19684772 perhitungan struktur gudang baja

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 ) :

25 0

20 ≤ 20

kg/m2

a =

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

Perencanaan Atap

Merencanakan Pola Beban

Data Perencanaan

Perencanaan Dimensi Gording

Perencanaan Gording Ujung

Perencaan Penggantung Gording

Perencanaan Ikatan Angin

Merencanakan

Pola Beban

Beban Mati Beban Hidup Beban Angin

Beban Penutup

AtapBeban

Terbagi RataBeban Profil Beban

Pengikat dllBeban

Terpusat

Beban Tekanan

Angin

Beban Angin Hisap

ambil q = 20

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 : 30

= 0.1

Angin Tekan = C x W = 3

Angin Hisap = 0.4 x W = 12

1.2 Data - Data perencanaan

Data AtapJenis : Asbes GelombangTebal : 5 mmBerat : 10.3 kg/m2Lebar Gelombang : 110 mmKedalaman Gelombang : 57 mmJarak Miring Gording : 110 cmJarak Kuda-Kuda (L) : 400 cm

Sudut Kemiringan Atap : 0.436 rad = 25 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 100 x 50 x 5 x 7

A = 11.85 tf = 7 mm Zx = 41.8

W = 9.3 kg/m Ix = 187 Zy = 8.94

a = 100 mm Iy = 14.8 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

1.3.2 Perencanaan Pembebanan1.3.2.1 Perhitungan BebanBeban MatiBerat Gording = 9.3 kg/mBerat Asbes Gelombang = w x l

= 10.3 x 1.1 = 11.33 kg/mBerat Total = 20.63 kg/m

alat Pengikat dll 10 % = 0.1 x 20.63 = 2.063 kg/m

= 22.69 kg/m

Beban Hidup

40 - 20 = 20

q = 20

= 0.997 x 20.00 = 19.94 kg/m

= 100 kg

Beban Angin

Tekanan Angin = 30

Angin Tekan = 3

Angin Hisap = 12 (menentukan = q)0.997 x 12.00 = 12 kg/m

Beban Mati + Beban Hidup > dari Beban Angin Hisap : 22.69 + 19.94 > 12

Beban Angin Hisap tidak perlu diperhitungkan ==> 3 kg/m

1.3.2.2 Perhitungan Momen Akibat Beban thp Sbx dan SbyBeban Mati

0.125 x ( 22.69 x 0.906 x 16 ) = 41.13 kgm

0.125 x ( 22.69 x 0.423 x 1.78 ) = 2.131 kgm

Beban Hidup Terbagi Rata

0.125 x ( 19.94 x 0.906 x 16 ) = 36.25 kgm

0.125 x ( 19.94 x 0.423 x 1.78 ) = 1.873 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 0.906 x 4 ) = 90.63 kgm

0.25 x ( 100 x 0.423 x 1.33 ) = 14.09 kgm

Beban Angin Terbagi Rata

= 0.125 x 3 x 16 = 6 kgm

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

41.134 kgm 2.131 kgm

36.252 kgm 1.873 kgmMw = 6 kgm

1.2 x 41.134 + 1.6 x 36.252 + 0.8 x 6 = 112 kgm

1.2 x 2.1312 + 1.6 x 1.8726 + 0.8 x 0 = 5.554 kgm

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

41.134 kgm 2.131 kgm

90.631 kgm 14.09 kgmMw = 6 kgm

1.2 x 41.134 + 1.6 x 90.631 + 0.8 x 6 = 199 kgm

1.2 x 2.1312 + 1.6 x 14.087 + 0.8 x 0 = 25.1 kgm

1.3.3 Kontrol Kekuatan Profil1.3.3.1 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

1.3.3.2 Kontrol Lateral Buckling

500 mm = 50 cm

1.76 x xEfy

= 1.76 x 1.12 x200000

= 56.90 cm240

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 = 41.8 x 2400 = 1003 KgmMny = Zy ( satu sayap ) * fy

=

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

1.3.3.3 Persamaan IterasiMux

+Muy

≤ 1

Beban Mati , Beban Angin dan Beban Hidup Terbagi Rata112.164

+5.554

≤ 10.9 x 1003 0.9 x 94.50.124 + 0.065 ≤ 1

0.19 ≤ 1 OK

Beban Mati , Beban Angin dan Beban Hidup Terpusat199.170

+25.097

≤1

0.9 x 1003 0.9 x 1050.198 + 0.266 ≤ 1

0.464 ≤ 1 OK

1.3.3.4 Kontrol Lendutan Profil

Lendutan Ijin f =L

=400

= 2.2222 cm180 180

Lendutan Akibat Beban Merata (1)

fx =5

x =5 x 0.426 x 0.906 x 400 4

384 E x Ix 384 x 2000000 x 187 = 0.3444 cm

fy =5

x =5 x 0.426 x 0.423 x 133 4

384 E x Iy 384 x 2000000 x 14.8 = 0.025 cm

Lendutan Akibat Beban Terpusat (2)

fx =5

xP

=1 x 100 x 0.906 x 400 3

384 E x Ix 48 x 2000000 x 187 = 0.3231 cm

fy =5

xP

=1 x 30 x 0.423 x 133 3

384 E x Iy 48 x 2000000 x 14.8 = 0.0212 cm

Lendutan Akibat Beban Angin merata (3)

fx =5

x =5 x 0.03 x 0.906 x 400 4

384 E x Ix 384 x 2000000 x 187 = 0.0242 cm

fy =5

x =5 x 0.03 x 0.423 x 133 4

384 E x Iy 384 x 2000000 x 14.8 = 0.0018 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

=

= ( 0.3444 + 0.3231 + 0.024 0.025 + 0.021 + 0.002

0.6934 cm < 2.222 cm OK

1.4 Perencanaan Penggantung Gording

1.4.1 Data Penggantung Gording

Jarak Kuda - Kuda (L) = 400 cmJumlah Penggantung Gording = 2 buahJumlah Gording = 9 buahJarak Penggantung gording = 133 cm

1.4.2 Perencanaan PembebananBeban MatiBerat Sendiri Gording = 9.3 kg/mBerat Asbes gelombang = 11.33 kg/m

= 20.63 kg/mAlat Pengikat dll 10 % = 0.1 x 20.63 = 2.063 kg/m

= 22.69 kg/m

x x L / 3 = 22.693 x 0.4226 x 1.333 = 12.787 kg

Beban Hidup

40 - 0.8 = 20

q = 20

= 0.997 x 20.00 = 19.94 kg/m

x x L / 3 = 19.939 x 0.4226 x 1.333 = 11.235 kg

= 100 kg

x = 100.0 x 0.4226 = 42.262 kg

Beban Angin

Angin Tekan = q = 3

= 0.997 x 3.00 = 2.991 kg/m

x x L / 3 = 2.9908 x 0.4226 x 1.333 = 1.685 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 12.8 + 1.6 x ( 11.2 + 42.3 ) + 0.8 x 1.7 = 102.29 kg

= 102.29 x 9 = 920.60 kg

1.4.3.1 Penggantung Gording Tipe B

panjang miring gording=

110= 0.825

L / 3 133.33

39.52

=920.60

= 1446.607 kgsin 39.52

1.4.4 Perencanaan Batang Tarik

Pu = 1446.607 kg

BJ 37 fu = 3700

fy = 2400

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

Ag perlu = Pu = 1446.607= 0.670j fy 0.9 x 2400

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

Ag perlu = Pu = 1446.607= 0.695j fu 0.75 0.75 x 3700 x 0.75

Menentukan

d =Ag x 4

=0.695 x 4

= 0.941 cmp p==> Pakai d = 10 mm

1.4.5 Kontrol KelangsinganJarak Penggantung Gording = 133 cm

Panjang Rb =

= 133.33 2 + 110 2= 173 cm

Cek :d >

Panjang Rb500

1 >172.85

5001 > 0.3457 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 = 30

= 0.9

= 0.4

300 cm 200 cm

a = 0.4363 rad = 25 0

1.5.2 Perhitungan Tinggi Ikatan Angin ( h )

9 m

9 + 2 x tg 0.436 = 9.933 m

9 + 4 x tg 0.436 = 10.87 m

9 + 6 x tg 0.436 = 11.8 m

9 + 9 x tg 0.436 = 13.2 m

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

0.50 x 30 x 0.9 x 1 x 9 = 121.5 kg

0.50 x 30 x 0.9 x 2 x 9.933 = 268 kg

0.50 x 30 x 0.9 x 2 x 10.87 = 293 kg

0.50 x 30 x 0.9 x 2.5 x 11.8 = 398 kg

0.50 x 30 x 0.9 x 3 x 13.2 = 534 kg

Rtotal = ( R1+R2+R3+R4+(R5/2)) = 121.5 + 268.18 + 293.4 + 398 + 267 = 1348.454 kg

1.5.4 Perencanaan Dimensi Ikatan Angin1.5.4.1 Menghitung gaya Normal

tg φ =2

= 0.54

φ = 26.565 0

R1 = 121.5 kgRtotal = 1348.454 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 1348.454

= 599.31 kg0.9

1.5.4.2 Menghitung gaya Pada Titik SimpulPada Titik Simpul A

ΣV = 0

===> S1 = - Rtotal ===> S1 = -1348 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

- ( - ) =

- ( 121.5 - -1348 ) cos j cos 26.565

-1643.458 kg


Pu = = -1643.46 x 1.6 x 0.75 = -1972.150 kg

BJ 37 fu = 3700

fy = 2400

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

Ag perlu = Pu = 1972.150= 0.913j fy 0.9 x 2400

Tidak Menentukan

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

Ag perlu = Pu = 1972.150= 0.948j fu 0.75 0.75 x 3700 x 0.75

Menentukan

d =Ag x 4

=0.948 x 4

= 1.098 cmp p==> Pakai d = 11 mm

1.5.6 Kontrol KelangsinganJarak kuda-kuda = 400 cm

= 400 2 + 110 2= 415 cm

Cek :d >

500

1.1 >414.85

5001.1 > 0.8297 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

= 199.170 x 0.75 = 149.377 kgm

= 25.097 x 0.75 = 18.823 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

1618.144 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 = 400 cm ==> λx =Lkx

=400

= 100.5ix 3.98

Ncrbx = =p 2 x 2000000 x 11.85

100.5 2= 23157.64 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=

1618.144= 0.153 < 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

= 1.0751 - (

1618.144)23157.64

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 = 1.075 > 1Sbx = 1.075

Muy = Mnty * SbySby =

Cmy≥ 1

1 - (Nu

)NcrbyUntuk elemen Beban Tranversal, ujung sederhana

Cmy = 1

Sby =1

= 1.0141 - (

1618.144)117366.49

Sby = 1.014 > 1Sby = 1.014

1.6.5 Perhitungan Momen Ultimate Sbx dan SbyMux = Sbx . Mntx = 1.0751 x 149.377 = 160.599 kgmMuy = Sby . Mnty = 1.0751 x 18.823 = 20.237 kgm

1.6.6 Perhitungan Persamaan InteraksiMnx = 1003 kgm Mny = 105 kgm

Pu+

Mux+

Muy≤ 12 x x Pn x Mnx x Mny

1618.144+

160.599+

20.237≤ 12 x 0.85 x 12437.136 0.9 x 1003 0.9 x 105

0.469 ≤ 1OK

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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


0.125 x 14.8 x 1.778 = 3.288 kg/m

Beban AnginLantai Dasar

Tekanan Angin = 30

Angin Tekan ( C = 0.9 ) = 0.9 x 30 = 27q = Angin Tekan x Jarak Gording = 27 x 1.25 = 33.75 kg/m

Angin Hisap ( C = 0.4 ) 0.4 x 30 = 12

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 = 12 x 1.25 = 15 kg/m

Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :

0.125 x 33.75 x 16 = 67.5 kgmN = q x Jarak Gording = 15 x 1.25 = 18.75 kg (Tarik)

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :

0.125 x 15 x 16 = 30 kgmN = q x Jarak Gording = 33.75 x 1.25 = 42.19 kg (Tekan)

Lantai 1

Tekanan Angin = 30

Angin Tekan ( C = 0.9 ) = 0.9 x 30 = 27q = Angin Tekan x Jarak Gording = 27 x 1 = 27 kg/m

Angin Hisap ( C = 0.4 ) 0.4 x 30 = 12q = Angin hisap x Jarak Gording = 12 x 1 = 12 kg/m

Akibat Beban Angin yg Tegak Lurus Dinding (tarik)

0.125 x 27 x 16 = 54 kgmN = q x Jarak Gording = 12 x 1 = 12 kg (Tarik)

Akibat Beban Angin yg Tegak Lurus Gevel (tekan)

0.125 x 12 x 16 = 24 kgmN = q x Jarak Gording = 27 x 1 = 27 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 67.5 + 0.5 x 0 + 0.5 x 0 = 87.75 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 18.75 + 0.5 x 0 + 0.5 x 0 = 24.38 kg

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :Mux = 1.2 x 0 + 1.3 x 30 + 0.5 x 0 + 0.5 x 0 = 39 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 42.19 + 0.5 x 0 + 0.5 x 0 = 54.84 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 54 + 0.5 x 0 + 0.5 x 0 = 70.2 kgm

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 )

Muy = 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 12 + 0.5 x 0 + 0.5 x 0 = 15.6 kg

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :Mux = 1.2 x 0 + 1.3 x 24 + 0.5 x 0 + 0.5 x 0 = 31.2 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 27 + 0.5 x 0 + 0.5 x 0 = 35.1 kg


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



500 mm = 50 cm

1.76 x xEfy

= 1.76 x 1.12 x200000

= 56.90 cm240


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

= 0.75 x 0.85 x Ag x fu


LP = iY

LB LP

1/4 x tf x bf2 x fy

lp

l =

f Nn = f .Ag . fy =

f Nn = f .Ae . fu =

λ p

= 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### > 54.84

OKLantai 1

> Nu### > 1404

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)Ncrbx

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

Lantai Dasar

Sbx =1

= 1.001 (Tarik)1 - (

24.38)23156.27

Sby =1

= 1.000 (Tarik)1 - (

24.38)117359.57

Sbx =1

= 1.002 (Tekan)1 - (

54.844)23156.27

Sby =1

= 1.000 (Tekan)1 - (

54.844)117359.57

Lantai 1

Sbx =1

= 1.001 (Tarik)1 - (

15.6)23156.27

Sby =1

= 1.000 (Tarik)1 - (

15.6)117359.57

Sbx =1

= 1.002 (Tekan)1 - (

35.1)23156.27

Sby =1

= 1.000 (Tekan)1 - (

35.1)117359.57

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

Nu=

24.375= 0.001008315 < 0.2 OK

24174

Nu+

Mux x Sbx+

Muy x Sby< 12 x f x Mnx x Mny

24.375+

87.8 x 1.001+

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

0.143 < 1OK

Lantai 1Nu

=15.6

= 0.000645321 < 0.2 OK24174

Nu+

Mux x Sbx+

Muy x Sby< 12 x f x Mnx x Mny

15.600+

70.2 x 1.001+

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

0.12 < 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=

54.84= 0.002268708 < 0.2 OK

###Nu

+Mux x Sbx

+Muy x Sby

< 12 x f x Mnx x Mny54.844

+39.0 x 1.002

+4.250 x 1.000

< 12 x 24174 0.9 x 1003 0.9 x 1050.089 < 1

OKLantai 1

Nu=

35.1= 0.001451973 < 0.2 OK

###Nu

+Mux x Sbx

+Muy x Sby

< 12 x f x Mnx x Mny35.100

+31.2 x 1.002

+3.945 x 1.000

< 12 x 24174 0.9 x 1003 0.9 x 1050.077 < 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



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


f . Nn

f . Nn fb

f . Nn

f . Nn fb

kg/cm2 =

kg/cm2 =

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


0.125 x 14.8 x 1 = 1.849 kg/m

Beban AnginLantai Dasar

Tekanan Angin = 30

Angin Tekan ( C = 0.9 ) = 0.9 x 30 = 27q = Angin Tekan x Jarak Gording = 27 x 1.25 = 33.75 kg/m

Angin Hisap ( C = 0.4 ) 0.4 x 30 = 12q = Angin hisap x Jarak Gording = 12 x 1.25 = 15 kg/m


0.125 x 33.75 x 9 = 38 kgmN = q x Jarak Gording = 15 x 1.25 = 18.75 kg (Tarik)


0.125 x 15 x 9 = 16.88 kgmN = q x Jarak Gording = 33.75 x 1.25 = 42.19 kg (Tekan)

Lantai 1

Tekanan Angin = 30

Angin Tekan ( C = 0.9 ) = 0.9 x 30 = 27q = Angin Tekan x Jarak Gording = 27 x 1 = 27 kg/m

Angin Hisap ( C = 0.4 ) 0.4 x 30 = 12q = Angin hisap x Jarak Gording = 12 x 1 = 12 kg/m


0.125 x 27 x 9 = 30.38 kgmN = q x Jarak Gording = 12 x 1 = 12 kg (Tarik)


0.125 x 12 x 9 = 13.5 kgmN = q x Jarak Gording = 27 x 1 = 27 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 38 + 0.5 x 0 + 0.5 x 0 = 49.36 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 18.75 + 0.5 x 0 + 0.5 x 0 = 24.38 kg

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :Mux = 1.2 x 0 + 1.3 x 16.88 + 0.5 x 0 + 0.5 x 0 = 21.94 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 42.19 + 0.5 x 0 + 0.5 x 0

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 )

= 54.84 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 30.38 + 0.5 x 0 + 0.5 x 0 = 39.49 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 12 + 0.5 x 0 + 0.5 x 0 = 15.6 kg

Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :Mux = 1.2 x 0 + 1.3 x 13.5 + 0.5 x 0 + 0.5 x 0 = 17.55 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 27 + 0.5 x 0 + 0.5 x 0 = 35.1 kg


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



500 mm = 50 cm

1.76 x xEfy

= 1.76 x 1.12 x200000

= 56.90 cm240


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

≤ 300Lk

=300

= 75.38 < 300 OK

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


LP = iY

LB LP

1/4 x tf x bf2 x fy

lp

l =

ix =3.98

= 75.38 < 300 OK

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### > 54.84

OKLantai 1

> Nu### > 1404

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

Ncrby =11.85 x 2400

= 117366.4931 kg

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 =0.492 2

= 117366.4931 kg

2.3.7.1 Komponen Struktur Ujung Sederhana Cm = 1

Sbx =Cmx

≥ 11 - (

Nu)Ncrbx

Lantai Dasar

Sbx =1

= 1.001 (Tarik)1 - (

24.38)41166.71

Sby =1

= 1.000 (Tarik)1 - (

24.38)117366.49

Sbx =1

= 1.001 (Tekan)1 - (

54.844)41166.71

Sby =1

= 1.000 (Tekan)1 - (

54.844)117366.49

Lantai 1

Sbx =1

= 1.000 (Tarik)1 - (

15.6)41166.71

Sby =1

= 1.000 (Tarik)1 - (

15.6)117366.49

Sbx =1

= 1.001 (Tekan)1 - (

35.1)41166.71

Sby =1

= 1.000 (Tekan)1 - (

35.1)117366.49

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

Nu=

24.375= 0.001008315 < 0.2 OK

24174Nu

+Mux x Sbx

+Muy x Sby

< 12 x f x Mnx x Mny24.375

+49.4 x 1.001

+2.390 x 1.000

< 12 x 24174 0.9 x 1003 0.9 x 1050.081 < 1

OKLantai 1

Nu=

15.6= 0.000645321 < 0.2 OK

24174Nu

+Mux x Sbx

+Muy x Sby

< 12 x f x Mnx x Mny15.600

+39.5 x 1.000

+2.219 x 1.000

< 1

f . Nn

f . Nn fb

f . Nn

f . Nn fb

2 x 24174 + 0.9 x 1003 + 0.9 x 105 < 1

0.068 < 1OK

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

Nu=

54.84= 0.002268708 < 0.2 OK

###Nu

+Mux x Sbx

+Muy x Sby

< 12 x f x Mnx x Mny54.844

+21.9 x 1.001

+2.390 x 1.000

< 12 x 24174 0.9 x 1003 0.9 x 1050.051 < 1

OKLantai 1

Nu=

35.1= 0.001451973 < 0.2 OK

###Nu

+Mux x Sbx

+Muy x Sby

< 12 x f x Mnx x Mny35.100

+17.6 x 1.001

+2.219 x 1.000

< 12 x 24174 0.9 x 1003 0.9 x 1050.044 < 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 27 = 81 kg/m

panjang x angin tekanTinggi Regel 5 = 7 m = 3 x 27 = 81 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 Pembebanan2.4.2.1 Beban Mati

f . Nn

f . Nn fb

f . Nn

f . Nn fb

qw regel 5 =

qw regel 2 =

m2

m2

m2

m2

Regel 5

= 7.5 x 20.63 = 155 kg

= 17.5 x 4.15 = 72.63 kg

= 7 x 9.3 = 65.1 kg

Regel 2

= 6 x 20.63 = 124 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 20 = 150 kg

Regel 2

= 6 x 20 = 120 kg

2.4.2.3 Beban AnginRegel 5

0.125 x 81 x 7 2 = 496 kgm

Regel 2

0.125 x 81 x 6 2 = 364.5 kgm

2.4.3 Syarat KekakuanRegel 5

Y =h

=700

= 3.5 cm200 200

Ix =5

xq x

384 E x Y

5x

4.961 x 7 4 = 2215.767384 0.02 x 3.5

===> Ix Profil yg Dipakai > 2215.767

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

= 155 + 138 + 281.4 = 574 kgNL Total = NL atap = 150 kg

Mw = 496 kgm

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 =

U = ( 1.2D + 1.6L+ 1.6W ) x 0.75Nu = ( 1.2 x 574 + 1.6 x 150 ) x 0.75 = 696 kg

Mntx = 1.6 x Mw x 0.75 = 1.6 x 496 x 0.75 = 595 kg

Regel 2

Y =h

=600

= 3 cm200 200

Ix =5

xq x

384 E x Y

5x

3.645 x 6 4 = 1025.156384 0.02 x 3

===> Ix Profil yg Dipakai > 1025.156

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

= 124 + 105.6 + 126.6 = 356 kgNL Total = NL atap = 120 kg

Mw = 364.5 kgmU = ( 1.2D + 1.6L+ 1.6W ) x 0.75Nu = ( 1.2 x 356 + 1.6 x 120 ) x 0.75 = 464 kg

Mntx = 1.6 x Mw x 0.75 = 1.6 x 364.5 x 0.75 = 437.4 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

22.83 2

L4

cm4

cm4

cm2 cm3

cm4 cm3

cm4

kg/cm2 =

kg/cm2 =

lc =lx

p2 . E . A

lx 2

lc =ly

p2 . E . A

ly 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

=696.465

= 0.007 < 0.20.85 x 122904

Pakai Rumus : Pu

+Mux

+Muy


Batang Dianggap Tidak Bergoyang Maka :

Sbx =Cmx

≥ 1 ;Cm = 11 - (

Nu)Ncrbx

Sbx =1

= 1.006 ≥ 11 - (

696.5)116040.87

Mux = Mntx . SbxMux = 595 x 1.006 = 599 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

=464.382

= 0.008 < 0.20.85 x 64416

Pakai Rumus : Pu

+Mux

+Muy


Batang Dianggap Tidak Bergoyang Maka :

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

Sbx =Cmx

≥ 1 ;Cm = 11 - (

Nu)Ncrbx

Sbx =1

= 1.008 ≥ 11 - (

464.4)56024.77

Mux = Mntx . SbxMux = 437.4 x 1.008 = 441 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


* Kontrol Lateral Buckling

1000 mm = 100 cm

1.76 x xEfy

= 1.76 x 100 x200000

= ### cm2400


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### ### ### ###### ### ### ###

### ###


Lateral Bracin Lb = 100 cm

Lateral Bracing = LB =

LP = iY

LB LP

1/4 x tf x bf2 x fy

b2tf

≤170

√ fyht≤

1680

√ fy

¿ ¿

¿ ¿

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! + 598.9448033 + 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 cmJarak Antar Gordng Horizont 100 cm

fc . Pn fb fb

Lp=1.76∗iy√ Efy

(1/4∗tf∗bf 2)∗fyx x x

x x x

x x x

2.5.2 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

arctgn β=β=β=

RB=R A

Sin β

x

KudaJarakKudaqRa = *

x

JarakGevelqRa *=

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 ###

= ### cm2

Ag=1 /4 πd2

d=√ Ag∗4π

√ x

d=√ Ag∗4π

Ag=1 /4 πd2

= ### 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

xxxxx

xxxxx

xxxxx

xxxxx

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

Mumax=18qu l

2 Dumax=12qu l

KONTROL LENDUTAN BALOKDimana Y ijin = L/360

Y max= 5

384(qD+qL)∗l4

EIx

Y mak < Y ijinPerbesar Profil

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

Sayap Badan

λ p=170

√ fyλ p=

1680

√ fy

ht

λr=370

√ f y− f rλr=

2550

√ f y

b2tf

b2tf

¿ λ pht¿ λ p

Profil KompakMnx = Mnp

b2tf

¿λ p¿ λr

Profil Tak Kompak

Mn=Mp−(Mp−Mr)λ−λpλr−λp

Profil Langsing

Mn=Mr( λr / λ)2Profil Langsing

λ p ¿ht¿ λr

Mn=Mr( λr / λ)2

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

Mumax=18qu l

2 Dumax=12qu l

PERHITUNGAN Ix PROFIL MINIMUMDimana Y ijin = L/360

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

Mn=Mp−(Mp−Mr)λ−λpλr−λp 2

KONTOL LATERAL BUCKLINGHitung λp dan λr daripada Lateral Buckling

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 p

Bentang PendekMnx = Mpx

λb¿λ p ¿ λr

Bentang Menengah

Mn=Cb(Mr+(Mr−Mp)(Lr−LLr−Lp

)≤Mp

Bentang Panjang

Mn=Mcr=CbπL √E¿GJ+(

πEL)2 I y I w≤Mp

Mnx Local Buckling > Mnx Lateral Buckling

Ambil Mnx Lateral Buckling Ambil Mnx Local Buckling

Jarak Lateral Bracing λb :λb = ?

KONTROL KUAT RENCANA GESERHitung h

tw

Hitung : 0.9 Mnx

0.9 Mnx > Mu maxPerbesar Profil

OK

KO

OK

KO

KO

OK

KONTROL KUAT RENCANA GESERHitung

htw

≤1100

√ fyVn = 0.6 fy Aw

1100

√ fy≤

htw

≤1370

√ fyVn=0. 6 f y Aw

1100 twh√ f y

Vn=900000Aw

(htw)2

Hitung 0.9 Vn

0.9 Vn > Vu Max

Profil Dapat Dipakai

Perbesar Profil


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


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


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


htw

≤1100

√ fy

4 Perencanaan Tangga Baja

4.1 Data PerencanaanTinggi tangga = 250 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 = 250 = 13.51351 buah18.5

= 14 buah

Jumlah injakan (n) = 14 buah

Lebar Bordes = 600 392 = 208 cmLebar Tangga = 200 20 = 180 cm

a = 32.5444 0 = 0.567719 rad

392 cm 208 cm

180 cm

180 cm

4.3 Perencanaan Pelat Tangga4.3.1 Perencanaan Tebal Pelat Tangga

Tebal Pelat Tangga = 4 mmBerat Jenis Baja = 7850 kg/m3

o o

−−

Tegangan Leleh Baja = 2400 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 2400 = 15552 kgcm

155.52 kgm

Syarat -> > Mu155.52 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



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

Zx = 6.14 cm3 ( Modulus Plastis )


MD

MD


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

φ Zx * fy = 0.9 6.14 2400 = 13262.4 kgcm

132.624 kgm

Syarat -> > Mu132.624 kgm > 50.16926 kgm

OK


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 = 8 mmBerat Jenis Baja = 7850 kg/m3Tegangan Leleh Baja = 2400 kg/m2Lebar Pelat Bordes = 2 m

4.5.2 Perencanaan Pembebanan Pelat BordesBeban Mati Berat Pelat = 0.008 2 7850 = 125.6 kg/m'Alat Penyambung (10 %) = 12.56 kg/m'

qD = 138.16 kg/m'Beban Hidup qL = 500 x 2 = 1000 kg/m'

= 0.125 138.16 0.480711 = 8.301881 kgm

φMn =

φMn =

φMn


MD

x x

Y max= 5384

(qD+qL)∗l4

EIx

x

x x

M D=18qDl

2

M L=18qL l

2

x x

x x

= 0.125 1000 0.480711 = 60.08889 kgm




= 0.25 200 0.64 = 32 cm3

φ Zx * fy = 0.9 32 2400 = 69120 kgcm

691.2 kgm

Syarat -> > Mu691.2 kgm > 106.1045 kgm

OK


f = L = 69.33333 = 0.192593360 360

= 0.083333 200 0.512 = 8.533333 cm4

Ix = 8.533333 cm4

= 5 ( 1.3816 10 ) 2.31E+07384 2100000 8.533333

= 0.191105 < 0.192593

OK

ambil Tebal Pelat Bordes = 8 mm

MD


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


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

40.6124.6.2 Perencanaan PembebananBeban MatiBerat Pelat = 0.008 0.693333 7850 = 43.54133 kg/m'Berat Profil I = = 17.2 kg/m'

= 60.74133 kg/m'Alat Penyambung ( 10 % ) = 6.074133 kg/m'

qD = 66.81547 kg/m'

= 0.125 66.81547 4.3264 = 36.1338 kgm

= 0.5 66.81547 4.3264 = 144.5352 kg

Beban Hidup qL = 500 x 0.693333 = 346.6667 kg/m'.

= 0.125 346.6667 4.3264 = 187.4773 kgm

= 0.5 346.6667 4.3264 = 749.9093 kg

4.6.3 Perhitungan Kombinasi Pembebanan

Mu = 1.4 36.1338 = 50.58733 kgmPu = 1.4 144.5352 = 202.3493 kgm

Tidak menentukan

Mu = 1.2 36.1338 + 1.6 187.4773 = 343.3243 kgmPu = 1.2 144.5352 + 1.6 749.9093 = 1373.297 kgm

Menentukan4.6.4 Kontrol Kekuatan Profil4.6.4.1 Penampang Profil fy = 2400 kg/m2


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 84 168016 15.49193 6 15.49193

6.25 10.97345 14 108.4435OK OK



Jarak Baut Pengikat : 250 mm = 25 cm

Lp = 125.4929 cm



== 0.25 3.2 100 2400 = 192000 kgcm= 1920 kgm


Zx = 84.184 cm3

φ Zx * fy = 0.9 84.184 2400 = 181837.4 kgcm

1818.374 kgm

Syarat -> > Mu1818.374 kgm > 343.3243 kgm

OK


f = L = 180 = 0.5360 360

Ix = 84.184 cm4

= 5 ( 0.668155 3.466667 ) 1.05E+09384 2100000 84.184

= 0.319696 < 0.5

φMn =

φMn =

φMn

ht≤

1680

√ fy

¿ ¿

¿ ¿



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 32.5444 = 15.05589 cm

4.7.2 Perencanaan Balok Induk Dengan Menggunakan Profil WF

250 x 125 x 5 x 8



63.7125

Syarat --> h > hmin25 > 15.05589

OK

4.7.3 Perencanaan Pembebanan

x

4.7.3.1 Perencanaan Pembebanan Anak TanggaBeban MatiBerat Pelat = 0.004 1.04 7850 = 32.656 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 32.5444 = 38.76035 kg/m'

115.9878 kg/m'Alat Penyambung (+ 10 %) = 11.59878 kg/m'

qD1 = 127.5866 kg/m'

Beban HidupqL1 = 500 1.04 = 520 kg/m'

Beban q1 Total = 1.2 qD + 1.6 qL= 1.2 127.5866 + 1.6 520= 985.1039 kg/m'

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

Berat Pelat Bordes = 0.008 1 0.693333 7850 = 43.54133 kgBerat Profil I = 17.2 1 = 17.2 kg

60.74133 kgAlat Penyambung (+ 10 %) = 6.074133 kg

Pd = 66.81547 kg

Beban HidupqL2 = 500 kg/m2 PL2 = 500 0.693333 1

= 346.6667 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 66.81547 + 1.6 346.6667= 634.8452 kg

x xx x

x

x x

x xxx

x x

x x

x x

4.7.4 Perhitungan Gaya - Gaya pada Tangga

Lab = 3.92 mLbc = 2.08 m

492.552 15.366 634.845 11.76 3.120 1728.147 1.040 3.92 RC6

Rc = 4264.476 kg

Rva = 985.10 3.92 830.84 2.08 1904.54 4264.48Rva = 3229.814 kg

B C

+

+

A

5092.1203 kgm

5294.716889 kgm

3229.813847

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

+ + ¿( () )

RAh = 0

Bidang M

Pers : Mx1 = RVA X1 - 0.5 q1

Mx1 = 3229.8138 X1 - 492.55193

dMx1

= 0 985.10386 X1 = 3229.8138

dX1 X1 = 3.2786531 m

X1 = 0 m MA = 0 Kgm

Xmax = 3.281 m Mmax= 5294.7169 Kgm tangga

X1 = 3.92 m MB = 5092.1203 Kgm tangga

B C

4.6501982

A a = 32.5444

Rav cos a

Rav sin a 3.92 m 2.08 m

Rav

X1 X2

-532.68369 kg

-2943.9981 -

2722.6517 kg -4264.4762 kg

Bidang D

Permisalan gayaDari kiri : searah jarum jam gaya dianggap positif

X= 0 m

DA = Rva cos a

= 3229.8138 cos 32.544= 2722.6517 kg

X= 3.92 m

Dbkiri = Rva cos a - q1 LAB

= -532.68369 kg

cos a

Dbkanan = P LBC - RC

= 634.84523 2.08 - 4264.4762

= -2943.9981 kg

X= 6 m

Dc = - RC

= -4264.4762 kg

X12

X12

x

x

x

x

x

x

x x

x

726.50807 kg

+

-

-1737.4881 kg

Bidang N

NA = -RVA sin a

= -3229.8138 sin 32.544= -1737.4881 kg

NBkiri = -RVA sin a + q1 L1

= 726.50807 kg

sin a

NBkanan -C = 0

4.7.5 Kontrol Kekuatan Profil4.7.5.1 Penampang Profil fy = 2400 kg/m2


125 170 210 168016 15.49193 5 15.49193

7.8125 10.97345 42 108.4435OK OK



Jarak Baut Pengikat : 250 mm = 25 cm

Lp = #VALUE! cm


b2tf

≤170

√ fyht≤

1680

√ fy

¿ ¿

¿ ¿



== 0.25 2.56 0.64 2400 = 983.04 kgcm= 9.8304 kgm


Zx = 310.445 cm3

φ Zx * fy = 0.9 310.445 2400 = 670561.2 kgcm

6705.612 kgm

Syarat -> > Mu6705.612 kgm > 5294.717 kgm

OK


f = L = 600 = 1.666667360 360

Ix = 3540 cm4

= 5 ( 1.532866 5.2 ) 1.30E+11384 2100000 3540

= 1.528344 < 1.666667

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

φMn =

φMn =

φMn


x x

Y max= 5384

(qD+qL)∗l4

EIx

x

5. Pembebanan

5.1 Perencanaan Beban Atap

Beban MatiBerat Gording = 9.3 4 = 37.2 kgBerat Asbes Gelombang = 11.33 4 = 45.32 kg

= 82.52 kg= 8.252 kg

Pm = 90.772 kgBerat Profil Kuda - Kuda = 0 cos 0.436332 = 0 kg

90.772Beban Hidup

Ph = 19.93877 4 = 79.75509 kg

108.9264 + 127.6081 = 236.5345 kg

5.2. Perencanaan Beban Angin (Gudang Tertutup)

W = 30 kg/m2

Beban Tekan Atap = 0.1 30 4 = 12 kg/m

Beban Sedot Atap = -0.4 30 4 = -48 kg/m

BebanTiup Kolom = 0.9 30 4 = 108 kg/m

Beban Sedot Atap = -0.4 30 4 = -48 kg/m

5.3 Perencanaan Beban Akibat Plat Lantai, Kolom Memanjang dan Melintang

Alat Pengikat dll (+10 %)

Pmtot

P Ultimate = 1.2 PD+ 1.6 PL =

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

0,9 - 0,4

P4 memanjang

xx

x

x x

x x

x x

x x

600

600

600

400 400 400. C

AB

A ( Balok Induk Melintang )0 x 0 x 0 x 0

A = 0 cm2 tf = 0 mm Zx = 0 cm3W = 0 kg/m Ix = 0 cm4 Zy = 0 cm3a = 0 mm Iy = 0 cm4 h = 0 mm

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

0B ( Balok Anak )

250 x 125 x 6 x 9



72.0225C ( Balok Induk Memanjang)

0 x 0 x 0 x 0



0Beban Mati- Bondex = 10.1 kg/m2

P3

P2

P1

P2

P3

P4

P5 P6 P6

mel

inta

ng

memanjang

B

II

II

III

IV

- Beton = 0.12 x 2400 = 288 kg/m2- Beban Finishing = 90 kg/m2

qM = 388.1 kg/m2

Beban Hidup

qL = 400 kg/m2

5.3.1 Perencanaan Pembebanan Portal Melintang

5.3.1.1 Beban P1Beban MatiBerat Pelat = 388.1 3 4 = 4657.2 kgBalok induk melintang = 0 3 = 0 kgBalok anak = 29.6 4 = 118.4 kg

Pm1 = 4775.6 kg

Beban HidupPh1 = 400 3 4 = 4800 kg

5.3.1.2 Beban P2Beban MatiBerat Pelat = 388.1 3 4 = 4657.2 kgBalok induk memanjang = 0 4 = 0 kgBalok induk melintang = 0 6 = 0 kg

Pm2 = 4657.2 kg


5.3.1.3 Beban P3Beban MatiBerat Pelat = 388.1 3 4 = 4657.2 kgBalok induk melintang = 0 6 = 0 kgBalok anak = 29.6 6 = 177.6 kg

Pm1 = 4834.8 kg


5.3.1.4 Beban P4Beban MatiBerat Pelat = 388.1 1.5 4 = 2328.6 kgBalok induk memanjang = 0 4 = 0 kgBerat Dinding = 4.15 4.5 4 = 74.7 kgBalok induk melintang = 6 0 = 0 kg

Pm4 = 2403.3 kg

Beban HidupPh4 = 400 1.5 4 = 2400 kg

5.3.2 Perencanaan Pembebanan Portal Memanjang

x xx

x

x x

x x

x xx

x

x x

x x

x xx

x

x xxx

x x

5.3.2.1 Beban P5 Beban Mati

Berat Pelat 388.1 6 2 = 4657.2 kgBalok induk melintang = 0 6 = 0 kgBalok induk memanjang = 0 2 = 0 kg

Pm5 = 4657.2 kg


5.3.2.2 Beban P6 Beban Mati

- Berat Pelat = 388.1 4 6 = 9314.4 kg- Balok induk memanjang = 0 4 = 0 kg- Balok induk melintang = 0 6 = 0 kg

Pm5 = 9314.4 kg Beban Hidup

Ph5 = 400 4 6 = 9600 kg

5.3.3 Beban Portal - Portal MelintangP1 Pm1 = 4775.6 kg

Ph1 = 4800 kg

P2 Pm3 = 4657.2 kgPh3 = 4800 kg

P3 Pm2 = 4834.8 kgPh2 = 4800 kg

P4 Pm4 = 2403.3 kgPh4 = 2400 kg

5.3.4 Beban - beban Portal MemanjangP5 Pm5 = 4657.2 kg

Ph5 = 4800 kg

P6 Pm6 = 9314.4 kgPh6 = 9600 kg

5.4 Perencanaan Beban Gempa ( Arah X )

24

4.5 9

5

F1

F2

x x

x x

x xxx

x xxx

6 6 6

Data Gempa:- Zone Gempa = 6- tanah lunakC = 0.95- I = 1

4

44

6 6 6

18

Kolom0 x 0 x 0 x 0



0

5.4.1 Perencanaan Beban Lantai (W1) 5.4.1.1 Beban Mati

Berat Plat = 388.1 18 4 = 27943.2 KgBalok Induk Memanjang = 0 4 = 0 KgBalok Induk Melintang = 0 18 = 0 KgBerat Dinding = 8.3 4 4.5 = 149.4 KgKolom = 0 4.5 2 = 0 Kg

0 4.5 1 = 0 Kg= 28092.6 Kg

5.4.1.2 Beban HidupBalok + Plat = Ph1 2 x Ph2 2 x Ph3 2 x Ph4

= 4800 9600 9600 4800= 28800 Kg

F1

x xxxxxx

xx

+ + ++ + +

= 28092.6 + 28800= 56892.6 Kg

5.4.2 Perencanaan Beban Atap ( W2)5.4.2.1 Beban MatiBerat Atap = 22.693 4 18 = 1802.805 KgBalok Kuda-kuda = 0 18 = 0 KgKolom = 0 2 2 = 0 KgBerat Dinding = 8.3 2 4 = 14.3 Kg

= 1817.105 Kg

5.4.2.2 Beban HidupPh1 = 19.93877 4 18 = 1435.592 kg

= 1817.105 + 1435.592= 3252.696 Kg

5.4.3 Berat Total Wt

= W1 + W2= 56892.6 + 3252.696= 60145.3 Kg

5.4.4 Perencanaan Gaya Gempa

T = 0.085 x

= 0.085 x 9= 0.441673

Tanah LunakC = 0.95 I = 1R = 4.5

V = (C x I x Wt )/R = 0.95 1 60145.3 4.5= 12697.34 Kg

Lantai = W1 x h1 = 56892.6 x 5= 284463 Kgm

Atap = W2 x H = 3252.696 x 9= 29274.27 Kgm

Σ Wi Hi = 313737.3 Kgm

F1 = W1 h1 x VS Wi hi

Beban Lantai ( W1tot)

Beban Atap ( W2tot)

Berat Total ( Wtot )

H3/4

3/4

CosαCosα

xx xx xx x

x x

x x

= 284463 x 12697.34313737.266726182

= 11512.57 Kg


= 29274.2667261815 x 12697.34313737.266726182

= 1184.766 Kg

5.4.5 Gambar Perencanaan Beban Akibat Gempa ( Arah X )

F21184.766 kg

4

F111512.57 kg

5

6 6 6

5.5 Perencanaan Beban Gempa ( Arah Y )

Data Gempa:- Zone Gempa = 6- tanah lunakC = 0.95- I = 1

Balok Kuda-Kuda0 x 0 x 0 x 0



0Kolom

0 x 0 x 0 x 0



0

5.5.1 Perencanaan Beban Lantai (W1) 5.5.1.1 Beban Mati

Berat Plat = 388.1 6 40 = 93144 KgBalok Induk Memanjang = 0 40 = 0 KgBalok Induk Melintang = 0 40 = 0 KgBerat Dinding = 8.3 6 4.5 = 224.1 KgKolom = 0 4.5 2 = 0 Kg

0 4.5 4.5 = 0 Kg= 93368.1 Kg

5.5.1.2 Beban HidupBeban Hidup Merata = 250 40 6 Kg

= 60000 Kg

= 93368.1 + 60000= 153368.1 Kg

5..2 Perencanaan Beban Atap ( W2)5.5.2.1 Beban MatiBerat Atap = 22.693 6 40 = 5446.32 KgBalok Kuda-kuda = 0 40 = 0 KgKolom = 0 2 2 = 0 KgBerat Dinding = 8.3 6 2 = 16.3 Kg

= 5462.62 Kg

5.5.2.2 Beban HidupPh1 = 19.93877 6 40 = 4785.305 kg

= 5462.62 + 4785.305= 10247.93 Kg

5.4.3 Berat Total Wt

= W1 + W2= 153368.1 + 10247.93= 163616 Kg

5.4.4 Perencanaan Gaya Gempa

T = 0.085 x

Beban Lantai ( W1tot)

Beban Atap ( W2tot)

Berat Total ( Wtot )

H3/4

x xxxxxx

xx

Cosαx xx xx x

x x

x x

x x

= 0.085 x 9= 0.441673

Tanah LunakC = 0.95 I = 1R = 4.5

V = (C x I x Wt )/R = 0.95 1 163616 4.5= 34541.16 Kg

Lantai = W1 x h1 = 153368.1 x 5= 766840.5 Kgm

Atap = W2 x H = 10247.93 x 9= 92231.33 Kgm

Σ Wi Hi = 859071.8 Kgm


= 766840.5 x 34541.16859071.826038838

= 30832.77 Kg


= 92231.3260388374 x 34541.16859071.826038838

= 3708.394 Kg

5.4.5 Gambar Perencanaan Beban Akibat Gempa ( Arah Y )

F23708.394 kg

F130832.77 kg

3/4

x xx x

P4 P3 P2 P1 P2 P3

Kombinasi Pembebanan

* Beban Mati + Beban Hidup

Pm = 90.772 kgPh = 79.75509 kg

6 6 6

Pm1 = 4775.6 kg Pm2 = 4657.2 kgPh1 = 4800 kg Ph2 = 4800 kg

Pm3 = 4834.8 kg Pm4 = 2403.3 kgPh3 = 4800 kg Ph4 = 2400 kg

* Beban Mati + Beban Hidup + Beban Angin

q = 12 kg/m q = -48 kg/m

200

P4 P3 P2 P1 P2 P3 P4

P4 P3 P2 P1 P2 P3 P4

q = 108 kg/m q = -48 kg/m6 6 6

* Beban Mati + Beban Hidup + Beban Gempa

1184.766 kg

11512.57 kg

6 6 6

P4 P3 P2 P1 P2 P3 P4

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


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


fu = 3700 kg/cm2fy = 2400 kg/cm2


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


Lateral Bracing Lb = 110 cm

Lp = 255.0503 cm


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

¿ ¿

¿ ¿



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 = 10123.25 Kgm

Nu = -28026.45 KgVu = 3834.452 kg

Max = 9049.01 KgmMbx = 10123.25 KgmMsx = 537.12 Kgm

Sby --> Muty = -8137 KgmNu = -27466 KgVu = 3149 kg

Max = 8137 KgmMbx = 7611.48 KgmMsx = 262.96 Kgm


300 x 300 x 10 x 15

A = 119.8 cm2 tf = 15 mm Zx = 1464.75 cm3W = 94 kg/m Ix = 20400 cm4 Zy = 652.5 cm3a = 300 mm Iy = 6750 cm4 h = 234 mm

bf = 300 mm tw = 10 mm Sx = 1360 cm3iy = 7.51 cm ix = 13.1 cm 1464.75


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

6.2.1 Kontrol Lendutan6.2.1.1 Kontrol Lendutan Arah X

f ijin = L = 500 = 1.388889 cm360 360

5 25.00 537.12 0.1 9049.01 10123.2548 200 20400

0.04114 cm

f <OK

6.2.1.2 Kontrol Lendutan Arah Yf ijin = L = 500 = 1.388889 cm

360 360

5 25.00 262.96 0.1 8137 7611.4848 200 6750

0.040588 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 = 0.8 (jepit-Sendi) L = 500 cmLkx = 400 cm

= 30.53435 cm

Ncrbx = 9.869022 2000000 119.8932.3466

Ncrbx = 2536200 kg

untuk arah y :kcy = 0.8 (jepit-Sendi) L = 500 cmLky = 400 cm

= 53.26232 cm (MENENTUKAN)

Ncrby = 9.869022 2000000 119.82836.874

Ncrby = 833529.2 kg

53.26232 24003.1415 2000000

0.587318

0.25 < < 1.2

ω = 1.43 1.1202541,67-0,67λc

Pn = Ag fy = 119.8 * 2400 = 256656.2 kgw 1.120254

28026.45 = 0.128469 < 0.20.85 256656.2

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


Cmx = 1Sbx = 1

1 - 28026.452536200

Sbx = 1.011174 < 1

Sbx = 1.011174

Mux = Mutx * SbxMux = 10123.25 1.011174 = 10236.37 kgm


Cmy = 1Sby = 1

1 - 28026.452536200

Sby = 1.011174 < 1

Sby = 1.011174

Muy = Muty * SbyMuy = -8137 1.011174 = -8227.923 kgm



Kgm

300 170 234 168030 15.49193 10 15.4919310 10.97345 23.4 108.4435

OK OK


b2tf

≤170

√ fyht≤

1680

√ fy

¿ ¿

¿ ¿

Sby=Cmy

1−( NuNcrby

)≥1

x

Sbx=Cmx

1−( NuNcrbx

)≥1

x

Lateral Buckling Lb = 500 cm

Lp = 381.5592 cm

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

76.5 cm4

Iw = Iy.((h^2)/4) = 1370672 cm6

197694.6 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) = 2312000 kgcmMp = Zx . fy = 3515400 kgcm < 1,5MyMy = Sx . fy = 3264000 kgcm

1,5 . My = 4896000 kgcm > Mp

Mnx = 7589364 kgcm > Mp

Mnx=Mp= 3515400 kgcm

Mny = Zy * fy= 652.5 2400= 1566000 kgcm= 15660 kgm


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

x1 = [π/s]*[sqrt((EGJA)/2)] =


x

Pu2ϕcPn

+Mux

ϕbMnx+

Muyϕ bMny

≤1

28026.45 + 10236.3676610317 + 8227.923212191221.7 256656.2 0.9 35154 0.9 15660

0.064234369004448 + 0.323540474642736 + 0.583789074229546

0.971564 < 1

OK


234 < 11002400 15.49193

0.0975 < 71.00469Plastis

Vn = 0.6 fy Aw= 0.6 2400 1 30= 43200 Kg

Vu < ФVn3834.452 < 0.9 432003834.452 < 38880

OK

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


340 x 250 x 9 x 14


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




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 =


Cmx = 1Sbx = 1

1 - 8457.0881186242

Sbx = 1.007181 < 1

Sbx = 1.007181


λc =

λc =

λc

ω =


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


Cmy = 1Sby = 1

1 - 8457.0881186242

Sby = 1.007181 < 1

Sby = 1.007181




Kgm

250 170 276 168028 15.49193 9 15.49193

8.928571 10.97345 30.66667 108.4435OK OK



Lp = 304.8409 cm


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

b2tf

≤170

√ fyht≤

1680

√ fy

¿ ¿

¿ ¿


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



Cb = 12,5 Mmax


Cb = 1.813


1,5 . My = 4608000 kgcm > Mp

Mnx = 4757518 kgcm > Mp


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


8457.088 + 15951.1003321742 + 01.7 158629.2 0.9 32640.58 0.9 5250

0.031360926501704 + 0.542988114349665 + 0

0.574349 < 1

OK


x1 = [π/s]*[sqrt((EGJA)/2)] =


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


300 x 200 x 8 x 12

htw

≤1100

√ fy


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




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 =


Cmx = 1Sbx = 1

1 - 28073620069.3

Sbx = 1.047421 < 1

Sbx = 1.047421

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



λc =

λc =

λc

ω =


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




Kgm

200 170 240 168024 15.49193 8 15.49193

8.333333 10.97345 30 108.4435OK OK



Lp = 239.3001 cm


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



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

x1 = [π/s]*[sqrt((EGJA)/2)] =

b2tf

≤170

√ fyht≤

1680

√ fy

¿ ¿

¿ ¿


x

Cb = 12,5 Mmax


Cb = 1.813


1,5 . My = 2775600 kgcm > Mp

Mnx = 2847728 kgcm > Mp


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


28073 + 6513.91060637561 + 01.7 88538.49 0.9 20245.25 0.9 2880

0.186512429941213 + 0.35750011318846 + 0

0.544013 < 1

OK


240 < 11002400 15.49193

0.1 < 71.00469Plastis


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

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

(Ruv

φRnv

)2+(Rut

Rnt

)2≤1

d1

d2d3

a

2T

2T

2T

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


Mu = 3240 kgm = 324000 kgcmPu = 1344.96 kg



Ruv = Pu = 1344.96 = 224.16 kgn 6


= 0.75 0.5 4100 1.1304 1= 1737.99 kg

a2


Pu

Mu


= 2.4 1.2 1 4100 0.75= 8856 kg


= 0.75 0.75 4100 1.1304= 2606.985 kg


224.16 + Rut < 11737.99 2606.985

Rut = T = 2270.745 kg


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


Σ T

a2

(Ruv

φRnv

)2+(Rut

Rnt

)2≤1

d1

d2d3

a

2T

2T

2T

= 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


Mu = 12614 kgm = 1261400 kgcmPu = 12203 kg



Ruv = Pu = 12203 = 1220.3 kgn 10


= 0.75 0.5 4100 2.5434 1= 3910.478 kg


= 2.4 1.8 1 4100 0.75= 13284 kg


= 0.75 0.75 4100 2.5434= 5865.716 kg


Pu

Mu


1220.3 + Rut < 13910.478 5865.716

Rut = T = 4035.266 kg


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


= 0.9 2500 0.651335 20 + 14661312

= 1480786 kgcm > 1261400 kgcmOK

Σ T

a2

(Ruv

φRnv

)2+(Rut

Rnt

)2≤1

d1 d2d3

d4 d5

a

2T

2T

2T

2T2T

7.4 Sambungan Balok dan Kolom ( Detail D )

57

60576 60

60

198141


Mu = 12976 kgm = 1297600 kgcmPu = 7343 kg



Ruv = Pu = 7343 = 734.3 kgn 10


= 0.75 0.5 4100 2.5434 1= 3910.478 kg


= 2.4 1.8 1 4100 0.75= 13284 kg


= 0.75 0.75 4100 2.5434


Pu

Mu

Pu

Mu

= 5865.716 kg


734.3 + Rut < 13910.478 5865.716

Rut = T = 4764.266 kg


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


= 0.9 2500 0.907929 20 + 17240522

= 1744480 kgcm > 1297600 kgcmOK

Σ T

a2

(Ruv

φRnv

)2+(Rut

Rnt

)2≤1

d1 d2d3

d4 d5

a

2T

2T

2T

2T2T

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 = +

= 68.62617 510.0183= 514.6147 kg/cm2

d2

fv2 fh2

2 + 2

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

f fn =

f fn

P

M

a1 a1a

c

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

7.5.2.2 Perhitungan Momen Yang Bekerja

1/6 B L2 2 =

s =

s maks = s min =

d1

L

c

b

b

c

B

1

2

3

~ 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

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 cm20.9 4100 #DIV/0!

- Putus: Pu =

s min s max

s min Bs min + s max

f fy Ag

f 0.75 fu Ag

P

M

C1/3x 1/3y

s max

s min

x y

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

Syarat tebal plat:a min = 3 mm

fu las E70 xx

a 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

19684772 perhitungan struktur gudang baja

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