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Pertemuan 9 Slope Deflection Method

Pertemuan 9 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.

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Pertemuan 9 Slope Deflection Method

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  1. Pertemuan 9Slope 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 : Penurunan Rumus Slope Deflection method

  4. P1 P2 A B C D E P1 P2 B A C D E Rumus Slope Deflection Method A = 0 D = 0 E = 0 Hanya ada B Free body :

  5. Moment Titik Momen batang Perjanjian tanda: Momen batang Momen titik + + - - Free body dengan anggapan momennya positip Moment Titik

  6. B MBA MBC MCB C MCE Dari free body kita tinjau titik B: MBD Syarat kesetimbangan MB=0 MBA + MBC + MBD = 0 ……….(1) Dari free body kita tinjau titik C : Dari kedua persamaan ini soal dapat diselesaikan Syarat kesetimbangan MC = 0 MCB + MCE = 0 ……….(2)

  7. MB’ A A B B MA’  P1 P2 B MAB A B A MBA l FEMBA FEMAB P1 P2 A B l Hubungan antara perputaran  dengan M  adalah positip bila searah jarum jam = + + A1 B1 A2 B2

  8. B1 A1 MA’ L MB’ B2 A2 MAB, MBA, A, B diambil tanda positip FEMAB, FEMBA diambil positip A = - A1 + A2 B = B1 - B2 A1 = M’A l/3EI B1 = MA’ l/6EI A2 = M’B l/6EI B2 = MB’ l/3EI A= -M’A.l/3EI + M’B.l/6EI x-2 -2A= 4M’A.l/3EI - 2M’B.l/6EI B= MA’.l/6EI - M’B.l/6EI B= MA’.l/6EI - M’B.l/6EI -2A- B= 3MA’.l/6EI MA’ = 2EI/l( -2 A- B) MB’ = 2EI/l( -2 B- A)

  9. MAB = FEMAB + 2EI/l ( -2 A - B) MBA = FEMBA + 2EI/l ( -2 B - A) FEMAB = momen primer dititik A dari batang AB FEMBA = momen primer dititik B dari batang AB Dapat disederhanakan menjadi: MAB = FEMAB + Kr AB ( -2 A - B) MBA = FEMBA + Kr BA ( -2 B - A)

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