5003 Lectures - Faculty of Engineering and Applied Science

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E5003 - Ship Structures I 44 © C.G. Daley where ff,fa are the distances from the to the centers of weight (fore and aft) gf,ga are the distances from the to the centers of buoyancy (fore and aft) The bending moment at midships is; or = W f − ∆ a f a f a = W f − ∆ g f a g f These are two ‘estimates’ of the maximum bending moment. We can combine the two, and increase our accuracy, by taking the average of the two; 1 ( W f + W f ) − ( ∆ g + ∆ g ) 1 = 2 a a f f 2 a a f f = BM W − BM B weight - buoyancy To find the buoyancy part, Murray suggested 1 ( ∆ g + ∆ g ) = ∆ ⋅ x 1 BM B = 2 a a f f 2 where x = average moment arm Murray suggested a set of values for x , as a function of the ship length, block coefficient and the ratio of draft to length; x L( a CB b) + ⋅ = where
E5003 - Ship Structures I 45 © C.G. Daley T/L a b .03 .209 .03 .04 .199 .041 .05 .189 .052 .06 .179 .063 This table for a and b can be represented adequately by the equation; a = . 239 − T / L b = . 1. 1T / L − . 003 Example using Murray’s Method Ship: Tanker L=278m, B=37m, CB=0.8 Assume wave bending moment is; WBMsag = 583800 t-m WBMhog = 520440 t-m The vessel weights, and weight bending moments are as follows; I TEM Weigh t lcg Moment (t) (m) ( t-m) Fwd car go 62000 40 2480000 fu el & wat er 590 116 68440 st eel 12000 55.6 667200 3,215,640 Aft car go 49800 37 1842600 machinery 3400 125 425000 ou tfit 900 120 108000 st eel 12000 55.6 667200 Σ 140690 t 3,042,800 To find the buoyancy moment we need the draft; W B = ∆ = C ⋅ L ⋅ B ⋅T ⋅γ BM w = 3,129,220
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