Plastic torque of solid shaft is 4/3

Elasto-plastic material: stress rises linearly with strain up to the yield stress in shear $\tau_y$ and then stays constant at $\tau_y$ (perfectly plastic).

Maximum elastic torque

The shaft is on the point of yielding when the surface stress reaches $\tau_y$. For a solid shaft of radius $R$ with $J = \dfrac{\pi R^4}{2}$:

$$T_e = \frac{\tau_y J}{R} = \frac{\tau_y}{R}\cdot\frac{\pi R^4}{2} = \frac{\pi\tau_y R^3}{2}$$

Fully plastic torque

When the whole section yields, the stress is $\tau_y$ at every radius. Take a ring of radius $r$, thickness $dr$, area $2\pi r\,dr$:

$$dT = \tau_y\,(2\pi r\,dr)\,r = 2\pi\tau_y r^2\,dr$$ $$T_p = \int_0^R 2\pi\tau_y r^2\,dr = 2\pi\tau_y\frac{R^3}{3} = \frac{2\pi\tau_y R^3}{3}$$

Ratio

$$\frac{T_p}{T_e} = \frac{2\pi\tau_y R^3/3}{\pi\tau_y R^3/2} = \frac{2}{3}\times 2 = \frac{4}{3}$$

Hence the plastic torque is $\dfrac{4}{3}$ times the maximum elastic torque.