BM, shear, thrust at any section, UDL on arch

Data: three-hinged symmetrical parabolic arch, span $L$, rise $f$, supports at the same level, origin at the left support. UDL $w$ per unit horizontal length over the whole span (if the load is given per m², multiply by the width of the arch to get $w$ per unit length). Self-weight neglected.

Arch equation and slope

$$y = \frac{4f\,x\,(L-x)}{L^2}, \qquad \tan\theta = \frac{dy}{dx} = \frac{4f\,(L-2x)}{L^2}$$

Reactions and thrust

$$R_A = R_B = \frac{wL}{2}, \qquad H = \frac{M_C}{f} = \frac{1}{f}\left[\frac{wL}{2}\cdot\frac{L}{2} - \frac{wL}{2}\cdot\frac{L}{4}\right] = \frac{wL^2}{8f}$$

Bending moment

$$M = \frac{wLx}{2} - \frac{wx^2}{2} - H\,y = \frac{w\,x\,(L-x)}{2} - \frac{wL^2}{8f}\cdot\frac{4f\,x\,(L-x)}{L^2} = 0$$

Radial shear

The simple-beam shear at $x$ is $V_b = w\left(\dfrac{L}{2} - x\right)$.

$$V = V_b\cos\theta - H\sin\theta = \cos\theta\left[w\left(\frac{L}{2}-x\right) - \frac{wL^2}{8f}\cdot\frac{4f\,(L-2x)}{L^2}\right] = \cos\theta\left[\frac{w\,(L-2x)}{2} - \frac{w\,(L-2x)}{2}\right] = 0$$

Normal thrust

$$N = V_b\sin\theta + H\cos\theta = H\tan\theta\sin\theta + H\cos\theta = \frac{H}{\cos\theta}$$ $$N = \frac{wL^2}{8f}\sqrt{1 + \frac{16f^2\,(L-2x)^2}{L^4}}$$

Results: $M = 0$, $V = 0$ and $N = \dfrac{H}{\cos\theta}$ at every section. $N$ is smallest at the crown ($N = H = \dfrac{wL^2}{8f}$) and largest at the springings, where $N = H\sqrt{1+\left(\dfrac{4f}{L}\right)^2}$.