FORMULAS FOR COLUMNS Machinery's Handbook, 31st Edition
285
Next determine Q , which is found as in Example 1, and will again be 0.0778. Assume Formula (2) applies: P As r Q 1 4 cr y 2 = − a k , . , . , . w d r d r d r 5400 40 000 1 4 00778 2 40 000 1 0 01945 40 000 2 5400 1 001945 2 2 2 2 2 # # # # = − = − = − a a a k k k As mentioned in Example 1 the least radius of gyration r of a rectangle is equal to 0.289 times the least dimension, d , in this case. Therefore, substituting for d the value r ÷ 0.289, , . . , . . . . . . r r r r = − = − = ` r 40 000 2 5400 # 0289 1 a j 001945 40 000 2 5400 0 289 0 289 # # # 001945 001945 0005638 00251 2 2 2 2 2 = − k Checking to determine if Q ÷ r 2 is greater or less than 2, . . . r Q 00251 00778 31 2 = = Therefore, Formula (1) should have been used to determine r and dimensions w and d . Using Formula (1), , . , . , . . . . Q d r r r r 5400 40 000 2 00778 40 000 2 0 289 2 2 2 2 4 # # # # # # # # = = = = ` j
. . 40 000 2 5400 0 0778 0 289 0 289 0 0004386 0289 0 145 0 50 inch . =
d
=
and w = 2 d = 1 inch as in the previous example. American Institute of Steel Construction.— For main or secondary compression mem bers with l / r ratios up to 120, safe unit stress = 17,000 − 0.485 l 2 / r 2 . For columns and brac
ing or other secondary members with l / r ratios above 120, For bracing and secondary members, safe unit stress, psi
, r l 1 18 000 18 000 , 2 + ⁄
=
2
For main members, safe unit stress, l ⁄ r 1.6 – Pipe Columns: Allowable concentric loads for steel pipe columns based on the above formulas are given in the table on page 286 . , r l 1 18 000 18 000 , 200 psi 2 2 # = + e o ⁄
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