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Slope Deflection Method

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Structural Analysis in University of The West of England

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Slope Deflection Method

  1. 1. Structural Design and Inspection- Deflection and Slope of Beams By Dr. Mahdi Damghani 2016-2017 1
  2. 2. Suggested Readings Reference 1 Chapter 16 2
  3. 3. Objective • To obtain slope and deflection of beam and frame structures using slope-deflection method 3
  4. 4. Introduction • Structural analysis method for beams and frames introduced in 1914 by George A. Maney • This method was later replaced by moment distribution method which is more advanced and useful (students are encouraged) to study this separately 4
  5. 5. Slope-Deflection Method • Sign convention: • Moments, slopes, displacements, shear are all in positive direction as shown 5 • Axial forces are ignored  xSM dx vd EI ABAB2 2          21 32 1 2 62 2 CxC x S x MEIv C x SxM dx dv EI ABAB ABAB
  6. 6. Slope-Deflection Method 6  AA vv dx dv x ,0@           AAABAB AABAB EIvxEI x S x MEIv EI x SxM dx dv EI   62 2 32 2  BB vv dx dv Lx ,@           AAABABB AABABB EIvLEI L S L MEIv EI L SLMEI   62 2 32 2 Solving for MAB and SAB       BABAAB vv LL EI M 3 2 2        BABAAB vv LL EI S 26 2        BABABA vv LL EI M 3 2 2        BABABA vv LL EI S 26 2 
  7. 7. Slope-Deflection Method • Last 4 equations obtained in previous slide are called slope-deflection equations • They establish force-displacement relationship • This method can find exact solution to indeterminate structures 7
  8. 8. Slope-Deflection Method • The beam we considered so far did not have any external loading from A to B 8 • In the presence of mid-span loading (common engineering problems) the equations become:   F ABBABAAB Mvv LL EI M      3 2 2    F ABBABAAB Svv LL EI S      26 2    F BABABABA Mvv LL EI M      3 2 2    F BABABABA Svv LL EI S      26 2 
  9. 9. Fixed End Moment/Shear • MAB F, MBA F are fixed end moments at nodes A and B, respectively. • Moments at two ends of beam when beam is clamped at both ends under external loading (see next slides) • SAB F, SBA F are fixed end shears at nodes A and B, respectively. • Shears at two ends of beam when beam is clamped at both ends under external loading (see next slides) 9
  10. 10. Fixed End Moment/Shear 10
  11. 11. Fixed End Moment/Shear 11
  12. 12. Example • Find support reactions. 12
  13. 13. Solution • This beam has 2 degrees of indeterminacy 13 1- Assume all beams are fixed & calculate FEM 2- Establish Slope- Deflection equations 3- Enforce boundary conditions & equilibrium conditions at joints 4- Solve simultaneous equations to get slopes/deflections
  14. 14. Solution 14 kNmMM F BA F AB 75.0 8 0.16    kNmMM F CB F BC 25.1 8 0.110    kNmMM F DC F CD 00.1 12 0.112 2   
  15. 15. Solution 15   F ijjijiij Mvv LL EI M      3 2 2    F jijijiji Mvv LL EI M      3 2 2  vi=0 and vj=0 for all cases
  16. 16. Solution • Equilibrium moments at the joints 16 MBA MBC MCB MCD 0ABM 0 0   BCBA B MM M 0 0   CDCB C MM M 0DCM
  17. 17. Solution • Substitution into slope deflection equations gives 4 equations and 4 unknown slopes. • By simultaneously solving the equations 17
  18. 18. Solution • Simply operation of substitution: 18
  19. 19. Solution • Now support reactions can easily be calculated as 19 15.485.16 85.1 0.1 15.15.06     BA AB R R 6 10 25.575.410 75.4 0.1 4.115.15.010     CB BC R R 12 6.44.712 4.7 0.1 4.15.012     DC CD R R
  20. 20. Solution 20 1.85 4.15 4.75 5.25 7.4 4.6 1.85 8.9 12.65 4.60
  21. 21. Solution 21
  22. 22. Example 2 • Obtain moment reaction at the clamped support B for 6m long beam if support A settles down by 5mm. EI=17×1012 Nmm2 22 A 4kN/m B
  23. 23. Solution 23   F ijjijiij Mvv LL EI M      3 2 2    F jijijiji Mvv LL EI M      3 2 2  A 4kN/m 5mm B        F ABBABAAB Mvv LL EI M 3 2 2           12 4000 0005.0 3 02 2 0 2 L LL EI M AAB  radL L EI A 4 3 1033.9 2 0015.0 24 4000          
  24. 24. Solution 24 A 4kN/m 5mm B   F ijjijiij Mvv LL EI M      3 2 2    F jijijiji Mvv LL EI M      3 2 2  radA 4 1033.9          F BABAABBA Mvv LL EI M 3 2 2            12 4000 0005.0 3 1033.90 2 2 4 L LL EI MBA mNMBA .3.18705
  25. 25. Case study 25
  26. 26. Case study 26 t=3.552 mm b=7.792 mm   F ijjijiij Mvv LL EI M      3 2 2 
  27. 27. Case study 27 Rotation and displacement can be obtained from the FEM Wire in composite beam
  28. 28. Case study 28 Rotation and displacement can be obtained from the FEM
  29. 29. Q1 • Determine the support reactions in the beam shown below. 29
  30. 30. Q2 • Calculate the support reactions in the beam shown below. 30
  31. 31. Q3 • Determine the end moments in the members of the portal frame shown. The second moment of area of the vertical members is 2.5I while that of the horizontal members is I. 31
  32. 32. Q4 • Analyze two span continuous beam ABC by slope deflection method. Then draw Bending moment & Shear force diagram. Take EI constant. 32

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