$M_n(++\frac{L-x}{x})=(w_{n}L)()(L-x)(\frac{1}{2}).$

Solving this equation for $M_n,$ taking $d\frac{M_n}{d}x=0,$ the value of $x$ can be calculated to equal $0\mathrm{.}414$ L $.$ This value is also applicable to uniformly loaded end spans of continuous beams with simple end supports, as will be illustrated in the next section.

FIGURE $8\mathrm{.}15$