Pure bending occurs when a beam is loaded with two equal and opposite [[Moment|moment]] couples along the same plane. Under the conditions of pure bending transverse planes stay straight and can be thought of as radial lines. Axial planes will either lengthen or contract, dependent on location, the axial plane that does not lengthen or contract is termed the neutral axis. ![[pure_bend.png|center|300]] The neutral axis radius of curvature, $\rho$, and the angle of curvature, $\theta$, can be used to determine the [[Strain|strain]]-state of a curved beam. $\varepsilon_x=−\dfrac{y}{\rho}$ Alternatively this can be expressed in terms of the maximum bending strain: $\varepsilon_x=−\dfrac{y}{y_{max}}\varepsilon_m$ Where the maximum strain is: $\varepsilon_m=\left|\dfrac{y_{max}}{\rho}\right|$ Using the relationship between [[Elastic|elastic]] [[Stress|stress]] and strain: $\sigma_x=E\varepsilon_x$ $\sigma_x=−\dfrac{y}{y_{max}}\sigma_m$ >[!proof]- >$\varepsilon_x=\dfrac{\delta}{L}$ >$\delta=(\rho-y)\theta-\rho\theta$ >$\delta=-y\theta$ >$L=\rho\theta$ >$\varepsilon_x = -\dfrac{y\theta}{\rho\theta}$ >$\varepsilon_x = -\dfrac{y}{\rho}$ The neutral axis must pass through the centroid of the beam cross-section, this provides a reliable way to determine the neutral axis. >[!proof]- >For a static beam, $F_x=\int\sigma_x dA=0$ >$\int\sigma_x dA = \int -\dfrac{y}{y_{max}}\sigma_m dA$ >$-\dfrac{\sigma_m}{y_{max}}\int ydA=0$ >$\int ydA=0$ >A reminder that $y$ is defined by the neutral axis, therefore the first moment about the neutral axis must be 0. In other words, the neutral axis must be the centroid of the beam cross-section. The stress can be related directly to the bending moment, $M$, and beam geometry by way of the cross-section [[Moment of Inertia|moment of inertia]], $I$. Bending moment is signed, it is positive when the positive $y$ axis is in compression due to bending. $\sigma_x=−\dfrac{My}{I}$ >[!proof]- >For a beam under bending moment, $M=\int -y\sigma_x dA$ >$M=\int -y\left(-\dfrac{y}{y_{max}}\sigma_m\right)dA$ >$M=\dfrac{\sigma_m}{y_{max}}\int y^2dA$ >$M=\dfrac{\sigma_m}{y_{max}}I$ >$\sigma_m=\dfrac{My_{max}}{I}$ >$\sigma_x=-\dfrac{y}{y_{max}}\dfrac{My_{max}}{I}$ >$\sigma_x=-\dfrac{My}{I}$ Finally, the radius of curvature can be determined solely by geometry and the materials Young’s modulus. $\rho=\dfrac{EI}{M}$ >[!proof]- >$\rho=\dfrac{y_{max}}{\varepsilon_m}$ >$\rho=\dfrac{y_{max}E}{\sigma_m}$ >$\rho=\dfrac{y_{max}EI}{My_{max}}$ >$\rho=\dfrac{EI}{M}$