HOW TO CALCULATE THE QUANTITIES OF CEMENT SAND AND BRICKS REQUIRED FOR 1ST CLASS BRICK WORK ? ALONG WITH FREE EXCEL SHEET DOWNLOAD.
There are many ways to calculate the size of bricks required
for construction of a brick wall, and I think most probably I have tried my
best to simplify this process as much as I can.
Along with this article I will provide you the excel sheet
for the ease of calculations and quick results.
If you want to directly skip the explanation you can skip
and at the end of the article I have attached the link from there you can
download the excel sheet of the topic.
1. We required some basic data about brick work like size of
bricks, mortar ratio and density of materials which may vary from country to
country,
Note: density (δ) of materials are only required if you want
to convert the values of sand in kg.
In this example lets take the size of bricks and other data as follows,
In this example lets take the size of bricks and other data as follows,
Size of bricks with mortar = 20 cm X 10 cm X 10 cm
= 0.2 m X 0.1 m X 0.1 m
Volume of one brick = 0.002 m3
Size of bricks without mortar = 19 cm X 9 cm X 9 cm
= 0.19 m X 0.09 m X 0.09 m
Volume of one brick = 0.0015 m3
Generally the thickness of mortar is assumed as 1 cm
Mortar ratio = 1: 4
Density (δ) of cement = 1400 kg/m3
Density (δ) of sand = 1450 kg/m3
2. At first you should know what’s your
requirement/specifications of brick work or if you are student and doing
calculation for study purpose then you have to assume certain base quantity on
the basis of that assumed quantity you can calculate the materials requirements very quickly.
In below example let’s assume the base volume as = 10 m3
3. moving to the calculation part lest first find the
quantity of bricks with mortar and without mortar,
To find the quantity of bricks (no. of bricks) with mortar
just divide the assumed base quantity with the size of brick with mortar\[ no.~ Of~ bricks~with~ mortar = {{base ~volume} \over size
~of ~brick ~with ~mortar} \]\[ no.~ Of~ bricks~with~ mortar = {10 \over (20~ X ~10 ~X
~10)}\]
\[ no.~ Of~ bricks~with~ mortar = {10 \over 0.002} \]
\[ no.~ Of~ bricks~with~ mortar = {5000~ nos.}\]
Note: the reason we
are finding the no. of bricks with mortar is because with the help of this
value we can find out the volume of mortar i.e. wet volume of mortar and the
volume of bricks in the assumed base quantity i.e. 10 m3.
4. For finding the volume occupied by the bricks, we have to
multiply the total no. of bricks with mortar with the volume of brick without
mortar.
To simplify this, the formula is give below\[ volume~ occupied~ by~ bricks ~= ~no. ~of ~bricks ~with
~mortar ~X \]\[~~~~~~~~~~~~~~~~~~~ volume~ of~ 1 ~brick ~without ~mortar \]
\[ volume~ occupied~ by~ bricks ~= ~5000 ~X~ 0.002 \]
\[ volume~ occupied~ by~ bricks ~= ~7.70~ m^3 \]
5. no we can simply find the volume occupied by mortar by simply deducting the above value from the assumed base volume i.e 10 m3
To simplify this, the formula is give below
\[ volume~ occupied~ by~ mortar ~= ~assumed ~base~ volume
~-\]\[~~~~~~~~~~~~~~~~~~~ volume~ volume ~occupied~ by~ bricks \]\[ volume~ occupied~ by~ mortar ~= ~10 ~-~ 7.70 \]\[ volume~ occupied~ by~ mortar ~= ~2.31 ~m^3\]6. now we have got the volume occupied by the mortar , but this value of mortar we got is without considering any wastage of mortar,
So to get the actual wet volume of mortar we have to
increase the volume of mortar by 15%
To simplify this, the formula is give below\[wet~ volume ~of ~mortar = {~ volume~ occupied~ by~ mortar} ~+ \]\[~~~~~~~~~~~~~~~~~~~ {~15 \%~ of~ volume~ occupied~ by~ mortar} \]
\[ wet~ volume ~of ~mortar = ~2.31 ~+ {~15\over 100}~ X ~2.31 \]\[ wet~ volume ~of ~mortar = ~2.31 ~+ 0.3465\]\[ wet~ volume ~of ~mortar = ~2.65 ~m^3\]7. Next step is to find the dry volume of mortar, and for that we have to increase the wet volume by 25%.
To simplify this, the formula is give below\[dry~ volume ~of ~mortar = ~ wet~ volume ~of ~mortar ~+ ~25\%~ of~ wet~ volume ~of ~mortar \]\[ dry~ volume ~of ~mortar = ~2.65 ~+ {~25\over 100}~ X ~2.65 \]\[ dry~ volume ~of ~mortar = ~3.31 ~m^3\]
8. After calculating the dry volume we have to calculate the volume of cement and sand.
To calculate the volume of cement we simply have to divide
the dry volume of mortar with addition of ration of cement, sand and then
multiply it with the part of cement on it,
\[~Volume~ of ~cement = {~ dry~ volume ~of ~mortar \over addition~ of~ cement ~ sand ~ratio }~ X ~ part~ of~ cement~ in~ the~ ratio. \]\[~Therefore, \]\[~Volume~ of ~cement = {~3.31 \over (1+4) }~ X ~ 1 \] \[~Volume~ of ~cement = {~3.31\over 5 }~ X ~ 1 \]\[~Volume~ of ~cement = 0.66 ~m^3\]
9. to find the volume of sand, we have to follow the same procedure , as we have find the volume of cement just you have to replace the part of cement with sand,
\[~Volume~ of ~sand = {~ dry~ volume ~of ~mortar \over addition~ of~ cement ~ sand ~ratio }~ X ~ part~ of~ sand~ in~ the~ ratio. \]\[~Therefore, \]\[dry~ volume ~of ~mortar = ~ wet~ volume ~of ~mortar ~+ ~25\%~ of~ wet~ volume ~of ~mortar \]\[ dry~ volume ~of ~mortar = ~2.65 ~+ {~25\over 100}~ X ~2.65 \]\[ dry~ volume ~of ~mortar = ~3.31 ~m^3\]
10. Now we have got the values of Volume of cement, sand in m3, but many a times what happens is we need the volume of cement in bags , and the volume of sand in brass or kg. for the purpose of ordering the quantity.
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
\[~density~ (δ) = {~mass ~(m) \over ~volume~ (v) }~ \] \[expressed ~in~ kg/m^3\]Therefore,\[~density~ of~ cement = {~mass~ of ~cement~ bag \over ~volume ~cement~bag }~ \] \[~1440 = {~50 \over ~volume ~cement~bag }~ \] \[~volume ~one ~cement ~bag = {~50 \over ~1440 }~ \] \[~volume~ of~cement~bag = {~0.035~ m^3 }~ \]
11. To find the no.of bags of cements we just have to divide
the volume of cement with the volume of one bags of cement.\[~ no.~of ~cement~ bags = {~ Volume~ of ~cement \over ~ volume ~one ~cement~bag }~ \]\[~ no.~of ~cement ~bags = {~ 0.66 \over ~0.035 }~ \]\[~ no.~of ~cement ~bags = {~ 19 ~ bags ~ }~ \]
12. To find the weight of sand and in kg we should know the density of sand and aggregate which is given below,
Density of sand = 1450 – 1500 kg /m3
therefore,\[~ Density of sand = {~ weight ~of ~sand~ in~ kg \over ~ volume~ of~ sand }~ \]\[~ 1450= {~ Weight~ of ~sand~ in ~kg \over ~ 2.65 }~ \]\[~ 1450~ X~ 2.65 = {~ Weight~ of ~sand ~in~ kg}~ \]\[~ Weight ~of~ sand / ~10~ m^3 = {~ 3842.5~ kg }~ \]\[~ Weight ~of~sand / ~1~ m^3 = {~ 384.3 ~kg }~ \]
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 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,\[ 1_{m^3} = 35.3147~ ft.^3~ (cubic~ft~)\]
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Great work, thanks a lot. This really helps!
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