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Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Recap – Shape Functions
This is a good place to stop and remind ourselves where we are in the process of formulating numerical solutions using finite element methods. For a component we are solving the global force displacement equation
for displacements, i.e.,
The key to doing this is formulating the global stiffness matrix [K] properly and finding its inverse. Once we have solved for the displacements AT THE NODES, we can interpolate displacements (u, v) across the element through the use of shape functions. For one dimensional line elements
where the coordinate axis was attached to the left end of the element.
dKF
1 KFd
x
xxx
xxx
d
dLx
LxdNdNx
Ldddu
2
12211
121 ˆ
ˆˆ,
ˆ1ˆˆˆ
ˆˆˆˆ
1
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
The linear shape functions for the one dimensional rod element are
relative to a local coordinate system attached to the left end.
For a two dimensional constant strain triangle, once the nodal displacements were determined, the displacements across the element can be interpolated again through the use of shape functions
and these shape functions are formulated using global coordinate axes (x, y). For the constant strain triangle the shape functions are linear in x and y, i.e.,
yxA
N
yxA
NyxA
N
mmmm
jjjjiiii
21
21
21
mmjjii uNuNuNyxu ,
mmjjii vNvNvNyxv ,
LxNˆ
11 LxNˆ
2
2
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
For the linear strain triangle element displacements were quadratic functions of position. Recall that
Here after the nodal displacements are determined recall that the coefficients above are determined through the following expression
and the displacements are interpolated across the element using
where now the shape functions are defined as
2
12112
10987
265
24321
,
,
yaxyaxayaxaayxv
yaxyaxayaxaayxu
da 1
dNdM
aMvu
1*
*
1* MN3
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
For the linear strain triangle element the displacements vary quadratically across the element and the shape functions must be able to interpolate the nodal displacements quadratically across the element. Thus for the example problem
bhxy
bx
bxN
hy
bhxy
hyN
bhxyN
hy
hyN
bx
bxN
hy
bhxy
bx
hy
bxN
444
444
4
2
2
242331
2
2
6
2
2
5
4
2
2
3
2
2
2
2
2
2
2
1
Note carefully that the shape functions are dependent upon a coordinate system whose origin was attached to the first corner node. Change the coordinate system and the shape functions change.
Tracking the shape function for each individual element relative to a single coordinate system now becomes problematic. This is not something one can do by hand or track easily in computer software.
We will begin the use a local coordinate system as we formulate higher order elements.
4
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Geometric Interpolation and the Concept of a Standardized Element
We just reviewed how shape functions are used to interpolate nodal displacements (a field quantity) across an element – once nodal displacements were known. Once displacements and how they vary across the element are known, derivatives of displacement can be taken to obtain strain. Once strain is computed we compute stress across the element.
If the nodal coordinates are available we can perform the same sort of interpolation to define the curved boundary (geometry) of an element. Consider the classic problem of a plate with a hole subject to a tensile stress boundary condition. We could use quadrilateral elements with four corner nodes and straight sides, i.e.,
But there is a loss in fidelity along the straight line edges of the element.
5
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
We could develop a quadrilateral element with mid side nodes just as we did with triangular elements. The boundary (geometry) between corner nodes could be curved. This would require a quadratic interpolation of the geometry based on the coordinates of the corners and the mid side node:
Here the coordinates of nodes are spatially located and a quadratic curve is interpolated through the nodes of the elements – the geometry is interpolated. Quadratic interpolation functions allow mid side nodes to be offset relative to a straight line connecting the corner nodes thus producing curved boundaries. This type of element has much more fidelity in modeling the geometry of this components. 6
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
s
t
1 111
s
t
y
x
s
t
1
1y
xs
t
Moving forward concepts are developed referencing the figures below and defining the mathematics going from left to right, i.e., from natural coordinates to global coordinates. However, inverses exist for these transformations such that one can define the geometry on the right hand side, transform the problem to the natural coordinates on the left hand side and perform calculations in the natural coordinate system. Calculations in the natural coordinate system are far easier. After a formal definition of isoparametric elements the mathematics associated with the mapping indicated below is presented.
Natural Coordinates
Global Coordinates
7
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Isoparametric FormulationDeveloping shape functions and element stiffness matrices for higher order elements in terms of a global coordinate system is difficult. Isoparametric formulations for finite elements alleviates a great deal of this complexity. In addition, the isoparametric formulation allows the development of elements that have curved sides in the actual component configuration.
A finite element is said to be isoparametric if the same interpolation functions define both the displacement shape functions and the geometric shape functions. Geometric shape functions define the transformation used to go back and forth from an x-y coordinate system to an s-t coordinate system for two dimensional elements. If the geometric interpolation functions are of lower order than the displacement shape functions the element is said to be subparametric. If the reverse holds, then the element is referred to as superparametric.
Isoparametric elements can and do have curved boundaries which make them more suitable in capturing actual geometry However, for the higher order elements considered it is necessary to employ numerical integration to evaluate the element stiffness matrix. Transformation to an s-t coordinate system, a natural coordinate system, facilitates integration. The methods are a necessity for quadratic elements as well as higher order elements. In commercial software the isoparametric formulation is used for both low order and higher order elements. 9
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
10
The concept of an isoparametric element is further clarified in the following figure. Here the isoparametric element is compared first relative to a superparametric element – an element where there are fewer nodes for the computation of displacements using the weak formulation of the solid mechanics problem then there are to describe the geometry of the element.
For a subparametric element there are more nodes for the computation of displacements then there are to describe the geometry.
Clearly for an isoparametric element the same interpolation functions can be used for field quantities and geometry.
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
For a two dimensional element the natural s-t coordinate system is defined by element geometry and by the element orientation in the global coordinate system. There is a transformation mapping for each element, and this transformation is used in element formulation.
The isoparametric formulation now will be discussed relative to a simple 4-node quadrilateral element. The formulation is general enough to extend to higher order elements, e.g., the 8-node quadrilateral element. The s-t coordinate system is attached to the center of the element, and need not be parallel or orthogonal to the x-y coordinate axes:
Rectangular Plane Stress Element
2
3
s
t
1 11
4
1
1
y 12
3
4x
s
t
),(),(
tsyytsxx
),(),(
yxttyxss
11
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
This quadrilateral element has eight degrees of freedom, i.e., two displacements at each node. The unknown nodal displacements are defined as
Here
4
4
3
3
2
2
1
1
vuvuvuvu
d
xyayaxaayxv
xyayaxaayxu
8765
4321
,,
4
1
xb b
2
h
h
y3
12
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
If we solve for the coefficients in the usual manner for the location of the coordinate axes given in the previous figure then the expressions for the displacements in the element are
or
]
[41,
43
21
uyhxbuyhxb
uyhxbuyhxbbh
yxu
]
[41,
43
21
vyhxbvyhxb
vyhxbvyhxbbh
yxv
dN
yxvyxu
,,
13
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
where
andbh
yhxbyxN
bhyhxbyxN
bhyhxbyxN
bhyhxbyxN
4))((),(
4))((),(
4))((),(
4))((),(
4
3
2
1
4
4
3
3
2
2
1
1
4321
4321
00000000
vuvuvuvu
NNNNNNNN
vu
14
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Again the element strains for this two dimensional element are
and
The {B} matrix can be found by taking appropriate derivatives of the shape functions. The resulting expression for strain will demonstrate that the strain in the x-direction is only dependent on y, the strain in the y-direction is only dependent on x, and the shear strain is dependent on both x and y, all in a linear fashion.
xv
yu
yvxu
xy
y
x
dB
15
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
y 12
3
4x
s
t
The shape functions defining displacements within the element have been defined in terms of the x-y coordinate system. We utilize the same shape functions to geometrically map the element from the natural coordinates (s-t) into the x-y coordinate system. That is let
sttsysttsx
8765
4321
]1111
1111[41
43
21
xtsxts
xtsxtsx
Solving for the coefficients using the nodal coordinate geometry in the figure yields
]1111
1111[41
43
21
ytsyts
ytsytsy
16
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Or in matrix notation
where
4)1)(1(),(
4)1)(1(),(
4)1)(1(),(
4)1)(1(),(
4
3
2
1
tstsN
tstsN
tstsN
tstsN
4
4
3
3
2
2
1
1
4321
4321
00000000
yxyxyxyx
NNNNNNNN
yx
17
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Note that
for both the x-y coordinate system and s-t coordinate systems. This a check for rigid body motion. If every node is subjected to the unit displacement, e.g.,
14321 NNNN
1111
1111
i
i
vu
then it is obvious that every point in the component has the same displacement. A scalar multiple of one produces the same result. This constitutes rigid body motion. Also recall that rigid body motion will produce zero strains throughout the component.
18
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
We now turn our attention to formulating the {B} matrix for the quadrilateral element. This formulation could be carried out in the x-y coordinate system but the computations are difficult to nearly impossible. It is tedious to execute these computations in the s-tcoordinate system, but it is doable.
To construct the element stiffness matrix we must have expressions for strains which are theoretically derived in terms of derivatives of the displacements with respect to the x-ycoordinate system. If you use the s-t coordinate system to find displacements, the displacements are functions of s and t, and not x and y. Therefore we need to apply the chain rule for differentiation. This means the derivatives of the displacements are
ty
yv
tx
xv
tv
sy
yv
sx
xv
sv
ty
yu
tx
xu
tu
sy
yu
sx
xu
su
19
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Focusing on
we solve this system of equations for
Similarly we solve
For
ty
yu
tx
xu
tu
sy
yu
sx
xu
su
xu
x
ty
yv
tx
xv
tv
sy
yv
sx
xv
sv
yv
y
20
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Using Cramer’s rule
where the determinants in the denominator is the determinant of the Jacobian matrix, i.e.,
ty
tx
sy
sx
ty
tu
sy
su
xu
x
ty
tx
sy
sx
tv
tx
sv
sx
yv
y
ty
tx
sy
sx
J
ty
tx
sy
sx
J
21
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
For the shear strain
These determinant expressions lead toty
tx
sy
sx
ty
tv
sy
sv
ty
tx
sy
sx
tu
tx
su
sx
xv
yu
xy
tu
sy
su
ty
Jxu
x1
sv
tx
tv
sx
Jyv
y1
22
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
and
In a matrix format
tv
sy
sv
ty
su
tx
sx
tu
J
xv
yu
xy
1
vu
tsy
sty
stx
tsx
stx
tsx
tsy
sty
J
yv
xu
yvxu
xy
y
x
0
0
1
23
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Substituting for displacements yields
or
4
4
3
3
2
2
1
1
4321
4321
00000000
0
0
1
vuvuvuvu
NNNNNNNN
tsy
sty
stx
tsx
stx
tsx
tsy
sty
Jxy
y
x
dND
dB
24
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
where we define the operator matrix
The matrix {B} is now expressed as a function of s and t, i.e.,
but because of J, the variables s and t appear in the numerator and the denominator of the components of the {B} matrix. This complicates integration to obtain the element stiffness matrix (refer to your Calculus text books for the integration of rational polynomials).
tsy
sty
stx
tsx
stx
tsx
tsy
sty
JD 0
0
1
4321
4321
00000000
0
0
1NNNN
NNNN
tsy
sty
stx
tsx
stx
tsx
tsy
sty
JB
25
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
A final note on computing the determinant of the Jacobian matrix. With
then
and
ii
i
ii
i
ytsNy
xtsNx
),(
),(
ii
i
ii
i
xt
tsNtx
xs
tsNsx
),(
),(
ii
i
ii
i
yt
tsNty
ys
tsNsy
),(
),(
26
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
ii
i
ii
i
ii
i
ii
i
yt
tsNxt
tsN
ys
tsNxs
tsNty
tx
sy
sx
J
),(),(
),(),(
Thus
and
ii
i
ii
i
ii
i
ii
i
ys
tsNxt
tsN
yt
tsNxs
tsNJ
),(),(
),(),(
27
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
One Dimensional Isoparametric MappingThe term isoparametric stems from the fact that we use the same shape function to interpolate the field quantities, e.g., the displacements
that we use for the geometry of the line element. The function
is used to describe the location in the transformed space of any a point on the line element in real space. Here a transformation is defined to take the natural coordinates into global coordinates. Isoparametric element equations are formulated using a natural coordinate system (s for a line element) that is defined by element geometry and not by the global coordinate system. The axial s-coordinate axis is attached to the line element and remains directed along the line element no matter how each individual line element is oriented with respect to the global coordinate system.
For a quadratic line element the functions would take the form
saau 21
saax 21
2321 sasaau 2
321 sasaax 29
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
1 2
x2x1 x
Shape functions
Consider a quadratic line element, i.e., a line element with 3 nodes. The s-coordinate axis is attached to the center of the element. Shape functions for this element are given below (reference Example 10.6 in Logan’s text book)
3
x3
23
2
1
1)(2
1)(
21)(
ssN
sssN
sssN
Isoparametric mapping
3
1)(
iii xsNx
32
21 12
12
1 xsxssxssx
1 23
s1 1
`
30
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Given a point in the natural coordinate system the corresponding mapped point in the global coordinates is defined using the isoparametric mapping equation
2
3
1
10
1
xxsxxsxxs
32
21 12
12
1 xsxssxssx
The shape functions are defined in the natural coordinate (s) and they are polynomials as they were before. In the global x-coordinate system the shape functions in general are not polynomials. Consider the following line element defined in the global coordinate system
460
3
2
1
xxx
1 2x
3
4 2
31
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
The isoparametric mapping, x(s), for this example is
which is a simple polynomial. The inverse mapping, s(x), is not simple
24253 xs
2
2
32
21
34
4162
102
1
12
12
1
ss
sssss
xsxssxssx
32
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
The shape functions in the global coordinates
Develop expressions for N1 and N3 as a homework exercise.
)(
42521021
24253
12
425321
21)(
2
2
xN
xx
xx
sssN
33
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Graphically the shape functions for node #2 plot as follows in the two coordinate systems
N2 (x) is slightly more complicated
but is painful to use with more than one element. Think about where to place the origin of the coordinate system.
1 2
s1 1
3
N2(s)
1 2x
3
4 2 11
N2(x)
N2 (s) is a simple polynomial
2
1)(2sssN
xxxN 42521021)(2
34
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Matrices for a Line Element
For a line element with quadratic shape functions
The strain displacement relationship is once again
dNuNuNuNu
332211
dB
ux
Nux
Nux
Nxu
33
22
11
35
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
E
Here
and stress is once again for a linear element is
x
Nx
Nx
NB 321 ,,
The only difference from before is that the shape functions are formulated in the natural coordinate system
and the derivatives above are expressed in the global x-coordinate system.
23
2
1
1)(2
1)(
21)(
ssN
sssN
sssN
36
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
We know that because we use an isoparametric mapping that
and we will use this expression to formulate the components of the {B} matrix which are derivatives of the shape functions. Using the chain rule from calculus
In Elasticity we derived the relationship
If J was greater than 1 we had volume expansion, between zero and 1 corresponds to volume contraction. Interpreting this relationship in terms of a line element we have
3
1)(
iii xsNx
dxds
ssN
xsN ii
)()(
dsJdx
odVJdV
37
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
or
From a computational standpoint
For a line element the calculation immediately above is made, inverted and used in the following expression:
Now the derivative of the shape functions are formulated in terms of the natural coordinate system.
3
1
)(i
ii x
dssdN
dsdxJ
dssdN
J
dxds
dssdN
xsN
i
ii
)(1
)()(
Jdxds 1
38
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
For the 3-noded line element
and the {B} matrix in the natural coordinate system is
321
3
1
22
122
12
)(
xsxsxs
xds
sdNJi
ii
sss
JB 2,
212,
2121
39
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
The element stiffness matrix is expressed as
The integral associated with ANY element in the global coordinates is transformed to an integral in the natural coordinate system where the integration will be from -1 to 1 in the local coordinates.
The Jacobean is a function of the s-coordinate in general and appears in the integrals. The specific form of J is determined by the values of x1, x2 and x3.
The components of the {B} matrix are polynomial functions in the s-coordinate system. In general Gaussian quadrature is used to evaluate the stiffness matrix.
Moreover, now think about the utility of the shape functions. Here their derivatives (the {B} matrix) are used to formulate the stiffness matrix in addition to their use in interpolating geometry (isoparametric elements), displacements through the elements, strains through the elements and stresses through the elements.
1
1
2
1
dsJABEB
dxABEBk
T
x
x
T
40
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Gauss QuadratureIn very simple terms a definite integral is defined as
where
Attention is now given to methods that can evaluate this definite integral numerically. A characteristic of a group of numerical integration, or quadrature, techniques known as Newton-Cotes equations are that integral estimates are based on evenly spaced values of the function. Consequently, the location of the evaluation points used in these types of numerical integration method are fixed, or predetermined.
41
aFbF
dxxfIb
a
dx
xdFxf
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Consider the trapezoidal rule which is the simplest method of the group. This method is based on taking the area under the straight line connecting the function values at the end of the integration interval. The formula for the trapezoidal rule is
Because the trapezoidal rule must use end point values of the function there are cases where the error associated with computation given above results in significant error. The integration error is quite noticeable with the function presented in the figure below.
42
2bfafab
dxxfIb
a
a b
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Next consider that the restraint of fixed base points is relaxed and one is free to evaluate the area under the straight line joining any two points on the curve. Note that the length of the base of the trapezoid is maintained.
By positioning the two points on the curve wisely, a straight line could be positioned that would balance the positive and negative errors. This is depicted in the following figure
Gauss quadrature is the name given to one class of techniques that implement this type of strategy. Before describing the approach and its use in deriving stiffness matrices for isoparametric elements we show how numerical integration formulas such as the trapezoidal rule can be found using the method of undetermined coefficients. This method will then be used to develop the Gauss quadrature formula. 43
a b
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
To illustrate the method of undetermined coefficients consider an alternative formulation for the trapezoidal rule
where c0 and c1 are constants. Realizing the trapezoidal rule must yield exact results when the function being integrated is a constant, or a linear function of x, then one can use these results to generate the trapezoidal rule. Consider that for f(x) = 1
n
iii
b
a
xfcbfcafc
bfafab
dxxfI
121
2
ab
dxcc
ab
ab
2
2
21 111
44
Undetermined Coefficients
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
And for f(x) = x
These two integrals yield two equations for the two unknown coefficients. Solving them simultaneously yields
which when substituted back into the original formulation for the integral of the function gives us back the trapezoidal rule, i.e.,
022
2
2
21
ab
ab
dxxabcabc
221
abcc
bfafab
bfabafabbfcafcI
2
22
21
45
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
The objective of the Gauss quadrature approach is to determine the unknown constants for the expression
However in contrast to the trapezoidal rule that used fixed end points a and b, the function arguments x1 and x2 are not fixed and treated as unknowns. Now we need four integral expressions to find the four unknowns, c1, c2 , x1 and x2.
We obtain these conditions by assuming the equation above produces the integral value exactly for a constant function and a linear function, i.e., equality holds for
keeping in mind that
2211 xfcxfcI
1
12211
1
12211 1
dxxxfcxfc
dxxfcxfc
46
1
1
1
1
0
21
dxx
dx
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
We use the “canonical” form of the integrals, i.e., the limits of integration are from -1 to 1 because these are the limits of integration we will use for isoparametric elements.
To arrive at two additional conditions we assume that the Gauss quadrature formulation yields exact results when the integrand is a polynomial of degree 3 or less.
On the previous overhead geometric arguments were made to make the point that equality holds exactly. Equality for these two expressions will be demonstrated momentarily.
If the two expressions above are exact, then
1
1
32222
1
1
22211
dxxxfcxfc
dxxxfcxfc
47
2211
1
1
3
2211
1
1
2
0
32
xfcxfcdxx
xfcxfcdxx
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
These two expressions along with the two previous expressions yields four equations in terms of four unknowns, i.e.,
which leads to (homework assignment)
48
0
32
0
211
322
3112211
222
2112211
22112211
212211
xcxcxfcxfc
xcxcxfcxfc
xcxcxfcxfc
ccxfcxfc
5773503.03
1
5773503.03
11
2
1
21
x
x
cc
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Consider the following integral with cubic polynomial integrand
Using the Gauss quadrature for this integral yields
One concludes from this brief example that the Gauss quadrature is exact for polynomials that are cubic or less. 49
1
1
32
32dxxxI
32
5773503.05773503.015773503.05773503.01 3232
32
222
31
211
2211
xxcxxc
xfcxfcI
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Beyond the two point formula described previously, three, four, five and six point versions of the Gauss quadrature approach have been used. The general form is
where n is the number of quadrature points. Values of c’s and x’s are summarized in the table to the right:
n
iii
nn
xfc
xfcxfcxfcI
1
2211
50
Number of gauss points Integration location, xi Weight, ci
1 x1 = 0 c1 = 2
2 x1 = ‐ 0.57735026918962 c1 = 1
x2 = + 0.57735026918962 c2 = 1
3 x1 = ‐ 0.77459666924148 c1 = 0.555556x2 = 0 c2 = 0.888889
x3 = + 0.77459666924148 c3 = 0.555556
4 x1 = ‐ 0.861136312 c1 = 0.3478548x2 = ‐ 0.339981044 c2 = 0.6521452x3 = + 0.339981044 c3 = 0.6521452x4 = + 0.861136312 c4 = 0.3478548
5 x1 = ‐0.906179846 c1 = 0.23692695x2 = ‐ 0.538469310 c2 = 0.4786287
x3 = 0.0 c3 = 0.5688889x4 = + 0.538469310 c4 = 0.4786287
x5 = +0.906179846 c5 = 0.23692695
6 x1 = ‐0.93246914 c1 = 0.1713245x2 = ‐ 0.661209386 c2 = 0.3607616x3 = ‐0.238619186 c3 = 0.4679139x4 = + 0.238619186 c4 = 0.4679139x5 = +0.661209386 c5 = 0.3607616x6 = +0.93246914 c6 = 0.1713245
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
If we want to extend this approach to integration over an area, say for an isoparametric element, then
In general, we do not have to use the same number of Gauss points in each direction, i.e., idoes not have to equal j, but in finite element analysis this is typically done.
n
ijiij
n
j
n
ijii
n
jj
n
iii
tsfcc
tsfcc
dttsfc
dtdstsfI
11
11
1
1 1
1
1
1
1
,
,
,
,
51
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Consider a four point Gauss integration which is shown in following figure
where for an arbitrary function of s and t
where all sampling points are +0.5773 or -0.5773 and all coefficients are equal to one. Hence the double integral is double summation technically, but really it is a single summation over four points in the element, i.e., the Gauss points.
2222121221211111
2
1
2
1
1
1
1
1
,,,,
,
,
tsfcctsfcctsfcctsfcc
tsfcc
dtdstsfI
n
ijiij
n
j
52
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
For a volume element we can easily extend the concepts as follows:
n
kkjikji
n
j
n
iztsfccc
dzdtdsztsfI
111
1
1
1
1
1
1
,,
,,
53
Section 10: ISOPARAMETRIC FORMULATION
Washkewicz College of Engineering
Element Stiffness MatrixIn general for a two dimensional element we have shown that
For a quadrilateral isoparametric element
If we use Gauss quadrature to evaluate the integral
A
TT dydxtBDBk
1
1
1
1
dtdstJBDBk TT
21212121
22222222
12121212
11111111
1
1
1
1
,,,
,,,
,,,
,,,
ccttsJtsBDtsB
ccttsJtsBDtsB
ccttsJtsBDtsB
ccttsJtsBDtsB
dtdstJBDBk
TTT
TTT
TTT
TTT
TT
54
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