Fatigue: S-N curve, endurance limit and Goodman/Soderberg lines

Fatigue under fluctuating stress: S-N curve, endurance limit and Marin factors, fatigue notch factor, Soderberg, Goodman and Gerber criteria and finite-life estimation.

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Why it matters

Most machine failures in service are fatigue failures: shafts, springs, gear teeth, bolts and robot links break after millions of load cycles at stresses well below the yield strength. A part that passes a static check can still crack at a keyway or fillet in a few weeks. Fatigue design tells you how large the alternating stress may be, given the mean stress, for the life you need.

Key ideas

Fluctuating stress. A cyclic stress between σ_max and σ_min is described by a mean σm and an amplitude (alternating component) σa. Special cases: fully reversed (σm = 0, R = −1), repeated or pulsating (σmin = 0, R = 0), and general fluctuating. The stress ratio is R = σmin/σmax.

Fatigue mechanism. A crack initiates at a stress raiser or surface defect, grows a little each cycle, and the remaining section finally fractures suddenly. The fracture surface shows smooth "beach marks" from the crack growth and a rough final-fracture zone. Failure is sudden even in ductile materials.

S-N (Wöhler) curve. Stress amplitude S plotted against cycles to failure N on log-log axes, obtained from rotating-beam tests of polished specimens under fully reversed bending.

  • Low-cycle fatigue: N < 10³. High-cycle fatigue: N > 10³.
  • Steels and titanium show a knee near 10⁶-10⁷ cycles; below the endurance limit Se′ they last practically indefinitely. For steels, Se′ ≈ 0.5·σut (for σut up to about 1400 MPa).
  • Aluminium, copper alloys and most non-ferrous metals have no true endurance limit; a fatigue strength at a stated life (e.g. 5×10⁸ cycles) is used instead.
  • For finite life between 10³ and 10⁶ cycles, a straight line on log-log axes is drawn from about 0.9·σut at 10³ cycles to Se at 10⁶ cycles (Basquin form S = a·N^b).

Endurance limit of the actual part. The real part differs from the specimen, so Se = ka·kb·kc·kd·ke·Se′ (Marin factors) for surface finish, size, reliability, temperature and miscellaneous effects. The values come from your data book charts; all are ≤ 1 (surface finish and size often reduce Se by 20-50%).

Notches in fatigue. The fatigue stress concentration factor Kf = 1 + q·(Kt − 1), where q (0 to 1) is the notch sensitivity from charts: q → 0 means the material is insensitive (Kf = 1), q → 1 means fully sensitive (Kf = Kt). Unlike static design of ductile parts, Kf must be applied in fatigue. The common textbook convention is to apply Kf to the alternating component (or equivalently divide Se by Kf); some texts also apply it to σm - follow your syllabus consistently.

Mean-stress criteria. On a plot of σa (vertical) against σm (horizontal), the failure line joins Se on the σa axis to a static strength on the σm axis:

  • Soderberg: straight line to σy. Most conservative; also guards against yielding.
  • Goodman (modified Goodman): straight line to σut. Widely used, slightly conservative compared with test data.
  • Gerber: parabola to σut. Fits test data best for ductile metals but is not conservative. A tensile mean stress reduces the allowable amplitude; a compressive mean stress is beneficial (this is why shot peening and pre-stressing help).

Formulas

σm = (σmax + σmin) / 2 · σa = (σmax − σmin) / 2 · R = σmin / σmax

  • stresses in Pa; R dimensionless.

Se′ ≈ 0.5·σut (steel, rotating-beam specimen, σut ≤ about 1400 MPa) Se = ka·kb·kc·kd·ke·Se′

  • Se′: specimen endurance limit (Pa); Se: endurance limit of the actual part (Pa); k-factors from charts, dimensionless.

Kf = 1 + q·(Kt − 1)

  • Kf: fatigue stress concentration factor; q: notch sensitivity (0 to 1); Kt: theoretical factor.

Soderberg: σa/Se + σm/σy = 1/n Goodman: σa/Se + σm/σut = 1/n Gerber: n·σa/Se + (n·σm/σut)² = 1

  • n: factor of safety; σa here includes Kf.

log S = log a + b·log N (finite life, 10³ ≤ N ≤ 10⁶), with S = 0.9·σut at N = 10³ and S = Se at N = 10⁶

  • S: fully reversed stress amplitude (Pa); N: cycles to failure; b: negative slope.

Worked examples

Example 1 (standard). A steel bar with a shoulder (Kt = 2.0, q = 0.8) carries an axial stress, on the nominal section, that fluctuates from 40 MPa to 200 MPa. σut = 600 MPa, σy = 400 MPa, corrected endurance limit Se = 250 MPa. Find n by Goodman and Soderberg, applying Kf to the alternating component.

  1. Components: σm = (200 + 40)/2 = 120 MPa; σa = (200 − 40)/2 = 80 MPa.
  2. Fatigue factor: Kf = 1 + q·(Kt − 1) = 1 + 0.8 × 1.0 = 1.8 → Kf·σa = 1.8 × 80 = 144 MPa.
  3. Goodman: 1/n = 144/250 + 120/600 = 0.576 + 0.200 = 0.776 → n = 1.29.
  4. Soderberg: 1/n = 144/250 + 120/400 = 0.576 + 0.300 = 0.876 → n = 1.14.

Example 2 (GATE level). A steel part with σut = 600 MPa and corrected Se = 250 MPa (at 10⁶ cycles) carries a fully reversed stress amplitude of 400 MPa. Estimate its life using the straight-line S-N model.

  1. End points: S₁ = 0.9 × 600 = 540 MPa at N₁ = 10³; S₂ = 250 MPa at N₂ = 10⁶.
  2. On log-log axes, log N = log N₁ + (log N₂ − log N₁) × (log S₁ − log S)/(log S₁ − log S₂).
  3. log 540 = 2.7324, log 400 = 2.6021, log 250 = 2.3979.
  4. log N = 3 + 3 × (2.7324 − 2.6021)/(2.7324 − 2.3979) = 3 + 3 × 0.3897 = 4.169.
  5. N = 10^4.169 ≈ 1.48 × 10⁴ cycles (about 14,800 cycles).

Common mistakes

  • Mixing up σa and σm, or using the stress range (σmax − σmin) as the amplitude.
  • Using the specimen endurance limit Se′ without the Marin factors.
  • Applying Kt instead of Kf, or forgetting the notch altogether because "the material is ductile".
  • Putting σut in Soderberg or σy in Goodman.
  • Assuming aluminium has an endurance limit.
  • Ignoring the static check: with a large σm, also verify σmax ≤ σy/n (Goodman alone does not prevent yielding).

For GATE ME

Expect numericals that give σmax and σmin (or a load range) and ask for n or a size by Goodman or Soderberg, questions on the S-N curve and endurance limit, finite-life estimation on the log-log line, and Kf from Kt and q. Also expect concept questions ranking Soderberg, Goodman and Gerber by conservativeness. Practise drawing the σa-σm diagram and locating the load line.

Quick check

  1. σmax = 300 MPa, σmin = −100 MPa. Find σm and σa.
  2. Kt = 2.5, q = 0.6. Find Kf.
  3. Which mean-stress line is the most conservative?
  4. Approximate Se′ for a steel with σut = 800 MPa? Answers: 1. σm = 100 MPa, σa = 200 MPa. 2. Kf = 1 + 0.6 × 1.5 = 1.9. 3. Soderberg. 4. About 400 MPa.

Interview questions

All Machine Design interview questions

Try answering each one aloud before you open it.

  1. 1.What is an S-N curve in the context of fatigue analysis?Concept

    An S-N curve, also known as a Wöhler curve, is a graphical representation of the relationship between the stress amplitude (S) and the number of cycles to failure (N) for a material under cyclic loading. It is used to predict the fatigue life of a material by showing how many cycles it can withstand at a given stress level before failure occurs.

  2. 2.Explain the concept of endurance limit in fatigue analysis.Concept

    The endurance limit, also known as the fatigue limit, is the maximum stress amplitude a material can withstand for an infinite number of cycles without failing. For ferrous materials, there is typically a distinct endurance limit, whereas non-ferrous materials do not have a clear endurance limit and may eventually fail at any stress level given enough cycles.

  3. 3.What is the Goodman line and how is it used in fatigue analysis?Concept

    On a plot of alternating stress σa against mean stress σm, the Goodman line joins the endurance limit Se on the σa axis to the ultimate strength σut on the σm axis: σa/Se + σm/σut = 1/n. Points below the line (scaled by the factor of safety n) are safe for infinite life. It is used to size parts carrying a fluctuating load, with Kf applied to the alternating component, and is usually paired with a static yield check.

  4. 4.Describe the Soderberg line and its application in fatigue analysis.Concept

    The Soderberg line is a conservative approach used in fatigue analysis to predict failure under combined mean and alternating stresses. It is plotted similarly to the Goodman line but uses the yield strength of the material instead of the ultimate tensile strength. The Soderberg line provides a more conservative estimate of the safe stress limits, making it useful in applications where safety is critical.

  5. 5.Why is the S-N curve important in the design of mechanical components?Application

    The S-N curve is crucial in mechanical design because it helps engineers predict the fatigue life of components subjected to cyclic loading. By understanding how a material behaves under different stress levels and cycles, designers can ensure that components will not fail prematurely, thus improving reliability and safety.

  6. 6.What happens if a material is used beyond its endurance limit?Application

    If a material is used beyond its endurance limit, it will eventually fail due to fatigue. The endurance limit represents the stress level below which a material can theoretically endure an infinite number of cycles without failure. Exceeding this limit means that the material will accumulate damage over time, leading to crack initiation and eventual failure.

  7. 7.How does the presence of a mean stress affect the fatigue life of a material?Application

    Mean stress does not change the stress range, but a tensile mean stress holds cracks open and lowers the alternating stress the part can survive, so for the same amplitude the life is shorter. Compressive mean stress closes cracks and improves life, which is why shot peening and surface rolling help. Goodman, Soderberg and Gerber lines quantify how the allowable amplitude falls as tensile mean stress rises.

  8. 8.Calculate the fatigue life of a steel component subjected to a stress amplitude of 250 MPa, given that its S-N curve follows the equation N = 10^12 / (σ^3).Numerical

    To calculate the fatigue life (N), use the given S-N curve equation: N = 10^12 / (σ^3). Substitute σ = 250 MPa: N = 10^12 / (250^3) = 10^12 / 15625000 = 64000 cycles. Therefore, the fatigue life of the component is 64,000 cycles.

  9. 9.A component made of aluminium alloy is subjected to a mean stress of 100 MPa and an alternating stress of 150 MPa. Its fatigue strength at the design life is 200 MPa and its ultimate tensile strength is 400 MPa. Using the Goodman relation, is the component safe?Numerical

    Goodman gives 1/n = σa/Sf + σm/σut = 150/200 + 100/400 = 0.75 + 0.25 = 1.0, so n = 1. The component sits exactly on the failure line with no margin, so it is not acceptable as a design. Note that aluminium has no true endurance limit, so a fatigue strength at the required number of cycles must be used, not a guessed fraction of σut.

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