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Question
how much quantity of cement and  sand required for
construction of 10 cum brick work( Brick size
3"*4"*9"(Standard size ) an d for 10 cum construction of SSM
(Size stone masonary) with Cm 1:6 for both.
 Question Submitted By :: Civil-Engineering
I also faced this Question!!     Answer Posted By  
 
Answer
# 1
Let the size of brick be 230mm x 115mm x 75mm
Vol. of brick( w/o mortar) = 0.00198 cum.
Thk. of mortar be 12mm.
vol. of brick with mortar = 242mm x 127mm x 87mm
= 0.00267 cum.
No. of bricks in 1 cum. with mortar = 1/0.00267
= 373.99
Say = 374 bricks.
Cement Mortar required = 1-( 374 x 0.00198)cum.
= 0.259 cum.
Say 0.26 cum

Add 15% for frog filling and wastage while execution.
Vol. of wet mortar = 0.299 cum.
Vol. of dry mortar( increase by 30% ) = 0.388 cum.

For CM of 1:6

Qty. of cement = 1/7 x (1440 x 0.388)
= 0.055 cum.
Qty. of sand = 0.055 x 6
= 0.330 cum.
 
Is This Answer Correct ?    97 Yes 31 No
B.s.dev
 
Answer
# 2
Let the size of brick be 230mm x 115mm x 75mm
Vol. of brick( w/o mortar) = 0.00198 cum.
Thk. of mortar be 12mm.
vol. of brick with mortar = 242mm x 127mm x 87mm
= 0.00267 cum.
No. of bricks in 1 cum. with mortar = 1/0.00267
= 373.99
Say = 374 bricks.
volume of 374 bricks= 374 x 0.00198 = 0.74052cum
Cement Mortar required for 1cum = 1 - 0.74052cum.
= 0.259 cum.
Say 0.26 cum

Add 15% for frog filling and wastage while execution.
.26 x 15/100 = 0.299 cum
Vol. of wet mortar = 0.299 cum.
Vol. of dry mortar( increase by 30% ) 0.299 x 30/100 =0.0897
0.299 + 0.0897 = 0.388 cum.

For CM of 1:6 (1 + 6 = 7)

Qty. of cement for 1cum = 1/7 x 0.388
= 0.055 cum.
Qty. of sand for 1cum = 0.055 x 6
= 0.330 cum.

Qty. of cement for 10cum =10 x 0.055
=0.55 cum.
Qty. of sand for 10cum = 10 x 0.330
= 3.30 cum.
 
Is This Answer Correct ?    42 Yes 15 No
Ar. G.s. Rehal
 
 
 
Answer
# 3
The size of brick be 230mm x 115mm x 75mm
Vol.of brick( w/o mortar) = 0.00198 cum.
Thk.of mortar be 12mm.
vol.of brick with mortar = 242mm x 127mm x 87mm = 0.00267 cum.
No.of bricks in 1 cum. with mortar = 1/0.00267 = 373.99
Say = 374 bricks.
Cement Mortar required = 1-(374 x 0.00198) cum= 0.259 cum.
Say 0.26 cum
Add 15% for frog filling and wastage while execution.
Vol. of wet mortar = 0.299 cum.
Vol. of dry mortar (increase by 30%) = 0.388 cum.
For CM of 1:6
Qty. of cement = 1/7 x (1440 x 0.388) = 0.055 cum.
Qty. of sand = 0.055 x 6 = 0.330 cum.
 
Is This Answer Correct ?    18 Yes 3 No
Uday Kiran
 
Answer
# 4
see CPWD ANALYSIS
480 BRICKS PER CUM I/C WASTAGE
1.24 BAGS CEMENT/CUM I/C WASTAGE
1.24*.035*6 = WILL BE THE QTY OF C.SAND REQD/ PER CUM
 
Is This Answer Correct ?    16 Yes 6 No
Sudhir
 
Answer
# 5
mortar required for 10 cu.m brick work

Dry mortar = 30% of brick work = 3 cu.m

one part of cement=3cu.m/sum of proportion

(3/1+6)=0.428cu.m

cement required in bags...0.428/vol. of one bag of cement

(0.428/0.035) =12.23 bags ..take 13 bags of cement

sand = 0.428*6= 2.57 cu.m
 
Is This Answer Correct ?    14 Yes 5 No
Jocky
 
Answer
# 6
The size of brick be 230mm x 115mm x 75mm
Vol.of brick( w/o mortar) = 0.00198 cum.
Thk.of mortar be 12mm.
vol.of brick with mortar = 242mm x 127mm x 87mm = 0.00267
cum.
No.of bricks in 1 cum. with mortar = 1/0.00267 = 373.99
Say = 374 bricks.
Cement Mortar required = 1-(374 x 0.00198) cum= 0.259 cum.
Say 0.26 cum
Add 15% for frog filling and wastage while execution.
Vol. of wet mortar = 0.299 cum.
Vol. of dry mortar (increase by 30%) = 0.388 cum.
For CM of 1:6
Qty. of cement = 1/7 x (1440 x 0.388) = 0.055 cum.
Qty. of sand = 0.055 x 6 = 0.330 cum
 
Is This Answer Correct ?    2 Yes 0 No
Arvind Vyas
 
Answer
# 7
Consumption & Rate analysis Brick Masnory 230mm 1:6
Superstructure With scaffolding

Providing and placing brick work 230mm 1:6 Superstructure
With scaffolding
1.3Ref: use 1..3 standard
Proportion of materials
Cement 1
Sand 6
Total 7
Std constant(1.5) / Total of proporttion(4)
0.186
1) Required volume of cement for 1 cum.

1 Proportion of cement 0.19
Add wastage(2.50%) 0.005
Vol of cement for 1 cum 0.190
0.035 is cum of per 50kg of bags 5.44.

As per standard we assumed that in 1cum of brick work ,the
quantity of mortor (Cement & Sand )is consider in
quantity. (24%) 1.28
Cost of cement per bags 325
Total cost of 1.34 bags 415.39

2) Required volume of Sand for 1 cum

2 Proportionof Sand 1.114
Add wastage(2.50%) 0.028
Vol of Sand for 1 cum 1.142
As per standard we assumed that in 1cum of brick work ,
the quantity of mortor (Cement & Sand )is consider in
quantity.(24%) 0.268
Cost of sand per brass 875
Total cost of 0.284 cum sand 234.85


3)Required brick for 1 cum of size 230x115x75mm

Actual Vol of one bricks is 0.00198cum,&
10 gap Allaround is become 0.00255
Nos of bricks in 1cum0.00255 1 392.16
Add wastage(10.00%) 39.22
Total Required brick for
1 cum 431.37
Cost of bricks 5.00
Total cost of brick 2156.86

A)Total material cost (1+2+3)
2807.10

4)Cost for Manpower ,Mixing ,Shifting ,
placing etc, Rs 15 / sft-Mumbai, 529.80

5) Cost for screeing,
cleaning ,of sand (10%) 333.69
6) Cost for Tools ,
Safety, curing etc(5%) 183.53
B) Cost of labour on
total material cost (4+5+6)
1047.02

C) Cost of Material & labour (A+B) 3854.12

5)Add over head(5%) 192.71
6)Add profit on(15%) 607.02


D- Total of (C+5+6),Over head & profit cost
P/A P230mm,1:6 Per Sqm. 4653.85
If the brick work is required to .


convert into Sqmt
1) for 230mm ; 0.23 1070.39
2) for 115mm ; 0.115 535.19
 
Is This Answer Correct ?    2 Yes 0 No
Bhupendra Potpose
 
Answer
# 8
lets the take size of brick is .230*.115*.075. now calculate
the quantity for 10 cum.

Area of one brick without mortar -0.00198375 Cum
No of brick in 10 cum= 10/0.00198375 (Area of bricks).
=5041 nos of bricks.

now calc. the bricks quantity in 10 cum with mortar.

lets take the mortar used 10mm in brickwork then
size of brick will be taken with mortar is =.240*.125*.085=
0.00255 (Area of one brick with mortar).

now in 10 cum brickwork no of bricks calculated with mortar
is=10/0.00255 =3922 nos

calculate the mortar in cum= bricks without mortar- bricks
with mortar= 5041-3922=1119 nos
Now area of brick without mortar is .230*.115*.075*1119=2.22cum

now 15% for frog filling and wastage taken in acc.

=2.22+2.22*15/100=2.55 cum

Dry volume is the 1/3rd of wet volume of mortar=
2.55+2.55/1/3=3.40 cum.

now the ratio is 1:6=7

3.40*14%=.486 cum cement
3.40*86%= 2.92 Cum Sand

=.486/.0347= 14.01 bags/10 Cum
= 14.01*6=84.06 Bags/10 Cum
 
Is This Answer Correct ?    0 Yes 0 No
Karan Bagga
 
Answer
# 9
No of bricks = 500 [5(Stredger face with 10mm CM)*10(Header face with 10mm CM)*10(Header face with 10mm CM)]
Volume of mortar required = 1-500*0.19*0.09*0.09
= 0.2305 cubic metre
Cement required for 1:6 = (0.2305*1440)/7
= 47.41 Kg (Approx. 48 Kg)
(Density of cement 1440 Kg/cu.m)
Sand required = 284.5 Kg (Approx. 288 Kg)
 
Is This Answer Correct ?    5 Yes 10 No
Muthu Kumaran
 
Answer
# 10
you are given 3x4x9 standard size of bricks
After placing mortar it became- 4x5x10 =
10.16cmx12.7cmx25.4cm
No of bricks are required=
10/.106x.127x.254=3069 nos
The size of brick without mortar= 3x4x9=
7.62cmx10.1cmx22.8cm
Then net mortar required =10-(3069*.076*.101*.228)
= 4.62cum
Due to frog filling, brick bonding course and
wastage add 15%
Volume of wet mortar= 4.62+4.62x15%=5.30cum
When dry increase this qty. by 1/3rd=
5.30+5.30x1/3=7.06cum
Ratio is =1+6=7 cement = 7.06/7= 1.06 = 29 bags
Sand= 7.06*6= 42.36cum
 
Is This Answer Correct ?    40 Yes 49 No
Sjtbehera
 

 
 
 
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