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Two-layer sand, plot stresses (bulk 18)

Short answer

At 7 m depth, total stress is 134 kPa, neutral stress is 29.43 kPa, and effective stress is 104.57 kPa.

A sand deposit consists of two layers. The top layer is 3 m [unit not printed] thick with bulk unit weight 18 kN/m³ and saturated unit weight of 21 kN/m³ and the bottom layer is 4 m thick with saturated density of 20 kN/m³. The ground water table is at a depth of 4 m below the ground surface and the zone of capillary saturation is 1 m above the water table. Calculate and plot the effective stress, total stress and neutral stress.

Given

Top layer 0 to 3 m: γ=18\gamma = 18 kN/m³ (bulk), γsat=21\gamma_{sat} = 21 kN/m³. Bottom layer 3 to 7 m: γsat=20\gamma_{sat} = 20 kN/m³. WT at 4 m; capillary saturation 1 m above the WT (3 m to 4 m). γw=9.81\gamma_w = 9.81 kN/m³.

The capillary zone (3 m to 4 m) lies in the bottom layer, so it is saturated with γsat=20\gamma_{sat} = 20 kN/m³. The top layer lies above the capillary zone, so it stays at its bulk weight of 18 kN/m³ (its saturated weight is not needed).

Calculation

Depth (m)σ\sigma (kPa)uu (kPa)σ′=σ−u\sigma' = \sigma - u (kPa)
0000
318×3=5418\times3 = 54−9.81-9.8163.81
4 (WT)54+20×1=7454 + 20\times1 = 74074.00
774+20×3=13474 + 20\times3 = 1349.81×3=29.439.81\times3 = 29.43104.57

Plot

 depth(m)  σ      u       σ'
   0       0      0       0
   3       54   -9.81    63.81
   4       74     0      74.0
   7      134   29.43   104.57

Plot each column against depth (downward), joining the points with straight lines. uu is zero above 3 m, negative between 3 m and 4 m and positive below the WT. Just above 3 m, σ′=54\sigma' = 54 kPa, so σ′\sigma' jumps by 9.81 kPa at the top of the capillary zone.

Answer: At 7 m, σ=134\sigma = 134 kPa, u=29.43u = 29.43 kPa, σ′=104.57\sigma' = 104.57 kPa.

✓ Answer checked by Claude Opus 5.5 (high effort)

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