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

3.6.2 · D3 · Physics › Spacecraft Structures & Systems Engineering › Structural design process — load cases, FOS (factor of safet

Shuru karne se pehle, do formulas yaad karo jinpe hum depend karte hain (from the parent note):

Definition Ek notation convention, jo neeche har jagah use hogi

Letter $S$ ek placeholder hai "wo quantity jise tum check karte ho" — ye ya to ek stress ya ek force ke liye stand kar sakta hai, bas condition ye hai ki tum $M S$ ratio ke dono sides par ek hi type consistently use karo. Ise explicit rakhne ke liye hum do physical kinds ke liye alag letters use karte hain aur unhe hamesha $S$ se map karte hain:

Forces ke liye $F$ ya $P$ (units: newtons, N). Jab problem forces mein pose ki gayi ho to hum set karte hain $S_{limit} \equiv F_{limit}$ , $S_{design} \equiv F_{design}$ , aur $S_{allow} \equiv F_{allow}$ (ek allowable force, jaise bolt ka proof load).
Stresses ke liye $σ$ (units: pascals, Pa = N/m²). Tab $S_{limit} \equiv σ_{limit}$ , etc. $M S$ formula dono cases mein identical hai — units bas ratio ke andar cancel ho jaate hain. Rule: numerator mein force aur denominator mein stress kabhi mix mat karo.

Ise dobara padho: $M S$ ek spare capacity ka fraction hai. $M S = 0$ matlab "bilkul enough, koi cushion nahi." $M S = 0.2$ matlab "factored demand se 20% zyada strong." $M S < 0$ matlab "fail."

Scenario matrix

Har FOS/load-case problem jo tumhe kabhi milegi wo in cells mein se kisi ek mein aati hai:

#	Case class	Kya ise alag banata hai	Example
A	Part size karo (area solve karo)	Unknown geometry hai; $M S = 0$ set karo	Ex 1
B	Part check karo ( $M S$ compute karo)	Geometry pata hai; kya $M S \geq 0$ hai?	Ex 2
C	Do alag FOS (yield aur ultimate)	Part ko dono checks pass karne chahiye	Ex 3
D	Simultaneous loads combine karo (RSS & law of cosines)	Vector add karo, scalar sum nahi	Ex 4
E	Zero / degenerate input	Zero load, zero margin, $S_{allow} = S_{design}$	Ex 5
F	Negative margin → redesign	$M S < 0$ ; nayi geometry dhundo	Ex 6
G	Real-world word problem (force-based $S$ )	Prose se numbers nikalo	Ex 7
H	Exam twist (kaun sa limiting hai?)	Do candidate failure modes; worst chuno	Ex 8

Neeche ke aath examples aathon cells ko hit karte hain. Har ek Verify step ke saath khatam hota hai, aur har number is page ke saath aane wale VERIFY block mein machine-checked hai.

Prerequisites jo tum open rakhna chahoge: Stress and Strain — Yield vs Ultimate Strength, Quasi-static Loads and Launch Environment.

Ex 1 — [Cell A] Ek strut size karo

Worked example Statement

Ek strut ek limit tensile force $P_{limit} = 20 kN$ carry karta hai. Material yield stress $σ_{y} = 300 MPa$ , $F O S_{yield} = 1.25$ . Minimum cross-sectional area $A$ nikalo.

Forecast: padhne se pehle guess karo — kya $A$ , $P_{limit} / σ_{y}$ se bada hoga ya chhota? (Ye bada hona chahiye, kyunki hum factored load ke against size karte hain, raw wale ke against nahi.)

Step 1 — demand factor karo. Yahan checked quantity $S$ ek force hai, isliye $P_{design} \equiv S_{design}$ set karo (general design load jo ek force ke liye specialised hai). Tab $P_{design} = F O S \times P_{limit} = 1.25 \times 20 = 25 kN$ . Ye step kyun? FOS load ko inflate karta hai, material ko nahi. Hume factored force survive karni hai.

Step 2 — stress limit likho. Stress = force / area, aur hum chahte hain ki ye yield se zyada na ho: $σ = \frac{P _{design}}{A} \leq σ_{y}$ Ye step kyun? Stress (force per unit area) wo quantity hai jo material actually "feel" karta hai. Yield se neeche → koi permanent bend nahi.

Step 3 — sabse chhota area solve karo ( $M S = 0$ set karo). $A \geq \frac{P _{design}}{σ _{y}} = \frac{25 000 N}{300 \times 1 0 ^{6} Pa} = 8.33 \times 1 0^{- 5} m^{2} = 83.3 mm^{2}$ Ye step kyun? Equality knife-edge hai ( $M S = 0$ ); koi bhi bada area positive margin deta hai.

Verify: $A = 83.3 mm^{2}$ wapas daalo: $σ = 25 000/ (83.3 \times 1 0^{- 6}) = 300 \times 1 0^{6}$ Pa $= σ_{y}$ . Exactly yield par → $M S = 0$ . ✓ Units: $N / m^{2} = Pa$ . ✓

Ex 2 — [Cell B] Ek bracket ka margin check karo

Worked example Statement

Ek bracket ka allowable stress $S_{allow} = σ_{allow} = 180 MPa$ hai. Limit load se worst-case stress $120 MPa$ hai, $F O S_{ult} = 1.4$ ke saath. Kya ye pass karta hai?

Forecast: $120$ , $180$ se kaafi neeche hai — safe lagta hai. Lekin hum abhi factor nahi kiya. $M S$ ka sign guess karo.

Step 1 — limit stress ko design par factor karo. $S_{design} = 1.4 \times 120 = 168 MPa$ . Ye step kyun? "Positive stress = positive margin" wali galti (parent note) yahan mar jaati hai: tum factored demand ke against compare karna chahte ho.

Step 2 — margin compute karo. $M S = \frac{180}{168} - 1 = 0.0714$ . Ye step kyun? $M S$ factor karne ke baad fractional spare capacity hai. Numerator aur denominator dono stresses (MPa) hain → units cancel ho jaate hain.

Result: $M S = + 0.07 \geq 0$ → pass, lekin sirf 7% cushion. Thin hai; ek engineer material add kar sakta hai.

Verify: $180 = 1.0714 \times 168$ ? $1.0714 \times 168 = 180$ . ✓ Aur $168 < 180$ confirm karta hai $M S > 0$ . ✓

Ex 3 — [Cell C] Yield AUR Ultimate dono

Worked example Statement

Ek fitting ek limit stress $S_{limit} = σ_{limit} = 100 MPa$ dekhti hai. Material: yield $σ_{y} = 250 MPa$ , ultimate $σ_{u} = 400 MPa$ . Factors: $F O S_{yield} = 1.1$ , $F O S_{ult} = 1.4$ . Kya ye dono checks pass karta hai?

Forecast: Do checks — tumhara guess kya hai, yield ya ultimate, dono mein se kaun tighter hoga?

Step 1 — yield margin. Design $= 1.1 \times 100 = 110 MPa$ ; $M S_{y} = \frac{250}{110} - 1 = 1.273$ . Ye step kyun? Yield permanent deformation rokta hai; ye lower strength ke against lower factor use karta hai.

Step 2 — ultimate margin. Design $= 1.4 \times 100 = 140 MPa$ ; $M S_{u} = \frac{400}{140} - 1 = 1.857$ . Ye step kyun? Ultimate rupture rokta hai; higher factor, higher strength.

Step 3 — part ko dono mein se worst pass karna chahiye. $min (1.273, 1.857) = 1.273$ → yield governs, phir bhi positive → dono pass. Ye step kyun? Dono alag physical requirements hain; chhota margin real wala hai.

Verify: $250/110 = 2.2727 \Rightarrow M S_{y} = 1.2727$ . $400/140 = 2.857 \Rightarrow M S_{u} = 1.857$ . Yield $<$ ultimate margin, dono $\geq 0$ . ✓ (Parent ki "dono ke liye same FOS" wali galti khatam karta hai.)

Ex 4 — [Cell D] Simultaneous loads combine karo — RSS aur general angle

Worked example Statement

Ek bracket par rakha $50 kg$ ka box usi waqt $9 g$ axial aur $4 g$ lateral QSL dekhta hai ( $g = 9.81 m/s^{2}$ ). (a) Resultant force nikalo jab dono $9 0^{\circ}$ par act karein (perpendicular). (b) Batao ki agar wo general angle $θ$ par milein to unhe kaise combine karein.

Forecast: Kya $9 g$ aur $4 g$ ko add karke $13 g$ milega, ya kuch chhota? Figure dekho — answer red diagonal ki length hai, do legs ko end-to-end nahi rakha.

Figure padhna: Blue arrow axial force $F_{a}$ hai jo right point kar raha hai; orange arrow lateral force $F_{l}$ hai jo upar point kar raha hai. Dono ek hi corner se shuru hote hain (chhota black right-angle square dikhata hai ki ye exactly $9 0^{\circ}$ apart hain). Corner se rectangle ke opposite corner tak red arrow resultant $F$ hai — iska length wahi hai jo hum compute karte hain. Neeche faint gray dotted line (galat) scalar sum $F_{a} + F_{l}$ hai, ye dikhane ke liye ki wo kitni lambi hai, aur isliye kitni over-conservative hai.

Step 1 — axial force. $F_{a} = m (9 g) = 50 \times 9.81 \times 9 = 4414.5 N$ . Step 2 — lateral force. $F_{l} = m (4 g) = 50 \times 9.81 \times 4 = 1962 N$ . Ye steps kyun? Force $=$ mass $\times$ acceleration; har $g$ -level ek acceleration hai.

Step 3a — Pythagoras se resultant (RSS), $9 0^{\circ}$ case. Kyunki dono forces figure mein right triangle ke perpendicular legs hain, resultant (hypotenuse, red) hai $F = F_{a}^{2} + F_{l}^{2} = 4414. 5^{2} + 196 2^{2} = 4831.6 N$ RSS kyun, $F_{a} + F_{l}$ nahi? Perpendicular vectors hypotenuse ke saath combine hote hain. Scalar sum $F_{a} + F_{l} = 6376.5 N$ (gray dotted path) orthogonal loads ke liye physically impossible hai aur mass waste karta hai.

Step 3b — general angle $θ$ (law of cosines). RSS sirf special case $θ = 9 0^{\circ}$ hai. Agar do forces apni directions ke beech interior angle $θ$ par milein, to resultant magnitude law of cosines se aata hai: $F = F_{a}^{2} + F_{l}^{2} + 2 F_{a} F_{l} cos θ$ Yahan cosine kyun? $cos θ$ measure karta hai ki do loads ek doosre ko kitna reinforce ya oppose karte hain. Limits check karo: $θ = 9 0^{\circ}$ par, $cos 9 0^{\circ} = 0$ aur middle term vanish ho jaata hai → RSS recover hota hai ( $4831.6 N$ ). $θ = 0^{\circ}$ par (same direction), $cos 0^{\circ} = 1$ → $F = F_{a} + F_{l} = 6376.5 N$ , full scalar sum. $θ = 18 0^{\circ}$ par (directly opposing), $cos 18 0^{\circ} = - 1$ → $F = ∣ F_{a} - F_{l} ∣ = 2452.5 N$ , sabse chhota possible. To RSS middle-ground case hai, aur har real combination $∣ F_{a} - F_{l} ∣$ aur $F_{a} + F_{l}$ ke beech hoti hai.

Verify: $4414. 5^{2} + 196 2^{2} = 1.9488 \times 1 0^{7} + 3.8494 \times 1 0^{6} = 2.3338 \times 1 0^{7}$ ; $= 4831.6 N$ . Cosine law $θ = 9 0^{\circ}$ par same deta hai, $θ = 0^{\circ}$ par $6376.5$ deta hai, $θ = 18 0^{\circ}$ par $2452.5$ deta hai. Aur $2452.5 < 4831.6 < 6376.5$ confirm karta hai ki RSS extremes ke beech hai. ✓ Units: $kg \cdot m/s^{2} = N$ . ✓

Ex 5 — [Cell E] Degenerate inputs (zero & knife-edge)

Worked example Statement

Teen quick sub-cases edges nail karne ke liye: (a) Ek part jis par zero applied load hai. (b) Ek part jahan $S_{allow}$ exactly design load ke barabar hai. (c) Ek part jisme zero-thickness (degenerate) member hai.

Forecast: $M S$ kya hoga jab koi load hi nahi?

Step 1 — (a) zero load, limit ki tarah treat karo. Jaise load shrink hota hai, $S_{limit} \to 0^{+}$ isliye $S_{design} = F O S \times S_{limit} \to 0^{+}$ . Hume $S_{allow} /0$ raw arithmetic mein nahi likhna chahiye (zero se division undefined hai). Iske bajaye $M S$ ko limit ki tarah padho $M S = lim_{S_{design} \to 0^{+}} (\frac{S _{allow}}{S _{design}} - 1) = + \infty$ Ye step kyun? Ratio without bound grow karta hai jaise denominator upar se zero approach karta hai. Physically: jo part (almost) kuch nahi carry kar raha uski us load ke against (almost) unbounded spare capacity hai — wo kabhi us se fail nahi ho sakta. Limit language statement ko rigorous rakhti hai.

Step 2 — (b) knife-edge. $S_{allow} = 168 MPa$ aur $S_{design} = 168 MPa$ maano. Tab $M S = \frac{168}{168} - 1 = 0$ . Ye step kyun? Ye acceptability ki exact boundary hai — "just barely OK," zero cushion. Estimate ke upar koi bhi scatter ab ise fail kar dega.

Step 3 — (c) zero area. Ex 1 se, $σ = P / A$ . Jaise $A \to 0^{+}$ , $σ \to + \infty$ , isliye ( $S_{allow}$ finite rakhen) $M S \to - 1$ (fully failed). Ye step kyun? Vanishing cross-section sabhi force ko koi material nahi mein concentrate kar deta hai → infinite stress. Isliye struts ka ek minimum area hota hai, kabhi zero nahi.

Verify: (b) $168/168 - 1 = 0$ . ✓ (a) $lim_{S_{design} \to 0^{+}} S_{allow} / S_{design} - 1 = + \infty$ . ✓ (c) $lim_{A \to 0^{+}} S_{allow} / (P / A) - 1 = - 1$ . ✓

Ex 6 — [Cell F] Negative margin → redesign

Worked example Statement

Ek plate ka $S_{allow} = σ_{allow} = 200 MPa$ , limit stress $S_{limit} = 160 MPa$ , $F O S_{ult} = 1.4$ hai. (a) Dikhao ki ye fail karta hai. (b) Kitne factor se area badhana hoga (equivalently stress kam karna) taaki $M S = 0$ mila sake?

Forecast: $160 < 200$ , isliye ye "theek lagta" hai — lekin pehle factor karo. Kya ye fail karega?

Step 1 — factor & check. $S_{design} = 1.4 \times 160 = 224 MPa$ ; $M S = \frac{200}{224} - 1 = - 0.107$ . Ye step kyun? $224 > 200$ → demand capacity se zyada → fail ~11% se. Ye exactly "positive stress but negative margin" trap hai.

Step 2 — fix target karo. Stress $1/ A$ ki tarah scale karta hai (force fixed). $M S = 0$ ke liye hume naya stress $S_{limit}^{'}$ chahiye jaise ki $1.4 S_{limit}^{'} = 200$ , matlab $S_{limit}^{'} = 142.9 MPa$ . Ye step kyun? Hum strength nahi add kar sakte (material fixed); area add karke stress kam karte hain.

Step 3 — area ratio. Kyunki $σ \propto 1/ A$ : $\frac{A _{new}}{A _{old}} = \frac{S _{limit}}{S _{limit}^{'}} = \frac{160}{142.9} = 1.12$ . Ye step kyun? Plate ko 12% mota karo knife-edge tak pahunchne ke liye; real cushion ke liye thoda aur add karo.

Verify: Naya design stress $= 224/1.12 = 200 MPa = S_{allow}$ → $M S = 0$ . ✓ Original $M S = 200/224 - 1 = - 0.1071$ . ✓

Ex 7 — [Cell G] Real-world word problem (allowable ek force hai)

Worked example Statement

"Liftoff ke dauran, ek $12 kg$ ka avionics unit ek shelf par bolted hai. Launch vehicle user manual ek quasi-static limit acceleration of $6 g$ axial quote karta hai ( $g = 9.81 m/s^{2}$ ). 4 identical bolts mein se har ek axial load equally share karta hai. Ek bolt ka proof (allowable) tensile force $F_{allow} = 3000 N$ hai. $F O S_{ult} = 1.4$ ke saath, kya bolts adequate hain?"

Notation par note: yahan checked quantity ek force hai, isliye hum set karte hain $S_{allow} \equiv F_{allow} = 3000 N$ aur $S_{design} \equiv F_{design}$ . $M S$ ratio ke dono sides forces (N) hain → units cancel, exactly stresses ki tarah.

Forecast: total factored load 4 bolts par split — kya har bolt ka share $3000 N$ ke neeche rahega?

Step 1 — total limit force. $F_{limit} = m (6 g) = 12 \times 9.81 \times 6 = 706.3 N$ . Ye step kyun? $g$ -level ko actual force mein convert karo $F = ma$ ke zariye; acceleration $6 g$ hai.

Step 2 — factor karo. $F_{design} = F O S_{ult} \times F_{limit} = 1.4 \times 706.3 = 988.8 N$ . Ye step kyun? Bolts ko factored ultimate demand survive karni hai, raw estimate nahi.

Step 3 — per-bolt share. $F_{bolt} = F_{design} /4 = 988.8/4 = 247.2 N$ . Ye step kyun? Ek symmetric 4-bolt pattern axial load equally split karta hai, isliye har bolt ek quarter dekhta hai.

Step 4 — bolt per margin. $M S = \frac{F _{allow}}{F _{bolt}} - 1 = \frac{3000}{247.2} - 1 = 11.14$ . Ye step kyun? Fractional spare capacity, force over force — units (N) ratio ke andar cancel ho jaate hain.

Result: $M S \approx + 11 \geq 0$ → axial QSL ke against bahut zyada adequate. Practice mein bolts akele axial QSL se rarely size hote hain; shear aur preload usually govern karte hain. Lekin ye axial check enormous margin ke saath cleanly pass karta hai.

Verify: $12 \times 9.81 \times 6 = 706.32$ ; $\times 1.4 = 988.85$ ; $/4 = 247.21$ ; $3000/247.21 - 1 = 11.14$ . ✓ Units: poore mein N, ratio dimensionless. ✓

Ex 8 — [Cell H] Exam twist — kaun sa mode governs?

Worked example Statement

Ek pressurized tank wall ek limit stress $S_{limit} = σ_{limit} = 150 MPa$ dekhti hai. Do candidate checks dono satisfy hone chahiye:

Yield: $σ_{y} = 280 MPa$ , $F O S_{yield} = 1.1$ .
Ultimate: $σ_{u} = 350 MPa$ , lekin kyunki part pressurized hai to user manual ek higher $F O S_{ult} = 2.0$ demand karta hai.

Kaun sa check governs karta hai, aur kya tank pass karta hai?

Forecast: ultimate strength ( $350$ ) yield strength ( $280$ ) se badi hai — lekin ultimate FOS ( $2.0$ ) bhi yield FOS ( $1.1$ ) se kaafi bada hai. Kaun sa effect jeetega?

Step 1 — yield margin. $S_{design, y} = 1.1 \times 150 = 165 MPa$ ; $M S_{y} = \frac{280}{165} - 1 = 0.697$ . Ye step kyun? Yield permanent deformation rokta hai; ye apne khud ke (lower) factor ko apni khud ki strength ke against use karta hai.

Step 2 — ultimate margin. $S_{design, u} = 2.0 \times 150 = 300 MPa$ ; $M S_{u} = \frac{350}{300} - 1 = 0.167$ . Ye step kyun? Ultimate rupture rokta hai; elevated pressurized factor factored demand ko bada bana deta hai.

Step 3 — worst (sabse chhota) margin chuno. $min (0.697, 0.167) = 0.167$ → ultimate governs. Bada pressurized factor higher raw strength ko overwhelm kar deta hai. Ye step kyun? Part ko har mode pass karna chahiye; sabse tight margin real wala hai. Dono ke liye ek FOS kabhi reuse mat karo (parent ki "same FOS" wali galti).

Result: dono margins $\geq 0$ hain → pass, lekin ye ultimate-limited hai $M S = + 0.17$ par. Trap ye assume karna hai ki badi raw strength ( $350$ ) matlab ultimate safest hai — factor ise flip kar deta hai.

Verify: $280/165 - 1 = 0.6970$ ; $350/300 - 1 = 0.1667$ . Ultimate margin $<$ yield margin → ultimate governs; dono $\geq 0$ → pass. ✓

Recall Har example kaun sa cell hit karta hai?

Ex1 sizing (A) ::: area ke liye solve karo, $M S = 0$ set karo Ex2 check margin (B) ::: $M S$ compute karo, positive lekin thin Ex3 two FOS (C) ::: yield aur ultimate dono checked, min lo Ex4 combine loads (D) ::: $9 0^{\circ}$ ke liye RSS, general angle ke liye law of cosines Ex5 degenerate (E) ::: zero load → $M S \to + \infty$ (limit ki tarah); zero area → $M S \to - 1$ Ex6 redesign (F) ::: negative $M S$ , area ratio se mota karo Ex7 word problem (G) ::: force-based $S$ , bolts par split Ex8 exam twist (H) ::: pressurized part par higher FOS ultimate ko govern karta hai

Aage explore karne ke liye related deep dives: Finite Element Analysis of Spacecraft Structures (jahan $S_{allow}$ actually aata hai), Mass Budget and Structural Efficiency (kyun hum over-design se nafrat karte hain), Qualification vs Acceptance Testing aur Random & Sine Vibration Testing (hardware mein margin prove karna).