Yield stress, ultimate stress — material behavior: Question bank

3.6.5 · D5 · Physics › Spacecraft Structures & Systems Engineering › Yield stress, ultimate stress — material behavior

Definition Is page par use hue symbols (ek baar padho, phir traps samajh aayenge)

Neeche sab kuch force-per-area (pascal, $Pa = N/m^{2}$ ) mein hai jab tak aur na bataya jaaye.

$F$ = part par kheenchne (ya dabaane) wali force, newtons $N$ mein.
$A_{0}$ = bar ka original cross-sectional area koi bhi load lagane se pehle, $m^{2}$ mein.
$A_{inst}$ = load lagey rehne ke dauran instantaneous (current) cross-sectional area — yeh tab sirkata hai jab bar stretch hoti hai aur neck karti hai. Isliye $A_{inst} \leq A_{0}$ .
Engineering stress $σ = F / A_{0}$ (fixed original area use karta hai). True stress $σ_{t} = F / A_{inst}$ (siraktey real area use karta hai).
$E$ = Young's modulus, elastic slope $σ = E ε$ , $Pa$ mein — yeh ek stiffness hai, dimensionally ek stress hai kyunki strain $ε$ dimensionless hai.
$σ_{y}$ = yield stress, $σ_{u}$ = ultimate stress (dono material properties hain, $Pa$ mein).

Definition Allowable stress, required FoS, aur margin formula (yahan define kiya, neeche use hoga)

$FoS_{req}$ = required factor of safety — woh safety multiplier jo standard maangta hai (spacecraft: yield ke against ~1.25, ultimate ke against ~1.5).
Allowable stress $σ_{allow}$ = sabse bada stress jo hum allow karte hain, jo material limit ko required FoS se divide karke milta hai: $\sigma_\text{allow}=\frac{\sigma_u}{\text{FoS}_\text{req}}\quad\text{(ultimate)}.$$ Required FoS *pehle se hi* $\sigma_\text{allow}$ ke andar hai — use dobaara divide mat karo.$
Margin of safety allowable ko applied stress $σ_{app}$ se compare karta hai: $MoS = \frac{σ _{allow}}{σ _{app}} - 1 \geq 0$

Yeh formula kyun hai: "safe" ka matlab hai applied stress allowable se zyada nahi honi chahiye, yaani $σ_{app} \leq σ_{allow}$ . Divide karne par $σ_{allow} / σ_{app} \geq 1$ milta hai; $1$ ghatane par "just safe" ko $0$ par re-centre kar deta hai, toh positive MoS spare capacity hai aur negative MoS fail hai. Kyunki required FoS pehle se $σ_{allow}$ mein bake ho chuka tha, denominator mein ek aur $FoS_{req}$ daalna safety factor ko double-count kar dega — yeh ek common aur mahenga galti hai.

True or false — justify

Ek material jiska Young's modulus

E

zyada ho (elastic slope,

Pa

mein) woh hamesha zyada strong hota hai.

False.

E

stiffness hai (elastic stretch ke against resistance), strength nahi. Do steels ek hi

E

share kar sakte hain phir bhi unke

σ_{y}

bilkul alag ho sakte hain; strength

σ_{y}, σ_{u}

mein hoti hai, elastic slope mein nahi.

Yield stress

σ_{y}

badal jaata hai agar aap rod ko mota kar dein.

False.

σ_{y}

ek material property hai (force per unit area). Ek moti rod yield par zyada force

F

(

N

mein) uthayegi (

F = σ_{y} A

), lekin stress limit khud unchanged rehti hai.

Ek baar part yield ho jaaye toh woh fail ho gaya aur alag ho gaya.

False. Yield permanent (plastic) deformation ki shuruaat hai — part muda hua hai lekin abhi bhi ek tukde mein hai. Separation baad mein hoti hai, fracture par,

σ_{u}

ke kaafi baad.

Engineering stress fracture tak true stress ke barabar rehta hai.

False. Woh tab agree karte hain jab area barely change hota hai. Necking ke baad real area

A_{inst}

tezi se sirkata hai, toh true stress (

σ_{t} = F / A_{inst}

) badhta hai jabki engineering stress (

σ = F / A_{0}

) girta hai.

Ultimate stress

σ_{u}

woh stress hai jo specimen ke exactly tootne ke waqt hota hai.

False.

σ_{u}

engineering curve ka peak hai, jo fracture se pehle pahuncha jaata hai. Peak ke baad specimen neck karta hai aur fracture par engineering stress actually

σ_{u}

se kam hoti hai.

Applied force ko double karne se yielding ka risk hamesha double ho jaata hai.

Iss sense mein True ki yeh stress ko double kar deta hai (

σ = F / A_{0}

, area fixed), yahi cheez yield compare karti hai — toh aap usi fixed

σ_{y}

ke against

σ

2 σ

par aa jaate ho.

Exactly

0

ka margin of safety matlab design fail ho gaya.

False.

MoS = 0

matlab applied stress exactly allowable ke barabar hai (jisme already required FoS shamil hai) — yeh pass hai, lekin zero spare ke saath. Negative MoS fail case hai.

Spacecraft yield aur ultimate ke liye same factor of safety use karte hain.

False. Yield typically

FoS_{req} \approx 1.25

use karta hai aur ultimate

\approx 1.5

. Bada ultimate FoS reflect karta hai ki fracture catastrophic aur non-recoverable hai, jabki yield sirf fit/spec kharaab karta hai.

Yield aur ultimate stresses ke same numbers hain chahe aap material ko kheechen ya dabaayen.

Generally false. Ductile metals ka yield tension aur compression mein roughly symmetric hota hai, lekin compression cracking ko rokta hai isliye brittle materials (glass, ceramics, concrete) compression mein tension ke mukable kahin zyada strong hote hain — dono limits ek order of magnitude tak differ kar sakti hain.

Spot the error

"Stress bas yeh hai ki aap kitna hard push kar rahe ho, isliye badi force matlab zyada stressed part."

Error: yeh area ignore karta hai. Stress ==

F / A_{0}

== hai, force per area. Usi force se ek patli rod par zyada stress hoga;

A

mota karne se load change kiye bina stress kam ho jaata hai.

MoS = σ_{allow} / (FoS_{req} \cdot σ_{app}) - 1

, extra safe rehne ke liye required FoS se dobaara divide kar rahe hain."

Error: safety factor ka double-counting.

σ_{allow}

pehle se hi

σ_{y} / FoS_{req}

ke barabar hai, toh FoS baked in hai. Sahi form hai

MoS = σ_{allow} / σ_{app} - 1

; ek aur

FoS_{req}

daalne se design actual se kahin zyada marginal lagta hai.

"0.2% offset line origin se shuru hoti hai, kisi bhi cheez ke parallel nahi."

Dono baaton mein galat. Offset line

ε = 0.002

par shuru hoti hai (origin par nahi) aur ==elastic slope

E

ke parallel== kheenchi jaati hai. Curve ke saath iska crossing hi

σ_{y}

define karta hai (figure "Why questions" ke neeche dekho).

σ_{y}

wahan hai jahan curve pehli baar non-linear hoti hai, isliye yeh proportional limit ke barabar hai."

Same point nahi hai. Proportional limit woh jagah hai jahan Hooke's law

σ = E ε

pehli baar toota. Yield woh jagah hai jahan permanent strain shuru hoti hai — yeh thoda upar hota hai, aur gradual-yielding alloys ke liye hum ise 0.2% offset se pin karte hain.

"Safe rehne ke liye hamen design allowables mein true stress use karni chahiye, kyunki yeh zyada physical hai."

Handbooks aur allowables almost hamesha engineering stress (

F / A_{0}

) mein hote hain. Engineering-stress formula mein true-stress value daalne se

σ_{u}

galat padh jaata hai, kyunki dono necking ke baad diverge ho jaate hain.

0.9 σ_{y}

par loaded part theek hai kyunki yield nahi hua."

Yield nahi hua, lekin yield ke liye check allowable

σ_{allow} = σ_{y} / FoS_{req}

ke against hota hai.

FoS_{req} = 1.25

ke saath allowable

0.8 σ_{y}

hai, toh

0.9 σ_{y}

margin fail karta hai chahe koi permanent set nahi aaya. Dekho Factor of safety and margins.

"Agar MoS yield ke liye positive hai, toh part automatically fracture ke against bhi safe hai."

Guaranteed nahi. Yield aur ultimate alag checks hain alag limits (

σ_{y}

σ_{u}

) aur alag required FoS ke saath. Dono ko independently clear karna zaroori hai.

Why questions

Engineers ek "failure" stress ki jagah do stress numbers kyun care karte hain?

Kyunki dono failures qualitatively alag hain: yield part ki shape/spec permanently kharaab karta hai (recoverable structure gone), jabki ultimate separation hai. Ek precision antenna yield par "fail" ho sakta hai kaafi pehle jab woh kabhi fracture bhi nahi karta.

Figure — Yield stress, ultimate stress — material behavior

0.2% offset line elastic slope ke parallel kyun kheenchi jaati hai, aur origin se nahi?

Figure dekho. Plastic region mein kisi bhi point se unloading ek line $E$ ke parallel neeche aati hai, aur jahan yeh land karti hai woh peeche rah jaana wala permanent strain hai. Toh

ε = 0.002

par ek parallel line exactly "

0.2%

permanent set" mark karti hai, aur curve ke saath iska crossing repeatable

σ_{y}

hai.

Stress–strain curve shuruaat mein straight kyun hoti hai?

Atomic bonds tiny springs ki tarah kaam karte hain; thoda sa stretch ek linear in displacement restoring force deta hai. Cross-section ke sabhi bonds par summing karne se macroscopic response linear ban jaata hai:

σ = E ε

. Dekho Young's modulus and elasticity.

Curve yield par straight line se peelaway kyun hoti hai?

Dislocations glide karne lagte hain — atoms ke planes irreversibly slip karte hain. Woh slip unloading par recover nahi hoti, isliye strain proportionally return nahi karta aur curve bend ho jaati hai.

Yield ke baad stress badhta kyun rehta hai flat rehne ki jagah (strain hardening)?

Dislocations pile up aur tangle ho jaate hain, ek doosre ko obstruct karte hain, isliye unhe chalate rehne ke liye aur stress chahiye. Material literally deform karna mushkil hota jaata hai jab tak

σ_{u}

par peak nahi aa jaata.

0.2% offset ki zaroorat hi kyun hai — yield directly kyun nahi padh lete?

Bahut saare alloys mein koi sharp yield point nahi hota; bend gradual hota hai, toh "kahan yield karta hai" ambiguous hai. Offset yield ko ek fixed 0.2% permanent strain par pin karta hai, labs mein ek repeatable number deta hai.

Ultimate point ke baad engineering stress kyun girta hai?

Ek local neck banta hai; wahan true cross-section

A_{inst}

tezi se sirkata hai. Kyunki engineering stress fixed original area

A_{0}

se divide karta hai, us unchanged

A_{0}

par chhota load girte hue stress ki tarah dikhta hai chahe neck wali jagah ka material kabhi se zyada stressed ho.

Ultimate ke liye yield se bada factor of safety kyun choose kiya jaata hai?

Kyunki fracture unrecoverable aur catastrophic hai, jabki yield ek intact (chahe deformed) part chhodta hai. Zyada consequence ⇒ zyada margin maanga jaata hai. Structural load cases and launch loads se link hai jahan worst-case loads kabhi fracture tak nahi pahunche chahiye.

Yield/ultimate margin checks mein stiffness (

E

) appear kyun nahi karta?

Margins stress limits ko applied stress se compare karte hain.

E

govern karta hai ki part kitna stretch karta hai, woh stress nahi jis par permanently deform ya toot ta hai — stiffness aur strength independent design levers hain. Material selection for spacecraft mein relevant hai.

Edge cases

Elastic region mein strain ka kya hoga agar applied stress exactly zero ho?

Strain exactly zero hai aur fully recoverable — "spring" aram mein hai. Yeh degenerate baseline hai:

ε = σ / E = 0

, koi stored energy nahi, koi permanent set nahi.

Ek perfectly brittle material (jaise glass) — curve par uska yield point kahan hai?

Essentially hota hi nahi hai. Yeh fracture tak linear-elastic rehta hai, isliye

σ_{y} \approx σ_{u}

aur plastic/strain-hardening region gayab ho jaata hai. "Every Young Ultimately Fractures" ki kahani Elastic→Fracture tak collapse ho jaati hai.

Kya material compression mein tension jaisi hi stress par yield aur toot ta hai?

Zaroori nahi. Compression cracks ko band karta hai, isliye brittle materials tensile stress se kahin zyada compressive stress carry karte hain; ductile metals zyada symmetric hain lekin unke design allowables phir bhi tension aur compression alag list karte hain. Yeh launch-load struts ke liye matter karta hai jo dono dekhte hain — dekho Structural load cases and launch loads.

Temperature aur kitni tezi se load karte hain yeh

σ_{y}

aur

σ_{u}

ko kaise badalta hai?

Metal ko garam karne se generally dono

σ_{y}

aur

σ_{u}

ghatt jaate hain (bonds aur dislocations aasaan se move karte hain), jabki tezi se load karna (high strain rate) ya thanda rehna dono ko badhata hai aur material ko zyada brittle banata hai. Handbook allowables isliye ek stated temperature aur rate par quote kiye jaate hain — kabhi bhi room-temperature values hot thruster bracket ya cryogenic tank par assume mat karo.

Agar load lagaya aur poori tarah hata liya jabki stress

σ_{y}

se neeche raha, toh kya bachta hai?

Kuch permanent nahi — part apni original length par wapas spring karta hai (zero residual strain). Yeh elastic region ka operational matlab hai: reversible.

Agar part yield se thoda aage load ho aur phir unload kiya jaaye, toh unloading path kaisa dikhega?

Yeh ek line ke saath neeche aata hai ==original elastic slope

E

ke parallel==, ek nonzero (permanent) strain par khatam hota hai. Woh bacha hua strain exactly woh plastic deformation hai jo offset method measure karta hai.

Agar applied stress exactly allowable stress ke barabar ho toh margin of safety kya hai?

MoS = σ_{allow} / σ_{app} - 1

σ_{app} = σ_{allow}

ke saath yeh ==

0

== hai — zero spare capacity ke saath pass; koi bhi extra load ise negative kar dega. (Denominator mein koi doosra FoS nahi — woh pehle se

σ_{allow}

ke andar hai.)

Ek cyclic load hamesha

σ_{y}

se neeche rehta hai — kya part indefinitely safe hai?

Zaroori nahi. Yield se neeche rehna static failure ko rokta hai, lekin repeated cycles phir bhi

σ_{y}

se neeche fatigue cracking cause kar sakte hain. Yeh ek alag limit hai — dekho Fatigue and fracture mechanics.

Necking ke instant par (peak

σ_{u}

), true area ka

A_{0}

ke mukable kya ho raha hai?

Local true area

A_{inst}

A_{0}

se neeche tezi se sirakna shuru ho jaata hai. Toh true stress (

F / A_{inst}

) upar jaata hai jabki engineering stress (

F / A_{0}

) neeche mud jaata hai — dono curves

σ_{u}

par hi split ho jaate hain.

Recall Quick self-test

Woh teen confusions batao jinhe mnemonic collapse karta hai. ::: Stiffness vs strength, yield vs fracture, engineering vs true stress. Kya $σ_{y}$ rod ke area par depend karta hai? ::: Nahi — material property; sirf failing force area ke saath scale karti hai. MoS denominator mein extra $FoS_{req}$ kyun nahi hota? ::: Kyunki $σ_{allow} = σ_{y} / FoS_{req}$ mein pehle se hai; dobaara divide karna safety factor ko double-count kar dega.