GTU 3151911 Dynamics of Machinery Summer 2026 Solved Paper & Solutions

D'Alembert's principle states that a body in motion can be treated as if it were in static equilibrium, if an imaginary force, called the inertia force, is added to the real forces. The inertia force has the magnitude $ma$ and acts opposite to the acceleration:

Significance: it turns a dynamic problem into a static one. The equations of equilibrium can then be used to find the forces and the torques in moving parts, such as the links of a mechanism, a slider-crank, a cam and a flywheel.

The piston of a slider-crank mechanism has a crank of radius $r$, a connecting rod of length $l$, and the ratio $n = l/r$. The crank turns at the angular speed $\omega$ (in rad/s), and its angle from the inner dead centre is $\theta$.

The acceleration of the piston is:

The inertia force of the reciprocating mass $m$ is:

It acts opposite to the acceleration. Its maximum is $m\omega^2 r\left(1 + \dfrac{1}{n}\right)$ at $\theta = 0$.

Example (assumed values): $m = 2$ kg, $r = 0.1$ m, $l = 0.4$ m, $N = 600$ rpm, $\theta = 30^\circ$.

A three-rotor system has three discs with the moments of inertia $I_1$, $I_2$ and $I_3$, joined by two shafts with the torsional stiffnesses $k_1$ and $k_2$. The angles of the discs are $\theta_1$, $\theta_2$ and $\theta_3$. It is a system with three degrees of freedom.

The shafts are taken as light (without mass) and elastic, the discs are taken as rigid, and the damping is neglected. The torsional stiffness of a shaft is $k = \dfrac{GJ}{l}$, where $G$ is the shear modulus, $J$ is the polar moment of area and $l$ is the length.

Equations of motion

Frequency equation

Put $\theta_i = \Theta_i \sin\omega t$, and $\lambda = \omega^2$. The solution exists only when the determinant of the coefficients is zero. It gives:

where

Natural frequencies

• $\omega_0 = 0$: the whole shaft turns as one rigid body, with no twist. This is not a vibration.
• The other two come from the quadratic:

Check: if $I_1 = I_2 = I_3 = I$ and $k_1 = k_2 = k$, then $S = 4k/I$ and $P = 3k^2/I^2$. So $\omega^2 = \dfrac{k}{I}$ and $\dfrac{3k}{I}$.

The lower frequency is the one-node mode, and the higher frequency is the two-node mode. A node is a section of the shaft that does not twist. In the one-node mode, the discs on the two sides of the node turn in opposite directions. In the two-node mode, the middle disc turns against the two end discs. The mode shape is found by putting each root back in the equations. For a system in which one end disc is fixed, the same method is used, with the frequency equation of the same form but with no zero root.

When a two-wheeler takes a turn, the spinning wheels are made to precess about a vertical axis. This produces a gyroscopic couple $C = I\omega\omega_p$, where $I$ is the moment of inertia of the wheel, $\omega$ is the spin speed and $\omega_p = v/R$ is the speed of precession.

The reaction couple on the vehicle tends to overturn it outwards from the turn. It adds to the overturning effect of the centrifugal force. To stay stable, the rider leans inwards, so that the weight gives a balancing couple. The effect grows with the speed and with the inertia of the wheels, and the lean angle must be larger.

The initial displacement of 20 mm does not change these three values. It only decides the size of the first amplitude.

Resonance in a high-speed rotor happens when the running speed is close to a critical speed. The design can be changed in these ways:

1. Change the stiffness or the mass, such as a thicker shaft, shorter bearing span or lighter disc, so that the critical speed is moved at least 20 to 30 per cent away from the running speed.
2. Add damping, with squeeze film dampers or flexible bearing supports, to limit the amplitude.
3. Balance the rotor accurately, to reduce the exciting force.
4. Design the rotor so that the machine passes quickly through the critical speed during start up.

The turning moment diagram plots the turning moment on the crankshaft against the crank angle.

The mean torque line is the same, but the area above it, which is the fluctuation of energy, is far smaller for a multi-cylinder engine.

The span is $L = 50 + 100 + 100 = 250$ mm. The shaft is simply supported at A and B. The load of 60 kg is 50 mm from A, and the load of 100 kg is 150 mm from A. The value of $E$ is not given, so take steel with $E = 200$ GPa. The mass of the shaft is neglected. The method is Dunkerley's method.

The logarithmic decrement $\delta$ is the natural logarithm of the ratio of two successive amplitudes of a damped free vibration:

For small damping, $\delta \approx 2\pi\zeta$.

Use: it is found from a recorded vibration, and then used to calculate the damping ratio $\zeta$ and the damping coefficient $c$ of a system, without a direct measurement of the damping.

In a horizontal rotating shaft, the critical (whirling) speed is the speed at which the shaft speed equals its natural frequency of bending. The effect of damping is as follows.

• The critical speed itself is almost unchanged by damping. It remains close to $\sqrt{k/m}$.
• Damping limits the whirl amplitude at the critical speed. Without damping the amplitude grows without limit, and with damping the peak is finite.
• The amplitude of the whirl is:

where $e$ is the eccentricity and $r$ is the ratio of the shaft speed to the critical speed.

• A higher damping gives a lower and broader peak. Above the critical speed, the shaft tends to self-centre, and the amplitude comes close to $e$.

Static balancing: the rotating masses are balanced when their centre of mass lies on the axis of rotation. The vector sum of their centrifugal forces is zero, $\sum F = 0$. The rotor then stays at rest in any position when it is placed on two horizontal knife edges. It is enough for a thin disc, such as a flywheel or a grinding wheel.

Dynamic balancing: both the resultant force and the resultant couple of the rotating masses are zero, $\sum F = 0$ and $\sum M = 0$. It is needed for a long rotor, with masses in different planes, such as the rotor of a motor or a crankshaft. A dynamically balanced rotor is also statically balanced.

In a reciprocating engine, the reciprocating mass $m$ gives an unbalanced force along the line of stroke. It has two parts:

• Primary unbalanced force:

It has the same frequency as the crank, which turns once in a revolution.

• Secondary unbalanced force:

Its frequency is twice the speed of the crank, and its maximum value is $1/n$ times that of the primary force, with $n = l/r$. So it is small.

Primary balancing is done by a balance mass on the crankshaft opposite to the crank, or, in a multi-cylinder engine, by arranging the cranks so that the primary forces cancel. Secondary balancing needs masses that turn at twice the crank speed, and it is usually done by the arrangement of the cylinders.

The mass to be balanced at the crank radius, for each crank, is the whole rotating mass and two thirds of the reciprocating mass:

Cylinder centres are 1.8 m apart, and the wheel centres are 1.5 m apart. So each cylinder is $\dfrac{1.8 - 1.5}{2} = 0.15$ m outside its wheel. Take crank 1 as the reference at $0^\circ$, and crank 2 at $90^\circ$ from it. The balance masses are at the radius $b = 0.8$ m, in the planes of the wheels A and B.

Planes and data (the plane A is the reference plane)

The wheel lifts when the maximum hammer blow, $B\omega^2$, equals the load of 30 kN on the wheel: