# Geotechnical Engineering

 Question 1
A consolidated drained (CD) triaxial test was carried out on a sand sample with the known effective shear strength parameters, $C^{\prime}=0$ and $\phi^{\prime}=30^{\circ}$. In the test, prior to the failure, when the sample was undergoing axial compression under constant cell pressure, the drainage valve was accidentally closed. At the failure, $360 \mathrm{kPa}$ deviatoric stress was recorded along with $70 \mathrm{kPa}$ pore water pressure. If the test is repeated without such error, and no back pressure is applied in either of the tests, what is the deviatoric stress (in $\mathrm{kPa}$, in integer) at the failure? _____
 A 11 B 200 C 350 D 500
GATE CE 2023 SET-2      Shear Strength of Soil
Question 1 Explanation:
We know that,
$\bar{\sigma}_{1 F}=\bar{\sigma}_{3 \mathrm{~F}} \tan ^{2}\left(45+\frac{\phi^{\prime}}{2}\right)+20 \tan \left(45+\frac{\phi^{\prime}}{2}\right)$
Assuming $c^{\prime}=0$ under both the conditions when drainage line in suddenly dosed.
$\left(\sigma_{\mathrm{cf}}+360-70\right)=\left(\sigma_{\mathrm{cy}}-70\right) \tan ^{2} 60+0\;\;\;...(i)$
When drainage is open and assuming the confining stress $\left(\sigma_{\mathrm{cf}}\right)$.
$\left[\left(\sigma_{\mathrm{cf}}+\sigma_{2}\right)-0\right]=\left[\sigma_{\mathrm{cf}}-0\right] \tan ^{2} 60^{\circ}+0 \;\;\;...(ii)$
From (i) and (ii), we get
\begin{aligned} 290-\sigma_{2} & =-70 \times \tan ^{2} 60^{\circ} \\ 290-\sigma_{d} & =-210 \\ \sigma_{d} &=500 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}
 Question 2
For the flow setup shown in the figure (not to scale), the hydraulic conductivities of the two soil samples, Soil 1 and Soil 2, are $10 \mathrm{~mm} / \mathrm{s}$ and $1 \mathrm{~mm} / \mathrm{s}$, respectively. Assume the unit weight of water as $10 \mathrm{kN} / \mathrm{m}^{3}$ and ignore the velocity head. At steady state, what is the total head (in $m$, rounded off to two decimal places) at any point located at the junction of the two samples? ____

 A 3.25 B 4.54 C 6.35 D 2.24
GATE CE 2023 SET-2      Seepage Analysis
Question 2 Explanation:

External pressure of $10 \mathrm{kPa}$ can be converted to equivalent hydrostatic, head $=\frac{10 \mathrm{kPa}}{10 \mathrm{kN} / \mathrm{m}^{3}}=1 \mathrm{~m}$
Total head at $A$,
$\mathrm{TH}_{\mathrm{A}}=$ Datum head + Pressure head
$\mathrm{TH}_{\mathrm{A}}=3 \mathrm{~m}+(1 \mathrm{~m}+1 \mathrm{~m})$
$\mathrm{TH}_{\mathrm{A}}=5 \mathrm{~m}$
Total head loss due to seepage through soil
\begin{aligned} & \mathrm{h}_{\mathrm{L}}=5 \mathrm{~m} \\ & Q=K_{e q} \mathrm{i} A\\ & =\left(\frac{L_1+L_2}{\frac{L_1}{K_1}+\frac{L_2}{K_2}}\right) \times \frac{h_L}{L} \times \mathrm{A} \\ & =\left(\frac{1+1}{\frac{1}{10}+\frac{1}{1}}\right) \times \frac{5}{2} \times \mathrm{A} \\ & =4.545 \mathrm{~A} \end{aligned}

Head loss through soil - 1 only
\begin{aligned} \mathrm{h}_{L_{1}} & =\frac{\mathrm{Q}}{\mathrm{K}_{1} \mathrm{~A}} \times \mathrm{L}_{1} \\ & =\frac{4.545 \mathrm{~A}}{10 \times \mathrm{A}} \times 1 \\ & =0.4545 \mathrm{~m} \end{aligned}
$\therefore$ Total head at junction $=\mathrm{TH}_{\mathrm{A}}-\mathrm{h}_{\mathrm{L} 1}$ $=5-0.4545=4.545 \mathrm{~m} \approx 4.54 \mathrm{~m}$

 Question 3
A circular pile of diameter $0.6 \mathrm{~m}$ and length $8 \mathrm{~m}$ was constructed in a cohesive soil stratum having the following properties: bulk unit weight $=$ $19 \mathrm{kN} / \mathrm{m}^{3}$; angle of internal friction $=0^{\circ}$ and cohesion $=25 \mathrm{kPa}$.
The allowable load the pile can carry with a factor of safety of 3 is ____ $\mathrm{kN}$ (round off to one decimal place).
[Adopt: Adhesion factor, $\alpha=1.0$ and Bearing capacity factor, $\left.\mathrm{N}_{\mathrm{C}}=9.0\right]$
 A 112.5 B 321.5 C 125.6 D 146.9
GATE CE 2023 SET-2      Deep Foundation
Question 3 Explanation:
End bearing capacity
$\mathrm{Q}_{\mathrm{pu}}=\mathrm{C}_{\mathrm{u}} \mathrm{N}_{\mathrm{c}} \mathrm{A}_{\mathrm{b}}$
where, $\quad C_{u}=25 \mathrm{kPa}$
$N_{c}=9$
$A_{b}=\frac{\pi}{4} \times 0.6^{2}=0.283 \mathrm{~m}^{2}$
$\therefore \quad \mathrm{Q}_{\mathrm{Pu}}=25 \times 9 \times 0.283$ $=63.62 \mathrm{kN}$

Skin friction capacity
$\mathrm{Q}_{\mathrm{s}}=\alpha \mathrm{C}_{\mathrm{u}} \mathrm{A}_{\mathrm{s}}$
Where,
\begin{aligned} \alpha & =1.0 \\ \mathrm{C}_{\mathrm{u}} & =25 \mathrm{kPa} \\ \mathrm{A}_{\mathrm{s}} & =\pi \mathrm{DL}=\pi \times 0.6 \times 8 \\ & =15.08 \mathrm{~m}^{2} \\ \therefore \quad Q_{\mathrm{s}} & =1 \times 25 \times 15.08=377 \mathrm{kN} \end{aligned}
$\therefore$ Ultimate pile capacity $=63.62+377$ $=440.62 \mathrm{kN}$
Allowable load $=\frac{440.62}{3}=146.87 \mathrm{kN}$
 Question 4
A square footing is to be designed to carry a column load of $500 \mathrm{kN}$ which is resting on a soil stratum having the following average properties: bulk unit weight $=19 \mathrm{kN} / \mathrm{m}^{3}$; angle of internal friction $=0^{\circ}$, and cohesion $=25 \mathrm{kPa}$. Considering the depth of the footing as $1 \mathrm{~m}$ and adopting Meyerhof's bearing capacity theory with a factor of safety of 3 , the width of the footing (in $\mathrm{m}$ ) is ____ (round off to one decimal place)
[Assume the applicable shape and depth factor values as unity; ground water level at greater depth.]
 A 1.2 B 2.4 C 3.4 D 4.5
GATE CE 2023 SET-2      Shallow Foundation and Bearing Capacity
Question 4 Explanation:
As per Meyerhoff's approach
$q_{u}=C N_{c}\left[i_{c} S_{c} d_{c}\right]+q N_{q}\left[{ }_{q} S_{q} d_{r}\right]+\frac{1}{2} \beta \gamma N_{r}\left[i_{r} S_{r} d_{r}\right]$
Given; shape factor $S_{c}, S_{q}, S_{r}=1$
depth factor $d_{c}, d_{v}, d_{r}=1$

$i_{c} i_{q} i_{r}=1$
For $\phi=0^{\circ}$
\begin{aligned} & \mathrm{N}_{\mathrm{c}}=5.14 \\ & N_{\mathrm{q}}=1 \\ & N_{r}=0 \\ & \therefore \quad \mathrm{q}_{\mathrm{u}}=25 \times 5.14 \times 1 \times 1 \times 1+19 \times 1 \times \\ & 1 \times 1 \times 1 \\ & =147.5 \mathrm{kPa} \\ & \mathrm{q}_{\mathrm{nu}}=\mathrm{q}_{\mathrm{u}}-\mathrm{rD}_{\mathrm{f}} \\ & =147.5-19 \times 1 \\ & =128.5 \mathrm{kPa} \end{aligned}
$q_{n s}=\frac{q_{n u}}{F O S}=\frac{128.5}{3}=42.83 \mathrm{kPa}$
Hence, $\frac{500}{B^{2}} \leq 42.83$
or $B \geq 3.416 m$
Hence $B=3.4 m$
 Question 5
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 A B B C C D D
GATE CE 2023 SET-2      Retaining Wall-Earth Pressure Theories
Question 5 Explanation:

There are 5 questions to complete.

### 1 thought on “Geotechnical Engineering”

1. Question No. 5 refers to Steel Structure