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Four wheel loads on 20 m beam: absolute max BM

Short answer

Place the 80 kN load 0.90 m from mid-span with the resultant opposite. Absolute maximum bending moment is 738.1 kN m under the 80 kN load.

Four-wheel loads of 30 kN, 80 kN, 60 kN and 30 kN as shown in figure move from right to left on a simply supported beam of span 20 m. Calculate absolute maximum bending moment that can be expected in the beam. [Figure: loads 30, 80, 60, 30 kN with spacings 3 m, 4 m, 3 m.]

Data: span L=20L = 20 m; loads 30, 80, 60, 30 kN with spacings 3 m, 4 m, 3 m (taken from the 30 kN at one end). Total W=200W = 200 kN. The absolute maximum moment does not depend on the direction of travel.

Position of the resultant

Measured from the first 30 kN load:

xˉ=30(0)+80(3)+60(7)+30(10)200=960200=4.8 m\bar{x} = \frac{30(0) + 80(3) + 60(7) + 30(10)}{200} = \frac{960}{200} = 4.8\ \text{m}

The 80 kN load (3 m from the first load) is the nearest to the resultant, at 1.8 m from it.

Absolute maximum BM

It occurs under a load when that load and the resultant are equidistant from the centre of the span. Take the 80 kN load as the critical one: it is placed 0.90 m from mid-span, and the resultant is on the other side of mid-span.

  30    80      60    30     (kN)
  |  3  |   4   |  3  |
  x=6.10 9.10   13.10  16.10 m from left support
  0.....................20

Positions from the left support: 30 kN at 6.10 m, 80 kN at 9.10 m, 60 kN at 13.10 m, 30 kN at 16.10 m; resultant at 10.90 m. All four loads are on the span.

RA=200×20−10.9020=91.0 kNR_A = 200\times\frac{20-10.90}{20} = 91.0\ \text{kN} Mmax=RA×9.10−30×3=91.0×9.10−90=738.1 kN mM_{max} = R_A\times 9.10 - 30\times 3 = 91.0\times 9.10 - 90 = 738.1\ \text{kN m}

Check: the other loads give a smaller moment (under the 60 kN load, M=702.1M = 702.1 kN m).

Answer: absolute maximum bending moment = 738.1 kN m, under the 80 kN load.

✓ Answer checked by Claude Opus 5.5 (high effort)

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