Project on Concrete Mix Design for various grades of Concrete
Introduction:Mix Design of Concrete is something where we find out the quantities of it's ingredients.It's ingredients comprise commonly the cement and aggregates.And we even find the quantity of any admixture if used along with the flyash if used.And to ultimately represent a ratio of cement to aggregates i.e.,cement:fine aggregate:coarse aggregate.So,this Mix Design is done before the construction and this Mix Design is tested by preparing the concrete cubes of 150 mm dimension and tested under the UTM.In such a way,the desired strength is checked.And if the desired strength is gained,the same Mix Design is usen for the construction process.If the required strength is not gained,the Mix Design is again carried out by changing the materials quantities and tried further.
M35 Grade Concrete by using some amount of Flyash.
Aim:To determine the required quantities of the materials to produce M35 grade concrete using the Mix Design.
Data Collection:Certain data is collected and used,we have nothing but just the grade of Concrete,so we assume the remaining data.
Grade of Concrete=M35
Cement=OPC 43
Nominal Size of Coarse Aggregate=20 mm
Workability=100 mm
Exposure=Severe
Method of Placing=Pumping
Degree of Supervision=Good
Admixture=Superplasticizer
Specific Gravity of Aggregates=2.74
Specific Gravity of Flyash=2.20
Zone of Fine Aggregate=Zone 2
Flyash Content=30% of cementious material
Specific Gravity of Admixture=1.145
Quantity of Admixture=2% of cementious material
Type of Concrete=Reinforced
Step 1:To calculate the Target Mean Strength of Concrete
Here,we determine the target mean strength of the concrete by using the formula present on the page no.1 of IS 10262:2009.
the formula be,
f'ck=fck+1.65s
where,
f'ck=target mean strength of the concrete
fck=compressive strength of concrete
s=standard deviation of the concrete present on Table 1 on page no.2 of IS 10262:2009
Then,
fck=35 N/(mm^2)
s=5 N/(mm^2)
So,
f'ck=35+1.65(5)=35+8.25=43.25 N/(mm^2)
Step 2:To determine the w/c ratio of Concrete
Here,we determine the w/c ratio of the concrete where we refer the Table 5 present on page no.20 of IS 456:2000.
from the above mentioned table,for reinforced concrete for severe exposure,
the
Minimum Cement Content=320 Kg/(m^3)
Maximum w/c ratio=0.45
So,we use the w/c ratio of 0.40,as lower the w/c ratio,the higher the strength of the concrete.
And,Minimum grade of concrete=M30
where,we use M35>M30
So,this condition is statisfied!
Step 3:To calculate the Water content required for Concrete Mixing
Here,we calculate the water content required for Concrete Mixing,which is done by rusing Table 2 present on page no.3 of IS 10262:2009,
from the above mentioned table,for nominal aggregate size of 20 mm,
the,
Maximum Water Content=186 Kg/(m^3)
but this is for workability of concrete of 50 mm,but we use the concrete of the workabilty of 100 mm,so,we have to increase the water content,this increment is done at a rate of 3% rise for every increase in 25 mm slump,so we have to increase the water content by 6% for the 50 mm rise of the slump value.
Now,
Water Content=186+(6/100)186=197.16 Kg/(m^3)
But we will using a admixture i.e.,Super Plasticizer,so we reduce the required water content by 29%
then,
Water Content=197.16-(29/100)197.16=140 Kg/(m^3)
Step 4:To calculate the Cement content of the Concrete
Here,we calculate the cement content required for the concrete,where we use the w/c ratio and water content.
w/c ratio=0.4
140/cement=0.4 => cement=140/0.4=350 Kg/(m^3) > 320 Kg/(m^3)
So,the condition is stastisfied.
As,we use flyash,we increase the cementious material by 10%.
So,cementious material=350+(10/100)350=385 Kg/(m^3)
where,we replace the 30% of the total cementious material with flyash,then
Flyash content=(30/100)*385=115 Kg/(m^3)
Cement Content=385-115=270 Kg/(m^3)
Step 5:To calculate the Admixture content
Here,we calculate the quantity of the Admixture which is equal to the 2% of the total cementious material.
Then,
Admixture Content=(2/100)* 385=7.7 Kg/(m^3)
Step 6:To calculate the Fine Aggregate and Coarse Aggregate content
Here,we calculate the quantities of both the aggregates i.e., fine aggregate quantity and coarse aggregate quantity by using the Table 3 present on page no.3 of IS 10262:2009
for the nominal size of the coarse aggregate of 20 mm for a zone 2 fine aggregate,
the coarse aggregate proportion with respective to the total volume of aggregates be,
COARSE AGGREGATE=0.62
but the above value is for w/c ratio=0.50,but we have w/c ratio=0.40 which means,as the w/c ratio decrease the cement content increase so,we have to increase the coarse aggregate content with repect to the total volume of the aggregates,which is done at a rate of 0.02 hike in 0.10 fall of w/c ratio.
So,COARSE AGGREGATE=0.62+0.02=0.64
But we are about to place the concrete by pumping,so we reduce the coarse aggregate content by 10% to facilitate the concrete for pumping and easy passage of the concrete through pipe.
COARSE AGGREGATE=0.64-(10/100)0.64=0.58
then,FINE AGGREGATE=1-0.58=0.42
Step 7:To determine the Mix Proportion of all the required ingredients
Here,we determine the quantities in terms of volume of all the materials for one cubic metre of concrete.
Let,volume of concrete=1 m^3
Now,
Volume of Cement=(Mass of Cement/Specific Gravity of Cement)*(1/1000)=(270/3.15)*(1/1000)=0.085 m^3
Volume of Flyash=(Mass of Flyash/Specific Gravity of Flyash)*(1/1000)=(115/2.2)*(1/1000)=0.052 m^3
Volume of Water=(Mass of Water/Speific Gravity of Water)*(1/1000)=(140/1)*(1/1000)=0.140 m^3
Volume of Admixture=(Mass of Admixture/Specific Gravity of Admixture)*(1/1000)=(7.7/1.145)*(1/1000)=0.006 m^3
Volume of Aggregate=1-Volume of Cement-Volume of Flyash-Volume of Water-Volume of Admixture=1-0.085-0.052-0.140-0.006=0.717 m^3
Then,
Mass of Coarse Aggregate=Volume of Total Aggregates*Proportion of Coarse Aggregates with respective to Total Aggregates Volume*Specific Gravity of Coarse Aggregate*1000=0.717*0.58*2.74*1000=1140 Kg/(m^3)
Mass of Fine Aggregate=Volume of Total Aggregates*Proportion of Fine Aggregates with respective to Total Aggregates Volume*Specific Gravity of Fine Aggregate*1000=0.717*0.42*2.74*1000=826 Kg/(m^3)
Step 8:To represent the Mix Ratio of the Concrete
Here,we represent the Mix Contents in Mix Ratio.
So,the contents for one cubic metre of the concrete are,
Cement=270 Kg/(m^3)
Fine Aggregate=826 Kg/(m^3)
Coarse Aggregate=1140 Kg/(m^3)
Flyash=115 Kg/(m^3)
Admixture=7.7 Kg/(m^3)
Water=140 Kg/(m^3)
Now,
Cement:Fine Aggregate:Coarse Aggregate=385:826:1140=1:2.14:2.96
M50 Grade Concrete without using Flyash.
Aim:To determine the required quantities of the materials to produce M50 grade concrete using the Mix Design.
Data Collection:Certain data is collected and used,we have nothing but just the grade of Concrete,so we assume the remaining data.
Grade of Concrete=M50
Cement=OPC 43
Nominal Size of Coarse Aggregate=20 mm
Workability=100 mm
Exposure=Extreme
Method of Placing=Pumping
Degree of Supervision=Good
Admixture=Superplasticizer
Specific Gravity of Aggregates=2.74
Specific Gravity of Flyash=2.20
Zone of Fine Aggregate=Zone 2
Specific Gravity of Admixture=1.145
Quantity of Admixture=2% of cementious material
Type of Concrete=Reinforced
Step 1:To calculate the Target Mean Strength of the Concrete
Here,we determine the target mean strength of the concrete by using the formula present on the page no.1 of IS 10262:2009.
the formula be,
f'ck=fck+1.65s
where,
f'ck=target mean strength of the concrete
fck=compressive strength of concrete
s=standard deviation of the concrete present on Table 1 on page no.2 of IS 10262:2009
Then,
f'ck=50+1.65(5)=50+8.25=58.25 N/(mm^2)
Step 2:To determine the w/c ratio of Concrete
Here,we determine the w/c ratio of the concrete where we refer the Table 5 present on page no.20 of IS 456:2000.
from the above mentioned table,for reinforced concrete for extreme exposure,
the,
Minimum Cement Content=360 Kg/(m^3)
Maximum w/c ratio=0.40
So,we use the w/c ratio of 0.35,as lower the w/c ratio,the higher the strength of the concrete.
And,Minimum grade of concrete=M40
where,we use M50>M40
So,this condition is statisfied!
Step 3:To calculate the water content required for Concrete Mixing
Here,we calculate the water content required for Concrete Mixing,which is done by rusing Table 2 present on page no.3 of IS 10262:2009,
from the above mentioned table,for nominal aggregate size of 20 mm,
the,
Maximum Water Content=186 Kg/(m^3)
but this is for workability of concrete of 50 mm,but we use the concrete of the workabilty of 100 mm,so,we have to increase the water content,this increment is done at a rate of 3% rise for every increase in 25 mm slump,so we have to increase the water content by 6% for the 50 mm rise of the slump value.
Now,
Water Content=186+(6/100)186=197.16 Kg/(m^3)
But we will using a admixture i.e.,Super Plasticizer,so we reduce the required water content by 29%
then,
Water Content=197.16-(29/100)197.16=140 Kg/(m^3)
Step 4:To calculate the Cement content of the Concrete
Here,we calculate the cement content required for the concrete,where we use the w/c ratio and water content.
w/c ratio=0.35
140/cement=0.35 => cement=140/0.35=400 Kg/(m^3) > 360 Kg/(m^3)
So,the condition is statisfied.
Step 5:To calculate the Admixture content
Here,we calculate the quantity of the Admixture which is equal to the 2% of the total cementious material.
Then,
Admixture Content=(2/100)* 400=8 Kg/(m^3)
Step 6:To calculate the Fine Aggregate and Coarse Aggregate content
Here,we calculate the quantities of both the aggregates i.e., fine aggregate quantity and coarse aggregate quantity by using the Table 3 present on page no.3 of IS 10262:2009
for the nominal size of the coarse aggregate of 20 mm for a zone 2 fine aggregate,
the coarse aggregate proportion with respective to the total volume of aggregates be,
COARSE AGGREGATE=0.62
but the above value is for w/c ratio=0.50,but we have w/c ratio=0.40 which means,as the w/c ratio decrease the cement content increase so,we have to increase the coarse aggregate content with repect to the total volume of the aggregates,which is done at a rate of 0.02 hike in 0.10 fall of w/c ratio.
So,COARSE AGGREGATE=0.62+0.03=0.65
But we are about to place the concrete by pumping,so we reduce the coarse aggregate content by 10% to facilitate the concrete for pumping and easy passage of the concrete through pipe.
COARSE AGGREGATE=0.65-(10/100)0.65=0.58
then,FINE AGGREGATE=1-0.58=0.42
Step 7:To determine the Mix Proportion of all the required ingredients
Here,we determine the quantities in terms of volume of all the materials for one cubic metre of concrete.
Let,volume of concrete=1 m^3
Now,
Volume of Cement=(Mass of Cement/Specific Gravity of Cement)*(1/1000)=(400/3.15)*(1/1000)=0.126 m^3
Volume of Water=(Mass of Water/Speific Gravity of Water)*(1/1000)=(140/1)*(1/1000)=0.140 m^3
Volume of Admixture=(Mass of Admixture/Specific Gravity of Admixture)*(1/1000)=(8/1.145)*(1/1000)=0.006 m^3
Volume of Aggregate=1-Volume of Cement-Volume of Water-Volume of Admixture=1-0.126-0.140-0.006=0.728 m^3
Then,
Mass of Coarse Aggregate=Volume of Total Aggregates*Proportion of Coarse Aggregates with respective to Total Aggregates Volume*Specific Gravity of Coarse Aggregate*1000=0.728*0.58*2.74*1000=1157 Kg/(m^3)
Mass of Fine Aggregate=Volume of Total Aggregates*Proportion of Fine Aggregates with respective to Total Aggregates Voulme*Specific Gravity of Fine Aggregate*1000=0.728*0.42*2.74*1000=838 Kg/(m^3)
Step 8:To represent the Mix Ratio of the Concrete
Here,we represent the Mix Contents in Mix Ratio.
So,the contents for one cubic metre of the concrete are,
Cement=400 Kg/(m^3)
Fine Aggregate=838 Kg/(m^3)
Coarse Aggregate=1157 Kg/(m^3)
Admixture=8 Kg/(m^3)
Water=140 Kg/(m^3)
Now,
Cement:Fine Aggregate:Coarse Aggregate=400:838:1157=1:2.09:2.89