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

Fluctuating stresses, the S-N curve and corrected endurance limit, fatigue notch factor, and the Gerber, Goodman and Soderberg criteria, with infinite-life and finite-life examples.

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Why it matters

Most automotive parts that break in service break by fatigue, not by a single overload: crankshafts, connecting rods, axle shafts, suspension springs and wheel studs all see millions of stress cycles. A part can fail at a stress far below its yield strength if the stress fluctuates, so cyclic loads must be designed against the endurance limit, with the effect of mean stress handled by a Goodman or Soderberg line.

Key ideas

Fatigue is progressive crack growth under fluctuating stress. A crack starts at a surface stress raiser (fillet, keyway, scratch, inclusion), grows a little each cycle (beach marks on the fracture face), and the remaining section finally breaks suddenly.

Fluctuating stress is described by the maximum and minimum stress, or by:

  • mean stress σ_m = (σ_max + σ_min)/2 and alternating (amplitude) stress σ_a = (σ_max − σ_min)/2;
  • stress ratio R = σ_min/σ_max (R = −1 fully reversed, R = 0 repeated, 0 < R < 1 fluctuating).

S-N (Wöhler) curve: alternating stress against cycles to failure on log-log axes, from rotating-beam tests (fully reversed, σ_m = 0). Steels show a knee at about 10⁶–10⁷ cycles below which life is effectively infinite; that stress is the endurance limit. Aluminium and most non-ferrous alloys have no true endurance limit, so a fatigue strength at a stated life (say 5 × 10⁸ cycles) is used instead.

For steels, a common estimate is S_e′ ≈ 0.5·S_ut for a polished rotating-beam specimen. The real part's endurance limit is lower: S_e = K_a·K_b·K_c·K_d·S_e′, with surface finish (K_a), size (K_b), reliability (K_c) and other effects (K_d) factors taken from the data book.

Notches in fatigue: the fatigue stress concentration factor K_f = 1 + q(K_t − 1), where the notch sensitivity q lies between 0 (no effect) and 1 (full K_t). In the method used in most Indian textbooks, K_f multiplies the alternating stress (or divides S_e); the mean stress is not magnified for ductile materials.

Mean-stress lines on the σ_m–σ_a diagram join the endurance limit S_e on the σ_a axis to a static strength on the σ_m axis:

  • Gerber parabola to S_ut: closest to test data, but nonlinear.
  • Goodman line to S_ut: linear, slightly conservative; widely used.
  • Soderberg line to S_yt: most conservative; also guards against yielding. Points below the chosen line are safe for infinite life. A tensile mean stress lowers the permissible amplitude; a compressive mean stress is beneficial, which is why shot peening and surface rolling help.

Finite life: for steel, the S-N line on log-log axes is often approximated between 0.9·S_ut at 10³ cycles and S_e at 10⁶ cycles.

Formulas

σ_m = (σ_max + σ_min) / 2 σ_a = (σ_max − σ_min) / 2 R = σ_min / σ_max S_e′ ≈ 0.5·S_ut (steel, rotating beam; estimate only) S_e = K_a·K_b·K_c·K_d·S_e′ K_f = 1 + q·(K_t − 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 Finite life: log N = 3 + 3·log(0.9·S_ut / σ_a) / log(0.9·S_ut / S_e) (10³ ≤ N ≤ 10⁶)

Symbols: σ_max, σ_min = extreme stresses (MPa); σ_m, σ_a = mean and alternating stress (MPa); S_e′ = specimen endurance limit, S_e = corrected endurance limit (MPa); S_ut, S_yt = ultimate and yield strength (MPa); K_a … K_d = modifying factors (–); K_t, K_f = theoretical and fatigue stress concentration factors (–); q = notch sensitivity (–); n = factor of safety (–); N = cycles to failure. In the formulas σ_a already includes K_f where there is a notch.

Worked examples

Example 1 (standard). A link sees stress fluctuating between 100 MPa and 300 MPa (tension). S_ut = 600 MPa, S_yt = 400 MPa, corrected S_e = 250 MPa. Find the factor of safety by Goodman and Soderberg.

  1. σ_m = (300 + 100)/2 = 200 MPa; σ_a = (300 − 100)/2 = 100 MPa.
  2. Goodman: 1/n = 100/250 + 200/600 = 0.400 + 0.333 = 0.733, so n = 1.36.
  3. Soderberg: 1/n = 100/250 + 200/400 = 0.400 + 0.500 = 0.900, so n = 1.11.
  4. Both exceed 1, so the link has infinite life; Soderberg gives the smaller margin.

Example 2 (GATE level). A steel part (S_ut = 600 MPa) has a fillet with K_t = 1.8 and notch sensitivity q = 0.85. Surface factor K_a = 0.80, size factor K_b = 0.85. The nominal stresses are σ_m = 120 MPa and σ_a = 60 MPa. Find the Goodman factor of safety.

  1. S_e′ = 0.5 × 600 = 300 MPa.
  2. S_e = 0.80 × 0.85 × 300 = 204 MPa.
  3. K_f = 1 + 0.85 × (1.8 − 1) = 1.68.
  4. Goodman with K_f on the alternating component: 1/n = (1.68 × 60)/204 + 120/600 = 0.494 + 0.200 = 0.694.
  5. n = 1.44.

Example 3 (finite life). Same steel with S_ut = 600 MPa and corrected S_e = 250 MPa, fully reversed σ_a = 300 MPa. Find the life.

  1. 0.9·S_ut = 540 MPa at 10³ cycles; 250 MPa at 10⁶ cycles.
  2. log N = 3 + 3 × log(540/300) / log(540/250) = 3 + 3 × 0.2553 / 0.3345 = 5.290.
  3. N ≈ 1.95 × 10⁵ cycles.

Common mistakes

  • Writing σ_a = σ_max − σ_min/2 without brackets; the whole range is halved.
  • Using S_e′ (polished specimen) instead of the corrected S_e of the actual part.
  • Applying K_f to the mean stress of a ductile part, or forgetting it on the alternating stress.
  • Mixing up which strength goes with which line: Goodman uses S_ut, Soderberg uses S_yt.
  • Saying a mean stress "adds to the amplitude". It does not change σ_a; it lowers the amplitude the material can tolerate.
  • Assuming aluminium has an endurance limit.

For GATE ME

Expect: computing σ_m, σ_a and R from loads; Goodman or Soderberg factor of safety; the permissible alternating stress for a given mean stress; diameter of a rod or shaft under fluctuating load; K_f from q and K_t; and finite life from a log-log S-N line. Practise drawing the σ_m–σ_a diagram and locating the load line.

Quick check

  1. Stress varies from −50 MPa to +150 MPa. Find σ_m, σ_a and R.
  2. Which mean-stress line is most conservative?
  3. K_t = 2.0 and q = 0.7. What is K_f?
  4. S_ut = 600 MPa, S_e = 250 MPa, σ_m = 150 MPa. What σ_a does the Goodman line allow (n = 1)?
  5. Does a compressive mean stress raise or lower fatigue strength?

Answers: 1. 50 MPa, 100 MPa, −0.33. 2. Soderberg. 3. 1.7. 4. 250 × (1 − 150/600) = 187.5 MPa. 5. Raises it.

Interview questions

All Design of Machine and Automotive Elements interview questions

Try answering each one aloud before you open it.

  1. 1.What is an S-N curve and what does it represent 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 cyclic stress amplitude (S) and the number of cycles to failure (N) for a material. It is used in fatigue analysis to predict the life of a material under cyclic loading. The curve typically shows a decreasing trend, indicating that as the stress amplitude decreases, the number of cycles to failure increases.

  2. 2.Define endurance limit and explain its significance in material fatigue.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. It is significant because it helps in designing components that are subjected to cyclic loading, ensuring they do not fail during their expected service life. Not all materials have a well-defined endurance limit; for example, ferrous metals typically do, while non-ferrous metals do not.

  3. 3.Explain the Goodman line and its application in fatigue analysis.Concept

    The Goodman line is a method used to predict the failure of materials under combined mean and alternating stresses. It is a linear relationship plotted on a graph with mean stress on the x-axis and alternating stress on the y-axis. The line connects the endurance limit on the y-axis to the ultimate tensile strength on the x-axis. It is used to determine safe stress levels for materials subjected to fluctuating loads.

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

    The Soderberg line is another method used in fatigue analysis, similar to the Goodman line, but it is more conservative. It uses the yield strength instead of the ultimate tensile strength to determine the safe stress levels. The Soderberg line connects the endurance limit on the y-axis to the yield strength on the x-axis. This approach is more conservative because it considers yielding as a failure criterion before fracture.

  5. 5.Why is the endurance limit important in the design of automotive components?Application

    The endurance limit is crucial in the design of automotive components because these components often experience cyclic loading during operation. By knowing the endurance limit, engineers can ensure that the components will not fail due to fatigue over the vehicle's expected lifespan. This is particularly important for components like crankshafts, connecting rods, and suspension parts, which are subjected to repeated stress cycles.

  6. 6.What could happen if the endurance limit is not considered in the design of a machine element?Application

    If the endurance limit is not considered in the design of a machine element, the component may fail prematurely due to fatigue. This can lead to unexpected breakdowns, increased maintenance costs, and potentially hazardous situations if the failure occurs in critical components. It is essential to design components to withstand the cyclic stresses they will encounter during their service life.

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

    Mean stress does not change the alternating stress, but it changes how much alternating stress the material can tolerate. A tensile mean stress opens cracks and lowers the permissible amplitude, which is what the Gerber, Goodman and Soderberg lines express: as σ_m rises toward the static strength, the allowable σ_a falls toward zero. A compressive mean stress keeps cracks closed and improves fatigue strength, which is why shot peening, cold rolling of fillets and nitriding, which leave compressive residual stress at the surface, are used on crankshafts and springs.

  8. 8.A steel component with S_ut = 800 MPa and a corrected endurance limit of 300 MPa carries a fully reversed stress of 400 MPa. Estimate its life using the usual log-log S-N line from 0.9·S_ut at 10³ cycles to S_e at 10⁶ cycles.Numerical

    Since 400 MPa exceeds the endurance limit, the life is finite. The line passes through 0.9 × 800 = 720 MPa at 10³ cycles and 300 MPa at 10⁶ cycles, so log N = 3 + 3·log(720/400)/log(720/300) = 3 + 3 × 0.2553/0.3802 = 5.014. That gives N ≈ 1.03 × 10⁵ cycles. In a real design the line constants come from test data or the data book, so this is an estimate.

  9. 9.A component is subjected to a mean stress of 100 MPa and an alternating stress of 150 MPa. Using the Goodman relation, determine if the component is safe if the ultimate tensile strength is 600 MPa and the endurance limit is 200 MPa.Numerical

    The Goodman criterion is σ_a/S_e + σ_m/S_ut = 1/n. Substituting, 150/200 + 100/600 = 0.750 + 0.167 = 0.917, so n = 1/0.917 = 1.09. The operating point lies just inside the Goodman line, so the component has infinite life, but with only about a 9 % margin, which is too small for most automotive parts.

  10. 10.What are the limitations of using the S-N curve for fatigue analysis?Application

    The S-N curve has several limitations: it is typically derived from tests on smooth, polished specimens, which may not represent real-world conditions with surface imperfections or notches. It also assumes constant amplitude loading, whereas real components often experience variable amplitude loading. Additionally, the S-N curve does not account for environmental factors such as temperature and corrosion, which can significantly affect fatigue life.

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