Imagine a material that, once undergoing [[Plastic|plastic]] deformation, ceases to increase in [[Stress|stress]] past the [[Yielding|yield]] stress. One outcome of such a material existing will be that it will cease to strain harden, the stress-[[Strain|strain]] curve will remain flat after the yield point. Such a material is referred to as perfectly plastic, due to the simple, known nature of perfectly plastic materials it is a great baseline for understanding plastic deformation in shafts. ![[torsion_plastic2.png]] ![[torsion_plastic1.png]] The above image shows a stress-radius diagram for a shaft undergoing perfectly plastic behaviour. The following should be noted: - Past a certain **yield torque** ($T_y$) the shaft will start yielding at a radius lower than $r_{max}$. - This lower radius, the radius of plastic deformation, is symbolized by $r_p$​. - The additional [[Moment|torque]] applied past the yield torque must be redistributed to parts of the shaft that can absorb more [[Force|load]], this is symbolized by the redistribution of the green triangle to the red triangle in the above figure. Torque causing yielding can be calculated using the procedure established for [[Elastic|elastic]] deformation. At yield torque: $T_y=\tau_y\dfrac{J}{r_{max}}$ Past yield torque: $T=\int_0^{r_p}r\tau dA + \int_{r_p}^{r_{max}}r\tau dA$ $T=2\pi\tau_y\left(\dfrac{c^3}{3}-\dfrac{r_p^3}{12}\right)$ >[!proof]- >$T=\int_0^{r_p}r\tau dA+\int_{r_p}^{r_{max}}r\tau dA$ >$T=\int_0^{r_p}r\tau_y\dfrac{r}{r_p}2\pi r dr+\int_{r_p}^{r_{max}}r\tau_y 2\pi rdr$ >$T = 2\pi\tau_y\dfrac{r_p^3}{4}+2\pi\tau_y\left(\dfrac{c^3}{3}-\dfrac{r_p^3}{3}\right)$ >$T=2\pi\tau_y\left(\dfrac{c^3}{3}-\dfrac{r_p^3}{12}\right)$