Project on Concrete Mix Design for various grades of Concrete
Calculate the Concrete Mix Design for M35 grade concrete with fly ash & M50 grade concrete without Fly ash
M35 grade concrete with fly ash
1)target mean strength
Ft= Fck+1.65s
Ft=target mean strength
Fck=charectarestic strength of concrete after 28days in this case 35N/mm²
S=standard deviation according to is 456 2000 table no 8 s is taken as 5N/mm² considaring good supervision on site
Ft=35+1.65×5
Ft=43.25N/mm²
2)water cement ratio
Considaring weather exposure or exposure condition as a severe according to the is 456 2000 table no 5 maximum free water cement ratio is 0.45 and minimum cement content is 320kg/m³
According to is 10262 1982 from fig no 1 for target strength at 28 day 43.25N/mm² free w/c is 0.34
According to is 10262 1982 from fig no 2 for target strength at 28 day 43.25N/mm² free w/c is 0.345
So select w/c as 0.34 <0.45 hence ok
3) SELECTION OF WATER CONTENT
As per is 10262 2019 table no 2 assume nominal size of aggregate is 20mm and aggregate is saturated surface dry condition Maximum Water Content per Cubic Metre of Concrete for Nominal Maximum Size of Aggregate is 186kg/m³
For 100 mm slumpe value water content can be increased by 3% for every increase in 25mm slumpe so for 100mm slumpe value
New Water content= 186+6/100×186=197.16kg/m³
We are using superplastisizer as well so we can decrease water content by 25%
New final water content= 197.16×0.75=147.87kg/m³.
4)CALCULATION OF CEMENT AND FLY ASH CONTENT
water cement ratio=0.34
Quantity of water/Quantity of cement+fly ash =0.34
Quantity of cement+fly ash=147.87/0.34=434.91kg/m³
434.91kg/m³>320kg/m³ hence ok
Assume Fly ash @ 30% of total cementitious material content= 434.91×0.3 =130.47kg/m³
So cement content is = 434.91-130.47=304.64kg/m³
5) PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE CONTENT
Assuming fine aggregate is belongs to zone 2 and from table no 3 of is 10262 table no 3 Volume of Coarse Aggregate per Unit Volume of Total Aggregate for Different Zones of Fine Aggregate
Assuming coarse aggregate is saturated surface dry condition and concrete is pumpable concrete so from clause 4.4.2 of is 10262 2009 quantity of coarse aggregate determine using table no 3 can be reduced upto 10% in our case from experience we will reduced quantity of coarse aggregate by 7%.
From table no 3 Volume of Coarse Aggregate per Unit Volume of Total Aggregate for water cement ratio 0.5 =0.64
In the present case water-cement ratio is 0.34. Therefore. volume of coarse aggregate is required to be increased to decrease the fine aggregate content. As the water-cement ratio is lower by 0.16. the proportion of volume of coarse aggregate is increased by 0.032 (at the rate of -/+ 0.01 for every ± 0.05 change in water-cement ratio) , Therefore. corrected proportion of volume of coarse aggregate for the water-cement ratio of 0.34 = 0.672
volume of coarse aggregate = 0.672×0.93=0.625
Volume of fine aggregate= 1-0.625=0.375
MIX CALCULATIONS
The mix calculations 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 ×1/1000=304.64/3.14×1/1000
=0.097m³
c) volume of fly ash= mass of fly ash/ specific gravity of fly ash ×1/1000
= 130.47/2.21×1/1000 =0.059m³
d)Volume of water= mass of water/specific gravity of water ×1/1000
= 147.87/1 ×1/1000=0.148m³
e)Volume of chemical admixture (superplasticizer) @ 2.0% by mass of cementitious material)
= Mass of admixture/ specific gravity of admixture×1/1000 = 8.69/1.145×1/1000=0.0075m³
f)Volume of all in aggregate= 1-[b+c+d+e]
=1-(0.097+0.059+0.0148+0.0075)
=0.8217m³
g)Mass of coarse aggregate=f x volume of coarse aggregate x Specific gravity of coarse aggregate × 1000
= 0.8217×0.625×2.75×1000 =1412kg/m³
h)Mass of fine aggregate= f×volume of fine aggregate ×specific gravity of fine aggregate ×1000
=0.8217×0.375×2.8×1000 =862.78kg/m³
M50 grade concrete without Fly ash
1)target mean strength
Ft= Fck+1.65s
Ft=target mean strength
Fck=charectarestic strength of concrete after 28days in this case 35N/mm²
S=standard deviation according to is 456 2000 table no 8 s is taken as 5N/mm² considaring good supervision on site
Ft=50+1.65×5
Ft=58.25N/mm²
2)water cement ratio
Considaring weather exposure or exposure condition as a very severe according to the is 456 2000 table no 5 maximum free water cement ratio is 0.45 and minimum cement content is 340kg/m³
According to is 10262 1982 from fig no 1 for target strength at 28 day 58.25N/mm² free w/c is 0.30
According to is 10262 1982 from fig no 2 for target strength at 28 day 58.25N/mm² free w/c is 0.32
So select w/c as 0.30 <0.45 hence ok
3) SELECTION OF WATER CONTENT
As per is 10262 2019 table no 2 assume nominal size of aggregate is 20mm and aggregate is saturated surface dry condition Maximum Water Content per Cubic Metre of Concrete for Nominal Maximum Size of Aggregate is 186kg/m³
For 150 mm slumpe value water content can be increased by 3% for every increase in 25mm slumpe so for 150mm slumpe value
New Water content= 186+12/100×186=208.32kg/m³
We are using superplastisizer as well so we can decrease water content by 25%
New final water content= 208.32×0.75=156.24kg/m³.
4)CALCULATION OF CEMENT
Mass of water/mass of cement =0.3
156.24/0.3 = mass of cement
Mass of cement= 520.8 kg/m³ >340kg/m³ hence ok
5) PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE CONTENT
Assuming fine aggregate is belongs to zone 2 and from table no 3 of is 10262 table no 3 Volume of Coarse Aggregate per Unit Volume of Total Aggregate for Different Zones of Fine Aggregate
Assuming coarse aggregate is saturated surface dry condition and concrete is pumpable concrete so from clause 4.4.2 of is 10262 2009 quantity of coarse aggregate determine using table no 3 can be reduced upto 10% in our case from experience we will reduced quantity of coarse aggregate by 10%.
From table no 3 Volume of Coarse Aggregate per Unit Volume of Total Aggregate for water cement ratio 0.5 =0.64
In the present case water-cement ratio is 0.30. Therefore. volume of coarse aggregate is required to be increased to decrease the fine aggregate content. As the water-cement ratio is lower by 0.20. the proportion of volume of coarse aggregate is increased by 0.04 (at the rate of -/+ 0.01 for every ± 0.05 change in water-cement ratio) , Therefore. corrected proportion of volume of coarse aggregate for the water-cement ratio of 0.30 = 0.68
volume of coarse aggregate = 0.68×0.90=0.612
Volume of fine aggregate= 1-0.612=0.388
MIX CALCULATIONS
The mix calculations 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 ×1/1000=520.8/3.14×1/1000 =0.1658
c)Volume of water= mass of water/specific gravity of water ×1/1000
= 156.24/1 ×1/1000=0.156m³
d)Volume of chemical admixture (superplasticizer) @ 2.0% by mass of cementitious material)
= Mass of admixture/ specific gravity of admixture×1/1000 = 10.42/1.145×1/1000=0.0090m³
e)Volume of all in aggregate= 1-[b+c+d]
= 1-[0.1658+0.156+0.0090] =0.6692m³
f)Mass of coarse aggregate=ex volume of coarse aggregate x Specific gravity of coarse aggregate × 1000
= 0.6692×0.612×2.75×1000 =1126.26kg/m³
g)Mass of fine aggregate= f×volume of fine aggregate ×specific gravity of fine aggregate ×1000
=0.6692×0.388×2.8×1000 =727kg/m³