Basics of Finite Element Analysis

What is FEA (Finite Element Analysis)?

A complex problem is divided into a smaller and simpler problems that can be solved by using the existing knowledge of mechanics of materials and mathematical tools

Why FEA ?
Modern mechanical design involves complicated shapes, sometimes made of different materials that as a whole cannot be solved by existing mathematical tools. Engineers need the FEA to evaluate their designs

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Basics of Finite Element Analysis

The process of dividing the model into small pieces is called meshing. The behavior of each element is well-known under all possible support and load scenarios. The finite element method uses elements with different shapes. Elements share common points called nodes.

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History of Finite Element Analysis

Finite Element Analysis (FEA) was first developed in 1943 by R. Courant, who utilized the Ritz method of numerical analysis and minimization of variational calculus. A paper published in 1956 by M. J. Turner, R. W. Clough, H. C. Martin, and L. J. Topp established a broader definition of numerical analysis. The paper centered on the "stiffness and deflection of complex structures".

By the early 70's, FEA was limited to expensive mainframe computers generally owned by the aeronautics, automotive, defense, and nuclear industries. Since the rapid decline in the cost of computers and the phenomenal increase in computing power, FEA has been developed to an incredible precision.
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Basics of Finite Element Analysis

FEA Applications Evaluate the stress or temperature distribution in a mechanical component. Perform deflection (Stiffness) analysis. Analyze the kinematics or dynamic response Perform vibration analysis Perform fatigue analysis Check Buckling failure Perform drop test
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The goal is to optimize for material

Basics of Finite Element Analysis

Consider a cantilever beam shown.

Finite element analysis starts with an approximation of the region of interest into a number of meshes (2D or 3D elements). Each mesh is connected to associated nodes (black dots) and thus becomes a finite element.

Node
Element
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Basics of Finite Element Analysis

After approximating the object by finite elements, each node is associated with the unknowns to be solved. For the cantilever beam the displacements in x and y directions would be the unknowns (2D mesh). This implies that every node has two degrees of freedom and the solution process has to solve 2n degrees of freedom, n is the number of nodes. Displacement
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Strain
Mechanical Engineering Dept

Stress
Stress & Strain relationship
7

Partial derivatives

Example a plate under load

Derive and solve the system of equations for a plate loaded as shown. Plate thickness is 1 cm and the applied load Py is constant
Py .

using two triangular elements,

Reaction forces

U1 thru U8, displacements in x and y directions

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Example a plate under load

Displacement within the triangular element (2D) with three nodes can be assumed to be linear. u = 1 + 2 x + 3 y v = 1 + 2 x + 3 y

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Example a plate under load

Displacement for each node,
Node 1
Node 2 Node 3

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Example a plate under load

Solve the equations simultaneously for and ,

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Example a plate under load

Substitute the location of the nodes: x1= 0, y1= 0 (node 1), x2=10, y2= 0 (node 2) and x3= 0, y3=4 (node 3) to obtain displacements u and v for element 1 Evaluate the constants a, b, and c (3)
10 4 0 0 Element 1 (1) (2)

Calculations: 2a = 40 a1 = 40, a2 = 0, a3 = 0 b1 = - 4, b2 = 4, b3 = 0
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c1 = -10, c2 = 0, c3 = 10
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Example
40 0
40 0

2a = 40 a1 = 40, a2 = 0, a3 = 0 b1 = - 4, b2 = 4, b3 = 0 c1 = -10, c2 = 0, c3 = 10 Calculations

Change of notations

1 = (1)U1

u 1 = U1, u2 = U3, u3 = U5, v1 = U2, v2 = U4, v3 = U6

2 = -(1/10)U1 + (1/10)U3 3 = -(1/4) U1+ (1/4) U5 1 = (1)U2 2 = -(1/10)U2 + (1/10) U4 3 = -(1/4) U2+ (1/4) U6
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Example
Substitute and to obtain displacements u and v for element 1.

1 = (1)U1 2 = -(1/10)U1 + (1/10)U3 3 = -(1/4) U1+ (1/4) U5 1 = (1)U2 2 = -(1/10)U2 + (1/10) U4 3 = -(1/4) U2+ (1/4) U6

u = 1 + 2 x + 3 y v = 1 + 2 x + 3 y

Calculation:

u1 = U1 + [-1/10 (U1) + (1/10) U3] x + [-(1/4) U1+ (1/4) U5 ] y v1 = U2 + [-1/10 (U2) + (1/10) U4] x + [-(1/4) U2+ (1/4) U6 ] y
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Example
Rewriting the equations in the matrix form, u1 = U1 + [-1/10 (U1) + (1/10) U3] x + [-(1/4) U1+ (1/4) U5 ] y v1 = U2 + [-1/10 (U2) + (1/10) U4] x + [-(1/4) U2+ (1/4) U6 ] y

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Example

Similarly the displacements within element 2 can be expressed as,

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Example
The next step is to determine the strains using 2D strain-

displacement relations,

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Example
Differentiate the displacement equation to obtain the strain
u1 = U1 + [-1/10(U1) + (1/10) U3] x + [-(1/4) U1+ (1/4) U5 ] y v1 = U2 + [-1/10(U2) + (1/10) U4] x + [-(1/4) U2+ (1/4) U6 ] y

1st element

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Example
Element 2 2nd element

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Normal & Shear components of stress (3D)

Normal stress is perpendicular to the cross section, (sigma). Shear stress is parallel to the cross section, (tau). y

First subscript indicates the axis that is perpendicular to the face

3D Element

yx yz zy zx
z

xy
xy
Second subscript indicates the positive direction of the shear stress

xz

x
x Due to equilibrium condition;

xy = yx zx = xz zy = yz

State of Stress
Three dimensional stress matrix

Two dimensional, Plane Stress

Stress & Strain Relationship

Uniaxial state of stress

x 0 , y = 0 , z = 0

x = (x / E ), y = - x ,

z = - x

Using the three dimensional (triaxial state of stress) stress strain relations for homogeneous, isotropic material and plane-stress,

x = (x / E ) - (y) - (z) = (x / E ) - (y / E ) - (z / E )
y = ( y / E ) - ( x ) - ( z ) = ( y / E ) - ( x / E ) - ( z / E ) z = (z / E ) - (x) - (y) = (z / E ) - (x / E ) - (y / E )
Stresses in terms of strains

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Stress & Strain Relationship

There are many practical problems where the stress in the z-direction is zero, this is referred to as the state of Plane Stress.

Shear stress xy = xy G

E G= 2(1 + )

Matrix form

FEA Results - Principal Stresses

Normal stresses on planes with no shear stresses are maximum and they are called principal stresses 1, 2, and 3, where 1 > 2 > 3

The three non-imaginary roots are the principal stresses

2 2 2 3 - (x + y + z) 2 + (x y + x z + y z - xy - xz - yz ) 2 2 2 (x y z - 2 xy xz yz - x yz - y xz - z xy )=0

3 - (x + y) 2 + (x y - xy) = 0
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Plane stress, two principal stresses, 3 = 0

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Displacement

Strain

Stress
Stress & Strain relationship

Partial derivatives

Material

Ductile Yield strength of the material is used in

designing components

Brittle Ultimate strength in tension and

compression is used in designing components

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Failure Theories Ductile Materials Maximum Shear Stress

Yield strength of a material is used to design components made of ductile material

Maximum shear stress theory (Tresca 1886)

(max )component > ( )obtained from a tension test at the yield point
= Sy

Failure

=
= Sy

Sy 2 To avoid failure

(max )component <

Sy
2

max =
=Sy

Sy 2n

n = Safety factor

Design equation
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Failure Theories Ductile Material von Mises Stress

Distortion energy theory (von Mises-Hencky)
Simple tension test (Sy)t

t
(Sy)h >> (Sy)t

Hydrostatic state of stress (Sy)h

h h h

Distortion contributes to failure much more than change in volume.

t
(total strain energy) (strain energy due to hydrostatic stress) = strain energy due to angular distortion > strain energy obtained from a tension test at the yield point failure
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von Mises Stress

3D case, to avoid failure (1 2)2 + (1 3)2 + (2 3)2
2

<

Sy

2D case (plane stress), 3 = 0 = (1 12 + 2 )

2 2

< Sy

Sy n

Design equation

Where is von Mises stress

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Failure Theories Brittle Materials Maximum Principal Stress

Characteristics of brittle materials;
1. Sut Syt 2. Suc >> Sut
3. Percent elongation < 5%

Perform two tests, one in compression and one in tension, draw the Mohrs circles for both tests. 2 1 Sut

Suc

Stress state

Tension test

Compression test

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Failure envelope The component is safe if the state of stress falls inside the 1 > 2 and 3 = 0 failure envelope.
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Failure Theories Brittle Materials

Modified Coulomb-Mohr theory

2 or 3
Sut
Safe Safe

2 or 3
Sut
Sut

Suc -Sut
Safe

1
-Sut

I
II III

Sut

Safe

Suc

Suc

Cast iron data

Three design zones

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Failure Theories Brittle Materials

Zone I
1 > 0 , 2 > 0 and 1 > 2

2
Sut

1 = n

Sut

Design equation

Sut

II
-Sut

Zone II
1 > 0 , 2 < 0 and 2 < Sut

III
Sut
-Suc

1 = n

Design equation

Zone III
1 > 0 , 2 < 0 and 2 > Sut
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1 (

1 Sut

1 Suc )

2
Suc

1 n

Design equation

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Formulation of the Finite Element Method

The classical finite element analysis code (h version)
The system equations for solid and structural mechanics problems are derived using the principle of virtual displacement and work (Bathe, 1982).

The method of weighted residuals (Galerkin Method)

weighted residuals are used as one method of finite element formulation starting from the governing differential equation.

Potential Energy and Equilibrium; The Rayleigh-Ritz Method

Involves the construction of assumed displacement field. Uses the total potential energy for an elastic body
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Formulation of the Finite Element Method

Lets denote the displacements of any point (x, y, z) of the object from the unloaded configuration as UT UT = [U(x, y, z) V(x, y, z) W(x, y,z)] The displacement U causes the strains

T = [x y z xy yz zx ]
and the corresponding stresses

T = [x y z xy yz zx ]
The goal is to calculate displacement, strains, and stresses from the given external forces.
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Formulation of the Finite Element Method

fxB f B= fyB fzB f S=

fxS fyS fzS f i=

fxi fyi fzi

f B Body forces (forces distributed over the volume of the body: (gravitational forces, inertia, or magnetic)
f S surface forces (pressure of one body on another, or hydrostatic pressure) f i Concentrated external forces
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Formulation of the Finite Element Method

Equilibrium condition and principle of virtual displacements

V dV =
T
Internal work

U f dV +
Work done by body forces

f dS +

iT

Work done by surface forces

Work done by external forces

The left side represents the internal virtual work done, and the right side represents the external work done by the actual forces as they go through the virtual displacement. The above equation is used to generate finite element equations. And by approximating the object as an assemblage of discrete finite elements, these elements are interconnected at nodal points Us denotes the displacement due to surface forces Ui denotes the displacement due to point forces
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Formulation of the Finite Element Method

Displacement interpolation matrix

The displacement at any point measured with respect to a local coordinate system for an element are assumed to be a function of the displacement at the nodes.

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H (m) is the displacement interpolation matrix

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Formulation of the Finite Element Method

strain-displacement matrix

B (m) is the rows of the strain-displacement matrix

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Formulation of the Finite Element Method

Elasticity matrix

Matrix form

C (m) is the elasticity matrix of element m and I(m) are the elements initial stresses. The elasticity matrix relates strains to stress.
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Displacement at any element

Displacement at any node

Displacement interpolation matrix

Strain-displacement matrix

Elasticity matrix
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Initial stress (residual stress)

39

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Formulation of the Finite Element Method

The formula for the principle of virtual displacements can be rewritten as the sum of integration over the volume and areas for each finite element,

Where m varies from 1 to the total number of elements

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Formulation of the Finite Element Method

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Formulation of the Finite Element Method

The equilibrium equation can be expressed using matrix notations for m elements.

where

B(m) Represents the rows of the strain displacement matrix C(m) Elasticity matrix of element m H(m) Displacement interpolation matrix U Vector of the three global displacement components at all nodes F Vector of the external concentrated forces applied to the nodes
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Formulation of the Finite Element Method

The above equation can be rewritten as follows,

The above equation describes the static equilibrium problem. K is the stiffness matrix.
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Continuing the example

B(m) - Represents the rows of the strain displacement matrix

C(m) - Elasticity matrix of element m

y x dx

dA = y dx
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y=4-

4 x 10 44

Example

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Example
Calculating the stiffness matrix for element 2.

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Example
The stiffness of the structure as a whole is obtained by combing the two matrices, K = K1 +K2

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Example
KU = R
The load vector R, equals Rc because only concentrated loads act on the nodes.

R=

where Py is the known external force and F1x, F1y, F3x, and F3y are the unknown reaction forces at the supports.
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Example
The following matrix equation can be solved for nodal point displacements

KU = R

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Example
The solution can be obtained by applying the boundary conditions

No deflection at the supports

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Example
The equation can be divided into two parts,

The first equation can be solved for the unknown nodal displacements, U3, U4, U7, and U8. And substituting these values into the second equation to obtain unknown reaction forces, F1x, F1y, F3x, and F3y

Once the nodal displacements have been obtained, the strains and stresses can be calculated.
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Finite Element Analysis

FEA is a mathematical representation of a physical system and the solution of that mathematical representation

FEA requires three steps

Pre-Processing Solving Matrix (solver) Post-Processing

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FEA Pre-Processing
Mesh Mesh is your way of communicating geometry to the solver, the accuracy of the solution is primarily dependent on the quality of the mesh. The better the mesh looks, the more accurate the solution is.

A good-looking mesh should have well-shaped elements (proportional), and the transition between densities should be smooth and gradual without skinny, distorted elements.
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FEA Pre-Processing
The mesh elements supported by most finite-element codes:

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FEA Pre-Processing Elements

Beam Elements
Beam elements typically fall into two categories; able to transmit moments or not able to transmit moments. Rod (bar or truss) elements cannot carry moments.

Entire length of a modeled component can be captured with a single element. This member can transmit axial loads only and can be defined simply by a material and cross sectional area.
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FEA Pre-Processing Elements

The most general line element is a beam.

(b) and (c) are higher order line elements.

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FEA Pre-Processing Elements

Plate and Shell Modeling
Plate and shell are used interchangeably and refer to surfacelike elements used to represent thin-walled structures.

A quadrilateral mesh is usually more accurate than a mesh of similar density based on triangles. Triangles are acceptable in regions of gradual transitions.
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FEA Pre-Processing Elements

Solid Element Modeling

Tetrahedral (tet) mesh is the only generally accepted means to fill a volume, used as automesh element by many FEA codes.
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10-node Quadratic
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FEA Pre-Processing - meshing

The mesh transition from .05 to .5 element size without control of transition (a) creates irregular mesh around the hole which will yield disappointing results.

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Effect of Mesh Element

Model 1 produces Von Mises stress of 18,000 psi. It uses a first-order solid tetrahedral element One element is placed across the thickness of the plate in bending, not able to handle the positive and negative bending. The elements are highly distorted

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Effect of Mesh Element

Model 2 produces maximum Von Mises stress of 32,000 psi. The mesh on model 2 is similar to that on model 1 but uses second-order solid tetrahedral elements. the mesh is too coarse to model stress distribution correctly or detect stress concentrations. Some elements are still highly distorted.

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Effect of Mesh Element

Model 3 produces maximum Von Mises stress of 49,000 psi This model uses second-order solid tetrahedral elements and has finer mesh to model stress distributions properly. Stress will increase with each mesh refinement. Thus, the process of mesh refining and solving the refined model must continue until the increase in stress between two consecutive iterations becomes sufficiently small. Only then can results be accepted as final.

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Max stress = 18,000 psi.

Max stress = 32,000 psi.

1st order tetrahedral coarse mesh

Max stress = 49,000 psi.

2nd order tetrahedral coarse mesh 2st order tetrahedral fine mesh
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CAD Modeling for FEA

CAD and FEA activities should be coordinated at the early stages of the design process to minimize the duplication of effort. There are four situations

CAD models prepared without consideration of FEA needs. Analytical geometry developed by or for analyst for sole purpose of FEA. CAD models prepared by the design group for eventual FEA. CAD models unsuitable for use in analysis due to the amount of rework required.
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CAD Modeling for FEA

Solid chunky parts (thick-walled, low aspect ratio)
parts mesh cleanly directly off CAD models.

Clean geometry
geometrical features must not prevent the mesh from being created. The model should not include buried features.

Parent-child relationships
parametric modeling allows defining features off other CAD features.

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CAD Modeling for FEA

Short edges and Sliver surfaces
Short edges and sliver surfaces usually accompany each other and on large faces can cause highly distorted elements or a failed mesh.

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CAD Modeling for FEA Sliver Surfaces

The rounded rib on the inside of the piston has a thickness of .30 and a radius of .145, as a result a flat surface of .01 by 2.5 is created. A mesh size of .05 is required to avoid distorted elements. This results in a 290,000 nodes. If the radius is increased to .15, a mesh size of .12 is sufficient which results in 33,500 nodes. Flat surface

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CAD Modeling for FEA

Sliver surface caused by misaligned features.

Fillet across shallow angle Sliver surface caused by a slightly undersized fillet
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CAD Modeling for FEA Sliver edge

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CAD Modeling for FEA Sliver Surfaces

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CAD Modeling for FEA Sliver Surfaces

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Guidelines for Geometry Planning

Delay inclusion of fillets and chamfers as long as possible. Where possible, try to use permanent datum as a reference to minimize dependencies. Avoid using fillet or draft edges as references for other features (parent-child relationship) Never bury a feature in your model. Delete or redefine unwanted or incorrect features.

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Guidelines for Part Simplification

In general, features listed below could be considered for suppression. But, consider the impact before suppression.

Outside corner breaks or rounds. Small inside fillets far from areas of interest. Screw threads or spline features unless they are specifically being studied. Small holes and slots outside the load path. Decorative or identification features. Large sections of geometry that are essentially decoupled from the behavior of interested section.
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Model with full detail

Model with details suppressed

Elements = 66727 No. of equations = 311421

Elements = 2565 No. of equations = 14889

Guidelines for Part Simplification

Fillet added to the rib

Holes removed

Fillet removed

Ribs needed for casting removed

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Ken Youssefi

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CAD Modeling for FEA Model Conversion

Try to use the same CAD system for all components in design. When the above is not possible, translate geometry through kernel based tools such as ACIS or Parasolids. Using standards based (IGES, DXF, or VDA) translations may lead to problem. Visually inspect the quality of imported geometry. Avoid modification of the imported geometry in a second CAD system.

Use the original geometry for analysis. If not possible, use a translation directly from the original model.
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Example of a solid model corrupted by IGES transfer

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FEA Pre-Processing
Material Properties
The only material properties that are generally required by an isotropic, linear static FEA are: Youngs modulus (E), Poissons ratio (v), shear modulus (G), and yield strength (or ultimate strength). Strength is needed if the program provides safety factor or performance result.

G = E / 2(1+v)
Provide only two of the three properties. Thermal expansion and simulation analysis require coefficient of thermal expansion, conductivity and specific heat values.

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FEA Pre-Processing
Nonlinear Material Properties
A multi-linear model requires the input of stress-strain data pairs to essentially communicate the stress-strain curve from testing to the FE model Highly deformable, low stiffness, incompressible materials, such as rubber and other synthetic elastomers require distortional and volumetric constants or a more complete set of tensile, compressive, and shear force versus stretch curve.

A creep analysis requires time and temperature dependent creep properties. Plastic parts are extremely sensitive to this phenomenon
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FEA Pre-Processing
Comments If you are selecting the property set from the codes library, be aware of the assumptions made with this selection. Their properties hold constant throughout the assigned entity. Average values are used (variation could be up to 15%).

Localized changes due to heat or other processing effects are not accounted for.
Any impurities present in the parent material are neglected.

The assumption is that there are no defects in the material

If possible, obtain material property values specific to the application under analysis.
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FEA Pre-Processing
Boundary Conditions (Loads and Constraints)

In FEA, the name of the game is boundary condition, that is calculating the load and figuring out constraints that each component experiences in its working environment.

garbage in, garbage out

The results of FEA should include a complete discussion of the boundary conditions.

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Boundary Conditions
Loads Loads are used to represent inputs to the system. They can be in the forms of forces, moments (torque), pressures, temperature, or accelerations. Constraints Constraints are used as reactions to the applied loads. Constraints can resist translational or rotational deformation induced by applied loads.

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Boundary Conditions
Linear Static Analysis
Boundary conditions are assumed constant from application to final deformation of system and all loads are applied gradually to their full magnitude.

Dynamic Analysis
The boundary conditions (Loads) vary with time.

Non-linear Analysis
The orientation and distribution of the boundary conditions vary as displacement of the structure is calculated.

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Boundary Conditions
Degrees of Freedom
Spatial DOFs refer to the three translational and three rotational modes of displacement that are possible for any part in 3D space. A constraint scheme must remove all six DOFs for the analysis to run. Elemental DOFs refer to the ability of each element to transmit or react to a load. The boundary condition cannot load or constrain a DOF that is not supported by the element to which it is applied.

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Boundary Conditions
Constraints and their geometric equivalent in classic beam calculation.
Fixed support

Pin support

Roller support

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Boundary Conditions
A solid face should always have at least three points in contact with the rest of the structure. A solid element should never be constrained by less than three points and only translational DOFs must be fixed.

Accuracy
The choice of boundary conditions has a direct impact on the overall accuracy of the model. Over-constrained model an overly stiff model due to poorly applied constraints.

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Boundary Conditions -Example

Excessive Constraints Model of the chair seat with patches representing the tops of the legs.

Patch 1 Patch 2 Patch 3

Patch 4

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Boundary Conditions -Example

It may appear to be acceptable to constrain each circular patch in vertical translation while leaving the rotational DOFs unconstraint. This causes the seat to behave as if the leg-toseat interfaces were completely fixed. A more realistic constraint scheme would be to pin the center point of each circular patch (translational), allowing the patch to rotate. Each point should be fixed vertically, and horizontal constraints should be selectively applied so that in-plane spatial rotation and rigid body translation is removed without causing excessive constraints.
Patch 1

Patch 2
Patch 3 Patch 4

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Boundary Conditions -Example

Constraining the center point of patch 1 in all 3 translational DOFs. Constraining x and y translations of the center point of patch 2. Constraining z and y translation of the center point of patch 3.

Constraining just the y translation of the center point of patch 4.

This scheme allows inplane translation induced by bending of the seat without rigid body translation or rotation.
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Patch 1 Patch 2 Patch 3 Patch 4

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Legs are fixed to seat

2000 N applied force distributed over the surface.

Use On Flat Face restraint

Fixed legs

In plane rotation is allowed Stress

Stress

Displacement Displacement

One leg is restrained in x,y,z, one in y, one in x,y, one in y,z

Stress Displacement
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All patches (legs) fixed

Displ. = .016 mm

All patches (legs) On Flat Face constrains, in plane rotation

Displ. = .06 mm

One leg is restrained in x,y,z, one in y, one in x,y, one in y,z

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Displ. = .02 mm

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Stress=11.6x107 N/m2

Stress=5.8x107 N/m2

All patches (legs) fixed

Stress=10.4x107 N/m2

All patches (legs) On Flat Face constrains, in plane rotation

One leg is restrained in x,y,z, one in y, one in x,y, one in y,z

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Summary of Pre-Processing
Build the geometry (CAD model for FEA) Prepare the model for meshing (simplify) Create the finite-element mesh Add boundary conditions; loads and constraints Select material or provide properties Specify analysis type (static or dynamic, linear or non-linear, thermal, etc.)
These activities are called finite element modeling.
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Solving the Model - Solver

Once the mesh is complete, and the properties and boundary conditions have been applied, it is time to solve the model. In most cases, this will be the point where you can take a deep breath, push a button and relax while the computer does the work for a change.

Multiple Load and Constraint Cases

In most cases submitting a run with multiple load cases will be faster than running sequential, complete solutions for each load case. Final Model Check

Ken Youssefi

Mechanical Engineering Dept

99

Post-Processing, Displacement Magnitude

Unexpectedly high or low displacements (by order of magnitude) could be caused by an improper definition of load and/or elemental properties.

Ken Youssefi

Mechanical Engineering Dept

100

Post-Processing, Displacement Animation

Animation of the model displacements serves as the best means of visualizing the response of the model to its boundary conditions.

Ken Youssefi

Mechanical Engineering Dept

101

Post-Processing, FEA of a connecting rod

Ken Youssefi

Mechanical Engineering Dept

102

Post-Processing, Stress Results

The magnitude of the stresses should not be entirely unexpected.

Second Mode (Twisting)

First Mode (Bending)

Ken Youssefi

Mechanical Engineering Dept

103

Post-Processing, thermal analysis

Deformation of a duct under thermal load

Ken Youssefi

Mechanical Engineering Dept

104

Deploy Mechanism Assembly Analysis

Displacement

Stress

Can crusher stress analysis

Use finer mesh size

Right click the Mesh icon and choose Failure Diagnostics

Add fillet to the slot edges (.1 in.)

Apply 200 N (45 lb)

Max stress (von Mises) = 43.9 MPa Sy = 96.5 MPa (Al 2014)

Safety factor n = 96.5/43.9 = 2.2 > 2.0

Max deflection 1.13 mm < 2 mm

Set gap to 5 in. Fix the back plate Design requirements Safety factor between 2.0-2.5 and deflection less than 2 mm

Mesh Quality
The ideal shape of a tetrahedral element is a regular tetrahedron with the aspect ratio of 1. Analogously, an equilateral triangle is the ideal shape for a shell element.

Sometimes, Irregular tetrahedral are created by the program. These distorted elements have high aspect ratio. An aspect ratio that is too high causes element degeneration, which in turn affects the quality of the results.

Aspect Ratio

Right click the Mesh icon and select Create Mesh Plot

Select Aspect ratio

Ken Youssefi Mechanical Engineering Dept.

114

View (animated) Displacements

Post-Processing
No

Does the shape of deformations make sense?

Yes

Review Boundary Conditions

View Displacement Fringe Plot Are magnitudes in line with your expectations? No
Yes

Review Load Magnitudes and Units

View Stress Fringe Plot Is the quality and mag. of stresses acceptable?
Yes

No

Review Mesh Density and Quality of Elements

View Results Specific To the Analysis

Ken Youssefi Mechanical Engineering Dept

115

FEA - Flow Chart

Ken Youssefi

Mechanical Engineering Dept

116

Last Comment
Finite Element Analysis makes a good engineer great

and a bad engineer dangerous !

Robert D. Cook, Professor of Mechanical Engineering University of Wisconsin, Madison