Structural Simulation Models

Filip C. Filippou

CEE Department

University of California, Berkeley

2001 PEER Annual Meeting

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PEER OpenSees Framework

Software framework for integrating

Material and component models. Emphasize degradation and failurebehaviors

Solution strategies: Static and dynamic for degrading and collapsingsystems

Performance evaluation based on simulated behavior

Utilize new computing resources

Engineering desktop workstations (SMP, distributed)

High-performance computing

Computational grids

Provide network communication mechanisms with scientificvisualization methods and databases

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M

P

Structural Beam-Column Models

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Axial Force-Flexure-Shear Behavior

Shear with degradation

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Beam-Column Joints

Joint Modeling: Shear-Bond Interaction (Lowes)

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Other issues

Parameter uncertainty - Sensitivity (DerKiureghian, Conte)

Shear Wall Models

Solution Strategies

Pull-out, bond deterioration

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Advances in Frame Element Formulations

Force based formulation for 1rst and 2nd order theory (exactinternal force distribution)

Large displacements with corotational formulation

Mixed force-displacement formulation for frame elements withcomplex interactions (composite action, pile-soil interaction)

Robust algorithms of state determination

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Push-Over of Two Story Frame

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Push-Over of Two Story Frame (Distributions)

Curvature Distribution of Two-Story Frame

Moment Distribution of Two Story Frame

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Tapered Beam - Curvature Distribution

Curvature distribution:

v sf x( )

v sd x( )

v s x( )

x0 20 40 60 80 100 120 140 160

2 10 4

1.5 10 4

1 10 4

5 10 5

0

5 10 5

1 10 4

force formulationdisplacement formulation with 1 elementdisplacement formulation with 2 elements

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Tapered Beam - Bending Moment Distribution

0 20 40 60 80 100 120 140 160200

100

0

100

200

300

400

force formulationdisplacement formulation with 1 elementdisplacement formulation with 2 elements

400

197.07

S sf x( )

S sd x( )

S s x( )

1440 x

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Advantages of Force Formulation

equilibrium is satisfied exactly along the element in every iteration;end compatibility is satisfied on convergence

distributed loads can be readily accommodated

a single element suffices for the entire member; no meshrefinement is necessary; localization problems are minimized

formulation is very robust in the presence of strength softening

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Low-Moehle Specimen No. 5 (EERC Report 1987-14)

-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3

Tip Displacement z (cm)

Tip

Disp

lacemen

t y (cm)

and Variable Axial Load

-3 -2 -1 0 1 20

10

20

30

40

50

60

70

80

90

100

Tip Displacement z (cm)

Co

mp

ressive Axial

N=89

N=2.2

Biaxial Bending

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Low-Moehle Specimen 5: Response in y

-3 -2 -1 0 1 2 3-40

-30

-20

-10

0

10

20

30

40

Tip Displacement y (cm)

Lo

ad y (kN

)

y

z

x P

51.44 cmp

z

x =44.48 kN

py

experiment

analysis

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Low-Moehle Specimen 5: Response in z

-3 -2 -1 0 1 2 3-40

-30

-20

-10

0

10

20

30

40

Tip Displacement z (cm)

Lo

ad z (kN

)

y

z

x P

51.44 cmp

z

x =44.48 kN

py

experiment

analysis

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-80 -60 -40 -20 0 20 40 60-2000

-1500

-1000

-500

0

500

1000

1500

2000

Fiber Strain (mil-cm/cm)

Mom

ent Mz (kN

-cm)

z

y

Low-Moehle Specimen 5: Reinforcing Steel Strain History

-2000

-1500

-1000

-500

0

500

1000

1500

2000

-40 -30 -20 -10 0 10 20 30 40Steel Strain (mil-cm/cm)

Mom

ent M

z (k

N-c

m)

z

y

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Correlation Studies for ISPRA columns (Bousias et al.)

12 Column Specimens with identical geometry and reinforcing S0

Uniaxial displacement cycles in x constant axial compression ~ 16% of axial capacity (?)

S1 Alternating uniaxial displacement cycles in x and y constant axial compression ~ 10% of axial capacity

S5, S7 Different biaxial displacement histories in x and y;

constant axial compression ~ 12% of axial capacity S9

Biaxial displacement history in x and y; two levels of axial compression ~ 3%->15% of axial capacity

S4 Displacement in x, Force in y constant axial compression

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ISPRA Specimen S1 - Lateral Displacement History

-100

-80

-60

-40

-20

0

20

40

60

80

100

0 5 10 15 20 25 30

Load Step

Late

ral D

ispl

acem

ent (

mm

)

X-direction

Y-direction

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ISPRA Specimen S1 - Flexural Response in x

-80

-60

-40

-20

0

20

40

60

80

-100 -80 -60 -40 -20 0 20 40 60 80 100

Displacement (mm)

She

ar F

orce

(kN

)

Experiment

Analysis

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ISPRA Specimen S1 - Flexural Response in y

-80

-60

-40

-20

0

20

40

60

80

-80 -60 -40 -20 0 20 40 60 80 100

Displacement (mm)

She

ar F

orce

(kN

)

Experiment

Analysis

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ISPRA Specimen S1 - Axial Displacement History

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

0 5 10 15 20 25 30

Load Step

Axi

al D

ispl

acem

ent (

mm

)

Experiment

Analysis

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ISPRA Specimen S5 - Lateral Displacement History

-150

-100

-50

0

50

100

150

-150 -100 -50 0 50 100 150

Displacement in x (mm)

Dis

plac

emen

t in

y (m

m)

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ISPRA Specimen S5 - Flexural Response in x

-80

-60

-40

-20

0

20

40

60

80

100

-150 -100 -50 0 50 100 150

Displacement in x (mm)

For

ce in

x (

kN)

Experiment

Analysis

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ISPRA Specimen S5 - Flexural Response in y

-60

-40

-20

0

20

40

60

80

-150 -100 -50 0 50 100 150

Displacement in y (mm)

For

ce in

y (

kN)

Experiment

Analysis

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ISPRA Specimen S5 - Axial Displacement History

-2

-1

0

1

2

3

4

5

0 5 10 15 20 25 30 35

Load Step

Axi

al D

ispl

acem

ent (

mm

)

Experiment

Analysis

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ISPRA Specimen S9 - Lateral Displacement History

-150

-100

-50

0

50

100

150

-150 -100 -50 0 50 100 150

Displacement in x (mm)

Dis

plac

emen

t in

y (m

m)

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ISPRA Specimen S9 - Flexural Response in x

-80

-60

-40

-20

0

20

40

60

80

-150 -100 -50 0 50 100 150

Displacement in x (mm)

For

ce in

x (

kN)

Experiment

Analysis

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ISPRA Specimen S9 - Flexural Response in y

-80

-60

-40

-20

0

20

40

60

80

-150 -100 -50 0 50 100 150

Displacement in y (mm)

For

ce in

y (

kN)

Experiment

Analysis

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ISPRA Specimen S9 - Axial Displacement History

-8

-6

-4

-2

0

2

4

6

0 5 10 15 20 25 30 35 40 45

Load Step

Axi

al D

ispl

acem

ent (

mm

)

Experiment

Analysis

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Second order analysis - Large displacements

The co-rotational formulationseparates rigid-body modes fromlocal deformations, using a singlecoordinate system thatcontinuously translates and rotatewith the element as thedeformation proceeds.$ , $Q q2 2

$y

X

Y$ , $Q q1 1

$ , $Q q3 3

$ , $Q q5 5

$ , $Q q4 4

$ , $Q q6 6

$x

Q q3 3,

Q q1 1,

Q q2 2,

x

y

basic system w/o rigid body modes

local systemw/ rigid body modes

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Lees Frame

E

E EH

y

=

=

=

70608

01

1020

MPa

MPa

.

σ

120 cm

P,w

120 cm

96 cm24 cm2 cm

3 cm

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Lee's Frame

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Parking Garage, 1994 Northridge Earthquake

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Shaking Table Specimen of Shahrooz-Moehle (1987)

6@36"=216"

FRAME A

FRAME B

FRAME C

2@75"=150"

FRAME 1

FRAME 2

FRAME 3

2@45"=90"

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Shaking Table Specimen El Centro 7.7

6th Floor Displacement Time History to EC7.7L

-0.30

-0.20

-0.10

0.00

0.10

0.20

0.30

0 2 4 6 8 10 12 14 16

Time (sec)

Measured Response

Calculated Response

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Shaking Table Specimen El Centro 49.3

6th Floor Displacement Time History to EC49.3L

-3.00

-2.50

-2.00

-1.50

-1.00

-0.50

0.00

0.50

1.00

1.50

2.00

2.50

0 2 4 6 8 10 12 14 16

Time (sec)

Meaasured Reponse

Calculated Repsonse