PE Civil (NCEES Principles and Practice of Engineering) — All Questions

Using the approximate 2:1 stress-distribution method, what is the vertical stress increase at 3 m depth below a 3 m x 3 m footing carrying 900 kN?

• a.100 kPa
• b.36 kPa
• c.50 kPa
• d.25 kPa✓

The 2:1 method spreads the load over an area that grows with depth: delta-sigma = Q/[(B + z)(L + z)] = 900/[(3 + 3)(3 + 3)] = 900/36 = 25 kPa. The stress increase drops rapidly with depth as the loaded area enlarges.

Using Boussinesq theory, the vertical stress directly beneath a point load is 0.477·Q/z^2. For Q = 100 kN at z = 2 m, what is the vertical stress increase?

• a.23.9 kPa
• b.25 kPa
• c.5.0 kPa
• d.11.9 kPa✓

Directly under a point load, delta-sigma = 0.477·Q/z^2 = 0.477 x 100 / 2^2 = 47.7/4 = 11.9 kPa. Boussinesq's elastic solution shows stress decreasing with the square of depth beneath a concentrated surface load.

Compared with a normally consolidated clay, an overconsolidated clay subjected to the same stress increase (staying below its preconsolidation pressure) will settle how much?

• a.The same amount
• b.More
• c.Less✓
• d.Zero

An overconsolidated clay loaded below its preconsolidation pressure follows the flatter recompression line, so it settles less than a normally consolidated clay following the steeper virgin compression line. The recompression index Cr is much smaller than Cc.

Increasing the compactive effort applied to a given soil generally has what effect on its compaction curve?

• a.Has no effect
• b.Raises the optimum moisture content
• c.Lowers the maximum dry density
• d.Raises the maximum dry density and lowers the optimum moisture content✓

Greater compactive effort (e.g., modified vs. standard Proctor) shifts the curve up and to the left: a higher maximum dry density at a lower optimum moisture content. The extra energy packs particles more tightly with less water needed for lubrication.

On a compaction (Proctor) plot, the zero-air-voids curve represents which condition?

• a.Zero water content (S = 0%)
• b.Maximum compactive effort
• c.The optimum moisture content
• d.Full saturation (S = 100%)✓

The zero-air-voids curve is the theoretical dry unit weight when all voids are filled with water (S = 100%). No compaction point can plot to its right because air cannot be fully removed; the curve is an upper bound the wet side approaches.

The modified Proctor test applies greater compaction energy than the standard Proctor test. Compared with the standard test, the modified test yields what?

• a.The same maximum dry density
• b.A lower maximum dry density
• c.A higher optimum moisture content
• d.A higher maximum dry density✓

The modified Proctor delivers more energy per unit volume, so it produces a higher maximum dry density at a lower optimum moisture content than the standard Proctor. It is specified when heavier loads (e.g., airfields, highways) demand denser fill.

Which soil type generally has the highest hydraulic conductivity (permeability)?

• a.Clay
• b.Gravel✓
• c.Silt
• d.Silty clay

Gravel has the highest hydraulic conductivity because its large, well-connected pore spaces let water flow freely. Clay has the lowest permeability (often 10^-9 m/s or less) owing to its tiny pores, which is why clays act as aquitards and consolidate slowly.

According to Terzaghi's principle, the shear strength and volume change of a saturated soil are governed by which stress?

• a.Pore water pressure alone
• b.Total unit weight
• c.Total stress
• d.Effective stress✓

Terzaghi's principle states that effective stress (sigma' = sigma - u) controls the strength and deformation of soil, because only the grain-to-grain contact stress mobilizes friction. This is why raising pore pressure (lowering sigma') can trigger failure.

Using the Mohr-Coulomb criterion, what is the shear strength of a soil with c = 10 kPa and phi = 30 degrees on a plane where the effective normal stress is 100 kPa?

• a.100 kPa
• b.77 kPa
• c.57.7 kPa
• d.67.7 kPa✓

Mohr-Coulomb: tau_f = c + sigma'·tan(phi) = 10 + 100 x tan(30) = 10 + 100 x 0.577 = 67.7 kPa. Shear strength combines a cohesion intercept with a frictional component that grows with effective normal stress.

A cohesionless sand (c = 0) has phi = 32 degrees. What is its shear strength on a plane where the effective normal stress is 150 kPa?

• a.93.7 kPa✓
• b.150 kPa
• c.75 kPa
• d.120 kPa

With c = 0, tau_f = sigma'·tan(phi) = 150 x tan(32) = 150 x 0.625 = 93.7 kPa. In cohesionless soil the shear strength is purely frictional and rises linearly with effective normal stress, passing through the origin.

In a total-stress (phi = 0) analysis of saturated clay under undrained loading, the shear strength is taken as which quantity?

• a.c + sigma·tan(phi) with phi = 30 degrees
• b.Always zero
• c.The undrained cohesion cu (constant)✓
• d.sigma·tan(phi)

Under undrained conditions saturated clay is analyzed with phi = 0, so its shear strength equals the constant undrained cohesion cu (half the unconfined compressive strength). Because no drainage occurs, strength does not increase with applied total stress in the short term.

In one-dimensional consolidation theory, what time factor Tv corresponds to 50% average degree of consolidation?

• a.0.848
• b.0.5
• c.0.197✓
• d.1.0

For 50% consolidation the dimensionless time factor is Tv = 0.197. (90% corresponds to Tv = 0.848.) These standard values from Terzaghi's consolidation solution let you convert between degree of consolidation and elapsed time.

If the drainage path length of a consolidating clay layer is doubled, the time required to reach a given degree of consolidation changes by what factor?

• a.4 times✓
• b.Unchanged
• c.2 times
• d.Half

From Tv = cv·t/H_dr^2, for a fixed Tv the time is proportional to the square of the drainage path: t is proportional to H_dr^2. Doubling H_dr multiplies the time by 4. This is why sand drains (shortening the drainage path) dramatically speed consolidation.

The Standard Penetration Test (SPT) N-value in sand correlates most directly with which soil property?

• a.Plasticity index
• b.Hydraulic conductivity
• c.Relative density and friction angle✓
• d.Organic content

The SPT blow count N reflects the relative density of granular soil, which in turn correlates with the friction angle and bearing capacity of sands. Correlations (with energy and overburden corrections to N60 and (N1)60) are widely used for foundation design in sand.

In bearing capacity theory, what is the value of the factor Nc for a soil with a friction angle of zero (saturated clay, undrained)?

• a.5.14✓
• b.1.0
• c.9.0
• d.0

For phi = 0 the bearing capacity factor Nc = 5.14 (equal to 2 + pi) for a strip footing, from Prandtl's plasticity solution. This underlies the undrained bearing capacity q_ult = 5.14·cu (plus surcharge) for saturated clay.

A deep (pile) foundation is generally selected instead of a shallow footing when:

• a.It is always cheaper than a footing
• b.The structure carries only light loads
• c.It is never structurally justified
• d.Near-surface soils are weak and load must be transferred to deeper competent strata✓

Deep foundations are used when the shallow soils are too weak or compressible to support the loads at acceptable settlement, so piles or drilled shafts carry the load down to firmer strata through end bearing and skin friction. They cost more, so they are chosen when shallow foundations are inadequate.