Solution to Problem 541 | Floor Framing

$w_o = (0.6 \, \text{lb/in}^2)(12 \, \text{in/ft})^2(s \, \text{ft})$

$w_o = 86.4s \, \text{lb/ft}$
 

$R_1 = R_2 = \frac{1}{2}(86.4s)(18)$

$R_1 = R_2 = 777.6s \, \text{lb}$
 

$M_{max} = 777.6s(9) - 86.4s(9)(4.5)$

$M_{max} = 3499.2s \, \text{lb}\cdot\text{ft}$
 

From Table B-7 in Appendix B of textbook: Properties of Wide-Flange Sections (W-Shapes): US Customary Units.
 

Designation	Section Modulus, S
W10 × 30	32.4 in3

 

$f_b = \dfrac{M}{S}$

$18\,000 = \dfrac{3499.2s(12)}{32.4}$

$s = 13.9 \, \text{ft}$           answer