The post Bar bending schedule of 2 way slab appeared first on Surveying & Architects.
]]>In this Article we will calculate the cutting length and prepare the B.B.S of the 2 way slab as shown below.
Let us redraw the above drawing with the dimensions for the calculation purpose as below.
Dimension of slab
Lx = five m.= 5000mm.
Ly = three m. = 3000mm.
Thickness = 125mm.
Main barOne = 10mm@ 150 c/c
Main barTwo or say distribution bar) = 10mm @ 150 c/c
Top extra barOne = 8mm @ 150 c/c
Top extra bar Two 8mm @ 150 c/c
Clear cover of the slab = 25mm.
Development length Ld = 45d.
1 The no. of the bars
a – Main bars.
The no. of main bars – One
= [Lx ÷ bar spacing ] + One
= [ 5000 ÷ 150] + One
= 33.33 + One
= 43.33 nos.
By rounding off = 44 nos.
b – Main bars – 2.
The no. of the main bars – Two ( or say distribution bar)
= [Ly ÷ bar spacing ] + One
= [ 3000 ÷ 150] + One
= 20 + One
= 21 nos.
c – Top extra bars One.
The no. of the top extra barsOne on one side
= [{(Ly ÷ Fore)÷ bar spacing} + One]
= [{ (3000 ÷ Fore) ÷ 150} +One]
= [{ 750 ÷ 150} +One]
= [{ 5} +One]
= Six nos.
The total no. of the top extra bars on both sides.
= [2 sides × Six nos.]
= 12 nos.
Note: The top extra bars are provided at the Ly/fore span of the main bar on the both sides.
c – Top extra bars Two .
The no. of the top extra bars Two on one side
= [{(Lx ÷ 4)÷ bar spacing} +One]
= [{ (5000 ÷ 4) ÷ 150} +One]
= [{ 1250 ÷ 150} +One]
= [{ 8.33} +One]
= 9.33 nos.
By rounding off = Ten nos.
The total no. of the top extra bars on both sides
= [Two sides × 10nos.]
= 20 nos.
Note: The top extra bars are provided at the Lx/fore span of the main bar on the both sides.
2 – Cutting length.
a – Main bar One.
Cutting length of the main bar One
= [clear span +(Two nos. × development length) + (extra crank length) – (Two nos. × 45° bend)]
= [ 3000mm. + ( Two nos. × Ld) + ( 0.42D ) – ( Two nos. × One d)]
Here, D = [slab thickness – ( top and bottom clear cover) + bar diameter ]
= [125mm – (Two nos × 25 mm.) +Ten mm ]
= [125mm – 60 mm]
D = 65 mm.
Cutting length of the main bar One.
= [ 3000mm. + ( Two nos. × 45 × Ten mm. ) + ( 0.42× 65mm. ) – ( Two nos. × One × 10mm)]
= [ 3000mm. + 900mm. + 27.3mm. – 20mm. ]
= 3907.30 mm. = 3.91m.
b – Main bar Two.
Cutting length of the main bar – Two
= [clear span +(Two nos. × development length) + (extra crank length) – (Two nos. × 45° bend)]
= [ 5000mm. + ( Two nos. × Ld) + ( 0.42D ) – ( Two nos. × One d)]
= [ 5000mm. + ( Two nos. × 45 × 10mm. ) + ( 0.42× 65mm. ) – ( Two nos. × One × 10mm)]
= [ 5000mm. + 900mm. + 27.3mm. – 20mm. ]
= 5907.30 mm. = 5.91m.
c – Top extra bar – One.
Cutting length of the top extra bar One
= [clear span Lx + (Two nos. × development length)]
= [5000mm. + ( Two nos. × 45d )]
= [ 5000mm. + ( Two nos. × 45 × 8mm. )]
= [ 5000mm. + 720mm.]
= 5720 mm. = 5.72m.
d – Top extra bar Two.
Cutting length of the top extra bar Two
= [clear span Ly + (Two nos. × development length)]
= [3000mm. + ( Two nos. × 45d )]
= [ 3000mm. + ( Two nos. × 45 × 8mm. )]
= [ 3000mm. + 720mm.]
= 3720 mm. = 3.72m.
Let us prepare the B.B.S table to calculate the weight of rebars.
Sl. No. 
Type of Bar 
Dia.in mm. 
Nos. 
Lengthin m. 
Total lengthin m. 
WeightKg/m. 
Totalbar wt.in kg. 
1. 
Main bar One 
10 
44 
3.91 
172.04 
0.617 
106.15 
2. 
Main bar Two 
10 
21 
5.91 
124.11 
0.617 
76.58 
3. 
Top extra bar One 
8 
12 
5.72 
68.64 
0.395 
27.11 
4. 
Top extra barTwo 
8 
20 
3.72 
74.40 
0.395 
29.39 
4. 
Total weight of the bars = 
239.23 

5. 
Add three % wastage = 
7.18 

6. 
The grand total wt. of the rebar’s = 
246.41 
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]]>The post Bar Bending Schedule B.B.S appeared first on Surveying & Architects.
]]>B.B.S The word B.B.S Plays a significant role in any Constructions of the High rise buildings. B.B.S refers to B.B.S. Well, What is the use of B.B.S ? Why we use B.B.S ? What is B.B.S ?
In this article we are majorly focused on how & where to start & what are the basics.
It’s a lengthy article, so brace yourself for at least 10 minutes. Let’s Get started.
First of all, Bar is any type of rebar which is used as a reinforcement in R.C.C. The bar may be a Mild Steel bar or H.Y.S.D bar or T.M.T Bar.
B.B.S is termed as ( Calculations of the total Steel required for the construction of a building ) We use steel to make concrete to be reinforced & for tension requirements. But how much steel required for constructing 15 floors building ? How much Steel I have to order ? All these questions are answered in B.B.S
In B.B.S, the bars are organized for each structural units ( Beams or columns or slabs or footings etc.) & detailed list is prepared which specifies the Bar location ( Bar in footings, slabs, beams or columns ), Bar Marking ( to identify the bar in accordance with the drawing ), Bar Size ( length of the bar used ), Quantity ( No. of Bars used ), Cutting length, Type of Bend & Shape of the bar in reinforcement drawings.
From 1950 to 2022 lots of modifications and enhancements happening in our world. In 1950, threestoriey buildings are high rise buildings now we are constructing a building with 200 floors+. There is a massive growth in the construction industry. Due to the vast increase in world population demands increased facilities, more need for space & more constructions.
Father of Estimation B.N. Data has given certain recommendations for the usage of steel in different components of the buildings. But, he didn’t mention any values if we use more bars in a single structural member.
Bar Bending Member 
Percentage 

Slab 
1% of total volume of concrete 
Beam 
2% of of total volume of concrete 
Column 
2.5% of total volume of concrete 
Footings 
0.8% of total volume of concrete 
At that time we’ve used only four bars in columns; now we are using 12+ bars in the columns based on load analysis. So, the percentage of the steel is increased in a column which reveals that the abovecited values are outdated. ( They are outdated “not wrong”) He wrote that book in 1950. Now we are in 2022. He gave recommendations according to the potentiality of the construction at that time. Now we are constructing 200 + floors in the small area.
Before dealing with the B.B.S, it’s very important to learn the basics of B.B.S. The belowmentioned table is a kickstart guide for learning B.B.S from scratch.
(If you are viewing the below table through mobile, scroll horizontally for a clear view)
S.No. 
Particulars 
Result 

1. 
Standard Length of the Steel Bar

12m or 40′ 
2. 
Weight of Barfor length = 1m 
D^{2}/162(were D = Dia of Bar) 
Ex: 
If length of bar is 12m with 10mm Dia then ,Weight of bar = D^{2}/162Therefore for length 1m = 1m x D^{2}/162= 1 x 10^{2}/162 = 0.61 KgsFor length 12m = 12 x 10^{2}/162= 7.40Kgs 
7.40Kgs 
3. 
Density of Steel 
7850Kg/m^{3} 
Below I am discussing the different concepts of B.B.S which are very useful while working with B.B.S. All these concepts are used in B.B.S design calculation of any structural member. So be familiar with the below concepts. To keep it clear, firstly the concepts are discussed and in the end, this post is closed with an example of B.B.S calculation of a member.
The hook length is commonly provided for stirrups in beams and ties in columns. In general, Hooks are added at the two ends of the rebar in stirrups or ties.
Hook Length = 9d (d is dia of the bar)
Below image makes you clear why the Hook length = 9d
From above fig, length of hook = [(Curved Portion) + 4d] = [(4d+d)+4d] = 9d
For clear understanding, look at the below image for calculation of the total length of stirrup the with two hooks at ends.
Total Cutting Length of stirrup or tie = Total length of Bar + 2 x Hook Length (Two hooks)
Total Cutting Length = L+2(9d)
Therefore Total Cutting length = L+18d (d is the Diameter of a bar)
Hope, now you are clear with the Hook length calculation.
The Bend length calculation is different for Cranked bars (bent up bars) and bends at corners.
The bars are usually cranked in Slabs and bars are bent at corners in Stirrups or ties.
As Shear stress is maximum at supports in Slab. To resist these stresses we usually crank the bars at the ends of supports in the slab. The below figure depicts the bent up bar in Slab. To calculate the bend length the below procedure is followed.
From the above figure as the bar is bent at an angle θ^{0} the additional length (l_{a}) is introduced.
Where, l_{a} = l_{1} – l_{2}–(i)
Tanθ = D/l_{2} ; Sinθ = D/l_{1}
Hence l_{1} = ^{D}/_{Sinθ} and l_{2} = ^{D}/_{tanθ}
Therefore from (i) : l_{a} = ^{D}/_{Sinθ} – ^{D}/_{tanθ}
Giving different θ values as 30^{0} , 45^{0}, 60^{0 }results different additional length l_{a} values as below.
θ^{0} 
^{D}/_{Sinθ} 
^{D}/_{tanθ} 
l_{a} =^{D}/_{Sinθ} – ^{D}/_{tanθ} 
30^{0} 
D/0.500 
D/0.573 
0.27D 
45^{0} 
D/0.707 
D/1.000 
0.42D 
60^{0} 
D/0.866 
D/1.732 
0.58D 
90^{0} 
D/1 
0 
1D 
135^{0} 
D/0.707 
D/1 
2.42D 
The additional length is added to the total length of the bar if the bars are cranked at a certain angle.
To keep the crank bar in position, an extra bar of length (L/4) is provided below the crank bar as shown in the below figure.
Therefore, the total length of bar = L+0.42D+0.42D+(L/4)+(L/4) = 1.5L+0.84D
The important standards used while calculating the bend length at corners
Here, ‘d’ = Diameter of bar
From above fig, There are 3 bends which are bent at an angle of 90^{0} and two bends are bent at an angle of 135^{0}
Total bend length = 3 x 90^{0} Bend length + 2 x 135^{0} Bend length = 3 x 2d + 2 x 3d = 12d = 12 x 8 = 96mm
Below table represents the total length of bar calculation for different types of bar shapes.
(If you are viewing the below table through mobile, scroll horizontally for a clear view)
Bar Shapes 
Total Length of Hooks 
Total Bend Length 
Total Length of Bar 

Straight Bar 
Two Hooks= 9d + 9d= 18d 
No bend 
l + 18D 
Bent Up at one End only 
Two Hooks= 9d + 9d= 18d 
One bendbent at anangle 45= 0.42D 
l + 18D + 0.42D 
Double Bent up Bar 
Two Hooks= 9d + 9d= 18d 
Two bendsbent at anangle 45^{0} 
l + 18D + 0.42D + 0.42D=l+18D+0.84D 
Overlap of bars 
Two Hooks= 9d + 9d= 18d 
No bends 
Overlap Length=(40d to 45d)+18d 
The standard length of the Rebar is 12m. Suppose the height of the column is 20 m. To purvey this requirement, two bars of length 12m and 8m are overlapped (joined) with overlap length.
Overlap Length for compression members (columns) = 50d
The Overlap Length for tension members (beams) = 40d
( d is the Diameter of the bar )
Have You seen the below picture on your top floor of the buildings ? We generally project some length of Bar on the last floor i.e., 50D. It is the used for further constructions purpose. (Constructing a new floor)
To understand clear, Here I am considering the below structural member of R.C.C Column and preparing a B.B.S for it.
BBS of Column 


Structural Member 
Column(3mx0.3mx0.3m) 
Bar Marking 
1. Main Bars2. Stirrups (Longitudinal bars) 
Dia of Bar 
1. Main Bars = 16mm ;2. Stirrups(Longitudinal bars) = 8mm 
No. of Bars used 
1. Main bars = 42. Stirrups = 30 
Cutting length 
1. Main bars = 3.16m2. Stirrups = 2.64m 
Total Length of bar 
1. Main bars = 18.4m2. Stirrups = 43.2m 
Weight of Steel bar 
1. Main bars = 29Kgs2. Stirrups =17Kgs 
Main bars = 4
To calculate the No. of longitudinal bars adopt spacing between bars is 0.1m
No. of Longitudinal bars = Length of column / Spacing = 3 / 0.1 = 30bars
Longitudinal bars = 30
= l +hook length + bend length = 0.3+0.3+0.3+0.3 +0.144+0.096=1.44m
Hence for Main Bars = 4.6m ; Longitudinal bars = 1.44m
Total weight of steel bar required to do BBS of above column = 46Kgs
1 – When B.B.S is available, cutting and bending of reinforcement can be done at the factory & transported to the site. This increases faster executions at site & reduces construction time and cost due to less requirement of workers for bar bending. Bar bending also avoids the wastage of steel reinforcement (5 to 10%) & thus saves project costs.
2 – Using a B.B.S when used for Fe500, saves 10% more steels reinforcement compared to fe415.
3 – It improves the quality control at site as reinforcement is provided as per B.B.S which is prepared using the provisions of respective detailing standard codes.
4 – It provides a better estimations of reinforcement steel requirement for each & every structural member which can be used to compute overall reinforcement requirement for entire project.
5 – It provides better stock management for reinforcement. The Steel requirement for the next phase of construction can be estimated with accuracy & procurement can be done. This prevents stocking of extra steel reinforcement at the site for a longer time, preventing corrosion of reinforcement in the case of coastal areas. It also the prevents a shortage of reinforcement for ongoing work by accurate estimation & thus concrete construction works can proceed smoothly.
6 – B.B.S is very much useful during auditing of reinforcement and provides checks on theft and pilferage.
7 – B.B.S can be used for reinforcement cutting, bending, and making a skeleton of a structural member before it can be placed at required position. Other activities such as excavations , P.C.C, etc.. can proceed parallel with this activities. So, overall project activities management becomes easy and reduces the time of construction. It becomes helpful in the preventing any damages due to construction time overrun.
8 – It provides benchmarks for the quantity & quality requirements for reinforcement & concrete works.
9 – B.B.S provides the steel quantity requirement much accurately and thus provides an option to optimize the design in case of cost overrun.
10 – It becomes easy for site engineers to verify and approve the bar bending and cutting length during inspection before placement of concrete with the use of B.B.S and helps in better quality control.
11 – It enables easy and fast preparation of bills of the construction works for clients and contractors.
12 – The quantity of reinforcement to be used is calculated using engineering formulas and standard codes, so there is no option for approximate estimation of steel reinforcement.
13 – With the use of B.B.S, mechanization of cutting and bending of reinforcement can be done, again reducing the cost & time of project and dependency on the skilled labor requirement. It also improves the reliability of the accuracy of bar cutting and bending.
14 – When mechanized bar cutting & bending is used, the cost of reinforced concrete work per unit reduces and helps in the cost optimization of construction projects.
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]]>The post Thumb Rules For Civil Engineers In Building Construction appeared first on Surveying & Architects.
]]>Thumb Rules For Civil Engineering is essential for any Civil Engineer, Site Engineers, or Civil Supervisor. They play a crucial role while taking quick decisions on site. There is some Civil Engineering Basic Knowledge that every Civil Engineers must know about this Riles.
The Thumb Rule of Civil Engineering or the thumb rule for construction helps you in finding out the solution using a simple mathematical formula & make smart decisions whenever needed.
But while using these thumb rules, u must remember that the thumb rule never gives the exact or accurate results, u just have used them for approximate results
There is a number of Thumb Rule For Civil Engineers that we used in construction work. So the following are some most frequently used Thumb rules on the Construction site.
The Thumb rule method is an approximate & comparing method. In thumb rules & the units are not the same when we compare to get the results. So ignore units while performing the thumb rule.
Following are thumb rules in Civil Engineering
The volume of concrete required = 0.038 m3/square feet area
Example:If Plan Area = 40 x 20 = 800 Sq. m.
So, for the plan area of 800 Sq. m. the area the total volume of concrete required
= 800 x 0.038m3 = 30.4m3
Following are some Important thumb rules for Steel Calculation for Slab, Beam, Column, and Footings.
Steel required in residential buildings = 4.5 Kgs – 4.75 Kgs / sq. Ft.
Steel required For Commercial buildings = 5.0 Kgs5.50 Kgs/Sq. Ft.
You can also use BN Data Recommendations for the more accurate result,
The following recommendations Thumb Rules For Civil Engineering are given in B N Data for the Steel quantity used in different members of the building.
Following are thumb rules for reinforcement in concrete members,
1) Slab – 1% of the total volume of concrete (Slab steel calculation thumb rule)
2) Beam – 2% of the total volume of concrete
3) Column – 2.5% of total volume of concrete
4) Footings – 0.8% of the total volume of concrete
How to calculate the steel quantity of slab having the Length, width, and depth of the slab is 5m x 4m x 0.15m
The Total Volume of Concrete for given Slab = 5 x 4 x 0.15
= 3m^{3}
As per the guidelines are given in the BN Dutta reference book the steel quantity of slab is 1% of the total volume of concrete utilized.
Thumb rule to estimate Steel quantity of above slab = Volume of Concrete x Density of Steel x % of Steel of Member
Steel weight required for above slab = 3 x 7850 x 0.01 = 235Kgs
For accurate estimation, you can refer to Bar Bending Schedule
Shuttering costs are taken as 1518% of the total construction of the building. Shuttering work is done to bring the concrete in Shape. The Thumb rule to estimate the shuttering required is 6 times the quantity of concrete or 2.4 times of the Plinth area.
For example, the concrete quantity is 0.5m^{3}, then
Area of Shuttering is 0.5 x 6 = 3m^{2}
Components of Shuttering
Shuttering Ply Quantity estimation
The Shuttering plate Ply, Battens, Nails are components of Shuttering.
Suppose, The Shuttering Ply has a length, width & depth of 2.44 x 1.22 x 0.012
The No. of Shuttering Ply Sheets = 0.22 times of Shuttering
Suppose, the Shuttering Area = 3m
Then Ply required for shuttering = 0.22 x 3 = 0.66m^{2}
Shuttering Batten usually has a length & width of 75mm x 40mm.
Batten Quantity = 19.82 x No. of Ply Sheets
If work requires 25 Ply sheets, the total quantity of Battens are 19.82 x 25 = 495 Battens
Nails & Binding Wire Quantity in Shuttering:
Approximately, 75 grams of Nails were used in the shuttering of the 1m^{2} area.
75gms of Binding wire is used for every 1m^{2} of Shuttering.
Thumb rule for Shuttering oil estimation :
Shuttering oil is applied on the shuttering plate surface used to deframe or deassemble from the concrete easily.
Total required Shuttering oil = 0.065 x Total Area of Shuttering
For every 15m^{2} of shuttering 1 liter of shuttering oil is consumed.
If, total area of shuttering is 15 m2, then Shuttering oil Consumption = 0.065 x 15 = 0.975.
Note: 1 bag of cement = 50Kgs
Thumb rule for Cement required in Brickwork, Cement Masonry & Plastering work in construction.
following are Thumb Rules for civil engineering for brickwork and cement quantity calculations.
Brickwork for 1m^{3}  Cement Qty in m^{3}  Cement Qty in Bags 
230 mm Brickwork  0.876m^{3}  25.4 Bags 
115 mm Brickwork  0.218m^{3}  6.32 Bags 
Cement Masonry Type & Mix  Cement Qty in Bags  Cement Qty in Kgs 
200mm in Cement Masonry work of Ratio 1:6 
0.124Bags/m^{2}  6.2Kgs/m^{2} 
150mm in Cement Masonry work of Ratio 1:6 
0.093Bags/m^{2}  4.65Kgs/m^{2} 
200mm in Cement Masonry work of Ratio 1:4 
0.206Bags/m^{2}  10.3Kgs/m^{2} 
150mm in Cement Masonry work of Ratio 1:4 
0.144Bags/m^{2}  7.2Kgs/m^{2} 
100mm in Cement Masonry work of Ratio 1:4 
0.103Bags/m^{2}  5.15Kgs/m^{2} 
Type of Plastering  Cement Qty in Bags  Cement Qty in Kgs 
Rough Plastering  0.09 Bags/m^{2}  4. 5Kgs/m^{2} 
Internal Wall Plastering  0.09 Bags/m^{2}  4.5 Kgs/m^{2} 
Duct Plastering  0.09 Bags/m^{2}  4.5 Kgs/m^{2} 
External Wall plastering  0.175 Bags/m^{2}  8.75 Kgs/m^{2} 
Stucco Plastering  0.175 Bags/m^{2}  8.75 Kgs/m^{2} 
Lathen Plastering  0.55 Bags/m^{2}  27.5 Kgs/m^{2} 
Following are Thumb Rule for reinforcement steel in concrete members,
1) Slab – 1% of the total volume of concrete (Slab steel calculation thumb rule)
2) Beam – 2% of the total volume of concrete
3) Column – 2.5% of the total volume of concrete
4) Footings – 0.8% of the total volume of concrete
Thumb Rule can be called a guideline, idea, or principle that helps you make quick decisions. “Arrive early” is an efficient good Thumb Rule for most appointments. It referred to builders who frequently use their thumb to estimate measurements. It is a helpful rule even being inexact.
The method for determining the widthdepth ratio of reinforced concrete beams is not specifically given in codes. However, a thumb rule can be used i.e. taking depth that is two and a half to three times the beam’s width. For longspan beams, it is generally economical to use deep and narrow sections.
Steel =3 to 5 kg / sqft.
Cement =0.5bags/ sqft.
RMC =0.05 m3/sqft.
Block =12.5 nos /sqm.
Electrical cast = Rs 133/sqft.
Plumbing cost = Rs 126/sqft.
Fire fighting cost = Rs 40/sqft.
External development = Rs 94.5/sqft.
To get more information, Visit Our Official website
Land Surveying & Architects
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]]>The post What is the Difference between main bars and distribution bars in slab appeared first on Surveying & Architects.
]]>Let us go through some of the FAQs to understand the main bars & distribution bars in a slab.
The reinforcement bars that are placed in the tension zone of the slab to resist the bending moment & to transfer the superimposed loads to the supports that are provided for the slab are called main bars.
Main bars are always placed at the bottom part of the slab reinforcement and distribution bars rest over the main bar as shown in the above drawing.
The diameter of the main bar should not be less than 8mm for the HYSD bar and 10mm for plain bars.
The reinforcement bars that are placed above the main bar in a oneway slab, to distribute the superimposed load to the main bar are called distribution bars.
As you can observe in the above drawing, distribution bars are provided in the longer span of the oneway slab & main bars are placed along the shorter span of the slab.
1. To transfer the superimposed loads to the supports.
2. To resist the bending moments.
3. To withstand shear stress in the tension zone.
4. To encounter shrinkage cracks due to temperature variation.
1. To distribute the superimposed load to the main bar.
2. To firmly hold the main bars in their given position.
3. To form a mesh with the main bar in concrete for better bonding strength.
4. To resist shrinkage stress due to climatic variation.
5. To resist the development of cracks due to external factors.
No. 
Main bar 
Distribution bar 
1. 
Used in shorter direction.

Used in longer direction of oneway slab. 
2. 
Placed in bottom part of slab reinforcement. 
Placed over the top of the main bar. 
3. 
Higher diameter bars are used. 
Lower diameter bars are used. 
4. 
Designed to resist bending moment. 
Not designed to resist bending moment. 
5. 
Transfers super imposed load to the supports 
Distributes super imposed load to the main bar. 
6. 
Resists shear stress at the tension zone. 
Resists shrinkage stress due to climatic variation. 
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]]>The post What Is The Standard Dimensional Requirements of building Components appeared first on Surveying & Architects.
]]>The standard dimension of all components of a building is given by the country’s concerned authorities, which defines the minimum size and other requirements to adhere to while designing a building.
These standards and specifications grant minimum safeguards to the workers during construction, the users’ health and comfort, and provide enough safety to the public.
The standard dimensional requirements for various components of buildings such as plinth, habitable room, kitchen, bathroom and water closet, ledge and loft, mezzanine floor, storeroom, garage, and staircase are given below.
The plinth or any component of a structure or outhouse must be positioned in relation to the surrounding ground level in order to provide proper drainage of the site. The plinth’s height from the surrounding ground level must be at least 450 mm.
Every internal courtyard must be elevated at least 150 mm above the determining ground level and adequately drained.
The height of rooms intended for human habitation shall not be less than 2.75 m measured from the floor’s surface to the lowest point of the ceiling (bottom of slab).
In the case of a pitched roof, the average height of rooms shall not be less than 2.75 m. The minimum clear headroom under beam folded plates or eaves shall be 2.4 m.
The abovementioned height requirements apply to residential, business, and mercantile buildings. In addition, for educational and industrial buildings, the following minimum requirements apply:
In the case where there is only one room with a minimum width of 2.4 m, the habitable room area shall not be less than 9.5 m2. In the case of two rooms, one of the rooms shall not be less than 9.5 m^{2} and the other room not less than 7.5 m^{2}, with a minimum width of 2.1 m.
A kitchen’s height from the floor’s surface to the lowest point in the ceiling (bottom slab) must not be less than 2.75 m, except for the portion to accommodate the floor trap of the upper floor.
The kitchen space when a separate eating room is offered must be at least 5.0 m^{2} in size and at least 1.8 m in width. When there is a separate shop, the kitchen size might be decreased to 4.5 m^{2}. A kitchen that is also intended to be used as a dining room must have a floor space of at least 7.5 m^{2} and a minimum width of 2.1 m.
A kitchen shall have:
The height of a bathroom or water closet should not be less than 2.1 m measured from the floor’s surface to the lowest point in the ceiling (bottom of slab).
A bathroom must have a minimum area of 1.8 m^{2} and a minimum width of 1.2 m. The water closet’s floor size must be 1.1 m^{2} with a minimum width of 0.9 m. If a bath and a water closet are integrated, the floor space must be at least 2.8 m^{2} with a minimum width of 1.2 m.
A bathroom or watercloset shall:
The minimum headroom of the ledge or tand/loft shall be 2.2 m. The maximum height of the loft shall be 1.5 m.
A ledge or TAND/loft in a livable room should not cover more than 25% of the floor area on which it is built and shall not, under any circumstances, interfere with the ventilation of the space.
The mezzanine floor shall have a minimum height of 2.2 m.
The mezzanine level must be at least 9.5 m^{2} in size to be utilized as a living space. The aggregate area of such a mezzanine floor in a building shall in no case exceed onethird of the plinth area of the building.
A mezzanine floor can be permitted over a room or a compartment, provided:
The height of a storeroom shall not be less than 2.2 m.
The size of a storeroom shall not be less than 3 m^{2}.
The height of a garage shall not be less than 2.4 m.
The size of garages shall be as given below:
The minimum width, minimum tread width, and maximum riser of staircases for buildings shall be as given below
The following minimum width shall be provided for staircases for respective occupancies,
The minimum width of tread without nosing shall be 300 mm. However, for one or twofamily dwellings, it may be reduced to not less than 250 mm.
The maximum height of the riser shall be 150 mm. However, for one or twofamily dwellings, it may be increased to not more than 190 mm. The number of risers shall be limited to 12 per flight.
The minimum clear headroom in any staircase or passage under the landing of a staircase shall be 2.2 m.
The height of rooms intended for human habitation shall not be less than 2.75 m measured from the floor’s surface to the lowest point of the ceiling (bottom of slab).
A bathroom must have a minimum area of 1.8 m^{2} and a minimum width of 1.2 m. The water closet’s floor size must be 1.1 m^{2} with a minimum width of 0.9 m. If a bath and a water closet are integrated, the floor space must be at least 2.8 m^{2} with a minimum width of 1.2 m.
The following minimum width shall be provided for staircases for respective occupancies,
Residential 1.00 to 1.25 m
Residential hotel 1.50 m
Assembly 2.00 m
Educational 1.50 m
Institutional 2.00 m
All other occupancies 1.50 m
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]]>The post What Is The Best Concrete Mix Ratio For Roof Slab, Beam & Column appeared first on Surveying & Architects.
]]>In this article we know about best mix ratio for concreting work like roof slab, column and beam for any type of residential and commercial building. As we know concrete mix ratio is design according to load acting on the structure.
Concrete is one of the most important building material used in construction industries for building a homes, low rise building, high rise building, commercial building, office building, apartment, industrial building, bridge, Dams, Channel, retaining walls and reservoir. Without concrete we can’t imaging any type of construction structure.
1) Plain cement concrete which is also known as PCC made of mixture of cement, sand, aggregate and water in definite and required proportion with desirable strength,
2) Reinforced cement concrete which is also known as RCC, if plain cement concrete provided with steel reinforcement known as rainforest cement concrete.
It all depends on type of structure, whether it is plain cement concrete or reinforced cement concrete. Best mix ratio for concrete is 1: 2: 4 (Cement, Sand, aggregate), in which one part is cement, 2 part is fine aggregate or sand and 3 part is coarse sand or aggregate. This is the safest best mix ratio for concrete or PCC for any type of construction where steel is not provided.
If concrete is provided with steel bar or reinforced, the safest best mix ratio for RCC concrete is 1: 1.5: 3 (Cement, Sand, aggregate), in which one part is cement, 1.5 part is fine aggregate or sand and 3 part is coarse sand or aggregate. This is the safest best mix ratio for rcc concrete for house construction where steel is provided.
Roof slab is horizontal RCC structure used to cover top of house, made of concrete provided with steel reinforcement. Generally, concrete mix ratio 1: 2: 4, 1: 1.5: 3 or 1: 1: 2 (Cement, Sand and aggregate) used for roof slab construction, but safest best concrete mix ratio for roof slab used is 1: 1.5: 3 (Cement, Sand and aggregate), in which one part is cement, 1.5 part is sand and 3 part is aggregate.
Beam is horizontal load bearing structural member of concrete fiber provided to resist the load coming from roof slab and transfer all coming load safely to column. Generally, concrete mix ratio 1: 1.5: 3 or 1: 1: 2 (Cement, Sand and aggregate) used for rcc beam construction, but, in house, typically, most common, safest best concrete mix ratio used for RCC beam is 1: 1.5: 3 (Cement, Sand and aggregate), in which one part is cement, 1.5 part is sand and 3 part is aggregate.
Column is vertical load bearing structure of concrete fiber provided with steel reinforcement to resist the load acting on it, it safely transfer all incoming load to bed of soil.
Generally, concrete mix ratio 1: 1.5: 3 or 1: 1: 2 (Cement, Sand and aggregate) used for rcc column construction, but, in house, typically, most common, safest best concrete mix ratio used for RCC column is 1: 1: 2 (Cement, Sand and aggregate), in which one part is cement, 1 part is sand and 2 part is aggregate.
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]]>The post What Is The Hidden Beam or Concealed Beam – Purpose & Advantages appeared first on Surveying & Architects.
]]>Hidden beams is defined as the beams whose depth is equal to the thickness of the slab. It also called as concealed beam provided within the depth of supporting slabs. It is popular and form an essential part of modern reinforced concrete framed structures
Hidden beams are generally inserted within the suspended slabs where slab thickness is considerable. The concept of concealed beam originated from flat slab concept. They are more applicable in commercial buildings.
Hidden beams are used for the following purposes
The hidden beam is a structural member, whose depth is equal to the thickness of the slab.
The hidden beam is also known as a concealed beam.
The hidden beam helps in enhancing the aesthetic appearance of a building by providing a larger flat ceiling, that gives a wider space look to the room.
The hidden beam provides a clear headroom height to work on the interior false ceilings, without any beam obstructions.
If the partition wall or brick wall comes over the slab, a hidden beam plays a major role in distributing the wall load, keeping the slab safe from the superimposed loads, that act over it.
Installation of the overhead tank ( larger capacity ) directly over the slab may develop cracks due to a higher concentrated load. If the water tank is rested over the hidden beam area, it helps to distribute the load to the adjacent supported beams.
Providing a hidden beam beneath the concentrated loads if any, that acts over the slab, helps to enhance the life span of the building.
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]]>The post Size Of Aggregate Used In RCC And PCC. appeared first on Surveying & Architects.
]]>All types of aggregate made of crushed of igneous rock granite rock sedimentary rock and Metamorphic rock, all type of aggregate unit in different purpose for building construction pavement, roads, Street and building and interior and exterior designing of house.
Some of these you will be familiar with that is fine aggregate and coarse aggregate containing different size and name like crushed stone, coarse sand, medium sand, fine sand ,silt ,clay and fine gravel ,medium gravel ,coarse gravel, pebbles cobbles and boulders.
When the aggregate is sieved through 4.75 mm size the aggregate is retained is known as coarse aggregate. Larger than 4.75mm size is coarse sand or aggregate, so coarse aggregate size arranging between 5mm to 256 mm or more.
1) fine gravel 4 to 8 mm
2) medium gravel 8 to 16 mm
3) coarse gravel 16 to 64 mm
4) pebbles 4 to 64 mm
5) cobbles 64 to 256 mm
6) boulder – more than 256 mm
Size of aggregate used in construction: it all depend on structure, whether it is plain cement concrete or reinforced cement concrete. Most commonly used aggregate size in construction is 20mm, 40mm and 75mm. Aggregate size 20mm used in RCC structure, 40mm aggregate size used in PCC structure or mass concreting and 75 mm and more size used in retaining wall construction.
Size of aggregate used in PCC: PCC is plain cement concrete made of mixture of cement, sand, aggregate and water without providing Steel, typically most common size of aggregate used in PCC is 40mm. This is best, perfect, ideal and standard size of aggregate used in PCC like construction of PCC pavement, road, walkway or driveway.
For the same strength or workability, concrete with large size aggregate will require lesser quantity of cement than concrete with a smaller size aggregate. In a mass concrete work or PCC the use of larger size aggregate will be useful due to the lesser consumption of cement, this will also reduce the heat of hydration and the corresponding thermal stresses and shrinkage cracks.
But in practice the size of aggregate cannot be increased to any limit on account of the limitation in the mixing, handling and placing equipment.
In large size aggregates surface area to be wetted per unit weight is less, and the water cement ratio is less which increases the strength. On the other hand in smaller size aggregates the surface area is increased which increases w/c ratio and lower strength is achieved.
In general for strength up to 200 kg/cm2 aggregates up to 40 mm may be used and for strength above 300 kg/cm2 aggregate up to 20 mm may be used.
Size of aggregate used in RCC: RCC is reinforced cement concrete made of mixture of cement, sand, aggregate and water provided with steel bar. Typically most common size of aggregate used in RCC structure like roof slab, column, beam and bridge deck is 20mm. This is best, perfect, ideal and standard size of aggregate used in RCC for construction of roof slab, column and beam.
Maximum size of coarse aggregate used in RCC: generally, 18mm, 20mm and 25 mm size of aggregate used in RCC work. 18mm is minimum size, 20mm is most common size used and maximum size of coarse aggregate used in RCC is 25mm.
Size of aggregate used in road construction: Road construction needs mass concreting work, where there is no limitation for concrete flow, most common size of aggregate used in road construction is 40mm. This is best, perfect, ideal, standard and maximum size of aggregate used in Road construction of pavement, walkway, driveway and etc.
Size of aggregate used in retaining wall: Retaining walls are relatively rigid walls used for supporting soil laterally so that it can be retained at different levels on the two sides. most common size of aggregate used for retaining wall construction is 75mm. This is best, perfect, ideal, standard and maximum size of aggregate used for retaining wall construction.
Size of aggregate used in Bridge construction: Bridge slab and their deck construction provided with reinforcement need mass concreting work, where there is no limitation for concrete floor. Most common size of aggregate used in bridge slab and their deck construction is 40 mm. This is best, perfect, ideal, standard and maximum size of aggregate used in bridge construction.
Size of aggregate used in concrete Dams: dam is heavily concrete, require a lot of mass concreting work, in Concrete Dams – 75mm aggregate size provided near the face where there steel reinforcement provided and 150 – 200mm maximum size of aggregate provided in the rest of the dam. Here concrete is poured by cranes from concrete buckets.
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]]>The post How To Calculate The Shuttering Area For Beam, Column And Slab appeared first on Surveying & Architects.
]]>Shuttering stands for a temporary structure formed to retain wet concrete that is poured in it to produce roofs and concreting walls in building structures.
It supports and retains the liquid concrete until it achieves the required strength and shape. When the structure becomes sufficiently strong to bear its load, this temporary structure is detached.
How to measure the shuttering area:
The calculation of shuttering is done with sq.m in the rate analysis of shuttering. To measure the area of shuttering, it is required to know the process for working out peripheral length (perimeter) of an any shape.
Peripheral length (Perimeter):
Perimeter belongs to the span around a two dimensional shape.
Given below, some vital formulas for measuring the shuttering area:
Perimeter of Square : 4S (S denotes Length of each side)
Perimeter of Rectangle : 2[L+B] (L & B denotes length & breadth)
Perimeter of Circle : 2πr (r denotes radius of circle)
Area of Rectangle = Length x Breadth
Area of Square = Side x Side
It should be kept in mind that each member in a structure ranging from slab or beam or column contains six sides (faces). The measurement for shuttering area is done with two methods. One is done with formula and other is by measuring the specific areas of faces.
The following formula is used :
Shuttering area = Peripheral length (Perimeter) x Depth
Computation of shuttering area of a column
Suppose, the column shuttering is performed on four sides.
Top face of the column is discarded for pouring concrete & the bottom face is secured to ground level. Here, the calculation should not done for top & bottom faces.
Toward column, shuttering is carried out for four sides and the other two sides.
The side of column remains in rectangular shape having side length as l and breadth as b.
Peripheral length of rectangular = l+b+l+b = 2l + 2b
Shuttering area = peripheral length (perimeter) x depth
Peripheral length = 2×0.8 + 2×0.6 = 1.6 + 1.2 = 2.8 sq.m
So, total area of shuttering for a column = 2.8 x 4 = 11.2 sq.m
When the shape of column is circular, then the following formula is used to work out the shuttering area.
2πr x Depth
Measurement of shuttering area for a beam
The shuttering is provided on five sides for a beam.
In beam, the top face is departed to fill concrete. Therefore, no shuttering is arranged on top face.
The calculation is done by determining the separate area of each faces as follow :
Face 1 : Area of rectangle = L x B = 0.8 x 4 = 3.2
Face 2 : Area of rectangle = L x B = 0.6 x 4 = 2.4
Face 3 : Area of rectangle = L x B = 0.8 x 4 = 3.2
Face 4 : Area of rectangle = L x B = 0.8 x 0.6 = 0.48
Face 5 : Area of rectangle = L x B = 0.8 x 0.6 = 0.48
Total Area of Shuttering = 3.2 + 2.4 + 3.2+ 0.48 +0.48 = 9.76.Sqm
Measurement of Shuttering Area of a Slab:
Slab is supported on beam, hence, shuttering is not required on four sides. The top of the slab is also departed to fill concrete and for curing. Hence, shuttering is arranged to the bottom of slab.
Shuttering area of Slab = Bottom area of slab = L x B
Bottom Area = 5 x 4 = 20 Sq.m
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]]>The post HOW TO CALCULATE STEEL QUANTITY FOR SLAB WITH BAR BENDING SCHEDULE appeared first on Surveying & Architects.
]]>One Way Slab  Ly/Lx > 2 
Two Way Slab  Ly/Lx < 2 
D = Depth of Slab – Top cover – Bottom cover
Length= 2950mm
Breadth= 6000mm
Beam size = 225x300mm
Main bars are 12 mm in diameter @ 150 mm centre to centre spacing
Distribution bars are 8 mm in diameter @ 150 mm centre to centre spacing. (Main Bar & Distribution Bar Difference)
Top and Bottom Clear Cover is 25 mm
Thickness of Slab – 150 mm
Therefore, To Find Its One Way Or Two Way Slab
Lx = 2950mm (shorter span)
Ly = 6000mm (Larger span)
Ly/Lx= 6000/2950 = 2.033
One Way Slab  Ly/Lx > 2 
Hence its oneway slab
So, We need to find
Distribution bars are placed on the top of the main bar.
Lower dimension bar is used as distribution bars.
Distribution bars are used to resist the shear stress, and cracks developed at the top of the slab
Find Cutting Length Of Main Bars For Floor Slab 1
D = 150 – (2 x 25) = 100mm
Inclined length = 0.42 x 100 = 42mm
45° bend = 1d
d = diameter of bar
45° bend = 1 x 12 = 12mm
Intermediate slab
L = 3100mm
No Of Bars Required For Cutting Main Bars
Opposite length of bar = 6000
Spacing = 150mm c/c as per given details
Total Length Of Bar Required For Main Bars Of Floor Slab 1
= no of bar x total length of bar
= 41 x 4.80 = 196.8m = 197m
Find Cutting Length Of Distribution Bars For Floor Slab 1
Length of distribution bars = (breadth of beam/2) + length of longer span + (breadth of beam/2)
Length of distribution bars = (225/2) + 6000 +(225/2) = 6225mm
Length of distribution bars = 6.225m
STEP 5
No Of Bars Required For Cutting Distribution Bars
Opposite length of bar = 2950
Spacing = 150mm c/c as per given details
Total Length Of Bar Required For Distribution Bars Of Floor Slab 1
= No Of Bar X Total Length Of Bar
= 21 x 6.225 = 130.725m
Number of top bars = (Lx/4) / spacing + 1
Number of top bars = ((2950/4) / 150) + 1 =5.91 = 6 nos of bars x 2 sides = 12 nos
Number of top bars = 12 nos of bars
Total Length Of Bar Required For Top Bars Of Floor Slab 1
Cutting Length Of top Bars = 6.225m
Total length of top bars = 6.225 x 10 = 27.25m
Length= 3100mm
Breadth= 6000mm
Lx shorter span = 3100mm
Ly larger span = 6000mm
Beam size = 225x300mm
Main bars are 12 mm in diameter @ 150 mm centre to centre spacing
Distribution bars are 8 mm in diameter @ 150 mm centre to centre spacing. (Main Bar & Distribution Bar Difference)
Top and Bottom Clear Cover is 25 mm
Consider Development length as 40 d
Thickness of Slab – 150 mm
Lx = 3100mm (shorter span)
Ly = 6000mm (Larger span)
Ly/Lx= 6000/3100 = 1.93
Two Way Slab 
Ly/Lx < 2 
So, We need to find
Find Cutting Length Of Main Bars For Floor Slab 2 (A2B2 direction)
D = 150 – (2 x 25) = 100mm
Inclined length = 0.42 x 100 = 42mm
d = diameter of bar
45° bend = 1 x 12 = 12mm
L = 3000mm
Length Of Main Bars (A2B2) = (300/2) + (225/2) + (1 x 42) – (12 x 2) + 3100 + 300 + (0.3 x 3000) = 4580.5
No Of Bars Required For Cutting Main Bars
Opposite length of bar = 6000
Spacing = 150mm c/c as per given details
Total Length Of Bar Required For Main Bars Of Floor Slab 2
= no of bar x total length of bar
= 41 x 4.6= 188.6m = 190m
Find Cutting Length Of Main Bars For Floor Slab 2 (C2D2 direction)
D = 150 – (2 x 25) = 100mm
Inclined length = 0.42 x 100 = 42mm
d = diameter of bar
45° bend = 1 x 12 = 12mm
L = 3000mm
Length Of Main Bars (C2D2) = (300/2) + (225/2) + (1 x 42) – (12 x 2) + 6000 + (225/2) = 6393m
No Of Bars Required For Cutting Main Bars
Opposite length of bar = 3100
Spacing = 150mm c/c as per given details
Total Length Of Bar Required For Main Bars Of Floor Slab 2
= no of bar x total length of bar
= 22 x 6.4= 140.8m = 141m
Number of top bars = (Ly/4) / spacing + 1
Number of top bars = ((6000/4) / 150) + 1 = 11 nos of bars x 2 sides = 22 nos
Number of top bars = (Lx/4) / spacing + 1
Number of top bars = ((3100/4) / 150) + 1 =6.16 = 7 nos of bars x 2 sides = 14 nos
Total Length Of Bar Required For Top Bars Of Floor Slab 2 (A2B2)
Cutting Length Of top Bars = 3.325m
Total length of top bars = 3.325 x 14 = 46.55m
Total Length Of Bar Required For Top Bars Of Floor Slab 2 (C2D2)
Cutting Length Of top Bars = 6.225m
Total length of top bars = 6.225 x 14 = 87.15m
RESULT,
FLOOR SLAB 1,
Main Bars = 185m Of 12mm Rod
Distribution Bars = 131m Of 8mm Rod
Top Bars = 63m Of 8mm Rod
Main Bars = 190m of 12mm rod
Main Bars = 141m Of 12mm Rod
Top Bars = 88 m of 8mm rod (A2B2)
Top Bars = 47m of 8mm rod (C2D2)
12mm rod required = 516m
8mm rod required = 329m
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