Title: Chapter 5, page 1
1Chapter 5 Two dimensional elements
- A primary advantage of the finite element method
is the ease with which it can be generalized (to
2-D, axisymmetric, etc.) - Also, the ease with which you can model different
materials and curved boundaries - We will discuss
- 2-D grids
- The linear triangular element
- The bi-linear rectangular element
- Continuous piecewise smooth equations
25.1 Two-Dimensional grids
- The linear triangular element
Node numbering must be counterclockwise.
- Use to model irregular shapes
- May have any orientation
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ccw
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Defined by
3- Bilinear Rectangular Element
Use for square or rectangular regions
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Defined by
Since there is no x2 and y2 terms in our defining
equation the rectangle must be parallel to x-y
coordinate system.
4- Making your grid
- 1. Divide the region into rectangular and
triangular subregions. - 2. For triangular elements
- Divide subregions into triangles
- Place same number of nodes along each side
- Connect nodes with straight lines
5- Place nodes at intersections.
- Number of elements in a subregion with n nodes
per side is (n-1)2 For example, the region shown
below will have (4-1)2 9 elements. - The nodes do not need to be equally spaced.
Varying the spacing will vary the element size.
6- Using triangular elements to model a curved
boundary - Place the same number of nodes along each side
- Connect the nodes using strait lines
- Note the more elements (nodes) you use the
closer you will get to the actual shape
7- Quadrilateral subregion
- Place the same number of nodes along opposite
sides - connect nodes on opposite sides using line
segments - Points of intersection are the interior nodes
- You can easily change a quadrilateral region into
triangular elements by drawing shortest diagonal
(this gives the closest approximation to an
equilateral triangle, which gives better results. - If the quadrilateral regions has n and m nodes
along adjacent sides, there will be a total of
2(n-1)(m-1) elements in the subregion.
8- The spacing between the boundary nodes can be
varied to alter element size. - You must have identical nodes between adjoining
subregions for continuity. - To improve results
- Smaller elements near a curved boundary
- More (smaller) elements where changes more
rapidly.
9- Triangular elements allow us to vary the element
size in a grid - Define a quadrilateral region
- Place 3 nodes on one side for every 2 nodes on
the other side.
10Number along the side with the fewest nodes. Why?
Because we want to keep the bandwidth to a
minimum. Recall, NBW maxBW1 max
difference between largest and smallest node
number1. This simplifies the solution. Example
grid generation -
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2 nodes per side
3 nodes per side
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3 nodes per side
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- 2 nodes same as opposite side.
12Connect opposite nodes
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13 14 15- Linear triangular element
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Ø Fj at (Xj, Yj)
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label nodes ccw
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163 equations, 3 unknowns
Solve for and rearrange to write based
on nodal values
17 18Where we defined coefficients a, b, c as shown
below and noted that the determinate of the
matrix shown below (with element coordinates) is
2element area.
19Then, for the linear triangular element, we can
write
Recall that the 1-D equations was
20The gradients are
21- Note
- Since the gradients are composed of constants (A,
a, b, c, and nodal values) the gradients will be
constants. - Therefore, the shape functions have a linear
slope in each coordinate direction. - This is why you need many small elements if the
gradient is large.
22- Properties of the shape functions
1.
2.
3. Each shape function varies linearly along the
sides between its node (i, j, or k) and the
other two nodes. Note this means that any line
of constant f is a strait line that intersects
two sides of the element. This makes it easy to
locate contour lines.
235. Ni varies linearly along ij and ik Nj
varies linearly along ji and jk Nk varies
linearly along ki and kj
24- varies linearly along ij and ik, jk
- (2) For constant the contour line
intersects 2 sides, straight lines. (unless
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25Example
- 1. Calculate at a point, given
and coordinates of nodes
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272. Find at point A where x 0.20 and y0.06
28Approximated to make the next calculations easier.
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0.06
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0.2
The contour line intersects sides i-j and j-k
29Side jk
30- Bilinear rectangular element
Solution plane
The element must be oriented with sides parallel
to the x,y axis, with node i in the lower left
hand corner. Nodes are labeled ccw.
Well use a local coordinate system (s, t) with
origin at node i because the shape functions are
easier to evaluate.
31- in terms of s-t coordinates
Linear in s along any line of constant t linear
in t along any line of constant s.
Could also use
Could also be written in terms of q-r system
(More on coordinate systems in the next chapter.)
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Note we define the length and width as 2b, 2a so
that we have a b when we use the q-r system.
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32Coefficients c1, c2, c3, c4 are determined by
using nodal coordinates in the s-t system.
Applying BCs
Solving for cs and writing as
33Solve for C1, C2, C3, and C4 and then using these
coefficients in the equation for f. Writing in
terms of the shape functions, we get
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34Shape Function Properties
- 1 at own node
- 0 at other nodes and along the sides it does not
touch.
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Find 3 values where the 50C contour line
intersects ij and km.
37Ø is a function of i and j only, along ij
38Along ij
Similarly, along km,
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along
40Global coordinates are (6.64, 4)
- If contour line were linear from side ij to mk,
then coordinate would be (6.60,4)
41A Continuous Piecewise Smooth Equation
Given 2 triangular elements and 2 bilinear
rectangular elements.
Denotes node i
Rectangle i is origin of st coordinate
system Triangle i can be any node
42Rectangle
Triangle
43Element 1
Shape functions only depend on global
coordinates for dimensions (length and width)
Global (x,y)
Local (s,t)
44For the triangle, the shape functions depend on
the global coordinates.
45So, for element (4)
46Assignment