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Level 5 — MasterySpacecraft Structures & Systems Engineering

Spacecraft Structures & Systems Engineering

⏱ 3 minutes100 marksprintable — key stays hidden on paper

Level 5 — Mastery (cross-domain: math + physics + coding) Time limit: 3 hours Total marks: 100

Instructions: Answer all THREE questions. Show all derivations. Numerical answers require units and appropriate significant figures. Where code is requested, pseudocode or Python (NumPy/SymPy) is acceptable.

Question 1 — Column Buckling, Fatigue & Reliability of a Strut (34 marks)

A pin-ended cylindrical aluminium strut in a launch adapter has length L=1.20 mL = 1.20\ \text{m}, outer diameter D=40 mmD = 40\ \text{mm}, wall thickness t=2.0 mmt = 2.0\ \text{mm}. Material: E=70 GPaE = 70\ \text{GPa}, yield σy=300 MPa\sigma_y = 300\ \text{MPa}, density ρ=2700 kg/m3\rho = 2700\ \text{kg/m}^3.

(a) Starting from the Euler–Bernoulli beam equation EI d2wdx2=−M=−PwEI\,\dfrac{d^2 w}{dx^2} = -M = -Pw, derive the critical buckling load for pin-ended boundary conditions, obtaining Pcr=π2EIL2P_{cr} = \dfrac{\pi^2 EI}{L^2}. State the buckled mode shape. (6)

(b) Compute the second moment of area II for the thin-walled tube (use I=π4(ro4−ri4)I = \tfrac{\pi}{4}(r_o^4 - r_i^4)), then PcrP_{cr}. Compute the axial stress at PcrP_{cr} and compare with σy\sigma_y: does the strut buckle or yield first? (6)

(c) During ascent the strut experiences a fluctuating axial load producing stress cycles. In one launch it sees the following blocks (an S–N curve of the form N⋅σm=CN \cdot \sigma^{m} = C with m=4m = 4, C=1.0×1022C = 1.0\times 10^{22} in SI-MPa units where σ\sigma in MPa):

Block σa\sigma_a (MPa) applied cycles nin_i
1 120 2000
2 80 15000
3 60 40000

Using Miner's rule, compute the cumulative damage per launch and the number of launches to failure. (8)

(d) The strut's failures follow an exponential model with MTTF=4000\text{MTTF} = 4000 launches (each launch = one "operation"). Write the reliability function R(n)R(n) and compute the reliability after the number of launches found in (c). If mission requires R≥0.99R \ge 0.99, is a single strut acceptable? (6)

(e) Two identical struts are placed in active (hot) redundancy — the load path survives if at least one survives. Derive the system reliability Rsys(n)R_{sys}(n) and compute it at n=40n = 40 launches. Write a short SymPy/Python snippet that returns RsysR_{sys}. (8)

Question 2 — Random Vibration & Modal Response of an Avionics Box (33 marks)

A single-degree-of-freedom model of an avionics box: mass m=5.0 kgm = 5.0\ \text{kg} on a mount of stiffness k=1.97×106 N/mk = 1.97\times 10^6\ \text{N/m}, damping ratio ζ=0.05\zeta = 0.05.

(a) Compute the natural frequency fnf_n in Hz. (4)

(b) The base is excited by a flat acceleration PSD W0=0.04 g2/HzW_0 = 0.04\ \text{g}^2/\text{Hz} across the relevant band. Using Miles' equation, Grms=π2 fn Q W0,G_{rms} = \sqrt{\frac{\pi}{2}\, f_n\, Q\, W_0}, derive/justify the equation from the SDOF transmissibility integral (state the resonant-approximation assumptions), then compute GrmsG_{rms} (in g) with Q=1/(2ζ)Q = 1/(2\zeta). (10)

(c) Using the 3-sigma rule, compute the peak acceleration and the peak inertial force on the box. If the mounting bracket has cross-section A=25 mm2A = 25\ \text{mm}^2 and σy=250 MPa\sigma_y = 250\ \text{MPa}, compute the factor of safety against yield (define FOS = σy/σapplied\sigma_y/\sigma_{applied}). Is FOS ≥1.25\ge 1.25 met? (9)

(d) Explain physically why increasing QQ (lower damping) increases GrmsG_{rms} but the dependence is only Q\sqrt{Q}, not linear. Then propose one design change to reduce GrmsG_{rms} by a factor of 2 and quantify what parameter change achieves it. (6)

(e) Write pseudocode for numerically integrating the SDOF response PSD ∫0∞∣H(f)∣2W0 df\int_0^\infty |H(f)|^2 W_0\, df (rather than using Miles) and explain how it converges to Miles' result. (4)

Part A — Thermal Stress (12 marks)

A titanium bracket (E=110 GPaE = 110\ \text{GPa}, α=8.6×10−6 K−1\alpha = 8.6\times10^{-6}\ \text{K}^{-1}) is rigidly constrained between two fixed points. It is cooled from +50∘C+50^\circ\text{C} to −70∘C-70^\circ\text{C} in eclipse.

(a) Derive the thermal stress for a fully constrained bar (εtotal=0\varepsilon_{total}=0) and compute σthermal\sigma_{thermal}. State whether it is tensile or compressive. (6)

(b) If σy=830 MPa\sigma_y = 830\ \text{MPa}, compute the margin of safety MS=σy/(FOS⋅∣σ∣)−1MS = \sigma_y/(FOS\cdot|\sigma|) - 1 with FOS=1.4FOS = 1.4. Is it positive? (6)

Part B — Link Budget (12 marks)

A LEO satellite downlink: transmit power Pt=5 WP_t = 5\ \text{W}, antenna gain Gt=6 dBiG_t = 6\ \text{dBi}, frequency f=8 GHzf = 8\ \text{GHz}, slant range d=2000 kmd = 2000\ \text{km}, ground station G/T=20 dB/KG/T = 20\ \text{dB/K}, data rate Rb=10 MbpsR_b = 10\ \text{Mbps}, required Eb/N0=8 dBE_b/N_0 = 8\ \text{dB}.

(c) Compute EIRP in dBW and free-space path loss (dB) using FSPL=20log⁡10(4πd/λ)\text{FSPL} = 20\log_{10}(4\pi d/\lambda). (6)

(d) Compute the received Eb/N0E_b/N_0 (use k=−228.6 dBW/K/Hzk = -228.6\ \text{dBW/K/Hz} for Boltzmann, and RbR_b in dB-Hz). Compute the link margin and state whether the link closes. (6)

Part C — Reliability Architecture (9 marks)

(e) A C&DH subsystem uses two processors in cold standby with a perfect switch. Each has failure rate λ=1×10−5 /hr\lambda = 1\times10^{-5}\ \text{/hr}. Derive the reliability Rcold(t)R_{cold}(t) for a cold-standby pair (from the two-term Poisson survival) and compute it for a t=8760t = 8760 hr (1 year) mission. Compare to a single unit's reliability. (9)

Answer keyMark scheme & solutions

Question 1

(a) [6] Governing equation EIw′′+Pw=0EIw'' + Pw = 0. Let k2=P/(EI)k^2 = P/(EI): w′′+k2w=0w'' + k^2 w = 0, solution w=Asin⁡kx+Bcos⁡kxw = A\sin kx + B\cos kx. BCs pin-ended: w(0)=0⇒B=0w(0)=0 \Rightarrow B=0; w(L)=0⇒Asin⁡kL=0w(L)=0 \Rightarrow A\sin kL = 0. Nontrivial A≠0⇒kL=nπA\ne0 \Rightarrow kL = n\pi. Lowest (n=1n=1): k=π/Lk = \pi/L, so Pcr=k2EI=π2EI/L2P_{cr} = k^2 EI = \pi^2 EI/L^2. (4) Mode shape w(x)=Asin⁡(πx/L)w(x) = A\sin(\pi x/L), half-sine. (2)

(b) [6] ro=0.020r_o = 0.020 m, ri=0.018r_i = 0.018 m. I=π4(0.0204−0.0184)=π4(1.6×10−7−1.04976×10−7)=π4(5.5024×10−8)=4.322×10−8 m4I = \tfrac{\pi}{4}(0.020^4 - 0.018^4) = \tfrac{\pi}{4}(1.6\times10^{-7} - 1.04976\times10^{-7}) = \tfrac{\pi}{4}(5.5024\times10^{-8}) = 4.322\times10^{-8}\ \text{m}^4. (2) Pcr=π2(70×109)(4.322×10−8)/(1.20)2=π2(70×109)(4.322×10−8)/1.44P_{cr} = \pi^2 (70\times10^9)(4.322\times10^{-8})/(1.20)^2 = \pi^2(70\times10^9)(4.322\times10^{-8})/1.44. Numerator π2⋅70×109⋅4.322×10−8=9.8696⋅3025.4=2.986×104\pi^2 \cdot 70\times10^9 \cdot 4.322\times10^{-8} = 9.8696\cdot 3025.4 = 2.986\times10^4. Divide by 1.44 → Pcr≈2.07×104P_{cr} \approx 2.07\times10^4 N ≈20.7\approx 20.7 kN. (2) Area A=π(ro2−ri2)=π(4×10−4−3.24×10−4)=π(7.6×10−5)=2.388×10−4 m2A = \pi(r_o^2-r_i^2) = \pi(4\times10^{-4}-3.24\times10^{-4}) = \pi(7.6\times10^{-5}) = 2.388\times10^{-4}\ \text{m}^2. Stress at PcrP_{cr}: σ=20700/2.388×10−4=8.67×107\sigma = 20700/2.388\times10^{-4} = 8.67\times10^{7} Pa =86.7= 86.7 MPa <σy=300< \sigma_y = 300 MPa. Buckling occurs first. (2)

(c) [8] NiN_i at each stress from N=C/σmN = C/\sigma^m, C=1022C=10^{22}, m=4m=4:

  • N1=1022/1204=1022/2.0736×108=4.823×1013N_1 = 10^{22}/120^4 = 10^{22}/2.0736\times10^8 = 4.823\times10^{13}
  • N2=1022/804=1022/4.096×107=2.441×1014N_2 = 10^{22}/80^4 = 10^{22}/4.096\times10^7 = 2.441\times10^{14}
  • N3=1022/604=1022/1.296×107=7.716×1014N_3 = 10^{22}/60^4 = 10^{22}/1.296\times10^7 = 7.716\times10^{14} (3)

Damage per launch D=∑ni/NiD = \sum n_i/N_i:

  • 2000/4.823×1013=4.147×10−112000/4.823\times10^{13} = 4.147\times10^{-11}
  • 15000/2.441×1014=6.145×10−1115000/2.441\times10^{14} = 6.145\times10^{-11}
  • 40000/7.716×1014=5.184×10−1140000/7.716\times10^{14} = 5.184\times10^{-11} D=1.548×10−10D = 1.548\times10^{-10} per launch. (3) Launches to failure =1/D≈6.46×109= 1/D \approx 6.46\times10^{9} launches. (Effectively infinite fatigue life — high-cycle safe.) (2)

(d) [6] R(n)=e−n/MTTF=e−n/4000R(n) = e^{-n/\text{MTTF}} = e^{-n/4000}. (2) Fatigue gives ~6.5×1096.5\times10^9 launches; using that nn gives R≈0R\approx0, but the physically meaningful check: at any realistic mission of tens of launches R≈1R\approx1. Grading: evaluate at the fatigue-life number → R=e−6.46×109/4000≈0R = e^{-6.46\times10^9/4000}\approx 0; since R≪0.99R\ll0.99 not acceptable for that many launches, but for typical missions (n∼40n\sim40): R=e−40/4000=e−0.01=0.990R=e^{-40/4000}=e^{-0.01}=0.990 — marginally acceptable. (4)

(e) [8] Hot/active redundancy, at least one of two survives: Rsys=1−(1−R)2=2R−R2R_{sys} = 1-(1-R)^2 = 2R - R^2. (3) At n=40n=40, R=e−0.01=0.99005R = e^{-0.01}=0.99005. Rsys=2(0.99005)−(0.99005)2=1.98010−0.98020=0.99990R_{sys} = 2(0.99005)-(0.99005)^2 = 1.98010 - 0.98020 = 0.99990. (3) Snippet: (2)

import numpy as np
def R_sys(n, mttf=4000):
    R = np.exp(-n/mttf)
    return 2*R - R**2
print(R_sys(40))   # 0.99990

Question 2

(a) [4] ωn=k/m=1.97×106/5.0=3.94×105=627.7 rad/s\omega_n = \sqrt{k/m} = \sqrt{1.97\times10^6/5.0} = \sqrt{3.94\times10^5} = 627.7\ \text{rad/s}. fn=ωn/2π=99.9≈100f_n = \omega_n/2\pi = 99.9 \approx 100 Hz. (4)

(b) [10] Transmissibility magnitude of SDOF base excitation ∣H(f)∣2=1+(2ζr)2(1−r2)2+(2ζr)2|H(f)|^2 = \dfrac{1+(2\zeta r)^2}{(1-r^2)^2+(2\zeta r)^2}, r=f/fnr=f/f_n. Response mean-square aˉ2=∫0∞∣H∣2W0 df\bar{a}^2 = \int_0^\infty |H|^2 W_0\,df. For light damping the integrand is sharply peaked at fnf_n; approximating W0W_0 flat and using the standard resonant integral ∫0∞∣H∣2df=πfnQ2\int_0^\infty |H|^2 df = \tfrac{\pi f_n Q}{2} gives Grms=π2fnQW0G_{rms}=\sqrt{\tfrac{\pi}{2}f_n Q W_0}. Assumptions: flat PSD over band, high-Q narrowband peak dominates, Q=1/(2ζ)Q=1/(2\zeta). (5) Q=1/(2⋅0.05)=10Q=1/(2\cdot0.05)=10. Grms=(π/2)(100)(10)(0.04)=(1.5708)(40)=62.83=7.93 gG_{rms}=\sqrt{(\pi/2)(100)(10)(0.04)} = \sqrt{(1.5708)(40)} = \sqrt{62.83} = 7.93\ \text{g}. (5)

(c) [9] 3σ3\sigma peak =3×7.93=23.78=3\times7.93 = 23.78 g. (2) Peak accel =23.78×9.81=233.3 m/s2= 23.78\times9.81 = 233.3\ \text{m/s}^2. Force F=ma=5.0×233.3=1166 NF = m a = 5.0\times233.3 = 1166\ \text{N}. (3) Stress σ=F/A=1166/(25×10−6)=4.665×107\sigma = F/A = 1166/(25\times10^{-6}) = 4.665\times10^7 Pa =46.65= 46.65 MPa. FOS =250/46.65=5.36≥1.25= 250/46.65 = 5.36 \ge 1.25. Yes, met. (4)

(d) [6] Grms∝QG_{rms}\propto\sqrt{Q} because random-vibration energy is a variance (power), and the area under the resonance peak scales with bandwidth ×\times peak: peak height ∝Q\propto Q but bandwidth ∝1/Q\propto 1/Q, so integrated power ∝Q\propto Q, and RMS ∝Q\propto\sqrt{Q}. (3) To halve GrmsG_{rms}, halve QQ ⇒ quadruple damping (since G∝QG\propto\sqrt Q, need Q→Q/4Q\to Q/4, i.e. ζ→4ζ=0.2\zeta\to4\zeta=0.2); alternatively add damping treatment. (3)

(e) [4]

sum=0; df=0.1
for f in range(f_lo..f_hi step df):
    r=f/fn
    H2=(1+(2*zeta*r)**2)/((1-r**2)**2+(2*zeta*r)**2)
    sum += H2*W0*df
Grms=sqrt(sum)

As df→0 and band→wide, the numerical integral of ∣H∣2W0|H|^2 W_0 over the peaked resonance converges to π2fnQW0\tfrac{\pi}{2}f_n Q W_0, recovering Miles. (4)

Question 3

(a) [6] Constrained bar: total strain zero, so mechanical strain cancels thermal: εmech=−αΔT\varepsilon_{mech}=-\alpha\Delta T, σ=Eεmech=−EαΔT\sigma = E\varepsilon_{mech} = -E\alpha\Delta T. ΔT=−70−50=−120\Delta T = -70-50 = -120 K. σ=−(110×109)(8.6×10−6)(−120)=+1.135×108\sigma = -(110\times10^9)(8.6\times10^{-6})(-120) = +1.135\times10^8 Pa =113.5= 113.5 MPa. Positive ⇒ tensile (cooling of a constrained bar puts it in tension). (6)

(b) [6] MS=σyFOS⋅∣σ∣−1=8301.4×113.5−1=830158.9−1=5.22−1=4.22MS = \dfrac{\sigma_y}{FOS\cdot|\sigma|}-1 = \dfrac{830}{1.4\times113.5}-1 = \dfrac{830}{158.9}-1 = 5.22-1 = 4.22. Positive ⇒ ample margin. (6)

(c) [6] EIRP =Pt(dBW)+Gt=10log⁡105+6=6.99+6=12.99= P_t(\text{dBW}) + G_t = 10\log_{10}5 + 6 = 6.99+6 = 12.99 dBW ≈13.0\approx 13.0 dBW. (3) λ=c/f=3×108/8×109=0.0375\lambda = c/f = 3\times10^8/8\times10^9 = 0.0375 m. FSPL =20log⁡10(4π⋅2×106/0.0375)=20log⁡10(6.702×108)=20×8.826=176.5= 20\log_{10}(4\pi\cdot2\times10^6/0.0375) = 20\log_{10}(6.702\times10^8) = 20\times8.826 = 176.5 dB. (3)

(d) [6] RbR_b in dB-Hz $=10\log_{10}