Yeh ek rapid-fire concept trap bank hai parent topic ke liye. Har line ek question ::: answer reveal hai. Answer ko cover karo, pehle apna verdict aur reason decide karo, phir check karo. Agar sirf "true/false" bol sakte ho lekin kyun nahi — toh samjha nahi abhi tak.
Neeche ke har trap mein kuch specific letters ka sahara liya gaya hai. Koi symbol use karne se pehle yeh batana aana chahiye ki uska matlab kya hai aur uske saath kaun si picture hai. Yeh rahe, har ek clearly:
Traps se pehle, do governing statements ko pin down karo taaki neeche ka har reveal kuch concrete point kar sake. Har ek earned hai, sirf assert nahi ki gayi.
Figure s01 — Rod ke along Temperature vs. position. Blue line true steady-state profile T(x) hai: yeh perfectly straight hai kyunki slope (yellow arrows, teen slices par identically drawn) kabhi nahi badalti, jo constant flux demand karta hai. Red dashed curve ek forbidden profile hai jiska slope vary karta hai; iske liye heat kuch slices mein pile up honi chahiye. Green dots fixed-temperature ends T1 (hot, left) aur T2 (cold, right) mark karte hain.
Figure s02 — Stress sign convention. Top (red): ek heated rod do rigid walls ke beech expand karna chahti hai; walls andar push karti hain (yellow arrows), isliye rod compression mein hai, σ<0. Bottom (blue): ek cooled rod shrink karna chahti hai; walls use stretched pakad ke rakhti hain (yellow arrows outward point karti hain), isliye yeh tension mein hai, σ>0. "Rigid walls" yahan matlab fixed-displacement (mechanical) ends.
Ek rod jo heat up hoti hai lekin dono ends par move karne ke liye free hai, zero thermal stress develop karti hai.
True — koi fixed-displacement constraint nahi, toh rod simply αLΔT se elongate hoti hai aur koi internal force build nahi hoti. Stress ke liye mechanical constraint chahiye, sirf temperature change nahi.
Agar ek bar ke dono ends fixed displacement par held hain aur use heat kiya jaaye, toh stress compressive hoga.
True — heating se material badhna chahta hai, walls length constant rakhne ke liye push back karti hain, aur andar squeeze hona compression hai (σ=−EαΔT<0 for ΔT>0).
Fixed displacement par dono ends held bar ko cool karne se tensile stress produce hota hai.
Fully constrained rod mein thermal stress rod ki length L par depend karta hai.
False — σ=−EαΔT mein koi L nahi; ek lambi rod absolute terms mein zyada expand karti hai lekin strainαΔT (aur isliye stress) length-independent hai.
1D steady state mein constant k ke saath, temperature profile ek straight line hoti hai.
True — constant k se d2T/dx2=0 milta hai, isliye slope constant hai aur T=C1x+C2 linear hai (figure s01 dekho).
Heat flux q, fixed-temperature ends wali 1D steady-state rod ke along vary karta hai.
False — steady state mein dq/dx=0, isliye q har cross section par same hai; jo ek face se enter karta hai woh doosre se exit karta hai.
Higher thermal conductivity k wale material mein hamesha higher thermal stress hota hai.
False — k yeh set karta hai ki heat kitni tezi se flow karti hai (temperature field), stress nahi; stress −EαΔTE, α aur ΔT par depend karta hai. High k gradients ko reduce bhi kar sakta hai aur isliye stress bhi.
CFRP dimensionally stable structures ke liye partly is liye prefer kiya jaata hai kyunki uska α near zero ho sakta hai.
True — α≈0 ke saath, term −EαΔT collapse ho jaata hai regardless of stiffness ya temperature swing, isliye thermal stress aur warping almost vanish ho jaate hain (dekho Composite Materials in Spacecraft).
Fourier's law q=−k∇T mein minus sign ka matlab hai ki heat temperature mein uphill flow kar sakti hai.
False — minus sign downhill flow enforce karta hai: yeh flux vector q ko gradient arrow ∇T ke opposite point karta hai, isliye heat hot se cold ki taraf jaati hai, second law follow karte hue.
"Beam 50 °C heat hoti hai toh 0.5 mm expand hoti hai; isliye yeh stressed hai."
Galat: free expansion displacement hai, stress nahi. Stress tab aata hai jab woh 0.5 mm ek fixed-displacement end se prevent ho. Pehle mechanical boundary condition check karo.
"Average temperature 25 °C hai, isliye main stress ke liye ΔT=25−T0 everywhere use karunga."
Non-uniform field ke liye galat: har slice ka apna T(x) hota hai, isliye stress σ(x)=−Eα[T(x)−T0]bar ke along vary karta hai. Average cold end par tension aur hot end par compression chhupa deta hai.
"Heat flux hai q=kdT/dx."
Minus sign missing hai. Uske bina q gradient ke upar point karega (cold→hot), physics violate karta hua. Correct form: q=−kdT/dx.
"Kyunki hot end expand karna chahta hai, isliye woh tension mein hona chahiye."
Ulta hai: held hote hue expand karna chahna matlab constraint dwara compressed hona → σ=−EαΔT<0. Tension tab milta hai jab woh shrink karna chahta hai (cooling).
"Rod stress hai σ=−EαΔTL."
Stray L galat hai; stress ki units pressure (Pa) hoti hain aur EαΔT already pascals deta hai. Length se multiply karna ek force-like quantity deta hai, stress nahi.
"d2T/dx2=0 tab bhi kaam karta hai jab k strongly temperature par depend karta ho."
Generally nahi — woh clean result constantk assume karta tha. True equation hai dxd(kdT/dx)=0; agar k=k(T) ho, toh k bahar nahi aa sakta, aur profile straight nahi rehti.
Steady state ka matlab hai ek slice mein energy ka koi accumulation nahi, isliye flux andar equals flux bahar — flux constant hai, necessarily zero nahi. Ek nonzero constant flux exactly woh rod hai jo steadily heat conduct kar rahi hai.
Thermal stress formula σ=−EαΔT mein E (stiffness) kyun hai?
Kyunki stress = stiffness × elastic strain, σ=Eεmech. Constraint εmech=−αΔT force karta hai; usse E se multiply karna us strain ko stress mein convert karta hai.
Usi formula mein minus sign kyun hai?
Constraint thermal strain cancel karta hai, isliye mechanical strain thermal one ka negative hai (εmech=−αΔT). Woh negative sign seedha Hooke's law ke through σ=−EαΔT mein flow karta hai, "heating → compression" encode karte hue.
Fatigue ke liye stress range Δσ=EαΔT (magnitude) kyun important quantity hai?
Fatigue cyclic loading se drive hoti hai — har orbit mein max aur min stress ke beech swing — absolute level se nahi. Ek badi swing ek part ko crack kar sakti hai even agar peaks moderate hoon (dekho Fatigue and Fracture Mechanics).
Ek strut ko permanent compression mein pre-stressing karna fatigue life kyun improve karta hai?
Agar part poore thermal cycle mein compressive (σ<0) rehta hai, toh yeh kabhi tensile reversals experience nahi karta jo cracks kholte aur badhate hain; stress ko zero ke ek side par rakhna fatigue crack propagation suppress karta hai.
Designers thermal stress reduce karne ke liye flexible (compliant) mounts kyun add karte hain?
Ek compliant mount fixed-displacement condition relax karta hai. Agar ends thoda move kar sakein, mechanical strain — aur isliye stress −EαΔT — free-expansion (zero-stress) case ki taraf drop ho jaata hai.
Low orbit mein ek spacecraft roughly har 90 minutes mein thermal cycle kyun dekhta hai?
Ek orbit use sunlight aur Earth's shadow se ek baar le jaata hai; shadow mein enter aur exit karna hot–cold swing drive karta hai, isliye cycles orbital period track karte hain (dekho Thermal Environment in Orbit).
Exactly zero — koi temperature change nahi matlab koi thermal strain nahi, isliye σ=−Eα⋅0=0 regardless of kitni rigidly clamped hai.
Agar α=0 ho (ideal zero-expansion material) toh stress kya hoga?
Kisi bhi temperature par zero thermal stress — jab kuch expand hi nahi karna chahta, constraint kuch nahi karta. Yahi near-zero-CTE composites ke peeche design goal hai.
Jab dono ends same fixed temperature T1=T2 par hoon, toh midpoint par temperature kya hai?
Poori rod us temperature par hai — linear profile mein zero slope hai, isliye T(x)=T1 everywhere aur heat flux zero hai.
Fixed-temperature lekin free-to-move (mechanically unconstrained) case mein thermal stress kya hai?
Uniform heating ke liye zero — free ends bar ko unimpeded expand karne dete hain, isliye koi mechanical strain force nahi hoti aur koi stress develop nahi hota, even though bar physically elongate hoti hai.
Agar k zero hota, toh steady-state flux kya hota?
Zero — ek perfect insulator koi heat carry nahi karta, isliye q=−kdT/dx=0; rod do-fixed-temperature steady state reach hi nahi kar sakti bina heat flow ke.
Linear profile T(x)=T1+LT2−T1x ke liye, magnitude mein thermal stress kahan sabse bada hoga?
Jis end par reference temperature T0 se sabse zyada door ho, kyunki σ(x)=−Eα[T(x)−T0] us temperature difference ke saath scale karta hai — aksar dono boundaries mein se ek.
Recall Quick self-test
Fixed-displacement rod, heated: stress ka sign? ::: Compressive, σ=−EαΔT<0 — expand karna chahti hai aur kar nahi sakti.
Kya thermal stress rod length par depend karta hai? ::: Nahi — σ=−EαΔT mein koi L nahi.
Fixed-temperature lekin free-to-move rod, heated: stress? ::: Zero — expansion unopposed hai.
Full steady-state conduction equation kya hai, aur constant k use kya banata hai? ::: ∇⋅(k∇T)=0; constant k ke saath yeh d2T/dx2=0 ban jaata hai, straight-line T(x) deta hai.
