5.3.2 Transfer matrix for a beam element with axial force

Differentiating the solution $w\left(x\right)$ of Eq. (5.46) to a homogeneous differential equation for the compressive axial force (Sign Convention 1)

$$w\left(x\right) = w_{0} - x \varphi_{0} + \left[\sin\left(\frac{\... ... \left(\frac{l}{\nu}\right)^{3} \frac{\left. Q_{z}\right\vert _{\circ}}{EI_{y}}$$

$$+ \left[\cos\left(\frac{\nu}{l}x\right) - 1\right] \left( \frac{l}{\nu}\right)^{2}\frac{\left. M_{y}\right\vert _{\circ}}{EI_{y}}$$ (5.50)

we get the expression for rotation:

$$\varphi_{x} = - w^{\prime}_{x}\left( x\right) = 0\cdot w_{0} - 1 ... ... \left(\frac{l}{\nu}\right)^{2} \frac{\left. Q_{z}\right\vert _{\circ}}{EI_{y}}$$

$$+ \left[\sin\left(\frac{\nu}{l}x\right)\right] \left(\frac{l}{\nu}\right) \frac{\left. M_{y}\right\vert _{\circ}}{EI_{y}}$$ (5.51)

Next we take the second derivative to find the bending moment $M_{y}\left(x\right)$:

$$M_{y}\left(x\right) = - EI_{y}w^{\prime\prime}_{x}\left(x\right) ... ...}{l}x \right)\right] \left(\frac{l}{\nu}\right)\left. Q_{z}\right\vert _{\circ}$$

$$+ \left[\cos\left(\frac{\nu}{l}x\right)\right] \left.M_{y}\right\vert _{\circ}$$ (5.52)

Taking a derivative from the bending moment $M_{y}\left(x\right)$, we obtain the shear force $Q_{z}\left(x\right)$:

$$Q_{z}\left(x\right) = - EI_{y}w^{\prime\prime\prime}_{x}\left(x\r... ...\left[\cos\left( \frac{\nu}{l}x \right)\right] \left. Q_{z}\right\vert _{\circ}$$

$$- \left[\sin\left(\frac{\nu}{l}x\right)\right]\frac{l}{\nu} \left.M_{y}\right\vert _{\circ}$$ (5.53)

For the constant normal force $N_{x}\left(x\right)$ and axial displacement $u\left(x\right)$ we take (Sign Convention 1)

$$u\left(x\right) = \left. u\right\vert _{\circ} + \frac{x}{NA}\left. N_{x}\right\vert _{\circ}$$ (5.54)

$$N_{x}\left(x\right) = \left. N_{x}\right\vert _{\circ}$$ (5.55)

Now consider the vectors

$$\mathbf{Z^{\left(N\right)}_{x}} = \mathbf{\left[ \begin{array}{c}... ...vert _{\circ} \\ \left.M_{y}\right\vert _{\circ} \end{array}\right]_{A}} \quad$$ (5.56)

the elements of which can be found from Eqs. (5.50)-(5.55).

These equations can now be written in matrix form:

$$\mathbf{Z^{\left(N\right)}_{x}} = \mathbf{U^{\left(N-1\right)}\cdot Z^{\left(N\right)}_{A}} % +$$ (5.57)

where $\mathbf{U^{\left(N-1\right)}}$ is the transfer matrix for the compressive axial force (Sign Convention 1).

$$\mathbf{U^{\left(N-1\right)}} = \left[ \begin{array}{cccccc} 1 & ... ...ight)\frac{l}{\nu} & \cos\left( \frac{\nu x}{l}\right) \end{array}\right] \quad$$ (5.58)

For the tensile axial force (Sign Convention 1), the transfer equations are given by

$$\mathbf{Z^{\left(N\right)}_{x}} = \mathbf{U^{\left(N+1\right)}\cdot Z^{\left(N\right)}_{A}} % +$$ (5.59)

where $\mathbf{U^{\left(N+1\right)}}$ is the transfer matrix:

$$\mathbf{U^{\left(N+1\right)}} = \left[ \begin{array}{cccccc} 1 & ... ...\frac{l}{\nu} & \mathrm{ch}\left(\nu\frac{x}{l}\right) \end{array}\right] \quad$$ (5.60)

For the compressive axial force (Sign Convention 2), the transfer equations are given by

$$\mathbf{Z^{\left(N\right)}_{x}} = \mathbf{U^{\left(N-2\right)}\cdot Z^{\left(N\right)}_{A}} % +$$ (5.61)

where $\mathbf{U^{\left(N-2\right)}}$ is the transfer matrix:

$$\mathbf{U^{\left(N-2\right)}} = \left[ \begin{array}{cccccc} 1 & ... ...ght)\frac{l}{\nu} & -\cos\left( \frac{\nu x}{l}\right) \end{array}\right] \quad$$ (5.62)

This transfer matrix has been obtained from Eqs. (5.58) and (5.60) by multiplying the 4th, 5th and 6th columns by $-1$ (see Sign Convention, Fig. 1.2).

For the tensile axial force (Sign Convention 2), the transfer equations are given by

$$\mathbf{Z^{\left(N\right)}_{x}} = \mathbf{U^{\left(N+2\right)}\cdot Z^{\left(N\right)}_{A}} % +$$ (5.63)

where $\mathbf{U^{\left(N+2\right)}}$ is the transfer matrix:

$$\mathbf{U^{\left(N+2\right)}} = \left[ \begin{array}{cccccc} 1 & ... ...rac{l}{\nu} & - \mathrm{ch}\left(\nu\frac{x}{l}\right) \end{array}\right] \quad$$ (5.64)

The transformation matrices of Eqs. (5.62) and (5.64) can be represented with the GNU Octave function ylfmII.m (p. ).