5.3.1 The set of solutions to a homogeneous differential equation

The fundamental set of solutions to the homogeneous differential equation

$$\frac{d^{4}w}{dx^{4}}{\,} {\pm}{\,} {\vert}\frac{S}{EI_{y}}{\vert}\frac{d^{2}w}{dx^{2}} = 0$$ (5.36)

for a compressive axial force with magnitude ${\vert}S{\vert}$ is

$$w_{1}^{\ast} = 1, \quad w_{2}^{\ast} = \frac{\nu}{l}x, \quad w_{3... ...rac{\nu}{l}x\right), \quad \quad w_{4}^{\ast} = \sin\left(\frac{\nu}{l}x\right)$$ (5.37)

and for a tensile axial force with magnitude

$$w_{1}^{\ast} = 1, \quad w_{2}^{\ast} = \frac{\nu}{l}x, \quad w_{3... ...}{l}x\right), \quad \quad w_{4}^{\ast} = \mathrm{sh}\left(\frac{\nu}{l}x\right)$$ (5.38)

where $\nu = l\sqrt{{S}/{EI_{y}}}$ is a characteristic parameter of a beam-column member and l is the length of the member.

Consider next the Wronski determinant $W\left(x\right)$ of the fundamental set of solutions from Eq. (5.37):

$$W\left(x\right) = \left\vert \begin{array}{cccc} 1 & \frac{\nu}{l... ...(\frac{\nu}{l}\right)^{3}\cos\left(\frac{\nu}{l}x\right) \end{array}\right\vert$$ (5.39)

Hence the Wronskian $W\left(x\right)$ value at $x=0$ is not 1:

$$W\left(0\right) = \left\vert \begin{array}{cccc} 1 & 0 & 1 & 0 \\... ...0 \\ 0 & 0 & 0 & -\left(\frac{\nu}{l}\right)^{3} \end{array}\right\vert \neq 1$$ (5.40)

Let us norm the Wronskian of the previous equation by

• subtracting the 1st column from the 3rd column and multiplying the result by $\left(-\left(\frac{l}{\nu}\right)^{2}\right)$
• subtracting the 2nd column from the 4th column and multiplying the result by $\left(-\left(\frac{l}{\nu}\right)^{3}\right)$
• multiplying the 2nd column by $\left(\frac{l}{\nu}\right)$.

Now the value of the Wronskian

$$W\left(x = 0\right) = \left\vert \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right\vert = 1$$ (5.41)

We make similar rearrangements with the fundamental set of solutions from Eq. (5.37) (with columns of Eq. (5.39)). Once more we

• subtract the 1st column from the 3rd column and multiply the result by $\left(-\left(\frac{l}{\nu}\right)^{2}\right)$:
  $- \left(\frac{l}{\nu}\right)^{2}\left[\cos\left(\frac{\nu}{l}x\right) - 1\right]$
• subtract the 2nd column from the 4th column and multiply the result by $\left(-\left(\frac{l}{\nu}\right)^{3}\right)$:
  $- \left(\frac{l}{\nu}\right)^{3}\left[\sin\left(\frac{\nu}{l}x\right) - \left(\frac{\nu}{l}x\right)\right]$
• multiply the 2nd column by $\left(\frac{l}{\nu}\right)$: x.

We get the normed fundamental set of solutions for the compressive axial force:

$$\begin{array}{ll} w_{1} = 1, & w_{2} = x, \\ w_{3} = - \left(\fr... ...sin\left(\frac{\nu}{l}x\right) - \left(\frac{\nu}{l}x\right)\right] \end{array}$$ (5.42)

The normed fundamental set of solutions for the tensile axial force is:

$$\begin{array}{ll} w_{1} = 1, & w_{2} = x, \\ w_{3} = \left(\frac... ...sh}\left(\frac{\nu}{l}x\right) - \left(\frac{\nu}{l}x\right)\right] \end{array}$$ (5.43)

There are two sign conventions (see Fig. (1.2)) for the internal reactions ^{5.4 5.5} (contact forces).

For the parameters $w_{0},\enspace w_{0}^{\prime}, \enspace w_{0}^{\prime\prime}$, and $w_{0}^{\prime\prime\prime}$ of the searchable function at $x = x_{0}$ (Sign Convention 1) we obtain

$$w_{0} = w_{0}, \enspace w_{0}^{\prime} = - \varphi_{0}, \enspace ... ...ac{M_{y}}{EI_{y}}, \enspace w_{0}^{\prime\prime\prime} = - \frac{Q_{z}}{EI_{y}}$$ (5.44)

and for Sign Convention 2

$$w_{0} = w_{0}, \enspace w_{0}^{\prime} = - \varphi_{0}, \enspace ... ...frac{M_{y}}{EI_{y}}, \enspace w_{0}^{\prime\prime\prime} = \frac{Q_{z}}{EI_{y}}$$ (5.45)

The complete solution for the compressive axial force (Sign Convention 1) is

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

$$+ \left.\frac{Q_{z}}{EI_{y}}\right\vert _{\circ}\left(\frac{l}{\n... ...frac{\nu}{l}x\right) - \left(\frac{\nu}{l}x\right)\right] + w_{e}\left(x\right)$$ (5.46)

and for the tensile axial force (Sign Convention 1)

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

$$- \left.\frac{Q_{z}}{EI_{y}}\right\vert _{\circ}\left(\frac{l}{\n... ...frac{\nu}{l}x\right) - \left(\frac{\nu}{l}x\right)\right] + w_{e}\left(x\right)$$ (5.47)

The complete solution for the compressive axial force (Sign Convention 2) is

$$w = w_{0} - \varphi_{0}x - \left.\frac{M_{y}}{EI_{y}}\right\vert ... ...\left( \frac{l}{\nu}\right)^{2}\left[\cos\left(\frac{\nu}{l}x\right) - 1\right]$$ (5.48)

and for the tensile axial force (Sign Convention 2)

$$w = w_{0} - \varphi_{0}x + \left.\frac{M_{y}}{EI_{y}}\right\vert ... ...frac{l}{\nu}\right)^{2}\left[ \mathrm{ch}\left(\frac{\nu}{l}x\right) - 1\right]$$ (5.49)

Relations between the 1st and 2nd sign conventions for a bending moment and shear force:

$\left.{M_{y}}\right\vert _{\circ \,\left( Sign \, Convention \, 1\right)}$ = $\left.{ - M_{y}}\right\vert _{\circ \,\left( Sign \, Convention \, 2\right)}$
$\left.{Q_{z}}\right\vert _{\circ \,\left( Sign \, Convention \, 1\right)}$ = $\left.{ - Q_{z}}\right\vert _{\circ \,\left( Sign \, Convention \, 2\right)}$