1 / 8

Pertemuan 10 Slope Deflection Method

Pertemuan 10 Slope Deflection Method. Matakuliah : S0114 / Rekayasa Struktur Tahun : 2006 Versi : 1. Learning Outcomes. Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu : Mahasiswa dapat menghitung struktur dengan metode slope deflection method. Outline Materi.

sybill-jordan
Download Presentation

Pertemuan 10 Slope Deflection Method

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Pertemuan 10Slope Deflection Method Matakuliah : S0114 / Rekayasa Struktur Tahun : 2006 Versi : 1

  2. Learning Outcomes Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu : • Mahasiswa dapat menghitung struktur dengan metode slope deflection method

  3. Outline Materi • Materi 1 : Analisa Balok Menerus

  4. 18 t 4 P = 20 t 6 t 3 t/m’ 2 t/m’ C B D A 10 I 3 I 2 I 12 24 12 3 Analisa BAlok Menerus Contoh: Balok menerus lihat gambar. Gambar diagram M, D. 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

  5. Pers. slope deflection MAB = 36 + 3 (-2A - B) = 36 - 6A - 3B MBA = - 36 + 3 (-2B - A) = - 36 - 3A - 6B MBC = 156 + 5 (-2B - C) = 156 - 10B - 5C MCB = -156 + 5 (-2C - B) = -156 - 5B - 10C MCD = 32 + 2 (-2C - D) = 32 - 4C - 2D MDC = -16 + 2 (-2D - C) = -16 - 2C - 4D

  6. Pers. MAB =0-6A- 3B =- 36 MBA+MBC=0-3A-16B-5C=-120 MCB+MCD=0-5B-14C-2D=-124 MDC+ 18=0-2C- 4D =- 2 (2  1) -240+32B+10C-3B = - 36 29B+10C=204 …...... (a) (4  3) - 1+C- 5B-14C-124 = 0 -5B-13C=125 ………(b) • 37.7B + 13C = 265.2 • -5B - 13C = 125 32,7B= 390,2 B= 11,93 C= -14,205 A= 0,03 D= 7,6

  7. =0 =0 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

  8. 107.7 20 73.6 18 6 3t/m’ 18 2t/m’ 6 6 12 18 18 34 34 4,6 8,9 8,9 1,4 +1,4 -4,6 6 35,4 32,6 1,4 16,6 9,1 26,9 35,4 + 11,42 9,1 + 16,6 + + - 6 - D - 1,4 26,9 8,58 173,74 32,6 + + - - - M 7,08 18 73,6 107,7 Gambar Diagram Gaya Dalam

More Related