Question

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.

Answer #1

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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