# Consolidation of Soils

 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
 A 0.35 B 0.78 C 0.87 D 1.28
GATE CE 2023 SET-1   Geotechnical Engineering
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}
 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) A 25 B 30 C 50 D 80
GATE CE 2022 SET-2   Geotechnical Engineering
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.
$=\frac{175-125}{225-125} \times 100=\frac{50}{100} \times 100=50$

 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)
 A 25 B 50 C 75 D 85
GATE CE 2022 SET-2   Geotechnical Engineering
Question 3 Explanation:
$\sigma _3=\sigma _c=100kPa , \sigma _d=\sigma _1-\sigma _3=150kPa$
For UU test Undrained shear strenth
$C_{uu}=\tau _f=\frac{\sigma _1-\sigma _3}{2}=\frac{150}{2}=75KPa$
 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)
 A 3525 B 1254 C 1736 D 2463
GATE CE 2022 SET-1   Geotechnical Engineering
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 ______________
 A 0.124 B 0.095 C 0.002 D 0.652
GATE CE 2021 SET-2   Geotechnical Engineering
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.

### 3 thoughts on “Consolidation of Soils”

1. In question 26 , there is no mention of void ratio and specific weight of clay .
From where does the value appear in the solution. Please explain

2. • 