Critical position of UDL shorter than span

A UDL shorter than the span gives maximum bending moment at a section when the load is placed so that the section divides the span and the load in the same ratio.

ILD for bending moment at section C

Simply supported beam $AB$, span $L$; section $C$ at $a$ from $A$, $b = L-a$. The ILD for $M_C$ is a triangle with peak $y_C = ab/L$ under $C$.

        A      c    C   (l-c)    B
        o------+----+-------+----o
  ILD:         y1  ab/L     y2

Ordinate left of $C$ at distance $x$ from $A$: $y = \dfrac{xb}{L}$. Right of $C$ at distance $x$ from $B$: $y = \dfrac{xa}{L}$.

Proof

Let a UDL of intensity $w$ and length $l$ (< $L$) have a length $c$ on the left of $C$ and $(l-c)$ on the right. Ordinates under its ends are

$$y_1 = \frac{(a-c)\,b}{L}, \qquad y_2 = \frac{\{b-(l-c)\}\,a}{L}$$

$M_C = w\times(\text{area under ILD below the load})$. Shift the load by a small distance $dx$ to the right. The left end loses a strip of height $y_1$ and the right end gains a strip of height $y_2$, so

$$\frac{dM_C}{dx} = w\,(y_2 - y_1)$$

$M_C$ is maximum when $dM_C/dx = 0$, i.e. $y_1 = y_2$:

$$\begin{aligned} (a-c)\,b &= (b-l+c)\,a \\ ab - cb &= ab - la + ca \\ c(a+b) &= la \Rightarrow \frac{c}{l} = \frac{a}{L} \end{aligned}$$

Hence the part of the load on the left of $C$ is to the whole load as the left part of the span is to the whole span:

$$\frac{c}{l} = \frac{a}{L}$$

Maximum moment

With $c = la/L$ the end ordinates are equal: $y_1 = y_2 = \dfrac{ab}{L}\left(1-\dfrac{l}{L}\right)$. The area under the load is two trapezoids:

$$M_{max} = w\,\frac{l}{2}\left(\frac{ab}{L} + y_1\right) = w\,l\,\frac{ab}{L}\left(1 - \frac{l}{2L}\right)$$

Result: place the UDL so that the section divides the load in the ratio $a : b$, the same as it divides the span.