Circular arch: BM, shear, thrust at D

Data: three-hinged circular arch, span $L = 60$ m, rise $f = 5$ m (crown hinge $C$ at mid-span). UDL 25 kN/m on the left half ($x = 0$ to 30 m); 500 kN vertical load on the right half, 15 m from $B$ ($x = 45$ m). Section $D$ is 15 m horizontally from $A$.

Step 1: Equation of the arch

$$y = \sqrt{R^2-(x-30)^2} - (R-f),\quad R = \frac{L^2/4+f^2}{2f} = \frac{60^2/4+5^2}{2\times 5} = 92.5\ \text{m}$$ $$\tan\theta = \frac{dy}{dx} = \frac{30-x}{\sqrt{R^2-(x-30)^2}}$$

At the section $x = 15$ m: $y = 3.776$ m, $\tan\theta = 0.1643$, so $\theta = 9.33^\circ$, $\sin\theta = 0.1622$, $\cos\theta = 0.9868$.

Step 2: Vertical reactions

Taking moments about $A$ for the whole arch:

$$R_B \times 60 = (25\times 30)\times 15 + 500\times 45 = 33750 \Rightarrow R_B = 562.5\ \text{kN}$$ $$R_A = 1250 - 562.5 = 687.5\ \text{kN}$$

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 $f = 5$ m):

$$H \times 5 = 687.5\times 30 - (25\times 30)\times 15 = 9375 \Rightarrow H = 1875\ \text{kN}$$

Step 4: Forces at the section (x = 15 m)

Simple-beam values at the section: $M_b = 7500$ kN m, $V_b = 312.5$ kN.

$$\begin{aligned} M &= M_b - H\,y = 7500 - 1875\times 3.776 = 420.6\ \text{kN m}\\ V &= V_b\cos\theta - H\sin\theta = 312.5\times 0.9868 - 1875\times 0.1622 = 4.31\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 312.5\times 0.1622 + 1875\times 0.9868 = 1900.86\ \text{kN} \end{aligned}$$

Answer: at $D$: bending moment $= 420.6$ kN m (sagging), radial shear $= 4.31$ kN, normal thrust $= 1900.86$ kN.