Question 6.P.5: A long steel tube, 7.5 cm internal diameter and 0.15 cm thic......
A long steel tube, 7.5 cm internal diameter and 0.15 cm thick, has closed ends, and is subjected to an internal fluid pressure of 3 MN/m². If E = 200 GN/m², and v = 0.3, estimate the percentage increase in internal volume of the tube.
Step-by-Step
The circumferential tensile stress is
\sigma_1 = \frac{pr}{t} = \frac{ (3 \times 10^6) (0.0383)}{(0.0015)} = 76.6 MN/m^2The longitudinal tensile stress is
\sigma_2 = \frac{pr}{2t} = 38.3 MN/m^2The circumferential strain is
\epsilon _1 = \frac{1}{E}(\sigma_1 – v\sigma_2)and the longitudinal strain is
\epsilon _2 = \frac{1}{E}(\sigma_2 – v\sigma_1)The volumetric strain is then
\begin{aligned} 2 \epsilon_1+\epsilon_2 & =\frac{1}{E}\left[2 \sigma_1-2 v \sigma_2+\sigma_2-v \sigma_1\right] \\ & =\frac{1}{E}\left[\sigma_1(2-v)+\sigma_2(1-2 v)\right] \\ \text{Thus} \\ 2 \epsilon_1+\epsilon_2 & =\frac{\left(76.6 \times 10^6\right)\left[(2-0.3)+\frac{1}{2}(1-0.6)\right]}{200 \times 10^9} \\ & =\frac{\left(76.6 \times 10^6\right)(1.9)}{\left(200 \times 10^9\right)}=0.727 \times 10^{-3} \end{aligned}The percentage increase in volume is therefore 0.0727%.
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