Answer:
[tex]\tau_{max} = 142.6[/tex] MPa
T = 1536.8 N m
Explanation:
Given data:
Torque = 3.5 k N m = 3.5*10^3 N.m
Diameter D = 5 cm = 0.05 m
a) from torsional equation we have
[tex]\frac[T}{J_{solid}} = \frac{\tau_{max}}{D/2}[/tex]
[tex]\frac{T}{\pi/32 D^4} = \frac{\tau_{max}}{D/2}[/tex]
solving for [tex]\tau_{max}[/tex]
[tex]\tau_{max} = \frac{16 T}{\pi D^3} =\frac{16 \times 3.5*10^3}{\pi 0.05^3}[/tex]
[tex]\tau_{max} = 142.6[/tex] MPa
B)
[tex]\tau = 37 MPa = 37 \times 10^6[/tex] Pa
[tex]D_i = 4.5 cm = 0.045[/tex] m
[tex]D_o = 6.5 cm = 0.065[/tex] m
[tex]\frac{T}{J_{hollow}} = \frac{\tau_{max}}{D_o /2}[/tex]
[tex]\frac{T}{(\pi/32) (0.065^4 - 0.045^4)} =\frac{37*10^6}{0.065/2}[/tex]
T = 1536.8 N m
