1 Cum Brick Work CM 1: 6 - How Many Bricks ,sand,Cement
Required? Brick Size 9" 4.5" 3"
Answers were Sorted based on User's Feedback
Answer / vivek yadav
Standard brick size 190*90*90
0.19*0.09*0.09= 0.001539m^3
1÷0.001539= 649.77
= 650 brick's (approx)
Actual brick size 230*115*76
0.23*0.115*0.076= 0.0020102m^3
1÷0.0020102= 497.46
= 498 brick's (approx)
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Answer / md.ali
Brick size: 9"X4.5"X3"--->convert to meters: 0.23X0.115X0.075
For No of Bricks = 0.75/(0.23X0.115X0.075)
= 378.09 No's
Cement =(dry materialXcement proportion)/sum of prop
=(0.35*1)/7
=0.05*1440
=72 kgs
sand =(dry material* sand proportion)/sum of prop
=(0.35*6)/7
=0.3*1850
=555 kgs or 0.555 tons
Is This Answer Correct ? | 0 Yes | 1 No |
Answer / nishanth
no. of bricks without mortor=504
no. of bricks with mortor=392
cement=58.75kg[1.17 bags]
sand=375kg[8.63ft cube]
Is This Answer Correct ? | 0 Yes | 1 No |
Answer / suvra sankha chattopadhyay
For 1 cum of brick work ,C.M 1:6
Brick size= 9" 4.5" 3"
= 0.2286 x 0.1143 x 0.0762 cum.
Volume of Brick= 1.991028276 x 1/1000 cum.
Number of bricks= cum of Brick work/Volume of 1 Brick
= 502.2530378 or 503 number of Brick.
For 1 cum of brick work 25 to 30% C.M required
i.e; 0.3 cum of C.M.
Dry cement concrete required; due to 60% void in sand;
=0.3 + 0.3 X 60%
=0.48 cum C.M.
As the C.M is in the ratio of 1:6;
Volume of Cement required =0.48/(1+6)
=0.06857 cum
Cement bags required = 0.06857/0.0347
=1.976 or 2 Bags of Cement.
Volume of sand required = (0.48 x 6)/(1+7)
= 0.411428 cum.
or 14.5304 cft.
Is This Answer Correct ? | 2 Yes | 4 No |
Answer / saiyyed
Number of Brick = 504
Quantity of cement = 0.43 cum = 13 Bags
Sand = 2.57 Cum
Is This Answer Correct ? | 0 Yes | 2 No |
Answer / mirza
M2.5 (1:6:12)
volume=1+6+12=19
total volume of ingredient for using=1.57
volume of cement=(1/19)*1.57=0.082cum
volume of cement=0.082*1440=118.08kg
1 bag of cement=50
actual bag of cements=118.08/50=2.37 bags of cement
volume of sand=(6/19)*1.57=0.495cum
volume of stone=(12/19)*1.57=0.991cum
no of brickswith mortar 10mm=405
no of bricks without mortar=500
Is This Answer Correct ? | 0 Yes | 2 No |
Answer / ashok sharma
Standard brick size in meter=.190*.090*.090 =.00154
add 10 mm mortar for ech sides=(.2*.1*.1)=(.002)
bricks required for 1 m3 =(1/(0.002)=500 nos
mortar required =(.002-.00154)*500=.23
add 10% for wastage = (.23*0.1+.23)=.253m3
add 25% for Dry Volumn = (.253*0.25+.253)=.316m3
Send is =(6/7)=0.857*0.316=0.271 cum = 9.571 cft
cement is =(1/7)=0.143*0.316=0.045*1440= 65 kg
Is This Answer Correct ? | 0 Yes | 2 No |
Answer / nabakumar
Without Mortar 1 brick volume = 0.23*0.11*0.070 = 0.0018
Brick Required for 1 Cum. = 1/0.0018 = 555 Nos
Cement Mortar Ratio - 1:6
cement required for 1 cum = 6+1 = 7
1/7 = 0.143 cum
0.143*1440 = 205.92 Kgs
Say 206 Kgs = 4.12 Bags
sand Required for 1 Cum = 0.143*6 = 0.858 cum.
Is This Answer Correct ? | 0 Yes | 3 No |
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