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)
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.