Arch 32 m: two rolling loads, max +/- moment at 8 m

Data: symmetrical parabolic three-hinged arch, $L = 32$ m, $f = 8$ m, $y = \dfrac{4\times 8\,x(32-x)}{32^2} = \dfrac{x(32-x)}{32}$. Two loads, 80 kN (leading) and 40 kN, 3 m apart, move from right to left (so the 40 kN load is 3 m to the right of the 80 kN load). Section at $x = 8$ m from the left support: $y = 6$ m, $\tan\theta = \dfrac{32-16}{32} = 0.5$, $\theta = 26.57^\circ$.

ILDs for the section at x = 8 m

Thrust ILD: $H = \dfrac{x'}{2f}$ (load left of the crown) or $\dfrac{L-x'}{2f}$ (right of the crown), peak $\dfrac{L}{4f} = 1$ at the crown. Moment ILD: $M = M_b - H\,y$.

The ILD of $M$ is zero at the neutral point, 12.8 m from $A$; it is positive on one side and negative on the other.

Maximum positive bending moment

The critical position is: 80 kN at 8 m, 40 kN at 11 m (measured from $A$).

$$M = \sum P\,y = 285\ \text{kN m}, \qquad H = \sum P\,y_H = 67.5\ \text{kN}$$

With $y = 6$ m, $\theta = 26.57^\circ$ and the simple-beam shear $V_b = 86.25$ kN at the section:

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 46.96\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 98.95\ \text{kN} \end{aligned}$$

Maximum negative bending moment

The critical position is: 80 kN at 16 m, 40 kN at 19 m (measured from $A$).

$$M = \sum P\,y = -225\ \text{kN m}, \qquad H = \sum P\,y_H = 112.5\ \text{kN}$$

With $y = 6$ m, $\theta = 26.57^\circ$ and the simple-beam shear $V_b = 56.25$ kN at the section:

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 0\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 125.78\ \text{kN} \end{aligned}$$

Answer: maximum positive moment $= +285$ kN m (80 kN at the section, 40 kN 3 m to its right); maximum negative moment $= -225$ kN m (80 kN at 16 m, i.e. at the crown, and 40 kN at 19 m).