Finite Element Analysis| FEA| ME8692 | UNIT-1| Part-1| Tamil

This video clearly explain to get a maximum mark in Finite Element Analysis (FEA) Unit -1 introduction to FINITE ELEMENT ANALYSIS (FEA).
FEA is a problematic subject for Anna university- 6th Semester-Mechanical engineering students. Unit-1 Important Question and Answers are discussed in this video
How to pass Finite element analysis subject in 30 min video is already uploaded in this Poriyaalan lecturer video channel.
FEA (Tamil)Full video link: • How to Pass Finite Ele...
FEA (English)Full video link: • How to Pass Finite Ele...
Unit wise Finite Element Analysis (FEA) Explanation
Unit:I
Part-1: • Finite Element Analysi...
Part-2: • Finite Element Analysi...
For more Details contact,
Prof.A.Aravindh:: 9952437464
FINITE ELEMENT ANALYSIS
UNIT I
INTRODUCTION
Historical Background - Mathematical Modeling of field problems in Engineering - Governing Equations - Discrete and continuous models - Boundary, Initial and Eigen Value problems- Weighted Residual Methods - Variational Formulation of Boundary Value Problems - Ritz Technique - Basic concepts of the Finite Element Method.
UNIT II
ONE-DIMENSIONAL PROBLEMS
One Dimensional Second Order Equations - Discretization - Element types- Linear and Higher order Elements - Derivation of Shape functions and Stiffness matrices and force vectors- Assembly of Matrices - Solution of problems from solid mechanics and heat transfer. Longitudinal vibration frequencies and mode shapes. Fourth Order Beam Equation -Transverse deflections and Natural frequencies of beams.
UNIT III
TWO DIMENSIONAL SCALAR VARIABLE PROBLEMS
Second Order 2D Equations involving Scalar Variable Functions - Variational formulation -Finite Element formulation - Triangular elements - Shape functions and element matrices and vectors. Application to Field Problems - Thermal problems - Torsion of Non circular shafts -Quadrilateral elements - Higher Order Elements.
UNIT IV
TWO DIMENSIONAL VECTOR VARIABLE PROBLEMS
Equations of elasticity - Plane stress, plane strain and axisymmetric problems - Body forces and temperature effects - Stress calculations - Plate and shell elements.
UNIT V
ISOPARAMETRIC FORMULATION
Natural co-ordinate systems - Isoparametric elements - Shape functions for iso parametric elements - One and two dimensions - Serendipity elements - Numerical integration and application to plane stress problems - Matrix solution techniques - Solutions Techniques to Dynamic problems - Introduction to Analysis Software.
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Sir practice sum la galarkins method Mattum podunga sir pls

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[@Poriyaalan]

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Sir Practice sumla least sqaure method la (-12ax^2+300x^2)^2 ku endha formula use pannanum

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Galerkins method la Y ku bathila }(Wi,R)dx = 0. Wi = neglect a¹, Wi = (x-x⁴) Panna 24.99 ~ 25 ans vanthurum

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how to assume trial function if not given sir..? in galerkin method

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Sir last step galerkin method... Doubt sir -10ax × 100 = -1000ax sir.. But ninga 100ax tha potu irukinga sir

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Sir, Trial function polynomial series yella problem thukku use pannalam except beam

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Trigonometry vechi weighted residual or Ritz edhachi onnu solve panna podhum sir

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Sir in trial function kudukalaa namma epdi assume panrathu sir. Athaa pathi oru ideaa kudunkaa sir

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point collocation method la antha 300 x square....(quadratic eqn a la varum aprm epdi kandupidika mudium R

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same doubt....got any answer ?

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Sir, I think "Finite element" means "countable/definite number of elements" not small particle.

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9:40

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Kastamana sum ethachum nadathuna paravala ella given um kuduthu keka maatagala Anna University la ...fea unit 1 part 1 boundary condition ilama or trail function ilama ethachum sum example solluga sir

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Good evening sir, lastla galarin methodla 100a/2(10) square potu iruku idhuku answer 10000dhana epadi sir 100000a/2 lakh vandhuchu

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Sir Galerkin method It would be nice if you could make it a little more understandable

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Galerkien's method multiple apo mistake pannitinga 1000 aa 100 nu potinga

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The galerkins method la , 100x100 =10000 , but neega 100000 nu pottu irukkinga

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Sir 2nd problem -25 or +25?

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Sir ennaku a1=- 25 varuthu But neenga a1=Plus25

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100*100=10000

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5/2

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