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## Review of Relative Density Principles

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**Review of Relative Density Principles**• Relative Density principles apply to compaction of relatively clean, coarse-grained soils. • Relatively clean usually taken to be less 12 % or less finer than the #200 sieve. • Important for compaction study of filters**Explain basic principles of compacting clean sands and**gravels Understand basic tests to obtain reference densities. Use 1 point compaction test in design and quality control Summarize minimum and maximum index density tests Detail the importance of water content in compacting clean sands and gravels Objectives**Review of Compaction Principles**• Compaction Tests are not commonly performed on soils with 12 % or fewer fines • Small percentage of fines means soils cannot easily hold water to examine range of water and effect on dry density**Review of Compaction Principles**• Compaction tests performed on clean sands may have this appearance Dr density w %**Compacting Clean Sands**• Clean sands are compacted most easily at either very dry or very wet water contents • At intermediate water contents, capillary stresses in voids resist compaction • Bulking is term for this phenomenon**Compacting Clean Sands**• Vibration most effective energy for sands • Use smooth-wheeled vibratory roller**Relative Density**• Alternative to traditional compaction test is relative density tests • Minimum Index Density • Maximum Index Density • Relative Density**Minimum Index Density**• Minimum index density of clean sand is that resulting from very loosely filling a steel mold. ASTM Method D4254 Sand dropped no more than 1”**Minimum Index Density**• After filling the mold, excess soil is carefully screed off. The volume of this mold is 0.1 ft3. Knowing the weight of soil in the mold, the dry density is easily computed**Maximum Index Density**• Example Minimum dry density = 96 pcf • Maximum index density of clean sand results from vibration at high amplitude on vibratory table for 10 minutes. ASTM D4253 • Example Maximum dry density = 117.5 pcf**Maximum Index Density**Weight on sample inside sleeve Vibratory table**Maximum Index Density**Weight on sample inside sleeve Vibratory table**Maximum Index Density**Sample densified by vibration Measure D height to determine new gd Plate on which weight sits during vibration**Void Ratio and Dry Density**• The void Ratio is calculated for each state of denseness of sample. • Maximum void ratio occurs at minimum index density - For Example Min.gd = 96.0 pcf • Minimum void ratio occurs at maximum index density For Example Maximum gd = 110.0 pcf**Minimum and Maximum Void Ratios**• First Calculate void ratio at Minimum gd • Next Calculate void ratio at Maximum gd**Relative Density Equation**Diagram below illustrates a relative density of about 40 % emeasured emax emin d measured dmax d min increasing density**Calculate Void Ratio of Compacted Sand**• Now, assume that the density of this sand was measured in a compacted fill and it was 102.5 pcf. Calculate a value for relative density of the fill. First, calculate the void ratio of the fill:**Compute Relative Density**• Now, use the values of void ratio in the relative density equation:**Compute Relative Density**• Relative Density Equation(rewritten in dry density terms) • Solve for Example: **Fort Worth Relative Density Study**• NRCS lab in Fort Worth studied 28 filter sands and used some published data • Minimum and Maximum Index Densities were performed on each sample • A 1 point dry Standard Proctor energy mold was also prepared for each sample. • Values of 50% and 70% relative density were plotted against the 1 point Proctor value**70 % Relative Density vs. 1 Point Proctor**• Conclusion is that the field 1 point Proctor dry test is about equal to 70 % relative density**50 % Relative Density vs. 1 Point Proctor**• Conclusion is that the 95 % of the field 1 point Proctor dry test is about equal to 50 % relative density**Relative Density Estimates from FW SML Study**gD70 = 1.075 x gd 1pt -9.61, for RD70 and gd 1pt in lb/ft3 gD50 = 1.07 x gd 1pt - 12.5, for RD50 and gd 1pt in lb/ft3**Relative Density Estimates from FW SML Study**• Example Relative Density Estimates • Given: 1 Point Proctor Testgd = 105.5 pcf • Estimate 70 % and 50% Relative Density • Given that measured gd is 98.7, evaluate state of compaction of sand.**Review of Relative Density**• Class Problem - Relative Density • A soil’s minimum index density is 96.5 pcf and its maximum index density is 111.5 pcf. The Gs value is 2.65 • Calculate the emin and emax • Compute the void ratio and dry density corresponding to a relative density value of 70 %**Class Problem Solution**• Given: Minimum index density is 96.5 pcf • Maximum index density is 111.5 pcf.**Class Problem Solution**• Now, substitue a value for RD of 70(%) in the relative density equation**Class Problem Solution**• Solving and Rearranging the equation:**Class Problem Solution**• Now, calculate a value for dry density at this void ratio: • Summary - The dry density corresponding to 70(%) relative density for this sample is 106.5 pcf**Other information on Relative Density**Chart is for silty sands (SM)**Class Problem**• Given that the water content of a silty sand that was obtained from a saturated zone of a channel bank measured 24.5 percent • What is the estimated relative density of the sand?**Class Problem Solution**• Reading from the chart, the estimated Rd value is about 42 percent.