Fixed Beam with Central Point Load

A point load in the middle of a clamped span splits its moment perfectly: PL/8 hogging at each end, PL/8 sagging under the load.

The equation

Mend = Mmid = PL/8
R = P/2 at each end
δmax = PL3/192EI at midspan
  • P is the concentrated load at midspan, in kN
  • L is the clear span, in meters
  • E and I are the modulus and second moment of area
  • δ is the vertical deflection
P = 20 kNL = 6 mBending moment, sagging plotted upPL/8 = 15 kNm at midspanPL/8 = 15 kNm hogging at each endcalculatgeek.com
Perfect symmetry: ends and midspan carry PL/8 alike, the deflection is a quarter of the pinned value, and the moment diagram crosses zero at the quarter points.

Worked example

Put 20 kN at the middle of a fixed 6 m span with EI = 16000 kNm2. The end and midspan moments are both 20 × 6/8 = 15 kNm, and the deflection is 20 × 63/(192 × 16000) = 1.4 mm, one quarter of the 5.6 mm the same beam deflects simply supported. The independent numerical solve of the clamped beam agrees.

When to use it

The fixity caveat applies with extra force here, because this case is unusually sensitive to it: lose the end restraint and the midspan moment doubles from PL/8 to PL/4 while the deflection quadruples. That swing is the whole argument for detailing the ends properly, or for designing to the pinned values when you cannot guarantee them.

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