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39079502-19684772-Perhitungan-Struktur-Gudang-Baja

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					       Pre - Eliminary Design
1 Perencanaan Atap

                     Perencanaan Atap


                 Merencanakan Pola Beban


                     Data Perencanaan


                                             Perencanaan Dimensi Gording


                                            Perencaan Penggantung Gording

                                              Perencanaan Gording Ujung


                                               Perencanaan Ikatan Angin



1.1 Merencanakan Pola Beban

   Pola Beban Diambil dari peraturan Pembebanan Indonesia untuk gedung 1983

                                                          Merencanak
                                                              an
                                                          Pola Beban




                  Beban Mati                                 Beban Hidup               Beban Angin




   Beban                                                                           Beban
  Penutup                             Beban            Beban            Beban                 Beban Angin
                 Beban Profil                                                     Tekanan
    Atap                            Pengikat dll    Terbagi Rata       Terpusat                  Hisap
                                                                                   Angin


   1.1.1 Merencanakan Beban Mati ( Berdasarkan Peraturan Pembebanan Indonesia Untuk Gedung )
        a. Atap
           Berat asbes        :      10.3 kg/m2
           Berat Profil       :     Menyesuaikan Perencanaan
           Berat Pengikat dll  :      10 % dari Berat Total

   1.1.2 Merencanakan Beban Hidup ( Berdasarkan Peraturan Pembebanan Indonesia Untuk Gedung )

       a. Beban Hidup Terbagi Rata ( Atap ) :
           α=      25 0
            q = (40 - 0.8 α) =        20    kg/m2     ≤       20   kg/m2

             ambil q =         20   kg/m2
       b. Beban Hidup Terpusat ( Atap )

           P=     100 kg

   1.1.3 Merencanakan Beban Angin ( Berdasarkan Peraturan Pembebanan Indonesia Untuk Gedung )

       a. Beban Tekanan Angin
          Bangunan Jauh dari Pantai -> asumsi Tekanan Angin :          30   kg/m2




         Koefisien Angin (C) tekan = (0.02 α - 0.4)   =     0.1

         Angin Tekan = C x W                     =    3    kg/m2

         Angin Hisap = 0.4 x W                   =    12   kg/m2

1.2 Data - Data perencanaan

   Data Atap
   Jenis                           :   Asbes Gelombang
   Tebal                           :      5 mm
   Berat                           :    10.3 kg/m2
   Lebar Gelombang                 :    110 mm
   Kedalaman Gelombang             :     57 mm
   Jarak Miring Gording            :    110 cm
   Jarak Kuda-Kuda (L)             :    400 cm
   Sudut Kemiringan Atap           :    0.44 rad     =      25     0



1.3 Perencanaan Dimensi Gording
   1.3.1 Perencanaan Profil WF untuk Gording Dengan ukuran :

    WF    100      x          50      x         5      x        7
     A = 11.85 cm2                      tf =      7   mm              Zx =    41.8 cm3
    W=       9.3 kg/m                  Ix =     187   cm4             Zy =    8.94 cm3
      a=    100 mm                     Iy =    14.8   cm4              h=       70 mm {=D - 2 x (tf + r)}
     bf =     50 mm                   tw =        5   mm
     iy =  1.12 cm                     ix =    3.98   cm

Mutu Baja =     BJ 37
     fu = 3700 kg/cm2 =            370 Mpa
      fy = 2400 kg/cm2 =           240 Mpa

  1.3.2 Perencanaan Pembebanan
  1.3.2.1 Perhitungan Beban
  Beban Mati
  Berat Gording                                                               =     9.3 kg/m
  Berat Asbes Gelombang      =                  w       x       l
                             =                 10.3     x      1.1            =    11.33 kg/m
                                                                Berat Total   =    20.63 kg/m
  alat Pengikat dll 10 % =         0.1          x     20.63                   =     2.06 kg/m
                                                                      qD      =    22.69 kg/m

  Beban Hidup
  Beban Terbagi Rata = (40 - 0.8 α) =                  40       -     20      =     20     kg/m2
                                                                       q      =     20     kg/m2
  qL = jarak gording horisontal x q             =     0.997     x    20.00    =    19.94 kg/m

  Beban Hidup Terpusat, PL                                                    =     100 kg

  Beban Angin
  Tekanan Angin                                                               =     30     kg/m2
  Angin Tekan                                                                 =      3     kg/m2
  Angin Hisap                                                                 =     12     kg/m2 (menentukan = q)
  q = jrk gording horisontal x angin hisap =          0.997     x    12.00    =    11.96   kg/m

  Beban Mati + Beban Hidup > dari Beban Angin Hisap :               22.69     +    19.94     >     11.96
  Beban Angin Hisap      tidak perlu diperhitungkan           ==>   qw =      3   kg/m

  1.3.2.2 Perhitungan Momen Akibat Beban thp Sbx dan Sby
  Beban Mati
  MXD = 1/8 (qD x cosα) L2 = 0.13   x ( 22.69     x    0.91                   x     16      )=     41.13 kgm
  MYD = 1/8(qDxsinα xL/3)2 =       0.13        x(     22.69     x    0.42     x    1.78     )=      2.13 kgm

  Beban Hidup Terbagi Rata
  MXLD = 1/8 (qL x cosα) L2 =      0.13        x(     19.94     x    0.91     x     16      )=     36.25 kgm
  MYL = 1/8(qLxsinαxL/3) =
                         2
                                   0.13        x(     19.94     x    0.42     x    1.78     )=      1.87 kgm

  Beban Hidup Terpusat
  MXL = 1/4 (qL x cosα) L =        0.25        x(     100       x    0.91     x      4      )=     90.63 kgm
  MYL = 1/4(qL x sinα)(L/3) =      0.25        x(     100       x    0.42     x    1.33     )=     14.09 kgm
Beban Angin Terbagi Rata
MXW = 1/8 x qw x L   =            0.13    x         3      x      16                  =     6   kgm


1.3.3.3 Besar Momen Berfaktor ( Mu = 1.2 M D + 1.6 ML + 0.8 MW )
* Mu Beban Mati ,Beban Angin dan Beban Hidup Terbagi Rata
Sumbu X                   Sumbu Y
MD =     41.13 kgm        MD =     2.13 kgm
ML =      36.25 kgm               ML =   1.87 kgm
Mw =          6 kgm

MUX =      1.2        x   41.13     +    1.6        x    36.25    +      0.8     x    6     =    ### kgm
MUY =      1.2        x   2.13      +    1.6        x    1.87     +      0.8     x    0     =    5.55 kgm


* Mu Beban Mati, Beban Angin dan Beban Hidup Terpusat
Sumbu X                   Sumbu Y
MD =   41.13 kgm          MD =    2.13 kgm
ML =      90.63 kgm               ML =   14.09 kgm
Mw =          6 kgm

MUX =      1.2        x   41.13     +    1.6        x    90.63    +      0.8     x    6     =    ### kgm
MUY =      1.2        x   2.13      +    1.6        x    14.09    +      0.8     x    0     =    25.1 kgm


1.3.3 Kontrol Kekuatan Profil
1.3.3.1 Penampang Profil
 Untuk Sayap                                    Untuk Badan
       bf             170                         h         1680
                 ≤                                      ≤
   2      tf           fy                        tw           fy
      50              170                        70         1680
                 ≤                                      ≤
   2      7           240                         5          240
     3.57        ≤   10.97                       14.0   ≤   108.4
                OK                                     OK

Penampang Profil Kompak, maka Mnx = Mpx

1.3.3.2 Kontrol Lateral Buckling
Jarak Baut Pengikat / pengaku lateral = L B =            500     mm       =     50   cm

                                             E
 LP =     1.76        x    iY       x
                                             fy
                                          200000
     =    1.76        x   1.12      x                     =      56.90   cm
                                            240

         Ternyata :        LB       <     LP    maka :   Mnx      =      Mpx

Mnx = Mpx = Zx . Fy =             41.8    x     2400      =      ### Kgm
Mny = Zy ( satu sayap ) * fy
    = 1/4 x tf x bf 2 x fy
    = 0.25        x        0.7      x          52          x     2400     =    10500 kgcm
    = 105 kgm
1.3.3.3 Persamaan Iterasi
            Mux                           Muy
                       +                                 ≤        1
          φb . Mnx                      φb . Mny

Beban Mati , Beban Angin dan Beban Hidup Terbagi Rata
      112.164                   5.554
                      +                        ≤      1
  0.9    x     ###         0.9    x    94.5
 0.12    +    0.07    ≤     1
              0.19    ≤     1    OK

Beban Mati , Beban Angin dan Beban Hidup Terpusat
     199.170                    25.097                                      1
                      +                        ≤
 0.9    x      ###         0.9     x   105
 0.2    +     0.27    ≤     1
              0.46    ≤     1    OK

1.3.3.4 Kontrol Lendutan Profil

                          L                    400
Lendutan Ijin f =                       =                =         2.22 cm
                         180                   180

Lendutan Akibat Beban Merata (1)
         5           qD + L cos α               L4                5        x    0.43   x    0.91   x    400 4
 fx =            x                                       =
        384            E      x                 Ix                        384     x    2000000     x    187
    =    0.34 cm

          5              qD + L       sin α (L/3)4                5        x    0.43   x    0.42   x    ### 4
  fy =           x                                       =
         384              E             x     Iy                          384     x    2000000     x    14.8
     =    0.03 cm

Lendutan Akibat Beban Terpusat (2)
         5             P   cos α                L3                1         x   100    x    0.91   x    400 3
 fx =            x                                       =
        384            E      x                 Ix                         48    x     2000000     x    187
    =    0.32 cm
         5             P    sin α               L3                1         x   30     x    0.42   x    ### 3
 fy =            x                                       =
        384            E      x                 Iy                         48    x     2000000     x    14.8
    =    0.02 cm

Lendutan Akibat Beban Angin merata (3)
         5            qW    cos α L4                              5        x    0.03   x    0.91   x    400 4
 fx =            x                                       =
        384            E      x     Ix                                    384     x    2000000     x    187
    =    0.02 cm

          5               qW          sin α (L/3)4                5        x    0.03   x    0.42   x    ### 4
  fy =               x                                   =
         384              E             x     Iy                          384     x    2000000     x    14.8
     =          0 cm


Lendutan total yang terjadi

ftot =   fx2 + fy2         =          (fx1 + fx2 + fx3)2 + (fy1 + fy2 + fy3)2

   = (   0.34        +   0.32           +      0.02 ) 2 + (      0.03       +   0.02   +     0     )2
                                  ≤
                                     ≤
     ftot =   0.69 cm           <        fijin =   2.22 cm          OK


1.4 Perencanaan Penggantung Gording

    1.4.1 Data Penggantung Gording

    Jarak Kuda - Kuda (L)                           =      400     cm
    Jumlah Penggantung Gording                      =       2      buah
    Jumlah Gording                                  =       9      buah
    Jarak Penggantung gording                       =      ###     cm




    1.4.2 Perencanaan Pembebanan
    Beban Mati
    Berat Sendiri Gording                                               =        9.3    kg/m
    Berat Asbes gelombang                                               =      11.33    kg/m
                                                                        =      20.63    kg/m
    Alat Pengikat dll 10 %      =        0.1        x      20.63        =      2.06     kg/m
                                                            qD          =      22.69 kg/m

     RD =      qD     x        sinα        x       L/3
        =     22.69   x          0.42      x        1.33     =          12.787          kg

    Beban Hidup
    Beban Terbagi Rata = (40 - 0.8 α) =                     40          -       0.8          =    20      kg/m2
                                                                                 q           =    20      kg/m2
    qL = jarak gording horisontal x q               =      0.997        x      20.00         =   19.94 kg/m
     RL =      qL     x        sinα        x       L/3
        =     19.94   x          0.42      x        1.33                =           11.235       kg

    Beban Terpusat =      PL               =        100 kg
     RL =      PL     x        sinα
        =     100.0   x          0.42      =         42.262        kg

    Beban Angin
    Angin Tekan = q                                                                          =        3   kg/m2
    qW = jarak gording horisontal x q               =      0.997        x       3.00         =   2.99 kg/m
    RW =      qW      x        sinα        x       L/3
       =      2.99    x          0.42      x        1.33     =              1.685       kg

    1.4.3 Perhitungan Gaya
    1.4.3.1 Penggantung Gording Tipe A
     RA = 1.2 RD + 1.6 RL + 0.8 RW
        =    1.2      x    12.8    +  1.6                   x (     11.2            +    42.3     ) +      0.8    x    1.7
        =      102.29    kg

    RA total = Ra x jumlah Gording         =         102.29             x           9        =        920.60      kg

    1.4.3.1 Penggantung Gording Tipe B
                          panjang miring gording                  110
      arctan β =                                           =                   =    0.83
                                L        /     3                 133.33
             β=          39.52 o
             RA                      920.60
     RB =                 =                        =        1446.607      kg
             sin β                 sin   39.52


    1.4.4 Perencanaan Batang Tarik

     Pu =    RB =         1446.607 kg
            BJ 37        fu = 3700 kg/cm2
                         fy = 2400 kg/cm2

    1.4.4.1 Kontrol Leleh
    Pu = φ . fy . Ag ; dengan φ = 0.9

     Ag perlu =               Pu          =              1446.607
                          ϕ                                                    =   0.670 cm2
                                    fy            0.9       x    2400
                                                                                   Tidak Menentukan
    1.4.4.2 Kontrol Putus
    Pu = φ . fu . 0,75 Ag ; dengan φ =            0.75

     Ag perlu =                    Pu              =                    1446.607
                          ϕ                                                                       =    0.7 cm2
                                   fu    0.75             0.75      x     3700       x     0.75
                                                                                                      Menentukan
    Ag perlu = 1/4 . π . d    2

                      Ag            x     4                0.7      x          4
      d       =                                    =                                 =     0.94 cm
                                    π                               π
     ==>      Pakai d =            10    mm


    1.4.5 Kontrol Kelangsingan
    Jarak Penggantung Gording =               ### cm

    Panjang Rb =         (jarak penggantung gording)2 + (panjang miring gording)2

                     =        133.33 2             +        110 2
                     =    ### cm

    Cek :                          Panjang Rb
               d          >
                                       500
                                  172.85
               1          >
                                   500
               1          >        0.35           OK


1.5 Perencanaan Ikatan Angin Atap

    1.5.1 Data Perencanaan Ikatan Angin Atap
    Tekanan Angin W         =    30 kg/m2
    Koefisien Angin C tekan =    0.9
    Koefisien Angin C hisap         =    0.4
     a1 =    300 cm                      a2 = 200 cm
     α=      0.44 rad               =       25 0
1.5.2 Perhitungan Tinggi Ikatan Angin ( h )
 h1 =     9   m
 h2 =     9      +           2        x    tg    0.44     =      9.93      m
 h3 =     9      +           4        x    tg    0.44     =      10.87     m
 h4 =     9      +           6        x    tg    0.44     =      11.8      m
 h5 =     9      +           9        x    tg    0.44     =      13.2      m

1.5.3 Perhitungan Gaya - Gaya yang Bekerja
R = 1/2 . W . C . a . h
 R1 = 0.50        x     30   x     0.9    x               1        x       9      =     121.5   kg
 R2 =    0.50    x           30       x    0.9    x       2        x      9.93    =     ###     kg
 R3 =    0.50    x           30       x    0.9    x       2        x     10.87    =     ###     kg
 R4 =    0.50    x           30       x    0.9    x      2.5       x      11.8    =     ###     kg
 R5 =    0.50    x           30       x    0.9    x       3        x      13.2    =     ###     kg

Rtotal = ( R1+R2+R3+R4+(R5/2)) = 121.5   +              268.18    +      293.36   +     ###     +    ###
                             =    1348.454                kg

1.5.4 Perencanaan Dimensi Ikatan Angin
1.5.4.1 Menghitung gaya Normal
                       2
         tg φ   =            =    0.5
                       4
           φ    =    26.57 0

        R1     =         121.5       kg
      Rtotal   =      1348.454 kg
Gaya Normal Gording Akibat Angin Dimana untuk angin tekan C =             0.9
                                       dan untuk angin hisap C =          0.4
             Chisap    x      Rtotal
  N     =
                     Ctekan
              0.4      x       1348.454
        =                                     =    599.31 kg
                          0.9

1.5.4.2 Menghitung gaya Pada Titik Simpul
 Pada Titik Simpul A
ΣV = 0
Rtotal + S1 = 0      ===> S1 = - Rtotal ===>            S1 =     ### kg

ΣH = 0
 S2 =     0

 Pada Titik Simpul B
EV = 0
R1 + S1 +S3 Cos ϕ = 0
 S3 =     -(     R1          -        S1   )              -(     121.5  -         ###    )
                                                  =
                cos          ϕ                                    cos 26.57
  S3 =   -1643.458      kg

1.5.5 Perencanaan Batang Tarik
  Pu = S3 x 1.6 x 0.75 =    -1643.46              x      1.6       x      0.75    =      -1972.150 kg
BJ 37    fu =   3700 kg/cm        2
               fy =     2400 kg/cm2

    1.5.5.1 Kontrol Leleh
    Pu = φ . fy . Ag ; dengan φ = 0.9

      Ag perlu =               Pu                =            1972.150
                           ϕ                                                    =       0.913 cm2
                                     fy                0.9       x    2400
                                                                                    Tidak Menentukan

    1.5.5.2 Kontrol Putus
    Pu = φ . fu . 0,75 Ag ; dengan φ =                 0.75

      Ag perlu =                    Pu                  =                    1972.150
                                                                                                          =   0.95 cm2
                           ϕ        fu          0.75           0.75    x       3700      x     0.75
                                                                                                              Menentukan
    Ag perlu = 1/4 . π . d2
                      Ag             x           4             0.95    x        4
      d       =                                         =                                =     1.1 cm
                                     π                                 π
     ==>        Pakai d =           11     mm

    1.5.6 Kontrol Kelangsingan
    Jarak kuda-kuda =     400 cm

     Panjang S3 =         (jarak kuda-kuda)2 + (jarak miring gording)2
                 =         400 2         +        110 2
                 =         ### cm

    Cek :                         Panjang S3
                 d         >
                                      500
                                 414.85
                1.1        >
                                  500
                1.1        >      0.83                 OK


1.6 Perencanaan Gording Ujung
     1.6.1 Perencanaan Pembebanan Mntx , Mnty dan Gaya Normal Akibat Angin
     Gording Ini adalah Balok Kolom. Akibat beban mati dan beban hidup Menghasilkan Momen Lentur
     Besaran Diambil Dari Perhitungan Gording

    Mntx = MUX (1.2 D + 1.6 L + 0.8 W) x 0.75           =        199.170        x       0.75    =         149.377   kgm
    Mnty = MUY (1.2 D + 1.6 L + 0.8 W) x 0.75 =         25.097      x                   0.75    =         18.823    kgm
    Nu = 1.6 x Rtotal (dari ikatan angin atap) x 0.75 =      1618.144               kg

    1.6.2 Perencanaan Profil Gording Ujung

     WF    100    x                 50       x    5             x     7
      A = 11.85 cm2                    tf = 7 mm                Zx = 41.8     cm3
     W = 9.3 kg/m                     Ix = 187 cm4              Zy = 8.9      cm3
       a = 100 mm                     Iy = 14.8 cm4              h = 70       mm    {=D - 2 x (tf + r)}
      bf = 50 mm                     tw = 5 mm
      iy = 1.12 cm                    ix = 3.98 cm

 Mutu Baja =    BJ 37
      fu = 3700 kg/cm2 =                        370 Mpa
  fy = 2400 kg/cm2                  =        240 Mpa

1.6.3 Kontrol Tekuk Profil

                                                     Lkx              400
Lkx =       400       cm        ==>          λx =               =               =      100.5
                                                      ix              3.98
                       π2 . E . A                          π2          x        2000000           x     11.85
Ncrbx       =                                    =
                         λx     2
                                                                      100.5 2
            =          23157.64             kg

                                                                Lkx            50
Lky =       50        cm        ==>                  λy =              =                    =   44.64
                                                                 iy           1.12
                       π2 . E . A                          π2          x        2000000           x     11.85
Ncrby       =                                    =
                         λy     2
                                                                      44.64 2
            =         117366.49             kg

Tekuk Kritis adalah arah X, Karena λx > λy                      ω=     2.29

        Ag x fy              11.85            x      2400
Pn =      ω            =                                        =     12437.136        kg
                                            2.29

       Pu                                    1618.144
                       =                                               =        0.15        <    0.2            (Pu = Nu)
  φ         Pn                  0.85          x    12437.136


Pakai Rumus =
                           Pu                                         Mux                               Muy
                                                      +                                     +                               ≤   1
                  2    x                φc . Pn                 φb     x      Mnx                φb      x       Mny

1.6.4 Perhitungan Faktor Pembesaran Momen
Gording dianggap tidak bergoyang, maka :

Mux = Mntx . Sbx                         Cmx
                             Sbx =                                              ≥      1
                                            Nu
                              1       - (        )
                                           Ncrbx
        Untuk elemen Beban Tranversal, ujung sederhana
        Cmx =     1
                              1
        Sbx =                                       =                         1.08
                              1618.144
                  1     - (               )
                              23157.64

        Sbx =         1.08          >            1
        Sbx =         1.08

Muy = Mnty * Sby                       Cmy
                             Sby =                                              ≥      1
                                           Nu
                             1       - (        )
                                          Ncrby
Untuk elemen Beban Tranversal, ujung sederhana
       Cmy =   1
                             1
      Sby =                                                            =      1.01
                            1618.144
               1      - (                )
                            117366.49

        Sby =         1.01          >            1
           Sby =   1.01

    1.6.5 Perhitungan Momen Ultimate Sbx dan Sby
    Mux = Sbx . Mntx =   1.08 x        149.377      =       160.599   kgm
    Muy = Sby . Mnty =   1.08 x         18.823      =        20.237   kgm


    1.6.6 Perhitungan Persamaan Interaksi
            Mnx = 1003 kgm                  Mny =   105   kgm

             Pu                              Mux                      Muy
                                 +                         +                       ≤    1
2     x     φc     x      Pn           φb    x     Mnx           φb    x    Mny
             1618.144                            160.599                   20.237
                                       +                         +                      ≤   1
2     x    0.85    x      12437.136          0.9    x    1003          0.9    x   105
                                            0.47    ≤      1
                                                   OK
      Pre - Eliminary Design
 2 Perencanaan Dinding
2.1 Data - Data perencanaan
    Data Dinding :
    Jenis                             :   Seng Gelombang
    Tebal                             :      4 mm
    Berat                             :    4.15 kg/m2
    Kedalaman Gelombang               :     25 mm
    Jarak Kolom Dinding (L)           :    400 cm
    Jarak Gording Lt Dasar            :    125 cm
    Jarak Gording Lt 1                :    100 cm

2.2 Perencanaan Regel Balok ( Dinding Samping )
    2.2.1 Perencanaan Profil WF untuk Regel Balok Dinding Dengan ukuran :

    WF    100      x        50       x      5     x       7
     A = 11.85 cm2            tf =      7 mm      Zx =    41.8 cm3
    W=      9.3 kg/m         Ix =     187 cm4     Zy =    8.94 cm3
      a = 100 mm             Iy =    14.8 cm4      h=       70 mm       {=D - 2 x (tf + r)}
     bf =    50 mm          tw =        5 mm      Sx =    37.5 mm
     iy = 1.12 cm            ix =    3.98 cm
                               r=         mm
Mutu Baja = BJ 37
     fu = 3700 kg/cm2 =              370 Mpa
      fy = 2400 kg/cm2 =             240 Mpa

   2.2.2 Perencanaan Pembebanan
   2.2.2.1 Perhitungan Beban
   Beban Mati
   Lantai Dasar
   Berat Gording                                                           =       9.3 kg/m
   Berat Seng Gelombang = 4.15              x     1.25                     =      5.19 kg/m
                                                         Berat Total       =     14.49 kg/m
   alat Pengikat dll 10 %       = 0.1       x    14.49                     =      1.45 kg/m
                                                       Berat Total         =     15.94 kg/m
   Myd = 1/8 x q x (L/3)2
                                = 0.13      x    15.94  x     1.78         =      3.54 kg/m

     Lantai 1
   Berat Gording                                                           =       9.3 kg/m
   Berat Seng Gelombang         = 4.15      x     1                        =      4.15 kg/m
                                                          Berat Total      =     13.45 kg/m
   alat Pengikat dll 10 %       = 0.1            13.45                     =      1.35 kg/m
                                                         Berat Total       =      14.8 kg/m
   Myd = 1/8 x q x (L/3)2       = 0.13      x    14.8     x     1.78       =      3.29 kg/m

   Beban Angin
   Lantai Dasar
   Tekanan Angin                                                           =     30     kg/m2
   Angin Tekan ( C = 0.9 )         =       0.9     x      30               =     27     kg/m2
   q = Angin Tekan x Jarak Gording =       27      x     1.25              =    33.75   kg/m
   Angin Hisap ( C = 0.4 )                 0.4     x      30               =     12     kg/m2
   q = Angin hisap x Jarak Gording =       12      x     1.25              =     15     kg/m
Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :
Mxw = 1/8 x q x (L)2   = 0.13      x    33.75       x   16   =      67.5 kgm
N = q x Jarak Gording = 15         x     1.25                =     18.75 kg       (Tarik)

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :
Mxw = 1/8 x q x (L)2   = 0.13      x      15      x     16   =        30 kgm
N = q x Jarak Gording = 33.75      x    1.25                 =     42.19 kg       (Tekan)

   Lantai 1
Tekanan Angin                                                =     30     kg/m2
Angin Tekan ( C = 0.9 )         =   0.9     x      30        =     27     kg/m2
q = Angin Tekan x Jarak Gording =   27      x       1        =     27     kg/m
Angin Hisap ( C = 0.4 )             0.4     x      30        =     12     kg/m2
q = Angin hisap x Jarak Gording =   12      x       1        =     12     kg/m

Akibat Beban Angin yg Tegak Lurus Dinding (tarik)
Mxw = 1/8 x q x (L)2   = 0.13      x     27       x     16   =          54 kgm
N = q x Jarak Gording = 12         x      1                  =          12 kg     (Tarik)

Akibat Beban Angin yg Tegak Lurus Gevel (tekan)
Mxw = 1/8 x q x (L)2   = 0.13      x      12    x       16   =          24 kgm
N = q x Jarak Gording = 27         x       1                 =          27 kg     (Tekan)

2.2.3 Kombinasi Pembebanan
Lantai Dasar
1. U = 1.4 D
       Muy = 1.4    x   3.54         =     4.96 kgm

2. U = 1.2D + 1.3W + λ L + 0.5 ( La atau Ha )
Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :
Mux = 1.2        x     0      +      1.3     x   67.5   +    0.5    x       0       +       0.5   x   0
     = 87.75 kgm
Muy = 1.2        x   3.54     +      1.3     x      0   +    0.5    x       0       +       0.5   x   0
     = 4.25 kgm
 Nu = 1.2        x     0      +      1.3     x  18.75   +    0.5    x       0       +       0.5   x   0
     = 24.38 kg

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :
Mux = 1.2      x       0    +     1.3     x       30    +    0.5    x       0       +       0.5   x   0
    = 39      kgm
Muy = 1.2      x     3.54   +     1.3     x        0    +    0.5    x       0       +       0.5   x   0
    = 4.25 kgm
 Nu = 1.2      x       0    +     1.3     x    42.19    +    0.5    x       0       +       0.5   x   0
    = 54.84 kg

   Lantai 1
1. U = 1.4 D
       Muy = 1.4       x     3.29     =    4.6 kgm
2. U = 1.2D + 1.3W + λ L + 0.5 ( La atau Ha )
Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :
Mux = 1.2        x     0      +      1.3     x    54    +    0.5    x       0       +       0.5   x   0
     = 70.2 kgm
Muy = 1.2        x   3.29     +      1.3     x      0   +    0.5    x       0       +       0.5   x   0
     = 3.95 kgm
 Nu = 1.2        x     0      +      1.3     x    12    +    0.5    x       0       +       0.5   x   0
     = 15.6     kg
Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :
Mux = 1.2       x      0    +     1.3     x       24      +      0.5     x     0    +   0.5   x   0
    = 31.2 kgm
Muy = 1.2       x    3.29   +     1.3     x        0      +      0.5     x     0    +   0.5   x   0
    = 3.95 kgm
 Nu = 1.2       x      0    +     1.3     x       27      +      0.5     x     0    +   0.5   x   0
    = 35.1     kg

2.2.4 Kontrol Kekuatan Profil
2.2.4.1 Penampang Profil
Untuk Sayap                          Untuk Badan
      bf             170               h                 1680
                ≤                            ≤
   2      tf          fy              tw                   fy
      50             170              70                 1680
                ≤                            ≤
   2      7          240               5                  240
     3.57       ≤   10.97             14.0 ≤             108.4
               OK                           OK
Penampang Profil Kompak, maka Mnx = Mpx

2.2.4.1 Kontrol Lateral Buckling
Jarak Baut Pengikat / pengaku lateral = L B =     500    mm       =     50    cm

                                         E
 LP =         1.76    x    iY    x
                                         fy
                                       200000
    = 1.76            x   1.12   x               =  56.90 cm
                                        240
             Ternyata :    LB    <    LP maka : Mnx   =   Mpx

Mnx = Mpx = Zx . Fy =      41.8       x      2400         =       ### Kgm
1.5 Myx = 1.5 Sx fy = 1.5   x        37.5      x  2400    =      1350 Kgm
===> Mnx         < 1.5 Myx

Mny = Zy ( satu sayap ) * fy
    = 1/4 x tf x bf 2 x fy
    = 0.25        x       0.7    x       52        x     2400     =    10500 kgcm
    = 105 kgm

2.2.5 Perhitungan Kuat Tarik
2.2.5.1 Kontrol Kelangsingan
                λp     ≤   300
         Lk          400
  λ=            =            =       100.5    <   300     OK
         ix          3.98
        λp




2.2.5.2 Berdasarkan Tegangan Leleh
φ Nn = φ .Ag . fy = 0.85   x   11.85          x   2400    = 24174 kg
                                                           Menentukan
2.2.5.3 Berdasarkan Tegangan Putus
φ Nn = φ .Ae . fu =   = 0.75 x 0.85 x Ag x fu
                      =    0.75     x  0.85        x      Ag      x     fu
                      =    0.75     x  0.85        x     11.85    x    3700
                      =      ### kg
                    Tidak Menentukan

2.2.5.4 Kontrol Kuat Tarik
Lantai Dasar
φ Nn      >   Nu
24174 >      54.84
        OK
   Lantai 1
φ Nn      >   Nu
24174 >      1404
        OK

2.2.6 Perhitungan Kuat Tekan
2.2.6.1 Kontrol Kelangsingan
         λp     ≤    200
         Lkx         400
λpx =           =           =   100.5    <      200         OK
          ix         3.98
         Lky          50
λpy =           =           =   44.64    <      200         OK
          iy         1.12

2.2.6.2 Berdasarkan Tekuk Arah X
          λx      fy      100.5            2400
 λc =                 =          x                           =     1.11
          π       E        3.14          2000000
 0.25     <       λc  <     1.2
             1.43                       1.43
  ω=                  =                                            =      1.67
        1.6 - 0.67 λc       1.6  -      0.67      x     1.11
                  fy                                    2400
 φ Nn = φAg ω         =    0.85  x      11.85     x                =      ###    kg
                                                        1.67

2.2.6.3 Berdasarkan Tekuk Arah Y
          λy      fy  =   44.64            2400
 λc =                            x                           =     0.49
          π       E        3.14          2000000
 0.25     <       λc  <     1.2
             1.43                       1.43
  ω=                  =                                            =      1.13
        1.6 - 0.67 λc       1.6  -      0.67      x     0.49
                  fy                                    2400
 φ Nn = φAg ω         =    0.85  x      11.85     x                =      ###    kg
                                                        1.13


2.2.7 Perhitungan Pembesaran Momen
               Ab     x      fy
Ncr      =
                        λc 2
              11.85   x     2400
Ncrbx    =                       =       23156.27       kg
                    1.108 2
              11.85   x     2400
Ncrby    =                       =      117359.57 kg
                    0.492 2

2.2.7.1 Komponen Struktur Ujung Sederhana Cm =          1
                Cmx
Sbx =                            ≥ 1
                    Nu
         1    - (       )
                  Ncrbx

Lantai Dasar
                     1
Sbx =                                    =      1.001        (Tarik)
                       24.38
        1      - (              )
                     23156.27
                     1
Sby =                                    =      1.000        (Tarik)
Sby =                                             =    1.000    (Tarik)
                           24.38
         1         - (             )
                         117359.57

                          1
Sbx =                                             =    1.002    (Tekan)
                           54.844
         1         - (             )
                          23156.27
                          1
Sby =                                             =    1.000    (Tekan)
                           54.844
         1         - (             )
                         117359.57

  Lantai 1
                           1
Sbx =                                             =    1.001    (Tarik)
                            15.6
         1         - (             )
                          23156.27
                          1
Sby =                                             =    1.000    (Tarik)
                            15.6
         1         - (             )
                         117359.57

                           1
Sbx =                                             =    1.002    (Tekan)
                            35.1
         1         - (             )
                          23156.27
                          1
Sby =                                             =    1.000    (Tekan)
                            35.1
         1         - (             )
                         117359.57

2.2.8 Kontrol Gaya Kombinasi
2.2.8.1 Angin Dari Arah Tegak Lurus Dinding (tarik)
Lantai Dasar
  Nu             24.38
         =                  =         0         <              0.2   OK
φ . Nn          24174

              Nu                       Mux        x    Sbx           Muy     x   Sby
                               +                               +                         <   1
     2   x         φ . Nn                φ         x    Mnx           φb     x    Mny
          24.375                       87.8        x   1.001         4.250   x   1.000
                               +                               +                         <   1
     2   x       24174                  0.9        x   1003           0.9    x    105
                                       0.14        <     1
                                                  OK
   Lantai 1
  Nu                15.6
          =                    =              0         <      0.2   OK
φ . Nn             24174

              Nu                       Mux        x    Sbx           Muy     x   Sby
                               +                               +                         <   1
     2   x       φ . Nn                  φ         x    Mnx           φb     x    Mny
          15.600                       70.2        x   1.001         3.945   x   1.000
                               +                               +                         <   1
     2   x       24174                  0.9        x   1003           0.9    x    105
                                       0.12        <     1
                                                  OK

2.2.8.2 Angin Dari Arah Tegak Lurus Gevel (tekan)
Lantai Dasar
  Nu          54.84
         =             =       0        <      0.2             OK
φ . Nn        24174
            Nu                   Mux    x     Sbx                    Muy     x   Sby
                            +                                  +                         <   1
     2 x         φ . Nn           φ     x     Mnx                    φb      x   Mny
              54.844                        39.0     x     1.002          4.250    x    1.000
                                  +                                +                            <   1
        2    x       24174                   0.9     x     1003            0.9     x     105
                                            0.09     <       1
                                                    OK
      Lantai 1
     Nu           35.1
             =             =            0           <       0.2    OK
   φ . Nn         24174
                Nu                          Mux      x     Sbx            Muy      x     Sby
                                  +                                +                            <   1
        2    x       φ . Nn                   φ      x      Mnx            φb      x     Mny
              35.100                        31.2     x     1.002          3.945    x    1.000
                                  +                                +                            <   1
        2    x       24174                   0.9     x     1003            0.9     x     105
                                            0.08     <       1
                                                    OK

2.3 Perencanaan Regel Horizontal Gevel
     2.3.1. Data - Data perencanaan tambahan
    Jarak Kolom Dinding (L)      :    300 cm
    Jarak Gording Lt Dasar       :    125 cm
    Jarak Gording Lt 1           :    100 cm

   2.3.2 Perencanaan Profil WF untuk Regel Horizontal Gevel Dengan ukuran :
    WF     100      x  50      x     5     x     7
      A = 11.85 cm2      tf =    7 mm     Zx = 41.8 cm3
     W=      9.3 kg/m   Ix = 187 cm4      Zy =      9 cm3
       a = 100 mm       Iy = 14.8 cm4       h=     70 mm
     bf =     50 mm    tw =      5 mm     Sx = 37.5 mm
      iy = 1.12 cm      ix = 3.98 cm             41.8
                          r=                     8.94
Mutu Baja = BJ 37
     fu = 3700 kg/cm2 =       370 Mpa
      fy = 2400 kg/cm2 =      240 Mpa

   2.3.3 Perencanaan Pembebanan
   2.3.3.1 Perhitungan Beban
   Beban Mati
   Lantai Dasar
   Berat Gording                                                           =        9.3 kg/m
   Berat Seng Gelombang = 4.15               x      1.25                   =       5.19 kg/m
                                                           Berat Total     =      14.49 kg/m
   alat Pengikat dll 10 %      = 0.1         x     14.49                   =       1.45 kg/m
                                                           Berat Total     =      15.94 kg/m
   Myd = 1/8 x q x (L/3)2      = 0.13        x     15.94    x      1       =       1.99 kg/m

     Lantai 1
   Berat Gording                                                           =        9.3 kg/m
   Berat Seng Gelombang        = 4.15        x      1                      =       4.15 kg/m
                                                            Berat Total    =      13.45 kg/m
   alat Pengikat dll 10 %      = 0.1         x     13.45                   =       1.35 kg/m
                                                           Berat Total     =       14.8 kg/m
   Myd = 1/8 x q x (L/3) 2
                               = 0.13        x     14.8     x      1       =       1.85 kg/m

   Beban Angin
   Lantai Dasar
   Tekanan Angin                                                           =       30   kg/m2
   Angin Tekan ( C = 0.9 )             = 0.9         x      30             =       27   kg/m2
q = Angin Tekan x Jarak Gording = 27        x     1.25       =     33.75 kg/m
Angin Hisap ( C = 0.4 )           0.4       x      30        =      12 kg/m2
q = Angin hisap x Jarak Gording = 12        x     1.25       =      15 kg/m

Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :
Mxw = 1/8 x q x (L)2   = 0.13      x    33.75       x    9   =     37.97 kgm
N = q x Jarak Gording = 15         x     1.25                =     18.75 kg      (Tarik)

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :
Mxw = 1/8 x q x (L)2   = 0.13      x      15      x      9   =     16.88 kgm
N = q x Jarak Gording = 33.75      x    1.25                 =     42.19 kg      (Tekan)

   Lantai 1
Tekanan Angin                                                =      30   kg/m2
Angin Tekan ( C = 0.9 )         =   0.9     x      30        =      27   kg/m2
q = Angin Tekan x Jarak Gording =   27      x       1        =      27   kg/m
Angin Hisap ( C = 0.4 )             0.4     x      30        =      12   kg/m2
q = Angin hisap x Jarak Gording =   12      x       1        =      12   kg/m

Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :
Mxw = 1/8 x q x (L)2   = 0.13      x     27         x    9   =     30.38 kgm
N = q x Jarak Gording = 12         x      1                  =        12 kg      (Tarik)

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :
Mxw = 1/8 x q x (L)2   = 0.13      x      12      x      9   =      13.5 kgm
N = q x Jarak Gording = 27         x       1                 =        27 kg      (Tekan)

2.3.3.2 Kombinasi Pembebanan
Lantai Dasar
1. U = 1.4 D
       Muy = 1.4       x     1.99     =    2.79 kgm
2. U = 1.2D + 1.3W + λ L + 0.5 ( La atau Ha )
Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :
Mux = 1.2        x     0      +      1.3     x  37.97    +   0.5    x      0       +       0.5   x   0
     = 49.36 kgm
Muy = 1.2        x   1.99     +      1.3     x      0    +   0.5    x      0       +       0.5   x   0
     = 2.39 kgm
 Nu = 1.2        x     0      +      1.3     x  18.75    +   0.5    x      0       +       0.5   x   0
     = 24.38 kg

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :
Mux = 1.2      x       0    +     1.3     x    16.88     +   0.5    x      0       +       0.5   x   0
    = 21.94 kgm
Muy = 1.2      x     1.99   +     1.3     x       0      +   0.5    x      0       +       0.5   x   0
    = 2.39 kgm
 Nu = 1.2      x       0    +     1.3     x    42.19     +   0.5    x      0       +       0.5   x   0
    = 54.84 kg

   Lantai 1
1. U = 1.4 D
       Muy = 1.4       x     1.85     =    2.59 kgm
2. U = 1.2D + 1.3W + λ L + 0.5 ( La atau Ha )
Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :
Mux = 1.2        x     0      +      1.3     x  30.38    +   0.5    x      0       +       0.5   x   0
     = 39.49 kgm
Muy = 1.2        x   1.85     +      1.3     x      0    +   0.5    x      0       +       0.5   x   0
    =   2.22   kgm
 Nu =    1.2     x      0       +    1.3        x   12      +      0.5     x     0    +   0.5   x   0
    =   15.6    kg

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :
Mux = 1.2       x      0    +     1.3     x     13.5        +      0.5     x     0    +   0.5   x   0
    = 17.55 kgm
Muy = 1.2       x    1.85   +     1.3     x        0        +      0.5     x     0    +   0.5   x   0
    = 2.22 kgm
 Nu = 1.2       x      0    +     1.3     x       27        +      0.5     x     0    +   0.5   x   0
    = 35.1     kg

2.3.4 Kontrol Kekuatan Profil
2.3.4.1 Penampang Profil
Untuk Sayap                          Untuk Badan
      bf             170               h                   1680
                ≤                            ≤
   2      tf          fy              tw                     fy
      50             170              70                   1680
                ≤                            ≤
   2      7          240               5                    240
     3.57       ≤   10.97             14.0 ≤               108.4
               OK                           OK
Penampang Profil Kompak, maka Mnx = Mpx

2.3.4.1 Kontrol Lateral Buckling
Jarak Baut Pengikat / pengaku lateral = L B =  500 mm     =               50    cm
                                        E
 LP = 1.76       x     iY     x
                                        fy
                                      200000
     = 1.76      x    1.12    x                 =  56.90 cm
                                       240
       Ternyata :      LB     <      LP maka : Mnx   =   Mpx

Mnx = Mpx = Zx . Fy =      41.8      x      2400            =       ### Kgm
1.5 Myx = 1.5 Sx fy = 1.5   x       37.5      x  2400       =      1350 Kgm
===> Mnx         < 1.5 Myx

Mny = Zy ( satu sayap ) * fy
    = 1/4 x tf x bf 2 x fy
    = 0.25        x       0.7   x          52        x     2400     =    10500 kgcm
    = 105 kgm

2.3.5 Perhitungan Kuat Tarik
2.3.5.1 Kontrol Kelangsingan
                λp     ≤   300
         Lk          300
  λ=            =            =      75.38       <   300     OK
         ix          3.98

2.3.5.2 Berdasarkan Tegangan Leleh
φ Nn = φ .Ag . fy = 0.85   x   11.85            x   2400    = 24174 kg
                                                             Menentukan
2.3.5.3 Berdasarkan Tegangan Putus
φ Nn = φ .Ae . fu =     = 0.75 x 0.85 x Ag x fu
                        =    0.75     x  0.85        x      Ag      x     fu
                        =    0.75     x  0.85        x     11.85    x    3700
                        =      ### kg
                      Tidak Menentukan
2.3.5.4 Kontrol Kuat Tarik
Lantai Dasar
φ Nn      >     Nu
24174 >       54.84
         OK
   Lantai 1
φ Nn      >     Nu
24174 >       1404
         OK

2.3.6 Perhitungan Kuat Tekan
2.3.6.1 Kontrol Kelangsingan
         λp     ≤    200
         Lkx         300
λpx =           =           =     75.38    <      200      OK
          ix         3.98
         Lky          50
λpy =           =           =     44.64    <      200      OK
          iy         1.12

2.3.6.2 Berdasarkan Tekuk Arah X
          λx      fy      75.38              2400
 λc =                 =          x                             =       0.83
          π       E        3.14            2000000
 0.25     <       λc  <     1.2
             1.43                         1.43
  ω=                  =                                                =      1.37
        1.6 - 0.67 λc       1.6  -        0.67      x     0.83
                  fy                                      2400
 φ Nn = φAg ω         =    0.85  x        11.85     x                  =      ###    kg
                                                          1.37

2.3.6.3 Berdasarkan Tekuk Arah Y
          λy      fy  =   44.64              2400
 λc =                            x                             =       0.49
          π       E          π             2000000
 0.25     <       λc  <     1.2
             1.43                         1.43
  ω=                  =                                                =      1.13
        1.6 - 0.67 λc       1.6  -        0.67      x     0.49
                  fy                                      2400
 φ Nn = φAg ω         =    0.85  x        11.85     x                  =      ###    kg
                                                          1.13

2.3.7 Perhitungan Pembesaran Momen
               Ab     x      fy
Ncr      =
                        λc 2
              11.85   x     2400
Ncrbx    =                       =         41166.71       kg
                    0.831 2
              11.85   x     2400
Ncrby    =                       =        117366.49 kg
                    0.492 2

2.3.7.1 Komponen Struktur Ujung Sederhana Cm =                     1
                Cmx
Sbx =                            ≥ 1
                    Nu
         1    - (       )
                  Ncrbx

Lantai Dasar
                      1
Sbx =                                      =      1.001        (Tarik)
                          24.38
         1      - (               )
        1      - (             )
                      41166.71
                      1
Sby =                                    =   1.000     (Tarik)
                        24.38
        1      - (             )
                     117366.49

                      1
Sbx =                                    =   1.001     (Tekan)
                       54.844
        1      - (             )
                      41166.71
                      1
Sby =                                    =   1.000     (Tekan)
                       54.844
        1      - (             )
                     117366.49

  Lantai 1
                     1
Sbx =                                    =   1.000     (Tarik)
                        15.6
        1      - (             )
                      41166.71
                      1
Sby =                                    =   1.000     (Tarik)
                        15.6
        1      - (             )
                     117366.49

                     1
Sbx =                                    =   1.001     (Tekan)
                        35.1
        1      - (             )
                      41166.71
                      1
Sby =                                    =   1.000     (Tekan)
                        35.1
        1      - (             )
                     117366.49

2.3.8 Kontrol Gaya Kombinasi
2.3.8.1 Angin Dari Arah Tegak Lurus Dinding (tarik)
Lantai Dasar
  Nu              24.375
          =                 =         0         <     0.2   OK
φ . Nn            24174
             Nu                  Mux     x     Sbx          Muy     x   Sby
                            +                         +                         <   1
     2 x          φ . Nn           φ     x     Mnx           φb     x    Mny
           24.375                49.4    x    1.001         2.390   x   1.000
                            +                         +                         <   1
     2 x          24174           0.9    x    1003           0.9    x    105
                                 0.08    <      1
                                        OK
   Lantai 1
  Nu               15.6
          =                 =         0         <     0.2   OK
φ . Nn            24174
             Nu                  Mux     x     Sbx          Muy     x   Sby
                            +                         +                         <   1
     2 x          φ . Nn           φ     x     Mnx           φb     x    Mny
           15.600                39.5    x    1.000         2.219   x   1.000
                            +                         +                         <   1
     2 x          24174           0.9    x    1003           0.9    x    105
                                 0.07    <      1
                                        OK

2.3.8.2 Angin Dari Arah Tegak Lurus Gevel (tekan)
Lantai Dasar
  Nu          54.84
         =             =       0        <      0.2    OK
φ . Nn        24174
            Nu                   Mux    x     Sbx           Muy     x   Sby
                            +                         +                         <   1
     2 x         φ . Nn           φ     x     Mnx           φb      x   Mny
             54.844                         21.9    x      1.001         2.390    x   1.000
                                    +                               +                            <   1
        2   x       24174                    0.9    x      1003           0.9     x    105
                                            0.05    <        1
                                                   OK
      Lantai 1
     Nu          35.1
             =            =             0              <    0.2    OK
   φ . Nn        24174
               Nu                           Mux        x    Sbx          Muy      x   Sby
                                    +                               +                            <   1
        2   x       φ . Nn                    φ     x       Mnx           φb      x    Mny
             35.100                         17.6    x      1.001         2.219    x   1.000
                                    +                               +                            <   1
        2   x       24174                    0.9    x      1003           0.9     x    105
                                            0.04    <        1
                                                   OK

2.4 Perencanaan kolom Gevel
    2.4.1 Data Perencanaan
    Panjang Beban Atap Regel 5      =        3     m       Panjang Cantilever =   1   m
    Panjang Beban Atap Regel 2      =        3     m       Jarak Kuda-kuda =      4   m

   Lebar Beban Atap Regel 5 = 2.5 m                        qw regel 5 = panjang x angin tekan
   Lebar Beban Atap Regel 2 = 2 m                                     =   3      x     27     =      81   kg/m
                                                           qw regel 2 = panjang x angin tekan
   Tinggi Regel 5 =         7   m                                     =   3      x     27     =      81   kg/m
   Tinggi Regel 2 =         6   m

     Regel 5
   Luas atap yg Dipikul oleh Regel 5 ( A1 ) = Lebar Beban Atap Regel 5 x Pjg Beban Atap Regel 5
                                            = 3        x    2.5
                                            = 7.5 m2
   Luas Dinding Regel 5 ( A2 )              = Pjg Beban Atap Regel 5 x Tinggi Regel 5
                                            = 2.5      x     7
                                            = 17.5 m2

     Regel 2
   Luas atap yg Dipikul oleh Regel 2 ( A3 ) =Lebar Beban Atap Regel 2 x Pjg Beban Atap Regel 2
                                            = 3        x     2
                                            = 6 m2
   Luas Dinding Regel 2 ( A4 )              = Pjg Beban Atap Regel 2 x Tinggi Regel 2
                                            = 2        x     6
                                            = 12 m2

   2.4.2 Perencanaan Pembebanan
   2.4.2.1 Beban Mati
      Regel 5
   ND atap = A1 x qD atap                              =    7.5     x   20.63     =   ### kg
   ND Dinding = A2 x qD Dinding                        =    17.5    x    4.15     =   72.63 kg
   ND Gording = Jml Gording . w Gording                =     7      x     9.3     =   65.1 kg

      Regel 2
   ND atap = A3 x qD atap                              =     6      x   20.63     =   ### kg
   ND Dinding = A4 x qD Dinding                        =    12      x    4.15     =   49.8 kg
   ND Gording = Jml Gording . w Gording                =     6      x     9.3     =   55.8 kg

   2.4.2.2 Beban hidup
      Regel 5
   NL atap = A1 x qL atap                               =      7.5      x       20     =   150 kg

      Regel 2
   NL atap = A2 x qL atap                               =       6       x       20     =   120 kg

   2.4.2.3 Beban Angin
      Regel 5
   Mw = 1/8 x qw x (h)2        =      0.13     x        81      x       7   2          =   ### kgm

     Regel 2
   Mw = 1/8 x qw x (h)2        =      0.13     x        81      x       6   2          =   364.5 kgm

   2.4.3 Syarat Kekakuan
      Regel 5
            h            700
     Y=            =                   =      3.5       cm
           200           200
            5              q            x    L4
    Ix =           x
           384            E             x     Y
            5            4.96           x     7 4               =      2215.767      cm4
                   x
           384           0.02           x    3.5
   ===> Ix Profil yg Dipakai >         2215.767 cm4

   Pakai Profil :
    WF     175         x    175         x     7.5       x       11
     A = 51.21      cm2        tf =    11    mm         Zx =   ###    cm3
     W = 40.2       kg/m      Ix =    2880   cm4        Zy =   ###    cm3
      a = 175       mm        Iy =    984    cm4         h=    175      -       2     x(    11      +   12    )
     bf = 175       mm       tw =      7.5   mm            =   136    mm
     iy = 4.38      cm        ix =     7.5   cm         Sx =   2050   mm
                                r=     12    cm
Mutu Baja = BJ 37
     fu = 3700 kg/cm2 =               370 Mpa
      fy = 2400 kg/cm2 =              240 Mpa

   Nd Profil  =    7     x     40.2    =  281.4 kg
   Nd total   = Nd atap + Nd (Dinding+Gording ) + Nd Profil
              = ###      +     ###     +  281.4     =     ### kg
  NL Total = NL atap = 150 kg
   Mw       =      ### kgm
  U = ( 1.2D + 1.6L+ 1.6W ) x 0.75
  Nu = ( 1.2       x    ###      +    1.6   x      150     ) x 0.75                    =   ### kg
  Mntx = 1.6       x    Mw       x   0.75   =      1.6      x  ###                     x   0.75   =     ###   kg

      Regel 2
            h            600
    Y=             =                   =       3        cm
          200            200
            5              q            x    L4
    Ix =           x
          384             E             x    Y
            5            3.65           x     6     4           =      1025.156      cm4
                   x
          384            0.02           x     3
   ===> Ix Profil yg Dipakai >         1025.156     cm4

   Pakai Profil :
    WF     150        x     100        x       6        x       9
     A=     26.84   cm2      tf = 9 mm        Zx =    ###    cm3
     W=      21.1   kg/m    Ix = 1020 cm4     Zy =   45.88   cm3
     D=      148    mm      Iy = 151 cm4       h=     150      -     2    x(      9       +    11      )
     Bf =    100    mm     tw = 6 mm             =    116    mm
     iy =    2.37   cm      ix = 6.17 cm      Sx =    138    mm
                              r = 11 cm
Mutu Baja = BJ 37
     fu = 3700 kg/cm2 =         370 Mpa
      fy = 2400 kg/cm2 =        240 Mpa

   Nd Profil  =    6     x     21.1    =  126.6 kg
   Nd total   = Nd atap + Nd (Dinding+Gording ) + Nd Profil
              = ###      +    105.6    +  126.6     =     ### kg
  NL Total = NL atap = 120 kg
   Mw       =    364.5 kgm
  U = ( 1.2D + 1.6L+ 1.6W ) x 0.75
  Nu = ( 1.2       x    ###      +    1.6   x      120     ) x 0.75       =      ### kg
  Mntx = 1.6       x    Mw       x   0.75   =      1.6      x  364.5      x      0.75   =      437.4   kg

   2.4.4 Kontrol Tekuk
      Regel 5
   untuk arah x :
   Lkx = 700       cm
           Lkx             700
    λx =            =                =   93.33
            ix              7.5
            λx      fy             93.33          2400
    λc =                     =             x                 =     1.03
            π       E                π          2000000
                    π2 . E . A               π2        x     2000000      x      51.21
   Ncrbx    =                        =
                       λx 2
                                                     93.33 2
            =      116040.87 kg
   untuk Arah y :
   Lky = 100       cm
           Lky             100
    λy =            =                =   22.83
            iy             4.38
            λy      fy             22.83          2400
    λc =                     =             x                 =     0.25
            π       E                π          2000000
                    π2 . E . A               π2        x     2000000      x      51.21
   Ncrby    =                        =
                       λy 2
                                                     22.83 2
            =     1939245.26 kg
   Tekuk Kritis Adalah Arah ====> X karena λx                >     λy
    0.25    <       λc       <      1.2
               1.43                            1.43
    ω=                       =                                      =     1.57
          1.6 - 0.67 λc             1.6    -   0.67    x    1.03
    Pn = Ag . fy =        51.21      x   2400   =      122904    kg
     Pu                       696.47
            =                                   =     0.01 <      0.2
   φ . Pn         0.85       x       122904
   Pakai Rumus :
                        Pu                           Mux                         Muy
                                           +                        +                           ≤      1
                2 x           φc . Pn           φb     x    Mnx           φb      x      Mny

   Batang Dianggap Tidak Bergoyang Maka :
                   Cmx
   Sbx =                            ≥ 1                      ;Cm =   1
                        Nu
            1     - (       )
         1      - (       )
                    Ncrbx
                      1
Sbx =                                   =        1.006     ≥     1
                        696.5
        1       - (             )
                     116040.87
Mux = Mntx . Sbx
Mux = ###      x    1.006     =   ### kgm

   Regel 2
untuk arah x :
Lkx = 600       cm
        Lkx             600
 λx =            =                =   97.24
         ix             6.17
         λx      fy             97.24           2400
 λc =                     =             x                  =     1.07
         π       E                π           2000000
                 π2 . E . A                π2        x     2000000      x      26.84
Ncrbx    =                        =
                    λx 2
                                                   97.24 2
         =       56024.77 kg
untuk Arah y :
Lky = 100       cm
        Lky             100
 λy =            =                =   42.19
         iy             2.37
         λy      fy             42.19           2400
 λc =                     =             x                  =     0.47
         π       E                π           2000000
                 π2 . E . A                π2        x     2000000      x      26.84
Ncrby    =                        =
                    λy 2                           42.19 2
         =      297583.57 kg                                                                 λ
                                                                                       0.25 <<
                                                                                             c 1.2



Tekuk Kritis Adalah Arah ====> X karena λx                 >     λy
 0.25    <       λc       <      1.2
            1.43                             1.43
 ω=                       =                                       =     1.62
       1.6 - 0.67 λc             1.6     -   0.67    x    1.07
 Pn = Ag . fy =        26.84      x   2400    =       64416    kg
  Pu                       464.38
         =                                    =     0.01 <      0.2
φ . Pn         0.85       x        64416
Pakai Rumus :
                     Pu                            Mux                         Muy
                                        +                         +                           ≤      1
             2 x           φc . Pn            φb     x    Mnx           φb      x      Mny

Batang Dianggap Tidak Bergoyang Maka :
                 Cmx
Sbx =                             ≥ 1                    ;Cm =   1
                     Nu
         1      - (       )
                    Ncrbx
                      1
Sbx =                                   =        1.008     ≥     1
                        464.4
         1      - (             )
                      56024.77
Mux = Mntx . Sbx
Mux = 437.4    x    1.008     =   ### kgm

2.4.5 Menentukan Mnx
   Regel 5
 * Penampang Profil
Untuk Sayap :                            Untuk Badan :
      bf            170                    h           1680
              ≤                                  ≤
   2     tf          fy                   tw            fy
        175                     170                     136            1680
                          ≤                                      ≤
    2          11               240                     7.5             240
        7.95              ≤    10.97                     18.1    ≤     108.4
                         OK                                     OK

 Penampang Profil Kompak, maka Mnx = Mpx

 * Kontrol Lateral Buckling
 Lateral Bracing = L B =    1000              mm         =      100 cm

                                                 E
  LP =      1.76          x     iY       x
                                                 fy
                                               200000
        = 1.76            x    100       x                       =     ###      cm
                                                2400

          Ternyata :            LB       <    LP      maka : Mnx        =      Mpx

 Mnx = Mpx = Zx . Fy =                   0     x         p       =      0      Kgm
 Mny = Zy ( satu sayap ) * fy
     = 1/4 x tf x bf 2 x fy
     = 0.25        x        0            x         02            x      p       =    0   kgcm
     = 0 kgm

    Regel 2
 Penampang Profil
 untuk Sayap                                 untuk Badan

      b 170                                  h 1680
         ≤                                     ≤
     2tf   fy                                t   fy

 #REF!                   170                 #REF!              1680
 #REF!            ≤ #REF!                    #REF!           ≤ #REF!
 #REF!            ≤ #REF!                    #REF!           ≤ #REF!
          #REF!                                       #REF!

 Penampang Profil Kompak, maka Mnx = Mpx

 Lateral BracingLb =           100      cm



                    E
 Lp = 1.76 * iy          Lp = #REF! cm
                    fy

          Ternyata Lp > Lb             maka Mnx = Mpx

Mnx = Mpx = Zx. Fy = #REF!              *  #REF! = #REF! Kgm
 Mny = Zy ( 1 flen ) * fy
     = (1 / 4 * tf * bf 2) * fy
     = 0.25 x       #REF! x     #REF! x
                                      #REF! = #REF! kgcm
     = #REF! kgm

 2.4.6 Persamaan Interaksi
                  Pu                                            Mux                      Muy
                                               +                                +               <   1
                                               +                        +                  <   1
                  2     x           φc . Pn         φb        x   Mnx       φb   x   Mny

      Regel 5

        #REF!           +         598.94       +        0
     0.17 x x                   0.9 x#REF!          0.9 x#REF!

       #REF!            +           #REF!      +

                                       #REF!   <    1

                                               OK


      Regel 2

        #REF!           +         #REF!        +        0
     0.17 x
          #REF!                 0.9 x
                                    #REF!           0.9 x#REF!

       #REF!            +           #REF!      +

                                       #REF!   <    1

                                               OK


2.5 Perencanaan Penggantung Gording Dinding Samping dan Gevel

    2.5.1 Data Penggantung Gording

    Jarak Kuda - Kuda       =      400 cm
                            =
    Jumlah Penggantung Gording       2 buah
    Jumlah Gording Gevel =           7 buah
    Jumlah Gording Dinding=          3 buah
                            =
    Jarak Penggantung gording ### cm
    Jarak antara Gevel      =      300 cm
                                   Dinding
    Jarak Antar Gordng Horizontal 125 cm =
                                   Gevel =
    Jarak Antar Gordng Horizontal 100 cm
    2.5.2 Perencanaan Pembebanan
    Dinding Samping
    Beban Mati
    Berat Sendiri Gording                =             0 kg/m
    Berat Seng Gelombang                 =          4.15 kg/m
                                         =          4.15 kg/m
    Alat Pengikat dll 10 = 0.1 x 4.15
                         %               =             0 kg/m
                                         =          4.15 kg/m

        Ra = q * JarakKuda − Kuda              =    16.6 kg

    Gevel
    Beban Mati
    Berat Sendiri Gording                      =       0 kg/m
    Berat Seng Gelombang                       =    4.15 kg/m
                                               =    4.15 kg/m
    Alat Pengikat dll 10 = 0.1 x 4.15
                         %                     =    0.42 kg/m
                                               =    4.57 kg/m
     Ra = q * JarakGevel                      =         13.7 kg

2.5.3 Perhitungan Gaya
2.5.3.1 Penggantung Gording Tipe A
Dinding Samping                                              `
                          Ra     =                     23.24 kg

                             R
Ra Total = Ra * jumlah Gording a              =        69.72 kg

Gevel                                                        `
                                   Ra         =        19.17 kg

                              R
Ra Total = Ra * jumlah Gording a        =      ### kg
2.5.3.2 Penggantung Gording Tipe B
Dinding Samping
                           arctgnβ = 0.94
                                   β = 0.75
                                   β = 43.14 o


        RA          Rb     =      69.72       =         ###   kg
RB =
       Sinβ                        0.68


Gevel
                                 arctgnβ =     1.4
                                         β=   0.95
                                         β = 54.44 o


        RA          Rb     =         ###      =         ###   kg
RB =
       Sinβ                         0.81

2.5.4 Perencanaan Batang Tarik
Dinding Samping
  Pu = ### kg
BJ 37 fu = 0 kg/cm2
   fy =    0 kg/cm3

Gevel
  Pu = ### kg
BJ 37 fu = 0 kg/cm2
   fy =     0 kg/cm3
2.5.4.1 Kontrol Leleh
Dinding Samping
Pu = φ fy Ag dengan φ = 0.9

                =
Ag perlu = Pu/φ fy         ###       =        ### cm2
                            0                  ###
        Ag * 4
d=
         π
Gevel
Pu = φ fy Ag dengan φ = 0.9
                =
Ag perlu = Pu/φ fy        ###       =      ### cm2
                           0                ###
         Ag * 4
  d=
          π
2.5.4.2 Kontrol Putus
Dinding Samping
Pu = φ fu 0.75 Ag dengan φ = 0.75

Ag Perlu = Pu               =       ###      =       ### cm2
         φ fu 0.75                    0              ###



Ag = 1 / 4π d=
             2    ### cm2



            =     ### x 4            d=      ### cm
                     3.14

Pakai d = 10 mm




Dinding Samping
Pu = φ fu 0.75 Ag dengan φ = 0.75

Ag Perlu = Pu               =       ###      =       ### cm2
         φ fu 0.75                    0              ###



Ag = 1 / 4π d=
             2    ### cm2



            =     ### x 4            d=      ### cm
                     3.14

Pakai d = 10 mm


2.5.5 Kontrol Kelangsingan
Dinding Samping
                      ### =
Jarak Penggantung Gordingcm

 PanjangRb = JrkPenggantungGording 2 + JrkantarGordingHorizontal 2
Panjang Rb = ###            +    15625

Panjang Rb =       ### cm
                                    1        >      ###
            PanjangRb                               500
         d≥
               500
                                    1        >     0.37
                                               OK

    Gevel
                         100 =
    Jarak Penggantung Gordingcm

     PanjangRb = JrkPenggantungGording 2 + JrkantarGordingHorizontal 2
    Panjang Rb = 10000          +    10000

    Panjang Rb =       ### cm
                                        1        >      ###
               PanjangRb                                500
            d≥
                  500
                                        1       >      0.28
                                               OK


2.6 Perencanaan Ikatan Angin Dinding

    2.6.1 Data Perencanaan Ikatan Angin Dinding
    Tekanan Angin W =0 kg/m2
                      =
    Koefisien Angin C0.9
     a1 =     300 cm          a2 =    200 cm
       α= 0         =       00

    2.6.2 Perhitungan Tinggi Ikatan Angin ( h )
     h1 =       9m
     h2 =    9     +      2 x tg x 0.44        =                9.93     m
     h3 =    9     +      4 x tg x 0.44        =               10.87     m
     h4 =    9     +      6 x tg x 0.44        =                11.8     m
     h5 =    9     +      9 x tg x 0.44        =                13.2     m

    2.6.3 Perhitungan Gaya - Gaya yang Bekerja
    R = 1/2 W C a h
      R1 = 0.50 x 0 x  #REF! x 1 x 9        =                  #REF!     kg
      R2 = 0.50 x 0 x  #REF! x 2 x 9.93     =                  #REF!     kg
      R3 = 0.50 x 0 x  #REF! x 2 x10.87     =                  #REF!     kg
      R4 = 0.50 x 0 x  #REF! x 2.5 x 11.8   =                  #REF!     kg
      R5 = 0.50 x 0 x  #REF! x 3 x 13.2     =                  #REF!     kg


                   Rtotal = ( R1+R2+R3+R4+(R5/2)) = #REF!                kg

    2.6.4 Perencanaan Dimensi Ikatan Angin
                                             tg φ      =        1
                                                                4
                                                       =      0.25

                                               φ       =      0.24 rad

                                              R1       =      #REF! kg
                                             Rtotal    =      #REF! kg

2.6.4.1 Menghitung gaya Pada Titik Simpul
Pada Titik Simpul A
ΣV = 0
Rtotal + S1 = 0
S1 = - Rtotal
  S1 = #REF! kg

ΣH = 0
S2 = 0


Pada Titik Simpul B
EV = 0
R1 + S1 +S3 Cos Φ = 0

             − ( R1 − S1 )
      S3 =
                Cosφ
  S3 = #REF! kg

2.6.5 Perencanaan Batang Tarik

 Pu = #REF! kg      Pu = S 3 *1.6 * 0.75
BJ 37 fu = 0 kg/cm2
  fy =     0 kg/cm3

2.6.5.1 Kontrol Leleh

Pu = φ fy Ag dengan φ = 0.9

                =
Ag perlu = Pu/φ fy           #REF!    =      #REF! cm2
                               0             #REF!

2.6.5.2 Kontrol Putus

Pu = φ fu 0.75 Ag dengan φ = 0.75

Ag Perlu = Pu                  =     #REF!     =      #REF!
         φ fu 0.75                       0            #REF!



Ag = 1 / 4π d=
             2          #REF! cm2



     Ag * 4
d=         =            #REF! x 4      d = #REF! cm
      π                    3.14
Pakai d = 12 mm
2.6.6 Kontrol Kelangsingan

                400
Jarak Kuda - Kuda = cm

PanjangS 3 = JrkantarKuda − Kuda 2 + Jrkantar 2 Re gelHorizontal 2

Panjang S3 =           0       +        0

Panjang S3 =               0 cm
                                       1.2        >       0
                       PanjangS 3                        500
                  d≥
                          500          1.2       >         0
                                                OK
                              Start




Masukkan Data - Data Perencanaan Bondex dan Balok Anak :
Panjang Bentang Beban Bondex Yang Dipikul Balok Anak = ?
                 Panjang Balok Anak = ?
                 Berat Sendiri Beton = ?
                Berat Sendiri Bondex = ?
              Berat Spesi per cm Tebal = ?
                     Berat Tegel = ?
                   Beban Berguna = ?




         Hitung Pembebanan terhadap Balok Anak :
                       Beban Mati
                      Beban Hidup




                 Hitung Tebal Lantai Bondex
 Tebal Lantai Bondex Dicari dengan Menggunakan Tabel yang
  ada dengan memperhitungkan Beban Berguna yang akan
Disalurkan Bondex ke Balok Anak sebagai Dasar Perencanaan.
                             T=?




            Hitung Luasan Tulangan Negatif Bondex
 Luasan Tulangan Negatif Bondex Dicari dengan Menggunakan
Tabel yang ada dengan memperhitungkan Beban Berguna yang
     akan Disalurkan Bondex ke Balok Anak sebagai Dasar
                        Perencanaan.
                            A=?




           Asumsikan Diamter Tulangan Negatif
                       Bondex :
                       φ = ? Mm




  Hitung Banyaknya Tulangan Yang Diperlukan Tiap 1 m :
                       A/As = ?
              Hasilnya Dibulatkan Keatas




               Hitung Jarak Tulangan Tarik :
 Jarak Tulangan Tarik = Jarak Tulangan yang Diperlukan ( 1
  m ) Dibagi dengan Banyaknya Tulangan yang diperlukan
         dengan Jarak yang Telah Ditetapkan Diatas




               Perencanaan Pembebanan
                      Beban Mati
                     Beban Hidup




            Hitung qU, Mu Max dan Du Max :
                 qU = 1.2 qD + 1.6 qL

                   1                         1
        Mu max =     qu l 2       Du max =     qu l
                   8                         2
                                                                    PERHITUNGAN Ix PROFIL MINIMUM
                                                                         Dimana Y ijin = L/360
                                                                              5 ( qD + qL ) * l 4
                                                                      Ix >
                                                                             384      EY



                                                Pilih Profil Baja Dimana Ix-nya Harus > Ix Minimum :
                                  A = ? ; W = ? ; a = ? ; bf = ? ; iy = ? ;tf = ? ; Ix = ? Iy = ? ; tw = ? ; Zx = ? ; Zy
                                                               = ? ; h = ? ; fu = ? ; Fy = ?




                                                           Perencanaan Pembebanan + Berat Profil
                                                                        Beban Mati
                                                                       Beban Hidup




                                              Hitung qU, Mu Max dan Du Max ( Berat Profil Dimasukkan ) :
                                                                qU = 1.2 qD + 1.6 qL

                                                                         1                            1
                                                           Mu max =        qu l 2          Du max =     qu l
                                                                         8                            2



                                                                      KONTROL LENDUTAN BALOK
                                                                        Dimana Y ijin = L/360
                                                                                     5 (qD + qL) * l 4
                                                                       Y max =
                                                                                    384     EIx


                                         KO
              Perbesar Profil                                                  Y mak < Y ijin


                                                                             OK


                                                               KONTROL LOKAL BUCKLING
                                               Hitung λp, λr Penampang Sayap dan λp, λr Penampang Badan :
                                                 Sayap                                          Badan
                                                         170                                                          1680
                                                  λp =                                                         λp =
                                                               fy                                                         fy
                                                           370                                                        2550
                                                λr =                                                           λr =
                                                          f y − fr                                                        fy

                                                          b                                                           h
                                                         2tf                                                          t




                                                                                       1                                            2
                                               b                                     Sayap                                     Badan            h ≤λ
                                                  ≤ λp                                                                                               p
                                              2tf                                                                                               t

       Profil Kompak                     OK                                                       Profil Kompak                          OK
        Mnx = Mnp                                                                                  Mnx = Mnp
                                                      KO                                                                                                    KO



                                                 b                                                                                                 h
                                        λp ≤           ≤ λr                                                                                λp ≤      ≤ λr
                                                2tf                                                                                                t


                                                                                                Profil Tak Kompak
                       OK                                                                                                                OK
                                                                                                                          λ − λp
                                                      KO                                   Mn = Mp − ( Mp − Mr )                                            KO
                                                                                                                          λr − λp
                                          Profil Langsing                                                                                Profil Langsing

                                         Mn = Mr (λr / λ ) 2                                                                             Mn = Mr (λr / λ ) 2




     Profil Tak Kompak
                        λ − λp                                                               Mnx Sayap > Mnx
Mn = Mp − ( Mp − Mr )                                                                        Badan
                        λr − λp                   2
                                                                                      OK                                            KO
                                                   Ambil Mnx Badan                                        Ambil Mnx Sayap
                                                    Local Buckling                                         Local Buckling




                                                                      KONTOL LATERAL BUCKLING
                                                                 Hitung λp dan λr daripada Lateral Buckling

                                                                                                E
                                                                              Lp = 1.76 * iy
                                                                                                fy

                                                                                   X1                2
                                                                       Lr = ry (      ) 1+ 1+ X 2 fL
                                                                                   fL
                                                                                           E
                                                                                   G=
                                                                                        2(1 + µ)

                                                                                      π     EGJA
                                                                               X1 =
                                                                                      Sx     2

                                                                                            S 2 Iw
                                                                               X 2 = 4(       )
                                                                                           GJ   Iy




                                                                        Jarak Lateral Bracing λb :
                                                                                 λb = ?




   KO                                                       KO
                         λ p ≤ λb ≤ λr                                             λb ≤ λ p


                         OK                                                                    OK
                                                                               Bentang Pendek
                                                                                 Mnx = Mpx
                       Bentang Menengah
                                        Lr − L
           Mn = Cb( Mr + ( Mr − Mp )(           ) ≤ Mp
                                        Lr − Lp


                  Bentang Panjang
                   π               πE
   Mn = Mcr = Cb     E _ I y GJ + ( ) 2 I y I w ≤ Mp
                   L                L




                                              Mnx Local Buckling > Mnx Lateral
                                              Buckling

                                           OK                                                  KO


                   Ambil Mnx Lateral Buckling                                                  Ambil Mnx Local Buckling


                                                            Hitung : 0.9 Mnx



                                KO
     Perbesar Profil                                     0.9 Mnx > Mu max


                                                            OK


                                                   KONTROL KUAT RENCANA GESER
                                                             Hitung h
                                                                         tw



                                                            h    10
                                                                10
                                OK                            ≤
Vn = 0.6 fy Aw                                             tw     fy



                                                            KO
                       1100        h 1370      OK                     1100t w
                              ≤      ≤              Vn = 0.6 f y Aw
                         fy       tw   fy                             h fy


                        KO


                                  900000 Aw
                         Vn =
                                   ( h )2
                                      tw




                         Hitung 0.9 Vn



                  KO
Perbesar Profil         0.9 Vn > Vu Max


                        OK


                        Profil Dapat Dipakai
                     Pre - Eliminary Design
 3 Perencanaan Bondex dan Balok Anak
3.1 Data - Data perencanaan
    Beban Hidup             :              400 Kg/m2
    Beban Finishing         :               90 Kg/m2
    Beban Berguna           :              490 Kg/m3

   Berat Beton Kering      :         2400 kg/m3
   Panjang Bentang Beban Bondex yang Dipikul Oleh Balok Anak            :     3      m
   Panjang Balok Anak      :        4      m

3.2 Perencanaan Pelat Lantai Bondex

   3.2.1 Data Perencanaan
   Berat Sendiri Beton                 =           2400 kg/m3
   Berat Sendiri Bondex                =            10.1 kg/m2
   Berat Spesi per cm Tebal            =              21 kg/m2
   Berat Tegel                         =              24 kg/m2

   3.2.2 Perencanaan Pembebanan
   Beban Mati
   Berat Beton            =     2400              *         0.12       =        288 Kg/m2
   Berat Bondex                                                        =       10.1 Kg/m2
   Berat Spesi 2 Cm       =      21               *           2        =         42 Kg/m2
   Berat Tegel 2 Cm       =      24               *           2        =         48 Kg/m2
                                                             qD        =      388.1 Kg/m2
   Beban Hidup
   Beban Hidup Lantai gudang                                           =          400 Kg/m2
   Beban Finishing                                                     =           90 Kg/m2
                                                             qL        =          490 Kg/m2


   3.2.3 Perencanaan Tebal Lantai Beton dan Tulangan Negatif
   3.2.3.1 Perencanaan Tebal Lantai
      qL =       490     kg/m2

   Beban Berguna yang Dipakai =       500    kg/m2
   Jarak Antar Balok            =        300 cm
   Jarak Kuda - Kuda            =        400 cm

   Dari Tabel Brosur ( Bentang Menerus dengan Tulangan Negatif ),didapat :
        t         =          12   mm
        A         =         3.57  cm2/m

   3.2.3.2 Perencanaan Tulangan Negatif
           Direncanakan Tulangan Dengan φ =             10 mm
                                       As =           0.79 mm2

        Banyaknya Tulangan Yang diperlukan Tiap 1 m =        A         =     3.57
                                                             As              0.79
                                                                        =    4.55    Buah
                                                                        =      5     Buah
                                 Jarak Tulangan Tarik =     200        cm
   Pasang Tulangan Tarik φ10 - 200

3.3 Perencanaan Dimensi Balok Anak
    3.3.1 Perencanaan Pembebanan
    Beban Mati ( D )
    Bondex                 =       3                     10.1                 =              30.3 kg/m
    Plat Beton             =       3                     0.12    2400         =               864 kg/m
    Tegel + Spesi          =       3                      90                  =               270 kg/m
                                                    qD                        =            1164.3 kg/m


   Beban Hidup ( L )
      qL         =              3       490               =           1470 kg/m

   3.3.3 Perhitungan qU , Mu Max dan Du Max
   qU = 1.2 qD + 1.6 qL
       qU         =        1.2     1164.3                1.6     1470         =          3749.16 Kg/m
                   1
   Mu max =          qu l 2     =       0.13        3749.16       16          =           7498.32 Kgm
                   8
                   1
   Du max =          qu l       =       0.5         3749.16           4       =          7498.32 Kg
                   2

   3.3.4 Perhitungan Ix Profil Yang Diperlukan
          Y=      L          =        400                 =       1.11
                 360                  360
                  5 ( qD + qL ) * l 4
            Ix >
                 384      EY
       Ix              >        5              (     11.64        14.7    )              2.56E+10
                               384                  2100000       1.11


       Ix              >      3763.29   cm4

   3.3.5 Perencanaan Profil WF untuk Balok Anak

                     250         x      125               x           6       x             9

            A=         37.66 cm2             tf =              9 mm               Zx =     351.86 cm3
            W=          29.6 kg/m           Ix =           4050 cm4               Zy =      72.02 cm3
             a=          250 mm             Iy =            294 cm4                h=         208 mm
            bf =         125 mm            tw =                6 mm                r=          12 mm
            iy =        2.79 cm             ix =            10.4 cm                        351.86
                                                                                            72.02
   Mutu Baja =               BJ 37
        fu =            3700 kg/cm2
         fy =           2400 kg/cm2

   3.3.6 Perencanaan Pembebanan + Beban Profil
   Beban Mati ( D )
   Bondex                 =        3       10.1                               =                 30.3 kg/m
   Plat Beton             =        3       0.12                  2400         =                  864 kg/m
   Tegel + Spesi          =        3        90                                =                  270 kg/m
   Berat Profil           =                                                   =                 29.6 kg/m
                                                    qD                          =      1193.9 kg/m


Beban Hidup ( L )
   qL         =             3           490               =         1470

3.3.7 Perhitungan qU , Mu Max dan Du Max ( Berat Profil Dimasukkan )
qU = 1.2 qD + 1.6 qL
    qU         =        1.2     1193.9     1.6       1470       =                    3784.68 Kg/m
            1
Mu max =      qu l 2        =           0.13        3784.68        16           =     7569.36 Kgm
            8
            1
Du max =      qu l          =           0.5         3784.68         4           =    7569.36 Kg
            2

3.3.8 Kontrol Lendutan Balok
       Y=      L        =               400               =       1.11
              360                       360
                      5 ( qD + qL) * l 4
      Y max =
                     384      EIx
                 =          5                   (    11.94        14.7     )         2.56E+10
                           384                      2100000       4050

                 =         1.04          <               1.11

                                        OK

3.3.9 Kontrol Kuat Rencana Momen Lentur
3.3.9.1 Kontrol Penampang

untuk Sayap                                                     untuk Badan
    b    170                                                    h 1680
       ≤                                                          ≤
   2tf     fy                                                   t   fy

   125                     170                                     208                1680
    18           ≤        15.49                                     6           ≤     15.49
   6.94          ≤        10.97                                   34.67         ≤      108.44
               OK                                                               OK

Penampang Profil Kompak, maka Mnx = Mpx

     Mp = fy * Zx
        = 2400         *               351.86
        = 844466.4 kgcm
        = 8444.66 kgm

3.3.9.2 Kontrol Lateral Buckling

Jrk Pengikat Lateral :       1000 mm                      =        100     cm



                     E
Lp = 1.76 * iy                  Lp =    141.75 cm
                     fy
                 E
Lp = 1.76 * iy
                 fy

            Ternyata          Lp > Lb         maka     Mnx = Mpx

          Mnx = Mpx = Zx. Fy = 351.86            *       2400          =      8444.66 Kgm
    Mny = Zy ( 1 flen ) * fy
        = (1 / 4 * tf * bf 2) * fy
        =     0.25 x        0.9   x 156.25 x    2400      =            84375 kgcm
        =      843.75 kgm

 0.9 Mp =      0.9        *       8444.66        =        7600.2 kgm

             0.9 Mp       >         Mu
             7600.2       >       7569.36
                         OK


3.3.9.3 Kontrol Kuat Rencana Geser

                  h    10
                      10
                    ≤
                 tw     fy
               208        <         1100
                6                   15.49

              34.67       <             71
                       Plastis

     Vn = 0.6 fy Aw
        =     0.6       2400            0.6      25
        =      21600 Kg

   Vu           <       ФVn
 7569.36        <       0.9        21600
 7569.36        <      19440
               OK
  4 Perencanaan Tangga Baja
4.1 Data Perencanaan
     Tinggi tangga                         =                 250 cm
     Lebar injakan (i)                     =                  28 cm
     Panjang Tangga                        =                 600 cm
     Lebar Pegangan Tangga                 =                  10 cm

4.2 Perencanaan Jumlah Injakan Tangga
     4.2.1 Persyaratan - Persyaratan Jumlah Injakan Tangga

            60 cm            <         ( 2t + I )        <                 65 cm
            25 o             <             a             <                 40 o

    Dimana : t = tinggi injakan (cm)
             i = lebar injakan (cm)
             a = kemiringan tangga

    4.2.2 Perhitungan Jumlah Injakan Tangga

    Tinggi tanjakan (t)      =            65         -                28            /       2
                             =              18.5 cm
    Jumlah Tanjakan          =           250         =                13.51 buah
                                         18.5
                             =            14       buah

    Jumlah injakan (n)       =            14            buah

    Lebar Bordes             =           600        −   392           =            208             cm
    Lebar Tangga             =           200        −    20           =            180             cm

                    a        =             32.54 0                =                     0.57 rad

                             392          cm                                       208             cm



180 cm




180 cm




4.3 Perencanaan Pelat Tangga
     4.3.1 Perencanaan Tebal Pelat Tangga

    Tebal Pelat Tangga     =        4        mm
    Berat Jenis Baja       =      7850      kg/m3
    Tegangan Leleh Baja    =      2400      kg/m2
    4.3.2 Perencanaan Pembebanan Pelat Tangga
Beban Mati
Berat Pelat            =                   0     x    1.8    x 7850            =        56.52 kg/m'
Alat Penyambung (10 %)                                                         =         5.65 kg/m'
                                                             qD                =        62.17 kg/m'
Beban Hidup
qL =        500                  x        1.8         =                900 kg/m'



4.3.3 Perhitungan M D dan ML
           1
 MD =        qDl 2
           8
   MD                =         0.13   x 62.17 x      0.08          =         0.61     kgm

          1
ML =        qLl 2
          8
   MD                =         0.13   x   900    x   0.08          =         8.82     kgm


4.3.4 Perhitungan Kombinasi Pembebanan M U

MU = 1.4 MD
    Mu =    1.4            x   0.61        =          0.85    kgm
                                               Tidak Menentukan
MU = 1.2 MD + 1.6 ML
    Mu =    1.2   x 0.61                   +          1.6    x 8.82            =      14.84 kgm
                                                                                    Menentukan


4.3.5 Kontrol Momen Lentur
                 1 2
          Zx =     bh            =        0.25   x   180     x    0.16         =       7.2      cm3
                 4
   φMn = φ Zx * fy               =        0.9    x    7.2    x 2400            =      15552     kgcm

   φMn =            155.52 kgm

Syarat ->         φMn                      >          Mu
                 155.52        kgm         >         14.84        kgm
                                          OK


4.3.6 Kontrol Lendutan

          f =        L           =         28         =           0.08
                    360                   360
                  1
          Ix =      bh 3         =        0.08   x   180     x    0.06         =      0.96      cm4
                 12
   Ix =             0.96       cm4


                   5 ( qD + qL) * l 4
          Y max =
                  384      EIx
                            5 ( qD + qL) * l 4
            Y max =
                           384      EIx

                       =          5                (     0.62         9     )            6.15E+05
                                 384                   2100000 x     0.96

                       =         0.04        <          0.08

                                            OK

    Ambil Pelat Tangga dengan Tebal =                     4     mm

4.4 Perencanaan Penyangga Pelat Injak

    4.4.1 Perencanaan Pembebanan
    Beban Mati
    Berat Pelat            =                0.14   x     0      x 7850           =               4.4 kg/m'
    Berat Baja Siku        =                 45    x     45     x  7             =               4.6 kg/m'
                                                                                                   9 kg/m'
    Alat Penyambung ( 10 % )                                                     =               0.9 kg/m'
                                                                qD               =               9.9 kg/m'

    Beban Hidup
    qL =        500                x        0.14                      =          70     kg/m'
                                                                .
    4.4.2 Perhitungan M D dan ML
             1
     MD =      qDl 2
             8
       MD              =         0.13   x   9.9    x    3.24          =         4.01       kgm

            1
    ML =      qLl 2
            8
       MD              =         0.13   x   70     x    3.24          =         28.35      kgm


    4.4.3 Perhitungan Kombinasi Pembebanan M U

    MU = 1.4 MD
        Mu =    1.4          x   4.01        =          5.61    kgm
                                                 Tidak Menentukan
    MU = 1.2 MD + 1.6 ML
        Mu =    1.2   x 4.01                 +           1.6    x 28.35          =        50.17 kgm
                                                                                        Menentukan

    4.4.4 Kontrol Momen Lentur
    Dari Perhitungan Sap 2000 Version 8.2.3 Didapat untuk Profil Siku 45x45x7 :
          Zx =       6.14   cm3     ( Modulus Plastis )

       φMn = φ Zx * fy             =        0.9    x    6.14    x 2400           =       13262.4       kgcm

       φMn =          132.62 kgm

    Syarat ->     φMn                        >           Mu
                 132.62          kgm         >          50.17        kgm
                                               OK


    4.4.5 Kontrol Lendutan

              f =        L           =         180          =          0.5
                        360                    360

    Dari Tabel Profil Baja Didapat :

       Ix =         10.42           cm4


                               5 ( qD + qL) * l 4
              Y max =
                              384      EIx

                         =           5               (     0.1      0.7      )           1.05E+09
                                    384                  2100000 x 10.42

                         =           0.5        <          0.5

                                               OK

    Ambil Profil Baja Siku Sama Kaki                       45     x    45    x    7

4.5 Perencanaan Pelat Bordes
     4.5.1 Perencanaan Tebal Pelat Bordes

    Tebal Pelat Tangga               =          8          mm
    Berat Jenis Baja                 =        7850        kg/m3
    Tegangan Leleh Baja              =        2400        kg/m2
    Lebar Pelat Bordes               =          2           m

    4.5.2 Perencanaan Pembebanan Pelat Bordes
    Beban Mati
    Berat Pelat            =      0.01 x      2                   x 7850          =         125.6 kg/m'
    Alat Penyambung (10 %)                                                        =         12.56 kg/m'
                                                                  qD              =        138.16 kg/m'
    Beban Hidup
    qL =        500                   x         2           =           1000 kg/m'

    4.5.3 Perhitungan M D dan ML
               1
     MD =        qDl 2
               8
        MD               =          0.13   x 138.16 x     0.48         =          8.3     kgm

              1
    ML =        qLl 2
              8
        MD               =          0.13   x 1000 x       0.48         =         60.09    kgm


    4.5.4 Perhitungan Kombinasi Pembebanan M U
    MU = 1.4 MD
        Mu =    1.4            x   8.3        =         11.62    kgm
                                                  Tidak Menentukan
    MU = 1.2 MD + 1.6 ML
        Mu =    1.2   x            8.3        +           1.6    x 60.09        =     106.1 kgm
                                                                                    Menentukan


    4.5.5 Kontrol Momen Lentur
                     1 2
              Zx =     bh           =       0.25    x     200    x   0.64       =      32       cm3
                     4
       φMn = φ Zx * fy              =        0.9    x     32     x 2400         =     69120     kgcm

       φMn =           691.2 kgm

    Syarat ->        φMn                      >           Mu
                     691.2         kgm        >          106.1       kgm
                                             OK


    4.5.6 Kontrol Lendutan

              f =      L            =       69.33          =         0.19
                      360                    360
                      1
              Ix =      bh 3        =       0.08    x     200    x   0.51       =     8.53      cm4
                     12
       Ix =          8.53          cm4


                             5 ( qD + qL) * l 4
              Y max =
                            384      EIx

                       =            5               (     1.38        10    )        2.31E+07
                                   384                  2100000 x    8.53

                       =           0.19       <          0.19

                                             OK

    ambil Tebal Pelat Bordes =                8           mm




4.6 Perencanaan Balok Bordes

    4.6.1 Perencanaan Balok Bordes dengan Profil I
               100           x        100            x                6             x            8

      A=             21.9 cm2                 tf =          8 mm                        Zx =     84.18 cm3
      W=             17.2 kg/m               Ix =        383 cm4                        Zy =     40.61 cm3
       a=             100 mm                 Iy =        134 cm4                         h=         84 mm
      bf =            100 mm                tw =            6 mm
      iy =           2.47 cm                 ix =        4.18 cm                                 84.18
                                                                                                 40.61
4.6.2 Perencanaan Pembebanan
Beban Mati
Berat Pelat =        0.01 x           0.69       x 7850                             =            43.54 kg/m'
Berat Profil I =                                                                    =             17.2 kg/m'
                                                                                    =            60.74 kg/m'
Alat Penyambung ( 10 % )                                                            =             6.07 kg/m'
                                                                qD                  =            66.82 kg/m'
              1
       MD =     q D L2       =        0.13       x 66.82 x           4.33           =          36.13     kgm
              8
              1
          PD = q D L         =            0.5    x 66.82 x           4.33           =          144.54    kg
              2
Beban Hidup               qL =        500            x               0.69           =           346.67 kg/m'
              1                                                 .
       ML =     q L L2       =        0.13       x 346.67 x          4.33           =          187.48    kgm
              8
              1
          PL = q L L         =            0.5    x 346.67 x          4.33           =          749.91    kg
              2
4.6.3 Perhitungan Kombinasi Pembebanan
MU = 1.4 MD
     Mu =     1.4  x 36.13      =        50.59    kgm
      Pu =    1.4  x 144.54     =       202.35    kgm
                                  Tidak menentukan
MU = 1.2 MD + 1.6 ML
     Mu =     1.2  x 36.13      +         1.6  x 187.48                             =        343.32 kgm
      Pu =    1.2  x 144.54     +         1.6  x 749.91                             =        1373.3 kgm
                                                                                           Menentukan
4.6.4 Kontrol Kekuatan Profil
4.6.4.1 Penampang Profil                                 fy =          2400 kg/m2

untuk Sayap                                                     untuk Badan

    b    170                                                    h 1680
       ≤                                                          ≤
   2tf     fy                                                   t   fy

   100                      170                                      84                        1680
    16           ≤         15.49                                     6              ≤          15.49
   6.25          ≤         10.97                                     14             ≤           108.44
               OK                                                                   OK

Penampang Profil Kompak, maka Mnx = Mpx

4.6.4.2 Kontrol Lateral Buckling

Jarak Baut Pengikat :            250 mm                    =                25 cm



                     E
Lp = 1.76 * iy
                     fy
                            E
    Lp = 1.76 * iy                      Lp =      125.49 cm
                            fy

                    Ternyata            Lp > Lb             maka      Mnx = Mpx

              Mnx = Mpx = Zx. Fy = 84.18                       *        2400           =        2020.42 Kgm
        Mny = Zy ( 1 flen ) * fy
            = (1 / 4 * tf * bf 2) * fy
            =     0.25 x        3.2   x 100 x                 2400       =            192000 kgcm
            =        1920 kgm

    4.6.5 Kontrol Momen Lentur

         Zx =           84.18      cm3

       φMn = φ Zx * fy              =             0.9   x 84.18 x 2400                 =      181837.44   kgcm

       φMn =          1818.37 kgm

    Syarat ->         φMn                          >           Mu
                     1818.37       kgm             >         343.32     kgm
                                                  OK


    4.6.6 Kontrol Lendutan

              f =       L           =          180             =         0.5
                       360                     360

       Ix =           84.18        cm4


                              5 ( qD + qL) * l 4
              Y max =
                             384      EIx

                        =           5                   (     0.67     3.47       )           1.05E+09
                                   384                      2100000 x 84.18

                        =          0.32           <           0.5

                                                  OK




4.7 Perhitungan Balok Induk Tangga

    4.7.1 Data - Data Perencanaan
          h min = I sin α =  28   x               sin        32.54       =            15.06      cm
4.7.2 Perencanaan Balok Induk Dengan Menggunakan Profil WF

             250       x       125         x          5      x          8

     A=       32.68 cm2             tf =       8 mm              Zx =   310.45 cm3
     W=        25.7 kg/m           Ix =    3540 cm4              Zy =    63.71 cm3
      a=        250 mm             Iy =     255 cm4               h=       210 mm
     bf =       125 mm            tw =         5 mm               r =       12         mm
     iy =      2.79 cm             ix =     10.4 cm                     310.45
                                                                         63.71

Syarat -->    h        >       hmin
             25        >        15.06
                      OK


4.7.3 Perencanaan Pembebanan




4.7.3.1 Perencanaan Pembebanan Anak Tangga
Beban Mati
Berat Pelat =           0   x 1.04 x 7850                    =           32.66 kg/m'
Berat Profil siku =    4.6  x   2    x 0.9        0.28       =           29.57 kg/m'
Berat Sandaran Besi                                          =              15 kg/m'
Berat Profil WF =     32.68     /       cos      32.54       =           38.76 kg/m'
                                                                              115.99 kg/m'
Alat Penyambung (+ 10 %)                                               =        11.6 kg/m'
                                                        qD1            =      127.59 kg/m'

Beban Hidup
  qL1       =            500     x   1.04         =       520         kg/m'

Beban q1 Total =       1.2 qD + 1.6 qL
               =          1.2    x 127.59         +        1.6    x   520
               =         985.1      kg/m'


4.7.3.2 Perencanaan Pembebanan Bordes
Beban Mati
Berat Profil WF =                                                      =      25.7        kg/m'

Berat Pelat Bordes =     0.01    x    1      x   0.69   x 7850         =       43.54 kg
Berat Profil I =         17.2    x    1                                =        17.2 kg
                                                                               60.74 kg
Alat Penyambung (+ 10 %)                                               =        6.07 kg
                                                           Pd          =       66.82 kg


Beban Hidup
  qL2 =            500 kg/m2                     PL2          =       500 x 0.69     x     1
                                                              =       346.67 kg
        jadi q2 total = 1.2 qD + 1.6 qL
                      =    1.2    x 25.7          +        1.6    x   500
                      = 830.84       kg/m'

         jadi P total = 1.2 PD + 1.6 PL
                      =    1.2   x 66.82          +        1.6    x 346.67
                      = 634.85        kg




4.7.4 Perhitungan Gaya - Gaya pada Tangga
                                                                                     Lab =              3.92 m
Σ Ma = 0                                                                             Lbc =              2.08 m
   1        2                                         1
  ( q 1 l ab ) + ( p (3l ab + 1.5l bc )) + (q 2 l cb ( l cb + l ab )) − ( Rc(l ab + l bc )) = 0
   2                                                  2
 492.552       15.366 + 634.845 ( 11.76 + 3.120 ) + 1728.147 ( 1.040 +                               3.92     )    RC
                                                      6
      Rc =      4264.48 kg

ΣV=0
Rva = q1l ab + q 2 l bc + 3P − Rc
     Rva = ( 985.10   3.92 ) +( 830.84                      2.08 ) + 1904.54 _ 4264.48
     Rva = 3229.81 kg




                                                                         B                 C
                                                                                 +

                                                                +
                                A
                                                                       5092.12       kgm
                                            5294.72       kgm

                                            3229.81




                RAh             =             0

Bidang M
  Pers :        Mx1 =        RVA x            X1                -       0.5   x         q1      x     X12
                Mx1 =       3229.81 x         X1                -      492.55 x         X12
  dMx1
                  =             0                           985.1        X1                =        3229.81
   dX1                                                                   X1                =         3.28           m

   X1             =            0        m                               MA =             0           Kgm
  Xmax            =           3.28      m                             Mmax=           5294.72        Kgm          tangga
   X1             =           3.92      m                               MB =          5092.12        Kgm          tangga

                                                                         B                 C

                                             4.65

                                A            a=             32.5444
                                        Rav cos a

                           Rav sin a                3.92 m                    2.08 m
                                Rav
                                              X1                                  X2



                                                                        -532.68   kg

                                                                         -2944    -
                             2722.65    kg                                            -4264.48 kg




 Bidang D
Permisalan gayaDari kiri : searah jarum jam gaya dianggap positif
        X=      0             m
                  DA =       Rva     x        cos a
                      = 3229.81           cos       32.544
                      = 2722.65           kg
        X=     3.92           m
               Dbkiri =      Rva       x      cos a               -                     q1    x     LAB
                      = -532.68           kg
                                                                                                          cos a
           Dbkanan =     P              x     LBC              -          RC
                   = 634.85                   2.08             -        4264.48
                   =   -2944                   kg
        X=    6          m
                Dc =     -                    RC
                   = -4264.48                 kg




                                                                        726.51          kg

                                                                    +



                                                       -



                                            -1737.49           kg




 Bidang N
               NA =            -RVA                  sin a
                =            -3229.81                sin 32.544
                =            -1737.49         kg

                NBkiri =       -RVA           sin          a              +             q1          L1
                    =        726.51           kg
                                                                                                         sin a
           NBkanan -C =         0

4.7.5 Kontrol Kekuatan Profil
4.7.5.1 Penampang Profil                                       fy =       2400 kg/m2

untuk Sayap                                                           untuk Badan

    b    170                                                          h 1680
       ≤                                                                ≤
   2tf     fy                                                         t   fy

   125                        170                                        210                   1680
    16             ≤         15.49                                        5            ≤       15.49
   7.81            ≤         10.97                                        42           ≤        108.44
                   OK                                                                  OK

Penampang Profil Kompak, maka Mnx = Mpx

4.7.5.1 Kontrol Lateral Buckling

Jarak Baut Pengikat :               250 mm                       =             25 cm



                        E
Lp = 1.76 * iy                      Lp =            0 cm
                        fy

                Ternyata            Lp > Lb          maka             Mnx = Mpx

          Mnx = Mpx = Zx. Fy = 310.45                      *            2400           =       7450.68 Kgm
    Mny = Zy ( 1 flen ) * fy
        = (1 / 4 * tf * bf 2) * fy
        =     0.25 x 2.56 x 0.64 x                     2400               =            983.04 kgcm
        =        9.83 kgm

4.7.5 Kontrol Momen Lentur

     Zx =          310.45    cm3

   φMn = φ Zx * fy              =             0.9   x 310.45 x 2400                    =     670561.2    kgcm

   φMn =          6705.61 kgm

Syarat ->         φMn                          >        Mu
                 6705.61     kgm               >      5294.72           kgm
                                              OK


4.7.6 Kontrol Lendutan

          f =       L           =          600             =            1.67
                   360                     360

   Ix =           3540       cm4
                    5 ( qD + qL) * l 4
       Y max =
                   384      EIx

               =          5               (      1.53       5.2   )       1.30E+11
                         384                   2100000 x   3540

               =         1.53        <          1.67

                                     OK
Profil yang Dipakai untuk Balok induk Adalah
Profil WF      250         x         125          x         5         x      8
5. Pembebanan
5.1 Perencanaan Beban Atap

    Beban Mati
    Berat Gording =        9.3         x       4                              =     37.2       kg
    Berat Asbes Gelombang =                  11.33 x    4                     =    45.32       kg
                                                                              =    82.52       kg
    Alat Pengikat dll (+10 %)                                                 =     8.25       kg
                                                                   Pm         =    90.77       kg
    Berat Profil Kuda - Kuda =                0         cos        0.44       =       0        kg
                                                                   Pmtot           90.77
    Beban Hidup
       Ph       =                19.94 x      4                               =    79.76       kg

    P Ultimate = 1.2 P D+ 1.6 PL =         108.93       +         127.61      =    236.53      kg

5.2. Perencanaan Beban Angin (Gudang Tertutup)


               (- 0,02α - 0,4)                                    - 0,4




                0,9                                                        - 0,4




        W             =               30 kg/m2

    Beban Tekan Atap              =          0.1    x   30    x      4        =     12      kg/m

    Beban Sedot Atap              =          -0.4   x   30    x      4        =     -48     kg/m

    BebanTiup Kolom               =          0.9    x   30    x      4        =     108     kg/m

    Beban Sedot Atap              =          -0.4   x   30    x      4        =     -48     kg/m



5.3 Perencanaan Beban Akibat Plat Lantai, Kolom Memanjang dan Melintang


                                        P4                        memanjang


    I
                                        P3
    I
      I
                                          P3
      I

600
                P5                        P2                       P6                                  P6

      I
      I                                   P1
600
                                                                                                                 melintang
                                          P2


    I
                                          P3
    I
600 I
                                          P4

      I
      V                                          B
                           400                    400                       400
                     .                                                                           C
                                          A
                                                                                  B

          A          ( Balok Induk Melintang )
                          0         x          0            x           0             x               0

              A=              0 cm2                  tf =       0 mm                      Zx =               0 cm3
              W=              0 kg/m                Ix =        0 cm4                     Zy =               0 cm3
               a=             0 mm                  Iy =        0 cm4                      h=                0 mm
              bf =            0 mm                 tw =         0 mm                       r=                0 mm
              iy =            0 cm                  ix =        0 cm                                         0
                                                                                                             0
          B          ( Balok Anak )
                        250         x          125          x           6             x               9

              A=         37.66 cm2                   tf =       9 mm                      Zx =        351.86 cm3
              W=          29.6 kg/m                 Ix =    4050 cm4                      Zy =         72.02 cm3
               a=          250 mm                   Iy =     294 cm4                       h=            208 mm
              bf =         125 mm                  tw =         6 mm                       r=             12 mm
              iy =        2.79 cm                   ix =     10.4 cm                                  351.86
                                                                                                       72.02
          C          ( Balok Induk Memanjang)
                          0         x         0             x           0             x               0

              A=              0 cm2                  tf =       0 mm                      Zx =               0 cm3
              W=              0 kg/m                Ix =        0 cm4                     Zy =               0 cm3
               a=             0 mm                  Iy =        0 cm4                      h=                0 mm
              bf =            0 mm                 tw =         0 mm                       r=                0 mm
              iy =            0 cm                  ix =        0 cm                                         0
                                                                                                             0
      Beban Mati
      - Bondex                                                                        =               10.1    kg/m2
      - Beton                    =             0.12         x      2400               =               288     kg/m2
      - Beban Finishing                                                               =                90     kg/m2
                                                                    qM                =              388.1    kg/m2
 Beban Hidup

    qL         =           400   kg/m2

 5.3.1 Perencanaan Pembebanan Portal Melintang

 5.3.1.1 Beban P1
 Beban Mati
 Berat Pelat =                   388.1 x     3     x    4    =   4657.2        kg
 Balok induk melintang =                     0     x    3    =      0          kg
 Balok anak =                    29.6    x   4               =    118.4        kg
                                                       Pm1   =   4775.6        kg

 Beban Hidup
   Ph1       =                   400     x   3     x    4    =   4800          kg

 5.3.1.2 Beban P2
 Beban Mati
 Berat Pelat =                   388.1 x     3     x    4    =   4657.2        kg
 Balok induk memanjang =                     0     x    4    =     0           kg
 Balok induk melintang =                     0     x    6    =     0           kg
                                                       Pm2   =   4657.2        kg

 Beban Hidup
   Ph2       =                   400     x   3     x    4    =   4800          kg

 5.3.1.3 Beban P3
 Beban Mati
 Berat Pelat =                   388.1 x     3     x    4    =   4657.2        kg
 Balok induk melintang =                     0     x    6    =      0          kg
 Balok anak =                    29.6    x   6               =    177.6        kg
                                                       Pm1   =   4834.8        kg

 Beban Hidup
   Ph3       =                   400     x   3     x    4    =   4800          kg

 5.3.1.4 Beban P4
 Beban Mati
 Berat Pelat =                   388.1 x     1.5   x    4    =   2328.6        kg
 Balok induk memanjang =                      0    x    4    =     0           kg
 Berat Dinding =                 4.15    x   4.5   x    4    =    74.7         kg
 Balok induk melintang =                      6    x    0    =     0           kg
                                                       Pm4   =   2403.3        kg

 Beban Hidup
   Ph4       =                   400     x   1.5   x    4    =   2400          kg


 5.3.2 Perencanaan Pembebanan Portal Memanjang

  5.3.2.1 Beban P5
  Beban Mati
* Beban Mati
  Berat Pelat                    388.1 x     6     x    2    =   4657.2   kg
  Balok induk melintang =                    0     x    6    =     0      kg
  Balok induk memanjang =                    0     x    2    =     0      kg
                                                       Pm5   =   4657.2   kg
     Beban Hidup
   * Beban Hidup
       Ph5       =                    400   x    6     x     2    =         4800          kg

     5.3.2.2 Beban P6
     Beban Mati
   * Beban Mati
     - Berat Pelat =                 388.1 x     4     x     6    =         9314.4   kg
     - Balok induk memanjang =                   0     x     4    =           0      kg
     - Balok induk melintang =                   0     x     6    =           0      kg
                                                            Pm5   =         9314.4   kg
     Beban Hidup
   * Beban Hidup
       Ph5       =                    400   x    4     x     6    =         9600          kg

    5.3.3 Beban Portal - Portal Melintang
        P1                  Pm1         =       4775.6 kg
                             Ph1        =         4800 kg

       P2                  Pm3         =        4657.2 kg
                           Ph3         =          4800 kg

       P3                  Pm2         =        4834.8 kg
                           Ph2         =          4800 kg

       P4                  Pm4         =        2403.3 kg
                           Ph4         =          2400 kg

    5.3.4 Beban - beban Portal Memanjang
        P5                 Pm5        =         4657.2 kg
                           Ph5        =           4800 kg

       P6                  Pm6         =        9314.4 kg
                           Ph6         =          9600 kg


5.4 Perencanaan Beban Gempa ( Arah X )




             F2
                                                                        2
            4

             F1                                                   4.5         9

            5



                             6              6          6

                        Data Gempa:
                        - Zone Gempa =           6
                        - tanah lunak
                        C             =         0.95
                        -I            =          1
                                                                               4
F
1                                                                                                    4
                                                                               4




                     6                   6                      6

                                       18



             Kolom
                 0        x      0            x             0       x                0

      A=             0 cm2             tf =         0 mm                Zx =                 0 cm3
      W=             0 kg/m           Ix =          0 cm4               Zy =                 0 cm3
       a=            0 mm             Iy =          0 cm4                h=                  0 mm
      bf =           0 mm            tw =           0 mm                 r=                  0 mm
      iy =           0 cm             ix =          0 cm                                     0
                                                                                             0

5.4.1 Perencanaan Beban Lantai (W1)
- Beban Mati:
5.4.1.1 Beban Mati
Berat Plat             =        388.1 x       18    x     4         =              27943.2       Kg
Balok Induk Memanjang =                        0    x     4         =                     0      Kg
Balok Induk Melintang  =                       0    x    18         =                     0      Kg
Berat Dinding          =         8.3           4    x    4.5        =                 149.4      Kg
Kolom                  =          0   x       4.5   x     2         =                     0      Kg
                                  0   x       4.5   x     1         =                     0      Kg
                                                                    =               28092.6      Kg

5.4.1.2 Beban Hidup
Balok + Plat              =     Ph1 + 2 x Ph2 + 2 x Ph3 + 2 x Ph4
                          =     4800 + 9600 + 9600 + 4800
                          =    28800    Kg

Beban Lantai ( W1 tot)    =   28092.6         +         28800
                          =   56892.6         Kg

5.4.2 Perencanaan Beban Atap ( W2)
5.4.2.1 Beban Mati
Berat Atap       =    22.69    Cosα           4     x    18         =                1802.8 Kg
Balok Kuda-kuda =       0   x Cosα x          18                    =                     0 Kg
Kolom               =   0   x    2  x         2                     =                     0 Kg
Berat Dinding     =    8.3 x     2  x         4                     =                  14.3 Kg
                                                                    =                1817.1 Kg
5.4.2.2 Beban Hidup
   Ph1         =                     19.94 x      4     x        18         =        1435.59   kg


Beban Atap ( W2 tot)      =         1817.1       +          1435.59
                          =         3252.7       Kg

5.4.3 Berat Total Wt

Berat Total ( W tot )     =          W1          +               W2
                          =        56892.6       +              3252.7
                          =        60145.3       Kg


5.4.4 Perencanaan Gaya Gempa

     T            =      0.09            x       H3/4
                  =      0.09            x              9 3/4
                  =         0.44

Tanah Lunak
    C             =      0.95                                     I         =          1
    R             =      4.5


V = (C x I x Wt )/R       =          0.95 x    1        x 60145.3           4.5
                          =        12697.34 Kg


  Lantai          =       W1             x       h1               =       56892.6      x       5
                                                                  =       284463      Kgm
   Atap           =       W2             x        H               =        3252.7      x       9
                                                                  =      29274.27     Kgm

                                              Σ Wi Hi             =      313737.27    Kgm




    F1            =            W1 h1              x               V
                               S Wi hi

                  =         284463                x         12697.34
                           313737.27

                  =     11512.57         Kg


    F2            =            W2 h2              x               V
                               S Wi hi

                  =        29274.27               x         12697.34
                           313737.27

                  =     1184.77          Kg
    5.4.5 Gambar Perencanaan Beban Akibat Gempa ( Arah X )


    F2
      1184.77 kg

                                                                                                       4

    F1
     11512.57 kg

                                                                                                       5



                             6                      6                      6



5.5 Perencanaan Beban Gempa ( Arah Y )

                         Data Gempa:
                         - Zone Gempa =                  6
                         - tanah lunak
                         C             =                0.95
                         -I            =                 1

                Balok Kuda-Kuda
                    0        x        0                  x             0       x          0

         A=             0 cm2                tf =              0 mm                Zx =       0 cm3
         W=             0 kg/m              Ix =               0 cm4               Zy =       0 cm3
          a=            0 mm                Iy =               0 cm4                h=        0 mm
         bf =           0 mm               tw =                0 mm                 r=        0 mm
         iy =           0 cm                ix =               0 cm                           0
                                                                                              0
                Kolom
                    0        x        0                  x             0       x          0

         A=             0 cm2                tf =              0 mm                Zx =       0 cm3
         W=             0 kg/m              Ix =               0 cm4               Zy =       0 cm3
          a=            0 mm                Iy =               0 cm4                h=        0 mm
         bf =           0 mm               tw =                0 mm                 r=        0 mm
         iy =           0 cm                ix =               0 cm                           0
                                                                                              0



    5.5.1 Perencanaan Beban Lantai (W1)
    - Beban Mati:
    5.5.1.1 Beban Mati
    Berat Plat             =        388.1 x              6     x   40          =          93144   Kg
    Balok Induk Memanjang =                              0     x   40          =              0   Kg
Balok Induk Melintang    =                         0     x        40        =              0      Kg
Berat Dinding            =            8.3          6     x        4.5       =          224.1      Kg
Kolom                    =             0      x   4.5    x         2        =              0      Kg
                                       0      x   4.5    x        4.5       =              0      Kg
                                                                            =        93368.1      Kg

5.5.1.2 Beban Hidup
Beban Hidup Merata        =          250 x        40     x         6       Kg
                          =         60000         Kg

Beban Lantai ( W1 tot)    =         93368.1       +              60000
                          =        153368.1       Kg

5..2 Perencanaan Beban Atap ( W2)
5.5.2.1 Beban Mati
Berat Atap       =    22.69 x     6           x   40                        =        5446.32 Kg
Balok Kuda-kuda =       0   x Cosα            x   40                        =              0 Kg
Kolom               =   0   x     2           x   2                         =              0 Kg
Berat Dinding     =    8.3  x     6           x   2                         =           16.3 Kg
                                                                            =        5462.62 Kg

5.5.2.2 Beban Hidup
   Ph1         =                     19.94 x       6     x        40        =       4785.31       kg


Beban Atap ( W2 tot)      =         5462.62       +          4785.31
                          =        10247.93       Kg

5.4.3 Berat Total Wt

Berat Total ( W tot )     =           W1          +            W2
                          =        153368.1       +          10247.93
                          =        163616.03      Kg


5.4.4 Perencanaan Gaya Gempa

     T            =      0.09          x          H3/4
                  =      0.09          x                 9 3/4
                  =         0.44


Tanah Lunak
    C             =      0.95                                      I        =         1
    R             =      4.5


V = (C x I x Wt )/R       =          0.95 x    1         x163616.03        4.5
                          =        34541.16 Kg


  Lantai          =      W1            x          h1               =     153368.1     x           5
                                                                   =     766840.5    Kgm
   Atap           =      W2            x           H               =     10247.93     x           9
                                                                   =     92231.33    Kgm
                                                  Σ Wi Hi        =     859071.83   Kgm




         F1        =             W1 h1                x          V
                                 S Wi hi

                   =            766840.5              x     34541.16
                                859071.83

                   =         30832.77      Kg


         F2        =             W2 h2                x          V
                                 S Wi hi

                   =            92231.33              x     34541.16
                                859071.83

                   =         3708.39       Kg




    5.4.5 Gambar Perencanaan Beban Akibat Gempa ( Arah Y )


    F2
      3708.39 kg



                        P4              P3            P2         P1         P2           P3
    F1
     30832.77 kg




Kombinasi Pembebanan

* Beban Mati + Beban Hidup

                                                            Pm         =            90.77 kg
                                                            Ph         =            79.76 kg


                                           200



    P4             P3              P2            P1         P2             P3            P4
      P4                    P3                P2         P1            P2                P3                  P4




                           6                                 6                           6



      Pm1 =                4775.6 kg                                       Pm2 =         4657.2 kg
      Ph1 =                  4800 kg                                       Ph2 =           4800 kg

      Pm3 =                4834.8 kg                                       Pm4 =         2403.3 kg
      Ph3 =                  4800 kg                                       Ph4 =           2400 kg



* Beban Mati + Beban Hidup + Beban Angin


                q =        12         kg/m                                         q =   -48        kg/m




           P4                    P3                P2            P1         P2               P3               P4




q =             108 kg/m                                                                      q =      -48    kg/m
                        6                                    6                           6




* Beban Mati + Beban Hidup + Beban Gempa




       1184.77 kg


                      P4                 P3             P2            P1            P2              P3               P4
11512.57 kg




              6   6   6
melintang
P4
 6 Perencanaan Dimensi Struktur Utama
6.1 Kontrol Dimensi Kuda -Kuda
    Dari Hasil Sap 2000 Diperoleh Mmax dan N max pada Frame 9:

                (U-G)         Beban Ultimate - Beban Gempa
                                 Mutx = -3679.74 Kgm
                                 Muty =          0 Kgm
                                   Nu = -2677.93 Kg
                                   Vu = -1184.35 kg

                                   Ma =     -3679.4 Kgm
                                   Mb =     -466.19 Kgm
                                   Ms =      301.67 Kgm


   6.1.1 Profil Baja yang Didapatkan Di SAP 2000 8.2.3 :

                  200              x         200             x           8      x            12

         A=        63.53 cm2                        tf =         12 mm              Zx =     513.15 cm3
         W=         49.9 kg/m                      Ix =       4720 cm4              Zy =     216.32 cm3
          a=         200 mm                        Iy =       1600 cm4               h=         150 mm
         bf =        200 mm                       tw =            8 mm              Sx =       37.5 mm
         iy =       5.02 cm                        ix =        8.62 cm                       513.15
                                                                                             216.32
   Mutu Baja =              BJ 37
        fu =           3700 kg/cm2
         fy =          2400 kg/cm2

   6.1.2 Kontrol Lendutan

                   f ijin =       L           =            993.2         =    2.76           cm
                                 360                        360

                       5 L2
                  f =       ( Ms − 0.1( Ma − Mb))
                      48 EI
          f=       5            98.64                      301.67    0.1     -3679.4       -466.19
                  48             200        4720

          f=           0.68 cm

                   f              <          f ijin
                                 OK




   6.1.3 Kontrol Tekuk

   untuk arah x :
     kcx =        1.2         (jepit-rol tanpa putaran sudut)
       L=       993.2             cm
     Lkx =        1191.84        cm
            Lkx
   λx =             =            138.26 cm                 (MENENTUKAN)
             ix
              π 2 EA
   Ncrbx =                   Ncrbx =          9.87     x 2000000 x 63.53
               λx 2                                    19117.07
                             Ncrbx =         65593.62 kg

   untuk arah y :
     kcy =        0.8        (jepit-Sendi)
       L=        110             cm
     Lky =        88             cm

              Lky   =             17.53 cm
   λy =
               iy
                  π 2 EA                               x 2000000 x 63.53
  Ncrby =                    Ncrby =          9.87
                   λy 2                                      307.3
                             Ncrby =                ### kg

   Maka dipakai λx karena λx > λy

              λx        fy      λc =         138.26         2400
     λc =                                     3.14         2000000
              π         E
                                λc =                1.52

       λc           >            1.2

   ω = 1.25 λc2                 ω=                  2.91


     Pn =          Ag fy          =           63.53            *       2400   =         52474.9 kg
                    w                                        2.91


     Pu
         =            2677.93                   =               0.06    <         0.2
    ϕcPn           0.85 x 52474.9

   Pakai Rumus =                  Pu   Mux   Muy
                                     +     +      ≤1
                                2ϕcPn ϕbMnx ϕbMny




X Batang Dianggap Tidak Bergoyang Maka :
                 Cmx
     Sbx =                 ≥1
                    Nu
              1− (       )
                   Ncrbx

     Cmx =          1
     Sbx =                        1
                    1             -          2677.93
                                  65593.62

     Sbx =     1.04        <         1

     Sbx =     1.04


  Mux = Mutx * Sbx
  Mux =      -3679.74 x   1.04       =       -3836.36 kgm


Y Batang Dianggap Tidak Bergoyang Maka :

                Cmy
     Sby =                ≥1
                   Nu
             1− (       )
                  Ncrby
    Cmy =          1
    Sby =                  1
                   1       -       2677.93
                                  65593.62

     Sby =     1.04        <         1

     Sby =     1.04


  Muy = Muty * Sby
  Muy =         0      x 1.04        =             0 kgm




  6.1.4 Menentukan Mnx

  6.1.4.1 Kontrol Penampang profil
  untuk Sayap                                        untuk Badan

       b    170                                      h 1680
          ≤                                            ≤                          Kgm
      2tf     fy                                     t   fy

     200                   170                          150             1680
      24           ≤      15.49                          8         ≤    15.49
     8.33          ≤      10.97                        18.75       ≤     108.44
               OK                                                  OK
Penampang Profil Kompak, maka Mnx = Mpx

Lateral Bracing             Lb =         110            cm



                      E
Lp = 1.76 * iy                   Lp =      255.05 cm
                      fy

           Ternyata              Lp > Lb           maka             Mnx = Mpx

          Mnx = Mpx = Zx. Fy = 513.15                    *            2400       =        12315.65 kgm
    Mny = Zy ( 1 flen ) * fy
        = (1 / 4 * tf * bf 2) * fy
        =     0.25 x        1.2   x 400 x               2400           =        288000 kgcm
        =        2880 kgm




6.1.5 Persamaan Interaksi
                         Pu   Mux   Muy
                            +     +      ≤1
                       2ϕcPn ϕbMnx ϕbMny



      2677.93                +                3836.36                  +              0
   1.7 x 52474.9                           0.9   x 12315.65                     0.9   x 2880

        0.03                 +                   0.35                  +              0

                                                             0.38      <         1

                                                                      OK



6.1.6 Kontrol Kuat Rencana Geser

                   h    10
                       10
                     ≤
                  tw     fy
                150          <          1100
               2400                     15.49

               0.06           <            71
                           Plastis

     Vn = 0.6 fy Aw
        =     0.6       2400               0.8           20
        =      23040 Kg

   Vu             <        ФVn
 1184.35          <        0.9          23040
    1184.35       <      20736
                 OK




6.2 Kontrol Dimensi Kolom
    Dari Hasil Sap 2000 Diperoleh Mmax dan N max pada Frame 13:

              (U-G)     Beban Ultimate - Beban Gempa
              Sbx -->      Mutx = 10123.25 Kgm
                             Nu = -28026.45 Kg
                             Vu = 3834.45 kg

                           Max = 9049.01 Kgm
                           Mbx = 10123.25 Kgm
                           Msx =   537.12 Kgm

              Sby -->      Muty =     -8137 Kgm
                            Nu =     -27466 Kg
                            Vu =       3149 kg

                           Max =       8137 Kgm
                           Mbx =    7611.48 Kgm
                           Msx =     262.96 Kgm
6.2.1 Profil Baja yang Didapatkan Di SAP 2000 8.2.3 :

              300            x          300               x          10       x            15

       A=      119.8 cm2                       tf =           15 mm               Zx =    1464.75 cm3
       W=         94 kg/m                     Ix =        20400 cm4               Zy =      652.5 cm3
        a=       300 mm                       Iy =         6750 cm4                h=         234 mm
       bf =      300 mm                      tw =             10 mm               Sx =       1360 cm3
       iy =     7.51 cm                       ix =          13.1 cm                       1464.75
                                                                                            652.5
Mutu Baja =             BJ 37
     fu =          3700 kg/cm2                   fr =         700 kg/cm2
      fy =         2400 kg/cm2

6.2.1 Kontrol Lendutan
6.2.1.1 Kontrol Lendutan Arah X
                f ijin =  L              =               500          =     1.39           cm
                         360                             360

                       5 L2
               f =          ( Ms − 0.1( Ma − Mb))
                      48 EI
        f=     5          25.00                         537.12      0.1    9049.01       10123.25
              48           200         20400

        f=         0.04 cm

               f              <         f ijin
                             OK

6.2.1.2 Kontrol Lendutan Arah Y
                f ijin =  L              =               500          =     1.39           cm
                         360                             360

                       5 L2
               f =          ( Ms − 0.1( Ma − Mb))
                      48 EI
        f=     5          25.00                         262.96      0.1     8137         7611.48
              48           200         6750

        f=         0.04 cm

               f              <         f ijin
                             OK

6.2.3 Kontrol Tekuk

untuk arah x :
  kcx =        0.8     (jepit-Sendi)
    L=        500          cm
  Lkx =       400          cm
       Lkx
λx =           =             30.53 cm
        ix
          π 2 EA
Ncrbx =
           λx 2
             π 2 EA
     Ncrbx =                    Ncrbx =         9.87    x 2000000 x 119.8
              λx 2                                         932.35
                                Ncrbx =              ### kg

     untuk arah y :
       kcy =        0.8         (jepit-Sendi)
         L=        500              cm
       Lky =       400              cm

               Lky     =             53.26 cm               (MENENTUKAN)
     λy =
                iy

                     π 2 EA                             x 2000000 x 119.8
     Ncrby =                    Ncrby =         9.87
                      λy 2                            2836.87
                                Ncrby =    833529.23 kg

     Maka dipakai λy karena λy > λx

                λy         fy      λc =         53.26        2400
       λc =                                      3.14       2000000
                π          E
                                   λc =              0.59

        0.25           <            λc           <                   1.2


ω=      1.43                       ω=                1.12
     1,67-0,67λc

       Pn =           Ag fy          =          119.8           *          2400   =         256656.17 kg
                       w                                      1.12


       Pu
           =             28026.45                =              0.13        <         0.2
      ϕcPn            0.85 x 256656.17

     Pakai Rumus =                   Pu   Mux   Muy
                                        +     +      ≤1
                                   2ϕcPn ϕbMnx ϕbMny




X Batang Dianggap Tidak Bergoyang Maka :
                   Cmx
       Sbx =                 ≥1
                      Nu
                1− (       )
                     Ncrbx

       Cmx =           1
       Sbx =                         1
                       1             -      28026.45
                                              ###

       Sbx =          1.01           <           1
     Sbx =        1.01


  Mux = Mutx * Sbx
  Mux =     10123.25 x           1.01            =        10236.37 kgm


Y Batang Dianggap Tidak Bergoyang Maka :

                 Cmy
     Sby =                 ≥1
                    Nu
              1− (       )
                   Ncrby
    Cmy =             1
    Sby =                         1
                      1           -          28026.45
                                               ###

     Sby =        1.01            <              1

     Sby =        1.01


  Muy = Muty * Sby
  Muy =        -8137           x 1.01            =        -8227.92 kgm

  6.2.4 Menentukan Mnx

  6.2.4.1 Kontrol Penampang profil
  untuk Sayap                                                     untuk Badan

       b    170                                                   h 1680
          ≤                                                         ≤                          Kgm
      2tf     fy                                                  t   fy

      300                         170                                234             1680
       30           ≤            15.49                                10        ≤    15.49
       10           ≤            10.97                               23.4       ≤     108.44
                   OK                                                           OK

  Penampang Profil Kompak, maka Mnx = Mpx

  Lateral Buckling               Lb =          500          cm


                          E
   Lp = 1.76 * iy                     Lp =      381.56 cm
                          fy

  Lr = iy . [x1/(fy-fr)] . Sqrt{1+[sqrt(1+(x2 . fl2))]}

  J = Σ(1/3).b.(t^3) =                76.5 cm4

  Iw = Iy.((h^2)/4)       =           ### cm6

  x1 = [π/s]*[sqrt((EGJA)/2)] =              197694.62 kg/cm2
x2 = 4.[(S/GJ)^] =                  4.01E-07 cm2/kg

   Lr =          1372.41 cm

    Lp            <           Lb      <            Lr       (INELASTIC BUCKLING)

Mnx = Cb.[Mr+[(Mp-Mr).((Lr-Lb)/(Lr-Lp))]]


                                            12,5 Mmax
                           Cb =
                                   2,5Mmax + 3Ma + 4Mb + 3Mc

                           Cb =      2.25




   Mr =     Sx(fy-fr) =    2312000 kgcm
   Mp =     Zx . fy =      3515400 kgcm       < 1,5My
   My =     Sx . fy =      3264000 kgcm
 1,5 . My            =     4896000 kgcm       > Mp

  Mnx =               ### kgcm     > Mp

Mnx=Mp=      3515400 kgcm

    Mny = Zy * fy
        = 652.5 x 2400
        = 1566000    kgcm
        =      15660  kgm



6.2.5 Persamaan Interaksi
                         Pu   Mux   Muy
                            +     +      ≤1
                       2ϕcPn ϕbMnx ϕbMny



     28026.45                 +         10236.37               +          8227.92
   1.7 x256656.17                     0.9  x 35154                     0.9 x 15660

          0.06                +             0.32               +            0.58

                                                     0.97      <        1

                                                              OK

6.2.6 Kontrol Kuat Rencana Geser

                    h    10
                        10
                      ≤
                   tw     fy
                  234       <         1100
                 2400                 15.49

                  0.1        <         71
                          Plastis

        Vn = 0.6 fy Aw
           =     0.6       2400         1            30
           =      43200 Kg

      Vu           <       ФVn
    3834.45        <       0.9        43200
    3834.45        <      38880
                  OK




6.3 Kontrol Dimensi Balok Melintang
    Dari Hasil Sap 2000 Diperoleh Mmax dan N max pada Frame 5 :

                (U-G)   Beban Ultimate - Beban Gempa
                           Mutx = -15837.38 Kgm
                           Muty =          0 Kgm
                             Nu = -8457.09 Kg
                             Vu = 9218.69 kg

                                Ma =    970.4 Kgm
                                Mb = 15837.88 Kgm
                                Ms = 11818.69 Kgm


   6.3.1 Profil Baja yang Didapatkan Di SAP 2000 8.2.3 :

                  340       x          250           x           9   x           14

         A=        101.5 cm2                  tf =       14 mm           Zx =   1360.02 cm3
         W=         79.7 kg/m                Ix =    21700 cm4           Zy =     439.2 cm3
          a=         340 mm                  Iy =     3650 cm4            h=        276 mm
         bf =        250 mm                 tw =          9 mm           Sx =      1280 cm3
         iy =          6 cm                  ix =      14.6 cm                  1360.02
                                                                                  439.2
   Mutu Baja =          BJ 37
       fu =        3700 kg/cm2                      fr =        700 kg/cm2
       fy =        2400 kg/cm2

6.3.1 Kontrol Lendutan

               f ijin =       L              =               500         =     1.39      cm
                             360                             360

                       5 L2
               f =          ( Ms − 0.1( Ma − Mb))
                      48 EI

         f=     5           25.00                          11818.69      0.1   970.4   15837.88
               48            200          21700

         f=         0.8 cm

               f              <            f ijin
                             OK

6.3.3 Kontrol Tekuk

untuk arah x :
  kcx =        1          (Sendi-Sendi)
    L=        600            cm
  Lkx =       600            cm
       Lkx
λx =           =               41.1
        ix
          π 2 EA
Ncrbx =                   Ncrbx =          9.87        x 2000000 x 101.5
           λx 2                                       1688.87
                          Ncrbx =                ### kg

untuk arah y :
  kcy =        1          (Sendi-Sendi)
    L=        600            cm
  Lky =       600            cm

         Lky   =                   100                     (MENENTUKAN)
λy =
          iy

        π 2 EA                                         x 2000000 x 101.5
Ncrby =                   Ncrby =          9.87
         λy 2                                        10000
                          Ncrby =        200341.15 kg

Maka dipakai λy karena λy > λx

          λy        fy       λc =          100              2400
 λc =                                      3.14            2000000
          π         E
                             λc =                   1.1

  0.25         <              λc             <                     1.2
ω=      1.43                  ω=           1.54
     1,67-0,67λc

       Pn =        Ag fy         =    101.5         *       2400     =         158629.17 kg
                    w                             1.54




       Pu
           =          8457.09          =             0.06        <       0.2
      ϕcPn         0.85 x158629.17

     Pakai Rumus =             Pu   Mux   Muy
                                  +     +      ≤1
                             2ϕcPn ϕbMnx ϕbMny




X Batang Dianggap Tidak Bergoyang Maka :
                  Cmx
       Sbx =                ≥1
                     Nu
               1− (       )
                    Ncrbx

      Cmx =          1
      Sbx =                      1
                     1           -   8457.09
                                       ###

       Sbx =       1.01          <     1

       Sbx =       1.01


     Mux = Mutx * Sbx
     Mux =     15837.38 x    1.01      =          15951.1 kgm


Y Batang Dianggap Tidak Bergoyang Maka :

                   Cmy
       Sby =                 ≥1
                      Nu
                1− (       )
                     Ncrby
      Cmy =          1
      Sby =                      1
                     1           -   8457.09
                                       ###

       Sby =       1.01          <     1

       Sby =       1.01


     Muy = Muty * Sby
     Muy =         0       x 1.01      =                 0 kgm
6.3.4 Menentukan Mnx

6.3.4.1 Kontrol Penampang profil
untuk Sayap                                                    untuk Badan

     b    170                                                  h 1680
        ≤                                                        ≤                           Kgm
    2tf     fy                                                 t   fy

   250                         170                                276              1680
    28            ≤           15.49                                9         ≤     15.49
   8.93           ≤           10.97                              30.67       ≤      108.44
                 OK                                                          OK

Penampang Profil Kompak, maka Mnx = Mpx

Lateral Buckling              Lb =         600           cm


                        E
Lp = 1.76 * iy                   Lp =      304.84 cm
                        fy

Lr = iy . [x1/(fy-fr)] . Sqrt{1+[sqrt(1+(x2 . fl2))]}

J = Σ(1/3).b.(t^3) =            53.32 cm4

Iw = Iy.((h^2)/4)       =    969768.5 cm6

x1 = [π/s]*[sqrt((EGJA)/2)] =           161407.01 kg/cm2

x2 = 4.[(S/GJ)^] =                       3.60E-09 cm2/kg

    Lr =         806.68 cm

    Lp              <          Lb            <           Lr    (INELASTIC BUCKLING)

Mnx = Cb.[Mr+[(Mp-Mr).((Lr-Lb)/(Lr-Lp))]]

                                                                        12,5 Mmax
                                                        Cb =
                                                               2,5Mmax + 3Ma + 4Mb + 3Mc

                                                        Cb =     1.81




   Mr =      Sx(fy-fr) = 2176000 kgcm
   Mp =      Zx . fy = 3264057.6 kgcm               < 1,5My
   My =      Sx . fy = 3072000 kgcm
 1,5 . My             =  4608000 kgcm               > Mp
  Mnx =             ### kgcm        > Mp

Mnx=Mp= 3264057.6 kgcm

    Mny = Zy ( 1 flen ) * fy
        = (1 / 4 * tf * bf 2) * fy
        =     0.25 x        1.4   x    625     x 2400        =    525000 kgcm
        =        5250 kgm




6.3.5 Persamaan Interaksi
                          Pu   Mux   Muy
                             +     +      ≤1
                        2ϕcPn ϕbMnx ϕbMny



      8457.09               +             15951.1            +          0
   1.7 x158629.17                      0.9   x 32640.58           0.9   x 5250

          0.03              +                 0.54           +          0

                                                      0.57   <     1

                                                             OK

6.3.6 Kontrol Kuat Rencana Geser

                   h    10
                       10
                     ≤
                  tw     fy
                  276       <         1100
                 2400                 15.49

                 0.12        <         71
                          Plastis

     Vn = 0.6 fy Aw
        =     0.6       2400           0.9           34
        =      44064 Kg

   Vu             <        ФVn
 9218.69          <         0.9       44064
 9218.69          <       39657.6
                 OK
6.3 Kontrol Dimensi Balok Memanjang
    Dari Hasil Sap 2000 Diperoleh Mmax dan N max pada Frame 5 :

                (U-G)         Beban Ultimate - Beban Gempa
                                 Mutx =       6219 Kgm
                                 Muty =          0 Kgm
                                   Nu =     -28073 Kg
                                   Vu =      -2663 kg

                                  Ma =       6219 Kgm
                                  Mb =       4435 Kgm
                                  Ms =        891 Kgm


   6.3.1 Profil Baja yang Didapatkan Di SAP 2000 8.2.3 :

                  300             x       200              x          8      x           12

         A=        72.38 cm2                     tf =         12 mm              Zx =   843.55 cm3
         W=         56.8 kg/m                   Ix =      11300 cm4              Zy =   248.32 cm3
          a=         300 mm                     Iy =       1600 cm4               h=       240 mm
         bf =        200 mm                    tw =            8 mm              Sx =      771 cm3
         iy =       4.71 cm                     ix =        12.5 cm                     843.55
                                                                                        248.32
   Mutu Baja =              BJ 37
        fu =           3700 kg/cm2                 fr =        700 kg/cm2
         fy =          2400 kg/cm2

   6.3.1 Kontrol Lendutan

                   f ijin =       L        =              500         =     1.39        cm
                                 360                      360

                           5 L2
                  f =           ( Ms − 0.1( Ma − Mb))
                          48 EI

          f=       5            25.00                     891        0.1    6219        4435
                  48             200     11300

          f=           0.08 cm

                   f              <       f ijin
                                 OK

   6.3.3 Kontrol Tekuk
     untuk arah x :
       kcx =        1         (Sendi-Sendi)
         L=        600           cm
       Lkx =       600           cm
            Lkx
     λx =            =                  48
             ix
               π 2 EA
     Ncrbx =                  Ncrbx =         9.87     x 2000000 x 72.38
                λx 2                                        2304
                              Ncrbx =        620069.3 kg

     untuk arah y :
       kcy =        1         (Sendi-Sendi)
         L=        600           cm
       Lky =       600           cm

              Lky    =           127.39                    (MENENTUKAN)
     λy =
               iy

             π 2 EA                                    x 2000000 x 72.38
     Ncrby =                  Ncrby =         9.87
              λy 2                                     16227.84
                              Ncrby =        88036.35 kg

     Maka dipakai λy karena λy > λx

               λy        fy      λc =        127.39         2400
      λc =                                    3.14         2000000
               π         E
                                 λc =                1.4

       0.25          <            λc            <                   1.2

ω=      1.43                     ω=                 1.96
     1,67-0,67λc

       Pn =         Ag fy         =           72.38            *          2400   =         88538.49 kg
                     w                                       1.96




       Pu
           =               28073                =              0.37        <         0.2
      ϕcPn          0.85     x88538.49

     Pakai Rumus =                Pu   Mux   Muy
                                     +     +      ≤1
                                2ϕcPn ϕbMnx ϕbMny




X Batang Dianggap Tidak Bergoyang Maka :
                  Cmx
       Sbx =                ≥1
                     Nu
               1− (       )
                    Ncrbx
                Cmx
     Sbx =                ≥1
                   Nu
             1− (       )
                  Ncrbx

    Cmx =          1
    Sbx =                      1
                   1           -    28073
                                   620069.3

     Sbx =       1.05          <      1

     Sbx =       1.05


  Mux = Mutx * Sbx
  Mux =        -6219    x 1.05        =       -6513.91 kgm


Y Batang Dianggap Tidak Bergoyang Maka :

                 Cmy
     Sby =                 ≥1
                    Nu
              1− (       )
                   Ncrby
    Cmy =          1
    Sby =                      1
                   1           -    28073
                                   620069.3

     Sby =       1.05          <      1

     Sby =       1.05


  Muy = Muty * Sby
  Muy =         0       x 1.05        =             0 kgm




  6.3.4 Menentukan Mnx

  6.3.4.1 Kontrol Penampang profil
  untuk Sayap                                         untuk Badan

       b    170                                       h 1680
          ≤                                             ≤                          Kgm
      2tf     fy                                      t   fy

     200                   170                           240             1680
      24           ≤      15.49                           8         ≤    15.49
     8.33          ≤      10.97                           30        ≤     108.44
                 OK                                                 OK

  Penampang Profil Kompak, maka Mnx = Mpx

  Lateral Buckling        Lb =       600       cm
                        E
Lp = 1.76 * iy                      Lp =       239.3 cm
                        fy

Lr = iy . [x1/(fy-fr)] . Sqrt{1+[sqrt(1+(x2 . fl2))]}

J = Σ(1/3).b.(t^3) =             53.32 cm4

Iw = Iy.((h^2)/4)       =     969768.5 cm6

x1 = [π/s]*[sqrt((EGJA)/2)] =              226284.21 kg/cm2

x2 = 4.[(S/GJ)^] =                          3.60E-09 cm2/kg

    Lr =           806.68 cm

    Lp              <           Lb             <             Lr   (INELASTIC BUCKLING)

Mnx = Cb.[Mr+[(Mp-Mr).((Lr-Lb)/(Lr-Lp))]]

                                                                           12,5 Mmax
                                                        Cb =
                                                                  2,5Mmax + 3Ma + 4Mb + 3Mc

                                                         Cb =       1.81




   Mr =       Sx(fy-fr) = 1310700 kgcm
   Mp =       Zx . fy = 2024524.8 kgcm                < 1,5My
   My =       Sx . fy = 1850400 kgcm
 1,5 . My              =  2775600 kgcm                > Mp

  Mnx =                 ### kgcm           > Mp

Mnx=Mp= 2024524.8 kgcm

     Mny = Zy ( 1 flen ) * fy
         = (1 / 4 * tf * bf 2) * fy
         =     0.25 x        1.2   x         400     x 2400          =       288000 kgcm
         =        2880 kgm




6.3.5 Persamaan Interaksi
                              Pu   Mux   Muy
                                 +     +      ≤1
                            2ϕcPn ϕbMnx ϕbMny



           28073                +                  6513.91           +             0
   1.7    x 88538.49               0.9    x 20245.25         0.9   x 2880

         0.19            +               0.36           +          0

                                                 0.54   <    1

                                                        OK

6.3.6 Kontrol Kuat Rencana Geser

                  h    10
                      10
                    ≤
                 tw     fy
                 240     <       1100
                2400             15.49

                0.1       <        71
                       Plastis

     Vn = 0.6 fy Aw
        =     0.6       2400       0.8          30
        =      34560 Kg

   Vu            <      ФVn
  2663           <      0.9      34560
  2663           <     31104
                OK
  7 Sambungan
                                              A


                 B




                     C                 D




7.1 Sambungan Kuda - Kuda ( Detail A )



                                                                                   60

                                                                 420               80

                                                                                  180
                                                                                  100

                                                                       200


    7.1.1 Data Perencanaan Sambungan
    Dari hasil perhitungan SAP diperoleh:

       Mu            =    1279.27      kgm        =      127927 kgcm
       Pu            =     1152         kg

    Baut Tanpa Ulir ( Bor ) Diameter φ =          8     mm
    Tebal Plat     =           10        mm

    7.1.2 Kontrol Kekuatan Baut
    7.1.2.1 Perhitungan Ruv yang Diterima Setiap Baut
       Ruv         =        Pu         =       1152          =           192 kg
                             n                   6
7.1.2.2 Perhitungan Kuat Geser Baut
   f Rnv       =       0.75      0.5                 fu         Ab                n
               =       0.75      0.5                4100        0.5               1
               =      772.44 kg

7.1.2.3 Perhitungan Kuat Tumpu Baut
   f Rn        =        2.4       d                  tp          fu              0.75
               =        2.4      0.8                  1         4100             0.75
               =       5904 kg

7.1.2.4 Perhitungan Kuat tarik Baut
   f Rnt       =       0.75       0.75               fu         Ab
               =       0.75       0.75                 4100           0.5
               =      1158.66 kg

7.1.2.5 Rumus Interaksi Geser dan Kuat Tarik Baut

          Ruv 2    R
     (        ) + ( ut ) 2 ≤ 1
         φRnv      Rnt
  192             +           Rut           <        1
 772.44                     1158.66

Rut = T =         870.66 kg

7.1.2.6 Kontrol Momen Sambungan

Letak Garis Netral a:
                                      2T
                           60

                           80         2T
                                                                            d3
                           180        2T                        d2
                                                           d1
                           100
                                                a
         200


   a =            ΣT             =      870.66      x            6
                 fy B                    2500       x            20
                                 =           0.1 cm

               d1 =                9.9 cm
               d2 =               27.9 cm
               d3 =               35.9 cm
                   Sdi =         73.69 cm
    f Mn =        0.9         fy          a2         B          +             S T di
                                                2
        =         0.9       2500         0.01        20         +         128311.88
                                                2
        =     128557.49 kgcm              >         127927 kgcm
                                         OK



7.2 Sambungan Kuda - Kuda dan Kolom ( Detail B )



                         Pu




                                    Mu                                                    60

                                                                    420                   80
                                                                                         180

                                                                                         100
                                                                            200



    7.2.1 Data Perencanaan Sambungan
    Dari hasil perhitungan SAP diperoleh:

       Mu          =       3240          kgm         =       324000 kgcm
       Pu          =      1344.96         kg

    Baut Tanpa Ulir ( Bor ) Diameter φ =             12    mm
    Tebal Plat     =           10        mm

    7.2.2 Kontrol Kekuatan Baut
    7.1.2.1 Perhitungan Ruv yang Diterima Setiap Baut
       Ruv         =        Pu         =      1344.96           =            224.16 kg
                             n                   6

    7.2.2.2 Perhitungan Kuat Geser Baut
       f Rnv       =        0.75     0.5             fu       Ab             n
                   =        0.75     0.5            4100     1.13            1
                   =      1737.99 kg

    7.2.2.3 Perhitungan Kuat Tumpu Baut
       f Rn        =        2.4       d              tp      fu             0.75
                   =        2.4      1.2              1     4100            0.75
                   =       8856 kg
    7.2.2.4 Perhitungan Kuat tarik Baut
       f Rnt       =       0.75       0.75                      fu       Ab
                   =       0.75       0.75                        4100     1.13
                   =      2606.99 kg

    7.2.2.5 Rumus Interaksi Geser dan Kuat Tarik Baut

              Ruv 2    R
         (        ) + ( ut ) 2 ≤ 1
             φRnv      Rnt
      224.16          +           Rut            <              1
     1737.99                    2606.99

    Rut = T =        2270.75 kg

    7.2.2.6 Kontrol Momen Sambungan

    Letak Garis Netral a:

                               60          2T

                               80          2T
                                                                                  d3
                               180         2T                            d2
                                                                    d1
                               100
                                                           a
             200


       a =            ΣT             =     2270.75     x                 6
                     fy B                   2500       x                 20
                                     =         0.27 cm

                   d1 =               9.73 cm
                   d2 =              27.73 cm
                   d3 =              35.73 cm
                       Sdi =         73.18 cm




    f Mn =            0.9            fy          a2             B        +             S T di
                                                       2
        =             0.9           2500        0.07           20        +        332357.74
                                                       2
        =          334028.37 kgcm                >             324000 kgcm
                                                OK

7.3 Sambungan Balok dan Kolom ( Detail C )
                         Pu


                                                                                 57
                                  Mu                                             60

                                                              576                60
                                                                                 60

                                                                                 198
                                                                                 141


                                                                           200

7.3.1 Data Perencanaan Sambungan
Dari hasil perhitungan SAP diperoleh:

   Mu          =       12614       kgm      =         1261400 kgcm
   Pu          =       12203        kg

Baut Tanpa Ulir ( Bor ) Diameter φ =        18       mm
Tebal Plat     =           10        mm

7.3.2 Kontrol Kekuatan Baut
7.3.2.1 Perhitungan Ruv yang Diterima Setiap Baut
   Ruv         =        Pu         =       12203          =         1220.3 kg
                         n                   10

7.3.2.2 Perhitungan Kuat Geser Baut
   f Rnv       =        0.75     0.5        fu          Ab           n
               =        0.75     0.5       4100        2.54          1
               =      3910.48 kg

7.3.2.3 Perhitungan Kuat Tumpu Baut
   f Rn        =        2.4       d         tp         fu           0.75
               =        2.4      1.8         1        4100          0.75
               =      13284 kg

7.3.2.4 Perhitungan Kuat tarik Baut
   f Rnt       =       0.75       0.75      fu            Ab
               =       0.75       0.75        4100          2.54
               =      5865.72 kg

7.3.2.5 Rumus Interaksi Geser dan Kuat Tarik Baut

         Ruv 2    R
    (        ) + ( ut ) 2 ≤ 1
        φRnv      Rnt
  1220.3       +        Rut         <       1
 3910.48              5865.72
    Rut = T =        4035.27 kg

    7.3.2.6 Kontrol Momen Sambungan


                                  57
                                              2T
                                  60
                                              2T
    576                           60
                                  60          2T
                                                                              d4   d5
                                              2T                         d3
                                  198
                                              2T                   d2
                                  141                         d1
                                                          a

                200

       a =            ΣT                =          4035.27     x              10
                     fy B                           2500       x              20
                                        =              0.81 cm

                d1    =                  13.29 cm
                d2    =                  33.09 cm
                d3    =                  39.09 cm
                d4    =                  45.09 cm
                d5    =                  51.09 cm
                          Sdi =         181.66



    f Mn =            0.9               fy           a2              B         +        S T di
                                                              2
          =           0.9              2500         0.65            20         +          ###
                                                              2
          =                 ### kgcm                  >            1261400 kgcm
                                                     OK




7.4 Sambungan Balok dan Kolom ( Detail D )


                                               Pu                   Pu
                                 Pu         Pu
                                                                                 57

                    Mu                                Mu                         60
                                                                    576          60

                                                                                 60

                                                                                198
                                                                                141




7.4.1 Data Perencanaan Sambungan
Dari hasil perhitungan SAP diperoleh:

   Mu          =         12976        kgm    =         1297600 kgcm
   Pu          =          7343         kg

Baut Tanpa Ulir ( Bor ) Diameter φ =         18       mm
Tebal Plat     =           10        mm

7.4.2 Kontrol Kekuatan Baut
7.4.2.1 Perhitungan Ruv yang Diterima Setiap Baut
   Ruv         =        Pu         =       7343            =         734.3 kg
                         n                   10

7.4.2.2 Perhitungan Kuat Geser Baut
   f Rnv       =        0.75     0.5         fu          Ab          n
               =        0.75     0.5        4100        2.54         1
               =      3910.48 kg

7.4.2.3 Perhitungan Kuat Tumpu Baut
   f Rn        =        2.4       d          tp         fu          0.75
               =        2.4      1.8          1        4100         0.75
               =      13284 kg

7.4.2.4 Perhitungan Kuat tarik Baut
   f Rnt       =       0.75       0.75       fu            Ab
               =       0.75       0.75         4100          2.54
               =      5865.72 kg

7.4.2.5 Rumus Interaksi Geser dan Kuat Tarik Baut

         Ruv 2    R
    (        ) + ( ut ) 2 ≤ 1
        φRnv      Rnt
  734.3        +        Rut            <     1
 3910.48              5865.72

Rut = T =    4764.27 kg
    7.4.2.6 Kontrol Momen Sambungan


                   57          2T
                   60
                               2T
                   60
                   60          2T
                                                             d4   d5
                               2T                       d3
                   198         2T
                                           d1     d2
                   141
                                      a
       200


       a =          ΣT          =     4764.27     x               10
                   fy B                2500       x               20
                                =         0.95 cm

              d1   =            13.15 cm
              d2   =            32.95 cm
              d3   =            38.95 cm
              d4   =            44.95 cm
              d5   =            50.95 cm
                       Sdi =   180.94



    f Mn =         0.9          fy          a2           B        +    S T di
                                                  2
        =          0.9         2500        0.91         20        +      ###
                                                  2
        =                ### kgcm           >          1297600 kgcm
                                           OK




7.5 Kontrol Kekuatan Sambungan Las
     7.5.1 Perencanaan Tebal Las Efektif pada Sambungan
     7.5.2 Perhitungan Tebal Las Efektif Pada Web
                aeff max Di Web           =            0.71       x               fu
                                                                            70          70.3
                                          =            0.71       x        3700          10
                                                                            70          70.3
                                          =            5.32       mm

                aeff max Di end plate     =            1.41       x               fu
                                                                            70          70.3
                                          =            1.41       x        4100          10
                                                                            70          70.3
                                          =             11.75     mm



7.5.3 Perhitungan Gaya yang Bekerja Pada Sambungan Las

Misal te =          1      cm

    A               =           72.2      +            34.8       =           107       cm2

     Sx =           d2           x        1             x         2
                    3

            =      72.2          +       34.8           x         1    x                       2
                                 3

            =     2544.22 cm3

         fv =      Pu            =              7343              =          68.63 kg/cm2
                   A                             107

         fh =      Mu            =            1297600             =         510.02 kg/cm2
                   Sx                         2544.22

  f total           =           fv2        +           fh2
                    =            68.63 +
                                       2
                                                       510.02 2
                    =           514.61 kg/cm2

7.5.4 Kontrol Kekuatan Las

φ fn =             0.75         0.6       70           70.3       =        2214.45 kg/cm2

   φ fn             >        f total
                   OK



7.5.5 Perhitungan Tebal efektif

te perlu            =        f total      =         514.61        =           0.23 cm
                               f fn                2214.45
   7.5.6 Perhitungan Lebar Perlu

   a perlu    =           0.87          =                1.23 mm              <          5.89 mm
                          0.71                                               OK

   7.5.7 Perhitungan Tebal Efektif Dengan Lebar Minimum

   a minimum =             4       mm
   te perlu =              4            x           0.71           =         2.83   mm



7.5 Sambungan Kolom Pondasi
     7.5.1 Data Perencanaan



                                   Rencana Panjang Plat Dasar kolom L                      40 cm
                                   Rencana Lebar Plat Dasar kolom B                        40 cm
                                   fc' beton                                               20 Mpa
                      P
                                   Momen yang bekerja pada dasar kolom Mu            1093100 kgcm
                                   Lintang yang bekerja pada dasar kolom Du              3854 kg
                                 M Normal yang bekerja pada dasar kolom Pu             22617 kg
                                   a                                                      300 mm
                                   a1                                                      50 mm
                                   b                                                    150.0 mm
                                   c                                                       50 mm
                                   d                                                      300 mm
                                   s pelat                                               1600 kg/cm2




             a1   a       a1


                                   c

                                   b
   d1                                   B
                                   b

                                   c

                  L

   7.5.2 Kontrol Pelat Landasan Beton ( Pondasi )

   fc' beton =             20     Mpa                =                 200 kg/cm2
   Pu yang bekerja =                        22617           kg
   Kekuatan nominal tumpu beton
                  Pn = 0.85 fc' A
                     A =       40       x         40              =            1600 cm2
             Pn =             0.85     200       1600             =             272000       kg

                 Pu           <=       f Pn
        22617                 <=        0.6        x                  272000
        22617                 <=            163200                       OK

7.5.2 Perencanaan Tebal Pelat Baja Pondasi
7.5.2.1 Perhitungan Tegangan Yang Bekerja Akibat Adanya Eksentrisitas

      e =    M                    =       1093100                 =            48.33 cm
             P                             22617
     A =     40                   x    40         =               1600 cm2
     W = 1/6 B L 2                =    1/6       40                 40 2 =           10666.67 cm3




      σ =        P                +     M
                 A                      W
         =            22617             +              1093100
                       1600                            10666.67
σ maks =                      116.61 kg/cm2
σ min =                        88.34 kg/cm2
 Jadi, q =                    116.61 kg/cm2




             2

                              3
             1

7.5.2.2 Perhitungan Momen Yang Bekerja
    ~ Daerah 1
       M = 1/2 q L2
         =     1/2    116.61         52
         = 1457.67 kgcm

   ~ Daerah 2
  a/b         =               30         /        15              =         2
   a1         =               0.1                 a2              =        0.05

   Ma            =               a1 q b2
                 =            0.1     116.61            15 2 =             2623.81 kgcm

   Mb            =               a2 q b2
                 =            0.05    116.61            15 2 =             1206.95 kgcm

   ~ Daerah 3
  a/b         =             5            /           30                =                0.17        <   0.5

   M3         =      1/2 q a12
              =          1/2       116.61                   52 =                   1457.67 kgcm

7.5.2.3 Perhitungan Tebal Pelat Baja
       s =    6 M                              t=                    6 M
               t2                                                    s plat

                                               =                     6 x         1457.67
                                                                              1600

                                               =        2.34           ~            3          cm




7.5.3 Perencanaan Diameter Angker
7.5.3.1 Perhitungan Tegangan Yang Bekerja Pada Angker


                                                            M


                                                        P




                                                     40
                                               x            y

                                                            M
                                                                20
                                σ min
                                                        P
                                                                           σ max

                                        1/3x        C           1/3y



  σ min       =        σ max
    x                   B-x

        x =       σ min B               =           88.34              x            40
                σ min + σ max                            88.34      +             116.61
                                               =           17.24 cm

       y =B - x                =              40              -        17.24           =            22.76 cm

  S min         =       1.5 d                  =       1.5 ( 2 x tf )
                =           1.5                2            x         1.2
                =                    3.6 cm

   1/3 x =          5.75 cm                    >       S min

   1/3 y =          7.59 cm                    >       S min

       r = 20 - 1/3 y          =              20              -        7.59            =            12.41 cm

      C =       40             -              7.59            -        5.75            =            26.67 cm

T =          M - P r           =                   1093100               -                 22617                12.41
                 C                                                             26.7

                               =                   30424.65       kg                   =           Pu




7.5.3.2 Perencanaan Diameter Angker
- Leleh:            Pu =           φ fy Ag

Ag perlu =                      30424.65                      =          8.25 cm2
                              0.9      4100                            #DIV/0!

- Putus:                Pu =                  φ 0.75 fu Ag

Ag perlu =                              30424.65                        =             #DIV/0! cm2
                              0.75        0.75                0                       #DIV/0!

- A baut perlu =                   30424.65                   =          27.16 cm2                 #DIV/0!
                                     1120

Untuk tiap sisi A baut perlu =   27.16     /        2 sisi       =      13.58                                cm2
Direncanakan menggunakan angker D30 ( A =               7.07 cm2 )
Jumlah angker dalam 1 sisi       =        #DIV/0!     /            7.07
                                 =        #DIV/0!     ~       #DIV/0! buah
Dipasang 3 D 30             A    =           21.2 cm2
                          Abaut  >      Aperlu
                                OK

7.5.4 Perencanaan Panjang Angker

Kekuatan Baut untuk menerima beban tarik pada tiap sisi adalah:
                30424.65          /        #DIV/0!       baut                          =           #DIV/0! kg
                      2 sisi

        l =                  #DIV/0!                      =          #DIV/0! cm
               4.47            2p           1.5

7.5.5 Perencanaan Sambungan Las




   t plat       =               3           cm

Profil Baja Bj 52                                  fu =                     0 kg/cm2
fu las E70 xx                   =           70     ksi
                                =           70            x          70.3    kg/cm3




Syarat tebal plat:
a min           =                3     mm
a max           =           t - 0.1         =             3           -            0.1
                                            =            2.9    cm


  af max        =           1.41 x fu elemen t
                                    fu las
                =             1.41         x              0           x            3
                               70          x             70.3
                =                   0 cm                  =                 0 mm

  Pakai          a              =              3 mm
                te              =           0.707 a                   =                2.12 mm
                 b              =            30 cm
                 d              =            30 cm
                Sx              =       b * d + ( d2 / 3)
                                =       30            30              +           900
                                                                                   3
                                =            1200 cm3

Akibat Pu :           fvp       =                 Pu
                                  2 ( 2 b + d ) te

                          =                      22617
                                        2         2    30          +       30         2.12
                          =        592.41 kg/cm2

Akibat Mu:       fhm      =        Mu         =            1093100
                                   Sx                        1200
                                              =          910.92 kg/cm2

   f las     =           0.75       x        0.6         x         70           x     70.3
             =          2214.45

  f total    =              fv2 + fh2
             =          592.41     2
                                      +     910.92   2

             =           ###      kg/cm2      <=     0.75 0.6 fu las
             =         1086.61    kg/cm2      <=      2214.45     kg/cm2            OK !!

				
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