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# Moment Distribution Method

Analysis of beams And Frames Using the moment Distribution Method ,, Depending on the Geometry and the material of the beam in consideration.

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### Moment Distribution Method

1. 1. DISPLACEMENT MEDTHOD OF ANALYSIS:MOMENT DISTRIBUTION ! Member Stiffness Factor (K) ! Distribution Factor (DF) ! Carry-Over Factor ! Distribution of Couple at Node ! Moment Distribution for Beams ! General Beams ! Symmetric Beams ! Moment Distribution for Frames: No Sidesway ! Moment Distribution for Frames: Sidesway 1
2. 2. Member Stiffness Factor (K) & Carry-Over Factor (COF) CB P w EI A B C D L1 /2 L1/2 L2 L3 Internal members and far-end member fixed at end support: 4 EI COF = 0.5 2 EI kCC = k DC = 4 EI L L K= L 1 C D COF = 0.5 B C K(BC) = 4EI/L2, K(CD) = 4EI/L3 2
3. 3. CB P w EI A B C D L1 /2 L1/2 L2 L3Far-end member pinned or roller end support: COF = 0 3EI 3EI k AA = 0 K= L L 1 C D K(AB) = 3EI/L1, K(BC) = 4EI/L2, K(CD) = 4EI/L3 3
4. 4. Joint Stiffness Factor (K) CB P w EI A B C D L1 /2 L1/2 L2 L3 K(AB) = 3EI/L1 K(BC) = 4EI/L2, K(CD) = 4EI/L3 Kjoint = KT = ΣKmember 4
5. 5. Distribution Factor (DF) CB P w EI A B C D L1 /2 L1/2 L2 L3 K DF = ΣK Notes: - far-end pined (DF = 1) - far-end fixed (DF = 0) A KAB/(K(AB) + K(BC) ) B KBC/(K(AB) + K(BC) ) C K(CD)/(K(BC) + K(CD) ) D DF 1 DFBA) DFBC DFCB DFCD 0 K(BC)/(K(BC) + K(CD) ) 5
6. 6. Distribution of Couple at Node CB (EI)1 (EI)2 (EI)3 A B C D L1 /2 L1/2 L2 L3 A B C D DF 1 DFBA DFBC DFCA DFCD 0 CB CO=0 CB(DFBC) CO=0.5 CB( DFBC) CB(DFBC) B CB( DFBC) 6
7. 7. 50 kN•m 2EI 3EI 3EI A B C D 4m 4m 8m 8m L1= L2 = L3 A B C DDF 1 0.333 0.667 0.5 0.5 0 50 kN•m CO=0 CO=0.5 50(.667) 50(.667) 16.67 50(.333) 50(.333) B 7
8. 8. Distribution of Fixed-End Moments P w (EI)1 M (EI)2 (EI)13 A F B C D L1 /2 L1/2 L2 L3 L1= L2 = 8 m, L3 = 10 m A B C D DF 1 DFBA DFBC DFCB DFCD 0 MF MF B MF 0 MF(DFBC) MF( DFBC) 0.5 0 MF(DFBC) B MF( DFBC) 8
9. 9. P w EI A 1.5PL1/8=30 B wL22/12=16 C wL32/12=25 D L1 /2 L1/2 L2 L3 L1= L2 = 8 m, L3 = 10 m A B C DDF 1 0.4 0.6 0.5 0.5 0 14 30 30 B 16 16 14 0 5.6 5.6 0.5 4.2 0 B 8.4 8.4 9
10. 10. Moment Distribution for Beams 10
11. 11. Example 1The support B of the beam shown (E = 200 GPa, I = 50x106 mm4 ).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 shape. 20 kN 3 kN/m 2EI 3EI A B C 4m 4m 8m 11
12. 12. 20 kN 3 kN/m 2EI 3EI 30 16 16 A B C CO 0.5 0 0.5 0 4m 4m 8m K1 = 3(2EI)/8 K2 = 4(3EI)/8 K1/(K1+ K2) K2/(K1+ K2) DF 1 0.333 0.667 0 [FEM]load -30 16 -16 Dist. 4.662 9.338 CO 4.669 Σ -25.34 25.34 -11.33 20 kN 24 kNA B B C 25.34 kN•m 11.33 kN•m6.83 kN 13.17 kN 13.75 kN 10.25 kN 12
13. 13. Note : Using the Slope 20 kN 3 kN/m Deflection 2EI 3EI 20+10 16 16 A B C 4m 4m 8m 3(2 EI ) M BA = θ B − 30 − − − (1) 8 4(3EI ) M BC = θ B + 16 − − − (2) 8 MBA MBC + ΣMB = 0: -MBA - MBC = 0 B (0.75 + 1.5)EIθB - 30 + 16 = 0 θB = 6.22/EI MBA = -25.33 kN•m, MBC = 25.33 kN•m 2(3EI ) M CB = θ B − 16 = −11.33 kN • m 8 13
14. 14. 20 kN 3 kN/m A C 11.33 B 6.83 kN 13.17 + 13.75 = 26.92 kN 10.25 kN 4m 4m 8m 13.75 6.83V (kN) + + x (m) - 4.58 m - -10.25 -13.17 27.32 6.13 M + +(kN•m) - - x (m) -11.33 -25.33Deflected shape x (m) 14
15. 15. Example 2From the beam 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 shape. 10 kN 50 kN•m 5 kN/m 2EI 3EI 3EI A B C D 4m 4m 8m 8m 15
16. 16. 10 kN 50 kN•m 5 kN/m 2EI 15 26.67 26.67 40 3EI A 0.5 B 0.5 C 0.5 D 4m 4m 8m 8m K1 = 3(2EI)/8 K2 = 4(3EI)/8 K3 = 3(3EI)/8 K1/(K1+ K2) K2/(K1+ K2) K2/(K2+ K3) K3/(K2+ K3) DF 1 0.333 0.667 0.571 0.428 1Joint couple -16.65 -33.35 50 kN•m CO=0 CO=0.5 50(.667) 50(.667) 50(.333) 50(.333) B 16
17. 17. 10 kN 50 kN•m 5 kN/m 2EI 15 26.67 26.67 40 3EI A 0.5 B 0.5 C 0.5 D 4m 4m 8m 8m K1 = 3(2EI)/8 K2 = 4(3EI)/8 K3 = 3(3EI)/8 K1/(K1+ K2) K2/(K1+ K2) K2/(K2+ K3) K3/(K2+ K3) DF 1 0.333 0.667 0.571 0.429 1Joint couple -16.65 -33.35 CO -16.675 FEM -15 26.667 -26.667 40 Dist. -3.885 -7.782 1.905 1.437 CO 0.953 -3.891 Dist. -0.317 -0.636 2.218 1.673 CO 1.109 -0.318 Dist. -0.369 -0.740 0.181 0.137 Σ -36.22 -13.78 -43.28 43.25 17
18. 18. 10 kN 50 kN•m 5 kN/m 2EI 36.22 13.78 43.25 43.25 3EI A B C D 4m 4m 8m 8m 10 kN 40 kN 36.22 kN•mA B 43.25 kN•m C DAy = 0.47 kN ByL = 9.53 kN CyR = 25.41 kN Dy = 14.59 kN 40 kN 13.78 kN•m 43.25 kN•m B C ByR = 12.87 kN CyL= 27.13 kN 18
19. 19. 10 kN 50 kN•m 5 kN/m A D 2EI B C 3EI 0.47 kN 14.59 kN 9.53+12.87=22.4 kN 27.13+25.41=52.54 kN 4m 4m 8m 8m 12.87 25.41 2.92 m V (kN) 0.47 x (m) 2.57 m -9.53 -14.59 30.32 -27.13 21.29 13.78M(kN•m) 1.88 x (m) -36.22 -43.25Deflected x (m) shape 19
20. 20. Example 3From the beam 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 shape. 2EI 40 kN 50 kN•m 10 kN/m 40 kN A D EI B C 3m 3m 9m 3m 20
21. 21. 120 kN•m 40 kN 2EI 40 kN 50 kN•m 10 kN/m A D 30 30 101.25 EI B C 0.5 0.5 3m 3m 9m 3m K1 = 4(2EI)/6 K2 = 3(EI)/9 K1/(K1+ K2) K2/(K1+ K2) DF 0 0.80 0.20 1Joint couple 40 10 -120 CO 20 -60 FEM 30 -30 101.25 Dist. Dist. -9 -2.25 CO -4.5 Σ 45.5 1 49 -120 21
22. 22. 2EI 40 kN 50 kN•m 10 kN/m 40 kN 1 A D 49 120 120 45.5 EI B C 3m 3m 9m 3m 40 kN 40 kN45.5 kN•m 1 kN•m A B 120 kN•m C D Ay = 27.75 kN ByL = 12.25 kN CyR = 40 kN 90 kN 49 kN•m 120 kN•m B C ByR = 37.11 kN CyL = 52.89 kN 22
23. 23. 2EI 40 kN 50 kN•m 10 kN/m 40 kN A D EI B C 27.75 kN 12.25+37.11 = 49.36 kN 52.89+40 = 92.89 kN 3m 3m 9m 3m 37.11 40 27.75 V (kN) + + + - x (m) -12.25 3.71 m - 37.75 -52.89 19.84M(kN•m) + 1 + x (m) - - - -45.5 -49 -120Deflected x (m) shape 23
24. 24. Example 4The support B of the beam shown (E = 200 GPa, I = 50x106 mm4 ) settles 10 mm.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 shape. 20 kN 15 kN•m12 kN•m 3 kN/m 2EI 3EI A B C 4m 4m 8m 24
25. 25. 20 kN 15 kN•m 12 kN•m 3 kN/m B A C 2EI 30 16 3EI 0.5 4 m 16 4m 10 mm 8 m 0.5 6(2 EI )∆ 6(2 EI )∆ 6(3EI ) ∆ A − = 9.375 6(3EI ) ∆ = 28.125 L 2 2L 2 = 28.125 L2 L2 ∆ [FEM]∆ ∆ B B6(2 EI )∆ C L2 K1 = 3(2EI)/8 K2 = 4(3EI)/8 K1/(K1+ K2) K2/(K1+ K2) DF 1 0.333 0.667 0 Joint couple -12 5 10 CO -6 5 [FEM]load -30 16 -16 [FEM]∆ 9.375 -28.125 -28.125 Dist. 12.90 25.85 CO 12.92 Σ -12 -8.72 23.72 -26.20 25
26. 26. Note : Using the 20 kN 18.75-9.375 15 kN•m 3 kN/mslope deflection 12 kN•m (9.375 )∆ B (28.125)∆ A C 2EI 3EI 28.125 (20+10)load (16)load 16 4m 4m 10 mm 8m 15 kN•m -12 MBA MBC 4(2 EI ) 2(2 EI ) M AB = θA + θ B + 20 − 18.75 − − − (1) 8 8 B 2(2 EI ) 4(2 EI ) M BA = θA + θ B − 20 + 18.75 − − − (2) 8 8 + ΣMB = 0: - MBA - MBC + 15 = 0( 2) − (1) 3(2 EI ) : M BA = θ B − 30 + 9.375 − 12 / 2 − − − (2a) 2 8 (0.75 + 1.5)EIθB - 38.75 - 15 = 0 4(3EI ) θB = 23.9/EI M BC = θ B + 16 − 28.125 − − − (3) 8 MBA = -8.7 kN•m, 2(3EI ) M CB = θ B − 16 − 28.125 − − − (4) 8 MBC = 23.72 kN•m 2(3EI ) M CB = θ B − 16 − 28.125 8 = −26.2 kN • m 26
27. 27. 20 kN 15 kN•m 12 kN•m 3 kN/m 2EI 8.725 23.7253EI 26.205 A B C 4m 4m 8m 20 kN 24 kN12 kN•m 8.725 kN•m 26.205 kN•m A B B C 23.725 kN•m Ay = 7.41 kN ByL = 12.59 kN ByR = 11.69 kN Cy = 12.31 kN 27
28. 28. 20 kN 15 kN•m 12 kN•m 3 kN/m A C 2EI B 3EI 26.205 7.41 kN 12.59+11.69 = 24.28 kN 12.31 kN 4m 4m 8m 7.41 11.69V (kN) + + x (m) - 3.9 m - -12.31 -12.59 41.64 12 + M(kN•m) x (m) -8.725 - -0.93 - -23.725 -26.205Deflected 10 mm θB = 23.9/EIshape x (m) 28
29. 29. Example 5For the beam shown, support A settles 10 mm downward, use the momentdistribution method to(a)Determine all the reactions at supports(b)Draw its quantitative shear, bending moment diagrams, and qualitativedeflected shape.Take E= 200 GPa, I = 50(106) mm4. 12 kN•m 6 kN/m B A 2EI C 1.5EI 10 mm 3m 3m 29
30. 30. 12 kN•m 4.5+(4.5/2) 6 kN/m = 6.75 B A 4.5 10 mm 2EI C 1.5EI 3m 0.5 3 m 0.5 100-(100/2) = 50 6(1.5 × 200 × 50)(0.01) 0.01 m C 32 MF∆ A = 100 kN • m K1 = 4(2EI)/3 K2 = 3(1.5EI)/3 K1/(K1+ K2) K2/(K1+ K2) DF 0 0.64 0.36 1Joint couple 12 CO 6 [FEM]load 6.75 [FEM]∆ 50 Dist. -40.16 -22.59 CO -20.08 Σ -20.08 -40.16 40.16 12 30
31. 31. 12 kN•m 6 kN/m B A 2EI C 1.5EI 3m 3m ΣΜ -20.08 -10.16 40.16 1220.08 kN•m 40.16 kN•m B C 40.16 + 20.08 20.08 kN = 20.08 kN 3 18 kN 12 kN•m 6 kN/m 40.16 kN•m C A 26.39 kN 8.39 kN 31
32. 32. 12 kN•m 6 kN/m B A 2EI C 1.5EI 10 mm 3m 3m20.08 kN•m 40.16 kN•m B C 20.08 kN 20.08 kN 12 kN•m 6 kN/m 40.16 kN•m C A 26.39 kN 8.39 kN V (kN) 26.39 + 8.39 x (m) - M (kN•m) -20.08 20.08 12 x (m)Deflected shape -40.16 x (m) 32
33. 33. Example 6For the beam shown, support A settles 10 mm downward, use the momentdistribution method to(a)Determine all the reactions at supports(b)Draw its quantitative shear, bending moment diagrams, and qualitativedeflected shape.Take E= 200 GPa, I = 50(106) mm4. 12 kN•m 6 kN/m B A 2EI C 1.5EI 10 mm 3m 3m 33
34. 34. • Overview 12 kN•m 6 kN/m B A C 10 mm 2EI 1.5EI 3m 3m 12 kN•m 6 kN/m B A 10 mm C R ∆ B ×C A R´ R + R C = 0 − − − (1*) 34
35. 35. • Artificial joint applied 12 kN•m 4.5+(4.5/2) 6 kN/m = 6.75 B A 4.5 10 mm 2EI C 1.5EI 3m 0.5 3 m 0.5 100-(100/2) = 50 6(1.5 × 200 × 50)(0.01) 0.01 m C 32 MF∆ A = 100 kN • m K1 = 4(2EI)/3 K2 = 3(1.5EI)/3 K1/(K1+ K2) K2/(K1+ K2) DF 0 0.64 0.36 1 Joint couple 12 CO 6 [FEM]load 6.75 [FEM]∆ 50 Dist. -40.16 -22.59 CO -20.08 Σ -20.08 -40.16 40.16 12 35
36. 36. 12 kN•m 6 kN/m B A 2EI C 1.5EI 3m 3m ΣΜ -20.08 -40.16 40.16 1220.08 kN•m 40.16 kN•m B C 40.16 + 20.08 18 kN 20.08 kN = 20.08 kN 12 kN•m 3 6 kN/m 40.16 kN•m C A 26.39 kN 8.39 kN 40.16 kN•m 40.16 kN•m 20.08 C 26.39 kN + ↑ ΣFy = 0 : − 20.08 − 26.39 + R = 0 R R = 46.47 kN 36
37. 37. • Artificial joint removed ∆ B A C 2EI 0.5 1.5EI 0.5 3m R´ 3m C C 75 B ∆ ∆ 6(1.5EI )( ) 100 EI = 75 75-(75/2) A 32 6(2 EI ) ∆ C 75 = 100 → ∆ = = 37.5 32 EI DF 0 0.64 0.36 1 [FEM]∆ -100 -100 +37.5 Dist. 40 22.5 CO 20 Σ -80 -60 6080 kN•m 60 kN•m 60 kN•m B C C A 46.67 kN 46.67 kN 20 kN 20 kN C 46.67 20 kN + ↑ ΣFy = 0 : R = 66.67 kN 37 R´
38. 38. • Solve equation Substitute R = 46.47 kN and R = 66.67 kN in (1*) 46.47 + 66.67C = 0 C = −0.6970 12 kN•m 6 kN/m B 20.08 kN•m A 10 mm 20.08 kN C R = 46.47 kN 8.39 kN ∆ B 80 kN•m A × C = −0.6970 46.67 kN 20 kN R´ = 66.67 kN 12 kN•m 6 kN/m 35.68 kN•m B A 2EI C 1.5EI 12.45 kN 5.55 kN 38
39. 39. 12 kN•m 6 kN/m35.68 kN•m B A 2EI C 1.5EI 10 mm 12.45 kN 5.55 kN 3m 3m V (kN) 12.45 0.925 m + x (m) -5.55 M (kN•m) 14.57 12 1.67 + x (m) - -35.68 Deflected shape x (m) 39
40. 40. Symmetric Beam • Symmetric Beam and Loading P P θ θ A B C D L´ L L´ real beamV´B L L V´C M L 2 2 + ΣMC´ = 0: − VB ( L ) + ( L)( ) = 0 EI 2 B´ C´ ML M VB = θ = M 2 EI L EI EI 2 EI conjugate beam M= θ L The stiffness factor for the center span is, therefore, 2 EI K= L 40
41. 41. • Symmetric Beam with Antisymmetric Loading P θ A D B θ C P L´ L L´ real beam 1 M L M 1 M L 2L ( )( ) + ΣMC´ = 0: − VB ( L) + ( )( )( ) = 0 2 EI 2 EI 2 EI 2 3V´B 2 ML L VB = θ = B´ 3 6 EI C´ 6 EI 1 M L V´C M= θ M ( )( ) L 2 EI 2 EI The stiffness factor for the center span is, therefore, conjugate beam 6 EI K= L 41
42. 42. Example 5aDetermine all the reactions at supports for the beam below. EI is constant. 15 kN/m A D B C 4m 6m 4m 42
43. 43. 15 kN/m A D wL2/15 = 16 B wL2/12 = 45 C wL2/15 = 16 4m 6m 4m 3EI 3EI 2 EI 2 EI K ( AB ) = = , K ( BC ) = = L 4 L 6 K ( AB ) K ( AB ) (3EI / 4) ( DF ) AB = = 1,( DF ) BA = = = 0.692, K ( AB ) K ( AB ) + K ( BC ) (3EI / 4) + ( 2 EI / 6) K ( BC ) (2 EI / 6) ( DF ) BC = = = 0.308 K ( AB ) + K ( BC ) (3EI / 4) + (2 EI / 6) DF 1.0 0.692 0.308 [FEM]load 0 -16 +45 Dist. -20.07 -8.93 ΣΜ -36.07 +36.07 30 kN 30 kN 8 90 kN 8 36.07 kN•m 36.07 kN•m 3 3 A B B C C D 4m 3m 3m 4m0.98 kN 29.02 kN 29.02 kN 0.98 kN 43 45 kN 45 kN
44. 44. 30 kN 30 kN 8 90 kN 8 36.07 kN•m 36.07 kN•m 3 3 A B B C C D 4m 3m 3m 4m0.98 kN 29.02 kN 29.02 kN 0.98 kN 45 kN 45 kN 15 kN/m A D B 74.02 kN C 74.02 kN 0.98 kN 0.98 kN 4m 6m 4m 45 29.02 V 0.98 (kN) x (m) -0.98 -29.02 -45 31.42 M + (kN•m) x (m) - - -36.07 -36.07 Deflected shape 44
45. 45. Example 5bDetermine all the reactions at supports for the beam below. EI is constant. 15 kN/m A C D B 15 kN/m 4m 3m 3m 4m 45
46. 46. Fixed End Moment 15 kN/m A C D wL2/15 = 16 B 11wL2/192 5wL2/192 = 30.938 = 14.063 5wL2/192 11wL2/192 = 14.063 = 30.938 wL2/15 = 16 A C D B 15 kN/m 15 kN/m 16.875 16 A C D 16 B 16.875 15 kN/m 46
47. 47. 15 kN/m 16.875 16 A C D 16 B 16.875 15 kN/m 4m 3m 3m 4m 3EI 3EI 6 EI 6 EI K ( AB ) = = = 0.75EI , K ( BC ) = = = EI L 4 L 6 0.75 1 ( DF ) AB = 1, ( DF ) BA = = 0.429, ( DF ) BC = = 0.571 0.75 + 1 0.75 + 1 DF 1.0 0.429 0.571 [FEM]load 0 -16 16.875 Dist. -0.375 -0.50 ΣΜ -16.375 16.375 8 30 kN 8 45 kN 3 16.375 kN•m 16.375 kN•m 4m 3 A B B C C D 4m5.91 kN 24.09 kN 27.96 kN 5.91 kN 27.96 kN 45 kN 24.09 kN 30 kN 47
48. 48. 8 30 kN 8 45 kN 3 16.375 kN•m 16.375 kN•m 4m 3 A B B C C D 4m5.91 kN 24.09 kN 27.96 kN 5.91 kN 27.96 kN 45 kN 24.09 kN 30 kN 15 kN/m A C D B 15 kN/m 5.91 kN 52.05 kN 52.05 kN 5.91 kN 27.96 27.96 V 5.91 5.91 (kN) x (m) -24.09 -24.09 M 16.375 (kN•m) x (m) -16.375 Deflected shape 48
49. 49. Moment Distribution Frames: No Sidesway 49
50. 50. Example 6From the frame shown use the moment distribution method to:(a) Determine all the reactions at supports(b) Draw its quantitative shear and bending moment diagrams,and qualitative deflected shape. 40 kN•m 48 kN 8 kN/m B C A 2.5EI 2.5EI 4m 3EI 5m D 5m 50
51. 51. 40 kN•m 48 kN 8 kN/m B C A 2.5EI 45 25 2.5EI 0.5 0.5 4mKAB = KBC = 3(2.5EI)/5 = 1.5 EI 3EI 0.5KBD = 4(3EI)/4 = 3EI 5m D 5m A B D C Member AB BA BC BD DB CB DF 1 0.25 0.25 0.5 0 1 Joint load -10 -10 -20 CO -10 FEM -45 25 Dist. 5 5 10 CO 5 Σ 0 -50 20 -10 -5 0 51
52. 52. 40 kN•m 48 kN 8 kN/m B CA 20 2.5EI 50 2.5EI 14 kN 10 16 kN 3EI 5 Member AB BA BC BD DB CB D Σ 0 -50 20 -10 -5 0 40 kN•m 58 kN 34 24 3.75 50 20 48 kN 3.75 40 kNA B 0 50 kN•m 58 10 B C 3.75 58 20 3.75 14 kN 34 kN 10 kN•m 16 kN 24 kN 3.75 kN 5 3.75 kN 58 52
53. 53. 40 kN•m 48 kN 8 kN/m B C A 2.5EI 20 2.5EI 14 kN 50 16 kN 3EI 10 4m 5 D 5m 5m 58 kN35 16 10 20 50 Moment diagram Deflected shape 5 53
54. 54. Moment Distribution for Frames: Sidesway 54
55. 55. Single Frames P P ∆ ∆ C C R´B C B R B x C1A D A D A D Artificial joint applied Artificial joint removed (no sidesway) (sidesway) 0 = R + C1R´ 55
56. 56. Multistory Frames ∆2 P3 P2 P4 P1 ∆1 P3 ∆´´ ∆´´P2 R2´´ R2 R2´ ∆´ ∆´ P4 x C1 x C2P1 R1´ R1 R1´´ 0 =R2 + C1R2´ + C2R2´´ 0 =R1 + C1R1´ + C2R2´´ 56
57. 57. Example 7From 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 shape.EI is constant. 16 kN 1m 4m B C 5m A D 57
58. 58. • Overview 16 kN 16 kN 1m 4m C C R´ B C B R B5m 5m = + x C1 A D A D A D artificial joint applied artificial joint removed (no sidesway) ( sidesway) R + C1R´ = 0 ---------(1) 58
59. 59. • Artificial joint applied (no sidesway) Fixed end moment: 16 kN b=1m a=4m C B R Pa2b/L2 Pb2a/L2 = 10.24 = 2.56 A D Equilibrium condition : + ΣF = 0: A + D + R = 0 x x x 59
60. 60. 16 kN 1m 4m C R A B C D B 0.5 DF 0 0.50 0.50 0.50 0.50 10.24 2.56 0 FEM 10.24 -2.565m 5m Dist. -5.12 -5.12 1.28 1.28 0.5 0.5 CO -2.56 0.64 -2.56 0.64 A D Dist. -0.32 -0.32 1.28 1.28 CO -0.16 0.64 -0.16 0.645.78 kN•m Dist. -0.32 -0.32 0.08 0.08 2.72 kN•m CO -0.16 0.04 -0.16 0.04 Dist. -0.02 -0.02 0.08 0.08B C Σ -2.88 -5.78 5.78 -2.72 2.72 1.32 5m 5m Equilibrium condition : + ΣF = 0: 1.73 - 0.81 + R = 0 xA Ax = 1.73 kN D Dx = 0.81 kN R = - 0.92 kN2.88 kN•m 1.32 kN•m 60
61. 61. • Artificial joint removed ( sidesway) Fixed end moment: Since both B and C happen to be displaced the same amount ∆, and AB and DC have the same E, I, and L so we will assume fixed-end moment to be 100 kN•m. ∆ ∆ B 5m C 100 kN•m R´ 100 kN•m 5m 5m 100 kN•m 100 kN•m A D Equilibrium condition : + ΣF = 0: A + D + R´ = 0 x x x 61
62. 62. ∆ ∆ B 5m C R´ A B C D 0.5 0.50 0.50 100 kN•m 100 kN•m DF 0 0.50 0.50 0 FEM 100 100 100 1005m 5m Dist. -50 -50 -50 -50 0.5 0.5 100 kN•m 100 kN•m CO -25.0 -25.0 -25.0 -25.0A Dist. 12.5 12.5 12.5 12.5 D CO 6.5 6.5 6.5 6.5 Dist. -3.125 -3.125 -3.125 -3.125 60 kN•m 60 kN•m CO -1.56 -1.56 -1.56 -1.56 Dist. 0.78 0.78 0.78 0.78 CO 0.39 0.39 0.39 0.39B C Dist. -0.195 -0.195 -0.195 -0.195 Σ 80 60 -60 -60 60 80 5m 5m Equilibrium condition: + ΣFx = 0:A Ax = 28 kN D Dx = 28 kN -28 - 28 + R´ = 0 80 kN•m 80 kN•m R´ = 56 kN 62
63. 63. Substitute R = -0.92 and R´= 56 in (1) : R + C1R´ = 0 -0.92 + C1(56) = 0 0.92 C1 = 56 16 kN 16 kNB C C R´ 1m 4m C R B B 5.78 2.72 60 60 4.79 3.71 2.72 60 60 3.71 5.78 4.79 + x C1 = 5m 5m 2.88 80 80 1.57 2.63 1.32 1.73 kN 0.81 28 28 1.27 kN A D A D A D 1.27 63
64. 64. 8.22 16 kN 3.71 1m 4m C B 4.79 4.79 3.71 3.71 4.79 4.79 3.71 5m 5m 1.57 2.63 1.27 kN 1.27 kN 1.57 2.63A D 2.99 kN Bending moment 13.01 kN diagram (kN•m) ∆ ∆ Deflected shape 64
65. 65. Example 8From 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, andqualitative deflected shape. 20 kN/m 3m B pin C 3EI 3m 2EI 4EI 4m A D 65
66. 66. • Overview 20 kN/mB 3m, 3EI B C B C C R R´ 3 m , 2EI 4m , 4EI = + x C1 A A A D D D artificial joint applied artificial joint removed (no sidesway) (sidesway) R + C1R´ = 0 ----------(1) 66
67. 67. • Artificial joint applied (no sidesway) 20 kN/m A B C DB 3m , 3EI C R 0.471 0.529 DF 0 1.00 1.00 0 0.5 15 FEM 15.00 -15.00 0.5 Dist. 7.065 7.935 0.5 CO 3.533 3 m , 2EI 4m, 4EI 15 Σ 18.53 -7.94 7.94 A D 7.94 kN•mKBA = 4(2EI)/3 = 2.667EI B C + ΣF = 0: xKBC = 3(3EI)/3 = 3EI 60 - 33.53 - 0 + R = 0 60KCD = 3(4EI)/4 = 3EI 3m 4m R = - 26.47 kN A Ax = 33.53 kN D Dx = 0 18.53 kN•m 0 67
68. 68. • Artificial joint removed ( sidesway) • Fixed end moment ∆ ∆ ∆ ∆ B 3m, 3EI C B 3m, 3EI C R´ R´100 kN•m6(2EI∆)/(3) 2 6(4EI)∆/(4) 2 100 kN•m 3 m, 2EI 100 kN•m 4m, 4EI 3 m, 2EI 4m, 4EI 6(2EI∆)/(3) 2 100 kN•m 3(4EI∆)/(4) 2 3(4EI)(75/EI)/(4) 2 A A = 56.25 kN•m D D Assign a value of (FEM)AB = (FEM)BA = 100 kN•m 6(2 EI ) ∆ 2 = 100 3 ∆AB = 75/EI 68
69. 69. ∆ ∆ A B C DB 3m C R´ 0.5 DF 0 0.471 0.529 1.00 1.00 0100 FEM 100 100 56.25 0.5 Dist. -47.1 -52.9 0 0.5 3m 4m CO -28.55 100 Σ 76.45 52.9 -52.9 56.25 56.25 A D 52.9. kN•m + ΣF = 0: C x B -43.12 - 14.06 + R´ = 0 3m 4m R´ = 57.18 kN A Ax = 43.12kN D 14.06 kN 76.45 kN•m 56.25 kN•m 69
70. 70. Substitute R = -26.37 and R´= 57.18 in (1) : R + C1R´ = 0 -26.47 + C1(57.18) = 0 26.47 C1 = 57.18 20 kN/mB B 52.9 kN•m B C C 3m R C 7.94 kN•m R´ 16.55 kN•m 7.94 kN•m 52.9 = 3m + 76.45 kN•m x C1 4m 18.53 kN•m 53.92 kN•m 33.53 kN 43.12 kN 53.49 kN A A 56.25 0 A 26.04 kN•m 5.52 kN kN 6.51 0 D D 14.06 D 5.52 kN 70
71. 71. 20 kN/m 16.55B 3m 16.55 C B C 16.55 kN•m 3m 53.92 kN•m 4m 53.92 53.49 kNA 26.04 kN•m A 5.52 kN kN 6.51 26.04 Moment diagram ∆ ∆ D D 5.52 kN B C A Deflected shape 71 D
72. 72. Example 8From 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 shape.EI is constant. 10 kN B C 4m A D 4m 3m 72
73. 73. • Overview 10 kN B C 4m A D 4m 3m R + C1R´ = 0 ---------(1) =10 kN B C B C R R´ + x C1 A D A D artificial joint applied artificial joint removed (no sidesway) (sidesway) 73
74. 74. • Artificial joint applied (no sidesway) 10 kN B C R 0 0 A D Equilibrium condition : + ΣF = 0: x 10 + R = 0 R = - 10 kN 74
75. 75. • Artificial joint removed (sidesway) • Fixed end moment 6EI∆BC/(4) 2 B R´ 6EI∆AB/(4) 2 C 3EI∆CD/(5) 2 100 kN•m 100 kN•m 6EI∆CD/(5) 2 4 m 6EI∆AB/(4) 2 A D 4m 3m Assign a value of (FEM)AB = (FEM)BA = 100 kN•m : 6 EI∆ AB = 100 42 ∆AB = 266.667/EI ∆ CD ∆BC B∆ C´ R´ ∆ ∆CD = ∆ / cos 36.87° = 1.25 ∆ = 1.25(266.667/EI) B´ C = 333.334/EI 36.87° C´ 4m ∆CD ∆BC = ∆ tan 36.87° = 0.75 ∆ 36.87° = 0.75(266.667/EI) A D C ∆AB = ∆ = 200/EI 4m 3m 75
76. 76. ∆BC= 200/EI, ∆CD = 333.334/EI 6EI∆BC/(4) 2 = 6(200)/42 = 75 kN•m B R´ 100 kN•m C 3EI∆CD/(5) 2 = 3(333.334)/52 = 40 kN•m 100 kN•m A D Equilibrium condition : + ΣF = 0: A + D + R´ = 0 x x x 76
77. 77. 75 A B C D B 75 R´ DF 0 0.50 0.50 0.625 0.375 1 0.5 C100 FEM 100 100 -75 -75 40 40 4 m Dist. -12.5 -12.5 21.875 13.125 0.5 0.5100 CO -6.25 10.938 -6.25A D Dist. -5.469 -5.469 3.906 2.344 4m 3m CO -2.735 1.953 -2.735 Dist. -0.977 -0.977 1.709 1.026KBA = 4EI/4 = EI, KBC = 4EI/4 = EI,KCD = 3EI/5 = 0.6EI Σ 91.02 81.05 -81.05 -56.48 56.4834.38 kN 34.38 kN 81.05 81.05 B C 56.48B 56.48 C 34.38 kN 34.38 kN + ΣF = 0: 39.91 kN A 43.02 kN x D 91.02 -43.02 - 39.91 + R´ = 0 34.38 kN R´ = 82.93 kN 34.38 kN 77
78. 78. Substitute R = -10 kN and R´= 82.93 kN in (1) : -10 + C1(82.93) = 0 R + C1R´ = 0 ---------(1) C1 = 10/82.93 81.0510 kN B C B C R R´ 81.05 56.48 x C1= 10/82.93 + 56.48 0 0 91.02 39.91 kN A 0 D A 43.02 kN D 0 = 34.38 kN 0 34.38 kN 9.77 10 kN B C 9.77 6.81 6.81 4m 10.98 4.81 kN A 5.19 kN D 4.15 kN 4.15 kN 4m 3m 78
79. 79. 9.77 10 kN B C 9.77 6.81 6.81 4m 10.98 A 5.19 kN 4.81 kN D 4.15 kN 4.15 kN 4m 3m 9.77 B B9.77 C C 6.81 A 10.98 A D D Bending moment diagram Deflected shape (kN•m) 79
80. 80. Example 9From 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,andqualitative deflected shape.EI is constant. 40 kN 20 kN B C 3EI 3m 4EI 4EI 4m A D 2m 3m 2m 80
81. 81. 40 kN• Overview 20 kN B C 3EI 3m 4EI 4EI 4m A D 2m 3m 2m R + C1R´ = 0 ----------(1) 40 kN = 20 kN B C B C R R´ + x C1 artificial joint applied artificial joint removed (no sidesway) (no sidesway)A A D D 2m 3m 2m 2m 3m 2m 81
82. 82. • Artificial joint applied (no sidesway) Fixed end moments: 40 kN PL/8 = 15 B C 20 kN R 15+(15/2) = 22.5 kN•m A D 2m 3m 2m Equilibrium condition : + ΣF = 0: A + D + R = 0 x x x 82
83. 83. A B C D 40 kN 20 kN B C DF 0 0.60 0.40 1.00 1.00 0 R 22.5 FEM 22.5 15 0.5 Dist. -13.5 -9.0 0.5 0.5 CO -6.75 Σ -6.75 -13.5 13.5 A D 2m 3m 2m KBA = 4(4EI)/3.6 = 4.444EI, KBC = 3(3EI)/3 = 3EI, 24.5 kN 15.5 kN 13.5 kN•m C + ΣF = 0: B x 40 kN 23.08 + 20 -7.75 + R´ = 0 13.5 B C R´ = - 35.33 kNA 7.75 kN 23.08 kN 24.5 kN 15.5 kN D 6.75 kN•m 024.5 kN 15.5 kN 83
84. 84. • Artificial joint removed (sidesway) Fixed end moments: 3m B C R´ m 3EI 06 4. 4 3.6 72 m 4EI 4EI A D 3(3EI)∆BC/(3) 2 6(3EI)∆BC/(3) 2 100 kN•m R´ 6(4EI)∆AB/(3.606) 2 B C 6(4EI)∆CD/(4.47) 2 100 kN•m 6(4EI)∆AB/(3.61) 2 3(4EI)∆CD/(4.472) 2 A DAssign a value of (FEM)AB = (FEM)BA = 100 kN•m : 6(4 EI ) ∆ AB = 100, ∆AB = 54.18/EI 3.612 84
85. 85. CC´ ∆ = ∆ABcos 33.69° = 45.08/EI B BB´ C C´ R´ C´ B´ ∆ CD .69 o C ∆ tan 26.57 = 22.54/EI 26.57° 26.57° 33 B∆ 33.69o AB =5 ∆ tan 33.69 = 30.05/EI 4.3A /E I D B´ ∆ BC = B C = 22.54 / EI + 30.05 / EI = 52.59 / EI ∆CD = ∆/cos 26.57°= 50.4/EI 3(3EI)∆BC/(3) 2 = 3(3EI)(52.59/EI) /(3) 2 = 52.59 kN•m R´100 kN•m B C 100 kN•m 3(4EI)∆CD/(4.472) 2 A = 3(4EI)(50.4/EI)/(4.472) 2 D = 30.24 kN•m 85
86. 86. A B C D 3EI B 52.59 C DF 0 0.60 0.40 1.00 1.00 0 R´ 100 0.5 FEM 100 100 -52.59 30.243m Dist. -28.45 -18.96100 0.5 0.5 4m CO -14.223 4EI Σ 85.78 71.55 -71.55 4EI 30.24 A 30.24 D 2m 3m 2m 23.85 kN 23.85 kN 71.55 kN•m + ΣF = 0: x C B -68.34 - 19.49 + R´ = 0 71.55 B C R´ = 87.83 kNA 68.34 kN 23.85 23.85 19.49 kN D 85.78kN•m 30.24 kN•m23.85 kN 23.85 kN 86