柱形杆的扭转和弯曲

9-1 扭转问题的位移解法

由 \((5-15)\) 式，设任意截面柱形杆的位移满足

\[ \begin{equation} \tag{9.1.1} \label{eq:torsion-displacement} \left \{ \begin{aligned} u &= - \alpha yz \\ v &= \alpha xz\\ w &= \alpha \phi(x,y) \end{aligned} \right. \end{equation} \]

其中 \(\phi(x,y)\) 为圣维南扭转函数，对应着横截面的翘曲。\(\alpha\) 是杆的单位长度扭转角。

将 \eqref{eq:torsion-displacement} 式代入以位移表示的平衡微分方程，前两个方程自动满足，第三个方程得到

\[ \begin{equation} \label{eq:phi-laplace} \tag{9.1.2} \nabla^2 \phi = 0 \end{equation} \]

通过位移解出应力分量：

\[ \begin{aligned} \sigma_x &= \lambda \theta + 2 G \frac{\partial u}{\partial x} = 0, \quad \sigma_y = \lambda \theta + 2 G \frac{\partial v}{\partial y} = 0 \\ \sigma_z &= \lambda \theta + 2 G \frac{\partial w}{\partial z} = 0, \quad \tau_{xy} = G \left( \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} \right) = 0 \\ \tau_{yz} &= \tau_{zy} = G \left( \frac{\partial w}{\partial y} + \frac{\partial v}{\partial z} \right) = \alpha G \left( \frac{\partial \phi}{\partial y} + x \right) \\ \tau_{zx} &= \tau_{xz} = G \left( \frac{\partial u}{\partial z} + \frac{\partial w}{\partial x} \right) = \alpha G \left( \frac{\partial \phi}{\partial x} - y \right) \end{aligned} \]

Remarks

• 在式 \eqref{eq:torsion-displacement} 的假设下，横截面内只有切应力分量 \(\tau_{zy}, \, \tau_{zx}\)，且与 \(z\) 无关，即在所有横截面上都相等。
• 如果翘曲函数 \(\phi(x,y) = 0\)，