A welded steel cylindrical drum made of a 10 mm thick plate has an internal diameter of 1.20 m. Find the change in diameter that would be caused by internal pressure of 1.5 MPa. Assume that Poisson's ratio is 0.30 and E = 200 GPa (longitudinal stress, σ_{y }= pD/4t circumferential stress, σ_{x} = pD/2t).

This question was previously asked in

BHEL ET Mechanical Held on May 2019

Option 2 : 0.459 mm

ISRO Scientist ME 2020 Paper

1464

80 Questions
240 Marks
90 Mins

__Concept:__

**Hoop strain in case of cylinder **is given by:

**\({\epsilon _{\rm{H}}} = \frac{{P \times {d}}}{{4 \times t \times E}} \times \left( {2 - \mu } \right)\)**

__Calculation:__

__Given:__

t = 10 mm = 10 × 10^{-3} m, Internal diameter (d) = 1.20 m, Internal pressure (P) = 1.5 MPa = 1.5 × 10^{6} Pa, Poisson’s ratio (μ) = 0.30

Modulus of elasticity (E) = 200 GPa = 200 × 10^{9} Pa

Hoop strain is:

\({\epsilon_H} = \frac{{{\rm{\Delta }}d}}{d} = \frac{{P \times {d}}}{{4 \times t \times E}} \times \left( {2 - \mu } \right)\)

\({\rm{\Delta }}d = \frac{{P \times {d^2}}}{{4 \times t \times E}} \times \left( {2 - \mu } \right)\)

\( {\rm{\Delta }}d= \frac{{1.5 \times {{10}^6} \times {{1.2}^2} \times \left( {2 - 0.3} \right)}}{{4 \times 10 \times {{10}^{ - 3}} \times 200 \times {{10}^9}}} = 4.59 \times {10^{ - 4}}\;m = 0.459\;mm\)