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ISRO MCF Technical Assistant Mechanical held on 23/06/2019

Option 1 : 100 MPa

SSC JE ME Full Test 4

5572

200 Questions
200 Marks
120 Mins

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