Updated for the new July 2020 FE Exam Topics

Free FE Practice Test# PrepFE™

## Free FE Other Disciplines Example Practice Problems

We've selected 10 diverse practice problems from our question bank that you can use to review for the Other Disciplines engineering FE exam and give you an idea about some of the content we provide.

◯ A.

1 1 1

◯ B.

1 1 2

◯ C.

1 2 1

◯ D.

1 2 2

◯ A.

1.1 W

◯ B.

3.0 W

◯ C.

7.1 W

◯ D.

21 W

◯ A.

0.51

◯ B.

2.18

◯ C.

1.93

◯ D.

1.20

◯ A.

$ (- \frac{1}{5},\frac{2}{5}), (0,0), (\frac{1}{7}, \frac{2}{7}) $

◯ B.

$ (- \frac{1}{5},\frac{2}{5}), (1,1), (\frac{1}{7}, \frac{2}{7}) $

◯ C.

$ (- \frac{1}{5},-\frac{2}{5}), (0,1), (\frac{1}{7}, \frac{2}{7}) $

◯ D.

$ (\frac{1}{5},\frac{2}{5}), (0,0), (\frac{1}{7}, \frac{2}{7}) $

$$ \begin{array}{c|c|c} \text{Bids Won} &0&1&2\\ \hline \text{\$ Earned} & -\$12,000 & \$88,000 & \$188,000 \\P(x) & 0.81 & 0.18 &0.01 \\ \end{array} $$

◯ A.

\$8,000

◯ B.

\$22,000

◯ C.

\$12,000

◯ D.

\$0

◯ A.

3.1 m/s

◯ B.

6.1 m/s

◯ C.

7.0 m/s

◯ D.

11 m/s

◯ A.

elastic behavior

◯ B.

Young's modulus

◯ C.

ultimate strength

◯ D.

proportional limit

◯ E.

plastic behavior

◯ A.

$\SI{49}{\micro\meter}$

◯ B.

0.31 mm

◯ C.

$\SI{265}{\micro\meter}$

◯ D.

230 nm

◯ A.

$\SI{240}{\kelvin}$

◯ B.

$\SI{234}{\kelvin}$

◯ C.

$\SI{108}{\kelvin}$

◯ D.

$\SI{124}{\kelvin}$

A.1 1 1

B.1 1 2

C.1 2 1

D.1 2 2

The correct answer is B.

Note that you could have solved for other variables besides b and still gotten to the same final solution.

A.1.1 W

B.3.0 W

C.7.1 W

D.21 W

The correct answer is B.

A.0.51

B.2.18

C.1.93

D.1.20

The correct answer is C.

The population is 50, but the sample is 9. Since we only have pressure values for 9 measurements out of the 50 total measurements, we must use the sample standard deviation formula. Ultimately, we must solve for: $$ s = \sqrt{ \frac{1}{(n-1)} \sum_{i=1}^n \left(X_i-\bar{X}\right)^2 } $$ Solve for the unknowns.

The total number of samples is $n=9$.

Calculate the arithmetic mean, $\bar{X}.$ $$ \bar{X}=(1/n)(X_1+X_2+...+X_n) \\ \bar{X}=(1/9)(20+20+20+22+ \\ 22+22+22+25+25) $$ $$\bar{X} = 22$$ Now that we have all the unknowns, solve for standard deviation. $$ s = \sqrt{ \frac{1}{9-1} \left[ 3(20-22)^2 + 4(22-22)^2 + 2(25-22)^2 \right] } $$ $$ \approx 1.93 $$

Note: You can use your TI-36X pro calculator's standard deviation function to solve this problem more quickly.

A.$ (- \frac{1}{5},\frac{2}{5}), (0,0), (\frac{1}{7}, \frac{2}{7}) $

B.$ (- \frac{1}{5},\frac{2}{5}), (1,1), (\frac{1}{7}, \frac{2}{7}) $

C.$ (- \frac{1}{5},-\frac{2}{5}), (0,1), (\frac{1}{7}, \frac{2}{7}) $

D.$ (\frac{1}{5},\frac{2}{5}), (0,0), (\frac{1}{7}, \frac{2}{7}) $

The correct answer is A.

$$ \begin{array}{c|c|c} \text{Bids Won} &0&1&2\\ \hline \text{\$ Earned} & -\$12,000 & \$88,000 & \$188,000 \\P(x) & 0.81 & 0.18 &0.01 \\ \end{array} $$

A.\$8,000

B.\$22,000

C.\$12,000

D.\$0

The correct answer is A.

First, what are the chances that we lose both bids ($x=0$). If winning has a 10% chance, then losing has a 90% chance. $$ P(0) = 0.9 \cdot 0.9 = 0.81 $$ What about when one bid loses and one bid wins? $$ P = 0.9 \cdot 0.1 = 0.09 $$

Once you work through the example, you see it's easier to understand than what the formal definition of

The expected value is also known as the expectation, mathematical expectation, mean, or first moment. It is the probability-weighted average of a discrete random variable's possible values.

A.3.1 m/s

B.6.1 m/s

C.7.0 m/s

D.11 m/s

The correct answer is A.

Ultimately, we must solve for the velocity at the outlet of the system, $v_2$. Use Bernoulli's equation to solve for $v_2. $ $$ \frac{p_2}{\gamma} +\frac{v_2^2} {2g} +z_2 = \frac{p_1}{\gamma} + \frac{v_1^2} {2g} +z_1 $$ Calculate the unknowns:

$p_1=0 \si{Pa}$ because the water is exposed to the atmosphere, which is 0 gauge pressure or 101.3 kPa absolute pressure

$\gamma_{water}=9.81\si{kN/m^3}$

$z_1=0.5\si{m}$

$g=9.81\si{m/s^2}$

$v_1=0$ since the water at $x_1$ is just sitting there, not moving

$v_2=?$

$p_2=0 \si{Pa}$ because the water is released to the atmosphere, which is 0 gauge pressure or 101.3 kPa absolute pressure

$z_2=0$ $$ \frac{0}{9.81\si{kN/m^3}} +\frac{v_2^2} {2(9.81\si{m/s^2})} +0 = \frac{0}{9.81\si{kN/m^3}} \\ + \frac{0^2} {2(9.81\si{m/s^2})} +0.5\si{m}; \text{solve for }v_2 $$ $$ v_2=3.1\si{m/s} $$

A. elastic behavior

B. Young's modulus

C. ultimate strength

D. proportional limit

E. plastic behavior

The correct answers are A and B.

Along A, a material's behavior is elastic - its deformation is not permanent. The slope of A is used to find Young's modulus.

A.$\SI{49}{\micro\meter}$

B.0.31 mm

C.$\SI{265}{\micro\meter}$

D.230 nm

The correct answer is C.

Ultimately, we must determine the increase in rod diameter after the rod is compressed. We can determine the increase in length of a member from its engineering strain. $$ \epsilon=\Delta L/L_0 $$ Which can be interpreted as $$ \epsilon_{lateral}=\Delta L_{lateral}/L_{0 lateral}=\Delta d/d_0 $$ Solve for the unknowns. Ultimately, we must solve for $\Delta d$.

$ d_0=12\si{mm}=0.012\si{m} \\ \Delta d=? \\ \epsilon_{lateral}=? $

Side note:

$ E= 70\times 10^9\si{Pa} \\ A=\pi/4\cdot d^2=\pi/4\cdot (0.012\si{m})^2=0.000113097 \si{m^2} \\ P=\text{internal loading}=500\si{kN}=500,000\si{N} $ $$ 70\times 10^9\si{Pa}=(-500,000\si{N}/0.000113097 \si{m^2})/\epsilon_{longitudinal} \\ \therefore \epsilon_{longitudinal}=-0.0631569 $$ Now we can go back and solve for lateral strain using Poisson's ratio. $$ 0.35=\frac{-\text{lateral strain}}{\text{longitudinal strain}}=\frac{-\epsilon_{lateral}}{-0.0631569} $$ $$ \therefore \epsilon_{lateral}=0.022104915 $$ Now we can go back to the first equation and solve for the change in diameter. $$ \epsilon_{lateral}=\Delta d/d_0 \\ 0.022104915=\Delta d/0.012\si{m};\text{solve for } \Delta d \\ \Delta d = 0.000265259\si{m}=\SI{265}{\micro\meter} $$ Some of the takeaways from this problem are:

Strain and deformation to a rod occur longitudinally (along the length of the rod) or laterally (along the cross section diameter length). Therefore, you can use any of the Hooke's Law equations and modify them to be in terms of lateral deformation or longitudinal deformation. In this problem, the rod got shorter and the diameter got larger as the two 500 kN pushed into the rod axially.

A.$\SI{240}{\kelvin}$

B.$\SI{234}{\kelvin}$

C.$\SI{108}{\kelvin}$

D.$\SI{124}{\kelvin}$

The correct answer is D.