Bending Shaft Diameter
Determine minimum shaft diameter under bending moment for solid or hollow shafts. k = inner/outer diameter.
Inputs
Formula
Solid d = ∛(32M/(π[σ])); Hollow dₒ = ∛(32M/(π(1−k⁴)[σ])) (M in N·mm)
Fundamentals
A shaft under transverse load bends; bending normal stress is the main controlling stress in shaft design.
Max bending stress $\sigma=M/W$, $W=\pi d^3/32$ the section modulus; simply-supported mid-load gives $M_\text{max}=FL/4$.
History
Beam bending and deflection theory was founded by Galileo and Euler-Bernoulli; the integration and area-moment methods matured in the 19th c.
Engineering applications
Used for stiffness checks of long shafts (leadscrews, reducer shafts) to avoid gear misload and early seal failure.
Glossary
| Bending moment $M$ | Algebraic sum of moments of forces on one side of a section. |
| Section modulus $W$ | Geometric resistance to bending, $W=I/c$. |
| Neutral axis | Layer unchanged in length during bending; zero stress. |
How to use
- Fill in Bending moment M, Allow. bend. [σ], Hollow ratio k (0=solid) in the Inputs section (watch the unit on each field).
- Click Calculate; the tool evaluates the formula shown above.
- Read Solid dia., Hollow outer (k>0) in the results area.
Formula notesBending M = F·L; solid d = ∛(32M/(π·σ)), hollow d = ∛(32M/(π(1−k⁴)σ)), k = d_i/d_o.
Formula · Worked Example · Knowledge
Formula
For a solid shaft under bending only, the strength condition on bending stress gives the minimum diameter:
$$d=\sqrt[3]{\frac{10M}{\sigma}}\ \text{(mm)}$$
Hollow shaft (inner/outer ratio $k=d_1/d_2$):
$$d_2=\sqrt[3]{\frac{10M}{(1-k^{4})\,\sigma}}\ \text{(mm)}$$
$M$ = max bending moment (N·mm), $\sigma$ = allowable bending stress (MPa).
$$d=\sqrt[3]{\frac{10M}{\sigma}}\ \text{(mm)}$$
Hollow shaft (inner/outer ratio $k=d_1/d_2$):
$$d_2=\sqrt[3]{\frac{10M}{(1-k^{4})\,\sigma}}\ \text{(mm)}$$
$M$ = max bending moment (N·mm), $\sigma$ = allowable bending stress (MPa).
Worked Example
Hollow shaft, $k=0.5$, allowable stress $\sigma=50\ \text{MPa}$, max moment $M=8.0\times10^{6}\ \text{N·mm}$.
$$d_2=\sqrt[3]{\frac{10\times8.0\times10^{6}}{(1-0.5^{4})\times50}}\approx119.5\ \text{mm}$$
Inner diameter $d_1=0.5\times119.5\approx59.8\ \text{mm}$.
$$d_2=\sqrt[3]{\frac{10\times8.0\times10^{6}}{(1-0.5^{4})\times50}}\approx119.5\ \text{mm}$$
Inner diameter $d_1=0.5\times119.5\approx59.8\ \text{mm}$.
Key Points
- For combined bending + torsion, compute diameter from each and take the larger (equivalent-stress method).
- Long shafts also need deflection and critical-speed (resonance) checks.
- Hollow shafts are lighter at equal strength — good for high-speed, weight-sensitive shafts.
Parameters
Inputs
| Parameter | Symbol | Unit | Default |
|---|---|---|---|
| Bending moment M (N·m) | M | N·m | 500 |
| Allow. bend. [σ] (MPa) | sigma | MPa | 80 |
| Hollow ratio k (0=solid) | k | 0 |
Outputs
| Result | Symbol | Unit |
|---|---|---|
| Solid dia. | d_solid | mm |
| Hollow outer (k>0) | d_hollow | mm |
Applications
- Common engineering use cases
FAQ
What formula does this tool use?
This tool computes per ISO / AGMA / ASME standard formulas: Solid d = ∛(32M/(π[σ])); Hollow dₒ = ∛(32M/(π(1−k⁴)[σ])) (M in N·mm)
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