Aircraft Structural Components Questions and Answers

Answer: C

Each of the four areas lies \( 0.25 \) m from the axis, so \( I = 4 \times 300 \times 10^{-6} \times 0.25^2 = 4 \times 3 \times 10^{-4} \times 0.0625 = 7.5 \times 10^{-5} \) m\( ^4 \). Booming off the stringers like this is the standard idealisation for a stiffened box.

Answer: D

Rearranging \( \sigma = pr/t \) gives \( p = \sigma t/r = 90 \times 10^6 \times 0.0012/1.8 = 60\,000 \) Pa. A differential of 60 kPa is representative of a transport cabin at cruise altitude, so this skin is about right for the duty.

Answer: A

The area is \( \pi \times 0.002^2 = 1.257 \times 10^{-5} \) m\( ^2 \), so the strength is \( 220 \times 10^6 \times 1.257 \times 10^{-5} = 2765 \) N. Rivet pitch in a seam follows directly from dividing the shear flow by a figure of this kind.

Answer: B

Each rivet must carry the flow over its own pitch, so \( p = 2800/140\,000 = 0.020 \) m. Pitches of around twenty to thirty millimetres are typical in a transport skin seam, and going much finer begins to weaken the sheet through the holes themselves.

Answer: C

Correct answer: 1800 kN m

\[ M = F \times d = 400 \times 4.5 = 1800\ \mathrm{kN\,m}. \] The centre of pressure is the point through which the resultant lift acts, so the moment about the root is simply that force times its distance. This is the load case that sizes the wing: the bending moment is greatest at the root and falls to zero at the tip, which is why spar caps taper outboard.

Answer: D

Correct answer: 1471 kN

Weight first, then the load factor: \[ 60{,}000 \times 9.81 = 588{,}600\ \mathrm{N} = 589\ \mathrm{kN}, \qquad 589 \times 2.5 = 1471\ \mathrm{kN}. \] The 588 kN offered is the 1 g weight with the load factor omitted, and 2943 kN is five times the weight. The ultimate load would be a further 1.5 times this, at about 2207 kN.

Answer: A

Civil practice uses an ultimate factor of 1.5 on limit load, so \( 1.5 \times 1471 = 2207 \) kN. The structure must carry limit load without detrimental permanent deformation and ultimate load for three seconds without failure.

Answer: B

Correct answer: 240 MPa

\[ \sigma = \frac{F}{A} = \frac{60{,}000\ \mathrm{N}}{250\ \mathrm{mm^2}} = 240\ \mathrm{N/mm^2} = 240\ \mathrm{MPa}. \] Newtons per square millimetre are megapascals directly, which makes this the convenient unit set for airframe work - no conversion is needed provided the force is in newtons and the area in square millimetres.

Answer: C

The margin of safety is the allowable divided by the applied, less one, so \( 480/240 - 1 = 1.0 \). A margin of zero means the member is exactly at its allowable, and a negative value means it fails, so margins are quoted to show how much is in hand.

Answer: D

The load is the pressure times the projected area, \( 55\,000 \times \pi \times 2.0^2 = 55\,000 \times 12.57 = 691 \) kN. This large force is reacted as tension in the fuselage shell, and reacting it in a light structure is why the bulkhead is domed rather than flat.