Elastic-plastic torsion of a circular shaft

Data

$R = 25$ mm, $\tau_y = 150$ MPa, $G = 80\times10^3$ MPa, $J = \pi R^4/2 = 613592$ mm$^4$.

(a) Torque at first yield

$$T_y = \frac{\tau_y J}{R} = \frac{\pi R^3\tau_y}{2} = \frac{150\times613592}{25} = 3.682\times10^6\ \text{N mm} = 3.68\ \text{kN m}$$

(b) Fully plastic torque

All fibres carry $\tau_y$:

$$T_p = \int_0^R \tau_y\,r\,(2\pi r\,dr) = \frac{2\pi}{3}\,\tau_yR^3 = \frac{2\pi}{3}\times150\times15625 = 4.909\ \text{kN m}$$ $$\frac{T_p}{T_y} = \frac43 = 1.333$$

so the shape factor of a solid circular shaft in torsion is 1.333.

(c) Elastic-plastic stage, core radius $r_c = 15$ mm

The stress is $\tau_y r/r_c$ for $r < r_c$ and $\tau_y$ for $r_c<r<R$:

$$T = \underbrace{\frac{\pi\tau_y r_c^3}{2}}_{\text{core}} + \underbrace{\frac{2\pi\tau_y}{3}\left(R^3 - r_c^3\right)}_{\text{plastic ring}} = \frac{\pi\tau_y}{6}\left(4R^3 - r_c^3\right)$$ $$T = \frac{\pi\times150}{6}\left(4\times15625 - 3375\right) = \frac{\pi\times150}{6}\times59125 = 4.644\ \text{kN m}$$

Twist per unit length is governed by the core yielding at $r_c$:

$$\theta = \frac{\tau_y}{G\,r_c} = \frac{150}{80\times10^3\times15} = 1.25\times10^{-4}\ \text{rad/mm} = 0.125\ \text{rad/m}$$

(At first yield $\theta_y = \tau_y/(GR) = 0.075$ rad/m, so the twist is $1.67$ times the yield twist.)

(d) Residual stresses after unloading from $T_p$

Unloading is elastic, so the recovery stress at radius $r$ is $\tau' = T_p r/J$.

$$\tau'(R) = \frac{4.909\times10^6\times25}{613592} = 200.0\ \text{MPa}$$

Residual stress $= \tau_y - \tau'$:

The residual stress reverses sign between the centre and the surface (zero at $r = 18.75$ mm, where $200r/25 = 150$). The torque from the residual stresses is zero, as the shaft is unloaded.

Answer: (a) $T_y = 3.68$ kN m; (b) $T_p = 4.91$ kN m; (c) $T = 4.64$ kN m, $\theta = 0.125$ rad/m; (d) residual shear stress $-50$ MPa at the surface and $+150$ MPa at the centre.