Combined Torsion & Bending Shaft Diameter

Calculate shaft diameter under simultaneous torsion and bending loads using equivalent moment methods.

Inputs

M

Bending Moment

N·mm

T

Torque

N·mm

\sigma

Allowable Bending Stress

MPa

\tau

Allowable Shear Stress

MPa

Formula Interpretation

Equivalent Torque

T_e = \sqrt{M^2 + T^2} \quad \text{(N}{\cdot}\text{mm)}

$M$ is the maximum bending moment (N·mm); $T$ is the maximum torque (N·mm); $T_e$ is the equivalent torque.

Equivalent Bending Moment

M_e = \dfrac{1}{2}(M + T_e) \quad \text{(N}{\cdot}\text{mm)}

$T_e$ is the equivalent torque; $M_e$ combines both loads into a single equivalent bending moment.

Solid Shaft — Bending Method

d = \sqrt[3]{\dfrac{10M_e}{\sigma}} \quad \text{(mm)}

Uses $M_e$ and allowable bending stress $\sigma$ (MPa).

Hollow Shaft — Bending Method

d_2 = \sqrt[3]{\dfrac{10M_e}{(1-k^4)\sigma}} \quad \text{(mm)}

$k$ is the inner-to-outer diameter ratio; $d_2$ is the outer diameter.

Solid Shaft — Torsion Method

d = \sqrt[3]{\dfrac{5T_e}{\tau}} \quad \text{(mm)}

Uses $T_e$ and allowable shear stress $\tau$ (MPa).

Hollow Shaft — Torsion Method

d_2 = \sqrt[3]{\dfrac{5T_e}{(1-k^4)\tau}} \quad \text{(mm)}

Uses $T_e$ , $k$ and allowable shear stress $\tau$ .

Knowledge: Combined Loading Design

Shaft cross-sections are usually circular. Because shafts are subject to both torque and bending moment, the design must ensure the shaft has sufficient strength to withstand bending, torsion, and shear loads. When determining shaft diameter, calculate separately for bending-only and torsion-only, then take the larger value.

Worked Example

Find the inner and outer diameters of a hollow shaft subjected simultaneously to a bending moment of $4{\times}10^5\,\text{N}{\cdot}\text{mm}$ and a torque of $6{\times}10^5\,\text{N}{\cdot}\text{mm}$ . The allowable bending stress is $50\,\text{MPa}$ , the allowable shear stress is $25\,\text{MPa}$ , and $k=0.5$ .

Step 1 — Equivalent torque Tₑ (Formula ④)

T_e = \sqrt{M^2 + T^2} = \sqrt{(4{\times}10^5)^2 + (6{\times}10^5)^2} \approx 7.21{\times}10^5 \text{ (N}{\cdot}\text{mm)}

Step 2 — Equivalent bending moment Mₑ (Formula ①)

M_e = \dfrac{1}{2}(M + T_e) = \dfrac{1}{2}(4{\times}10^5 + 7.21{\times}10^5) \approx 5.61{\times}10^5 \text{ (N}{\cdot}\text{mm)}

Step 3 — Outer diameter d₂ by torsion method (Formula ⑥)

d_2 = \sqrt[3]{\dfrac{5T_e}{(1-k^4)\tau}} = \sqrt[3]{\dfrac{5{\times}7.21{\times}10^5}{(1-0.5^4){\times}25}} \approx 53.6 \text{ (mm)}

Step 4 — Inner diameter d₁

d_1 = k \cdot d_2 = 0.5 \times 54 = 27 \text{ (mm)}

Therefore, outer diameter d₂ = $54\,\text{mm}$ and inner diameter d₁ = $27\,\text{mm}$ .

Extended Knowledge

•When the shaft material is a plastic material such as carbon steel, failure often occurs due to shear stress, so only the equivalent torque needs to be calculated.
•When the shaft material is a brittle material such as cast iron or hardened steel, failure by bending stress is more likely, so only the equivalent bending moment needs to be calculated.
•In practice, always take the larger of the two method results to ensure sufficient safety margin against both failure modes.

Combined Torsion & Bending Shaft Diameter

Calculate shaft diameter under simultaneous torsion and bending loads using equivalent moment methods.

Inputs

M

Bending Moment

N·mm

T

Torque

N·mm

\sigma

Allowable Bending Stress

MPa

\tau

Allowable Shear Stress

MPa

Formula Interpretation

Equivalent Torque

T_e = \sqrt{M^2 + T^2} \quad \text{(N}{\cdot}\text{mm)}

$M$ is the maximum bending moment (N·mm); $T$ is the maximum torque (N·mm); $T_e$ is the equivalent torque.

Equivalent Bending Moment

M_e = \dfrac{1}{2}(M + T_e) \quad \text{(N}{\cdot}\text{mm)}

$T_e$ is the equivalent torque; $M_e$ combines both loads into a single equivalent bending moment.

Solid Shaft — Bending Method

d = \sqrt[3]{\dfrac{10M_e}{\sigma}} \quad \text{(mm)}

Uses $M_e$ and allowable bending stress $\sigma$ (MPa).

Hollow Shaft — Bending Method

d_2 = \sqrt[3]{\dfrac{10M_e}{(1-k^4)\sigma}} \quad \text{(mm)}

$k$ is the inner-to-outer diameter ratio; $d_2$ is the outer diameter.

Solid Shaft — Torsion Method

d = \sqrt[3]{\dfrac{5T_e}{\tau}} \quad \text{(mm)}

Uses $T_e$ and allowable shear stress $\tau$ (MPa).

Hollow Shaft — Torsion Method

d_2 = \sqrt[3]{\dfrac{5T_e}{(1-k^4)\tau}} \quad \text{(mm)}

Uses $T_e$ , $k$ and allowable shear stress $\tau$ .

Knowledge: Combined Loading Design

Worked Example

Step 1 — Equivalent torque Tₑ (Formula ④)

T_e = \sqrt{M^2 + T^2} = \sqrt{(4{\times}10^5)^2 + (6{\times}10^5)^2} \approx 7.21{\times}10^5 \text{ (N}{\cdot}\text{mm)}

Step 2 — Equivalent bending moment Mₑ (Formula ①)

M_e = \dfrac{1}{2}(M + T_e) = \dfrac{1}{2}(4{\times}10^5 + 7.21{\times}10^5) \approx 5.61{\times}10^5 \text{ (N}{\cdot}\text{mm)}

Step 3 — Outer diameter d₂ by torsion method (Formula ⑥)

d_2 = \sqrt[3]{\dfrac{5T_e}{(1-k^4)\tau}} = \sqrt[3]{\dfrac{5{\times}7.21{\times}10^5}{(1-0.5^4){\times}25}} \approx 53.6 \text{ (mm)}

Step 4 — Inner diameter d₁

d_1 = k \cdot d_2 = 0.5 \times 54 = 27 \text{ (mm)}

Therefore, outer diameter d₂ = $54\,\text{mm}$ and inner diameter d₁ = $27\,\text{mm}$ .

Extended Knowledge

•When the shaft material is a plastic material such as carbon steel, failure often occurs due to shear stress, so only the equivalent torque needs to be calculated.
•When the shaft material is a brittle material such as cast iron or hardened steel, failure by bending stress is more likely, so only the equivalent bending moment needs to be calculated.
•In practice, always take the larger of the two method results to ensure sufficient safety margin against both failure modes.