A timing pulley pressed onto a shaft. It slipped under load after two months. The press fit was supposed to hold. I’d specified H7/p6 — a standard interference fit. But I’d never calculated the actual interface pressure. I just grabbed the fit from a table. This is the calculation I do now for every press joint that transmits torque.
Why press fits work
A press fit transmits torque through friction between the shaft and the hub. The interference (the shaft is bigger than the hole) creates a radial pressure at the interface. That pressure, multiplied by the friction coefficient and the contact area, gives the torque the joint can hold.
The interface pressure for a solid shaft in a hub is:
p = E × δ × (1 – d²/D²) / (2 × d)
Where:
- p — interface pressure in N/mm²
- E — Young’s modulus (206,000 for steel)
- δ — radial interference in mm (difference between shaft radius and hole radius)
- d — shaft diameter in mm
- D — hub outside diameter in mm
The example
The pulley was on a Ø20 mm shaft. The hub was 40 mm OD. I used H7/p6: the hole is Ø20 +0.000/+0.021, the shaft is Ø20 +0.035/+0.022. The minimum interference is 0.022 – 0.021 = 0.001 mm. The maximum is 0.035 – 0.000 = 0.035 mm. The minimum radial interference is δ_min = 0.001/2 = 0.0005 mm.
Interface pressure at minimum interference: p = 206,000 × 0.0005 × (1 – 20²/40²) / (2 × 20) = 206,000 × 0.0005 × 0.75 / 40 = 1.93 N/mm². That’s tiny. The maximum pressure (at max interference) is p = 206,000 × 0.0175 × 0.75 / 40 = 67.6 N/mm². The range is enormous — 1.9 to 67.6. At the low end, the joint slips. At the high end, the hub could crack.
The problem: H7/p6 has a huge range because the tolerances overlap. The minimum interference is barely positive. I should have specified a tighter fit or calculated the required interference from the torque.
Working backward from torque
The torque the joint must transmit is T = 5 N·m (a small timing pulley). The friction torque capacity is:
T = p × π × d × L × μ × (d/2)
Where L is the engagement length and μ is the friction coefficient (steel on steel, dry: 0.15). For L = 25 mm, d = 20 mm:
T = p × π × 20 × 25 × 0.15 × 10 = p × 2,356 N·mm.
For T = 5,000 N·mm (5 N·m): p = 5,000 / 2,356 = 2.1 N/mm².
Now back-calculate the required interference. From the pressure formula: δ = p × 2 × d / (E × (1 – d²/D²)) = 2.1 × 40 / (206,000 × 0.75) = 0.00054 mm radial. That’s 0.0011 mm diameter interference minimum.
But I need a safety factor of 2 against slip (the friction coefficient varies, there might be oil on the parts). So p = 4.2 N/mm², and δ_diameter = 0.0022 mm minimum. The H7/p6 minimum interference was 0.001 mm — half what I needed. The pulley slipped.
What I specified instead
I moved to H7/u6. The hole is still Ø20 +0.000/+0.021. The shaft is now Ø20 +0.064/+0.048. Minimum interference = 0.048 – 0.021 = 0.027 mm. Maximum = 0.064 – 0.000 = 0.064 mm. Minimum radial δ = 0.0135 mm. Interface pressure at minimum: p = 206,000 × 0.0135 × 0.75 / 20 = 104 N/mm². That gives T = 104 × 2,356 = 245,000 N·mm = 245 N·m. Way more than the 5 N·m needed. The joint won’t slip.
But wait — 104 N/mm² might yield the hub. The hub material is aluminum (E = 70,000, yield = 200 MPa). The hoop stress in the hub is σ = p × (D² + d²) / (D² – d²) = 104 × (1600 + 400) / (1600 – 400) = 173 N/mm². That’s under the 200 MPa yield but close. For a safety factor of 1.5, I’d need σ under 133 MPa. I went to a steel hub (or a slightly larger OD). The steel hub handles it easily.
The press force
When you press the parts together, the required force is:
F_press = p × π × d × L × μ
For p = 104 N/mm², d = 20, L = 25, μ = 0.15: F = 104 × π × 20 × 25 × 0.15 = 24,500 N = 2.5 tons. That needs a hydraulic press, not a bench vise. I tell the shop up front so they don’t try to hammer it in.
When to use a key instead
If the torque is high (over 50 N·m) or the joint sees reversing loads, I don’t rely on press fit alone. I add a Woodruff key or a keyway. The press fit handles the positioning; the key handles the torque. The press fit prevents fretting; the key prevents slip. They work together. A press fit without a key is fine for low torque (5 N·m). For high torque, the key is the backup.
The number I check: required interface pressure from the torque, back-calculated to minimum interference, with a 2x friction safety factor. Then verify the hub doesn’t yield under that pressure. Then calculate the press force. Don’t just pick H7/p6 from a table — calculate it. The pulley that slipped had H7/p6 with barely any interference. It was never going to hold.
