Yield stress, ultimate stress — material behavior: Worked examples

3.6.5 · D3 · Physics › Spacecraft Structures & Systems Engineering › Yield stress, ultimate stress — material behavior

Scenario matrix

Yield/ultimate stress ke baare mein har question in cells mein se kisi ek mein aata hai. Har example par "Which cell" tag batata hai ki woh kahan baithta hai.

#	Cell class	Kya special hai	Covered by
A	Elastic, safe	$σ_{a pp} < σ_{y}$ , margin positive	Ex 1
B	Exactly at yield	$σ_{a pp} = σ_{y}$ , $M o S = 0$ boundary	Ex 2
C	Past yield, below ultimate	plastic but not broken	Ex 3
D	Negative margin (fails)	$M o S < 0$ , part under-designed hai	Ex 4
E	Zero / degenerate input	$F = 0$ , ya $A \to 0$ (limiting stress $\to \infty$ )	Ex 5
F	0.2% offset geometry	bending curve se $σ_{y}$ padhna	Ex 6
G	Ultimate + necking	engineering vs true stress diverge karte hain	Ex 7
H	Real-world word problem	load survive karne ke liye area choose karo	Ex 8
I	Exam twist	area double karna / material change karna; trap question	Ex 9

Hum in symbols ka use karenge throughout (sab parent note mein define hain, yahan repeat kiye hain taaki guess na karna pade):

Definition Symbol cheat-sheet

$F$ = pulling force, newtons mein ( $N$ ).
$A_{0}$ = original (undeformed) cross-section area, $m^{2}$ mein.
$A_{in s t}$ = instantaneous (current) cross-section area test ke dauran — $A_{0}$ se chhota ho jaata hai jab specimen neck karta hai. Sirf true stress ke liye use hota hai.
$F_{f r a c}$ = woh force jo specimen fracture ke waqt carry kar raha hota hai.
$σ = F / A_{0}$ = engineering stress, pascals mein ( $Pa = N/m^{2}$ ). $1 MPa = 1 0^{6} Pa$ , $1 GPa = 1 0^{9} Pa$ .
$σ_{a pp}$ = engineering stress jo actually service mein ek part par apply hoti hai, yaani $σ_{a pp} = F_{a ppl i e d} / A_{0}$ . (Yeh sirf working load ke liye evaluate ki gayi $σ$ hai — hum $σ_{a pp}$ likhte hain jab kisi limit se compare karte hain.)
$ε = Δ L / L_{0}$ = strain, ek pure number (koi units nahi).
$E$ = Young's modulus (elastic slope), dekho Young's modulus and elasticity.
$σ_{y}$ = yield stress, $σ_{u}$ = ultimate stress.
$σ_{allow}$ = allowable stress = limit ( $σ_{y}$ ya $σ_{u}$ ) ko required factor of safety se divide karo, yaani $σ_{allow} = σ_{l imi t} / F o S_{r e q}$ . Yeh woh single number hai jiske against har example applied stress compare karta hai — isse hamesha explicitly form karo.
$F o S$ = factor of safety; $F o S_{r e q}$ = woh required factor of safety jo design ko meet karna chahiye (jaise $1.25$ yield ke liye, $1.5$ ultimate ke liye). $M o S$ = margin of safety — dekho Factor of safety and margins.

Woh master formula jis par hum lean karenge, seedha parent se. Notice karo ki ek baar jab humne $σ_{allow} = σ_{l imi t} / F o S_{r e q}$ bana liya, toh margin sirf "allowable over applied, minus one" hai: $M o S = \frac{σ _{allow}}{σ _{a pp}} - 1 = \frac{σ _{l imi t}}{F o S _{r e q} \cdot σ _{a pp}} - 1 (\geq 0 means "passes")$

Har example neeche pehle $σ_{allow}$ form karta hai, phir $σ_{a pp}$ se divide karta hai, isliye upar ke dono written forms hamesha same computation hain.

Yaha poora matrix ek stress–strain curve par draw kiya gaya hai, taaki compute karne se pehle tum dekh sako ki har cell kahan baithti hai.

Figure — Yield stress, ultimate stress — material behavior

Worked examples

Worked example Example 1 — Elastic aur safe (Cell A)

Ek titanium bracket ka area $A_{0} = 40 mm^{2}$ hai aur woh $F = 6000 N$ carry karta hai. Material: $σ_{y} = 830 MPa$ , $E = 114 GPa$ . Required $F o S_{y i e l d} = 1.25$ . Kya yeh safe hai, aur fractionally kitna stretch karta hai?

Forecast: $40 mm^{2}$ par $6000 N$ — andaaza lagao: kya $σ$ $100$ ke kareeb hai ya $1000$ MPa ke? Kya margin bada hoga ya chhota?

Area ko SI mein convert karo. $40 mm^{2} = 40 \times 1 0^{- 6} m^{2}$ . Yeh step kyun? Stress ko $N/m^{2}$ chahiye; $1 mm^{2} = 1 0^{- 6} m^{2}$ , isliye mm² mein chhod do toh answer ek million se badh jaayega.
Stress compute karo. $σ = \frac{6000}{40 \times 1 0 ^{- 6}} = 150 \times 1 0^{6} Pa = 150 MPa$ . Yeh step kyun? Stress woh cheez hai jo hum $σ_{y}$ se compare karte hain, force se nahi (dekho Stress and strain fundamentals).
Yield se compare karo. $150 < 830$ → elastic, isliye Hooke's law $σ = E ε$ valid hai. Yeh step kyun? Sirf yield ke neeche hum $E$ use kar sakte hain strain get karne ke liye.
Strain. $ε = \frac{σ}{E} = \frac{150}{114000} = 1.316 \times 1 0^{- 3}$ . Yeh step kyun? Hum Hooke's law $σ = E ε$ ko $ε = σ / E$ mein rearrange karte hain taaki fractional stretch milein jo question poochh raha hai; yeh valid hai sirf isliye kyunki step 3 ne confirm kiya ki hum elastic hain.
Allowable stress aur margin. $σ_{allow} = \frac{σ _{y}}{F o S _{r e q}} = \frac{830}{1.25} = 664 MPa$ , isliye $M o S = \frac{σ _{allow}}{σ _{a pp}} - 1 = \frac{664}{150} - 1 = 3.43$ . Yeh step kyun? Hum pehle de-rated allowable stress banate hain (woh single limit jo part actually reach karne ki permission hai), phir apply ki gayi $150$ MPa ko uske against compare karte hain — woh ratio minus one margin hai.

Verify: $M o S = 3.43 > 0$ → passes, bahut bada. Sanity: $150$ MPa yield ka $\approx 18%$ hai, isliye bada margin expected hai. Units: $MPa / MPa$ dimensionless hai ✓; strain dimensionless hai ✓.

Worked example Example 2 — Exactly at yield,

$M o S = 0$ boundary (Cell B) Ek aluminium rod ka $σ_{y} = 480 MPa$ hai. Koi factor of safety demand nahi ( $F o S_{r e q} = 1$ ), toh exactly $M o S = 0$ ke liye kaunsa applied stress chahiye? $M o S = 0$ ka physically kya matlab hai?

Forecast: $M o S = 0$ razor's edge hai. Aage padhne se pehle applied stress guess karo.

Allowable form karo aur $M o S = 0$ set karo. $F o S_{r e q} = 1$ ke saath, $σ_{allow} = σ_{y} /1 = 480 MPa$ , isliye $0 = \frac{σ _{allow}}{σ _{a pp}} - 1$ . Yeh step kyun? $M o S = 0$ exact pass/fail line hai, cell A aur cell D ki boundary; isse $σ_{allow}$ ke zariye likhna har example ko same footing par rakhta hai.
Solve karo. $\frac{σ _{allow}}{σ _{a pp}} = 1 \Rightarrow σ_{a pp} = σ_{allow} = 480 MPa$ . Yeh step kyun? Rearrange karne se woh applied stress isolate hoti hai jo just allowable ko touch karti hai.
Interpret karo. $σ_{a pp} = σ_{allow} = σ_{y}$ par (kyunki $F o S_{r e q} = 1$ ), applied stress exactly allowable ko reach karti hai: zero cushion. Yeh step kyun? Hum algebra ko physically interpret karte hain kyunki design review ko meaning chahiye ("koi cushion nahi bacha"), sirf ek number nahi — yeh engineer ko batata hai ki part permanent set ke bilkul onset par hai.

Verify: Plug back karo: $M o S = 480/480 - 1 = 1 - 1 = 0$ ✓. Matlab: part permanent deformation ke onset par hai — thoda aur load aur yeh yield karna shuru kar dega. Real spacecraft kabhi yahan nahi fly karte; isliye $F o S_{r e q} > 1$ hota hai.

Worked example Example 3 — Past yield lekin ultimate ke neeche (Cell C)

Ek test coupon ko deliberately $σ_{a pp} = 520 MPa$ tak load kiya gaya hai. Material: $σ_{y} = 480 MPa$ , $σ_{u} = 570 MPa$ , $E = 70 GPa$ . Kya yeh yield ho gaya hai? Kya toot gaya hai? Kya elastic strain formula abhi bhi valid hai?

Forecast: $520$ $480$ aur $570$ ke beech mein baithta hai. Predict karo ki usne dono limits mein se kaunsi cross ki hai.

Yield se compare karo. $520 > 480$ → yielded: permanent (plastic) deformation shuru ho gayi hai. Yeh step kyun? $σ_{y}$ cross karna plastic set ke onset ki definition hai.
Ultimate se compare karo. $520 < 570$ → fracture nahi hua: abhi bhi load carry kar raha hai, yield knee ke baad lekin peak se pehle. Yeh step kyun? Ultimate load-carrying peak hai; uske neeche part intact hai.
Strain rule check karo. Kyunki $520 > σ_{y}$ , Hooke's law $σ = E ε$ ab total strain nahi deta — strain ka kuch hissa ab permanent hai aur elastic slope ko follow nahi karta. Yeh step kyun? $E$ sirf springy part describe karta hai; agar yahan $ε = σ / E$ use karo toh total stretch underestimate ho jaayegi.

Verify: Order check: $480 < 520 < 570$ , yaani $σ_{y} < σ_{a pp} < σ_{u}$ ✓. Toh "yielded, not broken" consistent hai. Agar ab unload karo, toh coupon sirf partially spring back karega — ek permanent set chhod kar, jo cell C ki fingerprint hai.

Worked example Example 4 — Negative margin: part fail karta hai (Cell D)

Ek strut $σ_{a pp} = 420 MPa$ experience karta hai. Material $σ_{y} = 480 MPa$ , required $F o S_{y i e l d} = 1.25$ . $M o S$ compute karo aur decide karo.

Forecast: $420$ $480$ se neeche hai, isliye safe lagta hai. Lekin $1.25$ ka ek $F o S$ allowable ko kha raha hai. $M o S$ ka sign guess karo.

Allowable stress. $σ_{allow} = \frac{σ _{y}}{F o S _{r e q}} = \frac{480}{1.25} = 384 MPa$ . Yeh step kyun? Factor of safety shrink karta hai kitna stress hum allowed hain use karne ke liye, material ki actual yield nahi.
Compare karo. $σ_{a pp} = 420 > 384 = σ_{allow}$ → allowable se upar. Yeh step kyun? Pass hona hai $σ_{a pp} \leq σ_{allow}$ ; yahan nahi hai.
Margin. $M o S = \frac{σ _{allow}}{σ _{a pp}} - 1 = \frac{384}{420} - 1 = 0.9143 - 1 = - 0.0857$ . Yeh step kyun? Already-formed $σ_{allow}$ use karna arithmetic ko $σ_{y} / (F o S_{r e q} \cdot σ_{a pp}) - 1$ ke identical rakhta hai aur negative sign obvious banata hai.

Verify: $M o S = - 0.086 < 0$ → fails. Trap note karo: raw stress ( $420$ ) yield ( $480$ ) se neeche hai, isliye part abhi permanently deform nahi karega — lekin yeh required safety margin violate karta hai, jo design demand karta hai. Cell D exactly yahi "theek lagta hai, margin fail karta hai" case hai.

Worked example Example 5 — Degenerate inputs: zero force, vanishing area (Cell E)

$σ_{y} = 480 MPa$ aur required $F o S_{r e q} = 1.25$ throughout lo. (a) Ek strut $A_{0} = 50 mm^{2}$ par $F = 0 N$ carry karta hai. Uska stress aur margin kya hai? (b) Fixed force $F = 4000 N$ wala ek tie-rod machined thinner aur thinner hota jaata hai. Jaise jaise $A_{0} \to 0$ stress ka kya hota hai?

Forecast: Guess karo: zero force ke saath kya margin zero hoga, ya infinite? Jaise jaise rod razor-thin hoti hai, kya stress kisi number par settle hota hai ya blow up karta hai?

Zero-force stress. $σ = \frac{0}{50 \times 1 0 ^{- 6}} = 0 Pa$ . Yeh step kyun? Stress force per area hai; koi force nahi matlab koi stress nahi, geometry kuch bhi ho.
Zero-force margin. $σ_{allow} = σ_{y} / F o S_{r e q} = 480/1.25 = 384 MPa$ ke saath, margin hai $M o S = \frac{σ _{allow}}{σ _{a pp}} - 1 = \frac{384}{0} - 1$ : $σ_{a pp} = 0$ se divide karna undefined hai (zero par koi number ki koi value nahi hoti). Yeh step kyun? Hum honestly plug in karna zaroori hai taaki expose ho sake ki formula $σ_{a pp} = 0$ par break karti hai — tum yahan koi finite margin report nahi kar sakte.
Undefined result ko $+ \infty$ interpret karo. Limit dekho: load ko tiny lekin nonzero rakho aur shrink karte jaao, $σ_{a pp} \to 0^{+}$ . Phir $\frac{σ _{allow}}{σ _{a pp}} \to + \infty$ , isliye $M o S \to + \infty$ . Kyunki margin bina bound ke badh jaata hai jaise load vanish hota hai, hum define karte hain zero-load case ko $M o S = + \infty$ ("infinitely safe"). Yeh step kyun? Ek undefined algebraic expression sirf limiting process ke zariye meaningful banti hai; limit dikhati hai ki margin $+ \infty$ ki taraf jaata hai, jo physical statement hai "koi load nahi carry karta part kabhi fail nahi ho sakta," isliye hum convention se woh value adopt karte hain.
Vanishing area, fixed force. $σ (A_{0}) = \frac{4000}{A _{0}}$ . Jaise jaise $A_{0} \to 0^{+}$ , $σ \to + \infty$ . Yeh step kyun? Ever-less area se squeeze hoti force physical reason hai ki nick, crack, ya notch (tiny local area) kahan cheezein fail hoti hain — dekho Fatigue and fracture mechanics.

Verify: (a) $σ = 0$ , aur $F = σ_{y} A$ yield par $F = 0$ input ke saath $0$ deta hai ✓; limit argument $M o S = + \infty$ deta hai. (b) Numerically: $A_{0} = 10 mm^{2} \Rightarrow σ = 400 MPa$ ; $A_{0} = 1 mm^{2} \Rightarrow 4000 MPa$ ; $A_{0} = 0.1 mm^{2} \Rightarrow 40000 MPa$ — growing without bound ✓. Yahi limiting behaviour poora reason hai ki stress concentrations dangerous hain.

Worked example Example 6 — 0.2% offset geometry (Cell F)

Ek alloy ka elastic slope $E = 68 GPa$ hai aur koi sharp yield point nahi hai. 0.2% offset line hai $σ = 68000 (ε - 0.002)$ MPa. Yeh measured curve ko $ε = 0.0090$ par intersect karta hai. $σ_{y}$ find karo.

Forecast: Offset line $ε = 0.002$ par strain axis par start hoti hai. Guess karo ki kya $σ_{y}$ $E \times 0.002 = 136$ MPa ke kareeb hai ya bahut zyada badi.

Offset line likho. Slope $E$ , right shift $0.002$ se: $σ = E (ε - 0.002)$ . Yeh step kyun? Elastic region jaisi same stiffness $E$ (ek parallel line), lekin $0.2%$ permanent strain par pinned — isliye offset labs mein repeatable hai (parent note).
Intersection strain par evaluate karo. $σ_{y} = 68000 (0.0090 - 0.002) = 68000 \times 0.0070$ . Yeh step kyun? By definition $σ_{y}$ wahan hai jahan offset line material curve cross karti hai — intersection strain seedha andar jaata hai.
Compute karo. $68000 \times 0.0070 = 476 MPa$ .

Verify: Units: $MPa \times$ (dimensionless strain) $= MPa$ ✓. Geometry check (neeche figure dekho): elastic part $ε = 0.002$ par $\approx E \times 0.002 = 136$ MPa reach karta hai — lekin yield point bahut larger strain ( $0.009$ ) aur higher stress ( $476$ MPa) par baithta hai, curve ke bent-over part par, exactly as expected.

Neeche ki figure step by step yeh construction build karti hai: dashed white line elastic slope $E$ hai; yellow line woh same slope hai jo right shift karke $ε = 0.002$ par start karne ke liye (green dotted marker); jahan yellow blue material curve se milti hai woh red dot hai jo defines karta hai $σ_{y} = 476$ MPa. Yellow ko green marker se red dot tak trace karo "0.2% shift ke saath yeh steepness kis angle ki hai?" idea ko visible dekhne ke liye.

Worked example Example 7 — Ultimate + necking: engineering vs true stress (Cell G)

Ultimate point par ek specimen peak force $F = 28500 N$ original area $A_{0} = 50 mm^{2}$ par carry karta hai. Fracture se just pehle necked region instantaneous area $A_{in s t} = 30 mm^{2}$ tak shrink ho gayi hai jabki fracture force $F_{f r a c} = 21000 N$ carry kar rahi hai. Engineering ultimate stress $σ_{u}$ , engineering fracture stress, aur true fracture stress find karo. Kaunsa bada hai?

Forecast: Engineering $A_{0}$ use karta hai (cheat-sheet se original area); true shrunk area $A_{in s t}$ use karta hai. Guess karo: ultimate se fracture tak engineering stress upar jaata hai ya neeche? True stress?

Engineering ultimate. $σ_{u} = \frac{28500}{50 \times 1 0 ^{- 6}} = 570 \times 1 0^{6} = 570 MPa$ . Yeh step kyun? $σ_{u}$ engineering curve ka peak hai, hamesha $A_{0}$ par (design handbooks engineering stress use karte hain).
Engineering fracture stress. $σ_{f r a c, e n g} = \frac{F _{f r a c}}{A _{0}} = \frac{21000}{50 \times 1 0 ^{- 6}} = 420 MPa$ . Yeh step kyun? Neck form hone ke baad, force drop hoti hai lekin hum original area $A_{0}$ se divide karte rehte hain, isliye engineering stress $σ_{u}$ se gir jaati hai.
True fracture stress. $σ_{f r a c, t r u e} = \frac{F _{f r a c}}{A _{in s t}} = \frac{21000}{30 \times 1 0 ^{- 6}} = 700 MPa$ . Yeh step kyun? True stress real shrinking area $A_{in s t}$ se divide karta hai; neck same force ko less material se concentrate karta hai, isliye true stress badhta hai.

Verify: $σ_{f r a c, e n g} = 420 < σ_{u} = 570$ ✓ (engineering necking ke baad drop hoti hai). $σ_{f r a c, t r u e} = 700 > 570$ ✓ (true stress badhti rehti hai). Yahi parent ki warning hai: engineering aur true stress kabhi mix mat karo — yeh necking ke baad diverge karte hain.

Worked example Example 8 — Real-world word problem: strut size karo (Cell H)

Ek lander leg ko touchdown par $F = 9000 N$ carry karna hai (dekho Structural load cases and launch loads). Material $σ_{y} = 480 MPa$ , required $F o S_{y i e l d} = 1.25$ . Minimum cross-sectional area kya hai, aur kaunsa standard round-bar diameter (nearest mm) kaam karta hai?

Forecast: Zyada force ko zyada area chahiye. Guess karo ki answer kuch mm² hoga ya tens of mm².

Allowable stress. $σ_{allow} = \frac{σ _{y}}{F o S _{r e q}} = \frac{480}{1.25} = 384 MPa$ . Yeh step kyun? Leg ko sirf de-rated stress tak pahunchne ki permission hai, raw yield tak nahi — $σ_{allow}$ woh ceiling hai jo har subsequent step respect karna chahiye.
$σ = F / A$ se minimum area. Require karo $σ_{a pp} = \frac{F}{A} \leq σ_{allow}$ , isliye $A_{min} = \frac{F}{σ _{allow}} = \frac{9000}{384 \times 1 0 ^{6}}$ . Yeh step kyun? Stress definition rearrange karna ek stress limit ko geometry requirement mein turn karta hai — aise engineers sizes pick karte hain.
Compute karo. $A_{min} = 2.344 \times 1 0^{- 5} m^{2} = 23.44 mm^{2}$ . Yeh step kyun? mm² mein wapas convert karna ek number deta hai jo hum real bar stock se match kar sakte hain.
Round bar diameter. $A = \frac{π}{4} d^{2} \Rightarrow d = \frac{4 A}{π} = \frac{4 \times 23.44}{π} = 5.46 mm$ . $d = 6 mm$ choose karo (round up). Yeh step kyun? Area/diameter ko up round karna zaroori hai, kabhi down nahi — chhota bar allowable stress exceed kar dega.

Verify: $6 mm$ bar ka $A = \frac{π}{4} (6)^{2} = 28.27 mm^{2} > 23.44$ ✓. Actual stress $= 9000/ (28.27 \times 1 0^{- 6}) = 318 MPa < 384$ ✓, isliye $M o S = σ_{allow} / σ_{a pp} - 1 = 384/318 - 1 = 0.21 > 0$ passes. $5 mm$ tak round down karna ( $19.6 mm^{2}$ ) $459$ MPa dega $>$ allowable — fails, confirming ki hum up round karna zaroori hai. Concluding insight: hamesha pehle area ke liye solve karo, phir physical dimension up round karo, kyunki area woh cheez hai jis par stress actually depend karta hai.

Worked example Example 9 — Exam twist: area double karo, material change karo (Cell I)

Original design: area $A$ ka bar force $F$ carry karta hai exactly apni yield stress par $σ_{y} = 480 MPa$ (isliye $σ_{a pp} = 480$ MPa). $F o S_{r e q} = 1$ lo. Ab (a) area double karo $2 A$ tak usi same force $F$ ke saath; (b) ek aisi material mein swap karo jiska $σ_{y} = 480$ hai lekin Young's modulus $E$ double hai. Dono ke liye, (i) stress, (ii) yield stress, (iii) margin ka kya hota hai?

Forecast: Classic trap — log confuse karte hain $σ$ change karna (geometry se) $σ_{y}$ change karne ke saath (ek material property) aur $E$ (stiffness) ko strength se confuse karte hain. Aage padhne se pehle teeno predict karo.

(a) Area double, same force — stress. $σ_{n e w} = \frac{F}{2 A} = \frac{1}{2} \frac{F}{A} = \frac{σ _{y}}{2} = 240 MPa$ . Yeh step kyun? Stress $F / A$ hai; denominator double karne se stress half ho jaata hai. Geometry stress change karti hai.
(a) Yield stress. $σ_{y}$ ek material property hai → $480 MPa$ par unchanged, isliye $σ_{allow} = σ_{y} / F o S_{r e q} = 480$ MPa. Yeh step kyun? Koi nayi material introduce nahi hui; sirf shape change hua. Yield stress — aur isliye allowable — geometry change se move nahi kar sakta.
(a) Margin. $M o S = \frac{σ _{allow}}{σ _{n e w}} - 1 = \frac{480}{240} - 1 = 2 - 1 = 1$ . Yeh step kyun? Hum $M o S = 0$ (yield par) se comfortable $+ 1$ tak aa gaye sirf bar ko mota karke — wahi $σ_{allow}$ halved applied stress par.
(b) $E$ double — stress, yield, margin. Stress $σ = F / A$ unchanged hai (geometry same, force same). $σ_{y} = 480$ unchanged, isliye $σ_{allow} = 480$ unchanged, hence $M o S$ unchanged. Yeh step kyun? $E$ govern karta hai stiffness/kitna stretch hota hai, nahi strength/kab yield hota hai — parent ki "higher $E$ ≠ stronger" galti. Ek stiffer bar elastically kam deform hoti hai lekin same stress par yield karti hai aur same margin rakhti hai.

Verify: (a) $σ_{n e w} = 240$ MPa, $M o S = 1$ — plug back karo: $480/240 = 2$ , minus $1$ hai $1$ ✓. (b) $E$ double karna elastic strain half karta hai ( $ε = σ / E$ ) lekin $σ$ , $σ_{y}$ , $σ_{allow}$ aur $M o S$ untouched chhod deta hai ✓. Trap defused: geometry stress move karti hai, material strength move karta hai, aur $E$ na strength move karta hai na stress — sirf elastic strain.

Wrap-up: matrix coverage

Recall Kya humne har cell hit ki?

A (Ex 1), B (Ex 2), C (Ex 3), D (Ex 4), E (Ex 5), F (Ex 6), G (Ex 7), H (Ex 8), I (Ex 9) — saari nau cells worked. Margin ka har sign (positive, zero, negative), degenerate zero/vanishing inputs, necking limit, ek sizing word problem, aur double-area/double- $E$ exam trap — sab cover hai.

Prerequisite links jo revisit karo agar koi step shaky laga: Stress and strain fundamentals, Young's modulus and elasticity, Factor of safety and margins, Material selection for spacecraft.