Discussion
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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