Circular arch ILDs at 3 m, then internal forces

Data: three-hinged symmetrical circular arch, $L = 12$ m, $f = 4$ m. Radius $R = \dfrac{L^2/4+f^2}{2f} = \dfrac{36+16}{8} = 6.5$ m (centre 2.5 m below the springing level). Section $D$ at $x = 3$ m from $A$:

$$y = \sqrt{6.5^2-(3-6)^2} - 2.5 = 3.266\ \text{m}, \qquad \tan\theta = \frac{6-3}{5.766} = 0.5203, \ \theta = 27.49^\circ$$

Influence lines (unit load at $x'$)

• Thrust: $H = \dfrac{x'}{2f} = \dfrac{x'}{8}$ for $x'\le 6$; $\dfrac{12-x'}{8}$ for $x'\ge 6$ (peak $0.75$ at the crown).
• Bending moment: $M = M_b - H\,y$, with $M_b = \dfrac{x'(12-3)}{12}$ for $x'\le 3$ and $\dfrac{3(12-x')}{12}$ for $x'\ge 3$.
• Radial shear: $V = V_b\cos\theta - H\sin\theta$ with $V_b = -\dfrac{x'}{12}$ (load left of $D$) or $\dfrac{12-x'}{12}$ (load right of $D$).
• Normal thrust: $N = V_b\sin\theta + H\cos\theta$.

The M-ILD is positive near $D$ and negative towards the crown; the V-ILD has a jump of $\cos\theta = 0.8871$ at $D$; the N-ILD has a jump of $\sin\theta = 0.4615$ at $D$ (shown by the two rows at 3 m).

Loading

UDL 20 kN/m on the left half ($x' = 0$ to 6 m) and a 40 kN point load at 3 m from the right support ($x' = 9$ m).

Each quantity $= 20\times(\text{area of ILD from 0 to 6}) + 40\times(\text{ordinate at 9 m})$. The areas are obtained by the trapezoidal rule on each straight segment (using the values immediately left and right of $D$):

$$\begin{aligned} M &= 14.02\ \text{kN m}\\ V &= 7.79\ \text{kN}\\ N &= 71.69\ \text{kN}\\ H &= 60\ \text{kN} \end{aligned}$$

Check (direct statics): $R_A = 100$ kN, $R_B = 60$ kN, $H = 60$ kN; $M = 100\times 3 - 20\times 3\times 1.5 - 60\times 3.266 = 14.02$ kN m.

Answer: bending moment $14.02$ kN m (sagging), radial shear $7.79$ kN, normal thrust $71.69$ kN at the section 3 m from the left support.