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Solve 2nd order differential equations using finite element method

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Mohammad Tawfik
Mohammad Tawfik

How to create and solve finite element models? Application to 2nd Order Differential Equations! #WikiCourses #FEM https://wikicourses.wikispaces.com/TopicX+Element+Equations

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2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Introduction to the Finite Element Method 2nd order DE’s in 1-D
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Objectives • Understand the basic steps of the finite element analysis • Apply the finite element method to second order differential equations in 1-D
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com The Mathematical Model • Solve: • Subject to: Lx fcu dx du a dx d         0 0   00 ,0 Q dx du auu Lx        
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Step #1: Discretization • At this step, we divide the domain into elements. • The elements are connected at nodes. • All properties of the domain are defined at those nodes.
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Step #2: Element Equations • Let’s concentrate our attention to a single element. • The same DE applies on the element level, hence, we may follow the procedure for weighted residual methods on the element level! 21 0 xxx fcu dx du a dx d             21 2211 21 , ,, Q dx du aQ dx du a uxuuxu xxxx              
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Polynomial Approximation • Now, we may propose an approximate solution for the primary variable, u(x), within that element. • The simplest proposition would be a polynomial!
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Polynomial Approximation • Interpolating the values of displacement knowing the nodal displacements, we may write:   01 bxbxu    01111 bxbuxu    2 12 1 1 12 2 u xx xx u xx xx xu                    02122 bxbuxu 
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Polynomial Approximation        e ux u u uu u xx xx u xx xx xu                           2 1 212211 2 12 1 1 12 2
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Step #2: Element Equations (cont’d) • Assuming constant domain properties: • Applying the Galerkin method: 21 2 2 0 xxx fcu dx ud a             02 2       Domain jiiji i j dxfxuxxcu dx xd xa   
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Step #2: Element Equations (cont’d) • Note that: • And:     ee hdx xd hdx xd 1 , 1 21                         Domain ij x x i j Domain i j dx dx xd dx xd a dx xd xa dx dx xd xa     2 1 2 2
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Step #2: Element Equations (cont’d) • For i=j=1: (and ignoring boundary terms) • Which gives: 0 12 1 2 1 2 2 2                                  x x eee dx h xx fu h xx c h a 0 23 1        ee e fh u ch h a
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Step #2: Element Equations (cont’d) • Repeating for all terms: • The above equation is called the element equation.                                    1 1 221 12 611 11 2 1 ee e fh u uch h a
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com What happens for adjacent elements?
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Objectives • Learn how the finite element model for the whole domain is assembled • Learn how to apply boundary conditions • Solving the system of linear equations
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Recall • In the previous lecture, we obtained the element equation that relates the element degrees of freedom to the externally applied fields • Which maybe written:                                    1 1 221 12 611 11 2 1 ee e fh u uch h a                    2 1 2 1 43 21 f f u u kk kk
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Two–Element example                    1 2 1 1 1 2 1 1 1 4 1 3 1 2 1 1 f f u u kk kk                    2 2 2 1 2 2 2 1 2 4 2 3 2 2 2 1 f f u u kk kk                                            3 2 1 3 2 1 3 2 1 2 4 2 3 2 2 2 1 1 4 1 3 1 2 1 1 0 0 Q Q Q f f f u u u kk kkkk kk
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Illustration: Bar application 1. Discretization: Divide the bar into N number of elements. The length of each element will be (L/N) 2. Derive the element equation from the differential equation for constant properties an externally applied force:   02 2    xF x u EA 0 2 1 2                   x x ij ij e dxfu dx d dx d h EA  
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Performing Integration:                      1 1 211 11 2 1 e e e e fh u u h EA Note that if the integration is evaluated from 0 to he, where he is the element length, the same results will be obtained. 0 2 1 2                   x x ij ij e dxfu dx d dx d h EA  
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Two–Element bar example                      1 2 1 1 1 2 1 1 11 11 f f u u h EA e                      2 2 2 1 2 2 2 1 11 11 f f u u h EA e                                              0 0 1 2 1 2 110 121 011 3 2 1 R fh u u u h EA e e
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Applying Boundary Conditions
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Applying BC’s • For the bar with fixed left side and free right side, we may force the value of the left-displacement to be equal to zero:                                              0 0 1 2 1 2 0 110 121 011 3 2 R fh u u h EA e e
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Solving • Removing the first row and column of the system of equations: • Solving:                      1 2 211 12 3 2 e e fh u u h EA              4 3 2 2 3 2 EA fh u u e
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Secondary Variables • Using the values of the displacements obtained, we may get the value of the reaction force:                                                    0 0 1 2 1 2 2 4 2 3 0 110 121 011 R fh fh fh e e e
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Secondary Variables • Using the first equation, we get: • Which is the exact value of the reaction force. R fhfh ee  22 3 efhR 2
2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Summary • In this lecture, we learned how to assemble the global matrices of the finite element model; how to apply the boundary conditions, and solve the system of equations obtained. • And finally, how to obtain the secondary variables.

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Solve 2nd order differential equations using finite element method

  • 1. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Introduction to the Finite Element Method 2nd order DE’s in 1-D
  • 2. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Objectives • Understand the basic steps of the finite element analysis • Apply the finite element method to second order differential equations in 1-D
  • 3. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com The Mathematical Model • Solve: • Subject to: Lx fcu dx du a dx d         0 0   00 ,0 Q dx du auu Lx        
  • 4. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Step #1: Discretization • At this step, we divide the domain into elements. • The elements are connected at nodes. • All properties of the domain are defined at those nodes.
  • 5. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Step #2: Element Equations • Let’s concentrate our attention to a single element. • The same DE applies on the element level, hence, we may follow the procedure for weighted residual methods on the element level! 21 0 xxx fcu dx du a dx d             21 2211 21 , ,, Q dx du aQ dx du a uxuuxu xxxx              
  • 6. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Polynomial Approximation • Now, we may propose an approximate solution for the primary variable, u(x), within that element. • The simplest proposition would be a polynomial!
  • 7. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Polynomial Approximation • Interpolating the values of displacement knowing the nodal displacements, we may write:   01 bxbxu    01111 bxbuxu    2 12 1 1 12 2 u xx xx u xx xx xu                    02122 bxbuxu 
  • 8. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Polynomial Approximation        e ux u u uu u xx xx u xx xx xu                           2 1 212211 2 12 1 1 12 2
  • 9. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Step #2: Element Equations (cont’d) • Assuming constant domain properties: • Applying the Galerkin method: 21 2 2 0 xxx fcu dx ud a             02 2       Domain jiiji i j dxfxuxxcu dx xd xa   
  • 10. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Step #2: Element Equations (cont’d) • Note that: • And:     ee hdx xd hdx xd 1 , 1 21                         Domain ij x x i j Domain i j dx dx xd dx xd a dx xd xa dx dx xd xa     2 1 2 2
  • 11. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Step #2: Element Equations (cont’d) • For i=j=1: (and ignoring boundary terms) • Which gives: 0 12 1 2 1 2 2 2                                  x x eee dx h xx fu h xx c h a 0 23 1        ee e fh u ch h a
  • 12. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Step #2: Element Equations (cont’d) • Repeating for all terms: • The above equation is called the element equation.                                    1 1 221 12 611 11 2 1 ee e fh u uch h a
  • 13. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com What happens for adjacent elements?
  • 14. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Objectives • Learn how the finite element model for the whole domain is assembled • Learn how to apply boundary conditions • Solving the system of linear equations
  • 15. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Recall • In the previous lecture, we obtained the element equation that relates the element degrees of freedom to the externally applied fields • Which maybe written:                                    1 1 221 12 611 11 2 1 ee e fh u uch h a                    2 1 2 1 43 21 f f u u kk kk
  • 16. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Two–Element example                    1 2 1 1 1 2 1 1 1 4 1 3 1 2 1 1 f f u u kk kk                    2 2 2 1 2 2 2 1 2 4 2 3 2 2 2 1 f f u u kk kk                                            3 2 1 3 2 1 3 2 1 2 4 2 3 2 2 2 1 1 4 1 3 1 2 1 1 0 0 Q Q Q f f f u u u kk kkkk kk
  • 17. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Illustration: Bar application 1. Discretization: Divide the bar into N number of elements. The length of each element will be (L/N) 2. Derive the element equation from the differential equation for constant properties an externally applied force:   02 2    xF x u EA 0 2 1 2                   x x ij ij e dxfu dx d dx d h EA  
  • 18. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Performing Integration:                      1 1 211 11 2 1 e e e e fh u u h EA Note that if the integration is evaluated from 0 to he, where he is the element length, the same results will be obtained. 0 2 1 2                   x x ij ij e dxfu dx d dx d h EA  
  • 19. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Two–Element bar example                      1 2 1 1 1 2 1 1 11 11 f f u u h EA e                      2 2 2 1 2 2 2 1 11 11 f f u u h EA e                                              0 0 1 2 1 2 110 121 011 3 2 1 R fh u u u h EA e e
  • 20. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Applying Boundary Conditions
  • 21. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Applying BC’s • For the bar with fixed left side and free right side, we may force the value of the left-displacement to be equal to zero:                                              0 0 1 2 1 2 0 110 121 011 3 2 R fh u u h EA e e
  • 22. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Solving • Removing the first row and column of the system of equations: • Solving:                      1 2 211 12 3 2 e e fh u u h EA              4 3 2 2 3 2 EA fh u u e
  • 23. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Secondary Variables • Using the values of the displacements obtained, we may get the value of the reaction force:                                                    0 0 1 2 1 2 2 4 2 3 0 110 121 011 R fh fh fh e e e
  • 24. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Secondary Variables • Using the first equation, we get: • Which is the exact value of the reaction force. R fhfh ee  22 3 efhR 2
  • 25. 2nd order DE’s in 1-D Mohammad Tawfik #WikiCourses http://WikiCourses.WikiSpaces.com Summary • In this lecture, we learned how to assemble the global matrices of the finite element model; how to apply the boundary conditions, and solve the system of equations obtained. • And finally, how to obtain the secondary variables.