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

A simply supported beam of span 8 m carries a uniformly distributed load of 4 kN/m over its full length. What is the vertical reaction at each support?

• a.64 kN
• b.32 kN
• c.16 kN✓
• d.8 kN

The total load is w·L = 4 x 8 = 32 kN. By symmetry each support carries half: 32/2 = 16 kN, which also equals w·L/2. Symmetric loading always splits equally between the two supports.

For that same simply supported beam (span 8 m, uniform load 4 kN/m), what is the maximum bending moment?

• a.256 kN·m
• b.32 kN·m✓
• c.64 kN·m
• d.16 kN·m

The maximum moment for a simply supported beam under a uniform load occurs at midspan and equals w·L^2/8 = 4 x 8^2 / 8 = 4 x 64 / 8 = 32 kN·m. The moment diagram is parabolic, peaking at the center.

What is the moment of inertia about the centroidal axis of a rectangular cross-section 200 mm wide and 300 mm deep (bending about the strong axis)?

• a.1.8x10^9 mm^4
• b.9.0x10^8 mm^4
• c.2.25x10^8 mm^4
• d.4.5x10^8 mm^4✓

For a rectangle, I = b·h^3/12 where h is the dimension parallel to the load. I = 200 x 300^3 / 12 = 200 x 2.7x10^7 / 12 = 4.5x10^8 mm^4. Depth is cubed, so orienting the section deep-side up sharply increases stiffness.

What is the section modulus of a rectangular cross-section 200 mm wide and 300 mm deep about its strong axis?

• a.6.0x10^6 mm^3
• b.3.0x10^6 mm^3✓
• c.9.0x10^6 mm^3
• d.1.5x10^6 mm^3

Section modulus S = b·h^2/6 = 200 x 300^2 / 6 = 200 x 90,000 / 6 = 3.0x10^6 mm^3. Bending stress is M/S, so a larger section modulus lowers the stress for a given moment.

What is the specified minimum yield strength, Fy, of ASTM A36 structural steel?

• a.36 ksi✓
• b.29,000 ksi
• c.50 ksi
• d.60 ksi

A36 steel has a minimum yield strength of 36 ksi (about 250 MPa), the value in its designation. A992 steel is 50 ksi and Grade 60 rebar is 60 ksi; 29,000 ksi is steel's modulus of elasticity, not a strength.

In LRFD design, which load combination correctly factors dead load (D) and live load (L) for a basic gravity case?

• a.1.0D + 1.0L
• b.1.4D + 1.7L
• c.0.9D + 1.6L
• d.1.2D + 1.6L✓

The governing ASCE 7 strength combination for dead plus live load is 1.2D + 1.6L. The higher factor on live load reflects its greater uncertainty. The 1.4D + 1.7L form is the older ACI ultimate-strength combination, not current LRFD.

A simply supported beam of span 6 m carries a single concentrated load of 20 kN at midspan. What is the maximum bending moment?

• a.15 kN·m
• b.120 kN·m
• c.30 kN·m✓
• d.60 kN·m

For a central point load on a simply supported beam, the maximum moment is P·L/4 = 20 x 6 / 4 = 30 kN·m. The moment diagram is triangular, peaking under the load.

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.

Using the Rational Method, what is the peak runoff for a 20-acre watershed with runoff coefficient C = 0.5 and rainfall intensity i = 3 in/hr?

• a.3 cfs
• b.60 cfs
• c.30 cfs✓
• d.10 cfs

The Rational Method is Q = C·i·A, where in U.S. units i is in in/hr and A in acres, giving Q directly in cfs (the unit conversion factor is approximately 1). Q = 0.5 x 3 x 20 = 30 cfs. The method suits small urban drainage areas.

Water flows at 2 m/s through a channel with cross-sectional flow area 0.5 m^2. What is the discharge?

• a.4 m^3/s
• b.1 m^3/s✓
• c.0.5 m^3/s
• d.2 m^3/s

The continuity equation gives discharge Q = V·A = 2 x 0.5 = 1 m^3/s. For steady incompressible flow, discharge is constant along the channel even as area and velocity vary.

By Manning's equation, if the channel slope is doubled (all else equal), the flow velocity increases by approximately what factor?

• a.4
• b.2
• c.1
• d.1.41✓

Manning's velocity is V = (1/n)·R^(2/3)·S^(1/2), so velocity is proportional to the square root of slope. Doubling S multiplies velocity by sqrt(2) = 1.41. Slope has a weaker-than-linear effect on velocity.

The resultant hydrostatic force on a vertical rectangular gate holding back water acts at what location, measured from the bottom of the gate?

• a.2H/3 above the bottom
• b.H/3 above the bottom✓
• c.H/2 (the centroid)
• d.H/4 above the bottom

Because hydrostatic pressure increases linearly with depth, the pressure distribution is triangular and its resultant acts at the centroid of that triangle, which is H/3 above the bottom (or 2H/3 below the surface). This is the center of pressure.

In the Rational Method, the design storm duration is typically taken equal to what quantity so that the entire watershed contributes to peak flow?

• a.One hour
• b.The return period
• c.The time of concentration✓
• d.The storm frequency

Peak runoff occurs when the storm duration equals the time of concentration, the time for runoff to travel from the hydraulically most distant point to the outlet. At that duration the whole watershed contributes simultaneously, maximizing the peak.

What is the degree of curve (arc definition) for a horizontal curve with radius 1,000 ft?

• a.11.46 degrees
• b.5.73 degrees✓
• c.2.86 degrees
• d.1.0 degrees

By the arc definition, D = 5,729.58 / R = 5,729.58 / 1,000 = 5.73 degrees. The degree of curve is the central angle subtending a 100-ft arc; a larger radius gives a flatter (smaller-degree) curve.

Stopping sight distance (SSD) is defined as the sum of the brake-reaction distance and which other component?

• a.The braking (deceleration) distance✓
• b.The roadway grade
• c.The horizontal curve length
• d.The superelevation runoff

SSD = brake-reaction distance (driver perception-reaction time times speed) + braking distance (distance to decelerate to a stop). Both components grow with speed, so SSD increases sharply at higher design speeds.

Using the fundamental traffic-flow relationship, what is the flow rate for a density of 40 veh/mi and a space-mean speed of 60 mph?

• a.2,400 veh/hr✓
• b.1.5 veh/hr
• c.240 veh/hr
• d.100 veh/hr

The fundamental relationship is flow = density x speed: q = k·v = 40 x 60 = 2,400 veh/hr. This links the three primary traffic-stream variables and defines capacity at the flow's peak.

On a horizontal curve, superelevation (banking of the roadway) is provided primarily to counteract what?

• a.The stopping sight distance
• b.The lateral (centripetal) demand of vehicles rounding the curve✓
• c.The vertical grade
• d.The pavement thickness requirement

Superelevation tilts the roadway inward so that a component of the vehicle's weight, together with side friction, supplies the centripetal force needed to turn. This reduces reliance on tire friction and improves safety and comfort on curves.

A crest vertical curve is 200 ft long and connects grades that differ by an algebraic total of 4%. What is the rate of vertical curvature, K?

• a.4
• b.800
• c.50✓
• d.0.02

K = L / A, where L is the curve length in feet and A is the algebraic difference in grades in percent. K = 200 / 4 = 50 ft per percent. K expresses the horizontal distance needed to change the grade by 1% and is checked against sight-distance minimums.

A task has an earned value (EV) of $8,000 and an actual cost (AC) of $10,000. What is the Cost Performance Index (CPI)?

• a.0.8✓
• b.0.2
• c.1.25
• d.1.0

CPI = EV / AC = 8,000 / 10,000 = 0.8. A CPI below 1.0 means the work is over budget (you earned less value than you spent). CPI is a core earned-value metric for project cost control.

In CPM scheduling, the total float of an activity is correctly computed as:

• a.The activity duration
• b.Early finish minus early start
• c.Late finish minus late start
• d.Late finish minus early finish✓

Total float = Late Finish - Early Finish (equivalently Late Start - Early Start). It is the time an activity can slip without delaying project completion. Activities on the critical path have zero total float.

A soil has a bank (in-place) volume of 100 cubic yards and a swell of 25%. What is the loose (excavated) volume?

• a.80 cubic yards
• b.125 cubic yards✓
• c.75 cubic yards
• d.100 cubic yards

Loose volume = bank volume x (1 + swell) = 100 x 1.25 = 125 cubic yards. Excavating loosens soil so it occupies more volume; swell governs hauling quantities, while shrinkage governs compacted fill.

A concrete slab measures 20 ft by 20 ft by 0.5 ft thick. What volume of concrete is required, in cubic yards?

• a.27 cubic yards
• b.200 cubic yards
• c.7.4 cubic yards✓
• d.3.7 cubic yards

Volume = 20 x 20 x 0.5 = 200 cubic feet. Converting to cubic yards, divide by 27: 200 / 27 = 7.4 cubic yards. Ordering concrete requires this conversion because ready-mix is sold by the cubic yard.

For a given set of materials, reducing the water-cement ratio of a concrete mix generally does what to compressive strength?

• a.Has no effect on it
• b.Makes it unpredictable
• c.Increases it✓
• d.Decreases it

By Abrams' law, lower water-cement ratio yields higher compressive strength (down to the point where the mix becomes unworkable). Excess water creates capillary voids that weaken the hardened paste. Admixtures maintain workability at low water-cement ratios.

What is the nominal diameter of a #8 reinforcing bar (U.S. customary sizes)?

• a.0.375 in
• b.0.75 in
• c.0.5 in
• d.1.0 in✓

U.S. rebar bar numbers give the diameter in eighths of an inch: #8 = 8/8 = 1.0 in. Thus a #4 bar is 0.5 in and a #3 bar is 0.375 in. This rule lets you size reinforcement quickly from the bar number.

The slump test on fresh concrete is primarily a measure of what property?

• a.Workability (consistency)✓
• b.Compressive strength
• c.Water-cement ratio
• d.Air content

The slump test measures the consistency and workability of fresh concrete, that is, how easily it flows and can be placed. It does not directly measure strength; separate cylinder tests determine compressive strength, and other tests measure air content.

A cantilever beam 3 m long carries a single concentrated load of 10 kN at its free end. What is the maximum bending moment?

• a.10 kN·m
• b.30 kN·m✓
• c.45 kN·m
• d.15 kN·m

For a cantilever with a point load at the free end, the maximum moment occurs at the fixed support and equals P·L = 10 x 3 = 30 kN·m. The moment diagram is triangular, zero at the free end and maximum at the fixation.

A cantilever beam 4 m long carries a uniformly distributed load of 5 kN/m over its full length. What is the maximum bending moment?

• a.40 kN·m✓
• b.10 kN·m
• c.80 kN·m
• d.20 kN·m

For a cantilever under a uniform load, the maximum moment is at the fixed end and equals w·L^2/2 = 5 x 4^2 / 2 = 5 x 16 / 2 = 40 kN·m. Contrast with a simply supported beam, where the same load gives only w·L^2/8.

A simply supported beam of span 10 m carries a uniformly distributed load of 6 kN/m. What is the maximum shear force?

• a.30 kN✓
• b.15 kN
• c.60 kN
• d.45 kN

Maximum shear in a simply supported beam under a uniform load occurs at the supports and equals the reaction, w·L/2 = 6 x 10 / 2 = 30 kN. Shear varies linearly from +wL/2 at one support to -wL/2 at the other, crossing zero at midspan.

A simply supported beam AB spans 6 m. A single 12 kN point load acts 2 m from support A. What is the vertical reaction at A?

• a.6 kN
• b.4 kN
• c.8 kN✓
• d.12 kN

Taking moments about B: R_A·L = P·b, where b = 4 m is the distance from the load to B. R_A = P·b/L = 12 x 4 / 6 = 8 kN. The reaction is larger at the support nearer the load.

A simply supported beam (span 4 m) carries a 48 kN point load at midspan. Given E = 200 GPa and I = 100x10^6 mm^4, what is the maximum deflection?

• a.1.6 mm
• b.0.8 mm
• c.3.2 mm✓
• d.6.4 mm

Midspan deflection for a central point load is delta = P·L^3/(48·E·I) = 48,000 x 4000^3 / (48 x 200,000 x 100x10^6) = 3.072x10^15 / 9.6x10^14 = 3.2 mm. Keep units consistent (N and mm) so E is in MPa.

What is the radius of gyration about the strong axis of a rectangular section 200 mm wide and 300 mm deep?

• a.122 mm
• b.86.6 mm✓
• c.75 mm
• d.100 mm

Radius of gyration r = sqrt(I/A). For a rectangle bending about the strong axis this reduces to r = h/sqrt(12) = 300/3.464 = 86.6 mm. The radius of gyration measures how the area is distributed about the axis and governs column slenderness.

A beam section has a section modulus of 3.0x10^6 mm^3 and resists a bending moment of 90 kN·m. What is the maximum bending stress?

• a.30 MPa✓
• b.90 MPa
• c.15 MPa
• d.45 MPa

Bending stress sigma = M/S = 90x10^6 N·mm / 3.0x10^6 mm^3 = 30 MPa (N/mm^2). The flexure formula sigma = M·c/I = M/S gives the peak stress at the extreme fiber.

A pinned-pinned steel column is 15 ft long with I = 100 in^4 and E = 29,000 ksi. What is the Euler critical buckling load?

• a.1,766 kips
• b.442 kips
• c.883 kips✓
• d.221 kips

Euler's formula: P_cr = pi^2·E·I/(K·L)^2 with K = 1 for pinned ends and L = 15 ft = 180 in. P_cr = pi^2 x 29,000 x 100 / 180^2 = 2.862x10^7 / 32,400 = 883 kips. Capacity drops with the square of the length.

What is the theoretical effective-length factor K for a column fixed against rotation and translation at both ends (braced against sidesway)?

• a.2.0
• b.1.0
• c.0.7
• d.0.5✓

For a column fixed at both ends with no sidesway the theoretical K = 0.5. Pinned-pinned gives K = 1.0, fixed-pinned gives 0.7, and a cantilever (fixed-free) gives 2.0. Lower K means a shorter effective length and higher buckling capacity.

A pinned column (K = 1) is 4 m long and its cross-section has a radius of gyration of 50 mm. What is the slenderness ratio?

• a.200
• b.80✓
• c.20
• d.40

Slenderness ratio = K·L/r = (1 x 4000 mm)/50 mm = 80. This dimensionless ratio controls whether a column fails by inelastic or elastic (Euler) buckling; higher values mean a more slender, buckling-prone column.

What is the plastic section modulus Z of a rectangular section 200 mm wide and 300 mm deep about its strong axis?

• a.3.0x10^6 mm^3
• b.2.25x10^6 mm^3
• c.4.5x10^6 mm^3✓
• d.9.0x10^6 mm^3

The plastic section modulus of a rectangle is Z = b·h^2/4 = 200 x 300^2 / 4 = 200 x 90,000 / 4 = 4.5x10^6 mm^3. Z is used for the fully plastic moment M_p = Z·Fy, whereas the elastic S = b·h^2/6 governs first yield.

What is the shape factor (ratio of plastic to elastic section modulus) for a solid rectangular cross-section?

• a.1.5✓
• b.1.7
• c.2.0
• d.1.0

Shape factor = Z/S = (b·h^2/4)/(b·h^2/6) = 6/4 = 1.5 for a rectangle. It represents the reserve strength between first yield and full plastification; wide-flange shapes have a lower shape factor near 1.1-1.2.

A steel tension member has a gross area of 10 in^2 and is made of Fy = 50 ksi steel. What is the nominal tensile yielding strength on the gross section?

• a.250 kips
• b.450 kips
• c.1,000 kips
• d.500 kips✓

Nominal yielding strength P_n = Fy·Ag = 50 x 10 = 500 kips. This is the yielding-on-the-gross-section limit state; a separate rupture check uses Fu on the net effective area, and the lower of the two governs.

For that same tension member (P_n = 500 kips for yielding), what is the LRFD design tensile strength for the yielding limit state?

• a.500 kips
• b.333 kips
• c.550 kips
• d.450 kips✓

For tensile yielding the resistance factor is phi = 0.90, so phi·P_n = 0.90 x 500 = 450 kips. (Rupture uses phi = 0.75 on Fu·Ae.) The design strength must equal or exceed the factored load.

A singly reinforced concrete beam has As = 3.0 in^2, fy = 60 ksi, f'c = 4 ksi, and b = 12 in. What is the depth of the equivalent (Whitney) stress block, a?

• a.8.82 in
• b.3.53 in
• c.4.41 in✓
• d.2.21 in

From horizontal equilibrium of the Whitney block, a = As·fy/(0.85·f'c·b) = (3.0 x 60)/(0.85 x 4 x 12) = 180/40.8 = 4.41 in. The 0.85·f'c uniform stress replaces the true parabolic concrete compression distribution.

For that beam (As = 3.0 in^2, fy = 60 ksi, a = 4.41 in, d = 20 in), what is the nominal flexural strength M_n?

• a.240 k-ft
• b.300 k-ft
• c.267 k-ft✓
• d.180 k-ft

M_n = As·fy·(d - a/2) = 180 x (20 - 2.205) = 180 x 17.795 = 3,203 k-in = 267 k-ft. The internal moment is the steel tension force times the lever arm to the compression resultant.

For a tension-controlled RC beam with M_n = 267 k-ft, what is the LRFD design flexural strength phi·M_n?

• a.240 k-ft✓
• b.300 k-ft
• c.267 k-ft
• d.213 k-ft

For a tension-controlled section (steel strain >= 0.005) the flexural resistance factor is phi = 0.90, so phi·M_n = 0.90 x 267 = 240 k-ft. Designing tension-controlled ensures ductile behavior with ample warning before failure.

What is the nominal cross-sectional area of a #8 reinforcing bar (U.S. customary)?

• a.0.60 in^2
• b.0.44 in^2
• c.1.00 in^2
• d.0.79 in^2✓

A #8 bar has a 1.0 in diameter, so its area = pi/4 x 1.0^2 = 0.79 in^2. Standard tabulated bar areas (0.11, 0.20, 0.31, 0.44, 0.60, 0.79 in^2 for #3 to #8) are memorized shortcuts on the PE exam.

A beam requires As = 3.0 in^2 of tension steel. Using #8 bars (0.79 in^2 each), what is the minimum number of bars needed?

• a.3 bars
• b.6 bars
• c.5 bars
• d.4 bars✓

Number of bars = required area / area per bar = 3.0 / 0.79 = 3.8, rounded UP to 4 bars (providing 3.16 in^2). You must round up so the supplied steel meets or exceeds the requirement.

At a truss joint, a diagonal member at 45 degrees and a horizontal member meet, with a 10 kN vertical load applied downward at the joint. What is the axial force in the diagonal?

• a.10 kN
• b.7.07 kN
• c.20 kN
• d.14.14 kN✓

Only the diagonal has a vertical component, so by sum(Fy) = 0 its vertical component balances the 10 kN load: F·sin(45) = 10, giving F = 10/0.707 = 14.14 kN. The horizontal member carries no vertical force by the method of joints.