Question 4 of 7: Pore, Total and Effective Stress in a Layered Profile Before and After Rapid Dewatering

Question 4: Pore, Total and Effective Stress in a Layered Profile Before and After Rapid Dewatering (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.

Approach. All three points sit inside the sand layer, so a single dry/ saturated unit-weight pair applies throughout; compute σ, u and σ′ = σ − u at each point for the initial water table, then repeat for the water table relocated to C. Because sand is highly permeable, treat the "immediately after" and "long after" states as identical for these three points — there is no clay-style consolidation lag while the water table itself sits inside the free-draining sand.

1. a) Initial stresses (water table at 1 m). Point A is at the surface (σ = u = σ′ = 0). Point B (2.5 m depth) is 1.5 m below the water table: $$\sigma_B = 19(1) + 22(1.5) = 52.0\ \text{kPa}, \quad u_B = 9.81(1.5) = 14.72\ \text{kPa}, \quad \sigma'_B = \boxed{37.29\ \text{kPa}}$$ Point C (5 m depth, the interface) is 4 m below the water table: $$\sigma_C = 19(1) + 22(4) = 107.0\ \text{kPa}, \quad u_C = 9.81(4) = 39.24\ \text{kPa}, \quad \sigma'_C = \boxed{67.76\ \text{kPa}}$$
2. b) Stresses after dewatering to C. The new water table sits exactly at C, so all of the sand from the surface to C is now above the water table (dry unit weight), and pore pressure at and above C is zero. This state is reached almost immediately because sand drains freely — there is no low-permeability layer between the surface and C to delay re-equilibration, so "immediately after" and "long after" are the same answer here: $$\sigma_B = 19(2.5) = 47.5\ \text{kPa}, \quad u_B = 0, \quad \sigma'_B = \boxed{47.5\ \text{kPa}}$$ $$\sigma_C = 19(5) = 95.0\ \text{kPa}, \quad u_C = 0, \quad \sigma'_C = \boxed{95.0\ \text{kPa}}$$ Point A is unaffected (it was already above the original water table). Effective stress rises by 47.5 − 37.29 = 10.2 kPa at B and 95.0 − 67.76 = 27.2 kPa at C — the classic dewatering effect of removing buoyant support and converting saturated to dry unit weight through the drained zone.

c) Why a step, not a consolidation curve. In a low-permeability clay, lowering the water table (or applying a load) generates an excess pore pressure that dissipates only slowly, as water is squeezed out through a long drainage path — that produces the familiar gradual, asymptotic rise in σ′ over months or years. Sand's permeability is many orders of magnitude higher, so any excess pore pressure it develops drains away essentially instantaneously on an engineering timescale; there is no meaningful "immediately after" distinct from "long after" for point B. The sketch is therefore a flat line at 37.3 kPa, a near-vertical rise the moment the water table is drawn down, then a flat line at 47.5 kPa — the opposite behaviour from the slow, load-driven excess-pore-pressure dissipation seen when a surcharge is placed on a saturated clay.