SA51: Frame Analysis Example (Matrix Displacement Method)
Analyze the frame shown below using the Matrix Displacement Method.
The frame is subjected to two concentrated member loads. There is
a concentrated load of \( \small kN \)
acting on Member BC. It is located \( \small m \) to the right of joint B.
The other concentrated load has a magnitude of \( \small kN\), and it is
located \( \small m \) to the left of joint C.
Members AB and DC each has a length of \( \small m \).
The overall length of BC is \( \small m \).
The members are made of structural steel with a modulus of elasticity ( \( \small E\) ) of \( \small GPa\).
Members AB and CD each has a corss-sectional area ( \(\small A\) ) of \( \small mm^2 \) and
moment of inertia ( \( \small I\) ) of \( \small mm^4 \).
Member BC has a cross-sectional area of \( \small mm^2 \);
and a moment of inertia of \( \small mm^4 \).
Note: Move the cursor over an input field to see the valid range of values.
Note: The source of this problem can be found in this video: YOUTUBE.
Note: For a detail explanation of the Matrix Displacement Method, take our online course.
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.
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: