FEM Lecture 2

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



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



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



Collapse Analysis of the World Trade Center



Ultimate collapse capacity of jacket structure



Analysis of soil performance



Analysis of bridge response



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



Linear Elastic




Δ

2P

2P ε

σ

1

E

0.04ε <

Linear elastic (infinitesimal displacements)

L

L

//

P AE

L

σε σ

ε

==

Δ =

Infinitesimal Displacements



Material Nonlinearity onlySwiss Federal Institute of Technology Page 22



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



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



Large displacements, large strains




Large displacements, large rotations and large strains (linear or nonlinear material behavior)

Linear or Nonlinear material behavior



Change in BC for displacement




Δ

2P

Chang in boundary conditions

2P


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



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 =



In the beginning both Sections are elastic



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


Types of Response Diagrams

Complex Types

U

R


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].


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)


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



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.



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


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⎧⎨⎩


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 −= ×


Simple Bar Example - Revisited


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 = ×


Simple Bar Example - Revisited

The procedure is repeated and the results of successive iterations aretabulated in the accompanying table.


FEM Lecture 2

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