Finite element method - Chapter 4: Development of beam equations

ppt 44 trang Gia Huy 25/05/2022 2610
Bạn đang xem 20 trang mẫu của tài liệu "Finite element method - Chapter 4: Development of beam equations", để tải tài liệu gốc về máy bạn click vào nút DOWNLOAD ở trên

Tài liệu đính kèm:

  • pptfinite_element_method_chapter_4_development_of_beam_equation.ppt

Nội dung text: Finite element method - Chapter 4: Development of beam equations

  1. Ministry of Industry & Trade Industrial University of HCM City Chapter 4: DEVELOPMENT OF BEAM EQUATIONS
  2. 4.1 Beam stiffness A beam is a long, slender structural member generally subjected to transverse loading that produces significant bending effects as opposed to twisting or axial effects. This bending deformation is measured as a transverse displacement and a rotation 1
  3. 4.1 Beam stiffness At all nodes, the following sign conventions are used: 1. Moments are positive in the counterclockwise direction. 2. Rotations are positive in the counterclockwise direction. 3. Forces are positive in the positive y direction. 4. Displacements are positive in the positive y direction. 2
  4. 4.1 Beam stiffness 3
  5. 4.1 Beam stiffness Euler-Bernoulli Beam Theory Consider the beam shown in figure subjected to a distributed loading w(x) (force/length). 4
  6. 4.1 Beam stiffness The force equilibrium of a differential element of the beam 5
  7. 4.1 Beam stiffness The moment equilibrium of a differential element of the beam 6
  8. 4.1 Beam stiffness is the radius of the deflected curve shown in figure The curvature  of the beam is related to the moment by E is the modulus of elasticity I is the principal moment of inertia about the z axis 7
  9. 4.1 Beam stiffness The curvature for small slopes =dv/dx is given by The curvature  rewritten: Relation displacement and distributed loading 8
  10. 4.1 Beam stiffness Relation displacement and distributed loading For constant EI and only nodal forces and moments Step 1: Select the Element Type 9
  11. 4.1 Beam stiffness Step 2: Select a Displacement Function There are four total degrees of freedom We express  as a function of the nodal degrees of freedom 1, 2, 1, 2 and as follows 10
  12. 4.1 Beam stiffness  rewritten: In matrix form, we express equation as: N called the shape functions for a beam element 11
  13. 4.1 Beam stiffness d called the displacement for a beam element Where 12
  14. 4.1 Beam stiffness Step 3: Define the Strain/Displacement and Stress/Strain Relationships Assume the following axial strain/displacement relationship to be valid 13
  15. 4.1 Beam stiffness We relate the axial displacement to the transverse displacement by 14
  16. 4.1 Beam stiffness The axial strain rewitten: Also using Hooke’s law, we obtain the beam flexure or bending stress formula as The bending moment and shear force are related to the transverse displacement function 15
  17. 4.1 Beam stiffness Step 4: Derive the Element Stiffness Matrix and Equations We now relate the nodal and beam theory sign conventions for shear forces and bending moments 16
  18. 4.1 Beam stiffness In matrix form, Equation relate the nodal forces to the nodal displacements of beam element become: Where the stiffness matrix is then 17
  19. 4.1 Beam stiffness Step 5: Assemble the Element Equations to Obtain the Global Equations and Introduce Boundary Conditions The same chapter 2, 3 Assemblage of beam element stiffness matrices? 18
  20. 4.1 Beam stiffness The global stiffness matrices for the two elements: 19
  21. 4.1 Beam stiffness The governing equations for the beam are thus given by The boundary conditions: 20
  22. 4.1 Beam stiffness After applying conditions boudary, the governing equations for the beam: 21
  23. Example Determine the nodal displacements and rotations, global nodal forces, and element forces for the beam shown in figure. E=30x106 psi, I= 500 in4 22
  24. Solution We must have consistent units: 10ft = 120in We must have 4 beam element and 5 node The stiffness matrix of beam elements are same 12 6LL− 12 6 22 EI 6LLLL 4− 6 4 [][][][]k(1)= k (2) = k (3) = k (4) = e e e e L3 −12 − 6LL 12 − 6 22 6LLLL 4− 6 4 23
  25. We obtain the global stiffness matrix and the global equations as given in Eq 24
  26. Applying of the boundary conditions The resulting equation is We obtain: 25
  27. We can now back-substitute the results to determine the global nodal forces as 26
  28. All nodal displacements have now been determined Element 1: Element 2, 3, 4 (homeworks)? 27
  29. Homeworks Determine the nodal displacements and rotations and the global and element forces for the beam shown in figure. E=210Gpa, I=2x10-4 m4 28
  30. Determine the nodal displacements and rotations and the global and element forces for the beam shown in figure. E=210Gpa, I=2x10-4 m4 29
  31. 4.2 Distributed Loading Fixed-end reactions for the beam 32
  32. 4.2 Distributed Loading Beam subjected to a uniformly distributed loading The equivalent nodal forces to be determined 33
  33. 4.2 Distributed Loading The nodal moments and forces m1, m2, f1y, f2y T [][]F0= f 1yy m 1 f 2 m 2 34
  34. 4.2 Distributed Loading The formulation application for a general structure 35
  35. 4.2 Distributed Loading The formulation application for a general structure with distributed loading only in this section 36
  36. Example For the cantilever beam subjected to the uniform load w in figure, solve for the right-end vertical displacement and rotation and then for the nodal forces. Assume the beam to have constant EI throughout its length 37
  37. Solution We have 1 beam element and 2 node The beam element stiffness matrix 38
  38. Applying the boundary conditions 39
  39. The global nodal forces. 40
  40. The correct global nodal forces as 41
  41. Homeworks For the cantilever beam subjected to the uniform load w in figure, solve for the right-end vertical displacement and rotation and then for the nodal forces. Assume the beam to have constant EI throughout its length 42