Some useful tips to work out the cutting Length of Bent Up Bar In Slab

For a site engineer, it is essential to work out the cutting length of bars based on the slab dimensions and provided instructions to the bar benders.

If the construction work is intended for the small area, the reinforcement detailing can be transferred to the bar benders. They will deal with the cutting length. But be careful that it may not be perfect as they do not consider the bends and cranks. They may provide some additional inches to the bars for the bends which are fully imperfect. Therefore, to get rid of this issue, a site engineer should try to compute calculate the cutting length independently.

In this article, detailed explanation is given for working out the length for reinforcement bars of slab.

The calculation is made on the following dimensions :-

Diameter of the bar = 12 mm
Clear Cover = 25 mm
Clear Span (L) = 8000
Slab Thickness = 200 mm
Development Length(Ld) = 40d

Process for computation

Cutting Length = Clear Span of Slab + (2 x Development Length) + (2 x inclined length) – (45° bend x 4) – (90° bend x 2)
Inclined length = D/(sin 45°) – dD/ (tan 45°) = (D/0.7071) – (D/1)= (1D – 0.7071D)/0.7071= 0.42 D
There exist four 45°bends at the inner side (1,2,3 & 4) and two 90° bends ( a,b ).
45 ° = 1d; 90 ° = 2d
Cutting Length = Clear Span of Slab + (2 X Ld) +(2 x 0.42D) – (1d x 4) – (2d x 2) [BBS Shape Codes]

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Some useful tips to measure loads on column, beam and slab

In order to work out the total load on columns, Beam and Slab, there should be clear ideas on the types of loads enforcing on the column.

Different Loads operating on Column:

1) Column Self Weight X Number of floors
2) Beams Self Weight per running meter
3) Load of walls per running meter
4) Total load on Slab (Dead load + Live load + Self weight)

Apart from above loading, the columns are also susceptible to bending moments which should be taken into consideration in the final design.

For Colomn: The Self weight of Concrete remains approx 2400 kg/m3, that is similar to 240 kN and self weight of steel is approx 8000 kg/m3.

Therefore, if we consider a column size of 230 mm x 600 mm with 1% steel and 3 meters standard height, the self weight of column is approx 1000 kg per floor that is equivalent to 10 kN.

At the time of making calculation, self weight of columns is taken as 10 to 15 kN per floor.

For Beam: Similar method is also used for making calculations of beam. Suppose, each meter of beam contains dimensions of 230 mm x 450 mm without slab thickness. Therefore, the self weight should be approx 2.5 kN per running meter.

For Walls: The Density of bricks differs among 1500 to 2000 kg per cubic meter. For a brick wall with thickness 6 inch, height 3 meter a length 1 meter. The load / running meter should be equivalent to 0.150 x 1 x 3 x 2000 = 900 kg, that is identical to 9 kN/meter. This method is useful for working out the load of brick per running meter for any brick type.

For aerated concrete blocks and autoclaved concrete blocks similar to Aerocon or Siporex, the weight per cubic meter should remain 550 to 700 kg per cubic meter.

When these blocks are utilized for construction, the wall loads for each running meter should remain as low as 4 kN/meter, the cost of the project is decreased considerably with the use of this block.

For Slab: Suppose, the slab contains thickness of 125 mm.

Therefore, self weight of each square meter of slab should be = 0.125 x 1 x 2400 = 300 kg that is identical to 3 kN.

Now, If finishing load is taken to be 1 kN per meter and superimposed live load to be 2 kN per meter. Therefore, from above data, the load of slab can be calculated as 6 to 7 kN approximately per square meter.

Factor of Safety: At the end, once the total load on a column is computed, consider the factor of safety that is very crucial for any building design for safe and convenient performance of building during its design life cycle.

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Guidelines to provide concrete cover for reinforcement in slab, footing, beam & column

Concrete cover: Concrete Cover is arranged for the reinforcement in Reinforced Cement Concrete. Cover means the spacing among the exterior surface of the concrete to the inserted reinforcement.

Benefits of arranging Concrete Cover: The purpose of covering is to provide protection against erosion. Reinforcement is susceptible to erosion and fire for atmospheric conditions. In case of improper cover erosion and cracks may occur in hardened RCC.

Covering is arranged for each and every component of the building (Slabs, Beams, footings) where the reinforcement is applied. The covering blocks are utilized to retain the reinforcement in exact position as well as providing a covering for reinforcement.

Several Types of Concrete Cover Block: Depending on the type of materials applied, the following types of covering blocks are commonly found -

1. Wooden concrete cover Block
2. Steel concrete cover block
3. PVC Block
4. Cement Masonry concrete cover block
5. Aluminium Block
6. Stones

Conditions for Concrete Cover: Covering differs based on the dimensions of the components (Slab, beam, column, footings, etc.) The conditions for arranging covering in RCC are provided below -

Condition - Covering

When the length of the item is ≤ 0.3 1 - 1" or 25mm or 0.025mWhen the length of the item remains among 0.4m to 0.5m then - 2" or 50mm or 0.050m
When the length of the item remains ≥ - 0.6m then 4" or 100mm or 0.1m

From above, the maximum concrete cover remains 0.1m or 100cm

1. Concrete Cover in Columns / Beams: The length and width of the column should be 0.5m and 0.45m. The covering for reinforcement in the column should be 0.050m from all sides and similar reinforcement should be designed accordingly. The Dimensions of Reinforcement in the column should be 0.40m and 0.35m.

Suppose the length and width of the column are 0.40 and 0.25. Covering should be equal. Consider the minimum dimension from the two dimensions i.e. 0.25. For 0.25m the covering of 0.025m should be provided. So, the covering of 0.025m is arranged in all the sides. Therefore, dimensions of reinforcement is 0.35m and 0.20m.

Total Length of Stirrup is 2x [0.35+0.20]+ 9D x 2 (hook length)

2. Concrete Cover for Slabs: Suppose, the length and width of the slab are 1.3m and 1.0m. The covering of 0.1m is arranged when the length of the bar is in excess of 0.6m. Use the same condition as mentioned. The covering of 0.1m is arranged from all the sides of the slab.

3. Concrete cover for footings: Suppose, the dimensions of Footing are 0.7m and 0.6m. To length and width of Mesh (reinforcement) utilized in footings are acquired by subtracting the cover. Use the similar principle as above. As per the condition, a concrete cover of 0.1m is subtracted from all the sides. Therefore, the dimensions of reinforcement are 0.5m and 0.4m.

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Guidelines for making perfect structural design

This civil engineering article focuses on the least standards which should be maintained for the design of various RCC structural elements like the columns, beams, slab and foundation as well as the least safe standards for the reinforcing bars to be applied for making the design of the above mentioned structural elements.

Minimum cross-sectional dimension for a Column should be 9″x 12″ (225 MM x 300 MM). It is the minimum approved size.

It is always recommended to utilize M20 grade concrete for construction as per IS 456:2000. The least steel in a 9″ x 9″ column should be 4 bars of 12 MM with stirrups of 8 MM steel rings at a spacing of 150 MM centre to centre. In a 9″ x 12″ column, more bars (6 bars with 12 mm diameter) should be added to sustain the total efficiently.

Least RCC beam size should not be lower than 9″x 9″ (225MM X 225MM), with an supplementary slab thickness of 125 MM.

Normally, there should be minimum of 4 bars, with 2 bars having 12 MM thickness in the bottom of the beam, and 2 bars having 10 MM at the top of the beam.

A concrete cover of 40 MM should also be provided. It is suggested to utilize M20 grade of concrete (1 part cement : 1.5 parts sand : 3 parts aggregate : 0.5 parts water).

Minimum thickness of RCC slab should be 5″ (125MM) since a slab may comprise of electrical pipes which are implanted into them which could be 0.5″ or more for internal wiring and as a result the depth of slab is decreased at specific places that lead to cracking, weakening and water leakage throughout rains. Therefore, a least thickness of 5″ should be retained.

Minimum size of foundation for a single storey of G+1 building should be 1m x 1m, where safe bearing strength of soil is 30 tonnes per square meter, and the anticipated load on the column does not surpass 30 tonnes.

The depth of footing should be minimum 4′under ground level. It is suggested to get to depths up to had strata.

Minimum Reinforcing bar details:

1. Columns: 4 bars of 12mm steel rods FE 500.
2. Beams: 2 bars of 12 mm in the bottom and 2 bars of 10 mm on the top.
3. Slab

a) One Way Slab: Main Steel 8 MM bars @ 6″ C/C and Distribution Steel of 6 mm bars @ 6″ C/C
b) Two Way Slab: Main Steel 8 MM bars @ 5″ C/C and Distribution Steel of 8 mm bars @ 7″ C/C

4. Foundation: Initially, there should be 6″ of PCC layer. Over it, a tapered or rectangular footing with minimum 12″ thickness should be arranged. Steel mesh of 8 mm bars @ 6″ C/C should be placed. In a 1m X 1m footing, there should be 6 bars of 8 mm on both segments of the steel mesh.

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Common thumb rules for civil engineering works

Thumb Rules is very important for any civil engineer, Site engineer or civil supervisor to obtain instant decisions on the construction site. By applying thumb, the engineers can get the solution with a simple mathematical formula and take proper decisions wherever required. Before applying these thumb rules, it should be kept in mind that the thumb rule can only provide fairly accurate results never the correct results.

The following types of thumb rules for civil engineers are commonly used in construction work :-

Thumb rule for measuring the Concrete Volume relating to the area:
The volume of concrete necessary = 0.038 m3/square feet area.

As for instance, if Plan Area = 40 x 20 = 800 Sq. m., total necessary volume of concrete will be as follow :-
= 800 x 0.038m3 = 30.4m3

Thumb rule for Steel quantity necessary for Slab, Beams, Footings & Columns:
Essential quantity of steel in residential buildings = 4.5 Kgs – 4.75 Kgs / Sq. Ft.
Essential quantity of steel in commercial buildings = 5.0 Kgs-5.50 Kgs/Sq. Ft.

Thumb Rules For Civil Engineers recommended by B N Datta for the Steel quantity that will be applied for several members of the building :-

Proportions of Steel in Structural Members:

1) Slab – 1% of the total volume of concrete
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

As for instance, suppose the length, width and depth of the slab are 5m, 4m and 0.15m. Now, the quantity of steel for the slab will be computed as follow :-

Initially, it is required to work out the concrete volume.
The total volume of concrete for the slab = 5x4x0.15 = 3m3

Secondly, work out the quantity of steel with formula as follow :-
Based on the guidelines provided in B. N. Dutta reference book, the quantity of steel in slab is 1% of the total volume of concrete used.
Thumb rule to work out the quantity of steel in above slab = Volume of concrete x density of steel x % of steel member.

The weight of steel necessary for above slab = 3x7850x0.01 = 235 kgs

To make perfect calculation, use bar bending schedule.

To learn how thumb rules are applied to calculate the shuttering area and the quantity of cement, sand, course aggregate in several grades of concrete, click on the following link civiconcepts.com

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Detail guidelines on Structural Design Codes And Standards

Design of RCC Structural Components: In this civil engineering article, you will be familiar with the minimum standards necessary for the design of RCC structural elements of a structure like columns, beams, slab and foundation.

Besides, the information is also given on the minimum safe standards for the reinforcing bars which should be applied for the design of the above mentioned structural elements.

Minimum cross-sectional dimension for a Column: Minimum standard size should be 9″x 12″(225 MM x 300 MM). It is recommended to use M20 grade concrete for construction as per IS 456:2000. The minimum steel in a 9″ x 9″ column should contain 4 bars of 12 MM containing stirrups of 8 MM steel rings at a spacing of 150 MM centre to centre. In a 9″ x 12″ column, two more bars can be added to make the total to 6 bars of 12 MM diameter. This design will be secured for up to G+1 floors.

Minimum RCC beam size : It should not be under 9″x 9″(225MM X 225MM), containing a extra slab thickness of 125 MM.

A minimum of 4 bars can be used containing 2 bars of 12 MM thickness in the bottom of the beam, and 2 bars of 10 MM at the top of the beam. The required concrete cover should be 40 MM. It is recommended to utilize M20 grade of concrete (1 part cement : 1.5 parts sand : 3 parts aggregate : 0.5 parts water).

Minimum thickness of RCC slab : It should be minimum 5″ (125MM) as a slab may comprise of electrical pipes implanted into them with thickness 0.5″ or more for internal wiring, which in fact minimizes the depth of slab at specific areas. It leads to cracking, decay and water leakage during rains. So, there should be a minimum thickness of 5″.

Minimum size of foundation : A size of 1m x 1m should be ideal for a single storey of G+1 building, where the safe bearing strength of soil should be 30 tonnes per square meter, and the impending load on the column does not be in excess of 30 tonnes,. Minimum depth of footing should remain 4′ underneath ground level. It is recommended to retain deepness up to had strata.

Reinforcing bar details (minimum):

1. Columns: 4 bars of 12mm steel rods FE 500.
2. Beams: 2 bars of 12 mm in the bottom and 2 bars of 10 mm on the top.
3. Slab

a) One Way Slab: Main Steel 8 MM bars @ 6″ C/C and Distribution Steel of 6 mm bars @ 6″ C/C
b) Two Way Slab: Main Steel 8 MM bars @ 5″ C/C and Distribution Steel of 8 mm bars @ 7″ C/C

4. Foundation: Initially, 6″ of PCC layer should be provided. Over it, there should be a tapered or rectangular footing with minimum 12″ thickness. Steel mesh of 8 mm bars @ 6″ C/C should be placed. In a 1m X 1m footing, it should comprise of 6 bars of 8 mm on both portions of the steel mesh.



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Why reinforcement is provided in a column

Concrete is strong at bearing compressive stress. Plain concrete can sustain compressive loads capably. But it is always recommended to utilize the R.С.С. columns in place of plain concrete columns in modern day structures for the following reasons :-

Temperature stresses are formed in Concrete because of differences in weather. Cracks may happen for this stress. To get rid of the issue and reduce cracks, some steel is provided at the face of the concrete.

Columns specifically slender columns are prone to lateral loads and moments. Sometimes, tensile stress may also build up particularly in columns at exterior boundary of the building. Steel can deal with this tensile stress.

While making the design of RCC structures, the reinforcement is provided in beams to tie them securely with beams.

Reinforcement is provided so that the size of the columns is not increased.

Reinforcement steel improves the ductility of the member so that the structure gets the ability to withstand earthquake in a superior way.

In R.C.C. columns, less area is required with regards to a plain concrete column It is found that steel can bear load m-times that of concrete of the similar area. To deal with a specific load, the section of an R.C.C. column will be much finer as compared to that of plain concrete. By applying R.С.С. columns, huge space is saved since the size of the column will be less.

A minimum area of steel should be arranged in the column if any case it is necessary for bearing load or not. It is performed to withstand tensile stresses which occur because of eccentricity of loads.

Two types of reinforcements are arranged in a R.C.C. column.

a. Longitudinal reinforcement.
b. Transverse reinforcement.

Longitudinal Reinforcement: The longitudinal reinforcement comprises of steel bars which are arranged longitudinally in a column. It is also known as main steel. The properties of longitudinal reinforcement are given below:

i. To distribute the compressive loads along with concrete, consequently minimizing the size of the column on the whole and parting more usable area.
ii. To withstand tensile stresses which are formed because of any moment or accidental eccentricity.
iii. To yield ductility to the column.
iv. To lessen the impact of creep and shrinkage because of continuous constant loading applied for a long time.

Transverse Reinforcement: The transverse reinforcement is arranged along the lateral direction of the column in the shape of ties spirals which cover the main steel. The function of transverse steel are as following -

i. To retain the longitudinal bars in exact place.
ii. To resist buckling of the main longitudinal bars.
iii. To avoid diagonal tension that happens because of transverse shear formed due to any moment or load.
iv. To yield ductility to the column.
v. To resist longitudinal splitting or bulging from concrete by enclosing it in the core.



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How to design any one way simply supported slab

In this construction video tutorial, you will learn how to create the design of one way slab (simply supported slab) that is standing on the masonry wall.

You will also gather information on all checks for making design of the slab.

The requirements and materials of the slab are given below :-

Clear span = 3m
Live load = 4000 N/m2
M20 concrete and fe 415 steel are used here.

The calculation is made on the following dimensions :-

A bearing of 120 mm is provided at each end. The distance among centers of bearing = 3+0.12 = 3.12 m

Necessary effective depth = Span/20 x modification factor
= 3120/20 x 1.40 = 112 mm

The bars with 8 mm diameter are arranged with clear cover of 15 mm.
Effective cover = 15 + 4 = 19 mm

Overall necessary depth = 112 + 19 = 131 mm

It is necessary to arrange an overall depth of 135 mm

So, effective depth = 135-19 = 116 mm

The Dead load of the slab will be calculated as follow :-
25 x 135 = 3375 N/m2 (floor finished with C20 mm).

Live load = 4000 N/m2
Therefore, total load = 7855 N/m2
Factor load = Wµ = 1.50 x 7855 = 11782.5 N/m2

To get more details, go through the following video tutorial.

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Detailed process for measuring Cutting Length Of Bent Up Bar In Slab

This construction video is a part of technical lectures on Civil Engineering. The video will introduce you with the detailed processes for estimating cutting length of bent up bar in slab

If the reinforcing bars are fabricated with bends, the straight bar is primarily sliced to a length below the sum of the indicated dimensions of the bent bar. The variation among the detailed length and the cut length is defined as the "bend curvature deduction" and alternatively called as gain, creep, and gyp.

When the bent-up bars are supplied, their support to the shear resistance shall not be in excess of half that of the total shear reinforcement.

The general method of bending of a bar adjacent to a support is at an angle of 450. The angle of bend may set at 300 in shallow beams where actual depth is below 1.5 times its breadth. The objective of bend adjacent to a support is as below :-

To withstand negative bending moment happening at the region of the support to withstand shear force that is superior at the support.

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