$T = \dfrac{P}{2\pi f} = \dfrac{4.5(1\,000\,000)}{2\pi(3)}$

$T = 238\,732.41 \, \text{N}\cdot\text{m}$

Based on maximum allowable shearing stress:

$\tau_{max} = \dfrac{16T}{\pi d^3}$

$50 = \dfrac{16(238\,732.41)(1000)}{\pi d^3}$

$d = 289.71 \, \text{mm}$

Based on maximum allowable angle of twist:

$\theta = \dfrac{TL}{JG}$

$1^\circ \left( \dfrac{\pi}{180^\circ} \right) = \dfrac{238\,732.41(26d)(1000)}{\frac{1}{32}\pi d^4 (83\,000)}$

$d = 352.08 \, \text{mm}$

Use the larger diameter, thus, d = 352 mm. *answer*

## Comments

## im confused on why do we n

im confused on why do we need to use the larger diameter

## It is actually a good thing

It is actually a good thing to ponder. It will be enjoyable and at your best advantage to reason it out yourself.

## the outermost fiber of a

the outermost fiber of a material is where the shearing stress is at maximum that is why we always use the outside diameter of a shaft when dealing with torsion