how to calculate concrete qty for footing using M25 grade
1:1:2 for 731 cft.
Answers were Sorted based on User's Feedback
Answer / dharmesh patel
we can only calculate the quantity in m3 only after that we
can converet in c.ft.
1/4 x 1.54 x 28.8= 11.08 bags cement for 1 m3 for the ratio
of 1:1:2
s0 731 cft = 20.719 m3 * 11.08 = 229.570 bags cement used
1/4 x 1.54 = .385 m3 sand for 1 m3
so 20.719 m3 x .385 = 7.976 m3 sand required in 731 sq .ft
2/4 x 1.54 = 0.77 m3 chips ( aggregate ) for 1 m3
so 20.719 m3 x 0.77 = 15.953 m3 chips required in 731 sq .ft
Is This Answer Correct ? | 41 Yes | 18 No |
Answer / vijayababu`
M25 – 1:1:2
Total parts – 4
Cement Required for 1 cum =1/4X1.57X1440
= 0.393X1440 = 567kg/cum
= 11.34 bags/cum
Is This Answer Correct ? | 17 Yes | 3 No |
Answer / sumanth
we can only calculate the quantity in m3 only after that we
can converet in c.ft.
1/4 x 1.54 x 28.8= 11.08 bags cement for 1 m3 for the ratio
of 1:1:2
s0 731 cft = 20.719 m3 * 11.08 = 229.570 bags cement used
1/4 x 1.54 = .385 m3 sand for 1 m3
so 20.719 m3 x .385 = 7.976 m3 sand required in 731 cu .ft
2/4 x 1.54 = 0.77 m3 chips ( aggregate ) for 1 m3
so 20.719 m3 x 0.77 = 15.953 m3 chips required in 731 cu .ft
Is This Answer Correct ? | 15 Yes | 10 No |
Answer / chetan patel
for 1 cmt cont.of 1:1:2
cement:1.52/4=0.38/.035=10.8
Sand :0.38 cmt
Agg :0.38*2=0.76 cmt
So for 731/35.31=20.70 cmt
C :20.70*10.8=224 bag
S :20.70*0.38=7.86 cmt
A :20.70*0.76=15.73 cmt
Is This Answer Correct ? | 8 Yes | 4 No |
Answer / gurubardhan kumar
We have total 735cu.Ft
Ratio of m25= 1:1:2
Now calculate
Cement
1/4x1.54=.385
0.385x1440=554.4kg
554/50=11.08
731/35.314=20.7
20.7x11.08=230 bag
Sand
1/4x1.54=.385
0.385x20.7=7.969cuft
Aggregate
2/4x1.54=0.77
0.77x20.7=15.99cuft
#Ingredient=1.54
Is This Answer Correct ? | 1 Yes | 2 No |
Answer / er. tiwari karan
M25:: 1:1:2 Total parts is 4 Cement part is 1/4
Dry Volume of M25 is approx 1.57 times of total volume
Density of Cement is 1440
Cement Required for 1 cum =1/4X1.57X1440=565.2 approx cu. mtr.
for Bags (india cement bags) is of 50kg so 565.2 kg /50kg = 11.304 bags per cu. mtr.
So as with sand
dry volume if always 1.57 for required volume
and aggregate sole qty of aggregate if crushed to get 1 cu. mt we required 1.57 cu.mt
Is This Answer Correct ? | 0 Yes | 1 No |
Answer / s jaikumar
For 1m3 of 1:1:2
Cement 1.5/4=0.375/0.036=10.41
Sand 0.37
Agg 2 *0.37 =0.74
So for 731cft /35.28=20.71
Cement is used 10.41*20.71=215.59 bags
Sand is used 0.37*20.71=7.66m3
Aggregates is used 0.74*20.71=15.32m3
Is This Answer Correct ? | 1 Yes | 4 No |
Answer / pushparaj
How much cement bags required (m-25 grade) for 1000cft
And also how to calculate
Thank you
Is This Answer Correct ? | 0 Yes | 3 No |
Answer / pervaiz
we have total 731 cu.ft
Now calculating
Cement
1/4x1.54x731 = 281.43/1.25 = 225.14 bags
Sand
1/4x1.54x731 = 281.43 cuft
Crush
2/4x1.54x731=562.87 cu.ft
Now verifying
Is This Answer Correct ? | 5 Yes | 10 No |
what is Brick crushing strength(PSI)?
how we can design the design mix(m25)
What type of tensile strength is handling by crank in beam? Thanks.
How to identify the vitrified tile and ceramic tile by observation only after executing the job.
which "s the best wood for commercial purpose?
The corrosion resistance of a material is.
What is the Allowable silt content of sand for construction purposes?
what is project planning
# 1. size of the Bricks # 2. Minimum Crushing Strength of Bricks
mix design concrete m30 for elevated water tank
How do you compute the volume of airflow?
01. a. Diagram of a slump cone…………… 1 b. Size of Slump Cone…………… 1 c. Volume of a slump cone by av. Dia method…………… 1 d. Volume of a slump cone by av. area method…………… 1 e. Volume of a slump cone by frustrum method …………… 1 f. Angles of slump cone …………… 1 g. Slump measured by slump cone due to…………… 1 h. Slump does not measured by cylinder / cube due to …………… 1 i) ii) i. Slump cone is to be filled in……………………..Layers.