The idealisation of skin as shear-carrying only is least accurate when
Answer: A
The assumption depends on the stringers holding most of the direct-stress-carrying area. Where the skin is thick enough to carry a substantial share, lumping it into booms distorts the distribution and a fuller model is needed.
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Two idealisations of the same structure giving different answers indicates
Answer: B
Correct answer: One has discarded something that matters
Idealisation is the deliberate discarding of what does not matter, so a disagreement means one of the two models has thrown away something that does. The loads and material properties are inputs common to both and are not the explanation. The useful response is to identify what the two treat differently - a load path retained in one and omitted in the other - rather than to prefer the answer that looks better.
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Modelling a wing box for preliminary sizing typically uses
Answer: C
At preliminary stage the geometry is still changing and turnaround matters more than the last few per cent of accuracy. Booms and shear panels give section properties and load paths quickly, and detailed meshes follow once the layout has settled.
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A rigid body mode appearing in a finite element model indicates
Answer: D
An unrestrained direction leaves the model free to translate or rotate without strain, which the solver reports as a singularity or a zero-energy mode. Adding the missing restraint, rather than artificially stiffening the model, is the correct fix.
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The principal reason airframe stress work still uses hand idealisations alongside finite element models is
Answer: A
A hand idealisation forces an engineer to decide what carries what, and that understanding is what makes a finite element result interpretable. The two are complementary, and a result nobody can explain is a result nobody should trust.
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A panel \( 150 \) mm long and \( 2 \) mm thick carries equal stress at both ends. The boom area it contributes at each end is
Answer: B
The formula \( B = (t b/6)(2 + \sigma_2/\sigma_1) \) with a stress ratio of one gives \( (2 \times 150/6) \times 3 = 50 \times 3 = 150 \) mm\( ^2 \). That is half the panel area of 300 mm\( ^2 \), as it must be when the stress is uniform.
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The same panel now has equal and opposite stresses at its two ends. Each boom receives
Answer: C
With a stress ratio of minus one the bracket becomes \( 2 - 1 = 1 \), so \( B = 50 \times 1 = 50 \) mm\( ^2 \). Only a sixth of the panel area is credited to each end, because much of the panel's load cancels within its own length.
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For a panel of \( 2 \) mm thickness and \( 150 \) mm length with a stress ratio of \( 0.5 \), the boom area contributed at the higher-stressed end is
Answer: D
Correct answer: 125 mm²
The boom area contributed by a panel at its higher-stressed end is
\[ B = \frac{t_D b}{6}\left(2 + \frac{\sigma_2}{\sigma_1}\right) = \frac{2 \times 150}{6}(2 + 0.5) = 50 \times 2.5 = 125 \ \mathrm{mm}^2 \]The bracket runs from 1 when the far end carries the opposite stress, through 2 when it carries none, to 3 when the stress is uniform - so the panel contributes most where the stress does not vary along it.
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A \( 2 \) mm skin of \( E = 70 \) GPa is stressed to \( 200 \) MPa at the stiffener. Its effective width is nearest to
Answer: A
Correct answer: 71 mm
The effective width of a buckled panel is
\[ 2w = 1.9t\sqrt{\frac{E}{\sigma}} = 1.9 \times 2 \times \sqrt{\frac{70\,000}{200}} = 3.8 \times 18.71 \approx 71 \ \mathrm{mm} \]with the stresses in consistent units. Note the square root: quadrupling the stiffener stress halves the effective width, which is why the structure grows less efficient as it approaches its ultimate load.
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If the stiffener stress in that panel rises from \( 200 \) to \( 320 \) MPa, the effective width becomes nearest to
Answer: B
Since the width varies inversely with the square root of stress, the factor is \( \sqrt{200/320} = \sqrt{0.625} = 0.79 \), giving \( 71 \times 0.79 = 56 \) mm. The structure grows progressively less efficient as it approaches its ultimate load.
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