Two-hinged arch: BM at section x-x

Reading of the figure: span $L = 18$ m, rise $h = 3$ m; A is at the left support. The dimensions 2, 6, 8 and 2 m are read as: section x-x at 2 m from A; the UDL of 60 kN/m covers the first 8 m (2 m + 6 m); the 85 kN load acts 16 m from A (8 m after the UDL, 2 m before B). The arch is taken as $I = I_c\sec\theta$ (so $EI_c = 10000$ kNm$^2$ cancels in $H$).

Arch profile

$$y = \frac{4h\,x(L-x)}{L^2} = \frac{x(18-x)}{27}$$

Reactions of the equivalent simple beam

UDL resultant $= 60\times8 = 480$ kN at 4 m.

$$V_B = \frac{480(4) + 85(16)}{18} = 182.222\ \text{kN},\qquad V_A = 565 - 182.222 = 382.778\ \text{kN}$$

Free bending moment $M_0$ (kNm)

Horizontal thrust

$$\int_0^{18} y^2\,dx = \frac{8h^2L}{15} = 86.4$$ $$\int M_0\,y\,dx = 15836.2 + 15802.5 + 297.0 = 31935.6$$ $$H = \frac{\int M_0 y\,dx}{\int y^2\,dx} = \frac{31935.6}{86.4} = 369.63\ \text{kN}$$

Bending moment at section x-x ($x = 2$ m)

$$\begin{aligned} y &= \frac{2(16)}{27} = 1.1852\ \text{m} \\ M_0 &= 382.778(2) - 30(2)^2 = 645.56\ \text{kNm} \\ M_{xx} &= M_0 - Hy = 645.56 - 369.63(1.1852) = 207.48\ \text{kNm} \end{aligned}$$

Answer: $H = 369.63$ kN and the bending moment at x-x is $M = 207.48$ kNm (sagging).