Hoop stress in a cylindrical pipe of inner diameter 50 mm and wall thickness 2 mm, when subjected to the internal pressure of 1 MPa is

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ISRO IPRC Technical Assistant Mechanical held on 10/12/2016

Option 3 : 12.5 MPa

ST 1: Building Materials

1159

20 Questions
20 Marks
20 Mins

__Concept:__

For thin cylindrical vessel:

Hoop stress, \({\sigma _h} = \frac{{pd}}{{2t}}\)

where p is internal pressure, d is the inner diameter of the cylinder, t is the thickness of the wall of the cylinder.

__Calculation:__

__Given:__

d = 50 mm, t = 2 mm, p = 1 MPa

\({\sigma _h} = \frac{{pd}}{{2t}} = \frac{{1 \times 50}}{{2 \times 2}} = 12.5~MPa\)

__Additional Information__

For thin cylindrical vessel:

Longitudinal stress, \({\sigma _L} = \frac{{pd}}{{4t}}\)

Volumetric strain, \({\epsilon_v} = \frac{{pd}}{{4tE}}\left( {5 - 4\mu } \right)\)

For thin spherical vessel:

Hoop stress/Longitudinal stress:

\({\sigma _h} = {\sigma _L} = \frac{{pd}}{{4t}}\)

Hoop strain/longitudinal strain:

\({\epsilon_L} = {\epsilon_h} = \frac{{pd}}{{4tE}}\left( {1 - \mu } \right)\)

Volumetric strain:

\({\epsilon_v} = 3{\epsilon_h} = \frac{{3pd}}{{4tE}}\left( {1 - \mu } \right)\)