Project on Concrete Mix Design for various grades of Concrete

AIM: Concrete Mix Design for M35 grade concrete with Fly Ash

INTRODUCTION: Concrete Mix Design is the process of determining the proportions of cement, aggregates, water and sometimes admixtures needed to achieve the desired properties of concrete. The mix design is based on the specific requirement for strength durability, workability based on the project’s specifications and intended use of concrete. Use of Code Book of IS 10262:2019 and IS 456:2000.

STEP BY STEP PROCEDURE:

Stipulation for Proportioning

a) Grade designation	M35
b) Type of cement	OPC 43
c) Type of mineral admixture	Fly ash
d) Maximum nominal size of aggregate	20mm
e) Minimum cement content and maximum water-cement ratio to be adopted and exposure conditions as per table 3 and table 5 of IS 456	Very severe
f) Workability	100mm (slump)
g) Method of concrete placing	Pumping
h) Type of aggregate	Crushed angular aggregate

Test data for materials

a) Cement used	OPC 43 grade conforming to IS 269
b) Specific gravity of cement	3.15
c) Fly ash	Conforming to IS 3812 (Part 1)
d) Specific gravity of fly ash	2.2
e) Specific gravity of 1) Coarse aggregate (at SSD condition) 2) Fine aggregate (at SSD condition)	2.74 2.65
f) Water absorption 1) Coarse aggregate 2) Fine aggregate	0.5 percent 1.0 percent

1. The coarse and fine aggregates are wet and their total moisture content is 2 percent and 5 percent respectively. Therefore, the free moisture content in coarse and fine aggregate shall be as shown in (h) below
2. Free (surface) moisture

• Coarse aggregate : Free moisture= Total moisture content – Water absorption= 2.0-0.5 =1.5%
• Fine aggregate : Free moisture= Total moisture content – Water absorption= 5.0-1.0 =4.0%

Target Strength for Mix Proportioning

f’_ck = f_ck + 1.65 S or f’_ck = f_ck + X (from clause 4.2,pg.no 2 IS 10262:2019)

whichever is higher

where f’_ck = target average compressive strength at 28 days,

f_ck = characteristic compressive strength at 28 days,

S= standard deviation, and

X= factor based on grade of concrete

From table 2 pg. no. 3, Assumed Standard Deviation, S= 5.0 N/mm²

From table 1 pg. no. 3, Value of X = 6.5

Therefore, target strength using both equations, that is,

1. f’_ck = f_ck + 1.65 S= 35 + 1.65* 5 = 43.25 N/mm²
2. f’_ck = f_ck + 6.5= 35 + 6.5 = 41.50 N/mm²

The higher value is adapted

Approximate Air Content

From Table 3, the approximate amount of entrapped air to be expected in normal (non-air-entrained) concrete is 1.0 % for 20mm nominal maximum size of aggregate.

Selection of Water-Cement ratio

From Fig. 1 pg. no. 4. Curve 2 for expected 28 days compressive strength of between 43 and < 53 N/mm².

The free water-cement ratio required for the target strength of 43.25 N/mm² is 0.40. which is lower than the maximum Water-Cement Ratio for a very severe environmental exposure condition from Table 5 of IS 456:2000 is 0.45. i.e 0.40 < 0.45, hence O.K.

Selection of Water-Content

From Table 4, Water Content per cu.m of concrete for nominal maximum size of aggregate is 186 kg (for 50mm slump). Therefore water content for 100 mm slump (increasing at the rate of 3 percent for every 25mm slump) = 186 + (6/100)x186 = 197.16 kg

Calculation of Cement Content

Water- Cement ratio = 0.40

Cementitious content= 197.16/0.40 = 493 kg/m³

10% is increased to have a cementitious content mass

i.e cementitious material= 493 x 1.1 = 542.3 kg

Considering 30% cement is replaced by fly ash in the cementitious material 542 x (30/100) = 163kg

Therefore the cement (OPC) content= 542-163 = 379 kg

Also new revised Water-Cement Ratio= (197/542) = 0.36

Proportion of Volume of Coarse Aggregate and Fine Aggregate content

From the table 5 ( pg. no 6, IS 10262:2019), volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate (Zone II) for water-cement ratio of 0.50= 0.62

The present case water-cement ratio is 0.36. Therefore have to be increase the volume of coarse aggregate so have to decrease the amount of fine aggregate content, as the water-cement ratio is lowered by (0.5-0.36)= 0.14

The proportion of volume of coarse aggregate is increased at the rate of 0.01 for every 0.05 change in the water-cement ratio, therefore the volume of coarse aggregate is ((0.14/0.05)*0.01)= 0.028. The corrected proportion of coarse aggregate for the water-cement ratio of 0.36 = 0.62 + 0.028 = 0.648

For pumpable concrete these values may be reduced up to 10 % as per the clause 5.5.2.

The volume of coarse aggregate = 0.648* 0.9= 0.583 m³, volume of fine aggregate content= 1-0.58= 0.42m³

Mix Calculations

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

1. Total volume = 1 m³
2. Volume of entrapped air in wet concrete = 0.01 m³
3. Volume of cement=(mass of cement/specific gravity of cement)*(1/1000) =(379/3.15) *( 1/1000)= 0.12 m³
4. Volume of fly ash =(mass of fly ash/specific gravity of fly ash)*(1/1000) =(163/2.2) *( 1/1000)= 0.07 m³
5. Volume of fly ash =(mass of water/specific gravity of water)*(1/1000) =(197/1) *( 1/1000)= 0.197m³
6. Volume of all the aggregate = [a – (b+c+d+e)]=[1-0.01-0.12-0.07-.197]= 0.603
7. Mass of Coarse aggregate = (f) * volume of C.A * Specific gravity of C.A * 1000

= 0.603* 0.58* 2.74* 1000 = 958 kg

8. Mass of Fine aggregate = (f) * volume of F.A * Specific gravity of F.A * 1000

= 0.603* 0.42* 2.65* 1000 =671 kg

Mix Proportion

1. Cement = 379 kg
2. Fly ash = 163 kg
3. Water(Net mixing) =197 kg
4. Fine aggregate(SSD) =671 kg
5. Coarse aggregate(SSD)= 958 kg
6. Free water-cementitious materials ratio = 0.36

Adjustment of Water, Fine aggregate and Coarse aggregate (if the F.A & C.A are in wet condition)

1. Fine Aggregate (wet)

Mass of wet fine aggregate = mass of F.A in SSD condition * (1+(free surface moisture/100) )

= 671 * (1+(4/100) )

= 697.84 kg/m³

2. Coarse Aggregate (wet)

Mass of wet fine aggregate = mass of C.A in SSD condition * (1+(free surface moisture/100) )

= 958 * (1+(1.5/100))

= 972.37 kg/m³

Since the C.A and F.A are in wet condition, the materials would be already moisture so have to subtract the SSD water content with the wet materials absorbed weight of water

• Water content contributed by wet C.A

= Mass of wet C.A – Mass of SSD C.A

= 972- 958= 14 kg/m³

• Water content contributed by wet F.A

= Mass of wet F.A – Mass of SSD F.A

= 698- 671= 27 kg/m³

The requirement for water content = 197- 14- 27 = 156 kg/m³

Mix Proportions after adjustment for wet aggregates

1. Cement = 379 kg
2. Fly ash = 163 kg
3. Water (to be added) =156 kg
4. Fine aggregate(wet) =698 kg
5. Coarse aggregate(wet)= 972 kg
6. Free water-cementitious materials ratio = 0.36

AIM: Concrete Mix Design for M50 grade concrete without Fly Ash

INTRODUCTION: Concrete Mix Design is the process of determining the proportions of cement, aggregates, water and sometimes admixtures needed to achieve the desired properties of concrete. The mix design is based on the specific requirement for strength durability, workability based on the project’s specifications and intended use of concrete. Use of Code Book of IS 10262:2019 and IS 456:2000.

STEP BY STEP PROCEDURE:

Stipulation for Proportioning

a) Grade designation	M50
b) Type of cement	OPC 43
c) Type of mineral admixture	None
d) Maximum nominal size of aggregate	20mm
e) Minimum cement content and maximum water-cement ratio to be adopted and exposure conditions as per table 3 and table 5 of IS 456	severe
f) Workability	100mm (slump)
g) Method of concrete placing	Pumping
h) Type of aggregate	Crushed angular aggregate

Test data for materials

a) Cement used	OPC 43 grade conforming to IS 269
b) Specific gravity of cement	3.15
c) Fly ash	Conforming to IS 3812 (Part 1)
d) Specific gravity of 1) Coarse aggregate (at SSD condition) 2) Fine aggregate (at SSD condition)	2.74 2.65
e) Water absorption 1) Coarse aggregate 2) Fine aggregate	0.5 percent 1.0 percent

1. The coarse and fine aggregates are wet and their total moisture content is 2 percent and 5 percent respectively. Therefore, the free moisture content in coarse and fine aggregate shall be as shown in (h) below
2. Free (surface) moisture

• Coarse aggregate : Free moisture= Total moisture content – Water absorption= 2.0-0.5 =1.5%
• Fine aggregate : Free moisture= Total moisture content – Water absorption= 5.0-1.0 =4.0%

Target Strength for Mix Proportioning

f’_ck = f_ck + 1.65 S or f’_ck = f_ck + X (from clause 4.2,pg.no 2 IS 10262:2019)

whichever is higher

where f’_ck = target average compressive strength at 28 days,

f_ck = characteristic compressive strength at 28 days,

S= standard deviation, and

X= factor based on grade of concrete

From table 2 pg. no. 3, Assumed Standard Deviation, S= 5.0 N/mm²

From table 1 pg. no. 3, Value of X = 6.5

Therefore, target strength using both equations, that is,

1. f’_ck = f_ck + 1.65 S= 50 + 1.65* 5 = 58.25 N/mm²
2. f’_ck = f_ck + 6.5= 50 + 6.5 = 56.50 N/mm²

The higher value is adapted

Approximate Air Content

From Table 3, the approximate amount of entrapped air to be expected in normal (non-air-entrained) concrete is 1.0 % for 20mm nominal maximum size of aggregate.

Selection of Water-Cement ratio

From Fig. 1 pg. no. 4. Curve 3 for expected 28 days compressive strength of 53 N/mm² and above

The free water-cement ratio required for the target strength of 58.25 N/mm² is 0.34.

Selection of Water-Content

From Table 4, Water Content per cu.m of concrete for nominal maximum size of aggregate is 186 kg (for 50mm slump). Therefore water content for 100 mm slump (increasing at the rate of 3 percent for every 25mm slump) = 186 + (x186 = 197.16 kg

Calculation of Cement Content

Water- Cement ratio = 0.34

Cement content= 197.16/0.34= 597.88 kg/m³

Proportion of Volume of Coarse Aggregate and Fine Aggregate content

From the table 5 ( pg. no 6, IS 10262:2019), volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate (Zone II) for water-cement ratio of 0.50= 0.62

The present case water-cement ratio is 0.36. Therefore have to be increase the volume of coarse aggregate so have to decrease the amount of fine aggregate content, as the water-cement ratio is lowered by (0.5-0.34)= 0.16

The proportion of volume of coarse aggregate is increased at the rate of 0.01 for every 0.05 change in the water-cement ratio, therefore the volume of coarse aggregate is ((0.16/0.05)*0.01)= 0.032. The corrected proportion of coarse aggregate for the water-cement ratio of 0.36 = 0.62 + 0.032 = 0.652

For pumpable concrete these values may be reduced up to 10 % as per the clause 5.5.2.

The volume of coarse aggregate = 0.652* 0.9= 0.586 m³, volume of fine aggregate content= 1-0.586= 0.4132m³

Mix Calculations

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

1. Total volume = 1 m³
2. Volume of entrapped air in wet concrete = 0.01 m³
3. Volume of cement=(mass of cement/specific gravity of cement)*(1/1000) =(598/3.15) *( 1/1000)= 0.19 m³
4. Volume of fly ash =(mass of water/specific gravity of water)*(1/1000) =(197/1) *( 1/1000)= 0.197m³
5. Volume of all the aggregate = [a – (b+c+d)]=[1-0.01-0.19-.197]= 0.603
6. Mass of Coarse aggregate = (f) * volume of C.A * Specific gravity of C.A * 1000

= 0.603* 0.586* 2.74* 1000 = 968 kg

7. Mass of Fine aggregate = (f) * volume of F.A * Specific gravity of F.A * 1000

= 0.603* 0.4132* 2.65* 1000 =660 kg

Mix Proportion

1. Cement = 379 kg
2. Fly ash = 163 kg
3. Water(Net mixing) =197 kg
4. Fine aggregate(SSD) =660 kg
5. Coarse aggregate(SSD)= 968 kg
6. Free water-cementitious materials ratio = 0.34

Adjustment of Water, Fine aggregate and Coarse aggregate (if the F.A & C.A are in wet condition)

1. Fine Aggregate (wet)

Mass of wet fine aggregate = mass of F.A in SSD condition * (1+(free surface moisture/100) )

= 660 * (1+(4/100) )

= 686.4 kg/m³

2. Coarse Aggregate (wet)

Mass of wet fine aggregate = mass of C.A in SSD condition * (1+ (free surface moisture/100) )

= 968 * (1+ (1.5/100))

= 982.52 kg/m³

Since the C.A and F.A are in wet condition, the materials would be already moisture so have to subtract the SSD water content with the wet materials absorbed weight of water

• Water content contributed by wet C.A

= Mass of wet C.A – Mass of SSD C.A

= 983- 968= 15 kg/m³

• Water content contributed by wet F.A

= Mass of wet F.A – Mass of SSD F.A

= 686- 660= 26 kg/m³

The requirement for water content = 197- 15- 26 = 156 kg/m³

Mix Proportions after adjustment for wet aggregates

1. Cement = 379 kg
2. Fly ash = 163 kg
3. Water (to be added) =156 kg
4. Fine aggregate(wet) =686 kg
5. Coarse aggregate(wet)= 983 kg
6. Free water-cementitious materials ratio = 0.36