Free FE Mechanical practice problems

Ten real questions from our FE Mechanical bank, drawn from 10 of its 14 study topics, with fully worked solutions and the same figures our members get. No signup, no email. Try each one first, then reveal the solution.

8 with figures10 topicsFull worked solutions

1
medium
A. Electrical fundamentals & DC circuit analysis
A voltage divider made of $R_1=120\,\Omega$ and $R_2=300\,\Omega$ is supplied from a 24 V DC source, with its output taken across $R_2$ . A load $R_L=600\,\Omega$ is then connected across the output terminals, as shown. The voltage across the load is most nearly:
Figure — drag or pinch to move around
1. 15 V
2. 17.1 V
3. 20 V
4. 21.2 V
Show the answer & worked solution
Answer: A.15 V
The load is in parallel with $R_2$ , so the lower leg is $R_2\|R_L=\dfrac{300\cdot600}{300+600}=200\,\Omega$ , in series with $R_1$ (Resistors in Series and Parallel, FE Reference Handbook 10.6 p. 363). The source current is $I=\dfrac{24}{120+200}=0.075$ A, and Ohm's law across the lower leg gives $V_{out}=I\,(R_2\|R_L)=0.075\cdot200=15$ V. Distractors: the unloaded divider value, loading ignored ( $24\cdot300/420=17.1$ V); the load placed in series with $R_2$ instead of in parallel ( $24\cdot900/1020=21.2$ V); and the load taken as replacing $R_2$ rather than paralleling it ( $24\cdot600/720=20$ V).
2
medium
A. Time value of money & equivalence
An engineer borrows $200,000 to be repaid in 15 equal end-of-year payments at 6% per year. The annual payment is most nearly:
Figure — drag or pinch to move around
1. $8,593
2. $13,333
3. $20,593
4. $25,333
Show the answer & worked solution
Answer: C.$20,593
The capital recovery factor (FE Reference Handbook 10.6 p. 235) converts a present amount into equal end-of-year payments: $(A/P,\ 6\%,\ 15)=\dfrac{i(1+i)^n}{(1+i)^n-1}=\dfrac{0.06\,(1.06)^{15}}{(1.06)^{15}-1}=0.10296$ , printed as 0.1030 in the handbook's 6% factor table. The annual payment is $200,000 × 0.10296 = $20,593. Distractors: $P/n$ , spreading the principal evenly and ignoring interest ($13,333); the sinking-fund factor $(A/F)$ read in place of $(A/P)$ ($8,593), which treats the loan as a future sum; and simple interest on the full principal for all 15 years, spread evenly, $P(1+in)/n$ ($25,333), which charges interest on money already repaid.
3
medium
A. Fluid statics, energy/momentum & internal/external flow
Water flows steadily through a pipe that contracts from a diameter $D_1 = 200$ mm to $D_2 = 150$ mm. The upstream velocity is $V_1 = 1.5$ m/s. The velocity in the smaller section is most nearly:
Figure — drag or pinch to move around
1. 0.844 m/s
2. 2 m/s
3. 2.67 m/s
4. 4.74 m/s
Show the answer & worked solution
Answer: C.2.67 m/s
Continuity (FE Reference Handbook 10.6 p. 185): the same flow $Q=Av$ passes both sections, so $A_1V_1=A_2V_2$ . With $A=\pi D^2/4$ the $\pi/4$ cancels, leaving $V_2=V_1\left(\dfrac{D_1}{D_2}\right)^2=1.5\left(\dfrac{200}{150}\right)^2=2.67$ m/s, faster in the smaller section, as it must be. Distractors: the diameter ratio used un-squared, $V_1D_1/D_2$ (2 m/s), forgetting that area goes as $D^2$ ; the area ratio inverted, $V_1(D_2/D_1)^2$ (0.844 m/s), which would have the water slow down in the smaller section; and the ratio squared twice, $V_1(D_1/D_2)^4$ (4.74 m/s), the area-ratio squaring applied a second time.
4
medium
A. Conduction, convection, radiation, transient & heat exchangers
A brick wall 200 mm thick with thermal conductivity $k=0.8$ W/(m·K) has a face area of 10 m². In steady operation its inside surface is at $T_1=20$ °C and its outside surface at $T_2=-4$ °C. The rate of heat loss through the wall is most nearly:
Figure — drag or pinch to move around
1. 640 W
2. 960 W
3. 1500 W
4. 11900 W
Show the answer & worked solution
Answer: B.960 W
The temperature difference across the wall is $T_1-T_2=20-(-4)=24$ K (a difference is the same in °C and K), and $L=0.2$ m. Conduction through a plane wall (FE Reference Handbook 10.6 p. 209): $\dot Q=\dfrac{kA\,(T_1-T_2)}{L}=\dfrac{0.8\cdot10\cdot24}{0.2}=960$ W. Distractors: dropped the minus sign of the outside temperature, taking $\Delta T=20-4=16$ K (640 W); built the resistance as $kL/A$ instead of the handbook's $R=L/(kA)$ , putting the conductivity where a resistivity belongs ( $A\Delta T/(kL)=1500$ W); and converted only the inside temperature to kelvin, $\Delta T=293-(-4)=297$ K (11900 W), where converting both faces, or neither, gives the same 24 K.
5
medium
A. Machine-element stress, failure theories, springs & deflection
At the critical point of a loaded machine element the plane-stress components are $\sigma_x=80\,\text{MPa}$ , $\sigma_y=30\,\text{MPa}$ , and $\tau_{xy}=25\,\text{MPa}$ (see the stress element). The distortion-energy (von Mises) effective stress is most nearly:
Figure — drag or pinch to move around
1. 74.3 MPa
2. 82.3 MPa
3. 95.8 MPa
4. 108 MPa
Show the answer & worked solution
Answer: B.82.3 MPa
The FE Handbook's distortion-energy theory (Mechanical Engineering — Static Loading Failure Theories, p. 452) gives the effective (von Mises) stress for a biaxial state as $\sigma'=\sqrt{\sigma_x^2-\sigma_x\sigma_y+\sigma_y^2+3\tau_{xy}^2}=\sqrt{80^2-(80)(30)+30^2+3(25)^2}=\sqrt{6775}$ = 82.3 MPa. Distractors: dropped the factor $3$ on the shear term ( $+\tau_{xy}^2$ ), 74.3 MPa; dropped the cross term $-\sigma_x\sigma_y$ (used $\sigma_x^2+\sigma_y^2+3\tau_{xy}^2$ ), 95.8 MPa; and added the cross term ( $+\sigma_x\sigma_y$ ) instead of subtracting it, 108 MPa.
6
medium
A. Stress & strain, bending, torsion, transformation, buckling & failure
A beam of solid rectangular cross-section, width $b=50\,\text{mm}$ and depth $h=100\,\text{mm}$ , carries an internal bending moment $M=5\,\text{kN}\cdot\text{m}$ . The maximum bending (flexural) stress is most nearly:
Figure — drag or pinch to move around
1. 10 MPa
2. 15 MPa
3. 40 MPa
4. 60 MPa
Show the answer & worked solution
Answer: D.60 MPa
Maximum bending stress (Mechanics of Materials — Stresses in Beams): $\sigma=\dfrac{Mc}{I}$ , with the centroidal $I=\dfrac{bh^3}{12}=\dfrac{(50)(100)^3}{12}=4.167\times10^{6}\ \text{mm}^4$ and $c=h/2=50\ \text{mm}$ , which for a rectangle is $\sigma=\dfrac{6M}{bh^2}$ . With $M=5\ \text{kN·m}=5\times10^{6}\ \text{N·mm}$ : $\sigma=\dfrac{6(5\times10^{6}\ \text{N·mm})}{(50\ \text{mm})(100\ \text{mm})^2}=60$ MPa. Distractors: used $I=bh^3/3$ , the area table's moment of inertia about the BASE, instead of the centroidal $bh^3/12$ (¼ of the true stress); dropped the factor 6 (used $\sigma=M/bh^2$ ); and divided by the plastic section modulus $Z=bh^2/4$ instead of the elastic $S=bh^2/6$ (two-thirds of the true stress).
7
medium
A. Force systems, equilibrium, trusses, friction & inertia properties
A simply-supported beam of span $L=6$ m carries a uniform load $w=4$ kN/m over its entire length plus a point load $P=12$ kN located 2 m from the left support (A). The pin reaction at A is most nearly:
Figure — drag or pinch to move around
1. 10 kN
2. 16 kN
3. 18 kN
4. 20 kN
Show the answer & worked solution
Answer: D.20 kN
Replace the uniform load by its resultant, $wL=(4\ \text{kN/m})(6\ \text{m})=24$ kN acting at midspan, 3 m from each support. Then take moments about the roller B, which removes $R_B$ (Statics — Equilibrium Requirements, $\Sigma M=0$ ): $R_A(6\ \text{m})=(12\ \text{kN})(4\ \text{m})+(24\ \text{kN})(3\ \text{m})$ , so $R_A=\dfrac{48+72}{6}=20$ kN. Check with $\Sigma F_y=0$ : $R_B=12+24-20=16$ kN. Distractors: the point-load moment arm taken as $a$ instead of $(L-a)$ — moments summed about the wrong support — yielding $R_B$ (16 kN); the point load split evenly regardless of its position ( $P/2$ , giving 18 kN); and the load intensity $w=4$ kN/m used as if it were the distributed load's resultant, instead of $wL=24$ kN, so A gets only $w/2$ from it (10 kN).
8
medium
A. Probability distributions & expected value
The life of a carbide cutting insert turning steel at a fixed cutting speed is normally distributed with mean $\mu=50$ min and standard deviation $\sigma=8$ min. The probability that a randomly selected insert lasts between 46 min and 62 min is most nearly:
Figure — drag or pinch to move around
1. 0.2417
2. 0.4332
3. 0.6247
4. 0.6915
Show the answer & worked solution
Answer: C.0.6247
Standardize each limit (Normal Distribution, $z=(x-\mu)/\sigma$ ): $z_a=(46-50)/8=-0.50$ and $z_b=(62-50)/8=1.50$ . From the Unit Normal Distribution table, $F(1.50)=0.9332$ and $F(-0.50)=1-F(0.50)=1-0.6915=0.3085$ (the table lists only positive $x$ ). So $P(46<X<62)=F(z_b)-F(z_a)=0.9332-0.3085=0.6247$ . Distractors: the minus sign on $z_a=-0.50$ dropped, so the table value $F(0.50)=0.6915$ was subtracted in place of $F(-0.50)=0.3085$ (0.2417); only the area from the mean to 62 min counted, $F(1.50)-0.5$ , leaving out the part between 46 min and the mean (0.4332); and the upper limit ignored, reporting $P(X>46)=1-F(-0.50)$ (0.6915).
9
medium
A. Kinematics & kinetics of particles and rigid bodies
A 1200 kg car skids with its wheels locked along a level road, slowing from 20 m/s to 8 m/s. Neglecting air resistance, the magnitude of the work done on the car by friction during the skid is most nearly:
1. 86.4 kJ
2. 202 kJ
3. 240 kJ
4. 278 kJ
Show the answer & worked solution
Answer: B.202 kJ
Principle of work and energy (FE Reference Handbook 10.6 p. 107): $T_2=T_1+U_{1\to2}$ on a level road, with $T=\tfrac12 mv^2$ . $T_1=\tfrac12\cdot1200\cdot20^2=240$ kJ and $T_2=\tfrac12\cdot1200\cdot8^2=38.4$ kJ, so $U_{1\to2}=T_2-T_1=-201.6$ kJ: friction does 202 kJ of negative work. Distractors: the speeds subtracted before squaring, $\tfrac12 m(\Delta v)^2=\tfrac12\cdot1200\cdot12^2$ (86.4 kJ), which is not the change in kinetic energy; the final speed ignored, taking the whole initial kinetic energy $T_1$ (240 kJ) as if the car had stopped; and the two kinetic energies added instead of subtracted, $T_1+T_2$ (278 kJ), the sign of the work term mishandled.
10
medium
A. Codes of ethics, public protection & licensure
An engineer preparing the switchgear specification for a hospital project owns a 20 percent interest in one of the three manufacturers likely to bid on the equipment. Under the NCEES Model Rules of Professional Conduct, the engineer should:
1. sell the ownership interest quietly after the contract is awarded to eliminate the conflict
2. continue normally, since the bids will be evaluated against objective criteria the client controls
3. disclose the ownership interest to the client in writing before continuing with the specification
4. write the specification around generic performance requirements so the ownership has no practical effect
Show the answer & worked solution
Answer: C.disclose the ownership interest to the client in writing before continuing with the specification
Model Rule B.6 (FE Reference Handbook 10.6 p. 5, Obligations of Licensees to Employer and Clients) requires a licensee to disclose to the client all known or potential conflicts of interest that could influence or appear to influence their judgment. A stake in a likely bidder is exactly that, and the specification is where it could act, so the client must hear of it before the work goes further; putting it in writing makes the disclosure a record, and the informed client then decides how to proceed. Distractors: drafting around the interest manages the conflict privately, and the client still does not know of it; objective bid criteria do not cure it, because the stake can shape the specification itself before any bid is scored; selling the interest after award keeps the conflict hidden through the very period in which it could operate.

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