Finite Element Analysis - UNIT-5

TO
By
Dr.G.PAULRAJ
Professor & Head
Department of Mechanical Engineering
Vel Tech (Owned by R S Trust)
Avadi, Chennai-600062.

Dr.G.PAULRAJ,
Professor&Head(Mech.)
VTRS,Avadi, Chennai.
UNIT-V ISOPARAMETRIC
FORMULATION

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,

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,
Concept of mapping in Iso-parametric elements

Dr.G.PAULRAJ,
Concept of mapping in Iso-parametric elements

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,

 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,

Dr.G.PAULRAJ,
Mapping of rectangular element in natural local coordinate
system to global Cartesian coordinate system

 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,

 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,

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,

 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,

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,

 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
J11 J12
=
J21 J22
Dr.G.PAULRAJ,

Strain- displacement relationship matrix
J22 -J12 0 0
 [B] = 1/ |J| 0 0 -J21 J11 x (1/4)
-J21 J11 J22 -J12
-(1-η) 0 (1- η) 0 (1+ η) 0 -(1+ η) 0
-(1-ε) 0 -(1+ε) 0 (1+ε) 0 (1-ε) 0
x 0 -(1-η) 0 (1- η) 0 (1+ η) 0 -(1+ η)
0 -(1-ε) 0 -(1+ε) 0 (1+ε) 0 (1-ε)
Dr.G.PAULRAJ,

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,
.
P (ξ, η) .
P (x, y)
Parent Element Isoparametric Quadrilateral Element

 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,

 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,

 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,
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

2(1-η)+2(1+η) (1+η) -(1+η) 1 0 J11 J12
J= (1/4) = =
-2(1+ξ)+2(1-ξ) (1+ξ)+(1-ξ) 0 ½ J21 J22
For this rectangular element, we find that J is a constant matrix. Now Strain
displacement relationship matrix
J22 -J12 0 0 -(1-η) 0 (1- η) 0 (1+ η) 0 -(1+ η) 0
 [B] =1/|J| 0 0 -J21 J11 x(1/4)x -(1- ξ) 0 -(1+ ξ) 0 (1+ ξ) 0 (1- ξ) 0
-J21 J11 J22 -J12 0 -(1-η) 0 (1- η) 0 (1+ η) 0 -(1+ η)
0 -(1-ε) 0 -(1+ε) 0 (1+ε) 0 (1-ε)
Dr.G.PAULRAJ,

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,

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,
Y
O
2
1
3
P(L1,L2,L3)
Natural coordinates for
triangular element

 The shape functions are,
 N1 = (L1) (2L1-1)
 N2 = (L2) (2L2-1)
 N3 = (L3) (2L3-1)
 N4= 4L1 L2
 N5 = 4L2 L3
 N6 = 4L3 L1
Dr.G.PAULRAJ,
Y
O
2 (0,1,0)
1 (1,0,0)
(0,0,1)3
5(0,1/2, 1/2)
A Six-noded triangular element
4(1/2, 1/2,0)
(1/2,0,1/2) 6

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,
(-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

 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,
i=1 i=1
i=1 i=1
8 8
8 8

Shape function for eight-noded quadrilateral element
 N1/8 Node = ¼ (1-ξ) (1-η) (-1- ξ-η)
 N2/8 Node = ¼ (1+ξ) (1-η) (-1+ ξ-η)
 N3/8 Node = ¼ (1+ξ) (1+η) (-1+ ξ+η)
 N4/8 Node = ¼ (1-ξ) (1+η) (-1- ξ+η)
 N5 = ½ (1-ξ2) (1-η)
 N6 = ½ (1+ξ) (1-η2)
 N7 = ½ (1-ξ2) (1+η)
 N8 = ½ (1-ξ) (1-η2)
Dr.G.PAULRAJ,

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,
Natural Coordinates in 1D
BB’
x
(x=x1)
A
A’ ξ
P’Q’
ξ=1ξ = -1
PQ
(x=x2)

 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,

 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,

 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,
Quadratic transformation between x and ξ
BB’
x
(x=x1)
AA’ ξC’
ξ=1ξ = -1
C
(x=x2)ξ = 0 (x=x3)

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,

 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,
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

 For n=1, the degree of polynomial should be (2n-1)
 i.e., 2x1-1 = 1 degree
Dr.G.PAULRAJ,
Gauss Integration

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,

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,

 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,

 w2 f(x2) = w2(x2
4 + 3x2
3 – x2) = 0.88888(04+3x04-0) =0
 w3 f(x3) = w3(x3
4 + 3x3
3 – x3)
 = 0.55555[(-0.774597)4 + 3(-0.774597)3- (0.774597) = - 0.1443
 Hence the above equation =>
1
 I =∫f(x3
4 + 3x3
3 – x) dx = 0.5443 + 0 – 0.1443 = 0.4
−1
Result:
1
∫f(x3
4 + 3x3
3 – x) dx = 0.4
-1
Dr.G.PAULRAJ,

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,

 For 2 point scheme, the equation is
1 1
∫∫f(x, y) dx dy= w1
2(x1, y1) + w1w2f(x1, y2) + w2w1 f(x2, y1) +w2
2f(x2, y2)
-1 -1
 For 2 points scheme,
 w1 = w2 = 1.0
 (x1, y1) = (+0.57735, +0.57735)
 (x2, y2) = (-0.57735, -0.57735)
Dr.G.PAULRAJ,

 Now, w1
2 f(x1, y1) = w1
2 (2x1
2 + 3x1y1 + 4y1
2) = 3.0
 w1 w2 f(x1, y2) = w1 w2 (2x1
2 + 3x1y2 + 4y2
2) = 1.0
 w2 w1 f(x2, y1) = w2 w1 (2x2
2 + 3x2y1 + 4y1
2) = 1.0
 w2
2 f(x2, y2) = w2
2 (2x2
2 + 3x2y2 + 4y2
2) = 3.0
1 1
 Hence, ∫ ∫ (2x2 + 3xy + 4y2) dx dy = 3 + 1 + 1+ 3 = 8.0
-1 -1
Result:
1 1
 Hence, ∫ ∫ (2x2 + 3xy + 4y2) dx dy = 8.0
-1 -1
Dr.G.PAULRAJ,

BEST OF LUCK
Dr.G.PAULRAJ,

Finite Element Analysis - UNIT-5

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