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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

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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

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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

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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

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M=∫ (fD + fI ) y dy = MD+ MI (N-m) = 802,884+2809,28 =3612,164 [N-m]