Read Microsoft PowerPoint - Structural Analysis Moment Distribution Lecture 9 [Compatibility Mode] text version

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WL - 65.0 × 3.0 × 3.0 = = -48.75 kNm 12 12

FEM BA (being clockwise) = +48.75 kNm

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FEM BC (being anti - clockwise) =- WL - 32.0 × 3.6 × 3.6 = = -34.56 kNm 12 12

FEM CB (being clockwise) = +34.56 kNm

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5000 × 10 4 K AB = = 16.67 ×103 3.0 ×103 5000 ×10 4 K BC = = 13.89 ×103 3.6 × 103 16.67 × 103 = 0.545 DFAB = 3 3 16.67 × 10 + 13.89 × 10 13.89 × 103 DFBC = = 0.455 3 3 16.67 ×10 + 13.89 × 10

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0.545 × -14.19 = -7.73 kNm being added to the end BA 0.455 × -14.19 = -6.46 kNm being added to the end BC

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carry - over process 0.5 × -7.73 = -3.865 kNm transferred to the end A 0.5 × -6.46 = -3.23 kNm transferred to the end C

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3 8500 ×10 4 3 K AB = × = 14.17 × 103 4 4.5 ×103 4 6500 × 10 4 K BC = = 17.33 × 103 3.75 × 103 3 3 5500 ×10 4 K CD = × = 11.00 × 103 4 4 3.75 ×103

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14.17 × 103 = 0.45 = 3 3 17.73 × 10 + 14.17 × 10

K BC 17.73 × 103 = = 0.612 = 3 3 3 17.73 × 10 + 11.00 × 10 K BC + K CD 4 = 1 - 0.612 = 0.388

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3PL 3 ×120 × 4.5 = 101.25 kNm = 16 16 2 PL 2 × 70 × 3.75 = = -58.33 kNm FEM BC = - 9 9 FEM CB = +58.33 kNm FEM BA = + FEM CD = - WL 220 × 3.75 = = -103.13 kNm 8 8

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Distribution BA = (+ 101.25 - 58.33)× 0.45 = 19.31 Distribution BC = (+ 101.25 - 58.33)× 0.55 = 23.61

Distribution CB = -14.8 × 0.612 = 9.06 Distribution CD = -14.8 × 0.388 = 5.74

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