工程有限元分析英文课件：工程中的有限元法
工程有限元分析英文课件：Formulation of Isoparametric
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Formulation of Isoparametric Finite Element Matrices
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Formulation of Isoparametric Finite Element Matrices 3.4 Formulation of Isoparametric Finite Element Matrices
for Plane Elasticity (平面弹性问题)
The interpolation of the element coordinates and element displacements using the same interpolation functions, which are defined in a natural coordinate system, is the basis of the isoparametric finite element formulation.
有限元介绍第一部分优秀课件
更精确，因而应用的范围更广泛。
但是，弹性力学也有其固有的弱点。由于研究对 象的变形状态较复杂，处理的方法又较严谨，因而解 算问题时，往往需要冗长的数学运算。但为了简化计 算，便于数学处理，它仍然保留了材料力学中关于材 料性质的假定：
2 有限元的应用范围 2-1 固体力学，包括强度、稳定性、震 动和瞬态问题的分析，线性和非线性 分析
2-2 传热学 2-3 电磁场 2-4 流体力学 2-5 金属成形过程的分析 2-6 焊接残余应力分析 2-7 热处理过程的分析
3 有限元的基本求解原理
3-1 材料力学与弹性力学 3-2 应力的概念 3-3 位移及应变，几何方程，刚体位移 3-4 应力应变关系，物理方程 3-5 虚功原理及虚功方程 3-6 单元刚度矩阵 3-7 整体分析 3-8 整体刚度矩阵的形式 3-9 支承条件的处理 3-10 求解方程
有限元介绍第一部 分
1 有限元的概念
有限元分析（FEA，Finite Element Analysis）是将求解域看成是由许多称为 有限元的小的互连子域组成，对每一单元 假定一个合适的近似解，然后推导求解这 个域总的满足条件，从而得到问题的解。 这个解不是准确解，而是近似解，因为实 际问题被较简单的问题所代替。由于大多 数实际问题难以得到准确解，而有限元不 仅计算精度高，而且能适应各种复杂形状， 因而成为行之有效的工程分析手段。
体所有各点的位移都远小于物体的原有尺寸，因而应变和转角 都远小于1，这样，在考虑物体变形以后的平衡状态时，可以 用变形前的尺寸来代替变形后的尺寸，而不致有显著的误差； 并且，在考虑物体的变形时，应变和转角的平方项或乘积项都 可以略去不计，这就使得弹性力学中的微分方程都成为线性方 程。
工程有限元分析英文课件：Shape Functions for 3D Elements
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有限元分析经典课件
有限元分析经典课件1. 简介有限元分析（Finite Element Analysis, FEA）是一种以数值模拟方法为基础，通过离散化处理求解结构力学问题的工程方法。
本课件将介绍有限元分析的基本原理和常用的应用领域。
2. 有限元分析的基本原理2.1 有限元方法概述有限元方法（Finite Element Method, FEM）是有限元分析的基础理论和计算方法。
本部分将介绍有限元方法的基本概念、基本步骤、离散化处理等内容。
2.2 有限元网格划分有限元网格划分是有限元分析的关键步骤，它将结构离散化为有限个小单元。
本部分将介绍有限元网格划分的方法、常用网格类型以及网格质量评价的方法。
2.3 有限元方程与加载有限元方程是描述结构力学问题的关键方程。
本部分将介绍有限元方程的推导过程，以及加载条件的处理方法。
2.4 有限元解与后处理有限元解是通过有限元分析得到的结构响应结果。
本部分将介绍有限元解的计算方法以及后处理方法，包括位移、应力、应变等结果的计算和可视化展示。
3. 有限元分析的应用案例3.1 结构力学分析结构力学分析是有限元分析的主要应用之一。
本部分将通过实例演示有限元分析在结构力学分析中的具体应用，包括静力学分析、动力学分析等。
3.2 热力学分析热力学分析是有限元分析的另一个重要应用领域。
本部分将通过实例演示有限元分析在热力学分析中的具体应用，包括热传导、热稳定性等问题的分析。
3.3 流体力学分析流体力学分析是有限元分析的扩展应用领域之一。
本部分将通过实例演示有限元分析在流体力学分析中的具体应用，包括流体流动、压力分布等问题的分析。
4. 有限元分析软件的介绍有限元分析软件是进行有限元分析的工具，市场上有多种成熟的有限元分析软件可供选择。
本部分将介绍一些常用的有限元分析软件，包括Ansys、Abacus等。
5. 总结有限元分析作为一种重要的数值模拟方法，已广泛应用于不同领域的工程问题。
本课件从理论原理到实际应用都进行了全面的介绍，相信对有限元分析的学习和应用都有很大帮助。
英文有限元方法Finite element method讲义 (1)
MSc in Mechanical Engineering Design MSc in Structural Engineering LECTURER: Dr. K. DAVEY(P/C10)Week LectureThursday(11.00am)SB/C53LectureFriday(2.00pm)Mill/B19Tut/Example/Seminar/Lecture ClassFriday(3.00pm)Mill/B192nd Sem. Lab.Wed(9am)Friday(11am)GB/B7DeadlineforReports1 DiscreteSystems DiscreteSystems DiscreteSystems2 Discrete Systems. Discrete Systems. Tutorials/Example I.Meshing I.Deadline 3 Discrete Systems Discrete Systems Tutorials/Example IIStart4 Discrete Systems. Discrete Systems. DiscreteSystems.5 Continuous Systems Continuous Systems Tutorials/Example II. Mini Project6 Continuous Systems Continuous Systems Tutorials/Example7 Continuous Systems Continuous Systems Special elements8 Special elements Special elements Tutorials/Example III.Composite IIDeadline *9 Special elements Special elements Tutorials/Example10 Vibration Analysis Vibration Analysis Vibration Analysis III Deadline11 Vibration Analysis Vibration Analysis Tutorials/Example12 VibrationAnalysis Tutorials/Example Tutorials/Example13 Examination Period Examination Period14 Examination Period Examination Period15 Examination Period Examination Period*Week 9 is after the Easter vacation Assignment I submission (Box in GB by 3pm on the next workingday following the lab.) Assignment II and III submissions (Box in GB by 3pm on Wed.)CONTENTS OF LECTURE COURSEPrinciple of virtual work; minimum potential energy.Discrete spring systems, stiffness matrices, properties.Discretisation of a continuous system.Elements, shape functions; integration (Gauss-Legendre).Assembly of element equations and application of boundary conditions.Beams, rods and shafts.Variational calculus; Hamilton’s principleMass matrices (lumped and consistent)Modal shapes and time-steppingLarge deformation and special elements.ASSESSMENT: May examination (70%); Short Lab – Holed Plate (5%); Long Lab – Compositebeam (10%); Mini Project – Notched component (15%).COURSE BOOKSBuchanan, G R (1995), Schaum’s Outline Series: Finite Element Analysis, McGraw-Hill.Hughes, T J R (2000), The Finite Element Method, Dover.Astley, R. J., (1992), Finite Elements in Solids and Structures: An Introduction, Chapman &HallZienkiewicz, O.C. and Morgan, K., (2000), Finite Elements and Approximation, DoverZienkiewicz, O C and Taylor, R L, (2000), The Finite Element Method: Solid Mechanics,Butterworth-Heinemann.IntroductionThe finite element method (FEM) is a numerical technique that can be applied to solve a range of physical problems. The method involves the discretisation of the body (domain) of interest into subregions, which are known as elements. This enables a continuum problem to be described by a finite system of equations. In the field of solid mechanics the FEM is undoubtedly the solver of choice and its use has revolutionised design and analysis approaches. Many commercial FE codes are available for many types of analyses such as stress analysis, fluid flow, electromagnetism, etc. In fact if a physical phenomena can be described by differential or integral equations, then the FE approach can be used. Many universities, research centres and commercial software houses are involved in writing software. The differences between using and creating code are outlined below:(A) To create FE software1. Confirm nature of physical problem: solid mechanics; fluid dynamics; electromagnetic; heat transfer; 1-D, 2-D, 3-D; Linear; non-linear; etc.2. Describe mathematically: governing equations; loading conditions.3. Derive element equations: convert governing equations into algebraic form; select trial functions; prepare integrals for numerical evaluation.4. Assembly and solve: assemble system of equations; application of loads; solution of equations.5. Compute:6. Process output: select type of data; generate related data; display meaningfully and attractively.(B) To use FE software1. Define a specific problem: geometry; physical properties; loads.2. Input data to program: geometry of domain, mesh generation; physical properties; loads-interior and boundary.3. Compute:4. Process output: select type of data; generate related data; display meaningfully and attractively.DISCRETE SYSTEMSSTATICSThe finite element involves the transformation of a continuous system (infinite degrees of freedom) into a discrete system (finite degrees of freedom). It is instructive therefore to examine the behaviour of simple discrete systems and associated variational methods as this provides real insight and understanding into the more complicated systems arising from the finite element method.Work and Strain energyFLuxConsider a metal bar of uniform cross section, A , fixed at one end (unrestrained laterally) and subjected to an axial force, F , at the other.Small deflection theory is assumed to apply unless otherwise stated.The work done, W , by the applied force F is .a ()∫′′=uau d u F WIt is worth mentioning at this early stage that it is not always possible to express work in this manner for various reasons associated with reversibility and irreversibility. (To be discussed later)The work done, W , by the internal forces, denoted strain energy , is se22200se ku 21u L EA 2121EAL d EAL d AL W ==ε=ε′ε′=ε′σ=∫∫εεwhere ε=u L and stiffness k EA L=.The principle of virtual workThe principle of virtual work states that the variation in strain energy is equal to the variation in the work done by applied forces , i.e.()u F u u d u F du d W u ku u ku 21du d ku 21W u0a 22se δ=δ⎟⎟⎠⎞⎜⎜⎝⎛′′=δ=δ=δ⎟⎠⎞⎜⎝⎛=⎟⎠⎞⎜⎝⎛δ=δ∫()0u F ku =δ−⇒Note that use has been made of the relationship δf dfduu =δ where f is an arbitrary functional of u . In general displacement u is a function of position (x say) and it is understood that ()x u δ means a change in ()u x with xfixed. Appreciate that varies with from zero to ()'u F 'u ()u F F = in the above integral.Bearing in mind that δ is an arbitrary variation; then this equation is satisfied if and only if F , which is as expected. Before going on to apply the principle of virtual work to a continuous system it is worth investigating discrete systems further. This is because the finite element formulation involves the transformation of a continuous system into a discrete one. u ku =Spring systemsConsider a single spring with stiffness independent of deflection. Then, 2F21u1F1u2k()()⎟⎟⎠⎞⎜⎜⎝⎛⎥⎦⎤⎢⎣⎡−−=−=2121212se u u k k k k u u 21u u k 21W()()()⎟⎟⎠⎞⎜⎜⎝⎛⎥⎦⎤⎢⎣⎡−−δδ=δ−δ−=δ21211212se u u k k k k u u u u u u k W()⎟⎟⎠⎞⎜⎜⎝⎛δδ=δ+δ=δ21212211a F F u u u F u F W , where ()111u F F = and ()222u F F =.Note here that use has been made of the relationship δ∂∂δ∂∂δf f u u f u u =+1122, where f is an arbitrary functional of and . Observe that in this case is a functional of 1u u 2W se u u u 2121=−, so()()(121212*********se se u u u u k u u ku 21du d u du dW W δ−δ−=−δ⎟⎠⎞⎜⎝⎛=δ=δ).The principle of virtual work provides,()()()()0F u u k u F u u k u 0W W 21221121a se =−−δ+−−−δ⇒=δ−δand since δ and δ are arbitrary we have. u 1u 2F ku ku 11=−2u 2 F ku k 21=−+represented in matrix form,u F K u u k k k k F F 2121=⎟⎟⎠⎞⎜⎜⎝⎛⎥⎦⎤⎢⎣⎡−−=⎟⎟⎠⎞⎜⎜⎝⎛=where K is known as the stiffness matrix . Note that this matrix is singular (det K k k =−=220) andsymmetric (K K T=). The symmetry is a result of the fact that a unit deflection at node 1 results in a force at node 2 which is the same in magnitude at node 1 if node 2 is moved by the same amount.Could also have arrived at equation above via()⎟⎟⎠⎞⎜⎜⎝⎛⎥⎦⎤⎢⎣⎡−−=⎟⎟⎠⎞⎜⎜⎝⎛⇒=⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛⎥⎦⎤⎢⎣⎡−−−⎟⎟⎠⎞⎜⎜⎝⎛δδ=δ−δ2121211121a se u u k k k k F F 0u u k k k k F F u u W WBoundary conditionsWith the finite element method the application of displacement constraint boundary conditions is performed after the equations are assembled. It is an interest to examine the implications of applying and not applying the displacement boundary constraints prior to applying the principle of virtual work. Consider then the single spring element above but fixed at node 1, i.e. 0u 1=. Ignoring the constraint initially gives()212se u u k 21W −=, ()()1212se u u u u k W δ−δ−=δ and 2211a u F u F W δ+δ=δ.The principle of virtual work gives 2211ku ku ku F −=−= and 2212ku ku ku F =+−=, on applicationof the constraint. Note that is the force required at node 1 to prevent the node moving and is the reaction force.21ku F −=21ku F =−Applying the constraint straightaway gives 22se ku 21W =, 22se u ku W δ=δ and 22a u F W δ=δ. The principle of virtual work gives with no information about the reaction force at node 1.22ku F =Exam Standard Question:The spring-mass system depicted in the Figure consists of three massless springs, which are attached to fixed boundaries by means of pin-joints at nodes 1, 3 and 5. The springs are connected to a rigid bar by means of pin-joints at nodes 2 and 4. The rigid bar is free to rotate about pivot A. Nodes 2 and 4 are distances and below pivot A, respectively. Each spring has the same stiffness k. Node 2 is subjected to an external horizontal force F 2/l 4/l 2. All deflections can be assumed to be small.(i) Write expressions for the extension of each spring in terms of the displacement of node 2.(ii) In terms of the degrees of freedom at node 2, write expressions for the total strain energy W of the spring-mass system. In addition, specify the variation in work done se a W δ resulting from the application of the force.2F (iii) Use Use the principle of virtual work to find a relationship between the magnitude of and the horizontal components of displacement at node 2.2F (iv) Use the principle of virtual work to show that the net vertical force imposed by the springs on the rigid-bar at node 2 is zero.Solution:(i) Directional vectors for springs are: 2112e 21e 23e +=, 2132e 21e 23e +−= and 145e e =. Extensions for bottom springs are: 221212u 23u e =⋅=δ, 223232u 23u e −=⋅=δ.Note that 2u u 24=, so 2u245−=δ.(ii)()2222222245232212se ku 87u 212323k 21k 21W =⎟⎟⎠⎞⎜⎜⎝⎛⎟⎠⎞⎜⎝⎛+⎟⎟⎠⎞⎜⎜⎝⎛−+⎟⎟⎠⎞⎜⎜⎝⎛=δ+δ+δ=, 222u F W δ=δ(iii) 2222a 22se ku 47F u F W u ku 47W =⇒δ=δ=δ=δ(iv) Need additional displacement degree of freedom at node 2. Let 22122e v e u u += and note that2221212v 21u 23u e +=⋅=δ and 2223232v 21u 23u e +−=⋅=δ.()⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛+−+⎟⎟⎠⎞⎜⎜⎝⎛+=δ+δ=222222232212se v 21u 23v 21u 23k 21k 21W ⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛δ+δ−⎟⎟⎠⎞⎜⎜⎝⎛+−+⎟⎟⎠⎞⎜⎜⎝⎛δ+δ⎟⎟⎠⎞⎜⎜⎝⎛+=δ22222222se v 21u 23v 21u 23v 21u 23v 21u 23k W Setting and gives0v 2=0u 2=δ2vert 222222se v F v 0v 21u 23v 21u 23k W δ=δ=⎟⎟⎠⎞⎜⎜⎝⎛⎟⎠⎞⎜⎝⎛δ⎟⎟⎠⎞⎜⎜⎝⎛−+⎟⎠⎞⎜⎝⎛δ⎟⎟⎠⎞⎜⎜⎝⎛=δ hence . 0F vert 2=Method of Minimum PotentialConsider the expression,()()F u u u TT 21212121c se K 21F F u u u u k k k k u u 21W W P −=⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛⎥⎦⎤⎢⎣⎡−−=−=where W F and can be considered as a work term with independent of . u F u c =+1122F i u iThe approach of minimising P is known as the method of minimum potential .Note that,()()u F 0F -u u =F u u u +u u K K K K 21W W P T T T T c se =⇒=δδ−δδ=δ−δ=δwhere use has been made of the fact that δδu u =u u T TK K as a result of K 's symmetry.It is useful at this stage to consider the minimisation of an arbitrary functional ()u P where()()3T T O H 21P P u u u u u δ+δδ+∇δ=δand the gradient ∇=P P u i i ∂∂, and the Hessian matrix coefficients H P u u ij i j=∂∂∂2.A stationary point requires that ∇=, i.e.P 0∂∂Pu i=0.Moreover, a minimum point requires that δδu u TH >0 for all δu ≠0 and matrices that possess this property are known as positive definite .Setting P W W K se c T=−=−12u u u F T provides ∇=−=P K u F 0 and H K =.It is a simple matter to check that with u 10= (to prevent rigid body movement) that K is positive definite and this is a property commonly associated with FE stiffness matrices.Exam Standard Question:The spring system depicted in the Figure consists of four massless unstretched springs, which are attached to fixed boundaries by means of pin-joints at nodes 1 to 4. The springs are connected to a slider at node 5. Theslider is constrained to move in a frictionless channel whose axis is to the horizontal. Each spring has the same stiffness k. The slider is subjected to an external force F 0453 whose direction is along the axis of the frictionless channel.(i)The deflection of node 5 can be represented by the vector 25155v u e e u +=, where and areunit orthogonal vectors which are shown in the Figure. Write the components of deflection and in terms of , where is the magnitude of , i.e. e 1e 25u 5v 5U 5U 5u 25U 5u =. Show that the extensions of eachspring, in terms of , are: 5U ()22/31U 515+=δ, ()22/31U 525−=δ, and2/U 54535−=δ−=δ.(ii) In terms of k and write expressions for the total strain energy W of the spring-mass system. Inaddition, specify the variation in work done 5U se a W δ resulting from the application of the force . 5F (iii) Use the principal of virtual work to find a relationship between the magnitude of and thedisplacement at node 5.5F 5U (iv) Use the principal of virtual work to determine an expression for the force imposed by the frictionless channel on the slider.(v)Form a potential energy function for the spring system. Assume here that nodes 1, 3 and 4 are fixed and node 5 is restricted to move in the channel. Use this function to determine the reaction force at node 2.Solution:(i) Directional vectors for springs and channel are: ()2115e e 321e +=, ()2125e e 321e +−=, 135e e −=, 45e e = and (21c 5e e 21e +=). Deflection c 555e U u =, so 2U v u 555==. Extensions springs are: ()3122U u e 551515+=⋅=δ, ()3122U u e 552525−=⋅=δ, 2Uu e 553535−=⋅=δand 2Uu e 554545=⋅=δ(ii)()()()252522245235225215se kU U 83131k 8121k 21W =⎟⎠⎞⎜⎝⎛+−++=δ+δ+δ+δ=, 55a U F W δ=δ(iii)5555a 55se kU 2F U F W U kU 2W =⇒δ=δ=δ=δ(iv) Need additional displacement degree of freedom at node 3. A unit vector perpendicular to the channel is(21p 5e e 21e +−=) and let p 55c 555e V e U u += and note that()()3122V3122U u e 5551515−++=⋅=δ and ()()3122V3122U u e 5552525++−=⋅=δ, 2V 2U u e 5553535+−=⋅=δ and 2V 2U u e 5554545−=⋅=δ()()()()()()()()⎟⎠⎞⎜⎝⎛−+++−+−++=δ+δ+δ+δ=255255255245235225215se V U 831V 31U 31V 31U k 8121k 21W ()()()()()555se V 0V 831313131kU 81W δ=δ−+−+−+=δ, where variation is onlyconsidered and is set to zero. Principle of virtual work .5V δ3V 0F V F V 0W p 55p 55se =⇒δ=δ=δ(v)()3122U u e 551515+=⋅=δ, ()()5552525V 3122Uu u e −−=−⋅=δ, where 2522e V u =. ()()()223333223232233245235225215V F U F U V 3122U 3122U k 21V F U F k 21P −−⎟⎟⎠⎞⎜⎜⎝⎛+⎥⎦⎤⎢⎣⎡−−+⎥⎦⎤⎢⎣⎡+=−−δ+δ+δ+δ=and ()0F V 3122U k V P 2232=−⎥⎦⎤⎢⎣⎡−−−=∂∂, which on setting 0V 2= gives ()⎥⎦⎤⎢⎣⎡−−=3122U k F 32.The reaction is .2F −System AssemblyConsider the following three-spring system 2F 21u 1F 1u 2kF 3F 4u 3u 4k 1k2334()()()234322322121se u u k 21u u k 21u u k 21W −+−+−=,()()()()()()343432323212121se u u u u k u u u u k u u u u k W δ−δ−+δ−δ−+δ−δ−=δ,44332211a u F u F u F u F W δ+δ+δ+δ=δ,and δδ implies that,W W se a −=0u F K u u u u k k 0k k k k 00k k k k 00k k F FF F 43213333222211114321=⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−+−−+−−=⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛=where again it is apparent that K is symmetric but also it is banded, i.e. the non-zero coefficients are located around the principal diagonal. This is a property commonly associated with assembled FE stiffness matrices and depends on node connectivity. Note also that the summation of coefficients in individual rows or columns gives zero. The matrix is singular and 0K det =.Note that element stiffness matrices are: , and where on examination of K it is apparent how these are assembled to form K .⎥⎦⎤⎢⎣⎡−−1111k k k k ⎥⎦⎤⎢⎣⎡−−2222k k k k ⎥⎦⎤⎢⎣⎡−−3333k k k kIf a boundary constraint is imposed then row one is removed to give:0u 1=u F K u u u k k 0k k k k 0k k k F F F 432333322221432=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−+−−+=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛=. If however a boundary constraint (say) is imposed then row one is again removed but a somewhatdifferent answer is obtained: 1u 1=u F K u u u k k 0k k k k 0k k k F F k F 4323333222214312=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−+−−+=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛+=)Direct FormulationIt is possible to formulate the stiffness matrix directly by moving one node and keeping the others fixed and noting the reactions.The above system can be solved for u , once possible rigid body motion is prevented, by setting u (say) to give 10=⇒=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−+−−+=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛=u F K u u u k k 0k k k k 0k k k F F F 432333322221432⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−+−−+=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛−4321333322221432F F F k k 0k k k k 0k k k u u uThe inverse stiffness matrix, K −1, is known as the flexibility matrix and, for this example at least, can be assembled directly by noting the system response to prescribed forces.In practice K −1is never calculated and the system K u F = is solved using a modern numerical linear system solver.It is a simple matter to confirm thatu u K 21u u u u k k 0k k k k 00k k k k 00k k u u u u 21W T 4321333322221111T4321se =⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−+−−+−−⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛= with F u T4321T4321a F F F F u u u u W δ=⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛δδδδ=δThus,()u F F u u K 0K W W Ta se =⇒=−δ=δ−δExample:k1F 2u23k2u3F321With use a direct method to find the assembled stiffness and flexibility matrices.0u 1=Solution:The equations of interest are of the form: 3232222u k u k F += and 3332323u k u k F +=.Consider and equilibrium at nodes 2 and 3. At node 2, 0u 3=()2212u k k F += and at node 3,.223u k F −=Consider and equilibrium at nodes 2 and 3. At node 2, 0u 2=322u k F −= and at node 3, . 323u k F =Thus: , , 2122k k k +=223k k −=232k k −= and 233k k =.For flexibility the equations of interest are of the form: 3232222F c F c u += and . 3332323F c F c u +=Consider and equilibrium at nodes 2 and 3. At node 2, 0F 3=122k F u = and at node 3,1223k F u u ==.Consider and equilibrium at nodes 2 and 3. At node 2, 0F 2=122k F u = and at node 3,()2133k 1k 1F u +=.Thus: 122k 1c =, 123k 1c =, 132k 1c = and 2133k 1k 1c +=.Can check that ⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡+⎥⎦⎤⎢⎣⎡−−+1001k 1k 1k 1k 1k 1k k k k k 2111122221 as required,It should be noted that the direct determination requires boundary constraints to be applied to ensure that the flexibility matrix exists, which requires the stiffness to be non-singular. However, the stiffness matrix always exists, so boundary conditions need not be applied prior to constructing the stiffness matrix with the direct approach.Large deformation theory for spring elementsThus far small deflection theory has been applied where the strains are measured using the Cauchy strainxu11∂∂=ε. A conjugate stress can be obtained by differentiating with respect the expression for strain energy density (energy per unit volume) 11ε211E 21ε=ω, i.e. 111111E ε=ε∂ω∂=σ, where E is Young’s Modulusand is the Cauchy stress (sometimes referred to as the Euler stress). 11σIn the case of large deformation theory we will restrict our attention to hyperelastic materials which are materials that possess an expression for strain energy density Ω (say) that is analytical in strain.The strain used in large deformation theory is Green’s strain (see Appendix II) which for a uniformly loadeduniaxial bar is 211x u 21x u E ⎟⎠⎞⎜⎝⎛∂∂+∂∂=.An expression for strain energy density (energy per unit volume) 211EE 21=Ω and the derived stress is 111111EE E S =∂Ω∂=, where E is Young’s Modulus and is known as the 211S nd Piola-Kirchoff stress . 2F21u1F1u2kBar subject to longitudinal deformationConsider a bar of length L and cross sectional area A represented by a spring element and subject to nodal forces and . 1F 2FThe strain energy is∫∫∫∫⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛∂∂+∂∂==Ω=Ω=212121x x 22x x 211x x V se dx x u 21x u EA 21dx E EA 21dx A dV WConsider further a linear displacement field of the form ()21u L x u L x L x u ⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛−= and note thatL u u xu 12−=∂∂. ()()221212x x 221212se u u L 21u u L EA 21dx L u u 21L u u EA 21W 21⎥⎦⎤⎢⎣⎡−+−=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛−+−=∫ ()()()⎥⎦⎤⎢⎣⎡−+−+−=4122312212se u u L 41u u L 1u u k 21W()()()(12312221212se u u u u L 21u u L 23u u k W δ−δ⎥⎦⎤⎢⎣⎡−+−+−=δ) and 2211a u F u F W δ+δ=δ.The principle of virtual work gives()()()⎥⎦⎤⎢⎣⎡−+−+−−=3122212121u u L 21u u L 23u u k F and()()(⎥⎦⎤⎢⎣⎡−+−+−=3122212122u u L 21u u L 23u u k F ), represented in matrix form as()()()()()⎟⎟⎠⎞⎜⎜⎝⎛⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−+−−−−−−−+−+⎥⎦⎤⎢⎣⎡−−=⎟⎟⎠⎞⎜⎜⎝⎛21121212121221u u L 3u u 1L 3u u 1L 3u u 1L 3u u 1L 2u u k 3k k k k F Fwhich is of the form[]u F G L K K += where is called the geometrical stiffness matrix and is the usual linear stiffnessmatrix. G K L KA common approximation used, depending on the magnitude of L /u u 12−, is⎟⎟⎠⎞⎜⎜⎝⎛⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡−−+⎥⎦⎤⎢⎣⎡−−=⎟⎟⎠⎞⎜⎜⎝⎛2121u u 1111L 2P 3k k k k F F where ()12u u k P −=.The fact that is non-linear (even in its approximate form) means that iterative solution procedures are required to be employed to determine the unknown displacements. G KNote that the approximate form is arrived at using the following strain energy expression()()⎥⎦⎤⎢⎣⎡−+−=312212se u u L 1u u k 21WExample:The strain energies for the springs in the above system (fixed at node 1) are k 1 F 2u 23k 2u3F321⎥⎦⎤⎢⎣⎡+=1322211seL u u k 21W and ()()⎥⎦⎤⎢⎣⎡−+−=323222322se u u L 1u u k 21WUse the principle of virtual work to obtain the assembled linear and geometrical stiffness matrices.()()()3322a 2322322322122212se1sese u F u F W u u u u L 23u u k u L 2u 3u k W W W δ+δ=δ=δ−δ⎥⎦⎤⎢⎣⎡−+−+δ⎥⎦⎤⎢⎣⎡+=δ+δ=δThus ()(⎥⎦⎤⎢⎣⎡−+−−⎥⎦⎤⎢⎣⎡+=2232232122212u u L 23u u k L 2u 3u k F ) and ()()⎥⎦⎤⎢⎣⎡−+−=22322323u u L 23u u k F⎟⎟⎠⎞⎜⎜⎝⎛⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡αα−α−α+α+⎥⎦⎤⎢⎣⎡−−+=⎟⎟⎠⎞⎜⎜⎝⎛32222212222132u u k k k k k F F where 1211L 2u k 3=α and ()23222u u L 2k 3−=α.Note that the element stiffness matrices are[][]111k K α+= and ⎥⎦⎤⎢⎣⎡αα−α−α+⎥⎦⎤⎢⎣⎡−−=222222222k k k k Kand it is evident how these should be assembled to form the assembled linear and geometrical stiffness matrices.2v21u 1v1u 2kxBar subject to longitudinal and lateral deflectionConsider a bar of length L and cross sectional area A represented by a spring element and subject to longitudinal and lateral displacements u and v, respectively.The normal strain is 2211x v 21x u 21x u E ⎟⎠⎞⎜⎝⎛∂∂+⎟⎠⎞⎜⎝⎛∂∂+∂∂= and the associated strain energy∫∫∫∫⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛∂∂+⎟⎠⎞⎜⎝⎛∂∂+∂∂==Ω=Ω=212121x x 22x x 211x x V se dx x v 21x u 21x u EA 21dx E EA 21dx A dV W ∫⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛∂∂⎟⎠⎞⎜⎝⎛∂∂+⎟⎠⎞⎜⎝⎛∂∂+⎟⎠⎞⎜⎝⎛∂∂≈21x x 232se dx x v x u x u x u EA 21WConsider further a linear displacement field of the form ()21u L x u L x L x u ⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛−= and()21v L x v L x L x v ⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛−=, and note thatL u u x u 12−=∂∂ and L v v x v 12−=∂∂. ()()()()⎥⎦⎤⎢⎣⎡−−+−+−=L v v u u L u u u u L EA 21W 21212312212se()()()()()()()1212121221221212se v v L v v u u k u u L 2v v L 2u u 3u u k W δ−δ⎦⎤⎢⎣⎡−−+δ−δ⎥⎦⎤⎢⎣⎡−+−+−=δ2v 22h 21v 11h 1a v F u F v F u F W δ+δ+δ+δ=δ and the principle of virtual work gives()()()⎥⎦⎤⎢⎣⎡−+−+−−=L 2v v L 2u u 3u u k F 21221212h1and ()()⎥⎦⎤⎢⎣⎡−−−=L v v u u k F 1212v1 ()()()⎥⎦⎤⎢⎣⎡−+−+−=L 2v v L 2u u 3u u k F 21221212h2and ()()⎥⎦⎤⎢⎣⎡−−=L v v u u k F 1212v2()()⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−−−+⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−+⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−=⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛22111212v 2h 2v 1h 1v u v u 101005.105.1101005.105.1Lu u k 1010000010100000L2v v k 0000010100000101k F F F FExam Standard Question:The spring system depicted in the Figure consists of two massless springs of equal length , which are attached to fixed boundaries by means of pin-joints at nodes 1 and 2. The springs are connected to a slider atnode 3. The slider is constrained to move in a frictionless channel whose axis is 45 to the horizontal. Each spring has the same stiffness . The slider is subjected to an external force F 1L =0L /EA k =3 whose direction is along the axis of the frictionless channel.FigureAssume the springs have strain density ⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛∂∂⎟⎠⎞⎜⎝⎛∂∂+⎟⎠⎞⎜⎝⎛∂∂+⎟⎠⎞⎜⎝⎛∂∂=Ω232x v x u x u x u E 21.(i) Write expressions for the longitudinal and lateral displacements for each spring at node 3 in terms of thedisplacement along the channel at node 3.(ii) In terms of displacement along the channel at node 3, write expressions for the total strain energy W of thespring-mass system. In addition, specify the variation in work done se a W δ resulting from the application of the force .3F (iii) Use the principle of virtual work to find a relationship between the magnitude of and the displacementalong the channel at node 3. 3FSolution:(i) Directional vectors for springs and channel are: ()2113e e 321e +=and ()2123e e 321e +−= and (21c 3e e 21e +=). Perpendicular vectors are: ()2113e 3e 21e +−=⊥and ()2123e 3e 21e +=⊥Deflection c 333e U u =, so 2U v u 333==.Longitudinal displacement: ()3122U u e 331313+=⋅=δ, ()3122U u e 332323−=⋅=δ.Lateral displacement: ()3122U u e 331313+−=⋅=δ⊥⊥, ()3122U u e 332323+=⋅=δ⊥⊥(ii) The strain energy density for element 1 is ⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛δ⎟⎠⎞⎜⎝⎛δ+⎟⎠⎞⎜⎝⎛δ+⎟⎠⎞⎜⎝⎛δ=Ω⊥21313313213L L L L E 21 The strain energy density for element 2 is ⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛δ⎟⎠⎞⎜⎝⎛δ+⎟⎠⎞⎜⎝⎛δ+⎟⎠⎞⎜⎝⎛δ=Ω⊥22323323223L L L L E 21 The total strain energy with substitution of 1L = gives()()()()[]()()()()[][]3322312232332322321313313213se U Uk 21k 21k 21W α+α=δδ+δ+δ+δδ+δ+δ=⊥⊥where and are constants determined on collecting up terms on substitution of and .1α2α231313,,δδδ⊥⊥δ2333a U F W δ=δ.(iii) The principle of virtual work gives⎥⎦⎤⎢⎣⎡α+α=⇒δ=δ=δ⎥⎦⎤⎢⎣⎡α+α=δ32133332323231se U 23kU F U F W U U 23U k WPin-jointed structuresThe example above is a pin-jointed structure. A reasonable good approximation reported in the literature for strain energy density, commonly used with pin-jointed structures, is⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛∂∂⎟⎠⎞⎜⎝⎛∂∂+⎟⎠⎞⎜⎝⎛∂∂=Ω22x v x u x u E 21This arises from strain-energy approximation 211x v 21x u E ⎟⎠⎞⎜⎝⎛∂∂+∂∂=. Can be used when 22x v x u ⎟⎠⎞⎜⎝⎛∂∂<<⎟⎠⎞⎜⎝⎛∂∂.。
工程有限元分析英文课件：Lagrange polynomials (拉格朗日多项式)
Lagrange Multiplier Function in dimensionless form
Lni ( )
n k 0
( (i
k ) k )
k i
( 0 )( 1) ( i1)( i1) ( n ) (i 0 )(i 1) (i i1)(i i1) (i n )
where i denotes the natural coordinate of node i.
Table 3.2 Lagrange polynomials in dimensionless form
-1
0 1, 1 1
origin
1
n=1
L10 ( )
1
1 1
1 2
(1 )
L11( )
(1)
1 (1)
1 2
(1 )
Lni
(
)
( 0 )( (i 0 )(i
1) 1)
( i1)( i1) ( n ) (i i1)(i i1) (i n )
7
Formulation of Isoparametric Finite Element Matrices
3.3 Family of Isoparametric Elements（等参元族）
A whole family of two – dimensional isoparametric elements can be formulated but the quadratic element （二次单元） is considered to be the best. Elements of an order higher than three are not really practical since it is time consuming to generate stiffness matrices.
工程有限元分析英文课件：Shape Functions for 3D Elements
L1 L2 L3 L4 1
(5.2)
The relation between volume coordinates and Cartesian
coordinates can be given as follows:
1 Li 6V (ai bi x ci y di z) i 1, 2, 3, 4
在右手坐标中，要使得右手螺旋在按照1-2-3的转向转动时是向4的方向前进。3
Volume CAonoardlyisnisatoefsT(Nhraeteu–r体aDl积iCm坐oe标onrs(d自ioin然naa坐lteP标sro))blems
To develop the shape functions for a tetrahedron element, we make
L3
V3 V
VolP412 , Vol1234
L2
V2 V
VolP341 Vol1234
L4
V4 V
VolP123 Vol1234
(5.1)
where Vi - the volume of the tetrahedron formed by the point P
and the nodes other than the node i (i = 1,2,3,4)
8
AVnoalluysmiseoCf Tohorredein–aDteims ensional Problems
4
DefinAitnioanlyosifsVoofluTmhereCeo–oDrdiimnaetnessional Problems
Four internal tetrahedrons having a common apex P (公共锥顶)
《有限元分析及应用》课件
受垂直载荷的托架
31
体单元
•线性单元 / 二次单元 –更高阶的单元模拟曲面的精度就越高。
低阶单元
更高阶单元
32
有限元分析的作用
复杂问题的建模简化与特征等效 软件的操作技巧（单元、网格、算法参数控制） 计算结果的评判 二次开发 工程问题的研究 误差控制
36
第二章 有限元分析的力学基础
(3) 研究的基本技巧
采用微小体积元dxdydz的分析方法（针对任意变
形体）
40
2.2 弹性体的基本假设
为突出所处理的问题的实质，并使问题简单化和抽 象化，在弹性力学中，特提出以下几个基本假定。
物质连续性假定: 物质无空隙，可用连续函数来描述 ；
物质均匀性假定: 物体内各个位置的物质具有相同特 性；
0.02 0.04 0.06 0.08
0.1
0.12
X
0.056
0.058
X
0.06
28
Y
Y
0 -0.02 -0.04 -0.06 -0.08
0
-0.001
-0.002
-0.003 0.054
-0.1 0
0.02 0.04 0.06 0.08
0.1
0.12
X
0.056
0.058
X
0.06
29
30
y
dy zy
1 2
zy
z
dz
0
略去微量项，得 yz zy
MY 0 zx xz
MZ 0
xy yx
剪切力互等定律
53
二维问题: 平衡微分方程
x yx X 0
x y xy y Y 0 x y
剪切力互等定律
有限元法及应用课件
载荷
节点: 空间中的坐标位置，具有 一定相应，相互之间存在物理 作用。 单元: 节点间相互作用的媒介， 用一组节点相互作用的数值矩阵 描述（称为刚度或系数矩阵)。
载荷
有限元模型由一些简单形状的单元组成，单 元之间通过节点连接，并承受一定载荷。
14
对于一个具体的工程结构，单元的划分越小， 求解的结果就越精确，同时，其计算工作量也就越 大。 梯子的有限元模型不到100个方程；
34
3）非线性边界 在加工、密封、撞击等问题中，接触和摩擦 的作用不可忽视，接触边界属于高度非线性边界。 平时遇到的一些接触问题，如齿轮传动、冲 压成型、轧制成型、橡胶减振器、紧配合装配等， 当一个结构与另一个结构或外部边界相接触时通 常要考虑非线性边界条件。 实际的非线性可能同时出现上述两种或三种 非线性问题。
10
2.几个基本概念 1）单元（element） 将求解的工程结构看成是 由许多小的、彼此用点联结的 基本构件如杆、梁、板和壳组 成的，这些基本构件称为单元。 在有限元法中，单元用一 组节点间相互作用的数值和矩 阵（刚度系数矩阵）来描述。
11
单元具有以下特征：
每一个单元都有确定的方程来描述在一定载荷 下的响应； 模型中所有单元响应的“和”给出了设计的总 体响应； 单元中未知量的个数是有限的，因此称为“有
限单元”。
12
2）节点（node） 单元与单元之间的联结点，称为节点。在有 限元法中，节点就是空间中的坐标位置，它具有 物理特性，且存在相互物理作用。 3）有限元模型（node） 有限元模型真实系统理想化的数学抽象。由 一些形状简单的单元组成，单元之间通过节点连 接，并承受一定载荷。 每个单元的特性是通过一些线性方程式来描 述的。作为一个整体，所有单元的组合就形成了 整体结构的数学模型。
《有限元分析概述》课件
PART 05
有限元分析的未来发展与 挑战
新技术与新方法的探索
人工智能与机器学
习
利用人工智能和机器学习技术， 自动构建有限元模型、优化求解 过程和提高分值算法和 求解技术，提高有限元分析的稳 定性和精度。
多物理场耦合
探索多物理场耦合的有限元分析 方法，以解决复杂工程问题中的 多物理场耦合问题。
边界条件的处理
在有限元分析中，边界条件的处理是重要的环节。边界条件通常通过在边界节点上施加约束或加载来实现，以模拟实际系统 的边界条件。
边界条件的处理方式需要根据具体问题进行分析和设定，以确保求解结果的准确性和可靠性。
求解与后处理
求解是有限元分析的核心步骤，涉及到建立方程组、求解方程组并得到离散化模型的结果。常用的求 解方法包括直接法、迭代法和优化算法等。
优化设计
03
根据计算结果，对结构进行优化设计，提高其性能或降低成本
。
PART 04
有限元分析的优缺点
有限元分析的优缺点
• 有限元分析（FEA）是一种数值 分析方法，用于解决各种工程问 题，如结构分析、热传导、流体 动力学等。它通过将复杂的物理 系统离散化为有限数量的简单单 元（或称为“有限元”）来模拟 系统的行为。这些单元通过节点 相互连接，形成一个离散化的模 型，可以用来预测系统的性能和 行为。
2023-2026
ONE
KEEP VIEW
有限元分析概述
REPORTING
CATALOGUE
目 录
• 有限元分析简介 • 有限元分析的基本原理 • 有限元分析的实现过程 • 有限元分析的优缺点 • 有限元分析的未来发展与挑战
PART 01
有限元分析简介
定义与背景
工程有限元分析英文课件：REVIEW
1 2
(1 )(1 2 )
9
Review
Corresponding to corner nodes:
N1
1 4
(1 )(1)
2 3
N5
1 3
N6
1 2
N7
N2
1 4
(1
)(1 )
1 3
N5
2 3
N6
N3
1 4
(1 )(1)
N4
1 4
(1
)(1)
1 2
N7
Thus, it is easy to show that
Such elements whose shape (or geometry) and field variables are
described by the same interpolation functions of the same order are
known as “isoparametric elements”.
N2
Nˆ 2
1 2
N4
L2
1
2L3
N3
Nˆ 3
1 2
N4
L3
1
2L2
and
4
Ni L1 L2 L3 L4 1
i 1
(0,1/2,1/2)
7
Review
Example 1: Construct the shape functions for the element shown in Fig.1
2
Review
For beam, plate and shell elements require that not only the displacements but also the slops must be continuous across the element boundaries.
有限元分析课件
02
1960年， R.W. Clough在他的名为“The finite element in plane stress analysis”的论文中首次提出了有限元（Finite Element）这一术语
03
从固体力学的角度来看，桁架结构与分割成有限个分区后的连续体在结构上存在相似性。
数学家们则发展了微分方程的近似解法，包括有限差分方法，变分原理和加权余量法。 在1963年前后，经过J. F. Besseling, R.J. Melosh, R.E. Jones, R.H. Gallaher, T.H.H. Pian（卞学磺）等许多人的工作，认识到有限单元法就是变分原理中Ritz近似法的一种变形，发展了用各种不同变分原理导出的有限元计算公式。
有限单元法的数学基础（2）
1965年和（张佑启）发现只要能写成变分形式的所有场问题，都可以用与固体力学有限单元法的相同步骤求解。
1969年和指出可以用加权余量法特别是Galerkin法，导出标准的有限元过程来求解非结构问题。
02
01
陈伯屏（结构矩阵方法） 钱令希（余能原理） 钱伟长（广义变分原理） 胡海昌（广义变分原理） 冯康（有限单元法理论） 20世纪60年代初期，冯康等人在大型水坝应力计算的基础上，独立于西方创造了有限元方法并最早奠定其理论基础。--《数学辞海》第四卷
应力
内力
把外载荷集中到节点上 把第i单元和第i+1单元重量的一半，集中到第i+1结点上
01
对于第i+1结点，由力的平衡方程可得：
02
令
建立结点的力平衡方程
根据约束条件，
01
对于第n+1个结点，第n个单元的内力与 第n+1个结点上的外载荷平衡，
工程有限元分析英文课件：Mesh Division (网格划分)
General Procedure of Finite Element Method
In structure analysis, the nodal degrees of freedom called nodal - displacement parameters (节点位移参量), normally refer to the displacements at each node.
(xi , yi ), (x j , y j ) and (xm , ym ) . 10
General Procedure of Finite Element Method
( um, vm )
(ui, vi )
( u, v )
(uj , vj)
For a 2 - D triangular element, there are two
General Procedure of Finite Element Method
Substituting
the
nodal
coordinates
into
u 1 2x 3 y
v
4
5
x
6
y
one
after the other, we have
ui 1 2 xi 3 yi u j 1 2 x j 3 y j
(1)The continuum is divided into two- or three- dimensional (二维 或三维) finite elements, which are separated by straight or curved lines (直线或曲线) (two-dimensional) or by flat or curved surfaces (平面或曲面) (three-dimensional).
英文有限元方法Finite element method讲义 (4)
(ii)
and u 3 are identical (i.e. u 3 (0 ) = u 2 (0 ) and u 3 (0 ) = u 2 (0 ) ), show that u 3 = u 2 . Also show by means of Lagrange’s equations of motion that the response of the system depicted in the Figure is governed by the equation M u + Ku = F , where
U1 U2
1
1 0 U1 0 1 U2
1 1
T
1
1 0 1 0 1 −1
The zero frequency is associated with the translation deformation mode. The zero occurs when the stiffness matrix K is singular, which happens when insufficient boundary conditions are specified to prevent bulk modes of movement.
Solution
2 2 1 1 2 2 m 1u 1 + m2u 2 k1 u1 − u 0 + k 2 u 2 − u1 + k 2 u 3 − u1 2 + m 3 u 3 , Wse = 2 2 2 2 2 1 1 2 2 L = T − Wse = m1 u 1 + m2u 2 k1 u1 − u 0 + k 2 u 2 − u1 + k 2 u 3 − u1 2 + m3u 3 − 2 2
工程中的有限元方法
工程中的有限元方法
有限元方法（Finite Element Method, FEM）是一种常见的工程分析方法，广泛应用于各种工程领域。
下面是其中一些常见的应用。
1. 结构力学分析：有限元方法在工程中最常见的应用之一是结构力学分析。
通过将结构分割成有限个小的单元，并在每个单元内使用简单的数学模型描述其行为，可以对结构进行力学性能的计算和预测。
这种方法可以用于分析各种类型的结构，如桥梁、航空器、建筑物等。
2. 热传导分析：有限元方法还可以应用于热传导问题的数值计算。
通过将热传导区域划分为有限个小的单元，并在每个单元内使用热传导方程进行模拟，可以计算和预测材料内部的温度分布和热流。
这种方法在热交换器设计、电子元器件散热等领域有广泛应用。
3. 流体力学分析：有限元方法也可以用于模拟和分析流体的运动和行为。
通过将流体域划分为有限个小的单元，并在每个单元内使用流体力学方程进行模拟，可以计算流体的速度、压力和流量。
这种方法在流体动力学、气动学和水动力学等领域有广泛应用。
4. 电磁场分析：有限元方法还可以用于模拟和分析电磁场的行为和效应。
通过将电磁场区域划分为有限个小的单元，并在每个单元内使用麦克斯韦方程组进行模拟，可以计算电场、磁场和电流。
这种方法在电力系统、电磁感应和电磁兼容
性等领域有广泛应用。
除了上述应用，有限元方法还可以用于声学和振动分析、优化设计、材料力学分析等各种工程问题的模拟和分析。
它有较强的灵活性和适应性，能够适用于各种复杂的工程情况，并且能够提供较为准确的数值解。
然而，它也需要充分的理论基础和严密的数值计算方法才能获得可靠的结果。
工程有限元分析英文课件：Element Load Vectors
Jacobian matrix is very simple, i.e.,
x y
a 0
J
ab
x y 0 b
4
Formulation of Isoparametric Finite Element Matrices
and
Q3
1 1
1 1
(1 i )(1i )
d d
1 2
2i
1
1
1 2
2i
1
1
4
Therefore, the equivalent nodal forces for gravity loading at a corner
node i can be given as
Piy
W 16
2
Formulation
P e
of
1I1so11p aFr(Ba)mtdetrdic
Finite
and
Element
F (B ) N
Matrices
T Pv J
By virtue of Eqs. (3.77) and (3.78), we have
P e
1 1
1 N
1
T
0 Gy
Q2
1 1
1 (1
1
i
)(1
i
)i
d d
4 3
i2
4 3
since i 1 for all corner nodes.
6
Piy
W 16
1
1
11(F1ormi )u(1latioin)oifI(s1opair)(a1metri )icFi in(1iteEi )l(e1meni )t
工程有限元分析英文课件：Constitutive Relations
(6.143)
or
C
(6.144)
Here,S and C are called the compliance matrix(柔度矩阵)
and stiffness matrix, respectively. It is clear that
C S 1
(6.145)
8
Finite Element Analysis for Plates and Shells
Fig. 6.19 Positive rotation(正向旋转) of principal material
axes(材料主轴) from x – y axes
10
Finite Element Analysis for Plates and Shells
Transformation relations for stress and strain are defined from the tensor theory for second rank tensors(二阶张量). We observe that strains do transform with the same transformation as stresses if the tensor definition of shear strain is used(which is equivalent to dividing the engineering shear strain by two).
0 0
0 0
3 12
13
0
0
0
0
S55
0
13
23 0 0 0 0 0 S66 23
(6.141)
3
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Introduction to Finite Element Method
One of the main reasons for the popularity of the FE method in
different fields of engineering is that once a well known commercial FEM software package（软件包）（such as ABAQUS, CATIA, ANSYS, NASTRAN and so on）is established, it can be
The finite element method was first developed in 1956 for the analysis of aircraft structures. Thereafter, within the past decades, the potentialities of the method for the solution of different types of applied science and engineering problems were recognized.
small, interconnected subregions
called finite elements ( 单 元 ）
which are so small that the shape
Chinese Idiom: Practice makes progress. Review leads to deeper understanding.
How can we achieve “understanding”？
(1)based on knowledge (2)processed by thinking
• Discretization of Problems (问题离散化)
• Formulation of Finite Element Matrices (有限元矩阵公式的建立)
• Structure Analysis (结构分析)
2
Introduction to Finite Element Method
various engineering systems.
7
Introduction to Finite Element Method
Subject Objective
• The purpose of this subject is to present the necessary concepts, fundamental principles and effective techniques of the FEM， so that you can use any age comfortably to solve engineering problems.
1. Homework 2. Final Exam（Open Book）
50% 50%
9
Introduction to Finite Element Method
1.1 Basic Concept of Finite Element Method
(1) A structure is divided into many
UNDERSTANDING !!! 3
Introduction to Finite Element Method What can we learn in class?
What should we learn from engineering application?
4
Introduction to Finite Element Method
《Finite Element Analysis in Engineering》 Cui Deyu Xu Yuanming (北航出版社，2013)
5
Reading is important for you!!
Introduction to Finite Element Method
Brief History of the FEM
Major Reference Books
《Finite Element Procedures in Engineering Analysis》 K. J. Bathe (Butterworth - Heinemann, 1999)
《The Finite Element Method》 O. C. Zienkiewicz (Butterworth - Heinemann, 2000)
• The general philosophy is to introduce the FEM
insightful but simple;
informative but concise;
and theoretical but applicable.
8 8
Introduction to Finite Element Method COURSE ASSESSMENT
used for the solution of any problem simply by changing the input
data. At present the Finite Element Method is developed into a key
technology in the modelling and simulation (建模及仿真) of
Introduction to Finite Element Method
The Finite Element Method (FEM) in Engineering
工程中的有限元法
1
Introduction to Finite Element Method
Key Words
• Virtual Work and Variational Principles (虚功和变分原理)