This set of Automotive Engine Components Design Multiple Choice Questions & Answers (MCQs) focuses on “IC Engine – Big End Cap and Bolts”.
1. Which force is experienced by the connecting rod cap?
a) Gravity force
b) Magnetic force
c) Axial force
d) Inertia force
View Answer
Explanation: The maximum force experienced by the connecting rod cap and bolts consists only of inertia force at the top dead Centre on the exhaust stroke. Hence, the inertia force is experienced by the connecting rod cap.
2. At which position of the piston, connecting rod cap experience inertia force?
a) Center
b) Side dead center
c) Top dead center
d) Bottom dead center
View Answer
Explanation: The maximum force experienced by the connecting rod cap and bolts consists only of inertia force at the top dead center on the exhaust stroke. Hence, connecting rod cap experience inertia force at the top dead center.
3. Inertia force acting on the connecting rod cap is given by?
a) Pi=mrω2 r(θ+\(\frac{cos2\theta}{n_1})\)
b) Pi=mrω2 r(cosθ+\(\frac{sin2\theta}{n_1})\)
c) Pi=mrω2 r(cosθ+\(\frac{cos2\theta}{n_1})\)
d) Pi=mrω2 r(secθ+\(\frac{cos2\theta}{n_1})\)
View Answer
Explanation: Pi=mrω2 r(cosθ+\(\frac{cos2\theta}{n_1})\)is the right answer where Pi is the inertia force, mr is the mass of reciprocating parts, ω is the angular velocity of the crank, r is the crank radius, n1 is the ratio of L/r and θ is the angle of inclination.
4. Calculate the mass of reciprocating parts when the mass of piston assembly is 0.6Kg and mass of the connecting rod is 0.4Kg?
a) 0.73Kg
b) 0.85Kg
c) 0.6Kg
d) 0.43Kg
View Answer
Explanation: mr= mass of piston assembly+ \(\frac{1}{3}\)mass of connecting rod
= 0.6+\(\frac{1}{3}\) (0.4)
= 0.73Kg
5. Find the crank radius when the length of the stroke is 80mm?
a) 10mm
b) 40mm
c) 30mm
d) 50mm
View Answer
Explanation: Crank radius r=\(\frac{l}{2}\) =\(\frac{80}{2}\) =40mm
6. Find the angular velocity of the crank rotating at 1200RPM?
a) 100.86rad/sec
b) 112.54rad/sec
c) 125.65rad/sec
d) 134.89rad/sec
View Answer
Explanation: ω=\(\frac{2\times \pi\times N}{60}\)
=\(\frac{2\times \pi\times 1200}{60}\)
= 125.66 rad/sec
7. In the formula (Pi)max=mrω2r(1+\(\frac{1}{n_1})\) what is mr?
a) Mass of reciprocating parts
b) Mass of all parts
c) Mass of cylinder
d) Mass of crankshaft
View Answer
Explanation: In this formula, (Pi)max is the maximum inertia force, mr is the mass of reciprocating parts, ω is the angular velocity of the crank, r is the crank radius, n1 is the ratio of L/r.
8. Calculate the inertia force where the diameter of bolts is 10mm and σt=35N/mm2?
a) 5555.5N
b) 6107.8N
c) 6478.6N
d) 5497.7N
View Answer
Explanation: (Pi)max=2(\(\frac{\pi\times d_c^2}{4})\)σt
=2\((\frac{\pi\times 10^2}{4})\)× 35
=5497.7N
9. Calculate the bending moment on the cap when (Pi)max=400N and span length is 80mm.
a) 3567.9N-mm
b) 4789N-mm
c) 5333.3N-mm
d) 6893.89N-mm
View Answer
Explanation: Mb=\(\frac{(P_i)_{max}l}{6}\)
=\(\frac{400×800}{6}\)
=5333.3N-mm
10. Calculate the thickness of cap when σb=100N/mm2, Mb=90853.95N-mm, y=\(\frac{t_c}{2}\) and I=6.25tc3?
a) 7.65mm
b) 8.53mm
c) 9.7mm
d) 6.23mm
View Answer
Explanation: σb=\(\frac{M_b \times y}{I}\)
100=\(\frac{90853.95\times \frac{t}{2}}{6.25t^3}\)
tc=8.53mm
Sanfoundry Global Education & Learning Series – Automotive Engine Components Design.
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