Tag: civil formula

How to Calculate Quantities for Painting Quantity for Room Internal Walls From Drawing?

How to Calculate Quantities for Flooring Work from Drawing? How to calculate Tile, Marble, Granite Quantity for floor from drawing?

Flooring Area is a horizontal Area usually different type of materials are used in flooring work like (concrete flooring, Tile, Granite, marble etc.…

Above drawings is a reference,

Size of Bed Room              =        2800 x 2700 mm / 2.8 x 2.7 m

Size of Hall & Dining         =        3000 x 2700 mm / 3.0 x 2.7 m

Size of Kitchen                   =        2000 x 2700 mm / 2.0 x 2.7 m

Size of Bath Room             =        1500 x 1700 mm / 1.5 x 1.7 m

Formula
Area of Flooring = Length of Room  X  Breadth of Room (L x B)

1. Bed Room Floor Area Which is in Grid (A-B) / (2-3)

Bedroom Floor Area = 2.8 x 2.7 = 7.56 m2

2. Hall & Dining Floor Area Which is in Grid (B-C) / (2-3)

Hall & Dining Area = 3.0 x 2.7 = 8.10 m2

3. Kitchen Area Floor Which is in Grid (C-D) / (2-3)

Kitchen Floor Area = 2.0 x 2.7 = 5.40 m2

4. Bath Room Floor Area Which is in Grid (A-B) / (1-2)

Bath Room Floor Area = 1.5 x 1.7 = 2.55 m2

Total Floor Area = Floor Area of Bedroom + Floor Area of Hall & Dining + Floor Area of Kitchen + Floor Area of Bath Room

Total Floor Area = 7.56 + 8.10 + 5.40 + 2.55 = 23.61 m2

For Example Tiles are used in this Flooring Work, if Size of Tile is 400 x 400 mm / 0.4 x 0.4 m

So, Nos of Tile = Total Floor Area / Area of One Tile

Area of one Tile = 0.4 x 0.4 = 0.16 m2

Total No of Tile = 23.61 / 0.16 = 147.56 / Say 148 No’s

Usually Wastages considers 5-10 % in Floor Tile / Granite / Marble Work
If Wastage is 5 %

Total No of Tile =     (148) + 5% of Total No’s of Tile

Total No of Tile =     (148) + (148*5%) =155.4 /156 No’s of Tiles Requires

How to calculate Quantities for Flooring Area? How to calculate Tile, Marble, Granite area for floor

Calculation for Flooring Work
Si.No Description Length (m) breadth / Width (m) Height (m) Area (m2) (length x Breadth)
1 Bed Room 2.80 2.70 7.56
2 Hall & Dining 3.00 2.70 8.10
3 Kitchen 2.00 2.70 5.40
4 Bath Room 1.50 1.70 2.55
Total Quantity for Floor Area 23.61

Floor Area Number of Tile Calculation

Floor Area Number of Tile Calculation -Tile Size 0.4 x 0.4 m = 0.16 m2
Si.no Description Floor Area Tile Area (0.4 x 0.4) = 0.16 m2 Total No of Tile Wastage (%) Total No of Tile With Wastage
1 Bed Room 7.56 0.16 47 5% 50
2 Hall & Dining 8.10 0.16 51 5% 53
3 Kitchen 5.40 0.16 34 5% 36
4 Bath Room 2.55 0.16 16 5% 17
Total Quantity of Tile 148
Total No of Tile Requires (Without Wastage) 148 Nos
Total No of Tile Required (With 5% of Wastage) 156 Nos

Basic Unit Conversion Table

Basic Unit Conversion Table

Basic Unit Conversion Table :- The SI base units are the standard units  of measurement defined by the International system of units (SI) , some of basic units with conversion table stated below.

A basic unit conversion table for length includes commonly used units such as meter, millimeter, centimeter, kilometer, foot/feet, inch, yard, and mile. This table provides a quick reference for converting measurements between these units, facilitating seamless transitions between different systems of measurement. Users can easily determine the equivalent values by consulting the conversion factors listed in the table.

For instance, they can convert lengths from meters to feet, inches to centimeters, or kilometers to miles, ensuring accuracy and consistency in various applications such as construction, engineering, and everyday measurements. This simple tool aids in efficient and precise unit conversions, making it a valuable resource for a wide range of fields.

Basic Unit Conversion Table

Si. No Unit  Equivalent  Units  Unit  Equivalent  Units 
meter (m)  unit Conversion 
1 1 meter 3.28083 foot  1 m 3.28083 ft 
2 1 meter 1000 millimeter 1 m 1000 mm
3 1 meter 100 centimeter 1 m 100 cm 
4 1 meter 39.3701 inch 1 m 39.3701 in
5 1 meter 0.001 kilometer 1 m 0.001 km
6 1 meter 1.09361 yard 1 m 1.09361 yd
7 1 meter 0.000621371 mile 1 m 0.000621371 mi
foot / feed  (ft)  unit Conversion
1 1 foot / feet  0.3048 meter 1 ft 0.3048 m
2 1 foot / feet  304.8 millimeter 1 ft 304.8 mm
3 1 foot / feet  30.48 centimeter 1 ft 30.48 cm
4 1 foot / feet  12 inch 1 ft 12 in
5 1 foot / feet  0.0003048 kilometer 1 ft 0.0003048 km
6 1 foot / feet  0.333333 yard 1 ft 0.333333 yd
7 1 foot / feet  0.000189394 mile 1 ft 0.000189394 mi
millimeter (mm)  unit Conversion 
1 1 millimeter 0.001 meter 1 mm 0.001 m
2 1 millimeter 0.00328084 foot  1 mm 0.00328084 foot 
3 1 millimeter 0.10 centimeter 1 mm 0.10 cm
4 1 millimeter 0.0393701  inch 1 mm 0.0393701  in
5 1 millimeter 1e-6 kilometer 1 mm 1e-6 km
6 1 millimeter 0.00109361 yard  1 mm 0.00109361 yd 
7 1 millimeter 6.2137e-7 mile 1 mm 6.2137e-7 mi
centimeter  (cm)  unit Conversion
1 1 centimeter 0.01 meter 1 cm 0.01 m
2 1 centimeter 0.0328084 foot  1 cm 0.0328084 foot 
3 1 centimeter 10 millimeter 1 cm 10 mm
4 1 centimeter 0.393701 inch 1 cm 0.393701 in
5 1 centimeter 1e-5 kilometer 1 cm 1e-5 km
6 1 centimeter 0.0109361 yard 1 cm 0.0109361 yd
7 1 centimeter 6.2137e-6 mile 1 cm 6.2137e-6 mi
inch  (in)  unit Conversion
1 1 inch 0.0254 meter 1 in 0.0254 m
2 1 inch 0.0833333 foot  1 in 0.0833333 foot 
3 1 inch 25.4 millimeter 1 in 25.4 mm
4 1 inch 2.54 centimeter 1 in 2.54cm
5 1 inch 2.54e-5 kilometer 1 in 2.54e-5 km
6 1 inch 0.0277778 yard 1 in 0.0277778 yd
7 1 inch 1.5783e-5 mile 1 in 1.5783e-5 mi
kilometre  (km)  unit Conversion
1 1 kilometer 1000 meter 1 km 1000 m
2 1 kilometer 3280.84 foot  1 km 3280.84 foot 
3 1 kilometer 1e+6 millimeter 1 km 1e+6 mm
4 1 kilometer 100000 centimeter 1 km 100000 cm
5 1 kilometer 39370.1 inch 1 km 39370.1 in
6 1 kilometer 1093.61 yard 1 km 1093.61 yd
7 1 kilometer 0.621371 mile 1 km 0.621371 mi
yard   (yd)  unit Conversion
1 1 Yard 0.9144 meter 1 yd
2 1 Yard 3.00 foot  1 yd 3.00 foot 
3 1 Yard 914.4 millimeter 1 yd 914.4 mm
4 1 Yard 91.44 centimeter 1 yd 91.44 cm
5 1 Yard 36 inch 1 yd 36 in
6 1 Yard 0.0009144 kilometer 1 yd 0.0009144 km
7 1 Yard 0.000568182 mile 1 yd 0.000568182 mi
mile   (mi)  unit Conversion
1 1 mile 1609.34 meter 1 mi 1609.34 m
2 1 mile 5280 foot  1 mi 5280 foot 
3 1 mile 1.609e+6 millimeter 1 mi 1.609e+6 mm
4 1 mile 160934 centimeter 1 mi 160934 cm
5 1 mile 63360 inch 1 mi 63360 in
6 1 mile 1.60934 kilometer 1 mi 1.60934 km
7 1 mile 1760 yard 1 mi 1760 yd

How to Calculate Quantities for Flooring Work from Drawing? How to calculate Tile, Marble, Granite Quantity for floor from drawing?

Flooring Area is a horizontal Area usually different type of materials are used in flooring work like (concrete flooring, Tile, Granite, marble etc.…

How to calculate Quantities for Flooring Work ? How to calculate Tile, Marble, Granite area for floor

Above drawings is a reference,
Size of Bed Room              =        2800 x 2700 mm / 2.8 x 2.7 m

Size of Hall & Dining         =        3000 x 2700 mm / 3.0 x 2.7 m

Size of Kitchen                   =        2000 x 2700 mm / 2.0 x 2.7 m

Size of Bath Room             =        1500 x 1700 mm / 1.5 x 1.7 m

Formula
Area of Flooring = Length of Room  X  Breadth of Room (L x B)

1. Bed Room Floor Area Which is in Grid (A-B) / (2-3)
Bedroom Floor Area = 2.8 x 2.7 = 7.56 m2

    2. Hall & Dining Floor Area Which is in Grid (B-C) / (2-3)
    Hall & Dining Area = 3.0 x 2.7 = 8.10 m2

    3. Kitchen Area Floor Which is in Grid (C-D) / (2-3)
    Bedroom Floor Area = 2.0 x 2.7 = 5.40 m2

    4. Bath Room Floor Area Which is in Grid (A-B) / (1-2)
    Bath Room Floor Area = 1.5 x 1.7 = 2.55 m2

    Total Floor Area = Floor Area of Bedroom + Floor Area of Hall & Dining + Floor Area of Kitchen + Floor Area of Bath Room

    Total Floor Area = 7.56 + 8.10 + 5.40 + 2.55 = 23.61 m2

    For Example Tiles are used in this Flooring Work, if Size of Tile is 400 x 400 mm / 0.4 x 0.4 m

    So, No of Tile = Total Floor Area / Area of One Tile
    Area of one Tile = 0.4 x 0.4 = 0.16 m2
    So, Total No of Tile = 23.61 / 0.16 = 147.56 / Say 148 No’s

    Usually Wastages considers 5-10 % in Floor Tile / Granite / Marble Work
    If Wastage is 5 %
    Total No of Tile =     (148) + 5% of Total No’s of Tile
    Total No of Tile =     (148) + (148*5%) =155.4 /156 No’s of Tiles Requires

    How to calculate Quantities for Flooring Area? How to calculate Tile, Marble, Granite area for floor

    Calculation for Flooring Work
    Si.No Description Length (m) breadth / Width (m) Height (m) Area (m2) (length x Breadth)
    1 Bed Room 2.80 2.70 7.56
    2 Hall & Dining 3.00 2.70 8.10
    3 Kitchen 2.00 2.70 5.40
    4 Bath Room 1.50 1.70 2.55
    Total Quantity for Floor Area 23.61

    Floor Area Number of Tile Calculation

    Floor Area Number of Tile Calculation -Tile Size 0.4 x 0.4 m = 0.16 m2
    Si.no Description Floor Area Tile Area (0.4 x 0.4) = 0.16 m2 Total No of Tile Wastage (%) Total No of Tile With Wastage
    1 Bed Room 7.56 0.16 47 5% 50
    2 Hall & Dining 8.10 0.16 51 5% 53
    3 Kitchen 5.40 0.16 34 5% 36
    4 Bath Room 2.55 0.16 16 5% 17
    Total Quantity of Tile 148
    Total No of Tile Requires (Without Wastage) 148 Nos
    Total No of Tile Required (With 5% of Wastage) 156 Nos

    How to Calculate Number of blocks or Hollow blocks Quantities in block wall ?

    How to Calculate Number of blocks or Hollow blocks Quantities in block wall ?

    Hollow Block :- is a concrete block used in construction to build an internal & External Walls , there are different size of hollow blocks are used in construction, Hollow Concrete blocks are made of Portland cement, fine aggregate and sand. Usually fly ash or bottom ash is mixed with the concrete mixture.

    How to calculate Hollow Block Quantities in Wall

    For Example Wall Size is

    Length (L) -1m, Height (H) – 1m and Width (W) is-1m  

    Standard block Size –400 x 200 x 200 mm / 0.40 x 0.20 x 0.20 m (Without Mortar)

    Mortar Thickness is -10mm, 10mm mortar used in all the four sides

    Block Size with Mortar – 410 x 210 x 210 mm / 0.41x 0.21x 0.21 m

    Formula to Find the Number of Block Quantity in Wall is

    Formula = Volume of Wall / Volume of One Block

    Volume of Wall   = Length (L) x Width (W) x Height (H)

    Volume of Wall   =  1 x 1 x 1 = 1.0 mᶟ

    Volume of one Block without Mortar = 0.40 x 0.20 x 0.20 = 0.0160 mᶟ

    Volume of one Block with Mortar       = 0.41x 0.21x 0.21 = 0.0181 mᶟ

    So, No of Block Required in 1mᶟ of Wall is

    No of Block without Mortar  =    1.0 / 0.0160 = 62.50 /say 63 No’s

    No of Block with Mortar        =    1.0 / 0.0181 = 55.31 /say 56 No’s

    Usually Wastages considers 5% in block Work

    No of Block with Mortar        =     (1.0 / 0.0181)+5% of Total No’s of Blocks

    No of Block with Mortar        =     (56)+(56*5%) = 59 No’s of Blocks Requires

    1. Block Without Mortar (400 x 200 x 200) mm Size  Required in 1mᶟ = 63 Nos

    2. Block With Mortar (410 x 210 x 210 ) mm Size  Required in 1mᶟ = 56 Nos  (Without wastage)

    3. Block With Mortar     (410 x 210 x 210) mm Size Required in 1mᶟ = 59 Nos (With wastage 5%) 

    How to Calculate Number of Brick Quantity in brick wall

    Brick is a construction material it’s made up with clay, composition of brick is Lime 2 to 5% Alumina-clay 20 to 30%, Silica-Sand – 50 to 60%, Bricks are rectangular blocks of backed clay which is used to construct a walls and payment and other some masonry in construction.

    Brick Masonry is constructed by placing of brick with mortar which is called as brick masonry Brick masonry is a highly durable form of construction

    How to calculate brick quantities in brick masonry Wall

    For Example Wall Size is

    Length (L) -1m, Height (H) – 1m and Width(W) is-1m

    Standard brick Size – 190 x 90 x 90 mm / 0.19 x 0.09 x 0.09 m (Without Mortar)

    Mortar Thickness is -10mm

    Brick Size with Mortar – 200 x 100 x 100 mm / 0.20 x 0.10 x 0.10 m

    Formula to Find the Number of Brick Quantity in Wall is

    Formula = Volume of Wall / Volume of One Brick 

    Volume of Wall   = Length (L) x Width (W) x Height (H)

    Volume of Wall   =  1 x 1 x 1 = 1.0 mᶟ

    Volume of one Brick without Mortar = 0.19 x 0.09 x 0.09 = 0.00154 mᶟ

    Volume of one Brick with Mortar       = 0.20 x 0.10 x 0.10 = 0.00200 mᶟ

    So, No of brick required in 1mᶟ of Wall is

    No of Brick without Mortar  =    1.0 / 0.00154 = 650 Nos

    No of Brick with Mortar        =    1.0 / 0.0020 = 500 Nos

    Usually Wastages considers 5% in brick Work 

    No of Brick with Mortar        =  (1.0 / 0.0020)+5% of Total Nos of Bricks

    No of Brick with Mortar        =  (500)+(500*5%) = 525 Nos of Bricks Requires

    1. Brick Without Mortar (190 x  90 x 90) mm Size  Required in 1mᶟ = 650 Nos

    2. Brick With Mortar (200 x 100 x 100 ) mm Size  Required in 1mᶟ = 500 Nos  (Without wastage)

    3. Brick With Mortar (200 x 100 x 100 ) mm Size  Required in 1mᶟ = 525 Nos (With wastage 5%) 

    How to Calculate Cement, Sand, Quantities for Plastering Work ?

    Plastering: – This work activity means a thin layer of mortar, applied over the Wall & ceiling masonry, which it acts like a damp-proof coat over the masonry work

    Mortar: – Is a mixture of Cement (or) Lime, sand & water, which is used in masonry work in construction to fill the gaps between the bricks and blocks.

    How to Calculate Cement, and Sand Quantities in Plastering Work

    For Example size of Wall

    Length of Wall (L) – 50m, Height of Wall (H) – 4.0m,

    Thickness of Plaster (T) – 12 mm (Inner Wall)

    STANDART PLASTER THICKNESS

    Plastering Area Thickness of Plaster
    Inner Plaster Work 10-12 mm
    Outer Plaster Work 15-20 mm
    Ceiling Plaster Work 6-8 mm


    Plaster Area =Length × Height

    Plaster Area =50×4 =200.00m2

    Volume of Mortar = Plaster Area x Plaster Thickness in meter

    Volume of Mortar = 200.00 x 0.012 = 2.40 mᶟ

    If Ratio of Plaster is – 1:6 , Which Means 1% of Cement & 6 % of Sand

    FORMULA TO FIND THE QUANTITIES OF CEMENT, SAND  IS

    Formula = \frac{Dry Volume of mortar  *  Volume of mortar *  Ratio of (Cement /Sand) }{Total Ratio (Cement + Sand)}  

    If it is Outer Wall Dry Volume of  mortar -1.33mᶟ

    Volume of Mortar -2.4 mᶟ

    Ratio of Cement = 1, Ratio of Sand =6, ,  Total Ratio (1+ 6) = 7

    Required Quantity of Cement & Sand in 2.4 mᶟ Plastering Work  is

    Cement = \frac{1.33 * 2.4 * 1}{ 1 +  6}   = 0.456 mᶟ

    Sand= \frac{1.33 * 2.4* 6}{ 1 +  6}   = 2.736 mᶟ

    Which Means 1:6 mortar Ratio of Plastering Works  Required,

    Cement – 0.456 mᶟ / 1mᶟ of cement Equal to 29 bags of Cement,

    So, 0.456x 29 = 13.224 Bags of Cement is required

    1 Bag of cement to is Equal to 50Kg, So 50 x 13.224  =661.20 kg Cement Required

    Sand – 2.736 mᶟ / 1mᶟ equal to 35.31 cu.ft / cubic foot / ftᶟ

    So, 2.736 x 35.31 = 96.608 cu.ft / cubic foot Sand is required

    How to Calculate Cement, Sand, and Aggregate Quantities in Concrete?

    CONCRETE:-

    Concrete it’s a construction component, a mixture of cement, sand, and course or fine aggregate with water which is called us concrete, which is used in construction in different structures, In Construction industry concrete are used to mix in following ways as follows

    1.Hand Mix Concrete 2. Machine mix Concrete 3. Ready Mix Concrete

    How to Calculate Cement, Sand, and Aggregate Quantities in Concrete

    For Example size of slab
    Length (L) – 1m, Width (B) – 1m, Thickness /Depth (D) -1.0 m

    SO, Total Volume of concrete = (L x B x D) = 1 x 1 x 1 = 1.0 mᶟ

    if Grade of Concrete -M7.5 ,  Ratio of M 7.5 Concrete – 1 : 4 : 8

    Which Means 1% of  Cement, 4% of  Sand & 8% of Aggregate

    FORMULA TO FIND THE QUANTITIES OF CEMENT, SAND, AGGREGATE IS

    Formula = \frac{Dry Volume of Concrete  *  Volume of Concrete  *  Ratio of (Cement / Sand /Aggregate) }{Total Ratio (Cement + Sand + Aggregate)}  

    Dry Volume of Concrete -1.54 mᶟ  , Volume of Concrete -1.0 mᶟ

    Ratio of Cement = 1,  Ratio of Sand =4,  Ratio of Aggregate =8 ,  Total Ratio (1+ 4+ 8) = 13

    Required Quantity of Cement , Sand & Aggregate in 10 mᶟ Concrete is

    Cement = \frac{1.54 * 1 * 1}{ 1 +  4 +  8}   = 0.118 mᶟ

    Sand= \frac{1.54 * 1 * 4}{ 1 +  4 +  8}   = 0.474 mᶟ

    Aggregate = \frac{1.54 * 1 * 8}{ 1 +  4 +  8}   = 0.948 mᶟ

    Which Means 1.0 mᶟ of M7.5 Grade Concrete Required,

    Cement – 0.118 mᶟ / 1mᶟ of cement Equal to 29 bags of Cement,

    So, 1.184 x 29 = 3.43 Bags of Cement is required

    1 Bag of cement to is Equal to 50Kg, So 50 x 3.43 = 171.169 kg Cement Required (Without Any Wastage)

    Sand – 0.474 mᶟ / 1mᶟ equal to 35.31 cu.ft / cubic foot / ftᶟ

    So, 0.474 x 35.31 = 16.732 cu.ft / cubic foot Sand is required (Without Any Wastage)

    Aggregate – 0.948 mᶟ / 1mᶟ equal to 35.31 cu.ft / cubic foot / ftᶟ

    So, 0.948 x 35.31 = 33.463 cu.ft / cubic foot Aggregate is required (Without Any Wastage)

    Bill of Quantity (BOQ) for Substructure Water Proofing Work

    Waterproofing Work

    Buildings is the method of forming a barrier over surfaces of foundations, roofs, walls, bathrooms and other structural members of buildings to prevent water penetrations through these surfaces. In summary, building surfaces are made water-resistant and waterproof.

    There are different type of water proofing method is used in Roofing work , Following BOQ sample for Roof is Torch applied bituminous Water proofing.

    A Bill of Quantities (BOQ) for Substructure Waterproofing Work is a detailed document used in construction projects to enumerate and quantify the various resources required for waterproofing the substructure of a building. This comprehensive document is crucial for cost estimation, tendering, and project management, providing a clear breakdown of materials, labor, and other essential elements

    “Here is a sample of the bill of quantity for Substructure Water Proofing Work , formatted as follows

    Click here below download button for to download this BOQ

    BILL OF QUANTITY FOR SUBSTRUCTURE WATER PROOFING WORK

    BILL OF QUANTITY FOR SUBSTRUCTURE WATER PROOFING WORK
    Si.No Description Unit Quantity Unit Rate Amount Remark
    Bill of Quantity for Water Proofing Work- Torch applied (or) Self adhesive membrane
    Horizontal Area Water Proofing-Basemen, Raft, Footing ,Foundation
    1 Supply and apply 50-100 mm thick plain cement concrete on Horizontal Surface as per project specification m2
    2 Surface Preparation on horizontal area before application of water proofing work m2
    3 Supply and apply one (or) two coat approved bituminous primer, as per project specification m2
    4 Supply and apply one (or) Two Layer Approved Torch applied (or) Self adhesive bituminous water proofing membrane over the primer surface, as per project specification m2
    5 Supply and apply one layer Geotextile (or) Terram Sheet as a separation layer over the membrane surface, as per project specification m2
    6 Supply and Apply 50-100mm thick protection screed over the separation layer , as per project specification m2
    Vertical Area Water Proofing- Retaining Wall , Column , Footing
    7 Supply and apply one (or) two coat approved bituminous primer, as per project specification m2
    8 Supply and apply one (or) Two Layer Approved Torch applied (or) Self adhesive bituminous water proofing membrane over the primer surface, as per project specification m2
    9 Supply and apply one layer bituminous (or) polypropylene protection board over the membrane surface, as per project specification m2
    10 Supply and apply aluminum flashing on Wall or Column Termination area , as per project specification lm
    Bill of Quantity for Water Proofing Work- Polyvinyl Chloride (PVC) Membrane
    Horizontal Area Water Proofing-Basemen, Raft, Footing ,Foundation
    1 Supply and apply 50-100 mm thick plain cement concrete on Horizontal Surface, as per project specification m2
    2 Surface Preparation on horizontal area before application of water proofing work m2
    3 Supply and Apply one layer geotextile as a base layer over the prepared area, as per project specification m2
    4 Supply and apply one layer PVC membrane over the geotextile, joints treated with glue and melting, as per project specification m2
    5 Supply and Apply one layer geotextile as a top layer over the PVC Membrane , as per project specification m2
    6 Supply and Apply 50-100mm thick protection screed over the separation layer , as per project specification m2
    Vertical Area Water Proofing- Retaining Wall , Column , Footing
    7 Supply and Apply one layer geotextile as a base layer over the prepared area, as per project specification m2
    8 Supply and apply one layer PVC membrane over the geotextile, joints treated with glue and melting, as per project specification m2
    9 Supply and Apply one layer geotextile as a top layer over the PVC Membrane, as per project specification m2
    10 Supply and apply one layer polypropylene protection board over the geotextile, as per project specification m2
    11 Supply and apply aluminum flashing on Wall or Column Termination area , as per project specification lm
    Total Value of Work Done For Substructure Water proofing Work

    Concrete Grades in Construction: What They Mean & How They’re Used

    Concrete is a mixture of cement, sand, aggregate and water, Grades of Concrete is defined as concrete mix proportion and the minimum strength of concrete at the end of curing period of 28days.

    The Concrete grade is can be known by calculating compressive strength of concrete, compaction factor test is used to find the compressive strength of concrete , unit of compressive strength is known as N/mm²

    Compressive Strength For Various Grade of Concrete

    Compressive Strength For Various Grade of Concrete
    Si.No Character of Concrete Concrete Grade MIX Ratio Compressive Strength
    1 Lean Concrete M -05 1 : 5 : 10 5 N/mm² 725 psi
    2 M -7.5 1 : 4 : 08 7.5 N/mm² 1087 psi
    3 Ordinary Concrete M -10 1 : 03 : 06 10 N/mm² 1450 psi
    4 M -15 1 : 02 : 04 15 N/mm² 2175 psi
    5 M -20 1 : 1.5 : 03 20 N/mm² 2900 psi
    6 M -25 1 : 1 : 02 25 N/mm² 3625 psi
    7 Standard Concrete M -30 Design Mix 30 N/mm² 4350 psi
    8 M -35 Design Mix 35 N/mm² 5075 psi
    9 M -40 Design Mix 40 N/mm² 5800 psi
    10 M -45 Design Mix 45 N/mm² 6525 psi
    11 High Strength Concrete M -50 Design Mix 50 N/mm² 7250 psi
    12 M -55 Design Mix 55 N/mm² 7975 psi
    13 M -60 Design Mix 60 N/mm² 8700 psi
    14 M -65 Design Mix 65 N/mm² 9425 psi
    15 M -70 Design Mix 70 N/mm² 10150 psi

    What are the Types of Levels used in Construction?

    There are different levels used in the construction, these levels are used to make the construction process easy and understandable, some of basic level as follows.

    LEVELS IN CONSTRUCTION

    LEVELS IN CONSTRUCTION
    BOB Bottom of Blinding
    TOB Top of Blinding
    BOF Bottom of Footing
    TOF Top of Footing
    PL Plinth Level
    GL Ground level
    EGL Excisting Ground Level
    NGL Natural Ground Level
    FGL Finished Ground Level
    SL Sill Level
    LL Lintel Level
    SSL Structural Slab Level / Slab Sill Level
    FFL Floor Finish Level / Finished Floor Level
    SFL Structural Floor Level / Structural Finish Level
    FG Finished Grade
    FL Finished Level
    GF Ground Floor
    MF Mezzanine Floor
    FF First Floor
    SF Second Floor
    RF Roof Floor
    TOW Top of Wall
    BP Bottom of Pool
    TP Top of Pool
    TB Top of Bench
    TK Top of Kerb
    NC Neck column


    How to Calculate Backfilling Quantity (or) Volume of Backfilling Quantity?

    Backfilling means the process used refilling soil (or) sand an excavated area, backfilling quantity usually calculating in m³ (Cubic meter), basically There are two type of backfilling method used in construction work.

    1.Handmade Backfilling – The Soil is refilling by Humans.

    2. Machinery made Backfilling – The Soil is refilling machinery.

    For Example:–

    Size of Footing

    Length (L) – 3.00 m,
    Width (B) – 2.50 m,
    Thickness /Depth (D) -0.75 m

     Size of Column

    Length (L) – 0.50 m
    Width (B) – 0.50 m
    Thickness /Depth (D) – 3.00 m

    Size of PCC

    Length (L) – 3.20 m,
    Width (B) – 2.70 m,
    Thickness /Depth (D) -0.10 m

    So, Volume of Backfilling is,

    Formula = Total Volume of Excavation Area – (Total Volume of PCC + Total Volume of Footing + Total Volume of Column) +10% of total Backfilling Quantity

    What is 10% – While doing backfilling compaction need 10% of More Soil for backfilling, so 10% of total Quantity need to add in calculation for backfilling quantity.

    Volume of Excavation = (3.20 +0.6+0.6) X (2.70+0.6+0.6) X (0.1+0.75+3.0) = 66.07 m³

               Total Volume of Excavation = 66.07 m³

    Volume of PCC – Length X Width X Depth = 3.20 X 2.70 X 0.10 = 0.864 m³

    Volume of Footing – Length X Width X Depth = 3.00 X 2.50 X 0.75 = 5.625 m³

    Volume of Column – Length X Width X Depth = 0.50 X 0.50 X 3.00 = 0.75 m³

    So, Total Volume of back filling is = (66.70 – (0.864+ 5.625+0.75)) + 10%

    = (66.07-7.329) + 10%

    = 58.831 + 10% of Total Back Filling Area

    = 58.831 + (58.831 X 10%)

    = 64.71 m³ Total Quantity of Soil (or) Sand required For Backfilling

    So, TOTAL VOLUME OF BACK FILLING IS = 64.71 m³

    How to Calculate Excavation Work (or) Volume of Excavation Quantity?

    Excavation means work involving the removal of soil or sand or rock from earth, basically two types of excavation methods used in construction

    1.Handmade excavation – The Soil or sand or rock is excavating by Humans.

    2. Machinery made excavation -The Soil or sand or rock is excavating by machineries.

    Volume of Excavation is,

    For Example:-

    Size of Footing
    Length (L) – 3.00 m,
    Width (B) – 2.50 m,
    Thickness /Depth (D) -0.75 m

     Size of Column Length (L) – 0.50 m
    Width (B) – 0.50 m
    Thickness /Depth (D) – 3.00 m

    Size of PCC Length (L) – 3.20 m,
    Width (B) – 2.70 m,
    Thickness /Depth (D) -0.10 m

    So, Volume of Excavation is

    Formula = (Length of PCC + Working Space) X (Width of PCC + Working Space) X (Height of PCC + Height of Footing + Height of Column)

    What is working space and description of working space given below this article

    Volume of Excavation = (3.20 +0.6+0.6) X (2.70+0.6+0.6) X (0.1+0.75+3.0)

    Total Volume of Excavation = 66.07 m³

    What is Working Space: – Working space is the additional space provided between both edges of the excavation of the footing, which is used for the labor movement, using machinery, removing shuttering boards, pouring of concrete, etc.

    Working spaces are considered depending on structures varying between shallow and deep foundations.

    Shallow Foundation: – if the depth of excavation is less than 0.6m which is called the shallow foundation, for this condition working space is not required.

    Deep Foundation: – if the depth of excavation is up to 3.0m or above which is called a deep foundation.

    If the depth of the deep foundation is up to 3.0 m working space provides 600mm of all the four sides of the footings.

    If the depth of the deep foundation is more than 3.0 m working space to be provided 600+50mm (for every 1 meter) shall be added on all four sides of the footing edge. 

    How to Calculate Weight of Reinforced steel Rod

    Reinforcement Steel

    Usually 6mm, 8mm, 10mm, 12mm, 16mm, 20mm, 25mm, 32mm, 40mm dia size of reinforced steel bar used in construction, in this bars 6mm & 8mm are called as plain bar and 12mm, 16mm, 20mm, 25mm, 32mm, 40mm bars are called as rebar (or) Torsion bar , usually standard length of one reinforced steel bar is 12m.

    Plain bar:-  The steel bar is don’t have rip which is called as plain bar

    Torsion / Rebar:-  The steel bar having a rip which is called as torsion or rebar

    Formula to find weight of reinforced steel rod / unit weight of steel bars.

             Formula = D²/162, ( D-Diameter of steel bar )

    Unit Weight calculation of Steel bar @ 1m Length

    Unit Weight of 6 mm Steel bar  =  6² /162    =   0.22 kg/m

    Unit Weight of 8 mm Steel bar  =  8² /162    =   0.395 kg/m

    Unit Weight of 10 mm Steel bar =  10² /162  =  0.617 kg/m

    Unit Weight of 12 mm Steel bar =  12² /162  =   0.889 kg/m

    Unit Weight of 16 mm Steel bar =  16² /162  =   1.580 kg/m

    Unit Weight of 20 mm Steel bar =  20² /162 =   2.469 kg/m

    Unit Weight of 25 mm Steel bar =  25² /162 =    3.858 kg/m

    Unit Weight of 32 mm Steel bar =  32² /162 =   6.321 kg/m

    Unit Weight of 40 mm Steel bar =  40² /162 =   9.877 kg/m

    Standard length of one reinforced steel bar is 12m so,

    Unit Weight Of Steel Bars

    Unit Weight Of Steel Bars
    Dia of Steel Bar (mm) Type Unit Weight 1m Length Unit Weight for 12m Length Steel Bar Unit
    6 mm Plain Bar 0.222 2.664 Kg
    8 mm 0.395 4.741 Kg
    10 mm Torsion (or) Rebar 0.617 7.404 Kg
    12 mm 0.889 10.67 Kg
    16 mm 1.580 18.96 Kg
    20 mm 2.469 29.63 Kg
    25 mm 3.858 46.30 Kg
    32 mm 6.321 75.85 Kg
    40 mm 9.877 118.52 Kg
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