Project on Concrete Mix Design for various grades of Concrete
Solution 1:-
Concrete Mix Design for M35 grade concrete with fly ash is as follows:-
1 Stipulations
a Grade designation: M35
b Type of cement: opc 43 grade conforming to IS 8112
c Type of mineral admixture: Flyash conforming to IS 3812( PART 1)
d Max nominal size of aggregate: 20mm
e Min cement content: 320kg/m3 from Table 5 IS 456 2000 condition severe.
f Max w/c ratio : 0.45 from Table 5 IS 456 2000 condition severe.
g Workability: 100mm ( slump for pumping)
h Exposure condition:- Severe ( for RCC).
i Method of concrete placing:- pumping
j Degree of supervision:- Good
k Type of aggregate:- Crush angular aggregate
l Max cement content:- 450 kg/m3
m Chemical type:- Superplasticizer.
2 Test data for material
a Cement used: OPC 43 grade conforming to IS 8112.
b Specific gravity of cement:- 3.15
c Flyash: conforming to IS 3812( PART 1)
d Specific gravity of flyash: 2.2
e Chemical admixture:- Superplasticizer conforming to IS 9103
f Specific gravity :- CA -2.74
FA - 2.74
g Water absorption:- CA -0.5%
FA- 1.0%
h Free surface moisture: CA- NIL
FA- NIL
i Sieve analysis - CA - Table 2 IS 383
FA - Table 4 IS 383 ( ZONE I)
3 Target strength for mix proportioning
f'ck= fck+ 1.65S From IS 10262 2019 table no2
f'ck= 35+ 1.65x 5= 43.25N/mm2
4 Selection of w/c ratio
w/c= 0.45 from IS 456 2000 condition severe
w/c= 0.5 from IS 10262 2019 fig 1
So w/c= 0.45 , which is less then 0.5
5 Selection of water content
Max water content for 20mm agg= 186 litre ( for 25 to 50mm slump) from IS 10262 2019 table no 4
So Max water content for 100mm slump= (186+(6/100)x186)=197 litre from IS 10262 2019 clauce 5.3
As superplasticizer is used , the water content may be reduced upto 20% and above based on trails water reduction of 29% is achieved with plasticizer.
So water content = 197x 0.71= 140 litre.
6 Calculation of cement content
w/c= 0.45
cement + flyash= 140/0.45= 311.11 kg/m3
So cement content= 320kg/m3 From IS 456 2000 table no 5 condition severe.
So cement content = 320kg/m3 which is greater than 311.11kg/m3
Proportioning containing flyash is as follows
The increase in cement content is based on experience this example is based on 10% increase.
So cement content = 320 x 1.1= 352kg/m3
Flyash @ 30% of total cement content= 352 x 0.3= 105.6kg/m3= 106kg/m3
So cement content= 352 - 106 = 246kg/m3
saving of cement = 352-246= 106 kg/m3
7 Proportioning of vol of CA and FA
Vol of CA corresponds to 20mm agg and FA ( zone -I) for W/C of 0.5= 0.6 from IS 10262 2019 table 5
But calculated w/c= 0.45
As w/c is lower by 0.05
So vol of CA needs to be increased by 0.01 from IS 10262 2019 clause 5.5.1
So vol of CA for w/c 0.45= 0.61
Note If agg is not angular then vol of agg may be required to increased.
for pumpable concrete values of Vol of CA is reduced by 10%
So vol of CA = 0.61X 0.9= 0.55
Vol of FA= 1-0.55=0.45
8 Mix Calulations:-
A Vol of concrete = 1m3
B Vol of cement= (246/3.15)x(1/1000)= 0.0780m3
C Vol of flyash= (106/2.2)x(1/1000)=0.048m3
D Vol of water= (140/1)x(1/1000)= 0.14m3
E Vol of admkixture= (7.6/1.145)x(1/1000)=0.006 mass of admixture = (320x0.024)= 7.6kg
E Vol of agg= (A-(b+c+d+e))= (1-( 0.0780+0.048+ 0.14+0.006 ))=0.728m3
F Mass of CA = E x Vol of CAx SG of CA X 1000= 0.728x 0.55x 2.74x 1000=1098kg/m3
G Mass of FA = E x Vol of FAx SG of FA X 1000= 0.728x 0.45x 2.74x 1000=898kg/m3
9 Mix proportion
Cement= 246kg/m3
Flyash= 106kg/m3
water= 140kg/m3
FA = 898kg/m3
CA= 1098kg/m3
chemical admixture= 7.6kg/m3
w/c= 0.45
Trail mix ratio =1: 2.55:3.12
The concrete mix design for M50 grade concrete is as follows:-
1 Stipulations
a Grade designation: M50
b Type of cement: opc 43 grade conforming to IS 8112
c Max nominal size of aggregate : 20mm
d Min cement content: 320kg/m3 from Table 5 IS 456 2000 condition severe.
e Max w/c ratio : 0.45 from Table 5 IS 456 2000 condition severe.
f Workability: 100mm ( slump for pumping)
g Exposure condition:- Severe ( for RCC).
h Method of concrete placing:- pumping
i Degree of supervision:- Good
j Type of aggregate:- Crush angular aggregate
k Max cement content:- 450 kg/m3
l Chemical type:- Superplasticizer.
2 Test data for material
a Cement used: OPC 43 grade conforming to IS 8112.
b Specific gravity of cement:- 3.15
c Chemical admixture:- Superplasticizer conforming to IS 9103
d Specific gravity of cement :- CA -2.74
FA - 2.74
e Water absorption:- CA -0.5%
FA- 1.0%
f Free surface moisture: CA- NIL
FA- NIL
G Sieve analysis - CA - Table 2 IS 383
FA - Table 4 IS 383 ( ZONE I)
3 Target strength for mix proportioning
f'ck= fck+ 1.65S From IS 10262 2019 table no2
f'ck= 50+ 1.65x 5= 58.25N/mm2
4 Selection of w/c ratio
w/c= 0.45 from IS 456 2000 condition severe
w/c= 0.5 from IS 10262 2019 fig 1
So w/c= 0.45 , which is less then 0.5
5 Selection of water content
Max water content for 20mm agg= 186 litre ( for 25 to 50mm slump) from IS 10262 2019 table no 4
So Max water content for 100mm slump= (186+(6/100)x186)=197 litre from IS 10262 2019 clauce 5.3
As superplasticizer is used , the water content may be reduced upto 20% and above based on trails water reduction of 20% is achieved with plasticizer.
So water content = 197x 0.8= 158 litre.
6 Calculation of cement content
w/c= 0.45
cement content= 150/0.45= 359 kg/m3
So cement content= 320kg/m3 From IS 456 2000 table no 5 condition severe.
So cement content = 359kg/m3 which is greater than 320kg/m3
7 Proportioning of vol of CA and FA
Vol of CA corresponds to 20mm agg and FA ( zone -I) for W/C of 0.5= 0.6 from IS 10262 2019 table 5
But calculated w/c= 0.45
As w/c is lower by 0.05
So vol of CA needs to be increased by 0.01 from IS 10262 2019 clause 5.5.1
So vol of CA for w/c 0.45= 0.632
Note If agg is not angular then vol of agg may be required to increased.
for pumpable concrete values of Vol of CA is reduced by 10%
So vol of CA = 0.632X 0.9= 0.568
Vol of FA= 1-0.568=0.432
8 Mix Calulations:-
A Vol of concrete = 1m3
B Vol of cement= (359/3.15)x(1/1000)= 0.114m3
C Vol of water= (158/1)x(1/1000)= 0.158m3
D Vol of admixture= (7.6/1.145)x(1/1000)=0.006m3 mass of admixture = (359x0.021)= 7.6kg
E Vol of agg= (A-(b+c+d))= (1-(0.114+0.158+0.006))=0.722m3
F Mass of CA = E x Vol of CAx SG of CA X 1000= 0.722x 0.568x 2.74x 1000=1123kg/m3
G Mass of FA = E x Vol of FAx SG of FA X 1000= 0.722x 0.432x 2.74x 1000=821kg/m3
9 Mix proportion
Cement= 359kg/m3
water= 158kg/m3
FA = 821kg/m3
CA= 1123kg/m3
chemical admixture= 7.6kg/m3
w/c= 0.45
Trail mix ratio =1: 2.28:3.12