Q21GATE 2013MCQ

Two different soil types (Soil 1 and Soil 2) are used as backfill behind a retaining wall as shown in the figure, where $\gamma _{t}$ is total unit weight, and c' and $\phi$ ' are effective cohesion and effective angle of shearing resistance. The resultant active earth force per unit length (in kN/m) acting on the wall is:

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} k_{a 1} &=\frac{1-\sin 30^{\circ}}{1+\sin 30^{\circ}}=\frac{1}{3}=0.33 \\ k_{a_{2}} &=\frac{1-\sin 40^{\circ}}{1+\sin 40^{\circ}}=0.217 \\ P_{a_{1}} &=\frac{1}{2} k_{a_{1}} \gamma_{1} H_{1}^{2}=\frac{1}{2} \times \frac{1}{3} \times 15 \times 2^{2} \\ &=10 \mathrm{kN} / \mathrm{m} \end{aligned}

\begin{aligned} P_{a_{2}} &=k_{a_{2}} \gamma_{1} H_{1} \cdot H_{2}=0.217 \times 15 \times 2 \times 2 \\ &=13.02 \mathrm{kN} / \mathrm{m} \\ P_{a_{3}} &=\frac{1}{2} \mathrm{k}_{02} \gamma_{2} \mathrm{H}_{2}^{2}=\frac{1}{2} \times 0.217 \times 20 \times 2^{2} \\ &=8.68 \mathrm{kN} / \mathrm{m} \\ P_{a} &=P_{a_{1}}+P_{\mathrm{a}_{2}}+P_{a_{3}}=10+13.02+8.68 \\ P_{\mathrm{a}} &=31.7 \mathrm{kN} / \mathrm{m} \end{aligned}

Q22GATE 2012MCQ

A smooth rigid retaining wall moves as shown in the sketch causing the backfill material to fail. The backfill material is homogeneous and isotropic, and obeys the Mohr-Coulomb failure criterion. The major principal stress is

🚀 Solve in Practice Mode

Q23GATE 2010MCQ

If $\sigma _{h}, \sigma_{v}, \sigma' _{h} \; and \; \sigma' _{v}$ represent the total horizontal stress, total vertical stress, effective horizontal stress and effective vertical stress on a soil element, respectively, the coefficient of earth pressure at rest is given by

🚀 Solve in Practice Mode

📖 Explanation

Coefficient of earth pressure at rest is the ratio of intensity of earth pressure at rest to the vertical stress at a specified depth. $K_{0}=\frac{\sigma_{h}}{\sigma_{v}}$ But for a saturated soil, $K_{0}=\frac{\alpha_{h}^{\prime}}{\sigma_{v}^{\prime}}$

Q24GATE 2008MCQ

When a retaining wall moves away from the backfill, the pressure exerted on the wall is termed as

🚀 Solve in Practice Mode

Q25GATE 2006MCQ

Figure given below shows a smooth vertical gravity retaining wall with cohesionless soil backfill having an angle of internal friction $\phi$ . In the graphical representation of Rankine's active earth pressure for the retaining wall shown in figure, length OP represents

🚀 Solve in Practice Mode

📖 Explanation

\begin{array}{l} O P=\text {Major stress}=\sigma_{1} \\ O Q=\text {Mirror stress}=\sigma_{3} \end{array}

Active condition: OP = Vertical stress at base $=\sigma_{v} \cos \beta=\gamma+\cos \beta$ OQ = Lateral earth pressure at base $=\mathrm{K}_{\mathrm{a}} \cdot \sigma_{\mathrm{v}} \cos \beta$ Note: QR and QR' are failure plane for active case.

Q26GATE 2005MCQ

A 3 m high retaining wall is supporting a saturated sand (saturated due to capillary action) of bulk density 18 $kN/m^{3}$ and angle of shearing resistance 30 $^{\circ}$ . The change in magnitude of active earth pressure at the base due to rise in ground water table from the base of the footing to the ground surface shall ( $\gamma _{w}=10 kN/m^{3}$ )

🚀 Solve in Practice Mode

📖 Explanation

When the water tabel is at the base of footing the active earth pressure will be

\begin{aligned} P_{1} &=K_{a} \gamma R=\left(\frac{1-\sin \phi}{1+\sin \phi}\right) \times \gamma \times z \\ &=\frac{1-\sin 30^{\circ}}{1+\sin 30^{\circ}} \times 18 \times 3 \\ &=1 / 3 \times 18 \times 3=18 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

When the water table rises upto the surface, the thrust due to water will also be added to the active earth pressure.

\begin{aligned} P_{2} &=K_{2} \gamma z+\gamma_{w} z \\ &=\frac{1}{3} \times (18-10) \times 3+10 \times 3 \\ &=8+30=38 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

$\therefore$ change in active earth pressure

\begin{array}{l} =P_{2}-P_{1}=38-18 \\ =20 \mathrm{kN} / \mathrm{m}^{2}(\text { increase }) \end{array}

Note: In strict terms $\gamma_{\text {sat }}$ should be taker

Q27GATE 2004MCQ

Using $\phi _{u}=0$ analysis and assuming planar failure as shown, the minimum factor of safety against shear failure of a vertical cut of height 4 m in a pure clay having $c_{u}=120 kN/m^{2}$ and $\gamma _{sat}=20 kN/m^{3}$ is

🚀 Solve in Practice Mode

📖 Explanation

The maximum height of unsupported cut $=\frac{4 c}{\gamma \sqrt{K_{A}}}$ For $\phi_{u}=0 ; K_{A}=\frac{1-\sin 0^{\circ}}{1+\sin 0^{\circ}}=1.0$ $\therefore$ Maximum height of unsupported cu $=\frac{4 \times 120}{20}=24 \mathrm{m}$ Actual height of cut =4m Factor of safety $=\frac{24}{4}=6$

Q28GATE 2004MCQ

An unsupported excavation is made to the maximum possible depth in a clay soil having $\gamma _{t}=18 kN/m^{3}$ , $c=100 kN/m^{2}$ , $\phi =30^{\circ}$ . The active earth pressure, according to Rankine's theory, at the base level of the excavation is

🚀 Solve in Practice Mode

📖 Explanation

The excavation is unsupported which means the maximum possible depth upto which the clay soil can be excavated is $2Z_{\mathrm{c}}$ . The active earth pressure at this depth is

\begin{aligned} P_{A} &=2 c \sqrt{K_{A}} \quad\left[\because K_{A}=\frac{1-\sin \phi}{1+\sin \phi}\right] \\ &=2 \times 100 \times \sqrt{\frac{1-\sin 30^{\circ}}{1+\sin 30^{\circ}}} \\ &=115.47 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Alternatively: $Z_{c}=\frac{2 c}{\gamma \sqrt{K_{A}}} \Rightarrow 2 Z_{c}=\frac{4 c}{\gamma \sqrt{K_{A}}}$ $\therefore$ Active pressure at a depth, $2Z_{c}$

\begin{array}{l} =\gamma_{1} \times 2 Z_{c} \times K_{A}-2 c \sqrt{K_{A}} \\ =18 \times \frac{4 \times 100 \times \sqrt{3}}{18} \times \frac{1}{3} \\ \quad-2 \times 100 \times \frac{1}{\sqrt{3}} \\ =230.94-115.47=115.47 \mathrm{kN} / \mathrm{m}^{2} \end{array}

Q29GATE 2004MCQ

A retaining wall of height 8 m retains dry sand. In the initial state, the soil is loose and has a void ratio of 0.5, $\gamma _{d}=17.8 kN/m^{3} \; and \; \phi =30^{\circ}$ . Subsequently, the backfill is compacted to a state where void ratio is 0.4, $\gamma _{d}=18.8 kN/m^{3} \; and \; \phi =35^{\circ}$ . The ratio of initial passive thrust to the final passive thrust, according to Rankine's earth pressure theory, is

🚀 Solve in Practice Mode

📖 Explanation

Passive earth pressure for loose sand is

\begin{aligned} K_{p} &=\frac{1+\sin \phi}{1-\sin \phi}=\frac{1+\sin 30^{\circ}}{1-\sin 30^{\circ}}=3 \\ \text { Now } P_{P_{1}} &=\frac{1}{2} K_{P} \gamma H^{2}=\frac{1}{2} \times 3 \times 17.8 \times 8^{2} \\ &=1708.8 \mathrm{kN} / \mathrm{m} \end{aligned}

Passive earth pressure for compacted sand is

\begin{aligned} K_{P}&=\frac{1+\sin \phi}{1-\sin \phi}=\frac{1+\sin 35^{\circ}}{1-\sin 35^{\circ}}=3.69\\ \text { Now } P_{B} &=\frac{1}{2} K_{p} \gamma H^{2} \\ &=\frac{1}{2} \times 3.69 \times 18.8 \times 8^{2} \\ &=2219.90 \mathrm{kN} / \mathrm{m} \end{aligned}

Ratio of $P_{P_{1}}$ and $P_{P_{2}}=\frac{1708.8}{2219.90}=0.77$

Q30GATE 2003MCQ

A braced cut, 5 m wide and 7.5 m deep is proposed in a cohesionless soil deposit having effective cohesion c'=0 and effective friction angle, ${\phi}' =36^{\circ}$ . The first row of struts is to be installed at a depth of 0.5 m below ground surface and spacing between the struts should be 1.5 m. If the horizontal spacing of struts is 3 m and unit weight of the deposit is 20 $kN/m^{3}$ , the maximum strut load will be

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} K_{a} &=\frac{1-\sin \phi^{\prime}}{1+\sin \phi^{\prime}} \\ &=\frac{1-\sin 36^{\circ}}{1+\sin 36^{\circ}}=0.2596 \end{aligned}

$\therefore$ Uniform pressure on side of cut

\begin{aligned} &=0.65 \mathrm{K}_{\mathrm{a}} \gamma \mathrm{H} \\ &=0.65 \times 0.2596 \times 20 \times 7.5 \\ &=25.31 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Area covered by strut

\begin{aligned} 1 &=3 \times (0.5+0.75) \\ &=3.75 \mathrm{m}^{2} \end{aligned}

Area covered by strut 2 and downward struts $=3 \times (0.75+0.75)=4.5 \mathrm{m}^{2}$ $\therefore$ Maximum force will be carried by strut 2 and downward struts

\begin{aligned} &=25.31 \times 4.5=113.89 \mathrm{kN} \\ &\simeq 113.90 \mathrm{kN} \end{aligned}