Intro to Soil Mechanics: the What, Why &
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Intro to Soil Mechanics: the what, why & how José E. Andrade, Caltech Thursday, June 23, 2011 The What? Thursday, June 23, 2011 What is Soil Mechanics? erdbaumechanik The application of the laws of mechanics (physics) to soils as engineering materials Karl von Terzaghi is credited as the father of erdbaumechanik Thursday, June 23, 2011 sands & gravels clays & silts Thursday, June 23, 2011 The Why? Thursday, June 23, 2011 Sandcastles what holds them up? Thursday, June 23, 2011 Palacio de Bellas Artes Mexico, DF uniform settlement Thursday, June 23, 2011 The leaning tower of Pisa differential settlement Thursday, June 23, 2011 ! Teton dam dam failure Thursday, June 23, 2011 Niigata earthquake liquefaction Thursday, June 23, 2011 Katrina New Orleans levee failure Thursday, June 23, 2011 MER: Big Opportunity xTerramechanics Thursday, June 23, 2011 MER: Big Opportunity xTerramechanics Thursday, June 23, 2011 The How? Thursday, June 23, 2011 Topics in classic Soil Mechanics • Index & gradation • Soil classification • Compaction • Permeability, seepage, and effective stresses • Consolidation and rate of consolidation • Strength of soils: sands and clays Thursday, June 23, 2011 Index & gradation Definition: soil mass is a collection of particles and voids in between (voids can be filled w/ fluids or air) solid particle fluid (water) Each phase has volume and mass gas (air) Mechanical behavior governed by phase interaction Thursday, June 23, 2011 solid Index & gradation water+air=voids Key volumetric ratios Key mass ratio void ratio water content Vv Mw e = [0.4,1] sand w = <1 for most soils V M s [0.3,1.5] clays s >5 for marine, organic Key link mass & volume Vv porosity η = Vt [0,1] ρ = M/V moist, solid, water, dry, etc. Vw saturation S = Vv [0,1] ratios used in practice to characterize soils & properties Thursday, June 23, 2011 Gradation & classification Grain size is main classification feature sands & clays & gravels silts • can see grains • cannot see grains • mechanics~texture • mechanics~water • d>0.05 mm • d<0.05 mm Soils are currently classified using USCS (Casagrande) Thursday, June 23, 2011 Fabric in coarsely-grained soils “loose packing”, high e Vv relative e = “dense packing”, low e Vs emax greatest possible, loosest packing emin lowest possible, densest packing emax e ID = e e− relative density max− min strongly affects engineering behavior of soils Thursday, June 23, 2011 Typical problem(s)xb in Soil Mechanics • Compact sand fill • Calculate consolidation of clay • Calculate rate of consolidation • Determine strength of sand SAND FILL • Calculate F.S. on sand (failure?) • Need: stresses & matl behavior PISA CLAY ROCK (UNDERFORMABLE, IMPERMEABLE) Thursday, June 23, 2011 Modeling tools Thursday, June 23, 2011 Theoretical framework • continuum mechanics • constitutive theory js W W • computational inelasticity 0 x X • nonlinear finite elements jf f x2 W0 x1 Thursday, June 23, 2011 Theoretical framework • continuum mechanics • constitutive theory balance of mass • computational inelasticity p˙ φ + ∇·v = −∇ · q Kf • nonlinear finite elements ∇·σ + γ = 0 balance of momentum Thursday, June 23, 2011 Theoretical framework • continuum mechanics = h • constitutive theory q k ·∇ darcy ! ep σ˙ = c : "˙ hooke • computational inelasticity k permeability tensor • nonlinear finite elements controls fluid flow ep c mechanical stiffness controls deformation Thursday, June 23, 2011 Theoretical framework • continuum mechanics t3 • constitutive theory tr tn+1 Fn+1 tn+1 F • computational inelasticity n tn • nonlinear finite elements t1 t2 Thursday, June 23, 2011 Theoretical framework • continuum mechanics • constitutive theory • computational inelasticity • nonlinear finite elements Displacement node Pressure node Thursday, June 23, 2011 Finite Element Method (FEM) • Designed to approximately solve PDE’s • PDE’s model physical phenomena • Three types of PDE’s: ) & ) '(1 '(1 '(0 '(0 '(/ '(/ '(. '(. "!#$ '(- '(- ! • Parabolic: fluid flow % '(, '(, '(+ '(+ '(* '(* '() '() ' & ' ' '() '(* '(+ '(, '(- '(. '(/ '(0 '(1 ) !!"!#$ • Hyperbolic: wave eqn • Elliptic: elastostatics Thursday, June 23, 2011 FEM recipe Strong from Weak form Galerkin form Matrix form Thursday, June 23, 2011 Multi-D deformation with FEM ∇·σ + f = 0 in Ω equilibrium u = g on Γg e.g., clamp σ · n = h on Γh e.g., confinement Γg Constitutive relation: Ω given u get σ Γ h e.g., elasticity, plasticity Thursday, June 23, 2011 Modeling Ingredients 1. Set geometry 2.Discretize domaiin H 3. Set matl parameters 4.417 4. Set B.C.’s 5. Solve B Thursday, June 23, 2011 Modeling Ingredients 1. Set geometry 2. Discretize domain 3. Set matl parameters 4. Set B.C.’s 5. Solve Thursday, June 23, 2011 Modeling Ingredients 1. Set geometry 2. Discretize domain 3. Set matl parameters 4. Set B.C.’s 5. Solve Thursday, June 23, 2011 Modeling Ingredients !a 1. Set geometry 2. Discretize domain !r 3. Set matl parameters 4. Set B.C.’s 5. Solve Thursday, June 23, 2011 Modeling Ingredients !a 1. Set geometry 2. Discretize domain !r 3. Set matl parameters 4. Set B.C.’s 5. Solve Thursday, June 23, 2011 FEM Program TIME STEP LOOP ITERATION LOOP ASSEMBLE FORCE VECTOR AND STIFFNESS MATRIX ELEMENT LOOP: N=1, NUMEL GAUSS INTEGRATION LOOP: L=1, NINT constitutive CALL MATERIAL SUBROUTINE model CONTINUE CONTINUE CONTINUE T = T + !T Thursday, June 23, 2011 Material behavior: shear strength • Void ratio or relative density • Particle shape & size Engineers have developed models to account for most • Grain size distribution of these variables • Particle surface roughness • Water Elasto-plasticity framework of choice • Intermediate principal stress • Overconsolidation or pre-stress Thursday, June 23, 2011 A word on current characterization methods Direct Shear Triaxial Pros: cheap, simple, fast, Pros: control drainage & stress good for sands path, principal dir. cnst., Cons: drained, forced failure, more homogeneous non-homogeneous Cons: complex Thursday, June 23, 2011 Material models for sands should capture • Nonlinearity and irrecoverable deformations v • Pressure dependence A • Difference tensile and p compressive strength Dv C • Relative density dependence B • Nonassociative plastic flow log -p ’ Thursday, June 23, 2011 Material models for sands should capture • Nonlinearity and irrecoverable deformations s’ • Pressure dependence 200 vc 191 kPa 241 310 • Difference tensile and 380 compressive strength (kPa) 100 q • Relative density dependence 0 100 200 300 • Nonassociative plastic flow -p’ (kPa) Thursday, June 23, 2011 Material models for sands should capture • Nonlinearity and irrecoverable deformations t’3 Von Mises • Pressure dependence Mohr Coulomb • Difference tensile and compressive strength • Relative density dependence t’ t’ • Nonassociative plastic flow 1 2 Loose sand Dense sand Thursday, June 23, 2011 Material models for sands should capture • Nonlinearity and irrecoverable deformations 1000 (kPa) 800 | • Pressure dependence 600 sa ar 400 |s -s • Difference tensile and 200 s compressive strength r 0 0 2 4 6810 ea (%) • Relative density dependence 4 3 • Nonassociative plastic flow 2 Dense Sand 1 Loose Sand (%) 0 v e -1 Thursday, June 23, 2011 Material models for sands should capture • Nonlinearity and irrecoverable deformations Yield Function q Plastic Potential • Pressure dependence Flow vector • Difference tensile and compressive strength • Relative density dependence • Nonassociative plastic flow -p’ Thursday, June 23, 2011 Elasto-plasticity in one slide Hooke’s law σ˙ = cep : ˙ Additive decomposition of strain ˙ = ˙ e + ˙ p Convex elastic region F (σ, α)=0 Non-associative flow ˙ p = λ˙ g, g := ∂G/∂σ K-T optimality λ˙ F =0 λ˙ H = ∂F/∂α α˙ − · Elastoplastic constitutive tangent 1 cep = ce ce : g f : ce, χ = H g : ce : f − χ ⊗ − Thursday, June 23, 2011 Examples Thursday, June 23, 2011 Example of elasto-plastic model !3 "=1 q "=7/9 N=0 N=0.5 $ # i p' M ! ! 1 2 v CSL CSL (a) (b) v1 Figure 3: Th~ree invariant yield surface on (a) deviatorF ic=planeFat d(iffeσrentv,aluπes of) and (b) l i yi meridian plane at different values of N. vc y G = G(σ, π¯ ) v2 with i M [1 + ln (πi/p)] if N = 0 η = (2.4) ln-p’ N/(1 N) M/N 1 (1 N) (p/πHi) − = Hif N(>p0., πi, ψ) -p’ − − -pi The function ζ controls the cross-sectional shape of the yield function on a deviatoric plane Thursday, June 23, 2011 Figure 2: Geometricasrepra fuesenctionntationofof ththe stateLode’sparameangle.ter ψW. e adopt the shape function proposed by Gudehus and where the stored energy is an isotropiArgyc furisnction[18,of19]thebelasticecausstraie ofnitstensormatheme. Thateictotalal simplistresscity i.e., tensor σ is decomposed according to Terzaghi’s well known expression for fully saturated soils (1 + ) + (1 ) cos 3θ i.e., σ = σ pδ, where p is the pore fluid pressure and is negative underζcom(θ)pres= sion, following − (2.5) − 2 continuum mechanics convention. The tensor δ is the second-order identity. Let us define three independentwhereinvarianastsshoforwnthe ineffeFiguctiverestr3,essthetensorellipσticit, y constant controls the form of the cross-section going from perfectly circular 3= 1 to convex triangular for = 7/9. The shape function 1 2 1 tr s p = tr σ, q = s , cos 3θ = 3 (2.2) 3 with <31 reflects√a6 classical featurχ e in geomaterials, which exhibit higher strength in triaxial where s = σ p δ is the deviatoriccomcomppresonsion.ent of thFigue effreectiv3 ealsostressreteflenctssor,theand geχ =ome√trtrsical2. interpretation for parameters M and N − The invariant p is the mean normalwhiceffectivh goe vstresernstheandslopis asseuofmedthtoe CbeSLnegatiandveththre coughurvaturout.