Neutral point in ILD of arch

Neutral point

For a section $D$ of a three-hinged arch, the neutral point is the position of a unit load on the span for which the bending moment at $D$ is zero. In the ILD of $M_D$ it is the point where the influence line crosses the base line (ordinate zero). A load on one side of it produces a positive moment and on the other side a negative moment at $D$; this is why a rolling load must be put on different sides of the neutral point for the maximum positive and negative moments.

Graphical determination (section D in the left half)

When a load $P$ acts on the left half of the arch (on the segment $DC$), the right portion $CB$ carries no load, so it is a two-force member: the reaction at $B$ acts along the line $BC$. For $M_D = 0$ the reaction at $A$ must act along the line $AD$ (its moment about $D$ is then zero). The load $P$, $R_A$ and $R_B$ are in equilibrium, so the three lines must be concurrent.

1. Join $B$ to the crown hinge $C$ and extend it.
2. Join $A$ to the section $D$ and extend it.
3. Let the two lines meet at $E$.
4. The vertical through $E$ meets the span at the neutral point $N$.

            C
         .-' '-.        E  (intersection of AD and BC)
      D-'       '-.   .'
     /             '-B.
    A    . . . N . . . . 
         neutral point (vertical below E)

Derivation of the position

Let the span be $L$, the rise $f$, and let the section be at $x_D = a$ with ordinate $y_D$. Take a unit load at a distance $s$ from $A$ with $a < s < L/2$.

$$R_A = \frac{L-s}{L}, \qquad H = \frac{M_C}{f} = \frac{R_B\,(L/2)}{f} = \frac{s}{2f}$$ $$M_D = R_A\,a - H\,y_D = \frac{(L-s)\,a}{L} - \frac{s\,y_D}{2f}$$

Putting $M_D = 0$:

$$a - s\left(\frac{a}{L} + \frac{y_D}{2f}\right) = 0 \Rightarrow s_0 = \frac{a}{\dfrac{a}{L}+\dfrac{y_D}{2f}} = \frac{2f}{\dfrac{2f}{L}+\dfrac{y_D}{a}}$$

This is the same as the intersection of the lines $y = \dfrac{y_D}{a}x$ (line $AD$) and $y = \dfrac{2f}{L}(L-x)$ (line $BC$).

For a parabolic arch, $y_D = \dfrac{4fa(L-a)}{L^2}$, so

$$s_0 = \frac{L^2}{3L-2a}$$

Example: $L = 100$ m, $f = 10$ m, $a = 30$ m: $s_0 = \dfrac{100^2}{300-60} = 41.67$ m from $A$. A load to the left of $s_0$ gives positive $M_D$, a load to the right gives negative $M_D$ (this is confirmed by the ILD ordinates $+8.4$ m at $D$ and $-6$ m at the crown).

For the radial shear at $D$ the neutral point is found in the same way: it is the load position for which $V_D = R_A\cos\theta - H\sin\theta = 0$, i.e. where the reaction at $A$ is parallel to the tangent at $D$ (line through $A$ parallel to the tangent at $D$ meets line $BC$ at the point vertically above the neutral point).