91Ó°ÊÓ

To calculate the Flächenmoment der zweiten Ordnung (I) for a rectangular cross-section, we can use the formula $$ I = \frac{b \cdot h^3}{12}, $$ where b is the width, and h is the height of the cross-section. Plugging in the given values \(b=0.06\mathrm{~m}\) and \(h=0.1\mathrm{~m}\), we get $$ I = \frac{0.06 \cdot (0.1)^3}{12} = 5 \times 10^{-6} \mathrm{~m}^4. $$

The given differential equation is $$ E I w^{\prime \prime \prime \prime}(x) = q, \quad x \in\left(-\frac{L}{2}, \frac{L}{2}\right). $$ We can rewrite this equation as $$ w^{\prime \prime \prime \prime}(x) = \frac{q}{EI}. $$ Integrating the equation four times with respect to x, we'll get the expression for w(x). 1. First integration: $$ w^{\prime \prime \prime}(x) = \frac{q}{EI} x + C_1. $$ 2. Second integration: $$ w^{\prime \prime}(x) = \frac{q}{2EI} x^2 + C_1x + C_2. $$ 3. Third integration: $$ w^{\prime}(x) = \frac{q}{6EI} x^3 + \frac{C_1}{2}x^2 + C_2x + C_3. $$ 4. Fourth integration: $$ w(x) = \frac{q}{24EI} x^4 + \frac{C_1}{6}x^3 + \frac{C_2}{2}x^2 + C_3x + C_4. $$ Now, we need to find the values of the constants \(C_1\), \(C_2\), \(C_3\), and \(C_4\) using the provided boundary conditions.

We are given the following boundary conditions: $$ \begin{aligned} w\left(-\frac{L}{2}\right) &= w\left(\frac{L}{2}\right) = 0 \mathrm{~m} \\ w^{\prime \prime}\left(-\frac{L}{2}\right) &= w^{\prime \prime}\left(\frac{L}{2}\right) = 0 \mathrm{~m}^{-1} \end{aligned} $$ Applying the boundary conditions to w(x) and w''(x), we get a system of four linear equations: 1. \(w\left(-\frac{L}{2}\right) = 0\): $$ \frac{q}{24EI} \left(-\frac{L}{2}\right)^4 + \frac{C_1}{6}\left(-\frac{L}{2}\right)^3 + \frac{C_2}{2}\left(-\frac{L}{2}\right)^2 + C_3\left(-\frac{L}{2}\right) + C_4 = 0 $$ 2. \(w\left(\frac{L}{2}\right) = 0\): $$ \frac{q}{24EI} \left(\frac{L}{2}\right)^4 + \frac{C_1}{6}\left(\frac{L}{2}\right)^3 + \frac{C_2}{2}\left(\frac{L}{2}\right)^2 + C_3\left(\frac{L}{2}\right) + C_4 = 0 $$ 3. \(w^{\prime \prime}\left(-\frac{L}{2}\right) = 0\): $$ \frac{q}{2EI} \left(-\frac{L}{2}\right)^2 + C_1\left(-\frac{L}{2}\right) + C_2 = 0 $$ 4. \(w^{\prime \prime}\left(\frac{L}{2}\right) = 0\): $$ \frac{q}{2EI} \left(\frac{L}{2}\right)^2 + C_1\left(\frac{L}{2}\right) + C_2 = 0 $$ By solving this system of linear equations for \(C_1, C_2, C_3, C_4\), we get \(C_1 = 0\), \(C_2 = -\frac{qL^2}{8EI}\), \(C_3 = 0\), and \(C_4 = \frac{qL^4}{64EI}\). Thus, the final expression for the deflection (w) is: $$ w(x) = \frac{q}{24EI}x^4- \frac{qL^2}{16EI}x^2 + \frac{qL^4}{64EI}. $$