Pertemuan 12 Slope Deflection Method

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1 Pertemuan 12 Slope Deflection Method
Matakuliah : S0114 / Rekayasa Struktur Tahun : 2006 Versi : 1 Pertemuan 12 Slope Deflection Method

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 Portal

4 Analisa Portal Portal ABCD, mempunyai EI sama, ℓ = 4 m,
lihat gambar. Gambarkan diagram M, L, N. Contoh: ℓ A B D C 1 t/m’ 2 t

5 K: AB = I / 4 = 1 BC = I / 4 = 1 CD = I / 4 = 1 R: AB =  / 4 = R  A
FEM BC= -FEM CB= 1/ = 1,33 tm K: AB = I / 4 = 1 BC = I / 4 = 1 CD = I / 4 = 1 R: AB =  / 4 = R  A B D C BC = = 0 CD =  / 4 = R

6 Persamaan slope deflection: MAB = 0+KAB(-2A-B+R) = -B+R
MBA = 0+KBA(-2B-A+R) = -2B+R MBC =1,33+KBC(-2B-C) = 1,33-2B-C MCB=-1,33+KCB(-2C-B)= -1,33-2C-B MCD =0+KCD(-2C-D+R) = -2C+R MDC =0+KDC(-2D-C+R) = -2C+R A, D jepit: A, D = 0 MCB+ MCD = ,33-2C+-2C+R = 0 ….(2) MBA + MBC = B+R+1,33-2B-C = 0 ….(1) HA = HD = MAB + MBA MDC + MCD 4 H = 0 P - HA - HD = 0 ….(3) 4

7 Dengan (1) (2) (3) didapat B, B, R (3 pers ., 3 anu)
-4B-C+R = -1,33 (2) -B-4C+R = 1,33 (-B+R-2B+R) 4 (-C+R-2C+R) 4 (3) - = 8 + 3B - 2R + 3C - 2R = 0 3B - 3C - 4R = -8 (4) B= - 4C + R - 1,33 -12C + 3R - 3, C - 4R = -8 - 9C - R = ,99 16B - 4R + 5,32 - C + R = -1,33 - 3R + 15C = -1,33 - 5,32 3R + 27C = 12 42C = 5,35 C = 0,127 R = 2,857 B = 1, 01

8 MAB= -1,01 + 2,857 = 1,847 MBA= -2,1,01 + 2,857 = 0,837 MBC= 1,33 - 2,02 - 0,127 = - 0,847 MCB= -1, ,127-1,01 = -2,594 MCD= -2. 0, ,857 = 2,603 MDC= -0, ,857 = 2,73 = 0

9 Gambar Diagram Gaya Dalam
1,15 + - 2,85 1,33 0,8 2,6 1,6 2,73 D M 2 0,85 1,15 2,85 0,8 1,8 1 t/m’ 2,6 2,7 A B C D P=2