Worked examples — Fracture mechanics — stress intensity factor K, toughness K_IC

3.6.9 · D3 · Physics › Spacecraft Structures & Systems Engineering › Fracture mechanics — stress intensity factor K, toughness K_

Upar diye kisi bhi symbol ko hum bina ek-line meaning ke use nahi karenge.

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The scenario matrix

Fracture problems ko ek grid ki tarah socho. Har row ek tarah ka sawaal hai jo universe (ya exam) pooch sakta hai. Agar tum har row se ek example solve kar sako, to tum is topic ka koi bhi problem solve kar sakte ho.

#	Cell (case class)	Kya alag hai	Example
A	Edge crack, $K_I$ nikalo	$Y=1.12$, poori length $a$ use karo	Ex 1
B	Centre crack, $K_I$ nikalo	$Y=1.0$, lekin half-length $a$ use karo	Ex 2
C	Failure decision ($K_I$ vs $K_{IC}$)	compare karo, output "safe / breaks"	Ex 3
D	Critical crack size $a_{\text{max}}$ solve karo	formula ko invert karo	Ex 4
E	Critical stress ${\sigma }_c$ solve karo	doosri taraf se invert karo	Ex 5
F	Degenerate / limiting input	$a\to 0$, huge $\sigma$, bahut brittle material	Ex 6
G	Mixed-mode ($K_I$ aur $K_{II}$)	$K_{\text{eff}}$ se combine karo	Ex 7
H	Real-world word problem (inspection interval)	fracture + crack growth in time	Ex 8
I	Exam-style twist (units / doubling trap)	scaling, unit landmines	Ex 9

Hum har cell neeche cover karte hain. Har example tumse pehle answer ka ballpark forecast karwata hai — guessing karna intuition ko reading se kahin zyada sharpen karta hai.

Map figure kaise padhein

Yeh figure kaise padhein. Do identical plates hain jinhein top-aur-bottom se yellow stress arrows $\sigma$ se kheecha ja raha hai — woh vertical pull hi horizontal crack ko open karta hai. Left mein, red line ek edge crack hai: yeh plate ki free left surface se shuru hoti hai aur uski labelled length wahi poori $a$ hai jo tum plug in karte ho, $Y=1.12$ ke saath. Right mein, blue line ek centre crack hai: yeh beech mein dabi hui hai, uska total span $2a$ hai (green arrows), aur tumhe half-length $a$ (blue label) plug in karni hai $Y=1.0$ ke saath. Yeh left-vs-right split hi #1 cheez hai jise students mix up karte hain — jab bhi problem mein "edge" ya "centre" aaye, yahan glance karo. Hum left panel Ex 1 aur Ex 3 mein use karte hain, right panel Ex 2 aur Ex 5 mein.

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Cell A — Edge crack, $K_I$ nikalo

Forecast: $\sigma$ parent note ke $150\text{MPa}$ example se thoda bada hai, aur crack bhi thodi lambi hai, toh expect karo $K_I$ kahin high-teens $\text{MPa}\sqrt{\text{m}}$ mein hoga.

1. Crack length ko metres mein convert karo. $a=3\text{mm}=0.003\text{m}$. Yeh step kyun? $K$ ki unit mein $\sqrt{\text{m}}$ tab hi kaam karta hai jab $a$ metres mein ho. Millimetres use karne par answer $\sqrt{1000}\approx 31.6\times$ bahut chota aata.
2. $\sqrt{\pi a}$ compute karo. $\pi a=3.1416\times 0.003=0.009425$, toh $\sqrt{0.009425}=0.09708{\text{m}}^{1/2}$. Yeh step kyun? Yeh "size factor" hai — sirf crack ki length akele stress ko kitna amplify karti hai, geometry se pehle.
3. Sab kuch multiply karo. $K_I=1.12\times 180\times 0.09708=19.6\text{MPa}\sqrt{\text{m}}$. Yeh step kyun? $K_I=Y\sigma \sqrt{\pi a}$ — yahi poora formula hai, assembled.

$\boxed{K_I=19.6\text{MPa}\sqrt{\text{m}}}$

Verify: Units: $[\text{MPa}]\times [{\text{m}}^{1/2}]=\text{MPa}\sqrt{\text{m}}$ ✓. Ballpark forecast (high-teens) se match kiya ✓. Aluminium ke $K_{IC}=35$ ke against, yeh hull safe hai (19.6 < 35).

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Cell B — Centre crack, $K_I$ nikalo (halving trap)

Forecast: Unhone total length $10\text{mm}$ di, lekin centre cracks half-length use karte hain. Agar bhool ke $0.010\text{m}$ use karoge toh answer $\sqrt{2}\approx 1.41\times$ zyada aayega. Guess ~$27\text{MPa}\sqrt{\text{m}}$.

1. Width ko halve karo. Total $2a=10\text{mm}\Rightarrow a=5\text{mm}=0.005\text{m}$. Yeh step kyun? Formula ka $a$ centre se ek tip tak measure hota hai. Poora crack symmetric hai, aur $K_I$ ek tip ke field ko describe karta hai. Poori width use karna double-count karna hai.
2. Size factor. $\sqrt{\pi \times 0.005}=\sqrt{0.015708}=0.12533{\text{m}}^{1/2}$. Yeh step kyun? Pehle wali hi size factor hai — yeh geometry-independent core hai.
3. Assemble karo. $K_I=1.0\times 220\times 0.12533=27.6\text{MPa}\sqrt{\text{m}}$. Yeh step kyun? $Y=1.0$ isliye kyunki buried crack ke paas koi free surface nahi hoti jo use open hone mein help kare.

$\boxed{K_I=27.6\text{MPa}\sqrt{\text{m}}}$

Verify: Agar galti se $a=0.010$ use karte toh $220\times \sqrt{\pi \cdot 0.010}=39.0$ milta — poora $41\%$ zyada. Halving matter karta hai. Ti ke $K_{IC}=55$ ke against, phir bhi safe hai ✓.

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Cell C — Kya yeh tootega? (the decision)

Forecast: Ceramics bahut brittle hote hain (tiny $K_{IC}$). Moderate stress ke under chota sa crack bhi $4$ ko cross kar sakta hai. Guess: yeh toot jayega.

1. Driving force $K_I$ compute karo. $a=0.001\text{m}$, $\sqrt{\pi \cdot 0.001}=0.05605{\text{m}}^{1/2}$; $K_I=1.12\times 60\times 0.05605=3.77\text{MPa}\sqrt{\text{m}}$. Yeh step kyun? Compare karne se pehle, woh number chahiye jo material feel kar raha hai.
2. Toughness se compare karo. Kya $K_I\ge K_{IC}$? $3.77$ vs $4.0$ → $3.77<4.0$. Yeh step kyun? Yahi inequality hai failure criterion — sirf yahi test survival decide karta hai.
3. Verdict batao. $K_I<K_{IC}$, toh tile abhi survive karti hai — lekin margin almost nahi hai.

$\boxed{K_I=3.77<K_{IC}=4.0\Rightarrow \text{no fracture (barely)}}$

Verify: Sirf $4.0/3.77=1.06$ ka margin hai, matlab $6\%$ stress increase (ya crack ka $\sim 1.1\text{mm}$ tak badhna) ise failure mein daal deta hai. Isliye ceramics ko Non-destructive testing (NDT) aur generous safety factors wali Damage tolerance philosophy ki zaroorat hoti hai ✓.

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Cell D — Critical crack size $a_{\text{max}}$ solve karo

Forecast: High stress + safety factor of 2 se $a_{\text{max}}$ kuch millimetres tak squeeze hoga.

1. Safe condition likho. $K_I\le K_{IC}/n$, yaani limit par $Y\sigma \sqrt{\pi a}=K_{IC}/n$. Yeh step kyun? $K_{IC}$ ko $n$ se divide karna margin andar build karta hai — hum design karte hain jaise material aadha tough ho.
2. Square root isolate karo. Dono sides ko $Y\sigma$ se divide karo: $\sqrt{\pi a}=\frac{K_{IC}}{Y\sigma n}$ Yeh step kyun? Hume $a$ akela chahiye, toh pehle $\sqrt{\pi a}$ block ko multiply kar rahi sari cheezein hata dete hain.
3. Dono sides ko square karo square root khatam karne ke liye: $\pi a={\left(\frac{K_{IC}}{Y\sigma n}\right)}^2$ Yeh step kyun? $a$ square root ke andar fansa hai; squaring bilkul wahi inverse operation hai jo ise azaad karta hai — aur right-hand side bhi square ho jata hai.
4. $\pi$ se divide karo $a$ ko isolate karne ke liye: $a_{\text{max}}=\frac{1}{\pi }{\left(\frac{K_{IC}}{Y\sigma n}\right)}^2$ Yeh step kyun? $\pi$ left side par $a$ ko multiply kar raha tha, toh woh right side par neeche chala jata hai — isliye "$1/\pi$" aata hai.
5. Numbers plug in karo. Inner fraction: $\frac{50}{1.12\times 300\times 2}=\frac{50}{672}=0.07440$. Square: $0.005535$. $\pi$ se divide: $a_{\text{max}}=0.001762\text{m}=1.76\text{mm}$. Yeh step kyun? Numbers design equation ko inspection spec mein badal dete hain.

$\boxed{a_{\text{max}}=1.76\text{mm}}$

Verify: $a=0.001762$ back-substitute karo: $K_I=1.12\times 300\times \sqrt{\pi \cdot 0.001762}=1.12\times 300\times 0.07440=25.0\text{MPa}\sqrt{\text{m}}=K_{IC}/n=50/2$ ✓. Inspection (ultrasonic ~1 mm) ko $1.76\text{mm}$ se chhoti cracks reliably dhundhni chahiye ✓.

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Cell E — Critical stress ${\sigma }_c$ solve karo

Forecast: Chota crack + decent toughness → high critical stress, likely kuch sau MPa.

1. Fracture par $K_I=K_{IC}$ set karo. $Y{\sigma }_c\sqrt{\pi a}=K_{IC}$. Yeh step kyun? Fracture tab hota hai jab exactly driving force toughness ke barabar ho — yahi woh boundary hai jis par hum solve karte hain.
2. ${\sigma }_c$ isolate karo. ${\sigma }_c=\frac{K_{IC}}{Y\sqrt{\pi a}}$. Yeh step kyun? Yahan $\sigma$ unknown hai, $a$ nahi — wahi equation, alag variable free hua.
3. Compute karo. $\sqrt{\pi \cdot 0.002}=0.07927{\text{m}}^{1/2}$; ${\sigma }_c=\frac{25}{1.0\times 0.07927}=315.4\text{MPa}$. Yeh step kyun? Yeh woh load hai jo is crack ke present rehte boom par kabhi nahi aana chahiye.

$\boxed{{\sigma }_c\approx 315\text{MPa}}$

Verify: ${\sigma }_c$ back feed karo: $1.0\times 315.4\times 0.07927=25.0=K_{IC}$ ✓. Griffith energy criterion ki intuition se match karta hai — chota crack bada stress tolerate karta hai ✓.

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Cell F — Degenerate & limiting inputs

Forecast: (a) zero hona chahiye; (b) tiny toughness → tiny critical stress; (c) unbounded.

1. (a) $a=0$. $K_I=1.12\times 100\times \sqrt{\pi \cdot 0}=0$. Yeh step kyun? Koi crack nahi wale body mein koi stress-intensity singularity nahi hoti — fracture mechanics apply hi nahi hoti, aur $K_I\to 0$ confirm karta hai ki formula gracefully degrade karta hai. Failure phir ordinary yield-strength checks par revert ho jati hai.
2. (b) Critical stress. ${\sigma }_c=\frac{K_{IC}}{Y\sqrt{\pi a}}=\frac{2}{1.12\times \sqrt{\pi \cdot 0.001}}=\frac{2}{1.12\times 0.05605}=31.9\text{MPa}$. Yeh step kyun? Dikhata hai ki low-$K_{IC}$ material kitna brutally limited hota hai — 1 mm flaw stress ko ~32 MPa par cap kar deta hai, jo typical structural loads se kahin neeche hai.
3. (c) $\sigma \to \infty$. Kyunki $K_I$, $\sigma$ mein linear hai, $K_I\to \infty$: koi bhi crack sufficiently high stress par fail ho jata hai. Yeh step kyun? Confirm karta hai ki koi infinitely strong cracked body nahi hoti — har crack ka ek finite critical stress hota hai ${\sigma }_c=K_{IC}/(Y\sqrt{\pi a})$.

$\boxed{(a)K_I=0(b){\sigma }_c=31.9\text{MPa}(c)K_I\to \infty }$

Verify: (b) back-check: $1.12\times 31.9\times 0.05605=2.00=K_{IC}$ ✓. Limiting behaviours monotone aur sign-consistent hain ✓.

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Cell G — Mixed-mode fracture

Forecast: Koi bhi mode akele $22$ exceed nahi karta, lekin combined kar ke kar sakta hai. Guess: unsafe.

1. Present modes identify karo. Opening $K_I=18$, sliding $K_{II}=10$, tearing $K_{III}=0$ (callout mein define kiye). Yeh step kyun? Combine karne se pehle tumhe pata hona chahiye ki har $K$ kaunsi physical loading represent karta hai.
2. $K_{\text{eff}}$ se combine karo. $K_{\text{eff}}=\sqrt{K_I^2+K_{II}^2+K_{III}^2}=\sqrt{18^2+10^2+0^2}=\sqrt{324+100}=\sqrt{424}=20.6\text{MPa}\sqrt{\text{m}}$. Yeh step kyun? Modes perpendicular components ki tarah add hote hain — root-sum-of-squares.
3. Compare karo. $K_{\text{eff}}=20.6<K_{IC}=22$ → safe, lekin sirf $1.07\times$ margin. Yeh step kyun? Mixed-mode ko pure Mode I ki tarah same toughness limit ke against judge kiya jata hai.

$\boxed{K_{\text{eff}}=20.6\text{MPa}\sqrt{\text{m}}<22\Rightarrow \text{safe (thin margin)}}$

Verify: $18^2+10^2=424$, $\sqrt{424}=20.59$ ✓. Note karo $K_{\text{eff}}>K_I$: Mode II ignore karne par falsely $22/18=1.22$ margin declare hota — mixed-mode analysis optional nahi hai ✓.

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Cell H — Real-world word problem (inspection interval)

Forecast: Gap ~$1.16\text{mm}$ hai at $0.25\text{mm/yr}$ → roughly 4–5 saal critical tak; pehle inspect karo.

1. Growth allowance nikalo. $\Delta a=a_{\text{max}}-a_0=1.76-0.60=1.16\text{mm}$. Yeh step kyun? Crack pehle se part-grown hai; sirf bacha hua margin time deta hai.
2. Critical tak time. $t=\frac{\Delta a}{\text{rate}}=\frac{1.16}{0.25}=4.64\text{years}$. Yeh step kyun? Constant growth rate ise ek simple distance/speed calculation bana deta hai. (Real fatigue Fatigue crack growth (Paris law) follow karta hai, jo accelerate karta hai — toh yeh linear estimate optimistic hai.)
3. Interval set karo. Time-to-critical ka aadha inspect karo: $4.64/2=2.32\text{years}$, round down karke 2-year inspection interval. Yeh step kyun? Halving guarantee karta hai ki kam se kam ek inspection crack ke $a_{\text{max}}$ tak pahunchne se pehle hogi — standard Damage tolerance philosophy "two-lives" rule.

$\boxed{t_{\text{crit}}\approx 4.6\text{yr}\Rightarrow \text{re-inspect every 2 years}}$

Verify: 4 saal baad: $a=0.6+4\times 0.25=1.6\text{mm}<1.76$ abhi safe; 5 saal baad: $1.85>1.76$ fail — toh critical year 4 aur 5 ke beech hai, $4.64$ yr se consistent ✓.

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Cell I — Exam-style twist (scaling & unit traps)

Forecast: (a) $a$ double karne par $\sqrt{2}\approx 1.41$ ka factor aayega; (b) 531 absurdly bada hai — unit error hai.

1. (a) Scaling. $K_I\propto \sqrt{a}$. $a$ double → factor $\sqrt{2}$: $K_I^{\prime }=12\times \sqrt{2}=16.97\text{MPa}\sqrt{\text{m}}$. Yeh step kyun? Kyunki $a$ square root ke andar hai, $K$ sub-linearly badhta hai — yeh ek crucial aur frequently-tested fact hai. $K$ ko double karne ke liye $4\times$ crack chahiye.
2. (b) Error spot karo. Student ne $a$ millimetres mein use kiya. Sahi: $a=0.004\text{m}$. Yeh step kyun? $\sqrt{\text{m}}$ vs $\sqrt{\text{mm}}$ mein $\sqrt{1000}\approx 31.6$ ka difference hai — bilkul wahi factor $531/16.8\approx 31.6$ jisne answer inflate kiya.
3. (b) Sahi value. $K_I=150\times \sqrt{\pi \times 0.004}=150\times 0.11210=16.8\text{MPa}\sqrt{\text{m}}$. Yeh step kyun? Physically sane number, real toughness values se comparable.

$\boxed{(a)K_I^{\prime }=17.0(b)K_I=16.8\text{MPa}\sqrt{\text{m}}}$

Verify: (a) $12\sqrt{2}=16.97$ ✓. (b) Student ka number $150\sqrt{\pi \cdot 4}=531.7$ sahi $16.8$ se $\sqrt{1000}=31.6\times$ zyada hai — yeh metre/millimetre ka tell-tale signature hai ✓. Stress concentration factors se related hai, aur $J$-integral ise J-integral mein generalize karta hai.

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Recall Self-test (guess karne ke baad reveal karo)

Edge crack $a=2$mm mein kaun sa Y use hoga? ::: $Y=1.12$ Total width $2a=6$mm wale centre crack mein metres mein kya $a$ use hoga? ::: $a=0.003$m (6 mm ka aadha) Crack length double karne par $K_I$ kitne se multiply hoga? ::: $\sqrt{2}\approx 1.41$ Fracture kab hota hai? ::: $K_I\ge K_{IC}$ $K_I,K_{II}$ wale do modes kaise combine hote hain? ::: $K_{\text{eff}}=\sqrt{K_I^2+K_{II}^2}$ $K_I\approx 500\text{MPa}\sqrt{\text{m}}$ milna usually matlab? ::: crack length mm mein rahi, m mein convert nahi ki