Mechanical Design and Analysis practice questions

Answer: B) 50 N/mm

• AIncorrect. That's the parallel value.
• BCorrect. \(1/k = 1/100 + 1/100\).
• CIncorrect. Series springs are softer.
• DIncorrect. Recheck the formula.

Answer: C) 10 teeth/in.

• AIncorrect. That's inverted.
• BIncorrect. Divide, don't multiply.
• CCorrect. P = N/D = 40/4.
• DIncorrect. Divide, don't add.

Answer: D) 200 kJ

On level ground with losses ignored, the brakes convert all of the kinetic energy into heat: ½(1,000 kg)(20 m/s)² = 200 kJ.

• AIncorrect. This value comes from ½mv = 0.5(1,000)(20) = 10,000, which forgets to square the velocity.
• BIncorrect. This value comes from mv² = 1,000(20²) = 400,000 J, which leaves out the factor of ½ in the kinetic-energy formula.
• CIncorrect. This value comes from mv = 1,000(20) = 20,000. That is the momentum (in kg·m/s), not an energy, and it neither squares v nor applies the ½.
• DCorrect. The brakes must absorb the vehicle's kinetic energy: KE = ½mv² = 0.5(1,000)(20)² = 200,000 J = 200 kJ.

Answer: C) The pinion has too few teeth

• AIncorrect. Lubrication doesn't cause interference.
• BIncorrect. Too few teeth cause it.
• CCorrect. Minimum tooth counts depend on pressure angle.
• DIncorrect. Speed isn't the cause.

Answer: C) 90% of identical bearings will reach

• AIncorrect. It's 90%.
• BIncorrect. It's a 90% reliability life.
• CCorrect. 10% may fail sooner.
• DIncorrect. It's at the rated load.

Answer: A) 95.5 N·m

• ACorrect. \(T = P/\omega = 10{,}000/104.7\).
• BIncorrect. Convert rpm to rad/s.
• CIncorrect. Recheck the conversion.
• DIncorrect. Recheck the powers of ten.

Answer: D) 400 rpm

• AIncorrect. Recheck the ratio.
• BIncorrect. The larger gear turns slower.
• CIncorrect. Apply the gear ratio.
• DCorrect. 1,200 × 20/60.

Answer: D) 14.1 MPa

Fillet-weld shear stress is based on the throat area, 0.707hL. Here τ = 10,000/(0.707 × 10 × 100) ≈ 14.1 MPa.

• AIncorrect. This is ten times the correct value and comes from a unit or decimal error, for example using a throat area of 70.7 mm² instead of 707 mm².
• BIncorrect. This uses the leg area (10 mm × 100 mm = 1000 mm²) instead of the throat area. The throat of a 45° fillet weld is 0.707h, which is smaller than the leg.
• CIncorrect. This comes from multiplying the leg-area stress by 0.707 instead of dividing by the throat area. A smaller throat area gives a higher stress, not a lower one.
• DCorrect. Throat t = 0.707 × 10 = 7.07 mm, so the throat area is 7.07 × 100 = 707 mm² and τ = 10,000 N / 707 mm² ≈ 14.1 MPa.

Answer: A) 20°

20° is the most common modern spur-gear pressure angle. 14.5° is the older standard, and 25° appears in some heavy-duty designs.

• ACorrect. 20° is today's dominant standard. It balances tooth strength, resistance to undercutting with small pinions, and bearing loads.
• BIncorrect. 14.5° was a common older standard. It is now mostly found on legacy or replacement gears because its weaker teeth are more prone to undercutting.
• CIncorrect. 25° is used in some heavy-duty applications that need stronger teeth. It is less common than 20° because it raises separating forces and bearing loads.
• DIncorrect. 30° is a common pressure angle for involute splines, not the general standard for power-transmitting spur gears.

Answer: C) 150 N·m

With no losses, power is conserved, so torque scales inversely with speed. A 3:1 reduction triples the torque: 3 × 50 = 150 N·m.

• AIncorrect. Torque stays the same only at a 1:1 ratio. A speed reduction changes torque by the gear ratio.
• BIncorrect. This divides by the ratio, which is what happens to speed. With power conserved (Tω = constant), torque rises when speed falls.
• CCorrect. For an ideal train, T_in·ω_in = T_out·ω_out. Output speed is 1/3 of input speed, so T_out = 3 × 50 = 150 N·m.
• DIncorrect. This adds the ratio to the torque (50 + 3). The ratio scales torque by multiplication, not addition.