19684772 perhitungan struktur gudang 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 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
1.5.5 Perencanaan Batang Tarik
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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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 = 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
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
= 0.75 x 0.85 x Ag x fu
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 =
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
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/malat Pengikat dll 10 % = 0.1 x 13.45 = 1.345 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 =
Berat Total = 14.8 kg/m
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
Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :
0.125 x 33.75 x 9 = 38 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 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
Akibat Beban Angin yg Tegak Lurus Dinding (tarik) :
0.125 x 27 x 9 = 30.38 kgmN = q x Jarak Gording = 12 x 1 = 12 kg (Tarik)
Akibat Beban Angin yg Tegak Lurus Gevel (tekan) :
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
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
≤ 300Lk
=300
= 75.38 < 300 OK
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
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
≤ 12 x x Mnx x Mny
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
≤ 12 x x Mnx x Mny
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
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 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
2.6.5 Perencanaan Batang Tarik
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
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 = 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
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
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 = 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
4.5.3 Perhitungan MD dan ML
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
Mu = 1.4 8.301881 = 11.62263 kgmTidak Menentukan
Mu = 1.2 8.301881 + 1.6 60.08889 = 106.1045 kgmMenentukan
4.5.5 Kontrol Momen Lentur
= 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
4.5.6 Kontrol Lendutan
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
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 = 40.612 cm3a = 100 mm Iy = 134 cm4 h = 84 mm
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
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 84 168016 15.49193 6 15.49193
6.25 10.97345 14 108.4435OK OK
Penampang Profil Kompak, maka Mnx = Mpx
4.6.4.2 Kontrol Lateral Buckling
Jarak Baut Pengikat : 250 mm = 25 cm
Lp = 125.4929 cm
Ternyata Lp > Lb maka Mnx = Mpx
Mnx = Mpx = Zx. Fy = 84.184 * 2400 = 2020.416 KgmMny = Zy ( 1 flen ) * fy
== 0.25 3.2 100 2400 = 192000 kgcm= 1920 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 > 343.3243 kgm
OK
4.6.6 Kontrol Lendutan
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
¿ ¿
¿ ¿
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 32.5444 = 15.05589 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 > 15.05589
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
untuk Sayap untuk Badan
125 170 210 168016 15.49193 5 15.49193
7.8125 10.97345 42 108.4435OK OK
Penampang Profil Kompak, maka Mnx = Mpx
4.7.5.1 Kontrol Lateral Buckling
Jarak Baut Pengikat : 250 mm = 25 cm
Lp = #VALUE! cm
Ternyata Lp > Lb maka Mnx = Mpx
b2tf
≤170
√ fyht≤
1680
√ fy
¿ ¿
¿ ¿
Lp=1. 76∗iy√ Efy
Mnx = Mpx = Zx. Fy = 310.445 * 2400 = 7450.68 KgmMny = Zy ( 1 flen ) * fy
== 0.25 2.56 0.64 2400 = 983.04 kgcm= 9.8304 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 > 5294.717 kgm
OK
4.7.6 Kontrol Lendutan
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
(1/4∗tf∗bf 2)∗fyx x x
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
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
72.0225C ( Balok Induk Memanjang)
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
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
Beban HidupPh2 = 400 3 4 = 4800 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
Beban HidupPh3 = 400 3 4 = 4800 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
Beban HidupPh5 = 400 6 2 = 4800 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
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
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
F2 = W2 h2 x VS Wi hi
= 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
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
0Kolom
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
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
F1 = W1 h1 x VS Wi hi
= 766840.5 x 34541.16859071.826038838
= 30832.77 Kg
F2 = W2 h2 x VS Wi hi
= 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
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
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 = 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
6.2.1 Profil Baja yang Didapatkan Di SAP 2000 8.2.3 :
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
652.5Mutu Baja = BJ 37
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
X Batang Dianggap Tidak Bergoyang Maka :
Cmx = 1Sbx = 1
1 - 28026.452536200
Sbx = 1.011174 < 1
Sbx = 1.011174
Mux = Mutx * SbxMux = 10123.25 1.011174 = 10236.37 kgm
Y Batang Dianggap Tidak Bergoyang Maka :
Cmy = 1Sby = 1
1 - 28026.452536200
Sby = 1.011174 < 1
Sby = 1.011174
Muy = Muty * SbyMuy = -8137 1.011174 = -8227.923 kgm
6.2.4 Menentukan Mnx
6.2.4.1 Kontrol Penampang profiluntuk Sayap untuk Badan
Kgm
300 170 234 168030 15.49193 10 15.4919310 10.97345 23.4 108.4435
OK 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 = 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
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
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
6.2.6 Kontrol Kuat Rencana Geser
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
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
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
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
(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
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
(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
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 Bautf Rnt = 0.75 0.75 fu Ab
= 0.75 0.75 4100 2.5434
Baut Tanpa Ulir ( Bor ) Diameter f =
Pu
Mu
Pu
Mu
= 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
(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
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