MAXSURF | MOSES | SACS | OpenWindPower - Calculating Radius of Gyration

Introduction

The radius of gyration is a useful property of structural systems which describes the distribution of mass about its axes. It is frequently used in floating systems to determine the behavior of vessels in conjunction with their cargo. This post describes the method by which one can obtain the radius of gyration for any SACS model.

Tow Analysis

A Tow analysis must be performed to calculate the radius of gyration in order to calculate the distribution of mass for the modeled steel any additional loading. This post is using a modified model from Sample 09 which is attached for reference.

Jacket Model

SACS Input

Any valid tow analysis may be used to obtain the radius of gyration. In order to report the necessary information to obtain the radius of gyration, the rigid body mass matrix must be turned on. This can be done by adding the RBMASS line to the end of the tow input file.

See an excerpt the sample tow input file below:

TOWOPT  ENEC    MP  WPOR         115.   0.05   20.0XYZ

...

RBMASS

END

SACS Output

After the Tow analysis is performed, the following output may be obtained from the tow listing file:

MASS MATRIX ABOUT SACS CENTER OF GRAVITY

IN SACS COORDINATES

(SLUGS AND FT)

	DX	DY	DZ	RX	RY	RZ
DX	0.309480E+05	0.000000E+00	0.000000E+00	0.000000E+00	0.000000E+00	0.000000E+00
DY	0.000000E+00	0.309480E+05	0.000000E+00	0.000000E+00	0.000000E+00	0.000000E+00
DZ	0.000000E+00	0.000000E+00	0.309480E+05	0.000000E+00	0.000000E+00	0.000000E+00
RX	0.000000E+00	0.000000E+00	0.000000E+00	0.599640E+08	0.113685E+06	0.115139E+08
RY	0.000000E+00	0.000000E+00	0.000000E+00	0.113685E+06	0.250398E+09	0.118995E+06
RZ	0.000000E+00	0.000000E+00	0.000000E+00	0.115139E+08	0.118995E+06	0.257185E+09

Calculating the Radius of Gyration

Radius of Gyration about the Center of Gravity

The radius of gyration about the center of gravity can be calculated using the following equation:

$K_{axis} = \sqrt{I_{axis}/m}$

where

$I_{axis}$ is the mass moment of inertia

and

$m$ is the mass

These values can be obtained for each axis with the following equations:

$K_x = \sqrt{I_{x}/m_x} =\sqrt{M_{4,4}/M_{1,1}}$

$K_y = \sqrt{I_{y}/m_y} =\sqrt{M_{5,5}/M_{2,2}}$

$K_z = \sqrt{I_{z}/m_z} =\sqrt{M_{6,6}/M_{3,3}}$

Note that typically the mass does not vary for each axis so $m_x=M_{1,1}$ is typically suitable for each equation.

Using the sample problem values from the tow listing file the following radii are calculated:

$K_x=\sqrt{0.599640E+08/0.309480E+05}=44.018$

$K_y=\sqrt{0.250398E+09/0.309480E+05}=89.950$

$K_z=\sqrt{0.257185E+09/0.309480E+05}=91.160$

Radius of Gyration about a Point

These values are for the radius of gyration about the center of gravity of the model. If the radius of gyration is required about a certain point (e.g. the center of motion), the parallel axis theorem may be applied to modify the radius of gyration about the center of gravity. The radius of gyration about the center of gravity can be calculated using the following equation:

$K_{axis} = \sqrt{(I_{axis}+m*r^2)/m}$

where

$r$ is the distance from the center of gravity to the point in question

Using the center of motion from the example file we can calculate the distances from the tow listing file:

The center of gravity taken from the tow listing file:

TOTAL CENTER OF GRAVITY X ......    112.49 FT        
                        Y ......      0.05 FT        
                        Z ......     41.37 FT

The center of motion taken from the tow listin file:

CENTER OF MOTION X .............    115.00 FT        
                 Y .............      0.05 FT        
                 Z .............     20.00 FT

The calculated distances are as follows:

$r_x=115.00-112.49=2.51$

$r_y=0.05-0.05=0.00$

$r_z=41.37-20=-21.37$

and the calculated radii are:

$K_x=\sqrt{(0.599640E+08+0.309480E+05*2.51^2)/0.309480E+05}=44.089$

$K_y=\sqrt{(0.250398E+09+0.309480E+05*0.00^2)/0.309480E+05}=89.950$

$K_z=\sqrt{(0.257185E+09+0.309480E+05*21.37^2)/0.309480E+05}=93.632$

Note that the radius of gyration does not change for $K_y$ since the distance between the two points is zero.

Sample Files

communities.bentley.com/.../radius-of-gyration.zip