Contoh Soal Analisa Matriks (Portal) Dari Tugas Mahasisa

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  • CONTOH SOAL ANALISA MATRIKS METODE KEKAKUAN BIASAPORTAL BIDANG (2D)UNIT : kN -m

    GAMBAR PORTAL

    Data properties penampang balok Data Beban

    Mutu beton fc' = 25 M.Pa P1 = 48 kNP2 = 26.87 kN

    Dimensi penampang q1 = 18 kN/mM1 = 98 kN.m

    Lebar balok 1, b1 = 0.25 m sudut P1 = 45 deg diukur dari sb x (sb global struktur)Tinggi balok 1, h1 = 0.3 mLuas pnp balok 1, A1 = 0.075 m2

    Lebar balok 2, b2 = 0.3 mTinggi balok 2, h2 = 0.6 mLuas pnp balok 2, A2 = 0.18 m2

    Lebar balok 3, b3 = 0.3 mTinggi balok 3, h3 = 0.4 mLuas pnp balok 3, A3 = 0.12 m2

    Lebar balok 4, b4 = 0.3 mTinggi balok 4, h4 = 0.4 mLuas pnp balok 4, A4 = 0.12 m2

    Panjang bentang

    Panjang bentang 1, L1 = 1.8 m

    Panjang bentang 2, L2 = 4.8 m

    Panjang bentang 3, L3 = 3.8 m

    Panjang bentang 4, L4 = 4.8 m

    Modulus Elastisitas, Ec = 23500000

    Inersia

    Inersia balok 1, = 1/12 x b1 x h1^3= 0.001

    Inersia balok2, = 1/12 x b2 x h2^3= 0.005

    Inersia balok3, = 1/12 x b3 x h3^3= 0.002

    Inersia balok4, = 1/12 x b4 x h4^3= 0.002

    I1

    I2

    I3

    I4

  • Penyusunan matriks-matriks

    GAMBAR 1. PORTAL DIKEKANGBEBAN HILANGKAN, GBRKAN DOF (misalkan DOF positif semua)

    GAMBAR 2 . GAMBAR PORTAL DIKEKANGBEBAN HILANGKAN, KECUALI BEBAN SELARAS DOF (BAIK BERUPA GAYA ATAU MOMEN)

    Susun matrik AD, ---> gaya luar yang selaras DOFDari perletakan didapat DOF = 6

    D1 0 0D2 0 0D3 0 0D4 0 0D5 0 0

    AD = D6 AD = M1 = -98D7 0 0D8 0 0D9 0 0D10 33.941D11 -33.941D12 0 0

    Selanjutnya menghitung reaksi ujung batang akibat gaya luar dan kekangan

    GAMBAR PORTAL DIKEKANGBEBAN DI MASUKAN, KECUALI YG SELARAS DOF,

    GAMBARKAN REAKSI PERLETAKAN AKIBAT KEKANGAN (ARL DAN ADL)

    Catatan : - ARL adalah reaksi perletakan semula, akibat beban primer - ADL adalah rekasi perletakan akibat kekangan pada posisi DOF atau gaya akibat beban terjepit yg selaras dg DOF - Penomoran ARL = Penomoran Reaksi Perletakan Struktur Awal - Penomoran ADL = Penomoran DOF

    Bh --> orde matriks d x 1 = AD4x1

    P1.sin - P1.cos

    YG SELARAS DOF DIHILANGKAN (P1 dan M1 DIKELUARKAN DAHULU),

  • Freebody A- E (Bentang 1) :

    GAMBAR FREEBODY SESUAI BENTANGAN YG DIKEKANGBESERTA REAKSI AKIBAT KEKANGAN (ARL DAN ADL) ---> urutan penomoran ARL & ADL sesuai sb lokal atau sb global ? Jawaban semenatara sb global karena penomoran ADL beradasarkan DOF dan DOF berdasarkan sb global

    Karena tidak ada gaya luar disepanjang bentang, maka reaksi perletakan tidak ada

    ARL1 = 0 = 0 kN

    ARL2 = 0 = 0 kN

    ADL1 = 0 = 0 kN.m

    ADL10 = 0 = 0 kN

    ADL11 = 0 = 0 kN

    ADL12 = 0 = 0 kN.m

    Freebody B- D (Bentang 2) :

    GAMBAR FREEBODY SESUAI BENTANGAN YG DIKEKANGBESERTA REAKSI AKIBAT KEKANGAN (ARL DAN ADL) ---> urutan penomoran ARL & ADL sesuai sb lokal atau sb global ? Jawaban semenatara sb global karena penomoran ADL beradasarkan DOF dan DOF berdasarkan sb global

    ARL 3 = 0 = 0 kN tanda (+) or (-) berdasarkan sb global tanda kuning = tidak sesuai uruatan sb globalapakah bisa ?

    ADL 2 = P2./ 2 = -13.435 kN tanda (+) or (-) berdasarkan sb global

    ADL 3 = P2.L2 / 8 = 16.122 kN.m tanda (+) or (-) berdasarkan sb global

    ADL 7 = P2./ 2 = -13.435 kN tanda (+) or (-) berdasarkan sb global

    ADL 8 = 0 = 0 kN tanda (+) or (-) berdasarkan sb global

    ADL 9 = P2.L2 / 8 = -16.122 kN.m tanda (+) or (-) berdasarkan sb global

    Freebody D - C (Bentang 3) :

    GAMBAR FREEBODY SESUAI BENTANGAN YG DIKEKANGBESERTA REAKSI AKIBAT KEKANGAN (ARL DAN ADL) ---> urutan penomoran ARL & ADL sesuai sb lokal atau sb global ? Jawaban semenatara sb global karena penomoran ADL beradasarkan DOF dan DOF berdasarkan sb global

    Karena tidak ada gaya luar disepanjang bentang, maka reaksi perletakan tidak ada

    ADL 7 = 0 = 0 kN

    ADL 8 = 0 = 0 kN

    ADL 9 = 0 = 0 kN.m

    ADL 4 = 0 = 0 kN

    ADL 5 = 0 = 0 kN

    ADL 6 = 0 = 0 kN.m

    Freebody E - D (Bentang 4) :

    GAMBAR FREEBODY SESUAI BENTANGAN YG DIKEKANGBESERTA REAKSI AKIBAT KEKANGAN (ARL DAN ADL) ---> urutan penomoran ARL & ADL sesuai sb lokal atau sb global ? Jawaban semenatara sb global karena penomoran ADL beradasarkan DOF dan DOF berdasarkan sb global

    ADL 10 = 0 = 0 kN tanda (+) or (-) berdasarkan sb global

    ADL 11 = q1.L4/2 = 43.2 kN tanda (+) or (-) berdasarkan sb global

    ADL 12 = = 34.56 kN.m tanda (+) or (-) berdasarkan sb global 1/12 x q1.L42

  • ADL 7 = 0 = 0 kN tanda (+) or (-) berdasarkan sb global

    ADL 8 = q1.L4/2 = 43.2 kN tanda (+) or (-) berdasarkan sb global

    ADL 9 = = -34.56 kN.m tanda (+) or (-) berdasarkan sb global

    Gabungkan seluruh reaksi ujung batang akibat gaya luar dan kekangan sehingga dapat disusun matriks ADL dan ARL

    Pada joint E reaksi gaya ujung batang yg dijumlahkan adalah sbb : urutan penomoran berdasarkan sb global

    ADL 10 Btg1 + ADL 10 Btg4 = 0 kN

    ADL 11 btg1 + ADL 11 btg4 = 43.2 kN

    ADL 12 btg1 + ADL 12 btg4 = 34.56 kN.m

    Pada joint D reaksi gaya ujung batang yg dijumlahkan adalah sbb : urutan penomoran berdasarkan sb global

    ADL 7 btg2 + ADL 7 btg3 + ADL 7 btg4 = -13.435 kN

    ADL 8 btg2 + ADL 8 btg3 + ADL 8 btg4 = 43.2 kN

    ADL 9 btg2 + ADL 9 btg3 + ADL 9 btg4 = -50.682 kN.m

    Susun matriks ADL dan ARL sbb :

    ARL 1 ARL 1 0ARL = ARL 2 = ARL 2 = 0

    ARL 3 ARL 3 0

    ADL 1 ADL 1 0ADL 2 ADL 2 -13.435ADL 3 ADL 3 16.122ADL 4 ADL 4 0ADL 5 ADL 5 0

    ADL = ADL 6 = ADL 6 = 0ADL 7 ADL 7 btg2 + ADL 7 btg3 + ADL 7 btg4 -13.435ADL 8 ADL 8 btg2 + ADL 8 btg3 + ADL 8 btg4 43.2ADL 9 ADL 9 btg2 + ADL 9 btg3 + ADL 9 btg4 -50.682ADL 10 ADL 10 btg1 + ADL 10 Btg4 0ADL 11 ADL 11 btg1 + ADL 11 Btg4 43.2ADL 12 ADL 12 btg1 + ADL 12 Btg4 34.56

    Susun matriks kekakuan [S] dan matriks reaksi perletakan [ARD]

    1/12 x q1.L42

  • Hitung matriks akibat displacement / perpindahan yaitu matriks kekakuan [S] dan matriksreaksi perletakan semula akibat displacement [ARD]

    Perpindahan 1 ---> yaitu akibat D1

    GAMBAR DISINI

    UNTUK MENDAPATKAN KEKAKUAN

    GAMBAR SEMUA FREEBODYGAMBAR FREEBODY BESERTA PERPINDAHANNYA DAN REAKSI AKIBATPERPINDAHAN TSB

    GAMBAR DISINI

    ARD 1.1 = 0 = 0 kN

    ARD 2.1 = - 6.Ec.Ix = -24479.1667 kN

    S 1.1 = 4. Ec.Ix = 29375 kN.mL1

    S 10.1 = S 10.1 btg1 + S 10.1 btg4 = = 24479.16667 kN

    S 11.1 = S 11.1 btg1 + S 11.1 btg4 = 0 kN

    S 12.1 = S 12.1 btg1 + S 12.1 btg4 = = 14687.5 kN.mL1

    ARD 3.1 = 0 = 0 kN

    S 2.1 = 0 = 0 kN

    S 3.1 = 0 = 0 kN.m

    S 7.1 = S 7.1 btg2 + S 7.1 btg3 + S 7.1 btg4 = 0 kN

    S 8.1 = S 8.1 btg2 + S 8.1 btg3 + S 8.1 btg4 = 0 kN

    S 9.1 = S 9.1 btg2 + S 9.1 btg3 + S 9.1 btg4 = 0 kN.m

    S 4.1 = 0 = 0 kN

    S 5.1 = 0 = 0 kN

    S 6.1 = 0 = 0 kN

    GAMBAR PERPINDAHAN 1 SATUAN (PUTARAN ROTASI) PD TITIK A

    L12

    6.Ec.Ix + 0L12

    2. Ec.Ix + 0

  • Perpindahan 2 ---> yaitu akibat D2

    GAMBAR DISINI

    UNTUK MENDAPATKAN KEKAKUAN

    GAMBAR SEMUA FREEBODYGAMBAR FREEBODY BESERTA PERPINDAHANNYA DAN REAKSI AKIBATPERPINDAHAN TSB

    GAMBAR DISINI

    ARD 1.2 = 0 = 0 kN

    ARD 2.2 = 0 = 0 kN

    S 1.2 = 0 = 0 kN.m

    S 10.2 = S 10.2 btg1 + S 10.2 btg4 = 0 kN

    S 11.2 = S 11.2 btg1 + S 11.2 btg4 = 0 kN

    S 12.2 = S 12.2 btg1 + S 12.2 btg4 = 0 kN.m

    ARD 3.2 = 0 = 0 kN

    S 2.2 = 12.Ec.Ix = 13769.53125 kN

    S 3.2 = = -33046.875 kN.m

    S 7.2 = S 7.2 btg2 + S 7.2 btg3 + S 7.2 btg4 = -12.Ec.Ix = -13769.5313 kN

    S 8.2 = S 8.2 btg2 + S 8.2 btg3 + S 8.2 btg4 = 0 kN

    S 9.2 = S 9.2 btg2 + S 9.2 btg3 + S 9.2 btg4 = = -33046.875 kN.m

    S 4.2 = 0 = 0 kN

    S 5.2 = 0 = 0 kN

    S 6.2 = 0 = 0 kN.m

    Perpindahan 3 ---> yaitu akibat D3

    GAMBAR DISINI

    GAMBAR PERPINDAHAN 1 SATUAN (TRANSLASI) PD TITIK B

    L23

    - 6.Ec.Ix L22

    L23

    - 6.Ec.Ix L22

  • UNTUK MENDAPATKAN KEKAKUAN

    GAMBAR SEMUA FREEBODYGAMBAR FREEBODY BESERTA PERPINDAHANNYA DAN REAKSI AKIBATPERPINDAHAN TSB

    GAMBAR DISINI

    ARD 1.3 = 0 = 0 kN

    ARD 2.3 = 0 = 0 kN

    S 1.3 = 0 = 0 kN.m

    S 10.3 = S 10.3 btg1 + S 10.3 btg4 = 0 kN

    S 11.3 = S 11.3 btg1 + S 11.3 btg4 = 0 kN

    S 12.3 = S 12.3 btg1 + S 12.3 btg4 = 0 kN.m

    ARD 3.3 = 0 = 0 kN

    S 2.3 = - 6.Ec.Ix = -33046.875 kN

    S 3.3 = = 105750.000 kN.mL2

    S 7.3 = S 7.2 btg2 + S 7.2 btg3 + S 7.2 btg4 = = 33046.875 kN

    S 8.3 = S 8.2 btg2 + S 8.2 btg3 + S 8.2 btg4 = 0 kN

    S 9.3 = S 9.2 btg2 + S 9.2 btg3 + S 9.2 btg4 = = 52875 kN.mL2

    S 4.3 = 0 = 0 kN

    S 5.3 = 0 = 0 kN

    S 6.3 = 0 = 0 kN.m

    Perpindahan 4 ---> yaitu akibat D4

    GAMBAR DISINI

    GAMBAR PERPINDAHAN 1 SATUAN (ROTASI) PD TITIK B

    L22

    4.Ec.Ix

    6.Ec.Ix L22

    2.Ec.Ix

  • UNTUK MENDAPATKAN KEKAKUAN

    GAMBAR SEMUA FREEBODYGAMBAR FREEBODY BESERTA PERPINDAHANNYA DAN REAKSI AKIBATPERPINDAHAN TSB

    GAMBAR DISINI

    ARD 1.4 = 0 = 0 kN

    ARD 2.4 = 0 = 0 kN

    S 1.4 = 0 = 0 kN.m

    S 10.4 = S 10.4 btg1 + S 10.4 btg4 = 0 kN

    S 11.4 = S 11.4 btg1 + S 11.4 btg4 = 0 kN

    S 12.4 = S 12.4 btg1 + S 12.4 btg4 = 0 kN.m

    ARD 3.4 = 0 = 0 kN

    S 2.4 = 0 = 0 kN

    S 3.4 = 0 = 0 kN.m

    S 7.4 = S 7.4 btg2 + S 7.4 btg3 + S 7.4 btg4 = - Ec.A = -742105.263 kNL3

    S 8.4 = S 8.4 btg2 + S 8.4 btg3 + S 8.4 btg4 = 0 kN

    S 9.4 = S 9.4 btg2 + S 9.4 btg3 + S 9.4 btg4 = 0 kN.m

    S 4.4 = Ec.A = 742105.2632 kNL3

    S 5.4 = 0 = 0 kN

    S 6.4 = 0 = 0 kN.m

    Perpindahan 5 ---> yaitu akibat D5

    GAMBAR DISINI

    UNTUK MENDAPATKAN KEKAKUAN

    GAMBAR SEMUA FREEBODYGAMBAR FREEBODY BESERTA PERPINDAHANNYA DAN REAKSI AKIBATPERPINDAHAN TSB

    GAMBAR PERPINDAHAN 1 SATUAN (TRANSLASI) PD TITIK C

    GAMBAR PERPINDAHAN 1 SATUAN (TRANSLASI) PD TITIK C

  • GAMBAR DISINI

    ARD 1.5 = 0 = 0 kN

    ARD 2.5 = 0 = 0 kN

    S 1.5 = 0 = 0 kN.m

    S 10.5 = S 10.5 btg1 + S 10.5 btg4 = 0 kN

    S 11.5 = S 11.5 btg1 + S 11.5 btg4 = 0 kN

    S 12.5 = S 12.5 btg1 + S 12.5 btg4 = 0 kN.m

    ARD 3.5 = 0 = 0 kN

    S 2.5 = 0 = 0 kN

    S 3.5 = 0 = 0 kN.m

    S 7.5 = S 7.5 btg2 + S 7.5 btg3 + S 7.5 btg4 = = 0 kN

    S 8.5 = S 8.5 btg2 + S 8.5 btg3 + S 8.5 btg4 - 12.Ec.Ix = -8222.773 kN

    S 9.5 = S 9.5 btg2 + S 9.5 btg3 + S 9.5 btg4 = - 6.Ec.Ix = -15623.2687 kN.m

    S 4.5 = 0 = 0 kN

    S 5.5 = 12.Ec.Ix = 8222.772999 kN

    S 6.5 = - 6.Ec.Ix = -15623.2687 kN.m

    Perpindahan 6 ---> yaitu akibat D6

    GAMBAR DISINI

    UNTUK MENDAPATKAN KEKAKUAN

    GAMBAR SEMUA FREEBODYGAMBAR FREEBODY BESERTA PERPINDAHANNYA DAN REAKSI AKIBATPERPINDAHAN TSB

    GAMBAR DISINI

    L33

    L32

    L33

    L32

    GAMBAR PERPINDAHAN 1 SATUAN (ROTASI) PD TITIK C

  • ARD 1.6 = 0 = 0 kN

    ARD 2.6 = 0 = 0 kN

    S 1.6 = 0 = 0 kN.m

    S 10.6 = S 10.6 btg1 + S 10.6 btg4 = 0 kN

    S 11.6 = S 11.6 btg1 + S 11.6 btg4 = 0 kN

    S 12.6 = S 12.6 btg1 + S 12.6 btg4 = 0 kN.m

    ARD 3.6 = 0 = 0 kN

    S 2.6 = 0 = 0 kN

    S 3.6 = 0 = 0 kN.m

    S 7.6 = S 7.6 btg2 + S 7.6 btg3 + S 7.6 btg4 = = 0 kN

    S 8.6 = S 8.6 btg2 + S 8.6 btg3 + S 8.6 btg4 6.Ec.Ix = 15623.2687 kN

    S 9.6 = S 9.6 btg2 + S 9.6 btg3 + S 9.6 btg4 = 2.Ec.Ix = 19789.47368 kN.mL3

    S 4.6 = 0 = 0 kN

    S 5.6 = - 6.Ec.Ix = -15623.2687 kN

    S 6.6 = 4.Ec.Ix = 39578.94737 kN.mL3

    Perpindahan 7 ---> yaitu akibat D7

    GAMBAR DISINI

    UNTUK MENDAPATKAN KEKAKUAN

    GAMBAR SEMUA FREEBODYGAMBAR FREEBODY BESERTA PERPINDAHANNYA DAN REAKSI AKIBATPERPINDAHAN TSB

    GAMBAR DISINI

    ARD 1.7 = 0 = 0 kN

    L32

    L32

    GAMBAR PERPINDAHAN 1 SATUAN (TRANSLASI) PD TITIK D

  • ARD 2.7 = 0 = 0 kN

    S 1.7 = 0 = 0 kN.m

    S 10.7 = S 10.7 btg1 + S 10.7 btg4 = 0 - Ec.A4 = -587500 kNL4

    S 11.7 = S 11.7 btg1 + S 11.7 btg4 = 0 kN

    S 12.7 = S 12.7 btg1 + S 12.7 btg4 = 0 kN.m

    ARD 3.7 = 0 = 0 kN

    S 2.7 = = -13769.53125 kN

    S 3.7 = 6.Ec.Ix = 33046.875 kN.m

    S 7.7 = S 7.7 btg2 + S 7.7 btg3 + S 7.7 btg4 = Ec.A4 = 1343374.794 kNL3 L4

    S 8.7 = S 8.7 btg2 + S 8.7 btg3 + S 8.7 btg4 = 0 kN

    S 9.7 = S 9.7 btg2 + S 9.7 btg3 + S 9.7 btg4 = 0 + 0 = 33046.875 kN.m

    S 4.7 = - Ec.A3 = -742105.2632 kNL3

    S 5.7 = 0 = 0 kN

    S 6.7 = 0 = 0 kN.m

    Perpindahan 8 ---> yaitu akibat D8

    GAMBAR DISINI

    UNTUK MENDAPATKAN KEKAKUAN

    GAMBAR SEMUA FREEBODYGAMBAR FREEBODY BESERTA PERPINDAHANNYA DAN REAKSI AKIBATPERPINDAHAN TSB

    GAMBAR DISINI

    ARD 1.8 = 0 = 0 kN

    ARD 2.8 = 0 = 0 kN

    S 1.8 = 0 = 0 kN.m

    - 12.Ec.Ix L23

    L22

    12.Ec.Ix + Ec.A3 +L23

    6.Ec.Ix +L22

    GAMBAR PERPINDAHAN 1 SATUAN (TRANSLASI) PD TITIK D

  • S 10.8 = S 10.8 btg1 + S 10.8 btg4 = 0 kN

    S 11.8 = S 11.8 btg1 + S 11.8 btg4 = 0 - 12.Ec.Ix = -4079.861111 kN

    S 12.8 = S 12.8 btg1 + S 12.8 btg4 0 - 6.Ec.Ix = -9791.666667 kN.m

    ARD 3.8 = S 8.8 = -881250 kN

    S 2.8 = 0 = 0 kN

    S 3.8 = 0 = 0 kN.m

    S 7.8 = S 7.8 btg2 + S 7.8 btg3 + S 7.8 btg4 = = 0 kN

    S 8.8 = S 8.8 btg2 + S 8.8 btg3 + S 8.8 btg4 = = 893552.6341 kNL2

    S 9.8 = S 9.8 btg2 + S 9.8 btg3 + S 9.8 btg4 = 0 + = 5831.602031 kN.m

    S 4.8 = 0 = 0 kN

    S 5.8 = - 12.Ec.Ix = -8222.772999 kN

    S 6.8 = 6.Ec.Ix = 15623.2687 kN.m

    Perpindahan 9 ---> yaitu akibat D9

    GAMBAR DISINI

    UNTUK MENDAPATKAN KEKAKUAN

    GAMBAR SEMUA FREEBODYGAMBAR FREEBODY BESERTA PERPINDAHANNYA DAN REAKSI AKIBATPERPINDAHAN TSB

    GAMBAR DISINI

    ARD 1.9 = 0 = 0 kN

    ARD 2.9 = 0 = 0 kN -33046.875 -6.Ec.Ix

    L43

    L42

    Ec.A2 + 12.Ec.Ix + 12.Ec.Ix +L33 L43

    6.Ec.Ix - 6.Ec.Ix +L32 L42

    L33

    L32

    GAMBAR PERPINDAHAN 1 SATUAN (ROTASI) PD TITIK D

  • S 1.9 = 0 = 0 kN.m

    S 10.9 = S 10.9 btg1 + S 10.9 btg4 = 0 kN

    S 11.9 = S 11.9 btg1 + S 11.9 btg4 = 6.Ec.Ix = 9791.666667 kN

    S 12.9 = S 12.9 btg1 + S 12.9 btg4 = 2.Ec.Ix = 15666.66667 kN.mL4

    ARD 3.9 = S 8.9 = 0 kN 5831.60203139

    S 2.9 = -6.Ec.Ix = -33046.875 kN

    S 3.9 = 2.Ec.Ix = 52875 kN.mL2

    S 7.9 = S 7.9 btg2 + S 7.9 btg3 + S 7.9 btg4 = 6.Ec.Ix + 0 +0 = 33046.875 kN

    S 8.9 = S 8.9 btg2 + S 8.9 btg3 + S 8.9 btg4 = 0 + 6.Ec.Ix = 5831.602031 kN

    S 9.9 = S 9.9 btg2 + S 9.9 btg3 + S 9.9 btg4 = = 176662.2807 kN.mL2 L3 L4

    S 4.9 = 0 = 0 kN

    S 5.9 = -6.Ec.Ix = -15623.2687 kN

    S 6.9 = 2.Ec.Ix = 19789.47368 kN.mL3

    Perpindahan 10 ---> yaitu akibat D10

    GAMBAR DISINI

    UNTUK MENDAPATKAN KEKAKUAN

    GAMBAR SEMUA FREEBODYGAMBAR FREEBODY BESERTA PERPINDAHANNYA DAN REAKSI AKIBATPERPINDAHAN TSB

    GAMBAR DISINI

    L22

    L42

    L22

    L22

    6.Ec.Ix -L32 L42

    4.Ec.Ix + 4.Ec.Ix + 4.Ec.Ix +

    L32

    GAMBAR PERPINDAHAN 1 SATUAN (TRANSLASI) PD TITIK E

  • ARD 1.10 = 0 = 0 kN

    ARD 2.10 = -12.Ec.Ix = -27199.07407 kN

    S 1.10 = 6.Ec.Ix = 24479.16667 kN.m

    S 10.10 = S 10.10 btg1 + S 10.10 btg4 = Ec.A4 = 614699.0741 kNL4

    S 11.10 = S 11.10 btg1 + S 11.10 btg4 0 = 0 kN

    S 12.10 = S 12.10 btg1 + S 12.10 btg4 = 0 = 24479.16667 kN.m

    ARD 3.10 = S 8.10 = 0 kN

    S 2.10 = 0 = 0 kN

    S 3.10 = 0 = 0 kN.m

    S 7.10 = S 7.10 btg2 + S 7.10 btg3 + S 7.10 btg4 = 0 + 0 - Ec.A4 = -587500 kNL4

    S 8.10 = S 8.10 btg2 + S 8.10 btg3 + S 8.10 btg4 = 0 kN

    S 9.10 = S 9.10 btg2 + S 9.10 btg3 + S 9.10 btg4 = 0 kN.m

    S 4.10 = 0 = 0 kN

    S 5.10 = 0 = 0 kN

    S 6.10 = 0 = 0 kN.m

    Perpindahan 11 ---> yaitu akibat D11

    GAMBAR DISINI

    UNTUK MENDAPATKAN KEKAKUAN

    GAMBAR SEMUA FREEBODYGAMBAR FREEBODY BESERTA PERPINDAHANNYA DAN REAKSI AKIBATPERPINDAHAN TSB

    GAMBAR DISINI

    L13

    L12

    12.Ec.Ix +L13

    6.Ec.Ix +L12

    GAMBAR PERPINDAHAN 1 SATUAN (TRANSLASI) PD TITIK E

  • ARD 1.11 = = -979166.667 kN DISINI SALAHNYA PERHATIKAN .......L1

    ARD 2.11 = 0 = 0 kN

    S 1.11 = 0 = 0 kN.m

    S 10.11 = S 10.11 btg1 + S 10.11 btg4 = 0 kN

    S 11.11 = S 11.11 btg1 + S 11.11 btg4 = 12.Ec.Ix = 983246.5278 kNL1

    S 12.11 = S 12.11 btg1 + S 12.11 btg4 = 0 + 6.Ec.Ix = 9791.666667 kN.m

    ARD 3.11 = S 8.11 = 0 kN DISINI SALAHNYA PERHATIKAN .......

    S 2.11 = 0 = 0 kN

    S 3.11 = 0 = 0 kN.m

    S 7.11 = S 7.11 btg2 + S 7.11 btg3 + S 7.11 btg4 = 0 kN

    S 8.11 = S 8.11 btg2 + S 8.11 btg3 + S 8.11 btg4 = 0 + 0 - 12.Ec.Ix = -4079.861111 kN

    S 9.11 = S 9.11 btg2 + S 9.11 btg3 + S 9.11 btg4 = 0 + 0 + 6.Ec.Ix = 9791.666667 kN.m

    S 4.11 = 0 = 0 kN

    S 5.11 = 0 = 0 kN

    S 6.11 = 0 = 0 kN.m

    Perpindahan 12 ---> yaitu akibat D12

    GAMBAR DISINI

    UNTUK MENDAPATKAN KEKAKUAN

    GAMBAR SEMUA FREEBODYGAMBAR FREEBODY BESERTA PERPINDAHANNYA DAN REAKSI AKIBATPERPINDAHAN TSB

    -Ec.A1 +

    Ec.A1 +L43

    L42

    L43

    L42

    GAMBAR PERPINDAHAN 1 SATUAN (ROTASI) PD TITIK E

  • GAMBAR DISINI

    ARD 1.12 = 0 = 0 kN

    ARD 2.12 = -6.Ec.Ix = -24479.16667 kN

    S 1.12 = 2.Ec.Ix = 14687.5 kN.mL1

    S 10.12 = S 10.12 btg1 + S 10.12 btg4 = = 24479.16667 kN

    S 11.12 = S 11.12 btg1 + S 11.12 btg4 = 0 + 6.Ec.Ix = 9791.666667 kN

    S 12.12 = S 12.12 btg1 + S 12.12 btg4 = 4.Ec.Ix = 60708.33333 kN.mL1 L4

    ARD 3.12 = 0 = 0 kN

    S 2.12 = 0 = 0 kN

    S 3.12 = 0 = 0 kN.m

    S 7.12 = S 7.12 btg2 + S 7.12 btg3 + S 7.12 btg4 = 0 kN

    S 8.12 = S 8.12 btg2 + S 8.12 btg3 + S 8.12 btg4 = 0 + 0 - 6.Ec.Ix = -9791.666667 kN

    S 9.12 = S 9.12 btg2 + S 9.12 btg3 + S 9.12 btg4 = 0 + 0 + 2.Ec.Ix = 15666.66667 kN.mL4

    S 4.12 = 0 = 0 kN

    S 5.12 = 0 = 0 kN

    S 6.12 = 0 = 0 kN.m

    Matriks kekakuan sbb :

    S 1.1 S 1.2 S 1.3 S 1.4 S 1.5 S 1.6 S 1.7 S 1.8 S 1.9 S 1.10 S 1.11 S 1.12S 2.1 S 2.2 S 2.3 S 2.4 S 2.5 S 2.6 S 2.7 S 2.8 S 2.9 S 2.10 S 2.11 S 2.12S 3.1 S 3.2 S 3.3 S 3.4 S 3.5 S 3.6 S 3.7 S 3.8 S 3.9 S 3.10 S 3.11 S 3.12S 4.1 S 4.2 S 4.3 S 4.4 S 4.5 S 4.6 S 4.7 S 4.8 S 4.9 S 4.10 S 4.11 S 4.12S 5.1 S 5.2 S 5.3 S 5.4 S 5.5 S 5.6 S 5.7 S 5.8 S 5.9 S 5.10 S 5.11 S 5.12

    = S 6.1 S 6.2 S 6.3 S 6.4 S 6.5 S 6.6 S 6.7 S 6.8 S 6.9 S 6.10 S 6.11 S 6.12S 7.1 S 7.2 S 7.3 S 7.4 S 7.5 S 7.6 S 7.7 S 7.8 S 7.9 S 7.10 S 7.11 S 7.12S 8.1 S 8.2 S 8.3 S 8.4 S 8.5 S 8.6 S 8.7 S 8.8 S 8.9 S 8.10 S 8.11 S 8.12S 9.1 S 9.2 S 9.3 S 9.4 S 9.5 S 9.6 S 9.7 S 9.8 S 9.9 S 9.10 S 9.11 S 9.12S 10.1 S 10.2 S 10.3 S 10.4 S 10.5 S 10.6 S 10.7 S 10.8 S 10.9 S 10.10 S 10.11 S 10.12S 11.1 S 11.2 S 11.3 S 11.4 S 11.5 S 11.6 S 11.7 S 11.8 S 11.9 S 11.10 S 11.11 S 11.12S 12.1 S 12.2 S 12.3 S 12.4 S 12.5 S 12.6 S 12.7 S 12.8 S 12.9 S 12.10 S 12.11 S 12.12

    L12

    6.Ec.Ix + 0L12

    L42

    4.Ec.Ix +

    L42

    S d x d

  • 29375 0 0 0 0 0 0 0 0 24479.1667 0 14687.50 13769.53125 -33046.875 0 0 0 -13769.5313 0 -33046.875 0 0 00 -33046.875 105750.000 0 0 0 33046.875 0 52875 0 0 00 0 0 742105.263 0 0 -742105.263 0 0 0 0 00 0 0 0 8222.77299898 -15623.268698 0 -8222.77299898 -15623.2687 0 0 0

    = 0 0 0 0 -15623.2686981 39578.947368 0 15623.2686981 19789.47368 0 0 00 -13769.5313 33046.875 -742105.26 0 0 1343374.794 0 33046.875 -587500 0 00 0 0 0 -8222.77299898 15623.268698 0 893552.63411 5831.602031 0 -4079.86111111 -9791.66666670 -33046.875 52875 0 -15623.2686981 19789.473684 33046.875 5831.60203139 176662.2807 0 9791.66666667 15666.6666667

    24479.167 0 0 0 0 0 -587500 0 0 614699.074 0 24479.16666670 0 0 0 0 0 0 -4079.86111111 9791.666667 0 983246.527778 9791.66666667

    14687.5 0 0 0 0 0 0 -9791.66666667 15666.66667 24479.1667 9791.66666667 60708.3333333

    Invers matriks kekakuan

    0.00018 -0.00030 -0.00002 -0.00020 -0.00008 -0.00002 -0.00020 0.00000 -0.00002 -0.00020 0.00000 0.00004-0.00030 0.00193 0.00033 0.00047 0.00092 0.00024 0.00047 0.00000 0.00024 0.00047 0.00000 -0.00018-0.00002 0.00033 0.00008 0.00004 0.00016 0.00004 0.00004 0.00000 0.00004 0.00004 0.00000 -0.00002-0.00020 0.00047 0.00004 0.00029 0.00014 0.00004 0.00029 0.00000 0.00004 0.00029 0.00000 -0.00008-0.00008 0.00092 0.00016 0.00014 0.00111 0.00035 0.00014 0.00000 0.00016 0.00014 0.00000 -0.00008

    = -0.00002 0.00024 0.00004 0.00004 0.00035 0.00014 0.00004 0.00000 0.00004 0.00004 0.00000 -0.00002-0.00020 0.00047 0.00004 0.00029 0.00014 0.00004 0.00029 0.00000 0.00004 0.00029 0.00000 -0.000080.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000-0.00002 0.00024 0.00004 0.00004 0.00016 0.00004 0.00004 0.00000 0.00004 0.00004 0.00000 -0.00002-0.00020 0.00047 0.00004 0.00029 0.00014 0.00004 0.00029 0.00000 0.00004 0.00029 0.00000 -0.000080.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.000000.00004 -0.00018 -0.00002 -0.00008 -0.00008 -0.00002 -0.00008 0.00000 -0.00002 -0.00008 0.00000 0.00004

    Matriks reaksi perletakan (ARD) karena displacement (akibat beban translasi & rotasi 1 satuan) sbb :

    ARD 1.1 ARD 1.2 ARD 1.3 ARD 1.4 ARD 1.5 ARD 1.6 ARD 1.7 ARD 1.8 ARD 1.9 ARD 1.10 ARD 1.11 ARD 1.12= ARD 2.1 ARD 2.2 ARD 2.3 ARD 2.4 ARD 2.5 ARD 2.6 ARD 2.7 ARD 2.8 ARD 2.9 ARD 2.10 ARD 2.11 ARD 2.12

    ARD 3.1 ARD 3.2 ARD 3.3 ARD 3.4 ARD 3.5 ARD 3.6 ARD 3.7 ARD 3.8 ARD 3.9 ARD 3.10 ARD 3.11 ARD 3.12

    0 0 0 0 0 0 0 0 0 0 -979166.666667 0= -24479.17 0 0 0 0 0 0 0 0 -27199.074 0 -24479.166667

    0 0 0 0 0 0 0 -881250 0 0 0 0

    Menghitung Displacement dari DOF

    FDL = ADL = ARL jika pada reaksi perletakan AD = FD ADL = FDL

    0.00018 -0.00030 -0.00002 -0.00020 -0.00008 -0.00002 -0.00020 0.00000 -0.00002 -0.00020 0.00000 0.00004 0 0-0.00030 0.00193 0.00033 0.00047 0.00092 0.00024 0.00047 0.00000 0.00024 0.00047 0.00000 -0.00018 0 -13.435-0.00002 0.00033 0.00008 0.00004 0.00016 0.00004 0.00004 0.00000 0.00004 0.00004 0.00000 -0.00002 0 16.122-0.00020 0.00047 0.00004 0.00029 0.00014 0.00004 0.00029 0.00000 0.00004 0.00029 0.00000 -0.00008 0 0-0.00008 0.00092 0.00016 0.00014 0.00111 0.00035 0.00014 0.00000 0.00016 0.00014 0.00000 -0.00008 0 0

    D = -0.00002 0.00024 0.00004 0.00004 0.00035 0.00014 0.00004 0.00000 0.00004 0.00004 0.00000 -0.00002 X -98 - 0-0.00020 0.00047 0.00004 0.00029 0.00014 0.00004 0.00029 0.00000 0.00004 0.00029 0.00000 -0.00008 0 -13.4350.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0 43.2-0.00002 0.00024 0.00004 0.00004 0.00016 0.00004 0.00004 0.00000 0.00004 0.00004 0.00000 -0.00002 0 -50.682-0.00020 0.00047 0.00004 0.00029 0.00014 0.00004 0.00029 0.00000 0.00004 0.00029 0.00000 -0.00008 33.941125 00.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 -33.94113 43.20.00004 -0.00018 -0.00002 -0.00008 -0.00008 -0.00002 -0.00008 0.00000 -0.00002 -0.00008 0.00000 0.00004 0 34.56

    0.00018 -0.00030 -0.00002 -0.00020 -0.00008 -0.00002 -0.00020 0.00000 -0.00002 -0.00020 0.00000 0.00004 0 -0.013611-0.00030 0.00193 0.00033 0.00047 0.00092 0.00024 0.00047 0.00000 0.00024 0.00047 0.00000 -0.00018 13.435 0.0374005-0.00002 0.00033 0.00008 0.00004 0.00016 0.00004 0.00004 0.00000 0.00004 0.00004 0.00000 -0.00002 -16.122 0.0037115-0.00020 0.00047 0.00004 0.00029 0.00014 0.00004 0.00029 0.00000 0.00004 0.00029 0.00000 -0.00008 0 0.0200732-0.00008 0.00092 0.00016 0.00014 0.00111 0.00035 0.00014 0.00000 0.00016 0.00014 0.00000 -0.00008 0 -0.007115

    D = -0.00002 0.00024 0.00004 0.00004 0.00035 0.00014 0.00004 0.00000 0.00004 0.00004 0.00000 -0.00002 X -98 = -0.006803-0.00020 0.00047 0.00004 0.00029 0.00014 0.00004 0.00029 0.00000 0.00004 0.00029 0.00000 -0.00008 13.435 0.02007320.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 -43.2 -8.3E-005-0.00002 0.00024 0.00004 0.00004 0.00016 0.00004 0.00004 0.00000 0.00004 0.00004 0.00000 -0.00002 50.682 0.0031017-0.00020 0.00047 0.00004 0.00029 0.00014 0.00004 0.00029 0.00000 0.00004 0.00029 0.00000 -0.00008 33.941125 0.0200275

    S d x d

    S-1

    ARDr x d

    ARDr x d

    D = S-1 .(AD - ADL) = S-1 . (FD - FDL)

  • 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 -77.14113 -4.8E-0050.00004 -0.00018 -0.00002 -0.00008 -0.00008 -0.00002 -0.00008 0.00000 -0.00002 -0.00008 0.00000 0.00004 -34.56 -0.006158

    -0.0140.0370.0040.020-0.007

    D = -0.0070.020

    -0.00010.0030.020

    -0.00005-0.006

    HITUNG REAKSI PERLETAKAN MATRIKS AR

    AR = ARL + ARD.D-0.014

    0.03740052440.00371148850.0200732322-0.007114549

    0 0 0 0 0 0 0 0 0 0 0 -979166.66667 0 -0.00680258240 + -24479.1667 0 0 0 0 0 0 0 0 -27199.0740741 0 -24479.167 x 0.02007323220 0 0 0 0 0 0 0 -881250 0 0 0 0 -8.282081E-005

    0.00310167290.020027496

    0 47.35528677 47.3552868 -4.836285E-0050 + -60.8111255 = -60.811125 -0.00615798840 72.98583873 72.9858387

    SB GLOBAL

  • CONTOH SOAL ANALISA MATRIKS METODE KEKAKUAN LANGSUNGPORTAL BIDANG (2D)UNIT : kN -m

    Data properties penampang balok Data Beban

    Mutu beton fc' = 25 M.Pa P1 = 48 kNP2 = 26.87 kN

    Dimensi penampang q1 = 18 kN/mM1 = 98 kN.m

    Lebar balok 1, b1 = 0.25 m sudut P1 = 45 deg diukur dari sb x (sb global struktur)Tinggi balok 1, h1 = 0.3 mLuas pnp balok 1, A1 = 0.075 m2

    Lebar balok 2, b2 = 0.3 mTinggi balok 2, h2 = 0.6 mLuas pnp balok 2, A2 = 0.18 m2

    Lebar balok 3, b3 = 0.3 mTinggi balok 3, h3 = 0.4 mLuas pnp balok 3, A3 = 0.12 m2

    Lebar balok 4, b4 = 0.3 mTinggi balok 4, h4 = 0.4 mLuas pnp balok 4, A4 = 0.12 m2

    Panjang bentang

    Panjang bentang 1, L1 = 1.8 mJarak beban, a1 = 0.9 mJarak beban, b1 = 0.9 m

    Panjang bentang 2, L2 = 4.8 mJarak beban, a2 = 2.4 mJarak beban, b2 = 2.4 m

    Panjang bentang 3, L3 = 3.8 mJarak beban, a3 = 1.9 mJarak beban, b3 = 1.9 m

    Panjang bentang 4, L4 = 4.8 mJarak beban, a4 = 2.4 mJarak beban, b4 = 2.4 m

  • Modulus Elastisitas, Ec = 23500000

    Inersia

    Inersia balok 1, = 1/12 x b1 x h1^3= 0.001

    Inersia balok2, = 1/12 x b2 x h2^3= 0.005

    Inersia balok3, = 1/12 x b3 x h3^3= 0.002

    Inersia balok4, = 1/12 x b4 x h4^3= 0.002

    Hitung matriks kekakuan batang 1 (SM1)

    Akibat displacement arah 1 --> D1 (translasi arah sb lokal 1)

    SM11 = Ec.A1 = 979166.667 SM41 = - Ec.A1 = -979166.67L1 L1

    SM21 = 0 = 0 SM51 = 0 = 0

    SM31 = 0 = 0 SM61 = 0 = 0

    Akibat displacement arah 2 --> D2 (translasi arah sb lokal 2)

    SM12 = 0 = 0 SM42 = 0 = 0

    SM22 = = 27199.0741 SM52 = = -27199.074

    SM32 = = 24479.1667 SM62 = = 24479.1667

    Akibat displacement arah 3 --> D3 (rotasi arah sb lokal 3)

    SM13 = 0 = 0 SM43 = 0 = 0

    SM23 = = 24479.1667 SM53 = = -24479.167

    SM33 = = 29375 SM63 = = 14687.5L1 L1

    Akibat displacement arah 4 --> D4 (translasi arah sb lokal 4)

    I1

    I2

    I3

    I4

    12.Ec.I 1 - 12.Ec.I 1

    L13 L13

    6.Ec.I 1 6.Ec.I 1

    L12 L12

    6.Ec.I 1 - 6.Ec.I 1

    L12 L12

    4.Ec.I 1 2.Ec.I 1

  • SM14 = - Ec.A1 = -979166.67 SM44 = Ec.A1 = 979166.667L1 L1

  • SM24 = 0 = 0 SM54 = 0 = 0

    SM34 = 0 = 0 SM64 = 0 = 0

    Akibat displacement arah 5 --> D5 (translasi arah sb lokal 5)

    SM15 = 0 = 0 SM45 = 0 = 0

    SM25 = = -27199.074 SM55 = = 27199.0741

    SM35 = = -24479.167 SM65 = = -24479.167

    Akibat displacement arah 6 --> D6 (rotasi arah sb lokal 6)

    SM16 = 0 = 0 SM46 = 0 = 0

    SM26 = = 24479.1667 SM56 = = -24479.167

    SM36 = = 14687.5 SM66 = = 29375L1 L1

    Susunan matriks kekakuan batang 1 (SM1) : --> catatan matriks kekakuan batang dalam arah sumbu lokal

    SM11 SM12 SM13 SM14 SM15 SM16 979166.666667 0 0 -979166.666667 0 0SM1 = SM21 SM22 SM23 SM24 SM25 SM26 SM1 = 0 27199.0740741 24479.1666667 0 -27199.0740741 24479.1666667

    SM31 SM32 SM33 SM34 SM35 SM36 0 24479.1666667 29375 0 -24479.1666667 14687.5SM41 SM42 SM43 SM44 SM45 SM46 -979166.666667 0 0 979166.666667 0 0SM51 SM52 SM53 SM54 SM55 SM56 0 -27199.0740741 -24479.1666667 0 27199.0740741 -24479.1666667SM61 SM62 SM63 SM64 SM65 SM66 0 24479.1666667 14687.5 0 -24479.1666667 29375

    Sudut batang 1 : = 90 deg

    Matriks transformasi rotasi :

    0 0 0 0 0 1 0 0 0 0R = 0 0 0 0 R = -1 0 0 0 0 0

    0 0 1 0 0 0 0 0 1 0 0 00 0 0 0 0 0 0 0 1 00 0 0 0 0 0 0 -1 0 00 0 0 0 0 1 0 0 0 0 0 1

    Transpose matrik Rotasi [R]

    0.00 -1 0 0 0 0

    = 1 0.00 0 0 0 00 0 1 0 0 0

    -12.Ec.I 1 12.Ec.I 1

    L13 L13

    - 6.Ec.I 1 - 6.Ec.I 1

    L12 L12

    6.Ec.I 1 - 6.Ec.I 1

    L12 L12

    2.Ec.I 1 4.Ec.I 1

    cos 1 sin 1 - sin 1 cos 1

    cos 1 sin 1 - sin 1 cos 1

    RT

  • 0 0 0 0.00 -1 00 0 0 1 0.00 00 0 0 0 0 1

    Matriks kekakuan struktur menjadi :

  • 6.123234E-017 -1 0 0 0 0 979166.666667 0 0 -979166.666667 0 0 0 1 01 6.123234E-017 0 0 0 0 0 27199.0740741 24479.1666667 0 -27199.0740741 24479.1666667 -1 0 0

    SMS 1 = 0 0 1 0 0 0 x 0 24479.1666667 29375 0 -24479.1666667 14687.5 x 0 0 10 0 0 6.123234E-017 -1 0 -979166.666667 0 0 979166.666667 0 0 0 0 00 0 0 1 6.123234E-017 0 0 -27199.0740741 -24479.1666667 0 27199.0740741 -24479.1666667 0 0 00 0 0 0 0 1 0 24479.1666667 14687.5 0 -24479.1666667 29375 0 0 0

    5.995667E-011 -27199.0740741 -24479.1666667 -5.995667E-011 27199.0740741 -24479.1666667 6.123234E-017 1 0 0 0 0 27199.07 0.00 -24479.17979166.666667 1.665463E-012 1.498917E-012 -979166.666667 -1.665463E-012 1.498917E-012 -1 6.123234E-017 0 0 0 0 0.00 979166.67 0.00

    SMS 1 = 0 24479.1666667 29375 0 -24479.1666667 14687.5 x 0 0 1 0 0 0 = -24479.17 0.00 29375.00-5.995667E-011 27199.0740741 24479.1666667 5.995667E-011 -27199.0740741 24479.1666667 0 0 0 6.123234E-017 1 0 -27199.07 0.00 24479.17-979166.666667 -1.665463E-012 -1.498917E-012 979166.666667 1.665463E-012 -1.498917E-012 0 0 0 -1 6.123234E-017 0 0.00 -979166.67 0.00

    0 24479.1666667 14687.5 0 -24479.1666667 29375 0 0 0 0 0 1 -24479.17 0.00 14687.50

    SM11 SM12 SM13 SM14 SM15 SM16 SM22 SM21 SM23SM1 = SM21 SM22 SM23 SM24 SM25 SM26 SMS1 = SM12 SM11 SM13

    SM31 SM32 SM33 SM34 SM35 SM36 Setelah tranfromasi menjadi SM32 SM31 SM33SM41 SM42 SM43 SM44 SM45 SM46 SM52 SM51 SM53SM51 SM52 SM53 SM54 SM55 SM56 SM42 SM41 SM43SM61 SM62 SM63 SM64 SM65 SM66 SM62 SM61 SM63

    Hitung matriks kekakuan batang 2 (SM2)

    Akibat displacement arah 1 --> D1 (translasi arah sb lokal 1)

    SM11 = Ec.A2 = 881250 SM41 = - Ec.A2 = -881250L2 L2

    SM21 = 0 = 0 SM51 = 0 = 0

    SM31 = 0 = 0 SM61 = 0 = 0

    Akibat displacement arah 2 --> D2 (translasi arah sb lokal 2)

    SM12 = 0 = 0 SM42 = 0 = 0

    SM22 = = 13769.5313 SM52 = = -13769.531

    SM32 = = 33046.875 SM62 = = 33046.875

    SMS = RT. SM . R

    12.Ec.I 2 - 12.Ec.I 2

    L23 L23

    6.Ec.I 2 6.Ec.I 2

    L22 L22

  • Akibat displacement arah 3 --> D3 (rotasi arah sb lokal 3)

    SM13 = 0 = 0 SM43 = 0 = 0

    SM23 = = 33046.875 SM53 = = -33046.875

    SM33 = = 105750 SM63 = = 52875L2 L2

    Akibat displacement arah 4 --> D4 (translasi arah sb lokal 4)

    SM14 = - Ec.A2 = -881250 SM44 = Ec.A2 = 881250L2 L2

    SM24 = 0 = 0 SM54 = 0 = 0

    SM34 = 0 = 0 SM64 = 0 = 0

    6.Ec.I 2 - 6.Ec.I 2

    L22 L22

    4.Ec.I 2 2.Ec.I 2

  • Akibat displacement arah 5 --> D5 (translasi arah sb lokal 5)

    SM15 = 0 = 0 SM45 = 0 = 0

    SM25 = = -13769.5 SM55 = = 13769.5

    SM35 = = -33046.9 SM65 = = -33046.9

    Akibat displacement arah 6 --> D6 (rotasi arah sb lokal 6)

    SM16 = 0 = 0 SM46 = 0 = 0

    SM26 = = 33046.875 SM56 = = -33046.875

    SM36 = = 52875 SM66 = = 105750L2 L2

    Susunan matriks kekakuan batang 2 (SM2) : --> catatan matriks kekakuan batang dalam arah sumbu lokal

    SM11 SM12 SM13 SM14 SM15 SM16 881250 0 0 -881250 0 0SM2 = SM21 SM22 SM23 SM24 SM25 SM26 SM2 = 0 13769.53125 33046.875 0 -13769.53125 33046.875

    SM31 SM32 SM33 SM34 SM35 SM36 0 33046.875 105750 0 -33046.875 52875SM41 SM42 SM43 SM44 SM45 SM46 -881250 0 0 881250 0 0SM51 SM52 SM53 SM54 SM55 SM56 0 -13769.53125 -33046.875 0 13769.53125 -33046.875SM61 SM62 SM63 SM64 SM65 SM66 0 33046.875 52875 0 -33046.875 105750

    Sudut batang 2 : = 90 deg

    Matriks transformasi rotasi :

    0 0 0 0 0 1 0 0 0 0R = 0 0 0 0 R = -1 0 0 0 0 0

    0 0 1 0 0 0 0 0 1 0 0 00 0 0 0 0 0 0 0 1 00 0 0 0 0 0 0 -1 0 00 0 0 0 0 1 0 0 0 0 0 1

    Transpose matrik Rotasi [R]

    6.123234E-017 -1 0 0 0 0

    = 1 6.123234E-017 0 0 0 00 0 1 0 0 00 0 0 6.123234E-017 -1 00 0 0 1 6.123234E-017 00 0 0 0 0 1

    -12.Ec.I 2 12.Ec.I 2

    L23 L23

    - 6.Ec.I 2 - 6.Ec.I 2

    L22 L22

    6.Ec.I 2 - 6.Ec.I 2

    L22 L22

    2.Ec.I 2 4.Ec.I 2

    cos 1 sin 1 - sin 1 cos 1

    cos 1 sin 1 - sin 1 cos 1

    RT

  • Matriks kekakuan struktur menjadi :

    6.123234E-017 -1 0 0 0 0 881250 0 0 -881250 0 0 6.123234E-017 1 01 6.123234E-017 0 0 0 0 0 13769.53125 33046.875 0 -13769.53125 33046.875 -1 6.123234E-017 0

    SMS 2 = 0 0 1 0 0 0 x 0 33046.875 105750 0 -33046.875 52875 x 0 0 10 0 0 6.123234E-017 -1 0 -881250 0 0 881250 0 0 0 0 00 0 0 1 6.123234E-017 0 0 -13769.53125 -33046.875 0 13769.53125 -33046.875 0 0 00 0 0 0 0 1 0 33046.875 52875 0 -33046.875 105750 0 0 0

    5.396100E-011 -13769.53125 -33046.875 -5.396100E-011 13769.53125 -33046.875 6.123234E-017 1 0 0 0 0 13770 0 -33047881250 8.431406E-013 2.023537E-012 -881250 -8.431406E-013 2.023537E-012 -1 6.123234E-017 0 0 0 0 0 881250 0

    SMS 2 = 0 33046.875 105750 0 -33046.875 52875 x 0 0 1 0 0 0 = -33047 0 105750-5.396100E-011 13769.53125 33046.875 5.396100E-011 -13769.53125 33046.875 0 0 0 6.123234E-017 1 0 -13770 0 33047

    -881250 -8.431406E-013 -2.023537E-012 881250 8.431406E-013 -2.023537E-012 0 0 0 -1 6.123234E-017 0 0 -881250 00 33046.875 52875 0 -33046.875 105750 0 0 0 0 0 1 -33047 0 52875

    Hitung matriks kekakuan batang 3 (SM3)

    Akibat displacement arah 1 --> D1 (translasi arah sb lokal 1)

    SM11 = Ec.A3 = 742105.263 SM41 = - Ec.A3 = -742105.26L3 L3

    SM21 = 0 = 0 SM51 = 0 = 0

    SM31 = 0 = 0 SM61 = 0 = 0

    Akibat displacement arah 2 --> D2 (translasi arah sb lokal 2)

    SM12 = 0 = 0 SM42 = 0 = 0

    SM22 = = 8222.773 SM52 = = -8222.773

    SM32 = = 15623.2687 SM62 = = 15623.2687

    Akibat displacement arah 3 --> D3 (rotasi arah sb lokal 3)

    SM13 = 0 = 0 SM43 = 0 = 0

    SM23 = = 15623.2687 SM53 = = -15623.269

    SM33 = = 39578.9474 SM63 = = 19789.4737L3 L3

    SMS = RT. SM . R

    12.Ec.I 3 - 12.Ec.I 3

    L33 L33

    6.Ec.I 3 6.Ec.I 3

    L32 L32

    6.Ec.I 3 - 6.Ec.I 3

    L32 L32

    4.Ec.I 3 2.Ec.I 3

  • Akibat displacement arah 4 --> D4 (translasi arah sb lokal 4)

    SM14 = - Ec.A3 = -742105.26 SM44 = Ec.A3 = 742105.263L3 L3

    SM24 = 0 = 0 SM54 = 0 = 0

    SM34 = 0 = 0 SM64 = 0 = 0

    Akibat displacement arah 5 --> D5 (translasi arah sb lokal 5)

    SM15 = 0 = 0 SM45 = 0 = 0

    SM25 = = -8222.773 SM55 = = 8222.773

    SM35 = = -15623.269 SM65 = = -15623.269

    Akibat displacement arah 6 --> D6 (rotasi arah sb lokal 6)

    SM16 = 0 = 0 SM46 = 0 = 0

    SM26 = = 15623.2687 SM56 = = -15623.269

    SM36 = = 19789.4737 SM66 = = 39578.9474L3 L3

    Susunan matriks kekakuan batang 3 (SM3) : --> catatan matriks kekakuan batang dalam arah sumbu lokal

    SM11 SM12 SM13 SM14 SM15 SM16 742105.263158 0 0 -742105.263158 0 0SM3 = SM21 SM22 SM23 SM24 SM25 SM26 SM3 = 0 8222.77299898 15623.2686981 0 -8222.77299898 15623.2686981

    SM31 SM32 SM33 SM34 SM35 SM36 0 15623.2686981 39578.9473684 0 -15623.2686981 19789.4736842SM41 SM42 SM43 SM44 SM45 SM46 -742105.263158 0 0 742105.263158 0 0SM51 SM52 SM53 SM54 SM55 SM56 0 -8222.77299898 -15623.2686981 0 8222.77299898 -15623.2686981SM61 SM62 SM63 SM64 SM65 SM66 0 15623.2686981 19789.4736842 0 -15623.2686981 39578.9473684

    Sudut batang 3 : = 0 deg

    Matriks transformasi rotasi :

    0 0 0 0 1 0 0 0 0 0R = 0 0 0 0 R = 0 1 0 0 0 0

    0 0 1 0 0 0 0 0 1 0 0 00 0 0 0 0 0 0 1 0 00 0 0 0 0 0 0 0 1 00 0 0 0 0 1 0 0 0 0 0 1

    -12.Ec.I 3 12.Ec.I 3

    L33 L33

    - 6.Ec.I 3 - 6.Ec.I 3

    L32 L32

    6.Ec.I 3 - 6.Ec.I 3

    L32 L32

    2.Ec.I 3 4.Ec.I 3

    cos 1 sin 1 - sin 1 cos 1

    cos 1 sin 1 - sin 1 cos 1

  • Transpose matrik Rotasi [R]

    1 0 0 0 0 0

    = 0 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1

    Matriks kekakuan struktur menjadi :

    1 0 0 0 0 0 742105.263158 0 0 -742105.263158 0 0 1 0 00 1 0 0 0 0 0 8222.77299898 15623.2686981 0 -8222.77299898 15623.2686981 0 1 0

    SMS 3 = 0 0 1 0 0 0 x 0 15623.2686981 39578.9473684 0 -15623.2686981 19789.4736842 x 0 0 10 0 0 1 0 0 -742105.263158 0 0 742105.263158 0 0 0 0 00 0 0 0 1 0 0 -8222.77299898 -15623.2686981 0 8222.77299898 -15623.2686981 0 0 00 0 0 0 0 1 0 15623.2686981 19789.4736842 0 -15623.2686981 39578.9473684 0 0 0

    742105.263158 0 0 -742105.263158 0 0 1 0 0 0 0 0 742105.263158 0 00 8222.77299898 15623.2686981 0 -8222.77299898 15623.2686981 0 1 0 0 0 0 0 8222.77299898 15623.2686981

    SMS 3 = 0 15623.2686981 39578.9473684 0 -15623.2686981 19789.4736842 x 0 0 1 0 0 0 = 0 15623.2686981 39578.9473684-742105.263158 0 0 742105.263158 0 0 0 0 0 1 0 0 -742105.263158 0 0

    0 -8222.77299898 -15623.2686981 0 8222.77299898 -15623.2686981 0 0 0 0 1 0 0 -8222.77299898 -15623.26869810 15623.2686981 19789.4736842 0 -15623.2686981 39578.9473684 0 0 0 0 0 1 0 15623.2686981 19789.4736842

    Hitung matriks kekakuan batang 4 (SM4)

    Akibat displacement arah 1 --> D1 (translasi arah sb lokal 1)

    SM11 = Ec.A4 = 587500 SM41 = - Ec.A4 = -587500L4 L4

    SM21 = 0 = 0 SM51 = 0 = 0

    SM31 = 0 = 0 SM61 = 0 = 0

    Akibat displacement arah 2 --> D2 (translasi arah sb lokal 2)

    SM12 = 0 = 0 SM42 = 0 = 0

    SM22 = = 4079.86111 SM52 = = -4079.8611

    SM32 = = 9791.66667 SM62 = = 9791.66667

    RT

    SMS = RT. SM . R

    12.Ec.I 4 - 12.Ec.I 4

    L43 L43

    6.Ec.I 4 6.Ec.I 4

    L42 L42

  • Akibat displacement arah 3 --> D3 (rotasi arah sb lokal 3)

    SM13 = 0 = 0 SM43 = 0 = 0

    SM23 = = 9791.66667 SM53 = = -9791.6667

    SM33 = = 31333.3333 SM63 = = 15666.6667L4 L4

    Akibat displacement arah 4 --> D4 (translasi arah sb lokal 4)

    SM14 = - Ec.A4 = -587500 SM44 = Ec.A4 = 587500L4 L4

    SM24 = 0 = 0 SM54 = 0 = 0

    SM34 = 0 = 0 SM64 = 0 = 0

    Akibat displacement arah 5 --> D5 (translasi arah sb lokal 5)

    SM15 = 0 = 0 SM45 = 0 = 0

    SM25 = = -4079.8611 SM55 = = 4079.86111

    SM35 = = -9791.6667 SM65 = = -9791.6667

    Akibat displacement arah 6 --> D6 (rotasi arah sb lokal 6)

    SM16 = 0 = 0 SM46 = 0 = 0

    SM26 = = 9791.66667 SM56 = = -9791.6667

    SM36 = = 15666.6667 SM66 = = 31333.3333L4 L4

    Susunan matriks kekakuan batang 4 (SM4) : --> catatan matriks kekakuan batang dalam arah sumbu lokal

    SM11 SM12 SM13 SM14 SM15 SM16 587500 0 0 -587500 0 0SM4 = SM21 SM22 SM23 SM24 SM25 SM26 SM4 = 0 4079.86111111 9791.66666667 0 -4079.86111111 9791.66666667

    SM31 SM32 SM33 SM34 SM35 SM36 0 9791.66666667 31333.3333333 0 -9791.66666667 15666.6666667SM41 SM42 SM43 SM44 SM45 SM46 -587500 0 0 587500 0 0SM51 SM52 SM53 SM54 SM55 SM56 0 -4079.86111111 -9791.66666667 0 4079.86111111 -9791.66666667SM61 SM62 SM63 SM64 SM65 SM66 0 9791.66666667 15666.6666667 0 -9791.66666667 31333.3333333

    6.Ec.I 4 - 6.Ec.I 4

    L42 L42

    4.Ec.I 4 2.Ec.I 4

    -12.Ec.I 4 12.Ec.I 4

    L43 L43

    - 6.Ec.I 4 - 6.Ec.I 4

    L42 L42

    6.Ec.I 4 - 6.Ec.I 4

    L42 L42

    2.Ec.I 4 4.Ec.I 4

  • Sudut batang 4 : = 0 deg

    Matriks transformasi rotasi :

    0 0 0 0 1 0 0 0 0 0R = 0 0 0 0 R = 0 1 0 0 0 0

    0 0 1 0 0 0 0 0 1 0 0 00 0 0 0 0 0 0 1 0 00 0 0 0 0 0 0 0 1 00 0 0 0 0 1 0 0 0 0 0 1

    Transpose matrik Rotasi [R]

    1 0 0 0 0 0

    = 0 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1

    Matriks kekakuan struktur menjadi :

    1 0 0 0 0 0 587500 0 0 -587500 0 0 1 0 00 1 0 0 0 0 0 4079.86111111 9791.66666667 0 -4079.86111111 9791.66666667 0 1 0

    SMS 4 = 0 0 1 0 0 0 x 0 9791.66666667 31333.3333333 0 -9791.66666667 15666.6666667 x 0 0 10 0 0 1 0 0 -587500 0 0 587500 0 0 0 0 00 0 0 0 1 0 0 -4079.86111111 -9791.66666667 0 4079.86111111 -9791.66666667 0 0 00 0 0 0 0 1 0 9791.66666667 15666.6666667 0 -9791.66666667 31333.3333333 0 0 0

    587500 0 0 -587500 0 0 1 0 0 0 0 0 587500 0 00 4079.86111111 9791.66666667 0 -4079.86111111 9791.66666667 0 1 0 0 0 0 0 4079.86111111 9791.66666667

    SMS 4 = 0 9791.66666667 31333.3333333 0 -9791.66666667 15666.6666667 x 0 0 1 0 0 0 = 0 9791.66666667 31333.3333333-587500 0 0 587500 0 0 0 0 0 1 0 0 -587500 0 0

    0 -4079.86111111 -9791.66666667 0 4079.86111111 -9791.66666667 0 0 0 0 1 0 0 -4079.86111111 -9791.666666670 9791.66666667 15666.6666667 0 -9791.66666667 31333.3333333 0 0 0 0 0 1 0 9791.66666667 15666.6666667

    cos 1 sin 1 - sin 1 cos 1

    cos 1 sin 1 - sin 1 cos 1

    RT

    SMS = RT. SM . R

  • Susun MATRIKS KEKAKUAN STRUKTUR

    D1 D10 D11 D121 2 3 13 14 15

    27199 0 -24479 -27199 0 -24479 10 979167 0 0 -979167 0 2

    SMS 1 = -24479 0 29375 24479 0 14688 3 D1-27199 0 24479 27199 0 24479 13 D10

    0 -979167 0 0 979167 0 14 D11-24479 0 14688 24479 0 29375 15 D12

    D2 D3 D7 D8 D9 D1 D2 D3 D4 D5 D6 D74 5 6 10 11 12 ==> PENOMORAN SB GLOBAL

    13770 0 -33047 -13770 0 -33047 4 D2 D1 29375 0 0 0 0 0 00 881250 0 0 -881250 0 5 D2 0 13770 -33047 0 0 0 -13770

    SMS 2 = -33047 0 105750 33047 0 52875 6 D3 D3 0 -33047 105750 0 0 0 33047-13770 0 33047 13770 0 33047 10 D7 D4 0 0 0 742105 0 0 -742105

    0 -881250 0 0 881250 0 11 D8 D5 0 0 0 0 8223 -15623 0-33047 0 52875 33047 0 105750 12 D9 D6 0 0 0 0 -15623 39579 0

    D7 0 -13770 33047 -742105 0 0 1343375D8 0 0 0 0 -8223 15623 0

    D7 D8 D9 D4 D5 D6 D9 0 -33047 52875 0 -15623 19789 3304710 11 12 7 8 9 D10 24479 0 0 0 0 0 -587500

    742105.263158 0 0 -742105.263158 0 0 10 D7 D11 0 0 0 0 0 0 00 8222.77299898 15623.2686981 0 -8222.77299898 15623.2686981 11 D8 D12 14688 0 0 0 0 0 0

    SMS 3 = 0 15623.2686981 39578.9473684 0 -15623.2686981 19789.4736842 12 D9-742105.263158 0 0 742105.263158 0 0 7 D4

    0 -8222.77299898 -15623.2686981 0 8222.77299898 -15623.2686981 8 D50 15623.2686981 19789.4736842 0 -15623.2686981 39578.9473684 9 D6

    D10 D11 D12 D7 D8 D913 14 15 10 11 12

    587500 0 0 -587500 0 0 13 D100 4079.86111111 9791.66666667 0 -4079.86111111 9791.66666667 14 D11

    SMS 4 = 0 9791.66666667 31333.3333333 0 -9791.66666667 15666.6666667 15 D12-587500 0 0 587500 0 0 10 D7

    0 -4079.86111111 -9791.66666667 0 4079.86111111 -9791.66666667 11 D80 9791.66666667 15666.6666667 0 -9791.66666667 31333.3333333 12 D9

    29375 0 0 0 0 0 0 0 0 24479 0 146880 13770 -33047 0 0 0 -13770 0 -33047 0 0 00 -33047 105750 0 0 0 33047 0 52875 0 0 00 0 0 742105 0 0 -742105 0 0 0 0 00 0 0 0 8223 -15623 0 -8223 -15623 0 0 0

    SMS = 0 0 0 0 -15623 39579 0 15623 19789 0 0 00 -13770 33047 -742105 0 0 1343375 0 33047 -587500 0 00 0 0 0 -8223 15623 0 893553 5832 0 -4080 -97920 -33047 52875 0 -15623 19789 33047 5832 176662 0 9792 15667

    24479 0 0 0 0 0 -587500 0 0 614699 0 244790 0 0 0 0 0 0 -4080 9792 0 983247 9792

    14688 0 0 0 0 0 0 -9792 15667 24479 9792 60708

  • 0.0002 -0.0003 0.0000 -0.0002 -0.0001 0.0000 -0.0002 0.0000 0.0000 -0.0002 0.0000 0.0000

    -0.0003 0.0019 0.0003 0.0005 0.0009 0.0002 0.0005 0.0000 0.0002 0.0005 0.0000 -0.0002

    0.0000 0.0003 0.0001 0.0000 0.0002 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

    -0.0002 0.0005 0.0000 0.0003 0.0001 0.0000 0.0003 0.0000 0.0000 0.0003 0.0000 -0.0001

    -0.0001 0.0009 0.0002 0.0001 0.0011 0.0004 0.0001 0.0000 0.0002 0.0001 0.0000 -0.0001

    = 0.0000 0.0002 0.0000 0.0000 0.0004 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000-0.0002 0.0005 0.0000 0.0003 0.0001 0.0000 0.0003 0.0000 0.0000 0.0003 0.0000 -0.0001

    0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

    0.0000 0.0002 0.0000 0.0000 0.0002 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

    -0.0002 0.0005 0.0000 0.0003 0.0001 0.0000 0.0003 0.0000 0.0000 0.0003 0.0000 -0.0001

    0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

    0.0000 -0.0002 0.0000 -0.0001 -0.0001 0.0000 -0.0001 0.0000 0.0000 -0.0001 0.0000 0.0000

    Batang 10 1

    0 2

    Gbr AM 1 = = 0 3

    0 4

    0 5

    0 6

    SB LOKAL

    Batang 20 1

    13.435 2

    AM 2 = = = 16.122 3

    Gbr 0 4

    13.435 5

    -16.122 6

    SB LOKAL

    Batang 30 1

    0 2

    Gbr AM 3 = = 0 3

    0 4

    0 5

    0 6

    SB LOKAL

    Batang 40 1

    43.2 2

    Gbr AM 4 = = = 34.56 3

    SMS-1

    Hitung rekasi di ujung batang / freebody (AML) ---> Beban dimasukkan kecuali beban di joint

    AML 1

    AML 2

    AML 3

    AML 4

    AML 5

    AML 6

    AML 1

    AML 2

    AML 3

    AML 4

    AML 5

    AML 6

    AML 1

    AML 2

    AML 3

    AML 4

    AML 5

    AML 6

    AML 1

    AML 2

    AML 3

  • 0 4

    43.2 5

    -34.56 6

    SB LOKAL

    Transformasi dari sumbu lokal ke sumbu global struktur

    Batang 10 10 2

    AM 1 = 0 30 40 50 6

    SB LOKAL

    0.00 -1.00 0.00 0.00 0.00 0.00 0.00 -1.00 0.00 0.00 0.00 0.001.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00

    = 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 x

    0.00 0.00 0.00 0.00 -1.00 0.00 = 0.00 0.00 0.00 0.00 -1.00 0.000.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.000.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 1.00

    0 10 2

    = 0 3 D10 13 D100 14 D110 15 D12

    SB GLOBAL

    Batang 20 1

    13.435 2AM 2 = 16.122 3

    0 413.435 5-16.122 6

    SB LOKAL

    0.00 -1.00 0.00 0.00 0.00 0.00 0.00 -1.00 0.00 0.00 0.00 0.001.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00

    = 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 x

    0.00 0.00 0.00 0.00 -1.00 0.00 = 0.00 0.00 0.00 0.00 -1.00 0.000.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.000.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 1.00

    AML 4

    AML 5

    AML 6

    AMS 1 = RT . AML 1

    RT

    AMS 1

    AMS 1

    AMS 2 = RT . AML 2

    RT

    AMS 2

  • -13.435 4 D28.22656E-016 5

    = 16.122 6 D3-13.435 10 D7

    8.22656E-016 11 D8-16.122 12 D9

    SB GLOBAL

    Batang 30 10 2

    AM 3 = 0 30 40 50 6

    SB LOKAL

    1.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.000.00 1.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00

    = 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 x

    0.00 0.00 0.00 1.00 0.00 0.00 = 0.00 0.00 0.00 1.00 0.00 0.000.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 1.00 0.000.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 1.00

    0 10 D70 11 D8

    = 0 12 D90 7 D40 8 D50 9 D6

    SB GLOBAL

    Batang 40 1

    43.2 2AM 4 = 34.56 3

    0 443.2 5

    -34.56 6

    SB LOKAL

    1.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.000.00 1.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00

    = 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 x

    0.00 0.00 0.00 1.00 0.00 0.00 = 0.00 0.00 0.00 1.00 0.00 0.000.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00

    AMS 2

    AMS 3 = RT . AML 3

    RT

    AMS 3

    AMS 3

    AMS 4 = RT . AML 4

    RT

    AMS 4

  • 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 1.00

    0 13 D743.2 14 D8

    = 34.56 15 D90 10 D4

    43.2 11 D5-34.56 12 D6

    SB GLOBAL

    BEBAN EKIVALEN GABUNGAN (AE)BTG 1 BTG 2 BTG 3 BTG 4

    1 0 0 0 0 0 0 12 0 0 0 0 0 0 2

    D1 3 0 0 0 0 0 0 3 D1D2 4 0 -13.435 0 0 -13.435 13.435 4 D2

    5 0 8.22656E-016 0 0 8.22656E-016 0.00 5D3 6 0 16.122 0 0 16.122 -16.122 6 D3

    - AE = D4 7 0 + 0 + 0 + 0 = 0 AE = 0 7 D4D5 8 0 0 0 0 0 0 8 D5D6 9 0 0 0 0 0 0 9 D6D7 10 0 -13.435 0 0 -13.435 13.435 10 D7D8 11 0 8.22656E-016 0 43.2 43.2 -43.2 11 D8D9 12 0 -16.122 0 -34.56 -50.682 50.682 12 D9D10 13 0 0 0 0 0 0 13 D10D11 14 0 0 0 43.2 43.2 -43.2 14 D11D12 15 0 0 0 34.56 34.56 -34.56 15 D12

    SB GLOBAL SB GLOBAL

    Beban pada joint langsung (Aj)

    1 0 12 0 2

    D1 3 0 3 D1D2 4 0 4 D2

    5 0 5D3 6 0 6 D3

    Aj = D4 7 = 0 7 D4D5 8 0 8 D5D6 9 -98 9 D6D7 10 0 10 D7D8 11 0 11 D8D9 12 0 12 D9

    D10 13 33.941125497 13 D10D11 14 -33.941125497 14 D11D12 15 0 15 D12

    SB GLOBAL

    AMS 4

  • BEBAN JOINT GABUNGAN (Ac)

    Ac = Aj + AE

    1 0 0 0 1 0 3 D12 0 0 0 2 13.435 4 D23 0 0 0 3 D1 -16.122 6 D34 0 13.435 13.435 4 D2 0 7 D45 0 -8.22656E-016 -8.22656E-016 5 0 8 D5

    6 0 -16.122 -16.122 6 D3 -98 9 D6Ac = 7 0 + 0 = 0 7 D4 13.435 10 D7

    8 0 0 0 8 D5 -43.2 11 D89 -98 0 -98 9 D6 50.682 12 D9

    10 0 13.435 13.435 10 D7 33.941125497 13 D1011 0 -43.2 -43.2 11 D8 -77.141125497 14 D1112 0 50.682 50.682 12 D9 -34.56 15 D1213 33.941125497 0 33.941125497 13 D1014 -33.941125497 -43.2 -77.141125497 14 D1115 0 -34.56 -34.56 15 D12

    DEFORMASI STRUKTUR

    0.0002 -0.0003 0.0000 -0.0002 -0.0001 0.0000 -0.0002 0.0000 0.0000 -0.0002 0.0000 0.0000 0 -0.014 3

    -0.0003 0.0019 0.0003 0.0005 0.0009 0.0002 0.0005 0.0000 0.0002 0.0005 0.0000 -0.0002 13.435 0.037 4

    0.0000 0.0003 0.0001 0.0000 0.0002 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 -16.122 0.004 6-0.0002 0.0005 0.0000 0.0003 0.0001 0.0000 0.0003 0.0000 0.0000 0.0003 0.0000 -0.0001 0 0.020 7-0.0001 0.0009 0.0002 0.0001 0.0011 0.0004 0.0001 0.0000 0.0002 0.0001 0.0000 -0.0001 0 -0.007 8

    = 0.0000 0.0002 0.0000 0.0000 0.0004 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 x -98 = -0.007 9-0.0002 0.0005 0.0000 0.0003 0.0001 0.0000 0.0003 0.0000 0.0000 0.0003 0.0000 -0.0001 13.435 0.020 100.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 -43.2 0.000 11

    0.0000 0.0002 0.0000 0.0000 0.0002 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 50.682 0.003 12

    -0.0002 0.0005 0.0000 0.0003 0.0001 0.0000 0.0003 0.0000 0.0000 0.0003 0.0000 -0.0001 33.941125497 0.020 13

    0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 -77.141125497 0.000 140.0000 -0.0002 0.0000 -0.0001 -0.0001 0.0000 -0.0001 0.0000 0.0000 -0.0001 0.0000 0.0000 -34.56 -0.006 15

    GAYA-GAYA DALAM UJUNG ELEMEN

    Batang 1Deformasi struktur dalam arah SUMBU GLOBAL di batang 1, sbb :

    0 10 2

    = -0.014 3 D10.020 13 D100.000 14 D11-0.006 15 D12

    Beban joint yang selaras DOF (AFC)

    AFC =

    DF = SFF-1.AFC SFF

    -1 = SMS -1

    DF

    DMS 1

  • Deformasi struktur dalam arah SUMBU LOKAL di batang 1, sbb :

    0 1 0 0 0 0 0 0 1-1 0 0 0 0 0 0 0 2

    = 0 0 1 0 0 0 x -0.013610586 = -0.014 3 D10 0 0 0 1 0 0.020027496 0.000 4 D110 0 0 -1 0 0 -4.83628E-005 -0.020 5 D100 0 0 0 0 1 -0.006157988 -0.006 6 D12

    Gaya ujung batang

    0 979166.666667 0 0 -979166.666667 0 0 0 0 47.355 47.355 10 0 27199.0740741 24479.1666667 0 -27199.0740741 24479.1666667 0 0 60.811 60.811 2

    AM 1 = 0 + 0 24479.1666667 29375 0 -24479.1666667 14687.5 x -0.014 = 0 + 0.000 = 0.000 30 -979166.666667 0 0 979166.666667 0 0 0.000 0 -47.355 -47.355 40 0 -27199.0740741 -24479.1666667 0 27199.0740741 -24479.1666667 -0.020 0 -60.811 -60.811 50 0 24479.1666667 14687.5 0 -24479.1666667 29375 -0.006 0 109.460 109.460 6

    Batang 2Deformasi struktur dalam arah SUMBU GLOBAL di batang 2, sbb :

    0.037 4 D20 5

    = 0.004 6 D30.020 10 D70.000 11 D80.003 12 D9

    Deformasi struktur dalam arah SUMBU LOKAL di batang 2, sbb :

    0 1 0 0 0 0 0.0374005244 0 1-1 0 0 0 0 0 0 -0.037 2 D2

    = 0 0 1 0 0 0 x 0.0037114885 = 0.004 3 D30 0 0 0 1 0 0.0200732322 0.000 4 D80 0 0 -1 0 0 -8.28208E-005 -0.020 5 D70 0 0 0 0 1 0.0031016729 0.003 6 D9

    Gaya ujung batang

    0 881250 0 0 -881250 0 0 2.29012E-018 0 72.986 72.986 113.435 0 13769.53125 33046.875 0 -13769.53125 33046.875 -0.037 13.435 -13.435 0.000 2

    AM 2 = 16.122 + 0 33046.875 105750 0 -33046.875 52875 x 0.004 = 16.122 + -16.122 = 0.000 30 -881250 0 0 881250 0 0 0.000 0 -72.986 -72.986 4

    M 1 = R . DMS 1

    M 1

    AM 1 = AML 1 + SM 1 . M 1

    DMS 2

    M 2 = R . DMS 2

    M 2

    AM 2 = AML 2 + SM 2 . M 2

  • 13.435 0 -13769.53125 -33046.875 0 13769.53125 -33046.875 -0.020 13.435 13.435 26.870 5-16.122 0 33046.875 52875 0 -33046.875 105750 0.003 -16.122 -48.366 -64.488 6

    Batang 3Deformasi struktur dalam arah SUMBU GLOBAL di batang 3, sbb :

    0.020 10 D70.000 11 D8

    = 0.003 12 D90.020 7 D4-0.007 8 D5-0.007 9 D6

    Deformasi struktur dalam arah SUMBU LOKAL di batang 3, sbb :

    1 0 0 0 0 0 0.020 0.020 1 D70 1 0 0 0 0 0.000 0.000 2 D8

    = 0 0 1 0 0 0 x 0.003 = 0.003 3 D90 0 0 1 0 0 0.020 0.020 4 D40 0 0 0 1 0 -0.007 -0.007 5 D50 0 0 0 0 1 -0.007 -0.007 6 D6

    Gaya ujung batang

    0 742105.263158 0 0 -742105.263158 0 0 0.020 0 0.000 0.000 10 0 8222.77299898 15623.2686981 0 -8222.77299898 15623.2686981 0.000 0 0.000 0.000 2

    AM 3 = 0 + 0 15623.2686981 39578.9473684 0 -15623.2686981 19789.4736842 x 0.003 = 0 + 98.000 = 98.000 30 -742105.263158 0 0 742105.263158 0 0 0.020 0 0.000 0.000 40 0 -8222.77299898 -15623.2686981 0 8222.77299898 -15623.2686981 -0.007 0 0.000 0.000 50 0 15623.2686981 19789.4736842 0 -15623.2686981 39578.9473684 -0.007 0 -98.000 -98.000 6

    Batang 4Deformasi struktur dalam arah SUMBU GLOBAL di batang 4, sbb :

    0.020 13 D100.000 14 D11

    = -0.006 15 D120.020 10 D70.000 11 D80.003 12 D9

    Deformasi struktur dalam arah SUMBU LOKAL di batang 4, sbb :

    1 0 0 0 0 0 0.020 0.020 1 D100 1 0 0 0 0 0.000 0.000 2 D11

    DMS 3

    M 3 = R . DMS 3

    M 3

    AM 3 = AML 3 + SM 3 . M 3

    DMS 4

    M 4 = R . DMS 4

  • = 0 0 1 0 0 0 x -0.006 = -0.006 3 D120 0 0 1 0 0 0.020 0.020 4 D70 0 0 0 1 0 0.000 0.000 5 D80 0 0 0 0 1 0.003 0.003 6 D9

    Gaya ujung batang

    0 587500 0 0 -587500 0 0 0.020 0 -26.870 -26.870 143.2 0 4079.86111111 9791.66666667 0 -4079.86111111 9791.66666667 0.000 43.2 -29.786 13.414 2

    AM 4 = 34.56 + 0 9791.66666667 31333.3333333 0 -9791.66666667 15666.6666667 x -0.006 = 34.56 + -144.020 = -109.460 30 -587500 0 0 587500 0 0 0.020 0 26.870 26.870 4

    43.2 0 -4079.86111111 -9791.66666667 0 4079.86111111 -9791.66666667 0.000 43.2 29.786 72.986 5-34.56 0 9791.66666667 15666.6666667 0 -9791.66666667 31333.3333333 0.003 -34.56 1.048 -33.512 6

    M 4

    AM 4 = AML 4 + SM 4 . M 4

  • 0 0 00 0 00 0 00 1 0-1 0 00 0 1

    -27199.07 0.00 -24479.170.00 -979166.67 0.00

    24479.17 0.00 14687.5027199.07 0.00 24479.17

    0.00 979166.67 0.0024479.17 0.00 29375.00

    SM25 SM24 SM26SM15 SM14 SM16SM35 SM34 SM36SM55 SM54 SM56SM45 SM44 SM46SM65 SM64 SM66

  • 0 0 00 0 00 0 0

    6.123234E-017 1 0-1 6.123234E-017 00 0 1

    -13770 0 -330470 -881250 0

    33047 0 5287513770 0 33047

    0 881250 033047 0 105750

  • 0 0 00 0 00 0 01 0 00 1 00 0 1

    -742105.263158 0 00 -8222.77299898 15623.26869810 -15623.2686981 19789.4736842

    742105.263158 0 00 8222.77299898 -15623.26869810 -15623.2686981 39578.9473684

  • 0 0 00 0 00 0 01 0 00 1 00 0 1

    -587500 0 00 -4079.86111111 9791.666666670 -9791.66666667 15666.6666667

    587500 0 00 4079.86111111 -9791.666666670 -9791.66666667 31333.3333333

  • D8 D9 D10 D11 D12

    0 0 24479 0 146880 -33047 0 0 00 52875 0 0 00 0 0 0 0

    -8223 -15623 0 0 015623 19789 0 0 0

    0 33047 -587500 0 0893553 5832 0 -4080 -9792

    5832 176662 0 9792 156670 0 614699 0 24479

    -4080 9792 0 983247 9792-9792 15667 24479 9792 60708

  • 00

    0

    000

    013.435

    16.122

    013.435-16.122

  • 00

    0

    000

    043.2

    34.56

    043.2

  • -34.56

  • D1D2D3D4D5

    D6D7D8D9D10D11D12

  • D1D11D10D12

    D2D3D8

  • D7D9

    D7D8D9D4D5D6

  • D10D11D12D7D8D9

  • Data bebanP1 = 48 kNP2 = 26.87 kNq1 = 18 kN/mM1 = 98 kN.msudut P1 = 45 deg

    Properties penampangBahan / material :

    E = 2.35E+07 kN/m2

    Batang 1 :0.25 m

    0.30 m7.50E-02 m2

    E = 2.35E+07 kN/m2

    5.63E-04 m41.8 m

    90 degrees

    Elemen dalam matriks kekakuan batang

    EA / L 4.EI / L 2.EI / L 6.EI / L^29.79E+05 2.94E+04 1.47E+04 2.45E+04

    Batang 3 :0.30 m0.40 m

    1.20E-01 m2E = 2.35E+07 kN/m2

    1.60E-03 m4

    3.8 m0 degrees

    Elemen dalam matriks kekakuan batang

    EA / L 4.EI / L 2.EI / L 6.EI / L^27.42E+05 3.96E+04 1.98E+04 1.56E+04

    b1 =h1 =

    A1 =

    I1 =

    L1 =theta1 =

    b3 =h3 =A3 =

    I3 =L3 =

    theta3 =

  • Rekap Elemen dalam Matriks Kekakuan BatangBatang EA / L 4.EI / L 2.EI / L

    1 9.79E+05 2.94E+04 1.47E+042 8.81E+05 1.06E+05 5.29E+043 7.42E+05 3.96E+04 1.98E+044 5.88E+05 3.13E+04 1.57E+04

    Tentukan parameter yang dibutuhkan :

    a. Sumbu global struktur pada tiap elementb. DOF sesuai sumbu global strukturc.

    a. Sumbu global struktur pada tiap element

    Element

    1 1 2 32 4 5 63 10 11 124 13 14 15

    b. DOF sesuai sumbu global struktur

    Global axis 1 2 3DOF 1 2 d1

    c.

    Element 1:R = 0

    00 0 10 0 00 0 00 0 0

    Element 2:

    R = 00

    0 0 1

    0 0 00 0 00 0 0

    Matriks Transformasi [R] dan [RT] tiap element

    Matriks Transformasi [R] dan [RT] tiap element

    cos 1 sin 1 - sin 1 cos 1

    cos 1 sin 1 - sin 1 cos 1

  • Element 3:

    R = 00

    0 0 1

    0 0 00 0 00 0 0

    Element 4:R = 0

    00 0 10 0 0

    0 0 00 0 0

    ===> DALAM SUMBU LOKALBatang = 1

    90 degreesEA / L 4.EI / L 2.EI / L 6.EI / L^29.79E+05 2.94E+04 1.47E+04 2.45E+04

    Cx = cos theta = 6.123234E-017Cy = sin theta = 1

    Local axis 1 2 3

    1 9.79E+05 0 02 0 2.72E+04 2.45E+043 0 2.45E+04 2.94E+04

    4 -9.79E+05 0 05 0 -2.72E+04 -2.45E+046 0 2.45E+04 1.47E+04

    ===> DALAM SUMBU LOKALBatang = 2

    90 degreesEA / L 4.EI / L 2.EI / L 6.EI / L^28.81E+05 1.06E+05 5.29E+04 3.30E+04

    Cx = cos theta = 6.123234E-017Cy = sin theta = 1

    cos 1 sin 1 - sin 1 cos 1

    cos 1 sin 1 - sin 1 cos 1

    Hitung matriks kekakuan batang 1 [SM1]

    theta1 =

    [SM1] =

    Hitung matriks kekakuan batang 2 [SM2]

    theta2 =

  • Local axis 1 2 31 8.81E+05 0 02 0 1.38E+04 3.30E+04

    3 0 3.30E+04 1.06E+054 -8.81E+05 0 0

    5 0 -1.38E+04 -3.30E+046 0 3.30E+04 5.29E+04

    ===> DALAM SUMBU LOKALBatang = 3

    0 degreesEA / L 4.EI / L 2.EI / L 6.EI / L^27.42E+05 3.96E+04 1.98E+04 1.56E+04

    Cx = cos theta = 1Cy = sin theta = 0

    Local axis 1 2 3

    1 7.42E+05 0 02 0 8.22E+03 1.56E+043 0 1.56E+04 3.96E+04

    4 -7.42E+05 0 05 0 -8.22E+03 -1.56E+046 0 1.56E+04 1.98E+04

    ===> DALAM SUMBU LOKALBatang = 4

    0 degreesEA / L 4.EI / L 2.EI / L 6.EI / L^25.88E+05 3.13E+04 1.57E+04 9.79E+03

    Cx = cos theta = 1Cy = sin theta = 0

    Local axis 1 2 31 5.88E+05 0 02 0 4.08E+03 9.79E+03

    3 0 9.79E+03 3.13E+044 -5.88E+05 0 05 0 -4.08E+03 -9.79E+036 0 9.79E+03 1.57E+04

    [SM2] =

    Hitung matriks kekakuan batang 3 [SM3]

    theta3 =

    [SM3] =

    Hitung matriks kekakuan batang 4 [SM4]

    theta4 =

    [SM4] =

  • Urutkan sesuai sumbu global/struktur dengan DOF yang didahulukan

    d1 d2 d33 4 6

    3 29375 0 04 0 13769.53125 -33046.8756 0 -33046.875 1057507 0 0 08 0 0 09 0 0 0

    10 0 -13769.53125 33046.87511 0 -5.31179E-011 -2.02354E-01212 0 -33046.875 5287513 24479.166667 0 014 -1.49892E-012 0 015 14687.5 0 01 -24479.166667 0 02 1.498917E-012 0 05 0 5.311786E-011 2.02354E-012

    Hitung Matriks Beban Joint/ NodalNodal forces:

    - Akibat beban di sepanjang bentang (kedua ujung dikekang)

    Element: 100

    = 0000

    Element: 20

    13.435= 16.122

    013.435-16.122

    Element: 300

    = 0

    MATRIKS KEKAKUAN STRUKTUR [SMS]

    [SMS] =

    [AML]1

    [AML]2

    [AML]3

  • 000

    Element: 40

    43.2

    = 34.560

    43.2-34.56

    Susun matriks beban ekivalen

    Joint Sb Global Element 1

    1 1 01 2 01 3 0

    -Ae = 2 4 02 5 02 6 03 7 03 8 03 9 04 10 04 11 04 12 05 13 05 14 05 15 0

    - Akibat Beban di joint (kedua ujung dikekang)

    Joint Sb Global

    1 1 01 2 01 3 02 4 0

    Aj = 2 5 02 6 03 7 03 8 03 9 -984 10 04 11 04 12 05 13 33.941125497

    [AML]4

  • 5 14 -33.9411254975 15 0

    - Beban kombinasi atau gabungan (Ae + Aj)

    1 02 0

    d1 0d2 13.4355 -8.22656E-016

    d3 -16.122Ac = d4 0 +

    d5 0d6 0d7 13.435d8 -43.2d9 50.682d10 0d11 -43.2d12 -34.56

    Displacement:d1 = -0.0136105859d2 = 0.0374005244d3 = 0.0037114885d4 = 0.0200732322d5 = -0.007114549d6 = -0.0068025824d7 = 0.0200732322d8 = -8.28208E-005d9 = 0.0031016729d10 = 0.020027496d11 = -4.83628E-005d12 = -0.0061579884

    Reactions:

    GAYA-GAYA DALAM UJUNG ELEMEN

    Batang 1Deformasi struktur dalam arah SUMBU GLOBAL di batang 1, sbb :

    Tentukan no. Sb global dahulu1 1 02 2 0

    3 d1 -0.013610585913 d10 0.02002749614 d11 -4.83628E-00515 d12 -0.0061579884

    Gaya ujung batang

    [DMS]1 =

  • 0 979166.666670 00 + 00 -979166.666670 00 0

    0 979166.666670 00 00 -979166.666670 00 0

    - - - - - -

    Batang 2Deformasi struktur dalam arah SUMBU GLOBAL di batang 2, sbb :

    tentukan no. Sb global dahulu4 d2 0.03740052445 5 0

    6 d3 0.003711488510 d7 0.020073232211 d8 -8.28208E-00512 d9 0.0031016729

    Gaya ujung batang

    0 88125013.435 0

    16.122 + 00 -881250

    13.435 0-16.122 0

    Batang 3

    AM 1 = AML 1 + SM 1 . M 1

    [AM1] =

    [DMS]2 =

    AM 2 = AML 2 + SM 2 . M 2

    [AM2] =

  • Deformasi struktur dalam arah SUMBU GLOBAL di batang 3, sbb :tentukan no. Sb global dahulu

    10 d7 0.020073232211 d8 -8.28208E-00512 d9 0.00310167297 d4 0.02007323228 d5 -0.0071145499 d6 -0.0068025824

    Gaya ujung batang

    0 742105.263160 00 + 00 -742105.263160 00 0

    Batang 4Deformasi struktur dalam arah SUMBU GLOBAL di batang 4, sbb :

    tentukan no. Sb global dahulu13 d10 0.02002749614 d11 -4.83628E-005

    15 d12 -0.006157988410 d7 0.020073232211 d8 -8.28208E-00512 d9 0.0031016729

    Gaya ujung batang

    0 58750043.2 0

    34.56 + 00 -587500

    43.2 0-34.56 0

    [DMS]3 =

    AM 3 = AML 3 + SM 3 . M 3

    [AM3] =

    [DMS]4 =

    AM 4 = AML 4 + SM 4 . M 4

    [AM4] =

  • diukur dari sb x (sb global struktur)

    Batang 2 :

    E =

    Elemen dalam matriks kekakuan batang

    12.EI / L^3 EA / L2.72E+04 8.81E+05

    Batang 4 :

    E =

    Elemen dalam matriks kekakuan batang

    12.EI / L^3 EA / L8.22E+03 5.88E+05

    b2 =h2 =

    A2 =

    I2 =

    L2 =theta2 =

    b4 =h4 =A4 =

    I4 =L4 =

    theta4 =

  • 6.EI / L^2 12.EI / L^3 theta (deg) Cx Cy2.45E+04 2.72E+04 90 6.12323E-017 13.30E+04 1.38E+04 90 6.12323E-017 11.56E+04 8.22E+03 0 1 09.79E+03 4.08E+03 0 1 0

    13 14 1510 11 127 8 910 11 12

    4 5 6 7 8d2 5 d3 d4 d5

    0 0 0 R =0 0 00 0 0

    00

    0 0 1

    0 0 0 R =0 0 00 0 0

    00

    0 0 1

    cos 1 sin 1 - sin 1 cos 1

    cos 1 sin 1 - sin 1 cos 1

  • 0 0 0 R =0 0 00 0 0

    00

    0 0 1

    0 0 0 R =

    0 0 00 0 0

    0

    00 0 1

    ===> DALAM SUMBU LOKAL

    12.EI / L^32.72E+04

    4 5 6

    -9.79E+05 0 00 -2.72E+04 2.45E+040 -2.45E+04 1.47E+04

    9.79E+05 0 00 2.72E+04 -2.45E+040 -2.45E+04 2.94E+04

    ===> DALAM SUMBU LOKAL

    12.EI / L^31.38E+04

    cos 1 sin 1 - sin 1 cos 1

    cos 1 sin 1 - sin 1 cos 1

  • 4 5 6-8.81E+05 0 0

    0 -1.38E+04 3.30E+04

    0 -3.30E+04 5.29E+048.81E+05 0 0

    0 1.38E+04 -3.30E+040 -3.30E+04 1.06E+05

    ===> DALAM SUMBU LOKAL

    12.EI / L^38.22E+03

    4 5 6

    -7.42E+05 0 00 -8.22E+03 1.56E+040 -1.56E+04 1.98E+04

    7.42E+05 0 00 8.22E+03 -1.56E+040 -1.56E+04 3.96E+04

    ===> DALAM SUMBU LOKAL

    12.EI / L^34.08E+03

    4 5 6-5.88E+05 0 0

    0 -4.08E+03 9.79E+03

    0 -9.79E+03 1.57E+045.88E+05 0 0

    0 4.08E+03 -9.79E+030 -9.79E+03 3.13E+04

  • d4 d5 d6 d7 d87 8 9 10 11

    0 0 0 0 00 0 0 -13769.53125 -5.3117859E-0110 0 0 33046.875 -2.0235375E-012

    742105.26316 0 0 -742105.26316 00 8222.772999 -15623.268698 0 -8222.7729989790 -15623.268698 39578.947368 0 15623.26869806

    -742105.26316 0 0 1343374.7944 5.31178590E-0110 -8222.772999 15623.268698 5.31179E-011 893552.63411010 -15623.268698 19789.473684 33046.875 5831.6020313940 0 0 -587500 00 0 0 0 -4079.86111111110 0 0 0 -9791.6666666670 0 0 0 00 0 0 0 00 0 0 -5.31179E-011 -881250

    - Akibat beban di sepanjang bentang (kedua ujung dikekang)Transformasi sb lokal ke global ==>

    Local axis1 6.12323400E-0172 13 = 04 05 06 0

    Transformasi sb lokal ke global ==>Local axis

    1 6.12323400E-0172 13 = 04 05 06 0

    Transformasi sb lokal ke global ==>Local axis

    1 12 0

    3 = 0

    [AMS]1

    [AMS]2

    [AMS]3

  • 4 05 06 0

    Transformasi sb lokal ke global ==>Local axis

    1 12 0

    3 = 04 05 06 0

    Element 2 Element 3

    0 00 00 0

    -13.435 08.22656E-016 0

    16.122 0+ 0 + 0 +

    0 00 0

    -13.435 08.22656E-016 0

    -16.122 00 00 00 0

    DOF

    12d1d25d3d4d5d6d7d8d9

    d10

    [AMS]4

  • d11d12

    Cek:0 0 - 0 0 - 0 0 - 0 13.435 - 0 -8.22656E-016 - 0 -16.122 - 0 = 0 - 0 0 -

    -98 -98 - 0 13.435 - 0 -43.2 - 0 50.682 -

    33.941125497 33.941125497 - -33.941125497 -77.141125497 -

    0 -34.56 -

    Deformasi struktur dalam arah SUMBU LOKAL di batang 1, sbb :Deformasi struktur dalam arah SUMBU GLOBAL di batang 1, sbb :

    6.12323400E-017-1

    0000

    M 1 = R . DMS 1

    M 1

  • 0 0 -979166.66667 0 027199.074074 24479.166667 0 -27199.074074 24479.1666666724479.166667 29375 0 -24479.166667 14687.5

    0 0 979166.66667 0 0-27199.074074 -24479.166667 0 27199.074074 -24479.1666666724479.166667 14687.5 0 -24479.166667 29375

    0 0 -979166.66667 0 027199.074074 24479.166667 0 -27199.074074 24479.1666666724479.166667 29375 0 -24479.166667 14687.5

    0 0 979166.66667 0 0-27199.074074 -24479.166667 0 27199.074074 -24479.1666666724479.166667 14687.5 0 -24479.166667 29375

    - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    Deformasi struktur dalam arah SUMBU LOKAL di batang 2, sbb :Deformasi struktur dalam arah SUMBU GLOBAL di batang 2, sbb :

    6.12323400E-017-1

    0000

    0 0 -881250 0 013769.53125 33046.875 0 -13769.53125 33046.875

    33046.875 105750 0 -33046.875 528750 0 881250 0 0

    -13769.53125 -33046.875 0 13769.53125 -33046.87533046.875 52875 0 -33046.875 105750

    Deformasi struktur dalam arah SUMBU LOKAL di batang 3, sbb :

    M 2 = R . DMS 2

    M 2

  • Deformasi struktur dalam arah SUMBU GLOBAL di batang 3, sbb :

    100000

    0 0 -742105.26316 0 08222.772999 15623.268698 0 -8222.772999 15623.2686980615623.268698 39578.947368 0 -15623.268698 19789.47368421

    0 0 742105.26316 0 0-8222.772999 -15623.268698 0 8222.772999 -15623.2686980615623.268698 19789.473684 0 -15623.268698 39578.94736842

    Deformasi struktur dalam arah SUMBU LOKAL di batang 4, sbb :Deformasi struktur dalam arah SUMBU GLOBAL di batang 4, sbb :

    10

    0000

    0 0 -587500 0 04079.8611111 9791.6666667 0 -4079.8611111 9791.666666667

    9791.6666667 31333.333333 0 -9791.6666667 15666.666666670 0 587500 0 0

    -4079.8611111 -9791.6666667 0 4079.8611111 -9791.6666666679791.6666667 15666.666667 0 -9791.6666667 31333.33333333

    M 3 = R . DMS 3

    M 3

    M 4 = R . DMS 4

    M 4

  • 0.30 m

    0.60 m1.80E-01 m22.35E+07 kN/m2

    5.40E-03 m44.8 m

    90 degrees

    Elemen dalam matriks kekakuan batang

    4.EI / L 2.EI / L 6.EI / L^2 12.EI / L^31.06E+05 5.29E+04 3.30E+04 1.38E+04

    0.30 m0.40 m

    1.20E-01 m22.35E+07 kN/m21.60E-03 m4

    4.8 m0 degrees

    Elemen dalam matriks kekakuan batang

    4.EI / L 2.EI / L 6.EI / L^2 12.EI / L^33.13E+04 1.57E+04 9.79E+03 4.08E+03

  • 9 10 11 12 13d6 d7 d8 d9 d10

    6.1232340E-017 1 0 0 0-1 6.1232339957E-017 0 0 00 0 1 0 00 0 0 6.1232339957E-017 10 0 0 -1 6.1232339957E-0170 0 0 0 0

    6.1232340E-017 1 0 0 0-1 6.1232339957E-017 0 0 00 0 1 0 0

    0 0 0 6.1232339957E-017 10 0 0 -1 6.1232339957E-0170 0 0 0 0

  • 1 0 0 0 00 1 0 0 00 0 1 0 0

    0 0 0 1 00 0 0 0 10 0 0 0 0

    1 0 0 0 0

    0 1 0 0 00 0 1 0 00 0 0 1 0

    0 0 0 0 10 0 0 0 0

    1 2 d1Global axis 1 2 3

    1 27199.074074 5.8291203258E-011 -24479.16666666672 5.82912E-011 979166.666666667 1.50E-123 -24479.16667 1.50E-12 2.94E+04

    13 -27199.07407 -5.829120326E-011 24479.166666666714 -5.8291E-011 -979166.666666667 -1.498916655E-01215 -24479.16667 1.50E-12 1.47E+04

    [SMS1] =

  • d2 5 d3Global axis 4 5 6

    4 13769.53125 5.3117858969E-011 -33046.8755 5.31179E-011 881250 2.02E-12

    6 -33046.875 2.02E-12 1.06E+0510 -13769.53125 -5.311785897E-011 33046.875

    11 -5.3118E-011 -881250 -2.023537485E-01212 -33046.875 2.02E-12 5.29E+04

    d7 d8 d9Global axis 10 11 12

    10 742105.26316 0 011 0 8222.7729989795 1.56E+0412 0 1.56E+04 3.96E+04

    7 -742105.2632 0 08 0 -8222.7729989795 -15623.26869806099 0 1.56E+04 1.98E+04

    d10 d11 d12Global axis 13 14 15

    13 587500 0 014 0 4079.8611111111 9.79E+03

    15 0 9.79E+03 3.13E+0410 -587500 0 011 0 -4079.8611111111 -9791.666666666712 0 9.79E+03 1.57E+04

    [SMS2] =

    [SMS3] =

    [SMS4] =

  • d9 d10 d11 d12 112 13 14 15 1

    0 24479.1666666667 -1.4989E-012 14687.5 -24479.1666666667-33046.875 0 0 0 0

    52875 0 0 0 00 0 0 0 0

    -15623.2686981 0 0 0 019789.47368421 0 0 0 0

    33046.875 -587500 0 0 05831.602031394 0 -4079.861111 -9791.6666666667 0176662.2807018 0 9791.6666667 15666.6666666667 0

    0 614699.074074074 5.82912E-011 24479.1666666667 -27199.07407407419791.666666667 5.8291203258E-011 983246.52778 9791.6666666667 -5.829120326E-01115666.66666667 24479.1666666667 9791.6666667 60708.3333333333 -24479.1666666667

    0 -27199.0740740741 -5.8291E-011 -24479.1666666667 27199.07407407410 -5.829120326E-011 -979166.6667 1.4989166552E-012 5.8291203258E-011

    2.0235375E-012 0 0 0 0

    -1 0 0 0 06.1232340E-017 0 0 0 0

    0 1 0 0 00 0 6.12323E-017 -1 00 0 1 6.1232339957E-017 00 0 0 0 1

    -1 0 0 0 06.1232340E-017 0 0 0 0

    0 1 0 0 00 0 6.12323E-017 -1 00 0 1 6.1232339957E-017 00 0 0 0 1

    0 0 0 0 01 0 0 0 0

    0 1 0 0 0

    AMS 1 = RT . AML 1

    AMS 2 = RT . AML 2

    AMS 3 = RT . AML 3

  • 0 0 1 0 00 0 0 1 00 0 0 0 1

    0 0 0 0 01 0 0 0 0

    0 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1

    Element 4

    0 00 00 00 -13.435 Ae =0 8.22656E-0160 16.1220 = 00 00 00 -13.435

    43.2 43.2-34.56 -50.682

    0 043.2 43.2

    34.56 34.56

    AMS 4 = RT . AML 4

  • Tata ulang Ac :d1 0d2 13.435d3 -16.122d4 0d5 0d6 -98

    Ac = d7 13.435d8 -43.2d9 50.682

    d10 33.941125497d11 -77.141125497d12 -34.56

    1 02 05 -8.226564873E-016

    Deformasi struktur dalam arah SUMBU LOKAL di batang 1, sbb :

    1 0 0 0 06.1232340E-017 0 0 0 0

    0 1 0 0 00 0 6.12323E-017 1 00 0 -1 6.1232339957E-017 00 0 0 0 1

  • 0 00 0

    x -0.0136105859 = 0 +-4.836284606E-005 0

    -0.020027496 0-0.0061579884 0

    0 00 0

    -0.0136105859 0 +-4.836284606E-005 0

    -0.020027496 0-0.0061579884 0

    - - - - - - -

    Deformasi struktur dalam arah SUMBU LOKAL di batang 2, sbb :

    1 0 0 0 06.1232340E-017 0 0 0 0

    0 1 0 0 00 0 6.12323E-017 1 00 0 -1 6.1232339957E-017 00 0 0 0 1

    2.2901216257E-018 0-0.0374005244 13.435

    x 0.0037114885 = 16.122 +-8.282080990E-005 0

    -0.0200732322 13.4350.0031016729 -16.122

    Deformasi struktur dalam arah SUMBU LOKAL di batang 3, sbb :

  • 0 0 0 0 01 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1

    0.0200732322 0-8.282080990E-005 0

    x 0.0031016729 = 0 +0.0200732322 0-0.007114549 0

    -0.0068025824 0

    Deformasi struktur dalam arah SUMBU LOKAL di batang 4, sbb :

    0 0 0 0 01 0 0 0 0

    0 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1

    0.020027496 0-4.836284606E-005 43.2

    x -0.0061579884 = 34.56 +0.0200732322 0

    -8.282080990E-005 43.20.0031016729 -34.56

  • 14 15d11 d12

    0 6.12323E-017 -10 1 6.1232339957E-0170 0 00 0 00 0 01 0 0

    0 6.12323E-017 -10 1 6.1232339957E-0170 0 0

    0 0 00 0 01 0 0

    RT =

    RT =

  • 0 1 00 0 10 0 0

    0 0 00 0 01 0 0

    0 1 0

    0 0 10 0 00 0 0

    0 0 01 0 0

    d10 d11 d1213 14 15 atau dg menggunakan matriks transformasi

    -27199.074074 -5.829120E-011 -24479.1666667-5.82912E-011 -979166.666667 1.50E-12

    24479.1666667 -1.498917E-012 1.47E+04

    2.72E+04 5.8291203E-011 2.45E+045.829120E-011 9.79E+05 -1.50E-12

    2.45E+04 -1.498917E-012 2.94E+04

    RT =

    RT =

  • d7 d8 d910 11 12 atau dg menggunakan matriks transformasi

    -13769.53125 -5.311786E-011 -33046.875-5.31179E-011 -881250 2.02E-12

    33046.875 -2.023537E-012 5.29E+041.38E+04 5.3117859E-011 3.30E+04

    5.311786E-011 8.81E+05 -2.02E-123.30E+04 -2.023537E-012 1.06E+05

    d4 d5 d67 8 9 atau dg menggunakan matriks transformasi

    -742105.26316 0 00 -8222.77299898 1.56E+040 -15623.2686981 1.98E+04

    7.42E+05 0 0.00E+000 8.22E+03 -1.56E+04

    0.00E+00 -15623.2686981 3.96E+04

    d7 d8 d910 11 12 atau dg menggunakan matriks transformasi-587500 0 0

    0 -4079.86111111 9.79E+03

    0 -9791.66666667 1.57E+045.88E+05 0 0.00E+00

    0 4.08E+03 -9.79E+030.00E+00 -9791.66666667 3.13E+04

  • 2 52 5

    1.498917E-012 0 3 d10 5.3117859E-011 4 d20 2.0235375E-012 6 d30 0 7 d40 0 8 d50 0 9 d60 -5.311786E-011 10 d70 -881250 11 d80 2.0235375E-012 12 d9

    -5.82912E-011 0 13 d10-979166.66667 0 14 d111.498917E-012 0 15 d125.829120E-011 0 1 1979166.666667 0 2 2

    0 881250 5 5

    Global axis0 1 00 2 0

    x 0 = 3 00 13 00 14 00 15 0

    Global axis0 4 -13.435

    13.435 5 8.2265648733E-016x 16.122 = 6 16.122

    0 10 -13.43513.435 11 8.2265648733E-016-16.122 12 -16.122

    Global axis0 10 00 11 0

    x 0 = 12 0

  • 0 7 00 8 00 9 0

    Global axis0 13 0

    43.2 14 43.2

    x 34.56 = 15 34.560 10 0

    43.2 11 43.2-34.56 12 -34.56

    Cek:0 1 - 0 2 - 0 d1 -

    13.435 d2 - -8.22656E-016 5 -

    -16.122 d3 - 0 d4 - 0 d5 - 0 d6 -

    13.435 d7 - -43.2 d8 -

    50.682 d9 - 0 d10 -

    -43.2 d11 - -34.56 d12 -

  • Cek: - - - - - - - - - - - -

    local axis0 0 10 0 2

    x -0.0136105859 = -0.013610586 30.020027496 -4.8363E-005 4

    -4.836285E-005 -0.020027496 5-0.0061579884 -0.006157988 6

  • Cek:47.3552867689 47.3552867689 47.355286768960.811125497 60.811125497 60.811125497

    0 = 0 1.4210854715E-014-47.355286769 -47.3552867689 -47.3552867689-60.811125497 -60.811125497 -60.811125497109.460025895 109.4600258945 109.4600258945

    47.3552867689 - 60.811125497 -

    1.421085E-014 - -47.355286769 - -60.811125497 -

    109.460025895 -

    47.3552867689 47.355286768960.811125497 60.811125497

    1.421085E-014 0-47.355286769 -47.3552867689-60.811125497 -60.811125497

    109.460025895 109.4600258945

    local axis0.0374005244 2.29012E-018 1

    0 -0.037400524 2

    x 0.0037114885 = 0.0037114885 30.0200732322 -8.2821E-005 4

    -8.282081E-005 -0.020073232 50.0031016729 0.0031016729 6

    Cek:72.985838728 72.985838728 72.985838728

    -13.435 0 0

    -16.122 = 2.9842795E-013 0-72.985838728 -72.985838728 -72.985838728

    13.435 26.87 26.87-48.366 -64.488 -64.488

  • local axis0.0200732322 0.0200732322 1

    -8.282081E-005 -8.2821E-005 2x 0.0031016729 = 0.0031016729 3

    0.0200732322 0.0200732322 4-0.007114549 -0.007114549 5

    -0.0068025824 -0.006802582 6

    Cek:0 0 00 0 098 = 98 980 0 00 0 0

    -98 -98 -98

    local axis0.020027496 0.020027496 1

    -4.836285E-005 -4.8363E-005 2

    x -0.0061579884 = -0.006157988 30.0200732322 0.0200732322 4

    -8.282081E-005 -8.2821E-005 50.0031016729 0.0031016729 6

    Cek:-26.87 -26.87 -26.87

    -29.785838728 13.414161272 13.414161272

    -144.02002589 = -109.460025895 -109.460025894526.87 26.87 26.87

    29.785838728 72.985838728 72.9858387281.048 -33.512 -33.512

  • 0 0 0 00 0 0 01 0 0 00 6.12323399574E-017 -1 00 1 6.123E-017 00 0 0 1

    0 0 0 00 0 0 01 0 0 0

    0 6.12323399574E-017 -1 00 1 6.123E-017 00 0 0 1

  • 0 0 0 00 0 0 01 0 0 0

    0 1 0 00 0 1 00 0 0 1

    0 0 0 0

    0 0 0 01 0 0 00 1 0 0

    0 0 1 00 0 0 1

    atau dg menggunakan matriks transformasi

    SMS = RT. SM . R 5.99566662083E-011 -27199.074979166.666666667 1.665E-012

    0 24479.1667

    -5.9956666208E-011 27199.0741-979166.666666667 -1.665E-012

    0 24479.1667

    [SMS1] =

  • atau dg menggunakan matriks transformasiSMS = RT. SM . R 5.39609995874E-011 -13769.531

    881250 8.431E-013

    0 33046.875-5.3960999587E-011 13769.5313

    -881250 -8.431E-0130 33046.875

    atau dg menggunakan matriks transformasi

    SMS = RT. SM . R 742105.263157895 00 8222.7730 15623.2687

    -742105.263157895 00 -8222.7730 15623.2687

    atau dg menggunakan matriks transformasiSMS = RT. SM . R 587500 0

    0 4079.86111

    0 9791.66667-587500 0

    0 -4079.86110 9791.66667

    [SMS1] =

    [SMS1] =

    [SMS1] =

  • Cek:29375 0 0 0 0

    0 13770 -33047 0 00 -33047 105750 0 00 0 0 742105 00 0 0 0 82230 0 0 0 -156230 -13770 33047 -742105 00 0 0 0 -82230 -33047 52875 0 -15623

    24479 0 0 0 00 0 0 0 0

    14688 0 0 0 0

    - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

  • 21

    d1d11d10d12

  • - - (0.00) (0.00) - - -

    5d2

    d3d8d7d9

    - 0

    2.98427949019E-013 0.00 -

    00

  • d7d8d9d4d5d6

    - - - - - -

    d10d11

    d12d7d8d9

    - 0 - 0

    - 0 - 0 - 0 - 0

  • -24479.167 -5.996E-011 27199.0741 -24479.167 6.12E-017 11.499E-012 -979166.667 -1.665E-012 1.499E-012 -1 6.123E-017

    29375 0 -24479.167 14687.5 x 0 0

    24479.1667 5.996E-011 -27199.074 24479.1667 0 0-1.499E-012 979166.667 1.665E-012 -1.499E-012 0 0

    14687.5 0 -24479.167 29375 0 0

  • -33046.875 -5.396E-011 13769.5313 -33046.875 6.12E-017 12.024E-012 -881250 -8.431E-013 2.024E-012 -1 6.123E-017

    105750 0 -33046.875 52875 x 0 033046.875 5.396E-011 -13769.531 33046.875 0 0

    -2.024E-012 881250 8.431E-013 -2.024E-012 0 052875 0 -33046.875 105750 0 0

    0 -742105.263 0 0 1 015623.2687 0 -8222.773 15623.2687 0 139578.9474 0 -15623.269 19789.4737 x 0 0

    0 742105.263 0 0 0 0-15623.269 0 8222.773 -15623.269 0 019789.4737 0 -15623.269 39578.9474 0 0

    0 -587500 0 0 1 09791.66667 0 -4079.8611 9791.66667 0 1

    31333.3333 0 -9791.6667 15666.6667 x 0 00 587500 0 0 0 0

    -9791.6667 0 4079.86111 -9791.6667 0 015666.6667 0 -9791.6667 31333.3333 0 0

  • 0 0 0 0 24479 0 146880 -13770 0 -33047 0 0 00 33047 0 52875 0 0 00 -742105 0 0 0 0 0

    -15623 0 -8223 -15623 0 0 039579 0 15623 19789 0 0 0

    0 1343375 0 33047 -587500 0 015623 0 893553 5832 0 -4080 -979219789 33047 5832 176662 0 9792 15667

    0 -587500 0 0 614699 0 244790 0 -4080 9792 0 983247 97920 0 -9792 15667 24479 9792 60708

    - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

  • 0 0 0 0 27199.0741 5.829E-0110 0 0 0 5.829E-011 979166.6671 0 0 0 = -24479.167 1.499E-012

    0 6.123E-017 1 0 -27199.074 -5.829E-0110 -1 6.123E-017 0 -5.829E-011 -979166.6670 0 0 1 -24479.167 1.499E-012

    - - - - - - - - - - - -

  • 0 0 0 0 13769.5313 5.312E-0110 0 0 0 5.312E-011 881250

    1 0 0 0 = -33046.875 2.024E-0120 6.123E-017 1 0 -13769.531 -5.312E-011

    0 -1 6.123E-017 0 -5.312E-011 -8812500 0 0 1 -33046.875 2.024E-012

    - -

    - - - - - - - - - -

    0 0 0 0 742105.263 00 0 0 0 0 8222.7731 0 0 0 = 0 15623.2687

    0 1 0 0 -742105.26 00 0 1 0 0 -8222.7730 0 0 1 0 15623.2687

    - - - - - - - - - - - -

    0 0 0 0 587500 00 0 0 0 0 4079.86111

    1 0 0 0 = 0 9791.666670 1 0 0 -587500 00 0 1 0 0 -4079.861110 0 0 1 0 9791.66667

  • - - - - - - - - - - - -

  • -24479.167 -27199.074 -5.829E-011 -24479.167 R =1.499E-012 -5.829E-011 -979166.667 1.499E-012

    29375 24479.1667 -1.499E-012 14687.5

    24479.1667 27199.0741 5.829E-011 24479.1667-1.499E-012 5.829E-011 979166.667 -1.499E-012

    14687.5 24479.1667 -1.499E-012 29375

    - - - - - - - - - - - - - - - - - - - - - - - -

  • -33046.875 -13769.531 -5.312E-011 -33046.875 R =2.024E-012 -5.312E-011 -881250 2.024E-012

    105750 33046.875 -2.024E-012 5287533046.875 13769.5313 5.312E-011 33046.875

    -2.024E-012 5.312E-011 881250 -2.024E-01252875 33046.875 -2.024E-012 105750

    - - - -

    - - - - - - - - - - - - - - - - - - - -

    0 -742105.26 0 0 R =15623.2687 0 -8222.773 15623.268739578.9474 0 -15623.2687 19789.4737

    0 742105.263 0 0-15623.269 0 8222.773 -15623.26919789.4737 0 -15623.2687 39578.9474

    - - - - - - - - - - - - - - - - - - - - - - - -

    0