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? ____
3.25 | |
4.54 | |
6.35 | |
2.24 |
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 _____________
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 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 20kN/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 _____
11.1 | |
15.3 | |
16.8 | |
8.9 |
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}
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
1.67 | |
2.5 | |
3.4 | |
4 |
Question 4 Explanation:
N_{f}=7 \quad N_{d}=24 \quad H=6 \mathrm{m}
Critical Hydraulic Gradient,
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}
Critical Hydraulic Gradient,
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________
1.1 | |
0.8 | |
1.6 | |
2.4 |
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
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.
Q1 is of SOM…. Kindly make appropriate corrections! Thanks for all of ur efforts!
Thanks for ur effort