Stress and strain — σ = F - A, ε = ΔL - L, Young's modulus E: Visual walkthrough

3.6.3 · D2 · Physics › Spacecraft Structures & Systems Engineering › Stress and strain — σ = F - A, ε = ΔL - L, Young's modulus E

Hum is ek line tak pahunchenge aur isme har letter ko samjhenge:

$Δ L = \frac{F L}{A E}$

Lekin pehle hum har letter ko earn karte hain.

Step 1 — Bar, pull, aur "force" actually hai kya

KYA HAI. Sabse simple cheez se shuru karte hain: ek solid bar (socho ek aluminium strut). Hum dono ends ko pakad ke bahar ki taraf kheenchte hain. Is "khinchaav" ki matra ek force hai, jise hum $F$ likhte hain, newtons (N) mein measure kiya jaata hai. Ek newton roughly ek chote seb ka weight hota hai jo tumhare haath mein rakha ho.

YAHAN SE KYUN shuru karein. Aage jo bhi hoga wo sab is pull ka response hai. Agar hum pull ko clearly picture nahi kar sakte, toh baaki kuch bhi samajh nahi aayega. Pull ki ek direction hoti hai (bar ke saath-saath) aur ek size (N mein).

PICTURE. Neeche bar dekho. Do pink arrows force $F$ hain — same size, opposite directions, taaki bar kahin ud nahi jaata; woh bas wahan baitha rehta hai, khichta hua.

Figure — Stress and strain — σ = F - A, ε = ΔL - L, Young's modulus E

$kitna hard pull karte hain, newtons mein F ⟷ a bar of length original length, metres mein L$

$F$ — pull. Bada $F$ = zyada strong tug.
$L$ — bar kitni lambi hai pehle hum use touch karein. Yeh word "pehle" yaad rakhna; baad mein kaam aayega.

Spacecraft Load Paths and Struts dekho yeh jaanne ke liye ki launch ke dौraan yeh forces kahan se aate hain.

Step 2 — Imaginary cut: bar ke andar force

KYA HAI. Ek chaaku lo aur imagine karo ki bar ko seedha across, pull ke perpendicular, kaat rahe ho. Sach mein mat kaato — bas cut face picture karo. Socho: dono halves ko saath kya rokta hai?

KYUN. Hum stress ko bahar se nahi dekh sakte. Trick yeh hai ki andar dekho. Agar bar still baith rahi hai (equilibrium), toh left half ko right half ko exactly $F$ se wapas kheenchna chahiye, aur vice versa. Yeh Newton's third law hai: material ke internal bonds wahi force $F$ us cut ke across carry karte hain.

PICTURE. Dashed blue line imaginary cut hai. Blue arrows internal force dikhate hain — same size $F$ jaise baahri pull, ab cut face ko cross karte hue reveal ho raha hai.

Step 3 — Crowding: force ko area se divide karne par stress milta hai

KYA HAI. Woh internal force $F$ poori cut face par spread hai. Us face ki area ko $A$ kaho (square metres mein, m²). Ab stress define karo:

$σ = \frac{F}{A}$

Area se kyun divide karein — aur yahi operation kyun, koi aur kyun nahi? Hum ek aisa number chahte hain jo material ko describe kare, na ki kisi particular chunky-ya-skinny bar ko. Ek aur sharp question pucho: cut face par bonds mein force kitna crowded hai? Ek wide face $F$ ko bahut saare bonds mein share karta hai (har ek thoda carry karta hai); ek narrow face $F$ ko thode bonds mein cramm karta hai (har ek bahut carry karta hai). "Ek area mein share ki gayi matra" exactly wahi hai jo area se division compute karta hai — isliye $F / A$ hai, na $F \times A$ ya $F - A$ .

PICTURE. Do faces, same $F$ . Left: moti face, arrows spread thin. Right: patli face, arrows packed tight. Pink arrows ki density hi stress hai.

Step 4 — Stretching: stretch ko length se divide karne par strain milta hai

KYA HAI. Pull ke neeche bar $L$ se $L + Δ L$ tak grow karta hai. Woh chota extra bit $Δ L$ (Greek $Δ$ ka matlab sirf "mein change" hai) stretch hai. Strain define karo:

$ε = \frac{Δ L}{L}$

Original length se kyun divide karein? Same logic jaise stress: hum ek material number chahte hain, na ki ek part number. Kya 1 mm stretch bahut hai? 1 mm chip par yeh enormous hai; 2 m strut par yeh kuch bhi nahi. "Stretch as compared to yeh kitna bada tha pehle" exactly original length $L$ se division hai. Isliye $Δ L / L$ hai, aur isliye hum original $L$ use karte hain (Step 1 ka "pehle wala" length), naya wala nahi.

PICTURE. Do bars same $Δ L$ se stretch hui hain. Chhoti wali badly deformed hai (bada strain); lambi wali barely notice karti hai (tiny strain). Yellow bracket $Δ L$ ko har bar ki apni length ke against dikhata hai.

Bar bhi kyun patla ho raha hai jab woh stretch hota hai? Yeh isi idea ka ek sideways cousin hai — doosre page ki story hai.

Step 5 — Experiment: stress aur strain saath-saath badhte hain

KYA HAI. Ab measure karo. Zyada se zyada weight latkaao, $σ$ (Step 3 se) aur $ε$ (Step 4 se) record karo, aur ek ko doosre ke against plot karo. Choti pulls ke liye ek straight line through the origin milti hai.

Unhe ek doosre ke against plot kyun karein? Kyunki hum suspect karte hain ki woh related hain, aur ek graph relationship dekhne ka sabse clean tarika hai. Ek straight line ka matlab hai: stress double karo → strain double. Symbols mein, $σ$ proportional to $ε$ hai, likha jaata hai $σ \propto ε$ .

PICTURE. Straight sloped chalk line: strain horizontal axis par, stress vertical par. Har point ek experiment hai. Line ki steepness Step 6 ki poori kahani hai.

Yeh straight part Hooke's Law ka domain hai. Yeh hamesha nahi rehta — Step 8 dekho.

Step 6 — Slope ka naam: Young's modulus $E$

KYA HAI. Origin se guzarti straight line ki equation $y = (slope) \times x$ hoti hai. Yahan $y = σ$ aur $x = ε$ hai. Slope ko ek naam do, $E$ :

$σ = E ε$

$E$ kyun invent karein? Slope ek single number hai jo poori line capture karta hai — material ki stiffness. Steep line = stiff material (bahut saara stress sirf thoda strain khareedta hai). Shallow line = floppy material. Woh ek number engineer ko aluminium, steel, aur composite ko ek nazar mein compare karne deta hai.

PICTURE. Same axes par do lines: steep blue (steel, bada $E$ ) aur shallow yellow (aluminium, chhota $E$ ). Har ek par rise-over-run triangle hi $E = σ / ε$ hai.

Step 7 — Master formula assemble karna: $Δ L = F L / (A E)$

KYA HAI. Ab hum Steps 3 aur 4 ko Step 6 mein substitute karte hain aur stretch $Δ L$ ke liye solve karte hain — woh cheez jo ek engineer actually predict karna chahta hai.

KYUN. Hum $F$ , geometry ( $A$ , $L$ ), aur material ( $E$ ) jaante hain. Hum deflection chahte hain. Toh hum definitions ko chain karte hain aur rearrange karte hain.

PICTURE. Substitution ka flow: $σ = E ε$ jisme $σ$ aur $ε$ expand kiye gaye hain, phir $Δ L$ akela bahar nikala gaya.

Step 6 se shuru karo aur dono sides expand karo:

$σ \frac{F}{A} = E ε \frac{Δ L}{L}$

Dono sides ko $L$ se multiply karo (Step 4 ke "divide by $L$ " ko undo karte hue) aur $E$ se divide karo:

$Δ L = \frac{F hard pull \to zyada stretch L lambi bar \to zyada stretch}{moti \to kam stretch A stiffer \to kam stretch E}$

Formula ko ek sentence ki tarah padho: stretch pull $F$ aur length $L$ (upar) ke saath badhta hai, aur area $A$ aur stiffness $E$ (neeche) ke saath ghatta hai. Upar har arrow tumhari physical intuition se match karta hai — yahi ek formula ke sach mein samajhe jaane ki nishani hai.

Step 8 — Edge cases: har letter zero TAK AUR infinity TAK

KYA HAI. Ek formula sirf tabhi trustworthy hota hai jab woh har letter ki dono extremes par sensibly behave kare. $F, A, E, L$ mein se har ek ko hum ab $0$ aur $\infty$ par bhejte hain, aur picture check karte hain.

KYUN. Real struts zero load, huge loads, near-rigid materials, hair-thin wires, aur long booms dekhte hain. Agar formula kisi bhi limit par nonsense deta, toh hum beech mein bhi trust nahi kar sakte. $Δ L = F L / (A E)$ padhte hue: upar wala letter ( $F, L$ ) badhne par $Δ L$ ko upar le jaata hai; neeche wala letter ( $A, E$ ) badhne par $Δ L$ ko neeche le jaata hai.

PICTURE. Panels low-stretch limits (top row) aur runaway-stretch limits (bottom row) dikhate hain, jisme har extreme mein bar draw ki gayi hai.

Stretch zero ho jaati hai ( $Δ L \to 0$ ) jab:

$F \to 0$ (koi pull nahi): koi force nahi, koi stretch nahi. ✓
$A \to \infty$ (infinitely moti bar): ek pahaad nahi kheechti jab tum use tug karo. ✓
$E \to \infty$ (perfectly rigid, "unobtainium"): infinite stiffness ka matlab zero deflection — woh idealisation jise engineers rigid body kehte hain. ✓
$L \to 0$ (wafer-thin slice): stretch karne ke liye almost kuch bhi nahi, isliye total stretch vanish ho jaata hai.

Stretch blow up ho jaati hai ( $Δ L \to \infty$ ) jab:

$F \to \infty$ (unlimited pull): top letter bhaag jaata hai — infinite force (formally) infinite stretch deta hai. Reality mein material yield karta hai aur phir snap ho jaata hai bahut pehle; math warn karta hai ki deflection unbounded hai.
$L \to \infty$ (endless boom): top letter bhaag jaata hai — same fractional strain jo ek ever-longer bar par apply hoti hai ever-larger total stretch accumulate karti hai.
$A \to 0$ (hair-thin wire, ya crack tip): bottom letter vanish ho jaata hai, isliye $Δ L \to \infty$ . Physically force almost koi material mein crammed ho jaata hai — stress $σ = F / A$ explode ho jaata hai, wire necks karti hai aur fail ho jaati hai. Yeh ek failure mode ki mathematics hai.
$E \to 0$ (ek floppy gel jisme koi stiffness nahi): bottom letter vanish ho jaata hai, isliye $Δ L \to \infty$ . Ek material jo koi resistance offer nahi karta woh kisi bhi pull ke neeche without limit stretch ho jaata hai.

Ek-picture summary

YEH FIGURE KYA KARTA HAI. Upar sab kuch, ek single board par compress karke taaki tum poori derivation ek glance se rebuild kar sako. Ise apna recall trigger use karo: agar tum yeh picture narrate kar sako, toh yeh page tumhara hai.

ISSE KAISE PADHEN, left se right. Pink arrows se shuru karo — woh Step 1 ka pull $F$ hai. Bar mein dashed blue cut follow karo — woh Step 2 ka internal force hai, jise humne Step 3 mein stress $σ = F / A$ mein convert kiya. Bar ki stretch uski length ke upar strain $ε = Δ L / L$ hai Step 4 ki. Right par, yellow line Step 5 ka experiment hai; iski slope $E$ hai (Step 6). Aur neeche payoff hai, $Δ L = F L / (A E)$ (Step 7), pink "up: $F, L$ " aur blue "down: $A, E$ " tags ke saath jo tumhe remind karte hain ki har letter stretch ko kis taraf push karta hai (Step 8).

Recall Feynman: plain words mein poora walkthrough

Humne ek bar pakdi aur use force $F$ se kheencha. Yeh samajhne ke liye ki andar kya ho raha hai, humne imagine kiya ki use kaatenge — do halves ek doosre ko usi $F$ se wapas kheenchte hain. Ab, kya woh force badi deal hai? Depend karta hai ki kitna material use share karta hai: moti face par yeh chill hai, patli face par yeh fierce hai. Toh humne force ko area se divide kiya aur crowding ko stress kaha.

Phir bar thoda sa stretch hua, $Δ L$ . Kya woh stretch badi deal hai? Depend karta hai ki bar pehle kitni lambi thi — ek millimetre chip ke liye huge hai, girder ke liye kuch nahi. Toh humne stretch ko original length se divide kiya aur us fraction ko strain kaha.

Humne experiment kiya: stress ko strain ke against plot karo aur, gentle pulls ke liye, ek straight line milti hai. Us line ki steepness ek number hai jo batata hai ki material kitna stubborn hai — woh hai Young's modulus $E$ . Steep line, stubborn stuff. Aur "gentle" matter karta hai: line tabhi straight rehti hai jab strain small ho, zyada se zyada kuch percent.

Aakhir mein humne sab joda. Likho "stress equals $E$ times strain," dono sides ko $F / A = E Δ L / L$ mein kholo, aur jugaao jab tak stretch akela na baith jaaye: $Δ L = F L / (A E)$ . Ise ek story ki tarah padho — zyada hard pull karo ya lambi bar use karo toh zyada stretch hogi; ise mota ya stiffer banao toh kam stretch hogi. Push karo pull ki jagah aur har sign flip ho jaata hai: bar chhoti ho jaati hai. Aur humne corners dono ends par check kiye: koi force nahi, ya infinite fatness, ya infinite stiffness zero stretch dete hain; jabki infinite force, endless boom, hair-thin wire, ya floppy zero-stiffness gel sab stretch ko infinity ki taraf race karwa dete hain — exactly jaisa common sense demand karta hai. Woh poori chain ek idea hai: part ka size hatao, aur jo bachta hai woh material hai jo bol raha hai.

Recall

Master deflection formula aur har letter ka effect ::: $Δ L = F L / (A E)$ ; $F$ aur $L$ se upar, $A$ aur $E$ se neeche. Force ko area se divide karke stress kyun lete hain ::: Yeh measure karne ke liye ki internal force bonds mein kitna crowded hai, jo ek material property hai, na ki part property. Stress–strain line ke slope ko kya kehte hain ::: Young's modulus $E$ — material ki stiffness. Compressive (negative) force ke neeche $Δ L$ ka kya hota hai ::: Woh negative ho jaata hai — bar chhoti ho jaati hai; formula automatically sign flip kar leta hai. Kya strain length $L$ par depend karta hai ::: Nahi — $ε = F / (A E)$ mein $L$ nahi hai; sirf total stretch $Δ L = ε L$ length ke saath scale karta hai.