Arch 160 m: rolling UDL, section 50 m from right

Data: three-hinged symmetrical parabolic arch, $L = 160$ m, $f = 25$ m, $y = \dfrac{4\times 25\,x(160-x)}{160^2} = 0.00390625\,x\,(160-x)$. UDL 30 kN/m, length 60 m, rolls from left to right. The section is 50 m from the right support, i.e. at $x = 110$ m: $y = 21.484$ m, $\tan\theta = 0.00390625\,(160-220) = -0.2344$, so the tangent slopes downward to the right ($\theta = -13.19^\circ$).

For a section in the right half, the neutral point lies between the crown and the section; a UDL on the part of the span to its left gives a hogging (negative) moment, and one on the part to its right gives a sagging (positive) moment, as the ILD below shows.

ILDs for the section at x = 110 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 $\dfrac{L}{4f} = 1.6$ 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, 92.63 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 = 94.53 m to 154.53 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.

$$\begin{aligned} M &= w\times(\text{area of ILD}) = 30\times(429.016) = 12870.5\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 30\times(42.564) = 1276.92\ \text{kN} \end{aligned}$$

At the section: $y = 21.484$ m, $\theta = -13.19^\circ$, simple-beam shear $V_b = -65.06$ kN.

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = (-65.06)\times0.9736 - 1276.92\times(-0.2282) = 228.04\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = (-65.06)\times(-0.2282) + 1276.92\times0.9736 = 1258.08\ \text{kN} \end{aligned}$$

Maximum negative bending moment

The UDL covers the region from x = 28.18 m to 88.18 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.

$$\begin{aligned} M &= w\times(\text{area of ILD}) = 30\times(-380.327) = -11409.8\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 30\times(68.4778) = 2054.33\ \text{kN} \end{aligned}$$

At the section: $y = 21.484$ m, $\theta = -13.19^\circ$, simple-beam shear $V_b = -654.52$ kN.

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = (-654.52)\times0.9736 - 2054.33\times(-0.2282) = -168.48\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = (-654.52)\times(-0.2282) + 2054.33\times0.9736 = 2149.49\ \text{kN} \end{aligned}$$

Answer: at 50 m from the right support, the maximum bending moment is $+12870$ kN m (UDL covering about 94.5 m to 154.5 m), with normal thrust $1258.1$ kN and radial shear $228$ kN; the maximum negative moment is $-11410$ kN m (UDL covering about 28.2 m to 88.2 m), with normal thrust $2149.5$ kN and radial shear $-168.5$ kN.