Structural design process — load cases, FOS (factor of safety): Exercises

3.6.2 · D4 · Physics › Spacecraft Structures & Systems Engineering › Structural design process — load cases, FOS (factor of safet

Prerequisites Structural design process — load cases, FOS (factor of safety) mein hain. Agar koi term unfamiliar lage, toh wahan define kiya gaya hai. Hum poore note mein yeh use karte hain:

Hum poore note mein $g = 9.81 m/s^{2}$ use karte hain. Yaad rakho $1 MPa = 1 0^{6} Pa = 1 N/mm^{2}$ , aur $1 kN = 1000 N$ .

Level 1 — Recognition

Kya tum identify kar sakte ho ki kaun si quantity kaun si hai, aur formula mein ek baar plug in kar sakte ho?

L1.1 — Pieces ko naam do

Ek datasheet kehta hai: material yield stress $300 MPa$ ; service mein worst expected stress $150 MPa$ ; required factor of safety $1.25$ . Har number ko allowable, limit, ya FOS label karo, phir batao ki $F O S$ kis cheez ko multiply karta hai.

Recall Solution L1.1

Allowable $S_{allow} = 300 MPa$ (jo material le sakta hai — uski yield).
Limit $S_{limit} = 150 MPa$ (worst load expected, yani demand).
$F O S = 1.25$ .

$F O S$ limit (demand) ko multiply karta hai. Toh design stress hai $1.25 \times 150 = 187.5 MPa$ . Hum demand ko inflate karte hain, material strength ko nahi. (Yeh exactly parent note ka "Fear inflates the FORCE, not the METAL" rule hai.)

L1.2 — Sign padho

Do brackets analyse kiye gaye hain. Bracket A: $M S = + 0.30$ . Bracket B: $M S = - 0.05$ . Kaun pass karta hai, kaun fail, aur jo pass karta hai usmein roughly kitni spare strength hai?

Recall Solution L1.2

Ek structure acceptable hai jab $M S \geq 0$ .

Bracket A ( $M S = + 0.30$ ) pass karta hai — usmein $30%$ spare strength hai already-factored (design) load ke beyond.
Bracket B ( $M S = - 0.05$ ) fail karta hai — yeh factored load se bhi $5%$ short hai. Redesign karo (thicken karo, rib add karo, ya material change karo).

L1.3 — One-step margin

Ek fitting ka allowable $S_{allow} = 250 MPa$ hai. Limit stress $160 MPa$ aur $F O S_{ult} = 1.4$ hai. Margin of safety compute karo.

Recall Solution L1.3

Step 1 — design stress (WHAT: limit ko factor karo; WHY: humein inflate ki hui demand se survive karna hai): $S_{design} = 1.4 \times 160 = 224 MPa .$ Step 2 — margin (WHAT: fractional spare capacity; WHY: " $- 1$ " ka matlab hai "exactly enough" zero read hota hai): $M S = \frac{250}{224} - 1 = 0.116.$ $M S = + 0.12 \geq 0$ → pass, lagbhag $12%$ spare ke saath.

Level 2 — Application

Multi-step: ek part size karo, ya ek poora load FOS chain se le jaao.

L2.1 — Tension mein ek strut size karo

Ek strut limit tensile force $P_{limit} = 30 kN$ carry karta hai. Material yield stress $σ_{y} = 275 MPa$ , $F O S_{yield} = 1.25$ . Minimum cross-sectional area $A$ ( $mm^{2}$ mein) nikalo taaki yield ke against $M S \geq 0$ ho.

Figure — Structural design process — load cases, FOS (factor of safety)

Recall Solution L2.1

Step 1 — design load (WHAT: demand ko factor karo): $P_{design} = 1.25 \times 30 = 37.5 kN = 37500 N .$ Step 2 — stress condition (WHY: stress force per area hai; permanent deformation bachane ke liye ise yield ke neeche rakho, jaise figure mein hai): $σ = \frac{P _{design}}{A} \leq σ_{y} .$ Step 3 — smallest $A$ ke liye solve karo ( $M S = 0$ minimum deta hai): $A \geq \frac{P _{design}}{σ _{y}} = \frac{37500}{275 \times 1 0 ^{6}} = 1.364 \times 1 0^{- 4} m^{2} = 136.4 mm^{2} .$ Koi bhi badi area $M S > 0$ deti hai. Answer: $A_{m i n} \approx 136 mm^{2}$ .

L2.2 — Full chain: limit → yield → ultimate

Ek panel limit stress $S_{limit} = 100 MPa$ dekhta hai. Yield allowable $= 240 MPa$ , ultimate allowable $= 320 MPa$ . $F O S_{yield} = 1.1$ , $F O S_{ult} = 1.4$ use karo. Dono margins compute karo aur batao ki panel acceptable hai ya nahi.

Recall Solution L2.2

Do independent checks — part ko dono pass karne honge.

Yield check: $M S_{yield} = \frac{240}{1.1 \times 100} - 1 = \frac{240}{110} - 1 = 1.182 - 1 = 0.182.$ Ultimate check: $M S_{ult} = \frac{320}{1.4 \times 100} - 1 = \frac{320}{140} - 1 = 2.286 - 1 = 1.286.$ Dono $\geq 0$ → acceptable. Driving (chhota) margin yield ka hai $+ 0.18$ par; ultimate comfortable hai. Practice mein chhota margin woh hai jise dekhna chahiye.

L2.3 — Allowable back out karo jo chahiye

$12 kN$ ka ek limit force cross-section $A = 50 mm^{2}$ wale fitting par lag raha hai. $F O S_{ult} = 1.4$ ke saath, material ko $M S = 0$ ke liye minimum kitna ultimate stress provide karna hoga?

Recall Solution L2.3

Step 1 — design load: $P_{design} = 1.4 \times 12 = 16.8 kN = 16800 N$ . Step 2 — design stress (di gayi area par force): $σ_{design} = \frac{16800}{50 mm ^{2}} = \frac{16800}{50 \times 1 0 ^{- 6} m ^{2}} = 336 \times 1 0^{6} Pa = 336 MPa .$ Step 3 — required allowable: $M S = 0$ ke liye humein $S_{allow} = σ_{design}$ chahiye. Answer: material ko $\geq 336 MPa$ ultimate supply karna hoga.

Level 3 — Analysis

Simultaneous loads combine karo, cases compare karo, decide karo ki design ko kya drive karta hai.

L3.1 — Quasi-static loads ka vector combination

Ek $40 kg$ ka avionics box ek bracket par mounted hai. Quasi-static envelope hai $\pm 8 g$ axial aur $\pm 3 g$ lateral, simultaneously acting. Resultant force nikalo, aur ise (galat) scalar sum se compare karo.

Recall Solution L3.1

Step 1 — axial force (WHAT: mass × acceleration; $8 g$ ka matlab hai $8 \times g$ ): $F_{a} = m \cdot (8 g) = 40 \times 8 \times 9.81 = 3139.2 N .$ Step 2 — lateral force: $F_{l} = m \cdot (3 g) = 40 \times 3 \times 9.81 = 1177.2 N .$ Step 3 — resultant (WHY Pythagoras: axial aur lateral perpendicular hain, toh woh right triangle ke do legs ki tarah combine hote hain — figure mein blue triangle dekho): $F = F_{a}^{2} + F_{l}^{2} = 3139. 2^{2} + 1177. 2^{2} = 3352.7 N .$ Compare karo naive scalar sum $3139.2 + 1177.2 = 4316.4 N$ se. Scalar sum $29%$ bada hai — yeh pretend karta hai ki dono loads same direction mein hain. Iske liye design karna mass waste karta hai, aur mass space mein dushman hai. Answer: resultant $\approx 3353 N$ (root-sum-square), $4316 N$ nahi.

L3.2 — Kaun sa load case design drive karta hai?

Same box/bracket ko teen alag stress states survive karne honge (har ek pehle se apne environment ka worst hai): quasi-static $95 MPa$ deta hai, random vibration $130 MPa$ deta hai, thermal $60 MPa$ deta hai. Yeh alag times par hote hain. Allowable $= 200 MPa$ , $F O S_{ult} = 1.4$ . Kaun sa case bracket ko size karta hai, aur uska margin kya hai?

Recall Solution L3.2

Load cases alag times par hote hain, toh hum envelope (worst single case) ke liye design karte hain, unke sum ke liye nahi.

QSL: design stress $= 1.4 \times 95 = 133 MPa$ .
Random: design stress $= 1.4 \times 130 = 182 MPa$ .
Thermal: design stress $= 1.4 \times 60 = 84 MPa$ .

Random vibration sabse bada hai → yeh design drive karta hai. Uska margin: $M S = \frac{200}{1.4 \times 130} - 1 = \frac{200}{182} - 1 = 0.099.$ Answer: random vibration drive karta hai, $M S = + 0.10$ (lagbhag $10%$ spare — patla hai, dekhte rehna chahiye).

L3.3 — Reverse: allowable $g$ nikalo

Ek bracket $5000 N$ resultant force withstand kar sakta hai $M S = 0$ par $F O S_{ult} = 1.4$ ke baad. Yeh $60 kg$ ka box carry karta hai. Agar lateral load $2 g$ par fixed hai, toh mount survive kar sakta hai maximum kitna axial $g$ ?

Recall Solution L3.3

Step 1 — allowed limit resultant (FOS undo karo: $5000 N$ design load hai, toh limit chhota hai): $F_{limit} = \frac{5000}{1.4} = 3571.4 N .$ Step 2 — lateral limit force: $F_{l} = 60 \times 2 \times 9.81 = 1177.2 N .$ Step 3 — axial leg ke liye right triangle solve karo: $F_{a} = F_{limit}^{2} - F_{l}^{2} = 3571. 4^{2} - 1177. 2^{2} = 3371.8 N .$ Step 4 — $g$ mein convert karo: $n_{a} = \frac{F _{a}}{m g} = \frac{3371.8}{60 \times 9.81} = 5.73.$ Answer: lagbhag $5.7 g$ axial tak ( $2 g$ lateral ke saath present).

Level 4 — Synthesis

Poora design loop, ya ek iterate-until-it-passes redesign.

L4.1 — Ek strut par full loop

Ek strut do load cases dekhta hai (alag times): QSL limit tension $18 kN$ ; sine-vibration limit tension $22 kN$ . Material: yield $250 MPa$ , ultimate $350 MPa$ . $F O S_{yield} = 1.1$ , $F O S_{ult} = 1.4$ use karo. $8 mm$ diameter ka ek round rod propose kiya gaya hai. Kya yeh dono checks pass karta hai? Agar nahi, toh minimum diameter nikalo.

Recall Solution L4.1

Step 1 — envelope: worst limit load $= max (18, 22) = 22 kN$ (sine drive karta hai). Step 2 — design loads:

Yield: $1.1 \times 22 = 24.2 kN$ .
Ultimate: $1.4 \times 22 = 30.8 kN$ .

Step 3 — $8 mm$ rod ki area: $A = \frac{π d ^{2}}{4} = \frac{π ( 8 ) ^{2}}{4} = 50.27 mm^{2} .$ Step 4 — dono stresses check karo:

Yield stress $= \frac{24200}{50.27 mm ^{2}} = 481.4 MPa > 250 MPa$ → fails (negative margin).
Ultimate stress $= \frac{30800}{50.27} = 612.7 MPa > 350 MPa$ → fails.

Step 5 — har check ke liye minimum area ( $M S = 0$ ):

Yield ko $A \geq \frac{24200}{250} = 96.8 mm^{2}$ chahiye.
Ultimate ko $A \geq \frac{30800}{350} = 88.0 mm^{2}$ chahiye.

Badi requirement (yield, $96.8 mm^{2}$ ) govern karti hai. Step 6 — minimum diameter: $d = \frac{4 A}{π} = \frac{4 \times 96.8}{π} = 11.10 mm .$ Answer: $8 mm$ rod dono fail karta hai; minimum diameter $\approx 11.1 mm$ (yield-driven).

L4.2 — Target margin ke liye redesign karo

L4.1 continue karo. Management minimum $M S_{yield} = + 0.20$ chahta hai (sirf $0$ nahi). Kaun sa diameter isko meet karta hai? Nearest whole mm tak round up karo.

Recall Solution L4.2

Humein $\frac{σ _{y}}{F O S _{yield} S _{limit-stress}} - 1 \geq 0.20$ chahiye, yani actual stress $\leq \frac{σ _{y}}{1.20 \times F O S _{yield}}$ hona chahiye... seedha areas mein kaam karo:

Step 1 — required area. $M S = \frac{σ _{y}}{P _{design,yield} / A} - 1 = 0.20$ se: $A \geq \frac{P _{design,yield}}{σ _{y}} \times (1 + 0.20) = 96.8 \times 1.20 = 116.16 mm^{2} .$ Step 2 — diameter: $d = \frac{4 \times 116.16}{π} = 12.16 mm \Rightarrow 13 mm (rounded up) .$ $d = 13 mm$ ke saath check karo: $A = \frac{π ( 13 ) ^{2}}{4} = 132.7 mm^{2}$ , yield stress $= \frac{24200}{132.7} = 182.4 MPa$ , toh $M S_{yield} = \frac{250}{182.4} - 1 = 0.371 \geq 0.20 ✓ .$ Answer: $13 mm$ diameter ( $M S_{yield} \approx + 0.37$ deta hai, $+ 0.20$ target se comfortably upar).

Level 5 — Mastery

Subtle traps, degenerate cases, aur process ke baare mein reasoning.

L5.1 — Zero / negative margin boundary

Ek part mein $S_{allow} = 168 MPa$ hai, limit stress $120 MPa$ , $F O S = 1.4$ . (a) $M S$ kya hai? (b) Limit stress ko sirf $5%$ upar re-estimate kiya gaya (to $126 MPa$ ). $M S$ dobara compute karo. Thin margins ke baare mein yeh kya sikhata hai?

Recall Solution L5.1

(a) Design stress $= 1.4 \times 120 = 168 MPa$ . Phir $M S = \frac{168}{168} - 1 = 0.$ Exactly zero — allowable design load ke barabar hai, just barely OK, koi spare nahi. (b) New design stress $= 1.4 \times 126 = 176.4 MPa$ : $M S = \frac{168}{176.4} - 1 = - 0.0476.$ Sirf $5%$ load re-estimate ne part ko "pass" se fail ( $- 4.8%$ ) mein flip kar diya. Lesson: $M S$ jo $0$ ke paas ho woh fragile hai — koi bhi load-estimate scatter (exactly woh ignorance jise $F O S$ cover karne ke liye hai) ise negative push kar sakta hai. Engineers ek working margin rakhte hain (jaise $\geq + 0.10$ – $0.20$ ) precisely taaki chhote updates redesign cause na karein.

L5.2 — Degenerate load combination

Ek mount $6 g$ axial aur $0 g$ lateral (pure axial event) dekhta hai. Yeh alag time par $0 g$ axial aur $6 g$ lateral (pure lateral event) bhi dekhta hai. Ek junior engineer inhe RSS karke $6^{2} + 6^{2} = 8.49 g$ banata hai. Kya yeh sahi hai?

Recall Solution L5.2

Nahi. RSS sirf un loads ke liye hai jo simultaneously act karte hain. Yahan do $6 g$ events alag times par hote hain, toh tum envelope lo: har event akela $6 g$ resultant hai (kyunki doosra axis $0 g$ hai). Structure ko $6 g$ case survive karna hai, $8.49 g$ nahi. $F_{event 1} = m (6 g), F_{event 2} = m (6 g), design to 6 g, not 8.49 g .$ Blindly non-simultaneous cases ko RSS karna load ko $2 \approx 1.41 \times$ se inflate karta hai — ek bada, unjustified mass penalty. Sahi governing load $6 g$ hai.

L5.3 — Ek untested tank ke liye FOS choose karna

Do parts: (A) ek machined metallic bracket, qualification tak ground-tested; (B) ek composite pressure vessel jo full level par proof-tested nahi hoga. Dono same limit load dekhte hain. Kaun sa higher $F O S$ pata hai, aur kyun — argue karo ki $F O S$ physically kya represent karta hai.

Recall Solution L5.3

(B), untested composite tank, higher $F O S$ pata hai. $F O S$ ek paid-for cushion hai ignorance ke against: manufacturing scatter, analysis error, material variability, aur untested failure modes. Qualification tak ground testing (part A) seedha ignorance remove karta hai — tumne physically demonstrate kiya hai ki metallic bracket hold karta hai, toh chhota cushion justified hai ( $F O S_{ult} \approx 1.25$ ). Part B mein zyada unknowns stack hain: composites metals se zyada scatter karte hain, delamination inspect karna mushkil hai, aur koi full proof test nahi ka matlab hai ki demonstrated capability uncertain hai. Zyada unknowns → bada cushion (pressurized/composite hardware often $F O S_{ult} \approx 1.5$ – $2.0$ use karta hai). Number track karta hai kitna nahi pata tumhe, is par nahi ki part paper pe kitna strong dikhta hai.

L5.4 — Two-axis stress aur margin saath mein

Ek lug simultaneously $F_{a} = 3000 N$ axial aur $F_{l} = 4000 N$ lateral carry karta hai. Uska cross-section $A = 30 mm^{2}$ hai. Allowable $= 200 MPa$ , $F O S_{ult} = 1.4$ . Resultant limit stress aur margin of safety nikalo.

Recall Solution L5.4

Step 1 — resultant limit force (RSS, simultaneous & perpendicular): $F = 300 0^{2} + 400 0^{2} = 9000000 + 16000000 = 5000 N .$ (Famous $3$ - $4$ - $5$ triangle.) Step 2 — limit stress: $S_{limit} = \frac{5000}{30 mm ^{2}} = \frac{5000}{30 \times 1 0 ^{- 6} m ^{2}} = 166.7 MPa .$ Step 3 — margin: $M S = \frac{200}{1.4 \times 166.7} - 1 = \frac{200}{233.3} - 1 = - 0.143.$ $M S = - 0.14 < 0$ → fails. Lug bada karna hoga: $M S = 0$ ke liye minimum area $A \geq \frac{1.4 \times 5000}{200} = 35 mm^{2}$ chahiye. Answer: $S_{limit} = 166.7 MPa$ , $M S = - 0.14$ (fails); $\geq 35 mm^{2}$ chahiye.

Recall Quick self-check ladder (khatam karne ke baad reveal karo)

L1 jaanta hai kaun sa number kaun sa hai. L2 FOS chain ke through ek part size karta hai. L3 load cases combine/compare karta hai (RSS vs envelope). L4 poora loop chalata hai aur target margin ke liye redesign karta hai. L5 kyun ke baare mein reason karta hai — thin-margin fragility, degenerate combinations, aur ignorance se $F O S$ choose karna.

Dekho bhi: Stress and Strain — Yield vs Ultimate Strength, Quasi-static Loads and Launch Environment, Random & Sine Vibration Testing, Finite Element Analysis of Spacecraft Structures, Mass Budget and Structural Efficiency, Qualification vs Acceptance Testing.