Now put a torque on that section. The stress it produces is $\tau = \dfrac{T r}{J}$ — read aloud tau equals T r over J, with $\tau$ the Greek letter tau, the symbol shear stress always wears. $\tau$ is the torsional shear stress in megapascals (MPa, which is exactly N/mm²); $T$ is the applied torque; $r$ is the radius of the point you are asking about, measured from the centre line; and $J$ is the polar moment of area from the last lesson.

That $r$ is the sentence's most important word. The stress is not one number for the whole section — it is zero at the axis and climbs in a straight line to its maximum at the surface, where $r = d/2$. Half the radius, half the stress. This is also why a hollow shaft is such a good bargain: the metal you bore out of the middle was barely working anyway.

Unit discipline, stated once for the whole chapter and then assumed. Work in the millimetre system: torque in N·mm, lengths in mm, J in mm⁴, and the stress falls out in N/mm², which IS the megapascal. Every nameplate and every job sheet quotes torque in newton-metres, so the very first thing you do is multiply by 1000. Skip it and your answer is a thousand times light — a 40 MPa shaft reported at 0.04 MPa, which looks so safe that nobody questions it.

Read backwards, the same relation rates a shaft: $T = \dfrac{\tau J}{r}$, with $\tau$ now the material's allowable shear stress and the answer the greatest torque the section may carry. Push the units through it every time. N/mm² × mm⁴ ÷ mm is N·mm — a torque, as promised. If the units refuse to cancel into the answer's units the rearrangement is wrong, no appeal; units that do cancel never prove you right. The check runs one way only.