FINITE ELEMENT METHODS COURSE FILE

Transcription

1 FINITE ELEMENT METHODS COURSE FILE

2 GEETHANJALI COLLEGE OF ENGINEERING AND TECHNOLOGY DEPARTMENT OF Mechanical Engineering Name of the Subject : Finite Element Methods (JNTU CODE ) : A60330 Programme : UG Branch: ME Version No : 0 Year: II Updated on : 06/06/05 Semester: II No. of pages : Classification status (Unrestricted / Restricted ) Distribution List : Prepared by : ) Name : V Rajasekhar ) Sign : ) Sign : 3) Design : Assit. professor 3) Design 4) Date : 08//05 4) Date : Verified by : ) Name : ) Sign : 3) Design : 4) Date : * For Q.C Only.) Name : ) Sign : 3) Design : 4) Date : Approved by : (HOD ) ) Name : ) Sign : 3) Date :

3 . Contents.. Cover page.. contents.3. Syllabus copy.4. Vision of the department.5. Mission of the department.6. PEOs and POs.. Course objectives and outcomes.8. Brief note on the importance of the course.9. Prerequisites.0. Instructional Learning outcomes.0 Course mapping with POs. Class Time Table. Individual Time Table.3 Lecture schedule with methodology being used.4 Detailed Notes.5 Additional Topics.6 University Question Papers. Question Bank.8 Assignment questions.9 Unit wise quiz questions 3.0 Tutorial problems 3. References, Websites and E links 3. Quality measurement sheet 3.3 Student List 3.4 Group wise student list for discussion topics

4 UNIT:I JNTU Syllabus Introduction of FEM for solving field problems. Stress and equilibrium. Boundary conditions. General description, comparison of FEM with other methods. Basic equations of elasticity, Strain displacement relations, Stress strain relations for D and 3D Elastic problems. One dimensional problems: Finite Element modeling coordinates and shape functions. Stiffness equations for axial bar element using potential energy approach. Assembly of global stiffness matrix and load vector. Finite Element equations. Quadratic shape functions. UNIT II Analysis of Truss: Stiffness equations for a truss bar element oriented in D plane Methods of assembly Plane truss element Space truss element Finite element analysis of trusses Analysis of Beams: Element stiffness matrix for two noded, two degree of freedom per node beam element and simple problems. UNIT III D-problems, Finite element modeling of two dimensional stress analysis with CST, and treatment of boundary conditions. Estimation of load vector, Stresses. Finite element modeling of Axi symmetric solids subjected to Axisymmetric loading with triangular elements. Two dimensional four nodded isoparametric elements UNIT IV Steady state Heat transfer Analysis: One dimensional analysis of Slab, fin and two dimensional analysis of thin plate. Analysis of uniform shaft subjected to torsion. UNIT V Dynamic analysis: Formulation of finite model, Element mass matrices, equations of Eigen values and Eigen vectors for a stepped bar, truss. Finite element Formulation of 3D problems in stress analysis, convergence requirements, mesh generation, techniques such as semi automatic and fully Automatic use of software s such as ANSYS,NISA,NASTRAN

5 .3 Vision of the Department: To develop a world class program with excellence in teaching, learning and research that would lead to growth, innovation and recognition.4 Mission of the Department: The mission of the Mechanical Engineering Program is to benefit the society at large by providing technical education to interested and capable students. These technocrats should be able to apply basic and contemporary science, engineering and research skills to identify problems in the industry and academia and be able to develop practical solutions to them.5 PEOs and PO s: The Mechanical Engineering Department is dedicated to graduating mechanical engineers who: Practice mechanical engineering in the general stems of thermal/fluid systems, mechanical systems and design, and materials and manufacturing in industry and government settings. Apply their engineering knowledge, critical thinking and problem solving skills in professional engineering practice or in non-engineering fields, such as law, medicine or business. Continue their intellectual development, through, for example, graduate education or professional development courses. Pursue advanced education, research and development, and other creative efforts in science and technology. Conduct them in a responsible, professional and ethical manner. Participate as leaders in activities that support service to and economic development of the region, state and nation..6 Course objectives and Outcomes: CO) To understand the theory of elasticity including strain/displacement and Hooke s law relationships.

6 CO) To analyze solid mechanics problems using classical methods and energy methods; CO3) To solve torsion problems in bars and thin walled members. CO4) To solve for stresses and deflections of beams under unsymmetrical loading; CO5) To analyze the maximum and minimum principal stresses using analytical and graphical (mohr s circle)methods. CO6) To obtain stresses and deflections of beams on elastic foundations; CO) To Understand the fundamental concepts of stress and strain and the relationship between both through the strain-stress equations in order to solve problems for simple tridimensional elastic solids Calculate and represent the stress diagrams in bars and simple structures. CO8) To apply various failure criteria for general stress states at points. Brief note on the importance of the course: COURSE DESCRIPTION: Finite Element Analysis (FEA) is a tool used for the evaluation of structures and systems, providing an accurate prediction of a component's response subjected to thermal and structural loads. Structural analyses include all types of steady or cyclic loads, mechanical or thermal. Thermal analyses include convection, conduction, and radiation heat transfer, as well as various thermal transients and thermal shocks. FEA is used to analyze complex geometries, whereas very simple ones (for example, a beam) can be analyzed using hand calculations. For a structure subjected to a load condition (thermal, mechanical, vibratory, etc.) its response (deflection, stress, etc.) can be predicted and measured against acceptable defined limits. In the most simplest terms, this is a factor of safety, which is the ratio of the stress in a component, to the allowable stress of the material. If a factor of safety is too small, the possibility of failure becomes unacceptably large; on the other hand, if the factor is unnecessarily large, the result is a uneconomical or nonfunctional design. For the majority of structural and machine applications, factors of safety are specified by design specifications or codes written by committees of experienced engineers, such as the American Institute of Steel Construction (design & construction of structural steel for buildings) and the American Concrete Institute (building codes requirements for reinforced concrete). Pre-requisites.The trainees should have a basic knowledge of mathematics, engineering mechanics and mechanics of solids.. It is assumed that the student has knowledge about basic calculus and differential equations. 3.It is also assumed that the student has some experience with Python (or is willing to learn)

7 Course Objectives of the Finite Element Methods Course Goals To learn basic principles and skills of finite element modeling and analysis. To learn the theory and characteristics of finite elements that represent engineering structures. To learn and apply finite element solutions to problems in mechanical engineering. To develop the knowledge and skills needed to effectively evaluate finite element analyses performed by others. * Student Learning Objectives Upon successful completion of this course, the student should be able to: Describe the Finite Element Analysis (FEA) procedure. Identify the application and characteristics of FEA elements such as bars, beams, planar elements, and common 3-D elements. Develop the stiffness equation for common FEA elements, and assemble element stiffness equations in to a global equation. Identify and apply suitable boundary conditions to a global structural equation, and reduce it to a solvable form. Specify appropriate figure(s)-of-merit for engineering functionality, and optimize using FEA software. Interpret results obtained from FEA software solutions, not only in terms of conclusions but also awareness of limitations. Write a comprehensive project report based on applied finite element analysis, and critically evaluate analysis reports written by peers. * Explain the underlying concepts behind variational methods and weighted residual methods in FEM. * Explain how the finite element method expands beyond the structural domain, for problems involving dynamics, heat transfer, and fluid flow. * Instructional learning out Comes Students will learn advanced topics and techniques in finite element methods and how to implement and apply these techniques to solve nonlinear systems of ordinary and partial differential equations.

8 Mapping on to Programme Educational Objectives and Programme Out Comes: Mapping of course out comes with program outcomes: Course PO PO PO3 PO4 PO5 PO6 PO PO8 PO9 PO0 PO outcomes CO CO CO3 CO4 CO5 CO6 CO CO8 CO9 CO0 Relationship of the course to Programme out comes: a b c d e f g Graduates will demonstrate knowledge of mathematics, science and engineering applications Graduates will demonstrate ability to identify, formulate and solve engineering problems Graduates will demonstrate an ability to analyze, design, develop and execute the programs efficiently and effectively Graduates will demonstrate an ability to design a system, software products and components as per requirements and specifications Graduates will demonstrate an ability to visualize and work on laboratories in multidisciplinary tasks like microprocessors and interfacing, electronic devices and circuits etc. Graduates will demonstrate working in groups and possess project management skills to develop software projects. Graduates will demonstrate knowledge of professional and ethical responsibilities

9 h i j k Graduates will be able to communicate effectively in both verbal and written Graduates will show the understanding of impact of engineering solutions on society and also be aware of contemporary issues like global waste management, global warming technologies etc. Graduates will develop confidence for self education and ability for life long learning. Graduates can participate and succeed in all competitive examinations and interviews. Relationship of the course to the program educational objectives : PEO PEO PEO 3 PEO 4 PEO 5 Our graduates will apply their knowledge and skills to succeed in a computer engineering career and/or obtain an advanced degree. Our graduates will apply basic principles and practices of computing grounded in mathematics and science to successfully complete hardware and/or software related engineering projects to meet customer business objectives and/or productively engage in research. Our graduates will function ethically and responsibly and will remain informed and involved as fully in their profession and in our society. Our graduates will successfully function in multi-disciplinary teams. Our graduates will communicate effectively both orally and in writing. Program Educational Objectives: PEO: Our graduates will apply their knowledge and skills to succeed in a computer engineering career and/or obtain an advanced degree. PEO: Our graduates will apply basic principles and practices of computing grounded in mathematics and science to successfully complete hardware and/or software related engineering projects to meet customer business objectives and/or productively engage in research. PEO3: Our graduates will function ethically and responsibly and will remain informed and involved as fully in their profession and in our society. PEO4: Our graduates will successfully function in multi-disciplinary teams. PEO5: Our graduates will communicate effectively both orally and in writing. Outcomes: a. Graduates will demonstrate knowledge of mathematics, science and engineering applications b. Graduates will demonstrate ability to identify, formulate and solve engineering problems c. Graduates will demonstrate an ability to analyse, design, develop and execute the programs efficiently and effectively

10 d. Graduates will demonstrate an ability to design a system, software products and components as per requirements and specifications e. Graduates will demonstrate an ability to visualize and work on laboratories in multi-disciplinary tasks like microprocessors and interfacing, electronic devices and circuits etc. f. Graduates will demonstrate working in groups and possess project management skills to develop software projects. g. Graduates will demonstrate knowledge of professional and ethical responsibilities h. Graduates will be able to communicate effectively in both verbal and written i. Graduates will show the understanding of impact of engineering solutions on society and also be aware of contemporary issues like global waste management, global warming technologies etc. j. Graduates will develop confidence for self education and ability for life long learning. k. Graduates can participate and succeed in all competitive examinations and interviews. GEETHANJALI COLLEGE OF ENGINEERING AND TECHNOLOGY CHEERYAL (V), KEESARA (M), R.R. Dist DEPARTMENT OF MECHANICAL ENGINEERING Year/Sem/Sec: III B.Tech II-Sem,Sec: B ROOM NO :LH 39 Time Period 3 4 Acad Yr : CLASS INCHARGE: Mr. B Bhasker LUNCH Monday AE ACS /HT LAB HT R&AC Tuesday DMM II ACS /HT LAB HVPE FEM Wednesday AE R&AC FEM HVPE DMM II DMM II* Thursday R&AC HVPE HT HT DMM II AE* Friday FEM DMM II R&AC AE HT* HVPE Saturday FEM* AE CRT CRT

11 GEETHANJALI COLLEGE OF ENGINEERING & TECHNOLOGY CHEERYAL (V), KEESARA (M), R.R. DIST DEPARTMENT OF MECHANICAL ENGINEERING INDIVIDUAL TIME TABLE Name of the faculty: V Rajasekhar Load = Rev: w.e.f.: Time Period LUNCH Monday Tuesday FEM Wednesday Thursday FEM Friday FEM Saturday FEM Teaching/Learning Methodology A mixture of lectures, tutorial exercises, and case studies are used to deliver the various topics. Some of these topics are covered in a problem-based format to enhance learning objectives. Others will be covered through directed study in order to enhance the students ability of learning to learn. Some case studies are used to integrate these topics and thereby demonstrate to students how the various techniques are inter-related and how they can be applied to real problems in an industry. LECTURE SCHEDULE: Total Teaching aids Sl Unit Reg./ No. of Topics to be covered used LCD. No No. Additional Periods OHP.BB I 6 Introduction to FEM Regular OHP,BB Re ma rks

12 Basic concepts, historical background, Regular OHP,BB 3 application of FEM General description, Regular OHP,BB 4 comparison of FEM with other methods Basic Regular OHP,BB equations of elasticity 5 Stress strain relation. 6 Strain displacement relations Regular BB Rayleigh-Ritz method Regular 8 Weighted residual method Regular BB 9 One dimensional problems Regular OHP,BB 0 Stiffness equations for axial bar element using Regular BB potential energy approach. Virtual energy principle, Finite element analysis of uniform bar Regular OHP,BB Finite element analysis of stepped bar and tapered Regular BB bar subjected to mechanical and thermal loads 3 Assembly of global stiffness matrix and load Regular BB vector 4 Quadratic shape functions, properties of stiffness Regular OHP,BB matrix. 5 II 4 Stiffness equations for a truss bar element oriented Regular BB in D plane 6 Finite element analysis of trusses Regular Plane truss element Regular OHP,BB 8 Space truss element Regular OHP,BB 9 Methods of assembly Regular BB 0 problems Regular BB Analysis of beams Regular OHP,BB Hermite shape functions Regular BB 3 Element stiffness matrix Regular 4 Load vector Regular 5 problems Regular OHP,BB 6 III 6 D-problems, CST, stiffness matrix and load Regular vector Iso parametric representation, shape functions, convergence requirements, problems Regular OHP,BB 8 Two dimensional four noded isoparametric Regular elements 9 Numerical integration Regular 30 Finite element modeling of Axisymmetric solids Regular OHP,BB subjected to Axisymmetric loading with triangular elements 3 3-D problems Regular LCD,OHP,BB 3 Tetrahedran element Regular OHP,BB 33 IV 6 Scalar field problems Regular BB 34 -D heat conduction Regular OHP,BB 35 -D fin elements, -D heat conduction, analysis of Regular BB thin plates 36 Composite slabs and problems Regular BB 3 V 5 Dynamic analysis Regular OHP,BB 38 Dynamic equations Regular OHPBB 39 Lumped and consistent matrices, Eigen values and Eigen vectors Regular OHP,BB

13 40 Mode shapes, analysis for bars and beams Regular OHP,BB 4 3D Stress analysis Regular LCD,BB 4 Mesh generation Regular LCD,BB 43 ANSYS and NASTRAN Regular LCD,BB Micro Plan: Sl. No Unit No. Total No. of Periods Date Topic to be covered in One Lecture Reg/ Teaching aids used Additio LCD/OHP/ nal BB Regular OHP,BB I 0 Introduction to FEM, Basic concepts, historical background 0 application of FEM General Regular OHP,BB description, 3 0 comparison of FEM with other Regular OHP,BB methods Basic equations of elasticity Tutorial class- Regular OHP,BB 4 0 Stress strain relation. Regular BB 5 0 Strain displacement relations BB 6 0 Rayleigh-Ritz method Regular OHP,BB 0 Weighted residual method Regular BB 8 0 Tutorial Class- Regular OHP,BB Solving University papers Regular BB Assignment test- Regular BB 9 0 One dimensional problems Addition al OHP,BB 0 0 Stiffness equations for axial bar BB element using potential energy approach. 0 Virtual energy principle, Finite element analysis of uniform bar BB 0 Finite element analysis of stepped bar and tapered bar subjected to mechanical and thermal loads 3 0 Assembly of global stiffness matrix and Regular OHP,BB load vector 4 0 Quadratic shape functions, Regular OHP,BB 5 0 properties of stiffness matrix. Regular BB 6 0 Numerical problems. Regular OHP,BB Tutorial Class-3 Regular OHP,BB Solving University papers Regular BB Assignment test- Regular BB II 0 Stiffness equations for a truss bar Regular OHP,BB element oriented in D plane 8 0 Finite element analysis of trusses Regular OHP,BB Re ma rks 9 0 Plane truss element Regular BB

14 0 0 Space truss element Regular LCD,OHP,BB 0 Methods of assembly Regular OHP,BB 0 Numerical problems Regular BB 3 0 Tutorial Class-5 Addition al OHP,BB Solving University papers Regular OHP,BB Assignment test-3 Regular OHP,BB 4 0 Analysis of beams Regular BB 5 0 Hermite shape functions Regular OHP,BB 6 0 Hermite shape functions Regular OHP,BB 0 Element stiffness matrix Regular BB 8 0 Load vector Regular OHP,BB 9 0 Problems Regular OHP,BB 30 0 Problems 3 0 Tutorial Class-6 Regular BB Solving University papers Regular OHP,BB Assignment test-4 Regular OHP,BB st Mid Examinations 3 III 0 D-problems, CST, Regular BB 33 0 stiffness matrix and load vector Regular OHPBB 34 0 Iso parametric representation, Regular OHP,BB 35 0 convergence requirements, problems Regular OHP,BB 36 0 shape functions, Regular BB 3 0 Problems BB 0 Problems Tutorial Class- Regular OHP,BB Solving University papers Regular OHP,BB Assignment test-5 Regular OHP,BB 38 IV 0 Two dimensional four noded Regular OHP,BB isoparametric elements 39 0 Numerical integration Regular BB 40 0 Finite element modeling of Addition BB Axisymmetric solids subjected to al Axisymmetric loading with triangular elements 4 0 Finite element modeling of Axisymmetric solids subjected to Axisymmetric loading with triangular elements Regular OHP,BB 4 0 Problems, Regular OHP,BB D problems Regular BB D problems Regular OHP,BB 45 0 Problems Regular OHP,BB Tutorial Class-8 Regular BB Solving University papers Regular OHP,BB Assignment test-6 Regular OHP,BB 46 IV 0 Scalar field problems Regular OHP,BB 4 0 -D heat conduction Regular OHP,BB D fin elements Regular BB D heat conduction BB 50 0 analysis of thin plates Regular OHP,BB 5 0 Composite slabs and problems Regular OHP,BB

15 5 0 Problems Regular BB 0 Problems 53 0 Tutorial Class-9 Regular BB Solving University papers Assignment test- Regular BB Addition OHP,BB al 54 V 0 Dynamic analysis 55 0 Dynamic equations 56 0 Lumped and consistent matrices, 5 0 Eigen values and Eigen vectors 58 0 Mode shapes, 59 0 analysis for bars and beams 60 0 Problems 6 0 Tutorial Class-8 Regular OHP,BB 6 0 Solving University papers Regular OHP,BB Assignment test-8 Mid Test-II Regular BB Regular OHP,BB DETAILED NOTES: Constant Strain Triangle For any element, the displacement components u( x, y ) and v( x, y ) are unknown. Following a Rayleigh-Ritz type solution, we assume a solution for each. The simplest assumption that can be made in this case is to assume that the displacement varies linearly over the element. Hence, we assume: where the 's and 's for the triangular element: u( x, y) x y 3 v( x, y) x y 3 are constants. These constants can be related to nodal displacements

16 Assume the corners of the triangle (nodes) are numbered CCW, and have coordinates ( x, y ), etc. as shown. At each node (i=,,3), assume the nodal displacements are given by ( u, v ). i i We can now write 6 "boundary conditions" as follows: For u(x,y): u u( x, y ) x y At node : 3 u u( x, y ) x y At node : 3 u u( x, y ) x y At node 3: For v(x,y): v v( x, y ) x y At node : 3 v v( x, y ) x y At node : 3

17 v v( x, y ) x y At node 3: We can now solve for the constants in terms of nodal displacements. Eqs. (0.) can be written in matrix form as Solution is: where x y u x y u x3 y3 3 u 3 ( a u a u a u ) /( A) 3 3 ( b u b u b u ) /( A) 3 3 ( c u c u c u ) /( A) a x y x y, a x y x y, a x y x y b y y, b y y, b y y c x x, c x x, c x x and (0.6) A x y ( area of triangle) x x y y Substituting (0.5) into (0.) and rearranging, u(x,y) can be written

18 u( x, y) [( a b x c y) u ( a b x c y) u A ( a b x c y) u ] (0.) Note that the a's, b's and c's are constants and depend only upon the nodal coordinates (x,y) of the 3 corner nodes. Defining the coefficients of where u as i N, equation (0.) becomes: i 3 u( x, y) Niui i N i ( x, y) ( a ) i b i x c i y A A similar result is obtained for v(x,y): 3 v( x, y) Nivi i The quantities Ni ( x, y ) are called shape functions. Note that the same shape functions apply for both u( x, y ) and v( x, y ). We can now obtain the strains by substituting displacement functions (0.8) and (0.0) into strain expressions Error! Reference source not found. to obtain:

19 3 3 u Ni bi xx ui ui x i x i A 3 3 v Ni ci yy vi vi y i y i A u v Ni Ni ci bi xy ui vi ui vi y x i y i x i A i A (0.) The last 3 equations for strains can be put into matrix notation as: u v xx b 0 b 0 b3 0 u yy 0 c 0 c 0 c 3 A v xy c b c b c 3 b 3 u3 Or, more compactly as (for any element "e"): (0.) e e e { } [ B ]{ q } v 3 where b 0 b 0 b3 0 e [ B ] 0 c 0 c 0 c 3 A c b c b c 3 b 3

20 and u v { e u q } v u v e Since the terms in [ B ] are constant for an element, the strains { e } 3 3 element; hence the name "constant strain triangle" or CST. We can now evaluate the internal strain energy U. Substituting (0.3) into Error! Reference source not found. gives: are constant within an e e T e T e e e V U { q } [ B ] [ D ][ B ]{ q } dv e T e T e e e = { q } [ B ] [ D ][ B ] dv { q } V (0.6) e The quantity in parentheses can be identified as the element stiffness matrix [ k ] and (0.6) can be written as: e where the element stiffness matrix [ k ] is defined by: e e T e e U { q } [ k ]{ q }

21 e e T e e [ k ] [ B ] [ D ][ B ] dv If the element has a constant thickness t e, then dv=tda. Assuming that E is constant over the element and noting that the terms in B are constants, then V e e e e T e e [ k ] t A [ B ] [ D ][ B ] e Note that the element stiffness matrix [ k ] is a 6x6 matrix, i.e., we have a 6 degree-of-freedom (dof) element. Note that the general form for the strain energy (0.) can be written in index notation also: 6 6 e e T e e { } [ ]{ } e e e ij i j i j U q k q k q q e e Because [ D ] is symmetric, the stiffness matrix [ ] symmetric matrix (always the case). (0.0) k defined by either (0.8) or (0.9) is a The stiffness matrix for the CST defined by (0.9) can be written in sub-matrix notation as: where each of the k k k e [ k ] k k k k k k (6x6) k is a (x) sub-matrix defined by ij

22 [ k ] ij (x) ( b ibjd cic jd33 ) ( bi c jd cib jd33 ) 4 A ( cib jd bic jd33 ) ( cic jd bi bjd33 ) (0.) D are material properties ( ij E, ) defined by Error! Reference source not found. and b and c are geometry parameters (x-y coordinates of nodes) defined by (0.6). where the the i i To define the external potential energy V, we have to define the external load. Suppose we have a uniform traction (pressure) p applied on the element edge defined by nodes and. The external potential then becomes: e L V [ u( s) p cos v( s) psin ] tds Note that 0 p cos is the component of p in the x direction. Displacements u and v on boundary - must be written as functions of position s on the boundary: u ( s) ( s / L) u ( s / L) u v ( s) ( s / L) v ( s / L) v Substituting u(s) and v(s) into V, and integrating over the boundary, gives:

23 cos sin cos V ptl u ptl v ptl u The last result can be written in matrix notation as e { e T } { 6 e e e } i i i V q F F q where e { F } e pt Lcos e pt Lsin e pt Lcos e pt Lsin 0 0

24 The matrix {F} represents the equivalent generalized nodal force vector due to pressure load on boundary -, i.e., we have replaced the pressure p on boundary - by the nodal forces {F} at nodes and. Note that the total force due to p on boundary - is (ptl) and divides equally between nodes and. = Another set of forces exists on the boundary of any element. These are due to surrounding elements that apply forces due to contact with the element in question, i.e., surrounding elements are being deformed and hence they try to deform the element in question and thereby put forces on this element. Additionally, where a node is at a support or "fixed," there will be a reaction force on the element node. Call these reaction forces { S }.

25 S 4 S 3 S 6 S S Si S 5 3 reactions from adjacent elem e e T e The ext. pot. energy due to reactions is V { q } { S } We can determine the equations of equilibrium for the element. Using Error! Reference source not found. and noting that U and V are functions of nodal e i displacements q, i,...,6, we have e e Since q i 0, then 6 e e e e ( U V ) e e i i qi ( U V ) 0 q e e ( U V ) q e i 0 for i,,...,6 Substituting U (0.0) and V [(0.3) and (0.5)] into (0.) gives the equilibrium equation for any element.

26 e Note that [ K ] is (6x6) and { F e } &{ e } e e e e [ k ]{ q } { F } { S } S are (6x) matrices.

27 Equations (0.6)-(0.8) provide the equilibrium equation for a single element. Suppose we look at a collection of elements (i.e., a complete structure). Then the total energy of the structure is given by the sum of internal and potential energy of all the elements ( N ): el U U q k q k q q Nel Nel N 6 6 el e e T e e { } [ ]{ } e e e str ij i j e e e i j (0.9) and str N N N e e T e e T e { } { } { } { } el el el V V q F q S e e e Nel 6 Nel 6 e e e e Fi qi Si qi e i e i (0.30) The principle of minimum potential energy for the structure requires that M ( Ustr Vstr ) str str i i qi ( U V ) 0 q where { } 0, the last equation requires that q i (0.3) q contains the M degrees of freedom for the structure (NOT dof for each element). For ( U V ) str q i str 0 for i,,..., M

28 Substituting U (0.9) and str str V (0.30) into (0.3) gives q Nel N el N el e T e e e T e e T e { q } [ k ]{ q } { q } { F } { q } { S e e e i for i,,..., M (0.33) Problem: The energy terms for each element are in terms of the element dof, but in order to obtain the equations of equilibrium for the structure (above equation), we have to take the partial derivatives with respect to the global structural dof. In order to complete the above, the element degrees of e q. For freedom { q } must be written in terms of the M global structural degrees of freedom { } any element, we can write a transformation between element local and global dof (called the localglobal transformation): e { q } [ T ] { q} (6x) (6 xm ) ( Mx) e The transformation will be nothing more then 's and 0's. As an example, suppose we have the following element and structural node numbering: y x q q 0 q 8 Consider element. Suppose we place element node at global node 6. p y 9 q q 4 q 3 3 q q 0 x q 6 q

29 q e q e 3 q 6 e q 4 e q 5 e q 3 e 6 q q q 4 q q 3 q Element nodes and local dofs Structural nodes and global dofs We see that for element, there is a correspondence between the 6 element local dofs at element nodes, and 3, and the 6 structural global dofs at nodes 6, and. We see that local (element) node corresponds to global node 6, local (element) node corresponds to global node, and local node 3 corresponds to global node. We can write this local to global transformation e e { q } [ T ]{ q} as: global node # [0] [0] [0] [0] [0] [0] [0] [0] [0] [0] 0 e q, 0 q3,4 [0] [0] [0] [0] [0] [0] [0] [0] [0] [0] 0 q 5,6 0 [0] [0] [0] [0] [0] [0] [0] [0] [0] [0] 0 Each [0] is a (x). The above says that for element, local (element) node corresponds to global

30 node 6, i.e., local dofs, correspond to global dofs,; local node corresponds to global node, i.e., local dofs 3,4 correspond to global dofs,, etc. e e Now transform U and V from local to global dof by substituting (0.34) into (0.9) and (0.30) to obtain e e T e e T e T e e T U { q } [ k ]{ q } { q} [ T ] [ k ][ T ]{ q} { q} [ e T e T e T e T e T e g V { q} [ T ] { F } { q} [ T ] { S } { q} { F } { q} (0.35) Now we can define the following element matrices in global dof (instead of local element dof): e e T e e [ Kg ] [ T ] [ k ] [ T ] ( MxM ) ( Mx6) (6x6) (6 xm ) e e T e Fg T F ( Mx) ( Mx6) (6x) { } [ ] { } e e T e g { S } [ T ] { S } To see what an element stiffness and force matrix written in global dof looks like, consider element again. We obtain for g [ ] K and Element Each block is a (x) sub-matrix g { } F : 6 6 k k k k k k k k k F F F 3

31 3 4 5 g [ K ] 6 k k 3 k 3 k 33 k k 3 { } F g k k 3 k Now the internal and external potential energy is given by Nel Nel N el { } [ ]{ } M M e T e e Ustr U q k g q k gijqiq j e e e i j (0.3) Nel Nel Nel e T e T e str g g e e e { } { } { } { } V V q F q S Nel M Nel M e Fgiqi e i e i e Sgiqi (0.38) Now we can substitute (0.3) and (0.38) into (0.3) to obtain:

32 q N N N el el el e e e T e T e { } [ ]{ } { } { } T e q k { } { } g q q Fg q Sg i for i,,..., M (0.39) which gives a system of M equations in terms of the structural displacements: or N N N el e el e el e kg q Fg Sg e e e [ ]{ } { } { } {0} (0.40) [ ] { } { } { } N N N el e el e el e kg q Fg Sg e e e When all the element contributions have been summed, we simply write [ K]{ q} { Q} { S} Note that when the element stiffness and force matrices are written in terms of structural displacements (using local to global transformation), they become additive [see eq. (0.4)]; i.e., to get the structural stiffness matrix [K], we sum the contributions for all elements.

33 Assemblage of Elements A single element by itself is useless. We must determine the equilibrium equations for an assemblage of elements that comprise the entire structures. Consider the following structure (only a few elements are taken to simplify the discussion) with a uniformly pressure p on the right boundary and fixed on the left boundary (assume a constant thickness t). We number the structural nodes from to as shown. We also number the elements from to as shown (in any order). For each global node of the structure, we can specify the (x,y) coordinates: x, y, i=,,,. i i y x p Each node of the structure will have two degrees of freedom (dof). We label these structural (global) degrees of freedom in order as shown to the right. Note that the structural nodal displacements are written without the superscript "e." The nodal displacement vector is written as { q } and is (4x) for this problem. q q 0 q 8 y 9 q q 4 q 3 q 6 q q q 0 x q 8 q q 4 q 3 We note that the left side is fixed (nodes, 5 and 9). Hence, displacement boundary conditions will require that q q q9 q0 q q8 0.

34 Note: we do not have to number the dof consistently and in sequence with the structural nodes. However, this makes the bookkeeping much, much simpler! For each element, we can construct a table called the element connectivity that specifies which structural (global) nodes are connected by an element. Hence, for the problem above, we have the following element connectivity table: Element No. Element Node Element Node Element Note that for the CST, element nodes MUST be given as CCW. Element node can be attached with any global node of the element. Note that if we are careful in numbering the nodes and choosing the element connectivity in a "systematic" manner, there will be a pattern to the element connectivity table (see above). An automatic mesh generator, like the one in FEMAP, tries to follow this pattern. Note that the global node numbers for the structure are somewhat arbitrary, i.e., we could number them in any order. However, it will turn out that there are optimum ways to number nodes (for a given structure and mesh) in order to reduce the bandwidth of the structural stiffness matrix [K] - this saves time solving the equations. For the mesh above, it would be optimum to number downward and left-to-right, as opposed to left-to-right and downward. We'll discuss that later. Likewise, the element numbering is arbitrary, but again there may be optimum approaches. An automatic mesh generator tries to do the numbering in an optimum fashion.

35 Note that for this structure, we have global nodes. There are degrees of freedom (dof) at each node (u and v). Hence, the structure has 4 dof and the structural stiffness matrix [K] will be (4x4). The structural equilibrium equations can be written as: [ K] { q} { Q} { S} (4x4) (4x) (4x) where [K]=structural stiffness matrix, {Q}=structural forces matrix (due to applied tractions and body forces) {S}=structure reaction forces due to boundary conditions Lets see how each element contributes to global matrices. Take element to start with. Note that we can use sub-matrix notation to divide the element matrices as following. Use a superscript of on the k terms to indicate element.

36 k k k 3 F (x) (x) [ k ] k k k3, { F } F (6x6) k3 k3 k33 F3 We now look at element and note that element node numbers,, 3 correspond to global node numbers, 5, (from the drawing of the mesh, or from the element connectivity table). We can indicate this information on the stiffness and force matrices as follows: 5 k k k 3 F (x) (x) [ k ] 5 k k k3, { F } 5 F (6x6) (6x) k3 k3 k33 F3 Hence, we see that element contributes stiffness and forces to global nodes, 5 and. Placing these contributions into the global stiffness matrix gives: Element only K {q} k k 3 k 3 k 33 k q, k q 3 3,4

37 3 4 5 k k 3 q 5,6 q,8 k q 9,0 6 6 k k k k k k k k k F F F 3 q 3,4 ** remember, each block is a (x) sub-matrix Now take element. Element only K {q} q,

38 3 4 5 q 3,4 q 5,6 q,8 q 9, q 3,4 ** remember, each block is a (x) sub-matrix Note that the distributed pressure load p is applied only to the right boundary of elements 0 and. Hence {F} for all elements except 0 and will be zero. For elements 0 and, we will have 0 { F } 0 0 ptl 0 ptl { F } 0 0 ptl 0 ptl where L4 is the length between global nodes 4 and 8, etc. 8

39 If we assemble all element stiffness matrices [k] and forces matrices {F} to the global equilibrium equations, we have the following result: Structural Equations of Equilibrium K {q} X X X X X X X X 3 X X X X X 4 X X X X 5 X X X X X X q, q 3,4 q 5,6 q,8 q 9,0 6 X X X X X X X X X X X X X X 8 X X X X 9 X X X 0 X X X X X X X X X X X X X X q 3,4 X means that one or more elements have contributed here ** remember, each block is a (x) sub-matrix Note that [K] is symmetric; also it is banded (semi-bandwidth=).

40 In the previous page, each X means that one or more elements have contributed to that (x) sub-matrix. For example, we note that node will have stiffness from elements, and 5. Hence, the, position of the global stiffness matrix will be equal to (note: you have to refer to the element connectivity to see which element node for each element corresponds to global node ): [ K ] [ k ] [ k ] [ k ] each sub-matrix is (x) The global node -6 coupling term [ K 6] will have contributions from elements and 5 since only these elements share the boundary between nodes and 6: [ K ] [ k ] [ k ]. Global node 6 will have stiffness contributions from elements, 3, 5, 6,, and 8: [ K ] [ k ] [ k ] [ k ] [ k ] [ k ] [ k ].

41 Question? What happened to the reactions {S} for each element? Why don't they show up in the structural stiffness matrix? Simple. It is equilibrium. Recall that when we make a free-body, in this case take a single finite element as the free-body, we will have equal and opposite reactions where the cut is made though the body. Consider elements and below: S 4 3 S 3 S 6 S 5 S S 3 S 6 S 5 S 4 S 3 S At the boundary between elements and, the reactions are equal and opposite. Hence, we add S S3 0 6 S6 0 them up we have: S, S S4 0, S 5 S5 0, and. Hence, all the reactions between elements sum to zero and do not have to be put into the structural equilibrium equations. OK, but what about the boundary where there are supports? What happens to the reactions there? For example, the cantilever plate example above: They don't disappear and should be included in the structural stiffness matrix. We know that there will be unknown reactions at global nodes, 5 and 9. We could call y R R 0 R 8 R R R 0 x p

42 R, R, R, 9 R, 0 R and 8 these reactions So we have the free body of the structure: R (consistent with global displacements). Structural Equations of Equilibrium with Support Reactions K {q} X X X X X X X X 3 X X X X X 4 X X X X 5 X X X X X X q, q 3,4 q 5,6 q,8 q 9,0 6 X X X X X X X X X X X X X X 8 X X X X 9 X X X 0 X X X X X X X X X X X X X X q 3,4 X means that one or more elements have contributed here ** remember, each block is a (x) sub-matrix OK, now one last step. We have to apply displacement boundary conditions. The structure is fixed at nodes, 5 and 9; thus, q q q9 q0 q q8 0. The easiest way to apply boundary conditions to any system of equations is as follows:. Zero out the row and column on the left side matrix (the [K] matrix) corresponding to each B.C., and zero out the row of the right side (the {Q} matrix) corresponding to each B.C.

43 . Place a on the diagonal of the left side matrix (the [K] matrix) corresponding to each B.C. You will notice that every dof that has a B.C. also corresponds to a dof where a support reaction (R) occurs. Applying B.C. as described above will thus eliminate the reactions from the equilibrium equations. A theoretical reason why we don t have to worry about reactions in structural equations of equilibrium? Because these support reactions R do no work (displacement is zero at support) and hence do not affect equilibrium of the structure!!! Structural Equations of Equilibrium with B.C. Applied K {q} X X , X q q 3,4 3 X X X X X 4 X X X X q 5,6 q,8 q 9,0 6 X X X X X X X X X X X X X 8 X X X X X X X X X X X X X X X X q 3,4 X means that one or more elements have contributed here ** remember, each block is a (x) sub-matrix

44 The structural equations with B.C. may now be solved for the unknown displacements. Note that when we solve the system of equations, the solution will give q q q9 q0 q q8 0, i.e, the st equation simply says () q 0, etc. Element Strains and Stresses Now we are ready to solve for the element strains and stresses. For each element, we can substitute the 6 global displacements corresponding to that element into (0.3): e e e { } [ B ]{ q } e=,,, no. of elements (3x) (3x6) (6x) The stresses for each element can then be obtained by substituting the strains for that element into Error! Reference source not found.: e e e { } [ D ]{ } e=,,, no. of elements (3x) (3x3) (3x) Evaluation of stress results based on stress components in the Cartesian coordinates directions (,,,.) leaves something to be desired. Why? Stresses in these directions xx yy xy etc may not necessarily represent the largest stresses and we need these in order to consider yielding or failure. You already know that you can calculate principal stresses and maximum shear stress using stress transformation equations or Mohr's Circle. Hence, stress results (stress components) are often represented in two additional ways: Principal stresses and maximum shear stress, and von Mises stress. Principal stresses can, as noted above, be obtained by either stress transformation equations or through the use of Mohr's Circle. An alternate approach to define principal stresses is to write:

45 xx p xy xy yx yy p yz zx zy zz p 0 Expansion of the determinant provides a cubic equation that can be solved for the three principal stresses. Comparing principal stresses to a tensile yield stress provides some measure of p evaluation; however, one has to keep in mind that comparing the principal stress (obtained from a three-dimensional stress state) to a yield stress obtained from a uniaxial tension test is risky at best. The von Mises stress provides a means to extrapolate uniaxial tensile test data (for yield stress) to a three-dimensional stress state. In effect, the von Mises stress provides an "equivalent" uniaxial stress approximation to the three-dimensional stress state in a body through the following equation: or VM xx yy yy zz zz xx ( ) ( ) ( ) 6 xy 6 yz 6 zx (0.45) ( ) ( ) ( ) VM p p p p3 p3 p where p p p3 (0.46) (,, ) are the principal stresses. Given the stress components ( xx, yy, xy, etc. ) or principal stresses, one can compute the von Mises stress. This representation has been used quite successfully to model the onset of yielding in ductile metals and collaborates well with experiment. It is widely used in industry. For a material to remain elastic,

46 VM (for no yielding) y Equation (0.46) forms an ellipsoid in 3-D (ellipse in -D) when the stresses are plotted in principal stress space. As long as the stress state represented by the principal stresses is inside the ellipse (the yield surface), the material is elastic. Element Libraries (or, choose the right element for a structural component and loading, in order to maximize potential for correct results with the least amount of computation) Many, many finite elements have been developed for use in modern FEM software. Choosing the correct element for a particular structural is paramount. For example, if a structural member behaves like a beam in bending, we should choose a beam element to model it, if a structural member behaves like a thin plate in plane stress, we should choose an appropriate element to model it,

47 if a structural member looks like a shell of revolution, we should use a thin shell of revolution element, if a structural member will experience a three-dimensional stress state, we have to choose an element that models that behavior, etc. Here are some examples of the types of elements available: Truss element (-D and 3-D) Beam bending element (-D and 3-D; straight and curved) dof at each node Membrane element (no bending; flat and curved) Triangular, Quad (both straight and curved sides) Planes Stress and Plane Strain elements Triangular and Quadrilateral shapes (both straight and curved sides). Plane stress requires that the only non-zero stresses occur in the plane of the element (however, strain does occur normal to plane). Generally applicable to thin geometries. Two displacement dof per node (NO rotational dof). Plane strain requires that the only non-zero strains occur in the plane of the element (strain is zero normal to plane, but stress is not zero). Long constrained geometries (for example, a long pipe, a dam). Elements with curved boundaries will always have 3 or more nodes per edge.

48 Plate and shell bending elements (bending and in-plane stresses; flat and curved elements) Triangular, Quad (both straight and curved sides) Plate and shell bending elements are characterized as being thin compared to other dimensions, and having no stress normal to the plate (similar to plane stress). Plate and shell bending elements will have in-plane and normal and displacements ( u, v, w)

49 rotations (, ) about the two axes in the plane of the plate/shell. No stiffness about the norm x y 8. Subject Contents 8.. Synopsis page for each period (6 pages) 8.. Detailed Lecture notes containing: a. Ppts b. Ohp slides c. Subjective type questions (approximately 5 t0 8 in no) d. Objective type questions (approximately 0 to 30 in no) e. Any simulations 8.3. Course Review (By the concerned Faculty): (i)aims (ii) Sample check (iii) End of the course report by the concerned faculty GUIDELINES: Distribution of periods: No. of classes required to cover JNTU syllabus : 60 No. of classes required to cover Additional topics : Nil No. of classes required to cover Assignment tests (for every units test) : 4 No. of classes required to cover tutorials : No. of classes required to cover Mid tests : No of classes required to solve University Question papers : Total periods 0 UNIVERSITY QUESTIONS. Derive strain-displacement relations for a 3-D elastic body.. (a)what are the merits and the demerits of Finite Element Methods? (b) If a displacement field is described as follows:

50 u=( x +y +6xy)and v=(3x+6y y )0 4, Determine the strain components xx, yy, and xy at the point x=; y=0. 3. Explain the following: (a)variational method and (b) Importance of Boundary conditions. 4. What are different engineering field applications of finite element method? Explain them with suitable examples. 5. (a) Write the steps involved with finite element analysis of a typical problem. (b) Describe Rayleigh- Ritz method. 6) (a) State properties of global stiffness matrix. ) Determine the local and global stiffness matrices of a truss element. 8) Compare FEM with FDM (Finite difference method). 9) Using finite element method to calculate displacements and stresses of the bar shown in fig. 0) For the stepped bar shown in figure, determine the nodal displacements, element stresses and Support reactions. Take P=300kN, Q=500 kn, E=x0 N/m. A=50mm, A=500mm, A=000 mm ) Determine the displacements and the support reactions for the uniform bar shown in Fig.. GivenP=300KN

51 ) Determine the nodal displacements, element stresses and support reactions for the bar as shown in fig. 3). Determine the stiffness matrix, stresses and support reactions for the truss structure as shown in fig. 4). Calculate the nodal displacements, stresses and support reactions for the truss shown in fig.

52 QUIZ QUESTIONS Choose the correct alternative:. The solution by FEM is [ ] (a) Always exact (b) mostly approximate (c) sometimes exact (d) never exact. Primary variable in FEM structural analysis is [ ] (a) displacement (b) force (c) stress (d) strain 3. The art of subdividing a structure into convenient number of smaller components is known as [ ] (a) global stiffness matrix (b) force vector (c) discretization (d) none 4. is/are the phase/s of finite element method [ ] (a) Preprocessing (b) Solution (c) Post Processing (d) a, b & c 5. The characteristics of the shape functions is/are [ ] (a) the shape function has unit value at one nodal point and zero value at the other nodes (b) the sum of the shape function is equal to one (c) a & b (d) none 6.) The points in the entire structure are defined using coordinates system is known as [ ] (a) local coordinates (b) natural coordinates (c) global coordinate system (d) none. A plane truss element has a stiffness matrix of order [ ] (a) x (b) 4 x 4 (c) 6 x 6 (d) x 8. Determinant of assembled stiffness matrix before applying boundary conditions is [ ] (a) < 0 (b) = 0 (c) > 0 (d) depends on the problem 9. Each node of a -D beam element has degrees of freedom [ ] (a) (b) (c) 3 (d) 4 0. The -D beam element should have continuity. [ ] (a) C3 (b) C (c) C (d) C0 II Fill in the blanks. A small units having definite shape of geometry and nodes is called. Each kind of finite element has a specific structural shape and is inter- connected with the adjacent element by 3. is the variation method. 4. is defined as the ratio of the largest dimension of the element to the smallest dimension. 5. are used to express the geometry or shape of the element. 6. The boundary condition which in terms of the field variables is known as. The Strain- Displacement matrix of -D bar element is given by [B] = 8. The joints are used to join the truss members. 9. The element stiffness matrix for -D beam element is given by [K] =

53 0. The displacement and at each end of the beam element are treated as the ANSWERS. b. A 3. C 4. D 5. C 6. C. B 8. B 9. B 0. c II Fill in the blanks. Finite element.. Nodes 3. Ritz method or Ray-Leigh Ritz method 4. Aspect ratio 5. Shape functions 6. Primary boundary condition. [ ] = /L[ ] lb 8. Pin 0) Transverse or rotaion Choose the correct alternative:. What is the traction force of a D body? [ ] a) Force per unit area b) force per unit length c) force per unit volume d) all of these. For an Ax symmetric triangular element what is the size of the Jacobean Matrix [ ] a) 4 x 4 b) x c) x 4 d) 4 x 3. The governing equation for convection process is [ ] a) q = h A T s b) q = h A[ T h - T s ] c) q = h A T h d) q = h A [ T s - T h ] 4. Ax symmetric solids subjected to axisymmetric loading, the stress-strain relations are [ ] a) σ = D b) σ = D / c) σ = / D d) σ = D - 5. The stiffness matrix for a triangular element in a two dimensional problem is often derived Using [ ] a) area coordinates b) Surface coordinates c) volume coordinates d) mass coordinates 6. A constant term in the displacement function ensures [ ] a) Constant mode b) zero stress c) rigid body mode d) zero deformation. Number of shape functions the quadrilateral plane stress elements are [ ] a) 8 b) 4 c) 3 d) 8. A 3 noded simply supported beam gives number of frequencies [ ] a) 3 b) c) 4 d) 5 9. A linear term in the displacement function ensures [ ] a) rigid body mode b) zero deformation c)zero stress d) constant mode 0. Conductance matrix is the equivalent of stiffness matrix in analysis [ ] a) dynamic b) fluid flow c)thermal d) static structural II Fill in the blanks. Only matrix is different in case of plane strain and plane stress.. occurs when there is a temperature difference within a body or between a body and its surrounding medium. 3. When fewer nodes are used to define the geometry than are used to define the shape function, the element is termed as 4. Units for convection heat transfer coefficient is 5. The consistent mass matrix size for beam element is 6. In a D steady state heat transfer problem, the shape function matrix is. The consistent mass matrix size for bar element is 8. Thermal conductivity K x =K y =K z in case of material. 9. The shape functions of a D element in terms of area co-ordinates is 0. A fin is an external surface which is added on to a surface to increase the ANSWERS