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Section 8: ISOPARAMETRIC FORMULATION
Lagrange Interpolation - Review
In data analysis for engineering designs we are frequently presented with a series of data values where the need arises to interpolate values between the given data points. Recall linear interpolation used extensively to find intermediate tabular values. Another common approach is using higher order polynomials to “curve fit” a function betweencommon approach is using higher order polynomials to curve fit a function between data values. The polynomial usually takes the form:
( ) nn xaxaxaaxf L2
210 ++=
For (n+1) data points, there is only one polynomial of order n that passes through all the values. For example, there is only one straight line (a first order polynomial) that passes h h d i i il l l b l f h d i
( ) n xaxaxaaxf 210
through two data points. Similarly, only one parabola connects a set of three data points. Polynomial interpolation consists of determing the unique nth-order polynomial that fits (n+1) data points. This polynomial then provides a formula to compute intermediate values. va ues.
Section 8: ISOPARAMETRIC FORMULATION
We have been doing this all semester without the formal definition given on the previous slide. Although there is only one nth-order polynomial that fits (n+1) data
i h i f h d h b ili d b i h fi l f f hpoints, there are a variety of methods that can be utilized to obtain the final form of the interpolating polynomial. These methods include (but are not limited to)
• Newton’s Divided Difference Approach
• The Method of Lagrange Polynomials
• Regression Analysis (linear and non-linear)
S li• Splines
Here we focus on Lagrange interpolating polynomials because the method leads directly to the formulation of shape functions for higher order elements with an appropriate number of internal nodesappropriate number of internal nodes.
The Lagrange interpolating polynomial is a reformulation of the Newton polynomial, but avoids the computation of divided differences. The Lagrange polynomial has the form:form:
( ) ( ) ( )i
n
iin xfxHxf ∑
=
=0
Section 8: ISOPARAMETRIC FORMULATION
where
( ) ( )( )∏−
=n
jxxxH
For example, the linear version (n=1) is
( ) ( )∏≠= −
=
ijj ji
i xxxH
0
( ) ( )( ) ( ) ( )
( ) ( ) ( )( ) ( )1
01
00
10
1
0 01 xf
xxxxxf
xxxxxf
xxxx
xf i
n
i
n
ijj ji
j
−−
+−−
=
−
−= ∑ ∏
= =
and the second order version is
ij ≠
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )102021 fxxxxfxxxxfxxxxf −−−−−−
One can begin to see the usefulness of Lagrange polynomials by realizing that each term H (x) will be 1 at x = x and 0 at all other data points This is the quality we are looking
( ) ( )( )
( )( ) ( ) ( )
( )( )( ) ( ) ( )
( )( )( ) ( )2
12
1
02
01
21
2
01
00
20
2
10
12 xf
xxxxxf
xxxxxf
xxxxxf
−−+
−−+
−−=
Hi(x) will be 1 at x = xi and 0 at all other data points. This is the quality we are looking for in a shape function, i.e., the shape function for a particular node is 1 at the node, and zero at all other nodes.
Section 8: ISOPARAMETRIC FORMULATION
In two dimensions the interpolation function has the form
where
( ) ( ) ( ) ( ) ( )
= ∑∑
==i
m
iii
n
iip ygyVxfxHyxf
00,
Three dimensional Lagrange interpolation has the form
mnp =
where
( ) ( ) ( ) ( ) ( ) ( ) ( )
= ∑∑∑
===i
l
iii
m
iii
n
iip zqyQygyVxfxHzyxf
000,,
where
lmnp =
Section 8: ISOPARAMETRIC FORMULATION
Recap – Shape Functions
This is a good place to stop and remind ourselves where we are in the process ofThis is a good place to stop and remind ourselves where we are in the process of formulation solutions using finite element methods. For a component we are solving the global force displacement equation
{ } [ ]{ }dKF =
for displacements, i.e.,
{ } [ ]{ }dKF =
{ } { }[ ] 1−= KFd
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
{ } { }[ ]= KFd
p , pdisplacements (u, v) across the element through the use of shape functions. For one dimensional elements
− xxx dxxdNdNddd 112
ˆˆˆ1ˆˆˆ
ˆˆˆˆ
where the coordinate axis was attached to the left end of the element.
−=+=
+=x
xxx
xxx
dLLdNdNx
Ldu
2
12211
121 ˆ
,1
Section 8: ISOPARAMETRIC FORMULATION
The linear shape functions for the one dimensional rod element are
xNˆ
11 −=xNˆ
2 =
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
h l b i l d i h h h f h f i
LN 11 L
N2
across the element can be interpolated again through the use of shape functions
( ) mmjjii uNuNuNyxu ++=,
( ) vNvNvNyxv ++=but 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.,
1
( ) mmjjii vNvNvNyxv ++=,
( )
( )yxA
N
yxA
N
jjjj
iiii
γβα
γβα
++
=
++
=
21
21
( )yxA
N
A
mmmm
jjjj
γβα ++
=
21
2
Section 8: ISOPARAMETRIC FORMULATION
For the linear strain triangle element displacements were quadratic functions of position. Recall that
H ft th d l di l t d t i d ll th t th ffi i t b
( )( ) 2
12112
10987
265
24321
,
,
yaxyaxayaxaayxv
yaxyaxayaxaayxu
+++++=
+++++=
Here after the nodal displacements are determined recall that the coefficients above are determined through the following expression
{ } [ ] { }da 1−= χand the displacements are interpolated across the element using
{ }{ }aMvu
=
*
where now the shape functions are defined as
{ }[ ] { } { }{ }dNdM
v
==
−1* χ
{ } { }[ ] 1* −= χMN
Section 8: ISOPARAMETRIC FORMULATION
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
d i ll h l h f h h l blquadratically across the element. Thus for the the example problem
hy
bhxy
bx
hy
bxN 242331 2
2
2
2
1 +++−−=
bx
bxN
hbhbhb
2
2
2
2
2
2 +−= Note carefully that the shape functions are dependent upon a coordinate system whose origin was attached to the first corner node Change the
bhxyN
hy
hyN
4
2
4
23
=
+−=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
hy
bhxy
hyN
bh444
2
2
5
4
−−=
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.
bhxy
bx
bxN 444
2
2
6 −−= We will begin the use a local coordinate system as we formulate higher order elements.
Section 8: ISOPARAMETRIC FORMULATION
Isoparametric Formulation
As we just saw developing shape functions as well as element stiffness matrices in terms j p g pof the global coordinate system for higher order elements will become enormously difficult. Isoparametric formulations for finite elements is a way around this complexity. The isoparametric formulation allows the development of elements that are non-rectangular and have curved sidesrectangular and have curved sides
A finite element is said to be isoparametric if the same interpolation functions define both the displacement shape functions and the geometric shape functions - the geometric transformation functions used to go back and forth from an x y coordinate system to an s ttransformation functions used to go back and forth from an x-y coordinate system to an s-tcoordinate system for quadrilateral 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 tend to have curved boundaries which make them more suitable in capturing geometric boundary conditions. However stress-strain relationships are complicated in that displacements are specified relative to a local coordinate system (s-t)whereas differentiation is made with respect to the global x-y coordinate system. It also becomes necessary to employ numerical integration to evaluate the element stiffness matrix.
Section 8: ISOPARAMETRIC FORMULATION
Geometric InterpolationWe just reviewed how shape functions are used to interpolate nodal displacements across
l t d l di l t k O k h di l tan element – once nodal displacements were known. Once we know how displacements vary across the element we can take derivatives of displacement 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 toIf the nodal coordinates are available we can perform the same sort of interpolation to define the boundary (geometry) of the 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
But there is a loss in fidelity along the straight line edges of the element ifline edges of the element if the component edge is actually curved.
Section 8: ISOPARAMETRIC FORMULATION
We could develop a quadrilateral element with mid side nodes just as we did with triangular elements. The boundary between corner nodes could have a curved shape based on a quadratic interpolation of the coordinates of the corners and the mid side b sed o qu d c e po o o e coo d es o e co e s d e d s denode:
Note that the coordinates of nodes are specified and the software must fit geometry through location of the nodes. Quadratic interpolation functions allows for nodes to be moved out of line relative to one another producing curved boundaries. This type of element would have much more fidelity in modeling the geometry of this componentis.
Section 8: ISOPARAMETRIC FORMULATION
The concepts are developed by looking at 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 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 execute calculations in the natural coordinate system. Calculations in the natural coordinate system are far easier, but still difficult.
t
1 11t
s1 1
1s
y
xt
Natural Coordinates t
1y
tGlobal Coordinates
s1
y
xs
Section 8: ISOPARAMETRIC FORMULATION
I l lIn class example
Section 8: ISOPARAMETRIC FORMULATION
One Dimensional Isoparametric MappingThe term isoparametric stems from the fact that we use the same shape function to interpolate the field quantities, i.e, displacemnts
that we use for the geometry of the line element The functionsaau 21 +=
that we use for the geometry of the line element. The function
is used to describe the location of any a point on the line element. Here a transformation is d fi d k h l di i l b l di I i l
saax 21 +=
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 gno 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 formq
2321 sasaau ++= 2
321 sasaax ++=
Section 8: ISOPARAMETRIC FORMULATION
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 were just derived.
x2x1 x1 2
p j
3 x3
1 2s1 1 3
`
Local (isoparametric) coordinates
( )1 ss −
Isoparametric mapping
3( )
( )2
1
21)(
21)(
sssN
sssN
+=
−−= ∑
=
=3
1)(
iii xsNx
( ) ( )11 +2
3 1)(2ssN −=
( ) ( ) ( ) 32
21 12
12
1 xsxssxssx −++
+−
−=
Section 8: ISOPARAMETRIC FORMULATION
Given a point in the natural coordinate system the corresponding mapped point in the global coordinates is defined using the isoparametric mapping equation
( ) ( ) ( ) 32
21 12
12
1 xsxssxssx −++
+−
−=
2
3
1
10
1
xxsxxsxxs
=====−=
2
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
l i l C id th f ll i li l t d fi d i th l b l di t tpolynomials. Consider the following line element defined in the global coordinate system
601 =
xx
x4 2
46
3
2
==
xx
1 23
Section 8: ISOPARAMETRIC FORMULATION
The isoparametric mapping, x(s), for this example is
( ) ( ) ( )( ) ( ) ( ) ( ) ( )( )2
32
21
416101
12
12
1
sssss
xsxssxssx
++
+−
=
−++
+−
−=
( ) ( ) ( ) ( ) ( )( )234
4162
02
ss
s
−+=
−++−=
which is a simple polynomial. The inverse mapping, s(x), is not simple
24253 xs −−
=2
Section 8: ISOPARAMETRIC FORMULATION
The shape functions in the global coordinates
( )( )
42531
425312
1)(2
xx
sssN
−−
+
−−
+=
( )42521021
21
22
xx −−−=
+
=
Do N1 and N3
)(2
2 xN=
Section 8: ISOPARAMETRIC FORMULATION
Graphically the shape functions for node #2 plot as follows in the two coordinate systems
N2(x)
1 23
N2(s)
x4 2 11
N2(x)
s1 1 1 23
N2(x) is slightly more complicated N2(s) is a simple polynomial
( )1)( sssN += ( )xxxN 4252101)( −−−=
but is painful to use with more than one element. Think about where to
2)(2 sN = ( )xxxN 425210
2)(2 =
o e e e e t. about w e e toplace the origin of the coordinate system.
Section 8: ISOPARAMETRIC FORMULATION
Matrices for a Line Element
F li l i h d i h f iFor a line element with quadratic shape functions
{ }{ }dNuNuNuNu ++= 332211
The strain displacement relationship is once again
{ }{ }dN=
xu∂∂
=ε
{ }{ }dB
ux
Nux
Nux
N
=∂∂
+∂∂
+∂∂
= 33
22
11
Section 8: ISOPARAMETRIC FORMULATION
Here
{ }
∂∂∂ NNNB 321
and stress is once again for a linear element is
{ }
∂∂∂=
xxxB 321 ,,
εσ E=
The only difference from before is that the shape functions are formulated in the natural coordinate systemnatural coordinate system
( )
( )1 2
1)( sssN −−=
( )
23
2
1)(2
1)(
ssN
sssN
−=
+=
and the derivatives above are expressed in the global x-coordinate system.
Section 8: ISOPARAMETRIC FORMULATION
We know that because we use an isoparametric mapping that
∑=3
)( xsNx
and we will use this expression to formulate the components of the {B} matrix which are d i ti f th h f ti U i th h i l f l l
∑=
=1
)(i
ii xsNx
derivatives of the shape functions. Using the chain rule from calculus
dxds
ssN
xsN ii
∂∂
=∂
∂ )()(
In Elasticity we derived the relationship
If J was greater than 1 we had volume expansion between zero and 1 corresponds to volumeodVJdV =
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
dsJdx =
Section 8: ISOPARAMETRIC FORMULATION
or
Jdxds 1
=
From a computational standpoint
=dsdxJ
For a line element the calculation immediately above is made, inverted and used in the
∑=
=3
1
)(i
ii x
dssdN
following expression:
=
∂∂
dxds
dssdN
xsN ii )()(
=
∂
dssdN
J
dxdsx
i )(1
Now the derivative of the shape functions are formulated in terms of the natural coordinate system.
Section 8: ISOPARAMETRIC FORMULATION
For the 3-noded line element
3 )(sdN
( ) 321
1
21212
)(
xsxsxs
xds
sdNJi
ii
−
+
+
−
=
= ∑
=
and the {B} matrix in the natural coordinate system is
( ) 321 22xsxx
{ }
−
+−= sss
JB 2,
212,
2121
Section 8: ISOPARAMETRIC FORMULATION
The element stiffness matrix is expressed as
[ ] { } { }∫= 2 dxABEBkx T
The integral associated with ANY element in the global coordinates is transformed to
[ ] { } { }
{ } { }∫
∫
−=
1
1
1
dsJABEB T
x
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-coordinateThe 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.
Section 8: ISOPARAMETRIC FORMULATION
F di i l l h l di i d fi d b l
Rectangular Plane Stress Element
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 g q yto the center of the element, and need not be parallel or orthogonal to the x-y coordinate axes:
2 t 2t
2
s1 1
11
1s
y 12
3),(),(
tsyytsxx
==
)( yxss =
3 41
4x
s),(),(
yxttyxss
==
Section 8: ISOPARAMETRIC FORMULATION
This quadrilateral element has eight degrees of freedom, i.e., two displacements at each node. The unknown nodal displacements are defined as
1
1
vu
4y
3
{ }
=3
2
2
uvu
d xb bh
h
4
3
vuv 1 2
Here
4v
( ) xyayaxaayxu +++( )( ) xyayaxaayxv
xyayaxaayxu
8765
4321
,,
+++=+++=
Section 8: ISOPARAMETRIC FORMULATION
If we solve for the coefficients in the usual manner the expressions for the displacements in the element are
( ) ( )( ) ( )( )
( )( ) ( )( ) ]
[41, 21
hbhb
uyhxbuyhxbbh
yxu −++−−=
( )( ) ( )( ) ]43 uyhxbuyhxb +−++++
( ) ( )( ) ( )( )[41, 21 vyhxbvyhxbbh
yxv −++−−=
or
( ) ( )( ) ( )( )
( )( ) ( )( ) ]
[4
43
21
vyhxbvyhxb
yybh
y
+−++++
( )( ) { }{ }dN
yxvyxu
=
,,
Section 8: ISOPARAMETRIC FORMULATION
where
bhyhxbyxN
4))((),(1
−−=
yhxbbh
yhxbyxN
bh
))(()(
4))((),(
4
2
++
−+=
bhyhxbyxN
bhyhxbyxN
4))((),(
4))((),(
4
3
+−=
++=
uandbh4
2
1
1
uvu
=
3
3
2
4321
4321
00000000
vuv
NNNNNNNN
vu
4
4
3
vu
Section 8: ISOPARAMETRIC FORMULATION
Again the element strains for this two dimensional element are
∂u
∂∂∂
=
yvx
y
x
εε
and
∂∂
+∂∂
xv
yuxyγ
and
{ } { }{ }dB=ε
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 d d b h d ll i li f hidependent on both x and y, all in a linear fashion.
Section 8: ISOPARAMETRIC FORMULATION
The shape functions defining displacements within the element have been defined in terms of the x-y coordinate system. We utilize the same procedure to map the element h b l i f i l di i h di h ishown below in terms of its natural coordinates into the x-y coordinate system. That is
letstatasaax 4321 +++=
4 t
1 13
1statasaay 8765 +++=
1
s1 1
2
1
1Solving for the coefficients using the geometry in the figure yields
1 2
( )( ) ( )( )
( )( ) ( )( ) ]1111
1111[41
21
xtsxts
xtsxtsx
+−++++
−++−−=
( )( ) ( )( ) ]1111 43 xtsxts +++++
( )( ) ( )( )1111[41
21 ytsytsy −++−−=
( )( ) ( )( ) ]11114
43 ytsyts +−++++
Section 8: ISOPARAMETRIC FORMULATION
Or in matrix notation
1
1
yx
=
2
2
4321
00000000
xyx
NNNNNNNN
yx
where
4
3
34321 0000
xyxNNNNy
)1)(1(4
)1)(1(),(1tstsN −−
=
4y
)1)(1(),(
4)1)(1(),(
3
2
tstsN
tstsN
++=
−+=
4)1)(1(),(
4),(
4
3
tstsN +−=
Section 8: ISOPARAMETRIC FORMULATION
Note that14321 =+++ NNNN
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.,
( )1111( )( )1111
1111==
i
i
vu
th it i b i th t i tthen it is obvious that every point in the component has the same displacement. A scalar multiple of one produces the same result. pThis constitutes rigid body motion. Also recall that rigid body motion will produce zero strains throughout thestrains throughout the component.
Section 8: ISOPARAMETRIC FORMULATION
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
yuxuusy
yu
sx
xu
su
∂∂+
∂∂=
∂∂∂
∂∂
+∂∂
∂∂
=∂∂
sy
yv
sx
xv
sv
tytxt
∂∂
∂∂
+∂∂
∂∂
=∂∂
∂∂+
∂∂=
∂
ty
yv
tx
xv
tv
sysxs
∂∂
∂∂
+∂∂
∂∂
=∂∂
∂∂∂∂∂
Section 8: ISOPARAMETRIC FORMULATION
Focusing on
sy
yu
sx
xu
su
∂∂
∂∂
+∂∂
∂∂
=∂∂
we solve this system of equations for
ty
yu
tx
xu
tu
∂∂
∂∂
+∂∂
∂∂
=∂∂
we solve this system of equations for
xu
x ∂∂
=ε
Similarly we solve
sy
yv
sx
xv
sv
∂∂
∂∂
+∂∂
∂∂
=∂∂
Forty
yv
tx
xv
tv
sysxs
∂∂
∂∂
+∂∂
∂∂
=∂∂
∂∂∂∂∂
yv
y ∂∂
=ε
Section 8: ISOPARAMETRIC FORMULATION
Using Cramer’s rule
yu ∂∂ vx ∂∂
ty
tu
sy
su
u ∂∂
∂∂
∂∂
∂∂
∂ tv
tx
sv
sx
v ∂∂
∂∂
∂∂
∂∂
∂
yxsy
sx
ttxx
∂∂∂∂
∂∂
∂∂=∂
=ε
yxsy
sx
ttyv
y
∂∂∂∂
∂∂
∂∂=∂∂
=ε
where the determinants in the denominator is the determinant of the Jacobian matrix, i.e.,
ty
tx
∂∂
∂∂
ty
tx
∂∂
∂∂
sy
sx
J ∂∂
∂∂
=
ty
tx
J
∂∂
∂∂
=
Section 8: ISOPARAMETRIC FORMULATION
For the shear strainyvux ∂∂∂∂
ty
tv
ss
tu
tx
ss
vuxy ∂∂
∂∂
∂∂
∂∂
+∂∂∂∂
∂∂
∂∂
=∂∂
+∂∂
=γ
yxsy
sx
yxsy
sxxyxy
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂∂∂
γ
These determinant expressions lead totttt ∂∂∂∂
∂∂∂∂∂ uyuyu 1
∂∂
∂∂
−∂∂
∂∂
=∂∂
=tu
sy
su
ty
Jxu
x1ε
∂∂∂∂∂ vxvxv 1
∂∂
∂∂
−∂∂
∂∂
=∂∂
=sv
tx
tv
sx
Jyv
y1ε
Section 8: ISOPARAMETRIC FORMULATION
and
∂∂ vu
∂∂
∂∂
−∂∂
∂∂
+∂∂
∂∂
−∂∂
∂∂
=
∂∂
+∂∂
=
vyvyuxxuJ
xv
yu
xy
1
γ
In a matrix format
∂∂∂∂∂∂∂∂ tsstststJ
∂∂∂∂∂∂
∂∂
−∂∂
∂∂
=
∂∂∂
=
uxx
tsy
sty
vxu
x
0
0
1εε
∂∂
∂∂
−∂∂
∂∂
∂∂
∂∂
−∂∂
∂∂
∂∂−
∂∂=
∂∂
+∂∂∂
=
v
tsy
sty
stx
tsx
sttsJ
yv
xu
yxy
y 0
γ
ε
y
Section 8: ISOPARAMETRIC FORMULATION
Substituting for displacements yields
∂∂∂∂
1
vu
∂∂
∂∂
−∂∂
∂∂
∂∂
∂∂
−∂∂
∂∂
=
2
2
1
4321
00000000
0
0
1uvuv
NNNNNNNN
stx
tsx
tsy
sty
Jy
x
εε
∂∂
∂∂
−∂∂
∂∂
∂∂
∂∂
−∂∂
∂∂
∂∂∂∂
4
3
34321 0000
vuvuNNNN
tsy
sty
stx
tsx
sttsJxyγ
or
4v
{ } { }{ }dB{ } { }{ }{ }{ }{ }dND
dB′=
=ε
Section 8: ISOPARAMETRIC FORMULATION
where we define the operator matrix
∂∂
∂∂
−∂∂
∂∂
tsy
sty 0
{ }
∂∂
∂∂
−∂∂
∂∂
∂∂
∂∂
−∂∂
∂∂
∂∂
∂∂
−∂∂
∂∂
=′
yyxxst
xts
xJ
D 01
The matrix {B} is now expressed as a function of s and t, i.e., ∂∂∂∂∂∂∂∂ tsststts
∂∂
−∂∂ 0yy
{ }
∂∂∂∂∂∂∂∂∂∂
∂∂
−∂∂
∂∂
∂∂∂∂
=4321
4321
00000000
0
0
1NNNN
NNNNst
xts
xtsst
JB
but because of J, the variables s and t appear in the numerator and the denominator of the
∂
∂∂∂
−∂∂
∂∂
∂∂
∂∂
−∂∂
∂∂
tsy
sty
stx
tsx
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).
Section 8: ISOPARAMETRIC FORMULATION
A final note on computing the determinant of the Jacobian matrix. With
xtsNx )(∑=
ii
i
ii
i
ytsNy
xtsNx
),(
),(
∑
∑=
=
then
∑ ∂∂
=∂∂
ii
i xs
tsNsx ),(
and
∑ ∂∂
=∂∂
ii
i xt
tsNtx ),(
∑∂∂
∂∂
=∂∂
ii
i
N
ys
tsNsy
)(
),(
∑ ∂∂
=∂∂
ii
i yt
tsNty ),(
Section 8: ISOPARAMETRIC FORMULATION
∂∂ yx
This leads to
[ ]
∂∂
∂∂
∂∂=
ty
tx
ssJ
∂∂∂
∂∂
∂
=∑∑
∑∑ii
ii
i
ii
i
ytsNxtsN
ys
tsNxs
tsN
),(),(
),(),(
∂∂ ∑∑
ii
ii y
tx
t
thus
∂
∂
∂
∂
∂
∂=
∑∑
∑∑i
ii
ii
i
tsNtsN
yt
tsNxs
tsNJ
)()(
),(),(
∂
∂
∂
∂− ∑∑
ii
i
ii
i ys
tsNxt
tsN ),(),(
Section 8: ISOPARAMETRIC FORMULATION
Gauss QuadratureA characteristic of a group of numerical integration, or quadrature, techniques known as g p g , q , qNewton-Cotes equations are integral estimates 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. Consider the trapezoidal rule which is the simplest method of the gro p This method is based on taking the area nder the straight linesimplest 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
( )
( ) ( )
= ∫ dxxfIb
a
( ) ( ) ( )
+
−=2
bfafab
Section 8: ISOPARAMETRIC FORMULATION
Because the trapezoidal rule must use end point values of the function there are cases where the error associated with computation defined on the previous slide results in i ifi C id h f i d i h f ll i fisignificant error. Consider the function presented in the following figure
Next consider that the restraint of fixed base points is relaxed and one is free to evaluateNext 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.
Section 8: ISOPARAMETRIC FORMULATION
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 methodtrapezoidal rule can be found using the method of undetermined coefficients. This method will then be used to develop the Gauss quadrature formula.
Section 8: ISOPARAMETRIC FORMULATION
To illustrate the method of undetermined coefficients consider an alternative formulation for the trapezoidal rule bp
( )
( ) ( ) ( )bfafab
dxxfIb
a
+
= ∫
where c and c are constants Realizing the trapezoidal rule must yield exact results when
( ) ( ) ( )
( ) ( )bfcafc
ab
10
2+=
−=
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
( ) ( ) ( )dxcc
ab
ab
=+ ∫−
−−
2
2
10 111
( )ab −=2
Section 8: ISOPARAMETRIC FORMULATION
And for f(x) = x2
−
+ −
∫−ab
dabab
0
222
10
=
=
+
− ∫−
−ab
dxxcc
These two integrals yield two equations for the two unknown coefficients. Solving them simultaneously yields
−
==210
abcc
which when substituted back into the original formulation for the integral of the function gives us back the trapezoidal rule, i.e.,
2
( ) ( )
( ) ( )bfabafabbfcafcI
−
+ −
=
+= 10
( ) ( )bfaf
+
22
Section 8: ISOPARAMETRIC FORMULATION
The objective of the Gauss quadrature approach is to determine the unknown constants for the expression
( ) ( )xfcxfcI +=
However in contrast to the trapezoidal rule that used fixed end points a and b, the function arguments x0 and x1 are not fixed and treated as unknowns. Now we need four integral
i t fi d th f k d
( ) ( )1100 xfcxfcI +=
expressions to find the four unknowns, c0, c1 , x0 and x1.
( ) ( ) ( ) 211
1100 ==+ ∫ dxxfcxfc
We obtain these conditions by assuming the equation above produces the integral value exactly for a constant function and a linear ( ) ( ) ( )
( ) ( ) ( ) 01
11100
11100
==+ ∫
∫
−
−
dxxxfcxfc
ffexactly for a constant function and a linear function. To arrive at two additional conditions the reasoning on the previous overhead is extended and we assume that the
i l fi h i l f b l( ) ( ) ( )
( )32
1
1
1
21100 ==+ ∫
−
dxxxfcxfcequation exactly fits the integral of a parabola (y = x2) and the integral y = x3 exactly. By doing this we determine all four unknowns and in the bargain obtain a two point
( ) ( ) ( ) 01
31100 ==+ ∫
−
dxxxfcxfcg p
integration formula that is exact through cubic polynomials. The four equations are (to the right)
Section 8: ISOPARAMETRIC FORMULATION
These four equations can be solve simultaneously (homework) for
110 == cc
1
5773503.03
10
10
−=−=x
Thus
5773503.03
11 ==x
11
which yields the interesting result that the simple addition of the function values evaluated at the points above (x x ) yields an integral estimate that is third order accurate
+
−=3
13
1 ffI
the points above (x0, x1) yields an integral estimate that is third order accurate.
Also notice that the integration limits on the previous page were from -1 to +1. This is convenient for isoparametric elements.
Section 8: ISOPARAMETRIC FORMULATION
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 below:
( ) ( ) ( )111100 −−+++= nn xfcxfcxfcI L
Section 8: ISOPARAMETRIC FORMULATION
In general for Gauss quadrature we can write
( )∑=n
xfcI
where n is the number of quadrature points for a given variable. If we want to extend this two integration over an area, say for an isoparametric element, then
( )∑=
=i
ii xfcI1
( )
∫
∫ ∫− −
=
n
dtdstsfI
1
1
1
1
1
,
( )
( )∑∑
∫ ∑− =
=
=
nn
iii
tsfcc
dttsfc1 1
,
( )
( )∑∑
∑∑==
=
=
n
ijiij
n
j
ijii
jj
tsfcc
tsfcc
11
11
,
,
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.
== ij 11
Section 8: ISOPARAMETRIC FORMULATION
Consider a four point Gauss integration which is shown in following figure
where for an arbitrary function of s and t1 1
( )
( )22
1 1
,
tsfcc
dtdstsfI
nn
=
=
∑∑
∫ ∫==
− −
( )( ) ( ) ( ) ( )2222121221211111
11
,,,,
,
tsfcctsfcctsfcctsfcc
tsfcci
jiijj
+++=
= ∑∑==
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.
Section 8: ISOPARAMETRIC FORMULATION
For a volume element we can easily extend the concepts as follows:
∫ ∫ ∫1 1 1
( )
( )∑∑∑
∫ ∫ ∫− − −
=
=
n
kjikji
nn
ztsfccc
dzdtdstsfI1 1 1
,,
,
( )∑∑∑=== k
kjikjiji
ztsfccc111
,,
Section 8: ISOPARAMETRIC FORMULATION
Element Stiffness MatrixIn genera for a two dimensional element we have shown thatg
For a quadrilateral isoparametric element
[ ] [ ] [ ] [ ]∫∫=A
TT dydxtBDBk
If we use Gauss quadrature to evaluate the integral
[ ] [ ] [ ] [ ]∫ ∫− −
=1
1
1
1
dtdstJBDBk TT
If we use Gauss quadrature to evaluate the integral
[ ] [ ] [ ] [ ]1
1
1
1
dtdstJBDBk TT= ∫ ∫− −
[ ] ( ) [ ] [ ] ( ) ( )[ ] ( ) [ ] [ ] ( ) ( ) 12121212
11111111
1 1
,,,
,,,
ccttsJtsBDtsB
ccttsJtsBDtsBTTT
TTT
+
=
[ ] ( ) [ ] [ ] ( ) ( )[ ] ( ) [ ] [ ] ( ) ( ) 21212121
22222222
,,,
,,,
ccttsJtsBDtsB
ccttsJtsBDtsBTTT
TTT
+
+
Section 8: ISOPARAMETRIC FORMULATION
I Cl E lIn Class Example
Section 8: ISOPARAMETRIC FORMULATION
Element Aspect RatioThe aspect ratio of an element is defined as the ratio of the longest dimension to the shortest p gdimension in a quadrilateral element. As the aspect ratio increases results can (not will) become less accurate. To illustrate the issue consider a cantilever beam subjected to a parabolic shear stress distribution at the free end of the beam:
Section 8: ISOPARAMETRIC FORMULATION
Four node quadrilateral elements are used to analyse the beam. Four aspect ratios are examined:
Section 8: ISOPARAMETRIC FORMULATION
If we look at the displacement of point A (which is not on the nuetral axis of the beam, the axis for which displacement equations are typically published in text books) we see that the
h i i h f h di d di l d d l ias the aspect ratio increases, the accuracy of the predicted displacement degrades relative to the exact solution:
Section 8: ISOPARAMETRIC FORMULATION
These results are presented in tabular form below:
Note that the previous results were obtained using four node, plane stress quadrilateral elements.
Section 8: ISOPARAMETRIC FORMULATION
The effects of aspect ratios on the performance of an element is not the same from element to element. The instructor uses higher order brick (volume) elements without any
l f fid li h i d i f i lapparent loss of fidelity when aspect ratios produce warning messages from commercial finite element software.
Still, it is good practice to maintain aspect ratios close to unity. The current generation of fi it l t ft h hi biliti th t ill t ti ll fi t ifinite element software has remeshing capabilities that will automatically fix geometric problems such as the examples shown below: