Question 1 |

The figure shows a vertical retaining wall with backfill consisting of cohesive-frictional soil and a failure plane developed due to passive earth pressure. The forces acting on the failure wedge are: P as the reaction force between the wall and the soil, R as the reaction force on the failure plane, C as the cohesive force along the failure plane and W as the weight of the failure wedge. Assuming that there is no adhesion between the wall and the wedge, identify the most appropriate force polygon for the wedge.

A | |

B | |

C | |

D |

Question 1 Explanation:

Question 2 |

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.

\sqrt{\frac{k_{H}}{k_{V}}} | |

\sqrt{\frac{k_{V}}{k_{H}}} | |

1 | |

\sqrt{\frac{k_{H}}{k_{V}}} |

Question 2 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}}}.

\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}}}.

Question 3 |

A smooth vertical retaining wall supporting layered soils is shown in figure. According to Rankine's earth pressure theory, the lateral active earth pressure acting at the base of the wall is ____ \mathrm{kPa} (round off to one decimal place)

45.2 | |

35.4 | |

54.2 | |

56.2 |

Question 3 Explanation:

\mathrm{K}_{\mathrm{a} 2}=\frac{1-\sin 25^{\circ}}{1+\sin 25^{\circ}}

Active earth pressure at the base of the wall.

\begin{aligned} &(\mathrm{Pa})_{\text {Base }}=\mathrm{Ka}_{2}\left(\mathrm{q}+\gamma_{1} \mathrm{z}_{1}+\gamma_{2} \mathrm{z}_{2}\right)-2 \mathrm{C} \sqrt{\mathrm{Ka}_{2}} \\ &=0.405+8(20+18 \times 3+19 \times 4)-2 \times 20 \times \sqrt{0.4058} \\ &= 35.39 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Active earth pressure at the base of the wall.

\begin{aligned} &(\mathrm{Pa})_{\text {Base }}=\mathrm{Ka}_{2}\left(\mathrm{q}+\gamma_{1} \mathrm{z}_{1}+\gamma_{2} \mathrm{z}_{2}\right)-2 \mathrm{C} \sqrt{\mathrm{Ka}_{2}} \\ &=0.405+8(20+18 \times 3+19 \times 4)-2 \times 20 \times \sqrt{0.4058} \\ &= 35.39 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Question 4 |

As per Rankine's theory of earth pressure, the inclination of failure planes is (45+\frac{\phi }{2})^{\circ} with respect to the direction of the minor principal stress.

The above statement is correct for which one of the following options?

The above statement is correct for which one of the following options?

Only the active state and not the passive state | |

Only the passive state and not the active state | |

Both active as well as passive states | |

Neither active nor passive state |

Question 4 Explanation:

Question 5 |

A concentrated vertical load of 3000 kN is applied on a horizontal ground
surface. Points P and Q are at depths 1 m and 2 m below the ground,
respectively, along the line of application of the load. Considering the ground to
be a linearly elastic, isotropic, semi-infinite medium, the ratio of the increase in
vertical stress at P to the increase in vertical stress at Q is ________. (in integer)

12 | |

36 | |

8 | |

4 |

Question 5 Explanation:

Increase in vertical stress,

\begin{aligned} \sigma _z&=\frac{KQ}{z^2}\\ \sigma _z&\propto \frac{1}{z^2}\\ \frac{\sigma _P}{\sigma _Q}&=\left ( \frac{z_Q}{z_P} \right )^2\\ &=\left ( \frac{2}{1} \right )^2=4 \end{aligned}

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