Inelastic Behavior

7/21/2019 Inelastic Behavior

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Inelastic Behavior of

Materials and Structures

Inelastic Behavior 1Revised 3/10/06


2/78

Inelastic Behavior of

Materials and Structures

Illustrate Inelastic Behavior of Materials and Structures

Explain Why Inelastic Response May be Necessary

Explain the Equal Displacement Concept

Introduce the Concept of Inelastic Design Response Spectra

Explain How Inelastic Behavior is Built into ASCE 7-05

Inelastic Behavior 2Revised 2/29/04


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Importance in Relation to ASCE 7-05

Derivation and Explanation of the Response

Reduction FactorR

Derivation and Explanation of the Displacement

Amplification FactorC

d

Derivation and Explanation of the OverstrengthFactor o

Inelastic Behavior 3


4/78

Inelastic Behavior of Structures

From Material

to Cross Section

to Critical Region

to Structure



5/78

Idealized Inelastic Behavior

From Material..


Strain

Stress

y

y u

u

= u

y


6/78

Stress-Strain Relationships for Steel



7/78

Stress-Strain Relationships for Steel



8/78

Unconfined Concrete

Stress-Strain Behavior


0

2000

4000

6000

800010000

12000

14000

16000

18000

20000

0 0.001 0.002 0.003 0.004

Strain, in./in.

Stres

s,psi

4500 psi

8800 psi

13,500 psi

17,500 psi


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Confined Concrete

Stress-Strain Behavior


0

1000

2000

3000

4000

5000

6000

7000

8000

0 0.01 0.02 0.03 0.04

Average strain on 7.9 in. gauge length

Stress,psi

no confinement

4.75 in.

3.5 in.

2.375 in.

1.75 in.

Pitch of in. dia.

spiral

Tests of

6 in. x 12 in.

cylindersHuge increase in Strain Capacity !


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Confinement by Spirals or Hoops

Asp

ds

fyhAsp

fyhAsp

Confinementfrom square

hoop

Forces actingon 1/2 spiral or

circular hoop

Confinementfrom spiral or

circular hoop



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Concrete Confinement

Rectangular hoops

with cross tiesConfinement by

transverse bars

Confinement by

longitudinal bars



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Idealized Stress-Strain Behavior of

Confined Concrete

Kent and Park Model

0

500

1000

1500

20002500

3000

3500

4000

4500

0 0.004 0.008 0.012 0.016

Strain, in./in.

Stress,psi

No Hoops

4 in.

6 in.9 in.

12 in.

Pitch of

in. dia.

spiral


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Unconfined

Confined


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Benefits of Confinement

Spiral Confinement

Virtually NO Confinement

Olive View Hospital, 1971 San Fernando Valley Earthquake



15/78

IDEALIZED INELASTIC BEHAVIOR

To Section..


Strain Stress Moment

y y My

y

Strain Stress Moment

u u Mu

u

ELASTIC

INELASTIC


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To Section..

Curvature


Moment

My

y u

Mu

=

u

y

NOTE:


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Software for Moment - Curvature Analysis

XTRACT



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IDEALIZED INELASTIC BEHAVIORTo Critical Region and Member..


LP

Curvature

My

Mu

y u

Moment

y

u

Area u

Area y

ELASTIC INELASTIC



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IDEALIZED INELASTIC BEHAVIORTo Critical Region and Member..


Rotation

Moment

My

y u

Mu

= u

y

NOTE:

Critical Region Behavior of a Steel Girder


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Critical Region Behavior of a Steel Girder


Strength Loss

Due to Flange

Buckling



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IDEALIZED INELASTIC BEHAVIORTo Structure..


Force

Displacement

= u

y

Note:


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Loss of Ductility through Hierarchy

Strain = 100

Curvature = 12 to 20

Rotation= 8 to 14

Displacement = 4 to 10



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Ductility and Energy Dissipation Capacity

System ductility of 4 to 6 is required foracceptable seismic behavior.

Good hysteretic behavior requiresductile materials. However, ductility in

itself is insufficient to provide acceptableseismic behavior.

Cyclic energy dissipation capacity is abetter indicator of performance.



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Response Under Reversed

Cyclic Loading

(t)

+

Time

Earthquakes impose DEFORMATIONS. Internal Forces Develop

as a result of the deformations.


Laboratory Specimen under Cyclic Deformation Loading


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Load Cell

Displacement

Transducer

(Controls )

Hydraulic

Ram

Records (F)

+

Time

1

8

7

6

5

4

3

2

Moment ArmL

Deformations are cyclic,

inelastic, and reversed




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Measured Hysteretic

Behavior

Force

Deformation

Load Cell

Hydraulic

Ram

Records (F)

Displacement

Transducer

(Controls ) Moment ArmL


Hysteretic Behavior


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Hysteretic Behavior

Deformation Deformation

Force Force

PINCHED

(Good)ROBUST

(Excellent)


Hysteretic Behavior


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Hysteretic Behavior

ForceForce


Deformation

PINCHED (with strength loss)

Poor

Deformation

x

BRITTLE

Unacceptable

Hysteretic Behavior


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Hysteretic Behavior

Deformation Deformation

ForceForceAREA=Energy Dissipated

ROBUST PINCHED (No Strength Loss)



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The structure should be able to sustain severalcycles of inelastic deformation without significant

loss of strength.

Some loss of stiffness is inevitable, but excessive

stiffness loss can lead to collapse.

The more energy dissipated per cycle, withoutexcessive deformation, the better the behaviorof the structure.



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The art of seismic resistant design is in the details.

With good detailing, structures can be designed

for force levels significantly lower than would berequired for elastic response.


Why is Inelastic Response Necessary?


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Why is Inelastic Response Necessary?

Compare the Wind and Seismic Design of a

Simple Building

Building Properties:

Moment Resisting Frames

Density= 8 pcf

Period T = 1.0 sec

Damping= 5%

Soil Site Class B

120

90

62.5

Wind:

100 MPH Exposure C

Earthquake:

Assume SD1=0.48g


Wi d


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Wind:

90

62.5

120

120 mph 3-Sec GustExposure C

Velocity pressure q= 36.8 psfGust factor G = 0.85

Pressure coefficient Cp= 1.3

Load and Directionality Factor for Wind = 1.6x0.85=1.36

Total wind force on 120 face:

VW120= 62.5*120*36.8*0.85*1.3*1.36/1000 = 415 kipsTotal wind force on 90 face:

VW90 = 62.5*90*36.7*0.85*1.3*1.36/1000 = 310 kips


E th k


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Earthquake:

90

62.5Building Weight W=

120*90*62.5*8/1000 = 5400

kips 120

V C WEQ S=

C S

T R IS

D= = =1 0 48

1 0 1 0 1 00480

( / )

.

. ( . / . ).

Total ELASTIC earthquake force (in each direction):

VEQ = 0.480*5400 = 2592 kips



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Comparison: Earthquake vs. Wind

120

2592 6.2415

EQ

W

VV

= =90

2592 8.4310

EQ

W

VV

= =

ELASTIC Earthquake forces 6.2 to 8.4 times wind!

Virtually impossible to obtain economical design



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How to Deal with Huge Earthquake Force?

Isolate structure from ground (Base Isolation) Increase Damping (Passive Energy Dissipation)

Allow Controlled Inelastic Response

Historically, Building Codes use Inelastic Response Procedure.

Inelastic response occurs through structural damage (yielding).

We must control the damage for the method to be successful.


Assume Frame is Designed for Wind

P h A l i P di t St th 500 k


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Force, kips

Displacement, inches

4.0

400

200

1.0 2.0 3.0

600

5.0

Pushover Analysis Predicts Strength = 500 k

Actual

Idealized

How will Frame Respond During

0 4g El Centro Earthquake?


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0.4g El Centro Earthquake?

Force, kips

4.0

600


400

Fy=500 k

K1=550 k/in

K2=11 k/in

200

2.0 3.0 5.01.0

Idealized SDOF Model

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0 2 4 6 8 10 12 14 16 18 20

T im e , Secon ds

Acceleration,g

Displacement, in.

El Centro Ground Motion

Response Computed by NONLIN:


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Maximum

Displacement:

Maximum

Shear Force:

Number ofYield Events:

4.79

542 k

15




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Yield Displacement = 500/550 = 0.91 inch


= =Maximum Displacement

Yield Displacement

4 79

0 915 26

.

..Ductility Demand

Interim Conclusion (The Good News)


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Interim Conclusion (The Good News)

The frame, designed for a wind force which is 15%of the ELASTIC earthquake force, can survive the

earthquake if:

It has the capability to undergo numerous cycles ofINELASTIC deformation

It has the capability to deform at least 5 to 6times the yield deformation

It suffers no appreciable loss of strength

REQUIRES ADEQUATE DETAILING


Interim Conclusion (The Bad News)


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Interim Conclusion (The Bad News)

As a result of the large displacements associated with

The inelastic deformations, the structure will suffer

considerable structural and nonstructural damage.

This damage must be controlled byadequate detailing and by limiting

structural deformations (drift)


Development of Equal Displacement


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Development of Equal Displacement

Concept of Seismic Resistant Design.

Concept used by:

UBC/IBC

NEHRP

ASCE-7

FEMA 273

In association with Force Based

design concept. Used to predict

design forces and displacements

In association with Static Push Over

Analysis. Used to predict displacementsat various performance points.


Th E l Di l t C t


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The Equal Displacement Concept

The displacement of an inelastic system, with

stiffness K and strength Fy, subjected to a particularground motion, is approximately equal to the displacement

of the same system responding elastically.

(The displacement of a system is independent of the

yield strength of the system)


Repeat Analysis for Various Yield Strengths

(All other parameters stay the same)


45/78


0

1

2

3

4

5

6

7

0 500 1000 1500 2000 2500 3000 3500 4000

Yield Strength (K)

MaximumD

isplacement(in)

(All other parameters stay the same)

0

1

2

3

4

5

6

0 500 1000 1500 2000 2500 3000 3500 4000

Yield Streng th (K)

DuctilityDemand

Avg. = 5.0 in.

Inelastic

Elastic

Elastic = 5.77

Constant Displacement Idealization


46/78

p

of Inelastic Response

0

500

1000

1500

2000

2500

3000

3500

0 2 4 6 8

Displacement, Inches

Forc

e,

Kips

ACTUAL BEHAVIOR

0

500

1000

1500

2000

2500

3000

3500

0 2 4 6 8


Forc

e,

Kips

IDEALIZED BEHAVIOR


Equal Displacement Idealization


47/78

q p

of Inelastic Response

For design purposes, it may be assumed that

inelastic displacements are equal to thedisplacements that would occur during an

elastic response.

The required force levels under inelastic responseare much less than the force levels required for

elastic response.


Equal Displacement Concept


48/78

0

500

1000

1500

2000

2500

3000

3500

0 2 4 6 8


Force,

Kips

Elastic

Inelastic

Design for This Force Design for this Displacement

of Inelastic Design

5.77


Equal Displacement Concept

of Inelastic Design


49/78

0

500

1000

1500

2000

2500

3000

3500

0 2 4 6 8


Force,

Ki

ps

Elastic

Inelastic

du=5.77dy=0.91

Ductility Supply MUST BE > Ductility Demand =


5 77

0 91

6 34.

.

.=

of Inelastic Design

3500 FE


50/78

0

500

1000

1500

2000

2500

3000

0 2 4 6 8


Force,

Kips

Using Response Spectrum, Estimate ELASTIC Force Demand FE

FI

dR dI

Estimate Ductility Supply, , and determine INELASTIC ForceDemand FI = FE / Design Structure for FI


Compute Reduced Displacement dR, and multiply by to obtain true

Inelastic Displacement dI.

Check Drift using di

ASCE-7 Approach


51/78

pp

Use basic elastic spectrum, but for strength,

divide all pseudoacceleration values by R,

a response modification factor that accounts for:

Anticipated ductility supply

Overstrength Damping (if different than 5% critical) Past performance of similar systems

Redundancy


Ductility/Overstrength

FIRST SIGNIFICANT YIELD


52/78

FIRST SIGNIFICANT YIELD

First

Significant

Yield

Force

Design

Strength


Displacement


53/78

First Significant Yieldis the level of force which causes complete

plastification of at least the most critical regionof the structure (e.g. formation of the first

plastic hinge)

The Design Strength of a Structure is equal

to the resistance at first significant yield.


Overstrength (1)


54/78

Force

Design

Strength


Displacement

Overstrength (2)


55/78

Force

Design

Strength


Displacement

Overstrength (3)


56/78

Force

ApparentStrength

Over StrengthDesign

Strength


Displacement

Sources of Overstrength


57/78

Sources of Overstrength

Sequential Yielding of Critical Regions

Material Overstrength (Actual vs Specified Yield) Strain Hardening

Capacity Reduction ( ) Factors Member Selection


Definition of Overstrength Factor


58/78

Overstrength Factor =Apparent Strength

Design Strength

Force

Apparent Strength

Over Strength

Design Strength

Displacement


Definition of Ductility Reduction Factor RdForce


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Elastic StrengthDemand


Design Strength

Apparent Strength

DisplacementElastic

Displacement

Demand

Strength

DemandReduction due

To Ducti lity

Over Strength

Definition of Ductility Reduction Factor

Elastic Strength Demand


60/78


Ductility Reduction Rd= Apparent Strength


DesignStrength

Apparent

Strength

Force

Displacement

Elastic

StrengthDemand

Elastic

Displacement

Demand

Strength

Demand

Reduction dueTo Ducti lity

Over Strength

Definition of Response Modification

Coefficient R


61/78

Overstrength Factor =Apparent Strength

Design Strength

Ductility Reduction Rd

=Elastic Strength Demand

Apparent Strength


Design Strength =RdR

=


Definition of Response Modification

Coefficient R


62/78

Force

Reduced (Design) Strength

R x Design Strength

DisplacementElastic

Displacement

Demand

x Design Strength

ANALYSIS DOMAIN


Definition of Deflection Amplification Factor Coefficient Cd

Elastic Strength


63/78

Elastic Strength

Demand


Design Strength

Apparent Strength

Elastic

Displacement

DemandE

Actual Inelastic

Displacement

DemandI

Computed Design

Displacement

DemandD

ASCE 7 Approach for Displacements


64/78

Determine Design Forces V=CsW, where C

sincludes

Ductility/Overstrength Reduction Factor R.

Distribute forces vertically and horizontally, and

compute displacements using linear elastic analysis.

Multiply computed displacements by Cd

to obtain

estimate of true inelastic response.


EXAMPLES of Design Factors

for Steel Structures


65/78


ASCE 7-05R o Rd Cd

Special Moment Frame 8 3 2.67 5.5

Intermediate Moment Frame 4.5 3 1.50 4.0Ordinary Moment Frame 3.5 3 1.17 3.0

Eccentric Braced Frame 8 2 4.00 4.0

Eccentric Braced Frame (Pinned) 7 2 3.50 4.0

Special Concentric Braced Frame 6 2 3.00 5.0

Ordinary Concentric Braced Frame 3.25 2 1.25 3.25

Not Detailed 3 3 1.00 3.0

Note: Rd

is Ductil ity Demand ONLY IF o is Achieved.

EXAMPLES of Design Factors

for Reinforced Concrete Structures


66/78

o e o ced Co c ete St uctu es

ASCE 7-05

R o Rd Cd

Special Moment Frame 8 3 2.67 5.5Intermediate Moment Frame 5 3 1.67 4.5

Ordinary Moment Frame 3 3 1.00 2.5

Special Reinforced Shear Wall 5 2.5 2.00 5.0Ordinary Reinforced Shear Wall 4 2.5 1.60 4.0

Detailed Plain Concrete Wall 2 2.5 0.80 2.0

Ordinary Plain Concrete Wall 1.5 2.5 0.60 1.5

Note:Rd is Ductility Demand ONLY IF o is Achieved.


ASCE-7 Elastic Spectra as

Adjusted for Ductility and Overstrength


67/78

j y g


C SR I

SDS=/

C S

T R IS

D= 1

( / )0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 1 2 3 4

Period, Seconds

Acc

eleration,g

R=1

R=2

R=3

R=4

R=6

R=8

SDS=1.0g SD1=0.48g

1 2

Using Modified ASCE-7 Spectrum

to Determine Force Demand


68/78


0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 1.0 2.0 3.0 4.0 5.0

Period, Seconds

Pseudoacceleration,g

R=1

R=4

0.15V=0.15W

Using Modified ASCE-7 Spectrum

to Determine Displacement Demand


69/78


0

5

10

15

20

25

0 1 2 3 4 5

Period, Seconds

Displa

cement,inches.

R=1

R=4

(R=4)*Cd

Cd=3.5

Displacements must be Multiplied by factor CdBecause Displacements Based on Reduced


70/78


Force would be too low INELASTIC d REDUCED ELASTICC=

0

5

10

15

20

25

0 1 2 3 4 5

Period, Seconds

Displacemen

t,inches.

R=1

R=4

(R=4)*Cd

Cd=3.5

inINELASTIC 65.3=


71/78

Equal Displacement approach may not beapplicable at very low Period Values


Equal Energy Concept(Applicable at Low Periods)


72/78

ELASTIC ENERGY

FE

EE

FI

F

F

E

I

y

E F F

F

F

FE E

E

I

y yE

I

= =0 5 0 52

. .


Equal Energy Concept

(Applicable at Low Periods)


73/78

INELASTIC ENERGY

FI

y u

EI

( )5.05.0 == yIyIuII FFFE


Equal Energy Concept

(Applicable at Low Periods)


74/78

Assuming EE= EI :

FE

FI

F

F

E

I= 2 1


100

Elastic:F

Newmark Inelastic Spectrum (for Psuedoacceleration)


75/78


0.1

1

10

0.01 0.1 1 10

Period, Seconds

Pseud

oVelocity,In/Sec

Equal Energy

Elastic:

Inelastic:

EqualDisp.

Transition

EI

FF =

12 =

EI

F

F

Newmarks


76/78

Inelastic Design Response Spectrum

To obtain Inelastic DISPLACEMENTSpectrum, multiply the Spectrum shown in

Previous slide by (for all Periods)


100

Newmarks Inelastic Design Response Spectrumfor Acceleration and Displacement


77/78


0.1

1

10

0.01 0.1 1 10Period, Seconds

PseudoV

elocity,

In/Sec

Disp.

Accel.

At very low periods, ASCE-7 Spectrum does not reduce to


78/78

ground acceleration, so this partially compensates forerror in equal displacement assumption at low Period

Values.

Period, T

Cs

Note: FEMA 273 has explicit modifies for computing Target

at Low Periods


Inelastic Behavior

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