Arch equation given: BM, thrust, shear at 5 m

Data: $y = x - \dfrac{x^2}{40}$ with the origin at $A$, span $L = 40$ m. Comparing with $y = \dfrac{4f\,x\,(L-x)}{L^2} = \dfrac{4f}{L}x - \dfrac{4f}{L^2}x^2$: $\dfrac{4f}{L} = 1 \Rightarrow f = 10$ m (the arch is parabolic with a rise of 10 m). UDL 30 kN/m on the right half ($x = 20$ to 40 m); the section is 5 m from $B$, i.e. at $x = 35$ m.

Step 1: Equation and slope

$$y = x - \frac{x^2}{40}$$ $$\tan\theta = \frac{dy}{dx} = 1 - \frac{x}{20}$$

At the section $x = 35$ m: $y = 4.375$ m, $\tan\theta = -0.75$, so $\theta = -36.87^\circ$, $\sin\theta = -0.6$, $\cos\theta = 0.8$.

Step 2: Vertical reactions

Taking moments about $A$ for the whole arch:

$$R_B \times 40 = (30\times 20)\times 30 = 18000 \Rightarrow R_B = 450\ \text{kN}$$ $$R_A = 600 - 450 = 150\ \text{kN}$$

Step 3: Horizontal thrust

The bending moment at the crown hinge $C$ is zero. Taking moments of the forces on the left of $C$ about $C$ (rise $f = 10$ m):

$$H \times 10 = 150\times 20 = 3000 \Rightarrow H = 300\ \text{kN}$$

Step 4: Forces at the section (x = 35 m)

Simple-beam values at the section: $M_b = 1875$ kN m, $V_b = -300$ kN.

$$\begin{aligned} M &= M_b - H\,y = 1875 - 300\times 4.375 = 562.5\ \text{kN m}\\ V &= V_b\cos\theta - H\sin\theta = (-300)\times 0.8 - 300\times (-0.6) = -60\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = (-300)\times (-0.6) + 300\times 0.8 = 420\ \text{kN} \end{aligned}$$

Answer: at 5 m from the right support: $M = 562.5$ kN m, normal thrust $N = 420$ kN, radial shear $V = -60$ kN.