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Problem 2. Plot (Excel plot) the nominal capacity interaction diagram of the column shown below by calculation (hand calculat
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23 24 25 26 27 SEPTEMBER 14 2019 SATURDAY bearing [1.5 16 fle=4801 Ry = 6opsi 6# 10 jebay 1.5 in clear cover. 13 B = 0.85-8.0SEPTEMBE 8 21 201 SEPTEMBE SATURDAY 2019 Cs: 0.003 AS Es (crd ) Cone C-1.5 = 0.003 * 7.36 x 29 000 X 2 320:16 ( [-2.5 Tot CTension Controlled CEU = 0.005, ECE EA SEPTEMBE 19 6 2011 THURSDAY NA S ed ( d = 1415 ed 0.00 3 + 0.005 5.487 0.003 a Calcul0.00257 18 SEPTEMBER 2019 WEDNESDAY Is = As Ry 220.8 K. 785.7 g R 6.5-0.85x7.66 N= 120.87 354.19 4220.8 = 0 to concrete leverSEPTEMBER 23 2019 MONDAY Moment [kipin)/ axial- (kip Point A Pure compressim Bizero tersim Balance point D Tensim control Pur3 SEPTEMBER 2019 16 MONDAY compression steel, cs Asry 7.36 x 60 = 220.8 R. 2 Concrete Cc = 0.85B, PE cb - 0.85 x 0.85 X 4 XBER 19 SEPTEMBER 2019 20 FRIDAY -003) 231.58 + 251.41 - 220.8k 262.19 R M concrete lever arm y ce upt - Bic 2 = 8- 0-85X5-437SEPTEMBE 2010 SUNDAY SEPTEMBE 2019 (8 4 3 + Cc Yet Csyls 15 Moment at this point m=0 Rin Plastic centroid ERI ER COMPLE UptSEPTEMBER 2019 22 SUNDAY 10 Concrete lever arm. Yec = ypt BIL = 8 -0.85 x 2.32 C - 7.014 o ouro Top steel lever arm = y es17 SEPTEMBE 2019 TUESDAY SE 2 3) Balance point mu NA = (= d (0003784 a C = 14,5 / 0.003 3 + 2.68x10-3 7 66 11, 14.5*(0.00 3)

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