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

3.6.3 · D3 · Physics › Spacecraft Structures & Systems Engineering › Stress and strain — σ = F - A, ε = ΔL - L, Young's modulus E

Cell	Ise kya alag banata hai	Example
A. Tension (kheenchna)	$F > 0$ , bar lambi hoti hai, $Δ L > 0$	Ex 1
B. Compression (dhakelna)	$F < 0$ , bar choti hoti hai, $Δ L < 0$ — sign flip hoti hai	Ex 2
C. Unknown geometry solve karna	$Δ L$ diya hai, $A$ dhundho	Ex 3
D. Zero-input / degenerate	$A \to 0$ , $Δ L = 0$ , ya $F = 0$ — formula "kya kehta hai"	Ex 4
E. Limiting behaviour	load yield ki taraf creep karta hai; Hooke's law kahan khatam hota hai?	Ex 5
F. Do materials / ratios	same load, compare karo — shared quantity cancel karo	Ex 6
G. Real-world word problem	messy units, MPa↔Pa, mm↔m, mass→force	Ex 7
H. Exam-style twist	area aur length dono ek saath badlo; net effect predict karo	Ex 8

Cell A — Tension

Worked example Ex 1 — Ek strut jo kheenchi ja rahi hai (Cell A)

Ek steel tie ( $E = 200$ GPa) $L = 1.5$ m lambi hai, cross-section $A = 2 \times 1 0^{- 4}$ m², $F = + 8 \times 1 0^{4}$ N se kheenchi ja rahi hai. $σ$ , $ε$ , aur $Δ L$ nikalo.

Forecast: steel stiff hai ( $E$ bahut bada hai), toh guess karo ki stretch sub-millimetre hogi. $Δ L$ ka sign? Positive — hum kheench rahe hain.

Stress: $σ = F / A = \frac{8 \times 1 0 ^{4}}{2 \times 1 0 ^{- 4}} = 4 \times 1 0^{8}$ Pa $= 400$ MPa. Yeh step kyun? Force per area material-level ka "crowding" hai — pehla normalized quantity.
Strain: $ε = σ / E = \frac{4 \times 1 0 ^{8}}{2 \times 1 0 ^{11}} = 2 \times 1 0^{- 3}$ . Yeh step kyun? Rearranged Hooke's law $σ = E ε$ — strain, stress ko stiffness se divide karna hai.
Stretch: $Δ L = ε L = 2 \times 1 0^{- 3} \times 1.5 = 3 \times 1 0^{- 3}$ m $= 3$ mm. Yeh step kyun? Hum woh "divide by $L$ " undo kar rahe hain jo strain define karta tha.

Verify: direct route $Δ L = F L / (A E) = \frac{8 \times 1 0 ^{4} \cdot 1.5}{2 \times 1 0 ^{- 4} \cdot 2 \times 1 0 ^{11}} = 3 \times 1 0^{- 3}$ m ✓. Forecast ki tarah positive. (3 mm — meri "sub-mm" guess se thoda bada, kyunki load heavy hai.)

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

Cell B — Compression (sign flip)

Worked example Ex 2 — Landing-leg compression mein (Cell B)

Ex 1 wala same steel member ( $E = 200$ GPa, $L = 1.5$ m, $A = 2 \times 1 0^{- 4}$ m²) ab $F = - 8 \times 1 0^{4}$ N (compression) se dhaka ja raha hai. $Δ L$ nikalo.

Forecast: load ka magnitude identical hai → length change ka magnitude bhi identical hoga, lekin minus sign ke saath. Guess karo $- 3$ mm.

Signed stress: $σ = F / A = \frac{- 8 \times 1 0 ^{4}}{2 \times 1 0 ^{- 4}} = - 4 \times 1 0^{8}$ Pa. Negative = compressive. Yeh step kyun? Sign rakhne se ek formula push aur pull dono cover karta hai.
Signed strain: $ε = σ / E = - 2 \times 1 0^{- 3}$ . Yeh step kyun? Hooke's law linear aur odd hai — input sign flip karo, output bhi flip hota hai.
Length change: $Δ L = ε L = - 3 \times 1 0^{- 3}$ m $= - 3$ mm (yeh chota hota hai). Yeh step kyun? Same undo-step jaise pehle; sign physics carry karta hai.

Verify: magnitude Ex 1 ke barabar hai (3 mm), sign negative hai ✓.

[!mistake] Compression ko stretch ki tarah report karna Compressed bar ka $Δ L < 0$ hota hai. Agar tum sign drop kar do toh tum accidentally ek aisi member mein length add kar doge jo actually shrink ho rahi hai — clearances check karte waqt yeh dangerous hai. Sign ko bilkul last sentence tak rakho.

Cell C — Geometry ke liye solve karna

Worked example Ex 3 — Stretch budget ke liye strut size karna (Cell C)

Ek aluminium strut ( $E = 70$ GPa) $L = 2$ m lambi hai aur $F = 1.4 \times 1 0^{4}$ N ke under $1$ mm se zyada stretch nahi karni chahiye. Minimum cross-section $A$ kya hai?

Forecast: kam allowed stretch ⇒ humein zyada area chahiye (stiffer path). Expect karo ek chota-lekin-finite $A$ jo roughly ek cm² order ka ho.

$Δ L = \frac{F L}{A E}$ se shuru karo aur $A$ ke liye solve karo: $A = \frac{F L}{E Δ L}$ . Yeh step kyun? Unknown ab geometry hai, toh $A$ isolate karo.
SI units mein plug karo ( $Δ L = 1 \times 1 0^{- 3}$ m): $A = \frac{1.4 \times 1 0 ^{4} \cdot 2}{70 \times 1 0 ^{9} \cdot 1 \times 1 0 ^{- 3}} = \frac{2.8 \times 1 0 ^{4}}{7 \times 1 0 ^{7}} = 4 \times 1 0^{- 4} m^{2} .$ Yeh step kyun? Koi bhi mismatch (mm vs m) answer corrupt kar dega — pehle convert karo.
Yeh $4 \times 1 0^{- 4}$ m² $= 4$ cm² hai — ek modest strut.

Verify: $A$ wapas daalo: $Δ L = \frac{1.4 \times 1 0 ^{4} \cdot 2}{4 \times 1 0 ^{- 4} \cdot 7 \times 1 0 ^{10}} = 1 \times 1 0^{- 3}$ m $= 1$ mm ✓. Pure-strength design se zyada area — yeh stiffness constraint ka bite karna hai, bilkul jaisa Spacecraft Load Paths and Struts warn karta hai.

Cell D — Zero aur degenerate inputs

Worked example Ex 4 — Formulas apni boundaries par kya kehte hain (Cell D)

Har formula ko uski boundary par literally padho.

Forecast: guess karo kaun sa "blow up" karta hai aur kaun sa quietly zero pe ja jaata hai.

$F = 0$ : $σ = 0/ A = 0$ , toh $ε = 0$ , toh $Δ L = 0$ . Koi load nahi, koi stretch nahi — ek healthy sanity check. Yeh step kyun? Ek correct model ko "kuch nahi hota" dena chahiye jab kuch bhi nahi kheenchta.
$Δ L = 0$ with $F \neq = 0$ : possible tabhi jab $E \to \infty$ (perfectly rigid) ya $A \to \infty$ . Real materials hamesha thoda stretch karte hain; truly zero stretch ek idealization hai. Yeh step kyun? Ideal rigid body ko real elastic behaviour se alag karta hai.
$A \to 0$ (ek baal-jaisi patli wire): $σ = F / A \to \infty$ . Stress diverge karta hai — force kisi material mein nahi crowd ho rahi, toh yeh hamesha break karta hai. Yahi wajah hai ki patli members pehle fail karti hain. Yeh step kyun? Dikhata hai ki formula real-world mein patli sections ki failure predict karta hai — stress concentration.
$L \to 0$ (ek bahut chota block): $ε = Δ L / L$ — fixed $σ$ ke liye, strain fixed hai, toh $Δ L = ε L \to 0$ . Choti cheezein high strain par bhi barely move karti hain. Yeh step kyun? Strain (intensive) ko displacement (extensive) se alag karta hai.

Verify (numeric): $F = 8 \times 1 0^{4}$ N ko $A = 1 0^{- 8}$ m² se lo: $σ = 8 \times 1 0^{12}$ Pa $= 8$ TPa — kisi bhi metal ki strength se thousands of times zyada, matlab instant fracture ✓. Aur $F = 0$ exactly $Δ L = 0$ deta hai ✓.

Cell E — Yield ki taraf limiting behaviour

Worked example Ex 5 — Hum yield ke kitne paas hain? (Cell E)

Aluminium 6061-T6 ka yield stress $σ_{y} = 270$ MPa aur $E = 70$ GPa hai. Ek strut $σ = 240$ MPa carry karta hai. Kya Hooke's law abhi bhi valid hai, aur yield ke against Factor of Safety kya hai?

Forecast: $240$ aur $270$ close hain — expect karo ek factor jo barely 1 se upar ho, matlab uncomfortably marginal.

Elasticity check: kya $σ < σ_{y}$ hai? $240 < 270$ ✓, toh hum abhi bhi linear region mein hain aur $σ = E ε$ valid hai. Yeh step kyun? Yield ke baad Hooke's law use karne se galat (bahut chota) strain milta hai.
Elastic strain: $ε = σ / E = \frac{2.4 \times 1 0 ^{8}}{7 \times 1 0 ^{10}} = 3.43 \times 1 0^{- 3}$ . Yeh step kyun? Valid hai kyunki step 1 ne confirm kiya ki hum yield ke neeche hain.
Factor of safety: $FoS = σ_{y} / σ = 270/240 = 1.125$ . Yeh step kyun? FoS, failure ceiling aur actual load ka ratio hai.

Verify: $ε = 3.43 \times 1 0^{- 3}$ back-check: $E ε = 7 \times 1 0^{10} \times 3.43 \times 1 0^{- 3} = 2.4 \times 1 0^{8}$ Pa $= 240$ MPa ✓. FoS $= 1.125$ — patla margin, forecast confirm hua; ek engineer ise aur strong banata.

Cell F — Do materials compare karo (ratios)

Cell G — Real-world word problem (messy units)

Worked example Ex 7 — Ek hanging equipment box (Cell G)

Ek $340$ kg ka instrument ek single titanium rod ( $E = 110$ GPa) se hang ho raha hai jiska length $L = 800$ mm aur diameter $d = 6$ mm hai. $g = 9.81$ m/s² lo. MPa mein stress aur mm mein stretch nikalo.

Forecast: chota rod, few-kN load → tens of MPa, sub-millimetre stretch.

Mass se force: $F = m g = 340 \times 9.81 = 3335.4$ N. Yeh step kyun? Weight hi load hai; koi bhi stress calculate karne se pehle mass ko newtons mein convert karo.
Diameter se area: $A = π (d /2)^{2} = π (3 \times 1 0^{- 3})^{2} = 2.827 \times 1 0^{- 5}$ m². Yeh step kyun? Stress ko area chahiye, diameter nahi — aur $d$ metres mein hona chahiye.
Stress: $σ = F / A = \frac{3335.4}{2.827 \times 1 0 ^{- 5}} = 1.180 \times 1 0^{8}$ Pa $= 118.0$ MPa. Yeh step kyun? MPa mein report karo kyunki yeh engineering habit hai — lekin Pa mein compute karo.
Stretch: $Δ L = \frac{σ L}{E} = \frac{1.180 \times 1 0 ^{8} \times 0.8}{1.1 \times 1 0 ^{11}} = 8.58 \times 1 0^{- 4}$ m $= 0.858$ mm. Yeh step kyun? $L$ metres mein (0.8, 800 nahi); yahan mm mix karna classic trap hai.

Verify: direct route $Δ L = F L / (A E) = \frac{3335.4 \times 0.8}{2.827 \times 1 0 ^{- 5} \times 1.1 \times 1 0 ^{11}} = 8.58 \times 1 0^{- 4}$ m ✓. 118 MPa aur 0.86 mm — dono forecast band mein hain.

Cell H — Exam-style twist

Worked example Ex 8 — Area AUR length ek saath badlo (Cell H)

Ek strut ko redesign kiya jaata hai taki uski length double ho jaaye ( $L \to 2 L$ ) aur cross-section triple ho jaaye ( $A \to 3 A$ ), same force aur material ke saath. (a) stress aur (b) stretch $Δ L$ ka kya hoga?

Forecast: stress sirf $A$ par depend karta hai, toh $A$ triple karne se stress ek-third ho jaana chahiye. Stretch dono $L$ (upar) aur $A$ (neeche) par depend karta hai — guess karo yeh $2/3$ se badlega.

Stress: $σ_{new} = \frac{F}{3 A} = \frac{1}{3} σ_{old}$ — length appear hi nahi karta, toh stress ke liye irrelevant hai. Yeh step kyun? Stress, force ÷ area hai; $L$ badlne se yeh touch nahi hota. Ek common exam trap yeh sochna hai ki lambi bars "zyada stressed" hoti hain.
Stretch: $Δ L_{new} = \frac{F ( 2 L )}{( 3 A ) E} = \frac{2}{3} \cdot \frac{F L}{A E} = \frac{2}{3} Δ L_{old}$ . Yeh step kyun? Dono factors $Δ L = F L / (A E)$ mein enter karte hain: lamba ⇒ zyada stretch ( $\times 2$ ), mota ⇒ kam ( $\div 3$ ), net $\times \frac{2}{3}$ .

Verify: $F = 8 \times 1 0^{4}$ , $A = 2 \times 1 0^{- 4}$ , $L = 1.5$ , $E = 2 \times 1 0^{11}$ (Ex 1 ke numbers) ke saath old $Δ L = 3$ mm. New: $A = 6 \times 1 0^{- 4}$ , $L = 3$ → $Δ L = \frac{8 \times 1 0 ^{4} \cdot 3}{6 \times 1 0 ^{- 4} \cdot 2 \times 1 0 ^{11}} = 2 \times 1 0^{- 3}$ m $= 2$ mm $= \frac{2}{3} (3 mm)$ ✓. Stress old $= 400$ MPa, new $= 133.3$ MPa $= \frac{1}{3}$ ✓.

Recall Kaun sa cell kaun sa tha?

Tension (positive stretch) ::: Ex 1 (Cell A) Compression (negative stretch) ::: Ex 2 (Cell B) Unknown area ke liye solve karna ::: Ex 3 (Cell C) Formula kya kehta hai jab $A \to 0$ ya $F = 0$ ::: Ex 4 (Cell D) Check karna ki hum abhi bhi yield ke neeche hain ::: Ex 5 (Cell E) Same stress, do materials ko ratio se compare karna ::: Ex 6 (Cell F) Messy real units (mass→force, mm→m) ::: Ex 7 (Cell G) Area aur length ek saath badalna ::: Ex 8 (Cell H)

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

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