Tubes Baja

Anota Hamsyah M. Khalis K

(201510340311120) (201510340311133 (201510340311133 )

BAB II PERHITUNGAN DAN PERENCANAAN

Direncanakan Bangunan Gedung Industri berbahan baja serta gambar kerjanya dengan data-data sebagai berikut : 1. Bentang Kuda-kuda (L)

: 20 m

2. Profil Kuda-kuda

: Siku Ganda

3. Jumlah Kuda-kuda (n)

:5

4. Jarak antar Kuda-kuda (l)

:4m

5. H1

:5m

6. H2

: 0,5 m

7. H3

: 0,5 m

8. Sudut α

: 25º

9. Beban Angin

: 40 kg/cm2

10. Jenis Atap

: Zincalum

11. Dinding Samping

: Terbuka

12. Ikatan Angin Dinding

: Portal Bracing

13. Mutu Baja

: A36

14. Jenis Sambungan

: Baut (A325) 1


Tinggi Atap

(201510340311120) (201510340311133 (201510340311133 )

 

= tan α. .b

 

= tan 25o. . 20 = 4,66 m Panjang sisi miring

= =

√  +   1010 +4,66

= 11,03 m

Jarak Antar Gording

= Sisi miring / jumlah gording = 11,03/ 9 = 1.22 m = 1.25 m

2


(201510340311120) (201510340311133 (201510340311133 )

Cos 250 = 0.906 Sin 250 = 0.423

Perhitungan Beban 1. Beban Mati yang Bekerja Digunakan penutup atap ( Zincalume R 700 700 ) tebal 0,5 mm Produksi PT. Gunung Raja Paksi

Berat sendiri profil atap per m2

= 4,10 kg/m2

Berat Gording

= 6,37 kg/m

Jarak antar Gording

= 1,25 m

Berat atap ( 1,25 x 4,10 kg/m2 )

= 5.13 kg/m

Berat gording

= 6.37 kg/m

qD

= 11.5 kg/m

3


a.

Beban Mati Tegak Lurus Bidang Atap

q = 11.5 kg/m

R Dy Dy R Dy Dy MDx MDx MDy

R Dx Dx R Dx Dx MDy MDy MDx MDx

∝

 ∝    06) ( 4 ) = 20.84 kg = 11.5  (0.906)    = q Cos ∝ (L)    06) (4 m) = 20,84 kg m = 11.5  (0.906)   = q Cos L

2

2

=0

b. Beban Mati Sejajar Bidang Atap

q = 11.5 kg/m

: qD = q Cos

(201510340311120) (201510340311133 (201510340311133 )

: qD = q Sin

∝

 ∝ L    = 11.5  (0,423) ( 2 ) = 4,86 kg    = q Sin ∝ (L)    23) (2 m) = 2,43 kg m = 11.5  (0.423)    = q Cos ∝ (L)    06) (2 m) = 5,21 kg m = 11.5  (0.906)   = q Sin

2

2

2

2

4


2. Beban Hidup a.

Beban Hidup Tegak Lurus Bidang Atap : PL = P Cos

P = 100 kg

R Ly R Ly MLx MLx MLy

b.

R Lx R Lx MLy MLy MLx MLx

∝

 ∝   = (100 kg)(0.906)   = P Cos ∝ (L)   = (100 kg)(0.906) (4 m)  = P Cos

=0

Beban Hidup Sejajar Bidang Atap

P = 100 kg

(201510340311120) (201510340311133 )

: PL = P Sin

 ∝   = (100 kg)(0.432)   = P Sin ∝ (L)   = (100 kg)(0.432) (2 m)   = P Cos ∝ (L)   = (100 kg)(0.906) (2 m) 

= 45,30 kg

= 90,60 kg m

∝

= P Sin

= 21,60 kg

= 21,60 kg m

= 55,19 kg m

5


(201510340311120) (201510340311133 )

3. Beban Angin P = 40 kg/m2 G = 0,85 ( Didapat dari pasal 26.9.1 SNI-1727-2013)

Dari hasil interpolasi di dapat : Arah angin, ᵞ = 0O, 180O Sudut atap

Kasus beban

25O

C NW

= -0,87

C NL

= -0,87

Aliran angin terhalang C NW

C NL

A

-0,87

-0,87

B

-0,40

-1,3

Tekanan Angin Datang

6


q

(201510340311120) (201510340311133 )

= P x G x C N = 40 x 0,85 x (-0,87) = -29,58 kg/m2 x 1,25 = - 36,98

a. Beban Mati Tegak Lurus Bidang Atap

q = -36,98 kg/m

R wy Mwx Mwx Mwy

=0

    = −36,98  (4 m)   = q (L)2

2

= -73,96 kg m

=0

b. Beban Mati Sejajar Bidang Atap

q = -36,98 kg/m

R wx R wx Mwy Mwx Mwx

    = −36,98 (2 ) = -36,98 kg   = q L

=

0

    = −36,98  (2 m) = - 18,49 kg m   = q (L)2

2

Tekanan Angin Pergi 7


q

(201510340311120) (201510340311133 )

= P x G x C N = 40 x 0,85 x (-0,87) = -29,58 kg/m2 x 1,25 = - 36,98

a. Beban Mati Tegak Lurus Bidang Atap

q = -36,98 kg/m

R wy Mwx Mwx Mwy

=0

    = −36,98  (4 m) = -73,96 kg m   = q (L)2

2

=0

c. Beban Mati Sejajar Bidang Atap

q = -36,98 kg/m

R wx R wx Mwy Mwx Mwx

    = −36,98 (2 ) = -36,98 kg   = q L

=

0

    = −36,98  (2 m) = - 18,49 kg m   = q (L)2

2

8


(201510340311120) (201510340311133 )

Kombinasi Pembebanan Tabel Hasil Perhitungan Kombinasi Reaksi

Keterangan

Beban

Beban Tetap

Beban Sementara 1

Beban Sementara 2

Beban Sementara 3

Angin Datang

Angin Datang

Angin Datang

Angin Pergi

Angin Pergi

Angin Pergi

Reaksi

L= 4 m

RDy

20,84

RLy

45,30 0

(1,2.Dx + 1,6.Lx)

97,49 kg

(1,2.Dx + 0,5.Lx + 1,0.w)

47,66 kg

47,66 kg

(1,2.Dx + 1,6.Lx + 0,5.w)

97,49 kg

97,49 kg

(0,9.Dx + 1,0.w)

18,76 kg

18,76 kg

RWy 0

L= 2 m

RDx

4,86

RLx

21,60 -36,98

RWx

(1,2.Dy + 1,6.Ly)

(1,2.Dy + 0,5.Ly)

(1,2.Dy + 1,6.Ly)

(0,9.Dy)

40,39 kg

16,63 kg

40,39 kg

19,44 kg

-36,98

9


(201510340311120) (201510340311133 )

Tabel Hasil Perhitungan Kombinasi Momen

Keterangan

Beban Tetap

Beban

Beban Sementara 1 Angin Datang

Angin Pergi


Angin Pergi


Angin Pergi

Momen

L=4m

MDx

20,84

MLx

90,6

MWx

-73,96

5,21

MLx

55,19

L=2m

-18,49

70,308

70,308

(1,2.Dx + 1,6.Lx + 0,5.w)

169,97

169,97

(0,9.Dx + 1,0.w)

18,756

18,756

(1,2.Dy + 1,6.Ly) 94,56

(1,2.Dy + 0,5.Ly)

33,847

33,847

(1,2.Dy + 1,6.Ly)

94,56

94,56

(0,9.Dy)

4,689

4,689

-18,49

MDy

2,43

MLy

21,6

MWy

169,97

(1,2.Dx + 0,5.Lx + 1,0.w)

-73,96

MDx

MWx

(1,2.Dx + 1,6.Lx)

0

(1,2.Dy + 1,6.Ly) 37,48

(1,2.Dy + 0,5.Ly)

13,716

13,716

(1,2.Dy + 1,6.Ly)

37,48

37,48

(0,9.Dy)

2,187

2,187

0

10


(201510340311120) (201510340311133)

 Perencanaan Gording Atap

Diambil profil C-Channel 100x50x20 dengan tebal 3,0 mm yang diproduksi oleh PT. Gunung Raja Paksi.

q

=

5,19 kg/m

t

=

3,00 mm

Cy

=

1.86 cm

Xo

=

4,41 cm

A

=

6,16 cm 2

Ix=101 cm4

J

=

1982 cm4

Iy=23 cm4

Cw

=

603 cm6

=

20,2 cm3

Zy

=

7,4 cm3

r x

=

3,91 cm

r y

=

1,88 cm

Zx

Mutu Baja = A36 = 36 x 1ksi = 36 x 6,875 = 247,5 MPa = 2475 kg/cm 2 Momen Nominal Penampang C-Channel

  ℎ   ℎ     ℎ  =

2

+

( – ) + ( − 2 ) (

− )

 

= . 10 .0,3 2 + 2 x 0,3 (10 - 2) + 0,3 (5 – 2 . 0,3) (10 – 0,3) = 17,829 cm 3

11


(201510340311120) (201510340311133)

 ℎ                =

( –

)+2

( − –

) + t( − )2 + ( − − )2

 

= 10 . 0,3 (1,86 - 0,3) + 2 . 2 . 0,3 (5 – 1,860,3(5 – 0,3 – 1,86) 2

 0,3) + 0,3(1,86 – 0,3) +  2

= 11.868 cm 3 Mnx = Zx f y = 17,829 x 2475 kg/cm 2 = 44126,78 kg.cm Mny = Zyf y = 11,868 x 2475 kg/cm 2 = 29372,71 kg.cm Persyaratan Momen Biaxial L= 4m Mux

Ø x Mnx

< 1,0

.   = 0,43 < 1,0…(OK) .  , L= 2m Mux

Ø x Mnx

+

Muy

Ø x Mny

< 1,0

.   + .   = 0,38< 1,0…(OK) .  , .  , Lendutan Pada Profil C - Channel ∆max

= =

qD

PL

L

240 400 240

=

1.67 cm

=

11.5 kg/m

=

100 kg

=>

=>

qDx

=

(11.5 kg/m).0,423

= 4.86 kg/m

qDy

=

(11.5).0,906

= 10.42 kg/m

PDx =

(100).0,423

PDy =

(100).0,906

= 42,3 kg = 90,6 kg

12


(201510340311120) (201510340311133)

Lendutan terhadap sumbu x (L=4m) : qx

=

4,86 kg/m

Px

=

42,3 kg

∆x

=

= =

4

5

q x . L

.

384

E . I x

+

1

3

P x . L

. 48 E . I x

(0,0486).(4004 )

5

(42,3).(4003 )

1

 . . 384 (2000000).(101) 48 (2000000).(101)

0,36 cm

∆x = 0,36 cm < ∆max = 1,67 cm ...(OK) Lendutan terhadap sumbu x dan y (L= 2m): qy

=

10,42 kg/m

q x = 4,86 kg/m

Py

=

90,6 kg

P x = 42,3 kg

∆y

=

1

4

5

q y . L

.

384

E . I y

+

3

.

48

P y . L

E . I y

(0,1042).(2004 )

5

=

. 384 (2000000).(23)

=

0,38 cm



1

(90,6).(2003 )

. 48 (2000000).(23)

∆y = 0,38 cm < ∆max = 1,67 cm ...(OK) ∆x

=

5

.

384

5

1

4

q x . L

E . I y

+

48

3

.

Px. L

E . I y

(0,0486).(200 4 )

=

. 384 (2000000).(101)

=

0,04 cm



1

(42,3).(2003 )

. 48 (2000000).(101)

∆y = 0,04 cm < ∆max = 1,67 cm ...(OK)

∆x +∆y √ 0.04 +0.38

∆ =

=

= 0.3821 cm < ∆max = 1,67 cm ...(OK)

Jadi gording dengan profil C 100x50x20, dengan tebal 3,0 mm dapat digunakan karena telah memenuhi persyaratan.

13


(201510340311120) (201510340311133)

 Perencanaan Sagrod Atap

Digunakan profil rod, fy = 240 Mpa Dari Profil c-channel 100 x 50 x 20 dengan t = 3 mm r x = 3,91 cm r y = 1,88 cm

Cek K elangsingan gording 

= =

lk r y 400 1.88

= 212,78 < 240 (Ok)

Gaya yang bekerja pada Sagrod

q.Sin250.L/2

P. Sin 250

a. B eban Mati = 11,5 kg/m x sin25 x2 m = 9,72 kg b. Beban Hidup = 100 kg x sin25 = 42,26 kg

c. Beban Angin Ry total

=0

= 1,2 qD+ 1,6 P L x = 1,2 x 9,72 + 1,6 x 42,26 = 79.28 kg

Pu

= 9.R y = 9 x 79.28 kg = 713.52 kg

14


φPn



φ x A x fy

Pu



φ x A x fy

A



A



d2



d

Pu φ x fy

1





(201510340311120) (201510340311133)

Pu

71.52 0,9 x 2400



0,33 cm 2

πd2

4 4xA π



4 x 0,33 3,14



0,41 cm 2

0,41  0,64 cm  0,8 cm



Digunakan sagrod dengan diameter 8 mm : 1

Imin =

I A

=

 .d

4

.

64 1

=

 .d

2

1

2

.d

16

=

1

.d

4

=

1 4

.0,8 = 0,2

4

Cek Kelangsingan Sagrod lk i min

< 300 =

125 0,2

= 625 > 300 (Tidak Aman)

Digunakan sagrod dengan diameter 18 mm : 1

Imin =

I A

=

 .d

4

64 1

.

=

 .d

2

1 16

2

.d

=

1

.d

4

=

1 4

.1,8 = 0,45

4

Cek Kelangsingan Sagrod lk i min

< 300 =

125 0, 45

= 277,78 < 300 (OK)

Maka sagrod dengan Ø18 dapat digunakan karena memenuhi persyaratan.

15


(201510340311120) (201510340311133)

Perencanaan Ikatan Angin Atap

Luas bidang angin, A = 2 . 0,5 . 11,03 = 11,03 m 2 Ditaksir iktan angin atap = 11,03 m 2

C NW

= -0,87

qangin = 40 kg/m2

C NL

= -0,87

A

G

= 0,85 ( Didapat dari pasal 26.9.1 SNI-1727-2013)

Tekanan Angin, q

= qh x G x C N = 40 x 0,85 x (0,87) = 29,58 kg/m 2

P

=Axq = 11,03 x 29,58 = 326,28 Kg

Tiap simpul menerima beban sebesar P

   = . 326,28 Kg 

= . P

= 54,38 Kg

16


Tan α =

(201510340311120) (201510340311133)

 = 1.25 

α = 51.3O Panjang bracing

cos α = 4r 4 = 6.40  r = cos 51. 3°

A >

Menghitung Reaksi ΣMA (54,38 kg) (5 m) R B R A

=0 = R B . (4 m) = 67,98 kg = -67,98 kg

ΣH P + S . cos (51.3o) S S S

=0 =0 = P / cos(51.3o) = 54,38 kg / cos 51,3 = 86,97

   

=

, = ,  

0,04 cm2

Menghitung Dimensi Rod A 0,04 cm2

  

    =      , = 0,34 cm =  =

=

2

√0,051

= 0,23 cm ( maka digunakan Rod dengan Ø 12

mm)

17


(201510340311120) (201510340311133)

Perencanaan Ikatan Angin Dinding

Beban angin samping dianggap tidak ada karena tidak terdapat dinding disisi-sisi bangunan sehinggu Pu = 0. Maka digunakan Ikatan Angin dengan Rod Ø 18 mm.

18


(201510340311120) (201510340311133)

PERHITUNGAN BEBAN PADA KUDA-KUDA  Beban mati ˗

Beban Rangka Kuda – Kuda Profil yang digunakan Double Siku 40x40x3

-

Berat sambungan 5% kuda-kuda

= 14,47

kg

Berat total kuda-kuda

= 289,45

kg

Jumlah Gording

= 9 x 2 = 18 Simpul

Beban yang diterima tiap gording

= 16,08

kg

Beban Atap Berat penutup atap

= 4,10 kg/m2

Luas atap

= 11,03 x 4 = 44,12 m 2

Beban atap

= 4,10 kg/m2 x 44,12 m 2 = 180,89 kg

P atap

=

,  

= 20,01 kg -

-

Beban gording Berat gording

= 5,19 kg/m

P gording

= 5,19 kg/m x 4 m = 20,76 kg

Total Beban Mati pada tiap gording PDL

= 16,08 kg + 20,01 kg + 20,76 kg = 56,85 kg

Gambar Beban Mati Pada Atap

19


(201510340311120) (201510340311133)

 Beban H idup Beban pekerja (PL)

= 100 kg

Gambar Beban Hidup Pada Atap.

 Beban Angin Angin Datang Rwx tepi

= -36,98 kg

Rwx

= -36,98 sin 25ᴼ

= -15,63 kg

Rwy

= -36,98 cos 25ᴼ

= -33,52 kg

Rwx tengah

= -36,98x 2

= -73,96 kg

Rwx

= -73,96 sin 25ᴼ

= -31,26 kg

Rwy

= -73,96 cos 25ᴼ

= -67,04 kg

Angin pergi Rwx tepi

= -36,98 kg

Rwx

= -36,98 sin 25ᴼ

= -15,63 kg

Rwy

= -36,98 cos 25ᴼ

= -33,52 kg

Rwx tengah

= -36,98x 2

= -73,96 kg

Rwx

= -73,96 sin 25ᴼ

= -31,26 kg

Rwy

= -73,96 cos 25ᴼ

= -67,04 kg

20


Arah Kiri

Arah Kanan

Tepi

Tepi

(201510340311120) (201510340311133)

Rwx

=

-15,63 kg

Rwx

=

-15,63 kg

Rwy

=

-33,52 kg

Rwy

=

-33,52 kg

Tengah

Tengah

Rwx

=

-31,26 kg

Rwx

=

-31,26 kg

Rwy

=

-67,04 kg

Rwy

=

-67,04 kg

Gambar Beban Angin Pada Atap Dengan Dinding Terbuka.

21


(201510340311120) (201510340311133)

Hasil Analisa STAAD Pro :

Gambar Penomoran Batang Pada Staad-Pro.

Gambar Beban Mati

Gambar Beban Hidup

22


(201510340311120) (201510340311133)

Gambar Beban Angin

Gambar Batang Tarik dan Tekan

Kombinasi Beban yang digunakan : a. 1,2 D + 1,6 L b. 1,2 D + 0,5 L + 1,0 W c. 1,2 D + 1,6 L + 0,5 W Setelah dianalisa dengan Staad-Pro gaya terbesar yaitu menggunakan kombinasi 1,2D + 1,6L.

23


(201510340311120) (201510340311133)

Gaya-G aya Batang H asil Perhitungan Staad-Pro Batang

Tepi Atas

Tepi Bawah

No. Batang 1 31 32 33 34 35 36 37 38 40 70 71 72 73 74 75 76 77 22 23 24 25 26 27 28 29 30 61 62 63 64 65 66 67 68 69

Panjang (m) 1.034 1.250 1.250 1.250 1.250 1.250 1.250 1.250 1.250 1.034 1.250 1.250 1.250 1.250 1.250 1.250 1.250 1.250 1.250 1.250 1.250 1.250 1.250 1.250 1.250 1.250 1.034 1.250 1.250 1.250 1.250 1.250 1.250 1.250 1.250 1.034

Tekan (kg)

Tarik (kg) -1430,00

1870,00 3170,00 3900,00 4050,00 3640,00 2660,00 1100,00 -1020,00 -1430,00 1870,00 3170,00 3900,00 4050,00 3640,00 2660,00 1100,00 -1020,00 3200,00 1330,00 31,21 -697,07 -854,84 -442,02 541,39 2100,00 4220,00 3200,00 1330,00 31,21 -697,07 -854,84 -442,02 541,39 2100,00 4220,00

24


Batang

Tegak

Diagonal

No. Beam 2 3 4 5 6 7 8 9 10 20 41 42 43 44 45 46 47 48 49 59 79 80 11 12 13 14 15 16 17 18 19 21 39 50 51 52 53 54 55 56 57 58 60 78

Panjang (m) 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 1.133 1.133 1.133 1.133 1.133 1.133 1.133 1.133 0.939 0.939 1.034 1.133 1.133 1.133 1.133 1.133 1.133 1.133 1.133 0.939 0.939 1.034

Tekan (kg)

(201510340311120) (201510340311133)

Tarik (kg)

941,86 747,75 519,52 123,11 63,11 -165,13 -393,36 -621,56 -849,82 -495,44 941,86 747,75 519,52 123,11 63,11 -165,13 -393,36 -621,56 -849,82 -495,44 3430,00 1208,00 -1690,00 -1180,00 -660,25 -143,03 374,25 891,55 1410,00 1930,00 1030,00 1300,00 -2150,00 -1690,00 -1180,00 -660,25 -143,03 374,25 891,55 1410,00 1930,00 1030,00 1300,00 -2150,00

25


(201510340311120) (201510340311133)

Cek Gaya Pada Titik Buhul Kondisi 1

ΣH

=0

S1 x Cos 25 0 – S40 x Cos 25 0

=0

1430 x Cos 25 0 - 1430 x Cos 25 0

=0

0 kg ΣV

= 0 kg…..(OK)

=0

S1 x Sin 25 0 + S40 x Sin 25 0 – S80

=0

1430 x Sin 25 0 + 1430 x Sin 25 0 – 1208

=0

0 kg

= 0 kg…..(OK)

Kondisi 2

ΣH

=0

S34 x Cos 25 0 – S33 x Cos 25 0 – S14

=0

4050 x Cos 25 0 - 3900x Cos 25 0 – 143,03

=0

0 ΣV

kg

= 0 kg…..(OK)

=0

S34 x Sin 250 - S33 x Sin 25 0 – S5

=0

4050 x Sin 25 0 - 3900 x Sin 25 0 – 123,11

=0

0 kg

= 0 kg…..(OK)

PERENCANAAN KUDA-KUDA 26


(201510340311120) (201510340311133)

1. Perencanaan Batang Tepi Atas Untuk No Beam 34

- Pmax = 4050 kg (Tekan) - L = 1,25 m = 125 cm

ax C

- Dicoba profil Double L 40 x 40 x 3 Untuk profil tunggal : = 2,336 cm2

A0

B

Ix0 = Iy0 = 3,53 cm4

Cy

Cx = Cy = 1,09 cm ix = iy

= 1,23 cm

I min

= 1,46 cm4

i min

= 0,79 cm

t 5mm

Untuk profil ganda :



A gab

= 2 (A0) = 2 (2,336) = 4,672 cm 2

ax

= Cx +

Ix gab

= 2 (Ix0) + 2 (A0) (ax )2 = 2 (3,53) + 2 (2,336) (1,34) 2 = 15,45 cm 4

ay

=0

Iy gab

= 2 (Iy0) + 2 (A0) (ay )2 = 2 (3,53) + 2 (2,336) (0) 2 = 7,06 cm 4

ix gab

=

iy gab

=

, = 1,09 + 0,25 = 1,34 cm 

    ,,     ,, =

= 1,82 cm

=

= 1,23 cm

Cek kelangsingan λ

= = =

L   ,

101,68

< 200 < 200 < 200 (OKE)

27


Batas langsing penampang =

= =

b    13,33

< 0,45

< 0,45

(201510340311120) (201510340311133)

    .

> 12,79 → Penampang Langsing

Maka digunakan rumus sebagai berikut :

   λ    (,)

Fe = =

Q

Maka, Fcr

> 0,44.Q.Fy

190,71

> 107,12

= = =



= 0,98

Fe

=

  ,  

= 1.34 – 0,76. 13,33

= 190,71 Mpa

Karena,

= 1,34 – 0,76(b/t)

(Q.) e

.0,658 F (,., [0,98.0,658 , ]247,5 142,67 Mpa

1426,7 kg/cm2

Cek kekuatan penampang Pn

> Pmax

 ×Ag×Fcr

> Pmax

Pn =  Pn = Pn =

0,9×4,672 ×1426,7 5999,14 kg

> Pmax > 4050 kg

(OKE)

28


(201510340311120) (201510340311133)

Untuk No Beam 1

- Pmax = 1430 kg (Tarik) - L = 1,034 m = 103 cm

ax C


A0

B

Ix0 = Iy0 = 3,53 cm4

Cy


= 1,23 cm

I min

= 1,46 cm4

i min

= 0,79 cm

t 5mm

Untuk profil ganda : A gab

= 2 (A0) = 2 (2,336) = 4,672 cm 2

ax

= Cx +

Ix gab

= 2 (Ix0) + 2 (A0) (ax )2 = 2 (3,53) + 2 (2,336) (1,34) 2 = 15,45 cm 4

ay

=0

Iy gab

= 2 (Iy0) + 2 (A0) (ay )2 = 2 (3,53) + 2 (2,336) (0) 2 = 7,06 cm 4

ix gab

=

iy gab

=

, = 1,09 + 0,25 = 1,34 cm 

    ,,     ,, =

= 1,82 cm

=

= 1,23 cm

 Cek kelangsingan

λ

L   ,

= =

83,78

=

< 300 < 300 < 300 (OKE)

 Cek kekuatan penampang

Pn

 x Ag x Fy

> Pmax

Pn = 

> Pmax

Pn =

> Pmax

Pn =

0,9×4,672 ×2475 10406,88 kg

> 1430 kg

(OKE)

29


(201510340311120) (201510340311133)

2. Perencanaan Batang Tepi Bawah Untuk No Beam 30


ax C


A0

B

Ix0 = Iy0 = 3,53 cm4

Cy


= 1,23 cm

I min

= 1,46 cm4

i min

= 0,79 cm

t 5mm




A gab

= 2 (A0) = 2 (2,336) = 4,672 cm 2

ax

= Cx +

Ix gab

= 2 (Ix0) + 2 (A0) (ax )2 = 2 (3,53) + 2 (2,336) (1,34) 2 = 15,45 cm 4

ay

=0

Iy gab

= 2 (Iy0) + 2 (A0) (ay )2 = 2 (3,53) + 2 (2,336) (0) 2 = 7,06 cm 4

ix gab

=

iy gab

=

, = 1,09 + 0,25 = 1,34 cm 

    ,,     ,, =

= 1,82 cm

=

= 1,23 cm

Cek kelangsingan λ

= = =

L   ,

83,78

< 200 < 200 < 200 (OKE)

30



= =

b    13,33

< 0,45

< 0,45

(201510340311120) (201510340311133)

    .



   λ    (,)

Fe = =

Q

Maka, Fcr

Fe

> 0,44.Q.Fy

280,87

> 107,12

= = = =



  ,  

= 1.34 – 0,76. 13,33

= 280,87 Mpa > Fy= 247,5 Mpa

Karena,

= 1,34 – 0,76(b/t)

= 0,98

(Q.) e

.0,658 F (,., [0,98.0,658 , ]247,5 161,28 Mpa

1612,8 kg/cm2


> Pmax

 ×Ag×Fcr

> Pmax

Pn =  Pn = Pn =

0,9×4,672 ×1612,8 6781,53 kg

> Pmax > 4220 kg

(OKE)

31


(201510340311120) (201510340311133)

Untuk No Beam 26

- Pmax = 854,84 kg (Tarik) - L = 1,25 m = 125 cm

ax C


A0

B

Ix0 = Iy0 = 3,53 cm4

Cy


= 1,23 cm

I min

= 1,46 cm4

i min

= 0,79 cm

t 5mm


= 2 (A0) = 2 (2,336) = 4,672 cm 2

ax

= Cx +

Ix gab

= 2 (Ix0) + 2 (A0) (ax )2 = 2 (3,53) + 2 (2,336) (1,34) 2 = 15,45 cm 4

ay

=0

Iy gab

= 2 (Iy0) + 2 (A0) (ay )2 = 2 (3,53) + 2 (2,336) (0) 2 = 7,06 cm 4

ix gab

=

iy gab

=

, = 1,09 + 0,25 = 1,34 cm 

    ,,     ,, =

= 1,82 cm

=

= 1,23 cm


λ

L   ,

= =

101.68

=

< 300 < 300 < 300 (OKE)


Pn

 x Ag x Fy

> Pmax

Pn = 

> Pmax

Pn =

> Pmax

Pn =

0,9×4,672 ×2475 10406,88 kg

> 854.84 kg

(OKE)

32


(201510340311120) (201510340311133)

3. Perencanaan Batang Tegak Untuk No Beam 79


ax C


A0

B

Ix0 = Iy0 = 3,53 cm4

Cy


= 1,23 cm

I min

= 1,46 cm4

i min

= 0,79 cm

t 5mm




A gab

= 2 (A0) = 2 (2,336) = 4,672 cm 2

ax

= Cx +

Ix gab

= 2 (Ix0) + 2 (A0) (ax )2 = 2 (3,53) + 2 (2,336) (1,34) 2 = 15,45 cm 4

ay

=0

Iy gab

= 2 (Iy0) + 2 (A0) (ay )2 = 2 (3,53) + 2 (2,336) (0) 2 = 7,06 cm 4

ix gab

=

iy gab

=

, = 1,09 + 0,25 = 1,34 cm 

    ,,     ,, =

= 1,82 cm

=

= 1,23 cm

Cek kelangsingan λ

= = =

L   ,

40,67

< 200 < 200 < 200 (OKE)

33



= =

b    13,33

< 0,45

< 0,45

(201510340311120) (201510340311133)

    .



   λ    (,)

Fe = =

Q

Maka, Fcr

Fe

> 0,44.Q.Fy

1191,93

> 107,12

= = = =



  ,  

= 1.34 – 0,76. 13,33

= 1191,93 Mpa > Fy= 247,5 Mpa

Karena,

= 1,34 – 0,76(b/t)

= 0,98

(Q.) e

.0,658 F (,., [0,98.0,658 , ]247,5 161,28 Mpa

1612,8 kg/cm2


> Pmax

 ×Ag×Fcr

> Pmax

Pn =  Pn = Pn =

0,9×4,672 ×1612,8 6781,53 kg

> Pmax > 3430 kg

(OKE)

34


(201510340311120) (201510340311133)

Untuk No Beam 10

- Pmax = 849.82 kg (Tarik) - L = 0.5 m = 50 cm

ax C


A0

B

Ix0 = Iy0 = 3,53 cm4

Cy


= 1,23 cm

I min

= 1,46 cm4

i min

= 0,79 cm

t 5mm


= 2 (A0) = 2 (2,336) = 4,672 cm 2

ax

= Cx +

Ix gab

= 2 (Ix0) + 2 (A0) (ax )2 = 2 (3,53) + 2 (2,336) (1,34) 2 = 15,45 cm 4

ay

=0

Iy gab

= 2 (Iy0) + 2 (A0) (ay )2 = 2 (3,53) + 2 (2,336) (0) 2 = 7,06 cm 4

ix gab

=

iy gab

=

, = 1,09 + 0,25 = 1,34 cm 

    ,,     ,, =

= 1,82 cm

=

= 1,23 cm


λ

L   ,

= =

40,67

=

< 300 < 300 < 300 (OKE)


Pn

 x Ag x Fy

> Pmax

Pn = 

> Pmax

Pn =

> Pmax

Pn =

0,9×4,672 ×2475 10406,88 kg

> 849.82 kg

(OKE)

35


(201510340311120) (201510340311133)

4. Perencanaan Batang Diagonal Untuk No Beam 18

- Pmax = 1930 kg (Tekan) - L = 1,13 m = 113cm

ax C


A0

B

Ix0 = Iy0 = 3,53 cm4

Cy


= 1,23 cm

I min

= 1,46 cm4

i min

= 0,79 cm

t 5mm




A gab

= 2 (A0) = 2 (2,336) = 4,672 cm 2

ax

= Cx +

Ix gab

= 2 (Ix0) + 2 (A0) (ax )2 = 2 (3,53) + 2 (2,336) (1,34) 2 = 15,45 cm 4

ay

=0

Iy gab

= 2 (Iy0) + 2 (A0) (ay )2 = 2 (3,53) + 2 (2,336) (0) 2 = 7,06 cm 4

ix gab

=

iy gab

=

, = 1,09 + 0,25 = 1,34 cm 

    ,,     ,, =

= 1,82 cm

=

= 1,23 cm

Cek kelangsingan λ

= = =

L   ,

91,92

< 200 < 200 < 200 (OKE)

36



= =

b    13,33

< 0,45

< 0,45

(201510340311120) (201510340311133)

    .



   λ    (,)

Fe = =

= 233,36 Mpa

Karena,

Maka, Fcr

233,36

> 107,12

= =

  ,  

= 0,98

> 0,44.Q.Fy

=

= 1,34 – 0,76(b/t)

= 1.34 – 0,76. 13,33

Fe

=



Q

(Q.) e

.0,658 F (,., [0,98.0,658 , ]247,5 157,30 Mpa

1573kg/cm2


> Pmax

 ×Ag×Fcr

> Pmax

Pn =  Pn = Pn =

0,9×4,672 ×1573 6614,52 kg

> Pmax > 1930 kg

(OKE)

37


(201510340311120) (201510340311133)

Untuk No Beam 39

- Pmax = 2150 kg (Tarik) - L = 1,03 m = 103 cm

ax C


A0

B

Ix0 = Iy0 = 3,53 cm4

Cy


= 1,23 cm

I min

= 1,46 cm4

i min

= 0,79 cm

t 5mm


= 2 (A0) = 2 (2,336) = 4,672 cm 2

ax

= Cx +

Ix gab

= 2 (Ix0) + 2 (A0) (ax )2 = 2 (3,53) + 2 (2,336) (1,34) 2 = 15,45 cm 4

ay

=0

Iy gab

= 2 (Iy0) + 2 (A0) (ay )2 = 2 (3,53) + 2 (2,336) (0) 2 = 7,06 cm 4

ix gab

=

iy gab

=

, = 1,09 + 0,25 = 1,34 cm 

    ,,     ,, =

= 1,82 cm

=

= 1,23 cm


λ

L   ,

= =

83,78

=

< 300 < 300 < 300 (OKE)


Pn

 x Ag x Fy

> Pmax

Pn = 

> Pmax

Pn =

> Pmax

Pn =

0,9×4,672 ×2475 10406,88 kg

> 2150 kg

(OKE)

38


(201510340311120) (201510340311133)

PERENCANAAN SAMBUNGAN Contoh Perhitungan: Mutu Baut A325 : Fu = 825 Mpa = 8250 kg/cm2 Mutu Pelat A572 : Fu = 414 Mpa = 4140 kg/cm2 Digunakan profil baja Siku Ganda 40 x 40 x 3 dengan Ag = 4,672 cm2 Direncanakan pelat buhul dengan t = 5 mm  Rencana Baut

Alubang

≤ 15 % Aprofil

d lub . t

≤ 15 % (4,672 cm2)

d lubang

≤

d lubang

≤ 1,17 cm

% ., / 2 ,

maka diambil dlubang = 1,1 cm dan Ø baut = 1 cm = 10 mm  Geser

Vd = 2 (Øf . r1 . fub . Ab) = 2 (0,75 . 0,5 . 8250 kg/cm2 . (1/4 π 12) cm2 = 4857 kg  Tumpu Pelat

Rd = 2,4 (Øf . db . tp . fup) = 2,4 (0,75 . 1 cm . 0,5 cm . 4140 kg/cm2) = 3726 kg Dipilih nilai terkecil antara tumpu baut (geser) dan tumpu pelat yaitu 3726 kg. Jumlah Baut N =

 =  = 1,13 ≈ 2 baut  

*Pu yang digunakan dicoba Pu maks

 Kontrol Terhadap Profil

Profil double siku 40 x 40 x 3 dengan Ag = 4,672 cm 2 Ø baut = 1 cm Alubang

≤ 15 % Aprofil

N .d . t

≤ 15 % (4,672 cm2)

(2) (1 cm) (0.3 cm)

≤ 0,7 cm2

0,6 cm2

≤ 0,7 cm2 → Ok

39


(201510340311120) (201510340311133)

 Perencanaan Jarak Baut Searah Dengan Batang

Jarak Baut : 3 db < s < 15 tp 3(10 mm) < s < 15 (5 mm) 30 mm

< s < 75 mm

Diambil jarak baut yaitu S=50 mm

Jarak Tepi Baut : 1,5 db < s` < 12 tp 1,5 (10 mm) < S1 < 12 (5 mm) 15 mm < S1 < 60 mm Diambil jarak tepi baut yaitu S1= 25 mm

40

Tubes Baja

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