= Solution
By vertical <force> balance and reflection <symmetry>, each support exerts an upward <force> $W/2$. Use the paper's convention in which the sagging <bending moment> is negative. For a cut at $0<x<L/2$, the left support's moment about that cut gives
$$
B(x)=-\frac W2x.
$$
For $L/2<x<L$, include the central downward <force> when taking moments of all loads to the left:
$$
B(x)=-\frac W2x+W\left(x-\frac L2\right)=-\frac W2(L-x).
$$
Thus
$$
\boxed{B(x)=\begin{cases}-Wx/2,&0<x<L/2,\\-W(L-x)/2,&L/2<x<L.\end{cases}}
$$
The <bending moment> is continuous at the load, with $B(L/2)=-WL/4$, and vanishes at both freely rotating supports. The jump is in its first <derivative>, reflecting the concentrated transverse <force>.
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