Dynamic structural analysis:Response history

L. Driemeier

R. Moura

M. Alves

• given a known load (in time and space) obtain the structure response (in time and

space)

• or, obtain

– displacement,

– velocity

– acceleration

of a loaded structure

Source: Concepts and applications of finite element

analysis, RD Cook, DS Malkus, ME Plesha, RJ Witt

Problem:

2

Response history

Modal methods:

It is necessary to

solve an eigen-

problem

Component mode

synthesis

(substructuring)

Ritz vectors:

More efficient than

modal methods

Response spectra

Usually no good

for non-linear response

3

Response history

Explicit methods

Direct integration

methods

Implicit direct

integration

algorithms

. ..

1 1{ } ({ } ,{ } ,{ } ,{ } )n n n n nD f D D D D

.. . ..

1 1 1{ } ({ } ,{ } ,{ } ,{ } ,{ } )n n n n n nD f D D D D D

int[ ]{ } [ ]{ } { } { }

[ ]{ } [ ]{ } [ ]{ } { }

ext

n n n n

ext

n n n n

M D C D R R

M D C D K D R

4

Explicit methods Implicit methods

•Conditionally stable (a critical

time step must not be exceeded

to avoid instability)

•Matrices can be made diagonal

(uncoupling)

•Low cost per time step but

many steps

•Wave propagation problem:

•Blast and impact loads

•High modes are important

•Response spans over

small time interval

•Unconditionally stable

(calculation remains stable

regardless of time step but

accuracy may suffer)

•Matrices cannot be made

diagonal (coupling)

•High cost per time step but few

steps

•Structural dynamics problem:

•Response dominated by

lower modes, eg structure

vibration, earthquacke

•Response spans over

many fundamental periods

•Number of multiplications per

time step implicit/explicit:

•2 for 1D

•15-150 2D

•4000 3D

•Because in implicit, matrix

becomes less narrowly banded

•Implicit requires more storage

5

Explicit direct integration

2

1

2

1

{ } { } { } { }2

{ } { } { } { }2

n n n n

n n n n

tD D t D D

tD D t D D

1

2

1 1 1 1

1 12

1{ } { } { } { } { } 2 { }

2

1{ } { } 2{ } { }

n n n n n n

n n n n

D D D D D t Dt

D D D Dt

1

1

2

2-

+

int[ ]{ } [ ]{ } { } { }ext

n n n nM D C D R R

1{ }nD6

𝑛𝑛 − 1 𝑛 + 1∆𝑡−∆𝑡

7

Use this in the bottom eq. slide 6

to obtain:

8

THIS IS EXPLICIT: TO BE IMPLEMENTED IN A PROGRAM

SEE EXERCISE I

Remarks on explicit integration

• [M] being diagonal uncouples the system [only if [C] is diagonal]

• The critical time step in the previous equation is

• Internal forces can be calculated from either

int int int1

{ } { } { } [ ] { }elsN

T

n n n nii

R r r B dV

int{ } [ ]{ }n nR K D

2

max

max

21

damping ratio

largest possible "calculated" frequency

t

It is not

necessary to

form [K]

Minimize n. of integration

points to 1: be carefull with

stress calculation though!9

Half-step central differences

10

0 0 1/2 1/2 0 0

1

0 0 0 0

2

1 0 0 0

1{ } { } { } { } { } { }

/ 2 2

{ } [ ] { } [ ]{ } [ ]{ }

{ } { } { } { }2

ext

tD D D D D D

t

D M R K D C D

tD D t D D

11

To start the processes

int[ ]{ } [ ]{ } { } { }ext

n n n nM D C D R R

Critical time step

• consistent mass matrix gives higher frequencies than

lumped mass matrix

• lumped mass increases ∆t

• higher order elements have higher frquencies than lower

order ones. Use the latter

/

mesh meshcr

L Lt

c E actual

n

cr

tC

t

12

Exercise I

13

14

Implicit direct integration

(Newmark family methods)

• Most of these methods are unconditionable stable: large time keeps the

solution stable, although may compromise accuracy

1

0

( )

n n

t

t t t

u u t

Average acceleration method

Integração

com

aceleração

cte.

t t+t

td

tt d

ttt dd 2

1

t0

15

Linear acceleration method

IMPLICIT

Integração

com

aceleração

linear

At

step

n+1

(=t)

16

Newmark relations

17

1

2

1 1 1 1[ ]{ } [ ]{ } [ ]{ } { }ext

n n n nM D C D K D R 1

0 0 0 0{ } [ ] { } [ ]{ } [ ]{ }extD M R K D C D

18

Implicit: to implement

De 2

De 1

Frequência Natural

19

Como calcular T?

int[ ]{ } [ ]{ } { } { }

[ ]{ } [ ]{ } [ ]{ } { }

ext

n n n n

ext

n n n n

M D C D R R

M D C D K D R

𝐷 = 𝑋sin(𝑡)

1. Atribua dimensões e material realistas à ponte de pedestres

abaixo e calcule suas frequências naturais e modos de vibrar.

Em seguida, adicione o carregamento indicado ao nó 1 e

calcule a amplitude da força para que o deslocamento do nó 2

seja L/10. Use os métodos implícitos e explícitos. Plote a

resposta do nó 2 no intervalo [0..4T ].

2

1

L

t

T=metade do primeiro

período natural da estrutura

Função seno+

+

Exercise II

20

Entrega de

Exercícios I e II

via e-disc até

22/03 22hs