Spacecraft Structures & Systems Engineering

Question 1 — Euler Buckling [12]

(a) Consider a buckled pin-pinned column under axial load $P$. Bending moment at position $x$ for deflection $w$: $M(x) = -P\,w$ (compressive load produces restoring moment). [1]

Substitute into $EI\,w'' = M$: $EI\,\frac{d^2w}{dx^2} = -Pw \implies w'' + k^2 w = 0,\quad k^2 = \frac{P}{EI}$ [2]

General solution $w(x) = A\sin kx + B\cos kx$. [1] BCs: $w(0)=0 \Rightarrow B=0$; $w(L)=0 \Rightarrow A\sin kL = 0$. [2] Non-trivial ($A\neq0$) requires $\sin kL = 0 \Rightarrow kL = n\pi$. [1] Lowest mode $n=1$: $k = \pi/L$, so $P_{cr} = k^2 EI = \dfrac{\pi^2 EI}{L^2}$. [1]

(b) Second moment of area for thin annulus, $D_o = 0.040$ m, $t=0.002$ m, $D_i = 0.036$ m: $I = \frac{\pi}{64}(D_o^4 - D_i^4) = \frac{\pi}{64}(0.040^4 - 0.036^4)$ $0.040^4 = 2.560\times10^{-6}$, $0.036^4 = 1.6796\times10^{-6}$; difference $=8.804\times10^{-7}$. $I = 0.049087 \times 8.804\times10^{-7} = 4.322\times10^{-8}\ \text{m}^4$. [2] $P_{cr} = \frac{\pi^2 (70\times10^9)(4.322\times10^{-8})}{1.5^2} = \frac{2.987\times10^{4}}{2.25} \approx 1.33\times10^{4}\ \text{N} \approx 13.3\ \text{kN}$ [2]

Question 2 — Random Vibration & Miner [12]

(a) The response PSD of a SDOF to white-noise base input is amplified by the transmissibility. The mean-square response is $G_{RMS}^2 = \int_0^\infty W_0 |H(f)|^2\,df$ For a lightly damped SDOF the integral of $|H|^2$ over frequency evaluates to $\dfrac{\pi}{2} f_n Q$. [3] Hence Miles' equation: $G_{RMS} = \sqrt{\frac{\pi}{2} f_n Q\, W_0}$ [3]

(b) $G_{RMS} = \sqrt{\frac{\pi}{2}(200)(20)(0.04)} = \sqrt{\frac{\pi}{2}\times160} = \sqrt{251.3} = 15.85\ g$. [2] $3\sigma$ peak $= 3 \times 15.85 = 47.5\ g$. [1]

(c) Miner's rule: cumulative damage $D = \sum_i \dfrac{n_i}{N_i}$ where $n_i$ = applied cycles at stress level $i$, $N_i$ = cycles to failure at that level (from S–N curve). [1] Failure predicted when $D = 1$. [1] Designers adopt $D < 1$ (e.g. 0.3–0.5) because Miner's rule ignores load-sequence effects and scatter, so a margin guards against premature fatigue. [1]

Question 3 — Fracture Mechanics [10]

(a) $K$ measures the crack-tip stress-field intensity; failure occurs when $K = K_{IC}$. Set $Y\sigma\sqrt{\pi a_c}=K_{IC}$: $a_c = \frac{1}{\pi}\left(\frac{K_{IC}}{Y\sigma}\right)^2 = \frac{1}{\pi}\left(\frac{55}{1.12\times400}\right)^2$ [2] $\frac{55}{448} = 0.12277$; squared $=0.015078$; $/\pi = 4.799\times10^{-3}$ m $\approx 4.80$ mm. [3]

(b) Margin $= a_c/a_{detect} = 4.80/1.0 = 4.8$. [2] Since detectable flaws are ~5× smaller than critical, the design is damage-tolerant (cracks are caught by NDI well before reaching critical size). [1]

(c) Yield-based FOS keeps operating stress below yield/ultimate by a factor; fracture/damage-tolerance design instead assumes pre-existing cracks and ensures they stay sub-critical and grow slowly enough between inspections. [2]

Question 4 — Reliability & Redundancy [12]

(a) Constant hazard rate: $\dfrac{dR}{dt} = -\lambda R$. Solve: $R(t) = e^{-\lambda t}$ (with $R(0)=1$). [2] $MTTF = \int_0^\infty R(t)\,dt = \int_0^\infty e^{-\lambda t}dt = 1/\lambda$. [2]

(b) Active parallel: system fails only if both fail. Unit unreliability $F = 1-e^{-\lambda t}$. $R_{sys}(t) = 1 - (1-e^{-\lambda t})^2$ [2] $\lambda t = 5\times10^{-5}\times8760 = 0.438$; $e^{-0.438}=0.6453$; $F = 0.3547$; $F^2 = 0.12581$. $R_{sys} = 1 - 0.12581 = 0.8742$. [2] (single unit alone would be 0.6453.)

(c)

def system_reliability(lam, t, n):
    from math import exp
    F = 1 - exp(-lam * t)      # single-unit unreliability
    return 1 - F**n            # active parallel: all n must fail

[4] (1 for exp model, 1 for unreliability, 1 for F**n, 1 for return of complement.)

Question 5 — Link Budget [8]

(a) $P_t = 5$ W $= 10\log_{10}5 = 6.99$ dBW. EIRP $= P_t + G_t = 6.99 + 6 = 12.99$ dBW $\approx 13.0$ dBW. [3]

(b) $P_r = \text{EIRP} - L_{path} + G_r - L_{sys} = 13.0 - 165 + 40 - 3 = -115\ \text{dBW}$. [3]

(c) $G/T$ is the receiver figure of merit = antenna gain divided by system noise temperature (referenced to same point); units dB/K. [2]

Question 6 — Explain Out Loud [6]

(a) V-model: left arm = decomposition (concept → requirements → design → component build); right arm = integration & verification (unit → subsystem → system → validation). Each level on the left maps horizontally to its verification activity on the right (e.g. requirements ↔ acceptance test). Verification demonstrates the built item meets the corresponding-level requirement. [3]

(b) TID (Total Ionising Dose) is cumulative charge trapping in oxides causing gradual parametric drift/failure — mitigated by shielding and rad-hard/rad-tolerant parts. SEE (Single Event Effects) are instantaneous events from a single ion (upset, latch-up) — mitigated by EDAC/TMR, watchdog resets, and current-limiting/latch-up protection. [3]

[
{"claim":"Euler Pcr for hollow strut approx 13.3 kN","code":"import math\nE=70e9; L=1.5; Do=0.040; Di=0.036\nI=math.pi/64*(Do**4-Di**4)\nPcr=math.pi**2*E*I/L**2\nresult = abs(Pcr-13300)<400"},
{"claim":"Miles GRMS approx 15.85 g","code":"import math\nG=math.sqrt(math.pi/2*200*20*0.04)\nresult = abs(G-15.85)<0.1"},
{"claim":"critical crack length approx 4.80 mm","code":"import math\nac=(1/math.pi)*(55/(1.12*400))**2\nresult = abs(ac*1000-4.80)<0.05"},
{"claim":"active parallel reliability at 1 yr approx 0.8742","code":"import math\nlam=5e-5; t=8760\nF=1-math.exp(-lam*t)\nR=1-F**2\nresult = abs(R-0.8742)<0.001"},
{"claim":"link received power = -115 dBW","code":"import math\nPt=10*math.log10(5); EIRP=Pt+6\nPr=EIRP-165+40-3\nresult = abs(Pr-(-115))<0.1"}
]