Fatigue — S-N curves, Miner's rule: Worked examples

3.6.8 · D3 · Physics › Spacecraft Structures & Systems Engineering › Fatigue — S-N curves, Miner's rule

Shuru karne se pehle, ek reminder ki har symbol ka plain words mein kya matlab hai (hum koi bhi symbol use nahi karte jo humne earn nahi kiya ho):

Definition Symbols ki cast

$S$ = stress amplitude — ek wiggle kitni zaroor se khinchta hai, MPa (megapascals) mein. Dekho Stress and Strain.
$N$ = cycles to failure agar woh amplitude akele kaam kare.
$σ_{f}^{'}$ = fatigue strength coefficient — woh (extrapolated) stress jo part ko ek hi cycle mein tod deta hai ( $N = 1$ ); socho "log–log line ka y-intercept".
$b$ = Basquin exponent — line kitni steeply droops karti hai; ek chota sa positive number, ~0.05–0.12.
$n_{i}$ = kitne cycles tum actually apply karte ho level $i$ par.
$D$ = accumulated damage, ek dimensionless "fraction of life used". $D = 1$ par failure.
$σ_{ma x}, σ_{min}$ = ek cycle ke andar reach hone wala peak aur trough stress (MPa) — wiggle ka upar aur neeche.
$σ_{m} = \frac{1}{2} (σ_{ma x} + σ_{min})$ = mean stress — woh mid-level jiske around wiggle hilta hai; yeh crack ko thoda sa khula rakhta hai (parent note se).
$σ_{U T S}$ = ultimate tensile strength — ek-pull breaking stress; neeche diye mean-stress rules iske against de-rate karte hain.
$S_{e}$ = endurance limit — ek stress jiske neeche ek true-endurance-limit material (plain-carbon / low-alloy steel) practically forever chalta hai; uska S–N curve yahan flat ho jaata hai. Aluminium aur titanium mein yeh hota nahi.

Definition "Fully reversed" ka matlab kya hai (R = −1)

Is page par use ki gayi Basquin line $S = σ_{f}^{'} N^{- b}$ fully reversed loading assume karti hai: wiggle zero ke around symmetrically swing karta hai, isliye $σ_{ma x} = + S$ aur $σ_{min} = - S$ , mean stress $σ_{m} = 0$ deta hai. Ratio $R = \frac{σ _{min}}{σ _{ma x}} = - 1$ iska shorthand hai. Agar $σ_{m} \neq = 0$ ho (wiggle offset ho, jaise ek bolt jo hamesha tension mein ho), toh wahi amplitude $S$ zyada damaging ho jaata hai, isliye real $N$ plain Basquin value se chhoti hoti hai. Engineers iske liye ek mean-stress rule se correct karte hain — jaise Goodman (ek straight de-rating line) ya Gerber (ek parabola) — jo S–N curve mein enter karne se pehle allowable amplitude ko shrink kar deta hai. Is page par har example $R = - 1$ hai sivaaye Example 9 ke, jahan hum $σ_{m}$ explicitly compute karte hain aur Goodman correction karte hain.

Scenario matrix

Is topic mein jo bhi aata hai woh in cells mein se kisi ek mein fit hota hai. Neeche ke examples cell label ke saath hain jo woh cover karte hain.

Cell	Kya cheez khaas hai	Covered by
A. High stress, low cycles	$S$ near $σ_{f}^{'}$ , isliye $N$ chhoti hai	Ex 1
B. Low stress, huge cycles	$S ≪ σ_{f}^{'}$ , isliye $N$ explode karta hai	Ex 2
C. Degenerate: $S = 0$	koi swing nahi → Basquin kya kehta hai?	Ex 3
D. Limiting: below / exactly at endurance limit	curve flat ho jaati hai → $N \to \infty$ (sirf steel); boundary $S = S_{e}$	Ex 4
E. Mixed spectrum, survives	$\sum n_{i} / N_{i} < 1$	Ex 5
F. Mixed spectrum, fails; find where	$\sum$ mid-block mein 1 cross karta hai	Ex 6
G. Remaining-life / allowable cycles	unknown $n$ ke liye solve karo given used-up budget	Ex 7
H. Real-world word problem	launch + on-orbit thermal spectrum	Ex 8
I. Exam twist: mean-stress + "sum the stresses" trap	non-zero mean, Goodman correction, illegal shortcut	Ex 9

Worked Examples

Worked example Example 1 — Cell A: high stress, low cycles

Ek aluminium fitting mein $σ_{f}^{'} = 900$ MPa aur $b = 0.10$ hai. Isse amplitude $S = 450$ MPa par cycle kiya jaata hai (fully reversed, $R = - 1$ ). $N$ nikalo.

Forecast: $S$ , $σ_{f}^{'}$ ki aadhi hai — kaafi high. Kya tum hazaaron, laakhon, ya sirf kuch sau cycles expect karte ho?

Basquin invert karo: $N = (\frac{S}{σ _{f}^{'}})^{- 1/ b}$ . Yeh step kyun? Humein $S$ diya gaya hai aur $N$ chahiye; Basquin unhe tie karta hai, isliye jo chahiye uske liye solve karo.
Ratio: $\frac{S}{σ _{f}^{'}} = \frac{450}{900} = 0.5$ . Kyun? Formula sirf applied stress aur coefficient ke ratio ki parwah karta hai.
Exponent: $- 1/ b = - 1/0.10 = - 10$ . Toh $N = 0. 5^{- 10} = 2^{10} = 1024$ . Kyun? $0. 5^{- 10} = (1/2)^{- 10} = 2^{10}$ — do ki ek clean power.

Answer: $N \approx 1024$ cycles.

Verify: Plug back karo: $S = 900 \cdot 102 4^{- 0.10}$ . Kyunki $1024 = 2^{10}$ , $102 4^{- 0.10} = 2^{- 1} = 0.5$ , deta hai $900 \times 0.5 = 450$ MPa. ✓ Low-life expectation confirmed.

Worked example Example 2 — Cell B: low stress, huge cycles

Wohi aluminium fitting ( $σ_{f}^{'} = 900$ MPa, $b = 0.10$ ), ab gently $S = 90$ MPa par cycle ki gayi (fully reversed). $N$ nikalo.

Forecast: Humne stress Example 1 ke mukable 5 ke factor se drop kiya. Kyunki exponent $- 10$ hai, life sirf thodi nahi badhni chahiye — isse explode karna chahiye. Kitne zeros?

Ratio: $\frac{90}{900} = 0.1$ . Kyun? Wohi inversion, naya ratio.
Power: $N = 0. 1^{- 10} = 1 0^{10}$ . Kyun? $0.1 = 1 0^{- 1}$ , aur $(1 0^{- 1})^{- 10} = 1 0^{10}$ .

Answer: $N = 1 0^{10}$ cycles.

Verify: $900 \cdot (1 0^{10})^{- 0.10} = 900 \cdot 1 0^{- 1} = 90$ MPa. ✓

Neeche figure dekho (Example 1 aur Example 2 dono points uspar marked hain). Lavender curve Basquin line hai. Butter-coloured double arrow notice karo: linear stress ko 450 se 90 MPa drop karna (ek modest vertical step) tumhe log- $N$ axis par enormous distance right ki taraf slide karta hai — coral dot se $N \approx 1 0^{3}$ par mint dot $N = 1 0^{10}$ tak. Yahi reason hai ki $N$ logarithmically plot kiya jaata hai: sirf ek log axis dono lives ko ek picture mein rakh sakta hai.

Figure — Fatigue — S-N curves, Miner's rule

Worked example Example 3 — Cell C: degenerate input

$S = 0$ Part mount hai lekin kabhi load nahi hua — stress amplitude exactly $S = 0$ hai. Basquin kya predict karta hai, aur kya yeh physical hai?

Forecast: Koi wiggle nahi. Kya part forever chalna chahiye, ya formula misbehave karta hai?

$N$ ke liye rearrange karo: $N = σ_{f}^{'}^{1/ b} \cdot S^{- 1/ b}$ — yaani $N$ , $S^{- 1/ b}$ ke proportional hai, aur $- 1/ b < 0$ . Yeh step kyun? Hum dekhna chahte hain ki $S \to 0$ hone par $N$ kaise behave karta hai, isliye $S$ -dependence isolate karo.
Limit: jaise $S \to 0^{+}$ , $S^{- 1/ b} \to + \infty$ , isliye $N \to \infty$ . Kyun? Shrinking number ki negative power blow up karti hai. Koi swing nahi ⇒ koi crack opening nahi ⇒ infinite life.
Nonsense se bachao: literally $S = 0$ $0^{- 10}$ ko undefined banata hai (zero se division). Isliye hum $S = 0$ ko limit "infinite life" ki tarah treat karte hain, formula mein $0$ plug karke nahi. Kyun? Physics kehta hai "koi load nahi, koi fatigue nahi"; algebra ko sirf limit interpretation chahiye, raw substitution nahi.

Answer: $N \to \infty$ (unloaded part kabhi fatigue nahi hota). $0^{- 1/ b}$ ko directly evaluate mat karo.

Verify: ek shrinking sequence try karo $S = 90, 9, 0.9$ MPa with $σ_{f}^{'} = 900, b = 0.10$ : $N = 1 0^{10}, 1 0^{20}, 1 0^{30}$ — without bound badhta hua. ✓

Worked example Example 4 — Cell D: below AND exactly at the endurance limit

Ek plain-carbon steel bracket mein true endurance limit $S_{e} = 200$ MPa hai (upar cast of symbols mein define kiya gaya). Teen orbit stress levels consider karo: (a) $S = 150$ MPa, (b) $S = 200$ MPa (exactly $S_{e}$ ), (c) $S = 250$ MPa. Kaunse infinite life denge?

Forecast: Ek clearly flat line ke neeche hai, ek clearly upar hai, aur ek exactly corner par baitha hai. Corner par kya hota hai?

Case (a) $S = 150 < S_{e}$ : endurance limit ke neeche ⇒ flat branch par ⇒ $N \to \infty$ (design-safe, effectively $> 1 0^{7}$ ). Yeh step kyun? Steels ke liye sloping Basquin line sirf $S_{e}$ ke upar rehti hai; neeche curve flat hai, isliye damage per cycle essentially zero hai.
Case (b) $S = 200 = S_{e}$ (the boundary): endurance limit defined hai highest stress ke roop mein jo infinite life deta hai, isliye equality included hai — exactly $S_{e}$ par part indefinitely survive karta hai. Yeh corner point hai jahan flat line slope se milti hai; convention se $S = S_{e}$ "infinite life" count karta hai, lekin zero margin ke saath, isliye koi bhi chota overshoot tumhe sloping (finite-life) branch par le jaata hai. Kyun? Boundary ko explicitly resolve karna zaroori hai: " $\leq S_{e}$ survives, $> S_{e}$ finite hai". $S = S_{e}$ ko finite treat karna ek safe design ko galat taur par throw away kar deta; isse comfortably safe treat karna yeh ignore karta ki koi cushion nahi hai.
Case (c) $S = 250 > S_{e}$ : limit se upar ⇒ sloping Basquin branch par ⇒ finite $N$ , exactly Example 1 ki tarah compute kiya gaya. Kyun? $S_{e}$ se upar endurance shortcut apply nahi hota; sloped curve read karni padti hai.

Answer: (a) infinite, (b) infinite lekin zero-margin (boundary $S = S_{e}$ included), (c) finite.

Verify (contrast trap): agar yeh titanium hoti, toh koi true endurance limit nahi hota, isliye $150$ MPa bhi finite $N$ deta — " $\leq S_{e}$ " rule ek steel-only privilege hai. ✓ (parent note ke mistake box ke saath consistent)

Worked example Example 5 — Cell E: mixed spectrum that survives

Ek strut mission mein do blocks dekhta hai: $n_{1} = 3 \times 1 0^{4}$ cycles ek level par jahan $N_{1} = 2 \times 1 0^{5}$ , aur $n_{2} = 6 \times 1 0^{4}$ cycles jahan $N_{2} = 3 \times 1 0^{5}$ . Kya yeh survive karta hai?

Forecast: Har block "life budget" ka ek slice use karta hai. Kya do slices milkar 100% ke andar rahenge?

Block 1 se damage: $D_{1} = \frac{n _{1}}{N _{1}} = \frac{3 \times 1 0 ^{4}}{2 \times 1 0 ^{5}} = 0.15$ . Yeh step kyun? Miner's rule: level $i$ par ek cycle budget ka $1/ N_{i}$ kharach karta hai, isliye $n_{i}$ cycles $n_{i} / N_{i}$ kharach karte hain.
Block 2 se damage: $D_{2} = \frac{6 \times 1 0 ^{4}}{3 \times 1 0 ^{5}} = 0.20$ . Kyun? Wohi rule, independent contribution.
Sum: $D = 0.15 + 0.20 = 0.35 < 1$ . Kyun? Failure sirf $D \geq 1$ par; hum fractions sum karte hain, kabhi stresses nahi.

Answer: survives, apni fatigue life ka $35%$ use karke.

Verify: do fractions dimensionless hain (cycles ÷ cycles), aur $0.35 < 1$ . ✓

Worked example Example 6 — Cell F: mixed spectrum that fails; find where

Ek joint repeated identical missions run karta hai. Har mission mein $n_{1} = 2 \times 1 0^{4}$ cycles at $N_{1} = 1 \times 1 0^{5}$ aur $n_{2} = 1 \times 1 0^{4}$ cycles at $N_{2} = 4 \times 1 0^{4}$ hote hain. Kis mission par $D$ pehli baar 1 reach karta hai?

Forecast: Har mission damage ka ek fixed chunk add karta hai. Budget khatam hone tak kitne missions?

Damage per mission: $D_{mission} = \frac{2 \times 1 0 ^{4}}{1 \times 1 0 ^{5}} + \frac{1 \times 1 0 ^{4}}{4 \times 1 0 ^{4}} = 0.20 + 0.25 = 0.45$ . Yeh step kyun? Dono blocks ke fractions sum karo per-mission life used pane ke liye.
$D = 1$ tak missions: $\frac{1}{0.45} = 2.22$ missions. Kyun? Total budget 1, har mission mein kharchya budget se divided.
Fraction interpret karo: 2 full missions ke baad $D = 0.90$ ; failure 3rd ke dauran hota hai, jab cumulative $D$ 1 cross karta hai. Kyun? Tum fractional mission nahi fly kar sakte — failure mission 3 ke beech mein land karta hai.

Answer: 3rd mission ke dauran fail hota hai (predicted $D = 1$ at $2.22$ missions).

Verify: $2 \times 0.45 = 0.90 < 1$ (2 survive karta hai), $3 \times 0.45 = 1.35 \geq 1$ (3 tak fail). ✓

Worked example Example 7 — Cell G: remaining life / allowable cycles

Ek part pehle se $D = 0.70$ accumulate kar chuka hai. Ek new load level mein $N_{4} = 5 \times 1 0^{4}$ hai. Failure se pehle woh kitne aur cycles $n_{4}$ le sakta hai?

Forecast: Budget ka sirf $30%$ bacha hai. Kya $n_{4}$ , $N_{4}$ se kaafi chhota hoga?

Remaining budget: $D_{left} = 1 - 0.70 = 0.30$ . Yeh step kyun? Failure $D = 1$ par; jo already spent ho chuka hai use subtract karo.
New damage = remaining budget set karo: $\frac{n _{4}}{N _{4}} = 0.30$ . Kyun? Hum new level par sirf tab tak cycles apply kar sakte hain jab tak woh leftover budget exhaust kar den.
Solve karo: $n_{4} = 0.30 \times 5 \times 1 0^{4} = 1.5 \times 1 0^{4}$ cycles. Kyun? Dono sides ko $N_{4}$ se multiply karo.

Answer: $n_{4} = 15, 000$ aur cycles.

Verify: $0.70 + \frac{1.5 \times 1 0 ^{4}}{5 \times 1 0 ^{4}} = 0.70 + 0.30 = 1.00$ exactly. ✓ Aur $n_{4} = 1.5 \times 1 0^{4} < N_{4} = 5 \times 1 0^{4}$ , sensible. ✓

Worked example Example 8 — Cell H: real-world word problem (launch + orbit)

Ek bracket ko ek launch aur 5-year orbit life survive karna hai. Do damage sources, Launch Loads & Environments aur Thermal Cycling on Orbit se:

Launch vibration (dekho Random Vibration & PSD): $n_{1} = 8 \times 1 0^{3}$ cycles ek level par jahan $N_{1} = 5 \times 1 0^{4}$ .
Thermal cycling on orbit: har 90 min mein ek cycle. Har thermal cycle mein $N_{2} = 1 \times 1 0^{5}$ hai. Poora mission kitni life use karta hai, aur kya yeh survive karta hai?

Forecast: 90-minute cycles ke 5 saal bahut zyada thermal cycles hain. Kya slow thermal drip ya violent launch zyada damage dominate karega?

Thermal cycles count karo: cycles per year $\approx \frac{365 \times 24 \times 60}{90} = \frac{525600}{90} = 5840$ ; 5 saalon mein, $n_{2} = 5 \times 5840 = 29200$ . Yeh step kyun? Miner ko ek applied cycle count chahiye; "har 90 min for 5 years" ko ek number mein convert karo.
Launch damage: $D_{1} = \frac{8 \times 1 0 ^{3}}{5 \times 1 0 ^{4}} = 0.16$ . Kyun? Launch block se spent life ka fraction.
Thermal damage: $D_{2} = \frac{29200}{1 \times 1 0 ^{5}} = 0.292$ . Kyun? Orbital block ke liye wohi rule.
Total & margin: $D = 0.16 + 0.292 = 0.452 < 1$ . Kyun? Survives; $D = 1$ ke against margin ko Safety Factors & Margins of Safety formalise karta hai (yahan hame $\approx 2.2 \times$ life margin hai).

Answer: fatigue life ka $\approx 45.2%$ use karta hai ⇒ mission survive karta hai. Yahan thermal cycling ( $0.292$ ) launch ( $0.16$ ) se thoda zyada damage karta hai.

Verify: $525600/90 = 5840$ exactly; $5 \times 5840 = 29200$ ; $0.16 + 0.292 = 0.452 < 1$ . ✓

Worked example Example 9 — Cell I: mean stress, Goodman correction, aur "sum the stresses" trap

Ek titanium fitting mein $σ_{U T S} = 950$ MPa, $σ_{f}^{'} = 1800$ MPa, $b = 0.10$ hai. Isme $σ_{ma x} = 260$ MPa aur $σ_{min} = - 60$ MPa wala cycle aata hai. (i) Amplitude $σ_{a}$ aur mean $σ_{m}$ nikalo. (ii) Kyunki $σ_{m} \neq = 0$ , equivalent fully-reversed amplitude $S_{eq}$ pane ke liye Goodman correction apply karo. (iii) Phir, do blocks $n_{1} = 1 \times 1 0^{4}$ at $N_{1} = 4 \times 1 0^{4}$ aur $n_{2} = 3 \times 1 0^{4}$ at $N_{2} = 9 \times 1 0^{4}$ given, total damage nikalo. Trap: ek classmate kehta hai "bas $S_{1} + S_{2}$ add karo aur strength se compare karo."

Forecast: Kya offset cycle same amplitude ki fully-reversed cycle se zyada ya kam damaging hai? Aur kya MPa mein do stresses add karna legal move bhi hai?

Amplitude aur mean: $σ_{a} = \frac{260 - ( - 60 )}{2} = \frac{320}{2} = 160$ MPa; $σ_{m} = \frac{260 + ( - 60 )}{2} = \frac{200}{2} = 100$ MPa. Yeh step kyun? Amplitude = peak-to-trough swing ki aadhi (woh crack kholti hai); mean = mid-level (ise khula rakhta hai). Kyunki $σ_{m} = 100 \neq = 0$ , yeh fully reversed nahi hai, isliye Basquin raw use nahi ho sakta.
Goodman correction: $S_{eq} = \frac{σ _{a}}{1 - σ _{m} / σ _{U T S}} = \frac{160}{1 - 100/950} = \frac{160}{1 - 0.10526} = \frac{160}{0.89474} = 178.8$ MPa. Kyun? Tensile mean crack ko khula rakhti hai, isliye offset cycle ek bade fully-reversed ki tarah bite karta hai. Goodman $160 \to 178.8$ MPa inflate karta hai; woh amplitude hai jo tum S–N curve mein le jaate — naive 160 MPa ke suggest se chhoti life deta hai. (Gerber, $(σ_{m} / σ_{U T S})^{2} = 0.01108$ use karte hue, gentler $S_{eq} = 160/0.98892 = 161.8$ MPa deta hai.)
Trap — "stresses add karo" kyun illegal hai: classmate $S_{1} + S_{2}$ compute karta (do amplitudes MPa mein add karke) aur strength se compare karta. Yeh meaningless hai: MPa values life-usages nahi hain, separate loading blocks mein sum nahi ho sakti, aur ek aisa number deti hain jinke paas koi failure criterion nahi. Miner's rule dimensionless fractions $n_{i} / N_{i}$ sum karta hai, kabhi stresses nahi. Kyun? Elastic superposition stresses tab add karta hai jab woh ek hi waqt ek hi point par kaam kar rahe hon — lekin ye time ke saath alag alag blocks hain, isliye sirf unke damage fractions add hote hain.
Correct damage sum: $D = \frac{n _{1}}{N _{1}} + \frac{n _{2}}{N _{2}} = \frac{1 \times 1 0 ^{4}}{4 \times 1 0 ^{4}} + \frac{3 \times 1 0 ^{4}}{9 \times 1 0 ^{4}} = 0.25 + 0.333 \dots = 0.583$ . Kyun? Pehle har block ko life fraction mein convert karo, phir add karo — yahi ek legal accumulation hai.

Answer: $σ_{a} = 160$ MPa, $σ_{m} = 100$ MPa; Goodman-equivalent $S_{eq} \approx 178.8$ MPa (mean ise zyada damaging banata hai); total damage $D = 7/12 \approx 0.583 < 1$ ⇒ survives. "Add the stresses" method invalid hai.

Verify: $(260 - (- 60)) /2 = 160$ ; $(260 + (- 60)) /2 = 100$ ; Goodman $160/ (1 - 100/950) = 178.82$ MPa $> 160$ (correctly inflated); $0.25 + 1/3 = 7/12 \approx 0.5833 < 1$ . ✓

Recall

$N$ kab infinite hota hai versus finite? $N$ tab finite hota hai jab tum Basquin's sloping line par ho (koi bhi Ti/Al, koi bhi steel $S_{e}$ se upar). $N$ sirf tab infinite hota hai jab (a) $S = 0$ limit ke roop mein, ya (b) ek steel ko at or below uski endurance limit par stress kiya gaya ho, yaani $S \leq S_{e}$ (equality included, lekin zero margin ke saath).

Recall Miner's rule: tum kya sum karte ho, aur failure number kya hai?

Dimensionless damage fractions $n_{i} / N_{i}$ (cycles applied ÷ cycles-to-failure) sum karo. Failure tab jab sum $D = 1$ reach kare. Kabhi MPa mein stresses mat summate karo.

Recall Basquin line yahan mean stress ke bare mein kya assume karti hai, aur non-zero mean ko tum kaise fix karte ho?

Fully reversed loading, $R = - 1$ , isliye $σ_{m} = 0$ . Agar $σ_{m} \neq = 0$ ho, pehle equivalent fully-reversed amplitude mein convert karo Goodman $S_{eq} = σ_{a} / (1 - σ_{m} / σ_{U T S})$ se (ya gentler Gerber parabola se), phir S–N curve mein enter karo — true life chhoti hoti hai.

Is trap ko convert karo

S_{1} + S_{2}

MPa mein add karna illegal hai; pehle har block ko

n_{i} / N_{i}

mein convert karo, phir sum karo.

Steel ke liye at or below

S_{e}

par,

N =

Infinite (design-safe); boundary

S = S_{e}

included hai lekin zero margin hai.

S = 0

N =

Limit mein infinite; kabhi

0

0^{- 1/ b}

mein plug mat karo (undefined).

Non-zero mean fix?

Goodman

S_{eq} = σ_{a} / (1 - σ_{m} / σ_{U T S})

phir S–N curve mein enter karo; true life chhoti hoti hai.