Circular arch: two rolling loads 20 kN and 30 kN

Data: symmetrical circular three-hinged arch, $L = 100$ m, $f = 10$ m, $R = 130$ m. Two loads, 20 kN (leading) and 30 kN, 5 m apart, roll from left to right (the 30 kN load is 5 m behind the 20 kN load). Section at $x = 25$ m: $y = \sqrt{130^2-25^2} - 120 = 7.574$ m.

ILDs for the section at x = 25 m

Thrust ILD: $H = \dfrac{x'}{2f}$ (load left of the crown) or $\dfrac{L-x'}{2f}$ (right of the crown), peak $\dfrac{L}{4f} = 2.5$ at the crown. Moment ILD: $M = M_b - H\,y$.

The ILD of $M$ is zero at the neutral point, 39.77 m from $A$; it is positive on one side and negative on the other.

Maximum positive bending moment

The critical position is: 20 kN at 25 m, 30 kN at 20 m (measured from $A$).

$$M = \sum P\,y = 408.46\ \text{kN m}, \qquad H = \sum P\,y_H = 55\ \text{kN}$$

With $y = 7.574$ m, $\theta = 11.09^\circ$ and the simple-beam shear $V_b = 9$ kN at the section:

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = -1.74\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 55.7\ \text{kN} \end{aligned}$$

Maximum negative bending moment

The critical position is: 20 kN at 55 m, 30 kN at 50 m (measured from $A$).

$$M = \sum P\,y = -308.82\ \text{kN m}, \qquad H = \sum P\,y_H = 120\ \text{kN}$$

With $y = 7.574$ m, $\theta = 11.09^\circ$ and the simple-beam shear $V_b = 24$ kN at the section:

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 0.48\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 122.38\ \text{kN} \end{aligned}$$

Answer: maximum positive moment at 25 m $= +408.5$ kN m with $N = 55.7$ kN and $V = -1.74$ kN; maximum negative moment $= -308.8$ kN m with $N = 122.4$ kN and $V = 0.48$ kN.