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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, r, E, g – cross section area, length, material density,
Young’s modulus, gravity acceleration
Fig.1 Problem solved
STEP 1 – MESH GENERATION
1D continuum is divided into a set of elements and
nodes, according to Fig.2.
Fig.2 FE mesh
STEP 2 – DISPLACEMENT APPROXIMATION
We concentrate now on the element no.1. Its displacement u(x) is approximated by
linear shape functions
u( x )  N. 
Where
N  [ N1 , N 2 ]
is the shape function matrix,
  [u1 , u2 ]T
matrix of deformation parameters, which have a
simple physical interpretation of displacements of mesh nodes - see Fig.3.
x1
x2
u1
u2
Lp
Fig.3 Bar element
Linear shape functions N1 
x2  x
,
x2  x1
N2 
x  x1
,
x2  x1
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
u( x)  N1 ( x).u1  N2 ( x).u2
as can be seen in Fig.4
Fig.4 Approximation of displacement
STEP 3 – ELEMENT STIFFNESS MATRIX
From displacements, the approximation of strain and stress can be obtained,

d
(N.δ)  B.δ
dx
  E.B.δ  δT .BT .E
and
where B is the matrix obtained from N by derivation
B
dN
1

[1, 1]
dx x2  x1
B
or
1
[1 , 1]
Lp
,
if we denote the length of element as Lp  x2  x1 .
Inserting stress and strain into the strain energy W1 
we obtain after some manipulations
x2

1 T 
1
W1  δ . ES  BT Bdx.δ  δT .k.δ

2  x1
2

where k is the element stiffness matrix
k
ES  1  1
Lp  1 1 
x2

x1
1
2
 Sdx
STEP 4 – ELEMENT LOAD MATRIX
By analogy, load matrix can be obtained inserting the displacement
x2
approximation into the potential of external forces
which results in P1  δ .f .
x1
T
The load matrix f 
P1   ur gS dx ,
1
1
rgSLp  represents total external loading of
2
1
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
energy of the 1st element can be written
W1 
1 T
U .K 1.U
2
U  [u1, u2 , u3 , u4 ]T , strain
where
 1 1

ES  1 1
K1 
Lp  0 0

0 0
0
0
0
0
0
0
0

0
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
0 0 0 0
0 0 0 0 




ES 0 1  1 0
ES 0 0 0 0 
.
K2 
,
K3 
L p 0  1 1 0
L p 0 0 1  1




0 0 0 0
0 0  1 1 
As the total strain energy is a sum of element contributions, we can write it as
3
1
1
W  Wi  UT .K1  K 2  K 3 .U  UT .K.U ,
2
2
i 1
where K is the global stiffness matrix of the structure
 1 1 0 0 


ES  1 2  1 0 
K
L p  0  1 2  1


 0 0 1 1 
.
By the same procedure, global load matrix F can be obtained as a sum of
element contributions to the total potential of external forces :
3
P   Pi  UT .F1  F2  F3   UT .F ,
i 1
1 
 2
1
F  rgSLp   .
 2
2
 
1 
STEP 6 – RESULTING SYSTEM OF EQUATIONS
Using previous matrices K, F, total potential energy can be expressed as
1
  UT .K.U  UT .F
2
From the stationary condition

0
U
we obtain the system of four linear algebraic equations for the four deformation
parameters u1, u2, u3, u4 :
K .U  F
.
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:
 2 1 0 
ES 
,
K

1
2

1

Lp 
 0  1 1 
u2 
U  u3 ,
u4 
 2
1
F  rgSLp 2
2
1 
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 EQUATIONS
Solution 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
FE
M
Fig.6 Stress in the bar