Circular arch 50 m: rolling UDL, 15 m section

Data: symmetrical circular three-hinged arch, $L = 50$ m, $f = 10$ m, $R = \dfrac{25^2+10^2}{2\times 10} = 36.25$ m. Rolling UDL 50 kN/m, 10 m long, from left to right. Section at $x = 15$ m: $y = \sqrt{36.25^2-10^2} - 26.25 = 8.593$ m.

ILDs for the section at x = 15 m

The ILD of the horizontal thrust is $H = \dfrac{x'}{2f}$ for a unit load at $x'$ left of the crown and $\dfrac{L-x'}{2f}$ right of it (peak $\dfrac{L}{4f} = 1.25$ at the crown). The ILD of the bending moment at the section is $M = M_b - H\,y$.

The ILD of $M$ crosses zero at the neutral point, 20.56 m from $A$ (the vertical through the intersection of line $A$–$D$ extended and line $B$–$C$ extended). The ordinates are positive on one side of it and negative on the other.

Maximum positive bending moment

The UDL covers the region from x = 7.7 m to 17.7 m (the whole load is on the span); this lies in the positive region of the ILD, with equal ordinates at the two ends of the load.

$$\begin{aligned} M &= w\times(\text{area of ILD}) = 50\times(30.687) = 1534.3\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 50\times(6.352) = 317.58\ \text{kN} \end{aligned}$$

At the section: $y = 8.593$ m, $\theta = 16.01^\circ$, simple-beam shear $V_b = 8.13$ kN.

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 8.13\times0.9612 - 317.58\times0.2759 = -79.79\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 8.13\times0.2759 + 317.58\times0.9612 = 307.50\ \text{kN} \end{aligned}$$

Maximum negative bending moment

The UDL covers the region from x = 23.49 m to 33.49 m (the whole load is on the span); this lies in the negative region of the ILD, with equal ordinates at the two ends of the load.

$$\begin{aligned} M &= w\times(\text{area of ILD}) = 50\times(-26.912) = -1345.6\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 50\times(10.641) = 532.05\ \text{kN} \end{aligned}$$

At the section: $y = 8.593$ m, $\theta = 16.01^\circ$, simple-beam shear $V_b = 215.1$ kN.

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 215.1\times0.9612 - 532.05\times0.2759 = 59.98\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 215.1\times0.2759 + 532.05\times0.9612 = 570.74\ \text{kN} \end{aligned}$$

Answer: maximum positive bending moment $= +1534.3$ kN m (radial shear $-79.8$ kN, normal thrust $307.5$ kN); maximum negative bending moment $= -1345.6$ kN m (radial shear $59.98$ kN, normal thrust $570.7$ kN).