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Chapter 5 · 3 hours

Soil Compaction

IOE past exam questions

Past questions and answers

20 questions set from this chapter, 3 of them more than once; 2 are most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 3 of 12 exams
  • Asked 3 times
  • 2079 Asoj · 2 marks
  • 2078 Poush · 2 marks
  • 2075 Bhadra · 2 marks

Explain the effect of compaction on the engineering behavior (properties) of soil.

Answer

Compaction is the process of densifying soil by reducing air voids using mechanical energy (rolling, ramming or vibration), at constant water content.

Effects on engineering behaviour

  1. Shear strength increases: particles interlock more closely and dry density rises.
  2. Compressibility and settlement decrease: a dense, stiffer soil settles less under the load.
  3. Permeability decreases: voids are reduced, so seepage through dams and embankments is lower.
  4. Bearing capacity increases: safe load for foundations and pavements is higher.
  5. Swelling and shrinkage decrease (when compacted wet of optimum), and volume change from water absorption is lessened.
  6. Liquefaction potential is reduced for sands due to the denser state.
  7. Slope stability improves: because of higher strength and lower pore pressure.
  8. Water absorption and frost damage decrease.

Compacting wet of optimum gives lower permeability and a more flexible soil (good for dam cores); compacting dry of optimum gives higher strength and stiffness (good for road subgrade) but more swelling.

  • Most repeated · 3 of 12 exams
  • Asked 3 times
  • 2078 Chaitra · 3 marks
  • 2078 Poush · 1 mark
  • 2074 Bhadra · 2 marks

Discuss the factors that affect the compaction of soil.

Answer

The dry density achieved by compaction depends on:

  1. Water content: at low water content the soil is stiff and resists compaction. Water lubricates the particles, so density increases up to the optimum moisture content (OMC) where γd\gamma_d is maximum. Beyond OMC, water occupies space and density falls.
  2. Compaction effort (energy): more effort (more blows, heavier rammer, more passes) gives a higher maximum dry density and a lower OMC. Increasing the effort beyond a limit gives little gain.
  3. Type of soil: well-graded coarse soils give a high γd,max\gamma_{d,max} and low OMC with a steep curve; clays have a low γd,max\gamma_{d,max}, high OMC and a flat curve. Poorly graded sands give a low density.
  4. Method of compaction: impact (hammer), kneading, vibration, and static pressure give different densities. Vibration is best for sands; kneading for clays.
  5. Lift thickness and number of passes in the field.
  6. Type of roller and speed: sheep-foot for clay, vibratory for sand.
  7. Admixtures: lime, cement or salts change OMC and density.
  8. Temperature, and initial moisture condition.
  • Asked 2 times
  • 2079 Asoj · 1 mark
  • 2078 Chaitra · 1 mark

What is relative compaction?

Answer

Relative compaction (RC) is the ratio of the dry unit weight (density) achieved in the field to the maximum dry unit weight obtained in the laboratory compaction test (Proctor test), expressed as a percentage:

RC=γd,fieldγd,max,lab×100RC = \frac{\gamma_{d,field}}{\gamma_{d,max,lab}} \times 100

It is used as the field control of compaction. Specifications normally require RC of 95% (standard Proctor) for embankments and 98 - 100% for pavement layers or the top of fills. Not to be confused with relative density DrD_r, which is for cohesionless soils.

  • 2079 Asoj · 3 marks

The maximum dry unit weight of a compacted soil mass is found to be 17 kN/m³ with optimum water content being 16%. Determine the void ratio and degree of saturation of this soil after compaction. Also, find the value of the maximum dry unit weight on the zero air void line at that optimum water content. Take specific gravity of soil solids as 2.65.

Similar questions: Compacted soil 18 kN/m³, porosity and ZAV (2074 Bhadra)

Answer

Given

γd=17\gamma_{d} = 17 kN/m³, w=16%w = 16\% (OWC), G=2.65G = 2.65, γw=9.81\gamma_w = 9.81 kN/m³.

Void ratio

e=Gγwγd−1=2.65×9.8117−1=0.529e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.65 \times 9.81}{17} - 1 = 0.529

Degree of saturation

S=wGe=0.16×2.650.529=0.801=80.1%S = \frac{wG}{e} = \frac{0.16 \times 2.65}{0.529} = 0.801 = 80.1\%

Zero air void unit weight at w=16%w = 16\%

γd,zav=Gγw1+wG=2.65×9.811+0.16×2.65=25.991.424=18.26 kN/m3\gamma_{d,zav} = \frac{G\gamma_w}{1 + wG} = \frac{2.65 \times 9.81}{1 + 0.16 \times 2.65} = \frac{25.99}{1.424} = 18.26\ \text{kN/m}^3

Answer: e=0.529e = 0.529, S=80.1%S = 80.1\%, γd,zav=18.26\gamma_{d,zav} = 18.26 kN/m³.

  • 2074 Bhadra · 3 marks

The maximum dry unit weight of a compacted soil mass is found to be 18 kN/m³ with optimum water content being 15%. Find the values of porosity and degree of saturation of this compacted soil. Also, find the value of the maximum dry unit weight on the zero air void line at that optimum water content. Take specific gravity of soil solid as 2.7.

Similar questions: Compacted soil 17 kN/m³, OWC 16% (2079 Asoj)

Answer

Given

γd=18\gamma_{d} = 18 kN/m³, w=15%w = 15\%, G=2.7G = 2.7, γw=9.81\gamma_w = 9.81 kN/m³.

Void ratio and porosity

e=Gγwγd−1=2.7×9.8118−1=0.4715e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.7 \times 9.81}{18} - 1 = 0.4715 n=e1+e=0.47151.4715=0.320=32.0%n = \frac{e}{1+e} = \frac{0.4715}{1.4715} = 0.320 = 32.0\%

Degree of saturation

S=wGe=0.15×2.70.4715=0.859=85.9%S = \frac{wG}{e} = \frac{0.15 \times 2.7}{0.4715} = 0.859 = 85.9\%

Zero air void unit weight

γd,zav=Gγw1+wG=2.7×9.811+0.15×2.7=26.491.405=18.85 kN/m3\gamma_{d,zav} = \frac{G\gamma_w}{1 + wG} = \frac{2.7 \times 9.81}{1 + 0.15 \times 2.7} = \frac{26.49}{1.405} = 18.85\ \text{kN/m}^3

Answer: n=32.0%n = 32.0\%, S=85.9%S = 85.9\%, γd,zav=18.85\gamma_{d,zav} = 18.85 kN/m³.

  • 2078 Poush · 3+2 marks

The following results were obtained from a standard compaction test.
Test No.123456
Water Content (%)11.012.112.813.614.616.3
Mass of compacted soil (gm)1920.52051.52138.52147.02120.02081.5
The specific gravity of solids is 2.7 and the volume of the compaction mould is 1000 cm³. A field compacted soil sample showed water content of 35% and unit weight of 2.318 Mg/m³. (i) Draw the compaction curve and determine the maximum dry unit weight and OMC. (ii) Find the relative compaction (RC).

Similar questions: Compaction test, RC and saturation (2073 Magh)

Answer

(i) Compaction curve

Mould volume =1000= 1000 cm³ = 1 litre, so bulk density ρ=m/1000\rho = m/1000 g/cm³ (Mg/m³), and

ρd=ρ1+w,γd=ρd×9.81\rho_d = \frac{\rho}{1+w}, \qquad \gamma_d = \rho_d \times 9.81
Testww (%)ρ\rho (Mg/m³)ρd\rho_d (Mg/m³)γd\gamma_d (kN/m³)ρd,zav\rho_{d,zav} (Mg/m³)
111.01.92051.73016.972.082
212.12.05151.83017.952.035
312.82.13851.89618.602.007
413.62.14701.89018.541.975
514.62.12001.85018.151.937
616.32.08151.79017.561.875

ZAV: ρd,zav=Gρw1+wG\rho_{d,zav} = \dfrac{G\rho_w}{1+wG} with G=2.7G = 2.7.

Plot ρd\rho_d (or γd\gamma_d) against ww and draw a smooth curve through the points, which rises to a peak between tests 3 and 4 and then falls (the curve stays below the ZAV values).

 rho_d
 1.90 |         .-*-.
 1.85 |       *      *
 1.80 |    *            *
 1.75 | *
      +---+---+---+---+---+-- w%
       11  12  13  14  15  16

From the smoothed curve:

  • Maximum dry density ≈1.90\approx 1.90 Mg/m³, i.e. maximum dry unit weight ≈18.6\approx 18.6 kN/m³
  • OMC ≈13.2%\approx 13.2\%

(ii) Relative compaction

Field dry density:

ρd,field=2.3181+0.35=1.717 Mg/m3 (γd=16.84 kN/m3)\rho_{d,field} = \frac{2.318}{1 + 0.35} = 1.717\ \text{Mg/m}^3\ (\gamma_d = 16.84\ \text{kN/m}^3) RC=ρd,fieldρd,max×100=1.7171.90×100=90.4%RC = \frac{\rho_{d,field}}{\rho_{d,max}} \times 100 = \frac{1.717}{1.90} \times 100 = 90.4\%

Answer: MDD ≈1.90\approx 1.90 Mg/m³ (18.6 kN/m³), OMC ≈13.2%\approx 13.2\%, relative compaction ≈90%\approx 90\% (below the usual 95% requirement).

  • 2073 Magh · 6 marks

The following results were obtained from a standard compaction test.
Test No.123456
Water content (%)11.012.112.813.614.616.3
Mass of compacted soil (gm)1920.52051.52138.52147.02120.02081.5
The specific gravity of solids is 2.7 and the volume of the compaction mould is 1000 cm³. A field compacted soil sample showed water content of 35% and unit weight of 2.318 Mg/m³. i) Draw the compaction curve and determine the maximum dry unit weight and OMC. ii) Find the relative compaction (RC). iii) Find the degree of saturation at the maximum dry unit weight.

Similar questions: Compaction test: curve, MDD, OMC, RC (2078 Poush)

Answer

Dry unit weight of each test is found from the bulk density of the 1000 cm³ mould, γd=γbulk1+w\gamma_d = \dfrac{\gamma_{bulk}}{1+w}.

Computation of dry density

γbulk=mass (g)1000 cm3\gamma_{bulk} = \dfrac{\text{mass (g)}}{1000\ \text{cm}^3} (in Mg/m³ = g/cm³)

Testw (%)Mass (g)γbulk\gamma_{bulk} (Mg/m³)γd\gamma_d (Mg/m³)
111.01920.51.92051.730
212.12051.52.05151.830
312.82138.52.13851.896
413.62147.02.14701.890
514.62120.02.12001.850
616.32081.52.08151.790

i) Compaction curve, γd,max\gamma_{d,max} and OMC

Plot γd\gamma_d (y-axis) against ww (x-axis) and draw a smooth curve through the six points. The dry density rises to a peak between 12.8% and 13.6% and then falls. A parabola through the three points around the peak gives the crest at about w=13.1%w = 13.1\% with γd≈1.90\gamma_d \approx 1.90 Mg/m³.

 γd
 1.90 |         . * .
 1.85 |      *         * 
 1.80 |    *             
 1.75 |  *                 *
 1.70 | *
      +----------------------- w %
        11  12  13  14  15  16
  • Maximum dry unit weight, γd,max≈1.90\gamma_{d,max} \approx 1.90 Mg/m³ (≈18.6\approx 18.6 kN/m³)
  • Optimum moisture content, OMC ≈13%\approx 13\%

ii) Relative compaction

Field dry unit weight:

γd,field=2.3181+0.35=1.717 Mg/m3\gamma_{d,field} = \frac{2.318}{1+0.35} = 1.717\ \text{Mg/m}^3 RC=γd,fieldγd,max×100=1.7171.90×100=90.4%RC = \frac{\gamma_{d,field}}{\gamma_{d,max}} \times 100 = \frac{1.717}{1.90} \times 100 = 90.4\%

iii) Degree of saturation at γd,max\gamma_{d,max}

e=Gγwγd−1=2.7×11.90−1=0.421e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.7 \times 1}{1.90} - 1 = 0.421 S=wGe=0.13×2.70.421=0.834S = \frac{wG}{e} = \frac{0.13 \times 2.7}{0.421} = 0.834

Answer: γd,max≈1.90\gamma_{d,max} \approx 1.90 Mg/m³, OMC ≈13%\approx 13\%; RC ≈90.4%\approx 90.4\%; S≈83%S \approx 83\% at maximum dry density.

The field soil has 35% water content, far wetter than OMC, so its RC is low and the sample is not properly compacted.

  • 2078 Chaitra · 1 mark

Describe placement water content.

Answer

Placement water content is the water content at which the soil is placed and compacted in the field. It is usually specified relative to the OMC (for example OMC ±2%\pm 2\%). Dry-of-optimum placement gives higher strength and stiffness but higher permeability and swelling potential; wet-of-optimum placement gives lower permeability, lower strength and better flexibility, as needed in the core of an earth dam.

  • 2078 Chaitra · 1 mark

Describe theoretical maximum dry density.

Answer

The theoretical maximum dry density is the dry density that a soil would have if all air were removed from the voids, i.e. at S=100%S = 100\% for the given water content. It is represented by the zero air void (ZAV) line:

γd,max(theory)=Gγw1+wG\gamma_{d,max(theory)} = \frac{G\gamma_w}{1 + wG}

It is an upper limit of dry density: the actual compaction curve always lies below it, since in practice 2 - 5% air remains in the soil.

  • 2074 Bhadra · 1 mark

What is zero air void (ZAV)?

Answer

The zero air void (ZAV) line (or saturation line) is the curve showing the relation between dry unit weight and water content when the degree of saturation is 100% (Va=0V_a = 0):

γd=Gγw1+wG\gamma_{d} = \frac{G\gamma_w}{1 + wG}

It is the theoretical maximum dry unit weight for each water content. The compaction curve lies to its left and never crosses it, because air cannot be fully expelled by compaction.

  • 2076 Baisakh · 1+3+2 marks

Define zero air void. What are the necessary precautions needed during field compaction in different environments, for homogeneous earth dams and subgrades for highways? Would you prefer to compact the soil on the dry side of OMC or the wet side of OMC?

Answer

Zero air void

The zero air void (ZAV) line is the plot of dry unit weight against water content for a fully saturated soil (S=100%S=100\%, no air):

γd=Gγw1+wG\gamma_{d} = \frac{G\gamma_w}{1 + wG}

It is the upper limit of the compaction curve.

Precautions during field compaction

General: use the correct roller, lift thickness (150 - 300 mm), number of passes; control the water content near OMC; check density with field tests (sand replacement, core cutter); remove stones and organic matter; maintain a uniform moisture by mixing; avoid over-compaction.

Homogeneous earth dam:

  • Compact on the wet side of OMC (about OMC + 1 to 2%) to get a flexible, low-permeability fill that can adjust to settlement without cracking.
  • Use sheep-foot or pneumatic rollers; bond each layer by scarifying the surface; keep a uniform water content; compact in thin layers; provide a filter drain to control seepage.
  • Avoid over-wetting, which would cause pore pressure build-up and slope instability during construction.

Highway subgrade:

  • Compact on the dry side or at OMC to get high strength and stiffness (CBR); the top 300 mm to be compacted to 95 - 100% of the standard/modified Proctor MDD.
  • Protect from rain and keep drainage working, because wetting later can cause swelling and strength loss; moisture can be adjusted to OMC by sprinkling or aeration.
  • Use vibratory rollers for granular soils.

Dry side or wet side?

For earth dams and canal linings (impermeability and flexibility), prefer the wet side of OMC. For highway subgrade and where strength is important, prefer the dry side up to the OMC, as the soil has higher strength, but there is risk of swelling and water absorption if the soil gets wet later.

  • 2078 Poush · 1 mark

Is it practically possible to maintain the optimum moisture content during compaction at field? Give reason.

Answer

It is not fully possible to maintain the exact optimum moisture content in the field, but it can be kept within about ±2%\pm 2\% of the OMC.

Reasons:

  • Water content changes after sprinkling, because of evaporation by sun and wind or addition by rain during hauling, spreading and rolling.
  • The borrow soil is non-uniform, so moisture is variable and cannot be mixed perfectly.
  • The field OMC differs from the laboratory OMC, since the field compaction energy and method are different.
  • Water content is checked only in selected spots, and the test results take time.
  • 2079 Jestha · 1+2+3 marks

What is compaction? How does it differ from consolidation? Describe briefly different methods of compaction with their relative merits and demerits.

Answer

Compaction

Compaction is the process of increasing the density of soil by reducing the air voids using mechanical energy such as rolling, ramming or vibration, with no change in water content.

Compaction versus consolidation

BasisCompactionConsolidation
CauseMechanical energy (instant)Sustained static load
ChangeAir is expelledWater is squeezed out
SoilPartly saturatedSaturated (fine) soils
TimeImmediateSlow (years in clay)
Water contentNo changeDecreases
PurposeImprove soil properties in fillNatural settlement under loads

Methods of compaction

MethodUsed forMeritsDemerits
Tamping (hand or mechanical rammers)Small areas, trenches, near structuresSimple, reaches confined placesSlow, uneven, small output
Smooth wheel rollerGravel, sand, road baseSmooth surface, good for finishingPoor for clay, thin depth
Sheep-foot rollerClayey soilsGood kneading, bonds the layersNot for sand; leaves rough surface
Pneumatic-tyred rollerSand, clay, generalVersatile, adjustable pressureCostly, complex
Vibratory rollerGranular soilsDeep compaction, efficient for sandsNot effective for clay, noisy
Impact / dynamic compaction (heavy weight drop)Loose deep fillsDeep improvementVibration to nearby structures
  • 2075 Bhadra · 2 marks

Write down the names of the different methods of compaction carried out in the field. Draw compaction curves for the Standard Proctor Test and Modified Proctor Test.

Answer

Field compaction methods

  1. Tamping (rammers)
  2. Smooth-wheel (static) rolling
  3. Sheep-foot (tamping foot) rolling
  4. Pneumatic-tyred rolling
  5. Vibratory rolling
  6. Dynamic compaction (heavy tamping), vibroflotation (deep)

Compaction curves: Standard and Modified Proctor

The Modified Proctor test uses more energy (heavier hammer and greater drop, 5 layers), giving a higher MDD and lower OMC than the Standard Proctor test (3 layers).

 gd
  |        .-"-.   Modified
  |       / MDD2 \
  |      /  .-"-. \ Standard
  |     /  / MDD1\ \
  |    /  /       \ \  ZAV line
  |   /  /         \ \ (above)
  +-----|---|------------ w
       OMC2 OMC1

Notes: The peak of the Modified Proctor curve is higher and to the left (OMC2_2 < OMC1_1).

  • 2073 Bhadra · 3 marks

Draw a compaction curve for a soil showing maximum dry density, optimum water content, zero-air void line, dry side and wet side of optimum water content.

Answer

A compaction curve plots dry unit weight (γd\gamma_d) against water content (ww) for a fixed compaction effort. The dry unit weight rises with ww up to a peak, then falls.

 gd
  |                    ZAV line
  |         MDD  .-"-.  .
  |  . - - - - -/     \   .
  |            /       \    .
  |          /          \
  |  dry   /      wet     \
  | side  /       side
  +------+----------------------- w
       OMC

Features:

  • Maximum dry density (MDD): the peak value of γd\gamma_d at the top of the curve.
  • Optimum water content (OWC or OMC): the water content corresponding to MDD.
  • Dry side of optimum: the left of the peak, where w<w < OMC. The soil is stiff, and has flocculated structure.
  • Wet side of optimum: the right of the peak, where w>w > OMC. The soil is dispersed with lower strength; the curve approaches the ZAV line.
  • Zero air void (ZAV) line: γd=Gγw/(1+wG)\gamma_d = G\gamma_w/(1+wG) for S=100%S = 100\%; the compaction curve lies always below and to the left of it.

The ZAV line is drawn using the specific gravity of the solids; it lies above the compaction curve as air always remains.

  • 2073 Bhadra · 3 marks

Compare the compaction characteristic curve for sand and clay.

Answer

BasisSand (cohesionless)Clay (cohesive)
Shape of curveFlat, with a small "dip" at intermediate water content, and a rise at very dry and saturatedDistinct peak at OMC
Sensitivity to waterVery lowHigh
Maximum dry densityHigh (about 18 - 21 kN/m³)Low (about 14 - 19 kN/m³)
OMCLow (about 6 - 10%)High (about 12 - 30%)
ReasonMoisture is not essential; bulking at low water due to capillary tension; in saturation water drains outWater lubricates the particles; excess water reduces density
Best methodVibrationKneading (sheep-foot)
 gd
  |  Sand ___.-~~~~~-.___
  |  (flat, dips)
  |
  |        Clay .-"-.
  |            /     \
  +-------------------- w

In sand, the highest density is obtained when it is either completely dry or fully saturated; the minimum occurs at about 4 - 8% water content because of bulking (capillary tension holds particles apart). In clay the density follows the typical bell-shaped curve with a clear optimum.

  • 2075 Bhadra · 2 marks

The maximum dry density of a compacted soil mass is found to be 18 kN/m³ with optimum water content being 15%. Find the degree of saturation of this compacted soil if the specific gravity of soil solids is given as 2.65. What will be the value of the maximum dry density it can be further compacted to?

Answer

Given

γd=18\gamma_{d} = 18 kN/m³, w=15%w = 15\%, G=2.65G = 2.65, γw=9.81\gamma_w = 9.81 kN/m³.

Degree of saturation

e=Gγwγd−1=2.65×9.8118−1=0.444e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.65 \times 9.81}{18} - 1 = 0.444 S=wGe=0.15×2.650.444=0.895=89.5%S = \frac{wG}{e} = \frac{0.15 \times 2.65}{0.444} = 0.895 = 89.5\%

Further compaction

The maximum value the dry unit weight can reach at the same water content is when all air is expelled (S=100%S = 100\%), on the ZAV line:

γd,max=Gγw1+wG=2.65×9.811+0.15×2.65=25.991.3975=18.60 kN/m3\gamma_{d,max} = \frac{G\gamma_w}{1 + wG} = \frac{2.65 \times 9.81}{1 + 0.15 \times 2.65} = \frac{25.99}{1.3975} = 18.60\ \text{kN/m}^3

Answer: S=89.5%S = 89.5\%; the soil can at most be compacted to 18.60 kN/m³ (in practice slightly less, since some air always remains).

  • 2078 Poush · 3 marks

A cylindrical specimen of a cohesive soil of 10 cm diameter and 20 cm height was prepared by compaction in a mold. Taking the specific gravity of soil solid as 2.65 and the wet weight of this specimen as 30 kN [as printed] and water content as 15%, find the following: (i) dry unit weight, void ratio and degree of saturation of this cylindrical specimen; (ii) if 95% of relative compaction is to be achieved in the field, what should be the dry unit weight of compacted soil of the same soil specimen in the field.

Answer

Given

D=10D = 10 cm, H=20H = 20 cm, G=2.65G = 2.65, w=15%w = 15\%, γw=9.81\gamma_w = 9.81 kN/m³.

Assumption: the weight "30 kN" is a misprint for 30 N, since 30 kN in this small mould would give an impossible unit weight of about 19,000 kN/m³.

Volume and unit weights

V=π4(0.10)2(0.20)=1.571×10−3 m3V = \frac{\pi}{4}(0.10)^2(0.20) = 1.571 \times 10^{-3}\ \text{m}^3 γ=30 N1.571×10−3=19 099 N/m3=19.10 kN/m3\gamma = \frac{30\ \text{N}}{1.571 \times 10^{-3}} = 19\,099\ \text{N/m}^3 = 19.10\ \text{kN/m}^3 γd=γ1+w=19.101.15=16.61 kN/m3\gamma_d = \frac{\gamma}{1+w} = \frac{19.10}{1.15} = 16.61\ \text{kN/m}^3

Void ratio

e=Gγwγd−1=2.65×9.8116.61−1=0.565e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.65 \times 9.81}{16.61} - 1 = 0.565

Degree of saturation

S=wGe=0.15×2.650.565=0.703=70.3%S = \frac{wG}{e} = \frac{0.15 \times 2.65}{0.565} = 0.703 = 70.3\%

(ii) Field requirement

Taking the specimen's dry unit weight as the maximum (laboratory) dry unit weight:

γd,field=0.95×γd,max=0.95×16.61=15.78 kN/m3\gamma_{d,field} = 0.95 \times \gamma_{d,max} = 0.95 \times 16.61 = 15.78\ \text{kN/m}^3

Answer: (i) γd=16.61\gamma_d = 16.61 kN/m³, e=0.565e = 0.565, S=70.3%S = 70.3\%; (ii) field γd≥15.78\gamma_d \ge 15.78 kN/m³.

  • 2078 Baisakh · 6 marks

Evaluate the construction of the embankment if the required degree of compaction is 95% and the dry density of the embankment was found as 1.78 g/cc. The result of the compaction test performed in the laboratory for the same material using a 950 cc mould are as follows.
w %7.711.514.617.519.721.2
Mass of wet soil (kg)1.701.892.031.991.961.92

Answer

Degree of compaction (relative compaction) is RC=γd,field/γd,maxRC = \gamma_{d,field}/\gamma_{d,max}. First find γd,max\gamma_{d,max} from the lab test, then compare with the field value.

Lab dry density (mould volume = 950 cm³)

γbulk=M950 cm3,γd=γbulk1+w\gamma_{bulk} = \frac{M}{950\ \text{cm}^3}, \qquad \gamma_d = \frac{\gamma_{bulk}}{1+w}
w (%)Mass (kg)γbulk\gamma_{bulk} (g/cc)γd\gamma_d (g/cc)
7.71.701.7891.662
11.51.891.9891.784
14.62.032.1371.865
17.51.992.0951.783
19.71.962.0631.724
21.21.922.0211.668

Plotting γd\gamma_d against ww gives a peak at

  • γd,max≈1.865\gamma_{d,max} \approx 1.865 g/cc, OMC ≈14.6%\approx 14.6\%

Check of the embankment

Required dry density =0.95×1.865=1.77= 0.95 \times 1.865 = 1.77 g/cc.

RC=1.781.865×100=95.4%RC = \frac{1.78}{1.865} \times 100 = 95.4\%

Answer: RC = 95.4% (> 95%), field dry density 1.78 g/cc > required 1.77 g/cc. The embankment construction is satisfactory.

The margin is small, so the result is acceptable but only just; the moisture content of the field fill should also be checked to be near the OMC.

  • 2075 Baisakh · 6 marks

In the construction of a road, the compaction specification required was 95% of Proctor maximum dry density at a field moisture content within 2% of the optimum moisture content. The maximum dry density and optimum moisture content obtained in the laboratory from the Standard Proctor test were 1.95 Mg/m³ and 13.5% respectively. A site engineer conducted sand cone test at two locations and obtained the following results.
Location No.Mass of soil removed (gm), WetMass of soil removed (gm), DryMass of sand used (gm)
143.8638.4639.51
237.3832.2132.39
The density of sand used was 1.86 Mg/m³. Check whether the specification was satisfied or not.

Answer

Sand cone test: volume of the hole = mass of sand filling it / density of sand. Then the field density and water content follow. The specification requires both

  • γd≥0.95 γd,max=0.95×1.95=1.8525\gamma_d \ge 0.95\,\gamma_{d,max} = 0.95 \times 1.95 = 1.8525 Mg/m³
  • ww within 13.5±2%13.5 \pm 2\%, that is 11.5% to 15.5%

Location 1

V=39.511.86=21.24 cm3V = \frac{39.51}{1.86} = 21.24\ \text{cm}^3 γbulk=43.8621.24=2.065 Mg/m3,γd=38.4621.24=1.811 Mg/m3\gamma_{bulk} = \frac{43.86}{21.24} = 2.065\ \text{Mg/m}^3, \qquad \gamma_d = \frac{38.46}{21.24} = 1.811\ \text{Mg/m}^3 w=43.86−38.4638.46×100=14.0%w = \frac{43.86 - 38.46}{38.46} \times 100 = 14.0\% RC=1.8111.95×100=92.8%RC = \frac{1.811}{1.95} \times 100 = 92.8\%

Location 2

V=32.391.86=17.41 cm3V = \frac{32.39}{1.86} = 17.41\ \text{cm}^3 γbulk=37.3817.41=2.147 Mg/m3,γd=32.2117.41=1.850 Mg/m3\gamma_{bulk} = \frac{37.38}{17.41} = 2.147\ \text{Mg/m}^3, \qquad \gamma_d = \frac{32.21}{17.41} = 1.850\ \text{Mg/m}^3 w=37.38−32.2132.21×100=16.05%w = \frac{37.38 - 32.21}{32.21} \times 100 = 16.05\% RC=1.8501.95×100=94.9%RC = \frac{1.850}{1.95} \times 100 = 94.9\%

Comparison

Locationγd\gamma_d (Mg/m³)RC (%)Required RCw (%)Allowed w
11.81192.89514.011.5 to 15.5
21.85094.99516.0511.5 to 15.5

Answer: The specification is not satisfied. Location 1 fails on density (RC 92.8% < 95%) though its water content is acceptable. Location 2 is just below 95% (94.9%) and its water content (16.05%) is above the 15.5% limit. Both areas need re-rolling (and drying at location 2).

The masses are taken as grams, as given; the result depends only on their ratios.

Questions from Old Question Collection (CE 552) (IOE BCE Soil Mechanics (CE552) papers from 2073 Bhadra to 2079 Asoj). Answers are written for this site; check them against your class notes.

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