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Introduction to Finite Element Analysis
(FEA) or Finite Element Method (FEM)

Finite Element Analysis (FEA) or Finite
Element Method (FEM)
The Finite Element Analysis (FEA) is a numerical method for solving problems of engineering and mathematical physics.
Useful for problems with complicated geometries, loadings, and material properties where analytical solutions can not be obtained. The Purpose of FEA
Analytical Solution




Stress analysis for trusses, beams, and other simple structures are carried out based on dramatic simplification and idealization:
– mass concentrated at the center of gravity
– beam simplified as a line segment (same cross-section)
Design is based on the calculation results of the idealized structure & a large safety factor (1.5-3) given by experience.

FEA


Design geometry is a lot more complex; and the accuracy requirement is a lot higher. We need
– To understand the physical behaviors of a complex object (strength, heat transfer capability, fluid flow, etc.)
– To predict the performance and behavior of the design; to calculate the safety margin; and to identify the weakness of the design accurately; and
– To identify the optimal design with confidence

Brief History
Grew out of aerospace industry
Post-WW II jets, missiles, space flight
Need for light weight structures
Required accurate stress analysis
Paralleled growth of computers

Common FEA Applications
Mechanical/Aerospace/Civil/Automotive
Engineering
Structural/Stress Analysis



Static/Dynamic
Linear/Nonlinear

Fluid Flow
Heat Transfer
Electromagnetic Fields
Soil Mechanics
Acoustics
Biomechanics

Discretization
Complex Object

Simple Analysis

(Material discontinuity,
Complex and arbitrary geometry)

Real
Word

Simplified
(Idealized)
Physical
Model

Mathematical
Model

Discretized
(mesh)
Model

Discretizations
Model body by dividing it into an equivalent system of many smaller bodies or units (finite elements) interconnected at points common to two or more elements
(nodes or nodal points) and/or boundary lines and/or surfaces.

Elements & Nodes - Nodal Quantity

Feature
Obtain a set of algebraic equations to solve for unknown (first) nodal quantity
(displacement).
Secondary quantities (stresses and strains) are expressed in terms of nodal values of primary quantity

Object

Elements

Nodes

Displacement

Stress
Strain

Examples of FEA – 1D (beams)

Examples of FEA - 2D

Examples of FEA – 3D

Advantages
Irregular Boundaries
General Loads
Different Materials
Boundary Conditions
Variable Element Size
Easy Modification
Dynamics
Nonlinear Problems (Geometric or Material)
The following notes are a summary from “Fundamentals of Finite Element Analysis” by David V. Hutton

Principles of FEA
The finite element method (FEM), or finite element analysis
(FEA), is a computational technique used to obtain approximate solutions of boundary value problems in engineering.
Boundary value problems are also called field problems. The field is the domain of interest and most often represents a physical structure. The field variables are the dependent variables of interest governed by the differential equation.
The boundary conditions are the specified values of the field variables (or related variables such as derivatives) on the boundaries of the field.

For simplicity, at this point, we assume a two-dimensional case with a single field variable φ(x, y) to be determined at every point P(x, y) such that a known governing equation (or equations) is satisfied exactly at every such point.

-A finite element is not a differential element of size dx × dy.
- A node is a specific point in the finite element at which the value of the field variable is to be explicitly calculated.

Shape Functions
The values of the field variable computed at the nodes are used to approximate the values at non-nodal points (that is, in the element interior) by interpolation of the nodal values. For the three-node triangle example, the field variable is described by the approximate relation φ(x, y) = N1(x, y) φ1 + N2(x, y) φ2 + N3(x, y) φ3 where φ1, φ2, and φ3 are the values of the field variable at the nodes, and
N1, N2, and N3 are the interpolation functions, also known as shape functions or blending functions.
In the finite element approach, the nodal values of the field variable are treated as unknown constants that are to be determined. The interpolation functions are most often polynomial forms of the independent variables, derived to satisfy certain required conditions at the nodes.
The interpolation functions are predetermined, known functions of the independent variables; and these functions describe the variation of the field variable within the finite element.

Degrees of Freedom
Again a two-dimensional case with a single field variable φ(x, y). The triangular element described is said to have 3 degrees of freedom, as three nodal values of the field variable are required to describe the field variable everywhere in the element (scalar).

φ(x, y) = N1(x, y) φ1 + N2(x, y) φ2 + N3(x, y) φ3
In general, the number of degrees of freedom associated with a finite element is equal to the product of the number of nodes and the number of values of the field variable (and possibly its derivatives) that must be computed at each node.

A GENERAL PROCEDURE FOR
FINITE ELEMENT ANALYSIS
• Preprocessing







Define the geometric domain of the problem.
Define the element type(s) to be used (Chapter 6).
Define the material properties of the elements.
Define the geometric properties of the elements (length, area, and the like).
Define the element connectivities (mesh the model).
Define the physical constraints (boundary conditions). Define the loadings.

• Solution



computes the unknown values of the primary field variable(s) computed values are then used by back substitution to compute additional, derived variables, such as reaction forces, element stresses, and heat flow.

• Postprocessing


Postprocessor software contains sophisticated routines used for sorting, printing, and plotting selected results from a finite element solution.

Stiffness Matrix
The primary characteristics of a finite element are embodied in the element stiffness matrix. For a structural finite element, the stiffness matrix contains the geometric and material behavior information that indicates the resistance of the element to deformation when subjected to loading. Such deformation may include axial, bending, shear, and torsional effects. For finite elements used in nonstructural analyses, such as fluid flow and heat transfer, the term stiffness matrix is also used, since the matrix represents the resistance of the element to change when subjected to external influences.

LINEAR SPRING AS A FINITE ELEMENT
A linear elastic spring is a mechanical device capable of supporting axial loading only, and the elongation or contraction of the spring is directly proportional to the applied axial load. The constant of proportionality between deformation and load is referred to as the spring constant, spring rate, or spring stiffness k, and has units of force per unit length. As an elastic spring supports axial loading only, we select an element coordinate system (also known as a local coordinate system) as an x axis oriented along the length of the spring, as shown.

Assuming that both the nodal displacements are zero when the spring is undeformed, the net spring deformation is given by δ= u2 − u1 and the resultant axial force in the spring is f = kδ = k(u2 − u1)
For equilibrium, f1 + f2 = 0 or f1 = − f2,
Then, in terms of the applied nodal forces as f1 = −k(u2 − u1) f2 = k(u2 − u1) which can be expressed in matrix form as or where

Stiffness matrix for one spring element is defined as the element stiffness matrix in the element coordinate system (or local system), {u} is the column matrix (vector) of nodal displacements, and { f } is the column matrix (vector) of element nodal forces.

with

known

{F} = [K] {X}

unknown

The equation shows that the element stiffness matrix for the linear spring element is a 2 × 2 matrix. This corresponds to the fact that the element exhibits two nodal displacements (or degrees of freedom) and that the two displacements are not independent (that is, the body is continuous and elastic).
Furthermore, the matrix is symmetric. This is a consequence of the symmetry of the forces (equal and opposite to ensure equilibrium).
Also the matrix is singular and therefore not invertible. That is because the problem as defined is incomplete and does not have a solution: boundary conditions are required.

SYSTEM OF TWO SPRINGS

These are external forces

Free body diagrams:

These are internal forces

Writing the equations for each spring in matrix form:
Superscript refers to element

To begin assembling the equilibrium equations describing the behavior of the system of two springs, the displacement compatibility conditions, which relate element displacements to system displacements, are written as:

And therefore: Here, we use the notation f ( j )i to represent the force exerted on element j at node i.

Expand each equation in matrix form:

Summing member by member:

Next, we refer to the free-body diagrams of each of the three nodes:

Final form:

(1)
Where the stiffness matrix:

Note that the system stiffness matrix is:
(1) symmetric, as is the case with all linear systems referred to orthogonal coordinate systems; (2) singular, since no constraints are applied to prevent rigid body motion of the system; (3) the system matrix is simply a superposition of the individual element stiffness matrices with proper assignment of element nodal displacements and associated stiffness coefficients to system nodal displacements.

(first nodal quantity)

(second nodal quantities)

Example with Boundary Conditions
Consider the two element system as described before where Node 1 is attached to a fixed support, yielding the displacement constraint U1 = 0, k1= 50 lb/in, k2= 75 lb/in,
F2= F3= 75 lb for these conditions determine nodal displacements U2 and U3.

Substituting the specified values into (1) we have:
Due to boundary condition

Example with Boundary Conditions
Because of the constraint of zero displacement at node 1, nodal force F1 becomes an unknown reaction force. Formally, the first algebraic equation represented in this matrix equation becomes:
−50U2 = F1 and this is known as a constraint equation, as it represents the equilibrium condition of a node at which the displacement is constrained. The second and third equations become which can be solved to obtain U2 = 3 in. and U3 = 4 in. Note that the matrix equations governing the unknown displacements are obtained by simply striking out the first row and column of the 3 × 3 matrix system, since the constrained displacement is zero (homogeneous). If the displacement boundary condition is not equal to zero (nonhomogeneous) then this is not possible and the matrices need to be manipulated differently (partitioning).

Truss Element
The spring element is also often used to represent the elastic nature of supports for more complicated systems. A more generally applicable, yet similar, element is an elastic bar subjected to axial forces only. This element, which we simply call a bar or truss element, is particularly useful in the analysis of both two- and threedimensional frame or truss structures. Formulation of the finite element characteristics of an elastic bar element is based on the following assumptions:
1.The bar is geometrically straight.
2.The material obeys Hooke’s law.
3.Forces are applied only at the ends of the bar.
4.The bar supports axial loading only; bending, torsion, and shear are not transmitted to the element via the nature of its connections to other elements.

Truss Element Stiffness Matrix
Let’s obtain an expression for the stiffness matrix K for the beam element. Recall from elementary strength of materials that the deflection δ of an elastic bar of length L and uniform cross-sectional area A when subjected to axial load P :

where E is the modulus of elasticity of the material. Then the equivalent spring constant k:

Therefore the stiffness matrix for one element is:

And the equilibrium equation in matrix form:

Truss Element Blending Function
An elastic bar of length L to which is affixed a uniaxial coordinate system x with its origin arbitrarily placed at the left end. This is the element coordinate system or reference frame. Denoting axial displacement at any position along the length of the bar as u(x), we define nodes 1 and 2 at each end as shown and introduce the nodal displacements: u1=u (x=0) and u2 = u(x = L).
Thus, we have the continuous field variable u(x), which is to be expressed
(approximately) in terms of two nodal variables u1 and u2. To accomplish this discretization, we assume the existence of interpolation functions N1(x) and N2(x)
(also known as shape or blending functions) such that u(x) = N1(x)u1 + N2(x)u2

Truss Element Blending Function
To determine the interpolation functions, we require that the boundary values of u(x)
(the nodal displacements) be identically satisfied by the discretization such that: u1=u (x=0) and u2 = u(x = L). lead to the following boundary (nodal) conditions:
N1(0) = 1 N2(0) = 0
N1(L) = 0 N2(L) = 1
As we have two conditions that must be satisfied by each of two one-dimensional functions, the simplest forms for the interpolation functions are polynomial forms:
N1(x) = a0 + a1x
N2(x) = b0 + b1x

Truss Element Blending Function
Where the polynomial coefficients are to be determined via satisfaction of the boundary (nodal) conditions. Application of conditions yields a0 = 1, b0 = 0 , therefore a1 = −(1/L) and b1 = x/L. Therefore, the interpolation functions are
N1(x) = 1 − x/L
N2(x) = x/L

total interpolation

Therefore the final expression of the blending function is: u2 u(x) = (1 − x/L)u1 + (x/L)u2
And in matrix form:

u1 u1 contribution u2 contribution

This is the displacement field in terms of nodal variables.

Truss Element Example
Figure depicts a tapered elastic bar subjected to an applied tensile load P at one end and attached to a fixed support at the other end. The cross-sectional area varies linearly from A0 at the fixed support at x = 0 to A0/2 at x = L. Calculate the displacement of the end of the bar (a) by modeling the bar as a single element having cross-sectional area equal to the area of the actual bar at its midpoint along the length, (b) using two bar elements of equal length and similarly evaluating the area at the midpoint of each, and compare to the exact solution.

Truss Element Example Solution a)
For a single element, the cross-sectional area is 3A0/4 and the element
“spring constant” and element equation are:

and

Applying the constraint condition U1 = 0, we find for U2 as the displacement at x = L

Truss Element Example Solution b)
Two elements of equal length L/2 with associated nodal displacements. For element
1, A1 = 7A0/8 so

while for element 2, we have

Since no load is applied at the center of the bar, the equilibrium equations for the system of two elements is:

Applying the constraint condition U1 = 0 results in

Truss Element Example Solution b)
Adding the two equations gives

and substituting this result into the first equation results in

Comparing the displacement for the three solution at x = L:
a)
b)
c) Exact solution

Truss Element Example Solution Comparison
Deflection

u(x)

Node

Shape function for interpolated values: u(x) = (1 − x/L)u1 + (x/L)u2

Truss Element Example Solution Comparison
Stress
Node

For stress results are much different, discontinuous for FEA and highly dependent on number of nodes

Beam Element
The usual assumptions of elementary beam theory are applicable here:
1. The beam is loaded only in the y direction.
2. Deflections of the beam are small in comparison to the characteristic dimensions of the beam.
3. The material of the beam is linearly elastic, isotropic, and homogeneous. The beam is prismatic and the cross section has an axis of symmetry in the plane of bending.

Beam Element

The equation shows that the element stiffness matrix for the beam element is a 4 ×
4 matrix. This corresponds to the fact that the element exhibits four degrees of freedom and that the displacements are not independent (that is, the body is continuous and elastic).
Furthermore, the matrix is symmetric. This is a consequence of the symmetry of the forces and moments (equal and opposite to ensure equilibrium).
Also the matrix is singular and therefore not invertible. That is because the problem as defined is incomplete and does not have a solution: boundary conditions are required.

Beam Element Shape Function and
Stiffness Matrix
Shape function:

With
And the Stiffness Matrix:

Way of Stacking Blocks/Elements
• Compatibility requirement: ensures that the
“displacements” at the shared node of adjacent elements are equal.
• Equilibrium requirement: ensures that elemental forces and the external forces applied to the system nodes are in equilibrium.
• Boundary conditions: ensures the system satisfy the boundary constraints and so on.

Limitations of Regular
FEA
Software
• Unable to handle geometrically nonlinear - large deformation problems: shells, rubber, etc.

Interpolation Functions for General
Element Formulation
In finite element analysis, solution accuracy is judged in terms of convergence as the element “mesh” is refined.
There are two major methods of mesh refinement.
In the first, known as h-refinement, mesh refinement refers to the process of increasing the number of elements used to model a given domain, consequently, reducing individual element size.
In the second method, p-refinement, element size is unchanged but the order of the polynomials used as interpolation functions is increased.
The objective of mesh refinement in either method is to obtain sequential solutions that exhibit asymptotic convergence to values representing the exact solution.

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...in this guide may not be included in your vehicle. For easy reference, keep this guide with your Owner Manual in your glove box. F denotes optional equipment www.chevy.com INSTRUMENT PANEL Turn Signal Lever/Driver Information Center Controls Driver Information Center Windshield Wipers Lever Audio Steering Wheel ControlsF Exterior/Interior Lamps Control Cruise Control Hood Release Tilt/Telescopic Buttons Lever Steering Wheel Lever Safety Belt Reminder Check Engine Low Tire Pressure Brake System Warning Antilock Brake System Warning 2 Charging System Oil Pressure Airbag Readiness Security Fog Lamps Cruise Control Set Traction Control Off StabiliTrak/Traction Control Off StabiliTrak/Traction Control Warning Lights On Climate Controls/ Heated Seat ButtonsF Audio System Hazard Warning Flashers Button/ Power Door Locks Button Auxiliary GaugesF StabiliTrak Button Accessory Power Outlet* * The accessory power outlet does not operate once the key is removed from the ignition and the driver’s door is opened. Note: Refer to your Owner Manual to learn about the information being relayed by the lights, gauges and indicators of the instrument cluster, as well as what to do to ensure safety and prevent damage. See Instruments and Controls in your Owner Manual. 3 REMOTE KEYLESS ENTRY TRANSMITTER Unlock Press to unlock the driver’s door only or all doors. The unlock setting can be changed in the Vehicle Settings menu by using the audio system...

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