# Force

Point load $W$ at position $a$

From Roark$^{10}$, the end rotations for a point load are

\begin{aligned} \theta_{0} &= - \frac{W}{6EI}\frac{a}{l}(2l - a)(l - a) \\ \theta_{1} &= \frac{W}{6EI}\frac{a}{l}\left( l^{2} - a^{2} \right) \end{aligned}

Letting the distance from end 1 be $b$ these equations can be rewritten

\begin{aligned} \theta_{0} &=-\frac{W}{6 E I} \frac{a b(l+b)}{l} \\ \theta_{1} &=\frac{W}{6 E I} \frac{a b(l+a)}{l} \end{aligned}

Varying load from $\lbrack a:b\rbrack$ with intensity $w_{a}$ and $w_{b}$.

Using the equations above from Roark the end rotations for a point load at $x$ are

\begin{aligned} &\theta_{0}=-\frac{W}{6 E I} \frac{x}{l}(2 l-x)(l-x) \\ &\theta_{1}=\frac{W}{6 E I} \frac{x}{l}\left(l^{2}-x^{2}\right) \end{aligned}

Using these and integrating over the element gives

\begin{aligned} &\theta_{0}=-\frac{1}{6 E I} \int_{a}^{b} w(x) \frac{x}{l}(2 l-x)(l-x) d x &(1)\\ &\theta_{1}=\frac{1}{6 E I} \int_{a}^{b} w(x) \frac{x}{l}\left(l^{2}-x^{2}\right) d x &(2) \end{aligned}

The load intensity is a linear function in the range $\lbrack a:b\rbrack$

\begin{aligned} w(x) &=w_{a}+\left(w_{b}-w_{a}\right) \frac{(x-a)}{(b-a)} \\ &=\frac{w_{a}(b-a)}{(b-a)}+\frac{\left(w_{b}-w_{a}\right) x}{(b-a)}-\frac{\left(w_{b}-w_{a}\right) a}{(b-a)} \\ &=\frac{w_{a} b-w_{a} a-w_{b} a+w_{a} a}{(b-a)}+\frac{\left(w_{b}-w_{a}\right) x}{(b-a)} \\ &=\frac{w_{a} b-w_{b} a}{(b-a)}+\frac{\left(w_{b}-w_{a}\right) x}{(b-a)} \end{aligned}

Or

$w(x) = w_{p} + w_{q}x \quad (3)$

where

\begin{aligned} &w_{p}=\frac{w_{a} b-w_{b} a}{(b-a)} \\ &w_{q}=\frac{\left(w_{b}-w_{a}\right)}{(b-a)} \end{aligned}

Substituting equation 3 in 1 for end 0

\begin{aligned} \theta_{0} &=-\frac{1}{6 E I} \int_{a}^{b}\left(w_{p}+w_{q} x\right) \frac{x}{l}(2 l-x)(l-x) d x \\ &=-\frac{w_{p}}{6 E I} \int_{a}^{b} \frac{x}{l}(2 l-x)(l-x) d x-\frac{w_{q}}{6 E I} \int_{a}^{b} \frac{x^{2}}{l}(2 l-x)(l-x) d x \\ &=-\frac{w_{p}}{6 E I l} \int_{a}^{b} x\left(2 l^{2}-3 l x+x^{2}\right) d x-\frac{w_{q}}{6 E I} \int_{a}^{b} x^{2}\left(2 l^{2}-3 l x+x^{2}\right) d x \\ &=-\frac{w_{p}}{6 E I l} \int_{a}^{b}\left(2 l^{2} x-3 l x^{2}+x^{3}\right) d x-\frac{w_{q}}{6 E I l} \int_{a}^{b}\left(2 l^{2} x^{2}-3 l x^{3}+x^{4}\right) d x \end{aligned}

so

\begin{aligned} \theta_{0}=-\frac{w_{p}}{6 E I l} & {\left[l^{2} x^{2}-l x^{3}+\frac{x^{4}}{4}\right]_{a}^{b}-\frac{w_{q}}{6 E I l}\left[\frac{2}{3} l^{2} x^{3}-\frac{3}{4} l x^{4}+\frac{x^{5}}{5}\right]_{a}^{b} } \\ =-\frac{w_{p}}{6 E I l} & {\left[l^{2}\left(b^{2}-a^{2}\right)-l\left(b^{3}-a^{3}\right)+\frac{\left(b^{4}-a^{4}\right)}{4}\right] } \\ &-\frac{w_{q}}{6 E I l}\left[\frac{2}{3} l^{2}\left(b^{3}-a^{3}\right)-\frac{3}{4} l\left(b^{4}-a^{4}\right)+\frac{\left(b^{5}-a^{5}\right)}{5}\right] \end{aligned}

Substituting equation 3 in 2 for end 1

\begin{aligned} \theta_{1} &=-\frac{1}{6 E I} \int_{a}^{b}\left(w_{p}+w_{q} x\right) \frac{x}{l}\left(l^{2}-x^{2}\right) d x \\ &=-\frac{w_{p}}{6 E I} \int_{a}^{b} x\left(l^{2}-x^{2}\right) d x-\frac{w_{q}}{6 E I l} \int_{a}^{b} x^{2}\left(l^{2}-x^{2}\right) d x \\ &=-\frac{w_{p}}{6 E I l} \int_{a}^{b}\left(l^{2} x-x^{3}\right) d x-\frac{w_{q}}{6 E I l} \int_{a}^{b}\left(l^{2} x^{2}-x^{4}\right) d x \end{aligned}

so

\begin{aligned} \theta_{1}=-\frac{w_{p}}{6 E I l} & {\left[\frac{l^{2} x^{2}}{2}-\frac{x^{4}}{4}\right]_{a}^{b}-\frac{w_{q}}{6 E I l}\left[\frac{l^{2} x^{3}}{3}-\frac{x^{5}}{5}\right]_{a}^{b} } \\ =-\frac{w_{p}}{6 E I l} & {\left[\frac{l^{2}\left(b^{2}-a^{2}\right)}{2}-\frac{\left(b^{4}-a^{4}\right)}{4}\right] } \\ &-\frac{w_{q}}{6 E I l}\left[\frac{l^{2}\left(b^{3}-a^{3}\right)}{3}-\frac{\left(b^{5}-a^{5}\right)}{5}\right] \end{aligned}

$^{10}$ Roark Formulas for Stress and Strain, Table 3 (1.e)