The Equations of Equilibrium balance the forces acting on each a node so that all the applied loads and forces from the elements attached to the nodes sum to zero.
The Equations of Compatibility set defections for all members attached to a node to be equal.
It is a simple, determinate structure of three rods pinned together at one end, and the other ends fastened to fixed points in space (visualize a wall).
We apply a skewed load to the common joint, which we will resolve into component vectors in the global coordinate system.
For this example, we will not employ a skew matrix, as that is usually only done for pressure vessels (such as pressurized aircraft, pressure vessels, and submersible vessels.)
1) Set up the global coordinate system
2) Idealize the structure into nodes and rods
3) Label each node and rod
4) Define X, Y, Z for each node
5) Define the ith and jth node for each rod
6) Define E and A for each rod.
Material is steel pipe (all rods)
Modulus of Elasticity = E = 30x106
Cross-sectional Area = A = 1 in2
7) Calculate length (L) for each rod
LA = sqrt(602+602+02) = 84.85
LA = sqrt(602+02+302) = 67.08
LA = sqrt(602+02+(-302)) = 67.08
Steps 6) and 7) above are not needed immediately.
| node | X | Y | Z |
| 1 | 60 | 0 | 0 |
| 2 | 0 | 60 | 0 |
| 3 | 0 | 0 | 30 |
| 4 | 0 | 0 | -30 |
| element | i | j | E | A | L |
| A | 1 | 2 | 30x106 | 1.0 | 84.85 |
| B | 1 | 3 | 30x106 | 1.0 | 67.08 |
| C | 1 | 4 | 30x106 | 1.0 | 67.08 |
L = [(Xj - Xi)2 + (Yj - Yi)2 + (Zj - Zi)2]1/2
To the far right is a depiction of the degrees of freedom at each node.
These are the component X, Y, Z directions that all the forces and displacements can act on our example structure.
Normally, there are six degrees of freedom at each node. However, pinned rods can only take translations, and cannot take moments -- thus there are no rotations about the X, Y, Z axes.
On Node 2 are the X, Y, Z translation degrees of freedom..
On Node 3 are the X, Y, Z translation degrees of freedom.
On Node 4 are the X, Y, Z translation degrees of freedom.
In the figure to the far right are depicted the component nodal forces that can be applied to the example structure. Some of these forces (in magenta) will be reactions, and some of them (in white) are applied loads.
Each of these nodal forces is labeled with the node number and the degree of freedom (direction) represented.
We label all these external forces in such manner so that we can identify them correctly and process them in the equations we are developing here.
The matrix form of the Equations of Equilibrium are
depicted to the right.
Above the matrix, in blue font, are column
headings. These are not normally displayed, but here serve as a guide
for the novice. The column headings over the central matrix represent
the degrees of freedom for each element, at each end of the element, in the
X, Y,
and Z global coordinate system.
To the extreme left are the Applied component Forces (some
will be reactions) in the X,
Y, Z,
global coordinate system, at the Nodes for the example structure.
To the extreme right are each element's component forces in the
X, Y,
and Z global coordinate system.
In the center is the Equilibrium Matrix,
E. This is the form we
use in Matrix Structural Analysis, It is the matrix that links
External Forces (and Reactions) at the nodes to Internal forces in the
elements at each end of the element. In this form, it is drop-in ready for
insertion into our matrix solution.
To the right is the abbreviated matrix form of the Equations of Equilibrium.
F =
E
f
EQUATIONS OF COMPATIBILITY
In the figure to the right are
depicted the component nodal and element displacements in the
X, Y,
and Z global coordinate system that can
be experienced by the example structure.
Some of these displacements (in magenta) will eventually be set to zero,
tying the structure down so it cannot move as a free body through space.
But we do not set that constraint in the compatibility equation set.
Each of these possible displacements is labeled with the node number and the
degree of freedom (direction) represented. A upper case letter
represents a nodal displacement, while a lower case letter represents an
element end displacement that is to be connected to a node.
We label all these displacements in such manner so that we can
identify them correctly and process them in the equations we are developing
here.
The figure above pictorially displays the Equations of Compatibility
In the figure to the right is the matrix form of the Equations of
Compatibility for the structure.
Above the matrix to the right, in blue font, are column headings.
These are not normally displayed, but here serve as a guide for the novice.
The column headings over the central matrix represent the degrees of freedom
for each node in the X,
Y, and Z
global coordinate system.
To the extreme left are all the possible component
displacements in the X,
Y, Z
global coordinate system for the ends of each element in the example
structure.
To the extreme right are the nodal component displacements in the
X, Y,
and Z global coordinate system.
In the center is the Compatibility Matrix,
C. This is the form we
use in Matrix Structural Analysis, It is the matrix that sets the
internal displacements of the elements ends in the
X, Y,
Z, global coordinate system equal to the
external displacements at the nodes in the X,
Y, Z,
global coordinate system In this form, it is drop-in ready for
insertion into our matrix solution.
To the right is the abbreviated matrix form of the Equations of Equilibrium.
u =
C
U
Because of this relation, it is not necessary to develop both the Equilibrium and Compatibility matrices separately -- when one is developed, the other is merely its transform. We develop both here in this example for understanding the general structural relations, and so the neophyte can verify that one is indeed the transform of the other.
Generally it is easier to create the Compatibility matrix, which is mostly all zeroes, except for a single "1" in each row -- and that "1" connects the element degree of freedom to its corresponding nodal degree of freedom.
It is the Compatibility and Equilibrium matrices which link the individual elements to the nodes and mathematically creates the structural system.
E = CT and C=ET
The STACKED element matriX EQUATION
We developed a general rod element's globalized stiffness matrix equation in the prior section. But we have yet to apply that specifically to each element, and assemble all those globalized stiffness matrix equations into a single stacked element matrix equation.
We need to do that now, before we can go further. To the right is the stacked element matrix equation for the three-rod example structure in condensed format. And below is the expanded format fully displaying all the individual cells.
The Stacked Element Matrix Equation in condensed format
The Stacked Element Matrix Equation (expanded):
The above expanded stacked element matrix equation demonstrates why few people try hand solutions with Matrix Structural Analysis -- Its only three simple elements, and the the matrices are already almost too large an unwieldy to manipulate by hand, or even put on a piece of paper.
But the Matrix Structural Analysis is relatively easy to accomplish with a computer, being a very regular and straight forward process -- easily programmed, especially with matrix software like MATLAB, or even with a spreadsheet like Excel.
We show the expanded equation above, with the variables, as a learning tool -- so the novice and student can see the processes, including what the elements of a condensed format look like when they are expanded. But in actual use and practice, Matrix Structural Analysis is a numerical solution, and we would normally only be processing the equations with numerical data -- in a computer.
We have now all the pieces to assemble the unsupported stiffness matrix. To the immediate right are the condensed matrix equations which we have developed to this point. This is the complete matrix relationship, from which we can extract a solution.
We simply the equation by substituting CT for E, and (l C)T for CT lT. Whenever possible, we use the transpose of a matrix because it is just a re-indexing process, and not a calculation process. every time we perform a matrix multiplication operation, it introduces calculation error, so it behooves us to avoid it if we can.
The actual process of assembling the unsupported stiffness matrix is:
1) Calculate direction cosines l
2) Assemble the Stacked Transformation Matrix l
3) Assemble the Compatibility Matrix C
4) Calculate L = l C
5) Assemble the Stacked Element Stiffness Matrix k
6) Calculate the Unsupported Stiffness Matrix K = LT k L
F = E f u = C U Equilibrium and Compatibility equations
f = lT k l u Stacked Element Matrix Equation
F = E lT k l C U Combine the above equations
E = CT Recognize the transpose relation
F = CT lT k l C U Substitute the transpose
F = (l C)T k (l C) U Collect the Transposed matrices
L = l C Multiply the terms
F = LT k L U Simplify the equation
K = LT k L The Unsupported Stiffness Matrix
F = K U Unsupported Stiffness Matrix Equation
which we do in the next section, appropriately titled SOLUTION