Perhitungan Nilai Cb Cb Value Calculation untuk Balok dan Kolom Lentur

7/23/2019 Perhitungan Nilai Cb Cb Value Calculation untuk Balok dan Kolom Lentur

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PerhitunganNilai Cb Balok dan Kolom Terlentur

IR. THAMRIN NST.

(Semoga bermanfaat )

2015

thamrinnst.wordpress.com



Perhitungan Nilai C b 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 C b for Simply Supported Beams

LoadLateral Bracing

Along SpanC b

NoneLoad at midpoint

At load point

NoneLoads at third points

At load pointsLoads symmetrically

placed

NoneLoads at quater points

At load pointsLoads at quater points

None

At midpoint

At third points

At quaterpoints

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

X X

1,14

X X

1,30 1,30

X

X X

1,45 1,45X

1,01X

X X

1,52 1,52

X

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

M M M M

M Cb

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

= 11/288 W L2.

M B = (1/2 W L) x 2/12 L – (2/12 W L) x 1/12 L

= 5/72 W L2.

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

= 3/32 W L2.

M maks = (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

LW LW LW LW

LW Cb

Cb 1,460

1/8 W L

W

1

L

A

B CMmaks

2

1/12 L

X X

1/3 L1/3 L1/12 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

M M M M

M Cb

Dimana,

R1 = R2 = 1/2 W L

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

M B = 1/8 W L2.

M C = M A = 35/288 W L2.

M maks = 1/8 W L2.

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

)8/1(5,122222

2

LW LW LW LW

LW Cb

Cb 1,014

1/8 W L2

W

1

L

A B C

Mmaks

2

1/12 L

X X

1/3 L1/3 L1/12 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

M M M M

M Cb

Dimana,

M1 ≠ M2 R1 = R2 = 1/2 W L (hanya akibat muatan terbagi rata).

M A = (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).

M B = 1/8 W L2 – (1/2 M1 + 1/2 M2)

M C = 3/32 W L2 – (1/4 M1 + 3/4 M2)

Momen M maks ada 3 kemungkinan, yaitu :

M maks = M B = 1/8 W L2 – (1/2 M1 + 1/2 M2), atau

M maks = M1, atau

M maks = M2

Kejadian khusus,M1 = M2 = 1/12 W L2

Maka,

M A = M C = 3/32 W L2 – 1/12 W L2 = 1/96 W L2

M B = 1/8 W L2 – 1/12 W L2 = 1/24 W L2

M maks = 1/12 W L2

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

)12/1(5,122222

2

LW LW LW LW

LW Cb

Cb 2,381

1/8 W L

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

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 1/8 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).

M B = (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).

M C = (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)


M maks = 1/8 W L2 – (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)

= 7/128 W L2 – (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)

= 3/32 W L2 – (3/4 M2 + 1/4 M1).

M C = (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)


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

M maks = M2.

1/8 W L

W

1

L

M1 M2

AB

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,

M A = 7/128 W L2 – 1/12 W L2 = – 11/384 W L2

M B = 3/32 W L2 – 1/12 W L2 = 1/96 W L2

M C = 15/128 W L2

– 1/12 W L2

= 13/384 W L2

M maks = M1 = M2 = 1/12 W L2

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

)12/1(5,122222

2

LW LW LW LW

LW Cb

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

M M M M

M Cb

Dimana, M1 ≠ M2



M A = (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).

M B = (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).

M C = (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)

Momen M maks

ada 1 kemungkinan, yaitu :

M maks = M1

1/8 W L

W

1

L

M1 M2

A B

CMmaks

2

1/12 L

X X

1/3 L1/3 L1/12 L 1/12 L 1/12 L




7


M A = (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).

M B = (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).

M C = (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)


M maks = M2

Kejadian khusus,


Maka,

M A = 11/288 W L2 – 1/12 W L2 = – 13/288 W L2

M B = 5/72 W L2 – 1/12 W L2 = – 1/72 W L2

M C = 3/32 W L2 – 1/12 W L2 = 1/96 W L2

M maks = M1 = M2 = 1/12 W L2

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

)12/1(5,122222

2

LW LW LW LW

LW Cb

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

M M M M

M Cb

1/8 W L

W

1

L

M1 M2

A B C

2X

Mmaks

X

1/3 L1/12 L

1/3 L1/12 L1/12 L1/12 L




8

Dimana,

M1 ≠ M2


M A = (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).

M B = (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).

M C = (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)


M maks = M B

Kejadian khusus,


Maka,

M A = 35/288 W L2 – 1/12 W L2 = 11/288 W L2

M B = 1/8 W L2 – 1/12 W L2 = 1/24 W L2 M C = 35/288 W L2 – 1/12 W L2 = 11/288 W L2

M maks = M B = 1/24 W L2

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

)24/1(5,122222

2

LW LW LW LW

LW Cb

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 L

W

1

L

M1 M2

A B C

2X

Mmaks

X

1/4 L1/16 L

1/4 L1/16 L

1/16 L

1/16 L

X

1/4 L




9

CBAmax

max

3435,2

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

= 15/512 W L2 – (15/16 M1 + 1/16 M2).

M B = (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).

M C = (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)


M maks = M1

Tinjauan disebelah kanan : M A = (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).

M B = (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).

M C = (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)


M maks = M2

Kejadian khusus,

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

M A = 39/512 W L2 – 1/12 W L2 = – 3/419 W L2

M B = 7/128 W L2 – 1/12 W L2 = – 11/384 W L2

M C = 39/512 W L2 – 1/12 W L2 = – 3/419 W L2

M maks = M1 = 1/12 W L2

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

)12/1(5,122222

2

LW LW LW LW

LW Cb

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



M A = (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).

M B = (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).

M C = (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).


M maks = 1/8 W L2 – (1/2 M1 + 1/2 M2)


M A = (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).

M B = (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). M C = (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,

M A = 55/512 W L2 – 1/12 W L2 = 2/83 W L2

M B = 15/128 W L2 – 1/12 W L2 = 13/384 W L2

M C = 63/512 W L2 – 1/12 W L2 = 11/277 W L2

M maks = M1 = 1/12 W L2

1/8 W L

W

1

L

M1 M2

A B C

2X

Mmaks

X

1/4 L

1/16 L

1/4 L1/16 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

LW LW LW LW

LW Cb

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

M M M M

M Cb

Dimana,

M1 ≠ M2

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


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 C = (3/2 P) x 3/16 L – (13/16 M1 + 3/16 M2)= 9/32 P L – (13/16 M1 + 3/16 M2)


M maks = M1, atau

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


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)

1/2 P L

P

1

L

M1 M2

A BC

2X

Mmaks

X

1/4 L1/16 L

1/4 L1/16 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)


M maks = M2, atau

M maks = (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,

M A = 3/32 P L – 5/16 P L = – 7/32 P L

M B = 6/32 P L – 5/16 P L = – 4/32 P L

M C = 9/32 P L – 5/16 P L = – 1/32 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

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

M M M M

M Cb

Dimana,

M1 ≠ M2

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


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

= 15/32 P L – (11/16 M1 + 5/16 M2).

M B = (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

2X

Mmaks

X

1/4 L1/16 L

1/4 L1/16 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).

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

= 21/32 P L – (19/16 M1 + 7/16 M2)


M maks = 3/16 P L


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

= 21/32 P L – (19/16 M2 + 7/16 M1)


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 C = 21/32 P L – 5/16 P L = 11/32 P L


M maks = 3/16 P L

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

)16/3(5,12

L P L P L P L P

L P Cb

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

M M M M

M Cb

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

M A = 1/2 M

M B = 0

M C = 1/2 M

Momen M maks = M

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

)(5,12

M M M

M Cb

Cb 2,273

1 M1

A

B

C

2M2

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

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

M B = 0,5/4 x (M2) = 0,1250 M2

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

Momen M maks = M2

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

)2(5,12

M M M M

M Cb

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 C = 1/4 M

Momen M maks = M

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

)(5,12

M M M M

M Cb

Cb 1,667

0,75 M2

M A 3

4

M B

M C

M2

Perhitungan Nilai Cb Cb Value Calculation untuk Balok dan Kolom Lentur

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Transcript of Perhitungan Nilai Cb Cb Value Calculation untuk Balok dan Kolom Lentur