Geotechnical Engineering is 10–15 questions on soil as a material and as a foundation: phase relations, classification, effective stress, settlement, strength, bearing capacity, retaining structures and slopes. The problems below mix short calculations with concept questions.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Civil
questions on the exam10–15
problems here18
of 14 sections12
Checked
What the section covers
The July 2020 specification lists eleven geotechnical topics:
Index properties and soil classification.
Phase relations.
Laboratory and field tests.
Effective stress.
Stability of retaining structures: active, passive and at-rest pressure.
Shear strength.
Bearing capacity.
Foundation types: spread and wall footings, mats, deep foundations.
Consolidation and differential settlement.
Slope stability of fills, embankments, cuts and dams.
Soil stabilization with chemical additives or geosynthetics.
18 geotechnical problems
0 / 18 answered · 0 correct
Question 1 of 18 · Geotechnical Engineering · 10–15 items
Match each project scenario below with the correct shear-strength parameter selection and drainage condition assumption: (A) Rapid excavation in saturated clay (short-term stability), (B) Long-term earth slope subject to seasonal wetting, (C) Sliding on a previously sheared clay seam in an old landslide.
Solution
Short term in saturated clay → total stress, su (φ = 0). Long term → effective stress c′, φ′. Reactivated slide plane → residual φr. FE Reference Handbook, Geotechnical — shear strength.
Seasonal wetting of an intact slope calls for drained parameters, not residual; and (C), the pre-sheared seam, is the case that needs residual.
(A) and (B) are right, but a previously sheared seam has lost its peak strength, so peak φ′ for (C) is unconservative.
Rapid loading of saturated clay is undrained, a long-term slope is drained (effective stress), and an old slip surface is at residual strength.
Residual strength belongs to the pre-sheared seam (C), and the short-term cut (A) needs undrained strength; every pairing here is shifted.
The drainage cases are swapped: the rapid cut is undrained and the long-term slope is drained.
Question 2 of 18 · Geotechnical Engineering · 10–15 items
A uniform soil deposit has a total unit weight of 120 lb/ft³ both above and below the groundwater table, which lies 10 ft below the ground surface. What is the vertical effective stress at a depth of 20 ft? (Unit weight of water = 62.4 lb/ft³)
Given
Total unit weight γ = 120 lb/ft³ (above and below the water table)
Depth to groundwater table = 10 ft
Depth of point z = 20 ft
Unit weight of water γw = 62.4 lb/ft³
Find
Vertical effective stress σ′ (lb/ft²)
Solution
σ = γz = 120 × 20 = 2,400 lb/ft²
u = γw(z − zw) = 62.4 × 10 = 624 lb/ft²
σ′ = σ − u = 1,776 lb/ft²
FE Reference Handbook, Geotechnical — effective stress.
σ = 120 × 20 = 2400 lb/ft² and u = 62.4 × 10 = 624 lb/ft², so σ′ = 2400 − 624 = 1776 lb/ft².
1152 lb/ft² uses the buoyant weight (120 − 62.4) over the full 20 ft, as if the water table were at the surface.
624 lb/ft² is the pore water pressure u = 62.4 × 10, not the effective stress.
576 lb/ft² counts only the buoyant weight of the 10 ft below the water table and leaves out the 10 ft above it.
Question 3 of 18 · Geotechnical Engineering · 10–15 items
According to Rankine's theory, what is the active earth pressure coefficient (Ka) for a cohesionless soil with an internal friction angle (φ) of 30°?
Given
Friction angle φ = 30°
Cohesionless soil (c = 0)
Find
Rankine active earth pressure coefficient Ka
Solution
Ka = (1 − sin φ)/(1 + sin φ) = tan²(45° − φ/2) = 0.333
FE Reference Handbook, Geotechnical — Rankine earth pressure.
Ka = (1 − sin 30°)/(1 + sin 30°) = 0.5/1.5 = 0.333, equal to tan²(45° − φ/2).
0.500 is sin 30° itself (or K0 = 1 − sin φ), not the Rankine active coefficient.
0.866 is cos 30°; it does not enter the Rankine coefficient for a level backfill.
3.000 is the passive coefficient Kp = 1/Ka; the active and passive states are swapped.
Question 4 of 18 · Geotechnical Engineering · 10–15 items
A soil sample has a wet unit weight of 120 pcf and a water content of 15%. What is its dry unit weight?
Given
Wet (moist) unit weight γ = 120 pcf
Water content w = 15 %
Find
Dry unit weight γd (pcf)
Solution
γd = γ/(1 + w) = 120/(1 + 0.15) = 104.35 pcf
FE Reference Handbook, Geotechnical — phase relationships.
18 pcf is 0.15 × 120, the water fraction alone; it is not a unit weight of the soil skeleton.
γd = γ/(1 + w) = 120/1.15 = 104.35 pcf, because water content is a fraction of the dry weight.
Subtracting 15 pcf (treating 15 % as 15 pcf) gives 105; water content is not a unit weight.
138 pcf is 120 × 1.15; the moist weight was multiplied by (1 + w) instead of divided.
Question 5 of 18 · Geotechnical Engineering · 10–15 items
A clay seam along a bedding plane in an old landslide has been previously sheared. Laboratory direct-shear tests show peak shear strength with φpeak = 32° and residual shear strength with φr = 18°. The seam dips at 25° and is subject to long-term groundwater seepage. For a planar sliding stability analysis, should peak or residual friction angle be used, and how does the reduction from φpeak to φr affect the factor of safety?
Solution
Infinite/planar slope, c′ = 0, seepage ignored for the ratio:
FS ∝ tan φr/tan β; FSr/FSpeak = tan 18°/tan 32° = 0.325/0.625 = 0.52
With φr = 18° < β = 25°, the dry slope is already below FS = 1. FE Reference Handbook, Geotechnical — slope stability.
A lower friction angle cannot raise FS, and a pre-sheared seam has lost essentially all cohesion, so neither half holds.
On a residual seam c′ ≈ 0, so friction governs and dropping from 32° to 18° changes FS directly.
Averaging peak and residual has no basis; the seam is already at residual.
A reactivated slip surface is at residual strength; with c′ ≈ 0, FS ∝ tan φ/tan β, so it falls by about tan 18°/tan 32° ≈ 0.52.
Peak strength is gone once the seam has been sheared; using 32° overstates resistance on a reactivated plane.
Question 6 of 18 · Geotechnical Engineering · 10–15 items
In the context of foundation engineering, what does the term 'negative skin friction' refer to?
Solution
Downdrag adds load above the neutral plane: Qmax = Qapplied + Qnsf. FE Reference Handbook, Geotechnical — deep foundations.
Bitumen coating is a remedy that reduces downdrag; it is not what the term means.
When surrounding soil settles more than the pile (fill over soft clay), friction acts downward and adds load instead of resisting it.
Frost heave pushes upward through adfreeze; negative skin friction is a downward drag.
Scour removes soil around a foundation; it is a different mechanism.
Question 7 of 18 · Geotechnical Engineering · 10–15 items
A highway subgrade is a high-plasticity clay (LL = 62, PI = 38, 87 % passing the No. 200 sieve, natural moisture 24 %, free swell 9 %) 30 in deep. The site sees freeze–thaw cycles; lime and cement are readily available, fly ash is not. The design calls for a soaked CBR of 12 after stabilization. Which stabilization method fits, and what must the trial sections verify?
Solution
Selection by soil type: PI > about 20 with high fines → lime; low PI / granular → cement. Lime trial sections check soaked CBR, swell, durability and depth of mix. FE Reference Handbook, Geotechnical — soil classification (AASHTO/USCS).
Blending aggregate does not cure the swell or plasticity of a PI-38 clay; it suits granular or low-plasticity soils.
Lime reacts with high-PI clay (cation exchange plus pozzolanic gain), lowering PI and swell; trials must confirm strength, swell and durability.
Fly ash is not reliably available here, so a design that depends on it is not the fit for this site.
Cement works best in granular and low-PI soils; in a PI-38 clay it is hard to mix uniformly and does little for swell.
Geogrid adds tensile reinforcement but leaves the clay's swell and moisture sensitivity untreated.
Question 8 of 18 · Geotechnical Engineering · 10–15 items
A uniformly loaded square footing, 8 ft × 8 ft, is placed at 4 ft depth on a single 12-ft-thick saturated clay layer. The water table is at the ground surface. An instantaneous stress increase of 1,500 psf is imposed by the footing. The clay layer rests on an impervious bedrock. The coefficient of consolidation for the clay is cv = 0.32 ft2/day. Time factor for 90% consolidation is Tv90 = 0.848. Determine the time in days required to reach 90% primary consolidation.
Given
Clay layer thickness H = 12 ft (saturated)
Water table at the ground surface; base of the layer: impervious bedrock
Coefficient of consolidation cv = 0.32 ft²/day
Time factor Tv at 90 % = 0.848
Find
Time to 90 % primary consolidation t (days)
Solution
t = TvHdr²/cv
Hdr = 12 ft (one-way drainage to the top)
t90 = 0.848 × (12 ft)²/(0.32 ft²/day) = 382 days
FE Reference Handbook, Geotechnical — consolidation, time rate.
32 days comes from TvHdr/cv; the drainage path must be squared.
89 days uses Tv = 0.197, the time factor for 50 % consolidation, not 90 %.
95 days takes Hdr = 6 ft (double drainage); impervious bedrock leaves only the top boundary draining.
170 days takes Hdr = 8 ft, subtracting the 4-ft footing depth from the 12-ft layer; the drainage path is the full layer thickness.
Single drainage gives Hdr = 12 ft, so t = 0.848 × 12²/0.32 = 382 days.
Question 9 of 18 · Geotechnical Engineering · 10–15 items
A three-layer soil profile consists of: (1) 5-ft sand layer (Es=3 000 psi), (2) 10-ft soft clay (Cc = 0.40, e0 = 1.0, Cα = 0.015), and (3) 8-ft stiff clay (Cc = 0.15, e0 = 0.60). A sudden load of 2,500 psf is applied. The water table is at 3 ft depth. For a 10-year settlement estimate, which settlement components should be included and in what order?
Solution
Stotal = Si + Sc + Ss
Ss = CαH/(1 + ep)·log(t/tp) — soft clay only
FE Reference Handbook, Geotechnical — consolidation and secondary compression.
Secondary compression of the stiff, lightly compressible clay is negligible and no Cα is given for it.
The 5-ft sand compresses elastically as soon as the load is applied, so immediate settlement is part of the total.
Sand drains almost instantly, so its settlement is immediate (elastic), not primary consolidation.
Over 10 years the soft clay with Cα = 0.015 keeps creeping after primary ends; leaving secondary out underestimates settlement.
Elastic settlement in the sand, consolidation in both saturated clays, and creep in the soft clay where Cα is given and significant.
Question 10 of 18 · Geotechnical Engineering · 10–15 items
In a standard penetration test (SPT), the blow count (N-value) is defined as the number of blows required to drive the split-spoon sampler:
Solution
ASTM D1586: 140-lb hammer, 30-in drop, three 6-in increments; N = blows(2nd) + blows(3rd). N60 corrects for hammer energy. FE Reference Handbook, Geotechnical — SPT.
N is the sum of blows for the second and third 6-in increments; the first increment only seats the sampler.
The full 18 in includes the first 6-in seating increment, which is excluded because it passes through disturbed soil.
The first 6 in is the seating drive and is discarded, so N cannot start there.
The standard SPT drive is 18 in, made of three 6-in increments, not 24 in.
Question 11 of 18 · Geotechnical Engineering · 10–15 items
In a direct shear test on sand, what happens to the peak friction angle as the relative density increases?
Solution
φpeak ≈ φcs + dilatancy contribution, which grows with relative density Dr. FE Reference Handbook, Geotechnical — shear strength of soils.
Denser sand dilates more when sheared, which raises, not lowers, the peak angle.
A friction angle near zero describes undrained clay (φ = 0 analysis), not a sand at any density.
Dense sand must dilate to shear, so it mobilizes a higher peak φ; loose sand shows little peak above the critical-state angle.
Peak φ depends strongly on packing; only the critical-state angle is roughly independent of density.
Question 12 of 18 · Geotechnical Engineering · 10–15 items
Under cyclic loading, saturated sands are prone to liquefaction. What behavior during a cyclic triaxial test is a primary indicator of impending liquefaction?
Solution
Pore pressure ratio ru = Δu/σ′3c → 1 marks initial liquefaction; σ′ = σ − u → 0. FE Reference Handbook, Geotechnical — effective stress.
Cyclic loading of loose saturated sand builds excess pore pressure until effective stress approaches zero and the specimen softens: initial liquefaction.
Strength rising with cycles is dilative, dense behavior, the opposite of liquefaction.
Liquefaction is defined by pore-pressure buildup; a test without it is not liquefying.
Stable loops mean the specimen is not approaching liquefaction; liquefaction shows growing strains and collapsing loops.
Question 13 of 18 · Geotechnical Engineering · 10–15 items
A contractor performed trial stabilization sections and obtained the results below. The project acceptance criteria are: soaked CBR ≥15, swell ≤1.0%, and field density ≥95% of maximum Proctor. Identify which dosage(s) meet all three acceptance criteria and recommend the next step.
Check soaked CBR ≥ 15 and swell ≤ 1.0 % on each row: 4 % fails all three; 5 %, 6 % and 7 % pass all. FE Reference Handbook, Geotechnical — compaction.
Field density is 95.7–95.8 % of Proctor for 5–7 %, so compaction passes for those trials.
At 4 % the soaked CBR is 12 (< 15), swell 1.4 % (> 1.0) and compaction 93.9 % (< 95), so it fails all three.
95.7 % meets a ≥ 95 % limit; a pass is a pass, and rejecting it adds a criterion that is not in the specification.
The 5 % and 6 % trials also pass all three criteria, so 7 % is not the only acceptable dosage.
5 % gives soaked CBR 16, swell 0.9 % and 112/117 = 95.7 %; 6 % and 7 % pass too, so 5 % is the lowest passing dose.
Question 14 of 18 · Geotechnical Engineering · 10–15 items
A falling-head permeability test is performed on a soil sample. Using the laboratory data below, calculate the estimated discharge (Q) through a field aquifer of this soil with a cross-sectional area of 200 m2 under a hydraulic gradient of 0.02.
6,000 L/min leaves out the field gradient: Q = kA is the discharge at i = 1, not at i = 0.02.
Question 15 of 18 · Geotechnical Engineering · 10–15 items
The Boussinesq equation is commonly used in geotechnical engineering to calculate:
Solution
Δσz = 3Pz³/(2πR5) for a point load P (R = distance to the point)
FE Reference Handbook, Geotechnical — stress distribution.
Rate of consolidation comes from Terzaghi's theory (Tv, cv), not from Boussinesq.
Rotational slope stability uses limit-equilibrium methods (Bishop, Fellenius), not an elastic stress distribution.
Wall pressures come from Rankine or Coulomb theory; Boussinesq-type solutions only add surcharge effects.
Boussinesq's elastic half-space solution gives Δσz at depth under a point or area load, used for settlement analysis.
Question 16 of 18 · Geotechnical Engineering · 10–15 items
In unsaturated soils, the effective stress concept is modified to include matric suction. How does an increase in matric suction affect the shear strength of the soil?
Suction is why a moist sand can stand in a vertical cut; it clearly changes strength.
Suction affects both volume change and strength; 'only compressibility' is too narrow.
Question 17 of 18 · Geotechnical Engineering · 10–15 items
Depth (ft)
N (blows/ft)
N60 (blows/ft)
Notes
10
15
18
Standard sampler, ER = 72 %
15
22
26
Standard sampler, ER = 72 %
20
18
22
Standard sampler, ER = 72 %
A boring log table reports SPT results for each sample depth: raw blow counts N measured with a hammer whose energy ratio is ER = 72 %, and the same counts normalized to 60 % energy, N60. A standard split-spoon sampler was used in a 4-in borehole. A bearing-capacity correlation needs energy-corrected SPT resistance. Which column should be used, and what check is needed first?
Solution
N60 = N·CECBCSCR, with CE = ER/60
Borehole (CB) and sampler (CS) corrections are 1.0 only for the standard 2.5–4.5 in borehole and standard sampler. FE Reference Handbook, Geotechnical — SPT corrections.
Bearing-capacity and settlement correlations are calibrated to N60; raw counts vary with hammer energy.
Interpolation is pointless when a measured value exists at the design depth.
The correction is N60 = N·ER/60 = 1.2N, not 0.72N, and the table has already applied it in the N60 column.
SPT correlations are written in blows per foot (per 300 mm); no unit conversion is required.
The values are already energy-corrected; what remains is confirming the borehole diameter and sampler type are the standard ones the correlation assumes.
Question 18 of 18 · Geotechnical Engineering · 10–15 items
A design note computes allowable bearing pressure as qa = qu/FS = 720/2.5 = 288 kPa, where qu = 720 kPa is the gross ultimate bearing capacity and the overburden pressure at the footing base is q0 = 30 kPa. Using the net convention (qnult = qu − q0, then qa = qnult/FS), what is the correct net allowable bearing pressure?
Given
Gross ultimate bearing capacity qu = 720 kPa
Overburden pressure at the footing base q0 = 30 kPa
Factor of safety FS = 2.5
Find
Net allowable bearing pressure qa (kPa)
Solution
qnult = qu − q0 = 720 − 30 = 690 kPa
qa,net = qnult/FS = 690/2.5 = 276 kPa
FE Reference Handbook, Geotechnical — bearing capacity.
258 kPa subtracts q0 after dividing by FS (288 − 30); the overburden is removed from the ultimate value before the factor of safety.
272 kPa takes q0 = 40 kPa; the overburden at the base is given as 30 kPa.
qnult = 720 − 30 = 690 kPa and qa = 690/2.5 = 276 kPa; the gross method overstates the allowable by applying FS to soil weight that is already there.
288 kPa is the original gross calculation, which divides qu by FS without removing the overburden.
345 kPa uses FS = 2.0 on the net value; the specified factor of safety is 2.5.
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Common traps
Void ratio vs porosity. e is voids over solids, n is voids over total volume; mixing them ruins every phase problem. Convert with n = e/(1 + e).
The phase identity. When three of saturation, water content, specific gravity and void ratio are given, the fourth follows from Gs·w = S·e.
Two classification systems. USCS and AASHTO name the same soil differently; read which one the question asks for.
Normally vs overconsolidated clay. Below the preconsolidation pressure settlement follows the recompression index, above it the compression index. Check σ′p first.
Water behind a wall. Earth-pressure coefficients apply to effective stress; hydrostatic pressure is added separately and is not reduced by K.
Seepage vs aquifer flow. Seepage under a dam or into an excavation is a geotechnical problem; aquifer properties and drawdown are taught on the FE Environmental pages.
Sources: FE Reference Handbook, Civil Engineering chapter; Das, Principles of Geotechnical Engineering.
Total stress below the water table
Strength, settlement and earth pressure all depend on effective stress. Below the water table, subtract pore pressure from total stress, or use the buoyant unit weight γ′ for each layer under water.
σ′ = σ − u u = γw·hw
How it connects
Pavement subgrades meet this section on transportation; the structure the foundation carries is structural. Aquifer hydraulics, well drawdown and contaminated soil belong to the FE Environmental exam: groundwater, soils and sediments. Earthwork volumes and compaction on a job are on construction and surveying.
Questions people ask
Which geotechnical topics come up most on the FE Civil?
NCEES does not rank topics inside a section. Effective stress runs through consolidation, strength, bearing capacity and earth pressure, which makes it the topic to master first.
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