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
| Oxygen gas at a pressure of 20 MPa is stored in a thin cylinder of thickness 2.5 mm and a mean diameter of 50 mm. The longitudinal stress in the cylinder is

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

\({\epsilon_L} = \frac{{Pd}}{{4tE}}\left( {1 - 2\mu } \right) = \frac{{{\sigma _2}}}{E}\left( {1 - 2\mu } \right)\)