Thermal analysis — conduction in structures, thermal stress: Worked examples

3.6.14 · D3 · Physics › Spacecraft Structures & Systems Engineering › Thermal analysis — conduction in structures, thermal stress

Yeh page ek workout hai. Parent note ne tumhe do engines diye the:

Conduction: $T (x) = T_{1} + \frac{T _{2} - T _{1}}{L} x$ (do siron ke beech linear temperature), aur heat flux $q = - k \frac{T _{2} - T _{1}}{L}$ .
Thermal stress: $σ = - E α Δ T$ jab expansion block ho jaye (compression negative hoti hai).

Yahan hum har tarah ka input test karenge jo yeh engines receive kar sakte hain — har sign, zero cases, "ek end free" case, limiting cases, ek real word problem, aur ek exam twist. Har solution se pehle, Forecast: answer cover karo aur sign aur rough size guess karo. Wahi guess hai jahan learning hoti hai.

Recall Symbols ka matlab kya hai phir se? (kholna ho toh kholein)

$k$ ::: thermal conductivity, W/(m·K) — material mein se heat kitni aasaani se flow karti hai
$α$ ::: coefficient of thermal expansion, 1/K — ek 1 m ki rod 1 K warm hone par kitni stretch hoti hai
$E$ ::: Young's modulus, Pa — stiffness; material ko 100% strain karne ke liye kitna stress chahiye
$Δ T$ ::: stress-free reference $T_{0}$ se temperature change
$σ$ ::: stress, Pa. Positive = tension (alag kheencha gaya), negative = compression (dabaaya gaya)

Scenario matrix

Is topic ka har problem is table ki ek row hai. Humara kaam hai sabko hit karna.

Cell	Kya cheez ise alag banati hai	Example
A. Heating, dono ends fixed	$Δ T > 0$ → badhna chahta hai → blocked → compression	Ex 1
B. Cooling, dono ends fixed	$Δ T < 0$ → sihrna chahta hai → blocked → tension	Ex 2
C. Non-uniform $T (x)$ , bar constrained	ek average $Δ T$ ek single constant $σ$ set karta hai (equilibrium)	Ex 3
D. Zero / degenerate input	$Δ T = 0$ , ya ek end free, ya $α = 0$ (CFRP) → stress = 0	Ex 4
E. Fixed–free (partial constraint)	expansion partially allowed → stress 0 aur full value ke beech	Ex 5
F. Limiting behaviour	$E$ , $α$ , ya $L$ ko bahut bada / chhota karo — kya dominate karta hai?	Ex 6
G. Real-world word problem	conduction + stress chained, units, kW/m²	Ex 7
H. Exam twist: fatigue cycle	jo matter karta hai woh hai range $Δ σ$ ek cycle mein, na ki ek value	Ex 8

Related deep topics: Fatigue and Fracture Mechanics, Material Selection for Spacecraft, Composite Materials in Spacecraft, Thermal Environment in Orbit.

Example 1 — Cell A: heating, fixed–fixed

Step 1 — $Δ T$ nikalo. $Δ T = 70 - 20 = 50 K$ Yeh step kyun? Stress sirf stress-free reference $T_{0}$ se change ki parwah karta hai, absolute temperature ki nahi.

Step 2 — Fixed–fixed formula lagao. $σ = - E α Δ T = - (70 \times 1 0^{9}) (23 \times 1 0^{- 6}) (50)$ $σ = - 80.5 \times 1 0^{6} Pa = - 80.5 MPa$ Yeh step kyun? Dono ends held hain → total strain force ho ke zero → elastic strain ko thermal strain exactly cancel karni padti hai, jisse $σ = - E α Δ T$ milta hai.

Verify: sign negative hai → compression, forecast se match karta hai. Units: $Pa \cdot \frac{1}{K} \cdot K = Pa$ . ✓

Example 2 — Cell B: cooling, fixed–fixed

Step 1 — $Δ T$ nikalo. $Δ T = - 100 - 20 = - 120 K$ Yeh step kyun? Cooling ek negative $Δ T$ deta hai; us sign ko sahi se carry karna hi poora game hai.

Step 2 — Formula lagao. $σ = - E α Δ T = - (70 \times 1 0^{9}) (23 \times 1 0^{- 6}) (- 120) = + 193.2 MPa$ Yeh step kyun? Do minus signs (formula sign aur cooling sign) multiply hokar plus dete hain → tension, bilkul waise jaisa physics kehti hai.

Verify: positive → tension. ✓ Ex 1 se bada hai kyunki $∣Δ T ∣$ bada hai (120 vs 50). Ek thanda bracket aksar dangerous case hota hai — tension cracks kholti hai. Dekho Fatigue and Fracture Mechanics.

Example 3 — Cell C: non-uniform temperature, ek constant stress

Step 1 — $T (x)$ likho. $T (x) = T_{1} + \frac{T _{2} - T _{1}}{L} x = 100 - 150 x$ Yeh step kyun? Steady-state conduction straight line deta hai (parent note, $d^{2} T / d x^{2} = 0$ ). Average compute karne ke liye chahiye.

Step 2 — Equilibrium ek constant axial force force karta hai. Bar ko kahin se bhi kaato. Axial force $F = σ A$ jo do cut faces par pull kar raha hai, balance hona chahiye (koi aur axial loads nahi act karte). $A$ constant hone par, $σ$ har jagah same hai — yeh $T (x)$ ko slice by slice track nahi kar sakta. Yeh step kyun? Yeh "har slice independently blocked" wale naive idea ka correction hai: individual slices independent nahi hain; woh ek saath welded hain aur ek force share karni padti hai.

Step 3 — Average temperature ke saath compatibility. Total elongation zero honi chahiye (ends fixed). Free thermal elongation average rise use karta hai: $⟨ Δ T ⟩ = \frac{1}{L} \int_{0}^{L} [T (x) - T_{0}] d x = \frac{T _{1} + T _{2}}{2} - T_{0} = \frac{100 + ( - 50 )}{2} - 25 = 0 K$ $ε_{total} = \frac{σ}{E} + α ⟨ Δ T ⟩ = 0 ⟹ σ = - E α ⟨ Δ T ⟩$ Yeh step kyun? Bar ka net length change wahi hai jise fixed ends veto karte hain — aur net length change average temperature par depend karta hai, kisi ek point par nahi.

Step 4 — Calculate karo. $σ = - (70 \times 1 0^{9}) (23 \times 1 0^{- 6}) (0) = 0 MPa$

Verify: kyunki $T_{0} = 25$ , $100$ aur $- 50$ ka exactly average hai, average rise zero hai, toh net axial stress zero hai. Neeche figure dekho. ✓

Figure — Thermal analysis — conduction in structures, thermal stress

Figure — kyun axial stress ek number hai, curve nahi. Dashed lavender line $T (x)$ hai jo $+ 10 0^{\circ}$ C se $- 5 0^{\circ}$ C tak girta hai; iska average (mint level) $+ 2 5^{\circ}$ C par baitha hai, $T_{0}$ ke equal. Kyunki average rise zero hai, uniform axial stress (coral band) $0$ MPa par ek flat value hai — na ki tempting slice-by-slice curve. Takeaway: ek uniform constrained bar ke liye, sirf average temperature axial stress mein enter karta hai. (Temperature ke shape se point-to-point bending stresses alag story hain jo Finite Element Analysis handle karta hai.)

Example 4 — Cell D: zero aur degenerate inputs

Step 1 — Case (a): free end. $Δ L = α L Δ T \neq = 0, σ = 0$ Kyun? Kuch bhi growth ko block nahi karta, toh strain puri thermal hai, koi elastic nahi. Displacement without stress — parent note ka classic Mistake 1 trap.

Step 2 — Case (b): koi temperature change nahi. $σ = - E α (0) = 0$ Kyun? Koi $Δ T$ nahi, koi thermal strain nahi jo fight kare. Trivially zero.

Step 3 — Case (c): near-zero CTE. $σ = - E α Δ T = - (E) (0) (200) = 0$ Kyun? Chahe ends fully fixed hon aur $Δ T$ bahut bada ho, agar material barely expand kare toh constrain karne ke liye kuch nahi hai. Isi liye Composite Materials in Spacecraft aur low-CTE alloys choose kiye jaate hain — dekho Material Selection for Spacecraft.

Verify: teenon $0$ dete hain, lekin sirf (c) design story batata hai: $α$ kill karo aur temperature swing chahe kuch bhi ho, thermal stress kill ho jaati hai. ✓

Example 5 — Cell E: fixed–free (partial constraint)

Step 1 — Full fixed–fixed stress (ceiling). $σ_{full} = - E α Δ T = - (110 \times 1 0^{9}) (8.6 \times 1 0^{- 6}) (150) = - 141.9 MPa$ Yeh step kyun? Yeh maximum possible hai — woh case jahan zero expansion allowed hai. Heating compression deta hai (negative), hamare convention se consistent. Baki sab iska ek fraction hai.

Step 2 — Allowed strain elastic (stress-producing) strain ko reduce karta hai. Free thermal strain: $ε_{th} = α Δ T$ . Agar 60% release ho jaaye, toh sirf 40% elastic hai: $ε_{elastic} = - 0.40 α Δ T$ Yeh step kyun? Stress sirf us strain se aata hai jo constraint forbid karta hai. Kuch expansion release karo → less forbidden strain → less stress. Minus compression ko negative rakhta hai.

Step 3 — Stress. $σ = E ε_{elastic} = - 0.40 E α Δ T = 0.40 \times (- 141.9) = - 56.76 MPa (compression)$ Yeh step kyun? Forbidden (elastic) strain ko stiffness $E$ se multiply karo strain ko stress mein convert karne ke liye — wahi $σ = E ε$ jo humne ceiling ke liye use ki, bas reduced strain ke saath.

Verify: $∣ - 141.9 ∣ \geq ∣ - 56.76 ∣ \geq 0$ — magnitude free (0) aur fully-fixed (141.9) bounds ke beech hai, exactly waise jaisa ek partial constraint hona chahiye, aur sign compressive rehta hai. ✓ Isi liye designers "compliance" (flexible mounts) add karte hain — dekho Deployable Structures aur Structural Dynamics.

Example 6 — Cell F: limiting behaviour

Step 1 — Aluminium. $σ_{a} = - E α Δ T = - (70 \times 1 0^{9}) (23 \times 1 0^{- 6}) (100) = - 161 MPa, ∣ σ_{a} ∣ = 161 MPa$

Step 2 — Invar. $σ_{i} = - (140 \times 1 0^{9}) (1.2 \times 1 0^{- 6}) (100) = - 16.8 MPa, ∣ σ_{i} ∣ = 16.8 MPa$ Yeh step kyun? Stress magnitude product $E α$ ke saath scale karti hai. Invar ka $E α = 1.68 \times 1 0^{5}$ hai aluminium ke $1.61 \times 1 0^{6}$ ke comparison mein — factor of ~9.6 chhota.

Step 3 — Limit $α \to 0$ . $lim_{α \to 0} (- E α Δ T) = 0$ Yeh step kyun? $E$ finite hai aur $Δ T$ fixed hai, toh product $α$ ke saath vanish ho jaata hai. Stiffness tumhe save nahi kar sakti agar koi expansion hi nahi hai — aur yeh tumhe doom bhi nahi kar sakti.

Verify: $∣ σ_{i} ∣/∣ σ_{a} ∣ = 16.8/161 = 0.104$ , yaani Invar aluminium ka lagbhag 10 % stress carry karta hai; dono compressive (negative) rehte hain. Low $α$ high $E$ ko beat karta hai. ✓

Example 7 — Cell G: real-world chained problem

Step 1 — $x = 1.5$ m par temperature. $T (x) = 120 + \frac{- 40 - 120}{2} x = 120 - 80 x, T (1.5) = 120 - 120 = 0^{\circ} C$ Yeh step kyun? Linear conduction profile; bas plug in karo.

Step 2 — Heat flux aur power. $q = - k \frac{T _{2} - T _{1}}{L} = - 200 \cdot \frac{- 40 - 120}{2} = - 200 \cdot (- 80) = 16, 000 W/m^{2} = 16 kW/m^{2}$ $P = q A = 16, 000 \times 4 \times 1 0^{- 4} = 6.4 W$ Yeh step kyun? Flux per unit area hai; actual watts nikalne ke liye $A$ se multiply karo jo boom se flow kar rahe hain.

Step 3 — Average rise se uniform axial stress. $⟨ Δ T ⟩ = \frac{T _{1} + T _{2}}{2} - T_{0} = \frac{120 + ( - 40 )}{2} - 40 = 0 K$ $σ = - E α ⟨ Δ T ⟩ = - (70 \times 1 0^{9}) (23 \times 1 0^{- 6}) (0) = 0 MPa$ Yeh step kyun? Same equilibrium logic jaisi Example 3 mein: ek uniform constrained bar ek constant axial stress carry karta hai, jo average temperature rise se set hoti hai — yahan exactly zero.

Verify: flux positive ⇒ heat $+ x$ direction mein hot se cold flow karta hai. ✓ $T (1.5) = 0^{\circ}$ C dono ends ke beech baitha hai. ✓ Average rise zero ⇒ net axial stress zero (halaanki temperature shape local bending stresses drive karta hai jo Finite Element Analysis handle karta hai). ✓ Poori thermal picture ke liye dekho Thermal Control Subsystems aur Thermal Environment in Orbit.

Example 8 — Cell H: exam twist (fatigue cycle)

Step 1 — Temperature swing. $Δ T_{swing} = 50 - (- 100) = 150 K$ Yeh step kyun? Cycle is poori range mein span karta hai; stress iske across swing karta hai.

Step 2 — Stress range. $Δ σ = E α Δ T_{swing} = (110 \times 1 0^{9}) (8.6 \times 1 0^{- 6}) (150) = 141.9 MPa$ Yeh step kyun? Peak tension (cold, $σ = + E α ∣Δ T_{cold} ∣$ ) se peak compression (hot, $σ = - E α Δ T_{hot}$ ) tak, do extreme stresses ka difference $E α Δ T_{swing}$ hai — ek positive range, jo fatigue analysis ko chahiye.

Step 3 — 10 saalon mein cycle count. $\frac{1440 min/day}{90 min/cycle} = 16 cycles/day$ $16 \times 365 \times 10 = 58, 400 cycles$ Yeh step kyun? Har orbit ek poora thermal cycle hai; mission ke liye multiply karo.

Verify: $Δ σ = 141.9$ MPa parent note ke Example 2 se match karta hai. ✓ Lagbhag 6×10⁴ cycles ~142 MPa par ek real Fatigue and Fracture Mechanics driver hai — low-CTE materials ya pre-load se mitigate karo. Detailed life prediction Finite Element Analysis use karta hai. ✓

Recall Self-test

Ek fixed–fixed rod ko cool karna kaun sa sign ka stress deta hai? ::: Positive — tension. Ek rod jiska ek end free hai aur $Δ T = 80$ K hai, uska stress kya hai? ::: Zero — yeh freely expand karta hai. Do materials, same $Δ T$ : less thermal stress kaun deta hai, low $α$ ya low $E$ ? ::: Jo chhota product $E α$ deta hai; low $α$ usually jeetta hai. Ek uniform bar ke liye dono ends fixed aur varying $T (x)$ ke saath, axial stress kya set karta hai? ::: Average temperature rise $⟨ Δ T ⟩$ , ek constant $σ = - E α ⟨ Δ T ⟩$ deta hai.