We are creating cement test samples in construction materials class and our group is to use a .55 water/cement ratio. Our test cylinder is 4quot; diameter by 8quot; height. I know that this is 0.058 ft?. I am having trouble trying to figure out a formula that i can figure out how much water/cement/aggregate that i should put in this cylinder in order to make 0.058 ft?.
component of a cylnder it really is largely like 2 same circles and a rectangle rectangle = radious of the top multiplyes by capacity of the intensity of the cylinder = 2 Pi R or Pi D prolonged by capacity of the deapth of the cylinder component of both circles = (Pi R^2)*2 (multiply it by capacity of two because there's a circle on each and every end (assuming it really is totally closed))
First you need to determine the specific density of the cement usually 3100kg/m3. The aggregate and sand will be about 2600kg/m3. Next determine the ratio of cement to fine aggregate (sand) to coarse aggregate. A fairly low strength concrete would have a ratio by weight of 1:2:4. The density of the finished concrete will be about 2400kg/m3. So start with 100kg of OPC (Ordinary Portland Cement) then fine aggregate is 200kg and coarse aggregate is 400kg. The water is 55kg (0.55 w/c ratio). Add up total weight and obtain volume by using overall density of mixture (2600kg/m3). This volume is called the yield. By using the ratio of the quantity that you require against the yield using 100kg of OPC you can calculate the amount of OPC, sand, Aggregate and water that you need.
