Reminder: ![[Pasted image 20261014145250.png|right|375]] $ \frac{dV}{dx}=p(x)\qquad \frac{dM}{dx}=V\qquad M=\frac{EI}{\rho} $ Combining these equations gives: $ \frac{d^2M}{dx^2}=\frac{dV}{dx}=p(x)\implies \frac{d^2}{dx^2}\left( \frac{EI}{\rho} \right)=p(x) $ Also: $ \tan\theta\approx\theta\qquad M(x)=\frac{EI}{\rho}\approx EI \frac{d\theta}{dx} $ - $V(x)$: internal shear force in the beam at position $x$ - $v(x)$: vertical deflection (displacement) of the beam at position $x$ >[!info] Shear/Moment/Deflection Eqns > $V(x)=V(x_{1})+\int_{x_{1}}^x p(s)ds$ >$M(x)=M(x_{1})+\int_{x_{1}}^x V(s)ds$ >$ \theta(x)=\theta(x_{1})+ \frac{1}{EI} \int_{x_{1}}^x M(s)ds$ > $v(x)=v(x_{1})+\int_{x_{1}}^x \theta(s)ds$ > > [!info] These results can be used two alternate ways for determining the deflection of a beam >> 1. **Fourth order approach**: start w/ loading $p(x)$ and perform the four integrations to obtain $v(x)$ > >2. **Second-order approach:** determine bending moment distribution $M(x)$ through FBDs and equilibrium analysis, then use that to find $v(x)$ | Type of support | Outcome | | --------------------------------------- | ------------------------------- | | Fixed | $v=0 \\ \theta=\frac{dv}{dx}=0$ | | Constrained rotation | $\theta=\frac{dv}{dx}=0 \\ V=0$ | | Free end | $ V = 0 \\ M = 0$ | | Pin/roller/double roller/roller in slot | $ v = 0 \\ M = 0 $ |