The post How to calculate the self weight of RCC footing And plinth beam appeared first on Surveying & Architects.
]]>In this Article we will discuss How To Calculate the Self weight or dead load of the R.C.C footing as shown in the drawing.
Length at the base = (L) = 750-mm.
Breadth at the base = (B) = 600-mm.
Total height (H) = 500-mm.
Top surface length = (l) = 450-mm.
Top surface width = (b) = 300-mm.
The self weight of any R.C.C structure is calculated by the formula
[ Unit weight × volume ]As you know, the unit weight of the R.C.C structure = (D) = 2500-kg / cum.
The self-weight of R.C.C footing
= ( 2500-kg / m³ × 0.1904-m³ )
= 476-kg.
The span of the plinth beam-1 ( PB-1 ) = 4500-mm.=4.5-m.
The span of the plinth beam-2 ( PB-2 ) = 3000-mm.=3.0-m.
Sectional dimension of the plinth beams = 230-mm. × 450-mm.
= 0.23-m. × 0.45-m.
The self weight of the (PB-2)
= [ Unit weight × (volume) ]
= [ 2500 kg / m³ × ( 3-m × 0.23-m. × 0.45-m) ]
= 776.25-kg.
The self weight of the (PB-1)
= [ Unit weight × (volume) ]
= [ 2500-kg/m³ × ( 4.5-m × 0.23-m. × 0.45-m) ]
= 1164.38-kg
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]]>The post How To Calculate The Cement Bags Required For 2000 sq ft House appeared first on Surveying & Architects.
]]>Hello guys in this article we know about quantity of bags of cement is required for 2000 square feet house. Actual quantity of cement calculation is based on as per design but if design is not given then cement quantity calculation is based on Thumb Rule on experience basis and how many cement bags used in construction of 2000 square feet house.
1) type of structure (Load Bearing/Framed)
2) Purpose of structure (Load depends on the utility)
3) Type of grade of concrete used
4) Size height of column and length of beam (in terms of height, as requirement varies according to construction i.e. whether horizontal or vertical. )
5) Most important is foundation type(depends on soil bearing conditions)
For general residential low rise building of ground floor house we have to calculate cement bags. We use different types of Thumb Rule for cement quantity calculation in all type of RCC structure footing, column, beam & RCC slab, brickwork, flooring & plastering.
1) thumb rule for cement in RCC concrete is 0.301 cement bags/ sq. ft. built up area of low rise ground floor residential building
2) thumb rule for cement bags in surface ground flooring is 0.089 cement bags/ sq. ft. of built up area of low rise ground floor residential building.
3) thumb rule for cement bags in external wall brickwork is 1.26 cement bags/m3 of brickwork of 9 inch ( cement sand ratio 1:6)
4) thumb rule for cement bags in internal wall brick work is 1.28 cement bags/m3 of brickwork of 4.5 inch ( cement sand ratio 1:4)
5) thumb rule for cement bags for external plastering is 0.017017 cement bags/ sq. ft. plastering
6) thumb rule for cement bags for internal plastering is 0.0086 cement bags/ sq. ft. plastering
7) thumb rule for cement bags for ceiling plastering is 0.0051 cement bags/ sq. ft. plastering
For the construction of residential building of ground floor we have RCC structure like footing, column, beam and slab. Taking one rough design of 2000 square feet house having different dimension of RCC structure like footing, column, beam and slab are following:
● volume of RCC concrete calculation
1) footing size = 1500 mm × 1500mm × 300 mm
Length of footing = 1500 mm
Width of footing = 1500 mm
Height of footing = 300 mm
Depth of earth excavation = 1500 mm
Concrete size in footing = 1200mm×1200mm×300mm
No of footing = 20
Wet volume of concrete in all 20 no of footing = 1.2 m × 1.2 m × 0.3 m × 20 = 8.64 m3
2) column size = 300 mm × 400 mm
No of column = 20
Height above DPC = 3000 mm
Height above RCC slab = 900 mm
Depth of earth excavation = 1500 mm
Total height of column = 5400 mm
Wet volume of concrete for all 20 column
= 0.3 × 0.4× 5.4 × 20 = 12.96 m3
3) plinth beam size = 300mm × 300 mm
Width of beam = 300 mm
Depth of beam = 300 mm
Overall length of all beam = 400 ft. = 120000 mm
Wet volume of concrete = 0.3 × 0.3 × 120 = 10.8 m3
4) door beam size for external wall = 225mm × 150 mm
Overall length for external beam = 180 ft. = 54000 mm
Volume = 0.225 × 0.15 × 54 = 1.823 m3
Door beam size for internal wall = 125 mm × 150 mm
Overall length for internal beam = 220 ft = 66000 mm
Volume = 0.125 × 0.15 × 66 = 1.238 m3
Total volume of concrete = 1.823 + 1.238 = 3.061 m3
5) roof wall beam size for external wall = 225 mm × 200 mm
Overall length for external beam = 180 ft. = 54000 mm
Volume = 0.225 × 0.2 × 54 = 2.43 m3
Roof wall beam size for internal wall = 125 mm × 200 mm
Overall length for internal wall beam = 220 ft. = 66000 mm
Volume = 0.125 × 0.2 × 66 = 1.65 m3
Total volume = 2.43 + 1.65 = 4.08 m3
6) RCC roof slab = 50′ × 40’×6″
Length of slab = 50 feet
Width of slab = 40 feet
Thickness of slab = 6 inch = 0.5 feet
Volume = 50′ × 40’× 0.5′ = 1000 cubic feet
1 m3 = 35.3147 cu ft.
Volume = 1000/35.3147 = 28.32 m3
Now Total wet volume of concrete required for 2000 square feet house
Footing = 8.64 m3
Column = 12.96 m3
Plinth beam = 10.8 m3
Door beam = 3.06 m3
Roof beam = 4.08 m3
Roof slab = 28.32 m3
Total wet volume of concrete = 67.86 m3
Dry volume of concrete = 67.86 ×1.54 = 104.5 m3
If we take M20 grade of concrete in which cement sand and aggregate ratio is 1:1.5:3
Cement quantity = (1/5.5)× 104.5m3 × 1440 kg/m3
Cement weight = 27361 kg
1 cement bag weight = 50 kg
no of cement bags = 27361/50 = 547.2
547.2 no of cement bags required for RCC concrete structure footing, beam, column & slab for 2000 sq ft house of low rise building of ground floor.
1) cement consumption in external brick wall
External brick wall length = 2(50+40) = 180 ft
Height of brick wall = 10 feet
Height of brick wall above roof slab = 3 feet
Depth of brick wall below plinth = 2 feet
Total height = 15 feet
Area of external brick wall = 180×15 = 2700 sq. ft.
Volume = 2700 sq. ft. × 9 inch = 2025 cu ft.
Volume = 2025/35.3147 = 57.43 m3
Volume of brickwork = 57.34 m3
Thumb rule for cement bags in external wall brickwork is 1.26 cement bags/m3 of brickwork of 9 inch ( cement sand ratio 1:6)
no of cement bags = 57.34 × 1.26 = 72.25 bags
2) Cement consumption in internal brick wall
brick wall length = 50+50+40+40 + 40= 220 ft.
Height of brick wall = 10 feet
Depth of brick wall below plinth = 2 feet
Total height = 12 feet
Area of external brick wall = 220×12 = 2640 sqft
Volume = 2640 sq. ft. × 4.5 inch = 990 cu ft.
Volume = 990/35.3147 = 28 m3
Volume of brickwork = 28 m3
Thumb rule for cement bags in internal wall brickwork is 1.28 cement bags/m3 of brickwork of 4.5 inch ( cement sand ratio 1:4)
no of cement bags = 28 × 1.28 = 36 bags
Total no of cement bags = 72.25 + 36 = 108.25
108.25 no of cement bags required for brickwork of 2000 square feet house
In ground flooring we use plain cement concrete m15 grade of concrete in which mix ratio of cement sand and aggregate is 1:2:4 and floor thickness is 6 inch ( 4 inch concrete + 2 inch paltering for flooring + tile flooring)
Ground flooring concrete slab = 50′ × 40’×6″
Length of slab = 50 feet
Width of slab = 40 feet
Thickness of slab = 6 inch = 0.5 feet
Volume = 50′ × 40’× 0.5′ = 1000 cubic feet
1 m3 = 35.3147 cu ft.
Volume = 1000/35.3147 = 28.317 m3
Wet volume = 28.317 m3
Dry volume = 28.317 × 1.54 = 43.60 m3
no of cement bags = (1/7) 43.60 × (1440/50)
no of cement bags = 179 bags
179 no of cement bags required for ground flooring of concrete, plastering of ground flooring and tile flooring of 2000 square feet house
1) External plastering of brick wall
Length total = 180 ft.
Height = 10 ft.
Area of plastering = 180 × 10 = 1800 sq. ft.
thumb rule for cement bags for external plastering is 0.017017 cement bags/ sq. ft. plastering
no of cement bags = 1800 × 0.017017 = 30.6 bags
2) internal plastering of brick wall
Length total = 620 ft.
Height = 10 ft.
Area of plastering = 620 × 10 = 6200 sq. ft.
thumb rule for cement bags for external plastering is 0.0086 cement bags/ sq. ft. plastering
no of cement bags = 6200 × 0.0086 = 53.32 bags
3) ceiling plastering of roof slab
Length total = 2000 ft.
thumb rule for cement bags for ceiling plastering is 0.0051 cement bags/ ft. plastering
no of cement bags = 2000 × 0.0051 = 10.2 bags
Total no of cement bags = 30.6+ 53.32 + 10.2 = 94
94 no of cement bags required for external wall plastering, internal wall plastering and ceiling plastering of 2000 square feet house
1) R.C.C Concrete = 547.2 bags
2) Brickwork = 108.25 bags
3) Flooring + Septic tank + stair = 179 bags
4) Plastering = 94 bags
Total no of cement bags = 928.45 = 929
929 (+ 5% and _ 5℅ shift) no of cement bags ( 547.2 cement bags for RCC concrete, 108.25 cement bags for brickwork, 179 cement bags for flooring, septic tank including stair & 94 cement bags for plastering) are required for construction of 2000 square feet house of low rise ground floor building.
This is a general guidelines and actual quantity of steel may vary for special design.
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]]>The post Calculate The Cost Of Filling A Plot With Construction Soil appeared first on Surveying & Architects.
]]>Let us consider a Plot or Site as shown in the below drawing for the calculation purpose…
Given data:
Length of the plot = 80 ft.
The breadth of the plot = 50 ft.
Depth of the plot to be filled at front end = 2 ft.
Depth of the plot to be filled at back end = 3.5 ft.
The volume of the plot to be filled with construction soil.
= [length × breadth × (average depth of the plot to be filled)]
= [ 80 ft. × 50 ft. × {( 2ft. + 3.5 ft.) ÷ 2}]
= [ 80 ft. × 50 ft. × 2.75 ft.]
= 11,000 cu ft.
When we fill the plot with the loose soil, we need an extra volume of soil for settlement & compaction.
As you know,
Loose soil = 1.25 × compacted soil.
So, the volume of soil required to fill the plot
= 1.25 × 11,000 cu ft.
= 13, 750 cu ft.
= 389.36 cum.
( As 1 cum. = 35.3147 cu ft.)
The cost of filling:
The cost of construction soil falls in a range of INR 300/- to 500/– per cum.
Let us consider an average of INR 400/– for the calculation purpose.
The cost of filling the plot with soil
= [vol. of soil required in cum. × cost / cum.]
= [ 389.36 cum. × 400/- per cum.]
= INR 1,55,744/-
The soil required in no. of trucks:
The vol. of a truck container = 500 cu ft.
The no. of truck required
= [ the vol. of soil required in cu ft. ÷ the vol. of a truck]
= [ 13,750 cu ft. ÷ 500 cu ft.]
= 27.5 nos.
The soil required in no. of tractor-trolley:
The vol. of a trolley = 70 cu ft.
The no. of trolley required
= [ the vol. of soil required in cu ft. ÷ the vol. of a trolley]
= [ 13,750 cu ft. ÷ 70 cu ft.]
= 196.43 nos. say 197 nos.
Note:
1. The cost of construction soil varies according to the regional market rate. Replace the soil cost to get the correct results.
2. Here, the compaction factor of 1.25 is taken, by considering the natural compaction by rainwater & vehicle movements. If the roller or mechanical compactor is used, the multiplication factor will be 1.35.
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]]>The post What-is-the-1.54-Constant-for-Concrete-in-Quantity-Survey appeared first on Surveying & Architects.
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For example
We need to calculate dry mortar for 1cum concrete
We have 1m x 1m x 1m = 1 cum (cubic meter)
1 cum is wet volume
Now we need to convert 1 cum wet volume into 1cum dry volume
Dry volume = wet volume + 54% of wet volume
So
Dry volume = 1 + ((55/100) x 1)
Dry volume = 1 + (.54 x 1)
Dry volume = 1 + .54
Dry volume = 1.54 cum
when you need to calculate
Dry volume of concrete than just multiply wet volume with 1.54 to calculate dry morter
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The post What-is-the-1.54-Constant-for-Concrete-in-Quantity-Survey appeared first on Surveying & Architects.
]]>The post How to Calculate Quantities Cement, Sand, Aggregate and water in 1 m3 Concrete Work appeared first on Surveying & Architects.
]]>Well, this is the fundamentals things of Building Estimation that every civil Engineer should go through it. Concrete comprises cement, fine aggregates, coarse aggregates, and water in mix proportion.
Let us calculate the quantities
Note: for Concrete work: Dry volume =wet volume x 1.54
For mortar work : Dry Volume = wet volume x 1.33
Consider,
Grade of concrete (M15) =1:2:4
Where 1= part of cement, 2 = part of sand, 4 = parts of aggregates
Total ratio = 1+2+4 =7
Volume of concrete =1m3 (wet volume)
Dry volume = 1x 1.54 m3 = 1.54 m^{3}
Let us know with the help of the above information I will calculate the quantities of cement, sand, and aggregates.
Quantity of cement in m^{3} = Dry Volume x (part of cement / total ratio)
= 1.54 x (1/7) =0. 22 m^{3}
Quantities of cement in kg = 1.54 x (1/7) x1440 = 316.8 kg (Density of cement =1440 kg/ m^{3})
Quantities of cement in bags = (316.8/ 50) =6.336 bags (1 bags = 50 kg)
Quantity of sand in m^{3} =1.54 x (2/7) =0.44 m^{3}
Quantity of sand in kg =( 0.44 x 1450) =638 kg (Density of sand =1450-1500)
Quantity of sand in ft^{3} = 0.44 x 35.3147 = 15.54 ft^{3} (1 m^{3} = 35.3147 ft^{3})
Quantity of aggregates in m^{3} =1.54 x (4/7) =0.88 m^{3}
Quantity of aggregates in kg = 0.88×1500 =1320 kg (Density of aggregates =1450 -1550)
Quantity of aggregates in ft^{3} =0.88 x 35.3147 = 31.076 ft^{3}
Quantity of water in liter depends upon the water-cement ration and quantity of cement.
In the above examples, I have calculated
Weight of cement in m3 = 0.22 m^{3}
Consider water-cement ration = 0.45
Amount of water required in 1 m^{3} concrete work = Water cement ratio x cement weight
Now, Water = 0.45 x0.220 =0.0999 m^{3}
Amount of water required in liter= 0.0999 x 1000 = 99.9 liters.
Therefore, For 1 m^{3} concrete work, 6.336 bags, 15.54 ft^{3} sand, 31.076 ft^{3}
And 99.9 liters required having 1:2:4 proportions.
Let us take the example slab having a length of 5 m, a width of 3 m, and a thickness of 150 mm.
Length of slab = 5 m
Width of slab = 3 m
Thickness of slab =0.15 m
And, M15 =1:2:4 where 1 parts cement, 2 part sand, 4 part aggregates
Volume of slab = length x width x thickness
= 5 x 3 x 0.15 = 2.25 m^{3}
We got Wet Volume = 2.25 m^{3}
Therefore, Dry Volume =2.25 x1.54 = 3.465 m^{3}
Volume of cement in m^{3} = 3.465 x (1/7)=0.495 m^{3}
Volume of cement in bags = (0.495 x 1440) /50 = 14.25 bags
Volume of sand in m^{3} = 3.465 x(2/7) = 0.99 m^{3}
Volume of aggregate in m^{3} = 3.465 x(4/7) =1.98 m^{3}
Quantity of water in m^{3} = w/c x volume of cement in m^{3}
= 0.45 x 0.495 = 0.22275 m^{3}
Quantity of water in liter = 0.22275 x 1000 =222.75 liters ( 1 m^{3} = 1000 liters)
Therefore for 2.25 m^{3} of slab, 14.25 bags, 0.99 m^{3} sand, 1.98 m^{3} coarse aggregates and 222.75 liters are required.
The post How to Calculate Quantities Cement, Sand, Aggregate and water in 1 m3 Concrete Work appeared first on Surveying & Architects.
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