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Data E = 2100 ton/cm 2 P2 = 4 ton H = 3m L = 4m P = 4 ton M = -4 ton-m Batang 1 A = 120 cm 2 I = 4000 cm 4 Batang 2 A = 120 cm 2 PENYELESAIAN : I = 4000 cm 4 1). DEGREE OF FREEDOM (D.O.F) 2). Matrik Kekakuan Dalam SUMBU LOKAL a. Batang 1 (AB) E = 21000000 t/m 2 Batang 1 = 5 m sin a = 0.602 A = 0.012 m 2 a = 37.000 cos 2 a = 0.638 I = 0.00004 m 4 cos a = 0.799 sin 2 a = 0.362 50400 0 0 -50400 0 0 0 80.640 201.600 0 -80.640 201.600 0 201.600 672.000 0 -201.600 672.000 [k] 1 = -50400 0 0 50400 0 0 0 -80.640 -201.600 0 80.640 -201.600 0 201.600 336.000 0 -201.600 336.000 Dbx Dby 2 1 Rbz L H P 2 1 C B A M P

Tgs Matrix Kd3

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  • DataE = 2100 ton/cm2 P2 = 4 tonH = 3 mL = 4 mP = 4 tonM = -4 ton-m

    Batang 1A = 120 cm2

    I = 4000 cm4

    Batang 2A = 120 cm2

    PENYELESAIAN : I = 4000 cm4

    1). DEGREE OF FREEDOM (D.O.F)

    2). Matrik Kekakuan Dalam SUMBU LOKALa. Batang 1 (AB)

    E = 21000000 t/m2 Batang 1 = 5 m sin a = 0.602A = 0.012 m2 a = 37.000 cos2 a = 0.638I = 0.00004 m4 cos a = 0.799 sin2 a = 0.362

    50400 0 0 -50400 0 00 80.640 201.600 0 -80.640 201.6000 201.600 672.000 0 -201.600 672.000

    [k]1= -50400 0 0 50400 0 00 -80.640 -201.600 0 80.640 -201.6000 201.600 336.000 0 -201.600 336.000

    DbxDby

    2

    1

    Rbz

    L

    H

    P

    21

    C

    B

    A

    M

    P

  • b. Batang 2 (BC)E = 21000000 t/m2 batang 2 = 3 m sin a = 1A = 0.012 m2 a = 90 cos2 a = 4E-33I = 0.00004 m4 cos a = 6E-17 sin2 a = 1

    84000 0 0 -84000 0 00 373.333 560 0 -373.333 5600 560 1120 0 -560 1120

    [k]2= -84000 0 0 84000 0 00 -373.333 -560 0 373.333 -5600 560 560 0 -560 560

    3). Matrik Kekakuan SUMBU GLOBAL [kg] = [T]T.[kl].[T]a. Batang 1 (AB)

    a = 90 cos a = 0.00sin a = 1.00

    6.12574E-17 1 0 0 0 0-1 6.12574E-17 0 0 0 0

    [T]1= 0 0 1 0 0 00 0 0 6.12574E-17 1 00 0 0 -1 6.12574E-17 00 0 0 0 0 1

    [kg] = [T]T.[kl].[T] :

    80.640 0.000 -201.600 -80.640 0.000 -201.6000.000 50400.000 0.000 0.000 -50400.000 0.000

    [Kg]1= -201.600 0.000 672.000 201.600 0.000 672.000-80.640 0.000 201.600 80.640 0.000 201.6000.000 -50400.000 0.000 0.000 50400.000 0.000

    -201.600 0.000 336.000 201.600 0.000 336.000

    b. Batang 2 (BC)a = 90

    cos a = 0sin a = 1

    0 1 0 0 0 0-1 0 0 0 0 0

    [T]2= 0 0 1 0 0 00 0 0 0 1 00 0 0 -1 0 00 0 0 0 0 1

  • Sehingga [kg] = [T]T.[kl].[T] :

    373.333 0.0E+00 -560 -373.333 0.0E+00 -560- 84000 0.0E+00 0.0E+00 -84000 0.0E+00

    [Kg]2= -560 0.0E+00 1120 560 0.0E+00 1120-373.3333 0.0E+00 560 373.3333 0.0E+00 5600.0E+00 -84000 0.0E+00 0.0E+00 84000 0.0E+00

    -560 0.0E+00 560 560 0.0E+00 560

    4). Perakitan Matriks Kekakuan Struktur Global Tereduksi

    80.640 0.000 -201.600 -80.640 0.000 -201.600 0 0 00.000 50400.000 0.000 0.000 -50400.000 0.000 0 0 0

    -201.600 0.000 672.000 201.600 0.000 672.000 0 0 0[Kg]s = -80.640 0.000 201.600 453.973 0.000 -358.400 -373.333 0.000 -560.000

    0.000 -50400.000 0.000 0.000 134400.000 0.000 0 -84000.000 0-201.600 0.000 336.000 -358.400 0.000 1456.000 560.000 0 1120.000

    0 0 0 -373.333 0 560.000 373.333 0 560.0000 0 0 0 -84000.000 0 0 84000.000 00 0 0 -560.000 0 560.000 560.000 0 560.000

    SETELAH DIRDUKSI :

    453.973 0.000 -358.400[Kg]s = 0.000 134400.000 0.000

    -358.400 0.000 1456.000

    5). Vektor Perpindahan {D} =[Kg]SR-1 . [{F} + {Fo}]

    4 453.973 0.000 -358.400 Dbx4 = 0.000 134400.000 0.000 x Dby-4 -358.400 0.000 1456.000 Rbz

    Dbx 0.0027341 0.0000000 0.0006730 4Dby = 0.0000000 0.0000074 0.0000000 x 4Rbz 0.0006730 0.0000000 0.0008525 -4

    Dbx 0.0082 mDby = 0.0000 mRbz -0.0007 m

    6). Vektor perpindahan bagi masing-masing elemen {d}i = [T]i . {D}a. Batang 1

    00

    {d}1 = 00.000030-0.008244

  • -0.000718

  • b. Batang 2

    0.000030-0.008244

    {d}2 = -0.000718000

    7). Element Forces : {f} = [k]i {d}i - {fo}ia. Batang 1 (AB)

    f1 50400 0 0 -50400 0 0 0.000000 0f2 0 81 202 0 -81 202 0.000000 0f3 = 0 202 672 0 -202 672 x 0.000000 - 0f4 -50400 0 0 50400 0 0 0.000030 4f5 0 -81 -202 0 81 -202 -0.008244 4f6 0 202 336 0 -202 336 -0.000718 -4

    f1 -1.500 0 -1.500 tonf2 0.520 0 0.520 tonf3 = 1.180 - 0 = 1.180 ton-mf4 1.500 4 -2.500 tonf5 -0.520 4 -4.520 tonf6 1.421 -4 5.421 ton-m

    b. Batang 2 (BC)

    f1 84000 0 0 -84000 0 0 0.00003 4 Dbxf2 0 373 560 0 -373 560 -0.00824 4 Dbyf3 = 0 560 1120 0 -560 1120 x -0.00072 - -4 Rbzf4 -84000 0 0 84000 0 0 0.00000 0 Dcxf5 0 -373 -560 0 373 -560 0.00000 0 Dcyf6 0 560 560 0 -560 560 0.00000 0 Rcz

    f1 2.500 4 -1.500 tonf2 -3.480 4 -7.480 tonf3 = -5.421 - -4 = -1.421 ton-mf4 -2.500 0 -2.500 tonf5 3.480 0 3.480 tonf6 -5.019 0 -5.019 ton-m

  • 8). Reaksi Tumpuan {F} = [Kg]s.{D} - ({F}+{Fo})

    80.640 0.000 -201.600 -80.640 0.000 -201.600 0 0 0 0 00.000 50400.000 0.000 0.000 -50400.000 0.000 0 0 0 0 0

    -201.600 0.000 672.000 201.600 0.000 672.000 0 0 0 0 0[F] = -80.640 0.000 201.600 453.973 0.000 -358.400 -373.333 0.000 -560.000 0.00824435 4

    0.000 -50400.000 0.000 0.000 134400.000 0.000 0.000 -84000.000 0.000 x 2.9762E-05 - 4-201.600 0.000 336.000 -358.400 0.000 1456.000 560.000 0.000 1120.000 -0.0007179 -4

    0 0 0 -373.333 0.000 560.000 373.333 0.000 560.000 0 00 0 0 0.000 -84000.000 0.000 0.000 84000.000 0.000 0 00 0 0 -560.000 0.000 560.000 560.000 0.000 560.000 0 0

    -0.520 0 -0.520 ton-1.500 0 -1.500 ton1.180 0 1.180 ton-m

    [F] = 4.000 - 4 = 0 ton4.000 4 0 ton-4.000 -4 0 ton-m-3.480 0 -3.480 ton-2.500 0 -2.500 ton-5.019 0 -5.019 ton-m

    8). GAMBAR

    -2.500 5.421

    -1.500 -7.480 -1.421-4.520

    0.520-1.500 1.180

    -2.500 3.480-5.019

    NFD SFD BMD