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Transcript of Tugas Ujian Khusus Analisa Struktur 1 - Handi Erfani (h1a109079)

TUGAS UJIAN KHUSUS ANALISA STRUKTUR I

TUGAS UJIAN KHUSUS ANALISA STRUKTUR IHSKK 412

q =1 t/m3 mStruktur 1

2 EI5 EI4 m4 m

METODE LEAST WORK CASTIGLIANO

Hcq =1 t/m3 m

Vc2 EI5 EI4 m4 m

Redundant Vc & Hc Bentang C- B (0 s 5)Ms - Vc sin .s - Hc cos .s + 0,4.s2Ms = Vc sin .s + Hc cos .s - 0,4.s2Ms = 0,8 Vc.s + 0,6 Hc.s + 0,4.s2 Bentang B - A (0 x 4)Ms 4Vc 3Hc + 0,4.x - VC.x + 0,8 Ms = 4Vc + 3Hc - 0,4.x + VC.x - 0,8

Eliminasi persamaan (1) & (2) )x 14,7 = 635,51628 Vc + 358, 68 Hc = 486,60528 ) x 24,4 = 514,535 Vc + 358,68 Hc = 555,4881 _ 120,98128 Vc = -68,88282 Vc = -0,57 t Vc = 0,57 t ( )

Substitusi Vc Ke persamaan 1

Va Q Vc = 0Va 0,4 -0,57 = 0Va = 0,97 t ( )

Ha P Hc = 0Ha 4,79 2,37 = 0Ha = 2,42 t ( )

Hc(3) Vc(8) + Q(6) +MA = 0-7,11 4,56 +2,4 + MA = 0MA= 9,27 tm ( )

GAMBAR BIDANG MOMEN, LINTANG & NORMAL

2 tmP1 = 3 tq = 1 t/mSTRUKTUR 2

2 t3,79m 3m 3m 1m 1m2.EI2.EI3.EI

h=1/8 qL2=1,8h=(P.a.b)/L=2,25METODE DALIL 3 MOMEN

A1 = 2/3 h.L = 4,5481 = 1,895A2 = h.L = 3,3752 = 2A3 = h.L = 1,1253 = 0,67Tinjau Bentang A-B-CMA(LAB/EIAB) + 2MB {(LBA/EIBA)+(LBC/EIBC)} + Mc (LAB/EIAB)= -(6A1a1/EIAB.LAB) -(6AXaX/EIXX.LXX) + {6(h1/LAB) + (h2/LBC)}0 + 2 MB (1,26 + 1) + 1,5 Mc = - 4,5484,53 MB + 1,5 Mc = - 4,548 .....(1)

Tinjau Bentang A-B-CMB(LBC/EIBC) + 2MC {(LCB/EICD)+(LBC/EIBC)} + MD (LCD/EICD)= -(6A2a2/EICB.LCB) -(6A3a3/EICD.LCD) + {6(h1/LBC) + (h2/LCD)}1,5 MB + 7 MC + 4 = - 5,0625 0,6281251,5 MB + 7 MC = - 9,69 .....(2)

Eliminasi (1)&(2) )x 7 = 31,71 MB + 10,5 MC = - 31,85 ) x 1,5 = 2,25 MB + 10,5 MC = -14,2825 _ 29,46 MB = - 17,315 MB = - 0,59 MB = 0,59 tm ( )Substitusi MB Ke persamaan 1 REAKSI PERLETAKAN

VA = 1,74 tVB = 2,05 0,22 = 1,83 tVC = 0,22 + 0,5625 = 0,78 tVD = 2,44 t

VA Q +VB + VC - 3 + VD = 01,7 3,79 + 1,83 +0,78 -3 + 2,44 = 0 0= 0 (ok!!) Ha 2 = 0Ha = 2 t ( )

GAMBAR BIDANG MOMEN, LINTANG & NORMAL

NAMA : HANDI ERFANINIM : H1A109079Page 1