1 / 40

IN THE NAME OF ALLAH CHAPTER 8 Beams and Frames

IN THE NAME OF ALLAH CHAPTER 8 Beams and Frames. 1/40. 8.1 : introduction Here, Beams with cross section that are symmetric with respect to plane of loading are considered. In this part we don’t have axial load. . 2/40.

rupert
Télécharger la présentation

IN THE NAME OF ALLAH CHAPTER 8 Beams and Frames

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. IN THE NAME OF ALLAHCHAPTER 8Beams and Frames 1/40

  2. 8.1 : introduction Here, Beams with cross section that are symmetric with respect to plane of loading are considered. In this part we don’t have axial load. 2/40

  3. For small deflection, we recal from elementery beams theory that: 3/40

  4. Beam Element For the node i : Q2i-1 :transverse displacement (υi) Q2i : slope or rotation (θi) ζ υi θi υj θj q = [υiθi υjθj ]T 4/40

  5. Hermite shape functions Since nodal values and nodal slop are involved we define Hermite shape functions which satisfy nodal value and slope continuity requirement. Hi(ζ) = ai + biζ + ciζ2 + diζ3 i=1,2,3,4 υ1×1 = N1×4 q4×1 deflection H = [ H1 H2 H3 H4 ] slope 5/40

  6. The condition given in the following table must be satisfied: The coefficient ai bicidican be easily obtained by imposing the above conditions. 6/40

  7. Thus: H1 = ¼ (1-ζ)2(2+ ζ) = ¼ (2-3 ζ +ζ3) H2 = ¼ (1-ζ)2(ζ+1) = ¼ (1-ζ-ζ2+ζ3) H3 = ¼ (1-ζ)2(2-ζ) = ¼ (2+3ζ-ζ3) H4 = ¼ (1+ζ)2(ζ-1) = ¼ (-1-ζ+ζ2+ζ3) The Hermit shape function can be used to write υin the form: υ(ζ) = H1υ1 + H2 + H3υ3 + H4 7/40

  8. The chain rule gives us: thus: υ = H1υ1 + H2θ1 + H3υ2 + H4θ2 We know that υ1and θ1 evaluated at node 1 are q1 and q2. We know that υ2and θ2 evaluated at node 2 are q3 and q4. 8/40

  9. Thus: Which may be denoted as: υ = N.q Thus: 9/40

  10. Stiffness Matrix Then substituting υ = N.q , we obtain: 11/40

  11. Substituting and we obtain: 12/40

  12. Each term in the matrix needs to be integrated: 13/40

  13. Point Load • Point load P on node m F2m-1 =P • Couple M on node k F2k =M

  14. Force Vector for Distributed load w(x) 14/40

  15. 15/40

  16. Example 8.1 For rhe beam and loading shown in figure determine the slopes at 2 and 3 and the vertical deflection at the midpoint of the distributed load. 16/40

  17. 18/40

  18. KQ =F Use elimination approach: 19/40

  19. Slope at 2: θ2 = Q4 = -2.679 × 10-4 rad Slope at 3: θ3 = Q6 = 4.464 × 10-4 rad To get vertical deflection at the midpoint of the element, use υ=Nqatζ=0 for element 2: = 0 + (½) (¼) (-2.679× 10-4) + 0 + (½) (¼) (-2.679× 10-4) υ = -8.93 × 10-5 20/40

  20. Example: K(1) , K(2) , K(3) , K(4) 21/40

  21. K . Boundary condition: q1 = q9 = 0 use penalty or Elimination approach for solve. 22/40

  22. stress in midpoint of element 2 23/40

  23. Beam on Elastic Point Supports 1. Single-row ball bearing can be considered as a linear spring on a node n and then adding the linear spring stiffness (bearing stiffness) KL to the diagonal location K(2n-1) (2n-1) . 2. Rotational (moment) stiffness can be considered for roller bearings and journal bearings as a rotational spring on node m and then adding the rotational spring stiffness (bearing stiffness) Kθto the diagonal location K(2m) (2m) . 24/40

  24. Beam on Elastic Supports • 3. In wide journal bearins and winkler foundation we use stiffness per unit lengthKsof the supporting medium. 25/40

  25. For elements supported on an elastic foundation this stiffness Ks has to be added to the element stiffness K . 26/40

  26. Frame Element ν'j X’j Θ’j ν'i Θ’i X’i

  27. νj νj X’j X’j Θ’j Θ’j νi + = νi X’i yj Θ’i Θ’i x'i Xj θj yi Xi θi 27/40

  28. Plane frames We consider here plane structures With rigidly connected members. This members will be similar to the beams except that axial loads and axial deformations are present. We have two displacements and a rotationaldeformation for each node. 27/40

  29. Recording that qʹ3 = q3and qʹ6 = q6which rotations with respect to the body we obtain the local-global transformation qʹ = L.q Where 29/40

  30. qʹ2 , qʹ3 , qʹ5 ,qʹ6:the beam degrees of freedom qʹ1 ,qʹ4: the truss element degrees of freedom 30/40

  31. Combining the two stiffnesses and arranging in proper locations, we get the element stiffness for a frame element: KʹFrame = Kʹonedimentional + KʹBeam 31/40

  32. Recognize that the element strain energy is given by: If there is distributed load on a member, we have: Where F = LT Fʹ 32/40

  33. The nodal loads due to the distributed load P are given by: F = LT Fʹ The value of F are added to the global load vector. Note here that positive P is in the yʹ direction. The point loads and couples are simply added to the global load vector. On gathering stiffnesses and loads we get the system of equations: KQ = F 33/40

  34. Example: for the portal frame shown in this figure Determine the displacements of the joints 34/40

  35. Element 1: Using noting that K1 = Kʹ1 , we find: Element 2 and 3: we get element stiffnesses for elements 2 and 3: 35/40

  36. For elements 2 and 3 which are oriented similarly with respect to the x , y axes we have l=0 , m=1 then: Noting that K2 = LT Kʹ2 L , we get: 36/40

  37. The global stiffness matrix is given by: 37/40

  38. From figure the load vector can easily be written as: The set of equations is given by: KQ=F On solving we get: 38/40

  39. Axial displacements of one dimensional element (Truss) Nodal displacements of beam element 39/40

  40. Stress in midpoint of element: 40/40

More Related