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NCEES Engineering Licensure

FE Other Disciplines Practice Exam

FE Other Disciplines exam practice — the NCEES discipline exam, your first step toward the PE.

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Built for FE Other Disciplines candidates

  • NCEES FE Other Disciplines topic mix: the broad fundamentals — statics, dynamics, materials, fluids, thermodynamics, electricity & magnetism, and chemistry
  • Answerable from the NCEES FE Reference Handbook + first principles
  • Mix of calculation and conceptual items, in SI and US customary units
  • Step-by-step solutions on every question (first 2 free; all with Premium)
  • Follow-up study chat to dig into any concept that trips you up

The FE Other Disciplines exam is one of the seven NCEES Fundamentals of Engineering discipline exams — a 110-question, computer-based test, taken near graduation, that is the first step toward becoming a licensed Professional Engineer (PE). Every question is meant to be answered from the FE Reference Handbook and engineering fundamentals, so practice that drills the right Other Disciplines topic mix, in the right units, with worked solutions is exactly how you build exam-day speed.

Crameleon builds your practice exam on demand — pick the pathway and module you're studying, and we generate a fresh paper with step-by-step solutions (the first 2 free; all of them with Premium). Every question comes with a follow-up study chat, so you can dig into any concept that trips you up without leaving the page. The generation takes under 2 minutes and your first exam is free.

Sample FE Other Disciplines exam questions

Three original questions on FE Other Disciplines exam topics, each with a worked solution.

Question 1Multiple choiceFluid Mechanics

A large open tank of water has a short, smooth, well-rounded nozzle in its side wall with an exit diameter of 50 mm. The centreline of the nozzle is 4.0 m below the free surface, which is held at a constant level. The jet discharges to the atmosphere. Neglect losses and the velocity of the free surface, and take g=9.81 m/s2g = 9.81\,\text{m/s}^2. What is the volumetric flow rate from the nozzle?

  1. A.

    12.3 L/s

  2. B.

    17.4 L/s

  3. C.

    69.6 L/s

  4. D.

    154 L/s

Worked solution
Answer:B

Apply the Bernoulli equation along a streamline from the free surface (point 1) to the jet just outside the nozzle (point 2). Both points are at atmospheric pressure, the free-surface velocity is negligible, and the elevation difference is h=4.0h = 4.0 m:

p1γ+V122g+z1=p2γ+V222g+z2⇒h=V222g\frac{p_1}{\gamma} + \frac{V_1^2}{2g} + z_1 = \frac{p_2}{\gamma} + \frac{V_2^2}{2g} + z_2 \quad \Rightarrow \quad h = \frac{V_2^2}{2g}

This is Torricelli's result: the jet velocity is that of free fall through the head hh:

V2=2gh=2(9.81)(4.0)=8.86 m/sV_2 = \sqrt{2gh} = \sqrt{2(9.81)(4.0)} = 8.86\,\text{m/s}

The nozzle area uses the diameter, so the radius is 25 mm:

A=πd24=π(0.050)24=1.963×10−3 m2A = \frac{\pi d^2}{4} = \frac{\pi(0.050)^2}{4} = 1.963 \times 10^{-3}\,\text{m}^2

Continuity gives the flow rate:

Q=AV2=(1.963×10−3)(8.86)=0.0174 m3/s=17.4 L/sQ = AV_2 = (1.963 \times 10^{-3})(8.86) = 0.0174\,\text{m}^3\text{/s} = 17.4\,\text{L/s}

Q ≈ 17.4 L/s (option B)

The most tempting wrong answer is A (12.3 L/s), which uses V=ghV = \sqrt{gh} and drops the factor of 2 that comes from the kinetic-energy term V2/2gV^2/2g in Bernoulli. Option C (69.6 L/s) uses πd2\pi d^2 for the area instead of πd2/4\pi d^2/4, the classic radius-versus-diameter slip. Option D (154 L/s) forgets the square root and uses V=2ghV = 2gh.

Question 2Short answerBasic Electrical Engineering

An ideal 24 V DC source drives a circuit consisting of a 6 Ω resistor in series with a parallel combination of a 12 Ω resistor and a 4 Ω resistor. That is, the current leaves the source, passes through the 6 Ω resistor, splits between the 12 Ω and 4 Ω resistors, and returns to the source. Determine the power dissipated in the 4 Ω resistor.

Worked solution
Answer:

P = 16 W

Reduce the network to find the source current, then work back to the branch of interest.

The two parallel resistors combine into a single equivalent resistance:

Rp=12×412+4=4816=3 ΩR_p = \frac{12 \times 4}{12 + 4} = \frac{48}{16} = 3\,\Omega

This is in series with the 6 Ω resistor, so the total resistance seen by the source is Req=6+3=9 ΩR_{eq} = 6 + 3 = 9\,\Omega, and by Ohm's law the source current is

I=249=2.667 AI = \frac{24}{9} = 2.667\,\text{A}

The voltage across the parallel pair is the current times the parallel equivalent (the same voltage appears across both branches because they are in parallel):

Vp=IRp=2.667(3)=8 VV_p = IR_p = 2.667(3) = 8\,\text{V}

The power dissipated in the 4 Ω resistor is then

P4=Vp2R=824=16 WP_4 = \frac{V_p^2}{R} = \frac{8^2}{4} = 16\,\text{W}

The current-divider rule gives the same result: the 4 Ω branch carries I4=2.667×12/(12+4)=2.0I_4 = 2.667 \times 12/(12 + 4) = 2.0 A, and P4=I42R=(2.0)2(4)=16P_4 = I_4^2 R = (2.0)^2(4) = 16 W. The source supplies 24×2.667=6424\times 2.667 = 64 W in total, of which 42.7 W goes to the 6 Ω resistor and 5.3 W to the 12 Ω resistor.

P = 16 W in the 4 Ω resistor

Question 3Worked problemStrength of Materials

A solid steel rod of diameter 20 mm and length 1.5 m is fitted snugly between two rigid, immovable walls at 20°C with zero initial stress. The steel has a modulus of elasticity of 200 GPa and a coefficient of thermal expansion of 12×10−612\times 10^{-6} per °C. The rod is then heated uniformly to 70°C. Assume the material remains linearly elastic and neglect buckling.

(a) Determine the axial stress induced in the rod, and state whether it is tensile or compressive.

(b) Determine the force the rod exerts on each wall.

(c) If instead one wall were removed so that the rod could expand freely, determine the elongation of the rod for the same temperature rise.

Worked solution
Answer:

(a) σ = 120 MPa, compressive; (b) F ≈ 37.7 kN; (c) δ = 0.90 mm

(a) Thermal stress. The rod wants to lengthen when heated, but the rigid walls hold its length fixed, so the total strain must be zero. The total strain is the free thermal strain plus the mechanical strain produced by the wall force:

εtotal=α ΔT+σE=0\varepsilon_{\text{total}} = \alpha\,\Delta T + \frac{\sigma}{E} = 0

Solve for the stress, with ΔT=70−20=50\Delta T = 70 - 20 = 50°C:

σ=−Eα ΔT=−(200×109)(12×10−6)(50)=−120×106 Pa\sigma = -E\alpha\,\Delta T = -(200 \times 10^9)(12 \times 10^{-6})(50) = -120 \times 10^6\,\text{Pa}

The negative sign means the stress is compressive: the walls push back on a rod that is trying to expand. Note that the rod length does not appear; thermal stress in a fully restrained bar is independent of length.

(b) Wall force. The force is the stress times the cross-sectional area:

A=πd24=π(0.020)24=3.14×10−4 m2A = \frac{\pi d^2}{4} = \frac{\pi(0.020)^2}{4} = 3.14 \times 10^{-4}\,\text{m}^2 F=σA=(120×106)(3.14×10−4)=37.7×103 N=37.7 kNF = \sigma A = (120 \times 10^6)(3.14 \times 10^{-4}) = 37.7 \times 10^3\,\text{N} = 37.7\,\text{kN}

By Newton's third law the rod pushes on each wall with 37.7 kN, and each wall pushes back on the rod with the same force.

(c) Free expansion. With one wall removed there is no restraint, so the stress is zero and the rod simply undergoes its free thermal elongation:

δ=α ΔT L=(12×10−6)(50)(1.5)=9.0×10−4 m=0.90 mm\delta = \alpha\,\Delta T\,L = (12 \times 10^{-6})(50)(1.5) = 9.0 \times 10^{-4}\,\text{m} = 0.90\,\text{mm}

This is the amount the walls had to suppress in part (a); the same 0.90 mm shortening produced by the 37.7 kN force follows from δ=FL/(AE)\delta = FL/(AE).

σ = 120 MPa (compressive); F ≈ 37.7 kN on each wall; free elongation δ = 0.90 mm

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