lectuer 1- sami vahid-1

7/28/2019 Lectuer 1- Sami Vahid-1

http://slidepdf.com/reader/full/lectuer-1-sami-vahid-1 1/38

Finite Element Method

Lecture 1

School of Mechanical and Automotive Engineering

Kingston University London

2012-2013



Lecture 1: Finite Element Method



Design analysis: hand calculations, experiments, and

computer simulations

FEA is the most widely applied computer simulation

method in engineering

Closely integrated with CAD/CAM applications

Why Finite Element Method?



Applications of FEM in

Engineering

• Mechanical/Aerospace/Civil/Automobile Engineering

• Structure analysis (static/dynamic, linear/nonlinear)

• Thermal/fluid flows

• Electromagnetics

• Geomechanics

•

Biomechanics• …..



FEM in Structural Analysis

• Divide structure into pieces (elements with nodes)

• Describe the behaviour of the physical quantities on each

element

• Connect (assemble) the elements at the nodes to form an

• approximate system of equations for the whole structure

• Solve the system of equations involving unknown

• quantities at the nodes (e.g., displacements)

• Calculate desired quantities (e.g., strains and stresses) at

selected elements



Example



Computer Implementations

• Preprocessing (build FE model, loads and constraints)

• FEA solver (assemble and solve the system of equations)

• Postprocessing (sort and display the results)



• ANSYS (General purpose, PC and workstations)

• SDRC/I-DEAS (Complete CAD/CAM/CAE package)

• NASTRAN (General purpose FEA on mainframes)

• ABAQUS (Nonlinear and dynamic analyses)

• COSMOS (General purpose FEA)

• ALGOR (PC and workstations)

• PATRAN (Pre/Post Processor)• HyperMesh (Pre/Post Processor)

• Dyna-3D

Available Commercial FEM

Software Packages



Structural analysis is based on the following four principles:

(1) Equilibrium of forces

(2) Constitutive equations (material properties)

(3) Compatibility of displacements(4) Boundary conditions

Application of (1) – (3) will lead in general to a second order

partial differential equation that can be solved using finite

elements or other techniques by applying (4).

Structural Background



Example:

Consider the equilibrium of a uniform beam under the action of

axial forces, see Figure 1



The element has end nodes 1 and 2 and is of length L. The

displacement at any point along the length of the element is

defined by the function u.



Over a small length x of the element (see Figure 2), the variation

of stress will be given by

x x

Hence, the resultant internal force in the x-direction will be given

by A x

x

.

If the element is subjected to an external force F per unit length,

then for equilibrium0

x

x A F

0

F

x A



But, (constitutive equation).

Over a small length, the longitudinal strain is defined as the rate

of change of longitudinal displacement i.e.

Hence the equilibrium equation becomes

E

0 F x

AE

x

u

.02

2

F

x

u AE



The equilibrium equation is thus a second order partial

differential equation which can be solved by the finite element

method once the appropriate boundary conditions e.g.

displacements at nodes 1 and 2 are known.



In the finite element method, a structure is divided into a number

of elements (1, 2 or 3 dimensional as appropriate) which are

linked together by nodes (see Figure 3). A linear element will

generally have two nodes per side, a quadratic element three

nodes per side, a cubic element four nodes per side etc.

Figure 3



Application of the finite element method results in a loaded

structure being represented by the matrix equation

where F is a vector of the external nodal applied load, is a

vector of the nodal displacements and K is the overall (global)stiffness matrix which is formed from an assemblage of the

individual element stiffness matrices K e.

F = K



The ties and struts that form a pin-jointed framework can be

considered to be effectively stiff springs. Each end of the tie or

strut will be able to move due to both the displacement of the

structure and the deformation of the member itself. This type of

spring element is shown in Figure 4.

Framework Analysis using Spring

Analogy



The element has two nodes 1 and 2. The displacement and applied

force at node 1 are u1 and F1 and the displacement and applied

force at node 2 are u2 and F2. The element has length L and cross-

sectional area A.

The stiffness of the element will be given by where E isYoung’s modulus.

If externally applied forces F1 and F2 induce displacements u1 and

u2 at nodes 1 and 2 respectively then

L

AE

k

1

21112111

uk uk uuk F

21111212

uk uk uuk F



For convenience, these equations can be written in matrix form as

Or in shorthand notation as

Where is the external load vector

is the element stiffness matrix

And is the nodal displacement vector.

2

1

11

11

2

1

u

u

k k

k k

F

F

eee K F

2

1

F

F F e

11

11

k k

k k

K e

2

1

u

ue



Now consider two elements joined together as shown in Figure 5.

A force-displacement equation for each element can be written asfollows:

Element:

Element:

2

1

11

11

2

1

u

u

k k

k k

F

F

3

2

22

22

3

2

u

u

k k

k k

F

F

Figure 5



Expanding both matrices so that they are in an equivalent form

gives:

and

The overall force-displacement matrix can be obtained by adding

these two matrices together. Thus:

3

2

1

22

22

3

2

1

0

0

000

u

u

u

k k

k k

F

F

F

3

2

1

11

11

3

2

1

000

0

0

u

u

u

k k

k k

F

F

F

3

2

1

22

2211

11

3

2

1

0

0

u

u

u

k k

k k k k

k k

F

F

F



Or

where FG is the global force vector, K G is the global stiffness

matrix and G is the global displacement vector.

GGG K F



Example 1:

Three dissimilar materials are friction welded together and

mounted between rigid end supports as shown in Figure Q1.If forces of 100 kN and 50 kN are applied as indicated, find the

movement of the interfaces between the materials and the forces

on the end supports. Material properties are given in Table Q1.

Aluminium Brass Steel

Area (mm2) 400 200 70

Length (mm) 280 100 100

E (GPa) 70 100 200

Table Q1

Figure Q1



Solution

Spring stiffness will be k 1 = 100 kN/mm, k 2 = 200 kN/mm and

k 3 = 140 kN/mm.



Force-displacement equations for each element will be:

Element :

Element :

Element :

1 1

2 2

100 100

100 100

F u

F u

2 2

3 3

200 200

200 200

F u

F u

3 4

4 4

140 140

140 140

F u

F u



Hence the global force-displacement equations will be given by:

The applied forces are F2 = -50 kN and F3 = 100 kN and the

boundary conditions are u1 = 0 mm and u4 = 0 mm.

Substitution of these values gives the following set of equations:

1 1

2 2

3 3

4 4

100 100 0 0

100 300 200 0

0 200 340 140

0 0 140 140

F u

F u

F u

F u

1 2

2 3

2 3

4 3

100 1

50 300 200 2

100 200 340 3

140 4

F u

u u

u u

F u



Solving equations (2) and (3) gives u2 = 0.048 mm and u3 = 0.323

mm.

Substitution in equations (1) and (4) gives F1 = -4.8 kN andF4 = -45.2 kN.



This is the method generally used in most FEA (Finite Element

Analysis) packages. In this method, the displacements at the

nodal points are chosen as the basic variables. Alternative

methods involve using the stresses (equilibrium method) or a

combination of stresses and displacements (mixed or hybrid

method) as the basic variables. These methods are more complex

mathematically.

The Finite Element Displacement

Method



The displacement variables are termed generalised co-ordinates

since the components involved may not be physically interpreted

as displacements. These variables are formed into an element.

Displacement vector e given by

n

i

e

:

:

3

2

1



where i is the vector of the displacements at node i.

Associated with the generalised displacement vector is a

generalised force vector Fe defined by

where Fi is the vector of the forces at node i.

n

i

e

F

F

F

F

F

F

:

:

3

2

1



e.g. The element shown in Figure 6 can be used in two-

dimensional elasticity

etc.

i

ii

v

u

j

j

j v

u

j

i

e

yi

xi

i F

F F



e.g. The element shown in Figure 7 can be used in plate bendingtheory

In stress analysis, the product of the generalised forces and thegeneralised displacements must give work.

yi

xi

i

i

w

p

j

i

e

.

yi

xi

i

i

M

M

F

F

p

j

i

e

F

F

F

F .



Shape functions are used to define the assumed variation of a

particular variable e.g. displacement over the whole domain of an

element.

Shape Functions



The element shown in Figure 8, which lies along the x-axis, has

two nodes i and j with co-ordinates xi and x j respectively. A

quantity is assumed to vary linearly across the element and has

values i and j at nodes i and j.

Since is assumed to vary linearly, it will be represented by theequation where a1 and a2 are constants.

and

Since = i at x = xi, and = j at x = x j, then

L

x x

x x

x xa

i j ji

i j

i j ji

1 L x xa

i j

i j

i j

2



Hence

In matrix form = N where N = and

The functions and are the shape functions

and show the variation of i and j respectively over the domain

of the element.

j jii ji

i

ji ji j ji N N

L

x x

L

x x x

L L

x x

ji N N

j

i

L

x x N

j

i

L

x x

N i

j



Example 2: The nodal co-ordinates xi and x j of an element are

3.0 cm and 4.5 cm respectively. The temperatures at these nodes

are found to be 27°C and 33°C respectively. What is the

temperature at x = 3.6 cm?

and

Hence at x = 3.6 cm,

4.5

1.5

j

i

x x x

N L

3.0

1.5 j

x N

4.5 3.6 3.6 3.027 33

1.5 1.5

0.6 27 0.4 33

29.4

T

C



The shape functions and are shown graphically

in Figure 9.

L

x x N

j

i

L

x x N i

j



Note that: at node i, Ni = 1 and N j = 0

at node j, Ni = 0 and N j = 1

And Ni + N j = 1 over the domain of the element.

Thus, each shape function has value 1 at its own node and value

0 at all other nodes. In addition, the derivatives of the shapefunctions with respect to x sum to zero.

The properties of shape functions given above can be used to

define the shape functions of an element.

lectuer 1- sami vahid-1

Documents