Solution to Problem 511 | Flexure Formula

By symmetry:
$R_1 = R_2 = \frac{1}{2}(80L)$

$R_1 = R_2 = 40L$
 

$(\,f_b\,)_{max} = \dfrac{Mc}{I}$
 

Where

$(\,f_b\,)_{max} = 3000 \, \text{psi}$

$M = 10L^2 \, \text{lb}\cdot\text{ft}$

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

$I = \dfrac{bh^3}{12} = \dfrac{2(4^3)}{12} = \frac{32}{3} \, \text{in}^4$
 

 

$3000 = \dfrac{10L^2(12)(2)}{32/3}$

$L = 11.55 \, \text{ft}$           answer