Numerical Modeling in Geotechnical Engineering 1 October ... · 3 Main References FLAC, Fast...
Transcript of Numerical Modeling in Geotechnical Engineering 1 October ... · 3 Main References FLAC, Fast...
PRESENTED BY: Siavash Zamiran
Ph.D. Student, Research and Teaching Assistant, Department of
Civil Engineering, Southern Illinois University Carbondale
Email: [email protected]
Website: www.zamiran.net
Linkedin: www.linkedin.com/in/zamiran
Numerical Modeling in Geotechnical Engineering 1
October 17, 18 2014
Workshop Schedule 2
Date Time Session Room
October 17 , 2014 9 AM to 10:30 AM Session 1 EB 0160
October 17 , 2014 10:45 AM to 12:15 PM Session 2 EB 0160
October 17 , 2014 2 PM to 3:30 PM Session 3 EB 1140
October 17 , 2014 3:45 PM to 5:15 PM Session 4 EB 1140
October 18 , 2014 9 AM to 10:30 AM Session 5 EB 1140
October 18 , 2014 10:45 AM to 12:15 PM Session 6 EB 1140
October 18 , 2014 2 PM to 3:30 PM Session 7 EB 1140
October 18 , 2014 3:45 PM to 5:15 PM Session 8 EB 1140
Main References 3
FLAC, Fast Lagrangian Analysis of Continua Manual, Itasca Inc., 2013
Steven F. Bartlett, Numerical Methods in Geotechnical Engineering,
The University of Utah, 2012
Outline 5
•Introduction to Numerical Modeling in Geotechnical
Engineers
• Numerical Modeling Using FLAC
•Foundation, Stone Column Modeling
•Slope Stability Analysis
•Structural Elements
•Seismic Considerations
Numerical Techniques 11
Numerical Techniques
•Finite Element
•Finite Difference
Commercially Available Software Packages
○ FLAC (Fast Lagrangian Analysis of Continua) (General FDM)
○ ABAQUS (FEM) (General FEM with some geotechnical relations)
○ ANSYS (FEM) (Mechanical/Structural)
○ PLAXIS (FEM) (Geotechnical)
○ SIGMA/W (FEM) (Geotechnical)
○ SEEP/W (FEM) (Seepage Analysis)
○ MODFLOW (FEM) (Groundwater Modeling)
FLAC and PLAXIS are the most commonly used by advanced
geotechnical consultants
Numerical Modeling Procedure 12
Selection of representative cross-section
• Idealize the field conditions into a design X-section
• Plane strain vs. axisymmetrical models
Choice of numerical scheme and constitutive relationship
• FEM vs FDM
• Elastic vs Mohr-Coulomb vs. Elastoplastic models
Characterization of material properties for use in model
• Strength
• Stiffness
• Stress - Strain Relationships
Grid generation
• Discretize the Design X-section into nodes or elements
Assign of materials properties to grid
Assigning boundary conditions
Calculate initial conditions
Determine loading or modeling sequence
Run the model
Obtain results
Interpret of results
Idealize Field Conditions to Numerical Modeling
13
• Complex site situation should be simplified reasonably for modeling
• Many 3D problems can be reduced to 2D problems by selection of
the appropriate X-sections.
Note for plane strain conditions to exist all strains are in the x-y coordinate system
(i.e., x-y plane). There is no strain in the z direction (i.e., out of the paper
direction). This usually implies that the structure or feature is relatively long, so
that the z direction and the balanced stresses in this direction have little influence
on the behavior within the selected cross section.
Plain Strain Numerical Modeling Examples
14
Deformation analysis of slopes Deformation analysis of tunnels
20
Flow to an injection and/or pumping well
Point Load on Soil
Axisymmetrical Numerical Modeling Examples
Constitutive Relationships 25
Constitutive Models: (i.e., stress-strain laws)
• Elastic: can be described by the linear elasticityequations such as
Hooke's law
•Viscoelastic: These are materials that behave elastically, but also
have damping: when the stress is applied and removed, work has to
be done against the damping effects. This implies that the material
response has time-dependence.
•Elaso-plastic: Materials that behave elastically generally do so
when the applied stress is less than a yield value. When the stress is
greater than the yield stress, the material behaves plastically and
does not return to its previous state. That is, deformation that occurs
after yield is permanent.
Itasca Consulting Group, Inc. 28
•Itasca is a global, employee-owned, engineering consulting and
software firm
•Are of concentration: mining, civil engineering, oil & gas,
manufacturing and power generation
•Since 1981
• Provides:
FLAC
FLAC3D
Flac/ Slope
PFC
Kubrix
3DEC
UDEC
Field Equations in FLAC 35
Solution of solid-body: equations of motion and constitutive relations
Heat-transfer: Fourier’s Law for conductive heat transfer
Fluid-flow problems: Darcy’s Law
Motion and Equilibrium
Bulk Modulus 40
where P is pressure, Vis volume, and ∂P/∂Vdenotes the partial derivative of
pressure with respect to volume.
Modeling Steps 54
•Generate a grid for the domain
•Assign Constitutive Model
•Assign material properties
•Assign boundary/loading conditions
•Solve for the system of algebraic equations using the initial
conditions and the boundary conditions (This usually done
by time stepping in an explicit formulation.)
•Implement the solution in computer code to perform the
calculations.
General solution procedure:
Start: COMMAND keyword value . . . <keyword value . . . > . . .
; comments
grid icol jrow
grid 10 10
model elastic
grid 20,20
model elas
gen 0,5 0,20 20,20 5,5 i=1,11
gen same same 20,0 5,0 i=11,21
grid 20,20
m e
gen 0,0 0,100 100,100 100,0 rat 1.25 1.25
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gr 10,10
m e
gen -100,0 -100,100 0,100 0,0 rat .80,1.25
Creating a circular hole in a grid
new
grid 20,20
m e
gen circle 10,10 5
model null region 10,10
model null region 10,10
Moving gridpoints with the INITIAL command
new
grid 5 5
model elastic
gen 0,0 0,10 10,10 10,0
ini x=-2 i=1 j=6
ini x=12 i=6
58
BAD GEOMETRY
(1) the area of the quadrilateral must be positive; and
(2) each member of at least one pair of triangular subzones which comprise the
quadrilateral must have an area greater than 20% of the total quadrilateral area
59
Assigning Material Models
Elastic Model MODEL elastic and MODEL mohr-coul require that material properties be
assigned via the PROPERTY
command. For the elastic model, the required properties are
(1) density;
(2) bulk modulus; and
(3) shear modulus.
60
Mohr-Coulomb plasticity model (1) density;
(2) bulk modulus;
(3) shear modulus;
(4) friction angle;
(5) cohesion;
(6) dilation angle; and
(7) tensile strength.
grid 10,10
model elas j=6,10
prop den=2000 bulk=1e8 shear=.3e8 j=6,10
model mohr j=1,5
prop den=2500 bulk=1.5e8 shear=.6e8 j=1,5
prop fric=30 coh=5e6 ten=8.66e6 j=1,5
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Sign Conventions
DIRECT STRESS – Positive stresses indicate tension; negative stresses indicate
compression.
SHEAR STRESS
63
PORE PRESSURE – Fluid pore pressure is positive in compression. Negative
pore pressure
indicates fluid tension.
GRAVITY – Positive gravity will pull the mass of a body downward (in the
negative y-direction).
Negative gravity will pull the mass of a body upward.
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;Example
grid 10 10
mod el
fix x i=1
fix x i=11
fix y j=1
app press = 10 j=11
ini sxx=-10 syy=-10
hist unbal
hist xvel i=5 j=5
hist ydisp i=5 j=11
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grid 10 10
mod el
prop d=1800 bulk=1e8 shear =.3e8
fix x i=1
fix x i=11
fix y j=1
app pres=1e6 j=11
hist unbal
hist ydisp i=5 j=11
ini sxx=-1e6 syy=-1e6 szz=-1e6
set gravity=9.81
step 900
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Performing Alterations
FLAC allows model conditions to be changed at any point in the solution
process. These changes
may be of the following forms.
• excavation of material
• addition or deletion of gridpoint loads or pressures
• change of material model or properties for any zone
• fix or free velocities for any gridpoint
;Example
grid 10,10
model elastic
gen circle 5,5 2
plot hold grid
gen adjust
plot hold grid
prop s=.3e8 b=1e8 d=1600
set grav=9.81
fix x i=1
fix x i=11
fix y j=1
solve
ini sxx 0.0 syy 0.0 szz 0.0 region 5,5
prop s .3e5 b 1e5 d 1.6 region 5,5
;mod null region 5,5
solve
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Excavate and fill in stages
grid 10,10
m e
prop s=5.7e9 b=11.1e9 d=2000
fix x i=1
fix y j=1
fix x i=11
apply syy -20e6 j=11
ini sxx -30e6 syy -20e6 szz -20e6
his unbal
his xdis i=4 j=5
solve
mod null i 4,7 j 3,6
solve
mod mohr i 4,7 j 3,6
prop s=.3e8 b=1e8 fric=30 i=4,7 j=3,6
mod null i=1,3 j=3,6
mod null i=8,10 j=3,6
ini xd=0 yd=0
his reset
his unbal
his xdis i=4 j=5
step 1000
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Foundation and stone column modeling
74
Example 1:
;------------STAGE 1-----------------
config
grid 80 40
model elastic i=1,80 j=1,20
gen 0,0 0,5 20,5 20,0 i=1,81 j=1,21
prop b=8.3e6 s= 3.8e6 d=1800 i=1,80
j=1,20
fix x i=1
fix x i=81
fix x y j=1
set g=9.81
history unbal
;solve
ini xdisp=0 ydisp=0
;
Foundation and stone column modeling
75
;------------STAGE 2-----------------
model mohr i=1,80 j=1,20
apply press=100e3 i=38,45 j=21
prop b=8.3e6 s= 3.8e6 d=1800 c=30e3
f=10 i=1,80 j=1,20
set large
;solve
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;Example 2:
;------------STAGE 1-----------------
config
grid 80 40
model elastic i=1,80 j=1,20
gen 0,0 0,5 20,5 20,0 i=1,81 j=1,21
fix x i=1
fix x i=81
fix x y j=1
set g=9.81
history unbal
group layer1 j=1,8
group layer2 j=9,16
group layer3 j=17,20
prop b=1.6e7 s= 7.6e6 d=1800 group layer1
prop b=1.5e6 s= 6.9e5 d=2000 group layer2
prop b=1.6e7 s= 7.6e6 d=2000 group layer3
solve
Foundation and stone column modeling
77
ini xdisp 0 ydisp=0
;
;------------STAGE 2-----------------
group cap j=20 i=32,48
group stonecolumn i=33 j=12,19
group stonecolumn i=47 j=12,19
group stonecolumn i=40 j=12,19
group stonecolumn i=36 j=12,19
group stonecolumn i=44 j=12,19
prop b=1.6e8 s= 7.6e7 d=2200 group
stonecolumn
prop b=1.6e8 s= 7.6e7 d=2200 group cap
solve
ini xdisp 0 ydisp=0
;
Foundation and stone column modeling
78
;------------STAGE 3-----------------
model mohr group layer1
model mohr group layer2
model mohr group layer3
prop b=1.6e7 s= 7.6e6 d=1800 c=35e3
f=10 group layer1
prop b=1.5e6 s= 6.9e5 d=2000 f=25 group
layer2
prop b=1.6e7 s= 7.6e6 d=2000 c=35e3
f=12 group layer3
apply press=80e3 i=32,48 j=21
solve
Foundation and stone column modeling
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Example 3:
;------------STAGE 1-----------------
config
grid 80 40
model elastic i=1,80 j=1,20
gen 0,0 0,5 20,5 20,0 i=1,81 j=1,21
fix x i=1
fix x i=81
fix x y j=1
set g=9.81
history unbal
group layer1 j=1,8
group layer2 j=9,16
group layer3 j=17,20
prop b=1.6e7 s= 7.6e6 d=1800 group layer1
prop b=1.5e6 s= 6.9e5 d=2000 group layer2
prop b=1.6e7 s= 7.6e6 d=2000 group layer3
solve
Foundation and stone column modeling
81
ini xdisp 0 ydisp=0
;
;------------STAGE 2-----------------
group cap j=20 i=32,48
prop b=1.6e8 s= 7.6e7 d=2200 group cap
solve
ini xdisp 0 ydisp=0
;------------STAGE 3-----------------
model mohr group layer1
model mohr group layer2
model mohr group layer3
prop b=1.6e7 s= 7.6e6 d=1800 c=35e3
f=10 group layer1
prop b=1.5e6 s= 6.9e5 d=2000 f=25 group
layer2
prop b=1.6e7 s= 7.6e6 d=2000 c=35e3
f=12 group layer3
apply press=80e3 i=32,48 j=21
solve
Foundation and stone column modeling
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0
1
2
3
4
5
6
-10 -8 -6 -4 -2 0
He
igh
t (m
)
Settlement (cm)
Settlement (cm)
Settlement (With Stone Column) (cm)
Foundation and stone column modeling
Slope stability analysis – Dry 86
config ats
grid 20,10
m m
prop s=.3e8 b=1e8 d=1500 fri=20 coh=1e10
ten=1e10
; warp grid to form a slope :
gen 0,0 0,3 20,3 20,0 j 1,4
gen same 9,10 20,10 same i 6 21 j 4 11
mark i=1,6 j=4
mark i=6 j=4,11
model null region 1,10
; displacement boundary conditions
fix x i=1
fix x i=21
fix x y j=1
; apply gravity
set grav=9.81
; soil properties -- note large
cohesion to force initial elastic
; behavior for determining
initial stress state. This will
prevent ; slope failure when
initializing the gravity stresses
Slope stability analysis – Dry 87
; displacement history of slope
his ydis i=10 j=10
; solve for initial gravity stresses
solve
;
;*** BRANCH: DRY ****
;... STATE: SL2 ....
; reset displacement components to zero
ini xdis=0 ydis=0
; set cohesion to 0
prop coh=0
; use large strain logic
set large
step 200
;save sl2.sav
step 800
solve
; Change cohesion to c=10e3
; soil properties -- note large
cohesion to force initial elastic
; behavior for determining
initial stress state. This will
prevent ; slope failure when
initializing the gravity stresses
Slope stability analysis – Water table 88
ini xdis=0.0 ydis=0.0
;
; Update Densities for saturated part ;*** BRANCH: WATER TABLE ****
; install phreatic surface in slope
water table 1 den 1000
table 1 (0,5) (6.11,5) (20,9)
;Change density for saturated
apply press 2e4 var 0 -2e4 from 1,4 to 6,6
step 6000
Slope stability analysis – Groundwater 90
config gw ats
grid 20,10
; Type "new" to start a new problem
;Mohr-Coulomb model
m m
prop s=.3e8 b=1e8 d=1500 fri=20 coh=1e10 ten=1e10
; warp grid to form a slope :
gen 0,0 0,3 20,3 20,0 j 1,4
gen same 9,10 20,10 same i 6 21 j 4 11
mark i=1,6 j=4
mark i=6 j=4,11
model null region 1,10
prop perm 1e-10 por .3
water den 1000 bulk 1e4
Slope stability analysis – Groundwater 91
; displacement boundary conditions
fix x i=1
fix x i=21
fix x y j=1
; pore pressure boundary conditions
apply pp 9e4 var 0 -9e4 i 21 j 1 10
apply pp 5e4 var 0 -3e4 i 1 j 1 4
ini pp 2e4 var 0 -2e4 mark i 1 6 j 4 6
fix pp mark
; apply gravity
set grav=9.81
hist gwtime
set mech off
solve
Slope stability analysis – Groundwater 92
;... STATE: SLGW3 ....
set flow off mech on
app press 2e4 var 0 -2e4 from 1 4 to 6 6
water bulk 0.0
; displacement history of slope
hist reset
his ydis i=10 j=10
solve
;... STATE: SLGW4 ....
ini xdis 0.0 ydis 0.0
prop coh 1e4 ten 0.0
set large
solve
Nail and Anchor modeling 95
t is annulus thickness of the shear zone and is
considered equal to 0.004 m.
Codes 97
struct node 1501 0.0 0.0
struct node 1500 0.0 -32.0
struct beam beg node 1500 end node 1501 prop 1001 seg 32
struct prop 1001
int 101 as from 31,35 to 31,67 bs from node 1500 to node 1501
int 102 as from 32,67 to 32,35 bs from node 1501 to node 1500
struct prop 1001 density 2400.0
struct cable begin node 1533 end node 1534 seg 1 tension 768000.0 prop 2001
struct cable begin node 1535 end node 1533 seg 6 prop 2002
struct prop 2002
struct prop 2001 spac 2.3 e 2.1E11 area 0.0015 yie 1e10 kb 0 sb 0
struct prop 2002 spac 2.3 e 2.1E11 area 0.0015 yie 1e10 kb 1.0E8 sb 1e8
Codes 102
config dyn
set dyn off
set dyn on
apply xquiet j=1
apply ff
apply nquiet squiet j=1 ;(bottom)
;----------------------damping
set dy_damp struc rayl 0.05 1.64 mass
set dy_damp rayl 0.05 1.64
apply sxy 4.8e5 hist table 1 j 1
set dytime 0
hist reset
hist dytime
his 10 xacc i=50 j=2
his 11 xacc i=50 j=26
solve dytime 4
table 1 0 0
table 1 0.02 -0.146
table 1 0.04 -0.374
table 1 0.06 -0.574
table 1 0.08 -0.761
table 1 0.1 -0.973
table 1 0.12 -1.223
table 1 0.14 -1.473
table 1 0.16 -1.682
table 1 0.18 -1.842
table 1 0.2 -1.97
table 1 0.22 -2.138
table 1 0.24 -2.389
table 1 0.26 -2.693
table 1 0.28 -2.971
table 1 0.3 -3.189
table 1 0.32 -3.345
table 1 0.34 -3.434