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Geosciences LibreTexts

1.3: Failure of Rocks

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  • Let's relate the previously discussed stresses to fracture and faulting. Failure can occur by frictional sliding or fracture. Byerlee's law defines the shear stress needed to cause sliding, essentially it defines the point of failure given an applied normal stress. The law tells us that shear stress (\(\tau\)), increases about linearly with the normal stress. This is modeled with the equation \(\tau=0.85\sigma_{n}\). Using this calculated shear stress, we can apply this to both frictional sliding and fracture. In the case of failure by frictional sliding, failure is determined by the equation

    \[\sigma_{s-fric} = C_{o} +\mu_{s}\sigma_{n},\]

    where \(\sigma_{s-fric}\) is shear stress, \(C_{o}\) is the cohesion of fractured rock, \(\mu_{s}\) is the friction coefficient, and \(\sigma_{n}\) is the normal stress. Failure occurs when \(\sigma_{s}\) on the fault plane is bigger than \(\sigma_{s-fail}\), or \(\sigma_{s}>\sigma_{s-fric}\). Friction (\(\mu_{s}\)) is typically \(\cong\) 0.6-0.8 for most rocks, but can be as low as 0.2 for talc and serpentine. The Mohr-Coulomb Fracture Criteria states that failure by fracture is determined by the equation 

    \[\sigma_{s-frac} = C_{1} +\mu_{o}\sigma_{n}.\]

    where \(\sigma_{s-frac}\) is the shear stress, \(C_{1}\) is the cohesion of unfractured rock, and \(\sigma_{n}\) is the fracture coefficient. Fracture occurs when \(\sigma_{s} > \sigma_{s-frac}\).

    1.4 Friction and Fracture Graph.png

    Figure \(\PageIndex{1}\): Failure by Fracture Graph 


    Calculating Failure Point

    1.4 Fault Example.png

    Figure \(\PageIndex{2}\): Calculating Failure

    Applying what we've learned so far, calculate if the above fault is at or below the failure point. We are given that \(\sigma_{n}\)=200 MPa, \(\sigma_{s}\)=100 MPa, C=10 MPa, and \(\mu_{s}\)=0.8.

    \[ \begin{align*} \sigma_{s-fric} &= C_{o} +\mu_{s}\sigma_{n} \\[4pt] &=10 \,MPa + 0.8(200\, MPa) \\[4pt] &=10 + 160 \\[4pt] &=170\, MPa \end{align*}\]

    Since \(\sigma_{s-fric} > \sigma_{s}\), where \(\sigma_{s-fric}\) is the stress needed for failure, and \(\sigma_{s}\) is the applied stress, 170>100, thus the fault is below the failure point.