Oxygen gas at a pressure of 20 MPa is stored in a thin cylinder o
![Oxygen gas at a pressure of 20 MPa is stored in a thin cylinder o](https://storage.googleapis.com/tb-img/production/19/06/Railways_Solution%20Improvement_Satya_10%20June_Madhu%28Dia%29.png)
A. 100 MPa
B. 20 MPa
C. 200 MPa
D. 150 MPa
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
Concept:
By thin-walled cylinder, we mean that the thickness ‘t' is very much smaller than the radius Ri and we may quantify this by stating that the ratio t / Ri of the thickness of radius should be less than 0.1.
In a thin shell circumferential stress is \({\sigma _c} = \frac{{Pd}}{{2t}}\;\) and
Longitudinal stress will be half of the circumferential stress i.e. \({\sigma _l} = \frac{{Pd}}{{4t}}\).
where p is the pressure, t is the thickness of cylinder, d is the diameter of the cylinder
Calculation:
Given:
p = 20 MPa, t = 2.5 mm and d = 50 mm.
Longitudinal stress
\({\sigma _l} = \frac{{Pd}}{{4t}}\)
\({\sigma _l} = \frac{{20\times 50}}{{4\times 2.5}} =100~MPa\)
Hoop stress \({\sigma _h} = \frac{{Pd}}{{2t}} = {\sigma _1}\) |
Hoop strain \({\epsilon_h} = \frac{{Pd}}{{4tE}}\left( {2 - \mu } \right)\) |
Longitudinal stress \({\sigma _L} = \frac{{Pd}}{{4t}} = {\sigma _2} = \frac{{{\sigma _1}}}{2}\) |
Longitudinal strain \({\epsilon_L} = \frac{{Pd}}{{4tE}}\left( {1 - 2\mu } \right) = \frac{{{\sigma _2}}}{E}\left( {1 - 2\mu } \right)\) |