# Solution to Problem 519 | Flexure Formula

**Problem 519**

A 30-ft beam, simply supported at 6 ft from either end carries a uniformly distributed load of intensity w_{o} over its entire length. The beam is made by welding two S18 × 70 (see appendix B of text book) sections along their flanges to form the section shown in Fig. P-519. Calculate the maximum value of w_{o} if the flexural stress is limited to 20 ksi. Be sure to include the weight of the beam.

**Solution 519**

Designation | S18 × 70 |

S | 103 in^{3} |

$(\,f_b\,)_{max} = \dfrac{M}{S}$

$20 = \dfrac{M}{2(103)}$

$M = 4120 \, \text{kip}\cdot\text{in}$

$M = \frac{1030}{3} \, \text{kip}\cdot\text{ft}$

**From the moment diagram:**

$M = 22.5w$

$\frac{1030}{3} = 22.5w$

$w = 15.26 \, \text{kip/ft}$

$w = \text{dead load, } DL + \text{live load, } w_o$

$15.26(1000) = 2(70) + w_o$

$w_o = 15\,120 \, \text{lb/ft}$

$w_o = 15.12 \, \text{ kip/ft}$ *answer*