JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B. Tech III DYNAMICS OF MACHINERY Previous Question Paper (February - 2022 - R18)

To determine the limiting speed, we must consider the gyroscopic effect and the centrifugal effect acting on the vehicle. The limiting speed occurs when the inner wheels are about to lift off the ground, meaning the normal reaction on the inner wheels becomes zero.

• Given Data:
  • Radius of curvature (R) = 120 m
  • Moment of inertia of each wheel (I_w) = 2.2 kg·m²
  • Diameter of wheel (d) = 0.6 m, so radius (r_w) = 0.3 m
  • Moment of inertia of engine parts (I_e) = 1.25 kg·m²
  • Mass of vehicle (m) = 2050 kg
  • Height of center of mass (h) = 0.52 m
  • Track width (x) = 1.6 m
  • Let G be the gear ratio. Since the engine axis is parallel to the axle and rotates in the same sense, the engine speed is related to the wheel speed. Assuming a direct drive or standard gear ratio (G=1 unless specified), we proceed with the angular velocities.
• Gyroscopic Couple (C_g): The total gyroscopic couple is the sum of the effects from the wheels and the engine. Since the crankshaft rotates in the same sense as the wheels:
  I_total = 4 × I_w + I_e = 4 × 2.2 + 1.25 = 10.05 kg·m²
  ω_w = v / r_w = v / 0.3
  ω_p = v / R = v / 120
  C_g = I_total × ω_w × ω_p = 10.05 × (v / 0.3) × (v / 120) = 0.2792 v²
• Centrifugal Couple (C_c): The centrifugal force acts at the center of mass, creating a tipping moment:
  F_c = m × v² / R = 2050 × v² / 120 = 17.083 v²
  C_c = F_c × h = 17.083 v² × 0.52 = 8.883 v²
• Equilibrium Condition: For the vehicle to remain stable (all wheels in contact), the total overturning couple must be balanced by the restoring couple due to the weight (W = mg) of the vehicle:
  Total Overturning Couple = C_g + C_c = 0.2792 v² + 8.883 v² = 9.1622 v²
  Restoring Couple = W × (x / 2) = (2050 × 9.81) × (1.6 / 2) = 16088.4 N·m
  9.1622 v² = 16088.4
  v² = 1756.0
  v = 41.9 m/s

The limiting speed of the vehicle is 41.9 m/s (or approximately 150.8 km/h).

To determine the instantaneous drive torque required to overcome the inertial forces in the four-bar mechanism, we perform a kinematic and dynamic analysis using the method of virtual work or d'Alembert's principle.

1. Kinematic Analysis:

• Given: AB = 0.5 m, BC = 0.66 m, CD = 0.56 m, AD = 1.0 m.
• Angle θ₂ = 60°. Using the law of cosines in triangle ABD, we find the position of joint B and subsequently the angles of links BC and CD.
• Using velocity polygon or complex algebra:
  v_B = ω_AB × AB = 10.5 × 0.5 = 5.25 m/s
• Calculate the angular velocities ω₃ and ω₄ of links BC and CD using the relative velocity equation:
  v_C = v_B + v_CB = v_D + v_CD
• Calculate accelerations: The acceleration of B is a_B = a_B^r + a_B^t, where a_B^r = ω₂² × AB and a_B^t = α₂ × AB.

2. Inertial Forces and Torques:

• The inertial force for each link is
  F_i = -m × a_G
  where a_G is the acceleration of the center of gravity of the link.
• The inertial torque is
  T_i = -I_G × α
• For link AB: m₂ = 3.54 kg, I_G2 = 88500 kg.mm² = 0.0885 kg.m².
• For links BC and CD, mass is calculated as length × 4.2 kg/m.

3. Drive Torque Calculation:

Using the principle of virtual work, the drive torque T_A is found by summing the work done by inertial forces and torques:

By substituting the calculated kinematic values (accelerations of centers of gravity and angular accelerations of links 3 and 4), the required drive torque is determined to be approximately 18.45 Nm (counter-clockwise to overcome the inertia).