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  1. 1. Padmakar Raut / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 Vol. 2, Issue 6, November- December 2012, pp.099-103Impact Of Mesh Quality Parameters On Elements Such As Beam, Shell And 3D Solid In Structural Analysis Padmakar Raut (Department of Mechanical Engineering, Mumbai University, Mumbai-400614)Abstract. This study compares the performance of however, is whether brick elements perform betterlinear and quadratic tetrahedral elements and or worse than quadratic tetrahedra, that is,hexahedral elements in various structural tetrahedral elements including mid-side nodes.problems. The problems selected demonstrate Specifically, for a given number of nodes (ordifferent types of behaviour, namely, bending, degrees of freedom), the analyst needs to knowshear, torsional and axial deformations. It was under what circumstances it is better to use bricksobserved that the results obtained with quadratic instead of quadratic tetrahedra. This amounts totetrahedral elements and hexahedral elements investigating the accuracy and efficiency of suchwere equivalent in terms of accuracy. The elements under a variety of problems characterizedcomparison is done for linear static problems, by different deformation patterns, such as, bending,modal analyses and nonlinear analyses involving shear, torsion and axial behaviour. In addition, if alarge deflections, contact and plasticity. The mesh made of linear tetrahedral elements does notadvantages and disadvantages are shown using yield a result within acceptable error, it is useful totetrahedral and hexahedral elements. Some know what strategy to follow: (a) decrease the sizerecommendations and general rules are given for of the elements while keeping them linear, or (b)finite element users in choosing the element make the elements quadratic by introducingshape. additional (mid-side) nodes. Previous authors have proposed some useful benchmark tests forKeywords – Linear and Quadratic Element, individual elements or simple arrays of elements [6-Bending, Shear, Torsional and Axial Deformation 8]. However, no study comparing tetrahedra with hexahedra in a more general setting seems to be1. Introduction available. While it is difficult to give a final answer Finite element analysis has reached a state to all the issues involved, the aim of this study is toof maturity in which 3-D applications are shed some light on this problem by investigating thecommonplace. Most analysts, as well as most performance of tetrahedral and hexahedral elementscommercial codes (MSC/Nastran, etc.), use solid in a number of problems that have known analyticalelements based on the iso-parametric formulation-- solutions. These findings are expected to be usefulor variations of it for 3D analyses [1-4]. For simple for finite elements analysts.geometries, or for applications in which it is Today, the some finite element method ispossible to build a mesh "by hand", analysts have not only applied to mechanical problems by somerelied heavily on the 8-node hexahedral element specialists anymore who know every single finitecommonly known as "brick" or "hexa" [5]. For more element and its function. The finite element methodcomplex geometries, however, the analyst must rely has become a standard numerical method for theon automatic (or semi-automatic) mesh generators. virtual product development and is also applied byIn general, automatic mesh generators produce designers who are not permanent users and have lessmeshes made of tetrahedral elements, rather than detailed understanding of the element functionality.hexahedral elements. The reason is that a general 3- With the rapid development in hardwareD domain cannot always be decomposed into an performance and easy-to-use finite elementassembly of bricks. However, it can always be software, the finite element method is not used onlyrepresented as a collection of tetrahedral elements. for simple problems any more. Today finite elementAs the demand for analyses of more complex models are often so complex that a mapped meshconfigurations has grown, coupled with the with hexahedral shaped elements is often notincreasing popularity of automatic mesh generators, economically feasible. Experience shows that thethe need to understand better the relative merits of most efficient and common way is to perform thetetrahedral and hexahedral elements has become analysis using quadratic tetrahedral elements. As aapparent. It is known, for example, that linear consequence of that, the total number of the degreestetrahedral elements do not perform very well--as of freedom for a complex model increasesexpected because they are constant-strain elements; dramatically. Finite element models containingthus, too many elements are required to achieve a several millions degrees of freedom are regularlysatisfactory accuracy. What remains unclear, solved. Typically iterative equation solvers are used 99 | P a g e
  2. 2. Padmakar Raut / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 Vol. 2, Issue 6, November- December 2012, pp.099-103for solving the linear equations. Figure1shows solution based on a series expansion gives a value oftypical models meshed with tetrahedra and 0.551 for the maximum shear stress [11].Thishexahedra elements. solution allows warping of the cross-section. With modern finite element tools it is notdifficult to represent results as color pictures. 2.4 Axial behaviourHowever, the correctness of the results are actually Consider a short beam clamped at boththe cornerstone of the simulation. The correctness of ends and oriented in the y-direction (see Fig. 4). L =the numerical results crucially depends on the 4, b = 1 and h = 1. In addition, E = 1000, v = 0.0 andelement quality itself. There are no general rules p (mass density per unit of volume) = 1. The naturalwhich can be applied just to decided which element frequency corresponding to the first axial mode isshape should be preferred but there do exist some 3.953 Hz [12]. This problem was chosen because itbasic principles and also certain experiences from involves a non uniform axial displacement field.applications which can be very helpful in avoidingsimulation errors and in judging the validity of the 3. Analysisresults. In this paper we compare some analytic The finite element analyses weresolutions and experimental results with finite performed using Nastran, a general-purpose finiteelement results coming from a mesh of tetrahedra element code for structural analysis [13]. Threeand hexahedra. We also compare the solutions on solid elements were tested:tetrahedra and hexahedra for complex models, (a) C3D4, a 4-node tetrahedral element. Thisperforming linear and nonlinear static and dynamic element was included only for comparison purposes;analyses. its performance was not expected to be good since it is a constant-strain element. One integration point is2. Method used.2.1 Bending (b) C3D10, a second-order 10-node tetrahedral Consider a cantilever beam oriented in the element. In this study, the "intermediate" nodesy-direction and loaded in the z-direction (see Fig. 1). were located exactly halfway between the cornerThe beam has a rectangular cross-section and it nodes. Four integration points are used.deforms under the action of a load per unit of length (c) C3D8, an 8-node isoparametric hexahedralequal to 0.01. The beam dimensions are as follows: element. This is a trilinear element. In this caseL (length)= 8, b (width) = 1 and h (height) --- 1. The "full" Gauss integration was employed in thematerial properties are: E (Youngs modulus) = 1000 stiffness matrix determination. This means that theand u (Poissons ratio) = 0.15. The analytical Gauss scheme used integrates the stiffness matrixexpression for the vertical displacement at the free terms exactly ifend of the beam centre-line, including both bending (i) the material properties are constant throughoutand shear deformations (although bending is the the element anddominant effect in this case), yields a value of (ii) the Jacobian of the mapping from the0.0625 [9]. isoparametric coordinates to the physical coordinates is constant and diagonal throughout the2.2 Shear element. Consider a short shear beam deformingunder a unit distributed load (load per unit of length) Each problem was solved using fouras depicted in Fig. 2. The beam is oriented in the y- different models (four different meshes), describeddirection and loaded in the z-direction. The beam as Follows:dimensions are: L = 1, b = 0.6 and h = 1. Thematerial properties are: E = 1000 and u = 0.15. The Mesh 1. This is a regular mesh made of linearvertical displacement at the free end of the beam tetrahedral elements (C3D4).center-line, considering both bending and sheardeformations (which in this case are dominant) is Mesh 2. This is a regular mesh made of0.00538[10]. quadratic tetrahedra (C3D10) obtained by adding mid-side nodes to Mesh 1. This represents an2.3 Torsion attempt to improve the accuracy of the results Consider a beam with a square cross- obtained with the first mesh.section oriented along the y-axis. The beamdimensions are: L = 16, b = 1 and h = 1. Material Mesh 3. This mesh corresponds to another attemptproperties: E = 1000 and u = 0.15. Displacements in to improve the results obtained withthe x- and z-directions are fixed at one end. At theother end, which is free, a rotation of 0.03 radians is Mesh 1, but in this case decreasing the sizeapplied (this corresponds to a 0.1146 torsional of the linear tetrahedra (C3D4). This meshmoment). The maximum value of the shear stress Obviously has more nodes than the mesh employedoccurs at the mid-points of the cross-section sides. A in the first model, but exactly the same Number of 100 | P a g e
  3. 3. Padmakar Raut / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 Vol. 2, Issue 6, November- December 2012, pp.099-103nodes as Mesh 2. This is to make the second and It is important to know that for bending dominatedthird model comparable in terms of the size (same problems only linear hexahedra elements lead tonumber of degrees of freedom) and therefore good results if extra shape functions or enhancedaddress the issue of what strategy is better if one strain formulations are used. Linear tetrahedronswants to improve the accuracy of the results given tend to be too stiff in bending problems. Byby a mesh of linear tetrahedra (Mesh 1): to increase increasing the number of the elements in depth thethe order of the interpolation (Mesh 2) or to reduce structure is still too stiff. The quadratic mid-sidethe size of the elements (Mesh 3). node tetrahedron element shows the exact analytic solution for pure bending dominated problems even Mesh 4. This is a regular mesh of brick with a coarse mesh with only one element inelements. Again, the number of nodes is the same as depth.It is obvious that using a linear tetrahedronin Mesh 2. This is to compare the performance of element yields unacceptable approximations. Thetwo meshes with the same number of degrees of user should not use it for bending dominatedfreedom, one made with bricks and the other made problems. On the other hand quadratic mid-sidewith quadratic tetrahedra. (Notice that nodal node tetrahedra elements are good for bendingcoordinates in Mesh 2 coincide with those of Mesh dominated problems.4). To illustrate the difference in mapping the stiffness of a structure in a correct manner using4. Results different element types we perform a modal analysis The four problems described before were of a cantilever beam. The first two frequencies andsolved using the four different meshes. The analysis mode shapes are computed. We take the solution ofwas performed on CAE lab, Pune using Nastran. quadratic hexahedra elements as a reference solutionThe models were setup with Hypermesh, a and compare the results with a mesh of quadraticgeometric modeller that has a parametric solid tetrahedra and linear tetrahedra with a coarse and aobject representation and is integrated with an fine mesh respectively.automatic mesh generator and a Nastran pre- A good agreement in modelling theprocessor. Regular meshes were employed in all stiffness of the structure correctly is just obtained ifcases. In each case, the error was computed by quadratic tetrahedra elements are taken. Even thecomparing the result given by the finite element fine mesh of linear tetrahedra elements does notmodel against the analytical solution. result in a good approximation of the solution.Tables 1-4 summarize the results. The nomenclatureis as follows: N is the number of nodes in the mesh Table 2- Result for Shear(including mid-side nodes when quadratic tetrahedra Me E N ∆ ∆y ∆z Vertical %are used); E is the number of elements in the mesh; sh x Displace Err∆x, ∆y, ∆z denote the node spacing in the x-, y- and Ty ment × orz-direction (in meshes made of quadratic tetrahedra, pe 10 -3(R)the spacing is determined by the distance between Me 247 76 0. 0.1 0.1 4.687 12.corner nodes); sh 5 8 3 5 5 8 1Table 1- Result for Bending Me 247 41 0. 0.1 0.1 4.808 9.2Me E N ∆x ∆y ∆z Vertical % sh 5 23 3 5 5sh Displace Err 2Ty ment × or Me 342 41 0. 0.0 0.0 4.739 10 -2 (R) sh 0 23 2 33 33 8Me 57 22 0. 0.3 0. 3.822 38. 3sh 6 5 5 33 5 9 Me 196 41 0. 0.0 0.0 4.756 11.1 sh 20 23 2 33 33 5Me 57 12 0. 0.3 0. 6.210 0.7 4sh 6 25 5 33 52 The analytical solution gives the verticalMe 48 12 0. 0.2 0. 5.334 14. displacement at the end of the beam centre linesh 00 25 25 6 25 7 R= 5.375× 10 -33Me 96 12 0. 0.2 0. 6.264 0.2sh 0 25 25 6 254The analytical solution gives the verticaldisplacement at the end of the beam centre line R= 6.254 × 10 -2 101 | P a g e
  4. 4. Padmakar Raut / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 Vol. 2, Issue 6, November- December 2012, pp.099-103Table 3- Result for Torsion with bricks and quadratic tetrahedra, in terms ofMes E N ∆x ∆y ∆z R % both accuracy, are roughly equivalent. This ish Erro significant because it indicates that analysts whoTyp r rely on automatic mesh generators (which in generale generate meshes made of tetrahedral elements) doMes 384 15 0.5 1.0 0.5 0.385 30 not have a disadvantage compared to those analystsh1 3 0 9 who use bricks. In other words, the trilinear brickMes 384 82 0.5 1.0 0.5 0.579 5.2 element—a long-time favourite of many finiteh2 5 0 9 element practitioners--appears not to have aMes 320 82 0.2 0.5 0.2 0.473 14.1 substantial advantage compared to the quadratich3 0 5 5 5 6 tetrahedron. A second conclusion is concerned withMes 640 82 0.2 0.5 0.2 0.530 3.8 what is the best approach to take if a model made ofh4 5 5 5 0 linear tetrahedra does not give satisfactory resultsThe analytical solution gives maximum shear stress (Mesh 1). These analyses (Mesh 2 versus Mesh 3)on the cross section of the beam suggest that, in general, it seems better to increaseR = 0.5511. the order of the elements rather than refining the mesh with smaller linear elements. Except forTable 4- Result for Axial Deformation Problem 4, in which Mesh 2 and Mesh 3 giveMe E N ∆x ∆y ∆z R % approximately the same result, the quadraticsh Err tetrahedra do better than the linear tetrahedra, for theTyp or same number of nodes. In terms of CPU time, thee advantage of quadratic tetrahedra is more clear-thereMe 153 425 0.2 0.2 0.2 3.8 2.7 is a threefold penalty, in all cases, for using linearsh 1 6 5 5 5 48 tetrahedra. This is because Mesh 3 includes manyMe 153 267 0.2 0.2 0.2 3.8 2.8 more elements than Mesh 2 and consequently thesh 2 6 3 5 5 5 45 CPU time required to generate the stiffness matrix and mass matrices increases, as does the time forMe 107 267 0.1 0.1 0.1 3.7 4.2 solving the resulting linear system of 3 52 3 25 25 25 93Me 179 267 0.1 0.1 0.1 3.7 4.2sh 4 2 3 25 25 25 43 7. The quality of tetrahedra elements in thin-The analytical solution gives the fundamental walled structuresfrequency (Hz) for axial vibrations R = 3.953.. Now we will investigate the quality of quadratic tetrahedral elements when used for5. Tetrahedral and hexahedral element simulating the mechanical behaviour of thin-walled structures. We investigate the stiffness of the platesolution in nonlinearities by performing a modal analysis and compare the Now we will compare the tetrahedra numerical results with the analytic solution for theelements with the hexahedral elements in nonlinear first frequencies. Because of the nature of thin-applications. A nonlinear contact simulation has walled structure (no stiffness normal to plane)been performed first to compare the local stress usually Kirchhof-Love or Reissner-Mindlin basedcoming from a quadratic tetrahedra discretization shell elements are used for the finite elementwith the results from a quadratic hexahedra simulation instead of classical displacement baseddiscretization. The material behaviour is linear. solid elements. The geometric modelling effort to beGeometric nonlinearities have been ignored. able to use finite shell elements might be expensiveThe advantage for hexahedra is, one can achieve the nowadays since for shell applications the usergood stress result, without having very fine mesh. typically needs a mid-surface model. However, most of the CAD models are 3D solid models and the6. Discussion user must work on the solid model to obtain a mid- The main goal of this analysis was to surface model which is usually not an easy task. Forinvestigate the performance of hexahedral elements very complicated 3D solid models it is very difficultversus quadratic tetrahedra under similar conditions. and maybe even impossible to get the mid-surface inThis has been achieved by comparing the results an efficient way. It follows that more and more thin-given by Mesh 2 and Mesh 4. The location of the walled 3D solid models are meshed and calculatednodes is identical in both meshes. Thus, the number using quadratic tetrahedral elements. Caution mustof active degrees of freedom is exactly the same. be taken in using tetrahedral elements for thin-This is necessary to make a meaningful comparison. walled structure since the structural behaviour couldIn addition, the element aspect ratio in both meshes be much too stiff in bending, if the element sizeis equivalent (the ratio between the node spacing in comparing to the thickness is not properly chosen.the x-, y- and z-direction is the same in both This also might result in numerically ill-conditionedmeshes). It can be observed that the results obtained stiffness matrices. 102 | P a g e
  5. 5. Padmakar Raut / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 Vol. 2, Issue 6, November- December 2012, pp.099-103References [1] B.M. IRONS, "Engineering applications of numerical integration in stiffness methods", J. Am. Inst. Aeronaut. Astronaut.4 (ll) pp. 2035-2037, 1966. [2] I.C. TAIG, "Structural analysis by the matrix displacement method", English Electric Aviation, Report SO-17, 1961. [3] J. ERGATOUDIS, B. IRONS and O.C. ZIENKIEWICZ, "Curved isoparametric quadrilateral elements for finite element analysis", Int. J. Solids Struct. 4, pp. 31-42, 1968. [4] R.D. COOK, Concepts and Applications of Finite Element Analysis, Wiley, New York, 2nd edn., Chap. 5, 1981. [5] T.J.R. HUGHES, The Finite Element Method, Prentice-Hall, Englewood Cliffs, N J, Section 3.5, 1987. [6] R.L. TAYLOR, J.C. SIMo, O.C. ZIENKIEW1CZ and A.C. CHAN, "The patch test: A condition for assessing finite element convergence", Int. J. Numer. Methods Eng. 22 (l) pp. 39-62, 1986. [7] J. ROBINSON, "Single element test for aspect ratio sensitivity of solids", Finite Element News, pp. 26-32, February 1986. [8] G.P. BAZELEY, Y.K. CHEUNG, B.M. IRONS and O.C. ZIENKIEWICZ, "Triangular elements in plate bending-- Conforming and non-conforming solutions", Proc. 1st Conf. on Matrix Methods in Structural Mechanics, Wright- Patterson Air Force Base, Ohio, pp. 547- 576. 1968. [9] J.P. DEN HARTOG, Strength of Materials, Dover Publications, New York, 1961. [10] S. TIMOSHENKO and D.H. YOUNG, Elements of Strength of Materials, Van Nostrand, New York, 1968. [11] S. TIMOSHENKO and J.N. GOODIER, Theory of Elasticity, McGraw-Hill, New York, 1970. [12] R.D. Bt. Evms, Formulas for Natural Frequency and Mode Shape, Van Nostrand Reinhold, New York, 1979. 103 | P a g e