Project on Concrete Mix Design for various grades of Concrete
Ans . Mix design for M35 grade of concrete without flyash.
A. Stipulations for proportion
1. Grade designation M35
2. Types of cement OPC 43 GRADE
3. Maximum nominal size of aggregate 20 mm
4. Maximum water cement ratio 0.45
5. Minimum cement content 340 kg/m^3
6. Workability 100 mm ( slump)
7. Exposure condition severe
8. Method of placing concrete Pumping
9. Degree of supervision Good
10. Types of aggregate angular
11. Maximum cement content 450 kg /m^3
12. Chemical admixture Types Superplasticizer
Data collection
Specific gravity of cement OPC 43 Grade - 3.15
Specific gravity of coarse and fine aggregate - 2.74
Specific gravity of water and chemical admixture- 1.0
a. Strength for proportion
Ft = FCK + (1.65 *S)
S = Standard deviation for M35 = 5
35 +(1.65*5)= 43.25 N/mm^2
b. Selection of water cement ration
From is 456- 2000 maximum water cement ratio will be 0.45 based on experience and from graph 1 it will be = 0.38 approximately
0.38 < 0.45 hence ok
c. Selection of water content
Maximum water content = 186 kg /m^3 (for slump 100 mm)
Estimated water = 186 +(0 .6 * 186 /100)
= 197.16 lt.
As superplaticizer is water content can be reduced upto 29% and above based in trails
= 197.16 * 0.71 = 140 lts.
Calculation of water cement ratio = 0.38
d. Cement content = 140 / 0.38
= 368.42 kg /m^3
320 kg <368.4kg < 450 kg (Okey )
e. Coarse and fine aggregate calculation
From table 3 of is 456: 2000 volume of coarse aggregate to 20 mm size aggregate and fine aggregate Zone 1. For water cement 0.5 = 0.6
In present case the water cement ratio = 0.38
Therefore volume of coarse aggregate is increased to decrease the fine aggregate content
Coarse aggregate for the water cement ratio of 0.38 = 0.624
Reduce by 10 percent for pumping concrete
Coarse aggregate = 0.624 * 0.9 = 0.5616
Fine aggregate = 1- 0.5616 = 0.4386
1. Mix proportion
A. Volume of concete = 1 cum
B. Volume of cement = (368.42 / 3.15)/ 1000 = 0.117 cum or 368.42 kg
C. Volume of water = 140 / 1000 =.14 cum. Or 140 kg
D. Volume of chemical admixture = 7/1000 =0.007 cum or 7 kg
E. Volume of aggregate = (1- (0.117+0.14+0.007))
= 0.736 cum
Volume of coarse aggregate = 0.736*0.5616*2.74 * 1000 = 1132.55 kg/m^3
Volume of fine aggregate = 0.736*0.4384*2.74 * 1000 = 884.0 kg/m^3
Ans . Mix design for M35 grade of concrete without flyash.
A. Stipulations for proportion
1. Grade designation M35
2. Types of cement OPC 43 GRADE
3. Maximum nominal size of aggregate 20 mm
4. Maximum water cement ratio 0.45
5. Minimum cement content 340 kg/m^3
6. Workability 100 mm ( slump)
7. Exposure condition severe
8. Method of placing concrete Pumping
9. Degree of supervision Good
10. Types of aggregate angular
11. Maximum cement content 450 kg /m^3
12. Chemical admixture Types Superplasticizer
Data collection
Specific gravity of cement OPC 43 Grade - 3.15
Specific gravity of coarse and fine aggregate - 2.74
Specific gravity of water and chemical admixture- 1.0
a. Strength for proportion
Ft = FCK + (1.65 *S)
S = Standard deviation for M35 = 5
35 +(1.65*5)= 43.25 N/mm^2
b. Selection of water cement ration
From is 456- 2000 maximum water cement ratio will be 0.45 based on experience and from graph 1 it will be = 0.38 approximately
0.38 < 0.45 hence ok
c. Selection of water content
Maximum water content = 186 kg /m^3 (for slump 100 mm)
Estimated water = 186 +(0 .6 * 186 /100)
= 197.16 lt.
As superplaticizer is water content can be reduced upto 29% and above based in trails
= 197.16 * 0.71 = 140 lts.
Calculation of water cement ratio = 0.38
d. Cement content = 140 / 0.38
= 368.42 kg /m^3
320 kg <368.4kg < 450 kg (Okey )
e. Coarse and fine aggregate calculation
From table 3 of is 456: 2000 volume of coarse aggregate to 20 mm size aggregate and fine aggregate Zone 1. For water cement 0.5 = 0.6
In present case the water cement ratio = 0.38
Therefore volume of coarse aggregate is increased to decrease the fine aggregate content
Coarse aggregate for the water cement ratio of 0.38 = 0.624
Reduce by 10 percent for pumping concrete
Coarse aggregate = 0.624 * 0.9 = 0.5616
Fine aggregate = 1- 0.5616 = 0.4386
1. Mix proportion
A. Volume of concete = 1 cum
B. Volume of cement = (368.42 / 3.15)/ 1000 = 0.117 cum or 368.42 kg
C. Volume of water = 140 / 1000 =.14 cum. Or 140 kg
D. Volume of chemical admixture = 7/1000 =0.007 cum or 7 kg
E. Volume of aggregate = (1- (0.117+0.14+0.007))
= 0.736 cum
Volume of coarse aggregate = 0.736*0.5616*2.74 * 1000 = 1132.55 kg/m^3
Volume of fine aggregate = 0.736*0.4384*2.74 * 1000 = 884.0 kg/m^3
Ans . Mix design for M35 grade of concrete without flyash.
A. Stipulations for proportion
1. Grade designation M35
2. Types of cement OPC 43 GRADE
3. Maximum nominal size of aggregate 20 mm
4. Maximum water cement ratio 0.45
5. Minimum cement content 340 kg/m^3
6. Workability 100 mm ( slump)
7. Exposure condition severe
8. Method of placing concrete Pumping
9. Degree of supervision Good
10. Types of aggregate angular
11. Maximum cement content 450 kg /m^3
12. Chemical admixture Types Superplasticizer
Data collection
Specific gravity of cement OPC 43 Grade - 3.15
Specific gravity of coarse and fine aggregate - 2.74
Specific gravity of water and chemical admixture- 1.0
a. Strength for proportion
Ft = FCK + (1.65 *S)
S = Standard deviation for M35 = 5
35 +(1.65*5)= 43.25 N/mm^2
b. Selection of water cement ration
From is 456- 2000 maximum water cement ratio will be 0.45 based on experience and from graph 1 it will be = 0.38 approximately
0.38 < 0.45 hence ok
c. Selection of water content
Maximum water content = 186 kg /m^3 (for slump 100 mm)
Estimated water = 186 +(0 .6 * 186 /100)
= 197.16 lt.
As superplaticizer is water content can be reduced upto 29% and above based in trails
= 197.16 * 0.71 = 140 lts.
Calculation of water cement ratio = 0.38
d. Cement content = 140 / 0.38
= 368.42 kg /m^3
320 kg <368.4kg < 450 kg (Okey )
e. Coarse and fine aggregate calculation
From table 3 of is 456: 2000 volume of coarse aggregate to 20 mm size aggregate and fine aggregate Zone 1. For water cement 0.5 = 0.6
In present case the water cement ratio = 0.38
Therefore volume of coarse aggregate is increased to decrease the fine aggregate content
Coarse aggregate for the water cement ratio of 0.38 = 0.624
Reduce by 10 percent for pumping concrete
Coarse aggregate = 0.624 * 0.9 = 0.5616
Fine aggregate = 1- 0.5616 = 0.4386
1. Mix proportion
A. Volume of concete = 1 cum
B. Volume of cement = (368.42 / 3.15)/ 1000 = 0.117 cum or 368.42 kg
C. Volume of water = 140 / 1000 =.14 cum. Or 140 kg
D. Volume of chemical admixture = 7/1000 =0.007 cum or 7 kg
E. Volume of aggregate = (1- (0.117+0.14+0.007))
= 0.736 cum
Volume of coarse aggregate = 0.736*0.5616*2.74 * 1000 = 1132.55 kg/m^3
Volume of fine aggregate = 0.736*0.4384*2.74 * 1000 = 884.0 kg/m^3
Ans . Mix design for M35 grade of concrete without flyash.
A. Stipulations for proportion
1. Grade designation M35
2. Types of cement OPC 43 GRADE
3. Maximum nominal size of aggregate 20 mm
4. Maximum water cement ratio 0.45
5. Minimum cement content 340 kg/m^3
6. Workability 100 mm ( slump)
7. Exposure condition severe
8. Method of placing concrete Pumping
9. Degree of supervision Good
10. Types of aggregate angular
11. Maximum cement content 450 kg /m^3
12. Chemical admixture Types Superplasticizer
Data collection
Specific gravity of cement OPC 43 Grade - 3.15
Specific gravity of coarse and fine aggregate - 2.74
Specific gravity of water and chemical admixture- 1.0
a. Strength for proportion
Ft = FCK + (1.65 *S)
S = Standard deviation for M35 = 5
35 +(1.65*5)= 43.25 N/mm^2
b. Selection of water cement ration
From is 456- 2000 maximum water cement ratio will be 0.45 based on experience and from graph 1 it will be = 0.38 approximately
0.38 < 0.45 hence ok
c. Selection of water content
Maximum water content = 186 kg /m^3 (for slump 100 mm)
Estimated water = 186 +(0 .6 * 186 /100)
= 197.16 lt.
As superplaticizer is water content can be reduced upto 29% and above based in trails
= 197.16 * 0.71 = 140 lts.
Calculation of water cement ratio = 0.38
d. Cement content = 140 / 0.38
= 368.42 kg /m^3
320 kg <368.4kg < 450 kg (Okey )
e. Coarse and fine aggregate calculation
From table 3 of is 456: 2000 volume of coarse aggregate to 20 mm size aggregate and fine aggregate Zone 1. For water cement 0.5 = 0.6
In present case the water cement ratio = 0.38
Therefore volume of coarse aggregate is increased to decrease the fine aggregate content
Coarse aggregate for the water cement ratio of 0.38 = 0.624
Reduce by 10 percent for pumping concrete
Coarse aggregate = 0.624 * 0.9 = 0.5616
Fine aggregate = 1- 0.5616 = 0.4386
1. Mix proportion
A. Volume of concete = 1 cum
B. Volume of cement = (368.42 / 3.15)/ 1000 = 0.117 cum or 368.42 kg
C. Volume of water = 140 / 1000 =.14 cum. Or 140 kg
D. Volume of chemical admixture = 7/1000 =0.007 cum or 7 kg
E. Volume of aggregate = (1- (0.117+0.14+0.007))
= 0.736 cum
Volume of coarse aggregate = 0.736*0.5616*2.74 * 1000 = 1132.55 kg/m^3
Volume of fine aggregate = 0.736*0.4384*2.74 * 1000 = 884.0 kg/m^3
Ans . Mix design for M35 grade of concrete without flyash.
A. Stipulations for proportion
1. Grade designation M35
2. Types of cement OPC 43 GRADE
3. Maximum nominal size of aggregate 20 mm
4. Maximum water cement ratio 0.45
5. Minimum cement content 340 kg/m^3
6. Workability 100 mm ( slump)
7. Exposure condition severe
8. Method of placing concrete Pumping
9. Degree of supervision Good
10. Types of aggregate angular
11. Maximum cement content 450 kg /m^3
12. Chemical admixture Types Superplasticizer
Data collection
Specific gravity of cement OPC 43 Grade - 3.15
Specific gravity of coarse and fine aggregate - 2.74
Specific gravity of water and chemical admixture- 1.0
a. Strength for proportion
Ft = FCK + (1.65 *S)
S = Standard deviation for M35 = 5
35 +(1.65*5)= 43.25 N/mm^2
b. Selection of water cement ration
From is 456- 2000 maximum water cement ratio will be 0.45 based on experience and from graph 1 it will be = 0.38 approximately
0.38 < 0.45 hence ok
c. Selection of water content
Maximum water content = 186 kg /m^3 (for slump 100 mm)
Estimated water = 186 +(0 .6 * 186 /100)
= 197.16 lt.
As superplaticizer is water content can be reduced upto 29% and above based in trails
= 197.16 * 0.71 = 140 lts.
Calculation of water cement ratio = 0.38
d. Cement content = 140 / 0.38
= 368.42 kg /m^3
320 kg <368.4kg < 450 kg (Okey )
e. Coarse and fine aggregate calculation
From table 3 of is 456: 2000 volume of coarse aggregate to 20 mm size aggregate and fine aggregate Zone 1. For water cement 0.5 = 0.6
In present case the water cement ratio = 0.38
Therefore volume of coarse aggregate is increased to decrease the fine aggregate content
Coarse aggregate for the water cement ratio of 0.38 = 0.624
Reduce by 10 percent for pumping concrete
Coarse aggregate = 0.624 * 0.9 = 0.5616
Fine aggregate = 1- 0.5616 = 0.4386
1. Mix proportion
A. Volume of concete = 1 cum
B. Volume of cement = (368.42 / 3.15)/ 1000 = 0.117 cum or 368.42 kg
C. Volume of water = 140 / 1000 =.14 cum. Or 140 kg
D. Volume of chemical admixture = 7/1000 =0.007 cum or 7 kg
E. Volume of aggregate = (1- (0.117+0.14+0.007))
= 0.736 cum
Volume of coarse aggregate = 0.736*0.5616*2.74 * 1000 = 1132.55 kg/m^3
Volume of fine aggregate = 0.736*0.4384*2.74 * 1000 = 884.0 kg/m^3
Mix design for M50 grade of concrete with flyash.
A. Stipulations for proportion
1. Grade designation M50
2. Types of cement OPC 43 GRADE
3. Maximum nominal size of aggregate 20 mm
4. Maximum water cement ratio 0.45
5. Minimum cement content 340 kg/m^3
6. Workability 100 mm ( slump)
7. Exposure condition severe
8. Method of placing concrete Pumping
9. Degree of supervision Good
10. Types of aggregate angular
11. Maximum cement content 450 kg /m^3
12. Chemical admixture Types Superplasticizer
Data collection
Specific gravity of cement OPC 43 Grade - 3.15
Specific gravity of coarse and fine aggregate - 2.74
Specific gravity of water and chemical admixture- 1.0
a. Strength for proportion
Ft = FCK + (1.65 *S)
S = Standard deviation for M35 = 5
50+(1.65*5)= 58.25 N/mm^2
b. Selection of water cement ration
From is 456- 2000 maximum water cement ratio will be 0.45 based on experience and from graph 1 it will be = 0.32 approximately
0.32 < 0.45 hence ok
c. Selection of water content
Maximum water content = 186 kg /m^3 (for slump 100 mm)
Estimated water = 186 +(0 .6 * 186 /100)
= 197.16 lt.
As superplaticizer is water content can be reduced upto 29% and above based in trails
= 197.16 * 0.71 = 140 lts.
Calculation of water cement ratio = 0.32
d. Cement content = 140 / 0.32
= 437.5kg /m^3
320 kg <437.5 kg < 450 kg (Okey )
Total cementitious contents =437.5 kg
Assume fly ash content 30 percentage of cementitious content= ( 437.5 *30 )/100 = 131.25 kg
Cement content = 306.25 kg
e. Coarse and fine aggregate calculation
From table 3 of is 456: 2000 volume of coarse aggregate to 20 mm size aggregate and fine aggregate Zone 1. For water cement 0.5 = 0.6
In present case the water cement ratio = 0.32
Therefore volume of coarse aggregate is increased to decrease the fine aggregate content
Coarse aggregate for the water cement ratio of 0.32 = 0.636
Reduce by 10 percent for pumping concrete
Coarse aggregate = 0.636* 0.9 = 0.5724
Fine aggregate = 1- 0.5724 = 0.4276
1. Mix proportion
A. Volume of concete = 1 cum
B. Volume of cementitious content = (437.5/ 3.15)/ 1000 = 0.139cum or 437.5 kg
Cement OPC Grade 43 = 306.25 kg
Fly ash content =131.25 kg
C. Volume of water = 140 / 1000 =.14 cum. Or 140 kg
D. Volume of chemical admixture = 7/1000 =0.007 cum or 7 kg
E. Volume of aggregate = (1- (0.139+0.14+0.007))
= 0.714 cum
Volume of coarse aggregate = 0.714*0.5724*2.74 * 1000 = 1119.82kg/m^3
Volume of fine aggregate = 0.714*0.4276*2.74 * 1000 = 836.50 kg/m^3