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Proctor Test (AASHTO T 99*, T 224 & T 272 Topic J Topic J * Replacement Procedure is no longer applicable

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Page 1: Proctor Test (AASHTO T 99*, T 224 & T 272 Topic J * Replacement Procedure is no longer applicable

Proctor Test (AASHTO T 99*, T 224 & T 272

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Topic J

* Replacement Procedure is no longer applicable

Page 2: Proctor Test (AASHTO T 99*, T 224 & T 272 Topic J * Replacement Procedure is no longer applicable

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AASHTO T-99 – Methods A and C

• Compact soil in mold in three equal lifts.• Each layer is compacted with a hammer that

weighs 5.5 lbs with a free fall of 12 inches.• Each layer is compacted 25 times with the

hammer.• When method B or D are required each layer

is compacted 56 times with the hammer.

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Page 5: Proctor Test (AASHTO T 99*, T 224 & T 272 Topic J * Replacement Procedure is no longer applicable

Sample Preparation (AASHTO T 99)

• Dry Sample below optimum moisture content.• % retained on No. 4 sieve (Method A or B)• % retained on ¾ inch sieve (Method C or D)

– Used for determining rock correction factor.

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Proctor - Procedures• Mix sample with water• Compact specimen in mold• Remove extension collar and trim soil• Weigh mold and wet soil• Subtract mold weight and multiply by 30 cf• Remove soil and test for moisture content• Repeat four more times at greater moisture

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Page 9: Proctor Test (AASHTO T 99*, T 224 & T 272 Topic J * Replacement Procedure is no longer applicable

Student Problem (J-3)

• The Moisture/Density relationship of clay is 105.9 pcf at 22.9% moisture. Calculate.

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Determining Percent Compaction/Moisture

• Percent Compaction = (Measured Dry Density/Maximum Dry Density) * 100%

• Percent Optimum Moisture Content = (Measured Moisture Content/Optimum Moisture Content)*100%

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Page 11: Proctor Test (AASHTO T 99*, T 224 & T 272 Topic J * Replacement Procedure is no longer applicable

Student Problem (J-3)

• The Moisture/Density relationship of clay is 105.9 pcf at 22.9% moisture. Calculate.

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92.7%82.5%

100.3%100.0%

84.2%131.1%

Page 12: Proctor Test (AASHTO T 99*, T 224 & T 272 Topic J * Replacement Procedure is no longer applicable

Student Problem (J-5)Sample Preparation

• The weight of the total sample in grams is 6356. The actual moisture of the soil is 2.0%. The desired moisture content to be achieved is 7.0%. Calculate the milliliters of water needed to obtain the desired moisture content?

• One milliliter of water equals 1 gram

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Student Problem (J-5) Sample Preparation

• Weight of sample (W) = 6356 grams• Actual moisture of soil (A) = 2.0%• Desired moisture content (OM) =7.0%• Calculate ml of water needed?

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Student Problem (J-5)

• D= Difference in moisture content

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D = OM - A

D = 7.0 – 2.0 = 5.0

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Student Problem (J-5)

• F= Percent moisture factor

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F = D/100

F = 5.0/100 = 0.05

Page 16: Proctor Test (AASHTO T 99*, T 224 & T 272 Topic J * Replacement Procedure is no longer applicable

Student Problem (J-5)

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Water needed = W x F

Water needed = 6356 g x 0.05 = 317.8g

1 gram = 1 milliliter

317.8 g = 317.8 ml

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Determining Moisture Content (w)From Lab Test

• Moisture content is the ratio of weight of water in soil sample to weight of solid particles in a given soil sample.

• w = (Ww /Ws) * 100%

• Ww = Weight of Water

• Ws = Weight of Solids

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Page 18: Proctor Test (AASHTO T 99*, T 224 & T 272 Topic J * Replacement Procedure is no longer applicable

Determining Moisture Content (w)From Lab Test

A = Weight of container and wet soil = 100 gramsB = Weight of container and dry soil = 75 gramsC = Weight of container = 20 gramsw = Moisture Content (%)

w = [(A-B)/(B-C)] x 100%

Weight of Water = A-BWeight of Solids = B-C

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Determining Moisture Content (w)From Lab Test

w = [(A-B)/(B-C)] x 100%

w = [(100-75)/(75-20)] x 100%

w = [25/55] x 100%

w = 45.4%

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Page 20: Proctor Test (AASHTO T 99*, T 224 & T 272 Topic J * Replacement Procedure is no longer applicable

Determining Moisture Content (w)From Lab Test

• See AASHTO page T-99-10

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Determining Dry Density Proctor Test

• Wet Density (W1) = Weight/Volume

• Dry Density (W) = Dry Weight of Soil/Volume

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Determining Dry Density Example

• Wet Weight of Soil in Mold = 14.81 lbs• Weight of Mold = 10.27 lbs• Volume of Mold = 1/30 cubic foot• Moisture Content of Sample (w) = 5.5%• Dry Density (W) = ? (pounds per cubic feet (pcf))

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Determining Dry Density Example

• W = [W1/(w+100)] x 100

• W = [136.2/(5.5+100)] x 100

• W = 129.1 pcf

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Zero Air Voids Curve

• Theoretical Maximum Dry Density (zav) plotted against corresponding Moisture Content

• zav = w /(w + 1/Gs)

• wpcf

• Gs = 2.65

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Zero Air Voids Curve (Specific Gravity = 2.65)Moisture Content Zero Air Voids Dens (pcf)

1% 161.092% 157.043% 153.184% 149.515% 146.016% 142.677% 139.498% 136.449% 133.52

10% 130.7211% 128.0412% 125.4613% 122.9914% 120.6115% 118.3316% 116.1217% 114.0018% 111.9619% 109.9820% 108.0821% 106.2422% 104.4623% 102.7424% 101.0825% 99.46

Page 27: Proctor Test (AASHTO T 99*, T 224 & T 272 Topic J * Replacement Procedure is no longer applicable

One Point Proctor/Family of Curves

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One Point Proctor/Family of Curves

• Plot all proctors of likes soils on one graph• Draw curve connecting all of the points of

maximum dry density/optimum moisture content.

• Find point on each proctor that represents 80% of optimum moisture content.

• Draw curve that connects 80% points.

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One Point Proctor/Family of Curves

• Perform One Point Proctor (AASHTO T-272)• Plot point obtained for Dry Density and

Moisture Content directly on Family of Curves plot.

• This point must fall within band from 80% to 100% of optimum moisture content.

• If not repeat one point proctor until meeting this requirement, then plot on Family of Curves.

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One Point Proctor/Family of Curves• Draw new proctor curve on Family of Curves that

connects through the one point, making the complete curve essentially parallel to the adjacent curves.

• New proctor curve can be used to determine percent compaction and percent optimum moisture for representative density test only.

• If the one point proctor doesn’t yield accurate point/curve then a full proctor must be performed.

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