ILD and BM at mid span for triangular load

Data: simply supported beam $AB$ of span $L$; $C$ at mid-span. A triangular load acts on the right half $CB$, zero at $C$ and rising to $W$ kN/m at $B$.

ILD for bending moment at C

A triangle with peak $\dfrac{ab}{L} = \dfrac{(L/2)(L/2)}{L} = \dfrac{L}{4}$ at $C$, 0 at $A$ and $B$. At a distance $u$ from $B$ (in the right half) the ordinate is

$$y = \frac{L/4}{L/2}\,u = \frac{u}{2}$$

   A o--------------C--------------o B
                    /\  L/4
   ILD M_C:      0 /    \ 0
   Load on CB:   0 at C ........ W at B

Load intensity

At distance $u$ from $B$: $w(u) = W\left(1 - \dfrac{2u}{L}\right)$, so $w = W$ at $B$ ($u=0$) and 0 at $C$ ($u = L/2$).

Bending moment

$M_C = \int w\,y\,du$ over the loaded length:

$$\begin{aligned} M_C &= \int_0^{L/2} W\left(1-\frac{2u}{L}\right)\frac{u}{2}\,du = \frac{W}{2}\left[\frac{u^2}{2} - \frac{2u^3}{3L}\right]_0^{L/2} \\ &= \frac{W}{2}\left[\frac{L^2}{8} - \frac{L^2}{12}\right] = \frac{W}{2}\times\frac{L^2}{24} = \frac{WL^2}{48} \end{aligned}$$

Check by statics

Total load $=\tfrac12\cdot W\cdot\dfrac{L}{2} = \dfrac{WL}{4}$, acting at $\dfrac{2}{3}\cdot\dfrac{L}{2} = \dfrac{L}{3}$ from $C$, i.e. $\dfrac{5L}{6}$ from $A$. $R_A = \dfrac{WL}{4}\times\dfrac{L/6}{L} = \dfrac{WL}{24}$.

$$M_C = R_A\times\frac{L}{2} = \frac{WL}{24}\times\frac{L}{2} = \frac{WL^2}{48}$$

Answer: $M_C = \dfrac{WL^2}{48}$ (sagging).