Question 2 of 6: Effective Stress under Undrained and Drained Loading

Question 2: Effective Stress under Undrained and Drained Loading (10 marks)

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.

Given. A fully saturated soil element ($S = 100\ \%$) confined in a rigid container, already carrying a total stress $\sigma$; the standpipe registers the static pore-water pressure $u_s$. A further total-stress increment $\Delta\sigma$ is then applied to the top platen.

Find. The effective stress $\sigma'$ immediately after loading (undrained) and after full dissipation (drained), expressed in the symbols $\sigma,\ \Delta\sigma,\ u_s,\ u_e$, with comment.

[Figure not reproduced: Figure 1 (redrawn) — saturated soil in a rigid container. The standpipe reads the pore-water pressure; the applied total stress is σ before loading and σ + Δσ after. See the official exam paper.]

Approach. Apply Terzaghi’s effective-stress principle $\sigma' = \sigma - u$ at each stage, and determine $u$ from the pore-pressure parameter of a saturated soil ($B = 1$) immediately after loading and from the drainage boundary condition after consolidation.

1. Establish the initial state. Before the increment is applied the pore water is in hydrostatic equilibrium with the standpipe, so the pore pressure is the static value $u = u_s$ and $$\sigma_0' = \sigma - u_s$$ This is the effective stress that the grain skeleton is actually carrying, and it is the stress that controls both the strength and the volume of the element.
2. Apply the increment with no drainage allowed. Water and mineral grains are some three orders of magnitude stiffer than the soil skeleton, so in a saturated soil that cannot change volume the skeleton cannot compress and cannot take any of the new load. Skempton’s pore-pressure parameter for a saturated soil is $B = \Delta u / \Delta\sigma_3 = 1$, hence the excess pore pressure generated is $$u_e = B\,\Delta\sigma = \Delta\sigma$$ and the total pore pressure becomes $u = u_s + u_e$.
3. Evaluate the undrained effective stress. Substituting into $\sigma' = \sigma - u$ with the new total stress $\sigma + \Delta\sigma$, $$\sigma_{\text{und}}' = (\sigma + \Delta\sigma) - (u_s + u_e) = (\sigma + \Delta\sigma) - (u_s + \Delta\sigma)$$ so that $$\boxed{\;\sigma_{\text{und}}' = \sigma - u_s = \sigma_0'\;}$$ The effective stress is completely unchanged; the whole of $\Delta\sigma$ is carried by the pore water and the standpipe level jumps instantaneously by $\Delta\sigma/\gamma_w$.
4. Allow full drainage. Given enough time water is expelled through the standpipe and the excess pore pressure decays to zero, $u_e \to 0$, so the pore pressure returns to its static value $u = u_s$ and the standpipe returns to its original level.
5. Evaluate the drained effective stress. With $u_e = 0$, $$\sigma_{\text{dr}}' = (\sigma + \Delta\sigma) - u_s = \sigma_0' + \Delta\sigma$$ The whole of the increment has transferred from the pore water to the grain skeleton, and the element has consolidated by the corresponding volume change.
6. Describe the intermediate states. At any partial degree of consolidation $U$ the pore water still carries the fraction $(1-U)$ of the increment, so $u_e = (1-U)\Delta\sigma$ and $\sigma' = \sigma_0' + U\,\Delta\sigma$. The path from step 3 to step 5 is exactly Terzaghi one-dimensional consolidation, with the time factor $T_v = c_v t / d^2$ setting the rate.

Comments on the results. Three consequences follow directly and are what the question is really testing. First, no volume change and therefore no settlement can occur during the undrained stage, because volume change in a soil is driven by effective stress alone. Second, the soil gains no shear strength during that stage: since $\tau_f = c' + \sigma'\tan\phi'$ and $\sigma'$ has not moved, the immediate stability of a clay foundation must be checked with the undrained strength $s_u$ and a total-stress ($\phi = 0$) analysis. Third, the long-term condition is the more heavily loaded one for the skeleton but the safer one for stability, so an embankment on soft clay is critical at end of construction while an excavation in clay is critical in the long term, when negative excess pore pressures have dissipated. Staged construction exploits exactly this: load, wait for $u_e$ to decay, gain strength, then load again.