Project on Concrete Mix Design for various grades of Concrete

Q.

Answer- Concrete mix design for M35 grade of concrete with fly ash-

Step-1 Data required-

Grade designation- M35

Type of cement – OPC 43 grade belongs to IS 8112

Minimum nominal size of aggregate -20mm

Minimum cement content – 320 kg for severe exposure condition taken from table-5 of IS 456-2000

Maximum water cement ratio = 0.45 (from table-5 of IS code 456-2000)

Workability of concrete – 75mm slump value

Exposure condition – severe (for reinforced concrete)

Method of concrete placing – by pumping

Type of aggregate used- crushed angular aggregate

Maximum cement content- 450 kg/m³

Chemical admixture- super plasticizer

Step-2 Test data for material

Cement used- OPC 43 grade it is confirmed corresponding to IS 8112

Specific gravity of cement- 3.15

Fly ash confirming to IS 3812 (part-1)

Specific gravity of fly ash – 2..2
specific gravity of coarse aggregate and fine aggregate- 2.74

Water absorption- coarse aggregate- 0.5% and fine aggregate – 1%

Step-3 calculate the target strength of mix proportioning-

f`ck = fck+1.65S

So S = 5 N/mm² (where S is standard deviation which is taken from table-1 of IS 10262-2019)

Therefore target mean strength = 35 + 1.65*5 = 43.25 N/mm²

Step-4 Selection of water-cement ratio-

w/c ratio = 0.45

but we can take w/c ratio = 0.40 which is less than 0.45

so w/c ratio= 0.40

(for severe weather exposure condition which is taken from table-5 of IS 456-2000)

Step-5 selection of water content-

Maximum water content for 20 mm aggregate is 186 kg or liter for slump range 25 to 50 mm

(it is taken from table-2 of IS code 10262-2019)

Therefore maximum water content for 75mm slum is

= 186 + 186*3/100 = 191.58 liter

Note- for every 25mm extra slump above 50mm we add 3% extra water content

If we are using super plasticizer then water content will be reduced upto 30% but we take for plasticizer which reduced the water content 29%

Therefore calculated water content on the basis of plasticizer is =

= 191.58 * 0.71 = 136.0218 liter

Step-6 calculation of cement content-

We calculated w/c ratio 0.40

So cementious material (cement+ fly ash) content =

= 136.02/0.40 = 340.05 kg/m³

Minimum cement content required for severe exposure condition = 320 kg/m³

(It is taken from table-5 of IS 456-2000)

So 340.05 kg/m³ > 320 kg/m³ hence ok

Step -7 Cementious material like fly ash-

It is increasing 10% cementious content which is calculate according to experiment

Therefore cementious material = 340.05 * 1.10 = 374.055 kg/m³

Water content = 136.02

So water-cement ratio = 136.02/374.055 = 0.363 < 0.40

Hence lower value of w/c value taken

Quantity of fly ash in total cementious material = 30% of total cementious material

= 374.055 * 0.30 = 112.21 kg/m³

Cement quantity = 374.055 – 112.21 = 261.845 kg/m³

Step-8 - proportion of volume of fine aggregate and coarse aggregate-

From table-3 of IS 10262-2019 the volume of coarse corresponding to 20mm size of aggregate and fine aggregate (zone-1) for water cement ratio 0.50 = 0.60

But we have the value of water cement ratio is 0.40

Water- cement ratio difference is equal to 0.10 and we have to increase the volume of coarse aggregate and decrease the volume of fine aggregate

So coarse aggregate increase by 0.002

(it increases at a rate of 0.01 for every 0.05 change in water-cement ratio)

Therefore the corrected volume of coarse aggregate is 0.62 for water-cement ratio of 0.40

If we are taking pumpable concrete (slump 75mm to 100mm) these values are reduced by 10%

So volume of coarse aggregate = 0.62 * 0.9 = 0.56

Volume of fine aggregate = 1- 0.56 = 0.44

Step-9 mix calculation-

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

1. a) Volume of concrete = 1m³
2. Volume of cement = mass of cement/specific gravity of cement *1/1000

= 261.845/3.15 *1/1000 = 0.083 m³

(Here specific gravity of cement is 3.15)

1. c) Volume of fly ash = mass of fly ash/specific gravity of fly ash * 1/1000

= 112.21/2.2 *1/1000 = 0.051 m³

1. Volume of water = mass of water/specific gravity of water * 1/1000

= 136.02/1 *1/1000 = 0.136 liter

1. Volume of chemical admixture = mass of chemical admixture/specific gravity of chemical admixture *1/1000

= 7.6/1.145 *1/1000 = 0.006 m³

1. So the volume of total aggregate(F.A&C.A) =

= (a-(b+c+d)) = (1-(0.083+0.051+0.136+0.006) = 0.67 m³

1. Mass of coarse aggregate = volume of total aggregate * volume of coarse aggregate * specific gravity of coarse aggregate * 1000

= 0.67 *0.56 * 2.74 * 1000 = 1028.048 kg/m³

1. h) Volume of fine aggregate = volume of total aggregate * volume of fine aggregate * specific gravity of fine aggregate * 1000

= 0.67 * 0.44 * 2.74 * 1000 = 807.752 kg/m³

Step-10 calculation of mix proportion-

Cement = 261.845 kg/m³

Fly ash = 112.21 kg/m³

Water = 136.02 kg/m³

Fine aggregate = 807.752 kg/m³

Coarse aggregate = 1028.048 kg/m³

Chemical admixture = 7 kg/m³

Water-cement ratio = 0.363

Total mix ratio = 1 : 2.16 : 2.75 corresponding to cement : fine aggregate : coarse aggregate

Q.

Answer- Mix design of M50 grade of concrete-

Step-1 Data required-

Grade designation- M50

Type of cement – OPC 43 grade belongs to IS 8112

Minimum nominal size of aggregate -20mm

Minimum cement content – 340 kg for very severe exposure condition taken from table-5 of IS 456-2000

Maximum water cement ratio = 0.45 (from table-5 of IS code 456-2000)

Workability of concrete – 50mm slump value

Exposure condition – very severe (for reinforced concrete)

Method of concrete placing – by dumpers and trucks

Type of aggregate used- crushed angular aggregate

Maximum cement content- 450 kg/m³

Chemical admixture- water reducer admixture

Step-2 Test data for material

Cement used- OPC 43 grade it is confirmed corresponding to IS 8112

Specific gravity of cement- 3.15
specific gravity of coarse aggregate and fine aggregate- 2.74

Water absorption- coarse aggregate- 0.5% and fine aggregate – 1%

Step-3 Target mean strength for mix proportioning-

F`ck = fck + 1.65*S

f`ck = 50 + 1.65*5 = 58.25 N/mm²

S that is standard deviation taken from table-1 of IS 10262-2019

Step-3 Selection of water cement ratio –

w/c = 0.45 for very severe condition taken from table-5 of IS456-2000

for better strength we can take w/c ration 0.40

0.40 < 0.45 hence it`s ok

Step-4 selection of water content-

Maximum water content for 20 mm aggregate is 186 kg or liter for slump range 25 to 50 mm

(it is taken from table-2 of IS code 10262-2019)

Therefore maximum water content for 50mm slum is

= 186 lite

We use chemical admixture like water reducing admixture about 10%

= 186 * 0.90 = 167.4 liter

Step-5 calculation of cement content-

We calculated w/c ratio 0.40

So cement content =

= 167.4 /0.40 = 418.5 kg/m³

Minimum cement content required for severe exposure condition = 340 kg/m³

(It is taken from table-5 of IS 456-2000)

So 410.5 kg/m³ > 340 kg/m³ hence ok

Water content = 167.4

so water-cement ratio = 167.4 = 0.407 > 0.40

Hence lower value of w/c value taken 0.40

Step-6 - proportion of volume of fine aggregate and coarse aggregate-

From table-3 of IS 10262-2019 the volume of coarse corresponding to 20mm size of aggregate and fine aggregate (zone-1) for water cement ratio 0.50 = 0.60

But we have the value of water cement ratio is 0.40

Water- cement ratio difference is equal to 0.10 and we have to increase the volume of coarse aggregate and decrease the volume of fine aggregate

So coarse aggregate increase by 0.002

(it increases at a rate of 0.01 for every 0.05 change in water-cement ratio)

Therefore the corrected volume of coarse aggregate is 0.62 for water-cement ratio of 0.40

Step-7 mix calculation-

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

1. a) Volume of concrete = 1m³
2. Volume of cement = mass of cement/specific gravity of cement *1/1000

= 418.5/3.15 *1/1000 = 0.132 m³

(Here specific gravity of cement is 3.15)

1. Volume of water = mass of water/specific gravity of water * 1/1000

= 167.4/1 *1/1000 = 0.167 liter

1. Volume of chemical admixture = mass of chemical admixture/specific gravity of chemical admixture *1/1000

= 7.6/1.145 *1/1000 = 0.006 m³

1. So the Volume of total aggregate(F.A&C.A) =

= (a-(b+c+d)) = (1-(0.132+0.167+0.006) = 0.695 m³

1. Mass of coarse aggregate = volume of total aggregate * volume of coarse aggregate * specific gravity of coarse aggregate * 1000

= 0.695 *0.62 * 2.74 * 1000 = 1180.666 kg/m³

1. h) Volume of fine aggregate = volume of total aggregate * volume of fine aggregate * specific gravity of fine aggregate * 1000

= 0.695 * 0.38 * 2.74 * 1000 = 723.634 kg/m³

Step-8 calculation of mix proportion-

Cement = = 418.5 kg/m³

Water = 167.4 kg/m³

Fine aggregate = 723.634 kg/m³

Coarse aggregate = 1180.666 kg/m³

Chemical admixture = 2.5 kg/m³

Water-cement ratio 0.40

Total mix ratio = 1 : 1.729 : 2.82 corresponding to cement : fine aggregate : coarse aggregate