Answer to Question #176307 in Mechanical Engineering for Sinazo

Question #176307

 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


1
Expert's answer
2021-04-01T01:14:25-0400

"\\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"


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