Section 4.4 kinematic analysis of simple rigid body systems
Subsection 4.4.1 simply supported beam
Perform the kinematic analysis of the following beam of length equal to l\text{.} By โreadingโ the applyed constraints the following equations can be written
\begin{gather*}
\vec{u}_A = \vec0\,,\\
\vec{u}_B \cdot \vec{n}_B = 0\,,
\end{gather*}
where \vec{n}_B = \transp{[0 \; 1]}\text{.} Previous equations can be also expressed by the following 3 scalar equations (m=3):
\begin{gather*}
u_{1A} = 0\,,\\
u_{2A} = 0\,,\\
u_{2B} = 0\,.
\end{gather*}
For the only rigid body in question, choose point A as the pole, therefore the 3 Lagrangian parameters (n = 3 ) that can be used for the kinematic description of the system are: Using the equation (4.1.5) the constraint conditions can be expressed with respect to the chosen Lagrangian coordinates
\begin{gather*}
u_{1A} = 0\,,\\
u_{2A} = 0\,,\\
u_{2A} + \varphi_{A} l = 0\,.
\end{gather*}
Conditions which give
\begin{equation*}
\underbrace{\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 1 & l
\end{array}\right]}_{\mat{A}}
\underbrace{\left[\begin{array}{c} u_{1A} \\ u_{2A} \\ \varphi_{A} \end{array}\right]}_{\vec{q}} =
\underbrace{\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]}_{\vec{d}}.
\end{equation*}
Before proceeding with the solution of the system, it is possible to classify the system by calculating the rank of the kinematic matrix using the following MATLABยฎ instructions.
\begin{equation*}
u_{2B} = l/10\,.
\end{equation*}
Therefore on the system of equations to be analyzed, the only effect is constituted by the modification of the vector of the known terms and there is no effect on the classification of the system which always remains kinematically determined (the matrix A has not changed). Therefore the system of equations becomes
\begin{equation*}
\underbrace{\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 1 & l
\end{array}\right]}_{\mat{A}}
\underbrace{\left[\begin{array}{c} u_{1A} \\ u_{2A} \\ \varphi_{A} \end{array}\right]}_{\vec{q}} =
\underbrace{\left[\begin{array}{c} 0 \\ 0 \\ l/10 \end{array}\right]}_{\vec{d}}.
\end{equation*}
The solution of the linear system provides the new values assumed by the Lagrangian parameters u_{1A} \text{,} u_{2A} and \varphi_{A} \text{.}
Subsection 4.4.2 L shaped beam with 3 roller supports
The assigned system is shown in the figure below. The proposed scheme does not present any particular novelty with respect to the previous scheme and the following MATLABยฎ instructions allow to obtain the desired solution.Subsection 4.4.3 system with three hinges
Perform the kinematic analysis of the following system. The โreadingโ of the assigned constraint conditions gives
\begin{gather*}
\vec{u}_A = \vec{0}\,,\\
\vec{u}_B = \vec{u}_{B'}\,,\\
\vec{u}_C = \vec0\,,
\end{gather*}
which can be also expressed by the following scalar equations (m=6):
\begin{gather*}
u_{1A} = 0\,,\\
u_{2A} = 0\,,\\
u_{1B} = u_{1B'}\,,\\
u_{2B} = u_{2B'}\,,\\
u_{1C} = 0\,,\\
u_{2C} = 0\,.
\end{gather*}
The rigid bodies under examination are two: for the body AB the point A is chosen as the pole and for the second body the point B' is chosen. Therefore the 6 Lagrangian parameters (n = 6 ) adopted for the kinematic description of the system are: Using the equation (4.1.5) the constraint conditions can be expressed with respect to the chosen Lagrangian coordinates, in particular:
\begin{gather*}
u_{1A} = 0\,,\\
u_{2A} = 0\,,\\
u_{1A} = u_{1B'}\,,\\
u_{2A} + \varphi_{A} 2l = u_{2B'}\,,\\
u_{1B'} + \varphi_{B'} l = 0\,,\\
u_{2B'} + \varphi_{B'} 2l = 0\,.
\end{gather*}
or, in matrix format,
\begin{equation*}
\underbrace{\left[\begin{array}{cccccc}
1 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & -1 & 0 & 0\\
0 & 1 & 2l & 0 & -1 & 0\\
0 & 0 & 0 & 1 & 0 & l\\
0 & 0 & 0 & 0 & 1 & 2l
\end{array}\right]}_{\mat{A}}
\underbrace{\left[\begin{array}{c} u_{1A} \\ u_{2A} \\ \varphi_{A} \\ u_{1B'} \\ u_{2B'} \\ \varphi_{B'} \end{array}\right]}_{\vec{q}} =
\underbrace{\left[\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right]}_{\vec{d}}.
\end{equation*}
In order to classify the system, the rank of the kinematic matrix is calculated.
\begin{equation*}
\underbrace{\left[\begin{array}{cccccc}
1 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & -1 & 0 & 0\\
0 & 1 & 2l & 0 & -1 & 0\\
0 & 0 & 0 & 1 & 0 & l\\
0 & 0 & 0 & 0 & 1 & 2l
\end{array}\right]}_{\mat{A}}
\underbrace{\left[\begin{array}{c} u_{1A} \\ u_{2A} \\ \varphi_{A} \\ u_{1B'} \\ u_{2B'} \\ \varphi_{B'} \end{array}\right]}_{\vec{q}} =
\underbrace{\left[\begin{array}{c} l/4 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right]}_{\vec{d}}\,.
\end{equation*}
The solution can be calculated using the following MATLABยฎ instructions.
Subsection 4.4.4 mechanism example
Perform the kinematic analysis of the following system. The constraint conditions assigned can be expressed as follows
\begin{gather*}
\vec{u}_A = \vec{0}\,,\\
\vec{u}_C = \vec{u}_{C'}\,,
\end{gather*}
The corresponding set of scalar equations is (m=5):
\begin{gather*}
u_{1A} = 0\,,\\
u_{2A} = 0\,,\\
u_{1C} = u_{1C'}\,,\\
u_{2C} = u_{2C'}\,,\\
u_{2E} = 0\,.
\end{gather*}
For the two rigid bodies under examination, points B and D are chosen as poles, thus the kinematics is described through the following 6 Lagrangian parameters (n=6): Always using the plane motion model represented by the equation (4.1.5) the constraint conditions can be expressed with respect to the chosen Lagrangian coordinates obtaining the following result,
\begin{gather*}
u_{1B} + \varphi_{B} 2l = 0\,,\\
u_{2B} = 0\,,\\
u_{1B} = u_{1D}\,,\\
u_{2B} + \varphi_{B} 2l = u_{2D}-\varphi_{D} l\,,\\
u_{2D} = 0\,.
\end{gather*}
Using a matrix format, the system of constraint equations can be written as
\begin{equation*}
\underbrace{\left[\begin{array}{cccccc}
1 & 0 & 2l & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & -1 & 0 & 0\\
0 & 1 & 2l & 0 & -1 & l\\
0 & 0 & 0 & 0 & 1 & 0
\end{array}\right]}_{\mat{A}}
\underbrace{\left[\begin{array}{c} u_{1B} \\ u_{2B} \\ \varphi_{B} \\ u_{1D} \\ u_{2D} \\ \varphi_{D} \end{array}\right]}_{\vec{q}} =
\underbrace{\left[\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right]}_{\vec{d}}.
\end{equation*}
It is possible to proceed with the calculation of the rank of the matrix as follows.
\begin{equation*}
\left[\begin{array}{ccccc}
1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
1 & 0 & -1 & 0 & 0\\
0 & 1 & 0 & -1 & l\\
0 & 0 & 0 & 1 & 0
\end{array}\right]
\left[\begin{array}{c} u_{1B} \\ u_{2B} \\ u_{1D} \\ u_{2D} \\ \varphi_{D} \end{array}\right] =
\left[\begin{array}{c} -2l\,\varphi_{B} \\ 0 \\ 0 \\ -2l\,\varphi_{B} \\ 0 \end{array}\right].
\end{equation*}
The following MATLABยฎ instructions allow to calculate the solution we are looking for.
\begin{gather*}
u_{1B} = -\varphi_{B} 2l \,,\\
u_{2B} = 0\,,\\
u_{1D} = -\varphi_{B} 2l\,,\\
u_{2D} = 0\,,\\
\varphi_{D} = - 2\varphi_{B} \,.
\end{gather*}
One of these can be represented as follows