Ahmed Elgamal. Stiffness Coefficients for a Flexural Element

Transcription

1 September 18, 00 Stiffness Coefficients for a Flexural Element 1

2 September 18, 00 Stiffness coefficients for a flexural element (neglecting axial deformations), Appendix 1, Ch. 1 Dynamics of Structures by Chopra. u u 4 u u 1 Positive directions degrees of freedom 41 1

3 September 18, 00 To obtain coefficients in 1 st column of stiffness matrix, move u 1 = 1, u = u = u 4 = 0, and find forces and moments needed to maintain this shape. u 1 1.0

4 September 18, 00 These are (see structures textboo) 6EI 1EI 6EI 1EI Note that Σ Forces = 0 Σ Moments = 0 1EI Σ M = 1EI - = 0 i.e. remember 1EI, and you can find other forces & moments Positive directions

5 September 18, 00 = ij =, where i is row number and j is column number = EI

6 September 18, 00 u = 1, u 1 = u = u 4 = u 6

7 September 18, 00 1EI 6EI u = 1 6EI Σ M = 0, Σ F = 0 1EI = EI Positive directions 1 4 7

8 September 18, 00 u = 1, u 1 = u = u 4 = u = 1 8

9 September 18, 00 EI 4EI 6EI Σ M = 0, Σ F = 0 6EI = - 6EI 6EI 4EI EI Positive directions 1 4 9

10 September 18, 00 u 4 = 1, u 1 = u = u = 0 u 4 =

11 September 18, 00 4EI EI 6EI Σ M = 0, Σ F = 0 6EI = - 6EI 6EI EI 4EI Positive directions

12 September 18, 00 Example: Water Tan u 1 u m m is lumped at a point & does not contribute in rotation u g u above was u in the earlier section of these notes 1

13 September 18, 00 Example: Water Tan (continued) u 1 u m u g m && u 0 && u 1 1EI + 6EI 6EI u 4EI u 1 m = && u 0 g Note Symmetry Rotational (used to be u ) 1

14 September 18, 00 Example: Water Tan (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 nd equation gives or 6EI 4EI u1 + u = 0 6EI 6 u = 4EI 4 u 1 = u1 u1 = * 14

15 September 18, 00 Example: Water Tan (continued) Substitute * into Equation 1 or, or, m & u m & u m & u 1EI + 6EI & 1 u1 mu g 4EI 18EI + 1 u1 mu g EI & 1 + u1 = mu g = = & Now, solve for u 1 and u can be evaluated from Equation * above. Static condensation can be applied to large MDOF systems of equations, the same way as shown above. 15

16 September 18, 00 Example: Water Tan (continued) m & u EI & 1 + u1 = mu g of water tan as we were given earlier. or of column if EI b = 0 I b 16

17 September 18, 00 Mandatory Reading Example 9.4 page 6-64 Example 9.8 page bending beam Sample Exercises: 9.5, 9.8, &

18 September 18, 00 Example (a) (b) Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN

19 September 18, 00 (c) (d) Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN

20 September 18, 00 (e) (f) Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN

21 September 18, 00 (g) (h) Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN

22 September 18, 00 (i) (j) Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN

23 September 18, 00 therefore, = EI , M = m 1 m 0 0 Sample Exercise: For the above cantilever system, write equation of motion and perform static condensation to obtain a DOF system. Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN

24 September 18, 00 Column Stiffness (lateral vibration) bending beam m EI =, ω = m beam with EI beam = 0 EI = 4

25 September 18, 00 u Case of EI b = (rigid roof) = 1EI 1EI = h 1 h 1E I 1 1 = + h1 1E h I 5

26 September 18, 00 h = h (See example 1.1 in Dynamics of Structures by Chopra) 96EIc = if EI b = EIc 7h beam column = 4EI h c 1ρ + 1 1ρ + 4, Ib ρ = & E c = E b = E 4I c 6

27 September 18, 00 Obtained by static condensation of x system u 1 force θ θ 1 f s call it u call it u u u u 1 = fs 0 0 Use to represent u and u in terms of u 1 & plug bac into and get f s = u 1 Technique can also be used for large systems of equations (See example 1.1 in Dynamics of Structures by Chopra) 7

28 September 18, 00 Draft Example Neglect axial deformation u u I b u 1 I c I c 8

29 September 18, 00 u 1 = 1 u = u = 0 6EI c 6EI c 1EI c ( ) 9

30 September 18, 00 u = 1 u 1 = u = 0 EI b 4EI b 4EI c 6EI c 0

31 September 18, 00 u = 1 u 1 = u = 0 6EI c 4EI b 4EI c EI b = E 4Ic 6Ic 6Ic 4 ( I + I ) b 6I I b c c 4 6I I b c ( ) Ib + Ic 1

32 September 18, 00 If frame is subjected to lateral force f s Then (for simplicity, let I c = I b = I) EI u u 8 u 1 = fs 0 0

33 September 18, 00 Static condensation: From nd and rd equations, u u 8 = = u u Note matrix inverse: a c b d 1 1 d = ad cb c b a 6 = u 1

34 September 18, 00 Substitute into 1 st equation EI EI 10 4 u 1 = fs = u 1 or = 168EI 10 (chec this result) 4

35 September 18, 00 Draft Example u u u 1 5

36 September 18, 00 u 1 = 1 u = u = 0 6EI c 6EI c 1EI c ( ) 6

37 September 18, 00 u = 1 u 1 = u = 0 EI b 4EI b 4EI c 6EI c 7

38 September 18, 00 u = 1 u 1 = u = 0 6EI c 4EI b 4EI c EI b 8

39 9 September 18, = c b b c b c b c c c c I I 4 I 6I I I I 4 6I 6I 6I 4I E = 0 0 f u u u EI s 1 For simplicity, let I b = I c

40 40 September 18, 00 Static condensation: 1 1 u u u = u = 1 u u = =

41 September 18, 00 Substitute in 1 st Equation EI u 1 = fs or, f = s 96 7 EI u 1 96EI or, = Same as in Example 1.1, 7 Dynamics of Structures by Chopra 41

42 September 18, 00 Sample Exercise 1) 1.1 Derive stiffness matrix for EI b h 1 EI c1 EI c h 1. For the special case of I c1 = I c = I b, h 1 = h = h and = h, find lateral stiffness of the frame. 4

43 September 18, 00 Sample Exercise ) Derive equation of motion for: m EI b Flexural rigidity of beams and columns h h EI c EI c m EI b EI c EI c use 600 lb/ft E = 9,000 si, Columns W8x4 sections with I c = 8.4 in 4 h = 1 ft h I b = ½ I c 4

44 September 18, 00 Sample Exercise (Optional) ) Derive lateral of system (need to use computer to invert x matrix) I b I b h = 1 ft I c I c I c 4 ft 4 ft E = 9,000 si, I c = 8.4 in 4 W8x4 sections I b = ½ I c 44