例題集

連続体の振動(2)

理解レベル   難易度: ★★
長さ$l$の両端支持はりのたわみ曲線を以下のように仮定したとき,レーリー法を用いて$1$次固有振動数$\omega$を求めよ. たわみ曲線 : $w(x)= W\left\{3\frac{x}{l} - 4\left( \frac{x}{l} \right)^3 \right\}$  $(0 \leqq x \leqq \frac{l}{2})$ 運動エネルギ : $T= \frac{1}{2} \int \rho A \left(\frac{\partial w}{\partial t} \right)^2 dx $ ひずみエネルギ : $U= \frac{1}{2} \int EI \left(\frac{d^2 w}{d x^2} \right)^2 dx $
変位を$w(x) \cdot \sin\omega t$とおくと,変位の最大値は$w(x) \cdot \omega$となることから, \[ \begin{align} T_{max} &=\frac{1}{2} \int_{0}^{\frac{l}{2}} \rho A \left[ W \left\{3 \frac{x}{l}-4 \left( \frac{x}{l} \right)^3 \right\} ・\omega\right]^2 dx \times 2 \\ &=\frac{1}{2} \rho A (W \omega)^2 \times 2 \int_{0}^{\frac{l}{2}} \left\{ 9 \frac{x^2}{l^2} -24 \frac{x^4}{l^4} -16 \frac{x^6}{l^6} \right\} dx \\ \end{align} \] \[ \begin{align} \int_{0}^{\frac{l}{2}} \left\{ 9 \frac{x^2}{l^2} -24 \frac{x^4}{l^4} -16 \frac{x^6}{l^6} \right\} dx &=\left[ \frac{9^3}{l^2} \cdot \frac{x^3}{3}-\frac{24}{l^4}\cdot \frac{x^5}{5}+ \frac{16}{l^6}\cdot \frac{x^7}{7} \right]_0^{\frac{l}{2}} \\ &=\frac{3}{l^2} \left( \frac{l}{2} \right)^3- \frac{24}{5 l^4} \left( \frac{l}{2} \right)^5 + \frac{16}{7l^6} \left( \frac{l}{2} \right)^7 \\ &=\frac{3 l^3}{8l^2}- \frac{24l^5}{5\cdot 32l^4}+ \frac{16 l^7}{7\cdot 128l^6} \\ &= \left( \frac{3}{8}-\frac{3}{20}+\frac{1}{56} \right)l \\ &= \left( \frac{21}{56}+\frac{1}{56}-\frac{3}{20} \right)l \\ &= \left( \frac{22}{58}-\frac{3}{20} \right)l \\ &= \left( \frac{55}{140}-\frac{21}{140} \right)l \\ &=\frac{34}{140}l \\ &=\frac{17}{70}l \end{align} \] \[ \begin{align} T_{max} &=\frac{1}{2} \rho A (W \omega)^2 \times 2 \times \frac{17}{70}l \\ &=\frac{1}{2} \rho A (W \omega)^2 \cdot \frac{17}{35}l \\ \end{align}\] \[ \begin{align}\frac{d^2 w}{dx^2} &=\frac{d^2}{dx^2} \left[W\left\{ 3\cdot \frac{x}{l}-4 \left( \frac{x}{l} \right)^3 \right\}\right] \\ &=\frac{d}{dx} \left[ W \left\{ \frac{3}{l}- \frac{4}{l^3} \cdot 3x^2 \right\}\right] \\ &=W \left\{ -\frac{12}{l^3}\cdot 2x \right\} \\ &=-W\cdot \frac{24}{l^3}x \end{align} \] \[ \begin{align} U_{max} &=\frac{1}{2} \int_{0}^{\frac{l}{2}} EI \left(-W \cdot \frac{24}{l^3}x\right)^2 dx \cdot 2 \\ &=\frac{1}{2} EI W^2\left( \frac{24}{l^3} \right)^2\times 2\int_{0}^{\frac{l}{2}} x^2dx \\ &=\frac{1}{2}EI W^2\left( \frac{24}{l^3} \right)^2 \times 2\left[ \frac{x^3}{3} \right]^{\frac{l}{2}}_0 \\ &=\frac{1}{2} EI W^2 \left( \frac{24}{l^3} \right)\times 2 \left\{ \frac{1}{3} \cdot \left( \frac{l}{2} \right)^3 \right\} \\ &=\frac{1}{2}EI W^2 \frac{24^2}{l^6}\times 2 \times \frac{l^3}{3\times 8} \\ &=\frac{1}{2} EI W^2 \cdot \frac{48}{l^3} \end{align} \] $T_{max}=U_{max}$より \[ \frac{1}{2} \rho A l W^2 \omega^2 \cdot \frac{17}{35} =\frac{1}{2}EI W^2 \cdot \frac{48}{l^3} \] \[ \begin{align} \omega^2 &= \frac{48EI \frac{1}{l^3}}{\frac{17}{35} \cdot \rho A l} \\ \therefore m_b &= \rho A l\\ \omega &= \sqrt{\frac{48 \times 35 \times EI}{17m_b l^3}} \\ &=9.941 \sqrt{\frac{EI}{m_b l^3}} \end{align} \\ \] \[ \therefore \omega = 9.94 \sqrt{\frac{EI}{m_b l^3}} \]