ILD for bending moment at section D of arch

Data: three-hinged circular arch, $L = 60$ m, $f = 5$ m (radius $R = \dfrac{900+25}{10} = 92.5$ m), crown hinge $C$ at 30 m. Section $D$ in the left half at $x = 15$ m from $A$ (as in the loaded arch of the companion problem): $y_D = 3.776$ m.

Principle

For a unit load at $x'$:

• $R_A = \dfrac{L-x'}{L}$, thrust $H = \dfrac{M_C}{f}$ with $M_C = \dfrac{x'}{2}$ (load on the left half) or $\dfrac{L-x'}{2}$ (right half).
• $M_D = R_A\,x_D - H\,y_D$ for a load right of $D$, and $M_D = R_B\,(L-x_D) - H\,y_D$ for a load left of $D$.

$$H = \frac{x'}{2f} = \frac{x'}{10} \ (x'\le 30), \qquad \frac{60-x'}{10}\ (x'\ge 30)$$

Ordinates

For a load at $D$ itself ($x'=15$): $M = \dfrac{15\times 45}{60} - \dfrac{15}{10}\times 3.776 = 5.586$ m. At the crown ($x'=30$): $M = 0.5\times 15 - 3\times 3.776 = -3.827$ m.

Neutral point

$$s_0 = \frac{2f}{\dfrac{2f}{L}+\dfrac{y_D}{x_D}} = \frac{10}{\dfrac{10}{60}+\dfrac{3.776}{15}} = 23.9\ \text{m from } A$$

Shape of the ILD

  ILD of M_D (m), straight between the points:
   A = 0  ->  D = +5.586  ->  N = 0  ->  C = -3.827  ->  B = 0

The ILD of $M_D$ is a straight line from 0 at $A$ to $+5.586$ m at $D$, then a straight line crossing zero at the neutral point (23.9 m), reaching $-3.827$ m at the crown and returning in a straight line to 0 at $B$. A load between $A$ and the neutral point produces positive (sagging) moment at $D$; a load between the neutral point and $B$ produces negative moment.