Soil: Screened soil falling under the following parameters:
|Gravel||< 5%||Falling under the USDA soil classifications loam, silt loam orclay loam|
|Coarse to medium sand||< 5%|
|Very fine sand||10-30%|
Hydrogel: Gelscape(r) or approved equal
B. Calculate ratio of stone to soil
|a. Take a small representative sample of soil from the stock pile (40-50g) (1.4-1.8 oz)||48 grams (1.69 oz)|
|b. Dry sample to a constant weight (about 48 hours) at 60C or 110 F||--|
|c. Weight the dry soil||40 grams (1.41 oz)|
|d. Subtract the dry soil weight from the wet soil weight to obtain the weight of moisture lost.||48 - 40 = 8 (0.28 oz)|
|e. Divide the water lost by the dry weight for the weight percentage of soil moisture in the stock pile.||8/40 = .20 or 20%|
iii. If the mix is for 15% soil, 85%stone, determine the amount of soil to add to meet the dry weight soil needed.
|Testing batch recommended stone weight||Example:|
|Target mixture of 85% stone||25kg (55.1 lb) stone|
|Determine soil dry weight needed||25kg stone ÷ 0.85 = 29.41 kg (64.8 lb) mixture|
|Determine moist soil needed to reach target mixture ration||29.41kg mix - 25 kg stone = 4.41kg (9.7 lb) soil|
|If the mix is for 18% soil, 82 stone, determine the amount of soil to add to meet the dry weight soil needed.||4.41 kg dry soil x 1.20 (soil plus moisture content calculated above)
= 5.29 kg (11.64 lb) moist soil
IF 25 kg stone = 82% of mix 5.5 kg (12.1 lb) dry soil is needed 6.6 kg (14.6 lb) moist soil is needed
25 kg stone/.82 = 30.49 kg mix 30.5 kg mix - 25 kg stone = 5.5 kg dry soil 5.5 kg dry soil x 1.2 = 6.6 kg moist soil
|7.5 grams (0.27 oz) of hydrogel in both mixtures|
NOTE: It has been our experience that the soil
works well in mixing if it is moist enough to form a ball when lightly
packed in a person's hands and breaks into 2-4 pieces when dropped from
around 12 inches into a hard surface. If it slumps it is too wet, if it
shatters, it will require a greater amount of additional water in the
Send the 3 sample material blends to a soils engineering laboratory
the following diagnostic tests.
When the results come back:
15% should give you acceptable strength.
IF 18% is at CBR = 50, reduce the soil in mixture to 15% (to allow for mixing variations)
20% may or may not give you acceptable strength. IF 20% is at CBR = 50, then reduce soil to 18% (again to account for mixing variation)
-or- graph the results of CBR against soil percentage in mix. Choose 2% less soil than level where line crosses CBR = 50
Scale the mixing process to match the equipment available.
Mixing "on the flat" on a paved surface with a front end loader (shovel)
|Spread a known weight of dry stone onto the flat surface in a thin layer. This requires knowing the unit weight of stone per loader bucket.||1000kg (2204.6 lb) stone in layer|
|Spread dry hydrogel evenly over the stone layer at the 30 units hydrogel per 100,000 units stone rate||300grams (10.58 oz) hydrogel|
|Converting weight to volume of soil component:||A) 176 kg (388 lb) dry soil is needed for a 15% soil mix|
|Weigh one bucket load of soil moist||B) 176 kg (388 lb) dry soil is needed for a 15% soil mix|
|Take 5 representative samples from the soil stock pile 2 days prior to mixing and tarp soil pile to minimize changes in the moisture content.||C) 105 kg (231.5 lb) moist soil per bucket in this example|
|Determine average soil moisture content as described earlier.
NOTE: a Speedy(r) moisture kit could provide on site moisture content measurement Available through many soil testing supply companies.
|D) soil moisture was calculated at 20% in the previous example
100x(105 ÷ (100+%moisture in bucket load)) = weight of dry soil per bucket
|Spread appropriate amount of screened topsoil over the stone-hydrogel layer, accounting for the moisture in the soil during weight measurement for blending.||105 kg /120 = 0.875
0.875 x 100 = 87.5 kg (193 lb) dry soil and 17.5 kg (38.5 lb) water in bucket
87.5 kg (193 lb) soil per bucket x 2= 175 kg (386 lb) dry soil per 2 buckets
close enough to mix ratio target of 176 kg (388 lb).
So for this example, use 2 buckets of soil per 1000 kg (2204.6 lb) stone for a 15% soil mix.
For a 20% mix, 2.8 buckets would be needed. 1000÷ 1.2 = 1250 so 250 kg soil needed.
250 kg ÷ 87.5 kg soil per bucket = 2.86 buckets, round down for safety to 2.8.
Mix the layered material until the soil is uniformly distributed within the stone adding just enough moisture to the system to prevent the soil from falling away from the stone and the mixture begins to hold together as it is packed in a ball in your hands. In any case, the material should be mixed at or below optimum moisture level for mixture compaction as determined by the Proctor density curve once the curve is established (during the bulk mixing process).
For immediate installation, the material could be mixed at the desired optimum moisture level for compaction. To do this, an accounting of the moisture in the soil is necessary.
In example, suppose the targeted moisture content of the stone-soil mixture is 9% moisture by weight as determined from the laboratory compaction tests.
For a 15% soil mixture with the soil added having a 20% moisture content, the soil component accounts for a baseline 3% moisture content prior to any addition of water during the mixing process. 0.15 x 0.20 = 0.03
Add water as a spray during the mixing procedure. It is advisable to have a metered source of water, or a known rate of flow to gauge the moisture content of the mixing batch. Prevent excessive moisture in the mixing process.
|Assuming the same material example:||1000 kg (2204.6 lb) stone + 175 kg (386 lb) soil (2 buckets of soil 87.5 kg dry soil per bucket) = 1175 kg (2590.6 lb) dry mix|
|targeted 9% moisture content||1175 kg x1.09=1281 kg (2823.8 lb)|
|moisture total in system||1281 kg-1175 kg = 106 kg (233.2 lb) water|
|water added in soil||17.5 kg (38.5 lb)(per bucket *2 buckets = 35 kg (77 lb) water|
|water needed to hit target moisture||106 kg - 35 kg = 71 kg water ----- 71 liters water
156.5 lb water ----- 18.75 gallons
If mixing the material in large quantities is expected, stockpile mixed material in 200-400 cyd batches.
For quality control, take a minimum of three 5 kg (11 lb) samples to be tested for:
|To quickly measure ratio stone to soil
Collect sample and obtain moist weight
6.00 kg (13.2 lb)
|Dry the stockpile sample and obtain sample dry weight||5.50 kg (12.1 lb)|
|Determine mixture moisture content
Submerge the sample in water and wash the stone free of soil
Pass the material through a #40 seive, while washing any residual soil
away. Retain and collect all stone from the sample
|6 - 5.5 = 0.5 (13.2-12.1 = 1.1)
0.5 ÷ 5.5 = 0.091 = moisture content
|dry the stone and obtain sample stone dry weight||4.68 kg (10.3 lb) stone|
|calculate the percentage of stone in the stone
Remember 95% of the stone should be retained on a 0.75 inch seive, now could be a good time to double check if stone is in question
|4.68 kg stone ÷ 5.5 kg sample dry weight = 0.851
(10.3÷12.1 = 0.851)
0.851 x 100 = 85.1
Target mixture @ 85.0
Place the material into construction area and compacted to not less than 95% Proctor peak density in 6 inch lifts. Verify using sand cone densimeter method.
It is suggested that the material be covered if it is to be stored for any period of time. If not covered, the material stockpile should be turned before use to eliminate any surface washing of the outer layer of stone during a rain event. Moisture content should be measured and adjusted to optimum compaction moisture prior to installation.
Total volume estimation and mix calculation
trench of 40 ft long x 20 ft wide x 3ft deep = 2400cuft
2400 cuft ÷ 27 = 89 cyds
(if optimum compacted density is est. @ 1950 kgm-3 ÷ 1328.94 = 1.467
tons / cyd)
(if optimum compacted density is est. @ 2000 kgm-3 ÷ 1328.94 = 1.505 tons / cyd)
89 cyds x 1.5 ton / cyd gross est. of compacted density = 133.5 tons material dry if 82% stone mix
133.5 x .82 = 109.5 tons stone
133.5 tons total - 109.5 tons stone = 24 tons dry soil
if soil @ 10% moisture by weight 24 x 1.1 = 26.4 tons soil
if soil @ 20% moisture by weight 24 x 1.2 = 28.8 tons soil