JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II DYNAMICS OF MACHINERY Previous Question Paper (November/December - 2020 - R16)

Given: Total mass M = 2000 kg, gauge w = 1.6 m, radius R = 30 m, velocity v = 54 km/h = 15 m/s, banking angle θ = 8°, wheel diameter D = 0.7 m (radius r = 0.35 m), mass of each axle pair m = 200 kg, radius of gyration k = 0.3 m, height of CG h = 1 m. Angular velocity of wheels ω_w = v / r = 15 / 0.35 = 42.86 rad/s. Angular velocity of precession ω_p = v / R = 15 / 30 = 0.5 rad/s. Moment of inertia of each axle pair I = m * k^2 = 200 * 0.3^2 = 18 kg·m^2.

1. Centrifugal Force (F_c): F_c = M * v^2 / R = 2000 * 15^2 / 30 = 15000 N. This force acts at the CG, creating a overturning moment M_c = F_c * h * cos(θ) = 15000 * 1 * cos(8°) = 14854 N·m.

2. Gyroscopic Couple (C_g): Total moment C_g = 2 * I * ω_w * ω_p = 2 * 18 * 42.86 * 0.5 = 771.48 N·m.

3. Static Load: Total weight W = M * g = 2000 * 9.81 = 19620 N. Static load on each rail (assuming symmetry) = W / 2 = 9810 N. The component of weight perpendicular to the track is W * cos(θ) = 19620 * cos(8°) = 19429 N. The vertical reaction due to weight per rail is 9714.5 N.

4. Rail Pressure Calculation: Let R_1 and R_2 be the reactions on the inner and outer rails. The total overturning moment M_total = M_c + C_g = 14854 + 771.48 = 15625.48 N·m. The change in reaction ΔR due to this moment is M_total / w = 15625.48 / 1.6 = 9765.9 N.

5. Final Reactions: The outer rail reaction R_2 = (W * cos(θ) / 2) + (M * g * h * sin(θ) / w) + ΔR = 9714.5 + (19620 * 1 * sin(8°) / 1.6) + 9765.9 = 9714.5 + 1707.5 + 9765.9 = 21187.9 N. The inner rail reaction R_1 = (W * cos(θ) / 2) - (M * g * h * sin(θ) / w) - ΔR = 9714.5 - 1707.5 - 9765.9 = -1758.9 N. The negative value indicates the inner wheel is about to lift off the rail.

Given: Weight of rod W = 700 N (m = 71.36 kg), radius of gyration k = 0.4 m, length l = 1 m, distance of CG from crank pin center b = 0.35 m, distance of CG from crosshead pin center a = l - b = 0.65 m, crank radius r = 0.25 m, speed N = 300 rpm (ω = 31.42 rad/s), crank angle θ = 135°. The ratio n = l/r = 1/0.25 = 4.

First, calculate the kinematics of the slider-crank mechanism at θ = 135°:

1. Determine the angle of the connecting rod (φ): sin φ = (sin θ) / n = (sin 135°) / 4 = 0.1768, so φ = 10.18°.
2. Calculate acceleration of the piston (a_p): a_p = ω²r(cos θ + (cos 2θ)/n) = (31.42)² * 0.25 * (cos 135° + (cos 270°)/4) = -173.5 m/s².
3. Calculate angular acceleration of the connecting rod (α): α = (ω² * sin θ) / (n * cos φ) = (31.42² * sin 135°) / (4 * cos 10.18°) = 178.5 rad/s².

To find the inertia force, replace the rod with a dynamically equivalent system of two masses (m_A at crosshead pin, m_B at crank pin):

The inertia force on the rod is the resultant of the inertia forces of the two masses. The inertia force of mass m_A is F_A = m_A * a_p = 24.98 * (-173.5) = -4334 N. The inertia force of mass m_B is F_B = m_B * ω² * r = 46.38 * (31.42)² * 0.25 = 11448 N. The torque on the crankshaft is calculated by taking the moments of these inertia forces about the crank pin center, including the torque due to the angular acceleration of the rod (I * α):

Substituting the values, the magnitude of the inertia force is approximately 12,250 N acting at an angle determined by the vector sum of the components, and the corresponding crankshaft torque is approximately 1,650 Nm.