ILD as system-specific; focal point recurrence

ILD as a system-specific diagram

An influence line shows how one response (a reaction, shear or moment at one fixed location) varies as a unit load moves over the structure. It depends only on the system: the spans, support types, flexural rigidities and the chosen response, and not on the loads. So:

• the ILD of $R_B$ for a propped cantilever (zero at A, 1 at B, curved between) is quite different from that of a simply supported beam of the same length, and a beam with spans 6 m + 8 m has different ILDs from one with 8 m + 8 m;
• once the ILD for a system is drawn, it can be used for any loading on that same system: $Q = \sum P_i y_i$ for point loads, $Q = w\times(\text{area of ILD under the load})$ for a UDL.

By contrast, a BMD or SFD is load-specific: it shows one response along the whole beam for one given loading. An ILD is system-specific and response-specific.

Recurrence formula of the left focal ratio (loading on right spans)

Consider a continuous beam with spans numbered from the left; span $n$ lies between supports $n-1$ and $n$, with length $L_n$ and moment of inertia $I_n$. Let the load be on a span to the right of span $n+1$ (all spans from the left end up to $n+1$ unloaded), and let moments be taken with one sign convention (hogging or sagging) at all supports.

Three-moment equation at support $n$ for the unloaded spans $n$ and $n+1$ (the load terms are zero):

$$M_{n-1}\frac{L_n}{I_n} + 2M_n\left(\frac{L_n}{I_n}+\frac{L_{n+1}}{I_{n+1}}\right) + M_{n+1}\frac{L_{n+1}}{I_{n+1}} = 0$$

Because the moment in an unloaded span is linear, the moments at its two ends have opposite signs and a fixed ratio. Define the left focal ratio of span $n$:

$$k_n = -\frac{M_{n-1}}{M_n}$$

and similarly for span $n+1$: $M_n = -k_{n+1}M_{n+1}$.

Substitute $M_{n-1} = -k_nM_n$ in the equation:

$$M_n\left[(2-k_n)\frac{L_n}{I_n} + 2\frac{L_{n+1}}{I_{n+1}}\right] = -M_{n+1}\frac{L_{n+1}}{I_{n+1}}$$

so that $M_n/M_{n+1} = -k_{n+1}$ gives the recurrence formula

$$\boxed{\,k_{n+1} = \frac{L_{n+1}/I_{n+1}}{(2-k_n)\,L_n/I_n + 2\,L_{n+1}/I_{n+1}}\,}$$

For equal spans with constant $EI$ this becomes $k_{n+1} = \dfrac{1}{4-k_n}$.

Starting value: if the left end of the beam is hinged, $M_0 = 0$ so $k_1 = 0$; if it is fixed, the slope there is zero, $2M_0 + M_1 = 0$, so $k_1 = \tfrac12$.

Focal point: the straight BMD of unloaded span $n$ is $M(x) = M_n\left[1 - (1+k_n)\dfrac{x}{L_n}\right]$, with $x$ measured from support $n$, so it crosses the beam axis at $x = \dfrac{L_n}{1+k_n}$ from support $n$. This point is the left focal point, and it is the same for any loading on the spans to the right of span $n$.