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Institute of Structural Engineering Page 1
Method of Finite Elements I
Chapter 4
Isoparametric Elements
[O] V1/Ch3/77-95[F] Ch16
Institute of Structural Engineering Page 2
Method of Finite Elements I30-Apr-10
Today’s Lecture Contents
• The Shape Function: Reminder from the previous lecture
• Looking for a uniform mapping across element types
• Isoparametric elements
• 1D Demonstration: Bar elements
Institute of Structural Engineering Page 3
Method of Finite Elements I
In the previous lecture, we defined an approximation for the displacement field:
This also defined the first derivative of the approximation (strain)
Strain field
Displacement field
The Galerkin Method
Note: this is just a vector of thecoefficients of the linear approximation function.
u1, u2 do not correspond to displacements at the nodes
Institute of Structural Engineering Page 4
Method of Finite Elements I
{ } [ ]{ }( )u N x d=
where u1,u2 are so far random coefficients. Instead, we can choose to write the same relationship using a different basis N(x):
Vector of degrees of freedom (DOFs) at the element nodes
Shape Function Matrix
Displacement field
In this case we express u in terms of the degrees of freedom, i.e., the displacements at the ends of the bar:
𝑑𝑑 = 𝑢𝑢(𝑥𝑥 = 0)𝑢𝑢(𝑥𝑥 = 𝐿𝐿) =
𝑢𝑢0𝑢𝑢𝐿𝐿 = 1 0
1 𝐿𝐿𝑢𝑢1𝑢𝑢2 ⇔
𝑢𝑢1𝑢𝑢2 = 1 0
1 𝐿𝐿−1 𝑢𝑢0
𝑢𝑢𝐿𝐿
The Galerkin Method
u0, uL correspond to the displacements at the nodes (nodal DOFs), which are contained in vector {d}. So here, we expressed the coefficients in terms of the nodal DOFs.
Institute of Structural Engineering Page 5
Method of Finite Elements I
{ } [ ]{ }( )u N x d=
where u1,u2 are so far random coefficients. Instead, we can choose to write the same relationship using a different basis N(x):
Vector of degrees of freedom at the
element nodes
Shape Function Matrix
Displacement field
In this case we express u in terms of the degrees of freedom, i.e., the displacements at the ends of the bar:
Substituting in our initial displacement approximation we obtain:
𝑑𝑑 = 𝑢𝑢(𝑥𝑥 = 0)𝑢𝑢(𝑥𝑥 = 𝐿𝐿) =
𝑢𝑢0𝑢𝑢𝐿𝐿 = 1 0
1 𝐿𝐿𝑢𝑢1𝑢𝑢2 ⇔
𝑢𝑢1𝑢𝑢2 = 1 0
1 𝐿𝐿−1 𝑢𝑢0
𝑢𝑢𝐿𝐿
𝑢𝑢 = 1 𝑥𝑥 1 01 𝐿𝐿
−1 𝑢𝑢0𝑢𝑢𝐿𝐿 = 𝐿𝐿 − 𝑥𝑥
𝐿𝐿𝑥𝑥𝐿𝐿
𝑢𝑢0𝑢𝑢𝐿𝐿 ⇒ 𝑁𝑁(𝑥𝑥) = 𝐿𝐿 − 𝑥𝑥
𝐿𝐿𝑥𝑥𝐿𝐿
The Galerkin Method
shape functions
Institute of Structural Engineering Page 6
Method of Finite Elements I30-Apr-10
The Shape Functions for the generic bar (truss) Element
Truss Element Shape Functions
(in global coordinates)
( ) [ ] 11 2
2
uu x N N
u
=
Institute of Structural Engineering Page 7
Method of Finite Elements I
[ ]{ } [ ]{ }( )
d N xd B d
dxε = =
Then:
The weak form also involves the first derivative of the approximation
Strain field
[ ] [ ] [ ]( ) 1 1 1d N x
Bdx L
= = −where
Strain Displacement Matrix
Displacement field
[ ]{ }( )u N x d=Vector of degrees of freedom at the
element nodes
The Galerkin Method
Institute of Structural Engineering Page 8
Method of Finite Elements I30-Apr-10
Strain-Displacement relation
defineThe truss element
stress displacement matrix
Constant Variation of strains along the element’s length
Institute of Structural Engineering Page 9
Method of Finite Elements I
EA depends on material and cross-sectional properties
We call the nodal force vector for distributed loads
The Galerkin Method
{ } [ ] ( )∫L
T
0
f = N -αx dx
This is defined as: { } ( )( )
0f xf x L
= =
f = for concentrated loads* at the nodes
The following FUNDAMENTAL FEM expression is derived
*Notice the placement of the concentrated load on the corresponding degree of freedom where it applies, i.e., x = 0 → Node 1, x = L → Node 2
Institute of Structural Engineering Page 10
Method of Finite Elements I
where
The Galerkin Method
[ ]{ } { }fK d =
The FEM equation to solve is (as in the DSM):
for a bar element with only two nodes(2-noded bar element)[ ] 1 1
1 1EAKL
− = −
We are however not restricted to only using 2-noded elements!
EduApp1
EduApp2
Institute of Structural Engineering Page 11
Method of Finite Elements I
Generalized ElementsWe are not restricted to only using 2-noded elements!
In order to generalize to
• elements of higher order (more nodes) and
• generalized geometries
we need to generalize the shape functions
Higher order
Why?
[O] Ch3§2, Fig. 3.2
Institute of Structural Engineering Page 12
Method of Finite Elements I30-Apr-10
Therefore: Shape functions will be defined as interpolation functions which relate the variables in the finite element with their values in the element nodes. The latter are obtained through solving the problem using finite element procedures.
In general we may write:
where: is the function under investigation (for example: displacement field)is the shape functions matrix
is the vector of unknowns in the nodes (for example: displacements)
Revisiting the Shape Function
( )( ) i ii
u x N x u u= =∑ N
( )u x
( )xN
[ ]1 2T
Nu u u u=
Institute of Structural Engineering Page 13
Method of Finite Elements I30-Apr-10
Regardless of the dimension of the element used, we have to bear in mind that Shape Functions need to satisfy the following constraints:
• in node i has a value of 1 and in all other nodesassumes a value of 0.
• Furthermore we have to satisfy the continuity between the adjoining elements.
Example: rectangular element with 6 nodes
The Shape Function
( )i xN
H4 H5
Why??
Institute of Structural Engineering Page 14
Method of Finite Elements I30-Apr-10
Usually, polynomial functions are used as interpolation functions, for example:
( )0
Ni
n ii
P x a x=
=∑
where n is the order of the polynomial; is equal to the number of unknowns in the nodes (degrees of freedom).
In the MFE we use three different polynomials: Lagrange Serendipity and Hermitian polynomials.
The Shape Function
Institute of Structural Engineering Page 15
Method of Finite Elements I30-Apr-10
Lagrange Polynomials
The Shape Function
A function φ(x) can be approximated by a polynomial of order n and the values of φi in those n+1 points:
( ) ( ) ( )1
1
ni
ii
x x L xϕ ϕ ϕ+
=
≈ =∑where the normalized Lagrange polynomial of order n-1 isdefined by n points, as follows:
( ) ( )( ) ( )( )( ) ( )
1 2
1 1 2
nj nn
ij i j i i i ni j
x x x x x x x xL x
x x x x x x x x=≠
− − − −= =
− − − −∏
Note that ( ) ( )0 & 0 for n ni i i jL x L x i j≠ = ≠
niL
Institute of Structural Engineering Page 16
Method of Finite Elements I30-Apr-10
Note that
( ) ( )0 & 0 for n ni i i jL x L x i j≠ = ≠
The Shape FunctionLagrange Polynomials
Institute of Structural Engineering Page 17
Method of Finite Elements I30-Apr-10
Lagrange Polynomials
The Shape Function
The shape function Ni of a Lagrange element with n nodes coincideswith the normalized Lagrange polynomial of degree n-1, i.e.
( ) ( )ni iN x L x=
This explains why C0 continuous 1D elements are also called Lagrange elements.
Institute of Structural Engineering Page 18
Method of Finite Elements I30-Apr-10
Lagrange Polynomials – 2D Example
( ) ( )( ) ( )
1 2 1 21 2 1 1 2 2
4 3 4 33 4 1 3 2 4
L x L x
L x L x
ϕ ϕ ϕ
ϕ ϕ ϕ
− −−
− −−
= +
= +
( )1 11 ,x y
( )2 22 ,x y
( )4 44 ,x y ( )3 33 ,x yLet us break the process into steps:
Interpolate along borders 1-2, 4-3:
11L 1
2L
x
y
( ) ( )
( ) ( )
1 2 1 22 11 2
2 1 2 1
4 3 4 3 341 2
4 3 4 3
,
,
x x x xL x L xx x x x
x xx xL x L xx x x x
− −
− −
− −= =
− −−−
= =− −
Therefore:
Step 1: Interpolate along x
Institute of Structural Engineering Page 19
Method of Finite Elements I30-Apr-10
Lagrange Polynomials – 2D Example
( )1 11 ,x y
( )2 22 ,x y
( )4 44 ,x y ( )3 33 ,x y
11L 1
2L
( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
1 1 2 2 3 4
1 4 1 2 2 3 1 2 2 3 4 3 1 4 4 31 1 1 1 2 2 2 1 3 2 2 4
L y L y
L y L x L y L x L y L x L y L x
ϕ ϕ ϕ
ϕ ϕ ϕ ϕ− −
− − − − − − − −
= + =
= + + +
2. Interpolate along borders 1-2, 4-3: (one possible way is below)
x
y
Step 2: Interpolate along y
Step 1: Interpolate along x
( ) ( )( ) ( )
1 2 1 21 2 1 1 2 2
4 3 4 33 4 1 3 2 4
L x L x
L x L x
ϕ ϕ ϕ
ϕ ϕ ϕ
− −−
− −−
= +
= +
Interpolate along borders 1-2, 4-3:
Institute of Structural Engineering Page 20
Method of Finite Elements I30-Apr-10
Lagrange Polynomials – 2D Example
1 11: u ,v
2 22 : u ,v
4 44 : u ,v3 33 : u ,v
To interpolate the displacement field for this 2D element, both• the horizontal u(x,y) and • vertical displacements v(x,y)should be interpolated
x
y
we can use the following approximation:
( ) ( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
2 4 1 3 4 2 3 11 2 3 4
2 1 4 1 2 1 3 2 4 3 3 2 4 3 4 1
,x x y y x x y y x x y y x x y y
u x y u u u ux x y y x x y y x x y y x x y y
− − − − − − − −= + + +
− − − − − − − −
( ) ( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
2 4 1 3 4 2 3 11 2 3 4
2 1 4 1 2 1 3 2 4 3 3 2 4 3 4 1
,x x y y x x y y x x y y x x y y
x y v v v vx x y y x x y y x x y y x x y y
v− − − − − − − −
= + + +− − − − − − − −
ui, vi indicate the horizontal/vertical displacements at each one of the 4 nodes (i) of this element
Institute of Structural Engineering Page 21
Method of Finite Elements I30-Apr-10
Lagrange Polynomials – 2D Example
1 11: u ,v
2 22 : u ,v
4 44 : u ,v3 33 : u ,v
The polynomials we used in this simple example are of order 1
x
y
If polynomials of higher order are to be used, further nodes should be typically introduced within the elements.
the nine-node biquadraticQuadrilateral (Lagrangian) [F
] Ch1
6§8
Institute of Structural Engineering Page 22
Method of Finite Elements I30-Apr-10
Serendipity PolynomialsThese functions are similar to Lagrangian polynomials, but are incomplete(the have missing terms). Due to this fact we do not need to introduceadditional inner nodes, as for Lagrangian polynomials of higher order.
Serendipity Polynomials
Lagrange Polynomials
Institute of Structural Engineering Page 23
Method of Finite Elements I30-Apr-10
Serendipity PolynomialsThese functions are similar to Lagrangian polynomials, but are incomplete(the have missing terms). Due to this fact we do not need to introduceadditional inner nodes, as for Lagrangian polynomials of higher order.
Serendipity PolynomialsLagrange Polynomials
the nine-node biquadraticQuadrilateral (Lagrangian)
the eight-node “serendipity” quadrilateral.
[F] Ch16§8
Institute of Structural Engineering Page 24
Method of Finite Elements I30-Apr-10
Hermitian PolynomialsLagrangian polynomials and serendipity functions provide a C0
continuity.
If we additionally need continuity of the first derivatives between thefinite elements we use Hermitian polynomials.A Hermitian polynomial of the order n, Hn(x), is a 2n+1 order polynomial.For example a Hermitian polynomial of the first order is actually a thirdorder polynomial.The first few Hermitian polynomials:
( )( )( )( )( )
0
1
22
33
4 24
1
2
4 2
8 12
16 48 12...
x
x x
x x
x x x
x x x
Η =
Η =
Η = −
Η = −
Η = − +
Institute of Structural Engineering Page 25
Method of Finite Elements I30-Apr-10
Hermitian Polynomials
Let us consider a bar element with nodes on its ends. Unknowns are values ofthe function φ at the end nodes 1 and 2, φ1 and φ2, and the first derivatives ofφ with respect to x , φ1,x and φ2,x
Remember: The 1st derivative of displacement, corresponds to rotation
( ) ( )1 2
1, 2,,x xx x x x
d x d xdx dxϕ ϕ
ϕ ϕ= =
= =
11 ( )x
( )xϕ
12 ( )x
2ϕ1ϕ
1,xϕ2,xϕ
Question: What well-known type of element will need to be represented by Hermitian Polynomials?
Institute of Structural Engineering Page 26
Method of Finite Elements I30-Apr-10
Hermitian PolynomialsHermitian shape functions relate not only the displacements at nodes to displacements within the elements, but also to the first order derivatives (e.g. rotational DOFs for a beam element).
( ) ( )2
0 11
( )( ) ii i i
i
u xu x N x u N xx=
∂ = + ∂ ∑
( )( )( )( )
0
0
1
1
1 at node i and 0 at other nodes
0 at all nodes
0 at all nodes
1 at node i and 0 at other nodes
i
i
i
i
N x
N x
N x
N x
=
′ =
=
′ =
Shape function ofthe derivative u´(x)
Shape functionof u(x)
01N 02N
11N12N
Institute of Structural Engineering Page 27
Method of Finite Elements I30-Apr-10
next lecture & lab
So far we discussed the form of Shape Functions using a formulation of the element in global coordinates
This element formulation may be extended to any type of element however, but suffers from the following limitations:
1 The construction of shape functions for high order elements with curved geometry is highly complicated.
2 The evaluation of integrals for element stiffness matrix and force vector is not possible in closed form.
These limitations may be overcome by combining isoparametric representation & numerical quadrature.
Isoparametric Elements
current lecture
Institute of Structural Engineering Page 28
Method of Finite Elements I30-Apr-10
Isoparametric Elements
Natural coordinates
Shapefunctions
Geometrydescription
Displacement interpolation
,x yu u
,x y
( , , )iN r s t, ,r s t
In general, we would like to be able to represent any element in a standardized manner – introducing a transformation between a set of standardized (natural) coordinates and the real (global) coordinates
[F] Ch16§2
Institute of Structural Engineering Page 29
Method of Finite Elements I30-Apr-10
In general, we would like to be able to represent any element in a standardized manner – introducing a transformation between a set of standardized (natural) coordinates and the real (global) coordinatesDifferent schemes exist for establishing such transformations:
1 sub parametric representations (less nodes for geometric than for displacement representation)
2 isoparametric representations (same nodes for both geometry and displacement representation)
3 super parametric representations (more nodes for geometric than for displacement representation)
Isoparametric ElementsH
ere
we
will
use
this
one
Institute of Structural Engineering Page 30
Method of Finite Elements I30-Apr-10
Displacement fields as well as the geometrical representation of the finite elements are approximated using the same approximating functions – shape functions
x
y
r
s
1, -1
1, 1-1, 1
-1, -1
1 1ˆ ˆ,u v
2 2ˆ ˆ,u v
3 3ˆ ˆ,u v
4 4ˆ ˆ,u v
1
4
2
3
This transformation allows us to refer to similar elements (eg. truss, beams, 2D elements) in a standard manner, using the natural coordinates (r,s), without every time referring to the specific global coordinate system (x,y).
Isoparametric Elements
Institute of Structural Engineering Page 31
Method of Finite Elements I30-Apr-10
Isoparametric elements can be one-, two- or three-dimensional:
The principle is to assure that the value of the shape function Ni is equal to one in node i and equals zero in other nodes.
1 1 1( , , ) ; ( , , ) ; ( , , )
n n n
i i i i i ii i i
x N r s t x y N r s t y z N r s t z= = =
= = =∑ ∑ ∑
1 1 1( , , ) ; ( , , ) ; ( , , )
n n n
i i i i i ii i i
u N r s t u v N r s t v w N r s t w= = =
= = =∑ ∑ ∑
Isoparametric Elements
same approximating functions – shape
functions for
Geometric interpolation
Displacement Interpolation
Institute of Structural Engineering Page 32
Method of Finite Elements I30-Apr-10
Consider the bar element with two end nodes at points x1, x2 defined on the Cartesian axis x.
Bar Element
Truss Element Shape Functions
(in global coordinates)
( ) [ ] 11 2
2
uu x N N
u
=
Institute of Structural Engineering Page 33
Method of Finite Elements I30-Apr-10
Strain-Displacement relation
define The truss element stress displacement matrix
Constant Variation of strains along the element’s length
Institute of Structural Engineering Page 34
Method of Finite Elements I30-Apr-10
Consider the bar element with two end nodes at points x1, x2 defined on the Cartesian axis x.
Is there a function M that can map each point on the x axis to a point on the ξaxis, so that:
Bar Element
and now consider the following standardtruss element defined on the natural axis ξ
ξ = -1 ξ = 1
ξ
ξ = 0
( )( )
1
2
1. @ 0 1
2. @ 1
x x
x L x
ξ
ξ
= = = −
= = =
MM
Institute of Structural Engineering Page 35
Method of Finite Elements I30-Apr-10
The relation between the x-coordinate and the r-coordinate is given as:
1u 2u
y
x
1x2x
0r =1r = − 1r =
( ) ( )2
1 21
1 1ˆ ˆ ˆ(1 ) (1 )2 2 i i
iu u u N uξ ξ ξ ξ
=
= − + + =∑
( )2
1 21
1 1(1 ) (1 )2 2 i i
ix x x N xξ ξ ξ
=
= − + + =∑
Bar Element
ξ = -1 ξ = 1ξ = 0
Then, relation between the displacement u and the nodal displacements are defined in the same way:
Institute of Structural Engineering Page 36
Method of Finite Elements I30-Apr-10
Finite Element Idealization – Isoparametric Coordinates
Truss Element Shape Functions
(in isoparametric “natural” coordinates)
, 2A l =ξ
1ξ = − 1ξ =0ξ =
( ) ( )11 12
N ξ ξ= −
-1 1
( ) ( )21 12
N ξ ξ= +
-1 1
( ) ( ) ( ) 11 2
2
uu N N
uξ ξ ξ
=
Institute of Structural Engineering Page 37
Method of Finite Elements I30-Apr-10
1 2
1 1 1(1 ) (1 )2 12 2
1
2 22
x x x x xd dx dx L Jx dx d d
d Jdx L
ξ ξξ ξξ ξ
ξ
−= − + +
−
−∂= = → = = = ∂
⇒ = =
Why?
Isoparametric formulation:
Strain Displacement Matrix
( ) ( ) ( ) 11 2
2
uu N N
uξ ξ ξ
=
( ) ( )11 12
N ξ ξ= − ( ) ( )21 12
N ξ ξ= +
( ) ( ) ( ) [ ]1 1 1 1N x N N
B Jx x L
ξ ξξξ ξ
−∂ ∂ ∂∂= = = = −
∂ ∂ ∂ ∂
, 2A l =ξ
1ξ = − 1ξ =0ξ =
Chair rule of differentiation
This is termed the Jacobian, J
Institute of Structural Engineering Page 38
Method of Finite Elements I30-Apr-10
Why?
Isoparametric formulation:
Strain Displacement Matrix
Constant Variation of strains along the element’s length
( ) ( ) ( ) 11 2
2
uu N N
uξ ξ ξ
=
( ) ( )11 12
N ξ ξ= − ( ) ( )21 12
N ξ ξ= +
( ) ( ) ( ) [ ]1 1 1 1N x N N
B Jx x L
ξ ξξξ ξ
−∂ ∂ ∂∂= = = = −
∂ ∂ ∂ ∂
, 2A l =ξ
1ξ = − 1ξ =0ξ =
Chair rule of differentiation
Institute of Structural Engineering Page 39
Method of Finite Elements I30-Apr-10
Let us now consider the derivation of the stiffness matrix K, using the iso-parametric expression. *The strain-displacement matrix is the same for the case of the bar element for both global and isoparametric coordinates:
We write up the integral for calculating the stiffness matrix:
Why? det T T
L L
EA dx EA dξ= =∫ ∫K B B B B J
Isoparametric Elements
Because dxdx d dd
ξ ξξ
= = J
[ ] 1
2
ˆ1 1 1 with ˆuuL
ε
= − =
B B
Institute of Structural Engineering Page 40
Method of Finite Elements I30-Apr-10
[ ] 1
2
ˆ1 1 1 with ˆuuL
ε
= − =
B B
Bar ElementThe strain-displacement matrix in this case remains unchanged:
[ ]1
21
1
21
11 1 ,
1 2
1 1 1 1
1 1 1 12
AE dx LdrL dr
AE L AErL L
−
−
− = − = = ⇒
− − = ⇒ = − −
∫K J J
K K
and the stiffness matrix is calculated as:
Institute of Structural Engineering Page 41
Method of Finite Elements I30-Apr-10
As pointed out, in order to evaluate the stiffness matrix we must differentiate the displacements with respect to the coordinates (x, y, z).
The shape functions will be expressed in natural coordinates. Therefore the coordinate transformation (Jacobian) between natural and cartesiancoordinate system should be used.
In the 3D domain we can express the derivative of a function which is expressed in terms of coordinates (x,y,z), with respect to another set ofcoordinates (r,s,t) as:
x y zr x r y r z r
x y zs x s y s z s
x y zt x t y t z t
ϕ ϕ ϕ ϕ
ϕ ϕ ϕ ϕ
ϕ ϕ ϕ ϕ
∂ ∂ ∂ ∂ ∂ ∂ ∂= + +
∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂
= + +∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂
= + +∂ ∂ ∂ ∂ ∂ ∂ ∂
Chain rule of differentiation!
Isoparametric Elements in 3D
𝜑𝜑,
Institute of Structural Engineering Page 42
Method of Finite Elements I30-Apr-10
Written in matrix notation:
x y zr x r y r z r
x y zs x s y s y s
x y zt x t y t z t
φ φ φ φ
φ φ φ φ
φ φ φ φ
∂ ∂ ∂ ∂ ∂ ∂ ∂= + +
∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂
= + + ⇒∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂
= + +∂ ∂ ∂ ∂ ∂ ∂ ∂
x y zr r r r x
x y zs s s s y
x y zt t t t z
φ φ
φ φ
φ φ
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
Isoparametric Elements
Institute of Structural Engineering Page 43
Method of Finite Elements I30-Apr-10
This is the Jacobian operator J:
1
−
∂= = ⇒
∂∂
⇒ =∂
x Jr x r x
Jx r
∂ ∂ ∂∂ ∂ ∂
∂∂
x y zr r r r x
x y zs s s s y
x y zt t t t z
φ φ
φ φ
φ φ
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
Isoparametric Elements
J
which relates the natural coordinate derivatives to the Cartesian coordinate derivatives.
Institute of Structural Engineering Page 44
Method of Finite Elements I30-Apr-10
Isoparametric Elements
T
V
dV= ∫K B CB
The expression of element stiffness matrix
The strain-displacement matrix B will be expressed in terms of the natural coordinates (r,s,t) and the volume differential dV refers to the Cartesian coordinate system.
Therefore: detdV dr ds dt= J
*The Jacobian determinant is used when making a change of variables and evaluating a multiple integral of a function over a region within its domain. To accommodate for the change of coordinates the magnitude of the Jacobian determinant arises as a multiplicative factor within that integral.
Institute of Structural Engineering Page 45
Method of Finite Elements I30-Apr-10
Isoparametric Elements
T
V
dV= ∫K B CB
The expression of element stiffness matrix
The strain-displacement matrix B will be expressed in terms of the natural coordinates and the volume differential dV refers to the Cartesian coordinate system.
Therefore: detdV dr ds dt= J
detT
V
dr ds dt= ∫K B CB J
Institute of Structural Engineering Page 46
Method of Finite Elements I30-Apr-10
Isoparametric Elements
T
V
q dV= ∫f N
The expression of the equivalent nodal force vector is
The strain-displacement matrix N will be expressed in terms of the natural coordinates (r,s,t) and the volume differential dV refers to the Cartesian coordinate system.
Therefore: detdV dr ds dt= J
detT
V
q dr ds dt= ∫f N J
Institute of Structural Engineering Page 47
Method of Finite Elements I30-Apr-10
Example: Quadratic Bar Element with 3 nodes
Based on what we so far discussed, how can we define the equations for a quadratic bar element with 3 nodes?Define the higher order element by using the Lagrange Polynomial for 3 nodes, in isoparametric coordinates ξ
( ) ( ) ( )
( ) ( )( )( )( ) ( ) ( )
( ) ( )( )( )( ) ( )
( ) ( )( )( )( ) ( ) ( )
1 2
3
1 2
3
1 2
3
3
1, 02 31 11
1 2 1 3
1, 01 3 22 21
2 1 2 3
1, 01 23 31
3 1 3 2
1 12
1
1 12
ni i iN L L
N N
N N
N N
ξ ξξ
ξ ξξ
ξ ξξ
ξ ξ ξ
ξ ξ ξ ξξ ξ ξ ξ
ξ ξ ξ ξ
ξ ξ ξ ξξ ξ ξ
ξ ξ ξ ξ
ξ ξ ξ ξξ ξ ξ ξ
ξ ξ ξ ξ
=− ==
=− ==
=− ==
= = ⇒
− −= → = −
− −
− −= → = −
− −
− −= → = +
− −
Institute of Structural Engineering Page 48
Method of Finite Elements I30-Apr-10
Example: Quadratic Bar Element with 3 nodes
Self Study TaskDerive the stiffness matrix and the equivalent nodal force vector of the quadratic 3-noded bar element!
Solution: [O] Ch3§3.3
Stiffness Matrix
Nodal force vector
Institute of Structural Engineering Page 49
Method of Finite Elements I
Higher order bar Elements
[O] Ch3§2, Fig. 3.2
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