panduan pembebanan dalam merancang gudang berbasis rangka baja

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Pre - Eliminary Design 1 Perencanaan Atap 1.1 Merencanakan Pola Beban Pola Beban Diambil dari peraturan Pembebanan Indonesia untuk gedung 1983 1.1.1 Merencanakan Beban Mati ( Berdasarkan Peraturan Pembebanan Indonesia Untuk Ge a. Atap Berat asbes : 10.3 Berat Profil : Menyesuaikan Perencanaan Berat Pengikat dll : 10 % dari Berat Total 1.1.2 Merencanakan Beban Hidup ( Berdasarkan Peraturan Pembebanan Indonesia Untuk G a. Beban Hidup Terbagi Rata ( Atap ) : 35 0 12 20 kg/m 2 a = q = (40 - 0.8 a) kg/m 2 kg/m 2 Merencanak an Pola Beban Beban Mati Beban Hidup Beban Angin Beban Penutup Atap Beban Terbagi Rata Beban Profil Beban Pengikat dll Beban Terpusat Beban Tekanan Angin Beban Angin Hisap Perencanaan Atap Merencanakan Pola Beban Data Perencanaan Perencanaan Dimensi Gording Perencanaan Gording Ujung Perencaan Penggantung Gording Perencanaan Ikatan Angin

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Panduan dalam mengerjakan tugas besar bangunan baja oleh mahasiswa S1 Teknik Sipil ITS Institut Teknologi Sepuluh Nopember Surabaya. Terapat perhitungan dan lainnya untuk menentukan spesifikasi baja yang akan dipakai agar bangunan tetap aman.

TRANSCRIPT

Pre - Eliminary Design

1 Perencanaan Atap

1.1 Merencanakan Pola Beban

Pola Beban Diambil dari peraturan Pembebanan Indonesia untuk gedung 1983

1.1.1 Merencanakan Beban Mati ( Berdasarkan Peraturan Pembebanan Indonesia Untuk Gedung )a. Atap

Berat asbes : 10.3Berat Profil : Menyesuaikan PerencanaanBerat Pengikat dll : 10 % dari Berat Total

1.1.2 Merencanakan Beban Hidup ( Berdasarkan Peraturan Pembebanan Indonesia Untuk Gedung )

a. Beban Hidup Terbagi Rata ( Atap ) :

35 0

12 ≤ 20

kg/m2

a =

q = (40 - 0.8 a) = kg/m2 kg/m2

Merencanakan

Pola Beban

Beban Mati Beban Hidup Beban Angin

Beban Penutup

AtapBeban

Terbagi RataBeban ProfilBeban

Pengikat dllBeban

Terpusat

Beban Tekanan

Angin

Beban Angin Hisap

Perencanaan Atap

Merencanakan Pola Beban

Data Perencanaan

Perencanaan Dimensi Gording

Perencanaan Gording Ujung

Perencaan Penggantung Gording

Perencanaan Ikatan Angin

ambil q = 12

b. Beban Hidup Terpusat ( Atap )

P = 100 kg

1.1.3 Merencanakan Beban Angin ( Berdasarkan Peraturan Pembebanan Indonesia Untuk Gedung )

a. Beban Tekanan Angin

Bangunan Jauh dari Pantai -> asumsi Tekanan Angin : 50

= 0.3

Angin Tekan = C x W = 15

Angin Hisap = 0.4 x W = 20

1.2 Data - Data perencanaan

Data AtapJenis : Asbes GelombangTebal : 5 mmBerat : 10.3 kg/m2Lebar Gelombang : 110 mmKedalaman Gelombang : 57 mmJarak Miring Gording : 171 cmJarak Kuda-Kuda (L) : 630 cm

Sudut Kemiringan Atap : 0.611 rad = 35 0

1.3 Perencanaan Dimensi Gording

kg/m2

kg/m2

Koefisien Angin (C) tekan = (0.02 a - 0.4)

kg/m2

kg/m2

1.3.1 Perencanaan Profil WF untuk Gording Dengan ukuran :

WF 150 x 75 x 5 x 7

A = 17.85 tf = 7 mm Zx = 98.2

W = 14 kg/m Ix = 666 Zy = 20.44

a = 150 mm Iy = 49.5 h = 120 mm {=D - 2 x (tf + r)}bf = 75 mm tw = 5 mmiy = 1.66 cm ix = 6.11 cm

r 8Mutu Baja = BJ 37

fu = 4100 410 Mpa

fy = 2500 250 Mpa

1.3.2 Perencanaan Pembebanan1.3.2.1 Perhitungan BebanBeban MatiBerat Gording = 14 kg/mBerat Asbes Gelombang = w x l

= 55 x 1.71 = 94.05 kg/mBerat Total = 108 kg/m

alat Pengikat dll 10 % = 0.1 x 108 = 10.81 kg/m

= 119 kg/m

Beban Hidup

40 - 28 = 12

q = 12

= 1.40 x 12.00 = 16.81 kg/m

= 100 kg

Beban Angin

Tekanan Angin = 50

Angin Tekan = 15

Angin Hisap = 20 (menentukan = q)1.700 x 20.00 = 34 kg/m

Beban Mati + Beban Hidup > dari Beban Angin Hisap : 119 + 16.81 > 34

Beban Angin Hisap tidak perlu diperhitungkan ==> 15 kg/m 25.5

1.3.2.2 Perhitungan Momen Akibat Beban thp Sbx dan SbyBeban Mati

0.125 x ( 119 x 0.819 x 39.69 ) = 483 kgm

0.125 x ( 119 x 0.574 x 4.41 ) = 37.58 kgm

Beban Hidup Terbagi Rata

0.125 x ( 16.81 x 0.819 x 39.69 ) = 48.77 kgm

0.125 x ( 16.81 x 0.574 x 4.41 ) = 5.315 kgm

cm2 cm3

cm4 cm3

cm4

kg/cm2 =

kg/cm2 =

qD

Beban Terbagi Rata = (40 - 0.8 a) = kg/m2

kg/m2

qL = jarak gording horisontal x q

Beban Hidup Terpusat, PL

kg/m2

kg/m2

kg/m2

q = jrk gording horisontal x angin hisap =

qw =

MXD = 1/8 (qD x cosa) L2 =

MYD = 1/8(qDxsina xL/3)2 =

MXLD = 1/8 (qL x cosa) L2 =

MYL = 1/8(qLxsinaxL/3)2 =

Beban Hidup Terpusat

0.25 x ( 100 x 0.819 x 6.3 ) = 129 kgm

0.25 x ( 100 x 0.574 x 2.10 ) = 30.11 kgm

Beban Angin Terbagi Rata

= 0.125 x 25.5 x 39.69 = 127 kgm

* Mu Beban Mati ,Beban Angin dan Beban Hidup Terbagi RataSumbu X Sumbu Y

483 kgm 37.58 kgm

48.768 kgm 5.315 kgmMw = 126.51 kgm

1.2 x 483 + 1.6 x 48.768 + 0.8 x 127 = 759 kgm

1.2 x 37.58 + 1.6 x 5.3147 + 0.8 x 0 = 53.6 kgm

* Mu Beban Mati, Beban Angin dan Beban Hidup TerpusatSumbu X Sumbu Y

483 kgm 37.58 kgm

129 kgm 30.11 kgmMw = 126.51 kgm

1.2 x 483 + 1.6 x 129.02 + 0.8 x 127 = 887 kgm

1.2 x 37.58 + 1.6 x 30.113 + 0.8 x 0 = 93.28 kgm

1.3.3 Kontrol Kekuatan Profil1.3.3.1 Penampang ProfilUntuk Sayap Untuk Badan

bf≤

170 h≤

16802 tf fy tw fy

75≤

170 120≤

16802 7 250 5 250

5.36 ≤ 10.75 24.0 ≤ 106.3OK OK

Penampang Profil Kompak, maka Mnx = Mpx

1.3.3.2 Kontrol Lateral Buckling

500 mm = 50 cm

1.76 x xEfy

= 1.76 x 1.66 x200000

= 82.64 cm250

Ternyata : < maka : Mnx = Mpx

MXL = 1/4 (qL x cosa) L =

MYL = 1/4(qL x sina)(L/3) =

MXW = 1/8 x qw x L

1.3.3.3 Besar Momen Berfaktor ( Mu = 1.2 MD + 1.6 ML + 0.8 MW )

MD = MD =

ML = ML =

MUX =

MUY =

MD = MD =

ML = ML =

MUX =

MUY =

Jarak Baut Pengikat / pengaku lateral = LB =

LP = iY

LB LP

Mnx = Mpx = Zx . Fy = 98.2 x 2500 = 2455 KgmMny = Zy ( satu sayap ) * fy

=

= 0.25 x 0.7 x 7.5 2 x 2500 = ### kgcm= 246.09 kgm

1.3.3.3 Persamaan IterasiMux

+Muy

≤ 1

Beban Mati , Beban Angin dan Beban Hidup Terbagi Rata758.873

+53.600

≤ 10.9 x 2455 0.9 x 2460.343 + 0.242 ≤ 1

0.59 ≤ 1 OK

Beban Mati , Beban Angin dan Beban Hidup Terpusat887.270

+93.276

≤1

0.9 x 2455 0.9 x 2460.361 + 0.421 ≤ 1

0.782 ≤ 1 OK

1.3.3.4 Kontrol Lendutan Profil

Lendutan Ijin f =L

=630

= 3.5 cm180 180

Lendutan Akibat Beban Merata (1)

fx =5

x =5 x 1.357 x 0.819 x 630 4

384 E x Ix 384 x 2000000 x 666 = 1.7113 cm

fy =5

x =5 x 1.357 x 0.574 x 210 4

384 E x Iy 384 x 2000000 x 49.5 = 0.199 cm

Lendutan Akibat Beban Terpusat (2)

fx =1

xP

=1 x 100 x 0.819 x 630 3

48 E x Ix 48 x 2000000 x 666 = 0.3204 cm

fy =1

xP

=1 x 100 x 0.574 x 210 3

48 E x Iy 48 x 2000000 x 49.5 = 0.1118 cm

Lendutan Akibat Beban Angin merata (3)

fx =5

x =5 x 0.255 x 0.819 x 630 4

384 E x Ix 384 x 2000000 x 666 = 0.3217 cm

fy =5

x =5 x 0.255 x 0.574 x 210 4

384 E x Iy 384 x 2000000 x 49.5 = 0.0374 cm

1/4 x tf x bf2 x fy

fb . Mnx fb . Mny

qD + L cos a L4

qD + L sin a (L/3)4

cos a L3

sin a L3

qW cos a L4

qW sin a (L/3)4

Lendutan total yang terjadi

=

= ( 1.7113 + 0.3204 + 0.322 0.199 + 0.112 + 0.037

2.3789 cm < 3.5 cm OK

1.4 Perencanaan Penggantung Gording

1.4.1 Data Penggantung Gording

Jarak Kuda - Kuda (L) = 630 cmJumlah Penggantung Gording = 2 buahJumlah Gording = 9 buahJarak Penggantung gording = 210 cm

1.4.2 Perencanaan PembebananBeban MatiBerat Sendiri Gording = 14 kg/mBerat Asbes gelombang = 94.05 kg/m

= 108 kg/mAlat Pengikat dll 10 % = 0.1 x 108 = 10.81 kg/m

= 119 kg/m

x x L / 3 = 118.85 x 0.5736 x 2.1 = 143.162 kg

Beban Hidup

40 - 0.8 = 20

q = 20

= 1.700 x 20.00 = 34 kg/m

x x L / 3 = 34 x 0.5736 x 2.1 = 40.953 kg

= 100 kg

x = 100.0 x 0.5736 = 57.358 kg

Beban Angin

Angin Tekan = q = 15

= 1.700 x 15.00 = 25.5 kg/m

x x L / 3 = 25.5 x 0.5736 x 2.1 = 30.715 kg

1.4.3 Perhitungan Gaya1.4.3.1 Penggantung Gording Tipe A

ftot = fx2 + fy2 (fx1 + fx2 + fx3)2 + (fy1 + fy2 + fy3)2

) 2 + ( ) 2

ftot = fijin =

qD

RD = qD sina

Beban Terbagi Rata = (40 - 0.8 a) = kg/m2

kg/m2

qL = jarak gording horisontal x q

RL = qL sina

Beban Terpusat = PL

RL = PL sina

kg/m2

qW = jarak gording horisontal x q

RW = qW sina

¿

= 1.2 x 143.2 + 1.6 x ( 41.0 + 57.4 ) + 0.8 x 30.7 = 353.66 kg

= 353.66 x 9 = 3182.98 kg

1.4.3.1 Penggantung Gording Tipe B

panjang miring gording=

171= 0.814

L / 3 210

39.16

=3182.98

= 5040.932 kgsin 39.16

1.4.4 Perencanaan Batang Tarik

Pu = 5040.932 kg

BJ 37 fu = 4100

fy = 2500

1.4.4.1 Kontrol LelehPu = φ . fy . Ag ; dengan φ = 0.9

Ag perlu = Pu = 5040.932= 2.240j fy 0.9 x 2500

Menentukan1.4.4.2 Kontrol PutusPu = φ . fu . 0,75 Ag ; dengan φ = 0.75

Ag perlu = Pu = 5040.932= 2.186j fu 0.75 0.75 x 4100 x 0.75

Tidak Menentukan

d =Ag x 4

=2.24 x 4

= 1.689 cmp p==> Pakai d = 17 mm

1.4.5 Kontrol KelangsinganJarak Penggantung Gording = 210 cm

Panjang Rb =

= 210 2 + 171 2= 271 cm

Cek :d >

Panjang Rb500

1.7 >270.82

5001 > 0.5416 OK

RA = 1.2 RD + 1.6 RL + 0.8 RW

RA total = Ra x jumlah Gording

arctan b =

b = o

RB =RA

sin b

RB =

kg/cm2

kg/cm2

cm2

cm2

Ag perlu = 1/4 . p . d2

(jarak penggantung gording)2 + (panjang miring gording)2

1.5 Perencanaan Ikatan Angin Atap

1.5.1 Data Perencanaan Ikatan Angin Atap

Tekanan Angin W = 50

= 0.9

= 0.4

300 cm 200 cm

a = 0.6109 rad = 35 0

1.5.2 Perhitungan Tinggi Ikatan Angin ( h )

9 m

9 + 2 x tg 0.611 = 10.4 m

9 + 4 x tg 0.611 = 11.8 m

9 + 6 x tg 0.611 = 13.2 m

9 + 9 x tg 0.611 = 15.3 m

1.5.3 Perhitungan Gaya - Gaya yang BekerjaR = 1/2 . W . C . a . h

0.50 x 50 x 0.9 x 1 x 9 = 202.5 kg

0.50 x 50 x 0.9 x 2 x 10.4 = 468 kg

0.50 x 50 x 0.9 x 2 x 11.8 = 531 kg

0.50 x 50 x 0.9 x 2.5 x 13.2 = 743 kg

0.50 x 50 x 0.9 x 3 x 15.3 = 1033 kg

Rtotal = ( R1+R2+R3+R4+(R5/2)) = 202.5 + 468.02 + 531 + 743 + 516 = 2460.564 kg

1.5.4 Perencanaan Dimensi Ikatan Angin1.5.4.1 Menghitung gaya Normal

tg φ =2

= 0.54

φ = 26.565 0

R1 = 202.5 kgRtotal = 2460.564 kg

Gaya Normal Gording Akibat Angin Dimana untuk angin tekan C = 0.9 dan untuk angin hisap C = 0.4

N =x

=0.4 x 2460.564

= 1093.6 kg0.9

1.5.4.2 Menghitung gaya Pada Titik SimpulPada Titik Simpul A

ΣV = 0

===> S1 = - Rtotal ===> S1 = -2461 kg

ΣH = 0

0

kg/m2

Koefisien Angin Ctekan

Koefisien Angin Chisap

a1 = a2 =

h1 =

h2 =

h3 =

h4 =

h5 =

R1 =

R2 =

R3 =

R4 =

R5 =

Chisap Rtotal

Ctekan

Rtotal + S1 = 0

S2 =

Pada Titik Simpul BEV = 0

- ( - ) =

- ( 202.5 - -2461 ) cos j cos 26.565

-2977.396 kg

1.5.5 Perencanaan Batang Tarik

Pu = = -2977.40 x 1.6 x 0.75 = -3572.875 kg

BJ 37 fu = 4100

fy = 2500

1.5.5.1 Kontrol LelehPu = φ . fy . Ag ; dengan φ = 0.9

Ag perlu = Pu = 3572.875= 1.588j fy 0.9 x 2500

Menentukan

1.5.5.2 Kontrol PutusPu = φ . fu . 0,75 Ag ; dengan φ = 0.75

Ag perlu = Pu = 3572.875= 1.549j fu 0.75 0.75 x 4100 x 0.75

Tidak Menentukan

d =Ag x 4

=1.588 x 4

= 1.422 cmp p==> Pakai d = 15 mm

1.5.6 Kontrol KelangsinganJarak kuda-kuda = 630 cm

= 630 2 + 171 2= 653 cm

Cek :d >

500

1.5 >652.79

5001.5 > 1.3056 OK

1.6 Perencanaan Gording Ujung1.6.1 Perencanaan Pembebanan Mntx , Mnty dan Gaya Normal Akibat AnginGording Ini adalah Balok Kolom. Akibat beban mati dan beban hidup Menghasilkan Momen LenturBesaran Diambil Dari Perhitungan Gording

= 887.270 x 0.75 = 665.453 kgm

= 93.276 x 0.75 = 69.957 kgm

R1 + S1 +S3 Cos j = 0

S3 = R1 S1

S3 =

S3 x 1.6 x 0.75

kg/cm2

kg/cm2

cm2

cm2

Ag perlu = 1/4 . p . d2

Panjang S3 = (jarak kuda-kuda)2 + (jarak miring gording)2

Panjang S3

Mntx = MUX (1.2 D + 1.6 L + 0.8 W) x 0.75

Mnty = MUY (1.2 D + 1.6 L + 0.8 W) x 0.75

2952.677 kg

1.6.2 Perencanaan Profil Gording Ujung

WF 100 x 50 x 5 x 7A = 11.85 cm2 tf = 7 mm Zx = 41.8 cm3W = 9.3 kg/m Ix = 187 cm4 Zy = 8.9 cm3a = 100 mm Iy = 14.8 cm4 h = 70 mm {=D - 2 x (tf + r)}

bf = 50 mm tw = 5 mmiy = 1.12 cm ix = 3.98 cm

Mutu Baja = BJ 37

fu = 3700 370 Mpa

fy = 2400 240 Mpa

1.6.3 Kontrol Tekuk Profil

Lkx = 630 cm ==> λx =Lkx

=630

= 158.3ix 3.98

Ncrbx = =p 2 x 2000000 x 11.85

158.3 2= 9335.40 kg

Lky = 50 cm ==> λy =Lkx

=50

= 44.64iy 1.12

Ncrby = =p 2 x 2000000 x 11.85

44.64 2= 117366.49 kg

Tekuk Kritis adalah arah X, Karena λx > λy 2.2867

Pn =Ag x fy

=11.85 x 2400

= 12437.136 kgw 2.2867

Pu=

2952.677= 0.279 < 0.2 (Pu = Nu)f Pn 0.85 x 12437.136

Pakai Rumus =Pu

+Mux

+Muy

≤ 12 x x Mnx x Mny

1.6.4 Perhitungan Faktor Pembesaran MomenGording dianggap tidak bergoyang, maka :

Mux = Mntx . SbxSbx =

Cmx≥ 1

1 - (Nu

)NcrbxUntuk elemen Beban Tranversal, ujung sederhanaCmx = 1

Sbx =1

= 1.4631 - (

2952.677)9335.40

Nu = 1.6 x Rtotal (dari ikatan angin atap) x 0.75 =

kg/cm2 =

kg/cm2 =

p2 . E . A

lx 2

p2 . E . A

ly 2

w =

fc . Pn fb fb

Sbx = 1.463 > 1Sbx = 1.463

Muy = Mnty * SbySby =

Cmy≥ 1

1 - (Nu

)NcrbyUntuk elemen Beban Tranversal, ujung sederhana

Cmy = 1

Sby =1

= 1.0261 - (

2952.677)117366.49

Sby = 1.026 > 1Sby = 1.026

1.6.5 Perhitungan Momen Ultimate Sbx dan SbyMux = Sbx . Mntx = 1.4626 x 665.453 = 973.294 kgmMuy = Sby . Mnty = 1.4626 x 69.957 = 102.320 kgm

1.6.6 Perhitungan Persamaan InteraksiMnx = 2455 kgm Mny = 246.09 kgm

Pu+

Mux+

Muy≤ 12 x x Pn x Mnx x Mny

2952.677+

973.294+

102.320≤ 12 x 0.85 x 12437.136 0.9 x 2455 0.9 x 246

1.042 > 1KO

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Pre - Eliminary Design

2 Perencanaan Dinding2.1 Data - Data perencanaan

Data Dinding :Jenis : Seng GelombangTebal : 4 mm

Berat : 4.15Kedalaman Gelombang : 25 mmJarak Kolom Dinding (L) : 400 cmJarak Gording Lt Dasar : 125 cmJarak Gording Lt 1 : 100 cm

2.2 Perencanaan Regel Balok ( Dinding Samping ) 2.2.1 Perencanaan Profil WF untuk Regel Balok Dinding Dengan ukuran :

WF 100 x 50 x 5 x 7A = 11.85 cm2 tf = 7 mm Zx = 41.8 cm3W = 9.3 kg/m Ix = 187 cm4 Zy = 8.938 cm3a = 100 mm Iy = 14.8 cm4 h = 70 mm {=D - 2 x (tf + r)}

bf = 50 mm tw = 5 mm Sx = 37.5 mmiy = 1.12 cm ix = 3.98 cm

r = mmMutu Baja = BJ 37

fu = 3700 370 Mpa

fy = 2400 240 Mpa

2.2.2 Perencanaan Pembebanan2.2.2.1 Perhitungan BebanBeban MatiLantai DasarBerat Gording = 9.3 kg/mBerat Seng Gelombang = 4.15 x 1.25 = 5.188 kg/m

Berat Total = 14.49 kg/malat Pengikat dll 10 % = 0.1 x 14.49 = 1.449 kg/m

Berat Total = 15.94 kg/m

0.125 x 15.94 x 1.778 = 3.541 kg/m

Lantai 1Berat Gording = 9.3 kg/mBerat Seng Gelombang = 4.15 x 1 = 4.15 kg/m

Berat Total = 13.45 kg/malat Pengikat dll 10 % = 0.1 13.45 = 1.345 kg/m

Berat Total = 14.8 kg/m

0.125 x 14.8 x 1.778 = 3.288 kg/m

Beban AnginLantai Dasar

Tekanan Angin = 50

Angin Tekan ( C = 0.9 ) = 0.9 x 50 = 45q = Angin Tekan x Jarak Gording = 45 x 1.25 = 56.25 kg/m

Angin Hisap ( C = 0.4 ) 0.4 x 50 = 20

kg/m2

kg/cm2 =

kg/cm2 =

Myd = 1/8 x q x (L/3)2 =

Myd = 1/8 x q x (L/3)2 =

kg/m2

kg/m2

kg/m2

q = Angin hisap x Jarak Gording = 20 x 1.25 = 25 kg/m

Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :

0.125 x 56.25 x 16 = 112.5 kgmN = q x Jarak Gording = 25 x 1.25 = 31.25 kg (Tarik)

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :

0.125 x 25 x 16 = 50 kgmN = q x Jarak Gording = 56.25 x 1.25 = 70.31 kg (Tekan)

Lantai 1

Tekanan Angin = 50

Angin Tekan ( C = 0.9 ) = 0.9 x 50 = 45q = Angin Tekan x Jarak Gording = 45 x 1 = 45 kg/m

Angin Hisap ( C = 0.4 ) 0.4 x 50 = 20q = Angin hisap x Jarak Gording = 20 x 1 = 20 kg/m

Akibat Beban Angin yg Tegak Lurus Dinding (tarik)

0.125 x 45 x 16 = 90 kgmN = q x Jarak Gording = 20 x 1 = 20 kg (Tarik)

Akibat Beban Angin yg Tegak Lurus Gevel (tekan)

0.125 x 20 x 16 = 40 kgmN = q x Jarak Gording = 45 x 1 = 45 kg (Tekan)

2.2.3 Kombinasi PembebananLantai Dasar1. U = 1.4 D

Muy = 1.4 x 3.541 = 4.958 kgm

Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :Mux = 1.2 x 0 + 1.3 x 112.5 + 0.5 x 0 + 0.5 x 0 = 146 kgmMuy = 1.2 x 3.541 + 1.3 x 0 + 0.5 x 0 + 0.5 x 0 = 4.25 kgmNu = 1.2 x 0 + 1.3 x 31.25 + 0.5 x 0 + 0.5 x 0 = 40.63 kg

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :Mux = 1.2 x 0 + 1.3 x 50 + 0.5 x 0 + 0.5 x 0 = 65 kgmMuy = 1.2 x 3.541 + 1.3 x 0 + 0.5 x 0 + 0.5 x 0 = 4.25 kgmNu = 1.2 x 0 + 1.3 x 70.31 + 0.5 x 0 + 0.5 x 0 = 91.41 kg

Lantai 11. U = 1.4 D

Muy = 1.4 x 3.288 = 4.603 kgm

Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :Mux = 1.2 x 0 + 1.3 x 90 + 0.5 x 0 + 0.5 x 0

Mxw = 1/8 x q x (L)2 =

Mxw = 1/8 x q x (L)2 =

kg/m2

kg/m2

kg/m2

Mxw = 1/8 x q x (L)2 =

Mxw = 1/8 x q x (L)2 =

2. U = 1.2D + 1.3W + λ L + 0.5 ( La atau Ha )

2. U = 1.2D + 1.3W + λ L + 0.5 ( La atau Ha )

= 117 kgmMuy = 1.2 x 3.288 + 1.3 x 0 + 0.5 x 0 + 0.5 x 0 = 3.945 kgmNu = 1.2 x 0 + 1.3 x 20 + 0.5 x 0 + 0.5 x 0 = 26 kg

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :Mux = 1.2 x 0 + 1.3 x 40 + 0.5 x 0 + 0.5 x 0 = 52 kgmMuy = 1.2 x 3.288 + 1.3 x 0 + 0.5 x 0 + 0.5 x 0 = 3.945 kgmNu = 1.2 x 0 + 1.3 x 45 + 0.5 x 0 + 0.5 x 0 = 58.5 kg

2.2.4 Kontrol Kekuatan Profil2.2.4.1 Penampang ProfilUntuk Sayap Untuk Badan

bf≤

170 h≤

16802 tf fy tw fy

50≤

170 70≤

16802 7 240 5 240

3.57 ≤ 10.97 14.0 ≤ 108.4OK OK

Penampang Profil Kompak, maka Mnx = Mpx

2.2.4.1 Kontrol Lateral Buckling

500 mm = 50 cm

1.76 x xEfy

= 1.76 x 1.12 x200000

= 56.90 cm240

Ternyata : < maka : Mnx = Mpx

Mnx = Mpx = Zx . Fy = 41.8 x 2400 = 1003 Kgm1.5 Myx = 1.5 Sx fy = 1.5 x 37.5 x 2400 = 1350 Kgm===> Mnx < 1.5 Myx

Mny = Zy ( satu sayap ) * fy

=

= 0.25 x 0.7 x 5 2 x 2400 = ### kgcm= 105 kgm

2.2.5 Perhitungan Kuat Tarik2.2.5.1 Kontrol Kelangsingan

≤ 300Lk

=400

= 101 < 300 OKix 3.98

2.2.5.2 Berdasarkan Tegangan Leleh

0.85 x 11.85 x 2400 = ### kgMenentukan

2.2.5.3 Berdasarkan Tegangan Putus

Jarak Baut Pengikat / pengaku lateral = LB =

LP = iY

LB LP

1/4 x tf x bf2 x fy

lp

l =

f Nn = f .Ag . fy =

λ p

= 0.75 x 0.85 x Ag x fu= 0.75 x 0.85 x Ag x fu= 0.75 x 0.85 x 11.85 x 3700= ### kg

Tidak Menentukan

2.2.5.4 Kontrol Kuat TarikLantai Dasar

> Nu### > 91.41

OKLantai 1

> Nu### > 2340

OK

2.2.6 Perhitungan Kuat Tekan2.2.6.1 Kontrol Kelangsingan

≤ 200Lkx

=400

= 101 < 200 OKix 3.98Lky

=50

= 44.64 < 200 OKiy 1.12

2.2.6.2 Berdasarkan Tekuk Arah Xfy

=101

x2400

= 1.108p E 3.142 20000000.25 < < 1.2

1.43=

1.43= 1.6681.6 - 0.67 x 1.108

fy= 0.85 x 11.85 x

2400= ### kgw 1.668

2.2.6.3 Berdasarkan Tekuk Arah Yfy = 44.64

x2400

= 0.492p E 3.142 2000000

0.25 < < 1.21.43

=1.43

= 1.1261.6 - 0.67 x 0.492fy

= 0.85 x 11.85 x2400

= ### kgw 1.126

2.2.7 Perhitungan Pembesaran Momen

Ncr =Ab x fy

2

Ncrbx =11.85 x 2400

= 23156.27164 kg1.108 2

Ncrby =11.85 x 2400

= 117359.5703 kg0.492 2

2.2.7.1 Komponen Struktur Ujung Sederhana Cm = 1

Sbx =Cmx

≥ 11 - (

Nu)

f Nn = f .Ae . fu =

f Nn

f Nn

lp

lpx =

lpy =

lc =lx

lc

w = 1.6 - 0.67 lc

f Nn = fAg

lc =ly

lc

w = 1.6 - 0.67 lc

f Nn = fAg

lc

1 - ( Ncrbx )

Lantai Dasar

Sbx =1

= 1.002 (Tarik)1 - (

40.63)23156.27

Sby =1

= 1.000 (Tarik)1 - (

40.63)117359.57

Sbx =1

= 1.004 (Tekan)1 - (

91.406)23156.27

Sby =1

= 1.001 (Tekan)1 - (

91.406)117359.57

Lantai 1

Sbx =1

= 1.001 (Tarik)1 - (

26.0)23156.27

Sby =1

= 1.000 (Tarik)1 - (

26.0)117359.57

Sbx =1

= 1.003 (Tekan)1 - (

58.5)23156.27

Sby =1

= 1.000 (Tekan)1 - (

58.5)117359.57

2.2.8 Kontrol Gaya Kombinasi2.2.8.1 Angin Dari Arah Tegak Lurus Dinding (tarik)Lantai Dasar

Nu=

40.625= 0.001680525 < 0.2 OK

24174

Nu+

Mux x Sbx+

Muy x Sby< 12 x f x Mnx x Mny

40.625+

146.3 x 1.002+

4.250 x 1.000< 12 x 24174 0.9 x 1003 0.9 x 105

0.208 < 1OK

Lantai 1Nu

=26

= 0.001075536 < 0.2 OK24174

Nu+

Mux x Sbx+

Muy x Sby< 12 x f x Mnx x Mny

26.000+

117.0 x 1.001+

3.945 x 1.000< 12 x 24174 0.9 x 1003 0.9 x 105

0.172 < 1OK

f . Nn

f . Nn fb

f . Nn

f . Nn fb

2.2.8.2 Angin Dari Arah Tegak Lurus Gevel (tekan)Lantai Dasar

Nu=

91.41= 0.00378118 < 0.2 OK

###Nu

+Mux x Sbx

+Muy x Sby

< 12 x f x Mnx x Mny91.406

+65.0 x 1.004

+4.250 x 1.001

< 12 x 24174 0.9 x 1003 0.9 x 1050.119 < 1

OKLantai 1

Nu=

58.5= 0.002419955 < 0.2 OK

###Nu

+Mux x Sbx

+Muy x Sby

< 12 x f x Mnx x Mny58.500

+52.0 x 1.003

+3.945 x 1.000

< 12 x 24174 0.9 x 1003 0.9 x 1050.101 < 1

OK

2.3 Perencanaan Regel Horizontal Gevel 2.3.1. Data - Data perencanaan tambahanJarak Kolom Dinding (L) : 300 cmJarak Gording Lt Dasar : 125 cmJarak Gording Lt 1 : 100 cm

2.3.2 Perencanaan Profil WF untuk Regel Horizontal Gevel Dengan ukuran :WF 100 x 50 x 5 x 7

A = 11.85 cm2 tf = 7 mm Zx = 41.8 cm3W = 9.3 kg/m Ix = 187 cm4 Zy = 9 cm3a = 100 mm Iy = 14.8 cm4 h = 70 mm

bf = 50 mm tw = 5 mm Sx = 37.5 mmiy = 1.12 cm ix = 3.98 cm 41.8

r = 8.938Mutu Baja = BJ 37

fu = 3700 370 Mpa

fy = 2400 240 Mpa

2.3.3 Perencanaan Pembebanan2.3.3.1 Perhitungan BebanBeban MatiLantai DasarBerat Gording = 9.3 kg/mBerat Seng Gelombang = 4.15 x 1.25 = 5.188 kg/m

Berat Total = 14.49 kg/malat Pengikat dll 10 % = 0.1 x 14.49 = 1.449 kg/m

Berat Total = 15.94 kg/m

0.125 x 15.94 x 1 = 1.992 kg/m

Lantai 1Berat Gording = 9.3 kg/mBerat Seng Gelombang = 4.15 x 1 = 4.15 kg/m

Berat Total = 13.45 kg/m

f . Nn

f . Nn fb

f . Nn

f . Nn fb

kg/cm2 =

kg/cm2 =

Myd = 1/8 x q x (L/3)2 =

alat Pengikat dll 10 % = 0.1 x 13.45 = 1.345 kg/mBerat Total = 14.8 kg/m

0.125 x 14.8 x 1 = 1.849 kg/m

Beban AnginLantai Dasar

Tekanan Angin = 50

Angin Tekan ( C = 0.9 ) = 0.9 x 50 = 45q = Angin Tekan x Jarak Gording = 45 x 1.25 = 56.25 kg/m

Angin Hisap ( C = 0.4 ) 0.4 x 50 = 20q = Angin hisap x Jarak Gording = 20 x 1.25 = 25 kg/m

Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :

0.125 x 56.25 x 9 = 63.28 kgmN = q x Jarak Gording = 25 x 1.25 = 31.25 kg (Tarik)

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :

0.125 x 25 x 9 = 28.13 kgmN = q x Jarak Gording = 56.25 x 1.25 = 70.31 kg (Tekan)

Lantai 1

Tekanan Angin = 50

Angin Tekan ( C = 0.9 ) = 0.9 x 50 = 45q = Angin Tekan x Jarak Gording = 45 x 1 = 45 kg/m

Angin Hisap ( C = 0.4 ) 0.4 x 50 = 20q = Angin hisap x Jarak Gording = 20 x 1 = 20 kg/m

Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :

0.125 x 45 x 9 = 50.63 kgmN = q x Jarak Gording = 20 x 1 = 20 kg (Tarik)

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :

0.125 x 20 x 9 = 22.5 kgmN = q x Jarak Gording = 45 x 1 = 45 kg (Tekan)

2.3.3.2 Kombinasi PembebananLantai Dasar1. U = 1.4 D

Muy = 1.4 x 1.992 = 2.789 kgm

Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :Mux = 1.2 x 0 + 1.3 x 63.28 + 0.5 x 0 + 0.5 x 0 = 82.27 kgmMuy = 1.2 x 1.992 + 1.3 x 0 + 0.5 x 0 + 0.5 x 0 = 2.39 kgmNu = 1.2 x 0 + 1.3 x 31.25 + 0.5 x 0 + 0.5 x 0 = 40.63 kg

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :Mux = 1.2 x 0 + 1.3 x 28.13 + 0.5 x 0 + 0.5 x 0 = 36.56 kgmMuy = 1.2 x 1.992 + 1.3 x 0 + 0.5 x 0 + 0.5 x 0 = 2.39 kgm

Myd = 1/8 x q x (L/3)2 =

kg/m2

kg/m2

kg/m2

Mxw = 1/8 x q x (L)2 =

Mxw = 1/8 x q x (L)2 =

kg/m2

kg/m2

kg/m2

Mxw = 1/8 x q x (L)2 =

Mxw = 1/8 x q x (L)2 =

2. U = 1.2D + 1.3W + λ L + 0.5 ( La atau Ha )

Nu = 1.2 x 0 + 1.3 x 70.31 + 0.5 x 0 + 0.5 x 0 = 91.41 kg

Lantai 11. U = 1.4 D

Muy = 1.4 x 1.849 = 2.589 kgm

Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :Mux = 1.2 x 0 + 1.3 x 50.63 + 0.5 x 0 + 0.5 x 0 = 65.81 kgmMuy = 1.2 x 1.849 + 1.3 x 0 + 0.5 x 0 + 0.5 x 0 = 2.219 kgmNu = 1.2 x 0 + 1.3 x 20 + 0.5 x 0 + 0.5 x 0 = 26 kg

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :Mux = 1.2 x 0 + 1.3 x 22.5 + 0.5 x 0 + 0.5 x 0 = 29.25 kgmMuy = 1.2 x 1.849 + 1.3 x 0 + 0.5 x 0 + 0.5 x 0 = 2.219 kgmNu = 1.2 x 0 + 1.3 x 45 + 0.5 x 0 + 0.5 x 0 = 58.5 kg

2.3.4 Kontrol Kekuatan Profil2.3.4.1 Penampang ProfilUntuk Sayap Untuk Badan

bf≤

170 h≤

16802 tf fy tw fy

50≤

170 70≤

16802 7 240 5 240

3.57 ≤ 10.97 14.0 ≤ 108.4OK OK

Penampang Profil Kompak, maka Mnx = Mpx

2.3.4.1 Kontrol Lateral Buckling

500 mm = 50 cm

1.76 x xEfy

= 1.76 x 1.12 x200000

= 56.90 cm240

Ternyata : < maka : Mnx = Mpx

Mnx = Mpx = Zx . Fy = 41.8 x 2400 = 1003 Kgm1.5 Myx = 1.5 Sx fy = 1.5 x 37.5 x 2400 = 1350 Kgm===> Mnx < 1.5 Myx

Mny = Zy ( satu sayap ) * fy

=

= 0.25 x 0.7 x 5 2 x 2400 = ### kgcm= 105 kgm

2.3.5 Perhitungan Kuat Tarik2.3.5.1 Kontrol Kelangsingan

≤ 300

2. U = 1.2D + 1.3W + λ L + 0.5 ( La atau Ha )

Jarak Baut Pengikat / pengaku lateral = LB =

LP = iY

LB LP

1/4 x tf x bf2 x fy

lp

Lk=

300= 75.38 < 300 OKix 3.98

2.3.5.2 Berdasarkan Tegangan Leleh

0.85 x 11.85 x 2400 = ### kgMenentukan

2.3.5.3 Berdasarkan Tegangan Putus

= 0.75 x 0.85 x Ag x fu= 0.75 x 0.85 x Ag x fu= 0.75 x 0.85 x 11.85 x 3700= ### kg

Tidak Menentukan

2.3.5.4 Kontrol Kuat TarikLantai Dasar

> Nu### > 91.41

OKLantai 1

> Nu### > 2340

OK

2.3.6 Perhitungan Kuat Tekan2.3.6.1 Kontrol Kelangsingan

≤ 200Lkx

=300

= 75.38 < 200 OKix 3.98Lky

=50

= 44.64 < 200 OKiy 1.12

2.3.6.2 Berdasarkan Tekuk Arah Xfy

=75.38

x2400

= 0.831p E 3.142 2000000

0.25 < < 1.21.43

=1.43

= 1.3711.6 - 0.67 x 0.831fy

= 0.85 x 11.85 x2400

= ### kgw 1.371

2.3.6.3 Berdasarkan Tekuk Arah Yfy = 44.64

x2400

= 0.492p E p 2000000

0.25 < < 1.21.43

=1.43

= 1.1261.6 - 0.67 x 0.492fy

= 0.85 x 11.85 x2400

= ### kgw 1.126

2.3.7 Perhitungan Pembesaran Momen

Ncr =Ab x fy

2

Ncrbx =11.85 x 2400

= 41166.70515 kg0.831 2

l =

f Nn = f .Ag . fy =

f Nn = f .Ae . fu =

f Nn

f Nn

lp

lpx =

lpy =

lc =lx

lc

w = 1.6 - 0.67 lc

f Nn = fAg

lc =ly

lc

w = 1.6 - 0.67 lc

f Nn = fAg

lc

Ncrby =11.85 x 2400

= 117366.4931 kg0.492 2

2.3.7.1 Komponen Struktur Ujung Sederhana Cm = 1

Sbx =Cmx

≥ 11 - (

Nu)Ncrbx

Lantai Dasar

Sbx =1

= 1.001 (Tarik)1 - (

40.63)41166.71

Sby =1

= 1.000 (Tarik)1 - (

40.63)117366.49

Sbx =1

= 1.002 (Tekan)1 - (

91.406)41166.71

Sby =1

= 1.001 (Tekan)1 - (

91.406)117366.49

Lantai 1

Sbx =1

= 1.001 (Tarik)1 - (

26.0)41166.71

Sby =1

= 1.000 (Tarik)1 - (

26.0)117366.49

Sbx =1

= 1.001 (Tekan)1 - (

58.5)41166.71

Sby =1

= 1.000 (Tekan)1 - (

58.5)117366.49

2.3.8 Kontrol Gaya Kombinasi2.3.8.1 Angin Dari Arah Tegak Lurus Dinding (tarik)Lantai Dasar

Nu=

40.625= 0.001680525 < 0.2 OK

24174Nu

+Mux x Sbx

+Muy x Sby

< 12 x f x Mnx x Mny40.625

+82.3 x 1.001

+2.390 x 1.000

< 12 x 24174 0.9 x 1003 0.9 x 1050.117 < 1

OKLantai 1

Nu=

26= 0.001075536 < 0.2 OK

24174Nu

+Mux x Sbx

+Muy x Sby

< 12 x f x Mnx x Mny

f . Nn

f . Nn fb

f . Nn

f . Nn fb

26.000+

65.8 x 1.001+

2.219 x 1.000< 12 x 24174 0.9 x 1003 0.9 x 105

0.097 < 1OK

2.3.8.2 Angin Dari Arah Tegak Lurus Gevel (tekan)Lantai Dasar

Nu=

91.41= 0.00378118 < 0.2 OK

###Nu

+Mux x Sbx

+Muy x Sby

< 12 x f x Mnx x Mny91.406

+36.6 x 1.002

+2.390 x 1.001

< 12 x 24174 0.9 x 1003 0.9 x 1050.068 < 1

OKLantai 1

Nu=

58.5= 0.002419955 < 0.2 OK

###Nu

+Mux x Sbx

+Muy x Sby

< 12 x f x Mnx x Mny58.500

+29.3 x 1.001

+2.219 x 1.000

< 12 x 24174 0.9 x 1003 0.9 x 1050.057 < 1

OK

2.4 Perencanaan kolom Gevel2.4.1 Data PerencanaanPanjang Beban Atap Regel 5 = 3 m Panjang Cantilever = 1 mPanjang Beban Atap Regel 2 = 3 m Jarak Kuda-kuda = 4 m

Lebar Beban Atap Regel 5 = 2.5 m panjang x angin tekanLebar Beban Atap Regel 2 = 2 m = 3 x 45 = 135 kg/m

panjang x angin tekanTinggi Regel 5 = 7 m = 3 x 45 = 135 kg/mTinggi Regel 2 = 6 m

Regel 5Luas atap yg Dipikul oleh Regel 5 ( A1 ) = Lebar Beban Atap Regel 5 x Pjg Beban Atap Regel 5

= 3 x 2.5

= 7.5Luas Dinding Regel 5 ( A2 ) = Pjg Beban Atap Regel 5 x Tinggi Regel 5

= 2.5 x 7

= 17.5

Regel 2Luas atap yg Dipikul oleh Regel 2 ( A3 ) = Lebar Beban Atap Regel 2 x Pjg Beban Atap Regel 2

= 3 x 2

= 6Luas Dinding Regel 2 ( A4 ) = Pjg Beban Atap Regel 2 x Tinggi Regel 2

= 2 x 6

= 12

2.4.2 Perencanaan Pembebanan

f . Nn

f . Nn fb

f . Nn

f . Nn fb

qw regel 5 =

qw regel 2 =

m2

m2

m2

m2

2.4.2.1 Beban MatiRegel 5

= 7.5 x 108 = 810 kg

= 17.5 x 4.15 = 72.63 kg

= 7 x 9.3 = 65.1 kg

Regel 2

= 6 x 108 = 648.3 kg

= 12 x 4.15 = 49.8 kg

= 6 x 9.3 = 55.8 kg

2.4.2.2 Beban hidupRegel 5

= 7.5 x 12 = 90 kg

Regel 2

= 6 x 12 = 72 kg

2.4.2.3 Beban AnginRegel 5

0.125 x 135 x 7 2 = 827 kgm

Regel 2

0.125 x 135 x 6 2 = 607.5 kgm

2.4.3 Syarat KekakuanRegel 5

Y =h

=700

= 3.5 cm200 200

Ix =5

xq x

384 E x Y

5x

8.269 x 7 4 = 3692.944384 0.02 x 3.5

===> Ix Profil yg Dipakai > 3692.944

Pakai Profil :WF 175 x 175 x 7.5 x 11

A = 51.21 tf = 11 mm Zx = 360

W = 40.2 kg/m Ix = 2880 Zy = 170

a = 175 mm Iy = 984 h = 175 - 2 x ( 11 + 12 )bf = 175 mm tw = 7.5 mm = 136 mmiy = 4.38 cm ix = 7.5 cm Sx = 2050 mm

r = 12 cmMutu Baja = BJ 37

fu = 3700 370 Mpa

fy = 2400 240 Mpa

Nd Profil = 7 x 40.2 = 281.4 kgNd total = Nd atap + Nd (Dinding+Gording ) + Nd Profil

= 810 + 138 + 281.4 = 1230 kgNL Total = NL atap = 90 kg

ND atap = A1 x qD atap

ND Dinding = A2 x qD Dinding

ND Gording = Jml Gording . w Gording

ND atap = A3 x qD atap

ND Dinding = A4 x qD Dinding

ND Gording = Jml Gording . w Gording

NL atap = A1 x qL atap

NL atap = A2 x qL atap

Mw = 1/8 x qw x (h)2 =

Mw = 1/8 x qw x (h)2 =

L4

cm4

cm4

cm2 cm3

cm4 cm3

cm4

kg/cm2 =

kg/cm2 =

Mw = 827 kgmU = ( 1.2D + 1.6L+ 1.6W ) x 0.75Nu = ( 1.2 x 1230 + 1.6 x 90 ) x 0.75 = 1215 kg

Mntx = 1.6 x Mw x 0.75 = 1.6 x 827 x 0.75 = 992 kg

Regel 2

Y =h

=600

= 3 cm200 200

Ix =5

xq x

384 E x Y

5x

6.075 x 6 4 = 1708.594384 0.02 x 3

===> Ix Profil yg Dipakai > 1708.594

Pakai Profil :WF 150 x 100 x 6 x 9

A = 26.84 tf = 9 mm Zx = 150

W = 21.1 kg/m Ix = 1020 Zy = 45.88

D = 148 mm Iy = 151 h = 150 - 2 x ( 9 + 11 )Bf = 100 mm tw = 6 mm = 116 mmiy = 2.37 cm ix = 6.17 cm Sx = 138 mm

r = 11 cmMutu Baja = BJ 37

fu = 3700 370 Mpa

fy = 2400 240 Mpa

Nd Profil = 6 x 21.1 = 126.6 kgNd total = Nd atap + Nd (Dinding+Gording ) + Nd Profil

= 648.3 + 105.6 + 126.6 = 880.5 kgNL Total = NL atap = 72 kg

Mw = 607.5 kgmU = ( 1.2D + 1.6L+ 1.6W ) x 0.75Nu = ( 1.2 x 880.5 + 1.6 x 72 ) x 0.75 = 879 kg

Mntx = 1.6 x Mw x 0.75 = 1.6 x 607.5 x 0.75 = 729 kg

2.4.4 Kontrol TekukRegel 5

untuk arah x :Lkx = 700 cm

λx =Lkx

=700

= 93.33ix 7.5fy

=93.33

x2400

= 1.029p E p 2000000

Ncrbx = =p 2 x 2000000 x 51.21

93.33 2= 116040.87 kg

untuk Arah y :Lky = 100 cm

λy =Lky

=100

= 22.83iy 4.38fy

=22.83

x2400

= 0.252p E p 2000000

Ncrby = =p 2 x 2000000 x 51.21

L4

cm4

cm4

cm2 cm3

cm4 cm3

cm4

kg/cm2 =

kg/cm2 =

lc =lx

p2 . E . A

lx 2

lc =ly

p2 . E . A

Ncrby = =22.83 2

= 1939245.26 kgTekuk Kritis Adalah Arah ====> X karena >

0.25 < < 1.21.43

=1.43

= 1.5711.6 - 0.67 x 1.029Pn = Ag . fy = 51.21 x 2400 = 122904 kgPu

=1214.55

= 0.012 < 0.20.85 x 122904

Pakai Rumus : Pu

+Mux

+Muy

≤ 12 x x Mnx x Mny

Batang Dianggap Tidak Bergoyang Maka :

Sbx =Cmx

≥ 1 ;Cm = 11 - (

Nu)Ncrbx

Sbx =1

= 1.011 ≥ 11 - (

1214.6)116040.87

Mux = Mntx . SbxMux = 992 x 1.011 = 1003 kgm

Regel 2untuk arah x :Lkx = 600 cm

λx =Lkx

=600

= 97.24ix 6.17fy

=97.24

x2400

= 1.072p E p 2000000

Ncrbx = =p 2 x 2000000 x 26.84

97.24 2= 56024.77 kg

untuk Arah y :Lky = 100 cm

λy =Lky

=100

= 42.19iy 2.37fy

=42.19

x2400

= 0.465p E p 2000000

Ncrby = =p 2 x 2000000 x 26.84

42.19 2= 297583.57 kg

Tekuk Kritis Adalah Arah ====> X karena >

0.25 < < 1.21.43

=1.43

= 1.6221.6 - 0.67 x 1.072Pn = Ag . fy = 26.84 x 2400 = 64416 kgPu

=878.85

= 0.016 < 0.20.85 x 64416

Pakai Rumus : Pu

+Mux

+Muy

≤ 12 x x Mnx x Mny

ly 2

λx λy

lc

w = 1.6 - 0.67 lc

f . Pn

fc . Pn fb fb

lc =lx

p2 . E . A

lx 2

lc =ly

p2 . E . A

ly 2

λx λy

lc

w = 1.6 - 0.67 lc

f . Pn

fc . Pn fb fb

0 .25<λc<1 .2

Batang Dianggap Tidak Bergoyang Maka :

Sbx =Cmx

≥ 1 ;Cm = 11 - (

Nu)Ncrbx

Sbx =1

= 1.016 ≥ 11 - (

878.8)56024.77

Mux = Mntx . SbxMux = 729 x 1.016 = 741 kgm

2.4.5 Menentukan MnxRegel 5

* Penampang ProfilUntuk Sayap : Untuk Badan :

bf≤

170 h≤

16802 tf fy tw fy

175≤

170 136≤

16802 11 240 7.5 240

7.95 ≤ 10.97 18.1 ≤ 108.4OK OK

Penampang Profil Kompak, maka Mnx = Mpx

* Kontrol Lateral Buckling

1000 mm = 100 cm

1.76 x xEfy

= 1.76 x 100 x200000

= ### cm2400

Ternyata : < maka : Mnx = Mpx

Mnx = Mpx = Zx . Fy = 0 x p = ### KgmMny = Zy ( satu sayap ) * fy

=

= 0.25 x ### x 0 2 x p = ### kgcm= ### kgm

Regel 2Penampang Profiluntuk Sayap untuk Badan

### 170 ### 1680### ### ### ###### ### ### ###

### ###

Penampang Profil Kompak, maka Mnx = Mpx

Lateral Bracing = LB =

LP = iY

LB LP

1/4 x tf x bf2 x fy

b2tf

≤170

√ fyht≤

1680

√ fy

¿ ¿

¿ ¿

Lateral Bracin Lb = 100 cm

Lp = ### cm

Ternya Lp > Lb maka Mnx = Mpx

Mnx = Mpx = Zx. Fy = ### * ### = ### KgmMny = Zy ( 1 flen ) * fy

== 0.25 ### ### ### = ### kgcm= ### kgm

2.4.6 Persamaan InteraksiPu

+Mux

+Muy

< 12 x x Mnx x Mny

Regel 5

#REF! + 1002.745305 + 00.17 x 0.9 ### 0.9 ###

#VALUE! + #REF! +

### < 1

OK

Regel 2

#REF! + #REF! + 00.17 ### 0.9 ### 0.9 ###

#REF! + #REF! +

### < 1

OK

2.5 Perencanaan Penggantung Gording Dinding Samping dan Gevel

2.5.1 Data Penggantung Gording

Jarak Kuda - Kuda = 400 cmJumlah Penggantung = 2 buahJumlah Gording Geve = 7 buahJumlah Gording Dind = 3 buahJarak Penggantung g = 133 cmJarak antara Gevel = 300 cmJarak Antar Gordng Horizont 125 cm

fc . Pn fb fb

Lp=1.76∗iy√ Efy

(1/4∗tf∗bf 2)∗fyx x x

x x x

x x x

Jarak Antar Gordng Horizont 100 cm2.5.2 Perencanaan PembebananDinding SampingBeban MatiBerat Sendiri Gording = 0 kg/mBerat Seng Gelombang = 4.15 kg/m

= 4.15 kg/mAlat Pengikat = 0.1 4.15 = 0 kg/m

= 4.15 kg/m

= 16.6 kg

GevelBeban MatiBerat Sendiri Gording = 0 kg/mBerat Seng Gelombang = 4.15 kg/m

= 4.15 kg/mAlat Pengikat = 0.1 4.15 = 0.415 kg/m

= 4.565 kg/m

= 13.7 kg

2.5.3 Perhitungan Gaya2.5.3.1 Penggantung Gording Tipe ADinding Samping `

Ra = 23.24 kg

Ra Total = Ra * jumlah Gord Ra = 69.72 kg

Gevel `Ra = 19.17 kg

Ra Total = Ra * jumlah Gord Ra = 134 kg2.5.3.2 Penggantung Gording Tipe BDinding Samping

0.938

0.753

43.14

Rb = 69.72 = 102 kg0.684

Gevel1.4

0.951

54.44

Rb = 134 = 165 kg0.814

o

o

KudaJarakKudaqRa = *

JarakGevelqRa *=

arctgn β=β=β=

RB=R A

Sin β

x

x

arctgn β=β=β=

RB=R A

Sin β

2.5.4 Perencanaan Batang TarikDinding Samping

Pu = 102 kgBJ 37 f 0 kg/cm2

fy = 0 kg/cm3

GevelPu = 165 kg

BJ 37 f 0 kg/cm2fy = 0 kg/cm3

2.5.4.1 Kontrol LelehDinding SampingPu = φ fy Ag dengan φ = 0.9

Ag perlu = Pu/ = 102 = ### cm20 ###

GevelPu = φ fy Ag dengan φ = 0.9

Ag perlu = Pu/ = 165 = ### cm20 ###

2.5.4.2 Kontrol PutusDinding SampingPu = φ fu 0.75 Ag dengan φ = 0.75

Ag Per Pu = 102 = ### cm2φ fu 0.75 0 ###

= ### cm2

= ### 4 d = ### cm3.1415

Pakai d 10 mm

Dinding SampingPu = φ fu 0.75 Ag dengan φ = 0.75

Ag Per Pu = 165 = ### cm2φ fu 0.75 0 ###

Ag=1 /4 πd2

d=√ Ag∗4π

√ x

RB=R A

Sin β

d=√ Ag∗4π

Ag=1 /4 πd2

= ### cm2

= ### 4 d = ### cm3.1415

Pakai d 10 mm

2.5.5 Kontrol KelangsinganDinding SampingJarak Penggantung G 133 cm

Panjang Rb = ### + ###

Panjang Rb = 183 cm1 > 183

500

1 > 0.366OK

GevelJarak Penggantung G 100 cm

Panjang Rb = ### + ###

Panjang Rb = 141 cm1 > 141

500

1 > 0.283OK

2.6 Perencanaan Ikatan Angin Dinding

2.6.1 Data Perencanaan Ikatan Angin DindingTekanan Angi 0 kg/m2Koefisien Ang 0.9a1 = 300 cm a2 = 200 cm

0 = 0 0

2.6.2 Perhitungan Tinggi Ikatan Angin ( h )h1 = 9 mh2 = 9 + 2 tg 0.436 = 9.933 mh3 = 9 + 4 tg 0.436 = 10.87 mh4 = 9 + 6 tg 0.436 = 11.8 mh5 = 9 + 9 tg 0.436 = 13.2 m

d≥PanjangRb500

Ag=1 /4 πd2

√ x

PanjangRb=√JrkPenggantungGording2+JrkantarGordingHorizontal2

d≥PanjangRb500

PanjangRb=√JrkPenggantungGording2+JrkantarGordingHorizontal2

α=

xxxx

xxxx

2.6.3 Perhitungan Gaya - Gaya yang BekerjaR = 1/2 W C a h

R1 = 0.50 0 ### 1 9 = ### kgR2 = 0.50 0 ### 2 9.933 = ### kgR3 = 0.50 0 ### 2 10.87 = ### kgR4 = 0.50 0 ### 2.5 11.8 = ### kgR5 = 0.50 0 ### 3 13.2 = ### kg

Rtotal = ( R1+R2+R3+R4+(R5/2)) = ### kg

2.6.4 Perencanaan Dimensi Ikatan Angin

tg φ = 14

= 0.25

φ = 0.245 rad

R1 = ### kgRtotal = ### kg

2.6.4.1 Menghitung gaya Pada Titik SimpulPada Titik Simpul AΣV = 0Rtotal + S1 = 0S1 = - Rtotal

S1 = ### kg

ΣH = 0S2 = 0

Pada Titik Simpul BEV = 0R1 + S1 +S3 Cos Φ = 0

S3 = ### kg

S3=−(R1−S1 )

Cosφ

x x

xxxxx

xxxxx

xxxxx

xxxxx

2.6.5 Perencanaan Batang Tarik

Pu = ### kgBJ 37 f 0 kg/cm2

fy = 0 kg/cm3

2.6.5.1 Kontrol Leleh

Pu = φ fy Ag dengan φ = 0.9

Ag perlu = Pu/ = ### = ### cm20 ###

2.6.5.2 Kontrol Putus

Pu = φ fu 0.75 Ag dengan φ = 0.75

Ag Per Pu = ### = ###φ fu 0.75 0 ###

= ### cm2

= ### 4 d = ### cm3.1415

Pakai d 12 mm2.6.6 Kontrol Kelangsingan

Jarak Kuda - 400 cm

Panjang S3 = 0 + 0

Panjang S3 = 0 cm1.2 > 0

500

1.2 > 0OK

Ag=1 /4 πd2

d=√ Ag∗4π

Pu=S3∗1.6∗0 . 75

d≥PanjangS 3

500

√ x

PanjangS3=√ JrkantarKuda−Kuda2+Jrkantar2RegelHorizontal2

Start

Masukkan Data - Data Perencanaan Bondex dan Balok Anak :Panjang Bentang Beban Bondex Yang Dipikul Balok Anak = ?

Panjang Balok Anak = ?Berat Sendiri Beton = ?

Berat Sendiri Bondex = ?Berat Spesi per cm Tebal = ?

Berat Tegel = ?Beban Berguna = ?

Hitung Pembebanan terhadap Balok Anak :Beban Mati

Beban Hidup

Hitung Tebal Lantai BondexTebal Lantai Bondex Dicari dengan Menggunakan Tabel yang ada dengan memperhitungkan Beban Berguna yang akan Disalurkan

Bondex ke Balok Anak sebagai Dasar Perencanaan.T = ?

Hitung Luasan Tulangan Negatif BondexLuasan Tulangan Negatif Bondex Dicari dengan Menggunakan

Tabel yang ada dengan memperhitungkan Beban Berguna yang akan Disalurkan Bondex ke Balok Anak sebagai Dasar

Perencanaan.A = ?

Asumsikan Diamter Tulangan Negatif Bondex :φ = ? Mm

Hitung Banyaknya Tulangan Yang Diperlukan Tiap 1 m :A/As = ?

Hasilnya Dibulatkan Keatas

Hitung Jarak Tulangan Tarik :Jarak Tulangan Tarik = Jarak Tulangan yang Diperlukan ( 1 m ) Dibagi dengan Banyaknya Tulangan yang diperlukan dengan

Jarak yang Telah Ditetapkan Diatas

Perencanaan PembebananBeban Mati

Beban Hidup

Hitung qU, Mu Max dan Du Max :qU = 1.2 qD + 1.6 qL

Mumax=18qu l

2 Dumax=12qu l

KO

OK

Sayap Badan

OK OK

KO KO

OK OK

KO KO

Pilih Profil Baja Dimana Ix-nya Harus > Ix Minimum :A = ? ; W = ? ; a = ? ; bf = ? ; iy = ? ;tf = ? ; Ix = ? Iy = ? ; tw = ? ; Zx = ? ; Zy

= ? ; h = ? ; fu = ? ; Fy = ?

Perencanaan Pembebanan + Berat ProfilBeban Mati

Beban Hidup

Hitung qU, Mu Max dan Du Max ( Berat Profil Dimasukkan ) :qU = 1.2 qD + 1.6 qL

KONTROL LENDUTAN BALOKDimana Y ijin = L/360

Y mak < Y ijinPerbesar Profil

KONTROL LOKAL BUCKLINGHitung λp, λr Penampang Sayap dan λp, λr Penampang Badan :

Sayap Badan

Profil KompakMnx = Mnp

Profil Tak Kompak

Profil Langsing Profil Langsing

1 2

Mnx Sayap > Mnx Badan

Profil KompakMnx = Mnp

Profil Tak Kompak

Hitung qU, Mu Max dan Du Max :qU = 1.2 qD + 1.6 qL

PERHITUNGAN Ix PROFIL MINIMUMDimana Y ijin = L/360

Mumax=18qu l

2 Dumax=12qu l

Y max= 5384

(qD+qL)∗l4

EIx

λ p=170

√ fyλ p=

1680

√ fy

ht

λr=370

√ f y− f rλr=

2550

√ f y

b2tf

b2tf

¿ λ pht¿ λ p

b2tf

¿λ p¿ λr

Mn=Mp−(Mp−Mr)λ−λpλr−λp

Mn=Mr( λr / λ)2

λ p ¿ht¿ λr

Mn=Mr( λr / λ)2

Mumax=18qu l

2 Dumax=12qu l

Ix> 5384

( qD+qL)∗l4

EY

OK KO

KO KO

OK OK

OK KO

KO

OK

Mnx Sayap > Mnx Badan

Ambil Mnx BadanLocal Buckling

Ambil Mnx SayapLocal Buckling

Profil Tak Kompak

2

KONTOL LATERAL BUCKLINGHitung λp dan λr daripada Lateral Buckling

Bentang PendekMnx = Mpx

Bentang Menengah

Bentang Panjang

Mnx Local Buckling > Mnx Lateral Buckling

Ambil Mnx Lateral Buckling Ambil Mnx Local Buckling

Jarak Lateral Bracing λb :λb = ?

KONTROL KUAT RENCANA GESERHitung

Hitung : 0.9 Mnx

0.9 Mnx > Mu maxPerbesar Profil

Mn=Mp−(Mp−Mr)λ−λpλr−λp

Lp=1.76∗iy√ Efy

Lr=r y(X1

f L)√1+√1+X 2 f

L2

G=E

2(1+μ )

X 1=πSx √ EGJA

2

X 2=4 (SGJ

)2IwIy

λb ¿ λ pλb¿λ p ¿ λr

Mn=Cb(Mr+(Mr−Mp)(Lr−LLr−Lp

)≤Mp

Mn=Mcr=CbπL √E¿GJ+(

πEL

)2 I y I w≤Mp

htw

OK

KO

OK

KO

KO

OK

KONTROL KUAT RENCANA GESERHitung

Vn = 0.6 fy Aw

Hitung 0.9 Vn

0.9 Vn > Vu Max

Profil Dapat Dipakai

Perbesar Profil

htw

≤1100

√ fy

1100

√ fy≤

htw

≤1370

√ fyVn=0. 6 f y Aw

1100 twh√ f y

Vn=900000Aw

(htw)2

Pre - Eliminary Design

3 Perencanaan Bondex dan Balok Anak3.1 Data - Data perencanaan

Beban Hidup : 400 Kg/m2Beban Finishing : 90 Kg/m2Beban Berguna : 490 Kg/m3

Berat Beton Kering : 2400 kg/m3Panjang Bentang Beban Bondex yang Dipikul Oleh Balok Anak : 3 mPanjang Balok Anak : 4 m

3.2 Perencanaan Pelat Lantai Bondex

3.2.1 Data PerencanaanBerat Sendiri Beton = 2400 kg/m3Berat Sendiri Bondex = 10.1 kg/m2Berat Spesi per cm Tebal = 21 kg/m2Berat Tegel = 24 kg/m2

3.2.2 Perencanaan PembebananBeban MatiBerat Beton = 2400 * 0.12 = 288 Kg/m2Berat Bondex = 10.1 Kg/m2Berat Spesi 2 Cm = 21 * 2 = 42 Kg/m2Berat Tegel 2 Cm = 24 * 2 = 48 Kg/m2

qD = 388.1 Kg/m2Beban HidupBeban Hidup Lantai gudang = 400 Kg/m2Beban Finishing = 90 Kg/m2

qL = 490 Kg/m2

3.2.3 Perencanaan Tebal Lantai Beton dan Tulangan Negatif3.2.3.1 Perencanaan Tebal Lantai

qL = 490 kg/m2

Beban Berguna yang Dipakai = 500 kg/m2Jarak Antar Balok = 300 cmJarak Kuda - Kuda = 400 cm

Dari Tabel Brosur ( Bentang Menerus dengan Tulangan Negatif ),didapat :t = 12 mmA = 3.57 cm2/m

3.2.3.2 Perencanaan Tulangan Negatif10 mm

As = 0.785 mm2

Banyaknya Tulangan Yang diperlukan Tiap 1 m = A = 3.57As 0.785

= 4.547771 Buah= 5 Buah

Direncanakan Tulangan Dengan φ =

Jarak Tulangan Tarik = 200 cm

3.3 Perencanaan Dimensi Balok Anak3.3.1 Perencanaan PembebananBeban Mati ( D )Bondex = 3 10.1 = 30.3 kg/mPlat Beton = 3 0.12 2400 = 864 kg/mTegel + Spesi = 3 90 = 270 kg/m

qD = 1164.3 kg/m

Beban Hidup ( L )qL = 3 490 = 1470 kg/m

3.3.3 Perhitungan qU , Mu Max dan Du MaxqU = 1.2 qD + 1.6 qL

qU = 1.2 1164.3 1.6 1470 = 3749.16 Kg/m

= 0.125 3749.16 16 = 7498.32 Kgm

= 0.5 3749.16 4 = 7498.32 Kg

3.3.4 Perhitungan Ix Profil Yang DiperlukanY = L = 400 = 1.111111

360 360

Ix > 5 ( 11.643 14.7 ) 2.56E+10384 2100000 1.111111

Ix > 3763.286 cm4

3.3.5 Perencanaan Profil WF untuk Balok Anak

250 x 125 x 6 x 9

A = 37.66 cm2 tf = 9 mm Zx = 351.861 cm3W = 29.6 kg/m Ix = 4050 cm4 Zy = 72.0225 cm3a = 250 mm Iy = 294 cm4 h = 208 mm

bf = 125 mm tw = 6 mm r = 12 mmiy = 2.79 cm ix = 10.4 cm 351.861

72.0225Mutu Baja = BJ 37

fu = 3700 kg/cm2fy = 2400 kg/cm2

3.3.6 Perencanaan Pembebanan + Beban ProfilBeban Mati ( D )

Pasang Tulangan Tarik φ10 - 200

Mumax=18qu l

2

Dumax=12qu l

Ix> 5384

( qD+qL)∗l4

EY

Bondex = 3 10.1 = 30.3 kg/mPlat Beton = 3 0.12 2400 = 864 kg/mTegel + Spesi = 3 90 = 270 kg/mBerat Profil = = 29.6 kg/m

qD = 1193.9 kg/m

Beban Hidup ( L )qL = 3 490 = 1470

3.3.7 Perhitungan qU , Mu Max dan Du Max ( Berat Profil Dimasukkan )qU = 1.2 qD + 1.6 qL

qU = 1.2 1193.9 1.6 1470 = 3784.68 Kg/m

= 0.125 3784.68 16 = 7569.36 Kgm

= 0.5 3784.68 4 = 7569.36 Kg

3.3.8 Kontrol Lendutan BalokY = L = 400 = 1.111111

360 360

= 5 ( 11.939 14.7 ) 2.56E+10384 2100000 4050

= 1.044053 < 1.111111

OK

3.3.9 Kontrol Kuat Rencana Momen Lentur3.3.9.1 Kontrol Penampang

untuk Sayap untuk Badan

125 170 208 168018 15.49193 6 15.49193

6.944444 10.97345 34.66667 108.4435OK OK

Penampang Profil Kompak, maka Mnx = Mpx

Mp = fy * Zx= 2400 * 351.861= 844466.4 kgcm= 8444.664 kgm

3.3.9.2 Kontrol Lateral Buckling

Mumax=18qu l

2

Dumax=12qu l

Y max= 5384

(qD+qL)∗l4

EIx

b2tf

≤170

√ fyht≤

1680

√ fy

¿ ¿

¿ ¿

Jrk Pengikat Lateral : 1000 mm = 100 cm

Lp = 141.751 cm

Ternyata Lp > Lb maka Mnx = Mpx

Mnx = Mpx = Zx. Fy = 351.861 * 2400 = 8444.664 KgmMny = Zy ( 1 flen ) * fy

== 0.25 0.9 156.25 2400 = 84375 kgcm= 843.75 kgm

0.9 Mp = 0.9 * 8444.664 = 7600.198 kgm

0.9 Mp > Mu7600.198 > 7569.36

OK

3.3.9.3 Kontrol Kuat Rencana Geser

208 < 11006 15.49193

34.66667 < 71.00469Plastis

Vn = 0.6 fy Aw= 0.6 2400 0.6 25= 21600 Kg

Vu < ФVn7569.36 < 0.9 216007569.36 < 19440

OK

Lp=1. 76∗iy√ Efy

(1/4∗tf∗bf 2)∗fyx x x

htw

≤1100

√ fy

4 Perencanaan Tangga Baja

4.1 Data PerencanaanTinggi tangga = 275 cmLebar injakan (i) = 28 cmPanjang Tangga = 600 cmLebar Pegangan Tangga = 10 cm

4.2 Perencanaan Jumlah Injakan Tangga4.2.1 Persyaratan - Persyaratan Jumlah Injakan Tangga

60 cm < ( 2t + I ) < 65 cm

25 < a < 40

Dimana : t = tinggi injakan (cm)i = lebar injakan (cm)a = kemiringan tangga

4.2.2 Perhitungan Jumlah Injakan Tangga

Tinggi tanjakan (t) = 65 - 28 / 2= 18.5 cm

Jumlah Tanjakan = 275 = 14.86486 buah18.5

= 15 buah

Jumlah injakan (n) = 15 buah

Lebar Bordes = 600 420 = 180 cmLebar Tangga = 200 20 = 180 cm

a = 33.47028 0 = 0.58387 rad

420 cm 180 cm

180 cm

180 cm

4.3 Perencanaan Pelat Tangga4.3.1 Perencanaan Tebal Pelat Tangga

Tebal Pelat Tangga = 4 mm

o o

−−

Berat Jenis Baja = 7850 kg/m3Tegangan Leleh Baja = 2500 kg/m24.3.2 Perencanaan Pembebanan Pelat TanggaBeban Mati Berat Pelat = 0.004 1.8 7850 = 56.52 kg/m'Alat Penyambung (10 %) = 5.652 kg/m'

qD = 62.172 kg/m'Beban Hidup qL = 500 x 1.8 = 900 kg/m'

= 0.125 62.172 0.0784 = 0.609286 kgm

= 0.125 900 0.0784 = 8.82 kgm

Mu = 1.4 0.609286 = 0.853 kgmTidak Menentukan

Mu = 1.2 0.609286 + 1.6 8.82 = 14.84314 kgmMenentukan

4.3.5 Kontrol Momen Lentur

= 0.25 180 0.16 = 7.2 cm3

φ Zx * fy = 0.9 7.2 2500 = 16200 kgcm

162 kgm

Syarat -> > Mu162 kgm > 14.84314 kgm

OK

4.3.6 Kontrol Lendutan

f = L = 28 = 0.077778360 360

= 0.083333 180 0.064 = 0.96 cm4

4.3.3 Perhitungan MD dan ML

MD

MD

4.3.4 Perhitungan Kombinasi Pembebanan MU

MU = 1.4 MD

MU = 1.2 MD + 1.6 ML

φMn =

φMn =

φMn

x x

M D=18qDl

2

M L=18qL l

2

x x

x x

x

x x

x xZx=14bh2

x x

Ix=1

12bh3

x x

Ix = 0.96 cm4

= 5 ( 0.62172 9 ) 6.15E+05384 2100000 0.96

= 0.038197 < 0.077778

OK

Ambil Pelat Tangga dengan Tebal = 4 mm

4.4 Perencanaan Penyangga Pelat Injak

4.4.1 Perencanaan PembebananBeban MatiBerat Pelat = 0.14 0.004 7850 = 4.396 kg/m'Berat Baja Siku = 45 45 7 = 4.6 kg/m'

8.996 kg/m'Alat Penyambung ( 10 % ) = 0.8996 kg/m'

qD = 9.8956 kg/m'

Beban Hidup qL = 500 x 0.14 = 70 kg/m'

.

= 0.125 9.8956 3.24 = 4.007718 kgm

= 0.125 70 3.24 = 28.35 kgm

Mu = 1.4 4.007718 = 5.610805 kgmTidak Menentukan

Mu = 1.2 4.007718 + 1.6 28.35 = 50.16926 kgmMenentukan

4.4.4 Kontrol Momen LenturDari Perhitungan Sap 2000 Version 8.2.3 Didapat untuk Profil Siku 45x45x7 :

Zx = 6.14 cm3 ( Modulus Plastis )

4.4.2 Perhitungan MD dan ML

MD

MD

4.4.3 Perhitungan Kombinasi Pembebanan MU

MU = 1.4 MD

MU = 1.2 MD + 1.6 ML

Y max= 5384

(qD+qL)∗l4

EIx

x

x xx x

M D=18qDl

2

M L=18qL l

2

x x

x x

x

x x

x

Ix=1

12bh3

x x

φ Zx * fy = 0.9 6.14 2400 = 13262.4 kgcm

132.624 kgm

Syarat -> > Mu132.624 kgm > 50.16926 kgm

OK

4.4.5 Kontrol Lendutan

f = L = 180 = 0.5360 360

Dari Tabel Profil Baja Didapat :

Ix = 10.42 cm4

= 5 ( 0.098956 0.7 ) 1.05E+09384 2100000 10.42

= 0.499074 < 0.5

OK

Ambil Profil Baja Siku Sama Kaki 45 45 7

4.5 Perencanaan Pelat Bordes4.5.1 Perencanaan Tebal Pelat Bordes

Tebal Pelat Tangga = 7 mmBerat Jenis Baja = 7850 kg/m3Tegangan Leleh Baja = 2500 kg/m2Lebar Pelat Bordes = 1.8 m

4.5.2 Perencanaan Pembebanan Pelat BordesBeban Mati Berat Pelat = 0.007 1.8 7850 = 98.91 kg/m'Alat Penyambung (10 %) = 9.891 kg/m'

qD = 108.801 kg/m'Beban Hidup qL = 500 x 1.8 = 900 kg/m'

= 0.125 108.801 0.36 = 4.896045 kgm

φMn =

φMn =

φMn

4.5.3 Perhitungan MD dan ML

MD

x x

Y max= 5384

(qD+qL)∗l4

EIx

x

x x

M D=18qDl

2

x x

x x

= 0.125 900 0.36 = 40.5 kgm

Mu = 1.4 4.896045 = 6.854463 kgmTidak Menentukan

Mu = 1.2 4.896045 + 1.6 40.5 = 70.67525 kgmMenentukan

4.5.5 Kontrol Momen Lentur

= 0.25 180 0.49 = 22.05 cm3

φ Zx * fy = 0.9 22.05 2400 = 47628 kgcm

476.28 kgm

Syarat -> > Mu476.28 kgm > 70.67525 kgm

OK

4.5.6 Kontrol Lendutan

f = L = 60 = 0.166667360 360

= 0.083333 180 0.343 = 5.145 cm4

Ix = 5.145 cm4

= 5 ( 1.08801 9 ) 1.30E+07384 2100000 5.145

= 0.15756 < 0.166667

OK

ambil Tebal Pelat Bordes = 7 mm

MD

4.5.4 Perhitungan Kombinasi Pembebanan MU

MU = 1.4 MD

MU = 1.2 MD + 1.6 ML

φMn =

φMn =

φMn

M L=18qL l

2

x x

x

x x

x xZx=14bh2

x x

Y max= 5384

(qD+qL)∗l4

EIx

x

Ix=1

12bh3

x x

4.6 Perencanaan Balok Bordes

4.6.1 Perencanaan Balok Bordes dengan Profil I

100 x 100 x 6 x 8

A = 21.9 cm2 tf = 8 mm Zx = 84.184 cm3W = 17.2 kg/m Ix = 383 cm4 Zy = 40.432 cm3a = 100 mm Iy = 134 cm4 h = 64 mm

bf = 100 mm tw = 6 mmiy = 2.47 cm ix = 4.18 cm 84.184

40.4324.6.2 Perencanaan PembebananBeban MatiBerat Pelat = 0.007 0.6 7850 = 32.97 kg/m'Berat Profil I = = 17.2 kg/m'

= 50.17 kg/m'Alat Penyambung ( 10 % ) = 5.017 kg/m'

qD = 55.187 kg/m'

= 0.125 55.187 3.24 = 22.35074 kgm

= 0.5 55.187 3.24 = 89.40294 kg

Beban Hidup qL = 500 x 0.6 = 300 kg/m'.

= 0.125 300 3.24 = 121.5 kgm

= 0.5 300 3.24 = 486 kg

4.6.3 Perhitungan Kombinasi Pembebanan

Mu = 1.4 22.35074 = 31.29103 kgmPu = 1.4 89.40294 = 125.1641 kgm

Tidak menentukan

Mu = 1.2 22.35074 + 1.6 121.5 = 221.2209 kgmPu = 1.2 89.40294 + 1.6 486 = 884.8835 kgm

Menentukan4.6.4 Kontrol Kekuatan Profil4.6.4.1 Penampang Profil fy = 2500 kg/m2

untuk Sayap untuk Badan

MU = 1.4 MD

MU = 1.2 MD + 1.6 ML

x x

M D=18qDL

2

PD=12qDL

M L=18qL L

2

PL=12qL L

x x

x x

x x

x x

xx

xx

xx

b2tf

≤170

√ fyht≤

1680

√ fy

100 170 64 168016 15.81139 6 15.81139

6.25 10.75174 10.66667 106.2525OK OK

Penampang Profil Kompak, maka Mnx = Mpx

4.6.4.2 Kontrol Lateral Buckling

Jarak Baut Pengikat : 250 mm = 25 cm

Lp = 122.9574 cm

Ternyata Lp > Lb maka Mnx = Mpx

Mnx = Mpx = Zx. Fy = 84.184 * 2500 = 2104.6 KgmMny = Zy ( 1 flen ) * fy

== 0.25 2.205 100 2500 = 137812.5 kgcm= 1378.125 kgm

4.6.5 Kontrol Momen Lentur

Zx = 84.184 cm3

φ Zx * fy = 0.9 84.184 2400 = 181837.4 kgcm

1818.374 kgm

Syarat -> > Mu1818.374 kgm > 221.2209 kgm

OK

4.6.6 Kontrol Lendutan

f = L = 180 = 0.5360 360

Ix = 84.184 cm4

= 5 ( 0.55187 3 ) 1.05E+09384 2100000 84.184

= 0.274623 < 0.5

φMn =

φMn =

φMn

b2tf

≤170

√ fyht≤

1680

√ fy

¿ ¿

¿ ¿

Lp=1. 76∗iy√ Efy

(1/4∗tf∗bf 2)∗fyx x x

x x

Y max= 5384

(qD+qL)∗l4

EIx

x

OK

4.7 Perhitungan Balok Induk Tangga

4.7.1 Data - Data Perencanaanh min = I sin α = 28 sin 33.47028 = 15.4352 cm

4.7.2 Perencanaan Balok Induk Dengan Menggunakan Profil WF

250 x 125 x 5 x 8

A = 32.68 cm2 tf = 8 mm Zx = 310.445 cm3W = 25.7 kg/m Ix = 3540 cm4 Zy = 63.7125 cm3a = 250 mm Iy = 255 cm4 h = 210 mm

bf = 125 mm tw = 5 mm r = 12 mmiy = 2.79 cm ix = 10.4 cm 310.445

63.7125

Syarat --> h > hmin25 > 15.4352

OK

4.7.3 Perencanaan Pembebanan

x

420 180

4.7.3.1 Perencanaan Pembebanan Anak TanggaBeban MatiBerat Pelat = 0.004 0.9 7850 = 28.26 kg/m'Berat Profil siku = 4.6 2 0.9 0.28 = 29.57143 kg/m'Berat Sandaran Besi = 15 kg/m'Berat Profil WF = 32.68 / cos 33.47028 = 39.1689 kg/m'

112.0003 kg/m'Alat Penyambung (+ 10 %) = 11.20003 kg/m'

qD1 = 123.2004 kg/m'

Beban HidupqL1 = 500 0.9 = 450 kg/m'

Beban q1 Total = 1.2 qD + 1.6 qL= 1.2 123.2004 + 1.6 450= 867.8404 kg/m'

4.7.3.2 Perencanaan Pembebanan BordesBeban MatiBerat Profil WF = = 25.7 kg/m'

Berat Pelat Bordes = 0.007 0.9 0.6 7850 = 29.673 kgBerat Profil I = 17.2 0.9 = 15.48 kg

45.153 kgAlat Penyambung (+ 10 %) = 4.5153 kg

Pd = 49.6683 kg

Beban HidupqL2 = 500 kg/m2 PL2 = 500 0.6 0.9

= 270 kgjadi q2 total = 1.2 qD + 1.6 qL

= 1.2 25.7 + 1.6 500= 830.84 kg/m'

jadi P total = 1.2 PD + 1.6 PL= 1.2 49.6683 + 1.6 270= 491.602 kg

x xx x

x

x x

x xxx

x x

x x

x x

4.7.4 Perhitungan Gaya - Gaya pada TanggaP P

420 180Lab = 4.2 mLbc = 1.8 m

433.920 17.640 491.602 5.4 4.800 1495.512 0.900 4.2 RC6

Rc = 3382.634 kg

Rva = 867.84 4.20 830.84 1.80 983.20 3382.63Rva = 2741.012 kg

B C

+

+

A

3857.8968 kgm

4328.644573 kgm

Σ Ma = 0

Σ V = 0

+ + + +( () )

(12q 1 l

ab2 )+( p(3 lab+1 .5 lbc ))+(q2 lcb(

12lcb+lab))−(Rc ( lab+lbc ))=0

Rva=q1 lab+q2 lbc+3 P−Rc

+ + ¿( () )

zz 2741.011786

RAh = 0

Bidang M

Pers : Mx1 = RVA X1 - 0.5 q1

Mx1 = 2741.0118 X1 - 433.92022

dMx1

= 0 867.84044 X1 = 2741.0118

dX1 X1 = 3.1584283 m

X1 = 0 m MA = 0 Kgm

Xmax = 3.1584283 m Mmax= 4328.6446 Kgm tangga

X1 = 4.2 m MB = 3857.8968 Kgm tangga

B C

5.0349327

A a = 33.4703

Rav cos a

Rav sin a 4.2 m 1.8 m

Rav

X1 X2

-754.17076 kg

-2497.7504 -

2286.4754 kg -3382.634 kg

Bidang D

Permisalan gayaDari kiri : searah jarum jam gaya dianggap positif

X= 0 m

DA = Rva cos a

= 2741.0118 cos 33.470= 2286.4754 kg

X= 4.2 m

Dbkiri = Rva cos a - q1 LAB

= -754.17076 kg

cos a

Dbkanan = P LBC - RC

= 491.60196 1.8 - 3382.634

= -2497.7504 kg

X= 6 m

Dc = - RC

= -3382.634 kg 1053.4693

X12

X12

x

x

x

x

x

x

x x

x

897.72598 kg

+

-

-1511.6799 kg

Bidang N

NA = -RVA sin a

= -2741.0118 sin 33.470= -1511.6799 kg

NBkiri = -RVA sin a + q1 L1

= 897.72598 kg

sin a

NBkanan -C = 0

4.7.5 Kontrol Kekuatan Profil4.7.5.1 Penampang Profil fy = 2500 kg/m2

untuk Sayap untuk Badan

125 170 210 168016 15.81139 5 15.81139

7.8125 10.75174 42 106.2525OK OK

Penampang Profil Kompak, maka Mnx = Mpx

4.7.5.1 Kontrol Lateral Buckling

Jarak Baut Pengikat : 250 mm = 25 cm

Lp = 138.8871 cm

Ternyata Lp > Lb maka Mnx = Mpx

b2tf

≤170

√ fyht≤

1680

√ fy

¿ ¿

¿ ¿

Lp=1. 76∗iy√ Efy

Mnx = Mpx = Zx. Fy = 310.445 * 2500 = 7761.125 KgmMny = Zy ( 1 flen ) * fy

== 0.25 2.56 0.64 2500 = 1024 kgcm= 10.24 kgm

4.7.5 Kontrol Momen Lentur

Zx = 310.445 cm3

φ Zx * fy = 0.9 310.445 2400 = 670561.2 kgcm

6705.612 kgm

Syarat -> > Mu6705.612 kgm > 4328.645 kgm

OK

4.7.6 Kontrol Lendutan

f = L = 600 = 1.666667360 360

Ix = 3540 cm4

= 5 ( 1.489004 4.5 ) 1.30E+11384 2100000 3540

= 1.359489 < 1.666667

OKProfil yang Dipakai untuk Balok induk AdalahProfil WF 250 x 125 x 5 x 8

φMn =

φMn =

φMn

(1/4∗tf∗bf 2)∗fyx x x

x x

Y max= 5384

(qD+qL)∗l4

EIx

x

275

60250

540

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5. Pembebanan

5.1 Perencanaan Beban Atap

Beban MatiBerat Gording = 9.3 5.2 = 48.36 kg/mBerat seng Gelombang = 4.16 5.2 / cos 20 = 23.02029 kg/m

= 71.380 kg/m= 7.138029 kg/m

Pm = 78.51832 kg/mBerat Profil Kuda - Kuda = 65.7 /cos 20 = 69.91648 kg/m

148.435 kg/mBeban Hidup

Ph = 20 5.2 = 104 kg/m

178.1218 + 166.4 = 344.5218 kg

5.2. Perencanaan Beban Angin (Gudang Tertutup)

W = 50 kg/m2

Beban Tekan Atap = 0 50 5.2 = 0 kg/m

Beban Sedot Atap = -0.4 50 5.2 = -104 kg/m

BebanTiup Kolom = 0.9 50 5.2 = 234 kg/m

Beban Sedot kolom = -0.4 50 5.2 = -104 kg/m

Alat Pengikat dll (+10 %)

Pmtot

P Ultimate = 1.2 PD+ 1.6 PL =

(- 0,02a - 0,4) - 0,4

0,9 - 0,4

x

x

x x

x x

x x

x x

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5.3 Perencanaan Beban Akibat Plat Lantai, balok Memanjang dan Melintang

780

640

780

5.2 m 5.2 m

A ( Balok Induk Melintang )500 x 200 x 9 x 14

A = 0 cm2 tf = 14 mm Zx = 1862.064 cm3W = 79.5 kg/m Ix = 0 cm4 Zy = 279.433 cm3a = 500 mm Iy = 0 cm4 h = 0 mm

bf = 200 mm tw = 9 mm r = 0 mmiy = 0 cm ix = 0 cm 1862.064

279.433B ( Balok Anak )

250 x 250 x 8 x 13

A = 82.06 cm2 tf = 13 mmW = 64.4 kg/m Ix = 9930 cm4a = 248 mm Iy = 3350 cm4 h = 192 mm

bf = 249 mm tw = 8 mm r = 16 mmiy = 6.29 cm ix = 10.8 cm Zx= 859.263 cm3

Zy= 405.6625 cm3C ( Balok Induk Memanjang)

500 x 200 x 9 x 14

A = 0 cm2 tf = 14 mm Zx = 1862.064 cm3W = 79.5 kg/m Ix = 0 cm4 Zy = 288.991 cm3a = 500 mm Iy = 0 cm4 h = 472 mm

bf = 200 mm tw = 9 mm r = 0 mmiy = 0 cm ix = 0 cm 1862.064

288.991

II

II

III

IV

P1

P1

P2

P2

P3

P3

P3

P3

P4

P4

P5 P6 P6 P6

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Beban Mati- Bondex = 10.1 kg/m2- Beton = 0.1 x 2400 = 240 kg/m2- Beban Finishing = 90 kg/m2

qM = 340.1 kg/m2

Beban Hidup

qL = 400 kg/m2

5.3.1 Perencanaan Pembebanan Portal Melintang

5.3.1.1 Beban P1Beban MatiBerat Pelat = 340.1 6.4/3 5.2 = 3772.843 kgBalok induk melintang = 79.5 6.4/3 = 169.6 kgBalok anak = 64.4 5.2 = 334.88 kg

Pm1 = 4277.323 kg

Beban HidupPh1 = 400 6.4/3 5.2 = 4437.333 kg

5.3.1.2 Beban P2Beban MatiBerat Pelat = 340.1 (6.4/6)+(7.8/6) 5.2 = 4185.497 kgBalok induk memanjang = 79.5 5.2 = 413.4 kgBalok induk melintang = 79.5 7.8/3 = 206.7 kg

Pm2 = 4805.597 kg

Beban HidupPh2 = 400 (6.4/6)+(7.8/6) 5.2 = 4922.667 kg

5.3.1.3 Beban P3Beban MatiBerat Pelat = 340.1 7.8/3 5.2 = 4598.152 kgBalok induk melintang = 79.5 7.8/3 = 206.7 kgBalok anak = 64.4 5.2 = 334.88 kg

Pm1 = 5139.732 kg

Beban HidupPh3 = 400 7.8/3 5.2 = 5408 kg

5.3.1.4 Beban P4Beban MatiBerat Pelat = 340.1 7.8/6 5.2 = 2299.076 kgBalok induk memanjang = 79.5 5.2 = 413.4 kgBerat Dinding = 250 10.5 5.2 = 13650 kgBalok induk melintang = 7.8/3 79.5 = 206.7 kg

Pm4 = 16569.18 kg

Beban HidupPh4 = 400 7.8/6 5.2 = 2704 kg

x xx

x

x x

x x

x xx

x

x x

x x

x xx

x

x x

xx

x x

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5.3.2 Perencanaan Pembebanan Portal Memanjang

5.3.2.1 Beban P5 Beban Mati

Berat Pelat 340.1 7.8/3 2.6 = 2299.076 kgBalok induk melintang = 79.5 7.8/3 = 206.7 kgBalok induk memanjang = 79.5 2.6 = 206.7 kg

Pm5 = 2712.476 kg

Beban HidupPh5 = 400 7.8/3 2.6 = 2704 kg

5.3.2.2 Beban P6 Beban Mati

- Berat Pelat = 340.1 5.2 7.8/3 = 4598.152 kg- Balok induk memanjang = 79.5 5.2 = 413.4 kg- Balok induk melintang = 79.5 7.8/3 = 206.7 kg

Pm5 = 5218.252 kg Beban Hidup

Ph5 = 400 5.2 7.8/3 = 5408 kg

5.3.3 Beban Portal - Portal MelintangP1 Pm1 = 4277.323 kg

Ph1 = 4437.333 kg

P2 Pm3 = 4805.597 kgPh3 = 4922.667 kg

P3 Pm2 = 5139.732 kgPh2 = 5408 kg

P4 Pm4 = 16569.176 kgPh4 = 2704 kg

5.3.4 Beban - beban Portal MemanjangP5 Pm5 = 2712.476 kg

Ph5 = 2704 kg

P6 Pm6 = 5218.252 kgPh6 = 5408 kg

5.4 Perencanaan Beban Gempa ( Arah X )

2.55

5.25 10.5

5.5

F1

F2

x x

x x

x xxx

x xxx

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Page 61

7.8 6.4 7.8

Data Gempa:- Zone Gempa = 3- tanah lunak

5.2

5.25.2

7.8 6.4 7.8

22

Kolom400 x 400 x 16 x 24

A = 255 cm2 tf = 24 mm Zx = 4105.216 cm3W = 200 kg/m Ix = 0 cm4 Zy = 1939.456 cm3a = 400 mm Iy = 0 cm4 h = 352 mm

bf = 400 mm tw = 16 mm r = 0 mmiy = 0 cm ix = 0 cm 4105.216

1939.456

5.4.1 Perencanaan Beban Lantai (W1) 5.4.1.1 Beban Mati

Berat Plat = 340.1 22 5.2 = 38907.44 KgBalok Induk Memanjang = 4 79.5 5.2 = 1653.6 Kgbalok anak memanjang 6 64.4 5.2 = 2009.28Balok Induk Melintang = 1 79.5 22 = 1749 KgBerat Dinding = 500 5.2 5.25 = 13650 KgKolom = 200 5.25 2 = 2100 Kg

200 5.25 1 = 1050 Kg= 61119.32 Kg

F1

x xx

xxxx

xx

xxx

x

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Page 62

5.4.1.2 Beban Hidupbeban hidup lantai = 400 22 5.2

== 45760 Kg

= 61119.32 + 45760= 106879 Kg

5.4.2 Perencanaan Beban Atap ( W2)5.4.2.1 Beban MatiBerat Atap = 78.518 22 = 1727.403 KgBalok Kuda-kuda = 65.7 22 = 1538.163 KgKolom = 200 2.5 2 = 1000 KgBerat Dinding = 500 2.5 5.2 = 507.7 Kg

= 4773.266 Kg

5.4.2.2 Beban HidupPh1 = 20 5.2 22 = 2288 kg

= 4773.266 + 2288= 7061.266 Kg

5.4.3 Berat Total Wt

= W1 + W2= 106879 + 7061.266= 113941 Kg

5.4.4 Perencanaan Gaya Gempa

T = ( 0.0853 x H^(3/4) )= ( 0.0853 x 10.5= 0.4976

Tanah LunakC = 0.75 I = 1

R = 4.5 (SRPMB)

V = (C x I x Wt )/R = 0.75 1 113940.6 4.5= 18990.1 Kg

Lantai = W1 x h1 = 106879.3 x 5.5= 587836.3 Kgm

Atap = W2 x H = 7061.266 x 10.5= 74143.29 Kgm

Σ Wi Hi = 661979.5 Kgm

Beban Lantai ( W1tot)

Beban Atap ( W2tot)

Berat Total ( Wtot )

(SNI 03-1726-2002 zona gempa 3)

Cosαxx

x xx x

x x

x x

xx

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Page 63

F1 = W1 h1 x VS Wi hi

= 587836.26 x 18990.1661979.54939674

= 16863 Kg

F2 = W2 h2 x VS Wi hi

= 74143.289396739 x 18990.1661979.54939674

= 2126.9 Kg5.4.5 Gambar Perencanaan Beban Akibat Gempa ( Arah X )

F22127 kg

5

F116863 kg

5.5

7.8 6.4 6

5.5 Perencanaan Beban Gempa ( Arah Y )

Data Gempa:- Zone Gempa = 3- tanah lunak

Balok Kuda-Kuda200 x 200 x 10 x 16

A = 83.7 cm2 tf = 16 mm Zx = 659.36 cm3W = 65.7 kg/m Ix = 0 cm4 Zy = 323.4 cm3a = 200 mm Iy = 0 cm4 h = 168 mm

bf = 200 mm tw = 10 mm r = 0 mmiy = 0 cm ix = 0 cm

Kolom400 x 400 x 16 x 24

A = 255 cm2 tf = 24 mm Zx = 4105.216 cm3W = 200 kg/m Ix = 0 cm4 Zy = 1939.456 cm3a = 400 mm Iy = 0 cm4 h = 352 mm

bf = 400 mm tw = 16 mm r = 0 mmiy = 0 cm ix = 0 cm

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5.5.1 Perencanaan Beban Lantai (W1) 5.5.1.1 Beban Mati

Berat Plat = 340.1 7.8 5.2 = 13794.46 KgBalok Induk Memanjang = 79.5 5.2 = 413.4 KgBalok Induk Melintang = 79.5 22 = 1749 KgBerat Dinding = 500 7.8 5.25 = 20475 KgKolom = 200 5.25 2 = 2100 Kg

200 5.25 4.5 = 4725 Kg= 43256.86 Kg

5.5.1.2 Beban HidupBeban Hidup Merata = 400 46.8 7.8 Kg

= 146016 Kg

= 43256.86 + 146016= 189273 Kg

5..2 Perencanaan Beban Atap ( W2)5.5.2.1 Beban MatiBerat Atap = 78.518 7.8 46.8 = 28662.33 KgBalok Kuda-kuda = 65.7 7.8 = 545.3485 KgKolom = 200 2.5 2 = 1000 KgBerat Dinding = 500 7.8 2.5 = 510.3 Kg

= 30717.98 Kg

5.5.2.2 Beban HidupPh1 = 20 7.8 46.8 = 7300.8 kg

= 30718 + 7300.8= 38018.78 Kg

5.4.3 Berat Total Wt

= W1 + W2= 189273 + 38018.78= 227292 Kg

5.4.4 Perencanaan Gaya Gempa

T = ( 0.0853 x H^(3/4) )= ( 0.0853 x 10.5= 0.4976

Tanah LunakC = 0.75 I = 1R = 4.5

V = (C x I x Wt )/R = 0.75 1 227291.6 4.5

Beban Lantai ( W1tot)

Beban Atap ( W2tot)

Berat Total ( Wtot )

x xxxxxx

xx

Cosαx xx xx x

x x

x x

x x

x x

x x

01+000STRUKTUR BANGUNAN BAJA

Page 65

= 37881.94 Kg

Lantai = W1 x h1 = 189272.9 x 5.5= 1041001 Kgm

Atap = W2 x H = 38018.78 x 10.5= 399197.2 Kgm

Σ Wi Hi = 1440198 Kgm

F1 = W1 h1 x VS Wi hi

= 1041000.708 x 37881.941440197.8680145

= 27382 Kg

F2 = W2 h2 x VS Wi hi

= 399197.16001453 x 37881.941440197.8680145

= 10500 Kg

5.4.5 Gambar Perencanaan Beban Akibat Gempa ( Arah Y )

F210500.198 kg

F127381.741 kg

Kombinasi Pembebanan

* Beban Mati + Beban Hidup

P4 P3 P2 P1 P2 P3 P4

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Pm = 78.51832 kgPh = 104 kg

7.8 6.4 6

Pm1 = 4277.32 kg Pm2 = 4805.597 kgPh1 = 4437.33 kg Ph2 = 4922.667 kg

Pm3 = 5139.73 kg Pm4 = 16569.18 kgPh3 = 5408 kg Ph4 = 2704 kg

* Beban Mati + Beban Hidup + Beban Angin

q = 0 kg/m q = -104 kg/m

q = 234 kg/m q = -104 kg/m7.8 6.4 7.8

200

P4 P3 P2 P1 P2 P3 P4

P4 P3 P2 P1 P2 P3 P4

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* Beban Gempa

2126.9363 kg

16863.161 kg

7.8 6.4 7.8

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6 Perencanaan Dimensi Struktur Utama6.1 Kontrol Dimensi Kuda -Kuda

Dari Hasil Sap 2000 Diperoleh Mmax dan N max pada Frame 9:

( U - G ) Beban Ultimate - Beban GempaMutx = -3679.739 KgmMuty = 0 Kgm

Nu = -2677.929 KgVu = -1184.348 kg

Ma = -3679.4 KgmMb = -466.19 KgmMs = 301.67 Kgm

6.1.1 Profil Baja yang Didapatkan Di SAP 2000 8.2.3 :

200 x 200 x 8 x 12

A = 63.53 cm2 tf = 12 mm Zx = 513.152 cm3W = 49.9 kg/m Ix = 4720 cm4 Zy = 216.32 cm3a = 200 mm Iy = 1600 cm4 h = 150 mm

bf = 200 mm tw = 8 mm Sx = 37.5 mmiy = 5.02 cm ix = 8.62 cm 513.152

216.32Mutu Baja = BJ 37

fu = 3700 kg/cm2fy = 2400 kg/cm2

6.1.2 Kontrol Lendutan

f ijin = L = 993.2 = 2.758889 cm360 360

5 98.64 301.67 0.1 -3679.4 -466.1948 200 4720

0.678128 cm

f <OK

6.1.3 Kontrol Tekuk

untuk arah x :

f =

f =

f ijin

f= 5 L2

48EI(Ms−0 .1(Ma−Mb ))

kcx = 1.2 (jepit-rol tanpa putaran sudut) L = 993.2 cmLkx = 1191.84 cm

= 138.2645 cm (MENENTUKAN)

Ncrbx = 9.869022 2000000 63.5319117.07

Ncrbx = 65593.62 kg

untuk arah y :kcy = 0.8 (jepit-Sendi) L = 110 cmLky = 88 cm

= 17.52988 cm

Ncrby = 9.869022 2000000 63.53307.2967

Ncrby = 4080610 kg

138.2645 24003.1415 2000000

1.524629

> 1.2

2.905618

Pn = Ag fy = 63.53 * 2400 = 52474.9 kgw 2.905618

2677.929 = 0.060038 < 0.20.85 52474.9

Pakai Rumus =

X Batang Dianggap Tidak Bergoyang Maka :

Maka dipakai lx karena lx > ly

λc =

λc =

λc

ω = 1.25 λc2 ω =

λx=Lkxix

λc= λxπ √ fyE √

Puϕ cPn

=x

Pu2ϕcPn

+Mux

ϕbMnx+

Muyϕ bMny

≤1

λy=Lkyiy

Ncrbx=π2EAλx2 x x

Ncrby=π2EAλy2

x x

Sbx=Cmx

1−( NuNcrbx

)≥1

Cmx = 1Sbx = 1

1 - 2677.92965593.62

Sbx = 1.042564 < 1

Sbx = 1.042564

Mux = Mutx * SbxMux = -3679.739 1.042564 = -3836.363 kgm

Y Batang Dianggap Tidak Bergoyang Maka :

Cmy = 1Sby = 1

1 - 2677.92965593.62

Sby = 1.042564 < 1

Sby = 1.042564

Muy = Muty * SbyMuy = 0 1.042564 = 0 kgm

6.1.4 Menentukan Mnx

6.1.4.1 Kontrol Penampang profiluntuk Sayap untuk Badan

Kgmb

2tf≤

170

√ fyht≤

1680

√ fy

Sby=Cmy

1−( NuNcrby

)≥1

x

x

200 170 150 168024 15.49193 8 15.49193

8.333333 10.97345 18.75 108.4435OK OK

Penampang Profil Kompak, maka Mnx = Mpx

Lateral Bracing Lb = 110 cm

Lp = 255.0503 cm

Ternyata Lp > Lb maka Mnx = Mpx

Mnx = Mpx = Zx. Fy = 513.152 * 2400 = 12315.65 kgmMny = Zy ( 1 flen ) * fy

== 0.25 1.2 400 2400 = 288000 kgcm= 2880 kgm

6.1.5 Persamaan Interaksi

2677.929 + 3836.36256538448 + 01.7 52474.9 0.9 12315.65 0.9 2880

0.030019161210477 + 0.34611455870202 + 0

0.376134 < 1

OK

6.1.6 Kontrol Kuat Rencana Geser

150 < 11002400 15.49193

0.0625 < 71.00469Plastis

ht≤

1680

√ fy

¿ ¿

¿ ¿

Lp=1. 76∗iy√ Efy

(1/4∗tf∗bf 2)∗fyx x x

Pu2ϕcPn

+Mux

ϕbMnx+

Muyϕ bMny

≤1

x x x

htw

≤1100

√ fy

Vn = 0.6 fy Aw= 0.6 2400 0.8 20= 23040 Kg

Vu < ФVn1184.348 < 0.9 230401184.348 < 20736

OK

6.2 Kontrol Dimensi KolomDari Hasil Sap 2000 Diperoleh Mmax dan N max pada Frame 13:

( U - G ) Beban Ultimate - Beban GempaSbx --> Mutx = 1290.96 Kgm

Nu = -27466 KgVu = -309.42 kg

Max = -410.86 KgmMbx = 440.05 KgmMsx = 1290.96 Kgm

Sby --> Muty = 6216.43 KgmNu = -26505.54 KgVu = 1624.88 kg

Max = 6216.43 KgmMbx = 1748.02 KgmMsx = -27210.39 Kgm

6.2.1 Profil Baja yang Didapatkan Di SAP 2000 8.2.3 :

400 x 400 x 10 x 15

A = 218.7 cm2 tf = 15 mm Zx = 2652.25 cm3W = 172 kg/m Ix = 66600 cm4 Zy = 1170 cm3a = 400 mm Iy = 22400 cm4 h = 334 mm

bf = 400 mm tw = 10 mm Sx = 1360 cm3iy = 9.75 cm ix = 17.5 cm 2652.25

1170Mutu Baja = BJ 41

fu = 4100 kg/cm2 fr = 700 kg/cm2fy = 2500 kg/cm2

6.2.1 Kontrol Lendutan6.2.1.1 Kontrol Lendutan Arah X

f ijin = L = 550 = 1.527778 cm360 360

5 30.25 410.86 0.1 410.86 1748.0248 200 66600

0.012883 cm

f <OK

6.2.1.2 Kontrol Lendutan Arah Yf ijin = L = 183.3333 = 0.509259 cm

360 360

5 3.36 1290.96 0.1 -410.86 440.0548 200 22400

0.010754 cm

f <OK

f =

f =

f ijin

f =

f =

f ijin

f= 5 L2

48EI(Ms−0 .1(Ma−Mb ))

f= 5 L2

48EI(Ms−0 .1(Ma−Mb ))

6.2.3 Kontrol Tekuk

untuk arah x :kcx = 0.65 (jepit-Sendi) L = 550 cmLkx = 357.5 cm

= 20.42857 cm

Ncrbx = 9.869022 2000000 218.7417.3265

Ncrbx = 10343724 kg

untuk arah y :kcy = 0.8 (jepit-Sendi) L = 183.3333 cmLky = 146.6667 cm

= 15.04274 cm

Ncrby = 9.869022 2000000 218.7226.2839

Ncrby = 19076526 kg

15.04274 25003.1415 2000000

0.169295

0.25 > <

ω = 1 11

Pn = Ag fy = 218.7 * 2400 = 524880 kgw 1

27466 = 0.061563 < 0.20.85 524880

Pakai Rumus =

Maka dipakai ly karena ly > lx

λc =

λc =

λc

ω =

λx=Lkxix

λc= λyπ √ fyE √

Puϕ cPn

=x

Pu2ϕcPn

+Mux

ϕbMnx+

Muyϕ bMny

≤1

λy=Lkyiy

Ncrbx=π2EAλx2 x x

Ncrby=π2EAλy2

x x

X Batang Dianggap Tidak Bergoyang Maka :

Cmx = 1Sbx = 1

1 - 2746610343724

Sbx = 1.002662 < 1

Sbx = 1.002662

Mux = Mutx * SbxMux = 1290.96 1.002662 = 1294.397 kgm

Y Batang Dianggap Tidak Bergoyang Maka :

Cmy = 1Sby = 1

1 - 26505.5410343724

Sby = 1.002569 < 1

Sby = 1.002569

Muy = Muty * SbyMuy = 1290.96 1.002569 = 1294.277 kgm

6.2.4 Menentukan Mnx

6.2.4.1 Kontrol Penampang profiluntuk Sayap untuk Badan

Kgm

400 170 334 168030 15.81139 10 15.81139

13.33333 10.75174 33.4 106.2525OK OK

Penampang Profil Kompak, maka Mnx = Mpx

b2tf

≤170

√ fyht≤

1680

√ fy

¿ ¿

¿ ¿

Sby=Cmy

1−( NuNcrby

)≥1

x

Sbx=Cmx

1−( NuNcrbx

)≥1

x

Lateral Buckling Lb = 500 cm

Lp = 485.3581 cm

Lr = iy . [x1/(fy-fr)] . Sqrt{1+[sqrt(1+(x2 . fl2))]}

76.5 cm4

Iw = Iy.((h^2)/4) = 4548600 cm6

#VALUE! kg/cm2

x2 = 4.[(S/GJ)^] = 4.01E-07 cm2/kg

Lr = 1372.41 cm

Lp < Lb < Lr (INELASTIC BUCKLING)

Mnx = Cb.[Mr+[(Mp-Mr).((Lr-Lb)/(Lr-Lp))]]

Cb = 12,5 Mmax

2,5Mmax + 3Ma + 4Mb + 3Mc

Cb = 2.251

Mr = Sx(fy-fr) = #VALUE! kgcmMp = Zx . fy = 6630625 kgcm < 1,5MyMy = Sx . fy = #VALUE! kgcm

1,5 . My = #VALUE! kgcm > Mp

Mnx = #VALUE! kgcm > Mp

Mnx=Mp= 6630625 kgcm

Mny = Zy * fy= 1170 2500= 2925000 kgcm= 29250 kgm

6.2.5 Persamaan Interaksi

J = S(1/3).b.(t^3) =

x1 = [π/s]*[sqrt((EGJA)/2)] =

Lp=1. 76∗iy√ Efy

x

Pu2ϕcPn

+Mux

ϕbMnx+

Muyϕ bMny

≤1

27466 + 1294.39705133537 + 1294.276552131.7 524880 0.9 66306.25 0.9 29250

0.030781265409685 + 0.021690548717929 + 0.049165301125546

0.101637 < 1

OK

6.2.6 Kontrol Kuat Rencana Geser

334 < 11002500 15.81139

0.1336 < 69.57011Plastis

Vn = 0.6 fy Aw= 0.6 2400 1 40= 57600 Kg

Vu < ФVn#VALUE! < 0.9 57600#VALUE! < 51840

#VALUE!

6.3 Kontrol Dimensi Balok MelintangDari Hasil Sap 2000 Diperoleh Mmax dan N max pada Frame 5 :

( U - G ) Beban Ultimate - Beban GempaMutx = -15837.38 KgmMuty = 0 Kgm

x x x

htw

≤1100

√ fy

Nu = -8457.088 KgVu = 9218.689 kg

Ma = 970.4 KgmMb = 15837.88 KgmMs = 11818.69 Kgm

6.3.1 Profil Baja yang Didapatkan Di SAP 2000 8.2.3 :

340 x 250 x 9 x 14

A = 101.5 cm2 tf = 14 mm Zx = 1360.024 cm3W = 79.7 kg/m Ix = 21700 cm4 Zy = 439.2 cm3a = 340 mm Iy = 3650 cm4 h = 276 mm

bf = 250 mm tw = 9 mm Sx = 1280 cm3iy = 6 cm ix = 14.6 cm 1360.024

439.2Mutu Baja = BJ 37

fu = 3700 kg/cm2 fr = 700 kg/cm2fy = 2400 kg/cm2

6.3.1 Kontrol Lendutan

f ijin = L = 500 = 1.388889 cm360 360

5 25.00 11818.69 0.1 970.4 15837.8848 200 21700

0.798377 cm

f <OK

6.3.3 Kontrol Tekuk

untuk arah x :kcx = 1 (Sendi-Sendi) L = 600 cmLkx = 600 cm

= 41.09589

Ncrbx = 9.869022 2000000 101.51688.872

Ncrbx = 1186242 kg

untuk arah y :

f =

f =

f ijin

f= 5 L2

48EI(Ms−0 .1(Ma−Mb ))

λx=Lkxix

Ncrbx=π2EAλx2 x x

kcy = 1 (Sendi-Sendi) L = 600 cmLky = 600 cm

= 100 (MENENTUKAN)

Ncrby = 9.869022 2000000 101.510000

Ncrby = 200341.2 kg

100 24003.1415 2000000

1.10269

0.25 < < 1.2

ω = 1.43 1.5356571,67-0,67λc

Pn = Ag fy = 101.5 * 2400 = 158629.2 kgw 1.535657

8457.088 = 0.062722 < 0.20.85 158629.2

Pakai Rumus =

X Batang Dianggap Tidak Bergoyang Maka :

Cmx = 1Sbx = 1

1 - 8457.0881186242

Sbx = 1.007181 < 1

Sbx = 1.007181

Maka dipakai ly karena ly > lx

λc =

λc =

λc

ω =

λc= λyπ √ fyE √

Puϕ cPn

=x

Pu2ϕcPn

+Mux

ϕbMnx+

Muyϕ bMny

≤1

λy=Lkyiy

Ncrby=π2EAλy2

x x

Sbx=Cmx

1−( NuNcrbx

)≥1

Mux = Mutx * SbxMux = 15837.38 1.007181 = 15951.1 kgm

Y Batang Dianggap Tidak Bergoyang Maka :

Cmy = 1Sby = 1

1 - 8457.0881186242

Sby = 1.007181 < 1

Sby = 1.007181

Muy = Muty * SbyMuy = 0 1.007181 = 0 kgm

6.3.4 Menentukan Mnx

6.3.4.1 Kontrol Penampang profiluntuk Sayap untuk Badan

Kgm

250 170 276 168028 15.49193 9 15.49193

8.928571 10.97345 30.66667 108.4435OK OK

Penampang Profil Kompak, maka Mnx = Mpx

Lateral Buckling Lb = 600 cm

Lp = 304.8409 cm

Lr = iy . [x1/(fy-fr)] . Sqrt{1+[sqrt(1+(x2 . fl2))]}

53.315 cm4J = S(1/3).b.(t^3) =

b2tf

≤170

√ fyht≤

1680

√ fy

¿ ¿

¿ ¿

Lp=1. 76∗iy√ Efy

Sby=Cmy

1−( NuNcrby

)≥1

x

x

Iw = Iy.((h^2)/4) = 969768.5 cm6

161407 kg/cm2

x2 = 4.[(S/GJ)^] = 3.60E-09 cm2/kg

Lr = 806.682 cm

Lp < Lb < Lr (INELASTIC BUCKLING)

Mnx = Cb.[Mr+[(Mp-Mr).((Lr-Lb)/(Lr-Lp))]]

Cb = 12,5 Mmax

2,5Mmax + 3Ma + 4Mb + 3Mc

Cb = 1.813

Mr = Sx(fy-fr) = 2176000 kgcmMp = Zx . fy = 3264058 kgcm < 1,5MyMy = Sx . fy = 3072000 kgcm

1,5 . My = 4608000 kgcm > Mp

Mnx = 4757518 kgcm > Mp

Mnx=Mp= 3264058 kgcm

Mny = Zy ( 1 flen ) * fy== 0.25 1.4 625 2400 = 525000 kgcm= 5250 kgm

6.3.5 Persamaan Interaksi

8457.088 + 15951.1003321742 + 01.7 158629.2 0.9 32640.58 0.9 5250

0.031360926501704 + 0.542988114349665 + 0

0.574349 < 1

OK

6.3.6 Kontrol Kuat Rencana Geser

x1 = [π/s]*[sqrt((EGJA)/2)] =

(1/4∗tf∗bf 2)∗fyx x x

Pu2ϕcPn

+Mux

ϕbMnx+

Muyϕ bMny

≤1

x x x

276 < 11002400 15.49193

0.115 < 71.00469Plastis

Vn = 0.6 fy Aw= 0.6 2400 0.9 34= 44064 Kg

Vu < ФVn9218.689 < 0.9 440649218.689 < 39657.6

OK

6.3 Kontrol Dimensi Balok MemanjangDari Hasil Sap 2000 Diperoleh Mmax dan N max pada Frame 5 :

( U - G ) Beban Ultimate - Beban GempaMutx = 6219 KgmMuty = 0 Kgm

Nu = -28073 KgVu = -2663 kg

Ma = 6219 KgmMb = 4435 KgmMs = 891 Kgm

6.3.1 Profil Baja yang Didapatkan Di SAP 2000 8.2.3 :

300 x 200 x 8 x 12

htw

≤1100

√ fy

A = 72.38 cm2 tf = 12 mm Zx = 843.552 cm3W = 56.8 kg/m Ix = 11300 cm4 Zy = 248.32 cm3a = 300 mm Iy = 1600 cm4 h = 240 mm

bf = 200 mm tw = 8 mm Sx = 771 cm3iy = 4.71 cm ix = 12.5 cm 843.552

248.32Mutu Baja = BJ 37

fu = 3700 kg/cm2 fr = 700 kg/cm2fy = 2400 kg/cm2

6.3.1 Kontrol Lendutan

f ijin = L = 500 = 1.388889 cm360 360

5 25.00 891 0.1 6219 443548 200 11300

0.082112 cm

f <OK

6.3.3 Kontrol Tekuk

untuk arah x :kcx = 1 (Sendi-Sendi) L = 600 cmLkx = 600 cm

= 48

Ncrbx = 9.869022 2000000 72.382304

Ncrbx = 620069.3 kg

untuk arah y :kcy = 1 (Sendi-Sendi) L = 600 cmLky = 600 cm

= 127.3885 (MENENTUKAN)

Ncrby = 9.869022 2000000 72.3816227.84

Ncrby = 88036.35 kg

f =

f =

f ijin

f= 5 L2

48EI(Ms−0 .1(Ma−Mb ))

λx=Lkxix

λy=Lkyiy

Ncrbx=π2EAλx2 x x

Ncrby=π2EAλy2

x x

127.3885 24003.1415 2000000

1.404701

0.25 < < 1.2

ω = 1.43 1.9619941,67-0,67λc

Pn = Ag fy = 72.38 * 2400 = 88538.49 kgw 1.961994

28073 = 0.373025 < 0.20.85 88538.49

Pakai Rumus =

X Batang Dianggap Tidak Bergoyang Maka :

Cmx = 1Sbx = 1

1 - 28073620069.3

Sbx = 1.047421 < 1

Sbx = 1.047421

Mux = Mutx * SbxMux = -6219 1.047421 = -6513.911 kgm

Y Batang Dianggap Tidak Bergoyang Maka :

Maka dipakai ly karena ly > lx

λc =

λc =

λc

ω =

λc= λyπ √ fyE √

Puϕ cPn

=x

Pu2ϕcPn

+Mux

ϕbMnx+

Muyϕ bMny

≤1

Sby=Cmy

1−( NuNcrby

)≥1

Sbx=Cmx

1−( NuNcrbx

)≥1

x

Cmy = 1Sby = 1

1 - 28073620069.3

Sby = 1.047421 < 1

Sby = 1.047421

Muy = Muty * SbyMuy = 0 1.047421 = 0 kgm

6.3.4 Menentukan Mnx

6.3.4.1 Kontrol Penampang profiluntuk Sayap untuk Badan

Kgm

200 170 240 168024 15.49193 8 15.49193

8.333333 10.97345 30 108.4435OK OK

Penampang Profil Kompak, maka Mnx = Mpx

Lateral Buckling Lb = 600 cm

Lp = 239.3001 cm

Lr = iy . [x1/(fy-fr)] . Sqrt{1+[sqrt(1+(x2 . fl2))]}

53.315 cm4

Iw = Iy.((h^2)/4) = 969768.5 cm6

226284.2 kg/cm2

x2 = 4.[(S/GJ)^] = 3.60E-09 cm2/kg

Lr = 806.682 cm

Lp < Lb < Lr (INELASTIC BUCKLING)

Mnx = Cb.[Mr+[(Mp-Mr).((Lr-Lb)/(Lr-Lp))]]

J = S(1/3).b.(t^3) =

x1 = [π/s]*[sqrt((EGJA)/2)] =

b2tf

≤170

√ fyht≤

1680

√ fy

¿ ¿

¿ ¿

Lp=1. 76∗iy√ Efy

x

Cb = 12,5 Mmax

2,5Mmax + 3Ma + 4Mb + 3Mc

Cb = 1.813

Mr = Sx(fy-fr) = 1310700 kgcmMp = Zx . fy = 2024525 kgcm < 1,5MyMy = Sx . fy = 1850400 kgcm

1,5 . My = 2775600 kgcm > Mp

Mnx = 2847728 kgcm > Mp

Mnx=Mp= 2024525 kgcm

Mny = Zy ( 1 flen ) * fy== 0.25 1.2 400 2400 = 288000 kgcm= 2880 kgm

6.3.5 Persamaan Interaksi

28073 + 6513.91060637561 + 01.7 88538.49 0.9 20245.25 0.9 2880

0.186512429941213 + 0.35750011318846 + 0

0.544013 < 1

OK

6.3.6 Kontrol Kuat Rencana Geser

240 < 11002400 15.49193

0.1 < 71.00469Plastis

(1/4∗tf∗bf 2)∗fyx x x

Pu2ϕcPn

+Mux

ϕbMnx+

Muyϕ bMny

≤1

x x x

htw

≤1100

√ fy

Vn = 0.6 fy Aw= 0.6 2400 0.8 30= 34560 Kg

Vu < ФVn2663 < 0.9 345602663 < 31104

OK

Ix = cm4 ix =Iy = cm4 iy =Sx = cm3 A =Sy = 444 cm3 q =

cm Zx = 1280 cm3 b = 400 mmcm Zy = 292 cm3 tf = 21 mmcm2 d = 400 mm tw = 13 mmkg/m h = 314 mm r = 22 mm

7 Sambungan

7.1 Sambungan Kuda - Kuda ( Detail A )

60

420 80

180100

200

7.1.1 Data Perencanaan SambunganDari hasil perhitungan SAP diperoleh:

Mu = 1279.27 kgm = 127927 kgcmPu = 1152 kg

8 mmTebal Plat = 10 mm

7.1.2 Kontrol Kekuatan Baut7.1.2.1 Perhitungan Ruv yang Diterima Setiap Baut

Ruv = Pu = 1152 = 192 kgn 6

Baut Tanpa Ulir ( Bor ) Diameter f =

A

B

C D

7.1.2.2 Perhitungan Kuat Geser Bautf Rnv = 0.75 0.5 fu Ab n

= 0.75 0.5 4100 0.5024 1= 772.44 kg

7.1.2.3 Perhitungan Kuat Tumpu Bautf Rn = 2.4 d tp fu 0.75

= 2.4 0.8 1 4100 0.75= 5904 kg

7.1.2.4 Perhitungan Kuat tarik Bautf Rnt = 0.75 0.75 fu Ab

= 0.75 0.75 4100 0.5024= 1158.66 kg

7.1.2.5 Rumus Interaksi Geser dan Kuat Tarik Baut

192 + Rut < 1772.44 1158.66

Rut = T = 870.66 kg

7.1.2.6 Kontrol Momen Sambungan

Letak Garis Netral a:

60

80

180

100

200

a = = 870.66 x 6fy B 2500 x 20

= 0.104479 cm

d1 = 9.895521 cmd2 = 27.89552 cmd3 = 35.89552 cm

Sdi = 73.68656 cm

Σ T

d1

d2d3

a

2T

2T

2T

(Ruv

φRnv

)2+(Rut

Rnt

)2≤1

f Mn = 0.9 fy B + S T di2

= 0.9 2500 0.010916 20 + 128311.92

= 128557.5 kgcm > 127927 kgcmOK

7.2 Sambungan Kuda - Kuda dan Kolom ( Detail B )

60

420 80180

100200

7.2.1 Data Perencanaan SambunganDari hasil perhitungan SAP diperoleh:

Mu = 3240 kgm = 324000 kgcmPu = 1344.96 kg

12 mmTebal Plat = 10 mm

7.2.2 Kontrol Kekuatan Baut7.1.2.1 Perhitungan Ruv yang Diterima Setiap Baut

Ruv = Pu = 1344.96 = 224.16 kgn 6

7.2.2.2 Perhitungan Kuat Geser Bautf Rnv = 0.75 0.5 fu Ab n

= 0.75 0.5 4100 1.1304 1= 1737.99 kg

a2

Baut Tanpa Ulir ( Bor ) Diameter f =

Pu

Mu

7.2.2.3 Perhitungan Kuat Tumpu Bautf Rn = 2.4 d tp fu 0.75

= 2.4 1.2 1 4100 0.75= 8856 kg

7.2.2.4 Perhitungan Kuat tarik Bautf Rnt = 0.75 0.75 fu Ab

= 0.75 0.75 4100 1.1304= 2606.985 kg

7.2.2.5 Rumus Interaksi Geser dan Kuat Tarik Baut

224.16 + Rut < 11737.99 2606.985

Rut = T = 2270.745 kg

7.2.2.6 Kontrol Momen Sambungan

Letak Garis Netral a:

60

80

180

100

200

a = = 2270.745 x 6fy B 2500 x 20

= 0.272489 cm

d1 = 9.727511 cmd2 = 27.72751 cmd3 = 35.72751 cm

Sdi = 73.18253 cm

f Mn = 0.9 fy B + S T di2

Σ T

a2

d1

d2d3

a

2T

2T

2T

(Ruv

φRnv

)2+(Rut

Rnt

)2≤1

= 0.9 2500 0.07425 20 + 332357.72

= 334028.4 kgcm > 324000 kgcmOK

7.3 Sambungan Balok dan Kolom ( Detail C )

5760

576 6060

198141

200

7.3.1 Data Perencanaan SambunganDari hasil perhitungan SAP diperoleh:

Mu = 12614 kgm = 1261400 kgcmPu = 12203 kg

18 mmTebal Plat = 10 mm

7.3.2 Kontrol Kekuatan Baut7.3.2.1 Perhitungan Ruv yang Diterima Setiap Baut

Ruv = Pu = 12203 = 1220.3 kgn 10

7.3.2.2 Perhitungan Kuat Geser Bautf Rnv = 0.75 0.5 fu Ab n

= 0.75 0.5 4100 2.5434 1= 3910.478 kg

7.3.2.3 Perhitungan Kuat Tumpu Bautf Rn = 2.4 d tp fu 0.75

= 2.4 1.8 1 4100 0.75= 13284 kg

7.3.2.4 Perhitungan Kuat tarik Bautf Rnt = 0.75 0.75 fu Ab

= 0.75 0.75 4100 2.5434= 5865.716 kg

Baut Tanpa Ulir ( Bor ) Diameter f =

Pu

Mu

7.3.2.5 Rumus Interaksi Geser dan Kuat Tarik Baut

1220.3 + Rut < 13910.478 5865.716

Rut = T = 4035.266 kg

7.3.2.6 Kontrol Momen Sambungan

5760

576 6060

198141

200

a = = 4035.266 x 10fy B 2500 x 20

= 0.807053 cm

d1 = 13.29295 cmd2 = 33.09295 cmd3 = 39.09295 cmd4 = 45.09295 cmd5 = 51.09295 cm

Sdi = 181.6647

f Mn = 0.9 fy B + S T di2

= 0.9 2500 0.651335 20 + 14661312

= 1480786 kgcm > 1261400 kgcmOK

Σ T

a2

d1 d2d3

d4 d5

a

2T

2T

2T

2T2T

(Ruv

φRnv

)2+(Rut

Rnt

)2≤1

7.4 Sambungan Balok dan Kolom ( Detail D )

57

60576 60

60

198141

7.4.1 Data Perencanaan SambunganDari hasil perhitungan SAP diperoleh:

Mu = 12976 kgm = 1297600 kgcmPu = 7343 kg

18 mmTebal Plat = 10 mm

7.4.2 Kontrol Kekuatan Baut7.4.2.1 Perhitungan Ruv yang Diterima Setiap Baut

Ruv = Pu = 7343 = 734.3 kgn 10

7.4.2.2 Perhitungan Kuat Geser Bautf Rnv = 0.75 0.5 fu Ab n

= 0.75 0.5 4100 2.5434 1= 3910.478 kg

7.4.2.3 Perhitungan Kuat Tumpu Bautf Rn = 2.4 d tp fu 0.75

= 2.4 1.8 1 4100 0.75= 13284 kg

7.4.2.4 Perhitungan Kuat tarik Baut

Baut Tanpa Ulir ( Bor ) Diameter f =

Pu

Mu

Pu

Mu

f Rnt = 0.75 0.75 fu Ab= 0.75 0.75 4100 2.5434= 5865.716 kg

7.4.2.5 Rumus Interaksi Geser dan Kuat Tarik Baut

734.3 + Rut < 13910.478 5865.716

Rut = T = 4764.266 kg

7.4.2.6 Kontrol Momen Sambungan

5760

6060

198141

200

a = = 4764.266 x 10fy B 2500 x 20

= 0.952853 cm

d1 = 13.14715 cmd2 = 32.94715 cmd3 = 38.94715 cmd4 = 44.94715 cmd5 = 50.94715 cm

Sdi = 180.9357

f Mn = 0.9 fy B + S T di2

= 0.9 2500 0.907929 20 + 17240522

= 1744480 kgcm > 1297600 kgcmOK

Σ T

a2

d1 d2d3

d4 d5

a

2T

2T

2T

2T2T

(Ruv

φRnv

)2+(Rut

Rnt

)2≤1

7.5 Kontrol Kekuatan Sambungan Las7.5.1 Perencanaan Tebal Las Efektif pada Sambungan7.5.2 Perhitungan Tebal Las Efektif Pada Web

aeff max Di Web = 0.707 x fu70 70.3

= 0.707 x 3700 1070 70.3

= 5.315789 mm

aeff max Di end plate = 1.41 x fu70 70.3

= 1.41 x 4100 1070 70.3

= 11.74761 mm

7.5.3 Perhitungan Gaya yang Bekerja Pada Sambungan Las

Misal te = 1 cm

A = 72.2 + 34.8 = 107 cm2

Sx = x 1 x 23

= 72.2 + 34.8 x 1 x 23

= 2544.222 cm3

fv = Pu = 7343 = 68.62617 kg/cm2A 107

fh = Mu = 1297600 = 510.0183 kg/cm2Sx 2544.22222222222

f total = +

d2

fv2 fh2

= 68.62617 510.0183= 514.6147 kg/cm2

7.5.4 Kontrol Kekuatan Las

0.75 0.6 70 70.3 = 2214.45 kg/cm2

> f totalOK

7.5.5 Perhitungan Tebal efektif

te perlu = f total = 514.6147 = 0.232389 cmf fn 2214.45

7.5.6 Perhitungan Lebar Perlu

a perlu = 0.87 = 1.230552 mm < 5.890469 mm0.707 OK

7.5.7 Perhitungan Tebal Efektif Dengan Lebar Minimum

a minimum = 4 mmte perlu = 4 x 0.707 = 2.828 mm

7.5 Sambungan Kolom Pondasi7.5.1 Data Perencanaan

Rencana Panjang Plat Dasar kolom L 40 cmRencana Lebar Plat Dasar kolom B 40 cmfc' beton 20 MpaMomen yang bekerja pada dasar kolom Mu 1093100 kgcmLintang yang bekerja pada dasar kolom Du 3854 kgNormal yang bekerja pada dasar kolom Pu 22617 kga 300 mma1 50 mmb 150.0 mmc 50 mmd 300 mms pelat 1600 kg/cm2

2 + 2

f fn =

f fn

P

M

a1 a1a

7.5.2 Kontrol Pelat Landasan Beton ( Pondasi )

fc' beton = 20 Mpa = 200 kg/cm2Pu yang bekerja = 22617 kgKekuatan nominal tumpu beton

Pn = 0.85 fc' AA = 40 x 40 = 1600 cm2

Pn = 0.85 200 1600 = 272000 kg

Pu <= f Pn22617 <= 0.6 x 27200022617 <= 163200 OK

7.5.2 Perencanaan Tebal Pelat Baja Pondasi7.5.2.1 Perhitungan Tegangan Yang Bekerja Akibat Adanya Eksentrisitas

e = M = 1093100 = 48.3309 cmP 22617

A = 40 x 40 = 1600 cm2

W = = 1/6 40 40 10666.67 cm3

P + MA W

= 22617 + 10931001600 10666.6666666667

116.6138 kg/cm288.3425 kg/cm2

Jadi, q = 116.6138 kg/cm2

1/6 B L2 2 =

s =

s maks = s min =

a1 a1a

d1

L

c

b

b

c

B

1

2

3

7.5.2.2 Perhitungan Momen Yang Bekerja ~ Daerah 1

M =

= 1/2 116.6138 5= 1457.672 kgcm

~ Daerah 2a / b = 30 / 15 = 2a1 = 0.1 a2 = 0.046

Ma =

= 0.1 116.6138 15 2623.809 kgcm

Mb =

= 0.046 116.6138 15 1206.952 kgcm

~ Daerah 3a / b = 5 / 30 = 0.166667 < 0.5

M3 =

= 1/2 116.6138 5 1457.672 kgcm

7.5.2.3 Perhitungan Tebal Pelat Bajas = 6 M t = 6 M

s plat

= 6 x 1457.6721600

= 2.338005 ~ 3 cm

7.5.3 Perencanaan Diameter Angker7.5.3.1 Perhitungan Tegangan Yang Bekerja Pada Angker

1/2 q L2

2

a1 q b2

2 =

a2 q b2

2 =

1/2 q a12

2 =

t2

1

P

M

√√

40

20

=x B - x

x = = 88.3425 x 4088.3425 + 116.6138

= 17.24124 cm

y = B - x = 40 - 17.24124 = 22.75876 cm

S min = 1.5 d = 1.5 ( 2 x tf )= 1.5 2 x 1.2= 3.6 cm

1/3 x = 5.74708 cm > S min

1/3 y = 7.586253 cm > S min

r = 20 - 1/3 y = 20 - 7.586253 = 12.41375 cm

C = 40 - 7.586253 - 5.74708 = 26.66667 cm

T = M - P r = 1093100 - 22617 12.41375C 26.7

= 30424.6549739822 kg = Pu

7.5.3.2 Perencanaan Diameter Angker- Leleh: Pu =

Ag perlu = 30424.6549739822 = 8.245164 cm2

s min s max

s min Bs min + s max

f fy Ag

P

M

C1/3x 1/3y

s max

s min

x y

0.9 4100 #DIV/0!

- Putus: Pu =

Ag perlu = 30424.6549739822 = #DIV/0! cm20.75 0.75 0 #DIV/0!

- A baut perlu = 30424.6549739822 = 27.16487 cm2 #DIV/0!1120

Untuk tiap sisi A baut perlu = 27.16487 / 2 sisi = 13.58244 cm2Direncanakan menggunakan angker D30 ( A = 7.065 cm2 )Jumlah angker dalam 1 sisi = #DIV/0! / 7.065

= #DIV/0! ~ #DIV/0! buahDipasang 3 D 30 A = 21.195 cm2

Abaut > AperluOK

7.5.4 Perencanaan Panjang Angker

Kekuatan Baut untuk menerima beban tarik pada tiap sisi adalah:30424.6549739822 / #DIV/0! baut = #DIV/0! kg

2 sisi

l = #DIV/0! = #DIV/0! cm4.472136 2 p 1.5

7.5.5 Perencanaan Sambungan Las

t plat = 3 cm

Profil Baja Bj 52 fu = 0 kg/cm2

= 70 ksi= 70 x 70.3 kg/cm3

f 0.75 fu Ag

fu las E70 xx

Syarat tebal plat:a min = 3 mma max = t - 0.1 = 3 - 0.1

= 2.9 cm

af max = 1.41 x fu elemen t fu las

= 1.41 x 0 x 370 x 70.3

= 0 cm = 0 mm

Pakai a = 3 mmte = 0.707 a = 2.121 mmb = 30 cmd = 30 cm

Sx == 30 30 + 900

3= 1200 cm3

Akibat Pu : fvp = Pu2 ( 2 b + d ) te

= 226172 2 30 + 30 2.121

= 592.4092 kg/cm2

Akibat Mu: fhm = Mu = 1093100Sx 1200

= 910.9167 kg/cm2

f las = 0.75 x 0.6 x 70 x 70.3= 2214.45

f total =

= 592.4092 910.9167= 1180718 kg/cm2 <= 0.75 0.6 fu las= 1086.608 kg/cm2 <= 2214.45 kg/cm2 OK !!

b * d + ( d2 / 3)

fv2 + fh2

2 + 2