Project on Concrete Mix Design for various grades of Concrete

M35 Mix design using Flyash:

Aim:

Our aim is to calculate the mix design for M35 grade of concrete using flyash.

Introduction:

• Mix design are calculated to find the strength of the concretes with the help of ingredients like cement, water,fine and coarse aggregates.
• Flyash can also be used as a external material with additional cement to find compressive strength.
• water and cement content can be calculated with the help of codes IS 10262-2019.
• water plays a important role in the mixdesign.
• M35 grade concrete is a high strength concrete where high concretes are used in dams,bridges etc,.

Procedure:

1. Target mean strength:

F'ck=fck+1.65S

here fck=compressive strength of concrete of after 28 days

S=standard deviation (can be taken from table-1 of code IS 10262-2019)

for M35 grade, value of S is 5N/mm2.

F'ck=35+1.65x5

=43.25 N/mm2

ii) Calculation of water cement ratio:

We can calculate w/c ratio from table-5 of IS 456-2000

here we are assuming the climate condition as severe.

from the table, the max water/cement ratio m35 grade of reinforced concrete is 0.45.

from the graph of IS 10262-2019,we are taking w/c ratio as 0.38.

iii) Calculation of Water content:

Here we are assuming size of coarse aggregates 20mm and grading of fine aggregates from zone-1 and slump of 100mm(assume).

From table-2 of IS 10262-2019 the max water content for the 20mm grade of aggregates is 186kgs and for every 25mm of increase of slump add 3%.

=186+(186x0.06)

=197.16kgs approx 197 kgs.

if plasticizers are used in mix design then w/c ratio can be reduce upto 20%.

hence water content=197x0.8

= 157.6 approx of 158litres.

iv) Calculation of Cement content:

w/c ratio=0.42

cement content=158/0.38

= 415.78 approx 416 kgs/cu.m.

From the code IS 456-2000 min cement content is 320kg/cu.m

416 kg/m3 > 320 kg/m3.

here fly ash is using so 10-15% increase in the cementious materials.

1. cementious material content=416x1.15=478.4 approx 478 kgs/cu.m
2. water content=158litres
3. w/c ratio=158/478 = 0.33
4. fly ash @ 30% to cementious material=0.3x478=143.4 approx 143 kgs/cu.m.
5. cement=478-143=335 kgs/cu.m
6. saving of cement from fly ash=416-335=81 kgs/m3.
7. flyash=143 kg/cu.m

v) Proportions of Volume of Coarse and Fine aggregates:

from the table-3 of IS 10262-2019

coarse aggregates-20mm

fine aggregates- zone-1

for water cement ratio of 0.5=0.6

but in our case w/c is 0.33

therefore volume of coarse aggregates needs to increase to decrease fine aggregates,for every plus or minus 0.05 change in w/c ration then add 0.01.

= (0.5-0.33)x0.01/0.05 + 0.6

=0.634

Assume pumpable concrete then these values are reduced by 10%

volume of coarse aggregates=0.634x0.9

=0.57

volume of fine aggregates=1-volume of coarse aggregates

=1-0.57

=0.43

vi) Mix calculations:

1)volume of concrete= 1 cu.m

2)volume of cement= mass of cement/sp.gravity of cement * 1/1000

= 335/3.15 * 1/1000

= 0.106 cu.m

3)volume of flyash= mass of flyash/sp.gravity of flyash * 1/1000

= 143/2.2 * 1/1000

= 0.065 cu.m.

4)volume of water= 158/1000

=0.158 cu.m.

5)volume of admixture = 7.6/1.146 * 1/1000

= 0.006 cu.m.

6) volume of aggregates= 1-(2+3+4+5)

= 1-(0.106+0.065+0.158+0.006)

= 1-0.335

=0.665 cu.m.

7) mass of coarse aggregates= volume of aggregates xvolume of coarse x sp.gravity of coarse aggregate x 1000

= 0.665 x 0.57 x 2.74 x 1000

= 1038.59 kg/cu.m

= approx 1039 kg/cu.m.

8) mass of fine aggregates= volume of aggregates x volume of fine aggregates x sp.gravity of fine aggregates x 1000

= 0.665 x 0.43 x 2.74 x 1000

= 783.5 kg/cu.m

= approx 784 kg/cu.m

Mix Proportions:

cement= 335 kgs

flyash= 143 kg/cu.m

water= 158 kg/cu.m

fine aggregate= 784 kg/cu.m

coarse aggregates= 1039 kg/cu.m

water/cement= 0.33

here i have assumed some data.

2)M50 Mix design:

Aim:

Our aim is to calculate the mix design for M50 grade of concrete.

Introduction:

• Mix design are calculated to find the strength of the concretes with the help of ingredients like cement, water,fine and coarse aggregates.
• water and cement content can be calculated with the help of codes IS 10262-2019.
• water plays a important role in the mixdesign.
• M50 grade concrete is a high strength concrete where high concretes are used in dams,bridges etc,.

Procedure:

Target mean strength:

F'ck=fck+1.65S

here fck=compressive strength of concrete after 28 days

S=standard deviation (can be taken from table-1 of code IS 10262-2019)

for M50 grade, value of S is 5 N/mm2.

F'ck=50+1.65x5

=58.25 N/mm2.

ii) Calculation of water cement ratio:

We can calculate w/c ratio from table-5 of IS 456-2000

here we are assuming the climate condition as extreme.

from the table, the max water/cement ratio for reinforced concrete for extreme is 0.40.

from the graph of IS 10262-2019,we are taking w/c ratio as 0.33.

iii) Calculation of Water content:

Here we are assuming size of coarse aggregates 20mm and grading of fine aggregates from zone-2 and slump of 75mm(assume).

From table-2 of IS 10262-2019 the max water content for the 20mm grade of aggregates is 186kgs and for every 25mm of increase of slump add 3%.

=186+(186x0.03)

=191.58 kgs approx 192 kgs.

if plasticizers are used in mix design then w/c ratio can be reduce upto 20%.

hence water content=192x0.8

= 153.6 approx of 154 litres.

iv) Calculation of Cement content:

from the graph- w/c is 0.33

cement content=154/0.33

=466.66 kg/cu.m approx 467 kg/cu.m

v) Calculation of Volume of Coarse and Fine aggregates:

from table-5 10262-2019

from the table-3 of IS 10262-2019

coarse aggregates-20mm

fine aggregates- zone-2

for water cement ratio of 0.5=0.62

but in our case w/c is 0.33

therefore volume of coarse aggregates needs to increase to decrease fine aggregates,for every plus or minus 0.05 change in w/c ration then add 0.01.

= (0.5-0.33)x0.01/0.05 + 0.62

=0.654

Assume pumpable concrete,then these values are reduced by 10%

volume of coarse aggregates=0.654x0.9

=0.588 approx 0.59

volume of fine aggregates=1-volume of coarse aggregates

=1-0.59

=0.41

vi) Mix calculations:

1)volume of concrete= 1 cu.m

2)volume of cement= mass of cement/sp.gravity of cement * 1/1000

= 467/3.15 * 1/1000

= 0.148 cu.m.

3)volume of water= mass of water/sp.gravity of water * 1/1000

=154/1000

=0.154 cu.m.

4)volume of chemical admixture= 7/1.145 * 1/1000

= 0.006 cu.m

5)volume of aggregates= 1-(2+3+4)

= 1-(0.148+0.154+0.006)

= 1-0.362

= 0.638

6) mass of coarse aggregates= volume of aggregates x volume of coarse aggregates x sp.gravity of coarse aggregates x 1000

= 0.638 x 0.59 x 2.74 x 1000

= 1031.39 kg/cu.m

approx 1031 kg/cu.m.

7) mass of fine aggregates= volume of aggregates x volume of coarse aggregates x sp.gravity of coarse aggregates x 1000

= 0.638 x 0.41 x 2.74 x 1000

= 716.73 kg/cu.m

approx 717 kg/cu.m.

Mix Proportions:

cement= 467 kgs

water= 154 kg/cu.m

fine aggregate= 717 kg/cu.m

coarse aggregates= 1031 kg/cu.m

water/cement= 0.33

here i have assumed some data.