Predicting the Density of Concrete

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SUMMARY

The density of concrete is approximately 2400 kg/m³, influenced by its component materials: water (998 kg/m³), cement (1506 kg/m³), aggregate (1602 kg/m³), and sand (1600 kg/m³). The apparent discrepancy in density arises from the presence of air in these materials, which is removed during mixing. To calculate the density of a concrete and steel mixture, one must consider the volumes of both components in the final product. This understanding allows for accurate density predictions when incorporating steel into the aggregate mix.

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smithson1984
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Hello Everyone,

I have a problem which is flummoxing me concerning concrete!

The density of concrete is approx. 2400 kg/m^3 (depending on mix) however the component parts which make up concrete are as follows:

Water (density 998 kg/m^3)
Cement (density 1506 kg/m^3)
Aggregate (density 1602 kg/m^3)
Sand (density 1600 kg/m^3)

My question is how do all of these relatively light materials combine to form a relatively dense material like concrete?

The reason for my question is that I wish to use some steel in the aggregate mix and thus increase the density of the aggregate but I cannot calculate what density of concrete this will produce without understanding the above.

Any help and/or simple estimate formulas would be greatly appreciated!

Thanks a lot in advance,

Ian
 
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The densities you are quoting (except for the water) contain a significant quantity of air. For example,the density of quartz is about 2650 kg/m^3, and the reason sand is so much less is that it is a mixture of quartz particles and air. The same for the aggregate - typical rock densities are 2000-3000 kg/m^3. When you mix these materials to create concrete, the air is removed, so you get a density closer to the bulk materials without the air. For your problem, why don't you just accept the final concrete density of 2400 kg/m^3, and calculate the density of a mixture of concrete and steel? Presumably you know the volumes of concrete and steel in the final product.
 
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Hi phyzguy,

Thanks very much for the response! I think you've cracked the problem and it now makes sense. The calculations I'm coming out with are now making much more sense.

It's obvious when you know how!

Cheers,

Ian
 
Glad I could help. Good luck with you project.
 
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