A clay layer 3.66 m thick rests beneath a deposit of submerged sand 7.92 m thick. The top of the sand is located 3.05 m below the surface of a lake. The saturated unit weight of the sand is 19.62 $kN / m ^{3} \text { and of the clay is } 18.36 kN / m ^{3}$ .

Compute (a) the total vertical pressure, (b) the pore water pressure, and (c) the effective vertical pressure at mid height of the clay layer (Refer to Fig. Ex. 5.4).

Question

A clay layer 3.66 m thick rests beneath a deposit of submerged sand 7.92 m thick. The top of the sand is located 3.05 m below the surface of a lake. The saturated unit weight of the sand is 19.62 $kN / m ^{3} \text { and of the clay is } 18.36 kN / m ^{3}$ .

Compute (a) the total vertical pressure, (b) the pore water pressure, and (c) the effective vertical pressure at mid height of the clay layer (Refer to Fig. Ex. 5.4).

Accepted Answer

(a) Total pressure

The total pressure σ, over the midpoint of the clay is due to the saturated weights of clay and sand layers plus the weight of water over the bed of sand, that is

[latex]\sigma_{t}=\frac{3.66}{2} imes 18.36+7.92 imes 19.62+3.05 imes 9.81=33.6+155.4+29.9=218.9 kN / m ^{2}[/latex]

(b) Pore water pressure is due to the total water column above the midpoint. That is

[latex]u_{w}=\frac{3.66}{2} imes 9.81+7.92 imes 9.81+3.05 imes 9.81=125.6 kN / m ^{2}[/latex]

(c) Effective vertical pressure

[latex]\sigma_{t}-u_{w}=\sigma^{\prime}=218.9-125.6=93.3 kN / m ^{2}[/latex]

Question 5.4: A clay layer 3.66 m thick rests beneath a deposit of submerg...

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