Bolt Diameter
Find the minimum required thread major diameter for an axial tensile load using [σ]=σ_s/S, rounded up to the nearest standard metric size.
Inputs
Formula
A = F/[σ]; d = √(4A/π); [σ] = σ_s / S
Fundamentals
Bolt diameter is the nominal thread size needed to carry the axial service load — the first step in joint design.
Tension check: $d\ge\sqrt{4F/(\pi[\sigma]n)}$, with $F$ load, $[\sigma]$ allowable stress, $n$ safety factor. Transverse loads also need preload and friction locking.
History
Thread standards were unified by Whitworth (1841, inch) and the metric ISO thread in the 1960s; property classes made joint design quantifiable.
Engineering applications
Used for flanges, steel structures and engine cylinder heads where threaded joints carry preload and fatigue.
Glossary
| Nominal dia $d$ | Major thread diameter, the primary size parameter. |
| Allowable $[\sigma]$ | Material limit stress divided by safety factor. |
| Safety factor $n$ | Margin for load fluctuation and failure consequence. |
How to use
- Fill in Axial load F, Yield strength σ_s, Safety factor S in the Inputs section (watch the unit on each field).
- Click Calculate; the tool evaluates the formula shown above.
- Read Min diameter, Standard size M in the results area.
Formula notesTensile bolt dia d = √(4F / (π·σ·n)); F = load, σ = allowable stress, n = safety factor.
Formula · Worked Example · Knowledge
Formula
Under pure axial tension, stress at the minor diameter is $\sigma=F/A$; the strength condition gives the required major diameter:
$$d=\sqrt{\frac{4F}{\pi\,\sigma\,n}}\ \text{(mm)}$$
$F$ = tensile load (N), $\sigma$ = allowable stress (MPa), $n$ = safety factor; minor diameter $d_1\approx0.8d$. With combined tension + preload torsion the factor becomes $8/3$.
$$d=\sqrt{\frac{4F}{\pi\,\sigma\,n}}\ \text{(mm)}$$
$F$ = tensile load (N), $\sigma$ = allowable stress (MPa), $n$ = safety factor; minor diameter $d_1\approx0.8d$. With combined tension + preload torsion the factor becomes $8/3$.
Worked Example
Bolt under tensile load $F=30000\ \text{N}$, allowable stress $\sigma=60\ \text{MPa}$, $n=1$.
$$d=\sqrt{\frac{4\times30000}{\pi\times60\times1}}\approx25.2\ \text{mm}$$
Round up to the standard size M27.
$$d=\sqrt{\frac{4\times30000}{\pi\times60\times1}}\approx25.2\ \text{mm}$$
Round up to the standard size M27.
Key Points
- Bolt size is usually found per single load case (tension, tension+torsion, shear), then the governing case is taken.
- Thread size is given by the major diameter $d$ of the external thread.
- Minor diameter $d_1\approx0.8d$ is used to check thread shear/bending.
Parameters
Inputs
| Parameter | Symbol | Unit | Default |
|---|---|---|---|
| Axial load F (N) | F | N | 10000 |
| Yield strength σ_s (MPa) | sigma_s | MPa | 400 |
| Safety factor S | S | 2.5 |
Outputs
| Result | Symbol | Unit |
|---|---|---|
| Min diameter | d_req | mm |
| Standard size M | d_std | mm |
Applications
- Common engineering use cases
FAQ
What formula does this tool use?
This tool computes per ISO / AGMA / ASME standard formulas: A = F/[σ]; d = √(4A/π); [σ] = σ_s / S
How accurate are the results?
Results match input precision, based on SI units and common engineering approximations; for critical duty re-check with a safety factor.
Where is it used?
Common engineering use cases