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.

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.

For that same joint (diagonal force 14.14 kN at 45 degrees), what is the axial force in the horizontal member?

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

By sum(Fx) = 0, the horizontal member balances the diagonal's horizontal component: F·cos(45) = 14.14 x 0.707 = 10 kN. Resolving the diagonal into components at the joint gives both unknown member forces.

Which ASCE 7 LRFD load combination governs when wind acts to counteract (uplift) the dead load?

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

The uplift-check combination is 0.9D + 1.0W, using the minimum (0.9) dead-load factor so gravity does not mask net uplift from wind. Using 1.2D here would unconservatively rely on more dead load resisting the wind.

A beam fixed at both ends carries a uniform load of 12 kN/m over a 6 m span. What is the maximum negative moment at the supports?

• a.54 kN·m
• b.18 kN·m
• c.36 kN·m✓
• d.27 kN·m

For a fixed-fixed beam under uniform load, the support (negative) moment is w·L^2/12 = 12 x 6^2 / 12 = 36 kN·m. End fixity shifts the peak moment to the supports, reducing the midspan moment compared with a simple span.

For that fixed-fixed beam (w = 12 kN/m, L = 6 m), what is the positive moment at midspan?

• a.36 kN·m
• b.24 kN·m
• c.18 kN·m✓
• d.9 kN·m

The midspan positive moment of a fixed-fixed beam under uniform load is w·L^2/24 = 12 x 36 / 24 = 18 kN·m, exactly half the support moment of w·L^2/12. Fixity roughly balances positive and negative moments.

A propped cantilever (fixed at one end, simply supported at the other) of span 4 m carries a uniform load of 10 kN/m. What is the maximum moment (at the fixed end)?

• a.40 kN·m
• b.20 kN·m✓
• c.10 kN·m
• d.13.3 kN·m

For a propped cantilever under uniform load the maximum moment is at the fixed end, M = w·L^2/8 = 10 x 4^2 / 8 = 20 kN·m. The prop reaction is 3wL/8 and the fixed-end reaction is 5wL/8.

A rectangular beam 200 mm x 300 mm carries a shear force of 60 kN. What is the maximum transverse shear stress?

• a.1.5 MPa✓
• b.1.0 MPa
• c.2.0 MPa
• d.3.0 MPa

For a rectangular section the maximum shear stress at the neutral axis is 1.5 times the average: tau_max = 1.5·V/A = 1.5 x 60,000 / (200 x 300) = 1.5 x 60,000/60,000 = 1.5 MPa. The parabolic VQ/Ib distribution peaks at the centroid.

A rectangular area 100 mm x 20 mm (I about its own centroid = 66,667 mm^4) has its centroid located 50 mm from a reference axis. What is its moment of inertia about that reference axis?

• a.66,667 mm^4
• b.2.5x10^6 mm^4
• c.5.07x10^6 mm^4✓
• d.5.0x10^6 mm^4

By the parallel-axis theorem, I = I_c + A·d^2 = 66,667 + (2000)(50^2) = 66,667 + 5,000,000 = 5.07x10^6 mm^4. The A·d^2 transfer term dominates when the area is far from the reference axis.

A short member carries an axial compression giving P/A = 10 MPa plus a bending moment giving M/S = 20 MPa. What is the maximum compressive stress on the section?

• a.30 MPa✓
• b.20 MPa
• c.10 MPa
• d.-10 MPa

For combined axial and bending the stress is sigma = P/A +/- M/S. On the fiber where both add, sigma_max = 10 + 20 = 30 MPa compression. Superposition of the uniform axial stress and the linear bending stress gives the extreme-fiber value.

For that same member (axial P/A = 10 MPa compression, bending M/S = 20 MPa), what is the stress on the opposite extreme fiber?

• a.30 MPa compression
• b.10 MPa compression
• c.30 MPa tension
• d.10 MPa tension✓

On the far fiber the bending stress opposes the axial stress: sigma = P/A - M/S = 10 - 20 = -10 MPa, i.e., 10 MPa tension. When bending exceeds the axial stress, part of the section goes into tension despite the net compressive load.

What is the modulus of elasticity of structural steel?

• a.29,000 ksi✓
• b.10,000 ksi
• c.11,200 ksi
• d.3,600 ksi

Structural steel has E = 29,000 ksi (about 200 GPa), a nearly constant value independent of grade. Aluminum is about 10,000 ksi and normal-weight concrete about 3,600 ksi; 11,200 ksi is steel's shear modulus G.

What is the modulus of elasticity of normal-weight concrete with f'c = 4,000 psi, using the ACI formula Ec = 57,000·sqrt(f'c)?

• a.3,605,000 psi✓
• b.1,800,000 psi
• c.57,000 psi
• d.4,000 psi

Ec = 57,000·sqrt(f'c) = 57,000 x sqrt(4,000) = 57,000 x 63.25 = 3,605,000 psi (about 3,605 ksi). Concrete stiffness rises with the square root of strength, not linearly.

For steel (Es = 29,000 ksi) embedded in concrete with Ec = 3,605 ksi, what is the modular ratio n?

• a.6
• b.29
• c.10
• d.8✓

The modular ratio n = Es/Ec = 29,000/3,605 = 8.0. It converts steel area to an equivalent concrete area in transformed-section analysis of cracked reinforced concrete.

What maximum usable concrete compressive strain does ACI assume at the extreme fiber for flexural strength design?

• a.0.005
• b.0.003✓
• c.0.001
• d.0.002

ACI 318 assumes the concrete crushes at a maximum usable strain of 0.003. The 0.005 value is the tensile-strain threshold that defines a tension-controlled section, not the concrete crushing strain.