Problem 10.8.Natural frequency of a hinged plate

Percentage error, [%]=

Step 1.  Give the analytical solution of the natural frequency.

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See Table 10.6 first row or Eq.10-99)

$$\begin{gathered} f^2=f_x^2+f_y^2+f_t^2=\frac{{\pi }^{2}D}{4m{L}^4}+\frac{{\pi }^{2}D}{4m{L}^4}+\frac{2{\pi }^{2}D}{4m{L}^4}=\frac{{\pi }^{2}D}{m{L}^4} \\ f=\pi \sqrt{\frac{D}{m{L}^4}} \end{gathered}$$

Step 2.  Give the approximate solution of the natural frequency.

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Deflection at the middle of the square isotropic plate is

Table 10.3.

$w=4.06\frac{mg{L}^4}{1000D}$

The natural frequency is approximated as

Eq.(8-57)

$f_{\mathrm{approx}}=\frac{18}{\sqrt{w}}=\frac{18}{\sqrt{4.06{\displaystyle \frac{mg{L}^4}{D}}}}=\frac{18}{\sqrt{4.06\times 9.81}}\sqrt{\frac{D}{m{L}^4}}=2.85\sqrt{\frac{D}{m{L}^4}}$

Step 3.  Compare the above solutions. Calculate the percentage error of the approximate solution.

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The analytical solution of the natural frequency is

See Table 10.6 first row or Eq.(10-99)

$$\begin{gathered} f^2=f_x^2+f_y^2+f_t^2=\frac{{\pi }^{2}D}{4m{L}^4}+\frac{{\pi }^{2}D}{4m{L}^4}+\frac{2{\pi }^{2}D}{4m{L}^4}=\frac{{\pi }^{2}D}{m{L}^4} \\ f=\pi \sqrt{\frac{D}{m{L}^4}} \end{gathered}$$

The natural frequency can be approximated also from the deflection of the plate.Deflection at the middle of the square isotropic plate is

Table 10.3.

$w=4.06\frac{mg{L}^4}{1000D}$

The natural frequency is approximated as

Eq.(8-57)

$f_{\mathrm{approx}}=\frac{18}{\sqrt{w}}=\frac{18}{\sqrt{4.06{\displaystyle \frac{mg{L}^4}{D}}}}=\frac{18}{\sqrt{4.06\times 9.81}}\sqrt{\frac{D}{m{L}^4}}=2.85\sqrt{\frac{D}{m{L}^4}}$

The above solutions are compared. The percentage error of the approximate solution is calculated:

$\frac{\mathit{f}\mathbf{–}{\mathit{f}}_{\mathbf{approx}}}{\mathit{f}}=\frac{{\displaystyle \pi \sqrt{\frac{D}{m{L}^4}}–2.85\sqrt{\frac{{\displaystyle D}}{{\displaystyle m{L}^4}}}}}{\pi \sqrt{\frac{D}{m{L}^4}}}=\frac{{\displaystyle \pi –2.85}}{\pi }=0.0928\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{21}\mathbf{\%}$