Loads on Aircraft Structures Questions and Answers

Answer: A

Correct answer: 2.0

In a steady level turn the vertical component of lift must equal the weight, so \[ n = \frac{1}{\cos\phi} = \frac{1}{\cos 60^\circ} = \frac{1}{0.5} = 2.0. \] The relation depends only on bank angle - not on speed, weight or aircraft type - and it rises steeply beyond 60 degrees.

Answer: B

Stalling speed rises with the square root of the load factor, so \( 60 \times \sqrt{2} = 60 \times 1.414 = 84.9 \) m/s. A forty per cent rise for a sixty-degree bank is why accelerated stalls catch pilots out in steep turns near the ground.

Answer: C

Correct answer: 4000 N/m^2

\[ \frac{480{,}000\ \mathrm{N}}{120\ \mathrm{m}^2} = 4000\ \mathrm{N/m}^2. \] Wing loading is weight per unit wing area, and it governs stalling speed, turning performance and, as the gust items here show, how sharply the aircraft responds to a vertical gust.

Answer: D

Correct answer: 1.45

The gust adds its lift increment to the 1 g the wing already carries in level flight: \[ n = 1 + 0.45 = 1.45. \] Reading the increment itself as the load factor gives 0.45, which would mean the aircraft was accelerating downwards - the check is that a positive upward gust must always give \( n>1 \).

Answer: A

Kinetic energy goes as the square of velocity, so twice the sink rate means four times the energy. This is why design sink rates are specified so carefully and why a heavy landing is treated as an inspection-triggering event.

Answer: B

Correct answer: 225 kJ

\[ E = \tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 50{,}000 \times 3^2 = 225{,}000\ \mathrm{J} = 225\ \mathrm{kJ}. \] The undercarriage must absorb this in a stroke of well under a metre, which is why oleo-pneumatic struts are used rather than simple springs - a spring returns the energy, whereas an oleo dissipates it.

Answer: C

Multiplying by the square root of the limit factor, \( 40 \times \sqrt{6} = 40 \times 2.449 = 98 \) m/s. The higher structural limit pushes the corner speed much further out than a transport's, which is exactly what makes tight aerobatic manoeuvres possible.

Answer: D

The moment is force times arm, \( 150 \times 5 = 375 \) kN m per side using the semi-span lift. Inertia relief from the wing structure, fuel and engines would reduce this, which is why a rigid estimate of this kind is conservative.

Answer: A

The inertia moment acts against the lift moment, so the net value is \( 375 - 120 = 255 \) kN m. This is inertia relief: the weight of the wing, its fuel and any engines hung on it all reduce the bending the root must carry, which is why mass placed outboard is structurally welcome.

Answer: B

Correct answer: Twice that of the light aircraft

The gust load increment varies inversely with wing loading: a heavily loaded wing has more mass per unit lifting area, so the same change in angle of attack produces a smaller change in load factor per unit weight. Halving the increment from 0.62 to 0.31 therefore means doubling the wing loading - and it is why a transport rides through turbulence that would throw a light aeroplane about.