Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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
Question 5 |
The seepage occurring through an earthen dam is represented by a flow net comprising of 10 equipotential drops and 20 flow channels. The coefficient of permeability of the soil is 3 mm/min and the head loss is 5 m. The rate of seepage (expressed in cm^{3}/s per m length of the dam) through the earthen dam is ___.
500 | |
50 | |
55.5 | |
555 |
Question 5 Explanation:
\begin{aligned} q&=\frac{KH.N_{f}}{N_{d}}\\ K&=\frac{3 \mathrm{mm}}{\mathrm{min}}=\frac{3 \times 10^{-3} \mathrm{m}}{60} \mathrm{m} / \mathrm{s}\\ H&=5 \mathrm{m}, N_{\mathrm{f}}=20, \quad N_{\mathrm{d}}=10\\ q&=\frac{3 \times 10^{-3}}{60} \times 5 \times \frac{20}{10} \mathrm{m}^{3} / \mathrm{sec} \text { per } \mathrm{m} \text { length of dam }\\ q&=\frac{3 \times 10^{-3} \times 5 \times 20}{60 \times 10} \times 10^{6} \mathrm{cm}^{3} / \mathrm{sec} \text { per } \mathrm{m} \text { length of dam }\\ q&=500 \mathrm{cm}^{3} / \mathrm{sec} \text { per } \mathrm{m} \text { length of } \mathrm{dam} \end{aligned}
Question 6 |
The relationship between the specific gravity of sand (G) and the hydraulic gradient (i) to initiate quick condition in the sand layer having porosity of 30% is
G=0.7i+1 | |
G=1.43i-1 | |
G=1.43i+1 | |
G=0.7i-1 |
Question 6 Explanation:
\begin{array}{l} i_{c}=\frac{G-1}{1+e}=(G-1)(1-n) \quad \because \frac{1}{1+e}=1-n \\ i_{c}=(G-1)(1-0.3)=(G-1) \times 0.7 \\ G=\frac{i_{c}}{0.7}+1=1.43 i_{c}+1 \end{array}
Question 7 |
Which of the following statement is TRUE for the relation between discharge velocity and seepage velocity?
Seepage velocity is always smaller than discharge velocity | |
Seepage velocity can never be smaller than discharge velocity | |
Seepage velocity is equal to the discharge velocity | |
No relation between seepage velocity and discharge velocity can be established. |
Question 7 Explanation:
Seepage velocity is given by
V_{s}=\frac{V}{n}
where, V= discharge velocity
\begin{aligned} n&=\frac{V_{V}}{V}<1 \\ \text{Hence,}\quad V_{s}&>V \end{aligned}
V_{s}=\frac{V}{n}
where, V= discharge velocity
\begin{aligned} n&=\frac{V_{V}}{V}<1 \\ \text{Hence,}\quad V_{s}&>V \end{aligned}
Question 8 |
Water is flowing at a steady rate through a homogeneous and saturated horizontal soil strip of 10 m length. The strip is being subjected to a constant water head(H) of 5 m at the beginning and 1 m at the end. If the governing equation of flow in the soil strip is \frac{d^{2}H}{dx^2}=0 (where x is the distance along the soil strip), the value of H (in m) at the middle of the strip is ________.
3 | |
2 | |
0.5 | |
4 |
Question 8 Explanation:
Given, equation of the flow of soil strip is
\begin{aligned} \frac{d^{2} H}{d x^{2}}&=0 \\ \Rightarrow \quad \frac{\mathrm{d} \mathrm{H}}{\mathrm{d} x}&=\mathrm{C}_{1} \\ \Rightarrow \quad H&=C_{1} x+C_{2} \\ \text{at }x=0, H=&5 \mathrm{m} \\ \Rightarrow \quad C_{2}&=5 \\ \Rightarrow \quad H&=C_{1} x+5 \\ \text{at }x&=10, H=1 \mathrm{m} \\ \Rightarrow \quad 1&=C_{1} \times 10+5 \\ \Rightarrow \quad 10 C_{1}&=-4 \\ \Rightarrow \quad C_{1}&=-\frac{2}{5} \\ \Rightarrow \quad H&=-\frac{2}{5} x+5 \\ \text{at }x&=5 \mathrm{m} \\ H&=-\frac{2}{5} \times 5+5 \\ H&=3 \mathrm{m} \end{aligned}
\begin{aligned} \frac{d^{2} H}{d x^{2}}&=0 \\ \Rightarrow \quad \frac{\mathrm{d} \mathrm{H}}{\mathrm{d} x}&=\mathrm{C}_{1} \\ \Rightarrow \quad H&=C_{1} x+C_{2} \\ \text{at }x=0, H=&5 \mathrm{m} \\ \Rightarrow \quad C_{2}&=5 \\ \Rightarrow \quad H&=C_{1} x+5 \\ \text{at }x&=10, H=1 \mathrm{m} \\ \Rightarrow \quad 1&=C_{1} \times 10+5 \\ \Rightarrow \quad 10 C_{1}&=-4 \\ \Rightarrow \quad C_{1}&=-\frac{2}{5} \\ \Rightarrow \quad H&=-\frac{2}{5} x+5 \\ \text{at }x&=5 \mathrm{m} \\ H&=-\frac{2}{5} \times 5+5 \\ H&=3 \mathrm{m} \end{aligned}
Question 9 |
The flow net constructed for the dam is shown in the figure below. Taking the coefficient ofpermeability as 3.8\times 10^{-6}m/s, the quantity of flow (in cm^{3}/s) under the dam per meter of dam is______________
2.85 | |
6.25 | |
7.18 | |
9.24 |
Question 9 Explanation:
Quantity of flow,
\begin{aligned} Q&=\mathrm{kH} \frac{\mathrm{N}_{\mathrm{f}}}{\mathrm{N}_{\mathrm{d}}}\\ \text{Here, } \quad N_{f}&=\text{ No. of flow channels =3 }\\ N_{d}&=\text{ No. of equipotential}\\ \text { drops }=10 \\ \text{Given,}\quad k&=3.8 \times 10^{-6} \mathrm{m} / \mathrm{s}\\ \text{and }\quad H&=6.3 \mathrm{m} \\ Q &=3.8 \times 10^{-6} \times 6.3 \times \frac{3}{10} \times 1 \\ &=7.182 \times 10^{-6} \mathrm{m}^{3} / \mathrm{s} \\ &=7.182 \times 10^{-6} \times 10^{6} \mathrm{cm}^{3 / \mathrm{s}} \\ \mathrm{Q} &=7.182 \mathrm{cm}^{3} / \mathrm{s} \end{aligned}
\begin{aligned} Q&=\mathrm{kH} \frac{\mathrm{N}_{\mathrm{f}}}{\mathrm{N}_{\mathrm{d}}}\\ \text{Here, } \quad N_{f}&=\text{ No. of flow channels =3 }\\ N_{d}&=\text{ No. of equipotential}\\ \text { drops }=10 \\ \text{Given,}\quad k&=3.8 \times 10^{-6} \mathrm{m} / \mathrm{s}\\ \text{and }\quad H&=6.3 \mathrm{m} \\ Q &=3.8 \times 10^{-6} \times 6.3 \times \frac{3}{10} \times 1 \\ &=7.182 \times 10^{-6} \mathrm{m}^{3} / \mathrm{s} \\ &=7.182 \times 10^{-6} \times 10^{6} \mathrm{cm}^{3 / \mathrm{s}} \\ \mathrm{Q} &=7.182 \mathrm{cm}^{3} / \mathrm{s} \end{aligned}
Question 10 |
Laplace equation for water flow in soils is given below.
\frac{\partial^2 H}{\partial x^2}+\frac{\partial^2 H}{\partial y^2}+\frac{\partial^2 H}{\partial z^2}=0
Head H does not vary in y and z directions.
Boundary conditions are: at x = 0, H = 5; and \frac{dH}{dx}=-1.
What is the value of H at x = 1.2? __________
\frac{\partial^2 H}{\partial x^2}+\frac{\partial^2 H}{\partial y^2}+\frac{\partial^2 H}{\partial z^2}=0
Head H does not vary in y and z directions.
Boundary conditions are: at x = 0, H = 5; and \frac{dH}{dx}=-1.
What is the value of H at x = 1.2? __________
1.2 | |
3.8 | |
5.6 | |
8.2 |
Question 10 Explanation:
\frac{\partial^{2} H}{\partial x^{2}}=0
Integrating both sides, we get
\frac{\partial H}{\partial x}=C_{1} \quad \ldots(i)
Integrating again
\begin{aligned} H&=C_{1} x+C_{2}\\ \text{At }\quad x&=0, H=5 \\ \therefore \quad 5&=C_{2} \\ \text{At }\quad x&=0, \frac{d H}{d x}=-1 \end{aligned}
From eq. (i)
\begin{aligned} \therefore C_{1}&=-1 \\ \therefore \quad H&=-x+5 \\ \text { At } x &=1.2 \mathrm{m} \\ \therefore \quad \mathrm{H}&=5-1.2=3.8 \mathrm{m} \end{aligned}
Integrating both sides, we get
\frac{\partial H}{\partial x}=C_{1} \quad \ldots(i)
Integrating again
\begin{aligned} H&=C_{1} x+C_{2}\\ \text{At }\quad x&=0, H=5 \\ \therefore \quad 5&=C_{2} \\ \text{At }\quad x&=0, \frac{d H}{d x}=-1 \end{aligned}
From eq. (i)
\begin{aligned} \therefore C_{1}&=-1 \\ \therefore \quad H&=-x+5 \\ \text { At } x &=1.2 \mathrm{m} \\ \therefore \quad \mathrm{H}&=5-1.2=3.8 \mathrm{m} \end{aligned}
There are 10 questions to complete.
Q1 is of SOM…. Kindly make appropriate corrections! Thanks for all of ur efforts!