Three-span continuous beam under uniformly distributed load

Figure 1: Diagram of problem
Figure 1: Diagram of problem

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.

Figure 2: Superposition of simpler cases
Figure 2: Superposition of simpler cases

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):

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