Vibration of Continuous Systems

Wave equation

A string of length $l$, tension $P$ (constant) and mass per unit length $\rho$ vibrates with small transverse displacement $w(x,t)$.

  P  ^ (theta at x)        ^ P (theta at x+dx)
      \                   /
  ----- \_______________/ -----
        x          x+dx

For an element $dx$, the net transverse force from the tension is

$$P\left[\frac{\partial w}{\partial x}\right]_{x+dx} - P\left[\frac{\partial w}{\partial x}\right]_x = P\frac{\partial^2 w}{\partial x^2}dx$$

By Newton's law, this equals $\rho\,dx\,\dfrac{\partial^2 w}{\partial t^2}$. Hence

$$\frac{\partial^2 w}{\partial t^2} = c^2\frac{\partial^2 w}{\partial x^2}, \qquad c = \sqrt{\frac{P}{\rho}}$$

$c$ is the wave speed.

Solution

Let $w(x,t) = W(x)\,T(t)$. Separating variables:

$$\frac{W''}{W} = \frac{\ddot T}{c^2T} = -\frac{\omega^2}{c^2} \;\Rightarrow\; W = A\cos\frac{\omega x}{c} + B\sin\frac{\omega x}{c}$$

Boundary conditions $W(0) = 0 \Rightarrow A = 0$ and $W(l) = 0 \Rightarrow \sin\dfrac{\omega l}{c} = 0$ (frequency equation):

$$\omega_n = \frac{n\pi c}{l}, \qquad f_n = \frac{n}{2l}\sqrt{\frac{P}{\rho}}, \qquad n = 1, 2, 3,\ldots$$

Mode shape: $W_n(x) = B_n\sin\dfrac{n\pi x}{l}$. The $n$th mode has $n-1$ nodes between the supports.

Numerical values

$$c = \sqrt{\frac{70}{0.0006}} = 341.6\ \text{m/s}$$ $$f_1 = \frac{c}{2l} = \frac{341.6}{2 \times 0.65} = 262.7\ \text{Hz}, \quad f_2 = 2f_1 = 525.5\ \text{Hz}, \quad f_3 = 3f_1 = 788.2\ \text{Hz}$$

Answer: $f_1 = 262.7$ Hz, $f_2 = 525.5$ Hz, $f_3 = 788.2$ Hz.

(a) Longitudinal vibration of a rod

For a rod of area $A$, modulus $E$ and density $\rho$, let $u(x,t)$ be the axial displacement. Axial force $P = EA\dfrac{\partial u}{\partial x}$. For an element $dx$:

$$\left(P + \frac{\partial P}{\partial x}dx\right) - P = \rho A\,dx\,\frac{\partial^2 u}{\partial t^2} \;\Rightarrow\; \frac{\partial^2 u}{\partial t^2} = c^2\frac{\partial^2 u}{\partial x^2}, \qquad c = \sqrt{\frac{E}{\rho}}$$

With $u = U(x)\sin\omega t$, $\;U = A_1\cos\dfrac{\omega x}{c} + B_1\sin\dfrac{\omega x}{c}$.

Fixed at $x = 0$, free at $x = l$:

• Fixed end: $U(0) = 0 \Rightarrow A_1 = 0$.
• Free end (no stress): $\dfrac{dU}{dx}(l) = 0 \Rightarrow \cos\dfrac{\omega l}{c} = 0$.

$$\frac{\omega_nl}{c} = \frac{(2n-1)\pi}{2} \;\Rightarrow\; f_n = \frac{(2n-1)c}{4l}, \quad n = 1, 2, 3, \ldots$$

Mode shape: $U_n = B\sin\dfrac{(2n-1)\pi x}{2l}$.

(For fixed-fixed or free-free ends, $f_n = nc/(2l)$.)

(b) Numerical values

Longitudinal:

$$c = \sqrt{\frac{200\times10^9}{7850}} = 5047.5\ \text{m/s}$$ $$f_1 = \frac{c}{4l} = \frac{5047.5}{4 \times 2} = 630.9\ \text{Hz}, \quad f_2 = 3f_1 = 1892.8\ \text{Hz}, \quad f_3 = 5f_1 = 3154.7\ \text{Hz}$$

Torsional shaft: The equation is the same with $E \to G$ and $u \to \theta$, wave speed $c_t = \sqrt{G/\rho}$:

$$c_t = \sqrt{\frac{80\times10^9}{7850}} = 3192.3\ \text{m/s}$$ $$f_1 = \frac{c_t}{4l} = \frac{3192.3}{4 \times 1.2} = 665.1\ \text{Hz}, \qquad f_2 = 3f_1 = 1995.2\ \text{Hz}$$

The torsional frequencies are independent of the shaft diameter, because $I_p$ cancels out from the equation.

Answer: Longitudinal: $630.9$, $1892.8$, $3154.7$ Hz. Torsional: $665.1$ and $1995.2$ Hz.

(a) Euler-Bernoulli beam equation

Consider an element $dx$ of a beam with bending stiffness $EI$, mass per unit length $\rho A$, deflection $y(x,t)$, shear force $V$ and bending moment $M$.

     V+dV
 M ->|====|<- M+dM
     V

Vertical equilibrium: $\;V - (V + dV) = \rho A\,dx\,\dfrac{\partial^2 y}{\partial t^2}$, that is $-\dfrac{\partial V}{\partial x} = \rho A\dfrac{\partial^2 y}{\partial t^2}$.

Moments (neglecting rotary inertia and shear deformation): $V = \dfrac{\partial M}{\partial x}$ and $M = EI\dfrac{\partial^2 y}{\partial x^2}$. Hence

$$EI\frac{\partial^4 y}{\partial x^4} + \rho A\frac{\partial^2 y}{\partial t^2} = 0$$

Put $y = Y(x)\sin\omega t$ and $\beta^4 = \dfrac{\rho A\omega^2}{EI}$:

$$\frac{d^4Y}{dx^4} - \beta^4Y = 0 \;\Rightarrow\; Y = C_1\cos\beta x + C_2\sin\beta x + C_3\cosh\beta x + C_4\sinh\beta x$$

Simply supported ends: $Y(0) = Y''(0) = 0 \Rightarrow C_1 = C_3 = 0$. At $x = l$, $Y(l) = Y''(l) = 0$ gives $C_4 = 0$ and

$$\sin\beta l = 0 \;\Rightarrow\; \beta_nl = n\pi$$ $$\omega_n = (n\pi)^2\sqrt{\frac{EI}{\rho Al^4}} = \frac{n^2\pi^2}{l^2}\sqrt{\frac{EI}{\rho A}}, \qquad Y_n = \sin\frac{n\pi x}{l}$$

(b) Numerical

$$I = \frac{bh^3}{12} = \frac{0.05 \times 0.1^3}{12} = 4.1667e-06\ \text{m}^4, \qquad A = bh = 0.005\ \text{m}^2$$ $$\sqrt{\frac{EI}{\rho A}} = \sqrt{\frac{200\times10^9 \times 4.1667e-06}{7850 \times 0.005}} = 145.710\ \text{m}^2/\text{s}$$ $$\omega_1 = \frac{\pi^2}{3^2}\times 145.710 = 159.79\ \text{rad/s}, \qquad f_1 = 25.43\ \text{Hz}$$ $$f_2 = 4f_1 = 101.72\ \text{Hz}, \qquad f_3 = 9f_1 = 228.88\ \text{Hz}$$

Answer: $f_1 = 25.43$ Hz, $f_2 = 101.72$ Hz, $f_3 = 228.88$ Hz.