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150 Appendix CwhereΩ A (θ) = −2Ω B (θ) = 2∞∑ (−1) mm=1∞∑ (−1) nn=1a msin(a m v) exp(−a 2 mθ), (C.8)b ncos(b n w) exp(−κb 2 nθ), (C.9)and a m = mπ, b n = (2n − 1)π/2. The load required to sustain the deflection follows fromintegration of the total stress:W (t) = − 2a2 bz∫ 1 ∫ 1−1 −1E p( ˆb3K pP + ε z)vdvdw,(C.10)where the strain is given byε z = − 24azv ∆. (C.11)L 3Working out the integrals in Equation C.10 yieldsW (θ) = c 1 ∆(θ) + c 2where∞∑∞∑m=1 n=11a 2 mb 2 n∫ θexp[−θ(a 2 m + κb 2 n)]0exp[θ ′ (a 2 m + κb 2 n)] d∆dθ ′ dθ′ , (C.12)c 1 = 192a3 bE p3L 3 , (C.13)c 2 = 256a3 bE 2 pˆb 23µL 3 K p. (C.14)An analytical solution is obtained when the deflection follows an exponential decay,for example[(∆ = ∆ 0 H(θ) (1 − h) + h exp − θτ )]R, (C.15)τ Swhere H is the Heaviside step function, h reflects the amount of slip (0 ≤ h ≤ 1, and h= 0 implies no slip) and τ S is a decay rate. Substituting Eq. C.15 in Eq. C.12 yieldsW (θ) = c 1 ∆(θ) + c 2∞∑×( [exp −θ∞∑∆ 0{exp(−θΓ nm ) +a 2 m=1 n=1 mb 2 n( )]τR− Γ nm − exp[−θΓ nm ]τ Shτ Rτ R − τ S Γ nm)}, (C.16)where Γ nm = a 2 m + κb 2 n.It can be shown that Eq. C.16 reduces to the usual hydrodynamic relaxation expressionwhen h = 0. In case of slip, i.e. h > 0, the decreasing deflection results in an inflection inthe load curve similar to that of the hydrodynamic contribution. Some model calculationswere performed for an elastic porous plate filled with glycerol and having a permeability of