Loads on Aircraft Structures Questions and Answers

Answer: D

The internal pressure at yield is approximately 20 MPa.

In a thin spherical shell the membrane stresses in every tangential direction are equal, and the radial stress is negligible:

\[ \begin{aligned} \sigma_{1} &= \sigma_{2} = \frac{pr}{2t} \\ \sigma_{3} &\approx 0 \end{aligned} \]

For this biaxial state the von Mises stress is \(\sqrt{\sigma_{1}^{2} - \sigma_{1}\sigma_{2} + \sigma_{2}^{2}} = \sigma_{1}\), so yield occurs when \(\sigma_{1} = \sigma_{Y}\):

\[ \begin{aligned} p_{Y} &= \frac{2t\sigma_{Y}}{r} = \frac{2 \times 0.010 \times 500}{0.5} \\ &= 20\ \text{MPa} \end{aligned} \]

The larger options equate the pressure itself with or close to the material's yield stress; the thin wall (r/t = 50) means the stress in the wall is 25 times the pressure, so the vessel yields at a much lower pressure.

Answer: C

Yielding occurs at about the same pressure \(p_{Y}\).

Under external pressure the membrane stresses have the same magnitude as before but are compressive: \(\sigma_{1} = \sigma_{2} = -pr/(2t)\). The von Mises criterion depends only on squares and products of the principal stresses, so reversing every sign leaves the equivalent stress unchanged, and the material has the same yield stress in compression as in tension.

So, with buckling ruled out by the assumption of elastic stability, yielding begins at the same pressure magnitude, about 20 MPa. Half or double the pressure would need a material weaker or stronger in compression, and 'never occurs' would only follow for a purely hydrostatic stress, which a thin shell does not carry.

Answer: A

Load factor compares the lift the wings are producing with the aircraft's weight, so it is one in steady level flight and rises in a manoeuvre or a gust. Every structural load case is expressed through it, which is why the V-n diagram is the starting point of airframe sizing.

Answer: B

The manoeuvre envelope shows load factor against equivalent airspeed, bounded below the corner speed by the stall and above it by the structural limit. Using equivalent rather than true airspeed keeps the diagram valid at every altitude, since the aerodynamic loads follow dynamic pressure.

Answer: C

Below the corner speed the wing cannot generate enough lift to reach the limit load factor, so it stalls first and the boundary is a parabola. Above that speed the structure becomes the limiting factor and the boundary is a horizontal line.

Answer: D

At the corner speed a full-deflection pull produces exactly the limit load factor and the aircraft stalls at the same moment. Below it the aircraft cannot be over-stressed by elevator alone, and above it full deflection would exceed the structural limit, which makes it the fastest speed for a maximum-rate turn.

Answer: A

Limit load is the maximum anticipated in service, and the structure must carry it and return to shape with no detrimental permanent set. Ultimate load, at a factor above it, is what the structure must survive without collapsing, though permanent deformation there is acceptable.

Answer: B

Civil requirements use an ultimate factor of 1.5 on limit load, and the structure must carry ultimate load for three seconds without failure. The figure is small compared with civil engineering practice because weight matters so much and the loads are known far more precisely.

Answer: C

Transport aircraft are certificated to about 2.5 g positive and minus 1 g negative, which is ample for the manoeuvres and gusts they meet. Aerobatic aircraft are built to around plus six and minus three, and fighters higher still, with a corresponding weight penalty.

Answer: D

A vertical gust adds a velocity component normal to the flight path, changing the effective angle of attack and so the lift. The resulting load-factor increment is what the gust envelope captures, and for large transports it frequently exceeds the manoeuvre case.