Project on Concrete Mix Design for various grades of Concrete
I)The Concrete mix design for M35 grade concrete with fly ash;
(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-I).
(d) Max. Nominal size of aggregate = 20mm.
(e) Min. Cement content = 320 kg/m3 (from IS 456: 2000 Table 5)
(f) Maximum water cement ratio: 0.45 (from table 5 of IS 456 - 2000)
(g) Workability: 100mm
(h) Exposure condition: severe (for R.C.C)
(i) method of concrete: Pumping
(j) degree of supervision: Good
(k) type of aggregate: Crushed angular.
(l) maximum cement content: 450 kg/m3
(m) chemical admixture: Super plasticizer
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 I
(d) specific gravity of fly Ash: 2.2
(e) chemical admixture: Super plasticizer conforming to IS 9103.
(f) S.P.G. of C.A = 2.74,
F.A = 2.74
(g) water absorption;
- Coarse aggregate: 0.5%
- Fine aggregate: 1.0%
(h) sieve analysis
- C.A = Zone - I conforming of IS 383 of Table 2.
- F.A = Zone - I conforming of IS 383 of Table 2.
Target strength of mix proportioning;
F’ck = fck + 1.65 S
fck = 35 N/mm2
S= 5 (from table - I of IS 10262: 2009)
Therefore, 35 + 1.65 (5) = 43.25 N/mm2
Selection of water cement ratio;
From table 5 of IS 456: 2000 severe exposure
Maximum w/c ratio = 0.45
According to graph 1 w/c ratio= 0.34
From graph 2 of IS 4031- 1965 w/c ratio = 0.42
Based on experience
Minimum w/c ratio = 0.34
0.34 < 0.45, hence ok.
Selection of water content;
From table to of IS 10262: 2009 maximum water content for 20mm = 186 liter.
(for25mm to 50 mm slump range)
Estimated water content for 100 mm slump= 186 + 6/100 × 186
= 197 liters.
We have to add 3% for 25mm slump after 50 mm slump ratio.
as superplasticizer is used, the water content can be reduced up to 30% based on trials
water content reduction of 29% is achieved with plasticizer.
Hence the arrived water content = 197 - 57
= 140 liters.
Calculation of cement content;
water Cement ratio = 0.34
cementitious material (cement + fly ash) content = 140/ 0.34
= 412 kg/ m3.
From table 5 of IS 456 minimum cement content for sever exposure condition = 320 kg/m3
412 kg /m3 > 320 kg/ m3, hence ok
Now to proportion the mix containing fly ash the following;
Increase cementitious material by 10%
Cementitious material content = 412 + 10/100 × 412
= 453 kg/m3
Water content = 140 kg/m3
So, water cement ratio = 140/453
= 0.309
Fly ash @ 30% total cementitious material content = 30% of 453
= 136 kg/m3.
Cement (OPC) = 453 - 136
= 314 kg/m3
Saving of cement while using fly ash = 412- 314
= 98 kg/m3
Fly ash being utilized = 136 kg/m^3.
Proportion of volume of C.A And F.A
From table 3 of IS 10262 – 2009, Volume of CA corresponding to 20mm size of aggregate and FA (zone1) for water cement ratio 0.5 = 0.6.
In the present case water cement ratio is 0.34 = 0.57
For pumpable concrete the value reduced by 10%
Therefore, the volume of coarse aggregate = 0.57-10/100x0.57
=0.51
Volume of Fine aggregate = 1-0.51 = 0.49
Mix design calculation
a) Volume of concrete = 1m^3
b)Volume of cement = mass of cement/specific gravity of cement x 1/1000.
= 314/3.15 x 1/1000
=99.68 x 0.001
= 0.0996m^3.
c)Volume of fly ash:
=mass of fly ash/ specific gravity of fly ash x 1/1000.
=136/2.2 x 0.001
= 0.061m^3.
d)Volume of water = mass of water/specific gravity of water x 1/1000.
=140/1 x 0.001
=0.140m^3.
e)Volume of admixture:
= mass of admixture/specific gravity of admixture x 1/1000.
= 6.28/1.145 x 0.001
= 0.00054m^3.
f)Volume of aggregate FA & CA = (a-(b+c+d+e)).
= 1-(0.0996+0.06+0.140+0.0054)
=1-0.302
=0.697m^3.
g) Mass of CA = f x volume of CA x specific gravity of CA x 1000.
= 0.697x0.51x2.74x1000
=974kg/m^3.
h) Mass of FA = f x volume of FA x specific gravity of FA x 1000.
=0.697x0.49x2.74x1000.
=936kg/m^3.
Mix proportions;
Cement = 314 kg/m^3.
Fly ash = 136 kg/m^3.
Water = 140 kg/m^3.
Fine aggregate = 936 kg/m^3.
Coarse aggregate = 974 kg/m^3
Admixture = 6.28 kg/m^3.
Water cement ratio = 0.34
Mix ratio; 1 : 2.98 :3.10.
II) M50 GRADE DESIGN MIX
Data collection;
Grade of concrete: M50
Type of cement: OPC 43 Grade.
Maximum nominal size of aggregate: 20mm.
Minimum cement content: 320 kg/m^3.
Maximum water cement ratio: 0.40
Workability: 100mm.
Exposure: severe.
Method of concreting: pumping.
Degree of supervision: good.
Type of cement: crushed angular
Maximum cement content: 450kg/m^3.
Admixture: superplasticizer.
Test data for materials:
Cement used: OPC 43 grade.
Specific gravity of cement: 3.15.
Specific gravity of coarse aggregate and fine aggregate: 2.74
Water absorption: C.A – 0.5%
F.A – 10%
Free surface: C.A & F.A - NIL.
Target mean strength of concrete:
f’ ck = fck+1.65 S.
fck=50
standard deviation S = 5
f’ck = 50+1.65(5) = 58.25 N/mm2.
Selection of water cement ratio:
From table 2 of IS 10262:2009. Maximum water content for 20mm aggregate = 186 liters.
Estimated water content for 100mm slump = 186+6/100x186 = 197 liters.
As superplasticizer 20% of water content reduced,
Therefore, the arrived water content = 158 liter.
Calculation of cement content:
Water cement ratio = 0.40
Cement content= 158/0.40 = 395kg/m^3.
Hence, 395kg/m^3 > 320kg/m^3.hence ok,
Proportion of volume of coarse aggregate of 20mm size:
in present water cement ratio= 0.40.
from table 3, of IS 10262:2009 volume of coarse aggregate of 20mm and fine aggregate zone 1 for water cement ratio = 0.6
therefore, increase in volume of coarse aggregate and decrease in volume of fine aggregate.
The corrected volume of coarse aggregate for the water cement ratio of 0.40=0.62.
For pumpable concrete, these values are reduced by 10%
The volume of coarse aggregate = 0.62x0.9 = 0.56.
Volume of fine aggregate = 1- 0.56 = 0.44
Mix calculation;
a) Volume of concrete = 1m^3.
b) Volume of cement = mass of cement/ specific gravity of cement x 1/1000 =395/3.15x1/1000
= 0.125m^3
c) Volume of water = mass of water /specific gravity of water x 1/1000.
= 158/1 x 1/1000.
=0.158m^3
d) Volume of admixture= mass of admixture/specific gravity of admixture x 1/1000.
=7.9/1.145 x 1/1000
=0.0068m^3
e) Volume of C.A. & F.A = (a-(b+c+d)).
= 1-(0.125+0.158+0.0068)
=1-0.289
= 0.711m^3.
f) Mass of C.A = e x volume of C.A x specific gravity x 1000.
=0.711x0.56x2.74x1000.
=1091kg/m^3.
g) Mass of F.A = e x volume of F.A x specific gravity x 1000.
=0.711x0.44x2.74x1000.
=857kg/m^3.
Mix proportion:
Cement = 395kg/m^3.
Water = 158kg/m^3.
Fine aggregate = 857kg/m^3.
Coarse aggregate = 1091kg/m^3.
Admixture = 7.9kg/m^3.
Water cement ratio = 0.40.
Mix ratio = 1 : 2.16 : 2.76.