I was reviewing a welded frame last week. The engineer had used a 50×50×3 mm square tube as a crossbeam spanning 1.5 m, carrying a 100 kg axis at center. He said “it’s square tubing, it’ll be fine.” I did the stress calculation. It wasn’t fine. The stress was 80% of yield. At any bump or overload, it would bend permanently. This is how I check a beam before it goes to the shop.
The bending stress formula
For a simply supported beam with a center load:
σ = M / Z
Where M is the bending moment and Z is the section modulus. For a center load on a simply supported beam, M = F × L / 4. The section modulus Z = I / y, where y is the distance from neutral axis to the outer fiber (half the beam height for a symmetric section).
For the 50×50×3 mm tube: I = (50⁴ – 44⁴) / 12 = (6,250,000 – 3,748,096) / 12 = 208,492 mm⁴. Z = I / 25 = 8,340 mm³.
Load F = 100 kg × 9.8 = 980 N. M = 980 × 1500 / 4 = 367,500 N·mm. σ = 367,500 / 8,340 = 44 N/mm².
Wait — that’s 44 MPa. For steel (yield 250 MPa), that’s only 18% of yield. That’s fine. But the engineer wasn’t carrying just the 100 kg axis. He was also carrying the cutting forces (500 N from a milling head) and the acceleration (the axis moves, so dynamic load is 1.5x). The real M is higher. Let me redo it.
The real load
Static: 980 N at center. Dynamic (1.5x): 1,470 N. Cutting force: 500 N at center (worst case). Total F = 1,970 N. M = 1,970 × 1,500 / 4 = 738,750 N·mm. σ = 738,750 / 8,340 = 88.6 N/mm². That’s 35% of yield. Still fine for static, but for a machine that cycles every 10 seconds, fatigue is the concern. Fatigue strength of steel is about 100 MPa. At 89 MPa, we’re close to the fatigue limit. After a few million cycles, it could crack.
But I also need to check deflection. The customer needs the beam to deflect less than 0.2 mm at center. Deflection δ = F × L³ / (48 × E × I) = 1,970 × 1500³ / (48 × 206,000 × 208,492) = 6,646,875,000,000 / 2,059,754,112,000 = 3.2 mm. That’s 16x over the 0.2 mm target. The beam was strong enough (stress OK) but too flexible. It would sag 3 mm under load. The axis would move with the beam, not the other way around.
What I changed
To get deflection under 0.2 mm, I need I ≥ F × L³ / (48 × E × δ) = 1,970 × 3,375,000,000 / (48 × 206,000 × 0.2) = 6,646,875,000,000 / 1,977,600 = 3,361,000 mm⁴. The 50×50 tube has I = 208,492. I need 16x more. That means going to a bigger section.
| Section | I (mm⁴) | Deflection (mm) | Stress (MPa) |
|---|---|---|---|
| 50×50×3 tube | 208,492 | 3.2 | 89 |
| 60×60×5 tube | 556,000 | 1.2 | 36 |
| 80×80×5 tube | 1,365,000 | 0.49 | 15 |
| 100×100×6 tube | 3,560,000 | 0.19 | 6.5 |
The 100×100×6 tube gives 0.19 mm deflection and 6.5 MPa stress. That’s overkill on stress but meets the deflection target. Alternatively, I could add a middle support post. If the span drops from 1.5 m to 0.75 m, deflection drops by 8x (L³). The 50×50 tube would then give 0.4 mm — still over 0.2 mm. I’d need 60×60 with a middle post: 0.15 mm. That’s cheaper than going to 100×100.
The lesson
Strength and stiffness are different things. The 50×50 tube was strong enough (stress was fine). It was too flexible. For machine frames, deflection is usually the limiting factor, not stress. A frame that doesn’t break but flexes 3 mm won’t hold tolerance. I always check both: stress against yield (with a 2x safety factor) and deflection against the required tolerance.
The numbers I check: bending stress σ = M/Z must be under yield/3. Deflection δ = FL³/(48EI) must be under the location tolerance. For a 1.5 m span, 50×50 tube is too flexible for precision work. Add a middle post or go to a bigger section. The beam that “was fine” was strong but floppy — strength isn’t the same as stiffness.
