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Engineering · Statics

Beam Deflection Calculator

Maximum deflection of a beam under a central or end point load.

Load case
N
Point load on the beam.
m
Span or cantilever length.
GPa
Material stiffness, entered in GPa.
m⁴
Section stiffness; scientific notation OK.
Worked examples — tap to load
Maximum deflection (δ)
0.08333mm

That is 0.00008333 m for a simply supported beam. Stiffer E or I means less deflection.

Deflected shape along the beam (mm)
Deflected shape of the simply supported beam, deflection in millimetres plotted against position along its length0.08333 mm002 m

Maximum beam deflection under a point load is δ = F·L³ ÷ (48·E·I) for a simply supported beam and δ = F·L³ ÷ (3·E·I) for a cantilever. A 1 kN load at the centre of a 2 m steel beam (E 200 GPa, I 1×10⁻⁵ m⁴) deflects 0.0833 mm.

What beam deflection is

Deflection is how far a beam bends out of its straight line when a load is applied. For a single point load, the largest deflection sits under the load on a simply supported beam and at the free end of a cantilever. It grows with the cube of the length, so doubling the span makes a beam bend eight times as much — and it shrinks as the material stiffness E or the section stiffness I rises.

δ = F·L³ ÷ (48·E·I)

Simply supported, central point load · F = load (N), L = span (m), E = Young’s modulus (Pa), I = second moment of area (m⁴)

Worked example

A simply supported steel beam spans 2 m and carries a 1 kN load at midspan. Steel has E = 200 GPa = 200×10⁹ Pa, and the section has I = 1×10⁻⁵ m⁴.

  1. 1
    Pick the load case. Central point load on a simply supported beam uses the divisor 48; an end load on a cantilever uses 3.
  2. 2
    Put everything in SI units. F in newtons, L in metres, E in pascals (200 GPa = 200×10⁹ Pa), and I in m⁴.
  3. 3
    Substitute into δ = F·L³ ÷ (48·E·I). δ = 1000 × 2³ ÷ (48 × 200×10⁹ × 1×10⁻⁵) = 8000 ÷ 9.6×10⁷.
  4. 4
    Read the deflection. δ = 8.33×10⁻⁵ m = 0.0833 mm. The same beam as a cantilever would deflect 8000 ÷ 6×10⁶ = 1.333 mm.

Point-load deflection formulas

Both give the maximum deflection δ. A stiffer beam — larger E or I — deflects less, so raising I is the most effective fix.

Load caseMaximum deflectionWhere it occurs
Simply supported, central loadδ = F·L³ ÷ (48·E·I)Under the load, at midspan
Cantilever, end loadδ = F·L³ ÷ (3·E·I)At the free end

Assumptions and limits

These formulas assume linear-elastic behaviour (the material obeys Hooke’s law and springs back), small deflections relative to the span, and a uniform, prismatic cross-section along the length. They cover a single concentrated point load only — distributed or multiple loads use different constants.

The section stiffness I depends entirely on the cross-section shape: a tall rectangle beam has I = b·h³ ÷ 12, so orienting the section with its depth vertical dramatically reduces deflection. Beyond the elastic limit, or once deflections get large, these results no longer hold and a full structural analysis is needed.

What is the second moment of area (I)?
I measures how a cross-section’s material is distributed about the bending axis, in m⁴. Material far from the neutral axis contributes most, so a deep section resists bending far better than a shallow one of the same area. For a rectangle, I = b·h³ ÷ 12.
What is the difference between a simply supported and a cantilever beam?
A simply supported beam rests on a support at each end and bends most under a central load, using the divisor 48. A cantilever is fixed at one end and free at the other, so the same load at its tip bends it 16 times as much — its divisor is only 3.
Why does length matter so much?
Deflection scales with L³. Doubling the span increases deflection eightfold, which is why long, unsupported beams sag and why adding an intermediate support helps so much.
What units should I use?
Work in SI: load in newtons, length in metres, E in pascals, and I in m⁴, giving δ in metres. The calculator lets you enter E in GPa and converts it (×10⁹) for you, then reports δ in both mm and m.
How do I reduce a beam’s deflection?
Increase the second moment of area I (a deeper section is most effective since I grows with depth cubed), use a stiffer material with higher E, shorten the span, or add supports. Reducing the load helps proportionally.
Does this account for the beam’s own weight?
No — it models a single concentrated point load only. Self-weight acts as a distributed load, which uses different formulas (for example 5wL⁴ ÷ 384EI for a simply supported beam), so combine cases by superposition if self-weight matters.