Project on Concrete Mix Design for various grades of Concrete
1 Answer )
Aim of the Project :
To calculate the Concrete Mix Design for M35 grade concrete with fly ash.
The mix design calculation of M35 Grade Concrete with fly ash and Poly Carboxylite type superplasticizer as per Indian Standard Codebook IS10262(2009) – Guidelines for Concrete Mix Proportioning.
The following equations need to be considered :
f’ck= fck+1.65s
here f’ck is the mean target strength for concrete mix design
fck is characteristic compressive strength at 28 days
s is the standard deviation
The volume of cement =1m^3
The volume of cement = (mass of cement /specific gravity of cement ) *1/1000
The volume of water = (mass of water/specific gravity of water )*1/1000
The volume of chemical admixture = (mass of chemical admixture/specific gravity of admixture) * *1/1000
The volume of aggregates = (volume of concrete –(volume of cement +volume of fly ash+volume of water +volume of admixture
Mass of coarse aggregate = volume of aggregate *volume of coarse aggregate *specific gravity of coarse aggregate *1000
Mass of fine aggregate = volume of aggregate * volume of fine aggregate *specific gravity of fine aggregate *1000
Objective :
The objective of this project is to produce M35 grade concrete economically.
To achieve desired workability, durability, and minimum strength of M35 grade concrete.
Here we use IS 456:2000 and IS 10262:2000 as a standard codebook for reference.
Procedure and calculation :
- Target mean strength is
f’ck=fck+1.65s
= 35 N/mm^2
From table 1 of IS 10262:2009
S=5N/mm^2
- Calculation of water-cement ratio :
Assuming extreme weather conditions
From table 5 of IS 456:2000
For extreme conditions w/c ratio =0.40
Minimum cement content from table 5 of IS 456:2000 is 360 kg/m^3
- Selection of water-cement ratio :
Assuming the nominal size of aggregate i.e 20mm
Therefore from table 2 of IS 10262:2009
Minimum water content = 186 liters
Slump range from 25 to 50mm
As superplasticizers reduce 20% water content
= 186*(100-20)/100 = 148.8
Approximately 149 liters
- Calculation of cementitious material :
Water cement ratio = 0.40
Water content W= 149 kg
Cementatious content = 149/0.40= 372.5=370kg/m^3
From table 5 we can see that the minimum cement content to be 360 kg/m^3
Therefore 360<370
So we increase by 10%
370*0.10 +370= 407 kg/m^3
w/c = 149/407
= 0.366
Fly ash at 30 % total cementitious material content = 407*30/100= 122kg/m^3
Cement (OPC) = 407-122= 285 kg/m^3
Saving of cement while using fly ash = 370-285= 85 kg/m^3
Fly ash = 122kg/m^3
- The volume of Coarse aggregates and Fine aggregates :
Fine aggregates are sieved through Zone 1 table 4 of IS 383
Coarse aggregates are sieved through table 2 of IS 383
According to table 3 of IS 10262:2009
The nominal size of aggregates is 20mm.
Therefore the volume of coarse aggregate is 0.60
Water cement ratio =0.4
(+/-0.01 for every +/-0.005)
So, the volume of coarse aggregate is 0.62
Volume of fine aggregate =1-0.62 = 0.38
- Mix calculations :
a.Volume of concrete = 1m^3
b.Volume of cement = (285/3.15)*1/1000= 0.090m^3
c.Volume of fly ash = ( 122/2.2)*1/1000= 0.055m^3
d.Volume of water = (149/1)*1/1000 = 0.149m^3
e.Volume of admixture = (7.6/1.145)*1/1000= 0.006m^3
- Volume of corse aggregate and fine aggregate = a-(b+c+d+e)
= 1-(0.090+0.055+0.149+0.006)
= 0.7m^3
- Mass of coarse aggregate = 0.7*0.62*2.74*1000= 1189.16 kg/m^3
- Mass of fine aggregate = 0.7*0.38*2.74*1000=728kg/m^3
- Results and calculation :
Cement = 285kg /m^3
Flyash = 122kg/m^3
Water = 149kg
Coarse aggregate =1189kg/m^3
Fine aggregate =728kg/m^3
Chemical admixture = 7.6 kg/m^3
Water cement ratio = 0.366
Therefore trial mix ratio = 285:728:1189 = 1.2.55:4.17 ( C:FA:CA )
We have to check for durability and strength of this mix if not again a trial ratio mix needs to be done.
2 Answer )
Aim of the project :
To calculate M50 grade concrete design without using fly ash.
The following equations need to be considered :
f’ck= fck+1.65s
here f’ck is the mean target strength for concrete mix design
fck is characteristic compressive strength at 28 days
s is the standard deviation
The volume of cement =1m^3
The volume of cement = (mass of cement /specific gravity of cement ) *1/1000
The volume of water = (mass of water/specific gravity of water )*1/1000
The volume of chemical admixture = (mass of chemical admixture/specific gravity of admixture) * *1/1000
The volume of aggregates = (volume of concrete –(volume of cement +volume of fly ash+volume of water +volume of admixture
Mass of coarse aggregate = volume of aggregate *volume of coarse aggregate *specific gravity of coarse aggregate *1000
Mass of fine aggregate = volume of aggregate * volume of fine aggregate *specific gravity of fine aggregate *1000
Objective :
The objective of this project is to produce M35 grade concrete economically.
To achieve desired workability, durability, and minimum strength of M35 grade concrete.
Here we use IS 456:2000 and IS 10262:2000 as a standard codebook for reference.
Procedure and calculation :
- Target mean strength is
f’ck=fck+1.65s
fck = 50N/mm^2
From table 1 of IS 10262
S=5N/mm^2
F’ck=50+(1.65*5)
= 58.25 N/mm^2
- Selection of water-cement Ratio :
Assuming exposure condition: Extreme
Therefore from table 5 IS 456:2000
Extreme w/c ratio = 0.40
Lower the w/c ratio better the workability.
Minimum cement content = 360kg/m^3
- Selection of water content :
Assuming nominal size of aggregates = 20mm
From table 2 of IS 10262:2009
MinimumWater content = 186 liter
As superplasticizers would reduce the water content by 20%
= 186*(100-20)/100 = 148.8
Water content = 149 liter
- Calculation of cement content :
Water cement ratio = w/c ratio = 0.40
Water content = w= 149kg
Cement content c= 149/0.40=372.5 = 373 kg/m^3
360<373kg/m^3
Hence okay.
- The volume of coarse aggregate and fine aggregate:
According to table 3 IS 102062:2009
The nominal size of aggregate is 20mm
Therefore the volume of coarse aggregate = 0.60
Water cement ratio =0.4
For every ( +/-0.01for every +/-0.005)
So the volume of coarse aggregate =0.62
Volume of fine aggregate = 1.0.62= 0.38
- Mix calculation :
- Volume of concrete = 1m^3
- Volume of cement = (373/3.15)*1/1000= 0.118m^3
- Volume of water = (149/1)*1/1000 = 0.149m^3
- Volume of admixture = (7.6/1.145)*1/1000 = 0.006
- Volume of coarse aggregate and fine aggregate = a-(b+c+d) = 1-( 0.118+0.19+0.006)= 0.727m^3
- Mass of coarse aggregate = 0.727*0.62*2.74*1000= 1235.02kg/m^3
- Mass of fine aggregate =0.727*0.38*2.74*1000= 756.9kg/m^3
7.Results and calculations :
cement = 373kg/m^3
water = 149 kg
coarse aggregate = 1235.02kg/m^3
Fine aggregate = 756.9kg/m^3
Chemical admixture = 7.6kg/m^3
Water cement ratio = 0.40
Therefore trial mix ratio : 373:756.9:1235.02 = 1:2.02:3.31 ( C: FA : CA )
Therefore thid trial mix should satisfy the strength and durability of M50concrete grade mix.
If doesn't met the required conditions we have to increase the details by 10% and should be checked.