Steps in the Finite Element Method Chung Hua University Department of Mechanical Engineering Dr. Ching I Chen
General Idea Engineers are interested in evaluating effects such as deformations, stresses, temperature, fluid pressure, and fluid velocities caused by forces, such as applied loads or pressures and thermal and fluid fluxes. The nature of distribution of the effects in a body depends on the characteristics of the force system and of the body itself. Our aim is to find this distribution of the effects.
Solid Mechanics: Example 1: Truss deformation under the external forces y x
Solid Mechanics: Example 1 : deformation results
Solid Mechanics : Example 2 : plate stress distribution under the external forces
Solid Mechanics : Example 2 : stress distribution results
Heat Transfer : Example 3 : temperature distribution in a short solid cylinder T Top = 40 T wall =T bot = 0
Heat Transfer : Example 3 : temperature distribution results
Fluid Mechanics : Example 4 : acceleration of a tank fluid a
Fluid Mechanics : Example 4 : height distribution
Fluid Mechanics : Example 5 : plane poiseuille flow P 1 =0.1 P 2 = 0
Fluid Mechanics : Example 5 : velocity distribution
Electrical Analysis : Example 6 : electric current flowing in a network
Electrical Analysis : Example 6 : results comparison V 1, volts V 2, volts V 3, volts V 4, volts I 2-1, amps I 3-1, amps I 2-3, amps I 4-2, amps I 4-3, amps I 1-4, amps Target 0.0 28.0 19.0 100.0 1.4 1.9 1.0 2.4 0.9 3.3 ANSYS 0.0 28.0 19.0 100.0 1.4 1.9 1.0 2.4 0.9 3.3
From previous examples we know the effects may be deformation, pressure, velocity, temperature, voltage, current, etc. depending on the type of the problem. Let s called these effects distribution as a function of u(x,y,z). For 2-D illustration: y Distribution of u u(x,y) (T,p) for entire body 2-D Body to be considered x
For convenience, we temporality use displacements or deformations in place of effects. We assume that it is difficult to find the distribution of u(x,y,z) by using conventional methods and decide to use the concept of discretization -- finite element method. We divide the body into a number of smaller regions called finite elements.
The consequence of such subdivision is that the distribution of displacements (effects) is also discretized into corresponding subzones. The subdivided elements are now easier to examine as compared to the entire body and distribution of u over it. y u (T,p) Distribution of u e (x,y) over a generic element e Finite element x
Steps of Finite Element Method Formulation and application of the finite element method are considered to consist of eight basic steps : Discretize and Select Element Configuration Select Approximation Models or Functions Define Gradient Unknown and Constitution Relations Derive Element Equations Assemble Element to Obtain Global System Solve for the Primary Unknowns Solve for Derived or Secondary Quantities Interpretation of Results
Step 1 : Discretize and Select Element Configuration How to discretize the body? Node 1. A point processes coordinate and number to describe its location in the structure body. N3 N2 N4 N1 N1 = (10, 0) N2 = ( 5,10) N3 = ( 0, 5) N4 = ( 8, 2)
Totally, there are 71 nodes
Totally, there are 723 nodes
Node (continued) 2. Having degree of freedom (DOF) to specify the load effects. - A degree of freedom can be defined as an independent (unknown) displacement that can occur at a point. - The solution is solved only for those nodes that we created.
Example 1: Truss deformation under the external forces dof = displacements at x, y direction (UX, UY) y 8 6 7 UY x 3 UX
Example 2 : plate stress distribution under the external forces dof = displacements at x, y direction (UX, UY) UY UX
Example 3 : temperature distribution in a short solid cylinder dof = temperature (T) T
Example 4 : acceleration of a tank fluid dof = displacements at x, y,z direction (UX, UY, UZ)
Example 5 : plane poiseuille flow dof = velocity at x, y direction (VX, VY) P 1 =0.1 P 2 = 0 VY VX
Example 6 : electric current flowing in a network dof = voltage (Volt) V
Node (continued) 3. Using to create the elements - The creation of element should be based on node to node connection.
Element 1. Elements are created by node to node connection. 2. The connection between node to node is straight line. Therefore the actual outside contour is approximation for curve line and surface boundary. How to approach the actual contour?
3-D course element to fine element
Element (continued) 3. What type of element should be used? Depending on the problem defined, However, geometrically, 1 D element 2 D element 3 D element
1 D element Cantilever Beam 1 2 3 1 2 3 4 T1 T2 T1 1 2 3 T2 1 2 3 4 Heat transfer
2 D element The shape of 2-D plane element quadrilateral triangle
2 D element P P h This is plane stress problem, neglect the thickness (a unit).
2 D element t Real Beam 2-D beam 2-D beam finite element
2 D element The effects of x-y plane are to be found. Y X This is plane strain problem, neglect z direction length.
3 D element The shape of 3-D solid element hexahedron prism tetrahedron
3 D element The shape of 3-D solid element tetrahedron hexahedron
3 D element The shape of 3-D solid element hexahedron
Element Configuration Globally, the configuration (location) of each element is not same. 9 1 2 3 5 5 5 y 6 1 15 x x element #1 at 0 to 5 element #2 at 5 to 10 element #3 at 10 to 15 element #1 in the range 0 x 5, 0 y 5 element #6 in the range 5 x 10, 6 y 9
Element Configuration (continued) In order to generally describe the displacements of each element, the local coordinate system should be used s s s each element has its own coordinate system, called local coordinate system, such that 0 s 5 1 2 3 5 5 5 x global coordinate system
Element Configuration (continued) Based on the local coordinate system, the generalization to the governing equation of each element cab be reached. No matter how the element domain is!! 1-D element L 1 L 2 s L 3 s s Each element can be different length. The domain will be 0 s Li
Element Configuration (continued) 2-D element t y t s x The local coordinate system can be selected by your convenience. s
Step 2 : Select Approximation Models or Functions 1. Choose a pattern or shape for the distribution of the unknown quantity u e in the element. 2. The nodal points of the element provide strategic for writing mathematical functions to describe the shape of the distribution of the unknown quantity over the domain of the element. distribution pattern or shape of u e (s,t) over a generic element e e
Step 2 : Select Approximation Models or Functions (continued) 3. A number of mathematical functions such as polynomials and trigonometric series can be used for this purpose, especially polynomials because of the ease and simplification they provide in the finite element formulation. What is f(x) x Interpolation function To find a function f(x) to pass the given points
Step 2 : Select Approximation Models or Functions (continued) 4. If an plane element has 4 unknown (u 1, u 2, u 3, u 4 ), the polynomial interpolation function can be expressed as u 3 u 4 e u 2 t u 1 s ust (, ) = N( stu, ) + N( stu, ) + 1 1 2 2 N ( s, t) u + N ( s, t) u 3 3 4 4 where u i are the values of the unknowns at the nodal points
Step 2 : Select Approximation Models or Functions (continued) 5. Generally, if we denote u as the unknown, the polynomial interpolation function can be expressed as u = Nu + N u + N u + L + N u 1 1 2 2 3 3 m m where u 1,u 2,u 3,,u m are the values of the unknowns at the nodal points N 1,N 2,N 3,,N m are the interpolation functions
Step 2 : Select Approximation Models or Functions (continued) 6. Remark One should realize that the solution in each step will be in terms of the unknowns only at the nodal points. In the finite element method, one should find the solution as the values of the unknown u at all the nodes. To initiate action toward obtaining the solution, one has assumed a priori or in advance a shape or pattern that one hope will satisfy the conditions, laws, and principles of the problem at hand.
Step 3 : Define Gradient - Unknown and Constitution Relations 1. This step depends on the type of the problem. Various disciple processes different unknown attribution. Solid mechanics Fluid mechanics Heat transfer (Thermal)
Solid mechanics x direction of displacement du Gradient Unknown ε = dx x strain-displacement σ = E ε x x x Constitution Relations Hook s law stress-strain
Fluid mechanics Fluid flow (poiseuille flow) in one dimension φ : fluid head or potential = p/γ + z k x : coefficient of permeability d φ g = Gradient Unknown dx x gradient-fluid head v=-kg x x x Constitution Relations Darcy s law velocity-gradient
Heat transfer Heat flow in one dimension k x : thermal conductivity q x : rate of heat flow (Btu/W or W) dt T = Gradient Unknown dx x gradient-temperature q = -kat x x x Constitution Relations Newton s cooling law Rate of heat flow-gradient
Step 4 : Derive Element Equations A number of alternatives are available for the derivation of element equations. Two two most commonly used are : The energy methods Principle of stationary potential energies Principle of stationary complementary energies Reissner s mixed principle Hybrid formulations Weighted residual methods Collocation Method Subdomain Method Galerkin s method Least-squares Method
The energy methods - These procedures are based on the ideas of finding consistent states of bodies or structures associated values of a scalar quantity assumed by the loaded bodies. In engineering, usually this quantity is a measure of energy or work. - In case of stress-deformation analysis, we define the function F to be the potential energy in a body under load. - The potential energy is defined as the sum of the internal strain energy U and the potential of the external loads W p. Π = U W p
Principle of minimum potential energy - Stationary value : the term stationary can imply a maximum, minimum point of a function F(x). F local maximum F(x) local minimum x - to find the point of stationary value, we have df dx = 0
Principle of minimum potential energy - The potential energy is function of unknown nodes values Π=Π ( u u Lu ),, n 1 2 - to find the point of stationary value, we have Π Π Π = 0 = 0 L = 0 u1 u2 u n
Weighted residual methods - The weighted residual methods are based on assuming an approximate solution for the governing differential equations. - The assumed solution must satisfy the initial and boundary conditions of the given problem. - Because the assumed solution is not exact, substitution of the solution into the differential equations will lead to some residuals or errors. - Each residual method requires the error to vanish over some selected interval or at some points.
Weighted residual methods Define 1-D column problem with A 1 = 0.25 in, A 2 = 0.125 in, L = 10 in, thickness = 0.125 in, P = 1000 lb, E = 10.4 10 6 lb/in 2 y A 1 L The governing differential equations P du EA( y) = 0 dy u(y) A 2 P Assume an approximate solution uy ( ) = Cy+ Cy + Cy 2 3 1 2 3
Weighted residual methods (continued) Since A A = + L du dy = + + 2 1 Ay ( ) A1 y t 2 ( C1 2C2y 3 C3y ) The residual error function will be 2 Re s/ E= (0.25-0.0125 y)( C1+ 2Cy 2 + 3 Cy 3 ) - 96.154 10-6
Collocation Method The error function is force to be zero at as many points as there are unknowns coefficients. We select y = L/3, y = 2L/3, y = L 2 10 10 10 6 0.25 0.0125 C1+ 2C2 + 3C3 96.154 10 = 0 3 3 3 2 20 20 20 6 0.25 0.0125 C1+ 2C2 + 3C3 96.154 10 = 0 3 3 3 ( ) 1 2 3 ( ( )) C C ( ) C ( ) 2 6 0.25 0.0125 10 + 2 10 + 3 10 96.154 10 = 0 which leads to uy ( ) = (423.08y + 21.65 10 y + 1.154 y ) 10 9 2 3 6
Subdomain Method The integral of the error function over some selected subintervals is forced to be be zero. The number of subintervals chosen must equal the number of unknowns coefficients. In this case we assumed three integrals: L /3 0 2 L /3 L /3 L 2 L /3 Res dy = 0 Res dy = 0 Res dy = 0
Subdomain Method (continued) After manipulation L /3 0 2 L /3 L /3 L 2 L /3 ( )( ) 1 2 3 + + = 2 6 0.25 0.0125y C 2Cy 3Cy 96.154 10 dy 0 ( )( ) 1 2 3 + + = 2 6 0.25 0.0125y C 2C y 3C y 96.154 10 dy 0 ( )( ) 1 2 3 + + = 2 6 0.25 0.0125y C 2Cy 3Cy 96.154 10 dy 0 which leads to uy ( ) = (391.35y + 6.08y + 809.61 10 y ) 10 2 3 3 6
Galerkin Method This Method requires the error to be orthogonal to some weighting functions Φ i, according to the integral of the domain b a Φ i Res dy = 0 i=1,2,3...n The weighting functions are chosen to be members of the approximation solution. uy ( ) = Cy+ Cy + Cy The weighting function 2 3 1 2 3 Φ ( y) = y Φ ( y) = y Φ ( y) = y 2 3 1 2 3
0 Galerkin Method (continued) This leads to L y ( y )( C Cy Cy ) dy 1+ 2 + 3 = 2 6 0.25 0.0125 2 3 96.154 10 0 L y 2 ( y )( C Cy Cy 2) 6 dy 0 1 2 3 0.25 0.0125 + 2 + 3 96.154 10 = 0 ( )( ) 1 2 3 L y 3 y C C y C y 2 6 dy 0 0.25 0.0125 + 2 + 3 96.154 10 = 0 which leads to uy ( ) = (400.64y + 4.01y + 0.935 y ) 10 2 3 6
Least-Squares Method This method requires the error to be minimized with respect to the unknowns coefficients in the assumed solution of the domain, according to the relationship minimize ( b 2 ) Res dy a which leads to b a Res Res dy = 0 C i
Least-Squares Method (continued) Since, the residual error function is Re s/ E= (0.25-0.0125 y)( C+ 2Cy+ 3 Cy ) - 96.154 10 1 2 3 2-6 0 L ( y)( C1 Cy 2 Cy 3 ) 2 6 + + 0.25 0.0125 2 3 96.154 10 ( ) 0.25 0.0125y dy = 0 L 0 0 L ( y)( C ) 1 C2y C3y 2 6 + + 0.25 0.0125 2 3 96.154 10 ( y)( C ) 1 Cy 2 Cy 3 ( y) 0.25 0.0125 2ydy = 0 2 6 + + 0.25 0.0125 2 3 96.154 10 ( y) 2 0.25 0.0125 3y dy 0 =
Least-Squares Method (continued) which leads to uy ( ) = (389.73y + 6.44y + 0.789 y ) 10 2 3 6
The Exact Solution The differential equation is P du A2 A1 EA( y) = 0 and ( ) 1 dy L The exact solution in obtained by integrating A y = A + y t u y Pdy du = 0 0 EA( y) or y Pdy PL A A uy ( ) = = ln A+ y ln A 0 EA( y) Et ( A2 A 1) L 2 1 1 1
Comparison of weighted residual solution Location along the bar Exact method Collect method Subdomain method Galerkin method Least square method y = 0 0 0 0 0 0 y = 2.5 0.001027 0.001076 0.001029 0.001041 0.001027 y = 5 0.002213 0.002259 0.002209 0.002220 0.002208 y = 7.5 0.003615 0.003660 0.003618 0.003624 0.003618 y = 10 0.005333 0.005384 0.005330 0.005342 0.005331
The element equation Use of either of the forgoing methods will leads to equation describing the behavior of an element, which are commonly expressed as [ k]{ u } = { f} [k] = element property matrix, stiffness matrix {u} = vector of unknowns at the element nodes, node displacements {f} = vector of element forcing parameters, nodal forces
Step 5 : Assemble Element to Obtain Global System Once the element equations are established for a generic element in step 4, it will recursively add them together to find the global equations. This assembling process is based on the law of compatibility continuity
Compatibility : plane problem Slope may not be equal i j Equal displacement The displacements of two adjacent or consecutive points must have identical values
Compatibility : bending problem (more severely) Equal slopes or gradients i j Equal displacement Not only the continuity of displacements but also the slopes or the first derivative of displacements are also continuity at adjacent or consecutive points.
The assemblage equations are or = [K]{U}={F} [K] = assemblage property matrix {U} = assemblage vector of nodal unknowns {F} = assemblage vector of nodal forcing parameters
Boundary conditions The global equations [K] {U} = {F} represent the properties of a body or structure. It tell us about the capabilities of the body to withstand applied forces.
How the body or structure will perform its engineering duties will depend on the surrounding and the problems it faces. These aspects can be called constrains or boundary conditions. Boundary conditions are the physical constraints or supports that must exist so that the structure or body can stand in space uniquely.
Boundary conditions are categorized as Essential, forced or geometric boundary conditions: This type of condition is commonly specified in terms of known values on the unknowns on a part of the body or structure. Such as the fixed point without slope or displacement in the body. Natural boundary conditions: This type of condition is commonly specified in terms of the first or second derivative of displacement. Such as the free or simply support end of the beam imply the moment is zero.
Example fluid head Is specified essential B.C. temperature Is specified cylinder Impervious to fluid fluid flux is proportion to the first derivative of fluid head natural B.C. heat flux is proportion to the first derivative of temperature Insulated against heat
The final assemblage equations are = or [ K ]{ U } = { F }
Step 6 : Solve for the Primary Unknowns Once the global system equation is derived, this is a set of linear (or nonlinear) simultaneous algebraic equations, which can be written in a standard familiar form as K U + K U + L+ K U = F 11 1 12 2 1n n 1 K U + K U + L+ K U = F 21 1 22 2 2n n 2 M M K U + K U + L+ K U = F n1 1 n2 2 nn n n
Step 6 : Solve for the Primary Unknowns These equations can be solved using any numerical method. At the end of this step, we have solved for the unknowns (displacements) U 1, U 2, U 3,,U n. These are called primary unknowns because they appear as the first quantities sought in the basic. Also, they are the degree of freedom defined in each node at the beginning - in solid mechanics displacements - in heat transfer temperature - one dimensional flow fluid head
Step 7 : Solve for Derived or Secondary Quantities Very often additional or secondary quantities must be computed from the primary quantities Stress-deformation problems - strains, stresses, moments and shear forces Flow problem - velocities, flow flux Thermal problem - heat flux
Step 8 : Interpretation of Results The results are usually obtained in the form of printed output from the computer. - Plot the values of displacement and stresses along the domain. - Tabulate the results.