Discussion
A boom of \( 800 \) mm\( ^2 \) lies \( 200 \) mm from the neutral axis of a section with \( I = 8 \times 10^7 \) mm\( ^4 \), carrying a shear of \( 40 \) kN. The step in shear flow at that boom is
Answer: B
Correct answer: 80 N/mm
The step at a boom is \[ \Delta q = \frac{V B y}{I} = \frac{40{,}000 \times 800 \times 200}{8 \times 10^{7}} = 80\ \mathrm{N/mm}. \] The step is proportional to the boom’s first moment of area about the neutral axis, so a boom lying on that axis produces no step at all.
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