Need help

[cyz1654]

I want to simulate a consolidated undrained cyclic test,first it is consolidated with 100KPA pressur,then a sine motion is assigned .I made some changes of other's code.But the results weren't prpper.Because the pore pressure was not accumulated with time,could someone have a look at it ?

wipe
logFile log.txt
set fmass 1 ;# fluid mass density
set smass 2 ;# saturated soil mass density
set G 5.5e4 ;
set B 1.5e5 ;
set bulk 2.2e6 ;#fluid-solid combined bulk modulus
set vperm 5.e-4 ;#vertical permeability (m/s)
set hperm 5.e-4 ;#horizontal permeability (m/s)
set accGravity 9.81 ;#acceleration of gravity
set vperm [expr $vperm/$accGravity/$fmass] ;#actual value used in computation
set hperm [expr $hperm/$accGravity/$fmass] ;#actual value used in computation
set press 0. ;# isotropic consolidation pressure on quad element(s)
set loadBias 0.07 ;# Static shear load, in percentage of gravity load (=sin(inclination angle))

set accMul 2. ;# acc. multiplier
set accNam whatever.acc ;# file name for input acc. record
set accDt 0.0166 ;# dt of input acc. record
set period 1.0 ;# Period for applied Sine wave
set deltaT 0.01 ;# time step for analysis
set numSteps 2400 ;# number of time steps
set gamma 0.6 ;# Newmark integration parameter

set massProportionalDamping 0. ;
set InitStiffnessProportionalDamping 0.002;



# ################################
# create the modelbuilder
# #################################

model BasicBuilder -ndm 3 -ndf 4

set g 9.81
set ld -8.0
# ################################
# build the model
# #################################

node 1 1.0 1.0 0.0
node 2 0.0 1.0 0.0
node 3 0.0 0.0 0.0
node 4 1.0 0.0 0.0
node 5 1.0 1.0 1.0
node 6 0.0 1.0 1.0
node 7 0.0 0.0 1.0
node 8 1.0 0.0 1.0

#triaxial test boundary
fix 1 0 0 1 1
fix 2 0 0 1 1
fix 3 0 0 1 1
fix 4 0 0 1 1
fix 5 0 0 0 1
fix 6 0 0 0 1
fix 7 0 0 0 1
fix 8 0 0 0 1


nDMaterial PressureDependMultiYield 1 3 $smass $G $B 29 .1 80 0.5\
29 0.21 0 0 10 0.02 1.

element brickUP 1 1 2 3 4 5 6 7 8 1 2.2e6 1 0 0 $vperm 0 0 0


updateMaterialStage -material 1 -stage 0

#stage 1
#===========================================================
set p 100
set np -100
#isotropic load
pattern Plain 1 "Constant" {
load 1 $np $np $p 0
load 2 $p $np $p 0
load 3 $p $p $p 0
load 4 $np $p $p 0
load 5 $np $np $np 0
load 6 $p $np $np 0
load 7 $p $p $np 0
load 8 $np $p $np 0
}


integrator Newmark 0.6 0.3025
numberer RCM
constraints Penalty 1e12 1e12
test NormDispIncr 1.0e-6 50 1
algorithm Newton
system UmfPack
analysis Transient

analyze 1 1

#stage 2
#===========================================================
#Axial loading
# or
# Simple Shear
#wipeAnalysis
#set previous load constant
remove sp 1 4
remove sp 2 4
remove sp 3 4
remove sp 4 4
remove sp 5 4
remove sp 6 4
remove sp 7 4
remove sp 8 4

wipeAnalysis
loadConst time 0
equalDOF 5 6 3 3
equalDOF 5 7 3 3
equalDOF 5 8 3 3
#
#




setTime 0.0 ;# reset time, otherwise reference time is not zero for time history analysis


# This is for the second stage of triaxial load, change for simple shear)
eval "recorder Node -file pwp2 -time -node 5 6 -dof 4 -dT 0.01 vel"

updateMaterialStage -material 1 -stage 1

pattern UniformExcitation 2 3 -accel "Sine 0 100 1 -factor 4"

integrator Newmark $gamma [expr pow($gamma+0.5, 2)/4]
rayleigh $massProportionalDamping 0.0 $InitStiffnessProportionalDamping 0.0
constraints Penalty 1.e18 1.e18 ;# can't combine with test NormUnbalance
test NormDispIncr 1.0e-6 25 0 ;# can't combine with constraints Lagrange
#algorithm Newton ;# tengent is updated at each iteration
algorithm KrylovNewton ;#
system ProfileSPD ;# Use sparse solver. Next numberer is better to be Plain.
numberer RCM ;# method to map between between equation numbers of DOFs
analysis VariableTransient ;# splitting time step requires VariableTransient


############# perform the Analysis and record time used #############
set startT [clock seconds]
analyze 5000 0.01 [expr 0.01/64] 0.01 15
set endT [clock seconds]
puts "Execution time: [expr $endT-$startT] seconds."

wipe

[kyonten]

I don't think you are simulating a cyclic CU test if you are applying motion at the base. You should be applying cyclic deviatoric stress at the top surface of the specimen after the isotropic consolidation step. The consolidation step is done under drained condition while the shearing step is done under undrained condition so appropriate pore pressure boundary conditions should be used.

[cyz1654]

[quote="kyonten"]I don't think you are simulating a cyclic CU test if you are applying motion at the base. You should be applying cyclic deviatoric stress at the top surface of the specimen after the isotropic consolidation step. The consolidation step is done under drained condition while the shearing step is done under undrained condition so appropriate pore pressure boundary conditions should be used.[/quote]


Hello,kyonten,could you tell me how to assign appropriate pore pressure boundary conditions in a cyclic CU test ?By the way ,could you tell me your Email?

[kyonten]

In the laboratory triaxial test during the consolidation stage drainage is allowed from the top and bottom of the specimen (two-way drainage) by opening the valves. The excess pore pressure is allowed to dissipate. During the shearing stage (say CU test) the valves are closed before applying the deviatoric stress. Excess pore pressure is generated.

Now for one element simulation we usually model one-quarter of the specimen and apply appropriate symmetric boundary conditions. Using symmetry we need to apply boundary condition for pore pressure only at the top surface, i.e. pervious during consolidation, and impervious during undrained shearing.

[cyz1654]

Hello,kyonten.As you suggested,I used 1/8 the element and applied cyclic deviatoric stress at the top surface of the specimen。Just as I expected,the pore pressure generated ,but the displacement seemed abnormal as it was almost zero .Could you please have a look at it and give me some advices? Thank you!


logFile log.txt
wipe

set massDen 2.0 ;# solid mass density
set fluidDen 1.0 ;# fluid mass density
set massProportionalDamping 0.0 ;
set stiffnessProportionalDamping 0.001 ;
set fangle 35 ;#friction angle
set ptangle 26.50 ;#phase transformation angle
set E 90000.0 ;#shear modulus
set poisson 0.40 ;
set G [expr $E/(2*(1+$poisson))] ;
set B [expr $E/(3*(1-2*$poisson))] ;
set press 0.0 ;# isotropic consolidation pressure on quad element(s)
set deltaT 0.010 ;# time step for analysis
set numSteps 2000 ;# Number of analysis steps
set gamma 0.600 ;# Newmark integration parameter
set period 1 ;# Period of applied sinusoidal load
set pi 3.1415926535 ;
set inclination 0 ;
set unitWeightX [expr ($massDen-$fluidDen)*9.81*sin($inclination/180.0*$pi)] ;# unit weight in X direction
set unitWeightY [expr -($massDen-$fluidDen)*9.81*cos($inclination/180.0*$pi)] ;# unit weight in Y direction
set ndm 3 ;# space dimension
set perm 5.1e-5

model BasicBuilder -ndm 3 -ndf 4

nDMaterial PressureDependMultiYield 1 3 $massDen $G $B $fangle .1 80 0.5 \
$ptangle 0.17 0.4 10 10 0.015 1.0

node 1 0.00000 0.0000 0.00000
node 2 0.00000 0.0000 0.500000
node 3 0.00000 0.50000 0.00000
node 4 0.00000 0.50000 0.500000
node 5 0.500000 0.0000 0.00000
node 6 0.500000 0.0000 0.500000
node 7 0.500000 0.50000 0.00000
node 8 0.500000 0.50000 0.500000


element brickUP 1 1 5 7 3 2 6 8 4 1 2.2e6 1 $perm $perm $perm 0 0 -9.81
updateMaterialStage -material 1 -stage 0

fix 1 1 0 1 0
fix 2 1 0 0 0
fix 3 1 1 1 0
fix 4 1 1 0 0
fix 5 0 0 1 0
fix 6 0 0 0 0
fix 7 0 1 1 0
fix 8 0 1 0 0

pattern Plain 1 "Constant" {
sp 2 4 0
sp 4 4 0
sp 6 4 0
sp 8 4 0

}

set load 12.5
pattern Plain 2 "Constant" {
load 1 0 $load 0 0
load 5 -$load $load 0 0
load 7 -$load 0 0 0
load 6 -$load $load -$load 0
load 8 -$load 0 -$load 0
load 4 0 0 -$load 0
load 2 0 $load -$load 0
}



# Define Analysis
system UmfPack
constraints Penalty 1e12 1e12
test NormUnbalance 1.0e-6 25 1
integrator Newmark 0.5 0.25
algorithm Newton
numberer RCM
analysis Transient
analyze 10 1


wipeAnalysis

remove loadPattern 1

loadConst -time 0.0

set SineSeries "Sine 0 1000 1 "
pattern Plain 3 $SineSeries {
load 2 0 0.0 -2.5 0
load 6 0 0.0 -2.5 0
load 8 0 0.0 -2.5 0
load 4 0 0.0 -2.5 0

}
eval "recorder Node -file pwp -time -node 2 -dT 0.125 -dof 4 vel"
eval "recorder Node -file disp -time -node 2 -dof 3 -dT 0.125 disp"

updateMaterialStage -material 1 -stage 1

system ProfileSPD
constraints Penalty 1e12 1e12
test NormDispIncr 1.0e-4 25 0
algorithm Newton
numberer RCM
integrator Newmark 0.5 0.25
analysis Transient
analyze 1000 0.01


wipe #flush ouput stream

[mzwygart]

Liquefaction of saturated sands was studied by means of pulsating loading triaxial tests on isotropically consolidated undrained laboratory samples. An analysis indicates that this type of test can be made to simulate idealized loading conditions on elements of soil in the field during earthquakes. The test results indicate that the danger of liquefaction of a saturated sand is determined by the following factors: (1) Void ratio; the higher the void ration the more easily liquefaction will occur. (2) Confining pressure; the lower the confining pressure the more easily liquefaction will occur, and (3) magnitude of cycle stress or strain; the larger the cyclic stress or strain the fewer the number of cycles required to induce liquefaction. Data from large shaking table experiments, and from observations of liquefaction during recent earthquakes lend qualitative support to the conclusions drawn from these laboratory cyclic loading triaxial tests.

[Aimeeing]

the laboratory triaxial test during the consolidation stage drainage is allowed from the top and bottom of the specimen (two-way drainage) by opening the valves. The excess pore pressure is allowed to dissipate. During the shearing stage (say CU test) the valves are closed before applying the deviatoric stress. Excess pore pressure is generated.

[koloki]

excessive pore pressure is allowed to dissipate. During the shearing stage (say CU test) the valves are closed before applying the deviatoric stress. Excess pore pressure is generated

[AlbertSmith]

Nice discussions are going on..