Fault shear stress from a plate-integrated force ($\bar{\tau }=F\sin (2\delta )/(2a)$)

A horizontally compressive force per width $F$, carried uniformly through thickness $a$, gives additional horizontal stress $F/a$. Resolving its traction on a fault dipping at angle $\delta$ gives the displayed mean shear stress. An isotropic background pressure adds only normal traction. The integrated force alone does not determine the actual depth-dependent stress distribution or whether a fault fails.