Basic Illustration of Finite Element Analysis (2013)

7/30/2019 Basic Illustration of Finite Element Analysis (2013)

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January 2013

Finite element analysis example.We set up a triangle with 6 internal nodes,as shown.

In this grid we have three diff erent triangular elements. One is the isoscelesright triangle treated by Jackson (Fig 2.16)

1



We also have a right triangle with sides 1,2 and√

5 and area 1. ( Nodes 6,7,8form this triangle, and also nodes 5,12,13.)

For this triangle we have

1 =1

2 (23 − 32) =1

2 (0− 2) = −1

1 =1

2(2 − 3) =

1

2(1 − 0) =

1

2

1 = −1

2(2 − 3) = −

1

2(0 − 2) = 1

2 =1

2(21 − 13) =

1

2(0− 0) = 0

2 =1

2(3 − 1) =

1

2(0 − 0) = 0

2 = −1

2(3 − 1) = −

1

2(2 − 0) = −1

3 = 12

(12 − 21) = 12

(0 − 0) = 0

3 =1

2(1 − 2) =

1

2(0− 1) = −

1

2

3 = −1

2(1 − 2) = −

1

2(0− 0) = 0

2



Note: Jackson claims thatP

= 1 but this result depends on the orientation of the triangle, which explains why I have

−1. But the all depend quadratically

on the s and s, and so are independent of their signs. For this triangle,

11 = (21 + 21) =1

4+ 1 =

5

4

12 = (12 + 12) =

µ1

2

¶(0) + 1× (−1) = −1

13 = (13 + 13) =1

2×

µ−

1

2

¶+ 1× 0 = −

1

4

22 = (22 + 22) = 0 + 1 = 1

23 = (23 + 23) = 0 + 0 = 0

33 = (23 + 23) =1

4+ 0 =

1

4

Thus we have:

The third triangle is formed by nodes 5,13,14 and also 6,7,16 and looks like

this:

3



This triangle also has area 1 and its shape functions have coefficients:

1 =1

2 (23 − 32) =1

2 (0 − 2× 1) = −1

1 =1

2(2 − 3) =

1

2(1− 2) = −

1

2

1 = −1

2(2 − 3) = −

1

2(0− 2) = 1

2 =1

2(21 − 13) =

1

2(0− 0) = 0

2 =1

2(3 − 1) =

1

2(2 − 0) = 1

2 = −1

2(3 − 1) = −

1

2(2 − 0) = −1

3 = 12

(12 − 21) = 12

(0 − 0) = 0

3 =1

2(1 − 2) =

1

2(0− 1) = −

1

2

3 = −1

2(1 − 2) = −

1

2(0− 0) = 0

And then

11 = (21 + 21) =1

4+ 1 =

5

4

12 = (12 + 12) =

µ−

1

2

¶(1) + 1× (−1) = −

3

2

13 = (13 + 13) =µ−1

2¶×µ−1

2¶

+ 1×

0 =

1

4

22 = (22 + 22) = 1 + 1 = 2

23 = (23 + 23) = 1

µ−

1

2

¶+ 0 = −

1

2

33 = (23 + 23) =1

4+ 0 =

1

4

4



Now we have to compute the matrix coefficients for the six internal nodes. Eightidentical triangles meet at node 1, so

11 = 8

µ1

2

¶= 4

Similarly we have 22 = 4 (1) = 4 = 33

55 = 66 = 2 × 1 + 2 + 1 = 5

Line 12 is a side of two identical triangles, so

12 = 2

µ−

1

2

¶= −1 = 13 = 14

Similarly 45 = 36 = 2

µ−

1

2

¶= −1

and these are the only non-zero elements. Thus

=

⎛⎜⎜⎜⎜⎜⎜⎝

4 −1 −1 −1 0 0−1 4 0 0 0 0−1 0 4 0 0 −1−1 0 0 4 −1 00 0 0 −1 5 00 0 −1 0 0 5

⎞⎟⎟⎟⎟⎟⎟⎠

,

with inverse: −1

=

⎛⎜⎜⎜⎜⎜⎜⎝

76

245

19

245

4

49

4

49

4

245

4

24519

245

66

245

1

49

1

49

1

245

1

2454

49

1

49

265

931

20

931

4

931

53

931

449

149

20931

265931

53931

4931

4

245

1

245

4

931

53

931

984

4655

4

46554

245

1

245

53

931

4

931

4

4655

984

4655

⎞⎟⎟⎟⎟⎟⎟⎠

or −1 =⎛⎜⎜⎜⎜⎜⎜⎝

0310 2 7755 × 10−2 8163 × 10−2 8163 × 10−2 1633× 10−2 1633× 10−2

7755× 10−2 0269 39 2041× 10−2 2041× 10−2 4082× 10−3 4082× 10−3

8163× 10−2 2041× 10−2 0284 6 2148× 10−2 4296× 10−3 5693× 10−2

8163× 10−2 2041× 10−2 2148× 10−2 0284 64 5693× 10−2 4296× 10−3

1633× 10−2 4082× 10−3 4296× 10−3 5693× 10−2 0211 39 8593× 10−4

1633× 10−2 4082× 10−3 5693× 10−2 4296× 10−3 8593× 10−4 021139

⎞⎟⎟⎟⎟⎟⎟⎠

Now we set up boundary conditions with the sloping side grounded and theother two sides at unit potential. Then the source vector includes the nodes on

the boundaries:

−1 =16X

=7

1 = 199+ 11111

+ 11414+ 11515

+ 11616

5



The only nonzero is 115 but 15= 0 so 1 = 0 Similarly:

−2 = 299 + 21010 + 21111

= 3× 2(−1

2)× 1 = −3

−3 = 399+ 388

+ 31616

= 2(−1

2)× 1 +

µ−

1

2+ 0

¶× 1 +

µ−

1

2+ 0

¶× 0

= −1 −1

2= −

3

2= −4

−5 = 51212+ 51313

+ 51414

= µ−1

2 −1¶× 1 + µ0

−

1

2¶× 1 + 0

= −2 = −6

Then to solve the problem, we multiply⎛⎜⎜⎜⎜⎜⎜⎝

76

245

19

245

4

49

4

49

4

245

4

24519

245

66

245

1

49

1

49

1

245

1

2454

49

1

49

265

931

20

931

4

931

53

9314

49

1

49

20

931

265

931

53

931

4

9314

245

1

245

4

931

53

931

984

4655

4

46554

245

1

245

53

931

4

931

4

4655

984

4655

⎞⎟⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎜⎜⎜⎝

03

151522

⎞⎟⎟⎟⎟⎟⎟⎠

=

⎛⎜⎜⎜⎜⎜⎜⎝

0542 860885 710642 860642 860528 570528 57

⎞⎟⎟⎟⎟⎟⎟⎠

The major uncertainty has to do with the value of the potential in the corners(nodes 13 and 7) where the insulating strip runs. Should we use zero or 1, orperhaps 1/2? If we use zero, we get diff erent values for 5 = 6 :

−5 = 51212 + 51313 + 51414

=

µ−

1

2− 1

¶× 0 +

µ0 −

1

2

¶× 1 + 0

= −1

2= −6

and then

⎛⎜⎜⎜⎜⎜⎜⎝

76

245

19

245

4

49

4

49

4

245

4

24519

245

66

245

1

49

1

49

1

245

1

2454

49

1

49

265

931

20

931

4

931

53

9314

49

1

49

20

931

265

931

53

931

4

9314

245

1

245

4

931

53

931

984

4655

4

46554

245

1

245

53

931

4

931

4

4655

984

4655

⎞⎟⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎜⎜⎜⎝

03

151555

⎞⎟⎟⎟⎟⎟⎟⎠

=

⎛⎜⎜⎜⎜⎜⎜⎝

0493880873470551020551020210202102

⎞⎟⎟⎟⎟⎟⎟⎠

or choosing 1/2

−5 = 51212+ 51313

+ 51414

=

µ−

1

2− 1

¶×

1

2+

µ0 −

1

2

¶× 1 + 0

= −5

4= −6

6



⎛

⎜⎜⎜⎜⎜⎜⎝

76

245

19

245

4

49

4

49

4

245

4

24519

245

66

245

1

49

1

49

1

245

1

245

449

149

265931

20931

4931

53931

4

49

1

49

20

931

265

931

53

931

4

9314

245

1

245

4

931

53

931

984

4655

4

46554

245

1

245

53

931

4

931

4

4655

984

4655

⎞

⎟⎟⎟⎟⎟⎟⎠

⎛

⎜⎜⎜⎜⎜⎜⎝

03

1515

125125

⎞

⎟⎟⎟⎟⎟⎟⎠=

⎛

⎜⎜⎜⎜⎜⎜⎝

051837087959

059694059694036939036939

⎞

⎟⎟⎟⎟⎟⎟⎠

The last choice gives results that come closest to the values from the relax-ation method I used as a check.⎛⎜⎜⎜⎜⎜⎜⎝

051837087959059694059694036939036939

⎞⎟⎟⎟⎟⎟⎟⎠

versus

⎛⎜⎜⎜⎜⎜⎜⎝

051508750608060804020402

⎞⎟⎟⎟⎟⎟⎟⎠

or a diff erence of

⎛⎜⎜⎜⎜⎜⎜⎝

065437052457−1 8191−1 8191−8 1119−8 1119

⎞⎟⎟⎟⎟⎟⎟⎠

%

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Basic Illustration of Finite Element Analysis (2013)