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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?
