Tectonics
Lecture 12
Earthquake Faulting
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Plane strain
Strain occurs only in a
plane. In the third
direction strain is zero.
σ3
σ1
ε2 = 0
3
2
1
Assumption of plane
strain for faulting
e.g., reverse fault: footwall
moves down. No strain in 2direction
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Plane strain
Local coordinates
σz
Poisson’s ratio
σx
εy = 0
υ= - εx / εz
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:
υ
1
υ
εy = − σx + σ y − σz = 0
E
E
E
or
for plane strain
σ y = υ (σ x +σ z )
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2D stress field
remote principal stress
remote principal stress
σ1
fault plane
normal stress
σN
σ2
σS
shear stress
σ1
P
θ
σ2
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 σN and shear stress σS
σ N = 1 2 (σ 1 + σ 2 ) − 1 2 (σ 1 − σ 2 )cos 2ϑ
σ S = 1 2 (σ 1 − σ 2 )sin 2θ
Note that: ½ (σ1 + σ2) = σm = mean stress
Local stresses on fault: σ1 > σ2 > σ3 compression positive
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Construction of Mohr stress circle:
shear stress vs. normal stress
σS axis
Maximum shear stress = ½ (σ1 - σ2) when θ = 45o
σS max
P
σS
(σ1 - σ2)/2
2θ
σ2
σN
σm
σ1
σN axis
Any point on circle has coordinates (σN, σS)
where:
(σ1 + σ2)/2
σ N = 1 2 (σ 1 + σ 2 ) − 1 2 (σ 1 − σ 2 )cos 2ϑ
σ S = 1 2 (σ 1 − σ 2 )sin 2θ
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Simple failure criteria
(a) Friction – Amonton’s Law
1st: Friction is proportional normal load (N)
Hence: F = µ N
- µ is the coefficient of friction
2nd: Friction force (F) is independent of the areas in contact
So in terms of stresses: σS = µ σN = σN tanφ
May be simply represented on a Mohr diagram:
σS
µ
e
p
slo
φ
µ= tan φ
φ is the “angle of friction”
σN
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Friction on pre-existing fault
σS
Amonton’s Law: criterion for frictional sliding
σS = µ σN on fault angle θ
sliding
slides when expanding Mohr circle
(with increasing applied stress)
touches/crosses friction criterion line
σS
P
φ
φ σ2
σN
2θ
σm
below line, won’t slide
σN
σ1
σ
sliding on
most
1
favourably
oriented
θ fault
sliding
P
σ2
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Friction on pre-existing fault: more usual situation
σS
Amonton’s Law: any point above line
will already have occurred
sliding
sliding
P2
now
sliding on
less
favourably
oriented
faults
P1
φ
φ
σ2
2θ1
2θ2
σ1
σN
θ1
σ1
θ2
σ2
P2
P1
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Simple failure criteria
(b) Faulting – Coulomb’s Law
σS = C + µi σN = σN tanφi
C is a constant – the cohesion
µi is the coefficient of “internal” friction
Tensile fracture
Shear fracture
σS
(σ2 = -σT)
σT – tensile strength
C
µi
e
p
slo
µi = tan φi
φi
σN
φi is the “angle of internal friction”
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Coulomb faulting criterion
Coulomb failure criterion
σS
unstable
Amonton friction criterion
failure
failure
σ1
stable
P1
failure
tension
σ2
tension
θ
compression σN
σ1
2θ
2θ
P2
σ2
P1
uniaxial
compression
θ
triaxial
compression
φi
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P2
Simple failure criteria
porous rock
(c) Effect of pore fluid pressure - pf
weight of water generates hydrostatic pressure
weight of rock generates lithostatic pressure
pressurized fluid
the effect is to reduce normal stress by pf
e.g. (σ1, σ2, σ3) becomes (σ1 - pf, σ2 - pf, σ3 - pf)
σS
Coulomb failure criterion
Mohr circle for
rock with pore
pressure
Failure
pf
Mohr circle for
dry rocks
σN
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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.
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dip β
Reverse or thrust faulting
compression positive
σy
β
σx
σz
Lithostatic stress: σz = ρ g z = p (pressure)
To produce the thrust a compressive tectonic stress is required: σxtect > 0
The total horizontal stress in the x-direction is:
σx = p + σxtect = ρgz + σxtect
this exceeds the vertical lithostatic stress:
σx > σz
The total horizontal stress in the y-direction is:
σy = ρgz + σytect = ρgz + ν σxtect (plane strain)
So:
σx > σy > σz (as υ is less that 1)
The vertical lithostatic stress is the minimum for reverse faulting
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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.
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Normal faulting
σy
dip β
σx
compression positive
β
σz
To produce the normal a tensile tectonic stress is required: σxtect < 0
The total horizontal stress in the x-direction is:
σx = ρgz + σxtect
this is less than the vertical lithostatic stress: σz > σx
The total horizontal stress in the y-direction is:
σy = ρgz + ν σxtect (plane strain)
So:
σz > σy > σx (as υ is less that 1)
So the vertical lithostatic stress is the maximum for normal faulting
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Strike slip faulting
σy
σz
compression positive
σx
Strike slip faulting requires one compressional and one tensional tectonic
stress : σxtect > 0
σytect < 0
For this case:
σx > σz > σy
So the vertical lithostatic stress is always the intermediate stress for strike
slip faulting
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Anderson faulting
dip β =
σz
60o
β
σx
σx > σz > σy
σx
σy
σz > σy > σx
dip β = 30o
σz
β
σx
σx > σy > σz
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Anderson faulting
Recall:
Lithostatic stress: σz = ρgz
σz
Total stress in x-direction:
σx = ρgz + σxtect
normal stress
σN
σx
σS
shear stress
β
θ
σxtect is positive for thrust faulting
and negative for normal faulting
Normal stress σN and shear stress σS
σ N = 1 2 (σ x + σ z ) − 1 2 (σ x − σ z )cos 2ϑ
σ S = 1 2 (σ x − σ z )sin 2θ
In terms of the lithostatic and
tectonic stresses:
σ tect
σN = ρgz +
(1 + cos 2θ )
2
σS = −
σ tect
2
sin 2θ
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Anderson faulting
σS = µ σN
Amonton’s Law:
In the presence of pore fluid:
So:
σS = µ (σN – pf)
σ tect
⎛
⎞
σ tect
±
sin 2θ = µ ⎜⎜ ρ g z − p f +
(1 + cos 2θ ) ⎟⎟
2
2
⎝
⎠
90
400
normal fault
thrust fault
σtect MPa
60
β
30
0
µ
thrust fault
0
1.0
normal fault
0
µ
-100
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1.0
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).
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