12 Earthquake Faulting - UCL · 2006-02-21 · (b) Faulting – Coulomb’s Law σ S = C + µ i σ...

GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD

Earthquake Faulting

Lecture 12

Tectonics


Plane strain

σ3

σ1

ε2 = 0

Assumption of plane strain for faulting

Strain occurs only in a plane. In the third direction strain is zero.

e.g., reverse fault: footwall moves down. No strain in 2-direction

1

32


Plane strain

σz

σx

εy = 0

01=−+−= zyxy EEE

συσσυε

From our statement of Hooke’s Law:

for plane strain

or )( zxy σσυσ +=

Local coordinates

Poisson’s ratio

υ= - εx / εz

so it is saying how much something expands in one direction if it is squeezed in the other


2D stress field

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:

( ) ( )θσσσ

ϑσσσσσ2sin)(21

2cos2121

21

2121

−=−−+=

S

N

Note that: ½ (σ1 + σ2) = σm = mean stress

fault plane

Normal stress σN and shear stress σS

σ1

σ2 σ2

σ1

σN

σS

θnormal stress

shear stress

remote principal stressre

mot

e pr

inci

pal s

tress P

Local stresses on fault: σ1 > σ2 > σ3 compression positive


Construction of Mohr stress circle: shear stress vs. normal stress

σS axis

σN axisσ1σ2 σm

(σ1 + σ2)/2

2θ

σS

σN

σS max

(σ1 - σ2)/2

P

Maximum shear stress = ½ (σ1 - σ2) when θ = 45o

Any point on circle has coordinates (σN, σS) where:

( ) ( )θσσσ


2cos2121

21

2121

−=−−+=

S

N


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

σN

φµ= tan φ

φ is the “angle of friction”slope µ


Friction on pre-existing faultσS

σNσ1σ2 σm

2θ

σS

σN

sliding

P

Amonton’s Law: criterion for frictional sliding σS = µ σN on fault angle θ

φφ

sliding

slides when expanding Mohr circle (with increasing applied stress) touches/crosses friction criterion line

below line, won’t slideσ1

σ2P

θ

sliding on most favourably oriented fault


Friction on pre-existing fault: more usual situationσS

σNσ1σ2

sliding

P2

Amonton’s Law: any point above line will already have occurred

φφ

sliding

σ1

σ2

P2

θ2

now sliding on less favourably oriented faults

P1

2θ1

2θ2

θ1

P1



(b) Faulting – Coulomb’s Law

σS = C + µi σN = σN tanφi

C is a constant – the cohesion µi is the coefficient of “internal” friction

µi = tan φi

φi is the “angle of internal friction”

σS

σN

φi

slope µ i

Tensile fracture

(σ2 = -σT)

Shear fracture

CσT – tensile strength


Coulomb faulting criterionσS

σN

σ1σ2

2θ

failure

P1

φi

failureσ1

σ2

θ

P2

failure

tension

uniaxial compression

triaxial compression

tension compression

stable

unstableCoulomb failure criterion

Amonton friction criterion

2θ

θ

P1 P2



(c) Effect of pore fluid pressure - pf

weight of water generates hydrostatic pressure

weight of rock generates lithostatic pressure

the effect is to reduce normal stress by pf

e.g. (σ1, σ2, σ3) becomes (σ1 - pf, σ2 - pf, σ3 - pf)

σS

σN

porous rock

pressurized fluid

pf

Coulomb failure criterion

Mohr circle for dry rocks

Mohr circle for rock with pore

pressureFailure


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.


Reverse or thrust faulting

β σx

σz

σy

Lithostatic stress: σz = ρ g z = p (pressure)To produce the thrust a compressive tectonic stress is required: σx

tect > 0

The total horizontal stress in the x-direction is:

σx = p + σxtect = ρgz + σx

tect

this exceeds the vertical lithostatic stress: σx > σz

The total horizontal stress in the y-direction is:

σy = ρgz + σytect = ρgz + ν σx

tect (plane strain)

So: σx > σy > σz (as υ is less that 1)

The vertical lithostatic stress is the minimum for reverse faulting

compression positivedip β


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 faultingσx

σz

σy

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

β

compression positivedip β


Strike slip faulting

σz

σx

σy

Strike slip faulting requires one compressional and one tensional tectonic stress : σx

tect > 0 σytect < 0

For this case: σx > σz > σy

So the vertical lithostatic stress is always the intermediate stress for strike slip faulting

compression positive


Anderson faulting σx > σz > σy

σz > σy > σx

σx > σy > σz

σz

σx

dip β = 60o

σz

σx

β

β

dip β = 30o

σx

σy


Anderson faultingσz

σx

σN

σS

θ

( ) ( )θσσσ


2cos2121

zxS

zxzxN

−=−−+=

normal stress

shear stress

Recall:

Normal stress σN and shear stress σS

β

Lithostatic stress: σz = ρgz

Total stress in x-direction:

σx = ρgz + σxtect

σxtect is positive for thrust faulting

and negative for normal faulting

In terms of the lithostatic and tectonic stresses:

θσσ

θσρσ

2sin2

)2cos1(2

tect

S

tect

N zg

−=

++=


Anderson faultingAmonton’s Law: σS = µ σN

In the presence of pore fluid: σS = µ (σN – pf)

So:⎟⎟⎠

⎞⎜⎜⎝

⎛++−=± )2cos1(

22sin

2θσρµθσ tect

f

tect

pzg

σtec

tM

Pa

µ0 1.0

-100

400

thrust fault

normal faultµ0 1.0

β

0

90

30

60normal fault

thrust fault


Coulomb stress model – Toda, Stein & KingIf 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).

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.

12 Earthquake Faulting - UCL · 2006-02-21 · (b) Faulting – Coulomb’s Law σ S = C + µ i σ...

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