2 NAEIM 2009 Nonlinear Response

7/29/2019 2 NAEIM 2009 Nonlinear Response

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Farzad Naeim Structural Dynamics for Practicing Engineers 1of 71

(Last Revision Date: 5-18-2009)

Part II:Analysis of Nonlinear

Structural Response

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

J ohn A. Martin & Associates, Inc.J ohn A. Martin & Associates, Inc.



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

Static Equilibrium:

Dynamic Equilibrium:

Three Simplifying Assumptions

for SDOF:1. Mass concentrated at the roof2. Roof is Rigid3. Axial Deformation of Columns

Neglected

kvp

)()()()( tvtktvctvmtp

STATIC AND DYNAMIC EQUILIBRIUM


2/36



Single-degree-of-freedom systemsubjected to base motion.

)(tgmtvtktvtctvm

)(tgm

Response of a SDOFsystem to earthquakeground motion:

STATIC AND DYNAMIC EQUILIBRIUM

In reality, if parts of thestructure collapses, orthere is pounding withanother structure, evenmass could be a

function of time.



TWO TYPES OF NONLINEARITY

MaterialMaterial

GeometricGeometric

(caused by large deformations)(caused by large deformations)


3/36



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

Analogous to the case of SDOF systems:

)(}]{[}]{[}]{[ tgMvKvCvM Influence vector

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

MM

M

M

n

n

n

n

1

2

1

1

2

1

.

.

0

...

....

00..

00..0

Symmetric

Symmetric

Symmetric

Sym

metric

Actually, in general 3-D analysis, each element of the above matrices could be a6x6 matrix.



MDOF SYSTEMS:

ORTHOGONALITY OF MODES

Bettisreciprocal work theorem can be used to develop two orthogonalityproperties of vibration mode shapes

and

It is further assumed for convenience that

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

n

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

n

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

n


4/36



As we will see, orthogonalityreduces:

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

.

.

ornset of independent equations.

This is a monumental achievement which drastically reduces the necessarycomputational efforts.



DAMPING IN NONLINEAR ANALYSIS

For linear systems we used modesuperposition method.

In mode superposition method thedamping ratio was defined for eachmode of vibration.

This is not possible for a nonlinearsystem because it has no truevibration modes.

A useful way to define the dampingmatrix for a nonlinear system is toassume that it can be represented asa linear combination of the mass andstiffness matrices of the initialelastic system.

This is called the Rayleighdamping.

][][][ KMC

An Example of Rayleigh Damping Functions


5/36



DAMPING IN NONLINEAR ANALYSIS

and are scalar multipliers whichmay be selected so as to provide agiven percentage of criticaldamping at any two periods ofvibrations.

If damping at the two selected

periods are1 and2, then:

][][][ KMC

An Example of Rayleigh Damping Functions

2

1

21

22

21

12

12

112



Basic Idealizations of

Nonlinear Behavior


6/36



BASIC IDEALIZATIONS OF NONLINEAR BEHAVIOR

Source: FEMA 440: Improvement of Nonlinear Static Seismic Analysis Procedures, Feb., 2005.



TWO TYPES OF STRENGTH DEGRADATION

Strength and stiffness degrading model

-600

-400

-200

0

200

400

600

-400 -300 -200 -100 0 100 200 300 400

Displacement

F

orce


7/36



MEASURED VERSUS CALCULATED RESPONSE

Comparison with of calculated versus experimental results

(a) a moment-critical column (b) a shear-critical column

Source: Kaul R., and Deierlein, G.G. (2004), Object oriented development of strength and stiffness degrading models for reinforced concrete structures,



Brute Force

Calculation of

Nonlinear Response


8/36



RESPONSE OF NONLINEAR SDOF SYSTEMS

)()()()( ttpttfttfttf sdi

1

1

1

)()(

)()()(

)()()()(

)(

)(

n

i

iit

n

i

tiis

d

i

tvtkr

tvttvtv

tvtkrtvtkf

ttvcf

ttvmf

)()()( ttgmttpttp e

For response to ground motions:

For response to ground motions:

-0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

interstorydrift

baseshear/baseshearatyield

)()()( ttgmvkttvcttvm ii and

Hysteretic Loops



NUMERICAL INTEGRATION

Various methods exist for integration ofVarious methods exist for integration ofequation of motion, including:equation of motion, including:

The Central Difference MethodThe Central Difference Method

TheThe HouboltHoubolt MethodMethod

TheThe NewmarkNewmark-- MethodsMethods The WilsonThe Wilson-- MethodMethod In each case the timeIn each case the time--step chosen (step chosen (t) must bet) must be

small enough to capture the variation andsmall enough to capture the variation anddetails in input ground motions and hystereticdetails in input ground motions and hysteretic

models.models.

See Bathe,See Bathe, Finite Element Procedures in Engineering Analysis,Finite Element Procedures in Engineering Analysis,Prentice Hall, 1982, for more details.Prentice Hall, 1982, for more details.


9/36



THE CENTRAL DIFFERENCE METHOD

t

vv

v

ii

i 2

11

112112

12

1

2

11

iiiiiiiii

i vvvtt

vv

t

vv

tt

vv

v

assuming1iv

v

t

iv

1iv

t t

t

vvz iii

11 and using the equation of motion

22

1

1

c

t

mzfp

c

t

mz isii

or: 11 iii ztvv

To start, given the initial conditions, Taylor series is used:

00012

vt

vtvv

For numerical stability:

Tt

10

1

5

1 Much smaller values are used for dynamic response toearthquake ground motions (i.e., 0.005 to 0.02 sec.)



THE NEWMARK- METHODS

1iv

v

t

iv

1iv

t t 1

22

1

11

1111

2

1

22

iiiii

iiii

iiii

vtvtvtvv

vt

vt

vv

pkvvcvm

(1)

(2)

(3)

Knowing the three unknowns:are found by solving the above three simultaneousequations.

iii vvv ,, 111 ,, iii vvv

4

10

)2.0forstablely(numericalMethodNewmarkExplicit0

)3.0forstablely(numericalMethodGoodman&Fox12

1

)4.0forstablely(numericalnamenobutUsed,8

1)5.0forstablely(numericalMethodonAcceleratiLinear6

1

stable)ynumericall(alwaysMethodonAcceleratiConstant4

1

Tt

Tt

Tt

Tt

Much smaller values

are used for dynamicresponse toearthquake groundmotions (i.e., 0.005to 0.02 sec.)


10/36



NUMERICAL INTEGRATION EXAMPLE

-1

0

1

0 1 2 3 4 5

Time (sec.)

GroundA

cceleration(m/s/s)

K=1000 kN/m

m =100 kN.s2/m

-0.05, -50

1, 50

1.1, 20

0, 0

0.05, 50

-1.1, -20

-1, -50-60

-40

-20

0

20

40

60

-2 -1 0 1 2

Displacement (m)

F

orce(kN)

Example 1:

The SDOF structure shown is excited by horizontalground acceleration as shown below. Assume 0%damping. Find the maximum force and displacementexperienced by the structure if:

1. The structure is elastic and has an infinite amount ofstrength.

2. The structure has an elastic-plastic force-displacement property as shown below and begins tocollapse when displacement exceeds 1.0 m.

Source: Modified fromArmouti, N. S., Earthquake Engineering, Theory and Implementation,Printed in Jordan, 2004,Pages 113-116.



NUMERICAL INTEGRATION EXAMPLE (CONTINUED)

Use the Central Difference Method to obtain:


50.00.04617.80.0183.8019

50.0-0.06416.50.0173.6018

50.0-0.17724.60.0253.4017

50.0-0.28746.90.0473.2016

50.0-0.38582.40.0823.0015

24.0-0.462124.90.1252.8014

-31.3-0.518149.50.1502.6013

-50.0-0.536138.30.1382.4012

-50.0-0.51987.70.0882.2011

-50.0-0.47410.10.0102.0010

-50.0-0.408-71.6-0.0721.809

-50.0-0.331-132.6-0.1331.608

-50.0-0.250-156.6-0.1571.407

-50.0-0.173-142.0-0.1421.206

-50.0-0.107-102.5-0.1031.005

-50.0-0.062-62.1-0.0620.804

-28.8-0.029-28.8-0.0290.603

-8.0-0.008-8.0-0.0080.402

0.00.0000.00.0000.201

0.00.0000.00.0000.000

Force (kN)Inelastic Disp. (m)Force (kN)Elastic Disp. (m)Time (sec.)Step No

Inelastic ResponseElastic Response


11/36





-22.80.26011.80.0126.2031

-45.50.23819.30.0196.0030

-50.00.23319.00.0195.8029

-35.30.24911.20.0115.6028

-5.70.278-1.1-0.0015.4027

26.20.310-13.0-0.0135.2026

47.70.332-19.7-0.0205.0025

50.00.334-18.5-0.0184.8024

50.00.316-9.9-0.0104.6023

50.00.2792.70.0034.4022

50.00.22114.10.0144.2021

50.00.14420.00.0204.0020

Force (kN)Inelastic Disp. (m)Force (kN)Elastic Disp. (m)Time (sec.)Step No

Inelastic ResponseElastic Response




-200

-100

0

100

200

-1.0 -0.5 0.0 0.5 1.0

Displacement (m)

RestoringF

orce(kN)

Elastic Force (kN)

Inelastic Force (kN)


-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

0.4

0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00

Time (sec.)

Displacement(m)

Inelastic Disp. (m)

Elastic Disp. (m)

Sd


12/36



The Concepts of

Nonlinear Response

and Design Spectra



Elastoplastic Force-Deformation Relation

yy u

u

f

fR 00

Source: Chopra, A., Earthquake Dynamics of Structures A Primer, 2nd Edition, EERI, 2005.


13/36




14/36




15/36




16/36




17/36




INELASTIC DESIGN SPECTRA

Have you ever seen or used aHave you ever seen or used a

nonlinear or inelastic responsenonlinear or inelastic response

spectrum before?spectrum before?

NLSPECTRUM

YES, YOU HAVE.YES, YOU HAVE.


18/36




The Concept of

Equivalent

Linearization


19/36



THE CONCEPT OF EQUIVALENT LINEARIZATION

Consider Free Vibration of a Linear System:

k1

v

fs + fd

A-A

0.1

sin1coscos

sinsinsin

22

2

2

2

A

v

Ac

f

tActAcvcftAv

A

vttkAkvftAv

d

d

s

v

fd

A-A

+

k1

v

fs

A-A



THE CONCEPT OF EQUIVALENT LINEARIZATION

ntDisplacemeElasticMaximum

ForceElasticMaximum

max, 2 AfAreaDE s

cAAcAArea 2

Now Consider Free Vibration of a Simple

Nonlinear System:

keq1

v

F

A-A k0

Actual

Equivalent

Area of the ellipse Energy Dissipated in one Cycle

kmcm

k

m

c

2;

2

MAXMAX EE

eqF

ED

2

1


20/36



BASIC LINEARIZATION EXAMPLE

K

m =1000 kN.s2/mExample 2:

The SDOF structure shown below is subjected to horizontalground excitation represented by the pseudo-velocity response

spectrum shown. Stiffness and strength properties of the systemare shown on the hysteresis loop below. Assume zero systemviscous damping and calculate the following.

1. The elastic strength demand and the correspondingdisplacement.

2. The maximum inelastic displacement demand using basic

equivalent linearization technique.


0.0

0.1

0.20.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.5 1.0 1.5 2.0 2.5 3.0 3.5

Period (sec.)

SV(m/sec.)

"0% Damping"

"25% Damping"

"50% Damping"FORCE

DISP.

Fy= 1600 kN.

y= +20 mm



EXAMPLE 2 (CONTINUED)

rad/s8.941000

000,800

m

k

m/s56.00;sec.7.02

vST

ELASTIC RESPONSE

kN/m000,8002.0

16000

y

yFk

kN006,5006.5000,1

m062.0

m/s006.556.094.8S

m062.094.8

56.0

2

a

aE

dE

v

vd

mSF

S

S

SS


21/36




m062.0:Assume max E

%43

062.01600

02.02062.0216002

2

1222

2

1

2

1

sec.24.12

rad/s08.5000,1

806,25

kN/m806,25062.0

1600

max

max

MAXMAXMAXMAX EE

yy

EE

eq

eq

eq

eq

eq

y

eq

F

F

F

ED

T

m

K

FK

INELASTIC RESPONSE

ITERATION 1:

m/s45.0%43;sec.24.1 vST

needediterationAnotherm062.0m089.008.5

45.0

vd

SS




m/s6.14.04 2aS yad

FmSF

S

kN600,16.1000,1

m1.0

max

max

m10.0:Assume max E

%51

1.01600

02.021.0216002

2

1222

2

1

2

1

sec.57.12

rad/s4000,1

000,16

kN/m000,1610.0

1600

max

max

MAXMAXMAXMAX EE

yy

EE

eq

eq

eq

eq

eq

y

eq

F

F

F

ED

T

m

K

FK

INELASTIC RESPONSE

ITERATION 2:

m/s4.0%51;sec.57.1 vST

acheived.eConvergencm1.0m1.04

4.0

vd

SS


22/36



The Concept of

Nonlinear Static or

Push-Over Analysis



DESIGN SPECTRA REPRESENTATIONS

Ordinary DesignOrdinary Design

Period

V/W(Acceleration)

DESIGN SPECTRUM

Spectral or Roof-top Displacement

V/W(Acce

leration)

ConstantPeriodLines

ELASTIC DEMAND SPECTRUM

PushPush--Over AnalysisOver Analysis

Composite or ADRSComposite or ADRSPlotPlot


23/36



PUSH-OVER CURVE OR CAPACITY SPECTRUM

Using simple modal analysis

equations, spectraldisplacement and roof-topdisplacement may be convertedto each other.

Roof-top Displacement

V/W(Acceleration)

Low-Strength; Low-Stiffness; Brittle

Moderate Strength and Stiffness; Ductile

High-Strength; High-Stiffness; Brittle



ATC-40 CAPACITY SPECTRUM METHOD


24/36



ASCE-41 COEFFICIENTS METHOD



NSP OR PUSH-OVER ANALYSIS

V/W(Acceleration)


5% damped elastic spectrum

This is an iterative procedure involving severalThis is an iterative procedure involving several

analyses.analyses.

e

For each analysis an effective period for anFor each analysis an effective period for an

equivalent elastic system and aequivalent elastic system and a

corresponding elastic displacement arecorresponding elastic displacement are

calculated.calculated.

This displacement is then divided by a dampingThis displacement is then divided by a damping

factor to obtain an estimate of real displacementfactor to obtain an estimate of real displacement

at that step of analysis.at that step of analysis.

Teff

T0

e/B

ATCATC--4040


25/36




V/W(Acceleration)


5% damped elastic spectrum

capacity spectrum

e

Here an estimate of elastic displacementHere an estimate of elastic displacement

is obtained first.is obtained first.

This displacement is then multiplied by aThis displacement is then multiplied by a

set of modification factors to arrive at anset of modification factors to arrive at an

estimate of the target inelasticestimate of the target inelastic

displacement.displacement.

ASCE 41ASCE 41



ATC-40 CAPACITY SPECTRUM METHOD

This is inherently an iterative procedure.

65.1

100ln41.031.2

12.2

100ln68.021.3

1

11205.0

10

eff

V

eff

A

effeq

eq

SR

SR

TT


26/36



MODIFIED CAPACITY SPECTRUM METHOD PER FEMA-440

This is inherently an iterative procedure.







27/36










28/36



COEFFICIENTS METHOD AS MODIFIED PER FEMA-440


MODIFIEDC1COEFFICIENT: g

TSCCCC e

at 2

2

3210 4




gT

SCCCC eat 2

2

32104


MODIFIEDC1COEFFICIENT:

y

a

F

mSR


29/36




gT

SCCCC eat 2

2

3210 4


C3COEFFICIENT ELIMINATED AND REPLACED

WITH A MINIMUM STRENGTH REQUIREMENT: X

motionsgroundfield-nearfor8.0

motionsgroundfield-farfor2.0

ln15.01

4

2

max

PPe

t

e

y

d

y

a

Tt

RF

mSR




Push-over analysis is in reality the extension of responsespectrum analysis in order to perform approximatenonlinear analysis.


30/36



INCLUSION OF P-DELTA EFFECTS AND GRAVITY FRAMING ARE VITAL!

ROOF DRIFT ANGLE vs. NORMAL IZED BASE SHEAR

Pushover: LA 20-Story, Pre-Northridge, Model M2, =0%, 3%, 5%, 10%

0.00

0.05

0.10

0.15

0.20

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

Roof Drift Angle

NormalizedBaseShear(V/W)

Strain-Hardening =0%




Source: Krawinkler, H. (2005), A few comments on P-Delta, Overstrength, Drift, Deformation Capacity, Collapse Capacity, and other issues,

Presentation at the LATBSDC Invitational Workshop, September 22, Los Angeles.

Typical push-over analysis curves when P- effects are properly considered



Modeling Nonlinear

Behavior for Dynamic

Analysis


31/36


(Last Revision Date: 5-18-2009)BASIC INGREDIENTS

Nonlinear material model for theNonlinear material model for theelements and connections must beelements and connections must be

defineddefined

Backbone CurvesBackbone Curves

Hysteretic BehaviorHysteretic Behavior

This could be done usingThis could be done using

Results of experimental researchResults of experimental research

Publications such as FEMAPublications such as FEMA--356, ATC356, ATC--6262

Journal papers and proceedings ofJournal papers and proceedings ofTechnical SeminarsTechnical Seminars



BACKBONE CURVE EXAMPLE

Source: British Columbia Schools Retrofit Project, Draft Guideline, 2006.

Backbone curve for blocked OSB/plywood shear wall system


32/36



HYSTERETIC PROPERTIES EXAMPLE

Source: Naeim, Mehrain and Alimoradi, External Peer Review of British Columbia Schools Retrofit Project, Draft Guideline, 2006.

The effect of strengthdegradation on theresponse of the first

floor is evident.



BACKBONE CURVE AND HYSTERETIC LOOPS EXAMPLE

Source: British Columbia Schools Retrofit Project, Draft Guideline, 2006.

Backbone curve for a tension-only CBF


33/36




Source: Naeim, Mehrain and Alimoradi, External Peer Review of British Columbia Schools Retrofit Project, Draft Guideline, 2006.

The effect of strengthdegradation on theresponse of the first

floor is evident.




Source: ATC-62 Draft, 2006.

Springs 1a & b: Gravity system in buildings

Figure 1. Backbone and hysteretic behavior of Spring 1a.

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14

Interstory Drift Ratio

F / Fy

-0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

interstorydrift


Figure 2. Backbone and hysteretic behavior of Spring 1b.

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14

Interstory Drift Ratio

F / Fy

-0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

interstorydrift



34/36






Spring 2a & 2b: Non-ductile moment resisting frame

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14

InterstoryDrift Ratio

F / Fy


-0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08-1

-0.5

0

0.5

1

1.5

interstorydrift


0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14

InterstoryDrift Ratio

F / Fy

-0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08-1

-0.5

0

0.5

1

1.5

interstorydrift






Spring 3a & 3b: Ductile moment resisting frame



0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14-0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

interstorydrift


0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14-0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08-1

-0.5

0

0.5

1

1.5

interstorydrift

baseshear/b

aseshearatyield


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Spring 4a & 4b: Nonductilebrace frames



0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14-0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08-1

-0.5

0

0.5

1

1.5

interstorydrift


0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 -0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08-1.5

-1

-0.5

0

0.5

1

1.5

interstorydrift






Spring 5a & b : Infill Walls



0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.5

1

1.5

interstorydrift


-0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08-1.5

-1

-0.5

0

0.5

1

1.5ForcevsDisp

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.5

1

1.5

interstorydrift

baseshear/bas

eshearatyield

-0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08-1.5

-1

-0.5

0

0.5

1

1.5

interstorydrift

baseshear/base

shearatyield


36/36



Time for a break!

Thank You.Thank You.

2 NAEIM 2009 Nonlinear Response

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Transcript of 2 NAEIM 2009 Nonlinear Response