Free FE Mechanical Practice Problems with Solutions [2026]
A. Togay Koralturk
Last updated on October 11, 2026
28 min read
In this guide, we have created a set of 30 free FE Mechanical practice problems for you, organized by knowledge area and taken directly from our full-length practice exams. If you would like to go beyond these 30 questions, check out our NCEES-style FE Mechanical practice exams, covering every piece of testable knowledge in the FE Reference Handbook and built in the exact same format as the actual exam.
If you'd like to take the practice exam, without the exam simulator, here are the free practice questions.
FE Mechanical Mechanical Design and Analysis practice problems
Problem 1 Mechanical Design and Analysis
A bracket is attached to a column by four identical bolts at the corners of a 120-mm × 80-mm rectangle, as shown. A vertical load of 12 kN acts downward in the plane of the bracket at 200 mm to the right of the centroid of the bolt group. The shear force on the most heavily loaded bolt is most nearly:
Show the solution
Correct answer: D. 10.9 kN
Reference: FE Reference Handbook 10.6, Mechanical Engineering — Fastener Groups in Shear
Every bolt carries the direct share P/n in the direction of the load plus a moment force Mr/Σr² perpendicular to its radius from the centroid, and the two are added as vectors; the most heavily loaded bolts are those whose moment force has a component in the direction of the direct force.
Direct force on every bolt: F₁ = 12/4 = 3.0 kN downward
M = 12(200) = 2,400 kN·mm
Σr² = 4(60² + 40²) = 20,800 mm²
Moment-force components on every bolt: vertical (M/Σr²)(60) = 6.92 kN
Horizontal (M/Σr²)(40) = 4.62 kN — on the right-hand pair the vertical component points downward, with the load
F = √[(3.0 + 6.92)² + 4.62²] = √(9.92² + 4.62²) = 10.9 kN
Problem 2 Mechanical Design and Analysis
The square-thread power screw shown is single-start with a mean thread diameter of 24 mm. A torque of 24 N·m applied to the screw raises a 5-kN load against thread friction (coefficient 0.11) and collar friction (collar mean diameter 36 mm, coefficient 0.07). The lead (mm) of the thread is most nearly: __________
Show the solution
Correct answer: B. 14
Reference: FE Reference Handbook 10.6, Mechanical Engineering — Power Screws
The torque to raise a load on a square-thread power screw is T_R = (Fd_m/2)(l + πµd_m)/(πd_m − µl) + Fµ_c d_c/2, the second term being the collar friction. The collar term is removed first, and the thread term is then solved for the lead.
Collar: Fµ_c d_c/2 = 5,000(0.07)(0.036)/2 = 6.3 N·m
Thread: 24 − 6.3 = 17.7 N·m = (Fd_m/2)(l + πµd_m)/(πd_m − µl)
(l + πµd_m)/(πd_m − µl) = 2(17.7)/(5,000 × 0.024) = 0.295
l + πµd_m = 0.295(πd_m − µl) → l(1 + 0.295 × 0.11) = πd_m(0.295 − 0.11) → l = 0.07540(0.185)/1.0325 = 0.01351 m = 13.5 mm, most nearly 14 mm
Problem 3 Mechanical Design and Analysis
A flat belt drives a 300-mm-diameter pulley at 600 rpm and transmits 4.5 kW, as shown. The belt wraps 160° around the pulley and the coefficient of friction is 0.30. If the belt is on the point of slipping, the tension (N) in the tight side is most nearly:
Show the solution
Correct answer: C. 842 N
Reference: FE Reference Handbook 10.6, Mechanical Engineering — Power Transmission; Statics — Belt Friction
The transmitted torque fixes the difference between the two belt tensions through the pulley radius, and impending slip fixes their ratio through the belt-friction relation with the wrap angle in radians. Solving the two together gives the tight-side tension.
ω = 600(2π/60) = 62.8 rad/s
T = P/ω = 4,500/62.8 = 71.6 N·m
F₁ − F₂ = 2T/D = 2(71.6)/0.300 = 477 N
F₁/F₂ = e^(µθ) = e^[0.30(160π/180)] = e^0.838 = 2.31
F₁ = 477(2.31)/(2.31 − 1) = 842 N
FE Mechanical Statics practice problems
Problem 4 Statics
The band brake shown holds a drum of 6-in radius against a torque of 1,500 lbf·in that tends to turn the drum in the direction indicated. The band wraps 240° of the drum with a coefficient of static friction of 0.26. One end of the band is anchored at the lever pivot O and the other end is attached to the lever 3 in from O; the operator pushes the lever 15 in from O. The operating force F needed to hold the drum is most nearly:
Show the solution
Correct answer: A. 25.4 lbf
Reference: FE Reference Handbook 10.6, Statics — Belt Friction
With the indicated torque, the drum's surface moves along the wrap from the lever-end tangent toward the anchored-end tangent, so friction drags the band toward O and the tension builds from P toward O: the end anchored at O is the tight end and the end attached to the lever at P is the slack end; the belt-friction law relates the two tensions through the wrap angle in radians, the torque fixes their difference, and the lever's moment balance about O gives the operating force.
e^(µθ) = e^(0.26 × 240π/180) = e^(1.089) = 2.972
T₁ − T₂ = M/r = 1,500/6 = 250 lbf
T₂(2.972 − 1) = 250, so T₂ = 126.8 lbf and T₁ = 376.8 lbf
ΣM_O = 0 on the lever: F(15) = 126.8(3), so F = 25.4 lbf
Problem 5 Statics
Four coplanar forces act on the ring shown: 520 N along the positive x axis; 400 N upward to the right on a 3 horizontal : 4 vertical slope; 600 N downward to the left at 30° below the negative x axis; and 350 N downward to the left on a 5 horizontal : 12 vertical slope. The magnitude of the resultant of the four forces is most nearly:
Show the solution
Correct answer: A. 321 N
Reference: FE Reference Handbook 10.6, Statics — Resultant (Two Dimensions); Resolution of a Force
Each force is resolved into x and y components from its slope triangle or angle with the sense read from the figure, the components are summed, and the resultant magnitude is the root of the sum of the squared component sums — the four forces nearly cancel, so the senses of the two down-left forces decide the answer.
ΣF_x = 520 + 400(3/5) − 600 cos 30° − 350(5/13) = 520 + 240 − 519.6 − 134.6 = 105.8 N
ΣF_y = 400(4/5) − 600 sin 30° − 350(12/13) = 320 − 300 − 323.1 = −303.1 N
R = [(105.8)² + (−303.1)²]^½ = 321 N
Problem 6 Statics
The crate shown rests on a 20° incline, with μs = 0.20 between the crate and the incline. A rope from the crate runs parallel to the incline, passes over a fixed drum at the top of the incline (μs = 0.30 between rope and drum), and hangs vertically to a worker who pulls on it with a force of 50 N. The largest crate mass (kg) that the worker can keep from sliding down the incline is most nearly:
Show the solution
Correct answer: C. 58.8 kg
Reference: FE Reference Handbook 10.6, Statics — Belt Friction (Friction)
The rope arrives at the drum along the incline and leaves it vertically, so its angle of contact is 90° + 20° = 110° = 1.920 rad. With the crate about to slide down, impending motion is toward the crate, so the crate side of the rope is the larger force F₁ and the worker's pull is F₂.
F₁ = F₂ e^(μθ) = 50 e^(0.30 × 1.920) = 50(1.779) = 88.9 N
On the incline the crate is on the point of sliding down, so friction acts up the slope at its limiting value and the rope tension makes up the difference:
T = W sin 20° − μs W cos 20° = W(0.342 − 0.20 × 0.940) = 0.154 W → W = 88.9/0.154 = 577 N
m = 577/9.81 = 58.8 kg
FE Mechanical Dynamics, Kinematics, and Vibrations practice problems
Problem 7 Dynamics, Kinematics, and Vibrations
The 5-kg block shown is released from rest and slides down the incline into an undeformed spring aligned with the incline. The coefficient of kinetic friction between the block and the incline is 0.20. The maximum compression of the spring (m) is most nearly:
Show the solution
Correct answer: B. 0.3 m
Reference: FE Reference Handbook 10.6, Dynamics — Principle of Work and Energy
Between release and maximum compression the block is momentarily at rest at both ends, so the kinetic energy terms vanish and the balance runs between gravity work, friction work, and spring energy over the full travel d + x.
Driving force along the incline, net of friction: mg(sin 30° − 0.20 cos 30°) = 5 × 9.81 × (0.5 − 0.1732) = 16.03 N
Energy balance at maximum compression: 16.03 × (2.0 + x) = ½ × 800 × x²
400x² − 16.03x − 32.06 = 0 → x = [16.03 + √(16.03² + 4 × 400 × 32.06)]/800 = 0.304 m
The friction term enters with a negative sign — it dissipates energy — and the block keeps working against it during the compression itself, which is why the travel is d + x rather than d alone.
Problem 8 Dynamics, Kinematics, and Vibrations
The wheel shown, of radius 0.30 m, rolls without slipping to the right with an angular velocity of 4.0 rad/s. A rod AB of length 1.25 m is pinned to the wheel at A, the point of the rim level with the centre at the front of the wheel, and its other end B slides on the smooth floor ahead of the wheel. At the instant shown, the angular velocity of the rod is most nearly:
Show the solution
Correct answer: B. 0.99 rad/s
Reference: FE Reference Handbook 10.6, Dynamics — Instantaneous Center of Rotation (Instant Centers); Plane Circular Motion
The wheel's instant centre is its contact point, which fixes the velocity of the pin A in magnitude and direction; the rod's own instant centre lies where the perpendicular to A's velocity through A meets the perpendicular to B's velocity through B, and the rod's angular velocity is A's speed divided by its distance from that centre.
Pin A is level with the centre, so its distance from the contact point is 0.30√2 = 0.424 m
v_A = 4.0(0.424) = 1.697 m/s, directed at 45° below the horizontal toward the front
B moves along the floor, so the rod's instant centre lies on the vertical through B; the perpendicular to v_A through A rises at 45° toward the front
B is √(1.25² − 0.30²) = 1.214 m ahead of A, so the instant centre is 1.214 m above A
1.214√2 = 1.716 m from A
ω_rod = v_A/1.716 = 1.697/1.716 = 0.99 rad/s
Problem 9 Dynamics, Kinematics, and Vibrations
A uniform solid cylinder is released from rest on a 30° incline, as shown. The coefficients of friction between the cylinder and the incline are μs = 0.15 and μk = 0.10. The acceleration (m/s²) of the cylinder's center is most nearly:
Show the solution
Correct answer: C. 4.06 m/s²
Reference: FE Reference Handbook 10.6, Dynamics — Plane Motion of a Rigid Body (Equations of Motion) (Mass Moment of Inertia)
Whether the cylinder rolls or slips is not given; it is decided by the friction rolling would require. Assuming rolling, the force and moment equations with a = rα give the acceleration and the friction force; that friction is then compared with the static limit μs N. If it exceeds the limit, the cylinder slips and the kinetic friction μk N governs instead.
Rolling assumed: ΣF: mg sin θ − f = m a; ΣM_G: f r = (m r²/2)(a/r) → f = m a/2 → a = (2/3)g sin θ = 3.27 m/s², f = mg sin θ/3 = 0.167 mg
Static limit: μs N = 0.15 mg cos 30° = 0.130 mg < 0.167 mg → the cylinder slips
Slipping: a = g(sin θ − μk cos θ) = 9.81(0.500 − 0.10 × 0.866) = 4.06 m/s²
FE Mechanical Mechanics of Materials practice problems
Problem 10 Mechanics of Materials
A 300-mm steel bar [α = 11.7 × 10⁻⁶/°C, E = 200 GPa, cross-sectional area 150 mm²] and a 120-mm aluminum bar [α = 23.6 × 10⁻⁶/°C, E = 69 GPa, cross-sectional area 650 mm²] are fixed to opposite rigid walls with their free ends facing each other across a 0.16-mm gap at 20 °C, as shown. The temperature of both bars rises uniformly. The temperature (°C) at which the compressive stress in the aluminum bar reaches 35 MPa is most nearly:
Show the solution
Correct answer: B. 91
Reference: FE Reference Handbook 10.6, Mechanics of Materials — Thermal Deformations; Material Properties
Until the bars meet each grows freely by δ_t = αL(T − T_o), and the gap closes when the two free growths sum to 0.16 mm (at 45.2 °C). Above that temperature the total length of the two bars can no longer increase, so the bars push on each other with one compressive force F and each is shortened elastically by FL/(AE). Compatibility: the free thermal growth of both bars equals the gap plus the elastic shortening of both bars.
F = σ_a A_a = (35)(650) = 22,750 N
Free growth per degree: (11.7 × 10⁻⁶)(300) + (23.6 × 10⁻⁶)(120) = 0.006342 mm/°C
Elastic shortening: F[L_s/(E_s A_s) + L_a/(E_a A_a)] = 22,750[300/(200,000 × 150) + 120/(69,000 × 650)] = 22,750(1.268 × 10⁻⁵) = 0.288 mm
0.006342 ΔT = 0.16 + 0.288 → ΔT = 70.7 °C
T = 20 + 70.7 = 90.7 °C, most nearly 91 °C (the steel bar then carries 22,750/150 = 152 MPa, still elastic)
Problem 11 Mechanics of Materials
A steel bar with an 18 mm × 35 mm cross section and a length of 1.1 m is pinned at both ends to rigid walls, as shown, and is stress-free at room temperature (α = 11.7 × 10⁻⁶ per °C, E = 200 GPa, proportional limit 250 MPa). The walls prevent any change in the bar's length. The temperature rise at which the bar buckles is most nearly:
Show the solution
Correct answer: C. 18.8 °C
Reference: FE Reference Handbook 10.6, Mechanics of Materials — Columns; Thermal Deformations
The walls hold the bar's length fixed, so a temperature rise produces a compressive stress equal to the modulus times the suppressed thermal strain; the bar buckles about the axis with the smaller radius of gyration when that stress reaches the Euler critical stress, and the modulus cancels from the equality.
Weak-axis radius of gyration: r = h/√12 = 18/√12 = 5.196 mm, so ℓ/r = 1,100/5.196 = 211.7 with K = 1.0 for pinned ends
Euler critical stress: σ_cr = π²E/(ℓ/r)² = π²(200 × 10⁹)/211.7² = 44.0 MPa, below the 250-MPa proportional limit, so the Euler formula applies
Restrained thermal stress: σ = EαΔT
EαΔT = π²E/(ℓ/r)², so ΔT = π²/[(ℓ/r)²α] = 9.870/(44,820 × 11.7 × 10⁻⁶)
ΔT = 18.8 °C
Problem 12 Mechanics of Materials
The simply supported steel beam shown (E = 200 GPa) has a solid square cross section and carries the distributed load and the midspan concentrated load shown. If the midspan deflection must not exceed 1/360 of the span, the minimum side (mm) of the square section is most nearly:
Show the solution
Correct answer: C. 127 mm
Reference: FE Reference Handbook 10.6, Mechanics of Materials — Deflection of Beams (Principle of Superposition)
Both loads act on the same simply supported span, so the midspan deflection is the superposition of the two table entries: 5wL⁴/(384EI) for the uniform load and PL³/(48EI) for the midspan point load. Setting the sum equal to the allowable deflection gives the required moment of inertia, and the square section follows from I = s⁴/12.
δ_allow = L/360 = 4,500/360 = 12.5 mm
EI·δ = 5wL⁴/384 + PL³/48 = 5(6,000)(4.5)⁴/384 + 11,000(4.5)³/48 = 32,000 + 20,900 = 52,900 N·m³
I_req = 52,900/[(200 × 10⁹)(0.0125)] = 2.12 × 10⁻⁵ m⁴ = 2.12 × 10⁷ mm⁴
s = (12I)^¼ = [12(2.12 × 10⁷)]^¼ = 126.2 mm → the minimum side is 127 mm (rounded up, since 126 mm would exceed the limit)
FE Mechanical Fluid Mechanics practice problems
Problem 13 Fluid Mechanics
Water (ρ = 1,000 kg/m³) flows at 0.060 m³/s through the reducing bend shown, which turns the flow through the angle shown in a horizontal plane. The inlet diameter is 100 mm and the outlet diameter is 60 mm. The gauge pressure is 300 kPa at the inlet and 90 kPa at the outlet. The magnitude of the resultant force (kN) exerted by the water on the bend is most nearly:
Show the solution
Correct answer: D. 2.4 kN
Reference: FE Reference Handbook 10.6, Fluid Mechanics — Pipe Bends, Enlargements, and Contractions (The Impulse-Momentum Principle)
Continuity across the reduction sets the two velocities: v₁ = 0.060/(π × 0.05²) = 7.64 m/s and v₂ = 0.060/(π × 0.03²) = 21.2 m/s
The x-momentum balance on the fluid, with pressure forces on both ports: Fx = P₁A₁ − P₂A₂ cos 60° − Qρ(v₂ cos 60° − v₁) = 2,356 − 127 − 178 = 2,051 N
The y-momentum balance (the bend lies in a horizontal plane, so the fluid weight acts out of plane): Fy = P₂A₂ sin 60° + Qρ v₂ sin 60° = 220 + 1,103 = 1,323 N
F = √(2,051² + 1,323²) = 2,440 N = 2.44 kN
The force the water exerts on the bend is equal and opposite to the force the bend exerts on the fluid, so its magnitude is the same 2.4 kN.
Problem 14 Fluid Mechanics
A thin plate 0.5 m × 0.2 m in plan is drawn horizontally between two fixed surfaces at the constant speed shown. The oil films above and below the plate have the thicknesses shown, and the oil has a specific gravity of 0.88 and a kinematic viscosity of 1.2 × 10⁻⁴ m²/s. The power (W) required to move the plate is most nearly:
Show the solution
Correct answer: C. 2.82 W
Reference: FE Reference Handbook 10.6, Fluid Mechanics — Stress, Pressure, and Viscosity (Definitions); Dynamics — Power and Efficiency
Each face of the plate shears a film of oil, so the drag is the sum of the two film forces, and with a linear profile each shear stress is µv/δ with its own gap. The dynamic viscosity comes from the kinematic viscosity through the oil's density, ρ = SG × ρ_water. The power is the drag force times the plate speed.
µ = νρ = (1.2 × 10⁻⁴)(0.88 × 1,000) = 0.106 Pa·s
τ₁ = µv/δ₁ = 0.106(0.5)/0.0015 = 35.2 Pa
τ₂ = µv/δ₂ = 0.106(0.5)/0.0025 = 21.1 Pa
F = (τ₁ + τ₂)A = (35.2 + 21.1)(0.5 × 0.2) = 5.63 N
P = Fv = 5.63(0.5) = 2.82 W
Problem 15 Fluid Mechanics
A nozzle bolted to the flanged end of a horizontal 100-mm-diameter pipe reduces the diameter to 50 mm and discharges a free jet of water (ρ = 998 kg/m³) to the atmosphere at 12 L/s, as shown. Friction in the nozzle is negligible. The force that the flange bolts must exert to hold the nozzle on the pipe is most nearly:
Show the solution
Correct answer: C. 82.3 N
Reference: FE Reference Handbook 10.6, Fluid Mechanics — Pipe Bends, Enlargements, and Contractions; The Impulse-Momentum Principle; Bernoulli Equation
The jet leaves at atmospheric pressure, so the gauge pressure at the nozzle inlet follows from the Bernoulli equation and the two velocities; the momentum balance on the nozzle in the flow direction then sets the pressure force at the inlet against the bolt force and the fluid's gain of momentum.
v₁ = Q/A₁ = 0.012/[π(0.050²)] = 1.528 m/s
v₂ = Q/A₂ = 0.012/[π(0.025²)] = 6.112 m/s
Bernoulli between the inlet and the free jet, both at the same elevation: p₁ = ρ(v₂² − v₁²)/2 = 998(6.112² − 1.528²)/2 = 17,470 Pa gauge
Pressure force at the inlet: p₁A₁ = 17,470[π(0.050²)] = 137.2 N
Momentum balance on the nozzle, flow direction: p₁A₁ − F = ρQ(v₂ − v₁) = 998(0.012)(6.112 − 1.528) = 54.9 N
F = 137.2 − 54.9 = 82.3 N
FE Mechanical Thermodynamics practice problems
Problem 16 Thermodynamics
A steam power plant operates on the Rankine cycle shown, with the state enthalpies tabulated below.
| State | Enthalpy (kJ/kg) |
|---|---|
| 1, pump inlet (saturated liquid) | 191.8 |
| 2, pump exit | 199.9 |
| 3, turbine inlet | 3,398.3 |
| 4s, turbine exit for isentropic expansion | 2,130.0 |
The turbine has an isentropic efficiency of 88% and the plant delivers a net power of 25 MW. The rate at which heat is rejected in the condenser is most nearly:
Show the solution
Correct answer: C. 47,166 kW
Reference: FE Reference Handbook 10.6, Thermodynamics — Rankine Cycle; Special Cases of Steady-Flow Energy Equation
The actual turbine work is the isentropic efficiency times the isentropic enthalpy drop, which fixes the actual turbine exit enthalpy; the net work per unit mass is the turbine work less the tabulated pump work, the mass flow is the net power over it, and the condenser rejects the mass flow times the enthalpy drop from the actual turbine exit to the pump inlet.
w_t = η_t(h₃ − h₄s) = 0.88(3,398.3 − 2,130.0) = 1,116.1 kJ/kg
h₄ = h₃ − w_t = 3,398.3 − 1,116.1 = 2,282.2 kJ/kg
w_net = w_t − (h₂ − h₁) = 1,116.1 − 8.1 = 1,108.0 kJ/kg
ṁ = 25,000/1,108.0 = 22.563 kg/s
Q̇_out = ṁ(h₄ − h₁) = 22.563(2,282.2 − 191.8) = 47,166 kW
Problem 17 Thermodynamics
An ideal air-standard Brayton cycle, shown schematically, operates with a compressor pressure ratio of 12, a compressor inlet temperature of 300 K, and a turbine inlet temperature of 1,500 K; air has k = 1.4 and constant specific heats. The fraction of the turbine work that is consumed by the compressor is most nearly:
Show the solution
Correct answer: C. 0.41
Reference: FE Reference Handbook 10.6, Mechanical Engineering — Brayton Cycle (Steady-Flow Cycle); Thermodynamics — Ideal Gas
In the ideal Brayton cycle the compressor and turbine are isentropic, so T₂/T₁ = r_p^((k−1)/k) and T₄/T₃ = r_p^(−(k−1)/k). With constant specific heats the compressor work is c_p(T₂ − T₁) and the turbine work is c_p(T₃ − T₄), so the back-work ratio is (T₂ − T₁)/(T₃ − T₄).
r_p^((k−1)/k) = 12^0.2857 = 2.034
T₂ = 300(2.034) = 610.2 K
T₄ = 1,500/2.034 = 737.5 K
w_c/w_t = (610.2 − 300)/(1,500 − 737.5) = 310.2/762.5 = 0.407, most nearly 0.41
Problem 18 Thermodynamics
Steam flows through the condenser shown at 2.0 kg/s, entering at an enthalpy of 2,400 kJ/kg and leaving as liquid at 180 kJ/kg. The condenser pressure is 50 kPa. Cooling water (cp = 4.18 kJ/(kg·K)) enters at 15°C and leaves at 55°C. The required cooling-water mass flow rate (kg/s) is most nearly:
Show the solution
Correct answer: A. 27 kg/s
Reference: FE Reference Handbook 10.6, Thermodynamics — Boilers, Condensers, Evaporators, One Side in a Heat Exchanger (Heat Exchangers)
The steam side rejects heat at the rate: Q̇ = ṁs(hi − he) = 2.0 × (2,400 − 180) = 4,440 kW
The cooling water absorbs the same rate: Q̇ = ṁcw cp(Tout − Tin)
ṁcw = 4,440/(4.18 × (55 − 15)) = 4,440/167.2 = 26.6 kg/s
The condenser pressure sets the saturation state (81.3°C at 50 kPa) but plays no role once both enthalpies are given.
FE Mechanical Heat Transfer practice problems
Problem 19 Heat Transfer
A 25-mm-diameter rod 0.35 m long spans an air duct, as shown, and dissipates 340 W; its ends are insulated. Air at 25 °C flows across the rod at 8 m/s. Assume the air properties ρ = 1.2 kg/m³, μ = 1.8 × 10⁻⁵ Pa·s, k = 0.026 W/(m·K), and Pr = 0.71 apply. The steady surface temperature (°C) of the rod is most nearly:
Show the solution
Correct answer: C. 220
Reference: FE Reference Handbook 10.6, Heat Transfer — Cylinder of Diameter D in Cross Flow; Newton's Law of Cooling
Find the Reynolds number, pick the C and n of the cylinder correlation for that range, form the Nusselt number and h, then apply Newton's law of cooling over the rod's lateral surface to get the temperature rise above the air.
Re_D = ρvD/μ = 1.2(8)(0.025)/(1.8 × 10⁻⁵) = 13,300 → C = 0.193, n = 0.618
Nu_D = 0.193(13,300)^0.618(0.71)^(1/3) = 61.0
h = Nu_D k/D = 61.0(0.026)/0.025 = 63.4 W/(m²·K)
T_w = T∞ + Q̇/(hπDL) = 25 + 340/[63.4π(0.025)(0.35)] = 25 + 195 = 220 °C
Problem 20 Heat Transfer
One face of a 0.12-m² aluminum plate (k = 200 W/(m·K)) is held at 120 °C and carries 300 aluminum pin fins, each 8 mm in diameter and 80 mm long, as shown. Air at 20 °C flows over the fins and the exposed plate with a convection coefficient of 80 W/(m²·K). The total rate of heat transfer from the finned face is most nearly:
Show the solution
Correct answer: B. 4,341 W
Reference: FE Reference Handbook 10.6, Heat Transfer — Fins; Basic Heat-Transfer Rate Equations
The finned face loses heat through the fins and through the plate area left bare between them; each pin fin follows the printed fin relation with its corrected length, and the bare base convects directly.
P = πD = 0.0251 m
A_c = πD²/4 = 5.03 × 10⁻⁵ m²
m = √(hP/kA_c) = √[80(0.0251)/(200 × 5.03 × 10⁻⁵)] = 14.1 m⁻¹
L_c = L + A_c/P = 0.080 + 0.002 = 0.082 m
mL_c = 1.16
tanh(1.16) = 0.821
q_fin = √(hPkA_c)(T_b − T∞)tanh(mL_c) = √(80 × 0.0251 × 200 × 5.03 × 10⁻⁵)(100)(0.821) = 11.671 W per fin
Q̇_fins = 300(11.671) = 3,501.4 W
Q̇_base = 80[0.12 − 300(5.03 × 10⁻⁵)](100) = 839.4 W
Q̇_total = 3,501.4 + 839.4 = 4,341 W
FE Mechanical Electricity and Magnetism practice problems
Problem 21 Electricity and Magnetism
The network shown has two branches, a 30-V source with 10 Ω and a 10-V source with 15 Ω, feeding terminals a and b, with a 30-Ω resistor bridging the terminals. An adjustable resistor R_L will be connected between a and b. The largest power R_L can absorb is most nearly:
Show the solution
Correct answer: C. 16.8 W
Reference: FE Reference Handbook 10.6, Electrical and Computer Engineering — Maximum Power-Transfer Theorem; Thévenin Equivalent Circuit
The network seen from a–b is replaced by its Thévenin equivalent: the open-circuit terminal voltage follows from the node equation with the three branches in parallel, and the Thévenin resistance is the parallel combination of the three resistors with both sources replaced by short circuits; maximum power transfer then occurs with R_L = R_Th and equals V_Th²/(4R_Th).
V_Th = (30/10 + 10/15)/(1/10 + 1/15 + 1/30) = 3.667/0.200 = 18.33 V
R_Th = 1/(1/10 + 1/15 + 1/30) = 5.00 Ω
P_max = V_Th²/(4R_Th) = 18.33²/[4(5.00)] = 16.8 W
Problem 22 Electricity and Magnetism
A 20-Ω resistor, a 60-mH inductor, and an 18-µF capacitor are connected in series across a 110-V (rms) sinusoidal source at 100 Hz, as shown. The rms voltage (V) across the inductor is most nearly:
Show the solution
Correct answer: C. 76
Reference: FE Reference Handbook 10.6, Electrical and Computer Engineering — AC Circuits
In a series RLC branch the impedance is Z = R + j(X_L − X_C) with X_L = ωL and X_C = 1/(ωC); the current is I = V/|Z| and the inductor voltage is V_L = IX_L.
ω = 2π(100) = 628.3 rad/s; X_L = 628.3(0.060) = 37.7 Ω; X_C = 1/[628.3(18 × 10⁻⁶)] = 88.4 Ω
|Z| = √[20² + (37.7 − 88.4)²] = √(400 + 2,570) = 54.5 Ω
I = 110/54.5 = 2.02 A
V_L = 2.02(37.7) = 76 V
FE Mechanical Material Properties and Processing practice problems
Problem 23 Material Properties and Processing
Four wide plates, lettered A to D in the figure, are each loaded in uniform tension normal to a single crack. Plates A and D are 4340 steel; plates B and C are 2024-T3 aluminum. Plate A has an edge crack 6.8 mm long, plate B an interior crack 12 mm from tip to tip, plate C an edge crack 4.5 mm long, and plate D an interior crack 13 mm from tip to tip. The plate that fractures at the lowest applied stress is:
Show the solution
Correct answer: A. Plate A
Reference: FE Reference Handbook 10.6, Materials Science/Structure of Matter — Representative Values of Fracture Toughness
Fracture occurs when the stress intensity reaches the fracture toughness, K_IC = Yσ√(πa): an exterior (edge) crack takes Y = 1.1 with a = the crack length, an interior crack takes Y = 1 with a = half the tip-to-tip length. The fracture stress of each plate is σ = K_IC/(Y√(πa)), with K_IC = 46 MPa·m^½ for 4340 steel and 44 MPa·m^½ for 2024-T3 aluminum.
Plate A (steel, edge, a = 6.8 mm): σ = 46/[1.1√(π × 0.0068)] = 286 MPa
Plate B (aluminum, interior, a = 6 mm): σ = 44/√(π × 0.006) = 320 MPa
Plate C (aluminum, edge, a = 4.5 mm): σ = 44/[1.1√(π × 0.0045)] = 336 MPa
Plate D (steel, interior, a = 6.5 mm): σ = 46/√(π × 0.0065) = 322 MPa
Plate A fractures first: the edge factor 1.1 and the full crack length outweigh its slightly higher toughness
Problem 24 Material Properties and Processing
A heat-sink slug must have a heat capacity of at least 900 J/K and a mass of no more than 1.5 kg. It is to be made of aluminum, copper, lead, or magnesium, whichever gives the smallest slug volume. The minimum volume (cm³) of the slug is most nearly:
Show the solution
Correct answer: B. 370
Reference: FE Reference Handbook 10.6, Materials Science/Structure of Matter — Properties of Metals; Thermal Properties
A slug of mass m and specific heat c_P has heat capacity mc_P, so holding 900 J/K within 1.5 kg needs c_P ≥ 900/1.5 = 600 J/(kg·K): aluminum (895.9) and magnesium (1,046.7) qualify; copper (389.4) and lead (129.8) cannot reach 900 J/K under 1.5 kg.
The slug's volume at the floor is V = 900/(ρc_P): aluminum 2,698(895.9) = 2.42 × 10⁶ J/(m³·K); magnesium 1,738(1,046.7) = 1.82 × 10⁶ J/(m³·K) → aluminum is smaller
V = 900/(2.42 × 10⁶) = 3.72 × 10⁻⁴ m³ = 372 cm³ (mass 1.0 kg, within the limit)
Copper would make the smallest slug of all, but at 2.3 kg it breaks the mass limit; the lightest metal, magnesium, stores less heat per cubic metre and needs a third more volume. The two requirements must be screened together before the volume is ranked.
FE Mechanical Measurements, Instrumentation, and Controls practice problems
Problem 25 Measurements, Instrumentation, and Controls
Two resistors are connected in parallel as shown to form a current shunt. Their resistances were measured as 1.5 ± 0.02 Ω and 3.9 ± 0.12 Ω, and the two uncertainties are independent. The uncertainty (Ω) of the equivalent resistance of the pair is most nearly:
Show the solution
Correct answer: A. 0.0139
Reference: FE Reference Handbook 10.6, Instrumentation, Measurement, and Control — Measurement Uncertainty; Electrical and Computer Engineering — Resistors in Series and Parallel
The uncertainty of a result is the square root of the sum of the squares of each measurement's uncertainty times the partial derivative of the result with respect to that measurement, and for two resistors in parallel R = R₁R₂/(R₁ + R₂).
∂R/∂R₁ = [R₂/(R₁ + R₂)]² = (3.9/5.4)² = 0.5216
∂R/∂R₂ = [R₁/(R₁ + R₂)]² = (1.5/5.4)² = 0.07716
w_R = √[(0.5216 × 0.02)² + (0.07716 × 0.12)²] = √(1.088 × 10⁻⁴ + 0.857 × 10⁻⁴)
w_R = 0.0139 Ω
FE Mechanical Mathematics practice problems
Problem 26 Mathematics
The flowchart shown initializes n and a counter c, then repeats a loop in which a decision on n selects one of two updates before the counter is incremented and tested. The value of n printed by the output step is:
Show the solution
Correct answer: D. 45
Reference: FE Reference Handbook 10.6, Electrical and Computer Engineering — Flowchart Definition
Each pass first tests the parity of n: an even n is halved, an odd n becomes 2n + 3. The counter is then incremented and the loop exits only when c ≥ 5, so the sequence is traced through five updates.
Pass 1: n = 12 even → 6, c = 1
Pass 2: n = 6 even → 3, c = 2
Pass 3: n = 3 odd → 9, c = 3
Pass 4: n = 9 odd → 21, c = 4
Pass 5: n = 21 odd → 45, c = 5 (c ≥ 5? yes → output n = 45)
Problem 27 Mathematics
A force F₁ of magnitude 20 N makes an angle of 60° with the positive x axis and 45° with the positive y axis, and its z-component is negative. A second force is F₂ = 6i − 4j + 3k N. The magnitude (N) of F₁ + F₂ is most nearly:
Show the solution
Correct answer: B. 20.2 N
Reference: FE Reference Handbook 10.6, Mathematics — Vectors
The three direction cosines of a vector satisfy cos²α + cos²β + cos²γ = 1, which fixes the size of the missing cosine; its sign comes from the stated direction of the z-component. The components of F₁ then follow from its magnitude, and the resultant is formed component by component before its length is taken.
cos²γ = 1 − cos²60° − cos²45° = 1 − 0.25 − 0.5 = 0.25 → cos γ = −0.5 (z-component negative)
F₁ = 20(0.5i + 0.7071j − 0.5k) = 10i + 14.14j − 10k N
F₁ + F₂ = 16i + 10.14j − 7k N
|F₁ + F₂| = √(16² + 10.14² + 7²) = 20.2 N
FE Mechanical Probability and Statistics practice problems
Problem 28 Probability and Statistics
Five observations of a response y at settings x are tabulated below.
| x | 2 | 4 | 5 | 7 | 9 |
|---|---|---|---|---|---|
| y | 10 | 15 | 14 | 21 | 24 |
A straight line is fitted to the data by least squares. The residual (observed minus predicted) at x = 7 is most nearly:
Show the solution
Correct answer: B. 0.95
Reference: FE Reference Handbook 10.6, Engineering Probability and Statistics — Linear Regression and Goodness of Fit; Least Squares
The printed least-squares line is ŷ = a + bx with b = S_xy/S_xx and a = ȳ − b x̄, where S_xy = Σxy − (1/n)(Σx)(Σy) and S_xx = Σx² − (1/n)(Σx)² are the corrected sums; the residual at a point is the observed value minus the line's prediction there.
n = 5, Σx = 27, Σy = 84
Σxy = 20 + 60 + 70 + 147 + 216 = 513
Σx² = 4 + 16 + 25 + 49 + 81 = 175
S_xy = 513 − (27)(84)/5 = 513 − 453.6 = 59.4
S_xx = 175 − 27²/5 = 175 − 145.8 = 29.2
b = 59.4/29.2 = 2.0342
a = 84/5 − 2.0342(27/5) = 16.8 − 10.985 = 5.815
ŷ(7) = 5.815 + 2.0342(7) = 20.055
Residual = 21 − 20.055 = 0.945, most nearly 0.95
FE Mechanical Ethics and Professional Practice practice problems
Problem 29 Ethics and Professional Practice
An engineer is planning a manufacturing plant. To cut cost, the owner asks that the environmental review cover only the construction phase and that the plant's visual effect on the surrounding landscape be left out of the review. Which two of the following actions are consistent with sustainability and life-cycle principles? (Select two.)
Show the solution
Correct answer: A,D
Reference: FE Reference Handbook 10.6, Ethics and Professional Practice — Societal Considerations
The handbook states that sustainable approaches adopted during planning, design, and construction carry forward throughout a project's operation and maintenance to end-of-life, and that life-cycle analysis (cradle to grave) assesses environmental consequences from design and development through utilization and disposal. Its sustainable principles include resource allocation and non-renewable resources, and the concerns engineers must address include landscape aesthetics, protection of ecosystems, resource conservation, pollution, emissions, and waste.
Extending the review through utilization and disposal, and evaluating non-renewable-resource consumption and resource conservation, follow those principles. Limiting the review to construction and excluding landscape aesthetics contradict them.
FE Mechanical Engineering Economics practice problems
Problem 30 Engineering Economics
Two machines are being compared at an interest rate of 10 % per year. Machine A costs $60,000, lasts 5 years, has a $10,000 salvage value and costs $9,000 per year to operate and maintain. Machine B lasts 10 years with no salvage value and costs $6,000 per year to operate and maintain. The first cost of machine B at which the two machines have the same equivalent uniform annual cost is most nearly:
Show the solution
Correct answer: D. $105,655
Reference: FE Reference Handbook 10.6, Engineering Economics — Capital Recovery; Sinking Fund
Alternatives with different lives are compared by equivalent uniform annual cost. Machine A's annual cost is the capital recovery of its first cost less the sinking-fund equivalent of its salvage, plus its O&M; machine B's is the capital recovery of its unknown first cost plus its O&M. Setting them equal gives the breakeven first cost.
EUAC_A = 60,000(A/P, 10 %, 5) − 10,000(A/F, 10 %, 5) + 9,000 = 60,000(0.2638) − 10,000(0.1638) + 9,000 = $23,190
EUAC_B = P_B(A/P, 10 %, 10) + 6,000 = 0.1627 P_B + 6,000
0.1627 P_B + 6,000 = 23,190
P_B = 17,190/0.1627 = $105,655
FE Mechanical Practice Exams
- Covers the complete exam scope by testing every piece of testable knowledge the FE Mechanical exam draws from the FE Reference Handbook.
- Built in the exact same format as the actual FE Mechanical exam — including realistic exhibits.
- Provides detailed step-by-step solutions.
FE Mechanical Practice Exams
- Covers the complete exam scope by testing every piece of testable knowledge the FE Mechanical exam draws from the FE Reference Handbook.
- Built in the exact same format as the actual FE Mechanical exam — including realistic exhibits.
- Provides detailed step-by-step solutions.
Download the free FE Mechanical practice problems as a PDF
All 30 problems, diagrams, and step-by-step solutions are also available as a single print-friendly PDF — free, with no email or account required. Print it, work the problems away from a screen, and keep the solutions for review.
Download the free FE Mechanical practice problems PDF and pair it with your own copy of the FE Reference Handbook, since the PDF's solutions cite the same handbook sections the on-page versions do.
What the 30 free FE Mechanical practice problems cover
The set mirrors the structure of the actual exam, with at least one problem from every one of the 14 knowledge areas in the NCEES FE Mechanical specification, weighted toward the heavy areas the same way the real 110-question exam is. The table shows each area's share of the real exam alongside its problems below.
| Knowledge area | Questions on the real exam | Free problems below |
|---|---|---|
| Mechanical Design and Analysis | 10–15 | 3 |
| Statics | 9–14 | 3 |
| Dynamics, Kinematics, and Vibrations | 10–15 | 3 |
| Mechanics of Materials | 9–14 | 3 |
| Fluid Mechanics | 10–15 | 3 |
| Thermodynamics | 10–15 | 3 |
| Heat Transfer | 7–11 | 2 |
| Electricity and Magnetism | 5–8 | 2 |
| Material Properties and Processing | 7–11 | 2 |
| Measurements, Instrumentation, and Controls | 5–8 | 1 |
| Mathematics | 6–9 | 2 |
| Probability and Statistics | 4–6 | 1 |
| Ethics and Professional Practice | 4–6 | 1 |
| Engineering Economics | 4–6 | 1 |
Work each problem with the NCEES FE Reference Handbook open — you can download your free copy through MyNCEES — an approved calculator in hand, and a target pace of about three minutes per question. Every solution cites the exact handbook section it draws from.
Where to get full-length FE Mechanical practice exams
These 30 questions came from our FE Mechanical Practice Exams — 10 full-length, timed exams in the same simulator, 1,100 questions with the same style of worked solutions, backed by our passing guarantee. A passing score is built on volume: seeing every question style at realistic difficulty, against the clock, until nothing on exam day is new — our guide on how to study for the FE exam lays out the schedule that gets you there. Thirty problems show you the standard; the full set is where the practice happens.
Frequently Asked Questions
How many questions is the FE Mechanical exam?
The FE Mechanical exam has 110 questions in a 6-hour appointment, of which 5 hours and 20 minutes is exam time, per NCEES. The questions span 14 knowledge areas, from mathematics through mechanical design and analysis, and the exam is computer-based with the FE Reference Handbook supplied onscreen.
Are these free practice problems like the actual FE Mechanical exam?
Yes, deliberately so. All 30 are real questions from our full-length FE Mechanical practice exams: written to the current NCEES specification, drawn across all 14 knowledge areas, with exam-style diagrams and answer choices engineered from the mistakes real candidates make. We selected them from the harder end of our bank, so they represent the practice that builds readiness rather than the warm-up end.
What score do you need to pass the FE Mechanical exam?
NCEES does not publish a passing score, and there is no fixed percentage such as 70%. Your total correct answers convert to a scaled score judged against a criterion standard, and there is no penalty for wrong answers, so answering every question is always the right strategy. Our guide on the FE exam passing score explains the scoring model in full.
How hard is the FE Mechanical exam?
Per NCEES Squared 2025, 69% of first-time takers pass FE Mechanical and 37% of repeat takers do, one of the higher first-time rates among the FE disciplines. The difficulty is less about any single topic and more about breadth under time pressure: 14 subject areas at roughly three minutes per question, with dynamics, fluids, thermodynamics and mechanical design each carrying 10 to 15 questions.
How many practice problems should I solve before the FE Mechanical exam?
Enough that full-length, timed exams stop surprising you — for most candidates that means several hundred specification-mapped problems plus multiple complete 110-question simulations. The 30 problems on this page are a diagnostic starting point; the volume layer is what our 10 full-length practice exams provide.
Can I download these FE Mechanical practice problems as a PDF?
Yes. The free PDF contains all 30 problems with their diagrams and complete step-by-step solutions, and requires no email or account. It is print-friendly, so you can work the problems on paper and check the solutions afterward.
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