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)
Is This Answer Correct ? | 0 Yes | 0 No |
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 |
how to calculate concrete quantity for footing using M25 grade 1:1:2 for 730cft.
Structural steel: For a thumb rule check in field, the allowable weight which can be hung down for 10mm dia mild steel rod, of yield value of 0, 25 KN/mm2 and applying a factor of safety 2.00, will be how many kg?
What is the acceptance criteria to determine the "Core density" of "DBM or BC" course,for any grade as per "MORTH" guidelines in road construction work ?
what is a moment connection and shear connection ? where it provided ? why it provided ?
M.C.Q
3 Answers Mitsui OSK Lines, TechInfini,
What is bmc stands for in bmc software?
2. What are the grades of steel used in the concrete design?
WHAT GOAL DO YOU HAVE IN YOUR CARRIER?
why cylinder for split tensile strength test ?
How much 1 cum of M15, M20 & M25 concrete weigh?
Find out the Cutting Length of Closed Hoop 2 Legged Stirrups for Beam as mentioned. Beam Size :200x300 mm,Nominal cover for all sides :25 mm,8 mm dia. Fe415/500/550.
what formula used to find the length of triangle ring in column?