Question:

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 4 m. Take, water density = 1000 kg/m\(^3\) and acceleration due to gravity = 10 m/s\(^2\). \includegraphics[width=0.25\linewidth]{image37.png} \[ \text{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 \(\underline{\hspace{1cm}}\).} \]

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To avoid buckling, the weight of the water must be less than the critical buckling load. Use the formula for buckling load to find the maximum permissible water depth.
Updated On: Dec 20, 2025
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Correct Answer: 2.5

Solution and Explanation

The maximum permissible depth of the water is determined by the critical buckling load of the column. The buckling load for a column with an internal hinge is given by: \[ P_{\text{cr}} = \frac{\pi^2 E I}{L^2} \] Where: - \( E = 200 \, \text{GPa} = 200 \times 10^9 \, \text{N/m}^2 \), - \( I = \frac{\pi d^4}{64} \), with \( d = 75 \, \text{mm} = 0.075 \, \text{m} \), - \( L = 4 \, \text{m} \). First, calculate \( I \): \[ I = \frac{\pi (0.075)^4}{64} = 3.31 \times 10^{-6} \, \text{m}^4 \] Now, calculate the critical load \( P_{\text{cr}} \): \[ P_{\text{cr}} = \frac{\pi^2 \times 200 \times 10^9 \times 3.31 \times 10^{-6}}{4^2} = 4.94 \times 10^5 \, \text{N} \] The force due to the water column is \( F = \rho g A h \), where:
- \( \rho = 1000 \, \text{kg/m}^3 \) is the water density,
- \( g = 10 \, \text{m/s}^2 \) is the acceleration due to gravity,
- \( A = \frac{\pi d^2}{4} = \frac{\pi (0.075)^2}{4} = 4.42 \times 10^{-3} \, \text{m}^2 \).
Thus: \[ F = 1000 \times 10 \times 4.42 \times 10^{-3} \times h = 44.2 h \, \text{N} \] Setting the critical load equal to the water load: \[ 44.2 h = 4.94 \times 10^5 $\Rightarrow$ h = \frac{4.94 \times 10^5}{44.2} \approx 11.16 \, \text{m} \] Thus, the maximum permissible depth of water is \( \boxed{2.9} \, \text{m} \).
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