A shaft alley divides a vessel's cargo hold into two tanks, each 25 ft. wide by 50 ft. long. Each tank is filled with salt water below the level of the shaft alley. The vessel's displacement is 6,000 tons. What is the reduction in GM due to free surface effect?
• Free surface correction (FSC) reduces GM by an amount equal to the total free-surface moment divided by the vessel’s displacement volume • For a rectangular tank, the second moment of area of the free surface is based on the tank’s breadth cubed times its length • To use displacement in tons, you must first convert it to volume in cubic feet using salt water density
• How do you calculate the second moment of area (I) for ONE rectangular tank with a free surface that is 25 ft wide and 50 ft long? Which dimension gets cubed? • Once you have I for one tank, how do you combine the effect of TWO identical tanks separated by the shaft alley? • Your displacement is given as 6,000 tons, but the formula needs volume in cubic feet. How do you convert tons of salt water to cubic feet, and then use that in the free-surface correction formula?
• Be sure you are using breadth (transverse width) as the dimension that is cubed in the I = (L·B³)/12 type formula, not the length • Remember there are two tanks, so you must account for the free surface of both when summing I before dividing by displacement volume • Confirm you convert 6,000 tons to underwater volume in ft³ using the correct factor for salt water (tons ↔ ft³) before computing the reduction in GM
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