# Solution to Problem 518 | Flexure Formula

**Problem 518**

A cantilever beam 4 m long is composed of two C200 × 28 channels riveted back to back. What uniformly distributed load can be carried, in addition to the weight of the beam, without exceeding a flexural stress of 120 MPa if (a) the webs are vertical and (b) the webs are horizontal? Refer to Appendix B of text book for channel properties.

**Solution 518**

Designation | C200 × 28 |

Area | 3560 mm^{2} |

Width | 64 mm |

S_{X-X} |
180 × 10^{3} mm^{3} |

I_{Y-Y} |
0.825 × 10^{6} mm^{4} |

x | 14.4 mm |

**a. Webs are vertical**

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

$120 = \dfrac{M}{2(180 \times 10^3 )}$

$M = 43,200,000 \, \text{N}\cdot\text{mm}$

$M = 43.2 \, \text{kN}\cdot\text{m}$

From the figure:

$M = 4w(2)$

$M = 8w$

$43.2 = 8w$

$w = 5.4 \, \text{kN/m}$

$w = 550.46 \, \text{kg/m}$

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

$550.46 = 2(28) + LL$

$LL = 494.46 \, \text{ kg/m}$ *answer*

**b. Webs are horizontal**

$I_{back} = I_{Y-Y} + Ax^2$

$I_{back} = (0.825 \times 10^6) + 3560(14.4^2)$

$I_{back} = 1\,563\,201.6 \, \text{mm}^4$

$I = 2I_{back} = 2(1\,563\,201.6)$

$I = 3\,126\,403.2 \, \text{mm}^4$

$(\,f_b\,)_{max} = \dfrac{Mc}{I}$

$120 = \dfrac{M(64)}{3\,126\,403.2}$

$M = 5\,862\,006 \, \text{N}\cdot\text{mm}$

$M = 5.862 \, \text{kN}\cdot\text{m}$

From the figure:

$M = 4w (2)$

$M = 8w$

$5.862 = 8w$

$w = 0.732\,75 \, \text{kN/m}$

$w = 74.69 \, \text{kg/m}$

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

$74.69 = 2(28) + LL$

$LL = 18.69 \, \text{ kg/m}$ *answer*