If I remember right Cd is 1.4 across a diagonal width, and Cm is a mass of water equal to the volume of the member. What I can't figure is how you wound up with square legs. Probably that Cd will change them to circular.
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"Being GREEN isn't easy"
.... Kermit
Are you using Cd and Cm with Morison's Equation for forces on an offshore fixed structures? Because these are different than the coefficents for added mass due to turbulence? Also Cm is sometimes referred to as Ci.
For Morrison's Equation (without motion):
You can reference API Recommended Practice 2A (either WSD or LRFD) for a table with such factors. "C.3.2.10 Morison Equation"
where
Cm- inertia coefficent
Cd- drag coefficent
Values appropriate for the case of a steady current with negligible waves or the case of large waves with Umo*Tapp>30
Umo- maximum horizontal particle velocity at storm mean water level (2D , not diffraction)
Tapp- Apparent Wave Period
D- platform leg diameter
smooth Cd=0.65, Cm=1.6
rough Cd=1.05, Cm=1.2
If <30 than you need to modify with wake encounter
What type of project is this? I will answer anyones question with hard facts as long as they are honest and detailed about the purpose for the calculations.
and then I realized you wrote "square section legs" (right? , who uses those??? And why?)
Ciria Underwater Engineering Group: Report UR8, Dynamics of Marine Structures: Methods of calculating the dynamic response of fixed structures subject to wave and current action..
pg.152 Table 4.2 and Table 4.3
for a perfect square without rounded edges...
to be used with morison's equation (with or without motions)