fem lecture 3

7/27/2019 FEM Lecture 3

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The Finite Element Method for the Analysis ofNon-Linear and Dynamic Systems

Prof. Dr. Eleni Chatzi

Lecture 3 - 9 October, 2012

Institute of Structural Engineering Method of Finite Elements II 1


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Introduction to Nonlinear Analysis

Conclusion from the previous example:

The basic problem in general Nonlinear analysis is to find a state of

equilibrium between externally applied loads and element nodal forces

tR t F= 0tR=t RB+

t RS+t RC

tF=

m

tVm

tB(m)T t(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 2


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Types of Response Diagrams

Basic Types



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Types of Response Diagrams

Complex Types



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Solution Algorithms for NL equations

Root finding for single variable NL problems f(x) = 0

Bisection Method

Assumption: f[a, b] and continuous

If f(a)>0, f(b)


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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| ,


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Newtons method for FE

Assume the general case where Kis a nonlinear function of thedisplacement U:

K(U)U= F

In order to write it in the form f(x) = 0, we define the residual r(U):

r(U) =K(U)U F r(U) = 0

Then, the Newton iteration formula becomes:

Uk+1 = Uk T1(Uk)r(Uk)

The slope of the tangent (the tangent stiffness) is

T(Uk) = dr(Uk)

dU =

dK(U)U F

dU |U=Uk

T(Uk) =K(Uk) +dK(Uk)

dU Uk



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Incremental Analysis

The basic approach in incremental analysis is:Find a state of equilibrium between externally applied loads and

element nodal forces between successive time steps t

t+tR t+tF= 0

Assuming that t+tR is independent of the deformations we have

t+tR= tF + F

We know the solution tFat time t and F is the increment in thenodal point forces corresponding to an increment in the

displacements and stresses from time tto time t+ t. This we canapproximate by

F= tKU



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Newton-Raphson MethodAssume the tangent stiffness matrix:

tK= tFtU

We may now substitute the tangent stiffness matrix into theequilibrium relation

tKU= t+tR tF

which gives us a scheme for the calculation of the displacements:

t+tU= tU + U

The exact displacements at time t+ tcorrespond to the appliedloads at t+ t, however we only determined these approximately aswe used a tangent stiffness matrix thus we may have toiterate to

find the solution.Institute of Structural Engineering Method of Finite Elements II 9

I l A l i


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We may use the Newton-Raphson iteration scheme to find the

equilibrium within each load increment

t+tK(i1)U(i) = t+tR t+tF(i1)

(out of balance load vector)

t+tU(i) = t+tU(i1)

+ U(i)

with Initial Conditions

t+tU(0) = tU; t+tK(0) = tK; t+tF(0) = tF


M difi d N (R h )M h d


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Modified Newton (Raphson)Method

It may be expensive to calculate the tangent stiffness matrix. In theModified Newton-Raphsoniteration scheme it is only calculated in thebeginning of each new load step

In the quasi-Newtoniteration schemes the secant stiffness matrix is used

instead of the tangent matrixInstitute of Structural Engineering Method of Finite Elements II 11

S i l C id ti


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Special Considerations

Standard Newton-Raphon methods perform poorly for buckingproblems, where the slope at limit points is exactly equal to 0


S i l C id ti


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Special Considerations

The Arc-Length Method for Nonlinear Post-Buckling

Also called Modified Riks Method.

Control the size of the load step using aparameter .

Solve for both and U in each Newtoniteration.

Assume F = independent of geometry. Then, can be thought of as a normalized loadparameter and the residual is given by

r(U, ) =K(U)U F

The load increment is computed using

=

s2 U2n

where the reference arc length is

s20 = F

nloadstepInstitute of Structural Engineering Method of Finite Elements II 13

Si l B E l R isit d


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Simple Bar Example - Revisited


Simple Bar Example Revisited


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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:

( )

2 106.6667 10

1 110 ( )

10 5

Iteration 1 : ( 1)

6.6667 10

6.6667 10 < (elastic section!)

1.3333

a b a b

a Y

a

b

b

t

K K u R F F

u

i

u u u

u

L

u

L

=

+ =

= =

+

== + =

= =

= = 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

= =

+ = = 1 3

Convergence in one iteration!

6.6667 10u =




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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 10

1 110 ( )

10 5Iteration 1:( 1)

1.3333 10

1.3333 10 < (elastic section!)

a b a b

a Y

t

K K u R F F

u

i

u u u

=

+ =

= =

+

=

= + =

=

(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 Y

T

a b b Y Y

a b a b

F F E A

K K u R F F u

=

= = + =

+ = =




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


The Continuum Mechanics Incremental Equations


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The Continuum Mechanics Incremental Equations

The basic ProblemEstablish the solution using an incremental formulation. Two mainapproaches exist for establishing equilibrium

Lagrangian Formulation:Track the movement of all particles of the body (located in aCartesian coordinate system), in their motion from the original to thefinal configuration (pathline)

Eulerian Formulation:The motion of the material through a stationary control volume isconsidered (streamlines). Mainly used in fluid mechanics.


Lagrangian vs Eulerian Formulation - 1D Example


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Lagrangian vs. Eulerian Formulation - 1D Example

Spatial or Eulerian coordinates (x): These coordinates are used to locate apoint in space with respect to a fixed basis.Material or Lagrangian coordinates (X): These coordinates are used to label

material points. If we sit on a material point, the label does not change with time.Example: Assume that the motion is

x= (X, t) =X(1 + 2t+t2)

The inverse of the map gives us X in terms of x, i.e.,

X= 1

(x, t) = x

(1 + 2t+ t2)

Then, the displacement of the material point X is

u(X, t) =(X, t) (X, 0) =X(2t+ t2)

The velocity of the material point is (Langrangian Description)

v(X, t) = u

t= 2X(1 +t)

Alternatively we can express the velocity in terms ofx (Eulerian Description)

v(X, t) =v(1(x, t), t) = 1x(1 +t)

(1 + 2t+t2)


Lagrangian Formulation


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Lagrangian Formulation

In solids we use the Lagrangian approach as the solution process moves from time

t to t+ titeratively following elements of the body in their motion.


fem lecture 3

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