Tectonics Lecture 12 Earthquake Faulting
Plane strain 3 Strain occurs only in a plane. In the third direction strain is zero. 1 ε 2 = 0 3 2 Assumption of plane strain for faulting e.g., reverse fault: footwall moves down. No strain in 2- direction 1
z Plane strain Local coordinates Poisson s ratio x ε y = 0 υ= - ε x / ε z so it is saying how much something expands in one direction if it is squeezed in the other From our statement of Hooke s Law: ε y υ 1 υ = x + y z = 0 E E E for plane strain or = υ + ) y ( x z
remote principal stress 2D stress field remote principal stress 1 normal stress fault plane θ 2 2 1 N S shear stress P N S Resolution of forces and areas both parallel and perpendicular to the fault leads to the following equations for normal and shear stress on the fault plate: Normal stress N and shear stress S ( + ) 1 2( ) = 1 2 1 2 = 1 2( )sin 2θ 1 2 1 2 cos 2ϑ Note that: ½ ( 1 + 2 ) = m = mean stress Local stresses on fault: 1 > 2 > 3 compression positive
S axis Construction of Mohr stress circle: shear stress vs. normal stress Maximum shear stress = ½ ( 1-2 ) when θ = 45 o S P ( 1-2 )/2 S max 2θ N axis 2 N m 1 ( 1 + 2 )/2 Any point on circle has coordinates ( N, S ) where: N S ( + ) 1 2( ) = 1 2 1 2 = 1 2( )sin 2θ 1 2 1 2 cos 2ϑ
(a) Friction Amonton s Law Simple failure criteria 1 st : Friction is proportional normal load (N) Hence: F = µ N - µ is the coefficient of friction 2 nd : Friction force (F) is independent of the areas in contact So in terms of stresses: S = µ N = N tanφ May be simply represented on a Mohr diagram: S slope µ φ µ= tan φ φ is the angle of friction N
Friction on pre-existing fault S Amonton s Law: criterion for frictional sliding S = µ N on fault angle θ S sliding P slides when expanding Mohr circle (with increasing applied stress) touches/crosses friction criterion line φ φ 2 sliding N 2θ m below line, won t slide 1 N P 1 θ sliding on most favourably oriented fault 2
Friction on pre-existing fault: more usual situation S sliding Amonton s Law: any point above line will already have occurred sliding φ φ 2 P 1 2θ 1 2θ 2 P 2 1 N 1 now sliding on less favourably oriented faults θ 1 2 P 2 θ 2 P 1
(b) Faulting Coulomb s Law Simple failure criteria S = C + µ i N = N tanφ i C is a constant the cohesion Tensile fracture µ i is the coefficient of internal friction Shear fracture S slope µ i φ i ( 2 = - T ) C T tensile strength µ i = tan φ i φ i is the angle of internal friction N
unstable S Coulomb faulting criterion failure Coulomb failure criterion Amonton friction criterion failure stable 1 failure tension 2θ P 1 compression N θ θ tension 2 uniaxial compression 2θ P 2 1 2 P 1 P 2 triaxial compression φ i
Simple failure criteria porous rock (c) Effect of pore fluid pressure - p f weight of water generates hydrostatic pressure weight of rock generates lithostatic pressure the effect is to reduce normal stress by p f e.g. ( 1, 2, 3 ) becomes ( 1 -p f, 2 -p f, 3 -p f ) S Mohr circle for rock with pore pressure Failure Mohr circle for dry rocks pressurized fluid Coulomb failure criterion p f N
Reverse fault The Desert Peak thrust in the Newfoundland Mountains, northwest Utah This is a good illustration of a hanging wall ramp over a footwall ramp. Note the offset of the Oe (Ordovician Eureka Quartzite) Total slip on this fault is about 1 km.
dip β Reverse or thrust faulting β y x compression positive Lithostatic stress: z = ρ g z = p (pressure) To produce the thrust a compressive onic stress is required: x > 0 The total horizontal stress in the x-direction is: x = p + x = ρgz + x z this exceeds the vertical lithostatic stress: x > z The total horizontal stress in the y-direction is: y = ρgz + y = ρgz + ν x (plane strain) So: x > y > z (as υ is less that 1) The vertical lithostatic stress is the minimum for reverse faulting
Normal faulting Newfoundland Mountains, northwest Utah Domino normal faults. These faults offset a Pennsylvanian-Devonian unconformity in a top to the east (left) sense of shear. Note that the faults, although presently nearly horizontal, cut the steeply dipping bedding at high angles.
Normal faulting dip β y x compression positive β z To produce the normal a tensile onic stress is required: x < 0 The total horizontal stress in the x-direction is: x = ρgz + x this is less than the vertical lithostatic stress: z > x The total horizontal stress in the y-direction is: y = ρgz + ν x (plane strain) So: z > y > x (as υ is less that 1) So the vertical lithostatic stress is the maximum for normal faulting
Strike slip faulting y z compression positive x Strike slip faulting requires one compressional and one tensional onic stress : x > 0 y < 0 For this case: x > z > y So the vertical lithostatic stress is always the intermediate stress for strike slip faulting
Anderson faulting dip β = 60 o z x x > z > y x β y z > y > x dip β = 30 o z x β x > y > z
Recall: z Anderson faulting Lithostatic stress: z = ρgz Total stress in x-direction: x normal stress N β θ x = ρgz + x x is positive for thrust faulting and negative for normal faulting S shear stress Normal stress N and shear stress S = 1 2 cos N S ( + ) 1 2( ) x = 1 2( )sin 2θ x z z x z 2ϑ In terms of the lithostatic and onic stresses: N = ρ g z + (1 + cos 2θ ) 2 S = sin 2θ 2
Amonton s Law: Anderson faulting S = µ N In the presence of pore fluid: S = µ ( N p f ) So: 400 ± 2 sin 2θ = µ ρ g z p f + 2 (1 + cos 2θ ) 90 normal fault thrust fault 60 MPa β 30 thrust fault 0 µ 1.0 normal fault -100 0 µ 1.0 0
Coulomb stress model Toda, Stein & King From King et al. (1994). Dependence of the Coulomb stress change on the regional stress magnitude, for a given earthquake stress drop. If the earthquake relieves all of the regional stress (left panel), resulting optimum slip planes rotate near the fault. If the regional deviatoric stress is much larger than the earthquake stress drop (right panel), the orientations of the optimum slip planes are more limited, and regions of increased Coulomb stress diminish in size and become more isolated from the master fault. In this and subsequent plots, the maximum and minimum stress changes exceed the plotted colour bar range (in other words, the scale is saturated).