1.1 Fundamental relations of a frame element

The motion of a frame element is composed of rigid body translation, rigid body rotation, and strain producing deformations (see [PW94] p. 171). Let us now apply a local coordinate system (x, z) to the frame element shown in Fig. 1.1. (X, Z) is the global coordinate system. We consider the right-handed coordinates shown in Fig. 1.3 and Sign Convention 2 in Fig. 1.2.

Figure 1.1: Rigid body displacements and rotation

We determine first the rigid body translation and rigid body rotation at the ends of the frame element shown in Fig.1.1:

$$\left[\begin{array}{c} u_{L} \\ w_{L} \\ \varphi_{L} \end{array... ...ight] \left[\begin{array}{c} u_{A} \\ w_{A} \\ \varphi_{A} \end{array}\right]$$ (1.1)

where
$u_{A}$, $u_{L}$ - axial displacements at the beginning and at the end of the element, respectively;
$w_{A}$, $w_{L}$ - transverse displacements at the beginning and at the end of the element, respectively;
$\varphi_{A}$, $\varphi_{L}$ - rotation at the beginning and at the end of the element, respectively.

The rigid body displacements and rotations of the element shown in Fig. 1.1 can be large.

The basic coordinate system (x*, z*) in Fig. 1.5 rotates and translates together with the element [Suz00], [YSK08]. With respect to this basic coordinate system, only local strain producing deformations remain (see [PW94] p. 171).

Figure 1.2: Sign Convention

Figure 1.3: Right-handed and left-handed coordinates

The finger points in the positive direction of the respective axis. Each arrow is labeled by an axis letter in the sequence x, y, z or 1, 2, 3.

Figure 1.4: Moment diagrams for unit force Q and unit moment M

The deformations (internal displacements [KHMW10], [KHMW05]) $\Delta w$ and $\Delta \Theta$ in Fig. 1.5 can be determined by the unit load method using the material law relationships [PW94].

We proceed decomposing the deformations $\Delta w$ and $\Delta \Theta$ into the effects of the shear force and bending moment $M$:

$$\left[\begin{array}{c} \Delta w \\ \Delta \Theta \end{array}\rig... ...{Q} + \Delta w_{M} \\ \Delta \Theta_{Q} + \Delta \Theta_{M} \end{array}\right]$$ (1.2)

Figure 1.5: Beam element deformations

Figure 1.6: Moment diagrams for force Q and moment M

Next, we shall consider the moment diagrams in Figs. 1.4, 1.6, and 1.7. When applying the unit load method, it follows that

$$\Delta w_{Q} = \int^{l}_{0}\frac{M_{Q}m_{1}}{EI}ds +\int^{l}_{0}\frac{Q_{Q}q_{1}}{GA_{red}}ds = \frac{l^{3}Q_{L}}{3EI} + \frac{lQ_{L}}{GA_{red}}$$

$$\Delta \Theta_{Q} \int^{l}_{0}\frac{M_{Q}m_{2}}{EI}ds = -\frac{l^{2}Q_{L}}{2EI}$$ (1.3)

where
$EI$ - bending stiffness of the element,
$GA_{red}$ - shear stiffness of the element.

Now we consider again the moment diagrams in Figs. 1.4 and 1.6, and apply the unit load method for finding $\Delta w_{M}$ and $\Delta \Theta_{M}$:

$$\Delta w_{M} \int^{l}_{0}\frac{M_{M}m_{1}}{EI}ds = -\frac{l^{2}M_{L}}{2EI}$$

$$\Delta \Theta_{M} \int^{l}_{0}\frac{M_{M}m_{2}}{EI}ds = \frac{lM_{L}}{EI}$$ (1.4)

Substituting Eqs. (1.3), (1.4) into Eq. (1.2), we have

$$\left[\begin{array}{c} \Delta w \\ \Delta \Theta \end{array}\rig... ... }} {\normalfont { \left[\begin{array}{c} Q_{L} \\ M_{L} \end{array}\right] }}$$ (1.5)

Figure 1.7: Shear force diagrams for force Q and unit force q

The axial deformation of the element

$$\Delta u_{L} = \frac{lN_{L}}{EA}$$ (1.6)

where
$N_{L}$ - axial force at the element end,
l - length of the element,
EA - axial stiffness of the element.

All deformations of the frame element arranged in matrix form:

$$\left[\begin{array}{c} \Delta u_{L} \\ \Delta w \\ \Delta \Thet... ...malfont {\left[\begin{array}{c} N_{L} \\ Q_{L} \\ M_{L} \end{array}\right] }}$$ (1.7)

Now we consider a frame element (shown in Fig. 1.8) with no loads applied between the ends.

Figure 1.8: Bending of a frame element (Sign Convention 2)

Equilibrium equations of the frame element are written as

$$\left[\begin{array}{c} N_{L} \\ Q_{L} \\ M_{L} \end{array}\righ... ...ray}\right] \left[\begin{array}{c} N_{A} \\ Q_{A} \\ M_{A} \end{array}\right]$$ (1.8)

Using Eq. (1.8), we can rewrite equations (1.7) for all deformations of the frame element in the form

$$\left[\begin{array}{c} \Delta u_{L} \\ \Delta w \\ \Delta \Thet... ...malfont {\left[\begin{array}{c} N_{A} \\ Q_{A} \\ M_{A} \end{array}\right] }}$$ (1.9)

We shall now assemble Eqs. (1.1), (1.8), (1.9) into a matrix of basic equations

$$\mathbf{Z_{L}} = \mathbf{U}\cdot\mathbf{Z_{A}}$$ (1.10)

where $\mathbf{Z_{L}}$, $\mathbf{Z_{A}}$ are the vectors of displacements and forces at the end and at the beginning of the element, respectively,

$$\mathbf{Z_{L}} = \left[\begin{array}{c} u_{L} \\ w_{L} \\ \varp... ...{A} \\ \varphi_{A} \\ \ldots \\ N_{A} \\ Q_{A} \\ M_{A} \end{array}\right]$$ (1.11)

and $\mathbf{U}$ is the transfer matrix (Sign Convention 2):

$$\mathbf{U} = {\normalsize { \left[ \begin{array}{ccccccc} 1 & 0 &... ... 0 & - 1 & 0 \\ 0 & 0 & 0 & \vdots & 0 & - l & -1 \end{array} \right] }} \quad$$ (1.12)

This matrix of the basic equations has the following structure [PW94]:

$$\mathbf{U} = \left[ \begin{array}{ccc} Rigid \;\; body & \vdots &... ...gs, & \vdots & \\ foundations, \;\; etc.) & \vdots & \end{array} \right] \quad$$ (1.13)