24-Bld-A6 Geotechnical Materials and Analysis · Undated paper
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.
The four stages map directly onto Terzaghi's one-dimensional consolidation model. Stage 1 (at equilibrium, valve closed): before any new load is applied, the piston rests at its initial position, the spring carries the pre-existing effective stress $\sigma_v'$, and the water gauge reads the pre-existing (hydrostatic) pore pressure $u_0$ — matching a saturated clay under its existing overburden, long after any earlier loading has fully consolidated.
Stage 2 (under stress $\sigma_v+\Delta\sigma$, undrained, valve still closed): the instant the new load $\Delta\sigma$ is applied, the valve (representing the clay's low permeability) has not yet allowed any water out, so the piston has not moved and the spring's compression — and hence the effective stress it carries — is unchanged. The entire increment $\Delta\sigma$ is instead carried by a spike in pore pressure, $\Delta u_1\approx\Delta\sigma$, exactly as excess pore pressure equals the full applied stress increment at $t=0^+$ in an undrained clay.
Stage 3 (under the same stress, drained, valve open): once the valve opens, water bleeds out of the chamber at a rate controlled by the valve's resistance (the clay's coefficient of permeability $k$ and drainage path length). As water escapes, the piston sinks and the spring compresses further, so it now carries part of $\Delta\sigma$ as additional effective stress; correspondingly the excess pore pressure has partially dissipated to $\Delta u_2<\Delta u_1$. At any intermediate time, $\Delta\sigma=\Delta\sigma'+\Delta u$, with the split between spring and water continuously shifting toward the spring.
Stage 4 (equilibrium under $\sigma_v+\Delta\sigma$, valve open): once enough time has passed for all of the excess pore pressure to dissipate, the piston stops moving, the spring alone carries the entire new stress ($\Delta\sigma'=\Delta\sigma$), and the gauge reads the pore pressure back down to its original (hydrostatic) value $u_0$ — primary consolidation is complete. The piston's total downward travel from Stage 1 to Stage 4 is the analogue of the consolidation settlement $S_c$, and the valve's resistance is why that settlement takes real time to develop (governed by $T_v=c_vt/H_{dr}^2$) rather than occurring instantly, as it would in a free-draining sand.
