1 Instituto Tecnológico de Aeronáutica Prof. Maurício Vicente Donadon AE-256 Lecture notes: Prof. Maurício V. Donadon NUMERICAL METHODS IN APPLIED STRUCTURAL.

1

Instituto Tecnológico de Aeronáutica Prof. Maurício Vicente Donadon

AE-256

Lecture notes:

Prof. Maurício V. Donadon

NUMERICAL METHODS IN APPLIED STRUCTURAL

MECHANICS

2


AE-256

Non-linear static problems

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Introduction

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Sources of nonlinearities in structural analysis

• Geometrical non-linearity

• Non-linear material behaviour

• Non-linear boundary conditions

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Geometrical nonlinearities

)(2

1 2,

2,

2,, xxxxxx wvuu

)(2

1 2,

2,

2,, yyyyyy wvuv

)(2

1 2,

2,

2,, zzzzzz wvuw

Normal strain-displacement relationships

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yxyxyxyxxy wwvvuuuv ,,,,,,,,

zxzxzxzxxz wwvvuuuw ,,,,,,,,

zyzyzyzyyz wwvvuuvw ,,,,,,,,


Shear strain-displacement relationships

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Nonlinear material behaviour

p

EL

AS

TIC

EL

AS

TIC

EL

AS

TO

-PL

AS

TIC

EL

AS

TO

-PL

AS

TIC

EL

AS

TIC

+

EL

AS

TIC

+

MIC

RO

CR

AC

KIN

GM

ICR

OC

RA

CK

ING

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Nonlinear boundary conditions

Transient boundary problems:

Boundary conditions change during the analysis!!!

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Nonlinear boundary conditions

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Solution methods for non-linear

static problems

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• Incremental solutions

• Iterative solutions

• Combined incremental/iterative solutions

• Arc-length method

• Quasi-static solutions

Solution methods

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General form for a static problem

( ) ( ) 0i i i eg x F x F

K = K0 f(x)Fe

0

20

( ) 0

( ) 0

e

e

g x K x F

g x K x x F

• Example: Linear/Nonlinear Spring

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0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

Normalized displacement (m)

No

rma

lize

d f

orc

e

LinearNonlinear

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General form for a static problem


Trivial solution: Displacement control

Non-trivial solution: Load control which is commonly used in structural analyses!!!

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Incremental solution

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INCREMENTAL SOLUTION METHOD BASED ON THE EULER METHOD

( ) ( ) 0 ( )

i e i e

e i

g x F x F F x F

F dF

x dx

1

1

,

1

( ) ( )

e ein n ee

en n n i n n

n n

dFF F Fx F

dxg x F F x

x x x

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THE EULER METHOD ALGORITHM

1, n ex F

1

ie

dFx F

dx

1n nx x x

1

,

( ) ( )

e en n e

en n n i n n

F F F

g x F F x

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0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

Displacement

Mo

rma

lize

d F

orc

e

Exact solutionNumerical (Load increment =0.01)Numerical (Load increment=0.001)

• Example: Nonlinear Spring

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0.4 0.5 0.6 0.7 0.8 0.9 1-0.082

-0.081

-0.08

-0.079

-0.078

-0.077

-0.076

-0.075

-0.074

-0.073

Displacement

g(x

)

Load increment=0.01


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0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 13.4

3.6

3.8

4

4.2

4.4

4.6

4.8

5

5.2

5.4x 10

-3

Displacement

g(x

)

Load increment=0.001


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Iterative solution

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ITERATIVE SOLUTION BASED ON THE NEWTON RAPHSON METHOD

( ) ( ) 0i eg x F x F

2

21 11 2

1

2n n

n n

dg d gg g x x

dx dx

1

11 1 1 1 n

n n n n n

dgx g x x x

dx

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THE NEWTON RAPHSON ALGORITHM

1, , n ex F tol

1 1( )n ng x

1 1( ) ?n ng x tol

1

11 1

1 1

nn n

n n n

dgx g

dx

x x x

STOP

YES

NO

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0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

Displacement

No

rma

lize

d f

orc

e

Exact solutionN-R (10 different load levels)

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0 0.2 0.4 0.6 0.8 1-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1x 10

-4

Normalized force

g(x

)

Tolerance=1.0e-3


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0 0.2 0.4 0.6 0.8 10

2

4

6

8

10

12

14

Normalized force

Nu

mb

er

of

ite

rati

on

s

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Combined incremental/iterative

solutions

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COMBINED INCREMENTAL/ITERATIVE SOLUTIONS

( , ) ( ) 0

LOAD FACTORi eg x F x F

1

11 1 1 1 n

n n n n n

dgx g x x x

dx

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INCREMENTAL/ITERATIVE SOLUTION ALGORITHM

1, , n ex F tol

1 1( )n ng x

1 1( ) ?n ng x tol

1

11 1

1 1

nn n

n n n

dgx g

dx

x x x

STOP

YES

NO

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0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

Normalized displacement

No

rma

lize

d f

orc

e

Exact solutionLoad factor=0.5Load factor=0.1Load factor=0.05Load factor=0.01

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0 0.2 0.4 0.6 0.8 1-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1x 10

-4

Normalized load

g(x

)

Tol=1.0e-3

Load factor=0.5Load factor=0.1Load factor=0.05Load factor=0.01

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0 0.2 0.4 0.6 0.8 12

4

6

8

10

12

14

16

Normalized load

Nu

mb

er

of

ite

rati

on

s

Load factor=0.5Load factor=0.1Load factor=0.05Load factor=0.01

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Arc-length

Method

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Highly non-linear structural responses

eF

d

AB

C

Dsnap-through

sn

ap-b

ack

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Arc-length method: Single DOF

eF

d1nd nd

A

B CD

nd

1nF

1n n nF F F 1n n nF F F

L

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Arc-length method: Constraint equation

2 22 2 2n nd b F L

b: scale factor, scale forces to the same order of magnitude of the displacements

2 2 22 2, 0n n nf d d b F L

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Arc-length method: Residual force

1, ( ) 0n n n n i ng d F F F d

Equations to be solved simultaneously

1

2 2 22 2

, ( ) 0

, 0

n n n n i n

n n n

g d F F F d

f d d b F L

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Arc-length method

1 11 1

1 11 1

, , 0

, , 0

q qq q q q q q

n n n n n nn n

q qqq q q q q

n n nnn n

g gg d g d d

d

f ff d f d d

d

1

2 2 22 2

, ( ) 0

, 0

n n n n i n

n n n

g d F F F d

f d d b F L

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Arc-length method

1 11 1

1 11 1

, , 0

, , 0

q qq q q q q q

n n n n n nn n

q qqq q q q q

n n nnn n

g gg d g d d

d

f ff d f d d

d

11 11

1 11 2 1 2

,

2 , 2

qq qqi

T nn nn

q qq q

nn n

Fg gK F

d d

f fd b F

d

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Arc-length method

1 1 1 1 =cte, n n n n n nd d d d d d

1 11

1 2 1 2 1 1

,

2 2 ,

q qq qn nT n

q qq q qn n nn

g dK F d

d b F f d

Equations to be solved simultaneously

1

1

q q qn n n

q q qn n nd d d

Augmented stiffness matrix

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Arc-length method

Error function computation:

, ,,

, ,

Tq q q qn n n nq q

n n q q q qn n n n

g d g dE d tol

f d f d

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Arc-length method

Constant arc-length algorithm:

1 11 1 0 1 0 0 01

1

( )

with 1

n n T n i n T nd d K F F F d K g d

2 21 2 2 21

2 2 21 1 21

n

n

d b F L

L d b F

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Arc-length method

Variable arc-length algorithm:

1 11 1 0 1 0 0 01

1

( )

with 1

n n T n i n T nd d K F F F d K g d

2 22 2 2

22 21 2

n n

n n

d b F L

L d b F

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Spherical Arc-length method: b=0

• The computational cost associated with the inversion of the augmented stiffness matrix during the iterations is very high because the

augmented stiffness matrix is neither symmetric nor banded!

• The scale factor is unknown “a priori”

• Better solution: Spherical Arc-length!!!!

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Spherical Arc-length method

1 1

1 1, , 0q q

q q q q q qn n n n n n

n n

g gg d g d d

d

11 1

1,

qq qqi

T nn nn

Fg gK F

d d

1 1 1

1 11 11 1

= ,

,

q q q q qT n nn

q qq q q qn T n Tn n

K d g d F

d K g d K F

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1 1 1

1 11 11 1

1 1

11

1

1

= ,

,

q q q q qT n nn

q qq q q qn T n Tn n

q q q qn n n

q q q qn n n n n

qq i q qn n n n

i

K d g d F

d K g d K F

d d d

d d d d d

d d d d

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1 1

1

1

1

q q q qn n n

qq i q qn n n n

i

q q q

d d d

d d d d

2 2 22 2, 0n n nf d d b F L

2

1 2 3 0q qa a a

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1 1 21

1 1 1 2 12

21 1 1 1 2 1 23

2 2

q qn n

q q q qn n n

q q q q qn n n n

a d d b F F

a d d d b F F

a d d d d b F F L

2

1 2 3 0q qa a a

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1 1 1,

k=1 First Root

k=2 Second Root

q q q q qn k n n k nd d d d


2

1 2 3 0q qa a a

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1 1 1,

1,

1,

cos

q q q q qn k n n k n

q qn n k

k q qn n k

d d d d

d d

d d


Choosing the root

1

1qnd

,1qnd 2

1qnd ,2

qnd

Solution 1 Solution 2

1 1 1,

1 2

1 3 2

min cos ,cos

If 0

q q q q qn k n n k n

qk

qk

d d d d

a a a

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111

1

2

1 1 1 1

arc-length constraint equation with b=0

1 is positive defined

1 otherwise

q qp T p T n

qp p n

p p

pq q q qn n n n

T

T

d K F K F

d d

d d L

L Ls

d d d d

s K

s K


The predictor solution

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Quasi-static solutions

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DYNAMIC RELAXATION

( ) ( ) ( , , )

( , , , ) ( , , ) ( ) ( )e i

i e

Mx t F t F x x t

g x x x t F x x t Mx t F t

20( , , , )

( ) 0 e

g x x x t K x x

Cx Mx F t


K = K0 f(x)

M

Fe(t)

C

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1/ 2 1/ 21/ 2

11/ 2 1/ 21/ 2 , 1/ 2

( / 2) ( / 2)n nn n

n nn n n e n n

x x x t t x t tx v

t tv v

x a v tM F vt

Central difference method

DYNAMIC RELAXATION

, cL Et c

c

Critical time step computation

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Central difference method

DYNAMIC RELAXATION

t

0x 1x 2x 3x

t

4x

t

Displacement field

Velocity field

1/ 2v 3/ 2v 5/ 2v 7 / 2v

t

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Damping definition

DYNAMIC RELAXATION

( ) ( ) ( ) 0Mx t Cx t Kx t Critical damping

12 4 1.0nC M M T

C M K

Rayleigh damping

2

2

12

1high highhigh

low lowlow

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EXPLICIT TIME INTEGRATION ALGORITHM1. Initial conditions, v0, σ0, n=0, t=0, compute M

2. Compute acceleration an = M-1Fe,n

3. Update nodal velocities: vn+1/2 = vn+1/2-α + αΔtan

4. α = 1/2 if n=0

5. α = 1 if n>06. Update nodal displacements: un+1 = un+ Δtvn+1/2

7. Compute strains8. Compute stresses9. Compute internal forces10. Compute residual force vector: Fi - Fe

11. Update counter and time: n = n+1, t = t+Δt12. If simulation not complete go to step 2

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0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

Time (s)

Fo

rce

(N

)

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0 0.2 0.4 0.6 0.8 1 1.20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1


No

rma

lize

d L

oa

d

1 N/s0.1 N/s0.01 N/sExact solution

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0 0.2 0.4 0.6 0.8 10

0.05

0.1

0.15

0.2

0.25

Time

Internal energyKinetic energy

• Example: Nonlinear Spring – 1.0 N/s

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0 2 4 6 8 100

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

Time



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0 20 40 60 80 1000

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

Time


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• Example: Nonlinear Spring – Damping effect

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

Time (s)

No

rma

lize

d d

isp

lac

em

en

t

Tlow=0.22

Thigh=0.12

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C M K

2

2

12

1high highhigh

low lowlow

2

2

52.36 36.901 52.362

28.50 0.0251 28.50

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0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

Time (s)

No

rma

lize

d d

isp

lac

em

en

t

DampedUndamped


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0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1


No

rma

lize

d f

orc

e

Dynamic relaxationExact solution

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0 2 4 6 8 100

0.02

0.04

0.06

0.08

0.1

0.12

0.14

Time (s)

Internal energyKinetic energyDissipated energy


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Over damping effects in dynamic relaxation

•Over damping MUST BE AVOIDED in dynamic relaxation methods!

•Special care must be taken with over damping

• Over damping increases artificially the internal energy of the system!!!!

1 Instituto Tecnológico de Aeronáutica Prof. Maurício Vicente Donadon AE-256 Lecture notes: Prof. Maurício V. Donadon NUMERICAL METHODS IN APPLIED STRUCTURAL.

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