Question

Calculate the field weights of the ingredients for 0.047 m³ of concrete by using ACI Method of Mix Design. Job specifications dictate the followings:

Cement content: 330 kg/m³

w/c: 0.58 (by weight, from the strength point of view)

w/c: 0.54 (by weight, from durability point of view)

Air Content: 1,40 %

For Fine Aggregate (Crushed): SSD Bulk Specific Gravity: 2.48, Total Moisture: 5,0 % and Absorption Aggregate: 2,8 %.

For Coarse Aggregate (Crushed): SSD Bulk Specific Gravity: 2.60, Absorption Capacity: 2.0 %, Total Moisture: 1.7 % and Dry Rodded Unit Weight: 1483 kg/m³.

Cement: Sulphate Resisting Cement, Specific Gravity: 3.18.

Answer 1

Ans) Given,

Cement content = 330 kg/

Average Water cement ratio = (0.58 + 0.54) / 2 = 0.56

Water required = Cement x w/c ratio = 330 x 0.56 = 184.8 kg/

Let the volume of coarse aggregate be 0.68 , then

Amount of coarse aggregate = dry rodded density x volume = 1483 kg/ x 0.68 = 1008.44 kg

=> Hence, required amount of coarse aggregate = 1008.44 kg/

Volume of fine aggregate = Total volume - Volume of water, cement, coarse aggregate and air

Also, volume = Weight / (1000 x specific gravity)

=> Fine aggregate volume = 1 - [(184.8/1000) + (330 / 3.18 x 1000) + (1008.44 / 2.60 x 1000) + 0.014]

=> Fine aggregate volume = 1 -  0.1848 + 0.1038 + 0.3878 + 0.014 = 0.31

=> Amount of fine aggregate = Volume x specific gravity x 1000

Hence, required amount of fine aggragete = 0.31 x 2.48 x 1000 = 768.8 kg/

Now, since both aggregates has moisture and has absorption capacity , amount of mixing water needs to be corrected as shown :

Net Water absorbed by coarse aggregate = (0.02 - 0.017) x 1008.44 = 3.02 kg

Net Water absorbed by fine aggregate = (0.028 - 0.05) x 768.8 = - 16.91 kg

=> Actual amount of water to be added = 184.8 + 3.02 - 16.91 = 170.91 kg/

Now,

Since batch weight of 1 concrete is now known , batch weight for trial mix of 0.047 can be calculated as follows :

Water = 170.91 kg/ x 0.047 = 8.03 kg

Cement = 330 kg/ x 0.047 = 15.51 kg

Fine aggregate = 768.8 kg/ x 0.047 = 36.13 kg

Coarse aggregate = 1008.44 kg/ x 0.047 = 47.40 kg