[[Force|Unloading]] a shaft can be thought of as applying an equal and opposite unloading [[Moment|torque]] to the shaft. Application of an unloading torque will produce a net zero torque. Applying a torque in the opposite direction will necessarily apply a negative [[Shear Stress|shear stress]], this unloading shear stress can be calculated as follows: $\tau_{ul}=\tau_y\left(\dfrac{4}{3}-\dfrac{1}{3}\dfrac{r_p^3}{r_{max}^3}\right)$ >[!proof]- >$T=\int_0^{r_p}r\tau_y\dfrac{r}{r_p}2\pi rdr+\int_{r_p}^{r_{max}}r\tau_y2\pi rdr−\int_0^{r_{max}}r\tau_{ul}\dfrac{r}{r_{max}}2\pi rdr=0$ >$0 = \tau_y\left(\dfrac{c^3}{3}-\dfrac{r_p^3}{12}\right)-\tau_{ul}\dfrac{c^3}{4}$ >$\tau_{ul}=\tau_y\left(\dfrac{4}{3}-\dfrac{1}{3}\dfrac{r_p^3}{r_{max}^3}\right)$ When fully yielded: $\tau_{ul}=\dfrac{4}{3}\tau_y$ >[!proof]- >$T = \int_0^{r_{max}}r\tau_y 2\pi rdr -\int_0^{r_{max}} r\tau_{ul}\dfrac{r}{r_{max}}2\pi rdr=0$ >$0=2\pi\left(\tau_y\dfrac{r_{max}^3}{3}-\tau_{ul}\dfrac{r_{max}^3}{4}\right)$ >$\tau_{ul}=\dfrac{4}{3}\tau_y$ [[Stress|Stresses]] remain in the shaft due to unloading a [[Plastic|plastically]] deformed shaft, these are called residual stresses. There will also a resulting residual angle of twist after unloading. The below image shows the result of initial plastic loading combined with unloading. ![[torsion_unload.png]] Residual stresses can cause shafts to fail prematurely. However, compressive residual stresses can be used to strengthen particular materials. Regardless of what the reason is, residual stresses should always be accounted for in designs that will undergo plastic deformation.