FE Mechanical Exam: 10 Practice Problems with Detailed Solutions

Answer: B

For a simply supported beam, we use static equilibrium to find the reactions.

Step 1: Draw the free-body diagram. Support A is a pin (provides vertical reaction \(R_A\)), and support B is a roller (provides vertical reaction \(R_B\)). The 30 kN load acts 3 m from A (and therefore 5 m from B).

Step 2: Sum moments about point A to eliminate \(R_A\):

$$\sum M_A = 0: \quad R_B \times 8 - 30 \times 3 = 0$$

$$R_B \times 8 = 90$$

$$R_B = \frac{90}{8} = \mathbf{11.25 \text{ kN}}$$

Step 3: Verify with vertical equilibrium:

$$\sum F_y = 0: \quad R_A + R_B = 30$$

$$R_A = 30 - 11.25 = 18.75 \text{ kN} \checkmark$$

The vertical reaction at support B is 11.25 kN. Note that choice C (18.75 kN) is the reaction at A — a common trap if you sum moments about the wrong support.

Answer: B

Apply the work-energy theorem: the net work done on the block equals its change in kinetic energy.

$$W_{\text{net}} = \Delta KE = \frac{1}{2}mv^2 - 0$$

Step 1: The only force doing work along the incline is the component of gravity parallel to the slope (no friction). The height descended is:

$$h = d \sin\theta = 5 \sin(30°) = 5 \times 0.5 = 2.5 \text{ m}$$

Step 2: Apply conservation of energy (or equivalently, the work-energy theorem):

$$mgh = \frac{1}{2}mv^2$$

Mass cancels:

$$v = \sqrt{2gh} = \sqrt{2(9.81)(2.5)}$$

$$v = \sqrt{49.05} = \mathbf{7.00 \text{ m/s}}$$

Note that the mass does not matter for this frictionless case — it cancels from both sides. Choice A (4.95 m/s) is the result if you forget to take the square root of 2 (using \(v = \sqrt{gh}\) instead). Choice C (9.90 m/s) is what you get if you use the full 5 m as the height instead of the vertical component \(d\sin\theta\). Always resolve the geometry carefully on incline problems.

Answer: B

For a shaft transmitting power, we first find the torque, then use the torsion formula for maximum shear stress.

Step 1: Calculate the angular velocity:

$$\omega = \frac{2\pi N}{60} = \frac{2\pi(600)}{60} = 62.83 \text{ rad/s}$$

Step 2: Calculate the torque from the power equation:

$$P = T\omega \implies T = \frac{P}{\omega} = \frac{10{,}000}{62.83} = 159.15 \text{ N·m}$$

Step 3: Use the direct formula for maximum shear stress in a solid circular shaft:

$$\tau_{\max} = \frac{16T}{\pi d^3}$$

Step 4: Substitute values (with \(d = 0.040\) m):

$$\tau_{\max} = \frac{16 \times 159.15}{\pi (0.040)^3} = \frac{2546.4}{\pi \times 6.4 \times 10^{-5}}$$

$$\tau_{\max} = \frac{2546.4}{2.0106 \times 10^{-4}} = 12.66 \times 10^6 \text{ Pa}$$

$$\tau_{\max} \approx \mathbf{12.7 \text{ MPa}}$$

The key formulas for shaft torsion problems: \(P = T\omega\) to find torque from power, and \(\tau_{\max} = Tc/J = 16T/(\pi d^3)\) for maximum shear stress in a solid circular shaft. Note that shear stress varies linearly from zero at the center to maximum at the outer surface.

Answer: C

For an ideal Rankine cycle, the turbine expansion is isentropic: \(s_4 = s_3\).

Step 1: Find the quality at the turbine exit (state 4) using the entropy:

$$s_4 = s_3 = 5.89 \text{ kJ/(kg·K)}$$

$$s_4 = s_f + x_4 \cdot s_{fg}$$

$$5.89 = 0.649 + x_4(7.502)$$

$$x_4 = \frac{5.89 - 0.649}{7.502} = \frac{5.241}{7.502} = 0.6986$$

Step 2: Calculate the enthalpy at state 4:

$$h_4 = h_f + x_4 \cdot h_{fg} = 192 + 0.6986(2392)$$

$$h_4 = 192 + 1671.5 = 1863.5 \text{ kJ/kg}$$

Step 3: Calculate the turbine work output per unit mass:

$$w_{\text{turbine}} = h_3 - h_4 = 2784 - 1863.5 = 920.5 \text{ kJ/kg}$$

Step 4: Neglecting pump work, the heat input is:

$$q_{\text{in}} = h_3 - h_f = 2784 - 192 = 2592 \text{ kJ/kg}$$

(We approximate \(h_1 \approx h_f\) at the condenser pressure since pump work is negligible.)

Step 5: Calculate the thermal efficiency:

$$\eta = \frac{w_{\text{net}}}{q_{\text{in}}} = \frac{920.5}{2592} = 0.355$$

$$\eta \approx \mathbf{33\% \text{ to } 36\%}$$

This is closest to C) 33%. On the FE exam, Rankine cycle problems are very common. Always start by finding the quality at the turbine exit using the isentropic condition \(s_4 = s_3\), then use \(h = h_f + x \cdot h_{fg}\) to get the exit enthalpy.

Answer: B

The Darcy-Weisbach equation gives the head loss due to friction:

$$h_f = f \frac{L}{D} \frac{v^2}{2g}$$

Step 1: Calculate the head loss:

$$h_f = 0.020 \times \frac{200}{0.150} \times \frac{(3)^2}{2(9.81)}$$

$$h_f = 0.020 \times 1333.3 \times \frac{9}{19.62}$$

$$h_f = 0.020 \times 1333.3 \times 0.4587$$

$$h_f = 12.23 \text{ m}$$

Step 2: Convert head loss to pressure drop:

$$\Delta P = \rho g h_f = 1000 \times 9.81 \times 12.23$$

$$\Delta P = 119{,}976 \text{ Pa} \approx \mathbf{120 \text{ kPa}}$$

The Darcy-Weisbach equation is the standard method for calculating friction losses in pipe flow. On the FE exam, you will typically be given the friction factor (or enough information to read it from the Moody diagram in the reference handbook). Remember that the Darcy friction factor is four times the Fanning friction factor — always check which one your reference is using.

Answer: A

For steady-state conduction through a composite wall, the heat flux is:

$$q'' = \frac{\Delta T}{R_{\text{total}}}$$

where the total thermal resistance per unit area is the sum of the individual layer resistances.

Step 1: Calculate the thermal resistance of each layer:

$$R_1 = \frac{L_1}{k_1} = \frac{0.200}{1.5} = 0.1333 \text{ m}^2\text{·K/W}$$

$$R_2 = \frac{L_2}{k_2} = \frac{0.100}{0.08} = 1.250 \text{ m}^2\text{·K/W}$$

Step 2: Calculate the total resistance:

$$R_{\text{total}} = R_1 + R_2 = 0.1333 + 1.250 = 1.3833 \text{ m}^2\text{·K/W}$$

Step 3: Calculate the heat flux:

$$q'' = \frac{T_{\text{hot}} - T_{\text{cold}}}{R_{\text{total}}} = \frac{800 - 50}{1.3833} = \frac{750}{1.3833}$$

$$q'' = \mathbf{542 \text{ W/m}^2} \approx 540 \text{ W/m}^2$$

Notice that the insulation layer, despite being thinner, has a much larger thermal resistance than the firebrick because of its low thermal conductivity. This is why insulation is so effective — it dominates the total resistance. Thermal resistance analysis (the “resistance circuit” analogy) is one of the most commonly tested heat transfer concepts on the FE exam.

Answer: B

The modified Goodman criterion states:

$$\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} = \frac{1}{n}$$

where \(n\) is the factor of safety.

Step 1: Substitute the given values:

$$\frac{100}{250} + \frac{150}{600} = \frac{1}{n}$$

$$0.400 + 0.250 = \frac{1}{n}$$

$$0.650 = \frac{1}{n}$$

Step 2: Solve for the factor of safety:

$$n = \frac{1}{0.650} = \mathbf{1.538 \approx 1.5}$$

The Goodman criterion is the most commonly used fatigue failure criterion on the FE exam and is found in the reference handbook. It plots a straight line from the endurance limit on the alternating-stress axis to the ultimate tensile strength on the mean-stress axis. Any stress state inside the line is “safe.” The factor of safety tells you how far your operating point is from the failure line. A value of \(n = 1.5\) means the component has a 50% margin above the failure threshold.

Answer: B

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

$$E = \frac{\sigma}{\varepsilon}$$

Step 1: Substitute the given values:

$$E = \frac{70 \text{ MPa}}{0.001} = 70{,}000 \text{ MPa} = \mathbf{70 \text{ GPa}}$$

This is consistent with the well-known modulus of elasticity for aluminum alloys, which is approximately 69–72 GPa. For reference, steel is approximately 200 GPa (answer D), and titanium is approximately 110 GPa. Knowing the typical modulus values for common engineering materials helps you quickly eliminate unreasonable answer choices on the FE exam.

Answer: B

For an isentropic process with an ideal gas, the temperature and pressure are related by:

$$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(k-1)/k}$$

Step 1: Calculate the pressure ratio:

$$\frac{P_2}{P_1} = \frac{800}{100} = 8$$

Step 2: Calculate the exponent:

$$\frac{k-1}{k} = \frac{1.4 - 1}{1.4} = \frac{0.4}{1.4} = 0.2857$$

Step 3: Calculate the final temperature:

$$T_2 = T_1 \times \left(\frac{P_2}{P_1}\right)^{0.2857} = 300 \times 8^{0.2857}$$

Step 4: Evaluate \(8^{0.2857}\):

$$8^{0.2857} = e^{0.2857 \ln 8} = e^{0.2857 \times 2.0794} = e^{0.5941} = 1.811$$

$$T_2 = 300 \times 1.811 = \mathbf{543 \text{ K}}$$

Isentropic relations for ideal gases are among the most frequently tested thermodynamics formulas on the FE exam. They appear in gas power cycle analysis (Brayton, Otto, Diesel) and in compressor/turbine calculations. The key relationships are all in the reference handbook — just make sure you use absolute temperatures (Kelvin) and know the specific heat ratio \(k\) for the gas.

Answer: C

The power delivered to the fluid (water horsepower) is:

$$P_{\text{fluid}} = \rho g Q h$$

Step 1: Calculate the fluid power:

$$P_{\text{fluid}} = 1000 \times 9.81 \times 0.05 \times 25$$

$$P_{\text{fluid}} = 12{,}262.5 \text{ W} = 12.26 \text{ kW}$$

Step 2: Account for pump efficiency to find the required input (brake) power:

$$P_{\text{input}} = \frac{P_{\text{fluid}}}{\eta} = \frac{12.26}{0.80}$$

$$P_{\text{input}} = \mathbf{15.33 \text{ kW}} \approx 15.3 \text{ kW}$$

Pump power calculations are extremely common on the FE Mechanical exam. The key relationship is \(P_{\text{fluid}} = \rho g Q h\), and the input power is always higher than the fluid power because no pump is 100% efficient. Note that answer B (12.3 kW) is the fluid power before dividing by efficiency — a common trap. Always divide by efficiency for pump input power and multiply by efficiency for turbine output power.