Group of Earthquake Engineering and Structural Dynamics ... · Lumped Plasticity Models: Ruaumoko,...
Transcript of Group of Earthquake Engineering and Structural Dynamics ... · Lumped Plasticity Models: Ruaumoko,...
29.August.2006
Post-earthquake residual deformations and seismic performance assessment of buildingsPost-earthquake residual deformations and seismic performance assessment of buildings
Ufuk Yazgan and Alessandro DazioGroup of Earthquake Engineering and Structural Dynamics
Institute of Structural Engineering, ETH-Zurich
© ETH Zürich
229.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Motivation
The need to reduce the uncertainties in the assessment of seismic performance after an earthquake
The need for a direct consideration of post-earthquake residual displacements in the seismic design of structures
Management of EarthqaukeRisks using Condition Indicators
329.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Post-Earthquake Residual Displacements
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0 5 10 15 20 25 30Time (s)
Displacement (m)
-160
-120
-80
-40
0
40
80
120
160
-0.06 -0.04 -0.02 0 0.02 0.04 0.06Displacement (m)
Force (kN)
Maximum Displacement
Residual Displacement
429.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Residual Deformations at the Component Level
(Source: Lestuzzi P., Wenk. T & Bachmann T., (1999), “DynamischeVersuche an Stahlbetontragwänden auf dem ETH-Erdbebensimulator”, IBK IBK Bericht Nr. 240, April 1999 )
Courtesy National Information Service for Earthquake Engineering, University of California, Berkeley.
529.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Residual Displacements at the Structural Level
Courtesy National Information Service for Earthquake Engineering, University of California, Berkeley.
629.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Residual Displacements due to Ground Deformations
Courtesy National Information Service for Earthquake Engineering, University of California, Berkeley.
729.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Post-Earthquake Assessment
A Seismic Performance Assessment Method for buildings based on residual displacements is being developed.
A Seismic Performance Assessment Method for buildings based on residual displacements is being developed.
Damaged Building
Reconnaissance Survey
Recorded Ground Motion FEM Model
Measured Residual Displacements
Estimates of Maximum
Displacements
PostPost--Earthquake Earthquake VulnerabilityVulnerability
0
0.2
0.4
0.6
0.8
1
0 2 4 6 8 10 12 14Spectral Displacment - Sd (cm)
P(D
>DC
i)
P(D>DC0)P(D>DC1)P(D>DC2)P(D>DC3)P(D>DC4)
829.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Performance-Based Design
Provide essential information about the the post-earthquake:- Reparability/Usability- Vulnerability
Estimates of Maximum
Displacements
0
1
2
3
4
5
6
0.01 0.1 1 10T [s]
Se [m/s2]
ABCDE
Estimates ofResidual
Displacements
PSH Model
Seismic Performance
FE Model
Known to be well correlated with the attained damage
929.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Lumped Plasticity Models: Ruaumoko, DRAIN-2DX, OpenSees, SeismoStructMoment, M (kNm)
0
50
100
150
200
250
0 0.1 0.2 0.3 0.4
Curvature, 8 (m-1)
270
270
1500
400
400 8 φ 16
φ 8 / 100
-150
-100
-50
0
50
100
150
-0.06 -0.04 -0.02 0 0.02 0.04 0.06
Displacement (m)
Force (kN)
RuaumokoSeismoStruct (Lump.Plast.)DRAIN-2DXOpenSees (Lump.Plast.)
…… are practically the same. are practically the same.
Simulated largeSimulated large--cycle responses cycle responses ……
2.0=′
=cg fA
Nn %1=gρ
Difficulties related to simulation of response
1029.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Results: Lumped Plasticity ModelsPractically, the maxima are the samePractically, the maxima are the same
However, the residual displacements are significantly differentHowever, the residual displacements are significantly different
Maximum Displacement, um (m)
0
0.015
0.03
0.045
0.06
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Ground Motion
Maximum Displacement, um (m)
0
0.015
0.03
0.045
0.06
Mean StandardDeviation
Ruaumoko(26/10/04)
SeismoStruct(v.3.9.5)
DRAIN-2DX(v.1.10)
OpenSees (1.6.2e)
Residual Displacment , ur (m)
0
0.005
0.01
0.015
0.02
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Ground Motion
Residual Displacement, ur (m)
0
0.005
0.01
0.015
0.02
Mean StandardDeviation
!!
Ry = 4T = 0.5 [s] ζζ == 5%
1129.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Fiber Models: OpenSees, SeismoStruct, Rechenbrett-2D
70
10070
100140
140
200
280
400
1500
Stress (MPa)
-600
-400
-200
0
200
400
600
0 0.01 0.02 0.03 0.04 0.05
Strain
OpenSees(Steel02)SeismoStruct(stl_mp)
Stress (MPa) -40
-30
-20
-10
0-0.03 -0.02 -0.01 0
Strain
Reinforcement:Reinforcement: Concrete:Concrete:
-150
-100
-50
0
50
100
150
-0.06 -0.04 -0.02 0 0.02 0.04 0.06
Displacement (m)
Force (kN)
SeismoStruct (Fiber)
OpenSees (Fiber)
Rechenbrett-2D
OpenSees (Concrete02)SeismoStruct(con_cc)
Simulated largeSimulated large--cycle responses cycle responses ……
…… practically the samepractically the same
1229.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Results: Fiber ModelsAgain Again …… Practically, the maxima are the samePractically, the maxima are the same
AgainAgain…… the residual displacements are significantly differentthe residual displacements are significantly different
Maximum Displacement, um (m)
0
0.015
0.03
0.045
0.06
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Ground Motion
Maximum Displacement, um (m)
0
0.015
0.03
0.045
0.06
Mean StandardDeviation
OpenSees (1.6.2e)
SeismoStruct(v.3.9.5)
Rechenbrett-2D
Residual Displacement , ur (m)
0
0.002
0.004
0.006
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Ground Motion
Residual Displacement, ur (m)
0
0.002
0.004
0.006
Mean StandardDeviation
!!
1329.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Results: Fiber vs Lumped Plasticity
Residual Displacement, ur (m)
0
0.005
0.01
0.015
0.02
0 0.005 0.01 0.015 0.02Model with lumped plasticity elements
Mod
el w
ith fi
ber e
lem
ents
Maximum Displacement, um (m)
0
0.01
0.02
0.03
0.04
0.05
0.06
0 0.01 0.02 0.03 0.04 0.05 0.06Model with lumped plasticity elements
Mod
el w
ith fi
ber e
lem
nts
Maximum DisplacementsMaximum Displacements Residual DisplacementsResidual Displacements
Residual displacements are significantly influenced by the Residual displacements are significantly influenced by the adopted modeling approachadopted modeling approach !!
1429.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Comparison with the experimental dataDescription of the shake table test
Lestuzzi, P., T.Wenk, H. Bachmann (1999), “Dynamische Versuche an Stahlbetontragwänden auf dem ETH-Erdbebensimulator”, IBK Bericht Nr.240, Institut für Baustatik und Konstruktion, ETH Zürich
1529.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Rotation - z = 0.25 [m]
-0.01
-0.005
0
0.005
0.01
0.015
0 4 8 12 16 20
Time [s]
Bending Moment [kNm]
-150
-100
-50
0
50
100
150
-0.01 -0.005 0 0.005 0.01 0.015Rotation - z=0.25 [m]
Rotation - z = 0.25 [m]
-0.01
-0.005
0
0.005
0.01
0.015
0 4 8 12 16 20
Time [s]
Bending Moment [kNm]
-150
-100
-50
0
50
100
150
-0.01 -0.005 0 0.005 0.01 0.015Rotation - z=0.25 [m]
Rotation - z = 0.25 [m]
-0.01
-0.005
0
0.005
0.01
0.015
0 4 8 12 16 20
Time [s]
Bending Moment [kNm]
-150
-100
-50
0
50
100
150
-0.01 -0.005 0 0.005 0.01 0.015Rotation - z=0.25 [m]
Response history
Peak deformations
Points of unloading
Rotation - z = 0.25 [m]
-0.01
-0.005
0
0.005
0.01
0.015
0 4 8 12 16 20
Time [s]
Bending Moment [kNm]
-150
-100
-50
0
50
100
150
-0.01 -0.005 0 0.005 0.01 0.015Rotation - z=0.25 [m]
Residual deformation
1629.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
-150
-100
-50
0
50
100
150
0 2 4 6 8 10 12 14 16 18
Time [s ]
Moment [kNm ]
Measured
Simulated
FE model is subjected to the measured rotation history
-150
-100
-50
0
50
100
150
-0.01 -0.005 0 0.005 0.01 0.015
Rotation
Mom
ent [
kNm
]
1729.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
0 2 4 6 8 10 12 14 16 18 20-60
-40
-20
0
20
40
60
Time(s)
Top
disp
lace
men
t(mm
)
ExperimentOpenSees
1829.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Example applicationLocated in a region of high seismicityMw7.4 Kocaeli, Turkey 1999 EarthquakeAround ~7000 buildings - mostly RC - MRF Three decision alternatives:
strengthening the frame by structural wallsno actionpreventing liquefaction by stone columns
0 400 800100[m]
1234567
Number of Stories
Adapazari, Turkey
Performance of the structural stock subjected to a range of seismic events needs to be assessed.
(Bayraktarli Y., U. Yazgan, A. Dazio A. and M. Faber, 2006)
1929.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Example applicationSet of representative RC frame buildings
Yield ground story drift, θyPriestley (1998)
•Fundamental mode shape, φ1
•Yield displacement at roof, Δy,roof
Nstory=1,2,3,…,7
hstory = 2.4 [m]Lbay = 6 [m] εsy = 2.1 [%]
Astory=300 [m2]
•Modal participation factor, Γ1Mstory= 0.6 [t/m2]
0
0.2
0.4
0.6
0.8
1
1.2
0 0.2 0.4 0.6 0.8 1T [s]
Sa [g]Elastic spectrum
Design spectrum
TEC-98, ζ=5%, Z1, R=8
Yield displacement, Δ*y
Base shear capacity, V*y
Yield displacement, Δ*y
Base shear capacity, V*y
0
200
400
600
800
1000
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35D [m ]
V [kN ]
1
2
3
4
5
6
7
Nstory
Equivalent SDOF Systems for RC - MRF's
Dazio(2000)Modal Analysis
2029.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Example applicationRepresentative models for retrofitted buildings
H/L ≅ 4ρ = 1%
• Yield curvature, φy
• Yield displacement at the top, Δy,w
• Flexural yield strength, My
• Base shear at yielding Vy
Added Structural Walls
• Yield displacement, Δy,roof
• Base shear at yielding Vy
Wall +Frame System Dazio(2000)
0
500
1000
1500
2000
2500
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4
D [m]
V [kN]
1
2
3
4
5
6
7
Nstory
Equivalent SDOF Systems for RC MRF + Wall
2129.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Example applicationGround Motions
T, ζ, Ry
Substitute SDOF System
Simulated Response
Maximum Displacement, Δmax
Residual Displacement, Δres
Performance State
Vision2000(1995)
Which Δres ?
10-8 10-6 10-4 10-2 1000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1nstory=4
P(P
S>P
Si)
ures[m]
FOOLSNCC
Δres [m]
2229.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
ConclusionsResidual displacements are significantly influenced by the adopted modeling approach
It is possible to establish a performance assessment strategy taking into account residual displacements to update the uncertainties
Further study is needed to identify:- Which deformations would provide an effective measure?
- Which are feasible to measure?
2329.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Outlook / Open QuestionsRC frame and structural wall systems are being studied
Effective intensity measures related to residual displacements are being investigated
Efficient ways to relate the seismic performance to structural response parameters needs to be identified
Methods to include nonlinear soil deformation in the seismic performance assessment needs to be investigated
Effect of ground motion record processing schemes on the simulated residual displacements needs to be studied
2429.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
ThankThank youyou
Any Questions?Any Questions?
2529.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
-150
-100
-50
0
50
100
150
-0.04 -0.02 0 0.02 0.04 0.06
Displacement (m)
Force (kN)
-0.04
-0.02
0
0.02
0.04
0.06
0 5 10 15 20Time [s]
Dis
plac
emen
t [m
]
Lumped-plasticity model(Ruaumoko)Fiber-element model(OpenSees)
2629.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
-80
-40
0
40
80
120
160
0 0.015 0.03 0.045
Displacement (m)
Force (kN)
Stiffness is updated at the Stiffness is updated at the end of each stepend of each step
Stiffness is updated within Stiffness is updated within each stepeach step
-80
-40
0
40
80
120
160
-0.015 0 0.015 0.03 0.045
Displacement (m)
Force (kN)
dt=0.01
dt=0.005
dt=0.002
dt=0.0005
dt=0.0001
ImprtanceImprtance of of ““time integration step sizetime integration step size”” and proper updating of stiffnessand proper updating of stiffness
2729.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Maximum Displacement, um (m)
0
0.02
0.04
0.06
0.08
0 0.02 0.04 0.06 0.08With late stiffness updating
With
pro
per s
tiffn
ess
upda
ting
Residual Displacement, ur (m)
0
0.005
0.01
0.015
0.02
0 0.005 0.01 0.015 0.02With late stiffness updating
With
pro
per s
tiffn
ess
upda
ting
Properly updating stiffness is crucial for Properly updating stiffness is crucial for
estimating the residual displacements estimating the residual displacements !!
2829.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Force (kN)
-200
-150
-100
-50
0
50
100
150
200
-0.06 -0.04 -0.02 0 0.02 0.04 0.06
Displacement (m)
1-element
3-elements
5-elements
7-elements
Force (kN)
-150
-100
-50
0
50
100
150
-0.06 -0.04 -0.02 0 0.02 0.04 0.06
Displacement (m)
Stiffness-based approach
Flexibilty-based approach
Effect of mesh densityEffect of mesh density Effect of element formulationEffect of element formulation
2929.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
-500
-400
-300
-200
-100
0
100
200
300
400
500
-0.005 0 0.005 0.01 0.015 0.02
Strain
Stress (MPa)
without partial reloading
with partial reloading
26% increase in strain at the unloaded state
Partial reloading
-150
-100
-50
0
50
100
150
-0.04 -0.03 -0.02 -0.01 0 0.01 0.02 0.03 0.04
Displacement (m)
Force (kN)
without partial reloading
with partial reloading
25% increase in the displacement at the unloaded state
Partial reloading
Force Force -- DisplacementDisplacement Steel StressSteel Stress--StrainStrain
Limited memory problem of the steel Limited memory problem of the steel hysteresishysteresis::
Properly simulating the response of steel is Properly simulating the response of steel is
critical for estimating the residual displacements critical for estimating the residual displacements !!
3029.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Residual displacements are significantly influenced by the adopted modeling approach
Differences in the small-cycle response leads to notably different residual displacement estimates
“Time integration step size” can have a strong effect on the computed residual displacements
Meshing of the system may have a significant influence on the computed residual displacements
“Memory limitation” problem related to steel hysteresis response results inappropriate unloading paths
3129.August.2006 Ufuk Yazgan / Institute of Structural Engineering / [email protected]
Limit State DriftsCalculation of limit state displacementsMaximum Displacements Hstory= 2.4 [m] - Height of a story MRF
HN FO O LS NC C FO O LS NC C φ1,1 Γ FO O LS NC Cnstory [m] < > <[m] [m] [m] [m] [m]> nstory <[m] [m] [m] [m] [m]>
1 2.4 0.002 0.005 0.015 0.025 0.025 0.0048 0.012 0.036 0.06 0.06 1 1.000 1.000 4.80E-03 1.20E-02 3.60E-02 6.00E-02 6.00E-022 4.8 0.002 0.005 0.015 0.025 0.025 0.0048 0.012 0.036 0.06 0.06 2 0.618 1.171 6.63E-03 1.66E-02 4.97E-02 8.29E-02 8.29E-023 7.2 0.002 0.005 0.015 0.025 0.025 0.0048 0.012 0.036 0.06 0.06 3 0.445 1.220 8.84E-03 2.21E-02 6.63E-02 1.11E-01 1.11E-014 9.6 0.002 0.005 0.015 0.025 0.025 0.0048 0.012 0.036 0.06 0.06 4 0.347 1.241 1.11E-02 2.78E-02 8.35E-02 1.39E-01 1.39E-015 12 0.002 0.005 0.015 0.025 0.025 0.0048 0.012 0.036 0.06 0.06 5 0.285 1.252 1.35E-02 3.37E-02 1.01E-01 1.68E-01 1.68E-016 14.4 0.002 0.005 0.015 0.025 0.025 0.0048 0.012 0.036 0.06 0.06 6 0.241 1.258 1.58E-02 3.96E-02 1.19E-01 1.98E-01 1.98E-017 16.8 0.002 0.005 0.015 0.025 0.025 0.0048 0.012 0.036 0.06 0.06 7 0.209 1.262 1.82E-02 4.55E-02 1.36E-01 2.27E-01 2.27E-01
Vision 2000 (1995) Dazio(2000) Dazio(2000)
MRF + SW
φ1,1 Γ FO O LS NC Cnstory <[m] [m] [m] [m] [m]>
1 1.000 1.000 4.80E-03 1.20E-02 3.60E-02 6.00E-02 6.00E-022 0.321 1.197 1.25E-02 3.13E-02 9.38E-02 1.56E-01 1.56E-013 0.156 1.291 2.38E-02 5.94E-02 1.78E-01 2.97E-01 2.97E-014 0.093 1.347 3.85E-02 9.63E-02 2.89E-01 4.82E-01 4.82E-015 0.061 1.384 5.68E-02 1.42E-01 4.26E-01 7.10E-01 7.10E-016 0.043 1.410 7.84E-02 1.96E-01 5.88E-01 9.80E-01 9.80E-017 0.032 1.430 1.04E-01 2.59E-01 7.77E-01 1.30E+00 1.30E+00
Dazio(2000) Dazio(2000)
Residual Displacements MRF
HN FO O LS NC C FO O LS NC C φ1,1 Γ FO O LS NC Cnstory [m] < > <[m] [m] [m] [m] [m]> nstory <[m] [m] [m] [m] [m]>
1 2.4 0.001 0.002 0.005 0.025 0.025 0.0024 0.0048 0.012 0.06 0.06 1 1.000 1.000 2.40E-03 4.80E-03 1.20E-02 6.00E-02 6.00E-022 4.8 0.001 0.002 0.005 0.025 0.025 0.0024 0.0048 0.012 0.06 0.06 2 0.618 1.171 3.32E-03 6.63E-03 1.66E-02 8.29E-02 8.29E-023 7.2 0.001 0.002 0.005 0.025 0.025 0.0024 0.0048 0.012 0.06 0.06 3 0.445 1.220 4.42E-03 8.84E-03 2.21E-02 1.11E-01 1.11E-014 9.6 0.001 0.002 0.005 0.025 0.025 0.0024 0.0048 0.012 0.06 0.06 4 0.347 1.241 5.57E-03 1.11E-02 2.78E-02 1.39E-01 1.39E-015 12 0.001 0.002 0.005 0.025 0.025 0.0024 0.0048 0.012 0.06 0.06 5 0.285 1.252 6.74E-03 1.35E-02 3.37E-02 1.68E-01 1.68E-016 14.4 0.001 0.002 0.005 0.025 0.025 0.0024 0.0048 0.012 0.06 0.06 6 0.241 1.258 7.91E-03 1.58E-02 3.96E-02 1.98E-01 1.98E-017 16.8 0.001 0.002 0.005 0.025 0.025 0.0024 0.0048 0.012 0.06 0.06 7 0.209 1.262 9.09E-03 1.82E-02 4.55E-02 2.27E-01 2.27E-01
Vision 2000 (1995)
MRF + SW
φ1,1 Γ FO O LS NC Cnstory <[m] [m] [m] [m] [m]>
1 1.000 1.000 2.40E-03 4.80E-03 1.20E-02 6.00E-02 6.00E-022 0.321 1.197 6.26E-03 1.25E-02 3.13E-02 1.56E-01 1.56E-013 0.156 1.291 1.19E-02 2.38E-02 5.94E-02 2.97E-01 2.97E-014 0.093 1.347 1.93E-02 3.85E-02 9.63E-02 4.82E-01 4.82E-015 0.061 1.384 2.84E-02 5.68E-02 1.42E-01 7.10E-01 7.10E-016 0.043 1.410 3.92E-02 7.84E-02 1.96E-01 9.80E-01 9.80E-017 0.032 1.430 5.18E-02 1.04E-01 2.59E-01 1.30E+00 1.30E+00
Dazio(2000) Dazio(2000)
θ1st Story Δ1st Story
θ1st Story Δ1st Story
Δ*=Δ1stStory / ( Γφ1,1)
Δ*=Δ1stStory / ( Γφ1,1)
Δ*=Δ1stStory / ( Γφ1,1)