Project on Concrete Mix Design for various grades of Concrete
Aim: Calculate the Concrete Mix Design for M35 grade concrete with fly ash
Data is taken from the example
a) Grade designation: M35
b) Type of cement: OPC 43 grade conforming to IS 8112
c) Type of mineral admixture: Fly ash conforming to IS 3812 ( part 1)
d) Maximum nominal size of aggregate: 20 mm
e) Minimum cement content : 320 kg/m3 (From table 5 of IS 456:2000 - Exposure condition)
f) Maximum water-cement ratio: 0.45 (From table 5 of IS 456:2000 - Exposure condition)
g)Workability : 100 m ( slump value for pumpable concrete)
h) Exposure condition: severe ( RCC)
i) Method of concrete placing: Pumping
i) Degree of supervision: Good
j) Type of aggregate: Crushed angular aggregate
k) Maximum cement content : 450 kg/m3
l) Chemical admixture type: Superplasticizer
Test data for materials:
a) Cement used: OPC 43 grade conforming to IS 8112
b) Specific gravity of cement: 3.15
c) Fly ash: conforming to IS 3812 (part 1)
d) Specific gravity of fly ash: 2.2
e) Chemical admixture: Superplasticizer conforming to IS 9103
f) Specific gravity of:
1) Coarse aggregate: 2.74
2) Fine aggregate: 2.74
3) Chemical admixture: 1.145
g) Water absorption:
1) Coarse aggregate : 0.5 percent
2) Fine aggregate: 1.0 percent
e) free surface moisture:
1) Coarse aggregate: nil
2) Fine aggregate: nil
k) Sieve analysis data: taken from the example problem
first, we need to calculate the target mean strength of the mix proportion
fck1 = fck +1.65 * S
where fck1 = target mean compressive strength at 28 days
fck = characteristic compressive strength at 28 days
S = standard deviation
From table 2 of IS 10262:2019 , standard deviation s = 5 N/mm2
Therefore, target strength = 35 +1.65*5 =43.25 N/mm2
Selection of water cement ratio :
from the fig 1 of IS :10262:2019, the water cement ratio corresponding to the strength of 43.25 N/mm2 is 0.39
from table 5 of IS 456:2000 for severe exposure, maximum water-cement ratio =0.45
the water cement ratio = 0.39 ( lower of the both)
Selection of water content
from table 4 of IS 10262: 2019, maximum water content for 20 mm aggregate = 186 lit ( for 25-50 mm slump range)
for every 25 m increase in the slump te water content is raised to 3%
estimated water content for 100 mm slump = 186 + 6/100*186
= 197 lit
As a superplasticizer is used, the water content can be reduced up to 30%
based on trails water content reduction of 29% is achieved with a plasticizer.
Hence the arrived water content = 197 * 0.71 =140 lit
Calculation of cement content :
water-cement ratio =0.39
cementitious material ( fly ash + cement) content = 140/0.39 = 359 kg/m3
from table 5 of IS 456:2000 the minimum cement for "severe" exposure condition = 359 kg/m3
359 > 320 kg/m3
Now, to proportion a mix containing fly ash the following steps are suggested:
a) the percentage of fly ash needs to be decided based on project requirement and quality of materials
b) in some cases increase in cementitious material can be warranted. An increase in cementitious material content and its percentage may be based on experience and trial ( this example it is increased to 10%)
cementitious material content = 359 * 1.1 = 395 kg/m3
water content =140 kg/m3
water-cement ratio = 140/395 = 0.354
fly ash @ 30% of total cementitious material content = 395*0.3 = 119 kg/m3
cement ( OPC ) = 395-119 =276 kg/m3
Saving of cement while using fly ash = 359 -276 =83 kg/m3
fly ash utilized = 119 kg/m3
The proportion of the volume of coarse aggregate and fine aggregate content.
from table 5 of IS 10262: 2019, the volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (zone I) for the water-cement ratio of 0.5 = 0.6
In the present case, the water-cement ratio is 0.39
* therefore the volume of coarse aggregate needs to be increased to decrease the fine aggregate content.
* as the water-cement ratio is lower by 0.1, the proportion of the volume of coarse aggregate is increased by 0.02
therefore, the corrected volume of coarse aggregate for the water-cement ratio of 0.39 =0.6+0.022
= 0.622
for pumpable concrete, these values are reduced by 10%
therefore, volume of coarse aggregate = 0.622*0.9 =0.5598 =0.56
volume of fine aggregate content = 1 - 0.56 =0.44
The mix calculation 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)
= 276 /3.15 *(1/1000)
= 0.0876 m3
c) volume of flyash = (mass of fly ash / specific gravity of fly ash )*(1/1000)
= (119/2.2)*(1/1000)
= 0.054 m3
d) volume of water = (mass of water / specific gravity of water )*(1/1000)
= (140/1)*(1/1000)
= 0.14 m3
e) volume of chemical admixture ((superplasticizer) (@ 1.0 percentby mass of cementitious material)
= (mass of chemical admixture/ specific gravity of chemical admixture )*(1/1000)
= (2.76/1.145)*(1/1000)
= 0.0024 m3
f) volume of all in aggregate = [(a-(b+c+d+e)]
= [(1-(0.0876 + 0.054 + 0.14 + 0.0024)]
= 0.716 m3
g) mass of coarse aggregate = f × volume of coarse aggregate × Specific gravity of coarse aggregate × 1000
= 0.716 x 0.56 x 2.74 x 1000
= 1098.63 kg/m3
h) mass of fine aggregate = f × volume of fine aggregate × Specific gravity of coarse aggregate × 1000
= 0.716 x 0.44 x 2.74 x 1000
= 863.2 kg/m3
Mix proportions:
Cement = 276 kg/m3
Fly ash = 119 kg/m3
Water = 140 kg/m3
Fine aggregate = 863 kg/m3
Coarse aggregate = 1098 kg/m3
Chemical admixture = 2.76 kg/m3
Free water-cementitious materials ratio
= 0.354
Trail Mix Ratio = 1 : 2.18 : 2.78
Aim: Calculate the Concrete Mix Design for M50 grade concrete without fly ash
Data is taken from the example
a) Grade designation: M50
b) Type of cement: OPC 43 grade conforming to IS 8112
c) Maximum nominal size of aggregate: 20 mm
d) Minimum cement content: 320 kg/m3 (From table 5 of IS 456:2000 - Exposure condition)
e) Maximum water-cement ratio: 0.45 (From table 5 of IS 456:2000 - Exposure condition)
f)Workability : 100 m ( slump value for pumpable concrete)
g) Exposure condition: severe ( RCC)
h) Method of concrete placing: Pumping
i) Degree of supervision: Good
j) Type of aggregate: Crushed angular aggregate
k) Maximum cement content : 450 kg/m3
l) Chemical admixture type: Superplasticizer
Test data for materials:
a) Cement used: OPC 43 grade conforming to IS 8112
b) Specific gravity of cement: 3.15
c) Fly ash: conforming to IS 3812 (part 1)
d) Chemical admixture: Superplasticizer conforming to IS 9103
e) Specific gravity of:
1) Coarse aggregate: 2.74
2) Fine aggregate: 2.74
g) Water absorption:
1) Coarse aggregate : 0.5 percent
2) Fine aggregate: 1.0 percent
e) free surface moisture:
1) Coarse aggregate: nil
2) Fine aggregate: nil
k) Sieve analysis data: taken from the example problem
first, we need to calculate the target mean strength of the mix proportion
fck1 =fck +1.65 * S
where fck1 = target mean compressive strength at 28 days
fck= characteristic compressive strength at 28 days
S = standard deviation
From table 2 of IS 10262:2019 , standard deviation s = 5 N/mm2
Therefore, target strength = 50 +1.65*5 =58.25 N/mm2
Selection of water cement ratio :
from the fig 1 of IS :10262:2019, the water cement ratio corresponding to the strength of 58.25 N/mm2 is 0.34
from table 5 of IS 456:2000 for severe exposure, maximum water-cement ratio =0.45
the water cement ratio = 0.34 ( lower of the both)
Selection of water content
from table 4 of IS 10262: 2019, maximum water content for 20 mm aggregate = 186 lit ( for 25-50 mm slump range)
for every 25 m increase in the slump te water content is raised to 3%
estimated water content for 100 mm slump = 186 + 6/100*186
= 197 lit
As a superplasticizer is used, the water content can be reduced up to 30%
based on trails water content reduction of 29% is achieved with a plasticizer.
Hence the arrived water content = 197 * 0.71 =140 lit
Calculation of cement content :
water-cement ratio =0.34
cement content = 140/0.34 = 412 kg/m3
from table 5 of IS 456:2000 the minimum cement for "severe" exposure condition = 412 kg/m3
412 > 320 kg/m3
The proportion of the volume of coarse aggregate and fine aggregate content.
from table 5 of IS 10262: 2019, the volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (zone I) for the water-cement ratio of 0.5 = 0.6
In the present case, the water-cement ratio is 0.34
* therefore the volume of coarse aggregate needs to be increased to decrease the fine aggregate content.
* as the water-cement ratio is lower by 0.1, the proportion of the volume of coarse aggregate is increased by 0.02
therefore, the corrected volume of coarse aggregate for the water-cement ratio of 0.34 =0.6+0.032
= 0.632
for pumpable concrete, these values are reduced by 10%
therefore, volume of coarse aggregate = 0.632*0.9 =0.5688 =0.57
volume of fine aggregate content = 1 - 0.57 =0.43
The mix calculation 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)
= 412 /3.15 *(1/1000)
= 0.130 m3
c) volume of water = (mass of water / specific gravity of water )*(1/1000)
= (140/1)*(1/1000)
= 0.14 m3
d) volume of chemical admixture ((superplasticizer) (@ 1.0 percentby mass of cementitious material)
= (mass of chemical admixture/ specific gravity of chemical admixture )*(1/1000)
= (4.12/1.145)*(1/1000)
= 0.0036 m3
e) volume of all in aggregate = [(a-(b+c+d)]
= [(1-(0.130 + 0.14 + 0.0036)]
= 0.7264 m3
f) mass of coarse aggregate = f × volume of coarse aggregate × Specific gravity of coarse aggregate × 1000
= 0.7264 x 0.57 x 2.74 x 1000
= 1135 kg/m3
g) mass of fine aggregate = f × volume of fine aggregate × Specific gravity of coarse aggregate × 1000
= 0.7264 x 0.43 x 2.74 x 1000
= 856 kg/m3
Mix proportions:
Cement = 412 kg/m3
Water = 140 kg/m3
Fine aggregate = 856 kg/m3
Coarse aggregate = 1135 kg/m3
Chemical admixture = 4.12 kg/m3
water-cement ratio = 0.34
Trail Mix Ratio = 1 : 2.07 : 2.75