Structural loads — axial (thrust), bending (wind shear), dynamic (vibration, acoustics, shock): Worked examples

3.6.1 · D3 · Physics › Spacecraft Structures & Systems Engineering › Structural loads — axial (thrust), bending (wind shear), dyn

Ye parent topic ka drill page hai. Hum yahan formulas re-derive nahi karenge — parent note mein har symbol kama ke aaya hai. Yahan hum teen load families ke har tarah ke situation ko dhundhte hain aur ek-ek ko, zor se, solve karte hain.

Neeche har jagah use hone wale symbols (saare parent note mein define hain — koi bhi anjaan lage toh ek baar wapas dekho):

$m (x)$ = height $x$ par cut ke upar baitha mass (kg).
$g = 9.81 m/s^{2}$ = gravity. $n$ = axial load factor "g's" mein (dimensionless).
$F_{axial} = m n g$ , aur $σ = F / A$ (force cross-section area $A$ par spread hoti hai).
$M$ = bending moment (N·m). $y$ = neutral axis se kisi fiber ki doori (m), bending axis ke perpendicular measure kiya gaya — stretched side par positive, squashed side par negative. $c$ = outer skin par $y$ ki value, yaani sabse badi doori (m). $I$ = second moment of area (m⁴), $σ_{bend} = M c / I$ .
$θ$ = ring ke around angle (radians), $0$ se $2 π$ tak jab tum circular cross-section ke around ek chakkar lagate ho; angle $θ$ par ek fiber neutral axis se $y = R sin θ$ ki height par hota hai. Yeh sirf saare fibers ko jodne ka ek bookkeeping angle hai.
$f_{n}$ = natural frequency (Hz), $Q = 1/ (2 ζ)$ = resonance amplification, $ζ$ = damping ratio.
$g_{RMS} = \frac{π}{2} f_{n} Q W$ = Miles' random-vibration response, $W$ = PSD (g²/Hz).
$k$ = stiffness constant (N/m): kisi part ko ek metre deflect karne mein kitna force lagta hai — ek "spring rate". Jo structure load ke neeche muskil se bhi jhukti ho, uska $k$ bada hota hai. Yeh natural frequency ko $f_{n} = \frac{1}{2 π} k / m$ ke through set karta hai.
$x_{static}$ = steadily applied load ke neeche part ka deflection; $x_{max}$ = jab woh load resonant frequency par oscillate kare tab peak deflection. Unka ratio exactly $Q$ hai: $x_{max} \approx Q x_{static}$ .

Scenario matrix

Is topic ka har problem inhi cells mein se exactly ek mein aata hai. Hamara kaam hai sab cover karna.

Cell	Load family	Jo cheez test ho rahi hai	Kis Example se Cover Hoga
A1	Axial	ek cut par ordinary compression	Example 1
A2	Axial	sign flip → tension (upper stage coasting, $n < 0$ / drag)	Example 2
A3	Axial	degenerate: $n = 0$ (free-fall / engine cutoff)	Example 3
B1	Bending	ordinary flexure stress $M c / I$	Example 4
B2	Combined	max-Q par same fiber par axial + bending add	Example 5
B3	Bending	limiting geometry: thin-ring $I = π R^{3} t$ , $R$ double karne ka effect	Example 6
D1	Dynamic	Miles' equation, phir $3 σ$ design load	Example 7
D2	Dynamic	frequency-separation pass/fail decision	Example 8
D3	Dynamic	degenerate: $ζ \to 0$ (undamped), $Q$ aur $x_{max}$ ka kya hoga	Example 9
W/X	Any	word problem + exam twist (opposite fiber sign)	Example 10

Axial cases

Worked example Example 1 — Cell A1: ordinary compression

Ek booster ke aadhe raaste par cut lagaya gaya hai. Cut ke upar $m (x) = 1500 kg$ baitha hai. Load-bearing wall ki cross-section area $A = 0.015 m^{2}$ hai. Is waqt vehicle $n = 5$ g's khich raha hai. Axial stress nikalo aur batao compression hai ya tension.

Forecast: Kuch hazar kg × kuch g's ≈ 70–80 kN, ~1/60th m² ke upar → tens of MPa, compressive (thrust squeeze karta hai). Guess ~5 MPa? Dekhte hain.

Internal force $F_{axial} = m n g = 1500 \times 5 \times 9.81 = 73, 575 N$ . Yeh step kyun? Cut par structure ko upar ki sab cheez ka weight support karna hai aur usse accelerate bhi karna hai; $g + a = n g$ dono ko ek hi term mein bundle kar deta hai.
Stress $σ = F / A = 73, 575/0.015 = 4.905 \times 1 0^{6} Pa = 4.905 MPa$ . Yeh step kyun? Stress = force jo actually use carry karne wale material par spread hoti hai.
Sign: neeche ka engine upar ke mass ko upar dhakelta hai → wall squeeze hoti hai → compressive (hamare convention se positive). Yeh step kyun? Direction matter karta hai: compression ek thin shell ko buckle kar sakta hai, isliye hume iska naam lena hoga.

Verify: Units: $kg \cdot (m/s^{2}) / m^{2} = N/m^{2} = Pa$ ✓. Ballpark ~5 MPa forecast se match kiya ✓.

Worked example Example 2 — Cell A2: sign flip (tension)

Burnout ke baad stage coast karta hai aur aerodynamic drag ab use decelerate kar raha hai. Effective acceleration axis ke saath-saath neeche ki taraf hai, isliye upar ka mass aage badhta rehna chahta hai aur structure ko kheenchta hai. Ise $n = - 0.5$ model karo (axis ke saath net 0.5 g ka "negative" load). Wahi $m = 1500 kg$ , $A = 0.015 m^{2}$ . Stress aur uska sign nikalo.

Forecast: Same numbers lekin magnitude daswaan hissa aur opposite sign — toh ek chota tensile stress, well under 1 MPa.

Force $F_{axial} = m n g = 1500 \times (- 0.5) \times 9.81 = - 7357.5 N$ . Yeh step kyun? $n$ negative ho sakta hai jab axial inertial load reverse ho (drag deceleration, coast). Formula phir bhi valid hai; sign physics carry karta hai.
Stress $σ = F / A = - 7357.5/0.015 = - 4.905 \times 1 0^{5} Pa = - 0.4905 MPa$ . Yeh step kyun? Usi division se magnitude; negative sign yahan puri baat hai.
Sign interpret karo: negative $σ$ = tension (hamare master convention se) — wall stretch ho rahi hai. Yeh step kyun? Thin shells tension se nahi darte lekin compression se dab jaate hain, isliye tension mein hona reassuring hai.

Verify: Magnitude exactly Example 1 ka $1/10$ hai (kyunki $∣ - 0.5∣/5 = 1/10$ ) → $4.905/10 = 0.4905$ MPa ✓. Sign wada kiya tha waise flip hua ✓.

Worked example Example 3 — Cell A3: degenerate,

$n = 0$ (free fall) Engine cut off ho jaata hai aur, drag ignore karte hue, poora stage free fall mein hai: woh exactly $g$ se neeche accelerate karta hai, toh $a = - g$ aur $g + a = 0$ , yaani $n = 0$ . Wahi $m$ , $A$ . Axial stress kya hoga?

Forecast: Free fall mein sab cheez saath girती hai — kuch bhi kisi par push nahi karta. Guess: zero stress.

Force $F_{axial} = m n g = 1500 \times 0 \times 9.81 = 0 N$ . Yeh step kyun? Yeh degenerate check hai: jab $n = 0$ ho, upar ke mass ke free body mein gravity aur acceleration exactly cancel ho jaate hain.
Stress $σ = 0/0.015 = 0 Pa$ . Yeh step kyun? Zero force → zero stress. Free fall mein structure koi axial load carry nahi karta.

Verify: Physically, free fall mein koi cheez "weightless" hoti hai — scale zero read karta hai. Hamara formula $0$ deta hai ✓. Yeh compression ( $n > 0$ ) aur tension ( $n < 0$ ) ke beech ki boundary hai.

Bending & combined cases

Worked example Example 4 — Cell B1: ordinary flexure

Ek thin cylinder shell: radius $R = 1.2 m$ , wall $t = 4 mm = 0.004 m$ . Ek gust bending moment $M = 6 \times 1 0^{5} N\cdotm$ banata hai. Peak bending stress nikalo. (Thin-ring second moment: $I = π R^{3} t$ ; outer fiber $c = R$ .)

Forecast: $I$ scales like $R^{3} t \approx π (1.7) (0.004) \approx 0.02$ m⁴, aur $M c / I \approx 6 \times 1 0^{5} \times 1.2/0.02 \approx 3 \times 1 0^{7}$ → tens of MPa.

Figure padhna (alt text): Left mein round shell ka cross-section end-on dikhta hai. Center se guzarta horizontal dashed line neutral axis hai — material ki woh ek layer jo tube bend hone par na stretch hoti hai na squash. Magenta arrow top outer fiber ko point karta hai, jo stretched (tension) hai; orange arrow bottom outer fiber ko point karta hai, jo compressed hai. Dono neutral axis se $c = R$ ki doori par hain, koi bhi material jitna door ja sakta hai — isliye extreme stress wahan hi rehta hai. Right panel stress $σ = M y / I$ ko neutral axis se fiber ki doori $y$ ke against plot karta hai (yaad raho $y$ symbol list mein define hai): yeh origin se guzarti straight line hai, upar magenta (tension) aur neeche orange (compression). Picture ka key insight: stress centre par zero hai aur skin tak straight line mein $\pm M c / I$ tak badhta hai — isliye sirf outermost fiber ko hi check karna hota hai.

Second moment $I = π R^{3} t = π (1.2)^{3} (0.004) = π \times 1.728 \times 0.004 = 0.021714 m^{4}$ . Yeh step kyun? Thin ring ke liye saara material neutral axis se ~distance $R$ par hota hai. $\int y^{2} d A$ add karne ke liye hum ring ke around angle $θ$ use karke chalte hain (symbol list mein define, $0$ se $2 π$ tak): angle $θ$ par ek fiber ki height $y = R sin θ$ hoti hai, aur arithmetic collapse hokar $π R^{3} t$ ban jaata hai. $I$ measure karta hai ki material bending axis se kitna door spread hai — jitna door, utna stiffer.
Outer fiber $c = R = 1.2 m$ — sabse zyada stretched (aur sabse zyada compressed) fiber outer skin hai (figure mein: woh fiber jise magenta/orange arrows point karte hain, ring ke top aur bottom par). Yeh step kyun? Stress neutral axis se distance $y$ ke saath linearly badhta hai (figure ka right panel), isliye $y = c$ par maximum hota hai.
Peak stress $σ_{bend} = M c / I = 6 \times 1 0^{5} \times 1.2/0.021714 = 3.316 \times 1 0^{7} Pa = 33.16 MPa$ . Yeh step kyun? Flexure formula: bending moment divided by section ki material-spread, times lever arm.

Verify: Units $N\cdotm \cdot m / m^{4} = N/m^{2}$ ✓. ~33 MPa forecast band mein hai ✓.

Worked example Example 5 — Cell B2: same fiber par axial + bending (max-Q)

Example 4 waali shell hi ( $I = 0.021714 m^{4}$ , $c = 1.2$ m, toh bending $= 33.16$ MPa). max-Q par stage axial compression bhi carry karta hai: upar mass $m = 3000 kg$ , $n = 4$ , aur wall area $A = 2 π R t = 2 π (1.2) (0.004) = 0.030159 m^{2}$ hai. Windward fiber par total compressive stress nikalo.

Forecast: Axial part kuch MPa hai; bending ~33 MPa hai; yeh add hote hain → total high-30s MPa mein.

Axial force $F = m n g = 3000 \times 4 \times 9.81 = 117, 720 N$ . Yeh step kyun? Cut par axial load, load-factor form mein.
Axial stress $σ_{ax} = F / A = 117, 720/0.030159 = 3.904 \times 1 0^{6} Pa = 3.904 MPa$ (compressive, positive). Yeh step kyun? Thin-ring wall area = circumference × thickness $= 2 π R t$ .
Compressed fiber par add karo $σ_{total} = σ_{ax} + σ_{bend} = 3.904 + 33.16 = 37.06 MPa$ (dono positive → dono compression → add hote hain). Yeh step kyun? Windward/leeward compressed fiber par dono effects same direction mein push karte hain, isliye add hote hain. Yeh worst-case sizing point hai — max-Q par inhe alag-alag kabhi analyze mat karo.

Verify: Total > akele koi bhi piece ✓. Windward fiber (compression + compression) leeward fiber se bura hai (compression − tension = $3.904 - 33.16 = - 29.3$ MPa, yaani net tension), confirm karta hai ki humne critical side sahi chuni ✓.

Worked example Example 6 — Cell B3: limiting geometry (radius double karo)

Example 4 ki shell lo ( $R = 1.2$ m, $t = 4$ mm, $M = 6 \times 1 0^{5}$ N·m, $σ = 33.16$ MPa). Ek designer radius double karke $R^{'} = 2.4 m$ karta hai lekin $t$ aur $M$ same rakhta hai. Naya bending stress kya hoga? Purane se ratio kya hoga?

Forecast: $I \propto R^{3}$ aur $c \propto R$ , isliye $σ = M c / I \propto R / R^{3} = 1/ R^{2}$ . $R$ double karne se stress quarter ho jaana chahiye → lagbhag $33/4 \approx 8$ MPa.

Naya second moment $I^{'} = π R^{'3} t = π (2.4)^{3} (0.004) = π \times 13.824 \times 0.004 = 0.173712 m^{4}$ . Yeh step kyun? $I^{'} = 2^{3} I = 8 I$ — material ko double door bahar push karne se stiffness $8$ guna badh jaati hai.
Naya stress $σ^{'} = M R^{'} / I^{'} = 6 \times 1 0^{5} \times 2.4/0.173712 = 8.289 \times 1 0^{6} Pa = 8.289 MPa$ . Yeh step kyun? $c$ double hua (sirf $\times 2$ ) lekin $I$ $\times 8$ badha, toh net $\times 2/8 = \times 1/4$ .
Ratio $σ^{'} / σ = 8.289/33.16 = 0.25$ . Yeh step kyun? $1/ R^{2}$ scaling law confirm karta hai — yeh ek structures engineer ka sabse useful lever hai: zyada chodi tube bending mein dramatically stiffer hoti hai.

Verify: Predicted $1/ 2^{2} = 1/4$ ✓; $33.16 \times 0.25 = 8.29$ MPa match karta hai ✓.

Dynamic cases

Worked example Example 7 — Cell D1: Miles' equation + 3σ

Ek component ka natural frequency $f_{n} = 120 Hz$ hai, damping $ζ = 0.05$ hai, aur yeh apne resonance par $W = 0.05 g^{2} / Hz$ ke random-vibration PSD par ride karta hai. RMS response $g_{RMS}$ nikalo, phir $3 σ$ design load.

Forecast: $Q = 1/ (2 \times 0.05) = 10$ . Root ke andar: $\frac{π}{2} \times 120 \times 10 \times 0.05 \approx 94$ , root ≈ 9.7 g. Times 3 ≈ 29 g.

Amplification $Q = 1/ (2 ζ) = 1/ (2 \times 0.05) = 10$ . Yeh step kyun? Resonance $Q$ se amplify karta hai; Miles' equation se pehle hume yeh chahiye.
RMS response $g_{RMS} = \frac{π}{2} f_{n} Q W = \frac{π}{2} \times 120 \times 10 \times 0.05 = 94.25 = 9.708 g$ . Yeh step kyun? Miles' equation random vibration ke poore resonance-peak integral ko ek closed form mein collapse kar deta hai; $π /2$ SDOF peak ke neeche ka area hai.
Design load $3 σ = 3 \times 9.708 = 29.12 g$ . Yeh step kyun? Random peaks Gaussian hote hain; $3 σ$ unhe 99.7% cover karta hai, yeh standard qualification level hai.

Verify: $94.25 = 9.708$ aur $\times 3 = 29.12$ ✓. Units: PSD g²/Hz hai, times Hz → g², root → g ✓.

Worked example Example 8 — Cell D2: frequency-separation decision

Ek launch vehicle demand karta hai ki payload ki pehli lateral natural frequency 15 Hz se upar rahe aur pehli axial 30 Hz se upar. Analysis predict karta hai ki payload ka $f_{lat} = 12 Hz$ aur $f_{ax} = 42 Hz$ hai. Kya yeh pass karta hai? Agar nahi, toh kya structure ko stiffer banana (stiffness constant $k$ badhana, iss page ke top par defined force-per-metre-of-deflection) help karta hai, aur agar $m$ unchanged rahe aur $k$ quadruple ho jaaye toh $f$ par kitna effect?

Forecast: Lateral 12 < 15 → fail. Axial 42 > 30 → pass. Stiffening: $f_{n} \propto k$ , toh $\times 4$ stiffness → $\times 2$ frequency → 12 ban jaata hai 24 Hz → pass.

Axial check: $42 Hz > 30 Hz$ → pass. Yeh step kyun? Frequency separation payload ke resonance ko LV ke strong low-frequency drive band se bahar rakhta hai, isliye hum har predicted mode ko uske apne minimum se compare karte hain.
Lateral check: $12 Hz < 15 Hz$ → fail. Yeh mode launcher se zyada shake ho sakta hai. Yeh step kyun? Minimum se neeche hona dangerous direction hai — excitation energy wahan rehta hai, toh floor se neeche ka mode resonance ka risk leta hai.
Stiffening se fix karo: $f_{n} = \frac{1}{2 π} k / m$ , toh $m$ fixed hone par $f_{n} \propto k$ . $k$ quadruple karne se $f^{'} = 4 f = 2 \times 12 = 24 Hz > 15$ → ab pass hota hai. Yeh step kyun? Stiffening sirf isliye help karta hai kyunki yeh $f_{n}$ ko band se door le jaata hai — isliye nahi ki stiffness ( $k$ ) intrinsically "safer" hai (parent note ka mistake box dekho). Agar band upar hota, toh wohi stiffening tumhe band mein push kar sakti thi.

Verify: $4 = 2$ , $12 \times 2 = 24 \geq 15$ ✓. Axial $42 \geq 30$ ✓, lateral $12 < 15$ (fail) ✓.

Worked example Example 9 — Cell D3: degenerate,

$ζ \to 0$ (undamped) Ek perfectly undamped idealization mein $ζ = 0$ hai. $Q$ nikalo, aur $x_{max} \approx Q x_{static}$ use karo (dono symbols page ke top par define hain: $x_{static}$ steady load ke neeche deflection hai, $x_{max}$ resonance par peak deflection) yeh batane ke liye ki resonant deflection ka kya hoga. Phir ek realistic tiny damping $ζ = 0.01$ lo, recompute karo, aur $x_{static} = 0.1 mm$ wale part par numbers lagao.

Forecast: $Q = 1/ (2 ζ)$ jaise jaise $ζ \to 0$ blast ho jaata hai → infinite amplification (physically impossible — model toot jaata hai). $ζ = 0.01$ par, $Q = 50$ , ek enormous lekin finite factor, $0.1$ mm ko ~ $5$ mm bana deta hai.

Undamped limit $Q = 1/ (2 \times 0) = \infty$ . Yeh step kyun? Yeh limiting case expose karta hai ki zero damping worst case hai — amplification unbounded hai. Real structures mein hamesha kuch $ζ$ hota hai, aur wohi unhe bachata hai.
Us limit mein deflection $x_{max} \approx Q x_{static} \to \infty$ . Yeh step kyun? $Q$ infinite hone par, peak deflection $x_{max}$ (woh amplitude jis tak part resonance par actually swing karta hai) bhi runaway kar jaata hai — physical warning ki truly undamped resonance part ko taar-taar kar deti. Isliye $ζ > 0$ hamesha matter karta hai.
Realistic $ζ = 0.01 \Rightarrow Q = 1/ (2 \times 0.01) = 50$ . Yeh step kyun? 1% damping bhi amplification ko $50 \times$ par cap kar deta hai — phir bhi bahut bada, dikhata hai ki low $ζ$ (high nahi) danger hai.
Numbers lagao $x_{max} \approx Q x_{static} = 50 \times 0.1 mm = 5 mm$ . Yeh step kyun? Abstract $Q$ ko ek physical deflection mein badlo jo tum imagine kar sako: ek hairline static sag resonance par ek violent $5$ mm swing ban jaata hai.

Verify: Jaise $ζ \to 0^{+}$ , $1/ (2 ζ) \to + \infty$ ✓, toh $x_{max} \to \infty$ ✓. $ζ = 0.01 \Rightarrow Q = 50$ ✓, aur $50 \times 0.1 mm = 5 mm$ ✓.

Word problem + exam twist

Worked example Example 10 — Cell W/X: real mission, opposite-fiber twist

Word problem: Max-Q par ek rocket ke critical cross-section par axial compressive stress $6 MPa$ hai (positive, hamare convention se) aur bending stress ki magnitude $40 MPa$ hai. Material tension mein $35 MPa$ par yield karta hai aur compression mein $250 MPa$ par (buckling alag baat hai). Exam twist: kaunsa fiber pehle fail hoga — windward (compressed) wala ya leeward (stretched) wala?

Forecast: Windward: $6 + 40 = 46$ MPa compression, $250$ se kaafi neeche → safe. Leeward: axial compression $6$ minus bending tension $40$ = net $34$ MPa tension, $35$ MPa tensile limit se thoda neeche → leeward fiber near-failure wala hai. Twist: tension side, na ki big-number compression side, critical hai.

Windward (compressed) fiber: $σ = + 6 + 40 = + 46 MPa$ (dono compressive, positive, add hote hain). Yeh step kyun? Windward side par bending usi fiber ko compress karta hai jise thrust already compress kar raha hai; positive-plus-positive.
Compressive limit se compare karo: $∣ + 46 ∣ = 46 < 250$ → margin bahut bada → critical nahi. Yeh step kyun? Bada number ≠ danger; tum relevant allowable (yahan compressive wala) se compare karna padega.
Leeward (stretched) fiber: $σ = + 6 - 40 = - 34 MPa$ — negative, toh net tension (bending tension axial compression par bhari padti hai). Yeh step kyun? Leeward side par bending tension axial compression se ladti hai; jo jeete woh sign set karta hai, aur yahan tension jeetti hai ( $- 34$ ).
Tensile limit se compare karo: $∣ - 34∣ = 34 < 35$ → bachta hai lekin sirf $35/34 - 1 = 2.9%$ margin ke saath → yeh critical fiber hai. Yeh step kyun? Tensile allowable ( $35$ ) compressive ( $250$ ) se kaafi kam hai, isliye wahan ek chhota stress zyada dangerous hota hai — exam-twist ka punchline.

Verify: Windward $∣ + 6 + 40∣ = 46$ MPa, 250 ke against margin bada ✓. Leeward $+ 6 - 40 = - 34$ MPa (net tension), $∣ - 34∣ = 34 < 35$ (pass karta hai) lekin margin $35/34 - 1 = 0.0294 \approx 2.9%$ ✓.

Recall Matrix self-check

Kaun se cell mein $n < 0$ hai? ::: A2 — tension, coast/drag deceleration (Example 2). Kaun se cell mein $n = 0$ hai? ::: A3 — free fall, zero axial stress (Example 3). Tube radius double karne se bending stress kis factor se badalta hai? ::: $1/4$ (stress $\propto 1/ R^{2}$ ) — Example 6. Jab tensile allowable ≪ compressive allowable ho toh kaunsa fiber critical hai? ::: leeward (net-tension) fiber — Example 10. Jaise $ζ \to 0$ , $Q$ aur $x_{max}$ ka kya hota hai? ::: dono $\to \infty$ ; low damping danger hai — Example 9.

Related: Factor of Safety & Margins of Safety · Beam Bending & Second Moment of Area · Random Vibration & PSD · Modal Analysis & Natural Frequencies · Max-Q and Dynamic Pressure · Rocket Equation & Thrust · Pyrotechnic Separation Systems.