Project on Concrete Mix Design for various grades of Concrete
PROJECT ON CONCRETE MIX DESIGN FOR VARIOUS GRADES OF CONCRETE
DATA REQUIRED FOR CONCRET MIX DESIGN
- Grade of concrete
- Maximum size of aggregate
- Minimum cement content
- Maximum w/c ratio
- Workability in terms of slump
- Exposure condition
- Maximum temperature at the pouring point
- Early age strength
- Grading zone of fine aggregate
- Type of aggregate
- Maximum cement content
- Admixture used
- Specific gravity of all the material
- Moisture content
AIM :-
To Calculate the Concrete Mix Design for M35 grade concrete with fly ash & M50 grade concrete without Fly ash
INTRODUCTION :-
Concrete mix design is the method of finding the precise proportion of cement, sand and aggregates for concrete to achieve the target strength in structures, therefore concrete mix design can be stated as concrete mix in terms of cement: sand: aggregates. The design is done according to the requirements of concrete strength. So, we can achieve the desirable properties of concrete either it is in fresh stage or in hardened stage.The fresh concrete properties like workability, setting time and hardened concrete properties like compressive strength, durability etc. are attained surely by this method. Use of additives like admixtures, retarders etc. other than basic ingredients are used to improve the properties of mix. This project consist of two parts, the first part deals with the design of M35 grade cement with fly ash as a part of it and second part is the design of M50 grade concrete without fly ash.
OBJECTIVE :-
The objective of this project is to produce a mix design for M35 grade concrete containing fly ash and M50 grade concrete. The mix will be prepared as per IS METHOD: Concrete mix proportioning guidelines from IS 10262:2009 and recommented guide lines from IS 10262:1982
EQUATIONS USED :-
A ) Targetmean strength
F*t = fck + 1.65 S
Where F*t = Target average compressive strength of concrete at 28 days
Fck = compressive strength of concrete at 28 days
S = standard deviation takes from table 8 from IS 456 – 2000
B ) CEMENT CONTENT
Cement content = water concrete / water-cement ratio
C ) VOLUME CALCULATION
- Volume of concrete
- 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 all in aggregate = [a –(b+c+d)]
PART : 1 To Calculate the Concrete Mix Design for M35 grade concrete with fly ash
STEP 1 :- DATA REQUIRED
GRADE OF CONCRETE M35
MAXIMUM AGGREGATE SIZE 20mm
MINIMUM CEMENT CONTENT 320kg/m^3 ( TABLE 5 OF IS 456-2000 )
MAXIMUM W/C RATIO 0.45 ( TABLE 5 OF IS 456-2000 )
WORKABILITY IN TERMS OF SLUMP 100mm
EXPOSURE CONDITION SEVER ( for reinforced concrete )
GRADING ZONE OF FINE AGGREGATE ZONE II
TYPE OF AGGREGATE CRUSHED ANGULAR AGGREGATE
MAXIMUM CEMENT CONTENT 450 kg/m^3
ADMIXTURE USED SUPER PLASTICIZER
MAXIMUM WATER CONTENT FOR 20mm AGGREGATE 186 LITRES
TYPE OF CEMENT OPC 43
STEP 2 :- TEST DATA FOR MATERIALS
SPECIFIC GRAVITY OF CEMENT 3.15
SPECIFIC GRAVITY OF FLY ASH 2.2
SPECIFIC GRAVITY OF COARSE AGGREGATE 2.74
SPECIFIC GRAVITY OF FINE AGGREGATE 2.74
WATER ABSORPTION OF COARSE AGGREGATE 0.5%
WATER ABSORPYION OF FINE AGGREGATE 10%
STEP 3 :- TARGET MEAN STRENGTH FOR MIXING PROPORTION
TARGET MEAN STRENGTH
Ft = Fck + 1.65 S
Where Ft = target average compressive strength at 28 days
Fck = compressive strength at 28 days = 35 N/mm^2
S = standard deviation = 5 N/mm^2 ( takes from table 1 of 10262:2009)
Ft = 35 + 1.65 x 5 = 43.25 N/mm^2
Target mean strength = 43.25 N/mm^2
STEP 4 :- SELECTION OF WATER-CEMENT RATIO
THE MAXIMUM WATER-CEMENT RATIO TO DESIGN M35 = 0.43
ADOPTED MAXIMUM WATER-CEMENT RATIO = 0.41 (FROM TABLE 5 OF IS 456:2000)
ADOPTED W/C RATIO = 0.41
STEP 5 :- SELECTION OF WATER CONTENT
FROM TABLE 2 OF IS 10262:2009
MAXIMUM WATER CONTENT FOR 20mm AGGREGATE = 186 LITRES ( for 25mm to 50mm slump range )
CLAUSE 4.2 SUGGESTS INCREASING OF 3% OF WATER CONTENT FOR EVERY EXTRA 25mm SLUMP. SO 6% IS TO BE ADDED TO IT ESTIMATE WATER CONTENT FOR 10mm SLUMP
186 + (6/100) X 186 = 197.16 LITRES
BASED ON TRIALS WITH SUPER PLASTICIZER WATER CONTENT REDUCTION OF 20% HAS BEEN ACHIEVED
[NOTE :- IF SUPER PLASTICIZER IS USED , THE WATER CONTENT CAN BE REDUCED UPTO 20% ]
HENCE THE ARRIVED WATER CONTENT
197.16 X 0.8 = 157.72 LITRES
WATER REQUIRED = 158 LITRES
STEP 6:- CALCULATION OF CEMENT AND FLY ASH
ADOPTED W/C RATIO = 0.44
TOTAL CEMENTIOUS REQUIREMENT = WATER CONTENT / WATER-CEMENT RATIO
= 158/0.41
= 385.36 Kg/m^3
TOTAL CEMENTIOUS REQUIRED REQUIREMENT = 385 Kg/m^3
- The Cementious content has to be increased in order to attain good workability and strength
- The increment required is determined on the basis of experience and research
Here we are increasing cementious content by 10%
FINAL CEMENTIOUS CONTENT = 385 X 1.10 = 424 kg/m^3
From Table 5 of IS 456:2000
Minimum cement content for sieve exposure conditions = 320kg
424 kg/m^3 > 320 kg/m^3 hence ok
Cementious material content = 385 kg/m^3
New water cement ratio = 158/385 = 0.4
Fly ash content is taken as 30% of the total cementious content = 385 x 0.3 = 115.5 kg/m^3
Cement required = 385 – 115 = 269.5
Cement quantity saved due to fly ash addition = 350 – 269 = 81 kg/m^3
CEMENTIOUS MATERIAL = 385 Kg/m^3
CEMENT = 269 Kg/m^3
Fly ash = 115.5 kg /m^3
STEP 7 :- VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE CONTENT
From table-3 of ( IS 10262:2009) volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate for water-cement ratio of 0.50 is 0.62
In the present case water-cement ratio is 0.44 . therefore ,volume of coarse aggregates is required to be increased to decrease. The fine aggregate content as the water-cement ratio is lower by 0.06 .
The proportion of volume of coarse aggregate is increased by 0.02 ( at the rate of -/+ 0.01 for every -/+ 0.5 change in water-cement ratio )
Therefore , corrected proportion of volume of coarse aggregate for the water-cement ratio of 0.44 = 0.632
For pumpable concrete these values should be reduced upto 10%
Therefore , VOLUME OF COARSE AGGREGATE = 0.632 x 0.9
= 0.568
VOLUME OF FINE AGGREGATE = 1 - 0.568
= 0.431
STEP 8 :- MIX CALCULATION
The mix calculations as per unit volume of concrete shall be as follows :
a) Volume of concrete = 1 m3
b)Volume of cement = (mass of cement )/(specific gravity of cement) x 1/1000
= (269/3.15) x 1/1000
= 0.085 m3
c)Volume of fly ash = (mass of fly ash)/(specific gravity of fly ash) x 1/1000
= (115/2.2) x 1/1000
= 0.062 m3
d)Volume of water = (mass of water)/(specific gravity of water) x 1/1000
= (158/1) x 1/1000
= 0.158 m3
e)Volume of chemical admixture = (mass of admixture)/(specific gravity of admixture)
X 1/1000
= (7.6/1.145) x 1/1000
= 0.006 m3
f)Volume of all in aggregate = [a-(b+c+d+e)]
= [ 1-(0.085+0.062+0.158+0.006)]
= 0.699 m3
g)Mass of coarse aggregate = f x volume of coarse aggregate x specific gravity of fine
Aggregate x 1000
= 0.699 x 0.568 x 2.74 x 1000
= 1077 kg/m3
h)Mass of fine aggregate = f x volume of fine aggregate x specific gravity of fine
Aggregate x 1000
= 0.699 x 0.431 x 2.74 x 1000
= 822 kg/m3
STEP 9 :- MIX PROPORTION
Cement = 269 kg/m3
Fly ash = 115 kg/m3
Water = 158 kg/m3
Fine aggregate = 822 kg/m3
Coarse aggregate 20mm = 1077 kg/m3
Chemical admixture = 7.6 kg/m3
Water cement ratio = 0.4
Mix proportion by weight = 1: 2.08:2.72
PART 2 :- To Calculate the Concrete Mix Design for M50 grade concrete without fly ash
STEP 1 :- DATA REQUIRED
GRADE OF CONCRETE M50
MAXIMUM AGGREGATE SIZE 20mm
MINIMUM CEMENT CONTENT 320kg/m^3 ( TABLE 5 OF IS 456-2000 )
MAXIMUM W/C RATIO 0.45 ( TABLE 5 OF IS 456-2000 )
WORKABILITY IN TERMS OF SLUMP 100mm
EXPOSURE CONDITION SEVER ( for reinforced concrete )
GRADING ZONE OF FINE AGGREGATE ZONE II
TYPE OF AGGREGATE CRUSHED ANGULAR AGGREGATE
MAXIMUM CEMENT CONTENT 450 kg/m^3
ADMIXTURE USED SUPER PLASTICIZER
MAXIMUM WATER CONTENT FOR 20mm AGGREGATE 186 LITRES
TYPE OF CEMENT OPC 53
STEP 2 :- SPECIFIC GRAVITY OF CEMENT
SPECIFIC GRAVITY OF CEMENT 3.15
SPECIFIC GRAVITY OF FLY ASH 2.2
SPECIFIC GRAVITY OF COARSE AGGREGATE 2.74
SPECIFIC GRAVITY OF FINE AGGREGATE 2.74
WATER ABSORPTION OF COARSE AGGREGATE 0.5%
WATER ABSORPYION OF FINE AGGREGATE 10%
STEP 3 :- TARGET MEAN STRENGTH FOR MIXING PROPORTION
TARGET MEAN STRENGTH
Ft = Fck + 1.65 S
Where Ft = target average compressive strength at 28 days
Fck = compressive strength at 28 days = 35 N/mm^2
S = standard deviation = 5 N/mm^2 ( takes from table 1 of 10262:2009)
Ft = 50 + 1.65 x 5 = 58.25 N/mm^2
Target mean strength = 58.25 N/mm^2
STEP 4 :- SELECTION OF WATER-CEMENT RATIO
THE MAXIMUM WATER-CEMENT RATIO TO DESIGN M50 = 0.45
ADOPTED MAXIMUM WATER-CEMENT RATIO = 0.44 (FROM TABLE 5 OF IS 456:2000)
ADOPTED W/C RATIO = 0.44
STEP 5 :- SELECTION OF WATER CONTENT
FROM TABLE 2 OF IS 10262:2009
MAXIMUM WATER CONTENT FOR 20mm AGGREGATE = 186 LITRES ( for 25mm to 50mm slump range )
CLAUSE 4.2 SUGGESTS INCREASING OF 3% OF WATER CONTENT FOR EVERY EXTRA 25mm SLUMP. SO 6% IS TO BE ADDED TO IT ESTIMATE WATER CONTENT FOR 10mm SLUMP
186 + (6/100) X 186 = 197.16 LITRES
BASED ON TRIALS WITH SUPER PLASTICIZER WATER CONTENT REDUCTION OF 20% HAS BEEN ACHIEVED
[NOTE :- IF SUPER PLASTICIZER IS USED , THE WATER CONTENT CAN BE REDUCED UPTO 20% ]
HENCE THE ARRIVED WATER CONTENT
197.16 X 0.8 = 157.72 LITRES
WATER REQUIRED = 158 LITRES
STEP 6:- CALCULATION OF CEMENT
ADOPTED W/C RATIO = 0.44
TOTAL CEMENTIOUS REQUIREMENT = WATER CONTENT / WATER-CEMENT RATIO
= 158/0.44
= 359Kg/m^3
From table 5 of IS 456, minimum cement content for sever exposure condition is 320 kg/m3
359 kg/m3 > 320kg/m3 hence ok
STEP 7 :- VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE CONTENT
From table-3 of ( IS 10262:2009) volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate for water-cement ratio of 0.50 is 0.62
In the present case water-cement ratio is 0.44 . therefore ,volume of coarse aggregates is required to be increased to decrease. The fine aggregate content as the water-cement ratio is lower by 0.06 .
The proportion of volume of coarse aggregate is increased by 0.02 ( at the rate of -/+ 0.01 for every -/+ 0.5 change in water-cement ratio )
Therefore , corrected proportion of volume of coarse aggregate for the water-cement ratio of 0.44 = 0.632
For pumpable concrete these values should be reduced upto 10%
Therefore , VOLUME OF COARSE AGGREGATE = 0.632 x 0.9
= 0.568
VOLUME OF FINE AGGREGATE = 1 - 0.568
= 0.431
STEP 8 :- MIX CALCULATIONS
The mix calculations as per unit volume of concrete shall be as follows :
a)Volume of concrete = 1 m3
b)Volume of cement = (mass of cement )/(specific gravity of cement) x 1/1000
= (359/3.15) x 1/1000
= 0.113 m3
c)Volume of water = (mass of water)/(specific gravity of water) x 1/1000
= (158/1) x 1/1000
= 0.158 m3
d)Volume of chemical admixture = (mass of admixture)/(specific gravity of admixture)
X 1/1000
= (7.6/1.145) x 1/1000
= 0.006 m3
e)Volume of all in aggregate = [a-(b+c+d)]
= [ 1-(0.113+0.158+0.006)]
= 0.723 m3
f)Mass of coarse aggregate = f x volume of coarse aggregate x specific gravity of fine
Aggregate x 1000
= 0.723 x 0.568 x 2.74 x 1000
= 1125kg/m3
g)Mass of fine aggregate = f x volume of fine aggregate x specific gravity of fine
Aggregate x 1000
= 0.723x 0.431 x 2.74 x 1000
= 853.81kg/m3
STEP 9 :- MIX PROPORTION
Cement = 359 kg/m3
Water = 158 kg/m3
Fine aggregate = 853.81 kg/m3
Coarse aggregate 20mm = 1125 kg/m3
Chemical admixture = 7.6 kg/m3
Water cement ratio = 0.44
Mix proportion by weight = 1: 2.28:3.12
RESULTS
MIX PROPORTION BY WEIGHT OF M35 GRADE OF CONCRETE WITH FLY ASH 1:2.08:2.72 AND ITS WATER-CEMENT RATIO WAS TO FOUND TO BE 0.4
MIX PROPORTION BY WEIGHT OF M50 GRADE OF CONCRETE WITHOUT FLY ASH 1:2.28:3.12 AND ITS WATER-CEMENT RATIO TO BE FOUND 0.44
I CAN RECTIFING MY DOUBTS FROM PREVIOUS PROJECTS