Wing ribs are spaced more closely near the wing root chiefly because
Answer: A
Bending moment and shear both peak at the root, so the skin panels there carry the highest compressive stress and need the shortest unsupported length to avoid buckling. Rib pitch is therefore a direct response to the local stress rather than to geometry.
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The floor beams of a transport fuselage carry
Answer: B
Besides the everyday payload, the floor must survive a rapid decompression of the hold beneath it, which loads it downward across its whole area. Vent paths between cabin and hold are provided precisely so that this differential does not collapse the floor.
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A stringer run-out, where a stringer terminates, is a fatigue-critical detail because
Answer: C
Ending a stiffener forces its load to shear into the surrounding skin over a short length, producing a local peak stress. Tapering the run-out gradually and easing the fastener pattern are standard measures, and such details are routinely the origin of service cracking.
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The vertical fin's side load is reacted into the fuselage principally through
Answer: D
The fin acts as a cantilever whose root moment must be reacted by the rear fuselage, so heavy frames pick up the spar loads and the shell carries the resulting torsion forward. This is why the rear fuselage of a transport is so noticeably reinforced.
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An engine pylon on a transport aircraft is designed to
Answer: A
The pylon transmits thrust, engine weight and manoeuvre inertia into the wing, and it is designed so that in a severe overload the engine departs cleanly rather than tearing the wing structure. That deliberate weak-link behaviour is a certification consideration in its own right.
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A cylindrical fuselage of radius \( 2 \) m and skin thickness \( 1.6 \) mm carries a cabin differential of \( 55 \) kPa. The hoop stress is nearest to
Answer: B
The thin-shell hoop stress is \( pr/t = 55\,000 \times 2 / 0.0016 = 68.8 \times 10^6 \) Pa, that is 68.8 MPa. Against an aluminium yield near 300 MPa this leaves ample static margin, but it is the cyclic repetition of it that governs the design.
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For the same fuselage, the longitudinal stress from pressurisation is
Answer: C
The longitudinal stress is \( pr/2t \), exactly half the hoop value, so \( 68.8/2 = 34.4 \) MPa. That the hoop stress is twice the longitudinal is why pressurisation cracks in a fuselage skin characteristically run along the fuselage rather than around it.
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A wing box \( 0.4 \) m deep carries a bending moment of \( 900 \) kN m. Treating the caps as the only flanges, the force in each cap is about
Answer: D
Idealising the box as two flanges separated by the depth, the couple gives \( M/h = 900\,000/0.4 = 2250 \) kN, one cap in tension and the other in compression. The approximation ignores the skins' own contribution and so overstates the cap load, which makes it a conservative first estimate.
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A skin panel \( 1.6 \) mm thick carries a shear flow of \( 120 \) kN/m. The shear stress in it is
Answer: A
Shear stress is the shear flow divided by the thickness, so \( 120\,000/0.0016 = 75 \times 10^6 \) Pa. Working in shear flow rather than stress is the habit of thin-walled analysis, because the flow is constant along a wall of varying thickness.
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A closed wing box of enclosed area \( 0.9 \) m\( ^2 \) carries a torque of \( 180 \) kN m. The shear flow round it is
Answer: B
The Bredt-Batho relation gives \( q = T/(2A) = 180\,000/(2 \times 0.9) = 100 \) kN/m. Note that the enclosed area, not the material area, is what appears, which is why closing the cell is so much more effective than adding material to an open one.
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