Youngs modulus of Mooring line

Wind & Water Materials
1

Dear sir

Am trying to find the Young’s modulus of the mooring line from default OC3 Moordyn values

The lines below are from Moordyn:

C(6) = {‘Name Diam MassDen EA BA/-zeta Can Cat Cdn Cdt’};
C(7) = {‘(-) (m) (kg/m) (N) (N-s/-) (-) (-) (-) (-)’};
C(8) = {‘main 0.09 77.7066 384.243E6 -0.8 1 0.1 1 0.1’};

Given EA = 384e6

for the given diameter 0.09m the area(A) turns around 2pi.09*.09/4 = 0.0127m2.(By assuming studless chain)

On Dividing the EA with A ,E =3.01E10 N/m2, which is less than E of steel 2.1E11 N/m2.

Kindly guide me if I miss any assumptions in the calculation of Young’s of modulus of the mooring line.

2

Dear Krishnaveni,

Yes, your calculations look correct, but I can’t comment on why your calculated value of the Young’s modulus doesn’t match the textbook value.

Best regards,

3

Dear Jason

Thanks for your quick response.

Am trying to find how the Diam,MassDen and EA are calculated in the default OC3 line file.

C(6) = {‘Name Diam MassDen EA BA/-zeta Can Cat Cdn Cdt’};
C(7) = {‘(-) (m) (kg/m) (N) (N-s/-) (-) (-) (-) (-)’};
C(8) = ‘main 0.09 77.7066 384.243E6 -0.8 1 0.1 1 0.1’};

Here the MassDen = 77.7066kg/m

Diam , which is referred as the (in the manual)volumetric equivalent diameter = sqrt(77.70664/7850pi())= 0.112m , which is higher than the diameter used in default line.

If you or Dr.Matt could kindly clarify on the line properties calculations , would be greatly appreciated.

Thanks
Krishnaveni

4

Dear Kirshnaveni,

The OC3-Hywind specification of the mooring system was simplified from the data provided by Statoil to hide the details of the “delta” connection, which at the time was not information that Statoil wanted to make publicly available. As a result, some of the line properties may be slightly off what you think they should be using the basic formulas you are using. But I cannot share the actual properties with you due to confidentiality of the data.

Best regards,

5

Thank you for your response Dr,Jason.