第21篇
4.3.5 谐波激励响应
考虑图4.20所示梁的稳态振动，\(a\)在作用于质量\({m}_{2}\)的力\(f\left( t\right) = {F}_{0}\cos {\omega t}\)作用下。
图4.20
运动方程为
\[
{y}_{1} = - {\delta }_{11}{m}_{1}{\ddot{y}}_{1} + \left( {f - {m}_{2}{\ddot{y}}_{2}}\right) {\delta }_{12},
\]
\[
{y}_{2} = - {\delta }_{21}{m}_{1}{\ddot{y}}_{1} + \left( {f - {m}_{2}{\ddot{y}}_{2}}\right) {\delta }_{22}
\]
或
\[
{\delta }_{11}{m}_{1}{\ddot{y}}_{1} + {\delta }_{12}{m}_{2}{\ddot{y}}_{2} + {y}_{1} = {\delta }_{12}f\left( t\right) , \tag{4.97}
\]
\[
{\delta }_{21}{m}_{1}{\ddot{y}}_{1} + {\delta }_{22}{m}_{2}{\ddot{y}}_{2} + {y}_{2} = {\delta }_{22}f\left( t\right) .
\]
将稳态解
\[
{y}_{1}\left( t\right) = {Y}_{1}\cos {\omega t},\;{y}_{2}\left( t\right) = {Y}_{2}\cos {\omega t},
\]
代入方程(4.97)得到
\[
\left( {1 - {\omega }^{2}{\delta }_{11}{m}_{1}}\right) {Y}_{1} - {\omega }^{2}{\delta }_{12}{m}_{2}{Y}_{2} = {\delta }_{12}{F}_{0},
\]
\[
-{\omega }^{2}{\delta }_{21}{m}_{1}{Y}_{1} + \left( {1 - {\omega }^{2}{\delta }_{22}{m}_{2}}\right) {Y}_{2} = {\delta }_{22}{F}_{0}. \tag{4.98}
\]
解为
\[
{Y}_{1} = \frac{{\delta }_{12}}{1 - \left( {{\delta }_{11}{m}_{1} + {\delta }_{22}{m}_{2}}\right) {\omega }^{2} + {m}_{1}{m}_{2}\left( {{\delta }_{11}{\delta }_{22} - {\delta }_{12}^{2}}\right) {\omega }^{4}}{F}_{0},
\]
\[
{Y}_{2} = \frac{{\delta }_{22} - {m}_{1}{\omega }^{2}\left( {{\delta }_{11}{\delta }_{22} - {\delta }_{12}^{2}}\right) }{1 - \left( {{\delta }_{11}{m}_{1} + {\delta }_{22}{m}_{2}}\right) {\omega }^{2} + {m}_{1}{m}_{2}\left( {{\delta }_{11}{\delta }_{22} - {\delta }_{12}^{2}}\right) {\omega }^{4}}{F}_{0}. \tag{4.99}
\]
分母可识别为特征多项式(4.86)。
图4.21
振幅\({Y}_{1}\)和\({Y}_{2}\)的绝对值如图4.21所示。当激励频率等于任一固有频率时，振幅无限增大。系统存在两个共振，在频率响应曲线上表现为峰值。
当满足以下条件时，\({Y}_{2} = 0\)发生反共振
\[
{\omega }^{2} = {\omega }_{a}^{2} = \frac{{\delta }_{22}}{{m}_{1}\left( {{\delta }_{11}{\delta }_{22} - {\delta }_{12}^{2}}\right) }. \tag{4.100}
\]
若已知受迫响应的振幅，则可计算惯性力的振幅，从而得到作用于梁上的动态力振幅(图\({4.20}, b\))为
\[
{\Phi }_{1} = {m}_{1}{\omega }^{2}{Y}_{1},\;{\Phi }_{2} = {m}_{2}{\omega }^{2}{Y}_{2} + {F}_{0}. \tag{4.101}
\]
随后可绘制动态弯矩图(图\({4.20}, c\))，并计算由谐波力产生的动态应力。
例4.7
图4.22所示的无质量梁\(a\)，其直径为\(d = {40}\mathrm{\;{mm}},\ell = 1\mathrm{\;m}\)、\(E = {210}\mathrm{{GPa}}\)和\(m = {50}\mathrm{\;{kg}}\)。a) 计算固有频率；b) 确定由振幅为\({F}_{0} = {20}\mathrm{\;N}\)、频率为\({0.179}\mathrm{\;{Hz}}\)的谐波力产生的受迫振动振幅；c) 绘制静弯矩图并确定最大静应力；d) 绘制动态弯矩图并计算最大动态应力的振幅。
解:柔度系数为
\[
{\delta }_{11} = {\ell }^{3}/{6EI},{\delta }_{12} = {\delta }_{21} = - {\ell }^{3}/{8EI},{\delta }_{22} = 5{\ell }^{3}/{24EI}.
\]
记
\[
\lambda = \frac{24EI}{{\omega }^{2}m{\ell }^{3}},
\]
频率方程为
\[
\left| \begin{matrix} \lambda - 8 & 3 \\ 6 & \lambda - 5 \end{matrix}\right| = 0,\;{\lambda }^{2} - {13\lambda } + {22} = 0,
\]
其解为
\[
{\lambda }_{1} = {11},\;{\lambda }_{2} = 2.
\]
固有频率为
\[
{\omega }_{1} = {1.477}\sqrt{{EI}/m{\ell }^{3}} = {1.073}\mathrm{{rad}}/\mathrm{{sec}}\text{,}
\]
\[
{\omega }_{2} = {3.464}\sqrt{{EI}/m{\ell }^{3}} = {2.516}\mathrm{{rad}}/\mathrm{{sec}}.
\]
图4.22
对于给定的数值数据，激励频率对应于\(\lambda = {10}\)。由式(4.99)可得振动幅值
\[
{Y}_{1} = - \frac{3\lambda }{\left( {\lambda - {11}}\right) \left( {\lambda - 2}\right) }\frac{{F}_{0}}{m{\omega }^{2}} = {0.118}\mathrm{\;{mm}},
\]
\[
{Y}_{2} = - \frac{{22} - {5\lambda }}{\left( {\lambda - {11}}\right) \left( {\lambda - 2}\right) }\frac{{F}_{0}}{m{\omega }^{2}} = - {1.105}\mathrm{\;{mm}}.
\]
对于图4.22所示的静载荷\(b\)，其静弯矩图如图4.22\(c\)所示。最大弯矩为\({368}\mathrm{{Nm}}\)，最大静应力为\({\sigma }_{st} = {58.5}\mathrm{\;N}/{\mathrm{{mm}}}^{2}\)。
动态力的幅值(4.101)为
\[
{\Phi }_{1} = {2m}{\omega }^{2}{Y}_{1} = - \frac{6\lambda }{\left( {\lambda - {11}}\right) \left( {\lambda - 2}\right) }{F}_{0} = {150}\mathrm{\;N},
\]
\[
{\Phi }_{2} = m{\omega }^{2}{Y}_{2} + {F}_{0} = \left\lbrack {1 - \frac{{22} - {5\lambda }}{\left( {\lambda - {11}}\right) \left( {\lambda - 2}\right) }}\right\rbrack {F}_{0} = - {50}\mathrm{\;N}.
\]
对于图4.22所示的动载荷\(d\)，其动弯矩图如图4.22e所示。最大弯矩为\({87.5}\mathrm{\;{Nm}}\)，最大动应力为\({\sigma }_{d} = {14}\mathrm{\;N}/{\mathrm{{mm}}}^{2}\)。