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Moment Distribution Method.ppt

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VigneshVenkatasubram2

This document discusses methods for analyzing structural frames including fixed end moments, relative stiffness, distribution factors, and multi-stage solutions for frames with sway. Key steps and equations are presented for determining fixed end moments, relative stiffness ratios, distribution factors, distribution moments, and carry over moments. Worked examples are provided to demonstrate applying these methods to analyze beams and frames with and without sway.

Engineering
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1 Lecture No. : 5 ‫المحاضرة‬ ‫الخامسة‬
2 Fixed End Moment Relative Stiffness Distribution Factor Table : FEM RS DF Fixed End Moment Distribution Moment Carry Over Moment Distribution Moment Solution Steps :
3 Fixed End Moment FEM L P L w 8 L P 12 L w 2 L w L P a b w L2 20 w L2 30 P a b2 L2 P b a2 L2 Solution Steps :
4 Relative Stiffness RS K = L 3 EI K = 3/4 Ko K = 1/2 Ko Sym K = L 2 EI K = L 6 EI Anti-Sym K = 3/2 Ko K = L 4 EI K = Ko Solution Steps :
5 Distribution Factor DF 1 3 2 1 1+2+3 2 1+2+3 3 1+2+3 1 6 1 3 1 2 Solution Steps :
6 Table : Fixed End Moment ‫تمهيدية‬ ‫مرحلة‬ Solution Steps :
7 Table : Fixed End Moment Distribution Moment Carry Over Moment Distribution Moment Carry Over Moment Distribution Moment Carry Over Moment Distribution Moment ‫مرحلة‬ ‫أولية‬ ‫مرحلة‬ ‫متكررة‬ ‫مرحلة‬ ‫متكررة‬ ‫مرحلة‬ ‫متكررة‬ Solution Steps :
8 Table : DF FEM DM COM DM MFinal COM DM COM DM Solution Steps :
9 Frames Without Sway With Sway Symmetry With Support Solution Steps :
10 A D 2 t/m 8 t 8 t B C 12 I I 4I 6 6 Solution Steps :
11 Take it In first DM only Solution Steps :
12 Solution Steps :
13 Advanced Example in beams
14 Beams with settlement Fixed End Moment D
15 Using 3M equation method Apply 3M eqn. at A : L EI MA 0+ ( ) 0+ 2 L EI MB + ( ) = 6 D L 6EI MA 2 MB + = D L2 D A B
16 Apply 3M eqn. at B : L EI MB + ( ) 0+ 2 0 + L EI MA ( ) = 6 D L -6EI MA 2MB + = D L2 D A B
17 6EI MA 2 MB + = D L2 By addition MA 0 MB + = MA MB - = MA 2MB + = -6EI D L2 6EI MA= D L2 MB = -6EI D L2
18 D A B 6EI MA = D L2 MB = -6EI D L2 6EI D L2 6EI D L2 + _
19
20 6 I A B 8 2I C EI=4800 t.m2 2 cm FEM : Example 16 : 6EI D L2 6x4800x0.02 62 6x2x4800x0.02 82 -16 18 Draw B.M.D for the shown beam
21 RS : KAB : KBC 1 6 2 8 : 1 6 1 4 : : 2 3 6 I A B 8 2I C
22 DF : DFAB : DFBC 2 2+3 3 2+3 : : 0.4 0.6 KAB : KBC : 2 3 6 I A B 8 2I C
23 A B C DF 0.4 0.6 FEM - 16 -16 +18 +18 +2 DM 0 0 - -0.8 1.2 COM -0.4 -0.6 0 0 DM 0 0 0 0 MFinal -16.4 -16.8 16.8 17.4 16.4 16.8 17.4 16.8
24 16.4 16.8 17.4
25 Frames with Sway
26 A D 2 t/m 8 t B C 12 I I 2 I 6 Example 17 : Draw B.M.D & N.F.D and S.F.D for the shown frame
27 A D 2 t/m 8 t B C 12 6 FEM : M FBC 12 L w 2 = 12 12 2 2  = = - 24 tm 8 L P M FAB = = 8 6 8 = - 6 tm M FBA = 6 tm M FCB = 24 tm
28 A D B C 12 I I 2 I 6 RS : KAB : KBC 1 6 2 12 : : 1 1 : KCD 1 6 : : 1 DF : DFBA : DFBC 1 2 1 2 : DFCB : DFCD 1 2 1 2 :
29 6 I A B D 2 t/m 2I I 12 6 C 8 t DF 0.5 0.5 0.5 0.5 FEM DM COM DM - 6 +6 -24 +24 0 0 0 0 +9 +9 -12 -12 +4.5 0 +4.5 -6 -6 0 0 0 +3 +3 -2.25 -2.25 COM DM +1.5 0 +1.5 -1.13 -1.13 0 0 0 +0.57 +0.56 -0.75 -0.75
30 DF 0.5 0.5 0.5 0.5 FEM DM COM DM - 6 +6 -24 +24 0 0 0 0 +9 +9 -12 -12 +4.5 0 +4.5 -6 -6 0 0 0 +3 +3 -2.25 -2.25 COM DM +1.5 0 +1.5 -1.13 -1.13 0 0 0 +0.57 +0.56 -0.75 -0.75 MFinal +0.28 +18.76 +15.14 -15.14 -7.51 COM DM +0.28 0 +0.28 -0.38 -0.38 0 0 0 +0.19 +0.19 -0.14 -0.14 -18.76
31 MFinal +0.28 +18.76 +15.14 -15.14 -7.51 -18.76 0.28 18.76 18.76 15.14 15.14 7.51
32 2 t/m 12 6 18.76 0.28 7.51 15.14 18.76 15.14 8 t 4 t 4 t 19.04/6 19.04/6 7.17 t 0.83 t 3.17 12+? t 12-? t 3.78 t 3.78 t Free Body Diagram 24 t
33 2 t/m 7.17 t 0.83 t ? t ? t 3.78 t 3.78 t 8 t 7.17 t 3.78 t S Fx ≠ 0 Free Body Diagram
34 2 t/m ? t ? t 7.17 t 3.78 t S Fx ≠ 0 Reason Sway do not included in the solution Solution Take Sway into consideration Sway is unknown ?
35 A D 2 t/m 8 t B C Ro A D B C R1 d Mo M1 First Stage Second Stage Solution Stages
36 First Stage Second Stage Sway Moment Reaction Mo Ro 0 d R1 M1 2 d 2 R1 2 M1 3 d 3 R1 3 M1 D d D R1 D M1 . . . . . . . . . . . . Final Stage D d Mo + D M1 Ro - D R1
37 Sway Moment Reaction Final Case D d Mo + D M1 Ro - D R1 RFinal No support In original structure Ro - D R1 = 0 Ro = D R1 D R1 Ro = MFinal= Mo+ D M1 MFinal= Mo+ M1 R1 Ro = 0
38 Return to our problem
39 2 t/m 7.17 t 0.83 t ? t ? t 3.78 t 3.78 t 8 t 7.17 t 3.78 t S Fx ≠ 0 First Stage :
40 2 t/m 7.17 t 3.78 t S Fx = 0 Ro 7.17 - 3.78 - Ro = 0 Ro = 3.39
41 Second Stage : A D B C d FEM : 6EI d L2 A D B C 12 I I 2 I 6 6EI d 62
42 6EI d L2 6EI d 62 EI is unknown Assume : EI d = 60 6 x 60 62 = 10 M FAB = - 10 tm M FBA = - 10 tm M FCD = - 10 tm M FDC = - 10 tm d is unknown
43 Second Stage can be solved as previous DF 0.5 0.5 0.5 0.5 FEM DM - 10 - 10 0 0 - 10 - 10
44 solved as anti-symmetry problem A D B C Or
45 A D B C 12 I I 2 I 6 RS : KAB : KBC 1 6 2 12 : : 2 3 DF : DFBA : DFBC 2 5 3 5 : 3 2 x
46 DF A B D C 0.4 0.6 FEM DM COM DM - 10 -10 0 0 +4 +6 +2 0 0 0 0 0 MFinal - 8 - 6 + 6 + 6 - 6 - 8
47 MFinal 6 6 8 - 8 - 6 + 6 + 6 - 6 - 8 6 8 6 M1
48 12 6 6 8 14/6 14/6 6 14/6 14/6 14/6 14/6 R1 Free Body Diagram 8
49 14/6 14/6 R1 S Fx = 0 14/6 + 14/6 - R1 = 0 R1 = 14/3
50 Ro Ro = 3.39 R1 R1 =14 / 3 RFinal = Ro - D R1 = 0 Ro = D R1 D R1 Ro = 14 / 3 3.39 = 0.726 = MFinal= Mo+ 0.726 M1
51 0.28 18.76 18.76 15.14 15.14 7.51 6 6 8 6 8 6 M1 Mo MFinal= Mo+ 0.726 M1 - + - - - + + + - - +.28+.726x-8 +7.51+.726x+8 -18.76+.726x6 -15.14+.726x-6 - 5.53 13.32 - 14.40 - 19.50
52 - 5.53 13.32 - 14.40 - 19.50 5.53 14.40 13.32 19.50 14.40 19.50
53 A D 2 t/m 8 t B C 12 6 5.53 14.40 13.32 19.50 14.40 19.50 9.97 16.95 12 2.03 36 19.05 + + + - - - - -
54 2 t/m 8 t 12 6 14.40 5.53 13.32 19.50 14.40 19.50 4 t 4 t 8.87/6 8.87/6 5.48 t 2.52 t 1.47 5.47 t 32.82/6 32.82/6 5.48 t Free Body Diagram - 5.53 13.32 - 14.40 - 19.50
55 2 t/m 14.40 19.50 24 t 12 t 12 t 5.1/12 t 5.1/12 t 0.425 t 11.575 t 12.425 t
56 2 t/m 8 t 5.48 t 2.52 t 5.48 t 5.48 t 24 t 11.575 t 12.425 t 5.48 t 5.48 t 11.575 t 12.425 t
57 2.52 5.48 5.48 11.575 12.425 SFD - - + + + 2 t/m 8 t 5.48 t 2.52 t 5.48 t 5.48 t 24 t 11.575 t 12.425 t 5.48 t 5.48 t 11.575 t 12.425 t
58 2 t/m 8 t 5.48 t 2.52 t 5.48 t 5.48 t 24 t 11.575 t 12.425 t 5.48 t 5.48 t 11.575 t 12.425 t 11.575 5.48 NFD - - - 12.425
59 A D 16 t B C 12 I 2 I 2 I 6 Example 18 : Draw B.M.D for the shown frame 3
60 A D 16 t B C 12 6 3 First Stage :
61 FEM : M FBC= -27 tm A D 16 t B C 12 6 3 B C 16 t 9 3 16 x 3 x 92 122 16 x 9 x 32 122 - 27 + 9 M FCB= +9 tm
62 RS : KAB : KBC 1 6 2 12 : : 2 2 : KCD 2 6 : : 3 DF : DFBA : DFBC 1 2 1 2 : DFCB : DFCD 2 5 3 5 : A D B C 12 I 2 I 2 I 6 3 4 x
63 DF 0.5 0.5 0.4 0.6 FEM DM COM DM 0 0 -27 +9 0 0 0 0 +13.5 +13.5 -3.6 -5.4 +6.75 0 +6.75 - 1.8 0 0 0 0 +0.9 +0.9 -2.7 -4.05 COM DM +0.45 0 +0.45 -1.35 0 0 0 0 +0.67 +0.68 -0.18 -0.27 6 I A B D 2 t/m 2I 2I 12 6 C 8 t
64 DF 0.5 0.5 0.4 0.6 MFinal +7.54 +15.11 +9.92 -9.92 0 COM DM +0.34 0 +0.34 -0.09 0 0 0 0 +0.04 +0.05 -0.14 -0.20 -15.11 FEM DM COM DM 0 0 -27 +9 0 0 0 0 +13.5 +13.5 -3.6 -5.4 +6.75 0 +6.75 - 1.8 0 0 0 0 +0.9 +0.9 -2.7 -4.05 COM DM +0.45 0 +0.45 -1.35 0 0 0 0 +0.67 +0.68 -0.18 -0.27
65 MFinal 7.54 15.11 15.11 9.92 9.92 +7.54 +15.11 +9.92 -9.92 0 -15.11
66 12 6 15.11 7.54 9.92 15.11 9.92 22.65/6 22.65/6 3.78 9.92/6 1.65 t Free Body Diagram 9.92/6
67 3.78 t 3.78 t ? t ? t 1.65 t 1.65 t 3.78 t 1.65 t First Stage :
68 3.78 t 1.65 t S Fx = 0 Ro 3.78 - 1.65 - Ro = 0 Ro = 2.13
69 Second Stage : A D B C d FEM : 6EI d L2 6EI d 62 A D B C 12 I I 2 I 6
70 6EI d L2 6EI d 62 EI is unknown Assume : EI d = 60 6 x 60 62 = 10 M FAB = - 10 tm M FBA = - 10 tm M FCD = - 10 tm M FDC = - 10 tm d is unknown
71 DF 0.5 0.5 0.4 0.6 FEM DM COM DM -10 -10 0 0 -5 0 0 0 +5 +5 +2 +3 +2.5 0 +2.5 +1 0 0 0 0 -0.5-0.5 -1 -1.5 COM DM -0.25 0 -0.25 -0.5 0 0 0 0 +0.25 +0.25 +0.1 +0.15 A B D C -10 -10 0 0 -10 -10 +10 +5 FEM
72 DF 0.5 0.5 0.4 0.6 MFinal - 7.62 - 5.27 +3.43 - 3.43 0 COM DM +0.13 0 +0.13 +0.05 0 0 0 0 -0.02 -0.03 -0.05 -0.08 5.27 FEM DM COM DM -10 -10 0 0 -5 0 0 0 +5 +5 +2 +3 +2.5 0 +2.5 +1 0 0 0 0 -0.5-0.5 -1 -1.5 COM DM -0.25 0 -0.25 -0.5 0 0 0 0 +0.25 +0.25 +0.1 +0.15
73 MFinal 3.43 3.43 5.27 7.62 5.27 M1 - 7.62 - 5.27 +3.43 - 3.43 0 5.27
74 12 6 5.27 7.62 12.89/6 12.89/6 3.43 3.43/6 3.43/6 12.89/6 3.43/6 R1 Free Body Diagram
75 R1 S Fx = 0 12.89/6 + 3.43/6 - R1 = 0 R1 = 2.72 12.89/6 3.43/6
76 Ro Ro = 2.13 R1 R1 =2.72 RFinal = Ro - D R1 = 0 Ro = D R1 D R1 Ro = 2.72 2.13 = 0.783 = MFinal= Mo+ 0.783 M1
77 M1 Mo MFinal= Mo+ 0.783 M1 - - - - + + - - +7.54+.783x-7.62 -15.11+.783x5.27 -9.92+.783x-3.43 +1.57 - 10.98 - 12.61 7.54 15.11 15.11 9.92 9.92 3.43 3.43 5.27 7.62 5.27
78 +1.57 - 10.98 - 12.61 1.57 10.98 12.61 10.98 12.61
79 1.57 10.98 12.61 10.98 12.61 A D 16 t B C 12 6 3 ?
80 10.98 12.61 16 t B C 9 3 1.63 1.63/4=0.41 11.39 16 x 3 x 9 12 = 36 24.61
81 1.57 10.98 12.61 10.98 12.61 11.39 24.61 + + - - - - BMD
82 A D 16 t B C 12 I 2 I 2 I 6 Example 19 : Draw B.M.D for the shown frame 8 4 t
83 First Stage : A D 16 t B C 12 6 8 4 t
84 FEM : A D 16 t B C 12 6 8 4 t L P 8 L P 8 L P M FBC = = - 24 tm M FCB = 24 tm 16 x 12 8 =
85 RS : KAB : KBC 1 6 2 12 : : 2 2 : KCD 2 8 : : 3 DF : DFBA : DFBC 1 2 1 2 : DFCB : DFCD 2 5 3 5 : A D B C 12 I 2 I 2 I 6 8
86 DF 0.5 0.5 0.4 0.6 FEM DM COM DM 0 0 -24 +24 0 0 0 0 +12 +12 -9.6 -14.4 +6 0 +6 - 4.8 -7.2 0 0 0 +2.4 +2.4 -2.4 -3.6 COM DM +1.2 0 +1.2 -1.2 -1.8 0 0 0 +0.6 +0.6 -0.48 -0.72 6 I A B D 2I 2I 12 8 C
87 DF 0.5 0.5 0.4 0.6 MFinal +7.5 +15.12 +18.9 -18.9 9.36 COM DM +0.3 0 +0.3 -0.24 -0.36 0 0 0 +0.12 +0.12 -0.12 -0.18 -15.12 FEM DM COM DM 0 0 -24 +24 0 0 0 0 +12 +12 -9.6 -14.4 +6 0 +6 - 4.8 -7.2 0 0 0 +2.4 +2.4 -2.4 -3.6 COM DM +1.2 0 +1.2 -1.2 -1.8 0 0 0 +0.6 +0.6 -0.48 -0.72
88 7.5 15.12 15.12 18.9 18.9 MFinal +7.5 +15.12 +18.9 -18.9 -15.12 9.36 9.36
89 12 6 15.12 7.5 18.9 15.12 18.9 22.62/6 22.62/6 3.77 28.26/8 3.53 t Free Body Diagram 28.86/8 9.36 8
90 3.77 t 3.77 t ? t ? t 3.53 t 3.53 t 3.77 t 3.53 t First Stage : A D 16 t B C 4 t 4 t
91 3.77 t 3.53 t S Fx = 0 Ro 3.77 - 3.53 - 4 - Ro = 0 Ro = -3.76 4 t
92 Second Stage : A D B C d FEM : 6EI d L2 6EI d 62 6x2EI d 82 A D B C 12 I 2 I 2 I 6 8
93 6EI d L2 6EI d 62 EI is unknown Assume : EI d = 48 6 x 48 62 = 8 M FAB = - 8 tm M FBA = - 8 tm M FCD = - 9 tm M FDC = - 9 tm d is unknown 12 x 48 82 = 9 6x2EI d 82
94 DF 0.5 0.5 0.4 0.6 FEM DM COM DM - 8 - 8 0 0 - 9 - 9 0 0 +4 +4 +3.6 +5.4 +2 0 +2 +1.8 +2.7 0 0 0 -0.9 -0.9 -0.8 -1.2 COM DM -0.45 0 -0.45 -0.4 -0.6 0 0 0 +0.2 +0.2 +0.18 +0.27 A B D C
95 DF 0.5 0.5 0.4 0.6 MFinal - 6.35 - 4.75 +4.59 - 4.59 -6.76 COM DM +0.1 0 +0.1 +0.09 +0.14 0 0 0 -0.05 -0.04 -0.04 -0.06 4.75 FEM DM COM DM - 8 - 8 0 0 - 9 - 9 0 0 +4 +4 +3.6 +5.4 +2 0 +2 +1.8 +2.7 0 0 0 -0.9 -0.9 -0.8 -1.2 COM DM -0.45 0 -0.45 -0.4 -0.6 0 0 0 +0.2 +0.2 +0.18 +0.27
96 MFinal 4.59 4.59 4.75 6.35 4.75 M1 - 6.35 - 4.75 +4.59 - 4.59 -6.76 4.75 6.76
97 6 4.75 6.35 11.1/6 11.1/6 4.59 11.35/8 11.35/8 11.1/6 11.35/8 R1 Free Body Diagram 6.76 4.59 4.59 4.75 6.35 4.75 6.76
98 R1 S Fx = 0 11.1/6 + 11.35/8 - R1 = 0 R1 = 3.27 11.1/6 11.35/8
99 Ro Ro = 3.76 R1 R1 = 3.27 RFinal = Ro + D R1 = 0 Ro = - D R1 D = R1 - Ro 3.27 -3.76 = -1.15 = MFinal= Mo-1.15 M1
100 M1 Mo MFinal= Mo- 1.15 M1 - - - - + + - - +7.5-1.15x-6.35 -15.12-1.15x4.75 -18.9-1.15x-4.59 +14.8 - 20.58 - 13.62 4.59 4.59 4.75 6.35 4.75 6.76 7.5 15.12 15.12 18.9 18.9 9.36 + + +9.36-1.15x+6.76 1.83
101 14.8 20.58 13.62 20.58 13.62 - 20.58 - 13.62 +1.83 +14.8 1.83
102 14.8 20.58 13.62 20.58 13.62 1.83 8 A D 16 t B C 12 6 4 t 17.1 48 30.9 BMD + - + -
103 Advanced Idea
104 Beams with springs First Stage :
105 First Stage : Ro Second Stage : R1 1 cm
106 First Stage Second Stage Settlement Moment Reaction Mo Ro 0 1 cm R1 M1 2 cm 2 R1 2 M1 3 cm 3 R1 3 M1 D cm D R1 D M1 . . . . . . . . . . . . Final Stage Mo + D M1 Ro + D R1 D cm
107 Settlement Moment Reaction Final Stage Mo + D M1 Ro + D R1 D cm F = K d Ro + D R1 = K D MFinal= Mo+ D M1
108 Beams with Intermediate Hinge Determinate Part Indeterminate Part Determinate Part Determinate Part
109 Indeterminate Part Indeterminate Parts
110 Solution Steps : Ro R1 Mo M1 d
111 First Stage Second Stage Moment Reaction Mo Ro 0 d R1 M1 2 d 2 R1 2 M1 3 d 3 R1 3 M1 D d D R1 D M1 . . . . . . . . . . . . Final Stage D d Mo + D M1 Ro - D R1 Settlement
112 Sway Moment Reaction Final Case D d Mo + D M1 Ro - D R1 RFinal No support In original structure Ro - D R1 = 0 Ro = D R1 D R1 Ro = MFinal= Mo+ D M1 MFinal= Mo+ M1 R1 Ro = 0
113 Summary Frames Without Sway With Sway Symmetry With Support
114 Frames With Sway Summary
115 First Stage : Mo Ro S Fx = 0
116 Second Stage : A D B C d M1 R1 S Fx = 0
117 Sway Moment Reaction Final Case Mo + D M1 Ro - D R1 RFinal No support In original structure Ro - D R1 = 0 Ro = D R1 D R1 Ro = MFinal= Mo+ D M1 MFinal= Mo+ M1 R1 Ro = 0
118 Questions

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