FE Other Disciplines Exam: 10 Practice Problems with Detailed Solutions

Answer: A

We apply the method of joints at joint B. Assume tension (pulling away from the joint) as positive for each member.

Step 1: Write the equilibrium equations at joint B.

Sum of forces in the x-direction:

\[ \sum F_x = 0: \quad -F_{BA}\cos 30° + F_{BC}\cos 60° = 0 \]

Sum of forces in the y-direction:

\[ \sum F_y = 0: \quad F_{BA}\sin 30° + F_{BC}\sin 60° - 10 = 0 \]

Step 2: From the x-equation, solve for \( F_{BC} \) in terms of \( F_{BA} \):

\( F_{BC}\cos 60° = F_{BA}\cos 30° \)

\( F_{BC}(0.5) = F_{BA}(0.866) \)

\( F_{BC} = 1.732\, F_{BA} \)

Step 3: Substitute into the y-equation:

\( F_{BA}(0.5) + 1.732\, F_{BA}(0.866) = 10 \)

\( 0.5\, F_{BA} + 1.5\, F_{BA} = 10 \)

\( 2.0\, F_{BA} = 10 \)

\( F_{BA} = \textbf{5.0 kN} \)

Since the result is positive, the assumed direction (tension) is correct. Answer choice B (8.66 kN) is the value of \( F_{BC} \), not \( F_{BA} \) — a common mix-up when two unknowns are solved simultaneously.

Answer: C

Normal stress for an axially loaded member is given by:

\[ \sigma = \frac{F}{A} \]

Step 1: Calculate the cross-sectional area of the rod. For a circle with diameter \( d = 20 \text{ mm} = 0.020 \text{ m} \):

\( A = \frac{\pi d^2}{4} = \frac{\pi (0.020)^2}{4} = \frac{\pi \times 4 \times 10^{-4}}{4} = 3.1416 \times 10^{-4} \text{ m}^2 \)

Step 2: Calculate the stress:

\( \sigma = \frac{25{,}000 \text{ N}}{3.1416 \times 10^{-4} \text{ m}^2} = \textbf{79.6 MPa} \)

Answer choice A (31.8 MPa) results from using the radius instead of the diameter in the area formula but squaring it as if it were the diameter — effectively dividing the area by 4 incorrectly. Answer choice B (63.7 MPa) comes from using the circumference formula instead of the area formula. Always double-check your area calculation and unit conversions.

Answer: C

The Carnot efficiency represents the theoretical maximum efficiency of any heat engine operating between two thermal reservoirs. It depends only on the absolute temperatures:

\[ \eta_{Carnot} = 1 - \frac{T_C}{T_H} \]

Step 1: Verify that temperatures are in absolute units (Kelvin). Both 800 K and 300 K are already absolute — no conversion needed.

Step 2: Calculate the efficiency:

\( \eta = 1 - \frac{300}{800} = 1 - 0.375 = 0.625 = \textbf{62.5\%} \)

Answer choice A (37.5%) is the ratio \( T_C/T_H \) itself, not the efficiency. This is a common error — remember that efficiency equals one minus the temperature ratio. Answer choice D (75.0%) could result from accidentally using °C instead of K if the problem were stated in Celsius. Always use absolute temperatures (K or °R) for Carnot calculations.

Answer: B

In a series circuit, the same current flows through every element. The total resistance determines the current, and each resistor’s voltage drop is proportional to its resistance.

Step 1: Find the total series resistance:

\( R_{total} = R_1 + R_2 + R_3 = 10 + 20 + 30 = 60 \text{ } \Omega \)

Step 2: Apply Ohm’s Law to find the circuit current:

\( I = \frac{V}{R_{total}} = \frac{120}{60} = 2 \text{ A} \)

Step 3: Calculate the voltage drop across the 20 Ω resistor:

\( V_{20} = I \times R_2 = 2 \times 20 = \textbf{40 V} \)

You can verify by checking that all voltage drops sum to the source voltage: \( V_{10} + V_{20} + V_{30} = 20 + 40 + 60 = 120 \) V. Answer choice C (60 V) is the drop across the 30 Ω resistor, not the 20 Ω — read the question carefully.

Answer: C

Projectile motion separates into independent horizontal and vertical components. The ball is launched horizontally, so the initial vertical velocity is zero.

Step 1: Find the time to fall 20 m. Using the vertical free-fall equation with \( v_{0y} = 0 \):

\( h = \frac{1}{2}g t^2 \)

\( t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 20}{9.81}} = \sqrt{4.077} = 2.019 \text{ s} \)

Step 2: Calculate horizontal distance. The horizontal velocity is constant (no air resistance):

\( d = v_x \times t = 15 \times 2.019 = \textbf{30.3 m} \)

Answer choice A (20.2 m) is approximately \( 10 \times t \), which could result from using the wrong horizontal speed. Answer choice D (35.0 m) might come from rounding errors or using \( g = 8.0 \text{ m/s}^2 \). Always keep at least three significant figures in intermediate calculations to avoid rounding errors on the exam.

Answer: B

This problem combines the continuity equation with Bernoulli’s equation for incompressible, frictionless flow along a horizontal streamline.

Step 1: Use continuity to find \( v_2 \). For an incompressible fluid, \( A_1 v_1 = A_2 v_2 \). Since areas scale with the square of diameter:

\( v_2 = v_1 \left(\frac{d_1}{d_2}\right)^2 = 2 \times \left(\frac{100}{50}\right)^2 = 2 \times 4 = 8 \text{ m/s} \)

Step 2: Apply Bernoulli’s equation along a horizontal streamline (elevation terms cancel):

\( P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \)

Step 3: Solve for \( P_2 \):

\( P_2 = P_1 + \frac{1}{2}\rho\left(v_1^2 - v_2^2\right) \)

\( P_2 = 200{,}000 + \frac{1}{2}(1{,}000)(2^2 - 8^2) \)

\( P_2 = 200{,}000 + 500(4 - 64) \)

\( P_2 = 200{,}000 - 30{,}000 = 170{,}000 \text{ Pa} = \textbf{170 kPa} \)

As the pipe narrows, velocity increases and pressure decreases — a direct consequence of conservation of energy. Answer choice D (200 kPa) ignores the velocity change entirely. Answer choice A (140 kPa) could result from doubling the dynamic pressure term by mistake.

Answer: B

In the elastic region, stress and strain are related by Hooke’s Law:

\[ \sigma = E \varepsilon \quad \Longrightarrow \quad \varepsilon = \frac{\sigma}{E} \]

Step 1: Convert to consistent units. Yield strength \( \sigma_y = 250 \text{ MPa} = 250 \times 10^6 \text{ Pa} \). Elastic modulus \( E = 200 \text{ GPa} = 200 \times 10^9 \text{ Pa} \).

Step 2: Calculate the yield strain:

\( \varepsilon_y = \frac{250 \times 10^6}{200 \times 10^9} = \frac{250}{200{,}000} = \textbf{0.00125} \)

This is a dimensionless quantity (m/m). Note that answer choice A (0.000625) is exactly half the correct answer, which could result from using \( E = 400 \) GPa or misplacing a factor of 2. Answer choice C (0.00250) is double the correct value, and answer choice D (0.01250) is ten times too large — a likely decimal-place error. Unit conversion errors are one of the most common mistakes on the FE exam.

Answer: C

For one-dimensional steady-state conduction through a flat wall with no internal heat generation, Fourier’s Law gives the heat flux (energy per unit area per unit time):

\[ q = \frac{k \, \Delta T}{L} \]

Step 1: Determine the temperature difference:

\( \Delta T = T_{inside} - T_{outside} = 22 - (-5) = 27 \text{ K} \)

(A difference of 27°C is the same as 27 K.)

Step 2: Convert thickness to meters:

\( L = 200 \text{ mm} = 0.200 \text{ m} \)

Step 3: Calculate the heat flux:

\( q = \frac{1.4 \times 27}{0.200} = \frac{37.8}{0.200} = \textbf{189 \text{ W/m}^2} \)

Answer choice A (94.5 W/m²) is exactly half the correct answer, which could result from using \( \Delta T = 22 - 5 = 17 \) instead of \( 22 - (-5) = 27 \) — a sign error. Answer choice D (378 W/m²) doubles the correct value, possibly from using \( L = 100 \) mm. Always double-check your temperature difference sign and unit conversions.

Answer: C

OSHA uses a cumulative noise dose model. When a worker is exposed to different noise levels during a shift, the total dose is calculated by summing the ratio of actual exposure time to permissible exposure time at each level:

\[ D = \frac{C_1}{T_1} + \frac{C_2}{T_2} + \cdots \]

where \( C_i \) is the actual duration of exposure at level \( i \) and \( T_i \) is the permissible exposure duration at that level.

Step 1: Calculate each fractional dose:

At 95 dBA: \( \frac{C_1}{T_1} = \frac{4}{4} = 1.00 \)

At 85 dBA: \( \frac{C_2}{T_2} = \frac{4}{16} = 0.25 \)

Step 2: Sum the fractional doses:

\( D = 1.00 + 0.25 = \textbf{1.25} \)

A dose of 1.25 (125%) exceeds OSHA’s permissible limit of 1.0 (100%), so this worker’s exposure is non-compliant. Engineering or administrative controls would be required to reduce exposure. Answer choice B (1.00) would mean the worker is exactly at the limit, which underestimates the actual dose. This type of straightforward regulatory calculation appears frequently on the FE exam.

Answer: B

The NPV is the present value of all future cash flows minus the initial investment. For a uniform annual series, we use the Present Worth factor (P/A):

\[ NPV = -P + A \times (P/A, \, i, \, n) \]

where the uniform series present worth factor is:

\[ (P/A, \, i, \, n) = \frac{(1+i)^n - 1}{i(1+i)^n} \]

Step 1: Calculate the present worth factor with \( i = 0.08 \) and \( n = 6 \):

\( (1.08)^6 = 1.5869 \)

\( (P/A, \, 8\%, \, 6) = \frac{1.5869 - 1}{0.08 \times 1.5869} = \frac{0.5869}{0.12695} = 4.6229 \)

Step 2: Calculate the NPV:

\( NPV = -50{,}000 + 12{,}000 \times 4.6229 \)

\( NPV = -50{,}000 + 55{,}475 = \textbf{\$5{,}475} \)

Since NPV > 0, the project is economically justified at the 8% MARR — the present value of savings exceeds the initial cost. Answer choice C ($12,000) is simply the annual savings without any discounting, and answer choice D ($22,000) might result from summing undiscounted savings (\( 6 \times 12{,}000 = 72{,}000 \)) and subtracting the investment — a fundamental error that ignores the time value of money.