As you know, one of the most important tests to qualify the aggregates used in construction jobs such as Concrete, Pavement (Asphalt), Embankment at Roads and Yards and so on is sieve analysis which determines the particle-size distribution of an aggregate. We have so many tests to qualify the aggregates for instance, Alkali-Aggregate Reaction, Stripping (for chemical properties) and Gradation and size, Toughness and abrasion resistance (Los Angeles Abrasion Test), Durability and soundness, Particle shape and surface texture, Specific gravity, Cleanliness and deleterious materials (for physical properties).
What is different between grading test of aggregate and other tests?
When other tests do not obtain the appropriate results, we have to reject the aggregate. But if we have not a good sorting of sieve analysis which is not compatible with the standards, we will be able to improve the gradation of aggregates by mixing several types of them. As the matter of fact, in the real job, we cannot usually find the aggregate generated in river mining (especially fine aggregate for concrete) in which the particle size distribution of the aggregate (sorting) will be completely compatible with the standards. What can we do?
In this article, I have brought a simple calculation accompanied by an example to solve this problem. Why do I bring you this article? Because I am willing to establish a link between Engineering fields and Financial Management so that in my next article you will see to determine the risk minimizing combination (Risk Diversification), we can use the same method.
Here, I explain this method with an example in the field of Concrete Technology.
Assume we are willing to produce the fresh concrete at the site. One of the most important components of the concrete to gain high workability and economic savings is good sorting of fine aggregate. Two important factors are controlling the fine aggregate of the concrete:
1) Sandy Equivalent (SE) > 75%
2) Fineness Modulus (FM) that should be: 2.4 < FM < 3.2
If Fineness Modulus increases more than 3.2, we have to use more cement to reach our specific compressive strength of concrete accompanied by high workability (75mm < slump < 100mm) in which we will not have any economic savings.
The formula of FM is as follows:
FM = SUM (cumulative percentage on specified sieves) / 100
Where:
F.M =fineness modulus
As we can see, F.M is completely referred to sieve analysis of the fine aggregate.
Where:
F.M =fineness modulus
As we can see, F.M is completely referred to sieve analysis of the fine aggregate.
specified sieves
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=
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0.150 mm (No. 100), 0.30 mm (No. 50), 0.60 mm (No. 30), 1.18 mm (No. 16), 2.36 mm (No. 8), 4.75 mm (No. 4), 9.5 mm (0.375-in.)
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Now, let me bring an example of an available sample of fine aggregate (A) that the sieve analysis is as follows:
Size of sieves 3/8” No.4 No.8 No.16 No.30 No.50 No.100
Passing (%) 100 85 63 47 28 12 7
We can calculate: F.M = 3.58
According to ASTM C 33, fine - aggregate grating limits is below cited:
Size of sieves 3/8” No.4 No.8 No.16 No.30 No.50 No.100
Passing (%) 100 95-100 80-100 50-85 25-60 5-30 0-10
We can see that fine aggregate (A) is not compatible with ASTM C 33 and so F.M is outside of the limit.
Now, we want to mix two other fine aggregates of (B) and (C) with fine aggregate of (A) so that we will have a new fine aggregate which will cover all standard requirements. The sieve analysis of these two fine aggregates is as follows:
Sieve Analysis of Fine- aggregate (B):
Size of sieves 3/8” No.4 No.8 No.16 No.30 No.50 No.100
Passing (%) 100 93 85 70 43 23 12
Sieve Analysis of Fine- aggregate (C):
Size of sieves 3/8” No.4 No.8 No.16 No.30 No.50 No.100
Passing (%) 100 100 97 83 65 43 22
What will be the volume percentages of A, B and C for our mixed aggregate?
Now, we have three variables which are “x” as the volume percentage of A, “y” as the volume percentage of B and “z” as the volume percentage of C.
It is clear, we can easily find these variables by using of the solving aMatrix Inverse as follows:
Assume we have Matrixes of S, X and P as follows:
S = Matrix (m*n) Where: m = n
X = Matrix (m*1)
P = Matrix (m*1)
S.X = P
S’ * S * X = S’ * P
S’ * S = I
I * X = S’ * P
I * X = X
X = S’ * P
You can solve above function by useing of Excel (Please see my spreadsheet of Excel).
There are some conditions to solve the matrix extracted from our example (mixed aggregates) below cited:
1) The number of rows, which are the the size of sieves, should be equal to the number of the variables (x, y, z) or columns in matrix S. In fact,we have: m = n
The number of rows in matrix X and P are equal to matrix S.
The number of columns in matrix X and P are equal to 1.
2) In this example, “x” is related to aggregate type A and respectively “y” for B and “z” for C.
3) Elements of matrixes (aij) are the same the percentages of passing
4)We should have enough date issued by the standards such as ASTM, AASHTO and so on.
5) We should use from the best data issued by the standards as Index Data which are compatible with our requirements to solve this matrix. What is the meaning? What are the Index Data? Here, the Index Data is to choose the size of sieves. One the the most important conditions is to choose the best size of sieves to decrease the variances (Diversification) that it could be also used to reduce the risk of assets in the field of Financial Management. I will depict it in my next article.
For the better perception, let me continue the example as follows:
According to above mentioned, we have three variables and we should choose three size of sieves included in below table:
Passing (%)
Size of Sieves A (x) B (y) C (z) Average of Limits
3/8” 100 100 100 100
No.4 85 93 100 97.5
No.8 63 85 97 90
No.16 47 70 83 67.5
No.30 28 43 65 42.5
No.50 12 23 43 17.5
No.100 7 12 22 5
How can we choose the best Size of Sieves?
Definitely we should diversify and decrease the variances. I suggest you below method as follows:
V1 = [(85- 97.5) +(93 - 97.5) +(100- 97.5)] / 3
V2= ----------------------------------------------
V3 = -----------------------------------------------
V4 = -----------------------------------------------
V5 = -----------------------------------------------
V6 = ----------------------------------------------
Then we should find the least amount of SUM (Vi) which is reaching to zero.
(In formula of SUM (Vi), we should know that n = 3 because we have only three variables and should choose three Size of sieves among six Size of Sieves).
Of course, it will be very hard and we should obtain all permutations and combinations. I will explain this method completely in my next article about Risk Diversification of Assets in the field of Financial Management.
Now, the size of sieves: No.8, No.16 and No.100 have the least amount of
SUM (V) but when we try to find the variables, we find below percentages:
x = 141%
y = -113%
z = 72%
F.M = 3.3
As we can see, the results are not logic. Therefore, we should use from next SUM (V) which give us the size of sieves: No.8, No.16 and No.50
The results are as follows:
x = 96%
y = 4%
z = 0%
F.M = 3.5
It means that we should use 96% of aggregate type A and 4% aggregate type B. We do not need to use from aggregate type C.
In fact, to find the appropriate results, you can utilize from try and error method by using of my Excel spreadsheet.
Note:
“All spreadsheets and calculation notes are available.
The people, who are
interested in having my spreadsheets
of this method as a template for further
practice,
do not hesitate to ask me by sending an email to:
soleimani_gh@hotmail.com or call
me on my cellphone:
+989109250225.
Please be informed these spreadsheets are
not free of charge.”
