Fatigue, Endurance Limit & Fluctuating Loads

Fatigue is a signature GATE ME Machine Design topic (~2 marks) because most machine parts fail under fluctuating, not static, loads. The endurance limit and the Goodman/Soderberg criteria are the core.

Fatigue fundamentals

Repeated (cyclic) stress causes failure well below the static strength. The S–N curve plots stress amplitude vs cycles to failure. For steels, a horizontal endurance limit Se appears (~10⁶ cycles): below it, life is effectively infinite. Rough estimate: Se ≈ 0.5·σ_ut for steel (specimen value), corrected by factors:
Se = ka·kb·kc·kd·ke·Se' — surface (ka), size (kb), load (kc), temperature (kd), reliability (ke). Non-ferrous metals (aluminium) have no true endurance limit — strength keeps falling.

Stress terms: mean σm = (σmax + σmin)/2; amplitude σa = (σmax − σmin)/2; stress ratio R = σmin/σmax. Fully reversed → σm = 0, R = −1.

Stress concentration: fatigue strength reduction factor Kf = 1 + q(Kt − 1), where Kt = theoretical factor, q = notch sensitivity.

Fluctuating-load criteria (σa vs σm)

• Soderberg (safe, conservative): σa/Se + σm/σy = 1/n.
• Goodman (common, uses ultimate): σa/Se + σm/σut = 1/n.
• Gerber (parabolic, least conservative): σa/Se + (σm/σut)² = 1/n... plotted as a curve fitting data best.

Exam Tricks & Tips

• 🎯 Se ≈ 0.5σ_ut for steel; apply surface/size/reliability factors to get the corrected endurance limit.
• 🎯 Soderberg is the most conservative (uses σ_y), Gerber the least; Goodman sits between (uses σ_ut) — pick per the question.
• 🎯 Fully reversed loading → σm = 0, R = −1; then design directly against Se.
• 🎯 Kf = 1 + q(Kt − 1): notch sensitivity q blends the geometric Kt into the actual fatigue factor.
• 🎯 Aluminium has NO endurance limit — define fatigue strength at a stated cycle count (e.g. 5×10⁸) instead.
• ❌ Common mistake: using the uncorrected Se' = 0.5σ_ut directly — real parts need the ka·kb·… modifying factors, which lower Se substantially.

Expected exam pattern

A 2-mark Goodman/Soderberg factor-of-safety problem (given σa, σm, Se, σut/σy), or an endurance-limit correction. Mean/amplitude decomposition and the Kf notch factor appear as sub-steps or MCQs.

Quick recap

Fatigue fails below static strength; steel has an endurance limit Se ≈ 0.5σ_ut (corrected by ka·kb·kc…). σm and σa from σmax, σmin. Soderberg (σy, safest), Goodman (σut), Gerber (parabolic). Kf = 1 + q(Kt−1). Aluminium: no endurance limit.

Cover the answer, recall, then check. 12 cards on fatigue for GATE ME.

Q1. What is the endurance (fatigue) limit?
A1. The stress amplitude below which a steel part survives effectively infinite cycles (~10⁶); seen as a horizontal S–N asymptote.

Q2. Rough estimate of endurance limit for steel?
A2. Se ≈ 0.5·σ_ut (specimen value), then corrected by surface, size, load, and reliability factors.

Q3. Do aluminium alloys have an endurance limit?
A3. No — their S–N curve keeps falling, so fatigue strength is quoted at a specific cycle count (e.g. 5×10⁸ cycles).

Q4. Define mean stress and stress amplitude.
A4. σm = (σmax + σmin)/2; σa = (σmax − σmin)/2.

Q5. What is fully reversed loading?
A5. σm = 0 (σmax = −σmin), stress ratio R = −1 — the design is made directly against Se.

Q6. State the Soderberg criterion.
A6. σa/Se + σm/σy = 1/n. The most conservative line (uses yield strength).

Q7. State the Goodman criterion.
A7. σa/Se + σm/σut = 1/n. Uses ultimate strength; common in design.

Q8. How does Gerber differ from Goodman?
A8. Gerber is a parabola (σm/σut term squared) fitting experimental data best; it is the least conservative of the three.

Q9. Formula for the fatigue stress-concentration factor?
A9. Kf = 1 + q(Kt − 1), where Kt = theoretical factor and q = notch sensitivity (0 to 1).

Q10. Order the three criteria from safest to least safe.
A10. Soderberg (safest) → Goodman → Gerber (least conservative).

Q11. What do the modifying factors ka, kb, kc do?
A11. They reduce the specimen endurance limit Se' for real surface finish (ka), size (kb), and load type (kc), etc.

Q12. Why design against fatigue rather than static strength?
A12. Most parts see cyclic loads and fail by fatigue at stresses well below the static yield/ultimate strength.

Most machine parts fail not under a single overload but after millions of stress cycles at loads well below yield — fatigue. It is the leading cause of mechanical failure, so GATE gives it heavy weight. The skill is combining mean and alternating stresses with the endurance limit through the standard failure lines.

Core concept: repeated loading nucleates and grows cracks until sudden fracture; a material can endure infinitely only if its alternating stress stays below the endurance limit, and real parts must derate that limit for size, surface and stress concentration.

Deep explanation

Beginner — the S–N curve and endurance limit

Plotting stress amplitude S vs cycles to failure N (log scale) gives the S–N curve. For steels it flattens to a horizontal endurance limit S_e' (about 0.5 S_ut for steel) below which life is effectively infinite (>10⁶ cycles). Non-ferrous metals (aluminium) show no true endurance limit — strength keeps dropping, so life is quoted at a set number of cycles.

Intermediate — mean and alternating stress

A fluctuating load has:

• Mean stress σ_m = (σ_max + σ_min)/2,
• Alternating stress σ_a = (σ_max − σ_min)/2.
  Fully reversed loading has σ_m = 0. The corrected endurance limit derates the ideal value:
  S_e = k_a k_b k_c … × S_e', where k factors cover surface finish, size, load type, temperature and reliability. Stress concentration at notches is applied to σ_a via the fatigue stress-concentration factor K_f.

Advanced — the failure criteria (mean–alternating diagram)

Plot σ_a vs σ_m and draw a failure line from S_e (on σ_a axis) to a strength on the σ_m axis:

• Goodman line (to S_ut): σ_a/S_e + σ_m/S_ut = 1/FoS. Standard, slightly conservative — the exam default.
• Soderberg line (to S_y): σ_a/S_e + σ_m/S_y = 1/FoS. Most conservative (guards against yielding too).
• Gerber parabola (to S_ut): fits data best but non-linear.
  A design point inside the chosen line is safe. Combined with a yield check, these size parts against both fatigue and first-cycle yield.

Worked example

A steel part (S_ut = 600 MPa, corrected S_e = 250 MPa) sees σ_m = 150 MPa and σ_a = 100 MPa. Find the factor of safety by the Goodman criterion.
1/FoS = σ_a/S_e + σ_m/S_ut = 100/250 + 150/600 = 0.40 + 0.25 = 0.65.
FoS = 1/0.65 = 1.54. The part is safe against fatigue but the modest margin warns against any increase in load.

GATE relevance

Fatigue is one of the highest-yield Machine Design areas: endurance-limit estimation, Goodman/Soderberg factor of safety, and stress-concentration effects. The Goodman and Soderberg equations are near-mandatory to memorise, along with S_e' ≈ 0.5 S_ut for steel.

Exam tricks & shortcuts

• Goodman uses S_ut; Soderberg uses S_y — Soderberg is the more conservative (adds a yield safeguard).
• Estimate the raw endurance limit as 0.5 S_ut for steel when not given.
• Apply stress-concentration factor K_f to the alternating stress σ_a (the damaging component).
• Mnemonic: "Soderberg is Safest (yield), Goodman goes to Ultimate."

Key formulas

• Stress components: σ_mean = (σ_max + σ_min)/2; σ_amplitude = (σ_max − σ_min)/2.
• Endurance limit σ_e (≈ 0.5σ_ult for steel); corrected σ_e' = k_a k_b k_c … σ_e.
• Goodman line: σ_a/σ_e + σ_m/σ_ult = 1/FoS.
• Soderberg: σ_a/σ_e + σ_m/σ_y = 1/FoS.
• Stress ratio: R = σ_min/σ_max.

Fatigue design combines mean and alternating stress via Goodman/Soderberg criteria against the endurance limit.

Worked Example

Problem: A machine part experiences a stress that fluctuates between σ_min = 40 MPa and σ_max = 160 MPa. The material has endurance limit S_e = 200 MPa and yield strength S_y = 350 MPa. Using the Soderberg criterion, find the factor of safety against fatigue.

Solution:
Resolve the cyclic stress into mean and alternating components:
Mean stress: σ_m = (σ_max + σ_min)/2 = (160 + 40)/2 = 100 MPa.
Alternating stress: σ_a = (σ_max − σ_min)/2 = (160 − 40)/2 = 60 MPa.

The Soderberg fatigue criterion (which uses the yield strength for the mean term, the most conservative line):
1/FoS = σ_a/S_e + σ_m/S_y.

Substitute:
1/FoS = 60/200 + 100/350 = 0.300 + 0.2857 = 0.5857.

Therefore:
FoS = 1/0.5857 ≈ 1.71.

Answer: σ_m = 100 MPa, σ_a = 60 MPa, and the Soderberg factor of safety ≈ 1.71.