HOW TO CALCULATE THE QUANTITY OF CEMENT, SNAD AND AGGREGATE REQUIRED FOR PCC (PLAIN CEMENT CONCRETE) WITH FREE EXCEL SHEET, WITH UNIT CONVERSION
First of all the aim of this article is to calculate the
quantity of materials required to Produce PCC (plain cement concrete), of
certain volume.
Along with this article I will provide you the excel sheet
for the ease of calculation and quick results.
But first let me explain the steps which involves in the
process of taking out quantities for PCC (plain cement concrete)
1. At first you should know what your requirement of PCC is
or if you are student and doing calculation for study purpose then you have to
assume certain base quantity on the basis of which you can calculate the
materials requirement very quickly.
In below example let’s assume the base volume is = 10 m3
NOTE: here we are assuming the base volume as the final
volume of concrete (i.e. when the concrete is ready to use at site so basically
we have to increase the base volume of concrete by certain percentage)
2. After assuming the base volume you should know about the
grades of concrete to be used as PCC
So let’s assume the grade of concrete is M25
For M25 grade of concrete the ration of cement, sand and
aggregates is = 1:1.5:3
3. To get the dry volume of concrete we have to increase the
wet volume by 52 %
Therefore,
\[dry~volume ~of ~concrete = ~wet ~volume ~+ ~52\%~ of~ wet~ volume \]
\[dry~volume~of~concrete = 10 + {~52\over100}~ X ~10\]
\[~~~~~~~~~~~~~~~~~~~~~~~~= ~10 ~+~ 5.2 \]
\[ dry~
volume ~of ~concrete = 15.2 ~m^3 \]
4. After calculating the dry volume we have to calculate the
volume of cement sand and aggregate step by step.
5. To calculate the volume of cement we simply have to
divide the dry volume of concrete with addition of ration of cement, sand and
aggregate and the multiply it with the part of cement on it
To simplify this, the formula is give below
\[~Volume~of ~cement ={~dry~ volume\over addition~ of~ ratio }~ X ~ part~ of~ cement~ in~ the~ ratio\]
Therefore,
\[~Volume~of ~cement={~15.2\over (1+1.5+3) }~ X ~ 1 \]
\[~Volume~of ~cement={~15.2\over 5.5 }~ X ~ 1 \]
\[~Volume~of ~cement=2.76 ~m^3\]
6. to find the volume of sand and aggregate same procedure
is to be followed, as we have find the volume of cement just you have to
replace the part of cement with sand and aggregates,
To simplify this, the formula is give below
\[~Volume~of ~sand={dry~ volume\over addition~ of~ ratio }~ X ~ part~ of~ sand~in~ the~ ratio. \]
Therefore
\[~Volume~of ~sand={~15.2\over (1+1.5+3) }~ X ~ 1.5 \]
\[~Volume~of ~sand={~15.2\over 5.5 }~ X ~ 1.5 \]
\[~Volume~of ~sand=4.15 ~m^3\]
7. to find the volume of aggrgegates,
\[~Volume~
of ~aggregates ={~dry~ volume\over addition~ of~ ratio }~ X ~ part~ of~
aggregates~ in~ the~ ratio. \]
Therefore,
\[Volume of cement = 15.2/ (1+1.5+3) * 3\]\[~Volume~
of ~aggregates = {~15.2\over (1+1.5+3) }~ X ~ 3 \]\[~Volume~
of ~aggregates = {~15.2\over 5.5 }~ X ~ 3 \]\[~Volume~
of ~aggregates =8.29 ~m^3\]
8. Now we have got the values of Volume of cement, sand and aggregates is in m3, but many a
times what happens is we required the volume of cement in bags, and the volume of sand and aggregates in brass or kg.
9. To find the volume of cement in bags, you should know the
volume of one cement bag which is available in market in m3.
And to calculate the volume of one bag of cement we just
required the density of cement which is = 1440 kg / m3
Now everyone knows the formula for density which is mass /
volume so from this formula we can calculate the mass of one bag of cement in
m3
\[~desnity~(δ)={~mass ~(m) \over ~volume~ (v)}~\]
\[expressed~in kg/m^3\]
\[~desnity~of ~cement={~mass~ of ~cement~ bag\over ~volume ~cement~bag}~ \]
\[~1440={~50\over ~volume~cement~bag}~ \]
\[~volume~one ~cement ~bag={~50\over ~1440}~ \]\[Volume ~of ~one~ cement~ bag = 0.035~ m3\]
\[~volume~cement~bag={~0.035~ m^3}~ \]
10. To find the no.of bags of cements we just have to devide
the volume of cement with the volume of one bags of cement
\[~ no.of cement bags ={~ Volume~ of ~cement\over ~ volume ~one ~cement~bag}~ \]
\[~no.of cement bags ={~ 2.76\over ~0.035 }~ \]
\[~no.of cement bags ={~ 79 ~ bags }~ \]
11. To find the weight of sand and aggregate in kg we should
know the density of sand and aggregate which is given below,
Density of sand = 1450 – 1500 kg /m3
Density of aggregates = 1450 – 1600 kg /m3
Therefore,
\[~Density of sand ={~ weight of sand in
kg\over ~ volume of sand }~ \]
\[~
1450={~ Weight of sand in kg \over ~
4.15 }~ \]
\[~
1450~ X~ 4.15 ={~ Weight~ of ~sand ~in~
kg}~ \]\[~
Weight ~of~ sand / ~10~ m^3 ={~ 6017~
kg }~ \]\[~
Weight ~of~sand / ~1~ m^3 ={~ 601.7 ~kg
}~ \]
12. Density of Aggregates = weight of aggregates in kg /
volume of aggregates
\[~
Density~ of ~aggregates ={~ weight ~of~
aggregates ~in ~kg\over ~ volume ~of~
aggregates }~ \]\[~
1600 ={~ Weight ~of~ aggregates ~in~kg
\over ~ 8.29 }~ \]\[~
1600~X ~8.29 ={~ Weight ~of~aggregates ~in
~kg}~ \]\[~Weight ~of ~aggregates / ~10 ~m^3 = {~ 13,264~ kg }~ \]\[~Weight ~of~aggregates / 1 ~m^3 ={~ 1,326~ kg }~ \]
13. There is one more thing which confuse most of the people
about the conversion of m3 to cubic ft.
In India we calculate the quantity in m3 and at the time of
ordering the same we use different units so to find the volume of sand and
aggregates for ordering purpose,
You just have to know the conversion factor for m3 to cubic
ft. then we can easily convert the value to get the quantity in brass.
To convert the values from m3 to ft3 (cubic ft) you have to
multiply the values with 35.3147
Because ,
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