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

What is the total vertical stress at a depth of 5 m in a soil with unit weight 18 kN/m^3 (water table well below)?

• a.18 kPa
• b.45 kPa
• c.23 kPa
• d.90 kPa✓

Total vertical stress is the weight of overlying soil: sigma = gamma·h = 18 x 5 = 90 kPa. Stress accumulates linearly with depth, analogous to hydrostatic pressure in a fluid.

At a point 4 m below the water table, the saturated unit weight is 20 kN/m^3 and water unit weight is 10 kN/m^3. What is the effective vertical stress?

• a.80 kPa
• b.40 kPa✓
• c.120 kPa
• d.0 kPa

Total stress sigma = 20 x 4 = 80 kPa; pore pressure u = 10 x 4 = 40 kPa. Terzaghi's principle gives effective stress sigma' = sigma - u = 80 - 40 = 40 kPa. Effective stress, not total stress, controls soil strength and settlement.

Using Rankine theory, what is the active earth pressure coefficient Ka for a cohesionless soil with friction angle 30 degrees?

• a.3.0
• b.0.5
• c.1.0
• d.0.33✓

Ka = tan^2(45 - phi/2) = tan^2(45 - 15) = tan^2(30) = (0.577)^2 = 0.333. The active case (wall moving away from the soil) gives the minimum lateral pressure; the passive coefficient would be its reciprocal, 3.0.

What is the at-rest earth pressure coefficient K0 for a normally consolidated sand with friction angle 30 degrees (Jaky's equation)?

• a.0.5✓
• b.0.33
• c.1.0
• d.0.87

Jaky's equation gives K0 = 1 - sin(phi) = 1 - sin(30) = 1 - 0.5 = 0.5. The at-rest condition applies when the wall does not move, so K0 falls between the active and passive coefficients.

A water table that was at the ground surface is lowered by dewatering. What happens to the effective stress at a point that remains below the new water table?

• a.It increases✓
• b.It decreases
• c.It becomes zero
• d.It is unchanged

Lowering the water table reduces pore water pressure while total stress changes little, so effective stress sigma' = sigma - u increases. This rise in effective stress is what drives consolidation settlement when sites are dewatered.

A soil sample has a total (moist) unit weight of 20 kN/m^3 at a water content of 25%. What is its dry unit weight?

• a.15 kN/m^3
• b.16 kN/m^3✓
• c.20 kN/m^3
• d.25 kN/m^3

Dry unit weight gamma_d = gamma / (1 + w), where w is the decimal water content. gamma_d = 20 / (1 + 0.25) = 20 / 1.25 = 16 kN/m^3. The dry unit weight removes the weight of pore water and is used to assess compaction.

A soil profile has a 2 m top layer of unit weight 16 kN/m^3 over a layer of unit weight 19 kN/m^3. What is the total vertical stress at a depth of 6 m (water table well below)?

• a.114 kPa
• b.108 kPa✓
• c.96 kPa
• d.90 kPa

Total vertical stress is the sum of the weights of the layers above: sigma = 16 x 2 + 19 x 4 = 32 + 76 = 108 kPa. Stress accumulates layer by layer using each layer's own unit weight and thickness.

A profile has 3 m of sand (gamma = 17 kN/m^3) above the water table, then 4 m of saturated sand (gamma_sat = 20 kN/m^3). Using gamma_w = 10 kN/m^3, what is the effective vertical stress at 7 m depth?

• a.91 kPa✓
• b.131 kPa
• c.171 kPa
• d.40 kPa

Total stress sigma = 17 x 3 + 20 x 4 = 131 kPa. Pore pressure u = gamma_w x 4 = 10 x 4 = 40 kPa. Effective stress sigma' = sigma - u = 131 - 40 = 91 kPa (Terzaghi's principle). Only the depth below the water table generates pore pressure.

For that profile (water table at 3 m depth, gamma_w = 10 kN/m^3), what is the pore water pressure at 7 m depth?

• a.40 kPa✓
• b.91 kPa
• c.70 kPa
• d.131 kPa

Pore pressure is hydrostatic below the water table: u = gamma_w x h_w = 10 x (7 - 3) = 10 x 4 = 40 kPa. Only the 4 m of water column below the water table contributes; the soil above the table is treated as having no pore pressure here.

A saturated soil has a saturated unit weight of 20 kN/m^3. Using gamma_w = 10 kN/m^3, what is its buoyant (effective) unit weight?

• a.10 kN/m^3✓
• b.15 kN/m^3
• c.30 kN/m^3
• d.20 kN/m^3

Buoyant unit weight gamma' = gamma_sat - gamma_w = 20 - 10 = 10 kN/m^3. Below the water table, effective stress is computed using this submerged unit weight, which accounts for the buoyant uplift of the pore water.

Using Rankine theory, what is the active earth pressure coefficient Ka for a cohesionless soil with a friction angle of 36 degrees?

• a.0.33
• b.0.26✓
• c.0.50
• d.3.85

Ka = tan^2(45 - phi/2) = tan^2(45 - 18) = tan^2(27) = (0.5095)^2 = 0.26. A higher friction angle lowers the active coefficient, reducing the lateral thrust the wall must resist.

Using Rankine theory, what is the passive earth pressure coefficient Kp for a cohesionless soil with a friction angle of 30 degrees?

• a.0.5
• b.3.0✓
• c.0.33
• d.1.0

Kp = tan^2(45 + phi/2) = tan^2(45 + 15) = tan^2(60) = (1.732)^2 = 3.0. The passive coefficient is the reciprocal of the active coefficient (1/0.333 = 3.0); passive resistance is mobilized when the wall pushes into the soil.

For a normally consolidated sand with a friction angle of 35 degrees, what is the at-rest earth pressure coefficient K0 (Jaky's equation)?

• a.0.35
• b.0.43✓
• c.0.65
• d.0.574

Jaky's equation: K0 = 1 - sin(phi) = 1 - sin(35) = 1 - 0.574 = 0.43. The at-rest coefficient applies to a non-yielding wall and lies between the active (0.27) and passive (3.7) values for this soil.

A 5 m high retaining wall backfilled with cohesionless soil (gamma = 18 kN/m^3, Ka = 0.333) has a horizontal water table well below the base. What is the total active thrust per meter of wall?

• a.75 kN/m✓
• b.37.5 kN/m
• c.225 kN/m
• d.150 kN/m

Total active thrust is the area of the triangular pressure diagram: Pa = 0.5·Ka·gamma·H^2 = 0.5 x 0.333 x 18 x 5^2 = 0.5 x 0.333 x 18 x 25 = 75 kN/m. Pressure grows linearly with depth, so the resultant is a triangle.

For that 5 m wall, at what height above the base does the resultant active thrust act?

• a.1.67 m✓
• b.3.33 m
• c.2.5 m
• d.1.0 m

Because the active pressure distribution is triangular (zero at top, maximum at base), the resultant acts at the centroid of the triangle, H/3 = 5/3 = 1.67 m above the base. This lever arm is used when checking overturning about the toe.

A 5 m high wall retains cohesionless soil (gamma = 18 kN/m^3, Kp = 3.0). What is the total passive thrust per meter of wall?

• a.337.5 kN/m
• b.225 kN/m
• c.675 kN/m✓
• d.75 kN/m

Passive thrust Pp = 0.5·Kp·gamma·H^2 = 0.5 x 3.0 x 18 x 25 = 675 kN/m. Passive resistance is far larger than active (here 9 times) because Kp = 1/Ka; it resists sliding at the wall toe but requires large wall movement to develop fully.

A strip footing rests on saturated clay with undrained shear strength cu = 40 kPa. Neglecting surcharge, what is the ultimate bearing capacity (Nc = 5.14)?

• a.257 kPa
• b.103 kPa
• c.206 kPa✓
• d.411 kPa

For a strip footing on clay under undrained (phi = 0) conditions, q_ult = cu·Nc = 40 x 5.14 = 206 kPa. The bearing capacity factor Nc = 5.14 (equal to 2 + pi) applies to the total-stress analysis of saturated clay.

For that footing (q_ult = 206 kPa), what is the allowable bearing pressure using a factor of safety of 3?

• a.103 kPa
• b.206 kPa
• c.618 kPa
• d.69 kPa✓

Allowable bearing q_all = q_ult/FS = 206/3 = 69 kPa. A factor of safety of about 3 is typical for bearing capacity to guard against soil-strength variability and to limit settlement.

A 3 m thick normally consolidated clay (Cc = 0.3, e0 = 0.8) has its effective stress increased from 100 kPa to 200 kPa. What is the primary consolidation settlement?

• a.75 mm
• b.90 mm
• c.150 mm✓
• d.301 mm

Sc = (Cc·H/(1 + e0))·log10(sigma'_f/sigma'_0) = (0.3 x 3000/1.8) x log10(200/100) = 500 x 0.301 = 150 mm. Doubling the effective stress produces one log-cycle fraction of settlement scaled by the compression index.

A fully saturated soil has a specific gravity of solids Gs = 2.70 and a water content of 20%. What is the void ratio?

• a.0.54✓
• b.0.20
• c.0.74
• d.2.70

For a saturated soil, S·e = Gs·w with S = 1, so e = Gs·w = 2.70 x 0.20 = 0.54. This phase relationship links water content directly to void ratio only when the soil is fully saturated.

A soil has a void ratio of 0.54. What is its porosity?

• a.0.35✓
• b.0.46
• c.0.65
• d.0.54

Porosity n = e/(1 + e) = 0.54/1.54 = 0.35. Void ratio (voids/solids) and porosity (voids/total) describe the same pore space with different reference volumes; n is always less than e.

A soil has Gs = 2.70 and void ratio e = 0.54. Using gamma_w = 9.81 kN/m^3, what is its dry unit weight?

• a.17.2 kN/m^3✓
• b.26.5 kN/m^3
• c.14.5 kN/m^3
• d.20.6 kN/m^3

Dry unit weight gamma_d = Gs·gamma_w/(1 + e) = 2.70 x 9.81 / 1.54 = 26.49/1.54 = 17.2 kN/m^3. Only the solid particles contribute to dry weight, spread over the total volume that includes the voids.

A saturated soil has Gs = 2.70 and void ratio e = 0.54. Using gamma_w = 9.81 kN/m^3, what is its saturated unit weight?

• a.20.6 kN/m^3✓
• b.26.5 kN/m^3
• c.18.9 kN/m^3
• d.17.2 kN/m^3

Saturated unit weight gamma_sat = (Gs + e)·gamma_w/(1 + e) = (2.70 + 0.54) x 9.81 / 1.54 = 3.24 x 9.81 / 1.54 = 20.6 kN/m^3. Here all voids are filled with water, so the pore water adds to the solids' weight.

A soil has Gs = 2.70, void ratio e = 0.54, and water content 15%. What is its degree of saturation?

• a.50%
• b.40%
• c.75%✓
• d.100%

From S·e = Gs·w, the degree of saturation S = Gs·w/e = (2.70 x 0.15)/0.54 = 0.405/0.54 = 0.75 = 75%. Degree of saturation is the fraction of the void volume occupied by water.

A sand has Gs = 2.70 and void ratio e = 0.54. What is the critical (upward) hydraulic gradient that causes a quick (boiling) condition?

• a.2.70
• b.0.90
• c.1.10✓
• d.1.70

Critical gradient i_cr = (Gs - 1)/(1 + e) = (2.70 - 1)/1.54 = 1.70/1.54 = 1.10. At this gradient the upward seepage force cancels the buoyant weight, effective stress drops to zero, and the sand loses all strength (quicksand).

Water flows through a soil with hydraulic conductivity k = 1x10^-4 m/s under a hydraulic gradient of 0.5 through a cross-sectional area of 2 m^2. What is the discharge?

• a.5x10^-4 m^3/s
• b.1x10^-5 m^3/s
• c.1x10^-4 m^3/s✓
• d.2x10^-4 m^3/s

Darcy's law: q = k·i·A = 1x10^-4 x 0.5 x 2 = 1x10^-4 m^3/s. Darcy flow through the total (gross) cross-sectional area is valid for the laminar seepage typical in soils.

For that flow (k = 1x10^-4 m/s, i = 0.5) in a soil with porosity 0.33, what is the seepage (pore) velocity?

• a.5x10^-5 m/s
• b.1.5x10^-4 m/s✓
• c.1.65x10^-5 m/s
• d.1x10^-4 m/s

Discharge velocity v = k·i = 1x10^-4 x 0.5 = 5x10^-5 m/s. Seepage velocity v_s = v/n = 5x10^-5/0.33 = 1.5x10^-4 m/s. Water moves faster than the Darcy (discharge) velocity because it flows only through the pore space.

A flow net beneath a dam (k = 2x10^-5 m/s) has 4 flow channels (Nf) and 8 equipotential drops (Nd) under a total head of 6 m. What is the seepage per meter of dam?

• a.3x10^-5 m^3/s
• b.6x10^-5 m^3/s✓
• c.9.6x10^-4 m^3/s
• d.1.2x10^-4 m^3/s

Flow-net seepage: q = k·H·(Nf/Nd) = 2x10^-5 x 6 x (4/8) = 2x10^-5 x 6 x 0.5 = 6x10^-5 m^3/s per meter. The ratio of flow channels to equipotential drops scales the total head loss into a discharge.

A 4 m thick clay layer with double drainage has cv = 2 m^2/yr. What is the time factor after 1 year?

• a.0.5✓
• b.0.25
• c.1.0
• d.0.125

With double drainage the drainage path H_dr is half the layer thickness = 2 m. Time factor Tv = cv·t/H_dr^2 = 2 x 1 / 2^2 = 2/4 = 0.5. Double drainage quarters the time compared with single drainage because the path is halved and it is squared.

A soil has a liquid limit of 45 and a plastic limit of 20. What is the plasticity index?

• a.65
• b.25✓
• c.45
• d.20

Plasticity index PI = LL - PL = 45 - 20 = 25. The PI is the range of water content over which the soil behaves plastically; larger values indicate more clay-like, plastic soils.

A clay has LL = 45, PL = 20, and an in-situ water content of 30%. What is the liquidity index?

• a.0.40✓
• b.1.0
• c.0.60
• d.0.20

Liquidity index LI = (w - PL)/PI = (30 - 20)/(45 - 20) = 10/25 = 0.40. LI locates the natural water content within the plastic range; values near 0 indicate stiff soil and values near 1 indicate soft, near-liquid soil.

A clay has a preconsolidation pressure of 200 kPa and a current effective overburden stress of 100 kPa. What is the overconsolidation ratio?

• a.2.0✓
• b.1.0
• c.0.5
• d.100

OCR = sigma'_p/sigma'_0 = 200/100 = 2.0. An OCR greater than 1 means the soil was once loaded to a higher stress (e.g., by past overburden or glaciers) and is now overconsolidated, so it will settle less under new load.

For a dry cohesionless infinite slope with a friction angle of 30 degrees inclined at 20 degrees, what is the factor of safety against sliding?

• a.1.0
• b.2.0
• c.1.59✓
• d.0.63

For a dry cohesionless infinite slope, FS = tan(phi)/tan(beta) = tan(30)/tan(20) = 0.577/0.364 = 1.59. The slope is stable as long as the friction angle exceeds the slope angle; cohesion and seepage would modify this.

A sand has emax = 0.90, emin = 0.40, and an in-situ void ratio of 0.60. What is its relative density?

• a.50%
• b.40%
• c.75%
• d.60%✓

Relative density Dr = (emax - e)/(emax - emin) = (0.90 - 0.60)/(0.90 - 0.40) = 0.30/0.50 = 0.60 = 60%. Relative density expresses how dense a granular soil is between its loosest and densest states and correlates with strength.

A compacted fill achieves a field dry unit weight of 17.5 kN/m^3 against a standard Proctor maximum dry unit weight of 18.5 kN/m^3. What is the relative compaction?

• a.94.6%✓
• b.105%
• c.90%
• d.100%

Relative compaction = gamma_d,field/gamma_d,max = 17.5/18.5 = 0.946 = 94.6%. Earthwork specifications commonly require at least 90-95% of the Proctor maximum dry density to ensure adequate strength and low settlement.

A 2 m wide strip footing bears at 1 m depth in sand (gamma = 18 kN/m^3, c = 0). With Nq = 18.4 and N-gamma = 15.7, what is the Terzaghi ultimate bearing capacity?

• a.331 kPa
• b.614 kPa✓
• c.283 kPa
• d.440 kPa

With c = 0, q_ult = q·Nq + 0.5·gamma·B·N-gamma, where surcharge q = gamma·Df = 18 x 1 = 18 kPa. q_ult = 18 x 18.4 + 0.5 x 18 x 2 x 15.7 = 331.2 + 282.6 = 614 kPa. The surcharge and self-weight terms both contribute in cohesionless soil.

A footing founded 1.5 m deep in soil (gamma = 18 kN/m^3) applies a gross bearing pressure of 200 kPa. What is the net bearing pressure?

• a.150 kPa
• b.173 kPa✓
• c.200 kPa
• d.227 kPa

Net bearing pressure = gross pressure - overburden removed = 200 - (gamma·Df) = 200 - (18 x 1.5) = 200 - 27 = 173 kPa. The net pressure is the stress increase over the pre-existing overburden and is what drives settlement.

Primary consolidation settlement of a saturated clay is caused by which mechanism?

• a.Gradual dissipation of excess pore water pressure and expulsion of water✓
• b.Shear failure of the soil
• c.Instantaneous distortion
• d.Elastic rebound of the soil

Primary consolidation is the time-dependent volume reduction as excess pore water pressure dissipates and water is squeezed out of the clay, transferring load to the soil skeleton. Its slow rate is governed by the low permeability of clay.

The additional settlement that occurs after primary consolidation is complete, at essentially constant effective stress, is called what?

• a.Primary consolidation
• b.Secondary compression (creep)✓
• c.Swelling
• d.Elastic (immediate) settlement

Secondary compression, or creep, is the continued slow settlement caused by particle rearrangement after excess pore pressures have dissipated. It is characterized by the secondary compression index C-alpha and matters most for organic and highly plastic soils.

A normally consolidated clay is one whose current effective overburden stress equals its maximum past pressure, giving what overconsolidation ratio?

• a.Less than 1
• b.Equal to 1✓
• c.Greater than 1
• d.Equal to 0

A normally consolidated clay has never carried a higher effective stress than it does now, so OCR = sigma'_p/sigma'_0 = 1. Such soils are more compressible than overconsolidated clays because loading follows the steeper virgin compression line.

The plastic limit of a fine-grained soil is the water content at the boundary between which two states?

• a.Saturated and dry states
• b.Solid and liquid states
• c.Liquid and plastic states
• d.Plastic and semisolid states✓

The plastic limit marks the transition between the plastic and semisolid states, measured as the water content at which a 3 mm soil thread crumbles. The liquid limit marks the plastic-to-liquid boundary; together they define the plasticity index.

In the Unified Soil Classification System (USCS), what group symbol denotes a well-graded gravel?

• a.GP
• b.SW
• c.GW✓
• d.CL

GW denotes a well-graded gravel (G = gravel, W = well graded). GP is a poorly graded gravel, SW is a well-graded sand, and CL is a low-plasticity clay. The first letter is the primary size fraction, the second describes gradation or plasticity.

A soil is classified as fine-grained in the USCS when more than 50% by weight passes which sieve?

• a.No. 10 sieve
• b.No. 4 sieve
• c.No. 200 sieve✓
• d.No. 40 sieve

A soil is fine-grained when more than 50% passes the No. 200 sieve (0.075 mm), the boundary between sand and silt/clay sizes. The No. 4 sieve separates gravel from sand. Fine-grained soils are then classified by their Atterberg limits.

A sand has D60 = 0.6 mm and D10 = 0.1 mm. What is its coefficient of uniformity, Cu?

• a.6.0✓
• b.3.0
• c.0.6
• d.0.17

Coefficient of uniformity Cu = D60/D10 = 0.6/0.1 = 6.0. A larger Cu indicates a wider range of particle sizes (better graded); a sand generally needs Cu of at least 6 (and gravel at least 4) to be classified as well graded.

A soil has D10 = 0.1 mm, D30 = 0.3 mm, and D60 = 0.6 mm. What is its coefficient of curvature, Cc?

• a.1.0
• b.3.0
• c.1.5✓
• d.0.5

Coefficient of curvature Cc = (D30)^2/(D10·D60) = 0.3^2/(0.1 x 0.6) = 0.09/0.06 = 1.5. A well-graded soil requires Cc between 1 and 3, indicating a smoothly distributed gradation curve without gaps.

A pile develops 200 kN of skin friction and 300 kN of end bearing. What is its ultimate axial capacity?

• a.300 kN
• b.500 kN✓
• c.100 kN
• d.200 kN

Ultimate pile capacity Q_ult = Q_s + Q_p = 200 + 300 = 500 kN, the sum of the shaft (skin) friction and the end (tip) bearing resistance. The two components are added because both resist the downward load.

For that pile (Q_ult = 500 kN), what is the allowable capacity using a factor of safety of 2.5?

• a.500 kN
• b.200 kN✓
• c.125 kN
• d.250 kN

Allowable pile capacity Q_all = Q_ult/FS = 500/2.5 = 200 kN. A factor of safety of about 2 to 3 is used for piles to account for uncertainty in soil resistance and installation effects; higher when capacity is estimated only from soil data.

A 0.4 m diameter, 10 m long pile in clay (cu = 50 kPa) uses an adhesion factor alpha = 0.5. What is the skin-friction capacity (alpha method)?

• a.157 kN
• b.628 kN
• c.314 kN✓
• d.250 kN

Skin friction Q_s = alpha·cu·(perimeter x length) = 0.5 x 50 x (pi x 0.4 x 10) = 0.5 x 50 x 12.57 = 314 kN. The alpha method reduces the undrained strength by an adhesion factor to give the unit shaft resistance in clay.

A quick (boiling) condition in sand occurs when upward seepage reduces the effective stress to what value?

• a.Negative infinity
• b.Double its original value
• c.Its maximum
• d.Zero✓

A quick condition develops when the upward seepage force exactly offsets the submerged weight, driving the effective stress to zero. With sigma' = 0 the sand has no shear strength and behaves like a heavy fluid (quicksand).

For a heavily overconsolidated clay, the at-rest earth pressure coefficient K0 can have what value?

• a.Negative
• b.Greater than 1✓
• c.Exactly zero
• d.Always less than the active coefficient

In heavily overconsolidated soils, locked-in lateral stresses from past unloading can make K0 exceed 1, meaning horizontal stress is greater than vertical stress. Jaky's K0 = 1 - sin(phi) applies to normally consolidated soil; an OCR correction raises it.