Answer:
3180.86 Nm
Explanation:
Moment of inertia for shaft AB, [tex]I_{AB}= \frac {\pi d^{4}}{32}=\frac {\pi 0.06^{4}}{32}=1.27235\times 10^{-6} m^{4}[/tex]
Torque in solid shaft AB will be given by
[tex]T_{AB}=\frac {\tau I_{AB}}{r}=\frac {75\times 10^{6} \times 1.27235\times 10^{-6}}{0.03}=3180.862562 Nm\approx 3180.86 Nm[/tex]
Where [tex]\tau[/tex] is shear stress, [tex]I_{AB}[/tex] is polar moment of inertia for shaft AB, r is the radius of shaft B
The inner diameter of pipe CD can found considering that the thickness of pipe is 0.006 m hence diameter= 0.09-(2*0.006)= 0.078 m
Moment of inertia for shaft CD will be
[tex]I_{CD}=\frac {\pi (0.09^{4}-0.078^{4})}{32}=2.8073\times 10^{-6} m^{4}[/tex]
Torque for shaft CD will be
[tex]T_{CD} =\frac {\tau I_{CD}}{r}[/tex] and here r = 0.045 m
[tex]T_{CD}}=\frac {75\times 10^{6} \times 2.8073\times 10^{-6}}{0.045}=4678.837 Nm\approx 4678.84 Nm[/tex]
The minimum of the two torques is the largest torque that can be applied. Therefore, the torque to apply equals 3180.86 Nm
