Four loads on 50 m girder: BM at 20 m, absolute

Data: span $L = 50$ m. Four loads 100, 150, 200, 150 kN, 3 m apart, moving from right to left with the 100 kN load leading. Total $W = 600$ kN. Distances below are measured from the right support (where the train enters); the section $C$ is at $a = 20$ m from it, $b = 30$ m.

(i) Maximum BM at C

ILD for $M_C$: peak ordinate $\dfrac{ab}{L} = \dfrac{20\times 30}{50} = 12$ m. Trial: the 200 kN load at $C$. The loads between the entry support and $C$ are the 150 kN (17 m), so $W_L = 150$ kN without the 200 kN and $350$ kN with it.

$$\frac{350}{20} = 17.5 \ge \frac{600}{50} = 12 \ge \frac{150}{20} = 7.5$$

The condition is satisfied, so the 200 kN load at $C$ gives the maximum. Positions from the right support: 100 kN at 26 m, 150 kN at 23 m, 200 kN at 20 m (at $C$), 150 kN at 17 m.

$$M_{C,max} = 6510.0\ \text{kN m}$$

Check with other loads placed at $C$ (kN m): 100 kN at $C$: 5400.0; first 150 kN at $C$: 6180.0; last 150 kN at $C$: 6240.0; 200 kN at $C$: 6510.0. The largest is 6510.0 kN m.

(ii) Absolute maximum BM

Resultant from the leading load: $\bar x = \dfrac{150(3)+200(6)+150(9)}{600} = 5.0$ m behind the 100 kN load, i.e. 1.0 m from the 200 kN load. Place the 200 kN load and the resultant symmetrically about mid-span: 200 kN at 24.50 m, resultant at 25.50 m from the right support.

$$R_{right} = 600\times\frac{50-25.50}{50} = 294.0\ \text{kN}$$

Moment under the 200 kN load, taking the part of the girder between the right support and the 200 kN load (it carries one 150 kN load, 3 m from the 200 kN load):

$$M_{max} = R_{right}\times 24.50 - 150\times 3 = 6753.0\ \text{kN m}$$

All four loads lie on the girder (the train spans 9 m).

Answer: (i) maximum BM at 20 m from the right support = 6510.0 kN m; (ii) absolute maximum BM = 6753.0 kN m (under the 200 kN load, 0.5 m from mid-span).