Arch 100 m rise 15 m: rolling UDL 10 m, section 60 m

Data: parabolic three-hinged arch, $L = 100$ m, $f = 15$ m, $y = 0.006\,x(100-x)$. Rolling UDL 50 kN/m of length 10 m. Section $D$ at $x = 60$ m (in the right half): $y = 14.4$ m, $\tan\theta = 0.006\,(100-120) = -0.12$, so the tangent falls to the right ($\theta = -6.84^\circ$).

For a section in the right half, the neutral point lies between the crown and the section.

ILDs for the section at x = 60 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 $\dfrac{L}{4f} = 1.6667$ 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, 54.55 m from $A$ (the vertical through the intersection of line $A$–$C$ extended and line $B$–$D$ 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 = 58.8 m to 68.8 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.

$$\begin{aligned} M &= w\times(\text{area of ILD}) = 50\times(42.72) = 2136\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 50\times(12.0667) = 603.33\ \text{kN} \end{aligned}$$

At the section: $y = 14.4$ m, $\theta = -6.84^\circ$, simple-beam shear $V_b = 121$ kN.

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 121\times0.9929 - 603.33\times(-0.1191) = 192.02\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 121\times(-0.1191) + 603.33\times0.9929 = 584.62\ \text{kN} \end{aligned}$$

Maximum negative bending moment

The UDL covers the region from x = 40.83 m to 50.83 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.

$$\begin{aligned} M &= w\times(\text{area of ILD}) = 50\times(-36.333) = -1816.7\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 50\times(15.2537) = 762.69\ \text{kN} \end{aligned}$$

At the section: $y = 14.4$ m, $\theta = -6.84^\circ$, simple-beam shear $V_b = -229.15$ kN.

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = (-229.15)\times0.9929 - 762.69\times(-0.1191) = -136.65\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = (-229.15)\times(-0.1191) + 762.69\times0.9929 = 784.55\ \text{kN} \end{aligned}$$

Answer: at 60 m from the left support the maximum bending moments are $+2136$ kN m (radial shear $192$ kN, normal thrust $584.6$ kN) and $-1817$ kN m (radial shear $-136.6$ kN, normal thrust $784.6$ kN), with the 10 m UDL in the positions stated above.