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Approximate Analysis of Multistoried frame by Portal method
- 1. Sanjivani Rural EducationSociety's Sanjivani College of Engineering, Kopargaon 423603. Department of Civil Engineering Course Title –Advance Analysis of Structures T. Y.BTech Civil, Semester –II(CE313) Unit-V – Approximate Analysis of Multistoried Frames. Topic- Numerical on Analysis of frame by Portal method Unit-V – Approximate Analysis of Multistoried Frames. Topic- Numerical on Analysis of frame by Portal method By, Prof. Santosh. R. Nawale (Assistant Professor)
- 2. Unit-V – ApproximateAnalysis of Multistoried Frames. Topic- 2) Numerical on Analysis of frame by Portal method. Determine the approximate values of moment, shear and Axial forces In member of frame loaded and supported as shown in figure using Portal Method of Analysis.
- 3. Unit-V – ApproximateAnalysis of Multistoried Frames. Topic- 2) Numerical on Analysis of frame by Portal method. Assume point of contra flexure at centre of Beam and columns.
- 4. Unit-V – ApproximateAnalysis of Multistoried Frames. Topic- 2) Numerical on Analysis of frame by Portal method. Consider Upper storey and Releasing frame from nodes 4,5,6,7. The Horizontal shear is divided among all the columns on the basis that each interior columns takes twice as much as exterior column. F =0. X H+2H+2H+H=30 H=5KN.
- 5. Unit-V – ApproximateAnalysis of Multistoried Frames. Topic- 2) Numerical on Analysis of frame by Portal method. ( +ve & -ve) - F =0. X H 5 +30=0 ( +ve & -ve) F =0. X H +25-10=0 1 1 1 1 +Ve & Anticlockwise -Ve - H 5 +30=0 H = 25KN. M =0. @I V 1.75+5 1.5=0 V =4.28KN. Clockwise F =0. Y 4 4 ( +ve& -ve) +4.28 =4.28 -V =0 V KN. 2 2 2 2 +Ve & Anticlockwise -Ve H +25-10=0 H =15KN. M =0. @J 4.28 1.75 V 1.75 10 1.5=0 V =4.29KN. Clockwise F =0. Y 5 5 ( +ve& -ve) +4.29-4.28 =0.01 V =0 V KN.
- 6. Unit-V – ApproximateAnalysis of Multistoried Frames. Topic- 2) Numerical on Analysis of frame by Portal method. ( +ve & -ve) F =0. X 15 10 H =0 F =0. Y ( +ve& -ve) -4.28 V =0 V KN. 3 3 3 3 +Ve & Anticlockwise -Ve 15 10 H =0 H =5KN. M =0. @K 4.29 1.75 V 1.75 10 1.5=0 V =4.28KN. Clockwise F =0. Y 6 6 ( +ve& -ve) -4.29+4.28 =0.01 -V =0 V KN. 7 7 F =0. Y ( +ve& -ve) -4.28 =4.28 V =0 V KN.
- 7. Unit-V – ApproximateAnalysis of Multistoried Frames. Topic- 2) Numerical on Analysis of frame by Portal method. Consider lower storey and Releasing frame from nodes 11,12,13,14. The Horizontal shear is divided among all the columns on the basis that each interior columns takes twice as much as exterior column. ( v e ; - v e ) F = 0 . X H + 2 H + 2 H + H = 2 5 + 5 0 6 H = 9 0 . H = 1 5 K N .
- 8. Unit-V – ApproximateAnalysis of Multistoried Frames. Topic- 2) Numerical on Analysis of frame by Portal method. ( +ve & -ve) = F =0. X 60 15 5 H 0 ( +ve & -ve) F =0. X 50 10 30 H =0 8 8 8 8 +Ve & Anticlockwise -Ve = 60 15 5 H 0 H = 50KN. M =0. @E 15 1.5 V 1.75+5 1.5=0 V = 17.14KN. Clockwise F =0. Y 11 11 ( +ve& -ve) +4.28+17.14 =21.42 -V =0 V KN. 9 9 9 9 +Ve & Anticlockwise -Ve 50 10 30 H =0 H =30KN. M =0. @F 17.14 1.75 V 1.75 10 1.5 30 1.5=0 V = 17.14KN. 0.01 17.14 Clockwise F =0. Y 12 12 ( +ve& -ve) +17.14 =0.01 V =0 V KN.
- 9. Unit-V – ApproximateAnalysis of Multistoried Frames. Topic- 2) Numerical on Analysis of frame by Portal method. ( +ve & -ve) = F =0. X 30 10 30 H 0 F =0. Y ( +ve& -ve) 10 10 10 10 +Ve & Anticlockwise -Ve = 30 10 30 H 0 H = 10KN. M =0. @G 17.14 1.75 30 1.5 V 1.75+10 1.5=0 V = 17.14KN. 17.14 Clockwise F =0. Y 13 13 ( +ve& -ve) -0.01+17.14 =0.01 V =0 V KN. 17.14 F =0. Y 14 14 ( +ve& -ve) -4.28 =21.42 V =0 V KN.
- 10. Unit-V – ApproximateAnalysis of Multistoried Frames. Topic- 2) Numerical on Analysis of frame by Portal method. Bending Moment Diagram
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