Q5GATE 2023MCQ

A vertical sheet pile wall is installed in an anisotropic soil having coefficient of horizontal permeability, $k_{H}$ and coefficient of vertical permeability, $k_{V}$ . In order to draw the flow net for the isotropic condition, the embedment depth of the wall should be scaled by a factor of _____, without changing the horizontal scale.

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📖 Explanation

We know that, for 2D flow. $\mathrm{K}=\mathrm{K}_{\mathrm{x}} \frac{\partial^{2} \mathrm{~h}}{\mathrm{dx}}+\mathrm{K}_{\mathrm{z}} \frac{\partial^{2} \mathrm{~h}}{\mathrm{dz}}=0$ When $\mathrm{K}_{1}=\mathrm{K}_{\text {horizontal }}$ and $\mathrm{K}_{2}=\mathrm{K}_{\text {vertical }}$ as shown in figure. $\frac{\partial^{2} \mathrm{~h}}{\frac{\mathrm{K}_{\mathrm{z}}}{\mathrm{K}_{\mathrm{x}}} \partial \mathrm{x}^{2}}+\frac{\partial^{2} \mathrm{~h}}{\partial \mathrm{z}^{2}}=0$ From the above equation we can note that, when soil is anisotropic with respect to permeability i.e. $k_{x}=k_{z}$ , the flow and equipotential line are not necessarily to be orthogonal as because $\frac{\mathrm{K}_{\mathrm{z}}}{\mathrm{K}_{\mathrm{x}}}=1$ , and we cannot get the actual form of Laplace equation which is $\frac{\partial^{2} h}{\partial x^{2}}+\frac{\partial^{2} h}{\partial z^{2}}=0$ . But if we replace $x_{t}=x \sqrt{\frac{K_{z}}{\mathrm{~K}_{x}}}$ Then, $\quad \partial x_{t}^{2}=\partial x^{2} \frac{K_{z}}{\mathrm{~K}_{x}}$ $\frac{\partial^{2} h}{\partial x_{t}^{2}}+\frac{\partial^{2} h}{\partial z^{2}}=0$ Thus, new flow line and equipotential lines are drawn on transformed section with $\mathrm{x}$ -distance changed to $x \sqrt{\frac{\mathrm{K}_{\mathrm{z}}}{\mathrm{K}_{\mathrm{x}}}}$ while keeping the vertical dimension constant. If we scale up the depth by keeping the horizontal distance same the depth will be scaled by factor $\sqrt{\mathrm{K}_{\mathrm{x}} / \mathrm{K}_{\mathrm{z}}}$ .

Q9GATE 2021NAT

A retaining wall of height 10 m with clay backfill is shown in the figure (not to scale). Weight of the retaining wall is 5000 kN per m acting at 3.3 m from the toe of the retaining wall. The interface friction angle between base of the retaining wall and the base soil is $20^{\circ}$ . The depth of clay in front of the retaining wall is 2.0 m. The properties of the clay backfill and the clay placed in front of the retaining wall are the same. Assume that the tension crack is filled with water. Use Rankine's earth pressure theory. Take unit weight of water, $\gamma_{w}=9.81 \mathrm{kN} / \mathrm{m}^{3}$ . The factor of safety (round off to two decimal places) against sliding failure of the retaining wall after ignoring the passive earth pressure will be ________________

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📖 Explanation

Depth of tension crack,

\begin{aligned} Z_{C}&=\frac{2 C}{\gamma } \text { If } \phi=0 \\ &=\frac{2 \times 30}{17.2 }=3.488 \mathrm{~m} \end{aligned}

Tension crack develops only in the back fill (clay) The tension crack will not develop in the clay which is infront of the wall. The total active thrust on the wall, due to the backfill, and the water in the tension crack $F_a=k_a\cdot \frac{\gamma H^2}{2}-2C\sqrt{k_a}\cdot H+\frac{2C^2}{\gamma }+\frac{\gamma _wz_c^2}{2}$ $k_a=1 \text{ since }\phi =0$ $F_a=1 \times \frac{17.2 \times 10^2}{2}-2 \times 30\sqrt{1} \times 10+\frac{2 \times 30^2}{17.2 }+\frac{9.81 \times 3.488^2}{2}=424.326 \; kN$ Factor of safety against sliding neglecting passive earth pressure on front side, $F_s$ $F_s=\frac{\mu \cdot W}{F_a}$ $\mu =$ coefficient of friction at base $= \tan \delta$ $F_s=\frac{\tan \delta \cdot W}{F_a}=\frac{\tan 20^{\circ} \times 5000}{424.326}=4.2888\approx 4.29$

Q18GATE 2016NAT

The soil profile at a site consists of a 5 m thick sand layer underlain by a $c-\phi$ soil as shown in figure. The water table is found 1 m below the ground level. The entire soil mass is retained by a concrete retaining wall and is in the active state. The back of the wall is smooth and vertical. The total active earth pressure (expressed in $kN/m^{2}$ ) at point A as per Rankine's theory is _________.

🚀 Solve in Practice Mode

📖 Explanation

In $\mathrm{c}-\phi$ soil Earth pressure

\begin{array}{l} p_{\mathrm{a}}=k_{\mathrm{a}} \sigma_{\mathrm{v}}-2 \mathrm{c} \sqrt{\mathrm{k}_{\mathrm{a}}} \\ k_{\mathrm{a}}=\frac{1-\sin \phi}{1+\sin \phi}=\frac{1-\sin 24}{1+\sin 24}=0.4217 \end{array}

Note: Below water table, water pressure will not be multiplied by $k_{\mathrm{a}}$ at point A

\begin{array}{l} \sigma_{v}=1 \times \gamma_{b}+4 \gamma_{\text {sat }}+3 \gamma_{\text {sat }} \\ \sigma_{v}=1 \times \gamma_{b}+\left(4 \times \gamma^{\prime}+4 \times \gamma_{w}\right)+\left(3 \gamma^{\prime}+3 \gamma_{w}\right) \\ \sigma_{v}=\left(1 \times \gamma_{b}+4 \gamma^{\prime}+3 \gamma^{\prime}\right)+\left(4 \gamma_{w}+3 \gamma_{w}\right) \end{array}

Earth Pressure at 'A'

\begin{array}{l} p_{\mathrm{a}}=k_{\mathrm{a}} \sigma_{v}-2 c \sqrt{k_{\mathrm{a}}} \\ =k_{a}\left[\gamma_{b}+4 \gamma^{\prime}+3 \gamma^{\prime}\right]\$+4 \gamma_{w}+3 \gamma_{w}-2 \times c^{\prime} \sqrt{k_{a}} \\ =0.4217[16.5+4(19-9.81)+3(18.5-9.81)] \\ +7 \times 9.81-2 \times 25 \times \sqrt{0.4217} \\ p_{\mathrm{a}}=69.65 \mathrm{kN} / \mathrm{m}^{2} \end{array}

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