Solution to Problem 524 | Flexure Formula

$5R_1 = 3W(3.5) + W(1)$

$R_1 = 2.3W$
 

$\Sigma M_{R1} = 0$

$5R_2 = 3W(1.5) + W(4)$

$R_2 = 1.7W$
 

$2.3W - Wx = 0$

$x = 2.3 \, \text{m}$
 

$M_{max} = \frac{1}{2}x (2.3W)$

$M_{max} = \frac{1}{2}(2.3)(2.3W)$

$M_{max} = 2.645W$
 

From Appendix B, Table B-3 Properties of I-Beam Sections (S-Shapes): SI Units, of text book.
 

$(\,f_b\,)_{max} = \dfrac{Mc}{I}$

$120 = \dfrac{2.645W (1000)}{1\,060 \times 10^3}$

$W = 48\,090.74 \, \text{N}$           answer