3.6.8 · Physics › Spacecraft Structures & Systems Engineering
Ek spacecraft strut static load ko safely carry kar sakta hai, lekin launch ke dauran hazaron choti vibrations ke baad snap ho sakta hai . Kyun? Kyunki ek material damage ko yaad rakh sakta hai. Stress ka har cycle ek microscopic crack ko thoda sa kholta aur badhata hai — jaise paperclip ko baar baar aage-peeche modna jab tak woh toot na jaye, jabke ek baar modne se kuch nahi hota. Fatigue repeated loading se failure hai jo static strength se neeche hoti hai.
Fatigue woh progressive, localized, permanent structural damage hai jo tab hoti hai jab koi material cyclic (repeated) loading ke under hota hai, aur aakhirkar stress amplitudes par failure hoti hai jo ultimate tensile strength σ U T S se kaafi neeche hoti hain.
Spacecraft ke liye kyun important hai: Launch = intense random vibration + acoustic loading + hazaron stress cycles ke kuch minute. Orbit par = har ~90 min mein thermal cycling (din/raat), har cycle joints ko strain karta hai. Kuch bhi ek baar load nahi hota — sab kuch baar baar load hota hai.
Ek cycle describe karne wali key quantities:
σ m = 2 σ ma x + σ min ( mean ) , σ a = 2 σ ma x − σ min ( amplitude )
Intuition Amplitude damage kyun drive karti hai
Crack tab badhti hai jab use baar baar khींcha jaata hai. Swing σ a use kholti aur bandh karti hai; mean σ m use thoda khula rakhta hai taaki har swing zyada kaatey. Isliye dono matter karte hain, lekin amplitude king hai.
Stress amplitude S = σ a (y-axis) ka number of cycles to failure N (x-axis, log scale ) ke against plot. Ise Wöhler curve bhi kehte hain.
KAISE banta hai: Bahut saare identical specimens lo. Har ek ko fixed amplitude S i par load karo jab tak woh toot na jaye, N i record karo. Points plot karo. Zyada stress → kam cycles survived.
Empirically, log S vs log N plot karne par high-cycle regime mein ek straight line milti hai. Log–log mein straight line ka matlab hai power law. Ise build karte hain:
log S = log σ f ′ − b log N
Kyun? Ek line y = c − b x jahan y = log S , x = log N hai. Dono sides exponentiate karo:
S = σ f ′ N − b ⟺ N = ( σ f ′ S ) − 1/ b
σ f ′ = fatigue strength coefficient (N = 1 par stress).
b = Basquin exponent (slope magnitude, small positive number ~0.05–0.12).
Definition Endurance (fatigue) limit
S e
Ek endurance limit woh stress hai jiske neeche S–N curve flat ho jaata hai aur material effectively infinite cycles (> 1 0 6 –1 0 7 ) tak survive karta hai. Ek true, well-defined endurance limit mainly plain-carbon aur low-alloy steels (aur kuch doosre BCC ferrous alloys) mein dikhai deta hai.
Aluminium, zyaatar non-ferrous alloys, AUR zyaatar common titanium alloys (jaise Ti-6Al-4V) mein true endurance limit NAHI hota — unki S–N curves 1 0 7 cycles ke baad bhi slope karte rehti hain. Inke liye infinite-life stress ki jagah "N = 5 × 1 0 8 par fatigue strength" quote ki jaati hai.
Asli problem: Real loading amplitudes ka ek mix hota hai (n 1 cycles S 1 par, n 2 S 2 par, …). S–N curve sirf N i = life batata hai agar woh amplitude akele act kare . Hum unhe kaise combine karein?
Intuition Damage-fraction idea
Maano amplitude S 1 part ko N 1 cycles mein khatam kar dega. Toh ek cycle total life ka 1/ N 1 use karta hai — "budget" ka ek hissa. n 1 aise cycles lagao → tumne n 1 / N 1 budget kharch kar diya. Saare fractions add karo. Failure tab hogi jab budget 1 (100%) hit ho jaye.
Assume karo ki damage linearly accumulate hoti hai aur level i par har cycle D total unit damage ka equal share N i 1 contribute karta hai. n i cycles level i par:
D i = N i n i
Total damage sum hai (independent contributions):
D = ∑ i D i = ∑ i N i n i
Failure criterion: damage pura budget reach kar le.
perfect kyun nahi ho sakta
Miner's rule loading order aur interaction ko ignore karta hai. Reality mein ek bada overload compressive residual stress chhod sakta hai jo baad ki cracking ko slow kar deta hai (isliye real D f ai l 0.7–2.0 tak ho sakta hai). Engineers safety factors isliye use karte hain kyunki D = 1 ek average hai, guarantee nahi.
Worked example Example 1 — Basquin se life padhna
Ek Ti strut mein σ f ′ = 1800 MPa, b = 0.09 hai. Cyclic amplitude S = 300 MPa. Cycles to failure N nikalo.
Step 1: S = σ f ′ N − b ⇒ N = ( S / σ f ′ ) − 1/ b use karo.
Ye step kyun? Hum S jaante hain; N pane ke liye Basquin ko invert karo.
Step 2: S / σ f ′ = 300/1800 = 0.1667 .
N = 0.166 7 − 1/0.09 = 0.166 7 − 11.11 .
Kyun? − 1/ b = − 11.11 .
Step 3: ln N = − 11.11 ln ( 0.1667 ) = − 11.11 ( − 1.7918 ) = 19.9 ⇒ N = e 19.9 ≈ 4.4 × 1 0 8 cycles.
Kyun? Log power ko tractable banata hai; wapas exponentiate karo.
Worked example Example 2 — Miner's rule mixed spectrum
Ek bracket dekhta hai: n 1 = 1 0 4 cycles S 1 par (jahan N 1 = 1 0 5 ) aur n 2 = 2 × 1 0 4 cycles S 2 par (jahan N 2 = 5 × 1 0 4 ). Kya yeh fail hoga?
Step 1: D 1 = n 1 / N 1 = 1 0 4 /1 0 5 = 0.10 . Kyun? Level 1 par use ki gayi life ka fraction.
Step 2: D 2 = n 2 / N 2 = 2 × 1 0 4 /5 × 1 0 4 = 0.40 . Kyun? Wahi, level 2.
Step 3: D = 0.10 + 0.40 = 0.50 < 1 . Kyun? Fractions ka sum < budget → survives , fatigue life ka 50% use ho chuka hai.
Worked example Example 3 — Naye level par allowable cycles
Wahi bracket pehle se D = 0.50 par hai. N 3 = 8 × 1 0 4 wale level par failure se pehle aur kitne cycles n 3 ho sakte hain?
Step 1: Remaining budget = 1 − 0.50 = 0.50 . Kyun? Aadhi life pehle se khatam ho chuki hai.
Step 2: n 3 / N 3 = 0.50 ⇒ n 3 = 0.50 × 8 × 1 0 4 = 4 × 1 0 4 cycles. Kyun? Remaining damage ko available budget ke barabar set karo aur solve karo.
Common mistake "Yield stress se neeche → kabhi failure nahi."
Kyun sahi lagta hai: Statics kehta hai stress < yield ⇒ elastic ⇒ safe. Kyun galat hai: Fatigue cracks microscopic stress concentrators par nucleate hoti hain; cyclic loading unhe elastic range mein bhi badhata hai. Fix: Sirf static margin nahi, S–N life zaroor check karo.
Stresses add karo, fractions nahi."
Kyun sahi lagta hai: Elastic stresses ke liye superposition kaam karta hai. Kyun galat hai: Miner's rule damage fractions n i / N i sum karta hai, jo dimensionless life-usages hain, stresses nahi. Fix: Har level ke liye S–N curve par jao, N i nikalo, phir n i / N i sum karo.
Common mistake "Titanium mein steel jaisi endurance limit hoti hai."
Kyun sahi lagta hai: Dono strong aerospace metals hain, aur textbook S–N pictures mein ek flat asymptote dikhai deta hai. Kyun galat hai: True endurance limit mainly ek plain-carbon/low-alloy steel feature hai; common Ti alloys (Ti-6Al-4V) aur aluminium 1 0 7 cycles ke baad bhi slope karte rehte hain. Fix: Ti aur Al ke liye ek specified high-cycle strength tak design karo, kabhi "infinite life" tak nahi.
Common mistake "Loading ka order matter nahi karta (Miner exact hai)."
Kyun sahi lagta hai: Formula sum mein symmetric hai. Fix: Ye ek linear approximation hai; overloads aur sequence real life badal dete hain. Safety factors use karo; D f ai l tests mein ~0.3–3 tak range karta hai.
Recall 12-year-old ko samjhao (Feynman)
Ek paperclip ko aage-peeche moro. Ek baar modna = theek hai. Lekin har wiggle ek tiny crack ko thoda aur bada karta hai, aur kaafi wiggles ke baad — snap! — chahe tumne kabhi zyada zyada nahi khींcha. S–N curve ek chart hai jo kehta hai "ITNA zyada wiggle karo → ITNE wiggles tak chalega." Miner's rule ek phone battery ki tarah hai: har hard wiggle zyada battery drain karta hai, har soft wiggle kam. Jab battery khatam ho jaaye (100%), metal toot jaata hai.
"Small Nudges Break Metal" — S –N curve life batata hai; B asquin line hai; M iner's rule drained fractions ko 1 tak sum karta hai. Aur "Damage adds to ONE, or you're not DONE."
Fatigue failure kya hai? Cyclic loading se progressive damage aur failure jo static ultimate strength se neeche ki stresses par hoti hai.
S–N (Wöhler) curve ke do axes kya hain? Stress amplitude S (y) vs cycles-to-failure N log scale par (x).
Basquin's law state karo. S = σ f ′ N − b , yaani log–log mein ek straight line; b Basquin exponent hai.
Endurance limit kya hai aur kis material mein true endurance limit hoti hai? Ek stress jiske neeche life effectively infinite hoti hai; well-defined limit mainly plain-carbon/low-alloy steels mein hoti hai. Aluminium aur zyaatar common Ti alloys (Ti-6Al-4V) mein NAHI hoti.
Palmgren–Miner's rule state karo. Failure tab jab ∑ i n i / N i = 1 ; har cycle total life ka 1/ N i fraction use karta hai.
Miner's rule mein n i / N i kya represent karta hai? Stress level i par n i cycles dwara consume ki gayi total fatigue life ka fraction.
Mean vs amplitude stress formulas? σ m = ( σ ma x + σ min ) /2 , σ a = ( σ ma x − σ min ) /2 .
Miner's rule sirf approximate kyun hai? Ye load sequence/interaction ignore karta hai; real failure damage sum ~0.3–3 tak range karta hai, isliye safety factors use kiye jaate hain.
σ f ′ = 1800 MPa, b = 0.09 , S = 300 MPa diya hai, N nikalo.N = ( 300/1800 ) − 1/0.09 ≈ 4.4 × 1 0 8 cycles.
Spacecraft par cyclic load ke do sources? Launch vibration/acoustics aur on-orbit thermal cycling (~90-min din/raat).
Stress and Strain — cyclic stress amplitude driver hai
Random Vibration & PSD — launch loads S–N spectrum feed karte hain
Thermal Cycling on Orbit — main long-term fatigue source
Fracture Mechanics & Crack Growth (Paris' Law) — S–N ke peeche physical mechanism
Safety Factors & Margins of Safety — kyun hum D < 1 tak design karte hain
Launch Loads & Environments