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力学书籍/力学/Dynamics of Multibody Systems (Shabana A.A.) (Z-Library)/auto/Dynamics of Multibody Systems (Shabana A.A.) (Z-Library).md
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@ -6597,11 +6597,13 @@ $$
$$
Determine the global position of the tip point and the center of mass of the beam $C$ .
确定梁 $C$ 的尖端点和质心的全局位置。
Solution At this instant of time, the transformation matrix A of Eq. 5 is given by
$$
\mathbf{A}={\begin{bmatrix}{\cos\theta^{i}}&{-\sin\theta^{i}}\\ {\sin\theta^{i}}&{\cos\theta^{i}}\end{bmatrix}}={\left[\begin{array}{l l}{0.8660}&{-0.500{\big]}}\\ {0.500}&{0.8660}\end{array}\right]}
\mathbf{A}={\begin{bmatrix}{\cos\theta}&{-\sin\theta}\\ {\sin\theta}&{\cos\theta}\end{bmatrix}}={\left[\begin{array}{l l}{0.8660}&{-0.500{\big]}}\\ {0.500}&{0.8660}\end{array}\right]}
$$
The global position of point $A$ can then be written as
@ -6616,10 +6618,10 @@ $$
\bar{\mathbf{u}}_{o}=\left[\begin{array}{l}{l}\\ {0}\end{array}\right]=\left[\begin{array}{l}{0.5}\\ {0}\end{array}\right]
$$
The vector $\bar{\ensuremath{\mathbf{u}}}_{f}$ is the elastic deformation of point $A$ and can be evaluated, since $\xi=1$ at point $A$ , as
The vector $\bar{{\mathbf{u}}}_{f}$ is the elastic deformation of point $A$ and can be evaluated, since $\xi=1$ at point $A$ , as
$$
\bar{\mathbf{u}}_{f}=\left[\!\!\begin{array}{c c}{\xi}&{0}\\ {0}&{3(\xi)^{2}-2(\xi)^{3}\!\!}\end{array}\!\!\right]\left[\!\!\begin{array}{c}{q_{f1}}\\ {q_{f2}}\end{array}\!\!\right]=\left[\!\!\begin{array}{c c}{1}&{0}\\ {0}&{1}\end{array}\!\!\right]\left[\!\!\begin{array}{c}{0.001}\\ {0.01}\end{array}\!\!\right]=\left[\!\!\begin{array}{c}{0.001}\\ {0.01}\end{array}\!\!\right]
\bar{\mathbf{u}}_{f}=\left[\begin{array}{c c}{\xi}&{0}\\ {0}&{3(\xi)^{2}-2(\xi)^{3}}\end{array}\right]\left[\begin{array}{c}{q_{f1}}\\ {q_{f2}}\end{array}\right]=\left[\begin{array}{c c}{1}&{0}\\ {0}&{1}\end{array}\right]\left[\begin{array}{c}{0.001}\\ {0.01}\end{array}\right]=\left[\begin{array}{c}{0.001}\\ {0.01}\end{array}\right]
$$
and accordingly
@ -6628,36 +6630,55 @@ $$
\bar{\mathbf{u}}_{A}=\left[\begin{array}{l}{0.501}\\ {0.01}\end{array}\right]
$$
The position vector $\mathbf{r}_{A}$ can be then written as $\mathbf{r}_{A}={\left[\begin{array}{l}{1.0}\\ {0.5}\end{array}\right]}+{\left[\begin{array}{l l}{0.8660}&{-0.500}\\ {0.500}&{0.8660}\end{array}\right]}{\left[\begin{array}{l}{0.501}\\ {0.01}\end{array}\right]}={\left[\begin{array}{l}{1.4289}\\ {0.75916}\end{array}\right]}$ At point $C,\xi=0.5$ and $\bar{\mathbf{u}}_{o}=[(l/2)\,0]^{\mathrm{T}}=[0.25\,0]^{\mathrm{T}}.$ . The deformation vector
$\bar{\ensuremath{\mathbf{u}}}_{f}$ at $C$ is given by $\bar{\mathbf{u}}_{f}=\left[\begin{array}{l l}{0.5}&{0}\\ {0}&{3(0.5)^{2}-2(0.5)^{3}}\end{array}\right]\left[\begin{array}{l}{0.001}\\ {0.01}\end{array}\right]=\left[\begin{array}{l}{0.0005}\\ {0.005}\end{array}\right]$
and the local position of point $C$ is $\bar{\mathbf{u}}_{C}=\bar{\mathbf{u}}_{o}+\bar{\mathbf{u}}_{f}=\left[\begin{array}{c}{0.25}\\ {0}\end{array}\right]+\left[\begin{array}{c}{0.0005}\\ {0.005}\end{array}\right]=\left[\begin{array}{c}{0.2505}\\ {0.005}\end{array}\right]$
The global position $\mathbf{r}_{C}$ can then be determined as $\mathbf{r}_{C}={\left[\begin{array}{l}{1.0}\\ {0.5}\end{array}\right]}+{\left[\begin{array}{l l}{0.8660}&{-0.500}\\ {0.500}&{0.8660}\end{array}\right]}{\left[\begin{array}{l}{0.2505}\\ {0.005}\end{array}\right]}={\left[\begin{array}{l}{1.2144}\\ {0.62958}\end{array}\right]}$
The position vector $\mathbf{r}_{A}$ can be then written as
$$
\mathbf{r}_{A}={\left[\begin{array}{l}{1.0}\\ {0.5}\end{array}\right]}+{\left[\begin{array}{l l}{0.8660}&{-0.500}\\ {0.500}&{0.8660}\end{array}\right]}{\left[\begin{array}{l}{0.501}\\ {0.01}\end{array}\right]}={\left[\begin{array}{l}{1.4289}\\ {0.75916}\end{array}\right]}
$$
At point $C,\xi=0.5$ and $\bar{\mathbf{u}}_{o}=[(l/2)\,0]^{\mathrm{T}}=[0.25\,0]^{\mathrm{T}}.$ The deformation vector $\bar{{\mathbf{u}}}_{f}$ at $C$ is given by
$$
\bar{\mathbf{u}}_{f}=\left[\begin{array}{l l}{0.5}&{0}\\ {0}&{3(0.5)^{2}-2(0.5)^{3}}\end{array}\right]\left[\begin{array}{l}{0.001}\\ {0.01}\end{array}\right]=\left[\begin{array}{l}{0.0005}\\ {0.005}\end{array}\right]
$$
and the local position of point $C$ is
$$
\bar{\mathbf{u}}_{C}=\bar{\mathbf{u}}_{o}+\bar{\mathbf{u}}_{f}=\left[\begin{array}{c}{0.25}\\ {0}\end{array}\right]+\left[\begin{array}{c}{0.0005}\\ {0.005}\end{array}\right]=\left[\begin{array}{c}{0.2505}\\ {0.005}\end{array}\right]
$$
The global position $\mathbf{r}_{C}$ can then be determined as
$$
\mathbf{r}_{C}={\left[\begin{array}{l}{1.0}\\ {0.5}\end{array}\right]}+{\left[\begin{array}{l l}{0.8660}&{-0.500}\\ {0.500}&{0.8660}\end{array}\right]}{\left[\begin{array}{l}{0.2505}\\ {0.005}\end{array}\right]}={\left[\begin{array}{l}{1.2144}\\ {0.62958}\end{array}\right]}
$$
Velocity Equations Differentiating Eq. 7 with respect to time yields
速度方程
对公式 7 关于时间求导，得到：
Velocity Equations Differentiating Eq. 7 with respect to time yields
$$
\dot{\mathbf{r}}_{P}^{i}=\dot{\mathbf{R}}^{i}+\dot{\mathbf{A}}^{i}\bar{\mathbf{u}}^{i}+\mathbf{A}^{i}\dot{\bar{\mathbf{u}}}^{i}
$$
where $(\,\cdot\,)$ denotes differentiation with respect to time. Using Eq. 6, one can write $\dot{\mathbf{u}}^{i}$ in terms of the time derivatives of the elastic coordinates of body $i$ as
where $(\,\dot\,)$ denotes differentiation with respect to time. Using Eq. 6, one can write $\dot{\mathbf{u}}^{i}$ in terms of the time derivatives of the elastic coordinates of body $i$ as
其中 $(\,\dot\,)$ 表示对时间求导。利用公式 6，可以将 $\dot{\mathbf{u}}^{i}$ 表示为物体 $i$ 的弹性坐标的时间导数，即
$$
\dot{\bar{\mathbf{u}}}^{i}=\mathbf{S}^{i}\dot{\mathbf{q}}_{f}^{i}
$$
where $\mathbf{S}^{i}=\mathbf{S}^{i}(x_{1}^{i},x_{2}^{i},x_{3}^{i})$ is the body shape matrix, and $\dot{\mathbf{q}}_{f}^{i}$ is the vector of elastic generalized velocities of body $i.$ . Substituting Eq. 14 into Eq. 13 yields
其中，$\mathbf{S}^{i}=\mathbf{S}^{i}(x_{1}^{i},x_{2}^{i},x_{3}^{i})$ 为人体形状矩阵，$\dot{\mathbf{q}}_{f}^{i}$ 为物体 $i$ 的弹性广义速度矢量。将公式 14 代入公式 13 得到
$$
\dot{\mathbf{r}}_{P}^{i}=\dot{\mathbf{R}}^{i}+\dot{\mathbf{A}}^{i}\bar{\mathbf{u}}^{i}+\mathbf{A}^{i}\mathbf{S}^{i}\dot{\mathbf{q}}_{f}^{i}
$$
where the equation $\dot{\bar{\mathbf{u}}}_{o}^{i}=\mathbf{0}$ is used. To isolate velocity terms, the central term on the right-hand side of Eq. 15 can, in general, be written as
其中使用方程 $\dot{\bar{\mathbf{u}}}_{o}^{i}=\mathbf{0}$。为了分离速度项，方程 15 右侧的中心项，通常可以写成
$$
\dot{\mathbf{A}}^{i}\bar{\mathbf{u}}^{i}=\mathbf{B}^{i}\dot{\theta}^{i}
$$
where ${\dot{\Theta}}^{i}$ is the vector whose elements $\dot{\theta}_{k}^{i}$ are the time derivatives of the rotational coordinates of the body reference and $\mathbf{B}^{i}=\mathbf{B}^{i}(\theta^{i},\mathbf{q}_{f}^{i})$ is defined as
where ${\dot{\theta}}^{i}$ is the vector whose elements $\dot{\theta}_{k}^{i}$ are the time derivatives of the rotational coordinates of the body reference and $\mathbf{B}^{i}=\mathbf{B}^{i}(\theta^{i},\mathbf{q}_{f}^{i})$ is defined as
其中 ${\dot{\theta}}^{i}$ 是一个向量，其元素 $\dot{\theta}_{k}^{i}$ 是刚体参考系旋转坐标的时间导数，而 $\mathbf{B}^{i}=\mathbf{B}^{i}(\theta^{i},\mathbf{q}_{f}^{i})$ 定义如下：
$$
\mathbf{B}^{i}=\left[{\frac{\partial}{\partial\theta_{1}^{i}}}(\mathbf{A}^{i}{\bar{\mathbf{u}}}^{i})\cdot\cdot\cdot{\frac{\partial}{\partial\theta_{n_{r}}^{i}}}(\mathbf{A}^{i}{\bar{\mathbf{u}}}^{i})\right]