Transcript PPT - Math For College

Finite Difference Method
Civil Engineering Majors
Authors: Autar Kaw, Charlie Barker
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
4/9/2015
http://numericalmethods.eng.usf.edu
1
Finite Difference Method
http://numericalmethods.eng.usf.edu
Finite Difference Method
An example of a boundary value ordinary differential equation is
d 2u 1 du u

  0, u (5)  0.008731" , u (8)  0.0030769"
dr2 r dr r 2
The derivatives in such ordinary differential equation are substituted by finite
divided differences approximations, such as
dy yi 1  yi

dx
x
d 2 y yi 1  2 yi  yi 1

2
dx
x2
3
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Example
Take the case of a pressure vessel that is being tested in the laboratory to
check its ability to withstand pressure. For a thick pressure vessel of inner
radius a and outer radius b, the differential equation for the radial
displacement u of a point along the thickness is given by
d 2 u 1 du u

 2 0
2
r dr r
dr
The pressure vessel can be modeled as,
d 2u
dr2
du
dr


ui 1  2ui  ui 1
r 2
ui 1  ui
r
Substituting these approximations gives you,
ui 1  2ui  ui 1 1 ui 1  ui ui

 2 0
ri r
r 2
ri

 1
1 
2
1
1
1





u




u

2
 i 1  r 2 r r r 2  i r 2 ui 1  0
r

r



r
i
i


i 

4
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Solution
Step 1 At node i  0,
Step 2 At node i  1,
"
r0  a  5" u0  0.0038731
r1  r0  r  5  0.6  5.6"

 1

2
1
1 
1



u




u


2 0
2
2  1
 0.6 2 5.60.6 u 2  0




5
.
6
0
.
6
0.6
5.6  

 0.6
1
2.7778u0  5.8851u1  3.0754u2  0
Step 3 At node i  2, r2  r1  r  5.6  0.6  6.2"

 1

1
2
1
1 
1



u3  0
u




u


1
2
2
2
2 
2


6.20.6 6.2   0.6 6.20.6 
0.6
 0.6
2.7778u1  5.8504u 2 3.0466u3  0
5
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Solution Cont
Step 4 At node i  3, r3  r2  r  6.2  0.6  6.8"
 2
 1

1
1
1 
1
u






u



2
 0.62 6.80.6 6.82  3  0.62 6.80.6 u4  0
0.62




2.7778u2  5.8223u3  3.0229u4  0
Step 5 At node i  4, r4  r3  r  6.8  0.6  7.4"
 2
 1

1
1
1 
1


u




u



3
4
 0.62 7.40.6 u5  0
 0.62 7.40.6 7.42 
0.62




2.7778u3  5.7990u4  3.0030u5  0
Step 6 At node i  5, r5  r4  r  7.4  0.6  8
u5  u |r b  0.0030769
"
6
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Solving system of equations
0
0
0
0
0  u0   0.0038731
 1
2.7778  5.8851 3.0754
 u  

0
0
0
0
1

  

 0

2.7778  5.8504 3.0466
0
0  u2  
0


  

u
0
0
2
.
7778

5
.
8223
3
.
0229
0
0
3

  

 0

0
0
2.7778  5.7990 3.0030 u4  
0

  

0
0
0
0
1  u5  0.0030769
 0
u0  0.0038731
u3  0.0032743
u1  0.0036165
u4  0.0031618
u2  0.0034222
7
u5  0.0030769
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Solution Cont
du
dr
r a

u1  u0 0.0036165  0.0038731

r
0.6
 0.00042767
30106  0.0038731



 max 

0
.
3

0
.
00042767

  21307psi
2
1  0.3 
5

36  103
FS 
 1.6896
21307
Et  20538 21307  768 .59
t 
8
20538 21307
100  3.744%
20538
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Solution Cont
Using the approximation of
d 2 y yi 1  2 yi  yi 1

dx2
x2
and
dy yi 1  yi 1

dx
2x 
Gives you
ui 1  2ui  ui 1
r 2

1
1


 2r r  r 2
i

9
1 ui 1  ui 1 ui

 2 0
ri 2r 
ri


 1
2
1 
1 
u i 1   



u



 r 2 r 2  i  r 2 2r r u i 1  0
i



i 

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Solution Cont
Step 1 At node i  0, r0  a  5
u0  0.0038731
Step 2 At node i  1, r1  r0  r  5  0.6  5.6"


 1

1
1 
2
1 
1






u



u


 25.60.6 0.62  0  0.62 5.62  1  0.6 2 25.60.6 u 2  0






2.6297u0  5.5874u1  2.9266u2  0
Step 3 At node i  2, r2  r1  r  5.6  0.6  6.2

 1

1
1 
2
1 
1

 


u3  0

u



u



2  1
2
2  2
2

26.20.6 
6.2 
 0.6
 26.20.6 0.6 
 0.6
2.6434u1  5.5816u 2 2.9122u3  0
10
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Solution Cont
Step 4 At node i  3, r3  r2  r  6.2  0.6  6.8

 1

1
1 
2
1 
1

 



u



u




3
2  2
2
2
 0.6 2 26.80.6 u 4  0



2
6
.
8
0
.
6
0
.
6
0
.
6
6
.
8






2.6552u2  5.5772u3  2.9003u4  0
Step 5 At node i  4, r4  r3  r  6.8  0.6  7.4


 1

1
1 
2
1 
1


 


u5  0

u



u


2  3
2
2  4
2








2
7
.
4
0
.
6
2
7
.
4
0
.
6
0.6 
7.4   0.6


 0.6
2.6651u3  5.5738u4  2.8903u5  0
Step 6 At node i  5, r5  r4  r  7.4  0.6  8"
u5  u |r b  0.0030769"
11
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Solving system of equations
0
0
0
0
0  u0   0.0038731
 1
2.6297  5.5874 2.9266
 u  

0
0
0
0
1

  

 0

2.6434  5.5816 2.9122
0
0  u2  
0


  

0
2.6552  5.5772 2.9003
0  u3  
0
 0



 0



0
0
2.6651  5.5738 2.8903 u4
0

  

0
0
0
0
1  u5  0.0030769
 0
u0  0.0038731
u3  0.0032689
u1  0.0036115
u4  0.0031586
u2  0.0034159
12
u5  0.0030769
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Solution Cont
du
dr
r a

 3u0  4u0  u2  3  0.0038731 4  0.0036115 0.0034159

 0.0004925
2r 
2(0.6)
30106  0.0038731



 max 

0
.
3

0
.
0004925

  20666psi
2
1  0.3 
5

36  103
FS 
 1.7420
20666
Et  20538 20666 128
t 
13
20538 20666
100  0.62323%
20538
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Comparison of radial
displacements
Table 1 Comparisons of radial displacements from two methods
14
r
uexact
u1st order
|єt|
u2nd order
|єt|
5
0.0038731
0.0038731
0.0000
0.0038731
0.0000
5.6
0.0036110
0.0036165
1.5160×10−1
0.0036115
1.4540×10−2
6.2
0.0034152
0.0034222
2.0260×10−1
0.0034159
1.8765×10−2
6.8
0.0032683
0.0032743
1.8157×10−1
0.0032689
1.6334×10−2
7.4
0.0031583
0.0031618
1.0903×10−1
0.0031586
9.5665×10−3
8
0.0030769
0.0030769
0.0000
0.0030769
0.0000
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Additional Resources
For all resources on this topic such as digital audiovisual
lectures, primers, textbook chapters, multiple-choice
tests, worksheets in MATLAB, MATHEMATICA, MathCad
and MAPLE, blogs, related physical problems, please
visit
http://numericalmethods.eng.usf.edu/topics/finite_differ
ence_method.html
THE END
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