2. Complete the mix design table with the following mix design parameters w/cm 0
ID: 1713997 • Letter: 2
Question
2. Complete the mix design table with the following mix design parameters w/cm 0.50 b) total cementing materials content 400 kg per cubic meter c) Type GUb Blended cement-50% Type GU and 50% Slag d) SG of the Type GUb is 2.975 e) air content 4.0% f) Fly ash content (25% of total cementing materials by mass) g) mortar volume content 60% of the total volume (include the air in the mortar volume) Density (kg/m3) 2975 2400 2650 2650 1000 Volume Ingredient Cement (Type GUb) Fly ash Coarse Aggregate (ssd) Fine Aggregate (ssd) Water Air total Mass (kg) Plastic Density- Total SCM % by mass =Explanation / Answer
Let total volume of concrete mix be 1 m3
Volume of air = 4% = 0.04 m3
Mass of air =0
Total cementing material content = 400 kg
W/ cm = 0.5 thus water content = 0.5 * 400 = 200 kg
Volume of water = 200/ 1000 = 0.2 m3
Mass of fly ash = 25% of total weight of cementing material = 100 kg
Volume of fly ash = 100/ 2400 = 0.0416 m3
Mass of GUB cement = 400- weight of fly ash = 400-100 = 300 kg
Volume of GUB cement = 300/ 2975 = 0.1008 m3
Volume of mortar = 60% of total volume = 0.6 m3
But volume of mortar = volume of air + volume of cement + volume of fly ash + volume of water + volume of fine aggregates
Thus volume of fine aggregates can be calculated from above relationship as = 0.2176 m3
Thus mass of fine aggregates = 0.2176 * 2650 = 576.64 Kg
Now total volume = volume of mortar + volume of coarse aggregates
Thus volume of coarse aggregates = 1- 0.6 = 0.4 m3
Thus mass of coarse aggregates = 0.4 * 2650 = 1060 Kg
Thus plastic density of concrete = 300+ 100+1060+576.64 + 200 = 2236.64 Kg / m3
Total supplementary cementing material by mass =(100/2236.64 )* 100= 4.47 %
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