About Shaft Design Calculator (Combined Bending & Torsion)
The shaft design calculator finds the minimum diameter of a solid circular shaft carrying a combined bending moment M and torque T. It applies either the maximum-shear-stress (Tresca) theory or the distortion-energy (von Mises) theory to the design point, divides the yield strength by the chosen factor of safety, and solves the cubic stress equation for the required diameter d. All inputs use a consistent SI unit set: moments in N·mm, strengths in MPa, and diameters and stresses in mm and MPa.
Enter the bending moment, torque, yield strength, and factor of safety, pick a failure theory, and optionally add fatigue stress-concentration factors Kb (bending) and Kt (torsion) at a shoulder, keyway, or hole. The tool returns the required diameter and then re-evaluates the actual bending stress, torsional shear stress, and von Mises stress at that diameter, reporting the achieved factor of safety against yield so you can confirm the design closes.
How It Works
- Enter the bending moment M and torque T (N·mm), the yield strength Sy (MPa), and the design factor of safety N. Optionally enter Kb and Kt to account for stress concentrations (default 1).
- Choose a failure theory: maximum-shear-stress (Tresca) is more conservative, while distortion-energy (von Mises) gives a smaller, more economical diameter for ductile shafts.
- The calculator solves for the required diameter: max-shear uses d³ = (16/(π·tau_allow))·√((Kb·M)² + (Kt·T)²) with tau_allow = Sy/(2N); von Mises uses d³ = (32N/(π·Sy))·√((Kb·M)² + 0.75·(Kt·T)²).
- At the computed diameter it back-calculates the bending stress σ_b = 32·Kb·M/(π·d³), shear stress τ = 16·Kt·T/(π·d³), the von Mises stress, and the achieved factor of safety FS = Sy/σ_vm.
Worked Example
A solid shaft carries a bending moment M = 1.0×10⁶ N·mm and a torque T = 1.0×10⁶ N·mm in a steel with yield strength Sy = 400 MPa, designed to a factor of safety N = 2 using the maximum-shear-stress (Tresca) theory with Kb = Kt = 1. The allowable shear stress is tau_allow = Sy/(2N) = 400/(2·2) = 100 MPa. The combined-load term is √(M² + T²) = √(1×10¹² + 1×10¹²) = 1.4142×10⁶ N·mm. The required diameter cubed is d³ = (16/(π·100))·1.4142×10⁶ = 0.050930·1.4142×10⁶ = 72,026 mm³, so d = ∛72,026 = 41.6 mm. Re-evaluating at d = 41.6 mm gives a bending stress σ_b = 32·M/(π·d³) = 141.4 MPa and a torsional shear stress τ = 16·T/(π·d³) = 70.7 MPa, for a von Mises stress σ_vm = √(141.4² + 3·70.7²) = 187 MPa. Sizing the same case by the distortion-energy (von Mises) theory instead yields a slightly smaller d = 40.7 mm, with an achieved factor of safety of 2.0 against the 400 MPa yield.
Formulas
- Maximum-shear-stress (Tresca) required diameter
d^3 = (16 / (pi * tau_allow)) * sqrt((Kb*M)^2 + (Kt*T)^2) ; tau_allow = Sy / (2*N)- Distortion-energy (von Mises) required diameter
d^3 = (32 * N / (pi * Sy)) * sqrt((Kb*M)^2 + 0.75 * (Kt*T)^2)- Bending stress at the required diameter
sigma_b = 32 * Kb * M / (pi * d^3)- Torsional shear stress at the required diameter
tau = 16 * Kt * T / (pi * d^3)- von Mises stress and achieved factor of safety
sigma_vm = sqrt(sigma_b^2 + 3 * tau^2) ; FS = Sy / sigma_vm
Standards & References
- ASME B106.1M (shaft design)
- Shigley's Mechanical Engineering Design (combined loading)
- Maximum-shear-stress (Tresca) & distortion-energy (von Mises) theories
Frequently Asked Questions
What is the difference between the maximum-shear-stress and distortion-energy theories?
The maximum-shear-stress (Tresca) theory assumes yielding begins when the maximum shear stress reaches Sy/2, while the distortion-energy (von Mises) theory uses the distortion strain energy. Tresca is more conservative and gives a slightly larger required diameter; von Mises is a closer match to test data for ductile steels and is the usual default for shafts.
Why include the stress-concentration factors Kb and Kt?
Real shafts have shoulders, fillets, keyways, grooves, and cross-holes that locally raise stress above the nominal value. The fatigue stress-concentration factors Kb (bending) and Kt (torsion) multiply M and T inside the radical so the design diameter accounts for these local peaks. Set Kb = Kt = 1 for a plain prismatic shaft with no discontinuity.
Why are the moments entered in N·mm rather than N·m?
The formulas are dimensionally consistent when moments are in N·mm and strengths in MPa (N/mm²): N·mm divided by N/mm² gives mm³, so the diameter comes out directly in mm. If your moments are in N·m, multiply by 1000 before entering them so that all quantities stay in the same millimetre-and-newton unit set.
Does the achieved factor of safety match the design factor of safety?
For the distortion-energy theory the achieved factor of safety FS = Sy/σ_vm comes out essentially equal to the design value N, because the diameter is solved so the von Mises stress equals Sy/N. For the Tresca theory the achieved von-Mises FS is slightly higher, since Tresca sizes against the more conservative shear criterion.