Post on 14-Mar-2018
1
A one hour course onNonlinear Modeling of Structures
Filip C. FilippouProfessor of Structural Engineering
University of California, Berkeley
Computational framework for earthquake simulation
• OpenSees (OPEN Software for Earthquake Engineering Simulation)http://opensees.berkeley.edu (last release 2.2.0, August 2010)
• FEDEASMatLab for teaching and concept developmenthttp://fedeaslab.berkeley.edu (last release 3.1 July 2010)
at Berkeley
2
Element Selection in EQ Engineering Practice
• Criteria– Economy in model development and result interpretation considering parameter
sensitivity and multiple ground motions– Knowledge and experience of analysis team– Detail of response (global, regional or local) and accuracy
• Selection (in decreasing popularity and increasing cost and expectations)– Linear elastic elements of any type (1d, 2d, 3d)– Nonlinear beam and column elements with “plastic hinges”– Nonlinear beam and column elements with material response integration
(fiber, fiber-hinge, inelastic beam-column finite elements)– 2d and 3d finite elements (few robust constitutive models, few advanced
features when compared with expectations: e.g. bond-slip, buckling of reinforcement, large discrete cracks, shear sliding, local buckling, fatigue, tearing of steel, etc)
Beam-Column Models
• Concentrated plasticity models = one rotational spring at each end + elastic element– Advantages: relatively simple, good for interface effects (e.g. bar pull-out) – Disadvantages: force-deformation of rotational spring depends on geometry and
moment distribution; relation to strains requires “plastic hinge length”; interaction of axial force, moment good for “metallic” elements; more complex interaction questionable; numerical robustness difficult
• Distributed inelasticity models (FE model) = consistent integration of section response at specific “control” or monitoring points– Advantages: versatile and consistent; material response can be incorporated in
section response; interaction of axial force and moment (and shear and torsion) can be rationally developed, thus numerical robustness is possible
– Disadvantages: can be expensive (not clear), require good understanding of integration to determine validity of strains and local response
3
Structural Beam-Column Models
M
N
Distributed inelasticity models
• 2 monitoring points usually at element ends = inelastic zone model- Good for columns and girders with low gravity loads- Consistent location of integration point?- Value of fixed length of inelastic zone?- Good for softening response (Fenves/Scott, ASCE 2006)- Variable inelastic zone element (CLLee/FCF, ASCE 2009); good for hardening
response- >2 monitoring points
- Good for girders with significant gravity loads- 4-5 integration points are advisable (“good plastic hinge length value with
corresponding integration weights)- One element per girder -> avoid very high local deformation values
4
Beam-Column Models: Concentrated Inelasticity
• Concentrated plasticity models =one or more rotational springs at each end + elastic element– Advantages:
relatively simple, good(?) for interface effects(e.g. shear sliding, rotation due to bar pull-out)
– Disadvantages:properties of rotational spring depend on geometry and moment distribution; relation to strains requires “plastic hinge length”;interaction of axial force, moment and shear ????;generality??? numerical robustness???M
N
Beam-Column Models: Distributed Inelasticity
• Distributed inelasticity models (1d FE model) = consistent integration of section response at specific “control” or monitoring points– Advantages:
versatile and consistent;section response from integration of material response thus N-My-Mz interaction (Bernoulli)(shear and torsion?? Timoshenko, …)thus numerical robustness is possible
– Disadvantages:can be expensive (what is the price $$$$) with “wasted sections” for localized inelasticityinaccuracy of local response (localization)thus better understanding of theory for interpretation of local response and damage is necessary
x
y
z
5
“Economic” Distributed(?) inelasticity models for Columns
• 2 monitoring points at element ends =inelastic zone model- Good for columns and girders with low gravity loads- N-My-Mz interaction straightforward- shear and torsion ???- Consistent location of integration point?- Value of fixed length of inelastic zone?- Good for softening response
(Fenves/Scott, ASCE 2006)- Hardening response Spreading inelastic zone
element SIZE (CL Lee/FCF, ASCE 2009 to appear)without N-M interaction; is generalization possible??
x
y
z
Good Distributed inelasticity models for Girders
- For girders with significant gravity loads- 4-5 integration points are advisable (“good plastic hinge length value with
corresponding integration weights)- One element per girder -> avoid very high local deformation values
6
Section Response for Distributed Inelasticity Models
• Section resultant formulation based on plasticity theory, damage, etc, etc– Relatively economical, effects can be “lumped” in a composite section response– Generalization and extension to other than the calibrated cases may not be
straightforward; hardening is very difficult to incorporate let alone softening– Section geometry must always be accounted for (are limit capacities sufficient?)
• Integration of 1d, 1 ½ d, 2d, and 3d material response– For midpoint integration the name fiber model is used; but other integration
methods are possible– For many fibers it can be expensive; how many fibers should be used?
Section Response for “Distributed” Inelasticity Models
• Section resultant formulation based on plasticity theory, damage, etc, etc– Relatively economical, effects can be
“lumped” in a composite section response
– Generalization and extension to other than the calibrated cases may not be straightforward; hardening is difficult to incorporate, let alone softening
– Section geometry must always be accounted for (are limit capacities sufficient?)
– Hardening ???– RC section, softening ???
-1 -0.5 0 0.5 1-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1numerical solutionanalytical solutionpolynomial approximation
2 2 2 23.5 1.2 1x x y y+ + =
7
Section Response for “Distributed” Inelasticity Models
• Integration of 1d, 1 ½ d, 2d, and 3d material response– For midpoint integration the name
fiber model is used; are other integration methods better??
– how many fibers should be used and for what purpose?
z
ya) yb)
z
MID25y
z
y
z
8
Two key ideas for beam-column elements
• How to incorporate nonlinear geometry under large displacements?– Formulate the element in the basic reference system without rigid body modes– Use corotational formulation to transform basic variables to global system– Keep element force-deformation relation in basic system simple: use linear
geometry under small deformations– Use one element per structural member (see point below)– If second-order effects within structural member are significant, break structural
member into 2 elements by inserting middle node• How to formulate a robust frame element?
– Use Hu-Washizu functional with independent interpolation of forces (exact under certain conditions), displacements and section deformations(Taylor/Filippou/Saritas Comp Mech, 2003)
– No need for mesh refinement; keep integration points to 4, or 2 with variable integration weight (5 respectively 3 points for girders under large element loads)
Advantages of Mixed Formulation in Corotational Framework
• Nonlinear geometry is uncoupled from basic element response; thus, geometric transformation classes (e.g. large displacements, P-Δ, linear) can be implemented once for all user frame elements (OpenSees, FEDEASLab)
• Force interpolation functions are exact under certain conditions
• Element deformations arise in weak form and represent well the inelastic strain distribution (see next point)
• Without need for mesh and integration point refinement there is no danger of localization; local strains are relatively accurate as long as the “plastic hinge length” is accurate ( = integration weight of end points) – fiber hinge is a good choice for columns under plastic or softening response
• The effect of distributed element loads can be accounted for exactly with the force interpolation functions
9
PEER 2004 Annual Meeting
4u
5u
1u
2u
X
Y
xy
L
L
4 1xΔ = −u u u
i
j
5 2yΔ = −u u u
nL
3u
6u
β
Corotational formulation for large displacement geometry
2 3
3 6
ββ
= −= −
v uv u
1 nL Lv = −
Tgu=p a q
gu∂
=∂v au
Te g gu t gu= +k k a k a
PEER 2004 Annual Meeting
Nonlinear geometry with large displacements
-40 -20 0 20 40 60 80 100 120 140 160-40
-20
0
20
40
60
80
100
120
x
y
Lee`s Frame
E
E EH
y
=
=
=
70608
01
1020
MPa
MPa
.
σ
120
cm
P,w
120 cm
96 cm24 cm2 cm
3 cm
Frame by Lee et al. (1968)
10
PEER 2004 Annual Meeting
Lee Frame (Lee et al. 1968)
0 10 20 30 40 50 60 70 80 90 100-10
-5
0
5
10
15
20
linear elas tic
elas to-plas ticwith k inem atichardening
Lee`s Fram e
Displacem ent v (cm )
App
lied
Load
P (
kN)
proposed flex . form ul. - 3 elm tsCic hon (1983) - 10 elm ts
PEER 2004 Annual Meeting
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.5
1
1.5
2
2.5x 104
End Rotation
Axi
al F
orce
2 E lements 4 E lements 8 E lements16 E lements32 E lements
Force-displacement of eccentric column (basic element)
Linear element
P
P
11
PEER 2004 Annual Meeting
Brace buckling (Black and Popov 1980)
-2 .5 -2 -1 .5 -1 -0 .5 0 0 .5 1 1 .5 2-1 0 0
-5 0
0
5 0
1 0 0
1 5 0
V e rt ic a l D is p la c e m e n t
Ver
tical
For
ce
B u c k lin g B ra c e w ith P -δ
PEER 2004 Annual Meeting
Brace buckling (Black and Popov 1980)
12
PEER 2004 Annual Meeting
P. Uriz and S. MahinUC Berkeley
PEER 2004 Annual Meeting
Lateral Buckling: CST and Quad element in 3d space
General multi-step analysis Analysis with scripts Script files NR and its Variants Results
Newton-Raphson algorithm and its variants
The Newton-Raphson algorithm requires that a new tangent stiffness matrixbe assembled at every iteration of every load step. This means that the linearsystem of equations for the displacement correction needs to be solved fromscratch at every iteration of every load step. For very large structural modelsthis is a very expensive proposition.
In the modified Newton-Raphson method the tangent matrix is not updatedat every iteration, but only once at the beginning of each load step. In theinitial stiffness method, the initial stiffness is used throughout the incremen-tal analysis. Alternative strategies that update the stiffness matrix every sooften are also possible. If the stiffness matrix is not updated, the last de-composition of the stiffness matrix is used for the solution of the linearizedequilibrium equations and only the load changes at each iteration.
Finally, quasi-Newton methods do not use the tangent stiffness matrix ofthe structure but obtain secant stiffness approximations of the inverse ofthe stiffness matrix from the displacement vectors of previous iterations.Among the best known quasi-Newton methods is the BFGS method, whichwas originally developed for nonlinear optimization problems. For a briefdescription of the method consult Bathe’s 1982 book pp. 759-761.
c©Filip C. Filippou - UC Berkeley
Lecture 11 / page 14
General multi-step analysis Analysis with scripts Script files NR and its Variants Results
Response of 2-dof truss under 2 load increments with Δλ = 0.5
8
1
8
EA=25000k=5,000
a
b1
2
3
no s t ep = 2 ;Dlam0 = 0 . 5 ;t o l = 1 . e−16;max i t e r = 10 ;S I n i t i a l i z e S t a t ePf = ze ro s ( nf , 1 ) ;S S i m p I n i t i a l i z ef o r k=1: no s t ep
S t i fUpd t = ’ ye s ’ ;S S impIncrementS t i fUpd t = ’ ye s ’ ;S S imp I t e r a t ei f ( ConvFlag )
S Update Sta teend
end
0 0.05 0.1 0.15 0.2 0.25 0.30
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Vertical translation (downward)
Load
fact
or λ
numerically exact solutionNR stepsfinal result for algorithm
Figure: Newton-Raphson method
c©Filip C. Filippou - UC Berkeley
Lecture 11 / page 15
General multi-step analysis Analysis with scripts Script files NR and its Variants Results
Response of 2-dof truss under 2 load increments with Δλ = 0.5
8
1
8
EA=25000k=5,000
a
b1
2
3
no s t ep = 2 ;Dlam0 = 0 . 5 ;t o l = 1 . e−16;max i t e r = 10 ;S I n i t i a l i z e S t a t ePf = ze ro s ( nf , 1 ) ;S S i m p I n i t i a l i z ef o r k=1: no s t ep
S t i fUpd t = ’ ye s ’ ;S S impIncrementS t i fUpd t = ’ no ’ ;S S imp I t e r a t ei f ( ConvFlag )
S Update Sta teend
end
0 0.05 0.1 0.15 0.2 0.25 0.30
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Vertical translation (downward)
Load
fact
or λ
numerically exact solutionmodified NR stepsfinal result for algorithm
Figure: Modified Newton-Raphson method
c©Filip C. Filippou - UC Berkeley
Lecture 11 / page 16
General multi-step analysis Analysis with scripts Script files NR and its Variants Results
Response of 2-dof truss under 2 load increments with Δλ = 0.5
8
1
8
EA=25000k=5,000
a
b1
2
3
no s t ep = 2 ;Dlam0 = 0 . 5 ;t o l = 1 . e−16;max i t e r = 10 ;S I n i t i a l i z e S t a t ePf = ze ro s ( nf , 1 ) ;S S i m p I n i t i a l i z eS t i fUpd t = ’ ye s ’ ;f o r k=1: no s t ep
S SimpIncrementS t i fUpd t = ’ no ’ ;S S imp I t e r a t ei f ( ConvFlag )
S Update Sta teend
end
0 0.05 0.1 0.15 0.2 0.25 0.30
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Vertical translation (downward)
Load
fact
or λ
numerically exact solutionmodified NR stepsfinal result for algorithm
Figure: Initial stiffness method
c©Filip C. Filippou - UC Berkeley
Lecture 11 / page 17
General multi-step analysis Analysis with scripts Script files NR and its Variants Results
Response of 2-dof truss under 2 load increments with Δλ = 0.5
8
1
8
EA=25000k=5,000
a
b1
2
3
no s t ep = 2 ;Dlam0 = 0 . 5 ;t o l = 1 . e−16;max i t e r = 10 ;S I n i t i a l i z e S t a t ePf = ze ro s ( nf , 1 ) ;S S i m p I n i t i a l i z ef o r k=1: no s t ep
S t i fUpd t = ’ ye s ’ ;S S impIncrementS Update Sta te
end
0 0.05 0.1 0.15 0.2 0.25 0.30
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Vertical translation (downward)
Load
fact
or λ
numerically exact solutionstepsfinal result for algorithm
Figure: Incrementation without correction
c©Filip C. Filippou - UC Berkeley
Lecture 11 / page 18
General multi-step analysis Analysis with scripts Script files NR and its Variants Results
Response of 2-dof truss under 5 load increments with Δλ = 0.2
8
1
8
EA=25000k=5,000
a
b1
2
3
no s t ep = 5 ;Dlam0 = 0 . 2 ;t o l = 1 . e−16;max i t e r = 10 ;S I n i t i a l i z e S t a t ePf = ze ro s ( nf , 1 ) ;S S i m p I n i t i a l i z ef o r k=1: no s t ep
S t i fUpd t = ’ ye s ’ ;S S impIncrementS Update Sta te
end
0 0.05 0.1 0.15 0.2 0.25 0.30
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Vertical translation (downward)
Load
fact
or λ
numerically exact solutionstepsfinal result for algorithm
Figure: Incrementation without correction
c©Filip C. Filippou - UC Berkeley
Lecture 11 / page 19
General multi-step analysis Analysis with scripts Script files NR and its Variants Results
Response of 2-dof truss under 10 load increments with Δλ = 0.1
8
1
8
EA=25000k=5,000
a
b1
2
3
no s t ep = 10 ;Dlam0 = 0 . 1 ;t o l = 1 . e−16;max i t e r = 10 ;S I n i t i a l i z e S t a t ePf = ze ro s ( nf , 1 ) ;S S i m p I n i t i a l i z ef o r k=1: no s t ep
S t i fUpd t = ’ ye s ’ ;S S impIncrementS Update Sta te
end
0 0.05 0.1 0.15 0.2 0.25 0.30
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Vertical translation (downward)
Load
fact
or λ
numerically exact solutionstepsfinal result for algorithm
Figure: Incrementation without correction
c©Filip C. Filippou - UC Berkeley
Lecture 11 / page 20
LF control in incrementation Stiffness parameter Algorithm with LF Control LF control in iteration Examples
Response of 2-dof truss under 22 load increments with Δλ = 0.1
8
1
8
EA=25000k=5,000
a
b1
2
3
no s t ep = 23 ;Dlam0 = 0 . 1 0 ;t o l = 1 . e−16;max i t e r = 10 ;S I n i t i a l i z e S t a t e ;Pf = ze ro s ( nf , 1 ) ;S I n i t i a l i z ef o r k=1: no s t ep
S t i fUpd t = ’ ye s ’ ;LoadCt r l = ’ ye s ’ ;S Inc r ementS t i fUpd t = ’ ye s ’ ;S S imp I t e r a t ei f ( ConvFlag )
S Update Sta teend
end
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450
0.2
0.4
0.6
0.8
1
Vertical displacement (downward)
Load
fact
or λ
Figure: Load factor control during incrementation; only21 steps are able to complete; the last step ”flip-flops”
c©Filip C. Filippou - UC Berkeley
Lecture 12 / page 11
LF control in incrementation Stiffness parameter Algorithm with LF Control LF control in iteration Examples
Response of 2-dof truss under 120 load increments with Δλ = 0.1
8
1
8
EA=25000k=5,000
a
b1
2
3
no s t ep = 120 ;Dlam0 = 0 . 1 0 ;t o l = 1 . e−16;max i t e r = 10 ;S I n i t i a l i z e S t a t e ;Pf = ze ro s ( nf , 1 ) ;S I n i t i a l i z ef o r k=1: no s t ep
S t i fUpd t = ’ ye s ’ ;LoadCt r l = ’ ye s ’ ;S Inc r ementS t i fUpd t = ’ ye s ’ ;LoadCt r l = ’ ye s ’ ;S I t e r a t ei f ( ConvFlag )
S Update Sta teend
end
0 0.5 1 1.5 2 2.5-1.5
-1
-0.5
0
0.5
1
1.5
Vertical displacement (downward)
Load
fact
or λ
Figure: Response of 2-dof truss with load factor control
c©Filip C. Filippou - UC Berkeley
Lecture 12 / page 18
13
Correlation studies of analysis with experiment: important but …
• Many tests have been conducted and more are under way– Before understanding the behavior of assemblies one should understand the
behavior of the constituent parts; not always possible or available– Reduced scale models require attention to scaling laws (e.g. weld fractures,
bond-slip)– “Older” tests are not complete either for lack of enough channels of
measurement or for lack of reporting (lost data); it is hard to obtain funding to “repeat” old tests
– Tests may have experimental errors (these are not reported always)– Success or failure can be decided by looking at all experimental data, not a
suitable subset of them– We can learn from failure as much as we learn from success, even though this
is not accepted practice in research publications; better paradigm is necessary
A simple start …
RC columns with biaxial bending and variable axial force
14
Low-Moehle Specimen 5: Load-Displacement Response in y
-3 -2 -1 0 1 2 3-40
-30
-20
-10
0
10
20
30
40
Tip Displacement y (cm)
Load
y (k
N)
y
z
x P
51.44 cmpz
x =44.48 kNp
y
experimentanalysis
Low-Moehle Specimen 5: Load-Displacement Response in z
-3 -2 -1 0 1 2 3-40
-30
-20
-10
0
10
20
30
40
Tip Displacement z (cm)
Load
z (k
N)
y
z
x P
51.44 cmpz
x =44.48 kNp
y
experimentanalysis
15
Low-Moehle Specimen 5: Reinforcing Steel Strain History
-80 -60 -40 -20 0 20 40 60 80-2000
-1500
-1000
-500
0
500
1000
1500
2000
Fiber Strain (mil-cm/cm)M
omen
t Mz
(kN
-cm
) z
y
-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
Effect of reinforcing bar pull-out from base
Low-Moehle Specimen 5: Reinforcing Steel Strain History
-80 -60 -40 -20 0 20 40 60 80-2000
-1500
-1000
-500
0
500
1000
1500
2000
Fiber Strain (mil-cm/cm)
Mom
ent M
z (k
N-c
m)
z
y
-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
16
PEER 2004 Annual Meeting
Shear Link
Eccentrically braced steel frames
Eccentrically Braced Frame
PEER 2004 Annual Meeting
L=28 in
δ
fb 5.985 in=
d 17.88 in=
f
w
t 0.521 int 0.314 in==
W18 40×
Loading History
0 2 4 6 8 10-4
-2
0
2
4
Pseudo time
Dis
plac
emen
t, in
Imposed vertical displacement at the right end of the element.
Equal moments at member ends
Shear Link Experiment (Hjelmstad/Popov 1983)
17
PEER 2004 Annual Meeting
Shear Link Experiment (Hjelmstad/Popov 1983)
-4 -3 -2 -1 0 1 2 3 4-250
-200
-150
-100
-50
0
50
100
150
200
250
Imposed Displacement δ (in)
Shea
r For
ce (k
ips)
ExperimentAnalysis
PEER 2004 Annual Meeting
b
d
Monitoring sectionv vA f
2σ1σ
s
Monitoring point
Concrete shear model
18
PEER 2004 Annual Meeting
Vecchio/Shim (2004) Beams A1 and A2 (shear-compresion)P
L=3660 mm (Beam A1)L=4570 mm (Beam A2)L=6400 mm (Beam A3)
b=307mm
D5 ties
M10
Beam A2
M30 M25
b=307mm
h=561mm
Beam A3
D4 ties
64mmeach
M10
M30M25
b=307mm
M30
D5 ties
M10
64mmeach
Beam A1
M25
0 5 10 15 20 25 300
100
200
300
400
500
600
Midspan Deflection (mm)
Load
(kN
)
Point10
Point25
Point45
Point113
Experiment2D-FEM ModelProposed Model
0 5 10 15 20 25 30 35 40 450
100
200
300
400
500
600
Midspan deflection (mm)
Experiment2D-FEM ModelProposed ModelA1 A2
PEER 2004 Annual Meeting
Vecchio/Shim (2004) Beam A1 (shear-compresion failure)
Point 10
Point 25
Point 45
Point 113
0 5 10 15 20 25 300
100
200
300
400
500
600
Midspan Deflection (mm)
Load
(kN
)
Point10
Point25
Point45
Point113
Experiment2D-FEM ModelProposed Model
19
PEER 2004 Annual Meeting
Vecchio/Shim (2004) Beam A3 (compression failure)
0 10 20 30 40 50 60 70 80 900
100
200
300
400
500
600
Midspan deflection (mm)
Load
(kN
)
Point10
Point25
Point40 Point65
Experiment2D-FEM ModelProposed Model
Point 10
Point 25
Point 40
Point 65
One element per half span with 5 Lobatto integration points is usedThe cross-section is divided into 10 midpoint layersBasic concrete material parameters are taken from Vecchio-Shim
PEER 2004 Annual Meeting
Thomsen-Wallace (2004) Slender Shear Walls
-100 -50 0 50 100-200
-150
-100
-50
0
50
100
150
200
Top Displacement (mm)
144"
48"
A A
48"
4"
#2 bars @7.5"8- #3 bars
(Concrete cover 0.75")
Section A-A-100 -50 0 50 100
-200
-150
-100
-50
0
50
100
150
200
Top Displacement (mm)
She
ar F
orce
(kN
)
20
PEER 2004 Annual Meeting
Lefas-Kotsovos-Ambraseys (1990) Shear Walls
•Shear walls SW21 and SW22 have aspect ratio L/h=2.0
Loading Direction
d
L
0 5 10 15 20 250
20
40
60
80
100
120
140
160
180
Top Displacement,mm
She
ar F
orce
,kN
Experiment ν=0Analysis ν=0Experiment ν=0.1Analysis ν=0.1
PEER 2004 Annual Meeting
Point35
Column depth-0.2 0 0.2
0.2
0.4
0.6
0.8
1
1.2Point40
Column depth-0.2 0 0.2
0.2
0.4
0.6
0.8
1
1.2Point60
Column depth-0.2 0 0.2
0.2
0.4
0.6
0.8
1
1.2Point97
Column depth
Col
umn
axis
-0.2 0 0.2
0.2
0.4
0.6
0.8
1
1.2
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9 Point30
Column depth-0.2 0 0.2
0.2
0.4
0.6
0.8
1
1.2
Evolution of shear wall cracking – Tensile damage
a) Point 30 b) Point 35 c) Point 97
21
Paolo Martinelii, Politecnico di Milano, ItalyFCF, University of California, Berkeley
Analytical Model of
Reinforced Concrete Walls
Height ≈ 20 m
Weight ≈ 226 tons
Web Wall
Flange Wall
Table motion direction
Gravity Column Precast Column
Full scale 7 story wall building (Panagiotou, Restrepo, Conte, UCSD)
Paolo Martinelii, Politecnico di Milano, ItalyFCF, University of California, Berkeley
Analytical Model of
Reinforced Concrete Walls
Confined and unconfined zones in the web wall
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Paolo Martinelii, Politecnico di Milano, ItalyFCF, University of California, Berkeley
Analytical Model of
Reinforced Concrete Walls
Interstory drift and displacement envelopes
• Excellent agreement in terms of interstory drift and displacement
• Increasing damage due to increasing intensity motions is visible
Paolo Martinelii, Politecnico di Milano, ItalyFCF, University of California, Berkeley
Analytical Model of
Reinforced Concrete Walls
• Good agreement in terms of residual displacement
• Satisfactory results in terms of floor acceleration
Residual displacement and floor acceleration envelopes
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Paolo Martinelii, Politecnico di Milano, ItalyFCF, University of California, Berkeley
Analytical Model of
Reinforced Concrete Walls
• Good agreement also in terms of internal forces
• Maximum discrepancy during last motion at shear wall base
Story shear and overturning moment envelopes
Paolo Martinelii, Politecnico di Milano, ItalyFCF, University of California, Berkeley
Analytical Model of
Reinforced Concrete Walls
• Significant lengthening of fundamental period of specimen (> than 2.5 x)
• Damage evolution tracked with remarkable accuracy
f1 = 1.83 Hz (0.55 s)Beginning of EQ1
f1 = 0.67 Hz (1.43 s)End of EQ4
Experimental and analytical frequency spectrum
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Paolo Martinelii, Politecnico di Milano, ItalyFCF, University of California, Berkeley
Analytical Model of
Reinforced Concrete Walls
Top displacement time history
Conclusions
• Nonlinear Analysis is gradually going to become a designer’s tool for the evaluation of existing structures and the design of new important structures
• It can offer significant insights into the global, but particularly into the local response of structures and, thus serve for the identification of local failure mechanisms
• The current state of the art permits the simulation of the hysteretic behavior of structural elements with limited success; deeper understanding of material behavior under cyclic loading is, however, indispensable in order to arrive at failure mode prediction
• Thorough correlations between numerical models and specimens of increasing complexity are indispensable; limited progress has been made to date, particularly regarding 3d response
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Continuing Challenges
• Effect for shear, torsion and interaction with axial force and bending moment (3d and not just 2d analysis for shear)
• Effect of bond-slip, pull-out of reinforcing steel
• 3d beam-column joint model that is robust and efficient
• 3d constitutive model for concrete under large inelastic strains (damage, dilatation, …)
• Buckling of reinforcing steel (global and not local)
• Low cycle fatigue of structural steel; fracture
• Simulation of structural subassemblies and full-scale structures
• Many more: partitions, slab-wall-column interactions, cladding, infills …
Future outlook
• Much work needs to be done before nonlinear analysis can have widespread use, because of the complexity of nonlinear solution algorithms and the lack of training of modern engineers; thus researchers and educators need to redouble their efforts in clarifying the concepts
• I hope that with the contributions of all those present in an open forum (open platform) we will be able to experience significant progress in the nonlinear simulation of concrete structures in the future