# FE Mechanical Practice Problems | PrepFE (2023)

Free FE Practice Test

## Free FE Mechanical Example Practice Problems

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

### 1) Given the values below, what is nearest the total power dissipated for this circuit?

$R_1 = \SI{10}{\Omega}$, $R_2 = \SI{3}{\Omega}$, $R_3 = \SI{8}{\Omega}$
$V_1 = \SI{5}{V}$, $V_2 = \SI{8}{V}$

◯A.&nbsp

$1.1 \si W$

◯B.&nbsp

$3.0 \si W$

◯C.&nbsp

$7.1 \si W$

◯D.&nbsp

$21 \si W$

### 2) The following three lines form a triangle with vertices at what points?\begin{align}x+3y=1 \\2x+y=0 \\-2x+y=0\end{align}

◯A.&nbsp

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

◯B.&nbsp

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

◯C.&nbsp

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

◯D.&nbsp

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

### 3) A car accelerates at $a(t) = 2t^2 + 2$ with an initial velocity of 10 m/s. How fast is it going after 3 s?

◯A.&nbsp

(Video) What I Used to Study for the FE Exam (Mechanical)

34 m/s

◯B.&nbsp

97 m/s

◯C.&nbsp

60 m/s

◯D.&nbsp

50 m/s

### 4) The plot below represents the movement of a block on a rough surface due to an external force. Select all that represent the line segment $b$.

◯A.&nbsp

kinetic friction

◯B.&nbsp

dynamic equilibrium

◯C.&nbsp

static friction

◯D.&nbsp

work

### 5) An aluminum rod with diameter of $d = \SI{12}{\milli\meter}$ is loaded axially in compression as shown. Using a Young's modulus of $\SI{70}{\giga\pascal}$ and Poisson ratio of 0.35, what is most nearly the increase in diameter of the rod at the midspan of the rod? Assume the internal loading at the midspan of the rod is -500 kN.

◯A.&nbsp

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

◯B.&nbsp

0.31 mm

◯C.&nbsp

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

◯D.&nbsp

230 nm

### 6) The water velocity at $x_2$ as the water exits the system into the atmosphere is most nearly:

◯A.&nbsp

3.1 m/s

◯B.&nbsp

6.1 m/s

◯C.&nbsp

7.0 m/s

◯D.&nbsp

11 m/s

(Video) FE Exam Practice Problem - Fluid Mechanics

### 7) A parallel flow heat exchanger operates with water ($c_{p,w} = \SI{4.186}{\kilo\joule\per\kilogram\per\kelvin}$) and oil ($c_{p,o} = \SI{3.4}{\kilo\joule\per\kilogram\per\kelvin}$). Water, the cold fluid, enters the inner tube at $\SI{300}{\kelvin}$ and leaves at $\SI{400}{\kelvin}$. What is the log mean temperature difference if oil, the hot fluid, enters the heat exchanger at $\SI{550}{\kelvin}$ and leaves the system at $\SI{450}{\kelvin}$?

◯A.&nbsp

$\SI{240}{\kelvin}$

◯B.&nbsp

$\SI{234}{\kelvin}$

◯C.&nbsp

$\SI{108}{\kelvin}$

◯D.&nbsp

$\SI{124}{\kelvin}$

◯A.&nbsp

0.51

◯B.&nbsp

2.18

◯C.&nbsp

1.93

◯D.&nbsp

1.20

### 9) In the stress-strain diagram, select all that point A describes.

◯A.&nbsp

elastic behavior

◯B.&nbsp

Young's modulus

◯C.&nbsp

ultimate strength

◯D.&nbsp

Necking behavior

◯E.&nbsp

plastic behavior

## Solutions

### 1) Given the values below, what is nearest the total power dissipated for this circuit?

$R_1 = \SI{10}{\Omega}$, $R_2 = \SI{3}{\Omega}$, $R_3 = \SI{8}{\Omega}$
$V_1 = \SI{5}{V}$, $V_2 = \SI{8}{V}$

(Video) FE Exam Review: Mechanics of Materials (2019.09.11)

A.$1.1 \si W$

B.$3.0 \si W$

C.$7.1 \si W$

D.$21 \si W$

### Explanation:

Refer to the Power Absorbed by a Resistive Element section in the Electrical and Computer Engineering chapter of the FE Reference Handbook.

Use the equation,$$P = \frac{V^2}{R}$$The total power dissipated in the circuit is the sum of the powers consumed by each individual resistor. Write an expression for the power consumed on each individual resistor and sum them together,$$P_{\text{total}} = \frac{(V_2-V_1)^2}{R_3} +\frac{(V_1)^2}{R_1+R_2}$$$$= \frac{(\SI{8}{V}-\SI{5}{V})^2}{\SI{8}{\Omega}} +\frac{(\SI{5}{V})^2}{\SI{13}{\Omega}}$$$$\approx \SI{3.05}{W}$$Remember that the $V$ in the equation $\frac{V^2}{R}$ refers to the voltage drop across the resistor. For the resistor, $R_3$, this is the difference between $V_2$ and $V_1$.

For the resistors $R_1$ and $R_2$, their total power consumption can be found by adding them together and dividing them into the voltage drop across both of them, which is simply the voltage $V_1$.

Another way to look at this is that the resistors $R_1$ and $R_2$ are together in series which are then in parallel with the voltage $V_1$. The voltage drop across the series combination of resistors is $V_1 - 0 = V_1$ because all the voltage sources share a common reference. Thus, the term $V_1$ is all that ends up in that particular term of the total power equation.

### 2) The following three lines form a triangle with vertices at what points?\begin{align}x+3y=1 \\2x+y=0 \\-2x+y=0\end{align}

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

### Explanation:

Find the intersection for each pair of lines. Using the initial two equations, let's try substituting the second into the first by solving the second for y:\begin{align} y &= -2x \\\Rightarrow x+3(-2x)&=1 \\-5x &= 1 \\x &= - \frac{1}{5}\end{align}Then plug that back into either of the first two equations to find the y-coordinate of the intersection.$$y = -2(- \frac{1}{5}) = \frac{2}{5} \\\rightarrow (x,y)=(- \frac{1}{5},\frac{2}{5})$$Now, we do the same thing twice for the second and third equations and then the first and third equations. For the second and third equations, we see that $y=-y$, so the second intersection we know that only (0,0) will work. For the last pair:\begin{align}y &= 2x \\\Rightarrow x+3(2x)&=1 \\7x &= 1 \\x &= \frac{1}{7} \\\rightarrow (x,y) &= (\frac{1}{7}, \frac{2}{7})\end{align}We have:$$(- \frac{1}{5},\frac{2}{5}), (0,0), (\frac{1}{7}, \frac{2}{7})$$

A.34 m/s

B.97 m/s

C.60 m/s

D.50 m/s

### Explanation:

Refer to the Particle Rectilinear Motion section in the Dynamics chapter of the FE Reference Handbook.$$a = \mathrm{d}v/\mathrm{d}t$$$$v = \int a\,\mathrm{d}t$$$$v = \frac{2}{3}t^3 + 2t + v_0$$$$= \frac{2}{3}(3)^3 + 2(3) + 10$$$$= \SI{34}{m/s}$$Refer to the Indefinite Integrals section in the Mathematics chapter of the FE Reference Handbook to see a list of typical integration identities. Alternatively, you could solve this integral straight from the calculator using the integration feature in your TI-36X Pro calculator.

### 4) The plot below represents the movement of a block on a rough surface due to an external force. Select all that represent the line segment $b$.

A.kinetic friction

B.dynamic equilibrium

C.static friction

D.work

The correct answers are A, B, and D.

### Explanation:

The block begins moving and remains in motion during $b$ because $x(t)$ is increasing. This is where the coefficient of kinetic friction applies. Work is being done because a force is acting over a distance.

Also, because there is motion but no acceleration on the block, it's also in dynamic equilibrium.

### 5) An aluminum rod with diameter of $d = \SI{12}{\milli\meter}$ is loaded axially in compression as shown. Using a Young's modulus of $\SI{70}{\giga\pascal}$ and Poisson ratio of 0.35, what is most nearly the increase in diameter of the rod at the midspan of the rod? Assume the internal loading at the midspan of the rod is -500 kN.

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

B.0.31 mm

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

D.230 nm

(Video) FE Exam Review: Mechanics of Materials, Part 1 (2022.02.22)

### Explanation:

Refer to the Uniaxial Stress-Strain section in the Mechanics of Materials chapter of the FE Reference Handbook.

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: lateral strain refers to strain that occurs to the sides of the rod (cross section) and NOT strain that occurs along the length of the rod, that would be longitudinal strain. Therefore the equation above was set up in terms of lateral strain because this problem asks for the change in length laterally (at the cross section). Solve for lateral strain using Poisson's ratio. $$\nu=\frac{-\text{lateral strain}}{\text{longitudinal strain}}=0.35=\frac{-\text{lateral strain}}{\text{longitudinal strain}}$$Solve for longitudinal strain using Hooke's law, so we can then come back and solve for lateral strain. $$E=\sigma/ \epsilon_{longitudinal}\rightarrow E=(P/A)/\epsilon_{longitudinal}$$Solve for the unknowns.
$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}$$70\times 10^9\si{Pa}=(-500,000\si{N}/0.000113097 \si{m^2})/\epsilon_{longitudinal}\\\therefore \epsilon_{longitudinal}=-0.0631569$$We can go back and solve for lateral strain using Poisson's ratio.$$0.35=\frac{-\text{lateral strain}}{\text{longitudinal strain}}=\frac{-\epsilon_{lateral}}{-0.0631569}$$$$\therefore \epsilon_{lateral}=0.022104915$$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 can 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. ### 6) The water velocity at$x_2$as the water exits the system into the atmosphere is most nearly: A.3.1 m/s B.6.1 m/s C.7.0 m/s D.11 m/s The correct answer is A. ### Explanation: Refer to the Principles of One-Dimensional Fluid Flow section in the Fluid Mechanics chapter of the FE Reference Handbook. 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$$\frac{0}{9.81\si{kN/m^3}} +\frac{v_2^2} {2(9.81\si{m/s^2})} +0 = \frac{0}{9.81\si{kN/m^3}} \\+ \frac{0^2} {2(9.81\si{m/s^2})} +0.5\si{m}; \text{solve for }v_2$$$$v_2=3.1\si{m/s}$$ ### 7) A parallel flow heat exchanger operates with water ($c_{p,w} = \SI{4.186}{\kilo\joule\per\kilogram\per\kelvin}$) and oil ($c_{p,o} = \SI{3.4}{\kilo\joule\per\kilogram\per\kelvin}$). Water, the cold fluid, enters the inner tube at$\SI{300}{\kelvin}$and leaves at$\SI{400}{\kelvin}$. What is the log mean temperature difference if oil, the hot fluid, enters the heat exchanger at$\SI{550}{\kelvin}$and leaves the system at$\SI{450}{\kelvin}$? A.$\SI{240}{\kelvin}$B.$\SI{234}{\kelvin}$C.$\SI{108}{\kelvin}$D.$\SI{124}{\kelvin}$The correct answer is D. ### Explanation: According to the FE Reference Handbook, the log mean temperature difference for parallel flow heat exchangers is defined as$$\Delta T_{lm} = {(T_{Ho} - T_{Co}) - (T_{Hi} - T_{Ci}) \over ln \left ( T_{Ho} - T_{Co} \over T_{Hi} - T_{Ci} \right )}$$ where$T_{Hi}$the hot fluid inlet temperature,$T_{Ho}$the hot fluid outlet temperature,$T_{Ci}$the cold fluid inlet temperature and$T_{Co}$the cold fluid outlet temperature.In our problem water is the cold fluid and oil the hot one, therefore we have $$\Delta T_{lm} = {(T_{oil,out} - T_{water,out}) - (T_{oil,in} - T_{water,in}) \over ln \left ( T_{oil,out} - T_{water,out} \over T_{oil,in} - T_{water,in} \right )} \\ = {(\SI{450}{\kelvin} - \SI{400}{\kelvin}) - (\SI{550}{\kelvin} - \SI{300}{\kelvin}) \over ln \left (\SI{450}{\kelvin} - \SI{400}{\kelvin} \over \SI{550}{\kelvin} - \SI{300}{\kelvin} \right )} \\= \SI{124.3}{\kelvin}$$ ### 8) A total of 50 valve pressure measurements were taken but only 9 of those measurements will be utilized. What is most nearly the standard deviation of the sample measurements?$$\begin{array}{c|c}\text{Pressure} & \text{# of Measurements} \\\hline20 & 3 \\22 & 4 \\25 & 2 \\\end{array}$$ A.0.51 B.2.18 C.1.93 D.1.20 The correct answer is C. ### Explanation: Refer to the Dispersion, Mean, Median, and Mode Values section in the Engineering Probability and Statistics chapter of the FE Reference Handbook. 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 deviationfunction to solve this problem more quickly.

### 9) In the stress-strain diagram, select all that point A describes.

A.elastic behavior

B.Young's modulus

C.ultimate strength

D.Necking behavior

E.plastic behavior

The correct answers are A and B.

### Explanation:

The graph above shows a material's reaction to increasing axial tensile force.

From the origin until Point B, the material behaves elastically to tensile force. In elastic behavior, the material deformation is not permanent. The material will go back to its original shape once the tensile stress is removed. Additionally, the slope of A is used to find Young's modulus.

Plastic behavior occurs from Point B to Point D. The material's deformation is permanent and the material will not go back to its original shape if the tensile force is removed. Necking behavior occurs within the plastic behavior region. Necking occurs after the tensile force is so strong that it surpasses the ultimate strength of the material (Point C). Necking refers to when the material's cross-sectional area gets smaller and smaller until the material fractures (Point D).

## FAQs

### Is the FE practice exam enough to pass? ›

Although the questions on the NCEES Practice Exam are great, there just aren't enough of them for a good FE exam review. We recommend you add on with other sources that have more practice problems like PrepFE.

How many questions do I need to get right on the FE mechanical exam? ›

NCEES generally does not provide a specific number of questions required to answer correctly for one to pass. However after doing the research and referring to different articles listed above it has been said that out of 110 problems, to pass, you will need to answer approximately 50 to 60 percent of them correctly.

How many people pass the FE mechanical exam? ›

Pass rates
ExamVolumePass rate
FE Electrical and Computer73264%
FE Environmental39664%
FE Industrial and Systems8851%
FE Mechanical202965%
3 more rows

How to pass the mechanical FE exam? ›

It is to work hard and smart.
1. Review the list of FE Exam topics and classify the exam topics into buckets. Familiarize yourself with the published list of topics that could be asked on the Fundamentals of Engineering (FE) exam. ...
2. Develop a study schedule. Develop a study schedule that takes into account: ...
3. Practice. Practice.

Is the FE practice exam harder than the actual exam? ›

NCEES Practice Exam

This is one that I recommend everyone use. It comes in all disciplines, so it is suitable for all engineering students. Some students tell me that this exam is easier than the actual exam, but they still recommend it.

How hard is the FE mechanical exam? ›

A 54% pass rate is a hard exam, the hardest among the FE exams. The FE Mechanical and FE Environmental exams now have the highest pass rate at 66% (see our chart in the next section below), and a 34% failure rate is something we consider very demanding.

What is the easiest FE exam to pass? ›

Among all the NCEES® FE exams, the FE Other (General) exam is the easiest FE exam to pass, especially for those who are confused about choosing the right engineering domain.

What is the minimum score to pass Fe mechanical? ›

6. The FE Exam Passing Score. The passing score of the FE exam on average is 70%.

Does the FE exam get curved? ›

In short, no, the FE Exam is not curved. But it is definitely important to understand how it is scored – let's lay it out as it is defined for us by the NCEES. When you take the FE Exam, you will complete 110 questions. Of those 110 questions, you are going to get a certain number correct.

How many hours should I study for the FE mechanical exam? ›

In general, we recommend 12 to 14 weeks of serious study for a minimum of three hours per day. However, FE exam prep time varies depending on which FE exam you plan to take. Each exam has different questions and a different number of test-takers.

### How many people pass the FE on their first try? ›

To put it in perspective, the FE Civil Exam only has a 62% pass rate and continues to go down year after year. The FE is a comprehensive exam that tests your knowledge of all the fundamentals of engineering. You have to put in the hard work and dedication to pass.

Is the FE exam harder the second time? ›

A question we often get coming in from those who are getting back on the horse after failing is whether or not the FE Exam gets harder to pass the more you take it. The short answer is No. But, that isn't the whole story. The FE Exam actually can get harder to pass the more you attempt it.

Is it possible to pass the FE exam without studying? ›

Although it may seem daunting at first to prepare for the FE exam when you have not much time left, the chances are that you can still pass the FE exam with little to no studying at all.

Can I study for the FE in a month? ›

We recommend planning to study for at least 2 to 3 months before your FE exam date. Once you get started practicing problems from all of the different topics covered, you'll have a better idea if you need to spend more or less time preparing.

Can you bring a cheat sheet to FE exam? ›

Because the exam is administered via computer, you are not permitted to bring any books to the testing center. However, you will be provided with the NCEES FE Reference Handbook to search through during the exam.

Why do I keep failing the FE exam? ›

Failing to set aside enough time to prepare

The best way to pass the FE exam is proper preparation. Start studying for the test at least one month in advance. Buy reference materials and practice tests early to find enough time to prepare. Time yourself with practice tests so you can enter that test with confidence.

How many times can you fail the FE? ›

After all, you can retake the FE exam. How many times per year, exactly? Well, NCEES® policy states that you can attempt the test once per every testing window, up to three times in 12 months.

Is the FE harder than the PE? ›

The FE exam is actually harder than the Professional Engineer (PE) exam, simply because it is much more broad. It's best to take it as close to graduation as you can, with all that schooling still fresh in your mind.

What percentage of questions do you have to get right to pass Fe? ›

6. The FE Exam Passing Score. The passing score of the FE exam on average is 70%.

How long should you study before taking the FE exam? ›

How long you should study for the FE exam depends on you as an individual. In general, we recommend 12 to 14 weeks of serious study for a minimum of three hours per day.

### What percentage of people pass the FE exam on the first try? ›

Everyone has different study habits and test-taking skills, and what works for someone else may not work for you. To put it in perspective, the FE Civil Exam only has a 62% pass rate and continues to go down year after year.

Is Fe or PE harder? ›

The FE exam is actually harder than the Professional Engineer (PE) exam, simply because it is much more broad. It's best to take it as close to graduation as you can, with all that schooling still fresh in your mind.

How many times can you fail the FE exam? ›

After all, you can retake the FE exam. How many times per year, exactly? Well, NCEES® policy states that you can attempt the test once per every testing window, up to three times in 12 months.

Do you lose points for wrong answers on the FE exam? ›

There are no deductions for wrong answers. The score is then converted to a scaled score, which adjusts for any minor differences in difficulty across the different exam forms.

Does FE exam discipline matter? ›

Does it matter which FE exam I take? No. You may take any FE exam discipline to get EIT certification, and it does not matter for your PE license.

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