Focal point ratio: define and derive left ratio

Definition

In a continuous beam loaded on one span only, the bending moment diagram in every unloaded span is a straight line, and it cuts the beam axis at a fixed point called the focal point of that span. The focal point ratio is the ratio of the moments at the two ends of such an unloaded span (far-end moment to near-end moment, with opposite signs):

• Left focal ratio $k_n$: for an unloaded span to the left of the loaded span, $k_n = -\dfrac{M_{n-1}}{M_n}$ (moment at the left end divided by the moment at the right end).
• Right focal ratio $k'_n$: for an unloaded span to the right of the loaded span, $k'_n = -\dfrac{M_n}{M_{n-1}}$.

Derivation of the left focal ratio

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 $n$.

Numerical example

Hinged left end A, span 1: $L_1 = 6$ m, $I_1 = I$; span 2: $L_2 = 4$ m, $I_2 = 2I$.

• $k_1 = 0$ (hinged end).
• $k_2 = \dfrac{4/2}{(2-0)\times6/1 + 2\times4/2} = \dfrac{2}{12+4} = 0.125$.
• The focal point of span 2 lies $L_2/(1+k_2) = 4/1.125 = 3.56$ m from its right end (support 2), that is 0.44 m from support 1.