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September 18, 2002

September 18, 2002. Ahmed Elgamal. Stiffness Coefficients for a Flexural Element Ahmed Elgamal. September 18, 2002. Ahmed Elgamal. Stiffness coefficients for a flexural element (neglecting axial deformations), Appendix A, Ch. 1 Dynamics of Structures by Chopra. u 2. u 1. u 4. u 3.

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September 18, 2002

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  1. September 18, 2002 Ahmed Elgamal Stiffness Coefficients for a Flexural Element Ahmed Elgamal

  2. September 18, 2002 Ahmed Elgamal Stiffness coefficients for a flexural element (neglecting axial deformations), Appendix A, Ch. 1 Dynamics of Structures by Chopra. u2 u1 u4 u3 Positive directions k21 k11 k41 k31 4 degrees of freedom

  3. September 18, 2002 Ahmed Elgamal To obtain k coefficients in 1st column of stiffness matrix, move u1 = 1, u2 = u3 = u4 = 0, and find forces and moments needed to maintain this shape. u1 1.0 L

  4. September 18, 2002 Ahmed Elgamal These are (see structures textbook) Note that S Forces = 0 S Moments = 0 S M = = 0 - , and you i.e. remember Positive directions k21 k11 can find other forces & moments k41 k31

  5. September 18, 2002 Ahmed Elgamal , where i is row number and j is column number

  6. September 18, 2002 Ahmed Elgamal u2 = 1, u1 = u3 = u4 = 0 u2 1.0 L

  7. September 18, 2002 Ahmed Elgamal u2 = 1 S M = 0, S F = 0 Positive directions k22 k12 k42 k32

  8. September 18, 2002 Ahmed Elgamal u3 = 1, u1 = u2 = u4 = 0 u3 = 1 1.0 L

  9. September 18, 2002 Ahmed Elgamal S M = 0, S F = 0 Positive directions k23 k13 k43 k33

  10. September 18, 2002 Ahmed Elgamal u4 = 1, u1 = u2 = u3 = 0 1.0 u4 = 1 L

  11. September 18, 2002 Ahmed Elgamal S M = 0, S F = 0 Positive directions k24 k14 k44 k34

  12. September 18, 2002 Ahmed Elgamal Example: Water Tank u1 m m is lumped at a point & does not contribute in rotation u2 L ug u2 above was u3 in the earlier section of these notes

  13. u1 m u2 L ug September 18, 2002 Ahmed Elgamal Example: Water Tank (continued) k11 k12 k21 k22 Rotational (used to be u3) “Note Symmetry”

  14. September 18, 2002 Ahmed Elgamal Example: Water Tank (continued) Static Condensation: Way to solve a smaller system of equations by eliminating degrees of freedom with zero mass. e.g., in the above, the 2nd equation gives or ----- *

  15. September 18, 2002 Ahmed Elgamal Example: Water Tank (continued) Substitute * into Equation 1 or, or, Now, solve for u1 and u2 can be evaluated from Equation * above. Static condensation can be applied to large MDOF systems of equations, the same way as shown above.

  16. September 18, 2002 Ahmed Elgamal Example: Water Tank (continued) k of water tank as we were given earlier. or of column if EIb = 0 Ib

  17. September 18, 2002 Ahmed Elgamal Mandatory Reading bending beam Example 9.4 page 329-331 Example 9.8 page 335-336 Homework: 9.5, 9.8, & 9.9

  18. September 18, 2002 Ahmed Elgamal Example (a) (b) Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN 0-13-855214-2

  19. September 18, 2002 Ahmed Elgamal (c) (d) Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN 0-13-855214-2

  20. September 18, 2002 Ahmed Elgamal (e) (f) Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN 0-13-855214-2

  21. September 18, 2002 Ahmed Elgamal (g) (h) Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN 0-13-855214-2

  22. September 18, 2002 Ahmed Elgamal (i) (j) Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN 0-13-855214-2

  23. September 18, 2002 Ahmed Elgamal therefore, , Homework: For the above cantilever system, write equation of motion and perform static condensation to obtain a 2 DOF system. Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN 0-13-855214-2

  24. m bending beam L , beam with EIbeam = 0 September 18, 2002 Ahmed Elgamal Column Stiffness (lateral vibration)

  25. September 18, 2002 Ahmed Elgamal Case of EIb = u (rigid roof) L h1 h2

  26. September 18, 2002 Ahmed Elgamal (See example 1.1 in Dynamics of Structures by Chopra) h L = 2h if column beam , &

  27. September 18, 2002 Ahmed Elgamal Obtained by “static condensation” of 3x3 system u1 force fs q1 q2 call it u2 call it u3 Use to represent u2 and u3 in terms of u1 & plug back into Technique can also be used for large systems of equations and get fs = ku1 (See example 1.1 in Dynamics of Structures by Chopra)

  28. September 18, 2002 Ahmed Elgamal Draft Example Neglect axial deformation u3 u2 u1 Ib Ic Ic

  29. September 18, 2002 Ahmed Elgamal u1 = 1 u2 = u3 = 0 L L

  30. September 18, 2002 Ahmed Elgamal u2 = 1 u1 = u3 = 0

  31. September 18, 2002 Ahmed Elgamal u3 = 1 u1 = u2 = 0

  32. September 18, 2002 Ahmed Elgamal If frame is subjected to lateral force fs Then (for simplicity, let Ic = Ib = I)

  33. September 18, 2002 Ahmed Elgamal Static condensation: From 2nd and 3rd equations, Note matrix inverse:

  34. September 18, 2002 Ahmed Elgamal Substitute into 1st equation or (check this result)

  35. September 18, 2002 Ahmed Elgamal Draft Example 2 u3 u2 u1 L 2L

  36. September 18, 2002 Ahmed Elgamal u1 = 1 u2 = u3 = 0

  37. September 18, 2002 Ahmed Elgamal u2 = 1 u1 = u3 = 0

  38. September 18, 2002 Ahmed Elgamal u3 = 1 u1 = u2 = 0

  39. September 18, 2002 Ahmed Elgamal For simplicity, let Ib = Ic

  40. September 18, 2002 Ahmed Elgamal Static condensation:

  41. September 18, 2002 Ahmed Elgamal Substitute in 1st Equation or, Same as in Example 1.1, Dynamics of Structures by Chopra or,

  42. September 18, 2002 Ahmed Elgamal Homework 1) 1.1 Derive stiffness matrix for EIb EIc1 h1 EIc2 h2 L 1.2 For the special case of Ic1 = Ic2 = Ib, h1 = h2 = h and L = 2h, find lateral stiffness k of the frame.

  43. September 18, 2002 Ahmed Elgamal Homework 2) Derive equation of motion for: EIb Flexural rigidity of beams and columns EIc EIc h E = 29,000 ksi, m EIb Columns W8x24 sections use 600 lb/ft with Ic = 82.4 in4 EIc h EIc h = 12 ft Ib = ½ Ic 2h

  44. September 18, 2002 Ahmed Elgamal Homework(Optional) 3) Derive lateral k of system (need to use computer to invert 3x3 matrix) Ib Ib h = 12 ft Ic Ic Ic 24 ft 24 ft E = 29,000 ksi, Ic = 82.4 in4 W8x24 sections Ib = ½ Ic

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