THE SCIENCE AND APPLICATIONS OF ACOUSTICS - H. H. Arnold ...

126 6. Membrane and PlatesThe elastic resistance to flexing provides the restoring force that acts on thecircular plate. While there may be some temptation to consider the coefficient−κ 2 E/[ρ(1 − μ 2 )] 2 on the right-hand side of Equation (6.38) as being analogousto −κ 2 E/ρ in Equation (5.28) for the transverse vibration of a bar, this is notstrictly true because a sheet will curl up sideways as it is bend downward along itslength. This is the Poisson’s effect in which the curling occurs from the lateral expansionas the longitudinal compression ensues from the bending of the plate. Anincrease in the effective stiffness is thereby produced. The Poisson’s ratio μ givenin Equation (6.38) is the negative ratio of the lateral strain Mξ/Myto Mζ/Mx, i.e.,μ =− ∂ξ/∂y∂ς/∂xIn order to keep the Poisson’s ratio a positive number, it is necessary to introducethe minus sign to counteract the effect that a positive longitudinal strain gives riseto a negative lateral strain of compression. The value of μ, which is a property ofthe material, may be obtained from standard tables and is generally of the value0.3. In Equation (6.38), the factor (1 − μ 2 ) −1 embodies the effective increase inthe stiffness of the plate resulting from the curling.In solving Equation (6.38) it is assumed thatz = Ψ(r) iωtwhich is then substituted into that equation to givein which∇ 2 (∇ 2 Ψ) − K 4 Ψ = 0 (6.39)K 4 = ω2 ρ(1 − μ 2 )κ 2 EThe substitution of the Helmholtz equation 2 Ψ =−K 2 Ψ into Equation (6.39)indicates that if Ψ can satisfy the Helmholtz equation, it will also constitute asolution to Equation (6.39). The function Ψ in the relationship 2 Ψ = K 2 Ψ willalso satisfy Equation (6.39), so the complete solution of this equation must be thesum of four independent solutions to∇ 2 Ψ ± K 2 Ψ = 0 (6.40)Equation (6.40) with the positive sign is the Helmholtz equation with circular symmetry,which yields the solutions J 0 (Kr) and Y 0 (Kr). But the boundary conditionthat the displacement must be finite at r = 0 at the center of the plate requiresthat the latter solution must be scrapped. The solution of Equation (6.40) with thenegative sign yields J 0 (iKr) and Y 0 (iKr); the latter term is also discarded. Theterm J 0 (iKr) is a modified Bessel function of the first kind, generally written as 22 The modified Bessel functions I n (x) are solutions of the modified Bessel differential equationd 2 ydx 2 + 1 ( )dyx dx − 1 + n2x 2 y = 0