338 Structural elements Figure 6.11. Mode shape: n = 4, m = 1 Here the plate is found to buckle according to the first mode n = m = 1. It corresponds to the critical temperature increase: θ c = 1 6α(1 +ν)  πh L  2 [6.99] Figure 6.12 refers to a steel plate L = 1 m. The natural frequency of the (1,1) mode is plotted versus θ together with the variation of the critical temperature increase as a function of the plate thickness. 6.3.1.3 Modal density and forced vibrations near resonance Of course, the formula [6.96] also holds in the particular case of unstressed plates. The point here is to investigate a few consequences of the n, m dependency of the natural frequencies which is governed by the coefficient:  nπ L x  2 +  mπ L y  2 It is easy to check that it can take similar values for several pairs of mode indices. This is illustrated in Figure 6.13, where all the natural frequencies are plotted for m and n varying from 1 to 10. The results refer to a steel square plate L = 1m,h = 2 mm. It clearly shows that the plate can vibrate according to several flexure modes, whose natural frequencies are very close to each other. So, if a plate is excited by a ‘nearly resonant’ force, even if the excitation spectrum is Plates: out-of-plane motion 339 Figure 6.12. Thermal buckling of a square steel plate Figure 6.13. Natural frequencies of the flexure modes of a square plate 340 Structural elements limited to a fairly narrow frequency range, the plate response is not restricted to a single resonant mode but is made up of a linear superposition of all the nearly resonant modes whose frequencies lie within the frequency range of the excita- tion signal. Furthermore, the coefficients entering in the superposition are very sensitive to ‘small’ defects in the plate geometry, support conditions and mater- ial. As a consequence, the experimental determination of the modal properties of plates is often a difficult task. On the other hand, the multi modal response can explain the complexity of the figures obtained by Chladni, already evoked in subsection 6.2.2. As an example, the response of a square plate, hinged at the four edges and excited by a point harmonic force F 0 δ(x − x 0 ) ∩δ(y − y 0 )e iω 0 t is considered here. Using the modal expansion [4.51] of the transfer function, the Fourier transform of the response is found to be: Z(x, y, x 0 , y 0 ; ω 0 ) = F 0 M G ∞  n=1 ∞  m=1 sin(nπx/L) sin(mπy/L) sin(nπ x 0 /L) sin(mπy 0 /L) (ω 2 n,m − ω 2 0 + 2iω 0 ω n,m ς n,m ) where M G = ρhL 2 /4 Consider, for instance, the case of a resonant excitation at frequency f 1,2 = f 2,1 . If the excitation point lies on the nodal line of the mode (1,2) y = L/2, the response is marked by a nodal line x = L/2. Conversely, if it lies on the nodal line of the mode (2,1) x = L/2, the response is marked by a nodal line y = L/2. Finally, if the excitation is applied at x = y = L/4 the two modes are excited with the same efficiency and the resulting nodal line is x =−y. Much more complicated nodal patterns can be obtained by increasing the frequency of excitation, as illustrated in colour plate 11, where the red colour corresponds to vibration levels equal to or less than 10% of the maximum vibration mag- nitude, representing thus the zones where the sand would accumulate in a Chladni experiment. 6.3.1.4 Natural modes of vibration of a stretched plate As already mentioned in subsection 6.3.1.2, the modal problem cannot be solved analytically in closed form for various boundary conditions. This is the case for instance of a plate with hinged supports along the lateral edges and left free along the longitudinal edges. Such a configuration is of practical importance in many industrial applications, such as the rolling process of thin strips of paper or metal, see Figure 6.14. In so far as such strips are stretched uniformly in the longitudinal Plates: out-of-plane motion 341 Figure 6.14. Stretching of a rolled strip of metal direction, the modal problem is formulated as follows: D  ∂ 4 Z ∂x 4 + 2 ∂ 4 Z ∂x 2 ∂y 2 + ∂ 4 Z ∂y 4  − F (0) xx ∂ 2 Z ∂x 2 − ω 2 ρhZ = 0 lateral edges: Z(0, y) = Z(L, y) = 0; ∂ 2 Z ∂x 2     x=0,L = 0 longitudinal edges:  ∂ 2 Z ∂y 2 + ν ∂ 2 Z ∂x 2      y=±ℓ/2 = 0;  ∂ 3 Z ∂y 3 + (2 −ν) ∂ 3 Z ∂x 2 ∂y      y=±ℓ/2 = 0 [6.100] If an exact separate variables solution is attempted as in subsection 6.3.1.2, the longitudinal mode shapes ϕ n (x) = sin(nπx/L) arise necessarily as admissible functions. However, no suitable lateral mode shapes can be found. Then, an approx- imate solution can be attempted based on the Rayleigh–Ritz method described in Chapter 5 subsection 5.3.6.3. As trial functions for the mode shapes, it seems reas- onable and convenient to adopt a linear manifold of products of the natural modes of bending vibration of beams. Furthermore, because of the orthogonality properties of such functions, to approximate suitably the shape of the mode (n, m) a single product will suffice: φ n,m (x, y) = ϕ n (x)ψ m (y) [6.101] 342 Structural elements where ϕ n (x) stands for the beam modes complying with the lateral support con- ditions and ψ m (y) for the beam modes complying with the longitudinal support conditions: ϕ n (x) = sin nπx L ψ 1 (y) = 1, ψ 2 (y) = 2y ℓ − 1 m>2, ψ m (y) = a m sin  ̟ m y ℓ  + b m sinh  ̟ m y ℓ  +c m cos  ̟ m y ℓ  + b m cosh  ̟ n y ℓ  [6.102] where m = 1, 2 refer to the rigid modes of pure translation and of pure rotation. So, in this example the Rayleigh–Ritz procedure is reduced to that of Rayleigh’s quotient. Accordingly, the natural frequencies are evaluated by: ω 2 n,m =  E (p) n,m   E (κ) n,m  = K n,m M n,m [6.103] where the functional of potential and kinetic energies are calculated by using the postulated mode shapes. Since the trial functions [6.102] do not comply with the boundary conditions on the longitudinal edges of the plate, it is appropriate to calculate the functional of potential energy by starting from the symmetric form:  E (p) n,m  =  ℓ 0 dy  L 0  M nm : χ nm + F (0) xx  ∂φ nm ∂x  2  dx [6.104] By using the relations [6.66] and [6.102 ] it is found that:  E (p) n,m  = DL 2   nπ L  4 µ 1 − 2ν  nπ L  2 µ 2 + µ 3 + 2(1 −ν)  nπ L  2 µ 4  + F (0) xx L 2  nπ L  2 µ 1 where µ 1 =  ℓ 0 (ψ m (y)) 2 dy, µ 2 =  ℓ 0 ψ m (y)ψ n m (y)dy µ 3 =  ℓ 0 (ψ n m (y)) 2 dy, µ 4 =  ℓ 0 (ψ ′ m (y)) 2 dy [6.105] Plates: out-of-plane motion 343 the functional of kinetic energy is readily found to be:  E (κ) n,m  = ρhLµ 1 2 [6.106] As an example we consider a strip of steel L = 10 m, l = 1m,h = 0.7 mm subjected to a uniform tensile stress σ xx which is varied from 0 to 100 Mpa. In Figure 6.15 the natural frequencies of the modes (1,1), (1,2) and (1,3) are plotted versus σ xx . The values in full lines refer to the Rayleigh quotient method and those marked by upward triangles refer to the finite element method. Both kinds of results are found to agree with each other within a few percent. As expected, when σ xx is sufficiently large, most of the stiffness is provided by the prestress term and the natural frequencies of the three modes become essentially the same and equal to f 1 = 1 2L h ℓ  σ xx ρ On the other hand, Figure 6.16 shows the mode shapes, obtained by using the finite element method, as viewed from two distinct points of view (see the reference frames below the views). As expected, the mode shape (1,1) is very close to the first bending mode of the equivalent beam of length L. The mode shape (1,2) is very close to the first torsional mode. Figures 6.17 and 6.18 refer to a strip of steel of low aspect ratio, L = 1m, l = 1m,h = 0.7 mm. The mode shape (1,1) is clearly marked by an anticlastic Figure 6.15. Frequencies of the modes (1, 1), (1, 2) and (1, 3) versus σ xx 344 Structural elements Mode (1,1) Mode (1,2) Mode (1,3) Figure 6.16. Mode shapes: L = 10 m, l = 1 m, h = 0.7 mm Plates: out-of-plane motion 345 Mode (1,1) Mode (1,2) Mode (1,3) Figure 6.17. Mode shapes: L = 1 m, l = 1 m, h = 0.7 mm bending along the lateral direction (cf. Figure 6.7). Again, the Rayleigh quotient and the finite element results compare satisfactorily. As a final remark, it may be noted that the natural frequencies of the (1,1) and (1,2) modes can be also obtained by modelling the plates as an equivalent pinned- pinned beam. According to the results of Chapters 2 and 3, the beam equation for 346 Structural elements Figure 6.18. Frequencies of the modes (1, 1), (1, 2), (1, 3) versus σ xx the mode (1,1) is found to be: Eh 3 ℓ 12 ∂ 4 Z ∂x 4 − σhℓ ∂ 2 Z ∂x 2 − ω 2 ρhℓZ = 0; Z(0) = Z(L) = 0; ∂ 2 Z ∂x 2     x=0 = ∂ 2 Z ∂x 2     x=L = 0 and the beam model for the mode (1,2) is: −  GJ T + σ hℓ 3 12  ∂ 2 ψ x ∂x 2 − ω 2 ρJψ x = 0 where J T = 1 3  h 3 ℓ 3 ℓ 2 + h 2  h 3 ℓ 3 ψ x (0) = ψ x (L) = 0 the prestress term is obtained by determining the prestress potential related to ψ x . The transverse displacement induced by the rotation about the beam axis y = 0 is: Z(x, y) = yψ x (x) − ℓ/2 ≤ y ≤ ℓ/2 Consequently, the prestress potential is: E p = σh  ∂ψ x ∂x  2  −ℓ/2 −ℓ/2 y 2 dy  L 0  sin  nπx L  2 dx = σ hℓ 3 L 24  ∂ψ x ∂x  2 Plates: out-of-plane motion 347 6.3.1.5 Warping of a beam cross-section: membrane analogy In Chapter 2 subsection 2.2.3.7, it was established that the warping function  of the torsion theory of Barré de Saint-Venantis the solution of the following boundary problem (cf. equations [2.61]): (y, z) = 0 −−−→ grad ·n = zn y − yn z =n ×r ·  i ∀r ∈ (C) [6.107] The solution for a rectangular cross-section was established as the following series: (y, z) = yz − 8 a ∞  n=0 (−1) n k 3 n cosh(k n b/2) sin(k n y)sinh(k n z) [6.108] Here it is of interest to reconsider the problem by noticing the analogy between the problem [2.61] and that of the transverse displacement of a stretched membrane. Let us consider a rectangular plate stretched uniformly in the longitudinal and lateral direction by an in-plane force F (0) normal to the plate edges. Furthermore, the plate is assumed to be so thin that the flexure terms can be neglected in comparison with the prestressed terms. Thus the equation [6.92] reduces to: ρh ¨ Z −F (0)  ∂ 2 Z ∂x 2 + ∂ 2 Z ∂y 2  = 0 [6.109] The problem analogue to [6.107] is expressed as:                   ∂ 2 Z ∂x 2 + ∂ 2 Z ∂y 2  = 0 Z(y,0) = Z(0, z) = 0 ∂Z ∂x     x =a/2 = y; ∂Z ∂y     y =b/2 =−x ⇔                   ∂ 2 Z ∂x 2 + ∂ 2 Z ∂y 2  = yδ  x − a 2  − xδ  y − b 2  Z(y,0) = Z(0, z) = 0 [6.110] where use is made of the central symmetry of the problem to deal with a quarter of plate of length a/2 and width b/2. The solution of [6.110] can be expressed as [...]... components of the metric tensor at P are denoted Gα ; Gβ ; Gζ They are determined as follows: d P = d r + dς n + ς d n ⇒ (d P )2 = (d r )2 + (dς )2 + ς 2 (d n )2 + 2 d r · d n [7 .28 ] d r is in the tangent plane so 2 2 (d r )2 = gα (dα )2 + gβ (dβ )2 [7 .29 ] The determination of (d n )2 is obtained from the differentiation of n dn = ∂n ∂n dα + dβ ⇒ (d n )2 = ∂α ∂β ∂n ∂α 2 (dα )2 + ∂n ∂β 2 (dβ )2 [7.30] or dn · dn = 2. .. [6.68], D [ [Z]] = p0 ⇒ 1 2 2 1 ∂ + + 2 2 2 r ∂r ∂r r ∂θ ∂ 2Z p0 1 ∂ 2Z 1 ∂Z + + 2 2 = 2 r ∂r D ∂r r ∂θ [6. 124 ] The problem is independent of θ, so the corresponding derivatives are null and [6. 124 ] is reduced to the ordinary differential equation: 1 d d2 + r dr dr 2 1 dZ d 2Z + r dr dr 2 = p0 D [6. 125 ] It may be noticed that 1 dZ d 2Z 1 dZ + = r 2 r dr r dr dr then [6. 125 ] takes the form: d d r dr... 2 2 gβ gα (dα )2 + 2 (dβ )2 2 Rα Rβ The dot product d r · d n is expressed as: dr · dn = = ∂r ∂r dα + dβ · ∂α ∂β ∂n ∂n dα + dβ ∂α ∂β ∂r ∂n ∂r ∂n · (dα )2 + (dβ )2 + ∂α ∂α ∂β ∂β ∂r ∂n ∂r ∂n · + · dα dβ ∂α ∂β ∂β ∂α [7.31] With [7 .25 ] and [7 .26 ] it follows that: dr · dn = 2 2 gβ gα (dα )2 + (dβ )2 Rα Rβ The results [7 .29 ] to [7.31] being substituted into [7 .28 ], we arrive at: (d P )2 = gα 1 + ς Rα 2 (dα )2. .. expansion of the type: Z(x, y) = qn,m ϕn,m (x, y) n m where ϕn,m (x, y) are the mode shapes of the following modal problem: ∂ 2Z ∂ 2Z + 2 ∂x ∂y 2 + 2 Z = 0 Z(x, 0) = Z(0, y) = 0 ∂Z ∂x x=a /2 = 0; ∂Z ∂y y=b /2 =0 After some straightforward algebra, we arrive at: ϕn,m (x, y) = sin(kn x) sin(ηkm y) qn,m = kn = 16 (ηkm )2 − (kn )2 (−1)n+m ab (ηkm kn )2 (ηkm )2 + (kn )2 (2n + 1)π , n = 0, 1, 2 ; a km = (2m +... r dr dZ dr = p0 r D [6. 126 ] 3 52 Structural elements The successive integrations of [6. 126 ] lead to: r d dr 1 d r dr d dr r dZ dr r dZ dr = Z= 1 d r dr d dr = r dZ dr = a 1 d p0 r + ⇒ 2D r r dr r p0 r 2 +a 2D dZ dr = p0 r 2 + a ln r + b 4D [6. 127 ] br dZ p0 r 3 r r p0 r 3 + + ar ln r + br ⇒ = +a ln r − +c 4D dr 16D 2 4 2 r2 r2 lng r − 4 4 p0 r 4 +a 64D + br 2 + c ln r + d 4 [6. 128 ] As the plate is assumed... u1 is the unit azimuthal or circumferential vector 3 72 Structural elements Figure 7. 12 Midsurface of an axisymmetric shell From [7. 42] we get: ∂r ∂α (ds )2 = 2 + ∂z ∂α 2 (dα )2 + r 2 (dθ )2 [7.43] Thus, the components of the metric tensor are found to be: gαα = ∂r ∂α 2 + ∂z ∂α 2 1 /2 ; gθ θ = r [7.44] A local coordinate system at a current point M can be also used The unit vectors are the vector t tangential... here on the principle of minimum potential Arches and shells: string and membrane forces 357 Figure 7 .2 Circular cylindrical shell developed as an equivalent plate energy – Z0 is determined by solving the discretized equation: 1 2R Z0 H D 4 2 R sin 0 2 R = pH sin 0 2 x 2R dx x dx; 2R where D = Eh3 12( 1 − ν 2 ) The result is Z0 = − 4p(2R)4 768p(1 − ν 2 ) = πD πE R h 3 So the ratio of the two displacement... applied on the lower face of a circular plate of radius R and thickness h The Lamé parameters are readily found to be gα = 1, gβ = r In statics, equation [6.114] is written as: − 1 ∂(rMrr ) ∂ +2 r ∂r r∂θ Mθ r ∂Mθ r + ∂r r + ∂Mθ θ ∂ 2 Mθ θ − 2 ∂θ 2 r∂r r = p0 [6. 120 ] Plates: out -of- plane motion 351 Then [6.1 12] takes the form: χrr = − ∂ 2Z ; ∂r 2 χθ r = − χθ θ = − 1 ∂ r ∂θ 1 2 ∂Z ∂r 1 ∂ r ∂θ +r ∂ ∂r... ; r ∂r 1 ∂Z r 2 ∂θ [6. 121 ] The elastic moments are: Mrr = D(χrr + νχθ θ ) = −D ∂ 2Z +ν ∂r 2 1 ∂ 1 ∂Z + r ∂r r ∂θ 1 ∂Z r ∂θ 1 ∂ 1 ∂Z ∂ 2Z 1 ∂Z + +ν 2 r ∂r r ∂θ r ∂θ ∂r 1 ∂ ∂Z 1 ∂ 1 ∂Z = − D(1 − ν) +r 2 r ∂θ ∂r ∂r r 2 ∂θ Mθ θ = D(χθ θ + νχrr ) = −D Mrθ = D(1 − ν)χrθ [6. 122 ] and the shear forces are: Qrz = 1 ∂Mθ r Mrr − Mθ θ ∂Mrr + + ; ∂r r ∂θ r Qrz = ∂Mrθ 1 ∂Mθ θ 2Mrθ + + ∂r r ∂θ r [6. 123 ] Equation [6.115]... R 2 (1 + ν) − 16 R 2 2 (3 + ν) ; (1 + 3ν) ; Mrθ = 0 and finally the shear force is, Qrz (r) = − p0 r 2 2π ⇒ 0 Qrz (R)R dθ = −π R 2 p0 as appropriate to fulfil the condition of global equilibrium It is also of interest to evaluate the influence of the boundary conditions on the plate deflection For a hinged outer edge the result is: Z(R) = 0; Z R4 p0 r = R 64D r R Mrr (R) = 0 ⇒ 4 − 2( 3 + ν) r 1+ν R 2 . Q rz = ∂M rθ ∂r + 1 r ∂M θθ ∂θ + 2M rθ r [6. 123 ] Equation [6.115] takes a form similar to [6.68], D[[Z]]=p 0 ⇒  ∂ 2 ∂r 2 + 1 r ∂ ∂r + 1 r 2 ∂ 2 ∂θ 2  ∂ 2 Z ∂r 2 + 1 r ∂Z ∂r + 1 r 2 ∂ 2 Z ∂θ 2  = p 0 D [6. 124 ] The. equation [6. 92] reduces to: ρh ¨ Z −F (0)  ∂ 2 Z ∂x 2 + ∂ 2 Z ∂y 2  = 0 [6 .109 ] The problem analogue to [6 .107 ] is expressed as:                   ∂ 2 Z ∂x 2 + ∂ 2 Z ∂y 2  = 0 Z(y,0). sin(k n x)sin(ηk m y) q n,m = 16 ab (−1) n+m (ηk m ) 2 − (k n ) 2 (ηk m k n ) 2  (ηk m ) 2 + (k n ) 2  k n = (2n + 1)π a , n = 0, 1, 2 ; k m = (2m + 1)π a , m = 0, 1, 2 where η = a/b is the aspect ratio of the plate. Numerical evaluation of