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Basics of Finite Element Analysis

What is FEA ?

A complex problem is divided into a smaller and

simpler problems that can be solved by using theexisting knowledge of mechanics of materials and

mathematical tools

Ken Youssefi Mechanical Engineering Dept 1

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




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.




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 costof computers and the phenomenal increase in computing power,

FEA has been developed to an incredible precision.




FEA Applications

• Perform deflection analysis.• Evaluate the stress or temperature


.• Check Buckling failure

• Analyze the kinematics or dynamic response.

• Perform vibration analysis.



The goal is to optimize for material




Consider a cantilever beam shown.

Finite element analysis starts with an approximation of the region of


.

connected to associated nodes (black dots) and thus becomes a finite

element.




• 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 Strain

Partial derivatives

Stress

Stress & Strain

relationship



Example – a plate under load

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

Py

.


,

Reaction

forces

U 1 thru U 8,

displacements

in x and y

directions




Displacement within the triangular element (2D) with threenodes can be assumed to be linear.

u = α1 + α2 x + α3 y

v = β 1 + β 2 x + β 3 y





Displacement for each node,





Solve the equations simultaneously for α and β ,





Substitute 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

Element 1

1

(2)

(3)Evaluate the constants a, b, and c

10 4 0 0


2a = 40

Calculations:

a1 = 40, a2 = 0, a3 = 0

b1 = - 4, b2 = 4, b3 = 0

c1 = -10, c2 = 0, c3 = 10



Example

2a = 40

a1

= 40, a2

= 0, a3

= 0

b1

= - 4, b2

= 4, b3

= 0

c1

= -10, c2

= 0, c3

= 10

Calculationsu1 = U1, u2 = U3, u3 = U5,

v1 = U2, v2 = U4, v3 = U6

Change of notations

40 00

40


α1 = 1

α2 = -(1/10)U1 + (1/10)U3

α3 = -(1/4) U1+ (1/4) U5

β 1 = (1)U2

β 2 = -(1/10)U2 + (1/10) U4

β 3= -(1/4) U2+ (1/4) U6



Example

u = α1 + α2 x + α3 yv = β 1 + β 2 x + β 3 y

Substitute α and β to obtain displacements u

and v for element 1.

α1 = (1)U1

α2

= -(1/10)U1 + (1/10)U3

α3 = -(1/4) U1+ (1/4) U5

β 1= (1)U2

β 2 = -(1/10)U2 + (1/10) U4


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

Calculation:

β 3 = -(1/4) U2+ (1/4) U6



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) U

4] x + [-(1/4) U

2+ (1/4) U

6] y




Example

Similarly the displacements within

element 2 can be expressed as,




Example

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

displacement relations,




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(U

2) + (1/10) U

4] x + [-(1/4) U

2+ (1/4) U

6] y




Example

Element 2




Normal & Shear components of stress (3D)Normal stress is perpendicular to the cross section, σ σσ σ (sigma).

Shear stress is parallel to the cross section, τ ττ τ (tau).

3DElement

y σ σσ σ y

τ ττ τ yz

τ ττ τ yxSecond subscript

indicates the positive

τ ττ τ xy

First subscript indicates

the axis that is

perpendicular to the face

x

z

y

σ σσ σ x

τ ττ τ xz

σ σσ σ z

τ ττ τ zx

τ ττ τ zy

stress

Due to equilibrium condition;

τ ττ τ xy = τ ττ τ yx

τ ττ τ zy = τ ττ τ yz

τ ττ τ zx = τ ττ τ xz



State of StressThree dimensional stress matrix

Two dimensional,

Plane Stress



Stress & Strain Relationship

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

ε x = (σ x / E ) - ν (ε y) - ν (ε z ) = (σ x / E ) - ν (σ y / E ) - ν (σ z / E )

ε y = (σ y / E ) - ν (ε x ) - ν (ε z ) = (σ y / E ) - ν (σ x / E ) - ν (σ z / E )

Uniaxial state of stress

ε x = (σ x / E ), ε y = - ν ε x , ε z = - ν ε x

σ x , σ y = 0 , σ z = 0


ε z

= σ z

E - ν ε x

- ν ε y

= σ z

E - ν σ x

- ν σ y

E

Stresses in

terms of strains



Stress & Strain RelationshipThere are many practical problems where the stress in the z -direction is zero,

this is referred to as the state of Plane Stress.

G = E

2(1 + ν)= γ xy Gτ ττ τ xy

Shear stress

Matrix

form



FEA Results - Principal Stresses

Normal stresses on planes with no

shear stresses are maximum and they

are called principal stresses σ1, σ2,and σ3, where σ1 > σ2 > σ3


The three non-imaginary roots are the principal stresses

2

2

σ3 - (σ x + σ y + σ z ) σ

2 + (σ x σ y + σ x σ z + σ y σ z - τ xy - τ xz - τ yz )σ -

(σ x σ y σ z - 2 τ xy τ xz τ yz - σ x τ yz - σ y τ xz - σ z τ xy) = 02

22

2

σ3 - (σ x + σ y) σ

2 + (σ x σ y - τ xy)σ = 0

Plane stress, two principal stresses, σ3 = 0



Failure Theories – Maximum Shear Stress

• Maximum shear stress theory (Tresca 1886)

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

ductile material

Sσ = S

(τmax )component > (τ )obtained from a tension test at the yield point Failure


σ = Sy

τ2

=

σ =Sy

(τmax )component <Sy

2

To avoid failure

τmax =Sy

2 nn = Safety

factor

Design equation



Failure Theories – von Mises Stress

• Distortion energy theory (von Mises-Hencky)

σt

Simple tension test → (S y)t

(S y)t (S y)h >>

Hydrostatic state of stress → (S y)h

σh


σt

failure much more thanchange in volume.

(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

σh

σh



von Mises Stress

½

2D case, σ3 = 0

σ′ = (σ12 – σ1σ2 + σ2

2) < S

(σ1 – σ2)2

+ (σ1 – σ3)2

+ (σ2 – σ3)2

2

½< Sy

3D case, to avoid failure


Where σ′ is von Mises stress

σ′ =S y

n

Design equation



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





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




Let’s denote the displacements of any point ( x, y, z ) of the objectfrom the unloaded configuration as UT

U T = [U ( x, y, z ) V ( x, y, z ) W ( x, y, z )]

The displacement U causes the strains

ε T = [ε x ε ε z γ x γ z γ zx ]


and the corresponding stresses

The goal is to calculate displacement, strains, and stresses from

the given external forces.

σ T = [σ x σ y σ z τ xy τ yz τ zx ]




Equilibrium condition and principle of virtual displacements

The left side represents the internal virtual work done, and the

ri ht side re resents the external work done b the actual

σ

Work done by body forces

Work done bysurface forces

Work done byexternal forces


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

U s – denotes the displacement due to surface forces

Ui

– denotes the displacement due to point forces




Displacement interpolation matrix


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

displacement at the nodes.

H(m)

is the displacement interpolation matrix




strain-displacement matrix


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



Matrix

form


Elasticity matrix


C (m) is the elasticity matrix of element m and σ I (m) are the

element’s initial stresses. The elasticity matrix relates strains

to stress.




The formula for the principle of virtual displacements can berewritten 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




σ

σ σ





The equilibrium equation can be expressed using matrix

notations for m elements.

where


B

(m)

Represents the rows of the strain displacement matrixC(m) Elasticity matrix of element m

H(m) Displacement interpolation matrix

U Vector of the three global displacement

components at all nodesF Vector of the external concentrated forces

applied to the nodes

F l ti f th Fi it El t M th d




The above equation can be rewritten as follows,


The above equation describes the static equilibrium problem.

K is the stiffness matrix.



Continuing the example

C(m) - Elasticity matrix of element m

B(m) - Represents the rows of the strain

displacement matrix


dx x

y

y = 4 - x4

10dA = y dx



Example




Example

Calculating the stiffness matrix for element 2.




Example

The stiffness of the structure as a whole is obtained by combing

the two matrices, K = K 1 +K 2


Example



Example

KU = R The load vector R, equals R c because onlyconcentrated loads act on the nodes.


where P y is the known external force and F 1 x, F 1 y, F3 x, and F3 y

are the unknown reaction forces at the supports.

R =

Example



Example

The following matrix equation can be solved for nodal pointdisplacements

KU = R


Example



ExampleThe solution can be obtained by applying the boundary conditions


No deflectionat the supports

Example



Example

The equation can be divided into two parts,


The first equation can be solved for the unknown nodal displacements,

U 3, U 4, U 7, and U 8. And substituting these values into the secondequation to obtain unknown reaction forces, F 1 x, F 1 y, F3 x, and F3 y

Once the nodal displacements have been obtained, the strains

and stresses can be calculated.



Finite Element Analysis

FEA requires three steps

FEA is a mathematical representation of a physical system

and the solution of that mathematical representation


• Pre-Processing• Solving Matrix (solver)

• Post-Processing

FEA P P i



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 withoutskinny, distorted elements.

FEA Pre-Processing - meshing



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.


FEA Pre Processing



FEA Pre-Processing

The mesh elements supported by most finite-element codes:


FEA P P i El t



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.

FEA P P i El




The most general line element is a beam.


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

FEA P P i El t




Plate and Shell Modeling

Plate and shell are used interchangeably and refer to surface-

like 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.




FEA Pre Processing Elements

Solid Element Modeling


Tetrahedral (tet) mesh is the only generally

accepted means to fill a volume, used as auto-

mesh element by many FEA codes.10-node Quadratic

CAD Modeling for FEA




• CAD models prepared by the design group for

eventual 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.

• CAD models unsuitable for use in analysis due to

the amount of rework required.

• Analytical geometry developed by or for analyst

for sole purpose of FEA.





• Solid chunky parts (thick-walled, low aspect ratio)

parts mesh cleanly directly off CAD models.

• Clean geometrygeometrical 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.





Short edges and Sliver surfacesShort edges and sliver surfaces usually accompany each other and on

large faces can cause highly distorted elements or a failed mesh.


CAD Modeling for FEA – Sliver Surfaces



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

Flat surface


s or e e emen s. s

results in a 290,000nodes. If the radius is

increased to .15, a mesh

size of .12 is sufficient

which results in 33,500

nodes.




Sliver surface caused by

misaligned features.


Fillet across shallow angle

Sliver surface caused by a slightly

undersized fillet

G id li f G t Pl i



Guidelines for Geometry Planning

• Delay inclusion of fillets and chamfers as long as

possible.

• Try to use permanent datums as references where

possible to minimize dependencies.


• vo us ng e or ra e ges as re erences or

other features (parent-child relationship)

• Never bury a feature in your model. Delete or

redefine unwanted or incorrect features.

Guidelines for Part Simplification



• Outside corner breaks or rounds.

• Small inside fillets far from areas of interest.

• Screw threads or spline features unless they are

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


spec ca y e ng stu e .

• Small holes outside the load path.

• Decorative or identification features.

• Large sections of geometry that are essentiallydecoupled from the behavior of interested section.

Guidelines for Part Simplification



f p f

Fillet addedto the rib

Holes removed


Fillet

removed

Ribs needed

for casting

removed




Model Conversion• Try to use the same CAD system for allcomponents in design.

• When the above is not possible, translate

geometry through kernel based tools such asACIS or Parasolids. Using standards based(IGES, DXF, or VDA) translations may lead to


pro em.

• Visually inspect the quality of importedgeometry.

• 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.

Example of a solid model corrupted by



p f p y

IGES transfer


FEA Pre-Processing



g

Material Properties

The only material properties that are generally required by

an isotropic, linear static FEA are: Young’s modulus

(E), Poisson’s ratio (v), shear modulus (G), and yieldstrength (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 andspecific heat values.

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,


suc as ru er an o er syn e c e as omers requ re

distortional and volumetric constants or a more complete setof 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

FEA Pre-Processing



• Their properties hold constant throughout the assigned entity.

• Average values are used (variation could be up to 15%).

CommentsIf you are selecting the property set from the code’s library,

be aware of the assumptions made with this selection.


• Localized changes due to heat or other processing effects are

not accounted for.

• Any impurities present in the parent material are neglected.

If possible, obtain material property values specific to the

application under analysis.

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.

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 appliedloads. Constraints can resist translational or

rotational deformation induced by applied loads.

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


.

Non-linear Analysis

The orientation and distribution of the boundary

conditions vary as displacement of the structure is

calculated.

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 theanalysis to run.


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.

Boundary Conditions



Constraints and their geometric equivalent in classic beam calculation.

Fixed support


Roller support

Pin support

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.

Accurac


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.

Boundary Conditions -Example



Excessive ConstraintsModel of the chair seat with patches representing the tops of

the legs.


Patch 3

Patch 1

Patch 2

Patch 4


It may appear to be acceptable to constrain each circular patch



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-to-

seat 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,


Patch 3

Patch 1

Patch 2

Patch 4

that in-plane spatial rotation and rigid body translation isremoved without causing excessive constraints.


• Constraining the center point of patch 1 in all 3



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 in-

plane translation induced

by bending of the seatwithout rigid body

translation or rotation.

Patch 3

Patch 1

Patch 2

Patch 4



Legs are fixed to seat

Applied force = 2000 N

Use On Flat Face restraint



Fixed legs



In plane rotation is allowed

Summary of Pre-Processing



• Build the geometry (CAD model for FEA)

• Prepare the model for meshing

• Create the finite-element mesh• Add boundary conditions; loads and


cons ra n s

• Select material or provide properties

• Specify analysis type (static or dynamic,linear or non-linear, thermal, etc.)

These activities are called finite element modeling.

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 thecomputer does the work for a change.


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

Unexpectedly high or low displacements (by order of magnitude)

Post-Processing, Displacement Magnitude



p y g p ( y g )

could be caused by an improper definition of load and/or elemental

properties.


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.


Post-Processing, FEA of a connecting rod



Ken Youssefi Mechanical Engineering Dept84

Post-Processing, Stress Results



Second Mode (Twisting)

The magnitude of the stresses should not be entirely unexpected.

Ken Youssefi Mechanical Engineering Dept85

First Mode (Bending)

Post-Processing, thermal analysis



Deformation of a duct under thermal load

Ken Youssefi Mechanical Engineering Dept

86

Deploy Mechanism – Assembly Analysis



Displacement

Stress

Can crusher stress

l i



analysis



Use finer mesh size

Right click the

M h i d



Mesh icon andchoose Failure

Diagnostics



Add fillet to the slot

edges (.1 in.)

Max stress (von Mises) = 43.9 MPa

Sy = 96.5 MPa (Al 2014)

S f t f t

Apply 200 N

(45 lb)



n = 96.5/43.9 = 2.2 > 2.0

Max deflection 1.13 mm < 2 mm

Safety factor

Design requirements

Safety factor between2.0-2.5 and deflection

less than 2 mm

Set gap as 5 in.

Fix the back plate

Post-Processing View (animated)

Displacements



Does the shape of deformations make sense?

View Displacement

Fringe Plot

Yes

Review Boundary

Conditions

No

Are ma nitudes in line with our ex ectations?Review Load Magnitudes

No


View Stress

Fringe Plot

Yes

Is the quality and mag. of stresses acceptable?

and Units

Review Mesh Density

and Quality of Elements No

View Results Specific

To the Analysis

Yes

FEA - Flow Chart

plate anlaysis

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