What torque does an M20 × 2.5 bolt take?
Answer
An M20 × 2.5 in Class 8.8 takes 425.9 N·m — 314.2 lb·ft — dry, tightened to 75% of proof load. Oiled, it is far less.
How it is worked out
- Torque is only a proxy for tension
T = K × D × F - Tensile stress area of this bolt
244.79 mm² - Target 75% of proof load
106.49 kN of clamp force in Class 8.8 - Multiply out
0.20 × 0.02 m × 106,486 N = 425.9 N·m
T = K · D · F — and 85–90% of that torque is spent on friction
Worth knowing
- Dry, as received — K 0.20
- Class 8.8 425.9 N·m (314.2 lb·ft) · Class 10.9 609.5 N·m (449.6 lb·ft) · Class 12.9 712.4 N·m (525.4 lb·ft) · A2-70 stainless 330.5 N·m (243.7 lb·ft)
- Lightly oiled — K 0.15
- Class 8.8 319.5 N·m · Class 10.9 457.2 N·m · Class 12.9 534.3 N·m · A2-70 stainless 247.9 N·m
- Anti-seize or moly — K 0.12
- Class 8.8 255.6 N·m · Class 10.9 365.7 N·m · Class 12.9 427.4 N·m · A2-70 stainless 198.3 N·m
- Why lubrication changes it so violently
- Almost all the torque goes into friction rather than tension. Halving the friction roughly doubles the preload for the same torque — which is how a bolt gets snapped with a correctly set wrench and a dry-torque chart.
- Single shear capacity
- 117.5 kN in Class 8.8
- Can it be reused
- At 75% of proof, yes — the bolt stayed elastic. Torque-to-yield fasteners have already stretched permanently and cannot.
- The thread it goes into
- This sizes the bolt, not the hole. A steel bolt in aluminium strips the aluminium first unless engagement is one and a half diameters or more.
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