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

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.

What is the ACI strength-reduction factor phi for a tension-controlled flexural member?

• a.0.75
• b.0.65
• c.0.85
• d.0.90✓

Tension-controlled flexure uses phi = 0.90. Compression-controlled (tied column) sections use 0.65, and shear/torsion use 0.75. The higher factor rewards the ductile, well-warned failure mode of tension-controlled beams.

What is the ACI strength-reduction factor phi for shear and torsion?

• a.0.65
• b.0.90
• c.0.85
• d.0.75✓

Shear and torsion use phi = 0.75. Shear failures are more brittle and less predictable than flexural yielding, so ACI assigns a lower resistance factor than the 0.90 used for tension-controlled flexure.

A rectangular RC beam has b = 12 in, d = 20 in, and f'c = 4,000 psi. What is the nominal concrete shear strength Vc = 2·sqrt(f'c)·b·d?

• a.45 kips
• b.15.2 kips
• c.30.4 kips✓
• d.60.7 kips

Vc = 2·sqrt(f'c)·b·d = 2 x sqrt(4,000) x 12 x 20 = 2 x 63.25 x 240 = 30,357 lb = 30.4 kips. This simplified expression estimates the shear carried by the concrete before stirrups are required.

A simply supported beam of span 9 m carries two equal 15 kN loads placed at the third points (3 m from each support). What is the maximum bending moment?

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

Each reaction equals 15 kN. The moment is maximum and constant between the loads, M = R·a = 15 x 3 = 45 kN·m. Third-point loading produces a region of pure bending (constant moment, zero shear) at midspan, as used in flexural test setups.

For that third-point-loaded beam (two 15 kN loads, span 9 m), what is the maximum shear force?

• a.7.5 kN
• b.45 kN
• c.15 kN✓
• d.30 kN

Each support reaction is 15 kN, and the shear between a support and the nearer load equals that reaction, 15 kN. Between the two loads the shear is zero, which is why the moment is constant there.

For a simply supported beam under a symmetric uniformly distributed load, where does the maximum bending moment occur?

• a.At midspan✓
• b.At the third point
• c.At the quarter point
• d.At a support

The maximum moment occurs at midspan, where the shear diagram crosses zero. Since dM/dx = V, the bending moment is extreme exactly where the shear is zero, which for symmetric uniform load is the center.

In a beam, a point of contraflexure (inflection point) is a location where which quantity is zero?

• a.Bending moment✓
• b.Reaction
• c.Deflection
• d.Shear force

A point of contraflexure is where the bending moment changes sign and passes through zero, so the curvature reverses. Identifying inflection points is useful for locating rebar cutoffs and splice locations in continuous members.

In beam theory, the slope of the bending-moment diagram at any point equals which quantity?

• a.The curvature
• b.The shear force✓
• c.The deflection
• d.The distributed load

The differential relationship is dM/dx = V, so the shear force equals the slope of the moment diagram. Likewise dV/dx = -w. These relationships let you sketch shear and moment diagrams by integration and differentiation.

In beam theory, the slope of the shear-force diagram at any point equals the negative of which quantity?

• a.The slope of the beam
• b.The support reaction
• c.The bending moment
• d.The distributed load intensity✓

The relationship is dV/dx = -w, so the shear diagram's slope equals the negative of the distributed load intensity. Under uniform load the shear diagram is a straight sloping line; where there is no load, shear stays constant.

An overhanging beam has supports at A (x = 0) and B (x = 6 m) and a 2 m overhang to point C (x = 8 m) with a 9 kN downward load at C. What is the reaction at B?

• a.9 kN
• b.3 kN
• c.6 kN
• d.12 kN✓

Sum of moments about A: R_B x 6 = 9 x 8, so R_B = 72/6 = 12 kN. The far support carries more than the total load because the overhang load lever arm exceeds the support spacing, forcing an offsetting downward reaction at A.

For that overhanging beam (9 kN at the tip, R_B = 12 kN), what is the reaction at A?

• a.3 kN (upward)
• b.9 kN (upward)
• c.12 kN (upward)
• d.-3 kN (downward)✓

Sum of vertical forces: R_A + R_B = 9, so R_A = 9 - 12 = -3 kN, i.e., 3 kN acting downward. Overhang loads can produce uplift (a hold-down force) at the far support, which the anchorage must resist.

A solid circular shaft 100 mm in diameter (J = 9.82x10^6 mm^4) carries a torque of 5 kN·m. What is the maximum torsional shear stress?

• a.25.5 MPa✓
• b.100 MPa
• c.50.9 MPa
• d.12.7 MPa

Torsion formula: tau = T·r/J = (5x10^6 N·mm x 50 mm)/9.82x10^6 mm^4 = 25.5 MPa. Shear stress varies linearly from zero at the axis to a maximum at the outer radius r = 50 mm.

What is the polar moment of inertia J of a solid circular shaft 100 mm in diameter?

• a.19.6x10^6 mm^4
• b.4.91x10^6 mm^4
• c.9.82x10^6 mm^4✓
• d.1.23x10^6 mm^4

For a solid circle J = pi·d^4/32 = pi x 100^4 / 32 = pi x 1x10^8 / 32 = 9.82x10^6 mm^4. J governs torsional stiffness; the bending moment of inertia I = pi·d^4/64 is exactly half of J.

A fully restrained steel bar (E = 200 GPa, alpha = 12x10^-6 /degC) is heated by 30 degC. What thermal stress develops?

• a.36 MPa
• b.144 MPa
• c.24 MPa
• d.72 MPa✓

A fully restrained member cannot expand, so sigma = E·alpha·(delta T) = 200,000 x 12x10^-6 x 30 = 72 MPa (compression). The stress is independent of the member length; only the temperature change and material properties matter.

An unrestrained steel bar 10 m long (alpha = 12x10^-6 /degC) is heated by 25 degC. How much does it elongate?

• a.0.3 mm
• b.6.0 mm
• c.3.0 mm✓
• d.1.5 mm

Free thermal expansion: delta L = alpha·L·(delta T) = 12x10^-6 x 10,000 mm x 25 = 3.0 mm. Because the bar is unrestrained it expands freely with no thermal stress; stress only develops when expansion is prevented.

A steel rod (A = 1,000 mm^2, L = 2 m, E = 200 GPa) carries an axial tension of 100 kN. What is its elongation?

• a.0.5 mm
• b.0.25 mm
• c.2.0 mm
• d.1.0 mm✓

Axial deformation: delta = P·L/(A·E) = (100,000 x 2000)/(1000 x 200,000) = 2x10^8/2x10^8 = 1.0 mm. Elongation grows with load and length but shrinks with a larger area or stiffer material.

For that steel rod (A = 1,000 mm^2) under 100 kN axial tension, what is the axial normal stress?

• a.200 MPa
• b.100 MPa✓
• c.50 MPa
• d.10 MPa

Axial stress sigma = P/A = 100,000 N / 1,000 mm^2 = 100 MPa. This uniform normal stress over the cross-section is the basis for the elongation and strain calculations.

A steel member is stressed to 100 MPa within its elastic range (E = 200 GPa). What is the axial strain?

• a.5x10^-4✓
• b.2.5x10^-4
• c.5x10^-3
• d.1x10^-3

By Hooke's law, strain = sigma/E = 100/200,000 = 5x10^-4 (0.05%). Strain is the ratio of deformation to original length and is dimensionless; it stays linear with stress until yielding.

A steel bar undergoes an axial strain of 5x10^-4. With Poisson's ratio 0.3, what is the magnitude of the lateral (transverse) strain?

• a.1.5x10^-4✓
• b.5x10^-4
• c.1.67x10^-4
• d.3x10^-4

Lateral strain = -nu x axial strain = -0.3 x 5x10^-4 = -1.5x10^-4 (a contraction). Poisson's ratio relates the transverse contraction to the axial extension; for steel it is about 0.3.

A material with an ultimate stress of 400 MPa is used at an allowable working stress of 160 MPa. What is the factor of safety?

• a.2.5✓
• b.1.6
• c.4.0
• d.0.4

Factor of safety = ultimate (or yield) strength / allowable stress = 400/160 = 2.5. The factor of safety provides a margin against overload, material variability, and analysis uncertainty.

In AISC Allowable Strength Design (ASD), what safety factor Omega is applied to tensile yielding on the gross section?

• a.1.5
• b.1.0
• c.2.0
• d.1.67✓

ASD tensile yielding uses Omega = 1.67, so the allowable strength is P_n/1.67. This corresponds to the LRFD phi = 0.90 for the same limit state; tensile rupture uses the more conservative Omega = 2.0.

A single 20 mm diameter bolt (area 314 mm^2) carries a 20 kN shear force. What is the average shear stress in the bolt?

• a.127 MPa
• b.63.7 MPa✓
• c.100 MPa
• d.31.8 MPa

Average shear stress = V/A = 20,000 N / 314 mm^2 = 63.7 MPa. For a bolt in single shear the full force acts on one cross-sectional plane; in double shear the area would double and the stress would halve.

What is the effective throat thickness of an 8 mm equal-leg fillet weld?

• a.4.0 mm
• b.5.66 mm✓
• c.8.0 mm
• d.11.3 mm

The effective throat of an equal-leg fillet weld is 0.707 times the leg size: 0.707 x 8 = 5.66 mm. The throat is the shortest distance across the weld triangle and is the plane on which the weld's shear strength is computed.

In reinforced concrete flexure, the 'balanced' condition is defined as the state in which:

• a.A shear failure occurs first
• b.Only the concrete fails
• c.The tension steel yields just as the concrete reaches its crushing strain✓
• d.Only the steel fails

At the balanced condition the tension steel reaches its yield strain at exactly the same instant the extreme concrete fiber reaches 0.003. Designers keep sections under-reinforced (below balanced) so steel yields first, giving a ductile, well-warned failure.

Why does ACI require flexural members to be under-reinforced (tension-controlled) rather than over-reinforced?

• a.It provides higher strength
• b.It gives a ductile failure with visible warning✓
• c.It has no structural effect
• d.It gives a brittle failure

An under-reinforced (tension-controlled) beam yields the steel before the concrete crushes, producing large deflections and cracking that warn of impending failure. An over-reinforced beam fails by sudden concrete crushing with little warning, which is prohibited.

If the unbraced length of an ideal Euler column is doubled (all else equal), the critical buckling load changes by what factor?

• a.1/2
• b.4 times
• c.1/4✓
• d.2 times

Euler's load P_cr = pi^2·E·I/(K·L)^2 is inversely proportional to the square of the length. Doubling L multiplies the denominator by 4, so the capacity drops to one-quarter. Bracing to reduce unbraced length is a very efficient way to raise column capacity.

A simply supported beam (span 6 m) carries a uniform load of 8 kN/m. What is the shear force at the quarter point, 1.5 m from a support?

• a.0 kN
• b.12 kN✓
• c.24 kN
• d.18 kN

Reaction = wL/2 = 24 kN. Shear at distance x is V = wL/2 - w·x = 24 - 8 x 1.5 = 12 kN. Shear decreases linearly from the support reaction toward zero at midspan.

For that beam (8 kN/m over 6 m), what is the bending moment at the quarter point (1.5 m from a support)?

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

Moment at x for a uniformly loaded simple span is M = w·x·(L - x)/2 = 8 x 1.5 x (6 - 1.5)/2 = 8 x 1.5 x 4.5 / 2 = 27 kN·m. It rises parabolically from zero at the support to the midspan maximum.

For that same beam (8 kN/m over a 6 m simple span), what is the maximum bending moment at midspan?

• a.36 kN·m✓
• b.48 kN·m
• c.18 kN·m
• d.27 kN·m

Maximum moment for a uniformly loaded simple span is w·L^2/8 = 8 x 6^2 / 8 = 8 x 36 / 8 = 36 kN·m. This exceeds the quarter-point value (27 kN·m), consistent with the parabolic moment diagram peaking at the center.

For a single concentrated load moving across a simple span, the absolute maximum bending moment occurs when the load is located where?

• a.At a support
• b.At the quarter point
• c.At midspan✓
• d.At the far end

For one moving point load the maximum moment develops when the load sits at midspan, giving M = P·L/4. Influence-line reasoning shows the midspan ordinate is the largest, so that position produces the governing design moment.

A short tied RC column has Ag = 144 in^2, Ast = 4 in^2, f'c = 4 ksi, and fy = 60 ksi. What is the nominal axial capacity P_o = 0.85·f'c·(Ag - Ast) + fy·Ast?

• a.816 kips
• b.490 kips
• c.600 kips
• d.716 kips✓

P_o = 0.85 x 4 x (144 - 4) + 60 x 4 = 0.85 x 4 x 140 + 240 = 476 + 240 = 716 kips. The concrete carries 0.85·f'c over the net area while the steel carries fy; ACI then applies phi and a 0.80 (tied) reduction for accidental eccentricity.

For a wide-flange (W-shape) steel beam bent about its strong axis, which axis is the strong axis?

• a.The x-x (major) axis✓
• b.The y-y (minor) axis
• c.Any axis, they are equal
• d.The vertical axis of the web

The strong axis is the x-x (major) axis, about which the moment of inertia is largest because the flanges are far from that neutral axis. Orienting the web vertical so bending is about x-x maximizes flexural stiffness and strength.

Lateral-torsional buckling is a limit state that primarily affects which condition of a steel beam?

• a.A fully laterally braced beam
• b.A beam whose compression flange is laterally unsupported over a long length✓
• c.A short, stocky beam
• d.A purely axially loaded member

Lateral-torsional buckling occurs when a beam's compression flange is not braced and the beam twists and displaces sideways under load. Adequate lateral bracing of the compression flange raises the flexural capacity to the full plastic moment.

A common serviceability deflection limit for live load on a typical floor beam is which fraction of the span L?

• a.L/360✓
• b.L/240
• c.L/180
• d.L/500

L/360 is a widely used live-load deflection limit for floor members supporting brittle finishes such as plaster. Serviceability limits control comfort and finish cracking; they are checked with service (unfactored) loads, separate from strength design.

At a section, V = 50 kN, the first moment of the connected area Q = 1x10^6 mm^3, and I = 1x10^8 mm^4. What is the shear flow?

• a.5,000 N/mm
• b.250 N/mm
• c.500 N/mm✓
• d.50 N/mm

Shear flow q = V·Q/I = 50,000 x 1x10^6 / 1x10^8 = 500 N/mm. Shear flow gives the force per unit length that fasteners or welds must transfer between connected parts of a built-up member.

What unit weight is commonly assumed for normal-weight reinforced concrete?

• a.150 pcf✓
• b.100 pcf
• c.62.4 pcf
• d.490 pcf

Normal-weight reinforced concrete is taken as about 150 pcf (roughly 24 kN/m^3). Water is 62.4 pcf and structural steel is 490 pcf; the 150 pcf value is used to estimate concrete self-weight for dead loads.

A concrete beam has a 12 in x 24 in cross-section (concrete = 150 pcf). What is its self-weight per foot of length?

• a.200 lb/ft
• b.150 lb/ft
• c.300 lb/ft✓
• d.600 lb/ft

Cross-sectional area = 12 x 24 = 288 in^2 = 2.0 ft^2. Self-weight = area x unit weight = 2.0 ft^2 x 150 lb/ft^3 = 300 lb/ft. Member self-weight is a distributed dead load that must be added to the applied loads.