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### Lecture 4.pdf

• 1. CENG6504: Finite Element Methods Beam Member Dr. Tesfaye Alemu 1
• 2. Learning Objectives Development of Beam Equations To review the basic concepts of beam bending To derive the stiffness matrix for a beam element To demonstrate beam analysis using the direct stiffness method 2 To compare the finite element solution to an exact solution for a beam To show how the potential energy method can be used to derive the beam element equations To show how the potential energy method can be used to derive the beam element equations
• 3. Development of Beam Equations In this section, we will develop the stiffness matrix for a beam element, the most common of all structural elements. The beam element is considered to be straight and to have constant cross-sectional area. 3
• 4. Development of Beam Equations We will derive the beam element stiffness matrix by using the principles of simple beam theory. The degrees of freedom associated with a node of a beam element are a transverse displacement and a rotation. 4
• 5. Development of Beam Equations We will discuss procedures for handling distributed loading and concentrated nodal loading. We will include the nodal shear forces and bending moments and the resulting shear force and bending moment diagrams as part of the total solution. 5
• 6. Development of Beam Equations We will develop the beam bending element equations using the potential energy approach. Finally, the Galerkin residual method is applied to derive the beam element equations 6
• 7. Beam Stiffness Consider the beam element shown below. The beam is of length L with axial local coordinate x and transverse local coordinate y. The local transverse nodal displacements are given by vi and the rotations by ϕi. The local nodal forces are given by fiy and the bending moments by mi. 7
• 8. 3. Forces are positive in the positive y direction. Beam Stiffness At all nodes, the following sign conventions are used on the element level: 1. Moments are positive in the counterclockwise direction. 2. Rotations are positive in the counterclockwise direction. 4. Displacements are positive in the positive y direction. 8
• 9. At all nodes, the following sign conventions are used on the global level: 1. Bending moments m are positive if they cause the beam to bend concave up. 2. Shear forces V are positive is the cause the beam to rotate clockwise. Beam Stiffness 9
• 10. Beam Stiffness (+) Bending Moment (-) Bending Moment 10
• 11. Beam Stiffness (+) Shear Force (-) Shear Force 11
• 12. Beam Stiffness The differential equation governing simple linear-elastic beam behavior can be derived as follows. Consider the beam shown below. 12
• 13. Beam Stiffness The differential equation governing simple linear-elastic beam behavior can be derived as follows. Consider the beam shown below. Write the equations of equilibrium for the differential element: 0 right side M      0 y F     ( ) w x dx 2 dx         ( ) 2 dx M M dM Vdx w x dx             ( ) ( ) V V dV w x dx     0 13
• 14. Beam Stiffness From force and moment equilibrium of a differential beam element, we get: 0 0 right side M Vdx dM        0 0 y F wdx dV       d dM w dx dx         or dM V dx  or dV w dx   14
• 15. Beam Stiffness The curvature  of the beam is related to the moment by: 1 M EI     where  is the radius of the deflected curve, v is the transverse displacement function in the y direction, E is the modulus of elasticity, and I is the principle moment of inertia about y direction, as shown below. 15
• 16. Beam Stiffness The curvature,  for small slopes is given as: dv dx   2 2 d v dx   Therefore: 2 2 2 2 d v M d v M EI dx EI dx    Substituting the moment expression into the moment-load equations gives:   2 2 2 2 d d v EI w x dx dx         For constant values of EI, the above equation reduces to:   4 4 d v EI w x dx         16
• 17. Beam Stiffness Step 1 - Select Element Type We will consider the linear-elastic beam element shown below. 17
• 18. Beam Stiffness Step 2 - Select a Displacement Function Assume the transverse displacement function v is: 3 2 1 2 3 4 v a x a x a x a     The number of coefficients in the displacement function ai is equal to the total number of degrees of freedom associated with the element (displacement and rotation at each node). The boundary conditions are: 1 ( 0) v x v   1 0 x dv dx    2 x L dv dx    2 ( ) v x L v   18
• 19. Beam Stiffness Step 2 - Select a Displacement Function Applying the boundary conditions and solving for the unknown coefficients gives: Solving these equations for a1, a2, a3, and a4 gives: 1 4 (0) v v a   3 2 2 1 2 3 4 ( ) v L v a L a L a L a      1 3 (0) dv a dx    2 2 1 2 3 ( ) 3 2 dv L a L a L a dx          3 1 2 1 2 3 2 2 1 v v v x L L                 2 1 2 1 2 1 1 2 3 1 2 v v x x v L L                 19
• 20. Beam Stiffness Step 2 - Select a Displacement Function In matrix form the above equations are: where   [ ] v N d      1 1 1 2 3 4 2 2 [ ] v d N N N N N v                       3 2 3 3 2 2 3 1 2 3 3 1 1 2 3 2 N x x L L N x L x L xL L L           3 2 3 2 2 3 4 3 3 1 1 2 3 N x x L N x L x L L L      20
• 21. Beam Stiffness Step 2 - Select a Displacement Function N1, N2, N3, and N4 are called the interpolation functions for a beam element. -0.200 0.000 0.200 0.400 0.600 0.800 1.000 0.00 1.00 N1 -0.200 0.000 0.200 0.400 0.600 0.800 1.000 0.00 1.00 N2 -0.200 0.000 0.200 0.400 0.600 0.800 1.000 0.00 1.00 N3 -0.200 0.000 0.200 0.400 0.600 0.800 1.000 0.00 1.00 N4 21
• 22. Beam Stiffness Step 3 - Define the Strain/Displacement and Stress/Strain Relationships The stress-displacement relationship is: We can relate the axial displacement to the transverse displacement by considering the beam element shown below: where u is the axial displacement function.   , x du x y dx   22
• 23. Beam Stiffness Step 3 - Define the Strain/Displacement and Stress/Strain Relationships dv u y dx   23
• 24. Beam Stiffness Step 3 - Define the Strain/Displacement and Stress/Strain Relationships One of the basic assumptions in simple beam theory is that planes remain planar after deformation, therefore: Moments and shears are related to the transverse displacement as:   , x du x y dx     2 2 d v m x EI dx          3 3 d v V x EI dx        2 2 d v y dx         24
• 25. Beam Stiffness Step 4 - Derive the Element Stiffness Matrix and Equations Use beam theory sign convention for shear force and bending moment. V+ V+ M+ M+ 25
• 26. Beam Stiffness Step 4 - Derive the Element Stiffness Matrix and Equations Using beam theory sign convention for shear force and bending moment we obtain the following equations:   3 1 1 1 2 2 3 3 0 12 6 12 6 y x d v EI f V EI v L v L dx L            3 2 1 1 2 2 3 3 12 6 12 6 y x L d v EI f V EI v L v L dx L               2 2 2 1 1 1 2 2 2 3 0 6 4 6 2 x d v EI m m EI Lv L Lv L dx L              2 2 2 2 1 1 2 2 2 3 6 2 6 4 x L d v EI m m EI Lv L Lv L dx L          26
• 27. Beam Stiffness Step 4 - Derive the Element Stiffness Matrix and Equations In matrix form the above equations are: 1 1 2 2 1 1 3 2 2 2 2 2 2 12 6 12 6 6 4 6 2 12 6 12 6 6 2 6 4 y y f v L L m L L L L EI f v L L L m L L L L                                                  where the stiffness matrix is: 2 2 3 2 2 12 6 12 6 6 4 6 2 12 6 12 6 6 2 6 4 L L L L L L EI L L L L L L L                    k 1 1 1 1 2 2 2 2 y y f v m k f v m                                27
• 28. Beam Stiffness Step 5 - Assemble the Element Equations and Introduce Boundary Conditions Assume the EI to be constant throughout the beam. A force of 1,000 lb and moment of 1,000 lb-ft are applied to the mid- point of the beam. Consider a beam modeled by two beam elements 28
• 29. Beam Stiffness Step 5 - Assemble the Element Equations and Introduce Boundary Conditions The beam element stiffness matrices are: 1 1 2 2 2 2 (1) 3 2 2 12 6 12 6 6 4 6 2 12 6 12 6 6 2 6 4 v v L L L L L L EI L L L L L L L                      k 2 2 3 3 2 2 (2) 3 2 2 12 6 12 6 6 4 6 2 12 6 12 6 6 2 6 4 v v L L L L L L EI L L L L L L L                      k 29
• 30. 1 1 2 2 1 1 2 2 3 2 2 2 2 2 2 3 3 2 2 3 3 12 6 12 6 0 0 6 4 6 2 0 0 12 6 12 12 6 6 12 6 6 2 6 6 4 4 6 2 0 0 12 6 12 6 0 0 6 2 6 4 y y y F v L L M L L L L F v L L L L EI M L L L L L L L L L F v L L L L L L M                                                                               Beam Stiffness Step 5 - Assemble the Element Equations and Introduce Boundary Conditions In this example, the local coordinates coincide with the global coordinates of the whole beam (therefore there is no transformation required for this problem). The total stiffness matrix can be assembled as: Element 1 Element 2 30
• 31. Beam Stiffness Step 5 - Assemble the Element Equations and Introduce Boundary Conditions The boundary conditions are: 1 1 3 0 v v     1 1 2 2 1 1 2 2 3 2 2 2 2 2 2 3 3 2 2 3 3 12 6 12 6 0 0 6 4 6 2 0 0 12 6 12 12 6 6 12 6 6 2 6 6 4 4 6 2 0 0 12 6 12 6 0 0 6 2 6 4 y y y F v L L M L L L L F v L L L L EI M L L L L L L L L L F v L L L L L L M                                                                               2 2 3 0 0 0 v                     31
• 32. Beam Stiffness Step 5 - Assemble the Element Equations and Introduce Boundary Conditions By applying the boundary conditions the beam equations reduce to: 2 2 2 2 3 2 2 3 1,000 24 0 6 1,000 0 8 2 0 6 2 4 lb L v EI lbft L L L L L L                                   32
• 33. Beam Stiffness Step 6 - Solve for the Unknown Degrees of Freedom Solving the above equations gives: 2 2 3 3 2 2 2 875 375 125 625 125 125 12 4 L L L L L L v in rad rad EI EI EI           Step 7 - Solve for the Element Strains and Stresses   2 2 d v m x EI dx        1 2 1 2 2 2 v d N EI v dx                      The second derivative of N is linear; therefore m(x) is linear. 33
• 34. Beam Stiffness Step 6 - Solve for the Unknown Degrees of Freedom Solving the above equations gives: 2 2 3 3 2 2 2 875 375 125 625 125 125 12 4 L L L L L L v in rad rad EI EI EI            Step 7 - Solve for the Element Strains and Stresses   3 3 d v V x EI dx        1 3 1 2 2 2 v d N EI v dx                      The third derivative of N is a constant; therefore V(x) is constant. 34
• 35. Beam Stiffness   2 2 d v m x EI dx        1 2 1 2 2 2 v d N EI v dx                      Step 7 - Solve for the Element Strains and Stresses   2 2 1 1 1 2 2 3 6 4 6 2 EI m Lv L Lv L L         2 2 2 1 1 2 2 3 6 2 6 4 EI m Lv L Lv L L       Assume L = 120 in, E = 29x106 psi, and I = 100 in4: Element #1: 3,875lb ft   3,562.5lb ft   5 4 2 2 3 0.0433 7.758 10 5.586 10 v in rad rad            35
• 36. Beam Stiffness   2 2 d v m x EI dx        1 2 1 2 2 2 v d N EI v dx                      Step 7 - Solve for the Element Strains and Stresses   2 2 2 2 2 3 3 3 6 4 6 2 EI m Lv L Lv L L         2 2 3 2 2 3 3 3 6 2 6 4 EI m Lv L Lv L L       Assume L = 120 in, E = 29x106 psi, and I = 100 in4: Element #2: 2,562.5lb ft    0  5 4 2 2 3 0.0433 7.758 10 5.586 10 v in rad rad            36
• 37. Beam Stiffness Step 7 - Solve for the Element Strains and Stresses Assume L = 120 in, E = 29x106 psi, and I = 100 in4: Element #1: 743.75lb  743.75lb   5 4 2 2 3 0.0433 7.758 10 5.586 10 v in rad rad              3 3 d v V x EI dx        1 3 1 2 2 2 v d N EI v dx                        1 1 1 2 2 3 12 6 12 6 y EI f v L v L L         2 1 1 2 2 3 12 6 12 6 y EI f v L v L L        37
• 38. Beam Stiffness Step 7 - Solve for the Element Strains and Stresses Assume L = 120 in, E = 29x106 psi, and I = 100 in4: Element #2: 256.25lb   256.25lb  5 4 2 2 3 0.0433 7.758 10 5.586 10 v in rad rad              3 3 d v V x EI dx        1 3 1 2 2 2 v d N EI v dx                        2 2 2 3 3 3 12 6 12 6 y EI f v L v L L         3 2 2 3 3 3 12 6 12 6 y EI f v L v L L        38
• 39. Beam Stiffness Step 7 - Solve for the Element Strains and Stresses 3,875lb ft   3,562.5lb ft  2,562.5lb ft  743.75lb 256.25lb  1,000 F lb    1,000 M lb ft     39
• 40. Beam Stiffness Example 1 - Beam Problem Consider the beam shown below. Assume that EI is constant and the length is 2L (no shear deformation). The beam element stiffness matrices are: 1 1 2 2 2 2 (1) 3 2 2 12 6 12 6 6 4 6 2 12 6 12 6 6 2 6 4 v v L L L L L L EI L L L L L L L                      k 2 2 3 3 2 2 (2) 3 2 2 12 6 12 6 6 4 6 2 12 6 12 6 6 2 6 4 v v L L L L L L EI L L L L L L L                      k 40
• 41. Beam Stiffness Example 1 - Beam Problem The local coordinates coincide with the global coordinates of the whole beam (therefore there is no transformation required for this problem). The total stiffness matrix can be assembled as: 1 1 2 2 1 1 2 2 3 2 2 2 2 2 3 3 2 2 3 3 12 6 12 6 0 0 6 4 6 2 0 0 12 6 24 0 12 6 6 2 0 8 6 2 0 0 12 6 12 6 0 0 6 2 6 4 y y y F v L L M L L L L F v L L EI M L L L L L L F v L L L L L L M                                                                         Element 1 Element 2 41
• 42. Beam Stiffness Example 1 - Beam Problem The boundary conditions are: 2 3 3 0 v v     1 1 2 2 1 1 2 2 3 2 2 2 2 2 3 3 2 2 3 3 12 6 12 6 0 0 6 4 6 2 0 0 12 6 24 0 12 6 6 2 0 8 6 2 0 0 12 6 12 6 0 0 6 2 6 4 y y y F v L L M L L L L F v L L EI M L L L L L L F v L L L L L L M                                                                         1 1 2 0 0 0 v                     42
• 43. Beam Stiffness By applying the boundary conditions the beam equations reduce to: 1 2 2 1 3 2 2 2 12 6 6 0 6 4 2 0 6 2 8 P L L v EI L L L L L L L                                   Solving the above equations gives: 1 2 1 2 7 3 3 4 1 L v PL EI                                 43
• 44. Beam Stiffness Example 1 - Beam Problem The positive signs for the rotations indicate that both are in the counterclockwise direction. The negative sign on the displacement indicates a deformation in the -y direction. 7 1 3 2 2 1 2 2 2 2 2 3 2 2 3 12 6 12 6 0 0 6 4 6 2 0 0 3 12 6 24 0 12 6 0 4 6 2 0 8 6 2 1 0 0 12 6 12 6 0 0 0 6 2 6 4 0 L y y y F L L M L L L L F L L P M L L L L L L F L L L L L L M                                                                       44
• 45. Beam Stiffness Example 1 - Beam Problem The local nodal forces for element 1: 7 3 1 2 2 1 2 2 2 2 12 6 12 6 6 4 6 2 3 0 4 12 6 12 6 0 6 2 6 4 1 L y y f L L P m L L L L P f L L L P m L L L L PL                                                                  The local nodal forces for element 2: 2 2 2 2 3 2 2 3 12 6 12 6 0 1.5 6 4 6 2 1 4 12 6 12 6 0 1.5 6 2 6 4 0 0.5 y y f L L P m L L L L PL P f L L L P m L L L L PL                                                                45
• 46. Beam Stiffness Example 1 - Beam Problem The free-body diagrams for the each element are shown below. Combining the elements gives the forces and moments for the original beam. 46
• 47. Beam Stiffness Example 1 - Beam Problem Therefore, the shear force and bending moment diagrams are: 47
• 48. Beam Stiffness Example 2 - Beam Problem Consider the beam shown below. Assume E = 30 x 106 psi and I = 500 in4 are constant throughout the beam. Use four elements of equal length to model the beam. The beam element stiffness matrices are: 2 2 ( ) 3 2 2 ( 1) ( 1) 12 6 12 6 6 4 6 2 12 6 12 6 6 2 6 4 i v v i i i i L L L L L L EI L L L L L L L                        k 48
• 49. Beam Stiffness Example 2 - Beam Problem Using the direct stiffness method, the four beam element stiffness matrices are superimposed to produce the global stiffness matrix. Element 1 Element 2 Element 3 Element 4 49
• 50. Beam Stiffness Example 2 - Beam Problem The boundary conditions for this problem are: 1 1 3 5 5 0 v v v        50
• 51. Beam Stiffness Example 2 - Beam Problem The boundary conditions for this problem are: After applying the boundary conditions the global beam equations reduce to: 1 1 3 5 5 0 v v v        2 2 2 2 2 2 2 3 3 4 2 2 4 24 0 6 0 0 0 8 2 0 0 6 2 8 6 2 0 0 6 24 0 0 0 2 0 8 v L L L EI L L L L L L v L L L                                        10,000 0 0 10,000 0 lb lb                      51
• 52. Beam Stiffness Example 2 - Beam Problem Substituting L = 120 in, E = 30 x 106 psi, and I = 500 in4 into the above equations and solving for the unknowns gives: The global forces and moments can be determined as: 2 4 2 3 4 0.048 0 v v in          1 1 2 2 3 3 4 4 5 5 5 25 ꞏ 10 0 10 0 10 0 5 25 ꞏ y y y y y F kips M kips ft F kips M F kips M F kips M F kips M kips ft              52
• 53. Beam Stiffness Example 2 - Beam Problem The local nodal forces for element 1: The local nodal forces for element 2: 1 2 2 1 3 2 2 2 2 12 6 12 6 0 6 4 6 2 0 12 6 12 6 0.048 6 2 6 4 0 y y f L L m L L L L EI f L L L m L L L L                                                 2 2 2 2 3 3 2 2 3 12 6 12 6 0.048 6 4 6 2 0 12 6 12 6 0 6 2 6 4 0 y y f L L m L L L L EI f L L L m L L L L                                                 5 25 ꞏ 5 25 ꞏ kips k ft kips k ft                 5 25 ꞏ 5 25 ꞏ kips k ft kips k ft                   53
• 54. Beam Stiffness Example 2 - Beam Problem The local nodal forces for element 3: The local nodal forces for element 4: 3 2 2 3 3 4 2 2 4 12 6 12 6 0 6 4 6 2 0 12 6 12 6 0.048 6 2 6 4 0 y y f L L m L L L L EI f L L L m L L L L                                                 4 2 2 4 3 5 2 2 5 12 6 12 6 0.048 6 4 6 2 0 12 6 12 6 0 6 2 6 4 0 y y f L L m L L L L EI f L L L m L L L L                                                 5 25 ꞏ 5 25 ꞏ kips k ft kips k ft                 5 25 ꞏ 5 25 ꞏ kips k ft kips k ft                   54
• 55. Beam Stiffness Example 2 - Beam Problem Note: Due to symmetry about the vertical plane at node 3, we could have worked just half the beam, as shown below. Line of symmetry 55
• 56. Beam Stiffness Example 3 - Beam Problem Consider the beam shown below. Assume E = 210 GPa and I = 2 x 10-4 m4 are constant throughout the beam and the spring constant k = 200 kN/m. Use two beam elements of equal length and one spring element to model the structure. 56
• 57. Beam Stiffness Example 3 - Beam Problem The beam element stiffness matrices are: 2 2 (1) 3 2 2 1 1 2 2 12 6 12 6 6 4 6 2 12 6 12 6 6 2 6 4 v v L L L L L L EI L L L L L L L                      k 3 2 2 (2) 3 2 2 2 2 3 12 6 12 6 6 4 6 2 12 6 12 6 6 2 6 4 v v L L L L L L EI L L L L L L L                      k The spring element stiffness matrix is: 3 4 (3) v v k k k k          k 3 3 4 (3) 0 0 0 0 0 v v k k k k                k 57
• 58. Beam Stiffness Example 3 - Beam Problem Using the direct stiffness method and superposition gives the global beam equations. 1 1 2 2 1 1 2 2 2 2 2 2 2 3 3 3 2 2 3 3 4 4 12 6 12 6 0 0 0 6 4 6 2 0 0 0 12 6 24 0 12 6 0 6 2 0 8 6 2 0 0 0 12 6 12 ' 6 ' 0 0 6 2 6 4 0 0 0 0 0 ' 0 ' y y y y F v L L M L L L L F v L L EI M L L L L L L F v L k L k M L L L L F v k k                                                                                 3 ' kL k EI     Element 1 Element 2 Element 3 58
• 59. Beam Stiffness Example 3 - Beam Problem The boundary conditions for this problem are: 1 1 2 4 0 v v v      1 1 2 2 1 1 2 2 2 2 2 2 2 3 3 3 2 2 3 3 4 4 12 6 12 6 0 0 0 6 4 6 2 0 0 0 12 6 24 0 12 6 0 6 2 0 8 6 2 0 0 0 12 6 12 ' 6 ' 0 0 6 2 6 4 0 0 0 0 0 ' 0 ' y y y y F v L L M L L L L F v L L EI M L L L L L L F v L k L k M L L L L F v k k                                                                                 3 ' kL k EI     2 3 3 0 0 0 0 v                         59
• 60. Beam Stiffness Example 3 - Beam Problem After applying the boundary conditions the global beam equations reduce to: 2 2 2 2 3 3 3 2 2 3 3 8 6 2 0 6 12 ' 6 2 6 4 0 y M L L L EI F L k L v P L M L L L                                                   Solving the above equations gives: 2 2 3 3 3 3 2 3 1 12 7 ' 7 1 ' 12 7 ' 9 1 12 7 ' PL EI k PL kL v k EI k EI PL EI k                                                             60
• 61. Beam Stiffness Example 3 - Beam Problem Substituting L = 3 m, E = 210 GPa, I = 2 x 10-4 m4, and k = 200 kN/m in the above equations gives: Substituting the solution back into the global equations gives: 3 2 3 0.0174 0.00249 0.00747 v m rad rad         1 2 2 1 2 2 2 2 2 3 3 2 2 3 4 12 6 12 6 0 0 0 0 6 4 6 2 0 0 0 0 12 6 24 0 12 6 0 0 6 2 0 8 6 2 0 0.00249 0 0 12 6 12 ' 6 ' 0.0174 0 0 6 2 6 4 0 0.00747 0 0 0 0 ' 0 ' y y y y F L L M L L L L F L L EI M L L L L L rad L F L k L k m M L L L L rad F k k                                                              0                       61
• 62. Beam Stiffness Example 3 - Beam Problem Substituting L = 3 m, E = 210 GPa, I = 2 x 10-4 m4, and k = 200 kN/m in the above equations gives: Substituting the solution back into the global equations gives: 3 2 3 0.0174 0.00249 0.00747 v m rad rad         1 1 2 2 3 3 4 69.9 69.7 116.4 0 50 0 3.5 y y y y F kN M kN m F kN M F kN M F kN                                                  62
• 63. Beam Stiffness Example 3 - Beam Problem The variation of shear force and bending moment is: 69.7kNm 139.5kNm  69.9kN  46.5 kN 63
• 65. Beam Stiffness Distributed Loadings In general, fixed-end reactions are those reactions at the ends of an element if the ends of the element are assumed to be fixed (displacements and rotations are zero). Therefore, guided by the results from structural analysis for the case of a uniformly distributed load, we replace the load by concentrated nodal forces and moments tending to have the same effect on the beam as the actual distributed load. 65
• 66. Beam Stiffness Distributed Loadings The figures below illustrates the idea of equivalent nodal loads for a general beam. We can replace the effects of a uniform load by a set of nodal forces and moments. 66
• 67. Beam Stiffness General Formulation We can account for the distributed loads or concentrated loads acting on beam elements by considering the following formulation for a general structure: where F0 are the equivalent nodal forces, expressed in terms of the global-coordinate components. These forces would yield the same displacements as the original distributed load. If we assume that the global nodal forces are not present (F = 0) then: 0 F = Kd- F 0 F = Kd 67
• 68. Beam Stiffness General Formulation We now solve for the displacements, d, given the nodal forces F0. Next, substitute the displacements and the equivalent nodal forces F0 back into the original expression and solve for the global nodal forces. 0 F = Kd- F This concept can be applied on a local basis to obtain the local nodal forces in individual elements of structures as: 0 f = kd- f 68
• 69. Beam Stiffness Example 5 - Load Replacement Consider the beam shown below; determine the equivalent nodal forces for the given distributed load. The work equivalent nodal forces are shown above. Using the beam stiffness equations: 2 2 1 1 2 2 1 1 3 2 2 2 2 2 2 2 12 2 12 6 12 6 6 4 6 2 12 6 12 6 6 2 6 4 y y wL wL wL wL f v L L m L L L L EI f v L L L m L L L L                                                                          69
• 70. Beam Stiffness Example 5 - Load Replacement Apply the boundary conditions: 2 3 2 2 2 12 6 2 6 4 12 wL v L EI L L L wL                                4 2 3 2 8 6 wL v EI wL EI                         We can solve for the displacements 1 2 2 1 3 2 2 2 2 12 6 12 6 6 4 6 2 12 6 12 6 6 2 6 4 v L L L L L L EI v L L L L L L L                                   1 1 0 v    70
• 71. Beam Stiffness Example 5 - Load Replacement In this case, the method of equivalent nodal forces gives the exact solution for the displacements and rotations. To obtain the global nodal forces, we will first define the product of Kd to be Fe, where Fe is called the effective global nodal forces. Therefore: 4 3 1 2 2 1 3 8 2 2 2 6 2 12 6 12 6 0 6 4 6 2 0 12 6 12 6 6 2 6 4 e y e e wL EI y e wL EI L L F L L L L M EI L L L F L L L L M                                                  2 2 2 5 12 2 12 wL wL wL wL                         71
• 72. Beam Stiffness Example 5 - Load Replacement Using the above expression and the fix-end moments in: 0 F = Kd- F 2 2 0 0 wL wL                            2 2 1 1 2 2 2 2 2 2 5 12 12 2 2 12 12 y y wL wL F wL wL M F wL wL M wL wL                                                                 0 y F   1 0 M     1y F wL     2 2 2 wL L wL   72
• 73. Beam Stiffness Example 6 - Cantilever Beam Consider the beam, shown below, determine the vertical displacement and rotation at the free-end and the nodal forces, including reactions. Assume EI is constant throughout the beam. We will use one element and replace the concentrated load with the appropriate nodal forces. 73
• 74. Beam Stiffness Example 6 - Cantilever Beam The beam stiffness equations are: 1 1 2 2 1 1 3 2 2 2 2 2 2 2 8 2 8 12 6 12 6 6 4 6 2 12 6 12 6 6 2 6 4 y y P PL P PL f v L L m L L L L EI f v L L L m L L L L                                                                            Apply the boundary conditions: 1 2 2 1 3 2 2 2 2 12 6 12 6 6 4 6 2 12 6 12 6 6 2 6 4 v L L L L L L EI v L L L L L L L                                   1 1 0 v    74
• 75. Beam Stiffness Example 6 - Cantilever Beam The beam stiffness equations become: To obtain the global nodal forces, we begin by evaluating the effective nodal forces. 2 3 2 2 12 6 2 6 4 8 P v L EI PL L L L                                3 2 2 2 5 48 8 PL v EI PL EI                         3 2 1 2 2 1 3 5 48 2 2 2 8 2 12 6 12 6 0 6 4 6 2 0 12 6 12 6 6 2 6 4 e y e e PL EI y e PL EI L L F L L L L M EI L L L F L L L L M                                                  2 3 8 2 8 P PL P PL                         75
• 76. Beam Stiffness Example 6 - Cantilever Beam Using the above expression in the following equation, gives: 0 F = Kd- F 2 3 8 2 8 P PL P PL                         e Kd = F 0 F F 2 0 0 P PL                        2 8 2 8 P PL P PL                          1 1 2 2 y y F M F M               F 76
• 77. Beam Stiffness Example 6 - Cantilever Beam In general, for any structure in which an equivalent nodal force replacement is made, the actual nodal forces acting on the structure are determined by first evaluating the effective nodal forces Fe for the structure and then subtracting off the equivalent nodal forces F0 for the structure. Similarly, for any element of a structure in which equivalent nodal force replacement is made, the actual local nodal forces acting on the element are determined by first evaluating the effective local nodal forces for the element and then subtracting off the equivalent local nodal forces associated only with the element. 0 f = kd- f 77
• 78. Beam Stiffness Comparison of FE Solution to Exact Solution We will now compare the finite element solution to the exact classical beam theory solution for the cantilever beam shown below. Both one- and two-element finite element solutions will be presented and compared to the exact solution obtained by the direct double-integration method. Let E = 30 x 106 psi, I = 100 in4, L = 100 in, and uniform load w = 20 Ib/in. 78
• 79. Beam Stiffness Comparison of FE Solution to Exact Solution To obtain the solution from classical beam theory, we use the double-integration method: ( ) M x y EI         where the double prime superscript indicates differentiation twice with respect to x and M is expressed as a function of x by using a section of the beam as shown: 0 M     0 y F     2 2 ( ) 2 2 wx wL M x wLx      ( ) V x wL wx    79
• 80. Beam Stiffness Comparison of FE Solution to Exact Solution To obtain the solution from classical beam theory, we use the double-integration method: ( ) M x y EI         2 2 2 2 w x L y Lx dx dx EI            3 2 2 1 6 2 2 w x Lx xL y C dx EI             4 3 2 2 1 2 24 6 4 w x Lx x L y C x C EI             4 3 2 2 24 6 4 w x Lx x L y EI           (0) 0 (0) 0 y y    Boundary Conditions 80
• 81. Beam Stiffness Comparison of FE Solution to Exact Solution Recall the one-element solution to the cantilever beam is: 4 2 3 2 8 6 wL v EI wL EI                         Using the numerical values for this problem we get:         4 6 4 2 3 2 6 4 20 100 8 30 10 100 20 100 6 30 10 100 lb in lb in in psi in v in psi in                               0.0833 0.00111 in rad          81
• 82. Beam Stiffness Comparison of FE Solution to Exact Solution The slope and displacement from the one-element FE solution identically match the beam theory values evaluated at x = L. The reason why these nodal values from the FE solution are correct is that the element nodal forces were calculated on the basis of being energy or work equivalent to the distributed load based on the assumed cubic displacement field within each beam element. 82
• 83. Beam Stiffness Comparison of FE Solution to Exact Solution Values of displacement and slope at other locations along the beam for the FE are obtained by using the assumed cubic displacement function.     3 2 3 2 2 2 2 3 3 1 1 ( ) 2 3 v x x x L v x L x L L L       The value of the displacement at the midlength v(x = 50 in) is: ( 50 ) 0.0278 v x in in    Using beam theory, the displacement at v(x = 50 in) is: ( 50 ) 0.0295 v x in in    83
• 84. Beam Stiffness Comparison of FE Solution to Exact Solution In general, the displacements evaluated by the FE method using the cubic function for v are lower than by those of beam theory except at the nodes. This is always true for beams subjected to some form of distributed load that are modeled using the cubic displacement function. The exception to this result is at the nodes, where the beam theory and FE results are identical because of the work- equivalence concept used to replace the distributed load by work-equivalent discrete loads at the nodes. 84
• 85. Beam Stiffness Comparison of FE Solution to Exact Solution The beam theory solution predicts a quartic (fourth-order) polynomial expression for a beam subjected to uniformly distributed loading, while the FE solution v(x) assumes a cubic (third-order) displacement behavior in each beam all load conditions. The FE solution predicts a stiffer structure than the actual one. This is expected, as the FE model forces the beam into specific modes of displacement and effectively yields a stiffer model than the actual structure. However, as more elements are used in the model, the FE solution converges to the beam theory solution. 85
• 86. Beam Stiffness Comparison of FE Solution to Exact Solution For the special case of a beam subjected to only nodal concentrated loads, the beam theory predicts a cubic displacement behavior. The FE solution for displacement matches the beam theory solution for all locations along the beam length, as both v(x) and y(x) are cubic functions. 86
• 87. Beam Stiffness Comparison of FE Solution to Exact Solution Under uniformly distributed loading, the beam theory solution predicts a quadratic moment and a linear shear force in the beam. However, the FE solution using the cubic displacement function predicts a linear bending moment and a constant shear force within each beam element used in the model. 87
• 88. Beam Stiffness Comparison of FE Solution to Exact Solution We will now determine the bending moment and shear force in the present problem based on the FE method. ( ) M x EIy    2 2 d EI dx  Nd   2 2 d EI dx  N d    ( ) M x EI  B d   2 3 2 2 3 2 6 12 4 6 6 12 2 6 x x x x L L L L L L L L                                       B 1 1 2 3 2 6 12 4 6 x x M EI v L L L L                      2 2 2 3 2 6 12 2 6 x x v L L L L                      88
• 89. Beam Stiffness Comparison of FE Solution to Exact Solution We will now determine the bending moment and shear force in the present problem based on the FE method. Position MFE Mexact X = 0 -83,333 lb-in -100,000 lb-in X = 50 in -33,333 lb-in -25,000 lb-in X = 100 in 16,667 lb-in 0 89
• 90. Beam Stiffness Comparison of FE Solution to Exact Solution The plots below show the displacement, bending moment , and shear force over the beam using beam theory and the one-element FE solutions. 90
• 91. Beam Stiffness Comparison of FE Solution to Exact Solution The FE solution for displacement matches the beam theory solution at the nodes but predicts smaller displacements (less deflection) at other locations along the beam length. 91
• 92. Beam Stiffness Comparison of FE Solution to Exact Solution The bending moment is derived by taking two derivatives on the displacement function. It then takes more elements to model the second derivative of the displacement function. 92
• 93. Beam Stiffness Comparison of FE Solution to Exact Solution The shear force is derived by taking three derivatives on the displacement function. For the uniformly loaded beam, the shear force is a constant throughout the singIe-element model. 93
• 94. Beam Stiffness Comparison of FE Solution to Exact Solution To improve the FE solution we need to use more elements in the model (refine the mesh) or use a higher-order element, such as a fifth-order approximation for the displacement function. 94
• 95. Beam Stiffness Potential Energy Approach to Derive Beam Element Equations Let’s derive the equations for a beam element using the principle of minimum potential energy. The procedure for applying the principle of minimum potential energy is similar to that used for the bar element. The total potential energy p is defined as the sum of the internal strain energy U and the potential energy of the external forces : p U     95
• 96. Beam Stiffness Potential Energy Approach to Derive Beam Element Equations The differential internal work (strain energy) dU in a one- dimensional beam element is: 1 2 x x V U dV     For a single beam element, shown below, subjected to both distributed and concentrated nodal forces, the potential energy due to forces (or the work done these forces) is: 2 2 1 1 y iy i i i i i S T v dS P v m           96
• 97. Beam Stiffness Potential Energy Approach to Derive Beam Element Equations If the beam element has a constant cross-sectional area A, then the differential volume of the beam is given as: The differential element where the surface loading acts is given as: dS = b dx (where b is the width of the beam element). dV dA dx  Therefore the total potential energy is:   2 1 2 1 p x x y iy i i i i x A x dA dx T vb dx P v m             97
• 98. Beam Stiffness Potential Energy Approach to Derive Beam Element Equations The strain-displacement relationship is: We can express the strain in terms of nodal displacements and rotations as: 2 2 x d v y dx      2 2 3 3 3 3 12 6 6 4 12 6 6 2 x x L xL L x L xL L y d L L L L                   [ ] x y B d    2 2 3 3 3 3 12 6 6 4 12 6 6 2 [ ] x L xL L x L xL L B L L L L             98
• 99. Beam Stiffness Potential Energy Approach to Derive Beam Element Equations The stress-strain relationship in one-dimension is: where E is the modulus of elasticity. Therefore:     [ ] x x E        [ ][ ] x y E B d    The total potential energy can be written in matrix form as:           1 2 T T T p x x y x A x dA dx bT v dx d P             [ ] x y B d    99
• 100. Beam Stiffness Potential Energy Approach to Derive Beam Element Equations Use the following definition for moment of inertia: If we define, as a line load (load per unit length) in the y direction and the substitute the definitions of x and x the total potential energy can be written in matrix form as: y w bT            2 0 0 [ ] [ ] [ ] 2 L L T T T T T p A E y d B B d dAdx w d N dx d P       2 A I y dA   Then the total potential energy expression becomes:           0 0 [ ] [ ] [ ] 2 L L T T T T T p EI d B B d dx w d N dx d P       100
• 101. Beam Stiffness Potential Energy Approach to Derive Beam Element Equations The nodal forces vector is: The elemental stiffness matrix is: Differentiating the total potential energy with respect to the displacement and rotations (v1, v2, 1 and 2) and equating each term to zero gives:     0 0 [ ] [ ] [ ] 0 L L T T EI B B dx d w N dx P          0 [ ] L T f w N dx P      0 [ ] [ ] L T k EI B B dx   101
• 102. Beam Stiffness Potential Energy Approach to Derive Beam Element Equations Integrating the previous matrix expression gives:   2 2 3 2 2 12 6 12 6 6 4 6 2 12 6 12 6 6 2 6 4 L L L L L L EI k L L L L L L L                      0 [ ] [ ] L T k EI B B dx   102
• 103. Beam Stiffness Galerkin’s Method to Derive Beam Element Equations The governing differential equation for a one-dimensional beam is: 4 4 0 d v EI w dx         We can define the residual R as: 4 4 0 0 1, 2, 3, and 4 L i d v EI w N dx i dx                 If we apply integration by parts twice to the first term we get:           , , 0 0 0 L L L xxxx i xx i xx i xxx i x xx EI v N dx EI v N dx EI N v N v          where the subscript x indicates a derivative with respect to x 103
• 104. Beam Stiffness Galerkin’s Method to Derive Beam Element Equations or Therefore the integration by parts becomes: Since , then the second derivative of v with respect to x is:   [ ] v N d    2 2 3 3 3 3 12 6 6 4 12 6 6 2 xx x L xL L x L xL L v d L L L L               [ ] x x v B d          , , 0 0 0 [ ] 0 L L L i xx i i i x N EI B dx d N w dx NV N m d           1, 2, 3, and 4 i  104
• 105. Beam Stiffness Galerkin’s Method to Derive Beam Element Equations The interpolation function in the last two terms can be evaluated: The above expression is really four equations (one for each Ni) and can be written in matrix form as:   0 0 0 [ ] [ ] [ ] [ ] [ ] L L L T T T T x B EI B dx d N w dx N m N V           [ ] 0 [0 1 0 0] [ ] [0 0 0 1] x x N x N x L         [ ] 0 [1 0 0 0] [ ] [0 0 1 0] N x N x L     105
• 106. Beam Stiffness Galerkin’s Method to Derive Beam Element Equations Therefore, the last two terms of the matrix form of the Galerkin formulation become (see the figure below): 1 (0) 2 (0) i V i m     3 ( ) 4 ( ) i V L i m L     106
• 107. Beam Stiffness Galerkin’s Method to Derive Beam Element Equations When beam elements are assembled, as shown below: Two shear forces and two moments form adjoining elements contribute to the concentrated force and the concentrated moment at the node common to both elements. 107
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