The problem asks to determine the stiffness component and find the internal stresses of a given frame using the stiffness method. The frame consists of a combination of beams and columns. One horizontal beam is subjected to a uniformly distributed load of $4 kN/m$. The dimensions of the frame are given as $4.0 m$ horizontally and $4.0 m$ vertically. Some values near the connections are written as $\frac{40}{60}$ and $\frac{40}{40}$, and $\frac{40}{40}$. These values probably refer to fixed-end moments or stiffness factors at different locations.

1. Problem Description

The problem asks to determine the stiffness component and find the internal stresses of a given frame using the stiffness method. The frame consists of a combination of beams and columns. One horizontal beam is subjected to a uniformly distributed load of $4 kN/m$. The dimensions of the frame are given as $4.0 m$ horizontally and $4.0 m$ vertically. Some values near the connections are written as $\frac{40}{60}$ and $\frac{40}{40}$, and $\frac{40}{40}$. These values probably refer to fixed-end moments or stiffness factors at different locations.

2. Solution Steps

To solve this problem, we need to follow the steps of the stiffness method (also known as the displacement method):

Step 1: Determine the degrees of freedom (DOF).

Based on the geometry, let's assume that there are three degrees of freedom: rotation at a joint, horizontal displacement and vertical displacement. Since the structure is a frame, we'll consider rotations at the joints and translations. The fixed supports will not have any displacement or rotation.

Step 2: Develop the stiffness matrix for each element.

For each element (beam or column), we need to establish its local stiffness matrix. The local stiffness matrix relates the forces and moments at the ends of the element to the displacements and rotations at the ends. This will depend on element properties like $E$, $I$, and $L$.

For a beam element with two nodes $i$ and $j$, the local stiffness matrix $k$ (6x6) relating nodal forces $F_i, M_i, F_j, M_j$ to nodal displacements $u_i, \theta_i, u_j, \theta_j$ can be constructed. A simplified form (4x4 relating moments and rotations if axial deformations are neglected) is:

$k = \frac{EI}{L} \begin{bmatrix} 4 & 2 \\ 2 & 4 \end{bmatrix}$ (This is for a single beam element related to rotation only).

Step 3: Assemble the global stiffness matrix.

We combine the local stiffness matrices of all elements into a global stiffness matrix $K$. This involves transforming the local coordinates to global coordinates and adding the corresponding elements in the local matrices into the correct position in the global matrix.

Step 4: Apply the boundary conditions and loads.

We apply the boundary conditions (fixed supports have zero displacements) and the external loads to the structure. The external loads are in terms of forces and moments. In this case, the distributed load on the beam will be converted to equivalent nodal forces and moments.

For the beam subjected to a uniformly distributed load $w$ over a length $L$, the fixed-end moments are:

$M = \frac{wL^2}{12}$.

And the fixed-end forces are: $V = \frac{wL}{2}$.

In this case $w=4 kN/m$ and $L = 4m$, so the fixed end moment is $\frac{4 * 4^2}{12} = \frac{64}{12} = \frac{16}{3} kN.m$ and shear forces are $4 * 4/2 = 8 kN$.

Step 5: Solve for the unknown displacements.

We solve the system of linear equations $KU = F$ for the unknown displacements $U$, where $K$ is the global stiffness matrix, $U$ is the vector of unknown displacements, and $F$ is the vector of external forces.

Step 6: Calculate the element forces and moments.

Once we have the displacements, we can calculate the forces and moments in each element using the element stiffness matrices and the known displacements. The internal stresses can then be found from these forces and moments.

Without the exact EI values and a more detailed diagram, calculating exact numerical values for displacements and internal stresses is not possible. However, based on the image, the fixed-end moments are $40/60$ and $40/40$, which might indicate the distribution factors at the joints. If those are distribution factors, then the moment distribution method can be employed to find internal stresses.

3. Final Answer

It is not possible to provide a numerical solution without more information about the frame properties (EI values) and a clear diagram that identifies the degrees of freedom. However, the steps to solve this problem using the stiffness method have been outlined above. The general expressions for element stiffness matrices and fixed-end moments due to uniformly distributed loads are also provided. The internal stresses can be obtained from these moments and forces. Also, moment distribution could be an alternative method if the given numbers next to the connections are the distribution factors.