Arch 100 m rise 12 m: rolling UDL 60 m, section 10 m

Data: parabolic three-hinged arch, $L = 100$ m, $f = 12$ m, $y = 0.0048\,x(100-x)$. A UDL of 20 kN/m, 60 m long, rolls from left to right. Section at $x = 10$ m: $y = 4.32$ m, $\tan\theta = 0.0048\,(100-20) = 0.384$.

Horizontal thrust

$H$ is maximum when the load is symmetrical about the crown (20 m to 80 m): the ILD ordinates are $\dfrac{x'}{2f} = 0.833$ at 20 m and $\dfrac{L}{4f} = 2.083$ at the crown:

$$H_{max} = 20\times\left[2\times\frac{(0.833+2.083)}{2}\times 30\right] = 1750\ \text{kN}$$

ILDs for the section at x = 10 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 $\dfrac{L}{4f} = 2.0833$ 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, 35.71 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 35.71 m (the rest of the load is still off the span), which is exactly the region where the ILD is positive.

$$\begin{aligned} M &= w\times(\text{area of ILD}) = 20\times 128.571 = 2571.4\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 20\times 26.573 = 531.46\ \text{kN} \end{aligned}$$

At the section: $y = 4.32$ m, $\theta = 21.01^\circ$, simple-beam shear $V_b = 386.73$ kN.

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 386.73\times0.9335 - 531.46\times0.3585 = 170.51\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 386.73\times0.3585 + 531.46\times0.9335 = 634.78\ \text{kN} \end{aligned}$$

Maximum negative bending moment

The negative region (35.71 m to 100 m) is longer than the load, so the 60 m load is placed with equal ILD ordinates at its two ends: it covers $x = 36.67$ m to 96.67 m (the whole load is on the span).

$$\begin{aligned} M &= w\times(\text{area of ILD}) = 20\times(-128) = -2560\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 20\times 75.926 = 1518.52\ \text{kN} \end{aligned}$$

At the section: $y = 4.32$ m, $\theta = 21.01^\circ$, simple-beam shear $V_b = 400.0$ kN.

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 400.0\times0.9335 - 1518.52\times0.3585 = -170.94\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 400.0\times0.3585 + 1518.52\times0.9335 = 1560.99\ \text{kN} \end{aligned}$$

Answer: maximum thrust $= 1750$ kN; maximum bending moment at 10 m from the left end $= +2571$ kN m (load from the support up to the neutral point), with thrust $H = 531.5$ kN, and $-2560$ kN m (negative), with $H = 1518.5$ kN.