Structural Engineering is one of the three largest civil sections, 10–15 questions on analysis and on member design in steel and reinforced concrete. This page also covers the civil side of statics and mechanics of materials: trusses, frames and beams of buildings and bridges.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Civil
questions on the exam10–15
problems here18
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What the section covers
The July 2020 specification lists eight topics:
Analysis of statically determinate beams, columns, trusses and frames.
Deflection of determinate beams, trusses and frames.
Column analysis: buckling and boundary conditions.
Determinacy and stability of beams, trusses and frames.
Elementary statically indeterminate structures.
Loads, load combinations and load paths: dead, live and lateral loads, influence lines and moving loads, tributary areas.
Design of steel components: beams, columns, tension members, connections, and the design philosophies behind them.
Design of reinforced concrete beams and columns, and the design philosophies behind them.
18 structural problems
0 / 18 answered · 0 correct
Question 1 of 18 · Structural Engineering · 10–15 items
What is the plastic section modulus Z of a solid rectangular cross-section of width b and depth h, bent about its strong axis?
Shape factor Z/S = 1.5 for a rectangle. FE Reference Handbook, Structural Design — plastic moment.
bh²/8 has no meaning for a rectangle's section properties; the wL²/8 beam-moment pattern is a common mix-up.
At full plastification each half of the section, area bh/2, acts at h/4 from the neutral axis, so Z = 2(bh/2)(h/4) = bh²/4.
bh³/12 is the moment of inertia I of the rectangle, not a section modulus.
bh²/6 is the elastic section modulus S = I/c, which governs first yield, not the fully plastic moment.
Question 2 of 18 · Structural Engineering · 10–15 items
What is the primary purpose of shear reinforcement in reinforced concrete beams?
Solution
Shear reinforcement (stirrups) carries Vs across inclined diagonal-tension cracks:
φVn = φ(Vc + Vs)
FE Reference Handbook, Structural Design — reinforced concrete beams, shear.
Closed stirrups help with torsion in spandrels too, but the primary role of beam shear reinforcement is diagonal tension, so 'only' makes this wrong.
Shrinkage and temperature steel controls those cracks, mainly in slabs and walls, not stirrups in beams.
Flexural strength comes from the longitudinal tension steel; stirrups do not add to the moment capacity.
Stirrups cross the inclined cracks that shear and moment together produce near supports, carrying the diagonal tension the concrete alone cannot.
Question 3 of 18 · Structural Engineering · 10–15 items
What is the balanced reinforcement ratio in a reinforced concrete beam?
Solution
Balanced condition: εc = 0.003 and εs = fy/Es at the same time. Under-reinforced sections (ρ < ρb) fail by ductile steel yielding first. FE Reference Handbook, Structural Design — reinforced concrete.
Deflection depends on stiffness and span; no reinforcement ratio is defined as the deflection minimum.
Equal top and bottom steel describes a doubly reinforced layout, not the balanced strain condition.
At the balanced ratio ρb the tension steel reaches yield strain just as the extreme concrete fiber reaches 0.003.
Current ACI limits tension steel through net tensile strain, not by allowing the full balanced ratio; ρb itself is not the code maximum.
Question 4 of 18 · Structural Engineering · 10–15 items
In steel design, what is the primary factor that differentiates between compact, non-compact, and slender sections?
Solution
Compact if λ = b/t ≤ λp, noncompact if λp < λ ≤ λr, slender if λ > λr. FE Reference Handbook, Structural Design — steel, local buckling (AISC 360).
Fy enters the λ limits, but the steel grade alone does not classify the section; the element proportions do.
Unbraced length controls lateral-torsional buckling, not the local-buckling classification of the cross-section.
Depth by itself is not the criterion; a deep section can still be compact if its elements are stocky.
AISC classifies flanges and webs by b/t (or h/tw) against the λp and λr limits; that ratio decides compact, noncompact or slender.
Question 5 of 18 · Structural Engineering · 10–15 items
Per ACI 318, what is the minimum specified concrete cover for the ties and longitudinal bars of a cast-in-place column that is not exposed to weather or in contact with the ground?
Solution
ACI 318 specified cover, cast-in-place concrete not exposed to weather or ground: beams and columns 1.5 in; slabs and walls 0.75 in; cast against earth 3 in. FE Reference Handbook, Structural Design — reinforced concrete (ACI 318).
0.75 in is the cover for slabs, walls and joists not exposed to weather, not for beams and columns.
ACI 318 requires 1.5 in of cover to the primary reinforcement, ties and spirals of interior beams and columns.
No ACI cover category uses 2.5 in for interior columns; exposed members use 1.5 to 2 in and members cast against earth 3 in.
3 in is required only for concrete cast against and permanently in contact with earth, such as footings.
Question 6 of 18 · Structural Engineering · 10–15 items
For a short steel column whose cross-section elements are all nonslender, what is the primary failure mode under axial compression?
Question 7 of 18 · Structural Engineering · 10–15 items
What type of connection is created when a beam is attached to a column flange with angles connected to the beam web only?
Solution
Double-angle web connections are simple shear connections (AISC 360 §B3.4a): they transfer shear and are assumed free to rotate. FE Reference Handbook, Structural Design — steel connections.
Fixed implies full moment restraint, the opposite of a web-angle connection that is assumed free to rotate.
Web angles transfer the beam end shear while letting the end rotate, so the joint is a simple (shear) connection.
A moment connection must also connect the flanges to transfer the flange forces; web angles alone cannot.
A rigid connection keeps the beam-column angle fixed, which needs flange connection; web-only angles allow rotation.
Question 8 of 18 · Structural Engineering · 10–15 items
In plastic analysis of steel beams, at what location does a plastic hinge form in a simply supported beam with a central point load?
Solution
Mmax = PL/4 at midspan; collapse when PL/4 = Mp, so Pu = 4Mp/L
FE Reference Handbook, Structural Design — plastic analysis.
The moment diagram peaks at PL/4 under the load; the hinge forms there when PL/4 = Mp, giving Pu = 4Mp/L.
Moment at the quarter points is only half the midspan value under a central load, so the section there is not yet plastic.
Pinned or roller supports carry no moment, so a plastic hinge cannot form there.
One hinge at the maximum moment turns a simply supported beam into a mechanism; that is how it collapses plastically.
Question 9 of 18 · Structural Engineering · 10–15 items
In composite beam design, the effective bending stiffness is determined by transforming one material's area to an equivalent section using the modular ratio. Which material property is compared by the modular ratio?
Solution
n = Es/Ec; transformed concrete width btr = b/n
FE Reference Handbook, Mechanics of Materials — composite sections.
Thermal expansion matters for temperature stresses, not for the transformed-section stiffness of a composite beam.
Ultimate strength sets failure load; the transformed section is an elastic stiffness tool based on moduli.
The modular ratio n = Es/Ec converts one material's area into an equivalent area of the other with the same stiffness.
Yield strength sets capacity, but transformed-section stiffness depends on elastic moduli, not strength.
Question 10 of 18 · Structural Engineering · 10–15 items
In lateral-torsional buckling analysis for a simply supported I-beam under uniform load, how does reducing the unbraced length affect the beam's lateral-torsional buckling resistance?
Solution
AISC: Lb ≤ Lp → Mn = Mp; Lp < Lb ≤ Lr → linear reduction; Lb > Lr → elastic Mcr. FE Reference Handbook, Structural Design — steel beams, lateral-torsional buckling.
Mcr grows rapidly as Lb shrinks (roughly with 1/Lb to 1/Lb²); below Lp the beam reaches the full plastic moment.
Unbraced length is the main variable in the lateral-torsional buckling equations; it always matters.
Shorter unbraced length makes the compression flange harder to buckle sideways, so the critical moment rises, not falls.
Torsional rigidity GJ is a section property; the unbraced length changes how much of it and of EIy the beam can mobilize.
Question 11 of 18 · Structural Engineering · 10–15 items
When performing a P-delta analysis in columns, which parameter primarily amplifies the additional moment caused by secondary effects?
Solution
M2nd = P·Δ; amplified moment ≈ M1/(1 − P/Pe)
FE Reference Handbook, Structural Design — second-order effects (moment magnification).
Area sets axial stress and squash load, but the P-Δ moment comes from load times displacement.
The second-order moment is the axial load times the lateral displacement, P·Δ, so displacement is the amplifying parameter.
Connection fixity changes frame stiffness indirectly, but the additional moment itself is P times the lateral displacement.
Cover affects durability and fire protection, not the second-order moment.
Question 12 of 18 · Structural Engineering · 10–15 items
In the moment distribution method, what is the carry-over factor from the near end to the far end of a prismatic member whose far end is fixed?
Solution
Slope-deflection for a prismatic member with θfar = 0: Mnear = 4EIθ/L, Mfar = 2EIθ/L, so
COF = Mfar/Mnear = (2EIθ/L)/(4EIθ/L) = 0.5
FE Reference Handbook, Structural Analysis — moment distribution.
A carry-over factor of 0 applies when the far end is pinned or a roller, which cannot hold a moment.
0.25 is not a carry-over factor for a prismatic member; it may come from mixing the factor with a distribution factor.
Applying a moment M at the near end of a prismatic member with a fixed far end induces M/2 at the far end, so COF = 0.5.
A factor of 1.00 would carry the full moment across; the far fixed end develops only half of it.
Question 13 of 18 · Structural Engineering · 10–15 items
In structural steel design, what is the primary purpose of a gusset plate?
Solution
Gusset plates are checked for tension yielding, rupture, block shear and Whitmore-section buckling. FE Reference Handbook, Structural Design — steel connections.
A gusset plate is the connector plate at a truss or brace joint; member forces pass through it by bolts or welds.
Bracing members provide lateral support; a gusset only connects them to the frame.
Cover plates on flanges raise a beam's I; a gusset sits at a joint and does not stiffen the beam span.
Intermediate bracing shortens unbraced length; the gusset plate is a connection element, not a brace.
Question 14 of 18 · Structural Engineering · 10–15 items
What is the development length concept in reinforced concrete design?
Solution
ACI 318 gives ld as a function of fy, √f′c, bar diameter, cover and spacing; hooks shorten the required straight length. FE Reference Handbook, Structural Design — reinforced concrete, development length.
Reinforcing steel and concrete expand at nearly the same rate; no embedment length is set for thermal movement.
Splice length is related to ld, but it is the lap of two bars, not the spacing between splices.
Development length ld is the embedment over which bond stress builds the bar's force up to Abfy.
Mill lengths are a supply matter (often 60 ft) and have nothing to do with development length.
Question 15 of 18 · Structural Engineering · 10–15 items
Per ACI 318, what is the strength reduction factor φ for flexure of a tension-controlled reinforced concrete section?
Solution
φ = 0.90 for εt ≥ 0.005; φ varies linearly to 0.65 (tied) at εt = εty. FE Reference Handbook, Structural Design — reinforced concrete (ACI 318).
0.65 is φ for compression-controlled sections with ties, not for tension-controlled flexure.
0.75 is φ for shear and torsion (and for spiral compression-controlled members), not for flexure.
0.85 is the stress-block factor 0.85f′c (or β1 for low-strength concrete), not a strength reduction factor.
ACI 318 uses φ = 0.90 when the net tensile strain εt ≥ 0.005 (tension-controlled).
Question 16 of 18 · Structural Engineering · 10–15 items
A steel column has Fy = 50 ksi, E = 29,000 ksi and KL/r = 85. Using the AISC flexural-buckling equations, determine whether buckling is elastic or inelastic and the critical stress Fcr.
FE Reference Handbook, Structural Design — steel columns (AISC 360).
0.877Fe is the elastic-range equation; it applies only when KL/r exceeds 113, which 85 does not.
Setting Fcr = Fy ignores buckling entirely; that is only the limit as KL/r approaches zero.
39.6 ksi is the Euler stress Fe itself, with neither the inelastic reduction nor the 0.877 factor.
26.5 ksi is φFcr = 0.90 × 29.5; the question asks for the nominal critical stress, before the resistance factor.
KL/r = 85 is below 4.71√(E/Fy) = 113, so Fcr = 0.658Fy/Fe Fy = 0.6581.26 × 50 ≈ 29.5 ksi.
Question 17 of 18 · Structural Engineering · 10–15 items
A column has nominal compressive strength Pn = 450 kips. Compute the LRFD design strength φPn (φ = 0.90) and the ASD allowable strength Pn/Ω (Ω = 1.67).
Given
Nominal strength Pn = 450 kips
φ = 0.90
Ω = 1.67
Find
φPn and Pn/Ω (kips)
Solution
φPn = 0.90 × 450 = 405 kips
Pn/Ω = 450/1.67 = 269 kips
LRFD checks factored loads, ASD checks service loads, so the two numbers are not directly comparable. FE Reference Handbook, Structural Design — AISC LRFD/ASD.
225 kips is 450/2.00; the ASD safety factor for compression is Ω = 1.67.
ASD does not reuse φ; it divides by Ω = 1.67 and compares against unfactored loads.
338 kips is 0.75 × 450; φ for column compression in AISC 360 is 0.90.
The two results are swapped: 269 kips is Pn/Ω and 405 kips is φPn.
φPn = 0.90 × 450 = 405 kips and Pn/Ω = 450/1.67 = 269 kips; they compare against factored and service loads respectively.
Question 18 of 18 · Structural Engineering · 10–15 items
Per ACI 318, what is the minimum flexural tension reinforcement As,min for a nonprestressed reinforced concrete beam?
Solution
As,min = max(3√f′c/fy, 200/fy) · bwd (f′c, fy in psi)
0.75ρb was an old maximum reinforcement limit, not a minimum.
1 % of Ag is the minimum longitudinal steel for columns, not for beams.
0.0018Ag is the shrinkage and temperature steel for slabs with Grade 60 bars, not the beam flexural minimum.
ACI 318 sets As,min as the larger of (3√f′c/fy)bwd and (200/fy)bwd (psi units), so the cracked section is stronger than the uncracked one.
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Common traps
Effective length. Buckling depends on KL, not L. The factor K comes from the end conditions, so the same member can have several capacities:
Pcr = π2EI/(KL)2
Units in slenderness and deflection. Length in feet, radius of gyration and E in inches and ksi: convert before you divide.
Tributary width vs tributary area. A beam carries load per foot from its tributary width; a column carries load from its tributary area.
Determinacy count. For a plane truss, compare m + r with 2j before choosing a method: more unknowns than equations means the structure is indeterminate, fewer means it is unstable.
Net vs gross section. A steel tension member is checked for yielding on the gross area and rupture on the net area through the bolt holes.
Effective depth. Concrete flexure uses d, from the compression face to the centroid of the tension steel, not the full depth h.
Sources: FE Reference Handbook, Civil Engineering and Mechanics of Materials chapters; Hibbeler, Structural Analysis.
Service loads in a strength check
A strength check compares factored load effects with design strength. Dead and live loads straight from the problem are service loads; apply the load combination first, then compare. Deflection checks go the other way and use service loads.
Statics in civil structures
The civil exam has 8–12 statics questions. Those set in buildings and bridges belong here: reactions of a simply supported girder, member forces of a roof or bridge truss by joints or by sections, and the centroid and moment of inertia of a built-up section. Two habits save time: look for zero-force members before writing equations, and use the parallel-axis theorem for any section built from rectangles.
Mechanics of materials has 7–11 questions on the civil exam. In civil form they are shear and moment diagrams of floor beams and girders, bending and shear stress in a rolled or rectangular section, beam deflection, and combined stress in a column with an eccentric load. The change in moment between two points equals the area of the shear diagram between them, so a maximum moment sits where the shear crosses zero, at a support or under a concentrated moment.
σ = M·c/I τ = V·Q/(I·b)
Superposition of standard cases from the Handbook beats integrating the elastic curve. Shafts in torsion, springs and pressure vessels are on FE Mechanical.
How it connects
Statics and mechanics of materials are taught in machine settings on FE Mechanical statics and mechanics of materials; their civil versions are the two sections below. Dynamics, 4–6 questions on the civil exam, is taught on FE Mechanical dynamics. Concrete and steel as materials, rather than members, are on materials; foundations under the structure are geotechnical.
Questions people ask
Do I need the AISC manual or ACI 318 for the FE Civil?
No. The FE supplies no design standards, and you cannot bring them; you work from the NCEES Handbook alone.
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