Fatigue — S-N curves, Miner's rule: Exercises — Hinglish deep dive

3.6.8 · D4 · Physics › Spacecraft Structures & Systems Engineering › Fatigue — S-N curves, Miner's rule

Do constants baar baar aate hain, toh inhe ek baar clearly naam de lete hain:

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

Level 1 — Recognition

L1.1

S–N curve mein failure ke liye cycles ki number kaun se axis par hoti hai, aur woh linear hai ya logarithmic?

Recall Solution

Horizontal (x) axis $N$ hai, jo failure tak ke cycles ki number hai, aur yeh logarithmic scale par draw ki jaati hai. Vertical (y) axis stress amplitude $S = σ_{a}$ hai. Log kyun? Kyunki $N$ kai powers of ten tak range karta hai ( $1 0^{3}$ se $1 0^{9}$ tak); linear axis par sab kuch left edge par squeeze ho jaata. Figure s01 dekho — equally-spaced ticks $1 0^{3}, 1 0^{4}, 1 0^{5}, \dots$ hain, na ki $1, 2, 3$ .

L1.2

Ek part har cycle mein $σ_{ma x} = + 220$ MPa aur $σ_{min} = - 60$ MPa dekhta hai. Mean stress $σ_{m}$ aur amplitude $σ_{a}$ nikalo.

Recall Solution

Kya karein: do cycle formulas mein plug in karo. $σ_{m} = \frac{σ _{ma x} + σ _{min}}{2} = \frac{220 + ( - 60 )}{2} = \frac{160}{2} = 80 MPa$ $σ_{a} = \frac{σ _{ma x} - σ _{min}}{2} = \frac{220 - ( - 60 )}{2} = \frac{280}{2} = 140 MPa$ Kyun: mean swing ka middle point hai (crack ko thoda khula rakhta hai); amplitude half-swing hai (isse kholti aur band karti hai). Underlying $σ$ definitions ke liye Stress and Strain dekho.

L1.3

True ya false: aluminium aur Ti-6Al-4V dono mein plain-carbon steel ki tarah ek sachchi endurance limit hoti hai.

Recall Solution

False. Ek sachchi endurance limit — ek flat floor jiske neeche life practically infinite hai — mainly plain-carbon / low-alloy steel ki khasiyat hai. Aluminium aur common titanium alloys (Ti-6Al-4V) $1 0^{7}$ cycles ke baad bhi slope karte rehte hain, isliye engineers " $N = 5 \times 1 0^{8}$ par fatigue strength" quote karte hain.

Level 2 — Application

L2.1

Ek strut mein $σ_{f}^{'} = 1800$ MPa aur $b = 0.09$ hai. Use amplitude $S = 450$ MPa par cycle kiya jaata hai. Failure tak kitne cycles honge?

Recall Solution

Step 1 — tool chunna. Hume $S$ pata hai aur $N$ chahiye, toh Basquin invert karo: $N = (S / σ_{f}^{'})^{- 1/ b}$ . Step 2 — ratio. $S / σ_{f}^{'} = 450/1800 = 0.25$ . Step 3 — exponent. $- 1/ b = - 1/0.09 = - 11.11$ . Step 4 — logs se evaluate karo ( $0.2 5^{- 11.11}$ jaisa power $ln$ se aasaan hai): $ln N = - 11.11 \times ln (0.25) = - 11.11 \times (- 1.3863) = 15.40$ $N = e^{15.40} \approx 4.9 \times 1 0^{6} cycles$ Logs kyun? Woh " $- 11.11$ power tak raise karo" ko ek simple multiplication mein badal dete hain, phir hum wapas exponentiate karte hain.

L2.2

Wahi strut ( $σ_{f}^{'} = 1800$ MPa, $b = 0.09$ ). Design life $N = 1 0^{7}$ cycles hai. Kaun sa amplitude $S$ allowed hai?

Recall Solution

Step 1 — tool chunna. Hume $N$ pata hai, $S$ chahiye: Basquin seedha use karo, $S = σ_{f}^{'} N^{- b}$ . Step 2 — power. $N^{- b} = (1 0^{7})^{- 0.09} = 1 0^{- 0.63}$ . $1 0^{- 0.63} = e^{- 0.63 \times l n 10} = e^{- 1.4506} = 0.2344$ Step 3 — multiply karo. $S = 1800 \times 0.2344 \approx 422 MPa$ . Kyun: zyada lambi required life par allowed stress kam hogi — curve neeche slope karti hai. L2.1 se compare karo: life ko $4.9 \times 1 0^{6}$ se $1 0^{7}$ tak push karne par ~28 MPa ka nuksan hua. Is bare number ko use karne se pehle Safety Factors & Margins of Safety dekho.

L2.3

Ek bracket $n_{1} = 1.0 \times 1 0^{4}$ cycles dekhta hai jis level par $N_{1} = 1.0 \times 1 0^{5}$ hai, aur $n_{2} = 3.0 \times 1 0^{4}$ cycles us level par jahan $N_{2} = 6.0 \times 1 0^{4}$ hai. Total Miner damage $D$ calculate karo. Kya yeh survive karta hai?

Recall Solution

Step 1 — level-1 fraction. $D_{1} = n_{1} / N_{1} = 1 0^{4} /1 0^{5} = 0.10$ . Step 2 — level-2 fraction. $D_{2} = n_{2} / N_{2} = 3 \times 1 0^{4} /6 \times 1 0^{4} = 0.50$ . Step 3 — sum. $D = 0.10 + 0.50 = 0.60$ . Verdict: $0.60 < 1$ ⇒ survive kar leta hai, apni fatigue life ka $60%$ use karke.

Level 3 — Analysis

L3.1

Ek part pehle se $D = 0.60$ accumulate kar chuka hai (L2.3 se). Ek naya load level aata hai jahan $N_{3} = 8.0 \times 1 0^{4}$ hai. Is level par failure tak kitne cycles $n_{3}$ chalenge?

Recall Solution

Step 1 — remaining budget. $1 - 0.60 = 0.40$ life bachi hai. Step 2 — fraction = budget set karo. $n_{3} / N_{3} = 0.40 \Rightarrow n_{3} = 0.40 \times 8.0 \times 1 0^{4} = 3.2 \times 1 0^{4}$ cycles. Kyun: Miner budget ko ek single pool maanta hai; jo bhi level baaki $0.40$ khaali kare, wahi failure trigger karega.

L3.2

Do nominally identical Ti struts test kiye jaate hain. Strut A, $N = 1.0 \times 1 0^{4}$ cycles ke baad $S = 600$ MPa par fail hota hai; Strut B, $N = 4.0 \times 1 0^{6}$ cycles ke baad $S = 300$ MPa par fail hota hai. In do points se $b$ aur $σ_{f}^{'}$ nikalo.

Recall Solution

Step 1 — tool chunna. Log–log straight line par do points uski slope fix karte hain. Basquin $S = σ_{f}^{'} N^{- b}$ ka log lo: $lo g S = lo g σ_{f}^{'} - b lo g N$ Step 2 — slope $b$ = ( $lo g S$ mein drop)/( $lo g N$ mein rise): $b = - \frac{l o g S _{B} - l o g S _{A}}{l o g N _{B} - l o g N _{A}} = - \frac{l o g 300 - l o g 600}{l o g ( 4 \times 1 0 ^{6} ) - l o g ( 1 0 ^{4} )}$ $= - \frac{- 0.3010}{6.6021 - 4} = - \frac{- 0.3010}{2.6021} = 0.1157$ Step 3 — coefficient $σ_{f}^{'}$ point A se: $σ_{f}^{'} = S_{A} N_{A}^{b} = 600 \times (1 0^{4})^{0.1157}$ . $(1 0^{4})^{0.1157} = 1 0^{0.4628} = 2.903 \Rightarrow σ_{f}^{'} = 600 \times 2.903 \approx 1742 MPa$ Do points kyun kaafi hain: ek straight line ko exactly do anchors chahiye; slope $b$ deta hai, aur koi bhi ek point intercept $σ_{f}^{'}$ back-solve kar deta hai.

L3.3

L3.2 se fit use karke ( $b = 0.1157$ , $σ_{f}^{'} = 1742$ MPa), $S = 400$ MPa par life predict karo.

Recall Solution

$N = (S / σ_{f}^{'})^{- 1/ b} = (400/1742)^{- 1/0.1157} = (0.22962)^{- 8.643}$ $ln N = - 8.643 \times ln (0.22962) = - 8.643 \times (- 1.4713) = 12.716$ $N = e^{12.716} \approx 3.33 \times 1 0^{5} cycles$ Sanity check: $400$ MPa dono test stresses ( $300$ aur $600$ ) ke beech hai, aur $3.3 \times 1 0^{5}$ unki lives ( $4 \times 1 0^{6}$ aur $1 0^{4}$ ) ke beech hai — interpolation self-consistent hai.

Level 4 — Synthesis

L4.1

Ek launch profile ek bracket ko teen vibration levels se hit karta hai. S–N curve se (fit $σ_{f}^{'} = 1742$ MPa, $b = 0.1157$ ):

$n_{1} = 2.0 \times 1 0^{3}$ cycles at $S_{1} = 500$ MPa
$n_{2} = 1.5 \times 1 0^{4}$ cycles at $S_{2} = 400$ MPa
$n_{3} = 8.0 \times 1 0^{4}$ cycles at $S_{3} = 300$ MPa

Ek launch se total Miner damage compute karo. Bracket kitne launches survive kar sakta hai?

Recall Solution

Step 1 — har level par life $N = (S / σ_{f}^{'})^{- 1/ b}$ se, jahan $- 1/ b = - 8.643$ hai:

$N_{1} = (500/1742)^{- 8.643} = (0.28703)^{- 8.643}$ : $ln N_{1} = - 8.643 (- 1.2481) = 10.788 \Rightarrow N_{1} \approx 4.85 \times 1 0^{4}$
$N_{2} = (400/1742)^{- 8.643} = 3.33 \times 1 0^{5}$ (L3.3 se)
$N_{3} = (300/1742)^{- 8.643} = (0.17222)^{- 8.643}$ : $ln N_{3} = - 8.643 (- 1.7588) = 15.202 \Rightarrow N_{3} \approx 4.00 \times 1 0^{6}$ Step 2 — damage fractions: $D_{1} = \frac{2 \times 1 0 ^{3}}{4.85 \times 1 0 ^{4}} = 0.04124, D_{2} = \frac{1.5 \times 1 0 ^{4}}{3.33 \times 1 0 ^{5}} = 0.04505, D_{3} = \frac{8 \times 1 0 ^{4}}{4.00 \times 1 0 ^{6}} = 0.02000$ Step 3 — damage per launch: $D_{launch} = 0.04124 + 0.04505 + 0.02000 = 0.10629$ . Step 4 — launches to failure: $D = 1$ jab $# launches = \frac{1}{0.10629} \approx 9.4 \Rightarrow 9 full launches$ (neeche round karo — 10waan $D$ ko 1 se aage push kar dega). Neeche kyun round karo: tum sirf woh launch fly kar sakte ho jo $D \leq 1$ ke saath khatam ho; $9 \times 0.10629 = 0.957 < 1$ , lekin $10 \times = 1.063 > 1$ . Loads Launch Loads & Environments se aate hain; amplitude spectrum khud Random Vibration & PSD mein derive ki jaati hai.

Level 5 — Mastery

L5.1

Ek structure har orbit-day mein do damage sources dekhti hai: (a) launch ek one-time event hai jo $D_{launch} = 0.10629$ deposit karta hai (L4.1 se); (b) orbit par, har ~90-min thermal cycle $S_{th} = 180$ MPa impose karta hai. Wahi fit use karke ( $σ_{f}^{'} = 1742$ MPa, $b = 0.1157$ ), mission 5 saal chalti hai. Har 90 minute mein ek thermal cycle ke saath, kya structure survive karti hai? Ek fatigue safety factor apply karo jo $D \leq 0.5$ require karta hai.

Recall Solution

Step 1 — thermal-cycle life. $N_{th} = (180/1742)^{- 8.643} = (0.10333)^{- 8.643}$ . $ln N_{th} = - 8.643 \times ln (0.10333) = - 8.643 \times (- 2.2694) = 19.615 \Rightarrow N_{th} \approx 3.30 \times 1 0^{8}$ Step 2 — 5 saal mein thermal cycles count karo. Har 90 min mein ek = din mein 16. $n_{th} = 16 \times 365.25 \times 5 = 29, 220 cycles$ Step 3 — thermal damage. $D_{th} = \frac{29 , 220}{3.30 \times 1 0 ^{8}} = 8.85 \times 1 0^{- 5}$ Step 4 — total mission damage. $D_{total} = D_{launch} + D_{th} = 0.10629 + 0.0000885 = 0.10638$ Step 5 — safety factor apply karo. Requirement: $D \leq 0.5$ . Kyunki $0.10638 \leq 0.5$ ⇒ margin ke saath survive karti hai; allowed half-budget ka used-up fraction $0.10638/0.5 = 0.213$ hai, yaani derated life ka sirf $21%$ . Physical insight: launch teen orders of magnitude se dominate karta hai — kuch minute ka violent vibration 5 saal ki gentle thermal breathing se zyada damage karta hai, kyunki damage $S^{1/ b}$ ki tarah scale karta hai aur launch stresses kahin zyada hain. Thermal cycling low-cycle, high-strain joints ke liye sabse zyada matter karti hai; Thermal Cycling on Orbit dekho. Agar koi crack pehle se hoti, toh tum poora model badal kar Fracture Mechanics & Crack Growth (Paris' Law) use karte.

L5.2

Usi structure ke liye, sirf extra thermal cycles (koi aur launch nahi) kitne lagenge taaki derated failure budget $D = 0.5$ reach ho, $D_{total} = 0.10638$ se shuru karke?

Recall Solution

Step 1 — remaining budget derated limit tak: $0.5 - 0.10638 = 0.39362$ . Step 2 — thermal level par cycles: $n = 0.39362 \times N_{th} = 0.39362 \times 3.30 \times 1 0^{8} = 1.30 \times 1 0^{8}$ cycles. Step 3 — years mein convert karo 16 cycles/day par: $1.30 \times 1 0^{8} / (16 \times 365.25) \approx 2.22 \times 1 0^{4}$ years. Itna bada number kyun: thermal amplitude S–N curve ki shallow tail par kaafi neeche hai, isliye har cycle ek vanishingly chhota fraction spend karta hai. Thermal fatigue yahan life-limiter nahi hai — launch hai.