FE Mechanical Practice Exam

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CONTENTS Introduction to NCEES Exams ..... .... .. ..... ............ .... .. .... .. ......... ..... ........ .. 1

About NCEES Exam format Examinee Guide Scoring and reporting Updates on exam content and procedures Exam Specifications ........... .. ..................................... ... .. ........... ........ 3

Practice Exam .. ............ ........ ..... .. ............. ................. .. ....... .......... .... .. 9

Solutions .............. ........ ..... ............ ............ .... ... ....... ......... ...... ...... ... .. 65

iii

About NCEES NCEES is a nonprofit organization made up of the U.S. engineering and surveying licensing boards in all 50 states, U.S. territories, and the District of Columbia. We develop and score the exams used for engineering and surveying licensure in the United States. NCEES also promotes professional mobility through its services for licensees and its member boards. Engineering licensure in the United States is regulated by licensing boards in each state and territory. These boards set and maintain the standards that protect the public they serve. As a result, licensing requirements and procedures vary by jurisdiction, so stay in touch with your board (ncees.org/licensingboards).

Exam format The FE exam contains 110 questions and is administered year-round via computer at approved Pearson VUE test centers. A 6-hour appointment time includes a tutorial, the exam, and a break. You'll have 5 hours and 20 minutes to complete the actual exam. In addition to traditional multiple-choice questions with one correct answer, the FE exam uses common alternative item types such as • • • •

Multiple correct options-allows multiple choices to be correct Point and click-requires examinees to click on part of a graphic to answer Drag and drop--requires examinees to click on and drag items to match, sort, rank, or label Fill in the blank-provides a space for examinees to enter a response to the question

To familiarize yourself with the format, style, and navigation of a computer-based exam, view the demo on ncees .org/ExamPrep.

Examinee Guide The NCEES hxaminee Guide is the official guide to policies and procedures for all NCEES exams. During exam registration and again on exam day, examinees must agree to abide by the conditions in the Examinee Guide, which includes the CBT Examinee Rules and Agreement. You can download the Examinee Guide at ncees.org/exams. It is your responsibility to make sure you have the current version. Scoring and reporting Exam results for computer-based exams are typically available 7-10 days after you take the exam. You will receive an email notification from NCEES with instructions to view your results in your MyNCEES account. All results are reported as pass or fail. Updates on exam content and procedures Visit us at ncees.org/e.xams for updates on everything exam-related, including specifications, exam-day policies, scoring, and corrections to published exam preparation materials. This is also where you will register for the exam and find additional steps you should follow in your state to be approved for the exam.

1

EXAM SPECIFICATIONS

3

Fundamentals of Engineering (FE) MECHANICAL CST Exam Specifications Effective Beginning with the N/A 20XX Examinations

•

The FE exam is a computer-based test (CBT). It is closed book with an electronic reference.

•

Examinees have 6 hours to complete the exam, which contains 110 questions. The 6-hour time also includes a tutorial and an optional scheduled break.

•

The FE exam uses both the International System of Units (SI) and the U.S. Customary System (USCS). Number of Questions

Knowledge

1.

Mathematics A Analytic geometry B. Calculus (e.g., differential, integral, single-variable, multivariable) C. Ordinary differential equations (e.g., homogeneous, nonhomogeneous, Laplace transfom1s) D. Linear algebra (e.g., matrix operations, vector analysis) E. Numerical methods (e.g., approximations, precision limits, error propagation, Taylor's series, Newton's method) F. Algorithm and logic development (e.g., flowchaits, pseudocode)

6-9

2.

Probability and Statistics A Probability distributions (e.g., nonnal, binomial, empirical, discrete, continuous) B. Measures of central tendencies and dispersions (e.g., mean, mode, standard deviation, confidence intervals) C. Expected value (weighted average) in decision making D. Regression (lineai·, multiple), curve fitting, a11d goodness of fit (e.g., correlation coefficient, least squai·es)

4-6

3.

Ethics and Professional Practice A Codes of ethics (e.g., NCEES Model Law, professional and technical societies, ethical and legal considerations) B. Public health, safety, and welfare C. Intellectual property (e.g., copyright, trade secrets, patents, trademarks) D. Societal considerations (e.g., economic, sustainability, life-cycle analysis, environmental)

4-6

4.

Engineering Economics A Time value of money (e.g., equivalence, present worth, equivalent annual worth, future worth, rate of return, annuities) B. Cost types and breakdowns (e.g., fixed, vai·iable, incremental, average, sunk) C. Economic aiialyses (e.g., cost-benefit, break-even, minimum cost, overhead, life cycle)

4-6

4

5.

5-8

Electricity and Magnetism A. Electrical fundamentals (e.g., charge, cun-ent, voltage, resistance, power, energy, magnetic flux) B. DC circuit analysis (e.g., Kirchhoff's laws, Ohm's law, series, parallel) C. AC circuit analysis (e.g., resistors, capacitors, inductors) D. Motors and generators

6.

A. B. C. D. E. F.

7.

Resultants of force systems Concurrent force systems Equilibrium ofrigid bodies Frames and trusses Centroids and moments of inertia Static friction

10-15

Dynamics, Kinematics, and Vibrations A. B. C. D. E. F. G. H. I. J. K.

8.

9-14

Statics

Kinematics of paiticles Kinetic friction Newton's second law for particles Work-energy ofpatticles Impulse-momentum of particles Kinematics of rigid bodies Kinematics of mechanisms Newton's second law for rigid bodies Work-energy of rigid bodies Impulse-momentum of rigid bodies Free and forced vibrations

9-14

Mechanics of Materials A. B. C. D. E. F. G. H. I. J. K.

Shear and moment diagrams Stress transformations and Mohr's circle Stress and strain caused by axial loads Stress and strain caused by bending loads Stress and strain caused by torsional loads Stress and strain caused by sheai· Stress and strain caused by temperature changes Combined loading Defonnations Column buckling Statically indeterminate systems

5

9.

Material Properties and Processing A. Properties (e.g., chemical, electrical, mechanical, physical, thennal) B. Stress-strain diagrams C. Ferrous metals D. Nonferrous metals E. Engineered materials (e.g., composites, polymers) F. Manufacturing processes G. Phase diagrams, phase transfonnation, and heat treating H. Materials selection I. Corrosion mechanisms and control J. Failure mechanisms ( e.g., thermal failure, fatigue, fracture, creep)

10.

Fluid Mechanics A. Fluid properties B. Fluid statics C. Energy, impulse, and momentum D. Internal flow E. External flow F. Compressible flow (e.g. , Mach number, isentropic flow relationships, nonnal shock) G. Power and efficiency H. Perfonnance curves I. Scaling laws for fans, pumps, and compressors

11 . Thermodynamics

7-11

10-15

10-15

A. Properties of ideal gases and pure substances B Energy transfers C. Laws of thermodynamics D. Processes E. Perfonnance of components F . Power cycles G. Refrigeration and heat pump cycles H. Nonreacting mixtures of gases I. Psychrometrics J. Heating, ventilation, and air-conditioning (HV AC) processes K. Combustion and combustion products

12.

Heat Transfer A. Conduction B. Convection C. Radiation D. Transient processes E. Heat exchangers

13.

Measurements, Instrumentation, and Controls A. Sensors and transducers B. Control systems (e.g., feedback, block diagrams) C. Dynamic system response D. Measurement uncertainty (e.g., error propagation, accuracy, precision, significant figures)

7-11

6

5-8

14.

Mechanical Design and Analysis A. B. C. D. E. F. G. H. I. J. K. L. M.

10-15

Stress analysis of machine elements Failure theories and analysis Defonnation and stiffness Springs Pressure vessels and piping Bearings Power screws Power transmission Joining methods (e.g., welding, adhesives, mechanical fasteners) Manufacturability (e.g., limits, fits) Quality and reliability Components (e.g. , hydraulic, pneumatic, electromechanical) Engineering drawing interpretations and geometric dimensioning and tolerancing (GD&T)

7

PRACTICE EXAM

9

FE MECHANICAL PRACTICE EXAM The equation of a sphere with center at (0, 1, -2) and a radius of 9 is:

1.

A 0 B. 0

x2 + (y- 1)2 + (z + 2)2 = 81 x2 + (y + 1)2 + (z - 2)2 = 81

0

C.

(x + 1)2 + (y + 1)2 + (z + 2)2 = 81

0

D.

(x)2 + (y - 1)2 + (z + 2)2 = 9

What is the area of the region in the first quadrant that is bounded by the line y = 1, the curve x = y3 12 , and the y-axis?

2.

o

A

2/5

0

B.

3/5

o

C.

2/3

O D.

1

Suppose/({)= t2. The area under the curve for O:St :S 2, estimated by using the trapezoidal rule with lit= 0.5, is most nearly:

3.

4.00 2.25 .f(t) 1.00 0.25

t NOTTO SCALE 0 0 0 0

A B. C.

D.

4.00 2.75 2.67 1.33

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FE MECHANICAL PRACTICE EXAM Which of the following is the general solution to the differential equation and boundary condition shown below?

4.

dy +5y=0;y(0)=1 dt 0

A

est

0

B.

e-5t

0

C.

0

D.

5.

eHt 5e-5t

Which of the following is a unit vector perpendicular to the plane determined by the vectors A = 2i + 4j and B = i + j - k? 0

A

-2i + j-k

0

B.

1 - -(i + 2j)

0

C.

1 - -(-2i + j - k)

0

D.

_l (-2i-j-k)

./5 J6

J6

The flowchart for a computer program contains the following segment:

6.

VAR=O IF VAR< 5 THEN VAR= VAR+ 2 ELSE EXIT LOOP LOOP

[

What is the value of VAR at the conclusion of this routine?

o

A

0

B.

0 C. D.

o

0 2 4

6

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FE MECHANICAL PRACTICE EXAM 7.

Suppose the lengths of telephone calls form a normal distribution with a mean length of 8.0 min and a standard deviation of 2.5 min. The probability that a telephone call selected at random will last more than 15.5 min is most nearly: 0 0 0

A. B.

C. O D.

8.

0.9987

A series of measurements gave values of 11, 11, 11, 11, 12, 13, 13, 14, for which the arithmetic mean is 12. The population standard deviation is most nearly: O A. O B.

o

C. O D.

9.

0.0013 0.0026 0.2600

1.42 1.25 1.19 1.12

Consider a set of three values: 4, 4, and 7. Match each of the statistical quantities with the correct value.

Statistical Quantities

Mean

.1..

.1..

Variance

_]_

Median

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12

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FE MECHANICAL PRACTICE EXAM You wish to estimate the mean M of a population from a sample of size n drawn from the population. For the sample, the mean is x and the standard deviation is s. The probable accuracy of the estimate improves with an increase in:

10.

A 0 B. 0 C. 0

0

11.

D.

M n

s M+s

According to the Model Rules, Section 240.15, Rules of Professional Conduct, licensed professional engineers are obligated to:

O A

ensure that design documents and surveys are reviewed by a panel of licensed engineers prior to affixing a seal of approval

o

B.

express public opinions under the direction of an employer or client regardless of knowledge of subject matter

o

C.

practice by performing services only in the areas of their competence and in accordance with the current standards of technical competence

o

D.

offer, give, or solicit services directly or indirectly in order to secure work or other valuable or political considerations

12.

As the designer of a distinctively shaped automobile body style, you may protect your intellectual property with a: 0

A

trademark

0

B.

patent

0

C.

copyright

0

D.

industrial design right

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13

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FE MECHANICAL PRACTICE EXAM 13.

A professional engineer was originally licensed 30 years ago and is still currently licensed. The engineer's employer asks the engineer to evaluate a newly developed computerized control system for public transportation. The employer requires that a currently licensed engineer evaluate the system. The engineer may accept the assignment under which conditions? 0

A.

The engineer is competent in modern control systems.

0

B.

The engineer supervises two recent graduates who have passed the FE exam.

0

C.

The engineer studied transportation systems in college.

O D.

14.

The engineer has regularly attended meetings of a professional engineering society.

An engineering firm has overall design responsibility for a large multifaceted construction project. The state in which the work is being conducted requires civil, structural, environmental, electrical, and mechanical engineering plans to be prepared and stamped by a professional engineer. Which of the following statements are true? Select all that apply.

15.

D A.

Responsibility for the coordination of the entire project can be accepted by the firm's senior engineer, who is responsible for signing and sealing all the submitted plans.

D B.

Responsibility for coordination of the entire project can be accepted by the firm's senior engineer if each technical submittal is signed and sealed by the qualified engineer assigned to that segment.

D C.

The·foremost responsibility of the professional engineer is to safeguard the health, safety, and welfare of the public when performing services for clients and employers.

D D.

The technical competence of the professional engineers on the project is not as important as their responsibility to disclose to their employers or clients all known or potential conflicts of interest or other circumstances that could influence or appear to influence their judgment or the quality of their professional service or engagement.

D E.

Licensing laws and rnles governing engineering professional practice may vary in each of the jurisdictions in which a licensee practices.

A printer costs $900. Its salvage value after 5 years is $300. Annual maintenance is $50. If the interest rate is 8%, the equivalent uniform annual cost is - - - -Enter your response in the blank. Answer to the nearest integer.

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FE MECHANICAL PRACTICE EXAM 16.

Economic analysis will be used to compute an equivalent value today for an estimated constant monthly expense that is projected to occur each month over the next three years. Select the column in the factor table that contains the value that could be used for the analysis. Factor Table -i = 2.00%

17.

n

PIF

PIA

PIG

FIP

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 30 40 50 60 100

0.9804 0.9612 0.9423 0.9238 0.9057 0.8880 0.8706 0.8535 0.8368 0.8203 0.8043 0.7885 0.7730 0.7579 0.7430 0.7284 0.7142 0.7002 0.6864 0.6730 0.6598 0.6468 0.6342 0.6217 0.6095 0.5521 0.4529 0.3715 0.3048 0.1380

0.9804 1.9416 2.8839 3.8077 4.7135 5.6014 6.4720 7.3255 8.1622 8.9826 9.7868 10.5753 11.3484 12.1062 12.8493 13.5777 14.2919 14.9920 15.6785 16.3514 17.0112 17.6580 18.2922 18.9139 19.5235 22.3965 27.3555 31.4236 34.7609 43.0984

0.0000 0.9612 2.8458 5.6173 9.2403 13.6801 18.9035 24.8779 31.5720 38.9551 46.9977 55 .6712 64.9475 74.7999 85.2021 96.1288 107.5554 119.4581 131.8139 144.6003 157.7959 171.3795 185.3309 199.6305 214.2592 291.7164 461.9931 642.3606 823.6975 1,464.7527

1.0200 1.0404 1.0612 1.0824 1.1041 1.1262 11487 11717 1.1951 1.2190 1.2434 1.2682 1.2936 1.3195 1.3459 1.3728 1.4002 1.4282 1.4568 1.4859 1.5157 1.5460 1.5769 1.6084 1.6406 1.8114 2.2080 2.6916 3.2810 7.2446

FIA 1.0000 2.0200 3.0604 4.1216 5.2040 6.3081 7.4343 8.5830 9.7546 10.9497 12.1687 13.4121 14.6803 15 .9739 17.2934 18.6393 20.0121 21.4123 22.8406 24.2974 25.7833 27.2990 28.8450 30.4219 32.0303 40.5681 60.4020 84.5794 114.0515 312.2323

AIP

AIF

1.0200 0.5150 0.3468 0.2626 0.2122 0.1785 0.1545 0.1365 0.1225 0.1113 0.1022 0.0946 0.0881 0.0826 0.0778 0.0737 0.0700 0.0677 0.0638 0.0612 0.0588 0.0566 0.0547 0.0529 0.0512 0.0446 0.0366 0.0318 0.0288 0.0232

1.0000 0.4950 0.3268 0.2426 0.1922 0.1585 0.1345 0.1165 0.1025 0.0913 0.0822 0.0746 0.0681 0.0626 0.0578 0.0537 0.0500 0.0467 0.0438 0.0412 0.0388 0.0366 0.0347 0.0329 0.0312 0.0246 0.0166 0.0118 0.0088 0.0032

AIG 0.0000 0.4950 0.9868 1.4752 1.9604 2.4423 2.9208 3.3961 3.8681 4.3367 4.8021 5.2642 5.7231 6.1786 6.6309 7.0799 7.5256 7.9681 8.4073 8.8433 9.2760 9.7055 10.1317 10.5547 10.9745 13.0251 16.8885 20.4420 23 .6961 33.9863

A business owner wants to build a reserve fund. To have $50,000 in the fund at the end of 10 years, the amount the owner will need to invest annually, assuming an annual return of 5%, is most nearly: 0

A.

$951

0

B.

$1,535

0

C.

$2,500

0

D.

$3,975

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FE MECHANICAL PRACTICE EXAM 18.

Consider the information below about a company's project. Ignore any time-dependent factors such as interest, inflation, and depreciation. Initial capital investment: $50,000 Cost to produce each item: $25 Selling price for each item: $40 Production volume (equal distribution through the year): Year 1: 1,000 units Years 2-3: 750 units Years 4 and above: 500 units The year of production in which the company reaches the end of the payback period for the project IS

------

Enter your response in the blank.

19.

An ammeter, a voltmeter, and a wattmeter were installed in an ac circuit and read 15 Arms, 115 Vrms, and 1,500 W, respectively. The power factor of the circuit is most nearly: 0 0 0 0

20.

A B. C. D.

-0.87 -0.5 0.5 0.87

A 120-V heater is designed to use a 20-Q resistor. The power (W) consumed by the heater when . . muse 1s- - - - - Enter your response in the blank

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16

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FE MECHANICAL PRACTICE EXAM 21.

In the resistor circuit shown below, the equivalent resistance Re 9 (0) at Terminals a-b is most nearly:

20

3Q

a

r O A.

o B. 0

o

22.

C. D.

6Q

3Q

b

22 20 4 2

A 1,000-Q resistor is in series with a 2-mH inductor. An ac voltage source operating at a frequency of 100,000 rad/sis attached as shown in the figure. The impedance (Q) of the RL combination is most nearly:

1,000 Q

2mH

0

A.

200 + jl,000

0

B.

0

C.

1,000 + j200 38.4 + jl92

0

D.

I,000 - j200

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FE MECHANICAL PRACTICE EXAM 23.

A 4-pole induction motor operates with a line voltage frequency of 60 Hz and 0.09 slip. The rotational speed (rpm) of the motor shaft is _ _ _ __ Enter your response in the blank.

Beam AB has a distributed load as shown and supports at A and B. If the weight of the beam is negligible, the force RB (kN) is most nearly:

24.

8kN/m

A

B RA

0 0 0 0

25.

A. B. C. D.

RB

24 12 10 8

Three forces act as shown below. The magnitude (N) of the resultant of the three forces is most nearly: y

12~ 5

SON O A. O B. o C.

O D.

140 191 370 396

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18

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FE MECHANICAL PRACTICE EXAM 26.

The moment (N·m) of force F shown below with respect to Point Pis most nearly:

~ F - 500N

pi I...

- ---Urmm

0

A.

31.7 ccw

0

B.

31.7 cw

0

C.

43.3 cw

0

D.

43.3 ccw

27.

300mm

In the figure below, the force (kN) in Member BC is most nearly:

D

C

E

5m

5m

F

5

3kN

I 3kN

O A.

6

o o o

9 15 18

B. C. D.

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H 3kN

19

G 3kN

NEXT-+

FE MECHANICAL PRACTICE EXAM 28.

Mark all of the zero-force members of the simple truss below. D

6m

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20

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FE MECHANICAL PRACTICE EXAM 29.

Evaluate the center of gravity in each figure below. Arrange the figures in increasing order of y-axis center of gravity. All bars are 1 unit wide.

FIGUREA

L

l

- - - - -Lowest

FIGURER

8

_J

FIGUREC

FIGURED

_ _ _ _ _Highest

FIGUREE

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21

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FE MECHANICAL PRACTICE EXAM The ramp in the figure is rotated until the angle (a) is at 37 degrees, and then the block begins to slide. The coefficient of static friction is most nearly:

30.

~ 31.

0

A.

0.602

0

B.

0.625

0

C.

0.754

0

D.

0.798

The moment of inertia about the centroid axis (Y-Y) of the figure below is 606 in4. The moment of inertia (in4 ) about a parallel axis (Y'-Y') that is 1 in. higher is_ _ _ __ Enter your response in the blank.

All dimensions are in inches.

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22

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FE MECHANICAL PRACTICE EXAM The hoist shown is holding an engine with a mass of 100 kg. The tension in the left cable (N) is most nearly:

32.

r-----------2m-----------7

CABLES

ENGINE

o o o o

A B. C.

D.

490 566 849 981

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23

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FE MECHANICAL PRACTICE EXAM 33.

Two blocks, A and B, are arranged so that A rests on top of B and is attached to a vertical wall by an inextensible string. A force of 30 N is applied to Block B, which is sufficient to make it slide to the left. If µK = 0.2 between A and B, and if µK = 0.4 between B and the bottom surface, the acceleration ofB (m/s2 ) is most nearly:

BLOCK A (2 kg)1----IN_E_X_T_E_N_S_IB_L_E_S_T_RIN_G--v

F=30N -------;BLOCK B (2 kg)

0 0 0 0

A B.

5.2 7.2

C.

9.1

D.

15.0

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FE MECHANICAL PRACTICE EXAM 34.

The 2-kg block shown in the figure is accelerated from rest by force F along the smooth incline for 5 m until it clears the top of the ramp at a speed of 8 m/s. The value of F (N) is most nearly:

4.5m

zj3 4

O A. O B.

11.8 19.6

o o

C.

24.6

D.

69.4

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25

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FE MECHANICAL PRACTICE EXAM The piston and cylinder of an internal combustion engine are shown in the following figure. If ro = 377 rad.ls, the piston speed (mm/s) when 0 = 90° is most nearly:

35.

'

0

A.

O B. o C. o D.

/ /

0 10,500 18,850 24,300

A rigid wheel with radius r = 0.25 m rolls without slip along a horizontal surface, as shown in the figure. The velocity of the center of gravity is vo = 5.0 m/s. The magnitude of the total velocity (m/s) of Point A is most nearly:

36.

0

A.

3.5

0

B.

5.0

0

C.

8.5

0

D.

9.2

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//

A

45° VG

26

YL X

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FE MECHANICAL PRACTICE EXAM 37.

In the figure below, Block B is initially at rest and is attached to an unstretched spring. Block A travels to the right and hits Block B. Immediately after impact, the velocity of Block Bis 6 m/s to the right. The maximum acceleration (m/s2) of Block B after impact is most nearly: 4kg

VA ~

G 0 0 0 0

38.

A. B. C. D.

B

~

1.5 2.25 6.0 9.0

A 4-kg mass with a velocity of 10 m/s travels across a horizontal surface with negligible friction and impacts a stationary 2-kg mass. Assuming that the impact is perfectly elastic, the velocity (m/s) of the 2-kg mass after impact is most nearly: 0

A.

6.7

0

B.

8.3

0

C.

10.0

0

D.

13.3

A cylinder has a radius r = 0. l m and mass moment of inertia about Point O of lo= 0.0428 k·gm 2 and rotates about a frictionless axle. A mass m = 1 kg is connected by a massless cord wrapped around the cylinder without slip. The angular acceleration, a, of the cylinder (s- 2) is most nearly:

39.

0

A.

4.7

0

B.

9.4

0

C.

18.6

0

D.

23.2 m

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27

-p

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FE MECHANICAL PRACTICE EXAM An automobile starts from rest on a test track. Its acceleration is plotted in the figure . The speed (m/s) at time t = 5 s is_ _ _ __

40.

Enter your response in the blank.

1.5 1

-1

- 1.5

1

0

3

2

4

5

6

7

TIME (s)

41.

An automobile with a mass of 1,020 kg is traveling at 10 m/s. At time t = 0, the driver activates the emergency brake and the automobile begins to skid. The properties of the relevant materials are shown.

I Material 1 I Material 2 I Static Friction

I

Rubber

I

Asphalt

I

0.9

Kinetic Friction 0.4

At t = l s, the speed (mis) of the automobile is most nearly: 0

A.

8.8

0

B.

6.1

0

C.

3.9

0

D.

1.2

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FE MECHANICAL PRACTICE EXAM 42.

If a pendulum is released from rest at Position 1, the velocity (mis) of the mass at Position 2 is most nearly:

'

~60° '' ' '' ''

m= 1 kg POSITION 1

,

.... J ......

I

''

' '

' ,''

POSITION 2

o o

A. B. 0 C. o D.

5.0 7.0 9.8 12.7

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FE MECHANICAL PRACTICE EXAM A 0.25-m steel rod with a cross-sectional area of 1,250 mm 2 and a modulus of elasticity E of 200 GPa is subjected to a 5,000-N force as shown below. The elongation of the rod (µm) is most nearly:

43.

5,000 N....,_~_ _ _ _ _ __....5,000 N

f,---0.25m~ 0 0 0

A. B.

o

D.

44.

C.

2.4 4.4 5.0 9.6

The piston of a steam engine shown is 50 cm in diameter, and the maximum steam gauge pressure is 1.4 MPa. If the design stress for the piston rod is 68 MPa, its cross-sectional area (m2) should be most nearly: SHORT PISTON ROD STEAM--PRESSURE

0

A.

40.4

X

10-4

0

B.

98.8

X

10-4

0

C.

228.0

X

10-4

0

D.

323.Q

X

10-4

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PISTON

30

NEXT~

FE MECHANICAL PRACTICE EXAM In the figure below, the value of the maximum shear stress (MPa) in Segment BC is most nearly:

45.

RIGID SUPPORT 100-mm-DIAMETER SOLID ALUMINUM CYLINDER (G = 28 GPa) 75-mm-O.D. AND 50-mm-I.D. HOLLOW STEEL CYLINDER (G = 83 GPa)

o o

A. B.

O C. o D.

46.

15.0 30.0 37.7 52.7

A shaft of wood is to be used in a certain process. If the allowable shearing stress parallel to the grain of the wood is 840 kN/m 2, the torque (N·m) transmitted by a 200-mm-diameter shaft with the grain parallel to the neutral axis is most nearly: 0 0

A. B.

o o

D.

C.

500 1,200 1,320 1,500

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31

NEXT~

FE MECHANICAL PRACTICE EXAM 47.

The maximum in-plane shear stress (ksi) in the element shown below is most nearly: 20 ksi

10 ksi

40 ksi

O A. O B. O C.

10 14.1 44.1

O D.

316

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32

NEXT~

FE MECHANICAL PRACTICE EXAM 48.

A plane member having a uniform thickness of 10 mm is loaded as shown below. The maximum shear stress (MPa) at Point O is most nearly: 5kN

100mm 0

100mm

5kN 0 0 0 0

49.

A. B. C. D.

2.5 5.0 7.5 10.0

The Euler formula for columns deals with: O A.

relatively short columns

O B.

shear stress

o o

C.

tensile stress

D.

elastic buckling

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33

NEXT-+

FE MECHANICAL PRACTICE EXAM 50.

A 50-m length of rail is installed in a railroad switching station during summer. It is rigidly attached at each end while the ambient temperature is 30°C. During winter, the ambient temperature at this section of track drops to - l 0°C. Based on the properties listed in the following table, the axial stress (MPa) in the section of track is most nearly:

Density (mg/m 3) 7.85

200

O A.

96

o

B.

72

O C.

36

o

29

D.

Young's Yield Rigidity Modulus Strength (GPa) (GPa) (MPa)

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75

Ultimate Sample Poisson's Stress Elongation Ratio % (MPa)

250

400

34

30%

0.32

Coefficient of Thermal Expansion (10-6/oC)

12

NEXT--+

FE MECHANICAL PRACTICE EXAM 51.

A screw rod is fixed at its two ends, A and C, as shown in the figure. A nut initially in the center of the rod is supported by a spring. The spring constant k = 30 kN/cm and is initially in a neutral position. When the nut is screwed down 2 cm, the support force (kN) at the top point of A is most nearly: /

A

A

5cm 2cm

t

k = 30kN/cm //

0

A.

0

0

B.

18

0

C.

42

0

D.

60

Copyright © 2020 by NCEES

35

NEXT~

FE MECHANICAL PRACTICE EXAM 52.

The silver/copper binary phase diagram is shown below. The composition of Ag-Cu alloy that will be completely melted at the lowest temperature is most nearly: 1,100 1,000

u

900

0

14

I i;:.:i

~ i;:.:i

800 780°

700

r--1

600

a+~ 500 400

0

20

Ag

80

40 60

60 40

80

Cu

20

0

COMPOSITION, % BY WEIGHT

0

A.

8.0 wt% Cu

0

B.

8.8 wt% Cu

0

C.

28.1 wt% Cu

0

D.

71.9 wt% Cu

Copyright © 2020 by NCEES

36

NEXT.-+

FE MECHANICAL PRACTICE EXAM 53.

An alloy that is 70% copper by weight is fully melted and allowed to cool slowly. What phases are present at 850°C?

1,100 1,000

,-._

LIQUI

a+L

900

u

13+ L

0

~

~ ~ i:i.l ~

i:i.l E-


FE MECHANICAL PRACTICE EXAM A power plant operates on the following simple Rankine cycle. Water is the working fluid. Disregard pressure losses in the piping, steam boiler, and superheater, and neglect kinetic and potential energy effects. Assume steady-state, steady-flow conditions.

73.

T2 =179.91°C P = l MPa

2 r-+----iSUPERHEATER

2

h2 = 2,778.1 kJ/kg

BOILER

P3=l MPa T3=500°C h3 =3,478.5 kJ/kg U3 = 3,124.4 kJ/kg

P = 0.01 MPa 4 h44 = 2,584.7 kJ/kg U4 = 2,437.7 kJ/kg

m= 50 kg/s PUMP

COOLING WATER

P 1= 1.0 MPa h1 = 168.6 kJ/kg 1l = 0.7

For the thermodynamic conditions shown, the turbine power (MW) is most nearly:

o o

A B.

O C. O D.

34.3 44.7 52.0 161,000

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47

NEXT..+

FE MECHANICAL PRACTICE EXAM The enthalpies provided in the figure below apply to the refrigeration cycle using refrigerant HFC134a. The coefficient of performance (COP) for this cycle is most nearly:

74.

h1 = 394 kJ/kg

h2 = 438 kJ/kg h3 = 270 kJ/kg 2

T

0

A.

0.35

0

B.

0.74

0

C.

2.82

0

D.

3.82

A machine uses nozzles that can produce only 50 kg/s of an ideal gas mixture with an average molecular weight of 30. If the volumetric flow rate is 30 m 3/s at 350 Kand 420 kPa, the required number of nozzles is - - - - -

75.

Enter your response in the blank.

Conditioned air enters a room at 13°C and 70% relative humidity. The dew-point temperature of the air is most nearly:

76.

0

A.

5°c

0

B.

8°C

0

C.

10°c

0

D.

B 0c

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48

NEXT~

FE MECHANICAL PRACTICE EXAM A vapor-compression refrigeration cycle using HFC-134a as the refrigerant has the pressureenthalpy diagram shown below. The evaporator temperature is 0°C, and the condenser temperature is 40°C.

77.

p

h

The cooling achieved by the evaporator (kJ/kg) is most nearly: O A. o B. o C. o D.

28 143 169 210

Methanol (CH3OH) bums in 20% excess air. What is the combustion equation?

78.

o o o

A.

CHJOH + 1.2 02 + 1.2 (3 .76) N2-. CO2+ 1.2 H2O + 1.2 (3 .76) N2

B.

CHJOH + 1.5 02 + 1.5 (3.76) N2-. CO2+ 2 H2O + 1.5 (3.76) N2

C.

CH3OH + 1.8 02 + 3 (3.76) N2-. CO2+ 2 H2O + 1.8 (3.76) N2

O D.

CH3QH + 1.8 02 + 1.8 (3.76) N2-. CO2+ 2 H2O + 1.8 (3 .76) N2 + 0.3 02

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49

NEXT-+

FE MECHANICAL PRACTICE EXAM The heat flux (W/m2) through 1 m 2 of the steel/aluminum plate system shown is most nearly:

79.

STEAM

--------I-

T r

Q

Too ==300K hoo =100 W/(m2•K)

CARBONSTo/ (10 mm)

= 60 W/(m·K) kA = 240 W/(m·K) ks

O O O O

A. B. C. D.

\_ALUMINUM (IO mm) (IGNORE RADIATION LOSSES AND CONTACT RESISTANCE

BETWEEN THE CARBON STEEL AND ALUMINUM PLATES.)

17,800 19,800 21,000 153,000

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50

NEXT~

FE MECHANICAL PRACTICE EXAM Hot air at 200°C flows across a 50°C surface. If the heat-transfer coefficient is 72 W/(m 2·°C), the heat-transfer rate (W) over 2 m 2 of the surface is most nearly:

80.

A.

0 0

B.

o o

C. D.

300 5,625 11,250 21,600

An infinitely long, 3-cm water pipe is laid on the ground surface A3 as shown below. In order to calculate the radiative heat transfer between pairs of surfaces, you must know the shape factor (or view factor) FiJ between these surfaces. Assume infinitely large sky and ground surfaces. If the shape factor F13 between A1 and A3 is 1/2, the shape factor F12 between A1 and A2 is most nearly:

81.

PIPE A1

0 O A.

1/4

o

B.

1/2

0 C.

3/4

O D.

1

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GROUNDA3

51

NEXT-+

FE MECHANICAL PRACTICE EXAM An enclosure has Surfaces 1 and 2, each with an area of 4.0 m 2 . The shape factor Fi-2 is 0.275 . Surfaces 1 and 2 are black surfaces with temperatures of 500°C and 400°C, respectively. The net rate of heat transfer (kW) by radiation between Surfaces 1 and 2 is most nearly:

82.

2

T2 = 400°C A 2 =4 m2

I I

I I I I

I

I I

I I

/ / / /

>-- - ----- - --,,

/

1 Ti= 500°C Ai= 4 m 2

o o

A B. O C. O D.

2.30 9.47 22.3 34.4

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52

NEXT-.+

FE MECHANICAL PRACTICE EXAM 83.

The lumped capacitance model is valid for which of the following metal cubes? Select all that apply.

D A.

h = 10 (W/m2 •K) k = 1,000 (W/m·K) side= 2 m

D B.

h = 500 (W/m2 ·K) k = l,000 (W/m·K)

side= 6 m

D C.

h = 2000 (W/m 2 ·K) k = 400 (W/m·K) side= 1 m

D D.

h = 100 (W/m2 ·K)

k = 2,000 (W/m·K) side= 20 m

D E.

h = 30 (W/m2 ·K)

k= 3,000 (W/m·K) side= 3 m

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53

NEXT~

FE MECHANICAL PRACTICE EXAM Water at 29°C flows through the inner pipe of a counterflow double-pipe heat exchanger, as shown in the figure. Hot oil at 204 °C enters the annular space between the inner and outer pipes at a flow rate of 2.25 kg/s. The following physical data are known:

84.

Property

Specific heat, kJ/(kg·K) Viscosity, kg/(m·s) Density, kg/m3

Water

Oil

4.186 5.06 X 10-4

3.5 6.3 X IQ- 3

986.7

815.3

79°c

The mass flow rate of the water (kg/s) is most nearly: O A. O B. o C.

2.25 4.70 6.58

D.

19.7

o

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54

NEXT-+

FE MECHANICAL PRACTICE EXAM 85.

Heat is transferred from a stream of hot water to a stream of cold air in the heat-exchanger arrangement shown in the figure below. The heat exchanger is assumed to be insulated from its surroundings, and the specific heats of the two streams are assumed to be constant. If the exit temperature of the hot stream is 65°C and the exit temperature of the cold stream is 60°C, the logarithmic-mean temperature difference for the heat exchanger is most nearly: WATER (HOT STREAM) Cp

m

= 4.18 kJ/(kg·K) = 4,500 kg/hr AIR (COLD STREAM) Cp = 1 kJ/(kg·K) m = 22,900 kg/hr

Tc1 = 24°C 0

A.

28.5°C

0

B.

41.0°C

0

C.

45.4°c

0

D.

50.0°C

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55

NEXT-+

FE MECHANICAL PRACTICE EXAM 86.

A resistance temperature detector (RID) provides a resistance output that is related to temperature by:

R = Ro [ 1 + a( T - To)] where

n

R = resistance, Ro = reference resistance,

n

= coefficient, 0 c- 1 T = temperature, °C To = reference temperature, °C U

Consider an RTD with Ro= 100 0, a= 0.004°C-1, and To= 0°C. The change in resistance (0) of the RTD for a 10°C change in temperature is most nearly: O A.

O B. O C.

o

D.

0.04 0.4 4.0 100.4

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56

NEXT~

FE MECHANICAL PRACTICE EXAM 87.

A resistance temperature detector (RTD) provides a resistance output R that 1s related to temperature by:

R = Ro[l + a(T- To)] where R = resistance, n Ro = reference resistance, 100 n a= linear coefficient of resistance, 0.3925 x 10-31°c T = temperature, °C To = reference temperature, 0°C

The uncertainty, UR, may be calculated from the simplified Kline-McClintock equation:

uzR =("dR aT

U

r

)2

where UR and Ur are the uncertainties in variables Rand T, respectively. The resistance of the RTD, R, is 110 n, and this measured value has an uncertainty UR=± O. l The uncertainty in Tis most nearly:

88.

0

A.

±0.0026°C

o

B.

±0.0040°C

o

C.

±0.1°C

o

D.

±2.5°C

n.

A temperature probe has a time constant r = 3.0 s. The temperature indicated by the probe is given by the equation:

T(t) = To - (To - Ti)e-tlr where To and Ti are 140°C and 40°C, respectively. The time (s) required before the probe indicates a temperature within ±1 °C of the actual oil temperature is most nearly: O A. O B. 0 C. o D.

3.0 4.6 9.0 13.8

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57

NEXT-+

FE MECHANICAL PRACTICE EXAM 89.

An automatic controls block diagram is shown below:

C(s)

R(s)

The single element relating the input to the output is best represented by: Option A

R(s) - -

--C(s)

Option B

R(s) - -

--C(s)

Option C

R(s) - -

--C(s)

Option D

R(s)

0

A.

Option A

0

B.

Option B

0

C.

Option C

0

D.

OptionD

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I (G1G2)/(l + Gl + G1G2H1)

58

C(s)

NEXT-+

FE MECHANICAL PRACTICE EXAM 90.

The transfer function relating a step input to the output of a control system is: 16

s + 0.8s 2 + 16s 3

The natural frequency

91.

ffin

of the system and the damping ratio i; are most nearly:

0

A.

ffin

= 2 rad/s; i; = 0.1

0

B.

ffin

= 2 rad/s; i; = 0.2

0

C.

ffin

= 4 rad/s; i; = 0.1

0

D.

ffin

= 4 rad/s; i; = 0.2

Yielding is considered failure for the ductile beam shown. The following data apply:

Yielding first occurs at Point X. Sy= 34 ksi cr2 = 0 ksi

! - - - - - - - - - - - - - ~l---+----a.... p X

M

The table below shows various calculated values for CTI and 0"3. Select the columns that show the values at which failures will occur.

Stress CTI (ksi) 0"3 (ksi)

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A

B

C

D

E

F

35 -35

24

45 -35

60 24

18 -62

82 62

0

59

NEXT-+

FE MECHANICAL PRACTICE EXAM 92.

A helical compression spring has a wire diameter of 2.34 mm and an outside diameter of 15 mm. For a spring load of 150 N, the shear stress (MPa) in the spring is most nearly:

O A o B. O C. o D.

93.

342 377 412 577

A helical compression spring has a spring constant of 38.525 N/mm and a free length of 190 mm. The force (N) required to compress the spring to a length of 125 mm is most nearly:

o o

A

B. O C.

O D.

94.

1,500 2,500 4,800 6,500

The pressure gauge in an air cylinder reads 1,680 kPa. The cylinder is constructed of a 12-mm rolled-steel plate with an internal diameter of 700 mm. The tangential (hoop) stress (MP a) inside the tank is most nearly:

o

A

O B. 0

o

C. D.

25 50 77

100

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60

NEXT-+

FE MECHANICAL PRACTICE EXAM The figure below shows an unpressurized vessel. Material properties are given with the figure.

95.

- - - - - - 615 mm - - - - -

MATERIAL PROPERTIES:

E = 210 x 103 MPa V = 0.24

= 10.5 X 10-6/°C Sy = 400 MPa

7A

U

1,000 mm

_JB

VERTICAL-AXIS PRESSURE VESSEL SECTION

Assume the internal pressure is increased to P; such that the stresses in the wall between Locations A andB are: at = 46.2 MPa at = 23. l MPa CJr

= 0

The increase in length (mm), along the outer wall, of the distance between Locations A and B due to the increase in pressure is most nearly: 0

A.

o o o

B. C. D.

0.06 0.11 0.19 0.22

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61

NEXT~

FE MECHANICAL PRACTICE EXAM A reliability study is conducted for a three-component system which consists of one component connected in series to two additional components that are connected in parallel, as shown in the figure. The results of independent component testing are shown in the boxes and indicate the probability that the individual component is functioning. The reliability of the entire system is most nearly:

96.

0

A.

O B. O C. O D.

0.09 0.41 0.61 0.81

A steel pulley with a minimum room-temperature bore diameter of 100.00 mm is to be shrunk onto a steel shaft with a maximum room-temperature diameter of 100 .15 mm.

97.

Assume the following: Room temperature = 20°C Coefficient of linear expansion of steel = 11 x 1o-6/°C Required diametral clearance for assembly= 0.05 mm To shrink the pulley onto the room-temperature shaft with the desired diametral clearance, the pulley must be heated to a minimum temperature of most nearly: 0

A.

65°C

0

B.

136°C

0

C.

182°C

0

D.

202°c

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62

NEXT-+

FE MECHANICAL PRACTICE EXAM 98.

A solid steel rod with a round cross section vertically cantilevered in the ground has the following properties:

r 1,._p

--,---J..---1

Vmax = I cm

10 cm

l.Om

Modulus of elasticity Cross-sectional moment of inertia Length

£=200 GPa I= 1.60 x 10-8 m 4 L = 1.00 m

A force P is applied along the horizontal direction, 10 cm below the free end, as shown. To cause a maximum deflection to the end of the rod of 1 cm, the force (N) required is most nearly:

0 A. 0 B. o C. o D.

1.13 113 132 6,620

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63

NEXT~

FE MECHANICAL PRACTICE EXAM Two aluminum bars with semicircular ends are bolted together in single shear, as shown, using a 5-mm bolt. The tension in the bolt is 8 kN. The tension on the bars is 100 N. The average shear stress (MPa) at the interface of the overlapping bars is most nearly:

99.

15 mm

,

~

I

I I I

0

I \

',

lOON--100N

5mm

_J 0 0 0 0

100.

0 0 0 0

A B. C.

D.

5mm 1--smm

0.56 0.63 5.1 51

The positional tolerance of the hole documented below is most nearly:

050

+0.5 -0.3

A B.

-0.3 -0.1 0.2 0.5

C.

D.

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64

SOLUTIONS

65

FE MECHANICAL SOLUTIONS Detailed solutions for each question begin on the next page.

1

A

26

A

51

B

76

B

2

A

27

D

52

C

77

B

3

B

28

see solution

53

D

78

D

4

B

29

see solution

54

B

79

C

5

C

30

C

55

B

80

D

6

D

31

648

56

B

81

B

7

A

32

D

57

C

82

B

8

D

33

A

58

B

83

A,E

9

see solution

34

C

59

A

84

B

10

B

35

C

60

D

85

C

11

C

36

D

61

C

86

C

12

D

37

D

62

A

87

D

13

A

38

D

63

B

88

D

14

B,C,E

39

C

64

RegionB

89

D

15

224 or 225

40

3

65

B

90

C

16

see solution

41

B

66

C

91

A,C,D,E

17

D

42

B

67

D

92

C

18

5

43

C

68

C

93

B

19

D

44

A

69

B

94

B

20

720

45

A

70

A

95

A

21

C

46

C

71

B

96

D

22

B

47

B

72

D

97

D

23

1,638

48

C

73

B

98

B

24

C

49

D

74

C

99

B

25

D

50

A

75

3

100

C

66

FE MECHANICAL SOLUTIONS 1.

Refer to the Mathematics section of the FE Reference Handbook. (x - h) 2 + (y - k)2 + (z - m) 2 = r 2 with center at (h,k,m) (x - 0)2 + (y- 1)2 + (z - (-2)) 2 = r 2

x 2 + (y- 1) 2 + (z + 2)2 = 81

THE CORRECT ANSWER IS: A

2.

Define a differential strip with length (x - 0) and height dy.

(x-0)--.j y

X

1

1

5/2 l

0

0

5/20

fdA=f xdy= fy312dy=L

=2 5

THE CORRECT ANSWER IS: A 3.

Refer to the Numerical Integration section in the Mathematics chapter of the FE Reference Handbook. 05 2 Area= ~ [ 0 2 + 2(0.5) 2 + 2(1 .0) 2 + 2(1.5) + (2)2 ]

THE CORRECT ANSWER IS: B

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67

= 2. 75

FE MECHANICAL SOLUTIONS 4.

Refer to Differential Equations in the Mathematics chapter of the FE Reference Handbook. The characteristic equation for a first-order, linear, homogeneous differential equation is: r+S=O which has a root at r = - 5. The form of the solution is then: y = Ce-at where a= a

and

a = 5 for this problem C is determined from the boundary condition. 1 = ce-S(O)

C=l Then, y = e- 51 THE CORRECT ANSWER IS: B

5.

Refer to the Vectors section in the Mathematics chapter of the FE Reference Handbook. The cross product of vectors A and Bis a vector perpendicular to A and B. I

j

k

2 4 0 = i(-4) - j(-2-0)+k(2-4)=-4i+2j-2k 1 1 - 1

To obtain a unit vector, divide by the magnitude. 2

Magnitude =~(- 4) + 22 +(-2)2

=..fiA =2-/6

-4i+2j - 2k _ -2i + j-k 2-/6 J6 THE CORRECT ANSWER IS: C

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68

FE MECHANICAL SOLUTIONS 6.

Refer to the Software Syntax Guidelines section in the Electrical and Computer chapter of the FE Reference Handbook. Step 1 2

VAR 0 2

3 4 4 6 EXIT LOOP At the conclusion of the routine, VAR= 6. THE CORRECT ANSWER IS: D

7.

Refer to the Normal Distribution section in the Engineering Probability and Statistics chapter of the FE Reference Handbook. Meanµ= 8 Standard deviation cr = 2.5 X = 15 .5 Calculate Z

x-µ =- =3 .

Refer to the Unit Normal Distribution table. Use 3 for x in the table (3 standard deviations away). R(x) = 0.0013

THE CORRECT ANSWER IS: A

8.

Refer to the Dispersion, Mean, Median, and Mode Values section in the Engineering Probability and Statistics chapter of the FE Reference Handbook.

cr=

4(11-12)2 +1(12-12)2 + 2(13-12)2 +1(14-12)2 8

cr=l.118 THE CORRECT ANSWER IS: D

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69

FE MECHANICAL SOLUTIONS 9.

Refer to the Dispersion, Mean, Median, and Mode Values section in the Engineering Probability and Statistics chapter of the FE Reference Handbook:

x-x

X

-1

4 4 7

-1 2

(x -x) 2 1 1 4

k=6

k= 15

x= 15/3 = 5 = mean . is . the value o f ( -n +- I Median 2 Median= 4

)th item. . (-3 +-1) = 2 items . 2

N

n = 1,638 THE CORRECT ANSWER IS: 1,638

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75

FE MECHANICAL SOLUTIONS 24.

Refer to the Systems of Forces section in the Statics chapter of the FE Reference Handbook. The triangular force distribution can be replaced with a concentrated force F acting through the centroid of the triangle. The magnitude of Fis numerically equal to the area of the triangle.

F = 1/2 (base)(height) = 1/2 (3 m)(8 kN/m)

F= 12 kN Sum the moments about Point A so that the only unknown is RB.

'i.MA= 0

F= 12kN

6RB-5F=0

6RB - 5(12 kN) = 0 RB = lOkN

8kN/m

A

1-=-- 3

m-

l--•-- 3m - - --~ B

-~

RA

RB

THE CORRECT ANSWER IS: C

25.

Refer to the Resultant (Two Dimensions) section in the Statics chapter of the FE Reference Handbook. 12 3 R.y = L.Fy =-(260) + -(300)50 = 370 13 5

Rx ="i.Fx =- 2-(260) + i(300) = 140 13

5

y

12~

R=396N

5

~3 4 ---------------------x

SON 12~ 5

THE CORRECT ANSWER IS: D

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76

L::)3 4

FE MECHANICAL SOLUTIONS 26.

Refer to the Moments (Couples) section in the Statics chapter of the FE Reference Handbook.

L

250 Ff[= 500 cos 30° = 433

Fv = 500 sin 30° = 250

433

MP= 250(0.30)-433(0.10) =31.7 N·m ccw

P

1----~Eomm I,..

300mm

THE CORRECT ANSWER IS: A

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77

..,I

FE MECHANICAL SOLUTIONS 27.

Refer to Plane Truss: Method of Sections in the Statics chapter of the FE Reference Handbook. Place a hypothetical cut as shown below, exposing Member BC as an external force. Then sum the moments about the point so the FBc provides the only unknown moment.

IM1 = 0 = Mc - MH - MG-MF

IM1 = (5FBc)-(5x3) - (10x3)-(15x3)=0

O= 5Fnc -15-30 - 45 FBc = 3 + 6 + 9 FBc=18kN cut

A

B

D

C

E

F

5 ft

5 ft

3kN

3kN

3kN

THE CORRECT ANSWER IS: D

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78

3kN

FE MECHANICAL SOLUTIONS 28.

Refer to the Statically Determinate Truss section in the Statics chapter of the FE Reference Handbook. Where members are aligned as follows with no external loads, zero force members occur. A B 0

C That is, joints where two members are along the same line (OA and OC) and the third member is at some arbitrary angle create zero-force members. That member (OB) is a zero-force member because the forces in OA and OC must be equal and opposite. For this specific problem, we immediately examine Joints Band E:

C B:

D E:

J(

C

E F

G BG is a zero-force member

CE is a zero-force member

Now, examine Joint G. Since BG is zero-force member, the joint effectively looks like:

A

G

F

and, therefore, CG is another zero-force member. Finally, examine Joint C. Since both CG and CE are zero-force members, the joint effectively looks like:

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79

FE MECHANICAL SOLUTIONS 28.

(Continued) and, therefore, CF is another zero-force member. Thus, BG, CE, CG, and CF are the zero-force members. D

THE CORRECT ANSWERS ARE MARKED ABOVE.

29.

Refer to the Centroids of Masses, Areas, Lengths, and Volumes section in the Statics chapter of the FE Reference Handbook.

A. Determination of Y = 5.38 ((6 X 1)(3) + (8 X 1)(6.5) + (2

X

1)(8)]/16

X

1)(9.5)]/16

B. Determination ofY = 6.25 ((8 X 1)(4) + (8 X 1)(8.5)]/16 C. Determination ofY = 5.00 ((4 X 1)(0.5) + (8 X 1)(5) + (4

D. Detennination of Y = 2.75 ((8 X 1)(0.5) + (8 X 1)(5)]/16

E. Detennination of Y = 4.44 ((5 X 1)(0.50) + (8 X 1)(5.0) + (3

X

])(9.5)]/16

THE CORRECT ORDER IS: D, E, C, A, B

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80

FE MECHANICAL SOLUTIONS 30.

Refer to the Friction section in the Statics chapter of the FE Reference Handbook. The maximum friction force developed before the block slides is Fs:::; µ Fn. Formulate a free-body diagram of the block and ramp. Decompose the gravity force into a normal component and a friction component, and thenµ= tan a= 0.754.

0 (l

Fn Fg

~ THE CORRECT ANSWER IS: C

31.

Refer to the Moment of Inertia Parallel Axis Theorem section in the Statics chapter of the FE Reference Handbook.

9(103 -6 3 ) = 12

(1)(6)3 + 12

=588+18=606 d d1 b bo do

= 10 =6 =9 =1 =6

Parallel axis theorem: 4 1.v-y = 606 in dy = l in. A = (2)(9) + (1)(6) + (2)(9) A = 42 in2 Jy '-y' iy'-y'

= ly-y + d/A = (606 + 42) in4 = 648 in4

THE CORRECT ANSWER IS: 648 Copyright© 2020 by NCEES

81

FE MECHANICAL SOLUTIONS 32.

Refer to the Resolution of a Force section in the Statics chapter of the FE Reference Handbook. Sum of forces at E = 0 Using symmetry,

Tieft

=

Tright

Vertical components: mg = 2 T sin 30° T=

mg = 100 X 9. 81 2sin30° 2x0.5

= 100 X 9.81 =981N

mg

THE CORRECT ANSWER IS: D

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82

FE MECHANICAL SOLUTIONS 33.

Refer to the Friction section in the Dynamics chapter of the FE Reference Handbook. The upper frictional force due to Block A on Block B is =0.2 mAg

}u

=0.2(2)(9.81) =3.924 N The lower frictional force on Block B at the bottom surface is FL =0.4(mA + ms)g = 0.4(2 + 2)9.81 = 15.696 N

Then from the Kinetics section,

"'f.F =ma F-FL-Fu = ma 30-15.696-3.924 = 2a a= 5.19 ""5.2 m/s

2

THE CORRECT ANSWER IS: A

34.

Refer to the Principle of Work and Energy section in the Dynamics chapter of the FE Reference Handbook. Ti + U1 + W1~2 = T2 + [h

0 + 0 + Fs = l/2 mv2 + mgh SF= 1/2 (2)(8)2 + (2)(9.81)(3) F=24.6N THE CORRECT ANSWER IS: C

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83

FE MECHANICAL SOLUTIONS 35.

Refer to Plane Motion of a Rigid Body-Kinematics (Instantaneous Center of Rotation) in the Dynamics chapter of the FE Reference Handbook. The crank and rod are two rigid bodies. At the moment when 0 = 90°, vp is desired (piston speed). ve = 50 mm x 377 rad/s = 18,850 mm/s

Both points are on the rod. By the method of instantaneous centers, the center of rotation is located where the line at P, J_ to vp , intersects the line at C, J_ to ve. Ve is parallel to vp so these meet at infinity. Thus the rotation of rod PC is 0, or Wpc = 0.

Since there is no rotation at this instant, all points of the rod move with the same velocity and Vp

= ve = 18,850 mm/s because

vP

= vc

+ wPC x ~;c and ffi Pc

THE CORRECT ANSWER IS: C

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84

= 0.

FE MECHANICAL SOLUTIONS 36.

Refer to the Relative Motion section in the Dynamics chapter of the FE Reference Handbook.

vG =5.0i mis (0

- 5.0 mis 0.25 m

~

~

=-::-~-ks-I =-20kS-1

rGA =0.25 [ cos 45°t + sin 45°]] m Then

= 5.0i +3.54t -3 .54) mis VA= 8.54f - 3.54] mis Then jvA I = '-"8.54 2 + 3.54 2 =9.24 mis IvAl === 9.2 mis This problem could also be solved by vectorially adding the translational motion and the rotational motion or by using Point O as the reference. In that case, IvAI= lwl x

JroAJ =20 s- 1x

0.462 m == 9.2 mis

THE CORRECT ANSWER IS: D

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85

FE MECHANICAL SOLUTIONS 37.

Refer to the Free Vibration section in the Dynamics chapter of the FE Reference Handbook. mI+kx=O

:. x = C1 cos ( Wnt) + C2 sin ( ffint)

where ffin

=

ff

x(O) = 0 :. C1 = 0, and x

m

= C2 sin( wn1)

x = C2 wn cos ( runt) x(O) = 6 = C2 wn, solving for C2 6

Cz=wn

. ( ro,l ) x.. = - C'_, 2ron2 sm

x=-6ron sin ( (J)nt) x = -6J¾sin( ront)

x =-9 mls 2 sin ( ront) ..

:. Xmax

I 2 = 9 ms

Alternate solution: Use kinetic energy of Block B to compress the spring a distance of x = 4. F = Ma at maximum compression, and force gives

36 = 4a

a= 9 mls 2 THE CORRECT ANSWER IS: D

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86

FE MECHANICAL SOLUTIONS 38.

Refer to the Impact section in the Dynamics chapter of the FE Reference Handbook. Use conservation of momentwn during impact.

where ( )' indicates after impact.

4(10) =40 =4v; + 2v;

(1)

Use the coefficient of restitution

e=I= .

V ' -V' 2 i V -Q I

( 1)(10) = 10 = -v; + v;

(2)

Solve (1) and (2) to find = 13.3 mis

v:

THE CORRECT ANSWER IS: D

39.

Refer to the Kinetics of a Rigid Body section in the Dynamics chapter of the FE Reference Handbook. I,M0 = 10 a T(O. l)

= 0.0428 a

LFY=may mg-T=ma No slip a=ar

(1)

(2) (3) T

(3) ._ (2)

mg-T=mar (1 kg)(9.81 m/s 2) - T= (1 kg)(0.1 m) a T=9.81-0.la (4)

T

(4)- (1) [9.81 - 0.1 a] 0.1 = 0.0428 a a= 18.6 s-2

THE CORRECT ANSWER IS: C

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87

FE MECHANICAL SOLUTIONS 40.

Refer to the Non-constant Acceleration section in the Dynamics chapter of the FE Reference Handbook. The speed is the integral of velocity with respect to time: Speed= initial speed+

Jadt = O+ J2 adt+ J3 adt+ J5 adt I

0

0

2

3

By inspecting the graph: First segment is a triangle, with area= (1/2) x 1 x 2 = 1 Second segment is a rectangle, with area = 1 x 1 = 1 Third segment is a triangle, with area= ( 1/2) x 1 x 2 = 1 The speed will be l+ l+ 1=3 mis. THE CORRECT ANSWER IS: 3

41.

Refer to the Particle Kinematics section in the Dynamics chapter of the FE R(!ference Handbook. From the question, the automobile is sliding, so use kinetic friction. The friction force is:

FN= 1,020 kg

x

9.807 m/s 2 = 10,003 N, so Fi= 4,001 N.

The deceleration is calculated from F = ma, so a= (4,001)/(1,020) = 3.92 m/s 2. After 1 second, the speed will be 10- 3.92 = 6.08 mis. THE CORRECT ANSWER IS: B

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88

FE MECHANICAL SOLUTIONS 42.

Refer to the Principle of Work and Energy section in the Dynamics chapter of the FE Reference Handbook.

T2 + U2 = T1 + U1 + W1->2 W1~2=0 T1 =O 1 2 T2 = -mv2 2 U1 =mgh1 U2=0 2-mv~ 2

= mg(2.5) ~ v2 = -Jsi = ~5(9.81) = 7111/s hi =Ssin30°=2.5

THE CORRECT ANSWER IS: B 43.

From Uniaxial Loading and Deformation in the Mechanics of Materials section of the FE Reference Handbook, the uniaxial deformation is: PL Deformation= 8 = AE

=(

(5,000)(0.25) 6

1,250 X 10-

)(

200 X 10

9

) = 5.0 x 10- 6

m = 5.0 µm

THE CORRECT ANSWER IS: C 44.

Refer to the Uniaxial Loading and Defonnation section in the Mechanics of Materials chapter of the FE Reference Handbook.

LF= PA=( 14x106 ) ( n(O~S)2 J= Froa Frod

= 275 kN = oA = 68 x 10 6 A

A= 40.4 X 10-41112

THE CORRECT ANSWER IS: A

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P=l.4MPa ~ m

89

FE MECHANICAL SOLUTIONS 45.

Refer to the Torsion section in the Mechanics of Materials chapter of the FE Reference Handbook.

Tr J

"t' = -

where T

= torque at section of interest (N · mm)

r

= radius to point of interest (mm), routside for maximum shear, and r = d

J

=section (polar) moment of inertia ( mm 4 )

2

For Section BC T = l,OOON·m

75

r =-mm 2

4

4

1t(75 - 50 )

J

=- - - - - mm 4 32

Hence the maximum torsional shear stress is given by

a= (1,000,000N-mml(-¥mm J

rr( 75 4 - 504 )

- -- ---'-mm4

32

= 15MPa

THE CORRECT ANSWER IS: A

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90

FE MECHANICAL SOLUTIONS 46.

Refer to the Torsion section in the Mechanics of Materials chapter of the FE Reference Handbook.

Td Tr 2 16T t=-=--=J rrd4 rrd3

32 T

= nd\ = n(0.2)\840 x 103 ) 16

16

T=l,319 N·m THE CORRECT ANSWER IS: C

47.

Refer to Mohr's Circle section in the Mechanics of Material chapter of the FE Reference Handbook. From a constructed Mohr's Circle, the maximum in-plane shear stress is 'tmax = R.

R

= ( 40; 20 )' + 10,

R

=Jwo

R

=

14. l ksi

THE CORRECT ANSWER IS: B

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91

FE MECHANICAL SOLUTIONS 48.

Refer to the Mohr's Circle-Stress, 2D section in the Mechanics of Materials chapter of the FE

Reference Handbook. The stress at Point O may be represented as

j

(JV

-.

(JV

Load where cr = -Area

10,000 N

cr"=( 100 mm )( 10 mm )= - lOMPa cr = "

5,000 N =S MPa (100 mm)(l0 mm)

Since cr1i is compressive,

cr1 = 5 MPa, cr2 = 0 MPa, 0'3 = - 10 MPa, Thus

2 =7.5 MPa

THE CORRECT ANSWER IS: C

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92

FE MECHANICAL SOLUTIONS 49.

Refer to the Columns section in the Mechanics of Materials chapter of the FE Reference Handbook. The Euler fonnula is used for elastic stability of relatively long columns, subjected to concentric axial loads in compression. THE CORRECT ANSWER IS: D

50.

Refer to the Thermal Deformation and Engineering Strain sections in the Mechanics of Materials chapter of the FE Reference Handbook.

11.T =30 - ( -10) =40°C 8 = a 11.T L = (12 x 10-6 ) ( 40 )( 50) = 24 x 10-3 m 24 " = _Q_ L = 50

c

X

10-3

= 0 .48 X 10-3

a = E £ = (200 xl0 9 )x( 0.48xl0-3 ) = 96 MPa THE CORRECT ANSWER IS: A

51.

Refer to the Mechanical Springs section in the Mechanical Engineering chapter of the FE Reference Handbook. This is an indeterminate axial load question.

P = k (11.L) = 30 kN x 2 cm = 60 kN cm

FA A

·:j8ABI =l8scl => FALAs = FcLsc L

ABF .. C - Lsc A ·F, -

t

t

LBc=3 cm

FA +Fe =P=60kN

FA

f:c )

FA

=FA (

f;~ )=

C

i_Fe

60 kN

=(it )60 =18 kN

THE CORRECT ANSWER IS: B Copyright © 2020 by NCEES

7 cm

B

Force equilibrium equation in the Statics chapter

FA + (

LAB=

p

93

FE MECHANICAL SOLUTIONS 52.

Refer to the Binary Phase diagrams in the Material Science/Structure of Matter chapter of the FE Reference Handbook. The eutectic composition is the point at which the liquid phase transforms directly to two solid phases. For the Ag-Cu phase diagram, the L ~ a + p occurs only at 28.1 % Cu and 71.9% Ag. THE CORRECT ANSWER IS: C

53.

Refer to the Binary Phase diagrams in the Material Science/Structure of Matter chapter of the FE Reference Handbook. At 850°C,

p + L phases are present.

1,100 1,000

LIQUID

962° ,......_

u

900

0

'-'

~

~ ~

800 92

780° 700

p...

;E ~

r'

600 a+~ 500 400 0 Ag

20 80

40 60

60 40

COMPOSITION (WT¾)

THE CORRECT ANSWER IS: D

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94

80 20

Cu

0

FE MECHANICAL SOLUTIONS 54.

Refer to the hardenability curves for agitated oil in the Materials Science/Structure of Matter chapter in the FE Reference Handbook. Find the intersection of 40-mm diameter with the center cooling rate curve C. Read the distance from the quenched end as 12.5 mm. Then, go to the Jominy hardenability curves graph for a 4140 steel and a 12.5-mm distance from the quenched end. The Re hardness is 52. Therefore, 52 is the expected center hardness for a 4140 steel of diameter 40 mm when quenched in an agitated oil bath. THE CORRECT ANSWER IS: B

55.

Refer to the Electrochemistry section in the Chemistry and Biology chapter and to the Corrosion section in the Materials Science/Structure of Matter chapter of the FE Reference Handbook. Aluminum is anodic relative to copper and, therefore, will corrode to protect the copper. THE CORRECT ANSWER IS: B

56.

Refer to Stress Concentration in the Brittle Materials section of the Materials Science/Structure of Matter chapter of the FE Reference Handbook. Critical crack length: K 1 = ycr-J'i; Solving for critical a, yields ac

~

J!~L,:

4

( :;:

ac =4.3 mm THE CORRECT ANSWER IS: B

57.

Refer to the Thermal and Mechanical Processing section in the Materials Science/Structure of Matter chapter of the FE Reference Handbook. Cold-working decreases the recrystallization temperature, ductility, and slipping or twining takes place. During cold work, grains become elongated instead of equiaxed. THE CORRECT ANSWER IS: C

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95

FE MECHANICAL SOLUTIONS 58.

Refer to the Stresses in Beams section of the Mechanics of Materials chapter of the FE Reference Handbook. Maximum bending moment= 900 mmx 1,200 N = 1.08 x 10 6 N·mm 6

( 1.08x 10 N·mm)(15 mm) Citension

=

4

llxlO mm

= 147 MPa

4

THE CORRECT ANSWER IS: B

59.

Refer to the Stress, Pressure, and Viscosity section and the Density, Specific Volume, Specific Weight, and Specific Gravity section in the Fluid Mechanics section of the FE Reference Handbook. Units of absolute dynamic viscosity(µ) are kg/(m·s). Units of kinematic viscosity (u) are m 2/s. :. the relationship between the two is: 'l)

= µ/p

where p is the density in kg/m 3 Use the definition of specific gravity to compute the fluid density. U

= 1.5/1.263(1,000) = 0.001188

=

1.19 X 10-3

THE CORRECT ANSWER IS: A

60.

Refer to the Archimedes Principle and Buoyancy section in the Fluid Mechanics chapter of the FE Reference Handbook.

THE CORRECT ANS\VER IS: D

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96

FE MECHANICAL SOLUTIONS 61.

Refer to the Characteristics of a Static Liquid section in the Fluid Mechanics chapter of the FE Reference Handbook. Since atmospheric pressure acts above the liquid surface and on the non-wetted side of the submerged gate, the resultant force acting on the gate is: FRnet

= (pgyc sin 0) A

where Ye = 1.5m and 0 = 90°

FR,oi - [ (1,600

FRnet

l

~ )(9 807; )(lsm)(sin90°)]c3 OOm x LOOm{1~

= 70, 610 N

The point of application of the resultant force is the center of pressure.

lxc

Yep= Ye+ YcA

The moment of inertia about the centroidal x-axis for a rectangle (from the table in the Statics chapter) is

Thus, 4

Yep= 1.5m + (

2 25 ) = 2.0 m from fluid surface or 1.0 m above the hinge 1.5m 3.00mxl.00m

)( · m

A moment balance about the hinge gives

y

t

L-..,.x

+) LMhinge = 0 (lm)FRnet

F-

l

-(3m)F =0

(lm)(70,610 N) F=--'---'-'-----'3m

3m

FRuet-

F =23,537 N =24 kN

t

THE CORRECT ANSWER IS: C

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~1

97

FE MECHANICAL SOLUTIONS 62.

Refer to the Energy Equation section in the Fluid Mechanics chapter of the FE Reference Handbook.

;¥ V~=

+ Z1+

i¾ =

+ J{ +

;! N' +

2gzl

THE CORRECT ANSWER IS: A

63.

Refer to the Impulse-Momentum Principle section in the Fluid Mechanics chapter of the FE Reference Handbook. Q = A1v1 = (0.01 m 2)(30 mis) = 0.3 m3/s

Since the water jet is deflected perpendicularly, the force F must deflect the total horizontal momentum of the water.

F = pQv = (1,000 kg/m 3) (0.3 m 3/s) (30 mis)= 9,000 N = 9.0 kN THE CORRECT ANSWER IS: B

64.

Refer to the Normal Shock Relationships section in the Fluid Mechanics chapter of the FE Reference Handbook. Flow in a converging/diverging nozzle transitioning from subsonic to supersonic will have Mach = 1 at the throat. Region B is the correct answer. The flow is subsonic in the chamber, with Ma= Vic = 450/600

= 0.75, so Ma< 1 in Region A

By definition, in a normal shock the flow slows from supersonic to subsonic. Therefore, Region C has Ma > 1 and Region D has Ma < 1. Region E expands to ambient, so it must still be subsonic Ma< 1. THE CORRECT ANSWER IS: REGION B

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98

FE MECHANICAL SOLUTIONS 65.

Refer to the Continuity Equation in the Principles of One-Dimensional Fluid Flow section in the Fluid Mechanics chapter of the FE Reference Handbook. The cross-sectional area of the pipe is: 7t 2 = -7t ( 0.10 )2 =0.007854m 2 Ac =-D

4

4

The flow rate is:

Q = Ac Ve= 0.007854 (2.5)(60) = 1.178 m 3/min THE CORRECT ANSWER IS: B

66.

Refer to the pump power equation in the Fluid Mechanics chapter of the FE Reference Handbook.

W= Qyh = M 11

·Q

since M

= yh

11

M · Q (0.9 kPa)(3 .0 m 3 /s) kN kW· s 11 - - - - -'----'--'-- --'-. -2- - · - -

- W-

4.0 kW

m -kPa kN·m

= 0.675 THE CORRECT ANSWER IS: C

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99

FE MECHANICAL SOLUTIONS 67.

Refer to Special Cases of Steady-Flow Energy Equation in the Steady-Flow Systems section in the Thermodynamics chapter of the F'E Reference Handbook. The isentropic efficiency for a pump is llp

w

=

s Wactual

The isentropic pump work for a constant density fluid is ws= vM = (0.001010 m3/kg)(l,000 kPa-10 kPa) = 0.9999 kJ/kg

and the actual power required at the pump is

Wa = ri1w 8 I'llµ= (so kg/s)(0.9999 kJ/kg)/0.7 = 71 kW THE CORRECT ANSWER IS: D

68.

Refer to the Affinity Laws section in the Fluid Mechanics chapter of the FE Reference Handbook.

12/1,000 = l8/N2

THE CORRECT ANSWER IS: C

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100

FE MECHANICAL SOLUTIONS 69.

Refer to the Thermodynamics chapter of the FE Reference Handbook. Use the ideal gas formula:

PV = mRT P=mRT V

R= 8,314J kmol kmol·K 28kg

(100 kg)(297 -

=297 _J_ kg-K

1

-](343 K)

P=---'-_k_g_·K-'--100m3 J

=102 , 0003 m =102 000 N·m ' 3 m N =102 000,

m2

=102kPa THE CORRECT ANSWER IS: B

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101

FE MECHANICAL SOLUTIONS 70.

Refer to the State Functions and the Properties for Two-Phase (vapor-liquid) Systems sections in the Thennodynamics chapter of the FE Reference Handbook. As vapor escapes, the mass within the tank is reduced. With constant volume, the specific volume within the tank must increase. This can happen only if liquid evaporates.

t VAPOR VAPOR

LIQUID

THE CORRECT ANSWER IS: A

71.

Refer to the Throttling Valve section in the Thermodynamics chapter of the FE Reference Handbook. From the steam table at 250°C, the saturation pressure = 3.973 MPa Enthalpy h1 at 250°C = 1,085.36 kJ/kg Through a throttling valve, /11 = h2 = 1,085.36 kJ/kg At 0.2321 MPa, hf= 524.99 kJ/kg hfg = 2,188.5 kJ/kg

h1 = hf+ xh1g Substituting: 1,085.36 = 524.99 + X 2,188.5 x = 560/2,188.5

= 0.256, approximately 0.26

THE CORRECT ANSWER IS: B

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102

FE MECHANICAL SOLUTIONS 72.

Refer to the Steady-Flow Systems section in the Thermodynamics chapter of the FE Reference Handbook. An energy balance on the boiler gives:

Q = m( h2 -

h1 ) = (SO,OOO kgJ[(3,322 -167.6) kJ/kg] = 43,811 kW= 44 MW 3,600 s

THE CORRECT ANSWER IS: D

73.

Refer to the Steady-Flow Systems section in the Thermodynamics chapter of the FE Reference Handbook. Assuming steady-state, steady-flow conditions, a first law analysis of the turbine yields, wt = h4 -

hs

and with the enthalpy values provided for States 3 and 4, the turbine work per unit mass is w,

= 3,478.5- 2,584.7 = 893.8

kJ/kg

The power produced by the turbine is given by

W = mw, = (50 kg/s)(893.8 kJ/kg) = 44,690 kW

= 44.7 MW

THE CORRECT ANSWER IS: B

74.

Refer to the Basic Cycles (Coefficient of Performance) section in the Thermodynamics chapter of the FE Reference Handbook. COP = hi - h4 and h4

hz -hi

=h3

hz -hi = 438-394 = 44 kJ/kg hi -h4 = 124 kJ/kg = 2.82 hz -hi 44 THE CORRECT ANSWER IS: C

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103

FE MECHANICAL SOLUTIONS 75.

Refer to PVT Behavior section in the Thermodynamics chapter of the FE Reference Handbook.

(!iJr

Pv = RT =>v= M

p

R = 1miversal gas constant= 8.314 kPa·m 3/(kmol·K) . Q Q MPQ and m =p =-; = T ( R) Solve for mass flow rate: 3

Mass flow rate

(30kg/kmol)(420kPa)(30m /s) (350K)(8.314kPa-m3 /kmol·K)

= 378,000 =129 _9 kg/s 2909.9

Nozzles max produce 50 kg/s: 9 k.. . .;g:; ___ - -1-29-·/s = 2.6 nozzles 50 kg/s per nozzle The machine will require 3 nozzles.

THE CORRECT ANSWER IS: 3

76.

Refer to the psychrometric chart in the Thermodynamics chapter of the FE Reference Handbook.

= RELATIVE HUMIDITY Tdb= DRY-BULB TEMPERATURE

(j>

B 0c DRY-BULB TEMPERATIJRE

At the given state,

Tdh

= l 3°C, ~ = 70%, co= 6. 5 _g_

kgda

, da = dry air.

Follow the co= 6.5 line to the left until the saturation curve is reached. This point is the dew point. Read down to find the dew-point temperature of 7 .6°C.

THE CORRECT ANSWER IS: B

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104

FE MECHANICAL SOLUTIONS 77.

From the P-h Diagram for Refrigerant HFC-134a given in the Thermodynamics chapter of the FE Reference Handbook:

h1

= 400 kJ/kg

f1J =h4 =257 kJ/kg Evaporator cooling is q 41

= h1 -

h4

= 400~ kg

257 ~ = 143~ kg kg

THE CORRECT ANSWER IS: B

78.

Refer to Combustion in Excess Air in the Combustion Processes section in the Thermodynamics chapter of the FE Reference Handbook. Stoichiometric combustion of methanol is: CH3OH + 1.5 02 + 1.5 (3.76) N2 - CO2+ 2 H2O + 1.5 (3.76) N2 In 20% excess air, multiply the air components on the reactants side by 1.2. This creates an excess of 0.3 units of air to add in order to balance on the products side. CH3OH + 1.8 02 + 1.8 (3.76) N2 - CO2+ 2 H2O + 1.8 (3.76) N2 + 0.3 02

THE CORRECT ANSWER IS: D

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105

FE MECHANICAL SOLUTIONS 79.

Refer to the Thennal Resistance section in the Heat Transfer chapter of the FE Reference Handbook.

R" =_!_+Les + LAL + _l h; kcs kAL h,o R" =

1 + 0.01 m + 0.01 m + 1 2 700 W/m ·K 60 W/m·K 240 W/m·K 100 W/m 2 ·K 2

R" = 0.01164 m 1 U=-,;=

R

•K

w

1

W =85.93-20.01164 m ·K m ·K 2

w

Q = UA(T; - ~J Q =U(T-T) A I

(2

A

=(ss.93

0

';1

m·K

)cs44K-300K)

Q =20,968 W /m 2 A

THE CORRECT ANSWER IS: C

80.

Refer to the Conduction section in the Heat Transfer chapter of the FE Reference Handbook.

Q = hA/J.T Q = 72(2)(150) Q=

21,600W

THE CORRECT ANSWER IS: D

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106

FE MECHANICAL SOLUTIONS 81.

Refer to the Shape Factor Relations section in the Heat Transfer chapter of the FE Reference Handbook. Consider the figure shown. The ground plane A3 is moved downward to location A3 ' so that A1 is halfway between A2 and A3'. The shape factor F12 is the fraction of all the rays leaving A1 that arrive at A2. By symmetry, half the rays leaving A1 strike A2 and half strike A3'.

A2

00

00 ...

00

0 A1

.. 00

A3 00

00

A'3 Now the shape factor FB is the same as F13' because each ray that leaves A1 and strikes A3 ' must cross A3, and each ray that leaves A1 and strikes A3 will, if extended, also strike A3'. Then by the equations in the Shape Factor Relations section:

Fi 1+ Fi2 + 1)3 = 1 Fi2 = 1)3 Thus Fi2 + Fi2 =1

but

t11 = 0 since Surface A 1 cannot "see" itself

1

Fi2=2 THE CORRECT ANSWER IS: B 82.

Refer to the Radiation section in the Heat Transfer chapter of the FE Reference Handbook.

Q1- 2 = crA1Fi-2 (T/ -T/)

= 5.67x10-8 ( 4)(0.275)( 773 4 -673 4 ) = 9474 = 9.47 kW

THE CORRECT ANSWER IS: B

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107

FE MECHANICAL SOLUTIONS 83.

Refer to the Transient Conduction Using the Lumped Capacitance Model section in the Heat Transfer chapter of the FE Reference Handbook. Lumped capacitance model is valid if :;;_ « 1 s

v=a

3

As =6a 2

THE CORRECT ANSWERS ARE: A, E

84.

Refer to the Heat Exchangers section in the Heat Transfer chapter of the FE Reference Handbook. Subscript H indicates hot fluid. Subscript C indicates cold fluid.

=111ccPc , (Tcout-Tc.m )

Water is the cold fluid, so:

. . CpH(~TH)

mc=mH-· - cP,c ~Tc

=(2.25 k /s)(~)(204-79) g 4.186 79-29 12 =(2.25)(0.836{ 5;)=4.70 THE CORRECT ANSWER IS: B

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108

FE MECHANICAL SOLUTIONS 85.

Refer to the Heat Exchangers section in the Heat Transfer chapter of the FE Reference Handbook.

(TH !i'ftm =

0

-Tc.1 )-(TH-1 -Tc 0 ) (

ln

tiT.

TH.

]

for counterflow HE

-Tc-I TH-I - Tc 0 O

= (65°C-24°C)-(110°C -60°C)

In( 65oC - 240c )

Im

110°C - 60°C

!i'ftm

= 45.4°C

THE CORRECT ANSWER IS: C

86.

Refer to the Resistance Temperature Detector (RTD) section in the Instrumentation, Measurement, and Controls chapter of the FE Reference Handbook.

R = Ro [ 1 + a( T - To)] tiR = dR tiT dT

= Roa!iT = (100 Q) (0.004°c- 1) (10°C) =4.0Q THE CORRECT ANSWER IS: C

87.

Refer to the Resistance Temperature Detector (RTD) section in the Instrumentation, Measurement, and Controls chapter of the FE Reference Handbook. Since Uro =Ua =URo =0

cJR

Also -=Ra 0

' aT

so UR= 0.1 = (R0 a)UT

= [100 X (0.3925 X 10-3 )]UT and UT= 2.55°C

THE CORRECT ANSWER IS: D

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109

FE MECHANICAL SOLUTIONS 88.

Refer to the Resistance Temperature Detector (RTD) section in the Instrumentation, Measurement, and Controls chapter of the FE Reference Handbook. Rewriting the temperature equation,

T(t) - T T; - To

-th

0 --=e

For T(t)=139°C, 139 -140 40-140

-t/3

--- = e

or

t = 3 ln 100 = 13.8 s

THE CORRECT ANSWER IS: D

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110

FE MECHANICAL SOLUTIONS 89.

Refer to the Control Systems section in the Instrumentation, Measurement, and Controls chapter of the FE Reference Handbook. The solution requires a step-by-step reduction of the system loops. First, reduce the inner loop.

+

>--~-C(s)

R(s) ~..___G_1l(_1_+_G_1)_ _

1

H1

1--------~

Next, combine the forward blocks.

I---------,--

Finally, reduce the outer loop. ,___ _ .. C(s)

THE CORRECT ANSWER IS: D

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111

C(s)

FE MECHANICAL SOLUTIONS 90.

Refer to the Second-Order Control System Models section in the Instrumentation, Measurement, and Controls chapter of the FE Reference Handbook. 16 The equation of the system can be written as ( 2 ) which is in the form

s s +0.8s+l6

16

thus

ro/ = 16

or

THE CORRECT ANSWER IS: C

91.

Refer to the Static Loading Failure Theories section in the Mechanical Engineering chapter of the FE Reference Handbook. Maximum shear stress theory is used for the ductile material. Calculate 'tma...-:

= (cr1 -

Stress Calculated 'tmax 'tmax > 17?

cr 3 )/2 and compare with Sy /2. A 35 fails

B 12 12-ok

C

40 fails

THE CORRECT ANSWERS ARE: A, C, D, E

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112

D 18 fails

E

40 fails

F 10 10- ok

FE MECHANICAL SOLUTIONS 92.

Refer to the Mechanical Springs section in the Mechanical Engineering chapter of the FE Reference Handbook. 8FD

't=Ks--3

red

where d=2.34mm d 0 =15mm

D=d0 -d=l5 - 2.34=12.66 mm

= D = 12.66 = 5.4 10

C

d

2.34

= 2C + 1 = 2 ( 5.410) + 1 = 1. 0924

K

2C

s

't

= 1.0924

2 ( 5 .410)

(8)(150N)(12.661mn) rc(2.34)

3

= 412.3 Iv1Pa

THE CORRECT ANSWER IS: C

93.

Refer to the Mechanical Springs section in the Mechanical Engineering chapter of the FE Reference Handbook. The force required to displace a spring an amount ofrom its free length is F = ko, where k is the spring constant or rate. In this case:

o

free length - compressed length 190 mm-125 mm

=

65mm

The force required to deflect the spring this amount is: F = ko = (38.525 N/mm)(65 mm)= 2,504 N THE CORRECT ANSWER IS: B

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113

FE MECHANICAL SOLUTIONS 94.

Refer to the Cylindrical Pressure Vessel section in the Mechanics of Materials chapter of the FE Reference Handbook. The cylinder can be considered thin-walled if t < do/20. In this case, t ro = do/2 = 362 mm. Thus, (jf

12 mm and

Pr

=-If

where r = 'i +ro

2

= 350 + 362 =356 mm 2

at= (1.680 MPa)(356 mm)= 49 _8 MPa 121mn

THE CORRECT ANSWER IS: B

95.

Refer to the Hooke's Law section in the Mechanics of Materials chapter of the FE Reference Handbook. The formula for the total longitudinal strain without a temperature rise is: Eaxial

1

1

E

210 x l0 3 MPa

=-(cr1-v(cr1 +crr))=

(23.1MPa-0.24(46.2MPa+0))=5.72xlO

This must be converted to displacement using the following formula: Eaxial

= ~/, where l is the length of the section under consideration

Ol = eaxial

X /

= 5.72 x 10-6 x 1,000 mm = 0.0572 mm THE CORRECT ANSWER IS: A

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114

_6

FE MECHANICAL SOLUTIONS 96.

Refer to the Reliability section in the Industrial and Systems Engineering chapter of the FE Reference Handbook. For the figure given,

R total = R series x R parallel R = 0.9x[ 1-(1-0.75)(1-0.60)]

R=0.81 THE CORRECT ANSWER IS: D

97.

Refer to the Manufacturability section in the Mechanical Engineering chapter and the Thermal Deformations section of the Mechanics of Materials chapter of the FE Reference Handbook. Required diameter change: 8diameter

= dshaft -

dpulley + required assembly clearance

= 100.15-100.00 + 0.05 = 0.20 mm

Required temperature change of pulley diameter: ()diameter= ad(!l.T)

!l.T = odiameter ad

=

O·20

= 181. goc

(1 lxl0-6)(100)

Temperature to which pulley must be heated: T = Troom +AT= 20 + 182 = 202°C

THE CORRECT ANSWER IS: D

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115

FE MECHANICAL SOLUTIONS 98.

Refer to the Cantilevered Beam Slopes and Deflection section in the Mechanics of Material chapter of the FE Reference Handbook.

vmax

Pa 2 = 6EI (3L-a)

Solve for P

P= 6Elvmax a 2 (3L-a)

p = ( 6)( 200 X 109 ~ )(1.60 X 10-

8

(o.81

m4 )(0 01 m)

rn 2 )(2.1 rn)

P= 112.9N

~_,__7

r

10cm

I

Vmax

= }Cffi

P

1: I I

1.0m

/

THE CORRECT ANSWER IS: B

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116

FE MECHANICAL SOLUTIONS 99.

Use the shear stress equation from the Uniaxial Loading andDefonnation section in the Mechanics of Materials chapter of the FE Reference Handbook.

t= TIA where t = shear stress

T= load= 100 N A= area= n[(15 mrn/2) 2 - (5 mm/2)2] = 157.1 mm2

t=0.63 MPa THE CORRECT ANSWER IS: B

100.

Refer to the Geometric Dimensioning and Tolerancing (GD&T) section of the Mechanical Engineering chapter of the FE Reference Handbook. The 0.2 number refers to the positional tolerance of the hole. THE CORRECT ANSWER IS: C

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