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# EX11.2 - CEE 415 Example 11.2 Calculate points on the P n...

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Unformatted text preview: CEE 415 Example 11.2: Calculate points on the P n vs. M n interaction diagram for the column shown below. Use f c ' = 4000 psi, f y = 60 ksi. d 1 = 2.5 in. d 2 = 7.5 in. d 3 = 12.5 in. d 4 = 17.5 in. 12 No. 9 bars Column is 20” x 20” and A st = 12.0 in. 2 1. Pure axial load capacity and maximum permissible P n (max): P o = 0.85 f c ′ ( A g – A st ) + f y A st = 1320 kips + 720 kips = 2040 kips Recall, P n (max) = r P o = 0.80 (2040 k) = 1630 kips 2. Find P n and M n at balanced strain conditions (the balanced point) (Note: This is also the end of the compression-controlled region where φ is equal to 0.65.) 0.003 0.002 n.a. c(bal) d 4 = d t 1 CEE 415 The strain in layer 4 is assumed to be at tension yielding, so f s4 = 60 ksi (tens.) Depth of Whitney’s stress block, a = β 1 c(bal) = 0.85(10.5 in.) = 8.93 in. With this information, find all the forces in the concrete and the layers of reinforcement. C c = 0.85 f c ′ a b = 0.85 (4 ksi)(8.93 ″ )(20 ″ ) = 607 kips C s1 = A s1 ( f s1 – 0.85 f c ′ ) = (4.0 in....
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EX11.2 - CEE 415 Example 11.2 Calculate points on the P n...

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