Parabolic arch 80 m: inclined load, forces at D

Data and assumptions: parabolic three-hinged arch, span $L = 80$ m, rise $f = 8$ m, $y = \dfrac{4\times 8\,x\,(80-x)}{80^2} = 0.005\,x\,(80-x)$. The 250 kN inclined load acts at $x = 30$ m (10 m left of the crown), pointing downward and to the right at 30° to the horizontal (assumed). Its components: horizontal $250\cos30^\circ = 216.51$ kN (to the right), vertical $250\sin30^\circ = 125$ kN (downward). 100 kN vertical load at 15 m from $B$ ($x = 65$ m). Section $D$ at $x = 15$ m.

Heights of the loads: $y(30) = 7.5$ m, $y(65) = 4.875$ m; at $D$: $y = 4.875$ m, $\tan\theta = 0.005\,(80-30) = 0.25$, $\theta = 14.04^\circ$.

Let the thrusts be $H_A$ (to the right at $A$) and $H_B$ (to the left at $B$), and the vertical reactions $V_A$, $V_B$.

Reactions

Horizontal equilibrium:

$$H_A - H_B + 216.51 = 0 \Rightarrow H_B = H_A + 216.51$$

Moments about $A$:

$$V_B\times 80 = 100\times 65 + 125\times 30 + 216.51\times 7.5 \Rightarrow V_B = \frac{11873.8}{80} = 148.42\ \text{kN}$$ $$V_A = 225 - 148.42 = 76.58\ \text{kN}$$

Hinge condition (moment about $C$ of the forces on the left part, $C$ at $(40, 8)$):

$$V_A\times 40 - H_A\times 8 - 125\times 10 - 216.51\times(8-7.5) = 0$$ $$H_A = \frac{76.58\times 40 - 1250 - 216.51\times 0.5}{8} = 213.11\ \text{kN}, \qquad H_B = 429.61\ \text{kN}$$

Forces at D (left part, 15 m from A, no load on it)

$$\begin{aligned} M &= V_A\times 15 - H_A\times y_D = 76.58\times 15 - 213.11\times 4.875 = 109.77\ \text{kN m}\\ V &= V_A\cos\theta - H_A\sin\theta = 76.58(0.9701) - 213.11(0.2425) = 22.61\ \text{kN}\\ N &= V_A\sin\theta + H_A\cos\theta = 76.58(0.2425) + 213.11(0.9701) = 225.32\ \text{kN} \end{aligned}$$

Answer: at $D$: bending moment $= 109.77$ kN m (sagging), normal thrust $= 225.32$ kN, radial shear $= 22.61$ kN.