use 50 kips/floor
Required. Compute axial shortening of columns H.3-6 and G-6 at the roof due to gravity loads using the closed-form equation given earlier. Provide column length corrections at levels 8, 16, 24, and the roof.
Solution. The variation of axial loads and cross-sectional areas for the two columns are shown in Figs. 8.32i and 8.32j. Observe that Ab is the actual area of the column at the foundation level multiplied by a factor of 0.9.
Thus Ab for the interior column = 0.9 x 147 = 132.3 in.2 Ab for the exterior column = 0.9 x 170 = 153. in.2
The loads Pb for the exterior and interior columns at foundation level are:
Pb = 31 x 50 = 1550 kips (exterior column), and Pb = 31 x 77 = 2387 kips (interior column)
Load Pt = 50 kips for the exterior, and 77 kips for the interior column.
Column Length-Shortening Computations for Column G.6 (Interior Column)
Figure 8.32i. Example 2, interior column G-6: (1) axial load variation; (2) variation of cross-sectional areas.
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