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

3.6.3 · D4 · Physics › Spacecraft Structures & Systems Engineering › Stress and strain — σ = F - A, ε = ΔL - L, Young's modulus E

Toolbox jo hum poore raste use karenge:

Level 1 — Recognition

Problem 1.1 (L1)

Ek round strut ka diameter $d = 20 mm$ hai. Uska cross-section ek circle hai. Uska cross-sectional area $A$ $m^{2}$ mein kya hoga?

Recall Solution 1.1

WHAT chahiye: woh area jis par force phailta hai. WHY: stress force per area hai, isliye jab tak $A$ $m^{2}$ mein na ho, hum kuch bhi compute nahi kar sakte.

Radius $r = d /2 = 10 mm = 0.010 m$ . Circle ka area: $A = π r^{2} = π (0.010)^{2} = π \times 1 0^{- 4} \approx 3.14 \times 1 0^{- 4} m^{2} .$ Dhyan do: $20 mm = 0.020 m$ ; metres ko square karna, na millimetres ko — yahi SI ko sahi rakhta hai.

Problem 1.2 (L1)

Ek bar ki original length $L = 2.5 m$ hai aur woh $Δ L = 1.25 mm$ stretch hoti hai. Strain ko ek pure number ke roop mein aur microstrain mein likho ( $1 μ ε = 1 0^{- 6}$ ).

Recall Solution 1.2

WHAT: fractional stretch. WHY $L$ se divide karte hain: wahi $1.25 mm$ ek chhhoti bar par bahut zyada hoti hai aur lambi par kuch nahi — sirf ratio material ka description deta hai. $ε = \frac{Δ L}{L} = \frac{1.25 \times 1 0 ^{- 3}}{2.5} = 5 \times 1 0^{- 4} .$ Microstrain mein: $5 \times 1 0^{- 4} = 500 \times 1 0^{- 6} = 500 μ ε$ .

Problem 1.3 (L1)

Ek material ka $E = 70 GPa$ hai. Stress $σ = 35 MPa$ ke under (abhi bhi elastic hai), strain kya hogi?

Recall Solution 1.3

WHY yeh formula: $σ = E ε$ materials ke liye Hooke's Law hai; stress aur stiffness se strain nikalne ke liye rearrange karo: $ε = σ / E$ . $ε = \frac{35 \times 1 0 ^{6}}{70 \times 1 0 ^{9}} = 5 \times 1 0^{- 4} .$ GPa ko zaroor $70 \times 1 0^{9} Pa$ banana padega, warna answer $1000$ se galat ho jayega.

Level 2 — Application

Problem 2.1 (L2)

Ek aluminium tie ( $E = 70 GPa$ ), length $L = 1.5 m$ , area $A = 2 \times 1 0^{- 4} m^{2}$ , tensile force $F = 1.4 \times 1 0^{4} N$ carry karta hai. Stress, strain, aur extension $Δ L$ nikalo.

Recall Solution 2.1

Step 1 — stress (WHAT: force ki bheed). $σ = \frac{F}{A} = \frac{1.4 \times 1 0 ^{4}}{2 \times 1 0 ^{- 4}} = 7 \times 1 0^{7} Pa = 70 MPa .$ Step 2 — strain (WHY: Hooke's law se stiffness undo karo). $ε = \frac{σ}{E} = \frac{7 \times 1 0 ^{7}}{70 \times 1 0 ^{9}} = 1 \times 1 0^{- 3} .$ Step 3 — extension (WHY: woh "divide by $L$ " undo karo jo strain define karta tha). $Δ L = ε L = 1 0^{- 3} \times 1.5 = 1.5 \times 1 0^{- 3} m = 1.5 mm .$

Problem 2.2 (L2)

Ek steel rod ( $E = 200 GPa$ ) ki length $L = 0.8 m$ aur area $A = 5 \times 1 0^{- 5} m^{2}$ hai; yeh $0.2 mm$ se zyada stretch nahi honi chahiye. Sabse bada force kya ho sakta hai jo yeh carry kar sake (assume karo yeh elastic rahega)?

Recall Solution 2.2

WHAT solve karna hai: $F$ , $Δ L$ par ek cap diya gaya hai. WHY rearrange: deflection equation $Δ L = F L / (A E)$ mein $F$ hai; use isolate karo. $F = \frac{A E Δ L}{L} = \frac{( 5 \times 1 0 ^{- 5} ) ( 200 \times 1 0 ^{9} ) ( 0.2 \times 1 0 ^{- 3} )}{0.8} .$ Numerator: $5 \times 1 0^{- 5} \times 2 \times 1 0^{11} = 1 \times 1 0^{7}$ ; times $0.2 \times 1 0^{- 3}$ deta hai $2000$ . $0.8$ se divide karo: $F = \frac{2000}{0.8} = 2500 N .$

Problem 2.3 (L2)

Ek titanium cable ( $E = 110 GPa$ ) ko $L = 1.2 m$ par $F = 6600 N$ carry karte hue apna stretch $Δ L = 0.4 mm$ tak limit karna hai. Minimum cross-sectional area nikalo.

Recall Solution 2.3

Pehle forecast karo: kam stretch allowed $\Rightarrow$ humein zyada material chahiye, toh expect karo ek chhhota par na-zyada-chhhota $A$ . WHY: unknown geometry $A$ ke liye $Δ L = F L / (A E)$ solve karo.

= \frac{7920}{4.4\times10^{7}} = 1.8\times10^{-4}\ \text{m}^2 = 1.8\ \text{cm}^2.$$ **Verify:** $\Delta L = \dfrac{6600\times1.2}{1.8\times10^{-4}\times1.1\times10^{11}} = \dfrac{7920}{1.98\times10^{7}} = 4\times10^{-4}\ \text{m} = 0.4\ \text{mm}$ ✓.

Level 3 — Analysis

Problem 3.1 (L3)

Ek strut ko garam kiya jaata hai toh woh expand karna chahti hai, lekin uske ends rigidly clamp hain toh length change nahi ho sakti. Material ka $E = 70 GPa$ hai aur, agar free hota, toh fractional amount $ε_{free} = 8 \times 1 0^{- 4}$ se expand hota. Clamp kaun sa stress induce karta hai, aur woh tension hai ya compression? (Yeh ek Thermal Stress setup hai.)

Recall Solution 3.1

WHAT ho raha hai: heat bar ko $ε_{free}$ strain se lamaba karna chahti hai, lekin clamps net length change ko zero karne par majboor karte hain. Ek hone-wali expansion cancel karne ke liye, clamps bar ko wapas push karte hain — matlab use squeeze karte hain. Toh mechanical strain $- ε_{free}$ hai (ek compression), figure dekho.

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

WHY sign flip hoti hai: strain signed hoti hai. Stretch $+$ hai, squeeze $-$ hai. Kyunki wall expansion rok deti hai, material aisa behave karta hai jaise use utna hi compress kiya gaya jitna woh badhna chahta tha. $σ = E ε = (70 \times 1 0^{9}) (- 8 \times 1 0^{- 4}) = - 5.6 \times 1 0^{7} Pa = - 56 MPa .$ Minus sign matlab compression hai — magnitude $56 MPa$ . Agar ise thanda kiya jaata, toh bar shrink karne ki koshish karti, clamps use stretch kar ke rakhte, aur sign $+$ (tension) hoti.

Problem 3.2 (L3)

Do struts A aur B same aluminium se bane hain aur same force $F$ carry karte hain. Strut B ka diameter A ka double hai. (i) unke stresses aur (ii) unke strains compare karo.

Recall Solution 3.2

WHY area matter karta hai: stress $F / A$ hai, aur area diameter ke square ke saath scale karta hai, $A = π (d /2)^{2}$ . $d$ double karne se $A$ $4$ guna ho jaata hai.

Neeche do circles dekho: B ka circle A ke circle se $4 \times$ bada hai.

(i) Stress: $σ_{B} = F / A_{B} = F / (4 A_{A}) = σ_{A} /4$ . Strut B ko ek chauthai stress feel hota hai. (ii) Strain: same material ka matlab same $E$ hai, toh $ε = σ / E$ same tarah scale karta hai: $ε_{B} = ε_{A} /4$ . Toh mota strut kam crowded feel karta hai aur proportionally kam stretch bhi karta hai. Yahan sab kuch geometry ne kiya, material ne nahi.

Problem 3.3 (L3)

Ek steel wire ( $E = 200 GPa$ , yield stress $σ_{y} = 250 MPa$ ) jis ka area $A = 1 \times 1 0^{- 6} m^{2}$ hai, $F = 300 N$ carry karta hai. Kya wire abhi bhi elastic region mein hai jahan $σ = E ε$ valid hai? Agar haan, toh uski strain nikalo.

Recall Solution 3.3

WHY pehle yield check karo: Hooke's Law sirf yield se neeche hold karta hai; uske aage material plastic behaviour mein jaata hai aur hamare formulas galat ho jaate hain. $σ = \frac{F}{A} = \frac{300}{1 \times 1 0 ^{- 6}} = 3 \times 1 0^{8} Pa = 300 MPa .$ Compare karo: $σ = 300 MPa > σ_{y} = 250 MPa$ . Wire yield ho gayi hai — yeh elastic region mein nahi hai, toh $ε = σ / E$ valid nahi hai aur wire permanently deform ho gayi hai (ya toot rahi hai). Honest answer hai "yahan $σ = E ε$ use nahi kar sakte." Stress-Strain Curve dekho jahan straight line khatam hoti hai.

Level 4 — Synthesis

Problem 4.1 (L4)

Ek launch strut ko $F = 4.0 \times 1 0^{4} N$ ke under aluminium ( $E = 70 GPa$ , yield $σ_{y} = 240 MPa$ ) ke $L = 2.0 m$ mein teen rules ek saath satisfy karne hain: (a) stress Factor of Safety $2$ ke saath yield se neeche rehni chahiye (yaani $σ \leq σ_{y} /2$ ); (b) extension $2.0 mm$ se zyada nahi hona chahiye. Dono ko satisfy karne wala minimum area nikalo, aur batao kaun sa rule govern karta hai.

Recall Solution 4.1

WHY do rules: strength aur stiffness alag limits hain (ek ceiling vs ek slope), isliye hum har ek ke liye separately area compute karte hain, phir bada lete hain — yahi binding constraint hai.

Rule (a), strength. Allowed stress $σ_{allow} = σ_{y} /2 = 120 MPa = 1.2 \times 1 0^{8} Pa$ . Chahiye $σ = F / A \leq σ_{allow}$ : $A_{a} = \frac{F}{σ _{allow}} = \frac{4.0 \times 1 0 ^{4}}{1.2 \times 1 0 ^{8}} = 3.33 \times 1 0^{- 4} m^{2} .$

Rule (b), stiffness. Chahiye $Δ L = F L / (A E) \leq 2.0 \times 1 0^{- 3} m$ :

= \frac{8.0\times10^{4}}{1.4\times10^{8}} = 5.71\times10^{-4}\ \text{m}^2.$$ **Decide karo:** strut ko *dono* satisfy karne hain, toh bada lo: $A_{\min} = \max(A_a, A_b) = 5.71\times10^{-4}\ \text{m}^2$. **Stiffness govern karta hai** — stretch limit akele strength se zyada mota strut force karta hai. [[Spacecraft Load Paths and Struts]] dekho ki pointing-critical structures mein deflection kyun aksar win karti hai.

Problem 4.2 (L4)

Ek single load path mein, force $F = 1.0 \times 1 0^{4} N$ ek steel section (part 1: $E_{1} = 200 GPa$ , $L_{1} = 0.5 m$ , $A_{1} = 1 \times 1 0^{- 4} m^{2}$ ) aur phir ek aluminium section (part 2: $E_{2} = 70 GPa$ , $L_{2} = 0.5 m$ , $A_{2} = 1 \times 1 0^{- 4} m^{2}$ ) se series mein guzarta hai. Total extension nikalo.

Recall Solution 4.2

WHY add karte hain: series load path mein same $F$ har part se guzarta hai, aur total stretch har part ki stretch ka sum hota hai — dekho Spacecraft Load Paths and Struts.

= \frac{5000}{2\times10^{7}} = 2.5\times10^{-4}\ \text{m}.$$ $$\Delta L_2 = \frac{FL_2}{A_2 E_2} = \frac{10^{4}\times0.5}{10^{-4}\times7\times10^{10}} = \frac{5000}{7\times10^{6}} = 7.14\times10^{-4}\ \text{m}.$$ **Total:** $\Delta L = \Delta L_1 + \Delta L_2 = 2.5\times10^{-4} + 7.14\times10^{-4} = 9.64\times10^{-4}\ \text{m} \approx 0.96\ \text{mm}.$ Softer aluminium zyaadatar stretch contribute karta hai even though parts same size ke hain — kam $E$ matlab zyada give.

Level 5 — Mastery

Problem 5.1 (L5)

Ek composite tension member ek CFRP core ( $E_{c} = 150 GPa$ , $A_{c} = 2 \times 1 0^{- 4} m^{2}$ ) aur ek aluminium sleeve ( $E_{a} = 70 GPa$ , $A_{a} = 2 \times 1 0^{- 4} m^{2}$ ) se bana hai jo bond hain toh woh same length share karte hain aur isliye same strain bhi. Dono mil kar $F = 5.0 \times 1 0^{4} N$ carry karte hain. Nikalo (a) shared strain, (b) har material mein force, aur (c) har ek mein stress.

Recall Solution 5.1

WHY same strain, same stress nahi: dono side by side chipke hain aur saath stretch hote hain, isliye geometry (strain) common hai, aur har material apna khud ka stress develop karta hai apne $E$ se. Yeh 4.2 ke series case ka ulta hai.

Force balance set up karo. Har ek carry karta hai $F_{i} = σ_{i} A_{i} = E_{i} ε A_{i}$ same $ε$ ke saath. Woh total mein sum karte hain: $F = ε (E_{c} A_{c} + E_{a} A_{a}) .$ (a) Shared strain. Stiffness sum: $E_{c} A_{c} = 150 \times 1 0^{9} \times 2 \times 1 0^{- 4} = 3.0 \times 1 0^{7}$ ; $E_{a} A_{a} = 70 \times 1 0^{9} \times 2 \times 1 0^{- 4} = 1.4 \times 1 0^{7}$ ; total $= 4.4 \times 1 0^{7} N$ .

= 1.136\times10^{-3}.$$ **(b) Har ek mein force** ($F_i = E_i A_i \varepsilon$): $$F_c = 3.0\times10^{7}\times1.136\times10^{-3} = 3.41\times10^{4}\ \text{N},$$ $$F_a = 1.4\times10^{7}\times1.136\times10^{-3} = 1.59\times10^{4}\ \text{N}.$$ Check: $F_c + F_a = 5.0\times10^{4}\ \text{N}$ ✓ — stiffer CFRP bada share grab karta hai. **(c) Har ek mein stress** ($\sigma_i = F_i/A_i$): $$\sigma_c = \frac{3.41\times10^{4}}{2\times10^{-4}} = 1.705\times10^{8}\ \text{Pa} \approx 171\ \text{MPa},$$ $$\sigma_a = \frac{1.59\times10^{4}}{2\times10^{-4}} = 7.95\times10^{7}\ \text{Pa} \approx 80\ \text{MPa}.$$ Same strain, alag stress — bilkul isliye kyunki $E$ alag hai. Stiff phase sabse bhaari load carry karta hai, yeh ek key idea hai jab har phase ke liye [[Stress-Strain Curve]] padho.

Recall Paanchon levels ke baad ek-line self-test

Series load path mein kya share hota hai, aur parallel (bonded) mein kya? ::: Series mein force share hota hai (extensions add hote hain); parallel bonded mein strain share hoti hai (forces add hote hain).