Solution to Problem 513 | Flexure Formula

$\Sigma M_{R2} = 0$

$9R_1 = 6(450) + 3600$

$R_1 = 700 \, \text{lb}$
 

$\Sigma M_{R1} = 0$

$9R_2 + 3(450) = 3600$

$R_2 = 250 \, \text{lb}$
 

 

$(\,f_b\,)_{max} = \dfrac{Mc}{I}$
 

Where
$M = 2850 \, \text{lb}\cdot\text{ft}$

$c = h/2 = 3/2 = 1.5 \, \text{in}$

$I = \dfrac{bh^3}{12} = \dfrac{2(3^3)}{12} = 4.5 \, \text{in}^4$
 

$(\,f_b\,)_{max} = \dfrac{2850(12)(1.5)}{4.5}$

$(\,f_b\,)_{max} = 11400 \, \text{ psi}$   @ 3 ft from right support       answer