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Reference worksheet · uniform circular shaft

Documented Circular-Shaft Elastic-Torsion Worksheet

Calculate a mathematical outer-diameter lower bound from an externally approved nominal elastic surface-shear-stress limit and, optionally, an externally approved elastic twist-rate limit.

General torsion equationsNo material presetNo generated safety factor
Scope, not shaft certification. This is a general engineering Saint-Venant circular-shaft worksheet, not an ISO, API, ASME or DIN design/acceptance calculation. It does not choose a material, shear allowable, load factor, safety factor, fatigue method, keyway factor, standard size or safe machine verdict.

Documented inputs

Use a decimal point or decimal comma; signs, thousands separators, units inside number fields and scientific notation are intentionally rejected.

Reference results

Mathematical outer-diameter lower bound
Stress-criterion diameter dτ
Twist-criterion diameter dφ
Governing enabled criterion
Corresponding inner diameter
Polar second moment J at bound
Nominal surface shear stress at bound
Elastic twist rate at bound

Equations and definitions

Let dₒ be outer diameter, dᵢ inner diameter, k = dᵢ/dₒ, T the constant internal torque magnitude, τlim the externally approved nominal elastic surface-shear-stress limit, G the approved shear modulus, and φ′lim the approved twist angle per unit length in rad/m.

J = π(dₒ⁴ − dᵢ⁴)/32 = πdₒ⁴(1 − k⁴)/32
τmax = T(dₒ/2)/J = 16T/[πdₒ³(1 − k⁴)]
φ′ = dφ/dx = T/(GJ)

Solving the first criterion for dₒ gives:

dτ = {16T/[πτlim(1 − k⁴)]}^(1/3)

When the optional twist-rate criterion is enabled:

dφ = {32T/[πGφ′lim(1 − k⁴)]}^(1/4)
dₒ,lower = max(dτ, dφ)

With only the stress criterion enabled, dₒ,lower = dτ. Diameter is not automatically rounded: selecting a production size, tolerance and manufacturing allowance is an external design action.

Dimensional check

T/τ has dimension length³, so dτ has dimension length. T/(Gφ′) has dimension length⁴ because φ′ has dimension 1/length, so dφ has dimension length. These are general elastic torsion equations, not formulas issued by ISO.

Model boundary

  • Use the governing internal design torque for the analysed segment—not automatically motor nameplate torque, average torque or power/RPM torque. Startup, reversal, braking, shock, transient and duty-cycle effects must already be handled by the controlled load basis.
  • The worksheet does not derive τlim from tensile yield or ultimate strength. The University of Maryland example explicitly warns that tensile-test strengths cannot be directly compared with shear stress.
  • No material/shear-modulus table, 0.577 multiplier, 50–60% rule, 1.5–3.0 safety factor or keyway Kt range is generated. Those depend on material data, failure theory, loading, fatigue method, geometry and controlled design rules.
  • Combined bending and torsion are not replaced by one universal “equivalent torque” here. A compatible combined-stress/fatigue method must keep its stress normalization, load factors and allowables consistent.
  • Stress concentrations at keyways, splines, shoulders, grooves, holes and fits; contact/fretting; fatigue; yielding; buckling; critical speed; lateral vibration; thermal effects and bearing/coupling loads are outside the model.
  • The hollow/solid geometry ratios for the same dₒ are (1 − k⁴) for J and nominal torsional section modulus, and (1 − k²) for cross-sectional area. They are geometric comparisons, not a claim that two shafts have equal fatigue strength or mass after redesign.

Source traceability

ClaimClassificaçãoEvidence
T = GJ dφ/dx; τ = Tr/J; circular/cylindrical-tube scopeGeneral linear-elastic circular-shaft torsion; not an ISO formulaMIT OCW 16.01, M9 Shafts: Torsion of Circular Shafts, PDF pages 5–6
Solid circle τmax = 16T/(πd³); published T = 5400 lbf·in gives 27.50 ksi at d = 1.000 inPublished formula and numerical exampleUniversity of Maryland, Mechanics of Materials—101, Example 4.3
Published elastic benchmark: d = 40 mm, τ = 130 MPa, T = 1633.63 N·m, G inferred from τ/γ = 80 GPa, twist ≈ 4.66° over 1 mPublished numerical exampleMississippi State University, Pure Torsion Sample Test, problem 1(a)
1 in = 0.0254 m; 1 lbf = 4.4482216152605 N under standard gravityExact SI conversionNIST SP 811, Appendix B.8 and footnotes

Accessed: 15 July 2026. Every actual torque factor, material property, allowable, twist limit, stress-concentration/fatigue factor and acceptance decision remains NEEDS_LICENSED_SOURCE until the controlled project/OEM/code basis is supplied.

Published benchmark entry

Enter T = 1633.63 N·m, k = 0, τlim = 130 MPa, enable twist, G = 80 GPa and φ′lim = 4.66 deg/m. The stress criterion gives 40.0000149 mm and governs by the rounding of the published inputs; calculated twist at that diameter is 4.65528 deg/m, consistent with the source’s 40 mm and 4.66° over 1 m.

Perguntas

Why does the worksheet not offer steel grades and safety factors?

A grade name alone does not establish heat treatment, product form, temperature, statistical basis, shear allowable, fatigue condition or design code. The old presets mixed nominal tensile properties, guessed shear limits and a second safety-factor division. This version accepts one controlled limit only.

Is the larger lower-bound diameter a final drawing size?

No. It is the exact ideal-model lower bound for the enabled input criteria. A competent design must select a manufactured size/tolerance and verify every omitted feature, load case, strength/fatigue criterion, fit and dynamic requirement.

Can I use the old combined-bending “equivalent torque” formula?

Not without a controlled method. Expressions normalized to bending stress and expressions normalized to torsional shear stress are not interchangeable. Substituting one into the other can change a factor of two and under-size a shaft.

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