1 NAEIM 2009 Linear Dyamics

7/29/2019 1 NAEIM 2009 Linear Dyamics

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Farzad Naeim Structural Dynamics for Practicing Engineers 1 of 159

(Last Revision Date: 5-26-2009)

Linear Dynamic AnalysisLinear Dynamic Analysis

Farzad Naeim, Ph.D., S.E., Esq.Farzad Naeim, Ph.D., S.E., Esq.Vice President and General CounselVice President and General Counsel

John A. Martin & Associates, Inc.John A. Martin & Associates, Inc.



Module ContentsModule Contents

A.A. single degree of freedom (SDOF) dynamicssingle degree of freedom (SDOF) dynamics

B.B. response spectrum v. design spectrumresponse spectrum v. design spectrum

C.C. dampingdamping

D.D. multiple degree of freedom (MDOF) dynamicsmultiple degree of freedom (MDOF) dynamics

E.E. mode shapesmode shapes

F.F. mass participationmass participation

G.G. modal combinationsmodal combinationsH.H. earthquake response of SDOF and MDOF systemsearthquake response of SDOF and MDOF systems

I.I. using elastic design spectrum in SDOF and MDOF systemsusing elastic design spectrum in SDOF and MDOF systems


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References:References: Primary References:Primary References:

Naeim, F. (ed),Naeim, F. (ed), The Seismic Design HandbookThe Seismic Design Handbook, 2, 2ndnd Edition,Edition,

Chapters 2 and 4Chapters 2 and 4

FEMA 356FEMA 356

Secondary References:Secondary References: Chopra, A.,Chopra, A., Dynamics of StructuresDynamics of Structures, 2, 2ndnd Edition.Edition.

Chopra, A., EarthquakeChopra, A., Earthquake Dynamics of StructuresDynamics of StructuresA PrimerA Primer, 2, 2ndnd Edition, EERI,Edition, EERI,

2005.2005.

Paz, M. (ed.),Paz, M. (ed.), International Handbook of Earthquake Engineering,International Handbook of Earthquake Engineering, Chapter 2.Chapter 2.

Paz, M., Structural DynamicsPaz, M., Structural Dynamics

Naeim F. and Kelly, J.,Naeim F. and Kelly, J., Design of Seismic Isolated StructuresDesign of Seismic Isolated Structures, Chapter 7., Chapter 7.



Single-degree-of-freedom systemsubjected to time-dependent force.

Static Equilibrium:

Dynamic Equilibrium:

Three Simplifying Assumptions

for SDOF:1. Mass concentrated at the roof

2. Roof is Rigid

3. Axial Deformation of Columns

Neglected

kvp

)()()()( tvtktvctumtp

STATIC AND DYNAMIC EQUILIBRIUM


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Single-degree-of-freedom system

subjected to base motion.

)(tgmkvvcvm )(tgm

Response of a SDOF

system to earthquake

ground motion:

STATIC AND DYNAMIC EQUILIBRIUM



Rotating particle of mass

Mass Moment of

Inertia:

For a point mass

For a rigid body

dmI2

2mrI

ROTATIONAL PROPERTIES OF PARTICLES AND RIGID BODIES


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b

2

1

2

1

3

2

0

mll

lbtm

abtm 4

16

22

0

baml

2

2

0

mrl

2rtm lbtm

12

2

0

lml

Rigid-body mass and mass moment of inertia.

2

abtm

18

22

0

baml

12

22

0

baml

abtm

2

b

2

b

2

a

2

a

32b

3

b

3

a

2

2a




Example 1. (Determination of Mass

Properties)

Compute the mass and mass moment of

inertia for the rectangular plate

shown.

Translational mass: Rotational mass moment of inertia:



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SIMPLIFIED STIFFNESS PROPERTIES OFLATERAL FORCE RESISTING ELEMENTS



Free vibration occurs when a structure oscillates under theFree vibration occurs when a structure oscillates under the

action of forces that are inherent in the structure withoutaction of forces that are inherent in the structure without

any externally applied timeany externally applied time--dependent loads or grounddependent loads or ground

motions.motions.

These inherent forces arise from the initial velocity andThese inherent forces arise from the initial velocity and

displacement the structure has at the beginning of the freedisplacement the structure has at the beginning of the free--

vibration phase.vibration phase.

UNDAMPED STRUCTURES:UNDAMPED STRUCTURES:

FREE VIBRATION OF SDOF SYSTEMS

0)()( tkvtvm 0)()( 2 tvtvorormk/

2 wherewhere

tvtv

tv

cos)0(sin)0(

)(

Solution:Solution:


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Free-vibration response of an undamped SDOF system.

fk

mT

12

2

Natural vibration period of a SDOF system:


Several important concepts of oscillatory motion can be

illustrated with this result:1. The amplitude of vibration is constant, so that the

vibration would, theoretically, continue indefinitely

with time. This cannot physically be true, because free

oscillations tend to diminish with time, leading to the

concept of damping.

2. The time it takes a point on the curve to make one

complete cycle and return to its original position is

called the period of vibration, T. The quantity is the

circular frequency of vibration and is measured in

radians per second. The cyclic frequencyfis defined as

the reciprocal of the period and is measured in cycles

per second, or hertz.

3. These three vibration properties depend only on the

mass and stiffness of the structure and are related as

follows:



Damped Structures:Damped Structures:


wherewhere

This equation has the general solution:This equation has the general solution:

Percentage of Critical DampingPercentage of Critical Damping

0)()()( tkvtvctvm

tvt

vvetv dd

dt cos)0(sin

])0()0([)(

m

C

C

C

cr 2

andand

21 d Damped Circular FrequencyDamped Circular Frequency


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Example 2.Example 2.Construct an idealized SDOF

model for the industrial building

shown in the figure, and estimate

the period of vibration in the two

principal directions . Note that

vertical cross bracings are made of

1-inch-diameter rods, horizontal

cross bracing is at the bottom chord

of trusses, and all columns are

W8 X24.






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Free vibration response of a damped SDOF system.

m

C

C

C

cr 2

Percentage ofPercentage ofCriticalCritical

DampingDamping

2)1(

)(ln

iv

iv

2

21 d

FREE VIBRATION OF SDOF SYSTEMSLogarithmic Decrement



Some Recommended Damping Values


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ttk

gmtv

sinsin

1

12

0

2

1

1

1(DLF)FactorionAmplificatynamic

2

D

Undamped Damped

RESPONSE OF SDOF SYSTEMS TO HARMONIC LOADING

Undamped:



ResonanceResonance

responseresponse

RESPONSE OF SDOF SYSTEMS TO HARMONIC LOADING


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Bucharest (1977) ground acceleration.

RESPONSE OF SDOF SYSTEMS TO IMPULSE LOADING




The maximum response to an impulse load will generally beThe maximum response to an impulse load will generally beattained on the first cycle.attained on the first cycle.

For this reason, the damping forces do not have time to absorbFor this reason, the damping forces do not have time to absorbmuch energy from the structure.much energy from the structure.

Therefore, damping has a limited effect in controlling theTherefore, damping has a limited effect in controlling themaximum response and is usually neglected when consideringmaximum response and is usually neglected when consideringthe maximum response to impulse type loads.the maximum response to impulse type loads.

The rectangular pulse is a basic pulse shape. This pulse has aThe rectangular pulse is a basic pulse shape. This pulse has azero (instantaneous) rise time and a constant amplitude,zero (instantaneous) rise time and a constant amplitude, ppoo,,which is applied to the structure for a finite durationwhich is applied to the structure for a finite duration ttdd..

During the time period when the load is on the structure (During the time period when the load is on the structure ( t < tt < tdd))

the equation of motion has the form:the equation of motion has the form:

Thereafter:Thereafter:opkvvm

ttvttv

tvd

d

cos)(sin)(

)(

dttt


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Maximum elastic response, rectangular and triangular load pulses.




Maximum

elasto-plastic

response, rectangular

load pulse.



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EXAMPLE: ANALYSIS FOR IMPULSE BASEACCELERATION

Example 3. The three bay frame shown

in the figure is assumed to be pinned atthe base. It is subjected to a ground

acceleration pulse which has an

amplitude of 0.5g and a duration of 0.4

seconds. It should be noted that this

acceleration pulse is similar to one

recorded at the Newhall Fire Station

during the Northridge earthquake (1994).

The lateral resistance at ultimate load is

assumed to be elasto-plastic. The

columns are W10 x 54 with a clear

height of 15 feet and the steel is A36

having a nominal yield stress of 36 ksi.

Estimate the following:

(a) the displacement ductility demand(b) the maximum displacement and

(c) the residual displacement.





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in.1.80.37.2max yvv




Computed displacement time history



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Short duration rectangular impulse.

RESPONSE OF SDOF SYSTEMS TOGENERAL DYNAMIC LOADING



Differential impulse response.


dtegtv d

t

t

sin1

0

dte

gdtegtv d

t

t

d

t

t

sin1

cos0

20

dte

gdtegtv d

t

t

d

t

t

d

sin

1

21cos2

02

2

0

tvdtegtv dt

t

cos

0

tvdtegtvd

t

t

2

0

2 sin21

Can you tell how the maximum values are related?

For small damping:

Pseudo-Velocity

Pseudo-Acceleration


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tvtv 2


If the damping term can be neglected as contributing little to the equilibrium equation,

the total acceleration can be approximated as

The effective earthquake force is then given as

)()(2

tvmtQ

The above expression gives the value of the base shear in a SDOF (i.e., a single-story

structure) at every instant of time during the earthquake time history under consideration.

The overturning moment acting on the base of the structure can be determined by

multiplying the inertia force by the story height

)()(2

tvhmtM



THE CONCEPT OF RESPONSE SPECTRUM

Consider a SDOF system:

AND

a given earthquake ground

motion:


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COMBINED D-V-A RESPONSE SPECTRUM(Single Damping Value)



COMBINED D-V-A RESPONSE SPECTRA(Multiple Damping Values)


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DISPLACEMENT RESPONSE SPECTRA(Damping Values = 0%, 2%, 5%, 10% and 20%)



ACCELERATION RESPONSE SPECTRA OR BASE SHEARCOEFFICIENTS ( = 0%, 2%, 5%, 10% and 20%)


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FROM REPONSE SPECTRUM TO DESIGN SPECTRUM



AN EARLY ATTEMPT TO CONSTRUCT DESIGN SPECTRA


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A TYPICAL NEWMARK-HALL DESIGN SPECTRA



TYPICAL DESIGN SPECTRUM DISPERSIONS


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2001 CBC DESIGN SPECTRUM



IBC / CBC-2007/ ASCE 7 / ASCE-41 DESIGN SPECTRUM


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IBC / CBC-2007/ ASCE 7 / ASCE-41 DESIGN SPECTRUM



Spectral displacementSpectral displacement max)(tvSd

Maximum base shearMaximum base shear

Maximum overturningMaximum overturning

momentmoment

dSmQ2

max

dShmM2

max

Spectral pseudoSpectral pseudo--velocityvelocity

Spectral pseudoSpectral pseudo--accelerationacceleration

dpv SS

dpa SS2

RECAP: RESPONSE SPECTRAL ENTITIES (SDOF)


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Example 4:Example 4:Using this design spectrum and

5% damping, find the maximum

base shear of the building in

Example 2.

RESPONSE SPECTRUM APPLICATION EXAMPLE (SDOF)





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N-S:

T = 0.287 sec.= 21.8 rad/sec,

f = 3.48 HZ

Sd=v(t)

max=0.42 in.

Qmax

= mSd

(0.485)(21.8)2(0.42) = 96.8 kips

E-W:

T= 0.23 sec,

= 27.2 rad/sec, f= 4.3 Hz:

Sd

= 0.28 in.

Qmax

= mSd

(0.485)(21.8)2(0.28) = 64.5 kips




Generalized single-degree-of-freedom system

APPROXIMATE ANALYSIS BY REDUCTION OF MDOFSYSTEMS TO EQUIVALENT SDOF SYSTEMS


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factorionparticipatearthquake

i

iimL

i ii

ii i

w

wW

2

2

*)(

= Effective Weight

GENERALIZED COORDINATES METHOD

Lgp *



Effective Weight, Effective Mass andEffective Weight, Effective Mass and

Mass Participation FactorMass Participation Factor

i ii

ii ieff

m

mM

2

2)(

i ii

ii i

w

wW

2

2

*)(

= Effective Weight

= Effective Mass

i ii i

eff

w

W

m

M *= Mass (or weight) Participation Factor

EFFECTIVE WEIGHT, MASS, ANDWEIGHT/MASS PARTICIPATION FACTORS


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Possible shape functions based on aspect ratio.

GENERALIZED COORDINATES METHOD



Example 5 :Example 5 :Considering the four-story, reinforced-

concrete moment frame building

shown, determine the generalized

mass, generalized stiffness, and

fundamental period of vibration in

the transverse direction using the

following shape functions:

(a)

(b)

.

)2/sin()( Lxx

All beams are 12in. x 20 in. All columns are 14 in x 14 in.

fc =4000 psi,Ec = 3.6x106 . Fy = 60 ksi.

Floor weights (total dead load) = 390 kips at the roof, 445 kips at the fourth and third

levels, and 448 kips at the first level.

Live loads are 30 psf at the roof and 80 psf per typical floor level.

ASSUME BEAMS ARE RIGID RELATIVE TO COLUMNS

Lxx /)(

EXAMPLE: CALCUALTION OF GENERALIZED PARAMETERS


30/80



Building of Example 5.




Assumed shape of column

deformation.



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32/80






RAYLEIGHS METHOD

The idea here is to use the static deflected shape of the building given a lateral load distribution as

the deformed shape under dynamic loading. Everything else is the same as the Generalized

Coordinate Method.


33/80



RAYLEIGHS METHOD

In an undamped elastic system, the

maximum potential energy can be

expressed in terms of the externalwork done by the applied forces. In

terms of a generalized coordinate

this expression can be written as

22)(

*

max

Ypp

YPE ii

Similarly, the maximum kinetic

energy can be expressed in terms

of the generalized coordinate as

22)(

*222

22

max

mYm

YKE

i

ii

(According to the principle of

conservation of energy for anundamped elastic system, these two

quantities must be equal to each other

and to the total energy of the system.

(A)

(B)

Ym

p*

*

*

*

2p

YmT

ii

i ii

vpg

vwT

2

2

Code Method B Formula



Frame of Example 5.

EXAMPLES: APPLICATION OF RAYLEIGHS METHOD

Example 6:Use Rayleighs method to determine the

spatial shape function and estimate the

fundamental period of vibration in the

transverse direction for the reinforced-

concrete building given in Example 5.

We want to apply static lateral loads that are

representative of the inertial loads on the

building. Since the story weights are

approximately equal, it is assumed that the

accelerations and hence the inertial loads

vary linearly from the base to the roof.

Note that the magnitude of loads is irrelevant

and is chosen for ease of computation.


34/80






Example 7:Example 7:

k/in333,482

2 L

AEk

k/in258,432

1 L

AEk

k/in258,432

1 L

AEk

The building shown here is

subjected to vertical ground

acceleration. If the dead load

on all floors and roof is 2

kips/foot, calculate the

fundamental period of

vibration in the vertical

direction using the basic

Rayleighs Method.

5@12=60

W14x61

W14x61

W14x68

W14x68

W14x78

30

k/in333,482

2 L

AEk

k/in341,552

3 L

AEk



35/80



Solution:Solution:

000000BaseBase

0.4430.443

0.0140.014

0.0500.050

0.0920.092

0.1320.132

0.1550.155

mm22

0.2970.297

0.2720.272

0.2030.203

0.1520.152

0.0760.076

0.1550.155

0.1550.155

0.1550.155

0.1550.155

60/g=0.15560/g=0.155

mm

11,69811,6984881.64881.60.005420.005425534155341

0.2970.2970.005420.0054230030011

3575.93575.90.004970.004974833348333

0.5690.5690.010390.0103924024022

1991.81991.80.003720.003724833348333

0.7720.7720.014110.0141118018033

999.4999.40.002170.002174325843258

0.9240.9240.016880.0168812012044

249.9249.90.001390.001394325843258

1.001.000.018270.01827606055

KK 22vvkkPPSTORYSTORY




Solution:Solution:

sec.039.0698,11

443.022

*

*

k

mT

039.0

7.213

01827.0443.02

7.213297.060569.060

772.060924.0600.160

2

*

*

*

T

Pp

p

YmT

ii

OR



36/80



DisplacementsDisplacements

Base ShearBase Shear

ForcesForces

Overturning MomentOverturning Moment

*max

)()(

m

Sxxv d

L

i ii

ii i

w

wW

2

2

*)(

where

*

2

maxm

SQ

paL gSWQ pa/*

max or or CWQ max

Lii

i

mQq

max

iiiO qhM

RESPONSE SPECTRUM ANALYSIS USINGGENERALIZED COORDINATES



Example 8.

Using the design spectrum

given, the shape function

determined in Example 6, and

the reinforced-concrete moment

frame of Example 5, determine

the base shear in the transverse

direction, the corresponding

distribution of inertia forces

over the height of the structure,

and the resulting overturning

moment about the base of thestructure.

rad/sec.8.715

Hz,39.1/1sec.,721.0

TfT

From the design spectrum Spa = 0.185g.

EXAMPLE: SPECTRUM ANALYSIS OF GENERALIZEDSDOF SYSTEM


37/80



*

2

max m

S

Q

paL

kips84.88666.0

)4.386)(185.0()827.0( 2

Level mi i 2iim mii mii/L qmax Vmax

4 0.252 1.000 0.252 0.252 0.305 27.10

27.10

3 0.288 0.866 0.226 0.255 0.308 27.36

54.46

2 0.288 0.685 0.135 0.197 0.238 21.14

75.60

1 0.288 0.428 0.053 0.123 0.149 13.24

0.666 0.827 88.84

Lii

i

mQq

max




kipsft2716

)12(24.13)5.22(14.21

)33(36.27)5.43(10.27

oM

The displacement is

ddSSmv *)/(max L

where

in.50.0in.80.0

in.035.1in.168.1

168.1)941.0)(242.1(

242.1666.0

827.0

941.0

)715.8(

)4.386)(185.0(

/and/

12

34

2

*2

vv

vv

v

S

mSS

iii

d

pad

L



38/80



EXAMPLE 8A: 1991 Problem A-2 (Page 101 of your book)





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40/80








41/80








42/80


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A Crash Course on MatrixA Crash Course on Matrix

AlgebraAlgebra

http://numericalmethods.eng.usf.edu/matrixalgebrabook/frmMatrixDL.asp



ANALYSIS OF MULTI-DEGREE-OF-FREEDOM SYSTEMS (MDOF)

Analogous to the case of SDOF systems:

)(}]{[}]{[}]{[ tgMvKvCvM Influence vector to be introduced later

In general case:

n

n

nn

nnnn

nn

nn

n

n

nn

nnnn

nn

nn

n

n

n

n

v

v

v

v

K

KK

KKK

KKKK

v

v

v

v

C

CC

CCC

CCCC

v

v

v

v

M

M

M

M

1

2

1

111

21222

1111211

1

2

1

111

21222

1111211

1

2

1

1

2

1

.

.

...

....

..

..

.

.

...

....

..

..

.

.

0

...

....

00..

00..0

tg

M

M

M

M

n

n

n

n

1

2

1

1

2

1

.

.

0

...

....

00..

00..0

Symmetric

Symmetric

Symmetric

Symmetric

Actually, in general 3-D analysis, each element of the above matrices could be a

6x6 matrix.


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45/80



The stiffness matrix for this type of structure has the triThe stiffness matrix for this type of structure has the tri--diagonaldiagonal

form shown below:form shown below:


..

...

...

...

1

1

4323

3212

21

nnn

n

kkk

k

kkkk

kkkk

kk

Stiffness Matrix for a 2-D Shear Building

First Floor

Roof



The short version is that for all our applicationsThe short version is that for all our applications MDOFsMDOFs cancan

be converted to a series ofbe converted to a series of SDOFsSDOFs..

The process is very similar to Generalized CoordinatesThe process is very similar to Generalized Coordinates

approach.approach.

The difference is that you get one SDOF for each mode youThe difference is that you get one SDOF for each mode you

consider.consider.

In spectrum analysis, since maximum modal responses areIn spectrum analysis, since maximum modal responses are

not simultaneous and the time is lost we have to estimatenot simultaneous and the time is lost we have to estimate

the maximum total response.the maximum total response.

SRSS and CQC are two way to estimate maximum modalSRSS and CQC are two way to estimate maximum modal

responses.responses. Everything else is basically the same although it looks aEverything else is basically the same although it looks a

whole lot more complicated.whole lot more complicated.



46/80



MDOF SYSTEMS: MODE SHAPES AND FREQUENCIES

The equations of motion for undamped free vibration of a multiple-degree-of-

freedom (MDOF) system can be written in matrix form as}0{}]{[}]{[ vKvM (A)

Since the motions of a system in free vibration are simple harmonic, the

displacement vector can be represented as

tvv sin}{}{ (B)Differentiating twice with respect to time results in

}{}{2

vv (C)Substituting Equation (C) into Equation (A) results in a form of the eigenvalue

equation,

}0{}{][][ 2 vMK (D)In order to have a nontrivial solution, the determinant of the coefficient matrix mustbe zero:

}0{])[]det([ 2 MK (E)



MDOF SYSTEMS:MODE SHAPES AND FREQUENCIES CALCULATION EXAMPLE

Example 9:It is assumed that the response in the transverse

direction for the reinforced-concrete moment

frame of Example 5 can be represented in terms

of four displacement degrees of freedom which

represent the horizontal displacements of the

four story levels. Determine the stiffness matrix

and the mass matrix, assuming that the mass is

lumped at the story levels. Use these properties

to calculate the frequencies and mode shapes of

the four-degree-of-freedom system.

..

...

...

...

1

1

4323

3212

21

nnn

n

kkk

k

kkkk

kkkk

kk


47/80


48/80



MDOF SYSTEMS:MODE SHAPES AND FREQUENCIES CALCULATION EXAMPLE

Mode shapes are obtained by substituting the values ofBi, one at a time, into theequations}0{}]){[]([ 2 vMK

and determining N-1 components of the displacement vector in terms of the first

component, which is set equal to unity. This results in the modal matrix

92.024.105.147.0

75.175.078.074.0

78.107.120.091.0

00.100.100.100.1

][

Solution with ETABS:

92.024.105.147.0

75.175.078.074.0

78.107.120.091.0

00.100.100.100.1

][



MDOF SYSTEMS:ORTHOGONALITY OF MODES

Bettis reciprocal work theorem can be used to develop two orthogonality

properties of vibration mode shapes

and

It is further assumed for convenience that

)(}0{}]{[}{ nmM mT

n

)(}0{}]{[}{ nmK mT

n

)(}0{}]{[}{ nmC mT

n


49/80



As we will see, orthogonality reduces:

n

n

nn

nnnn

nn

nn

n

n

nn

nnnn

nn

nn

n

n

n

n

v

v

vv

K

KK

KKKKKKK

v

v

vv

C

CC

CCCCCCC

v

v

vv

M

M

MM

1

2

1

111

21222

1111211

1

2

1

111

21222

1111211

1

2

1

1

2

1

.

.

...

....

..

..

.

.

...

....

..

..

.

.

0

...

....

00..00..0

tg

M

M

M

M

n

n

n

n

1

2

1

1

2

1

.

.

0

...

....

00..

00..0

Symmetric

Symmetric

Symmetric

Symmetric

MDOF SYSTEMS:ORTHOGONALITY OF MODES

to:

n

n

n

n

n

n

n

n

n

n

n

n

v

v

v

v

K

K

K

K

v

v

v

v

C

C

CC

C

v

v

v

v

M

M

M

M

*

*

.

.

*

*

0..00

0..00

......

......

00..

00..0

*

*

.

.

*

*

0..00

0..00

......

......

0..0

00..0

*

*

.

.

*

*

0..00

0..00

......

.....

00..0

00..0

1

2

1

*

*

1

*

2

*

1

1

2

1

*

*

1

0

*

1

*

1

1

2

1

*

*

1

*

2

*

1

tg

n

n

L

L

L

L

1

2

1

.

.

or n set of independent equations.

This is a monumental achievement which drastically reduces the necessary

computational efforts.



Example 10:Example 10:The stiffness matrix and the shape of the first mode of

vibration of a two degree of freedom system are given

below. Find the shape of the second mode.

23

310

6.0

0.1

1

1K

b

a

MDOF SYSTEMS:ORTHOGONALITY OF MODES EXAMPLE


50/80



Solution:Solution:

56.4

0.156.4

8.1

2.80.1

08.12.8

0

08.12.8

0

0

23

3106.00.1

0

222

22

2

2

2

2

21

ba

ba

b

a

b

a

TK

MDOF SYSTEMS:ORTHOGONALITY OF MODES EXAMPLE



MDOF SYSTEMS:DYNAMIC RESPONSE USING MODAL SUPERPOSITION

Since any MDOF system having N degrees of freedom also has N independent

vibration mode shapes, it is possible to express the displaced shape of the

structure in terms of the amplitudes of these shapes by treating them as

generalized coordinates (sometimes called normal coordinates). Hence the

displacement at a particular location, vi, can be obtained by summing the

contributions from each mode as

tYtvN

n

nini

1

In a similar manner, the complete displacement vector can be expressed as

}]{[}{}{1

tYtYtvN

n

nn

Resulting in)}({}]{][[}]{][[}]{][[ tPYKYCYM

or

)}({}{}]{][[}{

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

tPT

nYKT

n

YCT

nYMT

n


51/80



MDOF SYSTEMS:EARTHQUAKE TIME HISTORY ANALYSIS

As in the case of SDOF systems

vector of influence coefficients of which component i represents the

acceleration at displacement coordinate i due to a unit ground acceleration at

the base. For the simple structural model in which the degrees of freedom are

represented by the horizontal displacements of the story levels, this vector

becomes a unity vector, {1}, since for a unit ground acceleration in the

horizontal direction all degrees of freedom have a unit horizontal acceleration.

)(}]{[)( tgMtPe

For earthquake ground motions:

)()(* tgtP nen L where }]{[}{ MTnnL



MDOF SYSTEMS:DYNAMIC RESPONSE USING MODAL SUPERPOSITION

Using the orthogonality conditions reduces this set of equations to the equation

of motion for a generalized SDOF system in terms of the generalized properties

for the nth mode shape and the normal coordinate Yn:

where the generalized properties for the nth mode are given as

)(**** tPYKYCYM nnnnnn

)}({}{loadingdgeneralize)(

}]{[}{

stiffnessdgeneralize2}]{[}{

dampingdgeneralize

}]{[}{massdgeneralize

n

*

*2

*

*

n

*

n

*

tPtP

MK

KMC

C

MM

T

n

nnn

T

n

n

nnnn

T

n

n

T

n


52/80




By substitution: *2 /)(2nnnnnnnn

MtgYYY L

Just as was the case for SDOF systems, response of each mode can be calculated

from

nn

nnn

M

tVtY

*

)()(

L

where t

n

t

n dtegtVnn

0

)()(sin)()(

The complete displacement of the structure at any time is then obtained by

superimposing the contributions of the individual modes

N

n

nn tYtYtv1

)}(]{[)(}{)}({

The resulting earthquake forces can be determined in terms of the effectiveaccelerations

*

2 )()()(n

nnnnnne

M

tVtYtY

L )(}{)}({ tYtv nenne




the corresponding effective modal earthquake force is given as

*/)(}]{[

)}(]{[)}({

nnnnn

nn

MtVM

tvMtq

L

The total earthquake force is obtained by superimposing the individual modal

forces

N

n

n tYMtqtq1

2 )(]][[)()(

The base shear can be obtained by summing the effective earthquake forces over

the height of the structure:

)(

)}({}1{)()( 1

tVM

tqtqtQ

nnen

H

in

T

inn

Where is the effective mass for the nth mode*2/ nnen MM L


53/80



MDOF SYSTEMS:THE CONCEPT OF EFFECTIVE MASS

The sum of the effective masses for all of the modes is equal to the total mass of

the structure.

This results in a means of determining the number of modal responses necessary

to accurately represent the overall structural response.

If the total response is to be represented in terms of a finite number of modes and

if the sum of the corresponding modal masses is greater than a predefined

percentage of the total mass, the number of modes considered in the analysis is

adequate.

If this is not the case, additional modes need to be considered.

Codes generally require that enough modes be considered so that 90% of total

mass is participating in response.

*2/nnen

MM L



The base shear for the nth mode, can also be expressed in terms of the effective

weight, Wen, as


)()( tVg

WtQ nn

enn

where

H

i ini

H

i ini

en

W

WW

1

2

2

1

The base shear can be distributed over the height of the building using modal

earthquake forces

n

nnn

tQMtq

L

)(}]{[)}({


54/80



The equations for the response of any mode of vibration are exactly equivalent to the

expressions developed for the generalized SDOF system. Therefore, the maximumresponse of any mode can be obtained in a manner similar to that used for the generalized

SDOF system.

MDOF SYSTEMS:RESPONSE SPECTRUM ANALYSIS

The maximum modal displacement:

dn

n

nn S

tVtY

maxmax

)()(

*

max / ndnnn MSY Lor

*maxmax

}{}{}{

n

dnnnnnn

M

SYv

L

Distribution of modal displacements throughout the structure:

Maximum effective modal earthquake forces:

*max

}]{[}{

n

pannn

nM

SMq

L

Maximum modal earthquake base shear: *2max / npannn MSQ L gSWQ panenn /max or

Maximum modal overturning moment: */ npannno MSMhM L



MDOF SYSTEMS: RESPONSE SPECTRUM ANALYSISCOMBINATION OF MODES

Take another look at aresponse or designspectrum.

We have achieved simplicityof engineering at a costbecause a significantparameter is missing on aresponse or design spectrumgraph.

What is that significant

parameter?


55/80




Because maximum modal responses rarely occur at the same timewe need to find a reasonable approach to combine modalresponses for engineering purposes.

A conservative approach would be to use the sum of the absolutevalues (SAV) of the modal responses:

A more reasonable approach based on probability theory is to usethe square-root-of-the-sum-of-the-squares (SRSS) method.

SRSS gives a good approximation of the response fortwo-dimensional structural systems.

N

n

nrr1

N

n

nrr

1

2




For three-dimensional systems, it has been shown that thecomplete-quadratic-combination (CQC) method may offer asignificant improvement in estimating the response of certainstructural systems. The complete quadratic combination isexpressed as

N

i

N

j

jiji rprr1 1

where2222

2/32

)1(4)1(

)1(8

ijp

and

cr

ij

cc/

/


56/80



MDOF SYSTEMS: EXAMPLE OFRESPONSE SPECTRUM ANALYSIS / COMBINATION OF MODES

Example 11:Example 11:A small two story frame structure is shown. Each

story is known to deflect 0.10 inches under a 10-

kips story shear. Plan dimensions are 20x20 and

the story heights are 12 ft. each. The first floor

weighs 92 kips and the roof weighs 86 kips.

Assume the structure has 7% damping and use a

two-degree of freedom mathematical model and

the design spectrum shown to calculate:

1. The mode shapes and frequencies of the

structure

2. Modal displacements, story shears and

base shears

3. SAV, SRSS and CQC values of the base

shear for the structure.

Source: Modified from an SE Exam Problem, 1990.




SOLUTION:

x2

x1m1

m2

k2

k1

x2

x1m1

m2

k2

k1

Mathematical

Model

}0{])[]det([ 2 MK

100100

100200

12

221

kk

kkkK

860

0921

0

0

2

1

gM

MM

B

B

gMK

86100100

10092200

860

092

100100

100200][][

22

0100008610092200 BB ; gB2

010000264007912 2 BB

0.436

2.901B

12.97

33.46

0.484

0.188T

Fundamental period


57/80




Mode Shape 1 (normalized at roof):

0][][2

12

1

x

xMK

00.186100100

10092200 1

2

2

x

B

B

0.1

625.01

Mode Shape 2 (normalized at roof):

x2= 1.0

x1 = 0.63

x2= 1.0

x1 = 0.63

0][][2

12

1

x

xMK

00.186100100

100922001

1

1

x

B

B

0.1

495.12




Modal Masses:

M1 1T

M 1 M1 0.316

M2 2T

M 2 M2 0.755

Modal Participation:

L1 1T

M L1 0.372

L2 2T M L2 0.579

Modal Displacements

v1 1 L1Sd

1

M1 v1

0.472

0.755

v2 2 L2Sd

2

M2 v2

0.123

0.082

Modal Accelerations:

a1 1 2 v1 a1 79.495127.192

a2 2 2

v2 a2137.128

91.724


58/80




Modal Story Shears:

q1 M a1 q118.947

28.338

V1 18.947 28.338 47.28

q2 M a2 q232.683

20.436

V2 32.683 20.436 53.11

SAV of Base Shear:

V 47.285 53.119 V 100.404

SRSS of Base Shear:

V 47.285( )2

53.119( )2

V 71.116




CQC of Base Shear:

0.0

1

1

2

2

1

1

1

0.388

2.574

1

p

8 2

1 0 0 0 0 1.5

1 0 0

2

2

4

2

0 0 1 0 0

2

8 2

1 1 0 1 0 1.5

1 1 0 2

2

4 2

1 0 1 1 0 2

8 2

1 0 1 0 1 1.5

1 0 1

2

2

4

2

0 1 1 0 1

2

8 2

1 1 1 1 1 1.5

1 1 1 2

2

4 2

1 1 1 1 1 2


59/80




p1

9.228 103

9.228 103

1

V V1 p0 0

V1 V1 p0 1 V2 V2 p1 0 V1 V2 p1 1 V2

V 71.441



MDOF SYSTEMS: EXAMPLES OFRESPONSE SPECTRUM ANALYSIS / COMBINATION OF MODES

Example 12:Example 12:The first two mode shapes and

frequencies of the structure shown to

the right are given. If the building is

subjected to an earthquake with the

response spectrum shown for motions

in the horizontal direction, compute

the R.M.S. of the maximum

displacement and moment at point B.

Neglect the effect of gravity. Sv

T0.6 1.6

1.6 ft/sec

1.2 ft/sec

Sv

T0.6 1.6

1.6 ft/sec

1.2 ft/sec

Sv

T0.6 1.6

1.6 ft/sec

1.2 ft/sec

1.5 L

L

L

M

M 3MvAvB

vC

T2 T1

6.8

0.32

n

25.608.0

58.733.0

0.10.1

in


60/80



EXAMPLE 12 CONTINUED

SOLUTION:

n

v

n

nnn

S

M

L MTnn L

83.13

41.0

1

1

0

25.658.700.3

08.033.000.3

1

1

0

00

00

003

25.658.71

08.033.01MM

M

M

M

L

n

T

nnMM

25.608.0

58.733.0

11

25.658.73

08.033.03

25.608.0

58.733.0

11

00

00

003

25.658.71

08.033.01M

M

M

M

Mn

52.990

012.3MMn

2.1

6.1v

S

; ;

6.80.32

nand



n

v

n

nnn

S

M

L


inM

MB 48.0

312.3

126.141.033.01

inSRSS B 19.517.548.022

*max

}]{[}{

n

vnnn

M

SMq

L

in

M

MB 17.5

6.852.99

122.183.1358.72

M

M

MMqA 11.13

12.3

126.1341.00.13max

1

M

M

MMqA 61.17

52.99

122.16.883.130.13

max

2

MLLMMB 67.195.111.13max

1

MLLMMB 42.265.161.17max

2

MLMLMSRSS B 94.3242.2667.1922


61/80



Review ProblemsReview Problems

Linear Dynamic AnalysisLinear Dynamic Analysis



EXAMPLE 13

Source: (ConstantinouM.C., Reinhorn A.M., and Whittaker A.S., Passive Energy Dissipation for Seismic/Wind Design and Retrofit Basic Principles, Pace Class Notes,

Multidisciplinary Center for Earthquake Engineering Research formerly NCEER)

36

w = 27 kN = 6000 lbs, Tel = 0.5 sec

36

w = 27 kN = 6000 lbs, Tel = 0.5 secDD uCF

DF

Du

DD uCF

DF

Du

DD uCF

DF

Du

1.1. Find the equivalent modal damping ratio, , for the portal frame below.

2.2. Does the addition of this damping device change the natural period of the system?

Assume C = 176.6 lbs/in/sec

SOLUTION:

mC

2 cos uuD

u

cos uuD

u

cosuCuCF DD

DF

cosDF

DF

cosDF

2coscos uCFF D


62/80



Therefore,

%30296.0

5.02

38660002

36cos6.176

2

36cos2

gW

C

95.030.11 22 d

Viscous damping does not affect natural frequency significantly.




The structural system shown below is modeled as a shear-building with three degrees of

freedom. The mass is distributed uniformly but the stiffness of columns is only partially

known. However, the first and the third mode shapes of vibration are given. Calculate

the second mode shape of this structure.

m

m

m=1.0 kips/in/sec2

k3 = 50 kips/in

m

m

m=1.0 kips/in/sec2

k3 = 50 kips/in

2.0

4.0

0.1

,

0.1

,

3.0

7.0

0.1

3

1

221

x

x

EXAMPLE 14


63/80



SOLUTION:

jiforM jT

i 0

100

010

001

M

0.02.04.00.10.0

0.03.07.00.10.0

1232

1221

xx

xxT

T

231.4385.0

000.1

2




EXAMPLE 15

A steel frame is shown below, together with its fundamental mode shape and a design

response spectrum. The frame has a fundamental period of vibration of 0.35 seconds.

Determine story forces, base shear, modal and mass participation factors.100 K

200 K

200 K

200 K0.3

0.6

0.8

1.0

0.0

0.1

0.1

0.2

0.2

0.3

0.3

0.4

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Period (sec.)

SpectralAcceleration(g)

Source: Modified from California SE B, 1973, Problem B1


64/80



SOLUTION:

Only first-mode response is needed.

2max

ii

ii

panii

niW

WS

g

WtqF

Floor W, kips W 2W iF , kips Story Shear

Roof 100 1.00 100 100 41.52 41.52

Forth 200 0.8 160 128 66.43 107.95

Third 200 0.6 120 72 49.82 157.77

Second 200 0.3 60 18 24.91 182.68

700 440 318 182.68At Roof,

52.41318

44030.000.1

100 g

gFi 41.52 k

66.43 k

49.82 k

24.91 k

Base Shear, V = 182.68 k

41.52 k

66.43 k

49.82 k

24.91 k

Base Shear, V = 182.68 k

Modal Participation Factor: 384.1318

4402

ii

ii

W

W

Mass / Weight Participation Factor:

87.0

700

318440/

2

*2

M

M

M

M nnen L




The figure below represents a three-story building with plan dimensions of 100x100.

The effective dead load on each floor is shown on the figure.

Assume the following matrices: 180 lb/ sq. ft.

180 lb/ sq. ft.

180 lb/ sq. ft.

000.1000.1000.1

610.1725.0950.1

387.0657.0860.2

Mode Shape Matrix:

Modal Frequency Matrix: sec/

7.61

5.38

1.15

rad

1. Determine the base shear and mass participation factor for each mode by using the

design response spectrum with 5% damping.

2. Determine the lateral load at each level at each mode.

3. What is the reasonable value for design base shear?

(

Source: Modified from California SE 1977 Problem A7

EXAMPLE 16


65/80



PSV for 0%, 2%, 5% and 10% Critical Damping

0

5

10

15

20

25

30

35

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2

Period (sec.)

PSV(in./sec)

Design Spectra:




SOLUTION:

22

ii

iivnii

ii

ii

panii

iW

W

g

SW

W

WS

g

WF

First Mode:

%5sec/25

.sec416.02

sec/1.15

forinS

Trad

vn

977.0

4.386

251.15

g

Svn

441.021086

92882

ii

ii

W

W

430.0441.0977.02

ii

iivn

W

W

g

S

430.0iii WF

Level W, kips W 2W iF , kips

Roof 1800 2.86 5148 14723 2215.4

Second 1200 1.95 2340 4563 1007.0

First 1800 1.0 1800 1800 774.6

4800 9288 21086 3997.0

2215.4 k

1007.0 k

774.6 k

First-Mode Base Shear, V = 3997.0 k

2215.4 k

1007.0 k

774.6 k



85.04800

210869288

/

2

*2

M

M

M

M nnen L



66/80



SOLUTION:

22

ii

iivnii

ii

ii

panii

i

W

W

g

SW

W

WS

g

WF

Second Mode:


14.04800

7.32074.1487

/

2

*2

M

M

M

M nnen L

%5sec/9

.sec163.02

sec/5.38

forinS

Trad

vn

897.0

4.386

95.38

g

Svn

464.07.3207

4.14872

ii

ii

W

W

416.0464.0897.02

ii

iivn

W

W

g

S

416.0iii WF


Roof 1800 -0.657 -1182.6 776.97 -491.9

Second 1200 0.725 870 630.75 361.9First 1800 1.000 1800 1800 748.7

4800 1487.4 3207.7 618.7

491.9 k

361.9 k

748.7 k

Second-Mode Base Shear, V = 618.7 k

491.9 k

361.9 k

748.7 k





SOLUTION:

22

ii

iivnii

ii

ii

panii

iW

W

g

SW

W

WS

g

WF

Third Mode:


01.04800

1.51806.564

/

2

*2

M

M

M

M nnen L

%5sec/5

.sec102.02

sec/7.61

forinS

Trad

vn

798.0

4.386

57.61

g

Svn

109.01.5180

6.5642

ii

ii

W

W

087.0109.0798.02

ii

iivn

W

W

g

S

087.0iii WF


Roof 1800 0.387 696.6 269.6 60.6

Second 1200 -1.61 -1932 3110.5 -168.0First 1800 1.000 1800 1800 156.6

4800 564.6 5180.1 49.2

60.6 k

168.0 k

156.6 k

Third-Mode Base Shear, V = 49.2 k

60.6 k

168.0 k

156.6 k




67/80



2215.4 k

1007.0 k

774.6 k


2215.4 k

1007.0 k

774.6 k


491.9 k

361.9 k

748.7 k


491.9 k

361.9 k

748.7 k


60.6 k

168.0 k

156.6 k


60.6 k

168.0 k

156.6 k


SRSS Base Shear:

kips9.40442.497.6183997 222




296 K

197 K

99 K

W=1000 kips

W=1200 kips

W=1200 kips

W=1200 kips

329 K

0.75

0.50

0.25

1.00

A structural steel plane rigid frame is shown below. The centers of gravity of story weights

are at the respective roof and floor levels. Lateral forces and corresponding total deflections

are shown on the figure. All story heights are 12 ft. Calculate the natural period of the frame

using the Rayleighs Method.

SOLUTION:

i

ii

i

ii

pg

w

T

2

2

Level i w, kipsi

p , kipsi

, in 2iiw iip

4 1000 329 1.000 1000.00 329.00

3 1200 296 0.75 675.00 222.00

2 1200 197 0.50 300.00 98.50

1 1200 99 0.25 75.75 24.752050.00 674.25

.sec557.0)25.674)(4.386(

20502 T

EXAMPLE 17

Source: Modified from California SE A, 1980, Problem A7


68/80



EXAMPLE 18

The figure below represents a three-story building. The effective dead loads are shown on

each floor. The following dynamic properties of the plane frame are given:

Eigenvectors,

984.0385.1572.0

697.1704.0220.1

714.0208.1690.1

IMT

Eigenvalues, sec/13.48

18.25

77.8

rad

1. Compute the mass participation factors

2. How can you check that your participation

factors are correct?3. Calculate the displacements of each floor

based on the spectra given above.

4. Calculate the interstory drift for each floorSource: Modified from California SE A, 1982, Problem A6.



SOLUTION:

1. Mass participation factors

Modal participation is:

M

MMPF

T

T

12000

0800

0080

4.386

1M

1

1

1

1

1

1

4.386120

4.386

804.386

80

984.0697.1714.0

385.1704.0208.1

572.022.169.1

M

MT

T



69/80




103.0

326.0

780.0

1

1

1

306.0351.0148.0

430.0146.0250.0

178.0253.0349.0

M

MT

T

84.072.0

780.0

4.3861208080

1780.0 2

2

1 MPF

15.0

72.0

326.02

2 MPF

00.072.0

103.02

3

MPF

2. Check Mass participation factors

The participation factors are in decreasing order with increasing mode shapes, and for allthree modes of the frame are considered:

00.199.00.015.084.0321 MPFMPFMPF




3. Floor Displacements

.sec

13.0

25.0

72.02

T dpa S

g

g

g

S 2

40.0

80.0

15.0

dn

n

T

n

nn SM

Mv

*max

ing

v

337.0

718.0

995.0

77.8

15.0)781.0(

572.0

220.1

690.1

2max1

ing

v

220.0

112.0192.0

18.25

80.0326.0

385.1

704.0208.1

2max2

ing

v

007.0

012.0

005.0

13.48

40.0102.0

984.0

697.1

714.0

2max3

inv SAV

564.0

842.0

192.1

invSRSS

403.0

728.0

013.1


70/80




4. Interstory Drifts

1 ifloorifloor vvISD

0066.0

0180.0

0162.0

220.0

108.0

304.0

337.0

382.0

276.0

321 ModeModeMode ISDISDISD

564.0

508.0

596.0

SAVISD

403.0

397.0

411.0

SRSSISD



EXAMPLE 19

A radio tower may be idealized as a weightless cantilever supporting three masses as shown.

The mode shapes and periods of the first two modes of vibration of the structure are as

follows:

Source:AnoldUSC/UCBExam

Problem.

8.01.0

4.14.0

0.10.1

.sec6.0

6.1

T

Spectral velocities of these two modes of vibration are as shown in the sketch for a given

earthquake and 5% of critical damping in each mode. Determine an approximation of the

maximum displacement at point c using the SRSS method of mode superposition and

considering the two modes of vibration.

a

b

c

.10

.20

.15

in

sk 2

in

sk 2

in

sk 2

Sv

T0.6 1.6

1.6 ft/sec

1.2 ft/sec

a

b

c

.10

.20

.15

in

sk 2

in

sk 2

in

sk 2

a

b

c

.10

.20

.15

in

sk 2

in

sk 2

in

sk 2

Sv

T0.6 1.6

1.6 ft/sec

1.2 ft/sec

Sv

T0.6 1.6

1.6 ft/sec

1.2 ft/sec


71/80




SOLUTION:

n

v

n

nnn

S

M

L 1MT

nn

L

27.0

18.0

1

1

1

16.021.01.0

02.006.01.0

1

1

1

20.000

015.00

0010.0

8.04.11

10.040.01L

n

T

nn

MM

8.01.0

4.14.0

11

16.021.01.0

02.006.01.0

8.01.0

4.14.0

11

2.0

15.0

1.0

8.04.11

10.040.01nM

522.00

0126.0nM

2.1

6.1v

ST

2

47.10927.3

;;



n

v

n

nnn

S

M

L


inc 698.0927.3126.0

126.1180.01.01

inc 569.047.10522.0

122.1270.08.02

inc 901.0569.0698.022


72/80



EXAMPLE 20

Source: An old USC / UCB Exam Problem.

A piece of process equipment in an oil refinery consists of a rigid cylinder 75 feetlong with concentrated

weights of50 kips at each end. For the purpose of this analysis it may be assumed that the cylinder itselfis weightless. It may further be assumed that the cylinder is supported by a frictionless pivot at a point

one third of its height above the base and that it is constrained against rotation by two lateral springs at

the base which have spring rate of100 kips/foot.

K K K = 100 kips/foot

50 kips

50 kips

50

25

K K K = 100 kips/foot

50 kips

50 kips

50

25

If the structure is

subjected to an

earthquake having a

spectral velocity of0.8

feet/secondat its

fundamental period,

determine the maximum

bending moment

developed at the pivot

point. Neglect damping.




SOLUTION:

= -1/2

= 1

= -1/2= -1/2

= 1= 1

ftkKK ii 50

2

1100

2

1100

2

1100

22

2*

94.125.12.32

50

2

11

2.32

502

22*

iiMM

08.594.1

50*

*

M

K 78.0

2

11

2.32

50

iiML

ftM

SY v 063.0

08.594.1

8.078.0*max

L

2max2

max sec64.1 ftYY 2maxmax sec

64.1 ftY

kipsMF 541.264.12.32

50max

kipsftMoment 12750541.2


73/80



EXAMPLE 21

Source: Chopra, A.K.,Dynamics of Structu res, Theory and App lications to Earthqua ke Engineering , Second Edition, Prentice Hall, Page 416.

Determine the natural periods and modes of vibration of the structure below:

b

y

xux

uy

u

e tu gy

d/2

d/2

Frame B

Frame C

FrameA

b

y

xux

uy

u

y

x

y

xux

uy

u

ee tu gy tu gy

d/2

d/2

Frame B

Frame C

FrameA

The structure is a one-story building. It consists of a roof, idealized as a rigid diaphragm,supported on three frames,A,B, and C, as shown. The roof weight is uniformly distributed

and has a magnitude of 100 lb/ft2. The lateral stiffnesses of the frames are:

Ky = 75 kips/ftfor frameA, and

Kx = 40 kips/ftfor framesB and C

The plan dimensions are b = 30

ftand d = 20 ft. the eccentricity

is e = 1.5 ft, and the height of the

building is 12 ft.




tuI

m

y

u

kek

ekk

u

u

I

mgy

o

y

y

yyy

o

0

1

b

y

xux

uy

u

e tugy

d/2

d/2

Frame B

Frame C

FrameA

b

y

xux

uy

u

y

x

y

xux

uy

u

ee tugy tugy

d/2

d/2

Frame B

Frame C

FrameA

12

22dbm

Io

xy k

dkek

2

22

kipsw 601002030

ftkipsgwm2

sec863.1

222sec825.201

12

ftkips

dbmIO

The lateral motion of the roof diaphragm in the x-direction is governed by:

02 xxx ukum

sec/553.6863.1

)40(22

radm

kxx

The coupled lateral (uy)-torsional ( u ) motion of the roof diaphragm is governed by:


74/80




kipseky 5.112755.1

ftkipskd

kek xy 75.81682

22

tuu

u

u

ugy

yy

0

1

825.201

863.1

75.81685.112

5.11275

825.201

863.1

With stiffness and mass matrices known, the eigenvalue problem for this two-DOF system

could be solved to obtain:

sec/878.51 rad sec/794.63 rad

0493.0

5228.01

0502.0

5131.03




0.523

0.049

0.513

0.0501.0

sec/878.51 rad sec/794.63 radsec/553.62 rad

0.523

0.0490.049

0.513

0.050

0.513

0.0501.01.0

sec/878.51 rad sec/794.63 radsec/553.62 rad


75/80



EXAMPLE 22 (SI UNITS)

Source: Modified from Armouti, N. S.,Earthquake Engineerin g, Theory and Implementation, Printed in Jordan, 2004, Pages 38-45.

The structure shown is idealized as a two degree of freedom system. Calculate the

maximum SRSS base shear and base moment if the structure is excited by a baseexcitation having the response spectrum shown below in the Z-Z direction.

L

2L

1.0

m m

v2

v1EI

Z

Z

EQ

M = 15 kN.S2/m

EI = 1500 kN.m2

L = 1 m

EI

0.447 -0.8

94

0

2

4

6

8

10

12

14

0.00 0.50 1.00 1.50 2.00 2.50

Period (sec)

SA(m/s/s)

Note that the members are considered to be axially

rigid but joint rotations are not ignored.




SOLUTION:

1. Mass Matrix

Apply unit acceleration to each DOF and calculate the

corresponding inertial forces. Note that the acceleration

of all other DOFs must be kept zero.

10

02mM

2. Stiffness Matrix

Stiffness matrix may be calculated in a variety of ways.

We use the Flexibility Method here.

Apply unit force in the direction of each DOF, then

find displacements accordingly using any convenient

structural analysis method.

2

1

2221

1211

2

1

F

F

ff

ff

v

v

fs1 = 1

f11

f21

EI

Lf

3

8 3

11

EI

Lf

3

21

2


76/80



Source: Chopra, A.K.,Dynamics of Structu res, Theory and App lications to Earthqua ke Engineering , Second Edition, Prentice Hall, Page 416.

fs2 = 1

f12

f22

EI

Lf

3

12

2

EI

Lf

3

7 3

22

The resulting flexibility matrix is:

76

68

3

3

EI

LF

The stiffness matrix is the inverse of the flexibility

matrix:

86

67

20

33L

EIK

3. Mode shapes and Frequencies

02 MK 010

02

86

67

20

3 23

m

L

EI

rad/sec58.12

77.3

258.1

377.03

mL

EI

35.285.0

1121





MTnn L nT

nn MM *

*max

}]{[}{

n

annnn

M

SMq

L

4. Modal Base Shear and Base Moment

where and

01.2894.0

447.0

10

021585.00.11

L

MODE 1:

84.4085.0

0.1

10

021585.00.1

*

1

M

m/sec2.2sec.67.12

77.3 11

11 aST

kN

37.1

23.3

84.40

2.201.285.0

0.1

10

0215

max1

q

3.23 kN

1.37 kN

M=7.83 kN.m

V=3.23 kN

kN.m7.831.37(1)3.23(2)M

kN23.3

V

Mode 1


77/80




85.44894.0447.0

10021535.20.12

L

MODE 2:

98.11235.2

0.1

10

021535.20.1*2

M

m/sec10sec.50.02

2258.12 22 aST

kN

140

120

98.112

1085.4435.2

0.1

10

0215

max2

q

kN.m100(1)401120(2)M

kN120

V

120 kN

140 kN

M=100 kN.m

V=120 kN

Mode 25. Maximum SRSS Base Shear and Base Moment

kN.m100.3110083.7MandkN04.12012023.3 22max22

max V



EXAMPLE 23: CALCULATION OF MASS AND STIFFNESS MATRICES

Source: Humar, L.H.,Dynamics of Structure s, 2ndEdition, A.A. Balkema Publishers, Lisse, 2002., pp 92-93.

The uniform rigid rectangular slab

of total massm shown in the Figure

is supported by three massless

columns rigidly attached to the slab

and fixed at the base. The columns

have a flexural rigidity ofEIabout

each of the two principal axes,

which are oriented so that they are

parallel to the adjacent sides of the

rectangle.

Evaluate the mass and stiffness

matrices for the system incoordinates v1, v2, and v3 defined at

the center of mass.

a

EI

v1

v2v3

LEI

EI

b

m


78/80



v1

v2

v3

k

k

k

k

k

3

12

L

EIk

SOLUTION:

EXAMPLE 23 (CONTINUED)

To obtain the first column of thestiffness matrix, we impose a unit

displacement in the direction of

coordinate 1 and identify the

internal spring forces that oppose

the displacement.

k11

k21

k31

11 11

k

k k

2222

0

3

31

21

11

ak

ak

akk

k

kk




Similarly, to obtain the second

column of the stiffness matrix, we

impose a unit displacement in the

direction of coordinate 2 and

identify the internal spring forces

that oppose the displacement.

k12

k22

k32

12

12

k

k k

2222

3

0

32

22

12

bkbkbkk

kk

k


79/80




To obtain the third column of the

stiffness matrix, we impose a unit

rotation in the direction of

coordinate 3 and identify the

internal spring forces that oppose

the rotation.

2222

33

23

13

4

3

23

23

2

2

baka

kb

kk

bkk

akk

13 2

ka

k

k13

k23

k33

2

b

2

b

2

a

2

a

2

kb

2

kb

2

ka

2

kb

2

ka




3

22

12where

4

3

22

230

203

L

EIk

baba

b

a

kK

12where00

00

0022

0

0

baI

I

m

m

M


80/80



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1 NAEIM 2009 Linear Dyamics

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