ce 620 ce 620 finite element method yogesh m desai
play

CE 620 CE 620 FINITE ELEMENT METHOD Yogesh M. Desai Department - PowerPoint PPT Presentation

CE 620 CE 620 FINITE ELEMENT METHOD Yogesh M. Desai Department of Civil Engineering Indian Institute of Technology Bombay Indian Institute of Technology Bombay Powai, Mumbai - 400076 Instructor Yogesh M. Desai Office Room No. 126 Civil


  1. CE 620 CE 620 FINITE ELEMENT METHOD Yogesh M. Desai Department of Civil Engineering Indian Institute of Technology Bombay Indian Institute of Technology Bombay Powai, Mumbai - 400076

  2. Instructor Yogesh M. Desai Office Room No. 126 Civil Engineering Dept. Phone No 7333 E-mail desai@civil.iitb.ac.in Lectures Lectures Tuesday Tuesday 14:00 - 15:25 14:00 15:25 (SLOT 10) Friday 14:00 - 15:25 Extra Lectures Wednesday 17:30 - 19:00 Consulting Hours C lti H Th Thursday d 15 00 15:00 - 17:00 17 00

  3. Introduction  Many problems in engineering and applied science are governed by differential or integral equations.  Due to complexities in geometry, properties ti and d boundary b d conditions diti i in most real-world problems, an exact solution cannot be obtained. solution cannot be obtained.

  4. Introduction Introduction Finite element method is an approximate numerical method for solving problems of engineering and mathematical sciences. Useful for problems with complicated geometries, external influences and g properties for which analytical solutions are not available.

  5. OBJECTIVES / LEARNING OUTCOMES OBJECTIVES / LEARNING OUTCOMES • Understanding of different semi-analytical / numerical methods to solve a variety of problems • Understanding of general steps of FEM • Understanding of finite element formulations • Ability to derive equations related to FEA of various1 D • Ability to derive equations related to FEA of various1-D , 2-D and 3-D problems

  6. OBJECTIVES / LEARNING OUTCOMES OBJECTIVES / LEARNING OUTCOMES • Understanding of advantages and disadvantages of the FEM • Exposure to computer implementation of the FEM Exposure to computer implementation of the FEM • Ability to do FE analysis independently with proper interpretation of results interpretation of results

  7. Course Contents  Introduction  Introduction  Overview of various methods to solve 5 – 8 Lectures integral and differential equations  Variational Calculus  Basics of Finite Element Methods  Local and Global Finite Element Methods  Application of FEM to solve various 1-D, 12 – 16 Lectures 2-D and 3-D Problems 2 D and 3 D Problems  C 0 Continnum  C 1 Continnum  C 1 Continnum  Convergence and Error Estimation  Iso-parametric formulation  Numerical integration

  8. Course Contents  Concept of Sub-structuring 3 – 5 Lectures  Conditions of Symmetry / Anti-symmetry  Computer Implementation of FEM  Application of FEM to Time Dependent 4 – 6 Lectures Problems Problems  Partial FEM  Exposure to Hybrid FEM Total Lectures ~ 28 28 (~ 42 Hrs)

  9. Assessment Scheme Assignments and Term Projects :20 % Mid - Term Exam :30 % (as per time table) End - Term Exam :50 % (as per time table) Notes : (1) Bring calculator to all the lecture sessions. ( ) g (2) 80% Attendance is required.

  10. Brief History It is difficult to document the exact origin of the FEM, because the basic concepts have evolved over a period of 150 or more years. i d f 150 Hrennikoff Hrennikoff [1941] [1941] – Framework Framework method method for for elasticity problems Courant [1943] - Variational form Levy [1947, 1953] - Flexibility and Stiffness L [1947 1953] Fl ibilit d Stiff Argyris [1955] - Energy Theorems and Structural Analysis Analysis Turner, Clough, Martin and Topp [1956] - Stiffness Method Cl Clough [1960] - Termed “Finite Elements” h [1960] T d “Fi it El t ”

  11. Brief History Brief History  In early 1960s, engineers used the method  In early 1960s, engineers used the method for approximate solution of problems in stress analysis, fluid flow, heat transfer, and other areas.  The first book on the FEM by Zienkiewicz and  Th fi t b k th FEM b Zi ki i d Chung was published in 1967.

  12. How can FEM Help ? • Can be applied to a variety of fields like structural mechanics, aerospace engineering, structural mechanics, aerospace engineering, geotechnical engineering, fluid mechanics, hydraulic and water resource engineering, mechanical engineering, nuclear engineering, electrical and electronics engineering, metall rgical metallurgical, chemical chemical and and en ironmental environmental engineering, meteorology and bioengineering, etc etc.

  13. • Easily applied to complex, irregular-shaped Easily applied to complex, irregular shaped objects composed of several different properties and having complex boundary conditions and external influences. • Applicable to steady-state (static), time A li bl t t d t t ( t ti ) ti dependent as well as characteristic value problems problems. • Applicable to linear as well as nonlinear pp problems.

  14. Fi it Finite Element Method is an Approximate Numerical El t M th d i A i t N i l Method to Solve Problems of Engineering and Mathematical Sciences. Any given problem reduces to

  15. CE 620 : FINITE ELEMENT METHOD Assignment No. 1 (Due on January 15, 2013) Q. 1 Matrices [K] 3  3, [T] 3  5, and {q} 5  1 are defined as        10 5 4 3 2 1 1 0.5         [K] = [K] T = 5 10 2 ; [ T] = 0.5 1 4 0 1               4 2 10 0.2 0 3 2 0 {q} = { 3 0 -2 1 -4} T Compute matrix [M 1 ] to [M 6 ] appearing in the following expression (a) [M 1 ] = [K] -1 (b) [M 2 ] = 3[K] -1 +4[K] (c) [M 3 ] = [K] [T] (d) [M 4 ] =5[K] -1 [T] {q} 1 {q} T [T] T [K][T]{q} (e) [M 5 ] = [T] T [K] [T] (f) [M 6 ] = 2 Q. 2 Given     2 2 d a t 3 t         a . Find [ ] b and [ ] c [ ] . a dt 3   dt 2 t 0 Q. 3 Given 1         T T    q [ K ] q q f where 2       k k k q f 11 12 13  1   1               [ K ] k k k ; q q ; f f   12 22 23 2 2             k k k q f 13 23 33 3 3          Derive equations arising from 0 ; 0 ; 0    q q q 1 2 3 Q. 4 Matrix equation [K] {q} = {f} is given where     6 7 9 2 13    7 18 6 10 6         [ K] = 9 6 42 15 11      2 10 15 8 6         13 6 11 6 24 {q} = [ q 1 q 2 q 3 q 4 q 5 ] T ; {f} = [100 70 -50 150 -35 ] T Compute {q} 5  1 by employing Gauss elimination algorithm.

Recommend


More recommend