Project on Concrete Mix Design for various grades of Concrete

AIM

To calculate the concrete mix design for M35 grade of concrete with fly ash

CODES FOR CONCRETE MIX DESIGN

IS 10262:2009

IS 456:2000

IS 8112

IS 9103

IS 3812 (part 1)

PROCEDURE AND CALCULATION

1. a) The concrete mix design for M35 grade concrete:

Stipulations:

Grade designation: M 35

Type of cement OPC 53 grades confirming to IS 8112

Maximum nominal size of aggregate: 20 mm

Minimum cement content: 320 kgm3

Maximum water-cement ratio: 0.45

Workability: 50-75 mm (slump)

Exposure condition: Severe (for reinforced concrete)

Method of concrete placing: Pumping

Degree of supervision: Good

Type of aggregate: Crushed angular aggregate

Maximum Cement Content: 450 kg/m3

Chemical Admixture Type: Super Plasticizer.

Test Data for Materials:

The specific gravity of cement: 3.15

The specific gravity of Coarse aggregate: 2.74

Specific Gravity of Fine aggregate: 2.67

water absorption, Coarse aggregate: 0.5%

Chemical admixture: Superplasticizer conforming to IS 9103.

Target Mean Strength:

ft = fck + 1.65 x S

ft = target average compressive strength at 28 days,

ick = characteristic compressive strength at 28 days, and S= standard deviation

The Standard deviation, S is taken 5.0 N/mm2 (Table | of IS 10262:2009)

Therefore, target strength = 35+ 1.65 x5,0 = 43.25 N/mm.

Selection of Water-Cement Ratio:

Given:

M35

Assume:

Exposure condition: Extreme (for reinforced concrete)

Table 5 - IS 456: 2000

Therefore,

Extreme= w/c ratio= 0.40

for M35, w/c ratio= 0.45

Lower the water cement ratio better the workability

0.40 < 0.45

Water cement ratio = 0.40

Minimum cement content = 360 kg/m^3

SELECTION OF WATER CONTENT

Assume:

Nominal size of aggregates= 20 mm

Table 2 -IS 10262:1009

Minimum water content = 186 litres or kg

Slump ranges from 25 to 50 mm

As superplasticizers (20 % reduced)

= 186 x80/100 = 148.8 (100-20 = 80)

Water content = 149 litres

CALCULATION OF CEMETITIOUS CONTENT (CEMENT + FLY ASH)

Water cement ratio = w/c= 0.40

Water content = W = 149 kg

Cementitious content = c=149/0.40 = 372.5 = 370 kg/m^3

Minimum cement content = 360 kg/m^3

360< 370 kg/m^3 (hence okay)

Increase by 10 %

370 x 0.10 + 370 = 407 kg/m^3

W= 149 kg

Water cement ratio = 149 407 = 0.366

Fly ash at 30 % total cementitious material content

= 407 x 0.3 = 122 kg/m^3

Cement (ordinary Portland cement) = 407-122 = 285 kg/m^3

Saving of cement while using fly ash = 370 - 285 = 85 kg/ m^3

Fly ash = 122 kg/m^3

VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES

Fine aggregates are sieved according to zone 1 of table 4 IS 383

Coarse aggregates are sieved according to table 2 IS 383

According,

Table 3 IS 10262: 2009

Nominal size of aggregates = 20 mm

Fine aggregates (zone 1)

Therefore.

Volume of coarse aggregate = 0.60

Water cement ratio = 0.4

(+/- 0.01 for every +/- 0.005)

So, volume of coarse aggregate = 0.62

Volume of fine aggregate = 1-0.62 = 0.38

MIX CALCULATION:

1. Volume of concrete = 1m^3 =a
2. Volume of cement = (285 / 3.15 x1/1000= 0.090 m^3 =b
3. Volume of fly ash = (122/2.2 x1/1000=0.055 m^3=c
4. Volume of water = (149/1)x1/1000=0.149 m^3 =d
5. Volume of admixture = (7.6/ 1.145x1/1000=0.006 m^3

6.Volume of coarse aggregate and fine aggregate = a - (b+ c+ d+ e) = 1- (0.090 + 0.055 +0.149+ 0.006) = 0.7 m^3

7.Mass of coarse aggregate = 0.7 x 0.62 x 2.74 x1000 = 1189.16 kg/m^3

8.Mass of fine aggregate = 0.7 x 0.38 x 2.74 x1000 = 728 kg/m^3

RESULTS AND CONCLUSION:

TRIAL MIX:

Cement = 285 kg/m^3

Fly ash = 122 kg/m^3

Water = 149 litres

coarse aggregate = 1189 kg/m13

Fine aggregate = 728 kg/m^3

Chemical admixture = 7.6 kg/m13

Water cement ratio = 0.366

Trial mix ratio: 285:728: 1189

= 1: 2.55: 4.17 (C: FA:CA)

AIM

To calculate the concrete mix design for M50 grade of concrete without fly ash

CODES FOR CONCRETE MIX DESIGN

IS 10262:2009

IS 456:2000

IS 8112

IS 9103

IS 3812 (part 1)

PROCEDURE AND CALCULATION

1. a) The concrete mix design for M50 grade concrete:

Stipulations for Concrete design mix

Grade designation: M50

Type of cement: OPC 43 grade

Maximum nominal size of aggregate:20mm

Minimum cement content:320 Kg (From Table 5 of IS 456:2000)

Maximum water-cement ratio: 0.45 (From Table 5 of IS 456:2000)

Degree of Workability: 100mm slump

Exposure condition: Sever

Method of concrete placing: Pumping

Degree of supervision: Good

Type of aggregate: Crushed angular aggregate

Maximum cement content: 450 kg/m^3

Chemical admixture type: Superplasticizer

Test Data for Material

Cement used: OPC 43 Grade conforming to IS 8112

Specific gravity of cement: 3.15

Chemical admixture: Super plasticizer conforming to IS 9103

Specific gravity of fly ash: 2.2

Specific gravity of coarse aggregate: 2.74

Specific gravity of fine aggregate: 2.74

Water absorption of coarse aggregate: 0.5%

Water absorption of fine aggregate: 1.0%

Sieve Analysis: Test for coarse aggregate should conforming to Table-2 of IS

Sieve Analysis: Test for fine aggregate should conforming to Grading of Zone of IS 383

Determination of Target Strength of concrete

Target mean strength f'ck= fck + (1.65*S)

Where,

f'ck= target average compressive strength at 28 days

fck = characteristics compressive strength at 28

S= standard deviation

From Table 1 of is 10262:2009 the standard deviation for M50 grade concrete is

S = 5 N/mm^2

Target mean strength f'ck= 50+ (1.65*5)

f'ck= 58.25 N/mm^2

Determination of water cement ratio

From Table 5 of IS 456-2000

Exposure condition is Severe

Maximum water-cement ratio: 0.45

Based on the experience, we have adopted water-cement ratio: 0.44

0.44<0.45, hence OK.

Selection of water content

From Table-2 of IS 10262:2009.

Nominal size of aggregates = 20 mm

maximum water content for 25 to 50 mm slump range =186 litre

Here slump value is 100mm

3% of water content must increase for every extra 25 mm slump.

Estimated water content for 100 mm slump =186+6/100 x186 =197 litres

As superplasticizer is used water content is reduced up to 20%

As trail it can be 20%

water content = 197*0.80 = 158 litres

Calculation of cement content

Water cement ratio =0.44

Water content = 158 litres

Cement content = 158/0.44 =359 kg/m^3

From Table 5 of IS 456, Minimum cement content for 'Severe' exposure conditions kg/m^3

359 kg/m^3>320 kg/ m13

Hence ok

Calculation of volume of coarse aggregate and fine aggregate

From Table 3 of IS: 10262-2009 volume of coarse aggregate corresponding tor 20mm aggregate and for fine aggregate (zone 2) water cement ratio 0.50 = 0.62

In the present case the water-cement ratio we have calculated is 0.44

So, the proportion of volume of coarse aggregate is increased by 0.02 (at the rate

for every ‡ 0.05 change in water-cement ratio)

So, the value is 0.64

For pumpable concrete the value is reduced by 10%

Volume of coarse aggregate =0.64*0.9 =0.56

Volume of fine aggregate= 1- 0.56 = 0.44

Mix calculation

1. Volume of concrete = 1 m^3
2. Volume of cement = Mass of cement/Specific gravity of cement * (1/1000)

= 359/3.15* (1/1000)

=0.114m^3

3. Volume of water = Mass of water/Specific gravity of water * (1/1000) =158/1* (1/1000) =0.158m^3
4. Volume of chemical admixtures = Mass of chemical admixture/Specific gravity of

water * (1/1000)

=7.18/1.145* (1/1000)

=0.0062m^3

1. e) volume of aggregate (FA+CA) =a- (b+c+d) =1-(0.114+.158+0.0062) =0.7218m^3
2. f) Mass of coarse aggregates = ex Volume of coarse aggregate × Specific gravity of

coarse aggregate × 100

=0.7218 x 0.568 x 2.74 x 1000 = 1123.35 kg/m^3

1. g) Mass of fine aggregates = e x Volume of fine aggregate × Specific gravity of fine

aggregate × 1000

=0.7218x 0.432x 2.74 x1000 =854.38 kg/m^3

RESULTS

Mix proportion

Cement =359kg/m13

Water =158kg/m13

Fine aggregate =854.38 kg/m13

Coarse aggregates =1123.35 kg/m13

Chemical admixtures =7.18 kg/m^3

Water cement ratio =0.40

Trail mix ratio =1:2.37:3.129