Project on Concrete Mix Design for various grades of Concrete
- Concrete Mix design for M35 grade concrete with fly ash
- AIM:
To Calculate Concrete Mix design for M35 grade concrete with Fly ash-based content as per code provision of IS 10262:2009 & IS 456:2000
- PROCEDURE
Given Data:
- Grade of concrete – M35
- Type of mineral admixture – Fly ash conforming to IS 3812 (Part-1)
Assuming the following points:
- Exposure condition – Severe
- Size of aggregate – 20 mm
- Slump value - 100 mm
- Admixture – Superplasticizer
- Coarse aggregate conforming to Table 2 of IS 383
- Fine aggregate – Zone 1 Table 1 of IS 383
Step -1: Target mean strength for mix proportioning
F’ck = Fck + 1.65 S
Where, F’ck – Target mean compressive strength at 28 days
Fck – Characteristics compressive strength at 28 days
S – Standard deviation
From Table 1 of IS 10262:2009
Standard deviation S = 5.0 N/mm^2
Therefore, F’ck = 35 + (1.65 * 5)
= 43.25 N/mm^2
STEP-2: Selection water-cement ratio
From Table 5 of IS 456:2000
Maximum water-cement ratio – 0.45
Based on experience adopt water-cement ratio – 0.45
Hence ok.
STEP-3: Selection of water content:
From Table 2 of IS 10262:2009
Maximum water content for 20 mm aggregate = 186 KG for 25 to 50 mm slump
Here we take slump value is 100 mm
So, we have to increase 3 % water content for every 25 mm slump
Estimated water content for 100 mm slump = 186 + (186 * 6/100)
= 197 Lit
Here, we use Superplasticizer as Admixture. So, water content can be reduced up to 30 percent
Based on trials with super plasticizer water content reduction is 20 % is achieved
Hence the calculated water content = 197 – (197 * 20%)
= 157.6 Lit
STEP-4: Calculation of Cement and Fly ash content
Water-cement ratio = 0.45
Cementitious material content = 157.6/0.45 = 350 KG
Since Fly ash is used the amount of cementitious content shall be increased to get good workability and strength. The percentage of cementitious content to be increased depends on the experience and research. Here we have increased the cementitious content by 10%.
Final cementitious content = 350 * 1.10
= 385 Kg
From Table 5 of IS 456:2000
Minimum cement content for severe exposure – 320 Kg
Min. cementitious content < Adopted cementitious content
Hence OK.
Fly ash content requirement is 30% replacement from cementitious content
= 385 * 0.30
= 115.5 KG
Cement content = cementitious content – fly ash content
= 385 – 115.5
= 269.5 KG
STEP-5: Volume of Coarse aggregate and Fine aggregate
Fine aggregate – Zone 1
Water-cement ratio – 0.45
Size of aggregate – 20 mm
From Table 3 of IS 10262:2009
Volume of coarse aggregate corresponding to fine aggregate zone 1 for water-cement ratio is 0.5 = 0.6
The present water-cement ratio is - 0.45
Water-cement ratio is lower by 0.15 times
So, for every 0.05 difference in W/C ratio, add 0.01 of volume of coarse aggregate.
Therefore, the volume of coarse aggregate = 0.63
As the slump is 100 mm and concrete placing is used by pump
So, reduced by 10% from volume of coarse aggregate
Finally, Ratio Volume of coarse aggregate = 0.63 * 90% = 0.56
Ratio of Volume of fine aggregate = 1-0.56 = 0.44
STEP-6: Mix calculations:
- Volume of concrete = 1 m^3
- Volume of cement = Mass of cement/ (Specific gravity of cement * 1000 )
= 269.5 / ( 3.15*1000)
= 0.0856 Cu. M
3.Volume of Fly ash = Mass of Flyash/ (Specific gravity of Flyash * 1000 )
= 115.5 / (2.2*1000)
= 0.0525 Cu. M
4.Volume of water = Mass of Water/ (Specific gravity of Water * 1000 )
= 157.6/(1 *1000)
= 0.157 Cu. M
5.Volume of Superplasticizer = Mass of Superplasticizer/ (Specific gravity of Superplasticizer * 1000 )
= 7/(1.145 * 1000)
= 0.006 Cu. M
6.Total aggregate requirement = A – (B+C+D+E)
= 1 – (0.0856+0.0525+0.157+0.006)
= 0.6989 Cu. M
7.Coarse aggregate = Total aggregate * C.A ratio * Specific gravity * 1000
= 0.6989 * 0.56 * 2.74 * 1000
= 1072.39 Kg
8.Fine aggregate = Total aggregate * F.A ratio * Specific gravity * 1000
= 0.6989 * 0.44 * 2.74 * 1000
= 842.59 Kg
STEP-7: Mix proportions:
CEMENT - 269.5 Kg
FLY ASH - 115.5 Kg
FINE AGGREGATE - 842.6 Kg
COARSE AGGREGATE - 1072.4 Kg
WATER CONTENT - 157.6 Lit
WATER-CEMENT RATIO - 0.45
SUPERPLASTICIZER - 7 Kg/m^3
- RESULT:
Therefore, Mix ratio of M35 grade concrete with fly ash based as per IS code is,
Cement: Fly ash: Fine aggregate: Coarse aggregate
1: 0.43: 3.13: 3.98
- Concrete Mix design for M50 grade concrete without Fly ash?
- AIM:
To calculate concrete mix design for M50 grade concrete as per IS 10262:2009 & IS 456:2000
- PROCEDURE:
Given data:
- Grade of concrete = M50
Assume the following data:
- Exposure condition – Moderate
- Size of aggregate – 20 mm
- Slump value – 100 mm
- Admixture - superplasticizer
- Coarse aggregate conforming of Table 2 IS 383
- Fine aggregate Zone 1 conforming of Table 1 IS 383
- STEP-1: Target mean strength for mix proportioning
F’ck = Fck + 1.65 S
Where, F’ck – Target mean compressive strength at 28 days
Fck – Characteristics compressive strength at 28 days
S – Standard deviation
From Table 1 of IS 10262:2009
Standard deviation S = 5.0 N/mm^2
Therefore, F’ck = 50 + (1.65 * 5)
= 58.25 N/mm^2
- STEP-2: Selection of water-cement ratio:
From Table 5 of IS 456:2000
Maximum water-cement ratio – 0.5
Based on experience adopt water-cement ratio – 0.45
Hence ok.
- STEP-3: Selection of water content:
From Table 2 of IS 10262:2009
Maximum water content for 20 mm aggregate = 186 KG for 25 to 50 mm slump
Here we take slump value is 100 mm
So, we have to increase 3 % water content for every 25 mm slump
Estimated water content for 100 mm slump = 186 + (186 * 6/100)
= 197 Lit
Here, we use Superplasticizer as Admixture. So, water content can be reduced up to 30 percent
Based on trials with super plasticizer water content reduction is 20 % is achieved
Hence the calculated water content = 197 – (197 * 20%)
= 157.6 Lit
- STEP-4: Calculation of Cement content
Water-cement ratio = 0.45
Cement content = 157.6/0.45 = 350 KG
From Table 5 of IS 456:2000
Minimum cement content for moderate exposure – 300 Kg
Min. cement content < Adopted cement content
Hence OK.
Calculation of super plasticizer = 7 Kg/m^3
STEP-5: Volume of Coarse aggregate and Fine aggregate
Fine aggregate – Zone 1
Water-cement ratio – 0.45
Size of aggregate – 20 mm
From Table 3 of IS 10262:2009
Volume of coarse aggregate corresponding to fine aggregate zone 1 for water-cement ratio is 0.5 = 0.6
The present water-cement ratio is - 0.45
Water-cement ratio is lower by 0.15 times
So, for every 0.05 difference in W/C ratio, add 0.01 of volume of coarse aggregate.
Therefore, the volume of coarse aggregate = 0.63
As the slump is 100 mm and concrete placing is used by pump
So, reduced by 10% from volume of coarse aggregate
Finally, Ratio of Volume of coarse aggregate = 0.63 * 90% = 0.56
Ratio of Volume of fine aggregate = 1-0.56 = 0.44
STEP-6: Mix calculations:
- Volume of concrete = 1 m^3
- Volume of cement = Mass of cement/ (Specific gravity of cement * 1000 )
= 350/ (3.15 * 1000)
= 0.111 Cu. M
3. Volume of water = Mass of water/ (Specific gravity of water * 1000 )
= 157.6 / ( 1* 1000)
= 0.157 Cu. M
4. Volume of Superplasticizer = Mass of superplasticizer/ (Specific gravity of superplasticizer * 1000 )
= 7 / ( 1.145 * 1000)
= 0.006 Cu. M
5.Total aggregate requirement = A – (B+C+D)
= 1 – (0.111+0.157+0.006)
= 0.726 Cu. M
6. Coarse aggregate = Total aggregate * C.A ratio * Specific gravity * 1000
= 0.726 * 0.56 * 2.74 * 1000
= 1114 Kg
7. Fine aggregate = Total aggregate * F.A ratio * Specific gravity * 1000
= 0.726 * 0.44 * 2.74 * 1000
= 875.26 Kg
STEP-7: Mix proportions:
CEMENT - 350 Kg
FINE AGGREGATE - 875.26 Kg
COARSE AGGREGATE - 1114 Kg
WATER CONTENT - 157.6 Lit
WATER-CEMENT RATIO - 0.45
SUPERPLASTICIZER - 7 Kg/m^3
- RESULT:
Therefore, Mix ratio of M50 grade concrete without fly ash based as per IS code is,
Cement: Fine aggregate: Coarse aggregate
1: 2.5: 3.18