CM - approfondimento BISCARINI GAI

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Approfondimento di Costruzioni Metalliche di Giulio Biscarini, A.A. 2013/14

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CM - approfondimento BISCARINI GAI

  1. 1. Facoltà di Ingegneria Civile e Industriale Corso di laurea magistrale in Ingegneria Civile COSTRUZIONI METALLICHE Plates and shells: buckling and nonlinear analysis Docente: Studenti: Prof. Ing. Franco Bontempi Giulio Biscarini Assistenti: Giordana Gai Ing. Francesco Petrini Ing. Paolo Emidio Sebastiani Anno Accademico 2013 – 2014
  2. 2. CONTENTS 1 BUCKLING OF THIN PLATES....................................................................................................................1 1.1 Critical load ..................................................................................................................................................1 1.1.1 Introduction.............................................................................................................................1 1.1.2 Simply supported plates compressed in one direction .............................................................1 1.1.3 Simply supported plates compressed along two opposite sides perpendicular to N .................3 1.1.3.1 Y = 0 simply supported ; Y = b free ...........................................................................4 1.1.3.2 Y = 0 built in ; Y = b free ...........................................................................................5 1.1.3.3 Y = 0 built in ; Y = b built in........................................................................................6 1.1.4 Simply supported plates compressed along two opposite sides parallel to N ...........................6 1.1.5 Simply supported plates under combined bending moment and compression .........................7 1.1.5.1 The factor k when Y = 0 and Y = b are built in............................................................8 1.1.5.2 The factor k when Y = 0 is simply supported and Y = b is built in................................8 1.1.6 Plates under the action of shearing stresses.............................................................................8 1.1.6.1 Combination of pure shear with longitudinal axial stresses.......................................9 1.1.6.2 Combination of pure shear with bending moment stresses.....................................10 1.2 Comparison between analytical and numerical code solution ......................................................11 1.2.1 Introduction ..........................................................................................................................11 1.2.2 Compressive forces................................................................................................................12 1.2.2.1 Case 1.....................................................................................................................13 1.2.2.2 Case 2.....................................................................................................................13 1.2.2.3 Case 3.....................................................................................................................14 1.2.2.3 Case 4.....................................................................................................................14 1.2.2.3 Case 5.....................................................................................................................15 1.2.2.3 Case 6.....................................................................................................................15 1.2.2.3 Case 7.....................................................................................................................16 1.2.2.3 Case 8.....................................................................................................................16 1.2.2.3 Case 9.....................................................................................................................17 1.2.2.3 Case 10...................................................................................................................17 1.2.2.3 Case 11...................................................................................................................18 1.2.2.3 Case 12...................................................................................................................18 1.2.2.13 Conclusion............................................................................................................19 1.2.3 Bending forces.......................................................................................................................20 1.2.3.1 Case 1.....................................................................................................................20 1.2.3.2 Case 2.....................................................................................................................22 1.2.3.3 Conclusion..............................................................................................................24 1.2.4 Shear forces...........................................................................................................................24 1.2.4.1 Case 1.....................................................................................................................25 1.2.4.2 Case 2.....................................................................................................................26 1.2.4.3 Case 3.....................................................................................................................26 1.2.4.4 Conclusion..............................................................................................................27 1.2.5 Compressive and shear forces................................................................................................28 1.2.5.1 Case 1.....................................................................................................................28 1.2.6 Bending moment and shear forces.........................................................................................29 1.2.6.1 Case 1.....................................................................................................................30 1.2.6.2 Conclusion..............................................................................................................31 1.3 Nonlinear static analysis ...............................................................................................................32 1.3.1 Elastic behavior .....................................................................................................................32
  3. 3. 1.3.1.1 Post - critical behavior ............................................................................................32 1.3.1.1 Geometric nonlinearity...........................................................................................33 1.3.2 Elastic - plastic behavior ........................................................................................................34 1.3.2.1 Material nonlinearity .............................................................................................34 1.3.2.2 Material nonlinearity and imperfection ..................................................................35 1.3.2.3 Material and geometric nonlinearity.......................................................................37 1.3.2.4 Comparison by varying thickness............................................................................38 1.4 Buckling of plates beyond the proportional limit ..........................................................................40 1.4.1 Introduction ..........................................................................................................................40 1.4.2 Large deflection of buckled plates .........................................................................................40 1.4.3 Approximate solution ............................................................................................................41 1.4.4 Approximate solution for very thin plate................................................................................43 1.4.5 Square plate simply supported...............................................................................................44 2 SHELLS........................................................................................................................................................53 2.1 Thin shells...................................................................................................................................................53 2.1.1 The membrane theory ...........................................................................................................53 2.1.2 Analysis of symmetrically loaded shell of revolution...............................................................55 2.1.3 Numerical examples...............................................................................................................56 2.1.3.1 Hemisphere............................................................................................................56 2.1.3.2 Toroidal tank ..........................................................................................................56 2.2 Thick shells.................................................................................................................................................62 2.2.1 The membrane theory ...........................................................................................................62 2.2.2 The differential equation of elastic the line ...........................................................................62 2.2.2.1 The stationary solution ..........................................................................................64 2.2.2.2 The general solution ..............................................................................................64 2.2.3 Cylindrical tank .....................................................................................................................65 2.3 Buckling of shells.......................................................................................................................................69 2.3.1 Buckling of cylindrical shells with circular section...................................................................69 2.3.1.1 Comparison by varying the ratio h/L ......................................................................70 2.3.1.2 Comparison by varying thickness............................................................................70 2.3.2 Nonlinear analysis..................................................................................................................72 2.3.3 Vary flat shells uniformly compressed....................................................................................73 2.3.3.1 Nonlinear analysis by varying thickness ..................................................................73 2.3.4 Cylindrical shell under the action of uniform axial compression .............................................75
  4. 4. 1 1-BUCKLING OF THIN PLATES 1.1 Critical load 1.1.1 Introduction There are various ways to calculate the critical buckling load for plates. - A first way is to consider, from the beginning, the plate with a small initial curvature (or with an applied lateral load): in this case, the critical load is the one at which the plate has lateral displacements that grow indefinitely. - Another way can be to consider that the plate buckles slightly under the action of a force, and that the value of the critical load is the one that the force must have in order to keep the plate in the deformed configuration. - The last way is the energy method: this is especially useful when the rigorous analytical solution is unknown or when we have plates reinforced by stiffeners of which we are interested only in an approximate value of the critical load. In this case we compare the work done by the forces acting in the middle plane of the plate (which cause only a lateral bending , without stretching ) with the strain energy of curvature associated with every form of lateral buckling: if the energy of bending is smaller , the equilibrium of the plate is stable , otherwise the plate is unstable and buckling occurs . We can also consider the stretching in the middle plane: in this case the flat condition of equilibrium of the plate becomes critical when the strain energy of stretching, released during buckling, is equal to the strain energy of bending of the plate. 1.1.2 Simply supported rectangular plates compressed in one direction We consider a rectangular plate, compressed in its middle plane by forces uniformly distributed along the sides x = 0 and x = a. Nx is the magnitude of the force per unit length of the edge: we search the value of Nx that causes buckling. In this case we can find the critical load by integration of the following equation, that represents the equilibrium of a plate with axial loads only. + 2 + = 1 + + 2 We can arrive at the same result also using the “energy method”. Figure 1.1
  5. 5. 2 In the case of a rectangular plate with simply supported edges, the deflection surface can be represented by the double trigonometric series = + The strain energy of bending can be written as ∆ = 8 + The work done by the compressive forces during buckling of the plate is 1 2 = 8 We find the critical value of compressive forces by equating the two terms. = ∑ ∑ + ∑ ∑ It’s shown that this expression has a minimum if all coefficients amn, except one, are taken equal to zero. = + So we can conclude that the plate buckles in a shape that can have several half-waves in the direction of compression but only one half wave in the perpendicular direction. The most interesting value is the smallest (obtained by taking n=1) , = + 1 Let us analyze this expression. The first term is the Euler load for a strip of length “a” and unit width, the second term represents in what proportion the stability of the continuous plate is greater than the stability of an isolated strip: this depends on the ratio a/b and on the number m, which indicates the number of half – waves into which the plate buckles. - If a is smaller than b, the second term is always smaller than the first and the minimum value of Ncr is obtained by taking m = 1 (assuming that the plate buckles in one half – wave ). The critical load is , = + - If a grows, we have various possibilities: we can write the critical load as , = where k is a numerical factor, that depends on the ratio a/b. As we can see in the figure 1.2, if we consider the curve m = 1, k is large for small values of a/b, then decreases (becomes a minimum when the plate is square) and then increases again. Assuming that the plate buckles in more half – waves , the plate has an inflection line dividing the plate in halves, and each half is in the condition of a simply supported plate of length a/(number of half – waves). The curves of m > 1 can be obtained by the curve m = 1 keeping the same ordinates and multiplying the abscissas by “m”.
  6. 6. 3 Figure 1.2 Short plates buckles in only one half – wave, while for long plates “m” can be very large . The critical value of the compressive stress is = ( − ) where E is the elastic modulus, ν is the Poisson coefficient, h is the thickness. a/b 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 k 27.0 13.2 8.41 6.25 5.14 4.53 4.20 4.04 4 a/b 1.1 1.2 1.3 1.4 1.41 k 4.04 4.13 4.28 4.47 4.49 Table 1.1 1.1.3 Simply supported plates compressed along two opposite sides perpendicular to N We consider the equation + 2 + = 1 + + 2 We assume that the plate buckles in m sinusoidal half – waves : the deflection can be written as = ( ) sin f(y) is a function of y alone, that depends on the boundary conditions along y=0 and y=b. We know the boundary conditions along x=0 and x=a = 0 + = 0 Substituting these equations in the first we obtain an ordinary differential equation for f(y), and the solution can be written in the following form: ( ) = ∝ + ∝ + cos + sin The four constants of integration must be determined by conditions of constraints along the sides y=0 e y=b; α and β are factors that depend on m, a, D, Nx.
  7. 7. 4 1.1.3.1 y=0 simply supported; y=b free Figure 1.3 = 0 − −→ = 0 + = 0 = − −→ ⎩ ⎪ ⎨ ⎪ ⎧ + = 0 + (2 − ) = 0 In this way we can write f(y) in the form: ( ) = sinℎ + sin Substituting f(y) in the condition y=b, we arrive at the buckled equation of equilibrium . − tanh = + tanh The smallest value of Nx is obtained by taking m=1: in this way, we assume that the buckled plate has only one half – wave. The critical compressive stress is , = , ℎ = ℎ where k is a numerical factor depending on the ratio a/b. a/b 0.5 1 1.2 1.4 1.6 1.8 2 2.5 3 k 4.40 1.440 1.135 0.952 0.835 0.755 0.698 0.610 0.564 a/b 4 5 k 0.516 0.506 Table 1.2
  8. 8. 5 1.1.3.2 y=0 built in; y=b free Figure 1.4 = 0 − −→ = 0 = 0 = − −→ ⎩ ⎪ ⎨ ⎪ ⎧ + = 0 + (2 − ) = 0 In this way we can write f(y) in the form: ( ) = cos − cosh ) + (sin – sinh Substituting f(y) in the condition y=b, we obtain two homogeneous linear equations in A and B. The critical load is determined by equating to zero the determinant of these equations and solving the transcendental equation that is obtained. , = , ℎ = ℎ k is a numerical factor depending on a/b : its minimum value is 1.328. Figure 1.5 a/b 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 k 1.70 1.56 1.47 1.41 1.36 1.34 1.33 1.33 1.34 a/b 1.9 2.0 2.2 2.4 k 1.36 1.38 1.45 1.47 Table 1.3
  9. 9. 6 1.1.3.3 y=0 built in; y=b built in = 0 − −→ = 0 = 0 = − −→ = 0 = 0 The critical value of compressive force is determined by solving the transcendental equation that is obtained. , = , ℎ = ℎ k is a numerical factor depending on a/b : the smallest value is for 0.6 < a/b < 0.7. In this case, a long compressed plate buckles in comparatively short waves. a/b 0.4 0.5 0.6 0.7 0.8 0.9 1.0 k 9.44 7.69 7.05 7.00 7.29 7.83 7.69 Table 1.4 1.1.4 Simply supported plates compressed along two opposite sides parallel to N In this case, the equation for the buckled plate is + 2 + = − We can use the same solution described in 1.1.3. The edge conditions at x=0 and x=a are the same of 1.1.3; if the force Ny remains parallel to the y axis after buckling, the bending moments vanish and the shearing force along y=-b/2 becomes . We find that the edge conditions for y=b/2 are: ⎩ ⎪ ⎨ ⎪ ⎧ + = 0 + (2 − ) = − We obtain a transcendental equation for calculating Nycr: there are two possible forms of buckling. For both cases we can write Nycr as , = We must use the curve that gives the smaller value of k. Considering ν=0.3, the first buckling form is antisymmetrical up to a value of b/a=1.316; then , for 1.316<b/a<2.632 the buckling form is symmetrical and then the value of k remain close to 2.31. The curve of antysimmetrical buckling can be used also for the case of a plate simply supported along 3 sides, with compressive force along the fourth, which is free to move (this plate is in the same conditions as the half of the antisymmetrical plate described before): in this case, to calculate k, it can be used that curve, using 2b/a instead of b/a.
  10. 10. 7 1.1.5 Simply supported plate under combined bending moment and compression Figure 1.6 We consider a plate simply supported along x=0 and x=a, loaded with distributed forces that follows the equation: = − where N0 is the intensity of the compressive force at the edge y=0, α is a numerical factor that determines different cases: = 0 − −→ = 2 − −→ 0 < < 2 − −→ The deflection of the buckled plate can be written as combination of double trigonometric series; the critical value of N0 is calculated by energy method. The calculations are more complicated (we must take account of the number of half-waves in which the plate can buckles, the various cases by changing η, the level of approximation for each case). The expression of the critical stress is the same of the other case, with different value of k. = a/b 0.4 0.6 0.75 0.8 1 1.5 η = 2 29.1 24.1 24.1 24.4 25.6 24.1 η = 4/3 18.7 12.9 11.5 11.2 11.0 11.5 η = 1 15.1 9.7 8.4 8.1 7.8 8.4 η = 4/5 13.3 8.3 7.1 6.9 6.6 7.1 η = 2/3 10.8 7.1 6.1 6.0 5.8 6.1 Table 1.5 Figure 1.7 Valid for η = 2
  11. 11. 8 In the case of pure bending the third approximation was used, while in the other cases the second. In the case of pure bending, the value of the critical stress becomes a minimum when a/b = 2/3: for the cases with η <2 the ratio a/b, at which σcr is minimum, increases, arriving at the unity when η =1 (that is the case of pure compression, at which the minimum value of Ncr is obtained when the plate is square). 1.1.5.1 The factor k when y=0 and y=b are built in a/b 0.30 0.35 0.40 0.45 0.47 0.48 0.50 0.60 0.70 η = 2 47.3 43.0 40.7 39.7 39.6 39.6 39.7 41.8 45.8 a/b 0.40 0.50 0.60 0.65 0.66 0.67 0.70 0.80 0.90 η = 1 17.7 14.7 13.7 13.56 13.57 13.58 13.65 14.3 15.4 Table 1.6 Comparing this table with Table 1.5, we see that when the edges are fixed, the critical stress increases (for long plates it can be larger by 75%, and the wave length diminishes). 1.1.5.2 The factor k when y=0 is simply supported and y=b is built in a/b 0.40 0.50 0.60 0.65 0.66 0.67 0.70 0.80 0.90 η = 2 29.5 26.0 24.65 24.48 24.48 24.48 24.6 25.3 26.6 Table 1.7 In this case, comparing this table with the one of the edges y=0 and y=b free, we see that fixing edges on the tension side has a really small effect on the critical stress and on the wave length . 1.1.6 Plates under the action of shearing stresses Figure 1.8 Now we consider a simply supported rectangular plate submitted to the action of shearing forces Nxy uniformly distributed along the four edges. To solve the problem it is used the energy method. As we did in the other cases, we start from the equation of the deflection w (double trigonometric series) and, by equaling the work done by external forces and the strain energy of bending, we find the minimum value of Nxy. In this case, the critical value of the stress does not depend on the sign of the applied forces. = where k is a numerical factor depending on a/b.
  12. 12. 9 a/b 1 1.2 1.4 1.5 1.6 1.8 2.0 2.5 3.0 k 9.34 8.0 7.3 7.1 7.0 6.8 6.6 6.1 5.9 Table 1.8 The deflection surface of the buckled plate follows the direction of the diagonal, which is the principal direction of compressive stresses. Figure 1.9 1.1.6.1 Combination of pure shear with longitudinal axial stresses Figure 1.10 This case presents different values of factor k. The compressive stress reduces the stability of the plate (already submitted to shear), while the tension stress increases the stability. Figure 1.11
  13. 13. 10 1.1.6.2 Combination of pure shear with bending moment stresses Figure 1.12 This case was studied using energy method: for a simply supported plate, we find that the numerical factor k depends not only on the ratio a/b, but also on the ratio τ / τcr , where τ is the actual shearing stress and τcr is the critical shearing stress determined in the case of “pure shear”. Figure 1.13 For τ / τcr <0.4 the effect of shearing stress on the critical value of bending stress is small (the shear action is small comparing to the one that takes the plate to buckling); when τ / τcr approaches unity, k decreases. This means that some bending stress can be added to pure shear case without a strong reduction of τcr. This problem was studied also for different boundary conditions: in the following tables the factor k is given. y = 0 and y = b are built in, x = 0 and x = a are simply supported a/b 1 1.5 2.0 2.5 3.0  k 12.28 11.12 10.21 9.81 9.61 8.99 Table 1.9 All four edges are built in a/b 1 1.5 2.0 2.5  k 14.71 11.50 10.34 10.85 8.99 Table 1.10
  14. 14. 11 1.2 Comparison between analytical and numerical code solution 1.2.1 Introduction We analyzed a square plate of side L = a = b = 50 cm and thickness t = 1 cm , under different load and constraint conditions. The analytical solution derives from the theory of Timoshenko (described in section 1.1) and is based on the use of numerical factor k, which is easily determined from the schedules as a function of the size of the plate (ratio a / b) depending on the load and constraint case that we consider. As calculation code we used both Straus 7 and Sap 2000 : the mesh used is 30x30 finite elements, for a total of 900 finite elements ISOP4. We will see that the two programs give results very close to each other . The comparison between the analytical solution and numerical solution is good too: differences % are contained in the majority of cases (we are on the order of 5% ) except when load conditions are more complicated , for example for the plate submitted to bending moment ( where the error rises to 10 % ) . In the following paragraphs we report the load cases studied: we develop each case by different boundary conditions. The pictures contained are extracted from Straus 7 . INITIAL DATA a = b = L 50 cm t 1 cm E 2E+8 KPa ν 0.30 Table 1.11 STRAUS 7 MODEL SAP 2000 MODEL Figure 1.14
  15. 15. 12 1.2.2 Compressive forces We used the following equations: = 12 (1 − ) = where Nl depends only on geometric data of the plate and on the material (steel). This value is fixed for all following cases. Nl [KN/m] 723.0 Ncr depends on boundary conditions through the numerical factor ki. We use the following schedule. Figure 1.15
  16. 16. 13 1.2.2.1 Case 1 : two edges simply supported, two edges free Figure 1.16 ANALYTICAL SOLUTION ki 1 Ncr 723.0 KN/m NUMERICAL CODE SOLUTION SAP 2000 736.6 KN/m STRAUS 7 688.8 KN/m DIFFERENCES % SAP 2000 - Analytical 1.8% STRAUS 7 - Analytical -5% Table 1.12 Figure 1.17 1.2.2.2 Case 2 : all four edges simply supported Figure 1.18 ANALYTICAL SOLUTION ki 4 Ncr 2892.2 KN/m NUMERICAL CODE SOLUTION SAP 2000 3043.4 KN/m STRAUS 7 2887.4 KN/m DIFFERENCES % SAP 2000 - Analytical 5 % STRAUS 7 - Analytical -0.2% Table 1.13 Figure 1.18
  17. 17. 14 1.2.2.3 Case 3 : two edges simply supported, two edges built in ANALYTICAL SOLUTION ki 6.5 Ncr 4699.8 KN/m NUMERICAL CODE SOLUTION SAP 2000 4867.0 KN/m STRAUS 7 4863.6 KN/m DIFFERENCES % SAP 2000 - Analytical 3.4% STRAUS 7 - Analytical 3.4% Table 1.13 1.2.2.4 Case 4 : three edges simply supported, one edge built in ANALYTICAL SOLUTION ki 5 Ncr 3615.2 KN/m NUMERICAL CODE SOLUTION SAP 2000 3830.2 KN/m STRAUS 7 3497.3 KN/m DIFFERENCES % SAP 2000 - Analytical 5.6% STRAUS 7 - Analytical -3.4% Table 1.14 Figure 1.19 Figure 1.20 Figure 1.22 Figure 1.21
  18. 18. 15 1.2.2.5 Case 5 : two edges simply supported, two edges built in ANALYTICAL SOLUTION ki 7.69 Ncr 5560.2 KN/m NUMERICAL CODE SOLUTION SAP 2000 5733.6 KN/m STRAUS 7 5536.9 KN/m DIFFERENCES % SAP 2000 - Analytical 3.0% STRAUS 7 - Analytical -0.4% Table 1.15 1.2.2.6 Case 6 : three edges simply supported, one edge built in ANALYTICAL SOLUTION ki 5.7 Ncr 4121.4 KN/m NUMERICAL CODE SOLUTION SAP 2000 4313.6 KN/m STRAUS 7 4141.7 KN/m DIFFERENCES % SAP 2000 - Analytical 4.5% STRAUS 7 - Analytical 0.5% Table 1.16 Figure 1.23 Figure 1.24 Figure 1.25 Figure 1.26
  19. 19. 16 1.2.2.7 Case 7 : three edges simply supported, one edge free ANALYTICAL SOLUTION ki 1.44 Ncr 1041.2 KN/m NUMERICAL CODE SOLUTION SAP 2000 1082.8 KN/m STRAUS 7 1013.3 KN/m DIFFERENCES % SAP 2000 - Analytical 3.8% STRAUS 7 - Analytical -2.8% Table 1.17 1.2.2.8 Case 8 : two edges simply supported, one edge built in, one edge free ANALYTICAL SOLUTION ki 1.7 Ncr 1229.2 KN/m NUMERICAL CODE SOLUTION SAP 2000 1276.6 KN/m STRAUS 7 1194.3 KN/m DIFFERENCES % SAP 2000 - Analytical 3.7% STRAUS 7 - Analytical -2.9% Table 1.18 Figure 1.27 Figure 1.28 Figure 1.29 Figure 1.30
  20. 20. 17 1.2.2.9 Case 9 : all four edges built in ANALYTICAL SOLUTION ki 10.2 Ncr 7375.1 KN/m NUMERICAL CODE SOLUTION SAP 2000 7255.2 KN/m STRAUS 7 7259.0 KN/m DIFFERENCES % SAP 2000 - Analytical -1.7% STRAUS 7 - Analytical -1.6% Table 1.19 1.2.2.10 Case 10 : two edges simply supported, two edges free ANALYTICAL SOLUTION ki 2.05 Ncr 1482.2 KN/m NUMERICAL CODE SOLUTION SAP 2000 1530.4 KN/m STRAUS 7 1466.9 KN/m DIFFERENCES % SAP 2000 - Analytical 3.1% STRAUS 7 - Analytical -1.0% Table 1.20 Figure 1.31 Figure 1.32 Figure 1.33 Figure 1.34
  21. 21. 18 1.2.2.11 Case 11 : three edges simply supported, one edge free ANALYTICAL SOLUTION ki 2.3 Ncr 1663.0 KN/m NUMERICAL CODE SOLUTION SAP 2000 1692.6 KN/m STRAUS 7 1708.9 KN/m DIFFERENCES % SAP 2000 - Analytical 1.7% STRAUS 7 - Analytical 2.7% Table 1.21 1.2.1.12 Case 12 :two edges simply supported, one edge built in,one edge free ANALYTICAL SOLUTION ki 2.3 Ncr 1663.0 KN/m NUMERICAL CODE SOLUTION SAP 2000 1708.7 KN/m STRAUS 7 1728.1 KN/m DIFFERENCES % SAP 2000 - Analytical 2.7% STRAUS 7 - Analytical 3.8% Table 1.22 Figure 1.35 Figure 1.36 Figure 1.37 Figure 1.38
  22. 22. 19 1.2.2.13 Conclusion We can order the 12 cases to find which boundary conditions give the maximum and the minimum critical load. It is easy to understand that the minimum value of critical compressive load is given by Case 1 (two edges simply supported), while the maximum is given by Case 9 (four edges built in). CASE Pcr [KN/m] 1 723.0 7 1041.2 8 1229.2 10 1482.2 11 1663.0 12 1663.0 2 2892.2 4 3615.2 6 4121.4 3 4699.8 5 5560.2 9 7375.1 Table 1.23 To understand the effect of boundary conditions, we can compare 3 cases: two edges (y=0 and y=b) are simply supported, while the other two (x=0 and x=a) are progressively stiffened. In the following table we see the critical load and the form of buckling: there is a progressive increase of the value of Pcr and a change of the way to buckle. CASE Pcr [KN/m] w [m]_XY view w [m]_YZ view 1 723.0 2 2892.2 5 5560.2 Table 1.25
  23. 23. 20 1.2.3 Bending forces The distribution of bending forces can be written as ( ) = − where N is the stress at y=0 and η is a coefficient that determines the type of load acting on the plate. = 2 − −→ = 1 − −→ The case of η = 0 is the same of section 1.2.2. The critical value of N is = where Nl is the same of the case of compressive stresses (section 1.2.2), depending only on geometric data of the plate and on the material (steel). Nl [KN/m] 723.0 k is a numerical factor depending on the ratio a/b, the case of load (pure bending or combination of bending and compression), and boundary conditions. 1.2.3.1 Case 1 : all four edges simply supported In this case we use the following schedule. PURE BENDING BENDING AND COMPRESSION Figure 1.39 Figure 1.40 Figure 1.41
  24. 24. 21 ANALYTICAL SOLUTION PURE BENDING BENDING+COMPRESSION ki 25.6 7.8 Ncr 18510.0 KN/m 5639.8 KN/m NUMERICAL CODE SOLUTION PURE BENDING BENDING+COMPRESSION SAP 2000 18537.5 KN/m 5655.2 KN/m STRAUS 7 18300.0 KN/m 5630.0 KN/m DIFFERENCES % PURE BENDING BENDING+COMPRESSION SAP 2000 - Analytical 0.15 % 0.27 % STRAUS 7 - Analytical 1.15 % -0.17 % Table 1.26 PURE BENDING Figure 1.42 Figure 1.43
  25. 25. 22 1.2.3.2 Case 2 : two edges simply supported, two edges built in In this case we use the following schedule. BENDING AND COMPRESSION PURE BENDING BENDING AND COMPRESSION Figure 1.44 Figure 1.45 Figure 1.46 Figure 1.47
  26. 26. 23 ANALYTICAL SOLUTION PURE BENDING BENDING+COMPRESSION ki 39.5 15.0 Ncr 28530.4 KN/m 10845.7 KN/m NUMERICAL CODE SOLUTION PURE BENDING BENDING+COMPRESSION SAP 2000 29064.5 KN/m 10662.7 KN/m STRAUS 7 26000 KN/m 10590 KN/m DIFFERENCES % PURE BENDING BENDING+COMPRESSION SAP 2000 - Analytical 1.73% -1.72% STRAUS 7 - Analytical -9.85% -2.41% Table 1.27 PURE BENDING BENDING AND COMPRESSION Figure 1.48 Figure 1.49
  27. 27. 24 1.2.3.2 Conclusion CASE BENDING BENDING AND COMPRESSION 1 k = 25.9 2 waves k = 7.8 1 wave 2 k = 39.5 2 waves k = 15.0 2 waves Table 1.28 1.2.4 Shear forces Now we consider uniformly distributed shearing stresses along the four edges. The critical value of T is = where Nl is the same of the case of compressive stresses (section 1.2.2), depending only on geometric data of the plate and on the material (steel). Nl [KN/m] 723.0 k is a numerical factor depending on the ratio a/b and on boundary conditions. Figure 1.50
  28. 28. 25 1.2.4.1 Case 1 : all four edges simply supported ANALYTICAL SOLUTION ki 9.7 Ncr 7013.6 KN/m NUMERICAL CODE SOLUTION SAP 2000 6576.6 KN/m STRAUS 7 6708.5 KN/m DIFFERENCES % SAP 2000 - Analytical 6.2% STRAUS 7 - Analytical 4.3% Table 1.29 Figure 1.53 Figure 1.51 Figure 1.52
  29. 29. 26 1.2.4.2 Case 2 : all four edges built in ANALYTICAL SOLUTION ki 15 Ncr 10845.7 KN/m NUMERICAL CODE SOLUTION SAP 2000 10512.9 KN/m STRAUS 7 10504.2 KN/m DIFFERENCES % SAP 2000 - Analytical 3.1% STRAUS 7 - Analytical 3.1% Table 1.30 1.2.4.3 Case 3 : two edges simply supported, two edges built in ANALYTICAL SOLUTION ki 12.3 Ncr 8893.5 KN/m NUMERICAL CODE SOLUTION SAP 2000 8921.9 KN/m STRAUS 7 9025.6 KN/m DIFFERENCES % SAP 2000 - Analytical -0.3% STRAUS 7 - Analytical -1.5% Table 1.31 Figure 1.57 Figure 1.56 Figure 1.54 Figure 1.55
  30. 30. 27 1.2.4.4 Conclusion CASE k w [m]_XY view 1 12.3 2 15.0 3 9.7 Table 1.32
  31. 31. 28 1.2.5 Compressive and shear forces In this case we have a combination of shearing and compressive forces. The critical value of T and N are given by the following expression, depending on the value of Tcr0 and Ncr0 calculated in section 1.2.4.1 and 1.2.2.2 , ≅ 1 − , or we can find them using the schedule, with curves that depends on the ratio a/b. 1.2.5.1 Case 1: all four edges simply supported Tcr0 (N=0) Ncr0 (T=0) SAP 2000 6576.6 KN/m 3043.4 KN/m STRAUS 7 6708.5 KN/m 2887.4 KN/m NUMERICAL CODE SOLUTION ANALYTICAL EXPRESSION Tcr = Ncr Ncr = Ncr0(1-Tcr/Tcr0) SAP 2000 2315.5 KN/m 2666.1 KN/m STRAUS 7 2452.4 KN/m 2887.2 KN/m Figure 1.58 Figure 1.59
  32. 32. 29 DIFFERENCES % SAP 2000 - Analytical 13% STRAUS 7 - Analytical 2% Table 1.33 1.2.6 Bending moment and shear forces In this case we have a combination of shearing and bending forces. The critical value of T and N are given by the following expression, depending on the value of Tcr0 and Ncr0 calculated in section 1.2.4.1 and 1.2.3.1 , ≅ 1 − , or we can find them using the schedule, with curves that depends on the ratio a/b. Figure 1.60 Figure 1.61 Figure 1.62
  33. 33. 30 1.2.6.1 Case 1: all four edges simply supported Tcr0 (N=0) Ncr0 (T=0) SAP 2000 7013.6 KN/m 18537.5 KN/m STRAUS 7 6708.5 KN/m 18300.2 KN/m NUMERICAL CODE SOLUTION ANALYTICAL EXPRESSION Tcr = Ncr Ncr = Ncr0(1-Tcr/Tcr0)0.5 SAP 2000 6103.1 KN/m 6906.5 KN/m STRAUS 7 6241.3 KN/m 6709.6 KN/m DIFFERENCES % SAP 2000 - Analytical 12% STRAUS 7 - Analytical 7% Table 1.34 Figure 1.63
  34. 34. 31 1.2.6.2 Conclusion CASE FORCES λcr w [m]_XY view SHEAR 6708.5 SHEAR AND BENDING 6241.3 SHEAR AND COMPRESSION 2452.4 Table 1.35
  35. 35. 32 1.3 Nonlinear static analysis 1.3.1 Elastic behavior 1.3.1.1 Post-critical behavior To read the post-buckling behavior we did a displacement control – nonlinear analysis on the plate. As boundary conditions we consider 2 sides simply supported and the other two free. a 50 cm b 50 cm s 1 cm Table 1.35 We used the numerical code Straus 7: we imposed an initial displacement of 1 mm and we run the incremental analysis (200 increments of 1mm, for a ultimate displacement of dx = 2cm). The model has a mesh 10x10, for a total of 100 finite elements ISOP4. The critical load was calculated in section 1.2.2.1: to make it comparable with the nodal reactions we put it in KN, by multiplying it for the length of the plate. = 688.5 ∗ 0.5 = . Figure 1.64
  36. 36. 33 The diagram is obtained in the following way: in the abscissas axis we put the nodal displacement imposed at the side y = b, while in the ordinate axis we put the sum of the nodal reaction of the side y = 0. The post-buckling behavior is stable, with hardening: there is an increase of the critical load with large displacements. 1.3.1.2 Geometric nonlinearity Now we want to take into account the possible geometric imperfections in the plate : to do this , we start from a plate not perfectly straight ( we draw the plate already embarked ) and we run the same analysis done in section 1.3.1.1. What we get is a curve that tends to theoretical behavior without imperfection. Figure 1.65 Figure 1.66
  37. 37. 34 1.3.2 Elastic-plastic behavior Until now we considered a purely elastic material: that is, we didn’t consider that the stresses associated with the critical buckling load may have exceeded the yield stress of the material. We proceed in this way: we insert the elastic-plastic perfect diagram into the properties of the material and run nonlinear static analysis. To highlight the different behaviors we study the same plate by variation of the thickness. In this discussion we consider a steel S235. We consider a little hardening, to avoid that the program gives problems reading the branch perfectly horizontal. σy 235 MPa εy 1.175*10-3 εu 7.5*10-2 Table 1.36 1.3.2.1 Material nonlinearity The first analysis is without imperfections: we consider a plate perfectly straight but we insert the σ–ε diagram. We find the yielding of the plate. We obtain the following diagram. Figure 1.67
  38. 38. 35 We do a simple control of the result. = 235000 ∗ 0.5 ∗ 0.01 = 1175 1.3.2.2 Material nonlinearity and imperfection We do the same analysis for a plate not perfectly straight: even if the imperfection is very small, we obtain that the limit of the yielding is really lower, as we can see in the following diagram. Figure 1.68 Figure 1.69
  39. 39. 36 Now we compare the buckling branch and the yielding branch obtained for a plate not perfectly straight. Figure 1.70 Figure 1.71
  40. 40. 37 1.3.2.3 Material and geometric nonlinearity Finally we do a nonlinear static analysis by including both geometric and material non linearity, for the plate with initial imperfection. The plate start to get the yielding stress in the lateral zone, but immediately it totally reach the yield stress. Von Mises – Combined Stresses (plane z-) White color indicates that σ > 235MPa Figure 1.72 Figure 1.73
  41. 41. 38 1.3.2.4 Comparison by varying thickness To understand better this coupled behavior (between yielding and buckling) we study two plates with different thickness: a thick plate with s = 3 cm and a thin plate with s = 2mm. We report the same diagram of Fig.1.72 for the two plate. THICK PLATE s = 3 cm Pcr buckling [KN] 9189.7 Fy _ straight plate [KN] 3525 Fy _ not perfectly straight plate [KN] 2571.8 Table 1.37 THIN PLATE s = 2 mm Pcr buckling [KN] 2.7 Fy _ straight plate [KN] 235 Fy _ not perfectly straight plate [KN] 14.1 Table 1.38 Figure 1.74 Figure 1.75
  42. 42. 39 To highline the difference of behavior between the 3 cases, we divide the values of the orange curve (calculated for an imperfect plate with both material and geometric non linearity) for the maximum value of the yield curve (green curve) , in order to adimensionalyze. We obtain the following diagram, that summarizes the analysis. Zoom Figure 1.76 Figura 1.77
  43. 43. 40 1.4 Buckling of plates beyond the proportional limit 1.4.1 Introduction We consider a case of a rectangular plate simply supported along all edges and uniformly compressed parallel to one side. As we know, the critical value of the compressive stress within the elastic limit is given by the formula: = 12 (1 − ) Experiments show that, when the compressive stress reaches the yield point of the material, which in this case is assumed equal to 235 Mpa, the plate buckles for any ratio b/h. This is shown in the following figure: Figure 1.78 To obtain such a curve we modelled with the software Straus7 a square plate with different value of ratio b/h and we report in a graph the variation of the critical value of the compressive stress in relation with the ratio b/h. As we can see in the Fig. 1.78, the Ncr remains constant for a low ratio b/h, because the collapse of the element is caused by the compressive stress that reaches the yield point, while varies for high value of the ratio b/h, because the collapse of the elements is produced by the buckling. We also can notice that this is exactly the same behaviour that we saw for the instability of inelastic bars. 1.4.2 Large deflections of buckled plates In this paragraph we will analyze the different behaviour of plates and struts when the stresses are beyond the proportional limit. The critical load for a strut is considered as the ultimate load, instead a thin buckled plate can carry much larger load than the critical load at which buckling begins. In order to investigate bending of plates when the stresses are above the critical value, we must consider the strain in the middle plane that we previously neglected it for the determination of the critical stresses for the different cases of buckling of plates.
  44. 44. 41 1.4.3 Approximate solution We analyze a simply supported plate that is compressed in the y direction and the lateral expansion in the x direction is prevented by a rigid frame (Figure.1.78). Figure.1.79 To get an approximate solution we use the solution method of minimum of strain energy 1) We assume an approximate expression for the deflection surface of the buckled plate = cos 2 cos 2 By assuming this expression for the deflection we are supposing that the deflection on the edges is zero: - The horizontal edges y = +/-a = cos 2 cos +/− 2 = 0 - The vertical edges x = +/-b = cos 2 cos +/− 2 = 0 2) The components v and u of the displacements in the middle plane are: = sin cos 2 − = sin cos 2 Where C1 , C2 and e are constants. By assuming these expressions for the components v and u of the displacements of the middle plate, we are supposing that the frame on the edge is absolutely rigid and that lateral displacements of the edges of the compressed plate (u) are entirely suppressed. - The vertical edges x = +/-b = sin(+/− ) cos 2 = 0 3) The components of the strain in the middle plane are: ⎩ ⎪⎪ ⎨ ⎪⎪ ⎧ = + 1 2 = + 1 2 = + + 4) By using the previous expressions we can determinate the strain energy U.
  45. 45. 42 5) For a given unit compression e of the plate, we can determinate the constants C1 , C2 and by using the condition that the strain energy U is a minimum. ⎩ ⎪⎪ ⎨ ⎪⎪ ⎧ = 0 = 0 = 0 By resolving the equations of the point 5 we determinate the critical compressive stress e = 0.632 h a → σ = 0.632t E (1 − υ )b The critical stress just calculated is equal to the critical compressive stress that we obtained in the paragraph 1.1.2 6) Determination of the stresses σx and σy : - We substitute the value of the constants f, C1 ad C2 in the equations of point 1 and 2 - We calculate the corresponding strain in the middle plane of the plate from equations of point 3 - We calculate the corresponding stresses by using the following expression: = 1 − + ( ) = 1 − ( ) + 7) I f we consider a compressive force n time greater than the buckling critical one, we can observe that the compressive stress have a non-uniform diagram (Figure.1.78) instead of having the classic uniform diagram of the compressive stress. This is why for a compressive force greater than the critical one, the vertical fibres instead of collapse ,as they would do if we analyzed a single strut, they can absorb a further load because their deflection out of the plane is hindered by the flexural rigidity of the horizontal fibres. For the restraints of the plate that we are considering, we know that the part of the horizontal fibres near of the edge is more rigid because is close to the restraints( that are considering having an infinite rigidity). So the vertical fibres near to the vertical edge are able to absorb a greater post-critical load because they can count on a greater flexural rigidity of the horizontal fibres. Instead of considering the non-uniform compressive stress diagram above-mentioned, we can analyze a plate characterized by a uniform compressive stress diagram by reducing the width of the plate (Figure.1.78). In order to calculate c we determinate the total compressive force taken by the plate at a unit compression n*ecr  n*σcr n*Ncr = −ℎ = 2 ℎ 0.623 + 0.377 = ℎ Where: = 2 0.623 + . Figure.1.80 c/2 c/2
  46. 46. 43 The factor c gives the relative diminishing of the resistance to compression of the plate caused by the buckling. The resistance reduction is represented by the diminishing of the width of the plate from c’ = 2ab to = 2 0.623 + . . We can state that this resistance is equivalent to that of a flat plate having a width equal to c (Figure.1.80). Conclusion: 1) A compressive plate simply supported is able to absorb a compressive force greater than the critical one 2) For plates under a compressive action greater than the critical one the compressive stress diagram is non-uniform 3) The buckling causes a diminishing of the resistance to compression of the plate 1.4.4 Approximate solution for very thin plate For thin plate as used in airplane constructions, the approximate solution that we above-calculated is not sufficiently accurate. The solution can be improved by changing the boundary conditions. In the previous case it was assumed that the frame is absolutely rigid and then the lateral displacements of the edges are entirely suppressed. Instead in this case we assume that the later bars of the frame keep the lateral edges of the plate straight but can move freely laterally. ⎩ ⎪ ⎨ ⎪ ⎧ = cos 2 cos 2 = sin cos 2 − = sin cos 2 + We resolve the problem in the same manner explained in the previous paragraph. The unique difference is the presence of the factor α that it is calculated from the condition that the sum of normal stresses along the vertical edge of the plate is equal to zero. In order to determinate the ultimate load that the plate can carry we can use the Tresca yield criterion. The complete failure of the plate occurs when the maximum shearing stress reaches the value τyp Where: = = 1 2 − By studying the diagram of the stresses σx and σy we observe that the maximum value of the shearing stress occurs at the points x = +/- a , y =0 By following the same steps of the previous paragraph we determinate the ultimate load of the plate: = 2ℎ = 2 ℎ 0.434 + 0.566
  47. 47. 44 1.4.5 Square plate simply supported In this paragraph we will analyze a square plate simply supported in order to determinate its behavior beyond the proportional limit. DATA a = b = 1000mm t = 10 mm E = 2E5 MPa Perfect elastic-plastic constitutive law σy = 235 MPa εy = 1.175E-3 Figure.1.81 Analytical solution: By following the steps described in the previous paragraph we determinate the equations that describe the components of strain in the middle plane of the plate: = cos cos 2 + 1 2 − 2 sin 2 cos 2 = cos cos 2 − + 1 2 − 2 sin 2 cos 2 We determinate the stresses: = 1 − + ( ) = 1 − ( ) + We chose two cross sections of the plate and we plot the stress diagram related to them 1) Section Y = a 2) Section X = a Figure.1.82 0 0.5 1 1.5 2 2.5 3 -1 -0.5 0 0.5 1 σx /σcr asse y/a Cross section x = a Analytical solution
  48. 48. 45 Figure.1.83 Numerical solution: 1) We modelled a square plate with the software Straus7 Boundary condition: Horizontal edges: - uy=0 - uz=0 Right vertical edge: - uz=0 - we create several links that connect every points on this edge in order to impose them to have the same displacement ux Left vertical edge: - uz = 0 - ux = 0 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 -1 -0.5 0 0.5 1 σy /σcr asse x/a Cross section y = a Analytical solution Figure.1.84 Y X Z
  49. 49. 46 2) We define the non linear cositutive law: Figure.1.85 We run the non linear static analysis by considering the non linear material: Figure.1.86 By solving this analysis we determinate the compressive force necessary to the stresses in order to reach the yield point of the material.
  50. 50. 47 Figure.1.87 Analytical solution Numerical solution σy 235 Mpa Step 4.7 Fcr 558 kN Fy 2350 kN Fy 2622.6 kN Error_Fy 10% Table 1.39 3) We consider the non linear geometry by applying a surface force to the center of the plate. Figure.1.88 We run the non linear static analysis by considering the non linear material and the non linear geometry For a load factor of 4.7 the compressive stresses reach the yield.
  51. 51. 48 Figure.1.89 In order to compare the results provided by this analysis we can proceed in the following manners: We compare the critical compressive force obtained by the software Strays7 with the critical one obtained by the analytical solution. Straus 7 Fcr = 1450.8 kN Analytical solution = − ∗ ∫ = ∗ ∗ ∗ = 554 kN 4) The analysis results : - We compare the compressive stresses diagram obtained by the analytical and numerical solution Figure.1.90 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 -1 -0.5 0 0.5 1 σy /σcr axis x/a Cross section y = a Straus7 Analytical solution
  52. 52. 49 Figure.1.91 If we extract the compressive stresses diagrams for different steps we can observe that for value of the compressive force minor than the critical compressive load the compressive stresses diagram is uniform while for compressive force greater than the critical one we observe a non-uniform diagram. Figure.1.92 Legend: Plate stress XX - Increment 4 - Load factor = 0.3 Plate stress XX - Increment 8 - Load factor = 0.7 Plate stress XX - Increment 10 - Load factor = 0.9 Plate stress XX - Increment 11 - Load factor = 1  This is the critical compressive load Plate stress XX - Increment 16 - Load factor = 1.5 Plate stress XX - Increment 21 - Load factor = 2 Plate stress XX - Increment 27 - Load factor = 2.6  The plate collapse -2 -1.5 -1 -0.5 0 0.5 1 1.5 -1 -0.5 0 0.5 1 σx /σcr axis y/a Cross section x = a Analytical solution Straus7 Uniform diagram Non-Uniform diagram
  53. 53. 50 5) In order to obtain a better solution, we repeat the analysis by considering different boundary conditions (Paragraph.1.3.4) Boundary condition: Horizontal edges: - uz=0 Right vertical edge: - uz=0 - we create several links that connect every points on this edge in order to impose them to have the same displacement ux Left vertical edge: - uz = 0 - ux = 0 Figure.1.93 By solving the analytical end the numerical solution we obtain the following graphs Figure.1.94 0 0.5 1 1.5 2 2.5 3 -1 -0.5 0 0.5 1 σy /σcr axis x/a Cross section y = a Straus7 Analytical solution Y X Z
  54. 54. 51 Figure.1.95 In addition we compare the results provided by the approximate and the performance analysis: Legend: P.A.S.10 Power analytical solution (thickness of the plate = 10 mm) A.S.10 analytical solution (thickness of the plate = 10 mm) P.S7.S.10 Power Straus7 solution (thickness of the plate = 10 mm) S7.S.10 Straus7 solution (thickness of the plate = 10 mm) Figura.1.96 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 -1 -0.5 0 0.5 1 σx /σcr axis y/a Cross section x = a Analytical solution Straus7 0 0.5 1 1.5 2 2.5 3 -1 -0.5 0 0.5 1 σx /σcr (asse x)/a C.S. x = a analytical solutions P.A.S.10 A.S.10
  55. 55. 52 Figura.1.97 Figura.1.96 Figura.1.97 -2 -1.5 -1 -0.5 0 0.5 -1 -0.5 0 0.5 1 σy/σcr (asse y)/a C.S. y = a analytical solutions P.A.S.10 A.S.10 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 -1 -0.5 0 0.5 1 σx /σcr (asse x)/a C.S. x = a Numerical solutions P.S7.S.10 S7.S.10 -1.5 -1 -0.5 0 0.5 1 1.5 -1 -0.5 0 0.5 1 σy /σcr (asse y)/a C.S. y = a Numerical solutions P.S7.S.10 S7.S.10
  56. 56. 53 2 - SHELLS 2.1 Thin shells 2.1.1 The membrane theory We consider a shell where the thickness of it is small in comparison to its other dimensions and we suppose that the resultant of the external forces that act on the element can be decomposed in two component: - One component is tangent to the meridian of the shell (X) - One component is normal to the surface of the shell (Z) By assuming the above-mentioned values of the external force, we obtain the following expressions for the components of stress: - σ1 : represented the component of stresses that is tangent to the meridian. This stresses are caused by the resultant of the external vertical forces. - σ2 : represented the component of stresses that is tangent to the parallel. This stresses are caused by the force Z and by the changing of direction of the stresses σ1. In fact, if we consider a circular cylindrical shell we can observed that the stresses σ1 , in order to remain in the same direction of the tangent of the parallel, change its direction. This implies that the stresses are not balanced and they generate a resultant radial force(Figure 3.1). This force is another cause of the creation of the σ2 stresses. Figure 2.1 The actions that are created by the above mentioned stresses are called membrane forces. These membrane forces are actually resultants of the normal stresses and the in-plane shear stresses that are uniformly distributed across the thickness. That means that the membrane is considered totally private of flexural and torsional rigidity . The corresponding theory of this membrane behaviour is called the membrane theory. The membrane theory can be explained by studying the deformation of the shell.
  57. 57. 54 In order to study the deformation of the shell, we subdivided the shell in elemental strips of unit width that follow the geometry of the meridians and parallels. If we observe the deformation of the strip, we can notice the following sentences: 1) Because the strips of the parallels are compressed or in traction they modify their radius but remaining circular . 2) In consequence at point 1, the strips of the meridians can deform and consequently their curvature can change in every point of the meridian. 3) The change of the curvature comports the creation of stresses that are not uniformly distributed across the thickness (σ’1). Then the stresses that act in the same direction of the tangent of the parallel is described by the following sum: = + ʹ 4) The stresses are not uniformly distributed as much as σ1’ is big in comparison to σ1. Because we considered a thin shell, the value of Ymax is very small and consequently the non uniformly stresses are very small in comparison to the uniformly stresses. Conclusion: This lead us to the assumption that for shell characterized by a very small thickness-to-minimal-radius ratio the stresses can be considered uniformly distributed across the thickness. There are different consequence of the above-mentioned assumption: 1) In each point of the membrane there is a plane stress state. The plane of the stress state is tangent to point of the element. 2) There aren’t bending or shear stresses in any point of the shell that is related with the assumption that the element doesn’t have flexural and torsional rigidity. In order to study a shell by using the membrane theory, there are some condition that the element must satisfy: 1) Boundary condition and deformation constraints must be compatible with the requirements of a pure membrane field (Figure.3.2). Figure 2.2 2) There must not be concentrated loads 3) There must not change the shell geometry If just one of the previous requirements would not be satisfy, this would create stresses that are not tangent to the middle surface and would contrast with the assumption that we previously made in the theory of the membrane.
  58. 58. 55 2.1.2 Analysis of a symmetrically loaded shell of revolution We consider a membrane described in Figure 3.3 This membrane was obtained by cutting a shell of revolution in two parts. In order to analyze just one part of the element, we replace the superior part by considering the force that the two parts exchange each other. Figure 2.3 S1 represents the integral of the stresses tangent to the meridians. In order to write the equation equilibrium for the vertical loads, we have to calculate the vertical component of the force S1. = sin Subsequently we cha write the equation equilibrium: sin 2 = → = Where: - Q is the resultant of the vertical loads By writing the equation equilibrium of the vertical loads we obtained the value of the stress S1 In order to determinate the value of the force S2 , that represent the integral of the stresses tangent to the parallels, we write the equation equilibrium of the element abcd of the membrane (figure.3.4). Figure 2.4 The equation equilibrium: ∗ ∗ + ∗ ∗ ∗ sin = ∗ ∗ ∗ ∗ sin ∗ ∗ + ∗ ∗ ∗ ∗ sin = ∗ ∗ ∗ ∗ sin ∗ ∗ + ∗ = ∗ ∗
  59. 59. 56 + = Because we determinated S1 with the previous equation equilibrium, we can obtain S2 by the expression: = − 2.1.3 Numerical examples In this paragraph we will solve some exercise by using the analytical formulation that we explained in the previously paragraph, subsequently we will compare the results with the results provided from a numerical analysis performed with the software Sap2000. 2.1.3.1 Hemisphere Question: Calculate the stresses of a hemisphere tank full of water DATA R = 3 m s = 0.1 m γW = 10 kN/m3 Analytical solution: Figure 2.5 Development: 1) We calculate the volume of the Hemisphere: Figure = 2 = 2 cos sin = 2 3 (1 − cos ) 2) By using the formula = we calculate S1 = 2 sin 3) By using the formula = − we calculate S2 In this case: - R1 = R2 = R - = cos
  60. 60. 57 Solution performed by Sap2000: We modelled with the software Sap2000 the cylinder and then we loaded the water tank by assigning a load pattern (Figure.3.5 and 3.6.) In order to fulfil the requirements of the membrane theory, we apply an uniform distributed load that varies linearly with the Z. In addition we define rollers as restraints for the edge of the structure in order to have a boundary condition that is compatible with the requirements of a pure membrane field . Figure 2.6 Figure 2.7
  61. 61. 58 Figure 2.8 Figure 2.9 SAP 2000 BELLUZZI step z θ S1 S2 S1 S2 Error S1 Error S2 m deg KN/m KN/m KN/m KN/m % % 1 0.000 0 45.53 44.65 45.00 45.00 -1.2% 0.8% 2 0.333 10 44.45 42.61 44.66 43.97 0.5% 3.1% 3 0.667 20 43.04 38.94 43.66 40.92 1.4% 4.8% 4 1.000 30 41.32 34.10 42.06 35.88 1.8% 5.0% 5 1.333 40 39.33 27.94 39.97 28.98 1.6% 3.6% 6 1.667 50 37.16 20.40 37.55 20.31 1.0% -0.5% 7 2.000 60 34.91 11.52 35.00 10.00 0.3% -15.1% 8 2.333 70 32.77 1.40 32.61 -1.83 -0.5% 176.1% 9 2.667 80 30.99 -9.84 30.77 -15.14 -0.7% 35.0% 10 3.000 90 29.90 -22.16 30.00 -30.00 0.3% 26.2% S2 S1 Table 2.1
  62. 62. 59 Figure 2.10 Conclusion: 1) As we can see in the figure 3.6 and 3.7, the stresses of the element are related only on the Z variable. 2) The S1 stress is always positive while the stress S2 is positive near the bottom of the element and negative near the top of it (Figure.3.6.). That means that the meridians are always in traction while the parallels are in traction near the top of the element and in compression near the bottom of it. This could be explained by analyzing the kinematic of a dome, that is characterized by the same geometry of the tank (Figure 3.9). Figure 2.11 -40 -30 -20 -10 0 10 20 30 40 50 0 20 40 60 80 stress[KN/m] θ [deg] Comparison S1 _ SAP S2 _ SAP S1 _analitica S2 _analitica
  63. 63. 60 2.1.3.2 Toroidal tank In this exercise we consider a tank characterized by the form of a torus that is full of water. Figure 2.12 Question: Calculate the stresses of the tank DATA Re = 4.5 m Ri = 3.5 m Rt = 0.5 m S = 0.05 m γW = 10 kN/m3 Analytical solution: Figure 2.13 In order to study an equation with a unique variable we subdivide the element in two portion (Figure...). The vertical load that is caused by the water in that part E of the element is absorbed by the external stress S1e while the vertical load that is caused by the water in the part I is absorbed by the internal stress. 1) By using the Guldino formulations we calculate the volume of the parts E and I. = 4 2 ( + 0.4244 ) = ∗ 4 2 ( − 0.4244 ) 2) by writing the equation equilibrium we determinate the stresses S1e and S1i = 4 + 0.4244 + = 4 − 0.4244 − 3) By using the formula = − we calculate S2e andS2i S1e S1i
  64. 64. 61 = − + = − − Solution performed by Sap2000: We modelled with the software Sap2000 the cylinder and then we loaded the water tank by assigning a load pattern. In order to fulfil the requirements of the membrane theory, we apply an uniform distributed load that varies linearly with the Z. In addition we define rollers as restraints for the edge of the structure in order to have a boundary condition that is compatible with the requirements of a pure membrane field . Figure 2.14 SAP 2000 BELLUZZI S1est S1int S2est S2int S1est S1int S2est S2int KN/m KN/m KN/m KN/m KN/m KN/m KN/m KN/m 1.73 2.135 -10.72 13.28 1.84 2.12 -16.54 14.87 Error S1est S1int S2est S2int % % % % 6% 0% 35% 11% Conclusion: 1) The stress S2 varies between a maximum positive value that is reached in the internal circumference and a minimum negative value that is reached in the external circumference . That means that the external circumference is in traction while the internal is in compression. Table 2.2
  65. 65. 62 2.2 Thick shells 2.2.1 Introduction Differently from the thin shells, these type of elements are characterized by having a thickness that we cannot ignore during the analytical analysis. By considering the thickness of the shell, we must consider bending and shear stresses in addition to the stresses that we regarded for the thin shells. The deformations that generate the bending and shear stresses are caused by two factors: 1) External forces or bending moment that are apply to the edge of the shell. In this case, the stresses quickly dampen and we can notice the presence of the bending and shear stresses just as a local effect around the edge. The dissipation of the bending and shear stresses is caused by the parallels that absorb part of these stresses because they are more rigid than the meridians. 2) External forces that are distributed on the surface of the element In this case, because the distributed forces deformed the shells, the bending and shear stresses are presented in the all element. However, these stresses are much smaller than the membrane stresses. Conclusion: In order to study the stress and deformation state of thickness shells, we can use the same theory that we described for thin shells. However if we want to study the local effects provided by External forces or bending moment that are apply to the edge of the shell we have to change theory. 2.2.2 The differential equation of the elastic line We consider a cylindrical tube where act a pressure from the internal to the external of the tube. The pressure have to be uniformly distributed on the same parallel and can varies along the meridian. In addition can be present external bending moment or radial forces in one or both of the cylinder edges. Figure 2.15 In these radial symmetry conditions, the elements deforms in the way that each parallel amplifies of reduces its radius but remains circular and the generatrix of the cylinder inflect on the radial plane. As we previously did, we subdivide the elements in strips of unit width and we assume that the pressure is supported by the longitudinal strips. The transversal strips contrast the deformation of longitudinal strips by reacting with radial forces that act on the longitudinal strips (Figure.4.2).
  66. 66. 63 These radial forces ρ are correlated with the variation of the parallels radius. In order to determinate this relation we perform the following steps: 1) Assume that the angle ω is very small = sin ≅ 2)Determinate ρ = 2 sin 2 → = → = 3) Determinate the dilatation of the generic parallel The perimeter of the circumference before the deformation is 2πR The perimeter of the circumference after the deformation is 2πw (w = R + ΔR) The dilatation is: = 2 2 = By using the elastic constitutive law : = = 1 4) Determinate the relation = ( ) = = 1 = = We can determinate the differential equation of the elastic line: + = Where = ( ) By solving the differential equation we can calculate the following values: - = - = − - = - = Figure 2.16
  67. 67. 64 2.2.2.1 The stationary solution By assuming = the differential equation become : + 4 = The equation solution is composed by a particular solution plus the general solution. The particular solution represent the effect of the pressure P that acts on the tube surface and don’t consider the external forces or bending moment that could act on the edge of the element. This solution represents the stationary solution because is correlated with the distributed load that act on the entire element. 2.2.2.2 The general solution In order to determinate the general solution we have to resolve the following equation: + 4 = 0 The solution is expressed by the equation: = sin + cos + sin + cos The displacement described by this equation represent the effect induced by the shear forces and the bending moments that are uniformly distributed on the edge of the element. By considering a tube that is very long in comparison with its other dimensions and assuming that the variable x starts where are applied the shear forces and bending moments the constant C1 and C2 can be considered equal to zero. = ∗ sin + cos If we consider = cos and = sin we can write the solution in a different formulation: = sin( + ) By using the formulation of the paragraph 4.1 we can obtain the following equations: - = sin( + ) - = √2 sin( + − 4⁄ ) - = 2 sin( + − 2⁄ ) - = −2√2 sin( + − 3 4⁄ ) As we can see, this equation represents a displacement that dissipates by going away from the loaded edge. The dissipation is regulated by the term that rapidly decrease with the increasing of the variable x. These expressions lead us to the observation that the bending moment (M) and the transverse shear (H) assume relevant value just in the part of the element near the loaded edge and can be ignored in the rest of the body (Figure.4.3 and 4.4); so they can be considered as local effects. Figure2.17 Figure 2.18 This is exactly what we declared in the previous paragraph by analyzing the effect of the action of the annular strips on the longitudinal strips.
  68. 68. 65 The dissipation of these local effects is fast as much as the wavelength is small and because the wavelength is related with the radius and the thickness of the element by ≅ 4.83 √ The dissipation of these local effects is fast as much as the radius and the thickness of the element are small. 2.2.3 Cylindrical tank Question: Calculate the following entities associated to one longitudinal strip - The displacement w - The transverse shear - The bending moment DATA R = 1 m L = 5 m h = 5 m s = 0.01 Material : Steel Constitutive law : linear elastic E = 2*10^8 kPa γt = 10 kN/m3 Figure 2.19 Analytical solution: The particular solution of the differential equation: = = (ℎ − ) The complete solution : = (ℎ − ) + ( sin + cos ) 1) We determinate the constant by using the surrounding conditions: Because the inferior edge is characterized by a fixed support the displacement w and the rotation φ are zero ⎩ ⎪ ⎨ ⎪ ⎧ ( = 0) = 0 → ℎ ∗ + = 0 = ʹ( = 0) = 0 → − + ( − ) = 0 → ⎩ ⎪ ⎨ ⎪ ⎧ = − ℎ − 1 = − ℎ By discretizing a longitudinal strip in 50 parts and using the formulation described in the previous paragraphs we determinate the displacemnt w the transverse shear T and the bending moment M.
  69. 69. 66 2) We determinate the following parameters α 13.13 1/m v 0.3 B 18.31 KN/m2 β 2.0E+06 kN/m3 a 12.85 C3 -2.46E-05 m C4 -2.50E-05 m 3) We determinate the displacements w and we consequently determinate the transverse shear and the bending moment Figure 2.20 Figure 2.21 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.000005 0.00001 0.000015 0.00002 0.000025 X [m] W [m] Displacement - W Analytical solution -2.00 -1.00 0.00 1.00 2.00 3.00 4.00 0 1 2 3 4 5 T [KN] X [m] Trasverse shear - T Analytical solution = ∗ ′′′ Table 2.3
  70. 70. 67 Figure 2.22 Solution performed by Sap2000: We modelled with the software Sap2000 the cylinder and then we loaded the water tank by assigning a load pattern. We lunch the analysis and then we export the displacements, the transverse shear and the bending moment values of a single meridian in order to compare the results provided by the both analysis (Figure4.10- 4.11-4.12). -0.05 0.00 0.05 0.10 0.15 0.20 0 1 2 3 4 5 M [KNm] X [m] Bending moment- M Analytical solution = ∗ ′′ Figure 2.23
  71. 71. 68 Figure 2.24 Figure 2.25 Figure 2.26 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.000005 0.00001 0.000015 0.00002 0.000025 0.00003 X [m] W [m] Displacement - W Sap2000 Analytical solution -2 -1 0 1 2 3 4 0 1 2 3 4 5 X TkN] X [m] Trasverse shear - T Sap2000 Analytical solution -0.05 -0.03 -0.01 0.01 0.03 0.05 0.07 0.09 0.11 0.13 0.15 0 1 2 3 4 5 M [kNm] X [m] Bending moment- M Sap2000 Analytical solution
  72. 72. 69 2.3 Buckling of shells 2.3.1 Buckling of cylindrical shells with circular section If we want to study the form of buckling of a shell, we must consider loads perpendicular to the middle plane: indeed, the behavior out of plane is the most important. From the analysis (that are described in section 2.1.2), we obtain that the form of buckling is a lateral movement of the entire shell. Figure 2.27 Figure 2.28 This results was found for all the cases described in the following sections 2.1.2 and 2.1.3. Undeformed shell Buckled shell
  73. 73. 70 2.3.1.1 Comparison by varying the ratio h/L To understand how rapidly the critical value of compressive force can vary, we studied a shell of fixed thickness and length L by varying the ratio h/L, where h is the height of the shell. We consider both hinged and built in ends. L [m] 1 s [m] 0.02 Table 2.4 Figure 2.29 We did the buckling analysis by using the numerical code Straus 7 : the mesh is a 10x10 finite elements ISOP4. From the diagram, we see that the critical load has a maximum when the ratio h/L approaches to 0.3. Before and after this value, the critical load decreases. 2.3.1.2 Comparison by varying thickness Now we did the same analysis by varying the thickness of the shell. Figure 2.30
  74. 74. 71 Figure2.31 Figure2.33Figure2.32
  75. 75. 72 2.3.2 Nonlinear analysis We chose one of the cases described before and we did nonlinear analysis, by using Straus 7. L [m] 1 h [m] 0.4 s [m] 0.02 Ends hinged dy – step [m] 0.006 Table 2.5 The results are summarized in the following diagram. We see that the blue line, obtained by displacement-control nonlinear analysis, has a maximum that coincides with the green line (buckling load obtained by buckling analysis). The red line is the force-control nonlinear analysis: it initially follows the blue line, then it has a snap, and arrives on the other positive side of the blue line: at this step the analysis fails and we can’t see the following increasing branch. This is a typical non-eulerian buckling: the behavior of this kind of structures is nonlinear even before the buckling limit. When the curve reaches the maximum, the shell buckles with a lateral form (see Fig. 2.1.2) and then flips over, going from an initial structure in analogy with “arches” to a final structure with “cables” behavior. At this point, the load may continue to increase, but the resistance mechanism is completely changed. Figure 2.34 Figure 2.35 Figure 2.36 Figure 2.37 Figure 2.38
  76. 76. 73 2.3.3 Very flat shells uniformly compressed In this case the axial stress, due to the action of the thrust H, becomes important: the buckled form is influenced by the presence of high axial stress and is not extensionless. 2.3.3.1 Nonlinear analysis by varying thickness We analyzed a shell with very low ratio h/L, by varying thickness. L [m] 1 h [m] 0.05 h / L 0.05 Ends hinged Table 2.6 -150 -100 -50 0 50 100 150 200 0 0.02 0.04 0.06 0.08 0.1 0.12 P[KN] Dy [m] s = 1 cm Force control Displacementcontrol Bucklinganalysis -800 -300 200 700 1200 1700 0 0.02 0.04 0.06 0.08 0.1 0.12 P[KN] Dy [m] s = 2 cm Force control Displacementcontrol Bucklinganalysis Figure 2.40 Figure 2.39
  77. 77. 74 We find the same behavior (snap through buckling), but we see that by increasing the thickness of the shell, the result of nonlinear analysis goes away from the ideal conditions obtained by buckling analysis. DIFFERENCES % s = 1 cm s = 2 cm s = 3 cm Buckling –force control NL analysis 6.4 % 12.9 % 26.5 % Table 2.7 In the following diagram we see that the thickness has a large influence on the critical value of compressive load. -2000 -1000 0 1000 2000 3000 4000 5000 6000 7000 0 0.02 0.04 0.06 0.08 0.1 0.12 P[KN] Dy [m] s = 3 cm Force control Displacementcontrol Bucklinganalysis 0 500 1000 1500 2000 2500 3000 3500 4000 0 0.02 0.04 0.06 0.08 0.1 P[KN] Dy [m] Force control - Nonlinear analysis s = 1 cm s = 2 cm s = 3 cm Figure 2.42 Figure 2.41
  78. 78. 75 2.3.4 Cylindrical shell under the action of uniform axial compression If a cylindrical shell is uniformly compressed in the axial direction buckling symmetrical occurs when the stresses compression reached the buckling value (Figure.5.1). The critical value of the compressive force Ncr per unit length of the edge of the shell can be obtained by using the energy method. As long as the shell remain remains cylindrical, the total strain energy is the energy of axial compression. However, when buckling begins, we have to consider: - The energy related to the strain of the middle surface in the circumferential direction - The energy related to the bending of the shell Thus the strain energy of the shell is increased; when the compression reached the critical value the increase of energy must be equal to the work done by the compressive load as the cylinder shortens owing the buckling. So in this paragraph we will determinate the two different increment of energy in order to determinate the critical buckling value. 1) We assume the radial displacement during the buckling described by the expression = − sin 2) We determinate the component f the deformation ε1 and ε2 after buckling by assuming that the axial compressive force during buckling remains constant. = − ℎ + = (1 − ) Observing that : = − − = − + sin We find that: = − sin + 3) By equating the expressions related to the increase energy during the buckling and the work done by the compressive forces during buckling we can determinate the different stresses buckling value: = ℎ = ℎ + This expression describes the autovettore related to the autovalore m. As we saw for the instability of monodimensional element the buckling value is the smallest fra tutti gli autovettori. This is obtained by the privies expression by equating: = ℎ = ℎ 3(1 − ) Example : We consider the same cylinder described in the previous paragraph DATA R = 1 m L = 5 m h = 5 m s = 0.01
  79. 79. 76 Material : Steel Constitutive law : linear elastic E = 2*10^8 kPa γt = 10 kN/m3 We determinate the buckling value by using two different method: 1) Analytical method = ℎ 3(1 − ) = 1210455 → = 12104.55 2) Numerical method We modelled the cylinder with the software Strauss7 Figure 2.43 Figure 2.44 The different between the two methods: Straus7 Ncr 12424 KN Analitica σcr 121045 KPa Ncr 12104.5 KN Error 3% Table 2.8 Ncr = 12424 kN
  80. 80. 77 BIBLIOGRAPHY  Timoshenko, Gere – “Theory of elastic stability”  Belluzzi – “Scienza delle costruzioni – Volume 3”  Bontempi, Sgambi, Malerba – “Sulla robustezza strutturale dei ponti a arco molto ribassato”

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