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Finite Element Analysis Sal Saad Al Deen Taher Assignment No: 1 ID: 20ME128

The document presents a method to solve a second order differential equation using finite difference and finite element methods. It uses a mesh spacing of 1/3 and applies the two methods to obtain solutions at grid points. The finite difference method derives equations representing the differential equation in finite form. The finite element method is also applied but not shown. Both methods are compared to the exact solution. Improving the accuracy by changing the mesh or shape functions is suggested.

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Sal Saad
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157 views8 pages

Finite Element Analysis Sal Saad Al Deen Taher Assignment No: 1 ID: 20ME128

The document presents a method to solve a second order differential equation using finite difference and finite element methods. It uses a mesh spacing of 1/3 and applies the two methods to obtain solutions at grid points. The finite difference method derives equations representing the differential equation in finite form. The finite element method is also applied but not shown. Both methods are compared to the exact solution. Improving the accuracy by changing the mesh or shape functions is suggested.

Uploaded by

Sal Saad
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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FINITE ELEMENT ANALYSIS SAL SAAD AL DEEN TAHER

ASSIGNMENT NO: 1 ID: 20ME128


Question: Solve the following differential equation by using the finite difference
1
method(solution) and the finite element method with∆ x= and compare both results with the
3
exact solutions.

d2 x d∅
dx 2
−∅ ( x ) =0 ,
dx ] x=0
=28 , ∅ (1 ) =0Exact Solution:

28
∅ ( x) = 2
( e x −e 2−x )
e +1

Improve the accuracy of the above calculation by changing mesh space or shape function.

Answer:

The given equation is

d2 x
−∅ ( x ) =0 ( 1 )
d x2

d∅ 1
which satisfies the boundary conditions
dx ] x=0
=28 , ∅ ( 1 )=0 A mesh spacing ∆ x=
3
is chosen

and the solution may then be found by-

a) Finite Difference Method


b) Finite Element Method

a) Finite Difference Method:

i = -1 i= 0 i=1 i=2 i=3

d2 x
Using Taylor Series Expansion can be expressed in finite difference form-
d x2

d 2 x ∅ ( x +∆ x )−2 ∅ ( x ) + ∅ ( x−∆ x )
=
d x2 ∆ x2

Substituting this value in equation no 1 we get,

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FINITE ELEMENT ANALYSIS SAL SAAD AL DEEN TAHER
ASSIGNMENT NO: 1 ID: 20ME128
∅ ( x +∆ x )−2 ∅ ( x ) + ∅ ( x−∆ x )
−∅ ( x )=0
∆ x2

≫ ∅ ( x +∆ x )−2 ∅ ( x ) −∆ x 2 ∅ ( x )+ ∅ ( x−∆ x ) =0(2)

An equation of this form arises at each of the interior grid points xi on the finite difference

mesh. Using this equation for i = -1 , i = 0, i = 1, i = 2, i = 3 and inserting the known boundary
d∅
conditions
dx ] x=0
=28 , ∅ ( 1 )=0 the governing equations are as below-

∅ −1−2 ∅0 −∆ x 2 ∅0 + ∅1=0(3)

∅ 0−2 ∅1−∆ x2 ∅ 1+ ∅ 2=0(4)

∅ 1−2 ∅ 2−∆ x 2 ∅ 2+ ∅ 3=0(5)

Using a central difference representation of the derivative boundary condition at x = 0,


d∅
dx ] x=0
=28 produces ,

∅ 1−∅−1
=28 ≫ ∅1 −∅−1=56 × ∆ x
2∆x

1
And at x = 1, ∅ ( 1 )=∅ 3=0, and putting ∆ x= in equations (3), (4) and (5) following equations
3
are formed-

56
−∅−1 + ∅1=
3

19
∅ −1− ∅ + ∅ =0
9 0 1

19
∅ 0− ∅ + ∅ =0
9 1 2

19
∅ 1− ∅ =0
9 2

Rearranging these equations in matrix form we get-

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FINITE ELEMENT ANALYSIS SAL SAAD AL DEEN TAHER
ASSIGNMENT NO: 1 ID: 20ME128

−1 0 1 0

| || || |
∅−1 56
−19
1 1 0
9 3
∅
−19 × 0 = 0
0 1 1 ∅1
9 0
−19 ∅2 0
0 0 1
9

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FINITE ELEMENT ANALYSIS SAL SAAD AL DEEN TAHER
ASSIGNMENT NO: 1 ID: 20ME128
Solving the above matrix using OCTAVE Program

Octave Code:

∅−1

{ }{ }
−31.4802
∅0 −20.9813
=
∅1 −12.8136
∅2 −6.0696

Using a forward difference representation of the derivative boundary condition at x=0,


d∅
dx ] x=0
=28 produces

∅ 1−∅ 0
=28 ≫ ∅1−∅ 0=28 × ∆ x
∆x

1
And at x = 1, ∅ ( 1 )=∅ 3=0, and putting ∆ x= following equations are formed-
3

28
−∅0 + ∅1=
3

19
∅ 0− ∅ + ∅ =0
9 1 2
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FINITE ELEMENT ANALYSIS SAL SAAD AL DEEN TAHER
ASSIGNMENT NO: 1 ID: 20ME128
19
∅ 1− ∅ =0
9 2

Rearranging these equations in matrix form we get-

−1 1 0

| |||| |
28
−19 ∅0
1 1 3
9 × ∅1 =
0
−19 ∅2
0 1 0
9

Solving the above matrix using OCTAVE Program

Octave Code:

∅ 0 −23.9755

{}{ }
∅ 1 = −14.6422
∅2 −6.9358

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FINITE ELEMENT ANALYSIS SAL SAAD AL DEEN TAHER
ASSIGNMENT NO: 1 ID: 20ME128
(b) Finite Element Method:

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FINITE ELEMENT ANALYSIS SAL SAAD AL DEEN TAHER
ASSIGNMENT NO: 1 ID: 20ME128
Exact Solution:

28
∅ ( x) = 2
( e x −e 2−x )
e +1

28 0 2−0
∅ 0=∅ ( 0 )= ( e −e ) =−21.3246
e2 +1
1 1
∅ 1=∅
1
()28
= 2
3 e +1
2−
(
e 3 −e 3 =−13.0132 )
2 2
∅ 2=∅ () 2 28
= 2
3 e +1
(
2−
e 3 −e 3 =−6.1611 )

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FINITE ELEMENT ANALYSIS SAL SAAD AL DEEN TAHER
ASSIGNMENT NO: 1 ID: 20ME128

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