Shearline
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Problem 6 · MMD May 2018

Cantilever, UDL plus tip load

A cantilever 6 m long carries a concentrated load of 45 kN at its free end and also a uniformly distributed load of 20 kN/m over its entire length. Find SF and BM at the wall and draw SF/BM diagrams.

Loading diagram

6 m20 kN/m45 kNABRA 165 kNMA -630 kN·m

RA

165 kN

0 m from A

MA

-630 kN·m

hogging at the wall

Max sagging BM

0 kN·m

at 6 m

Max hogging BM

-630 kN·m

at 0 m

SF and BM diagrams

x3 m|SF 105|BM -225
Shear force · kN-200-1000100200ABBending moment · kN·m-1000-50005001000AB

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

  1. Step 1

    Wall reactions

    Vertical reaction at the wall RA = 45 + 20×6 = 165 kN up. Fixing moment MA = 45×6 + 20×6×3 = 270 + 360 = 630 kN·m hogging.

  2. Step 2

    SF at the wall

    SF just inside the beam at the wall is +165 kN (left-up positive). It then falls linearly at 20 kN/m to +45 kN just before the tip, and to 0 after the 45 kN load.

  3. Step 3

    BM at the wall

    BM = −630 kN·m at the wall (hogging), a parabola that returns to 0 at the free end.