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In: Civil Engineering

Calculate the field weights of the ingredients for 0.047 m3 of concrete by using ACI Method...

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.

Solutions

Expert Solution

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


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