Solution to Problem 532 | Economic Sections

$S_{required} \ge 46.875 \, \text{in}^3$
 

From Appendix B, Table B-8 Properties of I-Beam Sections (S Shapes): US Customary Units, of text book: Use S15 × 42.9 with S = 59.6 in³.       answer
 

Checking:
$S_{resisting} \ge S_{live-load} + S_{dead-load}$

$S_{live-load} = 46.875 \, \text{in}^3$

$S_{live-load} = \dfrac{\frac{1}{8}(42.9)(25^2)(12)}{20\,000}$

$S_{live-load} = 2.011 \, \text{in}^3$
 

$S_{live-load} + S_{dead-load} = 46.875 + 2.011$

$S_{live-load} + S_{dead-load} = 48.886 \, \text{in}^3$

$(S_{resisting} = 59.6 \, \text{in}^3) \gt 48.886 \, \text{in}^3$   (okay!)
 

Actual bending moment:
$M = M_{live-load} + M_{dead-load}$

$M = (\frac{1}{8}w_oL^2)_{live-load} + (\frac{1}{8}w_oL^2)_{dead-load}$

$M = \frac{1}{8}(1000)(25^2) + \frac{1}{8}(42.9)(25^2)$

$M = 81,476.56 \, \text{lb}\cdot\text{ft}$
 

Actual stress:
$(\,f_b\,)_{max} = \dfrac{M}{S}$

$(\,f_b\,)_{max} = \dfrac{81\,476.56(12)}{59.6}$

$(\,f_b\,)_{max} = 16\,404.68 \, \text{psi}$

$(\,f_b\,)_{max} = 16.4 \, \text{ ksi}$           answer