The thin plate theory assumes the following deformational kinematics:

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1 MEG6007 (Fall, 2017) - Solutions of Homework # 8 (ue on Tuesay, 29 November 2017) 8.0 Variational Derivation of Thin Plate Theory The thin plate theory assumes the following eformational kinematics: u = u 0 (x, y) z x v = v 0 (x, y) z y w = w(x, y) (1) where u, v an w are the isplacements along the x, y an z coorinates; u 0 (x, y) an v 0 (x, y) represent the stretching the plate neutral surface; z is measure from the neutral surface of the plate, respectively. The strain-isplacement relations are obtaine as ɛ xx = u x = u 0 x z 2 w x 2 ɛ yy = v x = u 0 x z 2 w y 2 γ xy = v x + u y = v 0 x + u 0 y 2z x y γ xz = γ yz = 0 (2) For an isotropic material, the constitutive relations are moele by ɛ xx = 1 E (σ xx νσ yy ), ɛ yy = 1 E (σ yy νσ xx ), τ xy = Gγ xy (3) σ xx = E 1 ν (ɛ 2 xx + νɛ yy ), σ yy = E 1 ν (ɛ 2 yy + νɛ xx ), where σ xx, σ xx an τ xy are stress components; E an G are Young s moulus an shear moulus; an, ν is Poisson s ratio. Variational formulation of plate vibrations may be formulate employing the following expression: The bening strain energy of a plate can be expresse as δv b = V { E 1 ν 2 δɛ xx (ɛ xx +νɛ yy )+ E 1 ν 2 δɛ yy (ɛ yy +νɛ xx )+δγ xy G ɛ xy }V (4) 1

2 When the plate is sufficiently thin an the applie loas o no cause the neutral surface to stretch, we have u 0 = v 0 = 0, (5) 8.1 Carry out the necessary variation of the strain energy given in equation (4), using the expression D = Eh 3 12(1 ν 2 ) where D is calle plate bening rigiity, h is the plate thickness. an is the plate area. 8.2 Derive the kinetic energy an the external energy assuming that the applie force f(x, y, t) is istribute on the plate surface. 8.3 Obtain Hamilton s principle for a rectangular thin plate. n Ientify the bounary conitions from your variations. 8.4 Obtain the necessary characteristic equation for vibration frequencies assuming the following homogeneous solution form: w(x, y, t) = W (x, y)t (t), T (t) = e ωt W (x, y) = 1 sin αx sin γy + 2 sin αx cos γy + 3 cos αx sin γy + 4 cos αx cos γy (7) + 5 sinh α 1 x sinh γ 1 y + 6 sinh α 1 x cosh γ 1 y + 7 cosh α 1 x sinh γ 1 y + 8 cosh α 1 x cosh γ 1 y β 4 = ω2 m D, α 2 + γ 2 = β 2 α γ 2 1 = β 2 where m is the plate mass per unit area, m = ρh an obtain the characteristic equation for a simply supporte plate case. - Solution: (6) Substituting equations(1) -(3) into (4), carrying out the through-the-thickness 2

3 θy w y 4 b 3 θ x a 1 2 x Fig. 1. Plate geometry. Note that θ x = w y, θ y = w x spatial integration an imposing the thin plate assumption (5), one obtains: δv b = D { [ ( 2 w x + 2 w 2 y ) (1 ν) 2 w 2 y ] δ 2 w 2 x 2 + [ ( 2 w x + 2 w 2 y ) (1 ν) 2 w 2 x ] δ 2 w 2 y 2 + 2(1 ν) 2 w x y δ 2 w x y } D = Eh3 12(1 ν 2 ) (8) where spatial integration across the plate thickness z is carrie out, an where D is calle plate bening stiffness, h is the plate thickness, an is the plate area. 3

4 In carrying out the inicate variations, we make use of the ientities: G(w) δ x Γ = l G(w)δw Γ G(w) δ y Γ = m G(w)δw Γ G(w) x δw G(w) δw (9) y where l an m are the irectional cosines of the unit normal vector outwar from the surface contour Γ. Hence, δv b can be shown to be δv b = [l D( 2 w Γ x + ν 2 w ) + m (1 ν) D 2 w 2 y2 x y ] δ x Γ + [m D( 2 w Γ y + ν 2 w ) + l (1 ν) D 2 w 2 x2 x y ] δ y Γ [l D Γ x ( 2 w x + ν 2 w ) + l(1 ν) D 2 w ] δw Γ 2 y2 x y2 [m D Γ y ( 2 w y + ν 2 w ) + m(1 ν) D 2 2 x2 + D 4 w δw, 4 = 4 x x 2 y y 4 (11) w ] δw Γ (10) x 2 y In orer to ientify the natural an essential bounary conitions, we introuce the following efinitions: M xx = D( 2 w y 2 ) M yy = D( 2 w y 2 + ν 2 w x 2 ) M xy = D(1 ν) Q x = M xx x Q y = M yy y + M xy y + M xy x x y 2 (12) 4

5 so that (10) simplifies to δv b = Γ + + Γ { [l M xx + m M xy ] δ x + [m M yy + l M xy ] δ y } Γ { [l x M xx + l y M xy] + [m y M yy + m x M xy] } δw Γ D 4 w δw (13) In orer for the above expression to be vali for all possible bounary contour shapes incluing a circular plate, we introuce the following transformation (see Fig.(2)) x = l n m s y = m n + l, l = cos φ, m = sin φ (14) s Substituting the transformation (14) into (13) we obtain δv b = [ M n δ s n M ns δ s + Q n δw ] s + D 4 w δw M n =l 2 M xx + 2lm M xy + m 2 M yy M ns =lm (M yy M xx ) + (l 2 m 2 ) M xy Q n =l Q x + m Q y 2 = 2 n R n + 2 s, 1 2 R = φ s (15) Note that via (9) we have s M ns δ s s = M ns δw Γ s M ns δw s (16) s If the bounary contour Γ is smooth an close, the virtual work one by the ege twisting moment M ns shoul vanish. Otherwise, a gap along the contour woul exist. Remark: For rectangular plates, there exist iscontinuities at four corners, known as corner conitions, which states that M xy = 0. This conition must be satisfie either explicitly or implicitly. Thus, for a smooth close contour the variation of the strain energy becomes 5

6 Fig. 2. Tangential an normal irections at a plate bounary δv b = [ M n δ s n + (Q n + M ns s )δw ] s + D 4 w δw (17) The kinetic energy δt, the energy ue to the bounary unknown springs δv s, an the work performe by the applie forces δw can be expresse in a similar manner as for the case of beam in the form of t2 t 1 δt t = t2 m(x, y) ẅ(x, y, t) δw(x, t) t (18) t 1 where m(x, y) = ρh is the plate mass per unit plate area, an the kinetic energy ue to in-plane motions (u = zw xx, v = zw yy ) is neglecte as it is 6

7 associate with h 3 compare to the plate size. The variation of the potential energy of the unknown bounary forces an moments are given by δv s = s [ k w w δw + k θ n δ n ] s (19) The work performe by the applie force δw = f(x, y, t)δw(x, y, t) (20) Hamilton s principle for a continuum plate can be written as t2 t 1 + { s (D 4 w + mẅ f) δw [ (V n + k w w ) δw + (M n k θ n ) δ n ] s } t = 0 (21) where (V n, M n ) are given by V n = D n 2 w (1 ν) D M n = D 2 w + (1 ν) D( 1 R s ( 2 w n s 1 R s ) s + 2 w s ) 2 (22) The governing equation of motion for a plate can thus be obtaine as with the bounary conitions: D 4 w + mẅ = f (23) (V n + k w w ) = 0 (M n k θ n ) = 0 (24) Free Vibrations of a Rectangular Plate Free vibrations of a rectangular plate can be stuie by specializing the bounary conitions (24) accoring to the convention shown in Fig. (3). First, the solution of the homogeneous equation of motion for the plate with f = 0 in (23) may assume 7

8 y n = y, s = -x n = -x, s = -y b a n = x, s = y n = -y, s= x x Fig. 3. Bounary escriptions for a rectangular plate w(x, y, t) = W (x, y)e jωt ( 4 β 4 )W (x, y) = 0, β 4 = ω2 m D ( 2 + β 2 )( 2 β 2 )W (x, y) = 0 (25) 8

9 Hence, its solution consists of ( 2 + β 2 )W 1 = 0 W 1 = e jαx e jγy, α 2 + γ 2 = β 2 ( 2 β 2 )W 2 = 0 W 2 = e α1x e γ1y, α1 2 + γ1 2 = β 2 W = W 1 + W 2 W (x, y) = 1 sin αx sin γy + 2 sin αx cos γy + 3 cos αx sin γy + 4 cos αx cos γy + 5 sinh α 1 x sinh γ 1 y + 6 sinh α 1 x cosh γ 1 y + 7 cosh α 1 x sinh γ 1 y + 8 cosh α 1 x cosh γ 1 y (26) Specializing the bounary conition(24) to a rectangular case requires the conversion of (n, s) in terms of (x, y) along the four eges as shown in Fig. (3). The bounary conitions along the four eges are obtaine as follows.. long y = 0 we have (s = x, n = y): V y = D y 2 w + (1 ν) D 3 w [D y x 2 y 2 w + (1 ν) D 3 w y x + k ww ] 2 (x,0) = 0 M y = D( 2 w y 2 + ν 2 w x 2 ) [ D( 2 w y 2 + ν 2 w x 2 ) + k θ y ] (x,0) = 0 (27) B. long y = b we have (s = x, n = y): V y = [D y 2 w + (1 ν) D 3 w y x ] [D 2 y 2 w + (1 ν) D 3 w y x k ww ] 2 (x,b) = 0 M y = D( 2 w y 2 + ν 2 w x 2 ) [ D( 2 w y 2 + ν 2 w x 2 ) k θ y ] (x,b) = 0 (28) C. long x = 0 we have (s = y, n = x): V x = [D x 2 w + (1 ν) D 3 w x y ] [D 2 x 2 w + (1 ν) D M x = D( 2 w y 2 ) [ D( 2 w y 2 ) + k θ 3 w x y + k ww ] 2 (0,y) = 0 x ] (0,y) = 0 (29) 9

10 D. long x = a we have (s = y, n = x): V x = [D x 2 w + (1 ν) D 3 w x y 2 ] M x = D( 2 w y 2 ) [D x 2 w + (1 ν) D [ D( 2 w y 2 ) k θ 3 w x y k ww ] 2 (a,y) = 0 x ] (a,y) = 0 (30) where it is unerstoo that the istribute bounary springs (k w, k θ ) are efine along the four eges. Note that there are a total of eight bounary conitions for plate vibrations. Hence, the vibration moes W (x, y) must consist of eight unknown coefficients: Simply supporte along all four eges: The bounary conitions for this case is obtaine from (27) -(30) by setting k w an k θ = 0: long y = 0 : w (x,0) = 0 an M y = D( 2 w y 2 + ν 2 w x 2 ) (x,0) = 0 long y = b : w (x,b) = 0 an M y = D( 2 w y 2 + ν 2 w x 2 ) (x,b) = 0 long x = 0 : w (0,y) = 0 an M x = D( 2 w y 2 ) (0,y) = 0 long x = a : w (a,y) = 0 an M x = D( 2 w y 2 ) (a,y) = 0 (31) Note that along y = 0 we have w (x,0) = 0, which implies that w oes not vary along the plate ege (meaning with respect to x). Hence we have long y = 0 : w (x,0) = 0 implies x 2 (x,0) = 0 2 w y 2 (x,0) = 0 = 0 (32) pplying this observation to the three remaining eges, we arrive at the following bounary conitions for a simply supporte plate: 10

11 long y = 0 : w (x,0) = 0 an long y = b : w (x,b) = 0 an long x = 0 : w (0,y) = 0 an long x = a : w (a,y) = 0 an y 2 (x,0) = 0 y 2 (x,b) = 0 x 2 (0,y) = 0 x 2 (a,y) = 0 (33) Clearly, the above bounary conitions are satisfie by retaining 1 -term in (26): W (x, y) = 1 sin αx sin γy, α 2 + γ 2 = β 2, β 4 = ω2 m(x, y) D sin αa = 0, sin γb = 0 α m a = πm, m = 1, 2,... γ m a = πn, n = 1, 2,..., α 2 + γ 2 = β 2 β 2 mn = π 2 [( m a )2 + ( n b )2 ] (34) Therefore, the natural frequencies of a simply supporte plate are given by D ω mn = βmn 2 m(x, y) = π2 [( m a )2 + ( n D b )2 ] m(x, y) (35) Note that the frequencies of a simply supporte beam are obtaine from the above frequency expression by setting { b a 0, ν = 0} {D = Eh3 /12 = EI, m(x, y) = m(x)} ω n = ( mπ a )2 EI m(x, y) (36) Completely free along all four eges: 11

12 The bounary conitions for this case is obtaine from (27) -(30) by setting k w = 0 an k θ = 0: long y = (0, b) : y [ + (2 ν) y2 2 w x 2 ] = 0 y + ν 2 w 2 x 2 = 0 (37) long x = (0, a) : x [ + (2 ν) x 2 y 2 ] = 0 ( 2 w y 2 ) = 0 (38) It shoul be emphasize that, if an approximation is introuce for w(x, y), it must implicitly satisfy M ns = M xy = 0. Otherwise, the following must be explicitly enforce: M xy = D 2 w x y = 0 at the four corners (39) 12