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Answer Posted / sjtbehera(hari machines)
i write the way of desigin mix. you can easily design the
mix of every grade.
Aggregates of different sizes, grading, surface texture
shape and other characteristics and the quality & qty of
cement & water used will produce concretes of different
strengths. A judicious choice of these elements in a
particular proportion is necessary to ensure the
characteristic strength and durability of concrete.
Parameters involved:
Standard deviation of compressive strength of concrete.
Cement content
water cement ratio --workability
proportion of fine and coarse aggregates
Grading of aggregates
Size, surface texture and shape of aggregates
vibration
The standard Deviation
According to IS 456, the characteristic strength is defined
as that value below which not more than 5% (1 in 20)
results are expected to fall.
At site depending on the degree of control, the strength of
each cube varies. Some far below and some far above and
many scattered around the average strength. The degree of
scatter is measured by standard deviation. The greater the
value of standard deviation,
the greater is the scatter and less the control at
site.
It is usually decided at site based on the test strength of
samples. (IS 456)When sufficient test results for a
particular grade of concrete is not available, the value
can be assumed.
Grade of concrete Assumed standard
deviation
M10, M15 3.50
M20, M25 4.00
M30, M35, M40 M45, M50 5.00
These values apply when the concrete is produced in fully
controlled condition-with weigh batching, control on w/c
ratio & workability, proper Aggregate grading &mixing.
Where there is a deviation from eh above, increase the
values by 1
1.Target Strength of Concrete (at 28 days) as per IS
456,
If, Ft is the target strength, Fck the characteristic
strengths the standard deviation,
Ft = Fck+1.65X S
S can be taken from the table above.
For M20, Ft = 28.25 N/mm2 (Like this you may find out for
any grade)
2. Selection W/C ratio
a. From table 5 of IS 456, the maximum w/c ratio can be
ascertained
b. From fig.2 of IS 10262 -1982, find the limiting w/c
ratio for a compaction factor. Chose the lower of the 2
For M20, a gives a value of .55 and b gives 50. Hence fix
it as >50
3. Cement content
From Is 456, Table 5 the cement required is 300Kg.
(You may have to increase this weight if the required
strength is not obtained)
It can become 310 or 320 but not more generally)
4. Water in Kg = Cement in Kg X W/C ratio
Hence Qty.of water = 300x.5= 150 liters
5. Determination of fine and coarse aggregates:
The coarse aggregates should confirm to IS 380-1970
and fine aggregate either Zone two or three
The % of sand (from IS 10262) for 20mm aggregate is 0.35
The specific gravity of coarse aggregate is 3.15
The specific gravity of fine aggregate is 2.6
From table 3of IS 10262, for 20 mm aggregate, the amount
of entrapped air in wet concrete is 2%. Taking this into
account, apply the formula,
V= X 1/1000
Where, V=Volume of concrete
W=weight of water
C=weight of cement
Sc=specific gravity of cement
P=% of fine aggregate
Fa=weight of fine aggregate
Sfa=specific gravity of fine aggregate
Thus, 0.98= <150 + (300/3.15) + 1/.35XFa/2.6>X1/1000
From this, find out the value of Fa (around 650Kg)
Now taking up equation 2 from IS 10262,
V= X1/1000
(Where Ca is the weight of coarse aggregate & Sca its
specific gravity.)
From this equation, Sca works out to around 1200 kg.
It is important to decide the specific gravity of:
Sand
Coarse aggregate
and the quality of water going to be used for construction.
Also, it very important to fix the grading of the
aggregates to avoid voids. The lower the size of the coarse
aggregate, the higher will be air entrapping nature. Hence
a judicious mix is to be decided.
By adding a suitable admixture, the w/c ratio can be
reduced and correspondingly the cement content reduced.
| Is This Answer Correct ? | 21 Yes | 8 No |
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