Fatigue: endurance limit, Goodman and Soderberg criteria

Fatigue under fluctuating stress: S–N curve and endurance limit, correction and notch factors, and Goodman, Soderberg and Gerber design with Miner's rule.

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Why it matters

Most broken shafts, axles, springs, bolts and crankshafts fail by fatigue, often at stresses well below the yield strength, after millions of load cycles. Fatigue design, using a corrected endurance limit and a mean-stress criterion such as Goodman or Soderberg, is how you make rotating and vibrating parts last.

Key ideas

Fatigue failure. Under fluctuating stress a crack starts at a stress raiser (surface scratch, fillet, keyway, inclusion), grows a little with each cycle, and the part finally breaks suddenly when the remaining section can no longer carry the load. The fracture surface shows smooth beach marks from the slow growth and a rough final-fracture zone. There is little visible deformation, so the failure is sudden.

Fluctuating stresses. For a stress varying between σ_max and σ_min:

  • mean stress σ_m = (σ_max + σ_min)/2,
  • alternating (amplitude) stress σ_a = (σ_max − σ_min)/2,
  • stress ratio R = σ_min/σ_max. Completely reversed: σ_m = 0, R = −1 (rotating shaft in bending). Repeated: σ_min = 0, R = 0.

S–N curve and endurance limit. Fatigue strength falls with the number of cycles N. Steels show a knee at about 10⁶ to 10⁷ cycles; the stress below which they survive indefinitely is the endurance limit S_e′. For a polished rotating-beam steel specimen S_e′ ≈ 0.5 S_ut (for S_ut up to about 1400 MPa). Aluminium and copper alloys have no true endurance limit, so a fatigue strength at a stated life (e.g. 5 × 10⁸ cycles) is used.

Corrected endurance limit of the actual part. The specimen value is reduced by Marin factors for surface finish (k_a), size (k_b), reliability (k_c), temperature (k_d) and miscellaneous effects (k_e): S_e = k_a·k_b·k_c·k_d·k_e·S_e′. Take these factors from your data book; machined and forged surfaces and larger diameters lower S_e considerably.

Notch effect. The fatigue stress concentration factor K_f = 1 + q(K_t − 1), with notch sensitivity q between 0 (no effect) and 1 (full K_t). In fatigue, K_f is applied to the alternating stress (and for ductile materials usually not to the mean stress).

Mean-stress criteria. On a plot of σ_a against σ_m, a straight failure line joins S_e on the σ_a axis to a strength on the σ_m axis:

  • Goodman: to S_ut. Widely used, reasonably close to test data.
  • Soderberg: to S_yt. More conservative, also guards against yielding.
  • Gerber: a parabola to S_ut; closest to test data for ductile steels but non-linear. A design point inside the line is safe; dividing the strengths by N gives the design line.

Cumulative damage. For several stress levels, Miner's rule Σ nᵢ/Nᵢ = 1 estimates life, where nᵢ is the number of cycles applied at a level with life Nᵢ.

Formulas

σ_m = (σ_max + σ_min)/2, σ_a = (σ_max − σ_min)/2 — MPa.

S_e′ ≈ 0.5·S_ut (steel, rotating beam); S_e = k_a·k_b·k_c·k_d·k_e·S_e′.

K_f = 1 + q·(K_t − 1) — dimensionless; 0 ≤ q ≤ 1.

Goodman: σ_a/S_e + σ_m/S_ut = 1/N.

Soderberg: σ_a/S_e + σ_m/S_yt = 1/N.

Gerber: N·σ_a/S_e + (N·σ_m/S_ut)² = 1.

Miner: n₁/N₁ + n₂/N₂ + … = 1.

Symbols: S_ut ultimate tensile strength, S_yt yield strength, S_e corrected endurance limit (all MPa); N factor of safety. In the criteria, σ_a includes K_f where a notch exists.

Worked examples

Example 1 (standard). A machined steel part has S_ut = 600 MPa. Given surface factor k_a = 0.8 and size factor k_b = 0.85 (from a data book), other factors 1, find the corrected endurance limit.

  1. Specimen limit: S_e′ = 0.5 × 600 = 300 MPa.
  2. S_e = k_a·k_b·S_e′ = 0.8 × 0.85 × 300 = 204 MPa.
  3. S_e = 204 MPa

Example 2 (GATE level). A 20 mm diameter rod of the steel in Example 1 (S_yt = 450 MPa) carries an axial load that varies from 10 kN to 50 kN (tension). The rod has a notch with K_f = 1.6. Find the factor of safety by Goodman and by Soderberg.

  1. A = π × 20²/4 = 314.2 mm².
  2. σ_max = 50 000/314.2 = 159.2 MPa; σ_min = 10 000/314.2 = 31.8 MPa.
  3. σ_m = (159.2 + 31.8)/2 = 95.5 MPa; σ_a = (159.2 − 31.8)/2 = 63.7 MPa.
  4. Notch on the alternating part: K_f·σ_a = 1.6 × 63.7 = 101.9 MPa.
  5. Goodman: 1/N = 101.9/204 + 95.5/600 = 0.4993 + 0.1592 = 0.6585 → N = 1.52.
  6. Soderberg: 1/N = 101.9/204 + 95.5/450 = 0.4993 + 0.2122 = 0.7115 → N = 1.41.
  7. N ≈ 1.52 (Goodman), 1.41 (Soderberg). Soderberg is lower because it uses S_yt. (For axial loading some data books add a load factor below 1 to S_e; follow your data book.)

Common mistakes

  • Using S_e′ of a polished specimen for a real machined part without correction factors.
  • Swapping σ_m and σ_a, or using σ_max in place of σ_a.
  • Applying the static rule "ignore K_t for ductile materials" to fatigue.
  • Writing the Goodman equation equal to N instead of 1/N.
  • Assuming aluminium has an endurance limit.
  • Forgetting that Soderberg uses yield strength and Goodman ultimate strength.

For GATE PI

Expect σ_m and σ_a from a load cycle, factor of safety or allowable stress from Goodman or Soderberg, the endurance limit with given correction factors, K_f from K_t and q, and Miner's rule life estimates. Practise drawing the Goodman diagram and placing the load line.

Quick check

  1. Stress varies from −40 to +120 MPa. Find σ_m and σ_a.
  2. K_t = 2.0 and q = 0.8. What is K_f?
  3. σ_a = 90 MPa, σ_m = 60 MPa, S_e = 200 MPa, S_ut = 400 MPa. Goodman FoS?
  4. Which line is more conservative, Goodman or Soderberg?

Answers: 1. 40 MPa and 80 MPa; 2. 1.8; 3. 1.67; 4. Soderberg.

Interview questions

All Theory of Machines and Machine Design interview questions

Try answering each one aloud before you open it.

  1. 1.What is fatigue in the context of materials and machine design?Concept

    Fatigue refers to the weakening or failure of a material caused by repeatedly applied loads, typically below the material's ultimate tensile strength. It occurs over time as the material undergoes cyclic stress, leading to the initiation and growth of cracks, eventually resulting in failure.

  2. 2.Define the endurance limit in fatigue analysis.Concept

    The endurance limit is the completely reversed stress amplitude below which a material survives an indefinitely large number of cycles; on the S–N curve it is the level where the curve flattens, around 10⁶ to 10⁷ cycles for steels. For polished steel rotating-beam specimens it is roughly half the ultimate strength. The value for a real part is much lower after correcting for surface finish, size, reliability and temperature, and most non-ferrous alloys have no true endurance limit.

  3. 3.Explain the Goodman criterion in fatigue analysis.Concept

    The Goodman criterion is a method used to predict the failure of a material under combined mean and alternating stresses. It is represented by a linear relationship between the mean stress and the alternating stress, where the line connects the material's ultimate tensile strength on the mean stress axis and the endurance limit on the alternating stress axis.

  4. 4.What is the Soderberg criterion, and how does it differ from the Goodman criterion?Concept

    The Soderberg criterion is another method for evaluating fatigue failure, which uses a more conservative approach than the Goodman criterion. It considers the yield strength instead of the ultimate tensile strength, connecting the yield strength on the mean stress axis to the endurance limit on the alternating stress axis. This results in a safer design but may be overly conservative.

  5. 5.Why is the endurance limit important in the design of rotating machinery?Application

    The endurance limit is crucial in the design of rotating machinery because these components often experience cyclic loading. Designing below the endurance limit ensures that the machinery can operate indefinitely without fatigue failure, enhancing reliability and safety.

  6. 6.What happens if a component is designed above its endurance limit?Application

    If a component is designed above its endurance limit, it is likely to experience fatigue failure after a certain number of cycles. This can lead to unexpected breakdowns, increased maintenance costs, and potential safety hazards.

  7. 7.How would you apply the Goodman criterion to design a shaft subjected to both mean and alternating stresses?Application

    Find the mean and alternating stresses at the critical section from the load cycle, and multiply the alternating stress by the fatigue stress concentration factor K_f of any fillet or keyway. Correct the specimen endurance limit for surface, size, reliability and temperature to get S_e. Then require σ_a/S_e + σ_m/S_ut = 1/N for the chosen factor of safety N, and solve for the diameter; finally check that the maximum stress does not exceed the yield strength.

  8. 8.A steel rod has an ultimate tensile strength of 600 MPa and an endurance limit of 250 MPa. Calculate the allowable alternating stress if the mean stress is 150 MPa using the Goodman criterion.Numerical

    Using the Goodman criterion: (σ_a / σ_e) + (σ_m / σ_u) = 1, where σ_a is the alternating stress, σ_e is the endurance limit (250 MPa), σ_m is the mean stress (150 MPa), and σ_u is the ultimate tensile strength (600 MPa). Solving for σ_a: (σ_a / 250) + (150 / 600) = 1, σ_a = 250 * (1 - 0.25) = 187.5 MPa.

  9. 9.A component is subjected to a mean stress of 100 MPa and an alternating stress of 200 MPa. If the yield strength is 400 MPa and the endurance limit is 150 MPa, check the safety using the Soderberg criterion.Numerical

    Using the Soderberg criterion: (σ_a / σ_e) + (σ_m / σ_y) = 1, where σ_a is the alternating stress (200 MPa), σ_e is the endurance limit (150 MPa), σ_m is the mean stress (100 MPa), and σ_y is the yield strength (400 MPa). Calculating: (200 / 150) + (100 / 400) = 1.333 + 0.25 = 1.583. Since 1.583 > 1, the design is not safe.

  10. 10.Explain why non-ferrous materials often do not have a well-defined endurance limit.Concept

    Non-ferrous materials, such as aluminum and copper, typically do not exhibit a clear endurance limit because their fatigue strength decreases continuously with the number of cycles. Unlike ferrous materials, which show a plateau in their S-N curve, non-ferrous materials do not reach a point where the fatigue strength remains constant.

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