Tham khảo tài liệu 'fundamentals of structural analysis episode 2 part 2', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Beam and Frame Analysis Displacement Method Part I by S. T. Mau 100 kN _b_c__d 2EỈ I 4El I ĨẼĨ I El e El f 5 m 1 5 m 5 m i 20 m 10 m A three-span bridge frame. Solution. Symmetry of the structure calls for the decomposition of the load into a symmetric component and an anti-symmetric component. 50 kN i b a 50 kN c d a 50 kN ị b d 50 kN Symmetric and anti-symmetric loads. We shall solve both problems in parallel. 1 Preparation. Note the stiffness computation in steps c and d . a Only nodes a and b are free to rotate when we take advantage of the symmetry anti-symmetry. Furthermore if we use the modified stiffness for the hinged end situation in member ab then we need to concentrate on node b only. b FEM for member ab. The concentrated load of 50 kN creates FEMs at end a and end b. The formula for a single transverse load in the FEM table gives us tfob - P Length - 5 10 - kN-m P L4th 50 14 kN-m c Compute DF at b Symmetric Case DFba DFjc DFb. 3EKab 2EKbc 4EKbc 2 fA bc 4 EL bc A 1 4 10 20 5 10 20 5 195 Beam and Frame Analysis Displacement Method Part I by S. T. Mau 3 2 4 3 2 4 4 T T 4 T 5 5 5 9 9 9 d Compute DF at b Anti-symmetric Case DFba DFjc DFb. 3EKab 6EKbc 4EKbc _ 2EE 4EI EIX _ 6 . 24 4 10 a c 4 5 b 10 20 5 3 6 1 3 4 5 5 5 13 13 13 e Assign DFs at a and e DF is one at a and zero at e. 2 Tabulation. We need to include only nodes a b and e in the table below. Moment Distribution Table for a Symmetric Case and an Anti-symmetric Case Symmetric Case Anti-symmetric Case Node a b e a b e Member ab bc be ab bc be DF 1 0 1 0 MEM Mab Mba Mbc Mbe Meb Mab Mba Mbc Mbe Meb EAM FEM DM x x COM DM COM SUM The solution to the original problem is the superposition of the two solutions in the above table. Mab .
