Solution to Problem 512 | Flexure Formula

$\Sigma M_B = 0$

$4R_A = 2(400 \sin 60^\circ) + 2(200 \sin 30^\circ)$

$R_A = 223.2 \, \text{lb}$
 

 

$M = 2(223.2) - 2(400 \cos 60^\circ)$

$M = 46.4 \, \text{lb}\cdot\text{ft}$
 

$(\,f_b\,)_{max} = \dfrac{Mc}{I} = \dfrac{Mr}{\pi r^4 / 4}$

$(\,f_b\,)_{max} = \dfrac{4M}{\pi r^3} = \dfrac{4(46.4)(12)}{\pi (0.5^3)}$

$(\,f_b\,)_{max} = 5671.52 \, \text{ psi}$           answer