Vibration of Single Degree of Freedom Systems

Vibration is the repeated to-and-fro motion of a body or system about its equilibrium position.

Effects

• Harmful: excess stress and fatigue failure, loosening of joints, noise, wear of bearings, discomfort to people, loss of accuracy in machine tools, and resonance damage to structures.
• Useful: vibrating screens and conveyors, compactors, ultrasonic cleaning and testing, musical instruments, vibratory finishing.

Terms

• Period ($T$): time for one complete cycle.
• Frequency ($f$): number of cycles per second, $f = 1/T$ (Hz); circular frequency $\omega = 2\pi f$.
• Amplitude: maximum displacement from the equilibrium position.
• Natural frequency: frequency at which a system vibrates freely after an initial disturbance, with no external force.
• Resonance: when the forcing frequency equals a natural frequency, the amplitude becomes very large (limited only by damping).
• Degrees of freedom: the minimum number of independent coordinates needed to define the position of all parts of the system at any time (a spring-mass has 1, a two-mass system has 2).
• Free vibration: system disturbed and left to vibrate. Forced vibration: an external force acts continuously.

Elements of a vibrating system

Springs in parallel: $k_{eq} = k_1 + k_2$. Springs in series: $\dfrac{1}{k_{eq}} = \dfrac{1}{k_1}+\dfrac{1}{k_2}$.

For a conservative system with no damping, the total energy stays constant:

$$T + U = \text{constant}$$

Spring-mass (massless spring)

Let the mass $m$ be displaced by $x$ from the static equilibrium position and have velocity $\dot{x}$.

$$T = \tfrac12 m\dot{x}^2, \qquad U = \tfrac12 k x^2$$

(The change of potential energy of gravity is cancelled by the static spring force, so only the extra spring energy $\tfrac12kx^2$ remains.)

Differentiating $T + U =$ constant:

$$m\dot{x}\ddot{x} + kx\dot{x} = 0 \;\Rightarrow\; m\ddot{x} + kx = 0$$ $$\omega_n = \sqrt{\frac{k}{m}}, \qquad f_n = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2\pi}\sqrt{\frac{g}{\delta_{st}}}$$

Alternatively, with $x = X\sin\omega_n t$, the maximum kinetic energy $\tfrac12 m\omega_n^2X^2$ equals the maximum potential energy $\tfrac12 kX^2$, which gives the same result.

Effect of the spring's own mass

Let the spring have mass $m_s$ and length $l$. A small element $dy$ at distance $y$ from the fixed end moves with velocity $\dfrac{y}{l}\dot{x}$ (assuming a linear variation).

$$T_{spring} = \int_0^l \tfrac12\left(\frac{m_s}{l}dy\right)\left(\frac{y\dot{x}}{l}\right)^2 = \tfrac12\cdot\frac{m_s}{3}\dot{x}^2$$

Total kinetic energy $T = \tfrac12\left(m + \dfrac{m_s}{3}\right)\dot{x}^2$. Hence

$$\omega_n = \sqrt{\frac{k}{m + m_s/3}}$$

One-third of the spring mass is added to the suspended mass as an effective mass. The natural frequency decreases slightly; the correction is negligible when $m_s \ll m$.

Data: $m = 12$ kg, $k_1 = 15000$ N/m, $k_2 = 25000$ N/m (parallel), $k_3 = 20000$ N/m (in series with the pair).

Equivalent stiffness

$$k_p = k_1 + k_2 = 15000 + 25000 = 40000\ \text{N/m}$$ $$k_{eq} = \frac{k_p k_3}{k_p + k_3} = \frac{40000 \times 20000}{40000 + 20000} = 13333.3\ \text{N/m}$$

Static deflection

$$\delta_{st} = \frac{mg}{k_{eq}} = \frac{12 \times 9.81}{13333.3} = 0.00883\ \text{m} = 8.83\ \text{mm}$$

Natural frequency and period

$$\omega_n = \sqrt{\frac{k_{eq}}{m}} = \sqrt{\frac{13333.3}{12}} = 33.33\ \text{rad/s}$$ $$f_n = \frac{\omega_n}{2\pi} = 5.305\ \text{Hz}, \qquad T = \frac{1}{f_n} = 0.1885\ \text{s}$$

Response to the initial conditions

$x(t) = x_0\cos\omega_n t + \dfrac{v_0}{\omega_n}\sin\omega_n t = X\sin(\omega_n t + \phi)$

$$X = \sqrt{x_0^2 + \left(\frac{v_0}{\omega_n}\right)^2} = \sqrt{0.02^2 + \left(\frac{0.5}{33.33}\right)^2} = 0.0250\ \text{m} = 25.0\ \text{mm}$$ $$\phi = \tan^{-1}\left(\frac{\omega_n x_0}{v_0}\right) = \tan^{-1}\left(\frac{33.33 \times 0.02}{0.5}\right) = 53.1^\circ$$ $$v_{max} = \omega_n X = 33.33 \times 0.0250 = 0.833\ \text{m/s}, \qquad a_{max} = \omega_n^2 X = 27.78\ \text{m/s}^2$$

Answer: $k_{eq} = 13333.3$ N/m; $\delta_{st} = 8.83$ mm; $f_n = 5.305$ Hz; $T = 0.1885$ s; $X = 25.0$ mm; $v_{max} = 0.833$ m/s; $a_{max} = 27.78$ m/s².

Model

  [ A ]=========shaft (l, GJ)=========[ B ]
   I_A         k_t = GJ/l             I_B
   theta_A                            theta_B

Let $\theta_A$ and $\theta_B$ be the angular displacements of the rotors in opposite directions (as in the fundamental mode, the two rotors move opposite to each other).

Equations of motion

The shaft torque is $k_t(\theta_A - \theta_B)$ (relative twist):

$$I_A\ddot\theta_A + k_t(\theta_A - \theta_B) = 0, \qquad I_B\ddot\theta_B - k_t(\theta_A - \theta_B) = 0$$

Take harmonic motion $\theta = \Theta\sin\omega t$:

$$(k_t - I_A\omega^2)\Theta_A - k_t\Theta_B = 0, \qquad -k_t\Theta_A + (k_t - I_B\omega^2)\Theta_B = 0$$

For a non-trivial solution, the determinant is zero:

$$(k_t - I_A\omega^2)(k_t - I_B\omega^2) - k_t^2 = 0 \;\Rightarrow\; \omega^2\left[I_AI_B\omega^2 - k_t(I_A+I_B)\right] = 0$$

The root $\omega = 0$ is rigid-body rotation. The other root is the natural frequency:

$$\omega_n = \sqrt{\frac{k_t(I_A+I_B)}{I_AI_B}} = \sqrt{\frac{GJ}{l}\cdot\frac{I_A+I_B}{I_AI_B}}, \qquad f_n = \frac{\omega_n}{2\pi}$$

Position of the node

At the node the twist is zero. From the first equation, $\dfrac{\Theta_B}{\Theta_A} = \dfrac{k_t - I_A\omega_n^2}{k_t} = -\dfrac{I_A}{I_B}$, so the amplitudes are inversely proportional to the inertias and are of opposite sign.

If $l_A$ and $l_B$ are the distances of the node from A and B ($l_A + l_B = l$), the twist per unit length is uniform, so $\Theta_A/l_A = \Theta_B/l_B$:

$$l_A = \frac{I_B}{I_A+I_B}\,l, \qquad l_B = \frac{I_A}{I_A+I_B}\,l$$

The node is nearer the rotor with the larger inertia. Each rotor vibrates as if fixed to a shaft of length $l_A$ or $l_B$, with $\omega_n = \sqrt{\dfrac{GJ}{l_A I_A}} = \sqrt{\dfrac{GJ}{l_B I_B}}$.

Equation of motion

For mass $m$, spring $k$ and viscous damper $c$, with displacement $x$ from equilibrium:

$$m\ddot{x} + c\dot{x} + kx = 0$$

Assume $x = Ae^{st}$:

$$ms^2 + cs + k = 0 \;\Rightarrow\; s_{1,2} = -\frac{c}{2m} \pm \sqrt{\left(\frac{c}{2m}\right)^2 - \frac{k}{m}}$$

The critical damping coefficient is the value of $c$ making the root zero: $c_c = 2\sqrt{km} = 2m\omega_n$. The damping ratio is $\zeta = c/c_c$. Then

$$s_{1,2} = \omega_n\left(-\zeta \pm \sqrt{\zeta^2 - 1}\right)$$

Case 1: under-damped ($\zeta < 1$)

The roots are complex: $s = -\zeta\omega_n \pm i\omega_d$, with damped frequency $\omega_d = \omega_n\sqrt{1-\zeta^2}$.

$$x(t) = Xe^{-\zeta\omega_n t}\sin(\omega_d t + \phi)$$

The motion is oscillatory with exponentially decaying amplitude.

Case 2: critically damped ($\zeta = 1$)

Equal real roots $s = -\omega_n$:

$$x(t) = (A + Bt)\,e^{-\omega_n t}$$

No oscillation; the mass returns to equilibrium in the shortest time.

Case 3: over-damped ($\zeta > 1$)

Two distinct negative real roots:

$$x(t) = Ae^{(-\zeta + \sqrt{\zeta^2-1})\omega_n t} + Be^{(-\zeta - \sqrt{\zeta^2-1})\omega_n t}$$

No oscillation; the return to equilibrium is slower than in the critical case.

 x
 |\      under-damped (oscillates, decays)
 | \   /\
 |  \_/  \_/\__ ...
 |   \
 |    \__________ over-damped (slow)
 |     \_______   critical (fastest, no overshoot)
 +----------------------------> t

Data: $m = 25$ kg, $k = 20000$ N/m, $x_2/x_0 = 0.25$ after $n = 2$ cycles.

Logarithmic decrement

$$\delta = \frac{1}{n}\ln\frac{x_0}{x_n} = \frac{1}{2}\ln\frac{1}{0.25} = \frac{\ln 4}{2} = 0.6931$$

Damping ratio

$$\delta = \frac{2\pi\zeta}{\sqrt{1-\zeta^2}} \Rightarrow \zeta = \frac{\delta}{\sqrt{4\pi^2 + \delta^2}} = \frac{0.6931}{\sqrt{39.478 + 0.4805}} = 0.1097$$

Natural and damped frequency

$$\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{20000}{25}} = 28.28\ \text{rad/s}$$ $$\omega_d = \omega_n\sqrt{1-\zeta^2} = 28.28\sqrt{1 - 0.1097^2} = 28.11\ \text{rad/s}, \qquad f_d = \frac{\omega_d}{2\pi} = 4.474\ \text{Hz}$$

Damping coefficient

$$c_c = 2\sqrt{km} = 2\sqrt{20000 \times 25} = 1414.2\ \text{N·s/m}$$ $$c = \zeta c_c = 0.1097 \times 1414.2 = 155.1\ \text{N·s/m}$$

Cycles to reach 10%

$$\frac{x_0}{x_n} = e^{n\delta} = 10 \Rightarrow n = \frac{\ln 10}{\delta} = \frac{2.3026}{0.6931} = 3.32\ \text{cycles}$$

Answer: $\delta = 0.6931$; $\zeta = 0.1097$; $c = 155.1$ N·s/m; $f_d = 4.474$ Hz; about 3.32 cycles to fall to 10%.

Data: $M = 90$ kg, $k = 800000$ N/m, $\zeta = 0.12$, $m_0e = 0.25$ kg·m, $N = 1200$ rpm.

Frequencies

$$\omega_n = \sqrt{\frac{k}{M}} = \sqrt{\frac{800000}{90}} = 94.28\ \text{rad/s} \;(N_n = 900\ \text{rpm})$$ $$\omega = \frac{2\pi \times 1200}{60} = 125.66\ \text{rad/s}, \qquad r = \frac{\omega}{\omega_n} = 1.3329$$ $$c = 2\zeta\sqrt{kM} = 2 \times 0.12 \times \sqrt{800000 \times 90} = 2036\ \text{N·s/m}$$

(i) Steady-state amplitude

For the exciting force $F_0 = m_0e\,\omega^2$ rotating with the shaft:

$$X = \frac{(m_0e/M)\,r^2}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}$$ $$X = \frac{(0.25/90) \times 1.7765}{\sqrt{(1 - 1.7765)^2 + (2 \times 0.12 \times 1.3329)^2}} = 0.005876\ \text{m} = 5.876\ \text{mm}$$

(ii) Force transmitted

The spring and damper forces are in quadrature:

$$F_T = X\sqrt{k^2 + (c\omega)^2} = 0.005876 \times \sqrt{(800000)^2 + (2036 \times 125.66)^2} = 4935\ \text{N}$$

The exciting force amplitude is $F_0 = m_0e\,\omega^2 = 0.25 \times 125.66^2 = 3948$ N, so the transmissibility is $4935/3948 = 1.250$.

(iii) Resonance

Resonance occurs when $\omega = \omega_n$, i.e. $N = 900$ rpm. At $r = 1$:

$$X_{res} = \frac{m_0e/M}{2\zeta} = \frac{0.25/90}{2 \times 0.12} = 0.01157\ \text{m} = 11.57\ \text{mm}$$

Answer: (i) $X = 5.876$ mm; (ii) $F_T = 4935$ N; (iii) resonance at $900$ rpm with amplitude about $11.57$ mm.

(a) Transmissibility

Transmissibility $T_R$ is the ratio of the amplitude of the force transmitted to the foundation to the amplitude of the force applied on the machine (force transmissibility). The same ratio applies to motion (displacement transmissibility) for base excitation.

For $m\ddot{x} + c\dot{x} + kx = F_0\sin\omega t$, the steady-state amplitude is

$$X = \frac{F_0/k}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}, \qquad r = \frac{\omega}{\omega_n}$$

The force reaching the foundation is the sum of spring and damper forces (in quadrature):

$$F_T = X\sqrt{k^2 + (c\omega)^2} = kX\sqrt{1 + (2\zeta r)^2}$$ $$T_R = \frac{F_T}{F_0} = \sqrt{\frac{1 + (2\zeta r)^2}{(1-r^2)^2 + (2\zeta r)^2}}$$

Key points:

• $T_R > 1$ for $r < \sqrt2$ (the foundation sees more force than applied).
• $T_R = 1$ at $r = \sqrt2$ for every damping ratio.
• Isolation ($T_R < 1$) begins only for $r > \sqrt2$. Above this, more damping increases transmissibility, so light damping is best, while some damping is needed to limit the peak when passing resonance.

(b) Isolator design (undamped)

For $\zeta = 0$:

$$T_R = \frac{1}{r^2 - 1} \;(r > \sqrt{2})$$ $$0.2 = \frac{1}{r^2 - 1} \Rightarrow r^2 = 6 \Rightarrow r = 2.4495$$ $$\omega = \frac{2\pi \times 1200}{60} = 125.66\ \text{rad/s}, \qquad \omega_n = \frac{\omega}{r} = 51.30\ \text{rad/s}$$ $$k = m\omega_n^2 = 400 \times 51.30^2 = 1052758\ \text{N/m} = 1052.8\ \text{kN/m}$$ $$\delta_{st} = \frac{mg}{k} = \frac{g}{\omega_n^2} = 0.00373\ \text{m} = 3.73\ \text{mm}$$

Isolation efficiency $= 1 - T_R = 80\%$.

Answer: Spring stiffness $\approx 1052.8$ kN/m (total, all springs); static deflection $\approx 3.73$ mm.