Loads on Aircraft Structures Questions and Answers

Answer: C

In a level turn the load factor is the reciprocal of the cosine of the bank angle, and the cosine of 60 degrees is one half, giving n equal to two. The relationship explains why a steep turn feels heavy and why stalling speed rises so quickly with bank.

Answer: D

Lift must equal the load factor times the weight, and lift goes as the square of speed, so the stalling speed scales with the square root of the load factor. At 2 g it is about forty per cent higher than in level flight, which is the origin of the accelerated stall.

Answer: A

Transport aircraft are certificated to about minus one g, which is far less demanding than the positive case because sustained negative manoeuvres are neither expected nor comfortable. Aerobatic aircraft carry minus three or beyond and pay for it in structure weight.

Answer: B

The margin between cruise and dive speeds accommodates upsets, control failures and a descent at maximum thrust before the design limit is reached. The regulations prescribe minimum separations between the two, so it is a certification requirement rather than a preference.

Answer: C

Two separate envelopes are drawn and the structure is sized by the outer boundary of the pair at each speed. For a large transport the gust case frequently governs at cruise speeds while the manoeuvre case governs lower down.

Answer: D

As a swept wing bends upward, the outer sections twist nose-down and shed load inboard, which reduces the root bending moment below the rigid estimate. Ignoring the effect gives an unnecessarily heavy structure, so flexible load distributions are computed routinely.

Answer: A

Below VA the wing stalls before the structural limit is reached, so a full pull on the elevator cannot over-stress the aircraft. This is why turbulence penetration speeds are specified at or below VA, though the protection applies to a single axis and not to abrupt reversals.

Answer: B

Correct answer: 150 kN

\[ L = n W = 2.5 \times 60 = 150\ \mathrm{kN}. \] The load factor is by definition the lift divided by the weight, so at 2.5 g the wing is carrying two and a half times the aircraft’s weight - and everything the wing supports feels the same multiple.

Answer: C

Correct answer: 225 kN

\[ 150 \times 1.5 = 225\ \mathrm{kN}. \] The factor of safety between limit and ultimate load is 1.5 in essentially all civil airworthiness codes. The limit load must be carried without detrimental permanent deformation; the ultimate load must be carried for three seconds without failure.

Answer: D

Corner speed is the stalling speed times the square root of the limit load factor, \( 60 \times \sqrt{2.5} = 60 \times 1.581 = 94.9 \) m/s. Below this speed a full elevator pull stalls the wing before the structural limit is reached.