Arch 100 m: rolling UDL 60 m, section at 30 m

Data: three-hinged symmetrical parabolic arch, $L = 100$ m, $f = 10$ m, $y = \dfrac{4f\,x(L-x)}{L^2} = 0.004\,x\,(100-x)$. A UDL of 25 kN/m and 60 m length moves from left to right. Section $D$ at $x = 30$ m: $y_D = 8.4$ m, $\tan\theta = 0.004\,(100-60) = 0.16$, $\theta = 9.09^\circ$. (When the load covers only the first or the last part of the span, part of the 60 m load is off the span.)

ILDs for the section at x = 30 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 $\dfrac{L}{4f} = 2.5$ 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, 41.67 m from $A$ (vertically below the intersection of line $AD$ produced and line $BC$). The ordinates are positive on one side of it and negative on the other.

Maximum positive bending moment

The UDL covers the span from $x = 0$ to 41.67 m (the rest of the load is still off the span), which is exactly the region where the ILD is positive.

$$\begin{aligned} M &= w\times(\text{area of ILD}) = 25\times 175 = 4375\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 25\times 43.403 = 1085.07\ \text{kN} \end{aligned}$$

At the section: $y = 8.4$ m, $\theta = 9.09^\circ$, simple-beam shear $V_b = 74.65$ kN.

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 74.65\times0.9874 - 1085.07\times0.158 = -97.72\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 74.65\times0.158 + 1085.07\times0.9874 = 1083.24\ \text{kN} \end{aligned}$$

Maximum negative bending moment

The UDL covers the span from $x = 41.67$ m to 100 m (the rest of the load is off the span beyond $B$), which is exactly the region where the ILD is negative.

$$\begin{aligned} M &= w\times(\text{area of ILD}) = 25\times(-175) = -4375\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 25\times 81.597 = 2039.93\ \text{kN} \end{aligned}$$

At the section: $y = 8.4$ m, $\theta = 9.09^\circ$, simple-beam shear $V_b = 425.35$ kN.

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 425.35\times0.9874 - 2039.93\times0.158 = 97.72\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 425.35\times0.158 + 2039.93\times0.9874 = 2081.51\ \text{kN} \end{aligned}$$

Maximum horizontal thrust (for information)

The ILD of $H$ is positive everywhere; the thrust is greatest when the 60 m load is symmetrical about the crown (20 m to 80 m): $H_{max} = 25\times 105 = 2625$ kN.

Answer: maximum bending moment at 30 m from the left support is $+4375$ kN m (UDL from the left support up to the neutral point at 41.67 m), with normal thrust $1083.2$ kN, and $-4375$ kN m (UDL from the neutral point to the right support), with normal thrust $2081.5$ kN.