Introduction to Plastic Range Stress and Strain

Definitions

• Yield moment $M_y$: the moment at which the extreme fibre first reaches $\sigma_y$.
• Plastic moment $M_p$: the moment when the whole section has yielded (a plastic hinge forms).
• Shape factor $S = M_p/M_y = Z_p/Z$: depends only on the section shape.

Assumptions: elastic-perfectly plastic stress-strain curve, same behaviour in tension and compression, plane sections remain plane, and no axial force.

Derivation (rectangular section)

Yield moment

$$M_y = \sigma_y Z = \sigma_y\,\frac{bh^2}{6}$$

Elastic-plastic stage. Let the elastic core have half-depth $y_0$ ($y_0 \le h/2$). The stress is $\sigma_y y/y_0$ in the core and $\sigma_y$ in the outer zones.

$$M = \underbrace{\sigma_y\,\frac{b\,(2y_0)^2}{6}}_{\text{core}} + \underbrace{2\sigma_y\,b\,\left(\frac h2 - y_0\right)\left(\frac{h/2 + y_0}{2}\right)}_{\text{outer zones}} = \sigma_y\,b\left(\frac{h^2}{4} - \frac{y_0^2}{3}\right)$$

Fully plastic ($y_0 \to 0$):

$$M_p = \sigma_y\,\frac{bh^2}{4} = \sigma_y Z_p,\qquad Z_p = \frac{bh^2}{4}$$

Shape factor

$$S = \frac{M_p}{M_y} = \frac{bh^2/4}{bh^2/6} = 1.5$$

Numerical values ($b = 60$, $h = 100$, $\sigma_y = 250$ MPa)

$$\begin{aligned} M_y &= 250\times\frac{60\times100^2}{6} = 250\times10^5 = 25.0\ \text{kN m} \\ M_p &= 250\times\frac{60\times100^2}{4} = 250\times1.5\times10^5 = 37.5\ \text{kN m} \\ \end{aligned}$$

Elastic core 60 mm deep, so $y_0 = 30$ mm:

$$M = 250\times60\left(\frac{100^2}{4} - \frac{30^2}{3}\right) = 15000\times(2500 - 300) = 33\times10^6\ \text{N mm} = 33.0\ \text{kN m}$$

(A value between $M_y$ and $M_p$, as expected.)

Collapse load (simply supported, central load $W$, span $L = 3$ m)

At collapse a plastic hinge forms under the load, with $M_{max} = WL/4 = M_p$:

$$W_p = \frac{4M_p}{L} = \frac{4\times37.5}{3} = 50.0\ \text{kN},\qquad W_y = \frac{4M_y}{L} = 33.3\ \text{kN}$$

Answer: $M_y = 25$ kN m, $M_p = 37.5$ kN m, shape factor $= 1.5$, $M = 33$ kN m for a 60 mm elastic core, and the collapse load is $50$ kN (first yield at $33.3$ kN).

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.