Derive structural quantities using ILD

Let $F$ be a structural quantity (reaction, shear, moment or member force) at a fixed section. Let $y$ be its influence ordinate (value of $F$ for a unit load) at the load position. By the principle of superposition (linear elastic structure), the effect of several loads is the sum of their separate effects.

1. Concentrated loads

A single load $P$ at a point where the ILD ordinate is $y$ gives $F = P\,y$, because a unit load gives $y$ and the effect is proportional to the load. For loads $P_1, P_2, \dots, P_n$ at ordinates $y_1, y_2, \dots, y_n$:

$$F = P_1y_1 + P_2y_2 + \dots + P_ny_n = \sum P_i\,y_i$$

2. Uniformly distributed load

Consider a UDL of intensity $w$ per unit length between $x_1$ and $x_2$. An elementary strip $dx$ carries load $w\,dx$, acting as a concentrated load at ordinate $y$ (the ILD ordinate at $x$):

$$dF = w\,dx\;y \Rightarrow F = w\int_{x_1}^{x_2} y\,dx = w\times(\text{area of ILD between }x_1\text{ and }x_2)$$

Areas above the base line are positive and those below are negative. For the maximum positive effect, load only the positive portions.

3. Couple

A couple $M_0$ is a pair of equal and opposite loads $P$ at a small distance $d$ apart, with $M_0 = P\,d$. If the ordinates at the two points are $y_1$ and $y_2$ (the second load acting opposite to the first):

$$F = P\,(y_2 - y_1) = P\,d\;\frac{y_2-y_1}{d} = M_0\,\frac{dy}{dx}$$

So the effect of a couple equals the couple magnitude multiplied by the slope of the ILD at the point of application (the sign follows the sense of the couple: positive when clockwise, for an ILD drawn with upward positive ordinates).

 Summary:   P  ->  F = P . y
            w  ->  F = w . (area of ILD under load)
            M0 ->  F = M0 . (slope of ILD)