For Beams (B - 3) | Solution to Problem 542
For Beams (B - 3)
$M_{max} = \frac{1}{8}(20)(62)$
$M_{max} = 90 \, \text{kN}\cdot\text{m}$
$S_{required} = \dfrac{M_{max}}{f_b} = \dfrac{90(1000^2)}{120}$
$S_{required} = 750 \times 10^3 \, \text{mm}^3$
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