Nodal Coordinates Mapping Jacobian 1 0 0 2 1 4 3 5 5 4 0 5 x y J 0 at 5 10s 10t 0 ie s t 12 Constant t Constant s Invalid mapping POOR ELEMENTS ID: 544731
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Slide1
AN ILL-BEHAVED ELEMENT
Nodal CoordinatesMappingJacobian
1
(0, 0)
2
(1,
4)
3
(5, 5)
4(0, 5)
x
y
|J| = 0 at 5 – 10s + 10t = 0; i.e., s – t = 1/2 Slide2
Constant
t
Constant
s
Invalid mapping
POOR ELEMENTS
In general the element geometry
is invalid if the Jacobian is either
zero or negative anywhere in the
element.
Problems also arise when the
Jacobian matrix is nearly singular
either due to round-off errors or
due to badly shaped elements.
To avoid problems due to badly
shaped elements, it is suggested
that the inside angles in quadrilateral
elements be > 15˚ and < 165˚
> 15
o
< 165
oSlide3
Quiz-like problem
Consider the quadrilateral shown in the figure. If we move Node 3 along the line x=y, how far we can move it in either direction before we get an ill-conditioned element?Answer in notes page
1
(0, 0)
2
(1, 0)
3
(2, 2)
4
(0, 1)
x
ySlide4
INTERPOLATION
Displacement Interpolation (8-DOF)the interpolation is done in the reference coordinates (
s, t
)The behavior of the element is similar to that of the rectangular element because both of them are based on the bilinear Lagrange interpolationBut what is different?Slide5
STRAIN
Derivatives needed for strainSlide6
Adding the dependence on DOF
The expression of [B] is complicated because the matrix [A] involves the inverse of Jacobian matrix
The strain-displacement matrix [B] is not constant
STRAIN
cont
.
Strain-displacement matrixSlide7
EXAMPLE 6.8
{u1, v1,
u2, v
2, u3, v3, u4, v4} = {0, 0, 1, 0, 2, 1, 0, 2}
Displacement and strain at (s,t)=(1/3, 0)?Shape Functions
At (s,t)=(1/3, 0)
x
y
1 (0,0)
2 (3,0)
3 (3,2)
4 (0,2)
s
t
1 (-1,-1)
2 (1,-1)
3 (1,1)
4 (-1,1)Slide8
EXAMPLE cont.
Location at the Actual ElementDisplacement at (s,t) = (1/3,0)Slide9
EXAMPLE cont.
Derivatives of the shape functions w.r.t. s and t.
But
, we need the derivatives w.r.t. x and y. How to convert?Slide10
EXAMPLE cont.
Jacobian Matrix
Jacobian is positive, and the mapping is valid at this point
Jacobian matrix is constant throughout the elementJacobian matrix only has diagonal components, which means that the physical element is a rectangleSlide11
EXAMPLE cont.
Derivative of the shape functions w.r.t. x and y.StrainSlide12
Quiz-like problems
Consider the quadrilateral element in the figure, with a single non-zero displacement component u3=0.01. Calculate the displacements as functions of s and t.
Calculate the Jacobian and its inverse at Node 3
Calculate exx at Node 3Solution:Element was used in Example 6.7 in the previous lecture, we had
Then
Continued on notes page
1
(0, 0)
2
(1, 0)
3
(2, 2)
4
(0, 1)
xySlide13
FINITE ELEMENT EQUATION
Element stiffness matrix from strain energy expression
[
k
(e)] is the element stiffness matrixIntegration domain is a general quadrilateral shapeDisplacement–strain matrix [
B] is written in (s, t) coordinates
we can perform the integration in the reference element