5.3.3 The particular solution of a differential equation

The particular solution $w_{e}\left(x\right)$ (see Eq. (5.35)) of the non-homogeneous differential equation

$$\left[{w}^{\prime\prime}\right]^{\prime\prime} {\mp} {\vert} \fra... ...ght)/{EI_{y}} + \mathcal{M}_{x}{\delta}^{\prime}\left(t - x_{0}\right)/{EI_{y}}$$ (5.65)

can be obtained with the Cauchy^5.6 formula [Stp59]:

$$w_{e}\left(x\right) = \int_{x_{0}}^{x}K\left(x,t\right) f_{n}\left(t\right)dt$$ (5.66)

where $K\left(x,t\right)$ is the normed solution of the corresponding homogeneous differential equation. To be more precise,

$$w_{e}\left(x\right) = \int_{x_{0}}^{x}K_{4}\left(x,t\right)f_{4}\... ...3}\left(t\right)dt + \int_{x_{0}}^{x}K_{2}\left(x,t\right)f_{2}\left(t\right)dt$$ (5.67)

Using the normed fundamental set of solutions for the compressive axial force from Eq. (5.42), we obtain

$$K_{4}\left(x,t\right) = w_{4}\left(x - t\right) = - \left(\frac{l}{\nu} \right)^{3}\left[... ...\left(x - t\right)\right) - \left(\frac{\nu}{l}\left(x - t\right)\right)\right]$$ (5.68)

$$K_{3}\left(x,t\right) = w_{3}\left(x - t\right) = - \left(\frac{l}{\nu}\right)^{2}\left[\cos\left(\frac{\nu}{l}\left(x - t\right)\right) - 1\right]$$ (5.69)

$$K_{2}\left(x,t\right) = w_{2}\left(x - t\right) = {\left(x - t\right)}$$ (5.70)

and the load functions $f_{n}\left(t\right)$ are

$$f_{4}\left(t\right) = {q_{z}\left(t\right)}/{EI_{y}}, \quad f_{3}... ... \mathcal{F}_{z}/{EI_{y}}, \quad f_{2}\left(t\right) = \mathcal{M}_{x}/{EI_{y}}$$ (5.71)

For the particular solution $w_{e}\left(x\right)$ at a constant load $q_{z}$, we evaluate the following integral:

$$w_{e}\left(x\right) = - \left(\frac{l}{\nu}\right)^{3}\frac{q_{z}... ...eft(x - t\right)\right) - \left(\frac{\nu}{l}\left(x - t\right)\right)\right]dt$$ (5.72)

Let us start with the first term of Eq. (5.72):

$$\int_{x_{0}}^{x}\sin\left(\frac{\nu}{l}\left(x - t\right)\right)d... ...ac{l}{\nu}\left[1 - \cos\left(\frac{\nu}{l}\left(x - x_{0}\right)\right)\right]$$ (5.73)

For the second term of the equation (5.72) we get

$$\int_{x_{0}}^{x}\frac{\nu}{l}\left(x - t\right)dt = \frac{l}{\nu}\frac{\nu^{2}}{l^{2}}\frac{\left(x - x_{0}\right)^{2}}{2}$$ (5.74)

At the constant load $q_{z}$, the particular solution is

$$w_{e}\left(x\right) = \frac{1}{2\nu^{4}}\left[\frac{\nu^{2}\left(... ... \cos\frac{\nu}{l}\left(x - x_{0}\right)\right)\right]\frac{q_{z}l^{4}}{EI_{y}}$$ (5.75)

If we take $x - x_{0} = l$, this particular solution will be in agreement with those examined in [Krä91a], [Krä91b], and [Bor79b].

In case of the point load $\mathcal{F}_{z}$, the particular solution is

$$w_{e}\left(x\right) = -\left(\frac{l}{\nu}\right)^{2}\frac{\mathc... ..._{0}}^{x} \left[\cos\left(\frac{\nu}{l}\left(x - t\right)\right) - 1\right]dt =$$

$$= - \left[\sin\left(\frac{\nu}{l}\left(x - x_{0}\right)\right) - ... ...eft(x - x_{0}\right)\right)\right] \frac{\mathcal{F}_{z}l^{3}}{{\nu}^{3}EI_{y}}$$ (5.76)

The general solution of the non-homogeneous differential equation (5.65) is the sum of the solution of the homogeneous differential equation (5.48) (Sign Convention 2) and the particular solutions of Eqs. (5.75), (5.76), and (1.36).

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

$$\frac{1}{2\nu^{4}}\left[\frac{\nu^{2}\left(x - x_{0}\right)^{2}}{... ... \cos\frac{\nu\left(x - x_{0}\right)}{l}\right)\right]\frac{q_{z}l^{4}}{EI_{y}}$$

$$\left[\left(\frac{\nu}{l}\left(x - x_{0}\right)\right) - \sin\lef... ...eft(x - x_{0}\right)\right)\right] \frac{\mathcal{F}_{z}l^{3}}{{\nu}^{3}EI_{y}}$$ (5.77)

Let us take the derivatives of the displacement of Eq. (5.77) and apply these to the governing differential equations (5.3), (5.27), and (5.18).

We get the following beam governing equations in transfer matrix form (for the compressive axial force, Sign Convention 2):

$$\underbrace{\left[\begin{array}{c} w_{x} \\ \varphi_{x} \\ \ldo... ...\varphi_{A} \\ \ldots \\ Q_{A} \\ M_{A} \end{array}\right]}_{\mathbf{Z_{A}}}$$

$$%\hspace*{30pt}\nonumber \\ %%\normalsize {\normalsize { \underb... ...2}} \end{array} \right]}_{\mathbf{\stackrel{\rm\circ}{Z}} \hspace*{3pt} ...} }}$$

$$% {\normalsize { \underbrace{ + \left[\begin{array}{c} \left[ \fr... ...\stackrel{\rm\circ}{Z}}} }} \qquad \hspace*{6pt} %\qquad \hspace*{10pt} \quad$$ (5.78)

Employing a symbolic matrix notation, the above equations can be written as

$${\mathbf{Z_{x}}} = {\mathbf{U^{\left( C\right)}_{x}}}{\mathbf{Z_{A}}} + {\mathbf{\stackrel{\rm\circ}{Z}}}$$ (5.79)

Consider next the finding of the particular solution with the Cauchy formula of Eq. (5.66) where the normed fundamental set of solutions for the tensile axial force of Eq. (5.43) is

$$K_{4}\left(x,t\right) = w_{4}\left(x - t\right) = \left(\frac{l}{\nu}\right)^{3}\left[ \m... ...\left(x - t\right)\right) - \left(\frac{\nu}{l}\left(x - t\right)\right)\right]$$ (5.80)

$$K_{3}\left(x,t\right) = w_{3}\left(x - t\right) = \left(\frac{l}{\nu}\right)^{2}\left[ \mathrm{ch}\left(\frac{\nu}{l}\left(x - t\right)\right) - 1\right]$$ (5.81)

$$K_{2}\left(x,t\right) = w_{2}\left(x - t\right) = {\left(x - t\right)}$$ (5.82)

The load functions are

$$f_{4}\left(t\right) = {q_{z}\left(t\right)}/{EI_{y}}, \quad f_{3}... ... \mathcal{F}_{z}/{EI_{y}}, \quad f_{2}\left(t\right) = \mathcal{M}_{x}/{EI_{y}}$$ (5.83)

To obtain the particular solution $w_{e}\left(x\right)$ at the constant load $q_{z}$ (for the tensile axial force), we evaluate the following integral:

$$w_{e}\left(x\right) = \left(\frac{l}{\nu}\right)^{3}\frac{q_{z}}{... ...t(x - t\right)\right) - \left(\frac{\nu}{l}\left(x - t\right)\right)\right]dt =$$

$$- \left[1 + \frac{1}{2}\left(\frac{\nu \left(x - x_{0}\right) }{l... ...thrm{ch}\frac{\nu \left(x - x_{0}\right)}{l}\right]\frac{ql^{4}}{\nu^{4}EI_{y}}$$ (5.84)

In case of the point load $\mathcal{F}_{z}$, the particular solution is

$$w_{e}\left(x\right) = -\left(\frac{l}{\nu}\right)^{2}\frac{\mathc... ...} \left[ \mathrm{ch}\left(\frac{\nu}{l}\left(x - t\right)\right) - 1\right]dt =$$

$$= \left[\frac{\nu}{l}\left(x - x_{0}\right) - \mathrm{sh}{\,}\fra... ...}{l}\left(x - x_{0}\right) \right] \frac{\mathcal{F}_{z}l^{3}}{{\nu}^{3}EI_{y}}$$ (5.85)

The general solution of the non-homogeneous differential equation (5.65) in case of the tensile axial force is the sum of the solution of the homogeneous differential equation (5.49) (Sign Convention 2) and the particular solutions of Eqs. (5.84), (5.85), and (1.36).

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

$$- \left[1 + \frac{1}{2}\left(\frac{\nu \left(x - x_{0}\right) }{l}\... ...thrm{ch}\frac{\nu \left(x - x_{0}\right)}{l}\right]\frac{ql^{4}}{\nu^{4}EI_{y}}$$

$$+ \left[\frac{\nu}{l}\left(x - x_{0}\right) - \mathrm{sh}{\,}\frac{\nu}{l}\left(x - x_{0}\right) \right] \frac{\mathcal{F}_{z}l^{3}}{{\nu}^{3}EI_{y}}$$ (5.86)

Let us take the derivatives of the displacement of Eq. (5.86) and apply these to the governing differential equations (5.3), (5.27), and (5.18).

We get the beam governing equations (5.87) in transfer matrix form (for the tensile axial force, Sign Convention 2):

$$\underbrace{\left[\begin{array}{c} w_{x} \\ \varphi_{x} \\ \ldo... ...\varphi_{A} \\ \ldots \\ Q_{A} \\ M_{A} \end{array}\right]}_{\mathbf{Z_{A}}}$$

$$%\hspace*{30pt}\nonumber \\ %%\normalsize {\normalsize { \underb... ...}} \end{array} \right]} _{\mathbf{\stackrel{\rm\circ}{Z}} \hspace*{3pt} ...} }}$$

$$% { \normalsize { \underbrace{ + \left[\begin{array}{c} % \normal... ...\stackrel{\rm\circ}{Z}}} }} \qquad \hspace*{6pt} %\qquad \hspace*{10pt} \quad$$ (5.87)

In a symbolic matrix notation, these equations can be written as

$${\mathbf{Z_{x}}} = {\mathbf{U^{\left( T\right)}_{x}}}{\mathbf{Z_{A}}} + {\mathbf{\stackrel{\rm\circ}{Z}}}$$ (5.88)

Equations (5.78) and (5.87) should be complemented with the normal force $N$ and the displacement $u$. The axial deformation of a frame element

$$u_{x}\left( x\right) = \frac{x}{EA}\left . N\right\vert _{\circ} \qquad \mathrm{and} \qquad N\left( x\right) = \left . N\right\vert _{\circ}$$ (5.89)

where $\left . N\right\vert _{\circ}$ - normal force at the beginning of the element,
$EA$ - axial stiffness of the element.

The symbolic matrix transfer equation with the normal force and Sign Convention 2 is

$$\mathbf{Z_{x}} = \mathbf{U_{x}\cdot Z_{0}} + \mathbf{\stackrel{\rm\circ}{Z}}$$ (5.90)

where

$$\mathbf{Z_{x}} = \left[ \begin{array}{c} u \\ w \\ \varphi_{y} ... ... w \\ \varphi_{y} \\ N_{x} \\ Q_{z} \\ M_{y} \end{array} \right]_{0} \quad$$ (5.91)

and
* if the normal force stands for compression (Sign Convention 2), then the transfer matrix $\mathbf{U_{x}} \equiv \mathbf{U^{N-2}_{x}}$ is that given in Eq. (5.62) and the vector of applied loads, $\mathbf{\stackrel{\rm\circ}{Z^{+}}}$, is shown in Table C.5;
* if the normal force stands for tension (Sign Convention 2), then the transfer matrix $\mathbf{U_{x}} \equiv \mathbf{U^{N+2}_{x}}$ is that given in Eq. (5.64) and the vector of applied loads, $\mathbf{\stackrel{\rm\circ}{Z^{-}}}$, is shown in Table C.6.

The transfer matrices $\mathbf{U^{N-2}_{x}}$ and $\mathbf{U^{N+2}_{x}}$ [Krä91b] can be computed using the GNU Octave function ylfmII.m (p. ).

The loading vectors $\mathbf{\stackrel{\rm\circ}{{Z}}}{ }^{+}$ (5.92) and $\mathbf{\stackrel{\rm\circ}{ {Z}}}{ }^{-}$ (5.93) for the constant load $q_{z}$ in case of the compression and tension force $N$, respectively, are:

$$\mathbf{\stackrel{\rm\circ}{ {Z}}}{ }^{+} = \left[\begin{array}{c... ...- x_{0}\right)}{l} \right) - 1 \right]\frac{ql^{2}}{\nu^{2}} \end{array}\right]$$ (5.92)

$$\mathbf{\stackrel{\rm\circ}{ {Z}}}{ }^{-} = \left[\begin{array}{c... ...( x - x_{0}\right)}{l} \right) \right]\frac{ql^{2}}{\nu^{2}} \end{array}\right]$$ (5.93)

The loading vectors can be computed using the GNU Octave function ylqII.m (p. ).

The loading vectors $\mathbf{\stackrel{\rm\circ}{ {Z}}}{ }^{+}$ (5.94) and $\mathbf{\stackrel{\rm\circ}{ {Z}}}{ }^{+}$ (5.95) for the point load $\mathcal{F}_{z}$ in case of the compression and tension force $N$, respectively, are:

$$\mathbf{\stackrel{\rm\circ}{ {Z}}}{ }^{+} = \left[\begin{array}{c... ... x_{0}\right)}{l}\right) \right]\frac{\mathcal{F}_{z}l}{\nu} \end{array}\right]$$ (5.94)

$$\mathbf{\stackrel{\rm\circ}{{Z}}}{ }^{-} = \left[\begin{array}{c}... ...frac{\left(x - x_{0}\right)}{l}\right) \right]\frac{Fl}{\nu} \end{array}\right]$$ (5.95)

The loading vectors can be computed using the GNU Octave function ylffzII.m (p. ).