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

things they were not in quest of ’.