ILD by focal point method: moment at support 2

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Data and focal ratios

Beam 1-2-3: end 1 fixed, 2 roller, 3 hinged; $L_1 = 20$ m, $L_2 = 10$ m, constant $EI$.

For span n with the far end at moment ratio $k$, the unloaded span acts as a rotational spring at the joint with stiffness $S = \dfrac{6EI}{L(2-k)}$, where $k$ is its focal ratio ($|M_{far}/M_{near}|$).

• Span 1-2 seen from joint 2 (end 1 fixed): left focal ratio $k = \tfrac12$ (focal point at $L_1/3$ from 1), so $S_{21} = \dfrac{6EI}{20(1.5)} = 0.2EI$.
• Span 2-3 seen from joint 2 (end 3 hinged): right focal ratio $k = 0$ (focal point at 3), so $S_{23} = \dfrac{6EI}{10(2)} = 0.3EI$.

Distribution of an unbalanced joint moment at 2: share to 2-1 $= 0.2/0.5 = 0.4$, share to 2-3 $= 0.3/0.5 = 0.6$.

Ordinates of $M_2$

Unit load in span 1-2 at $x$ from 1 ($a = x$, $b = 20 - x$): the fixed-end moment at 2 is $F_{21} = \dfrac{a^2b}{L_1^2}$, and the part held by span 1-2 is 0.6 of it:

$$M_2 = -0.6\,\frac{x^2(20-x)}{400}$$

Unit load in span 2-3 at $y$ from 2: with the hinge at 3 released, the end moment at 2 is $F'_{23} = \dfrac{y(L_2-y)(2L_2-y)}{2L_2^2}$, and the part held is 0.4 of it:

$$M_2 = -0.4\,\frac{y(10-y)(20-y)}{200}$$

Sample: $x = 12$: $M_2 = -0.6\times\dfrac{144\times8}{400} = -1.728$. $y = 2$ (position 22 m): $M_2 = -0.4\times\dfrac{2\times8\times18}{200} = -0.576$.

(The values above were checked against the exact three-moment solution.)

ILD shape

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The ILD is entirely hogging. The maximum is about $-1.78$ at x = 13.3 m in span 1-2, and about $-0.77$ at 24.2 m in span 2-3; it is zero at 1, 2 and 3.