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

3.6.1 · D4 · Physics › Spacecraft Structures & Systems Engineering › Structural loads — axial (thrust), bending (wind shear), dyn

Intuition Is page ko kaise use karein

Ye problems ek ladder ki tarah chadhte hain. L1 sirf yeh pehchanne ke liye hai ki kaun sa load kaun sa hai. L5 tak aate-aate tum axial, bending, aur dynamic reasoning ko ek hi shot mein combine karte ho — bilkul wahi jo ek launch-vehicle stress engineer max-Q par karta hai. Pehle har ek problem ko Solution collapse karke try karo, phir open karo. Yahan har number machine-verified hai.

Recurring symbols (sab parent parent topic mein banaye gaye hain):

$F_{axial}$ = long axis ke saath andar ka push/pull (newtons, N).
$A$ = cross-section area jo us push ko carry karti hai (square metres, $m^{2}$ ).
$σ$ = stress = force jo area par spread hoti hai, pascals (Pa) ya megapascals (MPa = $1 0^{6}$ Pa) mein.
$n$ = axial load factor, ek plain number: "structure kitne g's feel karta hai".
$M$ = bending moment, ek sideways twist jo newton-metres (N·m) mein measure hota hai.
$I$ = second moment of area, material centre-line se kitna door spread hai ( $m^{4}$ ).
$c$ = centre-line se outermost fibre tak ki distance (m).
$f_{n}$ = natural frequency, wo "musical note" jis par part ring karna pasand karta hai (hertz, Hz).
$Q$ = amplification, resonance par shaking kitni baar badi ho jaati hai.

Level 1 — Recognition

Recall Solution L1.1

Yeh ek axial (thrust) load hai. Engine base par push karta hai aur upar ka saara mass speed up hone ka resistance deta hai, isliye lower structure dono taraf se push hoti hai → compressive. Kaise dikhta hai: socho jaise tum apne haath se ek soda can ko neeche dabaate ho — walls ek squeeze carry karti hain.

Recall Solution L1.2

(a) sideways gust → bending (slender body ek broomstick ki tarah flex karta hai jise beech mein push kiya jaaye).
(b) sound → dynamic (acoustics sub-family; frequencies par pressure waves).
(c) explosive bolt → dynamic (shock sub-family; ek bahut short, sharp transient).
(d) steady thrust → axial (axis ke saath ek slowly-varying, quasi-static push).

Level 2 — Application

Recall Solution L2.1

(i) Force. Kya: $F_{axial} = m n g$ use karo. Kyun: term $g + a$ (gravity plus engine acceleration) ko $n g$ mein bundle kiya gaya hai, isliye ek number $n$ dono effects carry karta hai. $F = 1500 \times 5 \times 9.81 = 7.358 \times 1 0^{4} N .$ (ii) Stress. Kya: $σ = F / A$ . Kyun: stress wo force hai jo us area par share hoti hai jo actually use carry karti hai. $σ = \frac{7.358 \times 1 0 ^{4}}{0.015} = 4.905 \times 1 0^{6} Pa = 4.9 MPa (compressive) .$

Recall Solution L2.2

Kya: $f_{n} = \frac{1}{2 π} k / m$ use karo. Kyun yeh tool: driving force ko zero set karo to motion simple harmonic hoti hai; $k / m$ angular frequency hai, aur $2 π$ se divide karne par "radians per second" "cycles per second" (Hz) mein convert hota hai.

= \frac{632.46}{2\pi} = 100.6\ \text{Hz}.$$ Dekho [[Modal Analysis & Natural Frequencies]] ke liye ki real parts mein aisi kai frequencies hoti hain.

Level 3 — Analysis

Recall Solution L3.1

(i) Kya: thin ring ke liye $I = π R^{3} t$ . Kyun: almost saara metal centre-line se $R$ distance par baitha hai, aur $I = \int y^{2} d A$ us material ko reward karta hai jo door placed ho. Figure s01 dekho — wall $R$ radius par ek thin band hai. $I = π (1.2)^{3} (0.004) = π \times 1.728 \times 0.004 = 0.021714 m^{4} .$ (ii) Kya: outer fibre ( $c = R = 1.2$ ) sabse zyada stretch/squeeze feel karta hai. Kyun $c = R$ : neutral axis se jitna door, utna zyada ek fibre strained hoti hai jab beam curve karti hai (figure s02). $σ_{bend} = \frac{6 \times 1 0 ^{5} \times 1.2}{0.021714} = 3.316 \times 1 0^{7} Pa = 33.2 MPa .$ $I$ ke baare mein aur detail Beam Bending & Second Moment of Area mein hai.

Figure — Structural loads — axial (thrust), bending (wind shear), dynamic (vibration, acoustics, shock)

Recall Solution L3.2

(i) Thin ring ki area. Kya: $A = 2 π R t$ (circumference × wall). Kyun: thin tube ko unroll karne par $2 π R$ length aur $t$ height ka ek rectangle milta hai. $A = 2 π (1.2) (0.004) = 0.030159 m^{2} .$ (ii) Axial stress. $σ_{axial} = \frac{5 \times 1 0 ^{5}}{0.030159} = 1.658 \times 1 0^{7} Pa = 16.6 MPa .$ (iii) Total. Kya: unhe add karo. Kyun: max-Q par dono loads ek hi fibre ko ek hi instant par squeeze karte hain, isliye unke stresses stack ho jaate hain (figure s03). $σ_{total} = 16.6 + 33.2 = 49.7 MPa (compressive) .$ Yeh worst-case sizing point hai — dekho Max-Q and Dynamic Pressure.

Level 4 — Synthesis

Recall Solution L4.1

(i) Kyun yeh formula: resonance par peak response static wale ka $1/ (2 ζ)$ times hota hai; low damping = high amplification. $Q = \frac{1}{2 ( 0.025 )} = 20.$ (ii) Miles' equation kya karta hai: yeh saare shaking spectrum ko ek RMS number mein collapse karta hai sharp resonance peak ko integrate karke; $π /2$ us peak ke neeche ke area se aata hai. $g_{RMS} = \frac{π}{2} \times 120 \times 20 \times 0.05 = 188.50 = 13.73 g .$ (iii) Kyun $3 σ$ : random peaks ek bell curve follow karte hain; $3 σ$ unka 99.7% cover karta hai, jo standard design envelope hai. $3 σ = 3 \times 13.73 = 41.2 g .$ Background: Random Vibration & PSD aur Factor of Safety & Margins of Safety.

Recall Solution L4.2

(i) $f_{n} = \frac{1}{2 π} k / m$ ko invert karo → $k = (2 π f_{n})^{2} m$ . $k_{1} = (2 π \times 90)^{2} \times 8 = (565.49)^{2} \times 8 = 2.558 \times 1 0^{6} N/m .$ (ii) $k_{2} = (2 π \times 140)^{2} \times 8 = (879.65)^{2} \times 8 = 6.190 \times 1 0^{6} N/m .$ (iii) Kyun frequency $k$ ke saath scale karti hai: $f_{n} \propto k$ , isliye $f_{n}$ ko $140/90$ se multiply karne ke liye $k$ ko us ratio squared se multiply karna hoga. $\frac{k _{2}}{k _{1}} = (\frac{140}{90})^{2} = 2.420.$ Isliye bracket ko requirement clear karne ke liye 2.42× stiffer banana hoga.

Level 5 — Mastery

Recall Solution L5.1

(i) Axial. $A = 2 π R t = 2 π (0.9) (0.006) = 0.033929 m^{2}$ .

= \frac{9.7120\times10^4}{0.033929} = 2.862\times10^6\ \text{Pa} = 2.86\ \text{MPa}.$$ **(ii) Bending.** $I = \pi R^3 t = \pi (0.9)^3 (0.006) = 0.013741\ \text{m}^4$, $c = R = 0.9$. $$\sigma_\text{bend} = \frac{Mc}{I} = \frac{4.5\times10^5 \times 0.9}{0.013741} = 2.948\times10^7\ \text{Pa} = 29.48\ \text{MPa}.$$ **(iii) Total.** Compressive fibre par add karo: $$\sigma_\text{total} = 2.86 + 29.48 = 32.34\ \text{MPa}.$$ **(iv) Margin.** *Kyun:* FoS demand ko inflate karta hai; MoS tumhe spare capacity fraction ke roop mein batata hai. $$\text{MoS} = \frac{300}{1.5 \times 32.34} - 1 = \frac{300}{48.51} - 1 = 6.184 - 1 = 5.18.$$ $\text{MoS} = +5.18 \ge 0$ → **comfortably pass** (bending driver hone se bahut bada margin hai). Dekho [[Factor of Safety & Margins of Safety]].

Recall Solution L5.2

(i) $Q = 1/ (2 ζ) = 1/ (2 \times 0.02) = 25.$ (ii) Kyun multiply karte hain: resonance par response static wale ka $Q$ times hoti hai. $σ_{peak} = 25 \times 40 = 1000 MPa .$ (iii) $MoS = \frac{500}{1.25 \times 1000} - 1 = \frac{500}{1250} - 1 = 0.4 - 1 = - 0.6.$ $MoS = - 0.6 < 0$ → FAILS. "Static" number safe lagta tha (40 MPa vs 500 MPa yield), lekin resonant amplification ne use 25× multiply kiya aur seedha yield ke paar chala gaya. Exactly isliye dynamic loads ko frequency ke hisaab se analyse kiya jaata hai, magnitude ke hisaab se nahi.

Recall Ek-line self-test

Axial aur bending stresses max-Q par kyun add hote hain lekin tum sirf bada wala kabhi nahi lete? ::: Kyunki yeh ek hi fibre ko ek hi moment par load karte hain, isliye unke stresses superpose hote hain: $σ_{total} = F / A + M c / I$ .