# Ex-12.1 - CEE 415 Example 12.1 Slenderness evaluation of a...

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CEE 415 1 Example 12.1: Slenderness evaluation of a column in a nonsway frame loaded only by axial loads. P D = 230 kips and P L = 140 kips (15% sustained); P u = 500 kips u = 12 ft (assume k = 1.0, conservative) f ´ c = 4000 psi, f y = 60 ksi. Aim for a square column with R-type reinforcement pattern and g ≈ 0.03 Preliminary design results in: Recall, P n (max) = r x P o , where r = 0.80 for a tied column. P o = 0.85 x (4 ksi) x (196 – 6.32) in. 2 + 60 ksi x 6.32 in. 2 P o = 645 + 379 = 1020 kips So, P n (max) = 0.65 x 0.80 x 1020 kips = 530 kips > P u (initially o.k.) Check for slenderness: k u / r = (1 x 144 in.)/ (0.3 x 14 in.) = 34.3 ACI Code limit for a short column is (no end moments, so take M 1 /M 2 = 1.0): k u / r < 34 - 12 ( M 1 / M 2 ) = 34 – 12(1.0) = 22 34.3 (slender column) 14 in. 14 in. 8 No. 8 bars A st = 6.32 in. 2 g = 0.0322

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CEE 415 2 1.0 1.79 500 1 1 0.75(1510) 0.75 m ns u c C P P Therefore, the column is slender and we need to magnify the design moment. Check the minimum moment, which is defined as: M 2 (min) = P u ( 0.6 in. + 0.03 h) = 500 k (1.02 in.) = 510 kip – in.
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Ex-12.1 - CEE 415 Example 12.1 Slenderness evaluation of a...

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