Solution to Problem 527 | Flexure Formula

$\Sigma M_{R2} = 0$

$12R_1 + 600x (x/2) = 6P$

$R_1 = 0.5P - 25x^2$
 

$12R_2 = 6P + 600x \, (12 + \frac{1}{2}x)$

$R_2 = 0.5P + 600x + 25x^2$
 

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

Refer to Solution 526 for values of c and I.
 

For moment at R₂:
$1200 = \dfrac{(300x^2)(6)(12)}{864}$

$x^2 = 48$

$x = 6.93 ~ \text{ft}$       answer
 

For moment under P:
$1200 = \dfrac{(3P - 150x^2)(6)(12)}{864}$

$14\,400 = 3P - 150x^2$

$14\,400 = 3P - 150(48)$

$P = 7\,200 ~ \text{lb}$           answer