ILD: moment at section 5, two-span beam

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.

Reading of the figure

The beam has two equal spans of $4@2 = 8$ m, hinged at A and on rollers at B and C. Section 5 is taken as the middle section of span BC (3 sections per span, numbered 1-3 in AB and 4-6 in BC), that is 12 m from A (4 m from B and 4 m from C).

Method

The ILD for the moment at a section in BC is the ILD of the simple span BC plus the effect of the support moment $M_B$:

$$\text{ILD}(M_5) = \text{ILD}_{\text{simple}} + \frac{4}{8}\,\text{ILD}(M_B)$$

Here $M_B = -\dfrac{x(L_1^2 - x^2)}{2L_1(L_1+L_2)}$ for a load in AB at $x$ from A, and $M_B = -\dfrac{y(L_2^2 - y^2)}{2L_2(L_1+L_2)}$ for a load in BC at $y$ from C. With $L_1 = L_2 = 8$ m: $M_B = -\dfrac{x(64-x^2)}{256}$.

Sample: load at the section ($y = 4$): simple ordinate $= \dfrac{4\times4}{8} = 2.0$; $M_B = -\dfrac{4(64-16)}{256} = -0.75$; $M_5 = 2.0 + 0.5(-0.75) = 1.625$. Sample: load at 4 m from A: $M_B = -0.75$, $M_5 = 0.5(-0.75) = -0.375$.

Ordinates (2 m interval)

ILD shape

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Positive (sagging) in BC with a peak of $+1.625$ under the section, and a small negative loop in AB (about $-0.385$ near x = 4.6 m; $-0.375$ at x = 4 m), zero at A, B and C.