Geodynamics. Brittle deformation and faulting Lecture Predicting fault orientations. Lecturer: David Whipp

Transcription

1 Geodynamics Brittle deformation and faulting Lecture Predicting fault orientations Lecturer: David Whipp david.whipp@helsinki.fi Geodynamics 1

2 Goals of this lecture Introduce Anderson s theory and how it can be applied to predict fault orientations 2

3 Faulting Beartooth Plateau, Wyoming, USA 3

4 Faulting Beartooth Plateau, Wyoming, USA 4

5 Anderson s theory Anderson formulated the Mohr-Coulomb criterion in terms of principal stresses and lithostatic pressure (σyy = ρgy) a 2 Anderson assumed σyy = σ3 for a reverse fault, σyy = σ1 for a normal fault and σyy = σ2 = (σ1 + σ3)/2 for a strike-slip fault 1 Normal fault 3 b 2 3 Reverse fault 1 c 2 1 Strike slip fault 3 Fig. 5.8, Stüwe,

6 Anderson s theory c = 0; fs = 0.85 Fig. 5.7, Stüwe, 2007 In terms of differential stress, Anderson s theory is Reverse: Normal: Strike-slip: d = 1 3 = 2(c + f s( yy w gy) p f 2 s +1 f s d = 1 3 = 2(c f s( yy w gy) p f 2 s +1+f s d = 1 3 = 2(c + f s( yy w gy) p f 2 s +1 6

7 Predicting fault orientation Now we ll look at how to apply Anderson s theory to predict the dip angle β of normal and reverse faults Assume principal stresses yy = gy xx = gy + xx where Δσxx is the tectonic deviatoric stress, which is positive for a reverse fault and negative for a normal fault 7

8 Predicting fault orientation Fig. 8.8, Turcotte and Schubert, 2014 We first need to relate σxx and σyy to σn and σs in order to apply Amonton s law by using the equation for the normal and shear stresses in a coordinate system rotated by angle θ with respect to the principal stresses (see Lecture set 3) n = 1 2 ( xx + yy )+ 1 2 ( xx yy) cos 2 s = 1 2 ( xx yy)sin2 Note that here, θ is with respect to vertical, θ = π/2 - β 8

9 Predicting fault orientation Fig. 8.8, Turcotte and Schubert, 2014 If we plug in the values for σxx and σyy, we find xx n = gy + (1 + cos 2 ) 2 Inserting the values above into the form of Amonton s law that includes pore fluid pressure, τ = fs(σn - pw), yields ± xx 2 s = xx 2 sin 2 = f s apple gy p w + sin 2 xx 2 (1 + cos 2 ) Note that the upper sign is for reverse faults (Δσxx > 0) and the lower for normal faults (Δσxx < 0) 9

10 Predicting fault orientation Fig. 8.8, Turcotte and Schubert, 2014 The previous expression can be rearranged to solve for Δσxx 2f s ( gy p w ) xx = ± sin 2 f s (1 + cos 2 ) If we assume that faulting will occur with the minimum tectonic stress, then Δσxx should be minimized By setting dδσxx/dθ = 0, we find tan 2 = 1 or tan 2 = ± 1 f s f s where the upper sign again applies to reverse faults and the 10 lower to normal faults

11 Predicting fault orientation Dip angle Figs. 8.9, 8.10, Turcotte and Schubert, 2014 Tectonic stress = 2700 kg m 3 p w = w gy w = 1000 kg m 3 y =5km Finally, the two equations from the previous slide can be combined to yield the tectonic stresses corresponding to angle θ or β xx = ±f s( gy p w ) (1 + f 2 s ) 1/2 f s where the upper sign again corresponds to reverse faults and the lower to normal faults 11

12 Let s see what you ve learned If you re watching this lecture in Moodle, you will now be automatically directed to the quiz! 12