Project on Concrete Mix Design for various grades of Concrete
1.Calculating the design mix for M35 grade concrete with fly ash, here is the some suitable data
Grade of concrete : M35
Type of cement:OPC 43
Maximum nominal size of coarse aggregates: 20mm
exposure condition: 'very severe'
min cementg content : 340 kg/m^3
max water cement ratio as per exposure condition:0.45
workability:100mm for reinforced comcrete
admixture :Fly ash
Target mean strength f'ck=fck=1.65*S =35=1.65*5= 43.25
Selection of water cement ratio from the table5 IS456 :2000 for very severe condition minimum water cement ratio =0.45 based on trails = 0.36. hence it is ok.
Selection of water content From the IS 10262:2009 maximum water content for 20mm size of aggregate = 186 liters or kg
estimated water content for 100mm slump = 186+6/100-186=197 ( every 25mm slump increase add 3 % of water content)
superplastisizers is used then the water content can be reduced up to 30%.
based on trails water reduced of 29% is acheived with plastisizers. where 197*0.71=140
Calculation of cementious content
water content/water cement ratio=140/0.36=389 kg/m^3
as per IS 456:2000 the minimum cement content=340kg/m^3
389kg/m^3>340kg/m^3 hence is ok.
Based on experience and trails 10% cementous content increased cementious content for add the fly ash.
cementious material content =389*110%=427.9kg/m^3
water cement ratio=140/428=0.327
30% Fly ash added
total cementious content =428*0.30=128
cement opc=428-128=300 kg/m^3
usage of Flyash=128
saving of cement while using fly ash=389-300=89kgs/m^3
Volume of coarse aggregate and fine aggregate content
as per IS 10262:2009 vol of coarse aggregate for 20 mm
fine aggregate for water cement ratio of 0.5 = 0.6 now w/c ratio is 0.36.
coarse aggregates increases to reduce the fine aggregate content. corrected vol of coarse aggregate 0.36=0.56
for pumpable concrete coarse aggregate reduced by 10%
then, vol of coarse aggregate= 0.56*10%=0.056
vol of fine aggregate content=1-0.56=0.044
Volume of Concrete
(a) Volume of concrete=1m^3
(b) Volume of cement content= 300/3.15*1/1000=0.095 m^3
(c)Volume of flyash =128/2.2*1/1000=0.058 m^3
(d) Volume of water =140/1*1/1000= 0.140 m^3
(e) volume of chemical admixtures =7.6/1.145*1/1000= 0.006m^3
(f) volume of total aggregates =(a-(b+c+d+e))=(1-(0.095+0.058+0.140+0.006)=0.701m^3
(g) mass of coarse aggregates = 0.701*0.56*2.74*1000=1076 kg/m^3
(h) mass of fine aggregates =0.701*0.44*2.74*1000=852 kg/m^3
Mix proportion for trail
Cement =300 kg/m^3
Flyash=128 kg/m^3
Water=140
Fine aggregates= 852 kg/m^3
Coarse aggregate =1076 kg/m^3
chemical admixture = 7 kg/m^3
water cement ratio = 0.327
Trail mix raio = 1: 1.99: 2.23.
2.Mix Design for M50 grade of concrete
Grade of concrete = M50
Type of cement = OPC 43
Maximum nominal size of aggregates = 20mm
Exposure condition: Moderate
Minimum cement content =300 kg/m^3
maximum water cement ratio = 0.50
Workability = 75 mm slump
Target mean strength
f'ck= fck+1.65*S = 50+1.65*5 = 58.25
Selection water cement ratio
Based on experience and trails water cement ratio 0.35 obtained from the graph as per IS 10262:2009
Selection of water content
For the nominal size of aggregates 20mm water content is 186 kg or liters
estimated water content for 75 mm slump = 186+3%*186= 191.58
As superplastisizers is used reduce the water content
Based on trails water reduction of 25% is achieved with plasticizers.
Hence the water content = 191.58*100-25%=144
Calculation of cement content
water content/water cement ratio = 144/0.35 = 411 kg/m^3
From the table5 IS 456:2000 minimum cement content for moderate condition = 300kg/m^3
411 kg/m^3 > 300kg/m^3 Hence it is ok.
Volume of total aggregates
Volume of coarse aggregates for nominal size of 20mm grade as per IS 10262:2009 is 0.66
then,
volume of fine aggregates is 1-0.66=0.34
Mix Calculations
(a) volume of concrete = 1m^3
(b) volume of cement = mass of cement/specific gravity of cement*1/1000= 411/3.15*1/1000= 0.130
(c) volume of water = mass of water /specific gravity of water*1/1000 = 144/1*1/1000 = 0.144
(d) volume of chemical admixture = 6.5/1.145* 1/1000= 0.0056
(e) Volume of aggregates = (1-(0.130+0.144+0.0056)) = 720 kg/m^3
(f) mass of coarse aggregates = e* volume of coarse aggregate*specific gravity of coarse aggregate* 1000
720*0.66*2.74*1000 = 1302 kg/m^3
(g) mass of fine aggregate = 720*0.34*2.74*1000=670 kg/m^3
Mix proportion for trail
cement = 411 kg/m^3
water = 144 kg/m^3
Fine aggregates = 670 kg/m^3
coarse aggregates = 1302 kg/m^3
chemical admixtures = 6.5 kg/m^3
Water cement ratio =0.35 kg/m^3
Trail mix ratio
1: 1.63: 3.16