The soil profile shown consists of dry sand (4-m thick) which overlies a layer of clay (3-m thick). Ground water table is located at the interface of the sand and clay.

How many meters must the ground water table rise to decrease the effective stress by 14 kPa, at the bottom of the clay layer?

If the water table rises to the top of the ground surface, what is the change in the effective stress (in kPa) at the bottom of the clay layer?

Compute the effective stress at the bottom of the clay layer in kPa.

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solutionGivene=0.49 and G_{s}=2.68( for sand )as it is dry state so the unit weigh of the dry soil\gamma_{d}=\frac{\gamma_{w} G}{1+e}=\frac{9.81(2.68)}{1+0.49}=\frac{17.644 K N}{m^{3}}for saturated \gamma_{\text {sat }}=\frac{\gamma_{w}(G+e)}{1+e}=\frac{9.81(2.68+0.49)}{1+0.49}=\frac{20.87 K N}{m^{3}}for clay e=0.9 and G_{s}=2.75 as water table is present so the degree of saturation will be 1\gamma_{s a t}=\frac{\gamma_{w}(G+e)}{1+e}=\frac{9.81(2.75+0.9)}{1+0.9}=\frac{18.84 K N}{m^{3}}The effective stress at the bottom of the clay layer\sigma^{\prime}=17.644(4)+(18.84-9.81)(3)=97.66 K p alet us assume the water table is rides to h so there is decrease in effective stress by14Kpa consider the effective stress at bottom of clay layer\begin{array}{l}97.66-14=17.644(4-h)+(20.87-9.81)(h)+(18.84-9.81) 3 \\83.66=97.666-6.584 h\end{array}h=2.127 \mathrm{mts} height form the water tableif the water table is at ground level the effective stress at bottom of clay us\sigma^{\prime}=(20.87-9.81) 4+(18.84-9.81) 3=71.33 \mathrm{Kpa} Thank you  please feel free to comment in case of any queries   ...