![ Fundamental Concept of FEM
• A continuous field of a certain domain having infinite degrees of
freedom is approximated by a set of piecewise continuous models
with a number of finite regions called elements. The number of
unknowns defined as nodes are determined using a given relationship
i.e.{F}=[K]*{d}.](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Finite_Element_Method analysis and theory of yeild line .ppt
- 1. G.H. RAISONI COLLEGE OF ENGINEERING & MANAGEMENT, PUNE. (An Autonomous Institute Affiliated to SPPU) Department of Civil Engineering TAE – 2 Name : Atif Malik Roll no : 05 Reg. no : 23ASTR2101007 Subject : FINITE ELEMENT METHOD PPT Topic : FINITE ELEMENT METHOD ANALYSIS Year : F.Y. SEM - II Course : MTECH STRUCTURAL ENGINEERING
- 2. Fundamental Concept of FEM • A continuous field of a certain domain having infinite degrees of freedom is approximated by a set of piecewise continuous models with a number of finite regions called elements. The number of unknowns defined as nodes are determined using a given relationship i.e.{F}=[K]*{d}.
- 3. Fundamental Concept of FEM Domain x x Domain with degrees of freedom •Red line-Continuous field over the entire domain. •Blue line-Finite number of linear approximations with the finite number of elements x 1 2 3 4 5 6 x Subdomain e Domain divided with subdomains with degrees of freedom
- 4. General Steps: 1) Discretize the domain a) Divide domain into finite elements using appropriate element types (1-D, 2-D, 3-D, or Axisymmetric) 2) Select a Displacement Function a) Define a function within each element using the nodal values 3) Define the Strain/Displacement and Stress/strain Relationships 4) Derive the Element Stiffness Matrix and Equations a)Derive the equations within each element
- 5. General Steps: 5) Assemble the Element Equations to Obtain the Global or Total Equations and Introduce Boundary Conditions a)Add element equations by method of superposition to obtain global equation 6) Solve for the Unknown Degrees of Freedom (i.e primary unknowns) 7) Solve for the Element Strains and Stresses 8) Interpret the Results
- 6. Applications: •Stress Analysis • Truss and frame analysis • Stress concentration •Buckling •Vibration analysis •Heat transfer •Fluid flow
- 7. Advantages of FEM : •Model irregularly shaped bodies •Compute General load conditions •Model bodies composed of different materials •Solve unlimited numbers and kinds of boundary conditions •Able to use different element sizes in places where loads or stresses are concentrated •Handle non-linear behavior using linear approximations •Reduce System Cost
- 8. FEM Packages •Large Commercial Programs • Designed to solve many types of problems • Can be upgraded fairly easily • Initial Cost is high • Less efficient •Special-purpose programs • Relatively short, low development costs • Additions can be made quickly • Efficient in solving their specific types of problems • Can’t solve different types of problems
- 10. Note on Stiffness matrix For a 1-D bar, the stiffness matrix is derived from the stress/strain relationship in Hooke’s law and the definitions of stress and strain. σx = Eεx. ; σx = P/A ; εx = du/dx = (d2x – d1x)/L By substitution: -f1x = EA (d2x – d1x) L f1x = EA (d1x – d2x) L Similarly for f2x: f2x = EA (d2x – d1x) L Combining into matrix form, the stiffness matrix is defined as [k] = EA L 1 -1 -1 1 f2 f1 d1x d2x L A E
- 11. Note on the displacement function •For a given set of nodes there exists a function that approximates the displacement at any position along the bar. •This function, called the displacement function, is derived from Pascal’s Triangle. •A new constant is introduced into the function for every node in the discretized domain. u(x) = a1 + a2x + a3x2 + … For 1-D
- 12. Note on the displacement function If a 1-D bar is broken into 2 elements, the displacement function would be u(x) = a1 + a2x + a3x2. Putting it into matrix notation: u(x) = [1 x x2] a1 a2 a3 By knowing the distances to the nodes and the displacements at those nodes, the equation becomes: u1 u2 u3 = 1 0 0 1 x2 x2 2 1 x3 x3 2 a1 a2 a3 , where x1 = 0, x2 and x3 are the distances to the nodes and u1, u2, and u3 are the displacements. The coefficients are found by solving the equation.
- 13. Example A A L L E 2E P Determine displacements of materials a and b if the load P is applied to the end of the bar given the above information. a b
- 14. SOLUTION: 1) Discretize the domain with appropriate elements. Element a 1 2 Element b 2 3 u1 u2 u2 u3 1 2 3 u1 u2 u3 f3 = P f1 f1 f3 = P f21 f22
- 15. SOLUTION: 2) Select a displacement function There will be new term for each element, and the terms are 1 2 3 u1 u2 u3 u(x) = a1 + a2x + a3x2 derived from Pascal’s triangle.
- 16. SOLUTION: 3) Define stress/displacement and stress/strain relationships σx = Eεx εx = du/dx 4) Derive the element stiffness matrix and element equations {F} = [k]{d} [k] = stiffness matrix [k] = EA L 1 -1 -1 1 {F} = Force {d} = displacement f1 f21 = L 1 -1 -1 1 EA u1 u2 f22 f3 = L 2 -2 -2 2 EA u2 u3 , a b
- 17. SOLUTION: 5) Construct Global equation and introduce boundary conditions and known variables. f1 f21+f22 f3 = L 1 -1 0 -1 3 -2 0 -2 2 EA u1 u2 u3 Global Equation B.C.: (x =0) u1 = 0 Known variables: f3 = P and f2 = f21+f22 = 0
- 18. SOLUTION: 6) Solve for unknowns. f1 0 P = L 1 -1 0 -1 3 -2 0 -2 2 EA 0 u2 u3 f1 = -EAu2 L 0 = EA(3u2-2u3) L P = EA(2u3-2u2) L u2 = PL EA u3 = 3PL 2EA f1 = -P
- 19. SOLUTION: 7) Solve for the element strains and stresses. εa = P = u2 EA L σa = Eεa = P A εb = 3P = u3 2EA L σa = 2E εb = P 3 A 8) Interpret the Results •After solving for the displacements, the coefficients of the displacement function can be determined.
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