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Beam Deflection Calculator

Max deflection, bending moment and stress for common beam cases

Updated · Free, no signup

m
kN or kN/m
cm⁴

From a section table (e.g. IPE 300 ≈ 8,356 cm⁴) or b·h³/12 for a rectangle.

cm³

I ÷ distance to extreme fibre; b·h²/6 for a rectangle.

Maximum deflection

5.39 mm

Maximum bending moment

30 kN·m

Maximum shear force

10 kN

Maximum bending stress

53.9 MPa

Span ÷ deflection

1,114

Higher is stiffer. L/360 is a common floor limit.

Allowable deflection at chosen limit

16.67 mm

Deflection check

OK — within L/360
  • Deflection is L/1,114; the L/360 limit allows 16.7 mm.
  • Bending stress of 53.9 MPa is 20% of the 275 MPa yield strength of S275 steel (before safety factors).

Deflected shape along the beam (mm, downward)

About the Beam Deflection Calculator

This beam deflection calculator gives the maximum deflection, bending moment, shear force and bending stress for the four textbook load cases: a simply supported beam with a central point load or a uniformly distributed load (UDL), and a cantilever with an end point load or a UDL. It also plots the deflected shape along the span.

It is useful for students checking homework, engineers doing a quick hand-check, and builders sanity-checking a joist, lintel or shelf bracket. Choose a material to fill in Young’s modulus, then enter the second moment of area (I) and elastic section modulus (S) from a steel section table or from b·h³/12 and b·h²/6 for a rectangle.

Results assume linear-elastic, small-deflection Euler–Bernoulli beam theory, a prismatic beam and no self-weight unless you include it in the load. Shear deformation, lateral-torsional buckling and connection details are not checked, so use it for preliminary sizing — a qualified engineer should verify anything structural.

With the default inputs, the maximum deflection is 5.39 mm. Change any value above to recalculate instantly.

How to use the beam deflection calculator

  1. 1Pick the support type and whether the load is a point load or spread evenly.
  2. 2Enter the span in metres and the load in kN (point) or kN/m (UDL).
  3. 3Choose a material or enter a custom Young’s modulus.
  4. 4Enter I and S for the section from a table or from b·h³/12 and b·h²/6.
  5. 5Compare the maximum deflection with the L/n limit and review moment and stress.

Formula and method

Simply supported, centre load: δ = PL³/48EI, M = PL/4
Simply supported, UDL: δ = 5wL⁴/384EI, M = wL²/8
Cantilever, end load: δ = PL³/3EI, M = PL
Cantilever, UDL: δ = wL⁴/8EI, M = wL²/2
Bending stress: σ = M/S

The formulas come from integrating the Euler–Bernoulli beam equation EI·y″ = M(x) with the boundary conditions of each support type. Deflection grows with the cube (point load) or fourth power (UDL) of the span, so doubling the span of a uniformly loaded beam makes it sag sixteen times as much — span is usually the dominant factor.

Stiffness EI combines the material (Young’s modulus E) and the cross-section shape (second moment of area I). Maximum bending stress is the peak moment divided by the elastic section modulus S. Units are converted internally to newtons, metres and pascals; deflection is reported in millimetres and stress in MPa (N/mm²). Self-weight is not added automatically.

δ
Maximum deflection
P
Point load (kN)
w
Uniformly distributed load (kN/m)
L
Span length (m)
E
Young’s modulus of the material (GPa)
I
Second moment of area of the section (cm⁴)
S
Elastic section modulus (cm³)
M
Maximum bending moment (kN·m)

Worked examples

6 m steel beam (IPE 300), 20 kN at mid-span

δ = 20,000 × 6³ ÷ (48 × 200×10⁹ × 8,356×10⁻⁸) ≈ 5.39 mm, or L/1114 — well inside L/360 (16.7 mm). The peak moment is 20 × 6 ÷ 4 = 30 kN·m, giving 30,000 ÷ 557×10⁻⁶ ≈ 53.9 MPa.

Same beam with a 10 kN/m uniform load

A 10 kN/m UDL over 6 m totals 60 kN. δ = 5 × 10,000 × 6⁴ ÷ (384 × EI) ≈ 10.1 mm (L/594). The moment is 10 × 6² ÷ 8 = 45 kN·m and each support carries 30 kN.

2 m steel cantilever (IPE 200), 5 kN at the tip

δ = 5,000 × 2³ ÷ (3 × 200×10⁹ × 1,943×10⁻⁸) ≈ 3.43 mm. The fixed end resists 5 × 2 = 10 kN·m, which on a 194 cm³ section is about 51.5 MPa.

3 m timber cantilever (45×195 mm), 2 kN/m

For a 45×195 mm joist, I = 0.045 × 0.195³ ÷ 12 ≈ 2,781 cm⁴. With E = 11 GPa, δ = 2,000 × 3⁴ ÷ (8 × EI) ≈ 66.2 mm, far beyond the 16.7 mm L/180 limit — this cantilever needs a deeper section or a back-span.

Frequently asked questions

What is the formula for beam deflection?+

For a simply supported beam with a central point load, δ = PL³ ÷ 48EI; with a uniform load, δ = 5wL⁴ ÷ 384EI. A cantilever deflects PL³ ÷ 3EI under an end load and wL⁴ ÷ 8EI under a uniform load.

What is an acceptable beam deflection?+

Building codes commonly limit live-load deflection to L/360 for floors with brittle finishes, L/240 for total load on floors and roofs, and L/180 for some roofs and cantilevers. Check the code and finishes that apply to your project.

How do I find the moment of inertia of a beam?+

For a solid rectangle, I = b·h³ ÷ 12 with h the depth in the bending direction. For steel I-beams, channels and hollow sections, use the Ix value from the manufacturer’s section table.

Why does a cantilever deflect so much more than a simply supported beam?+

A cantilever has only one support, so the whole span acts like a lever from the fixed end. For the same span and point load it deflects 16 times as much as a simply supported beam loaded at mid-span.

Does this include the beam’s own weight?+

No. Add the self-weight to the uniform load yourself — for example, an IPE 300 weighs about 42 kg/m, or roughly 0.41 kN/m.

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