Parabolic arch 90 m: BM, shear, thrust at D

Data: parabolic three-hinged arch, $L = 90$ m, $f = 6$ m. 180 kN vertical load at 25 m from $A$; UDL 6 kN/m on the right half ($x = 45$ to 90 m). Section $D$ at $x = 30$ m from $A$.

Step 1: Equation of the arch

$$y = \frac{4f\,x\,(L-x)}{L^2} = \frac{4\times 6\,x\,(90-x)}{90^2}$$ $$\tan\theta = \frac{dy}{dx} = \frac{4f\,(L-2x)}{L^2}$$

At the section $x = 30$ m: $y = 5.333$ m, $\tan\theta = 0.0889$, so $\theta = 5.08^\circ$, $\sin\theta = 0.0885$, $\cos\theta = 0.9961$.

Step 2: Vertical reactions

Taking moments about $A$ for the whole arch:

$$R_B \times 90 = 180\times 25 + (6\times 45)\times 67.5 = 22725 \Rightarrow R_B = 252.5\ \text{kN}$$ $$R_A = 450 - 252.5 = 197.5\ \text{kN}$$

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 $f = 6$ m):

$$H \times 6 = 197.5\times 45 - 180\times 20 = 5287.5 \Rightarrow H = 881.25\ \text{kN}$$

Step 4: Forces at the section (x = 30 m)

Simple-beam values at the section: $M_b = 5025$ kN m, $V_b = 17.5$ kN.

$$\begin{aligned} M &= M_b - H\,y = 5025 - 881.25\times 5.333 = 325\ \text{kN m}\\ V &= V_b\cos\theta - H\sin\theta = 17.5\times 0.9961 - 881.25\times 0.0885 = -60.59\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 17.5\times 0.0885 + 881.25\times 0.9961 = 879.34\ \text{kN} \end{aligned}$$

Answer: at $D$: bending moment $= 325$ kN m, radial shear $= -60.59$ kN, normal thrust $= 879.34$ kN.