Four loads on 100 m girder: BM at C, absolute

Data: girder $AB$, $L = 100$ m; section $C$ at $a = 35$ m from $A$, $b = 65$ m. Wheel loads from the rear: 100, 160, 180, 120 kN at spacings 4 m, 3 m, 2 m; the train moves left to right (assumed as drawn, so the 120 kN load leads). Total $W = 560$ kN.

(i) Maximum BM at C

ILD for $M_C$: peak $\dfrac{ab}{L} = \dfrac{35\times 65}{100} = 22.75$ m; ordinate $0.65x$ for $x\le 35$ and $0.35(100-x)$ for $x\ge 35$.

Place the 160 kN load at $C$. Load on the left of $C$ including it: $100 + 160 = 260$ kN.

$$\frac{260}{35} = 7.43 > \frac{560}{100} = 5.6 > \frac{100}{35} = 2.86$$

The condition for maximum is satisfied. Positions from $A$: 100 kN at 31 m, 160 kN at 35 m (at $C$), 180 kN at 38 m, 120 kN at 40 m.

$$M_{C,max} = 12081.0\ \text{kN m}$$

(If the train moves the other way, with the 100 kN load leading, the same method gives 12171.0 kN m.)

(ii) Absolute maximum BM

Resultant from the 120 kN (leading) load: $\bar x = \dfrac{100(9)+160(5)+180(2)}{560} = 3.679$ m behind it. For each load, place it and the resultant symmetrically about mid-span and find the moment under the load:

The largest is under the 180 kN load. With this load at 50.84 m from $A$, the loads are at 100 kN: 43.84 m, 160 kN: 47.84 m, 180 kN: 50.84 m, 120 kN: 52.84 m (all on the girder).

$$R_A = 284.7\ \text{kN}, \qquad M_{max} = R_A\times 50.84 - (\text{moments of the loads on its left})= 13293.9\ \text{kN m}$$

Answer: (i) $M_C = 12081.0$ kN m; (ii) absolute maximum BM = 13293.9 kN m under the 180 kN load.