Noesis
Level 2 — RecallSpacecraft Structures & Systems Engineering

Spacecraft Structures & Systems Engineering

⏱ 30 minutes50 marksprintable — key stays hidden on paper

Level 2 — Recall & Standard Problems

Time limit: 30 minutes Total marks: 50

Q1. (4 marks) Define stress (σ\sigma) and strain (ε\varepsilon), giving their formulas and SI units. State Young's modulus EE and its relation to σ\sigma and ε\varepsilon.

Q2. (5 marks) A titanium tie-rod of cross-sectional area A=200 mm2A = 200\text{ mm}^2 carries an axial tensile force F=60 kNF = 60\text{ kN}. The rod is 1.5 m1.5\text{ m} long with E=110 GPaE = 110\text{ GPa}. (a) Compute the stress σ\sigma. (2) (b) Compute the strain ε\varepsilon and elongation ΔL\Delta L. (3)

Q3. (6 marks) State the Euler buckling load formula for a pin-ended column. A solid circular aluminium strut of diameter d=20 mmd = 20\text{ mm}, length L=1.2 mL = 1.2\text{ m}, E=70 GPaE = 70\text{ GPa} is pin-ended. Compute its critical buckling load PcrP_{cr}. (Take I=πd4/64I = \pi d^4/64.)

Q4. (6 marks) Define factor of safety (FOS). A structural member has yield stress σy=250 MPa\sigma_y = 250\text{ MPa} and experiences an applied stress of σ=100 MPa\sigma = 100\text{ MPa} under design load. (a) Compute the FOS. (2) (b) State whether a design requiring FOS≥1.5\text{FOS} \geq 1.5 is met. (1) (c) Define the margin of safety (MoS) and compute it. (3)

Q5. (6 marks) State Miner's rule for cumulative fatigue damage. A component undergoes two loading blocks: n1=104n_1 = 10^4 cycles at a stress whose fatigue life is N1=5×104N_1 = 5\times10^4, and n2=2×104n_2 = 2\times10^4 cycles where N2=1×105N_2 = 1\times10^5. Compute the cumulative damage DD and state whether failure is predicted.

Q6. (5 marks) Define the stress intensity factor KK and fracture toughness KICK_{IC}. Write K=YσπaK = Y\sigma\sqrt{\pi a} and state the fracture criterion. For Y=1.12Y=1.12, σ=150 MPa\sigma = 150\text{ MPa}, crack length a=2 mma = 2\text{ mm}, compute KK.

Q7. (4 marks) Define MTBF and the exponential failure model reliability R(t)R(t). If a unit has failure rate λ=2×10−5 /hr\lambda = 2\times10^{-5}\text{ /hr}, compute its MTBF and the reliability at t=104 hrt = 10^4\text{ hr}.

Q8. (5 marks) In FMEA, define RPN in terms of Severity (S), Occurrence (O), and Detection (D). For a failure mode with S=8S=8, O=4O=4, D=5D=5, compute the RPN. Briefly state what a high RPN indicates.

Q9. (4 marks) List and briefly define the four standard verification methods used in spacecraft systems engineering.

Q10. (5 marks) Define RMS acceleration from a flat acceleration PSD. Given a flat PSD of W0=0.04 g2/HzW_0 = 0.04\text{ g}^2/\text{Hz} over a bandwidth from 20 Hz20\text{ Hz} to 2000 Hz2000\text{ Hz}, compute the GrmsG_{rms}.

End of paper

Answer keyMark scheme & solutions

Q1. (4 marks)

  • Stress: σ=F/A\sigma = F/A, force per unit area, units Pa (N/m²). (1)
  • Strain: ε=ΔL/L\varepsilon = \Delta L/L, fractional change in length, dimensionless. (1)
  • Young's modulus EE = ratio of stress to strain in the elastic region. (1)
  • Relation: σ=Eε\sigma = E\varepsilon; units of EE are Pa. (1)

Q2. (5 marks) (a) σ=F/A=60×103/(200×10−6)=3×108 Pa=300 MPa\sigma = F/A = 60\times10^3 / (200\times10^{-6}) = 3\times10^8\text{ Pa} = 300\text{ MPa}. (2) (b) ε=σ/E=300×106/110×109=2.727×10−3\varepsilon = \sigma/E = 300\times10^6 / 110\times10^9 = 2.727\times10^{-3}. (2) ΔL=εL=2.727×10−3×1.5=4.09×10−3 m=4.09 mm\Delta L = \varepsilon L = 2.727\times10^{-3}\times1.5 = 4.09\times10^{-3}\text{ m} = 4.09\text{ mm}. (1) Why: stress from definition, strain from Hooke's law, elongation from strain definition.

Q3. (6 marks)

  • Formula: Pcr=π2EIL2P_{cr} = \dfrac{\pi^2 EI}{L^2} (pin-ended, effective length = L). (2)
  • I=πd4/64=π(0.02)4/64=7.854×10−9 m4I = \pi d^4/64 = \pi (0.02)^4/64 = 7.854\times10^{-9}\text{ m}^4. (2)
  • Pcr=π2(70×109)(7.854×10−9)/(1.2)2P_{cr} = \pi^2 (70\times10^9)(7.854\times10^{-9})/(1.2)^2. (1)
  • =π2×549.8/1.44=5426.6/1.44=3768 N≈3.77 kN= \pi^2 \times 549.8 / 1.44 = 5426.6/1.44 = 3768\text{ N} \approx 3.77\text{ kN}. (1)

Q4. (6 marks) (a) FOS=σy/σ=250/100=2.5\text{FOS} = \sigma_y/\sigma = 250/100 = 2.5. (2) (b) 2.5≥1.52.5 \geq 1.5 → requirement met. (1) (c) MoS=σallowFOSreq⋅σ−1\text{MoS} = \dfrac{\sigma_{allow}}{\text{FOS}_{req}\cdot\sigma} - 1, or simply MoS=σyσ⋅FOSreq−1\text{MoS} = \dfrac{\sigma_y}{\sigma\cdot\text{FOS}_{req}} - 1. (1) MoS=250100×1.5−1=1.667−1=0.667\text{MoS} = \dfrac{250}{100\times1.5} - 1 = 1.667 - 1 = 0.667 (positive → adequate). (2)

Q5. (6 marks)

  • Miner's rule: D=∑niNiD = \sum \dfrac{n_i}{N_i}; failure predicted when D≥1D \geq 1. (2)
  • D=1045×104+2×104105=0.2+0.2=0.4D = \dfrac{10^4}{5\times10^4} + \dfrac{2\times10^4}{10^5} = 0.2 + 0.2 = 0.4. (3)
  • D=0.4<1D = 0.4 < 1 → no failure predicted. (1)

Q6. (5 marks)

  • KK = stress intensity factor; characterizes stress field magnitude at crack tip. (1)
  • KICK_{IC} = fracture toughness, material property; fracture when K≥KICK \geq K_{IC}. (2)
  • K=Yσπa=1.12×150×106×π×0.002K = Y\sigma\sqrt{\pi a} = 1.12\times150\times10^6\times\sqrt{\pi\times0.002}. (1)
  • π×0.002=6.283×10−3=0.07927\sqrt{\pi\times0.002} = \sqrt{6.283\times10^{-3}} = 0.07927; K=1.12×150×106×0.07927=1.332×107 Pam=13.3 MPamK = 1.12\times150\times10^6\times0.07927 = 1.332\times10^7\text{ Pa}\sqrt{\text{m}} = 13.3\text{ MPa}\sqrt{\text{m}}. (1)

Q7. (4 marks)

  • MTBF = mean time between failures; MTBF=1/λ\text{MTBF} = 1/\lambda. (1)
  • R(t)=e−λtR(t) = e^{-\lambda t}. (1)
  • MTBF=1/(2×10−5)=50000 hr\text{MTBF} = 1/(2\times10^{-5}) = 50000\text{ hr}. (1)
  • R=e−2×10−5×104=e−0.2=0.8187R = e^{-2\times10^{-5}\times10^4} = e^{-0.2} = 0.8187. (1)

Q8. (5 marks)

  • RPN=S×O×D\text{RPN} = S\times O\times D. (2)
  • =8×4×5=160= 8\times4\times5 = 160. (2)
  • High RPN indicates a high-priority risk requiring corrective/mitigation action. (1)

Q9. (4 marks) 1 mark each:

  • Analysis — verify by calculation/modelling/simulation.
  • Test — verify by applying stimuli and measuring the response on hardware.
  • Inspection — verify by visual/physical examination (dimensions, workmanship).
  • Demonstration — verify by operating the item to show functional performance.

Q10. (5 marks)

  • For a flat PSD: Grms=W0⋅ΔfG_{rms} = \sqrt{W_0 \cdot \Delta f}. (2)
  • Δf=2000−20=1980 Hz\Delta f = 2000 - 20 = 1980\text{ Hz}. (1)
  • Grms=0.04×1980=79.2=8.90 gG_{rms} = \sqrt{0.04\times1980} = \sqrt{79.2} = 8.90\text{ g}. (2)
[
{"claim":"Q2 stress 300 MPa and elongation ~4.09 mm","code":"sigma=60e3/200e-6; eps=sigma/110e9; dL=eps*1.5; result=(abs(sigma-300e6)<1e3) and (abs(dL-4.09e-3)<1e-5)"},
{"claim":"Q3 Euler buckling load ~3768 N","code":"import math; I=math.pi*(0.02)**4/64; Pcr=math.pi**2*70e9*I/1.2**2; result=abs(Pcr-3768)<5"},
{"claim":"Q5 Miner damage = 0.4","code":"D=1e4/5e4+2e4/1e5; result=abs(D-0.4)<1e-9"},
{"claim":"Q7 MTBF 50000 hr and R(1e4)=0.8187","code":"import math; MTBF=1/2e-5; R=math.exp(-2e-5*1e4); result=(abs(MTBF-50000)<1e-6) and (abs(R-0.8187)<1e-3)"},
{"claim":"Q10 Grms ~8.90 g","code":"import math; G=math.sqrt(0.04*1980); result=abs(G-8.90)<0.02"}
]