Solution to Problem 522 | Flexure Formula

$I_{NA} = 981.33 \, \text{in}^4$
 

$\Sigma M_{R1} = 0$

$12R_2 = 300(12)(6) + 9P$

$R_2 = 1800 + 0.75P$
 

$M = \frac{1}{2} \, [ \, (1800 + 0.25P) + (-900 + 0.25P) \, ] \, (9)$

$M = 4050 + 2.25P \, \text{lb}\cdot\text{ft}$
 

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

$1400 = \dfrac{(4050 + 2.25P)(6)(12)}{981.33}$

$P = 6680.63 \, \text{lb}$
 

Check if the shear at P is positive as assumed
$-900 + 0.25P = -900 + 0.25(6680.63)$

$-900 + 0.25P = 770.16 \, \text{ lb}$       (okay!)
 

Thus, P = 6680.63 lb.           answer