Given:

A simply supported beam of length $L$ subject to a force $F.$ The deflection the beam $y$ is characterized by the deflection equation

$y(x)=\{\begin{array}{cc}-x+\frac{R{x}^3}{6EI},& xa\\ -x+\frac{R{x}^3}{6EI}-\frac{F}{6EI}(x-a{)}^3,& x>a\end{array}$

where I is the moment of inertia of the cross section, $R$ is the reaction force on the beam at the left end, $$ is $t$ clockwise rotational angle of the beam at the left end, and $E$ is the Young's modulus of the beam's material. I, $$ and $E$ are all geometry$-$ or material$-$based constants. $E$ is found in a table $($MaterialElasticity$.$mat$),$ and th other values are found with the following $($already derived$)$ equations

$I=\frac{w{h}^3}{12}$

$R=\frac{F}{L}(L-a)$

$=\frac{Fa}{6\text{E}\text{I}\text{L}}(2L-a)(L-a)$

Eq$.2$

Eq$.3$

Eq$.4$

Since all the properties are either given or can be directly calculated, to find deflection at a point simply find th coefficients I,$R,$ and $E$ and then plug them into Equation $1.$

Case $1$: Single force

For the beam shown with width $w=0\mathrm{.}2[m],$ height $h=0\mathrm{.}2[m],$ and modulu $E={190}^{**}{10}^{{?}^{{?}^{?}}}9[Pa],$ calculate the beam deflection at $x=2[m]$ and $x=5[m]$

First, calculate the constants:

$I=\frac{w{h}^3}{12}=\frac{0\mathrm{.}2**{0\mathrm{.}2}^3}{12}=1\mathrm{.}33**{10}^{-4}[m^4]$

$R=\frac{F}{L}(L-a)=\frac{500}{10}(10-3)=350[N]$

$=\frac{Fa}{6\text{E}\text{I}\text{L}}(2L-a)(L-a)=\frac{(500)(3)}{6(190**{10}^9)(1\mathrm{.}33**{10}^{-4})(10)}(17)(7)=1\mathrm{.}682**{10}^{-5}[rad]$

For $)(a,$ use the first half of Eq$.1$:

$y(2)=-(2)+\frac{R(2^3)}{6EI}=-(1\mathrm{.}682**{10}^{-5})(2)+\frac{(350)(2^3)}{6(190**{10}^9)(1\mathrm{.}33**{10}^{-4})}-1\mathrm{.}517**{10}^{-5}[m]or0\mathrm{.}015[mm]$

For $)>(a,$ use the second half of Eq$.1$:

$y(2)=-(5)+\frac{R(5^3)}{6EI}-\frac{F}{6EI}(x-a{)}^3=-(1\mathrm{.}682**{10}^{-5})(5)+\frac{(350)(5^3)}{6(190**{10}^9)(1\mathrm{.}33**{10}^{-4})}-\frac{500}{6(190**{10}^9)(1\mathrm{.}33**{10}^{-4})}(2{)}^3$

$-1\mathrm{.}781**{10}^{-4}[m]or0\mathrm{.}18[mm]$

This application can be generalized with $x$ as a vector instead of a single value to find the deflection at all points on the beam.