Calculate the deflection of aluminium profiles
How far does a profile bend under load? Choose the support, load and length; deflection, bending stress and rating appear instantly.
Material and safety
Modulus of elasticity of aluminium 70,000 N/mm². Standard: AlMgSi 0.5 F25 with Rp0.2 = 200 N/mm².
Good The profile carries the load with little deflection.
- Deflection f
- 1.55 mmL/644 · Good
- Bending stress σ
- 25.9 N/mm²19 % of 133 · Good
- Maximum bending moment
- 124.9 N·m
- of which from dead weight
- f 0.04 mm · 18.5 N
Cut to L, in 0.5 mm steps. Add the bearing or clamping length if needed. To the profile page
How the calculator works
It is based on beam theory for slender beams with a constant cross-section. F is the load, L the length, E the modulus of elasticity (aluminium 70,000 N/mm²) and I the second moment of area about the bending axis. For a distributed load, q = F / L.
| Support | Load | Deflection f | Moment M |
|---|---|---|---|
| Simply supported | Point load, middle | F·L³ / (48·E·I) | F·L / 4 |
| distributed | 5·q·L⁴ / (384·E·I) | q·L² / 8 | |
| Fixed at one end | Point load, end | F·L³ / (3·E·I) | F·L |
| distributed | q·L⁴ / (8·E·I) | q·L² / 2 | |
| Fixed at both ends | Point load, middle | F·L³ / (192·E·I) | F·L / 8 |
| distributed | q·L⁴ / (384·E·I) | q·L² / 12 |
The dead weight acts as a distributed load with the same support and is added to the load. The bending stress is σ = M / W with the section modulus W. The permissible stress is σzul = Rp0.2 / safety factor.
Rating
- Deflection: up to L/500 good, up to L/300 borderline, above that too high.
- Stress: up to 80 % of σzul good, up to 100 % borderline, above that overloaded.
Limits of the calculation
- Only elastic deformation under static load. Impacts, vibrations and fatigue strength are not included.
- The flexibility of connectors, T-slot nuts and screw connections is not included. Real frames therefore deflect more; “fixed” is the most favourable limiting case.
- Shear deformation, torsion and buckling are not considered.
- The section values of the ProFix profiles are catalogue values of the nominal cross-section. Extrusion tolerances change them slightly. For profiles without catalogue values, choose “Other profile” and enter the values from the data sheet.
- The result is an estimate to help you choose. It does not replace a structural calculation or check.
Example: ProFix Profile 40 × 40 slot 8 Light with 50 kg
A ProFix Profile 40 × 40 slot 8 Light rests on two supports 1,000 mm apart. A load of 50 kg hangs in the middle, i.e. F = 490.5 N. The profile has I = 96,200 mm⁴ and W = 4,830 mm³.
- Deflection from the load: f = F·L³ / (48·E·I) = 490.5 · 1,000³ / (48 · 70,000 · 96,200) = 1.52 mm
- Dead weight: 1.89 kg/m gives 18.5 N and an additional 0.04 mm.
- Total: f = 1.55 mm, that is L/644.
- Bending stress: σ = M / W = 25.9 N/mm², permissible 133.3 N/mm² (19 % utilised).
Result: Deflection below L/500 and stress in the green range. The profile is sufficient for this case. Use the calculator above to check other lengths, loads and profiles.
Frequently asked questions
How do I calculate the deflection of an aluminium profile?
You need the support, load, span L, the modulus of elasticity of aluminium (70,000 N/mm²) and the second moment of area I of the profile about the bending axis. For a simply supported beam with a point load in the middle, f = F·L³ / (48·E·I). The calculator above uses the catalogue values of the ProFix profiles and adds the dead weight.
Which deflection is permissible?
That depends on the application. For frames and cladding, L/300 is the usual limit; for workbenches and guides rather L/500 or less. With a 1000 mm span, that is 3.3 mm and 2 mm respectively.
How do I convert kg to newtons?
Mass in kg times gravitational acceleration 9.81 m/s². 10 kg is roughly 98 N. The calculator converts automatically when you choose the unit kg.
What does upright instead of flat achieve?
The second moment of area grows with the cube of the height in the load direction. A 40 × 80 profile mounted upright deflects about four to five times less than flat.
Why is a fixed beam so much stiffer?
Fixing the ends prevents rotation at the support. With a point load in the middle, deflection drops to a quarter, with a distributed load to a fifth compared with a simply supported beam. Real profile connections are never completely rigid, so the result lies between the two cases.
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