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Element calculations and Assembly
In the last lecture, general expressions were developed for the shiftness matrix and the
force vector using the piecewise linear basis functions. These expressions are to be
evaluated as the sums of the integrals over all the elements. This lecture discusses the
systematic procedure of this evolution. First, the integrals are evaluated elementwise and
then they are assembled over all the elements. For the convenience of the evolution of the
integrals over the elements, local or element notation is introduced.
6.1Local or Element Notation
In the last lecture, we introduced a numbering scheme for the elements, nodes, nodal
coordinates and the basis functions. This is called a global numbering scheme. It is shown
in fig 6.1.
FIG 6.1- Global Numbering Scheme
For the convenience of the element wise evaluation of the shiftness matrix and the force
vector, at the element level , we introduce another scheme for the nodes, nodal
coordinates and the basis functions. This scheme is called the local or element numbering
scheme. For kth element, this numbering scheme is illustrated in fig 6.2
FIG 6.2 – Elemental Numbering Scheme for k-th element
For kth element, note that, the global numbers of the two end nodes are k and
k+1respectively (fig 6.1). Now we number them as 1 and 2 (fig 6.2). We call ie local or
element numbering scheme. Note that the global notation for the coordinated of the end
nodes is xk and xk+1 (fig 6.1). Now, we denote them as x1k and x2k (fig 6.2). This is called
local or element notation. The subscript here denotes the local node number. For every
element, the nodes will always be numbered as 1 and 2 according to the local numbering
scheme. Therefore to identify the element under consideration, a superscript is added to
the notation of the coordinates. The superscript denotes the element number. Figure 6.1
shows that the only non-zero basis functions for the k……… element are  k and k1 .
This is the global notation for the basis functions. Under the local or element notation, we
denote them as N1k and N2k . Here, the subscript denotes the local number of the node at
which the basis function attains the value unity and the superscript represents the element
number under consideration.
These functions are called as the shape or interpolating functions.
Using equation (5.3), the expressions for the shape functions can be written asx
x
N1k   k  k 1
hk
x k2  x

hk
x  xk
N k2   k 1 
hk
x  x1k

6.1
hk
When the expression (5.2) for the element size becomes
h k  x k 1  x k
 x k2  x1k
Consider the expression (5.5) for the approximate solution:
u h (x) 
N 1
 u j  j (x)
6.3
j 1
Note that, since
is unity at node j (fig 6.1), the unknown coefficient uj is the
unknown displacement at node j. The unknowns uj are called as the degrees of freedom
of the rod. Since, the subscript j denotes the global number of the node, they are called as
global degrees of freedom. For the k-th element, the only non zero basis functions are
k and k1 .Therefore, the expression for u h (x) over the k-th element becomes
u h  u k k  u k  1 k  1
6.4
For the convenience of element wise calculations of the shiftness matrix and force vector.
We introduce another notation for u k and u k 1 . We denote them as u1k and u k2 . Here,
the subscript denotes the local number of the node to which the displacement belongs and
the superscript represents the element under consideration. This notation is called the
local or element notation and (u1k , u k2 ) are called the local or element degrees of freedom
of the k-th element. Then the expression (6.4) under the element notation becomes
u h  u1k N1k  u k2 N k2
6.5
6.2 Expressions for Element Shiftness matrix and Force Vector
Now, we shall determine the row and column number of the matrix equation (5.14) to
which the k-th element makes the contribution. For this purpose, consider equation(5.13):
N 1
 k ij u j  Fi
for i 1,2,.... , N  1
6.6
j 1
Where the expressions for the shiftness matrix Kij and the force vector Fi are given by
equations (5.11) and (5.12):
K ij 
N
X K 1
 
EA
K 1 X K
di d j
dx
dx dx
for i  1,2,..., N  1;
j 1,2,...., N  1
and
Fi 
6.7
X k 1
N
  f i . dx   i |
xL
for i  1,2,..., N  1
K 1 X k
6.8
The above expression can be written as
k ij 
N
 k ij(k )
for i  1,2,..., N  1
k 1
j  1,2,...., N  1
k ij( k ) 
x k 1
di d j
EA
 dx  dx dx
6.9
for i  1,2,...., N  1
xk
j  1,2,...., N  1
6.10
and
Fi 
Fi( k )
N
 Fi(k )  i
k 1
x k 1

 fi  dx
xk
|
x L
for i  1,2,..., N  1
6.11
for i  1,2,..., N  1
6.12
Note th at
k ij( k )
and
Fi( k )
represent the contributions of the k-th element to the shiftness
matrix k ij and the force vector Fi . The second term of equation (6.11) is a contribution
from the last element only. This term needs to be treated separately and therefore not
included in the expression for Fi( k ) and the present discussion. Using the expressions
(6.10) and (6.12), the contribution to equation (6.6) of the k-th element can be written as
N 1
 Kij(k)u j  Fi(k )
for i  1,2,...., N  1
6.13
j1
Note that the values of I for which i ( x ) is nonzero over the k-th element are i = k and
i=k+1 . Therefore, the only nonzero rows of k ij( k ) and Fi( k ) are I=k and I=k+1. Thus, the
only nonzero equations of the set (6.13) are I=k and I=k+1. These expressions are as
follows:
ik:
N 1  x k 1

 x k 1

dk d j 


   EA dx dx dx u j    fk dx 
j1  x k

 xk

i  k 1:
6.14
N 1  x k 1

 x k 1

dk 1 d j 


   EA dx dx dx u j    fk 1dx 
j1  x k

 xk

6.15
Since, the only values of j for which  j (x) is nonzero over the interval (x k , x k 1) are j=k
and k+1. Therefore, the above equations become:
ik
 x k 1

 x k 1

 x k 1

dk dk 
dk dk 1 




  EA dx dx dx u k    EA dx dx dx u k 1    fk dx 
 x

 x

 x

 k

 k

 k

i  k 1
6.16
 x k 1

 x k 1

 x k 1

dk 1 dk 
dk 1 dk 1 




dx u k 1    fk 1dx 
  EA dx dx dx u k    EA dx
dx
 x

 x

 x

 k

 k

 k

6.17
In matrix form, this can be written as
[k ( k ) ]{u}  {F( k ) }
where
6.18
[k ( k ) ] 
j k
j  k 1
x k 1
 x k 1

dk dk
dk dk 1


EA
dx
EA
dx

dx dx
dx dx
 

xk
 xk

x k 1
 x k 1
dk 1 dk
dk 1 dk 1 
EA
dx
EA
dx 

i  k 1


dx dx
dx
dx
 x

xk
 k

ik
6.19
and
 x k1

i  k   fk dx 
 xk

{F( k ) } 
x k1


i  k  1 fk 1dx 


 xk

In elemental notation, the above quantities can be written as
[k ( k ) ] 
j k
j  k 1
 x2
dN k
  EA 1
dx
 x1k
 x k2
dN k2

EA
 k
dx
 x1
k
ik
i  k 1
dN1k dN k2 
k EA dx dx dx 
x1

x k2
dN1k
dN k2 dN k2 
dx  EA
dx 
dx
dx
dx
k
x1

dN1k
dx
dx
6.20
x k2
6.21
and
{F( k ) } 
 xk

 2 K 
  f N1 dx 
ik
 xk

 1

 k

 x2


i  k 1  f NK
dx
2
 

 x1k

6.22
Note that, the elements of k ij( k ) and Fi( k ) are zero when i  k or k+1 and j  k or k+1.
Thus,
k ij( k )  0
for i  k, k  1; j  k, k  1
and
Fij( k )  0
6.23
for i  k, k  1
6.24
Therefore, these quantities are often expressed as
x k
 2
dN1K
EA

dx
x k
(k )
[k ]   1k
x
 2
dN K
2
  EA
dx
x k
 1
and
dN1K
dx
dx
dN1K
dx
dx


dN1K dN K
2
 EA dx dx dx 

xk
1


xk
2
K

dN K
dN
2
2
 EA dx dx dx 

xk
1

xk
2
6.25
x k

 2 k 
  fN 1 dx 
x k



{f ( k ) }   1k

x 2

 fN k dx 
 2 
x1k

6.26
The matrix (6.25) is called as the element shiftness matrix whereas the vector (6.26) is
called as the element force vector. Note that the notation for the elemental quantities uses
small letters : small k for the shiftness matrix and small f for the force vector. The
notation for the corresponding full matrix or full vector uses capital letters : capital K for
the shiftness matrix and capital F for the force vector.
The expressions (6.25) and (6.26) help in the systematic evaluation of the global shiftness
matrix K ij and global force vector Fi of equation (6.6). The procedure can be described as
follows:
(i)
First, the shiftness matrix and the force vector of each element is evaluated
using the expressions (6.25) and (6.26).
This can be done using a DO loop. Thus, for k=1,…N, the elements shiftness matrix
k ij( k ) and element force vector f ij( k ) are evaluated using
k ijk 
f ik 
x k2
k
dN ik dN j
 EA dx dx dx
k
x1
x k2
for i  1,2; j  1,2
6.27
 fN i dx
k
for i  1,2
x1k
6.28
(ii)
k
(k )
Next, these quantities are expanded to the full size [K ] and {F } . This is
done using the relationship between the local and global node numbering
systems.
(iii)
Finally, the expanded matrices and vectors of all the elements are added to get
the global shiftness matrix [k] and global force vector {F}. At this stage, the
second term of the expression for {F} (equation 6.8)
also needs to be added.
The last 2 steps constitute what is called as the global assembly procedure. The
details of the procedure are discussed in section 6.4. Before that, section 6.3 discusses
an example on element calculation.
6.3 Example on calculations of element Shiftness matrix and Force vector.
Consider a typical emement K. The nodes, the degrees of freedom and the shape
functions of the element are shown in Fig 6.3
Fig 6.3: A typical element K
Assume that
EA(x) = EA, a constant
F(x) = fo, a constant
6.29
Using equation (6.1), the expressions for Nki , I=1,2 are:
N1k
x k2  x

hk
N k2
x  x1k

hk
6.30
where
h k  x k2  x1k
6.31
Substituting equations (6.29)-(6.30) into the expression (6.25) for [k ] ,we get
(k)
x k
 2
1
1
  EA ( h k )(  h k )dx
x k
[ k ( k ) ]   1k
 x2

1
1
  EA ( h k )(  h k )dx
 xk
 1
EA  1  1

h k  1 1 


 EA ( h1k )( h1k )dx 

x1k


x k2

1
1
 EA ( h k )( h k )dx 

x1k

x k2
6.32
Similarly, substituting equations (6.29)- (6.30) into the expression (6.26) for {f ( k ) } ,
we get
x k

xk

x
 2

2
  f o ( h k )dx 
x k



{f k }   1k

x 2

k
x  x1
 f (
)dx 
  o hk

x1k

f h 1
 o k
2 1
6.33
6.4 Assembly of shiftness Matrix and Force Vector
After the evaluation of element shiftness matrix and element force vector for all the
elements, these quantities need to be “assembled” to get the global shiftness matrix and
global force vector. As stated at the end of section 6.2, this procedure has two steps:
(i)
Expansion of the element shiftness matrix and element force vector to the full
size.
(ii)
Addition of the expanded matrices and vectors over all the elements. At this
stage, the second term of the expression for {F} (equation 6.8) also needs to
be added.
Let us first discuss the first step. Note that equations (6.25) and (6.26) are the
expressions for the element shiftness matrix [k ( k ) ] and the element force vector {f ( k ) }
while equations (6.21) and (6.22) are the expressions for their expanded versions
[K ( k ) ] and {F( k ) } . When we compare equation (6.25) with (6.21), we observe that
(1,1) element of [k ( k ) ] occupies the position (k,k) of the expanded matrix [K ( k ) ] . This
is because K is the global number of the local node 1 of the element k. thus the first
step involves:
(i)
(ii)
(iii)
Choose the element k ij( k ) , i  1, j  1
Find the global number of the local nodes I and j of the element k, let they be r
and s respectively.
Then the element k ij( k ) occupies the position in rth row and sth column of the
expanded matrix [K ( k ) ] . Thus, the element k ij( k ) goes to the location K (rsk ) in
the expanded matrix.
(iv)
Repeat the steps (i)-(iii) for the other values of I and j. The remaining
elements of [K ( k ) ] are made zero.
The first step can be expressed mathematically by introducing a matrix [C], called as the
connectivity matrix, which relates the local and global numbering systems. The number
of rows in the connectivity matrix is equal to the number of nodes per element and the
number of columns is equal to the number of elements. Thus, the row index of [C]
denotes the local node number and the column number of [C] represents the element
number. The elements of [C] are the corresponding global numbers. Thus, for the mesh
of Fig 6.1, the connectivity matrix becomes
[ c] 


2 
 1


 2
3 




 3
4 




6.34
 .
. 




k
k  1

th
k




row
. 
 .


. 
 .
 N
N  1
The first row of the connectivity matrix contains the global numbers of the first and
second nodes of element 1. The global numbers corresponding to the first and second
nodes of element 1 are written in the second row. Continuing in this way, the global
numbers of the first and second nodes of element k appear in the k-th row. The last row
contains the global numbers associated with the first and second nodes of the last, ie Nth
element. The expression (6.34), in the index notation, can be expressed as
r  c ki
6.35
It means the global number r of the local node I of the element k is obtained as the value
of the element of the connectivity matrix in kth row and ith column. As an example,
consider the case o k=3 and I=2. The expression (6.35) gives r  C32  4 . This means 4 is
the global number of the second node of the element 3. this can be verifies from fog 6.1.
Now, the first step of the assembly procedure can be expresses as follows. The expanded
matrix [K ( k ) ] is obtained from the element shiftness matrix [k ( k ) ] by the relation:
k (rsk )  k ij(k )
0
where r  Cki , s  Ckj;
6.36
otherwise
Similarly, to obtain the expanded vector {F( k ) } from the element force vector {f ( k ) } , we
can write
Frk  fik
0
where r  Cki ;
otherwise
6.37
Thus, we follow the following procedure:
(i)
Choose the element f ik , i  1
(ii)
Find the global number of the local node I from the connectivity matrix, Let it
be r.
(iii)
Then, the element f ik goes to the location Frk of the expanded matrix.
(iv)
Repeat the steps (i)-(iii) for the other values of i. The remaining elements of
{Fk } are made zero.
The second-step is straight-forward. After obtaining the expanded versions of the element
shiftness matrix and the element force vector for all the elements, they are added as
follows:
[k ] 
{F} 
N
 [k ( k ) ]
k 1
N
6.38
 {F(k ) }  {P}
k 1
6.39
The matrix {P} corresponds to the second term of equation (6.6). Note that, the only
basis function which is nonzero at x = L is  N1 .
For other, ie value at x = L is 1.Thus
i |  0
for i  1,2,...N
x L
P
for i  N  1
6.40
Therefore, the vector {P} can be written as
{P}   0 
 
 
 
0 
 
  
I
st
Nth
row
row
(N+1)th row
6.41
6.5 Example on Assembly of shiftness matrix and Force vector
As an example, consider the mesh of 6 element (N=6) and 7 nodes, shown in Fig 6.4
1
2
3
4
5
6
. . . . . .
Element number
2
3
4
6
7
Global node n o
x1 x 2
u1 u 2
x3
x 4 x5 x6
u 4 u5 u 6
x7
Coordinates in global notation
u7
Global degrees of freedom
.
1
u3
5
Fig 6.4 Mesh with 6 elements
The connectivity matrix for this mesh can be written as :
[ c] 
1
2

3

4
5

6
2
3
4

5
6

7 
6.42
Let
[k
(k )
k ( k )
]   11
k)
k (21
(k ) 
k12

k)

k (22
and
6.42
f ( k ) 
{f ( k ) }   1( k ) 
f 2 
6.44
be the element shiftness matrix and the element force vector of the elements k=1,2,3,4,5,6
Consider the element 1, ie k=1. Note that
(i) i  1, j  1 gives r  C ki  C11  1, s  C kj  C11  1;
(ii ) i  1, j  2 gives r  C ki  C11  1, s  C kj  C12  2;
(iii ) i  2, j  1 gives r  C ki  C12  2, s  C kj  C11  1;
(iv ) i  2, j  2 gives r  C ki  C12  2, s  C kj  C12  2;
6.45
Then as per equation (6.36), the element of the shiftness matrix of the element 1, [k (1) ] ,
occupy the following locations in the expanded matrix [k (1) ] :
(1)
(1)
(1)
(1)
1)
1)
1)
1)
k11
 K11
, k12
 K12
, k (21
 K (21
, k (22
 K (22
6.46
Similarly as per equation (6.37) the elements of the force vector of the element 1, {f ( k ) } ,
occupy the following locations in the expanded vector {F( k ) } :
f1(1)  F1(1) , f 2(1)  F2(1)
6.47
(1)
(k)
The remaining elements of the expanded matrix [k ] and the expanded vector {F } are
zero. Thus, the matrix [k (1) ] becomes:
[K (1) ] 
k (1)
 11
1)
k (21

 0
 0

 0
 0

 0
(1)
k12
1)
k (22
0
0
0
0
0
0 0 0 0 0

0 0 0 0 0

0 0 0 0 0
0 0 0 0 0

0 0 0 0 0
0 0 0 0 0

0 0 0 0 0
and the vector {F(1) } becomes :
{F(1) } 
 F (1) 
 1 (1) 
F2 
 0 


 0 


 0 
 0 


 0 
6.48
Similarly, we obtain the expanded versions of the element shiftness matrix [k ( k ) ] and the
element force vector {f ( k ) } for the remaining elements, ie for k=2,3,4,5,6. It can easily be
verified that, for the 3rd element (ie for k=3), the expanded matrix [K (3) ] and the
expanded vector [F( k ) ] are:
[ k ( 3) ] 
0
0

0

0
0

0
0

0
0
0
0
0
0
0 k
0 k
( 3)
11
( 3)
21
( 3)
k12
k (223)
0
0
0
0
0
0
0
0
0
0 0 0
0 0 0
0 0 0

0 0 0
0 0 0

0 0 0
0 0 0
6.49
and
{F( 3) } 
0
0
 
F13 
 3
F2 
0
 
0
 
6.50
0
This completes the first step.
Int the 2nd step, we add all the expanded matrices and vectors. Thus, equation (6.38) gives
the following expression for the global shiftness matrix:
[k ] 
6
 [k (k ) ]
k 1
k (1)
 11
(1)
k12

 0
 0

 0

 0

 0
1)
k (21
0
0
0
0
1)
( 2)
k (22
 k11
k (212)
0
0
0
( 2)
k12
(3)
k (222)  k11
(3)
k12
3)
k (21
3)
( 4)
k (22
 k11
0
0
0
0
0
0
0
0
k (214)
(5)
k (222)  k11
( 5)
k12
0
0
0
0
0
Similarly, the sum of the expanded force vector becomes:
0
5)
k (21
5)
( 6)
k (22
 k11
( 6)
k12
0 

0 

0 
0 

0 
6) 
k (21

( 6) 
k 22 
6.51
6
{F(k ) } 
k 1














f 2(1)  f1( 2) 

f 2( 2)  f1(3) 

f 2(3)  f1( 4) 

f 2( 4)  f1(5) 
f 2(5)  f1(6) 

f 2(6) 
f1(1)
6.52
However, before we get the global force vector {F}, we need to add the vector {P} to the
above expression. Since N(no. of elements) = 6, the (N+1)th element, ie the element of
the vector {P} will be  .The remaining elements will be zero as per equation (6.41).
Thus, {P} becomes:
{P} 
0 
0 
 
0 
 
0 
0 
 
0 
 
P 
6.53
Substituting the expressions (6.52) and (6.53) in equation (6.39) we get the following
expression for the global force vector {F}
| F |
6
 (F)(k )  (P)
k 1














f 2(1)  f1( 2) 

f 2( 2)  f1(3) 

f 2(3)  f1( 4) 

f 2( 4)  f1(5) 
f 2(5)  f1(6) 

f 2(6)   
f1(1)
6.54
Now, as a section 6.3, assume that EA and f (distributed force) are constant for the entire
bar. Further, assume that the length h k of each element is constant. Let us denote ie by h.
Then
hk  h  L
6.55
N
Then, equation (6.32) implies that the element shiftness matrix [K ( k ) ] is identical for
each element and is given by
[k ( k ) ] 
EA  1  1
h  1 1 
for k  1,2,3,4,5,6
6.56
Similarly, equation (6.33) implies that the element force vector {f ( k ) } is identical for each
element and is given by
f h 1
{f (k ) }  0  
2 1
for k  1,2,3,4,5,6
6.57
Substituting the expression (6.56) in equation (6.51), we get
0
0
0
0
0
 1 1
 1 1  1  1
0
0
0
0 

 0 1 11 1
0
0
0
EA 

[k ] 
0
0
1 11 1
0
0

h
0
0
0
1 1 1 1 0 


0
0
0
 1 1  1  1
0
0
0
0
0
0
 1 1 

 1 1 0 0 0 0 0 
 1 2  1 0 0 0 0 


 0 1 2 1 0 0 0 
EA 


0 0 1 2 1 0 0 

h
 0 0 0 1 2 1 0 


 0 0 0 0  1 2  1
 0 0 0 0 0 1 1 


6.58
Further, substituting the expression (6.57) in equation (6.54), we get
 1 
 2 


 2 

f h 
| F | 0  2 
2  2 


 2 
1  2P 
 f o h 
6.59
In actual calculations, the assembly procedure is appropriately modified to reduce the
computational time and storage requirements. When, the number of elements is large,
storing of the expanded matrices and vectors for each element needs a lot of storage
requirement. Therefore, the process is modified as follows:
 Once the expanded version [k (1) ] of the element shiftness matrix of the first
element (k=1) is obtained , the element shiftness matrices of other elements are
not expanded.
 Instead, the locations of the elements of the shiftness matrix [k ( 2) ] of the element
two (k=2) are determined using equation (6.36).
 From the connectivity (6.34), it is easy to see that
( 2)
k11
 (2,2) location of the exp anded matrix ,
( 2)
k12
 (2,3) location of the exp anded matrix ,
2)
k (21
 (3,2) location of the exp anded matrix ,
2)
k (22
 (3,3) location of the exp anded matrix .
6.60