Answer / basharat ali

1.54/sum of ratio
1.54/7=.22cum cement
.22/.0347=6.340bags cement (Say 7)
.22×2=.44cum :-15.5368cft :-19.047bags
.22×4=.88cum:- 31.07cft:-38.09bags

Answer / engr. daniel otoyo

Normal concrete such as 1:2:4 mix has a density of 2400kg/cm
1cm of concrete will weigh 2400x1kg = 2400kg
Sum of mix ratio = 1+2+4 = 7
Ratio (fraction) of cement weight = 1/7
Weight of cement (in 1cm concrete) is therefore = 1/7 x 2400 = 342.857kg
Cement in 50kg bags will give us 342.857/50 = 6.86kg
This approximates to 7bag/cm of 1:2:4 concrete if the desired strength is to be achieved.
Other constituent like fine and coarse aggregates can be solved likewise and their densities used to translate their weight into volumes.

Answer / ruatsanga

For 1 m3 of Concrete (1:2:4)
Required volume of concrete = 1 m3
Estimated volume of dry material = 1 x 1.54 (1.54 = constant)
= 1.54 m3
Mix total = 1 + 2 + 4 = 7 ( 1 : 2 : 4 )
Ingredients Volume = 1.54 x 1/7
= 0.22 m3 of cement
#Bags of cement = 0.22 / 0.035 (per 50 kg Bag)
= 6.289 bags of cement

Answer / gopalakrishna

cement: 1*1.33/7=0.19*1440=273.6/50=5.47bags say 6
sand :2*1.33/7=0.38 cum ,jelly 20mm =4*1.33/7=0.76cum

Answer / ngozi favour

no volume is given. Then it
just 1bag of cement to 100kg
of fine aggregate and 200kg
of coarse aggregate

Answer / syed mohd akram

1:2:4 MIX m 6.4bags lagte ha

castigliano's theorems are based on

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