MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 3 ppt

58 Structural elements hold, the dilatational (P ) waves are reflected without mode conversion. Again, it is stressed that sliding supports would be difficult to manufacture. Incidentally, if the solid is replaced by a non-viscous fluid, i.e. a medium in which the Poisson effect and the shear stresses do not exist, the present calculation takes on its full practical interest. In a fluid, the dilatation waves are called acoustic waves and the natural modes defined by the formulas [1.136], [1.137] are the acoustic modes of a rectangular enclosure provided with rigid and motionless walls. 1.4.7.3 Shear plane modes of vibration Starting from the plane shear waves [1.86], it is immediately verifiable that shear stresses σ xy = GdY /dx and σ xz = GdZ/dx are the only nonzero stress components. They both must vanish at the faces x = 0 and x = L x which are assumed to be free. Whence the boundary conditions: dY dx     0 = dY dx     L x = 0; dZ dx     0 = dZ dx     L x = 0 [1.139] From [1.77] it follows that the partial differential equation to be solved is: G  (Y (x)  j + Z(x)  k)  + ω 2 ρ 0 ((Y (x)  j + Zx  k)) = 0 [1.140] Applying the relation [A.2.14] in Appendix A.2 to calculate the Laplacian of the displacement vector, it is readily shown that the vector equation [1.140] can be split into two identical and uncoupled equations of the same type as those already encountered in the last examples: d 2 Y dx 2 +  ω c s  2 Y = 0; d 2 Z dx 2 +  ω c s  2 Z = 0 [1.141] Hence the waves oscillating in the Oy direction are found to be identical to those oscillating in the Oz direction. Such a result could have been anticipated, since the transverse sections parallel to the Oyz plane are not deformed and oscillate as a rigid body in the Oy and in the Oz directions, independently. As a consequence, the corresponding natural modes of vibration can be split into two families which are identical to each other, except that they vibrate into two orthogonal directions. The common modal quantities are: ϕ (T ) n (x) = cos  nπx L x  ; ω n = nπc s L x ; n = 0, 1, 2, [1.142] By using the principle of superposition, the physical mode shapes can be written as: ϕ (T ) n (x) = (a n  j + b n  k)ϕ (T ) n (x) [1.143] Solid mechanics 59 where the vector (a n  j +b n  k) specifies the direction of vibration, that is the polar- ization of the wave. Clearly, any transverse direction is possible and equivalent to the others. It can be also noted that modes at zero frequency exist, which are immediately recognized as a rigid displacement of the whole body in an arbitrary transverse direction. 1.5. From solids to structural elements 1.5.1 Saint-Venant’s principle The difficulty in obtaining the analytical solutions of elastodynamic problems is also encountered when attempting to solve static problems, as for instance that of determining the stresses in a three-dimensional body induced by forces applied on small portions of it. As an example, let us consider a cylindrical rod inserted in a tensile test machine, see Figure 1.23. If a numerical simulation of the experiment is performed, by using the finite element method (regarding the finite element method see Chapter 3, section 3.4), quite enlightening results are obtained concerning the displacement field and the stress distribution, especially sufficiently near the grips. Major information can be suitably summarized as follows: 1. Near the loaded parts of the rod, in a portion extending no more than a fraction of the rod radius in each direction, the displacement field, and even more conspicuously the stress field, are found to vary in all directions, in connection with the detailed space distribution of the contact forces exerted by the grips. Such fields characterize the so called local response of the system rod plus grips. Figure 1.23. Cylindrical rod set in a tensile test machine 60 Structural elements 2. In the current portion of the rod, sufficiently far from the grips, the stress field is found to reduce practically to only one non-vanishing component, namely the uniform axial stress σ zz (r, θ , z) = T/S, where cylindrical coordinates r, θ,zare used. Here, T stands for the resultant of the tensile force system exerted by the grips and S denotes the cross-sectional area of the rod. The displacement field is marked by an axial component denoted by W and a radial component denoted by U, which are independent of the polar angle θ. W is observed to increase in proportion to z, starting from the fixed grip, and U varies in proportion to r. All these fields are independent of the dis- tribution of the loading by the grips and characterize the global response of the rod. As can be expected from the simplicity of the response observed sufficiently far from the grips, the analytical solution of the problem becomes straightforward, provided it is formulated in a suitable way. A first idea is to adopt the model of a circular cylinder loaded at its bases by a uniform axial force density whose resultant is T . The problem can be thus solved by using a 2D model, since the θ dependency is removed. Further, if the interest is restricted to determining the axial displace- ment and the axial stress, a 1D model becomes sufficient. Finally, to determine σ zz only, it becomes unnecessary to model the rod as an elastic body because a single DOF system suffices. This elementary example emphasizes, if necessary, the importance of modelling mechanical systems in close relation to the nature ofthe information desired. In continuous systems, a major simplification occurs when the study of the local and of the global responses can be split into two distinct prob- lems. This is often the case in solid mechanics and this kind of simplification has been stated as a principle by Barré de Saint-Venant (1885), which is enunciated as follows: The elastic response induced by a local force system, whose resultant force and torque are both zero, become negligible far enough from the small loaded por- tion of the body. In other words, if sufficiently far from the loaded domain, the response depends solely upon the resultant force and torque of the actual loading system. Here ‘local’ means a part which is much smaller than the size of the whole body, or of the boundary in the case of contact forces, and ‘far enough’ means distances substantially greater than the length scales of the loaded part. As detailed further in the next chapters, the Saint-Venant principle may be con- sidered rightly as the cornerstone for modelling solids as structural elements, even if it is not explicitly invoked in most cases. Nevertheless, care has to be taken when using it, as its validity is not universal. In particular, an important exception is encountered in the case of thin shells of revolution loaded by concentrated forces. This can be easily realized by observing the deformation of a flexible pail filled of water, or sand, and held by its handle. Indeed, it can be observed that the reactions at Solid mechanics 61 the handle fasteners induce an ovalization of the pail cross-sections which extends far beyond the vicinity of the fasteners. The colour Plates 1 and 2 illustrate such results which are further discussed in Chapter 8, see also Plates 13 and 14. 1.5.2 Shape criterion to reduce the dimension of a problem In most engineering applications, use is made of structural elements which are characterized by a few generic particularities in their geometry. Such peculiarities can be advantageously used to simplify their mathematical modelling. In partic- ular, most structural elements can be characterized by the fact that one or two dimension(s) is/are much smaller than the other(s). Then, the corresponding strain components can be neglected and finally the dimensions connected to the small scales can be suppressed, as detailed in the rest of the book for several types of structures. A few simple ‘order of magnitude’ calculations can be used to support this simplifying statement. 1.5.2.1 Compression of a solid body shaped as a slender parallelepiped An elastic bar shaped as a slender parallelepiped is constrained on the face x = 0 and loaded on the opposite face x = L x by a compressive force  T ( e ) = T 0  i parallel to 0x, see Figure 1.24. The three-dimensional problem is not simple if the local response is needed. Indeed, it depends upon the stress distribution on the loaded face, and also on the boundary conditions in the vicinity of the opposite face. As already indicated in the example of the rod, the same problem is drastically simplified when restricted to the global response. According to Saint- Venant’s principle, the global solution may be expected to be valid in the largest portion of the bar, if L x is much larger than L y and L z . The actual loading exerted on the face x = L x is described by the force density t (e) x (y, z). However, for determining the global response, an equivalent load can be defined as the resultant T 0 =  (S) t (e) x (y, z) dy dz applied to the centre of the free end of the bar. For mathematical convenience, the face x = 0 is provided with a sliding support condition, i.e. (X(0, y, z) = 0). The other faces are free. With these hypotheses, the Figure 1.24. Bar shaped as a slender parallelepiped under compressive load 62 Structural elements equilibrium is obtained from an axial state of stress whose nonzero component is: σ xx = T 0 L y L z [1.144] Of course, this result fully agrees with the rod example of subsection 1.5.1. How- ever, the interesting point here is that by solving the same problem along the Oy and Oz axes instead of Ox, we find much lower stresses which are, σ yy = T 0 L x L z ; σ zz = T 0 L x L y ⇒ σ yy σ xx = L z L x ≪ 1; σ zz σ xx = L y L x ≪ 1 [1.145] On the other hand, as the strains increase with the stresses, so, for any load of same magnitude, the following inequality holds: ε xx ≫ ε yy , ε zz [1.146] Thus, the idea which arises naturally is to neglect the transverse strains ε yy , ε zz which are found to be much smaller than the axial strain. In other words, if the body extends much more in the axial direction than in the transverse direc- tions, it will be modelled as an equivalent body which is completely rigid in the transverse directions. The corresponding stresses needed to obtain the equilibrium of the solid are given by the Lagrange multipliers associated with the constraint conditions: ε zz = ε yy = 0 [1.147] 1.5.2.2 Shearing of a slender parallelepiped What was said just above in the case of a normal loading is also true in the case of a tangential loading. For instance, let us consider the two similar problems sketched in Figure 1.25. In the case (a), a load T 0  i is applied on the face z = L z , the opposite face being fixed in any direction. From the equilibrium conditions it is not difficult to show that normal stresses are identically zero. Hence div  X = 0 and the displacement field is necessarily in the longitudinal direction (Y = Z = 0). The shear stress and strain fields are: σ zx = T 0 L x L y ; ε zx = 1 2 ∂X ∂z [1.148] In the case (b), the load T 0  k is applied on x = L x . Of course, the same reasoning as in case (a) applies. The shear stress and strain fields are: σ xz = T 0 L y L z ; ε xz = 1 2 ∂Z ∂x [1.149] Solid mechanics 63 Figure 1.25. Shear deformation of a slender parallelepiped As the stresses are again increasing functions of the strains, it follows that: (σ zx ) (a) (σ xz ) (b) = L z L x ≪ 1 ⇒ ∂X ∂z ≪ ∂Z ∂x [1.150] From similar calculations related to the other faces of the parallelepiped, it can also be stated that: L x ≫ L y , L z ⇒ ∂X ∂z ; ∂X ∂y ; ∂Y ∂z ; ∂Z ∂y ≪ ∂Z ∂x ; ∂Y ∂x [1.151] A major conclusion arising from such orders of magnitude is that if L x is large enough in comparison with the other two dimensions and if the local effects due to the loads and the boundary conditions are negligible, the gradient of the displace- ment field is also negligible, if taken along the directions L y , L z . As an immediate corollary, the strain components in the planes parallel to Oyz can be discarded. In other words, the cross-sections of the parallelepiped parallel to Oyz can be modelled as rigid bodies. 1.5.2.3 Validity of the simplification for a dynamic loading The relations [1.136] and [1.137] defining the natural frequencies and mode shapes of a cubical solid are convenient to discuss the validity of the previous simplifications in the dynamic domain. Since we assume here that L x ≫ L y , L z , 64 Structural elements the first natural frequencies associated to the plane modes – indexed by the numbers (n,0,0) – are much lower than that of the other modes. Therefore, if the frequency range of the excitation spectrum is much smaller than the lowest frequency (f 1,1,0 or f 1,0,1 ) of the first non-plane mode, then the dynamic response can be obtained with sufficient accuracy by taking into account the plane modes only (cf. for instance [AXI 04], Chapters 7 and 9). In other words, the natural modes of vibration which are related to transversal strains (Oy, Oz directions) can be safely neglected if their frequency is much higher than the frequency range of the excitation. It may be also noted that, according to such an approximation, the frequency spectrum of the guided waves is restricted to the plane wave branches, which are the lowest branches of the complete 3D frequency spectrum. 1.5.2.4 Structural elements in engineering From the previous considerations, it is concluded that to make tractable the analysis of engineering structures, the first simplifying assumption is to replace the real three-dimensional continuous medium by an equivalent continuous medium of smaller dimension. In this way, structural elements can be defined, as cables or beams which are one-dimensional structures and as plates and shells, which are two-dimensional structures, see Figure 1.26. It is also worth emphasizing that if a dimension is changed, all the formulation of the equilibrium equations is deeply modified too, like the boundary conditions and space distribution of the loading. On the other hand, the advantages gained by using such simplified models in structural analysis are very important. The following chapters are devoted to work out such simplified models and to studying their properties both in the static and in the dynamic domains. However, the present book is not exhaustive by far and has been restricted, on purpose, to the structural elements which are the most commonly encountered in engineering. For mathematical convenience, we start by studying Figure 1.26. From solids to structural elements Solid mechanics 65 the straight beams, then the plates and finally, the curved elements such as thearches and the shells. The present order of presentation is appropriate to take into account progressively the increasing mathematical difficulties encountered in modelling such structures. However, from a logical point of view, an exactly inverse order of presentation should be more concise, as, starting from a 3D solid body, it is possible to deduce all the other models as particular cases, as indicated in Figure 1.26. Chapter 2 Straight beam models: Newtonian approach Beams are slender structural elements which are employed to support transverse as well as axial loads. They are of very common use as columns, masts, lintels, joists etc, or as constitutive parts of supporting frames of buildings, cars, airplanes etc. They may be considered as the simplest structural element because of the relative simplicity of the equilibrium equations arising from a suitable condensation of the space variables into a single one, namely the abscissa along the beam length. However, the approximations used to represent in an appropriate manner the 3D solid as an equivalent 1D solid involve several subtleties, and several degrees of approximation can be proposed to refine the beam model when necessary. The object of this first of four chapters devoted to beams is to introduce the basic ideas of the elementary theory of beams. For the sake of simplicity at least, it is found convenient to adopt here a Newtonian instead of a variational approach. The latter will be adopted in the next chapters to establish more refined models than those presented here. Straight beam models: Newtonian approach 67 2.1. Simplified representation of a 3D continuous medium by an equivalent 1D model 2.1.1 Beam geometry Let the dimensions of a structure be defined by three scale factors denoted L, D 1 , D 2 . If one dimension L, called the length, is much larger than the two others, called transverse dimensions (see Figure 2.1 in which D 1 ≈ D 2 ≈ D) it is said to fall within the class of ‘beams’. The ratio L/D is called the slenderness ratio. As indicated in Chapter 1, subsection 1.5.2, the purpose of the present chapter is to model a beam structure as a one-dimensional equivalent medium. Having this purpose in mind, a first step is to simplify the 3D expression of the strain and displacement fields in a suitable way. 2.1.2 Global and local displacements Let us cut a beam by a nearly transversal plane. The centre of the area of this section – which is supposed homogeneous and isotropic – is called the centroid denoted C. The line of centroids,orcentral axis is defined as the line that passes through the centroids along the beam. The cross-sections are then defined as the beam sections which are perpendicular to the central axis. For the sake of simplicity, Figure 2.1. Geometric representation of a beam [...]... approach 77 2. 1.6 .2 Equilibrium of the moments The balance of moments is calculated with respect to the centre -of- mass of the beam element, located at abscissa x It is written as: x+dL /2 x−dL /2 (S ) r× ρ ∂ 2 ∂ t1 − ∂x ∂t 2 x+dL /2 = x−dL /2 (S ) (r × f (e) dS dx [2. 20] (e) ) dS dx + M (x; t) dL Using the relations [2. 1], [2. 8], and [2. 9], the following one-dimensional equations are obtained: ρJ ∂Mx ∂ 2 ψx... of little interest in most engineering applications 2. 2.1 .2 General solution of the static equilibrium without external loading Here, the equation [2. 22] is reduced to its static and homogeneous form: − ∂ ∂x ∂X ∂x =0 [2. 23] dx +b ES ES [2. 24] The solution is: X(x) = a where a and b are two arbitrary constants If the beam is homogeneous and of constant cross-section, [2. 24] gives: X(x) = ax +b ES [2. 25]... respectively By substituting this elastic stress into the first force equation [2. 18], we arrive at: ρS ∂ 2X ∂ − 2 ∂x ∂t ES ∂X ∂x (e) = Fx (x; t) [2. 22] Equation [2. 22] was first established by Navier 1 824 , see [SOE 93] for a short historical survey of the vibration equations of structural elements As detailed in Chapter 4, equation [2. 22] governs the elastic vibrations in the axial direction, which can be... (Figure 2. 3) 70 Structural elements 2. 1 .3 Local and global strains The components of the local strains are obtained by substituting [2. 2] into the expression [1 .22 ] or [1 .25 ] of the local strain tensor The result is: εxx (x, y, z) = 1 2 ∂ξx ∂ξx + ∂x ∂x = ∂ψy ∂ψz ∂X +z −y ∂x ∂x ∂x εxy (x, y, z) = 1 2 ∂ξy ∂ξx + ∂y ∂x = ∂ψx 1 ∂Y −z −ψz + 2 ∂x ∂x εxz (x, y, z) = 1 2 ∂ξz ∂ξx + ∂z ∂x = ∂ψx 1 ∂Z +y +ψy + 2 ∂x... 2. 2 .2. 5 Intermediate supports An elastic support of stiffness Kz introduces the discontinuity condition: GS ∂Z ∂x x0+ − ∂Z ∂x x0− − Kz Z(x0 ; t) = 0 [2. 40] 2. 2 .3 Torsion mode of deformation 2. 2 .3. 1 Torsion without warping Figure 2. 18 shows a beam element shaped as a circular cylinder, in which the twisting of a few superficial fibres due to torsion are sketched to help visualize the deformation 2. 2 .3 .2. .. ignoring the con(e) tribution of the moment density M (x, t), which will be included later in subsection 2. 2.4 Integration of the local force densities on a beam element of infinitesimal length dL gives: x+dL /2 dx x−dL /2 ρ (S ) ∂ 2 ξ ∂ t1 − ∂x ∂t 2 x+dL /2 dS = dx x−dL /2 (S ) f (e) dS = F (e) (x; t) dL [2. 17] 76 Structural elements Using the relations [2. 1], [2. 6] and [2. 8], the following one-dimensional... implies a = a ′ = −1 /2 and the conditions at the spring provide the linear system: (1 + κξ0 ) −1 −1 1 + κ(1 − ξ0 ) 2 κ b ξ0 ′ = 2 b 2 1 − ξ0 86 Structural elements Then the solution is: 2 1 + κξ0 1 − ξ0 ξ X− 2 = − ; X0 2( 1 + κξ0 (1 − ξ0 )) 2 0 ≤ ξ ≤ ξ0 2 1 + κξ0 1 − ξ0 (ξ − 1) 1 − ξ 2 X+ = + ; X0 2( 1 + κξ0 (1 − ξ0 )) 2 ξ0 ≤ ξ ≤ 1 Limit cases: κ→∞ X− ξ(ξ0 − ξ ) = ; 0 ≤ ξ ≤ ξ0 ; X0 2 X+ (ξ − ξ0 )(1 −... analytical results already established in the case of the longitudinal mode of deformation to the shear case 2. 2 .2. 2 General solution without external loading x Z(x) = a 0 dx +b GS [2. 34 ] a and b are two arbitrary real constants 2. 2 .2. 3 Elastic boundary conditions If the end x = 0 is supported by a spring of stiffness coefficient Kz , the displacement of the beam is constrained by the reaction R = −Kz... 0 [2. 35 ] x=0 ∂Z ∂x =0 [2. 36 ] x=0 2 fixed end: Z(0, t) = 0 [2. 37 ] Qz (0, t) + R = 0 ⇒ GS The two limit cases are: 1 free end: 2. 2 .2. 4 Concentrated loads (e) If a transverse force denoted Fz (t) is applied on the cross-section located at x0 , the discontinuity condition is: Qz (x0+ ; t) − Qz (x0− ; t) = −Fz(e) (t) [2. 38 ] For an elastic beam it is written as: GS ∂Z ∂x x0+ − ∂Z ∂x x0− = −Fz(e) (t) [2. 39 ]... [1. 32 ] still holds and can be expressed either by using a vector or a simplified indicial notation, according to which the first index of the stress term is replaced by ‘x’, because (S) is normal to Ox In terms of local quantities it reduces to: ∂ 2 ∂ t1 = f (e) (x, y, z; t) − ∂x ∂t 2 ∂ 2 ξj ∂σxj (e) ρ 2 − = fj (x, y, z; t); ∂x ∂t ρ [2. 16] (j = 1, 2, 3) 2. 1.6.1 Equilibrium of forces For the sake of . Ozrespectively (Figure 2. 3) . 70 Structural elements 2. 1 .3 Local and global strains The components of the local strains are obtained by substituting [2. 2] into the expression [1 .22 ] or [1 .25 ] of the local. In terms of local quantities it reduces to: ρ ∂ 2  ξ ∂t 2 − ∂  t 1 ∂x =  f (e) (x, y, z; t) ρ ∂ 2 ξ j ∂t 2 − ∂σ xj ∂x = f (e) j (x, y, z; t); (j = 1, 2, 3) [2. 16] 2. 1.6.1 Equilibrium of forces For. strings of musical instruments. Study of the transverse equilibrium of beams subjected to tensile or compressive loads is postponed to Chapter 3, subsection 3 .2. 3 and Chapter 4, subsection 4 .2. 4. 2. 1.5