# Seepage Analysis

 Question 1
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   Geotechnical Engineering
Question 1 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 2
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 _____________
 A 2.2 B 2.7 C 1.5 D 3.4
GATE CE 2021 SET-2   Geotechnical Engineering
Question 2 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 3
A granular soil has a saturated unit weight of 20$kN/m^3$ and an effective angle of shearing resistance of 30$^{\circ}$. The unit weight of water is 9.81 $kN/m^3$. A slope is to be made on this soil deposit in which the seepage occurs parallel to the slope up to the free surface. Under this seepage condition for a factor of safety of 1.5, the safe slope angle (in degree, round off to 1 decimal place) would be _____
 A 11.1 B 15.3 C 16.8 D 8.9
GATE CE 2019 SET-1   Geotechnical Engineering
Question 3 Explanation:
\begin{aligned} \gamma _{sat}&=20kN/m^3\\ \gamma _w&=9.81kN/m^3\\ \therefore \; \gamma '&=\gamma _{sat}-\gamma _w=10.19kN/m^3\\ \therefore \; \phi '&=30^{\circ},\; F=1.5 \end{aligned}
For infnite slope, the F.O. safety for seepage parallel to the slope in a granular soil (C = 0) is given below
\begin{aligned} F&=\frac{\gamma '}{\gamma _{sat}}\frac{\tan \phi '}{\tan i}\\ 1.5&=\frac{10.19}{20}\frac{\tan 30}{\tan i}\\ i&=11.095 \; \text{say}\; 11.1^{\circ} \end{aligned}
 Question 4
A flownet below a dam consists of 24 equipotential drops and 7 flow channels. The difference between the upstream and downstream water levels is 6 m. The length of the flow line adjacent to the toe of the dam at exit is 1 m. The specific gravity and void ratio of the soil below the dam are 2.70 and 0.70, respectively. The factor of safety against piping is
 A 1.67 B 2.5 C 3.4 D 4
GATE CE 2018 SET-2   Geotechnical Engineering
Question 4 Explanation:
$N_{f}=7 \quad N_{d}=24 \quad H=6 \mathrm{m}$
$i_{c}=\frac{G-1}{1+e}=\frac{2.7-1}{1+0.7}=1$
Exit Gradient $(i_{\text{exit}})$
\begin{aligned} &=\frac{\Delta h}{l}=\frac{\left(\frac{H}{N_{d}}\right)}{l}=\frac{\left(\frac{6}{24}\right)}{1 m}=\frac{1}{4} \\ \text { F.O.S. } &=\frac{i_{c}}{i_{\text {exit }}}=\frac{1}{\left(\frac{1}{4}\right)}=4 \end{aligned}
 Question 5
A sheet pile has an embedment depth of 12 m in a homogeneous soil stratum. The coefficient of permeability of soil is $10^{-6}\, m/s$. Difference in the water levels between the two sides of the sheet pile is 4 m. The flow net is constructed with five number of flow lines and eleven number of equipotential lines. The quantity of seepage (in $\frac{cm^{3}}{s}$ per m, up to one decimal place) under the sheet pile is________
 A 1.1 B 0.8 C 1.6 D 2.4
GATE CE 2017 SET-2   Geotechnical Engineering
Question 5 Explanation:
$N_{f}=$ No. of flow lines $-1=5-1=4$
$N_{d}=$ No. of equipotential lines $-1=11-1=10$
$H=4 \mathrm{m}$
Discharge,
$q=\frac{K H N_{f}}{N_{d}}=10^{-6} \times 4 \times \frac{4}{10}=16 \times 10^{-7} \mathrm{m}^{3} / \mathrm{sec} / \mathrm{m}$
$q=16 \times 10^{-7} \times 10^{6} \mathrm{cm}^{3} / \mathrm{sec} / \mathrm{m}$
$q=1.6 \mathrm{cm}^{3} / \mathrm{s}$ per meter length of dam

There are 5 questions to complete.

### 2 thoughts on “Seepage Analysis”

1. Q1 is of SOM…. Kindly make appropriate corrections! Thanks for all of ur efforts!

2. 