Introduction• It is originated in the aerospace industry as a tool to study

stresses in complex air- frame structure.• It is numerical technique for obtaining approximate solution to

a wide variety of engineering problem.• Analytical methods provide accurate solution with

applications limited to simple geometries.• Experimental methods are used to test prototypes or full scale

models.• And they are costly and may not be feasible in some cases.• But Numerical methods are most efficient technique for

engineering analysis which can treat complex geometries also.• Among many FEA/FEM is the most versatile & compressive

numerical technique in the hands of engineers today.• FEM is used to obtain approximate solutions of boundary

value problems in engineering.

• Boundary value problems is a mathematical problems with one or more dependant variables satisfying different equation within the known domain of independent variable & satisfying specific conditions on the boundary of the domain.

• The boundary conditions are the specified values of the field variables (or related variables such as derivatives) on the boundaries of the field.

• Depending on the type of physical problem being analyzed, the field variables may include physical displacement, temperature, heat flux, fluid velocity, etc.

• FEM allows for descretizing the intricate geometries in to small fundamental volumes called finite elements.

• It is then possible to write the governing equations & material properties for these elements &this will be in terms of unknown values at points called nodes.

• These equations are then assembled by taking proper care to loads & constraints which results in a set of equation which are to be solved & the results describe the behavior of the original complex body being analyzed. The solution to these equations would provide an exact closed form solution to the particular problem being studied.

Advantages of FEMIncrease productivity (reduce uncertainty)Minimize physical prototyping (optimize performance before prototyping)Innovative products in less time & less costs (reduced testing and redesign costs thereby shortening the product development time)Applicable to linear and non linear problemEasily applied to complex, irregular shaped objects composed of several different materials and having complex boundary conditionsApplicable to steady state, time dependant and eigenvalue problemsOne method can solve a wide variety of problems, including problems in solid mechanics, chemical reaction, electromagnetics, biomechanics, heat transfer and acoustics.

Disadvantages of FEM Experience and judgment are needed in order to

construct a good finite element model A powerful computer and reliable FEM software are

essential. Input and output data may be large and tedious to

prepare and interpret. The FEM is applied to an approximation of the

mathematical model of a system(the source of so called inherited error)

Susceptible to user introduced modeling error:1. poor choice of elements types2. distorted elements3. geometry not adequately modeled

Application of FEM

FEM is not limited to mechanical system alone but to a range of engineering problems such as

• Stress analysis • Dynamic analysis • Deformation studies • Fluid flow analysis • Heat flow analysis • Magnetic flux studies And these are different types of commercial software

available in the market for the application of finite elements such as ALGOR, ANSYS, COSMOS/M, SAP 90/200., ABAQUS.

Stress Analysis

In the finite element modeling process there are Six Steps

Step 1 - Discretization: The problem domain is discretized into a collection of simple shapes, or elements.

Step 2 - Develop Element Equations: Developed using the physics of the problem, and typically Weighted Residual (Galerkin’s Method) or variational (Ritz) principles.

Step 3 - Assembly: The element equations for each element in the FEM mesh are assembled into a set of global equations that model the properties of the entire system.

Step 4 - Application of Boundary Conditions: Solution cannot be obtained unless boundary conditions are applied. They reflect the known values for certain primary unknowns. Imposing the boundary conditions modifies the global equations.

Step 5 - Solve for Primary Unknowns: The modified global equations are solved for the primary unknowns (Elemental Displacements) at the nodes.

Step 6 - Calculate Derived Variables (Elemental stress, strain, reaction forces) : Calculated using the nodal values of the primary variables.

And this six steps, categorized in the FEA program to three main stages as follows:

BASIC STEPS IN THE FINITE ELEMENT METHOD The basic steps involved in any finite element analysis consist of the following:

Preprocessing Phase

1. Create and discretize the solution domain into finite elements; that is, subdivide the problem into nodes and elements.

2. Assume a shape function to represent the physical behavior of an element; that is, an approximate continuous function is assumed to represent the solution of an element.

3. Develop equations for an element.

4. Assemble the elements to present the entire problem. Construct the global stiffness matrix.

5. Apply boundary conditions, initial conditions, and loading.

Solution Phase

6. Solve a set of linear or nonlinear algebraic equations simultaneously to obtain nodal results, such as displacement values at different nodes or temperature values at different nodes in a heat transfer problem.

Post processing Phase

7. Obtain other important information. At this point, you may be interested in values of principal stresses, heat fluxes, etc.

Typical elements commonly used in finite Element analysis

• The type of elements to be used for generating the mesh depends upon the actual geometry and the type of problem being considered. Some typical elements generally found in most of the FE software are shown below

a) One dimensional

2 node lines (BE2) 3 node quadratic (BE3) b) Two dimensional Triangle

3 - Node lines (TR3) 6 -node quadratic (TR6)

Quadrilateral

4-Node linear (QU4) 8 - node quadratic (QU8)c) Three dimensional

Tetrahedral

4 - Node linear

Modeling recommendations

1. Maintain compatibility : e.g. Do not attach quadratic quadrilateral elements to linear quadrilateral

2. Adopt fine mesh in regions of high stress 3. Reduce bandwidth by suitable numbering of

nodes.

Linear quadratic

Quadraticquadrilateral

Band width of matrix is the number of columns from first to the last non-zero column.

The element band width = max. Difference between the largest and smallest node Numbers of the element

The total number of band width = element band width +1

e.g.

Element band width = 5-4 = 1

Total number of band width = 1 + 1 = 2

e.g.

BW=21 - 10, Total No. BW= 11 + 1 = 12

4. Exploit symmetry of elements wherever possible methods of element refinement as per point 2 above

P -method: using size of element but of a higher order interpolating polynomial

h - method: using same element type, but of smaller size

1 2 3 4 5

6 10

14 2112

In general, there are several approaches to formulating finite element problems:

(1) Direct formulation, (2) The Minimum Total Potential Energy

Formulation, (3) Rayleigh – Ritz Method and (Variation Method)(4) Weighted Residual FormulationsAgain, it is important to note that the basic steps involved

in any finite element analysis, regardless of how we general the finite element model, will be tile same as those listed above.

Approaches in Finite Element Method

DIRECT FORMULATIONThe following problem illustrates the steps and the procedure involved in direct formulation.

EXAMPLE 1.0

Consider a bar with a variable cross section supporting a load P, as shown in Figure 1.0 The bar is fixed at one end and carries the load P at the other end. Let us designate the width of the bar at the top by w1, at the bottom by w2, its thickness by t, and its length by L. The bar's modulus of elasticity will be denoted by E. We are interested in determining how much the bar will deflect at various points along its length when it is subjected to the load P. We will neglect the weight of the bar in the following analysis, assuming that the applied load is considerably larger than the weight of the bar:

Figure 1.0 A bar under axial lodaingPreprocessing Phase

1. Discretize the solution domain into finite elements.

The given bar is modeled using four individual segments,with

each segment having a uniform cross section.The cross-

sectional area of each element is represented by an average

area of the cross sections at the nodes that define the

element.This model is shown in Figure2.

Figure 2.0 Subdividing the bar in to elements and nodes

2. Assume a solution that approximates the behavior of an

element.

In order to study the behavior of a typical element, let's

consider the deflection of a solid member with a uniform cross

section A that has a length ℓ when subjected to a force F, as

shown in Figure 3.

The average stress rr in the member is given

by AF (1)

The average normal strain e of the member is

defined as the change in length ∆ℓ per unit

original length ℓ of the member:

ll (2)

Over the elastic region, the stress and strain

are related by Hooke's Law, according to the

equation

E (3)

where E is the modulus of elasticity of the

material. Combining Eqs. (1), (2), and (1.3)

and simplifying, we have

ll

AEF

(4)

Note that Eq. (4) is similar to the equation for

a linear spring, F = kx. Therefore, a centrally

loaded member of uniform cross section may

be modeled as a spring with an equivalent

stiffness of

l

AEkeq (5)

Figure 3.0 A solid member of uniform crossection subjected to a force F

the bar is represented by a model consisting of four elastic springs (elements) in series, and the elastic behavior of an element is modeled by an equivalent linear spring according to the equation

f = keq (Ui+1 + Ui) = 111

1 2UUE

l

AAUU

l

EAi

iiii

avg

(6)

where the equivalent element stiffness is given

by

Keq = E

l

AA ii

21 (7)

Ai and Ai+l are the cross-sectional areas of the member at nodes i and i+1, respectively, and ℓ is the length of the element.

_

The free body diagram of nodes, which shows the

forces acting on nodes 1 through 5 of this model, is

depicted in Figure 4. Static equilibrium requires that the

sum of the forces acting on each node be zero. This

requirement creates the following five equations:

node 1: R1 -- kl(U2 - Ui) = 0

node2:kl(U2-Ul)-k2(u3-u2)=0 (8)

node 3: k2(u3-u2)-k3(u4-u3) = 0

node 4: k3(u4 - u3) - k4(u5 - u4) = 0

node 5: k4(us-u4) – P = O

Node 1

Node 2

Node 3

Node 4

Node 5

K4(U5-U4)

R1

P

K3(U4-U3)

K2(U3-U2)

K1(U2-U1)

Figure 4.0 Free body diagram of the nodes

Rearranging the equilibrium equations given by Eq. (8)

by separating the reaction force R1 and the applied

external force P from the internal forces, we have:

kl U1- klU2 = R1 .................................. (a)

klU2 - kl Ul - k2u3 + k2 u2 = 0 ....................(b)

(9)

k2u3- k2u2-k3u4+ k3u3 = 0 .........................(c)

k3u4 - k3u3 - k4u5 - k4u4 = 0 .......................(d)

k4us- k4u4 = P..............................................(e)

Presenting the equilibrium equations of Eq. (9) in a

matrix form, we have:

P

R

u

u

u

u

u

kk

kkkk

kkkk

kkkk

kk

0

0

0

1

5

4

3

2

1

44000

443300

033220

002211

00011

(10)

-

+

It is also important to distinguish between the reaction

forces and the applied loads in the load matrix.

Therefore, the matrix relation of Eq. (10) can be written

as:

0

0

0

0

1R

=

Pu

u

u

u

u

kk

kkkk

kkkk

kkkk

kk

0

0

0

0

5

4

3

2

1

44000

443300

033220

002211

00011

(11)

We can readily show that under additional nodal loads and other fixed boundary

conditions, the relationship given by Eq. (11) can be put into the general form

{R} = [K] {u} – {F} (12)

which stands for:

{reaction matrix} = [stiffness matrix] {displacement matrix} - {load matrix}

we find that because the bar is fixed at the top, the

displacement of node 1 is zero. Thus, the first row of the

system of equations given by Eq. (10) should read u1 = 0.

Thus, application of the boundary condition leads to the

following matrix equation:

Pu

u

u

u

u

kk

kkkk

kkkk

kkkk

0

0

0

0

5

4

3

2

1

44000

443300

033220

002211

00001

(13)

The solution of the above matrix yields the nodal

displacement values. In the next section, we will develop the

general elemental stiffness matrix and discuss the

construction of the global stiffness matrix by inspection.

Develop equations for an element. • Because each of the elements in Example 1.1 has

two nodes, and with each node we have associated a displacement, we need to create two equations for each element.

• These equations must involve nodal displacements and the element's stiffness. Consider the internally transmitted forces fi and fi+1 and the end displacements ui and ui+1 of an element, which are shown in Figure 5.

Static equilibrium conditions require that the sum

of fi and fi+1 be zero, so that fi and fi+1 are given

in the positive y-direction. Thus, we write the

transmitted forces at nodes i and i+1 ac-

cording to the following equations:

)(

)(

1

1

iieqii

iieq

uukf

uukfi

(14)

Equation (1.14) can be expressed in a matrix form by

11 i

i

eqeq

eqeq

i u

u

kk

kk

f

fi (15)

Figure 5.0 Internally transmitted forces through an arbitrary element.

Assemble the elements to present the entire

problem.

Applying the elemental description given by Eq. (15) to

all elements and assembling them (putting them

together) will lead to the formation of the global

stiffness matrix. The stiffness matrix for element (1) is

given by

11

11)1(

kk

kkk

and it position in the global fness matrix is given by

00000

00000

00000

00011

00011

1

kk

kk

k G

The nodal displacement matrix is shown alongside the

position of element 1 in the global stiffness matrix to

aid us to observe the contribution of a node to its

neighboring elements. Similarly, for elements (2), (3),

and (4), we have

22

22)2(

kk

kkk and its position in the global matrix

33

33)3(

kk

kkk and its position in the global matrix

00000

03300

03300

00000

00000

3

kk

kkk G

44

44)4(

kk

kkk and its position in the global matrix

44000

44000

00000

00000

00000

4

kk

kk

k G

The final global stiffness matrix is obtained by assembling, or adding, together each

element's position in the global stiffness matrix:

GGGGG kkkkk 4321

44000

443300

033220

002211

00011

kk

kkkk

kkkk

kkkk

kk

k G (16)

The gobal stiffness matrix obtained using elemental description is identical to the global

matrix we obtained earlier from the analysis of the FBD of the nodes, as given by the left

hand side of Eq. (10).

Apply boundary conditions and load.

The bar is fixed at the top, which leads to the

boundary condition u1 = 0..the external load P is

applied at node 5. Applying these conditions

results in the following set of linear equations.

Pu

u

u

u

u

kk

kkkk

kkkk

kkkk

0

0

0

0

5

4

3

2

1

44000

443300

033220

002211

00001

(17)

Solve a system of algebraic equations

simultaneously.

In order to obtain numerical values of the nodal

displacements, let us assume that E = 10.4 x 106 lb/in2

(aluminum), w1 = 2 in, w2 = 1 in, t = 0.125 in, L = 10

in, and P = 1000 lb.

TABLE 1 Properties of the elements in Example 1.0

Element Nodes Average cross-sectional area (in2)

Length(in) Modulus of elasticity (lb/in2)

Element's stiffness coefficient (lb/in)

1

2

3

4

1 2

2 3

3 4

4 5

0.234375

0.203125

0.171875

0.140625

2.5

2.5

2.5

2.5

10.4 x 106

10.4 x 106

10.4 x 106

10.4 x 106

975 x 103

845 x 103

715 x 103

585 x 103

The variation of the cross-sectional area of the bar in the y-

direction can be expressed by:

yytyL

wwwyA 0125.025.0)125.0(

10

)21(2

121

(18)

Using Eq. (18), we can compute the cross-sectional areas at each

node:

A1 = 0.25 in2

A2 = 0.25 - 0.0125(2.5) = 0.21875 in2

A3 = 0.25 - 0.0125(5.0) = 0.1875 in2

A4 = 0.25 - 0.0125(7.5) = 0.15625 in2

A5 = 0.125 in2

Next, the equivalent stiffness coefficient for each element is computed from the equations

E

AAkeq ii

21

in

bxx

k 36

1 109755.22

104.1025.021875.0

in

bxx

k 36

2 108455.22

104.1021875.01875.0

in

bxx

k 36

3 107155.22

104.101875.015625.0

in

bxx

k 36

4 105855.22

104.1015625.0125.0

and the elemental matrices are

975975

975975103

1

111

kk

kkk

845845

845845103

22

222

kk

kkk

715715

715715103

33.

333

kk

kkk

585585

585585103

44

444

kk

kkk

Assembling the elemental matrices leads to the generation of the global stiffness matrix:

585585000

58558571571500

07157158458450

00845845975975

000975975

103Gk

Applying the boundary condition u1 = 0 and the load P = 1000 lb, we get

3

5

4

3

2

1

3

10

0

0

0

585585000

585130071500

071515608450

008451820975

00001

10

u

u

u

u

u

Because in the second row, the -975 coefficient gets multiplied by u1 = 0, we need only to

solve the following 4 x 4 matrix:

35

4

3

2

3

10

0

0

0

58558500

58513007150

07151560845

008451820

10

u

u

u

u

Therfoer the displacement solution is: u1= 0, u2= 0.001026 in, u3= 0.002210 in, u4= 0.003608

in, and u5 = 0.005317 in.

Postprocessing Phase, Obtain other information.

For the given example 1.0, we may be interested in obtaining other information, such as the

average normal stresses in each element. These values can be determined from the equation

ii

avg

iiavg

avg

iieq

avg

uuE

A

uuEA

A

uuk

A

f

1

11 (19)

Since the displacements of different nodes are known, Eq. (1.19) could have been obtained

directly from the relationship between the stresses and strains,

ii uu

EE

1 (20)

Employing Eq. (20) , we compute the average normal stress for each element is

2

6121 4268

5.2

0001026.0104.10in

bxuuE

2

232 4925in

buuE

2

343 5816in

buuE

2

454 7109in

buuE

Figure 6.0 The internal forces

y

P

L

f=p

P p

f=p

In the above figure, we note that for the given problem,

regardless of where we cut a section through the bar,

the internal force at the section is equal to 1000 lb. So,

2

1 4267234375.0

1000in

bA

f

av

2

2 4923203125.0

1000in

bA

f

av

2

3 5818171875.0

1000in

bA

f

av

2

4 7111140625.0

1000in

bA

f

av

Ignoring the errors we get from rounding off our answers, we find that these

results are identical to the element stresses computed from the displacement

information. This comparison tells us that our displacement calculations are

good for this problem.

Reaction Forces

For the above problem, the reaction force may be computed in a

number of ways. First, referring to Figure 4, we note that the statics

equilibrium at node 1 requires

R1 = k1 (u2 - ul) = 975 x 103(0.001026 - 0) = 1000 lb

The statics equilibrium for the entire bar also requires that

R1 =P = 1000lb

As you may recall, we can also compute the reaction forces from the

general reaction equation

{R} = [K]{u} - (F}

{reaction malrlx} = [stiffness matrix] {displacemenl matrix} -

{load matrix}

General Differential Formulation of a Structural Problem in 1 D

Definition

Weighted Residual Statement

Ex. Cantilever Beam with Uniform Distributed Load

Temperature effect