Answer to Question #176307 in Mechanical Engineering for Sinazo

Question

 A thin steel cylinder, 400 mm in diameter with closed ends, is rotated at a speed of 5 000 r/min about a longitudinal axis through its centre. If it is simultaneously subjected to an internal pressure of 4 MPa, calculate the thickness of the cylinder, assuming that the maximum allowable tensile stress is limited to 170 MPa. The density of the steel is 7 800 kg/m

Answer 1

$\omega=\frac{2 \pi N}{60}=\frac{2 \pi \times 5000}{60}=523.598 rad/s$

$r= \frac{D}{2}=\frac{400}{2}=200 mm=0.2m$

Now the thickness of the cylinder is given as

$t=\frac{Pr}{\sigma - \rho r^2 \omega^2}$

$t=\frac{4 \times 10^6 \times0.2}{170 \times10^6 - 7800 \times r^2 \times 523.598^2}=0.0094715 m=9.4715 mm$