NONLINEAR DYNAMIC FINITE ELEMENT ANALYSIS IN ZSOIL :

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NONLINEAR DYNAMIC FINITE ELEMENT ANALYSIS IN ZSOIL : with application to geomechanics & structures Th.Zimmermann copyright zace services ltd

Far-field BC needed 2-phase medium Ground motion

For time being in Z_Soil: limited structural dynamics
a, or d t

with some extensions analysis by geomod

STATICS RECALL

STATIC EQUILIBRIUM STATEMENT, 1-PHASE
Boundary value problem displacement imposed on u Equilibrium 12 +(12 /x2)dx2 12 f1 traction imposed on  11 11+(11/x1)dx1 x2 x1 dx1 direction 1: (11/x1)dx1dx2+(12 /x2) dx1dx2+ f1dx1dx2=0 L(u)= ij/xj + fi=0 (Differential equation of equilibrium)

FORMAL DIFFERENTIAL PROBLEM STATEMENT
1-phase,linear or nonlinear) (equilibrium) (displ.boundary cond.) (traction bound. cond.) Incremental elasto-plastic constitutive equation: NB: Time is steps

Kd=F MATRIX FORM -DISCRETIZATION LEADS TO THE MATRIX FORM….
FOR LINEAR STATICS Kd=F ( K=stiffness matrix, F=vector of nodal forces d=vector of nodal displacements)

DYNAMICS

DYNAMIC EQUILIBRIUM STATEMENT, 1-PHASE
Boundary value problem displacement imposed on u Equilibrium traction imposed on  12 +(12 /x2)dx2 12 f1 11 11+(11/x1)dx1 x2 x1 dx1 direction 1: (11/x1)dx1dx2+(12 /x2) dx1dx2+ f1dx1dx2=0 L(u)= ij/xj + fi=0

FORMAL DIFFERENTIAL PROBLEM STATEMENT
Deformation(1-phase): (equilibrium) (displ.boundary cond.) (traction bound. cond.) (initial conditions) Incremental elasto-plastic constitutive equation: NB: Time is real

Kd=F Ma(t)+[Cv(t)]+Kd(t) =F(t) COMPARING MATRIX FORMS 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) optional

SOLUTION TECHNIQUES -MODAL ANALYSIS -FREQUENCY DOMAIN ANALYSIS both essentially restricted to linear problems -DIRECT TIME INTEGRATION appropriate for a fully nonlinear analysis

DIRECT TIME INTEGRATION (linear case)…a)
Using Newmark’s algorithm : At each time step, solve:

DIRECT TIME INTEGRATION (linear case)…b)

Ma(t)+Cv(t)+Kd(t)=F(t) >>>>
MATRIX FORMS 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=…………

NEWMARK IS A 1-STEP ALGORITHM
All information to compute solution at time tn+1, is in solution at time tn , restart is easy

NUMERICAL ( ALGORITHMIC) DAMPING CAN EXIST
and varies with parameters (γ ,β) Newmark(0.6,0.3025) HHT Newmark(0.5,0.25) IT MAY BE WANTED OR NOT

DISCRETIZATION APPROXIMATES HIGH FREQUENCIES
Exact sol.: Filtering of high frequencies may be desirable

HHT Hilber-Hughes-Taylor α method
HHT filters high frequencies without damping low frequencies

NUMERICAL ( ALGORITHMIC) DAMPING CAN EXIST
and varies with parameters (γ ,β) HHT(-0.3) IT MAY BE WANTED OR NOT

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

Lumped masses tend to lead to underestimate frequencies:
ILLUSTRATION

C=αM+βK is RAYLEIGH DAMPING
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 it maintains mode decoupling in modal analysis

RAYLEIGH DAMPING b): PARENTHESIS ON MODAL ANALYSIS

RAYLEIGH DAMPING d) COMPARING THE MODAL EQUATION
WITH THE 1DOF VISCOUSLY DAMPED OSCILLATOR YIELDS:

RAYLEIGH DAMPING e)

RAYLEIGH DAMPING f) this can be plotted

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

NONLINEAR DYNAMICS

CONSTITUTIVE MODEL: ELASTIC-PERFECTLY PLASTIC
1- dimensional E y this problem is non-linear

FROM LOCAL TO GLOBAL NONLINEAR RESPONSE

SOLUTION OF LINEARIZED PROBLEM, static case
Nonlinear problem to solve d Linearize at , w. Taylor exp. hence the following algorithm: i: iteration n: step

THE PROBLEM IS NONLINEAR & THEREFORE
NEEDS ITERATIONS tends to 0 Fn+1 Fn i: iteration n: step d

d Fn Fn+1 NEWTON- RAPHSON & al. ITERATIVES SCHEMES d Fn Fn+1 KTo 2.Constant stiffness,use KTo till i: iteration n: step 3.Modified NR, update KT opportunistically, each step e.g.,till 1.Full NR, update KT at each step & iteration, till 4. BFGS, “optimal”secant scheme

TOLERANCES ITERATIVE ALGORITHMS

Ma(t)+Cv(t)+N(d(t))=F(t)
MATRIX FORMS STATICS (nonlinear case) N(d)=F DYNAMICS (nonlinear case) Ma(t)+Cv(t)+N(d(t))=F(t) >>>> (e.g.)

DIRECT TIME INTEGRATION (nonlinear case)
Using Newmark’s algorithm (or Hilber’s): At each time step, solve:

Ma(t)+Cv(t)+N(d(t))=F(t) or Ma(t)+N(d,v)=F(t)
MATRIX FORMS STATICS (nonlinear case) N(d)=F DYNAMICS (nonlinear case) Ma(t)+Cv(t)+N(d(t))=F(t) or Ma(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=………… Like for linear case

SEISMIC INPUT a >>> equilibrium >>Fin+Fdamp+Fel = Fext

SEISMIC INPUT b yields

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