Hong Life CalculatorsUnderstanding the bolt tensile stress area and shear calculator — concept and principle
A threaded bolt is weakest where the thread is cut, so its strength is not based on the nominal diameter but on the tensile stress area A_s. This calculator finds A_s from diameter and pitch and divides the load by it to give the average tensile stress, or the average shear stress when the threads lie in the shear plane. Use it to check that a bolt size is in the right range before consulting a code or catalogue.
How it is calculated
- Tensile stress area: A_s = (π/4)·(d − 0.9382·P)², with d the nominal diameter and P the thread pitch, both in mm.
- Tensile stress: σ = F·1000 / A_s, with F in kN, giving N/mm² = MPa.
- Shear stress: τ = F·1000 / (A_s·n), where n is the number of shear planes the load passes through.
The two stresses are alternative load cases: pure axial pull or pure sideways shear.
The principle
ISO 898-1 defines A_s as the area of a circle whose diameter is the average of the pitch diameter d₂ and a slightly reduced root diameter d₃. For the ISO metric thread profile (fundamental triangle height H = 0.866P) that average works out to d − 0.9382P, which is why the formula needs only d and P. Real bolts break in tension at loads that match this area well, so it is the basis of proof loads. In shear, use A_s if the shear plane cuts the threads and πd²/4 if it cuts the plain shank; each extra shear plane shares the load.
Worked example
- Inputs: M12 coarse thread, d = 12 mm, P = 1.75 mm, load 20 kN, one shear plane.
- d − 0.9382 × 1.75 = 12 − 1.64185 = 10.35815 mm.
- A_s = (π/4) × 10.35815² = 0.785398 × 107.29127 = 84.27 mm² (the familiar 84.3 mm² table value).
- Tensile stress = 20,000 N ÷ 84.266 mm² = 237.34 MPa.
- Shear stress with threads in one plane = 20,000 ÷ (84.266 × 1) = 237.34 MPa.
| Nominal diameter (mm) | Tensile stress area (mm²) |
|---|---|
| 3 | 1.448723007 |
| 6 | 14.91743757 |
| 9 | 42.52331908 |
| 12 | 84.26636752 |
| 15 | 140.1465829 |
| 18 | 210.1639652 |
| 24 | 392.6102307 |
How to read the result
The property class encodes strength: 8.8 means a nominal tensile strength of 800 MPa and a yield of 0.8 × 800 = 640 MPa; 10.9 means 1,000 MPa and 900 MPa. 237 MPa is far below the yield of class 8.8, but codes apply safety factors and, for shear, use only a fraction of tensile strength (EN 1993-1-8 uses 0.6 or 0.5 of f_ub depending on class when threads are in the shear plane).
Common mistakes
- Using the nominal area πd²/4 for a threaded section, which overstates capacity by about 34% for M12.
- Mixing up fine and coarse pitch.
- Adding tensile and shear stresses directly instead of using an interaction check.
- Ignoring preload, prying and fatigue.
Frequently asked questions
What is the tensile stress area of M10 and M16 coarse bolts?
With pitches 1.5 mm and 2 mm the formula gives about 58.0 mm² and 156.7 mm², matching the ISO 898-1 tables (58.0 and 157 mm²).
Does tightening torque change the stress area?
No. Torque creates preload, adding tensile stress, but the area stays A_s.
How much does a fine thread help?
M12 × 1.25 has A_s ≈ 92.1 mm² against 84.3 mm² for the coarse thread.
Sources: ISO 898-1:2013, Mechanical properties of fasteners made of carbon steel and alloy steel — Part 1: Bolts, screws and studs · EN 1993-1-8, Eurocode 3: Design of steel structures — Design of joints