Hi civil engineers today I will share some
useful calculations . this calculations are fundamental for civil engineers.
therefore this calculations are used most of the times to calculate various
materials like cement and sand volume in mortar ,brick volume in brick word,
cement ,sand and aggregate in concrete, volume of steel in reinforced cement
concrete.
QUANTITY OF CEMENT ,FINE AGGREGATE, COARSE AGGREGATE REQUIRED FOR DIFFERENT GRADES OF CEMENT CONCRETE FOR ONE CUBIC METRE.
GRADE OF CONCRETE

RATIO

CEMENT BAGS

FINE AGGREGATE (SAND )
M^{3}

COARSE AGGREGATE
20 mm
M^{3}

COARSE AGGREGATE
40 mm
M^{3}

M_{7.5}

1:4:8

3.40

0.48



0.96

M_{10}

1:3:6

4.40

0.46

0.92



M_{15}

1:2:4

6.30

0.44

0.88



M_{20}

1:1.5:3

8.05

0.42

0.84



NOTE: ONE BAG CEMENT =
50 KG OR 0.034 CUB METRE

DRIVE HOW IT IS CALCULATED
for one m concrete for the ratio 1:2:4
Quantity of cement is one part
Quantity of sand is two parts
Quantity of coarse aggregate is four parts
total 1+2+4 = 7 unit
VOLUME OF CEMENT
Volume of cement = 1/7 x 1.5
=0.214 m^{3}
= 0.214 x
1440 (note for unit weight of
cement ( 1m cement ) = 1440 kg / m)
= 308.16
kg
=
308.16/50
= 6.30 bags
VOLUME OF SAND
Volume of sand = (2/7) x 1.5
= 0.428 m^{3}
Volume of coarse aggregate
Volume of coarse aggregate = (4/7 ) x 1.5 =
0.857 m^{3}
^{}
QUANTITY OF CEMENT ,SAND ( FINE AGGREGATE )
, COARSE AGGREGATE AND BRICKS REQUIRED FOR DIFFERENT RATIO OF MORTAR IN ONE
CUBIC METRE OF BRICK WORK.
MORTAR RATIO

CEMENT BAGS

SAND
(m^{3})

BRICKS ( nos )

1:2

3.45

0.24

500

1:3

2.40

0.25

500

1:4

1.75

0.25

500

1:5

1.49

0.26

500

1:6

1.20

0.25

500

1:7

1.00

0.25

500

DERIVE HOW IT IS CALCULATED
To find no of bricks in one cub.metre
Size of brick = 19 cm x 9 cm x 9 cm
Mortar thick = 1 cm
Volume of one brick without mortar = 0.19 x
0.09 x 0.09
= 0.00154 m^{3}
So size of brick with mortar thick = 20 cm
x 10 cm x 10 cm
Now volume of one brick = 0.3 x 0.1 x0.1
=
0.002 m^{3}
No of brick in one cub metre = 1 / 0.002
= 500 nos
TO FIND VOLUME OF MORTAR ( CEMENT ,SAND, )
REQUIRED FOR ONE CUBIC METRE
Volume covered by bricks in one cubic metre
without mortar = volume of one brick x no of brick
= 0.00154 x 500
= 0.76 m^{3}
Volume of mortar = total volume – volume of
bricks ( 500 nos )
=
1 – 0.76
=
0.24m^{3 }( wet volume )
Now we are take mortar ratio is 1:5
Total unit = 1+5 = 6
Volume of cement = (1x0.24 x 1.3 )/ 6
= 0.052 m^{3}
= 0.052 x
1440
= 74.88 kg
=78.44/ 50
= 1.49 bags
Volume of sand = (5 x 0.24 x 1.3 ) / 6
= 0.26 m^{3}
^{}
BREAK UP COST OF BUILDING
1.earth work excavation and filling = 1 % total cost
2.foundation concrete = 5 % total cost
3.damp proof course = 1% total cost
4.brick work =
34 % total cost
5. flooring = 16 % total cost
6. roofing
= 20 % total cost
7. joinery =
16 % total cost
8 plastering =
10 % total cost
9. white washing ,colour washing and
painting = 2 % total cost
10 .miscellaneous = 5 % total cost
PERCENTAGE OF REINFORCEMENT
Lintel and beam = 1 %
to 2.0 %
Column = 1 % to 5.0 %
Foundation =
0.5 % to 0.8 %
Roof slab = 0.7 % to 0.8 %
BINDING WIRE CALCULATION
Binding wire of ting wire is used to tieup
reinforcements. Gauge of binding wire 14 to 16 gauge or 2 mm dia. Binding wire is calculated based on total
volume of reinforcement . usually 1 kg
to 1.3 kg per quintal ( 100 kg ) of
taken as reinforcement.
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