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
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.