ILD: moment at section x-x, continuous 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

Two-span beam ABC: A hinged, B and C rollers, $AB = 6$ m, $BC = 8$ m, constant $EI$. Section x-x is 4 m from A (2 m from B).

Method

$$\text{ILD}(M_{x}) = \text{ILD}_{\text{simple}} + \frac{4}{6}\,\text{ILD}(M_B)$$

Simple-span ordinate for the section at $a = 4$, $L_1 = 6$: $x(6-4)/6 = x/3$ for $x \le 4$ and $4(6-x)/6$ for $x \ge 4$.

$$M_B = -\frac{x(L_1^2-x^2)}{2L_1(L_1+L_2)} = -\frac{x(36-x^2)}{168}\ (\text{load in AB}),\qquad M_B = -\frac{y(64-y^2)}{224}\ (\text{load in BC},\ y\text{ from C})$$

Sample: load at the section ($x = 4$): $\tfrac{4\times2}{6} = 1.333$; $M_B = -\dfrac{4(36-16)}{168} = -0.476$; $M = 1.333 + \tfrac46(-0.476) = 1.016$. Sample: load at the middle of BC ($y = 4$, x = 10): $M_B = -\dfrac{4(64-16)}{224} = -0.857$; $M = \tfrac46(-0.857) = -0.571$.

Ordinates (2 m interval)

ILD shape

             **
          ***  **
       ***       *
   ****           **
***-----------------**------------------------***
                      ***                *****
                         ****************


^                    ^                          ^

Peak $+1.016$ under the section (x = 4 m). The ILD is zero at A and B, becomes negative in BC with a largest value of about $-0.59$ near x = 9.4 m, and is zero at C.