Example

Determine the allowable axial load for the square tubular column shown in Figure 6.31. Assume that Fy = 40 ksi, KxLx = KyLy = 10 ft, and the dead to live load ratio is 1. Use the ASD and LRFD methods.

Solution

ASD method

Since the square tube is a doubly symmetric closed section, it will not be subject to torsional-flexural buckling. It can be designed by flexural buckling.

1. Sectional properties of full-section. w = 8 . 00 - 2(R + t ) = 7 . 415 in.

Ix = Iy = 2(0 .105)[(1/12)(7. 415)3 + 7 .415(4- 0 .105/2)2]+ 4(0.0396)(4.0 -0 .1373)2 = 33 .763 in. 4 rx = ry = y/lx/A = a/33 .763/3 .273 = 3 .212 in .

2. Nominal buckling stress, Fn. According to Equation 6.72, the elastic flexural buckling stress, Fe, is computed as follows:

KL 10 x 12

r 3 212

0 0

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