Time domain analysis by AQWA

Wind & Water Structural Analysis
1

Dear Jason
I am trying to model V-shaped semisubmersible in the ANSYS-AQWA, but unfortunately I am facing some problems in regard with Time-Domain analysis of the system.
I have got Time series of motions which are relative to COG(-30.6,0,-16) in six DoFs by AQWA. However, i want to get the motions in SWL（0,0,0）. I found the Tranmat martix in this Web. And the equation is q_cg=Transmatq_swl. In other words,q_swl=Transmat^-1q_cg.
Transmat=[1 0 0 0 16.2 0
0 1 0 -16.0 0 30.6
0 0 1 0 -30.6 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1]
Transmat^-1=[1 0 0 0 -16.2 0
0 1 0 16.0 0 -30.6
0 0 1 0 30.6 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1]
Then the displacement in Heave related to swl can be expressed Heave_swl=Heave_cg+30.6*Pitch_cg. In this paper, the aerodynamic force is simplified using a force-speed curve of NREL 5MW wind turbine. So the wind force can be converted to a steady thrust force in constat wind load case.
But after i calculate the heave motion related to SWl, the average heave motion is positive. However, the average motion in heave should be negative according to fig.1.
fig.1

draft.png
draft.png717×540 21 KB

fig.2
heave displacment.png
heave displacment.png1118×842 19.1 KB

I would really appreciate it if you can help me!

best regards
Lixian.Zhang

2

Dear Lixian,

Sorry for the confusion, but it looks like I had a typo in the forum post you linked to. Here are the corrected equations:

Consider the linear system:

(M_swl + A_swl)qdd_swl + B_swlqd_swl + C_swl*q_swl = F_swl

where q_swl = [ surge, sway, heave, roll, pitch, yaw ]^T are the platform displacements, and F_swl are the platform loads, relative to the platform centerline at still water level (SWL) (0,0,0). The same equations can be written in terms of displacements and loads about the center of gravity located at (x_cg, y_cg, z_cg) using the following transformation:

q_swl = TransMatq_cg
q_cg = TransMat^-1q_swl
F_cg = TransMat^T*F_swl

where,
TransMat =
[ [ 1 0 0 0 -z_cg y_cg ];
[ 0 1 0 z_cg 0 -x_cg ];
[ 0 0 1 -y_cg x_cg 0 ];
[ 0 0 0 1 0 0 ];
[ 0 0 0 0 1 0 ];
[ 0 0 0 0 0 1 ] ]

TransMat^-1 =
[ [ 1 0 0 0 z_cg -y_cg ];
[ 0 1 0 -z_cg 0 x_cg ];
[ 0 0 1 y_cg -x_cg 0 ];
[ 0 0 0 1 0 0 ];
[ 0 0 0 0 1 0 ];
[ 0 0 0 0 0 1 ] ]

and
TransMat^T =
[ [ 1 0 0 0 0 0 ];
[ 0 1 0 0 0 0 ];
[ 0 0 1 0 0 0 ];
[ 0 z_cg -y_cg 1 0 0 ];
[ -z_cg 0 x_cg 0 1 0 ];
[ y_cg -x_cg 0 0 0 0 1 ] ]

Thus, the transformed system is:

(M_cg + A_cg)qdd_cg + B_cgqd_cg + C_cg*q_cg = F_cg

where,
M_cg = TransMat^TM_swlTransMat
A_cg = TransMat^TA_swlTransMat
B_cg = TransMat^TB_swlTransMat
C_cg = TransMat^TC_swlTransMat

Thus, you have the signs backwards on x_cg and z_cg (y_cg = 0).

Best regards,

3

Dear Jason

Thanks for your help

yours
Lixian.Zhang