Balok dan Kolom Terlentur

  

Perhitungan

Nilai Cb Balok dan Kolom Terlentur IR. THAMRIN NST.

  (Semoga bermanfaat)

  2015 thamrinnst.wordpress.com

  

Balok Diatas Perletakan Sederhana

Tabel : Nilai Cb Balok Diatas Perletakan Sederhana.

  X X 1,67 1,67

  X X 1,12

  X ^W 1,00

  X X 1,56 1,56

  X X 1,06

  X ^W 1,06

  X 1,52 1,52

  X X

  X ^W 1,01

  X X 1,45 1,45

  X ^W

  X X 1,30 1,30

  X X 1,14 ^W

  P P P P P P W

  X 1,11

  X X 1,11

  

Table 3-1

Values for C b for Simply Supported Beams Load

  

None

At midpoint

At third points

At quater

points

At fifth points

  

Lateral Bracing

Along Span

C b

  

None

Load at midpoint

  

At load point

None

Loads at third points

  

At load points

Loads symmetrically

placed

  

None

Loads at quater points

  

At load points

Loads at quater points

  Sumber : AISC – 2005, 13 th Editon, Steel Construction Manual.

  X X 1,14

  X X 1,67 1,67

  X X 1,32

  X X

  X 1,14

  X X 1,00

  X X 1,67 1,67

  X 1,12 Balok Diatas Perletakan Sederhana

  C O N T O H : Muatan terbagi rata penuh, dengan pengaku lateral (lateral bracing) sebanyak 2 (dua) buah ditengah-tengah bentang.

  W

  1

  2 X

  X L A B C

  Mmaks ₂ 1/8 W L 1/3 L 1/3 L

  1/12 L 1/12 L 1/12 L 1/12 L Gambar : Balok memikul beban terbagi rata, tanpa bracing ditengah bentang.

  12 ,

  5 M

  max Cb 

  2 ,

  5 M 

  3 M 

  4 M 

  3 M

  max A B C

  Dimana, R1 = R2 = 1/2 W L

  M A = (1/2 W L) x 1/12 L

• – (1/12 W L) x 1/24 L

  2 = 11/288 W L .

  M B = (1/2 W L) x 2/12 L

• – (2/12 W L) x 1/12 L

  2 = 5/72 W L .

  M = (1/2 W L) x 3/12 L C – (3/12 W L) x 3/24 L

  2 = 3/32 W L .

  M maks = (1/2 W L) x 4/12 L

• – (4/12 W L) x 2/12 L

  2 = 1/9 W L .

  

2

  12 , 5 ( 1 /

  9 W L )

  Cb 

  2

  2

  2

  2

  2 , 5 ( 1 /

  9 W L )  3 ( 11 / 288 W L )  4 ( 5 /

  72 W L )  3 ( 3 /

  32 W L )

  Cb  1,460

• – (5/12 W L) x 5/24 L = 35/288 W L

  2

  1 ( / 288 4 ) 35 (

  3 ) 8 / 1 ( 5 ,

  2 ) 8 /

  1 ( 5 ,

  12

  2

  2

  2 .

  2

  2 L W L W L W L W

L W

Cb

     

   Cb 1,014 1/8 W L ² W

  1 L A B C Mmaks

  2 1/12 L

  X X 1/3 L 1/3 L

  ) 288 / 35 ( 3 ) 8 /

  M maks = 1/8 W L

  Muatan terbagi rata penuh, dengan pengaku lateral (lateral bracing) sebanyak 2 (dua) buah ditengah-tengah bentang.

     

  Gambar : Balok memikul beban terbagi rata, tanpa bracing ditengah bentang.

  C B A max max

  3

  4

  3 5 , 2 5 ,

  12 M M M M

  

M

Cb

  Dimana, R1 = R2 = 1/2 W L

  2 .

  M A

  = (1/2 W L) x 5/12 L

  2 .

  M B

  = 1/8 W L

  2 .

  M C = M A = 35/288 W L

  

1/12 L 1/12 L 1/12 L

• – (1/4 W L) x 1/8 L – (3/4 M1 + 1/4 M2) = 3/32 W L
• – (1/2 M1 + 1/2 M2)
• – (1/4 M1 + 3/4 M2) Momen M
• – (1/2 M1 + 1/2 M2), atau

  M B = 1/8 W L

  Mmaks

  2 1/4 L 1/4 L 1/4 L

  Kejadian khusus, M1 = M2 = 1/12 W L

  M maks = M1, atau M maks = M2

  2

  = 1/8 W L

  B

  = M

  M maks

  ada 3 kemungkinan, yaitu :

  maks

  2

  M C = 3/32 W L

  2

  2 – (3/4 M1 + 1/4 M2).

  1/8 W L ² W

  = (1/2 W L) x 1/4 L

  M A

  ≠ M2 R1 = R2 = 1/2 W L (hanya akibat muatan terbagi rata).

  Dimana, M1

     

  M Cb

  12 M M M M

  3 5 , 2 5 ,

  4

  3

  C B A max max

  

Gambar : Balok memikul beban terbagi rata, tanpa bracing ditengah bentang.

  1). Muatan terbagi rata penuh, tanpa pengaku lateral (lateral bracing) ditengah-tengah bentang.

  

Balok Diatas Perletakan Jepit-jepit

  1 L M1 M2 A

B

C

2 Maka,

  2 M B = 1/8 W L

• – 1/12 W L
• – 1/12 W L

  2

  2

  2

  = 1/96 W L

  2,381

   Cb

     

  2 L W L W L W L W L W Cb

  2

  2

  2

  12

  2

  1 ( 5 ,

  2 ) 12 /

  5 ,

  1 ( 3 ) 12 / 1 (

  24 / 1 ( 4 ) 96 /

  ) 96 / 1 ( 3 )

  2

  = 1/12 W L

  2 M maks

  = 1/24 W L

  2

  M A = M C = 3/32 W L

  2). Muatan terbagi rata penuh, dengan pengaku lateral (lateral bracing) ditengah-tengah bentang.

  

W

  1

  2 X L M1 M2 A

  Mmaks B ₂ C 1/8 W L

Mmaks

  1/8 L 1/8 L 1/8 L 1/8 L 1/2 L

Gambar : Balok memikul beban terbagi rata, dengan satu bracing ditengah bentang.

  12 ,

  5 M

  max Cb 

  2 ,

  5 M 

  3 M 

  4 M 

  3 M

  max A B C

  Dimana, M1

  ≠ M2 R1 = R2 = 1/2 W L (hanya akibat muatan terbagi rata). Tinjauan disebelah kiri :

  = (1/2 W L) x 1/8 L

  M A

• – (1/8 W L) x 1/16 L – (7/8 M1 + 1/8 M2)

  2

  = 7/128 W L – (7/8 M1 + 1/8 M2).

  M B = (1/2 W L) x 1/4 L

• – (1/4 W L) x 1/8 L – (3/4 M1 + 1/4 M2)

  2

  = 3/32 W L – (3/4 M1 + 1/4 M2).

  M C = (1/2 W L) x 3/8 L

• – (3/8 W L) x 3/16 L – (5/8 M1 + 3/8 M2)

  2

  = 15/128 W L

• – (5/8 M1 + 3/8 M2) Momen M maks ada 2 kemungkinan, yaitu :

  2 M maks = 1/8 W L

• – (1/2 M1 + 1/2 M2), atau M maks = M1.

  Tinjauan disebelah kanan :

  M A = (1/2 W L) x 1/8 L

• – (1/8 W L) x 1/16 L – (7/8 M2 + 1/8 M1)

  2

  = 7/128 W L – (7/8 M2 + 1/8 M1).

  M B = (1/2 W L) x 1/4 L

• – (1/4 W L) x 1/8 L – (3/4 M2 + 1/4 M1)

  2

  = 3/32 W L – (3/4 M2 + 1/4 M1).

  M = (1/2 W L) x 3/8 L C

• – (3/8 W L) x 3/16 L – (5/8 M2 + 3/8 M1)

  2

  = 15/128 W L

• – (5/8 M2 + 3/8 M1) Momen M maks ada 2 kemungkinan, yaitu :

  2 M maks = 1/8 W L

• – (1/2 M1 + 1/2 M2), atau M = M2.

  maks

• – 1/12 W L
• – 1/12 W L

  ≠ M2 R1 = R2 = 1/2 W L (hanya akibat muatan terbagi rata). Tinjauan disebelah kiri :

   Cb 2,381

  3a). Muatan terbagi rata penuh, dengan pengaku lateral (lateral bracing) sebanyak 2 (dua) buah ditengah-tengah bentang.

  

Gambar : Balok memikul beban terbagi rata, dengan dua bracing ditengah bentang.

  C B A max max

  3

  4

  3 5 , 2 5 ,

  12 M M M M

  M Cb

     

  Dimana, M1

  M A = (1/2 W L) x 1/12 L

  2 L W L W L W L W L W Cb

  2 – (11/12 M1 + 1/12 M2).

  M B = (1/2 W L) x 2/12 L

  2 – (10/12 M1 + 2/12 M2).

  M C = (1/2 W L) x 3/12 L

  2

  maks

  ada 1 kemungkinan, yaitu :

  M maks = M1 1/8 W L ² W

  1 L M1 M2 A B C

  Mmaks

  2 1/12 L

  X X 1/3 L

1/3 L

     

  2

  1/12 L 1/12 L 1/12 L

  2

  Kejadian khusus, M1 = M2 = 1/12 W L

  2

  ( hanya akibat beban terbagi rata penuh bentang ) Maka,

  M A = 7/128 W L

  2

  2

  =

  2 M B = 3/32 W L

  2

  2

  = 1/96 W L

  2 M C = 15/128 W L

  2

  2

  = 13/384 W L

  2 M maks

  = M1 = M2 = 1/12 W L

  2

  ) 384 / 13 ( 3 ) 96 /

  1 ( / 384 4 ) 11 (

  3 ) 12 / 1 ( 5 ,

  2 ) 12 /

  1 ( 5 ,

  12

  2

  2

• – (1/12 W L) x 1/24 L – (11/12 M1 + 1/12 M2) = 11/288 W L
• – (2/12 W L) x 1/12 L – (10/12 M1 + 2/12 M2) = 5/72 W L
• – (3/12 W L) x 3/24 L – (9/12 M1 + 3/12 M2) = 3/32 W L
• – (9/12 M1 + 3/12 M2) Momen M

Tinjauan disebelah kanan :

  M = (1/2 W L) x 1/12 L A

• – (1/12 W L) x 1/24 L – (11/12 M2 + 1/12 M1)

  2

  = 11/288 W L – (11/12 M2 + 1/12 M1).

  M B = (1/2 W L) x 2/12 L

  2

  = 5/72 W L – (10/12 M2 + 2/12 M1).

  M C = (1/2 W L) x 3/12 L

• – (3/12 W L) x 3/24 L – (9/12 M2 + 3/12 M1)

  2

  = 3/32 W L

• – (9/12 M2 + 3/12 M1) Momen M maks ada 1 kemungkinan, yaitu :

  M maks = M2

  Kejadian khusus,

2 M1 = M2 = 1/12 W L ( hanya akibat beban terbagi rata penuh bentang )

  Maka,

  2

  2

  2 M A = 11/288 W L =

• – 1/12 W L – 13/288 W L

  2

  2

  2 M B = 5/72 W L =

• – 1/12 W L – 1/72 W L

  2

  2

  2 M C = 3/32 W L = 1/96 W L

• – 1/12 W L

2 M maks = M1 = M2 = 1/12 W L

  2

  12 , 5 ( 1 /

  12 W L )

  Cb 

  2

  2

  2

  2

  2 , 5 ( 1 /

  12 W L )  3 ( 13 / 288 W L )  4 ( 1 /

  72 W L )  3 ( 1 /

  96 W L )

  Cb  2,419

  3b). Muatan terbagi rata penuh, dengan pengaku lateral (lateral bracing) sebanyak 2 (dua) buah ditengah-tengah bentang.

  

W

  1

  X L M1 ₂ M2 A B C 1/8 W L

  2 X

  

Mmaks

1/3 L 1/3 L

  

1/12 L 1/12 L 1/12 L 1/12 L

Gambar : Balok memikul beban terbagi rata, dengan dua bracing ditengah bentang.

  12 ,

  5 M

  max Cb 

  2 ,

  5 M 

  3 M 

  4 M 

  3 M

  max B A C

• – (6/12 W L) x 3/12 L – (1/2 M1 + 1/2 M2) = 1/8 W L
• – (7/12 W L) x 7/24 L – (5/12 M1 + 7/12 M2) = 35/288 W L
• – (5/12 M1 + 7/12 M2) Momen M maks ada 1 kemungkinan, yaitu :
• – 1/12 W L

  2

  = 11/288 W L

  2 M maks = M B = 1/24 W L

  2

  ) 288 / 11 ( 3 ) 24 /

  1 ( / 288 4 ) 11 (

  3 ) 24 / 1 ( 5 ,

  2 ) 24 /

  1 ( 5 ,

  12

  2

  2

  2

  2

  2 L W L W L W L W L W Cb

     

   Cb 1,042

  4a). Muatan terbagi rata penuh, dengan pengaku lateral (lateral bracing) sebanyak 3 (tiga) buah ditengah-tengah bentang.

  

Gambar : Balok memikul beban terbagi rata, dengan tiga bracing ditengah bentang.

  1/8 W L ² W

  1 L M1 M2 A B C

  2 X Mmaks

  X 1/4 L 1/16 L

  1/4 L 1/16 L 1/16 L 1/16 L

  2

  2 M C = 35/288 W L

  

X

1/4 L

  Kejadian khusus, M1 = M2 = 1/12 W L

  Dimana, M1

  ≠ M2 R1 = R2 = 1/2 W L (hanya akibat muatan terbagi rata).

  M A = (1/2 W L) x 5/12 L

  = 35/288 W L

  2 – (7/12 M1 + 5/12 M2).

  M B = (1/2 W L) x 6/12 L

  2 – (1/2 M1 + 1/2 M2).

  M C = (1/2 W L) x 7/12 L

  2

  M maks = M B

  2

  = 1/24 W L

  (

  hanya akibat beban terbagi rata penuh bentang

  ) Maka,

  M A = 35/288 W L

  2

  2

  = 11/288 W L

  2 M B

  = 1/8 W L

  2

  2

• – 1/12 W L
• – 1/12 W L

• – (1/16 W L) x 1/32 L – (15/16 M1 + 1/16 M2) = 15/512 W L
• – (2/16 W L) x 1/16 L – (14/16 M1 + 2/16 M2) = 7/128 W L
• – (3/16 W L) x 3/32 L – (13/16 M1 + 3/16 M2) = 39/512 W L
• – (13/16 M1 + 3/16 M2) Momen M maks ada 1 kemungkinan, yaitu :
• – (1/16 W L) x 1/32 L – (15/16 M2+ 1/16 M1) = 39/512 W L
• – (2/16 W L) x 1/16 L – (14/16 M2 + 2/16 M1) = 7/128 W L

• – (3/16 W L) x 3/32 L – (13/16 M2 + 3/16 M1) = 39/512 W L
• – (13/16 M2 + 3/16 M1) Momen M maks ada 1 kemungkinan, yaitu :
• – 1/12 W L
• – 3/419 W L
• – 1/12 W L
• – 11/384 W L
• – 1/12 W L
• – 3/419 W L

  ) Maka,

  M A

  = 39/512 W L

  2

  2

  =

  2 M B = 7/128 W L

  2

  2

  =

  2 M C = 39/512 W L

  2

  2

  2 M maks = M1 = 1/12 W L

  =

  2

  ) 419 / 3 ( / 384 3 ) 11 (

  / 419 4 ) 3 ( 3 ) 12 /

  1 ( 5 ,

  2 ) 12 /

  1 ( 5 ,

  12

  2

  2

  2

  2

  2 L W L W L W L W L W Cb

     

  hanya akibat beban terbagi rata penuh bentang

  (

  2

  = (1/2 W L) x 2/16 L

  C B A max max

  3

  4

  3 5 , 2 5 ,

  12 M M M M

  M Cb

     

  Dimana, M1

  ≠ M2 R1 = R2 = 1/2 W L (hanya akibat muatan terbagi rata). Tinjauan disebelah kiri :

  M A = (1/2 W L) x 1/16 L

  2 – (15/16 M1 + 1/16 M2).

  M B

  2 – (14/16 M1 + 2/16 M2).

  M1 = M2 = 1/12 W L

  M C = (1/2 W L) x 3/16 L

  2

  M maks

  = M1 Tinjauan disebelah kanan :

  M A = (1/2 W L) x 1/16 L

  2 – (15/16 M2 + 1/16 M1).

  M B

  = (1/2 W L) x 2/16 L

  2 – (14/16 M2 + 2/16 M1).

  M C = (1/2 W L) x 3/16 L

  2

  M maks

  = M2 Kejadian khusus,

   Cb 2,847

• – (5/16 W L) x 5/32 L – (11/16 M1 + 5/16 M2) = 55/512 W L
• – (6/16 W L) x 3/16 L – (10/16 M1 + 6/16 M2) = 15/128 W L
• – (7/16 W L) x 7/32 L – (9/16 M1 + 7/16 M2) = 63/512 W L
• – (1/2 M1 + 1/2 M2) Tinjauan disebelah kanan :
• – (5/16 W L) x 5/32 L – (11/16 M2 + 5/16 M1) = 55/512 W L

• – (6/16 W L) x 3/16 L – (10/16 M2 + 6/16 M1) = 15/128 W L
• – (7/16 W L) x 7/32 L – (9/16 M2 + 7/16 M1) = 63/512 W L
• – 1/12 W L
• – 1/12 W L
• – 1/12 W L

  hanya akibat beban terbagi rata penuh bentang

  ) Maka,

  M A = 55/512 W L

  2

  2

  = 2/83 W L

  2 M B

  = 15/128 W L

  2

  2

  = 13/384 W L

  2 M C = 63/512 W L

  2

  2

  = 11/277 W L

  2 M maks = M1 = 1/12 W L

  2 1/8 W L ² W

  1 L M1 M2 A B C

  2 X Mmaks

  X 1/4 L 1/16 L

  1/4 L 1/16 L 1/16 L 1/16 L

  (

  Kejadian khusus, M1 = M2 = 1/12 W L

  2

  M C = (1/2 W L) x 7/16 L

  4b). Muatan terbagi rata penuh, dengan pengaku lateral (lateral bracing) sebanyak 3 (tiga) buah ditengah-tengah bentang.

  

Gambar : Balok memikul beban terbagi rata, dengan tiga bracing ditengah bentang.

  Dimana, M1

  ≠ M2 R1 = R2 = 1/2 W L (hanya akibat muatan terbagi rata). Tinjauan disebelah kiri :

  M A

  = (1/2 W L) x 5/16 L

  2 – (11/16 M1 + 5/16 M2).

  M B = (1/2 W L) x 6/16 L

  2 – (10/16 M1 + 6/16 M2).

  2 – (9/16 M1 + 7/16 M2).

  2 – (9/16 M2 + 7/16 M1).

  Momen M

  maks

  ada 1 kemungkinan, yaitu :

  M maks = 1/8 W L

  2

  M A = (1/2 W L) x 5/16 L

  2 – (11/16 M2 + 5/16 M1).

  M B = (1/2 W L) x 6/16 L

  2 – (10/16 M2 + 6/16 M1).

  M C = (1/2 W L) x 7/16 L

  

X

1/4 L

  2

  12 , 5 ( 1 /

  12 W L )

  Cb 

  2

  2

  2

  2

  2 , 5 ( 1 /

  12 W L )  3 ( 2 /

  83 W L )  4 ( 13 / 384 W L )  3 ( 11 / 277 W L )

  Cb  1,946

  4c). Muatan terpusat, dengan pengaku lateral (lateral bracing) sebanyak 3 (tiga) buah ditengah-tengah bentang.

  P P P

  1

  2 X

  X X L M1

  M2 Mmaks 1/2 P L

  B A 3/16 P L

  C 1/16 L 1/16 L 1/4 L 1/4 L

  1/4 L 1/16 L 1/16 L

Gambar : Balok memikul terpusat, dengan tiga bracing ditengah bentang.

  12 ,

  5 M

  max Cb 

  2 ,

  5

  3

  4

  3 M  M  M  M

  max A B C

  Dimana, M1

  ≠ M2 R1 = R2 = 3/2 P (hanya akibat muatan terpusat P). Tinjauan disebelah kiri :

  M A = (3/2 P) x 1/16 L

• – (15/16 M1 + 1/16 M2) = 3/32 P L – (15/16 M1 + 1/16 M2).

  M B = (3/2 P) x 2/16 L

• – (14/16 M1 + 2/16 M2) = 6/32 P L – (14/16 M1 + 2/16 M2).

  M = (3/2 P) x 3/16 L C

• – (13/16 M1 + 3/16 M2) = 9/32 P L – (13/16 M1 + 3/16 M2)

  Momen M maks ada 2 kemungkinan, yaitu :

  M maks = M1, atau M maks = (3/2 P) x 4/16 L

• – (12/16 M1 + 4/16 M2) Tinjauan disebelah kanan :

  M A = (3/2 P) x 1/16 L

• – (15/16 M2 + 1/16 M1) = 3/32 P L – (15/16 M2 + 1/16 M1).

  M B = (3/2 P) x 2/16 L

• – (14/16 M2 + 2/16 M1) = 6/32 P L – (14/16 M2 + 2/16 M1).

  M C = (3/2 P) x 3/16 L

• – (13/16 M2 + 3/16 M1)
• – (13/16 M2 + 3/16 M1) Momen M

• – 5/16 P L = – 7/32 P L
• – 5/16 P L = – 4/32 P L
• – 5/16 P L = – 1/32 P L

  ≠ M2 R1 = R2 = 3/2 P (hanya akibat muatan terpusat P). Tinjauan disebelah kiri :

  C B A max max

  3

  4

  3 5 , 2 5 ,

  12 M M M M

  M Cb

     

  Dimana, M1

  M A = (3/2 P) x 5/16 L

  1,923 4d). Muatan terpusat, dengan pengaku lateral (lateral bracing) sebanyak 3 (tiga) buah ditengah-tengah bentang.

  M B

  = (3/2 P) x 6/16 L

  1/2 P L P

  1 L M1 M2 A B C

  2 X Mmaks

  X 1/4 L 1/16 L

  1/4 L 1/16 L 1/16 L 1/16 L

  

X

1/4 L

  

Gambar : Balok memikul beban terpusat, dengan tiga bracing ditengah bentang.

  Cb

  P P 3/16 P L

  M C = 9/32 P L

  = 9/32 P L

  maks

  ada 2 kemungkinan, yaitu :

  M maks = M2, atau M maks = (3/2 P) x 4/16 L

  Kejadian khusus, M1 = M2 = 5/16 P L ( hanya akibat beban terpusat )

  Maka,

  M A = 3/32 P L

  M B = 6/32 P L

  M maks

  

  = M1 = 5/16 P L

  ) 32 / 1 ( 3 )

  

32 /

4 ( 4 ) 32 /

  7 ( 3 ) 16 / 5 (

  5 ,

  2

)

16 / 5 (

  5 ,

  12 L P L P L P L P L P Cb

     

• – (11/16 M1 + 5/16 M2) = 15/32 P L – (11/16 M1 + 5/16 M2).
• – (10/16 M1 + 6/16 M2)

  = 18/32 P L – (10/16 M1 + 6/16 M2).

  M = (3/2 P) x 7/16 L C

• – (9/16 M1 + 7/16 M2) = 21/32 P L – (19/16 M1 + 7/16 M2)

  Momen M maks ada 1 kemungkinan, yaitu :

  M maks = 3/16 P L

  Tinjauan disebelah kanan :

  M A = (3/2 P) x 5/16 L

• – (11/16 M2 + 5/16 M1) = 15/32 P L – (11/16 M2 + 5/16 M1).

  M B = (3/2 P) x 6/16 L

• – (10/16 M2 + 6/16 M1) = 18/32 P L – (10/16 M2 + 6/16 M1).

  M = (3/2 P) x 7/16 L C

• – (9/16 M2 + 7/16 M1) = 21/32 P L – (19/16 M2 + 7/16 M1)

  Momen M maks ada 1 kemungkinan, yaitu :

  M maks = 3/16 P L

  Kejadian khusus, M1 = M2 = 5/16 P L ( hanya akibat beban terpusat )

  Maka,

  M A = 15/32 P L

• – 5/16 P L = 5/32 P L

  M B = 18/32 P L

• – 5/16 P L = 8/32 P L

  M = 21/32 P L C

• – 5/16 P L = 11/32 P L Momen M maks ada 1 kemungkinan, yaitu :

  M maks = 3/16 P L 12 , 5 ( 3 /

16 P L )

  Cb  2 , 5 ( 3 /

  16 P L )  3 ( 5 /

  32 P L )  4 (

8 /

  32 P L )  3 ( 11 /

  32 P L ) Cb  0,789

  

Kolom Jepit-jepit dan Jepit-Sendi

1). Kolom memikul momen distribusi linear, tanpa pengaku lateral.

  = M ) 2 /

  1/4H 1/4H 1/4H 1/4H

  2 H (b)

  1 M A B C

  (a)

  2 M2 H ha hb

  1 M1 A B C

   Cb 2,273

     

  M Cb

  12 M M M

  2 ) ( 5 ,

  2 / 1 ( ( 3 ) 5 ,

  1 ( ( 3 ) 4 )

  maks

  Gambar : Kolom memikul momen distribusi linear, tanpa pengaku lateral, (a) perletakan jepit, (b) perletakan jepit sendi .

  Momen M

  M A = 1/2 M M B = 0 M C = 1/2 M

  Maka, M1 = M2 = M, dan ha = hb = 1/2H

  Bila, M1/M2 = 1 ,

  M 2 M 1 

  hb M 2 ha M 1  , atau hb ha

  Gambar (a) : Dengan perbandingan garis pada segitiga, diperoleh

     

  M Cb

  12 M M M M

  3 5 , 2 5 ,

  4

  3

  C B A max max

  1/4H 1/4H 1/4H 1/4H Bila,

  0,75 M2

  M1/M2 = 0,75 = 3/4 , Maka,

  3 M _A

  M1 = 0,75 M2

  M B M = 1,25/3 x (0,75 M2) = 0,3125 M2

  A

  4 M C M B = 0,5/4 x (M2) = 0,1250 M2 M C = 2,25/4 x (M2) = 0,5625 M2

  Momen M maks = M2

  M2

  12 , 5 ( M 2 )

  Cb 

  2 , 5 ( M 2 )  3 ( , 3125 M 2 )  4 ( , 1250 M 2 )  3 ( , 5625 M 2 )

  Cb  2,222

  Selanjutnya lihat tabel berikut,

Sumber : Charles G. Salmon, Steel Structures Design and Behavior, 5th Edition, 2009, page 429.

  Gambar (b) :

  M A = 3/4 M M B = 1/2 M M = 1/4 M

  C

  Momen M maks = M 12 , 5 ( M )

  Cb 

  2 , 5 ( ) 3 ( 3 / 4 ) 4 ( 1 / 2 ) 3 ( 1 / 4 )

  M  M  M  M Cb 

  1,667