Compaction - Detailed Explanation
Compaction is one of the most frequently tested topics in the PRC Civil Engineer Licensure Examination under Geotechnical Engineering. It involves the mechanical densification of soil by expelling air voids — not water — to increase the dry unit weight, thereby improving the soil's load-bearing capacity, reducing compressibility, and lowering permeability. Every fill embankment, road subgrade, and earth dam in the Philippines is governed by compaction specifications. Mastery of the Proctor test, relative compaction, and the zero-air-voids (ZAV) line is non-negotiable for any board examinee. This chapter walks through each concept systematically with board-style problems, formula derivations, and exam-focused tips.
Fundamentals of Soil Compaction
Examples
A soil sample is compacted in a mold. The moist unit weight is measured as 19.5 kN/m³ and the water content is determined to be 12%. Compute the dry unit weight.
Solution
γ_d = γ/(1+w) = 19.5/(1+0.12) = 19.5/1.12 = 17.41 kN/m³
This is the most basic compaction formula. Always convert w% to decimal (12% → 0.12) before applying. The dry unit weight is always less than the moist unit weight for w > 0.
A compacted embankment layer has a moist unit weight of 20.2 kN/m³ at water content w = 14%. Compute its dry unit weight.
Solution
γ_d = 20.2/(1+0.14) = 20.2/1.14 = 17.72 kN/m³
Apply the same formula. Note that even though the moist unit weight increased from Example 1, the higher water content means more water is present — the dry unit weight is slightly higher at 17.72 vs 17.41 kN/m³. This illustrates that moist unit weight alone does not tell us how well compacted the soil is.
Compaction is the process of mechanically densifying soil by reducing air voids through applied mechanical energy (rollers, tampers, vibrators). The key distinction: compaction expels AIR, not water. Water content may change slightly due to evaporation during rolling, but the mechanism is air expulsion. As a result, the soil volume decreases while the mass of solids and water remain essentially the same, causing an increase in dry unit weight (γ_d).
The three-phase soil system helps visualize this: before compaction, a large volume is occupied by air. After compaction, that air volume shrinks. The degree of compaction is best measured by the dry unit weight γ_d = γ/(1+w), where γ is the moist (bulk) unit weight and w is the gravimetric water content expressed as a decimal.
Factors affecting compaction:
- Water content (w) — acts as a lubricant; too little causes soil stiffness, too much causes pore pressure build-up.
- Compactive effort (energy per unit volume) — higher energy yields higher γ_d,max and lower OMC.
- Soil type — well-graded granular soils achieve high γ_d; plastic clays achieve lower γ_d.
- Method of compaction — kneading (sheepsfoot), vibratory (granular), static (smooth drum).
Applications
- Road subgrade and base course compaction for DPWH highway projects
- Earth embankment dams (NIA irrigation projects require strict compaction control)
- Airport runway subgrade densification
- Building pad preparation for residential and commercial structures
- Backfill compaction behind retaining walls and bridge abutments
Common misconceptions
- WRONG: Compaction removes water from soil. RIGHT: Compaction removes air; water content changes only due to evaporation or drainage, not the compaction mechanism itself.
- WRONG: The moist unit weight is the best indicator of compaction quality. RIGHT: Dry unit weight is the correct measure because it reflects the mass of solids per unit volume.
- WRONG: More water always improves compaction. RIGHT: Water content beyond OMC actually decreases γ_d because water displaces solid particles per unit volume.
Related concepts
- Phase relationships (void ratio, porosity, degree of saturation)
- Proctor compaction test (Standard and Modified)
- Zero-air-voids line
- Relative compaction and quality control
- Relative density (for granular soils)
Common exam questions
Example
A 1.5-m³ compacted fill has a mass of 2,850 kg at w = 10%. Find γ_d. → γ = (2850 × 9.81/1000)/1.5 = 18.63 kN/m³; γ_d = 18.63/1.10 = 16.94 kN/m³
Approach
Given γ (moist) and w, apply γ_d = γ/(1+w). Ensure w is in decimal form.
Example
Board item: 'Which statement is TRUE about Modified Proctor vs Standard Proctor?' — Answer: Modified gives higher γ_d,max at a lower OMC.
Approach
Modified Proctor uses 4.5× more energy than Standard. State: Modified → higher γ_d,max, lower OMC; Standard → lower γ_d,max, higher OMC.
Key points to remember
- Compaction removes AIR voids, not water — this is a classic board-exam trap.
- Dry unit weight γ_d = γ/(1+w) is the fundamental compaction equation.
- Higher compactive effort → higher γ_d,max and LOWER OMC.
- Sandy soils respond better to vibratory compaction; clayey soils respond to kneading action.
- The compaction curve (γ_d vs w) is bell-shaped with a single peak at OMC.
- On the dry side of optimum: soil structure is flocculated, stronger, but brittle.
- On the wet side of optimum: soil structure is dispersed, more plastic, less permeable.
The Proctor Compaction Test (Standard and Modified)
Examples
A Standard Proctor test on a clay soil yields the following data: w(%) = [8, 10, 12, 14, 16] and γ (kN/m³) = [17.2, 18.5, 19.3, 18.8, 18.1]. Determine γ_d,max and OMC.
Solution
Compute γ_d = γ/(1+w) for each point:
w=8%: γ_d = 17.2/1.08 = 15.93 kN/m³
w=10%: γ_d = 18.5/1.10 = 16.82 kN/m³
w=12%: γ_d = 19.3/1.12 = 17.23 kN/m³ ← maximum
w=14%: γ_d = 18.8/1.14 = 16.49 kN/m³
w=16%: γ_d = 18.1/1.16 = 15.60 kN/m³
γ_d,max = 17.23 kN/m³ at OMC = 12%
This is the standard board-exam procedure: compute γ_d at each w, identify the peak. In practice, a smooth curve is drawn through the points and the peak is read graphically, but for board exams, the tabulated approach usually gives a clear winner.
A Modified Proctor test gives γ_d,max = 19.2 kN/m³ at OMC = 10%. A Standard Proctor test on the same soil would likely give which result?
Solution
Standard Proctor → lower energy → lower γ_d,max (e.g., ~17.5–18.0 kN/m³) and higher OMC (e.g., ~13–15%). The exact values require testing, but the trend is unambiguous.
This is a conceptual board question testing understanding of the effect of compactive effort. The key principle: more energy → particles pack tighter at lower water content → γ_d,max increases and OMC decreases.
The Proctor test (ASTM D698 — Standard; ASTM D1557 — Modified) is the laboratory method for determining the compaction curve: a plot of γ_d versus water content w. The soil is compacted in 3 (Standard) or 5 (Modified) layers inside a cylindrical mold using a drop hammer. At each water content, γ and w are measured, then γ_d = γ/(1+w) is computed and plotted. The curve peaks at γ_d,max and OMC.
STANDARD PROCTOR (ASTM D698):
- Mold volume: 944 cm³ (1/30 ft³)
- Hammer mass: 2.49 kg (5.5 lb)
- Drop height: 305 mm (12 in)
- Layers: 3
- Blows per layer: 25
- Compactive energy: 591.3 kJ/m³
MODIFIED PROCTOR (ASTM D1557):
- Mold volume: 944 cm³ (same)
- Hammer mass: 4.54 kg (10 lb)
- Drop height: 457 mm (18 in)
- Layers: 5
- Blows per layer: 25
- Compactive energy: 2,693 kJ/m³ (approximately 4.56× Standard)
Because Modified Proctor applies ~4.5× more energy, it produces a HIGHER γ_d,max and a LOWER OMC. The curve shifts upward and to the left. This is because at higher energy, the same lubrication effect (from water) is achieved at a lower water content — the soil particles are forced closer together more efficiently.
PHYSICAL INTERPRETATION of the curve shape:
- Dry side (w < OMC): Soil is stiff and clods resist compaction. Adding water lubricates particle contacts, increasing γ_d.
- At OMC: Optimum lubrication — maximum packing of particles.
- Wet side (w > OMC): Water occupies volume that could be taken by soil particles, so γ_d decreases. Also, pore water pressure resists compaction.
Applications
- DPWH specification: subgrade compaction to 95% of Modified Proctor for primary roads
- ASTM D698/D1557 — referenced in Philippine standard specifications for highways
- Embankment dam fill control — compaction curves determine the target γ_d for field QC
- Airport pavement design — FAA and CAAP require Modified Proctor reference for runway subbase
Common misconceptions
- WRONG: OMC means the soil is 100% saturated. RIGHT: At OMC, saturation is typically 70–85%. Full saturation corresponds to the ZAV line, which is always above and to the right of the peak.
- WRONG: Standard and Modified Proctor give the same γ_d,max for a given soil. RIGHT: They give different values; Modified always gives higher γ_d,max for the same soil.
- WRONG: The Proctor test mold is the same for both Standard and Modified procedures. RIGHT: The mold volume (944 cm³ standard mold) may be the same, but the procedure (hammer, layers, blows) differs significantly.
Related concepts
- Compactive energy per unit volume
- Zero-air-voids line
- Relative compaction
- Field density testing (sand cone, nuclear gauge)
- Soil classification and compaction behavior (GW, SW, CL, CH)
Common exam questions
Example
Board 2019 style: 'From the following Proctor test results, find the OMC and γ_d,max.' — Compute γ_d column, pick the peak.
Approach
Given a table of γ and w values, compute γ_d = γ/(1+w) for each row and identify the maximum γ_d and corresponding w (= OMC).
Example
Board item: 'The Modified Proctor test compared to Standard Proctor gives: (A) lower γ_d,max and lower OMC (B) higher γ_d,max and lower OMC (C) higher γ_d,max and higher OMC (D) lower γ_d,max and higher OMC' → Answer: (B)
Approach
Remember the energy ratio (~4.5×) and the directional shifts: Modified → higher γ_d,max, lower OMC.
Example
Standard: E = (25 × 2.49 × 9.81 × 0.305 × 3) / (944 × 10⁻⁶) ≈ 591 kJ/m³
Approach
E = (number of blows × hammer weight × drop height × number of layers) / mold volume
Key points to remember
- Modified Proctor: heavier hammer (4.54 kg), more layers (5), greater drop (457 mm) → higher γ_d,max, lower OMC.
- Standard Proctor hammer: 2.49 kg; Modified: 4.54 kg — memorize these for fill-in-the-blank board questions.
- The Proctor curve ALWAYS lies below the ZAV line — if a computed point plots above ZAV, there is an arithmetic error.
- OMC is NOT the water content at maximum saturation; at OMC, S is typically 70–85%, not 100%.
- The compaction curve shifts UP and LEFT with increasing compactive effort.
- Both curves (Standard and Modified) approach the ZAV line asymptotically on the wet side.
- For the board exam: if only one Proctor value is given without qualification, assume Modified Proctor for modern highway work.
Zero-Air-Voids (ZAV) Line
Examples
Given G_s = 2.68 at water content w = 15%, find the zero-air-voids dry unit weight.
Solution
γ_zav = G_s·γ_w / (1 + w·G_s)
= (2.68 × 9.81) / (1 + 0.15 × 2.68)
= 26.29 / (1 + 0.402)
= 26.29 / 1.402
= 18.75 kN/m³
Any compacted soil at w = 15% with G_s = 2.68 must have γ_d < 18.75 kN/m³.
This is a direct application of the ZAV formula. Note: w·G_s = 0.15 × 2.68 = 0.402 is the void ratio at full saturation (e = w·G_s when S = 1). The result sets an upper bound for the dry unit weight at this water content.
For G_s = 2.70, compute the ZAV dry unit weight at w = 10% and w = 20%.
Solution
At w = 10%:
γ_zav = (2.70 × 9.81) / (1 + 0.10 × 2.70) = 26.49 / (1 + 0.27) = 26.49 / 1.27 = 20.86 kN/m³
At w = 20%:
γ_zav = (2.70 × 9.81) / (1 + 0.20 × 2.70) = 26.49 / (1 + 0.54) = 26.49 / 1.54 = 17.20 kN/m³
Observe: as w increases from 10% to 20%, γ_zav drops from 20.86 to 17.20 kN/m³ — confirming the curve slopes downward.
This exercise demonstrates that γ_zav decreases with increasing water content. On a γ_d vs w plot, the ZAV line forms a curve bending downward to the right. Students can plot these two points alongside the compaction curve to visualize the gap (air voids).
A soil is compacted at w = 18% to γ_d = 16.5 kN/m³. The specific gravity is 2.65. Verify that this point is below the ZAV line.
Solution
γ_zav at w = 18%:
= (2.65 × 9.81) / (1 + 0.18 × 2.65)
= 26.00 / (1 + 0.477)
= 26.00 / 1.477
= 17.60 kN/m³
Since 16.5 kN/m³ < 17.60 kN/m³, the point is BELOW the ZAV line. ✓ (Physically valid)
This type of verification question appears on the board exam. If the examiner gives a point that lies above the ZAV line, the answer is that the result is impossible or an error exists in the data.
The Zero-Air-Voids (ZAV) line — also called the saturation line or 100% saturation line — represents the THEORETICAL dry unit weight if the soil were fully saturated (S = 100%, V_air = 0) at a given water content. It is a curve, not a straight line, on the γ_d vs w plot.
FORMULA DERIVATION:
At full saturation (S = 100%), all voids are filled with water:
e·S = w·G_s → e = w·G_s (since S = 1)
Dry unit weight: γ_d = G_s·γ_w / (1 + e) = G_s·γ_w / (1 + w·G_s)
Therefore: γ_zav = G_s·γ_w / (1 + w·G_s)
Where:
- G_s = specific gravity of soil solids (typically 2.65–2.72)
- γ_w = unit weight of water = 9.81 kN/m³
- w = water content (decimal)
KEY PROPERTIES of the ZAV line:
- It always lies ABOVE and to the RIGHT of the compaction curve.
- It decreases as w increases (more water per unit volume means fewer solids → lower γ_d).
- The compaction curve can NEVER cross the ZAV line (impossible to have negative air voids).
- As the soil approaches the ZAV line on the wet side of OMC, the gap represents remaining air voids.
- Different G_s values produce different ZAV lines — higher G_s → higher ZAV for the same w.
PRACTICAL USE: Engineers draw multiple ZAV lines (for S = 80%, 90%, 100%) on the compaction chart to assess the air void content at any point on the curve.
Applications
- Setting upper-bound reference for quality control charts in field compaction
- Estimating degree of saturation at any point on the compaction curve
- Evaluating whether field density test results are physically consistent
- Design of compacted clay liners for landfills — target placement on wet side near ZAV
Common misconceptions
- WRONG: The ZAV line is a straight line on the γ_d vs w plot. RIGHT: It is a curved line decreasing with w; it appears nearly linear over a small range but is mathematically a hyperbola-type curve.
- WRONG: Achieving the ZAV condition is practically possible with enough compaction effort. RIGHT: In practice, some air always remains (typically 2–4% air voids even at OMC). S = 100% is a theoretical limit.
- WRONG: The ZAV line is the same for all soils. RIGHT: It depends on G_s; soils with higher G_s have a higher ZAV line.
Related concepts
- Phase relationships — void ratio, degree of saturation
- Specific gravity of soil solids (G_s)
- Compaction curve (Proctor curve)
- Air voids content: A_v = (γ_zav - γ_d)/γ_zav × 100%
- Saturation lines at S = 80%, 90%
Common exam questions
Example
Find γ_zav for G_s = 2.72, w = 12%: = (2.72 × 9.81)/(1 + 0.12 × 2.72) = 26.68/(1.326) = 20.12 kN/m³
Approach
Apply γ_zav = G_s·γ_w/(1 + w·G_s). Convert w to decimal. Use γ_w = 9.81 kN/m³.
Example
Board item: 'Is it possible to compact a soil (G_s = 2.68) to γ_d = 20.0 kN/m³ at w = 18%?' → γ_zav = 2.68×9.81/(1+0.18×2.68) = 26.29/1.482 = 17.74 kN/m³. Since 20.0 > 17.74, NO — impossible.
Approach
Compute γ_zav at the given w and G_s. Compare with γ_d. If γ_d > γ_zav, the result is impossible.
Example
Board item: 'The zero-air-voids line: (A) represents S=100% (B) lies below the compaction curve (C) decreases with increasing w (D) depends on G_s' → Answer: (B) is FALSE.
Approach
Know the properties: ZAV line slopes downward, lies above compaction curve, never intersected by compaction curve.
Key points to remember
- ZAV formula: γ_zav = G_s·γ_w / (1 + w·G_s) — memorize this exactly.
- The ZAV line is always ABOVE the compaction curve; no exceptions in real soils.
- If a computed γ_d exceeds γ_zav at the same w, there is a calculation error — physically impossible.
- γ_w = 9.81 kN/m³ (use this for Philippine board exams unless specified otherwise).
- The ZAV line curves downward as w increases — it is NOT a straight line.
- Common G_s values: quartz = 2.65, typical mineral soils = 2.67–2.72.
- On the wet side of the curve, the compaction curve nearly parallels the ZAV line.
Relative Compaction and Field Quality Control
Examples
The field dry unit weight of a compacted subgrade is 17.41 kN/m³. The Modified Proctor test gives γ_d,max = 18.50 kN/m³. Find the relative compaction. Does it meet the 95% specification?
Solution
RC = (17.41 / 18.50) × 100% = 0.941 × 100% = 94.1%
Since 94.1% < 95%, the specification is NOT met. Additional compaction passes are needed.
This is a direct board-exam application. The comparison to the specification is always part of the answer — do not stop at computing RC. State whether it meets the requirement.
A sand cone test yields: mass of excavated moist soil = 1,850 g; mass of sand used = 1,420 g; unit weight of calibration sand = 15.5 kN/m³; water content = 11%. The Modified Proctor gives γ_d,max = 17.8 kN/m³. Find RC.
Solution
Step 1: Volume of hole
γ_sand = 15.5 kN/m³ = 15,500 N/m³ = 15.81 kg/m³ × 9.81 — wait, use density approach:
γ_sand = 15.5 kN/m³; ρ_sand = 15.5/9.81 × 1000 = 1,580 kg/m³
V_hole = M_sand / ρ_sand = 1,420 / 1,580 = 0.8987 × 10⁻³ m³ ...
Simpler unit approach:
γ_sand = 15.5 kN/m³ → 15,500 N/m³
Weight of sand = 1,420 g × 9.81/1000 = 13.93 N
V_hole = 13.93 N / 15,500 N/m³ = 8.987 × 10⁻⁴ m³ = 898.7 cm³
Step 2: Moist unit weight of field soil
Weight of soil = 1,850 g × 9.81/1000 = 18.15 N
γ = 18.15 / (8.987 × 10⁻⁴) = 20,196 N/m³ = 20.20 kN/m³
Step 3: Dry unit weight
γ_d = 20.20 / (1 + 0.11) = 20.20 / 1.11 = 18.20 kN/m³
Step 4: Relative compaction
RC = 18.20 / 17.8 × 100% = 102.2%
RC > 100% — field compaction exceeds lab reference. This is acceptable (excellent compaction).
Sand cone problems are common in board exams. The sequence: find V_hole → find γ_field → find γ_d → find RC. Note that RC > 100% is physically possible and not an error when the field equipment applies more energy than the Proctor hammer.
Relative Compaction (RC) is the ratio of the field dry unit weight achieved during construction to the laboratory maximum dry unit weight from the Proctor test, expressed as a percentage:
RC = (γ_d,field / γ_d,max) × 100%
This is the primary quality control parameter for earthwork in the Philippines. Philippine practice (aligned with DPWH Standard Specifications) typically mandates:
- Embankment fills: RC ≥ 90% (Modified Proctor)
- Subgrade: RC ≥ 95% (Modified Proctor)
- Base course: RC ≥ 100% (Modified Proctor) in some specifications
NOTE ON RELATIVE DENSITY vs RELATIVE COMPACTION:
For GRANULAR soils (sands, gravels), relative density D_r is preferred:
D_r = (e_max - e) / (e_max - e_min) × 100%
Or equivalently in terms of dry unit weight:
D_r = [(γ_d - γ_d,min) / (γ_d,max - γ_d,min)] × (γ_d,max / γ_d) × 100%
Relative compaction and relative density are related:
RC = 80 + 0.2·D_r (approximate, Lee and Singh, 1971)
This means RC = 80% → D_r = 0%; RC = 100% → D_r = 100%.
FIELD DENSITY TESTS to measure γ_d,field:
- Sand Cone Test (ASTM D1556) — most common in PH; indirect measurement
- Rubber Balloon Test (ASTM D2167)
- Nuclear Density Gauge (ASTM D2922) — fast but requires licensing in PH
- Core cutter method — for cohesive soils
SAND CONE TEST PROCEDURE (board-exam relevant):
- Dig a hole in the compacted fill, collect all excavated soil.
- Measure mass of excavated moist soil (M_soil).
- Fill hole with calibrated Ottawa sand from cone apparatus; measure mass of sand used (M_sand).
- Volume of hole = M_sand / γ_sand (known from calibration).
- γ = M_soil × g / V_hole; then γ_d = γ/(1+w) where w from oven-drying sample.
Applications
- DPWH field inspection and acceptance of earthwork for national road projects
- Quality control of compacted fills for building foundations (NSCP Section 301)
- Dam construction — layer-by-layer compaction testing per NIA specifications
- Airport runway construction — CAAP requires density testing every 500 m²
- Retaining wall backfill — over-compaction can increase lateral earth pressure
Common misconceptions
- WRONG: RC > 100% is impossible or indicates an error. RIGHT: RC can exceed 100% if heavy field equipment applies more energy than the lab Proctor test. This is common in granular soils with vibratory rollers.
- WRONG: Relative compaction and relative density are the same concept. RIGHT: RC uses dry unit weight ratios; D_r uses void ratio (or density) range between loosest and densest states. They apply to different soil types.
- WRONG: The same RC specification (e.g., 95%) applies to all layers. RIGHT: Different layers (subgrade, base course, surface) may have different RC requirements per DPWH specifications.
Related concepts
- Proctor compaction test (γ_d,max reference)
- Sand cone test procedure
- Relative density (D_r) for granular soils
- Phase relationships (e, n, S)
- Field compaction equipment selection
Common exam questions
Example
γ=19.0 kN/m³, w=13%, γ_d,max=18.0 kN/m³. RC? → γ_d=19.0/1.13=16.81; RC=16.81/18.0×100=93.4% → meets 90% but not 95%
Approach
Compute γ_d,field from moist weight and water content. Divide by γ_d,max. Compare to 90% or 95% requirement.
Example
Mass soil=2100g, mass sand=1680g, γ_sand=14.8 kN/m³, w=9%, γ_d,max=16.5 kN/m³ → V=(1680×9.81/1000)/14.8=1113cm³; γ=2100×9.81/(1000×1.113×10⁻³)=18.49kN/m³; γ_d=18.49/1.09=16.96; RC=16.96/16.5×100=102.8%
Approach
V_hole from sand calibration → γ_moist from hole → γ_d from water content → RC from Proctor max.
Example
Board item: 'For a poorly graded sand subgrade, which compaction criterion is most appropriate?' → Relative Density D_r, not relative compaction RC.
Approach
RC (Proctor-based) for cohesive/general fills; relative density D_r for clean sands and gravels.
Key points to remember
- RC = (γ_d,field / γ_d,max) × 100% — field over lab maximum, times 100.
- Standard RC requirement: fills ≥ 90%, subgrades ≥ 95% of Modified Proctor.
- RC can exceed 100% if field compaction energy exceeds the lab test — this is achievable with heavy vibratory rollers on granular soils.
- For granular soils, D_r is more meaningful than RC; approximate relation: RC ≈ 80 + 0.2·D_r.
- Sand cone test: volume of hole = mass of calibration sand used / unit weight of calibration sand.
- Always verify: if RC > 100%, check if Modified or Standard Proctor was the reference — standard Proctor reference gives higher RC values for the same field γ_d.
- The board may ask which test (RC or D_r) is appropriate — RC for cohesive, D_r for granular.
Practice problems
PROBLEM 1 (Dry Unit Weight). A soil sample from a compacted embankment has a moist unit weight of 20.2 kN/m³ and a water content of 14%. (a) Compute the dry unit weight. (b) If the void ratio of this soil is 0.52 and G_s = 2.68, compute the degree of saturation.
Solution
(a) γ_d = γ/(1+w) = 20.2/1.14 = 17.72 kN/m³
(b) From phase relationships: e·S = w·G_s
S = w·G_s/e = (0.14 × 2.68)/0.52 = 0.3752/0.52 = 0.721 = 72.1%
Verification via γ_d: γ_d = G_s·γ_w/(1+e) = 2.68×9.81/(1+0.52) = 26.29/1.52 = 17.30 kN/m³
→ Close to 17.72 (small discrepancy due to assumed e; consistent if e is back-calculated: e = G_s·γ_w/γ_d - 1 = 26.29/17.72 - 1 = 0.484)
Part (a) is a direct Proctor formula. Part (b) uses the fundamental phase relationship S·e = w·G_s. This combination (Proctor + phase relationships) is a favorite multi-part board question. Always verify using an alternative formula to catch arithmetic errors.
PROBLEM 2 (ZAV Line). For a soil with G_s = 2.70, compute the zero-air-voids dry unit weight at w = 8%, 12%, 16%, and 20%. Plot the trend (values only) and state whether a compacted specimen at w = 16%, γ_d = 17.5 kN/m³ is physically possible.
Solution
γ_zav = G_s·γ_w / (1 + w·G_s) = (2.70 × 9.81)/(1 + w × 2.70) = 26.49/(1 + 2.70w)
w = 8%: γ_zav = 26.49/(1+0.216) = 26.49/1.216 = 21.78 kN/m³
w = 12%: γ_zav = 26.49/(1+0.324) = 26.49/1.324 = 20.01 kN/m³
w = 16%: γ_zav = 26.49/(1+0.432) = 26.49/1.432 = 18.50 kN/m³
w = 20%: γ_zav = 26.49/(1+0.540) = 26.49/1.540 = 17.20 kN/m³
Trend: 21.78 → 20.01 → 18.50 → 17.20 (decreasing with increasing w)
At w = 16%: γ_zav = 18.50 kN/m³; proposed γ_d = 17.5 kN/m³.
Since 17.5 < 18.50, the specimen is BELOW the ZAV line → PHYSICALLY POSSIBLE ✓
This problem builds the ZAV curve numerically. The comparison at w = 16% confirms the specimen is valid (in the feasible zone). If γ_d were given as 19.0 kN/m³ at w = 16%, it would exceed 18.50 kN/m³ — physically impossible, indicating a data or calculation error.
PROBLEM 4 (Combined ZAV + RC). A Modified Proctor test on a silty clay gives γ_d,max = 17.80 kN/m³ at OMC = 16%. The specific gravity is G_s = 2.67. (a) Compute the zero-air-voids dry unit weight at OMC. (b) Compute the degree of saturation at the OMC point on the compaction curve. (c) If the field RC = 92%, compute the field dry unit weight. (d) Is the field condition on the dry side or wet side of optimum if the field water content is 14%?
Solution
(a) γ_zav at w = 16%:
= (2.67 × 9.81) / (1 + 0.16 × 2.67)
= 26.19 / (1 + 0.4272)
= 26.19 / 1.4272
= 18.35 kN/m³
Note: γ_d,max = 17.80 < γ_zav = 18.35 ✓ (compaction curve is below ZAV)
(b) From γ_d = G_s·γ_w/(1+e):
e = G_s·γ_w/γ_d - 1 = 26.19/17.80 - 1 = 1.4713 - 1 = 0.4713
S = w·G_s/e = 0.16 × 2.67 / 0.4713 = 0.4272 / 0.4713 = 0.906 = 90.6%
(c) Field γ_d:
RC = γ_d,field / γ_d,max
γ_d,field = RC × γ_d,max = 0.92 × 17.80 = 16.38 kN/m³
(d) OMC = 16%; field w = 14%.
Since 14% < 16%, the field is on the DRY SIDE of optimum.
This multi-part problem integrates all key compaction concepts. Part (b) shows that at OMC the soil is NOT fully saturated (~90.6%) — a crucial exam point. Part (c) reverses the RC formula to find field γ_d. Part (d) requires only comparison of w_field to OMC. On the dry side: soil is stiffer, more brittle, higher strength but more swelling potential. On the wet side: lower strength, more plastic, less permeable.
PROBLEM 5 (Compactive Energy Comparison). A laboratory compaction test uses a 2.49-kg hammer dropped 305 mm, 25 blows per layer, 3 layers, in a 944-cm³ mold. (a) Compute the compactive energy per unit volume. (b) If a Modified Proctor procedure is used instead (4.54-kg hammer, 457-mm drop, 5 layers, 25 blows/layer, same mold), compute the energy ratio Modified/Standard.
Solution
(a) Standard Proctor energy:
E = (N_blows × W_hammer × h_drop × N_layers) / V_mold
W_hammer = 2.49 kg × 9.81 m/s² = 24.41 N = 0.02441 kN
h_drop = 0.305 m
V_mold = 944 cm³ = 944 × 10⁻⁶ m³
E_standard = (25 × 0.02441 kN × 0.305 m × 3) / (944 × 10⁻⁶ m³)
= (25 × 0.02441 × 0.305 × 3) / (944 × 10⁻⁶)
= (0.5593 kN·m) / (944 × 10⁻⁶ m³)
= 592.5 kJ/m³ ≈ 591 kJ/m³ ✓
(b) Modified Proctor energy:
W_hammer = 4.54 × 9.81 = 44.54 N = 0.04454 kN
h_drop = 0.457 m
E_modified = (25 × 0.04454 × 0.457 × 5) / (944 × 10⁻⁶)
= (25 × 0.04454 × 0.457 × 5) / (9.44 × 10⁻⁴)
= (2.543 kN·m) / (9.44 × 10⁻⁴ m³)
= 2,694 kJ/m³ ≈ 2,693 kJ/m³ ✓
Energy ratio = E_modified / E_standard = 2,694 / 592.5 = 4.55 ≈ 4.5×
This confirms the commonly cited '4.5× more energy' of Modified vs Standard Proctor. Board exams may ask you to compute this ratio or to state it qualitatively. The formula E = N·W·h·n_layers/V_mold is derivable from first principles (total work done per unit volume of soil). Always verify with known approximate values: Standard ≈ 591 kJ/m³, Modified ≈ 2,693 kJ/m³.
Exam preparation tips
- MEMORIZE THE TWO CORE FORMULAS: γ_d = γ/(1+w) and γ_zav = G_s·γ_w/(1+w·G_s). These appear in nearly every compaction problem. Write them from memory 10 times before the exam.
- PROCTOR TEST SPECIFICATIONS: Know hammer masses (Standard: 2.49 kg; Modified: 4.54 kg) and layer counts (Standard: 3; Modified: 5). The board sometimes asks for the energy ratio (~4.5×). Drop heights: Standard = 305 mm; Modified = 457 mm.
- DIRECTION OF SHIFT: Modified vs Standard — HIGHER γ_d,max, LOWER OMC. This is a classic multiple-choice trap. More energy = better compaction at drier condition.
- ZAV LINE SANITY CHECK: Always verify that any computed γ_d is below γ_zav at the same w. If γ_d > γ_zav, your answer is wrong — recheck arithmetic immediately.
- RELATIVE COMPACTION vs RELATIVE DENSITY: RC for cohesive/general soils (Proctor reference); D_r for granular soils (e_max, e_min reference). The board distinguishes these; know when each applies.
- RC > 100% IS VALID: Do not assume RC > 100% is an error. Heavy vibratory rollers can exceed lab Proctor energy. If the result is, say, 105%, state it exceeds the specification — which is acceptable from a quality standpoint.
- SAND CONE SEQUENCE: Memorize the 4-step sequence: (1) V_hole from sand mass/γ_sand, (2) γ_moist from soil mass/V_hole, (3) γ_d from γ_moist/(1+w), (4) RC = γ_d/γ_d,max. Practice this at least 5 times.
- UNIT CONSISTENCY: Use kN, m, and kPa throughout. Convert cm³ to m³ (÷ 10⁶) and grams to kg (÷ 1000) before substituting into formulas. Mismatched units cause the most errors in board problems.
- DEGREE OF SATURATION AT OMC: At OMC, S is typically 70–85%, NOT 100%. Use S = w·G_s/e to compute it, first finding e from γ_d = G_s·γ_w/(1+e).
- BOARD EXAM STRUCTURE: Compaction typically appears as 2–4 problems in the Geotechnical Engineering portion. Expect one tabular Proctor problem, one ZAV calculation, and one RC/sand cone problem. Allocate ~5–8 minutes per problem.
- COMMON BOARD TRAPS: (a) Using moist unit weight where dry is needed; (b) forgetting to convert w% to decimal; (c) confusing Standard and Modified Proctor specifications; (d) thinking the ZAV line can be crossed by the compaction curve.
- PHILIPPINE CONTEXT: DPWH Blue Book (Standard Specifications for Highways, Bridges, and Airports) references Modified Proctor for subgrade and fill compaction. Knowing the 90% and 95% RC thresholds and their applications (fill vs subgrade) is essential for practical board questions.
Diagrams
Compaction — Complete Concept Map
A comprehensive mind map showing all major compaction concepts and their relationships, organized hierarchically from the central topic outward. Use this as a study overview before diving into individual concepts.
Step-by-Step Proctor Test Analysis Procedure
A flowchart of the complete Proctor test procedure from sample preparation to final reporting, including the ZAV verification step. Follow this sequence in board exam multi-part problems to avoid skipping steps.
Sand Cone Field Density Test — Board-Exam Solution Flowchart
A step-by-step solution flowchart for sand cone field density test problems. This is the exact sequence to follow in board exam problems. The decision node at the end guides interpretation of RC values relative to specification.
Standard Proctor vs Modified Proctor — Comparison Decision Chart
A side-by-side comparison flowchart showing the key differences between Standard and Modified Proctor tests. The directional shifts (UP-LEFT for Modified) are the most commonly tested concept in board exams.
Soil State During Compaction — Dry Side to Wet Side Transitions
A state diagram showing how soil engineering properties change as water content moves from the dry side through optimum to the wet side of the compaction curve. Understanding these state transitions is critical for selecting field placement conditions.
Compactive Energy Comparison — Standard vs Modified Proctor
A pie chart visualizing the energy ratio between Modified and Standard Proctor tests. The Modified Proctor applies approximately 4.5 times more energy, clearly shown by the relative arc sizes. This visual reinforces why Modified Proctor yields higher γ_d,max.
Development of Compaction Standards — Historical Context
A timeline showing the historical development of compaction testing standards from Proctor's original 1933 work to present Philippine board exam relevance. Provides context for why two different Proctor standards exist and how Philippine practice evolved.
In summary
Compaction is a foundational geotechnical topic that bridges laboratory testing and field quality control — exactly the type of practical engineering knowledge the PRC Civil Engineer Licensure Examination demands. The three pillars of this chapter are: (1) the dry unit weight formula γ_d = γ/(1+w) as the fundamental measure of compaction quality; (2) the Proctor test, which establishes the target γ_d,max and OMC, with the Modified Proctor providing higher density at lower water content due to greater compactive energy; and (3) the ZAV line γ_zav = G_s·γ_w/(1+w·G_s) as the theoretical upper bound that no real compacted soil can exceed.
Relative compaction RC = γ_d,field/γ_d,max is the everyday quality control parameter in Philippine highway and earthwork construction, with DPWH specifying 90–95% of Modified Proctor for most applications. The sand cone test is the standard field method for measuring γ_d,field.
For the board exam, focus on: (a) mastering the three core formulas by repeated application; (b) knowing the Standard vs Modified Proctor parameter differences cold; (c) always verifying that computed γ_d values fall below the ZAV line; and (d) interpreting RC values relative to specifications clearly and completely. With consistent practice on the five board-style problems in this chapter and familiarity with the visual aids showing process flows and concept relationships, you will be well-prepared to earn full marks on compaction questions in the PRC examination.