fem lecture 2
DESCRIPTION
2nd lecture in a series of lectures in ETH.TRANSCRIPT
The Finite Element Method for the Analysis ofNon-Linear and Dynamic Systems
Prof. Dr. Eleni Chatzi
Lecture 2 - 2 October, 2012
Institute of Structural Engineering Method of Finite Elements II 1
Non Linear FE - Background and Motivation
Linear Analysis Assumptions
Infinitesimally small displacements
Linearly elastic material
No gaps or overlaps occurring during deformations - Thedisplacement field is smooth
Nature of Boundary Conditions remains unchanged
Steady State Assumption
No dependence on time
Institute of Structural Engineering Method of Finite Elements II 2
Non Linear FE - Background and Motivation
What kind of problems are not steady state and linear?
Material behaves Nonlinearly
Geometric Nonlinearity (ex. p-∆ effects, follower force)
Contact Problems (Hertzian stress)
Loads vary fast compared to the eigenfrequencies of thestructure
Varying Boundary conditions (ex. freezing of water, welding)
General feature:
Response becomes load path dependent
Institute of Structural Engineering Method of Finite Elements II 3
Non Linear FE - Background and Motivation
What is the added value of being able to assess the Nonlinearnon-steady state response of structures?
Assessing the:
Structural response of structures to extreme events (rock-fall,earthquake, hurricanes)
Performance (failures and deformations) of soils
Verifying simple models
Institute of Structural Engineering Method of Finite Elements II 4
Non Linear FE - Background and Motivation
Collapse Analysis of the World Trade Center
Institute of Structural Engineering Method of Finite Elements II 5
Non Linear FE - Background and Motivation
Ultimate collapse capacity of jacket structure
Institute of Structural Engineering Method of Finite Elements II 6
Non Linear FE - Background and Motivation
Analysis of soil performance
Institute of Structural Engineering Method of Finite Elements II 7
Non Linear FE - Background and Motivation
Analysis of bridge response
Institute of Structural Engineering Method of Finite Elements II 8
Non Linear FE - Background and Motivation
Steady state problems (Linear/Nonlinear):
The response of the system does not change over time
KU = R
Propagation problems (Linear/Nonlinear):
The response of the system changes over time
MU(t) + CU(t) + KU(t) = R(t)
Eigenvalue problems:
No unique solution to the response of the system
Av(t) = λBvInstitute of Structural Engineering Method of Finite Elements II 9
Introduction to Nonlinear analysis
Classification of Nonlinear analyses
Swiss Federal Institute of Technology Page 20
Method of Finite Elements II
Introduction to non-linear analysis• Classification of non-linear analyses
Type of analysis Description Typical formulation used
Stress and strain measures used
Materially-nonlinear only
Infinitesimal displacements and strains; stress train relation is non-linear
Materially-nonlinear-only (MNO)
Engineering strain and stress
Large displacements, large rotations but small strains
Displacements and rotations of fibers are large; but fiber extensions and angle changes between fibers are small; stress strain relationship may be linear or non-linear
Total Lagrange (TL) Updated Lagrange (UL)
Second Piola-Kirchoff stress, Green-Lagrange strain Cauchy stress, Almansi strain
Large displacements, large rotations and large strains
Displacements and rotations of fibers are large; fiber extensions and angle changes between fibers may also be large; stress strain relationship may be linear or non-linear
Total Lagrange (TL) Updated Lagrange (UL)
Second Piola-Kirchoff stress, Green-Lagrange strain Cauchy stress, Logarithmic strain
Institute of Structural Engineering Method of Finite Elements II 10
Introduction to Nonlinear analysis
Linear Elastic
Swiss Federal Institute of Technology Page 21
Method of Finite Elements II
Introduction to non-linear analysis• Classification of non-linear analyses
Δ
2P
2P ε
σ
1
E
0.04ε <
Linear elastic (infinitesimal displacements)
L
L
//
P AE
L
σε σ
ε
==
Δ =
Infinitesimal Displacements
Institute of Structural Engineering Method of Finite Elements II 11
Introduction to Nonlinear analysis
Material Nonlinearity onlySwiss Federal Institute of Technology Page 22
Method of Finite Elements II
Introduction to non-linear analysis• Classification of non-linear analyses
Δ
2P
2P ε
σ
Materially nonlinear only (infinitesimal displacements, but nonlinear stress-strain relation)
L
L
/
0.04
Y Y
T
P A
E E
σσ σ σε
ε
=−
= +
<
/P A
1
E1 TEYσ
Infinitesimal Displacements, but Nonlinear Stress Strain relation
Institute of Structural Engineering Method of Finite Elements II 12
Introduction to Nonlinear analysis
Large displacements, small strainsSwiss Federal Institute of Technology
x
y
L
x′
y′
L
′Δ
ε′
0.04L
εε
′ <′ ′Δ =
Linear or Nonlinear material behavior
Institute of Structural Engineering Method of Finite Elements II 13
Introduction to Nonlinear analysis
Large displacements, large strains
Swiss Federal Institute of Technology Page 24
Method of Finite Elements II
Introduction to non-linear analysis• Classification of non-linear analyses
Large displacements, large rotations and large strains (linear or nonlinear material behavior)
Linear or Nonlinear material behavior
Institute of Structural Engineering Method of Finite Elements II 14
Introduction to Nonlinear analysis
Change in BC for displacement
Swiss Federal Institute of Technology Page 25
Method of Finite Elements II
Introduction to non-linear analysis• Classification of non-linear analyses
Δ
2P
Chang in boundary conditions
2P
Institute of Structural Engineering Method of Finite Elements II 15
Example: Simple Bar Structure
Material Nonlinearity only
Assumptions: Small displacements, strains, load is applied slowly.
Swiss Federal Institute of Technology
Section a Section b
10cmaL = 5cmbL =
tR
tu2Area 1cm=
ε
σ
1
E1 TEYσ
0.002Yε =
7 2
5 2
10 N / cm10 N / cm
: yield stress: yield strain
T
Y
Y
EEσε
=
=1
2
3
4
2 4 6
tR
t
⇒ Calculate the displacement at the point of load application.Institute of Structural Engineering Method of Finite Elements II 16
Example: Simple Bar Structure
Swiss Federal Institute of Technology
Section a Section b
10cmaL = 5cmbL =
tR
tu2Area 1cm=
,
(elastic region)
(plastic region)
t tt t
a ba b
t t tb a
tt
tt Y
YT
u uL L
R A A
E
E
ε ε
σ σ
σε
σ σε ε
= = −
+ =
=
−= +
ε
σ
1
E1 TEYσ
0.002ε =
7 2
5 2
10 N / cm10 N / cm
: yield stress: yield strain
T
Y
Y
EEσε
=
=
1
2
3
4
2 4 6
tR
t
(unloading)Eσε Δ
Δ =
Institute of Structural Engineering Method of Finite Elements II 17
Example: Simple Bar Structure
In the beginning both Sections are elastic
Institute of Structural Engineering Method of Finite Elements II 18
Example: Simple Bar Structure
Section A is elastic while Section B is plastic
Since the stress on Section B is higher, it will yield first at time t∗:
Unloading occurs before Section A yields.Institute of Structural Engineering Method of Finite Elements II 19
Introduction to Nonlinear Analysis
Conclusion from the previous example:
The basic problem in general Nonlinear analysis is to find a state ofequilibrium between externally applied loads and element nodal forces
tR−t F = 0tR =t RB +t RS +t RC
tF =∑m
∫tVm
tB(m)T tτ (m) tdV(m)
where RB : body forces, RS : surface forces, RC : nodal forces
We must achieve equilibrium for all time steps whenincrementing the loading
Very general approach
Includes implicitly also dynamic analysis!Institute of Structural Engineering Method of Finite Elements II 20
Types of Response Diagrams
Basic Types
F
U
Institute of Structural Engineering Method of Finite Elements II 21
Types of Response Diagrams
Complex Types
U
R
Institute of Structural Engineering Method of Finite Elements II 22
Solution Algorithms for NL equations
Root finding for single variable NL problems f (x) = 0
Bisection Method
Assumption: f [a, b] ∈ < andcontinuous
If f (a) > 0, f (b) < 0⇒ a ≤ x ≤b ⇒ f (x) = 0
Fixed Point Iteration
Write f (x) = 0 in the form f (x) = x − q(x),the solution x satisfies x = q(x)Recurrence relation: xk+1 = g(xk)
Convergence: If g ′(x) is defined over [a, b] and apositive constant K exists with |g(x)| ≤ K ,∀x ∈ [a, b] then g(x) has a unique fixed pointx ∈ [a, b].
Institute of Structural Engineering Method of Finite Elements II 23
Solution Algorithms for NL equations
Root finding for single variable NL problems f (x) = 0
Newton (Raphson) Method
Defined by the recurrence relation
xk+1 = xk − f(xk)f′(xk)
terminate when |xk+1 − xk | ≤ ε, ε <<Convergence: quadratic→ |x − xk+1| ≤ C |x − xk+1|2
Secant Method
Defined by the recurrence relation
xk+1 = xk − f(xk)xk−xk−1
f(xk)−f(xk−1)
Convergence: superlinear order
→ α = 1+√
52
(golden ratio)
Institute of Structural Engineering Method of Finite Elements II 24
Incremental Analysis
The basic approach in incremental analysis is:Find a state of equilibrium between externally applied loads andelement nodal forces
t+∆tR−t+∆t F = 0
Assuming that t+∆tR is independent of the deformations we have
t+∆tR =t F + F
We know the solution tF at time t and F is the increment in thenodal point forces corresponding to an increment in thedisplacements and stresses from time t to time t + ∆t. This we canapproximate by
F =t KU
Institute of Structural Engineering Method of Finite Elements II 25
Incremental Analysis
Newton-Raphson MethodAssume the tangent stiffness matrix:
tK =∂tF
∂tU
We may now substitute the tangent stiffness matrix into theequilibrium relation
tKU =t+∆t R−t F
which gives us a scheme for the calculation of the displacements:
t+∆tU =t U + U
The exact displacements at time t + ∆t correspond to the appliedloads at t + ∆t, however we only determined these approximately aswe used a tangent stiffness matrix thus we may have to iterate tofind the solution.
Institute of Structural Engineering Method of Finite Elements II 26
Incremental Analysis
We may use the Newton-Raphson iteration scheme to find theequilibrium within each load increment
t+∆tK(i−1)∆U(i) =t+∆t R−t∆t F(i−1)
(out of balance load vector)
t+∆tU(i) =t+∆t U(i−1) + ∆U(i)
with Initial Conditions
t+∆tU(0) =t U; t+∆tK(0) =t K; t+∆tF(0) =t F
Institute of Structural Engineering Method of Finite Elements II 27
Modified Newton (Raphson)Method
It may be expensive to calculate the tangent stiffness matrix. In theModified Newton-Raphson iteration scheme it is only calculated in thebeginning of each new load step
Pinching of hysteretic loops due to shear cracking and bond slip of the reinforcement is commonly observed in reinforced concrete structures during cyclic loading. This phenomenon is taken into account by introducing in the expression of Kz the “slip length” parameter a, and a function f (z):
1 a ( )=
+ ⋅ ⋅pin zz
z
KKf z K
(8)
In the above equation pinzK is the hysteretic parameter affected by pinching, Kz is the original
expression of the hysteretic parameter obtained from (3) and parameters a, f(z) are given by: 2
m2
(z-z )a ( ' 1) & f(z)=exp -= −ss
Az
µ (9)
where As is a control parameter which may be linked to the size of crack opening or reinforcement slip or both; µ is the normalized curvature attained at the load reversal prior the current loading circle; zsis the range where slip occurs. A non zero value of the parameter zm will shift the effective slip region so that it is symmetric about z=zm.
7 Method of Analysis
The computer program “Plastique” which is described herein, is capable to perform the different types of analysis, namely; Push-over analysis, Quasi-static analysis, Non-linear dynamic analysis and Eigenvalue analysis The first two analysis types, although significantly simplified, can lead to valuable conclusions concerning the behavior of the structure and the possible collapse mechanism. The applied procedure can be described in brief as follows. In the case of 2-D analysis the structure is assumed to consist of a finite number of nodes interconnected by a finite number of elements. The types of elements have been described in section 3. In the case of 3-D analysis the structure is assumed to consist of the aforementioned 2-D frames, assuming a rigid diaphragm assemblage of their horizontal dof’s per floor slab. Loads may be applied at the nodes or along the elements. In both cases though, they are transformed to nodal loads.
Figure 5. Modified Newton Raphson Method
After the formation of the stiffness matrix the equilibrium equations are solved by an efficient algorithm based on the Gaussian elimination method. The structure stiffness is stored in a banded form
c) Pinching or Slip
V.K. Koumousis et al.372
In the quasi-Newton iteration schemes the secant stiffness matrix is used
instead of the tangent matrixInstitute of Structural Engineering Method of Finite Elements II 28
Simple Bar Example - RevisitedSwiss Federal Institute of Technology
( ) ( 1) ( 1)
( ) ( 1) ( )
(0) (0) (0)
( ) ( )
with initial conditions;
;
if section is elastic if se
t t i t t t t i t t ia b a b
t t i t t i i
t t t t t t t t ta a b b
t tt t
a ba b
t
T
K K u R F F
u u u
u u F F F F
CA CAK KL L
EC
E
+Δ +Δ − +Δ −
+Δ +Δ −
+Δ +Δ +Δ
+ Δ = − −
= + Δ
= = =
= =
== ction is plastic⎧⎨⎩
Institute of Structural Engineering Method of Finite Elements II 29
Simple Bar Example - RevisitedSwiss Federal Institute of Technology
0 0 (1) 1 1 (0) 1 (0)
4(1) 3
7
1 (1) 1 (0) (1) 3
1 (1)1 (1) 4
1 (1)1 (1)
Load step 1: 1:( )
2 10 6.6667 101 110 ( )10 5
Iteration 1: ( 1)6.6667 10
6.6667 10 < (elastic section!)
1.3333
a b a b
a Ya
bb
tK K u R F F
u
iu u u
uL
uL
ε ε
ε
−
−
−
=
+ Δ = − −
⇓
×Δ = = ×
+
=
= + Δ = ×
= = ×
= = 3
1 (1) 3 1 (1) 4
0 0 (2) 1 1 (1) 1 (1)
10 < (elastic section!)
6.6667 10 ; 1.3333 10
( ) 0
Y
a b
a b a b
F F
K K u R F F
ε−×
= × = ×
+ Δ = − − = 1 3
Convergence in one iteration!6.6667 10u −= ×
Institute of Structural Engineering Method of Finite Elements II 30
Simple Bar Example - Revisited
Swiss Federal Institute of Technology
1 1 (1) 2 2 (0) 2 (0)
4 3 4(1) 3
7
2 (1) 2 (0) (1) 2
2 (1) 3
2
Load step 2: 2 :( )
(4 10 ) (6.6667 10 ) (1.333 10 ) 6.6667 101 110 ( )10 5
Iteration 1: ( 1)1.3333 10
1.3333 10 < (elastic section!)
a b a b
a Y
tK K u R F F
u
iu u uε ε
ε
−
−
−
=
+ Δ = − −
⇓
× − × − ×Δ = = ×
+
=
= + Δ = ×
= ×(1) 3
1 (1) 4 1 (1) 2 (1) 4
1 1 (2) 2 2 (1) 2 (1) (2) 3
2.6667 10 > (plastic section!)
1.3333 10 ; ( ( ) ) 2.0067 10
( ) 2.2 10
b YT
a b b Y Y
a b a b
F F E A
K K u R F F u
ε
ε ε σ
−
−
= ×
= × = − + = ×
+ Δ = − − ⇒ Δ = ×
Institute of Structural Engineering Method of Finite Elements II 31
Simple Bar Example - Revisited
The procedure is repeated and the results of successive iterations aretabulated in the accompanying table.
Institute of Structural Engineering Method of Finite Elements II 32