Respuesta :
Answer:
thickness 0.106 inc
Explanation:
outer radius of shaft is given as [tex]= \frac {d_o}{2} = 3/2 = 1.5 inch[/tex]
power = 150 hp = 150*550 lb- ft/s
angular velocity is given as \omega = [tex]\frac{2 \pi N}{60}[/tex]
[tex]\omega =\frac{2 \pi 1000}{60} = 104.71 rad/s[/tex]
we know that torque is given as [tex]T =\frac{P}{\omega}[/tex]
[tex]T = \frac{150*550}{104.71}[/tex]
T = 787.89 lb.in
T =9454.67 lb ft
minimum allowable shear stress is given as
[tex]\tau = \frac{Tr_0}{\frac{\pI}{2}(r_o^4 - r_i^4)}[/tex]
[tex]8 *10^{3}= \frac{9454.67*1.5}{\frac{\pI}{2}(1.5^4 - r_i^4)}[/tex]
[tex]r_1 = 1.39 inc[/tex]
[tex]thickness = r_o - r_i = 1.5 - 1.39 = 0.106 inc[/tex]
