ILD: bending moment at D, 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.

Method

Two-span beam: AB = 10 m, BC = 20 m (A hinged, B and C rollers), D at 5 m from A. The ILD for $M_D$ is the simple-span ILD for $M_D$ plus the contribution of the continuity moment at B:

$$\text{ILD}(M_D) = \text{ILD}_{\text{simple}}(M_D) + \frac{5}{10}\,\text{ILD}(M_B)$$

with

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

The simple-span ordinate is $x/2$ for $x \le 5$ and $(10-x)/2$ for $x \ge 5$.

Sample: $x = 5$: simple ordinate $= 2.5$; $M_B = -\dfrac{5(100-25)}{2\times10\times30} = -0.625$; so $M_D = 2.5 + 0.5(-0.625) = 2.1875$. Sample: $x = 15$ ($y = 15$): $M_B = -\dfrac{15(400-225)}{2\times20\times30} = -2.1875$; $M_D = 0.5\times(-2.1875) = -1.094$.

Ordinates of $M_D$ (2.5 m in AB, 5 m in BC)

ILD shape

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


^               ^                               ^

Maximum positive ordinate $+2.19$ at D; the ordinate is zero at B (x = 10 m) and the negative loop in BC reaches about $-1.28$ near x = 18.5 m (the tabulated value at the middle of BC, x = 20 m, is $-1.25$).