Finite Element Method

Many practical problems in engineering are either extremely difficult or impossible to solve by conventional analytical methods. Such methods involve finding mathematical equations which define the required variables. For example, the distribution of stresses and displacements in a solid component, or of pressure and velocities in the flow of a fluid, might be required.

In the past, the common practice was to simplify such problems to the point where an analytical solution could be obtained which, it was hoped, bore some resemblance to the solution of the real problem. Because of the uncertainties involved in such a procedure (over-simplification of the initial problem), large factors of safety would be introduced.

What is FEM (FEA)?

The basis of the finite element method (also commonly known as Finite Element Analysis) is that a body or structure may be divided into smaller elements of finite dimensions called Finite Elements. The original body or structure is then considered as an assemblage of these elements connected at a finite number of joints called Nodes or Nodal Points.

Steps involved in a Finite Element Analysis procedure

1. Discretization (subdivision of the body or structure into smaller elements)

2. Application of mathematical model to each element (formulation of the properties of the elements)

3. Combination of the elements to obtain the properties of the whole body/structure (assembly)

Thus, instead of solving the problem for the entire body/structure in one operation, in the method of finite elements, attention is mainly devoted to the formulation of properties of the constituent elements. A common procedure is then adopted for combining the elements.

Single Spring Element

When a pin-jointed truss element is part of a structure, its ends will be able to move due to (i). displacement of the structure (lump displacement) (ii). deformation of the member

This may be modelled by a spring element.

Fig. 1

The nodes of the spring element are indicated by points i and j in Fig. 1.

Let if = nodal force at node i iu = nodal displacement at node i

For this particular case of the spring element, we will consider only one degree of freedom (DOF) for each node of the spring. We will assume that displacements are possible only along the axis of the spring.

Spring element loaded by oppositely directed forces

Let us first evaluate the response of a spring element to which oppositely directed forces of magnitude N are applied.

If we now consider that a spring element is used to model a truss member of length L and cross-sectional area A, the stiffness 1k of the spring will be given by

1AEk L= [1]

where E is Young's Modulus for the material.

Using the sign convention that forces and displacements are positive in the positive x-direction, the forces may be related to the displacements by the following equations:

1 1 1 2 1 1 1 2f k u u k u k u = = [2] 2 1 2 1 1 1 1 2f k u u k u k u + = = [3]

In matrix form, the system of equations can be written as

1 1 1 122 1 1

f k k uuf k k

=

[4]

if = (element) nodal force vector ju = (element) nodal displacement vector

ijk = (element) "stiffness" matrix

local stiffness matrix

We shall denote by eiK

or eiK the element stiffness matrix for

the i numbered element. Therefore,

1 11 1

1 1

e ek k

K Kk k

= =

An important property of the stiffness matrix for an element (and as will be seen later, for a complete structure) is that it is symmetrical.

Assembly of Spring Elements

Consider a system of combined spring elements. For this particular configuration of linked spring elements, the assembly process is equivalent to finding the stiffness matrix for the whole system.

As a n example, let us consider a system consisting of two spring elements (Fig. 2). Note that 1F , 2F and 3F are externally applied loads.

The element nodal force components are denoted by ijf , where i is the element number, and j is the node number for each element.

Drawing free body diagrams for each spring element allows for the representation of internal nodal forces ( ijf ).

Using equation [4], the force-displacement relationship for each element may be written as

11 1 1 11 21 12

f k k uuk kf

=

221 2 2

2 32 22

uf k kuk kf

=

Fig. 2

Assembly Method 1

We expand the two matrix equations so that they are in equivalent form

11 1 11

1 21 123

00

0 0 00

k k ufuk kfu

=

12

1 2 2 22 32 22

0 0 0 000

u

f k k uuk kf

=

1 1

1 1 1

0 0

0 0 0e

k kK k k

=

2 2 2

2 2

0 0 0 0

0eK k k

k k

=

The matrix equation for the whole system is obtained by adding the matrix equation for each system (superposition)

(assembly of spring elements addition)

11 1 1 1

1 22 1 1 1 2 2 2

2 32 22

0

0

f k k uf f k k k k u

uk kf

+ +

=

[5]

1 2+e eK K K=

[6]

Note:

1) It is imperative that the element stiffness matrices are first expanded before proceeding with the addition operation.

2) The global stiffness matrix will have dimensions n n , where n is the number of degrees of freedom (DOF) of the system.

Assembly Method 2

An alternative way of assembling the stiffness matrix for the system is to consider the equilibrium of forces for each node of the system.

global stiffness matrix

Equilibrium of forces at node 1 For node 1, the element nodal force is equal to the externally applied force 1F :

11 1f F=

1 1 1 1 2 F k u k u =

Equilibrium of forces at node 2 The sum of element nodal forces applied to node 2 is equal to the externally applied force 2F :

1 22 1 2f f F+ =

( )2 1 1 1 2 2 2 3 F k u k k u k u+ = +

Equilibrium of forces at node 3 The element nodal force applied to node 3 is equal to the externally applied force 3F :

22 3f F=

3 2 2 2 3 F k u k u = +

Writing the equations for nodal equilibrium in matrix form, we obtain the same global stiffness as when we expand the element stiffness matrices and added them together.

Assembly Method 3

A 3rd method for performing the assembly of elementary stiffness matrices is to use a table of connectivity.

As an introduction to the use of the method, a sample connectivity table is given below. Details of the method will be introduced through a sample problem.

Element no. Local Element

Node 1 Node 2

1 1 2

2 2 3

Boundary and Load Conditions

Let us assume that:

1 0u = (node 1 is fixed) 2 3F F P= =

We have

11 1 11 1

1 22 1 2 1 1 2 2 2

2 33 2 22

0

0

f k kF uf f F k k k k u

uF k kf

+ +

= =

global node numbers

Therefore

1 11

1 1 2 2 232 2

0 0

0

k kFP k k k k uP uk k

+

=

which reduces to

21 2 232 2

uk k kPuP k k

+ =

and 1 1 2.F k u=

The unknowns are 23

uU

u

= and the reaction force 1F .

Solving the equations, we obtain the nodal displacements and the reaction force.

2 13

1 2

2

2

Pu ku P P

k k

=

+ 1 2F P=

Problem 1

For the spring system above,

1 100 /k N mm= , 2 200 /k N mm= , 3 100 /k N mm= 500P N= , 1 4 0u u= =

Find: a) the global stiffness matrix b) the displacements of nodes 2 and 3 c) the reaction forces at nodes 1 and 4 d) the force in the spring 2

Problem 2

For the above spring system, with arbitrarily numbered nodes and elements as shown, find the global stiffness matrix.

Solution :

Truss element

Note that the analysis is based on the fact that the elements used sustain only axial forcesrod or truss member

For this particular configurationa) each node of a b) the bar is no longer aligned with the global

axis

Local and Global Coordinates

The spring element can be used important engineering problemexample. Structures by frictionless hinges which imply that only tension or compression exists in the bars.shown below.

russ element having 2 DOF / NODE

Note that the analysis is based on the fact that the elements used sustain only axial forces (bars or rods). The terms

member can be used indifferently in what follows.

configuration, we consider the following:each node of a bar has 2 degrees of freedom

no longer aligned with the global x coordinate

Local and Global Coordinates

The spring element can be used as a simplified modelimportant engineering problems. A structure is one such

Structures normally consist of trusses, i.e. bars joined by frictionless hinges which imply that only tension or

ts in the bars. A typical truss structure is

Fig. 3

2 DOF / NODE

Note that the analysis is based on the fact that the elements The terms bar,

can be used indifferently in what follows.

e following:

x coordinate

model in many A structure is one such

normally consist of trusses, i.e. bars joined by frictionless hinges which imply that only tension or

A typical truss structure is

Analysis of truss structures provides an opportunity to discuss the concept of local and global coordinate systems.

In previous discussions, we considered only 1 DOF for each node of the spring element.

Let us now consider the more general configuration where each node of the spring element has 2 DOF. In addition, we suppose that the spring element is inclined at some angle to the horizontal x axis .

From the diagram, we see that it is necessary to consider 2 coordinate systems when dealing with the problem of springs having arbitrary orientations.

In previous sections, the spring elements were aligned with the x axis . The assembly process was facilitated in some respect, as all the spring elements had a common frame of reference. The elementary stiffness matrices only needed to be expanded before proceeding with the superposition.

In the case of trusses, for which the bar elements have a priori different orientations; assembly of individual elements will be possible only if a common reference is used for the different elements of the structure.

For systems consisting of non-aligned members, the Finite Element Analysis procedure would have to be modified in the following way:

1. Discretization 2. Formulation: model applied to individual elements (in

local coordinates) 3. Choice of a common reference system (global

coordinates) 4. Use of coordinate transformation to express the model

for individual elements in terms of global coordinates. (globalization)

5. Assembly of elements.

Notations (analysis of 1 spring element (no. 1))

Local coordinates: ( x , y )

11xu ,

11yu ,

12xu ,

12yu (nodal displacements)

11xf , 11yf , 12xf , 12yf (nodal forces) 1k (stiffness constant)

u

(nodal disp. vector) f (nodal force vector)

1eK (element stiffness matrix)

Global coordinates: ( X , Y )

11XU , 11YU , 12XU , 12YU (nodal displacements) 11XF ,

11YF ,

12XF ,

12YF (nodal forces)

U

(nodal disp. vector) F

(nodal force vector) 1GK

(element stiffness matrix)

Coordinate Transformations

1 1 11 1 11 1 11 1 1

.cos .sin.sin .cos

x X Y

y X Y

u U Uu U U

= +

= +

1 1 12 2 21 1 12 2 2

.cos .sin.sin .cos

x X Y

y X Y

u U Uu U U

= +

= +

Using the following convention: cos c = , sin s =

the relationship between the local and global nodal displacements is given by:

1 11 11 11 11 12 2

1122

0 00 0

0 00 0

x X

y Y

x X

Yy

u Uc s

u Us cc su Us c Uu

=

[7]

The 4 4 matrix that appears above is called a transformation matrix and will be denoted by T .

Similarly, we have a relationship between the local nodal forces and the global nodal forces.

Note that for elements that sustain only axial forces (tension/compression), the transverse local nodal forces are zero.

11 11 11

1 12 2

12

0 00 0 0

0 00 00

Xx

Y

x X

Y

Ff c sFs c

c sf Fs c F

=

[8]

Recall that the force-displacement relationship (written in local coordinate variables) for a spring element is given as

11 11 1 1 1

11 1

1 12 212

0 00 0 0 0 0

0 00 0 0 00

xx

y

x x

y

uf k ku

k kf uu

=

[9]

Substituting [7] & [8] in [9], we obtain

1 11 11 11 11 11 1

1 12 21 12 2

0 00 0 0 00 0 0 0 0 0 0 0

0 0 0 00 00 0 0 00 0 0 0

X X

Y Y

X X

Y Y

F Uk kc s c sF Us c s c

c s c sk kF Us c s cF U

=

Written in symbolic notation

. 1. .eT F K T U=

[10]

Pre-multiplying [10] by TT , we obtain

. . . 1. .T T eT T F T K T U=

Since T is an orthogonal matrix, . TT T I=

Therefore, we obtain the expression for the force-displacement relationship in terms of variables written in the global coordinate system;

. . 1 .T eF T K T U=

[11]

The stiffness matrix 1GK can be simplified as follows:

Stiffness matrix written in global

coordinates : 1GK

1 1

11 1

1

0 00 0 0 00 0 0 0 0 0 0 0

0 0 0 00 00 0 0 00 0 0 0

0 0 1 0 1 0 0 00 0 0 0 0 0 0 0

0 0 1 0 1 0 0 00 0 0 0 0 0

G

k kc s c ss c s cK

c s c sk ks c s c

c s c s

s c s ckc s

s c

=

=

0 0c s

s c

1 1

0 00 0 0 0 0 0

0 00 0 0 00 0

G

c s c s c s

s cK kc s c sc s

s c

=

2 2

2 2

2 2

2 2

1 1 G

c cs c cs

cs s cs sK kc cs c cs

cs s cs s

=

Example

Let us consider the case of a three member truss.

Find the displacements of the truss members given the load P .

Problem

Solution : Problem

Construct the stiffness matrix of the above given truss. Determine the nodal displacements and reaction forces. Find the normal stress in element 3.

1 2 2L L m= = 3 2 2L m= 1 2 3

280A A A mm= = = 200E GPa=

Solution :