UNIT 2 MOMENT DISTRIBUTION METHOD By Asst Prof

UNIT 2: MOMENT DISTRIBUTION METHOD By: Asst. Prof. Mithun Sawant Department of Civil Engineering

TYPE 1: BOTH ENDS FIXED Pr. No. 1: Analyze the Continuous beam loaded as shown in fig. by MDM. A 25 k. N 10 k. N/m 40 k. N D C 3 m 5 m B 1 m 2. 50 m

STEP 1 : FIXED END MOMENTS: Considering Anticlockwise +ve and Clockwise -ve

STEP 2 : DISTRIBUTION FACTOR: Joint Member B BA Stiffness Factor K 1. 80 EI BC C CB CD 0. 44 0. 55 1. 80 EI 0. 56 0. 44

Step 3 : Moment Distribution: Joint Member DF FEM A BC CB CD DC 0 0. 44 0. 56 0. 44 0 20. 83 -20. 83 4. 69 -14. 06 25 -25 +7. 10 +9. 04 -6. 13 -4. 82 -3. 06 +4. 52 +1. 72 -2. 53 -1. 265 +0. 86 +0. 71 -0. 482 -0. 241 +0. 355 +0. 11 +0. 135 -0. 199 -0. 156 -11. 71 +11. 71 -17. 67 +17. 65 +3. 55 +1. 35 0. 675 DM COM +0. 56 +0. 28 DM Final Moments D BA DM COM C AB DM COM B +25. 34 -2. 41 -1. 99 -1. 00 -0. 38 - 0. 19 -28. 60

STEP 3 : MOMENT DISTRIBUTION: Joint Member DF FEM A BC CB CD DC 0 0. 44 0. 56 0. 44 0 20. 83 -20. 83 4. 69 -14. 06 25 -25 +7. 10 +9. 04 -6. 13 -4. 82 -3. 06 +4. 52 +1. 72 -2. 53 -1. 265 +0. 86 +0. 71 -0. 482 -0. 241 +0. 355 +0. 11 +0. 135 -0. 199 -0. 156 -11. 71 +11. 71 -17. 67 +17. 65 +3. 55 +1. 35 0. 675 DM COM +0. 56 +0. 28 DM Final Moments D BA DM COM C AB DM COM B +25. 34 -2. 41 -1. 99 -1. 00 -0. 38 -0. 19 -28. 60

SFD Reaction Calculations A 25 k. N 10 k. N/m 40 k. N D C 3 m 5 m 10 k. N/m 25. 34 k. N. m B 11. 71 k. N. m 1 m 25 k. N 2. 50 m 17. 67 k. N. m 2. 50 m 40 k. N 28. 60 k. N. m

SFD 27. 73 k. N 17. 81 k. N 4. 76 k. N 0 0 22. 27 k. N 20. 24 k. N 22. 19 k. N

SUPERIMPOSED BMD 50 k. N. m 25. 34 k. N. m 31. 25 k. N. m 11. 71 k. N. m 18. 75 k. N. m 17. 67 k. N. m 28. 60 k. N. m

BMD ON TENSION SIDE 28. 60 k. N. m 25. 34 k. N. m 17. 67 k. N. m 11. 71 k. N. m 2. 60 k. N. m 12. 73 k. N. m 26. 87 k. N. m

TYPE 2: BOTH ENDS HINGED Pr. No. 2: Analyze the Continuous beam loaded as shown in fig. by MDM. 15 k. N/m 40 k. N 2 m 60 k. N 4 m 4 m A B 3 m C 3 m D

STEP 1 : FIXED END MOMENTS: Considering Anticlockwise +ve and Clockwise -ve

STEP 2 : DISTRIBUTION FACTOR: Joint Member B BA Stiffness Factor K 1. 50 EI BC C CB CD 0. 33 0. 67 1. 50 EI 0. 67 0. 33

STEP 3 : MOMENT DISTRIBUTION: Joint A Member AB BA BC CB CD DC 1 0. 33 0. 67 0. 33 1 +35. 56 -17. 78 +20. 00 -20. 00 +45. 00 -45. 00 DF FEM DM B D -35. 56 COM +45. 00 -17. 78 0. 00 DM DM +20. 00 -20. 00 +67. 50 +5. 13 +10. 43 -31. 83 -15. 67 -15. 92 +5. 22 +10. 67 -3. 50 -1. 75 +5. 34 +1. 17 -3. 58 -1. 79 +0. 59 +1. 20 -0. 40 -0. 19 +24. 01 -48. 16 +5. 25 COM DM +0. 58 COM DM 0. 00 +22. 50 -35. 56 COM Final Moments C 0. 00 -1. 72 -1. 76 0. 00 As support A and D are HINGED they are capable of having resisting moments. Therefore Moment @ A and D Should be equal to ZERO.

STEP 3 : MOMENT DISTRIBUTION: Joint A Member AB BA BC CB CD DC 1 0. 33 0. 67 0. 33 1 +35. 56 -17. 78 +20. 00 -20. 00 +45. 00 -45. 00 DF FEM DM B D -35. 56 COM +45. 00 -17. 78 0. 00 DM COM DM +20. 00 -20. 00 +67. 50 +5. 13 +10. 43 -15. 92 +10. 67 -31. 83 +5. 22 -3. 50 -15. 67 -1. 75 +5. 34 +1. 17 -3. 58 -1. 79 +0. 59 +1. 20 +24. 01 -0. 40 -48. 16 +5. 25 DM +0. 58 COM 0. 00 +22. 50 -35. 56 COM DM Final Moments C +0. 59 -24. 01 0. 00 -1. 72 -1. 76 -0. 19 +48. 16 0. 00 As support A and D are HINGED they are capable o having resisting moments. Therefore Moment @ A and D Should be equal to ZERO.

SFD Reaction Calculations 15 k. N/m 40 k. N 2 m 60 k. N 4 m 4 m A 3 m B 40 k. N 22. 67 k. N D C 15 k. N/m 24. 01 k. N. m 17. 34 k. N 3 m 23. 96 k. N 48. 16 k. N. m 36. 04 k. N 60 k. N 38. 03 k. N 21. 97 k. N

SFD 22. 67 k. N 17. 34 k. N 23. 96 k. N 38. 03 k. N 21. 97 k. N 17. 34 k. N 36. 04 k. N

SUPERIMPOSED BMD 90 k. N. m 53. 33 k. N. m 47. 96 k. N. m 24. 01 k. N. m 30 k. N. m

BMD ON TENSION SIDE 48. 16 k. N. m 24. 01 k. N. m 6. 09 k. N. m 45. 33 k. N. m 65. 92 k. N. m

TYPE 3: ONE END FIXED OTHER END HINGED Pr. No. 3: Analyze the Continuous beam loaded as shown in fig. by MDM. 10 k. N/m 40 k. N 1 m A 30 k. N 2 m B C 4 m D 1 m 1 m

STEP 2 : DISTRIBUTION FACTOR: Joint Member B BA Stiffness Factor K 2. 33 EI BC C CB CD 0. 57 0. 43 2. 50 EI 0. 40 0. 60

STEP 3 : MOMENT DISTRIBUTION: Joint Member DF FEM A B C D AB BA BC CB CD DC 0 0. 57 0. 43 0. 40 0. 60 1 +17. 78 -8. 89 +13. 33 -13. 33 +7. 50 -7. 50 DM +7. 50 COM +3. 75 +17. 78 DM COM -13. 33 +11. 25 -2. 53 -1. 91 +0. 83 +1. 25 +0. 42 -0. 96 -0. 18 +0. 38 +0. 19 -0. 09 -0. 11 -0. 08 +0. 04 +0. 05 -11. 77 +11. 77 -13. 13 +13. 13 -0. 24 -0. 12 DM Final Moments +13. 33 -1. 27 DM COM -8. 89 +16. 39 0. 00 +0. 58 0. 00

SFD Reaction Calculations 10 k. N/m 40 k. N 1 m 30 k. N 2 m A B C 4 m 40 k. N 16. 38 k. N. m 28. 20 k. N 1 m 10 k. N/m 11. 77 k. N. m 11. 80 k. N 19. 66 k. N D 1 m 13. 13 k. N. m 20. 34 k. N 30 k. N 21. 67 k. N 8. 34 k. N

SFD 28. 20 k. N 11. 80 k. N 19. 66 k. N 21. 67 k. N 8. 34 k. N 11. 80 k. N 20. 34 k. N

SUPERIMPOSED BMD 26. 67 k. N. m 20 k. N. m 16. 39 k. N. m 11. 77 k. N. m 13. 13 k. N. m 15 k. N. m

BMD ON TENSION SIDE 16. 39 k. N. m 13. 13 N. m 11. 77 k. N. m 11. 82 k. N. m 7. 55 k. N. m 8. 44 k. N. m

TYPE 4: OVERHANG Pr. No. 4: Analyze the Continuous beam loaded as shown in fig. by MDM. 40 k. N/m A 4 m 20 k. N C B 3 m D 2 m

STEP 2 : DISTRIBUTION FACTOR: Joint B Member Stiffness Factor K BA 2. 00 EI BC C 0. 50 CB CD 0. 50 1. 33 EI 0. 00 1. 00 0. 00 As C is considered as a simple support, its relative stiffness with respect to (w. r. t) joint B will And as it considered as Simple Support NO carry over moment (COM) will carried towards C from B.

STEP 3 : MOMENT DISTRIBUTION: Joint Member DF FEM A BA BC CB CD 0. 00 0. 50 1. 00 0. 00 +53. 33 -53. 33 +30. 00 -30. 00 +40. 00 +11. 67 -10. 00 -40. 00 +5. 02 DM COM Final Moments C AB DM COM B -5. 00 +2. 50 -39. 16 +1. 25 +60. 41 As C is considered as a simple support, its relative stiffness with respect to (w. r. t) joint B will be And as it considered as Simple Support NO carry over moment (COM) will carried towards C from B.

SFD Reaction Calculations 40 k. N/m A 20 k. N C B 4 m 3 m 60. 41 k. N. m 40 k. N/m 85. 31 k. N 39. 16 k. N. m 74. 69 k. N 40 k. N/m 59. 72 k. N D 2 m 40 k. N. m 60. 28 k. N 20. 00 k. N 20 k. N

SFD 85. 31 k. N 59. 72 k. N 20. 00 k. N 74. 69 k. N 60. 28 k. N 20. 00 k. N

BMD ON TENSION SIDE 60. 41 k. N. m 40. 00 k. N. m 39. 16 k. N. m 5. 42 k. N. m 30. 22 k. N. m

TYPE 5: SINKING Pr. No. 5: Analyze the Continuous beam loaded as shown in fig. by MDM. If support B sinks by 10 mm. 40 k. N 2 m A 10 k. N/m 2 m 4 m B 20 k. N 6 m D C 2 m W. r. t. Left Support always Observe Right Support while deciding Sink as Positive or Negative. 1. For Beam AB as w. r. t to A, B is downward. Therefore we consider as Negative. 2. And when we consider BC as w. r. t B, C is Upward. Therefore considered Positive. as

STEP 1 : FIXED END MOMENTS: Due to sink additional moment is created as

STEP 2 : DISTRIBUTION FACTOR: Joint B Member Stiffness Factor K BA 1. 50 EI BC C 0. 33 CB CD 0. 67 EI 0. 00 1. 00 0. 00 As C is considered as a simple support, its relative stiffness with respect to (w. r. t) joint B will And as it considered as Simple Support NO carry over moment (COM) will carried towards C from B.

STEP 3 : MOMENT DISTRIBUTION: Joint Member DF FEM A BA BC CB CD 0. 00 0. 67 0. 33 1. 00 0. 00 +95. 00 +55. 00 -3. 33 -63. 33 +40. 00 -34. 62 -17. 05 +23. 33 0. 00 -40. 00 +40. 00 -17. 31 DM COM Final Moments C AB DM COM B +11. 67 -7. 82 -3. 85 +12. 56 -3. 91 +73. 78 As C is considered as a simple support, its relative stiffness with respect to (w. r. t) joint B will be And as it considered as Simple Support NO carry over moment (COM) will carried towards C from B.

SFD Reaction Calculations 40 k. N 2 m A 10 k. N/m 2 m B 4 m 20 k. N D C 6 m 2 m 73. 78 k. N. m 40 k. N 38. 41 k. N 12. 56 k. N. m 1. 59 k. N 10 k. N/m 21. 24 k. N 40 k. N. m 38. 76 k. N 20. 00 k. N 20 k. N

SFD

BMD ON TENSION SIDE

TYPE 6: NON- SWAY FRAME Pr. No. 6: Analyze the Frame loaded as shown in fig. by MDM. 40 k. N/m B 4. 00 m 2. 50 m C 2. 50 m 20 k. N 2. 50 m A 2. 50 m D

NON-SWAY

STEP 1 : FIXED END MOMENTS: -

STEP 2 : DISTRIBUTION FACTOR: - Joint B Member BA Stiffness Factor K 2. 80 EI BC C CB CD 0. 29 0. 71 2. 80 EI 0. 71 0. 29

STEP 3 : MOMENT DISTRIBUTION: Joint Member DF FEM A B C AB BA BC CB CD DC 0. 00 0. 29 0. 71 0. 29 0. 00 +12. 50 -12. 50 +53. 33 -53. 33 +12. 50 -12. 50 40. 83 DM COM -11. 84 COM -14. 50 -10. 30 +5. 15 -3. 66 +1. 83 -1. 30 +0. 65 -0. 19 -0. 46 +0. 19 -30. 76 +30. 76 -1. 49 -0. 75 -0. 53 -0. 27 DM Final Moments +14. 50 -2. 11 DM COM +28. 99 -4. 21 DM +3. 45 -40. 83 -28. 99 -5. 92 DM COM D +11. 84 +5. 92 +4. 21 +2. 11 +1. 49 +0. 75 +0. 53 +0. 27 -3. 45

SFD Reaction Calculations 30. 76 k. N. m 40 k. N/m 30. 76 k. N. m 80 k. N 15. 46 k. N 20 k. N 3. 45 k. N. m 4. 54 k. N 80 k. N 3. 45 k. N. m 80 k. N

SFD 80 k. N 15. 46 k. N 80 k. N 4. 54 k. N

BMD ON TENSION SIDE 30. 76 k. N. m 7. 89 k. N. m 3. 45 k. N. m 7. 89 k. N. m 49. 24 k. N. m 3. 45 k. N. m

TYPE 6: NON- SWAY FRAME Pr. No. 7: Analyze the Frame loaded as shown in fig. by MDM. 20 k. N/m B 1. 00 m 6 k. N 3. 00 m A 4. 00 m C 2. 50 m 6 k. N 3. 00 m D

NON-SWAY: -

STEP 1 : FIXED END MOMENTS: -

STEP 2 : DISTRIBUTION FACTOR: - Joint B Member BA Stiffness Factor K 5 EI BC C CB CD 0. 20 0. 80 5 EI 0. 80 0. 20

STEP 3 : MOMENT DISTRIBUTION: Joint Member A B C D AB BA BC CB CD DC DF 0. 00 0. 20 0. 80 0. 20 0. 00 FEM +1. 13 -3. 38 +6. 67 -6. 67 +3. 38 -1. 13 +3. 29 DM COM -0. 66 COM -1. 32 -1. 06 +0. 53 -0. 42 +0. 21 -0. 17 +0. 09 -0. 02 -0. 07 +0. 02 -4. 47 +4. 47 -0. 11 -0. 06 -0. 04 -0. 02 DM Final Moments +1. 32 -0. 13 DM COM +2. 63 -0. 26 DM COM -2. 63 -0. 33 DM +0. 59 -3. 29 +0. 66 +0. 33 +0. 26 +0. 13 +0. 11 +0. 06 +0. 04 +0. 02 -0. 59

SFD Reaction Calculations 4. 47 k. N. m 20 k. N/m 4. 47 k. N. m 20 k. N 5. 47 k. N 6 k. N 0. 59 k. N. m 0. 53 k. N 20 k. N 0. 59 k. N. m 20 k. N

SFD 20 k. N 5. 47 k. N 20 k. N 0. 53 k. N

BMD ON TENSION SIDE 4. 47 k. N. m 1. 00 k. N. m 5. 53 k. N. m 0. 59 k. N. m

TYPE 6: NON- SWAY FRAME Pr. No. 7: Analyze the Frame loaded as shown in fig. by MDM. 2. 4 k. N/m 10 k. N 12 k. N

NON-SWAY: -

STEP 1 : FIXED END MOMENTS: - STEP 2 : DISTRIBUTION FACTOR: Joint B Member Stiffness Factor K BA 2. 33 EI BC BD 0. 43 0. 57 0. 00

STEP 3 : MOMENT DISTRIBUTION: Joint Member A B C AB BA BD BC CB DF 0. 00 0. 43 0. 00 0. 57 0. 00 FEM +3. 20 -3. 20 +15. 00 +4. 50 -4. 50 +16. 30 DM -7. 01 COM -3. 51 Final Moments -0. 31 0. 00 -9. 29 -4. 65 -10. 21 +15. 00 -4. 79 -9. 15

SFD Reaction Calculations 0. 31 k. N. m 2. 40 k. N/m 10. 21 k. N. m 10 k. N 15 k. N. m 2. 30 k. N 7. 43 k. N 1. 35 k. N 10 k. N 4. 79 k. N. m 12 k. N 10. 65 k. N 9. 15 k. N. m 17. 43 k. N

SFD

BMD ON TENSION SIDE 15. 00 k. N. m 10. 21 k. N. m 0. 31 k. N. m 4. 79 k. N. m 9. 15 k. N. m

TYPE 6: SWAY FRAME Pr. No. 7: Analyze the Frame loaded as shown in fig. by MDM. 10 k. N 3 k. N/m

NON-SWAY: -

STEP 1 : FIXED END MOMENTS: -

STEP 2 : DISTRIBUTION FACTOR: Joint B Member BA Stiffness Factor K 2. 60 EI BC C CB CD 0. 23 0. 77 2. 60 EI 0. 77 0. 23

STEP 3 : MOMENT DISTRIBUTION: - TABLE 7. 1 Joint Member A B C D AB BA BC CB CD DC DF 1. 00 0. 23 0. 77 0. 23 1. 00 FEM +6. 25 -6. 25 +5. 00 -5. 00 0. 00 DM -6. 25 COM 0. 00 -3. 13 AM 0. 00 -9. 38 +5. 00 -5. 00 0. 00 Imbalance -4. 38 DM +1. 01 +3. 37 +3. 85 +1. 93 +1. 69 -1. 49 -1. 30 -0. 65 -0. 75 +0. 50 +0. 58 +0. 29 +0. 25 -0. 07 -0. 22 -0. 19 -0. 06 -8. 73 +8. 73 -0. 87 +0. 87 COM DM -0. 44 COM DM +0. 15 COM DM Final Moments 0. 00 -5. 00 +1. 15 -0. 39 +0. 17 0. 00

STEP 4 : SWAY FORCE : Horizontal Reaction @ A: 3 k. N/m

CONTD. . . Horizontal Reaction @ D:

IMPORTANT NOTE: PURE SWAY ANALYSIS Due to Sway, these will be subjected to additional moments i. e. ; And since is unknown, these moments are also unknown. Ø . Both Ends of Frame Fixed: Therefore the ratio of initial equivalent moments at the tops of the columns, Ø Both Ends of Frame Hinged: Therefore the ratio of initial equivalent moments at the tops of the columns, Ø One End Fixed other End Hinged: Therefore the ratio of initial equivalent moments at the tops of the columns,

SWAY ANALYSIS: - TABLE 7. 2 Therefore, Joint Member Say 1: 1 A B C D AB BA BC CB CD DC DF 1. 00 0. 23 0. 77 0. 23 1. 00 FEM 0. 00 +1. 00 0. 00 Imbalance +1. 00 DM -0. 23 -0. 77 -0. 39 +0. 30 +0. 15 -0. 12 -0. 06 +0. 01 +0. 05 +0. 01 +0. 84 -0. 84 +0. 84 COM DM +0. 09 COM DM -0. 03 COM DM Final Moments 0. 00 +1. 00 -0. 23 +0. 09 -0. 03 0. 00

The moments obtained are due to Sway force S For a sway force of 0. 34 k. N, the sway moments as per Table 7. 2 Hence Actual Sway force of 9. 08 k. N, the actual sway moments will be : Joint A B C D Member AB BA BC CB CD DC Moments (Table 7. 2) 0. 00 +0. 84 -0. 84 +0. 84 0. 00 Actual Sway Moments 0. 00 +22. 43 -22. 43 +22. 43 0. 00 Non-Sway Moments 0. 00 -8. 73 +8. 73 -0. 87 +0. 87 0. 00 Final Moments 0. 00 +13. 70 -23. 30 +23. 30 0. 00