Question 5 of 6: Consolidation, seepage heads and effective stress

Question 5: Consolidation, seepage heads and effective stress (Value: 20 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.

(a) One-dimensional consolidation by the spring analogy

In the analogy the spring represents the soil skeleton and the water the pore fluid; the piston has a small orifice that meters how fast water can escape. (a) Before loading, the spring carries the seating load and the water is at its static pressure. (b) The instant a load increment $\Delta\sigma$ is applied with the orifice effectively closed, the near-incompressible water carries all of it: the excess pore pressure jumps to $u = \Delta\sigma$ while the spring is still uncompressed, so the effective stress increment is zero. This is the “end of construction”, undrained condition. (c) As water bleeds through the orifice the piston descends, the spring compresses and progressively takes load: $u$ decays and $\sigma'$ grows, always obeying $\sigma = \sigma' + u$. (d) When flow stops, all excess pore pressure has dissipated ($u = 0$) and the full increment is carried by the spring ($\sigma' = \Delta\sigma$). The time-dependent transfer of load from water to skeleton, governed by how quickly water can drain, is consolidation, and the settlement is the accumulated compression of the spring.

(b) Pressure, total head and elevation head in the seepage cell

Given. Upward flow through a 4 m soil column: the supply standpipe holds water 4 m above the container’s free surface, which itself stands 1 m above the top of the soil; point A is 1 m above the base of the soil.

Find. Elevation head, total head and pressure head (hence pore pressure $u=\gamma_w h_p$) at the entrance, at A and at the exit, and their distributions with height.

Approach. Take the datum at the base of the soil. Total head is fixed at each free-water surface and drops linearly through the soil at the hydraulic gradient $i=\Delta h/L$; pressure head is total head minus elevation head.

1. Hydraulic gradient. The two free surfaces differ by 4 m and all loss is across the 4 m of soil, so $$i = \frac{\Delta h}{L} = \frac{4}{4} = \boxed{1.0}\quad(\text{flow upward, base}\to\text{top}).$$
2. Total head. At the entrance (base) the total head equals the supply-standpipe level, $h = 9\ \text{m}$; at the exit (top of soil) it equals the container surface, $h = 5\ \text{m}$. It falls linearly at $1\ \text{m}$ of head per metre risen, so at A (1 m above base) $h_A = 9 - 1 = 8\ \text{m}$.
3. Elevation head. Directly from the datum: entrance $z = 0$, A $z = 1\ \text{m}$, exit $z = 4\ \text{m}$.
4. Pressure head and pore pressure. $h_p = h - z$, then $u = \gamma_w h_p$: entrance $h_p = 9-0 = 9\ \text{m} \Rightarrow u = 88.3\ \text{kPa}$; A $h_p = 8-1 = 7\ \text{m} \Rightarrow u = 68.7\ \text{kPa}$; exit $h_p = 5-4 = 1\ \text{m} \Rightarrow u = 9.8\ \text{kPa}$ (the 1 m of ponded water above the soil).

(c) Effective stresses at the top and bottom of the clay layer

Given. A three-layer profile with a perched water table in the upper sand and artesian pressure in the lower sand.

Find. The effective vertical stress $\sigma'$ at the top ($z=4$ m) and bottom ($z=8$ m) of the clay.

Approach. Build total stress by summing $\gamma\,\Delta z$ down each layer; take the pore pressure at the top of the clay from the upper water table and at the bottom of the clay from the artesian piezometric surface; then $\sigma' = \sigma - u$.

1. Total stress at the top of the clay (4 m). $\sigma = 16.5(2) + 19(2) = 33 + 38 = 71\ \text{kPa}$.
2. Pore pressure at the top of the clay. Hydrostatic below the water table (2 m depth): $u = \gamma_w(4-2) = 9.81\times 2 = 19.6\ \text{kPa}$. Hence $\sigma'_{\text{top}} = 71 - 19.6 = \boxed{51.4\ \text{kPa}}$.
3. Total stress at the bottom of the clay (8 m). Add the 4 m of clay: $\sigma = 71 + 20(4) = 71 + 80 = 151\ \text{kPa}$.
4. Pore pressure at the bottom of the clay. Governed by the artesian lower sand, whose piezometric surface is 4 m above GL, so the pressure head at 8 m depth is $8+4 = 12\ \text{m}$: $u = 9.81\times 12 = 117.7\ \text{kPa}$. Hence $\sigma'_{\text{bot}} = 151 - 117.7 = \boxed{33.3\ \text{kPa}}$.

The effective stress is lower at the bottom of the clay than at the top, because the artesian pressure buoys the base of the layer. This uplift reduces the confining stress on the lower part of the clay and, if the artesian head rose further, could approach zero effective stress (heave/boiling) at the base — a key design concern for excavations over artesian aquifers.