Calculate the hoop tension developed at the base of the wall of a circular water tank of diameter 12.6 m storing water to a depth of 4 m. Take the specific weight of water as 10 kN/m³.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 05/06/2024 (Morning)
Detailed Explanation
Water pressure pushes the wall of a circular tank outward. Like a thin cylinder under internal pressure, the wall resists by developing hoop tension around its circumference. Water pressure is largest at the base, so hoop tension is largest there too.
Pressure at the base: \( p = \gamma H = 10 \times 4 = 40\ \text{kN/m}^2 \)
Hoop tension per metre height: \[ T = \frac{pD}{2} = \frac{40 \times 12.6}{2} = 252\ \text{kN} \]
Why “252 kN” Is Correct
This is the cylinder hoop-force formula (force = pD/2 per unit length) applied to a 1 m high strip of wall at the base, giving 252 kN (per metre height).
Why the Other Options Are Wrong
- A. 200 kN: 200 kN does not follow from γHD/2.
- B. 152 kN: 152 kN does not follow from γHD/2.
- D. 504 kN: 504 kN uses pD instead of pD/2 (both halves of the ring instead of one).
Key Point
Hoop tension in a circular tank wall at the base: T = γHD/2 per metre height.
Key Concepts for Students
- Hoop tension falls linearly to zero at the water surface (free-base wall).
- The tank wall is designed with hoop steel to carry this tension (IS 3370).
- Hoop stress = T/(wall thickness × 1 m).
Practice more: this is question 1 in Thin Cylindrical and Spherical Shells MCQs – Part 1. Try the whole set and check every answer instantly.