Simply Supported Beam with Central Point Load

One load in the middle of one span: the case every lab bench test recreates and every quick check reaches for.

The equation

Mmax = PL/4 under the load
V = R = P/2
δmax = PL3/48EI at midspan
  • P is the concentrated load, in kN
  • L is the span between supports, in meters
  • E is the modulus of elasticity
  • I is the second moment of area
  • δ is the vertical deflection
P = 20 kNL = 6 mBending moment, sagging plotted upMmax = PL/4 = 30 kNmcalculatgeek.com
Concentration is expensive: the same total load spread along the span would produce half this moment, which is why point loads on joists and purlins deserve respect.

Worked example

A 6 m span with EI = 16000 kNm2 carries 20 kN at midspan. Reactions are 10 kN each, the moment under the load is 20 × 6/4 = 30 kNm, and the deflection is 20 × 63/(48 × 16000) = 0.0056 m, or 5.6 mm. Spread the same 20 kN evenly along the span instead and the moment halves to 15 kNm while the deflection drops to 3.5 mm: concentration is what hurts.

When to use it

The same elastic limits as the uniform case apply, plus one of its own: the load must actually be concentrated. A bearing plate a few hundred millimetres long still counts; a load spread over a third of the span does not, and sits somewhere between this case and the uniform one. Check local effects at the load point too, since crushing and web checks arrive before flexure on short spans.

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