Lecture 6; The Finite Element Method
1-dimensional spring systems
Lecture 6; The Finite Element Method
1
1-dimensional spring systems
(modified 2011-09-21)
With a 1-dimensional spring system we here mean
springs in series
where the individual springs are
linear, i.e. expose linear force-elongation relations
k1
k2
1
2
Note that the identity of a spring is indicated by a ring-symbol, while its
stiffness is referred to as k
Lecture 6; The Finite Element Method
2
1-dimensional spring systems
(modified 2011-09-21)
Why study (linear) spring systems?
• You are familiar with springs, since you have met them in a number of
courses
• By using what you already know (e.g. equilibrium), we may find a matrix
formulation which has the same structure as an FE-formulation for a
general elastostatic problem of Solid Mechanics
Lecture 6; The Finite Element Method
3
1-dimensional spring systems
(modified 2011-09-21)
Example problem
Let us first, in order to have something to compare with, in the
"traditional" way determine the spring forces and spring elongations in the
example below. We will later on solve the same problem by a structured
FE-treatment.
F
k
k
F
k
1
3
2
1
2
Lecture 6; The Finite Element Method
4
1-dimensional spring systems
(modified 2011-09-21)
3
4
Example problem; cont.
In order to solve our problem, we need to look at equilibrium,
constitutive relations and compatibility.
First, make free body diagrams of the nodes/connection points 2 and 3,
resp., and study equilibrium (T denotes a spring force)
F
k
k
F
k
1
3
2
1
4
2
3
F
T1
F T
2
2
 T2  F  T1  0
A statically indeterminate
problem
(2 eq. & 3 unknowns)
Lecture 6; The Finite Element Method
5
1-dimensional spring systems
(modified 2011-09-21)
T2
T3
3
 T3  F  T2  0
Example problem; cont.
The constitution tells us (linear springs)
F
k
k
F
k
1
3
2
1
2
3
T1  k1 , T2  k 2 , T3  k 3
where δ denotes the spring elongation. We now have 5 eq. for 6
unknowns, which we fix by compatibility
1   2   3  0
Let us now do some calculations (see the next page)
Lecture 6; The Finite Element Method
6
1-dimensional spring systems
(modified 2011-09-21)
4
Example problem; cont.
F
k
k
F
k
1
3
2
1
By equilibrium
4
2
3
By constitution and compatibility
T1  T3  T2  F
T1  T2  T3  0
With the left expressions inserted into the right one we get
2F
T2  
3
F
 T1  T3 
3
The elongations are finally found by inserting the obtained spring forces
into the constitutive relations (only a scaling with k)
Lecture 6; The Finite Element Method
7
1-dimensional spring systems
(modified 2011-09-21)
Structured analysis of spring systems
Let us now focus on a general structured FE approach for spring
systems, which contains the following steps
I.
Find, for each spring, a relation between the spring loading and the
spring displacements
II. Find, for the spring structure, a relation between the structural loading
and the structural displacements
III. Solve the structural problem for the given loading and restraints, and
calculate entities such as spring forces and spring elongations
Let us now see how this is done!
Lecture 6; The Finite Element Method
8
1-dimensional spring systems
(modified 2011-09-21)
Step I
Find, for each spring, a relation between spring loading and the
spring displacements
Lecture 6; The Finite Element Method
9
1-dimensional spring systems
(modified 2011-09-21)
Spring equilibrium
Let us now study equilibrium for an arbitrary spring in a spring system as
illustrated below, where the index e refers to the element number, where
the spring nodes/end points are labeled n1e and n2e, resp., and where
the spring loading is described by the forces f1e and f2e, resp.
f1e
ke
n1e
n2 e
f 2e
e
x
Note that the forces are defined positive in the positive x-direction!
Lecture 6; The Finite Element Method
10
1-dimensional spring systems
(modified 2011-09-21)
Spring equilibrium; cont.
Let us make an imaginary cut in the spring, insert the spring force Te, and
study equilibrium for each part of the spring
f1e
n1e

n2 e
Te
f1e  Te  0
  Te  f 2e  0
On matrix form we thus have
 f1e   1
     Te
 f 2e   1 
Lecture 6; The Finite Element Method
11
1-dimensional spring systems
(modified 2011-09-21)
f 2e
Spring equilibrium; cont.
Let us now proceed and look at the spring deformation, where the
elongation (as before) is labeled δe, while the displacement of its end
nodes are called d1e and d2e; resp.
d1e n1e
n2 e
Obviously
 e  d 2e  d1e
or, on matrix form
 d1e 
 e   1 1 
d 2 e 
Lecture 6; The Finite Element Method
12
1-dimensional spring systems
(modified 2011-09-21)
d 2e
Spring equilibrium; cont.
Since the springs are linear, we have
Te  ke  e
Lecture 6; The Finite Element Method
13
1-dimensional spring systems
(modified 2011-09-21)
Transformation diagram
We thus have got the following, so called, transformation diagram
 1 1T
 f e
Te
 f e
e
d e
 f1e 
 
 f 2e 
spring loading or
element forces
ke
1 1
d e
 d1e 
 
d 2e 
Lecture 6; The Finite Element Method
14
1-dimensional spring systems
(modified 2011-09-21)
spring displacements or
element displacements
Transformation diagram; cont.
A counter-clockwise trip in the transf. diag. will give us the so called
spring stiffness [k]e, which relates the element displacements to the
element forces
 1 1T
Te
 f e
 f e
ke
e
1 1
d e
 1
 1
 1
  Te   ke e   ke  1 1d e
1
1
1

k e
Lecture 6; The Finite Element Method
15
1-dimensional spring systems
(modified 2011-09-21)
Transformation diagram; cont.
The complete transf. diag. for the spring thus takes the form
 1 1T
Te
 f e
 f e  k e d e
k e
ke
e
1 1
d e
Lecture 6; The Finite Element Method
16
1-dimensional spring systems
(modified 2011-09-21)
 1  1
, k e  ke 

 1 1 
Step II
Find, for the spring structure, the relation between the structual
loading and the structural displacements
Lecture 6; The Finite Element Method
17
1-dimensional spring systems
(modified 2011-09-21)
Structural loading and structural displacements
For the spring structure, we introduce structural nodes, with associated
structural loads and structural displacements, according to the example
below
F1
F2
1
F3
3
2
D2
Lecture 6; The Finite Element Method
18
4
2
1
D1
F4
1-dimensional spring systems
(modified 2011-09-21)
3
D3
D4
Structural loading and structural displacements; cont.
We introduce the structural load matrix {F} and the structural
displacement matrix {D} as illustrated below
F1
F2
1
F3
3
2
D2
T


F  F1 F2 F3 F4
,
T


D  D1 D2 D3 D4
Lecture 6; The Finite Element Method
19
4
2
1
D1
F4
1-dimensional spring systems
(modified 2011-09-21)
3
D3
D4
The structural problem
What we now seek is a relation between the structural load matrix {F} and
the structural displacement matrix {D}, which will take the form
F  K D
F1
F2
1
F3
3
2
D1
F4
4
2
1
D2
3
D3
D4
Note that we here use lower case letters for spring related quantities, and
capital letters for structural quantities.
Lecture 6; The Finite Element Method
20
1-dimensional spring systems
(modified 2011-09-21)
The structural problem; cont.
In the transformation diagram we thus have to find
 1 1T
Te
 f e
?
F 
k e
ke
e
1 1
K  ?
d e
D
?
Lecture 6; The Finite Element Method
21
1-dimensional spring systems
(modified 2011-09-21)
The relation between spring disp. and structural disp.
We fix this by so called connectivity matrices (unique for each spring).
For spring 1 we get
f1e 1
 D1 
f 2e1
 
 d1e1  1 0 0 0  D2 


D 
d
0
1
0
0
 3 
 2e1  
 D4 
d   [C ] D
1
1
F1
d 2e1
F4
F3
3
2
4
2
1
D2
Lecture 6; The Finite Element Method
22
n2 e 1
d1e 1
F2
1
D1
n1e1
1-dimensional spring systems
(modified 2011-09-21)
3
D3
D4
The relation between spring disp. and structural disp.; cont.
For spring 2 we get
 D1 
 
 d1e 2  0 1 0 0  D2 


D 
d
0
0
1
0
 3 
 2e2  
 D4 
d 2  [C ]2 D
F1
n2e2
d 2e2
d1e2
F4
F3
3
2
4
2
1
D2
Lecture 6; The Finite Element Method
23
f 2e2
n1e  2
F2
1
D1
f1e2
1-dimensional spring systems
(modified 2011-09-21)
3
D3
D4
The relation between spring disp. and structural disp.; cont.
For spring 3 we get
 D1 
 
 d1e3  0 0 1 0  D2 


D 
d
0
0
0
1
 3 
 2 e 3  
 D4 
d 3  [C ]3 D
F1
n2e3
d 2 e 3
d1e3
F4
F3
3
2
4
2
1
D2
Lecture 6; The Finite Element Method
24
f 2 e 3
n1e 3
F2
1
D1
f1e 3
1-dimensional spring systems
(modified 2011-09-21)
3
D3
D4
The relation between spring disp. and structural disp.; cont.
Thus,
 1 1T
Te
 f e
?
F 
k e
ke
e
1 1
d e
K  ?
C e
D
Lecture 6; The Finite Element Method
25
1-dimensional spring systems
(modified 2011-09-21)
The relation between spring loads and structural loads
Let us now continue, and study the relation between spring loads and
structural loads. If looking closely at the structural nodes, and using
Newtons 3:rd law and the previously defined spring loads, we get the
following situation
1
f1e 1
1
f 2e1
2
f1e 2
F4
F3
F2
F1
f 2e2
2
3
f1e 3
3
f 2 e 3
4
Equilibrium implies
F1  f1e1  0
F3  f 2e  2  f1e3  0
F2  f 2e1  f1e  2  0
Lecture 6; The Finite Element Method
26
1-dimensional spring systems
(modified 2011-09-21)
F4  f 2e3  0
The relation between spring loads and structural loads; cont.
Thus
F1  f1e1
F2  f 2 e1  f1e  2
F3  f 2 e 2  f1e 3
F4  f 2 e3
or
 F1  1
 F  0
 2 
 
 F3  0
 F4  0
or, actually
0
0
1  f1e1  1

 
0  f 2e1  0
0
0

0
0
0  f1e2  0

 
1   f 2 e  2  1
0
0

no el
F    C Te  f e
e 1
Lecture 6; The Finite Element Method
27
1-dimensional spring systems
(modified 2011-09-21)
0
0  f1e3 


0   f 2 e 3 
1
The relation between spring loads and structural loads; cont.
Thus
 1 1
no el
 C e
T
Te
 f e
e 1
F 
k e
ke
e
1 1
d e
Lecture 6; The Finite Element Method
28
T
1-dimensional spring systems
(modified 2011-09-21)
K  ?
C e
D
The structural problem
By an anti-clockwise trip in the transf. diag. we get
no el
 1 1
T
 C e
T
Te
 f e
e 1
F 
no el
F    C Te k e C e D
k e
ke
e
1 1
d e
C e
D
Lecture 6; The Finite Element Method
29
1-dimensional spring systems
(modified 2011-09-21)
e 1

K 
The structural problem; cont.
external loads
internal loads
no el
 1 1
T
 C e
T
Te
 f e
e 1
k e
ke
e
1 1
d e
F   K D
F 
K 
C e
D
Lecture 6; The Finite Element Method
30
1-dimensional spring systems
(modified 2011-09-21)
K  
no el
T
 C e k e C e
e 1
 1  1
k e  ke 


1
1


Step III
Solve the structural problem for given loading and restraints,
and calculate entities such as spring forces and
spring elongations
Let us see what this looks like in our example!
Lecture 6; The Finite Element Method
31
1-dimensional spring systems
(modified 2011-09-21)
Example problem
Find the spring forces and spring elongations in the example below
F
k
k
F
k
1
3
2
1
2
With the connectivity matrices previously found, we get
(see the next page)
Lecture 6; The Finite Element Method
32
1-dimensional spring systems
(modified 2011-09-21)
3
4
Example problem; cont.
1
0
no el
T
K    C e k e C e  
0
e 1
0

0
1

0
0

0
1  1  1 1 0 0 0
k





0    1 1  0 1 0 0 
0
0
0 0 
0  1  1 0 1 0 0 0 0  1  1 0 0 1 0
k
k





1  1 1  0 0 1 0 1 0  1 1  0 0 0 1
0 1 
0


Lecture 6; The Finite Element Method
33
1-dimensional spring systems
(modified 2011-09-21)
Example problem; cont.
 1 1 0 0 
 1 2  1 0 

K   k 
 0  1 2  1
 0 0 1 1 


We thus have to solve
 1  1 0 0   D1 = 0  F1 = ?
 1 2  1 0   D = ?  F = F
  2    2 
k
 0  1 2  1  D3 = ?  F3 = - F
 0 0  1 1   D = 0  F = ?

 4   4 
Lecture 6; The Finite Element Method
34
1-dimensional spring systems
(modified 2011-09-21)
Example problem; cont.
The unknown displacements D2 and D3 are now found by solving the
equation system one gets when eliminating all rows and columns
associated with locked displacements
implying
 1  1 0 0   D1 = 0  F1 = ?
 1 2  1 0   D = ?  F = F
  2    2 
k
 0  1 2  1  D3 = ?  F3 = - F
 0 0  1 1   D = 0  F = ?

 4   4 
 D2  1 2 1  F   1  F
 2  1  D2   F 
k
     
  


 1 2   D3   F 
 D3  3k 1 2  F   1 3k
Lecture 6; The Finite Element Method
35
1-dimensional spring systems
(modified 2011-09-21)
Example problem; cont.
The spring elongations are then found by (c.f. the transformation
diagram)
 e   1 1C e D
0
which for instance gives
 
1 0 0 0  1  F F
 e1   1 1
  

0 1 0 0  1 3k 3k
 0 
Finally, the spring forces are found by (c.f. the transformation diagram)
Te  ke e
(  ke  1 1C e D )
Implying e.g. that
Te1 
F
3
Lecture 6; The Finite Element Method
36
1-dimensional spring systems
(modified 2011-09-21)
As before :)
Example problem; cont.
The way used for finding the structural problem described above
(assembling the spring stiffnesses by using connectivity matrices)
constitutes the general structured way to describe FEM.
In a hand calculation situation, it is often easier to use the procedure
described below (not using connectivity matrices).
Also when considering implementation in a computer code it is
generally too expensive to directly work with connectivity matrices (both
with respect to computational time and memory allocation). Instead one
saves the positions of the non-zero elements in the connectivity matrices.
Lecture 6; The Finite Element Method
37
1-dimensional spring systems
(modified 2011-09-21)
Example problem; cont.
The “direct identification” method
The contribution to [K] from element 1, which we here call [K]1, is
= D1 = D2
d1e 1 d 2e1
 1  1 d1e 1= D1
k 1  k 


1
1

 d 2e1= D2
D1
1
 1
K 1  k 
0
0

Lecture 6; The Finite Element Method
38
1-dimensional spring systems
(modified 2011-09-21)
D2
1
1
0
0
D3 D4
0 0
0 0

0 0
0 0
D1
D2
D3
D4
Example problem; cont.
The “direct identification” method; cont.
The contribution to [K] from element 2, which we here call [K]2, is
= D2 = D3
d1e 2 d 2 e  2
k 2
 1  1 d1e 2= D2
 k

 1 1  d 2 e =2 D3
K 2
Lecture 6; The Finite Element Method
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1-dimensional spring systems
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D1 D2
0 0
0 1
 k
0  1
0 0

D3 D4
0 0
 1 0

1 0
0 0
D1
D2
D3
D4
Example problem; cont.
The “direct identification” method; cont.
The contribution to [K] from element 3, which we here call [K]3, is
= D3 = D4
d1e3 d 2 e 3
 1  1 d1e3= D3
k 3  k 


1
1

 d 2 e =3 D4
D1 D2 D3 D4
0 0 0 0 
0 0 0 0 

K 3  k 
0 0 1  1
0 0  1 1 


Lecture 6; The Finite Element Method
40
1-dimensional spring systems
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D1
D2
D3
D4
Example problem; cont.
The “direct identification” method; cont.
The resulting structural stiffness is found by adding all contributions
(giving the same result as before)
3
K    K e
e 1
1
0
0
1
 1 (1  1)
1
0

 k
 1 (1  1)  1
0
0

0

1
1


Lecture 6; The Finite Element Method
41
1-dimensional spring systems
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Example problem; cont.
The “direct identification” method; cont.
After having solved the structural problem, we proceed in the following
way when calculating the spring forces (note that we in the “direct
identification” procedure have not set up any connectivity matrices).
no el
 1 1
T
 C e
T
Te
 f e
e 1
F 
k e
ke
e
1 1
d e
K 
C e
D
Lecture 6; The Finite Element Method
42
1-dimensional spring systems
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Te  ke  1 1d e
Example problem; cont.
The “direct identification” method; cont.
As an example, we get for element 1
no el
T1 T
e
 1 1
T
 C e
T
 f e
e 1
F 
k e
ke
=k
e
1 1
 d1e = D1
 
d 2 e = D2
K 
C e
D
Lecture 6; The Finite Element Method
43
1-dimensional spring systems
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 D1 
T1  k  1 1  
 D2 
 0 
 k  1 1

 F / 3k 
 F /3
As before :)
An alternative treatment of our example problem by
taking advantage of the symmetry
F
k
k
F
k
1
3
2
2
1
3

F
k
1
2
1
Lecture 6; The Finite Element Method
44
4
1-dimensional spring systems
(modified 2011-09-21)
2
Decreasing the
length by two,
increases the
2k
3 stiffness by two!
Example problem; cont.
The contribution to [K] from element 1, which we here call [K]1, is
D1 D2 D3
= D1 = D2
d1e 1 d 2e1
 1  1 d1e 1= D1
k 1  k 

 1 1  d 2e1= D2
 1  1 0 D1
K 1  k  1 1 0 D2
 0 0 0 D3
Lecture 6; The Finite Element Method
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1-dimensional spring systems
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Example problem; cont.
The contribution to [K] from element 2, which we here call [K]2, is
= D2 = D3
d1e 2 d 2 e  2
k 2
 1  1d1e 2= D2
 2k 


1
1

d 2 e  =2 D3
K 2
Lecture 6; The Finite Element Method
46
1-dimensional spring systems
(modified 2011-09-21)
D1 D2
0 0
 2 k 0 1

0  1
D3
0  D1
 1 D2

1  D3
Example problem; cont.
We thus have to solve
 1  1 0   D1 = 0  F1  = ?
   
k  1 3  2  D2 =? F2  = F


 0  2 2   D3 = 0  F3 = ?
 1  1 0   D1   F1 
   
k  1 3  2  D2    F2  = F


 0  2 2   D3   F3 
and gets
 D2  F / 3k
As before :)
Lecture 6; The Finite Element Method
47
1-dimensional spring systems
(modified 2011-09-21)
Some final comments
The spring stiffness matrix [k]e and the structural stiffness matrix [K] are
symmetric and singular (can not be inverted).
However, if we have prevented rigid body motions, we may solve the
structural problem (the reduced stiffness matrix obtained by removing
rows and columns associated with locked displacements is not singular).
One may finally also note that the sum of all components in a row or
column of [k]e or [K] is always zero.
Lecture 6; The Finite Element Method
48
1-dimensional spring systems
(modified 2011-09-21)