Visual walkthrough — Thermal analysis — conduction in structures, thermal stress

3.6.14 · D2 · Physics › Spacecraft Structures & Systems Engineering › Thermal analysis — conduction in structures, thermal stress

Yeh page parent result Thermal Stress ko ek-ek picture ke saath rebuild karti hai. Hum ek metal ki bar aur ek garam Sun se shuru karte hain, aur koi bhi symbol use nahi karte jab tak hum use draw na kar lein. Akhir tak aap dekh paoge ki ek garam, bolted-down beam kyun push back karta hai — aur kitni force se.

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Step 1 — "Expansion" ka matlab kya hai

KYA. Ek seedhi metal bar lo. Use uniformly garam karo. Yeh thodi lambi ho jaati hai. Hum extra length ko $\Delta L$ kehte hain (padho: "length mein change", triangle $\Delta$ ka matlab sirf change hota hai).

KYUN. Isse pehle ki hum forces ki baat karein, hume ek clean number chahiye ki bar kitni lambi hoti hai. Baaki sab kuch isi ek measurement par bana hai.

PICTURE. Upar wali bar thandi hai, length $L$ (original length). Neeche wali bar garam hai — uska right end $\Delta L$ se right ki taraf khisak gaya hai.

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Step 2 — Temperature kaise strain mein badal jaata hai

KYA. Experiment kehta hai: bar ko $\Delta T$ degrees garam karo aur uska strain seedhe proportion mein badhta hai. Double heating, double stretch. Proportionality constant material ka coefficient of thermal expansion hota hai, likha jaata hai $\alpha$ (Greek "alpha").

KYUN. Hume temperature (jo orbit deta hai) se strain (jo mechanics ko matter karta hai) tak ek bridge chahiye. $\alpha$ woh bridge hai, aur yeh har material ke liye ek fixed number hota hai — dekho Material Selection for Spacecraft.

PICTURE. Ek seedhi line: horizontal axis temperature change $\Delta T$ hai, vertical axis thermal strain ${\epsilon }_{\text{th}}$ hai. Jitni steep line, utna "twitchier" material. Aluminium ki line steep hai; carbon-fibre ki almost flat hai.

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Step 3 — Force kaise strain mein badal jaata hai (Hooke's Law)

KYA. Ek second ke liye heat bhool jao. Ek thandi bar lo aur use force se khiincho. Yeh stretch hoti hai. Push karo toh squeeze hoti hai. Elastic range mein, strain aapke apply kiye stress ke proportional hota hai.

KYUN. Step 4 mein bar ko thermally stretch karne se mana kiya jaayega — toh kuch aur us stretch ko cancel karna hoga. Woh "kuch aur" mechanical squashing hai. Ise use karne ke liye, hume force↔strain rule khud, clean chahiye.

PICTURE. Left: ek bar ko ek arrow se khincha ja raha hai (tension) — badhti hai aur teal ho jaati hai. Right: ek bar ko inward arrows se push kiya ja raha hai (compression) — sikodhti hai aur orange ho jaati hai. Unke paas graph: stress $\sigma$ upar axis par, strain $\epsilon$ across — origin se ek seedhi line jiska slope $E$ hai.

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Step 4 — Trap: bar ko move karne se mana karo

KYA. Ab bar ke dono ends ko rigid wall se bolt karo, phir use $\Delta T$ se garam karo. Bar badhna chahti hai ${\epsilon }_{\text{th}}=\alpha \Delta T$ se — lekin walls mana kar deti hain. Final length change: zero.

KYUN. Yahi parent note ka poora point hai. Ek free bar koi stress build nahi karti (neeche mistake box dekho). Stress tabhi aata hai jab wanted motion block ho jaata hai. Blocking hi woh cheez hai jo material mein force store karti hai.

PICTURE. Do walls (hatched) bar ko clamp karti hain. Ek faint "ghost" bar dikhata hai ki garam bar kahan pahunchna chahti thi (dashed, wall ke paar). Real bar $L$ length par stuck hai. Har wall par inward orange arrows dikhate hain ki material bahar dhakel raha hai aur wall waapis dhakel rahi hai.

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Step 5 — Dono strains add karo aur total ko zero set karo

KYA. Bar ek saath do strains experience karti hai: thermal stretch jo woh chahti hai (${\epsilon }_{\text{th}}$) aur ek elastic strain (${\epsilon }_{\text{el}}$) jo wall ke push se force ki jaati hai. Wall guarantee karti hai ki total length change zero hai: ${\epsilon }_{\text{total}}={\epsilon }_{\text{th}}+{\epsilon }_{\text{el}}=0.$

KYUN. Yeh single equation poori derivation ka hinge hai. "Total motion = 0" "walls ko move nahi karne deti" ka mathematical statement hai. Rearrange karo toh yeh kehta hai ki elastic strain ko thermal strain ko exactly undo karna hoga.

PICTURE. Strain ki ek number line. Right-pointing teal arrow of length ${\epsilon }_{\text{th}}=\alpha \Delta T$ (wanted growth). Left-pointing orange arrow of length ${\epsilon }_{\text{el}}$ (forced squash). Dono same length ke hain, tip-to-tail $0$ par land karte hain.

${\epsilon }_{\text{th}}+{\epsilon }_{\text{el}}=0$ se: ${\epsilon }_{\text{el}}=-{\epsilon }_{\text{th}}=-\alpha \Delta T.$

• Minus sign: elastic strain thermal wale ke opposite point karta hai. Heating (badhna chahti hai) squashing se jawab milta hai.

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Step 6 — Us forced strain ko stress mein convert karo

KYA. ${\epsilon }_{\text{el}}=-\alpha \Delta T$ ko Step 3 ke Hooke's Law $\sigma =E{\epsilon }_{\text{el}}$ mein daalo.

KYUN. Strain invisible hota hai; stress woh cheez hai jo metal crack karta hai, mirrors warp karta hai, aur joints ko fatigue karta hai. Hooke's Law hi ek aisa tool hai jo ek known strain ko actual stress mein badalta hai jis par hum design karte hain — isliye hi hum ne ise Step 3 mein alag kiya tha.

PICTURE. Ek flow strip: box "$\Delta T$" → (×$\alpha$) → box "${\epsilon }_{\text{th}}$" → (sign flip, wall) → box "${\epsilon }_{\text{el}}$" → (×$E$) → box "$\sigma$". Har arrow par operation label kiya gaya hai.

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Step 7 — Har sign cover karo (saare cases)

KYA. Formula mein ek aisa term hai jiska sign flip hota hai: $\Delta T$. Woh single sign poori physical story control karta hai. Teeno cases walk karo.

KYUN. Contract rule: reader ko kisi aisi situation mein mat chodo jo tumne draw nahi ki. Orbit ek strut ko sunlight se shadow tak le jaata hai — woh dono signs dekhega, aur crossing point bhi — aur isliye hum ne ise alag kiya tha.

PICTURE. Teen clamped bars stack ki hui hain.

• Top, $\Delta T>0$ (heated): badhna chahti hai → wall squeeze karti hai → compression (orange, inward arrows). $\sigma <0$.
• Middle, $\Delta T=0$: kuch nahi chahti → wall kuch nahi karti → zero stress. Degenerate case.
• Bottom, $\Delta T<0$ (cooled): sikodhna chahti hai → wall use waapis khinchti hai → tension (teal, outward arrows). $\sigma >0$.

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Step 8 — Jab bar ka ek end free ho (degenerate limit)

KYA. Ek bolt dhila karo. Ab bar badh sakti hai. Stress ka kya hota hai?

KYUN. Real spacecraft joints rock-solid se floppy tak range karte hain. Do extreme — perfectly clamped aur perfectly free — beech mein sab kuch bracket karte hain. Hume spectrum ke dono ends dikhane chahiye.

PICTURE. "Free" se "fixed" tak ek dial. Free par: ghost aur real bar coincide karte hain, arrows vanish hote hain, $\sigma =0$. Fixed par: ghost overshoot karta hai, full arrows, $\sigma =-E\alpha \Delta T$. Beech mein ek curve dikhata hai ki stress constraint stiffness ke saath badhta hai.

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Ek-picture summary

Upar sab kuch, ek canvas par: temperature bilkul left par, strain drive kar raha hai, wall se block ho raha hai, right par stress ban raha hai — neeche sign map ke saath.

Recall Feynman retelling — ise ek kahani ki tarah bolo

Ek metal bar basically ek spring hai jo andar se same stuff se bani hai. Use garam karo aur har atom thoda zyada wide jiggle karta hai, toh poori bar lambi hone ki koshish karti hai — har degree ke liye ek fraction $\alpha$, multiply by kitne degrees add kiye. Yahi hai wanting-to-grow. Ab dono ends ko wall se bolt karo. Bar badhne ki koshish karti hai, wall mana karti hai. Toh bar exactly utni squash ho jaati hai jitna badhna chahti thi — dono zero motion par cancel ho jaate hain. Ek squashed spring push back karta hai, aur kitna hard push karta hai yeh depend karta hai ki woh kitna stiff hai, jise hum $E$ kehte hain. Stiffness ko squash ki amount se multiply karo aur tumhe unit area per push milta hai: wahi stress hai, $\sigma =-E\alpha \Delta T$. Minus sign sirf bookkeeping hai — heat matlab push (compression), cold matlab pull (tension). Bar ko free karo aur squash gayab ho jaata hai, toh stress bhi gayab ho jaata hai. Woh last fact designer ka escape hatch hai: heat se ladho mat, bar ko move karne do.

Recall Quick self-test

Kaunsi single quantity ka sign decide karta hai tension vs compression? ::: $\Delta T$ — positive (heating) → compression; negative (cooling) → tension. Ek bar garam hoti hai lekin dono ends par free hai. Uska thermal stress kya hai? ::: Zero — koi expansion rok nahi raha. Stress formula mein $\Delta T$ ke saath kaun se do material numbers multiply hote hain? ::: Young's modulus $E$ aur expansion coefficient $\alpha$. Engineers flexible mounts kyun add karte hain? ::: Constraint fraction $c$ lower karne ke liye, jo stress seedhe proportion mein lower karta hai.

Prerequisite links: Structural Dynamics, Finite Element Analysis, Composite Materials in Spacecraft, Thermal Environment in Orbit.