Circular arch 120 m: two rolling loads, 30 m section

Data: three-hinged circular arch, $L = 120$ m, $f = 15$ m. Radius $R = \dfrac{60^2+15^2}{2\times 15} = 127.5$ m. Two loads, 8 kN (leading) and 12 kN, 10 m apart, roll from left to right (the 12 kN load is 10 m behind the 8 kN load). Section at $x = 30$ m: $y = \sqrt{127.5^2-30^2} - 112.5 = 11.42$ m.

ILDs for the section at x = 30 m

Thrust ILD: $H = \dfrac{x'}{2f}$ (load left of the crown) or $\dfrac{L-x'}{2f}$ (right of the crown), peak $\dfrac{L}{4f} = 2$ at the crown. Moment ILD: $M = M_b - H\,y$.

The ILD of $M$ is zero at the neutral point, 47.57 m from $A$; it is positive on one side and negative on the other.

Maximum positive bending moment

The critical position is: 8 kN at 30 m, 12 kN at 20 m (measured from $A$).

$$M = \sum P\,y = 177.27\ \text{kN m}, \qquad H = \sum P\,y_H = 16\ \text{kN}$$

With $y = 11.42$ m, $\theta = 13.61^\circ$ and the simple-beam shear $V_b = 4$ kN at the section:

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 0.12\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 16.49\ \text{kN} \end{aligned}$$

Maximum negative bending moment

The critical position is: 8 kN at 70 m, 12 kN at 60 m (measured from $A$).

$$M = \sum P\,y = -146.36\ \text{kN m}, \qquad H = \sum P\,y_H = 37.33\ \text{kN}$$

With $y = 11.42$ m, $\theta = 13.61^\circ$ and the simple-beam shear $V_b = 9.33$ kN at the section:

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 0.29\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 38.48\ \text{kN} \end{aligned}$$

Answer: at 30 m from the left support: maximum positive bending moment $=+177.27$ kN m with normal thrust $16.49$ kN and radial shear $0.12$ kN; maximum negative bending moment $=-146.36$ kN m with normal thrust $38.48$ kN and radial shear $0.29$ kN.