Project on Concrete Mix Design for various grades of Concrete
1. Design of M35 Grade of Concrete using Fly ash:-
A) Design data:-
Grade Designation:- M35
Type of Cement:- OPC53
Type of Admixture:- Fly ash
Maximum Nominal Size of aggregate:- 20 mm
Minimum cement Content:- 340 Kg, (As per table 5 : IS 456)
Maximum Cement content:- 450 kg, (as per IS code)
Partial replacement of cement by fly ash:- 30% (as per requirement)
Maximum Water cement ratio:- 0.45 (as per table 5: IS56)
Workability :- 100 mm
Chemical admixture:- PCS type.
B) test Data for Matrerials:-
Specific gravity of OPC53 grade cement:- 3.15 (As per IS : 2720)
Mineral admixture :- Fly ash (as per IS: 12089)
Specific graity of Fly Ash:- 2.90
Chemical admixture :- @0.6% (IS:9103)
Specific gravity of Coarse aggregate:- 2.74, ( as per IS:2386)
Specific gravity of Fine aggregtae:- 2.74
Specific gravity of PCe basede:- 1.12
Water absorption of coarse aggregate:- 0.5 % (IS: 2386)
Water absorption of fine aggregate:- 1 %
Sieve analysis :- (as per IS: 2386)
Coarse aggregate:- (conforming to table 2, IS:383)
Fine aggregate:- Zone II ( Conforming to table 4, IS:383)
C) Target Strength for M35 grade of concrete:-
Ft= Fck+ 1.65 S
Ft= 35+ 1.65*5 (table 1, IS:10262)
Ft= 43.25 N/mm2
D) Selection of water cement ratio:-
maximum W-C ratio = 0.45 (table 5, IS:456)
Adopted W-C ratio= 0.45
E) Selection of Water Content/cubic meter:-
maximum Water content = 186 L , for 25- 50 mm slump. ( table 2, IS: 10262).
hence for Slump value 100, we need to increase 3% water content for every 25 mm increase in slump.
hence, total 6% slump will be increased.
now, Total water Content= 186+ (0.06*186) = 197.16 L.
Since , we are using Super plasticezers, water content can be reduced upto 20 %.
hence, Water content = 197.16- (0.20*197.16) = 157.728 L, Assume, 160 L.
Adopted water Content= 160 L
F) calculatiuon of Cement and fly ash Content:-
W/C ratio= 0.45
Cement/Cementious material = 160/0.45= 355.55 kg/ cum, approx- 356 kg.
Since fly ash is used amount of cementious material can be increased to get good workiability and Strength. Hence we increse 10 % by weigth cementious content.
therefore, Final cement Content= 356*1.10= 391.6 kg.
as 391.6 kg > 340 kg (minimum cement content)
hence, Adopted Cement Content= 391.60 Kg/cum
Now replacement of Cement by Fly Ash, 30%= 391.60*0.30= 117.48 approx, 118.0 kg.
Final Cement Content Required= 391.6-117.48= 274.12 kg.
Superplasticezer @ 0.6% of cementious material= 391.6*0.006)= 2.35 kg.
G) Coarse and Fine aggregates:-
Coarse aggregate ratio= 0.62, ( table-3, IS: 10262, for W/C=0.50)
Coarse aggregate ratio (for W/C ratio, 0.45) = 0.63 (adding 0.01 for every decrease in 0.05 and vice versa)
Final fine aggregate ratio= (1.0.63)= 0.37.
H) Mix Calculation of M35 grade concrete:-
1. Volume of concrete= 1 cum.
2. total Volume of cement= Weigth of cement/ (SG*1000)= 274/(3.15*1000) =0.087 cum.
3. Total volume of fly Ash = weight of Fly Ash/( S.G. *1000)= 117.48/(2.90*1000)= 0.040 cum.
4. Volume of Water= Weight of water/(S.G*1000)= 160/(1*1000)= 0.160 cum.
5. Volume of Superplasticizer= weight/(SG*1000)= 2.35/(1.12*1000)= 0.002.
6. Total aggregate requirement= ( 1-(0.087+0.040+0.160+0.002)= 0.71 cum.
7. Coarse Aggregate =0.71*C A ratio* SG*1000= 0.71*0.63*2.74*1000= 1225.6 kg.
8. fine aggregate= 0.71*F A ratio* SG*1000= 0.71*.37*2.74*1000= 719.8 kg.
J) Total material required for 1 cum. M35 grade of Concrete:-
1. Cement= 275 kg
2. Fly ash= 118 kg
3 .Fine aggregate= 720 Kg
4. Coarse aggregate (20mm)= 1225.6 kg
5. Water= 160 L
6. PCE Based Super plasticezer= 2L
K) The Design Mix Ratio of M35 grade of concrete
Cement:Fly ash: FA: CA: Water= 1:0.43: 2.62: 4.46: 0.58
2. Design of M50 grade of concrete Without fly Ash
A) Design data:-
Grade Designation:- M50
Type of Cement:- OPC 53
Type of Admixture:- Non
Maximum Nominal Size of aggregate:- 20 mm
Minimum cement Content:- 340 Kg, (As per table 5 : IS 456, for sever condition)
Maximum Cement content:- 450 kg, (as per IS : 456)
Maximum Water cement ratio:- 0.45 (as per table 5, IS:456)
Workability :- 100 mm
Chemical admixture:- PCS type.
B) test Data for Matrerials:-
Specific gravity of OPC53 grade cement:- 3.15 (As per IS : 2720)
Chemical admixture :- @0.6% (IS:9103)
Specific gravity of Coarse aggregate:- 2.74, ( as per IS:2386)
Specific gravity of Fine aggregtae:- 2.74
Specific gravity of PCE based:- 1.12
Water absorption of coarse aggregate:- 0.5 % (IS: 2386)
Water absorption of fine aggregate:- 1 %
Sieve analysis :- (as per IS: 2386)
Coarse aggregate:- (conforming to table 2, IS:383)
Fine aggregate:- Zone II ( Conforming to table 4, IS:383)
C) Target Strength for M50 grade of concrete:-
Ft= Fck+ 1.65 S
Ft= 50+ 1.65*5 (table 1, IS:10262)
Ft= 58.25 N/mm2
D) Selection of water cement ratio:-
maximum W-C ratio = 0.45 (table 5, IS:456)
Adopted W-C ratio= 0.45
E) Selection of Water Content/cubic meter:-
maximum Water content = 186 L , for 25- 50 mm slump. ( table 2, IS: 10262).
hence for Slump value 100, we need to increase 3% water content for every 25 mm increase in slump.
hence, total 6% slump will be increased.
now, Total water Content= 186+ (0.06*186) = 197.16 L.
Since , we are using Super plasticezers, water content can be reduced upto 20 %.
hence, Water content = 197.16- (0.20*197.16) = 157.728 L, Assume, 160 L.
Adopted water Content= 160 L
F) calculatiuon of Cement and fly ash Content:-
W/C ratio= 0.45
Cement/Cementious material = 160/0.45= 355.55 kg/ cum, approx- 356 kg.
as 356 kg > 340 kg (minimum cement content)
hence, Adopted Cement Content= 356 Kg/cum
Superplasticezer @ 0.6% of cementious material= 356*0.006)= 2.14 kg.
G) Coarse and Fine aggregates:-
Coarse aggregate ratio= 0.62, ( table-3, IS: 10262, for W/C=0.50)
Coarse aggregate ratio (for W/C ratio, 0.45) = 0.63 (adding 0.01 for every decrease in 0.05 and vice versa)
Final fine aggregate ratio= (1.0.63)= 0.37.
H) Mix Calculation of M35 grade concrete:-
1. Volume of concrete= 1 cum.
2. total Volume of cement= Weigth of cement/ (SG*1000)= 356/(3.15*1000) =0.130 cum.
3. Volume of Water= Weight of water/(S.G*1000)= 160/(1*1000)= 0.160 cum.
4. Volume of Superplasticizer= weight/(SG*1000)= 2.14/(1.12*1000)= 0.0019.
5. Total aggregate requirement= ( 1-(0.130+0.160+0.0019)= 0.708 cum.
6. Coarse Aggregate =0.708*C A ratio* SG*1000= 0.708*0.63*2.74*1000= 1222.15 kg.
7. fine aggregate= 0.71*F A ratio* SG*1000= 0.708*.37*2.74*1000= 717.77 kg.
J) Total material required for 1 cum. M35 grade of Concrete:-
1. Cement= 356 kg
2 .Fine aggregate= 717.77 Kg
3. Coarse aggregate (20mm)= 1222.15 kg
4. Water= 160 L
5. PCE Based Super plasticezer= 2 L.
K) The Design Mix Ratio of M35 grade of concrete
Cement: FA: CA: Water= 1: 2.02: 3.43: 0.45