Last month I was building a weld fixture. The arm was 300 mm long, Ø25 mm steel, holding a 15 kg part at the tip. I drew it up, sent it to the shop, and when it came back the tip drooped 0.8 mm under load. The drawing said it was fine. The part didn’t locate. I’d skipped a deflection check. This is how I do it now.
The formula I actually use
For a cantilever with an end load, the tip deflection is:
δ = F · L³ / (3 · E · I)
I don’t memorize the derivation. I use the formula, plug in real numbers, and check the result. The variables are:
- F — force at the tip (Newtons). For a 15 kg part: F = 15 × 9.8 = 147 N.
- L — cantilever length in mm. 300 mm.
- E — Young’s modulus in N/mm². Steel is 206,000. Aluminum is 70,000. That’s the single biggest mistake I see — people use steel’s E for an aluminum part and wonder why it flexes.
- I — area moment of inertia in mm⁴. For a round shaft: I = π · d⁴ / 64.
Plugging in the numbers
For the Ø25 mm steel shaft:
I = π × 25⁴ / 64 = π × 390,625 / 64 = 19,175 mm⁴.
δ = 147 × 300³ / (3 × 206,000 × 19,175)
δ = 147 × 27,000,000 / (3 × 206,000 × 19,175)
δ = 3,969,000,000 / 11,857,650,000
δ = 0.335 mm.
Wait — that says 0.34 mm, but the fixture drooped 0.8 mm. Why? Because I forgot the weld. The shaft was welded to a base plate, and the weld wasn’t perfectly rigid. The weld itself deflected. The “fixed” end wasn’t actually fixed. It was more like a semi-fixed end, which doubles the deflection. So the real number is closer to 0.6-0.7 mm. Add the part weight plus the weld gap, and 0.8 mm makes sense.
The table I keep next to my desk
For solid round steel shafts, tip deflection at 100 N load (mm), cantilever length L:
| Shaft Ø | I (mm⁴) | L=100mm | L=200mm | L=300mm | L=500mm |
|---|---|---|---|---|---|
| Ø10 | 491 | 0.103 | 0.824 | 2.78 | 12.9 |
| Ø16 | 3,217 | 0.016 | 0.126 | 0.427 | 1.98 |
| Ø20 | 7,854 | 0.0066 | 0.053 | 0.179 | 0.830 |
| Ø25 | 19,175 | 0.0027 | 0.022 | 0.073 | 0.340 |
| Ø30 | 39,794 | 0.0013 | 0.010 | 0.035 | 0.164 |
| Ø40 | 125,664 | 0.0004 | 0.003 | 0.011 | 0.052 |
Read this table as: for a 100 N load, a Ø20 shaft at 300 mm cantilever bends 0.18 mm. That’s within my 0.2 mm tolerance. A Ø16 shaft at 300 mm bends 0.43 mm. Not acceptable. I’d go to Ø20 or shorter.
The three corrections I always make
1. Fixed end isn’t fixed. A welded or bolted joint isn’t 100% rigid. I multiply the calculated deflection by 1.5 to 2.0 to account for it. For a critical location, I design the support as a cantilever with a guided end (two supports), which reduces deflection by 8x.
2. Aluminum flexes 3x more than steel. Same geometry, aluminum gives δ that’s 206,000/70,000 = 2.94x the steel value. If the table says 0.2 mm for steel, aluminum is 0.59 mm. I don’t let people specify aluminum arms for precision fixtures.
3. Add the part’s own weight to the force. People size for the machining force (say 50 N) but forget the 15 kg part hanging on the end. The tip load is 147 N from the part alone. The cutting force is on top of that. If I’d calculated F = 50 N instead of 147 N, I’d have predicted 0.11 mm and been wrong by 7x.
What I changed on that fixture
I went from Ø25 to Ø30 steel. At 300 mm with 147 N: δ = 147 × 27,000,000 / (3 × 206,000 × 39,794) = 0.16 mm. With the weld correction (×1.7), that’s 0.27 mm. Still over my 0.2 mm target. So I also shortened the cantilever from 300 to 250 mm. δ = 147 × 15,625,000 / (3 × 206,000 × 39,794) = 0.094 mm. With correction: 0.16 mm. That works.
The alternative was going to Ø40, which added 3 kg to the arm and cost more material. Shortening the overhang was cheaper and stiffer. Sometimes the best design change is making the thing shorter, not thicker.
When I use a different formula
If the load is distributed along the beam (like a rail with brackets), it’s a different case. Uniform load deflection is δ = w · L⁴ / (384 · E · I) for simply supported, or δ = w · L⁴ / (8 · E · I) for cantilever. I use the point-load formula for most fixture arms, but for linear rails mounted on a support, I switch to distributed load. The L⁴ term means length matters enormously — go from 1 m to 1.2 m and deflection doubles.
The number I check before I send any arm to the shop: calculated tip deflection under actual load, times a 1.7x weld-joint factor, must be under the location tolerance. If it isn’t, I don’t redesign with a bigger diameter first — I look at whether I can shorten the overhang. Geometry beats material.
