- Parallel axis theorem: $I_{B}=I_{O }+Ad_{OB}^2$ ![[Pasted image 20261004174212.png|right|200]] - Reminder flexural stress $\sigma=-\frac{My}{I}$, shear stress $\tau=\frac{vA^*\bar{y}^*}{It}$ - Flexural stress $\sigma$ varies linearly with $y,$ equals zero at neutral axis - Shear stress $\tau$ depends on shape of cross section, maximum magnitude of $\tau$ occurs at or near the neutral axis - Direction of $\tau$ governed by direction of $v$ $v(x^+)=v(x^-)+P_{0},\quad v(x_{2})=v(x_{1})+\int_{x_{1}}^{x_{2}}p(\zeta)d\zeta$ $ M(x^+)=M(x^-)-M_{0},\quad M(x_{2})=M(x_{1})+\int_{x_{1}}^{x_{2}} v(\zeta)d\zeta $ ![[Pasted image 20261004174311.png|center|300]]