Higher Order Elements

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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 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 -1 1

In the above A1 and A2. Let us find the stress for this case shown in Fig. 121.(b) with the

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) (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


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 K = E/6l A1 1 -12

1 3 -4

-12 -4 +A2 16

3 1 -4

1 11 -12

-4 -12 16

Particular cases: (a) When A2 = 2A1

17 K = E/6l A1 3 -20

3 25 -28

-20 -28 48

(b) When A2 = A1=A (for uniform element with three nodes) 7 K = 2AE /6l 1 -8 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 equidistance. 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.


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 Pascals 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 =