pbl
Post on 21-Jan-2016
13 Views
Preview:
DESCRIPTION
TRANSCRIPT
1
Jurusan Teknik Kelautan FT. Kelautan ITS
UJIAN TENGAH SEMESTER GENAP 2006-2007 Mata kuliah : Perencanaan Bangunan Laut III Kode : Dosen : Ir. Murdjito, M.Sc.Eng Hari/tanggal : Selasa, 24 April 2007 Waktu ujian : 120 menit Sifat : Buku tertutup
Diketahui jacket platform dengan 4 kaki. Jarak antar kaki 75 m arah sumbu X dan 60 m arah sumbu Z, mempunyai 2 deck. Main deck berada pada elevasi 65 m dari seabed sedang cellar deck pada elevasi 60 m dari seabed. Dead load 2500 ton dan life load 1500 ton. Beban vertical dianggap bekerja secara merata dengan titik berat deck berada di tengah-tengah deck. Luas penampang angin masing-masing deck sebagai berikut:
Deck Luas penampang (m2) Ax Az
Main Deck 750 600 Cs main deck 1.0 1.50 Cellar Deck 940 750 Cs cellar deck 1.0 1.0
Kecepatan angin pada ketinggian 10 m dari MSL, V10 = 32.5 m/s dan masa jenis udara, ρair, pada suhu kamar sebesar 1.2 kg/m3. sedang 1/n factor adalah 1/8. Platform leg berupa caisson OD=2400 mm dan thickness of 100 mm. Hubungan kaki-deck dianggap fixed pada ketinggian 60 m dari seabed. Water depth 50 m. Modulus Elastisitas baja 210 GPa. Allowable tension/compression stress 165 MPa dan 100 MPa untuk shear stress, yield stress sebesar 260 MPa. Platform beroperasi pada daerah dengan tinggi gelombang maximum sebesar 4 m , periode gelombang 6.5 detik dan panjang gelombang 58.5 m. Teori gelombang linier Airy dengan CD=0.8 dan CM=1.6. Beban gelombang dapat dihitung dengan teori Morison. Beban arus diabaikan. Masa jenis air laut, ρ, sebesar 1.025 ton/m3.Tumpuan kaki jacket dengan seabed dianggap tumpuan fixed. Pertanyaan:
a. Tentukan besarnya gaya angin arah X, Z dan diagonal b. Tentukan gaya gelombang dan moment maksimum arah X (maksimum terjadi
pada sudut fase 345o atau ωt=6.0 rad) c. Tentukan gaya reaksi di masing-masing kaki akibat beban vertikal dan beban
lingkungan arah sumbu X. d. Tentukan tegangan maximum compression stress, bending stress dan shear stress
serta rasio tegangan kombinasi axial bending. e. Check tegangan struktur kaki terhadap kriteria tegangan struktur dengan API RP
2A WSD.
2
Catatan: Moment of inertia: Ixx = d3t/8, cross section area: A=dt;
Normal/compression stress: fa=P/A; Bending stress : fb=Mc/Ixx; Shear stress: v=V/(0,5A)
Rasio tegangan kombinasi: b
b
a F
f
F
faUC
Gaya angin:
windsair ACUFwind 2
2
Morison F=∫ (fD + fI ) dy = FD + FI (N) dan M=∫ (fD + fI ) y dy = MD+ MI (N-m) dengan asumsi: y=h+H/2 cos ωt
Nttkh
ky
kh
kyH
k
DCF DD
coscos
sinh
2
sinh
2sinh
32 22
2
Ntkh
kyH
k
DCF II
sin
sinh
sinh
82
2
mNttQHk
DCM D
D
coscos64
1
2
2
mNtQHk
DCM I
I
sin8
22
2
2
Dimana
kh
kykykykyQ
2
2
1sinh
1)(22cosh2sinh2
kh
kykykyQ
sinh
1coshsinh2
API stress criteria Axial tension: Fa=0.6 Fy Axial compression The allowable axila compresive stress, Fa, should be determined from the following AISC formulas for mebers with D/t ratio equal or less than 60:
3
2
2
8
)/(
8
)/(3
3
5
2
/1
Cc
rkL
Cc
rkL
FyCc
rkL
Fa
for kL/r < Cc and 2
2
)/(23
12
rkL
EFa
for kL/r Cc
Where 2
122
Fy
ECc
E= Young’s Modulus of elasticity, ksi (MPa) k= effective length factor ~1.0 L= unbraced length, in. (m); r= radius of gyration, in. (m)
3
For member with D/t > 60 then the local buckling criteria is used. Bending The allowable bending stress, Fb, should be determined from:
FyFb 75.0 for D/t 10,340/Fy and FyEt
FyDFb
74.184.0 for
Fyt
D
Fy
680,20340,10
FyEt
FyDFb
58.072.0 for 300
680,20
t
D
Fy
Where
D= outer diameter of the cylindrical member, in. (m) t= wall thickness, in. (m)
Shear stress The allowable shear stress, Fv, should be determined from:
FyFv 4.0
Combined stress for cylindrical members Cylindrical members subjected to combined compression and flexure should be proportioned to satisfy following requirement at all points along their length:
0.16.0
22
Fb
ff
Fy
fa bybx
When 15.0Fa
fathe following formula may be used:
0.122
Fb
fbyfbx
Fa
fa
Where:
fa = actual axial stress, ksi (MPa) fbx = in-plane bending, ksi (MPa) fby = out of plane bending, ksi (MPa)
oooooOOOOO SELAMAT BEKERJA OOOOOooooo
4
JAWAB: Luas penampang angin masing-masing deck sebagai berikut
Deck
Luas
penampang(m²)
Ax Az
Main deck 750 600
Cs main deck 1 1.5
cellar deck 940 750
Cs cellar deck 1 1
jarak antar kaki arah sumbu X = 75 m
jarak antar kaki arah sumbu Z= 60 m
Main deck = 65 m
cellar deck= 60 m V₁₀ = 32.5 m/s
1/n= 1/8
ρ= 1.2 kg/m3
water depth = 50 m
platform leg berupa coisson OD= 2400 mm
2.4 m
thickness= 100 mm
0.1 m
Modulus Elastisitas baja = 210 Gpa
allowable tension/compression stress= 165 MPa
allowable tension/shear stress= 100 MPa
yield stress= 260 MPa
gelombang max = 4 m
periode gelombang= 6.5 sekon
P. gelombang = 58.5 m
CD= 0.8
CM= 1.6
ρair laut= 1.025 ton/m3
5
Vh=32.5m/s
Cs platform = 1
1) KECEPATAN ANGIN
-MAIN DECK (Vmd) 36.6 m/s
-Cs main deck = 36.6 m/s
-Cellar Deck = 36.3 m/s
-Cs cellar deck = 36.3 m/s
2) gaya angin : windsair ACUFwind 2
2
Fx Fz
][125.1406875016.362
025.1NxxFmaindeck
][5.1125460016.362
025.1. NxxmaindeckF
][7575.18116.362
025.1. NxxCsmaindeckF ][1363.285.116.36
2
025.1. NxxCsmainF
[525.1748794013.362
025.1. NxxcellardeckF
[8.1395275013.362
025.1. NxxcellardeckF
][60375.18113.362
025.1.. NxxcellarCsF ][6038.18113..36
2
025.1.. NxxcellarCsF
Gaya angin pada sudut 3450 Main deck [N] Cellar deck [N] FѲx=10269.73
FѲx= 12765.89
FѲz = 8687.067 FѲz= 8557.717
6
Moment of inertia: Ixx = d3t/8, cross section area: A=dt;
Normal/compression stress: fa=P/A; Bending stress : fb=Mc/Ixx; Shear stress: v=V/(0,5A)
Rasio tegangan kombinasi: b
b
a F
f
F
faUC
y=50+4/2 cos 345 = 51.931 [m] k = 2 / ٨ = 0.2419
ω = = 1.051
Nttkh
ky
kh
kyH
k
DCF DD
coscos
sinh
2
sinh
2sinh
32 22
2
Nx
xx
x
xxx
x
xxFD 144.16345cos345cos
502419.0sinh
931.512419.02
502419.0sinh
931.512419.02sinh4051.1
2419.032
4.28.0025.122
2
Ntkh
kyH
k
DCF II
sin
sinh
sinh
82
2
Nx
xx
x
xxFI 77,58345sin
502419,0sinh
931,512419.0sinh4051,1
2419.08
4.26,1025.1 22
Morison F=∫ (fD + fI ) dy = FD + FI (N) = 16,144+58,77 = 74,914 N
kh
kykykykyQ
2
2
1sinh
1)(22cosh2sinh2 =
809,122502419,0sinh
1)931,512419,0(2931,512419,02cosh931,512419,02sinh931,512419,022
2
1
x
xxxxxxxQ
kh
kykykyQ
sinh
1coshsinh2
4459,18502419,0sinh
1931,512419,0cosh931,512419,0sinh931,512419,02
x
xxxQ
mNttQHk
DCM D
D
coscos64
1
2
2
mNxxx
M D 884,802345cos345cos809,1224051,12419,064
4,28,0025,1 2
2
mNtQHk
DCM I
I
sin8
22
2
2
mNx
xxM I 28,2809345sin4459,184051,1
2419,08
4,26,1025,1 2
2
2
7
M=∫ (fD + fI ) y dy = MD+ MI (N-m) = 802,884+2809,28 =3612,164 [N-m]
top related