158 T.Rumpel,K.Schweizerhof and M.HaBler 3.2 Hydrostatic Loading-Incompressible Fluids Under Gravity Loading The follower force part depends as in 3.1 from the change in the normal and of the gradient of the fluid under gravity loading.The latter comes into the formulation after partial integration resulting in 哈= u·[g:eeW+genW"y]△dEdn ou 0 w:w 6u, 0 dedn.(19) 0 △1n It is obvious that the first part is non-symmetric and disappears if g is set to zero.The interesting part is the volume conservation and its influence on the pressure in the linearized form.The linearized pressure is a function of the variation of the fluid level A u and the local structural deformation Au △9p=pg·(△°u-△u) (20) The volume change is zero thus the linearization is zero as well: ont·△°uddn=0. (21) Focusing on the fluid load part-the second part in(21)-we obtain based on the direction of the normal on the fluid level,which is identical to the direction of gravity,the components of the free fluid surface and the corresponding displacement ont =ont. g gl' (22) △°u=△u· (23) Thus the volume change due to the change in the fluid level can be written as g△°u· g dogdon (24) Obviously both quantities in the integral are scalars;in addition the fluid level displacement is uniform,thus we obtain (25)
158 T. Rumpel, K. Schweizerhof and M. Haßler 3.2 Hydrostatic Loading – Incompressible Fluids Under Gravity Loading The follower force part depends as in 3.1 from the change in the normal and of the gradient of the fluid under gravity loading. The latter comes into the formulation after partial integration resulting in δgΠ∆n lin = ρ 2 η ξ δu · [g · eξWξ + geηWη] ∆u dξdη + η ξ gpt 2 ⎛ ⎝ δu δu,ξ δu,η ⎞ ⎠ · ⎛ ⎝ 0 Wξ Wη WξT 0 0 WηT 0 0 ⎞ ⎠ ⎛ ⎝ ∆u ∆u,ξ ∆u,η ⎞ ⎠ dξdη. (19) It is obvious that the first part is non-symmetric and disappears if g is set to zero. The interesting part is the volume conservation and its influence on the pressure in the linearized form. The linearized pressure is a function of the variation of the fluid level ∆0u and the local structural deformation ∆u ∆gp = ρg · (∆ou − ∆u). (20) The volume change is zero thus the linearization is zero as well: ∆gv = η ξ ∗nt · ∆u dξdη + η ξ ont · ∆ou doξdoη = 0. (21) Focusing on the fluid load part – the second part in (21) – we obtain based on the direction of the normal on the fluid level, which is identical to the direction of gravity, the components of the free fluid surface and the corresponding displacement ont = ont · g |g| , (22) ∆ou = ∆o u · g |g| . (23) Thus the volume change due to the change in the fluid level can be written as ∆ov = η ξ ont · g |g| ∆ou · g |g| doξdoη. (24) Obviously both quantities in the integral are scalars; in addition the fluid level displacement is uniform, thus we obtain ∆ov = ∆ou η ξ ont · g |g| doξdoη = ∆ouSt. (25)
FE Modelling and Simulation of Gas and Fluid Supported Structures 159 S,is the size of the water surface,which can also be computed via a boundary integral over the enclosure of the fluid volume projected onto the direction of gravity nt' g dEdn. (26) Thus the change in the water level height can be written as A'u-5 1 St JoJ *nt·△udξdn (27） and the corresponding pressure change becomes △p=pg·△°u-pg·△u *nt·△udξdm-pg·△u (28) The linearized variational form of the gravity potential depending on the fluid level is then obtained as △pnt·6ud'dm *nt·△uddn u*ntg·△dedn. (29) Obviously the second part of this equation is a non-symmetric term.However, combining both non-symmetric parts ofanda symmetric ex- pression results for the complete sum 9nm=9+69+9n: =69Ⅱ：+ + u*ndd n·△udd Term I 6u·('nt图g+g8*nt)△ud'dm Term II ou 0 w:w Au 0 0 △u,e dfdn. Su:n 0 △un Term III (30)
FE Modelling and Simulation of Gas and Fluid Supported Structures 159 St is the size of the water surface, which can also be computed via a boundary integral over the enclosure of the fluid volume projected onto the direction of gravity St = η ξ ∗nt · g |g| dξdη. (26) Thus the change in the water level height can be written as ∆ou = ∆ov St = 1 St η ξ ∗nt · ∆u dξdη (27) and the corresponding pressure change becomes ∆p = ρg · ∆ou − ρg · ∆u = ρ |g| St η ξ ∗nt · ∆u dξdη − ρg · ∆u. (28) The linearized variational form of the gravity potential depending on the fluid level is then obtained as δgΠ∆p lin = η ξ ∆p∗nt · δu dξdη = ρ |g| St η ξ δu · ∗ nt dξdη η ξ ∗nt · ∆u dξdη −ρ η ξ δu · ∗ nt g · ∆u dξdη. (29) Obviously the second part of this equation is a non-symmetric term. However, combining both non-symmetric parts of δgΠ∆n lin and δgΠ∆p lin , a symmetric expression results for the complete sum δgΠlin = δgΠ∆n lin + δgΠ∆p lin + δgΠt = δgΠt + + ρ |g| St η ξ δu · ∗nt dξdη η ξ ∗nt · ∆u dξdη Term I − ρ 2 η ξ δu · ( ∗nt ⊗ g + g ⊗ ∗nt)∆u dξdη Term II + η ξ gpt 2 ⎛ ⎝ δu δu,ξ δu,η ⎞ ⎠ · ⎛ ⎝ 0 Wξ Wη WξT 0 0 WηT 0 0 ⎞ ⎠ ⎛ ⎝ ∆u ∆u,ξ ∆u,η ⎞ ⎠ dξdη. Term III (30)