1.2.3 The basic system of equations for a frame element

We consider a frame element with loads applied between the ends (Fig. 1.11).

Figure 1.11: Frame element (Sign Convention 2)
\includegraphics[width=90mm]{./joonised/talaREM.eps}

Here, we rewrite Eq. (1.40) in the form

$\displaystyle {\mathbf{I_{6\times 6}\cdot Z_{L}}} - {\mathbf{U_{x=l}}}{\mathbf{Z_{A}}} = {\mathbf{\stackrel{\rm\circ}{Z_{x=l}}}}$     (1.47)

or
$\displaystyle \mathbf{\widehat{IU}_{6\times 12}}\cdot\mathbf{\widehat{Z}} = \mathbf{\stackrel{\rm\circ}{Z}}$     (1.48)

where $\mathbf{\widehat{IU}}$ is the enlarged transfer matrix $\mathbf{U}$:

$\displaystyle \mathbf{\widehat{IU}} =\mathbf{I} - \mathbf{U},$     (1.49)
$\mathbf{\widehat{Z}}$ is the vector of displacements and forces (DaFs) of the element:

$\displaystyle \mathbf{\widehat{Z}} =
\left[\begin{array}{c}
\mathbf{Z_{L}} \\
\mathbf{Z_{A}}
\end{array}\right]$     (1.50)

$\mathbf{Z_{L}}$, $\mathbf{Z_{A}}$ are the displacement and force vectors at the end and at the beginning of the element, respectively,
$\mathbf{I}$ is a unit matrix (for the frame element $6 \times 6$ ),
$\mathbf{U}$ is the transfer matrix of Eq. (1.45) at $x=l$ (Sign Convention 2),
$\mathbf{\stackrel{\rm\circ}{Z}}$ is the loading vector of Eq. (1.46) at (Sign Conventions 1 and 2).

Note that Eq. (1.42) is a second-order equation and requires two boundary conditions, whereas Eq. (1.14) is a fourth-order equation requiring four boundary conditions. These (2+4=6) boundary conditions should be well posed. They are divided into static and kinematic boundary conditions.

The fixed-end displacements and rotation at the end and at the beginning of the element are respectively

$\displaystyle \mathbf{v_{L}} =
\left[\begin{array}{c}
u_{L} \\
w_{L} \\
\varp...
...A}} =
\left[\begin{array}{c}
u_{A} \\
w_{A} \\
\varphi_{A}
\end{array}\right]$     (1.51)

Figure 1.12: Internal reactions1.2 1.3(contact forces1.4) and displacements of joints
\begin{picture}(158,160)
%\{ tiny
{\small\thinlines
% drawpath\{4.0\}\{146.0\...
...nter}\end{minipage}}
% end\{picture\}
%\}
%%%%%%%
} % small % tiny
\end{picture}

The fixed-end forces and the moments $\mathbf{s_{L}}$, $\mathbf{s_{A}}$ at the end and at the beginning of the element, respectively, are shown in Eq. (1.52).

At unloaded joints (contacts) of elements, the sum of the fixed-end forces and moments is zero.

$\displaystyle \mathbf{s_{L}} =
\left[\begin{array}{c}
N_{L} \\
Q_{L} \\
M_{L}...
...bf{s_{A}} =
\left[\begin{array}{c}
N_{A} \\
Q_{A} \\
M_{A}
\end{array}\right]$     (1.52)

Sometimes the fixed-end forces and moments at joints are called internal reactions1.8 1.9 or joint contact forces1.10 [GN12]. We shall call them contact forces and contact moments.

A boundary value problem is a differential equation together with a set of boundary conditions. Solutions to the differential equations of a frame element are given with the basic system of equations (1.47). The boundary value problem should be well posed. The kinematic and static boundary conditions [KW90] of a frame element are shown in Figs. (1.10) and (1.12).



andres
2014-09-09