NONLINEAR DYNAMIC FINITE ELEMENT ANALYSIS IN ZSOIL :

with application to geomechanics & structures

Th.Zimmermann

copyright zace services ltd

Ground motion

Far-field BC needed

2-phase medium

For time being in Z_Soil: limited structural dynamics

t

a, or d

analysis by geomod

with some extensions

STATICS RECALL

STATIC EQUILIBRIUM STATEMENT, 1-PHASE

traction imposed on

displacement imposedon u

11 11+(11/x1)dx1

12 +(12 /x2)dx2

12

f1

direction 1:

(11/x1)dx1dx2+(12 /x2) dx1dx2+ f1dx1dx2=0

L(u)= ij/xj + fi=0

x1

x2

dx1

Equilibrium

Boundary value problem

(Differential equation of equilibrium)

FORMAL DIFFERENTIAL PROBLEM STATEMENT1-phase,linear or nonlinear)

Txonfijij 0,Txonuu ukk

Txontn tijjij ,

epD

Incremental elasto-plastic constitutive equation:

(equilibrium)

(displ.boundary cond.)

(traction bound. cond.)

tu

NB: Time is steps

MATRIX FORM

-DISCRETIZATION LEADS TO THE MATRIX FORM….

FOR LINEAR STATICS

Kd=F

( K=stiffness matrix, F=vector of nodal forcesd=vector of nodal displacements)

DYNAMICS

DYNAMIC EQUILIBRIUM STATEMENT, 1-PHASE

11 11+(11/x1)dx1

12 +(12 /x2)dx2

12

f1

direction 1:

(11/x1)dx1dx2+(12 /x2) dx1dx2+ f1dx1dx2=0

L(u)= ij/xj + fi=0

x1

x2

dx1

1u

1 1 2u dx dx

iu

Boundary value problem

Equilibriumdisplacement imposedon u traction imposed

on

FORMAL DIFFERENTIAL PROBLEM STATEMENT

, 0ij ji if o xTnu

Deformation(1-phase):

k k uu u on xT

,ij j j i tn t xon T

epDIncremental elasto-plastic constitutive equation:

(equilibrium)

(displ.boundary cond.)

(traction bound. cond.)

( ) ( ) ( )u tt t t

0 0( 0, ) ( ); ( 0, ) ( )u t x u x u t x u x (initial conditions)

NB: Time is real

STATICS(linear case)

Kd=F

We obtain(Linear system size:Ndofs=Nnodes x NspaceDim,-d=nodal displacements-F=nodal forces)

DYNAMICS(linear case)

Ma(t)+[Cv(t)]+Kd(t) =F(t)where

We obtain(Linear system size:Ndofs=Nnodes x NspaceDim,But 3xNdofs unknowns)

v = d; a = d

COMPARING MATRIX FORMS

optional

SOLUTION TECHNIQUES

-MODAL ANALYSIS-FREQUENCY DOMAIN ANALYSIS

both essentially restricted to linear problems

-DIRECT TIME INTEGRATIONappropriate for a fully nonlinear analysis

DIRECT TIME INTEGRATION (linear case)…a)

Using Newmark’s algorithm :At each time step, solve:

2

1.

2. ( / 2)[(1 2 ) 2 ]

3. [(1 )

:

: ,

: 0.

]

5, 0.25( lg .)

algorithmic

with

wh and areere

typically trape

paramet

zoidal

e

t t

s

a o

r

t

n 1 n 1 n 1 n 1

n 1 n n n n 1

n 1 n n n 1

Ma Cv Κd F

d d v a a

v v a a

1 ( 1)nt n t

DIRECT TIME INTEGRATION (linear case)…b)

2

[ , (........) ]

[ , (.

,

2. ( / 2)[(1 2 ) 2 ]

3. [(1 ) ]

..

1

.....) ]

.

* *

functionof

functionof

at t

t t

t

n 1

n 1 n n n n 1

n 1 n n n 1

n 1 n 1 n 1 n 1

n 1 n 1 n

n 1 n 1 n

d d v a a

a d

v v a a

Ma C

d

v Κ

v

d F

Κ d Fn 1 n 1

STATICS(linear case)

Kd=F

DYNAMICS(linear case)

Ma(t)+Cv(t)+Kd(t)=F(t)

>>>>at any tn+1

we have an equivalent static problem

K*dn+1=F*n+1

an+1=…………vn+1=…………

v = d; a = d

MATRIX FORMS

NEWMARK IS A 1-STEP ALGORITHM

All information to compute solution at time tn+1, is in solution at time tn , restart is easy

n n+1

NUMERICAL ( ALGORITHMIC) DAMPING CAN EXIST

● ● ● ● ●● ●

●

●

●

●

Newmark(0.5,0.25)

HHT

IT MAY BE WANTED OR NOT

and varies with parameters (γ ,β)

●

●

●

●

Newmark(0.6,0.3025)

Err>10%

fi Contin. Discrete N=…

Err>10%

fi Contin. Discrete N=…

;2

ii i

Ef i

L

; 2sin2 1 2N i

i N i

k if

m N

k

Continuous case Discrete case

L = 10 L = 10m

particular case: 1; / ; 1, ,1

LE k EA l E A l m Al

N

;2

ii i

Ef i

L

; 2sin2 1 2N i

i N i

k if

m N

k

Continuous case Discrete case

L = 10 L = 10m

particular case: 1; / ; 1, ,1

LE k EA l E A l m Al

N

DISCRETIZATION APPROXIMATES HIGH FREQUENCIES

Filtering of high frequencies may be desirable

Exact sol.:

HHT Hilber-Hughes-Taylor α method

HHT filters high frequencies without damping low frequencies

NUMERICAL ( ALGORITHMIC) DAMPING CAN EXIST

● ● ● ● ●● ●

●

●

●

●

HHT(-0.3)

IT MAY BE WANTED OR NOT

and varies with parameters (γ ,β)

●

●

●

●

Algorithmic data for Newmark …or HHT(under CONTROL/AN..

Mass can be CONSISTENT (as obtained by FEM)or LUMPED (concentrated at (some) nodes)

Only lumped masses are available in ZSOIL

Lumped masses tend to lead to underestimate frequencies

Err>10%

fi Contin. Discrete N=…

Err>10%

fi Contin. Discrete N=…

;2

ii i

Ef i

L

; 2sin2 1 2N i

i N i

k if

m N

k

Continuous case Discrete case

L = 10 L = 10m

particular case: 1; / ; 1, ,1

LE k EA l E A l m Al

N

;2

ii i

Ef i

L

; 2sin2 1 2N i

i N i

k if

m N

k

Continuous case Discrete case

L = 10 L = 10m

particular case: 1; / ; 1, ,1

LE k EA l E A l m Al

N

Lumped masses tend to lead to underestimate frequencies:ILLUSTRATION

RAYLEIGH DAMPING a)Recall: Ma(t)+Cv(t)+Kd(t)=F(t)

C=αM+βK is RAYLEIGH DAMPING

α,β:constants

This form of damping is not representative of physical reality, in general. Its success is due to the fact that itmaintains mode decoupling in modal analysis

2

2

0 1...

( )

i i ii i

i i

y c y y i n

c Rayleigh

+

modal transformation ,

modal matrix, generates a decoupled

system of modal equations;

for each mode we have:

with the

and the

u = Φy

Φ

C uM u+ + K u = 0

RAYLEIGH DAMPING b):PARENTHESIS ON MODAL ANALYSIS

RAYLEIGH DAMPING d)

COMPARING THE MODAL EQUATION

2 2

2

2

( ) 0 1...

2 0 1...

2 ( )

i i ii i

i i i

y y y i n

y y y i n

i

+

+

WITH THE 1DOF VISCOUSLY DAMPED OSCILLATOR

YIELDS:

RAYLEIGH DAMPING e)

20 1

0

1

2

2 ( ) ,

1/0.5

1/

2 ( , )

define α and β and the viscousdamping at any frequency

( ) / 2

i i i

i i i

j j j

k k k

from i one gets

therefore pairs

k

+

+

RAYLEIGH DAMPING f) this can be plotted

2 (ω,ξ) pairs are used to define α0,β0 in ZSOIL

NONLINEAR DYNAMICS

E

y

CONSTITUTIVE MODEL: ELASTIC-PERFECTLY PLASTIC 1- dimensional

this problem is non-linear

epE

FROM LOCAL TO GLOBAL NONLINEAR RESPONSE

SOLUTION OF LINEARIZED PROBLEM, static case

N(d)=F

....)/()()( 1

1

1 ddNdNdN i

n

i

n

/ in+1( N d)Δd = F - N(d )

i+1 in+1 n+1d = d + Δd

Tn+1K

hence the following algorithm:

Linearize at , w. Taylor exp.i

nd 1

Nonlinear problem to solveF

d

i

nd 1

i: iterationn: step

THE PROBLEM IS NONLINEAR & THEREFORE NEEDS ITERATIONS

... .TOL in+1 n+1

i( N/ d)Δd = F - N(d = ΔF)

i+1 in+1 n+1d = d + Δd

nd

i: iterationn: step

tends to 0

d

on+1K

1FFn

Fn+1

1

1nd

Fon+1K

NEWTON- RAPHSON & al.ITERATIVES SCHEMES

i: iterationn: step

1.Full NR, update KT at each step & iteration, till .TOLF i

ndd

1FFn

Fn+1

1

1nd

To

nK 11F

2Fnd

d

Fn

Fn+1iF

2.Constant stiffness,use KTo

till .TOLF i

3.Modified NR, update KT opportunistically, each step e.g.,till

KTo

.TOLF i

4. BFGS, “optimal”secant scheme

TOLERANCES ITERATIVE ALGORITHMS

STATICS(nonlinear case)

N(d)=F

DYNAMICS(nonlinear case)

Ma(t)+Cv(t)+N(d(t))=F(t)

>>>>

MATRIX FORMS

(e.g.)

DIRECT TIME INTEGRATION (nonlinear case)

Using Newmark’s algorithm (or Hilber’s):At each time step, solve:

2

algor

2. ( / 2)[(1 2 )

:

:

2 ]

3. [(1 ) ]

,

: 0

ithmic

.5, 0.25

t t

with

where

typical

and are paramete

t

ly

rs

n+1 n+1 n+1

n 1 n

n

n n n 1

n 1 n n 1

+

n

11. Ma +Cv + = F

d d v a a

v v a a

N(d )1 ( 1)nt n t

STATICS(nonlinear case)

N(d)=F

DYNAMICS(nonlinear case)

Ma(t)+Cv(t)+N(d(t))=F(t)orMa(t)+N(d,v)=F(t)>>>>at any tn+1,

we have an equivalent static problem

N*(dn+1)=F*n+1

an+1=…………vn+1=…………

MATRIX FORMS

Like for linear case

SEISMIC INPUT a

uG

FextFin+Fel+Fdamp

uG

uG

uR

uT

uG

FextFin+Fel+Fdamp

uG

FextFin+Fel+Fdamp

uG

uG

uR

uT

uG

uG

uR

uT

uG

FextFin+Fel+Fdamp

uG

uG

uR

uT

uG

FextFin+Fel+Fdamp

uG

FextFin+Fel+Fdamp

uG

uG

uR

uT

uG

uG

uR

uT

>>Fin+Fdamp+Fel = Fext

equilibrium>>>

T R R extM u +Cu + Ku = F

SEISMIC INPUT b

,with R T G R T G R T Gu u - u ,u u - u u = u - u

yields

....

( )

with input as BC or

with input as inertia forces under seismic

T T T ext G G

R R R ext G

M u +Cu + Ku = F +Cu + Ku

M u +Cu + Ku = F - Mu

T R R extM u +Cu + Ku = F

Time-history