EX11.1 - CEE 415 Example 11.1: Design a square tied column...

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CEE 415 1 2 212 . 0.85 ( 0.85 ) u g c g y c P A in r f f f       . 6 . 14 . 212 2 in in h b 2 1 0.85 3.47 . ( 0.85 ) u st g c yc P A A f in f f r  Example 11.1: Design a square tied column and a circular spiral column for concentric axial loads. Use f ´ c = 4000 psi, f y = 60 ksi and f yt = 60 ksi Assume the loads are: P D = 150 kips and P L = 200 kips Start design with g 0.02. P u = 1.2 P D + 1.6 P L = 500 kips Using class equation for A g for a tied column ( = 0.65 and r = 0.8): Use b = h = 15 in., then A g = 225 in. 2 Use second class equation to find total steel area, A st : Select eight No. 6 bars, so A st = 3.52 in. 2 (or 4 No. 9 bars, A st = 4.00 in. 2 ) Check (with eight No. 6 bars): P o = 0.85f ´ c (A g – A st ) + f y A st = 753 k + 211 k = 964 kips P n (max) = r P o = (0.65)(0.8)(964 k) = 501 kips > 500 kips (o.k.) Spacing of #3 ties: s 16 d b (longit.) = 16 (0.75) = 12 in. (governs) s 48 d b (tie) = 48 (0.375) = 18 in. s b & h = 15 in.
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CEE 415 2 2 173 in. 0.85 ( 0.85 ) u g c g y c P A r f f f       4 14.8 in. g A D  2 1 0.85 3.22 in. ( 0.85 ) u st g c yc P A A f f f r  Final design is given below: Using class equation for A
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This note was uploaded on 03/29/2011 for the course CEE 415 taught by Professor Wight during the Spring '11 term at University of Michigan.

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EX11.1 - CEE 415 Example 11.1: Design a square tied column...

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