How much Bricks, Cement, Sand is required for a one cubic meter of brick work?

How much cement is required for a one meter cube of brick work?
Answer:
The size of the modular brick is 190 X 90 X 90 (refer the picture)Mortar Thickness is 10mmAssume required brickwork volume is 1 cubic metre (m3)

Any brick walls consist of bricks and cement mortar.

So, first of all, we are going to find the volume of bricks with mortar thickness and then volume of bricks alone.

Volume of bricks with mortar

Volume of 1 brick with mortar = 200 X 100 X 100 ( 10 mm mortar thickness on all sides)

= 0.2 X 0.1 X 0.1

Volume of brick with mortar = 0.002 Cum (m3)

Therefore, Number of bricks required for 1 cubic metre = 1/0.002 = 500 No.s

Volume of bricks without mortar

Volume of 1 brick without mortar = 190 X 90 X 90

= 0.19 X 0.09 X 0.09

Volume of 1 brick without mortar = 0.001539 Cum (m3)

Volume of 500 bricks without mortar = 500 X 0.001539 Cum

Volume of bricks without mortar for 1 cum = 0.7695 Cum (m3)

Therefore,

Required amount of cement mortar = 1 Cum – Volume of bricks without mortar

= 1 – 0.7695

Required amount of cement mortar = 0.2305 Cum (m3) (Wet Condition)

Note – The above volume is in a wet condition that means we need 0.2305 cement mortar in mixed condition (after adding water). In order to find the dry volume, we need to multiply 33 % as bulkage of sand.

Dry volume of a mortar = 0.2305 cum X 1.33 = 0.306565 cum

We know the mortar ratio is 1:6 (1 part Cement & 6 Part Sand = 7 Part)

Required amount Cement quantity in brickwork = 0.306565 X 1/7 X 1440 kg

Density of cement = 1440 kg. The reason to multiply this density is, the above multiplication will give us only required amount of cement quantity in brickwork as a cubic metre. But we need cement in Kg. Therefore we are multiplying the 1440 kg density of cement to calculate the cement quantity.

Required amount Cement quantity = 63 Kg = 1.26 bags (50 Kg bag)

Required amount of Sand = 0.306565 X 6/7 = 0.26277 Cubic metre (m3)

Therefore, For 1 cum of brickwork, we need

500 Numbers of bricks
63 kg of cement (1.26 bags)
0.263 m3 of sand. (9.29~10cu.ft)

Estimate the Quantities and Cost of Building Materials for House Construction

                              In order to construct a good quality house in the amount you have budgeted, a thorough understanding of the quality parameters, cost and quantities of these building materials are required.The cost of construction depends majorly on the following factors:

1. Architectural Design opted (like Open Top, Sloped Roofs, terraces with add-on features etc.,)
2. Structural Design (depends upon type of strata available for foundation and numbers of floors / configurations (basement, stilt,G+2 etc.)
3. Specification of Building materials selected (Quality/Brand of materials used for painting, flooring, woodwork, Bathroom, Electrical etc.)
4. Exterior Finish chosen (i.e. front elevation design, stone cladding, facade, etc.)
5. Peripheral external developments (such as compound wall, driveway, landscape, hardscape, Gate etc,).

The Construction cost can be broadly split into Labor and Material Cost. The extremely increasing construction trends are considered the driving force behind this fast upraise of total building construction costs.

 

The major raw material, intermediately and finished construction materials contributing major pie to overall material cost are:

Requirements for Building Construction:

1. Reinforcing Bars / Steel

2. Cement

3. Sand/Fine Aggregates

4. Coarse Aggregates

5. Bricks/Blocks

6. Tiles

7. Paints

8. Fittings (Window, Door, CP Fittings, Sanitary wares, Plumbing and Electrical fittings)

1. Reinforcing Bars / Steel required for building construction:

Reinforcement Steel is the most important structural material in construction. Steel is used in RCC (Reinforcement cement concrete). Generally rebars available in the market are manufactured through Thermo mechanical treatment (TMT). Rebars comes in different grades (i.e, Fe415, Fe500, etc.,). Fe500 is generally recommended by structural designer for structural requirement fulfillment.
The approximate Steel consumption per sq.ft built up area (BUA) is 4 kg (for low rise construction i.e., less than 4 floors of construction). Steel contributes the most among all individual materials, about 25% of total material cost. So, a price rise of Rs.5 per kg can make big difference in the total cost of construction.

2. Cement required for building construction:

Cement is an important construction material and when mixed with materials like sand, aggregates (stone chips), and water, it binds them together. It is used in concrete, in brick masonry work, in tiling, and in plaster works.
Good quality cement should feel smooth when rubbed between fingers. If a small quantity of cement is thrown into a bucket of water it should sink and not float. Cement should always be kept free from moisture. Its storage should have finished floor raised to at least 300mm above ground level and should have airtight storage. Use of cement older than 2 months should be avoided as cement loses strength with increase in its shelf life.
OPC 53 grade is generally used for concrete works and blended cement (PPC & PSC) for masonry, tiling and plaster works.
The approximate cement consumption per Sq.ft built up area (bua) is 0.4 bags. Cement as a construction material contributes about 16% of total material cost.

3. Sand required for building construction:

Sand is used mainly in Concrete, Masonry, Plaster and Flooring. Good sand should be well graded i.e., particle size ranging from 10mm to 0.150 mm for concrete and masonry works, and 5mm to 0.150 for plaster. It should be free from slit/clay and organic matter.
Natural Sand (also called River Sand) is obtained from River Beds. Due to environmental impacts and stringent laws by the government, Natural sand is slowly and gradually being replaced by Crushed sand (for concrete and masonry works) & Plaster sand (for plaster works). Crushed Sand and Plaster Sand are manufactured from Quarry Stone using latest production technology.
Sand consumption per sq.ft built up area (bua) is 1.8 cft and contributes about 12% of total material cost for building construction.

4. Aggregate required:

Crushed rocks are used as coarse Aggregates and are generally used in making concrete. Coarse aggregates are normally available in two fractions 20mm and 10mm for concrete making.
Aggregates should be clean, dense & hard. The aggregate should be neither flaky nor elongated. Flaky and Elongated aggregates decrease the strength of the concrete and demands more cement. Aggregates should be stored properly and different fractions must not be intermixed. Both these aggregate fractions should be used invariably.
Coarse aggregate (chips/gravel) consumption per sq.ft built up area (bua) is 1.35 cft. Aggregate as a construction material contributes about 8% of total material cost.

5. Bricks required for House Construction:

Bricks in old days, were commonly made of clay and were known as burnt clay bricks. Now a days, bricks are made of other materials such as fly ash. But clay bricks are still widely used in low rise residential constructions today. Bricks are used for masonry wall construction. Other substitute materials to bricks are Concrete solid/hollow blocks, Autoclaves Aerated Concrete (AAC) Blocks and Cellular light weight concrete CLC Blocks.
The clay bricks should have uniform size, uniform copper color, plain (without undulated surfaces), rectangular surfaces with parallel sides and sharp straight edges. Well burnt brick should give a metallic sound when struck with other brick. Good bricks should not exceed +/- 3 mm tolerances in length and +/- 1.5 mm tolerances in width and height. Water absorption should not exceed 20% by weight.
Bricks approximately cost Rs.7000 per 1000 units (Nos). Bricks contribute to about 5% of total material cost and are consumed approximately at 1.45 brick per sqft of built up area (BUA).

6. Tiles:

Ceramic tiles are generally made from red or white clay fired in a kiln. They are finished with a durable glaze which carries the color and design. Ceramic tiles are manufactured for both wall and floor, having varying degrees of wear resistance and water absorption. High strength and Low water absorption ceramic floor tiles are commonly known as Vitrified tiles. Tiles prices vary according to their types and quality.
Tile should be easy to clean, strong, sturdy and stain resistant. Tiles in wet area like bathroom should be of anti-skid floor type.
Tiles consumption per sq.ft built up area (BUA) is 1.3 sq.ft. Tiles contribute about 8.0% of total material cost.

7. Paints:

Paints can be broadly classified into water based or solvent based. They come in thousands of shades and gives multiples finishes like Matt, satin and glossy finish. Certain Paints also have washables, anti-algae/fungal, crack bridging properties.
When selecting an interior paint, try choosing water-based paint instead of oil-based gloss paint. Water-based paints have less odor than conventional oil-based paints.They are much easier to clean up and are durable.
When selecting an external paint look for waterproofing, anti-algae, and dirt pick resistance properties.
Paints (Internal- Emulsion and external grade) consumption per sq.ft built up area (BUA) is 0.18 liter (0.14 liter for internal painting and 0.4 for external painting).
Paints contribute about 4.1% of total material cost.

The Finishers (Bricks, Tiles, and Paints) collectively contribute 16.5% of total material cost.

 

 

Plaster of Paris & Gypsum

What is Plaster of Paris?

                                       Plaster of Paris is obtained by pulverizing the gypsum (calcium sulphate hemihydrate CaSO₄ 0.5H₂O) which is heated to a temperature of 150 deg centigrade.
Upon addition of water, Plaster of Paris (POP) becomes regular gypsum (dihydrate) again causing the resultant material to harden. This hardening material can be used to create moulds for casting and in construction.

What is Gypsum?

                     Gypsum is a soft sulfate mineral composed of Calcium sulfate dihydrate (CaSO4:2H20). It is widely used as a fertilized, in mould, in sculptures and as plaster material.
Gypsum is a chalk like material and is very light in weight. It is available in crystalline form in nature.
In recent years, the construction sector has witnessed a number of new trends, technology advancements and innovations across applications, all aimed at making construction faster and delivering higher performance. Gypsum although a much older material than the cement & sand plaster has rarely been used extensively in construction industry. Now a days, Gypsum has proved to be a miraculous material aiding interior construction due to its properties.

Gypsum

What is Gypsum Plaster?

             When dry POP powder is mixed with water it hardens. This material which can be applied over brick, block or concrete surface to form a smooth surface is called gypsum plaster.

Earlier, a 6 mm coat of gypsum plaster (termed as POP punning) is usually applied on the top of cement plaster to give a smooth finish to it before painting. This is a two stage process and involves various elements like sand, cement and water which has to be mixed onsite. This process is slowly being replaced by a direct application of single coat of gypsum plaster. In gypsum plaster, readymade POP powder is mixed with water and applied directly on the wall.
Gypsum Plaster can be applied directly on any brick, solid or hollow blocks, AAC blocks and plaster boards. Gypsum plaster has good insulation properties, fire resistant and impact resistant. Also, gypsum saves a lot of time during construction and has superior finish. These properties have clearly drawn attention of real estate builders and contractors towards choosing gypsum plaster over traditional cement plaster.

Gypsum plastering in Walls

Technical Specification of Gypsum Plaster:

Colour of Finished surface: White
Setting Time: 25-30 Minutes
Coverage area (considering 12mm thickness): 21 sq.ft per 25 Kgs Bag
Compressive Strength: 60-70 kg/cm2
Shelf Life: 4 Months

How to Storage of Gypsum Plaster at Site?

Exposure to water or moisture reduces the setting time and strength of gypsum plaster. So gypsum has to be stored properly.

• Gypsum Plaster (POP) bags has to be stored on elevated surface (dry platform) made of bricks/timber/concrete at site.
• The minimum shelf life of Gypsum Plaster 3-4 months from the date of manufacture. But if properly stored gypsum can be use in excessive of 6 months subject to temperature and humidity.

Merits of Gypsum Plaster:

Contractors and Builders started preferring gypsum plaster instead of due to its superior finish and time saving attributes. Some of the advantages of gypsum plaster are:

1. Ease of Application (Workability): Gypsum can be directly applied over brick/block work without separate finishing. It is also very easy to apply and level gypsum plaster.
2. No Shrinkage Cracks: Gypsum reaction produces less heat as compared to cement reaction with water. So there are fewer Shrinkage cracks in gypsum plaster as compared to traditional cement plaster
3. Quick Setting Time: Gypsum sets quickly (i.e., within 25-30 mins). So painting could be started 72 hours after application of gypsum plaster. Plaster has to be dried up before painting.
4. No curing: Unlike Traditional Cement Plaster, Gypsum plaster doesn’t need any curing saving water and time during construction
5. High Productivity: Reduces time considerably when compared to conventional cement plaster
6. High Performance: Excellent high strength after drying, Durable and Light weight (Reduces dead load on structure)
7. Smooth Finish: Perfectly lined, levelled, smooth walls and perfect right angled corners
8. Reduced Supervision: Careful quality checking is required for cement plaster as cement and sand has to be properly proportioned. In contrast, gypsum plaster doesn’t require same amount of quality checks for application thus reducing supervision efforts.
9. Readily available raw materials: Gypsum is a ready available material. Natural Sand, which is a raw material used in Traditional cement plaster, is hard to obtain. It is also banned in multiple states in India
10. Fire resistant:Gypsum plaster is highly resistant to fire
11. Low thermal conductivity: Gypsum has low thermal conductivity. This saves electrical cost for heating and cooling rooms in a building.
12. Decorative application: It can be easily applied to decorative purposes also and can be mould into different shapes

Demerits of Gypsum Plaster:

• Gypsum plaster cannot be used for outside walls since they retain dampness. Also gypsum plastering cannot be done in areas which are continuously damp such as bathroom etc.,
• Gypsum plaster is costlier than traditional cement motar plaster (cement and sand) for same thickness of plaster. But in areas where natural/river sand is not available for construction, cement mortar plastering should be covered with a 6 mm gypsum layer making cement plaster more costly.

welcoming your feedback also....

Tips for Preparation of Bar Bending Schedule

What is Bar Bending Schedule?

Bar bending is a process of cutting and bending reinforcement steel into shapes as suggested by the structural engineer for various structural elements like like Slab, Beam, Coloum, Footing etc.,.
Bar Bending Schedule is a list details of bent reinforcement bars used in any given structural concrete (RCC) element. The list contains bar mark, its diameter, length, shape and weight.
Bar bending schedule is used by site engineers/home owners to estimate the quantities of various diameters of bars used in a construction, perform bar bending at site, and check the reinforcement work done by the contractor.

Advantages of bar bending schedule in construction:

The following are the advantages of bar bending schedule in construction.

1. Provides better estimation of reinforcement steel to be used in each structural concrete member. Also you can compute the overall reinforcement required for the project.
2. Helps you in procurement of materials and better stock management. Reinforcement Steel is the costliest material out of all building materials accounting for almost 25% of the project cost. It is not always advisable to purchase the whole steel lot in one go (for house construction projects of more than 2 Floors). Steel requirement has to be divided in phases based on the construction. Bar bending schedule makes it easy for ordering the exact quantities of steel required for a specific phase of construction work.
3. Bar Bending Schedule provides exact estimate of the quantities required. So wastages at site can be avoided by careful monitoring of the contractor and using bar bending schedule along with that
4. Using Bar Bending Schedule, reinforcement can be cut and bend at factory and can be directly transported to site. This will not only reduce the wastage of reinforcement steel at site but also will reduce on labor costs, thus helping in reducing the overall cost of the project
5. Bar Bending Schedule helps site engineers and quality inspectors to check and approve the reinforcement thus aiding in better quality control
6. Billing can be prepared real fast and easy using Bar bending schedule. It is also easy for clients to approve those bills for payment
7. Bar bending schedule can help the team cut and bend the bars at a separate location while other activities are performed in parallel, thus saving time during construction

Step by Step Preparation of a Bar Bending Schedule:

First, Download a Bar Bending Schedule excel sheet to start working with the steel calculations.

Now, Follow these steps for calculating the exact weight of the reinforcement required.

Step-1: Identification of Members (Steel bars)

Each type of bar is usually tagged with a number or bar mark in the drawing. Use that number or bar mark to identify the bar. If there are no such numbers, follow a bar mark system yourself. Now, List down all the shapes of bars you see in the drawing. Mark the diameter of the bar in the table.



The above example shows the bar marks like #5, #6, #9 etc., along with the diameter of the bar.

Step-2: Counting the number of each type of bar:

Count the number of bars of each of those shapes and note it in the bar bending schedule.
For small concrete shapes, you can manually count them in the drawing.
For big slab structure, to calculate number of bars is you need to check the spacing between two bars (centre to centre c/c distance) and the width of concrete in which the bars are distributed.
Number of bars = (Concrete width in the direction that these bars are distributed – 2 X cover)/ (Spacing between bars c/c distance)
For example: Suppose 12 mm bars are distributed in a slab of 10 mtrs at 300 mm c/c spacing. Considering a cover of 100 mm. Then the number of bars would be
Number of bars = (10,000 – 2 X 100mm)/(300) = 32.66 ~ 33 Bars of 12 mm dia

Step-3: Calculation of Length of Reinforcement Bars

Calculate the length of each of those bars. You can do that with the following formula.
Length of Steel Bar = Length of concrete (concrete direction in which the bar is placed) – (2 X Cover concrete) + Development length
Development length or Anchorage length is the L shaped leg provided to a bar

Step-4: Calculate unit weight of individual bar

Now fill the unit weight of each bar. Unit weight of bar = Volume of bar X density (7850 kg/m3)
Unit weight of bar with dia (D) = (D X D)/162
Aternatively, You can use the following table:

Dia of bar (mm)	Weight (Kg/mtr)
8	0.395
10	0.617
12	0.889
16	1.580
20	2.469
25	3.858
32	6.321

 Now after filling all those details, your table should look something like this.

Step-5: Calculate the total weight of Steel Required

Calculate the weight of bars of a particular shape
Weight of Steel bar (of a specific shape) = No of bars of that shape X Length of Bar X Unit weight of the bar
Adding the weight of the steel bars gives you get the total steel quantity in Kgs.
Error for rolling margin should be included in the calculations.

Note: Although this procedure looks simple, there are lots of Good Engineers who do mistakes by miscounting the number of bars, calculating the length incorrectly etc., So be careful while you calculate. Creating a complete Bar Bending schedule for steel calculation will always help you minimize mistakes. Also you can refer to the detailed calculation for any changes or mistake that you have committed.
Also, keep some thumb rules handy so as to double check your calculation. This way if there is any major deviation, you can check your bar bending schedule for mistakes.

Tips for Preparing Bar Bending Schedule:

1. Prepare bar bending schedule for different RCC elements separately (i.e., Divide the table into parts for Footing, Pedestal, Plinth Beam, Coloumn, Slab etc.,)
2. Keep the unit weights of bars handy. This will help you in designing the
3. Keep some thumb rules handy that will help you in double checking your calculation. Some of them are
   • The total quantity of steel for a low rise building structure varies from 3-4 Kgs/sqft of build-up area
   • Volume of steel reinforcement for slab is roughly 1.5-2% the volume of concrete (Volume of steel can be calculated by dividing the weight by density of steel i.e., 7850 kg/cum)
     

     RCC member	% Steel by volume of concrete
     Slab	1.5-2%
     Column	2.5-3%
     Beam	2.5-3%

     For ExampleFor 10 cum of concrete in a column considering 2.5% of steel, the quantity of steel would be10 cum concrete in a column will have 0.25 cum of steel (considering steel volume of 2.5%)
     0.25 cum will weigh , 0.25*7850 = 1962.5 kg (as density of steel 7850 kg/cum)
     For 10cum concrete the steel required is 1962.5 kg
     Per cum concrete steel requirement will be 196.25 kg.
     
     A positive rolling margin (weight higher than standard weight) shall fall short while placing at site.
     
     Keep an eye on Positive rolling %, to save on steel cost.
4. Rolling margin allowance should be given prime importance at site while preparing the BBS. As mentioned earlier BBS gives us the cutting length of a bar but we procure steel by weight (as the standard weight by calculation varies from actual weight of bar due to rolling margin).
5. Mentioning type of steel is also important sometimes. This depends on the project requirement. If more than one grade of bar is used. This helps procurement team in buying the right type of steel required for the work

Types of Foundations Used in Building Construction

            Foundation is the most essential part of the structure which transmits the loads acting on the structure and the self-weight of the structure, safely to the ground/subsoil.
The functional requirements of foundations are – strength and stability.

• Strength – The foundation should be strong enough to bear the combined dead, imposed and wind loads.
• Stability – The foundation should transmit all loads without causing any movements in the soil, which would compromise the stability of the structure. It should provide a level base for the superstructure.

Soils have bearing capacities; a more appropriate term would be ‘Safe Bearing Capacity (SBC)’. The area of the foundation should be sufficiently large enough, so that the pressure on the ground does not exceed the SBC. Swelling and shrinking in the soil causes ground movements which affect the stability of the building. So, they need to be taken into account.
If the soil directly beneath the ground surface is weak or is prone to swelling and shrinking, then foundation can be taken to a depth where the soil strata is stronger or where the soil is unaffected by changes in moisture.
In any case a soil investigation is a preferable, to determine the type of soil, moisture content and the presence of aggressive chemicals which may attack the concrete used in the foundation.
The factors which need to be considered for selecting the type of foundation and designing it are –

• Magnitude of the building loads – the depth or area of the foundation system needs to be increased proportional to the magnitude of the total load on the foundation system. Also, the pattern of the loads needs to be taken in account, to make the design more economical by providing foundation of lesser area/depth where magnitude of the loads is low.
• Subsoil and groundwater conditions – such as
• Soil type – as some soils have very low bearing capacity, in these cases, piles are more economical than conventional pad foundations, depending on the depth of hard strata from the ground surface, which can bear the load.
• Depth of ground water table – to take necessary precautions to prevent water-logging in the site during construction, plan for drainage of the water and to analyze up thrust pressure of water on building.
• Topography and the terrain of the site and its surroundings – Different terrains require different foundation systems. In hilly areas, the foundation system required I different form that used in plain areas.
• The presence of roads, and other buildings in the vicinity of the building site – the walls of the adjacent building and the roads can restrict the area that needs to be excavated, making combined foundations preferable over isolated pad foundations.
• Ease of construction and cost – Some foundation types like pile foundations require specialist construction machinery and manpower, making them costlier, and overall a cumbersome construction process.
• Requirements as prescribed by building codes (CPWD, BIS, etc.) – The foundation system needs to be designed keeping in mind the laws and standards as prescribed national and state agencies.

Various Types of foundations used in Building Construction:

Foundations can be broadly classified into two categories –

• Shallow Foundations – these are used when stable soil of adequate bearing capacity is available relatively near to the ground surface. They are built directly built below the lowest part of the building.
• Deep Foundations – these are used when the soil available relatively near to the ground surface is unstable and has inadequate bearing capacity. They transfer the loads to a soil or rock stratum which has adequate bearing capacity, present at a greater depth.

But more specifically they can be classified based on their shape and utility. are of following types

Strip foundations:

Strip foundations consist of a continuous longitudinal strip of concrete or masonry, which provide a firm and level base on which the walls can be built. They are more common in load-bearing structures.
The width (spread) of the foundation depends on the safe bearing capacity of the soil. The thickness of the foundation depends on the strength of the foundation material. A general rule is that the projection of the concrete strip each side of the wall should be no greater than the thickness of the concrete.
Strip foundations are suitable for continuous loads. The building load is evenly distributed along the length of the foundation.

Pad foundations:

The foundation to the columns of masonry, reinforced concrete and steel columns is usually in the form of a square or rectangular isolated pad of concrete to spread a concentrated load. The area of the pad foundation depends on the load on the foundation and the bearing capacity and shear strength of the soil. The depth of the foundation depends on the strength of the foundation material used (masonry/RCC).

In case the load on the foundation is high the thickness of the concrete pad is same as the projection of the concrete pad around the column. Pad foundations are used in combination with plinth beams. The plinth beams are used to support masonry walls. The plinth beams transfer this load to the pad foundation.
Pad foundations are preferred where the depth of the excavation is no more than a few meters below the ground surface.

Combined Foundations:

The foundations of adjacent columns are combined when a column is so close to the boundary of the site or the columns are closely spaced to each other. It is more economical and is easier to construct. The thickness of the concrete should be at least equal to the projection of the strip each side of the column. The combined continuous foundation is reinforced with re-bars to limit the depth of the foundation. Stirrups used are used position the reinforcement and provide resistance to shear.

The pad can be symmetrical (rectangular) or asymmetrical (trapezoidal) depending on the pattern of the loads.

Raft/Mat Foundations:

A raft foundation is a continuous slab of concrete usually covering an area equal to or greater than the base of the building which support the walls or lightly loaded columns and serve as a base for the ground floor. The word raft is used in the sense that the slab of the concrete floats on the surface as a raft/mat on the surface of water. These types of foundations are suitable when the bearing capacity of the soil is low, but the soil should be such that it can be properly & easily compacted and levelled.
Type of raft/mat foundations are-

1. Solid slab raft – these are reinforced using re-bars, in both directions
2. Beam and slab raft – they are suitable for heavier loads; the beams transfer the loads from the columns to the slab. The walls are built on the beams.
3. Cellular raft – they are very rigid, and are usually built in soils where settlement is not uniform, and is prone to shrinkage and swelling. The construction of basements also becomes easier. They are suitable for tall buildings.

On the compacted soil, a base course (minimum depth is 4” or 100mm) of gravel or crushed stones is provided to prevent the groundwater from rising to the surface. A layer of sand (minimum depth is 2” or 51mm) is also provided to absorb the excess water from the concrete slab during curing. A moisture barrier is provided between the layers of gravel and sand, usually sheets of polyethylene. Control joints and thickened edges are preferable in the raft foundation. Control joints should be provided so as to allow the concrete to crack along predetermined lines. And the thickened edges enhance rigidity and stability.

Pile Foundations:

Piles re usually columns, usually of RCC, driven or cast into the ground. The main function of the pile is to transmit loads to lower levels of the ground by a combination of friction along their sides and the end bearing at the stratum. Piles that transfer loads mainly by friction to clays and silts are termed as friction piles, and those mainly transfer loads by end bearing to compact gravel or rock are called end-bearing piles.

A pile foundation system consists of end-bearing or friction piles, pile caps, tie beams for transferring the building loads down to a suitable bearing stratum. Piles are usually driven in clusters of two or more. The pile cap joins the cluster of piles so that it can distribute the load from a column or plinth beam equally among the piles. The construction of pile foundation requires specialist machinery, which lift the piles in position and drive them into the ground.

Scoffolding - Types & its Details

Scaffolding work is explained simply in below images

How many bricks required for 1 cu.m wall ???

1.     How many bricks required for 1 cu.m wall of 1^1/2 brick thickness ???

ANSWER:

 

                                       For a 1⅟₂   brick thickiness wall (i.e. 30cm nominal thickness). Therefore nominal volume of wall of 20m length , 1m  height & 0.3m thickness = 20*5*.3 = 30 cu.m.

                          Normally mortar joint will be less than 1cm, but assuming mortar joint as   1cm thickness, then actual thickness of wall will be = 29cm. therefore , actual thickness = 20*5*.29 = 29 cu.m.

      Nominal Size of standard brick =  20cm*10cm*10cm.

 

       Number of std bricks = 29/(.20*.10*.10) =  14500

       Then,

           Number of bricks per cu.m(nominal) = 14500/30 =483.3333

                                                                                =483 bricks.

      Considering 5% breakages, wasteges,etc (assume)

                    So,Number of bricks required for 1 cu.m = 500 bricks

        RESULT:

        Therefore, number of bricks required for 1 cu.m = 500 bricks .

DESIGN OF A CONCRETE APPROACH ROAD AT EYYAKUNAM VILLAGE --- PROJECT REPORT

   Project available @ below link.....

                              This design project titled “Design of cement concrete approach road at Eyyakunam village’’ deals with the pavements in general, its types and design factors taking a case study of a road leading to a small village between Gingee and Tiruvannamalai. The concentration of the project report is keened on the area of design of rigid pavement which is the best suited pavement design for the road in consideration. The design of rigid pavement for the proposed road is illustrated and designed for the soil condition and traffic condition of the approach road to Eyyakunam village. Codal recommendations provided by the Indian Road Congress, “Guidelines for the design of flexible pavements for highways” (IRC 37-2001) are strictly followed in preparing this design report. Codal recommendations provided by the Indian Road Congress, “Guidelines for the design of plain jointed rigid pavements for highways” (IRC 58-2002) are strictly followed in preparing this design report.

The next part of the report, sandwiches the above manual pavement design with the help of a flowchart methodology. Computerisation of this process will reduce the consumption of design time, complexity in manual designing and increase the reliability.

For more details..............

Click here to download.

CHALLENGING TEST FOR CIVIL ENGINEERS & CHANCE TO WIN FREE RECHARGES

HI CIVILERS......

    NOW WE WILL CHECK UR KNOWLEDGE IN CIVIL SYLLABUS.... IF U ANSWERED 2 QUEZTIONS CORRECTLY WITH EXPLANATI0N, WE WILL RECHARGE UR MOBILE NO OF RS.20 FOR 10 LUCKY WINNERS..

DO
   U
     READY??

 QUEZ:1

 QUEZ:2

RESULTS WILL BE ANNOUNCED ON APRIL 2ND 2017.....


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Important Unit conversion for Civil Engineers

Unit conversion for Civil Engineers:

         Mostly useful for site Engineers...

kkkikk

HOW TO CALCULATE QUANTITIES OF CEMENT, SAND AND AGGREGATE FOR NOMINAL CONCRETE MIX (1:2:4)?

       Mix design is a process of determining the right quality materials and their relative proportions to prepare concrete of desired properties like workability, strength, setting time and durability.

While following a mix design is advised to optimise the material consumption, it is not possible at site to always come up with Mix design. Nominal mix concrete is prepared by approximate proportioning of cement, sand and aggregate to obtain target compressive strength.

Mix ratio of concrete defines the ratio of cement sand and aggregate by volume in that order. So a mix ratio of 1:2:4 represents Cement: Sand: Aggregate – 1:2:4 (by Volume)

Material requirement for producing 1 Cum of Nominal Concrete Mix

The following are the materials required to produce 1 Cum of Concrete of a given Nominal Mix Proportion.

Method-1: 

 
DLBD method to determine material requirement for Nominal Concrete Mix (M20 – 1:2:4)

The DLBD (Dry Loose Bulk Densities) method is an accurate method to calculate cement, sand and aggregate for a given nominal mix concrete. This gives accurate results as it takes into account the Dry Loose Bulk Densities of materials like Sand and Aggregate which varies based on the local source of the material

For calculation, We consider a nominal concrete mix proportion of 1:2:4 (~M20).

Step-1:
Calculated Volume of materials required.:-

01 bag (50 kg) of cement = 35 litres or 0.035 cubic meter (cum)

Since we know the ratio of cement to sand (1:2) and cement to aggregate (1:4)

Volume of Sand required would be = 0.035*2 = 0.07 cubic meter (cum)

Volume of Aggregate required would be = 0.035*4 = 0.14 cubic meter (cum)

Step-2:
Convert Volume requirement to weights:-

To convert Sand volume into weight we assume,  we need the dry loose bulk density (DLBD). This density for practical purposes has to be determined at site for arriving at the exact quantities. We can also assume the following dry loose bulk densities for calculation.

DLBD of Sand = 1600 kgs/cum

DLBD of Aggregate = 1450 Kgs/Cum

So, Sand required = 0.07*1600 = 112 kgs

and Aggregate required = 0.14*1450 =203 kgs

Considering water/cement (W/C) ratio of 0.55

We can also arrive at the Water required = 50*0.55 = 27.5 kg

So, One bag of cement (50 Kgs) has to be mixed with 112 kgs of Sand, 203 Kgs of aggregate and 27.5 kgs of water to produce M20 grade concrete.

Step-3:
Calculate Material requirement for producing 1 cum Concrete

From the above calculation, we have already got the weights of individual ingredients in concrete.

So, the weight of concrete produced with 1 Bag of cement (50 Kgs) =50 kg + 140 kg + 203 kg + 27.5 kg = 420.5 kg say 420 kgs

Considering concrete density = 2400 kg/cum,

One bag of cement and other ingredients can produce = 420/2400 = 0.175 Cum of concrete (1:2:4)

01 bag cement yield = 0.175 cum concrete with a proportion of 1:2:4

01 cum of concrete will require

Cement required = 1/0.175 = 5.72 Bags

Sand required = 140/0.175 = 800 Kgs

Aggregate required = 203/0.175 = 1160 kgs

Method-2:

 
Empirical method to determine material requirement for Nominal Concrete Mix:-

Although empirical method is easy to use in determining the materials requirement for Nominal Concrete mix, it sometimes doesnt give accurate results as it doesn’t take into factor the local variations in the materials.

Let’s design M20 grade concrete. Ratio for M20 concrete is 1 : 2 : 4

Step-1:
Calculate the Volumes of material required in 1 Cum concrete

The dry volume of concrete mixture is always greater than the wet volume. The ratio of dry volume to the wet volume of concrete is 1.54.

So 1.54 Cum of dry materials (cement, sand and aggregate) is required to produce 1 Cum of concrete

Volume of Cement required = 1/(1+2+4) X 1.54 = 1/7 X 1.54 = 0.22 Cum

Volume of Sand required = 2/7 X 1.54 = 0.44 Cum or 15.53 cft

Volume of Aggregate required = 4/7 X 1.54 = 0.88 Cum or 31.05 cft

Note: 1 cubic meter = 35.29 cubic feet.

Step-2:
Calculate the weights of materials required in 1 Cum concrete:-

Density of Cement (loose) = 1440 kgs/cum

So weight of cement required = 1440 X 0.22 = 316.2 Kgs or 6.32 bags

Density of Sand = 1600 Kgs/cum

Weight of Sand required  = 1600 X 0.44 = 704 kgs

Density of Aggregate = 1450 kgs/cum

Weight of aggregate required = 1450 X 0.88 = 1,276 Kgs

Types of beam based on Reinforcements

SINGLY REINFORCED SECTION :

When the beam us loaded it can suffer through two different type of beam(mode of failure):

1.Ductile Failure &

2.Brittle Failure

Now, in case of a brittle failure of beam when a beam gets over-stressed, then till the point of over-stressing any major cracks aren’t developed in beam, but as soon the point is passed, suddenly the beam fails.Well thus doesn’t seem good h..!! Hence we always design the beam to fail in the ductile fashion, consequently giving us more time to notice that the beam is over-stressed and it’s the time that we should do some repairs or evacuate the building.

Now, how do we make sure that the beam will fail in ductile manner. This can only possible when the strength of the steel will be less than compression strength of the concrete. This means that when the ultimate loads are reached, as the total strength of the steel will be  pretty less than the strength of concrete, steel will try to strain more than what the concrete will experience strain in compression. Now we all know that steel is ductile and able to handle higher strains without any kind of failure. So this is how we can get a ductile response of beam. In this case, Concrete is all in under control while steel is straining itself.

But suppose if  there are certain restrictions like we cannot use a section greater than say 25″ x 25″ & we have to deal with very high moments. Now this high moment will tend to increase the tension steel demand & consequently a bigger concrete block will be required in compression resulting in pushing the neutral axis further down. Now this is an important part. One might think that so what if neutral axis go further down, I have the whole concrete beam which can take the compression. But wait..!!

As a neutral axis starts shifting down, the strains in the extreme fiber starts to increase. And this will be our concern in case of the compression fiber. Now concrete is brittle, so at start/first it will not show that much impact. But this high value of the strain will cause concrete to crush itself which will lead to the ultimate failure and this will be sudden.  That’s why we can’t put more steel than a certain amount.

So, what decides this limit of steel ? Well, it’s all experimental based approach & to an extent we can prove it mathematically too. Balancing tension & compressive forces, take the depth of a compression block, now draw the strains and look if the strain in the concrete is more than the allowable strain. But in case of designing by us, then some codes specify the maximum r/f ratio in beam whereas some codes specify the maximum allowable neutral axis depth that concrete beam can achieve. And if the desing is under this, thenductile failure in a beam will occur and we are safe.

DOUBLY REINFORCED SECTION :

Now as we mentioned before that we are restricted to a 24″ x 24″ section & we have pretty heavy moment & we are exceeding the limits mentioned in the code. So now the only option is to provide thr compression reinforcement & add some tension steel. Now the extent to which we are adding the compression reinforcement will tell us whether the beam is brittle or ductile.

Suppose the limit for tension reinforcement for singly reinforced beam is X, & to resist this moment we have to add extra 0.5X of tension reinforcements. We also decide to include 0.8X of compression reinforcement. Hence,  now we have 1.5X(1X+.5X) of tension reinforcement & 0.8X of compression reinforcement provided along with compression stress block. So now what happens is that this 0.8X of compression r/f will tend to balance the effect of 0.5X of tension r/f. As we have added 0.8X of compression r/f, the beam doesn’t experience a very high compression strain compared to tension strain. Thus the compression bars will be under lower stress than its yield point. Thus to balance a fully stressed 0.5X of tension r/f we will need a higher amount of compression r/f . Now, the concrete block is responsible to resist the X amount of the tension r/f which is a limit for ductile behavior. Hence in this case the doubly reinforced beam will act as a ductile beam and so we can say that it is an under-reinforced beam/section. All good..!!

But suppose we decide to add only 0.4X of extra compression r/f. Well This will counter max of 0.4X of tension r/f in worst case scenario(it may be lower). So now, the plain concrete stress block will be responsible to resist 1.1X [1X + (.5X-.4X) ]of tension r/f and this is above the limit resulting in brittle failure. So in this case, even though the beam is doubly reinforced, it will experience a brittle failure i.e. it’s an over-reinforced section.