Portal Bidang 1 Diketahui Struktur portal bidang (plane frame ) seperti gambar di bawah : Data Batang :L1=

  2 ^{2 2 3 2 3 2 2 2 3 2 3}

  1

  3

  2

  1

  2

  1

  2

  1

  1

  6

  5

  4

  3

  2

  2

  4

  1

  2

  6

  4

  5

  3

  1

  2

  Stiffness Method )

  1 Analisislah struktur di atas dengan menggunakan metode Kekakuan Langsung (Direct

  1

  

O

) :

  1 Batang 2 (α = 0

  1

  5

  6

• − + +
• − + + − + = L EI L EI L

  1 =

  2

  12

  6

  12

  6

  6

  4

  6

  2

  12

  6

  12

  6

  6

  6

  2

  4

  EI L EI L EI L EI L EI L EI L EA L EA L EI L EI L EI L EI L EI L EI L EI L EI L EA L EA S M

                 

                 

  1 P [ ]

  2 L

  2 L

  1

  X Z

  9 Y

  7

  8

  3

  1

• − + + − + − −
• &mi
  • 8.000 0.000 0.000 8.000 0.000 0.000 4

  1

  1 [R T ]

  )/12 = 213333.33 cm

  12EI/L

  4 ) α ( O ) EA/L

  2

) I (cm

  2 ) A (cm

  Tabel Perhitungan : ( i ) ( j ) ( k ) L (cm) E (kg/cm

  Sistim Koordinat Global

  Penyelesaian : Penomoran Lokal Penomoran Global

  P = 2000 kg

  2 Data Beban :

  4 E = sama kg/cm

  )/12 = 720000 cm

  3

  4 I

2 = (40x60

  3

  6EI/L

  = (40x40

  1

  2 I

  = 40 x 60 = 2400 cm

  2

  2 A

  = 40 x 40 = 1600 cm

  1

  = 300 cm A

  2

  = 500 cm L

  1

  Data Batang : L

  Diketahui Struktur portal bidang (plane frame ) seperti gambar di bawah :

  3

  2

  ) :

  3 Batang 2 : 8.000 0.000 0.000 -8.000 0.000 0.000 1 0.000 0.320 0.160 0.000 -0.320 0.160 2 0.000 0.160 9600.000 0.000 -0.160 4800.000 3

  Batang 1 (α = 90

O

  6 Matriks Rotasi Transformasi [R T ] :

  5

  4

  3

  2

  1

  0.000 -0.320 -0.160 0.000 0.320 -0.160 5 0.000 0.160 4800.000 0.000 -0.160 9600.000 6

  = E

  2

  ]

  M

  [S

  2

  4EI/L

  1

  9

  8

  7

  0.000 -0.031 -0.015 0.000 0.031 -0.015 2 0.000 0.015 1280.000 0.000 -0.015 2560.000 3

  1 = E -4.800 0.000 0.000 4.800 0.000 0.000 1

  [S M ]

  4.800 0.000 0.000 -4.800 0.000 0.000 7 0.000 0.031 0.015 0.000 -0.031 0.015 8 0.000 0.015 2560.000 0.000 -0.015 1280.000 9

  Matriks Kekakuan Batang Lokal [S M ] : Batang 1 :

  1 2 300 1 2400 720000.0 8.000 0.320 0.160 9600.000 4800.000

  2

  3 1 500 1.5 1600 213333.3 90 4.800 0.031 0.015 2560.000 1280.000

  1

  2EI/L

• 1
• 1

  1 [R

  T

  ]

  2

  =

  1

  1

  1

• 1
• α + α + α − α + =

• α + α + α − α +
• 0.015 0.000 2560.000 0.015 0.000 1280.000 9 [S
• 0.015 0.000 1280.000 0.015 0.000 2560.000 3

  2

  1

  9

  8

  7

  5

  4

  ] = E

  J

  6 [S

  3

  2

  1

  9 Penataan Ulang Matriks Kekakuan Struktur [S J ] (Rearrangement ) :

  8

  7

  6

  5

  4

  3

  2

  1

  9

  3

  6

  4

  2 3 {A FC }

  R T T

  1 Cos Sin Sin Cos Cos Sin Sin Cos

  1

           

  [ ]          

  9 Untuk batang 2 tidak perlu ditransformasikan karena Sumbu Lokal sama dengan Sumbu Global. Hal ini juga berlaku untuk semua batang yang mempunyai Sumbu Lokal sama dengan Sumbu Global.

  8

  5 7 {A RC }

  4

  6 {A C } = + =

  1

  5

  } 2000 2000

  E

  } {A

  J

  {A

  = P = 2000 kg 2000 1

  C } = {A J } + {A E } : Beban Ekivalen {A E } = 0 Beban Joint {A J } : Titik 1 :

  9 Analisis Beban Joint Ekivalen {A

  8

  7

  8

  6

  7

  T

  MS

  0.031 0.000 -0.015 -0.031 0.000 -0.015 7 0.000 4.800 0.000 0.000 -4.800 0.000 8

  MS ] = [R T ] T [S M ] [R T ] Batang 1 :

  1 Matriks Kekakuan Batang Global [S

  1

  1

  =

  2

  T

  ]

  1 [R

  1

  1

  1

  

O

) :

  1 Batang 2 (α = 0

  1

  T 1 = -1

  1 [R T ]

  1

  ) :

  Matriks Rotasi Transformasi Transpose [R T ] T : Batang 1 (α = 90

O

  ]

  = E -0.031 0.000 0.015 0.031 0.000 0.015 1 0.000 -4.800 0.000 0.000 4.800 0.000 2

  5

  3

  ] = E

  J

  4 [S

  3

  2

  1

  Perakitan Matriks Kekakuan Struktur [S J ] :

  6 Catatan :

  5

  4

  2

  7

  1

  = E -8.000 0.000 0.000 8.000 0.000 0.000 4 0.000 -0.320 -0.160 0.000 0.320 -0.160 5 0.000 0.160 4800.000 0.000 -0.160 9600.000 6

  2

  ]

  MS

  [S

  3 Batang 2 : 8.000 0.000 0.000 -8.000 0.000 0.000 1 0.000 0.320 0.160 0.000 -0.320 0.160 2 0.000 0.160 9600.000 0.000 -0.160 4800.000 3

  2

  1

  9

  8

S FF S FR S RF S RR

• 1992.35 + -7.65 + 2000.00 =
• 2000.00 + 2000.00 =
• [S

  ML

  J

  ].{D

  T

  .[R

  i

  ]

  M

  i

  }

  = {A

  i Batang 1 :

  i

  }

  M

  {A

  i

  }

  M

  {D

  i

  }

  4.8 -4.8 0 7 0.03072 0.01536 -0.03072 0.01536 8

  M

  0.01536 1280 -0.01536 2560 3 -3207.6917 299194.76 7

  1

  1

  3

  2

  1

  9

  8

  7

  1088.48 2

  =

  {A

  4.8 -4.8 0 7 0.03072 0.01536 -0.03072 0.01536 8

  1 x 0 9

  1 0 8 x

  0 7

  0.01536 1280 -0.01536 2560 3

  =

  1

  }

  M

  ]

  i

  2

  0.00 {A

  0.00

  0.00

  0.00

  0.00

  0.00 0.00 -4.80

  0.00

  } = - 0.00 + -0.03 0.00 -0.02 0.00 *

  R

  0.00 0.00 -0.32 -0.16 -0.16

  0.00 0.00 -1992.35 -1992.35 4

  0.00 0.00 249.04

  0.00

  0.00 -8.00

  Matriks Reaksi Tumpuan {A R } = - [A RC } + [S RF ] {D F }

  } = 1/E 0.000 0.000 0.000 0.000 * = 1/E 0.00 3 (rad - cm) 0.000 0.000 0.000 0.000 0.00 6

  F

  {D

  0.125 0.000 0.000 0.000 2000 249.04 1 0.000 0.195 0.000 0.000 0.00 2

  Matriks Perpindahan {D F } = [S FF ] -1 {A FC }

  0.02 0.00 1280.00

  0.00 0.00 0.00 5 = - 0.00 + -7.65 = -7.65 7 (kg-cm)

  }

  0.00 0.00 0.00 8 0.00 -3825.31 -3825.31 9

  ML

  = {A

  i

  }

  M

  {A

  Gaya Ujung Batang :

  0.00 0 = 0 … Ok.

  0.00 1000000.00 + -1000000.00 =

  = 0 1000000.00 + -996174.69 + -3825.31 =

  9

  .500 + M

  4

  P.500 - H

  terhadap titik 3 = 0

  0.00 0 = 0 … Ok. Σ ΣΣ ΣM

  0.00

  4 + H 7 + P = 0

  Σ ΣΣ ΣH = 0 H

  Kontrol Reaksi :

• [S
• 0.01536 2560 -0.01536 1280 9
• 4.8 4.8 0 1
• 0.03072 -0.01536 0.03072 -0.01536 2
• 1
• 1 37036.19 1 1 -62332.24 2 1 -3207.69 3

• 0.01536 2560 -0.01536 1280 9 x
• 4.8 4.8 0 1 62332.2427
• 0.03072 -0.01536 0.03072 -0.01536 2 37036.1937
• 1088.48 8 = -4105276.47 9
• 299194.76 1
• 8211121.82 3
• 6 6 0 4 300000 -0.135 -0.0675 0.135 -0.0675 5
• 20000000 0.0675 3600 -0.0675 7200 6

• 62332.243 2 x
• 911.518273 4 300000 805.2350241 5
• 20000000 11065714.14 6

• 911.52 4 300805.24 5
• 8934285.86 6

  300805.24 10

  911.52 5 8934285.86 6 Catatan : Untuk Gambar Bidang M, D dan N dapat dicoba sendiri dengan cara Superposisi.

  2

  1

  3

  5

  4

  6

  2

  2

  1

• 0 0.0118519 2844.444444 0 -0.011852 1422.222222 12
• 5.33333333 0 5.333333333 0 4 0 -0.023704 -0.01185185 0 0.0237037 -0.01185185 5 0 0.0118519 1422.222222 0 -0.011852 2844.444444 6
• 1

  4

  5

  6

  10

  11

  12

  3

  1

  5.333333333 -5.33333333 0 10 0 0.0237037 0.011851852 0 -0.023704 0.011851852 11 =

  1 0 11 x

  1 x 0 12

  0 10

  Batang 2 :

  6 -6 0 1 300000 0.135 0.0675 -0.135 0.0675 2 {A

  M

  }

  2

  = 20000000 + 0.0675 7200 -0.0675 3600 3

  1 37036.1937 1

  1

  1 x -3207.6917 3 1 36884.27 4 1 -56400.98 5 1 3140.81 6

  911.5182725 1 300000 -805.235024 2 = 20000000 + -11788878.2 3

  911.52 1 299194.76 2

  = 8211121.82 3

  Batang 3 :

  5.333333333 -5.33333333 0 10 0 0.0237037 0.011851852 0 -0.023704 0.011851852 11 {A

  M

  }

  3

  =

  2

• 1 36884.27 4 1 -56400.98 5 1 3140.81 6
• 0 0.0118519 2844.444444 0 -0.011852 1422.222222 12 x
• 5.33333333 0 5.333333333 0 4 56400.9816 0 -0.023704 -0.01185185 0 0.0237037 -0.01185185 5 36884.274 0 0.0118519 1422.222222 0 -0.011852 2844.444444 6 3140.80619
• 911.52 11 = 3336220.47 12
• 300805.24 4