Project on Concrete Mix Design for various grades of Concrete
1.Calculate the Concrete Mix Design for M35 grade concrete with fly ash.
Aim:
To calculate the concrete mix design for M35 grade concrete with fly ash as given below.
Introduction:
concrete mix design is the process of finding right proportions of cement,sand and aggregate for concrete to achieve target strength in structure.
concrete mix = cement : sand : aggregate
procedure
1.cement
- Specific gravity of cement = 3.15
- Minimum Cement Content= 320kg
- Maximum Cement Content= 450kg
2.Site conditions
- Exopsure = Severe
- Workability = 100-120mm slump
- Method of placing = pumping
- Degree of supervision = Good
3.Admixture
- Mineral Admixture = Fly ash (30%)
- Sp.Gr admixture super plasticiers = 1.12
- Chemical Admixture = @0.6%
- Sp.Gr of fly ash = 2.50
4.Aggregates
- Sp.Gr of coarse aggregates = 2.74
- Sp.Gr of fine aggregates = 2.74
- water absorption of C.A=0.5%
- water absorption of F.A = 1%
- Type of aggregate = Crushed angular aggregate
- Maximum nominal size = 20mm
Target strength :(Table 2)
Ft = FCK + 1.65*S (where S from table (2) IS10262)
=35+1.65*5
= 43.25 N/MM^2
Selection of water cement ratio :( Table 5 )
The water/ cementitious ratio for target mean strenth and OPC 53 = 0.38
According to the table 5 IS 456:2000;Maximum water cement ratio for exposure = 0.45
Selection of water content: ( Table 4 IS10262 )
Maximum water content = 186 L
Water content for 100mm slump = 186 +(0.06+186)
=197.16 L
=197 L
Since we are use super plasticizers;
% of reduction in water content = 20%
=197 - (0.20 *197)
=157.6
Adopted water content = 158 L
Calculation of cementitious content:
Cementitious content = water content / water cement ratio
=158/0.38
=415.78 Kg
=416kg > 320kg/m^3 ( minimum cementitious )
Increased cementitious content = 415.78 * 1.1
=457.35
=458kg/m^3
Revised W/C ratio =158/458
=0.344
Fly ash content = 30% *458
=30/100 * 458
=137kg/m^3
Cement content =458 - 137
=321 kg/m^3 <450kg/m^3 ( maxmium cement content )
Super plasticiers @ 0.6% the mass = 457.35 @ 0.6%
=457.35 * 0.6/100
=2.74kg
Coarse and fine aggregate:
Coarse aggregate =0.64 ( Table 3 - IS10262 for W/C = 0.5 )
C.A ratio for 0.45W/C ratio = 0.65 ( adding 0.01 for every decrease 0.5 and vice verse )
Final F.A ratio = 1 - 0.65
=0.35
caluculation of volume :
(a) volume of concrete = 1cu.m
(b) Total volume of cement = cement/Sp.Gr *1000
=321/3.15*1000
=0.101cu.m
(c)Total volume of fly ash = Fly ash/Sp.Gr*1000
= 137/2.5*1000
= 0.054cu.m
(d)Volume of water = water/Sp.Gr *1000
=158/1*1000
=0.158cu.m
(e)Volume of superplasticizer = superplasticizer/Sp.Gr*1000
=2.74/1.12 * 1000
=0.002cu.m
(f)Total aggregates requirement = A - (B+C+D+E)
=1-(0.101+0.054+0.158+0.02)
=0.69 cu.m
(g)Coarse aggregates (C.A)=F*C.A ratio * Sp.Gr*1000
=0.69*0.65*2.74*1000
=1228.89 kg
(h)Fine aggregates(F.A) = F*F.A ratio * Sp.Gr * 1000
=0.69*0.35*2.74*1000
=661.71kg
Cement flyash sand aggregate water and super plasticizer needed for M35 grade concrete:
- Cement = 321kg
- Flyash = 118kg
- Fine aggregate =662kg
- Coarse aggregate =1229kg
- water =158L
- Super plasticiers=2L
Result:
The mix ratio of M35 Grade concrete by weight is cement : flyash :CA :FA :Water = 1 :0.4 :2.6 :4.46 :0.58
2.Concrete Mix Design for M50 grade without fkyash:
Aim:
To calculate concrete mix design for M50 grade of concrete witout flyash.
Introduction:
Concrete mix design is the process of finding right proportions of cement,sand and aggregate for concrete to achieve target strength in structures.
Concrete mix = cement :sand :aggregate
The concrete mix design involves various steps,calculations and laboratory testing to find mix propotions , this process is usually to adopted for structure which requires higher grades of concrete.
Given:
Grade = M50
Cement = opc 53
maximum nominal size of aggregate = 20mm
minimum cement content = 360 kg (Table 5 IS456 )
maximum cement content = 450 kg as per code
maximum water cement ratio =0. 45 ( Table 5 IS456 )
Chemical admixture type : super plasticizer.
degree of supervision :good
Procedure:
Test data for materials:
- Cement used opc53 (confirming to IS8112 )
- Specific gravity of cement =3.15 ( test IS:2720 )
- Chemical admixture @0.6% (IS9103 )
- *Specific gravity of CA = 2.74
- *Specific gravity of FA = 2.74
- *admixture pce based = 1.12
- Water absorption ( test IS 2386 )
- CA = .5%
- FA = 1%
- Seive analysis (test IS 2386 )
- CA = confirming to table 2 IS383
- FA = zone II confirming to table IS 383
Target strength:
Fck = fck + 1.65s ( where s from table 1 IS10262 )
=50 +1.65 * 5
=58.25 N/mm^2
PSelection of water content :
Maximum water cement ratio = 0.35 ( table 5 IS456 )
Adopted W/C ratio = 0.35
Selection of water content :
Maximum water content = 186 L (from table 2 IS 1026 )
Adopted water content is for 25 to 50mm slump, but we need 100mm slump, clause of 4.2 of IS 1026 suggests to increase 3% of water content to get an extra 25mm slump.
Hence, 6% to get additional 50mm slump ofour concrete.
Water content for 100mm slump = 186 + (0.06 +186 )
= 197.16 L
Based on trails with super plasticizer water content reduction of 27% has been acheived, hence the arrived water content
=197 - [197* 27/100 ]
=143 L
Calculation of cement content :
Water - cement ratio = 0.35
Cement content = 143/0.35 = 410 kg /m^3
From the table 5 of IS 456,minimum cement content for extreme exposure condition = 360 kg/m^3
Hence, 410kg/m^3 > 369kg/m^3 ok.
Proportion of volume of CA & FA content :
Coarse aggregate ratio = 0.62 ( from table 3 IS10262, for W/C ratio = 0.5 )
Coarse aggregate ratio for 0.35W/C = 0.65 ( adding 0.01 for every decrease .05 and vice versa )
Volume of fine aggregate = 1- 0.585 =0.415
Mix calculation for M50 grade concrete:
a. Volume of concrete = 1m^3
b. Volume of cement = ( mass of cement /SG )*(1/1000) = (410/3.12)*(1/1000) = 0.131m^3
c. Volume of water = ( mass of water / (SG*1000)) =(143/(1*1000) = 0.143m^3
d. Volumeof super plasticizer = ( mass of superplasticizer /(SG*1000) = (8.2/(1.145*1000)) = 0.007m^3
e.Volume of aggregate = (a-(b+c+d)) = 1- (0.131 + 0.143 +0.007) = 0.719m^3
e. Mass of coarse aggregate = volume of all aggregate * volume of CA *SG of CA * 1000
=0.719 * 0.585 * 2.74 * 1000
=1152.48kg
f.Mass of FA =Volume of all aggregate *volume of FA * SG of FA * 1000
= 0.719 * 0.415 * 256 *1000
=763.86kg
Result:
Concrete mix proportion:
cement = 410 kg/m^3
water =143kg/m^3
super plasticizer = 7kg/m^3
FA=763.865kg/m^3
CA=1152.48kg/m^3
w/c ratio = 0.35
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