95 0.05Equivalent Height 0.81 -0.31Roof Disp to SD ratio 1.17 -0.17

Effective Modal Mass 0.91 0.07 0.01Equivalent Height 0.75 -0.27 0.19Roof Disp to SD ratio 1.22 -0.28 0.06

Effective Modal Mass 0.89 0.08 0.02 0.00Equivalent Height 0.72 -0.25 0.16 -0.13Roof Disp to SD ratio 1.24 -0.33 0.12 -0.03

Effective Modal Mass 0.88 0.09 0.02 0.01 0.00Equivalent Height 0.70 -0.24 0.15 -0.12 0.10Roof Disp to SD ratio 1.25 -0.36 0.16 -0.06 0.02

Effective Modal Mass 0.87 0.09 0.03 0.01 0.00 0.00Equivalent Height 0.69 -0.24 0.15 -0.11 0.09 -0.09Roof Disp to SD ratio 1.26 -0.38 0.18 -0.09 0.04 -0.01

Effective Modal Mass 0.86 0.09 0.03 0.01 0.01 0.00 0.00Equivalent Height 0.68 -0.23 0.14 -0.11 0.09 -0.08 0.07Roof Disp to SD ratio 1.26 -0.39 0.20 -0.11 0.06 -0.02 0.01

Effective Modal Mass 0.86 0.09 0.03 0.01 0.01 0.00 0.00 0.00Equivalent Height 0.68 -0.23 0.14 -0.10 0.08 -0.07 0.07 -0.06Roof Disp to SD ratio 1.26 -0.40 0.21 -0.12 0.07 -0.04 0.02 0.00

Effective Modal Mass 0.85 0.09 0.03 0.01 0.01 0.00 0.00 0.00 0.00Equivalent Height 0.67 -0.23 0.14 -0.10 0.08 -0.07 0.06 -0.06 0.06Roof Disp to SD ratio 1.27 -0.40 0.22 -0.13 0.08 -0.05 0.03 -0.01 0.00

Effective Modal Mass 0.85 0.09 0.03 0.01 0.01 0.00 0.00 0.00 0.00 0.00Equivalent Height 0.67 -0.22 0.14 -0.10 0.08 -0.07 0.06 -0.06 0.05 -0.05Roof Disp to SD ratio 1.27 -0.41 0.23 -0.14 0.09 -0.06 0.04 -0.02 0.01 0.00

1-Story

2-Story

3-Story

4-Story

9-Story

10-Story

5-Story

6-Story

7-Story

8-Story

mode mode

mode

Base Overturning MomentEquivalent Height =

Base Sheari

ii

47

Base Shear Force

48

max0 Effective Modal Mass ( , | )i i i giV PSA T x &&

49

Mode Effective Modal Mass(kip-sec2/ft)

Effective Weight(kip)

Spectral Pseudo-Acceleration, SA(g)

1 18.74 603.4 0.96

2 1.96 63.1 0.84

3 0.62 20.0 0.75

4 0.26 8.4 0.65

5 0.11 3.5 -

6 0.04 1.3 -

7 0.01 0.3 -

Total : 700 kip 50

max01

max02

max03

max04

18.74 0.96 579.3 kip

1.96 0.84 53.0 kip (9% of first-mode response)

0.62 0.75 15.0 kip (3% of first-mode response)

0.26 0.65 5.5 kip

V g

V g

V g

V g

(1% of first-mode response)

Absolute Sum approach (Absolute combination)

max0 658 kipV

Square-root of Sum of Squares (SRSS) combination

max0 584 kipV

51

Modal base shear demand

52

53