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BLENDING OF AGGREGATES

BLENDING OF AGGREGATES. BLENDING OF AGGREGATES. BLENDING OF AGGREGATES. BLENDING OF AGGREGATES.

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BLENDING OF AGGREGATES

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  1. BLENDING OF AGGREGATES BLENDING OF AGGREGATES BLENDING OF AGGREGATES BLENDING OF AGGREGATES Warning: If your mind hasn’t yetblended with “Gradation of Aggregates” (posted on the instructor’s website) then you’ll probably struggle with this presentation. You should also have “Assignment 1 Example” at the ready.

  2. Not all the asphalt cement ends up coating the aggregate particles… What is asphalt concrete? …coated with asphalt cement… …some is absorbed into the water-permeable voids within the aggregate particles. …and then compacted. There are 3 basic elements to the compacted mix… Basically, its just aggregate… 1. Aggregate Particles 2. Asphalt Cement 3. Air Voids

  3. Voids in the Mineral Aggregate VMA There’s stone, and sand particles all coated with asphalt cement and there’s a few air voids (the white spaces). Well, VMA is one of the most important properties of an asphalt paving mix. If we removed all the asphalt cement but the aggregate stayed put it would look like this: Firstly, here’s a representation of a compacted paving mix. … “Volatile Mercenaries of Alberta”? In mathematical terms… This diagram shows the mix without the aggregate…there’s just the asphalt cement and air voids. The space occupied by the asphalt cement and air voids (not including the asphalt cement absorbed) is voids in the mineral aggregate. … so what’s it mean? … “Voluptuous Mama’s of America”? Any help? Of what…the “Virile Men’s Association”? Are you a member? Vvma = Va + Vbe Vvma is the Volume of Voids in the Mineral Aggregate Vmb is the Bulk Volume of the Mix Va is the Volume of Air Voids in the Mix Vbe is the Effective Volume of Asphalt Cement in the Mix

  4. Densely graded mixes are going to offer the greatest resistance to loads because they minimize the void space and therefore the movement paths within the mix. Mathematically, Fuller Grading Curves offer the maximum density and minimum voids: Once the mix has been compacted to its densest state then there’s nowhere for particles to go when subjected to traffic. VMA and Dense Graded Aggregate where pi = total % passing sieve size i di = width of opening of sieve size i D = largest size (sieve opening) in gradation

  5. This allows for a reduction in the number of finer sizes which will in turn allow an increase in voids The goal is to blend the stock aggregates to produce a blend that falls between the two limits These can be used to set grading limits (“specs”) as with the red and green dashed curves. Notice that the High Spec is above the ⅜” Fuller Curve below the #8 sieve But why would we want to increase the voids? When plotted they look like this: VMA and Dense Graded Aggregate

  6. Experience has shown that asphalt mixes need to have between 3% and 5% air voids VMA and Dense Graded Aggregate This is a compromise between stability (lower with higher voids) and durability. If the voids are too low, then asphalt cement can bleed to the surface in hot weather. Its analogous to Portland Cement Concrete and air entraining to match exposure to the environment. So minimum VMA values are specified for different Nominal Maximum Particle Sizes

  7. Normally, aggregate stocks come from different sources: 1. stone comes from crushed bedrock, 2. sand comes from natural deposits, 3. mineral filler comes from the bottom of the crusher (dust) Blending Stock Aggregates There are two basic approaches: Simultaneous equations (used on Assignment 1) Trial and Error (spreadsheet, used for lab)

  8. Two questions using simultaneous equations. 1. Produce a blend of the given aggregates that results in 28.0% passing the 0.6 mm sieve. Assignment 1 Example The proportion of dust is set at 2 %. The second condition is that all the proportions must sum to 1: The condition given is for the 0.6 mm sieve. Therefore, C = 100% - 50.51% - 2% = 47.49% Let C be the fraction of Cyclone A Crushed and N be the fraction of Newmarket Sand. When we apply the proportions to find the percent passing the 0.6 mm sieve for the blend, we get: Multiply equation 2 by 0.005 then subtract 2 from 1 0.005C + 0.51N + 1.00(.02) = 0.28 1 0.005C + 0.51N = 0.26 or 0.005C + 0.005N = 0.0049 C + N = 0.98 C + N + 0.02 = 1.0 2 (0.51 - 0.005)N = 0.26 - 0.0049 (0.505)N = 0.2551 N = 0.505149 ≈ 50.51%

  9. Now for the blended gradation Assignment 1 Example: Q 1 Cont’d That’s the one all right! So we’re good to go. Before we start to crank off the blend, let’s check the proportions by finding P0.6 This is the percent passing the 0.6 mm sieve given for condition 1. Now we do the same for all the other sizes… P16.0 = 0.4749x100.0%+0.5051x100.0%+0.02x100.0% = 100.0% P2.36 = 0.4749x3.2% + 0.5051x82.0% + 0.02x100.0% = 44.938% P2.36 ≈ 44.9% P9.5 = 0.4749x59.0% + 0.5051x100.0% + 0.02x100.0% = 80.529% P9.5 ≈ 80.5% P4.75 = 0.4749x16.0% + 0.5051x96.0% + 0.02x100.0% = 58.088% P4.75 ≈ 58.1% P12.7 = 0.4749x90.0%+0.5051x100.0%+0.02x100.0% = 95.251% P12.7 ≈ 95.3% P1.18 = 0.4749x2.0% + 0.5051x73.0% + 0.02x100.0% = 39.822% P1.18 ≈ 39.8% P0.3 = 0.4749x0.0% + 0.5051x36.0% + 0.02x100.0% = 20.184% P0.3 ≈ 20.2% P0.150 = 0.4749x0.0% + 0.5051x21.0% + 0.02x100.0% = 12.607% P0.150 ≈ 12.6% P0.6 = 0.4749x0.5% + 0.5051x51.0% + 0.02x100.0% = 27.9976% P0.6 ≈ 28.0% P19.0 = 0.4749x100.0%+0.5051x100.0%+0.02x100.0% = 100.0% P0.075 = 0.4749x0.0% + 0.5051x9.2% + 0.02x100.0% = 6.647% P0.075 ≈ 6.6% P25.4 = 0.4749x100.0%+0.5051x100.0%+0.02x100.0% = 100.0%

  10. Assignment 1 Example: Q1 Finale Now, how much Newmarket Sand will be used? The final step(s) is to determine the mass of asphalt cement required for the 6 asphalt contents listed. Now for the other Newmarket Sand masses: These values represent the cumulative mass (the scale reading) after adding the required mass of each stock sieve size Applying this formula to each %AC: And, how much DFC Dust will be used? Since we’re using % of total mix, we have to use a bit of algebra. (I won’t tell anyone if you don’t.) Let the Mass of AC required = Mb and Pb = %AC We can check these numbers by summing over each size and calculating the % passing which should agree with the blend gradation The table above is used to weigh out and blend the sieved stock for individual 1210 g specimens. Now for the other Cyclone A masses: Any sizes with 100% passing will retain none of the material. Hence 0 g are recorded for all these sizes. First, how much Cyclone A will be used? P19.0 = 100%x(1.00 – 0/1210) = 100.0% (100.0%) P9.5 = 100%x(1.00 – 235.6/1210) = 80.5% (80.5%) The 4.75 mm sieve will retain 100 – 96 or 4% of the mass of Newmarket Sand used: 0.04 x 611.17 = 24.4 g P2.36 = 100%x(1.00 – 666.2/1210) = 44.9% (44.9%) The 4.75 mm sieve will retain 100 – 16 or 84% of the mass of Cyclone A used: 0.84 x 574.63 = 482.7 g P12.7 = 100%x(1.00 – 57.5/1210) = 95.25% (95.3%) P4.75 = 100%x(1.00 – 507.1/1210) = 58.1% (58.1%) For a specimen with 1210 g of aggregate we need 2% of 1210 g for DFC Dust = 24.2 g P16.0 = 100%x(1.00 – 0/1210) = 100.0% (100.0%) The 9.5 mm sieve will retain 100 – 59 or 41% of the mass of Cyclone A used: 0.41 x 574.63 = 235.6 g The 2.36 mm sieve will retain 100 – 82 or 18% of the mass of Newmarket Sand used: 0.18 x 611.17 = 110.0 g The 2.36 mm sieve will retain 100 – 3.2 or 96.8% of the mass of Cyclone A used: 0.968 x 574.63 = 556.2 g The 12.7 mm sieve will retain 100 – 90 or 10% of the mass of Cyclone A used: 0.10 x 574.63 = 57.5 g The last column should (and does) sum to the total specimen mass (1210 g) For a specimen with 1210 g of aggregate we need 50.51% of 1210 g of all Newmarket Sand sizes = 611.17 g For a specimen with 1210 g of aggregate we need 47.49% of 1210 g of all Cyclone A sizes = 574.63 g

  11. 2. Produce a blend of the given aggregates that results in 74.0% passing the 9.5 mm sieve and 61.0% passing the 2.36 mm sieve. Assignment 1 Example : Q 2 Let G be the fraction of Gumshoe Gold, S be the fraction of Saltwater Sludge and H be the fraction of Homer’s Donut Crumbs. The conditions given are for the 9.5 mm & 2.36 mm sieves. Applying the first condition for P9.5: The third condition is that all the proportions must sum to 1: The second condition for P2.36: 0.34G + 0.89S + H = 0.74 1 0.02G + 0.48S + H = 0.61 2 G + S + H = 1.00 3 for instructions, visit http://math.mohawkcollege.ca/calc.asp Using the Sharp EL546W Calculator, Mode 21: G = 0.392098555 ≈ 39.21% S = 0.011045029 ≈ 1.10% and H = 0.596856414% ≈ 59.69%

  12. Now for the blended gradation Assignment 1 Example: Q 2 Cont’d Got ‘em both! So we’re good to go. Before we start to crank off the blend, let’s check the proportions by finding P9.5 and P2.36 Now we do the same for all the other sizes… P19.0 = 0.3921x100% + 0.0110x100% + 0.5969x100% = 100% P25.4 = 0.3921x100% + 0.0110x100.0% + 0.5969x100% = 100% P0.150 = 0.3921x0.0% + 0.0110x1.0% + 0.5969x5.0% = 2.9955% P0.150 ≈ 3.0% P0.6 = 0.3921x0.0% + 0.0110x8.0% + 0.5969x53.0% = 31.7237% P0.6 ≈ 31.7% P9.5 = 0.3921x34.0% + 0.0110x89.0% + 0.5969x100.0% = 74.0004% P9.5 ≈ 74.0% P4.75 = 0.3921x9.0% + 0.0110x67.0% + 0.5969x100.0% = 63.9559% P4.75 ≈ 64.0% P0.075 = 0.3921x0.0% + 0.0110x0.0% + 0.5969x2.0% = 1.1938% P0.075 ≈ 1.2% P12.7 = 0.3921x79.0% + 0.0110x98.0% + 0.5969x100 = 91.7439% P12.7 ≈ 91.7% P1.18 = 0.3921x0.0% + 0.0110x21.0% + 0.5969x100% = 59.921% P1.18 ≈ 59.9% P2.36 = 0.3921x2.0% + 0.0110x48.0% + 0.5969x100.0% = 61.0022% P2.36 ≈ 61.0% P0.3 = 0.3921x0.0% + 0.0110x1.5% + 0.5969x26.0% = 15.5359% P0.3 ≈ 15.5% P16.0 = 0.3921x98.0% + 0.0110x100% + 0.57969x100% = 99.2158% P16.0 ≈ 99.2%

  13. Assignment 1 Example: Q2 Finale Now, how much Saltwater Sludge will be used? Applying this formula to each %AC: The final step(s) is to determine the mass of asphalt cement required for the 6 asphalt contents listed. Since we’re using % of total mix, we have to use a bit of algebra. (I won’t tell anyone if you don’t.) Let the Mass of AC required = Mb and Pb = %AC Now for the other Saltwater Sludge masses: Now for the other Gumshoe Gold masses: The table above is used to weigh out and blend the sieved stock for individual 1200 g specimens. First, how much Gumshoe Gold will be used? How much Homer’s Donut Crumbs will be used? Any sizes with 100% passing will retain none of the material. Hence 0 g are recorded for all these sizes. The 9.5 mm sieve will retain 100 – 89 or 11% of the mass of Saltwater Sludge used: 0.11 x 13.2 = 1.5 g The 16.0 mm sieve will retain 100 – 98 or 2% of the mass of Gumshoe Gold used: 0.02 x 470.5 = 9.4 g The 4.75 mm sieve will retain 100 – 9 or 91% of the mass of Gumshoe Gold used: 0.91 x 470.5 = 428.2 g The 2.36 mm sieve will retain 100 – 2.0 or 98.0% of the mass of Gumshoe Gold used: 0.98 x 470.5 = 461.1 g P9.5 = 100%x(1.00 – 312.0/1200) = 74.0% (74.0%) P12.7 = 100%x(1.00 – 98.8/1200) = 91.8% (91.8%) The 12.7 mm sieve will retain 100 – 79 or 21% of the mass of Gumshoe Gold used: 0.21 x 470.5 = 98.8 g The 4.75 mm sieve will retain 100 – 67 or 33% of the mass of Saltwater Sludge used: 0.33 x 13.2 = 4.4 g For a specimen with 1200 g of aggregate we need 39.21% of 1200 g of all Gumshoe Gold sizes = 470.5 g P16.0 = 100%x(1.00 – 9.4/1200) = 99.2% (99.2%) P4.75 = 100%x(1.00 – 432.6/1200) = 64.0% (64.0%) The 9.5 mm sieve will retain 100 – 34 or 66% of the mass of Gumshoe Gold used: 0.66 x 470.5 = 310.5 g P2.36 = 100%x(1.00 – 468.0/1200) = 61.0% (61.0%) The last column should (and does) sum to the total specimen mass (1200 g) The 2.36 mm sieve will retain 100 – 48 or 52% of the mass of Saltwater Sludge used: 0.52 x 13.2 = 6.9 g For a specimen with 1200 g of aggregate we need 1.10% of 1200 g of all Saltwater Sludge sizes = 13.2 g P19.0 = 100%x(1.00 – 0.0/1200) = 100.0% (100.0%) We can check these numbers by summing over each size and calculating the % passing which should agree with the blend gradation For a specimen with 1200 g of aggregate we need 59.69% of 1200 g of all Homer’s Donut Crumb sizes = 716.3 g The 12.7 mm sieve will retain 100 – 98 or 2% of the mass of Saltwater Sludge used: 0.02 x 13.2 = 0.3 g These values represent the cumulative mass (the scale reading) after adding the required mass of each stock sieve size

  14. Before the third lab, each group must perform its blending calculations. • The first priority is making sure your sieve results are reasonable. • If you’ve mixed up any sizes, even if you have a sieving error of 0%, you’ll still be using garbage for data. Trial and Error • The instructor will provide you with a set that is free of calculation errors, based on the data you submitted after you submit the report for the first lab.

  15. The numbers shown are for illustration only (they weren’t that great) The stock proportions are found by multiplying each sieve result by the trial proportion. The gradation of the blended aggregate is found by adding the proportions of the materials: The PMF is calculated (100 – PCA – PFA) These are the trial proportions for CA (PCA)& FA (PFA). The last column shows the percent error if the blend percent passing is outside the spec range Where does the 2.7% come from? This section is for the sieve sizes and specs. • Once you have a decent set of data, start an Excel spreadsheet using the following format: This section is for your sieve analysis results. The error in P3/8” is 92.7% - 90% = 2.7% Eg.: 50.4% = 4.1% + 44.3% + 2.0% Eg.: 4.1% = 7.8% x 0.520 Eg.: 2.0% = 100% x 0.020 Eg.: 44.3% = 96.3% x 0.460 Trial and Error

  16. Of course a grading plot makes the numbers easier to understand: • What would it look like before specifying PCA and PFA? Trial and Error

  17. What would it look like before specifying PCA and PFA? • Now you’re looking at the grading curve for the mineral filler (PMF = 100%) • Typically you’d set the PMF to a reasonable number (say 4%) and set PCA = PFA (= say 48%) • If you specified 100% CA you’d see the CA grading curve, or 100% FA would get you the FA grading curve Trial and Error

  18. Typically you’d set the PMF to a reasonable number (say 4%) and set PCA = PFA (= say 48%) • This curve is a bit too far to the fine end indicating that more coarse is needed and less fine: try 4% more CA and 2% less FA to drop the No 200 to about midrange Trial and Error

  19. In this instance the CA P3/8” is too close to the upper spec • a coarser CA (1/2”) would be required to meet the spec • Adding more coarse would drop the curve below the lower spec for the No. 4 sieve but more coarse would be needed to get the curve below the high limit for the 3/8” sieve • After a bit more tinkering the error on the 3/8” sieve could not be reduced below 2.7%. This was the best I could do. • …and the final gradation curve... Trial and Error

  20. It takes a lot of work to get the gradation plot to look like the ones shown in this presentation • A step-by-step example of how this is done can be found in the “Graphing Standards” posted on my website home page Trial and Error

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