Free FE Mechanical practice problems

Ten real questions from our FE Mechanical bank — one from each knowledge area, with the same exam-realistic figures and fully worked solutions our members get. No signup, no email. Try each one first, then reveal the solution.

8 with figures10 knowledge areasFull worked solutions

1
medium
A. Electrical fundamentals & DC circuit analysis
A voltage divider across a 24 V DC source uses $R_1=200\,\Omega$ (series) and $R_2=300\,\Omega$ (shunt). A load $R_L=600\,\Omega$ is then connected across the output terminals (across $R_2$ ). The output voltage across the load is most nearly:
1. 9.6 V
2. 19.6 V
3. 14.4 V
4. 12 V
Show the answer & worked solution
Answer: D.12 V
Loaded bottom leg: $R_2\|R_L=\dfrac{300\cdot600}{900}=200\,\Omega$ . $V_{out}=V\dfrac{R_2\|R_L}{R_1+R_2\|R_L}=24\cdot\dfrac{200}{400}=12$ V. Distractors: the unloaded divider value (loading ignored), the inverted divider ratio $R_1/(R_1+R_2)$ , and the load wrongly placed in series with $R_2$ instead of in parallel.
2
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 $D_2 = 100$ mm. The upstream velocity is $V_1 = 2$ m/s. The velocity in the smaller section is most nearly:
1. 32 m/s
2. 4 m/s
3. 8 m/s
4. 1 m/s
Show the answer & worked solution
Answer: C.8 m/s
Continuity $A_1V_1=A_2V_2$ with $A\propto D^2$ gives $V_2=V_1(D_1/D_2)^2=2\,(200/100)^2=8$ m/s. Distractors: the diameter ratio raised to the FOURTH power ( $V_1(D_1/D_2)^4$ , the area-ratio squaring applied a second time); the DIAMETER ratio used un-squared ( $V_1 D_1/D_2$ , forgetting $A\propto D^2$ ); and the ratio both inverted and un-squared ( $V_1 D_2/D_1$ ).
3
medium
A. Conduction, convection, radiation, transient & heat exchangers
A plane wall of thermal conductivity $k=0.8$ W/m·K has a face area $A=10$ m² and thickness $L=0.2$ m, with its two faces held at a temperature difference $\Delta T=20$ K. By Fourier's law, the steady rate of heat conduction through the wall is most nearly:
1. 1600 W
2. 80 W
3. 800 W
4. 160 W
Show the answer & worked solution
Answer: C.800 W
Fourier's law: $q=\dfrac{kA\,\Delta T}{L}=\dfrac{0.8\cdot10\cdot20}{0.2}=800$ W. Distractors: used half the thickness — the distance to the mid-plane — in the denominator ( $kA\Delta T/(L/2)$ , which doubles the rate); dropped the area $A$ (reported the heat flux $k\Delta T/L$ as a rate); and omitted the thickness entirely ( $kA\Delta T$ , as if $L=1$ m).
4
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:
1. 82.3 MPa
2. 95.8 MPa
3. 74.3 MPa
4. 70.5 MPa
Show the answer & worked solution
Answer: A.82.3 MPa
$\sigma'=\sqrt{\sigma_x^2-\sigma_x\sigma_y+\sigma_y^2+3\tau_{xy}^2}=\sqrt{80^2-(80)(30)+30^2+3(25)^2}=82.3$ MPa. Distractors: dropped the cross term $-\sigma_x\sigma_y$ (used $\sigma_x^2+\sigma_y^2+3\tau^2$ ); used $3\tau_{xy}$ instead of $3\tau_{xy}^2$ (shear left un-squared); and dropped the factor $3$ on the shear term ( $+\tau_{xy}^2$ ).
5
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:
1. 120 MPa
2. 10 MPa
3. 15 MPa
4. 60 MPa
Show the answer & worked solution
Answer: D.60 MPa
$I=\dfrac{bh^3}{12}=4166667\,\text{mm}^4$ , $c=h/2=50\,\text{mm}$ ; $\sigma=\dfrac{Mc}{I}=\dfrac{6M}{bh^2}=\dfrac{6(5\times10^6)}{(50)(100)^2}=60$ MPa. Distractors: used $I=bh^3/3$ instead of $bh^3/12$ (¼ of the true stress); used $c=h$ instead of $h/2$ (twice the stress); and dropped the factor 6 (used $\sigma=M/bh^2$ ).
6
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:
1. 18 kN
2. 32 kN
3. 16 kN
4. 20 kN
Show the answer & worked solution
Answer: D.20 kN
ΣM about B: $R_A=\dfrac{P(L-a)}{L}+\dfrac{wL}{2}=\dfrac{12(4)}{6}+\dfrac{4\cdot6}{2}=20$ kN. Distractors: the entire distributed load assigned to A instead of half ( $+wL$ , giving 32 kN), the point-load moment arm taken as $a$ instead of $(L-a)$ — moments summed about the wrong support — yielding R_B (16 kN), and the point load split evenly regardless of its position ( $P/2$ , giving 18 kN).
7
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:
1. $12,000
2. $20,593
3. $13,333
4. $8,593
Show the answer & worked solution
Answer: B.$20,593
$(A/P,0.06,15)=\dfrac{i(1+i)^n}{(1+i)^n-1}=0.10296$ , so $A=$ $200,000 $\times\ 0.10296=$ $20,593. Distractors: $P/n$ ignoring interest, the interest-only payment $P\cdot i$ , and the sinking-fund factor (A/F) used in place of A/P.
8
medium
A. Probability distributions & expected value
The breakdown voltage of a batch of Zener diodes is normally distributed with mean $\mu=50$ V and standard deviation $\sigma=8$ V. Given $\Phi(0.50)=0.6915$ and $\Phi(1.50)=0.9332$ , the probability that a randomly selected diode breaks down between 54 V and 62 V is most nearly:
1. 0.7583
2. 0.6247
3. 0.9332
4. 0.2417
Show the answer & worked solution
Answer: D.0.2417
$z_1=(54-50)/8=0.50$ and $z_2=(62-50)/8=1.50$ . $P=\Phi(z_2)-\Phi(z_1)=0.9332-0.6915=0.2417$ . Distractors subtract the wrong lower tail ( $1-\Phi(z_1)$ instead of $\Phi(z_1)$ ), ignore the lower limit and report $\Phi(z_2)$ alone, and report the complement $1-P$ .
9
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
The Model Rules require prompt written disclosure of any business interest that could influence professional judgment; the informed client then decides how to proceed. Distractors: drafting around the interest conceals the conflict instead of disclosing it; objective bid criteria do not cure an undisclosed stake in a bidder who helped shape the spec; divesting after award removes the benefit only after the influence has already operated.
10
medium
A. Properties, phase diagrams, processing, corrosion & failure
A 2 kg block of aluminum is heated from 25 °C to 185 °C. Its specific heat is $c=900$ J/(kg·°C). The heat that must be added is most nearly:
1. 144 kJ
2. 288 kJ
3. 333 kJ
4. 824 kJ
Show the answer & worked solution
Answer: B.288 kJ
$Q=mc\,\Delta T=(2)(900)(160)=288000$ J $=288$ kJ. Distractors: the final temperature $T_2=185$ °C used in place of $\Delta T=160$ °C; the mass omitted ( $Q=c\,\Delta T$ ); and the absolute final temperature $T_2+273=458$ K used as the temperature change — a $\Delta T$ is identical in °C and K, so no 273 is added.

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