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Perhitungan Nilai Cb Balok dan Kolom Terlentur IR. THAMRIN NST.

(Semoga bermanfaat)

2015

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Perhitungan Nilai Cb Untuk Balok dan Kolom Lentur, 2015 Ir. Thamrin Nst.

1

Balok Diatas Perletakan Sederhana

Tabel : Nilai Cb Balok Diatas Perletakan Sederhana.

Table 3-1

Values for Cb for Simply Supported Beams

Load Lateral Bracing

Along Span Cb

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

None

At midpoint

At third points

At quater points

At fifth points

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

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 X

1,14

X X

1,11

X

X

1,67 1,67

X

1,11

P

P

P

P

P P

W

X X

1,14

W

X X

1,30 1,30

X

W

X X

1,45 1,45

X

W

1,01

X

X X

1,52 1,52

X

W

1,06

X

X

1,06

X X

1,56 1,56

X

W

1,00

X

X

1,12

X

1,12


2

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.

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

CBAmax

max

3435,2

5,12

MMMM

MCb

Dimana,

R1 = R2 = 1/2 W L

MA = (1/2 W L) x 1/12 L – (1/12 W L) x 1/24 L

= 11/288 W L2.

MB = (1/2 W L) x 2/12 L – (2/12 W L) x 1/12 L

= 5/72 W L2.

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

= 3/32 W L2.

Mmaks = (1/2 W L) x 4/12 L – (4/12 W L) x 2/12 L

= 1/9 W L2.

)32/3(3)72/5(4)288/11(3)9/1(5,2

)9/1(5,122222

2

LWLWLWLW

LWCb

Cb 1,460

1/8 W L2

W

1

L

A

B C

Mmaks

2

1/12 L

X X

1/3 L 1/3 L 1/12 L 1/12 L 1/12 L


3

Muatan terbagi rata penuh, dengan pengaku lateral (lateral bracing) sebanyak 2 (dua)



CBAmax

max

3435,2

5,12

MMMM

MCb

Dimana,

R1 = R2 = 1/2 W L

MA = (1/2 W L) x 5/12 L – (5/12 W L) x 5/24 L = 35/288 W L2.

MB = 1/8 W L2.

MC = MA = 35/288 W L2.

Mmaks = 1/8 W L2.

)288/35(3)8/1(4)288/35(3)8/1(5,2

)8/1(5,122222

2

LWLWLWLW

LWCb

Cb 1,014

1/8 W L2

W

1

L

A B C

Mmaks

2

1/12 L

X X

1/3 L 1/3 L 1/12 L 1/12 L 1/12 L


4

Balok Diatas Perletakan Jepit-jepit

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

bentang.


CBAmax

max

3435,2

5,12

MMMM

MCb

Dimana,

M1 ≠ M2

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

MA = (1/2 W L) x 1/4 L – (1/4 W L) x 1/8 L – (3/4 M1 + 1/4 M2)

= 3/32 W L2 – (3/4 M1 + 1/4 M2).

MB = 1/8 W L2 – (1/2 M1 + 1/2 M2)

MC = 3/32 W L2 – (1/4 M1 + 3/4 M2)

Momen Mmaks ada 3 kemungkinan, yaitu :

Mmaks = MB = 1/8 W L2 – (1/2 M1 + 1/2 M2), atau

Mmaks = M1, atau

Mmaks = M2

Kejadian khusus,

M1 = M2 = 1/12 W L2

Maka,

MA = MC = 3/32 W L2 – 1/12 W L

2 = 1/96 W L

2

MB = 1/8 W L2 – 1/12 W L

2 = 1/24 W L

2

Mmaks = 1/12 W L2

)96/1(3)24/1(4)96/1(3)12/1(5,2

)12/1(5,122222

2

LWLWLWLW

LWCb

Cb 2,381

1/8 W L2

W

1

L

M1 M2

A

B

C

Mmaks

2

1/4 L 1/4 L 1/4 L


5

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

bentang.

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

CBAmax

max

3435,2

5,12

MMMM

MCb

Dimana,

M1 ≠ M2


Tinjauan disebelah kiri :

MA = (1/2 W L) x 1/8 L – (1/8 W L) x 1/16 L – (7/8 M1 + 1/8 M2)

= 7/128 W L2 – (7/8 M1 + 1/8 M2).

MB = (1/2 W L) x 1/4 L – (1/4 W L) x 1/8 L – (3/4 M1 + 1/4 M2)

= 3/32 W L2 – (3/4 M1 + 1/4 M2).

MC = (1/2 W L) x 3/8 L – (3/8 W L) x 3/16 L – (5/8 M1 + 3/8 M2)

= 15/128 W L2 – (5/8 M1 + 3/8 M2)


Mmaks = 1/8 W L2 – (1/2 M1 + 1/2 M2), atau

Mmaks = M1.

Tinjauan disebelah kanan :

MA = (1/2 W L) x 1/8 L – (1/8 W L) x 1/16 L – (7/8 M2 + 1/8 M1)

= 7/128 W L2 – (7/8 M2 + 1/8 M1).

MB = (1/2 W L) x 1/4 L – (1/4 W L) x 1/8 L – (3/4 M2 + 1/4 M1)

= 3/32 W L2 – (3/4 M2 + 1/4 M1).

MC = (1/2 W L) x 3/8 L – (3/8 W L) x 3/16 L – (5/8 M2 + 3/8 M1)

= 15/128 W L2 – (5/8 M2 + 3/8 M1)


Mmaks = 1/8 W L2 – (1/2 M1 + 1/2 M2), atau

Mmaks = M2.

1/8 W L2

W

1

L

M1 M2

A B

C

Mmaks

2

1/8 L 1/2 L

X

1/8 L 1/8 L 1/8 L

Mmaks


6

Kejadian khusus,

M1 = M2 = 1/12 W L2 (hanya akibat beban terbagi rata penuh bentang)

Maka,

MA = 7/128 W L2 – 1/12 W L

2 = – 11/384 W L

2

MB = 3/32 W L2 – 1/12 W L

2 = 1/96 W L

2

MC = 15/128 W L2 – 1/12 W L

2 = 13/384 W L

2

Mmaks = M1 = M2 = 1/12 W L2

)384/13(3)96/1(4)384/11(3)12/1(5,2

)12/1(5,122222

2

LWLWLWLW

LWCb

Cb 2,381

3a). Muatan terbagi rata penuh, dengan pengaku lateral (lateral bracing) sebanyak 2 (dua)


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

CBAmax

max

3435,2

5,12

MMMM

MCb

Dimana,

M1 ≠ M2



MA = (1/2 W L) x 1/12 L – (1/12 W L) x 1/24 L – (11/12 M1 + 1/12 M2)

= 11/288 W L2 – (11/12 M1 + 1/12 M2).

MB = (1/2 W L) x 2/12 L – (2/12 W L) x 1/12 L – (10/12 M1 + 2/12 M2)

= 5/72 W L2 – (10/12 M1 + 2/12 M2).

MC = (1/2 W L) x 3/12 L – (3/12 W L) x 3/24 L – (9/12 M1 + 3/12 M2)

= 3/32 W L2 – (9/12 M1 + 3/12 M2)


Mmaks = M1

1/8 W L2

W

1

L

M1 M2

A B C Mmaks

2

1/12 L

X X

1/3 L 1/3 L 1/12 L 1/12 L 1/12 L


7


MA = (1/2 W L) x 1/12 L – (1/12 W L) x 1/24 L – (11/12 M2 + 1/12 M1)

= 11/288 W L2 – (11/12 M2 + 1/12 M1).

MB = (1/2 W L) x 2/12 L – (2/12 W L) x 1/12 L – (10/12 M2 + 2/12 M1)

= 5/72 W L2 – (10/12 M2 + 2/12 M1).

MC = (1/2 W L) x 3/12 L – (3/12 W L) x 3/24 L – (9/12 M2 + 3/12 M1)

= 3/32 W L2 – (9/12 M2 + 3/12 M1)


Mmaks = M2

Kejadian khusus,


Maka,

MA = 11/288 W L2 – 1/12 W L

2 = – 13/288 W L

2

MB = 5/72 W L2 – 1/12 W L

2 = – 1/72 W L

2

MC = 3/32 W L2 – 1/12 W L

2 = 1/96 W L

2

Mmaks = M1 = M2 = 1/12 W L2

)96/1(3)72/1(4)288/13(3)12/1(5,2

)12/1(5,122222

2

LWLWLWLW

LWCb

Cb 2,419

3b). Muatan terbagi rata penuh, dengan pengaku lateral (lateral bracing) sebanyak 2 (dua)


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

CBAmax

max

3435,2

5,12

MMMM

MCb

1/8 W L2

W

1

L

M1 M2

A B C

2 X

Mmaks

X

1/3 L 1/12 L

1/3 L 1/12 L 1/12 L 1/12 L


8

Dimana,

M1 ≠ M2


MA = (1/2 W L) x 5/12 L – (5/12 W L) x 5/24 L – (7/12 M1 + 5/12 M2)

= 35/288 W L2 – (7/12 M1 + 5/12 M2).

MB = (1/2 W L) x 6/12 L – (6/12 W L) x 3/12 L – (1/2 M1 + 1/2 M2)

= 1/8 W L2 – (1/2 M1 + 1/2 M2).

MC = (1/2 W L) x 7/12 L – (7/12 W L) x 7/24 L – (5/12 M1 + 7/12 M2)

= 35/288 W L2 – (5/12 M1 + 7/12 M2)


Mmaks = MB

Kejadian khusus,


Maka,

MA = 35/288 W L2 – 1/12 W L

2 = 11/288 W L

2

MB = 1/8 W L2 – 1/12 W L

2 = 1/24 W L

2

MC = 35/288 W L2 – 1/12 W L

2 = 11/288 W L

2

Mmaks = MB = 1/24 W L2

)288/11(3)24/1(4)288/11(3)24/1(5,2

)24/1(5,122222

2

LWLWLWLW

LWCb

Cb 1,042

4a). Muatan terbagi rata penuh, dengan pengaku lateral (lateral bracing) sebanyak 3 (tiga)


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

1/8 W L2

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

X

1/4 L


9

CBAmax

max

3435,2

5,12

MMMM

MCb

Dimana,

M1 ≠ M2



MA = (1/2 W L) x 1/16 L – (1/16 W L) x 1/32 L – (15/16 M1 + 1/16 M2)

= 15/512 W L2 – (15/16 M1 + 1/16 M2).

MB = (1/2 W L) x 2/16 L – (2/16 W L) x 1/16 L – (14/16 M1 + 2/16 M2)

= 7/128 W L2 – (14/16 M1 + 2/16 M2).

MC = (1/2 W L) x 3/16 L – (3/16 W L) x 3/32 L – (13/16 M1 + 3/16 M2)

= 39/512 W L2 – (13/16 M1 + 3/16 M2)


Mmaks = M1


MA = (1/2 W L) x 1/16 L – (1/16 W L) x 1/32 L – (15/16 M2+ 1/16 M1)

= 39/512 W L2 – (15/16 M2 + 1/16 M1).

MB = (1/2 W L) x 2/16 L – (2/16 W L) x 1/16 L – (14/16 M2 + 2/16 M1)

= 7/128 W L2 – (14/16 M2 + 2/16 M1).

MC = (1/2 W L) x 3/16 L – (3/16 W L) x 3/32 L – (13/16 M2 + 3/16 M1)

= 39/512 W L2 – (13/16 M2 + 3/16 M1)


Mmaks = M2

Kejadian khusus,


Maka,

MA = 39/512 W L2 – 1/12 W L

2 = – 3/419 W L

2

MB = 7/128 W L2 – 1/12 W L

2 = – 11/384 W L

2

MC = 39/512 W L2 – 1/12 W L

2 = – 3/419 W L

2

Mmaks = M1 = 1/12 W L2

)419/3(3)384/11(4)419/3(3)12/1(5,2

)12/1(5,122222

2

LWLWLWLW

LWCb

Cb 2,847


10

4b). Muatan terbagi rata penuh, dengan pengaku lateral (lateral bracing) sebanyak 3 (tiga)


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

Dimana,

M1 ≠ M2



MA = (1/2 W L) x 5/16 L – (5/16 W L) x 5/32 L – (11/16 M1 + 5/16 M2)

= 55/512 W L2 – (11/16 M1 + 5/16 M2).

MB = (1/2 W L) x 6/16 L – (6/16 W L) x 3/16 L – (10/16 M1 + 6/16 M2)

= 15/128 W L2 – (10/16 M1 + 6/16 M2).

MC = (1/2 W L) x 7/16 L – (7/16 W L) x 7/32 L – (9/16 M1 + 7/16 M2)

= 63/512 W L2 – (9/16 M1 + 7/16 M2).


Mmaks = 1/8 W L2 – (1/2 M1 + 1/2 M2)


MA = (1/2 W L) x 5/16 L – (5/16 W L) x 5/32 L – (11/16 M2 + 5/16 M1)

= 55/512 W L2 – (11/16 M2 + 5/16 M1).

MB = (1/2 W L) x 6/16 L – (6/16 W L) x 3/16 L – (10/16 M2 + 6/16 M1)

= 15/128 W L2 – (10/16 M2 + 6/16 M1).

MC = (1/2 W L) x 7/16 L – (7/16 W L) x 7/32 L – (9/16 M2 + 7/16 M1)

= 63/512 W L2 – (9/16 M2 + 7/16 M1).

Kejadian khusus,


Maka,

MA = 55/512 W L2 – 1/12 W L

2 = 2/83 W L

2

MB = 15/128 W L2 – 1/12 W L

2 = 13/384 W L

2

MC = 63/512 W L2 – 1/12 W L

2 = 11/277 W L

2

Mmaks = M1 = 1/12 W L2

1/8 W L2

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

X

1/4 L


11

)277/11(3)384/13(4)83/2(3)12/1(5,2

)12/1(5,122222

2

LWLWLWLW

LWCb

Cb 1,946

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

ditengah-tengah bentang.

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

CBAmax

max

3435,2

5,12

MMMM

MCb

Dimana,

M1 ≠ M2

R1 = R2 = 3/2 P (hanya akibat muatan terpusat P).


MA = (3/2 P) x 1/16 L – (15/16 M1 + 1/16 M2)

= 3/32 P L – (15/16 M1 + 1/16 M2).

MB = (3/2 P) x 2/16 L – (14/16 M1 + 2/16 M2)

= 6/32 P L – (14/16 M1 + 2/16 M2).

MC = (3/2 P) x 3/16 L – (13/16 M1 + 3/16 M2)

= 9/32 P L – (13/16 M1 + 3/16 M2)


Mmaks = M1, atau

Mmaks = (3/2 P) x 4/16 L – (12/16 M1 + 4/16 M2)


MA = (3/2 P) x 1/16 L – (15/16 M2 + 1/16 M1)

= 3/32 P L – (15/16 M2 + 1/16 M1).

MB = (3/2 P) x 2/16 L – (14/16 M2 + 2/16 M1)

= 6/32 P L – (14/16 M2 + 2/16 M1).

MC = (3/2 P) x 3/16 L – (13/16 M2 + 3/16 M1)

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

P

P

3/16 P L


12

= 9/32 P L – (13/16 M2 + 3/16 M1)


Mmaks = M2, atau

Mmaks = (3/2 P) x 4/16 L – (12/16 M2 + 4/16 M1)

Kejadian khusus,

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

Maka,

MA = 3/32 P L – 5/16 P L = – 7/32 P L

MB = 6/32 P L – 5/16 P L = – 4/32 P L

MC = 9/32 P L – 5/16 P L = – 1/32 P L

Mmaks = M1 = 5/16 P L

)32/1(3)32/4(4)32/7(3)16/5(5,2

)16/5(5,12

LPLPLPLP

LPCb

Cb 1,923

4d). Muatan terpusat, dengan pengaku lateral (lateral bracing) sebanyak 3 (tiga) buah

ditengah-tengah bentang.

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

CBAmax

max

3435,2

5,12

MMMM

MCb

Dimana,

M1 ≠ M2

R1 = R2 = 3/2 P (hanya akibat muatan terpusat P).


MA = (3/2 P) x 5/16 L – (11/16 M1 + 5/16 M2)

= 15/32 P L – (11/16 M1 + 5/16 M2).

MB = (3/2 P) x 6/16 L – (10/16 M1 + 6/16 M2)

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

P

P

3/16 P L


13

= 18/32 P L – (10/16 M1 + 6/16 M2).

MC = (3/2 P) x 7/16 L – (9/16 M1 + 7/16 M2)

= 21/32 P L – (19/16 M1 + 7/16 M2)


Mmaks = 3/16 P L


MA = (3/2 P) x 5/16 L – (11/16 M2 + 5/16 M1)

= 15/32 P L – (11/16 M2 + 5/16 M1).

MB = (3/2 P) x 6/16 L – (10/16 M2 + 6/16 M1)

= 18/32 P L – (10/16 M2 + 6/16 M1).

MC = (3/2 P) x 7/16 L – (9/16 M2 + 7/16 M1)

= 21/32 P L – (19/16 M2 + 7/16 M1)


Mmaks = 3/16 P L

Kejadian khusus,

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

Maka,

MA = 15/32 P L – 5/16 P L = 5/32 P L

MB = 18/32 P L – 5/16 P L = 8/32 P L

MC = 21/32 P L – 5/16 P L = 11/32 P L


Mmaks = 3/16 P L

)32/11(3)32/8(4)32/5(3)16/3(5,2

)16/3(5,12

LPLPLPLP

LPCb

Cb 0,789


14

Kolom Jepit-jepit dan Jepit-Sendi

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

Gambar : Kolom memikul momen distribusi linear, tanpa pengaku lateral,

(a) perletakan jepit, (b) perletakan jepit sendi.

CBAmax

max

3435,2

5,12

MMMM

MCb

Gambar (a) :

Dengan perbandingan garis pada segitiga, diperoleh

hb

M2

ha

M1 , atau

hb

ha

M2

M1

Bila,

M1/M2 = 1 ,

Maka,

M1 = M2 = M, dan

ha = hb = 1/2H

MA = 1/2 M

MB = 0

MC = 1/2 M

Momen Mmaks = M

)2/1(3)0(4)2/1(3)(5,2

)(5,12

MMM

MCb

Cb 2,273

1 M1

A

B

C

2 M2

H

ha

hb

(a)

1 M

A

B

C

2

H

(b)

1/4H

1/4H

1/4H

1/4H

1/4H

1/4H

1/4H

1/4H


15

Bila,

M1/M2 = 0,75 = 3/4 ,

Maka,

M1 = 0,75 M2

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

MB = 0,5/4 x (M2) = 0,1250 M2

MC = 2,25/4 x (M2) = 0,5625 M2

Momen Mmaks = M2

)25625,0(3)21250,0(4)23125,0(3)2(5,2

)2(5,12

MMMM

MCb

Cb 2,222

Selanjutnya lihat tabel berikut,

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

Gambar (b) :

MA = 3/4 M

MB = 1/2 M

MC = 1/4 M

Momen Mmaks = M

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

)(5,12

MMMM

MCb

Cb 1,667

0,75 M2

MA 3

4

MB

MC

M2

perhitungan nilai cb - thamrin nasution · perhitungan nilai cb untuk balok dan kolom lentur, 2015...

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