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
Answer:- Mix Design for M35 grade concrete
Concrete is designed or does mix design as per our requirement to achieve the desired compressive strength.
Here, In this question, I have to do a mix design for M35 grade concrete.
Here, 'M' refers to the Design mix & the number '35' refers 35Mpa compressive strength that has to be achieved in 28 days after casting of concrete.
- Mixed designs are designed by the structural engineer and the mix proportion is determined by the concrete producer.
- Mix design is the most intellectual & rational approach for the selection of specific materials with their proportion & more or less for their unique characteristics.
- In the design mix, I have to give the details about how much & in which quantity of concrete ingredients like coarse aggregate, fine aggregate, cement, fly ash, & water are required for the M35 grade of concrete.
- STIPULATION FOR PROPORTIONING
- a) Grade designation : M35
- b) Type of concrete : OPC 43 grade
- 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/m^3 ( From Table-5 of IS 456:200 as a severe condition)
- f) Maximum water cement ratio : 0.45 ( From Table-5 of IS 456:200 as a severe condition)
- g) Workability (slump) : 100 mm ( slump value for Pumpable condition)
- h)Exposure condition : Severe ( for RCC )
- i) Method of concrete placing : Pumping
- j) Type of aggregate : Crushed angular aggregate
- k) Degree of Supervision : Good
- l) Maximum cement content : 450 kg/m^3 ( as per clause 8.2.4.2 of IS 456:2000)
- m) Chemical admixture type : superplasticizer
2. Test date for materials
- a) Cement used : OPC 43 grade conforming IS 8112
- b) Specific Gravity of Cement : 3.15
- c) Fly ash : conforming IS 3812 part-1
- d) Specific Gravity of Fly ash : 2.2
- e) Specific Gravity of C.A & F.A : 2.74 & 2.74
- f) Water- absorption of C.A & F.A : 0.5% & 10%
Step 1 - Determination Of Target Strength of M35
formula,
f ' ck = fck + 1.65 x S
formula,
f ' ck = fck + 1.65 x S
here,
fck = 35 N/mm^2
S ( standard deviation in N/mm^2 ) = 5 N/mm^2 ( from table -1 of IS 10262- 2009)
Now,
The Target strength = 35 + 1.65 x 5 = 43.25 N/mm^2
Step 2 - Selection of water/cement ratio:-
From Table 5 of IS 456:2000,
Maximum water-cement ratio for Sever exposure condition = 0.45
Based on experience, adopt a water-cement ratio as 0.4.
0.4<0.45,
Hence ok.
Step 3 - Selection of Water Content
From Table 2 of IS 10262- 2009,
Maximum water content for 20 mm (Nominal maximum size of aggregate) = 186 Kg = 186 litre
(for slump 25mm to 50 mm range)
But we have to reach a slump of 100 mm then as per codel provision, we can increase/decrease water content by 3% for a change in a slump of 25mm
Therefore,
Estimated water content for 100mm slump = 186 + 6 x 186 / 100 = 197 liter.
As superplasticizer is used, the water content has to be reduced up to 30%
Based on experience water content reduction of 29% is achieved with a plasticizer.
Hence,
The arrived water content = 197 x 0.71 = 140 liter
Step 4 - Selection of Cement Content
Water-cement ratio = 0.4
Water-cement ratio = 0.4
Cementitious material content = 140 / 0.4 = 350 kg/m^3
From Table 5 of IS 456:2000,
Minimum cement Content for severe exposure condition = 320 kg/m^3
350 kg/m^3 > 320 kg/m^3,
hence, OK.
As per clause 8.2.4.2 of IS: 456:2000, Maximum cement content = 450 kg/m^3.
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 the project requirement & Quality of materials.
b) In some cases increase in cementation material can be warranted. increase in cementation material content & its percentage may be based on experience & Trial.
(This mix design is with an increase of 10% cementation material Content )
cementation material content = 350 x 1.10 = 385 kg/m^3
Water content = 140 kg/m^3
so, corrected water-cement ratio = 140/385 = 0.364
Fly ash @ 30% of total cementation material = 385 x 30 / 100 = 115 kg /m^3
Hence,
cement content (OPC 43) = 385 - 115 = 270 kg/m^3
Fly ash = 115 kg/m^3
( we are saving 350-270= 80 kg/m^3 of cement)
Step 5 - Estimation of Volume of Coarse Aggregate& Fine aggregate content
From Table 3 of IS 10262- 2009,
The volume of C.A corresponding to 20mm size aggregate and F.A ( Zone 1) for water-cement = 0.6
in the present case, the water-cement ratio is 0.4
then, The volume of C.A needs to be increased or decreased F.A content.
(For every ±0.05 change in w/c, the coarse aggregate proportion is to be changed by 0.01. If the w/c is less than 0.5 (standard value), the volume of coarse aggregate is required to be increased to reduce the fine aggregate content. If the w/c is more than 0.5, the volume of coarse aggregate is to be reduced to increase the fine aggregate content.)
corrected volume of C.A for w/c of 0.4 = 0.6 + 0.02 = 0.62
(for pumpable concrete values are reduced by 10%)
Volume of C.A = 0.62 x 0.9 = 0.56
Volume of F.A = 1 - 0.56 = 0.44
Step 6 - Estimation of the mix ingredients
a) Volume of concrete = 1 m^3
b) Volume of cement = (Mass of cement / Specific gravity of cement) x (1/100)
= (270 / 3.15) x (1/1000)
= 0.086 m^3
c) Volume of fly ash = (Mass of fly ash / Specific gravity of fly ash ) x (1/100)
= (115 / 2.2) x (1/1000)
= 0.052 m^3
d) Volume of water = (Mass of water / Specific gravity of water) x (1/1000)
= (140/1) x (1/1000)
= 0.14 m^3
e) Volume of Chemical admixture = (Mass of Chemical admixture / Specific gravity of Chemical admixture ) x (1/1000)
= (7/1.145) x (1/1000)
= 0.006 m^3
f) Volume of total aggregates (C.A + F.A ) = a – (b + c + d + e )
= 1 – ( 0.086 + 0.052 + 0.14 + 0.006)
= 0.716 m^3
g) Mass of coarse aggregates = f x Vol. of C.A x Specify gravity of C.A x 1000
= 0.716 x 0.56 x 2.74 x 1000
= 1098 kg/m^3
f) Mass of fine aggregates = f x Vol. of F.A x Specify gravity of F.A x 1000
=0.716 x 0.44 x 2.74 x 1000
= 863 kg/m^3
Hence,
Result
Mix proportion for M35 grade concrete is
Cement = 270 Kg/m^3
Fly ash = 150 Kg/m^3
Water = 140 Kg/m^3
Fine aggregate = 863 Kg/m^3
Coarse aggregate = 1098 Kg/m^3
Chemical Admixture = 7 Kg/m^3
water-cement ratio = 0.364
M35 mix ratio = 1 : 2.24 : 2.85
Calculation of Mix design of M50 grade concrete without fly ash
- STIPULATION FOR PROPORTIONING
- a) Grade designation : M50
- b) Type of concrete : OPC 43 grade
- c) Maximum Nominal Size of aggregate : 20 mm
- d) Minimum cement content : 360 kg/m^3 ( From Table-5 of IS 456:200 as a severe condition)
- e) Maximum water cement ratio : 0.40 ( From Table-5 of IS 456:200 as a severe condition)
- f) Workability (slump) : 150 mm ( slump value for Pumpable condition)
- g) Exposure condition : Severe ( for RCC )
- h) Method of concrete placing : Pumping
- i) Type of aggregate : Crushed angular aggregate
- j) Degree of Supervision : Good
- k) Maximum cement content : 450 kg/m^3 ( as per clause 8.2.4.2 of IS 456:2000)
- l) Chemical admixture type : superplasticizer
2. Test date for materials
- a) Cement used : OPC 43 grade conforming IS 8112
- b) Specific Gravity of Cement : 3.15
- c) Fly ash : conforming IS 3812 part-1
- d) Specific Gravity of Fly ash : 2.2
- e) Specific Gravity of C.A & F.A : 2.74 & 2.74
- f) Water- absorption of C.A & F.A : 0.5% & 10%
Step 1 - Determination Of Target Strength of M50
formula,
f ' ck = fck + 1.65 x S
formula,
f ' ck = fck + 1.65 x S
here,
fck = 50 N/mm^2
S ( standard deviation in N/mm^2 ) = 5 N/mm^2 ( from table -1 of IS 10262- 2009)
Now,
The Target strength = 50 + 1.65 x 5 = 58.25 N/mm^2
Step 2 - Selection of water/cement ratio:-
From Table 5 of IS 456:2000,
Maximum water-cement ratio for Sever exposure condition = 0.40
Based on experience, adopt a water-cement ratio as 0.35
0.35<0.40,
Hence ok.
Step 3 - Selection of Water Content
From Table 2 of IS 10262- 2009,
Maximum water content for 20 mm (Nominal maximum size of aggregate) = 186 Kg = 186 litre
(for slump 25mm to 50 mm range)
But we have to reach a slump of 100 mm then as per codel provision, we can increase/decrease water content by 3% for a change in a slump of 25mm
Therefore,
Estimated water content for 100mm slump = 186 + 12 x 186 / 100 = 208.32 liter.
As superplasticizer is used, the water content has to be reduced up to 30%
Based on experience water content reduction of 29% is achieved with a plasticizer.
Hence,
The arrived water content = 208.32 x 0.71 = 148 liter
Step 4 - Selection of Cement Content
Water-cement ratio = 0.35
Water-cement ratio = 0.35
Cementitious material content = 148 / 0.35 = 423 kg/m^3
From Table 5 of IS 456:2000,
Minimum cement Content for severe exposure condition = 360 kg/m^3
423 kg/m^3 > 360 kg/m^3,
hence, OK.
As per clause 8.2.4.2 of IS: 456:2000, Maximum cement content = 450 kg/m^3.
Step 5 - Estimation of Volume of Coarse Aggregate& Fine aggregate content
From Table 3 of IS 10262- 2009,
The volume of C.A corresponding to 20mm size aggregate and F.A ( Zone 1) for water-cement = 0.6
in the present case, the water-cement ratio is 0.35
then, The volume of C.A needs to be increased or decreased F.A content.
(For every ±0.05 change in w/c, the coarse aggregate proportion is to be changed by 0.01. If the w/c is less than 0.5 (standard value), the volume of coarse aggregate is required to be increased to reduce the fine aggregate content. If the w/c is more than 0.5, the volume of coarse aggregate is to be reduced to increase the fine aggregate content.)
corrected volume of C.A for w/c of 0.4 = 0.6 + 0.025 = 0.625
(for pumpable concrete values are reduced by 10%)
Volume of C.A = 0.625 x 0.9 = 0.5625
Volume of F.A = 1 - 0.5625 = 0.4375
a) Volume of concrete = 1 m^3
b) Volume of cement = (Mass of cement / Specific gravity of cement) x (1/100)
= ( 423 / 3.15) x (1/1000)
= 0.1343 m^3
c) Volume of water = (Mass of water / Specific gravity of water) x (1/1000)
= (148/1) x (1/1000)
= 0.148 m^3
d) Volume of Chemical admixture = (Mass of Chemical admixture / Specific gravity of Chemical admixture ) x (1/1000)
= (10/1.145) x (1/1000)
= 0.0087 m^3
e) Volume of total aggregates (C.A + F.A ) = a – (b + c + d )
= 1 – ( 0.1343 + 0.148 + 0.0087 )
= 0.709 m^3
f) Mass of coarse aggregates = e x Vol. of C.A x Specify gravity of C.A x 1000
= 0.709 x 0.5625 x 2.74 x 1000
= 1093 kg/m^3
g) Mass of fine aggregates = e x Vol. of F.A x Specify gravity of F.A x 1000
= 0.709 x 0.4375 x 2.74 x 1000
= 850 kg/m^3
Hence,
Result
Mix proportion for M50 grade concrete is
Cement = 423 Kg/m^3
Water = 148 Kg/m^3
Fine aggregate = 850 Kg/m^3
Coarse aggregate = 1093 Kg/m^3
Chemical Admixture = 10 Kg/m^3
water-cement ratio = 0.35
M35 mix ratio = 1 : 2.01 : 2.584