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MANE 4240 & CIVL 4240 Introduction to Finite Elements

MANE 4240 & CIVL 4240 Introduction to Finite Elements. Prof. Suvranu De. Numerical Integration in 2D. Reading assignment: Lecture notes, Logan 10.4. Summary: Gauss integration on a 2D square domain Integration on a triangular domain Recommended order of integration

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MANE 4240 & CIVL 4240 Introduction to Finite Elements

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  1. MANE 4240 & CIVL 4240Introduction to Finite Elements Prof. Suvranu De Numerical Integration in 2D

  2. Reading assignment: Lecture notes, Logan 10.4 • Summary: • Gauss integration on a 2D square domain • Integration on a triangular domain • Recommended order of integration • “Reduced” vs “Full” integration; concept of “spurious” zero energy modes/ “hour-glass” modes

  3. Integration point Weight 1D quardrature rule recap Choose the integration points and weights to maximize accuracy Newton-Cotes Gauss quadrature 1. ‘M’ integration points are necessary to exactly integrate a polynomial of degree ‘2M-1’ 2. Less expensive 3. Exponential convergence, error proportional to 1. ‘M’ integration points are necessary to exactly integrate a polynomial of degree ‘M-1’ 2. More expensive

  4. Example f(x) x -1 1 A 2-point Gauss quadrature rule is exact for a polynomial of degree 3 or less

  5. 2D square domain t 1 1 1 s 1 Using 1D Gauss rule to integrate along ‘t’ Using 1D Gauss rule to integrate along ‘s’ Where Wij=Wi Wj

  6. For M=2 Wij=Wi Wj=1 Number the Gauss points IP=1,2,3,4

  7. The rule Uses M2 integration points on a nonuniform grid inside the parent element and is exact for a polynomial of degree (2M-1) i.e., A M2 –point rule is exact for a complete polynomial of degree (2M-1)

  8. 1 1 s1=0, t1=0 W1= 4 1 s 1 CASE I: M=1 (One-point GQ rule) is exact for a product of two linear polynomials t

  9. CASE II: M=2 (2x2 GQ rule) t 1 1 1 s 1 is exact for a product of two cubic polynomials

  10. CASE III: M=3 (3x3 GQ rule) t 1 1 3 6 2 1 1 9 7 s 4 8 5 1 is exact for a product of two 1D polynomials of degree 5

  11. Examples If f(s,t)=1 A 1-point GQ scheme is sufficient If f(s,t)=s A 1-point GQ scheme is sufficient If f(s,t)=s2t2 A 3x3 GQ scheme is sufficient

  12. 2D Gauss quadrature for triangular domains Remember that the parent element is a right angled triangle with unit sides The type of integral encountered t 1 t s=1-t t s 1

  13. Constraints on the weights if f(s,t)=1

  14. Example 1. A M=1 point rule is exact for a polynomial t 1/3 1 1/3 s 1

  15. Why? Assume Then But Hence

  16. Example 2. A M=3 point rule is exact for a complete polynomial of degree 2 t 1/2 1 1 2 1/2 s 3 1

  17. Example 4. A M=4 point rule is exact for a complete polynomial of degree 3 t (0.2,0.6) (1/3,1/3) 1 2 1 3 4 s (0.2,0.2) (0.6,0.2) 1

  18. Recommended order of integration “Finite Element Procedures” by K. –J. Bathe

  19. “Reduced” vs “Full” integration Full integration: Quadrature scheme sufficient to provide exact integrals of all terms of the stiffness matrix if the element is geometrically undistorted. Reduced integration: An integration scheme of lower order than required by “full” integration. Recommendation: Reduced integration is NOT recommended.

  20. 1 s t 1 s t s2 st t2 Which order of GQ to use for full integration? To computet the stiffness matrix we need to evaluate the following integral For an “undistroted” element det (J) =constant Example : 4-noded parallelogram

  21. 1 s t s2 st t2 s2t st2 1 s t s2 st t2 Hence, 2M-1=2 M=3/2 Hence we need at least a 2x2 GQ scheme Example 2: 8-noded Serendipity element

  22. 1 s t s2 st t2 s3 s2t st2 t3 s4 s3t s2t2 st3 t4 Hence, 2M-1=4 M=5/2 Hence we need at least a 3x3 GQ scheme

  23. Reduced integration leads to rank deficiency of the stiffness matrix and “spurious” zero energy modes “Spurious” zero energy mode/ “hour-glass” mode The strain energy of an element Corresponding to a rigid body mode, If U=0 for a mode d that is different from a rigid body mode, then d is known as a “spurious” zero energy mode or “hour-glass” mode Such a mode is undesirable

  24. Example 1. 4-noded element y 1 1 Full integration: NGAUSS=4 Element has 3 zero energy (rigid body) modes Reduced integration: e.g., NGAUSS=1 1 x 1

  25. Consider 2 displacement fields y y x x We have therefore 2 hour-glass modes.

  26. Propagation of hour-glass modes through a mesh

  27. Example 2. 8-noded serendipity element y 1 1 Full integration: NGAUSS=9 Element has 3 zero energy (rigid body) modes Reduced integration: e.g., NGAUSS=4 1 x 1

  28. Element has one spurious zero energy mode corresponding to the following displacement field Show that the strains corresponding to this displacement field are all zero at the 4 Gauss points y x Elements with zero energy modes introduce uncontrolled errors and should NOT be used in engineering practice.

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