Yield stress, ultimate stress — material behavior: Exercises

3.6.5 · D4 · Physics › Spacecraft Structures & Systems Engineering › Yield stress, ultimate stress — material behavior

Neeche reuse hone wale constants (sab "engineering" quantities hain — force ko original area $A_{0}$ se divide karo):

Level 1 — Recognition

L1.1

Ek rod $F = 6000 N$ le jaati hai original area $A_{0} = 40 mm^{2}$ ke across. Stress ke liye kaun sa formula use hoga, aur woh MPa mein kya hai?

Recall Solution

KYA: stress matlab force ko area pe spread karna, $σ = F / A_{0}$ . WHY woh formula: parent note ki pehli galti ("stress sirf force hai") exactly wahi hai jo hum avoid karte hain — wahi force ek chhoti area pe zyada damaging hoti hai, isliye hum hamesha area se divide karte hain. Unit care — woh ek jagah jahan log slip karte hain: $1 mm^{2} = 1 0^{- 6} m^{2}$ , to $40 mm^{2} = 40 \times 1 0^{- 6} m^{2}$ . $σ = \frac{6000}{40 \times 1 0 ^{- 6}} = 1.5 \times 1 0^{8} Pa = 150 MPa .$

L1.2

Batao ki har quantity ek material property hai (alloy ke liye fixed) ya part ki shape/load pe depend karti hai: (a) $σ_{y}$ , (b) yield pe force $F_{y}$ , (c) $E$ , (d) $σ_{a pp}$ .

Recall Solution

KYA/KYU: ek material property metal ke atomic bonds aur dislocations mein buni hoti hai, isliye usse fark nahi padta ki tum bar kitni moti kaat rahe ho.

(a) $σ_{y}$ — material property. Yielding dislocation glide hai; woh ek set stress pe hota hai, set force pe nahi.
(b) $F_{y} = σ_{y} A_{0}$ — shape pe depend karta hai. Moti bar → same $σ_{y}$ lekin bada $F_{y}$ .
(c) $E$ — material property (elastic line ka slope).
(d) $σ_{a pp}$ — load aur shape dono pe depend karta hai (yeh tumhari loading ka $F / A_{0}$ hai).

Level 2 — Application

L2.1

Ek titanium strut hai jiska $A_{0} = 25 mm^{2}$ , $E = 114 GPa$ , $σ_{y} = 830 MPa$ hai. Yeh $F = 10000 N$ carry kar raha hai. Stress find karo, confirm karo ki yeh abhi bhi elastic hai, aur strain find karo.

Recall Solution

Step 1 — stress. $σ = F / A_{0} = 10000/ (25 \times 1 0^{- 6}) = 4.0 \times 1 0^{8} Pa = 400 MPa$ . Step 2 — elastic check (pehle KYU): Hooke's law $σ = E ε$ sirf neeche $σ_{y}$ ke legal hai. Yahan $400 < 830$ , to hum straight-line region mein hain — allowed hai. Step 3 — strain. $ε = σ / E = 400 MPa /114000 MPa = 3.51 \times 1 0^{- 3}$ . (Dono ko MPa mein rakho taaki units cancel hokar pure number bane.)

L2.2

Same strut. Required $FoS_{y i e l d} = 1.25$ hai. Yield ke against margin of safety compute karo.

Recall Solution

KYA: margin of safety, $MoS = \frac{σ _{a l l o w}}{FoS _{r e q} σ _{a pp}} - 1$ , jahan yield check ke liye $σ_{a l l o w} = σ_{y}$ hai. $MoS = \frac{830}{1.25 \times 400} - 1 = \frac{830}{500} - 1 = 1.66 - 1 = 0.66.$ KYU iska matlab "pass" hai: $MoS = 0.66 \geq 0$ — required 1.25 se already divide karne ke baad bhi hamare paas 66% headroom bacha hai.

Level 3 — Analysis

L3.1 (offset yield, geometric)

Ek stress–strain test elastic slope $E = 68 GPa$ deta hai. 0.2% offset line hai $σ = 68000 (ε - 0.002)$ (MPa). Yeh measured curve ko $ε = 0.0095$ par intersect karta hai. $σ_{y}$ find karo, aur neeche diye picture se padho ki offset construction ka matlab kya hai.

Figure — Yield stress, ultimate stress — material behavior

Recall Solution

KYA offset karta hai (figure dekho): kai alloys gradually bend karte hain, to koi ek obvious "yield point" nahi hota. Hum ek line draw karte hain elastic slope ke parallel (same $E$ ) lekin right mein $ε = 0.002$ se start karne ke liye shift karte hain. Jahan shifted line (teal, dashed) real curve (orange) ko cross karti hai woh $σ_{y}$ define karta hai — woh intersection plum dot hai. KYU 0.2%: yeh yield ko 0.002 ke fixed permanent strain se pin karta hai, taaki har lab ko same number mile. Compute: $σ_{y} = 68000 (0.0095 - 0.002) = 68000 \times 0.0075 = 510 MPa .$

L3.2 (engineering vs true stress reasoning)

Ek specimen apne ultimate point tak pahunchta hai. Wahan engineering stress $σ_{u} = 500 MPa$ hai (original area $A_{0}$ use karte hue). Thodi der baad, ek neck form hoti hai aur real load-bearing area $0.80 A_{0}$ tak drop ho jaati hai jabki force momentarily ultimate force pe hold karti hai. Us instant par true stress compute karo aur parent ke teesre mistake se explain karo ki handbook kaun sa number print karta hai.

Recall Solution

Step 1 — ultimate force recover karo. Engineering stress $A_{0}$ use karta hai: $F_{u} = σ_{u} A_{0} = 500 MPa \times A_{0}$ . Step 2 — true stress shrunken area use karta hai $A_{in s t} = 0.80 A_{0}$ : $σ_{t r u e} = \frac{F _{u}}{A _{in s t}} = \frac{500 A _{0}}{0.80 A _{0}} = 625 MPa .$ KYU yeh diverge karte hain: necking ke baad real area collapse ho jaati hai, to chhoti area se divide karne par true stress badhta rehta hai jab engineering stress girta hai. Handbook engineering value print karta hai ( $500 MPa$ ) — design allowables aur $σ_{u}$ tables sab $F / A_{0}$ hain.

Level 4 — Synthesis

L4.1 (do checks, governing margin choose karo)

Ek aluminium 7075 fitting: $A_{0} = 80 mm^{2}$ , $σ_{y} = 480 MPa$ , $σ_{u} = 570 MPa$ . Ek launch load case $F = 24000 N$ apply karta hai (dekho Structural load cases and launch loads). $FoS_{y i e l d} = 1.25$ aur $FoS_{u l t} = 1.5$ use karte hue, dono margins compute karo aur batao kaun sa govern karta hai (chhota wala jeetta hai).

Recall Solution

Step 1 — applied stress. $σ_{a pp} = 24000/ (80 \times 1 0^{- 6}) = 3.0 \times 1 0^{8} Pa = 300 MPa$ . Step 2 — yield margin. $MoS_{y} = \frac{480}{1.25 \times 300} - 1 = \frac{480}{375} - 1 = 1.28 - 1 = 0.28.$ Step 3 — ultimate margin. $MoS_{u} = \frac{570}{1.5 \times 300} - 1 = \frac{570}{450} - 1 = 1.2667 - 1 = 0.267.$ Step 4 — kaun govern karta hai (figure dekho): chhota margin binding wala hota hai — jab load badhta hai yeh pehle fail hota hai. $MoS_{u} = 0.267 < MoS_{y} = 0.28$ , to ultimate yahan govern karta hai (barely). Dono $\geq 0$ hain, to fitting pass karti hai.

L4.2 (backwards thickness design karo)

Same alloy aur load ( $F = 24000 N$ ), lekin ab tumhe guarantee karni hai $MoS_{y} \geq 0$ aur $MoS_{u} \geq 0$ same required factors ke saath. Minimum original area $A_{0}$ kya hai, aur isse ek square cross-section ki minimum side kya hogi?

Recall Solution

KYU backwards work karo: $MoS \geq 0$ ka matlab hai $σ_{a l l o w} \geq FoS_{r e q} σ_{a pp}$ , yaani applied stress allowable se zyada nahi honi chahiye. Har ek ko required area mein badlo. Yield need: $σ_{a pp} \leq σ_{y} / FoS_{y} = 480/1.25 = 384 MPa$ $\Rightarrow A_{0} \geq F /384 MPa = 24000/ (384 \times 1 0^{6}) = 6.25 \times 1 0^{- 5} m^{2} = 62.5 mm^{2}$ . Ultimate need: $σ_{a pp} \leq σ_{u} / FoS_{u} = 570/1.5 = 380 MPa$ $\Rightarrow A_{0} \geq 24000/ (380 \times 1 0^{6}) = 6.316 \times 1 0^{- 5} m^{2} = 63.16 mm^{2}$ . Governing (badi) area jeetti hai: $A_{0} = 63.16 mm^{2}$ . Square side: $s = 63.16 = 7.95 mm$ .

Level 5 — Mastery

L5.1 (elastic-se-permanent poori kahani)

Ek steel tie-rod: $L_{0} = 500 mm$ , $A_{0} = 100 mm^{2}$ , $E = 200 GPa$ , $σ_{y} = 350 MPa$ , $σ_{u} = 500 MPa$ . (a) Pehle yield kaun sa force cause karta hai? (b) Us force par rod kitni stretch hui hai (elastic elongation, mm mein)? (c) Ek fault case $F = 45000 N$ isse slam karta hai — kya rod abhi bhi elastic hai, permanently bent hai, ya toot gayi hai? (d) Kaun sa force use todega?

Recall Solution

(a) Pehle yield par force. Yield ek stress limit hai, to force hai $F_{y} = σ_{y} A_{0}$ : $F_{y} = 350 \times 1 0^{6} \times 100 \times 1 0^{- 6} = 35000 N .$ (b) Us force par elastic elongation. Yield par strain hai (abhi bhi elastic line par) $ε = σ_{y} / E = 350/200000 = 1.75 \times 1 0^{- 3}$ , to $Δ L = ε L_{0} = 1.75 \times 1 0^{- 3} \times 500 mm = 0.875 mm .$ Kyun yahan $A_{0}$ nahi $L_{0}$ : elongation strain × length hai, area apna kaam stress mein kar chuka. (c) Fault case $F = 45000 N$ . Stress $= 45000/ (100 \times 1 0^{- 6}) = 450 MPa$ . Landmarks se compare karo: $350 (σ_{y}) < 450 < 500 (σ_{u})$ . To yeh yield ke past lekin ultimate ke neeche hai — rod permanently deform ho gayi (plastically bent) lekin toot nahi gayi. (d) Breaking force. Fracture (engineering) $σ_{u}$ ko $A_{0}$ ke upar track karta hai: $F_{b r e ak} = σ_{u} A_{0} = 500 \times 1 0^{6} \times 100 \times 1 0^{- 6} = 50000 N .$

L5.2 (alloy choose karo)

Ek mass-critical bracket ke liye tum same load $σ_{a pp} = 250 MPa$ ke liye do alloys mein se koi ek choose kar sakte ho: Alloy A ( $σ_{y} = 300$ , $σ_{u} = 360 MPa$ ) ya Alloy B ( $σ_{y} = 380$ , $σ_{u} = 430 MPa$ ). Required $FoS_{y i e l d} = 1.25$ , $FoS_{u l t} = 1.5$ hain. Har alloy ke liye dono margins compute karo, pass/fail decide karo, aur alloy choose karo. (Ties into Material selection for spacecraft.)

Recall Solution

Alloy A: $MoS_{y} = 300/ (1.25 \times 250) - 1 = 300/312.5 - 1 = 0.96 - 1 = - 0.04$ → negative, yield FAIL. (Ultimate ke saath bother karne ki zaroorat nahi: ek negative margin design ko condemn kar deta hai.) Alloy B: $MoS_{y} = 380/ (1.25 \times 250) - 1 = 380/312.5 - 1 = 1.216 - 1 = 0.216$ → pass. $MoS_{u} = 430/ (1.5 \times 250) - 1 = 430/375 - 1 = 1.1467 - 1 = 0.147$ → pass. Decision: Alloy A fail karta hai (uska yield margin zero se neeche hai), isliye tumhe Alloy B choose karna hai, jiska governing margin $MoS_{u} = 0.147 \geq 0$ hai.