Soil investigation at a site gave the following information. Fine sand exists to a depth of 10.6 m and below this lies a soft clay layer 7.60 m thick. The water table is at 4.60 m below the ground surface.The submerged unit weight of sand yb is 10.4 kN/m^3, and the wet unit weight above the water

Question

Soil investigation at a site gave the following information. Fine sand exists to a depth of 10.6 m and below this lies a soft clay layer 7.60 m thick. The water table is at 4.60 m below the ground surface. The submerged unit weight of sand [latex]\gamma_{b}[/latex] is [latex]10.4 \mathrm{kN} / \mathrm{m}^{3}[/latex], and the wet unit weight above the water table is [latex]17.6 \mathrm{kN} / \mathrm{m}^{3}[/latex]. The water content of the normally consolidated clay[latex]w_{n}=40 \%[/latex], its liquid limit [latex]w_{l}=45 \%[/latex], and the specific gravity of the solid particles is 2.78. The proposed construction will transmit a net stress of 120[latex]120 \mathrm{kN} / \mathrm{m}^{2}[/latex] at the center of the clay layer. Find the average settlement of the clay layer.

Accepted Answer

For calculating settlement [latex]S_{t}=\frac{C_{s} H}{1+e_{0}} \log \frac{p_{0}+\Delta p}{p_{0}}[/latex] (7.15a)

[latex]S_{t}=\frac{C_{c}}{1+e_{0}} H \log \frac{p_{0}+\Delta p}{p_{0}} \quad ext { where } \Delta p=120 \mathrm{kN} / \mathrm{m}^{2}[/latex]

[latex]C_{c}=0.009\left(w_{l}-10 ight)[/latex] , (7.17) [latex]C_{c}=0.009\left(w_{l}-10 ight)=0.009(45-10)=0.32[/latex]

[latex]S=\frac{w G_{s}}{e} ext { or } e=\frac{w G_{s}}{S}[/latex] , [latex]e_{0}=\frac{w G}{S}=w G=0.40 imes 2.78=1.11 \quad ext { since } S=1[/latex]

[latex]\gamma_{b}[/latex], the submerged unit weight of clay, is found as follows

[latex]\gamma_{ ext {sat }}=\frac{\gamma_{w}\left(G_{s}+e_{0} ight)}{1+e_{0}}=\frac{9.81(2.78+1.11)}{1+1.11}=18.1 \mathrm{kN} / \mathrm{m}^{3}[/latex]

[latex]\gamma_{b}=\gamma_{\mathrm{sat}}-\gamma_{w}=18.1-9.81=8.28 \mathrm{kN} / \mathrm{m}^{3}[/latex]

The effective vertical stress [latex]p_{0}[/latex] at the mid height of the clay layer is

[latex]p_{0}=4.60 imes 17.6+6 imes 10.4+\frac{7.60}{2} imes 8.28=174.8 \mathrm{kN} / \mathrm{m}^{2}[/latex]

Now [latex]S_{t}=\frac{0.32 imes 7.60}{1+1.11} \log \frac{174.8+120}{174.8}=0.26 \mathrm{~m}=26 \mathrm{~cm}[/latex]

Average settlement = 26 cm.

Question 7.5: Soil investigation at a site gave the following information....

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