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Saturday, November 24, 2018

Concrete Mix Design


 Mix Design of Concrete (M-35) : 

1.       Characteristic Strength =35 N/mm2
2.       Target mean strength at 28 days as per Morth ,table 1700-3A =47 N/mm2
3.       Maximum Size of aggregate =20mm
4.       Type of Expose-Normal
5.       Degree of Quality Control-Good
6.       Minimum Cement content as per Morth,Table 1700-3A =400 kg/mm3
7.       Maximum water cement ratio as per morth,table 1700-3A---=0.45
8.       Air entrapped as per IS: 10262………=2.0%
9.       Grade of concrete-M35
10.   Workability required in terms of slump at site as per Morth, Table 1700-3 = 150 to 200 mm
11.   Cement Used Dalmia konarak OPC 53 Grade (IS: 12269-2013)

Coarse Aggregate:

As we will use Coarse Aggregate 20mm & 10mm (Normal Size), So we can have to satisfy 

combined grading criteria IS :383,Table -2. The proportions of Coarse aggregate and Fine 

aggregate are CA : FA=60:40 & CA1 (20mm) : CA2 (10mm) =50:50

20mm (Pakur Varity) Specific Gravity= 2.87

10mm (Pakur Varity) Specific Gravity= 2.83

       Specific Gravity  of Fine Aggregate=2.72

       Specific Gravity  of  Cement=3.15

       Water=1

CA:FA=60:40,CA1:CA2=50:50

W/C Ratio= 0.42

Cement =450 kg/m3

Water=450 x 0.42=189 kg/m3

For slump 175mm water content increased upto 15 %=189+15% of 189=217.35 lt/m3

Super Plasticizer (Admixture)  @ 0.8% was taken by weight of cement=450 x 0.8%=3.6 kg/m3

As super Plasticizer used the water content can be reduced  10%

Now water content is = 217.35 x 0.9=195.615 Lt



Now, Volume of Cement =450/3.15 x 1/1000 =0.142 m3

(Where, Weight of Cement=-450kg,Specific Gravity of  cement =3.15 )

Volume of water =195.615/1 X  1/1000 = 0.195 m3

Volume of Super Plasticizer = 3.6/1.11 X 1/1000 =0.003 m3

(Where, Weight of Super Plasticizer=3.6kg, specific gravity of S.P.=1.11 )

Volume of air Entrapped @ 2% =0.02 m3

Total Volume=(0.142+0.195+0.003+0.02)=0.360 m3

Volume of all in aggregate =(1 – 0.360)=0.64 m3

Mass of CA =0.64 x 2.85 x 0.60 x 1000=1094.40 Kg

 (Where, Vol of all in aggregate =0.64,Specific gravity of CA =2.85, CA : FA = 60: 40)

Mass of FA = 0.64 x 2.72 x 0.40 x 1000 = 696.32 kg

(Where, Vol of all in aggregate =0.64,Specific gravity of FA =2.72, CA : FA = 60: 40)

Now,

Cu.m  Before Moisture Correction,

Cement= 450kg

20 mm Aggregate= 547.20 kg

10 mm Aggregate= 547.20 kg

Fine aggregate= 696.32 kg

Water = 195.615 kg

Super plasticizer = 3.6 kg

Sunday, November 4, 2018

**15 TECHNICAL TERMS ABOUT REINFORCEMENT A CONSTRUCTION SUPERVISOR

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1: Concrete Cover for Reinforcement:-

Concrete cover is also called clear cover for reinforcement. and it is the distance b/w steel surface and nearest outer surface of concrete structure.

Concrete cover for different RCC members

Pile  = 75 mm

Pile cap =75 mm

Footing = 75 mm

Column below ground = 65 mm

Column above ground =  40 mm

Shear wall below ground = 65 mm

Shear wall above ground = 40 mm

Beam below ground =  65 mm

Beam above ground = 40 mm

Slab below ground = 50 mm

Slab above ground = 20 mm

Stair  = 20 mm

2: Unite Weight of Steel bar:-

Site supervisor should be know the unite weight of steel we have given weight of steel for different dia of steel

8 mm ø bar – 0.12 kg/ft

10 mm ø bar – 0.19 kg/ft

12 mm ø bar – 0.27 kg/ft

16 mm ø bar – 0.48 kg/ft

20 mm ø bar – 0.75 kg/ft

25 mm ø bar – 1.25 kg/ft

3: Lap in compression zone:-

Steel should be lapping for compression zone with 50d where ( d ) is dia of steel bar

4: Lapping for Tension:-


Steel should be lapping for tension zone with 40d where ( d ) is dia of steel bar

5: No Lapping:-

Lap should be not provide larger than 36mm dia of steel

6: Require bending wire:-

For binding one ton of MS bar 7 kg to 13 kg binding wire is required.

7: Standard length of Mild steel bar:-

Mild steel Standard length is 40′

8: Minimum Longitudinal Bar in a Column:-

Use Minimum 4 No’s of steel in Square column and 6 No’s for circular column

9: Placing Stirrup:-

Place 1st stirrup with the spacing of 2 inches.

10: Hook’s Angle of Stirrup:-

Stirrup’s hook angle is 135°.

11:  Bend for Vertical Bar of Column:-

Remember that vertical column bars are called longitudinal bars, and bend is provide at one end of longitudinal bar with the angle of 90 degree and then length should be not less than 18 inches.

12: End Hook for Longitudinal Bar of Beam:-

Column end hook angel should be with 90 degree

13: Hook’s Length of Stirrup:-

The stirrup hook length should be not less than 3 inches or use this formula = 6d where d is dia of steel bar which used for stirrup.

14: Maximum Lap in a Zone:-

Don’t use more than 50% lapping zone in any RCC structure.
15: Bent Up bar:-

Provide bent up  bar near at support of slab and beam.


Saturday, November 3, 2018

3bhk house plan


Watch the video about 3bhk 🏠 house plan

What is centre line method

* WHAT IS CENTRE LINE METHOD:-
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The center line method is suitable for wall cross sections. and the center line method the total length of centre line should be multiplying with breadth and depth of the any wall than it will give us the total quantity at the same time. and each length wall length should be measure to center line of the wall.

To fine out the quantity of wall or any construction building. And this method ( Centre line ) is useful for walls with equivalent cross sections. In order to obtain the total quantity instantly, the total centre line length is multiplied with breadth and depth of relevant item. When cross walls or partitions or verandah walls are attached with main wall, the length of the centre line decreases by half of breadth for each junction. Such junction or joints are analyzed cautiously at the time of estimating total centre line length.


*SOLUTION :- 
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(1)Total Center line length = ?

(2)Now Quantity of Excavation

(3)Volume of Excavation = Total Centre line length x Breadth of Foundation x Depth of Foundation

(4)Volume of Excavation = ?

(5)Quantity of concrete in foundation = Total Centre line length x Breadth of Foundation x Depth of concrete.

(6)Quantity of concrete in foundation  = ?

(7)Quantity of bricks in foundation = Total Centre line length x Breadth of Foundation x ( Total depth of foundation – concrete depth)

(8)Quantity of bricks in foundation =?

(9)Quantity of D.P.C = Total Centre line length x Breadth of Foundation x thickness of D.P.C  suppose (25mm)

(10)Quantity of D.P.C = ?

(11)Quantity of plinth beam = Total Centre line length x Breadth of the beam x Depth of beam

(12)Quantity of plinth beam = ?

(13)Brick work in super structure = Total centre line length x breadth of wall x height of wall

(14)Brick work in super structure = ?

(15)Quantity of wall plaster for 2 sides = Total centre line length x height of wall x thickness of plaster

(16)Quantity of wall plaster for 2 sides = ?

(17)Quantity of Paint for 2 sides of wall = Total centre line length x height of wall  x 2  = Area of paint in sft    ( multiply 2 for wall 2 sides of wall.

(18)We have two kind of wall in the figure > Long wall and short wall

* Note:- The centre line method is the one of the best method to find out the building quantity. and this method is used at the construction site. therefore we have to learn the centre line method with long wall and short wall method. with this method we can find out the exact quantity of any construction building.


Compound of road structures


*COMPOUND OF ROAD STRUCTURE:-

1: sub – Grade                2: Sub – Base                        3: Base Course               4: Binder Course

                       5: Surfacing

(1)SUB GRADE:-
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Sub-grade  is the the natural surface of the road, which is very important for road and this is lowest layer of the road.

Sub Grade construction:

this is the natural surface of the soil. therefore if the bearing capacity of soil is good so we have should not do any change in sub grade. but if the bearing capacity of soil is low than we have to up the sub grade level up to 60 cm (2ft) from the flood level. and we have to used the cheap materials in the sub grade just for making surface if the bearing capacity is good.

(2)SUB BASE :-
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Sub base is the one of the main load bearing layer of the road. which role is to spread the total load of the sub base to sub grade. and this is also the important layer of the road. And we have to used the good quality of materials in the sub base. because when the sub base is not good so the road will be not work for long time.

 construction Sub Base:

The sub base can constructed above the sub grade. and the thickness of the sub base should be from 3 inch to 6 inch or 7.5cm to 15cm.

(3)Base Course:-
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the base course the pavements layer of material in asphalt road. and this layer in between the sub base and binder course. when the sub base designed than we directly construct the base course on above this.

Construction of Base Course

we have to use the strong materials in this layer of the road. and the bottom layer essentially comprises of unbound mixture from course and fin aggregate , and also used to crush sand, to get the desire load bearing capacity and absorb the upcoming traffic load.and we have to use the good quality of materials in the base course construction.

(4)Binder Course :-
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 in road structure the binder course are the different combination of aggregate and binders. and the binder course is layer which bind all the materials in road structure. which are present in the base course , sub base and sub grade. in binder course we can bind the materials with asphalt (bitumen). and binder course is suitable  for a wide range of of used from heavy duty of road pavement to car parks and also for race road etc.

(5)SURFACING :-
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Surfacing  is the waterproof layer of road, which protect the underlying pavement. and also increase the skid resistance of road. the surfacing prevent the road from the cracks . and its anti- glare surface during the wet weather and surfacing can increase the reflexive surface for the night driving.
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**IRRIGATION TECHNICS Irrigation:- The science of artificial application of water to the land,in accordance with the 'Crop requiremen...

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