calculation of cement bags in 1:2:4 concrete mix
Answers were Sorted based on User's Feedback
Answer / akashkhutia
For 1 m3 of concrete (1:2:4)
Required volume of concrete = 1 m3
Estimated volume of dry material = 1 x 1.65 = 1.65 m3
Mix totals = 1+2+4 = 7 (1:2:4 cement:sand:gravel)
Ingredient Volumes: 1.65 x 1/7 = 0.235 m3 cement
# Bags of cement: 0.235 m3 cement / .0332 m3 per 50 Kg bag
= 7 bags of cement
| Is This Answer Correct ? | 216 Yes | 82 No |
Answer / uma
For 1 m3 of concrete (1:2:4)
Metal -0.92cum
Sand -0.46cum
Cement-0.23cum
Weight of cement per cum-1440kgs
For 1:2:4 mix- 0.23x1440=331.2kgs
IN BAGS-331.2/50 =6.6Bags
| Is This Answer Correct ? | 86 Yes | 23 No |
Answer / balvinder bhagat
For 1 m3 of concrete (1:2:4)
Mix totals=7 (1:2:4 cement:sand:gravel)
Cement volume: 1* 1/7=0.143 m3 cement
#Bags of cement: 0.143 m3 cement * 1(As 1m3=6.4bag in 1:2:4)
=6.4 bags
| Is This Answer Correct ? | 89 Yes | 50 No |
Answer / anil raut
For 1 m3 of concrete (1:2:4)
for Required volume of concrete 1.00
Add extra for additional voids in concrete ( 25% to 35%)
0.35
Total qty= 1.35 Cum
Mix total (1+2+4) 7
Therefore Quantity of cement = (1.35/7) 0.19
Cum
Qty of Cement in bags = ( 0.19/0.03) 6.4 Bags say(7
bag)
Qty of sand in bags = ( 0.19*2) 0.39 Cum
Qty of Aggregate = ( 0.19*4) 0.77 Cum
| Is This Answer Correct ? | 39 Yes | 10 No |
Answer / vishnutom
For 1 m3 of concrete (1:2:4)
for Required volume of concrete 1.00
Add extra for additional voids in concrete ( 25% to 35%)=Say
35%=35/100=0.35.
Total qty=1+0.35=1.35 Cum
Mix total (1+2+4)=7
Therefore Quantity of cement = (1.35/7)=0.19 Cum
Qty of Cement in Kgs = (0.19*1440)=273.6Kgs
(1cum=1440Kgs)
Qty of Cement in bags= 273.6/50=5.4 bags=Say 6 bags
(50kgs=1 bag)
Qty of sand in bags = ( 0.19*2)=0.39Cum
Qty of Aggregate = ( 0.19*4)=0.77Cum
| Is This Answer Correct ? | 25 Yes | 11 No |
Answer / a
For finding cement
1+2+4=7
1/7x1.52
=.217x35.28
=7.65/1.25
=6.12bag/per 1cum.
| Is This Answer Correct ? | 46 Yes | 36 No |
Answer / suresh
for i m3 of concrete required
0.9 m3 of metal
0.45 m3 of mix
cement
0.45x1440/2x50 = 6.48kg
| Is This Answer Correct ? | 37 Yes | 28 No |
Answer / daniel dandan
for 1 cu.m of concrete, (1:2:4)
1=cement; 2=sand; 4=gravel
add 25% for loose volume(to cover voids),
total concrete volume = 1.25 cu.m.
total parts = 1+2+4 = 7 parts
cement = (1/7)*1.25
= 0.1786 cu.m
density of cement = 1440kg/cu.m
thus, 0.1786 * 1440 = 257.184kg.
for 40kg per bag of cement:
bag of cement = 257.14kg/(40kg per bag)
= 6.4 bags....answer
| Is This Answer Correct ? | 18 Yes | 9 No |
Answer / goran
mix Proportion =(1:2:4)
then (1C+2S+4G)=7C
for one cu.meter of concrete
concrete Gamma =2240 m3/kg
then
7C=2240=
C = 320 kg
for 50 kg bag
C = 6.4 bags per one cu.meter
| Is This Answer Correct ? | 11 Yes | 2 No |
Answer / s.mohamed imthiyaz
1+2+4=7........ then total ingredient=1.57
1part of cement..... So 1 / 7 =0.142
Then 0.142*1.57=0.2229
Unit weight of concrete is 1440kg/m3
0.2229*1440=320.97kg
1cement bag=50kg
320.97/50= 6.41 bags................
| Is This Answer Correct ? | 9 Yes | 3 No |
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