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Moment

Point moment MMat position a

From Roark11^{11}, the end rotations for a point load at aa are

θ0=−M6EIl(2l2−6al+3a2)θ1=M6EI(l2−3a2)\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 bb these equations can be rewritten

θ0=M6EIl(l2−3b2)θ1=M6EI(l2−3a2)\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 [a:b]\lbrack a:b\rbrack with intensity mam_{a} and mbm_{b}.

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

θ0=−M6EIl(2l2−6xl+3x2)θ1=M6EI(l2−3x2)\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

θ0=−16EIl∫abm(x)(2l2−6xl+3x2)dx(4)θ1=16EI∫abm(x)(l2−3x2)dx(5)\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)=mp+mqx(6)m(x) = m_{p} + m_{q}x \quad(6)

where

mp=mab−mba(b−a)mq=(mb−ma)(b−a)\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

θ0=−16EIl∫ab(mp+mqx)(2l2−6xl+3x2)dx=−mp6EIl∫ab(2l2−6xl+3x2)dx−mq6EIl∫ab(2l2x−6x2l+3x3)dx\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

θ0=−mp6EIl[2l2x−3x2l+x3]ab−mq6EIl[l2x2−2x3l+3x44]ab=−mp6EIl[2l2(b−a)−3(b2−a2)l+(b3−a3)]−mq6EIl[l2(b2−a2)−2(b3−a3)l+3(b4−a4)4]\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

θ1=16EI∫ab(mp+mqx)(l2−3x2)dx=mp6EI∫ab(l2−3x2)dx+mq6EI∫ab(l2x−3x3)dx\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

θ1=mp6EIl[l2x−x3]ab+mq6EIl[l2x22−3x44]ab=mp6EIl[l2(b−a)−(b3−a3)]+mq6EIl[l2(b2−a2)2−3(b4−a4)4]\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^{11} Roark Formulas for Stress and Strain, Table 3 (3.e)