Free PE Civil Geotechnical Practice Test

Answer: A. 56.7%

RQD counts only sound pieces longer than 100 mm (4 in.), then divides by the full run length.

Qualifying pieces: 300 + 250 + 180 + 120 = 850 mm

RQD = 850 ÷ 1,500 × 100 = 56.7%

Why not the others:

• B (68.0%): Uses an 80 mm cutoff instead of 100 mm (1,020 ÷ 1,500).
• C (73.9%): Divides the qualifying length by the recovered length (850 ÷ 1,150) instead of the run length.
• D (76.7%): This is core recovery: every recovered piece over the run length (1,150 ÷ 1,500).

If you missed it: Write both definitions side by side: recovery uses all recovered core; RQD uses only sound pieces longer than 100 mm. Both divide by the run length.

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §3.6.5.4 (Recovery) and §3.6.5.5 (Rock Quality Designation), pp. 3-44 to 3-47, including the soundness criteria on p. 3-47.

Answer: 24.0 blows/ft (accepted range: 23.9 to 24.1 blows/ft)

Blow counts scale inversely with energy, so more energy per blow means fewer blows. To convert to the 60% reference, multiply by the energy ratio:

N60 = N × (ER ÷ 60) = 18 × (80 ÷ 60) = 24.0 blows/ft

Common wrong answers:

• 13.5: Multiplying by 60/80 runs the correction backward. A hammer that delivers more energy produces lower field counts, so N60 must be higher than the field N.
• 18.0: Skips the energy correction. Automatic hammers usually deliver more than 60% energy, which is why the correction exists.

If you missed it: Remember the direction: a more efficient hammer means the field N is too low compared with the 60% reference, so the correction raises it.

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §3.7.1 (Energy Efficiency of Hammers), pp. 3-54 to 3-55.

Answer: A. Install an observation well or piezometer in the relevant stratum and take repeated readings after the level stabilizes

Drilling fluid added during mud rotary drilling disturbs the water level in the hole, and a reading taken right after drilling may not have stabilized. A properly installed observation well or piezometer, read more than once after equilibrium, measures the water level or pore pressure in the stratum you care about.

Why not the others:

• B (Use the 6 ft reading, since it was measured directly in the borehole): A single reading in a hole just filled with drilling fluid reflects the drilling, not necessarily the groundwater.
• C (Estimate the water table from where the soil samples change color): Color changes can hint at past water levels or oxidation, but they don't measure the current level or pressure.
• D (Use the shallowest depth at which a sample was described as wet): "Wet" descriptions are qualitative and can come from drilling water or perched water. They aren't a substitute for a stabilized measurement.

If you missed it: List the ways a borehole reading can be wrong (drilling fluid, time to stabilize, perched water, multiple aquifers) and the instrument that fixes each.

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §3.13 (Groundwater Measurements), including §3.13.3 Observation Wells and §3.13.4 Water Level Measurements, pp. 3-78 to 3-82.

Answer: D. SW

Fines are under 5%, so no dual symbol is needed. Sand outweighs gravel in the coarse fraction, so the first letter is S.

Cu = D60 ÷ D10 = 1.8 ÷ 0.08 = 22.5

Cc = D30² ÷ (D60 × D10) = 0.45² ÷ (1.8 × 0.08) = 1.41

A well-graded sand needs Cu ≥ 6 and 1 ≤ Cc ≤ 3. Both are met, so the symbol is SW. (The full group name would add "with gravel" because gravel is at least 15%.)

Why not the others:

• A (GW): The first letter follows whichever coarse fraction is larger. Sand (70%) outweighs gravel (30%).
• B (SP): SP means poorly graded. That would need Cu < 6 or Cc outside 1 to 3, and neither applies here.
• C (SW-SM): Dual symbols such as SW-SM apply when fines are 5% to 12%. This soil has 3%.

If you missed it: Write the USCS fines thresholds (under 5%, 5–12%, over 12%) and the gravel and sand Cu limits (4 and 6) on one card.

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §4.2.1 (Unified Soil Classification System), Table 4-9 and its notes, pp. 4-15 to 4-16; Eqs. 4-1 and 4-2, pp. 4-18 to 4-19.

Answer: B. 76.8 kPa

Total stress adds up the weight of both layers:

σv = (2 × 18) + (4 × 20) = 116.0 kPa

The point is 4 m below the water table:

u = 4 × 9.81 = 39.24 kPa

Effective stress is total stress minus pore pressure:

σ′v = 116.0 − 39.24 = 76.8 kPa

Why not the others:

• A (39.2 kPa): This is the pore-water pressure alone.
• C (116.0 kPa): This is total stress. Pore pressure was never subtracted.
• D (155.2 kPa): Adds pore pressure instead of subtracting it.

If you missed it: Sketch three columns, total stress, pore pressure, effective stress, and fill them at the water table and at the point. Measure pore pressure from the water table, not the ground surface.

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §2.2 (Principle of Effective Stress, Eq. 2-13) and §2.3 (Overburden Pressure), pp. 2-19 to 2-21.

Answer: C. 128.9 pcf

For a saturated soil, S = 1, so e = wGs.

e = 0.22 × 2.70 = 0.594

γsat = (Gs + e)γw ÷ (1 + e) = (2.70 + 0.594)(62.4) ÷ 1.594 = 128.9 pcf

Why not the others:

• A (66.5 pcf): This is the buoyant (effective) unit weight, γsat − γw.
• B (105.7 pcf): This is the dry unit weight, Gs·γw ÷ (1 + e). It leaves out the water.
• D (149.4 pcf): Treats the void ratio as equal to the water content (e = 0.22). For a saturated soil, e = wGs.

If you missed it: Memorize one link, Se = wGs, and derive the unit weights from it rather than memorizing all of them.

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §2.1 (Basic Weight-Volume Relationships), Table 2-2.

Answer: A. Unconsolidated-undrained (UU) triaxial test

End of construction on soft saturated clay is a short-term, undrained case: the load goes on faster than the clay can drain. A UU test keeps drainage closed throughout, so it measures the undrained strength that controls at that moment.

Why not the others:

• B (Consolidated-drained (CD) triaxial test): CD tests give drained, long-term strength. That case matters later, after excess pore pressures dissipate.
• C (Drained direct shear test): Drained direct shear also gives long-term drained parameters, not the short-term undrained strength.
• D (Constant-head permeability test): Permeability tests measure flow, not strength.

If you missed it: Make a two-row table: short-term (undrained, total stress, UU) versus long-term (drained, effective stress, CD or CU with pore pressures).

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §5.5.2 (Strength Testing of Soils in the Laboratory) and Table 5-8 (shear strength issues for typical geotechnical features), pp. 5-43 and 5-46 to 5-47.

Answer: B. 3.5 × 10⁻³ cm/s

The falling-head equation comes from Darcy's law plus the drop in standpipe volume:

k = (aL ÷ At) · ln(h1 ÷ h2)

k = (1.0 × 30) ÷ (50 × 120) × ln(80 ÷ 40) = 0.005 × 0.693 = 3.5 × 10⁻³ cm/s

Why not the others:

• A (1.5 × 10⁻³ cm/s): Uses log base 10 instead of the natural log (0.005 × 0.301).
• C (5.0 × 10⁻³ cm/s): Leaves out the log term entirely.
• D (8.7 cm/s): Swaps the standpipe and specimen areas, so the area ratio is inverted.

If you missed it: Re-derive the falling-head equation once from Darcy's law. If you can derive it, you won't mix up the log base or the areas.

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §5.6 (Permeability), pp. 5-58 to 5-61. The equation shown is derived from Darcy's law and a volume balance on the standpipe.

Answer: C. 30.0°

At failure, σ′1 = 40 + 80 = 120 kPa. For a cohesionless soil, the Mohr circle touches a failure line through the origin:

sin φ′ = (σ′1 − σ′3) ÷ (σ′1 + σ′3) = 80 ÷ 160 = 0.5

φ′ = 30.0°

Why not the others:

• A (19.5°): Uses sin φ′ = σ′3 ÷ σ′1 (40 ÷ 120), which isn't the Mohr–Coulomb relationship.
• B (26.6°): Uses tan instead of sin: the circle's radius over its center gives sin φ′, not tan φ′.
• D (60.0°): This is the failure-plane angle, 45° + φ′/2, not the friction angle.

If you missed it: Draw one Mohr circle with its tangent line and label radius, center, and φ′. The geometry shows why it's sine.

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §5.5.2.2 (Triaxial Tests) and Appendix B (Mohr's circle).

Answer: A. 95.3%; passes

Relative compaction compares dry unit weights, so convert the field reading first:

γd = γ ÷ (1 + w) = 126 ÷ 1.12 = 112.5 pcf

RC = 112.5 ÷ 118 × 100 = 95.3%, which meets the 95% requirement.

This settles density only. Other requirements, such as a moisture window or lift thickness, still need their own check.

Why not the others:

• B (106.8%; passes): Divides the moist unit weight by the maximum dry unit weight, mixing wet and dry.
• C (94.0%; fails): Converts to dry unit weight by subtracting the water content, 126 × (1 − 0.12), instead of dividing by (1 + w).
• D (95.3%; fails): Computes correctly but misreads the criterion: 95.3% is above 95%.

If you missed it: Always convert the field reading to dry unit weight before comparing it with a lab maximum dry unit weight.

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §2.1 (weight-volume relationships) and §5.8.3 (relative compaction and field compaction specifications).

Answer: C. 1H:1V (45°)

Appendix A classifies cohesive soil with an unconfined compressive strength greater than 0.5 tsf but less than 1.5 tsf as Type B. Table B-1 in Appendix B allows a maximum slope of 1H:1V (45°) for Type B in excavations less than 20 ft deep.

The table gives a maximum: distress or surcharge loads can require a flatter actual slope.

Why not the others:

• A (Vertical): For a simple slope under Table B-1, vertical sides apply only to stable rock.
• B (¾H:1V (53°)): ¾H:1V applies to Type A, which needs at least 1.5 tsf.
• D (1½H:1V (34°)): 1½H:1V applies to Type C, including cohesive soil at 0.5 tsf or less, sand or loamy sand, and submerged or freely seeping soil.

If you missed it: Memorize the strength breakpoints (0.5 and 1.5 tsf) and the slopes (¾:1, 1:1, 1½:1). Also remember that excavations deeper than 20 ft need sloping designed by a registered professional engineer.

Source: 29 CFR 1926 Subpart P, Appendix A (Soil Classification), definitions of Type A, B, and C and paragraphs (c)(1)–(4); 29 CFR 1926 Subpart P, Appendix B (Sloping and Benching), Table B-1, including paragraph (c)(3) (actual slopes) and note 3 (excavations over 20 ft).

1 → b, 2 → a, 3 → c.

An inclinometer casing tilts as the ground moves sideways, so it profiles lateral movement with depth. A piezometer reads water pressure at its tip; comparison with baseline readings shows the excess pore pressure generated by the fill. A settlement plate rides down with the original ground surface, giving the amount and rate of settlement. Fill density (d) is a compaction test, not a monitoring instrument.

Why not the others:

• d: In-place density is checked with field compaction tests such as a nuclear gauge, not with these instruments.

If you missed it: For each instrument, say what physical quantity it senses and which construction decision it feeds, for example "piezometer, so I know whether to wait before the next lift."

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §7.9.1 (Embankment Construction Monitoring by Instrumentation), §7.9.1.2 (Types of Instrumentation), pp. 7-49 to 7-51.

Answer: B. 0.69

Cyclic stress ratio from the simplified procedure:

CSR = 0.65 · (amax ÷ g) · (σv ÷ σ′v) · rd

σv = 20 × 120 = 2,400 psf

u = 15 × 62.4 = 936 psf

σ′v = 2,400 − 936 = 1,464 psf

CSR = 0.65 × 0.30 × (2,400 ÷ 1,464) × 0.95 = 0.304

The earthquake magnitude matches the CRR basis (7.5), so no magnitude scaling is needed:

FS = CRR ÷ CSR = 0.21 ÷ 0.304 = 0.69. FS below 1 means liquefaction is predicted at this depth.

Why not the others:

• A (0.54): Measures pore pressure from the ground surface instead of the water table, so σ′v is too low and CSR too high.
• C (1.13): Leaves out the σv ÷ σ′v term, which understates the cyclic stress.
• D (1.45): Inverts the ratio, CSR ÷ CRR.

If you missed it: Write the CSR equation and say in words what each term does. Then check the magnitude: if it isn't 7.5, you need a magnitude scaling factor.

Source: FHWA-NHI-11-032, GEC 3: LRFD Seismic Analysis and Design of Transportation Geotechnical Features (2011), §6.3.4 (Evaluation of Liquefaction Potential), Eq. 6-10 on p. 6-30 for magnitude scaling, Eq. 6-16 on p. 6-36 for CSR, and p. 6-37 for the CRR/CSR ratio.

Answer: D. 1.06

Resolve the weight W and the horizontal force kh·W onto the slip plane.

Normal force: N = W cos β − kh W sin β

Driving force: T = W sin β + kh W cos β

FS = (cos β − kh sin β) tan φ′ ÷ (sin β + kh cos β)

= (0.940 − 0.15 × 0.342)(0.577) ÷ (0.342 + 0.15 × 0.940) = 1.06

The static factor of safety is tan 30° ÷ tan 20° = 1.59, so the earthquake load cuts the factor of safety by about a third and leaves almost no margin above 1.0.

Why not the others:

• A (1.59): This is the static factor of safety, with no seismic force at all.
• B (1.44): Subtracts kh from the static factor of safety, which isn't a valid shortcut.
• C (1.12): Adds kh to the driving force but forgets that it also reduces the normal force.

If you missed it: Draw the block with W, kh·W, N, and T and resolve each force onto the slip plane yourself. Don't memorize the final formula.

Source: FHWA-NHI-11-032, GEC 3: LRFD Seismic Analysis and Design of Transportation Geotechnical Features (2011), §6.2.2 (Limit Equilibrium Pseudo-Static Stability Analysis), p. 6-4. The infinite-slope expression is derived from force equilibrium on the stated model.

Answer: C. 240 m/s

Vs30 is 30 m divided by the time a shear wave takes to travel through the top 30 m.

t = 12 ÷ 150 + 18 ÷ 400 = 0.080 + 0.045 = 0.125 s

Vs30 = 30 ÷ 0.125 = 240 m/s

The slow upper layer controls because the wave spends most of its time there.

Why not the others:

• A (300 m/s): A thickness-weighted arithmetic average. It overstates the velocity by giving too little weight to slow layers.
• B (275 m/s): A simple arithmetic average of the two velocities.
• D (218 m/s): Averages the two velocities harmonically but ignores the layer thicknesses.

If you missed it: Practice with three layers and confirm you always sum travel times (thickness ÷ velocity) before dividing into 30 m.

Source: Thompson and others (2011), USGS publication on site-response mapping, Parkfield, California, abstract (Vs30 as 30 m divided by travel time to 30 m). See also FHWA-NHI-11-032, GEC 3: LRFD Seismic Analysis and Design of Transportation Geotechnical Features (2011), Chapter 4, for site characterization.

Answer: B. 0.82

The slope angle is tan β = 1 ÷ 2.5 = 0.40.

Dry factor of safety: tan φ′ ÷ tan β = 0.6745 ÷ 0.40 = 1.69

With parallel seepage at the surface, the factor drops by γ′ ÷ γsat:

γ′ = 122 − 62.4 = 59.6 pcf

FS = (59.6 ÷ 122) × 1.69 = 0.82

Seepage roughly halves the factor of safety, enough here to make the slope unstable.

Why not the others:

• A (0.40): Applies the γ′ ÷ γsat reduction twice.
• C (1.61): Uses γ′ ÷ γw instead of γ′ ÷ γsat.
• D (1.69): The dry-slope factor of safety, with seepage ignored.

If you missed it: Derive the seepage case once: the effective normal force uses buoyant weight, while the driving force uses saturated weight.

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §6.3 (Infinite Slope Analysis), §6.3.1 and §6.3.2 (c-φ soils with parallel seepage).

Answer: 102.1 mm (accepted range: 101.1 to 103.1 mm)

For normally consolidated clay:

Sc = [Cc · H ÷ (1 + e0)] · log10(σ′f ÷ σ′0)

Sc = [0.24 × 3,000 mm ÷ 1.80] × log10(180 ÷ 100) = 400 × 0.2553 = 102.1 mm

This is the final primary settlement only. It says nothing about how long it takes.

Common wrong answers:

• 235.1 mm: Uses the natural log instead of log base 10.
• −38.8 mm: Uses the stress increment (80 kPa) as if it were the final stress, which gives a log of a number below 1.

If you missed it: Write the equation with the units on every term, and check the ratio inside the log is final ÷ initial stress, not the increment.

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §7.5.2.1 (Normally Consolidated Soils), p. 7-24.

Answer: A and C

A and C. Consolidation time grows with the square of the drainage distance. Drains let water escape sideways to a nearby drain instead of traveling to the top or bottom of the layer, so settlement finishes much sooner. The amount of primary settlement depends on load and compressibility, which don't change.

Why not the others:

• B: Vertical drains carry water. They aren't reinforcement and don't reduce settlement magnitude.
• D: Cc is a soil property measured in a consolidation test. Drains don't change it.
• E: Piezometers are how you confirm that excess pore pressure is actually dissipating before adding more fill.

If you missed it: Separate the two questions in every consolidation problem: how much (magnitude) and how fast (rate). Then label which one each ground-improvement method changes.

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §7.7.2 (Reducing Settlement Time), §7.7.2.2 (Vertical Drains), pp. 7-46 to 7-47; §7.5.3 (Consolidation Rates).

1 → c (separation), 2 → a (filtration), 3 → b (reinforcement), 4 → d (drainage).

The difference between filtration and drainage is flow direction. In filtration, water crosses the fabric while soil stays behind. In drainage, water moves along the plane of the material to an outlet. Separation keeps two layers apart. Reinforcement carries tension.

If you missed it: For each function, name the direction of water flow (if any) and what failure it prevents: mixing, piping, sliding, or water build-up.

Source: FHWA NHI-05-037, Geotechnical Aspects of Pavements, Ch. 7 Part A, §7.2.9 (geocomposite drains), §7.2.10 (separator layers), §7.2.16 (geotextile separation and filter design); FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §6.9.2 (Reinforced Soil Slopes).

Answer: D. Restore the outlet pipes so water collected in the base has somewhere to go

A drainage system is a chain: base, collector, outlet. The base is fine. The chain is broken at the outlets, so water has no exit and backs up into the base. Crushed or clogged outlets are a common reason drainage systems fail, which is why they need inspection and maintenance.

Why not the others:

• A (Add asphalt thickness so less water reaches the base): Thicker asphalt doesn't remove water already trapped in the base, and it doesn't fix the outlets.
• B (Replace the base with a more permeable aggregate): The aggregate already meets its spec. More permeability can't help if water can't leave the system.
• C (Nothing; a permeable base drains itself regardless of outlets): A permeable base still needs a free path to an outlet.

If you missed it: Trace water from the surface to its final outlet in a pavement section, and list where the path can be blocked.

Source: FHWA NHI-05-037, Geotechnical Aspects of Pavements, Ch. 7 Part A, §7.2.4 to §7.2.6 (edge drains, outlets, and maintenance).

Answer: B. 0.26 m³/day per m

Each flow channel carries the same flow. Combining Darcy's law with the flow-net geometry gives:

q = k · H · (Nf ÷ Nd)

q = (2 × 10⁻⁶ m/s)(4.5 m)(4 ÷ 12) = 3.0 × 10⁻⁶ m³/s per m

× 86,400 s/day = 0.26 m³/day per m

Why not the others:

• A (0.026 m³/day per m): Uses 8,640 s/day instead of 86,400 s/day in the time conversion.
• C (0.78 m³/day per m): Ignores the flow-net shape factor Nf ÷ Nd.
• D (2.3 m³/day per m): Inverts the shape factor, using Nd ÷ Nf.

If you missed it: Use the number of flow channels divided by the number of equipotential drops. Check that the units give discharge per unit wall length, and use 86,400 seconds per day.

Source: USBR, Design Standards No. 13, Chapter 8: Seepage (2014), §8.4.5 (Flow Nets), pp. 8-53 to 8-56, especially Figures 8.4.5-1 and 8.4.5-2. This is a teaching reference, not one of the NCEES-listed exam standards.

Answer: B. 2.73

Upward seepage removes effective stress. It reaches zero at the critical gradient:

ic = γ′ ÷ γw = (122 − 62.4) ÷ 62.4 = 0.955

FS = ic ÷ iexit = 0.955 ÷ 0.35 = 2.73

Why not the others:

• A (0.37): Inverts the ratio, iexit ÷ ic.
• C (2.86): Assumes a critical gradient of exactly 1.0, a common rule of thumb, instead of computing γ′ ÷ γw = 0.955.
• D (5.59): Uses γsat ÷ γw instead of γ′ ÷ γw for the critical gradient.

If you missed it: Derive ic by setting effective stress to zero at the base of a soil column with upward flow. One line of algebra shows why it's γ′ ÷ γw.

Source: USBR, Design Standards No. 13, Chapter 8: Seepage (2014), §8.2.2.2 (High Exit Gradients in a Cohesionless Soil), pp. 8-5 to 8-6. This is a teaching reference, not one of the NCEES-listed exam standards.

Answer: A. It increases by about 29 kPa, which can cause new consolidation settlement under the building

Once pore pressures equilibrate, effective stress rises by however much the pore pressure falls. With total stress unchanged, lowering the water table 3 m gives:

Δu = −3 × 9.81 = −29.4 kPa

Δσ′v = Δσv − Δu = 0 − (−29.4) = +29.4 kPa

So σ′v rises by about 29 kPa. Clay compresses under added effective stress just as it would under a new load, so the building can settle even though nobody touched it. That's why dewatering plans often include settlement monitoring of neighboring structures.

Why not the others:

• B (It decreases by about 29 kPa, so the clay swells and the building heaves): Gets the sign backward. Lower pore pressure means higher effective stress.
• C (It doesn't change, because the building load hasn't changed): The soil's stress state changes even when the structural load doesn't.
• D (It increases by about 59 kPa, which doubles the building's bearing capacity): Doubles the pore-pressure change, and bearing capacity isn't the concern here; settlement is.

If you missed it: For any change in groundwater, write Δσ′ = Δσ − Δu and decide the sign of each term before calculating.

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §2.2 (Principle of Effective Stress, Eq. 2-13) and §7.5 (consolidation settlement).

Answer: A, B, and C

A, B, and C. Ice lenses grow when cold penetrates soil that can pull water up to the freezing front. Take away any one of the three, and heave won't develop. Cold weather alone isn't enough.

Why not the others:

• D: Earthquakes have nothing to do with ice-lens formation.
• E: A treated surface layer isn't a requirement for heave. Treatment is one way to manage the problem.

If you missed it: For each condition, name one way designers remove it: non-frost-susceptible fill, insulation or depth, and drainage or capillary breaks.

Source: FHWA NHI-05-037, Geotechnical Aspects of Pavements, Ch. 7 Part C, §7.5.6 (Frost-Susceptible Soils).

Answer: D. X shows swelling (expansive soil); Y shows wetting-induced collapse

Expansive clays take on water and grow in volume, which can lift slabs and pavements. Collapsible soils have an open structure held together by weak bonds, such as clay binders or soluble cements, that give way on wetting, so the soil drops suddenly. Same trigger, opposite movement.

Why not the others:

• A (Both samples are collapsible): Only Y collapsed. X gained volume.
• B (X shows collapse; Y shows swelling): Reverses the two behaviors.
• C (Both responses are ordinary primary consolidation): Primary consolidation is volume change from load as excess pore pressure dissipates. Both responses here were caused by wetting at constant load.

If you missed it: Make a two-column card: expansive versus collapsible, with typical soil, trigger, movement direction, and one mitigation for each.

Source: FHWA NHI-05-037, Geotechnical Aspects of Pavements, Ch. 7 Part C, §7.5.3 (Collapsible Soils) and §7.5.4 (Swelling Soils); FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §5.7 (Volume Change Phenomena).

Answer: 6,960 lb/ft (accepted range: 6,890 to 7,030 lb/ft)

Ka = (1 − sin 32°) ÷ (1 + sin 32°) ≈ 0.30726

Effective vertical stress: 6 × 120 = 720 psf at the water table; 720 + 10 × (125 − 62.4) = 1,346 psf at the base.

Soil force (effective stress × Ka):

• Top triangle: ½ × 0.30726 × 720 × 6 = 664 lb/ft
• Rectangle below the water table: 0.30726 × 720 × 10 = 2,212 lb/ft
• Lower triangle: ½ × 0.30726 × (1,346 − 720) × 10 = 962 lb/ft
• Soil subtotal = 3,838 lb/ft

Water force: ½ × 62.4 × 10² = 3,120 lb/ft

Using unrounded values, total = 3,837.66 + 3,120 = 6,957.66 lb/ft ≈ 6,960 lb/ft

Common wrong answers:

• 4,720 lb/ft: Ignores the water table: uses 120 pcf over the full 16 ft with no water force.
• 3,840 lb/ft: Computes the soil force correctly but leaves out water pressure.
• 4,800 lb/ft: Multiplies water pressure by Ka, as if water were soil. Water pressure acts equally in all directions.

If you missed it: Draw the pressure diagram in two parts, effective soil pressure and water pressure, before adding anything. Never multiply water pressure by Ka.

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §2.9 (Eq. 2-23 for Ka) and §2.9.1 (Distribution of Lateral Earth and Water Pressures); FHWA NHI-06-089, Soils and Foundations Reference Manual, Vol. II (2006), §10.2.2 and §10.3 (Lateral Pressures Due to Water).

Answer: B. 2.31 m

Soil pressure is triangular:

Psoil = ½ × (1/3) × 18 × 6² = 108 kN/m, acting at H/3 = 2.00 m

Surcharge pressure is rectangular (Ka × q):

Pq = (1/3) × 24 × 6 = 48 kN/m, acting at H/2 = 3.00 m

Take moments about the base:

ȳ = (108 × 2.00 + 48 × 3.00) ÷ 156 = 2.31 m

Why not the others:

• A (2.00 m): The location of the soil resultant alone, as if there were no surcharge.
• C (2.69 m): Swaps the centroids: puts the triangular soil force at H/2 and the rectangular surcharge force at H/3.
• D (3.00 m): The location of the surcharge resultant alone.

If you missed it: Treat each pressure shape (triangle, rectangle) as its own force at its own centroid, then combine with a moment balance.

Source: FHWA NHI-06-089, Soils and Foundations Reference Manual, Vol. II (2006), §10.2.2 (Active and Passive Lateral Earth Pressures) and §10.4 (lateral pressure from surcharge, Δph = K·qs).

Answer: B. 0.50 (at-rest, Ko)

Active pressure develops only when the wall moves away from the soil. A restrained wall stays near the at-rest condition:

Ko = 1 − sin φ′ = 1 − sin 30° = 0.50

Using Ka here would underestimate the load by about a third.

Why not the others:

• A (0.33 (active, Ka)): Ka assumes the wall moves enough to mobilize active pressure, which the slabs prevent.
• C (1.00): K = 1 is not the at-rest value from Jaky's relationship for this normally consolidated sand.
• D (3.00 (passive, Kp)): Passive pressure develops when the wall is pushed into the soil, the opposite case.

If you missed it: Sketch the K-versus-wall-movement curve: passive on one side, at rest in the middle, active on the other. Then ask how much the wall can actually move.

Source: FHWA NHI-06-089, Soils and Foundations Reference Manual, Vol. II (2006), §10.2 and Figure 10-4 (effect of wall movement on wall pressures), §10.2.1 (At-Rest Lateral Earth Pressure, Eq. 10-2).

Answer: D. FS ≈ 1.27; the wall does not meet the 1.5 sliding criterion

Sliding resistance is base friction:

FS = (N · tan δ) ÷ Pa = (9,000 × 0.45) ÷ 3,200 = 1.27

That's below the required 1.5, so the wall needs more resistance, such as a wider base or a shear key, or a reduced load.

Why not the others:

• A (FS ≈ 0.79; the wall fails in sliding): Inverts the ratio, driving over resisting.
• B (FS ≈ 1.27; the wall is fine because FS is above 1.0): Uses 1.0 as the criterion. The project requires 1.5.
• C (FS ≈ 1.90; the wall meets the criterion): Multiplies the result by 1.5 instead of comparing against it.

If you missed it: Say the check in words: resisting force over driving force, compared with the required minimum. Then compute.

Source: FHWA NHI-06-089, Soils and Foundations Reference Manual, Vol. II (2006), §10.5.5 (Evaluate Overturning and Sliding), p. 10-43, which lists FS ≥ 1.5 for sliding.

Answer: C. 44.9 kips

Apparent pressure:

p = 0.65 × 0.30 × 120 × 24 = 561.6 psf

Strut load = pressure × tributary area:

F = 561.6 × 8 × 10 = 44,928 lb = 44.9 kips

Why not the others:

• A (5.6 kips): Uses only the horizontal spacing and drops the tributary height.
• B (22.5 kips): Uses half the tributary height (4 ft).
• D (89.9 kips): Doubles the tributary height to 16 ft, counting the full strut spacing above and below.

If you missed it: For any support load from an apparent pressure diagram: pressure × tributary height × horizontal spacing. Check the tributary heights add up to the wall height.

Source: FHWA NHI-06-089, Soils and Foundations Reference Manual, Vol. II (2006), §10.2.6 (Semi-Empirical Lateral Earth Pressure Diagrams). The 0.65 Ka γ H envelope is supplied as an exercise input; use the diagram your reference specifies for the soil type.

Answer: A and C

A and C. An anchor resists movement only if its grip is in ground that isn't part of the moving soil wedge. The bond length goes beyond the critical failure surface, and the unbonded (free) length passes through the potentially moving zone without gripping it.

Why not the others:

• B: A bond zone inside the active wedge moves with the soil it's meant to hold back.
• D: Anchor loads come from the earth pressure on each anchor's tributary area, so spacing matters directly.
• E: Bonding inside the moving zone transfers load into soil that's moving, which defeats the purpose of the free length.

If you missed it: Sketch a wall, the critical failure surface, and one anchor. Label the free length and the bond length and explain each in one sentence.

Source: FHWA-IF-99-015, GEC 4: Ground Anchors and Anchored Systems (1999), Chapter 2 (anchor components, p. 5) and §5.3.3 (horizontal anchor loads from apparent pressure diagrams, pp. 65 to 67). GEC 4 is a teaching reference, not one of the NCEES-listed exam standards.

Answer: C. 3,200 psf

Shape factor: sc = 1 + 6/(5 × 6) = 1.20

qult = su · Nc · sc + γ · Df · Nq = 1,500 × 5.14 × 1.20 + 120 × 3 × 1.0 = 9,252 + 360 = 9,612 psf

qall = 9,612 ÷ 3 = 3,204 psf ≈ 3,200 psf

Why not the others:

• A (2,690 psf): Leaves out the shape factor.
• B (3,080 psf): Computes net allowable pressure, (qult − γDf) ÷ 3, instead of the requested gross allowable pressure.
• D (9,610 psf): The ultimate capacity, with no factor of safety.

If you missed it: Read the prompt for what the factor of safety applies to, gross or net, and whether depth and shape factors are included. Those words change the answer.

Source: FHWA NHI-06-089, Soils and Foundations Reference Manual, Vol. II (2006), §8.4.2 (bearing-capacity framework), Eq. 8-1, and Table 8-4 (shape correction factors), φ = 0 row, p. 8-27. The factor of safety and ignored depth factors are supplied as exercise inputs.

Answer: C. 125 kPa

Only the loaded dimension is reduced:

B′ = 3.0 − 2(0.30) = 2.4 m; L′ = 4.0 m

q = 1,200 ÷ (2.4 × 4.0) = 125 kPa

This is the equivalent uniform pressure used with effective dimensions. It is not the peak contact pressure at the toe.

Why not the others:

• A (160 kPa): This is the maximum toe pressure from the linear formula q(1 + 6e/B). It's a different quantity.
• B (147 kPa): Reduces both dimensions by 2e, though the load is centered along the 4 m side.
• D (100 kPa): Uses the full footing area and ignores the eccentricity.

If you missed it: For eccentric loads, check which direction the eccentricity acts and reduce only that dimension.

Source: FHWA NHI-06-089, Soils and Foundations Reference Manual, Vol. II (2006), §8.4.3.1 (Footing Shape: Eccentricity and Effective Dimensions), Eqs. 8-7 and 8-8, p. 8-25.

Answer: 617 psf (accepted range: 611 to 623 psf)

The 2:1 method spreads the load over an area that grows by z in each plan dimension:

Δσ = Q ÷ [(B + z)(L + z)] = 200,000 ÷ [(8 + 10)(8 + 10)] = 200,000 ÷ 324 = 617 psf

Common wrong answers:

• 3,125 psf: Uses the footing area (8 × 8) with no spreading.
• 1,183 psf: Spreads by only z/2 in each dimension (8 + 5 = 13 ft).

If you missed it: Measure z from the footing base, not the ground surface, and add z (not 2z) to each dimension.

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §2.4.2 (Approximate (2:1) Stress Distribution Concept), Figure 2-10.

Answer: B. 2.6 in.

The final stress passes the preconsolidation stress, so the curve has two pieces: recompression up to σ′p, then virgin compression beyond it.

Sc = [H ÷ (1 + e0)] · [Cr log10(σ′p ÷ σ′0) + Cc log10(σ′f ÷ σ′p)]

= (96 in. ÷ 1.95) × [0.05 log10(2,500 ÷ 1,500) + 0.35 log10(3,300 ÷ 2,500)]

= 49.2 × [0.0111 + 0.0422] = 2.6 in.

Why not the others:

• A (0.8 in.): Uses Cr for the whole stress range, ignoring that loading goes past σ′p.
• C (5.1 in.): Leaves out the (1 + e0) term, using the 96 in. thickness directly.
• D (5.9 in.): Uses Cc for the whole range, as if the clay were normally consolidated.

If you missed it: Before using any settlement equation, place σ′0, σ′p, and σ′f on one number line. That tells you which index applies to which segment.

Source: FHWA NHI-06-088, Soils and Foundations Reference Manual, Vol. I (2006), §7.5.2.2 (Overconsolidated Soils), Eq. 7-4, p. 7-26.

Answer: C. 205.0 kips

Perimeter = 4 × 14/12 = 4.667 ft; toe area = (14/12)² = 1.361 ft²

Shaft: Qs = α · su · perimeter · L = 0.8 × 1,000 × 4.667 × 50 = 186,700 lb

Toe: Qt = 9 · su,toe · A = 9 × 1,500 × 1.361 = 18,400 lb

Using unrounded values, Qu = 186.6667 + 18.3750 = 205.0417 kips ≈ 205.0 kips

Why not the others:

• A (186.7 kips): Shaft resistance only; the toe was left out.
• B (198.9 kips): Uses the shaft su (1,000 psf) at the toe instead of the toe stratum's 1,500 psf.
• D (251.7 kips): Uses α = 1.0, the full undrained strength, as shaft adhesion.

If you missed it: Write capacity as two lines, shaft and toe, each with its own area and strength. Most errors are a perimeter used as an area or the wrong stratum at the toe.

Source: FHWA NHI-06-089, Soils and Foundations Reference Manual, Vol. II (2006), §9.4, Eq. 9-2, p. 9-22; §9.5.2.1, Eq. 9-8 (α-method), p. 9-47, and Eq. 9-9 (toe resistance), p. 9-51; summation procedure, p. 9-52.

Answer: C. 44 kips

Unit negative shaft resistance: fs = β · σ′v = 0.25 × 1,250 = 312.5 psf

Perimeter = π × 1.5 ft = 4.712 ft

Drag load = 312.5 × 4.712 × 30 = 44,200 lb ≈ 44 kips

The drag load acts downward. It adds to the load the pile must carry. It doesn't help resist it.

Why not the others:

• A (1.5 kips): Stops at the drag per foot of pile (fs × perimeter) and never multiplies by the 30 ft length.
• B (22 kips): Uses the radius instead of the diameter in the perimeter (π × 0.75 ft).
• D (177 kips): Leaves out β and uses the full effective stress as shaft resistance.

If you missed it: Find the neutral plane first. Shaft friction above it acts downward (drag); below it acts upward (resistance). Keep the signs straight.

Source: FHWA NHI-06-089, Soils and Foundations Reference Manual, Vol. II (2006), §9.5.2.2, Eq. 9-12 (β-method), p. 9-53, and §9.8 (Downdrag or Negative Shaft Resistance), p. 9-87.

Answer: A. Neither alternative

Check each limit separately:

• Alternative 1: moment passes (45 ≤ 60), but deflection fails (16 > 10).
• Alternative 2: deflection passes (8 ≤ 10), but moment fails (70 > 60).

Neither meets both limits. Both the deflection limit and the bending-moment limit must be satisfied.

Why not the others:

• B (Alternative 1 only): Alternative 1 exceeds the 10 mm deflection limit.
• C (Alternative 2 only): Alternative 2 exceeds the 60 kN·m moment limit.
• D (Both alternatives): Each alternative fails one of the two limits.

If you missed it: List every acceptance criterion before reading results, then check each one. Don't stop at the first one that passes.

Source: FHWA NHI-06-089, Soils and Foundations Reference Manual, Vol. II (2006), §9.4, p. 9-22 (geotechnical and structural capacity; deformation results and performance criteria), and §9.7, p. 9-84 (general lateral-load analysis context). The limits and analysis results are supplied as exercise inputs.

Answer: D. 0.625 in.

The Davisson line is the reference elastic-compression line, PL/AE, plus a fixed offset:

Δ = PL ÷ AE + 0.15 in. + b ÷ 120

PL ÷ AE = (300 kips × 720 in.) ÷ (144 in² × 4,000 ksi) = 0.375 in.

Offset = 0.15 + 12/120 = 0.25 in.

Δ = 0.625 in. If the measured movement on the loading curve at 300 kips exceeds 0.625 in., the curve has reached the Davisson offset-limit criterion at or below 300 kips. This criterion does not give an allowable load by itself.

Why not the others:

• A (0.250 in.): The offset alone, without elastic shortening.
• B (0.375 in.): Elastic shortening alone.
• C (0.525 in.): Drops the b/120 term.

If you missed it: For piles less than 24 in. in diameter or width, write this Davisson line as "elastic plus 0.15 in. plus b/120" and check units: b and L in inches, E in ksi, P in kips.

Source: FHWA NHI-06-089, Soils and Foundations Reference Manual, Vol. II (2006), §§9.15.7.4–9.15.7.5, Eqs. 9-51 and 9-52, p. 9-167; Figure 9-68a, p. 9-166. The 0.15 in. + b/120 offset applies to the manual's less-than-24-in. pile case.

Answer: A and C

A and C. CSL times sound waves traveling between pairs of tubes, so it checks concrete integrity only along those paths. It does not measure the soil's strength or establish the shaft's axial geotechnical resistance.

Why not the others:

• B: Concrete outside the tube cage isn't on any tested path, so CSL doesn't evaluate it.
• D: Integrity and capacity are different questions. Capacity needs load testing or analysis.
• E: CSL signals travel through the shaft concrete, not the soil below the toe.

If you missed it: For each integrity or load test you know, write two lines: what it actually measures, and what people wrongly assume it proves.

Source: FHWA NHI-06-089, Soils and Foundations Reference Manual, Vol. II (2006), §9.14.1 (The Standard Crosshole Sonic Logging Test), pp. 9-146 to 9-148, and §9.14.3 (coverage limits), p. 9-152.