Solution to Problem 509 | Flexure Formula

$R_1 = R_2 = \frac{1}{2}(12)(w)$

$R_1 = R_2 = 6w$
 

$f_b = 10 \, \text{ksi} = 10,000 \, \text{psi}$

$M = 18w \, \text{lb}\cdot\text{ft}$

$c = 6$
 

Centroidal moment of inertia of one tube:
$A = \pi r^2 = 0.20$

$r = 0.2523 \, \text{ in}$   hollow portion of the tube was neglected
 

$\bar I_x = \dfrac{\pi r^4}{4} = \dfrac{\pi (0.2523^4)}{4}$

$\bar I_x = 0.0032 \, \text{in}^4$
 

Moment of inertia at the center of the section:
$d_1 = 6 \sin 30^\circ = 3 \, \text{in}$
 

$I_1 = \bar I_x + A{d_1}^2$

$I_1 = 0.0032 + 0.2(3^2)$

$I_1 = 1.8 \, \text{in}^4$
 

$I_2 = \bar I_x + A{d_2}^2$

$I_2 = 0.0032 + 0.2(6^2)$

$I_2 = 7.2 \, \text{in}^4$
 

$I = 4I_1 + 2I_2 = 4(1.8) + 2(7.2)$

$I = 21.6 \, \text{in}^4$
 

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

$10,000 = \dfrac{18w(12)(6)}{21.6}$

$w = 166.7 \, \text{ lb/ft}$           answer