easy step on how to solve slope deflection

3,149 views
2,890 views

Published on

Published in: Education, Technology, Business
0 Comments
9 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
3,149
On SlideShare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
256
Comments
0
Likes
9
Embeds 0
No embeds

No notes for slide

easy step on how to solve slope deflection

  1. 1. DISPLACEMENT METHOD OF ANALYSIS: SLOPE DEFLECTION EQUATIONS ! ! ! ! ! ! ! ! ! General Case Stiffness Coefficients Stiffness Coefficients Derivation Fixed-End Moments Pin-Supported End Span Typical Problems Analysis of Beams Analysis of Frames: No Sidesway Analysis of Frames: Sidesway 1
  2. 2. Slope – Deflection Equations P i w j k Cj settlement = ∆j Mij P i w j Mji θi ψ θj 2
  3. 3. Degrees of Freedom M θΑ B A 1 DOF: θΑ C 2 DOF: θΑ , θΒ L θΑ A P B θΒ 3
  4. 4. Stiffness kAA kBA 1 B A L k AA = 4 EI L k BA = 2 EI L 4
  5. 5. kBB kAB A B 1 L k BB = 4 EI L k AB = 2 EI L 5
  6. 6. Fixed-End Forces Fixed-End Moments: Loads P L/2 PL 8 L/2 PL 8 L P 2 P 2 w wL2 12 wL 2 L wL2 12 wL 2 6
  7. 7. General Case P i w j k Cj settlement = ∆j Mij P i w j Mji θi ψ θj 7
  8. 8. Mij P i j w Mji θi L θj ψ 4 EI 2 EI θi + θj = M ij L L Mji = 2 EI 4 EI θi + θj L L θj θi (MFij)∆ + (MFji)∆ settlement = ∆j + P (MFij)Load M ij = ( settlement = ∆j w (MFji)Load 4 EI 2 EI 2 EI 4 EI )θ i + ( )θ j + ( M F ij ) ∆ + ( M F ij ) Load , M ji = ( )θ i + ( )θ j + ( M F ji ) ∆ + ( M F ji ) Load 8 L L L L
  9. 9. Equilibrium Equations i P w j k Cj Cj M Mji Mji jk Mjk j + ΣM j = 0 : − M ji − M jk + C j = 0 9
  10. 10. Stiffness Coefficients Mij i j Mji L θj θi kii = 4 EI L k ji = 2 EI L ×θ i k jj = 4 EI L ×θ j 1 kij = 2 EI L + 1 10
  11. 11. Matrix Formulation M ij = ( 4 EI 2 EI )θ i + ( )θ j + ( M F ij ) L L M ji = ( 2 EI 4 EI )θ i + ( )θ j + ( M F ji ) L L  M ij  (4 EI / L) ( 2 EI / L) θ iI   M ij F   M  =   + (2 EI / L) ( 4 EI / L) θ j   M ji F    ji      kii k ji [k ] =  kij  k jj   Stiffness Matrix 11
  12. 12. P i Mij w j Mji θi L [ M ] = [ K ][θ ] + [ FEM ] θj ψ ∆j ([ M ] − [ FEM ]) = [ K ][θ ] [θ ] = [ K ]−1[ M ] − [ FEM ] Mij Mji θj θi Fixed-end moment Stiffness matrix matrix + (MFij)∆ (MFji)∆ [D] = [K]-1([Q] - [FEM]) + (MFij)Load P w (MFji)Load Displacement matrix Force matrix 12
  13. 13. Stiffness Coefficients Derivation: Fixed-End Support Mj θi Mi Real beam i j L Mi + M j L Mi + M j L/3 M jL 2 EI L Mj EI Conjugate beam Mi EI θι MiL 2 EI M j L 2L MiL L )( ) + ( )( ) = 0 2 EI 3 2 EI 3 M i = 2 M j − − − (1) + ΣM 'i = 0 : − ( + ↑ ΣFy = 0 : θ i − ( M L MiL ) + ( j ) = 0 − − − (2) 2 EI 2 EI From (1) and (2); 4 EI )θ i L 2 EI )θ i Mj =( L Mi = ( 13
  14. 14. Stiffness Coefficients Derivation: Pinned-End Support Mi θi i L Mi L 2L 3 Mi EI θi + ΣM ' j = 0 : ( MiL 2 EI M i L 2L )( ) − θ i L = 0 2 EI 3 ML θi = ( i ) 3EI θi = 1 = ( θj j Real beam Mi L Conjugate beam θj + ↑ ΣFy = 0 : ( MiL ML ) − ( i ) +θ j = 0 3EI 2 EI θj =( MiL 3EI ) → Mi = L 3EI − MiL ) 6 EI 14
  15. 15. Fixed end moment : Point Load P Real beam Conjugate beam A A B B L M EI M M M EI M EI ML 2 EI ML 2 EI P PL2 16 EI PL 4 EI M EI PL2 16 EI PL ML ML 2 PL2 + ↑ ΣFy = 0 : − − + = 0, M = 2 EI 2 EI 16 EI 8 15
  16. 16. P PL 8 PL 8 L P 2 P 2 P/2 PL/8 P/2 -PL/8 -PL/8 - -PL/8 -PL/16 - -PL/16 PL/4 + -PL/8 − PL − PL PL PL + = + 16 16 4 8 16
  17. 17. Uniform load w Real beam Conjugate beam A A L B B M EI M M M EI M EI ML 2 EI ML 2 EI w wL3 24 EI wL2 8 EI M EI wL3 24 EI wL2 ML ML 2 wL3 + ↑ ΣFy = 0 : − − + = 0, M = 2 EI 2 EI 24 EI 12 17
  18. 18. Settlements Mi = Mj Real beam Mj Conjugate beam L Mi + M j M L A M EI B ∆ ∆ M EI Mi + M j L M EI ML 2 EI ML 2 EI M M EI ∆ + ΣM B = 0 : − ∆ − ( ML L ML 2 L )( ) + ( )( ) = 0, 2 EI 3 2 EI 3 M= 6 EI∆ L2 18
  19. 19. Pin-Supported End Span: Simple Case P w B A L 2 EI 4 EI θA + θB L L 4 EI 2 EI θA + θB L L θA A θB + P B w (FEM)BA (FEM)AB A B = 0 = (4 EI / L)θ A + (2 EI / L)θ B + ( FEM ) AB − − − (1) M BA = 0 = (2 EI / L)θ A + (4 EI / L)θ B + ( FEM ) BA − − − ( 2) M AB 2(2) − (1) : 2 M BA = (6 EI / L)θ B + 2( FEM ) BA − ( FEM ) BA M BA = (3EI / L)θ B + ( FEM ) BA − ( FEM ) BA 2 19
  20. 20. Pin-Supported End Span: With End Couple and Settlement P w MA B A L ∆ 2 EI 4 EI θA + θB L L 4 EI 2 EI θA + θB L L θA A P θB w B (MF BA)load (MF AB)load (MF AB)∆ A A B B (MF BA) ∆ 4 EI 2 EI F F θA + θ B + ( M AB )load + ( M AB ) ∆ − − − (1) L L 2 EI 4 EI F F θA + θ B + ( M BA )load + ( M BA ) ∆ − − − (2) M BA = L L 2(2) − (1) 3EI 1 1 M F F F : M BA = θ B + [( M BA )load − ( M AB )load ] + ( M BA ) ∆ + A 2 2 2 2 L M AB = M A = E lim inate θ A by 20
  21. 21. Fixed-End Moments Fixed-End Moments: Loads P PL 8 L/2 L/2 PL 8 L/2 PL 1 PL 3PL + ( )[−(− )] = 8 2 8 16 P L/2 wL2 12 wL2 12 wL2 1 wL2 wL2 )] = + ( )[−( − 12 2 12 8 21
  22. 22. Typical Problem CB w P1 P2 C A B L1 P PL 8 PL 8 L2 wL2 12 L 0 PL 4 EI 2 EI M AB = θA + θB + 0 + 1 1 8 L1 L1 0 EI PL 2 EI 4 M BA = θA + θB + 0 − 1 1 8 L1 L1 0 2 P2 L2 wL2 4 EI 2 EI M BC = θB + θC + 0 + + L2 L2 8 12 0 2 − P2 L2 wL2 2 EI 4 EI θB + θC + 0 + − M CB = 8 12 L2 L2 w wL2 12 L 22
  23. 23. CB w P1 P2 C A B L1 L2 CB M BC MBA B M BA = M BC PL 2 EI 4 EI θA + θB + 0 − 1 1 8 L1 L1 P L wL 4 EI 2 EI = θB + θC + 0 + 2 2 + 2 L2 L2 8 12 2 + ΣM B = 0 : C B − M BA − M BC = 0 → Solve for θ B 23
  24. 24. P1 MAB CB w MBA A L1 B P2 MBC L2 C M CB Substitute θB in MAB, MBA, MBC, MCB 0 PL 4 EI 2 EI θA + θB + 0 + 1 1 M AB = L1 L1 8 0 EI PL 2 EI 4 M BA = θA + θB + 0 − 1 1 8 L1 L1 0 2 4 EI 2 EI P2 L2 wL2 θB + θC + 0 + + M BC = 8 12 L2 L2 0 2 − P2 L2 wL2 2 EI 4 EI θB + θC + 0 + − M CB = L2 L2 8 12 24
  25. 25. P1 MAB A Ay CB w MBA B L1 P2 MCB MBC L2 Cy C By = ByL + ByR B P1 MAB B A Ay L1 MBA ByL P2 MBC ByR L2 C MCB Cy 25
  26. 26. Example of Beams 26
  27. 27. Example 1 Draw the quantitative shear , bending moment diagrams and qualitative deflected curve for the beam shown. EI is constant. 10 kN 6 kN/m C A 4m 4m B 6m 27
  28. 28. 10 kN 6 kN/m C A 4m PL 8 4m P B 6m wL2 30 PL 8 wL2 20 w FEM MBA MBC [M] = [K][Q] + [FEM] 0 4 EI 2 EI (10)(8) θA + θB + 8 8 8 0 2 EI 4 EI (10)(8) M BA = θA + θB − 8 8 8 0 4 EI 2 EI (6)(6 2 ) θB + θC + M BC = 6 6 30 0 2 EI 4 EI (6)(6) 2 θB + θC − M CB = 6 6 20 M AB = B + ΣM B = 0 : − M BA − M BC = 0 4 EI 4 EI (6)(6 2 ) + =0 ( )θ B − 10 + 8 6 30 2.4 θB = EI Substitute θB in the moment equations: MAB = 10.6 kN•m, MBC = 8.8 kN•m MBA = - 8.8 kN•m, MCB = -10 kN•m 28
  29. 29. 10 kN 8.8 kN•m 10.6 kN•m A 6 kN/m C 8.8 kN•m 4m B 4m MAB = 10.6 kN•m, 6m MBC = 8.8 kN•m MBA = - 8.8 kN•m, 10 kN•m MCB= -10 kN•m 2m 18 kN 10 kN 6 kN/m 10.6 kN•m A B 8.8 kN•m B 10 kN•m 8.8 kN•m Ay = 5.23 kN ByL = 4.78 kN ByR = 5.8 kN Cy = 12.2 kN 29
  30. 30. 10 kN 10.6 kN•m 6 kN/m C A 4m 5.23 kN B 4m 6m 10 kN•m 12.2 kN 4.78 + 5.8 = 10.58 kN V (kN) 5.8 5.23 + + x (m) - 4.78 - 10.3 M (kN•m) -10.6 Deflected shape -12.2 + -8.8 2.4 θB = EI x (m) -10 x (m) 30
  31. 31. Example 2 Draw the quantitative shear , bending moment diagrams and qualitative deflected curve for the beam shown. EI is constant. 10 kN 6 kN/m C A 4m 4m B 6m 31
  32. 32. 10 kN 6 kN/m C A 4m PL 8 P 4m B 6m wL2 30 PL 8 w wL2 20 FEM 10 [M] = [K][Q] + [FEM] 0 4 EI 2 EI (10)(8) θA + θB + M AB = − − − (1) 8 8 8 10 2 EI 4 EI (10)(8) θA + θB − − − − (2) M BA = 8 8 8 4 EI 2 EI 0 (6)(6 2 ) θB + θC + − − − (3) M BC = 6 6 30 2 EI 4 EI 0 (6)(6) 2 θB + θC − − − − (4) M CB = 6 6 20 6 EI 2(2) − (1) : 2 M BA = θ B − 30 8 3EI θ B − 15 − − − (5) M BA = 8 32
  33. 33. MBA MBC B M BC 4 EI (6)(6 2 ) = θB + 6 30 M CB = 2 EI (6)(6) θB − 6 20 2 − − − (3) Substitute θA and θB in (5), (3) and (4): MBA = - 12.19 kN•m − − − (4) 3EI θ B − 15 − − − (5) 8 + ΣM B = 0 : − M BA − M BC = 0 M BA = MBC = 12.19 kN•m MCB = - 8.30 kN•m 3EI 4 EI (6)(6 2 ) + = 0 − − − (6 ) ( )θ B − 15 + 8 6 30 7.488 θB = EI 4 EI 2 EI θA + θ B − 10 8 8 − 23.74 θA = EI Substitute θ B in (1) : 0 = 33
  34. 34. 10 kN 12.19 kN•m A 4m 4m B 6 kN/m C 12.19 kN•m 6m 8.30 kN•m MBA = - 12.19 kN•m, MBC = 12.19 kN•m, MCB = - 8.30 kN•m 2m 10 kN A 18 kN B B 12.19 kN•m 6 kN/m C 8.30 kN•m 12.19 kN•m Ay = 3.48 kN ByL = 6.52 kN ByR = 6.65 kN Cy = 11.35 kN 34
  35. 35. 10 kN 6 kN/m C A 3.48 kN 4m 4m B 6m 11.35 kN 6.52 + 6.65 = 13.17 kN V (kN) 6.65 3.48 x (m) - 6.52 -11.35 14 M (kN•m) x (m) -12.2 Deflected shape − 23.74 θA = EI θB = 7.49 EI -8.3 x (m) 35
  36. 36. Example 3 Draw the quantitative shear , bending moment diagrams and qualitative deflected curve for the beam shown. EI is constant. 10 kN 4 kN/m C A 2EI 4m B 4m 3EI 6m 36
  37. 37. 10 kN 4 kN/m C A 2EI (10)(8)/8 (10)(8)/8 4m 4m 0 M AB 3EI B (4)(62)/12 6m (4)(62)/12 10 4(2 EI ) 2(2 EI ) (10)(8) = θA + θB + − − − (1) 8 8 8 10 M BA = 2(2 EI ) 4(2 EI ) (10)(8) θA + θB − 8 8 15 8 − − − (2) 2(2) − (1) 3(2 EI ) (3 / 2)(10)(8) : M BA = θB − − − − ( 2a ) 2 8 8 12 4(3EI ) (4)(6 2 ) θB + − − − (3) M BC = 6 12 37
  38. 38. 10 kN 4 kN/m C A 3EI 2EI (3/2)(10)(8)/8 B (4)(62)/12 4m 6m 4m (4)(62)/12 15 3(2 EI ) (3 / 2)(10)(8) M BA = θB − − − − ( 2a ) 8 8 12 2 4(3EI ) (4)(6 ) θB + − − − (3) M BC = 6 12 − M BA − M BC = 0 : 2.75EIθ B = −12 + 15 = 3 θ B = 1.091 / EI 3(2 EI ) 1.091 ( ) − 15 = −14.18 kN • m 8 EI 4(3EI ) 1.091 ( ) − 12 = 14.18 kN • m = 6 EI 2(3EI ) = θ B − 12 = −10.91 kN • m 6 M BA = M BC M CB 38
  39. 39. 10 kN A 2EI 14.18 4m 4 kN/m C 10.91 B 14.18 3EI 6m 4m MBA = - 14.18 kN•m, MBC = 14.18 kN•m, MCB = -10.91 kN•m 10 kN A 24 kN B 4 kN/m 14.18 kN•m 140.18 kN•m 10.91 kN•m C Ay = 3.23 kN ByL = 6.73 kN ByR = 12.55 kN Cy = 11.46 kN 39
  40. 40. 10 kN 4 kN/m C 10.91 kN•m A 2EI 3.23 kN 4m V (kN) 3.23 B 3EI 11.46 kN 6m 4m 6.77 + 12.55 = 19.32 kN 12.55 2.86 + -6.73 + x (m) -11.46 12.91 M (kN•m) 5.53 + + - -10.91 -14.18 Deflected shape - x (m) θB = 1.091/EI x (m) 40
  41. 41. Example 4 Draw the quantitative shear , bending moment diagrams and qualitative deflected curve for the beam shown. EI is constant. 10 kN 12 kN•m 4 kN/m C A 2EI 4m B 4m 3EI 6m 41
  42. 42. 10 kN 12 kN•m 4 kN/m C A 2EI 3EI 2/12 = 12 1.5PL/8 = 15 B wL wL2/12 = 12 4m 6m 4m M BA = 3(2 EI ) θ B − 15 − − − (1) 8 M BC = 4(3EI ) θ B + 12 − − − (2) 6 2(3EI ) θ B − 12 − − − (3) 6 12 kN•m MBA MBA M CB = B MBC MBC Jo int B : − M BA − M BC − 12 = 0 -3.273/EI 4(2 EI ) 2(3EI ) (10)(8) θA + θB + M AB = 8 8 8 7.21 θA = − EI 3.273 M BA = 0.75 EI ( − ) − 15 = −17.45 kN • m EI 3.273 M BC = 2 EI (− ) + 12 = 5.45 kN • m EI 3.273 M CB = EI (− ) − 12 = −15.27 kN • m EI 0 − (0.75 EI − 15) − (2 EIθ B + 12) − 12 = 0 θB = − 3.273 EI 42
  43. 43. 3.273 ) − 15 = −17.45 kN • m EI 3.273 M BC = 2 EI (− ) + 12 = 5.45 kN • m EI 3.273 M CB = EI ( − ) − 12 = −15.27 kN • m EI M BA = 0.75EI ( − 24 kN 10 kN 4 kN/m A 5.45 kN•m B 2.82 kN 17.45 kN•m 7.18 kN C 10.36 kN 10 kN 12 kN•m 2.82 kN B 17.54 kN 13.64 kN 4 kN/m C A 15.27 kN•m 15.27 kN•m 13.64 kN 43
  44. 44. 10 kN 12 kN•m 4 kN/m C 15.27 kN•m A 2.82 kN 4m 3EI 2EI 7.21 EI 3.273 θB = − EI B 17.54 kN 6m 4m 13.64 kN θ A = − 10.36 V (kN) 2.82 + + 3.41 m - - -7.18 M (kN•m) -13.64 11.28 7.98 + + - x (m) - -5.45 -15.27 -17.45 Deflected shape − 7.21 θA = EI x (m) x (m) θB = 3.273 EI 44
  45. 45. Example 5 Draw the quantitative shear, bending moment diagrams, and qualitative deflected curve for the beam shown. Support B settles 10 mm, and EI is constant. Take E = 200 GPa, I = 200x106 mm4. 10 kN 12 kN•m 6 kN/m B C A 3EI 2EI 4m 4m 10 mm 6m 45
  46. 46. 10 kN 12 kN•m 6 kN/m B C A 3EI 2EI 6 EI∆ L2 4m 4m A P B 6 EI∆ L2 [FEM]∆ PL 8 6 EI∆ L2 6m 6 EI∆ L2 ∆ PL 8 10 mm 2 wL 30 ∆ B w C wL2 30 [FEM]load -12 4(2 EI ) 2(2 EI ) 6(2 EI )(0.01) (10)(8) M AB = θA + θB + + − − − (1) 8 8 82 8 2(2 EI ) 4(2 EI ) 6(2 EI )(0.01) (10)(8) M BA = θA + θB + − − − − ( 2) 8 8 82 8 0 4(3EI ) 2(3EI ) 6(3EI )(0.01) (6)(6 2 ) M BC = θB + θC − + − − − (3) 6 6 62 30 0 2(3EI ) 4(3EI ) 6(3EI )(0.01) (6)(6) 2 M CB = θB + θC − − − − − (4) 2 6 6 6 30 46
  47. 47. 10 kN 12 kN•m 6 kN/m B C A 3EI 2EI 4m 4m 10 mm 6m 4(2 EI ) 2(2 EI ) 6(2 EI )(0.01) (10)(8) θA + θB + + − − − (1) 8 8 82 8 2(2 EI ) 4(2 EI ) 6(2 EI )(0.01) (10)(8) = θA + θB + − − − − ( 2) 8 8 82 8 M AB = M BA Substitute EI = (200x106 kPa)(200x10-6 m4) = 200x200 kN• m2 : 4(2 EI ) 2(2 EI ) θA + θ B + 75 + 10 − − − (1) 8 8 2(2 EI ) 4(2 EI ) = θA + θ B + 75 − 10 − − − (2) 8 8 16.5 M AB = M BA 2(2) − (1) 3(2 EI ) : M BA = θ B + 75 − (75 / 2) − 10 − (10 / 2) − 12 / 2 − − − (2a ) 8 2 47
  48. 48. 10 kN 12 kN•m 6 kN/m B C A 3EI 2EI 4m 4m 10 mm 6m MBA = (3/4)(2EI)θB + 16.5 MBC = (4/6)(3EI)θB - 192.8 + ΣMB = 0: - MBA - MBC = 0 MBA MBC B (3/4 + 2)EIθB + 16.5 - 192.8 = 0 θB = 64.109/ EI Substitute θB in (1): θA = -129.06/EI Substitute θA and θB in (5), (3), (4): MBA = 64.58 kN•m, MBC = -64.58 kN•m MCB = -146.69 kN•m 48
  49. 49. 10 kN 12 kN•m 6 kN/m B 146.69 kN•m C A 64.58 kN•m 4m 4m 64.58 kN•m 6m MBA = 64.58 kN•m, MBC = -64.58 kN•m 2m MCB = -146.69 kN•m 12 kN•m 18 kN 10 kN A 64.58 kN•m B 146.69 kN•m B Ay = 11.57 kN ByL = -1.57 kN 6 kN/m 64.58 kN•m ByR = -29.21 kN C Cy = 47.21 kN 49
  50. 50. 10 kN 12 kN•m 6 kN/m C A 2EI 11.57 kN 3EI B 4m 47.21 kN 6m 4m θA = -129.06/EI θB = 64.109/ EI 1.57 + 29.21 = 30.78 kN 11.57 + V (kN) 1.57 x (m) - -29.21 M (kN•m) 12 58.29 146.69 kN•m -47.21 64.58 + x (m) -146.69 Deflected shape x (m) θA = -129.06/EI 10 mm θB = 64.109/ EI 50
  51. 51. Example 6 For the beam shown, support A settles 10 mm downward, use the slope-deflection method to (a)Determine all the slopes at supports (b)Determine all the reactions at supports (c)Draw its quantitative shear, bending moment diagrams, and qualitative deflected shape. (3 points) Take E= 200 GPa, I = 50(106) mm4. 6 kN/m B 2EI 3m C 1.5EI 12 kN•m A 10 mm 3m 51
  52. 52. 6 kN/m B C 2EI 3m 6(32 ) = 4.5 12 6(1.5 × 200 × 50)(0.01) 32 = 100 kN • m 12 kN•m A 1.5EI 10 mm 3m 6 kN/m 4.5 C C MFw A 0.01 m MF∆ 100 kN • m A M CB = 4(2 EI ) θC 3 M CA = 4(1.5EI ) 2(1.5 EI ) θC + θ A + 4.5 + 100 − − − (2) 3 3 12 M AC = − − − (1) 2(1.5EI ) 4(1.5 EI ) θC + θ A − 4.5 + 100 − − − (3) 3 3 2(2) − (2) 3(1.5EI ) 3(4.5) 100 12 : M CA = θC + + + 3 2 2 2 2 − − − ( 2a ) 52
  53. 53. 6 kN/m B 2EI 3m C 1.5EI 12 kN•m A 10 mm 3m 4(2 EI ) θ C − − − (1) 3 3(1.5 EI ) 3(4.5) 100 12 = θC + + + 3 2 2 2 M CB = M CA MCB MCA − − − ( 2a ) • Equilibrium equation: M CB + M CA = 0 (8 + 4.5) EI 3(4.5) 100 12 θC + + + =0 C 3 2 2 2 − 15.06 θC = = −0.0015 rad EI 2(1.5 EI ) − 15.06 4(1.5 EI ) Substitute θC in eq.(3) 12 = ( )+ θ A − 4.5 + 100 − − − (3) EI 3 3 − 34.22 θA = = −0.0034 rad EI 53
  54. 54. 6 kN/m B C 2EI 12 kN•m 1.5EI 3m A 10 mm 3m − 15.06 − 34.22 = −0.0015 rad θ A = = −0.0034 rad EI EI 2(2 EI ) 2( 2 EI ) − 15.06 = θC = ( ) = −20.08 kN • m 3 3 EI 4(2 EI ) 4(2 EI ) − 15.06 = θC = ( ) = −40.16 kN • m 3 3 EI θC = M BC M CB 20.08 kN•m B 20.08 kN C 40.16 kN•m 40.16 + 20.08 = 20.08 kN 3 18 kN 6 kN/m 40.16 kN•m C 26.39 kN 12 kN•m A 8.39 kN 54
  55. 55. 12 kN•m 6 kN/m θ C = −0.0015 rad B θ A = −0.0034 rad C 2EI 3m B 40.16 kN•m C 20.08 kN 40.16 kN•m V (kN) 6 kN/m C 26.39 kN 26.39 12 kN•m A - Deflected shape -20.08 20.08 12 -40.16 θ C = 0.0015 rad 8.39 kN 8.39 x (m) + M (kN•m) 10 mm 3m 20.08 kN•m 20.08 kN A 1.5EI x (m) x (m) θ A = 0.0034 rad 55
  56. 56. Example 7 For the beam shown, support A settles 10 mm downward, use the slope-deflection method to (a)Determine all the slopes at supports (b)Determine all the reactions at supports (c)Draw its quantitative shear, bending moment diagrams, and qualitative deflected shape. Take E= 200 GPa, I = 50(106) mm4. 6 kN/m B 2EI 3m C 1.5EI 12 kN•m A 10 mm 3m 56
  57. 57. 12 kN•m 6 kN/m B C 2EI 3m C B ∆C 6( 2 EI )∆ C 4 EI∆ C = 2 3 3 M BC M CB M CA = 2(2 EI ) 4 EI = θC − ∆C 3 3 4(2 EI ) 4 EI = θC − ∆C 3 3 A 1.5EI 10 mm 3m EI∆ C C ∆C 4 EI∆ C 3 A 6 kN/m 4.5 100 − − − (1) C A C − − − (2) 0.01 m A 6(1.5 EI ) ∆ C = EI∆ C 2 3 6(32 ) = 4.5 12 6(1.5 × 200 × 50)(0.01) 32 = 100 kN • m 4(1.5 EI ) 2(1.5EI ) θC + θ A + EI∆ C + 4.5 + 100 − −(3) 3 3 12 2(1.5 EI ) 4(1.5 EI ) M AC = θC + θ A + EI∆ C − 4.5 + 100 − −(4) 3 3 2(3) − (4) 3(1.5 EI ) EI 3(4.5) 100 12 : M CA = θ C + ∆C + + + − − − (3a) 2 3 2 2 2 2 57
  58. 58. 12 kN•m 6 kN/m B C 2EI 3m A 1.5EI 3m 18 kN • Equilibrium equation: 6 kN/m MBC B By C (C y ) CB = −( MCB MCA C M BC + M CB (C y ) CA ) 3 MCB MCA (Cy)CB 10 mm C 12 kN•m A Ay M + 12 + 18(1.5) M CA + 39 = CA = 3 3 ΣM C = 0 : M CB + M CA = 0 − − − (1*) ΣC y = 0 : (C y ) CB + (C y ) CA = 0 − − − (2*) (Cy)CA Substitute in (1*) 4.167 EIθ C − 0.8333EI∆ C = −62.15 − − − (5) Substitute in ( 2*) − 2.5 EIθ C + 3.167 EI∆ C = −101.75 − − − (6) From (5) and (6) θ C = −25.51 / EI = −0.00255 rad ∆ C = −52.27 / EI = −5.227 mm 58
  59. 59. 6 kN/m B 2EI C 3m 1.5EI 12 kN•m A 10 mm 3m • Solve equation θC = − 25.51 = −0.00255 rad EI Substitute θC and ∆C in (1), (2) and (3a) M BC = 35.68 kN • m − 52.27 ∆C = = −5.227 mm EI M CB = 1.67 kN • m Substitute θC and ∆C in (4) θA = M CA = −1.67 kN • m − 2.86 = −0.000286 rad EI 6 kN/m 35.68 kN•m B B y = 18 − 5.55 = 12.45 kN 2EI 3m C 1.5EI 3m 12 kN•m A 18(4.5) − 12 − 35.68 6 = 5.55 kN Ay = 59
  60. 60. 6 kN/m 35.68 kN•m B C 2EI 12.45 kN 3m 12 kN•m A 1.5EI 10 mm 5.55 kN 3m V (kN) 12.45 θ C = −0.00255 rad + 0.925 m x (m) ∆ C = −5.227 mm -5.55 θ A = −0.000286 rad M (kN•m) 1.67 14.57 + 12 - x (m) -35.68 Deflected shape ∆ C = 5.227 mm θ C = 0.00255 rad x (m) θ A = 0.000286 rad 60
  61. 61. Example of Frame: No Sidesway 61
  62. 62. Example 6 For the frame shown, use the slope-deflection method to (a) Determine the end moments of each member and reactions at supports (b) Draw the quantitative bending moment diagram, and also draw the qualitative deflected shape of the entire frame. 10 kN 12 kN/m C 3m B 3EI 40 kN 3m 2EI A 1m 6m 62
  63. 63. 10 kN 36/2 = 18 12 kN/m • Equilibrium equations C B 3m 40 kN 3m A 1m 2EI 2 36 (wL /12 ) =36 PL/8 = 30 3EI PL/8 = 30 6m • Slope-Deflection Equations M AB = M BA = M BC 10 MBC MBA 10 − M BA − M BC = 0 − − − (1*) Substitute (2) and (3) in (1*) 10 − 3EIθ B + 30 − 54 = 0 − 14 − 4.667 = (3EI ) EI 2(3EI ) θ B + 30 − − − (1) 6 θB = 4(3EI ) θ B − 30 − − − (2) 6 Substitute θ B = 3( 2 EI ) θ B + 36 + 18 − − − (3) = 6 − 4.667 in (1) to (3) EI M AB = 25.33 kN • m M BA = −39.33 kN • m M BC = 49.33 kN • m 63
  64. 64. 10 kN 12 kN/m 20.58 C 3m B 49.33 39.33 3EI 40 kN 3m 25.33 A 1m 2EI 10 -39.3 -49.33 MAB = 25.33 kN•m 27.7 MBA = -39.33 kN•m -25.33 MBC = 49.33 kN•m 6m Bending moment diagram 12 kN/m B B C θB 49.33 39.33 27.78 kN θB θB = -4.667/EI 40 kN A 17.67 kN 25.33 Deflected curve 64
  65. 65. Example 7 Draw the quantitative shear, bending moment diagrams and qualitative deflected curve for the frame shown. E = 200 GPa. 25 kN 5 kN/m B 240(106) mm4 5m 180(106) C 120(106) mm4 60(106) mm4 A E D 3m 3m 4m 65
  66. 66. 25 kN PL/8 = 18.75 18.75 B 240(106) mm4 C 5m 5 kN/m 180(106) 6.667+ 3.333 120(106) mm4 (wL2/12 ) = 6.667 60(106) mm4 A E D 3m 3m 0 4(2 EI ) 2(2 EI ) M AB = θA + θB 5 5 0 2(2 EI ) 4(2 EI ) M BA = θA + θB 5 5 4( 4 EI ) 2(4 EI ) M BC = θB + θ C + 18.75 6 6 2(4 EI ) 4(4 EI ) M CB = θB + θ C − 18.75 6 6 3( EI ) θC M CD = 5 3(3EI ) M CE = θ C + 10 4 4m M BA + M BC = 0 8 16 8 ( + ) EIθ B + ( ) EIθ C = −18.75 − − − (1) 5 6 6 M CB + M CD + M CE = 0 16 3 9 8 ( ) EIθ B + ( + + ) EIθ C = 8.75 − − − (2) 6 5 4 6 From (1) and (2) : θ B = − 5.29 EI θC = 2.86 EI 66
  67. 67. Substitute θB = -1.11/EI, θc = -20.59/EI below 4(2 EI ) 0 2(2 EI ) M AB = θA + θB 5 5 0 2(2 EI ) 4(2 EI ) M BA = θA + θB 5 5 4( 4 EI ) 2(4 EI ) M BC = θB + θ C + 18.75 6 6 2(4 EI ) 4(4 EI ) θB + θ C − 18.75 M CB = 6 6 3( EI ) θC M CD = 5 3(3EI ) θ C + 10 M CE = 4 MAB = −4.23 kN•m MBA = −8.46 kN•m MBC = 8.46 kN•m MCB = −18.18 kN•m MCD = 1.72 kN•m MCE = 16.44 kN•m 67
  68. 68. MAB = -4.23 kN•m, MBA = -8.46 kN•m, MBC = 8.46 kN•m, MCB = -18.18 kN•m, MCD = 1.72 kN•m, MCE = 16.44 kN•m 20 kN 25 kN 16.44 kN•m 3m 3m C C 2.54 kN B 2.54 kN 2.54-0.34 =2.2 kN 8.46 kN•m 18.18 kN•m (20(2)+16.44)/4 (25(3)+8.46-18.18)/6 14.12 kN = 14.11 kN = 10.88 kN 10.88 kN B 8.46 kN•m (8.46 + 4.23)/5 = 2.54 kN 5m A E 2.2 kN 5.89 kN 14.12+14.11=28.23 kN 1.72 kN•m (1.72)/5 = 0.34 kN B 5m 2.54 kN A 0.34 kN 4.23 kN•m 10.88 kN 28.23 kN 68
  69. 69. 24.18 1.29 m 2.33 m 14.11 10.88 1.18 m + + -2.54 2.82 m - -14.12 - -2.54 + 0.78 m - -8.46 -5.89 -8.46 + - -18.18 0.34 4.23 Shear diagram 0.78 m + 3.46 1.72 + - 1.18 m -16.44 1.67m Moment diagram 1.29 m 2.33 m θB = −5.29/EI 1.18 m θC = 2.86/EI 1.67m Deflected curve 69
  70. 70. Example of Frames: Sidesway 70
  71. 71. Example 8 Determine the moments at each joint of the frame and draw the quantitative bending moment diagrams and qualitative deflected curve . The joints at A and D are fixed and joint C is assumed pin-connected. EI is constant for each member 3m 1m B 10 kN C 3m A D 71
  72. 72. • Overview • Unknowns C B 1m 10 kN θB and ∆ • Boundary Conditions 3m MAB Ax 3m A Ay θA = θD = 0 MDC D D • Equilibrium Conditions x - Joint B Dy B MBA MBC ΣM B = 0 : M BA + M BC = 0 − − − (1*) - Entire Frame + → ΣFx = 0 : 10 − Ax − Dx = 0 − − − (2*) 72
  73. 73. ∆ ∆ (0.375EI∆)∆ 1m B 10 kN (5.625)load MAB C (0.375EI∆)∆ 10 kN C B 4m 4m 3m (1.875)load A (0.375EI∆)∆ (1/2)(0.375EI∆)∆ (0.375EI∆)∆ A D Ax D Dx 3m MBA • Slope-Deflection Equations 5.625 0.375EI∆ 2( EI ) 10(3)(12 ) 6 EI∆ M AB = θB + + 2 − − − (1) 4 42 4 5.625 0.375EI∆ 2 4( EI ) 10(3 )(1) 6 EI∆ M BA = θB − + 2 − − − (2) 4 42 4 M BC = 3( EI ) θB 3 − − − (3) 1 M DC = 0.375EI∆ − 0.375EI∆ = 0.1875EI∆ − − − (4) 2 MDC + ΣM B = 0 : ( M AB + M BA ) 4 Ax = 0.375EIθ B + 0.1875EI∆ + 1.563 − − − (5) Ax = + ΣM C = 0 : Dx = M DC = 0.0468EI∆ − − − (6) 4 73
  74. 74. Equilibrium Conditions: M BA + M BC = 0 − − − (1*) 10 − Ax − Dx = 0 − − − (2*) • Solve equation Substitute (2) and (3) in (1*) 2EI θB + 0.375EI ∆ = 5.625 ----(7) Slope-Deflection Equations: Substitute (5) and (6) in (2*) 2( EI ) M AB = θ B + 5.625 + 0.375EI∆ − − − (1) − 0.375EIθ B − 0.235EI∆ = −8.437 − − − (8) 4 4( EI ) M BA = θ B − 5.625 + 0.375EI∆ − − − (2) From (7) and (8) can solve; 4 − 5.6 44.8 3( EI ) θB = ∆= M BC = θ B − − − (3) EI EI 3 − 5.6 44.8 M DC = 0.1875EI∆ − − − (4) Substituteθ B = and ∆ = in (1)to (6) EI EI Horizontal reaction at supports: Ax = 0.375EIθ B + 0.1875EI∆ + 1.563 − − − (5) Dx = 0.0468EI∆ − − − (6) MAB = 15.88 kN•m MBA = 5.6 kN•m MBC = -5.6 kN•m MDC = 8.42 kN•m Ax = 7.9 kN Dx = 2.1 kN 74
  75. 75. C B 1m 10 kN MAB = 15.88 kN•m, MBA = 5.6 kN•m, MBC = -5.6 kN•m, MDC = 8.42 kN•m, Ax = 7.9 kN, Dx = 2.1 kN, 5.6 3m 8.42 D 2.1 kN 15.88 7.9 kN 3m A 5.6 C B B 7.8 A ∆ = 44.8/EI C ∆ = 44.8/EI 5.6 15.88 D 8.42 Bending moment diagram A θB = -5.6/EI D Deflected curve 75
  76. 76. Example 9 From the frame shown use the slope-deflection method to: (a) Determine the end moments of each member and reactions at supports (b) Draw the quantitative bending moment diagram, and also draw the qualitative deflected shape of the entire frame. B C 2 EI 10 kN 2m pin 4m 2.5 EI EI A D 4m 3m 76
  77. 77. • Overview B ∆ 2EI B´ 10 kN 2m A EI MAB Ax Ay 4m ∆ CD C C´ ∆BC ∆ 2.5EI • Unknowns θB and ∆ MDC • Boundary Conditions Dx D 3m 4m θA = θD = 0 Dy • Equilibrium Conditions - Joint B B MBA MBC ΣM B = 0 : M BA + M BC = 0 − − − (1*) - Entire Frame + → ΣFx = 0 : 10 − Ax − Dx = 0 − − − (2*) 77
  78. 78. • Slope-Deflection Equation C´ B ∆ ∆ CD ∆BC 5 B´ 2EI C ∆ 10 kN 36.87° 4m 2.5EI EI 2m PL/8 = 5 D A 4m 3m 0.375EI∆ B ∆´ B´ = ∆ / cos 36.87° = 1.25 ∆ C´ ∆CD ∆BC = ∆ tan 36.87° = 0.75 ∆ 36.87° ∆ C 2( EI ) θ B + 0.375EI∆ + 5 − − − (1) 4 (6)(2.5EI)(1.25∆)/(5)2 = 0.75EI∆ 4( EI ) M BA = θ B + 0.375EI∆ − 5 − − − (2) C´ 4 ∆CD= 1.25 ∆ 3(2 EI ) C M BC = θ B − 0.2813EI∆ − − − (3) 4 M AB = M DC = 0.375EI∆ − − − (4) A D 0.75EI∆ 6EI∆/(4) 2= 0.375EI∆ 0.5625EI∆ B B´ (1/2) 0.5625EI∆ C´ C (1/2) 0.75EI∆ (6)(2EI)(0.75∆)/(4) 2 = 0.5625EI∆ ∆BC= 0.75 ∆ 78
  79. 79. • Horizontal reactions B 2 EI pin C 10 kN 2m 4m 2.5 EI EI A D 4m 3m MBC MBA B B 10 kN C MBC MBC 4 4 C A Ax = (MBA+ MAB-20)/4 -----(5) MAB D Dx= (MDC-(3/4)MBC)/4 ---(6) MDC MBC/4 79
  80. 80. Equilibrium Conditions: M BA + M BC = 0 − − − (1*) 10 − Ax − Dx = 0 − − − (2*) Slope-Deflection Equation: • Solve equations Substitute (2) and (3) in (1*) 2.5EIθ B + 0.0938 EI∆ − 5 = 0 − − − (7) Substitute (5) and (6) in (2*) 2( EI ) 6 EI∆ 0.0938EIθ B + 0.334EI∆ − 5 = 0 − − − (8) θB + 5 + 2 − − − (1) 4 4 From (7) and (8) can solve; 6 EI∆ 4( EI ) M BA = θB −5 + 2 − − − (2) 4 4 1.45 − 14.56 θB = ∆= 3(2 EI ) 3(2 EI )(0.75∆) EI EI M BC = θB − − − − (3) 2 4 4 1.45 − 14.56 3(2.5EI )(1.25∆) Substituteθ B = and ∆ = in (1)to (6) M DC = − − − (4) EI EI 52 Horizontal reactions at supports: MAB = 15.88 kN•m MBA = 5.6 kN•m ( M BA + M AB − 20) − − − (5) Ax = MBC = -5.6 kN•m 4 MDC = 8.42 kN•m 3 Ax = 7.9 kN M DC − M BC 4 Dx = − − − (6) Dx = 2.1 kN 4 80 M AB =
  81. 81. 1.91 1.91 B 2 EI A pin C 10 kN 2m MAB = 11.19 kN•m MBA = 1.91 kN•m MBC = -1.91 kN•m 2.5 EI EI 11.19 kN•m 8.27 kN 0.478 kN 4m 5.46 4m 1.73 D MDC = 5.46 kN•m Ax = 8.28 kN•m Dx = 1.72 kN•m 0.478 kN 3m ∆ ∆ B 1.91 1.91 B C C 5.35 A θB=1.45/EI θB=1.45/EI 11.19 5.46 D Bending-moment diagram A D Deflected shape 81
  82. 82. Example 10 From the frame shown use the moment distribution method to: (a) Determine all the reactions at supports, and also (b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected curve. 3m 20 kN/m B pin C 3EI 3m 2EI 4m 4EI A D 82
  83. 83. • Overview • Unknowns ∆ ∆ 3m 20 kN/m B 3EI 2EI θB and ∆ C • Boundary Conditions 4EI θA = θD = 0 4m • Equilibrium Conditions - Joint B A [FEM]load D B 3m MBA MBC ΣM B = 0 : M BA + M BC = 0 − − − (1*) - Entire Frame + → ΣFx = 0 : 60 − Ax − Dx = 0 − − − (2*) 83
  84. 84. • Slope-Deflection Equation B 20 kN/m B 3m 6(2EI∆)/(3) = 1.333EI∆ 2 C 3EI = 15 wL2/12 2EI 4EI 6(2EI∆)/(3) 2 = 1.333EI∆ A A [FEM]load D 0 M BA 1.5EI∆ 4m wL2/12 = 15 M AB ∆ C ∆ 3m (1/2)(1.5EI∆) 6(4EI∆)/(4) 2 = 1.5EI∆ [FEM]∆ D 4(2 EI ) 2(2 EI ) = θA + θ B + 15 + 1.333EI∆ = 1.333EIθ B + 15 + 1.333EI∆ ----------(1) 3 3 0 2( 2 EI ) 4(2 EI ) = θA + θ B − 15 + 1.333EI∆ = 2.667 EIθ B − 15 + 1.333EI∆ ----------(2) 3 3 3(3EI ) θ B = 3EIθ B ----------(3) 3 0 3( 4 EI ) = θ D + 0.75 EI∆ = 0.75 EI∆ ----------(4) 4 M BC = M DC 84
  85. 85. • Horizontal reactions MAB C B 1.5 m + ΣM B = 0 : 60 kN 4m 1.5 m M BA + M AB + 60(1.5) 3 Ax = 1.333EIθ B + 0.889 EI∆ + 30 − − − (5) Ax = A Ax MBA D Dx MDC + ΣMC = 0: Dx = M DC = 0.188EI∆ − − − (6) 4 85
  86. 86. Equilibrium Conditions M BA + M BC = 0 − − − (1*) 60 − Ax − Dx = 0 − − − (2*) Equation of moment M AB = 1.333EIθ B + 15 + 1.333EI∆ − − − (1) M BA = 2.667 EIθ B − 15 + 1.333EI∆ − − − (2) M BC = 3EIθ B − − − (3) M DC = 0.75 EI∆ − − − ( 4) Horizontal reaction at support Ax = 1.333EIθ B + 0.889 EI∆ + 30 − − − (5) Dx = 0.188EI∆ − − − (6) • Solve equation Substitute (2) and (3) in (1*) 5.667 EIθ B + 1.333EI∆ = 15 − − − (7) Substitute (5) and (6) in (2*) − 1.333EIθ B − 1.077 EI∆ = −30 − − − (8) From (7) and (8), solve equations; θB = − 5.51 EI ∆= 34.67 EI 34.67 − 5.51 and ∆ = EI EI M AB = 53.87 kN • m M BA = 16.52 kN • m M BC = −16.52 kN • m Substituteθ B = in (1)to (6) M DC = 26.0 kN • m Ax = 53.48 kN Dx = 6.52 kN 86
  87. 87. B 3m 20 kN/m C 3m M AB = 53.87 kN • m M BA = 16.52 kN • m M BC = −16.52 kN • m 53.87 kN•m 4m M DC = 26.0 kN • m 16.52 kN•m Ax = 53.48 kN 53.48 kN A D 5.55 kN 26 kN•m Dx = 6.52 kN 6.52 kN 5.55 kN ∆ 16.52 C B ∆ B 16.52 C 53.87 A Moment diagram 26. D A Deflected shape D 87

×