No bending moment under UDL in parabolic arch

Statement: a three-hinged parabolic arch carrying a UDL $w$ per unit horizontal length over the whole span has zero bending moment at every section.

Let the span be $L$ and the rise $f$, supports at the same level, origin at the left support.

Arch equation

$$y = \frac{4f\,x\,(L-x)}{L^2}$$

Reactions and thrust

By symmetry $R_A = R_B = \dfrac{wL}{2}$. The bending moment at the crown hinge is zero, so, taking moments of the left half about $C$:

$$H f = \frac{wL}{2}\times\frac{L}{2} - w\,\frac{L}{2}\times\frac{L}{4} = \frac{wL^2}{8} \Rightarrow H = \frac{wL^2}{8f}$$

Bending moment at a section at distance x

$$\begin{aligned} M_x &= R_A\,x - \frac{w\,x^2}{2} - H\,y \\ &= \frac{wLx}{2} - \frac{wx^2}{2} - \frac{wL^2}{8f}\cdot\frac{4f\,x\,(L-x)}{L^2} \\ &= \frac{w\,x\,(L-x)}{2} - \frac{w\,x\,(L-x)}{2} = 0 \end{aligned}$$

So $M_x = 0$ for every $x$. The beam bending moment $\dfrac{wx(L-x)}{2}$ is exactly cancelled by $H\,y$, because both are parabolic with the same shape.

Conclusion: the parabola is the funicular (equilibrium) shape for a UDL on the horizontal span. The arch is then in pure compression (thrust) with no bending and no radial shear.