2.016 Hydrodynamics Prof. A.H. Techet

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1 .01 Hydrodynics Reding #.01 Hydrodynics Prof. A.H. Techet Added Mss For the cse of unstedy otion of bodies underwter or unstedy flow round objects, we ust consider the dditionl effect (force) resulting fro the fluid cting on the structure when forulting the syste eqution of otion. This dded effect is dded ss. Most floting structures cn be odeled, for sll otions nd liner behvior, by syste eqution with the bsic for siilr to typicl ss-spring-dshpot syste described by the following eqution: x + bx + kx = f () t (.1) where is the syste ss, b is the liner dping coefficient, k is the spring coefficient, f(t) is the force cting on the ss, nd x is the displceent of the ss. The nturl frequency ω of the syste is siply k. (.) ω = In physicl sense, this dded ss is the weight dded to syste due to the fct tht n ccelerting or decelerting body (ie. unstedy otion: d 0 ) ust ove soe volue of surrounding fluid with it s it oves. The dded ss force opposes the otion nd cn be fctored into the syste eqution s follows: x + bx + kx = f () t x (.) where is the dded ss. Reordering the ters the syste eqution becoes: ( + ) x + bx + kx = f () t (.4) Fro here we cn tret this gin s siple spring-ss-dshpot syste with new ss = + such tht the nturl frequency of the syste is now ω = k k = (.5) + It is iportnt in ocen engineering to consider floting vessels or pltfors otions in ore thn one direction. Added ss forces cn rise in one direction due to otion in different direction, nd thus we cn end up with x trix of dded ss coefficients. version.0 updted 8/0/ A. Techet

2 .01 Hydrodynics Reding # ooking siply t body in two-diensions we cn hve liner otion in two directions nd rottionl otion in one direction. (Think of these coordintes s if you were looking down on ship.) Two diensionl otion with xis (x,y) fixed on the body. 1: Surge, : Swy, : Yw The unstedy forces on the body in the three directions re: = du 1 + F du + 1 du (.) du F = du + du (.7) du F = du + du (.8) Where, F, nd F, re the surge (x-) force, swy (y-) force nd yw oents respectively. It is coon prctice in Ocen Engineering nd Nvl Architecture to write the oents for roll, pitch, nd yw s F 4, F 5, nd F nd the ngulr otions in these directions s X 4, X 5, nd X. This set of equtions, (.)-(.8), cn be written in trix for, F = [ M ] u, du F = 1 du 1 du (.9) version.0 updted 8/0/ A. Techet

3 .01 Hydrodynics Reding # Considering ll six degrees of freedo the Force Mtrix is: u u u F = (.10) u u u We will often bbrevite how we write the Force trix given in (.10) using tensor nottion. The force vector is written s the ccelertion vector s F = F i, where i = 1 N,,, 4 N, 5,, (.11) iner Forces Moents u i = [, u u, u u 5, u ], (.1) u, 1 4, nd the dded ss trix [ ] s ij where i, j =1,,, 4, 5,. (.1) A good wy to think of the dded ss coponents, ij, is to think of ech ter s ss ssocited with force on the body in the i direction. th direction due to unit ccelertion in the j th For syetric geoetries the dded ss tensor siplifies significntly. For exple, figure shows dded ss vlues for circle, ellipse, nd squre. In the cse of the circle nd squre, oveent in the 1 nd directions yields siilr geoetry nd identicl dded ss coefficients ( = ). 11 version.0 updted 8/0/ A. Techet

4 .01 Hydrodynics Reding # Circle Ellipse Squre 11 = = ρd 11 = ρπb 11 = =151. πρ 4 = 0 = ρπ = 0. 4 πρ = ρ b Two diensionl dded ss coefficients for circle, ellipse, nd squre in 1: Surge, : Swy, : Yw sing these coefficients nd those tbulted in Newn s Mrine Hydrodynics on p.145 we cn deterine the dded ss forces quite siply. In three-diensions, for sphere (by syetry): 1 11 = = = ρ = A (.14) A OTHER ij TERMS ARE ZERO ( i j ). version.0 updted 8/0/ A. Techet

5 .01 Hydrodynics Reding # Generl DOF forces nd oents on Rigid body oving in fluid: Velocities: G Trnsltion Velocity : () t = (, 1, ) (.15) G Rottionl Velocity :() t = ( 1,, ) ( 4,5, ) (.1) All rottion is tken with respect to Origin of the coordinte syste (often plced t the center of grvity of the object for siplicity!). Forces: (force in the j th direction). ( i = 1, 45,,,, nd j, k, l = 1,, ) F j = i ij ε jkl i (.17) li Moents: ( i = 1, 45,,,, nd j, k, l = 1,, ) Einstein s sution nottion pplies! The lternting tensor ε jkl is siply M j = i j +,i ε jkl i l +,i ε jkl li (.18) k i 0; if ny j, k, l re equl ε jkl = 1; if jkl,, re in cyclic order (.19) 1; if jkl,, re in nti-cyclic order The full for of the force in the x-direction ( ) is sued over ll vlues of i: for kl, =1,,. F N 1 = N j=1 i=1 i= i= i=4 i=5 i= ε 1kl 1 l1 ε 1kl l ε l ε 1kl 1kl 4 k l 4 (.0) i=1 i= i= i=4 ε1kl 5 l 5 ε 1kl k l i=5 i= version.0 updted 8/0/ A. Techet

6 .01 Hydrodynics Reding # Next we cn choose the index k to cycle through. It is helpful to note tht the only ters where k plys role, contin ε jkl. Following the definition for ε jkl given in (.19) nd since j = 1, ll ters will be zero for k = 1. Therefore k cn only tke the vlue of or : = N N 1 j=1 i=1 i= i= i=4 i=5 i= ε 1 l 1 l1 ε 1 l l ε 1 l l ε 1 l 5 l 5 ε 4 l 4 ε 1 l 1 l l i=1 i= i= i=4 i=5 i= k = ε 1 l 1 l1 ε 1 l l ε 1 l l ε 1 l 4 l 4 ε 5 l5 ε 1 l 1 l l i =1 i= i= i=4 i=5 i= k = (.1) Finlly we cycle through the index l. Agin it is helpful to note tht the only ters where l plys role, contin ε jkl. Following the definition for ε jkl given in (.19) nd since j = 1, nd k = or, then ll ters will be zero for l = 1 nd soe zero for the cse l = nd others zero when l =. ike before l cn only tke the vlue of or such tht l k j : = N N 1 j=1 i=1 i= i= i=4 i=5 i= ε ε 1 ε ε ε 1 ε 1 1 i=1 i= i= i=4 i=5 i= k =; l = ε 1 ε ε ε ε ε 1 i =1 i= i= i=4 i=5 i= k =; l = (.) On the second row of the eqution bove, the indices of the lternting tensor, ε jkl, re in cyclic order jkl = 1 ( ε 1 =+1). In the third row, the indices re in nti (or reverse) cyclic order: ε 1 = 1 where jkl = 1. More thn likely you will never hve to write out ll six force equtions with ll the ters s the velocity nd ccelertion of the body will be zero in certin directions. However for full sekeeping nlysis of ship then one dy you just ight need to be ble to deterine ll the forces! version.0 updted 8/0/ A. Techet

7 .01 Hydrodynics Reding # Typicl Exple: For body oving in the fluid with velocity nd ccelertion K V = (1,0,1,0,0,1) = (,0,, 0, 0, ) = (,0,, 0, 0, ) (.) 1 1 K = (1,0,0,0,0,1) = (,0,0,0,0, 1 ) (.4) we cn find the force on the body in the X-direction. The force in the x-direction is so j=1. First substitute 1 for every instnce of j in eqution (.17) to get: F j=1 = = i i1 ε 1kl i li (.5) Next we need to cycle through the possible vlues for i (i = 1,,,4,5,). ooking t eqution (.5), it is cler tht the only i th ccelertions tht will tter re the non-zero ones fro (.4), thus 1 nd, nd the only i th velocities to consider re for i = 1,, nd [eqn (.)]. = ε 1kl 1 l1 ε 1kl k l ε 1kl l (.) N i=1 i= i=1 i= i= Now look t the k-index: ( k j k =, ) However = 0 nd 0 thus for k = ll ssocited ters will be zero, so we only hve to del with k =. Since j =1nd k = the only vlue left for l, tht could result in non-zero ters, is. = N i=1 i= ε ε ε 1 1 i=1 i= i= k =; l = (.7) If the body in question ws siple, syetricl sphere we could reduce this even further. sing the dded ss vlues fro (.14) nd trusting tht the off-digonl dded ss ters re zero (just for the sphere), the force in the x-direction on sphere, given (.) nd (.4), is = N 1 11 (.8) i=1 version.0 updted 8/0/ A. Techet

8 .01 Hydrodynics Reding # Deterining D Added Mss sing Slender Body Theory To forulte the dded ss of syste such s ship or subrine tht cn be odeled s slender body, we first need the two-diensionl sectionl dded ss coefficients. We will consider slender body to hve chrcteristic length in one direction tht is considerbly longer thn its length in the other two directions. For these slender bodies we cn use known D coefficients to find the unknown D dded ss coefficient for the body. The dded ss force cting on the body due to unstedy otion is F j = i ij ε jkl i (.9) li where ij is the dded ss in the i th direction due to unit ccelertion in the j th direction nd i,j = 1:. The dded ss tensor, ij, is syetric! To find the D dded ss coefficients consider siply the body geoetry, ignoring for now the ctul otions of the vessel. To strt, orient the 1-xis long the long xis of the slender body s shown in figure 1. The D dded ss coefficients will be found by suing (or integrting) the dded ss coefficients of the D cross-sectionl slices long the body. Slender body oriented with the long xis in the 1-direction. version.0 updted 8/0/ A. Techet

9 .01 Hydrodynics Reding # The sectionl dded ss coefficients re tbulted for siple geoetries. In generl, with the slender body ligned lengthwise long the 1-xis, the D cross-sectionl slice is ligned with the - plne, soe distnce x fro the origin (figure 1). This D slice is shown in figure. To find the D coefficients we need to know the D coefficient of ech section (strip) long the length of the vessel. For unifor dieter cylinder this is quite siple, but for ships with coplex geoetry there is bit ore work involved. D cross-sectionl slice of slender body. The D coefficients will be written s ij wheres the D coefficients re written s ij. Fro here on we will follow the bsic forultions used in the hndbook: Principles of Nvl Architecture Vol III., (1989) Soc. Nvl Arch. nd Mrine Engineers, p = = = = x = 5 = x = 4 = x 4 = x 55 = x version.0 updted 8/0/ A. Techet

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