Structural Engineering is worth 10 to 15 questions on the 110-question FE Civil exam. The items are far shorter than anything you did in a design course: a shear and moment diagram, a load combination, a column capacity check, a deflection from a standard loading case. Speed comes from recognizing the standard case, not from deriving it.
Below are five problems with full solutions that show the handbook table or equation used at each step. Structural is the area where candidates most often burn time re-deriving something the handbook already tabulates, so pay attention to which lookup each solution starts from. Work each one on paper before reading ahead.
Exam weight: NCEES lists Structural Engineering at 10-15 questions of the 110-question FE Civil exam. Work each problem below on paper first, then reveal the worked solution — reading a solution you have not attempted builds recognition, not recall.
What NCEES Tests in Structural Engineering
The NCEES outline for Structural Engineering spans analysis of statically determinate beams, trusses and frames; deflection of determinate structures; column analysis including buckling and slenderness; and structural determinacy and stability. Expect to read or sketch shear and moment diagrams, find reactions, and pull a deflection formula for a standard loading case straight from the handbook's beam tables rather than integrating anything.
It also covers design and detailing basics for steel, concrete, timber and masonry, plus loads and load combinations of the ASCE 7 factored form. Questions on materials stay at the code-concept level: which combination governs, what a reinforcement ratio or a compact section means, how a bolt or weld capacity is computed. Nothing requires a full code book.
5 Free Structural Engineering Practice Problems
Each structural engineering problem below comes from the FEExamClick FE Civil question bank, with worked solutions, and matches the style, difficulty and format of the real exam. Attempt each one under a three-minute limit — roughly the pace the exam demands.
Problem 1 — C. Column analysis (e.g., buckling, boundary conditions)
A column has cross-sectional area \( A = 50\,\mathrm{in}^{2} \) and radius of gyration \( r = 4\,\mathrm{in} \) about the weak axis. The moment of inertia \( I \) about that axis is most nearly:
Answer: D) \( 800\,\mathrm{in}^{4} \)
Step 1 – Identify given information: \(A = 50\,\text{in}^2\), \(r = 4\,\text{in}\) (weak axis).
Step 2 – Relation between \(I\), \(A\), and \(r\):
$$r = \sqrt{I/A} \implies I = A \cdot r^2$$
Step 3 – Substitute and solve:
$$I = 50 \times (4)^2 = 50 \times 16 = 800\,\text{in}^4$$
Step 4 – Answer: \(I = 800\,\text{in}^4\), matching option (D). This is the reverse of the typical \(r = \sqrt{I/A}\) calculation, which is useful when designing a section to meet a slenderness requirement.
Option (B) is what you get by treating the given radius of gyration as half the section depth and running the rectangle formula \(I = bh^{3}/12\), instead of applying \(I = Ar^{2}\) directly.
Correct answer: (D)
Problem 2 — F. Loads, load combinations, and load paths (e.g., dead, live, lateral, influence lines and moving loads, tributary areas)
Wind velocity pressure: \( q_z = 0.00256 K_z K_{zt} K_d V^2 \) (\( \mathrm{lb/ft}^{2} \)). Given \( V = 140\,\mathrm{mph} \), \( K_z = 0.85 \), \( K_{zt} = 1.0 \), and \( K_d = 0.85 \), what is \( q_z \) (\( \mathrm{psf} \)) most nearly?
Answer: B) 36.3
Step 1 – Identify given information:
\( V = 140\,\text{mph} \), \( K_z = 0.85 \), \( K_{zt} = 1.0 \), \( K_d = 0.85 \).
Step 2 – Velocity pressure (FE Handbook, p. 280):
\( q_z = 0.00256\,K_z K_{zt} K_d V^2 \), with \( V \) in mph and \( q_z \) in lb/ft\(^2\).
Step 3 – Solve:
$$q_z = 0.00256(0.85)(1.0)(0.85)(140)^2 = 0.00256(0.7225)(19{,}600) = 36.3\,\text{psf}$$
Step 4 – Answer:
\( q_z \approx 36.3\,\text{psf} \).
(C) \( 42 \) drops the directionality factor \( K_d = 0.85 \), which is the term most often forgotten because it is the same \( 0.85 \) for almost every structure.
(A) \( 31.8 \) and (D) \( 50.2 \) follow from no clean route.
Remember what \( q_z \) is and is not: it is a PRESSURE, and the design wind force needs pressure coefficients that the Handbook does not print. Note also the square on \( V \), which is why a ten per cent higher basic wind speed costs twenty-one per cent more pressure.
Correct answer: (B)
Problem 3 — H. Design of reinforced concrete components (e.g., codes and design philosophies, beams, columns)
A reinforced concrete beam has \( f'_c = 6{,}000\,\mathrm{psi} \). Using \( \beta_1 = 0.85 - 0.05(f'_c - 4000)/1000 \) (\( \mathrm{psi} \)), what is \( \beta_1 \)?
Answer: C) 0.75
Step 1 – Find the row in the table:
\( f'_c = 6{,}000\,\text{psi} \) falls in the middle range, \( 4{,}000 < f'_c < 8{,}000 \), where \( \beta_1 \) slopes.
Step 2 – Apply the expression (FE Handbook, p. 280):
$$\beta_1 = 0.85 - \dfrac{0.05\left(f'_c - 4{,}000\right)}{1{,}000}$$
Step 3 – Solve:
$$\beta_1 = 0.85 - \dfrac{0.05(6{,}000 - 4{,}000)}{1{,}000} = 0.85 - 0.10 = 0.75$$
Step 4 – Answer:
\( \beta_1 = 0.75 \).
(B) \( 0.8 \) is what the same expression gives at \( 5{,}000\,\text{psi} \), that is with the excess strength halved.
(A) \( 0.85 \) is the first row applied outside its range, and (D) \( 0.65 \) is the floor, which is not reached until \( 8{,}000\,\text{psi} \).
Read the expression as a rate: \( \beta_1 \) falls by \( 0.05 \) for every \( 1{,}000\,\text{psi} \) above \( 4{,}000 \), until it stops at \( 0.65 \).
Correct answer: (C)
Problem 4 — B. Deflection of statically determinant beams, trusses, and frames
Truss deflection by unit load method: You need the horizontal displacement at joint J. For two truss members, use \( \Delta = \sum(F_i f_i L_i / (A_i E)) \). Data: Member 1: \( F_1 = 50\,\mathrm{kN} \), \( f_1 = +0.60 \), \( L_1 = 3\,\mathrm{m} \), \( A_1 = 800\,\mathrm{mm}^{2} \). Member 2: \( F_2 = 30\,\mathrm{kN} \), \( f_2 = -0.40 \), \( L_2 = 4\,\mathrm{m} \), \( A_2 = 600\,\mathrm{mm}^{2} \). \( E = 200\,\mathrm{GPa} \). What is \( \Delta_J \) (most nearly)?
Answer: A) \( 0.16\,\mathrm{mm} \)
Step 1 – Identify given information:
| Member | \(F_i\) (kN) | \(f_i\) | \(L_i\) (m) | \(A_i\) (mm\(^2\)) |
|---|---|---|---|---|
| 1 | 50 | +0.60 | 3 | 800 |
| 2 | 30 | -0.40 | 4 | 600 |
Step 2 – Unit load method formula: \(\Delta_J = \displaystyle\sum \dfrac{F_i f_i L_i}{A_i E}\)
Step 3 – Evaluate each member contribution:
Member 1: \(\dfrac{50 \times 10^3 \times 0.6 \times 3}{800 \times 10^{-6} \times 200 \times 10^9} = \dfrac{90{,}000}{160{,}000} = 5.625 \times 10^{-4}\,\text{m}\)
Member 2: \(\dfrac{30 \times 10^3 \times (-0.4) \times 4}{600 \times 10^{-6} \times 200 \times 10^9} = \dfrac{-48{,}000}{120{,}000} = -4.000 \times 10^{-4}\,\text{m}\)
$$\Delta_J = 5.625 \times 10^{-4} - 4.000 \times 10^{-4} = 1.625 \times 10^{-4}\,\text{m} \approx 0.16\,\text{mm}$$
Step 4 – Answer: \(\Delta_J \approx 0.16\,\text{mm}\), matching option (A). The positive result indicates displacement in the assumed direction. Member 2 has a negative contribution because its real and virtual forces have opposite signs.
Option (B) is what you get by adding both member contributions as magnitudes, ignoring that the second member's virtual force is opposite in sign and must subtract.
Correct answer: (A)
Problem 5 — B. Deflection of statically determinant beams, trusses, and frames
A cantilever beam (\( L = 2.5\,\mathrm{m} \)) has an end moment \( M_0 = 4\,\mathrm{kN \cdot m} \) applied at the free end. \( E = 200\,\mathrm{GPa} \), \( I = 2.0 \times 10^{-6}\,\mathrm{m}^{4} \). What is the tip deflection magnitude (\( \mathrm{mm} \))?
Answer: D) 31.2
Step 1 – Identify given information:
| Length | \(L = 2.5\,\text{m}\) |
| End moment | \(M_0 = 4\,\text{kN}\!\cdot\!\text{m}\) |
| \(E\) | \(200\,\text{GPa}\) |
| \(I\) | \(2.0 \times 10^{-6}\,\text{m}^4\) |
Step 2 – Tip deflection formula for cantilever with end moment: From the FE Handbook:
$$\delta_{\text{tip}} = \dfrac{M_0 L^2}{2EI}$$
Step 3 – Substitute and solve:
$$\delta = \dfrac{4{,}000 \times (2.5)^2}{2 \times 200 \times 10^9 \times 2.0 \times 10^{-6}} = \dfrac{4{,}000 \times 6.25}{800{,}000} = \dfrac{25{,}000}{800{,}000} = 0.03125\,\text{m} = 31.25\,\text{mm}$$
Step 4 – Answer: \(\delta_{\text{tip}} \approx 31.2\,\text{mm}\), matching option (D). Option (A) \(0\) is incorrect because an applied moment causes curvature and deflection. Option (B) corresponds to a different load case.
Correct answer: (D)
Using the FE Reference Handbook for Structural Engineering
Find the beam deflection and shear-moment diagram tables in the Structural section early and know their location by feel. Nearly every determinate beam question resolves to one of those standard cases, sometimes superposed. Also flag the load combination list, since problems often supply dead, live, wind and snow values and simply ask which combination controls.
Four Mistakes That Cost Points
- Redrawing shear and moment diagrams from scratch The handbook tabulates reactions, maximum shear and maximum moment for the common simple and cantilever cases. If the loading matches, read the value. Deriving it costs three minutes on a question budgeted for two, and integration sign errors are a common way to lose an item you actually understood.
- Ignoring the effective length factor K Euler buckling problems state a support condition for a reason. Pinned-pinned K is 1.0, fixed-free is 2.0, fixed-fixed is 0.5. Using L instead of KL changes the critical load by a factor of four in the fixed-free case, and that wrong value is always sitting among the choices.
- Evaluating only one load combination Candidates compute a single combination, see a plausible number and move on. The question usually asks for the controlling factored load, which means checking every applicable combination. It is arithmetic rather than judgment, so the fix is discipline: list them, evaluate all of them, then take the maximum.
- Swapping section modulus and moment of inertia S equals I over c. Bending stress uses M over S while deflection uses I directly. Substituting one for the other produces a result with the wrong dimensions, but under time pressure people accept it. Check the units on your answer: stress in psi or MPa, deflection in inches or millimeters.
Frequently Asked Questions
How many structural questions are on the FE Civil exam?
NCEES specifies 10 to 15 questions for Structural Engineering out of the 110 on the exam. That is enough that skipping the area is not viable, but not so much that you need design-course depth. Aim to be fast and reliable on determinate analysis rather than deep on any single code.
Do I need to know AISC or ACI provisions for the FE?
Not the code books themselves. The exam supplies what you need in the NCEES FE Reference Handbook, and questions test the concepts a code applies rather than clause lookup: load combinations, capacity reduction factors, reinforcement limits, compact versus noncompact behavior. Being comfortable with the handbook's steel and concrete pages matters more than memorizing provisions.
Are shear and moment diagrams actually drawn on the exam?
You will not sketch anything on screen, but you will read them. Typical items give a beam and loading and ask for the maximum moment, the location of zero shear, or which of four diagrams matches. Sketching quickly on your scratch board is still the fastest route to picking the right choice.
What is the best way to practice structural problems for the FE?
Mixed sets, timed. Working ten deflection problems in a row teaches the formula but not the recognition skill the exam actually tests. Shuffle beams, trusses, columns and load combinations together so every problem starts with the question of what kind of item this is, which is exactly how it feels on exam day.
Keep Going
Done with structural engineering? Work through the other FE Civil knowledge areas with more free practice problems, review the full FE Civil exam topic breakdown, or plan your preparation with our FE exam study guide and study timeline. Or test yourself against the clock with the free FE Civil practice exam: 20 timed questions in the CBT format.