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Finite Element Method
Chapter 10
Isoparametric Formulation
Basic Principle of Isoparametric Elements
Definition:
The term isoparametric (same parameters) is derived from the use of
the same shape (interpolation) functions N to define the element’s
geometric shape as are used to define the displacements within the
element.
The basic principle of isoparametric elements is that the interpolation
functions for the displacements are also used to represent the
geometry of the element.
Alternatively:
4 4
1 1
4 4
1 1
,
,
i i i i
i i
i i i i
i i
u N u v N v
x N x y N y
Basic Principle of Isoparametric Elements
In this formulation, displacements are expressed in terms of the
natural (local) coordinates and then differentiated with respect to
global coordinates. Accordingly, a transformation matrix [J], called
Jacobian, is produced.
If the geometric interpolation functions are of lower order than the
displacement shape functions, the element is called
subparametric. If the reverse is true, the element is referred to as
superparametric.
The isoparametric formulation is generally applicable to 1-, 2-
and 3- dimensional stress analysis. The isoparametric family
includes elements for plane, solid, plate, and shell problems. Also,
it is applicable for nonstructural problems.
Basic Principle of Isoparametric Elements
The isoparametric formulation makes it possible to generate
elements that are nonrectangular and have curved sides. So it
can facilitate an accurate representation of irregular elements.
Numerous commercial computer programs have adopted this
formulation for their various libraries of elements.
Two-Noded Bar Isoparametric Element
Basic Principle of Isoparametric Elements in 2D
As shown in the figure, the local (natural) coordinate system (,) for
the two elements have their origins at the centroids of the elements,
with (,) varying form –1 to 1. The natural coordinate system needs
not to be orthogonal and neither has to be parallel to the x-y axes.
The coordinate transformation will map the point (,) in the master
element to x(,) and y(,) in the slave element.
s
t
1 11
1
s
t
y
x
Examples
s
t
1
1
y
x
s
t
Isoparametric Formulation of Two-
Noded Bar Element Stiffness Matrix
Isoparametric Formulation of the Bar
Element Stiffness Matrix
Step 1: Select the Element Type
We consider the bar element to have two degrees of freedom s.
For the special case when the s and x axes are parallel to each other,
the s and x coordinates can be related by
where xc is the global coordinate of the element centroid.
Using the global coordinates x1 and x2 in
1 2
2 1
2
2
x xs x
x x
2c
Lx x s
Step 2: Select Displacement Functions
We begin by relating the natural coordinate to the global coordinate by
This linear shape functions map the s coordinate of any point in the element
to the x coordinate
1 2
1
1 2 1 2
2
1 1
11 1
2
1 1,
2 2
s
x s x s x
x s sx N N N N
x
Step 2: Select Displacement Functions
The displacement function within the bar is now defined by the same
shape functions
Step 3: Define the Strain/Displacement and Stress/Strain Relationships
1
1 2 1 2
2
1 1,
2 2
u s su N N N N
u
Using Chain R le
2
1 2
u
x
x
u
x
u u x
s x s
u x Lu xJ
s sx s
u u u
x J s L s
Step 3: Define the Strain/Displacement and Stress/Strain Relationships
2 1
1
2
2
2 1 1
since
1 1
x
x
x
u uu
s
uu u
ux L s L L
B d
BL L
E E B d
Step 4 Derive the Element Stiffness Matrix
0
1
10
1 1
1 1
[ ] [ ][ ]
J:Jacobian Matrix
2
[ ] [ ] [ ][ ] [ ] [ ][ ]
1 1[ ]
2
1 1
L
T
L
T T
k B D B Adx
f x dx f s J ds
x LJ
s
Lk B D B A J d
EAk
s B D B A ds
L
Body Forces
1
0 1
1
1
[ ] { }
[ ] { } [ ] { }
1
12{ }
1 12 2
2
T
b b
V
L
T T
b b b
bb b
f N X dV
f A N X dx A N X J ds
s
ALXLf A X ds
s
Surface Forces
[ ] { }T
s x
S
f N T dS
Assuming the cross section is constant and the traction is uniform
over the perimeter and along the length of the element, we obtain
1
0 1
1
1
[ ] { } [ ] { } [ ] { }
1
12{ }
1 12 2
2
L
T T T
s s x x x
S
b x x
f f N T dS N T dx N T J ds
s
L Lf T ds T
s
Isoparametric Formulation of the
Three-Noded linear strain bar
Three-Noded linear strain isoparametric bar
Three-noded linear strain isoparametric bar
1
1 2 3 2
3
2
1 2 3
1 1, , 1
2 2
2
1 2
x
x
x N N N x
x
s s s sN N N s
u
x
u u x x LJ
s x s s
u u u
x J s L s
Three-noded linear strain bar isoparametric element
13
, 2
1
3
1
2
3
2 1 2 12
2 2
2 2 1 2 1 4
2 1 2 1 4
i s i
i
uu s s
N u s us
u
uu u s s s
ux L s L L L
u
s s sB
L L L
1
1[ ] [ ] [ ][ ]
2
Tk B D B A J ds
LJ
The Stiffness Matrix of three-noded linear strain bar
The Stiffness Matrix of three-noded linear strain bar
The Stiffness Matrix of Three-Noded linear strain bar
Numerical Integration
Gauss Quadrature
Numerical Integration
Gauss Quadrature
• Gauss quadrature implements a strategy of
positioning any two points on a curve to define a
straight line that would balance the positive and
negative errors.
• Hence, the area evaluated under this straight line
provides an improved estimate of the integral.
Two points Gauss-Legendre
Formula• Assume that the two Integration points are xo and x1 such that:
• c0 and c1 are constants, the function arguments x0 and x1 are
unknowns…….(4 unknowns)
1
0 0 1 1
1
( ) ( )I f x dx c f x c f x
Two points Gauss-Legendre
Formula
• Thus, four unknowns
to be evaluated require
four conditions.
• If this integration is
exact for a constant,
1st order, 2nd order, and
3rd order functions:
1
1
3
1100
1
1
2
1100
1
1
1100
1
1
1100
0)()(
3
2)()(
0)()(
21)()(
dxxxfcxfc
dxxxfcxfc
dxxxfcxfc
dxxfcxfc
Two points Gauss-Legendre Formula
• Solving these 4 equations, we can determine c0, c1, x0 and x1.
3
1
3
1ffI
0 1
0
1
The weighting factors are: 1
The Integration points are:
10.5773503
3
10.5773503
3
c c
x
x
Two points Gauss-Legendre Formula
• Since we used limits for the previous integration from –1 to 1 and the actual limits are usually from a to b, then we need first to transform both the function and the integration from the x-system to the xd-system
1ax
1bx
d2
abdx
2
ab
2
abx
f(x)
x
a b
f(xo)
f(x1)
xo x1
-1 1
f()
Higher-Points Gauss-Legendre Formula
1
11
( ) ( )n
i i
i
I f d W f
Weight Integration point
1
11
1 1 2 2
( ) ( )
( ) ( ) ( )
n
i i
i
n n
I f d W f
I W f W f W f
Multiple Points Gauss-LegendrePoints Weighting factor Function argument Exact for
2 1.0 -0.577350269=−1/ 3 up to 3rd
1.0 0.577350269= 1/ 3 degree
3 0.5555556=5/9 -0.774596669=−1/ 0.6 up to 5th
0.8888889=8/9 0.0 degree0.5555556=5/9 0.774596669= 1/ 0.6
4 0.3478548 -0.861136312 up to 7th
0.6521452 -0.339981044 degree0.6521452 0.3399810440.3478548 0.861136312
6 0.1713245 -0.932469514 up to 11th
0.3607616 -0.661209386 degree0.4679139 -0.2386191860.4679139 0.2386191860.3607616 0.6612093860.1713245 0.932469514
Example 1
Evaluate the stiffness matrix for the three-noded bar element using
two-point Gaussian quadrature.
Three–noded bar with two Gauss points
We have for the stiffness matrix
Using two-point Gaussian quadrature, we evaluate the stiffness matrix at the two
points shown :
Example 1
Example 1
Multiple Integration
dydxyxfI
d
cy
b
ax
),(
• Double integral:
AA
ddJfdydxyxf ),(),(
In General transformation to natural
coordinate:
1 1
1 1( , )I f d d
1 1
1 11 1
( , ) ( , )n n
ij i j
i j
I f d d W f
Where Wij =Wi Wj
2 21 1
1 11 1
( , ) ( , )
1 1 1 1 1 1 1 1( , ) ( , ) ( , ) ( , )
3 3 3 3 3 3 3 3
ij i j
i j
I f d d W f
f f f f
For n=2 Wij =Wi Wj=1
1
1
1 1
3
1,
3
1
3
1,
3
1
3
1,
3
1
3
1,
3
1
Shape function for Isoparametric 6-Nodes triangular Elements
1
2
3
4
5
6
2 2 1 1
2 1
2 1
4 (1 )
4
4 (1 )
N
N
N
N
N
N
Example 1. A n=1 point rule is exact for a polynomial
ts
tsf 1~),(
3
1,
3
1
2
1fI
s
t
1
11/3
1/3
Gauss Points for Isoparametric triangular Elements
Example 2. A M=3 point rule is exact for a complete polynomial of degree 2
22
1~),(
tsts
ts
tsf
2
1,0
6
10,
2
1
6
1
2
1,
2
1
6
1fffI
s
t
1
1
1/2
1/2
21
3
Gauss Points for Isoparametric triangular Elements
Isoparametric Formulation of the Plane
Quadrilateral Element Stiffness Matrix
Step 2: Select Displacement Functions
In other words, we look for shape functions that map the regular
shape element in isoparametric coordinates to the quadrilateral
in the x-y coordinates whose size and shape are determined by
the eight nodal coordinates x1, y1, x2, y2, ….., x4, y4.
8765
4321
),(
),(
aaaay
aaaax
Step 2: Select Displacement Functions
1 2 3 4
5 6 7 8
1
2
8
( , )
( , )
( , ) 1 0 0 0 0
( , ) 0 0 0 0 1
a a a ax
y a a a a
a
ax
y
a
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 11
1 1 1 1 1 1 1 14
1 1 1 1 1 1 1 1
x a a x
x a a x
x a a x
x a a x
Step 2: Select Displacement Functions
])1)(1()1)(1()1)(1()1)(1[(4
1
1111
1111
1111
1111
4
111),(
4321
4
3
2
1
4
3
2
1
xxxx
x
x
x
x
a
a
a
a
x
i
i
i
i
i
i
yN
xN
y
x
y
x
y
x
y
x
NNNN
NNNN
y
x4
1
4
1
4
4
3
3
2
2
1
1
4321
4321
0000
0000
),(
),(
Step 2: Select Displacement Functions
)1)(1(4
1,)1)(1(
4
1
)1)(1(4
1,)1)(1(
4
1
43
21
NN
NN
These shape functions are seen to map the (,) coordinates of
any point in the rectangular element in the above master element
to x and y coordinates in the quadrilateral (slave) element.
For example, consider the coordinates of node 1, where:
=-1,=-1 using the above equation, we get x=x1 , y=y1
Shape Function for 4-Nodes quadrilateral Elements
Shape Function for 4-Nodes quadrilateral Elements
Step 2: Select Displacement Functions
),,,2,1(,11
niNn
i
i
where n = the number of shape functions associated with
number of nodes
Shape Function for 4-Nodes quadrilateral Elements
Step 2: Select Displacement Functions
nodesotherallat
inodeatN i
0
1
Step 2: Select Displacement Functions
i
i
i
i
i
i
vN
uN
v
u
v
u
v
u
v
u
NNNN
NNNN
v
u4
1
4
1
4
4
3
3
2
2
1
1
4321
4321
0000
0000
),(
),(
][][ dNv
u
where u and v are displacements parallel to the global x and y
coordinates
Step 3: Define the Strain/Displacement and Stress/Strain
Relationships
y
y
fx
x
ff
y
y
fx
x
ff
ff x y
x
f x y f
y
Using Chain Rule
Step 3: Define the Strain/Displacement and Stress/Strain Relationships
1
N fN f
x xJ J
N N f f
y y
NN x y
x
N x y N
y
Can be
computed
We want to compute
these for the B matrix
This is known as the
Jacobian matrix (J) for the
mapping (,) → (x,y)
Define the Strain/Displacement and Stress/Strain Relationships
i
i
ii
i
i
i
i
ii
i
i
yN
xN
yN
xN
yx
yx
J4
1
4
1
4
1
4
1][
1 1[ ]
where
y y
Jx xJ
x y x yJ
1
1
1
1
N N
xJ
N N
y
N y y N
x
N x x NJ
y
N y N y N
x J
N x N x N
y J
Since:
Define the Strain/Displacement and Stress/Strain Relationships
{ }
( ) ( )0
1 ( ) ( )0
( ) ( ) ( ) ( )
{ } [ ] [
x
y
xy
x
y
xy
u
x
v
y
u v
y x
y y
ux x
vJ
x x y y
uD
v
][ ][ ]D N d
Define the Strain/Displacement and Stress/Strain Relationships
(3 8) (3 2) (2 8)
{ } [ ][ ][ ]
{ } [ ][ ]
( ) ( )0
1 ( ) ( )[ ] 0
( ) ( ) ( ) ( )
[ ] [ ] [ ]
D N d
B d
y y
x xD
J
x x y y
B D N
Define the Strain/Displacement and Stress/Strain Relationships
Derive the Element Stiffness Matrix and Equations
1 1
1 1
[ ] [ ] [ ][ ]
( , ) ( , )
[ ] [ ] [ ][ ]
T
A
A A
T
k B D B t dx dy
f x y dx dy f J d d
k B D B t J d d
The shape function are:
)1(4
1,)1(
4
1,)1(
4
1,)1(
4
1
)1(4
1,)1(
4
1,)1(
4
1,)1(
4
1
4321
4321
NNNN
and
NNNN
)1)(1(4
1,)1)(1(
4
1
)1)(1(4
1,)1)(1(
4
1
43
21
NN
NN
Derive the Element Stiffness Matrix and Equations
Their derivatives:
i
i
ii
i
i
i
i
ii
i
i
yN
xN
yN
xN
yx
yx
J4
1
4
1
4
1
4
1][
Derive the Element Stiffness Matrix and Equations
i
i
ii
i
i
i
i
ii
i
i
yNy
JxNx
J
yNy
JxNx
J
4
1
22
4
1
21
4
1
12
4
1
11
,
,
44
33
22
11
,4,3,2,1
,4,3,2,1
2221
1211][
yx
yx
yx
yx
NNNN
NNNN
JJ
JJJ
Derive the Element Stiffness Matrix and Equations
22 1 2 3 4
12 1 2 3 4
11 1 2 3 4
21 1 2 3 4
11 1 1 1
4
11 1 1 1
4
11 1 1 1
4
11 1 1 1
4
J y y y y
J y y y y
J x x x x
J x x x x
Derive the Element Stiffness Matrix and Equations
11 12
21 22
11 22 12 21
[ ]J J
JJ J
J J J J J
1
11 12 2
1 2 3 4
21 22 3
4
0 1 1
1 0 11
1 0 18
1 1 0
y
J J yJ x x x x
J J y
y
Explicit formulation for |J| for 4 node Element
Derive the Element Stiffness Matrix and Equations
][][][ NDB
22 12
1 2 3 4
11 21
1 2 3 4
11 21 22 12
( ) ( )0
0 0 0 01 ( ) ( )[ ] 0
0 0 0 0
( ) ( ) ( ) ( )
J J
N N N NB J J
N N N NJ
J J J J
,12,22,21,11
,21,11
,12,22
0
0
][
iiii
ii
ii
i
NJNJNJNJ
NJNJ
NJNJ
B
Derive the Element Stiffness Matrix and Equations
1 2 3 4
1B B B B B
J
Derive the Element Stiffness Matrix and Equations
22 , 12 ,
11 , 21 ,
11 , 21 , 22 , 12 ,
0
[ ] 0
i i
i i i
i i i i
J N J N
B J N J N
J N J N J N J N
Derive the Element Stiffness Matrix and Equations
Derive the Element Stiffness Matrix and Equations
22 1 2 3 4
12 1 2 3 4
11 1 2 3 4
21 1 2 3 4
11 1 1 1
4
11 1 1 1
4
11 1 1 1
4
11 1 1 1
4
J y y y y
J y y y y
J x x x x
J x x x x
In the our text Book
Derive the Element Stiffness Matrix and Equations
Derive the Element Stiffness Matrix and Equations
A
T
V
Tb dydxtXNdVXNf }{][}{][}{
The element body force matrix
1 1
1 1{ } [ ] { }
(8 1) (8 2) (2 1)
T
bf N X t J d d
b
b
Y
XX}{
Basic Principle of Isoparametric Elements
and Xb and Yb are the weight densities (body weight/unit volume)
in the x and y directions, respectively.
L
T
S
Ts dxtTNdSTNf }{][}{][}{
1
1{ } [ ] { }
(4 1) (4 2) (2 1)
T
sf N T t J d
1
143
43
4
4
3
3
200
00
dL
tp
p
NN
NN
f
f
f
fT
s
s
s
s
Note that for one-dimensional transformation | J | = L / 2. Also,
p, p are the pressure distributions in and , respectively.
The element surface force matrix
Basic Principle of Isoparametric Elements
We assumed surface loading at edge with overall length
L (Since N1 = 0 and N2 = 0 along edge =1, and hence,
no nodal forces exist at nodes 1 and 2,
1 13
4
12
1
1
Problem: Consider the following isoparamteric map
y
34
1 2(3,1) (6,1)
(6,6)(3,6)
xISOPARAMETRIC COORDINATES
GLOBAL COORDINATES
1 1 2 2 3 3 4 4
1 1 2 2 3 3 4 4
u N u N u N u N u
v N v N v N v N v
Displacement interpolation
Shape functions in isoparametric coordinate system
)1)(1(4
1,)1)(1(
4
1
)1)(1(4
1,)1)(1(
4
1
43
21
NN
NN
1 1 2 2 3 3 4 4
1 1 2 2 3 3 4 4
( , ) ( , ) ( , ) ( , )
( , ) ( , ) ( , ) ( , )
1 1(1 )(1 ) 3 (1 )(1 ) 6
4 4
1 1 3(3 )(1 )(1 ) 6 (1 )(1 ) 3
4 4 2
1 1(1 )(1 ) 1 (1 )(1 ) 1
4 4
1 1(1 )(1 ) 6
4 4
x N x N x N x N x
y N y N y N y N y
x
y
7 5
(1 )(1 ) 62
The isoparamtric mapping
The Jacobian matrix
NOTE: The diagonal terms are due to stretching of the sides along the
x-and y-directions. The off-diagonal terms are zero because the
element does not shear.
30
2
50
2
x y
Jx y
3(3 )
2
7 5
2
x
y
since
1 3 / 2 0 2 / 3 01 15
and0 5 / 2 0 2 / 515 / 4 4
J J
Hence, if I were to compute the columns of the B1 matrix along the
positive x-direction
Hence
22 1, 12 1,
1 11 1, 21 1,
11 1, 21 1, 22 1, 12 1,
0
[ ] 0
J N J N
B J N J N
J N J N J N J N
1
5 15 1 00 082 4
3 13 1[ ] 0 0 0
2 4 8
3 1 5 1 3 1 5 10 0
2 4 2 4 8 8
B
Example1
For the four-noded linear plane element shown with a uniform surface
traction along side 2–3, evaluate the force matrix by using the energy
equivalent nodal forces. Let the thickness of the element be h=0.1 in.
Example1
Shape functions N2 and N3 must be used,
as we are evaluating the surface traction
along side 2–3 (at s = 1).
Example1
evaluated along s = 1
Example1
Flowchart to evaluate k by four-point Gaussian quadrature
Evaluate the stiffness matrix for the
quadrilateral element shown in Figure
using the four-point Gaussian quadrature
rule.
Let E = 30×106 psi, = 0.25 and h=1 in.
Example 2
we evaluate the k matrix. Using the four-point rule, the four points are:
Solution
1 1
2 2
3 3
4 4
, 0.5773, 0.5773
, 0.5773,0.5773
, 0.5773, 0.5773
, 0.5773,0.5773
1 2 3 4 1.0W W W W
Example 2
Example 2
22 1, 12 1,
1 11 1, 21 1,
11 1, 21 1, 22 1, 12 1,
0
[ ] 0
J N J N
B J N J N
J N J N J N J N
22 1 2 3 4
22
12 1
1
1 21
,
1,
11 1 1 1
4
12 0.5773 1 2 1 0.5773 4 1 0.5773 4 1 0.5773 1.0
4
with similar computations used to obtain , . Also,
1 1(1 ) (1 0.577
and
3) 0.39434 4
1(1
4
J J J
J y y y y
J
N
N
1
) (1 0.5773) 0.39434
Example 2
Example 2
Example 3 For the rectangular element shown previous
Example, assume plane stress conditions
Let E = 30×106 psi, = 0.3 and displacements:
u1 = 0 , v1 = 0
u2 = 0.001 , v2 = 0.0015
u3 = 0.003 , v3 = 0.0016
u4 = 0 , v4= 0
Evaluate the stresses at s=0 , t=0
Solution
Example 3
22 , 12 ,
11 , 21 ,
11 , 21 , 22 , 12 ,
22 1 2 3 4
22
12 11 21
0
[ ] 0
11 1 1 1
4
12 0 1 2 1 0
Simili
4 1 0 4 1 0 14
0,arly 1, 0
i i
i i i
i i i i
J N J N
B J N J N
J N J N J N J N
J y y y y
J
J J J
1, 2, 3, 4,
1, 2, 3, 4,
1 1 1 1 1 1(1 ) (1 0) ,Similarly
Similarly
and4 4 4 4 4 4
1 1 1 1 1 1(1 ) (1 0) , and
4 4 4 4 4 4
N N N N
N N N N
Example 3
Higher-Order Shape Functions
In general, higher-order element shape functions can
be developed by adding additional nodes to the sides
of the linear element.
These elements result in higher-order strain variations
within each element, and convergence to the exact
solution thus occurs at a faster rate using fewer
elements.
Another advantage of the use of higher-order
elements is that curved boundaries of irregularly
shaped bodies can be approximated more closely than
by the use of simple straight-sided linear elements.
or, in compact index notation, we express
where i is the number of the shape function
Shape function of a quadratic isoparametric
element
Shape function of a cubic isoparametric element
HW: 10.6, 10.8, 10.15 and 10.21