Elastoplastic behavior during torsion

Plastic moment

Consider a bar with a thick-walled circular cross-section. The bar is modeled by an elastic-ideally-plastic material behavior, in shear, according to the figure below.

Elastic-ideally-plastic material behavior in shear. The stress in the material can not excedd the yield limit $\tau_\mathrm{y}$

Let us determine the moment corresponding to initial yielding. This is obtained by using the relations in the previous section and considering the case when $\tau_{\rm y}$ for $r=b$ (where the stress is largest).

$$\tau_\mathrm{y} = \frac{ \Mv_\mathrm{y} b}{\Kv} \quad \Rightarrow \quad \Mv_\mathrm{y}=
\frac{\tau_\mathrm{y} \, \Kv}{b}$$

We can also determine the plastic moment $\Mv_\mathrm{u}$ where the entire cross-section is in a plastic state. In this case the stress is $\tau=\tau_{\rm y}$ for all $r$ which gives

$$\Mv_\mathrm{u}=\int_A \tau_{\mathrm y} \, r \, {\mathrm d}A=
\tau_{\mathrm y} \int_a^b r \,2 \pi r \, \dr=
\tau_{\mathrm y} \, \frac{2 \pi}{3} (b^3-a^3)$$

An illustration of the corresponding stress state is shown in the figure below.

Note
The plastic moment $\Mv_\mathrm{u}$ is the limit load the cross-section can withstand before failing.

Plastic strengthening

The plastic strengthening is defined as:

$$\beta = \frac{\Mv_\mathrm{u} – \Mv_\mathrm{y}}{ \Mv_\mathrm{y} } =
\ldots =\frac{4b}{3} \cdot\frac{ b^3-a^3}{b^4-a^4}-1$$

and is a factor indicating how much the torque can be increased if one allows the cross-section to reach a fully plastic state, compared to initial yielding. For thin-walled cross-sections $\beta=0$, since the cross-section reaches a fully plastic state as soon as one point starts to yield.

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