MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 11 potx

378 Structural elements subsection 8.3.2.4, bending is highly confined in the vicinity of the interface and outside this zone of accommodation the radial displacements are very close to the asymptotic values given here. 7.3.7.3 Pressurized toroidal shell As sketched in Figure 7.16, we consider a toroidal shell of circular cross-section, loaded by an internal pressure p, assumed to be uniform. An approximate solution may be reasonably proposed, according to which the displacement field is the superposition of two distinct radial dilatations, one concerning the cross sectional (meridian) circles (radiusa) and the other the parallel circles of radius lying between R −a and R +a. The resulting displacement field is thus written as  ξ = q 1  i +q 2 n. Furthermore, if the aspect ratio R/a of the torus is sufficiently large it can be assumed that q 1 and q 2 are essentially constant. The problem can be then solved by using the Rayleigh–Ritz or Galerkin procedure, based again on the principle of minimum potential energy. However, as the trial function is constant in the local frame  i, n, it may be found more expedient to write directly the equilibrium equations in terms of q 1 and q 2 . Both methods are successively worked out below. First, the metrics of the shell is given by: ds 2 = g 2 θ dθ 2 + g 2 ϕ dϕ 2 = r 2 dθ 2 + a 2 dϕ 2 g θ = r = R +a sin ϕ, g ϕ = a Figure 7.16. Toroidal shell Arches and shells: string and membrane forces 379 The principal curvature radii are: 1 R ϕ = dϕ ds m = 1 a ⇒ R ϕ = a 1 R θ = sin ϕ r ⇒ R θ = R +a sin ϕ sin ϕ The displacement field is:  ξ = U n + W  t = q 2 n + q 1  i  i = cos ϕ  t + sin ϕ n U = q 2 + q 1 sin ϕ; W = q 1 cos ϕ The small strains are given by the relations [7.36] in which the following variables are in correspondence: α → ϕ; R α = R ϕ = a; X α = W β → θ; R θ = R + a sin ϕ sin ϕ ; X β = V ≡ 0 X ς → U V is a free rotation around the torus axis and may be discarded. Then: η ϕϕ = q 2 a , η θθ = q 1 + q 2 sin ϕ R + a sin ϕ The elastic stresses are:  N ϕϕ N θθ  = Eh 1 − ν 2  1 ν ν 1  η ϕϕ η θθ  N ϕϕ = Eh 1 − ν 2  q 2 a + ν q 1 + q 2 sin ϕ R + a sin ϕ  N θθ = Eh 1 − ν 2  q 1 + q 2 sin ϕ R + a sin ϕ + ν q 2 a  380 Structural elements The equilibrium equations follow: N ϕϕ a + N θθ sin ϕ R + a sin ϕ = p ∂[(R +a sin ϕ)N ϕϕ ] ∂ϕ − N θθ ∂[(R +a sin ϕ)] ∂ϕ = 0 ∂N ϕϕ ∂ϕ + a cos ϕ R + a sin ϕ (N ϕϕ − N θθ ) = 0 Generally, they are difficult to integrate; however if a/R ≪ 1, the following approximations can be made: N ϕϕ = Eh 1 − ν 2  q 2 a + ν q 1 + q 2 sin ϕ R + a sin ϕ  ≃ Eh 1 − ν 2  q 2 a  1 + ν a sin ϕ R  + ν q 1 R  N θθ = Eh 1 − ν 2  q 1 + q 2 sin ϕ R + a sin ϕ + ν q 2 a  ≃ Eh 1 − ν 2  q 1 R + ν q 2 a  1 + a sin ϕ R  If the terms dependent on a/R are neglected, the stresses become independent of ϕ. The corresponding approximate values of N ϕϕ , N θθ are: N ϕϕ = N θθ ∼ = Eh 1 − ν 2  q 2 a  (1 + ν) = Eh 1 − ν  q 2 a  where q 2 a = q 1 R ⇒ Rq 2 = aq 1 whence the approximate results: q 2 ∼ = a 2 p(1 −ν) Eh q 1 ∼ = aRp(1 − ν) Eh Turning now to the more refined Galerkin procedure, the Lagrangian of the system takes the form, L(q 1 , q 2 ) =−E s + W p Its variation gives the equilibrium equations: δL(q 1 , q 2 ) =−δE s + δW p = 0 First, the variation of the pressure work density is determined: δw p =−pn · δ  ξdS ⇒ δw p = p(δq 2 − δq 1  i ·n) dS dS = ra dθ dϕ, r = R +a sin ϕ,  i ·n =−sin ϕ Arches and shells: string and membrane forces 381 then, δW p =paδq 2  2π 0  2π 0 (R + a sin ϕ)dϕ dθ + paδq 1  2π 0  2π 0 (R + a sin ϕ) sin ϕdϕdθ δW p =4π 2 a 2 Rp  δq 2 a + δq 1 2R  Reduced displacements defined by ¯q 1 = q 1 /R; ¯q 2 = q 2 /a may be introduced and give: δW p = 4π 2 a 2 Rp  δ ¯q 2 + δ ¯q 1 2  The factor 1/2 present in the generalized force associated with the displacement ¯q 1 is associated with the term asinϕ in the curvature radius formula. This term has been neglected in the calculus made just above. Then, the strain energy is deduced from the stress and the strain expressions, η ϕϕ =¯q 2 , η θθ ∼ = ¯q 1 N ϕϕ = Eh 1 − ν 2 ( ¯q 1 + ν ¯q 2 ), N θθ = Eh 1 − ν 2 ( ¯q 2 + ν ¯q 1 ) which give: δe s = aR Eh 1 − ν 2 ( ¯q 1 δ ¯q 1 + ν(¯q 2 δ ¯q 1 +¯q 1 δ ¯q 2 +¯q 2 δ ¯q 2 ))  2π 0  2π 0 dϕ dθ = 4π 2 aR Eh 1 − v 2 ( ¯q 1 δ ¯q 1 + v( ¯q 2 δ ¯q 1 +¯q 1 δ ¯q 2 +¯q 2 δ ¯q 2 )) Finally the equilibrium equations are: Eh 1 − ν 2 ( ¯q 1 + ν ¯q 2 ) = a p 2 Eh 1 − ν 2 ( ¯q 2 + ν ¯q 1 ) = ap leading immediately to the following approximate solution: q 1 =  1 2 − ν  paR Eh ; q 2 =  1 − ν 2  pa 2 Eh 382 Structural elements These results are more accurate than those obtained by the direct approximate solution. 7.3.7.4 Spherical cap loaded by its own weight The problem is sketched in Figure 7.17. In local coordinates the loading is given by, f ς =−ρhgcos ϕ; f θ = 0; f ϕ = ρhg sin ϕ Because of the symmetries, the solution is independent of θ and V = 0. The local equilibrium is then described by, (N θθ − N ϕϕ ) cos ϕ − sin ϕ dN ϕϕ dϕ = ρhgR(sin ϕ) 2 N θθ + N ϕϕ =−ρhgR cos ϕ and N ϕϕ verifies: 2 cos ϕN ϕϕ + sin ϕ dN ϕϕ dϕ =−ρhgR which can be integrated (though integration is not obvious) to give: N ϕϕ = −ρhRg 1 + cos ϕ ⇒ N θθ = ρhRg  1 1 + cos ϕ − cos ϕ  A simpler manner to obtain N ϕϕ is to write the global equilibrium of the shell in the vertical direction. The area of an elementary strip of the shell surface delimited Figure 7.17. Spherical cap loaded by its own weight Arches and shells: string and membrane forces 383 by two horizontal planes at z and z +dz is readily found to be 2πRsin ϕR dϕ; the corresponding weight is −ρhg2πRsin ϕR dϕ. So, the total weight is:  P =−ρhg  ϕ 0 2πR sin ϕR dϕ =−2πR 2 ρhg(1 − cos ϕ) as it must be balanced by the vertical component of support reactions, we get: 2πR 2 ρhg(1 − cos ϕ) =−2πR(sin ϕ) 2 N ϕϕ (ϕ) ⇒ N ϕϕ (ϕ) =−ρhgR 1 − cos ϕ (sin ϕ) 2 = −ρhgR 1 + cos ϕ The normal stresses are plotted in Figure 7.18 versus ϕ. N ϕϕ is negative, hence compressive, whatever the ϕ value may be. However, N θθ becomes tensile in the range ϕ>52 ◦ . This less intuitive result caused many flaws and even failures in masonry domes because masonry resists compressive stresses much better than tensile stresses. For instance, the dome of St Peter basilica in Rome was found largely cracked during the seventeenth century. The empiricalsolution implemented by the architects – which was perfectly satisfactory – has been to reinforce the dome perimeter support by a masonry ring [COT 90]. Once the stresses are known, it is possible to calculate the strains and the dis- placements. The analytical solution is not very simple and requires one to assume specific boundary conditions. The interest of such analytical solutions, which hold for particular boundary conditions, is to obtain approximate solutions for various Figure 7.18. Normal stresses plotted versus the ϕ angle 384 Structural elements actual boundary conditions which are not too far from the assumed ones, provided Saint-Venant’s principle can be invoked. This point is illustrated below. If the material is linear elastic, the strains in the spherical dome are: η θθ = 1 E (N θθ − νN ϕϕ ) = W R cot ϕ + U R η ϕϕ = 1 E (N ϕϕ − νN θθ ) = 1 R dW dϕ + U R Eliminating U gives, η θθ − η ϕϕ = 1 R  W cot ϕ − dW dϕ  = ρhgR(1 + ν) E  2 1 + cos ϕ − cos ϕ  This equation has the general form W cot ϕ −dW/dϕ = f(ϕ)and the solution is W = sin ϕ  f(ϕ)/sin ϕdϕ+ Cste W = −ρhgR 2 (1 + ν) E sin ϕ  Log  1 + cos ϕ 1 + cos ϕ 0  + 1 1 + cos ϕ − 1 1 + cos ϕ 0  the radial displacement U is readily found to be: U = −ρhgR 2 E  1 + ν 2(1 + cos ϕ) − cos ϕ  + W cotgϕ In Figures 7.19, the reduced value of U is plotted as a function of ϕ for an hemisphere and a dome defined by a 3π/4 base angle. The tangential displacement along the meridians is set to zero but the other components are left free, which is Figure 7.19. Analytical displacements of two spherical caps Arches and shells: string and membrane forces 385 unrealistic. However, these plots show clearly that the radius of the parallel circles is reduced in the upper zone ϕ>52 ◦ thus inducing a compressive hoop stress and enlarged in the lower zone, thus inducing a tensile hoop stress. note. – Finite element solution A finite element solution which accounts for both the membrane and bending effects shows that a clamped boundary condition does not deeply modify the ana- lytical solution except in the vicinity of the clamping. In Figures 7.20a, 7.20b and 7.21a, 7.21b, the deformed shapes of a masonry dome with the base either on slid- ing or clamped support can be compared. The radius is 20 m, the thickness 10 cm and the weight 400 tons. To make the deformed shapes clearly visible, the actual displacements have been multiplied by the factor 20000. The left-hand side plot shows the deflection of a meridian and the right-hand side plot shows that of an Figure 7.20a. Hemispherical cap provided with sliding supports at the base Figure 7.20b. Membrane stresses N ϕϕ , N θθ 386 Structural elements Figure 7.21a. Hemispherical cap clamped at the base Figure 7.21b. Membrane stresses N ϕϕ , N θθ angular sector. Computed deflection in the case of a sliding support is found to be very close to the analytical solution which discards bending effects, except at the immediate vicinity of the base, see Figure 7.20a. This close agreement also holds so far as the membrane stresses are concerned, except near the top and the base of the dome, where bending stresses are present, see Figure 7.20b. As could be expected, the effect of clamping the dome at the base is to increase further the importance of bending near the base which becomes significant over a much larger zone than in the case of a sliding base, see Figure 7.21a. Outside this perturbed zone, the dome response remains essentially the same in both support configurations. 7.3.7.5 Conical shell of revolution loaded by its own weight The structure is shown in Figure 7.22. The shell is shaped as a conical frustum of revolution around the axis Oz; the cone half angle is denoted ψ, the shell thickness Arches and shells: string and membrane forces 387 Figure 7.22. Truncated cone loaded by its own weight is h and p = ρgh is the weight per unit area of the shell. The bottom of the frustum is supported in the Oz direction. We are interested in determining the membrane stresses. The position of a current point is defined by the two parameters θ and ξ = OP. The top and the bottom of the cone are defined by ξ = ξ 0 and ξ = ξ 1 respectively. The coordinates of a point are: r = ξ sin ψ; x = ξ sin ψ cos θ; y = ξ sin ψ cos θ; z = ξ cos ψ So, the coefficients of the metrics are given by: (ds) 2 =(dx) 2 + (dy) 2 + (dz) 2 (ds) 2 =  (sin ψ cos θ) 2 + (sin ψ sin θ) 2 + (cos ψ) 2  (dξ) 2 +  (sin θ) 2 + (cos θ) 2  (ξ sin ψdθ) 2 (ds) 2 =(dξ) 2 + (ξ sin ψdθ) 2 then, g ξ = 1; g θ = ξ sin ψ The principal curvatures are given by: χ ξξ = 1 R ξ = ∂ 2 z ∂ξ 2 ∂r ∂ξ − ∂ 2 r ∂ξ 2 ∂z ∂ξ = 0; χ θθ = 1 R θ = 1 r ∂z ∂ξ = cos ψ ξ sin ψ ⇒ R θ = ξtgψ [...]... as: 2 E= 0 L11 [Wn ] = η 2 n4 + L 22 [Rψn ] = 2 [Wn Rψn ] L11 [Wn ] + L 12 [Rψn ] dθ L21 [Wn ] + L 22 [Rψn ] n2 − ̟ 2 Wn ; 2( 1 + ν) n2 1 + 2 2(1 + ν) L 12 [νn ] = −η 2 n2 Rψn [8.69] − ̟ 2 Rψn ; L 12 [un ] = −nWn Bent and twisted arches and shells 411 So, the energy functional takes the form: 2 1 E= 2 1 = η 2 n4 + 2 − ̟ 2 αn + 2 n2 ; 2 (1 + ν) 2 2 2 2 2 − ̟ 2 βn − 2 2 c αn β n n2 1 ; + 2 2 (1 + ν) = 2 c... ν) = 2 c = n2 1 + η 2 2 [8.70] which leads to the following modal system: 2 1 − 2 − 2 c − 2 2 2 2 c 0 αn = 0 βn − 2 [8.71] Again there exist two modal branches, such that: 2 ̟(1)n = 2 1 + 2 2 2 + = √ ; 2 1 2 ̟ (2) n − 2 2 2 2 1 = +2 + 2 2 2 − √ ; 4 c [8. 72] In Figure 8 .11, the natural frequencies of the out -of- plane modes are plotted versus n, for a steel ring R = 1 m a/R = 0.1 The modes of the lower... as: 2 E= 0 L11 [un ] + L 12 [vn ] dθ L21 [un ] + L 22 [vn ] [un vn ] L11 [un ] = {1 + η 2 n4 − ̟ 2 }un L 12 [vn ] = n(1 + η 2 n2 )(bn cos(nθ) − an sin(nθ)) L21 [un ] = n(1 + η [8.37] 2 2 n )(αn sin(nθ) − βn cos(nθ)) L 22 [vn ] = {n2 (1 + η 2 ) − ̟ 2 }vn which leads to the quadratic and symmetrical form: 2 2 2 2 E = An αn + βn + Bn an + bn + 2Cn (bn αn − βn an ) An = 1 + η 2 n4 − ̟ 2 ; Bn = n2 (1 + η 2. .. − 2 R ∂ 4U ∂ 3V − ∂θ 4 ∂θ 3 ∂ 2V ∂U + ∂θ ∂θ 2 + EI + 4 R ES R2 ∂V +U ∂θ ∂ 3U ∂ 2V − 3 ∂θ ∂θ 2 − ρS 2 U = 0 [8.34] 2 − ρSω V = 0 which are rewritten in the following dimensionless matrix form: L11 L21 L11 = 1 + η 2 L 22 = −(1 + η 2 L 12 L 22 ∂4 − ̟ 2; ∂θ 4 un 0 = vn 0 L 12 = ∂ ∂3 − 2 3 ∂θ ∂θ 2 ∂ ∂3 ) 2 − ̟ 2 ; L21 = − + 2 3 ∂θ ∂θ ∂θ ωR ̟ = c [8.35] The mode shapes are of the following admissible type:... the following integrals: r0 = − rn>1 = − f0 2 f0 π f0 t1 = π tn>1 = − f0 π +π /2 −π /2 cos θ dθ = − f0 ; π r1 = − f0 π +π /2 −π /2  k  2f0 (−1) +π /2 cos(nθ) cos θ dθ = π(4k 2 − 1)  −π /2 0 +π /2 −π /2 (cos θ )2 dθ = − f0 2 if n = 2k if n = 2k + 1 [8.51] f0 (sin θ )2 dθ = 2 +π /2 −π /2  k  −4f0 k(−1) sin(nθ) sin θ dθ = π(4k 2 − 1)  0 if n = 2k if n = 2k + 1 [8. 52] In principle, the generalized displacements... hand, the shear elastic stresses are: σs1 = 2Gεs1 ; Mss = G ∂ψs ∂s 2 2 ς1 + 2 dS = GJ (S) G ∂X2 R ∂s Ms1 = σs2 = 2Gεs2 ⇒ (S) ∂ψs ; ∂s [8.66] 2 2 GI1 ∂X2 dS = R R ∂s Substituting [8.65] and [8.66] into the equilibrium equations [8.64], the vibration equations are written as: ψs ∂ 2 X2 + R ∂s 2 2 ¨ ρS X2 + 2 EI1 ∂s GI1 ∂X2 R 2 ∂s = F2 − ∂ − ∂s ψs ∂ 2 X2 + R ∂s 2 EI1 ¨ ρJ ψs + R (e) ∂ ∂s ∂ψs GJ ∂s ∂M1... 1+ εs2 = ς1 R ∂(Xs + 2 ψ1 − ς1 2 ) X1 − 2 ψs + ∂s R ∂ ∂ς1 −1 ∂(X1 − 2 ψs ) ∂s ∂ ∂ 2 −1 Xs + 2 ψ1 − ς1 2 (1 + ς1 /R) [8.6] ∂Xs + 2 ψ1 − ς1 2 (1 + ς1 /R) ∂(X2 + ς1 ψs ) ∂s In the case of slender arches the strains [8.6] can be simplified, reducing to: εss = ∂ψ1 ∂ 2 X1 − 2 ψs ∂Xs + 2 − ς1 + ∂s ∂s ∂s R εs1 = ∂ψs 1 (Xs + 2 ψ1 − ς1 2 ) ∂X1 + − 2 − 2 − 2 R ∂s ∂s εs2 = 1 ∂ψs ∂X2 + ς1 ψ1 + 2 ∂s... natural frequency is also specified on the plots The natural pulsations of the modes n > 1 are given by the roots of the equation: 2 1 where 4 ̟n − 2 2 1 ̟n + = (n2 + 1)(n2 η 2 + 1); 4 2 =0 = η 2 n2 (n2 − 1 )2 4 2 [8.44] As expected, there are two distinct families or branches defined as: 2 ̟n1 = 2 1 2 1− 1−4 4 2 4 1 ; 2 ̟n2 = 2 1 2 1+ 1−4 4 2 4 1 [8.45] As indicated in Figure 8.4, the lower modal branch corresponds... X1 ∂ 2 X1 + 2 2 ∂s R ⇒ M2 = −EI2 X1 ∂ 2 X1 + 2 2 ∂s R [8.17] 396 Structural elements The equation of transverse vibration is: ¨ ρS X1 + 2 ∂s 2 EI2 (e) X1 ∂ 2 X1 +2 2 ∂s R2 + ∂M2 EI2 (e) X1 = F1 − 4 ∂s R [8.18] Equation [8.18] differs from the straight beam case by two additional stiffness terms which arise as a consequence of the finite curvature of the neutral fibre 8.1.3 .2 Vibration modes of a circular... δX1 + 2 δs 2 R ∂X1 ∂s s2 ds = ∂M2 δX1 ∂s − s1 ∂ 2 M2 M2 + 2 ∂s 2 R δX1 ds s2 [8.14] s1 in which M2 designates the bending moment about the flexure axis n2 The equation of transverse equilibrium and the related boundary conditions are found to be: (e) ¨ ρS X1 − ∂M2 (s; t) ∂ 2 M2 M2 (e) − 2 = F1 (s; t) − 2 ∂s ∂s R M2 + M(e) − ∂M2 (e) + T1 ∂s s=s1 = 0; s=s1 = 0; −M2 + M(e) + s=0 ∂M2 (e) + T1 ∂s =0 s=0 . variation of strain energy is: −δ[E s ]=  s 2 s 1 M 2  ∂ 2 δX 1 δs 2 + δX 1 R 2  ds =  s 2 s 1  ∂ 2 M 2 ∂s 2 + M 2 R 2  δX 1 ds +  M 2 δ  ∂X 1 ∂s  − ∂M 2 ∂s δX 1  s 2 s 1 [8.14] in which M 2 designates. be: σ ss =−Eς 1  ∂ 2 X 1 ∂s 2 + X 1 R 2  ⇒ M 2 =−EI 2  ∂ 2 X 1 ∂s 2 + X 1 R 2  [8.17] 396 Structural elements The equation of transverse vibration is: ρS ¨ X 1 + ∂ 2 ∂s 2  EI 2  ∂ 2 X 1 ∂s 2 + 2  X 1 R 2  + EI 2 R 4 X 1 =. coefficients of the metrics are given by: (ds) 2 =(dx) 2 + (dy) 2 + (dz) 2 (ds) 2 =  (sin ψ cos θ) 2 + (sin ψ sin θ) 2 + (cos ψ) 2  (dξ) 2 +  (sin θ) 2 + (cos θ) 2  (ξ sin ψdθ) 2 (ds) 2 =(dξ) 2 +