Solution to Problem 523 | Flexure Formula

$\Sigma M_{R2} = 0$

$12R_1 = 600(12)(6) + 3P$

$R_1 = 3600 + 0.25P$
 

$\Sigma M_{R1} = 0$

$12R_2 = 600(12)(6) + 9P$

$R_2 = 3600 + 0.75P$
 

$M = \frac{1}{2} \, [ \, (3600 + 0.25P) + (-1800 + 0.25P) \, ](9)$

$M = 8100 + 2.25P \, \text{lb}\cdot\text{ft}$
 

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

$1400 = \dfrac{(8100 + 2.25P)(6)(12)}{981.33}$

$P = 4880.63 \, \text{lb}$
 

Check if the shear at P is positive as assumed
$-1800 + 0.25P = -1800 + 0.25(4880.63)$

$-1800 + 0.25P = -579.84 \, \text{ lb}$   (not okay!)
 

From the actual shear diagram:
$(3600 + 0.25P) - 600x = 0$

$x = \dfrac{3600 + 0.25P}{600}$
 

$M_{max} = \frac{1}{2} x (3600 + 0.25P)$

$M_{max} = \frac{1}{2} \left( \dfrac{3600 + 0.25P}{600} \right) (3600 + 0.25P)$

$M_{max} = \dfrac{(3600 + 0.25P)^2}{1200}$
 

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

$1400 = \dfrac{\dfrac{(3600 + 0.25P)^2}{1200}(6)(12)}{981.33}$

$22\,897\,700 = (3600 + 0.25P)^2$

$P = 4740.62 \, \text{ lb}$           answer