Simple Example - Uniaxial Strain

Lecture Notes 14 Simple Example: Uniaxial Strain Uniaxial strain is excellent approximation, but not exact. Sediment loading erosion: x1 x3 x2 Assume e22 = e33 = 0, e11 ≠ 0 τ 11 = (λ + 2µ )e11 τ 22 = τ 33 = λ e11 = τ 11 = λ λ + 2µ τ 11 = ν 1− ν τ 11 (1 − ν )Ee11 (1+ ν )(1 − 2ν ) ⎡ ⎢1 ⎢ τ = − ρ gd ⎢ 0 ⎢ ⎢ ⎢0 ⎣ 0 1 3 0 ⎤ 0⎥ ⎥ 0 ⎥ = τ uniaxial ⎥ 1⎥ ⎥ 3⎦ For γ = 0.25, τ 11 = −ρ gh ; −0.27 kbar/km (crustal rock) τ 22 = τ 33 = τ 11 3 ; −0.1 kbar/km ⇒ deviatoric stress ⎡−0.27 0 0 ⎤ ⎡−0.15 0 0 ⎤ ⎡−0.12 0 0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ −0.09 0 ⎥=⎢ 0 −0.15 0 ⎥+ ⎢ 0 −0.06 0 ⎥ ⎢ 0 ⎢⎣ 0 0 −0.09⎥⎦ ⎢⎣ 0 0 −0.15⎥⎦ ⎢⎣ 0 0 −0.06⎥⎦ Unloading: Assume initially lithostatic τ 11 = τ 22 = τ 33 = −ρ gd ⎡ ⎤ ⎡4 ⎢0 0 0 ⎥ ⎢9 ⎢ ⎥ ⎢ 2 ⎢ ⇒ 0 − ρ gd 0 ⎥ = ⎢0 ⎢ ⎥ ⎢ 3 ⎢ ⎥ ⎢ 2 ⎢0 0 − ρ gd ⎥ ⎢ 0 ⎣ ⎦ ⎣ 3 0 − 2 9 0 ⎡ 4 ⎤ 0 ⎥ ⎢− ⎢ 9 ⎥ 0 ⎥ ρ gd + ⎢ 0 ⎢ ⎥ ⎢ ⎥ 2 ⎢ 0 − ⎥ ⎣ 9⎦ ⇒ thrust faults popups give direction of compressional stress Assume σ 2 < σ 3 tectonic stress Type of faulting: (compression negative) x1 x3 (E-W) x2 (N-S) 0 − 4 9 0 ⎤ 0 ⎥ ⎥ 0 ⎥ ρ gd ⎥ ⎥ 4 − ⎥ 9⎦ σ 2 < σ 3 < σ 1 ⇒ thrusting on E-W plane σ 2 < σ 1 < σ 3 ⇒ strike-slip σ 1 < σ 2 < σ 3 ⇒ normal faulting on N-S plane Stress increment from burial ⎡ ⎢1 ⎢ τ = − ⎢0 ⎢ ⎢ ⎢0 ⎣ 0 1 3 0 ⎡5 ⎤ ⎢ 0⎥ ⎢9 ⎥ 0 ⎥ ρ gd = − ⎢ 0 ⎢ ⎥ ⎢ ⎥ 1 ⎢0 ⎥ ⎣ 3⎦ 0 5 9 0 ⎡ 4 ⎤ 0⎥ ⎢− ⎥ ⎢ 9 0 ⎥ ρ gd + ⎢ 0 ⎥ ⎢ ⎥ ⎢ 5⎥ ⎢ 0 ⎣ 9⎦ 0 2 9 0 ⎤ 0⎥ ⎥ 0 ⎥ ρ gd ⎥ ⎥ 2⎥ 9⎦ Near surface in Ventura Basin, thrusting occurs. Stress tensor could look something like: ⎡0 0 ⎢ ⎢0 −τ ⎢⎣0 0 0 ⎤ ⎥ 0 ⎥ −ατ ⎥⎦ τ ≡ "tectonic" stress 0 ≤α ≤1 1+ α p=− τ 3 Assume tectonic stress independent of depth. For simplicity, assume τ 1 α= ⇒ p=− 2 2 Total deviatoric stress is ⎡ 4 ⎢− ⎢ 9 ⎢ 0 ⎢ ⎢ ⎢ 0 ⎣ 0 2 9 0 ⎤ ⎡ 1 0⎥ ⎢− ⎥ ⎢ 2 0 ⎥ ρ gd − ⎢ 0 ⎥ ⎢ ⎥ ⎢ 2⎥ ⎢0 ⎣ 9⎦ 0 1 2 0 ⎤ 0⎥ ⎥ 0⎥ τ ⎥ ⎥ 0⎥ ⎦ Normal faulting occurs if τ 2 τ 2 4 − ρ gd + < ρ gd − < ρ gd 2 9 2 9 9 2 − ρ gd < −τ 3