Project on Concrete Mix Design for various grades of Concrete
Calculate the concrete mix design for m35 grade concrete with fly ash
Step1: Stipulations for proportioning
Grade designation : M 35
Type of cement : OPC 53 grade confirming to IS 8112
Maximum nominal size of aggregate : 20 mm
Minimum cement content : 340 kg/m3
Maximum water-cement ratio : 0.45
Workability : 50-100 mm (slump)
Exposure condition : Very Severe (for reinforced concrete)
Method of concrete placing : Pumping
Degree of supervision : Good
Type of aggregate : Crushed angular aggregate
Maximum cement content : 450 kg/m3
Chemical admixture type : Super Plasticizer
Step 2: Test Data for Materials
Type of cement : OPC 53 grade confirming to IS 8112
Specific gravity of cement : 3.12
Specific gravity of Coarse aggregate : 2.74
Specific gravity of Fine aggregate : 2.56
Step3: Target Strength for Mix Proportioning
ftarget = fck + 1.65 xS
'ftarget‘ is the target average compressive strength at 28 days,
‘fck‘ is the characteristic compressive strength at 28 days, and
‘s’ is the standard deviation
As per the table I of IS: 10262-2009 , standard deviation, s = 5.0 N/mm2
Therefore, target strength = 35 + 1.65 x 5.0 = 43.25 N/mm2
Step4: Selection of Water/Cement Ratio
From Table 5 of IS 456:2000, for servere exposure,Maximum water-cement ratio = 0.45
Step5: Selection of Water Content
From Table 2 of IS 456, maximum water content =186 litres (for 25 to 50 mm slump range) for 20 mm aggregate.
Estimated water content for 100 mm slump =186+(6/100) x 186 =197 liter.
( If Super plasticizer is used, the water content can be reduced upto 20% and above.)
Based on trials with Super plasticizer water content reduction of 25% has been achieved, Hence the arrived water content = 197-[197 x (25/100)] = 147 litres.
Step6: Calculation of Cement Content
Water-cement ratio =0.45C
Cementcontent = 147/0.45 = 350 kg/m3
From Table 5 of IS 456, minimum cement content for ‘Very Severe‘ exposure condition = 340 kg/m3
350 kg/m3 > 340 kg/m3, hence, O.K.
Now, to proportion a mix containing fly ash the following steps are suggested
Cementitions material content = 350 × 1.10 =385kg/m3
Water content = 147kg/m3
So water cement ratio = 147/385 = 0.381
Fly ash @20% total amenities material content= 385 × 20% = 77 kg/m3
Cement (OPC) = 385 -77 = 308kg/m3
Saving of cement while using fly ash = 350- 308= 42kg/m3
Fly ash being utilized = 77 kg/m3.
Step7: The Proportion of the Volume of Coarse Aggregate and Fine Aggregate Content
From Table 3 of IS 10262, volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone II) for a water-cement ratio of 0.50 = 0.62.
In the present case, the water-cement ratio is 0.45. Therefore, the volume of coarse aggregate is required to be increased to reduce the fine aggregate content.
As the water-cement ratio is lower by 0.10, the proportion of the volume of coarse aggregate is increased by 0.02 (at the rate of -/+ 0.01 for every ± 0.05 change in the water-cement ratio). Therefore. the corrected proportion of the volume of coarse aggregate for the water-cement ratio of 0.45 = 0.63.
NOTE – In case the coarse aggregate is not angular one. then also the volume of coarse aggregate may be required 10 be increased suitably, based on experience.
For pumpable concrete, these values should be reduced by 10 percent.
Therefore, volume of coarse aggregate = 0.63 x 0.9 = 0.567.
Volume of fine aggregate content =1 – 0.567 =0.433
Step8: Mix Calculations
a) Volume of concrete = 1m3
b) Volume of cement = (Mass of cement/Specific Gravity of Cement) X (1/1000)
= (350/3.12) X (1/1000)
= 0.112 m3
c) volume of fly ash = 77/2.2 × 1/1000
=0.035 m3
d) Volume of water = (Mass of water/specific gravity of water) X (1/1000)
= (147/1) X (1/1000)
= 0.147 m3
e) Volume of superplasticizer = (Mass of superplasticizer / Specific gravity of admixture) X (1/1000)
= (7/1.145) X (1/1000)
= 0.006 m3
f) Volume of all aggregate = (a-(b+c+d))
= 1 – (0.112 + 0.147+0.006)
= 0.735 m3
g) Mass of coarse aggregate = Volume of all aggregate X volume of coarse aggregate X Specific gravity of coarse aggregate X 1000
= 0.735 x 0.567 x 2.74 x 1000
= 1141.88 kg
h)Mass of fine aggregate = Volume of all aggregate X volume of fine aggregate X Specific gravity of fine aggregate X 1000
= 0.735 x 0.433 x 2.56 x 1000
= 814.73 kg
Step9: Concrete Mix Proportions
Cement = 350 kg/m3
Water = 147 kg/m3
Fly ash =77 kg/m3
Super plasticizer = 7 kg/m3
Fine aggregates = 814.73 kg/m3
Coarse aggregate = 1141.88 kg/m3
W/c = 0.45
Concrete Mix Ratio Obtained
The Mix Ratio of M35 as obtained is 1 : 2.32 : 3.26 (Cement : Fine aggregate : Coarse aggregate).
The concrete mix design for M50 grade without fly ash.
Step 1: Stipulations for proportioning
Grade designation : M 50
Type of cement : OPC 53 grade confirming to IS 8112
Maximum nominal size of aggregate : 20 mm
Minimum cement content : 360 kg/m3
Maximum water-cement ratio : 0.45
Workability : 50-75 mm (slump)
Exposure condition : Extreme (for reinforced concrete)
Method of concrete placing : Pumping
Degree of supervision : Good
Type of aggregate : Crushed angular aggregate
Maximum cement content : 450 kg/m3
Chemical admixture type : Super Plasticizer
Step 2: Test Data for Materials
Cement used : OPC 53 grade confirming to IS 8112
Specific gravity of cement : 3.12
Specific gravity of Coarse Aggregate : 2.74
Specific gravity of Fine Aggregate : 2.56
Step 3: Target Strength for Mix Proportioning
ftarget = fck + 1.65 x s
ftarget denotes the Target Average Compressive Strength at 28 days,
fck denotes the Characteristic Compressive Strength at 28 days,
‘s’ stands for Standard Deviation
According to the table 1 of IS: 10262-2009 ,
standard deviation, s = 5.0 N/mm2
Therefore, target strength = 50 + 1.65 x 5.0 = 58.25 N/mm2
Step 4: Selection of Water/Cement Ratio
From Table 5 of IS 456 : 2000,
Maximum water-cement ratio = 0.35
Step 5: Selection of Water Content
From Table 2 of IS 456, maximum water content =186 litres (for 25 to 50 mm slump range) for 20 mm aggregate.
Estimated water content for 100 mm slump =186+(6/100) x 186 =197 litres.
(Note: If Super plasticizer is used, the water content can be reduced up to 20% and above.)
Based on trials with Super plasticizer water content reduction of 27% has been achieved, Hence the arrived water content = 197 – [197 x (27/100)] = 143 litres.
Step6: Calculation of Cement Content
Water-cement ratio = 0.35
Cement content = 143/0.35 = 410 kg/m3
From Table 5 of IS 456, minimum cement content for ‘Extreme exposure’ condition = 360 kg/m3
410 kg/m3 > 360 kg/m3, hence, O.K..
Step 7: Proportion of Volume of Coarse Aggregate and Fine Aggregate Content
From Table 3 of IS 10262. volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone II) for a water-cement ratio of 0.50 =0.62.
In the present case, the water-cement ratio is 0.35. Therefore. the volume of coarse aggregate is required to be increased to decrease the fine aggregate content. As the water-cement ratio is lower by 0.10. the proportion of the volume of coarse aggregate is increased by 0.02 (at the rate of -/+ 0.01 for every ± 0.05 change in the water-cement ratio).
Therefore. the corrected proportion of the volume of coarse aggregate for the water-cement ratio of 0.35 = 0.65.
NOTE – In case the coarse aggregate is not angular one, then the volume of coarse aggregate may be increased by 10 percent based on experience.
For pumpable concrete, these values should be reduced by 10 percent.
Therefore, volume of coarse aggregate = 0.65 x 0.9 = 0.585.
Volume of fine aggregate content =1 – 0.585 =0.415 m3
Step 8: MixCalculations
a) Volume of concrete = 1m3
b) Volume of cement = (Mass of Cement/ Its Specific Gravity ) X (1/1000)
= (410/3.12) X (1/1000)
= 0.131 m3
c) Volume of Water = (Mass of water/specific gravity of water) X (1/1000)
= (143/1) X (1/1000)
= 0.143 m3
d) Volume of Superplasticizer = (Mass of Superplasticizer / Specific gravity of admixture) X (1/1000)
= (8.2/1.145) X (1/1000)
= 0.007 m3
e) Volume of all aggregate = (a – (b + c + d))
= 1 – (0.131 + 0.143+0.007)
= 0.719 m3
e) Mass of coarse aggregate = Volume of All Aggregate X Volume of Coarse Aggregate X Specific Gravity of Coarse Aggregate X 1000
= 0.719 x 0.585 x 2.74 x 1000
= 1152.48 kg
f) Mass of fine aggregate = Volume of all aggregate X volume of fine aggregate X Specific gravity of fine aggregate X 1000
= 0.719 x 0.415 x 2.56 x 1000
= 763.86 kg
Step 9: Concrete Mix Proportions
Cement = 410 kg/m3
Water = 143 kg/m3
Super plasticizer = 7 kg/m3
Fine aggregates = 763.86 kg/m3
Coarse aggregate = 1152.48 kg/m3
W/c = 0.35
Concrete Mix Ratio Obtained
The Mix Ratio of M50 as obtained is1 : 2.28 : 3.12 (Cement : Fine aggregate : Coarse aggregate).