--- title: "Soderberg design with reversed bending and steady torque" subject: "Machine Design I" chapter: 5 key: "semester_6/machine_design_i:soderberg-bending-torsion-shaft" source: https://www.bctengineer.com/q/semester_6/machine_design_i/soderberg-bending-torsion-shaft site: BCT Engineering --- # Soderberg design with reversed bending and steady torque Machine Design I · Chapter 5: Design of shafts - Asked in: year not given, 8 marks - Video: [Shaft Design for INFINITE LIFE and Fatigue Failure in Just Over 10 Minutes!](https://www.youtube.com/watch?v=Fp5-jjmaksM) — Less Boring Lectures - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/machine_design_i/soderberg-bending-torsion-shaft - Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-5 ## Question 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. ## 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. ## Answer The bending stress is fully reversed ($\sigma_m = 0$) and the torsion is steady ($\tau_a = 0$). By the distortion energy theory, the equivalent alternating and mean stresses are $$ \sigma_a' = \frac{32 K_f M}{\pi d^3}, \qquad \sigma_m' = \sqrt{3}\,\frac{16 K_{fs} T}{\pi d^3} $$ Substituting these in the Soderberg line $\dfrac{\sigma_a'}{S_e} + \dfrac{\sigma_m'}{S_y} = \dfrac{1}{n}$ gives $$ \frac{1}{n} = \frac{16}{\pi d^3}\left[\frac{2 K_f M}{S_e} + \frac{\sqrt{3}\,K_{fs} T}{S_y}\right] $$ #### Corrected endurance limit $$ \begin{aligned} S_e' &= 0.5 \times 700 = 350\ \text{MPa} \\ S_e &= 0.77 \times 0.85 \times 0.814 \times 350 = 186.5\ \text{MPa} \end{aligned} $$ #### Stress concentration terms $$ \begin{aligned} K_f M &= 1.8 \times 500 = 900\ \text{N.m} = 900 \times 10^3\ \text{N.mm} \\ K_{fs} T &= 1.5 \times 700 = 1050\ \text{N.m} = 1050 \times 10^3\ \text{N.mm} \end{aligned} $$ #### Substituting $$ \begin{aligned} \frac{K_f M}{S_e} &= \frac{900 \times 10^3}{186.5} = 4826.6 \\ \frac{K_{fs} T}{S_y} &= \frac{1050 \times 10^3}{420} = 2500 \\ \frac{2 K_f M}{S_e} + \frac{\sqrt{3}\,K_{fs} T}{S_y} &= 2(4826.6) + 1.732(2500) = 9653.2 + 4330.1 = 13\,983 \end{aligned} $$ $$ d^3 = \frac{16\,n}{\pi}\times 13\,983 = \frac{16 \times 2}{\pi}\times 13\,983 = 1.424\times10^5\ \text{mm}^3 $$ $$ d = 52.2\ \text{mm} $$ The nearest standard size above this is 55 mm. **Answer: d = 52.2 mm, so use a 55 mm diameter shaft.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the stress components for fully reversed bending and steady torque? A. Both bending and torsion alternating B. Bending steady and torsion alternating C. sigma_m = 0 (bending alternating) and tau_a = 0 (torsion steady) D. All stresses are zero Answer: C. Bending is alternating only; torsion is mean only. (Key point: Stress state) 2. What is Se? A. 350 MPa B. About 227 MPa C. About 120 MPa D. About 186.5 MPa Answer: D. 0.77 x 0.85 x 0.814 x 350 = 186.5 MPa. (Key point: Corrected endurance limit) 3. In the equation, the alternating bending term is divided by __ and the steady torsion term by __. A. Se; Sy B. Sy; Se C. Sut; Se D. Se; Sut Answer: A. Soderberg uses Se for alternating and Sy for mean. (Key point: Which strength is used) 4. What diameter results and what is adopted? A. 41.5 mm, adopt 42 mm B. 47.6 mm, adopt 50 mm C. 54.4 mm, adopt 55 mm D. 75.9 mm, adopt 80 mm Answer: B. d^3 = 16 x 2 x 10580/pi = 1.078e5 mm^3. 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