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Clay over sand, effective stress at depths

Short answer

Effective vertical stresses at 0, 1, 2, 4, and 7 m depths are 0, 29.81, 40.00, 60.38, and 87.95 kPa, respectively.

A soil profile consists of 4 m and 3 m thick clay and sand layers, respectively. The clay layer lies above the sand layer and the ground water table is seen at 2 m depth from the ground surface. Above the water table, there lies a 1 m thick capillary saturated zone. Determine the effective vertical stress at 0 m, 1 m, 2 m, 4 m and 7 m depths from the ground surface. Take the bulk unit weight and saturated unit weight of clay as 20 kN/m³ and 20 kN/m³ respectively. Take the saturated unit weight of sand as 19 kN/m³.

Given

Clay 0 to 4 m (above), sand 4 to 7 m. WT at 2 m. Capillary saturated zone 1 m thick above the WT (from 1 m to 2 m). Clay: γ=γsat=20\gamma = \gamma_{sat} = 20 kN/m³; sand γsat=19\gamma_{sat} = 19 kN/m³; γw=9.81\gamma_w = 9.81 kN/m³. Soil above the capillary zone (0 to 1 m) has u=0u = 0.

Calculation

Depth (m)σ\sigma (kPa)uu (kPa)σ′=σ−u\sigma' = \sigma - u (kPa)
0000
120×1=2020\times1 = 20−9.81×1=−9.81-9.81\times1 = -9.8129.81
220×2=4020\times2 = 40040.00
440+20×2=8040 + 20\times2 = 809.81×2=19.629.81\times2 = 19.6260.38
780+19×3=13780 + 19\times3 = 1379.81×5=49.059.81\times5 = 49.0587.95

Answer: Effective vertical stress at 0, 1, 2, 4 and 7 m = 0, 29.81, 40.00, 60.38 and 87.95 kPa respectively.

✓ Answer checked by Claude Opus 5.5 (high effort)

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