# Solution to Problem 509 | Flexure Formula

**Problem 509**

A section used in aircraft is constructed of tubes connected by thin webs as shown in Fig. P-509. Each tube has a cross-sectional area of 0.20 in^{2}. If the average stress in the tubes is no to exceed 10 ksi, determine the total uniformly distributed load that can be supported in a simple span 12 ft long. Neglect the effect of the webs.

**Solution 509**

$R_1 = R_2 = 6w$

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

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

$c = 6$

**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*