The amount of cement are specified in the C1293. And the W/C ratio can be determined by the results of the workability test.
The specification for coarse aggregate is that the 'dry mass of coarse aggregate per unit volume of concrete' should be 'equal to 0.70 +- 0.02 of its dry-rodded bulk density', which means that, for example, if the dry-rodded bulk density of the coarse aggregate is 1650 kg/m^3, then you will have to add around 0.70*1650=1155 kg coarse aggregates for each 1 m^3 of fresh concrete.
To find out the amount of fine aggregates needed, use the equation that the volumes of all the components per unit volume of fresh concrete ( that is the amount of cement, water, etc. per unit volume divided by their density respectively) sums up to 1 unit volume (e.g. 1 m^3).
And thus you can obtain the proportions of every component.
FOR CASTING A CUBE OF 150 *150*150 MM YOU NEED GENERALLY ABOUT 8-8.5 KG OF WET CONCRETE. COARSE AND FINE AGGREGATES MAKING ABOUT 70-75% OF ALL. COARSE AGGREGATES NEEDED ABOUT 45% OF ALL WHICH MAKES 3.6 - 3.8 KG .( WATER CEMENT RATIO 0.35 - 0.40)
If you know the density of your aggregates then you can easily calculate it. In a mixtures with 300kg cement /m³ and a w/c ratio of 0,5 (150kg/m³), you’ll have already taken 0,25% of the available volume. This means the other 75% needs to be filled with aggregates. Normal aggregates have a density of 2500 kg/m³, this leads to 1875kg of aggregates needed. Depending on the method you use to calculate the mixtures (Volume based or skeleton based) and the d/D of your aggregates, you’ll need something like 800-1000 of fine/coarse to fill one m³ of concrete. One cube 150 is 0,338% of a m³. This means 2,7-3,4kg of fine and coarse aggregates. I normally try to create 10% of additional concrete so I’m 100% sure I can fill the cubes. If you use lightweight aggregates, different amount of cement or different w/c-ratio this calculation will change a little, but the idea will stay the same.
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