Start with any of the 14 FE Mechanical topic sets below. Every problem is original, written against the current NCEES FE Mechanical exam specification, and comes with a step-by-step solution that cites where the equation lives in the FE Reference Handbook. There are 76 free problems in total — no account, no card, no email required.
The table is ordered by exam weight rather than alphabetically, because on a 110-question paper those weights decide where your study hours belong. The four heaviest areas — Dynamics, Kinematics, and Vibrations, Fluid Mechanics, Mechanical Design and Analysis and Thermodynamics — are worth 10 to 15 questions each. Together they can account for close to half the exam.
Try These Featured Visual Problems
These come straight from the PECivilClick FE Mechanical bank — the same clean, exam-style diagrams you get inside the platform. Pick an answer and see instantly whether you are right.
Statics — Force Components in 3D
A force of 500 N acts along a line from point A(1, 2, 3) to point B(4, 6, 3). What is the x-component of this force?
Direction vector from A to B:
$$\vec{d} = (4-1, 6-2, 3-3) = (3, 4, 0)$$
Magnitude: \(|\vec{d}| = \sqrt{3^2 + 4^2 + 0^2} = 5\)
Unit vector: \(\hat{u} = \frac{1}{5}(3, 4, 0) = (0.6, 0.8, 0)\)
Force component:
$$F_x = F \cdot u_x = 500 \times 0.6$$
$$\boxed{F_x = 300 \text{ N}}$$
Dynamics — Projectile Range
A projectile is fired at 30° above horizontal with initial velocity 50 m/s. What is the horizontal range? (g = 10 \(\text{m/s}^{2}\))
For projectile motion:
$$R = \frac{v_0^2 \sin(2\theta)}{g}$$
$$R = \frac{50^2 \sin(60^{\circ})}{10} = \frac{2500 \times 0.866}{10}$$
$$\boxed{R = 216.5 \text{ m}}$$
Thermodynamics — Polytropic Boundary Work
A gas in a piston-cylinder assembly undergoes a polytropic process (PV^1.3 = constant) from 150 kPa, 0.05 m\(^{3}\) to 700 kPa, 0.015 m\(^{3}\). The boundary work (kJ) is most nearly:
Refer to the Closed-System (Boundary) Work entry in the First Law section of the Thermodynamics chapter of the FE Reference Handbook.
Polytropic boundary work (\(PV^n\) = constant, \(n \ne 1\)) is the area under the curve on the P–V plane, and it reduces to the end-state \(PV\) products: \(W = \dfrac{P_2 V_2 - P_1 V_1}{1 - n}\)
Note the volume ratio has already been fixed by the two given states, so no property tables are needed — only the end-point \(PV\) products and the exponent \(n\).
End-state products (recall \(1\ \text{kPa}\cdot\text{m}^3 = 1\ \text{kJ}\)): \(P_1 V_1 = (150)(0.05) = 7.5\ \text{kJ}, \qquad P_2 V_2 = (700)(0.015) = 10.5\ \text{kJ}\)
$$W = \dfrac{10.5 - 7.5}{1 - 1.3} = \dfrac{3.0}{-0.3} = -10.0\ \text{kJ}$$
The negative sign is expected: the gas is compressed (V drops from 0.05 to 0.015 m³), so the surroundings do work on the gas.
Why the other options fail: • +10.0 kJ — correct magnitude, wrong sign (numerator reversed). Compression can never give positive boundary work. • −7.5 kJ — used the air value \(n = 1.4\) instead of the stated \(n = 1.3\): \(3.0/(1-1.4) = -7.5\). • −3.0 kJ — forgot the \(1/(1-n)\) factor and reported just \(P_2V_2 - P_1V_1\).
Correct answer: (B) −10.0 kJ
Fluid Mechanics — Pump NPSH (Cavitation Check)
NPSHA for a pump with suction tank at atmospheric pressure (101 kPa), 3 m below pump centerline, friction losses of 1.5 m, and water at 20°C (vapor pressure 2.34 kPa) is most nearly:
What NPSH available means.
\( NPSH_A \) is the head margin at the pump inlet above the fluid's vapor pressure — how far the suction is from boiling (cavitating). Referenced to the pump centerline:
$$NPSH_A = \dfrac{P_{atm}}{\rho g} + z - h_f - \dfrac{P_v}{\rho g}$$
where \( z \) is the height of the free surface relative to the pump centerline. The tank is below the pump (suction lift), so \( z = -3 \) m — the static lift SUBTRACTS.
Term by term (with \( \rho g = 9.81 \text{ kN/m}^3 \)):
$$\dfrac{P_{atm}}{\rho g} = \dfrac{101}{9.81} = 10.3 \text{ m}$$
$$z = -3 \text{ m} \quad (\text{lift, tank below pump})$$
$$h_f = 1.5 \text{ m} \quad (\text{friction along the suction line})$$
$$\dfrac{P_v}{\rho g} = \dfrac{2.34}{9.81} = 0.24 \text{ m}$$
Add them up:
$$NPSH_A = 10.3 - 3 - 1.5 - 0.24 = 5.6 \text{ m}$$
Why the other options fail: • 10.3 m is only the atmospheric head \( P_{atm}/\rho g \) — it ignores the lift, the friction, and the vapor pressure entirely. • 8.8 m keeps the atmospheric head and removes only the friction (\( 10.3 - 1.5 \)), forgetting that the tank is 3 m below the pump. • 3.2 m comes from a sign slip — e.g. subtracting the vapor pressure twice, or double-counting a term. The reliable guard is to write \( z = -3 \) once and add the four terms.
Physical check: a positive \( NPSH_A = 5.6 \) m means the pump has margin before cavitation. If the manufacturer's required \( NPSH_R \) exceeded 5.6 m, this arrangement would cavitate and you would have to lower the pump, shorten the suction line, or cool the water.
Correct answer: (B)
Heat Transfer — Fin Parameter
A rectangular fin (k = 200 W/m·K) has width 50 mm, thickness 5 mm, length 100 mm. With h = 25 W/m\(^{2}\)·K and base temperature 100°C above ambient, the fin parameter m is:
For a rectangular fin:
\(P = 2(w + t) = 2(0.05 + 0.005) = 0.11\) m \(A_c = w \times t = 0.05 \times 0.005 = 0.00025\) m\(^{2}\)
$$m = \sqrt{\frac{hP}{kA_c}}$$
$$m = \sqrt{\frac{25 \times 0.11}{200 \times 0.00025}} = \sqrt{\frac{2.75}{0.05}} = \sqrt{55}$$
$$\boxed{m = 7.42 \text{ m}^{-1}}$$
Mechanics of Materials — Load Sharing Between Two Rods
A rigid bar is supported by two vertical rods: a steel rod (\(E_s\) = 200 GPa, \(A_s\) = 300 mm\(^{2}\), \(L_s\) = 1 m) and an aluminum rod (\(E_a\) = 70 GPa, \(A_a\) = 600 mm\(^{2}\), \(L_a\) = 1 m). Both rods are attached at the same point. A vertical load of 100 kN is applied. The force (kN) carried by the steel rod is most nearly:
Refer to the Uniaxial Stress-Strain section in the Mechanics of Materials chapter of the FE Reference Handbook.
Step 1 – Compatibility. The two rods hang from the same ceiling and are joined at the same point, so they must stretch by the same amount: \(\delta_s = \delta_a \quad\Rightarrow\quad \dfrac{F_s L}{A_s E_s} = \dfrac{F_a L}{A_a E_a}\)
Step 2 – The split is by axial stiffness \(k = AE/L\), not by area alone: \(k_s = \dfrac{(300)(200{,}000)}{1000} = 60{,}000\ \text{N/mm}, \qquad k_a = \dfrac{(600)(70{,}000)}{1000} = 42{,}000\ \text{N/mm}\)
so \(F_a = F_s\,(42/60) = 0.7\,F_s\).
Step 3 – Equilibrium: \(F_s + F_a = 100 \quad\Rightarrow\quad 1.7\,F_s = 100 \quad\Rightarrow\quad F_s = 58.8\ \text{kN}\)
(and \(F_a = 41.2\) kN, which checks: \(58.8 + 41.2 = 100\) ✔).
Why the other options fail: • 50.0 kN assumes the two rods simply split the load equally. They do not — the stiffer rod attracts more force, and steel here is the stiffer of the two. • 33.3 kN splits the load by area only (\(300/900\)), ignoring that steel's modulus is nearly three times aluminium's. Area alone never governs; the product \(AE\) does. • 66.7 kN is the same area-only split (\(600/900\)) assigned to the wrong rod.
Sanity check: \(k_s > k_a\), so the steel rod must carry more than half of the 100 kN, but not as much as the 3:1 modulus ratio alone would suggest — aluminium has twice the area, which pulls the split back toward even. 58.8 kN sits exactly in that window.
Correct answer: (D)
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Free FE Mechanical practice problems by topic
| FE Mechanical topic | Free problems | Difficulty | Approx. time | Questions on the real exam |
|---|---|---|---|---|
| FE Mechanical Dynamics, Kinematics, and Vibrations | 5 | 2 easy · 2 med · 1 hard | ~15 min | 10-15 |
| FE Mechanical Fluid Mechanics | 5 | 2 easy · 2 med · 1 hard | ~15 min | 10-15 |
| FE Mechanical Mechanical Design and Analysis | 5 | 2 easy · 2 med · 1 hard | ~15 min | 10-15 |
| FE Mechanical Thermodynamics | 5 | 2 easy · 2 med · 1 hard | ~15 min | 10-15 |
| FE Mechanical Mechanics of Materials | 5 | 2 easy · 2 med · 1 hard | ~15 min | 9-14 |
| FE Mechanical Statics | 5 | 2 easy · 2 med · 1 hard | ~15 min | 9-14 |
| FE Mechanical Heat Transfer | 5 | 2 easy · 2 med · 1 hard | ~15 min | 7-11 |
| FE Mechanical Material Properties and Processing | 5 | 2 easy · 2 med · 1 hard | ~15 min | 7-11 |
| FE Mechanical Mathematics | 5 | 2 easy · 2 med · 1 hard | ~15 min | 6-9 |
| FE Mechanical Electricity and Magnetism | 5 | 2 easy · 2 med · 1 hard | ~15 min | 5-8 |
| FE Mechanical Measurements, Instrumentation, and Controls | 5 | 2 easy · 2 med · 1 hard | ~15 min | 5-8 |
| FE Mechanical Engineering Economics | 5 | 2 easy · 2 med · 1 hard | ~15 min | 4-6 |
| FE Mechanical Ethics and Professional Practice | 5 | 2 easy · 2 med · 1 hard | ~15 min | 4-6 |
| FE Mechanical Probability and Statistics | 5 | 2 easy · 2 med · 1 hard | ~15 min | 4-6 |
| Total free problems | 70 | — | ~3.5 hr | 110 on exam day |
How FE Mechanical differs from FE Civil
Both exams run 110 questions in 5 hours and 20 minutes of testing time, inside a six-hour appointment that also covers the nondisclosure agreement, the tutorial and a 25-minute scheduled break. The topic lists are where they diverge.
Four FE Mechanical knowledge areas have no counterpart at all on FE Civil: FE Mechanical Thermodynamics, FE Mechanical Heat Transfer, Measurements, Instrumentation and Controls, and FE Mechanical Mechanical Design and Analysis. Together they are worth roughly a third of the paper. Dynamics is also weighted far more heavily here, at 10-15 questions against 4-6 on the civil exam.
This is why a question bank shared between disciplines serves neither well. Our FE Mechanical exam prep and FE Civil exam prep banks are built and maintained separately.
How to use these problem sets
Work each problem on paper before revealing the solution. Reading a worked solution you have not attempted builds recognition, not recall, and recognition is precisely what fails under exam pressure. Keep the FE Reference Handbook open while you practise, because finding an equation quickly is a skill the exam tests just as directly as applying it.
Time yourself at about three minutes per question. That is the pace 110 questions in 5 hours 20 minutes demands, and it is the reason candidates who only ever practise untimed are surprised on exam day.
Where to start
If your exam is months away, work down the table from the top so your hours land on the heaviest areas first. If it is weeks away, start instead with FE Mechanical Ethics and FE Mechanical Engineering Economics: together they are 8-12 questions, they need no derivation, and they are the fastest points on the paper to secure.
For a full schedule, see how long to study for the FE exam and the week-by-week FE study plan. To rehearse under real conditions, the FE Mechanical platform includes full-length CBT simulations timed to the same 5 hr 20 min window.