building estimation methods
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Building Estimation Methods
Prepared by:-Er. Simarpreet SinghA.P, Civil Engg.ACET, Amritsar
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TABLE OF CONTENTS One Room Building
Concrete work in foundation (Center line and separate wall method)
Brickwork in footing (Center line and separate wall method)
Brickwork in walls (Center line and separate wall method) Two Rooms Building
Concrete work in foundation (Center line and separate wall method)
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One Room Building
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5m
4m
0.3m0.3m
0.5m0.6m0.9m
0.4m0.6m
3.5m
Section at A-A’ and B-B’ Thickness of footing and foundation is 30 cm each
A
A’
B B’
Plan
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0.9 m
5.3m
4.3 m
Total Center line = 5.3+5.3+4.3+4.3=19.2 m
Width = 0.9 m Total Qty of concrete in foundation= (19.2×0.9×0.3) cum
Center Line Method
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0.9 m
5.3m6.2m
4.3 m
3.4 m
Separate Wall Method
Width = 0.9 m Total Qty of concrete in fdn= (2×6.2×0.9×0.3) + (2×3.4×0.9×0.3)cum
Long Wall = 5.3+0.9=6.2 m Short Wall = 4.3-0.9=3.4 m
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0.6 m
5.3m
4.3 m
Total Center line = 5.3+5.3+4.3+4.3=19.2 m
Width = 0.6 m Total Qty of brickwork in Ist footing= (19.2×0.6×0.3) cum
Center Line Method
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0.6 m
5.3m
5.9 m
4.3 m
3.7 m
Width = 0.6 m
Separate Wall MethodLong Wall = 5.3+0.6=5.9 m Short Wall = 4.3-0.6=3.7 m
Total Qty of brickwork in Ist footing= (2×5.9×0.6×0.3) + (2×3.7×0.6×0.3)cum
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5.3m
4.3 m
0.3 m
Total Center line = 5.3+5.3+4.3+4.3=19.2 m
Width = 0.3 m Total Qty of brickwork in walls= (19.2×0.3×3.5) cum
Center Line Method
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5.6 m
5.3m
4.3 m
4 m
0.3 m
Width = 0.3 m
Separate Wall MethodLong Wall = 5.3+0.3=5.6 m Short Wall = 4.3-0.3=4.0 m
Total Qty of brickwork in walls= (2×5.6×0.3×3.5) + (2×4.0×0.3×3.5)cum
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Two Rooms Building
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5m
4m0.3m
0.3m
0.5m0.6m0.9m
0.4m0.6m
3.5m
Section at A-A’ and B-B’ Thickness of footing and foundation is 30 cm each
A
A’
B B’
Plan
![Page 13: Building estimation methods](https://reader033.vdocuments.us/reader033/viewer/2022061513/587813511a28ab971e8b7759/html5/thumbnails/13.jpg)
Center Line MethodTotal Center line = 5.3+5.3+5.3+5.3+4.3+4.3+4.3=34.1 m
Width = 0.9 m Total Qty of concrete in fdn= (34.1×0.9×0.3)
5.3m 5.3m
4.3 m
0.9 m
Intermediate wall
– (2×0.9×(0.9/2) ×0.3)cum
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Separate Wall MethodLong Wall = 5.3+5.3+0.9=11.5 m Short Wall = 4.3-0.9=3.4 m
Width = 0.9 m Total Qty of concrete in fdn= (2×11.5×0.9×0.3)
5.3m 5.3m
4.3 m
0.9 m
11.5 m
3.4m
Intermediate wall
+ (3×3.4×0.9×0.3)cum
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THANK YOU
Any Questions?