Shearline
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Problem 8

40 m beam on a level bed

A beam of uniform cross section, 40 m long weighing 320 t, is completely resting on a plain horizontal surface. It carries a UDL of 200 t each over 10 m of its length at the ends and a UDL of 800 t over 20 m of its mid-length. Draw the SF and BM diagrams of the beam.

  • Tonnes treated as the force unit used in the paper (marine practice).
  • The bed supplies a uniform upward reaction of total load / 40 m.

Loading diagram

40 m28 t/m48 t/m28 t/mAB

Max sagging BM

1000 t·m

at 20 m

SF and BM diagrams

x20 m|SF 0|BM 1000
Shear force · t-100-50050100ABBending moment · t·m-1000-50005001000AB

Drag across the plots to read values. Sagging BM is drawn positive.

  1. Step 1

    Bed reaction

    Total load = 320 + 200 + 200 + 800 = 1520 t. Uniform bed reaction = 1520/40 = 38 t/m up. Self-weight = 8 t/m.

  2. Step 2

    Net load

    0–10 m and 30–40 m: 8 + 20 − 38 = 10 t/m net up. 10–30 m: 8 + 40 − 38 = 10 t/m net down. A hogging-free sagging box — SF peaks ±100 t at the 10 m and 30 m changes, BM peaks 1000 t·m at mid-length.