Answer to Question #176291 in Mechanical Engineering for Sinazo

Question #176291

A thin rim with a mean diameter of 1.2 m and cross section of 15 mm ×200mm rotates at 800 r/min. Calculate change in diameter due to rotation if E=200GPa and density is 7800kg/m^3


1
Expert's answer
2021-04-01T01:14:42-0400

Radius, r "=\\frac{D}{2}=\\frac{1.2}{2}=0.6 m"

Area, A"= 15 \\times 2000=3000 mm^2= \\frac{3000}{1000000}=0.003 m^2"

"\\omega= \\frac{2 \\pi N}{60}=\\frac{2 \\times \\pi \\times800}{60}=83.775 rad\/s"

Velocity ",v=r\\omega=0.6 \\times 83.775 =50.2655 m\/s"

Magnitude of force, F=mv="\\rho A v^2=7800 \\times 0.003 \\times (50.2655)^2=59122.919 N"

Stress, "\\sigma =\\frac{F}{A}=\\frac{59122.919}{0.003}=19.708 MPa"

Change in diameter , "\\triangle d=\\frac{Fd}{AE}=\\frac{59122.919 \\times1.2}{0.003 \\times 200 \\times 10^9}=0.000118245=0.118245 mm"


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