Four loads: maximum BM at C

Data: simply supported beam $AB$, $L = 15$ m; $C$ at $a = 6$ m from $A$, $b = 9$ m. Loads 50, 70, 80, 60 kN with spacings 2 m, 2 m, 3 m (50 kN at one end). Total $W = 260$ kN.

ILD for $M_C$

Peak ordinate $\dfrac{ab}{L} = \dfrac{6\times 9}{15} = 3.6$ m. Ordinate $0.6x$ for $x\le 6$ and $0.4\,(15-x)$ for $x\ge 6$.

Critical load

Condition: $\dfrac{W_L + P_k}{a} \ge \dfrac{W}{L} \ge \dfrac{W_L}{a}$, with $\dfrac{W}{L} = \dfrac{260}{15} = 17.33$ kN/m.

For the 70 kN load at $C$ (train moving left to right, 60 kN leading): load on the left without it $=50$ kN, with it $=120$ kN.

$$\frac{120}{6} = 20 \ge 17.33 \ge \frac{50}{6} = 8.33$$

The condition holds, so the 70 kN load must be at $C$. Positions from $A$: 50 kN at 4 m, 70 kN at 6 m, 80 kN at 8 m, 60 kN at 11 m.

$$M_{C,max} = 692.0\ \text{kN m}$$

If the train moves in the opposite direction (50 kN leading), the 80 kN load goes to $C$ (positions 60 kN at 3 m, 80 kN at 6 m, 70 kN at 8 m, 50 kN at 10 m), and the same calculation gives 692.0 kN m, the same value.

Answer: maximum bending moment at $C$ = 692.0 kN m.