The generalized stiffness matrix for the beam-column element in the global coordinate system is:

\[ Equation00 \]

Note: K is symmetrical (the lower triangular part of the matrix is not shown.)

where \( \small \alpha = EA/L \), \( \small \beta = EI/L^3 \), \( \small c = cos \theta \), and \( \small s = sin \theta \).

A.1) For Members AB and DC:

\( Equation05 \)

\( Equation05 \)

\( Equation07 \)

\( Equation08 \)

\[ Equation09 \]

The stiffness matrix \( \small K \) relates member-end forces and displacements using equation: \( \small F = KD \) where \( \small F \) is the vector of member-end forces and \( \small D \) represents member-end displacements. The following diagrams show the elements of these two vectors.

A.2) For Member BC:

\( Equation010 \)

\( Equation011 \)

\( Equation012 \)

\( Equation013 \)

\[ Equation014 \]

The stiffness matrix \( \small K \) relates member-end forces and displacements using equation: \( Equation015 \) where \( \small P \) is the vector of fixed-end forces and \( \small D \) in the global coordinate system.

\[ Equation016 \]

where \( \small Q \) is the force transformation matrix and \( \small p \) is the fixed-end forces in the local coordinate system.

\[ Equation017 \]

To determine vector \( p \), we need to determine the fixed-end forces due to the applied loads. We know the fixed-end forces for a single concentrated load. If the load magnitude is \( \small R \), we can write:

For the left concentrated load, we get:

For the right concentrated load, we have:

Adding the two load cases, we get the total fixed-end forces:

The vector of fixed-end forces in the local coordinate system becomes:

\[ \]

In global coordinate system, this vector can be written as:

\[ \]

The structure has six (6) degrees of freedom: displacements and rotations at joints B and C.

The system stiffness matrix can be written as:

\[ \]

Making proper substitutions from the member stiffness matrices, we get:

\[ \]

BC is the only member subjected to applied loads. The member-end forces for BC are:

So, the equivalent joint loads due to the applied concentrated loads are:

Or,

In vector form, we can write the joint loads as:

\[ \]

The system of equations relating the joint vector to the joint displacement vector can be written as:

\[ \small KD = F \]

Since \( \small K \) and \( \small F \) were determined in the previous steps, we can write:

\[ \]

Solving the system of equations for the displacement vector, we get:

\[ \]

The horizontal and vertical displacements are given in meter. If we convert them to millimeters, we get:

\[ \]

Knowing these values, we can draw the general deformed shape of the frame.

(the drawing is not to scale)

Member-end forces can be calculated using the joint displacements and rotations \( \small D_1 \) through \( \small D_6 \).

F.1) Member AB

The member-end displacement vector for AB is:

\[ \]

The member stiffness matrix is:

\( \)

Then, we can write:

\( \)

Showing these forces graphically, we get:

F.2) Member BC

Given the system displacement vector, the member displacement vector becomes:

\[ \]

The member stiffness matrix is:

\[ \]

Then, we can write:

\[ \]

Showing these forces graphically, we get:

F.3) Member DC

The member-end displacement vector for DC is:

\[ \]

The member stiffness matrix is:

\[ \]

Then, we can write:

\[ \]

Showing these forces graphically, we get: