A wing skin carries \( 200 \) MPa direct stress and \( 60 \) MPa shear. Its larger principal stress is nearest to
Answer: C
The principal values are \( \sigma/2 \pm \sqrt{(\sigma/2)^2 + \tau^2} = 100 \pm \sqrt{10\,000 + 3\,600} = 100 \pm 116.6 \), giving 216.6 and minus 16.6 MPa. Note the compressive value appearing despite the applied direct stress being wholly tensile.
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For that same skin, the von Mises equivalent stress is nearest to
Answer: D
For combined direct and shear stress, \( \sigma_{vm} = \sqrt{\sigma^2 + 3\tau^2} = \sqrt{40\,000 + 10\,800} = \sqrt{50\,800} = 225 \) MPa. The factor of three on the shear term is what distinguishes the criterion from a naive combination.
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A wing carries a distributed load whose resultant is \( 300 \) kN acting \( 4 \) m from the root. The root bending moment is
Answer: A
Correct answer: 1200 kN m
The bending moment is the resultant times its distance from the section:
\[ M = 300 \times 4 = 1200 \ \mathrm{kN\,m} \]The root is where the accumulated moment is greatest, which is why it is normally the critical section for bending. Inertia relief from the wing’s own weight and from engines and fuel carried outboard reduces this figure, sometimes substantially.
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A fastener of \( 5 \) mm diameter bears on \( 2 \) mm sheet carrying \( 4 \) kN. The bearing stress is
Answer: B
Correct answer: 400 MPa
Bearing stress uses the projected area of the hole, that is diameter times thickness:
\[ \sigma_{br} = \frac{P}{dt} = \frac{4000}{0.005 \times 0.002} = \frac{4000}{10^{-5}} = 4 \times 10^8 \ \mathrm{Pa} = 400 \ \mathrm{MPa} \]The projected rectangle, not the hole’s circular area, is the convention, and it is what bearing allowables are quoted against.
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Applying a fitting factor of \( 1.15 \) to an ultimate load of \( 2000 \) N gives a design load of
Answer: C
Correct answer: 2300 N
The fitting factor multiplies the load the fitting must be shown to carry:
\[ 2000 \times 1.15 = 2300 \ \mathrm{N} \]It is applied in addition to the ultimate factor of 1.5 already in the load, so the fitting is designed for 1.725 times limit load. Dividing rather than multiplying gives 1739 N and would make the fitting weaker rather than stronger, which is the error the distractor tests.
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A lug is checked for tension, shear tear-out and bearing, giving allowables of \( 42 \), \( 38 \) and \( 55 \) kN. Its capacity is
Answer: D
Correct answer: 38 kN
The lug fails in whichever mode gives way first, so its capacity is the smallest of the three allowables:
\[ \min(42, 38, 55) = 38 \ \mathrm{kN} \]Here shear tear-out governs, which would point the designer to increasing the edge distance rather than the width or the thickness. Adding or averaging the three would credit the lug with strength it does not have.
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A four-row fastened joint carries \( 20 \) kN. If the outer rows each take \( 30 \) per cent and the inner rows share the rest equally, each inner row carries
Answer: A
Correct answer: 4 kN
The outer rows take 30 per cent each, that is 60 per cent between them, leaving 40 per cent for the two inner rows:
\[ \frac{0.40 \times 20}{2} = \frac{8}{2} = 4 \ \mathrm{kN} \]The uneven sharing is the point: the outer rows carry half as much again as the inner ones, because they are where the stiffness mismatch between the two sheets is greatest. Adding more rows in the middle does almost nothing, since the load will not reach them.
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A wing box section has an allowable of \( 240 \) MPa and an applied stress of \( 260 \) MPa. Its margin of safety is
Answer: B
Correct answer: -0.08
Applying the definition,
\[ MS = \frac{240}{260} - 1 = 0.923 - 1 = -0.077 \approx -0.08 \]The negative sign means the member is overloaded at its design condition and must be strengthened or the load reduced. However small, a negative margin cannot be accepted - the factor of safety is already in the applied load, so there is no further reserve behind it.
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A frame carries a radial load of \( 30 \) kN at a radius of \( 2 \) m. The order of magnitude of the bending moment in the ring is
Answer: C
Standard ring solutions give a peak moment of the order of the load times the radius times a coefficient below one, so \( 30 \times 2 \) gives 60 kN m as an upper bound and the real figure falls in the tens. An order-of-magnitude check like this catches gross errors quickly.
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The equivalent stress under pure shear of \( \tau \), by von Mises, is
Answer: D
Setting the direct stress to zero in \( \sqrt{\sigma^2 + 3\tau^2} \) leaves \( \sqrt{3}\tau = 1.73\tau \). It follows that the shear yield stress is the tensile yield divided by root three, about fifty-eight per cent of it.
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