Project on Concrete Mix Design for various grades of Concrete

1) Mix design for M35 grade concrete with fly ash

STIPULATIONS FOR PROPORTIONING

1. Grade Designation : M35
2. Type of cement : OPC 43 grade
3. Type of mineral admixture : Fly ash
4. Maximum Nominal size of aggregates : 20 mm
5. Minimum cement content : 320 kg/m3
6. Maximum water-cement ratio : 0.45
7. Workability : 100mm
8. Exposure condition : Severe
9. Method of concrete placing : Pumping
10. Degree of supervision : Good
11. Type of aggregate : Crushed Angular aggregate
12. Maximum cement content : 450 kg/m3
13. Chemical admixture type : Superplasticizers

TEST DATA FOR MATERIALS

1. Cement used : OPC 43 grade
2. Specific gravity of cement : 3.15
3. Fly ash : confirming to IS3812 (part1)
4. Specific gravity of fly ash : 2.2
5. Chemical admixture : superplasticizer
6. Specific gravity of 1) Coarse aggregate : 2.74 2) Fine aggregate : 2.74
7. Water absorption 1) Coarse aggregate : 0.5 2) Fine aggregate : 1
8. Free surface moisture 1) Coarse aggregate: NIL 2) Fine aggregate : NIL
9. Fine aggregate : Confirming to grading Zone I of Table 4 of IS 383

TARGET STRENGTH FOR MIX PROPORTIONING

F'ck = Fck + 1.65s

where, F'ck = target average compressive strength at 28 days

Fck = characteristic compressive strength at 28 days

s = standard deviation

F'ck = 35 + 1.65 x 5 = 43.25 N/mm2

SELECTION OF WATER-CEMENT RATIO

From Table 5 of IS 456, the maximum water-cement ratio = 0.45

SELECTION OF WATER CONTENT

From table 2 Maximum water content for 20 mm aggregate = 186 litre ( For 25 to 50 mm slump )

Estimated water content for 100 mm slump = 186 + 0.06 x 186 = 197 litre

As Superplasticizer is used, the water content can be reduced upto 20 percent and above

So, reduce 20 percent of water

Arrived water content = 0.78 x 186 = 145.08 litre

CALCULATIONS OF CEMENT CONTENT

Water-cement ratio = 0.45

Cementitious material (cement + fly ash) content = water content / w/c ratio =145.08 / 0.45 = 322.4 kg/m3

From Table 5 of IS 456, minimum cement content for severe exposure condition = 320 kg/m3

322.4 kg/m3 > 320 kg/m3, hence OK

Cementitious material content = 322.4 x 1.10 = 354.64 kg/m3

Water content = 145.08 kg/m3

Water-cement ratio = 145.08 / 354.64 = 0.409

Fly ash @ 30% of total cementitious material content = 0.3 X 354.64 = 106.392 Kg/m3

Cement( OPC ) = 354.64 - 106.39 = 248.25 Kg/m3

Saving of cement while using fly ash = 322.4 - 248.25 = 74.15 Kg / m3

Fly ash being utilized = 106.39 Kg/m3

PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE CONTENT

From table 3, volume of course aggregate corresponding to 20mm size aggregate and fine aggregate( Zone I) for water-cement ratio of 0.5=0.6

In the present case water cement ratio is 0.45. Therefore, volume of course aggregate is required to be increased to decrease the fine aggregate. As the water-cement ratio is lower by 0.05, the proportion of volume of course aggregate is increased by 0.01 ( at the rate of -/+ 0.01 for every +/- 0.05 change in water cement ratio). Therefore, corrected proportion of volume of course aggregate for the water-cement ratio of 0.45 = 0.61)

For Pumping concrete these calues should be reduced by 10%

Therefore, volume of course aggregate = 0.62 x 0.9 = 0.56

Volume of Fine aggregate = 1-0.56 = 0.44

MIX CALCULATIONS

a) Total volume = 1 m3

b) Volume of cement = Mass of cement / Specific gravity of cement x 0.001
= 248.25 / 3.15 *0.001 = 0.079 m3

c) Volume of Fly ash = Mass of fly ash / Specific gravity of fly ash x 0.001

= 106.39 / 2.2 *0.001 = 0.048 m3

d) Volume of water = Mass of water / Specific gravity of water x 0.001
= 145.08 /1 *0.001
= 0.145 m3

e) Volume of chemical admixture
(superplasticizer) (@ 2.0 percent
by mass of cementitious material) = Mass of chemical admixture x Specific gravity of admixture
= 7 / 1.145 *0.001
= 0.007 m3
f) Volume of all in aggregate = [(a-b)-(c+d+e)]
= [ 1 – (0.079+0.048+0.145+0.007)]
= 0.721 m3
g) Mass of coarse aggregate
= f × Volume of coarse aggregate × Specific gravity of coarse aggregate × 1000
= 0.721 x 0.56 x 2.74 × 1000
= 1106.30 kg
h) Mass of fine aggregate
= f × volume of fine aggregate × Specific gravity of fine aggregate × 1000
= 0.721 x 0.44 x 2.74 × 1000
= 869.23 kg

MIX PROPORTIONS FOR TRIAL NO. 1

1. Cement = 248 Kg/m3
2. Fly Ash = 106 Kg/m3
3. Water = 145 Kg/m3
4. Fine Aggregate = 869.23 Kg/m3
5. Coarse aggregate = 1106 Kg/m3
6. Chemical admixture = 7 Kg/m3
7. Water-cement ratio = 0.40

2) M50 grade concrete

STIPULATIONS FOR PROPORTIONING

1. Grade Designation : M50
2. Type of cement : OPC 43 Grade
3. Maximum Nominal size of aggregates : 20 mm
4. Minimum cement content : 320 Kg/m3
5. Maximum water-cement ratio : 0.45
6. Workability : 100 mm slump
7. Exposure condition : Severe
8. Method of concrete placing : Pumping
9. Degree of supervision : Good
10. Type of aggregate : Crushed angular aggregate
11. Maximum cement content : 450 Kg/m3
12. Chemical admixture type : Superplasticizer

TEST DATA FOR MATERIALS

1. Cement used : OPC 43 Grade
2. Specific gravity of cement : 3.15
3. Chemical admixture : Superplasticizer
4. Specific gravity of 1) Coarse aggregate : 2.74 2) Fine aggregate : 2.74
5. Water absorption 1) Coarse aggregate : 0.5 2) Fine aggregate : 1
6. Free surface moisture 1) Coarse aggregate : NIL 2) Fine aggregate : NIL
7. Fine aggregate : Confirming to grading Zone I of Table 4 of IS 383

TARGET STRENGTH FOR MIX PROPORTIONING

F'ck = Fck + 1.65s

where, F'ck = target average compressive strength at 28 days

Fck = characteristic compressive strength at 28 days

s = standard deviation

F'ck = 50 + 1.65 x 5 = 58.25 N/mm2

SELECTION OF WATER-CEMENT RATIO

From Table 5 of IS 456, the maximum water-cement ratio = 0.45

SELECTION OF WATER CONTENT

From table 2 Maximum water content for 20 mm aggregate = 186 litre ( For 25 to 50 mm slump )

Estimated water content for 100 mm slump = 186 + 0.06 x 186 = 197 litre

As Superplasticizer is used, the water content can be reduced upto 20 percent and above

So, reduce 20 percent of water

Arrived water content = 0.8 x 186 = 148.8 litre

CALCULATIONS OF CEMENT CONTENT

Water-cement ratio = 0.45

Cement content = water content / w/c ratio = 148.8 /0.45 = 330 Kg/m3

From Table 5 of IS 456, minimum cement content for severe exposure condition = 320 Kg/m3

PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE CONTENT

From table 3, volume of course aggregate corresponding to 20mm size aggregate and fine aggregate( Zone I) for water-cement ratio of 0.5=0.6

In the present case water cement ratio is 0.45. Therefore, volume of course aggregate is required to be increased to decrease the fine aggregate. As the water-cement ratio is lower by 0.05, the proportion of volume of course aggregate is increased by 0.01 ( at the rate of -/+ 0.01 for every +/- 0.05 change in water cement ratio). Therefore, corrected proportion of volume of course aggregate for the water-cement ratio of 0.45 = 0.61)

For Pumping concrete these calues should be reduced by 10%

Therefore, volume of course aggregate = 0.61 x 0.9 = 0.55

Volume of Fine aggregate = 1-0.55 = 0.45

MIX CALCULATIONS

a) Total volume = 1 m3

b) Volume of cement = Mass of cement / Specific gravity of cement x 0.001
= 330 / 3.15 *0.001 = 0.105 m3

c) Volume of water = Mass of water / Specific gravity of water x 0.001
= 148.8 /1 *0.001
= 0.149 m3

d) Volume of chemical admixture
(superplasticizer) (@ 2.0 percent
by mass of cementitious material) = Mass of chemical admixture x Specific gravity of admixture
= 6.6 / 1.145 *0.001
= 0.006 m3
e) Volume of all in aggregate = [a-(b+c+d)]
= [ 1 – (0.105+0.149+0.006)]
= 0.74 m3
f) Mass of coarse aggregate
= f × Volume of coarse aggregate × Specific gravity of coarse aggregate × 1000
= 0.74 x 0.55 x 2.74 × 1000
= 1115.18 kg
g) Mass of fine aggregate
= f × volume of fine aggregate × Specific gravity of fine aggregate × 1000
= 0.74 x 0.45 x 2.74 × 1000
= 912.42 kg

MIX PROPORTIONS FOR TRIAL NO. 1

1. Cement = 330 Kg/m3
2. Water = 149 Kg/m3
3. Fine Aggregate = 912 Kg/m3
4. Coarse aggregate = 1115 Kg/m3
5. Chemical admixture = 6.6 Kg/m3
6. Water-cement ratio = 0.45