Everything so far has been analysis: here is a shaft, how hard is it working. Now turn it round. You know the torque, you know what the steel is allowed to stand, and you need a diameter. That is design, and it has its own form: $d = \sqrt[3]{\dfrac{16 T}{\pi \tau}}$ — d equals the cube root of sixteen T over pi tau.

$d$ is the minimum solid shaft diameter in millimetres, $T$ is the torque the shaft must carry (in N·mm, so a nameplate's newton-metres get their ×1000 first), and $\tau$ is the allowable shear stress for the material in N/mm². Where did the 16 come from? Put $r = d/2$ and $J = \pi d^{4}/32$ into $\tau = Tr/J$ and the 32 halves into a 16. It is the same relation you already know, wearing working clothes — which is why reaching for 32 here quietly oversizes the bar by about 26%.

The cube root is the friendly part. Capacity goes as $d^{3}$, so eight times the torque needs only twice the diameter. That single fact is why shafts never look as heavy as the loads they carry would suggest, and it is worth having in your hands before you compute anything.

Then the last step, the one exams forget and shops never do: the answer is a minimum, and bar comes in stock sizes. You round UP to the next available size — never to the nearest one. Below the calculated diameter the stress is above the allowable, and the allowable is exactly where your safety margin already lives. You do not get to spend it twice. Nor do you jump to the biggest bar in the rack: oversize costs mass, inertia, bearings and couplings, and it moves the failure somewhere you did not choose.