Pertemuan 10 Slope Deflection Method Matakuliah : S0114 / Rekayasa Struktur Tahun : 2006 Versi : 1 Pertemuan 10 Slope Deflection Method
Learning Outcomes Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu : Mahasiswa dapat menghitung struktur dengan metode slope deflection method
Materi 1 : Analisa Balok Menerus Outline Materi Materi 1 : Analisa Balok Menerus
Analisa BAlok Menerus A B C D Contoh: Balok menerus lihat gambar. Gambar diagram M, D. 3 12 24 A B C D 3 I 10 I 2 I 4 2 t/m’ 3 t/m’ P = 20 t 18 t 6 t Kr AB = 3I / 12 = 3 Kr BC = 10I / 24 = 5 Kr CD = 2I / 12 = 2 FEMAB = - FEMAB = 1/12.3.122 = 36 FEMBC = - FEMCB = 1/12.2.242 +1/8.20.24 = 156 FEMCD = 18.4.82/122 = 32 FEMDC = -18.42.8/122 = -16
Pers. slope deflection MAB = 36 + 3 (-2A - B) = 36 - 6A - 3B MBA = - 36 + 3 (-2B - A) = - 36 - 3A - 6B MBC = 156 + 5 (-2B - C) = 156 - 10B - 5C MCB = -156 + 5 (-2C - B) = -156 - 5B - 10C MCD = 32 + 2 (-2C - D) = 32 - 4C - 2D MDC = -16 + 2 (-2D - C) = -16 - 2C - 4D
Pers. MAB =0-6A- 3B =- 36 MBA+MBC=0-3A-16B-5C=-120 MCB+MCD=0-5B-14C-2D=-124 MDC+ 18=0-2C- 4D =- 2 (2 1) -240+32B+10C-3B = - 36 29B+10C=204 …...... (a) (4 3) - 1+C- 5B-14C-124 = 0 -5B-13C=125 ………(b) 37.7B + 13C = 265.2 -5B - 13C = 125 32,7B= 390,2 B= 11,93 C= -14,205 A= 0,03 D= 7,6
MAB= 36 – 6 . 0,03 – 3 . 11,9 = 0=0=0 MBA=-36 – 6 . 11,9 – 3 . 0,03 = -107,7 MBC= 156 – 10 . 11,9 – 5 . 14 = 107,7 MCB=-156 – 10.14,2 – 5.11,9 = -73,62 MCD= 32 – 4 . –14 – 2 . 7,6 = 73,62 MDC= -16 – 4 .7,6 – 2 . –14,2 = -18 =0
Gambar Diagram Gaya Dalam 18 34 12 6 2t/m’ 3t/m’ 107.7 73.6 20 8,9 +1,4 4,6 1,4 -4,6 16,6 32,6 35,4 26,9 9,1 + - 8,58 11,42 107,7 173,74 73,6 7,08 D M