Module 9 (RESPONSE SPECTRUM MODAL ANALYSIS)

CE-409: Introduction to Structural Dynamics and Earthquake Engineering

MODULE 9

Prof. Dr. Akhtar Naeem Khan & Prof. Dr. Mohammad Javed [email protected] [email protected]

1

University of Engineering & Technology, Peshawar, Pakistan

RESPONSE SPECTRUM MODAL ANALYSIS

CE-409: MODULE 9 (Fall-2013) 2

Modal decoupling of the EOMs

It is already known that the equations of motion for a a MDOF with lumped mass system and undergoing only lateral displacement can be written as:

[ ]{ } [ ]{ } [ ]{ } { }p(t)ukucum =++


Let be the modal matrix (matrix of mode shapes) in which the nth column is the nth mode shape of vibration (i.e. each column represents a particular mode shape).

Recalling the results of Prob M8.2

[ ]Φ


[ ]

−−=

−−==Φ

445.0802.0000.1

000.1445.0804.0

802.0000.1446.0

3210.5550.8022.243

1.2470.4451.802

1.0001.001.000

φφφ

{ } [ ]{ }qu Φ=

{ } [ ]{ }qu Φ=

{ } [ ]{ }qu Φ=

Where {u} is displacement vector and {q} is the modal amplitude vector.

CE-409: MODULE 9 (Fall-2013)

tSinBtCosA)t( 1n1n ωω 111q +=0.446

0.804

1.00

*q1

0.804q1

0.446q1

0.50q1

0.402q1

0.223q1

a b c

0. 5q1q1

a b c d ? (ϕ & q)

ϕ1

d ?

u1=ϕ1*q1


Substituting values of from slide 3 in the:

[ ][ ]{ } [ ][ ]{ } [ ][ ]{ } { }p(t)qkqcqm =Φ+Φ+Φ


[ ]{ } [ ]{ } [ ]{ } { }p(t)ukucum =++ { } { } { }uuu &,

[ ]TΦ

[ ] [ ] [ ]{ } [ ] [ ] [ ]{ } [ ] [ ] [ ]{ } [ ] { } p(t)qkqcqm TTTT Φ=ΦΦ+ΦΦ+ΦΦ

[ ]{ } [ ]{ } [ ]{ } { }P(t)qKqCqM =++

Pre-multiply both sides by

Where = Modal mass matrix

= Modal stiffness matrix

= Modal damping matrix

= Modal (applied) forces vector

[ ]M

[ ]C[ ]K

{ })t(P



Because of orthogonally properties of mode shapes (i.e., each mode shape is independent of others) as shown below

0.445

0.802

1.00

1.00

0.445

0.802

0.446

0.804

1.000

First mode shape Second mode shape

Third mode shape

Mode shapes being normalized by taking greatest floor term taken as 1



Note that a matrix said to be orthogonal if

where [I] is an identity matrix in which diagonal terms are 1 and off

diagonal terms are 0 and therefore det [I]=1. are diagonal matrices (i.e., matrices in which off diagonal terms are zero)

[ ]A [ ] [ ] [ ]IAA T =

[ ] [ ] [ ] C and K , M

m1

m2

m3

k1

k2

k3

[ ]

=

3

2

1

m00

0m0

00m

m

m3

m2

m1


Modal decoupling of the EOMsSince are diagonal matrices so the N coupled equations replaces by N uncoupled equation for SDOF systems

[ ] [ ] [ ]K and C, M

(t)PqKqCqMnnnnnnn

=++

Where Mn= Generalized mass for the nth natural mode

Kn= Generalized stiffnes for the nth natural mode

Cn= Generalized damping for the nth natural mode

Pn(t)= Generalized force for the nth natural mode



The equation given on previous slide is for nth mode of MDOF of order N. All the independent equations for N modes in matrix form can be written as

Where M= Diagonal matrix of the generalized modal masses K= Diagonal matrix of the generalized modal stiffnesses C= Diagonal matrix of the generalized modal dampings

P(t) = Column vector of the generalized modal forces Pn(t)

(t)PKqqCqM =++

CE-409: MODULE 9 (Fall-2013) 10

Modal analysis for earthquake forces

The uncoupled equations of motions for earthquake excitations can be written as

{ι}= Influence vector (refer slide 15 for details) of size Nx1. {ι} = {1}Nx1 for structures where the dynamic degrees of freedom are displacements in the same direction as the ground motion

[ ]{ } [ ]{ } [ ]{ } { } [ ] { })(tp(t)PqKqCqM effT

eff Φ==++

{ } [ ]{ } )( tum(t)pWheregeffι−=

CE-409: MODULE 9 (Fall-2013) 11

Modal analysis for earthquake forces

Replacing

For nth mode

[ ] [ ]{ } { }LmT =Φ ι

[ ]{ } [ ]{ } [ ]{ } { } { } )(tuL(t)PqKqCqM geff −==++⇒

)(tuLqKqCqM gnnnnnnn −=++

)(tuM

Lq

M

Kq

M

M2q

M

Mg

n

nn

n

nn

n

nnnn

n

n −=++ ωζ

)( 2 tuM

Lqq2qor g

n

nnnnnnn −=++ ωωζ

L and m has same units

CE-409: MODULE 9 (Fall-2013) 12

Modal Participation factors

The term Ln/Mn has been given the name of participation factor for

the nth mode and is represented by Γn (capital Greek alphabet for

Gamma)

Γn is usually considered a measure of the degree to which the nth

mode participates in the response. This terminology is misleading, however, because Γn is not independent of how the mode is

normalized, nor a measure of the modal contribution to a response quantity.

The magnitude of the participation factor is dependent on the normalization method used for the mode shapes.

{ } [ ]{ }{ } [ ] { }n

Tn

Tn

n

nn

m

m

M

L

φφιφ

==Γ

CE-409: MODULE 9 (Fall-2013) 13

Participation factors

Once the modal amplitudes {q} have been found the

displacements of the structure are obtained from

The above mentioned equation to determine modal displacements cancel out the effect of normalization carried out to calculate q (slide 11).

The displacements associated with the nth mode are given by

{ } [ ]{ }qu Φ=

{ } [ ]{ }(t)q(t)u nnn φ =

CE-409: MODULE 9 (Fall-2013) 14

Effective weight of structure in nth mode, Wn

Effective weight of structure in nth mode=

It shall be noted that the sum of the all effective weights for an

excitation in a given direction ( i.e. for a given {ι}) should equal the total weight of the structure. Note, this may not be the case where rotational inertia terms also exist in the mass matrix.

Many building codes require that a sufficient number of modes be used in the analyses such that the sum of the effective weights is at least 90% of the weight of the structure. This provides a measure on the number of modes required in the analysis.

{ } [ ]{ } { } [ ]{ }{ } [ ] { }

( ) gM

L)g L(gL

M

Lg

m mW

n

2n

nnnn

n

nT

n

Tn

Tn

n =Γ===φφ

ιφιφm

CE-409: MODULE 9 (Fall-2013) 15

{ }

=1

1

1

ι

Influence (Direction) vector{ }ι{ι} = Influence vector= {1}Nx1 for structures where the dynamic degrees of freedom are displacements in the same direction as the ground motion

Direction of EQ is horizontal

u1

u2

u3

Direction of EQ is horizontal

u1

u2

u3

{ }

=0

1

1

ι

CE-409: MODULE 9 (Fall-2013) 16

Base Shear Force in the structure in the nth node, Vbn

The base shear in nth mode can be determined using relation

Where Wn = Effective weight of structures in nth mode

=

=

g

Ag

M

L

g

AWV

n

2n

nbnnn

CE-409: MODULE 9 (Fall-2013) 17

Distribution of nodal (joint) forces in the structure from the base shear

In many design codes the first step is to compute the modal base shear force and this is then distributed along the structures (shown on next slide) to each degrees of freedom.

The distributed loads are assumed to give the same displacements in the structure as those generated by the exciting base shear.

{ } [ ]{ }nn

bnn m

L

Vf φ=

CE-409: MODULE 9 (Fall-2013) 18

≡

Base shear acting in nth mode, Vbn

Vbn distributed along the structure in nth mode

∑=++= nnnnbn ffffV 321

Distribution of nodal (joint) forces in the structure from the base shear

Vbn

m1n

m2n

m3n

f1n

f2n

f3n

CE-409: MODULE 9 (Fall-2013) 19

Response spectrum model analysis: ExampleA 3 story R.C. building as shown below is required to be designed for a design earthquake with PGA=0.3g, and its elastic design spectrum is given by Fig 6.9.5 multiplied by 0.3). Carry out the dynamic analysis by using the above mentioned design spectrum. Take:

Story height = 10ft

Total stiffness of each story = 250 kips/in.

Weight of each floor = 386.4 kipsm1

m2

m3

k1

k2

k3

CE-409: MODULE 9 (Fall-2013) 20

Mass and stiffness matrices

[ ]

=

100

010

001

m

m=W/g = (386.4k) /(396.4 in/sec2)= 1.0 kip-sec2/in

[ ]

−−−

−=

2502500

250500250

0250500

k

[ ] [ ]

−−−−−

−−=−

2

2

2

2

2502500

250500250

0250500

n

n

n

n mk

ωω

ωω

CE-409: MODULE 9 (Fall-2013) 21

[ ] [ ][ ] 0det 2 =− mkSetting nω yields following values

sec/ 49.28

sec/ 69.19

sec/ 04.7

3

2

1

rad

rad

rad

n

n

n

===

ωωω

sec 22.0

sec 32.0

sec 89.0

3

2

1

===

n

n

n

T

T

T

Natural frequencies

CE-409: MODULE 9 (Fall-2013) 22

0

2502500

250500250

0250500

31

21

11

21

21

21

=

−−−−−

−−⇒

φφφ

ωω

ω

n

n

n

Normalized coordinates of first mode shape

[ ] [ ][ ]{ } 0 12

1 =− φω mk n

Substituting 1 and 56.94 11

2

1== φω

n

Mode shapes

CE-409: MODULE 9 (Fall-2013) 23

0

1

44.2002500

25044.450250

025044.450

31

21=

−−−

−⇒

φφ

First row gives 80.1 025044.450 2121 =⇒=− φφ

24.2 0250)80.1(44.450250

025044.450250

3131

3121

=⇒=−+−⇒=−+−

φφφφSecond row gives

{ }

+++

=

=

=

=⇒1.00

0.80

0.45

2.24/2.24

1.80/2.24

1.00/2.24

31

21

11

12.24

1.80

1.00

φφφ

φ

Mode shapes

CE-409: MODULE 9 (Fall-2013) 24

0

7.3872502500

2507.387500250

02507.387500

32

22

12

=

−−−−−

−−⇒

φφφ

7.3872

2=

nω

0

1

3.1372500

2503.112250

02503.112

32

22=

−−−−

−⇒

φφ

80.0 &

45.0

32

22

−==⇒

φφ

−++

=

0.80

0.45

1.00

32

22

12

φφφ

Mode shapes

CE-409: MODULE 9 (Fall-2013) 25

+−+

=

+−+

=

0.45

1.00

0.80

Similarly

33

23

31

0.56

1.25

1.00

φφφ

[ ]

+−+

−++

=

+−+

−++

==Φ

0.45

1.00

0.80

0.80

0.45

1.00

1.00

0.80

0.45

3210.56

1.25

1.00

0.80

0.45

1.00

2.24

1.80

1.00

φφφ

Mode shapes

CE-409: MODULE 9 (Fall-2013) 26

Modal mass, Mn and participation factor,Γn

{ } [ ]{ }{ } [ ] { }n

Tn

Tn

n

nn

m

m

M

L

φφιφ

==Γ

{ } [ ]{ }{ } [ ] { }11

1

11

1

φφφ

m

m

M

LT

T1 ==Γ

For structure given in problem{ι}={1}

{ } [ ]{ }{ } [ ] { }n

Tn

Tn

n

nn

m

m

M

L

φφφ

1==Γ∴

CE-409: MODULE 9 (Fall-2013) 27

=1

1

1

100

010

001

00.1

80.0

45.0T

1L

=1

1

1

1.00 0.80 .4501L

/ftsec-kip 72/in.sec-kip 25.2L 22 ==1

{ } [ ]{ }

==1

1

1

100

010

001

1

31

21

11

1

T

T1 mL

φφφ

φ


CE-409: MODULE 9 (Fall-2013) 28

{ } [ ] { }111 φφ mM T=

=00.1

80.0

45.0

100

010

001

00.1

80.0

45.0T

1M

=00.1

80.0

45.0

1.00 0.80 .4501M

/ft sec-kip 22.1/in.sec-kip 84.1 22 ==1M


CE-409: MODULE 9 (Fall-2013) 29

22.184.1

25.2

11 ===Γ

M

L1

Similarly 36.084.1

65.0

22 ===Γ

M

L2

14.084.1

25.0

33 ===Γ

M

L3

Participation factor, Γn

CE-409: MODULE 9 (Fall-2013) 30

Effective weight of structure participating in nth mode, Wn

{ } [ ]{ }{ } [ ]{ }{ } [ ] { } g

M

Lg

m

mmW

n

n

nT

n

Tn

Tn

n

2

==

φφιφιφ

/insec-kip 4.386 2=g( )

kips 1.10634.386*84.1

25.2 2

1

21

1 === gM

LW

( )kips 7.884.386*

84.1

65.0 2

2

22

2 === gM

LW

( )kips 1.134.386*

84.1

25.0 2

3

23

3 === gM

LW

CE-409: MODULE 9 (Fall-2013) 31

Mass of the structure participating in nth mode , PMn

Participating mass of the structure in nth mode= W

WPM n

n =*

%7.91917.04.386*3

1.10631*1 ====

W

WPM

%7.7077.04.386*3

7.882*2 ====

W

WPM

%13.10113.04.386*3

1.133*3 ====

W

WPM

00.1PM =Σ⇒Most of the code requires that such number of modes shall be considered so that ΣPM≥ 0.9. In our case, indeed, the consideration of just the first mode would have been sufficient as PM1≥ 0.9

CE-409: MODULE 9 (Fall-2013) 32

=

=

gW

gg

M

LV n

nn

n

nbn

AA2

Base shear in nth mode, Vbn

Mode 1: Tn1=0.89 sec

kipsg

gTg

WVn

1b 0.6453.0*

1*

8.1*1.1063.

A

111 =

=

=

kipsg

g

gWV 2

b 1.723.0*71.2

*7.883.0*A

:sec 0.32TFor 22n2 ==

==

kipsg

g

gWV 3

b 7.103.0*71.2

*1.133.0*A

:sec 0.22TFor 33n3 ==

==

Values of A for each Tn can be determined from Fig. 6.9.5 given on next slide

CE-409: MODULE 9 (Fall-2013) 33

0.22 sec

0.32 sec

0.89 sec

sec 22.0

sec 32.0

sec 89.0

3

2

1

===

n

n

n

T

T

T

CE-409: MODULE 9 (Fall-2013) 34

Nodal forces acting on the structure in nth mode, fn

{ } [ ]{ }nn

bnn m

L

Vf φ=

[ ] [ ]

=

⇒

=

31

21

11

1

1

31

21

11

3

2

1

1

1

3

2

1

φφφ

φφφ

mL

V

f

f

f

mL

V

f

f

fb

n

n

n

b

n

n

n

First mode

=

=

7.286

3.229

0.129

00.1

80.0

45.0

1200

0120

0012

)12*25.2(

645

31

21

11

f

f

f

CE-409: MODULE 9 (Fall-2013) 35

{ } [ ]{ }nn

bnn m

L

Vf φ=

[ ]

−++

=

=

0.80

0.45

1.00

1200

0120

0012

)12*65.0(

1.72

32

22

12

2

2

32

22

12

φφφ

mL

V

f

f

fb

Second mode

−++

=

6.88

8.49

8.110

32

22

12

f

f

f


CE-409: MODULE 9 (Fall-2013) 36

{ } [ ]{ }nn

bnn m

L

Vf φ=

[ ]

+−+

=

=

0.45

1.00

0.80

1200

0120

0012

)12*25.0(

7.10

33

23

13

3

3

33

23

13

φφφ

mL

V

f

f

fb

Third mode

+−+

=

3.19

8.42

2.34

33

23

13

f

f

f


CE-409: MODULE 9 (Fall-2013) 37

{ } { }

+−+

−++

==3.19

8.42

2.34

6.88

8.49

8.110

7.286

3.229

0.129

321 nnnn ffff


CE-409: MODULE 9 (Fall-2013) 38

645.0 kips

Mode 1

286.7 k

229.3 k

129.0 k

i1

j1

10.7 kips

Mode 3

19.3 k

42.8 k

34.2 k

i3

j3

Mode 2

72.0 kips

88.6 k

49.8 k

110.8 k

i2

j2


CE-409: MODULE 9 (Fall-2013) 39

Combination of Modal MaximaThe use of response spectra techniques for multi-degree of freedom structures is complicated by the difficulty of combining the responses of each mode.

It is extremely unlikely that the maximum response of all the modes would occur at the same instant of time.

When one mode is reaching its peak response there is no way of knowing what another mode is doing.

The response spectra only provide the peak values of the response, the sign of the peak response and the time at which the peak response occurs is not known.

CE-409: MODULE 9 (Fall-2013) 40

Combination of Modal Maxima

Therefore

and, in general

The combinations are usually made using statistical methods.

{ } [ ]{ } maxmax qu Φ≠

{ } [ ]{ } maxmax qu Φ≤

CE-409: MODULE 9 (Fall-2013) 41

Combined Response ro

Let rn be the modal response quantity (base shear, nodal

displacement, inter-storey drift, member moment, column stress etc.) for mode n .The r values have been found for all modes (or for as many modes that are significant).

Most design codes do not require all modes to be used but many do require that the number of modes used is sufficient so that the sum of the Effective Weights of the modes reaches, say, 90% of the weight of the building. Checking the significance of the Participation Factors may be useful if computing deflections and rotations only.

CE-409: MODULE 9 (Fall-2013) 42

Absolute sum (ABSSUM) method

The maximum absolute response for any system response quantity is obtained by assuming that maximum response in each mode occurs at the same instant of time. Thus the maximum value of the response quantity is the sum of the maximum absolute value of the response associated with each mode. Therefore using ABSSUM method

This upper bound value is too conservative. Therefore, ABSSUM modal combination rules is not popular is structural design applications

∑=

≤N

nnoo rr

1

CE-409: MODULE 9 (Fall-2013) 43

Square-Root-of-the Sum-of-the-Squares (SRSS) method

The SRSS rule for modal combination, developed in E.Rosenblueth’s PhD thesis (1951) is

The most common combination method and is generally satisfactory for 2-dimensional analyses is the square root of the sum of the squares method. The method shall not be confused with the root-mean-square of statistical analysis as there is no denominator.

( ) 2/1

1

2∑≅=

N

nnoo

rr

CE-409: MODULE 9 (Fall-2013) 44

This method was very commonly used in design codes until about1980. Most design codes up to that time only considered the earthquake acting in one horizontal direction at a time and most

dynamic analyses were limited to 2-dimensional analyses.

Square-Root-of-the Sum-of-the-Squares (SRSS) method

CE-409: MODULE 9 (Fall-2013) 45

Three Dimensional Structures In three-dimensional structures, different modes of free-

vibration in different directions may have very similar natural frequencies.

If one of these modes is strongly excited by the earthquake at a

given instant of time then the other mode, with a very similar natural frequency, is also likely to be strongly excited at the same instant of time. These modes are often in orthogonal horizontal directions but there may be earthquake excitation directions where both modes are likely to be excited.

In these cases the Root-Sum-Square or SRSS combination method has been shown to give non-conservative results for the likely maximum response. In such cases some other methods such as CQC, DSC are used

CE-409: MODULE 9 (Fall-2013) 46

Modal combination of responses

Using SRSS method

( ) 23i2

2i2

1iji AAAAA −++==

j3

Aj3

Mj3

i3

Mi3

Ai3

Aj1

j1

Mj1

i1

Mi1

Ai1

Aj2

j2

Mj2

i2

Mi2

Ai2

( ) 23i2

2i2

1ii MMMM −++=

( )23j2

2j2

1jj MMMM −++=

Mode 1 Mode 2 Mode 3

Consider nodes i & j of the frame for which R.S.modal analysis was carried out on previous slides

CE-409: MODULE 9 (Fall-2013) 47

Caution

It must be stressed that what ever response item r that the analyst or designer requires it must be first computed in each mode before the modal combination is carried out.

If the longitudinal stress is required in a column in a frame, then the longitudinal stress which is derived from the axial force and bending moment in the column must be obtained for each mode then the desired combination method is used to get the maximum likely longitudinal stress.

It is NOT correct to compute the maximum likely axial force and the maximum likely bending moment for the column then use these axial forces and bending moments, after carrying out their modal combinations, to compute the longitudinal stress in the column.

CE-409: MODULE 9 (Fall-2013) 48

Home Assignment No. M9

A 3 story R.C. building as shown below is required to be designed for a design earthquake with PGA=0.25g, and its elastic design spectrum is given by Fig 6.9.5 (Chopra’s book) multiplied by 0.25). It is required to carry out the dynamic modal analysis by using the afore mentioned design spectrum . Take:

• Story height = 10ft

•Total stiffness of first 2 stories = 2000 kips/ft.

• Total stiffness of top floor = 1500 kips/ft

• Mass of first 2 floors = 5000 slugs

• Mass of top floor = 6000 slugs

m1

m2

m3

k1

k2

k3

Module 9 (RESPONSE SPECTRUM MODAL ANALYSIS)

Technology

n n n t

n n qn

modal decoupling

n modes

q slide

n n kn ln qn

modal participation

modal displacements