Thermal analysis — conduction in structures, thermal stress: Exercises

3.6.14 · D4 · Physics › Spacecraft Structures & Systems Engineering › Thermal analysis — conduction in structures, thermal stress

Yeh ek self-testing page hai. Har problem ka ek clean statement diya gaya hai; worked solution ek collapsible callout mein chhupa hua hai — pehle khud try karo, phir reveal karo. Problems L1 Recognition (sahi formula dhundho) se lekar L5 Mastery (design-level synthesis) tak climb karti hain.

Yahan sab kuch parent note par build karta hai. Shuru karne se pehle, make sure karo ki tumhare paas woh chaar tools hain jo usne build kiye the:

Recall Chaar formulas jo tum use karoge

Fourier's law (1D): $q = - k \frac{d T}{d x}$ — heat flux, conductivity ka temperature graph ki steepness ke saath product hai, minus sign ke saath kyunki heat downhill hot se cold ki taraf jaati hai. Linear temperature profile: $T (x) = T_{1} + \frac{T _{2} - T _{1}}{L} x$ — dono end temperatures ke beech ek seedhi line. Thermal stress, full constraint: $σ = - E α Δ T$ — woh stress jo ek rod feel karti hai jab woh $α Δ T$ se expand karna chahti hai lekin use allow nahi kiya jaata. Free expansion: $Δ L = α L Δ T$ — ek free rod har degree mein kitni length gain karti hai.

Symbols: $k$ = thermal conductivity (W/m·K), $E$ = Young's modulus (Pa, "stiffness"), $α$ = coefficient of thermal expansion (per K, "heat ke saath kitna badhne ka mann karta hai"), $Δ T$ = temperature change (K), $L$ = length (m), $A$ = cross-section area (m²).

Ek quick sign compass haath mein rakho:

Figure — Thermal analysis — conduction in structures, thermal stress

Figure s01 — Sign compass. Left panel: ek red rod do black hatched walls ke beech clamp ki hui hai aur heated ki ja rahi hai; rod ke andar ke black arrows baahaar ki taraf point karte hain (woh grow karna chahti hai) jabki walls usse block karti hain, isliye rod compression (σ < 0) mein end hoti hai. Right panel: wahi clamped red rod cooled ki ja rahi hai; andar ke arrows andar ki taraf point karte hain (woh shrink karna chahti hai) jabki walls usse baahaar hold karti hain, isliye rod tension (σ > 0) mein hai. Take-away: ek held rod ko heat karo → negative stress, ek held rod ko cool karo → positive stress.

Kisi bhi flux problem se pehle, is sketch ke saath coordinate convention fix karo:

Figure s03 — Flux sign convention. Ek horizontal rod left (x = 0, hot, red dot) se right (x = L, cold) tak jaati hai. Black $x$ -axis rightward (+x hot end se cold end ki taraf) point karta hai. Ek red arrow jis par $q > 0$ likha hai same +x direction mein point karta hai, dikhata hai ki heat physically hot end se cold end ki taraf flow kar rahi hai. Upar ka temperature graph left se right tak girta hai, isliye iska slope $d T / d x$ negative hai; usse $q = - k d T / d x$ mein feed karne par negative slope ek positive $q$ ban jaata hai. Isi liye " $q$ positive" aur "heat flows in +x (hot → cold)" ka ek hi matlab hai.

Level 1 — Recognition

L1·Q1 — Flux sign padho

Ek rod mein $T_{1} = 8 0^{\circ} C$ hai $x = 0$ par aur $T_{2} = 2 0^{\circ} C$ hai $x = L$ par. $k$ ya $L$ ke liye kisi bhi number ke bina, heat kis direction mein flow karti hai, aur kya flux $q$ positive hai ya negative (jahan $+ x$ end 1 se end 2 ki taraf point karta hai)?

Recall Solution

Heat hamesha hot se cold ki taraf flow karti hai. Hot end $x = 0$ par hai, cold end $x = L$ par hai, isliye heat $+ x$ direction mein flow karti hai → flux positive hai (figure s03 se compare karo). Formula se check karo: $T$ ka graph $x$ badhne ke saath girta hai, isliye $\frac{d T}{d x} < 0$ . Tab $q = - k \frac{d T}{d x} = - k \cdot (negative) = positive$ . ✓

L1·Q2 — Formula chuno

Tumhe bataya gaya hai ki ek strut dono ends par rigidly bolted hai aur uniformly heat ki ja rahi hai. Tumhe stress find karna hai. Kaun sa single formula apply hota hai, aur answer ka sign tumhe kya batata hai?

Recall Solution

Dono ends fixed + uniform heating = full constraint, isliye $σ = - E α Δ T$ use karo. Uniform heating ka matlab $Δ T > 0$ hai, isliye $σ < 0$ → compression. Rod ko squeeze kiya ja raha hai kyunki woh expand nahi kar sakti (figure s01 ka left panel).

Level 2 — Application

L2·Q1 — Midpoint temperature

Ek copper bracket ki length $L = 0.4$ m hai, $T_{1} = 6 0^{\circ} C$ aur $T_{2} = - 4 0^{\circ} C$ hai. $x = 0.1$ m par temperature find karo.

Recall Solution

Linear profile $T (x) = T_{1} + \frac{T _{2} - T _{1}}{L} x$ use karo. Slope $= \frac{- 40 - 60}{0.4} = \frac{- 100}{0.4} = - 250 K/m$ . $T (0.1) = 60 + (- 250) (0.1) = 60 - 25 = 3 5^{\circ} C$ . Yeh kaisa dikhta hai: ek seedhi line jo steadily girte jaati hai; quarter raste par jaane ke baad humne total $10 0^{\circ}$ swing ka ek quarter drop kiya hai.

L2·Q2 — Heat flux magnitude

Wohi copper bracket ( $k = 400$ W/m·K, $L = 0.4$ m, $T_{1} = 6 0^{\circ} C$ , $T_{2} = - 4 0^{\circ} C$ , cross-section $A = 2 \times 1 0^{- 4}$ m²). Heat flux $q$ aur total heat power $P = q A$ find karo jo usme flow ho rahi hai.

Recall Solution

Fourier's law $q = - k \frac{T _{2} - T _{1}}{L}$ se shuru karo aur ek ek piece build karo taaki koi step chhupe nahi. Step 1 — temperature difference. $T_{2} - T_{1} = - 40 - 60 = - 100$ K. Negative kyun? End 2, end 1 se thanda hai, isliye $+ x$ mein jaate waqt temperature $100$ K drop karti hai. Step 2 — slope (gradient). $\frac{T _{2} - T _{1}}{L} = \frac{- 100}{0.4} = - 250$ K/m. Kyun? Us $- 100$ K drop ko length ke $0.4$ m par spread karo; har metre $250$ K drop karta, aur sign negative rehta kyunki temperature $x$ ke saath girti hai. Step 3 — $- k$ se multiply karo. $q = - 400 \times (- 250)$ . Do minus signs kyun? $- k$ Fourier's downhill rule hai (heat gradient ke neeche run karti hai); $- 250$ falling slope hai. Ek negative ka negative positive hota hai: $- 400 \times (- 250) = + 100, 000$ . Toh $q = 100, 000 W/m^{2} = 100 kW/m^{2}$ . Positive → heat $+ x$ mein flow karti hai (hot se cold), figure s03 se match karta hai. ✓ Step 4 — total power. $P = q A = 100, 000 \times 2 \times 1 0^{- 4} = 20 W$ . Area se multiply kyun? Flux power per square metre hai; cross-section $A$ se multiply karne par actual watts milti hain jo poori rod se pass ho rahi hain.

L2·Q3 — Free expansion length

Ek steel rail-jaisi fitting, $L = 1.2$ m, $α = 12 \times 1 0^{- 6}$ /K, $- 3 0^{\circ} C$ se $+ 9 0^{\circ} C$ tak warm hoti hai jabki dono ends par free hai. Yeh kitni lambi ho jaati hai, aur stress kya hai?

Recall Solution

$Δ T = 90 - (- 30) = 120$ K. $Δ L = α L Δ T = 12 \times 1 0^{- 6} \times 1.2 \times 120 = 1.728 \times 1 0^{- 3}$ m $= 1.728$ mm. Kyunki woh free hai, kuch bhi growth resist nahi karta → $σ = 0$ . Displacement hai, stress nahi.

Level 3 — Analysis

L3·Q1 — Gradient ke dono ends par stress

Ek aluminum beam ( $E = 70$ GPa, $α = 23 \times 1 0^{- 6}$ /K), fully constrained, $T_{1} = 10 0^{\circ} C$ se $T_{2} = - 5 0^{\circ} C$ tak linear profile carry karti hai. Reference temperature $T_{0} = 2 5^{\circ} C$ (stress-free assembly temperature) use karte hue, $σ (0)$ aur $σ (L)$ find karo.

Figure s02 — Beam ke saath temperature aur stress. Black seedhi line $T (x)$ hai, $x = 0$ par $+ 10 0^{\circ}$ C se $x = L$ par $- 5 0^{\circ}$ C tak girte hue. Black dashed horizontal line stress-free reference $T_{0} = 2 5^{\circ}$ C mark karti hai. Red dot wahan baitha hai jahan temperature line $T_{0}$ ko cross karti hai, $x = 0.5 L$ par — yahi exactly woh jagah hai jahan thermal stress zero hai. Red dot ke left mein material $T_{0}$ se zyada hot hai (expand karna chahta hai → compression); right mein material $T_{0}$ se thanda hai (shrink karna chahta hai → tension). Isliye stress linearly hot end par $- 121$ MPa se cold end par $+ 121$ MPa tak jaata hai, red dot par sign flip karte hue.

Recall Solution

Constrained slice ke liye local stress: $σ (x) = - E α [T (x) - T_{0}]$ . Hot end $x = 0$ : $T - T_{0} = 100 - 25 = 75$ K. $σ (0) = - 70 \times 1 0^{9} \times 23 \times 1 0^{- 6} \times 75 = - 120.75 \times 1 0^{6} \approx - 121$ MPa → compression (hot end grow karna chahta hai, squeeze kiya ja raha hai). Cold end $x = L$ : $T - T_{0} = - 50 - 25 = - 75$ K. $σ (L) = - 70 \times 1 0^{9} \times 23 \times 1 0^{- 6} \times (- 75) = + 120.75 \times 1 0^{6} \approx + 121$ MPa → tension (cold end shrink karna chahta hai, stretch kiya ja raha hai). Yeh kaisa dikhta hai (figure s02): stress ek seedhi line hai jo exactly wahan zero cross karti hai jahan $T (x) = T_{0} = 2 5^{\circ} C$ hai, yani yahan midpoint par.

L3·Q2 — Stress zero kahan hai?

L3·Q1 ki beam ke liye, kis position $x$ (aur temperature) par thermal stress exactly zero hai?

Recall Solution

$σ (x) = 0$ ke liye $T (x) = T_{0} = 2 5^{\circ} C$ chahiye. Profile: $T (x) = 100 + \frac{- 50 - 100}{L} x = 100 - \frac{150}{L} x$ . Set $= 25$ : $100 - \frac{150}{L} x = 25 \Rightarrow \frac{150}{L} x = 75 \Rightarrow x = \frac{75}{150} L = 0.5 L$ . Neutral point midpoint par baitha hai, aadhe raste mein, kyunki $2 5^{\circ}$ exactly $10 0^{\circ}$ aur $- 5 0^{\circ}$ ke beech mein hai. Ek taraf material compressed hai, doosri taraf stretched hai (figure s02 mein red dot).

L3·Q3 — Kaun sa end zyada dangerous hai?

Ek material tension mein $150$ MPa par fail hoti hai lekin $400$ MPa compression tolerate karti hai (ek common ceramic-jaisi asymmetry). L3·Q1 ke stresses use karte hue, kaun sa end failure ke zyada close hai, aur safety margin kya hai?

Recall Solution

Compression end: $∣ σ ∣ = 121$ MPa, $400$ MPa limit ke against → margin $= 400/121 \approx 3.3$ . Tension end: $σ = + 121$ MPa, $150$ MPa limit ke against → margin $= 150/121 \approx 1.24$ . Cold (tension) end kaafi zyada dangerous hai: uska margin sirf $1.24$ hai $3.3$ ke against. Brittle spacecraft materials cold end par tensile thermal stress se crack karti hain, hot end par nahi — pehli intuition ke ulta.

Level 4 — Synthesis

Definition Stress range

$Δ σ$ Jab ek structure ek cold state aur ek hot state ke beech cycle karta hai, toh uska stress ek most-tensile value aur ek most-compressive value ke beech swing karta hai. Stress range $Δ σ$ us swing ki total size hai — dono extremes ke beech peak-to-peak distance, hamesha positive magnitude ke roop mein li jaati hai. Fatigue theory isi ki parwah karti hai, kyunki woh repeated back-and-forth hi cracks badhata hai, na ki absolute sign. Kyunki do extremes sirf temperature swing $Δ T = T_{hot} - T_{cold}$ se differ karte hain, aur $σ = - E α T$ har state par, unhe subtract karne par minus sign aur reference $T_{0}$ cancel ho jaata hai: $Δ σ = ∣ σ_{hot} - σ_{cold} ∣ = E α Δ T$ . Hum minus isliye nahi drop karte kyunki hum careless hain, balki isliye ki ek range ek distance hai, aur distances positive hote hain.

L4·Q1 — Fatigue cycle count

Ek titanium strut ( $E = 110$ GPa, $α = 8.6 \times 1 0^{- 6}$ /K), fully constrained, har $90$ -minute orbit mein ek baar $- 10 0^{\circ} C$ aur $+ 5 0^{\circ} C$ ke beech cycle karti hai. Find karo (a) stress range $Δ σ$ , aur (b) ek $10$ -year mission mein cycles ki sankhya.

Recall Solution

(a) Stress range peak-to-peak swing hai, isliye woh sirf temperature swing par depend karta hai: $Δ T = 50 - (- 100) = 150$ K. $Δ σ = E α Δ T = 110 \times 1 0^{9} \times 8.6 \times 1 0^{- 6} \times 150 = 141.9 \times 1 0^{6} \approx 142$ MPa. (Yeh most-compressive se most-tensile tak swing ki magnitude hai; upar ki definition ke anusaar, ek range ek positive distance hai, isliye $σ = - E α Δ T$ ka minus sign drop ho jaata hai.) (b) Cycles per day: $\frac{1440 min/day}{90 min/cycle} = 16$ . Per year: $16 \times 365 = 5840$ . Ten years: $5840 \times 10 = 58, 400 \approx 5.8 \times 1 0^{4}$ cycles. Yeh ek real fatigue driver hai — dekho Fatigue and Fracture Mechanics.

L4·Q2 — Stress reduction ke liye material trade

L4·Q1 ki strut $142$ MPa range experience karti hai. Ek designer titanium ki jagah ek CFRP tube swap karta hai jisme $α_{CFRP} = 0.5 \times 1 0^{- 6}$ /K aur $E_{CFRP} = 130$ GPa hai. New stress range kya hai, aur yeh kitne factor se drop hua?

Recall Solution

$Δ σ_{CFRP} = 130 \times 1 0^{9} \times 0.5 \times 1 0^{- 6} \times 150 = 9.75 \times 1 0^{6} \approx 9.75$ MPa. Reduction factor $= \frac{142}{9.75} \approx 14.6 \times$ . Chahe CFRP stiffer hai (higher $E$ ), uska near-zero $α$ product $E α$ ko dominate karta hai. Isi liye Composite Materials in Spacecraft aur sochhe-samjhe Material Selection for Spacecraft thermal-stress ki ladaai jeet jaate hain — tum $α$ factor par attack karte ho, $E$ factor par nahi.

L4·Q3 — Partial constraint (fixed–spring)

Ek rod ( $E$ , $α$ , area $A$ , length $L$ ) $x = 0$ par fixed hai lekin $x = L$ par rigid wall ki jagah stiffness $k_{s}$ (N/m) ki ek soft spring se hold ki gayi hai. Uniform heating $Δ T$ par, free expansion $Δ L = α L Δ T$ ka sirf ek hissa allow hota hai. Dikhao ki rod force yeh hai: $F = \frac{E A α Δ T}{1 + \frac{E A}{k _{s} L}},$ aur do limits $k_{s} \to \infty$ aur $k_{s} \to 0$ check karo.

Recall Solution

Maan lo far end actually $δ$ baahaar move karta hai (free $Δ L$ se kam). Rod blocked amount $(Δ L - δ)$ se compress hoti hai, jo ek compressive force produce karti hai: $F = \frac{E A}{L} (Δ L - δ) .$ Wohi force spring ko $δ$ baahaar push karta hai, isliye $F = k_{s} δ \Rightarrow δ = F / k_{s}$ . Substitute karo: $F = \frac{E A}{L} (α L Δ T - \frac{F}{k _{s}}) = E A α Δ T - \frac{E A}{k _{s} L} F .$ $F$ collect karo: $F (1 + \frac{E A}{k _{s} L}) = E A α Δ T$ , se milta hai: $F = \frac{E A α Δ T}{1 + \frac{E A}{k _{s} L}} .$ Limit $k_{s} \to \infty$ (rigid wall): denominator $\to 1$ , isliye $F \to E A α Δ T$ , yaani stress $σ = F / A = E α Δ T$ — full-constraint result. ✓ Limit $k_{s} \to 0$ (free end): denominator $\to \infty$ , isliye $F \to 0$ — ek free rod koi stress carry nahi karti. ✓ Soft spring smoothly dono textbook cases ke beech interpolate karta hai — yahi real spacecraft situation hai, jahan mounts stiff hote hain lekin kabhi perfectly rigid nahi hote.

Intuition Edge cases jo tum clearly state kar sako

$Δ T = 0$ (koi temperature change nahi): har formula collapse ho jaata hai — $Δ L = 0$ , $σ = - E α \cdot 0 = 0$ , aur $Δ σ = 0$ . Bilkul bhi koi thermal loading nahi, chahe constraint kitni bhi stiff kyun na ho. Thermal stress completely temperature change se drive hota hai; $Δ T$ khatam karo aur stress khatam. Cycling ke under $k_{s} \to 0$ (perfectly compliant mounts): L4·Q3 se, $F \to 0$ , isliye stress range $Δ σ \to 0$ bhi ho jaata hai. Ek perfectly floppy mount koi force transmit nahi karta chahe rod hot-cold-hot cycle kare, isliye usse koi fatigue loading nahi milti — rod simply freely expand aur contract karti hai. Yahi flexure mounts aur thermal isolators ke peeche limiting ideal hai: thodi positional stiffness trade karo, cyclic stress mein badi drop ke liye. Catch yeh hai ki ek truly floppy mount structure ko jagah par hold bhi nahi kar sakta, isliye real designs ek finite $k_{s}$ choose karte hain jo pointing stiffness aur fatigue ke beech balance kare.

Level 5 — Mastery

L5·Q1 — Full design decision

Ek optical bench truss length $L = 0.5$ m ke struts use karta hai, fully constrained, ek orbital swing $Δ T = 150$ K par. Requirement: $1 0^{7}$ -cycle fatigue life ke liye stress range $Δ σ \leq 40$ MPa. Teen candidate materials:

Material	$E$ (GPa)	$α$ ( $1 0^{- 6}$ /K)
Aluminum 6061	69	23.6
Titanium 6Al-4V	114	8.6
CFRP (quasi-iso)	70	2.0

(a) Har ek ke liye $Δ σ = E α Δ T$ compute karo. (b) Kaun se $40$ MPa gate pass karte hain? (c) Jo fail karte hain unke liye, $40$ MPa gate ke under laane ke liye L4·Q3 se kaun sa compliance factor $R = \frac{1}{1 + E A / ( k _{s} L )}$ chahiye hoga?

Recall Solution

(a) $Δ σ = E α Δ T$ use karte hue $Δ T = 150$ K ke saath:

Aluminum: $69 \times 1 0^{9} \times 23.6 \times 1 0^{- 6} \times 150 = 244.3 \times 1 0^{6} \approx 244$ MPa.
Titanium: $114 \times 1 0^{9} \times 8.6 \times 1 0^{- 6} \times 150 = 147.1 \times 1 0^{6} \approx 147$ MPa.
CFRP: $70 \times 1 0^{9} \times 2.0 \times 1 0^{- 6} \times 150 = 21.0 \times 1 0^{6} = 21$ MPa. (b) Gate $40$ MPa hai. Sirf CFRP pass karta hai ( $21 < 40$ ). Aluminum aur titanium fail karte hain. (c) Compliance, constraint stress ko $R$ se multiply karta hai (L4·Q3 se, $σ_{eff} = R \cdot E α Δ T$ ). Humein $R \leq 40/Δ σ_{rigid}$ chahiye:
Aluminum: $R \leq 40/244 = 0.164$ — mount sirf rigid-constraint force ka $\sim 16%$ pass kare (bahut soft mounts).
Titanium: $R \leq 40/147 = 0.272$ — mount sirf $\sim 27%$ pass kare. Design call: CFRP gate bina kisi special mounting ke meet karta hai, isliye yeh lowest-risk choice hai: koi floppy flexures design nahi karne, koi positional-stiffness penalty nahi, aur margin bhi bacha ( $21$ vs $40$ MPa). Aluminum aur titanium dono ko deliberately compliant mounts (flexures / thermal isolators) chahiye hote un $R$ values ke size ke, jo mass aur complexity add karte jabki pointing stiffness erode karte. Yeh precision structures mein low- $α$ Composite Materials in Spacecraft ki textbook win hai; dekho Deployable Structures bhi jahan isi stress se thermal snap-through ek failure mode hai.

L5·Q2 — Coupled conduction + stress budget

Ek aluminum strut ( $k = 200$ W/m·K, $E = 70$ GPa, $α = 23 \times 1 0^{- 6}$ /K, $L = 0.8$ m, $A = 1 \times 1 0^{- 4}$ m²) ek $T_{1} = 12 0^{\circ} C$ fitting aur ek $T_{2} = - 14 0^{\circ} C$ radiator ko bridge karta hai, fully constrained, $T_{0} = 2 0^{\circ} C$ par stress-free. Find karo (a) conducted heat power, (b) cold end par tensile stress, aur (c) kya yeh ek $200$ MPa tensile allowable mein survive karta hai.

Recall Solution

(a) Flux $q = - k \frac{T _{2} - T _{1}}{L} = - 200 \cdot \frac{- 140 - 120}{0.8} = - 200 \cdot (- 325) = 65, 000 W/m^{2}$ . Power $P = q A = 65, 000 \times 1 \times 1 0^{- 4} = 6.5$ W hot→cold flow kar rahi hai. (b) Cold-end stress: $σ (L) = - E α [T_{2} - T_{0}] = - 70 \times 1 0^{9} \times 23 \times 1 0^{- 6} \times (- 140 - 20)$ $= - 70 \times 1 0^{9} \times 23 \times 1 0^{- 6} \times (- 160) = + 257.6 \times 1 0^{6} \approx + 258$ MPa (tension). (c) $258$ MPa $> 200$ MPa allowable → yeh fail karta hai cold end par. Conduction theek hai ( $6.5$ W easily radiated hai), lekin frigid end par tensile thermal stress limit exceed kar raha hai. Remedy: low- $α$ material swap karo, compliance add karo, ya fitting ko Thermal Control Subsystems ke zariye thermally isolate karo effective $T - T_{0}$ shrink karne ke liye.

Recall Self-test recall

Hot-at-x=0, cold-at-x=L ke liye heat flow ki direction, jab +x 0 se L tak ho ::: +x direction, flux positive. Ek held, uniformly heated rod mein thermal stress ka sign ::: negative (compression). Linear gradient mein thermal stress kahan zero hai ::: jahan $T (x)$ stress-free reference $T_{0}$ ke barabar ho. Kaun sa product shrink karna hai thermal stress cut karne ke liye, aur kaun sa factor usually dominate karta hai ::: $E α$ shrink karo; $α$ (expansion) factor usually choice dominate karta hai. $k_{s} \to \infty$ par fixed–spring formula se full-constraint stress ::: $σ \to E α Δ T$ . $Δ T = 0$ hone par thermal stress ka kya hota hai ::: bilkul khatam ho jaata hai; koi temperature change nahi matlab koi thermal loading nahi. Jab mounts perfectly compliant ho jaate hain ( $k_{s} \to 0$ ) toh fatigue stress range ka kya hota hai ::: range $Δ σ \to 0$ ; ek floppy mount koi cyclic force transmit nahi karta.