Project on Concrete Mix Design for various grades of Concrete
Aim :-
To Calculate The Concrete Mix Design For M35 Grade Concrete With Fly Ash
Equations :-
* Target Mean Strength : f'ck = fck+1.65 s
f'ck = Target Mean Strength
fck = Compressive Strength At 28 Days Of Concrete
s = Standard Deviation ( IS 10262:2009 , Table-1 )
* Volume Of Cement : ( Mass Of Cement / Specific Gravity Of Cement ) X 1/1000
* Volume Of Water : ( Mass Of Water / Specific Gravity Of Water ) X 1/1000
* Volume Of Chemical Admixture : ( Mass Of Chemical Admixture / Specific Gravity Of Chemical Admixture ) X
1/1000
* Volume Of Fly Ash : ( Mass Of Fly Ash / Specific Gravity Of Fly Ash ) X 1/1000
*Volume Of Concrete : 1m^3
*Volume Of Aggregate : ( Volume Of Concrete - ( Volume Of Cement + Volume Of Fly Ash + Volume Of
Water+Volume Of Admixture )
* Mass Of Coarse Aggregates : Volume Of Aggregate X Volume Of Coarse Aggregate X Specific Gravity Of Coarse
Aggregate X 1000
* Mass Of Fine Aggregate : Volume Of Aggregate X Volume Of Fine Aggregate X Specific Gravity Of Fine
Aggregate X 1000
Codes For Concrete Mix Design :-
* IS 10262:2009
* IS 456:2000
* IS 8112
* IS 9103
* IS 3812 ( Part 1 )
Objectives :-
* The Main Objective Is To Produce Concrete As Economically As Possible
* To Achieve Desired Workability , Durability And Minimum Strength Of M35 Grade Concrete
Procedure And Calculation :-
1. Target Mean Strength
F'ck = fck + 1.65 s
fck = 35N/mm^2 ( M35 )
Table 1 - Is 10262:2009
s= 5 N/mm^2
f'ck = 35 + ( 1.65 X 5 )
= 43.25 N/mm^2
2. Selection Of Water Cement Ratio :
Given :
Assume : M35
Exposure Condition : Extreme ( For Reinforced Concrete )
Table 5 - IS 456 : 2000
Therefore , Extreme = w/c Ratio = 0.40
M35 = w/c Ratio = 0.45
Lower The Water Cement Ratio Better The Workability
0.40 < 0.45
Water Cement Ratio = 0.40
Minimum Cement Content = 360 Kg/m^3
3. Selection Of Water Content :-
Assume : Nominal Size Of Aggregate = 20mm
Table 2 - IS 10262 : 2009
Minimum Water Content = 186 Litres Or Kg
Slump Ranges From 25 To 50 mm
As Superplasticizers ( 20 % Reduced )
= 186 X 80/100 = 148.8 Therefore , ( 100 - 20 = 80 )
Water Content = 149 Litres
4. Calculation Of Cemetitious Content ( Cement + Fly Ash )
Water Cement Ratio = w/c = 0.40
Water Content = W = 149 Kg
Cementitious Content = c = 149/0.40 = 372.5 = 373 Kg/m^3
Minimum Cement Content = 360 Kg/m^3
360 < 373 Kg/m^3 ( Hence Okay )
Increase By 10%
370 X 0.10 + 370 = 407 Kg/m^3
W = 149 Kg
Water Cement Ratio = 149/407 = 0.366
Fly Ash At 30% Total Cementitious Material Content
= 407 X 30/100 =122 Kg/m^3
Cement ( Ordinary Portland Cement ) = 407-122 = 285 Kg/m^3
Saving Of Cement While Using Fly Ash = 370 - 285 = 85 Kg/m^3
Fly Ash = 122 Kg/m^3
5. Volume Of Coarse Aggregate And Fine Aggregates
Fine Aggregate Are Seived According To Zone 1 Of Table 4 IS 383
Coarse Aggregate Are Sieved According To Table 2 IS 383
According ,
Table 3 IS 10262 : 2009
Nominal Size Of Aggregates = 20 mm
Fine Aggregates ( Zone 1 )
Therefore ,
Volume Of Coarse Aggregate = 0.60
Water Cement Ratio = 0.4 ( +/- 0.01 For Every +/- 0.005 )
So , Volume Of Coarse Aggregate = 0.62
Volume Of Fine Aggregate = 1-0.62 = 0.38
6. Mix Calculation :-
1. Volume Of Concrete = 1m^3
2. Volume Of Cement = ( 285/3.15) X 1/1000 = 0.090 m^3
3. Volume Of Fly Ash = ( 122/2.2 ) X 1/1000 = 0.055 m^3
4. Volume Of Water = ( 149/1 ) X 1/1000 = 0.149 m^3
5. Volume Of Admixture = ( 7.6/1.145 ) X 1/1000 =0.006 m^3
6. Volume Of Coarse Aggregate And Fine Aggregate = a - ( b + c + d + e ) = 1 - ( 0.090 + 0.055 + 0.149 +0.006 )
= 0.7 m^3
7. Mass Of Coarse Aggregate = 0.7 X 0.62 X 2.74 X 1000 = 1189.16 Kg/m^3
8. Mass Of Fine Aggregate = 0.7 X 0.38 X 2.74 X 1000 = 728 Kg/m^3
Results And Conclusion :-
Trial Mix :-
Cement = 285 Kg/m^3
Fly Ash = 122 Kg/m^3
Water = 149 Kg
Coarse Aggregate = 1189 Kg/m^3
Fine Aggregate = 728 Kg/m^3
Chemical Admixture = 7.6 Kg/m^3
Water Cement Ratio = 0.366
Trial Mix Ratio : 285 : 728 : 1189 = 1 : 2.55 : 4.17 ( C : FA : CA )
*This Trial Mix Ratio Should Match The Workability , Durability , Minimum Strength Of M35 Grade Concrete.
* It Is Does Not Satisfy The Above Criteria Two Or More Trials For Increase Of 10 % Should Be Done.
Aim :-
To Calculate M50 Grade Without Fly Ash
Equations :-
* Target Mean Strength : f'ck = fck + 1.65 s
f'ck = Target Mean Strength
fck = Compressive Strength At 28 Days Of Concrete
s = Standard Deviation ( IS 10262 : 2009 , Table -1 )
* Volume Of Cement : ( Mass Of Cement / Specific Gravity Of Cement ) X 1/1000
* Volume Of Water : ( Mass Of Water / Specific Gravity Of Water ) X 1/1000
* Volume Of Chemical Admixtures : ( Mass Of Chemical Admixture / Specific Gravity Of Chemical Admixture ) X
1/1000
* Volume Of Concrete : 1 m^3
* Volume Of Aggregate : ( Volume Of Concrete - ( Volume Of Cement + Volume Of Water + Volume Of Chemical
Admixture ))
* Mass Of Coarse Aggregate : Volume Of Aggregate X Volume Of Coarse Aggregate X Specific Gravity Of Coarse
Aggregate X 1000
* Mass Of Fine Aggregate : Volume Of Aggregate X Volume Of Fine Aggregate X Specific Gravity Of Fine
Aggregate X 1000
Codes For Concrete Mix Design :-
* IS 10262 : 2009
* IS 456 : 2000
* IS 8112
* IS 9103
* Is 3812 ( Part 1 )
Objectives :-
* The Main Objective Is To Produce Concrete As Economically As Possible
* To Achieve Desired Workability , Durability And Minimum Strength Of M50 Grade Concrete
Procedure And Calculation :-
1. Target Mean Strength
F'ck = fck + 1.65 s
fck = 50 N/mm^2 ( M50 )
Table 1-IS 10262:2009
s = 5 N/mm^2
f'ck = 50 + ( 1.65 X 5 )
= 58.25 N/mm^2
2. Selection Of Water Cement Ratio :-
Given :-
Assume : M50
Exposure Comdition : Extreme ( For Reinforced Concrete )
Table 5 - IS 456 : 2000
Therefore ,
Extreme = w/c Ratio = 0.40
M50 = w/c Ratio = 0.40
Lower The Water Cement Ratio Better The Workability 0.40
Water Cement Ratio = 0.40
Minimum Cement Content = 360 Kg/m^3
3. Selection Of Water Content :-
Assume : Nominal Size Of Aggregate = 20 mm
Table 2 - IS 10262 : 2009
minimum Water Content = 186 Litres Or Kg Slump Range From 25 To 50 mm
As Superplasticizers ( 20 % Reduced )
= 186 X 80/100 = 148.8 Therefore , ( 100 - 20 = 80 )
Water Content = 149 Litres
4. Calculation Of Cement Content :-
Water Cement Ratio = w/c = 0.40
Water Content = W = 149 Kg
Cement Content = c = 149/0.40 = 372.5 = 373 kg/m^3
Minimum Cement Content = 360 Kg/m^3
360<373 Kg/m^3 ( Hence Okay )
5. Volume Of Coarse Aggregates And Fine Aggregates
Fine Aggregates Are Sieved According To Zone 1 Of Table 4 Is 383
Coarse Aggregate Are Sieved According To Table 2 Is 383
According ,
Table 3 IS 10262:2009
Nominal Size Of Aggregate = 20 mm
Fine Aggregate ( Zone 1 )
Therefore ,
Volume Of Coarse Aggregate = 0.60
Water Cement Ratio = 0.4 ( +/- 0.01 For Every +/-0.005 )
So , Volume Of Coarse Aggregate = 0.62
Volume Of Fine Aggregate = 1 - 0.62 = 0.38
6. Mix Calculation :-
1. Volume Of Concrete = 1m^3
2. Volume Of Cement = ( 373/3.15 ) X 1/1000 = 0.118 m^3
3. Volume Of Water = ( 149/1 ) X 1/1000 = 0.149 m^3
4. Volume Of Admixture = ( 7.6/1.145 ) X 1/1000 = 0.006
5. Volume Of Coarse Aggregate And Fine Aggregate = a - ( b + c+ d ) = 1-( 0.118 + 0.149 + 0.006 ) = 0.727 m^3
6. Mass Of Coarse Aggregate = 0.727 X 0.62 X 2.74 X 1000 = 1235.02 Kg/m^3
7. Mass Of Fine Aggregate = 0.727 X 0.38 X 2.74 X 1000 = 756.9 Kg/m^3
Result And Conclusion :-
Trial Mix :-
Cement = 373 Kgm^3
Water = 149 Kg
Coarse Aggregate = 1235.02 Kg/m^3
Fine Aggregate = 756.9 Kg/m^3
Chemical Admixture = 7.6 Kg/m^3
Water Cement Ratio = 0.40
Trial Mix Ratio : 373 : 756.9 : 1235.02 = 1 : 2.02 : 3.31 ( C : FA : CA )
* This Trial Mix Ratio Should Match The Workability , Durability , Minimum Strength Of M50 Grade Concrete.
* If It Does Not Satisfy The Above Criteria Two Or More Trials For Increase Of 10 % Should Be Done.