Make cost estimation of RC building
TweetRCC Column specification
- Foundation are 2m long x 2m wide x 0.45m . deep
- PCC 1:4:8 is used in the foundation
- RCC 1:2:4 used in foot
- 350mm is the size from the top of the foot to the base level and RCC 1:2:4 is used
- RCC column quantity
The shoe sole Estimate the amount of RCC in the column Calculation of reinforcement in In foundation excavation work, estimate the excavation volume.
Foundations and columns Estimate the total cost of all jobs
Estimate the amount of PCC in the amount of foundation. Estimate the amount of RCC in
Estimate the amount of PCC in the amount of foundation.
From the cross-section, the depth of the PCC in the foundation is 80 mm. According to the plan, the length of the foundation is 2 m and the width is 2 m.
Conventional cement concrete, PCC 1:4:8 is used in the foundation.
The amount of PCC can be estimated by = 2m length x 2m width x 0.08m depth = 0.32 cum.
Estimate the amount of RCC in the sole RCC 1:2:4 is used in the footer.
From the cross section, the length of the foundation is 2m, the width of the foundation is 2m and the depth is 0.45m The amount of RCC in the foundation can be calculated as = 2 m x 2 m x 0.45 m = 1.8 cum.
Reinforcement amount in footing
Length of the footing is two m and width is two m, clean cover =40mm No. of bars =(clean period/spacing) +1 Clear period=general period-2 x clean cover=2 m-2 x 0.04m=1.ninety two m Total No.of bars alongside the period of footing=(1.ninety two/0.15)+1= 14 bars. Similarly, Total No.of bars alongside the width of footing=(1.ninety two/0.15)+1= 14 bars. The general period of metallic bar=period-2x clean cover +20 x dia.metallic=2 m -2 x 0.04+ 20×0.012=2.sixteen m. Unit Weight of metallic= (metallic dia x metallic dia) /162= 12 x 12/162=0.89 kg/m Total weight of metallic amount in footing may be calculated by=2 methods x No`s of bars x general metallic period x unit weight of metallic =2 x 14 no`s x 2.sixteen m x 0.89 kg/m =53.eighty two kgs
Main bars weight
No. of main bars =6 no’s Total length of bar from footing to slab bottom =0.45 m+0.3 m +bottom bent-clear cover+0.6 m+3.3 m+top bent=0.45+0.3+0.15-0.08+0.6+3.3+0.1=4.89 m Unit weight =dia x dia/162=16 x 16/162=1.58 kg/m Total weight of main bars =No.of bars x total length x unit weight=6No’s x 4.89 m x 1.58 kg/m=46.35 kgs
Length of the footing is two m and width is two m, clean cover =40mm No. of bars =(clean period/spacing) +1 Clear period=general period-2 x clean cover=2 m-2 x 0.04m=1.ninety two m Total No.of bars alongside the period of footing=(1.ninety two/0.15)+1= 14 bars. Similarly, Total No.of bars alongside the width of footing=(1.ninety two/0.15)+1= 14 bars. The general period of metallic bar=period-2x clean cover +20 x dia.metallic=2 m -2 x 0.04+ 20×0.012=2.sixteen m. Unit Weight of metallic= (metallic dia x metallic dia) /162= 12 x 12/162=0.89 kg/m Total weight of metallic amount in footing may be calculated by=2 methods x No`s of bars x general metallic period x unit weight of metallic =2 x 14 no`s x 2.sixteen m x 0.89 kg/m =53.eighty two kgs
Main bars weight
No. of main bars =6 no’s Total length of bar from footing to slab bottom =0.45 m+0.3 m +bottom bent-clear cover+0.6 m+3.3 m+top bent=0.45+0.3+0.15-0.08+0.6+3.3+0.1=4.89 m Unit weight =dia x dia/162=16 x 16/162=1.58 kg/m Total weight of main bars =No.of bars x total length x unit weight=6No’s x 4.89 m x 1.58 kg/m=46.35 kgs
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