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Mohr-Coulomb failure
Goal: To understand relationship between
stress, brittle failure, and frictional faulting
and to use this relationship to predict rock
behavior
Stress review
• Stress = Force/Area
• 3 principal vectors: σ1, σ2, and σ3 at right
angles to each other
• σ 1 ≥ σ2 ≥ σ3
• σ1 is the maximum principal stress
direction, σ2 is the intermediate principal
stress direction, and σ3 is the minimum
principal stress direction
We also define:
• Static stress as σ1 = σ2 = σ3
• Lithostatic stress as static stress
generated by mass of overlying rocks
• Differential stress (σd) as (σ1 - σ3)
• Confining pressure as σ2 = σ3 for the
conditions σ1 > σ2 = σ3
Shear stress and normal stress
For any plane in a stress field defined by σ1,
σ2, and σ3 with strike parallel with σ2:
σ1
θ
σ3
σ3
σ1
The stress is resolved into 2 components:
1. Shear stress (σs), acting parallel with the
plane
2. Normal stress (σn), acting perpendicular
to the plane
σ1
σ
n
θ
σ3
σ3
σs
σn
σ1
σs
Stress components are related by:
1. σs = ½(σ1 - σ3)sin(2θ)
2. σn = ½(σ1 + σ3) - ½(σ1 - σ3)cos(2θ)
where θ = angle between plane and σ1
σ1
σ
n
θ
σ3
σ3
σs
σn
σ1
σs
Mohr diagram for stress
Relationship between σ1, σ3, σs, and σn is
plotted graphically in Cartesian coordinates
σs
σn
Mohr circle for stress: circle with diameter =
σd plotted on mohr diagram
Center on the σn-axis at point = ½(σ1 + σ3)
σs
σ3
σ1
σn
½(σ1 + σ3)
Finding σs, and σn
Can use a Mohr circle to find σs, and σn for
any plane
σs
σ3
σ1
σn
Plot a line from center to edge of circle at
angle 2θ-clockwise from σn-axis
σs
2θ
σ3
σ1
σn
X- and y-coordinates of intersection of line
and circle define σs and σn for the plane
(σs, σn) of plane
σs
σ3
σ1
σn
Coulomb’s failure criterion
• Every homogeneous material has a
characteristic failure envelope for brittle
shear fracturing
• Combinations of σs and σn outside of the
envelope result in fracture
Determining failure envelope
Experimental rock deformation
The Coulomb envelope
Tensile
Fracture
σs
Shear
Fracture
Stable
2θ
σ3
σ1
Stable
Shear
Fracture
σn
Coulomb law of failure
σs
σc = σ0 + tan(φ)σn
φ
σ
0
σn
σc = σ0 + tan(φ)σn
Formula defines shear stress under which
rocks will fracture
σc = critical shear stress — σs at failure
σ0 = cohesive strength — σs when σn = 0
φ = angle of internal friction — φ ≈ 90 - 2θ
• For most rocks, angle of internal friction ≈ 30°
• Therefore, θ at failure is also ≈ 30°
• σs is greatest when θ = 45°
Failure
envelopes
for different
rocks
Slip on pre-existing fractures
Pre-existing fractures have no cohesive
strength, σ0 = 0
Failure envelopes for pre-existing fractures
derived experimentally
Envelope of sliding friction
σs
φf = angle of sliding friction
σn
Byerlee’s law
Describes frictional sliding envelope
σc = tan(φf)σn
φf ≈ 40° for low confining pressures and
≈ 35° for high confining pressures
Byerlee’s law for different rock types
Effect of pore-fluid pressure
Pore fluid pressure (Pf) effectively lowers the
stress in all directions
The effective stresses (σ1eff, σ2eff, and σ3eff) =
principal stresses - Pf
σ1eff = σ1 - Pf
σ2eff= σ2 - Pf
σ3eff = σ3 - Pf
σs
Stable stress conditions
σ1
σ3
σn
σs
Increase in pore fluid pressure can
drive faulting!!
σ1eff
σ3eff
σ1
σ3
σn