ILD: moment at support B, fixed-propped 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

Beam ABC: A fixed, B and C supports, $AB = 2L$, $BC = L$, constant $EI$. Use the moment-distribution (stiffness) form: rotational stiffness at B of AB is $4EI/2L = 2EI/L$ (far end fixed) and of BC is $3EI/L$ (far end C is a roller, moment zero). So the distribution factors at B are $DF_{BA} = 2/5 = 0.4$ and $DF_{BC} = 3/5 = 0.6$.

Unit load in AB at $x$ from A: $M_{BA}^F = \dfrac{x^2(2L-x)}{(2L)^2}$; support moment $M_B = (1 - 0.4)M_{BA}^F$:

$$M_B = -0.6\,\frac{x^2(2L-x)}{4L^2}$$

Unit load in BC at $y$ from B: the fixed-end moment at B with C released is $\dfrac{y(L-y)(2L-y)}{2L^2}$, and $M_B = (1-0.6)$ of it:

$$M_B = -0.4\,\frac{y(L-y)(2L-y)}{2L^2}$$

Sample: $x = L$: $M_B = -0.6\times\dfrac{L^2\cdot L}{4L^2} = -0.15L$. $y = 0.5L$: $M_B = -0.4\times\dfrac{0.5\cdot0.5\cdot1.5}{2}L = -0.075L$.

Ordinates (at 0.25 and 0.5 times each span length)

ILD shape

*****---------------------------*--------------**
     ****                     ** ****     *****
         ****                *       *****
             ****         ***
                 *********
^                               ^               ^

The ILD is hogging throughout; the largest ordinate in AB is about $-0.178L$ near $x = 1.33L$, and in BC about $-0.077L$ near $0.42L$ from B (mid-span value $-0.075L$).