Max BM at 10 m: UDL and two point loads

Data: span $L = 25$ m, section $C$ at $a = 10$ m from the left support, $b = 15$ m. The train is a 20 kN/m UDL over 4 m (= 80 kN), then a 10 kN load 2 m beyond the UDL and a second 10 kN load another 2 m further. The train moves left to right, so the 10 kN loads lead.

ILD for $M_C$

Peak ordinate under $C$: $y_C = \dfrac{ab}{L} = \dfrac{10\times 15}{25} = 6$ m. Ordinate at $x$ from $A$: $0.6x$ (left of $C$), $0.4(25-x)$ (right of $C$).

   A o----------C---------------o B
                ^ 6.0
   ILD: 0 -> 6.0 at C -> 0

Critical position

Try positions of the leading 10 kN load (distance $p$ from $A$):

The maximum is near $p = 16$ m. Then the loads are: 10 kN at 16 m, 10 kN at 14 m, and the UDL spans 8 m to 12 m (covering $C$).

Ordinates: at 16 m: $0.4\times 9 = 3.6$; at 14 m: $0.4\times 11 = 4.4$; under the UDL: $4.8$ at 8 m, $6.0$ at 10 m, $5.2$ at 12 m.

$$\begin{aligned} M_{max} &= 10(3.6) + 10(4.4) + 20\left[\tfrac{4.8+6.0}{2}\times 2 + \tfrac{6.0+5.2}{2}\times 2\right]\\ &= 36 + 44 + 20\,(10.8 + 11.2) = 36+44+440 = 520\ \text{kN m} \end{aligned}$$

Answer: maximum bending moment at the section = 520 kN m.