seismic nonlinear dr alsaadi

By: Dr. Balhan A. Alsaadi

P.U.P.R Professor

Evaluation of the Non linear

Behavior of Structural

Concrete Elements subjected

to Seismic loads

Evaluación del comportamiento

no lineal de elementos

estructurales de hormigón

sometido

a cargas sísmicas

1. Identify

or Revise Criteria

2. Generate Trial Design 3. Develop Model

6. Refine Design 5. Performance Evaluate

Structural Design 4. Analyze Model

High probability of “Failure”

High Uncertainty

Importance of Details

Earthquake Protective Design Philosophical Issues

Target Building Performance:

1. Structural Performance Level:

•

•

•

•

•

•

Immediate Occupancy Damage Control Range

Life Safety

Limited Safety Range

Collapse Prevention

Not Considered

Elastic vs. Inelastic Response

The red line shows the force and displacement that would be reached if the structure responded elastically.

The green line shows the actual force vs. displacement response of the structure

The pink line indicates the minimum strength required to hold everything together during inelastic behavior

The blue line is the force level that we design for.

We rely on the ductility of the system to prevent collapse.

5

From 1997 NEHRP Provisions

1. Linear Static Equivalent

Lateral force Method

2.Linear

Dynamic

Modal

Response

Spectrum

Analysis Method

3.Nonlinear Static

Including P- D

4. Nonlinear Time History

Seismic Load Structural Analysis Procedure

Analysis Options:

1.

Linear static - (considered the

least accurate)

Linear model subject to lateral

loading determined by ASCE 7-10.

Similar to IBC 2009 equivalent

lateral force method.

Allowed only for structures without

irregularities defined in ASCE 7).

Allowed only for some structures

which do not have any irregularities defined.

Plan Structural Irregularities

• 1a - Torsional Irregularity

• 1b - Extreme Torsional Irregularity

• 2 - Re-entrant Corners

• 3 - Diaphragm Discontinuity

• 4 - Out-of-plane Offsets

• 5 - Nonparallel Systems

8

Analysis Options:

2.

Linear Dynamic Response Spectrum Analysis Method (considered more accurate than the LS)

Linear model subjected to response spectral or time history loading.

Allowed only for some structures which do not have any irregularities defined for the NSP.

Design Response Spectrum

10

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 1 2 3 4 5 6 7TS T0

Period, seconds

Sp

ec

tra

l A

cc

ele

rati

on

, g

0.4SDS

Sa = SD1 / T

Sa = SDS(0.4 + 0.6 T/T0)

Sa = SD1 TL / T2

Drawn for SS = 1.0, Fa = 1.0 S1 = 0.4, Fv = 1.5 TL = 4

Analysis Options:

3.

Nonlinear static – Including P- D (considered accurate enough

for most structures)

Structural model which Include P- D effect. Structural model with nonlinear

material behavior assigned to structural

elements subjected to an earthquake

Stability: P-Δ Effects

12

P

Δ Deflection introduces P-Δ moment which increases deflection, which increases moment ….. Structure must be designed to prevent collapse due to P-Δ effects

Analysis Options:

Buildings with non-orthogonal

lateral system.

Building with a vertical stiffness

irregularity.

Building which has a torsional

stiffness irregularity in any story.

Any structure where the horizontal

dimension of any story exceeds that

of an adjacent story by 1.4.

Required if any of the following are true:

Analysis Options:

4.

Nonlinear time history -

(considered the most accurate)

Structural model with nonlinear

material behavior assigned to

structural elements subjected to an

earthquake time history loading.

Required for certain structures

including those when R> Rmax. R is a parameter related to the structures’ capacity / the seismic demand.

Permitted for all structures

Nonlinear Modeling : Nonlinear static With P- D

A model that considers material nonlinearity in all elements which comprise it including: 1.

2 3 4. 5. 6.

Likely plastic hinge regions modeled with

FEMA 356 nonlinear hinges. Member curvature and lateral drift Duration of loads and the effects of shrinkage and creep

Concrete elements behavior under Service loads

(Serviceability) which include Deflection control

and cracking control

Allowances for Moment Redistribution

Interaction with the supporting foundation

Uncracked

Section

Cracked

Section

F

Idealized Behavior

Actual Behavior

c) Load-displacement overall to laterl load bihavior of the member

M

Mn

My

Mcr

Fn

Fy Cracked Section Semi-cracked Section

Uncracked Section Fcr

a) RC cantilever subjected b) behavior at the section

Behavior of reinforced concrete element in flexure (a) member subjected to lateral load, (b) moment-curvature response, (c) load-deformation response

Service Actual Behavior

Lateral

Load Ultimate

Yield

Factored

Load versus deflection behavior of a reinforced concrete frame

Displacement

Ib = Icr/μb

ACI 10.10.4.1 Recommended EI (Option 1)

Compression Members:

Columns

Walls – Uncracked

– Cracked

Flexural Members:

Beams

Flat plates/slabs

0.70EcIg

0.70EcIg

0.35EcIg

0.35EcIg

0.25EIg

Compression Members:

EI = [ 0.80 + 25 Ast

Ag

] [ 1- Mu

Puh - 0.5

Pu

P0 ] E Ig

0.35 EIg EI 0.875E Ig ≤ ≤

Flexural Members:

EI= (0.10+ 25 r) [ 1.2- 0.2 bw

d ] EIg

0.25 EIg EI 0.5 EIg ≤ ≤

ACI 10.10.4.1 Recommended EI (Option 2)

EI

Inelastic Hinge Spring

EI

EI

M

My

Elastic

End rotation,

Rigid-Plastic Hinge

EI

y

a) Concentrated-Hinge Models

M

My

Curvature,

b) Spread-of-plasticity Model

Nonlinear beam-column element models for frame analysis (a) concentrated-

hinge type, (b) spread-of-plasticity type.

Definition of Drift

22

I

C xedx

Structural displacement,

where,

xe Elastic deflection calculated

from design forces

dC Deflection amplification factor

Importance factor I

P-Delta

• What if your analysis program “includes” P-Delta and you don’t want to make a second set of output?

• max must still be checked

• Compute * from displacements that include P-Delta, then

23

max*1

*

P-Delta

• What if your analysis program “includes” P-Delta and you don’t want to make a second set of output?

• max must still be checked

• Compute * from displacements that include P-Delta, then

24

max*1

*

Drift Ratio Limits

Structure Occupancy Category UBC

I or II III IV

4 stories, no masonry 0.025 0.020 0.015 0.025*

Masonry cantilever 0.010 0.010 0.010

Other masonry 0.007 0.007 0.007

All other 0.020 0.015 0.010 0.020*

25

Nonlinear of Concrete Structures Why?

• Improving our prediction of the expected

range of structural response by modeling

‘real behavior’.

• Reduce the uncertainties that we control.

• Understand those that we cannot.

• Develop our ‘model in the mind’.

Nonlinear of Concrete Structures Why?

• By exploring solutions inside the code

• Alternate Means of Compliance

• By reducing structural scope and cost

• By improving structural & seismic performance

for the same or lower scope/cost.

• By improving post-earthquake outcomes and

reducing life-cycle costs.

• While improving our understanding of structural

behavior to make us better Designers

INPUT DATA:

f’c, fy, b, d, h, Asi , di”from the compression fiber

Ec=57,000 *SQR(f’c), f’’c=0.9*f’c e o=1.71*f’c/Ec, Es=29,000,000 Psi

FOR e c=0 to .003

Z e cm /e0

a Ln(1+Z^2)/Z g =1- 2*(Z- tan^-1(Z))/( a * Z^2)

1

e si e cm * ( Xn-di)/Xn

fsi e si * Es

Tsi S Asi * fsi

Cc a * f’’c *b * Xn

Find Xn which develop

Cc=Ts

N o

Yes

M=Cc*(h/2- g *Xn) + Tsi *(h/2-di) F = e cm/Xn

NEXT e c

MOMENT –CURVATURE DIAGRAM EXACT APPROACH

by: Dr. Balhan A. Alsaadi

M-C DIAGRAM FOR MU+

0

20

40

60

80

100

120

140

0 0.2 0.4 0.6 0.8 1 1.2

Beam Section Mu +

(12* 20) in

3# 8 2# 8

INPUT DATA:

f’c, fy, b, d, h, Asi , di”from the compression fiber

Ec=57,000 *SQR(f’c), f’’c=0.9*f’c, P0 e eo=1.71*f’c/Ec, ecm=.003 , Es=29,000,000 Psi

FOR P =0 to P0

Z e cm /e0

a Ln(1+Z^2)/Z g =1- 2*(Z- tan^-1(Z))/( a * Z^2)

1

e si e cm * ( Xn-di)/Xn

fsi e si * Es

Tsi S Asi * fsi

Cc a * f’’c *b * Xn

Find Xn which develop

P+Cc=Ts

N o

Yes

Mn=Cc*(h/2- g *Xn) + Tsi *(h/2-di) F = e cm/Xn

NEXT P

IINTERACTION DIAGRAM EXACT APPROACH

by: Dr. Balhan A. Alsaadi

INTERACTION DIAGRAM

Exact -Approach by: Dr. Balhan A. Alsaadi

Pmax

et >=.005

6# 8

fc=3Ksi fy=50 Ksi

g=0.8

b=16 in

g h=24 in

0

200

400

600

800

1000

1200

0 1000 2000 3000 4000 5000

Pn [kips]

Pb

Mb =

eb

e m

in

Section tension - controlled

Transition zone

e y <= e t <.005

e t = e y

Balanced strain condition

Section compression - controlled

e t < e y

Maximum axial compression

e c=0.003

e c=0.003

e c=0.003

e c=0.003

e c=0.003

Mn (K. Inch)

Seismic Hazard & Ground Motions

Earthquake Ground Motion

Selection and Scaling

Source: GeologyCafe.com (Base map modified after the Geologic Map

of California by Jenning, C.W., 1997, California Dept. of Mines and

Geology)

ASCE 7- 2010 Seismic Provisions and

Determination of Seismic design Categories

By

IBC 2009

Seismic Design Categories

To be determined for every structure

function of: Occupancy Category

Spectral Response Accelerations SDS and SD1.

Used to determine analysis options, detailed requirements, height limitations, and other limits on usage.

Seismic Design Categories labeled A-F

36

Seismic Ground Motion Values

• Mapped Acceleration Parameters

– Ss = Mapped 5% damped, spectral response acceleration parameter at short periods

– S1 = Mapped 5% damped spectral response acceleration parameter at a period of 1 sec.

.

37

Site Classes

38

Compute SMS and SM1

• SMS = FaSS

– Fa from Table 11.4-1

• SM1= FvS1

– Fv from Table 11.4-2

39

Spectral Response Accelerations SDS and SD1

• SDS is the design, 5% damped, spectral response acceleration for short periods.

• SD1 is the design, 5% damped, spectral response acceleration at a period of 1 sec.

• SDS and SD1 are used in selecting the Seismic Design Category and in the analysis methods.

40

SDS = 2*SMS/3 SD1 = 2*SM1/3

Importance Factor, I

• See ASCE 7-05 Table 11.5-1

– Function of Occupancy Category

• Requirement for structures adjacent to occupancy category IV structures where access is needed to get to the category IV structure.

41

Seismic Design Category continued….

• SD1 = 0.351

• SDS = 0.535

• Use Seismic Design

Category D

42 Seismic Provisions Example – A

Beginner’s Guide to ASCE 7-05

Deformation Compatibility

Applies to

• SD Category D+

• All structural components not in SFRS

• Check capacity for gravity load combined with effects induced from design drift; rational analysis of restraint required

• ACI 318 Chap 21 acceptable alternate

43

Component resisting earthquake

effect

and

the ACI Section of Chapters to

Be Satisfied

Seismic Design Category SDS

A B C D, E, F

Energy Dissipation In nonlinear range of response

Ordinary Ordinary Intermediate Special

Chapters 1 to 19 and 22

None 21.1.1.4 21.1.1.5 21.1.1.6

Analysis and design

requirements

None

21.1.2 21.1.2 21.1.2 & 21.1.3

Materials None None 21.1.4 to 21.1.7

Frame members 21.2 21.3 21.5 & 21.6 & 21.7 & 21.8

Structural walls and

coupling beams

None

None 21.9

Precast structural walls 21.4 21.4 & 21.10

Structural diaphragms and

trusses

None

21.11

Foundations 21.12

Frame members not

proportioned to resist forces

induced by earthquake

motions

21.13

Anchors 21.1.8 21.1.8

ACI Table R21.1.1 SECTIONS OF CHAPTER 21 TO BE SATSFIED IN TYPICAL APPLICATINS

5. 1.2D + 1.0E + L + 0.2S

6. 0.9D + 1.0W

7. 0.9D + 1.0 E

1. 1.4D 2. 1.2D + 1.6L + 0.5(Lr or S or R) 3. 1.2D + 1.6(Lr or S or R) + (L or 0.5W) 4. 1.2D + 1.0W + L + 0.5(Lr or S or R)

E=Eh+Ev

E=Eh-Ev

E = seismic load effect Eh = effect of horizontal seismic forces Ev = effect of vertical seismic forces

Eh= ΩQE

EV=0.2*SDS*D

5. (1.2 + 0.2SDS)D + Ω QE + L + 0.2S 7. (0.9 – 0.2SDS)D + Ω QE + 1.6H

The load factor on H shall be set equal to zero in combination 7 if the structural action due to H counteracts that due to E Ω= Overstrength Factor (ASCE 7-10) Table 12.2-1

Required Strength (Factored Load) ACI-318-11 and ASCE 7-10

ORDINARY STRUCTURAL (SDC B ) ACI 21.2 Ordinary MOMENT FRAME

a.) Ordinary Moment Frames shall satisfy ACI 21.2 b.) Ordinary Reinforced Concrete Structural Walls need not satisfy ACI Chapter 21 provisions

ACI 21.2 Ordinary Moment Frames

ACI 21.2.2 Beams shall have at least two of the longitudinal bars continuous along top and bottom faces and must be developed at the face of support ACI 21.2.3 Columns having clear height (Lu<=5*C1) shall satisfy the ACI 21.3.3.2 for shear provisions.

Vu=(Mnt+Mnb)/Lu Column flexural strength shall be calculate for Pu

FVn>= Min resulting in the highest flexural strength

ACI 21.3.3.2 Vu= (1.2 + 0.2SDS)VD + 2 VQE +V L + 0.2VS

ACI 21.3 INTERMEDIATE MOMENT FRAMES( SDC C )

PU<= (AG*f`c/10) Beam Reinforcement Details ACI 21.3.4

PU> (AG*f`c/10) Column Reinforcement Details ACI 21.3.5

Two Way Slab system without beams ACI 21.3.6

Vu=(Mnt+Mnb)/Lu Column flexural strength shall be calculate for Pu

FVn>= Min resulting in the highest flexural strength

ACI 21.3.3.2 Vu= (1.2 + 0.2SDS)VD + 2 VQE +V L + 0.2VS

Vu=(Mnl+Mnr)/Ln + Wu*Ln/2

FVn>= Min Where : Wu=1.2D+1.0L+.2S

ACI 21.3.3.1 Vu= (1.2 + 0.2SDS)VD + 2 VQE +V L + 0.2VS

Beam Shear Strength

Column Shear Strength

-

Mn >= 1/3 Mn + -

Mn >= 1/5( Mn)max. face

Mn Mn _ _

ACI 21.3.4 Beams ( SDC C )

2h 2h

h

b

Smax=Min

d/4 8*db (longitudinal)

24*db(hoop)

12 inches Over length 2h From support

face

-

+

- -

+

+ +

Smax=d/2 < Spacing Limits in ACI 11.4.5

_

ACI 21.3.5 Column( SDC C )

L0

L0

L0

L0

8*db (Longitudinal) 24 db (hoop) C1 0.5 * Min C2 12 Inches

L0=Max

1/6*Lu C1

C2

18 Inches

Maximum spacing over length Lo from joint face S0=Min

s0

s0

C1

C2

Smax

< S

pac

ing

Lim

its

in A

CI 1

1.4

.5

_

Max

ACI 21.5 Beams

ACI 21.5.1.1 Pu < _ Ag*f`c /10

ACI 21.5.1.2 Ln > _ 4d

Min 0.3h

10”

Min C2 +2C2

C2+1.5C1

ACI 21.5.1.3 ACI 21.5.1.4

SPECIAL MOMENT FRAME (SDC DEF)

Bwmin = Bwmax =

Mn >= 0.5 Mn + -

Mn >= 0.25*( Mn ) Suup.face

Mn Mn _ _

ACI 21.5.2 Beams Longitudinal Reinforcement

+

-

+ +

- - -

Asmax=0.025 bw*d

Asmin= Max 3 f`c

fy bw*d

200 fy

bw*d

Lap Splices of flexural reinforcement shall be Enclosed over the lap length by hoops at maximum spacing:

Smax=Min d/4 4”

Lap splices shall no be used

Within the Joint

2h from the support face

Maximum flexural Yielding

ACI 21.5.3 Lap splices and Transverse reinforcement

Vu = (1.2 + 0.2SDS)VD + 2 VQE +V L + 0.2VS

Mpr1+Mpr2 ln 2

Ve = + Vureq =Max

(1.2WD+1.0WL+.02s)*ln

FVn=FVc+FVs

FVc=0 If both occur

Mpr1+Mpr2 ln

Pu< _ Ag*f`c / 20

> _ 0.5Vureq

2h 2h

h

b

Smax=Min d/4 6*db (longitudinal)

6 inches

Over length 2h From support face

ln Mpr2 Mpr1

ACI 21.5.4 Beams Shear Strength and Transverse reinforcement

Mpr=As*(1.25*fy)*(d-a/2)

Ast*(1.25*fy)

0.85f`c*b a =

Smax=d/2 < Spacing Limits in ACI 11.4.5 _

ACI 21.6 Columns

ACI 21.6.1 Pu> Ag f`c/10

ACI 21.6.1.1 C1 > _ 12”

C2 > _ 12”

ACI 21.6.1.2 C2

C1 > _ 0.4

Asmax=.06 Ag

Asmin=.01 Ag ACI 21.6.3.1

Strong column/weak beam design moments

Mn,l

_

Mn,l

+

Mn,top

col

Mn,l

_

Mn,r

+ Mn,l

+ Mn,r

_

Mn,bot

col

Mn,l

_

Mn,l

+

Mn,bot

col

Mn,bot

col

Mn,bot

col

Mn,bot

col

Mn,top

col

Mn,top

col

Mn,top

col

Mn,bot

col

M + M > (M +M ) col col

n bot n top _ + -

n n 6 _ 5

Nominal column moments must be checked at maximum and minimum axial forces.

ACI 21.6 Columns

L0=Max

1/6*Lu Max

C1 C2

18 Inches

Maximum Spacing over length Lo from joint face

S0Max

S0Max

L0

L0

L0

L0

C1

C2

6*db (Longitudinal) C1 0. 25 * Min C2

4”<=[S0=4+( )]<=6”

S0 Max=Min

3 14-hx

Ashmin = max

0.3 S*bc*f`c

fy [(

Ag

Ach )-1 ]

0.09 S*bc*f`c

fy

S<= Min

6db

6”

Joint of Special moment frames

Beam in both side of column Vuj =1.25 fy (As top+As bot) - Vcol

Beams in one side of column 1.25 fy (As top) - Vcol

1.25 fy (As bot) - Vcol

Vuj = Max

Joint configurations and strength coefficients

F Vn= g f’c Aj

ACI21.9.7 Coupling Beams

Ln/h >=4 Design as flexural member of special moment

frame for moment and shear

Ln/h <2

Vu> 4 f’c Acw

AND Two intersection groups of diagonally

placed bars symmetrical about mid span

2<=Ln/h <4

Vu> 4 f’c Acw

OR

Two intersection groups of diagonally

placed bars symmetrical about mid span

Design as flexural member of special moment

frame for moment and shear

OR

f’c f Vnmax= f *10 *

>= bw/5

Avd req = Vu

f *2 *fy * Sin a

Additional longitudinal and transverse reinforcement Shall be

distributed around the beam perimeter in each direction

AS additional l = 0.002 bw * S S=12”

PREGUNTAS?