# Moment

Point moment $M$at position a

From Roark$^{11}$, the end rotations for a point load at $a$ are

\begin{aligned} \theta_{0} &= - \frac{M}{6EIl}\left( 2l^{2} - 6al + 3a^{2} \right)\\ \theta_{1} &= \frac{M}{6EI}\left( l^{2} - 3a^{2} \right) \end{aligned}

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

\begin{aligned} \theta_{0} &= \frac{M}{6EIl}\left( l^{2} - 3b^{2} \right) \\ \theta_{1} &= \frac{M}{6EI}\left( l^{2} - 3a^{2} \right) \end{aligned}

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

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

\begin{aligned} \theta_{0} &= - \frac{M}{6EIl}\left( 2l^{2} - 6xl + 3x^{2} \right) \\ \theta_{1} &= \frac{M}{6EI}\left( l^{2} - 3x^{2} \right) \end{aligned}

Using these and integrating over the element gives

\begin{aligned} \theta_{0} &= - \frac{1}{6EIl}\int_{a}^{b}{m(x)\left( 2l^{2} - 6xl + 3x^{2} \right)dx} &(4) \\ \theta_{1} &= \frac{1}{6EI}\int_{a}^{b}{m(x)\left( l^{2} - 3x^{2} \right)dx} &(5) \end{aligned}

As with the forces the moment intensity can be written as

$m(x) = m_{p} + m_{q}x \quad(6)$

where

$\begin{gathered} m_{p}=\frac{m_{a} b-m_{b} a}{(b-a)} \\ m_{q}=\frac{\left(m_{b}-m_{a}\right)}{(b-a)} \end{gathered}$

Substituting equation 6 in 4 for end 0

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

so

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

Substituting equation 6 in 5 for end 1

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

so

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

$^{11}$ Roark Formulas for Stress and Strain, Table 3 (3.e)