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TO FIND THE GRADATION OF COARSE AGGREGATE
BY SIEVE ANALYSIS (ASTM C136-05).
THEORY & IMPORTANCE:
This experiment is conducted to find and check coarse aggregate gradation i.e. crush. The main bulky component of the concrete is crush; it is used together with its strength-giving properties to increase the volume of the concrete. it calculates its fineness module to find the water ratio to be used, along with the value of the specific crush gravity and its maximum grain size.
The maximum size of the coarse aggregate will be calculated from the modulus fineness table in this experiment. For example, if its maximum size is 3⁄4 inches, it means that all grain size is less than 3⁄4 inches. This can also be expressed by writing it as a crush: 3⁄4 inch down.
APPARATUS DETAILS:
For this gradation, the sieve used is 1 1⁄2 in, 1 in, 3⁄4 in, 1⁄2 in, 3/8 in, and #4 sieve. The distinction is that #4 sieve means that in one linear inch of the sieve there are four holes, while 1 1⁄2 inch sieve means the one side dimension of the sieve hole is 1 1⁄2 inches (not that in one linear inch of the sieve there are one and a half inches). The sieves are stacked with their lengths in ascending order. The crushed sample here is 3000 g.
APPARATUS:
- Sieve Apparatus or sieve set.
- Electronic Balance.
- Brittle brush.
- Empty plate.
 |
| Electronic Scale |
 |
| Sieve Set & shaker |
MATERIAL:
Sample of Coarse aggregate
PROCEDURE:
Ø Took the digital balance and set
the scale reading to zero
Ø Took and measured the weight of
pan.
Ø Put some crush in the pan and
measure 3000 grams of crush with the help of electronic balance.
Ø Put the crush in a sieve of 1. 5
inches and began shaking until no more seeds could escape it readily. Using the
electronic balance, measured the weight retained on the sieve and noted it in
the table.
Ø Some stone had fallen out of the
reduced sieve placed it in it and begins to shake the sieve for sufficient
time. Noted the sieves read of the retained weight.
Ø Similarly, placed the entire
remaining crush in the below sieve and began shaking for enough moment and
noticed the weight retained in all the sieves.
Ø Calculated the percentage of each
sieve's weight retained.
Ø The percentage of the weight that
had passed through each sieve was found in the next phase. The complete quantity entered will be 3000
grams for 1.5 inch sieve.
Ø But it won't be the same for the
1 inch sieve because the upper sieve had retained some weight, so the proportion of the passing was calculated relative to the quantity that enters
that sieve.
Ø For a sieve of 3⁄4 inches, the
total quantity entering the sieve will be less the value for the upper sieve,
similarly, for other sieves, the same method should be used to calculate the
percentage of the passed weight
Ø The cumulative proportion was
calculated in the next column. This was
the weight proportion that would be retained if the crush were placed directly
on the sieve. It will be the same as it
was for 1. 5 inch sieve, but for 1-inch sieve, it would be the sum of the
proportion retained by 1. 5 inch sieve plus the one retained. Similarly, the cumulative proportion was
calculated for other sieves, this was denoted by a1, a2, a3, etc
Ø Then
at the end to find the crush's fineness module add each sieve's cumulative
proportion and divide it by 100.
Observations and Calculations:
Total weight of
natural coarse aggregate = 3kg.
Sieve
No
|
Weight
Retained
on
Sieve
(kg)
|
Percentage
of
weight
Retained
(%)
|
Percentage
Of
weight
Passed
(%)
|
Cumulative
Percentage
of
Retained
(%)
|
1
in
|
0
|
0
|
100
|
0
|
3/4
in
|
0.3594
|
11.98
|
88.02
|
11.98
|
1/2
in
|
1.1526
|
38.42
|
49.6
|
50.40
|
3/8
in
|
0.9732
|
32.44
|
12.16
|
82.84
|
#4
|
0.489
|
16.30
|
0.86
|
99.14
|
Finess modulus formula= a1+a2+a3+a4+a5 / 100
(Note:- a is the Cumulative Percentage of Retained (%) )
Finess
modulus= 244.36/100=2.44
nice sir g
ReplyDeletevery informative sir
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