How we calculate the cement for the M 15 normal concrete with
aggregate size is 40 mm ?
Answers were Sorted based on User's Feedback
Answer / hari
for M15 proportion is 1:2:4
dry volume of concrete in 1 cum = 1.54
cement qty - 1.54/(1+2+4)*1 = 0.22cum
Fine Agg - 1.54/(1+2+4)*2 = 0.44cum
coarse Agg - 1.54/(1+2+4)*4 = 0.88 cum
check any proportion - (0.22+0.44+0.88) = 1.54cum
Is This Answer Correct ? | 8 Yes | 0 No |
Answer / r. dilli babu
M15 (1:2:4)
=((0.45/sand ratio)*1440)
Cement=324 kg , 6.48bags, 8.1cft
sand=16.2 Cft
Mattel=32.4 Cft
Is This Answer Correct ? | 4 Yes | 0 No |
Answer / mangesh
for M15 proportion is 1:2:4
assume wet volume 1 m3 of work
to produce 1 m3 of wet volume we need 1.54 dry volume
cement qty=1.54/1+2+4=0.22 m3
hence no of bag=0.22*1440/50=6.33 nos
sand qty=0.154*2=0.308 m3
agg qty=0.154*4= 0.616 m3
40 mm is max size of agg & it is the std value
Is This Answer Correct ? | 5 Yes | 6 No |
for M15 proportion is 1:3:6
assume wet volume 1 m3 of work
to produce 1 m3 of wet volume we need 1.54 dry volume
cement qty=1.54/1+3+6=0.154 m3
hence no of bag=0.154*1440/50=4.44 nos
sand qty=0.154*3=0.462 m3
agg qty=0.154*6=0.924 m3
40 mm is max size of agg & it is the std value
Is This Answer Correct ? | 4 Yes | 10 No |
Answer / raju
for M15 proportion is 1:3:6
assume wet volume 1 m3 of work
to produce 1 m3 of wet volume we need 1.54 dry volume
cement qty=1.54/1+2+4=0.22 m3
hence no of bag=0.22*1440/50=6.33 nos
sand qty=0.154*2=0.308 m3
agg qty=0.154*4= 0.616 m3
Is This Answer Correct ? | 1 Yes | 7 No |
How much quantity of rod binding work by a rod binder per day
what will be the quantity of following materials in P.C.C for the total area of (100'x70'x 0'4") 1)NO. OF CEMENT BAGS 2)SAND IN CU.FT 3)METTLE IN CU.FT
0 Answers Pratibha Industries,
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