BM at section under moving load train

For a train of loads $P_1,\dots,P_5$ at fixed spacings crossing a simply supported beam of span $L$, the bending moment at a section $C$ is found from the ILD of $M_C$ by trying the critical position of the train.

ILD for $M_C$

Let $C$ be at $a$ from the left support and $b = L-a$. The ILD is a triangle with peak $ab/L$ under $C$. Left slope $= b/L$, right slope $= -a/L$.

    A o-------------C-------------o B
                    ^ ab/L
    P1  P2  P3  P4  P5   ->  moving

$$M_C = \sum P_i\,y_i$$

Condition for maximum

Let $W$ be the total load on the span and $W_L$ the load on the left of $C$. Shifting the train to the right by $\delta$:

$$\Delta M_C = \frac{\delta}{L}\left(W_L\,b - W_R\,a\right) = \delta\left(W_L - \frac{W\,a}{L}\right)$$

$M_C$ is maximum when $\Delta M_C$ changes sign from positive to negative as a load $P_k$ crosses $C$:

$$\frac{W_L + P_k}{a} \ge \frac{W}{L} \ge \frac{W_L}{a}$$

where $W_L$ excludes $P_k$ (load just right of $C$) and $W_L + P_k$ includes it (just left of $C$). At such a position the average load on the left of $C$ equals the average load on the whole span, and $P_k$ stands over the section.

Steps

1. Place the heaviest load near $C$ and test the condition; move to the next load if it fails. Remember that loads leaving or entering the span change $W$.
2. At the critical position, read the ordinates under every load of the train from the ILD (similar triangles).
3. Compute $M_{C,max} = P_1y_1 + P_2y_2 + P_3y_3 + P_4y_4 + P_5y_5$.
4. If more than one load satisfies the test, compute for each and take the largest.
5. For the absolute maximum moment anywhere, put a load and the resultant of the loads on the span symmetric about mid-span.