Introduction to Finite Element Concepts

1. Introduction to Finite Element Concepts

2. Overview of the basic mechanical /stress terms

3. Review : Stress values • StressesNormal Stress : normal to cutting planeShear Stress : tangential to cutting planeNormal and Shear Stress values at Point Pdepend on :- Location of Point P on the cutting plane- Cutting direction=> Stress behavior at Point P can be described by the Stress tensor S

4. Review : Stress values Normal Stress values: xx = xyy = yzz = z The normal of a face and the normalstress vector have the same direction. z zz zy zx yz xz yy xx xy yx Shear Stress values:xy =yxxz =zx yz =zy y Shear Stresses in two perpendicular cutting directions at a point have the same value. x

5. Review : Stress values • Stress Tensor S :6 independent stress values from 3 perpendicularcutting planes at a point describe the stress behavior completely.Xxy xz S = xy yyz xz yz zThe stress tensor is symmetrical .

6. Review : Stress values • Example : Plane Stress= stress behavior in a plane membraneA plane membrane is under loading onlyin its plane.The thickness of the membrane is smallcompared to its two other dimensions.No forces in z-direction : xz =yz = z = 0xxy S = xy y

7. Review : Stress values • Coordinate Transformation (Plane Stress): = ½(x+ y) + ½(x- y)cos2 + xysin2 = ½(x+ y) - ½(x- y)cos2 - xysin2  = - ½(x- y)sin2 + xycos2  - Angle between x- and -Axis y   x

8. Review : Stress values • Invariant of Transformation (Plane Stress) :x+ y = +  = 1+ 2 Sum of normal stress valuesx2+ 2xy2 + y2=  2+ 22 +  2 = 12+ 22 • Special Case : Hydrostatic Stress behaviorx= y =  =  Same normal stress values in all cutting directions xy =  = 0 Shear stress values are equal Zero

9. Review : Stress values • Principal Stresses (Plane Stress) := Maximum values of Normal Stresses under the condition : d/d = 0 and d/d = 0 this leads to : - (x- y)sin2 + 2xycos2 = 0 2xy=> tan 2* = tan 2(* + /2) = x- y* and (* + /2) are two perpendicular cutting directions, called Principal Directions .

10. Review : Stress values • Trigonometric transformation :1 x- ycos2* = = [ 1 + tan22* ]½[ (x- y )2 + 4xy2]½ tan2* 2xysin2* = = [ 1 + tan22* ]½[ (x- y )2 + 4xy2 ]½ • Principal Stresses 1 and 2 : 1 >  21,2 = ½(x+ y)  ½[ (x- y )2 + 4xy2 ]½(*) = (* + /2) = 0 !! No Shear Stresses

11. Review : Stress values • Principal Shear Stress (Plane Stress) := Maximum values of Shear Stresses under the condition : d/d = 0 this leads to : - (x- y)cos2 - 2xysin2 = 0 x- y=> tan2** = tan2(** + /2) = - 2xy

12. Review : Stress values • Because : tan2** = - 1 / tan2*2** and 2* are oriented perpendicular to each other,this results in an angle of 45o between the direction of maximum shear stress ** and the direction of maximum normal stress *. • Principal shear stress max : max =  ½[ (x- y )2 + 4xy2 ]½or max=  ½(1- 2)Using ** in  and  results in a Non-zero value ofnormal stress value M : M(**) = ½(x + y) = ½(1 + 2)

13. Review : Stress values • Equivalent stresses V :There is a 3–dimensional stress behaviour in an arbitrary part. Experimental Data have been received from a tension specimen (1-dimensional stress behaviour). The equivalent stress is used to compare 3-dimensional stress behavior with the 1-dimensional stress behaviour of the tension test. Different strength hypotheses have been developed.

14. Review : Stress values • Equivalent stresses V : Normal stress hypothesis :Assumption : The maximum Normal stress value is responsible for the materialload V =1 Used for : brittle materials

15. Review : Stress values • Equivalent stresses V :Shear stress hypothesis :(1864 described by H.Tresca)Assumption : The the material load is characterized by the maximum stress value.Plane Stress:Max = ½(1 - 2 ) => V =1 - 2V = [(x - y)2+ 4xy2]½

16. Review : Stress values • Equivalent stresses V :von Mises stress hypothesis :(named after Huber (1872-1950), v. Mises (1883-1953)and Hencky (1885-1951))Assumption : The material load is characterized by the energy which is used for the change of the shape without a change in the volume of the part Plane Stress:V =[ 12 + 22-12 ]½= [ x2 + y2-xy + 3xy2]½Used for : ductile materials

17. Review : Strain values • Deformation in space :Kinematical relationStrains : x = u/x , y = v/y , z = w/zAngular distortions :xy = u/y + v/x , xz = u/z + w/x , yz = v/z + w/yStrain tensor V : symmetricx ½xy ½xz V = ½ xy y ½ yz ½ xz ½yz z

18. Review : Strain values Assumption : Small DeformationStrain :x = u/x , y = v/y Gliding or Shearing(Angular distortion) = u/y ,  = v/xxy =  +  = u/y + v/x u+(u/y)dx (u/y)dy R’ • Plane Strain : S’ v+(v/y)dy R S  Q’ (v/x)dx dy /2-xy Strain tensorx ½xy V = ½ xy y  P’ v+(v/x)dx v Q dx P u u+(u/x)dx y x

19. Review : Strain values • Coordinate transformation (Plane Strain) : = ½(x+ y) + ½(x- y)cos2 + ½xysin2 = ½(x+ y) - ½(x- y)cos2 - ½xysin2½ = - ½(x- y)sin2 + ½xycos2  - Angle between x- and - axis y   x

20. Review : Strain values • Principal strains (Plane Strain) := Maximum strain values under the condition : d/d = 0 and d/d = 0 this leads to : - (x- y)sin2* + xycos2* = 0 xy=> tan 2* = tan 2(* + /2) = x - y* and ( * + /2 ) are two perpendicular oriented cutting directions.

21. Review : Strain values • Principal strains (Plane Strain) 1 and 2 1,2 = ½(x+ y)  ½[ (x- y )2 + xy2 ]½(*) = (* + /2) = 0 !! No angular distortion • Maximum angular distortion : d/d = 0 - (x- y)cos2** - xysin2** = 0 x- y=> tan 2** = tan 2(** + /2) = - xy=> M(**) = ½(x + y) = ½(1 + 2) ( Strains  0 )

22. Introduction to FEM • FEM = Finite ElementMethod • FEM is a numerical approximation method,which is used for the calculation and optimization of the structural behavior of mechanical parts ! • Finite Element = Discrete structural description of continua withhelp of mathematical formulations

23. Historical Overview • 1950´s: Application of matrix methods for structural analysis • 1950´s: Introduction of the displacement- and stiffness method for complex aerospace structures • 1960: Birth date of the name “Finite Elements“ • 1970´s: FEM applications mainly in the aerospace and the automotive industry (NASA -> NASTRAN) • 1980´s: Introduction of powerful computer graphics • 1997: FEM is the standard tool for structural analysis

24. Seminar Contents • Basic Theory of FEM • Element Types and their Use • Geometric model • Finite Element model • Material Data • Element Properties • Loads • Boundary Conditions • Postprocessing

25. Introduction to FEM • Finite Elements have simple geometric shapes, like triangle, rectangle, cube. • The points at the corners are called Node or Grid point.Nodes connect the elements. • The description of the structural behaviour inside an element is done by the calculation of the nodal displacements (= discretization) in combination withspecial shape function (= mathematical functions usedfor integration across the element region)

26. Introduction to FEM • Analytical preparation Principle of virtual work=Minimal change of energy due to external load A static equilibrium between internal and external forces is reached when the Elastic Potential becomes minimal; this means the first derivation of  is Zero. /ui = 0 - System of equations with n equations for n unknown displacements ui - Displacement of the i-th degree of freedom

27. Introduction to FEM • Elastic Potential  (Potential Energy) = ½  T dV -  uTfK dV -  uTfV dV -  uTfA dA  Strain vector : x , y , z , xy , yz , zx Stress vector : x , y , z , xy , yz , zxu Vector of displacements : ux , uy , uz , rx , ry, rzfK Vector of external forces and moments on nodesfV Vector of inertial forces (Acceleration, Rotation)fA Vector of forces on faces (Pressure)(Thermal load is not considered here.)

28. Introduction to FEM • Continuous – Discrete displacement relations The continuous displacement field in an element u(x,y,z)will be described by Shape functions N(x,y,z) and the discrete nodal displacements of the element ui.u(x,y,z) = N(x,y,z) * ui

29. Introduction to FEM • Strain – Displacement relations Strains can be derived from the displacementaccording to the theory of linear elasticity by theuse of the Cauchy-Matrix D (Differential operator).(x,y,z) = D * u(x,y,z)Using the Continuous – Discrete displacement relation (x,y,z) = DN(x,y,z) * ui

30. Introduction to FEM • Stress – Strain relationFor a linear material the Hooke’s Matrix E statesthe relation between stress and strain.(x,y,z) = E * (x,y,z)Using the Strain – Displacement relation (x,y,z) = ED * u(x,y,z)Using the Continuous – Discrete displacement relation (x,y,z) = EDN(x,y,z) * ui

31. Introduction to FEM • Getting system of linear equationsAll preceding equations will be put into the equation of theelastic potential . Creating the derivates /ui = 0 weget the basic equation of the FE-method in the linear static. K ui = FGlobal stiffness matrix : Sum of elemental stiffness matrixesK =  (DN)TE(DN) dV (Maxwell´s Law : Matrix is symmetric)External nodal forces (inertial loads and loads on faces are been represented by equivalent nodal forces F =  NTfK dV +  NTfV dV +  NTfA dA

32. Introduction to FEM Y uy3 • Example : Triangular element (Plane stress) ux3 uy2 ux2 uy1 ux1 X

33. Introduction to FEM • Assumption : Linear shape functionux(x,y) = a0 + a1x + a2y uy(x,y) = b0 + b1x + b2yre-written in matrix form : a0 b0 a1 b1 ux(x,y) = [ 1 x y ] a2 uy(x,y) = [ 1 x y ] b2

34. Introduction to FEM • Using linear shape function for all 3 nodes :ux1 = a0 + a1x1 + a2y1 uy1 = b0 + b1x1 + b2y1ux2 = a0 + a1x2 + a2y2 uy2 = b0 + b1x2 + b2y2ux3 = a0 + a1x3 + a2y3 uy3 = b0 + b1x3 + b2y3re-written in matrix form : ux1 1 x1 y1 a0uy1 1 x1 y1 b0ux2 = 1 x2 y2 a1uy1 = 1 x2 y2 b1ux3 1 x3 y3 a2uy1 1 x3 y3 b2uxi = G * a uyi = G * b

35. Introduction to FEM • a and b as function of uxi anduyi :a = G-1 * uxib = G-1 * uyi • Getting the Continuous – Discrete displacement relations : ux1 ux2ux(x,y) = [ 1 x y ] G-1uxi = N(x,y) uxi = [ N1 N2 N3 ] ux3 uy1 uy2uy(x,y) = [ 1 x y ] G-1uyi = N(x,y) uyi = [ N1 N2 N3 ] uy3

36. Introduction to FEM • Strains (Plane stress) :x(x,y) = ux(x,y)/xy(x,y) = uy(x,y)/yxy(x,y) = ux(x,y)/y + uy(x,y)/xre-written in matrix form : x(x,y) /x 0 ux(x,y) y(x,y) = 0 /y uy(x,y) xy(x,y) /y /x  = D * u

37. Introduction to FEM • Using discrete nodal displacements ux1 ux2 ux3 x (x,y) N1,x N2,x N3,x 0 0 0 uy1 y (x,y) = 0 0 0 N1,y N2,y N3,y uy2 xy(x,y) N1,y N2,y N3,y N1,x N2,x N3,x uy3Linear Shape function Ni = Ni(x,y)Derivatives : Ni,x = const. , Ni,y = const. => constant strain value over the element region

38. Introduction to FEM • Stresses , isotropic Material x (x,y) E 1  0 x (x,y) y (x,y) =  1 0 y (x,y) xy(x,y) 1-2 0 0 (1- )/2 xy(x,y)  = H * Shear Modulus : G = E/ 2(1+  )Poisson Ratio : 0 <  < 0.5Because all strains (x,y) are constant values, all stress value are constant over the element region as well !!

39. Solve a number of simple problems, add them all up and get the answer of a complex problem • Divide a complex problem into simple ones • Divide complex geometry into simple objects which we can understand (Lines, Squares, Cubes) • Use the computer to do millions (and millions, …) of numerical operations • Use modern hardware equipment to present the results graphically Calculate the area of a circle How does FEM work?

40. u F Mathematical Model Formulation of simple elements K = Spring stiffness (Ea/L). U = Spring elongation F = Spring force K * U = F

41. Mathematical Model Dividing of the geometry into simple elements and assembling all elements [K] = Stiffness matrix of the part (Sum of all elements) {U} = Components of the displacements of the single nodes of the part {F} = Components of the loads of the single nodes of the part [K] *{U} = {F} Solving the matrix equation withthousands of unknowns

42. A given problem is discretized by dividing the original domain into simply shaped subdomains, the so called elements. • Each element it quite simple, and the program can figure out its mechanical properties quite easily • By summation of all the element contributions one gets the whole model behavior uy Y ux [ k ]e { u }e = { f }e X element level Theoretical Background

43. Terminology • Each element is connected to its neighbour only at a number of points, called nodes • Each node has 6 independent possibilities to move: 3 translational and 3 rotational • These independent possibilities to move are called degrees of freedom (DOF’s)

44. Y Translations Tx, Ty, andTz (1, 2, and 3) Rotations Rx, Ry, andRz (4, 5, and 6) Forces Fx, Fy, andFz Moments Mx, My, andMz Ry ,My Ty ,Fy Rx ,Mx Tz ,Fz X Tx ,Fx Rz ,Mz Cartesian Coordinate System Z Terminology

45. Steps in an FE Analysis Geometry Displacements Stresses Forces Strains Elements Loads Analysis Model Analysis Results Solver Contour Plots X-Y-Plots Listings Constraints Materials Properties

46. FEM Program algorithm • Represent continuous model as a collection of elements and connections • Formulate element stiffness matrices [k] • Assemble all element stiffness matrices to a global stiffness matrix [K] • Generate load vector [F] • Solve matrix [F]=[K][x] with respect to [x] • Calculate element stresses and strains We shall illustrate this using a simple example...

47. Continuous Model F = 1000 N A = 20mm2E = 210000 N/mm2 L = 50 mm A = 50mm2E = 210000 N/mm2 L = 50 mm

48. Discrete model We make one-dimensional elements with two degrees of freedom Node 3 A = 20mm2E = 210000 N/mm2 L = 50 mm Element 2 Node 2 A = 50mm2E = 210000 N/mm2 L = 50 mm Element 1 Node 1

49. Element stiffness matrices For one element, the stiffness relation is Each of the element matrices then becomes:

50. Global stiffness matrix The Global stiffness matrix is assembled by combining the matrices at the appropriate degrees of freedom for each node