Fuselage Design and Layout Questions and Answers

Answer: A

Multiplying, \( 6 \times 30 = 180 \) seats. Capacity, seat pitch and the number of rows together set the cabin length, which is why the interior layout is settled before the fuselage itself is sized. The diameter follows from seats abreast rather than from capacity.

Answer: B

Computing \( 30 \times 0.81 = 24.3 \) m. Galleys, lavatories and the extra space at exit rows all add to this, so the overall cabin is appreciably longer than the seating block alone, and the fuselage longer again once nose and tail are included.

Answer: C

Dividing length by diameter gives \( 38/4 = 9.5 \), which sits comfortably in the usual range of eight to twelve for a transport. Below about eight the afterbody cannot close gently enough and separates; above twelve the wetted area and bending weight begin to dominate.

Answer: D

Correct answer: 377 m^2

The curved surface of a cylinder is

\[ A = \pi d L = \pi \times 4 \times 30 = 377\ \mathrm{m^2} \]

The distractor 754 uses \( 2\pi d L \), doubling the circumference. Note that this covers the constant section only - the nose and tailcone add perhaps a third as much again in practice, and wetted area is what drives both skin friction drag and structural weight.

Answer: A

Correct answer: 55 kPa

The differential is the difference between cabin and ambient pressure:

\[ 0.75 - 0.20 = 0.55\ \mathrm{bar} = 55\ \mathrm{kPa} \]

since one bar is 100 kPa. This is the pressure the shell must contain, and it is what sizes the skin. A cabin altitude of about 2400 metres against a cruise altitude near 11,000 gives figures of this order on a transport.

Answer: B

Correct answer: 61 MPa

Hoop stress in a thin-walled cylinder is

\[ \sigma = \frac{pr}{t} = \frac{55 \times 10^{3} \times 2}{1.8 \times 10^{-3}} \]

\[ = 61.1 \times 10^{6}\ \mathrm{Pa} = 61\ \mathrm{MPa} \]

Keeping the units consistent is the whole difficulty: pascals and metres throughout. The longitudinal stress is half this, which is why fatigue cracks in a pressure cabin tend to run along the fuselage rather than around it.

Answer: C

Correct answer: Between 5 and 6 rows

Add the plugs and divide by the pitch:

\[ \frac{2.7 + 1.8}{0.81} = \frac{4.5}{0.81} = 5.56 \]

so the stretch buys five rows with something left over. In practice the plug lengths are chosen as whole multiples of the pitch precisely to avoid the awkward remainder, and they are placed fore and aft of the wing so that the centre of gravity moves as little as possible.

Answer: D

Multiplying, \( 5 \times 6 = 30 \) seats. Whether an extra exit pair is then required depends on which certification band the new total falls into, and adding a door is expensive enough that a stretch is sometimes sized deliberately to stay below the threshold.

Answer: A

Dividing, \( 44/4 = 11.0 \), still within the usual range though approaching its upper end. Beyond about twelve the added wetted area and the shrinking rotation angle both start to cost more than the extra capacity is worth, which is what limits how far a family can stretch.

Answer: B

Correct answer: 3.5 m

The tail swings on an arc about the main gear contact point, so the descent is the arm times the sine of the rotation angle:

\[ 20 \times \sin 10^\circ = 20 \times 0.1736 = 3.47\ \mathrm{m} \]

The available clearance must exceed this, which is why a stretch that lengthens the rear fuselage so directly threatens the rotation angle - a longer arm gives a larger descent for the same rotation.