Motion Analysis of Floating Structure [PDF]

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1.1. Motion Analysis of Floating Structure 1.1.1. Equation of Motion Floating structures have 6 degrees of freedoms. Its motions are divided into two big groups, those are translation mode and rotation mode. The 6 motions are (Journee & Massie, 2001):  Translations Mode 1. Surging, translation movement on the X axis 2. Sway, translation movement on the Y axis 3. Heave, translation movement on the Z axis  Rotation Mode 4. Rolling, rotation on the X axis 5. Pitching, rotation on the Y axis 6. Yawing, rotation on the Z axis



Figure 1. Six degrees of freedom in floating structures (Journee & Massie, 2001) The equation of motion of all six degrees of freedom can be written as: 6



∑ ( M jk + A jk ) S´ k + B jk S´k +C jk S k =F j eiωt k=1



(Equation 1)



Where : k : mode of motion takes 1,2,3,4,5 and 6 for the surge, sway, heave, roll, pitch and yaw, respectively j : mode of excitation and takes the values similar to k for the corresponding modes Mjk : mass matrix containing the mass, mass moment of inertia and products of inertia of the body Ajk : added mass matrix containing added mass and added moment of inertia per unit acceleration, which are frequency dependent



Bjk



: damping matrix containing damping force and moment of inertia per unit velocity : restoring matrix containing restoring force and moment matrix per unit displacement : complex motion displacement vector per unit wave amplitude, and : complex external force and moment vector per unit amplitude.



Cjk S Fj



Equation 1 can also be represented as action and reaction of forces in structures. The left part of equation is the term of reaction forces while the right part is the term of action forces (Djatmiko, 2012). Among 6 motions of floating structures, there are only 3 pure oscillatory motions (heave, roll and pitch) since these motions has restoring forces. The others (surge, sway and yaw) do not return to its equilibrium positions unless the exciting forces acting alternately from opposite direction (Bhattacharyya, 1978). Variables of added mas, damping coefficient and stiffness of submerged part in heaving, pitching and rolling motion and the solution of motion can be calculated by dividing structures into several sections. This method is widely known as strip theory (Bhattacharyya, 1978). 1.1.1.1. Heaving Motions Heaving g motions happens in the mode 3. The displacement of motion can be written as z. So that the equation of motion can be simplified into (Bhattacharyya, 1978): ω (¿ ¿ e t) ´ ( M + A 33 ) z + B33 ´z +C 33 z=F o cos ¿ Where: Fo



: Amplitude of excitation force (N)



ωe



: Frequency of wave encountering (rad/sec)



(M + A 33) ´z



: Inertial force (N)



B 33 ´z



: Damping force (N)



C33 z



: Restoring force (N)



ω (¿¿ e t ) Fo cos ¿



1.1.1.2.



(Equation 2)



Rolling Motions



: Exciting force (N)



Rolling motions happens in the mode 4. The displacement in the motion can be written as ϕ. So that the equation of motion can be simplified into (Bhattacharyya, 1978): ω (¿¿ e t) ( M + A 44 ) ∅´ + B44 ´∅+ C44 ∅=M o cos ¿ Where: Mo



(Equation 3)



: Amplitude of excitation Moment (N m)



ωe



: Frequency of wave encountering (rad/sec)



( M+ A 44 ) ∅´



: Inertial Moment (N m)



B 44 ∅´



: Damping Moment (N m)



C 44 ∅



: Restoring Moment (N m)



ω (¿ ¿ e t ) M o cos ¿



: Exciting Moment (N m)



1.1.1.3. Pitching Motions Pitching motions happens in the mode 5. The displacement in the motion can be written as ϕ. So that the equation of motion can be simplified into (Bhattacharyya, 1978): ω (¿¿ e t) ´ ´ ( M + A 55 ) θ +B 55 θ+C 55 θ=M o cos ¿ Where: Mo ωe



(Equation 4)



: Amplitude of excitation Moment (N m) : Frequency of wave encountering (rad/sec)



( M + A 55 ) θ´ : Inertial Moment (N m) B 55 θ´



: Damping Moment (N m)



C55 θ



: Restoring Moment (N m)



ω (¿ ¿ e t ) M o cos ¿



: Exciting Moment (N m)



MODELLING BUOY SIZING The shape of the buoy is cylindrical. Modelling of the buoy is done on Maxsurf Hull Modeller software. The size of the buoy is taken to be: Buoy Size Radius, R 2.00 Height, H 1.50 Draught, T 0.75



m m m



MASS DISTRIBUTION The mass of cylinder is assumed to be distributed uniformly in its volume. The draught of the floating structures is also assumed to be half of the height. So that the weight (displacement) can be shown as:



∆=∇ × ρwater Where:



∆ = Displacement (kg)



∇ = Volume displacement (kg) For Cylindrical shape 2



∇=π R T



ρwater



= Density of water (kg/m3)



Using the formula above, the value of displacement is found to be: Volume Displacement,V Displacement, Δ



9.42 9424. 78



m3 kg



The other parameter for mass distribution is its moment of inertia. The formula for uniformly distributed cylinder is:



1 I x= M R 2 4 1 I y = M R2 4 1 I z= M R 2 2 Where:



I x = Moment of Inertia of x axis (kg m2) I y = Moment of Inertia of y axis (kg m2) I z = Moment of Inertia of z axis (kg m2) M = Mass/ Displacement of cylinder (kg) R = Radius of cylinder (m) Using the equation above the moment of inertia can be found: Ixx Iyy Izz



9424.7 8 9424.7 8 18849. 56



kg m2 kg m2 kg m2



Center of Gravity Since the mass is distributed uniformly the center of gravity can be taken: VCG TCG LCG



0.75 0.00 0.00



m m m



(SEBAIKNYA ADA GAMBAR AXIS NYA MAS, JADI TERJAWAB 0.75 M ITU TERHADAP MANA, TP BELUM TAK BUAT MAS) Radii of Gyration The parameter of moment inertia in Maxsurf Motion is filled using radii of gyration. While the radii of gyration is defined:



R x=



√ √



R y=



Iy M



R z=



Iz M



√



Ix M



Where:



I x = Moment of Inertia of x axis (kg m2) I y = Moment of Inertia of y axis (kg m2) I z = Moment of Inertia of z axis (kg m2) M = Mass/ Displacement of cylinder (kg)



R x = Radi of gyration x axis (m) R y = Radi of gyration y axis (m) R z = Radi of gyration z axis (m) Using the equation above the radii of gyration of model can be found: Rxx Ryy Rzz



1.00 1.00 1.41



m m m



The analysis of motion is done in Maxsurf Motion. The model is stripped (divided) into:



Using strip theory in Maxsurf Motion the response amplitude operator is:



HEAVE MOTION 1.5 1



RAO (m/m)



.5 . 0,000 0,002 0,004 0,006 0,008 0,010 0,012 0,014



ω (rad/s) 0 Deg



30 Deg



60 Deg



90 Deg



45 Deg



ROLL MOTION 8 6



RAO (deg/m)



4 2 . 0,000 0,002 0,004 0,006 0,008 0,010 0,012 0,014



ω (rad/s) 0 Deg



30 Deg



60 Deg



90 Deg



45 Deg



PITCH MOTION 4 3



RAO (deg/m)



2 1 . 0,000 0,002 0,004 0,006 0,008 0,010 0,012 0,014



ω (rad/s) 0 Deg



30 Deg



60 Deg



90 Deg



45 Deg



The analysis of response of structure can be conducted in either: 1. Regular Waves 2. Irregular Waves For regular waves, we assumed the wave to be 1 m height and 3.817 rad/s frequency, the response will be: Headi ng (deg) 0.000 30.00 0 45.00 0 60.00 0 90.00 0



Heave (m) 0.063



Response Roll Pitch (deg) (deg) 0.000 0.232



0.079



0.478



0.277



0.059



0.677



0.266



0.038



0.829



0.244



0.028



0.957



0.198



Note: 







 



Analisa kali ini baru ngeluarin heave roll pitch aja mas, soalnya ini masih pake strip bukan pake panel mas (softwarenya) kalua pake panel nanti bisa ngeluarin 6 gerak semuanya mas tp agak lama ngerunningnya. Kalau memang butuh bisa aku runningin mas, tapi kalau butuh heave aja bisa pake ini mas. Asumsi bendanya beban merata ya mas hehehe, kalua ada penjelasan lebih lengkap mengenai distribusi bebannya bisa nanti tak coba hitung mas moment inersianya yang baru. Kalau dimensinya sudah fix nanti bisa diganti-ganti lagi kok mas, santaii Refrensi yang biasa dipake mas, di lab hidro bukunya udah tak print mas:



References Bhattacharyya, R., 1978. Dynamics of Marine Vehicles. New York: John Wiley & Sons. Djatmiko, E. B., 2012. Perilaku dan Operabilitas Bangunan Laut di Atas Gelombang Acak. Surabaya: ITS Press. Journee, J. & Massie, W., 2001. Offshore Hydromechanics. 1st ed. Delft : TU Delft.