3. UNIT V ISOPARAMETRIC
FORMULATION
Natural co-ordinate systems –
Isoparametric elements – Shape functions
for iso parametric elements – One and two
dimensions – Serendipity elements –
Numerical integration and application to
plane stress problems - Matrix solution
techniques – Solutions Techniques to
Dynamic problems – Introduction to
Analysis Software.
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
4. ISO-PARAMETRIC ELEMENTS
The Iso-parametric concept in the finite elements analysis to help
properly mapping the curved boundaries.
The concept of mapping regular triangular and rectangular elements in
natural coordinate system, to arbitrary shapes in global system as
shown in Figure.
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
Concept of mapping in Iso-parametric elements
6. CO-ORDINATE
TRANSFORMATION
The shape functions for defining deflection at any point interms of the
nodal displacement, also it is used for coordinate transformation form
natural local coordinate system to global Cartesian system.
And successfully achieved in mapping parent element to required shape
in global system.
Thus the Cartesian coordinate of a point in an element may be
expressed as
x = N1 x1 + N2 x2 + ... + Nn xn
y = N1 y1 + N2 y2 + ... + Nn yn
z = N1 z1 + N2 z2 + ... + Nn zn
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
7. or in matrix form
{x} = [N] {x}e
where N are shape functions and (x)e are the coordinates of nodal
points of the element.
The shape functions are to be expressed in natural coordinate system.
For example consider mapping of a rectangular parent element into a
quadrilateral element
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
9. The parent rectangular element shown in Figure has nodes 1, 2, 3 and 4
and their coordinates are (–1, –1), (–1, 1), (1, 1) and (1, –1).
The shape functions of this element are
N1 = [(1 − ξ) (1− η)]/4, N2 =[(1+ ξ ) (1− η)]/4
N3 = [(1+ ξ ) (1+ η)]/4 and N4 = [(1- ξ ) (1+ η)]/4
P is a point with coordinate (ξ, η). In global system the coordinates of
the nodal points are (x1, y1), (x2, y2), (x3, y3) and (x4, y4)
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
10. To get this mapping we define the coordinate of point P as
x = N1 x1 + N2 x2 + N3 x3 + N4 x4 and
y = N1 y1 + N2 y2 + N3 y3 + N4 y4
Noting that shape functions are such that at node i, Ni = 1 and all others
are zero, it satisfy the coordinate value at all the nodes.
Thus any point in the quadrilateral is defined in terms of nodal
coordinates.
Similarly other parent elements are mapped suitably when we do
coordinate transformation.
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
11. ISOPARAMETRIC,SUPERPARAMETRIC
AND SUBPARAMETRIC ELEMENTS
In the finite element analysis with iso-parametric elements, shape
functions are used for defining the geometry as well as displacements.
If the shape functions defining the boundary and displacements are the
same, the element is called as iso-parametric element. For example, in
Figure (a) all the eight nodes are used in defining the geometry and
displacement.
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
12. The element in which more number of nodes are used to define
geometry compared to the number of nodes used to define
displacement are known as superparametric element.
One such element is shown in Figure(b) in which 8 nodes are used to
define the geometry and displacement is defined using only 4 nodes. In
the stress analysis where boundary is highly curved but stress gradient
is not high, one can use these elements advantageously.
Figure (c) shows a subparametric element in which less number of
nodes are used to define geometry compared to the number of nodes
used for defining the displacements.
Such elements can be used advantageously in case of geometry being
simple but stress gradient high.
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
13. SHAPE FUNCTION FOR 4 NODED
RECTANGULAR ELEMENT
Figure shows the typical parent element and isoparametric quadrilateral
element.
For parent element, the shape functions are,
N1 =(1/4) (1 − ξ) (1− η), N2 = (1/4)(1+ ξ) (1− η)
N3 = (1/4) (1+ ξ) (1+ η) and N4 = (1/4) (1− ξ) (1+ η)
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
16. STIFFNESS MATRIX FOR
4 NODED RECTANGULAR
ELEMENT
Assembling element stiffness matrix for isoparametric element is a
tedious process since it involves co-ordinate transformation form
natural co-ordinate system to global co-ordinate system.
We know that, General element stiffness matrix equation is,
Stiffness matrix, [K] = [B]T [D] [B] dv
v
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
.
P (ξ, η) .
P (x, y)
Parent Element Isoparametric Quadrilateral Element
17. For Isoparametric quadrilateral element,
[K] =t [B]T [D] [B] ∂x ∂y
For natural co-ordinates,
1 1
[K] =t [B]T [D] [B] x |J| x ∂ε ∂η [∂x ∂y = |J| ∂ε ∂η ]
-1 -1
Where, t = Thickness of the element
|J|= Determinant of the Jacobian
ε , η = Natural coordinates
[B] = Strain- displacement relationship matrix
[D] = Stress-Strain relationship matrix
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
18. For two dimensional problems,
1 v 0
Stress-Strain relationship matrix, [D] = E/(1-v2) v 1 0
0 0 (1-v)/2
(For plane stress conditions)
1-v v 0
[D] = E/(1+v)(1-2v) v 1-v 0 (For plane strain conditions)
0 0 (1-2v)/2
Where, E = Young’s modulus
v = Poisson’s ratio
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
19. P1: Consider a rectangular element as shown in figure. Assume plane
stress condition E= 30x 106 N/m2, v=0.3 and q = [0. 0. 0.002, 0.003, 0.006,
0.0032, 0, 0 ]. Evaluate J, B and D at ξ=0 and η=0.
Solution:
Jacobian matrix J =
-(1-η)x1+(1-η)x2+(1+η)x3-(1+η)x4 -(1-η)y1+(1-η)y2+(1+η)y3 -(1+η)y4
(1/4)
-(1-ξ)x1-(1+ξ)x2+(1-ξ)x3+(1-ξ)x4 -(1-ξ)y1-(1+ξ)y2+(1-ξ)y3+(1-ξ)y4
J J
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
q2
+
C (1, 0.5)
q8
(2, 1)(0, 0)
(2, 1)
(0, 1)
q7
q6
q5
q1
Y
q3
q4
X
3
2
4
1
21. ½ 0 0 0 -¼ 0 ¼ 0 ¼ 0 - ¼ 0
[B] =1/½ 0 0 0 1 x 0 -½ 0 -½ 0 ½ 0 ½
0 0 ½ 0 -½ -¼ -½ ¼ ½ ¼ ½ -¼
The stresses at ξ=0 and η=0
1 v 0
Stress-Strain relationship matrix, [D] = E/(1-v2) v 1 0
(For plane stress conditions) 0 0 (1-v)/2
1 0.3 0
[D] = (30x106)/(1-0.09) 0.03 1 0
0 0 0.35
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
22. SHAPE FUNCTION FOR 6 NODED
TRIANGULAR ELEMENT
The natural coordinates for triangular elements are conveniently
defined as shown in figure. The three natural coordinates are defined by
the following expressions:
L1 = A1/A = Area of ∆P23/ Area of ∆123
L2 = A2/A = Area of ∆1P3/ Area of ∆123
L3 = A3/A = Area of ∆P12/ Area of ∆123
L1+L2+L3 =1
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
Y
O
2
1
3
P(L1,L2,L3)
Natural coordinates for
triangular element
24. SHAPE FUNCTION FOR 8 NODED
RECTANGULAR ELEMENT
Consider the parent eight node rectangular element in ξ-η frame as well
as the general quadrilateral element in the physical x-y space as shown
in figure.
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
(-1, 0) 8 6(1,0)
(-1, 1) 4
(-1, -1) 1
3(1,-1)
η
ξ
5(0,-1)
7 (0, 1)
X
Y
6(x6, y6)
(x8, y8) 8
3(x3, y3)(x7, y7) 7
(x4, y4)
4
(x1, y1) 1
5(x5, y5)
2(x2, y2)
O
General Eight-noded quadrilateral element
25. For eight-noded isoparametric element, the Jacobian can be written as,
Σ(∂Ni /∂ξ)xi Σ (∂Ni /∂ξ)yi
|J| =
Σ (∂Ni /∂η)xi Σ(∂Ni /∂η)yi
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
i=1 i=1
i=1 i=1
8 8
8 8
27. NATURAL COORDINATES AND
COORDINATE TRANSFORMATION
The natural coordinate used for 1-D case designated by ξ.
Consider a child line shown in figure with two points defined by their
coordinates A(x1) and B(x2).
The parent line indicating ξ varying from A’(-1) to B’(+1) is also shown
in figure.
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
Natural Coordinates in 1D
BB’
x
(x=x1)
A
A’ ξ
P’Q’
ξ=1ξ = -1
PQ
(x=x2)
28. An appropriate linear transformation between x and ξ is given by
x(ξ) = [ (x1 + x2)/2 ] + [ (x1 - x2)/2 ] ξ
= [ (1- ξ)/2] x1 + [ (1+ ξ)/2] x2
x(ξ) = N1x1 + N2x2
Where Ni = Interpolation function used for coordinate transformation.
If the shape function used for coordinate transformation are of lower
degree than those for unknown field variable called as Sub-parametric
elements.
If the shape function used for coordinate transformation are of higher
degree than those for unknown field variable called as Superparametric
elements.
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
29. Rewriting the above equation in the standard notation,
x1
x = [N1 N2]
x2
For point P’, ξ = 0.5, correspondingly,
x = [ (x1 + 3x2)/4 ]
Which is the point P.
For point Q,
x = [ x1 + (x2 + x1)/4 ]
correspondingly, ξ = -0.5,
Which is the point Q’.
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
30. In order to fit a Quadratic transformation between x and ξ, to need one
more point C as shown in figure.
The transformation as
x(ξ) = Σ Nixi = N1x1 + N2x2 + N3x3
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
Quadratic transformation between x and ξ
BB’
x
(x=x1)
AA’ ξC’
ξ=1ξ = -1
C
(x=x2)ξ = 0 (x=x3)
31. NUMERICAL INTEGRATION
In One Dimension:
There are several schemes available for the numerical evaluation of
definite integrals.
In finite element analysis, Gauss quadrature method is mostly preferred.
Let the one-dimensional integral to be evaluated be
1 n
I =∫f(x) dx = Σwi f(xi)
−1 i= 1
Where, wi is weight function.
f(xi) is values of the function at pre-determined sampling points.
Function f(xi) is calculated at several sampling points i.e., n= 1, 2, 3,…
and each value of f(xi) multiplied by weight function wi. Finally all the
terms are added, it gives the value of integration.
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
32. Table shows the location of Gauss sampling points f(xi) and
corresponding weight function wi for different number of point (n) and
also gives the Gauss points for integration from -1 to 1.
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
No. of
points, n
Location, xi Corresponding weights, wi
1 x1 = 0.00000000 W1 = 2.00000000
2 x 1,x 2 = ±0.57735027 W1 = W2 = 1.00000000
3 x1, x3 = ±0.77459667
x2 = 0.000
W1 = W3 = 0.55555556
W2 = 0.88888889
4 x1, x4 = ±0.86113631
x2, x3 = ±0.33998104
W1 = W4 = 0.34785485
W2 = W3 = 0.65214515
33. For n=1, the degree of polynomial should be (2n-1)
i.e., 2x1-1 = 1 degree
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
Gauss Integration
34. For two dimension:
For two dimensional integration, we can write
1 1 n n
I =∫∫f(x, y) dx dy= ΣΣwj wi f(x, y)
-1 -1 j=1 i=1
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
35. 1
P1: Evaluate the integral I=∫(x4 + 3x3 - x) dx
−1
Solution:
We know that as per Gaussian Quadrature method
1 n
I =∫f(x) dx = Σwi f(xi)
−1 i= 1
and also the integral function of (2n-1) degree can be evaluated most
accurately by n Gauss-points.
Hence to find the no. of Gauss-points n, equate (2n-1) with the degree of
integral function.
i.e., 2n-1 =4 => n= (4 + 1)/2 = 3 points.
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
36. That is , the integral can be evaluated by 3points scheme.
Now, the equation (1) can be written as,
1
I =∫f(x) dx = w1 f(x1) + w2 f(x2) + w3 f(x3)
−1
where f(x) = (x4 + 3x3 - x)
w1 = 5/9 = 0.55555 and x1 = 0.774597
w2 = 8/9 = 0.88888 and x2 = 0
w3 = 5/9 = 0.55555 and x3 = -0.774597
Now, w1 f(x1) = w1(x1
4 + 3x1
3 - x1)
= 0.55555[(0.774597)4 + 3(0.774597)3- (0.774597) = 0.5443
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
38. 1 1
P2: Evaluate the integral I =∫ ∫ (2x2 + 3xy + 4y2) dx dy
-1 -1
We know that as per Gaussian Quadrature method
1 1 n n
I =∫∫f(x, y) dx dy= ΣΣwj wi f(xi, yj)
-1 -1 j=1 i=1
Where f(x, y) = 2x2 + 3xy + 4y2
To find the no. of Gauss- points n,
2n-1 =2 => n = (2+1)/2 = 1.5 =2
Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.