Illustration of FE algorithm on the example of 1D problem

Problem:

Stress and displacement analysis of a one-dimensional bar, loaded only by its own weight, acting in the direction of the x-axis Given data: S, L, , E, g – cross section area, length, material density,

Young’s modulus, gravity acceleration

Fig.1 Problem solved

STEP 1 – MESH GENERATION1D continuum is divided into a set of elements and nodes, according to Fig.2.

Fig.2 FE mesh

STEP 2 – DISPLACEMENT APPROXIMATIONWe concentrate now on the element no.1. Its displacement u(x) is approximated by linear shape functions

Where is the shape function matrix,

matrix of deformation parameters, which have a simple physical interpretation of displacements of mesh nodes - see Fig.3.

Fig.3 Bar element

u x( ) .N

N [ , ]N N1 2

[ , ]u u T1 2

u1 u2

Lp

x1 x2

Linear shape functions

where x1, x2 are the nodal coordinates according to Fig.3.

Displacement of any point of the bar is thus determined by displacement of its nodes

as can be seen in Fig.4

Fig.4 Approximation of displacement

,,12

12

12

21 xx

xxN

xx

xxN

2211 ).().()( uxNuxNxu

STEP 3 – ELEMENT STIFFNESS MATRIXFrom displacements, the approximation of strain and stress can be obtained,

and

where B is the matrix obtained from N by derivation

or ,

if we denote the length of element as .

Inserting stress and strain into the strain energywe obtain after some manipulations

where k is the element stiffness matrix

δBδN .).( dx

d EE TT .... BδδB

]1,1[1

12

xxdx

dNB ]1,1[

1

pLB

12 xxLp

δkδδBBδ ..2

1..

2

1 2

1

1T

x

x

TT dxESW

11

11

pL

ESk

2

1

21

1

x

x

SdxW

STEP 4 – ELEMENT LOAD MATRIX

By analogy, load matrix can be obtained inserting the displacement

approximation into the potential of external forces ,

which results in .

The load matrix represents total external loading of

element (in this particular case its weight), localised into point forces acting in the nodes of the element.

STEP 5 – GLOBAL STIFFNESS AND LOAD MATRIX

Creating global matrix of deformation parameters , strain energy of the 1st element can be written

2

1

1

x

x

dxgSuP

fδ .1TP

1

1

2

1pgSLf

Tuuuu ],,,[ 4321U

UKU ..2

111

TW

where

is the stiffness matrix of the first element, enhanced by zero-element lines and columns to enable multiplication by the matrix U. By the same logic, stiffness matrices of the second and third element can be written as

.

As the total strain energy is a sum of element contributions, we can write it as

,

where K is the global stiffness matrix of the structure

.

0000

0000

0011

0011

1pL

ESK

1100

1100

0000

0000

,

0000

0110

0110

0000

32pp L

ES

L

ESKK

UKUUKKKU ..2

1..

2

13

1321

T

i

TiWW

1100

1210

0121

0011

pL

ESK

By the same procedure, global load matrix F can be obtained as a sum of element contributions to the total potential of external forces :

,

.

STEP 6 – RESULTING SYSTEM OF EQUATIONSUsing previous matrices K, F, total potential energy can be expressed as

From the stationary condition

FUFFFU .. 321

3

1

TT

iiPP

1

2

2

1

2

1pgSLF

FUUKU ...2

1 TT

0U

we obtain the system of four linear algebraic equations for the four deformation

parameters u1, u2, u3, u4 : .

This is the basic equation to be solved. We can see that the stiffness matrix K is the matrix of coefficients. Detailed inspection of our K matrix shows, nevertheless, that it is singular and that the equations have thus no unambiguous solution. This is caused by the fact that we have not specified any boundary condition yet. We should remember a very important fact:

In the static problems of displacement version of FEM, at least so many boundary conditions must be defined to prevent free motion of the analysed model. If this condition is not met, the resulting stiffness matrix is singular and solution of equations usually breaks down.

To meet this condition in our case, we prescribe zero displacement for the first unknown, u1 = 0. This means that from the system of equations we must leave out the first equation, which results in the following form of our global matrices:

FUK .

1

2

2

2

1,,

110

121

012

4

3

2

pp

gSL

u

u

u

L

ES FUK

If we look at the structure of K matrix now, we can see it is a symmetric, positive definite band matrix with dominant diagonal, which is very helpful for quick, stable and efficient computational solution of very large systems of equations.

STEP 7 – SOLUTION OF EQUATIONSSolution of the basic system of equations yields nodal displacements, from which the approximation of element displacement, strain and stress can be obtained according to the equations given above. Figs.5 and 6 show the difference between the analytical and numerical solution of this illustrative example. We should note, that the exact correspondence of numerical and analytical solution in nodal displacements is only a consequence of a very simple problem, it cannot be generalized. Typical for the elements with linear shape functions is, of course, linear approximation of displacements, and constant approximation of stress and strain on elements, with no continuity between elements - see Fig.6.

Analytic solution

FEM

Fig.5 Displacements of bar

Analytic solution

FEM

Fig.6 Stress in the bar