Nonlinear Frame Finite Elements in OpenSees Michael H. Scott Associate Professor School of Civil and Construction Engineering
OpenSeesDays Users Workshop Richmond, CA September ber 26, 2014
Types of Nonlinearity
Two sources of nonlinear frame element response: Material – yielding, strain hardening, crushing of concrete, etc. Geometry – loss of stability due to loads acting through large
displacements
An analysis can account for each source of nonlinearity separately, giving four possible approaches Geometry Geometry Linear Linear (GL) (GL) Geometry Geometry Nonlinea Nonlinearr (GN) (GN) Material Linear (ML) ML, GL ML, GN Material Nonlinear (MN) MN, GL MN, GN
Types of Nonlinearity
Two sources of nonlinear frame element response: Material – yielding, strain hardening, crushing of concrete, etc. Geometry – loss of stability due to loads acting through large
displacements
An analysis can account for each source of nonlinearity separately, giving four possible approaches Geometry Geometry Linear Linear (GL) (GL) Geometry Geometry Nonlinea Nonlinearr (GN) (GN) Material Linear (ML) ML, GL ML, GN Material Nonlinear (MN) MN, GL MN, GN
Steel Frame Pushover Analysis 150 kip
150 kip
1.33λ 33λ σ
300 kip
300 kip
15 ft
αE
σy
E
0.67λ 67λ ε
Beams: W18x76 Columns: W14x90
15 ft
σy =36 ksi E =30,000 ksi α=0.02
V b b = 2λ
30 ft
Simple steel frame model analyzed under four approahces Relatively large column axial loads will intensify both material and geometric nonlinear response for demonstration purposes
Steel Frame Pushover Analysis 1000 ) p i k ( r a e h S e s a B
200
ML, GL ML, GN
800
) 150 p i k ( r a e h 100 S e s a B 50
600 400 200 0 0
MN, GL MN, GN 10
20
Roof Displacement (in)
30
0 0
MN, GL
MN, GN
5
10
15
Roof Displaceme Displacement nt (in)
20
We observe the following: Material nonlinearity kicks in well before geometric nonlinearity Geometric nonlinearity allows for prediction of loss of stability for increasing displacement
Section Force-Deformation Response At each cross-section along a frame element, we must determine the section forces for any given section deformations Material nonlinearity eminates from the stress-strain response in each frame element Heuristic approach through stress-resultant section models, e.g., moment-curvature; or Integrate stress-strain response via “fiber section” approach y
z
x Ai ( y i , z i )
Commands for Section Definition
uniaxialMaterial modelName $tag ...
Define uniaxial stress-strain models for use in Bernoulli beam elements Elastic, Steel01, Steel02, Concrete01, Concrete02, etc. nDMaterial modelName $tag ...
Define multiaxial stress-strain models for use in Timoshenko beam elements ElasticIsotropic, J2Plasticity, ConcreteMCFT, etc.
Commands for Section Definition General definition of Bernoulli cross-section using patches and layers of fibers whose stress-strain response is defined by uniaxialMaterial objects section Fiber $tag { patch $type $matTag ... layer $type $matTag ... fiber $matTag ... ...
} Use NDFiber with nDMaterial objects instead of Fiber with uniaxialMaterial objects for Timoshenko beams Specific cross-sections obtained with “canned” models section WFSection2d $tag $matTag ... section RCSection2d $tag $matTag ...
Rectangular Steel Section Rectangular section with EPP uniaxial stress-strain response Compute moment-curvature response for increasing number of fibers Exact solution for M y = f y bd 2 /6 and M p = bd 2 /4 N fiber
σ
.. . i y i
σy
d
.. .
E
d /2
ε
2 1 b
Rectangular Steel Section x 10
N fiber =
4
2 4
) 3 n i p i k ( t 2 n e m o M 1
) 3 n i p i k ( t 2 n e m o M 1
Yield
Exact Computed
0 0
0.2
0.4
0.6
0.8
Curvature (1/in) x 10
N fiber =
4
1 x 10
8
x 10
) 3 n i p i k ( t 2 n e m o M 1
Exact Computed 0.6
Curvature (1/in)
0.8
1 x 10
−3
0.4
N fiber =
4
) 3 n i p i k ( t 2 n e m o M 1
0.4
0.2
0.6
0.8
Curvature (1/in) 4
0.2
Exact Computed
−3
Yield
4
Yield
0 0
4
0 0
N fiber =
4
x 10
4
0 0
1 −3
x 10
16
Yield
Exact Computed 0.2
0.4
0.6
Curvature (1/in)
0.8
1 −3
x 10
Reinforced Concrete Section EPP steel and Concrete01 concrete Using “canned” RCSection2d command Confined and unconfined concrete
′
f cc
24 in
c
f
2 in
2 in
, s s e r t S
′
f cu
εcc 24 in
εcu
Strain, εc
Reinforced Concrete Section Moment-curvature response for increasing levels of axial load With and without confining effects of transverse reinforcement Modify the Concrete01 input parameters for confined concrete 15000
12000
) n i p 10000 i k ( t n e m 5000 o M
) n i p 8000 i k ( t 6000 n e m 4000 o M
ν = 0.3 ν = 0.2 ν = 0.1
10000
ν = = 0.1 0.0 ν ν = 0.2
ν = 0.0
2000 0 0
ν = 0.3 0.5
1
Curvature (1/in)
1.5 −3
x 10
(c) Without Confining Effects
0 0
0.5
1
Curvature (1/in)
1.5 −3
x 10
(d) With Confining Effects
Numerical Integration of Element Response For most material nonlinear element formulations, cross-section response is integrated numerically along the frame element length in order to determine element force-deformation response Sections located at discrete points along the element length, each with a prescribed weight Highly accurate Gauss-based quadrature commonly used x 1
x 2
x 3
x 4
x 5
w 1
w 2
w 3
w 4
w 5
L
Displacement-Based Frame Element element dispBeamColumn $tag $ndI $ndJ $transfTag Legendre $secTag 2
Strict compatibility Linear axial and cubic Hermitian transverse displacement fields Constant axial deformation and linear curvature along element length
Weak equilibrium Equilibrium satisfied only at the nodes, not at every section along the element Two-point Gauss-Legendre integration along element length
Improve numerical solution by using more elements per member (mesh- or h-refinement)
Propped Cantilever Constant axial load and increasing moment applied at propped end Fiber-discretized section response with strain-hardening stress-strain Two Gauss-points per element Investigate refinement for increasing number of elements σ
M
σy N
θ
αE E
U
ε
100 in 10 web fibers 1
2
...
2 fibers each flange N ele -1
N ele
W14x90
Global Response N ele
) n 15000 i p i k ( 10000
=1
N ele
=1
15000 10000
M
, t n 5000 e m o 0 M 0
5000
Computed Exact 0.05
0.1 N ele
) n 15000 i p i k ( 10000
0.15
0.2
0 −0.4
−0.3
=2
−0.2 N ele
−0.1
0
−0.1
0
−0.1 U (in)
0
=2
15000 10000
M
, t n 5000 e m o 0 M 0
5000
0.05
0.1 N ele
) n 15000 i p i k ( 10000
0.15
0.2
0 −0.4
−0.3
= 4
−0.2 N ele
=4
15000 10000
M
, t n 5000 e m o 0 M 0
5000
0.05
0.1 Rotation, θ
0.15
(rad)
0.2
0 −0.4
−0.3
−0.2
Deflection,
Local Flexural Response N ele
−3
x 10
=1
10
) n i / 1 ( ) x (
) n i p i k ( ) x (
Computed Exact
5 0
1 0
M
−5 0
20
40
60
N ele
−3
x 10
80
−1 0
100
=2
20
40
60
N ele
4
x 10
10
80
100
80
100
80
100
=2
2
) n i p i k ( ) x (
5 0
κ
1 0
M
−5 0
20
40
60
N ele
−3
x 10
80
−1 0
100
=4
20
40
60
N ele
4
x 10
10
) n i / 1 ( ) x (
=1
2
κ
) n i / 1 ( ) x (
N ele
4
x 10
=4
2
) n i p i k ( ) x (
5 0
κ
1 0
M
−5 0
20
40 x
60
(in)
80
100
−1 0
20
40 x
60
(in)
Local Axial Response N ele 0 ) n i / −0.005 n i ( ) x ( −0.01 a
ε
−0.015 0
=1
N ele 0
) p i k ( ) −200 x (
Computed Exact 20
40 N ele
N
60
80
−400 0
100
40 N ele
0
) n i / −0.005 n i ( ) x −0.01 (
) p i k ( ) −200 x (
60
80
100
80
100
80
100
=2
N
a
ε
20
40 N ele
60
80
−400 0
100
20
=4
40 N ele
0
0
) n i / −0.005 n i ( ) x −0.01 (
) p i k ( ) −200 x (
60
=4
N
a
ε
−0.015 0
20
=2
0
−0.015 0
=1
20
40 60 x (in)
80
100
−400 0
20
40 60 x (in)
Steel Frame Pushover Analysis Investigate refinement of load-displacement response for increasing number of displacement-based elements per member 150 kip
150 kip
1.33λ σ
300 kip
300 kip
15 ft
αE
σy
E
0.67λ ε
Beams: W18x76 Columns: W14x90 V b = 2λ
30 ft
15 ft
σy =36 ksi E =30,000 ksi α=0.02
Steel Frame Pushover Analysis Coarse mesh over-predicts strength – unconservative Improved solution with refined mesh 200
N ele = 1 N ele = 2 N ele = 4
) 150 p i k ( r a e h 100 S e s a B 50
0 0
10
20
30
Roof Displacement (in)
40
Force-Based Frame Element element forceBeamColumn $tag $ndI $ndJ $transfTag Lobatto $secTag $Np
Average compatibility Nodal displacements are balanced by weighted integral of section deformations Complex state determination Use Gauss-Lobatto integration so that extreme flexural response captured at element ends
Strong equilibrium Equilibrum of nodal and section forces satisfied at all points along element Constant axial force and linear bending moment in absence of member loads Straightforward to include member loads
Improve numerical solution by using more integration points per element while maintaining mesh of one element per member
Propped Cantilever Constant axial load and increasing moment applied at propped end Fiber-discretized section response with strain-hardening stress-strain Investigate refinement for increasing number of Gauss-Lobatto point using one element σ
M
σy N
θ
αE E
U
ε
100 in 1
2
...
N p -1
N p
10 web fibers 2 fibers each flange W14x90
Global Response N p
) n 15000 i p i k ( 10000
= 3
N p
=3
15000 10000
M
, t n 5000 e m o 0 M 0
5000
Computed Exact 0.05
0.1 N p
) n 15000 i p i k ( 10000
0.15
0.2
0 −0.4
−0.3
= 4
−0.2 N p
−0.1
0
−0.1
0
−0.1 U (in)
0
=4
15000 10000
M
, t n 5000 e m o 0 M 0
5000
0.05
0.1 N p
) n 15000 i p i k ( 10000
0.15
0.2
0 −0.4
−0.3
=5
−0.2 N p
=5
15000 10000
M
, t n 5000 e m o 0 M 0
5000
0.05
0.1 Rotation, θ
0.15
(rad)
0.2
0 −0.4
−0.3
−0.2
Deflection,
Local Flexural Response =3
N p
−3
x 10 10
) n i / 1 ( ) x (
) n i p i k ( ) x (
Computed Exact
5 0
1 0
M
−5 0
20
40 N p
−3
x 10
60
80
−1 0
100
=4
20
40
60
N p
4
x 10
10
80
100
80
100
80
100
=4
2
) n i p i k ( ) x (
5 0
κ
1 0
M
−5 0
20
40 N p
−3
x 10
60
80
−1 0
100
=5
20
40
60
N p
4
x 10
10
) n i / 1 ( ) x (
=3
2
κ
) n i / 1 ( ) x (
N p
4
x 10
=5
2
) n i p i k ( ) x (
5 0
κ
1 0
M
−5 0
20
40 x
60
(in)
80
100
−1 0
20
40 x
60
(in)
Local Axial Response N p 0 ) n i / −0.005 n i ( ) x ( −0.01 a
ε
−0.015 0
=3
N p −190.8
) p i k ( ) −190.8 x (
Computed Exact 20
40 N p
N
60
80
−190.8 0
100
40 N p
60
80
100
80
100
80
100
=4
−190.8
) p i k ( ) −190.8 x (
N
a
ε
20
40 N p
60
80
−190.8 0
100
20
=5
40 N p
0
−190.8
) n i / −0.005 n i ( ) x ( −0.01
) p i k ( ) −190.8 x (
60
=5
N
a
ε
−0.015 0
20
=4
0 ) n i / −0.005 n i ( ) x ( −0.01 −0.015 0
=3
20
40 60 x (in)
80
100
−190.8 0
20
40 60 x (in)
Steel Frame Pushover Analysis Investigate refinement of load-displacement response for increasing number of Gauss-Lobatto integration points per element Maintain one element per member 150 kip
150 kip
1.33λ σ
300 kip
300 kip
15 ft
αE
σy
E
0.67λ ε
Beams: W18x76 Columns: W14x90 V b = 2λ
30 ft
15 ft
σy =36 ksi E =30,000 ksi α=0.02
Steel Frame Pushover Analysis Same yield point predicted in all cases Post-yield stiffness more flexible with fewer integration points 200 ) 150 p i k ( r a e h 100 S e s a B 50
0 0
=5 N N pp = 4 N p = 3
10
20
30
Roof Displacement (in)
40
Force-Based Plastic Hinge Frame Element element forceBeamColumn $tag $ndI $ndJ $transfTag HingeRadau $secTagI $lpI $secTagJ $lpJ $secTagE or element beamWithHinges $tag $ndI $ndJ $secTagI $lpI $secTagJ $lpJ $E $A $I $transfTag
Control integration weights at element ends Important for strain-softening section response x 1
x 2
x 3
x 4
x 5
x 6
I
J l pI
3l pI
L − 4(l pI + l pJ ) L
3l pJ
l pJ
Reinforced Concrete Bridge Pier P = 0.3f c Ag ′
550mm x 550mm square 12 bars, d b = 20 mm 40 mm clear cover
V , U m 5 6 . 1 = L
V = Base Shear
Tanaka and Park (1990) Specimen 7
Displacement-Based Elements Post-peak response is mesh-dependent Function of element length 800
1
) 600 N k ( r a e h 400 S e s a B200
0 0
36
20
40
60
Displacement (mm)
80
Force-Based Element Post-peak response depends on number of integration points Function of integration weight at base of column 800 ) 600 N k ( r a e h 400 S e s a B200
0 0
6 5 4
20
40
60
Displacement (mm)
80
Force-Based Plastic Hinge Element Post-peak response controlled by plastic hinge length l p = 0.22L from empirical equation 800 ) 600 N k ( r a e h 400 S e s a B200
0 0
20
40
60
Displacement (mm)
80
Drawback to Force-Based Plastic Hinge Element For strain-hardening section behavior, post-peak response is too flexible 15
10 d a o L
5 Computed Exact 0 0
5
10
Displacement
15
Modeling Recommendations There’s no silver bullet
Strain-Hardening Section Response
Use mesh of displacement-based elements Use one force-based elements with 4 to 6 Gauss-Lobatto points Plastic hinge element not recommended because post-peak response will be too flexible Strain-Softening Section Response
Use force-based plastic hinge element Response with displacement-based elements is mesh dependent Response with Gauss-Lobatto force-based element depends on number of integration points
Geometric Transformation of Element Response Element formulation of material nonlinearity inside the basic system (free or rigid body displacement modes) Element formulation of geometric nonlinearity outside the basic system Basic System
J
β u l 3
∆u ly
u b 2 I
u b 3 Ln
u l 6 u l 5
β L + ∆ u lx
u l 2 I
J L u l 1
Local Coordinate System
u l 4
Geometric Transformation geomTransf Linear $tag
Small displacement assumptions in local to basic transformation Linear transformation of forces and displacements geomTransf PDelta $tag
Small displacement assumption transformation of displacements Account for transverse displacement of axial load in equilibrium relationship geomTransf Corotational $tag
Fully nonlinear transformation of displacements and forces Exact in 2D but some approximations in 3D
Steel Frame Pushover Analysis Examine pushover response for different levels of gravity load λ50 kip λ50 kip V
λ100 kip
λ100 kip
12 ft
V
Beams: W18x76 Columns: W14x90 σy =36 ksi E =30,000 ksi α=0.02
12 ft
Steel Frame Pushover Analysis P − ∆ and Corotational – similar results for lateral displacement λ
= 1
λ
400
400
) 300 p i k ( r a e h 200 S e s a B 100
) 300 p i k ( r a e h 200 S e s a B 100
0 0
20
40
60
80
100
0 0
20
Roof Disp (in) λ
= 3
λ 400
) 300 p i k ( r a e h 200 S e s a B 100
) 300 p i k ( r a e h 200 S e s a B 100
20
40
60
Roof Disp (in)
60
80
100
80
100
Roof Disp (in)
400
0 0
40
= 2
80
100
0 0
20
40
= 4
60
Roof Disp (in)
Steel Frame Pushover Analysis “Exact” Corotational predicts change in vertical displacement Important for collapse prediction and post-buckling capacity λ
= 1
λ
400
400
) 300 p i k ( r a e 200 h S e s a B 100
) 300 p i k ( r a e 200 h S e s a B 100
0 0
P -∆ P -∆ Corotational 5
10
15
20
0 0
5
Roof Disp (in) λ
= 3
λ 400
) 300 p i k ( r a e 200 h S e s a B 100
) 300 p i k ( r a e 200 h S e s a B 100
5
10
Roof Disp (in)
10
15
20
15
20
Roof Disp (in)
400
0 0
= 2
15
20
0 0
5
= 4
10
Roof Disp (in)
Elastic Buckling Use mesh of corotational frame elements to simulate buckling Simply-supported W14x90, L=100 in, P cr =29579 kip L/r =16.26: short column, but demonstrates point Imperfection applied to nodes, u (t ) = 0.1 sin(π x /L) in 4
4
x 10
N ele =2
=4 N N eele le =8
) 3 p i k ( d a 2 o L l a i x A1
0 0
29579 kip
2
4
6
Axial Deflection (in)
8
Concept works well for inelastic buckling too
10
