The Linear Shaft That Sagged: Beam Deflection, by the Numbers
A Ø20 mm hardened linear shaft, 800 mm long, supported at both ends. A carriage carrying 20 kg rode on two linear bearings. At mid-span, the shaft sagged 0.3 mm under the carriage. The carriage tipped. The parts it positioned were off. This is the deflection calculation that should have been done before the shaft was ordered.
The formula
A simply supported beam (shaft supported at both ends, load at mid-span) deflects:
δ = F · L³ / (48 · E · I)
Where F is the load at mid-span (N), L is the support span (mm), E is Young’s modulus (200,000 N/mm² for steel), and I is the section moment of inertia.
For a solid circular shaft, I = π · d⁴ / 64. For Ø20 mm: I = π × 20⁴ / 64 = 7,854 mm⁴. F = 20 kg × 9.8 = 196 N. L = 800 mm.
δ = 196 × 800³ / (48 × 200,000 × 7,854) = 196 × 5.12×10⁸ / (7.54×10¹⁰) = 100,352,000,000 / 75,400,000,000 = 1.33 mm. We measured 0.3 mm because the load wasn’t exactly at mid-span — the carriage was at about 60% of the span. But the calculation shows the shaft was way too flexible.
What the calculation tells you
A 0.3 mm sag on an Ø20 shaft is expected. The question is: can the system tolerate it? If the carriage positions a part to ±0.1 mm, no. If it supports a non-critical conveyor bed, fine. The tolerance is what decides the shaft size, not the other way around.
| Shaft diameter | I (mm⁴) | Deflection at mid-span (800 mm, 20 kg) |
|---|---|---|
| Ø16 | 3,217 | 3.25 mm |
| Ø20 | 7,854 | 1.33 mm |
| Ø25 | 19,175 | 0.55 mm |
| Ø30 | 39,794 | 0.27 mm |
| Ø35 | 73,660 | 0.15 mm |
Going from Ø20 to Ø30 cuts deflection by a factor of 5. The shaft diameter to the fourth power — doubling the diameter makes it 16× stiffer. This is why upsizing the shaft is the cheapest fix for sag.
Support options
If upsizing the shaft is too expensive, add supports. A shaft supported at three points (mid-span support) cuts deflection dramatically:
- Simply supported (2 points): δ = 1.33 mm (our problem)
- Center support (3 points): deflection drops to about 0.2 mm — but the carriage has to cross the support, which requires a linear bearing that can pass over it (a “supported shaft” with a groove, or a profile rail instead)
- Fixed at both ends (clamped): δ drops by about 4× compared to simply supported. Clamping the ends reduces the effective span. This is the cheapest fix if the ends can be rigidly held.
The real fix
We replaced the Ø20 shaft with a Ø25. The deflection dropped to 0.55 mm. The customer’s tolerance was ±0.3 mm — close but not quite. We added a center support (a pillow block at mid-span) and switched to a profile linear rail (which is inherently supported along its length, not at two points). The sag disappeared. The carriage ran straight.
The other option was a profile rail from the start. Profile rails (HGH20, etc.) are machined flat along their length and bolted to a rigid extrusion. They don’t sag because the support is continuous. A round shaft is a beam; a profile rail is a straight edge. If precision matters, use the rail, not the shaft.
When to use what
| Application | Sag tolerance | Use |
|---|---|---|
| Light duty conveyor guide | ±1 mm | Round shaft, Ø20, 800 mm span |
| Pick-and-place carriage | ±0.2 mm | Round shaft, Ø30, or center support |
| Positioning axis (servo-driven) | ±0.05 mm | Profile linear rail, not round shaft |
| Vision stand / measuring | ±0.01 mm | Profile rail on machined surface |
The calculation rule: Do the deflection math before ordering the shaft. δ = F·L³/(48·E·I). If δ is bigger than your tolerance, upsizing the diameter (to the 4th power) or adding supports is cheaper than fixing the sag after installation. A sagging shaft isn’t a wear problem — it’s an undersized beam problem from day one.
