Answer / piyush singh

DIRECT ESTIMATION METHOD : =

M20 (1:1.5:3) for 10 m^3

1. cement content = 15.2 \ (1+1.5+3)
= 2.763 cum
= 82 bags of 50 kg

2. water = 2.763 cum * 1.5
3. ca + fa = 2.763 * 3 cum

* here 15.2 is a standard value worked out from experiments.
so for 10 m^3 u can divide the sum of ratio by 15.2 to get
amount of cement directly...

* 50 kg of cement (1 bag) = (1/30) m^3 so = 0.034 cum
2.763cum / 0.034 cum = 82 bags

similarly u can work it out for

M25 1:1:2
m15 1:2:4
m10 1:3:6

Answer / piyush singh

according to the IS method of design mix
1. select the grade of concrete
2. select the target strength using the equation
f(target) = Fck (selected grade) + 1.65 * sd
3. a) using the table given in IS select the maximum cement
. content
b) selecct the nominal size of the aggregate
c) water cement ration
4. then procced as code says.. and u will get actual amount
of cement , fa ,ca ,water needed for 1 M^3

Answer / hitenpatel

a) For 1 m3 of concrete Mix 1:2:4 (M15)
Add 50% for wet concrete = 1.50 m3
= 1.50/(1+2+4) = 0.214 m3
For 1 m3 = 30 bags of cement required (1440/50) say 30 bags
Cement = 0.214 x 30 = 6.42 bags
Sand = 0.214 x 2 = 0.428 m3
Aggregate = 0.214 x 4 = 0.856 m3

Answer / sridhar reddy

what piyash said is write but in place of water take fine aggreate

Answer / ansari asif ahemad

What Piyush said is right but in place of water take Sand. I think its typing mistake.

Answer / subbarao

by using constants of materials we have to find for 1cu.m and
then for 10cu.m

Answer / sanket bagade

a) For 1 m3 of concrete Mix 1:2:4 (M15)
Add 50% for wet concrete = 1.50 m3
= 1.50/(1+2+4) = 0.214 m3
For 1 m3 = 30 bags of cement required (1440/50) say 30 bags
Cement = 0.214 x 30 = 6.42 bags
Sand = 0.214 x 2 = 0.428 m3
Aggregate = 0.214 x 4 = 0.856 m3

Hai, I have applied for APPSC AEE (I&CAD) post.I need previous AEE papers of civil engineering.I hope u will kindly consider this and send.

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