Max height of unsupported clay excavation

For a vertical cut in clay with $\phi = 0$, cohesion $c$ and unit weight $\gamma$, Rankine's active pressure at depth $z$ is

$$p_a = \gamma z - 2c$$

since $K_a = 1$. The pressure is zero at $z_0 = 2c/\gamma$ and tensile above it.

For an unsupported (vertical) cut, the soil is stable up to the height at which the total active thrust over the full height is zero; i.e. the compressive force below balances the tension above, so no support is needed:

$$\begin{aligned} P_a &= \int_0^{H_c}(\gamma z - 2c)\,dz = \tfrac12\gamma H_c^2 - 2cH_c = 0 \\ H_c &= \frac{4c}{\gamma} \end{aligned}$$

Result: $H_c = 4c/\gamma$ (for $c = c_u$ in undrained condition, $H_c = 4c_u/\gamma$). Using a factor of safety, the safe height is $H = H_c/F$.

Note: because soil cannot carry tension, tension cracks of depth $2c/\gamma$ form; considering them, the practical safe height is smaller (the theoretical value with the cracks is $2.67c/\gamma$ by Terzaghi's analysis), so a factor of safety of 2 or more is used.