Solution to Problem 521 | Flexure Formula | Strength of Materials Review at MATHalino

Problem 521
A beam made by bolting two C10 × 30 channels back to back, is simply supported at its ends. The beam supports a central concentrated load of 12 kips and a uniformly distributed load of 1200 lb/ft, including the weight of the beam. Compute the maximum length of the beam if the flexural stress is not to exceed 20 ksi.

Solution 521

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Relevant data from Appendix B, Table B-9 Properties of Channel Sections: US Customary Units, of text book.

Designation	C10 × 30
S	20.7 in³

From the shear diagram:
$M_{max} = \frac{1}{2} \, [ \,(6 + 0.6L) + 6 \, ](L/2)$

$M_{max} = 3L + 0.15L^2$

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

$20(1000) = \dfrac{(3L + 0.15L^2)(1000)(12)}{2(20.7)}$

$0.15L^2 + 3L - 69 = 0$

$L = 13.66 \, \text{and} \, -33.66 \, \text{(meaningless)}$

Use $L = 13.66 \, \text{ ft}$ answer

Solution to Problem 521 | Flexure Formula

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