# Three-span continuous beam under uniformly distributed load

### Solution

\begin{align} a & = \frac{l}{L} \\ A & = \frac{1 - 2a^{2} + a^{3}}{12a - 16a^{2}} \\ M_{mid} & = \frac{wL}{8}(L - 8lA) \\ M_{pier} & = \frac{wl}{2} \bigg( l + L(2A - 1) \bigg) \end{align}

Symbol Description
$$M_{mid}$$ Midspan moment (positive sagging)
$$M_{pier}$$ Pier moment (positive hogging)
$$w$$ Uniformly distributed load (force per length)
$$l$$ Distance between abutment and pier
$$L$$ Total length of structure

### Derivation

Consider the problem by superposition of a simply supported beam under a uniformly distributed load and a simply supported beam under two symmetric point loads as shown in Figure 2.

We then note that the total deflection at $$x = l$$ is equal to $$0$$:

\begin{align} \frac{PL^{3}}{6EI}\Bigg[3\bigg(\frac{l}{L}\bigg)^{2}-4\bigg(\frac{l}{L}\bigg)^{3}\Bigg] &= \frac{w}{12EI}\Bigg[\frac{l^{4}}{2} - Ll^{3} + \frac{L^{3}l}{2}\Bigg] \\ \frac{PL^{3}}{24EI}\Bigg[12\bigg(\frac{l}{L}\bigg)^{2}-16\bigg(\frac{l}{L}\bigg)^{3}\Bigg] &= \frac{wL^{4}}{24EI}\Bigg[\bigg(\frac{l}{L}\bigg)^{4}-2\bigg(\frac{l}{L}\bigg)^{3}+\bigg(\frac{l}{L}\bigg)\Bigg] \end{align}

$\displaystyle \textrm{Let }a = \frac{l}{L}$

\begin{align} P[12a^{2} - 16a^{3}] &= wL[a^{4} - 2a^{3} + a] \\ 4Pa(3 - 4a) & = wLa(a^{3} - 2a^{2} + 1) \\ P &= \frac{wL(a^{3} - 2a^{2} + 1)}{4a(3 - 4a)} \\ P &= \frac{1 - 2a^{2} + a^{3}}{12a - 16a^{2}}wL \end{align}

$\displaystyle \textrm{Let }A = \frac{1 - 2a^{2} + a^{3}}{12a - 16a^{2}}$

$P = AwL$

Expression for the mid-span moment from superposition (positive sagging):

\begin{align} M_{mid} &= \frac{wL^{2}}{8} - Pl \\ &= \frac{wL^{2}}{8} - AwLl \\ &= wL \bigg( \frac{L}{8} - lA \bigg) \\ &= \frac{wL}{8}(L - 8lA) \end{align}

Expression for the moment over the pier from superposition (positive hogging):

\begin{align} M_{pier} &= \bigg( \frac{wl^{2}}{2} - \frac{wLl}{2} \bigg) - Pl \\ & = \frac{wl}{2}\bigg(l - L\bigg) + AwLl \\ & = wl\bigg(\frac{l}{2} - \frac{L}{2} + AL\bigg) \\ & = \frac{wl}{2} \bigg( l + L(2A - 1) \bigg) \end{align}