Project on Concrete Mix Design for various grades of Concrete

1)ans-

Grade designation : M35 Design mix

1. b) Type of cement:53 grade Ordinary Portland Cement conforming IS 12269
2. c) Maximum nominal size of coarse aggregate : 20 mm
3. d) Minimum amount of cement : 400 kg/m'as IS 2911-Part 1/Sec 2 & IRC 78-2014.
4. e) Maximum water-cement ratio :0.40 as per IRC 78 -2014,
5. f) Workability : 150-200 mm slump as per IRC 78-2014,
6. g) Exposure condition : Moderate (For Reinforced Concrete)
7. h) Method of concrete placing : Tremie
8. i) Degree of supervision : Good
9. k) Type of aggregate : Crushed Angular Aggregates

1) Chemical admixture type : Super Plasticizer Normet

Cement used: 53 grade Ordinary Portland cement conforming IS 12269

3. b) Specific gravity of cement :3.14
4. c) Chemical admixture : Super Plasticizer conforming to IS 9103
5. d) Specific gravity of

1) Coarse aggregate 20 mm: 2.799

2) Coarse aggregate 10 mm: 2.789

Combined Specific Gravity of aggregate ( 20 mm-45% & 10 mm 55% )=2.792

2) Fine aggregate : 2.517

1. e) Water absorption:

1) Coarse aggregate 20 mm: : 0.41 %

2) Coarse aggregate 10 mm :0.59

TEST DATA FOR MATERIALS

3) Fine aggregate : 1.87%

20. f) Aggregate Impact Value : 20.52%
21. g) Combined Flakiness & Elongation Index : 27.57 %
22. h) Sieve analysis:

1) Coarse aggregate: Conforming to all in aggregates of Table 2 of IS 383

2) Fine aggregate : Conforming to Grading Zone III of Table 4 of IS 383

TARGET STRENGTH FOR MIX PROPORTIONING

f'ck =fck + 1.65 s where

f'ck = average target compressive strength of concrete at 28 days,

fck = characteristics compressive strength of concrete at 28 days,

S = standard deviation. From table 1 of IS 456 & IS 10262 assumed Standard

Deviation, s = 5 N/mm²; therefore target strength of concrete = 35 + 1.65 x 5 =43.25 N/mm². Can adopt this value while doing the concrete mix design those who are working other than highway project

SELECTION OF WATER CEMENT RATIO

Based on the trial , adopted water cement ratio 0.40

From the Table 5 of IS 456 maximum Water Cement Ratio for moderate condition is 0.50

As per Section 709.1.9 of IRC 78, maximum water cement ratio is 0.40 so 0.40 is satisfying all codal provision whether it is IS 456 0.40 < 0.50 = 0.40 hence ok.

SELECTION OF WATER CONTENT

From Table 2 of IS 10262:2009, maximum water content for 20 mm aggregate = 186 liter (for

25 to 50 mm slump range)

but for an increase by about 3 percent for every additional 25 mm slump so here estimated

water content for 175 mm slump = 186+(15/100) x 186 = 213.9 or 214 liter.

Based on trials with Super plasticizer water content reduction of 21.50% has been achieved.

How it comes , see below calculation:

Water requirement , if we intended to keep cement 420 kg & wlc ratio 0.40 for our concrete

mix design calculated water will be 168.0 liter , so required reduction in water when using

superplaticizer = 100 - (168/214 x 100) = 21.50 %.

Net required water = 214 - (214 x 21.50) % = 214-54.0 = 167.99 liter say 168 liter.

Note :For 75 m…

CALCULATION OF CEMENT CONTENT

As discuss earlier adopted based on trial w/c Ratio = 0.40

Cement Content = 168.0/0.40 = 420 kg/m3

As per IS 2911-Part 1/Sec 2 minimum cement content for tremie concrete is 400 kg/m².

(As per Section 709.1.9 of IRC 78 - 2014 minimum cement content 400 kg & maximum

water coment ratio is 0.40 specified)

but we had taken 420 kg/m3 > 400 kg/m3 hence ok.

PROPORTION OF VOLUME OF COARSE AGGREGA AND FINE AGGREGATE CONTENT

As per table number 3 of IS 10262:2009 volume of coarse aggregate for 20 mm nominal size

aggregate and fine aggregate (Zone III) for having water-cement ratio of 0.50 =0.64

(a)In the present case water-cement ratio is 0.40. Therefore. volume of coarse aggregate is

required to be increased to decrease the fine aggregate content. As the water-cement ratio

is lower by 0.10. the proportion of volume of coarse aggregate is increased by 0.02 (at the

rate of -/+ 0.01 for every + 0.05 change in water-cement ratio).

= (0.10/0.05) = 2.0 times of 0.01 so 0.01 2.0= 0.020

(b)= a+b = 0.64 + 0.020 =0.66 (at the rate of -/+ 0.01 for every + 0.05 change in wlc ratio).

Therefore, corrected proportion of volume

MIX CALCULATIONS

The mix calculations per unit volume of concrete shall be as follows:

1. a) Volume of concrete = 1 m3
2. b) Volume of cement = [Mass of cement] / {[Specific Gravity of Cement] x 1000} = 400/{3.15 x 1000}

= 0.126 m3

1. c) Volume of water = [Mass of water] / {[Specific Gravity of water] x 1000}

= 168.01{1 x 1000} = 0.168 m

1. d) Base on trial we had kept admixture 1.0 percentage by weight of cement

Volume of admixture = [Mass of admixture ]/{[Specific Gravity of admixture ] x 1000}

= 4.2 1{1.090 x 1000}

= 0.0039 m

1. e) Volume of all in aggregate = [a-(b+c+d)]

= [1-(0.126+0.168+0.0039)]

= 0.702 m

1. f) Mass of coarse aggregate= e xVolume of Coarse Aggregate x Specific Gravity of coarse

Aggregate x 1000

= 0.702 x 0.59 x 2.792 x 1000

= ||43.97 kg/m

1. g) Mass of fine aggregate= e x Volume of Fine Aggregate x Specific Gravity of

Aggregate x 1000

= 0.702 x 0.41 x 2.517 x 1000

= 724.44 kg/m3

Cement = 420 kg/m3

Water = 168.0 l/m3

Fine aggregate = 724.44 kg/m3

Coarse aggregate 20 mm = 1143.97 x 45 %= 514.79 kg/m3

Coarse aggregate 10 mm = 1143.97 x 55%= 629.18 kg/m3

Chemical admixture = 4.2 kg/m3

Water-cement ratio = 0.40

1.Aggregate 20 mm = 514.79/ (1+(0.41/100)} = 513.0 kg

2. Aggregate 10 mm = 729.18/ (1+(0.59/100)} = 725.0 kg
3. Sand= 724.44/ (1+(1.87/100)} = 711.0 kg

4. Water= 2561-420-711-513-725 = 192.0 kg

Note-.Do the trial in dry position as in practical in batching plant you can cot use aggregate in SSD condition.

3.Do the number of trials with variation of £ 10 percent of water-cement ratio & different cement content,

4.Do the surface moisture correction whenever required.

Design mix of M50

Grade Designation = M-50
Type of cement = O.P.C-43 grade
Fine Aggregate = Zone-II

Sp. Gravity
Cement = 3.15
Fine Aggregate = 2.61
Coarse Aggregate (20mm) = 2.65
Coarse Aggregate (10mm) = 2.66

Minimum Cement (As per contract) =400 kg / m³
Maximum water cement ratio (As per contract) = 0.45

Mix Calculation: –

1. Target Mean Strength = 50 + ( 5 X 1.65 ) = 58.25 Mpa

2. Selection of water cement ratio:-
Assume water cement ratio = 0.35

3. Calculation of water: –
Approximate water content for 20mm max. Size of aggregate = 180 kg /m³ (As per Table No. 5 , IS : 10262 ).

Now water content = 180 X 0.8 = 144 kg /m³

4. Calculation of cement content:-
Water cement ratio = 0.35
Water content per cum of concrete = 144 kg
Cement content = 144/0.35 = 411.4 kg / m³
Say cement content = 412 kg / m³ (As per contract Minimum cement content 400 kg / m³ )
Hence O.K.

5. Calculation for C.A. & F.A.

Volume of concrete = 1 m³
Volume of cement = 412 / ( 3.15 X 1000 ) = 0.1308 m³
Volume of water = 144 / ( 1 X 1000 ) = 0.1440 m³
Volume of Admixture = 4.994 / (1.145 X 1000 ) = 0.0043 m³
Total weight of other materials except coarse aggregate = 0.1308 + 0.1440 +0.0043 = 0.2791 m³

Volume of coarse and fine aggregate = 1 – 0.2791 = 0.7209 m³
Volume of F.A. = 0.7209 X 0.33 = 0.2379 m³ (Assuming 33% by volume of total aggregate )

Volume of C.A. = 0.7209 – 0.2379 = 0.4830 m³

Therefore weight of F.A. = 0.2379 X 2.61 X 1000 = 620.919 kg/ m³

Say weight of F.A. = 621 kg/ m³

Therefore weight of C.A. = 0.4830 X 2.655 X 1000 = 1282.365 kg/ m³

Say weight of C.A. = 1284 kg/ m³

Considering 20 mm: 10mm = 0.55: 0.45
20mm = 706 kg .
10mm = 578 kg .
Hence Mix details per m³
Increasing cement, water, admixture by 2.5% for this trial

Cement = 412 X 1.025 = 422 kg
Water = 144 X 1.025 = 147.6 kg
Fine aggregate = 621 kg
Coarse aggregate 20 mm = 706 kg
Coarse aggregate 10 mm = 578 kg
Admixture = 1.2 % by weight of cement = 5.064 kg.

Water: cement: F.A.: C.A. = 0.35: 1: 1.472: 3.043