The problem asks to determine the vertical displacement at point I (which I assume is at the top of the structure), for the given structure, subjected to an external load of $9$ kN/m. The structure is a rectangular frame with fixed supports at A and B. The length of the horizontal member AB is given as $L$.

1. Problem Description

The problem asks to determine the vertical displacement at point I (which I assume is at the top of the structure), for the given structure, subjected to an external load of $9$ kN/m. The structure is a rectangular frame with fixed supports at A and B. The length of the horizontal member AB is given as $L$.

2. Solution Steps

Since the problem only provides the distributed load and the length of the base ($L$), and doesn't specify the height or material properties (Young's modulus, $E$, and moment of inertia, $I$), we can only express the vertical displacement at the top of the structure in terms of these unknown parameters.

First, we assume that the frame members are uniform and have the same material and cross-sectional properties ($E$ and $I$). We can use Castigliano's second theorem to determine the vertical displacement at the top. Castigliano's theorem states that the displacement in the direction of an applied force is the partial derivative of the strain energy with respect to that force.

$$\delta = \frac{\partial U}{\partial P}$$

where $\delta$ is the displacement, $U$ is the strain energy, and $P$ is the applied force.

For a frame structure, the strain energy due to bending is:

$$U = \int \frac{M^2}{2EI} dx$$

We'll need to introduce a dummy vertical load $P$ at the point where we want to find the displacement (top center). The distributed load is $q = 9$ kN/m.

Let's divide the structure into three parts: the left vertical member (1), the top horizontal member (2), and the right vertical member (3). Let the height of the frame be $h$.

Due to symmetry, the moments in the two vertical members will be equal.

Moment in the horizontal member (2) at a distance $x$ from the center:

$M_2(x) = \frac{P}{2} - \frac{q}{2} (\frac{L^2}{4} - x^2)$, where $-L/2 \le x \le L/2$

Moment in vertical member (1) at a distance $y$ from A:

$M_1(y) = \frac{PL}{4} + \frac{qL^2}{8}$

Vertical displacement at the top:

$\delta = \frac{\partial U}{\partial P} = \frac{1}{EI} \int M \frac{\partial M}{\partial P} dx$

Because of the symmetry of the structure:

$\delta = \frac{2}{EI} \int_0^h M_1(y) \frac{\partial M_1(y)}{\partial P} dy + \frac{1}{EI} \int_{-L/2}^{L/2} M_2(x) \frac{\partial M_2(x)}{\partial P} dx$

$\frac{\partial M_1}{\partial P} = \frac{L}{4}$

$\frac{\partial M_2}{\partial P} = \frac{1}{2}$

$\delta = \frac{2}{EI} \int_0^h (\frac{PL}{4} + \frac{qL^2}{8}) (\frac{L}{4}) dy + \frac{1}{EI} \int_{-L/2}^{L/2} (\frac{P}{2} - \frac{q}{2} (\frac{L^2}{4} - x^2)) (\frac{1}{2}) dx$

Since we are looking for the displacement due to the external load, we set $P=0$:

$\delta = \frac{2}{EI} \int_0^h (\frac{qL^2}{8}) (\frac{L}{4}) dy + \frac{1}{EI} \int_{-L/2}^{L/2} (- \frac{q}{2} (\frac{L^2}{4} - x^2)) (\frac{1}{2}) dx$

$\delta = \frac{2}{EI} (\frac{qL^3}{32})h + \frac{1}{EI} \int_{-L/2}^{L/2} (- \frac{q}{4} (\frac{L^2}{4} - x^2)) dx$

$\delta = \frac{qL^3h}{16EI} - \frac{q}{4EI} [\frac{L^2}{4}x - \frac{x^3}{3}]_{-L/2}^{L/2}$

$\delta = \frac{qL^3h}{16EI} - \frac{q}{4EI} [\frac{L^2}{4}(L) - \frac{L^3}{12}]$

$\delta = \frac{qL^3h}{16EI} - \frac{q}{4EI} [\frac{3L^3-L^3}{12}]$

$\delta = \frac{qL^3h}{16EI} - \frac{q}{4EI} [\frac{2L^3}{12}]$

$\delta = \frac{qL^3h}{16EI} - \frac{qL^3}{24EI}$

$\delta = \frac{qL^3}{EI} (\frac{h}{16} - \frac{1}{24}) = \frac{9L^3}{EI}(\frac{h}{16} - \frac{1}{24})$

3. Final Answer

The vertical displacement at the top is $\frac{9L^3}{EI}(\frac{h}{16} - \frac{1}{24})$, where $q = 9$ kN/m, $L$ is the length of the base, $h$ is the height of the frame, $E$ is the Young's modulus, and $I$ is the moment of inertia of the frame members.

Without values for $E$, $I$, and $h$, we can only provide the expression in terms of these unknowns.

Final Answer: $\delta = \frac{9L^3}{EI}(\frac{h}{16} - \frac{1}{24})$