**CALCULATION OF ONE CUBIC CONCRETE MATERIAL**

The calculation of material needed to produce one cubic concrete is an integral part of every engineer. Most especially, the civil engineer and the builders, Ignorant of this makes you worthless in the field as the field demand accuracy of such calculation. This article is targeted at given a quick and brief method of carrying out this calculation without stress.

As we all know, concrete is a mixture of
cement, sand and coarse aggregates. You will agree with that, when these
components are mixed together, there will be voids due to their coarse nature. On
these bases, the ancient civil engineers and scientist carried out some trial
mixt and come up with the conclusion that in every one cubic meter (1m^{3})
of a wet concrete, the volume decrease by 52% to 54% due to the void in the dry
state.

**ELIMINATION
OF THE VOIDS**

To eliminate this shortage, it is there said that, the volume of every one cubic meter (1m^{3}) Dry concrete should increase by 52% to %54% to cover the shortage. Therefore, to achieve one cubic meter (1m^{3}) of concrete by volume, the dry volume will be either (1.52m^{3}) or (1.54m^{3}) multiply by the wet volume (safety factor x wet volume of concrete). This number is therefore called shrinkage or safety factor of concrete. Good knowledge of this factor is what gives quick access to the calculation of quantity of materials needed for the production of a given volume of concrete.

**CALCULATION
OF MATERILS FOR (1M**^{3}) OF CONCRETE

^{3}) OF CONCRETE

To calculate the materials needed to produce one cubic meter, the following must be born in mind.

- Density of all the materials: cement =
1440kg/m
^{3}, sand = 1600kg/m^{3}, coarse aggregate = 1600kg/m^{3}. - Volume of a 5okg bag of cement = 0.0347 ~
0.035m
^{3} - Sand and coarse aggregate dumpers are measure in meter cube and tones. 1000kg = 1tone

Haven born the above points in mind, let’s go to the calculation of the material needed to produce (1m^{3}) of concrete.

**MATERIAL NEEDED**

Safety factor =1.52

Volume of wet concrete = (1m^{3})

dry concrete volume (safety factor x wet volume) 1.52 x 1 = 1.52m^{3}

Total volume wet concrete require = 1.52m^{3}

Mix ratio = 1: 1.5: 3

Sum of ration =1+1.5+3 = 5.5 therefor,

Volume of cement required = 1/5.5 x 1.52 =
0.1818 x 1.52 = 0.276m^{3}

Volume of one bag of 50kg of cement = 0.035

Quantity of cement require =0.276/0.035 = 7.88 ~ 8bages of cement.

Volume of sand require = 1.5/5.5 x 1.52 =
0.2727 x 1.52 = 0.415m^{3}

Density of sand = 1600kg/m^{3}

Mass = 1600 x 0.415 = 664kg = 0.7tones

Volume of coarse aggregate = 3/5.5 x 1.52 =
0.5454 x 1.52 = 0.829m^{3}

Density of aggregate = 1600kg/m^{3}

Mass =1600 x 0.829 = 1326.4kg = 1.3tones

**Conclusion
**

For every one cubic meter of the mix ratio 1: 1.5 :3

Quantity of cement = 8 bags = 400kg

Sand = 0.415m^{3} = 664kg = 0.7tones

Stones (coarse aggregate) = 0.829m^{3}
= 1326.4kg = 1.3tones

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