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
Concrete mix design for M35 grade concrete with fly ash:
I. Stipulations for proportioning:
- Grade designation= M35
- Type of cement= OPC 53 grade confirming to IS 8112.
- Type of mineral admixture= Fly ash confirming to IS 3812(part2)
- Maximum nominal size of aggregate=20mm
- Minimum cement content= 340kg/m3 ( from table 5 of IS 456:2000)
- Maximum water-cement ratio= 0.45 (From table 5 of IS 456:2000)
- Workability= 100mm (slimp value for pumpable concrete)
- 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= Superplasticizer.
II. Test data for materials:
- Cement used= OPC 53 confirming to IS 8112
- Specific gravity of cement=3.15
- Mineral admixture= Fly ash confirming to IS 12089
- Specific gravity of fly ash=2.9
- Chemical [email protected]% confirming to IS 9103
- Specific gravity of:
- Coarse aggregate= 2.74
- Fine aggregate= 2.74
- Admixture PCE based= 1.12
g.Water absorption:
- Coarse aggregate=0.5%
- Fine aggregate=1%
h.Sieve analysis:
- Coarse aggregate= confirming to table 2-IS 383
- Fine aggregate= Zone II (confirming to tablr 4-IS 383)
i.Free surface moisture:
- Coarse aggregate= Nil
- Fine aggreate=Nil
III. Target strength of mix proportioning:
F'ck=Fck+1.65S
F'ck= 35+(1.65X5)
F'ck=43.25 N/mm^2
IV. Selection of water-cement ratio:
From table 5 of IS 456:2000 for very severe exposure,the w/c ratio=0.45
Based on experience,adopt watercement ratio as 0.45
therefore, 0.45=0.45 hence ok
V. Selection of water content:
From table 2 of IS 10262:2000,maximum water content for 20mm aggregate=186 Litres(for 25 to 50mm slump range)
Estimated water content for 100mm slump= 186+0.6% of 186
= 186+ (0.6/100X186)
= 197.16 litres = 198 litres
Since we are using plasticizers, the water content can be reduced up to 20%
The amount of water content to be reduced is depend on the type of superplasticizer.
Therefore,Percentage reduction in water content=20%
Water content required= 198-(20% of 198)
=198-(20/100X198)
=158.4 Litres= 160 litres
VI. Calculation of cement content:
Water cement ratio=0.45
Cementitious material(cement+fly ash)content= 160/0.45
= 355.55 kg
Since fly ash is used the amount of cementitious content(cement+fly ash)shall be increased.
The percentage of cementitious content to be increased depends on the experience and research.
Here, we have increased the cementitous content by 10%.
Final cementitious content(cement+fly ash)=356X1.10
= 391.6 kg/m^3
From table 5 of IS 456, minimum cement content required for very severe exposure condition=340kg/m^3.
Adopted cementitious content=391.6 kg/m^3
391.6>340 kg/m^3 hence ok.
Fly ash @30% total cementitious material content= 391.6X0.30
=117.48kg/m^3=118kg/m^3
Cement content requirement= 391.6-117.48
= 274.12 kg/m^3
Calculation of superplasticizer @0.6% to the mass of cementitious content=391.6X0.6%
=391.6X(0.6/100)
= 2.34kg/m^3
VII. Proportion of volume of coarse and fine aggregate:
From table 3 of IS 10262:2009,volume of coarse aggregate corresponding to 20mm size of aggregate and fine aggregate(ZoneII) for water cement ratio=0.62
As w/c ratio is lower by 0.10,proportion of volume of coarse aggregate is increased by 0.02.(at the rate of +/- 0.01 for every 0.05 change in w/c ratio)
Therefore, Volume of fine aggregate content=1-0.63
= 0.37
VIII. Mix calculations:
- Volume of concrete=1m^3
- Volume of cement=(Mass of cement/Specific gravity of cement)X (1/1000)
=(275/3.15) X (1/1000)
= 0.087 cu.m
c. Volume of fly ash=(Mass of fly ash/Specific gravity of fly ash)X (1/1000)
= (118/2.5) X (1/1000)
0.047 cu.m
d. Volume of water=(Mass of water/Specific gravity of water)X (1/1000)
=(160/1)X(1/1000)
= 0.160cu.m
e. Volume of superplasticizer= (Mass of superplasticizer/Specific gravity of superplasticizer)X (1/1000)
= (2.35/1.12) X (1/1000)
= 0.002 cu.m
f. Volume of aggregate (CA & FA)= {a-[b+c+d+e]}
= {1-[0.087+0.047+0.160+0.002]}
= 1-0.296
=0.71 cu.m
g. Mass of coarse aggregate= fX Volume of coarse aggregate X specific gravity of coarse aggregate X 1000
= 0.71 X 0.63 X 2.74 X 1000
= 1225.6 kg/m^3
h. Mass of fine aggregate= f X Volume of fine aggregate X specific gravity of fine aggregate X 1000
= 0.71 X 0.37 X 2.74 X 1000
= 719.79 kg/m^3
XI. Mix proportions for trial number:
- Cement =275kg/m^3
- Fly ash =118kg/m^3
- Water =160 litres
- Fine aggregates =720 kg/m^3
- Coarse aggregate =1226kg/m^3
- Chemical admixture =2L
To find the design mix ratio divide the calculated value of all the materials by the weight of cement.
Mix ratio of M35 grade concrete by weight = Cement : Fly ash : Coarse aggregate : Fine aggregate : Water
= 1 : 0.4 : 4.46 : 2.6 : 0.58
Concrete mix design for M50 grade concrete without fly ash:
I. Stipulations for proportioning:
- Grade designation= M50
- Type of cement= OPC 53 grade confirming to IS 8112.
- Maximum nominal size of aggregate=20mm
- Minimum cement content= 360kg/m3 ( from table 5 of IS 456:2000)
- Maximum water-cement ratio= 0.35 (From table 5 of IS 456:2000)
- Workability= 50-75mm (slimp value for pumpable concrete)
- 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= Superplasticizer.
II. Test data for materials:
- Cement used= OPC 53 confirming to IS 8112
- Specific gravity of cement=3.12
- Specific gravity of:
- Coarse aggregate= 2.74
- Fine aggregate= 2.56
g.Water absorption:
- Coarse aggregate=0.5%
- Fine aggregate=1%
h.Free surface moisture:
- Coarse aggregate= Nil
- Fine aggreate=Nil
III. Target strength of mix proportioning:
F'ck=Fck+1.65S
F'ck= 50+(1.65X5)
F'ck=58.25 N/mm^2
IV. Selection of water-cement ratio:
From table 5 of IS 456:2000 for extreme exposure,the w/c ratio=0.35
Based on experience,adopt water cement ratio as 0.45
therefore, 0.35<0.45 hence ok
V. Selection of water content:
From table 2 of IS 10262:2000,maximum water content for 20mm aggregate=186 Litres(for 25 to 50mm slump range)
Estimated water content for 100mm slump= 186+0.6% of 186
= 186+ (0.6/100X186)
= 197 litres
Since we are using plasticizers, the water content can be reduced up to 20% or above.
The amount of water content to be reduced is depend on the type of superplasticizer.
Therefore,Percentage reduction in water content=27%
Water content required= 197-(27% of 198)
=197-(27/100X198)
=143 Litres
VI. Calculation of cement content:
Water cement ratio=0.35
Cementitious material(cement+fly ash)content= 160/0.35
= 410 kg/m^3
From table 5 of IS 456 minimum cement content for extreme exposure condition= 360kg/m^3
410kg/m^3.360kg/m^3, hence OK.
VII. Proportion of volume of coarse and fine aggregate:
From table 3 of IS 10262:2009,volume of coarse aggregate corresponding to 20mm size of aggregate and fine aggregate(ZoneII) for water cement ratio 0.50=0.62
In the present case water-cement ratio is 0.35. Therefore the volume of coarse aggregate is required to be increased to be decreased the fine aggregate.
As w/c ratio is lower by 0.10,proportion of volume of coarse aggregate is increased by 0.02.(at the rate of +/- 0.01 for every 0.05 change in w/c ratio)
Therefore, the corrected proportion of the volume of the coarse aggregate for water cement ratio is 0.35=0.65
For pumpable concrete,these values should be reduced by 10%.
Therefore, volume of coarse aggregate=0.65X0.9
=0.585.
Volume of fine aggregate content= 1-0.585
= 0.415
VIII. Mix calculations:
- Volume of concrete=1m^3
- Volume of cement=(Mass of cement/Specific gravity of cement)X (1/1000)
=(410/3.12) X (1/1000)
= 0.131 cu.m
c. Volume of water=(Mass of water/Specific gravity of water)X (1/1000)
=(143/1)X(1/1000)
= 0.143 cu.m
d. Volume of superplasticizer= (Mass of superplasticizer/Specific gravity of superplasticizer)X (1/1000)
= (8.2/1.145) X (1/1000)
= 0.007 cu.m
e. Volume of all aggregate (CA & FA)= {a-[b+c+d]}
= {1-[0.131+0.143+0.007]}
= 1-0.281
=0.719 cu.m
f. 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/m^3
g. 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/m^3
XI. Mix proportions for trial number:
- Cement =410 kg/m^3
- Water =143 litres
- Fine aggregates =736.86 kg/m^3
- Coarse aggregate =1152.48 kg/m^3
- Super plasticizer =7 kg/m^3
- Water cement ratio = 0.35
To find the design mix ratio divide the calculated value of all the materials by the weight of cement.
Mix ratio of M35 grade concrete by weight = Cement : Coarse aggregate : Fine aggregate : Water
= 1 : 2.8 : 1.79 : 0.34