- ![[Pasted image 20260927151820.png|right|250]] Distributed load acts upwards on the beam -> internal shear force causes a clockwise rotation of beam segments - Internal moments cause **compression in the top fibers** of the segments - **Shear force / bending moment equation**: ${ V=\frac{dM}{dx} }$ - **Axial stress/strain relation**: ${ \sigma_{x}=E\varepsilon_{x}=-\frac{My}{I} }$ - Slope of shear diagram equals the intensity of the distributed loading: $\frac{dV}{dx}=w(x)$ - Slope of moment diagram equals the shear: $\frac{dM}{dx}=V(x)$ $ \Delta V=\int w(x)dx\quad\Delta M=\int V(x)dx $ - A state of ==pure bending== exists $\int V(x)dx=0,$ zero applied shear - Under the action of equal & opposite negative bending couples at its ends, the top fibers of the beam stretch and the bottom fibers shorten - Plane that divides the region of compression from the region of stretch is called the ==neutral surface== of the beam - Perpendicular to the neutral surface is the ==plane of bending==. Loading and supports for the beam are assumed to be symmetrical about the plane of bending - Deformation of the initially straight beam axis is known as the ==deflection curve== of the beam - ![[Pasted image 20260927152830.png|right|350]] **Euler-Bernoulli Beam Theory** - Cross-sections which are plane and are perpendicular to the axis of the undeformed beam, remain plane and remain perpendicular to the deflection curve of the deformed beams - Deformation in the plane of the cross-section (i.e. transverse strains $\varepsilon_{y}$ and $\varepsilon_{z}$ may be neglected in deriving an expression for longitudinal strain $\varepsilon_{x}$ ![[Pasted image 20260927153203.png|center|550]] $ \varepsilon_{x}=\lim_{ \Delta x \to 0 } \frac{\Delta x^*-\Delta x}{\Delta x} $ $ A^*B^*=\Delta x=\rho\Delta\theta^* $ $ P^*Q^*=\Delta x^*=(\rho-y)\Delta\theta^* $ $ \varepsilon_{x}=\lim_{ \Delta x \to 0 } \frac{(\rho-y)\Delta\theta^*-\rho\Delta\theta}{\rho\Delta\theta} $ $ \boxed{ \varepsilon_{x}=-\frac{y}{\rho} }\qquad \boxed{ \sigma_{x}=E\varepsilon_{x}=-\frac{Ey}{\rho} } $ $ F_{x}=\int_{A}\sigma_{x}dA\implies M_{x}=-\int_{A}y\sigma_{x}dA=\frac{E}{\rho}\int_{A}y^2dA=\frac{EI}{\rho} $ $ \boxed{ M=\frac{EI}{\rho} }\qquad \boxed{ \sigma_{x}=-\frac{My}{I} } $ $\sigma_{max}=\frac{Mc}{I}$ where $c$ is the perpendicular distance from the neutral axis, where $\sigma_{max}$ acts Second moment of area $I=\int_{A} y^2dA$