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Jawab : a). Apakah f S e > 0,50 f p u, bila ya f P S = f p u ( 1 – 0,5 X A P S X f p u

175. a =269. 269. 785. 140. 320. Z1=6 97,5. ZII =650,5. 140. 460. Jawaban : B. Jawab : a). Apakah f S e > 0,50 f p u, bila ya f P S = f p u ( 1 – 0,5 X A P S X f p u b. d X f c’

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Jawab : a). Apakah f S e > 0,50 f p u, bila ya f P S = f p u ( 1 – 0,5 X A P S X f p u

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  1. 175 a=269 269 785 140 320 Z1=6 97,5 ZII =650,5 140 460 Jawaban : B Jawab : a). Apakah f S e > 0,50 f p u, bila ya f P S = f p u ( 1 – 0,5 X A P S X f p u b. d X f c’ f s e = 1.100 Mpa > 0,50 X 1.860 Mpa = 930 Mpa ( ya ! ) f p s = 1.860 ( 1 – ( 0,5 ) X 2.350 X 1.860 ) = 1.625 Mpa 460 X 785 X 48 b). Menentukan luas daerah tekan dan a. mm Bila a > tebal flen – tidak bisa dianggap bentuk persegi panjang T’ ( total ) = ( 2.350 ) ( 1.625 ) = 3.819 KN = F T’ = C’=3.819 KN = 3.819 X 103 N C’ = 0.85 x fc’x Ac Luas daerah tekan = Ac = C’ = 3.819 X 10 3 N / ( N / mm 2 ) 0,85 f c’ 0,85 X 48 Luas daerah tekan = Ac = 93,6 X 103 mm2 Luas daerah flen = 175 X 460 =(80,5 X 10 3mm2 Luas sisa ( luas yang ditanggung badan ) = 13,1 X 10 3 mm2 a = tebal flen + tinggi badan yang menanggung beban = 175 + 13,1 X 10 3 = 175 + 94 = 269 mm 140 a > tebal flen, 269 mm > 175 mm Tidak bisa dianggap berbentuk persegi panjang ( tetap bentuk profil I ). 30

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