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Machine Design
Standard Equation

ASME Combined Shaft Sizing Equation

Sizing standard for transmission shafting under combined torsional shear and reversed fatigue bending loads.

Primary Mathematical Expression

d = [ (32 * N_sf / (\pi * S_y)) * \sqrt{M^2 + T^2} ]^(1/3)

Design Schematic

Bearing ABearing BPulley 1F1Gear SeatF_mesh (Bending)KeywayTorque (T)Fillet Shoulders

Nomenclature & Variables

SymbolVariable NameMetric UnitImperial UnitDescription
dShaft diametermminThe minimum required outer diameter of the circular solid shaft.
N_sfSafety factordimensionlessdimensionlessDesign Factor of Safety (FOS) to account for uncertainties.
S_yMaterial yield strengthMPapsiThe material tensile yield strength limit.
MBending momentN·min·lbThe bending moment load acting on the shaft section.
TTorqueN·min·lbThe torsional torque load acting on the shaft section.

Step-by-Step Derivation

  1. 1

    Under combined bending and torsion, the shaft experiences maximum normal stress \sigma_x and shear stress \tau_{xy}.

  2. 2

    According to the Maximum Shear Stress theory (Tresca criteria), the maximum shear stress is: \tau_{max} = \sqrt{(\sigma_x/2)^2 + \tau_{xy}^2}.

  3. 3

    Substituting bending stress \sigma_x = 32M / (\pi * d³) and torsional shear stress \tau_{xy} = 16T / (\pi * d³): \tau_{max} = (16 / \pi * d³) * \sqrt{M^2 + T^2}.

  4. 4

    Set allowable shear stress to S_y / (2 * N_sf): S_y / (2 * N_sf) = (16 / \pi * d³) * \sqrt{M^2 + T^2}.

  5. 5

    Solving for diameter d yields the ASME combined loading sizing equation: d = [ (32 * N_sf / (\pi * S_y)) * \sqrt{M^2 + T^2} ]^(1/3).

Worked Example Calculation

Problem Statement

Determine the solid shaft diameter for structural steel (S_y = 250 MPa) subjected to a bending moment of 180 N·m and torque of 350 N·m. Use a safety factor N_sf of 2.0.

Calculation Steps
  • Identify input parameters: S_y = 250 MPa = 250 N/mm², M = 180 N·m = 180,000 N·mm, T = 350 N·m = 350,000 N·mm, N_sf = 2.0.
  • Apply the ASME Combined Sizing equation: d = [ (32 * N_sf / (\pi * S_y)) * \sqrt{M^2 + T^2} ]^(1/3).
  • Calculate root term: \sqrt{180,000^2 + 350,000^2} \approx 393,573 N·mm.
  • Compute: d³ = (32 * 2.0 / (\pi * 250)) * 393,573 \approx 0.081487 * 393,573 \approx 32,071 mm³.
  • Take the cube root: d = 32,071^(1/3) \approx 31.8 mm.
Final ResultShaft Diameter = 31.8 mm

Engineering Assumptions

  • Homogeneous, isotropic, elastic material behaving in accordance with Tresca maximum shear stress yield criteria.
  • Steady torque and fully reversed cyclic bending load.

Design Limitations

  • Does not account for dynamic fatigue stress concentration factors due to keyways, shoulders, or surface finishes (requires Gerber/Goodman fatigue criteria).

Academic References & Standards

ASME B106.1Mstandard

Design of Transmission Shafting

Standard code for sizing transmission shafting.

Shigley Ch 7textbook

Shigley's Mechanical Engineering Design, 11th Edition

Textbook covering rotating shaft design under fatigue and static loads.