tugas baja 2
DESCRIPTION
BAJA 2TRANSCRIPT
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α 30 0
5.40
4.40
2 4.40 5.40 4.40 2
L1
L2
L3
5.40 5.40 5.40
DITANYAKAN
- PERHITUNGAN GORDING DAN GAMBAR- PENGGANTUNG GORDING- IKATAN ANGIN- Gevel- LANTAI DAN BALOK ANAK- PERENCANAAN TANGGA- PERENCANAAN KANOPI
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BAB IPERHITUNGAN GORDING
1.1 DATA PERENCANAAN- JARAK MENDATAR GORDING = 1100- JARAK MIRING GORDING = 1170.21- JARAK PENGGANTUNG GORDING = 1800- KEMIRINGAN ATAP = 30 0- ATAP ASBES GELOMBANG ( GEL.BESAR T.76 250 X 110 ) DENGAN TEBAL 5 mm = 10.18
DIRENCAKAN DIMENSI GORDING PROFIL WF 100 X 50 X 5 X 7A = 11.85 Lx = 3.98ϑ = 9.30 Ly = 1.12I x = 18.7 cm 2 Zx = 42.00Iy = 14.8 cm 2 Zy = 4.375
1.2 PEMBEBANANA . Beban mati ( D )
Berat atap = 10.18 x 1.15 = 11.71Berat gording = 9.30
21.01Berat lem ± 20% = 4.20
25.21
= 25.21MDx = 1 1 25.21 cos 30 5.40 2 = 79.57
8 = 8
Mdy = 1 1 25.21 sin 30 1.80 2 = 5.108 = 8
3
B. Beban mati ( L )
Akibat air hujan ( H )
= 40 - 0.8 α ≤ 20 PPIUG 83.3.22-b= 40 - 0.8 30 = 16
Maka ϑH = 20
= 20 x Jarak antara gording= 20 x 1.15= 23
MHx = 1 1 23 cos 30 5.40 2 = 72.608 = 8
Mhy = 1 1 23 sin 30 1.80 2 = 4.668 = 8
3Akibat beban hidup terpusat ( La ) : p = 100 kg
MLax = 1 1 100 cos 30 5.40 = 116.914 = 4
Mlay = 1 1 100 sin 30 1.80 = 22.504 = 4
cm 2
kg/m 1
kg/m1
kg/m1
kg/m1
kg/m1
kg/m1
ϑo kg/m1
( ϑ0 cos α )L2
( ϑ0 cos α )L2
0≥
ϑH kg/m 2 kg/m 2
kg/m 2
kg/m 2
ϑH
kg/m 2
( ϑ0 cos α )L2
( ϑ0 sin α )L2
0≥
( P cos α )L
( P sin α )L
KG/M2
KG/M1
KG/M1
KG/M1
KG/M1
KG/M1
KG/M1
mmmm
mm
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3C. Beban Angin ( W )
Tekanan Angin w = 40 (PPIUG 83 PS 4.2 ( 2 ) )
0.02 0.4- 0.4
+ 0.9 - 0.4
α ≤ 65 0
Koefisien angin tekanc = 0.02 α - 0.4
= 0.02 30 - 0.4 = 0.2 ≈ 0
Koefisien angin hisapc = - 0.4ϑ w = c . w . b
= - 0.4 40 1.15 = -18.4
MWx = 1 ϑ w = 1 -18.4 5.40 2 = -67.078 8
Mwy = 0
D . Momen BerfaktorKombinasi I 1.2 D + 1.6 La atau H + L atau 0.8 w LRFD 6.2 - 3
MU = 1.2 MD + 1.6 MLa + 0.8 MwMux = 1.2 79.57 + 1.6 116.91 + 0.8 0
= 282.53719 Kg/mMuy = 1.2 5.10 1.6 22.50 0
= 42.125
Kombinasi II 1.2 D + 1.3 W + ƔL + 0.5 ( La atau H )Mu = 1.2 Md + 1.3 Mw + 0.5 MLaMux = 1.2 79.57 + 1.3 0 + 0.5 116.91
= 153.93619 Kg/mMuy = 1.2 5.10 + 0 + 0.5 22.50
= 17.38 Kg/m
1.3.1 Kontrol Profil ( LRFD Tabel 751Plat sayap
= 50 = 3.572 7
λ p =170
= 170 = 10.97Fy 240
< λ p
Plat badan
kg/m2
kg/m1
L 2 kg/m 2
bf
2 t f
bf
2 t f
α
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h = 100 - 2 x 7 - 2 x 8 = 14tw 5
h< Fy
λ p = 1680 = 1680 = 108.46fy 240
Karena Penampang kompak maka Mnx = Mpx = Zx.Fy1.3.2 Kontrol lateral buckling
Jarak antara 2 pengikat asbes gelombang atau jarak penahan lateral = 53.10Lb = 53.10 cmLp = 1.76 x Ly E
FyLb<Lp
= 1.76 x 1.12 2 10 6 = 56.91 cm2400
Karena Lb < Lp maka Mnx = Mpx = zx fyMnx - Mpx = Zx . Fy
= 42.00 2400 = 100800 Kg/cmMny = Mpy = = (zy .1 flens ) . y
== 1 0.7 5 2 2400
4= 10500 Kg/Mc
Kontrol interaksi Momen lentur
Mux + Muy ≤ 1ɸ Mny
282.54 + 17.38 ≤ 10.9 100800 0.9 10500
0.0031 + 0.002 ≤ 1 0.0050 < 1 OK
1.3.3 Kontrol Lendutan
Untuk balok biasa lendutan maksimum F = L240
dengan pembebanan tetap tidak berfaktor ( LRFD Tabel 6.4 -1 )Beban tetap adalah beban mati ( D ) dan beban hidup ( La dan H )
F = L = 550 = 2.29 Cm240 240
Lendutan yang terjadi
F = ≤ f = L240
F x = 5 +384 Eiy 48 Eix
Fy = 5 +384 Eiy 48 Eiy
Akibat beban mati ( D )ϑD = 25.21 Kg/MϑxD = 25.21 = 25.21 cos 30 = 21.832 kg/m
= 0.218 Kg/Cm
tw
( 1/4.tf.by 2 ) fy
ɸMnx
f x 2 + f y 2
ϑ x . Ly4 Px. Lx3
ϑ x . Ly4 Px. Ly3
cos α
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ϑyD = 25.21 = 25.21 sin 30 = 12.605 Kg/m= 0.12605 Kg/cm
Akibat beban hujan ( H )ϑh = 23 Kg/mϑxh = 23 = 23 cos 30 = 19.918 Kg/m
= 0.19918 Kg/Cmϑxh = 23 = 23 sin 30 = 11.5 Kg/m
= 0.115 Kg/cm
Akibat beban terpusat ( La )P = 100 kgPx = 100 = 100 cos 30 = 86.60 Kg/m
= 0.866 Kg/CmPy = 100 = 100 Sin 30 = 50 Kg/m
= 0.5 Kg/Cm
Lendutan arah sumbu x
Fx = 5 0.218 + 0.19918 5.40 0.866 5.40 3384 2 10 6 187 48 2 10 6 187
=
sin α
cos α
cos α
cos α
cos α
+
4
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BAB IIPENGGANTUNG GORDING
2.1. Perhitungan Pembebanan
a. Beban mati ( D ) ϑy = ϑD sin α
= 25.21 sin 30 0 = 12.61 Kg/m
= ϑy ( L/ 3 )= 8.25 1.8 = 14.85 KgJumlah gording yang di gantung sebanyak 11 batang
RI total = 14.85 x 11 = 163.35 Kg
b. Beban hidup1 . Beban air hujan ( H )
ϑy = ϑH sin α= 23 sin 30 0 = 11.5 Kg/m
R2 = ϑy ( L/ 3 )= 7.53 . 1.8 = 13.554 Kg
R2 total = 13.554 . 11 = 149.094 Kg2 . Beban Terpusat ( La )
ɸ y == 100 sin 30 0 = 50 Kg
2.2. Keseimbangan Gaya
RI
Pla sin α
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ɸ = arc . 1.15 = 0.639 01.8
1.15
ɸ R tot = 1.2 1.6 R2 tot= 1.2 163.35 + 1.6 149.094= 434.5704 Kg
R tot
1.8
T = R tot = 434.5704 = 869.141 Kgsin 30
2.3 Perencanaan batang tarik
Beban mati = 163.35 KgBeban hidup = 149.094 Kggaya tarik = 869.141 Kg
Pu = ɸ fy AgAg = Pu = 869.141 = 0.402 cm 2
ɸ fy 0.9 2400
d = 4 Ag = 4 0.402 = 0.513 cmπ 3.14
Dipakai batang bulat dengan diameter 5 mm
2.4 Kontrol batang tarik2.4.1 Kontrol leleh
Pu = ɸ. Fy . Ag= 0.9 2400 0.402= 869.141 Kg ≥ 869.141 Kg
2.4.2 Kontrol kelangsingan
L ≤ 500D
L = 1.15 = 2.30Sin 30
D ≥ 2.30 = 0.0046 cm<
1.0 cm Ok500
2.4.3 Kontrol putus
ɸ u = 0.75 ɸ Fu . Ag
˄ g = Pu0.75 ɸ . Fu
= 869.141
R1 tot +
Sin α
( 1/4 . π. 1.0 2)
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0.75 0.9 3700= 0.348
10 mm = 0.348
Bab IIIIkatan angin
3.1 Pembebanan
Ikatan angin dipasang untuk menahan beban angin dalam semua kemungkinan arah angin
9.80
2.20 2.20 2.70 2.70 2.20 2.20
Cm 2
Dipakai ɸ =
A1 A2 A3 A4
h1 h2 h3 h4 h5 h6 h7
RA RB
R1 R2 R3 R4
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Gambar 3.1 Pembebanan ikatan angin
3.2 Perhitungan beban ikatan angin3.2.1 Data perhitungan
- Tekanan tiup angin ( w ) 40 ( PPIUG 83. 4.2.2 )- Koefisien angin ( c ) 0.9 ( PPIUG 83. 4.3.1 a )- Luas bidang ( A )- Gaya yang bekerja ( R ) 0.5 W.C.A
3.2.2 Perhitungan luas bidang
= 0.00 m
= 0.00 1.625 tan 30 0 = 0.75725 m
= 0.00 3.250 tan 30 0 = 1.5145 m
= 0.00 4.875 tan 30 0 = 2.27175 m
= 0.00 6.500 tan 30 0 = 3.0290 m
= 0.00 8.1875 tan 30 0 = 3.815375 m
= 0.00 9.875 tan 30 0 = 4.60175 m
= 1/2 0.00 0.75725 1.625 = 0.615266
= 1/2 0.7573 1.5145 3.250 = 3.691594
= 1/2 1.515 3.029 3.3125 = 7.525172
= 1/2 3.8154 4.60175 1.6875 = 14.2039
3.2.3 Perhitungan gaya yang bekerja
= 1/2 40 0.9 0.615266 = 11.075 Kg
= 1/2 40 0.9 3.691594 = 66.449 Kg
= 1/2 40 0.9 7.525172 = 135.453 Kg
= 1/2 40 0.9 14.2039 = 255.670 KgR tot = 468.647 Kg
= 1/2 = 340.812 Kg
3.3 Perhitungan Dimensi Ikatan Angin3.3.1 Gaya pada ikatan angin
4.40 mO = arc tg 3.45
O 4.40= 32.09
3 1.15
z y = 0
0
TI = -Cos O
= 340.812 - 11.075Cos 32.09 0
= 389.300 Kg ( TARIK )3.3.2 Dimensi Ikatan Angin Atap
kg/m2
h1
h2
h3
h4
h5
h6
h7
A1 m2
A2 m2
A3 m2
A4 m2
R1
R2
R3
R4
RA R1 R2 R3 R4
RA
T1
R1
TI cos O + R1 - RA =
RA R1
+
++
+
+
+
x 2
+
+
+
+
+ + + +
x
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- Leleh
Pu = ɸ 0.75Δg = Pu
ɸ 0.75 Fy= 389.300
0.9 0.9 2400= 0.2003
- PutusPu = ɸ 0.75Δg = Pu
ɸ 0.75 Fu= 389.300
0.75 0.75 3700= 0.1871
Dipakai Δg = 0.2003
d = 4 0.2003 = 1.175 Cm2π
Dipakai d = 12.9 mm
- Kontrol kelangsingan
d > L500
L = 4.40 = 4.77 = 476.7 cmcos 22.69
d = 477 = 0.953 cm500
13 > 9.53 Ok
Fy Δg
Cm2
Fu Δg
Cm2
Cm2
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BAB IVPERENCANAAN GEVEL
4.10
5.40
4.40
2.20 2.20 2.70 2.70 2.20 2.20
Gambar. 4.1 Pembebanan Gevel
4.1 Data Perencanaan4.1.1 Perencanaan profil
Menggunakan profil wf 175 175 7.5 11
Δ = 51.21 ix = 7.50ϑ = 40.2 iy = 4.38Ix = 2880 zx = 360Iy = 984 zy = 117
4.1.2 Pembebanan4.1.2.1 Beban vetikal
- Berat sendiri Gevel = 8.1686 40.20 = 328.38+
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- Berat dinding setengah bata = 1/2 7.5944 8.1686 1.6875 2 = 26.60- Berat penutup atap = 1.785 2 1/2 5.5 10.18 2 = 172.63- Berat gording = 1/2 5.5 9.30 4 = 102.30
629.91
4.1.2.2 Beban HorisantalBeban merata akibat tekanan angin ( w )ϑx = c.w.a = 0.9 40 3.375 = 121.50 Kg/mMx= 1 121.5 8.172 124.1123 Kg/m
84.1.2.3 Beban Aksial ( Beban hidup )
NL= 1785 1 5.5 2.0 20 2 = 339.15 Kg2
4.1.2.4 Beban berfaktora. beban aksial ( Nu )Nu= 1.2 Nd + 1.6 NL
= 1.2 629.91 + 1.6 339.15 1298.526 Kgb. momen ( Mntx )Mntx = 1.6 Mx
= 1.6 124.112 = 198.58 Kg/m
4.2 Kontrol profil
8.17
3.42
4.75
x y
Gambar 4.2 tekuk pada level
4.2.1
Lkx = 1.0 8.17 = 8.17 m = 817.00 cmλx = Lkx = 817.00 = 108.93
Lx 7.50Ncrtx = = 2 10 6 = 18102.1 Kg
108.93
Lky = 1.0 4.75 = 4.75 m = 475.00 cmλy = Lky = 475.00 = 108.45
Ly 4.38
Ncrby = = 2 10 6 = 181831.8 kg108.45
λ = λx = 108.93λc= λx = fy
=108.93 2400
=1.2003
π E π 2 10 6λc > 1.2w = 1.25 = 1.25 1.2003 2 = 1.801
Nn= Δg.fy = 51.21 2400 = 68240.4 kgw 1.801
ND
1/8 ϑx Lx2 =
Menghitung nilai πc
π2E π2
λx2
π2 E π2
xy2
λc 2
++
=
+
=
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Nu = 1298.526159 = 0.022 < 0.2ɸc.Nn 0.85 68240.4
Digunakan rumus 2Nu < 0.2 maka Nu Mux + Muy ≤ 1.0 LRFD ( 7.4.3.3 )
ɸ.Nn 2ɸ.Nn ɸb Mnx ɸb Mny
4.2.2 Kontrol balok
Mux = ϑbx. Mntx + ϑsx . MLtxϑbx = Cm ≥ 1 cm = 1
1 NuNrcbx
= 1 = 1.00861 1557.324
181831.8Mux = 1.0086 x 198.5796
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cos 30 = 0.866sin 30 = 0.5tan 30 = 0.466cos 32.09 = 0.847
7.10 4.106.07
8.199
1.27 1.10
1.17
1.80
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BAB IPERHITUNGAN GORDING
29.16
24.72
KG/M2
KG/M1
KG/M1
KG/M1
KG/M1
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Lb<Lp
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5 0.0825 0.0753 5.50 4384 2 10 6 187
721.9843125 7687680000000
5.02718577665441000000000E-10 1.43616000000000000000000E+12
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BAB IIPENGGANTUNG GORDING
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149.094
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Bab IIIIkatan angin
m
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Gambar 3.1 Pembebanan ikatan angin
3.3125
1.6875
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Jangan di hapus0.2550.923
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BAB IVPERENCANAAN GEVEL
13.90
Gambar. 4.1 Pembebanan Gevel
328.38KgKg
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26.60172.63102.30
629.91
KgKgKg
Kg
Kg
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1.43616000000000000000000E+12
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0.40 2
0.404 4.40
5 5.40
DI SINI MUNGKIN STAMBUKNYA
PA REA ARE
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0.40 setelah memasukkan stambuk jangan lupa cari dulu akar dari nilai yang ditunjuk arah panah ke 1 menggunakan kalkulator kemudian masukkan nilainya di arah panah ke 2
Panah 1 0.2551045 1.175Panah ke 2
ini adalah hasil dari akar nilai di atas setelah menggunakan kalkulator
masukkan nilai dari gambar untuk tinggi rangka atap
DI SINI MUNGKIN STAMBUKNYA4.80
- JARAK MENDATAR GORDING =- JARAK MIRING GORDING =- JARAK PENGGANTUNG GORDING =
PA REA ARE 7.108.20
=
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dari nilai yang ditunjuk arah panah ke 1 menggunakan kalkulator kemudian masukkan nilainya di arah panah ke 2
ini adalah hasil dari akar nilai di atas setelah menggunakan kalkulator
masukkan nilai dari gambar untuk tinggi rangka atap
= 1086.72= 1150.00= 1830.00
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YAYASAN RAZAK POROSIUNIVERSITAS LAKIDENDE
FAKULTAS TEKNIK - JURUSAN TEKNIK SIPILJl. Sultan Hasanuddin No. 234 Telp. (0408) 21777 Unaaha
KABUPATEN KONAWE SULAWESI TENGGARA
LEMBAR ASISTENSINAMA : INDRASTAMBUK : 210 201 040TUGAS : BAJA II
DOSEN : EDY MANGKERE,ST
NO HARI/TANGGAL URAIAN PARAF
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