Fracture mechanics — stress intensity factor K, toughness K_IC: Question bank

3.6.9 · D5 · Physics › Spacecraft Structures & Systems Engineering › Fracture mechanics — stress intensity factor K, toughness K_

Shuru karne se pehle, is page mein use hone waale har symbol ko clearly define karte hain taaki koi bhi trap undefined notation ke peeche na chupi rahe.

Recall Do symbols jo yeh page baar baar test karta hai

$K_{I}$ hai stress intensity factor: ek number jo aap apne load aur crack se compute karte ho ( $K_{I} = σ π a Y$ ). $K_{I C}$ hai fracture toughness: ek number jo material ka apna hota hai, lab mein measure kiya jaata hai. Fracture tab hota hai jab computed wala owned wale tak pahunch jaata hai: $K_{I} \geq K_{I C}$ .

Geometry	$Y$ (approx.)	Kyun
Centre crack, infinite plate	$1.00$	Ideal reference case
Single edge crack, wide plate	$1.12$	Free edge crack ko zyaada aasaani se khulne deta hai
Edge crack, finite width (crack section ka zyaada hissa fill karta hai)	$> 1.2$ , badhta hua	Load carry karne ke liye kam material bachta hai
Embedded penny-shaped crack	$\frac{2}{π} \approx 0.64$	Surrounding material opening ko rokta hai

True or false — justify karein

TF1 — "

K_{I C}

crack ki property hai, isliye lambi crack ka

K_{I C}

zyaada hoga."

False.

K_{I C}

ek material property hai, kisi bhi alloy aur temperature ke liye fixed; crack length applied

K_{I}

ko change karti hai, material ki toughness

K_{I C}

ko nahi.

TF2 — "Agar applied stress yield strength se kam hai, toh part fracture nahi ho sakta."

False. Yield strength bulk ke plastic flow ko govern karti hai; fracture crack tip par

K_{I}

K_{I C}

se govern hota hai, jahan stress

1/ r

se amplify hota hai aur bonds toot sakte hain jab bulk abhi elastic hi ho.

TF3 — "Crack length double karne se

K_{I}

double ho jaata hai."

False.

K_{I} \propto a

hai, isliye

a

double karne se

K_{I}

sirf

2 \approx 1.41

se multiply hota hai, 2 se nahi — formula mein square root hi poora point hai.

TF4 — "Zyaada tough material (higher

K_{I C}

) ki yield strength bhi hamesha zyaada hoti hai."

False. Yeh aksar trade off karte hain: high-strength alloys aksar kam tough hote hain, aur isliye fracture mechanics ek alag check ke roop mein exist karta hai.

TF5 — "

K_{I}

ke units

MPa m

hain, iska matlab yeh ek tarah ka stress hai."

False. Real stress ke units sirf MPa hote hain; extra

m

K_{I}

ko ek field-scaling quantity batata hai jo

1/ r

singularity ki strength set karta hai, kisi ek point par stress nahi.

TF6 — "Crack-tip opening stress

σ_{y y} = K_{I} / 2 π r

infinite hai, isliye har cracked part turant fail ho jaata hai."

False. Infinity ek bilkul sharp tip ki mathematical idealisation hai; real materials tip ko plastic flow se blunt karte hain, aur failure is baat se decide hota hai ki poore field ka scale

K_{I}

K_{I C}

tak pahunche, na ki kisi literal infinite point se.

TF7 — "Total length

2 a

waale centre crack ke liye aap

K_{I} = σ π a Y

mein poora

2 a

daaltey ho."

False. Aap half-length

a

use karte ho; formula

a

ke terms mein likha hai, isliye 8 mm total length wale centre crack ke liye

a = 4

mm use hoga.

TF8 — "Har crack ke liye geometry factor

Y = 1

hota hai."

False.

Y = 1

sirf infinite plate mein ideal centre crack ke liye hota hai; edge crack ka

Y \approx 1.12

hota hai, aur finite-width ya curved geometries mein

Y

aur bhi badhta hai (upar

Y

table dekhein).

TF9 — "Safety factor

n

badhane se aap jitna bada crack tolerate kar sakte ho woh bhi badh jaata hai."

False. Kyunki

a_{m a x} = \frac{1}{π} (\frac{K _{I C}}{Y σ n})^{2}

(upar derive kiya), bada

n

actually

a_{m a x}

ghataata hai — aap zyaada safe rehne ke liye chhhote flaws detect karne ki demand karte hain.

TF10 — "Sirf Mode I hi failure cause kar sakta hai."

False. Mode II aur Mode III (sliding, tearing) bhi cracks drive karte hain; mixed loading unhe

K_{eff} = K_{I}^{2} + K_{I I}^{2} + K_{I I I}^{2}

ke zariye combine karta hai, lekin Mode I zyaadatar spacecraft cases mein dominate karta hai (neeche three-mode figure dekhein).

Error dhundhein

SE1 — "Aluminium 2024-T3 ka

K_{I C} = 35

MPa

m

hai, aur mere part par

σ = 35

MPa hai, toh main exactly fracture limit par hoon."

Error yeh hai ki ek stress (35 MPa) ko ek toughness (35 MPa

m

) se compare kiya ja raha hai — alag units hain. Pehle

K_{I} = σ π a Y

build karein aur usko

K_{I C}

se compare karein.

SE2 — "Koi visible crack nahi hai, isliye

K_{I} = 0

hai aur part bilkul safe hai."

Error yeh hai ki zero flaw size assume ki ja rahi hai. Damage tolerance yeh assume karta hai ki ek undetectable crack NDT detection limit par maujood hai; aap us chhupe crack ke liye

K_{I}

compute karte ho, na ki a=0 ke liye.

SE3 — "Angular part

cos \frac{θ}{2} (1 - sin \frac{θ}{2} sin \frac{3 θ}{2})

dikhata hai ki stress

θ = 0

par seedha aage sabse zyaada hai, jo

K_{I} / 2 π r

ke barabar hai; isliye crack sideways mud jaata hai."

θ = 0

par value sahi mein

K_{I} / 2 π r

hai aur yeh peak opening stress hai — aur

θ = 0

seedha aage point karta hai (polar-coordinate figure dekhein), toh yeh crack ko forward drive karta hai, sideways nahi; reasoning geometry ko galat padh raha hai.

SE4 — "

Y

cracks ko safer banata hai kyunki yeh ek 'correction' hai, isliye bada

Y

bada allowable crack deta hai."

Error yeh hai ki

Y

ko protective samjha ja raha hai. Bada

Y

usi crack ke liye

K_{I}

badhata hai, isliye yeh allowable size

a_{m a x} \propto 1/ Y^{2}

ko ghataata hai.

SE5 — "Kyunki

K_{I} \geq K_{I C}

matlab failure hai,

K_{I} \leq K_{I C} / n

use karna over-cautious double-counting hai."

Yeh double-counting nahi hai.

K_{I} = K_{I C}

physical failure edge hai;

n

se divide karna deliberately us edge se peeche hat jaata hai taaki loads, crack size, aur material scatter mein uncertainty absorb ho sake.

SE6 — "Ceramics ka

K_{I C}

bahut chhota hota hai (2–5), isliye inhe kabhi spacecraft par use nahi karna chahiye."

Error load context ko ignore karta hai. Thermal tiles bahut kam mechanical stress carry karti hain, isliye brittle, low-

K_{I C}

ceramic bhi

K_{I}

K_{I C}

se kaafi neeche rakh sakta hai; toughness sirf us stress ke relative matter karti hai jise survive karna hai.

SE7 — "

K_{I} = σ π a Y

mein crack length millimetres mein daalna theek hai jab tak

σ

MPa mein hai."

Error unit inconsistency hai. MPa

m

mein

K_{I}

ke liye

a

metres mein chahiye; 2 mm crack 0.002 m enter karni hogi, warna answer

1000 \approx 31.6

se off hoga.

Why questions

WHY1 — Crack-tip stress

1/ r

se kyun scale hota hai,

1/ r

se nahi?

Kyunki traction-free crack faces waala elasticity solution exactly yahi square-root singularity demand karta hai taaki tip ke across transmit hone wala total force finite rahe;

1/ r

blow-up infinite force carry karega.

WHY2 —

K_{I}

mein sirf

a

ki jagah

π a

kyun hai?

π

exact Westergaard solution mein elliptical crack front ke aaspaas stress field integrate karne se nikalti hai; yeh freely choose nahi ki gayi balki fixed hai taaki

K_{I}

real physical crack ke actual field se match kare.

WHY3 —

K_{I}

useful kyun hai agar tip stress technically sabke liye infinite hai?

Kyunki har sharp crack wahi

1/ r

shape share karta hai, isliye ek akela scaling number

K_{I}

poore field ko fully characterise karta hai; us ek number ko

K_{I C}

se compare karna ek impossible point-by-point stress comparison ko replace kar deta hai.

WHY4 — Damage-tolerance design

a_{m a x}

kyun compute karta hai instead of sirf "koi crack allowed nahi"?

Kyunki manufacturing, impacts, aur fatigue cracks guarantee karte hain; inhe forbid nahi kar sakte, isliye aap us sabse badi size set karte ho jo safe rahe aur demand karte ho ki inspection usse bada kuch bhi pakade.

WHY5 — High-strength alloys ko often extra fracture-mechanics scrutiny kyun chahiye?

Inki high yield strength bade stresses carry karne deti hai bina yielding ke, lekin inki toughness aksar kam hoti hai, isliye

K_{I}

un crack sizes par

K_{I C}

tak pahunch sakta hai jo strength check kabhi flag nahi karega.

WHY6 — Failure criterion

\geq

(ek threshold) kyun use karta hai instead of gradual "damage accumulates" law?

Jab

K_{I}

K_{I C}

tak pahunchta hai, bonds itni tezi se toot te hain ki plastic flow tip ko blunt nahi kar paata, isliye propagation unstable ho jaati hai — yeh bhaag jaati hai, isse event ek sharp threshold ban jaata hai, gradual creep nahi.

WHY6 — Inspectors ko NDT method choose karne se pehle

a_{m a x}

kyun jaanna chahiye?

Chosen method ko

a_{m a x}

se chhote flaws reliably detect karne chahiye; agar eddy-current/ultrasound 1–2 mm cracks dhundh leta hai aur

a_{m a x}

5.4 mm hai, toh real margin hai — lekin agar

a_{m a x}

0.5 mm hota, toh woh method useless hoti.

Edge cases

EC1 — Jab crack length

a = 0

ho (ek truly flawless part) toh

K_{I}

kya hoga?

K_{I} = σ π \cdot 0 Y = 0

, isliye fracture mechanics predict karta hai ki koi crack-driven failure nahi — part tab ordinary yield strength se limited hota hai.

EC2 — Jab applied stress

σ \to 0

ho toh

a_{m a x}

ka kya hoga?

a_{m a x} = \frac{1}{π} (K_{I C} / Y σ n)^{2} \to \infty

: essentially koi load nahi hai, toh koi bhi crack tolerable hai, jo intuition se match karta hai ki unloaded structures fracture nahi karte.

EC3 —

K_{I} = K_{I C}

exactly (equality, not

>

) physically kya represent karta hai?

Unstable propagation ka precise onset — crack knife-edge par hai, na safely arrested na abhi run kar raha; design kabhi yahan nahi baitta, isliye

n > 1

aapko neeche push karta hai.

EC4 — Bahut high-toughness material (

K_{I C} \to \infty

) ke liye structure ko kya limit karta hai?

Fracture side par kuch nahi —

a_{m a x} \to \infty

— isliye failure conventional yielding ya buckling par wapas aa jaata hai; toughness governing constraint nahi rehti.

EC5 — Pure Mode I loading mein

K_{I I}

aur

K_{I I I}

kya hain, aur

K_{eff}

kya ban jaata hai?

Dono zero hain, isliye

K_{eff} = K_{I}^{2} + 0 + 0 = K_{I}

; mixed-mode formula gracefully plain Mode I case mein reduce ho jaata hai.

EC6 — Crack plane ke along (

θ = 0

), angular factor

cos \frac{θ}{2} (1 - sin \frac{θ}{2} sin \frac{3 θ}{2})

kya evaluate hota hai, aur yeh kyun matter karta hai?

Yeh

cos 0 (1 - 0) = 1

ban jaata hai, isliye

σ_{y y} = K_{I} / 2 π r

seedha aage —

K_{I}

ko define aur interpret karne ki sabse clean jagah, kyunki messy angular terms vanish ho jaate hain (yeh polar figure mein

θ = 0

ray hai).