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Question
what is the ratio of m50 grade in concrete block
 Question Submitted By :: Civil-Engineering
I also faced this Question!!     Answer Posted By  
 
Answer
# 1
there is no ratio for m50 grade concrete. it is fully based
on design mix. following steps should be followed while you
consider design mix.

Grade Designation = M-50
Type of cement = O.P.C-43 grade
Brand of cement = Vikram ( Grasim )
Admixture = Sika [Sikament 170 ( H ) ]
Fine Aggregate = Zone-II

Sp. Gravity
Cement = 3.15
Fine Aggregate = 2.61
Coarse Aggregate (20mm) = 2.65
Coarse Aggregate (10mm) = 2.66

Minimum Cement (As per contract) =400 kg / m3
Maximum water cement ratio (As per contract) = 0.45

Mix Calculation: -

1. Target Mean Strength = 50 + ( 5 X 1.65 ) = 58.25 Mpa

2. Selection of water cement ratio:-
Assume water cement ratio = 0.35

3. Calculation of water: -
Approximate water content for 20mm max. Size of aggregate
= 180 kg /m3 (As per Table No. 5 , IS : 10262 ). As
plasticizer is proposed we can reduce water content by 20%.

Now water content = 180 X 0.8 = 144 kg /m3

4. Calculation of cement content:-
Water cement ratio = 0.35
Water content per cum of concrete = 144 kg
Cement content = 144/0.35 = 411.4 kg / m3
Say cement content = 412 kg / m3 (As per contract Minimum
cement content 400 kg / m3 )
Hence O.K.

5. Calculation for C.A. & F.A.: [ Formula's can be seen in
earlier posts]-

Volume of concrete = 1 m3
Volume of cement = 412 / ( 3.15 X 1000 ) = 0.1308 m3
Volume of water = 144 / ( 1 X 1000 ) = 0.1440 m3
Volume of Admixture = 4.994 / (1.145 X 1000 ) = 0.0043 m3
Total weight of other materials except coarse aggregate =
0.1308 + 0.1440 +0.0043 = 0.2791 m3

Volume of coarse and fine aggregate = 1 0.2791 = 0.7209 m3
Volume of F.A. = 0.7209 X 0.33 = 0.2379 m3 (Assuming 33%
by volume of total aggregate )

Volume of C.A. = 0.7209 0.2379 = 0.4830 m3

Therefore weight of F.A. = 0.2379 X 2.61 X 1000 = 620.919
kg/ m3

Say weight of F.A. = 621 kg/ m3

Therefore weight of C.A. = 0.4830 X 2.655 X 1000 = 1282.365
kg/ m3

Say weight of C.A. = 1284 kg/ m3

Considering 20 mm: 10mm = 0.55: 0.45
20mm = 706 kg .
10mm = 578 kg .
Hence Mix details per m3
Increasing cement, water, admixture by 2.5% for this trial

Cement = 412 X 1.025 = 422 kg
Water = 144 X 1.025 = 147.6 kg
Fine aggregate = 621 kg
Coarse aggregate 20 mm = 706 kg
Coarse aggregate 10 mm = 578 kg
Admixture = 1.2 % by weight of cement = 5.064 kg.

Water: cement: F.A.: C.A. = 0.35: 1: 1.472: 3.043

Observation: -
A. Mix was cohesive and homogeneous.
B. Slump = 120 mm
C. No. of cube casted = 9 Nos.
7 days average compressive strength = 52.07 MPa.
28 days average compressive strength = 62.52 MPa which is
greater than 58.25MPa
Hence the mix accepted.
 
Is This Answer Correct ?    3 Yes 0 No
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