Solution to Problem 526 | Flexure Formula

$12R_1 + 3(6w_o) = 6P$

$R_1 = 0.5P - 1.5w_o$
 

$\Sigma M_{R1} = 0$

$12R_2 = 6P + 15(6w_o)$

$R_2 = 0.5P + 7.5w_o$
 

$(\,f_b\,)_{max} = \dfrac{Mc}{I}$
 

Where:
$f_b = 1200 \, \text{psi}$

$c = \frac{1}{2}h = \frac{1}{2}(12) = 6 \, \text{in}$

$I = \dfrac{bh^3}{12} = \dfrac{6(12^3)}{12} = 864 \, \text{in}^4$
 

For moment at R₂:
$1200 = \dfrac{18w_o(6)(12)}{864}$

$w_o = 800 \, \text{lb/ft} \,\,$ answer
 

For moment under P:
$1200 = \dfrac{(3P - 9w_o)(6)(12)}{864}$

$14\,400 = 3P - 9w_o$

$14\,400 = 3P - 9(800)$

$P = 7200 \, \text{ lb}$           answer