Geodynamics - Special Cases

Lecture Notes 15 Special Cases In general τ ij = λ eiiδij + 2µeij Plane stress: (e.g., τ zz = 0, τ xz = 0 ) ⎡τ xx ⎢ τ = ⎢τ xy ⎢⎣ 0 p=− τ xy 0⎤ ⎥ τ yy 0⎥ 0 0⎥⎦ τ xx + τ yy 3 ⎡ 2τ xx − τ yy ⎢ 3 ⎢ ⎢ τ dev = ⎢ τ xy ⎢ ⎢ 0 ⎢⎣ τ zzdev = − τ xx + τ yy 3 τ xy 2τ yy − τ xx 3 0 ⎤ ⎥ ⎥ ⎥ 0 ⎥ ⎥ −τ xx − τ yy ⎥ ⎥⎦ 3 0 Solving for strains: 1 (τ xx − ντ yy ) E 1 eyy = (τ yy − ντ xx ) E exx = ezz = − exy = ν E (τ xx + τ yy ) τ xy 2µ Plane strain: (e.g., ezz = exz = 0 = τ zz dev ⇒ τ xx dev = −τ yy dev ) ⎡ ⎢τ ⎢ xx τ = ⎢τ xy ⎢ ⎢0 ⎣ τ xy τ yy 0 ⎤ 0 ⎥ ⎥ 0 ⎥ τ xx + τ yy ⎥ ⎥ 2 ⎦ p = −τ zz ⎡ τ xx − τ yy ⎢ ⎢ 2 ⎢ τ dev = ⎢ τ xy ⎢ ⎢ 0 ⎢⎣ τ xy τ yy − τ xx 2 0 1+ ν ⎡ ⎣τ xx (1 − ν ) − ντ yy ⎤⎦ E 1+ ν ⎡ eyy = ⎣τ yy (1 − ν ) − ντ xx ⎤⎦ E exx = exy = τ xy 2µ ⎤ 0⎥ ⎥ ⎥ 0⎥ ⎥ 0⎥ ⎥⎦ Note: Plate stress → E → ν → 2 λµ λ + 2µ Plane strain E 1−ν2 ν 1− ν λ ← Plane stress: (e.g., τ 33 = 0) �11 �22 Figure 15.1 1 (τ 11 − ντ 22 ) E 1 e22 = (τ 22 − ντ 11 ) E e11 = e33 = − ν E (τ 11 + τ 22 ) Tectonic Stress – Often plane stress is useful (3D → 2D). Assume mostly in plate stress Assume τ zz = −ρ gz (lithostatic) τ xx = −ρ gz + ∆τ xx τ yy = −ρ gz + ∆τ yy The non-lithostatic (as opposed to deviatoric) stress: τ zz ' = 0 ⎫ ⎪ τ xx ' = ∆τ xx ⎬ τ yy ' = ∆τ yy ⎪⎭ plane stress Assume for simplicity ∆τ xx = ∆τ yy = σ e11 = e22 = e33 = − (1 − ν ) σ ⇒ compression for σ compression E 2ν σ E ⇒ extension for σ compression dilation θ = (2 − 4ν ) σ E =− δρ ρ Look at the change in lithostatic pressure at a depth h → h + δh = h(1 − 2ν ρ gh → (ρ + δρ )g(h + δh ) Plane stress, σ 1 = σ 2 = σ �3 �2 h l �1 l Figure 15.2 σ E ) ε1 = ε 2 = ε3 = (1 − ν ) σ E −2ν σ E θ = ε1 + ε 2 + ε 3 = Now to conserve mass, (2 − 4ν ) 2(1 − 2ν ) σ= σ E E δρ δV =− =θ ρ V New lithostat: δτ zz = −(ρ + δρ )g(h + δh ) + ρ gh 2ν = (1 + θ )g(1 + σ )ρ h − ρ gh E = (θ + 2ν σ )ρ gh E δτ zz 2(1 − ν ) = ρ gh σ E Use ν = 0.25 E = 100 GPa = 1 Mbar h = 100 km ρ = 3 Mg/m 3 = 3 g/cm 3 δτ zz 2(1 − ν ) = ρ gh ; 5% σ E