Test d'entraînement FE Exam

• A. 8
• B. 10
• C. 12
• D. 16

Réponse : C. 12

The antiderivative of 3x^2 + 2 is x^3 + 2x. At x = 2 that is 8 + 4 = 12, and at x = 0 it is 0, so the definite integral is 12. A common slip is to forget the 2x term and stop at 8, so integrate every term before you plug in the limits.

• A. y = C1 e^(2x) + C2 e^(-2x)
• B. y = C1 cos(2x) + C2 sin(2x)
• C. y = C1 e^(4x) + C2 e^(-4x)
• D. y = C1 e^(2x) + C2 x e^(2x)

Réponse : A. y = C1 e^(2x) + C2 e^(-2x)

Try y = e^(rx). The characteristic equation is r^2 - 4 = 0, so r = 2 or r = -2, and the two real roots give e^(2x) and e^(-2x). The cosine and sine form would come from y'' + 4y = 0, the term x e^(2x) appears only for a repeated root, and e^(4x) mistakes r^2 = 4 for r = 4.

For the matrix A with rows [2, 1] and [4, 3], the determinant of A is _____ and the trace of A is _____.

Options pour le trou 1 : 2, 6, 10, 14

Options pour le trou 2 : 3, 5, 7, 10

Réponse : 2, 5

For a 2 by 2 matrix, the determinant is ad - bc = (2)(3) - (1)(4) = 6 - 4 = 2. The trace is the sum of the diagonal entries, 2 + 3 = 5. Mixing up the two is a common error: the determinant subtracts the cross products, and the trace only adds the diagonal.

• A. 4 kN
• B. 6 kN
• C. 10 kN
• D. 8 kN

Réponse : D. 8 kN

Take moments about the right support: the left reaction times 6 m equals 12 kN times 4 m, the distance from the load to the right support. That gives 48 / 6 = 8 kN. The right reaction is then 12 - 8 = 4 kN. The load sits closer to the left support, so the left support carries more of it.

• A. 20 N
• B. 196 N
• C. 490 N
• D. 1,226 N

Réponse : B. 196 N

The normal force on a horizontal surface equals the weight: 50 x 9.81 = 490.5 N. Maximum static friction is the coefficient times the normal force, 0.40 x 490.5 = 196.2 N. Choosing 490 N forgets the coefficient, and 1,226 N comes from dividing by 0.40 instead of multiplying.

Coche toutes les bonnes réponses.

• A. The two end forces are equal in magnitude and opposite in direction.
• B. The forces act along the line joining the two pins.
• C. The member carries shear force and a bending moment along its length.
• D. The member can be in tension or in compression.
• E. The two end forces must be perpendicular to each other.

Réponse : A. The two end forces are equal in magnitude and opposite in direction., B. The forces act along the line joining the two pins., D. The member can be in tension or in compression.

For a body loaded at two points to stay in equilibrium, the two forces must be equal, opposite and collinear, which means they act along the line between the pins. Whether they pull the ends apart or push them together decides tension or compression. With no loads along the length there is no shear or bending in the member, and the forces are collinear, not perpendicular.

• A. 4 s
• B. 9 s
• C. 6 s
• D. 16 s

Réponse : C. 6 s

Newton's second law gives the deceleration: a = F / m = 6,000 / 1,500 = 4 m/s^2. With constant deceleration, the time to stop is t = v / a = 24 / 4 = 6 s. The choice of 4 s takes the value of the deceleration as the time, so check that your final number has the units the question asks for.

A 2 kg mass moves at 6 m/s. Its kinetic energy is _____ J and its linear momentum is _____ kg*m/s.

Options pour le trou 1 : 12, 18, 36, 72

Options pour le trou 2 : 6, 12, 24, 36

Réponse : 36, 12

Kinetic energy is (1/2) m v^2 = 0.5 x 2 x 36 = 36 J. Momentum is m v = 2 x 6 = 12 kg*m/s. Energy depends on the square of the speed and momentum on the speed itself, which is why 72 is the typical wrong answer when the one-half is forgotten.

• A. 23.9 MPa
• B. 95.5 MPa
• C. 47.7 MPa
• D. 191 MPa

Réponse : B. 95.5 MPa

Stress is force divided by area. The area is pi x (10 mm)^2 = 314.2 mm^2, so the stress is 30,000 N / 314.2 mm^2 = 95.5 N/mm^2, which is 95.5 MPa. If you use the 20 mm diameter as the radius, the area is four times too big and you get 23.9 MPa. Always check whether a formula needs diameter or radius.

• A. 20 kN*m
• B. 10 kN*m
• C. 40 kN*m
• D. 80 kN*m

Réponse : A. 20 kN*m

For a simply supported beam with a uniform load, the maximum moment is at mid-span and equals w L^2 / 8 = 10 x 16 / 8 = 20 kN*m. The choice of 80 kN*m uses w L^2 / 2, which is the formula for a cantilever's fixed end, and 40 kN*m uses w L^2 / 4. Match the formula to the support conditions before you plug in numbers.

Coche toutes les bonnes réponses.

• A. In the linear elastic region, stress is proportional to strain.
• B. Young's modulus is the slope of the linear part of the stress-strain curve.
• C. The yield strength is always equal to the ultimate tensile strength.
• D. Necking begins at about the ultimate tensile strength.
• E. Stress stays proportional to strain after the metal yields.

Réponse : A. In the linear elastic region, stress is proportional to strain., B. Young's modulus is the slope of the linear part of the stress-strain curve., D. Necking begins at about the ultimate tensile strength.

Hooke's law holds in the linear elastic region, and Young's modulus is the slope there. On the engineering stress-strain curve, the ultimate tensile strength is the peak, and necking, the local thinning of the specimen, starts about there. Yield strength is lower than ultimate strength in a ductile metal, and once the metal yields, stress is no longer proportional to strain.

• A. 4.7 kPa
• B. 58.9 kPa
• C. 470.9 kPa
• D. 47.1 kPa

Réponse : D. 47.1 kPa

Hydrostatic gauge pressure is rho x g x h = 800 x 9.81 x 6 = 47,088 Pa, or about 47.1 kPa. The 58.9 kPa choice uses the density of water, 1,000 kg/m^3, and the others are off by a factor of ten. The pressure depends only on the density and the depth, not on the shape or width of the tank.

Water flows at 2 m/s in a pipe with a 100 mm diameter that narrows to a 50 mm diameter. The velocity in the narrow section is _____ m/s, and the flow rate is about _____ m^3/s.

Options pour le trou 1 : 4, 6, 8, 16

Options pour le trou 2 : 0.0039, 0.0157, 0.0628, 0.1571

Réponse : 8, 0.0157

By continuity the flow rate stays the same, so area times velocity is constant. Halving the diameter cuts the area to one quarter, so the velocity is four times larger: 8 m/s. The flow rate is area times velocity: pi x (0.05 m)^2 x 2 m/s = 0.0157 m^3/s. Doubling the speed is the common mistake, since area depends on the square of the diameter.

• A. Release the design, since the employer has the final say.
• B. Document her concern, raise it through the proper channels, and notify the appropriate authorities if the danger remains.
• C. Quietly resign and say nothing further about the design.
• D. Post her concerns about the design publicly right away.

Réponse : B. Document her concern, raise it through the proper channels, and notify the appropriate authorities if the danger remains.

The codes of ethics for engineers put the safety, health and welfare of the public first, above loyalty to an employer. That means writing down the concern, using the proper channels and, if the danger remains, informing the proper authorities. Staying silent or simply resigning leaves the danger in place, and going public first skips the channels the code expects you to use.

Coche toutes les bonnes réponses.

• A. Accept work only in areas where you are competent.
• B. Make public statements only in an objective and truthful way.
• C. Accept gifts from a contractor to speed up an approval.
• D. Disclose a conflict of interest to the affected parties.
• E. Sign and seal plans prepared by others that you have not reviewed.

Réponse : A. Accept work only in areas where you are competent., B. Make public statements only in an objective and truthful way., D. Disclose a conflict of interest to the affected parties.

Professional codes call for working within your competence, speaking honestly in public and disclosing conflicts of interest. Taking gifts to influence a decision is a conflict and can amount to bribery. Sealing work you have not reviewed is wrong because the seal means you take responsibility for it, so you must have direct control over the work.