Given :
Power, P = 20 kW
Speed, N = 430 rpm
Allowable shear stress, τ = 65 MPa
Torque in the shaft is given by :


T = 444.37 N.m
Diameter of the solid shaft is
![$d=\sqrt[3]{(16 T)/(\pi \tau)}](https://img.qammunity.org/2022/formulas/engineering/college/s57jr21n0r9v1iywsjsh93w3f8jxsy3zo2.png)
![$d=\sqrt[3]{(16 * 444.37)/(3.14 * 65)}](https://img.qammunity.org/2022/formulas/engineering/college/6pfs0bzpmdupslx86i2fk6zlult16st0sj.png)
![$d=\sqrt[3]{34.83} $](https://img.qammunity.org/2022/formulas/engineering/college/ejwe5gq66j9528efz8vq7c3hlj8c56ckr7.png)
d = 3.265 m
d = 326.5 mm
Internal diameter of the hollow shaft is :






mm
Percentage savings in the weight is given by :
Percentage saving =







= 99.01 %