At the final state of the slump test for yield stress, an axisymmetric deposit of height is just able to support itself. Hydrostatic pressure gives radial pressure gradient , while lubrication theory makes the magnitude of the basal shear stress
The marginally yielded final profile therefore satisfies
Integration gives
Using mass conservation, the known volume is
Solving for the yield stress produces the estimate
Thus one measures the final radius and inserts it with and . 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 , so the deposit is approximately a cone,
This constant-slope profile differs from the square-root edge of the yield-stress-fluid model.
Let be the downward velocity between the vertical plates. A steady vertical momentum balance gives
Symmetry requires , so the stress magnitude is
The wall stress first reaches the yield stress when . The onset measurement therefore gives
For , define . The central region is a plug flow of a yield-stress fluid, while the layers are yielded. The Herschel–Bulkley fluid law gives, for ,
Integrating from the no-slip wall gives
in the yielded layer, and the plug moves at
The volumetric flow rate per unit span is consequently
Near onset, the plug term dominates and
The observed exponent two therefore corresponds to , a linear post-yield constitutive law at small shear rate. Far above onset, the fully yielded contribution scales as
The observed exponent four corresponds to , a shear-thinning square-root law at high shear rate.
A plausible stress curve is therefore odd in , starts at the two yield-stress values , and has finite linear slope just after yield:
It then crosses smoothly to the concave high-rate behavior
with the negative-rate branch fixed by odd symmetry. This combines the two power laws inferred independently from the measured flux limits.

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