Solution to Problem 504 | Flexure Formula

$12R_1 = 9(2000)$

$R_1 = 1500 \, \text{lb}$
 

$\Sigma M_{R1} = 0$

$12R_2 = 3(2000)$

$R_2 = 500 \, \text{lb}$
 

Maximum fiber stress:
 

$(\,f_b\,)_{max} = \dfrac{Mc}{I} = \dfrac{4500(12)(2)}{\dfrac{2(4^3)}{12}}$

$(\,f_b\,)_{max} = 10,125 \, \text{ psi}$           answer
 

Stress in a fiber located 0.5 in from the top of the beam at midspan:

$\dfrac{M_m}{6} = \dfrac{4500}{9}$

$M_m = 3000 \, \text{lb}\cdot\text{ft}$
 

$f_b = \dfrac{My}{I}$

$f_b = \dfrac{3000(12)(1.5)}{\dfrac{2(4^3)}{12}}$

$f_b = 5,062.5 \, \text{ psi}$           answer