A finite element method for element
扩展有限元法(XFEM)及其应用12
• 1999年，以美国西北大学Belytschko教授为代表 的研究组首先提出了XFEM的思想[21]，2000年， 他们正式提出了XFEM术语[22]。XFEM是迄今为 止求解不连续力学问题的最有效的数值方法，它 在标准有限元框架内研究问题，不需要对结构内 存在的几何或物理界面进行剖分，保留了CFEM 的所有优点。XFEM与CFEM的最根本区别在于所 使用的网格与结构内部的几何或物理界面无关， 从而克服了在诸如裂纹尖端等高应力和变形集中 区进行高密度网格剖分所带来的困难，当模拟裂 纹扩展时也无需对网格进行重新剖分。XFEM在 处理裂纹问题包括以下三个方面[23]：
• 2. 单位分解法（PUM） • 2.1 单位分解法的基本概念 • 1996年Melenk和Babuska[24]及Duarte和 Oden[25]先后提出了单位分解法（PUM）， 其基本思想是任意函数ψ(x)都可以用域内一 组局部函数NI(x)ψ(x)表示，即 • ，, （1） • 其中，NI(x)为有限单元形状函数, 它形成一 个单位分解。 N ( x) 1 • , (2） • 基于此，可以对有限元形状函数根据需要 进行改进。
• 数值方法，如有限元、边界元、无单元法等，特别是有限 元法（FEM）已被广泛用于处理不连续问题。有限元法具 有其它数值方法无可比拟的优点，如适用于任意几何形状 和边界条件、材料和几何非线性问题、各向异性问题、容 易编程等，是数值分析裂纹问题的主要手段。这方面的工 作很多，无法一一列举。Oritz等[1]及Belytschko等[2]通过 使用多场变分原理，用可以横贯有限单元的“弱”（应变） 间断模拟剪切带。Dvorkin等[3]通过修改虚功原理表达式 考虑了“强”（位移）间断问题；Lotfi和Sheng[4]将HuWashizu变分原理推广至具有内部间断的物体中；通过考 虑软化本构律和界面上的面力－位移关系，Simo及其同 事[5，6]提出了分析强间断问题的统一框架，很多研究者 [7-12]将该法应用到变形局部化分析中。Borja[13]提出了 分析强间断问题的标准Galerkin公式，并证明它与假定改 进应变逼近等价。
有限元法单元英语
有限元法释义finit element method[计] 有限元法;finite element method有限元素法;点击人工翻译，了解更多人工释义实用场景例句全部The finite - element method is the most versatile.有限元法是最有用的方法.辞典例句The rectangular groove guide ( RGG ) is analyzed by finite element method ( FEM ) .本文用有限元法分析了矩形槽波导.互联网According to axi - symmetry of shaft, the semi - analytical finite element method is used.考虑到结构的轴对称性质, 分析时采用了半分析有限元法.互联网FEM FCT is used to solve three dimensional hypersonic inviscid flow.从Euler方程出发,利用流量修正有限元法(FEM?FCT)求解三维无粘流动的高速流场.互联网The result are compared with the finite element method's, and anastomosed. "本文计算结果与有限元法分析结果作了比较, 结果吻合较好.互联网Method: 3 - D finite element modeling was computed and analyzed.方法: 采用三维有限元法建立模型并计算、分析.互联网The cloth draping property is studied by using finite element method.以梁单元为模型,运用有限元法研究织物的悬垂性问题.互联网Therefore, the core question of rigid - viscoplastic finite element method has been solved.从而解决了刚粘塑性有限元法核心问题.互联网A permanent magnetic field for mono - crystal furnace was designed.采用有限元法设计了硅单晶炉用永磁磁体.互联网The edge finite element interpolation function of 1 - forms for prism is derived.就三梭柱单元导出了棱边有限元法的1 -形式线性插值基函数.互联网Methods: Three - dimensional finite element analysiswas adopted.方法: 采用三维有限元法.互联网A mixed method - NES FEM is systematically illustrated.提出了新型等效源法与有限元法的一种新耦合算法.互联网The finite element method was used for analysis of heat stress in chip on board ( COB ).本文采用有限元法分析了板上芯片( COB ) 的热应力分布.互联网Finite element method ( FEM ) occupies an important part in Computer Aided Engineering ( CAE ) methods.有限元法,也称有限单元法或有限元素法,在计算机辅助工程CAE 中占有重要的位置.互联网A finite element model for drawing process of high carbon wire with inner micro - defects was built.通过对高碳盘条内部缺陷的假设,采用有限元法研究了高碳盘条拉拔过程中,工艺参数对裂纹扩展情况的影响.互联网。
1.The finite element method-its____ basis and fundamentals
能源与建筑工程学院
1.The standard discrete system and origins of the finite element method
1.2 The structural element and the structural system in which Ke11 , Ke12 , etc., are submatrices which are again square and of the size l × l , where l is the
number of force and displacement components to be considered at each node. The element properties were
assumed to follow a simple linear relationship. In principle, similar relationships could be established for non-linear materials, but discussion of such problems will be postponed at this stage. In most cases considered in this volume the element matrices Ke will be symmetric.
能源与建筑工程学院
1.The standard discrete system and origins of the finite element method
1.1 Introduction The existence of a unified treatment of ‘standard discrete problems’ leads us to the first definition of the
土木工程专业英语词汇(整理版)
精品文档. 第一部分必须掌握，第二部分尽量掌握第一部分：1 Finite Element Method 有限单元法2 专业英语Specialty English3 水利工程Hydraulic Engineering4 土木工程Civil Engineering5 地下工程Underground Engineering6 岩土工程Geotechnical Engineering7 道路工程Road (Highway) Engineering8 桥梁工程Bridge Engineering9 隧道工程Tunnel Engineering10 工程力学Engineering Mechanics11 交通工程Traffic Engineering12 港口工程Port Engineering13 安全性safety17木结构timber structure18 砌体结构masonry structure19 混凝土结构concrete structure20 钢结构steelstructure21 钢-混凝土复合结构steel and concrete composite structure22 素混凝土plain concrete23 钢筋混凝土reinforced concrete24 钢筋rebar25 预应力混凝土pre-stressed concrete26 静定结构statically determinate structure27 超静定结构statically indeterminate structure28 桁架结构truss structure29 空间网架结构spatial grid structure30 近海工程offshore engineering31 静力学statics32运动学kinematics33 动力学dynamics34 简支梁simply supported beam35 固定支座fixed bearing36弹性力学elasticity37 塑性力学plasticity38 弹塑性力学elaso-plasticity39 断裂力学fracture Mechanics40 土力学soil mechanics41 水力学hydraulics42 流体力学fluid mechanics43 固体力学solid mechanics44 集中力concentrated force45 压力pressure46 静水压力hydrostatic pressure47 均布压力uniform pressure48 体力body force49 重力gravity 50 线荷载line load51 弯矩bending moment52 torque 扭矩53 应力stress54 应变stain55 正应力normal stress56 剪应力shearing stress57 主应力principal stress58 变形deformation59 内力internal force60 偏移量挠度deflection61 settlement 沉降62 屈曲失稳buckle63 轴力axial force64 允许应力allowable stress65 疲劳分析fatigue analysis66 梁beam67 壳shell68 板plate69 桥bridge70 桩pile71 主动土压力active earth pressure72 被动土压力passive earth pressure73 承载力load-bearing capacity74 水位water Height75 位移displacement76 结构力学structural mechanics77 材料力学material mechanics78 经纬仪altometer79 水准仪level80 学科discipline81 子学科sub-discipline82 期刊journal ,periodical83文献literature84 ISSN International Standard Serial Number 国际标准刊号85 ISBN International Standard Book Number 国际标准书号86 卷volume87 期number 88 专著monograph89 会议论文集Proceeding90 学位论文thesis, dissertation91 专利patent92 档案档案室archive93 国际学术会议conference94 导师advisor95 学位论文答辩defense of thesis96 博士研究生doctorate student97 研究生postgraduate98 EI Engineering Index 工程索引99 SCI Science Citation Index 科学引文索引精品文档.100ISTP Index to Science and Technology Proceedings 科学技术会议论文集索引101 题目title102 摘要abstract 103 全文full-text104 参考文献reference105 联络单位、所属单位affiliation 106 主题词Subject 107 关键字keyword108 ASCE American Society of Civil Engineers 美国土木工程师协会109 FHWA Federal Highway Administration 联邦公路总署110 ISO International Standard Organization 111 解析方法analytical method 112 数值方法numerical method 113 计算computation 114 说明书instruction 第二部分：岩土工程专业词汇1.geotechnical engineering 岩土工程2.foundation engineering 基础工程3.soil, earth 土4.soil mechanics 土力学cyclic loading 周期荷载unloading 卸载reloading 再加载viscoelastic foundation 粘弹性地基viscous damping 粘滞阻尼shear modulus 剪切模量5.soil dynamics 土动力学6.stress path 应力路径7.numerical geotechanics 数值岩土力学二. 土的分类 1.residual soil 残积土groundwater level 地下水位 2.groundwater 地下水groundwater table 地下水位3.clay minerals 粘土矿物 4.secondary minerals 次生矿物ndslides 滑坡 6.bore hole columnar section 钻孔柱状图7.engineering geologic investigation 工程地质勘察8.boulder 漂石9.cobble 卵石10.gravel 砂石11.gravelly sand 砾砂12.coarse sand 粗砂13.medium sand 中砂14.fine sand 细砂15.silty sand 粉土16.clayey soil 粘性土17.clay 粘土18.silty clay 粉质粘土19.silt 粉土20.sandy silt 砂质粉土21.clayey silt 粘质粉土22.saturated soil 饱和土23.unsaturated soil 非饱和土24.fill (soil)填土25.overconsolidated soil 超固结土26.normally consolidatedsoil 正常固结土27.underconsolidated soil 欠固结土28.zonal soil 区域性土29.soft clay 软粘土30.expansive (swelling) soil 膨胀土31.peat 泥炭32.loess 黄土33.frozen soil 冻土24.degree of saturation 饱和度25.dry unit weight 干重度26.moist unit weight 湿重度45.ISSMGE=International Society for Soil Mechanics and Geotechnical Engineering 国际土力学与岩土工程学会四. 渗透性和渗流1.Darcy ’ｓlaw 达西定律2.piping 管涌3.flowing soil 流土4.sand boiling 砂沸5.flow net 流网6.seepage 渗透（流）7.leakage 渗流8.seepage pressure 渗透压力9.permeability渗透性10.seepage force 渗透力11.hydraulic gradient 水力梯度12.coefficient ofpermeability 渗透系数五. 地基应力和变形1.soft soil 软土2.(negative) skin friction of driven pile 打入桩（负）摩阻力3.effective stress 有效应力4.total stress 总应力5.field vane shear strength 十字板抗剪强度6.low activity 低活性7.sensitivity 灵敏度8.triaxial test 三轴试验9.foundation design 基础设计10.recompaction 再压缩11.bearing capacity 承载力12.soil mass 土体13.contact stress (pressure)接触应力（压力）14.concentrated load 集中荷载15.a semi-infinite elastic solid 半无限弹性体16.homogeneous 均质17.isotropic 各向同性18.strip footing 条基19.square spread footing 方形独立基础20.underlying soil (stratum ,strata)下卧层（土）21.dead load =sustained load 恒载持续荷载22.live load 活载23.short –term transient load 短期瞬时荷载24.long-term transient load 长期荷载25.reduced load 折算荷载26.settlement 沉降27.deformation 变形28.casing 套管29.dike=dyke 堤（防）30.clay fraction 粘粒粒组31.physical properties物理性质32.subgrade 路基33.well-graded soil 级配良好土34.poorly-graded soil 级配不良土35.normal stresses 正应力36.shear stresses 剪应力37.principal plane 主平面38.major (intermediate, minor) principal stress 最大（中、最小）主应力39.Mohr-Coulomb failure condition 摩尔-库仑破坏条件40.FEM=finite element method 有限元法41.limit equilibrium method 极限平衡法42.pore water pressure 孔隙水压力43.preconsolidation pressure 先期固结压力44.modulus of compressibility 压缩模量45.coefficent of compressibility 压缩系数pression index 压缩指数47.swelling index 回弹指数48.geostatic stress 自重应力49.additional stress 附加应力50.total stress 总应力51.final settlement 最终沉降52.slip line 滑动线六. 基坑开挖与降水1 excavation 开挖（挖方）2dewatering （基坑）降水 3 failure of foundation 基坑失稳4 bracing of foundation pit 基坑围护5 bottom heave=basal heave （基坑）底隆起6 retaining wall 挡土墙7 pore-pressure distribution 孔压分精品文档.布8 dewatering method 降低地下水位法9 well pointsystem 井点系统（轻型）10 deep well point 深井点11vacuum well point 真空井点12 braced cuts 支撑围护13 braced excavation 支撑开挖14 braced sheeting 支撑挡板七. 深基础--deep foundation 1.pile foundation 桩基础1)cast –in-place 灌注桩diving casting cast-in-place pile 沉管灌注桩bored pile 钻孔桩special-shaped cast-in-place pile 机控异型灌注桩piles set into rock 嵌岩灌注桩rammed bulb pile 夯扩桩2)belled pier foundation 钻孔墩基础drilled-pier foundation 钻孔扩底墩under-reamed bored pier 3)precast concrete pile 预制混凝土桩4)steel pile 钢桩steel pipe pile 钢管桩steel sheet pile 钢板桩5)prestressed concrete pile 预应力混凝土桩prestressedconcrete pipe pile 预应力混凝土管桩 2.caisson foundation 沉井（箱）3.diaphragm wall 地下连续墙截水墙 4.friction pile 摩擦桩 5.end-bearing pile 端承桩 6.shaft 竖井；桩身7.wave equation analysis 波动方程分析8.pile caps 承台（桩帽）9.bearing capacity of single pile 单桩承载力teral pile load test 单桩横向载荷试验11.ultimate lateral resistance of single pile 单桩横向极限承载力12.static load test of pile 单桩竖向静荷载试验13.vertical allowable load capacity 单桩竖向容许承载力14.low pile cap 低桩承台15.high-rise pile cap 高桩承台16.vertical ultimate uplift resistance of single pile 单桩抗拔极限承载力17.silent piling 静力压桩18.uplift pile 抗拔桩19.anti-slide pile 抗滑桩20.pile groups 群桩21.efficiency factor of pile groups 群桩效率系数（η）22.efficiency of pile groups 群桩效应23.dynamic piletesting 桩基动测技术24.final set 最后贯入度25.dynamic load test of pile桩动荷载试验26.pile integrity test 桩的完整性试验27.pile head=butt 桩头28.pile tip=pile point=pile toe 桩端（头）29.pile spacing 桩距30.pile plan 桩位布置图31.arrangement of piles =pile layout桩的布置32.group action 群桩作用33.end bearing=tip resistance 桩端阻34.skin(side) friction=shaft resistance 桩侧阻35.pile cushion 桩垫36.pile driving(by vibration) （振动）打桩37.pile pulling test 拔桩试验38.pile shoe 桩靴39.pile noise 打桩噪音40.pile rig 打桩机九. 固结consolidation1.Terzzaghi ’ｓ consolidation theory 太沙基固结理论2.Barraon ’ｓ consolidation theory 巴隆固结理论3.Biot ’ｓ consolidation theory 比奥固结理论4.over consolidation ration (OCR)超固结比5.overconsolidation soil 超固结土6.excess pore water pressure 超孔压力7.multi-dimensional consolidation 多维固结8.one-dimensional consolidation 一维固结9.primary consolidation 主固结10.secondary consolidation 次固结11.degree of consolidation 固结度12.consolidation test 固结试验13.consolidation curve 固结曲线14.time factor Tv 时间因子15.coefficient of consolidation 固结系数16.preconsolidation pressure 前期固结压力17.principle of effective stress 有效应力原理18.consolidation under K0 condition K0固结十. 抗剪强度shear strength 1.undrained shear strength 不排水抗剪强度2.residual strength 残余强度3.long-term strength 长期强度4.peak strength 峰值强度5.shear strain rate 剪切应变速率6.dilatation 剪胀7.effective stress approach of shear strength 剪胀抗剪强度有效应力法8.total stress approach of shear strength 抗剪强度总应力法9.Mohr-Coulomb theory 莫尔－库仑理论10.angle of internal friction 内摩擦角11.cohesion 粘聚力12.failure criterion 破坏准则13.vane strength 十字板抗剪强度14.unconfined compression 无侧限抗压强度15.effective stress failure envelop 有效应力破坏包线16.effective stress strength parameter 有效应力强度参数十一. 本构模型--constitutive model 1.elastic model 弹性模型 2.nonlinear elastic model 非线性弹性模型3.elastoplastic model 弹塑性模型4.viscoelastic model 粘弹性模型5.boundary surface model 边界面模型6.Duncan-Chang model 邓肯－张模型7.rigid plastic model 刚塑性模型8.cap model 盖帽模型9.work softening 加工软化10.work hardening 加工硬化11.Cambridge model 剑桥模型12.ideal elastoplastic model 理想弹塑性模型13.Mohr-Coulomb yield criterion 莫尔－库仑屈服准则14.yield surface 屈服面15.elastic half-space foundation model 弹性半空间地基模型16.elastic modulus 弹性模量17.Winkler foundation model 文克尔地基模型十二. 地基承载力--bearing capacity of foundation soil 1.punching shear failure 冲剪破坏 2.general shear failure 整体剪切破化 3.local shear failure 局部剪切破坏 4.state of limit equilibrium 极限平衡状态5.critical edge pressure 临塑荷载6.stability of foundation soil 地基稳定性7.ultimate bearing capacity of foundation soil 地基极限承载力8.allowable bearing capacity of foundation soil 地基容许承载力十三. 土压力--earth pressure精品文档.1.active earth pressure 主动土压力2.passive earth pressure 被动土压力3.earth pressure at rest 静止土压力4.Coulomb ’ｓearth pressure theory 库仑土压力理论5.Rankine ’ｓ earth pressure theory 朗金土压力理论十四. 土坡稳定分析--slope stability analysis 1.angle of repose 休止角2.Bishop method 毕肖普法3.safety factor of slope 边坡稳定安全系数4.Fellenius method of slices 费纽伦斯条分法5.Swedish circle method 瑞典圆弧滑动法6.slices method 条分法十五. 挡土墙--retaining wall1.stability of retaining wall 挡土墙稳定性2.foundation wall 基础墙3.counter retaining wall 扶壁式挡土墙4.cantilever retaining wall 悬臂式挡土墙5.cantilever sheet pile wall 悬臂式板桩墙6.gravity retaining wall 重力式挡土墙7.anchored plate retaining wall 锚定板挡土墙8.anchored sheet pile wall 锚定板板桩墙十六. 板桩结构物--sheet pile structure 1.steel sheet pile 钢板桩2.reinforced concrete sheet pile 钢筋混凝土板桩3.steel piles 钢桩4.wooden sheet pile 木板桩5.timber piles 木桩十七. 浅基础--shallow foundation 1.box foundation 箱型基础 2.mat(raft) foundation 片筏基础 3.strip foundation 条形基础 4.spread footing 扩展基础 pensated foundation 补偿性基础 6.bearing stratum 持力层7.rigid foundation 刚性基础8.flexible foundation 柔性基础9.embedded depth of foundation 基础埋置深度 foundation pressure 基底附加应力11.structure-foundation-soil interaction analysis 上部结构－基础－地基共同作用分析十八. 土的动力性质--dynamic properties of soils1.dynamic strength of soils 动强度2.wave velocity method 波速法3.material damping 材料阻尼4.geometric damping 几何阻尼5.damping ratio 阻尼比6.initial liquefaction 初始液化7.natural period of soil site 地基固有周期8.dynamic shear modulus of soils 动剪切模量9.dynamic ma 二十. 地基基础抗震 1.earthquake engineering 地震工程2.soil dynamics 土动力学 3.duration of earthquake 地震持续时间 4.earthquake response spectrum 地震反应谱5.earthquake intensity 地震烈度 6.earthquake magnitude 震级7.seismic predominant period 地震卓越周期8.maximum acceleration of earthquake 地震最大加速度二十一. 室内土工实验 1.high pressure consolidation test 高压固结试验 2.consolidation under K0 condition K0固结试验 3.falling head permeability 变水头试验4.constant head permeability 常水头渗透试验5.unconsolidated-undrained triaxial test 不固结不排水试验(UU)6.consolidated undrained triaxial test 固结不排水试验(CU)7.consolidated drained triaxial test 固结排水试验(CD)paction test 击实试验9.consolidated quick direct shear test 固结快剪试验10.quick direct shear test 快剪试验11.consolidated drained direct shear test 慢剪试验12.sieve analysis 筛分析13.geotechnical model test 土工模型试验14.centrifugal model test 离心模型试验15.direct shear apparatus 直剪仪16.direct shear test 直剪试验17.direct simple shear test 直接单剪试验18.dynamic triaxial test 三轴试验19.dynamic simple shear 动单剪20.free （resonance ）vibration column test 自(共)振柱试验二十二. 原位测试 1.standard penetration test (SPT)标准贯入试验 2.surface wave test (SWT)表面波试验 3.dynamic penetration test(DPT)动力触探试验4.static cone penetration(SPT) 静力触探试验 5.plate loading test 静力荷载试验teral load test of pile 单桩横向载荷试验7.static load test of pile 单桩竖向荷载试验8.cross-hole test 跨孔试验9.screw plate test 螺旋板载荷试验10.pressuremeter test 旁压试验11.light sounding 轻便触探试验12.deep settlement measurement 深层沉降观测13.vane shear test 十字板剪切试验14.field permeability test 现场渗透试验15.in-situ pore water pressure measurement 原位孔隙水压量测16.in-situ soil test 原位试验。
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 ⎟⎠⎞⎜⎝⎛∂∂<<⎟⎠⎞⎜⎝⎛∂∂.。
边界元与有限元
边界元与有限元边界元法boundary element method定义：将力学中的微分方程的定解问题化为边界积分方程的定解问题，再通过边界的离散化与待定函数的分片插值求解的数值方法。
所属学科：水利科技(一级学科) ；工程力学、工程结构、建筑材料(二级学科) ；工程力学（水利）(三级学科)边界元法（boundary element method）是一种继有限元法之后发展起来的一种新数值方法，与有限元法在连续体域内划分单元的基本思想不同，边界元法是只在定义域的边界上划分单元,用满足控制方程的函数去逼近边界条件。
所以边界元法与有限元相比，具有单元个数少,数据准备简单等优点.但用边界元法解非线性问题时，遇到同非线性项相对应的区域积分，这种积分在奇异点附近有强烈的奇异性，使求解遇到困难。
简介边界元法是在有限元法之后发展起来的一种较精确有效的工程数值分析方法。
又称边界积分方程-边界元法。
它以定义在边界上的边界积分方程为控制方程，通过对边界分元插值离散，化为代数方程组求解。
它与基于偏微分方程的区域解法相比，由于降低了问题的维数，而显著降低了自由度数，边界的离散也比区域的离散方便得多，可用较简单的单元准确地模拟边界形状，最终得到阶数较低的线性代数方程组。
又由于它利用微分算子的解析的基本解作为边界积分方程的核函数，而具有解析与数值相结合的特点，通常具有较高的精度。
特别是对于边界变量变化梯度较大的问题，如应力集中问题，或边界变量出现奇异性的裂纹问题，边界元法被公认为比有限元法更加精确高效。
由于边界元法所利用的微分算子基本解能自动满足无限远处的条件，因而边界元法特别便于处理无限域以及半无限域问题。
边界元法的主要缺点是它的应用范围以存在相应微分算子的基本解为前提，对于非均匀介质等问题难以应用，故其适用范围远不如有限元法广泛，而且通常由它建立的求解代数方程组的系数阵是非对称满阵，对解题规模产生较大限制。
对一般的非线性问题，由于在方程中会出现域内积分项，从而部分抵消了边界元法只要离散边界的优点。
Introduction to Finite Element Method
1. Mathematical Model
(1) odeling
Physical Problems Mathematica l Model Solution
Identify control variables Assumptions (empirical law)
(2) Types of solution Sol. Eq.
(b) total number of element (mesh) 1D: 2D: 3D:
b. Select a shape function 1D line element: u=ax+b c. Define the compatibility and constitutive law
d. Form the element stiffness matrix and equations (a) Direct equilibrium method (b) Work or energy method (c) Method of weight Residuals
Continuous system Time-independent PDE Time-dependent PDE Discrete system Linear algebraic eq. ODE
(2) Discretization Modeling a body by dividing it into an equivalent system of finite elements interconnected at a finite number of points on each element called nodes.
(2) Analysis procedures of linear static structural analysis
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
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A finite element method for the dynamic analysis of automatic transmission ge...
N Zhang; A Crowther; D K Liu; J Jeyakumaran Proceedings of the Institution of Mechanical Engineers; 2003; 217, 6; ProQuest Science Journals pg. 461
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