Numerical Modeling in Geotechnical Engineering October 17,

Transcription

1 1 Numerical Modeling in Geotechnical Engineering October 17, PRESENTED BY: Siavash Zamiran Ph.D. Student, Research and Teaching Assistant, Department of Civil Engineering, Southern Illinois University Carbondale Website: Linkedin:

2 Workshop Schedule 2 Date Time Session Room October 17, AM to 10:30 AM Session 1 EB 0160 October 17, :45 AM to 12:15 PM Session 2 EB 0160 October 17, PM to 3:30 PM Session 3 EB 1140 October 17, :45 PM to 5:15 PM Session 4 EB 1140 October 18, AM to 10:30 AM Session 5 EB 1140 October 18, :45 AM to 12:15 PM Session 6 EB 1140 October 18, PM to 3:30 PM Session 7 EB 1140 October 18, :45 PM to 5:15 PM Session 8 EB 1140

3 3 Main References FLAC, Fast Lagrangian Analysis of Continua Manual, Itasca Inc., 2013 Steven F. Bartlett, Numerical Methods in Geotechnical Engineering, The University of Utah, 2012

4 4 Prerequisites Soil Mechanics Foundation Engineering Strength of Material

5 Outline 5 Introduction to Numerical Modeling in Geotechnical Engineers Numerical Modeling Using FLAC Foundation, Stone Column Modeling Slope Stability Analysis Structural Elements Seismic Considerations

6 6 Chapter 1 Introduction to Numerical Modeling in Geotechnical Engineers

7 7 Numerical modeling approach

8 8 Simple definition of modeling

9 9 Application of modeling

10 10 Numerical modeling procedure

11 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

12 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

13 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.

14 Plain Strain Numerical Modeling Examples 14 Deformation analysis of slopes Deformation analysis of tunnels

15 Plain Strain Numerical Modeling Examples 15 Dynamic analysis MSE walls

16 Plain Strain Numerical Modeling Examples 16 Retaining wall Embankment dam

17 Plain Strain Numerical Modeling Examples 17 Strip footing Roadway embankment

18 18 Axisymmetrical Conditions

19 Axisymmetrical Numerical Modeling Examples 19 Circular Footing Single Pile

20 Axisymmetrical Numerical Modeling Examples 20 Flow to an injection and/or pumping well Point Load on Soil

21 21 Finite Element vs Finite Difference

22 22 Finite Element Method

23 23 Finite Difference Method

24 24 Basic explicit calculation cycle

25 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.

26 26 Constitutive Relationships II

27 27 Chapter 2 Numerical Modeling Using FLAC

28 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

29 Download a demo version 29 szamira -as

30 30 FLAC program

31 31

32 32

33 33 Terminology

34 34 Special Features

35 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

36 36 Normal and Shear Stresses

37 37 Normal and Shear Strain

38 38 Strain - Displacement

39 39 Hooke s law

40 Bulk Modulus 40 where P is pressure, Vis volume, and P/ Vdenotes the partial derivative of pressure with respect to volume.

41 41 Elastic Correlations

42 Motion and Equilibrium 42 In a continuous solid body:

43 Constitutive Relation 43 strain rate is derived from velocity gradient as follows

44 Constitutive Model 44 The simplest example of a constitutive law is that of isotropic elasticity:

45 45 Finite Difference Zones

46 46 Zone and Gridpoint

47 47 Finite Difference Grid with 400 zones

48 48 Zone Numbers

49 49 Grid Point Numbers

50 50 Boundary Condition

51 51 Initital Conditions

52 52 Boundary Conditions

53 53 Applied Condition

54 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.

55 55 Numerical Flowchart

56 56 Grids

57 General solution procedure: Start: COMMAND keyword value... <keyword value... >... ; comments grid icol jrow grid 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 57 grid 20,20 m e gen 0,0 0, , ,0 rat

58 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

59 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

60 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

61 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 61

62 62 Applying Boundary and Initial Conditions

63 Sign Conventions DIRECT STRESS Positive stresses indicate tension; negative stresses indicate compression. SHEAR STRESS 63

64 64

65 65 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.

66 66

67 67

68 ;Example grid 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 68

69 69 grid 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

70 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 70

71 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

72 72 Saving/Restoring Problem State save file.sav rest file.sav save fill1.sav

73 73 Chapter 3 Foundation, Stone Column Modeling

74 Foundation and stone column modeling 74 Example 1: ; STAGE config grid 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 ;

75 Foundation and stone column modeling 75 ; STAGE 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

76 Foundation and stone column modeling 76 ;Example 2: ; STAGE config grid 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

77 Foundation and stone column modeling 77 ini xdisp 0 ydisp=0 ; ; STAGE 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 ;

78 Foundation and stone column modeling 78 ; STAGE 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

79 79 Foundation and stone column modeling

80 Foundation and stone column modeling 80 Example 3: ; STAGE config grid 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

81 Foundation and stone column modeling 81 ini xdisp 0 ydisp=0 ; ; STAGE 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 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

82 82 Foundation and stone column modeling

83 Height (m) Foundation and stone column modeling Settlement (cm) 2 Settlement (With Stone Column) (cm) Settlement (cm)

84 84 Chapter 4 Slope Stability Analysis

85 Groundwater simulation 85 Water table Steady state analysis Transient analysis

86 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

87 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

88 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

89 89 Permeability

90 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

91 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

92 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

93 93 Chapter 5 Structural elements

94 Structural element 94 Beam Pile Cable

95 Nail and Anchor modeling 95 t is annulus thickness of the shear zone and is considered equal to m.

96 96 Material properties

97 Codes 97 struct node struct node 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 struct cable begin node 1533 end node 1534 seg 1 tension 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 yie 1e10 kb 0 sb 0 struct prop 2002 spac 2.3 e 2.1E11 area yie 1e10 kb 1.0E8 sb 1e8

98 98 Chapter 6 Seismic Considerations

99 Important points 99 Loading Damping Boundary condition

100 100 Boundary conditions

101 101 Boundary conditions

102 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 mass set dy_damp rayl 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 table table table table table table table table table table table table table table table table table