Parabolic arch 100 m: UDL left half and 80 kN

Data: parabolic three-hinged arch, $L = 100$ m, rise $f = 10$ m at the crown $C$. UDL 50 kN/m on the left half ($x = 0$ to 50 m); 80 kN vertical load 25 m from $B$ ($x = 75$ m). Section $D$ at $x = 20$ m from $A$.

Step 1: Equation of the arch

$$y = \frac{4f\,x\,(L-x)}{L^2} = \frac{4\times 10\,x\,(100-x)}{100^2}$$ $$\tan\theta = \frac{dy}{dx} = \frac{4f\,(L-2x)}{L^2}$$

At the section $x = 20$ m: $y = 6.4$ m, $\tan\theta = 0.24$, so $\theta = 13.5^\circ$, $\sin\theta = 0.2334$, $\cos\theta = 0.9724$.

Step 2: Vertical reactions

Taking moments about $A$ for the whole arch:

$$R_B \times 100 = (50\times 50)\times 25 + 80\times 75 = 68500 \Rightarrow R_B = 685\ \text{kN}$$ $$R_A = 2580 - 685 = 1895\ \text{kN}$$

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):

$$H \times 10 = 1895\times 50 - (50\times 50)\times 25 = 32250 \Rightarrow H = 3225\ \text{kN}$$

Step 4: Forces at the section (x = 20 m)

Simple-beam values at the section: $M_b = 27900$ kN m, $V_b = 895$ kN.

$$\begin{aligned} M &= M_b - H\,y = 27900 - 3225\times 6.4 = 7260\ \text{kN m}\\ V &= V_b\cos\theta - H\sin\theta = 895\times 0.9724 - 3225\times 0.2334 = 117.66\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 895\times 0.2334 + 3225\times 0.9724 = 3344.82\ \text{kN} \end{aligned}$$

Answer: at $D$: bending moment $= 7260$ kN m, normal thrust $= 3344.82$ kN, radial shear $= 117.66$ kN.