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Left and right focal point ratios

Short answer

Left focal ratio k_n = -M_{n-1}/M_n is the far-to-near end moment ratio of an unloaded span when loading lies to the right; right focal ratio k'_n = -M_n/M_{n-1} applies for loading to the left. Recurrences are k_{n+1} = (L_{n+1}/I_{n+1})/[(2-k_n)L_n/I_n + 2L_{n+1}/I_{n+1}] and the analogous right ratio formula.

Define the term left and right focal point ratios. Also write their expressions.

Focal point

When a continuous beam is loaded in one span only, the bending moment diagram in every unloaded span is a straight line. For a given unloaded span this line always passes through the same point of the beam axis, whatever the load position, and this point is called the focal point of the span.

Left and right focal ratios

  • Left focal ratio (knk_n or knLk^L_n) of span nn: the ratio of the moment at the left (far) end to the moment at the right (near) end of an unloaded span, when the load lies on spans to its right. The two end moments have opposite signs, so the ratio is defined with a minus sign to make it positive: kn=−Mn−1/Mnk_n = -M_{n-1}/M_n. The left focal point of the span lies Ln/(1+kn)L_n/(1+k_n) from its right support, i.e. knLn/(1+kn)k_nL_n/(1+k_n) from its left support.
  • Right focal ratio (kn′k'_n or knRk^R_n) of span nn: the ratio of the moment at the right (far) end to the moment at the left (near) end of an unloaded span, when the load lies on spans to its left; kn′=−Mn/Mn−1k'_n = -M_n/M_{n-1}. The right focal point lies Ln/(1+kn′)L_n/(1+k'_n) from its left support.

Expressions

For span nn between supports n−1n-1 and nn, with Ln/InL_n/I_n the flexibility ratio of the span:

kn+1=Ln+1/In+1(2−kn) Ln/In+2 Ln+1/In+1(left ratios, from the left end)k_{n+1} = \frac{L_{n+1}/I_{n+1}}{(2-k_n)\,L_n/I_n + 2\,L_{n+1}/I_{n+1}}\qquad(\text{left ratios, from the left end}) kn−1′=Ln−1/In−1(2−kn′) Ln/In+2 Ln−1/In−1(right ratios, from the right end)k'_{n-1} = \frac{L_{n-1}/I_{n-1}}{(2-k'_n)\,L_n/I_n + 2\,L_{n-1}/I_{n-1}}\qquad(\text{right ratios, from the right end})

For equal spans with constant EIEI: kn+1=14−knk_{n+1} = \dfrac{1}{4-k_n}, kn−1′=14−kn′k'_{n-1} = \dfrac{1}{4-k'_n}.

Starting values: hinged end k=0k = 0; fixed end k=12k = \tfrac12. Example (equal spans, constant EIEI, hinged end): k1=0k_1 = 0, k2=0.25k_2 = 0.25, k3=0.2667k_3 = 0.2667, k4=0.2679k_4 = 0.2679.

✓ Answer checked by Claude Opus 5.5 (high effort)

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