HIGHER ORDER ELEMENTS UNIAXIAL ELEMENTS: A uniform uniaxial element is shown in the figure. (a) General form (b) Contracted form after imposing boundary condition Uniform Rod We know that for the structure shown in the figure ,the stress is constant at every section .Hence strain is constant. The property strain is constant leads to the conclusion that the displacement field is linear. In the figure (a), there are two degrees of freedom. Hence the polynomial describing the displacement field must have two generalised constants A 1 and A 2 . We saw this field as u = A 1 +A 2 x
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HIGHER ORDER ELEMENTS
UNIAXIAL ELEMENTS:
A uniform uniaxial element is shown in the figure.
(a) General form
(b) Contracted form after imposing boundary condition
Uniform Rod
We know that for the structure shown in the figure ,the stress is constant at every
section .Hence strain is constant. The property strain is constant leads to the conclusion
that the displacement field is linear. In the figure (a), there are two degrees of freedom.
Hence the polynomial describing the displacement field must have two generalised
constants A1 and A2 . We saw this field as
u = A1 +A2x
Upon evaluation of the stiffness matrix we found that it is the same one obtained from the
structural mechanics concept. Therefore the above polynomial is the exact representation
of the displacement field for a two noded uniform axial member.
Now let us consider a two noded uniformly varying axial element as shown in the figure.
(a) General form
(b) Contracted form after applying boundary condition
Fig. 121. Uniformly varying axial element.
In Fig. 121.(a),there are two degrees of freedom one at each node. The suitable
polynomial is
U = A1 + A2
Here A1 and A2 are generalized coordinates and not areas as shown in Fig. 121.(a)
The corresponding stiffness matrix is
K = E(A1 + A2)/2l 1 -1
-1 1
In the above A1 and A2. Let us find the stress for this case shown in Fig. 121.(b) with the
following values
A1 = 1 cm2
A2 = 3cm2
L = 1cm
F1 = 1000KN
Since there is only one element k =K assume that element axis and global axis coincide,
we get after deleting first row and first column
K =2E
R = Kr
1000/E = 2u2
u2 = 500/E units
Now assuming the shape function in terms of natural coordinates we get
U = N1u1 +N2u2 = N2u2 Since u1 = 0
U = (1-(x/l) u1 + (x/l) u2
= (du/dx) (1/l) 500/E = 500/E
Stress = E = 500E/E = 500
The solution indicates that the stress is constant throughout and is equal in 500KN/cm2.
Now the question is ,is it the correct answer? Obviously looking at Fig. 121(b), we infer
that the stress is continuously varying. It is not constant. Since the stress varies
continuously the strain also to vary. This shows that the assumed linear displacement
field is only approximate. How to improve the solution then?
One way is instead of assuming one element for the entire system, We can subdivide or
dicretize it into very many smaller elements as shown in Fig 122. As an example, Fig.
122(a) is divided in to three elements. The equivalent system is shown in the magnitude
of the stress differs in the three elements 1,2 and 3.This is an improvement over the first
consideration, that is treating the system as a single element. Whatever results we get
from strength of materials approach, the same results will be obtained by performing the
finite element analysis of the system shown in Fig. 122(a). Instead of three elements, the
elements can be increased further. Then the result will improve. Our knowledge of the
convergence requirements confirms this inference, that is when the elements are made
smaller and smaller, the strain in the elements becomes almost constant. A constant strain
situation in an axial element gives rise to a linear displacement field which we have
assumed for the element. For getting the exact answer, we require infinite elements in the
case of uniformly varying element. This is not so when the given system has constant
area of cross section. This we have seen in section 10.6. Thus we have seen in the case of
approximate displacement field we can improve the answer by resorting to finer
divisions.
On the other hand we can assume a polynomial displacement field having very many
generalized coordinates instead of two for the element shown in Fig 121(a). The question
is can we arbitrarily increase the generalized coordinates as we please? We know the
requirement that as many generalized coordinates should be present as there are degrees
of freedom. The degrees of freedom for the element can be increased beyond two only if
there are additional nodes. These additional nodes are placed as shown in Fig. 123
We have provided two additional nodes in the element as shown in Fig. 123. Note that we
have not reduced the size of the element. These additional nodes can be placed in any
member. However for computational convenience they are placed equidistant from each
other. The corresponding displacement field becomes
U = A1+A2x +A3x2 +A4x3
From the linear polynomial, it has changed in to cubic polynomial. This is known as third
order element.
Elements in which additional nodes are present more than the minimum are known as
Higher order elements
In axial elements, these additional nodes are known as internal nodes. Why do we call
them internal nodes? The two exterior nodes 1 and 2 are known as primary nodes. We
may also note the sequence of the numbered first and then the internal nodes
The structure shown in Fig. 124 consists of three elements. Element 1 and 3 are exact.
Therefore additional node is not going to improve the flexibility of the system whereas in
element 2,there are two additional nodes. Another point to be noted is that nodes 1 and 3
serve as connection purpose. These nodes are known as internal nodes whose only
purpose is to increase the flexibility of the element. Therefore theoretically in irregular
cross section axial element, we have infinite number of generalized coordinates
approaching the true flexibility of the system
If the element has got only one internal node then it is known as quadratic element
Example.
Using the isoparametric concept, set up the element stiffness matrix for the three noded
axial element shown in Fig. 125
Solution
Using the element and simple natural coordinate system we can express the cross
sectional area of the element as
A = [1/2(1-r) ½(1+r)] A1
A2
Isoparametric concept requires
X = N1x1 + N2x2 + N3x3
The alternative form is
R = (x –x3)/(l/2)
Dr=dx/(l/2)
Now we know
u = N1u1 + N2u2 + N3u3
u1
u = [r/2(1-r),r/2(1+r),(1-r2)] u3
u2
= du/dx = (du/dr)(dr/dx)
u1
= [(2r-1),(2r+1),-4r]*(1/2)*(2/l) u3
u2
Now k = BTdB dv
Where dv = Adx = [1/2(1-r)A1+1/2(1+r)A2]dx
Dv = [1/2(1-r)A1+1/2(1+r)A2]l/2 dr
1
K = BTdB[1/2(1-r)A1+1/2(1+r)A2]l/2 dr
-1
Substituting for B in Eq. (5) and observing d = E upon carefully integrating we get
11 1 -12 3 1 -4
K = E/6l A1 1 3 -4 +A2 1 11 -12
-12 -4 16 -4 -12 16
Particular cases:
(a) When A2 = 2A1
17 3 -20
K = E/6l A1 3 25 -28
-20 -28 48
(b) When A2 = A1=A (for uniform element with three nodes)
7 1 -8
K = 2AE /6l 1 7 -8
-8 -8 16
TWO DIMENSIONAL ELEMENTS
There is a presence of internal node in an uniaxial element. There are two types of
uniaxial elements, namely,
(1) Uniform cross-section element
(2) Tapering uniaxial element
In the case of uniform cross-section element, the presence of internal node does not have
any effect over the flexibility of the element. It does not improve the answer. On the other
hand, in tapering element presence of internal node has significant influence in unproving
the flexibility of the element. An infinite number of such nodes leads to the exact
flexibility of the element because each node corresponds to a degree of freedom which in
turn represents for one generalized coordinate in the polynomial of the displacement
field. Infinite number of degrees of freedom means the polynomial represents the true
displacement field of the element. For convenience the internal nodes are placed at equi-
distance.
In two dimensional structure, the system flexibility can be improved in two ways.
(a) By having more number of simple elements
(b) By having fewer numbers of higher order elements. Which of the two ways better?
No conclusive answer is available at this stage even through higher order elements
are used in the system in the places of holes, cut outs, etc., along with the simple
elements. For further aspects on this aspect, advanced texts may be referred. For two-
dimensional structures, we use triangular, rectangular and quadrilateral elements. In
this primer let us learn about the triangular and rectangular elements.
TRIANGULAR ELEMENT
In the case of uniaxial element, we can have arbitrarily any number of internal nodes.
However in triangular element for given size,even through theoretically we can have any
number of internal nodes, the criterion of the completeness or balanced polynomial of the
Pascal’s triangle plays an important role. Secondly, experts in this field are of the opinion
that the vertices and mid-points are convenient points for function evaluation. However,
the Pascal triangle consideration plays an important role. Let us consider section, the
sixth requirement where the Pascal triangle is shown. In the Pascal triangle, for a
complete quadratic model, there are six terms to be considered as shown below.
U = A1+A2x +A3y+A4x2+A5xy+A6y2
Similarly for v displacement six terms will be there.
This means an higher order quadratic model must have six nodes in all, of which three
are the nodes at the vertices. Where to place the remaining three nodes ? According to
experts opinion, the best place would be at the centre of each side as shown in the figure.
We could have placed three nodes internally. However it is unwritten law wherever
possible we must avoid the usage of internal node as far as possible. The three nodes
placed on the sides are known as Secondary External Nodes.
In any element involving primary, secondary, and internal nodes, the sequence of
numbering should be in the same order as mentioned.
SHAPE FUNCTIONS FOR THE QUADRATIC MODEL:
Node(1). The L1 coordinate for nodes 1,4,and 2 are (1,1/2,0) respectively. A lagrangian
shape function can be constructed as follows :
(s-s4)(s-s2)
N1 = --------------------------
(s1 –s4)(s1-s2)
where s4 =1/2
s2 =0
s1 =1
(s-1/2)(s-0)
N1 = -------------------------- = s(2s-1)
(1 –1/2)(1-0)
Let s =L1 = area coordinate of the system
N1 = L1(2L1 -1)
It can be seen that the value of this function at all other nodes except node 1 is zero. The
same is shown in the figure
Similarly by cylic rotation we can obtain for other primary nodes as
N2 = L2(2L2 -1)
N3 = L3(2L3 -1)
Now let us find the shape functios for the secondary nodes 4,5 and 6.
Node(4). As before L1 coordinates for nodes 1,4 and 2 are (1,1/2,0). Now
(s-s1) (s-s2)N4 = (s4-s1) (s4-s2)
s1 = 1
s2 = 0s4 = ½
(s-s1) (s-s2)N4 = = (s-s1)/(-1/4) (1-1/2) (1/2-0)
N4 = 4s(1-s)
S = L1
N4 = 4L1(1-L1)
Along the line containing the nodes 1,4 and 2
L1+L2+L3 = 1
L1+L2+0 = 1
L3 = 0 along this line
L2 = (1-L1)
N3 = L1L2
Similarly
N5 = 4 L2L3
N6 = 4 L3L1
SHAPE FUNCTIONS FOR THE CUBIC MODEL
Consideration of the Pascal triangle for a cubic model in the section shows that for the
cubic model, ten terms have to be considered as follows:
U = A1+A2x +A3y+A4x2+A5xy+A6y2+A7x3+ A8x2y+ A9xy2 + A10y3
Similarly v displacement will have 10 generalized coordinates A11 through A20.
Ten terms in the polynomial means ten nodes will have to be present in the model. As far
possible, the nodes should be placed on the sides in some orderly fashion so that shape
functions can be easily evaluated.
Cubic model
In the case of quadratic model there was no difficulty. We could accommodate all on the
boundaries without dificulty. In this model it is possible to place only nine nodes on the
sides as shown in the figure. The remaining one node at the convenient point as shown in
the figure.
CONSTRUCTION OF SHAPE FUNCTIONS:
Node 1: Consider the line having nodes1,4,5 and 2 then
(s-s4) (s-s5) (s-s2)N1 = (s1-s4) (s1-s5)(s1-s2)
s1 = 1
s2 = 0
s4 = 2/3s5 = 1/3
Substituting the above
(s-2/3) (s-1/3) (s-0)N1 = (1-2/3) (1-1/3)(1-0)
(3s –2)(3s-1)s = 2s =L1
(3L1 –2)(3L1-1)L1
N1 = 2
It can be verified that this is the shape function for node 1Similarly
(3L2 –2)(3L2-1)L2
N2 = 2
(3L3 –2)(3L3-1)L3
N3 = 2
Node 4:
Consider the line having nodes 4,10 and 7
(s-s10) (s-s7)S1 = (s4-s10) (s4-s7)
s4 =2/3
s10 =1/3 L1ordinates
s7 = 0
(s-1/3) (s-0) ( 3s –1)sS1 = = (2/3-1/3) (2/3-0) 2
Let s= L1
(3L1-1)L1
S1 = ----------------------
2
Similarly consider line having nodes 4 and 9
Then
(s-s9)
S2 = --------------
(s4 –s9)
where
s9 = 0
s4 = 1/3
(s-0)
S2 = ------------ = 3s
(1/3 –0)
Let
s = L2
Therefore
S2 =3L2
3L23L1(3L1 -1)
Now N4 =S1 x S2= ---------------------------------
2
9L1L2(3L1 -1)
N4 = ------------------
2
It may be verified that this function gives a function equal to one at node 4 zero at all
other nodes. Along similar lines shape functions for all other side nodes can be
constructed.
Node 10: (internal node)
At node 10,three lines are intersecting. For any one line shape function can be found and then multiplied as shown below.
Consider line 1,10,7 and consider L1 coordinates along thin line
(s-s4) (s-s7)S1 = (s10-s4) (s10-s7)
where
s4 = 2/3s10 =1/3s7 = 0
(s-2/3) (s-0)S1 = = -3s(3s-2) (1/3-2/3) (1/3-0)
s =L1
S1 = -3L1(3L1-2)
Similarly
S2 = -3L2(3L2-2)
S3 = -3L3(3L3-2)
N10 = -3*3*3*(3L1-2) (3L2-2) (3L3-2) L1 L2 L3
= -27(3L1-2) (3L2-2) (3L3-2) L1 L2 L3
At the node 10
(3L1-2) = -1
(3L2-2) = -1
(3L3-2) = -1
N10 = 27 L1 L2 L3
It must be verified that this is the shape function for node 10.
14.6. Rectangular Elements – Lagrangian Family
we have seen in example 12.5, Fig . 106, how to construct shape functions for a simple
four noded rectangular element. This element belongs to the Lagrangian Family. The
displacement field for this element is
u = A1+A2x+A3y+A4xy
This satisfies all the criteria for convergence. Let us see how the pattern is in the Pascal
triangle
1 1
x y
xy
It may be seen that all the elements bounded by arrows (diamond) are included in the displacement field
As a second example let us see the rectangular element shows in Fig. 132.
Fig. 132. Lagrangian element (quadratic model)
Shape function for any node can be found as the product of two Lagrangian shape
functions in two mutually perpendicular directions at that node.
The u and v displacement fields will each consists of nine generalized coordinates and the
corresponding polynomial variables will be as shown below
1
x y
x2 xy y2
x3 x2y xy2 y3
x4 x3y x2y2 xy3 y4
We may note that in the diamond there are nine terms
As a third example let us see the element shown in Fig. 133
1
x y
x2 xy y2
x3 x2y xy2 y3
x4 x3y x2y2 xy3 y4
In Fig. 133 there are sixteen nodes. Hence there will be sixteen generalized coordinates.
Within the diamond shown in the Pascal triangle, there are sixteen polynomial variables
in general the Pascal triangle for Lagrangian elements will be as shown in Fig. 134.
xn yn
xnyn
Fig. 134. General Pascal triangle for Lagrangian elements
The advantage of Lagrangian elements is that we can very easily generate shape
functions for any nodes in the element. The disadvantages are:
(a) In general internal nodes in the element is not preferable as far as possible
(b) Poor convergence properties of the higher order polynomials
(c) It will be noticed that the expressions of shape function will contain some very higher
order terms while omitting some lower ones.
RECTANGULAR ELEMENTS- ‘SERENDIPITY’ FAMILY:
In general it is easy to evaluate the function if the nodes are present on the sides. It is
usually most convenient to make the functions dependent on nodal values placed on the
element boundary. Let us see below four elements.
Linear element-Serendipity Type
(a) Linear element. The nodes are on the corners. The Pascal triangle requirement is
similar to what we have seen in the case of Lagrangian linear element. There is no
internal node in this element.
(b) Quadratic element. In this element also there is no internal node. On the boundary
the displacement field varies quadratically. There are eight nodes and there will be
eight generalised coordinates in the u and v displacement field. The geometric
isotropy is obtained as shown in the Pascal triangle. In the direction of arrows there
are eight terms which are balanced.
(c) Cubic Model.
Here there are twelve nodes. The Pascal triangle expansion is shown below. It may be
noted that that the geometric invariance is preserved here.
It may be seen that there are twelve terms shown by the arrows in the Pascal triangle.
(d) Quartic model. This is shown in the figure . In this model a central node is added so
that all terms of a complete fourth order expansion would be available.
All the elements in the fourth order expension (ie.,) 15 terms plus two terms from the
quintic expansion give seventeen terms.
Summary.
(a) In linear model we have considered all the terms plus one term from the quadratic
expansion.
(b) In quadratic model all the terms upto quadratic , ie., six terms plus two terms from the
succeeding cubic expansion.
(c) In cubic model all the terms upto cubic ,ie., ten terms plus two terms from the
succeeding fourth order expansion.
(d) In quadratic model all the terms upto quartic , ie., fifteen terms plus two terms from
the succeeding fifth order expansion.
The extra nodes placed on the sides other than the primirary nodes are called Secondary
External Nodes.
The primary advantage of Serendipity elements is the absence of internal nodes (except
quartic elements where one internal node is present) in the various elements. Because of
this , these elements are very efficient from computational point of view. However, the
disadvantage is generation of shape functions for these elements is not a straight forward
procedure. Advanced texts may be referred for further understanding of this aspect.
The shape functions have been originally derived by inspection. It becomes more and
more difficult as we progress further to higher elements. A certain ability and ingenuity is
required on the part of the analyst to evolve these functions. It is therefore befitting to call
these family of elements as “ Serendipity Elements” after the famous princes of Serendip
noted for their chance discoveries. Chamber’s dictionary gives meaning as follows:
Serendipity- the faculty of making happily chance finds.
Horace Walpole coined the word from the title of the fairy tale “ The three princes of
Serendipity ” whose heroes ‘were always making discoveries by accident and sagacity, of