Finite Difference Method
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Finite Difference Method - Numerical Methods - Lecture Slides, Slides of Mathematical Methods for Numerical Analysis and Optimization
Main points are: Finite Difference Method, Boundary Value, Ordinary Differential Equation, Differences Approximations, Pressure Vessel, Solving System of Equations, Comparison of Radial Displacements, Decimal Number to Binary
Typology: Slides
2012/2013
1 / 13
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Finite Difference Method
Finite Difference Method
An example of a boundary value ordinary differential equation is
2 1 2 0 , (^5 )^0.^008731 ", (^8 )^0.^0030769 "
2
+ − = u = u =
r
u
dr
du
dr r
d u
The derivatives in such ordinary differential equation are substituted by finite
divided differences approximations, such as
x
y y
dx
dy i i
≈ +^1 −
( )^2
1 1 2
x
y y y dx
d y i i i ∆
Solution
Step 1 At node^ i^ =^0 ,^ r 0 = a =^5 " u 0 =^0.^0038731 "
Step 2 At node i =^1 ,^ r 1 = r 0 +∆ r =^5 +^0.^6 =^5.^6 "
( ) ( ) (^ )(^ )^ ( ) (^ )(^ )^
2 0 2 2 1 2 ^2 =
+^ +
u + − − − u u
2. 7778 u 0 − 5. 8851 u 1 + 3. 0754 u 2 = 0
Step 3 At node i = 2 , r 2 =^ r 1 +∆ r =^5.^6 +^0.^6 =^6.^2 "
( )( ) ( )( )
2 1 2 2 2 2 ^3 =
+^ +
u + ^ − − − u u
2. 7778 u 1 − 5. 8504 u 2 + 3. 0466 u 3 = 0
Solution Cont
Step 4 At node i = 3 , r 3 =^ r 2 +∆ r =^6.^2 +^0.^6 =^6.^8 "
( )( ) ( )( ) 0
- 8 0. 6
1
- 6
1
- 8
1
- 8 0. 6
1
- 6
2
- 6
1 2 2 2 2 3 2 ^4 =
+^ +
u + ^ − − − u u
2. 7778 u 2 − 5. 8223 u 3 + 3. 0229 u 4 = 0
Step 5 At node
Step 6 At node
i = 4 , r 4 =^ r 3 +∆ r =^6.^8 +^0.^6 =^7.^4 "
( )( ) (^) ( ) ( )( ) 0
- 4 0. 6
1
- 6
1
- 4
1
- 4 0. 6
1
- 6
2
- 6
1 (^2 32242) ^5 =
+^ +
u + ^ − − − u u
2. 7778 u 3 − 5. 7990 u 4 + 3. 0030 u 5 = 0
i = 5 , r 5 =^ r 4 +∆ r =^7.^4 +^0.^6 =^8
u 5 = u | r = b = 0. 0030769 "
Solution Cont
r
u u
dr
du
r a ∆
=^100. 6
= (^) =− 0. 00042767
- 3 ( 0. 00042767 ) 21307 psi 5
2
6 max =
σ = ×
FS =^36 ×^103 =
E t = 20538 − 21307 =− 768. 59
∈ =^20538 −^21307 × =
t
Solution Cont
( )^2
1 1 2
x
y y y
dx
d y i i i
Using the approximation of
( x )
y y
dx
dy i i
and^11
( ) (^ )^
0 2
(^21) 2
1 1 2
(^1 1) − = ∆
−
∆
- −^ + − + − i
i i i i
i i i r
u r
u u r r
u u u
2 1222 ^1 =
− − i +
i
i i
i i
u
r r r
u
r r
u
r r r
Gives you
Solution Cont
Step 4 At node i = 3 , r 3 = r 2 +∆ r = 6. 2 + 0. 6 = 6. 8
( )( ) ( )( )
2 2 2 2 3 2 ^4 =
+^ +
+^ − −
(^) − + u u u
2. 6552 u 2 − 5. 5772 u 3 + 2. 9003 u 4 = 0
Step 5 At node
Step 6 At node
i = 4 , r 4 =^ r 3 +∆ r =^6.^8 +^0.^6 =^7.^4
( )( ) (^) ( ) ( )( )
2 3 2 2 4 2 ^5 =
+^ +
(^) − + u u u
2. 6651 u 3 − 5. 5738 u 4 + 2. 8903 u 5 = 0
i = 5 , r 5 =^ r 4 +∆ r =^7.^4 +^0.^6 =^8 "
u 5 = u | r = b = 0. 0030769 "
Solving system of equations
=
−
−
−
−
- 0030769
0
0
0
0
- 0038731
0 0 0 0 0 1
0 0 0 2. 6651 5. 5738 2. 8903
0 0 2. 6552 5. 5772 2. 9003 0
0 2. 6434 5. 5816 2. 9122 0 0
- 6297 5. 5874 2. 9266 0 0 0
1 0 0 0 0 0
5
4
3
2
1
0
u
u
u
u
u
u
u 0 = 0. 0038731
u 1 = 0. 0036115
u 2 = 0. 0034159
u 3 = 0. 0032689
u 4 = 0. 0031586
u 5 = 0. 0030769
Comparison of radial displacements