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Soderberg design with reversed bending and steady torque

Short answer

Corrected Se:0.77×0.85×0.814×350 =186.5 MPa. Using Soderberg with DE: σa' =32KfM/(πd³), σm' =√3·16KfsT/(πd³). Substituting into σa'/Se + σm'/Sy =1/n gives d³ = (16n/π)[2KfM/Se + √3 KfsT/Sy] =1.424×10⁵ mm³ → d=52.2 mm, adopt55 mm.

A rotating solid shaft carries a fully reversed bending moment of 500 N.m and a steady torque of 700 N.m at a section having a keyseat. The shaft material has Sut = 700 MPa and Sy = 420 MPa. Take the uncorrected endurance limit as 0.5 Sut, surface factor 0.77, size factor 0.85, reliability factor 0.814 (99 percent), and other factors 1. The fatigue stress concentration factors are Kf = 1.8 (bending) and Kfs = 1.5 (torsion). Using the Soderberg approach with the distortion energy theory and a factor of safety of 2, find the shaft diameter.

The bending stress is fully reversed (σm=0\sigma_m = 0) and the torsion is steady (τa=0\tau_a = 0). By the distortion energy theory, the equivalent alternating and mean stresses are

σa′=32KfMπd3,σm′=3 16KfsTπd3\sigma_a' = \frac{32 K_f M}{\pi d^3}, \qquad \sigma_m' = \sqrt{3}\,\frac{16 K_{fs} T}{\pi d^3}

Substituting these in the Soderberg line σa′Se+σm′Sy=1n\dfrac{\sigma_a'}{S_e} + \dfrac{\sigma_m'}{S_y} = \dfrac{1}{n} gives

1n=16πd3[2KfMSe+3 KfsTSy]\frac{1}{n} = \frac{16}{\pi d^3}\left[\frac{2 K_f M}{S_e} + \frac{\sqrt{3}\,K_{fs} T}{S_y}\right]

Corrected endurance limit

Se′=0.5×700=350 MPaSe=0.77×0.85×0.814×350=186.5 MPa\begin{aligned} S_e' &= 0.5 \times 700 = 350\ \text{MPa} \\ S_e &= 0.77 \times 0.85 \times 0.814 \times 350 = 186.5\ \text{MPa} \end{aligned}

Stress concentration terms

KfM=1.8×500=900 N.m=900×103 N.mmKfsT=1.5×700=1050 N.m=1050×103 N.mm\begin{aligned} K_f M &= 1.8 \times 500 = 900\ \text{N.m} = 900 \times 10^3\ \text{N.mm} \\ K_{fs} T &= 1.5 \times 700 = 1050\ \text{N.m} = 1050 \times 10^3\ \text{N.mm} \end{aligned}

Substituting

KfMSe=900×103186.5=4826.6KfsTSy=1050×103420=25002KfMSe+3 KfsTSy=2(4826.6)+1.732(2500)=9653.2+4330.1=13 983\begin{aligned} \frac{K_f M}{S_e} &= \frac{900 \times 10^3}{186.5} = 4826.6 \\ \frac{K_{fs} T}{S_y} &= \frac{1050 \times 10^3}{420} = 2500 \\ \frac{2 K_f M}{S_e} + \frac{\sqrt{3}\,K_{fs} T}{S_y} &= 2(4826.6) + 1.732(2500) = 9653.2 + 4330.1 = 13\,983 \end{aligned} d3=16 nπ×13 983=16×2π×13 983=1.424×105 mm3d^3 = \frac{16\,n}{\pi}\times 13\,983 = \frac{16 \times 2}{\pi}\times 13\,983 = 1.424\times10^5\ \text{mm}^3 d=52.2 mmd = 52.2\ \text{mm}

The nearest standard size above this is 55 mm.

Answer: d = 52.2 mm, so use a 55 mm diameter shaft.

✓ Answer checked by Claude Opus 5.5 (high effort)

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