Project on Concrete Mix Design for various grades of Concrete
Calculate the Concrete Mix Design for M35 grade concrete with fly ash :-
1)TARGET MEAN STRENGTH
f'ck=fck+1.65s
=35+1.65*5.0
=43.25 N/mm^2
2)Selection of water cement ratio:-
in IS 456:2000 water cement ratio-0.45
0.4<0.45 HENCE OK
3)Selecting of water content :-
20 mm size 186 liter ( for 50mm slump)
for 100mm slump 186
186+(6/100)*186
=197 liters
Add super plasticizar 20% reduce
197*0.8=158 l
Total water required 158 liters
4)calculatation of cement and fly ash required
adopted w/c ratio-0.44
total cementitious (cement+fly ash) ewquired =158/0.44=359 kg/m^3
the cementitious content has to bee increase in order toattin good workability and strength.the increment required is determined on the basis of experience and researce . Here we are increasing cementitious content by 10% =359*1.10=395 kg final cementitious content.
From Table 5 of IS456,minimum cement content for'severe' condition 320 kg
395kg/m^3>320 kg/m^3 hence ok.
Cementitious material content 395
New water cement Ratio 158/395 =0.4
Fly ash content is taken as30% of total cementitious content 395*.3=118.5 kg/m^3
cement required 395-118=277 kg/m^3
Cement quality saved due to fly ash addition 350-277=73 kg/m^3
5)volume of coarse aggregate and fine aggregate content
From Table 3 of IS 10262:2009 volume of coarse aggregate corresponding to 20 mm size aggregate (zoneII) for water cement ratio 0.50 in present case water cement ratio 0.44,volume of coarse aggregate is required to be increase and decrease the fine aggregate content.as cthe water cement ratio is lower by 0.06
Therefore,corrected proportion of volume of coarse aggregate for the water-cement ratio0.44=0.632
For pumpable concrete these values should be reduced up to 10% Therefore,volume of coarse aggregate =0.632*0.9=0.568
volume of fine aggregate 1-0.568=0.431
6)Mix calculation
a)Volume of concrete 1m^3
- b) Volume of cement
(Mass of cement/s.p.Gravity of cement)*1/1000
=(277/3.158)*1/100=0.087m^3
c)Volume of fly ash =(Mass of fly ash/ s.p. Gravity of fly ash )*1/1000
=(118/2.2)*1/1000
=0.053
d)Volume of water=(mass of water/s.p. Gravity of water )*1/1000
=(158/1)*1/1000=0.158M^3
e)volume of chemical admixture
=mass of admixture/s.p. gravity of admixture*1/1000
=(7.6/1.145)*(1/1000)=0.006
f)Volume of all aggregate=
a-(b+c+d+e)=1-(0.087+0.053+0.158+0.006)
=0.696m^3
g)Mass of concrete =
f*(volume of coarse aggregate*s.p. gravity of fine aggregate*1000)
=0.696*0.568*2.74*1000
=1077 kg/m^3
h)Mass of fine aggregate =f*volume of fine aggregate*s.p. gravity of fine aggregate*1000)
=0.696*0.431*2.74*1000
=822 kg/m^3
7)mix proportions
cement 277 kg/m^3
fly ash 118kg/m^3
water 158l/m^3
Fine aggregate 822 kg/m^3
coarse aggregate 1077 kg/m^3
chemical admixture 7.18 kg/m^3
water cement ratio 0.4
mix proportion by weight 1:2.9:3.8
M50 grade concrete without Fly ash:-
1)Target Mean Strength
f'ck=fck+1.65s
=50+1.65*5.0=58.25N/mm^2(from IS 10262:2009)
2)Selection of water cement ratio:-
in IS 456:2000 water cement ratio-0.45
0.4<0.45 HENCE OK
Adopted w/c ratio =0.44
3)Selecting of water content :-
20 mm size 186 liter ( for 50mm slump)
for 100mm slump 186
186+(6/100)*186
=197 liters
Add super plasticizar 20% reduce
197*0.8=158 l
Total water required 158 liters
4)Calculation of cement required
1158/0.44=359 kg/m^3
from Table 5 of IS 456,minimum cement content 'severe' exp[osure conditions 359kg/m^3>320kg/m^3 hence;ok
5)volume of coarse aggregate and fine aggregate content
From Table 3 of IS 10262:2009 volume of coarse aggregate corresponding to 20 mm size aggregate (zoneII) for water cement ratio 0.50 in present case water cement ratio 0.44,volume of coarse aggregate is required to be increase and decrease the fine aggregate content.as cthe water cement ratio is lower by 0.06
Therefore,corrected proportion of volume of coarse aggregate for the water-cement ratio0.44=0.632
For pumpable concrete these values should be reduced up to 10% Therefore,volume of coarse aggregate =0.632*0.9=0.568
volume of fine aggregate 1-0.568=0.431
6)Mix calculation
a)Volume of concrete 1m^3
- b) Volume of cement
(Mass of cement/s.p.Gravity of cement)*1/1000
=(359/3.158)*1/100=0.114 m^3
d)Volume of water=(mass of water/s.p. Gravity of water )*1/1000
=(158/1)*1/1000=0.158M^3
e)volume of chemical admixture
=mass of admixture/s.p. gravity of admixture*1/1000
=(7.6/1.145)*(1/1000)=0.006 m^3
f)Volume of all aggregate=
a-(b+c+d+e)=1-(0.114+0.158+0.006)
=0.722m^3
g)Mass of concrete =
f*(volume of coarse aggregate*s.p. gravity of fine aggregate*1000)
=0.722*0.568*2.74*1000
=1123 kg/m^3
h)Mass of fine aggregate =f*volume of fine aggregate*s.p. gravity of fine aggregate*1000)
=0.722*0.431*2.64*1000
=821 kg/m^3
7)mix proportions
cement 359 kg/m^3
water 158 l/m^3
Fine aggregate 821 kg/m^3
coarse aggregate 1123 kg/m^3
chemical admixture 7.18 kg/m^3
water cement ratio 0.44
mix proportion by weight 1:2.8:3.12