Question 1 |

The specific gravity of a soil is 2.60 . The soil is at 50 \% degree of saturation with a water content of 15 \%. The void ratio of the soil is

0.35 | |

0.78 | |

0.87 | |

1.28 |

Question 1 Explanation:

\begin{aligned}
\mathrm{G}_{\mathrm{s}} & =2.60 \\
\mathrm{~S} & =50 \% \\
\mathrm{w} & =15 \%
\end{aligned}

We know,

\begin{aligned} \mathrm{e}_{\mathrm{s}} & =\omega \mathrm{G}_{\mathrm{s}} \\ \mathrm{e} \times 0.5 & =0.15 \times 2.60 \\ \mathrm{e} & =\frac{0.15 \times 2.60}{0.5} \\ \mathrm{e} & =0.78 \end{aligned}

We know,

\begin{aligned} \mathrm{e}_{\mathrm{s}} & =\omega \mathrm{G}_{\mathrm{s}} \\ \mathrm{e} \times 0.5 & =0.15 \times 2.60 \\ \mathrm{e} & =\frac{0.15 \times 2.60}{0.5} \\ \mathrm{e} & =0.78 \end{aligned}

Question 2 |

A saturated compressible clay layer of thickness h is sandwiched between two
sand layers, as shown in the figure. Initially, the total vertical stress and pore
water pressure at point P, which is located at the mid-depth of the clay layer,
were 150 kPa and 25 kPa, respectively. Construction of a building caused an
additional total vertical stress of 100 kPa at P. When the vertical effective stress
at P is 175 kPa, the percentage of consolidation in the clay layer at P is
_____________. (in integer)

25 | |

30 | |

50 | |

80 |

Question 2 Explanation:

Initial effective stress at P

\sigma _{P_1}'=150-25=125kPa

After construction,

\sigma _{P_2}'=125+100=225kPa

100% consolidation will occur when effective stress at P reaches 225 kPa.

Percentage of consolidation when vertical effective stress is at P is 175 kPa.

=\frac{175-125}{225-125} \times 100=\frac{50}{100} \times 100=50

\sigma _{P_1}'=150-25=125kPa

After construction,

\sigma _{P_2}'=125+100=225kPa

100% consolidation will occur when effective stress at P reaches 225 kPa.

Percentage of consolidation when vertical effective stress is at P is 175 kPa.

=\frac{175-125}{225-125} \times 100=\frac{50}{100} \times 100=50

Question 3 |

In a triaxial unconsolidated undrained (UU) test on a saturated clay sample, the
cell pressure was 100 kPa. If the deviatoric stress at failure was 150 kPa, then
the undrained shear strength of the soil is _________ kPa. (in integer)

25 | |

50 | |

75 | |

85 |

Question 3 Explanation:

\sigma _3=\sigma _c=100kPa , \sigma _d=\sigma _1-\sigma _3=150kPa

For UU test

Undrained shear strenth

C_{uu}=\tau _f=\frac{\sigma _1-\sigma _3}{2}=\frac{150}{2}=75KPa

For UU test

Undrained shear strenth

C_{uu}=\tau _f=\frac{\sigma _1-\sigma _3}{2}=\frac{150}{2}=75KPa

Question 4 |

A raft foundation of 30 m x 25 m is proposed to be constructed at a depth of 8 m
in a sand layer. A 25 m thick saturated clay layer exists 2 m below the base of
the raft foundation. Below the clay layer, a dense sand layer exists at the site.
A 25 mm thick undisturbed sample was collected from the mid-depth of the clay
layer and tested in a laboratory oedometer under double drainage condition. It
was found that the soil sample had undergone 50% consolidation settlement in
10 minutes.

The time (in days) required for 25% consolidation settlement of the raft foundation will be ______. (round off to the nearest integer)

The time (in days) required for 25% consolidation settlement of the raft foundation will be ______. (round off to the nearest integer)

3525 | |

1254 | |

1736 | |

2463 |

Question 4 Explanation:

\begin{aligned} (T_v)_{50}&=C_v\frac{t}{d^2} \;\;\; (\text{from lab})\\ \frac{\pi}{4}(0.5)^2&=C_v \times \frac{10 min}{\left ( \frac{25}{2} \times 10^{-3} \right )^2}\\ (T_v)_{25}&=C_v\frac{t}{d^2} \;\;\; (\text{from field})\\ \frac{\pi}{4}(0.25)^2&=C_v \frac{t}{(12.5)^2}\\ \end{aligned}

Since the soil is same -> C_v same

\begin{aligned} \frac{\pi}{4}(0.25)^2&= \frac{\frac{\pi}{4}(0.5)^2 \times (12.5 \times 10^{-3})^2}{10} \times \frac{t}{(12.5)^2}\\ t&=1736 \;days \end{aligned}

Question 5 |

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 5 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}

There are 5 questions to complete.