Column Design

— David Wagner 2007/06/28 15:51

Columns are 10”x16”, with the longer dimension in the transverse direction.

Use 2” minimum cover (7.7.1b), and consider #4 ties and #8 main bars.

The moment created by the wind at the top of the column is as a uniform cantilevered load supported at the other. (The small amount of wind not blowing below ground level is not considered.)

M_{wind}={w l^2}/8 = {0.314*(12*18)}/{8} = 1831

The top end moment results in single curvature and is the sum of the main span moment and the moment from the wind load, less the small moment created by the cantilever (ignoring more complicated wind effects such as uplift on the cantilever): 3412 kip-in + 1831 -754 kip-in = 4489 kip-in = 374 kip-ft.

Take E as 1 (same concrete throughout)

Column: {EI}/{l_c} = {10*16^3}/{12*21}=162.54

Beam: {EI}/{l_c} = {10*16^3}/{12*26}=131.28

Psi_{top} = {0.7*162.54}/{0.35*131.28} =2.476

Psi_{pin}=infty approx 10

Assume this is a nonbraced sway frame. (No bracing is provided in the transverse direction.)

k=2.2

The bottom of the column is pinned, resulting in single curvature.

{k l_u}/r = {2.2*(21*12-8)}/{0.3*16}=111.8 » 40 → This is a slender column in a sway frame. Verify Q={Sigma P_u Delta_o}/{V_{us} l_c} > 0.05 just to be certain. Then it will likely be necessary to…

Start over with larger member dimensions.

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