1.BOOK PERHITUNGAN STRUKTUR BAJA H BEAM ATAU WF.pdf

1

Book1.xlsxF:\arsitektur test\8.STRUKTUR BAJA 1\Book1.xlsx

3

PEMBANGUNAN GEDUNG KANTOR 9 LANTAI 

Project Client

Engineer Name

 

Checked

Approved

GUSTI

Date

PERENCANAAN DIMENSI STRUKTUR UTAMA

1

TYPE FRAME

=

BALOK]

2

NO NOMOR FRAME

=

23

3

DIPEROLE DIPEROLEH H DARI SAP 2000

4 4A

4B

5

L

=

8

m

NOTE :APABILA MOMEN TERDAPAT MX DAN MY MAKA KEDUANYA DIPERHITUNGKAN..

MU TX

=

-3680

kgm

NU

=

-2678

kg

MUTY

=

0

kg m

VU

=

-1184

kg

Ma

=

-3679

kg m

φc

=

0.85

Mb

=

-466

kg m

φb

=

0.90

Ms

=

302

kgm

PROFIL PROFIL BAJA YANG DIGUNAKA DIGUNAKAN N TYPE

=

H BEAM

UK.

=

200

X

8

X

12

mm

A

=

63.53

cm2

tf

=

12.00

mm

2 00

X

Zx

=

513.15

c m3

W

=

49.87

kg/m

Ix

=

4720.00

cm4

Zy

=

216.32

c m3

a

=

200.00

mm

Iy

=

1600.00

cm4

h

=

150.00

mm

bf

=

2 00.00

mm

tw

=

8.00

mm

Sx

=

472.00

cm3

iy

=

5.02

cm cm

ix

=

8.62

cm

fu

=

3700.00

kg/cm2

fr

=

fy

=

24 2400.00

kg/cm2

F IJIN

=

2.22

cm

 f 

=

0.44

cm

 f 

<

F IJIN

0.44

<

2.22

M MU UTU T U BAJA BAJA

KONTROL LENDUTAN F IJIN

=

L

G

=

fl

=

700.00

kg/cm2 PENAMPANG ROL

1500.00

kg/cm2 PENAMPANG WELD

815772.97 kg/cm2 MODULUS GESER BAJA 17 1700.00

kg/cm2

360

 f 

=

2

5L

48EI 

(Ms - 0,1 (Ma - Mb)

OK 6 6A

KONTRO KONTROLL TEKUK TEKUK ARAH ARAH X KC X

=

0.80

L

=

800.00

JEPIT SENDI cm

LKX

=

640.00

cm cm

ʎx

=

LKX

ʎx

=

ix

Ncrbx 

=

2

π EA ʎx

74.25

cm

MENENTUKAN

2

Ncrbx

=

227477.03

kg

2

Book1.xlsxF:\arsitektur test\8.STRUKTUR BAJA 1\Book1.xlsx

3

PEMBANGUNAN GEDUNG KANTOR 9 LANTAI 

Project Client

Engineer Name

 

Checked

Approved

GUSTI

Date

6B

ARAH ARAH y KC Y

=

0.80

L

=

800.00

JEPIT SENDI cm

LKY

=

640.00

cm cm

ʎy

=

LKX

ʎy

=

iy NcrbY 

=

127.49

cm

MENENTUKAN

2

NcrbY

=

77149.03

fy

ʎc

=

1.41

E

1.41

>

1.20

π EA

kg

2

ʎy DIPERGUNAKAN

=

ʎc

=

127.49

cm

ʎ yang terpilih

π

OK 2

ω

=

1.25ʎc

Pn

=

Ag

x

fy

ω

=

2.47

Pn

=

61719.22

=

0.05

<

0.20

kg

ω

=

Pu 0.05

φc Pn

OK 7 7A

BATANG DIANGGAP TIDAK BERGOYANG (X) SbX

Cmx

=

Nu

1-

7B

Ncrbx 

Cmx

=

1

MUX

=

MU TX

SbY

=

8

x

CmY 1-

8B

> 1

CmY

=

1

MUY

=

MU TY

Nu

SbX

= <>

0.99 1.00

SbX

=

1.00

MUX

=

-3679.74 0.97

SbY

=

0.97

<>

1.00

SbY

=

1.00

SbY

MUY

=

0.00

170

8.33

<

10.97

> 1

NcrbY 

x

SbX 0.99

kgm

kgm

MENEN MENENTU TUKAN KAN MNX MNX KONTROL PENAMPANG PENAMPANG PROFIL SAYAP

=

b 2tf 

BADAN

=

h t 

<

<

DICOBA LATERAL BRACING

=

800.00

cm

Lp

=

1.76

x

fy

KOMPAK

1680

18.75

fy

iy

<

108.44

KOMPAK

E

Lp

=

255.05

fy

Lp

>

lb

255.05

>

800.00

MNX

><

MPX

cm

3

Book1.xlsxF:\arsitektur test\8.STRUKTUR BAJA 1\Book1.xlsx

Project

3

PEMBANGUNAN GEDUNG KANTOR 9 LANTAI 

Client Engineer Name

Checked

Approved

GUSTI

Date

9

CHECK KONDISI BATAS LATERAL LATERAL KONDISIONAL KONDISIONAL BUCKLING

Lr

=

iy

x

x1

x

1

x

t

+

1

+

x2 x2

2

x

fl

J

=

6.83

c m4

lw

=

90000.00

c m6

x1

=

123915.72

kg/cm2

x2

=

1.616E-06

kg/cm2

Lr

=

1634.10

cm

lb

<

Lr

800.00

<

1634.10

KONTROL

=

2.00

FL J

=

2

x

bf

3

3

lw

=

ly 2

x1

=

2

h

x

2

π

E

G

Sx

x2

=

4.00

J

A

2

Sx

2

x

GJ

lw ly

Lp

<

255.05

<

BENTANG MENENGAH

Cb 9A

BENTANG PENDEK MN

9B

LIHAT DI TABEL CB

=

MN

=

MP

=

Zx

x

fy

=

Cb

x

Mr

Mr

=

Sx

fy

-

Mp

=

Zx

x

My

=

Sx

x

MNY

11

1.14

MN

=

 

12315.65

kgm

802400.00

kgcm

BENTANG BENTANG MENENGA MENENGAH H MNX

10

=

=

0.25

x

+

Mp

-

Mr

-

lb

lr

-

lp

Mr

=

fy

Mp

=

 

1231564.80   kgcm

fy

My

=

 

1132800.00   kgcm

MNX

=

 

849182.03

  kgcm

MNY

=

 

288000.00

  kgcm

tf

fr

lr

x

2

bf 

fy

LIMIT STATE OF COMPRESSION COMPRESSION FLANGE LOCAL BUCKLING . SAYAP

=

KOMPAK

BADAN

=

KOMPAK

PERSAMAA PERSAMAAN N INTERAKS INTERAKSII =

PU 2φc Pn

+

MuX φb MnX

+

Muy φb MnY

<

1

 

0.51

=

 

0.51

<

 

1.00

OK