Transcript Document

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
Equilibrium
displacement imposed
on u
12 +(12 /x2)dx2
traction imposed
on 
f1

x2
dx1
11+(11/x1)dx1
11
x1
12
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)
 ij,j  fi  0 on  x T
(equilibrium)
u k  uk
on u x T
(displ.boundary cond.)
 ij,jn j  t i
on t x T
(traction bound. cond.)
  u  t
Incremental elasto-plastic constitutive equation:
  D 
ep
NB: Time is steps
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
 u1

x2

11+(11/x1)dx1
11
x1
dx1
f1
12
direction 1:
u1dx1dx2 (11/x1)dx1dx2+(12 /x2) dx1dx2+ f1dx1dx2=0
L(u)=
  ui 
ij/xj + fi=0
FORMAL DIFFERENTIAL PROBLEM STATEMENT
Deformation(1-phase):
(t )  u (t )  t (t )
ui   ij , j  fi  0 on  xT
(equilibrium)
uk  uk
on u xT
(displ.boundary cond.)
 ij , j n j  ti
on t x T
(traction bound. cond.)
u(t  0, x)  u0 ( x); u(t  0, x)  u0 ( x)
(initial conditions)
Incremental elasto-plastic constitutive equation:
  D 
ep
NB: Time is real
COMPARING MATRIX FORMS
STATICS
(linear case)
DYNAMICS
(linear case)
optional
Kd=F
Ma(t)+[Cv(t)]+Kd(t)
=F(t)
where
We obtain
v = d; a = d
(Linear system size:
We obtain
Ndofs=Nnodes x NspaceDim,
(Linear system size:
-d=nodal displacements
Ndofs=Nnodes x NspaceDim,
-F=nodal forces)
But 3xNdofs unknowns)
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, tn 1  (n  1)t
solve:
1. Man 1  Cvn 1  Κd n  1  Fn  1
with :
2. d n  1  d n  tvn  (t / 2)[(1  2  )an  2  an 1 ]
2
3. vn  1  vn  t[(1   )an   an 1 ]
where :  and  are algorithmic parameters,
typically :  0.5,   0.25(trapezoidal a lg o.)
DIRECT TIME INTEGRATION (linear case)…b)
at tn  1 ,
2. d n 1  d n  tvn  (t / 2)[(1  2  )an  2  an  1 ]
2
 an 1  function of [d n 1 , (........) n ]
3. vn  1  vn  t[(1   )an   an  1 ]
 vn 1  function of [d n 1 , (........) n ]
1. Man 1  Cvn 1  Κd n 1  Fn 1
*
 Κ d
*
F
n1 n1
MATRIX FORMS
STATICS
(linear case)
DYNAMICS
(linear case)
Kd=F
Ma(t)+Cv(t)+Kd(t)=F(t)
v = d; a = d
>>>>
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
n
n+1
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
Continuous case
Discrete case
L = 10
Exact sol.: 
i
fi 
2 L
E

; i  i
L = 10
k
fi 
N i
2
m
k

 i
;  N i  2sin 
 
m
 N 1 2 
particular case: E    1; k  EA / l; E  A  1, l 
fi Contin.
L
, m   Al
N 1
Discrete N=…
Err>10%
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
Continuous case
Discrete case
L = 10
fi 
i
E
; i  i
2 L 
fi Contin.
L = 10
k
fi 
N i
2
m
k

 i
;  N i  2sin 
 
m
 N 1 2 
Discrete N=…
particular case: E    1; k  EA / l; E  A  1, l 
Err>10%
L
, m   Al
N 1
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


M u+ C u+ K u = 0
with the modal transformation u = Φy ,
and Φ the modal matrix, generates a decoupled
system of modal equations;
for each mode we have:


y i  ci y i   yi  0
2
i
i  1...n
ci   + i2 ( Rayleigh)
RAYLEIGH DAMPING d)
COMPARING THE MODAL EQUATION


y i  ( + i2 ) y i  i2 yi  0
i  1...n
WITH THE 1DOF VISCOUSLY DAMPED OSCILLATOR


y  2  y   2 y  0
YIELDS:
2i i  ( + i2 )  i
i  1...n
RAYLEIGH DAMPING e)
from 2i i  ( 0 + 1i2 )  i, one gets
1/ i
0.5 
1/  j
i   0   i 
  

 j  1   j 
therefore 2 pairs ( ,  )
define α and β and the viscous damping at any frequency
 k  ( + k2 ) / 2k
k
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

y
  Eep 
E

this problem is non-linear
FROM LOCAL TO GLOBAL NONLINEAR RESPONSE
SOLUTION OF LINEARIZED PROBLEM, static case
Nonlinear problem to solve
F

N(d)=F
i
d
Linearize at
n 1 , w. Taylor exp.
d
i
n 1
N(din11 )  N(din1 )  (N / d)d  ....
hence the following algorithm:
i
( N / d)Δd = F - N(d n+1
)
T
K n+
1
i+1
i
d n+1
= d n+1
+ Δd
i: iteration
n: step
d
THE PROBLEM IS NONLINEAR & THEREFORE
NEEDS ITERATIONS
i
( N/ d)Δd = Fn+1 - N(d n+1
) = ΔF i ...  TOL.
o
K n+1
i+1
i
d n+1
= d n+1
+ Δd
Fn+1
 F1
F
Fn
tends to 0
o
K n+1
d
d n d1n1
i: iteration
n: step
NEWTON- RAPHSON & al.
ITERATIVES SCHEMES
Fn+1
Fn
K To
n1
dn d1n1
F i
Fn
KTo
dn
F1 F 2
 F1
Fn+1
d

i: iteration
n: step
d
1.Full NR, update KT at each
step & iteration, till Fi  TOL.
2.Constant stiffness,use KTo
till Fi  TOL.
3.Modified NR, update KT
opportunistically, each step
e.g.,till Fi  TOL.
4. BFGS, “optimal”secant
scheme
TOLERANCES
ITERATIVE ALGORITHMS
MATRIX FORMS
STATICS
(nonlinear case)
DYNAMICS
(nonlinear case)
N(d)=F
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, tn 1  (n  1)t
solve:
1. Man+1 + Cv n+1 + N(d n+ 1 ) = Fn+1
with :
2. d n  1  d n  tv n  (t / 2)[(1  2  )an  2  an  1 ]
2
3. vn  1  vn  t[(1   )an   an  1 ]
where :  and  are algorithmic parameters,
typically :  0.5,   0.25
MATRIX FORMS
STATICS
(nonlinear case)
DYNAMICS
(nonlinear case)
N(d)=F
Ma(t)+Cv(t)+N(d(t))=F(t)
or
Ma(t)+N(d,v)=F(t)
>>>>at any tn+1,
Like for linear case
we have an equivalent static problem
N*(dn+1)=F*n+1
an+1=…………
vn+1=…………
SEISMIC INPUT a
equilibrium
>>>
>>Fin+Fdamp+Fel = Fext
 M u + Cu + Ku = F
T
R
R
ext
SEISMIC INPUT b
 M uT + Cu R + Ku R = F ext
with u  u - u ,u  u - u , u = u - u
R
T
G
R
T
G
R
T
G
yields
M uT + CuT + KuT = F ext + CuG + KuG
with input as BC....or
R
R
R
M u + Cu + Ku = F
ext
- Mu
G
with input as inertia forces (under seismic )
Time-history