Question

An air vessel has circumferential stress of 267.75 N/mm²; longitudinal stress of 133.875 N/mm²; value of Young's modulus E = 2.1 x 10⁵ N/mm², Poisson's ratio m = 0.3: value of the circumferential strain will be

• A. 0.0005090
• B. 0.001084
• C. 0.0002050
• D. 0.002040

Answer 1

Answer:

B. 0.001084

Given: $\sigma_{h} = 267.75N / m m ^ 2$ $\sigma_{L} = 133.875N / m m ^ 2$ , $E =2.1\times 10 ^ 5 N / m m ^ 2$ $m = \mu =0.3$ $\Rightarrow$ circumferential strain is: $\epsilon_{h} = \epsilon_{l} = \frac{\sigma_1}{ E} - \mu\frac{\sigma_2}{ E} \Rightarrow \epsilon_{h} = \frac{267.75}{2.1 \times 10 ^ 5} - 0.3 \frac{133.875}{2.1 \times 10 ^ 5} = 0.001084$