> [!info] Torsion stresses in shafts >- **Strain:** the shear strain $\gamma$ varies linearly with radaius $\rho$ through the cross section of the shaft, regardless of the material makeup of the cross section >- **Stress:** across annular regions on the cross-section where the material makeup is a constant, the shear stress $\tau$ varies linearly with radius $\rho$ through the cross section of the shaft: $\tau=G\gamma=T\rho /I_{P}$ where $I_{P}$ is the polar area movement of the cross section >- **Angle of twist** $\Delta \phi=\phi_{D}-\phi_{B}=\int_{0}^L \frac{T}{GI_{P}}dx=\frac{TL}{GI_{P}}$ >[!info] Consider an axial torque $T$ acting on a shaft w/ a circular cross section. Solve using the **four-step plan** >1. Equilibrium >$ > (1)\quad\sum M=T_{2}+T-T_{1}=0 >$ >2. Torque/rotation >$ > (2)\quad\Delta \phi_{1}=\frac{T_{1}(2L)}{GI_{P}} >$ >$ > (3)\quad\Delta \phi_{2}=\frac{T_{2}L}{GI_{P}} >$ >3. Compatibility >$ >(4)\quad\phi_{C}=\phi_{B}+\Delta \phi_{1}=\Delta \phi_{1} >$ >$ > (5)\quad\phi_{D}=\phi_{C}+\Delta \phi_{2}=\Delta \phi_{1}+\Delta \phi_{2}=0 >$ >4. Solve >$ >(2),(3),(5)\implies 2 \frac{T_{1}L}{GI_{P}}+\frac{T_{2}L}{GI_{P}}=0\to T_{2}=-2T_{1}\quad (6) >$ >$ > (1),(6) \implies -2T_{1}+T-T_{1}=0\to T_{1}=\frac{T}{3},T_{2}=-\frac{2}{3}T >$