Fracture mechanics — stress intensity factor K, toughness K_IC: Exercises

3.6.9 · D4 · Physics › Spacecraft Structures & Systems Engineering › Fracture mechanics — stress intensity factor K, toughness K_

Shuru karne se pehle, plain words mein ek reminder taaki koi symbol mystery na rahe:

Figure — Fracture mechanics — stress intensity factor K, toughness K_IC

Figure dekho: wahi formula, do crack pictures. Left mein crack edge se shuru hoti hai aur depth $a$ tak pahunchti hai. Right mein crack beech mein hoti hai, total mouth-to-tip length $2 a$ , toh hum $a$ = uska aadha feed karte hain. Yeh "konsi length use karoon" sahi karna sabse common galti hai — figure har problem ke liye tumhara anchor hai.

L1 · Recognition

Exercise 1.1

$K_{I}$ ki units batao, aur ek sentence mein kaho ki $K_{I C}$ kya hai.

Recall Solution

$K_{I}$ ki units hain stress × √length, yani $MPa m$ . $K_{I C}$ material ki fracture toughness hai: $K_{I}$ ki critical value jis par ek Mode-I crack unstable (runaway) propagation shuru karta hai. Yeh measure ki jaati hai, calculate nahi, aur part ki shape par depend nahi karti.

Exercise 1.2

Ek wide plate mein centre crack ki total length $2 a = 10 mm$ hai. $K_{I} = σ π a Y$ mein $a$ ki kya value jaayegi, aur $Y$ kya hai?

Recall Solution

Hum kya karte hain: total length ko aadha karo, kyunki formula centre crack ke liye half-length use karta hai (figure, right panel dekho). $a = \frac{1}{2} (10 mm) = 5 mm = 0.005 m, Y = 1.0.$

L2 · Application

Exercise 2.1

Ek titanium panel ( $Y = 1.0$ , centre crack) $σ = 300 MPa$ ke centre crack of total length $2 a = 6 mm$ par carry karta hai. $K_{I}$ compute karo.

Recall Solution

Step 1 — half length: $a = 3 mm = 0.003 m$ . Step 2 — plug in: $K_{I} = 300 π \times 0.003 \times 1.0.$ $π a = 3.1416 \times 0.003 = 0.009425 m$ . $0.009425 = 0.09708 m^{1/2}$ . $K_{I} = 300 \times 0.09708 = 29.1 MPa m .$ (MPa mein kaam karne se answer directly $MPa m$ mein milta hai — pascals mein convert karne ki zaroorat nahi.)

Exercise 2.2

Ek hull skin mein hoop stress $σ = 180 MPa$ ke neeche depth $a = 2.5 mm$ ( $Y = 1.12$ ) ka edge crack hai. $K_{I}$ nikalo.

Recall Solution

Edge crack: poori depth use karo, $a = 0.0025 m$ . $π a = 3.1416 \times 0.0025 = 0.007854 m$ . $0.007854 = 0.08862 m^{1/2}$ . $K_{I} = 180 \times 0.08862 \times 1.12 = 17.9 MPa m .$

L3 · Analysis

Exercise 3.1

Aluminium 2024-T3 ka $K_{I C} = 35 MPa m$ hai. Ek pressure hull region $σ = 150 MPa$ ke saath edge crack, $Y = 1.12$ carry karta hai. Koi safety factor ignore karte hue ( $n = 1$ ), kaunsi crack depth $a_{c}$ par part fracture karega?

Recall Solution

Hum kya karte hain: applied $K_{I}$ ko material limit ke barabar set karo aur $a$ ke liye solve karo. $σ π a_{c} Y = K_{I C} \Rightarrow a_{c} = \frac{1}{π} (\frac{K _{I C}}{Y σ})^{2} .$ Yeh rearrangement kyun: hume crack size chahiye, toh $a$ ko isolate karo — $Y σ$ se divide karne ke baad dono sides ko square karo. $\frac{K _{I C}}{Y σ} = \frac{35}{1.12 \times 150} = \frac{35}{168} = 0.20833 m^{1/2} .$ $(0.20833)^{2} = 0.043403, a_{c} = \frac{0.043403}{3.1416} = 0.013815 m \approx 13.8 mm .$ ~13.8 mm se gehri crack 150 MPa par is hull ko fracture kar deti hai.

Exercise 3.2

Usi hull ke liye, safety factor $n = 2$ add karo. Inspectors ko maximum allowable crack size $a_{m a x}$ kya pakadni chahiye?

Recall Solution

Kya badalta hai: ab hum require karte hain $K_{I} \leq K_{I C} / n$ , toh $K_{I C}$ ki jagah $K_{I C} / n$ rakh do: $a_{m a x} = \frac{1}{π} (\frac{K _{I C}}{Y σ n})^{2} .$ $\frac{K _{I C}}{Y σ n} = \frac{35}{1.12 \times 150 \times 2} = \frac{35}{336} = 0.104167 m^{1/2} .$ $(0.104167)^{2} = 0.010851, a_{m a x} = \frac{0.010851}{3.1416} = 0.003454 m \approx 3.45 mm .$ Gaur karo $a_{m a x}$ , $a_{c}$ ka ek chauthai hai (13.8/3.45 ≈ 4). Yeh accident nahi hai — ek squared term ke andar factor $n = 2$ allowable crack ko $n^{2} = 4$ se chota kar deta hai.

L4 · Synthesis

Exercise 4.1

Ek titanium truss (Ti-6Al-4V, $K_{I C} = 55 MPa m$ ) mein edge crack $Y = 1.12$ hai aur $σ = 200 MPa$ , safety factor $n = 2$ par operate karta hai. (a) $a_{m a x}$ nikalo. (b) NDT (dekho Non-destructive testing (NDT)) reliably $1.5 mm$ tak ke cracks dhundhta hai. Kya safe margin hai?

Recall Solution

(a) $\frac{K _{I C}}{Y σ n} = \frac{55}{1.12 \times 200 \times 2} = \frac{55}{448} = 0.122768 m^{1/2} .$ $(0.122768)^{2} = 0.015072, a_{m a x} = \frac{0.015072}{3.1416} = 0.004797 m \approx 4.80 mm .$ (b) Sabse chhota detectable crack (1.5 mm) maximum tolerable crack (4.80 mm) se kaafi neeche hai, toh haan, margin hai — inspection roughly $4.80/1.5 = 3.2 \times$ chhote cracks pakadti hai danger size se. Yahi Damage tolerance philosophy ka essence hai: design aisa karo ki critical crack comfortably us size se bada ho jo inspection dekh sakti hai.

Exercise 4.2

Wahi truss crack fatigue se steady rate $0.4 mm/year$ se grow karti hai (ek Fatigue crack growth (Paris law) regime jo hum yahan constant treat karte hain). Yeh just-detected size $1.5 mm$ se shuru hoti hai. Kitne saal mein yeh $a_{m a x} = 4.80 mm$ tak pahunchegi, aur kaunsa inspection interval ek visit spare ke saath safe rakhega?

Recall Solution

Step 1 — remaining crack budget: $Δ a = 4.80 - 1.5 = 3.30 mm$ . Step 2 — $a_{m a x}$ tak pahunchne ka time: $t = \frac{Δ a}{rate} = \frac{3.30 mm}{0.4 mm/yr} = 8.25 years .$ Step 3 — inspection interval: yeh guarantee karne ke liye ki crack critical hone se pehle kam se kam ek inspection ho, us life ka aadha inspect karo: $8.25/2 \approx 4.1 years$ . Conservatism ke liye round down karo, har 4 saal mein inspect karo.

L5 · Mastery

Exercise 5.1

Tumhe $σ = 250 MPa$ , edge crack $Y = 1.12$ , safety factor $n = 1.5$ , aur NDT jo $2.0 mm$ tak ke cracks detect kar sakta hai, ke liye ek bracket ka material chunna hai. Do candidates:

Steel 4340: $K_{I C} = 50 MPa m$
CFRP: $K_{I C} = 25 MPa m$

Har ek ke liye $a_{m a x}$ compute karo. Kaun sa material genuine inspection margin deta hai (yaani $a_{m a x} > 2.0 mm$ )?

Recall Solution

$a_{m a x} = \frac{1}{π} (\frac{K _{I C}}{Y σ n})^{2}$ use karo jahan $Y σ n = 1.12 \times 250 \times 1.5 = 420$ .

Steel 4340: $\frac{50}{420} = 0.119048, (0.119048)^{2} = 0.014172, a_{m a x} = \frac{0.014172}{3.1416} = 0.004511 m \approx 4.51 mm .$ CFRP: $\frac{25}{420} = 0.059524, (0.059524)^{2} = 0.003543, a_{m a x} = \frac{0.003543}{3.1416} = 0.001128 m \approx 1.13 mm .$

Decision: Steel ka $a_{m a x} = 4.51 mm$ 2.0 mm detection floor se comfortably upar hai — margin ke saath safe. CFRP ka $a_{m a x} = 1.13 mm$ us size se chhota hai jo NDT dekh bhi sakta hai (2.0 mm) — ek critical crack exist kar sakti hai aur inspection mein undetected nikal jaaye. CFRP yahan damage-tolerance test fail karta hai aur use reject karna chahiye (ya $σ$ kam karke redesign karo).

Exercise 5.2 (mixed-mode capstone)

Ek aluminium fitting ( $K_{I C} = 35 MPa m$ ) mein ek crack combined loading experience karta hai: $K_{I} = 20 MPa m$ aur $K_{I I} = 12 MPa m$ (Mode III negligible). Parent note se effective stress intensity $K_{eff} = K_{I}^{2} + K_{I I}^{2} + K_{I I I}^{2}$ use karke, kya fitting fracture karega?

Recall Solution

Hum kya karte hain: mode contributions ko ek number mein combine karo aur toughness se compare karo. $K_{eff} = 2 0^{2} + 1 2^{2} + 0^{2} = 400 + 144 = 544 = 23.32 MPa m .$ Kyunki $K_{eff} = 23.3 < K_{I C} = 35$ hai, fitting fracture nahi karta — yeh toughness ka $23.3/35 = 67%$ par hai, yaani roughly $1.5$ ka effective safety factor.

Recall Self-test cloze

Stress intensity factor ka formula ::: $K_{I} = σ π a Y$ Total length $2 a$ ke centre crack ke liye, formula mein use ki jaane wali value hai ::: half-length $a$ Maximum allowable crack size safety factor ke saath scale karti hai ::: $(1/ n)^{2}$ ke hisaab se — $n$ double karne se crack budget quarter ho jaata hai Safe inspection interval zyada se zyada hona chahiye ::: crack ki remaining life ka aadha (two-inspection rule) Mixed mode ke liye effective stress intensity ::: $K_{eff} = K_{I}^{2} + K_{I I}^{2} + K_{I I I}^{2}$