 Solution

At the final state of the slump test for yield stress, an axisymmetric deposit of height $h(r)$ is just able to support itself. Hydrostatic pressure gives radial pressure gradient $\rho g{h}^{\prime }(r)$, while lubrication theory makes the magnitude of the basal shear stressThe marginally yielded final profile therefore satisfiesIntegration givesUsing mass conservation, the known volume isSolving for the yield stress produces the estimateThus one measures the final radius $R$ and inserts it with $V$ and $\rho$. The estimate assumes a thin deposit, negligible surface tension, complete initial yielding, and a spatially uniform yield stress.

Dry sand is a granular material governed primarily by frictional stability. Its final free surface reaches the angle of repose ${\theta }_r$, so the deposit is approximately a cone,This constant-slope profile differs from the square-root edge of the yield-stress-fluid model.