ILD: moment at middle support, 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.

Data

Beam ABC with pinned ends and a middle support B: $L_1 = 2$ m (stiffness $2EI$), $L_2 = 6$ m (stiffness $5EI$). Ends A and C carry no moment. The ILD for the moment at B is obtained from the three-moment equation with $L_1/I_1 = 2/2 = 1$ and $L_2/I_2 = 6/5 = 1.2$.

$$M_AL_1/I_1 + 2M_B\left(\frac{L_1}{I_1} + \frac{L_2}{I_2}\right) + M_CL_2/I_2 = -(\text{load terms}),\qquad 2\left(1 + 1.2\right) = 4.4$$

Formulas

• Unit load in AB at $a$ from A ($b = L_1 - a$): load term $= \dfrac{ab(L_1+a)}{I_1L_1}$, so

$$M_B = -\frac{ab(L_1+a)}{I_1L_1\times4.4} = -\frac{a(2-a)(2+a)}{2\times2\times4.4}\ \ (\text{with } I_1 = 2)$$

• Unit load in BC at $y$ from C: load term $= \dfrac{y(L_2^2-y^2)}{I_2L_2}$, so

$$M_B = -\frac{y(36-y^2)}{5\times6\times4.4} = -\frac{y(36-y^2)}{132}$$

Sample: $a = 1$: $M_B = -\dfrac{1\times1\times3}{17.6} = -0.1705$. Load 4 m from C ($y = 4$, i.e. $x = 4$ m from A): $M_B = -\dfrac{4\times20}{132} = -0.6061$.

(Note: ordinates are given at x measured from A; positions in BC are x = 2 + distance from B.)

Ordinates of $M_B$ (every 1 m, so the 2 m positions are included)

ILD shape

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

The ILD is negative (hogging) everywhere and zero at A, B and C. The largest ordinate is about $-0.63$ near $x = 4.5$ m (2.5 m from B, in the longer span); the short span AB has a maximum of only about $-0.175$ near $x = 1.15$ m.