Project on Concrete Mix Design for various grades of Concrete

1. Calculate the Concrete Mix Design for M35 grade concrete with fly ash

AIM: To calculate the concrete Mix Design for M35 grade concrete with fly ash

PROCEDURE:

Step-1

Required data for proportioning:

• Grade designation = M35
• Type of cement = OPC 43 grade conforming to IS-8112
• Maximum nominal size of aggregate = 20 mm
• Minimum cement content = 300 kg/`m^3` (From table 5 of IS 456-2000 codal provision)
• Maximum water-cement ratio = 0.50 (From table 5 of IS 456-2000 codal provision)
• Workability (Slump value) = 100mm
• Exposure condition = Moderate (For reinforced concrete)
• Method of concrete placing = Pumping
• Degree of supervision = Good
• Type of aggregate = Crushed angular aggregate
• Chemical admixture type = Superplasticizer

Step-2

Test Data for material:

• Cement used = OPC 43 grade conforming to IS 8112
• The specific gravity of cement = 3.15
• Fly ash = conforming to IS 3812 (Part-1)
• The specific gravity of fly ash = 2.2
• Chemical admixture type = Superplasticizer
• Conforming to IS 9103
• Specific gravity of coarse aggregate and fine aggregate = 2.74
• Water absorption of coarse aggregate = 0.5%
• Water absorption of fine aggregate = 1.0%
• Free surface moisture of coarse aggregate and fine aggregate is Nil.

Step-3

Target mean strength of concrete:

Target mean strength of concrete(f'ck) = fck+1.65 X S

Where:

• fck = characteristic compressive strength at 28 days.
• S = standard deviation (From IS 10262-2009, table-1 or IS 456-2000, Table-8)

• Given concrete grade is M35
• Compressive strength after 28 days is 35 N/`mm^2`
• Standard deviation is 5 N/`mm^2`.(IS 10262-2009, Table-1)

Now f'ck = fck+1.65 X S

=35+1.65 X 5

=43.25 N/`mm^2`

Step-4

Selection of water-cement ratio:

• From the table-5 of IS 456-2000 for moderate exposure, maximum water-cement ratio = 0.50
• Based upon the experience, adopt water-cement ratio = 0.45
• 0.45<0.50 Hence OK.

Step-5

Selection of water content:

• As per IS 10262-2009, table-2 The maximum nominal size of aggregate is 20 mm.
• The standard slump value is between 25 to 50 mm.
• The maximum water content for 20mm aggregate = 186 liters

• Estimated water content = 100 mm slump

=`186+6/100 X 186`

=197 liters

Then the superplastisizer is used, the water content is reduced by 30%. Based on the trail & errors, water content reduction of 29% is achived with plasticizer

Hence the arrived water content = `197X 71/100`= 139.87 =140 liters

Step-6

Calculation of Cement content:

• Water-cement ratio = 0.45
• water content = 140 liters
• Cementies material (Cement + fly ash) content

= `140/0.45`

= 311 kg/`mm^3`

From table-5 of IS 456-2000, the minimum cement content for moderate exposure condition = 300 kg/`mm^3`

Hence, 311 kg/`mm^3` > 300 kg/`mm^3`

Now the percentage of fly ash needs to be decided based on project requirements and quality of materials. In some cases, an increase in cementitious material can be warranted. The cementitious material content is increased by 10%.

• Cementitious material content = 311 X 1.10= 342 kg/`m^3`
• water content = 140 kg/`m^3`
• Water-cement ratio = 140/342 = 0.40
• Flyash @ 30% total cementitious material content = 342 X 30% =102.6 kg/`m^3`
• Cement (OPC) = 342-102.6 = 240 kg/`m^3`
• Saving of cement while using fly ash = 311-240 = 71 kg/`m^3`
• Fly ash being utilized = 102.6 kg/`m^3`

Step-7

The proportion of the volume of coarse aggregate and fine aggregate:

From table-3 of IS 10262-2009, the volume of coarse aggregate corresponding to 20 mm size and the fine aggregate (zone-1) for water-cement ratio = 0.6

• In the present case, the water-cement ratio is 0.5
• Therefore the volume of coarse aggregate needs to be increased to decrease the fine aggregate content.
• As the water-cement ratio is lower by 0.01 the proportion of the volume of coarse aggregate is increased by 0.02
• Therefore corrected volume of coarse aggregate for the water-cement ratio of 0.5=0.61
• For pumpable concrete, these values are reduced by 10%
• Therefore volume of coarse aggregate = 0.61-10% = 0.51
• Volume of fine aggregate content = 1-0.51 = 0.49

Step-8

Mix calculation:

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

• a) Volume of concrete = 1`m^3`
• b) Volume of cement:

`240/3.15 X 1/1000`= 0.076`m^3`

• c) Volume of Flyash

`102.6/2.2 X 1/1000`= 0.046 `m^3`

• d) Volume of water:

`140/1 X 1/1000 = 0.14 m^3`

• e) Volume of chemical admixture:

`7.6/1.145 X 1/1000 = 0.006 m^3`

• f) Volume of aggregate (CA & FA):

= [a-(b+c+d)]

= [1-(0.076+0.140+0.006)]

= 0.77 `m^3`

• g) Mass of coarse aggregate:

= f X volume of coarse aggregate X specific gravity of coarse aggregate X 1000

= 0.77 X 0.51 X 2.74 X 1000

= 1075.99 kg/`m^3`

• h) Mass of Fine aggregate:

= f X volume of fine aggregate X specific gravity of fine aggregate X 1000

= 0.77 X 0.49 X 2.74 X 1000

= 1033.80 kg/`m^3`

Step-9

Mix proportion:

• Cement : 240 kg/`m^3`
• Fly ash : 102.6 kg/`m^3`
• Water : 140 kg/`m^3`
• Fine aggregate : 1033.80 kg/`m^3`
• Coarse aggregate : 1075.99 kg/`m^3`
• Chemical admixture : 7 kg/`m^3`
• Water-cement ratio : 0.45

Mix ratio 1:4.3:4.48

Step-10

Result:

The mix ratio of the M35 grade of concrete with fly ash =1:4.3:4.48

The slump shall be measured and the water content and dosage of admixture shall be adjusted for achieving the required slump on the trail. The mix proportion shall be reworked for the actual water content and checked for durability requirements.

2. Concrete mix design for M50 grade concrete without fly ash

AIM: Concrete mix design for M50 grade concrete without fly ash.

Procedure:

Step-1:

Requirements for proportioning:

• Grade designation = M50
• Type of cement = OPC 43 grade conforming to IS 8112
• Maximum nominal size of aggregate = 20 mm
• Minimum cement content 320 kg/`m^3` (From table 5 of IS 456-2000)
• Maximum water-cement ratio = 0.45 (From table 5 of IS 456-2000 codal provision)
• Workability (Slump value) = 100mm
• Exposure condition = Severe (For reinforced concrete)
• Method of concrete placing = Pumping
• Degree of supervision = Good
• Type of aggregate = Crushed angular aggregate
• Chemical admixture type = Superplasticizer

Step-2:

Test Data for material:

• Cement used = OPC 43 grade conforming to IS 8112
• The specific gravity of cement = 3.15
• Chemical admixture type = Superplasticizer (conforming to IS 9103)
• Specific gravity of coarse aggregate and fine aggregate = 2.74
• Water absorption of coarse aggregate = 0.5%
• Water absorption of fine aggregate = 1.0%
• Free surface moisture of coarse aggregate and fine aggregate is Nil.

Step-3:

Target mean strength of concrete:

Target mean strength of concrete(f'ck) = fck+1.65 X S

Where:

• fck = characteristic compressive strength at 28 days.
• S = standard deviation (From IS 10262-2009, table-1 or IS 456-2000, Table-8)

• Given concrete grade is M50
• Compressive strength after 28 days is 50 N/`mm^2`
• The standard deviation is 5 N/`mm^2`.(IS 10262-2009, Table-1)

• NOW (f'ck) = fck+1.65 X S

= 50+1.65 X 5.0

= 58.25 `N/mm^2`

Step-4

Selection of water-cement ratio:

• From the table-5 of IS 456-2000 for moderate exposure, maximum water-cement ratio = 0.45
• Based upon the experience, adopt water-cement ratio = 0.40
• 0.40<0.45 Hence OK.

Step-5

Selection of water content:

• As per IS 10262-2009, table-2 The maximum nominal size of aggregate is 20 mm.
• The standard slump value is between 25 to 50 mm.
• The maximum water content for 20mm aggregate = 186 liters

• Estimated water content = 100 mm slump

=`186+6/100 X 186`

=197 liters

Then the superplasticizer is used, the water content is reduced by 30%. Based on the trial & errors, a water content reduction of 29% is achieved with a plasticizer

Hence the arrived water content = `197X 71/100`= 139.87 =140 liters

Step-6

Calculation of Cement content:

• Water-cement ratio = 0.40
• water content = 140 liters
• Cementies material

= 140/0.40

= 350 kg/`mm^3`

From table-5 of IS 456-2000, the minimum cement content for moderate exposure condition = 320 kg/`mm^3`

Hence, 350 kg/`mm^3` > 320 kg/`mm^3`.

Step-7

The proportion of the volume of coarse aggregate and fine aggregate:

From table-3 of IS 10262-2009, the volume of coarse aggregate corresponding to 20 mm size and the fine aggregate (zone-1) for water-cement ratio = 0.6

• In the present case, the water-cement ratio is 0.40
• Therefore the volume of coarse aggregate needs to be increased to decrease the fine aggregate content.
• As the water-cement ratio is lower by 0.01 the proportion of the volume of coarse aggregate is increased by 0.02
• Therefore corrected volume of coarse aggregate for the water-cement ratio of 0.40=0.62
• For pumpable concrete, these values are reduced by 10%
• Therefore volume of coarse aggregate = 0.62-10% = 0.52
• Volume of fine aggregate content = 1-0.52 = 0.48.

Step-8

Mix calculation:

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

• a) Volume of concrete = 1`m^3`
• b) Volume of cement:

`350/3.14 X 1/1000= 0.110``m^3`

• c) Volume of water:

`140/1 X 1/1000= 0.14``m^3`

• d) Volume of chemical admixture:

`7.6/1.145 X 1/1000 = 0.006 m^3`

• e) Volume of aggregate (CA & FA):

= [a-(b+c+d)]

= [1-(0.110+0.140+0.006)]

= 0.744 `m^3`

• f) Mass of coarse aggregate:

= e X volume of coarse aggregate X specific gravity of coarse aggregate X 1000

= 0.744 X 0.52 X 2.74 X 1000

=1060 kg/`m^3`

• g) Mass of Fine aggregate:

= e X volume of fine aggregate X specific gravity of fine aggregate X 1000

= 0.744 X 0.48 X 2.74 X 1000

= 978.50 kg/`m^3`

Step-9:

Mix proportion:

• Cement : 350 kg/`m^3`
• Water : 140 kg/`m^3`
• Fine aggregate : 978.50 kg/`m^3`
• Coarse aggregate : 1060 kg/`m^3`
• Chemical admixture : 7 kg/`m^3`
• Water-cement ratio : 0.40

Mix ratio : 1:2.79:3.02

Step-10:

Result:

The mix ratio of the M50 grade of concrete without fly ash = 1:2.79:3.02

The slump shall be measured and the water content and dosage of admixture shall be adjusted for achieving the required slump on the trail. The mix proportion shall be reworked for the actual water content and checked for durability requirements.