Four loads on 60 m girder: BM at C, absolute

Data: girder $AB$, $L = 60$ m; section $C$ at $a = 20$ m from $A$, $b = 40$ m. Loads from the rear: 170, 215, 150, 120 kN at spacings 3 m, 2 m, 3 m; the train moves left to right (120 kN leads, as drawn). Total $W = 655$ kN.

(i) Maximum BM at C

ILD for $M_C$: peak $\dfrac{ab}{L} = \dfrac{20\times 40}{60} = 13.33$ m; ordinate $\dfrac{2x}{3}$ for $x\le 20$ and $\dfrac{20(60-x)}{60}$ for $x\ge 20$.

Place the 215 kN load at $C$. Load on the left of $C$ including it: $170 + 215 = 385$ kN.

$$\frac{385}{20} = 19.25 > \frac{655}{60} = 10.9 > \frac{170}{20} = 8.5$$

The condition for maximum is satisfied. Positions from $A$: 170 kN at 17 m, 215 kN at 20 m (at $C$), 150 kN at 22 m, 120 kN at 25 m.

$$M_{C,max} = 8093.3\ \text{kN m}$$

(ii) Absolute maximum BM

Distance of the resultant behind the 120 kN load: $\bar x = \dfrac{170(8)+215(5)+150(3)}{655} = 4.405$ m. Trying each load with the resultant placed symmetrically about mid-span:

The largest value is under the 215 kN load, at 29.70 m from $A$. Then the loads are at: 170 kN: 26.70 m, 215 kN: 29.70 m, 150 kN: 31.70 m, 120 kN: 34.70 m (all on the girder).

$$R_A = 324.2\ \text{kN}, \qquad M_{max} = R_A\times 29.70 - (\text{moments of the loads on its left}) = 9121.0\ \text{kN m}$$

Answer: (i) $M_C = 8093.3$ kN m; (ii) absolute maximum BM = 9121.0 kN m under the 215 kN load.