Stress on a fault plane

Consider a fault plane with normal $\ensuremath{\mbox{\boldmath ${n}$}} = (n_1,n_2,n_3)$ in a homogeneous stress field $\underline{\sigma}$ as in Figure 2.

Figure: Decomposition of the traction vector ${t}$ on a plane into the normal stress ${\sigma_n}$ and shear stress ${\tau}$ components.

Performing all calculations in the principal stress system we can easily obtain the components of the normal, ${\sigma_n}$, and shear, ${\tau}$, stresses. The traction vector ${t}$ is given by

$$-\ensuremath{\mbox{\boldmath${t}$}}= \ensuremath{\mbox{\boldmath$... ...nsuremath{\sigma_{1}}n_1,\ensuremath{\sigma_{2}}n_2,\ensuremath{\sigma_{3}}n_3)$$ (16)

remembering that our sign convention dictates that compressive stress is positive and thus has the opposite direction to the plane normal. From the traction vector we calculate the normal and shear stresses on the plane

$$-\ensuremath{\mbox{\boldmath${\sigma_n}$}}= (\ensuremath{\mbox{\b... ...sigma_{2}}n_2^2+\ensuremath{\sigma_{3}}n_3^2)\ensuremath{\mbox{\boldmath${n}$}}$$ (17)

$$-\ensuremath{\mbox{\boldmath${\tau}$}}= \ensuremath{\mbox{\boldmath${t}$}}- \ensuremath{\mbox{\boldmath${\sigma_n}$}}$$ (18)

We find for the shear stress components

$$-\tau_1 = (\ensuremath{\sigma_{1}}- (\ensuremath{\sigma_{1}}n_1^2 + \ensuremath{\sigma_{2}}n_2^2 + \ensuremath{\sigma_{3}}n_3^2))n_1$$

$$-\tau_2 = (\ensuremath{\sigma_{2}}- (\ensuremath{\sigma_{1}}n_1^2 + \ensuremath{\sigma_{2}}n_2^2 + \ensuremath{\sigma_{3}}n_3^2))n_2$$

$$-\tau_3 = (\ensuremath{\sigma_{3}}- (\ensuremath{\sigma_{1}}n_1^2 + \ensuremath{\sigma_{2}}n_2^2 + \ensuremath{\sigma_{3}}n_3^2))n_3$$

and using the relation $n_1^2 + n_2^2 + n_3^2 = 1$

$$-\tau_1 = (n_2^2(\ensuremath{\sigma_{1}}- \ensuremath{\sigma_{2}}) + n_3^2(\ensuremath{\sigma_{1}}- \ensuremath{\sigma_{3}}))n_1$$

$$-\tau_2 = (n_2^2(\ensuremath{\sigma_{1}}- \ensuremath{\sigma_{2}}) + n_3^2(... ...nsuremath{\sigma_{3}}) - (\ensuremath{\sigma_{1}}- \ensuremath{\sigma_{2}}))n_2$$

$$-\tau_3 = (n_2^2(\ensuremath{\sigma_{1}}- \ensuremath{\sigma_{2}}) + n_3^2(... ...nsuremath{\sigma_{3}}) - (\ensuremath{\sigma_{1}}- \ensuremath{\sigma_{3}}))n_3$$

Finally, remembering that $R = (\ensuremath{\sigma_{1}}- \ensuremath{\sigma_{2}})/(\ensuremath{\sigma_{1}}- \ensuremath{\sigma_{3}})$ and using $K = n_3^2 + Rn_2^2$ we can write

$$-\ensuremath{\mbox{\boldmath${\tau}$}}= (\ensuremath{\sigma_{1}}-... ...left ( \begin{array}{c} Kn_1 \\ (K - R)n_2 \\ (K - 1)n_3 \end{array} \right )$$ (19)

and hence the magnitude of the shear stress is

$$\tau = (\ensuremath{\sigma_{1}}- \ensuremath{\sigma_{3}})\sqrt{n_3^2 + R^2n_2^2 - K^2}$$ (20)

Equation 19 shows that the direction of shear stress on a plane is independent of the absolute magnitude of the principal stresses and only depends on the orientation of the plane in the stress field and on the ratio $R$ [Bott, 1959]. Comparing Equation 19 with Equation 15 we see that the shear stress direction on a fault plane is determined by the reduced stress tensor and that neither adding an isotropic stress nor multiplying the tensor with a positive constant will modify that direction. This result is central to inversion schemes that estimate the principal stresses from earthquake focal mechanisms.

The maximum shear stress directions and magnitudes are obtained from the stationary points when Equation 20 is differentiated with respect to the coordinates of the fault plane. I will only state the results here, for a full treatment see e.g. Jaeger and Cook [1979]. There are three directions of maximum shear stress, all of which bisect the angles between the principal stresses. The greatest is

$$\tau = \frac{1}{2}(\ensuremath{\sigma_{1}}- \ensuremath{\sigma_{3}})$$ (21)

bisecting the directions of $\sigma_{1}$ and $\sigma_{3}$. The other stationary values are, as expected,

$$\tau = \frac{1}{2}(\ensuremath{\sigma_{1}}- \ensuremath{\sigma_{2... ...ace{0.5cm} \tau = \frac{1}{2}(\ensuremath{\sigma_{2}}- \ensuremath{\sigma_{3}})$$ (22)

The normal stresses corresponding to these directions are

$$\frac{1}{2}(\ensuremath{\sigma_{1}}+ \ensuremath{\sigma_{3}}) \hs... ...}) \hspace{0.5cm} \frac{1}{2}(\ensuremath{\sigma_{2}}+ \ensuremath{\sigma_{3}})$$ (23)