Parabolic arch 20 m: two loads and UDL, 5 m section

Data: parabolic three-hinged arch, $L = 20$ m, rise $f = 5$ m. Loads: 20 kN at 3 m and 30 kN at 7 m from $A$; UDL 25 kN/m on the right half ($x = 10$ to 20 m). Section at $x = 5$ m from $A$.

Step 1: Equation of the arch

$$y = \frac{4f\,x\,(L-x)}{L^2} = \frac{4\times 5\,x\,(20-x)}{20^2}$$ $$\tan\theta = \frac{dy}{dx} = \frac{4f\,(L-2x)}{L^2}$$

At the section $x = 5$ m: $y = 3.75$ m, $\tan\theta = 0.5$, so $\theta = 26.57^\circ$, $\sin\theta = 0.4472$, $\cos\theta = 0.8944$.

Step 2: Vertical reactions

Taking moments about $A$ for the whole arch:

$$R_B \times 20 = 20\times 3 + 30\times 7 + (25\times 10)\times 15 = 4020 \Rightarrow R_B = 201\ \text{kN}$$ $$R_A = 300 - 201 = 99\ \text{kN}$$

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):

$$H \times 5 = 99\times 10 - 20\times 7 - 30\times 3 = 760 \Rightarrow H = 152\ \text{kN}$$

Step 4: Forces at the section (x = 5 m)

Simple-beam values at the section: $M_b = 455$ kN m, $V_b = 79$ kN.

$$\begin{aligned} M &= M_b - H\,y = 455 - 152\times 3.75 = -115\ \text{kN m}\\ V &= V_b\cos\theta - H\sin\theta = 79\times 0.8944 - 152\times 0.4472 = 2.68\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 79\times 0.4472 + 152\times 0.8944 = 171.28\ \text{kN} \end{aligned}$$

Answer: at 5 m from the left support: $M = -115$ kN m, normal thrust $N = 171.28$ kN, radial shear $V = 2.68$ kN.