Project on Concrete Mix Design for various grades of Concrete
1) MIX DESIGN FOR M35 GRADE CONCRETE WITH FLYASH.
a) AIM:
TO determine the mix design for M35 grade concrete with flyash as a cementitious material to the concrete. to determine the content of different materials required for the concrete with respect to the ratio obtained from mix design
The data required for mix design is obtained from IS 10262 - 2019 code book.
b) INTRODUCTION:
The brief summary of materials required for the mix design
- M35 grade of concrete
- cementitious material= flyash
- grade of cement = 43
- Max size of aggregate = 20 mm
- Exposure condition= severe
- Min. cement content = 320 kg/m ^3 As per (IS 456 - 2000)
- Max. water cement ratio=0.37 As per (IS 10262 - 2019)
- slump value = 100 mm
- concrete placing= by pumping
- crushed angular aggregate
- Degree of supervision= Good
- Max. cement content = 450 kg/m^3 As per (IS 456 - 2000)
- Zone 1 sand
- flyash content 30 %
c) PROCEDURE FOR MIX DESIGN:
i) To determine target mean strength of concrete
ft=fck+1.65s
ft= target mean strength at 28 days of concrete curing
fck = compressive strength of concrete at 28 days of concrete curing
s = standard deviation according to grade of concrete as per (IS 10262 - 2019)
ft=35+1.65x5
ft=43.25 n/mm^2
ii) To determine water cement ratio as per (IS 10262 - 2019)
w/c = 0.37 (according to the graph of water cement ratio & 28 days compressive strength of concrete , curve 2 is taken into consideration as the grade of cement is 43 )
iii) To determine the water content as per (IS 10262 - 2019)
water content = 186 litres (according to the aggregate size & slump value is considered from IS 10262 - 2019 )
as 100 mm slump is considered there is a increase of +/- 3% for 25 mm slump
=186+(6/100x186)
= 197
as the chemical admixture is taken into consideration which is superplasticizer the water content is decreased by 29%
= 197x0.71
= 139.87
the final water content = 139.87 litres
iv) To determine the cement content from water cement ratio
w/c = 0.37 water content = 139.87
cement content = 139.87/0.37
cement content = 378.02 kg (according to IS 456 - 2000 value of cement content is more than min. cement content i.e 378.02>320 hence ok )
v) To determine the coarse aggregate & fine aggregate ( as per IS 10262 - TABLE NO. 5)
The value of coarse aggregate is mainly depends on zone of aggregate & size of aggregate
the table value mainly shows the reading for 0.5 water cement ratio so at the rate of +/- 0.01 for every +/- 0.05 change in water cement ratio
the value obtained is (the value obtained from table is mainly for the water cement ratio 0.5, as the water cement ratio is decreased the value of coarse aggregate should be increased that is 0.626)
volume of coarse aggregate for w/c ratio = 0.626
for pumpable concrete values are reduced by 10 % .
volume of coarse aggregate = 0.626x0.9=0.5634 m^3
volume of fine aggregate = 1-0.5634= 0.4366 m^3
vi) Mix calculation :
a) volume of concrete = 1 m^3
b) volume of cement =( mass of cement / specific gravity of cement ) x 1/1000
volume of cement = (378.02/3.15)x(1/1000)
volume of cement = 0.12 m^3
c) volume of water = ( mass of water / specific gravity of water ) x 1/1000
volume of water = (139.87/1)x(1/1000)
volume of water = 0.139 m^3
d) volume of chemical admixture = ( mass of chemical admixture / specific gravity of chemical admixture ) x 1/1000
mass of chemical admixture is 1% to 3% of cementitious material (as per IS 456 - 2000 )
lets assume 2% of cementitious material
mass of chemical admixture = 378.02x2/100
mass of chemical admixture = 7.56 kg
vol. of chemical admixture = (7.56/1.145)x(1/1000)
vol. of chemical admixture = 0.0066 m^3
e) volume of flyash
cementitious material = 378.02x1.10 (the value is increased as the flyash content is more than 20 % )
cementitious material = 415.822 kg/m^3
water content = 139.87 litres
w/c ratio = 139.87/415.822
w/c ratio = 0.336
flyash is considered as 30% of total cementitious material
flyash = 415.822x0.3
flyash = 124.74 kg/m^3
volume of flyash = (124.74/2.2)x(1/1000)
volume of flyash = 0.079 m^3
f) volume of coarse aggregate & fine aggregate =(a-(b+c+d+e))
volume of all aggregate = (1-(0.12+0.139+0.0066+0.079))
volume of all aggregate = 0.6554 m ^3
vii) mass of coarse aggregate = vol. of all aggregate x vol. of coarse aggregate x specific gravity of coarse aggregate x 1000
mass of coarse aggregate = 0.6554x0.5634x2.74x1000
mass of coarse aggregate = 1011.75 kg/m^3
mass of fine aggregate = 0.6554x0.4366x2.74x1000
mass of fine aggregate = 784.04 kg/m^3
D) FINAL RESULT
Mix proportion for M35 grade of concrete
CEMENT = 378.02 kg/m^3
FLYASH = 124.74 kg/m^3
WATER = 139.87 kg/m^3
CHEMICAL ADMIXTURE = 7.56 kg/m^3
FINE AGGREGATE = 784.04 kg/m^3
COARSE AGGREGATE =1011.75 kg/m^3
WATER CEMENT RATIO = 0.37
MIX RATIO = 1:2.07:2.67
2) MIX DESIGN FOR M50 GRADE CONCRETE WITHOUT FLYASH.
a) AIM:
TO determine the mix design for M50 grade concrete . To determine the content of different materials required for the concrete with respect to the ratio obtained from mix design.
The data required for mix design is obtained from IS 10262 - 2019 code book.
b) INTRODUCTION:
The brief summary of materials required for the mix design
- M50 grade of concrete
- grade of cement = 43
- Max size of aggregate = 20 mm
- Exposure condition= severe
- Min. cement content = 320 kg/m ^3 As per (IS 456 - 2000)
- Max. water cement ratio=0.3 As per (IS 10262 - 2019)
- slump value = 100 mm
- concrete placing= by pumping
- crushed angular aggregate
- Degree of supervision= Good
- Max. cement content = 450 kg/m^3 As per (IS 456 - 2000)
- Zone 1 sand
c) PROCEDURE FOR MIX DESIGN:
i) To determine target mean strength of concrete
ft=fck+1.65s
ft= target mean strength at 28 days of concrete curing
fck = compressive strength of concrete at 28 days of concrete curing
s = standard deviation according to grade of concrete as per (IS 10262 - 2019)
ft=50+1.65x5
ft=58.25 N/mm^2
ii) To determine water cement ratio as per (IS 10262 - 2019)
w/c = 0.3 (according to the graph of water cement ratio & 28 days compressive strength of concrete , curve 2 is taken into consideration as the grade of cement is 43 )
iii) To determine the water content as per (IS 10262 - 2019)
water content = 186 litres (according to the aggregate size & slump value is considered from IS 10262 - 2019 )
as 100 mm slump is considered there is a increase of +/- 3% for 25 mm slump
=186+(6/100x186)
= 197
as the chemical admixture is taken into consideration which is superplasticizer the water content is decreased by 29%
= 197x0.71
= 140
the final water content = 140 litres
iv) To determine the cement content from water cement ratio
w/c = 0.3 water content = 140
cement content = 140/0.3
cement content = 466.66 kg (according to IS 456 - 2000 value of cement content is more than min. cement content i.e 466.66>320 hence ok )
v) To determine the coarse aggregate & fine aggregate ( as per IS 10262 - TABLE NO. 5)
The value of coarse aggregate is mainly depends on zone of aggregate & size of aggregate
the table value mainly shows the reading for 0.5 water cement ratio so at the rate of +/- 0.01 for every +/- 0.05 change in water cement ratio
the value obtained is (value obtained from table is 0.6 but as the water cement ratio is less than 0.5 the value should be increased which is taken as 0.64)
volume of coarse aggregate for w/c ratio = 0.64
for pumpable concrete values are reduced by 10 % .
volume of coarse aggregate = 0.64x0.9=0.576 m^3
volume of fine aggregate = 1-0.576= 0.424 m^3
vi) Mix calculation :
a) volume of concrete = 1 m^3
b) volume of cement =( mass of cement / specific gravity of cement ) x 1/1000
volume of cement = (466.66/3.15)x(1/1000)
volume of cement = 0.148 m^3
c) volume of water = ( mass of water / specific gravity of water ) x 1/1000
volume of water = (140/1)x(1/1000)
volume of water = 0.14 m^3
d) volume of chemical admixture = ( mass of chemical admixture / specific gravity of chemical admixture ) x 1/1000
mass of chemical admixture is 1% to 3% of cementitious material (as per IS 456 - 2000 )
lets assume 2% of cementitious material
mass of chemical admixture = 466.66x2/100
mass of chemical admixture = 9.33 kg
vol. of chemical admixture = (9.33/1.145)x(1/1000)
vol. of chemical admixture = 0.0081 m^3
e) volume of coarse aggregate & fine aggregate =(a-(b+c+d))
volume of all aggregate = (1-(0.148+0.14+0.0081))
volume of all aggregate = 0.7039 m ^3
vii) mass of coarse aggregate = vol. of all aggregate x vol. of coarse aggregate x specific gravity of coarse aggregate x 1000
mass of coarse aggregate = 0.7039x0.576x2.74x1000
mass of coarse aggregate = 1110.92 kg/m^3
mass of fine aggregate = 0.7039x0.424x2.74x1000
mass of fine aggregate = 817.76 kg/m^3
D) FINAL RESULT
Mix proportion for M50 grade of concrete
CEMENT = 466.66 kg/m^3
WATER = 140 kg/m^3
CHEMICAL ADMIXTURE = 9.33 kg/m^3
FINE AGGREGATE = 817.76 kg/m^3
COARSE AGGREGATE =1110.92 kg/m^3
WATER CEMENT RATIO = 0.3
MIX RATIO = 1:1.75:2.38