Answer:
25 - [tex]\sqrt[4]{26.66*10^{-8} }[/tex] mm
Explanation:
Given data
steel tube : outer diameter = 50-mm
power transmitted = 100 KW
frequency(f) = 34 Hz
shearing stress ≤ 60 MPa
Determine tube thickness
firstly we calculate the ; power, angular velocity and torque of the tube
power = T(torque) * w (angular velocity)
angular velocity ( w ) = 2[tex]\pi[/tex]f = 2 * [tex]\pi[/tex] * 34 = 213.71
Torque (T) = power / angular velocity = 100000 / 213.71 = 467.92 N.m/s
next we calculate the inner diameter using the relation
[tex]\frac{J}{c_{2} } = \frac{T}{t_{max} }[/tex] = 467.92 / (60 * 10^6) = 7.8 * 10^-6 m^3
also
c2 = (50/2) = 25 mm
[tex]\frac{J}{c_{2} }[/tex] = [tex]\frac{\pi }{2c_{2} } ( c^{4} _{2} - c^{4} _{1} )[/tex] = [tex]\frac{\pi }{0.050} [ ( 0.025^{4} - c^{4} _{1} ) ][/tex]
therefore; 0.025^4 - [tex]c^{4} _{1}[/tex] = 0.050 / [tex]\pi[/tex] (7.8 *10^-6)
[tex]c^{4} _{1}[/tex] = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)
39.06 * 10^-8 - 12.402 * 10^-8 =26.66 *10^-8
[tex]c_{1} = \sqrt[4]{26.66 * 10^{-8} }[/tex] =
THE TUBE THICKNESS
[tex]c_{2} - c_{1}[/tex] = 25 - [tex]\sqrt[4]{26.66*10^{-8} }[/tex] mm
