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

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.