MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 12 pot

418 Structural elements and stress variables: W ←→ N zz V ←→ N zθ + M zθ R U ←→ ∂M zz ∂z + 2 ∂M zθ R∂θ ϕ θ ←→ M zz [8.86] Either the displacement or the stress variable is set to zero according to whether the degree of freedom concerned is fixed or free. As in the case of plates, the transverse shear forces at the boundaries differ from their expression at a current point within the shell: V zu = Q zu + ∂M zθ R∂θ ; V zθ = N zθ + M zθ R [8.87] V zu and V zθ are known as the Kirchhoff effective shear stress resultants of the first and of the second kind, respectively. 8.3.2 Elastic vibrations 8.3.2.1 Small elastic strain and stress fields The membrane components of global strains are: η zz = ∂W ∂z ; η θθ = ∂V R∂θ + U R ; η zθ = 1 2  ∂W R∂θ + ∂V ∂z  [8.88] The global strains in flexure and torsion are: χ zz =− ∂ψ θ ∂z ; χ θθ =− ∂ψ z R∂θ ; χ zθ = −1 2  ∂ψ θ R∂θ + ∂ψ z ∂z  [8.89] Flexure angles are related to translation displacements by: ψ θ =+ ∂U ∂z ; ψ z =− V R + ∂U R∂θ [8.90] Whence the following components of the elastic stress field: N zz = Eh 1 − ν 2  ∂W ∂z + ν  ∂V R∂θ + U R  N θθ = Eh 1 − ν 2  ∂V R∂θ + U R + ν ∂W ∂z  N zθ = Gh  ∂W R∂θ + ∂V ∂z  [8.91] Bent and twisted arches and shells 419 M zz =−D  ∂ 2 U ∂z 2 + ν R 2  ∂ 2 U ∂θ 2 − ∂V ∂θ  M θθ =−D  1 R 2  ∂ 2 U ∂θ 2 − ∂V ∂θ  + ν ∂ 2 U ∂z 2  M θz = −D(1 − ν) R  ∂ 2 U ∂θ∂z − 1 2 ∂V ∂z  where D = Eh 3 12(1 − ν 2 ) [8.92] 8.3.2.2 Equations of vibrations By substituting the elastic stresses [8.91], [8.92] into the Love equations [8.84], we arrive at the following system of vibration equations: 1. Axial equation: ρh ¨ W − Eh 1 − ν 2  ∂ 2 W ∂z 2 + 1 − ν 2 ∂ 2 W R 2 ∂θ 2 + 1 + ν 2 ∂ 2 V R∂θ∂z + ν ∂U R∂z  = f (e) z (θ, z;t) [8.93] 2. Tangential equation: ρh ¨ V − Eh 1 − ν 2           1 + h 2 12R 2  ∂ 2 V R 2 ∂θ 2 + 1 − ν 2 ∂ 2 V ∂z 2  + 1 + ν 2 ∂ 2 W R∂θ∂z + ∂U R 2 ∂θ − h 2 12R 2  ∂ 3 U R 2 ∂θ 3 + ∂ 3 U ∂z 2 ∂θ           = f (e) θ (θ, z;t) [8.94] 3. Radial equation: ρh ¨ U + Eh 1 − ν 2          U R 2 + h 2 12R 2  ∂ 4 U R 2 ∂θ 4 + 2 ∂ 4 U ∂z 2 ∂θ 2 + R 2 ∂ 4 U ∂z 4  +ν ∂W R∂z + ∂V R 2 ∂θ − h 2 12R 2  ∂ 3 V ∂z 2 ∂θ + ∂ 3 V R 2 ∂θ 3           = f (e) r (θ, z;t)+ ∂M (e) θ R∂θ + ∂M (e) z ∂z [8.95] 420 Structural elements In these equations, the terms proportional to h 2 /12R 2 are due to bending and twisting. More generally, to understand the relative importance of the various terms appearing in the coupled system [8.93] to [8.95], it is advisable to discuss first a few particular problems which can be solved by using further simplifying assumptions. 8.3.2.3 Pure bending model As for a circular ring, the flexure terms can be singled out by neglecting the membrane strains. So, turning back to equations [8.88], it is assumed that: ∂W ∂z = 0; ∂V R∂θ =− U R ; ∂W R∂θ =− ∂V ∂z [8.96] Accordingly, the bending and torsion stresses are: M zz =−D  ∂ 2 U ∂z 2 + ν R 2  ∂ 2 U ∂θ 2 + U  M θθ =−D  1 R 2  ∂ 2 U ∂θ 2 + U  + ν ∂ 2 U ∂z 2  M θz = −D(1 − ν) R  ∂ 2 U ∂θ∂z + ∂W R∂θ  [8.97] The shear forces are: Q zr =−D  ∂ 3 U ∂z 3 + 1 R 2  ∂ 3 U ∂z∂θ 2 +  1 + ν 2  ∂U ∂z  Q θr =−D  1 R 3 ∂ 3 U ∂θ 3 + 1 R ∂ 3 U ∂θ∂z 2 + 1 R 3 ∂U ∂θ  [8.98] Substituting [8.98] into the radial equation [8.84], an equation dealing with the sole radial displacement Uis obtained. In the absence of external load, it is written as: ρh ¨ U + D  ∂ 4 U ∂z 4 + 2 R 2 ∂ 4 U ∂z 2 ∂θ 2 + 1 R 4 ∂ 4 U ∂θ 4  +  (1 + 0.5ν) R 2 ∂ 2 U ∂z 2 + ∂ 2 U R 4 ∂θ 2  = 0 [8.99] As expected, the stiffness operator comprises plate and shell components which appear within the first and second brackets, respectively. A convenient way to assess the relative importance of these components is to calculate the natural modes of Bent and twisted arches and shells 421 vibration. Analytical solution is straightforward if the cylinder bases are hinged. The radial displacement can be written as: U n,m = sin  mπz H  sin(nθ) [8.100] where H denotes the cylinder height; n is the circumferential (or azimuth) index and m is the axial index. Substituting [8.100] into [8.99], the following natural pulsations are found: ω n,m = c R  12(1 −ν 2 ) h R       mπR H  2 + n 2   mπR H  2 + n 2 − 1  − ν 2  mπR H  2 [8.101] where c = √ E/ρ The ratio of the contribution of the shell terms over that of the plate terms to the modal stiffness is found to be: κ n,m =    K (S) n,m    K (P ) n,m  n 2 +  mπR H  2  −1 [8.102] Hence, κ n,m decreases rather quickly as the modal order increases, as a mere con- sequence of the wavelength shortening. If κ n,m is sufficiently small [8.101] may be replaced by: ω n,m = c R  12(1 − ν 2 ) h R   mπR H  2 + n 2  [8.103] which is identical to the result obtained for a rectangular plate L x = πR and L y = H hinged at the four edges, see formula [6.96]. The reason for L x = πR instead of L x = 2πR, as could be assumed at first sight, is the 2π periodicity which holds in the case of the cylinder instead of the π periodicity which holds in the case of the rectangular plate. 8.3.2.4 Constriction of a circular cylindrical shell Let us consider a circular cylindrical shell loaded by a constant radial force density f 0 = F/2πR along a parallel line located at height z 0 : f 0 δ(z − z 0 ) [8.104] 422 Structural elements The present problem is clearly independent of θ. Furthermore, the tangential displacement must be zero, because of the axial symmetry. Nevertheless, the hoop stress N θθ is of paramount importance. Thus the membrane strains reduce to: η zz = ∂W ∂z ; η θθ = U R ; η zθ = 0 [8.105] The non-zero stress components are: N zz = Eh 1 − ν 2  ∂W ∂z + ν U R  ; N θθ = Eh 1 − ν 2  U R + ν ∂W ∂z  M zz =−D ∂ 2 U ∂z 2 ; M θθ =−Dν ∂ 2 U ∂z 2 ; Q zr =−D ∂ 3 U ∂z 3 [8.106] In the static (or quasi-static) domain, the equilibrium equations simplify to: ∂N zz ∂z = 0 ⇒ Eh 1 − ν 2  ∂ 2 W ∂z 2 + ν R ∂U ∂z  = 0 ⇒ ∂W ∂z + ν R U = C [8.107] The radial equation is written as:  ∂Q zr ∂z +  + N θθ R = f 0 δ(z − z 0 ) ⇒ D ∂ 4 U ∂z 4 + Eh 1 − ν 2  U R 2 + ν R ∂W ∂z  = f 0 δ(z − z 0 ) [8.108] In equation [8.107], the constant Ccan be set to zero, which means that the shell is assumed to be free axially at both ends. The bending equation [8.108] is thus reduced to: D d 4 U dz 4 + Eh R 2 U = f 0 δ(z − z 0 ) [8.109] It can be noticed that the left-hand side of the equilibrium equation is the same as that of a bended straight beam resting on a uniform elastic foundation, see Figure 8.14, the beam rigidity being set to D = EI and the support stiffness coefficient per unit being set to K S = Eh/R 2 . As the problem is symmetric with respect to the plane z = z 0 , it can be formulated as: D d 4 U dz 4 + Eh R 2 U = 0 D d 3 U dz 3     z=z 0 = 1 2 f 0 ; dU dz     z=z 0 = 0 [8.110] where half a shell loaded by half the actual load is considered. Bent and twisted arches and shells 423 K Figure 8.14. Bending beam on an elastic foundation Another consequence of symmetry is the disappearance of the first derivative of U along the loaded parallel. Solving first for the characteristic equation, we arrive at: λ 4 =− 12(1 − ν 2 ) (Rh) 2 ⇒ λ 1 = k(1 + i); λ 2 = k(1 − i); λ 3 =−λ 1 ; λ 4 =−λ 2 where k = 4  3(1 − ν 2 ) (Rh) 2 [8.111] After some elementary trigonometric manipulations, the general solution of the differential equation [8.110] is expressed as: U(z) = e −kz (A 1 cos(kz) + B 1 sin(kz)) + e +kz (A 2 cos(kz) + B 2 sin(kz)) [8.112] A necessary condition is A 2 = B 2 = 0, otherwise the solution would increase exponentially when moving off the loaded parallel, which is unrealistic. The two other constants are determined by using the conditions at the loaded parallel. Finally, the shell radial deflection is found to be: U(z) = f 0 e −kz 8Dk 3 cos(k(z − z 0 )) [8.113] Figure 8.15 shows the shell constricted by a radial inward load. The results of the finite element computation are found to agree closely with the analytical result [8.113]. As a remarkable result it is found that the deflected portion of the shell is highly confined around the loaded parallel line, as U vanishes exponen- tially with the short characteristic length ℓ c ≃ √ Rh. Accordingly, as soon as H ≫ ℓ c , constriction is quite insensitive to the support conditions at the cylinder bases. 424 Structural elements Figure 8.15. Deformed shell (180 ◦ sector) Figure 8.16a. Membrane stress N θθ , N zz , see text for indentification. The stress distributions are plotted along a meridian in Figures 8.16a and 8.16b. It is noted that the profile of N θθ is marked by a sharp compressive peak, whose maximum is centred at the loaded parallel. Moving off this parallel, a secondary traction peak of much smaller magnitude is first observed, then the stress vanishes in an exponential way. As expected, N zz is practically zero. On the other hand, the most prominent component of the bending moment is M zz ; however M θθ is not negligible. Finally, the present problem gives us the opportunity to turn back to that of the pressurized vessel already studied in Chapter 7 subsection 7.3.7.2. Indeed, it is now possible to discuss the end effects at the interface between the cylindrical body and the hemispherical caps. Based on pure membrane effects, the radial displacement of the caps was found to be less than that of the cylinder by the ratio (1−ν)/(2−ν). As a consequence, the cylinder ends are constricted and bending is induced at the interface, which accounts for the condition of displacement continuity. According Bent and twisted arches and shells 425 Figure 8.16b. Bending moments M θθ , M zz , see text for indentification. R z F 0 R { () + ( – )} Figure 8.17. Cylindrical shell pinched by radial forces along two opposite meridian lines to the present analysis, it can be asserted that bending remains a local effect, closely confined within a characteristic length scale √ Rh from the interface. 8.3.2.5 Bending about the meridian lines A circular cylindrical shell is loaded along two opposite meridian lines by a radial force density, which is assumed to be uniform, as sketched in Figure 8.17. Because of the loading distribution, the deflection of the shell is independent from the axial coordinate z. It can be also expected that bending is the main deformation about the loaded meridian lines. Accordingly, the membrane effects are discarded, and the elastic energy is reduced to the bending term: E e = HR 2  2π 0 M θθ χ θθ dθ [8.114] The n-th harmonic component of the displacement field is first written as: U n cos(nθ); V n sin(nθ); W n ≡ 0 [8.115] 426 Structural elements However, the disappearance of the membrane hoop strain η θθ , implies: V n =− U n n [8.116] Substituting the field [8.115], [8.116] into equations [8.89], [8.92], the elastic energy contained in the harmonic n is found to be: E (n) e = πDH(n 2 − 1) 2 2R 3 U 2 n [8.117] The load is expanded as a Fourier series: F(θ) = F 0 {δ(θ) + δ(π −θ)}= 2F 0 π  1 2 + ∞  k=1 cos(2kθ)  [8.118] The equilibrium equation for the harmonic n = 2k is: πDH(4k 2 − 1) 2 R 3 U 2k = 2F 0 π ; k ≥ 1 [8.119] Thus the lateral deflection is given by the Fourier series: U(θ) = 24(1 − ν 2 )R 3 F 0 π 2 Eh 3 H ∞  k=1 cos(2kθ) (4k 2 − 1) 2 [8.120] The series converges very fast to about 0.11. Most of the deflection is pro- duced by the ovaling mode n = 2. However, the contribution of harmonics of a few higher ranks is also perceptible, see Figure 8.18. Finally, it is worth stressing that the solution [8.120] accounts for bending only. The harmonic n = 0ofthe external loading is expected to deform the shell according to a breathing mode of the type already described in Chapter 7 subsection 7.2.5.2 for a circular ring. The ring model agrees with equation [8.108], provided the z dependency is removed. However, owing to the large difference between the membrane and bending rigid- ities of a shell, the contribution of the n = 0 mode to the shell deformation can be safely neglected in comparison with the contribution of the first bending modes n = 2, 4, 8.3.2.6 Natural modes of vibration n = 0 In order to analyse the breathing modes, the Love equations are simplified by neglecting the membrane stresses N zz and N zθ , which are much smaller than the Bent and twisted arches and shells 427 Figure 8.18. Deflection of the shell, harmonic truncation: 2N = 20 and 2N = 2 hoop stress N θθ . Accordingly, the admissible field of displacement is: V = 0; W =− R ν ∂U ∂z [8.121] As the breathing modes are independent of θ , the strains of interest can be simplified by using [8.121] as: η θθ = U R ; χ zz =− d 2 U dz 2 χ θθ = ∂ψ θ R∂θ = 0; χ zθ = V R − ∂U R∂θ = 0 [8.122] The variation of elastic energy is: δE e =  H 0  2π 0 {N θθ δη θθ + M zz δχ zz }Rdθdz [8.123] That of kinetic energy is: δE κ = ρh  H 0  2π 0 ˙ Uδ ˙ UR dθ dz [8.124] Whence the modal equation: Eh R 2 (1 − ν 2 )  U + h 2 12R 2 ∂ 4 U ∂z 4  − ω 2 ρhU = 0 [8.125] [...]... [8. 128 ] 1 Membrane energy (n) Em(zz) = 2 πR 2 H π EhR 3 2( 1 − ν 2 )n4 (n) (n) Nzz ηzz dz = 0 0 d 2 Un dz2 1 − n2 R2 d 2 Un dz2 H 2 dz [8. 129 ] Bending energy about a tangential direction (n) Eb(zz) = = H πR 2 0 π RD 2 (n) M(n) χzz dz zz H 0 2 d 2 Un dz2 +ν Un dz [8.130] 3 Bending energy about a meridian line (n) Eb(θ θ ) = = πR 2 π RD 2 H (n) (n) Mθ θ χθ θ dz 0 H 0 1 − n2 R2 2 2 Un + ν 1 − n2 R2 d 2. .. algebraic system of the canonical form:          0 wn,m 1 0 0 K11 K 12 K13 2  2 c K21 K 22 K23  − ωn,m 0 1 0  vn,m  = 0 0 0 0 1 un,m K31 K 32 K33 [8.149] E where c2 = ρ(1 − ν 2 ) The stiffness coefficients appearing in [8.149] are given by: K11 = mπ H 2 + n R 2 1−ν 2 n R K 22 = (1 + κ) K33 = 1−ν 2 1 + κR 2 R2 n R + 2 + 2 2 mπ H mπ H 2 2 1+ν mπ n 2 R H ν mπ =− R H n 2 1 mπ = +n +κ + 2 R H... ∂β [A.3.4] Invariance of the length of an infinitesimal segment implies: ds 2 = dx 2 + dy 2 = = ∂x ∂α 2 + ∂x ∂x dα + dβ ∂α ∂β ∂y ∂α 2 dα 2 + 2 ∂y ∂y dα + dβ ∂α ∂β + ∂x ∂β 2 + ∂y ∂β 2 2 2 2 dβ 2 = gα dα 2 + gβ dβ 2 [A.3.5] Appendices 451 The result is suitably written in matrix form as: ds 2 = dx 2 + dy 2 = [dx dy] 2 ds = 2 gα dα 2 2 + gβ 1 0 0 1 dx dy g2 dβ = [dα dβ] α 0 2 0 2 gβ [A.3.6] dα dβ where... dψθ dz (n) (n) χθ θ = − nψz R (n) χθ z = − = 1 R =− d 2 Un ; dz2 Eh 1 − 2 (n) Nzz = M(n) = −D zz (n2 − 1) Un ; R2 dUn 1 −n ; n dz (n) (n) R d 2 Un n2 dz2 d 2 Un ν(1 − n2 )Un + dz2 R2 Mθ θ = −D − Mθ z = − − d 2 Un (n2 − 1) Un + ν R2 dz2 D(1 − ν) 1 −n + R 2n dUn dz [8. 128 ] 8.3.3 .2 Membrane and bending-torsion terms of elastic energy The elastic energy of the shell includes a membrane term together with... shell The following relative orders of magnitude are arrived at: (n) Eb(zz) (n) Em(zz) (n) Eb(θ θ ) (n) Em(zz) (n) Et(zθ) (n) Em(zz) 2 ∼ ε {1 + ν(1 − n2 ) 2 } = 6 2 ∼ ε {(1 − n2 )2 η4 + (1 − n2 ) 2 } = 6 ∼ = 2 6 (1 − n2 )(1 − 2n2 ) (n) where Em(zz) ∼ = 2n2 [8.134] 2 2 π EhUn 2( 1 − ν 2 )η3 1 n4 These ratios show that values of η, ε being fixed, the relative contribution of the membrane energy decreases... modal equation is: ER 2 1 d 4 Un Eh2 (n2 − 1 )2 Un 1 + − 2 ρ 1 + 2 2 n4 dz4 2 )R 4 1−ν 12( 1 − ν n Un = 0 [8.1 42] The mode shapes are the same as for straight beams and the natural frequencies can be easily determined by using the Rayleigh quotient; for instance, if the shell is hinged at its bases they are found to be: fn,m = c 2 R 1 (1 − ν 2 ) n2 +1 n2 mπ nη 4 + ε 2 (1 − n2 )2 ; 12 c= E ρ [8.143] Bent... following families of predominant deformation, mc 1 − ν h2 , 1+ 2H 2 12R 2 mc , w ≫ u; v ≡ 0 (2) Axial modes: fa ≃ 2H (1) Torsion modes: ft = (3) Breathing modes: fr ≃ c h2 1+ 2 R 12R 2 mπ R H v = 0; u≡w≡0 4 , u ≫ w; v≡0 438 Structural elements Figure 8 .22 Frequency plot of a simply supported circular cylindrical shell The expression given here for fr is identical to the result [8. 126 ] It is also noticed... Y = (x z −z x ) dt; Z = (y x −x y ) dt Forming the curl of this displacement field, it can be verified that: curl X = 2 [A .2. 12] Appendices A .2. 5 The Laplace operator ∇ · (∇ ) = ∇ · (∇ A) = A .2. 6 449 = 2 2 2 + + 2 2 ∂x ∂y ∂z2 [A .2. 13] A = ( Ax )i + ( Ay )j + ( Az )k [A .2. 14] Other useful formulas −− −→ curl grad =0 [A .2. 15] div curl A = 0 [A .2. 16] −− − −→ −−−− curl curl A = grad divA − A −− −→ div... ratios of the shell, as illustrated in Figures 8 .20 a and 8 .20 b 4 32 Structural elements Figure 8 .20 a Reduced elastic potential η = 1 Figure 8 .20 b Reduced elastic potential η = 10 Furthermore, by retaining the elastic terms arising from the two preponderant membrane and flexure terms only, the radial vibration equation is written as: Eh3 (n2 − 1 )2 Un 1 EhR 2 1 ∂ 4 Un + + ρh 1 + 2 2 n4 ∂z4 2 )R 4 1−ν 12( 1... [8.150] K 12 = K21 = − K13 = K31 K23 = K 32 where κ= 2 h2 12R 2 Numerical solution of the modal system [8.149] can be conveniently carried out by using a software such as MATLAB The results thus obtained are found to be in close agreement with those arising from a finite element model As expected, there exist three distinct families of modes, which can be identified by looking at the relative amplitude of . 419 M zz =−D  ∂ 2 U ∂z 2 + ν R 2  ∂ 2 U ∂θ 2 − ∂V ∂θ  M θθ =−D  1 R 2  ∂ 2 U ∂θ 2 − ∂V ∂θ  + ν ∂ 2 U ∂z 2  M θz = −D(1 − ν) R  ∂ 2 U ∂θ∂z − 1 2 ∂V ∂z  where D = Eh 3 12( 1 − ν 2 ) [8. 92] 8.3 .2. 2 Equations. ν 2  ∂ 2 W ∂z 2 + 1 − ν 2 ∂ 2 W R 2 ∂θ 2 + 1 + ν 2 ∂ 2 V R∂θ∂z + ν ∂U R∂z  = f (e) z (θ, z;t) [8.93] 2. Tangential equation: ρh ¨ V − Eh 1 − ν 2           1 + h 2 12R 2  ∂ 2 V R 2 ∂θ 2 + 1. + Eh 1 − ν 2          U R 2 + h 2 12R 2  ∂ 4 U R 2 ∂θ 4 + 2 ∂ 4 U ∂z 2 ∂θ 2 + R 2 ∂ 4 U ∂z 4  +ν ∂W R∂z + ∂V R 2 ∂θ − h 2 12R 2  ∂ 3 V ∂z 2 ∂θ + ∂ 3 V R 2 ∂θ 3           = f (e) r (θ,