Geotechnical Engineering
10–15 of 110 itemsNCEES draws between 10 and 15 of the 110 from this area on any legal paper. Its outline lists 11 sub-topics; the bank holds 57 verified templates against them, which expand to 1,590 questions.
Sub-topics, from the NCEES outline
- AIndex properties and soil classifications
- BPhase relations
- CLaboratory and field tests
- DEffective stress
- EStability of retaining structures (e.g., active/passive/at-rest pressure)
- FShear strength
- GBearing capacity
- HFoundation types (e.g., spread footings, deep foundations, wall footings, mats)
- IConsolidation and differential settlement
- JSlope stability (e.g., fills, embankments, cuts, dams)
- KSoil stabilization (e.g., chemical additives, geosynthetics)
5 worked problems
Real questions from the bank, at the depth the exam asks them. Pick an answer first: the reasoning opens when you do, along with why each wrong answer is tempting.
Question 1 of 5
A footing 4 ft wide bears on saturated sand with a unit weight of 130 pcf, and the water table sits at the base of the footing. With N-gamma = 15.7, the width term of the bearing capacity equation is most nearly:
Spread footing 4 ft wide bearing on saturated sand at 130 pcf, with the water table at the level of the footing base. Show the answer and the reasoning
Why C
Given: B = 4 ft, gamma_sat = 130 pcf, N-gamma = 15.7, water table at the footing base. Relation: below the water table the width term uses the buoyant unit weight: 0.5 x gamma' x B x N-gamma, gamma' = gamma_sat - 62.4. Substitute: gamma' = 67.6 pcf; 0.5(67.6)(4)(15.7). Result: 2,123 psf, roughly half the dry value.
Why A is tempting
Left the footing width out.
Why B is tempting
Used the unit weight of water alone rather than the buoyant unit weight of the soil.
Why D is tempting
Omitted the factor of one half.
Question 2 of 5
A soil sample has a specific gravity of solids of 2.7, a void ratio of 1.00 and a water content of 14%. The moist unit weight of the soil is most nearly:
Show the answer and the reasoning
Why B
Given: Gs = 2.7, e = 1.00, w = 0.14. Relation: gamma = Gs x gamma_w x (1 + w)/(1 + e). Substitute: gamma = 2.7(62.4)(1 + 0.14)/(1 + 1.00). Result: gamma = 96.03 pcf. Check: the dry unit weight is 84.2 pcf, and (1 + w) adds the pore water to it.
Why A is tempting
Reported the dry unit weight; the question asks for the moist weight, which includes the pore water.
Why C is tempting
Used the saturated unit weight. The sample is only partially saturated at this water content.
Why D is tempting
Left out the void ratio entirely, treating the soil as solid grains with no pore space.
Question 3 of 5
A soil profile has a water table 6 ft below the ground surface. Above it the soil has a moist unit weight of 120 pcf; below it the saturated unit weight is 127 pcf. The vertical effective stress 15 ft below the surface is most nearly:
Soil profile with the water table 6 ft below ground. Above it the soil is moist at 120 pcf; below it saturated at 127 pcf. The point of interest is 15 ft deep. Show the answer and the reasoning
Why C
Given: water table at 6 ft, 120 pcf above it, 127 pcf saturated for 9 ft below it. Relation: sigma' = gamma(dw) + (gamma_sat - 62.4)(depth below the water table). Substitute: sigma' = 120(6) + (127 - 62.4)(9) = 720 + 581. Result: sigma' = 1,301 psf. Check: total stress 1863 psf less pore pressure 562 psf gives the same.
Why A is tempting
Measured the pore pressure from the ground surface rather than from the water table.
Why B is tempting
Applied the buoyant unit weight above the water table as well. Only the submerged soil is buoyant.
Why D is tempting
Gave the total stress. Subtracting the pore water pressure is what makes it effective.
Question 4 of 5
A clay layer 20 ft thick drains from its upper face only. Its coefficient of consolidation is 15 in^2/day. For a time factor of Tv = 0.403, the elapsed time is most nearly:
Show the answer and the reasoning
Why B
Given: H = 20 ft, single drainage, cv = 15 in^2/day, Tv = 0.403. Relation: t = Tv x Hdr^2/cv, with Hdr = H for one drainage face. Substitute: Hdr = 240 in; t = (0.403)(240)^2/15. Result: t = 1,550 days.
Why A is tempting
Halved the result after already accounting for two-way drainage.
Why C is tempting
Doubled the time for one-way drainage on top of using the full thickness as the path.
Why D is tempting
Took the path as twice the thickness, as if water ran down and back up.
Source NAVFAC DM-7, Soil Mechanics and Foundations — DM-7.01 Chapter 5, rate of consolidation
Question 5 of 5
A cohesionless backfill has an effective friction angle of 28 degrees. The Rankine active earth pressure coefficient is most nearly:
Show the answer and the reasoning
Why A
Given: phi = 28 degrees, cohesionless. Relation: Ka = tan^2(45 - phi/2). Substitute: Ka = tan^2(31). Result: Ka = 0.361. Check: (1 - sin phi)/(1 + sin phi) gives the same, and active is always below 1.
Why B is tempting
Used the at-rest coefficient K0 = 1 - sin(phi).
Why C is tempting
Did not square the tangent.
Why D is tempting
Used a plus sign, which gives the passive coefficient. Active pressure is always less than one.
Source NAVFAC DM-7, Soil Mechanics and Foundations — DM-7.02 Chapter 3, lateral earth pressure
Four mistakes that cost the question
Each of these lands on an answer that is offered, so it costs the question outright. The minutes are our estimate of the time each one burns on top, against an average of 2.9 minutes a question (320 minutes for 110).
Costs the question and about 3 min
Using the total unit weight below the water table
Effective stress is total stress minus pore pressure, which below the water table is the buoyant unit weight times depth. Using the saturated unit weight with no pore pressure deducted, or deducting pore pressure above the water table too, both give answers that are offered.
Costs the question and about 3 min
Using one unit weight throughout the bearing capacity equation
With the water table at the footing base, the surcharge term above the base takes the moist unit weight and the width term below it takes the buoyant one. Using the same unit weight in both overstates the capacity.
Costs the question and about 2 min
Using the wrong drainage path in consolidation time
Time is the time factor times the drainage path squared, divided by the coefficient of consolidation. The drainage path is half the layer thickness when water can leave at the top and bottom, and the full thickness when it can leave from one face only. The wrong choice changes the time four times, and both are offered.
Costs the question and about 3 min
Taking water content as a fraction of total weight
Water content is the weight of water over the weight of solids, not over total weight. The phase relations, such as dry unit weight equals moist unit weight over (1 + w), all assume that definition; using total weight gives answers a few percent off, close enough to be offered.
Questions about Geotechnical Engineering
Is Geotechnical Engineering mostly calculation?
Yes. Phase relations, effective stress, consolidation, bearing capacity and earth pressure are all computation, and few of them come with a drawing: on the official NCEES practice exam we counted only one figure in the area. Building the soil profile from the stem, water table included, is the first minute of most of these questions.
Which bearing capacity equation should I use?
Terzaghi's, or the general bearing capacity equation, with the factors the question or the FE Reference Handbook gives. The pieces to get right are the shape factors, the effective unit weight when the water table is near the footing, and whether the question wants ultimate, allowable or net capacity.
Do I need to classify soils by USCS and AASHTO?
Index properties and soil classification are a sub-topic, so know how grain size, the coefficients of uniformity and curvature, and the Atterberg limits place a soil in each system. The classification charts are in the FE Reference Handbook; the work is computing their inputs correctly.
What earth pressure theory does the exam use?
Rankine for most items: active and passive coefficients from the friction angle, at-rest from 1 minus the sine of the friction angle for a normally consolidated soil, and a thrust equal to half the coefficient times unit weight times height squared, acting at one third of the height.
How many study hours this area is worth · Using the handbook under the clock