A hydraulic steel tube that burst at 150 bar. It was a 15 mm OD, 1.5 mm wall, seamless steel tube (ST37.4). The system pressure was 100 bar. The customer used schedule 40 tubing. It was rated for 200 bar by the supplier. But it burst at 150. The issue: the tube was bent to a 2D radius, and the bend thinned the wall. This is about thin-walled pressure tubing.
The thin-wall formula
The hoop stress in a thin-walled cylinder under internal pressure:
σ_h = P · D / (2 · t)
Where P is the pressure (bar), D is the mean diameter (mm), and t is the wall thickness (mm). For a 15 mm OD, 1.5 mm wall tube: D = 15 – 1.5 = 13.5 mm. At 150 bar (15 N/mm²): σ_h = 15 × 13.5 / (2 × 1.5) = 101 N/mm² = 101 MPa. The yield strength of ST37.4 is 235 MPa. The safety factor is 235/101 = 2.3. That should be fine. But the tube burst.
The bend thinning
When you bend a tube, the outer wall stretches. The wall thins at the bend. For a 2D bend radius (2 × OD = 30 mm), the wall thins by about 15%. At the bend: t = 1.5 × 0.85 = 1.275 mm. σ_h = 15 × 13.5 / (2 × 1.275) = 79.4 N/mm². Still below yield. But the bend also work-hardens the material. At the bend, the ductility drops. The burst pressure is about 20% lower than the straight tube. Effective burst pressure: 150 × 0.8 = 120 bar. At 150 bar working pressure, it bursts.
What I changed
1. Upsized to 2.0 mm wall. With 2.0 mm wall: σ_h at 150 bar = 15 × 13.5 / (2 × 2.0) = 50.6 MPa. Safety factor: 235/50.6 = 4.6. Even with 15% bend thinning (t=1.7 mm): σ_h = 59.6 MPa. SF = 3.9. Safe. The 2.0 mm wall tube costs 15% more but doesn’t burst.
2. Increased bend radius. I specified a 3D bend radius instead of 2D. The wall thinning drops from 15% to 8%. At 2.0 mm wall, the bend wall is 1.84 mm. σ_h = 55 MPa. The longer bend also reduces stress concentration.
3. Specified cold-drawn seamless tube. Welded tubing has a seam that’s weaker in fatigue. For hydraulic lines over 100 bar, I specify seamless cold-drawn tube (ST37.4 or E235). The supplier’s “schedule 40” rating was for straight tube, not bent. I add 30% margin for the bend.
The burst pressure formula
For a quick check:
P_burst = 2 · t · R_m / D
Where R_m is the tensile strength (360 MPa for ST37.4). For 1.5 mm wall, 13.5 mm ID: P_burst = 2 × 1.5 × 360 / 13.5 = 80 N/mm² = 800 bar. That’s the theoretical burst. But with safety factor 4 (for hydraulic): P_working = 200 bar. The supplier was right. But at the bend, P_burst drops 20% to 640 bar. With SF 4: P_working = 160 bar. At 150 bar, it’s close. The tube was at the edge.
The safety factor table
| Application | Minimum safety factor |
|---|---|
| Water, low pressure, static | 3 |
| Hydraulic, steady pressure | 4 |
| Hydraulic, shock/vibration | 6 |
| Steam/gas, hazardous | 8 |
The wall thickness I use: add 30% for bends. The burst tube wasn’t undersized for straight length — it was at a tight 2D bend that thinned the wall. Specify 3D bends or upsized wall thickness. Safety factor of 4 minimum, 6 for shock.
