Question

# Calculate the field weights of the ingredients for 0.039 m3 of concrete by using ACI Method...

Calculate the field weights of the ingredients for 0.039 m3 of concrete by using ACI Method of Mix Design. Job specifications dictate the followings:

Cement content: 300 kg/m3

w/c: 0.57 (by weight, from the strength point of view)

w/c: 0.53 (by weight, from durability point of view)

Fine Aggregate (SSD)/Coarse Aggregate (SSD)=0.45 (by weight)

Air Content: 1,30 %

For Fine Aggregate (Crushed): SSD Bulk Specific Gravity: 2.47, Total Moisture: 5,0 % and Absorption Aggregate: 2,7 %.

For Coarse Aggregate (Crushed): SSD Bulk Specific Gravity: 2.60, Absorption Capacity: 2.0 %, Total Moisture: 1.9 % and Dry Rodded Unit Weight: 1470 kg/m3.

Cement: Sulphate Resisting Cement, Specific Gravity: 3.17.

Ans) Given,

Cement content = 300 kg/

Average Water cement ratio = (0.57 + 0.53) / 2 = 0.55

Water required = Cement x w/c ratio = 300 x 0.55 = 165 kg/

Let the total weight of coarse aggregate be 'x' kg and fine aggregate be 'y' kg then

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

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

=> 1 - [(165/1000) + (300 / 3.17 x 1000) + (x/ 2.6 x 1000) + (y/ 2.47 x 1000) + 0.013]

=> 1 = 0.165 + 0.0946 + 0.000384 x + 0.000404 y + 0.013

=> 0.000384 x + 0.000404 y = 0.7274

Given, y / x = 0.45

=> y = 0.45 x

=>  0.000384 x + 0.000404 (0.45 x) = 0.7274

=> 0.0005658x = 0.7274

=> x = 1285.61 kg

Hence, required amount of coarse aggregate (x) = 1285.61 kg/

Hence, required amount of fine aggragete (y) = 0.45 x 1244.75 kg = 578.53 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 - 0.019) x 1285.61 = 1.28 kg

Net Water absorbed by fine aggregate = (0.027 - 0.05 ) x 578.53 = -13.30 kg

=> Actual amount of water to be added = 165 + 1.28 - 13.30 = 152.98kg/

Now,

Since batch weight of 1 concrete is now known , batch weight for trial mix of 0.039 can be calculated as follows :

Water = 152.98 kg/ x 0.039 = 5.96 kg

Cement = 300 kg/ x 0.039 = 11.70 kg

Fine aggregate = 578.53 kg/ x 0.039 = 22.56 kg

Coarse aggregate = 1285.61 kg/ x 0.039 = 50.14 kg

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