0- Numerical methods / Finite differences 1- Deformation and Estress, tensors 2- Basic equations of Solid Mechanics 3- Constitutive laws 4-

Transcription

1 0- Numerical methods / Finite differences 1- Deformation and Estress, tensors 2- Basic equations of Solid Mechanics 3- Constitutive laws 4- Orientation of faults on Earth (Anderson) 5- Critical shape of mountains... rôle of water

2 Finite Differences, with the diffusion equation dt d²t =κ 2 dt dx We resolve the equation of heat diffusion, in 1D. we need to discretize space (x), time (t) and temperature T. dt T i+1 T i T i+1 T i Progade differenciation: =. dx x i+1 x i Δx dt T i T i 1 dt T i+1 T i 1 Retrogade differenciation:, Central:. dx Δx dx 2Δx

3 T i+1 T i T i T i 1 dt d( ) T i+1 2T i +T i 1 d2 T dx Δx Δx Second order derivative: = = = dx Δx dx 2 Δx 2 dt d²t =κ 2 dt dx Δt Δt Explicit method: T i =(1 2κ 2 ). T i +(κ 2 ). (T i+1 +T i 1 ) Δx Δx + T i +κ Implicit method : T i = + Δt + + ( T +T i+1 i 1 ) 2 Δx Δt 1+2κ Δx 2 Mixte method: ( well known as Crank-Nicolson ) T i T i T 2T +T T 2T +T κ i+ 1 i i 1 i+ 1 i i 1 = +. 2 Δt 2 Δx 2 Δx + ( )

4 T 2 T Diffusion equation: =κ 2. t x T i T i T 2T +T i+1 i i 1 = κ 2. Δt Δx Δt therefore: T i =T +(κ ). (T 2T +T i i+1 i i 1 ). 2 Δx + + Stability of the aproximation depends on the value of (κ Algorithm : Δt ). 2 Δx

5 The critical time step for a finite difference method to work Ti t+1 - Ti t = k. t/ x² (Ti+1 t -2 Ti t + Ti-1 t) Ti t+1 = Ti t (1 2.k. t/ x²) + 2k. t/ x² (Ti+1 t + Ti-1 t)/2 If 2kDT/dx²=DD=0 => Ti t+1 = Ti t, nothing changes. If DD= ½ => Ti t+1 = (Ti+1 t + Ti-1 t)/2 : new T is the mean. If DD is between 0 and ½ => all good, but may take time to compute. If DD > ½ => Ti t+1 > (Ti+1 t + Ti-1 t)/2, instability (divergence). Stability Condition for the numerical scheme: t x² /(4.k)

6 Implementation with Matlab : We write a script to calculate the thermal state of an oceanic lithosphere as it is created at an oceanic ridge (mantle arrives at the surface at 1300 C), with the surface condition To=0 C. Thermal diffusion is given κ =10 ⁶m²/s. Prepocessing : - Spatial discretisation 1D, km : resolution... - Time discretisation Ma : numerical time-step... Processing : Resolve the equation step by step with an explicit method. Post-processing :Visualize he geothermal gradient with time. Plotting every 5 Ma for instance. * At which age is the temperature equal to 500 C at 20 km depth? * What is the critical time step, given a resolution of 1 km?

7 An elastic bar may be disccretized as a series of masses related by springs of rigidity k. Which system of equations allows to deduce the stresses and displacements of these masses? σ =ρ. g x u ε= x σ=e. ε 2 u ρ.g = E x2 1- Analytical solution: σ = ρgx u = ρ g x2/2e 2- finite difference solution: U1 U2 U3 U4 U5 σ i+1 =σ i +(x i+1 x i ). ρ. g 2 U i U i+1 2U i +U i 1 ρg = = 2 2 E x (x i+1 x i 1 )

8 FLAC :Example of a finite differences Lagrangian thermo-mechanical approach Equation of motion at the nodes: Strain-rates in elements: Temperature at nodes : t+ t Surface processes (erosion) : constitutive laws : Integration of forces at nodes : Boundary conditions -velocities: Displacements at nodes :

9 Numerical methods using finite element distinct elements, and finite differences

10 MESHES

11 Lagrangian vs. Eulerian formulations Coordonates of the Lagrangian nodes move with the material, and markers in an element move passively. The mesh distorsion becomes large and causes big problems to resolve equations, requiring remeshing, and numerical errors. Coordonates of the Eulerian nodes are fix in space, but internal markers transport their properties through the mesh. Therefore is is difficult to restore the history of materials (bad for elasticity). There is no distorsion of the mesh, but it must be much larger domain in order to account for the entire mvoement of the medium in time.

12 Example...

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14 Models acounting for the fluid behaiovr of the mantle, and account for phase transformation, but don't account for faults at the surface... Li & Gerya, 2012 Tackley, 2011

15 Subduction benchmarks -- Schmeling et al., Codes use different assumptions : - Upper boundary condition (fix/ free), -Presence/absence of elasticity -Algorithm of interpolation

16 Problems of numerical interpolation

17 Deformation

18 Deformation Linearised Lagrangian deformation: Can be decomposed in symetrical and anti-symetrical parts: εij symetrical Infinitesimal Lagrangian tensor ε ε ωij antisymetrical ωij describes rigid rotation ~0

19 Ejemplos de tensores de deformacion y x ay ay

20 Stress tensor (Cauchy) t n =t 1.n 1+t 2.n 2 +t 3.n 3=t i.n i T i.ni =σ ji.n j =σ ij.n j is a symetric tensor

21 The state of stress ni a vertical column in the field of gravity Weight of a column : F y= m.g. Mass m= ρ.g.h.δa Fy σyy=ρ.g.y State of stress lithostatic: σxx=σzz=σyy Tectonic stress (~deviatoric) σxx : σxx= ρ g y + σxx The horizontal force necessary to maintain this column is : F x = σxxdy = ρ g h2/2

22 Principal stresses and invariants σ2 y σ1 θ x [ σ xx σ xy σ xz σ xy σ yy σ yz ][ σ xz σ1 σ yz 0 σ zz σ2 0 0 σ3 ] Decomposed in an isotropic P ( for deformation), plus deviatoric components: σ = -P + σdev [ σ xx σ xy σ xz σ xy σ yy σ yz ][ ][ σ xz σ xx P σ xy P 0 0 σ yz = 0 P 0 + σ xy σ yy P 0 0 P σ zz σ xz σ yz σ xz σ yz σ zz P P is pressure: First invariant of the tensor P= tr (σ)/3 = σ.ι/3= Σ (σii)/3= σ1+σ2+σ3= σxx+σyy+σzz τii is the deviator = Second stress Invariant. ]

23 Principal stresses and Mohr circle representation σ 1,2 =P±τ II I (σ xx +σ yy ) P= = 3 2 ( σ xx σ yy ) 2 τ II = II = +τ xy 4 2 Their projection on the Mohr circle (in 2D): tan2θ= 2τ xy σ yy P=-p τ II=R 2 ( σ xx +σ yy ) ( σ xx σ yy ) σ 1,2 =P±τ II = ± +τ xy 2 4 2

24 Normal and Shear Stresses Principal stresses may in turn be decomposed in a referencial of any given plane defined by its angle θ with respect to the X axis, with a NORMAL and TANGENCIAL component. B θ σ1 σn τ O σ2 A ( σ 1 +σ 2 ) ( σ 1 σ 2 ) σ n= + sin2θ 2 2 ( σ 1 σ 2 ) τ= cos2θ 2

25 2- Equations of solid mechanics Equilibrium of forces on a rigid block A block of density ρ, of length l and height h, lies on a slope of angle with the X axis α. Write the 2 equations of force equilibrium in the X and Y directions, and express the shear stress τ, that prevents the block from moving. τ = f (ρ,g, h, α)? y l h R F x g α

26 2.A) Mass conservation M= ρ( x,t )dv V Mass flux through S: ρυ nds S With the Gauss theorem: ρυ n ds= ( ρυ ) dv S V Mass variation in time equals the mass flux that exits the volume M = ( ρυ ) dv t V M ρ y con = dv t V t With a uniform density in the volume: CONTINUITY EQUATION With constant density in time: INCOMPRESSIBILITY EQUATION V υi dρ +ρ =0 dt xi υi xi =0 [ ] ρ + ( ρυ ) dv= 0 t dρ +ρ υ= 0 dt υ= 0

27 2.B) Conservation of momentum) Newton's statement: The sum of the forces on a medium's surface ( stresses T) and of the volumic forces (gravity g) are equal to its acceleration γ. F =mγ d TdS+ ρgdv = dt ρvdv S V V T i =σ ij. n j σ ij dv i ( x +ρgi ) dv = ρ dt dv V j V σ ij xj +ρg i =ρ dv i dt In 2D and in static: σ xx τ xy + =0 x y σ yy τ xy + +ρg= 0 y y

28 Elementary equilibrium In 2D, over an element of size dx and dy, with stresses σxx, σyy,σxy acting on each face and accounting for the gravity field, one expresses the force equilibrium in both directions... σyy(y+dy σxy(y+dy) y+dy σxy(x) σxx(x+dx) σxx(x) y σxy(x+dx) x+dx σxy(x+dx) x σyy(y) a) Σ Fx: σxx(x).dy + σxy(y).dx +σxy(y+dy).dx + σxx(x+dx).dy = 0 Σ Fy=?... = 0 b) Dividing by the volume dxdydz... σ xx τ xy + =0 x y σ yy τ xy + +ρg= 0 y y

29 3- Constitutive laws Elasticity Hook,1666: σ = Ε. ε Rigidity of tectonic plates (effective thickness h), is involved in the flexural equations D= E.h3/(1-ν)

30 Young's modulus E is the ratio of a unixial stress and the corresponding strain (σ22=σ33=0). Poisson's coefficient ν gives the associated vertical contraction: ε 11 = ε 11 σ 11 ε 33 ν= σ E ε 22 ε 11 = ε 33 ε 11 ε 22 σ= [ D ] ε E ν σ ij = (ε + δ ε )with Young and Poisson's parameters 1 +ν ij 1 2ν ij kk σ ij =2μ ε ij +λ δ ij ε kk with Lamé's parameters μ,λ In plane strain, ε zz =ε xz =ε yz=0 [] [ 1 σ xx E ν σ yy = (1 ν )² 0 σ xy ][ ] ν 0 ε xx 1 0. ε yy 1 ν 0 ε xy 2

31 Exercise A crust is submitted to a unixial stress ( σ2= 3= 0 ). Assuming for crustal rock that ν=0.25 and E=50 GPa, a) how much strain εxx will result from the application of a lateral stress σxx=100mpa? b) If the crust is l =10km, h =3km, how much does it shorten (Ux) and thicken (Uy)? Uy Ux εxx= ³, Ux= 12 m, Uy= 0.9m.

32 SUMMARY:Solid mechanics equations υi dρ +ρ =0 dt xi A) 1 equation of continuity (mass conservation): B) 3 equations of motion (Newton) : σ ij xj +ρg i =ρ dv i dt C) 6 constitutive equations (behaviour): vis pl ε ij = ε el + ε + ε ij ij ij = 1 dτ ij 1 + τ ij+ λ τ ij 2G dt 2μvis 10 equations for 10 unknowns : ρ(1), vi(3), σ ij (6) ) OK!

33 4- FAULTS ORIENTATIONS ON EARTH (ANDERSON's THEORY, 1919) The surface of the Earth is free of stress (contact with the Atmosphere). Since the orientation of the principal stresses correspond to directions of ZERO tangential stress, they must be parallel and perpendicular to the earth surface. With a criteriion of failure that considers the friction angle of 30 of most rocks, one can predict the orientation of faults. This theory results from the friction property of rocks.

34 Coulomb, 1773 : τs = So - µ.σn, µ= tanφ τs σn τs

35 Criteria of rupture (yield enveloppe) The rutpure surface is defined as the function in the principal stresses referential that satisfies : f (σ 1,σ 2,σ 3 )=0 at the plastic state( σ=σ s ) f< 0 at the elastic state (σ<σ s ) The criteria of Tresca ( ~ 1866) no depends on σ 2 intermediate: For σ 1 σ 2 σ 3 ( σ 1 σ 3 )/ 2=k and f=σ 1 σ 3 2k The criteria of Mohr-Coulomb depends on pressure. f=σ 1 1 sin ϕ σ C 1+sin ϕ 3 El criterio de Drucker-Prager depende de la pression tambien...

36 THEORY OF PLASTICITY TO PREDICT THE ORIENTATION OF FAULTS - The medium is assumed to be at a state of failure everywhere, with a deviatoric stress set to k: 2 (σ σ ) xx yy τ 2 =k 2 = +σ xy deformation is incompresible (φ= ψ= 0, directions of principal stress // deformation: 2σ xy 2 ε pxy tan2( x,σ 1 )= = p p σ xx σ yy ε xx ε yy - Force Equilibrium provides a solution in the form of perpendicular sli-lines, with a geometry defined by the boundary conditions, pressure P, velocities u,v, and orientaction θ = ( slip line, x) P+2kθ, du vdθ= constant along α P 2kθ, dv+udθ= constant along β

37 EXamples of slip-lines predicted by the theory and obtained in experiments Nadai, 1960

38 ELASTICITY, PLASTICITY, solutions for the problem of the INDENTOR Elasticity Tapponnier et al., 1976 Plasticity : slip along rigid blocs

39 5- Accretionary prisms and mountains Buiter et al., 2012

40 THEORY OF THE CRITICAL TAPERED WEDGE (Chapple, 1978, Davis et al.,1983, Dahlen, 1984) Hipoteses to predict the orientation of faults in an accretionary prism: a) the whole medium is at the state of failure : τ = µ.σn, µ = tan φ b) condition for failure at the base (inclined at angle β): τ = µb.σn Solution of the force equilibirum in a vertical column For α<<1, β<<1: 1 sin φ, 1+sin φ σ 1 +σ 3 Surface Slope: α+β=( β+μb ) Fault geometries: sin ψ= 3(σ 1 σ 3 )

41 Adam & Reuther (2000): along the margen of North Chile, variation in faults styles are attributed to changes in β and µ

42 Subduction of seamounts (Dominguez et al., 1998)

43 Continental Scale prism : the effect of climate on mountains α=β ( 1 sin φ 1 sin φ 1)+μb. 1+sin φ 1+sin φ 4 km 1 km Lamb et Davis, 2003

44 Evolution of a deforming wedge in time : numerical approach 3 km 32 km 100 km Prism with β = 6, φ = φ b = 30, α(t=0) = 0. A shear bands form to accomodate the shortening and build the topography (α) / its geometry evolves in time. Normal faults accommodate the slope at some point (super-critical taper).

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46 6- Effect of water : the «pore pressure effect» Tensile fracture filled with fluid (vein) during dilation. Stress σ1 is parallel to ths teructure. The effective stress is very low, because high fluid pressure counteracts the normal (lithostatic) stress. Increasing pore fluid pressure favors failure! May lead to tensile failure deep in crust Effective stress = σ n fluid pressure

47 Friction at the plate interface and the effect of water Oleskevitch et al., 1999 Water has a fundamental role in plate tectonics... See next presentation.. Faregeng & Ellis, 2006