Circular arch 100 m: rolling UDL 25 m, 15 m section

Data: three-hinged symmetrical circular arch, $L = 100$ m, $f = 10$ m. Radius $R = \dfrac{50^2+10^2}{2\times 10} = 130$ m. Rolling UDL 50 kN/m, length 25 m, moving left to right. Section at $x = 15$ m: $y_D = \sqrt{130^2-35^2} - 120 = 5.2$ m.

Maximum horizontal thrust

The ILD of $H$ is a triangle with peak $\dfrac{L}{4f} = 2.5$ at the crown. $H$ is maximum when the load is placed symmetrically about the crown (from 37.5 m to 62.5 m), where the ordinates are $1.875$ at the ends and $2.5$ at the middle:

$$\text{Area} = 2\times\frac{(1.875+2.5)}{2}\times 12.5 = 54.6875\ \text{m}, \qquad H_{max} = 50\times 54.6875 = 2734.4\ \text{kN}$$

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 $\dfrac{L}{4f} = 2.5$ 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, 36.59 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 25 m load is shorter than the positive region, so it is placed with equal ILD ordinates at its two ends: it covers $x = 4.75$ m to 29.75 m (the whole load is on the span).

$$\begin{aligned} M &= w\times(\text{area of ILD}) = 50\times 145.66 = 7283\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 50\times 21.5625 = 1078.13\ \text{kN} \end{aligned}$$

At the section: $y = 5.2$ m, $\theta = 15.62^\circ$, simple-beam shear $V_b = 521.88$ kN.

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 521.88\times0.9631 - 1078.13\times0.2692 = 212.34\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 521.88\times0.2692 + 1078.13\times0.9631 = 1178.82\ \text{kN} \end{aligned}$$

Maximum negative bending moment

In the same way, the load is placed with equal end ordinates in the negative region: it covers $x = 44.71$ m to 69.71 m (the whole load is on the span).

$$\begin{aligned} M &= w\times(\text{area of ILD}) = 50\times(-110.388) = -5519.4\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 50\times 52.088 = 2604.38\ \text{kN} \end{aligned}$$

At the section: $y = 5.2$ m, $\theta = 15.62^\circ$, simple-beam shear $V_b = 534.86$ kN.

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 534.86\times0.9631 - 2604.38\times0.2692 = -186.07\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 534.86\times0.2692 + 2604.38\times0.9631 = 2652.22\ \text{kN} \end{aligned}$$

Absolute maximum bending moment

The absolute maximum is found by repeating the above for every section and every position of the UDL (the 25 m load fully on the span). The largest value occurs at a section about 22 m from the support, with the load covering about 7.9 m to 32.9 m of the span:

$$M_{abs} \approx 8061\ \text{kN m (sagging, at } x \approx 22.2\ \text{m})$$

(The largest hogging moment anywhere is about 6382 kN m, near $x \approx 24$ m, which is smaller.)

Answer: $H_{max} = 2734.4$ kN; $M_{max}$ at 15 m $=+7283$ kN m (and $-5519$ kN m); absolute maximum bending moment $\approx 8061$ kN m.