Question 1 |

In an aggregate mix, the proportions of coarse aggregate, fine aggregate and mineral filler are 55%, 40% and 5%, respectively. The values of bulk specific gravity of the coarse aggregate, fine aggregate and mineral filler are 2.55, 2.65 and 2.70, respectively. The bulk specific gravity of the aggregate mix (round off to two decimal places) is ____________

2.6 | |

3.21 | |

5.64 | |

6.25 |

Question 1 Explanation:

G_{m}=\frac{55+40+5}{\frac{55}{2.55}+\frac{40}{2.65}+\frac{5}{2.70}}=2.596

Question 2 |

The void ratio of a clay soil sample M decreased from 0.575 to 0.510 when the applied pressure is increased from 120 kPa to 180 kPa. For the same increment in pressure, the void ratio of another clay soil sample N decreases from 0.600 to 0.550. If the ratio of hydraulic conductivity of sample M to sample N is 0.125, then the ratio of coefficient of consolidation of sample M to sample N (round off to three decimal places) is ______________

0.124 | |

0.095 | |

0.002 | |

0.652 |

Question 2 Explanation:

\begin{aligned} m_{v}&=\frac{a_{V}}{1+e_{0}}=\frac{\Delta e}{\left(1+e_{0}\right) \times(\Delta \bar{\sigma})} & (\Delta \bar{\sigma} \text { is same for both } M \text { and } N)\\ m_{v 1}&=\frac{0.575-0.510}{(1+0.575) \times \Delta \bar{\sigma}}\\ m_{v 2}&=\frac{0.600-0.550}{(1+0.600) \times \Delta \bar{\sigma}}\\ \frac{C_{v_{1}}}{C_{V_{2}}} &=\frac{\frac{k_{1}}{m_{v_{1}} \gamma_{w}}}{\frac{k_{2}}{m_{v_{2}} \gamma_{w}}}=\frac{k_{1}}{k_{2}} \times \frac{m_{v_{2}}}{m_{v_{1}}} \\ &=0.125\left(\frac{1.575}{1.6}\right) \times\left(\frac{0.60-0.55}{0.575-0.510}\right) \\ &=0.0947 \simeq 0.095 \end{aligned}

Question 3 |

The soil profile at a road construction site is as shown in figure (not to scale). A large embankment is to be constructed at the site. The ground water table (GWT) is located at the surface of the clay layer, and the capillary rise in the sandy soil is negligible. The effective stress at the middle of the clay layer after the application of the embankment loading is 180 \mathrm{kN} / \mathrm{m}^{2}. Take unit weight of water, \gamma_{w}=9.81 \mathrm{kN} / \mathrm{m}^{3}.

The primary consolidation settlement (in m, round off to two decimal places) of the clay layer resulting from this loading will be ______

The primary consolidation settlement (in m, round off to two decimal places) of the clay layer resulting from this loading will be ______

0.12 | |

0.88 | |

0.45 | |

0.33 |

Question 3 Explanation:

Primary consolidation settlement

\Delta H=\frac{C_{c} H}{1+e_{0}} \log _{10}\left(\frac{\bar{\sigma}_{0}+\Delta \bar{\sigma}}{\bar{\sigma}_{0}}\right)

\bar{\sigma}_{0}+\Delta \bar{\sigma}=Effective stress at the center of clay layer after embankment loading

=180 \mathrm{kN} / \mathrm{m}^{2}

\bar{\sigma}_{0}= Effective stress at the centre of clay layer before embankment loading

\begin{aligned} \gamma_{\text {sub }} \text { of clay layer } &=\frac{G-1}{1+e} \gamma_{w}=\frac{(2.65-1)}{1+\frac{w G}{1}} \times 9.81 \\ &=\frac{1.65 \times 9.81}{1+0.45 \times 2.65}=7.383 \mathrm{kN} / \mathrm{m}^{3} \\ \bar{\sigma}_{0} &=(18.5 \times 2)+(7.383) \times 3=59.149 \mathrm{kNm} /{ } \\ e_{0} &=\frac{w G}{1}=0.45 \times 2.65=1.1925 \\ \Rightarrow \qquad \qquad \qquad \qquad \qquad & \Delta \mathrm{H}=\frac{0.25 \times 6}{1+1.1925} \log _{10}\left(\frac{180}{59.149}\right)=0.33 \mathrm{~m} \end{aligned}

\Delta H=\frac{C_{c} H}{1+e_{0}} \log _{10}\left(\frac{\bar{\sigma}_{0}+\Delta \bar{\sigma}}{\bar{\sigma}_{0}}\right)

\bar{\sigma}_{0}+\Delta \bar{\sigma}=Effective stress at the center of clay layer after embankment loading

=180 \mathrm{kN} / \mathrm{m}^{2}

\bar{\sigma}_{0}= Effective stress at the centre of clay layer before embankment loading

\begin{aligned} \gamma_{\text {sub }} \text { of clay layer } &=\frac{G-1}{1+e} \gamma_{w}=\frac{(2.65-1)}{1+\frac{w G}{1}} \times 9.81 \\ &=\frac{1.65 \times 9.81}{1+0.45 \times 2.65}=7.383 \mathrm{kN} / \mathrm{m}^{3} \\ \bar{\sigma}_{0} &=(18.5 \times 2)+(7.383) \times 3=59.149 \mathrm{kNm} /{ } \\ e_{0} &=\frac{w G}{1}=0.45 \times 2.65=1.1925 \\ \Rightarrow \qquad \qquad \qquad \qquad \qquad & \Delta \mathrm{H}=\frac{0.25 \times 6}{1+1.1925} \log _{10}\left(\frac{180}{59.149}\right)=0.33 \mathrm{~m} \end{aligned}

Question 4 |

A rectangular footing of size 2.8m x 3.5m is embedded in a clay layer and a vertical load is placed with an eccentricity of 0.8 m as shown in the figure (not to scale). Take Bearing capacity factors: N_{c}=5.14,N_{q}=1.0, and N_{\gamma}=0.0; Shape factors: s_{c}=1.16,s_{q}=1.0 and s_{\gamma}=1.0; Depth factors: d_{c}=1.1,d_{q}=1.0 and d_{\gamma}=1.0; and Inclination factors: i_{c}=1.0 and i_{q}=1.0 and i_{\gamma}=1.0.

Using Meyerhoff's method, the load (in kN, round off to two decimal places) that can be applied on the footing with a factor of safety of 2.5 is _________________

Using Meyerhoff's method, the load (in kN, round off to two decimal places) that can be applied on the footing with a factor of safety of 2.5 is _________________

441 | |

564 | |

876 | |

689 |

Question 4 Explanation:

\begin{aligned}& \text { Given data: } 2.8 \times 3.56, e=0.8, D_{f}=1.5 \mathrm{~m} \\ N_{c}&=5.14, \qquad N_{q}=1, \qquad N_{r}=0 \\ S_{c}&=1.16, \qquad S_{q}=1, \qquad S_{r}=1.0 \\ d_{c}&=1.1, \qquad d_{q}=1, \qquad d_{f}=1.0 \\ i_{c}&=1, \qquad \quad i_{q}=1, \qquad i_{r}=1.0 \\ \gamma&=18.2, \qquad c=40 \mathrm{kN} / \mathrm{m}^{2} \\ B^{\prime}&=B-2 e_{x}=2.8-2(0.8)=1.2 \mathrm{~m} \\ L^{\prime}&=L=3.5 \end{aligned}

Ultimate bearing capacity

\begin{aligned} q_{u} &=C N_{c} S_{c} d_{c} i_{c}+\gamma D_{p} N_{q} S_{q} d_{q} i_{q}+0.5 B^{\prime} \gamma N_{r} S_{r} d_{r} \\ q_{u} &=40 \times 5.14 \times 1.16 \times 1.1+18.2 \times 1.5 \times 1+0 \\ q_{n u} &=q_{u}-\gamma D_{f}=262.346 \mathrm{kN} / \mathrm{m}^{2} \\ q_{n s} &=\frac{q_{n u}}{F}=104.938 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

\begin{aligned} \text { Net safe load } &=q_{n s} \times A=\frac{q_{n u}}{F} \times\left(B^{\prime} \times L\right) \\ &=104.938 \times 1.2 \times 3.5 \mathrm{kN} \end{aligned}

Load applied on the footing = 440.740 kN

Ultimate bearing capacity

\begin{aligned} q_{u} &=C N_{c} S_{c} d_{c} i_{c}+\gamma D_{p} N_{q} S_{q} d_{q} i_{q}+0.5 B^{\prime} \gamma N_{r} S_{r} d_{r} \\ q_{u} &=40 \times 5.14 \times 1.16 \times 1.1+18.2 \times 1.5 \times 1+0 \\ q_{n u} &=q_{u}-\gamma D_{f}=262.346 \mathrm{kN} / \mathrm{m}^{2} \\ q_{n s} &=\frac{q_{n u}}{F}=104.938 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

\begin{aligned} \text { Net safe load } &=q_{n s} \times A=\frac{q_{n u}}{F} \times\left(B^{\prime} \times L\right) \\ &=104.938 \times 1.2 \times 3.5 \mathrm{kN} \end{aligned}

Load applied on the footing = 440.740 kN

Question 5 |

An elevated cylindrical water storage tank is shown in the figure. The tank has inner diameter of 1.5 m. It is supported on a solid steel circular column of diameter 75 mm and total height (L) of 4m.Take, water density=1000 \mathrm{~kg} / \mathrm{m}^{3} and acceleration due to gravity = 10 \mathrm{m/s}^{2}.

If elastic modulus (E) of steel is 200 GPa, ignoring self-weight of the tank, for the supporting steel column to remain unbuckled, the maximum depth (h) of the water permissible (in m, round off to one decimal place) is _____________

If elastic modulus (E) of steel is 200 GPa, ignoring self-weight of the tank, for the supporting steel column to remain unbuckled, the maximum depth (h) of the water permissible (in m, round off to one decimal place) is _____________

2.2 | |

2.7 | |

1.5 | |

3.4 |

Question 5 Explanation:

\begin{aligned} P_{cr}&=\frac{\pi ^2 EI}{L^2} =\text{Buckling load }=\text{weight of water} \\ \frac{\pi ^2 EI}{(2L)^2}&=(\rho gh) \times \frac{\pi}{4}d^2 =\rho g \times \text{volume of tank}\\ &\frac{\pi ^2 \times 200 \times 10^9 \times \pi (75 \times 10^{-3})^4}{(2 \times 4)^2 \times 64}=1000 \times 10 \times h \times \frac{\pi}{4} \times 1.5^2\\ h&=2.7\; m \end{aligned}

Question 6 |

From laboratory investigations, the liquid limit, plastic limit, natural moisture content and flow index of a soil specimen are obtained as 60%, 27%, 32% and 27%, respectively. The corresponding toughness index and liquidity index of the soil specimen, respectively, are

0.15 and 1.22 | |

0.19 and 6.60 | |

1.22 and 0.15 | |

6.60 and 0.19 |

Question 6 Explanation:

Flow index = 27

\begin{aligned} \text { Plasticity index, } \qquad I_{P}&=W_{L}-W_{P}=33 \\ \text { Toughness index, } \qquad I_{T}&=\frac{I_{P}}{I_{f}}=\frac{33}{27}=1.22 \\ \text { Liquidity index, } \qquad I_{L}&=\frac{W_{n}-W_{P}}{W_{L}-W_{P}}=0.151 \end{aligned}

\begin{aligned} \text { Plasticity index, } \qquad I_{P}&=W_{L}-W_{P}=33 \\ \text { Toughness index, } \qquad I_{T}&=\frac{I_{P}}{I_{f}}=\frac{33}{27}=1.22 \\ \text { Liquidity index, } \qquad I_{L}&=\frac{W_{n}-W_{P}}{W_{L}-W_{P}}=0.151 \end{aligned}

Question 7 |

A clay layer of thickness H has a preconsolidation pressure p_{c}
and an initial void ratio e_{0}.The initial effective overburden stress at the mid-height of the layer is p_{0}. At the same location, the increment in effective stress due to applied external load is \Delta p. The compression and swelling indices of the clay are C_{c}
and C_{s}, respectively. If p_{0} \lt p_{c} \lt \left(p_{0}+\Delta p\right), then the correct expression to estimate the consolidation settlement \left(S_{c}\right) of the clay layer is

s_{c}=\frac{H}{I+e_{0}}\left[C_{c} \log \frac{p_{c}}{p_{0}}+C_{s} \log \frac{p_{0}+\Delta p}{p_{c}}\right] | |

s_{c}=\frac{H}{I+e_{0}}\left[C_{s} \log \frac{p_{c}}{p_{0}}+C_{c} \log \frac{p_{0}+\Delta p}{p_{c}}\right] | |

s_{c}=\frac{H}{I+e_{0}}\left[C_{c} \log \frac{p_{0}}{p_{c}}+C_{s} \log \frac{p_{0}+\Delta p}{p_{c}}\right] | |

s_{c}=\frac{H}{I+e_{0}}\left[C_{s} \log \frac{p_{0}}{p_{c}}+C_{c} \log \frac{p_{0}+\Delta p}{p_{c}}\right] |

Question 7 Explanation:

As the soil is initially in over consolidate state (p_o \lt p_c), due to the external applied load (\Delta p) the soil initially undergoes re-compression upto (p_c) and then changes to virgin compression from (p_c) to (p_f)

The re-compression index (C_r) is also called Swelling index, (C_S).

s_{c}=\frac{H}{I+e_{0}} \left[C_{s} \log \frac{p_{c}}{p_{0}}+C_{c} \log \frac{p_{0}+\Delta p}{p_{c}}\right]

Question 8 |

As per the Unified Soil Classification System (USCS), the type of soil represented by 'MH' is

Inorganic silts of high plasticity with liquid limit more than 50% | |

Inorganic silts of low plasticity with liquid limit less than 50% | |

Inorganic clays of high plasticity with liquid limit less than 50% | |

Inorganic clays of low plasticity with liquid limit more than 50% |

Question 8 Explanation:

Question 9 |

The most appropriate triaxial test to assess the long-term stability of an excavated clay slope is

consolidated drained test | |

unconsolidated undrained test | |

consolidated undrained test | |

unconfined compression test |

Question 9 Explanation:

To assess the long term stability of clayey soil, the results of consolidated drained (CD) test are used.

Question 10 |

An unsupported slope of height 15 m is shown in the figure (not to scale), in which the slope face makes an angle 50^{\circ} with the horizontal. The slope material comprises purely cohesive soil having undrained cohesion 75 kPa. A trial slip circle KLM, with a radius 25 m, passes through the crest and toe of the slope and it subtends an angle 60^{\circ} at its center O. The weight of the active soil mass (W, bounded by KLMN) is 2500 kN/m, which is acting at a horizontal distance of 10 m from the toe of the slope. Consider the water table to be present at a very large depth from the ground surface.

Considering the trail slip circle KLM, the factor of the safety against the failure of slope under undrained condition (round off to two decimal places) is ___________

Considering the trail slip circle KLM, the factor of the safety against the failure of slope under undrained condition (round off to two decimal places) is ___________

1.96 | |

2.45 | |

8.25 | |

6.32 |

Question 10 Explanation:

\begin{aligned} &\begin{aligned} \mathrm{FOS} &=\frac{\text { Resisting moment }}{\text { Actuating moment }} \\ \mathrm{FOS} &=\frac{\mathrm{C}_{\mathrm{u}} l R}{w \bar{x}} \\ l &=\text { Length of ac } \mathrm{KLM} \\ \bar{x} &=\text { Distance of ' } \mathrm{w}^{\prime} \text { from toe } \\ \Rightarrow \qquad \qquad \quad \mathrm{FOS} &=\frac{75 \times 2 \pi \times 25 \times \frac{60}{36} \times 25}{2500 \times 10} \\ \Rightarrow \qquad \qquad \quad \mathrm{FOS} &=1.96 \end{aligned} \end{aligned}

There are 10 questions to complete.