Approximate Analysis of Multistored Frame by Cantilever method Numerical-II
- 1. Sanjivani Rural Education Society'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 Cantilever method Unit-V – Approximate Analysis of Multistoried Frames. Topic- Numerical on Analysis of frame by Cantilever method By, Prof. Santosh. R. Nawale (Assistant Professor) Mail Id- nawalesantoshcivil@sanjivani.org.in
- 2. Unit-V – Approximate Analysis of Multistoried Frames. Topic- 1) Numerical on Analysis of frame by Cantilever method. Determine the approximate values of moment, shear and Axial forces In member of frame loaded and supported as shown in figure using Cantilever Method of Analysis. Area of each exterior columns is one half of the area of the Interior Columns.
- 3. Unit-V – Approximate Analysis of Multistoried Frames. Topic- 1) Numerical on Analysis of frame by Cantilever method. Assume point of contra flexure at centre of Beam and columns.
- 4. Unit-V – Approximate Analysis of Multistoried Frames. Topic- 1) Numerical on Analysis of frame by Cantilever method. 1) Determination of the CG of the Frame: Exterior Column= A; Interior Column=2A. To find out CG of the frame. Taking moment of the area of columns at leftmost column axis. A1 x 0+A2 x 4.5+A3 x 10.5=(A1+A2+A3) 2A x4.5 +A x10.5=(A+2A+A) 19.5A=4A ; =4.875m. X X A1 x 0+A2 x 4.5+A3 x 10.5=(A1+A2+A3) 2A x4.5 +A x10.5=(A+2A+A) 19.5A=4A ; =4.875m. Consider Upper storey and Releasing frame from nodes 3,4,5. X X X
- 5. Unit-V – Approximate Analysis of Multistoried Frames. Topic- 1) Numerical on Analysis of frame by Cantilever method. H3+H4+H5=25 Taking Moment @ Point P ∑M @p=0. V3 x 4.875 +V4 x0.375 +V5x5.625=(H3 +H4 +H5)x1.5. As per Assumption: Axial forces in column is proportional to its distance from centroid and areas of the column group at that level. Ratio=Stress/Distance from C.G. As per Assumption: Axial forces in column is proportional to its distance from centroid and areas of the column group at that level. Ratio=Stress/Distance from C.G. V V V 3 5 4 A 2A A = = 4.875 0.375 5.625 V V V 3 5 4 = = 4.875 0.375 5.625 0.375V 3 V = =0.154V 4 3 0.5×4.875 5.625V 3 V = =1.154V 3 5 4.875 V3 x 4.875 +V4 x0.375 +V5x5.625=(H3 +H4+H5)x1.5. V3 x 4.875 +0.154 x V3 x0.375 +1.154 x V3 x5.625= 25x1.5. 4.875 V3 + 0.058 V3 +6.491 V3 =37.5 11.424 V3 =37.5. V3 =3.283 ; V4=0.506 ;V5=3.789.
- 6. Unit-V – Approximate Analysis of Multistoried Frames. Topic- 1) Numerical on Analysis of frame by Cantilever method. F =0. Y 1 ( +ve& -ve) -3.283 =3.283 V =0 V KN. F =0. Y 2 ( +ve& -ve) -3.283-0.506 =3.789 V =0 V KN. ( +ve & -ve) F =0. X 3 3 1 1 -Ve & Anticlockwise +Ve ( +ve & -ve) -4.925 M =0. @G 3.283 2.25-H 1.5=0 H =4.925KN. F =0. X H +25=0 H =20.076KN. Clockwise F =0. Y 1 1 ( +ve& -ve) -3.283 =3.283 V =0 V KN. 4 4 2 2 -Ve & Anticlockwise +Ve ( +ve & -ve) M =0. @H 3.283 2.25 H 1.5 3.789 3=0 H =12.503KN. F =0. X H +20.076-12.503=0 H =7.573KN. Clockwise F =0. Y 2 2 ( +ve& -ve) -3.283-0.506 =3.789 V =0 V KN. 5 5 ( +ve & -ve) + X H 7.573=0 H =7.573KN.
- 7. Unit-V – Approximate Analysis of Multistoried Frames. Topic- 1) Numerical on Analysis of frame by Cantilever method. Consider lower storey and Releasing frame from nodes 8,9,10. H8+H9+H10=25+50 Taking Moment @ Point P ∑M @p=0. V8 x 4.875 +V9 x0.375 +V5x5.625=(H8 +H9+H10)x 4.5- 50x3. As per Assumption: Axial forces in column is proportional to its distance from centroid and areas of the column group at that level. Ratio=Stress/Distance from C.G. As per Assumption: Axial forces in column is proportional to its distance from centroid and areas of the column group at that level. Ratio=Stress/Distance from C.G. V V V 8 9 10 A 2A A = = 4.875 0.375 5.625 V 0.5V V 8 9 10 = = 4.875 0.375 5.625 0.375V 8 V = =0.154V 9 8 0.5×4.875 5.625V 8 V = =1.154V 10 8 4.875 V8 x 4.875 +V9 x0.375 +V10x5.625=(H8 +H9+H10)x 4.5- 50x3. V8 x 4.875 +0.154 x V8 x0.375 +1.154 x V8 x5.625 = 75x4.5-50 x 3. 4.875 V8 + 0.058 V8 +6.491 V8 =187.5. 11.424 V8 =187.50. V8 =16.41 ; V9 =0.154x16.41=2.53; V10= 1.154 x 16.41=18.94.
- 8. Unit-V – Approximate Analysis of Multistoried Frames. Topic- 1) Numerical on Analysis of frame by Cantilever method.
- 9. Unit-V – Approximate Analysis of Multistoried Frames. Topic- 1) Numerical on Analysis of frame by Cantilever method. 6 3.283 F =0. Y ( +ve& -ve) -34.467= =31.184 V 0 V KN. 31.184 0.506 F =0. Y 7 ( +ve& -ve) -5.308 =35.986 V =0 V KN. ( +ve & -ve) F =0. X 6 6 8 8 6 6 -Ve & Anticlockwise +Ve ( +ve & -ve) 3.283 M =0. @F H 2.25-4.925 1.5 +31.184 2.25=0 H =27.901KN. F =0. X H +50-27.901+4.925=0 H =27.024KN. Clockwise F =0. Y ( +ve& -ve) -34.467= =31.184 V 0 V KN. 7 9 9 7 -Ve & Anticlockwise +Ve ( +ve & -ve) 31.184 0.506 M =0. @E 31.184 2.25 H 2.25 35.986 3-12.503 1.5=0 H =70.830KN. F =0. X 27.024 H -70.830+12.503 Clockwise F =0. Y 7 ( +ve& -ve) -5.308 =35.986 V =0 V KN. 7 =0 H =31.303KN. 10 10 ( +ve & -ve) +7.573 31.303 H =0 H =38.876KN.
- 10. Unit-V – Approximate Analysis of Multistoried Frames. Topic- 1) Numerical on Analysis of frame by Cantilever method. Bending Moment Diagram
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