This is an introductory lecture on the matrix displacement method
for analyzing statically indeterminate structures.
Our point of departure here is the slope-deflection formulation,
as it forms a basis for the matrix method.
But initially, we are going to limit our discussion to the analysis of continuous beams only,
we'll expand the formulation in future lectures for dealing with other types of structures.
To begin, please review the slope-deflection method presented in Lectures SA27 through SA33.
You also need to be familiar with basic matrix notation and operations,
as we are going to write our equations using matrices.
Consider this particular continuous beam.
If we were to analyze it using the slope-deflection method,
first we would have written the slope- deflection equations for each member.
Then, we would have used the joint equilibrium equations
to find the unknown rotations and displacements.
In this lecture, we are going to more or less follow the same process, but we'll do it using matrices.
First, we write an expanded version of the slope-deflection equations
for each structural member in matrix form.
Then, we combined the member matrices
to form the joint equilibrium equations, also in matrix form.
The solution for this system of linear equations
is the set of unknown slopes and deflections at the joints of the structure,
which can then be used to compute member-end forces.
Please keep the parallels between the slope- deflection method and the matrix method in mind
as we go through the formulation.
Consider a typical beam segment subjected to a uniformly distributed load.
Here are its slope-deflection equations.
They give us member-end moments in terms of member-end rotations, theta,
and the relative end displacement, Delta.
Let's number the equations 1 and 2.
We also want to represent the member-end shear forces in terms of theta and Delta.
For this free-body diagram we can write:
Sum of the forces in the Y-direction must be zero.
Sum of the moments about end I must be zero.
Solving these equations for V_ij and V_ji, we get:
We number these equations 3 and 4.
Here are our four equations:
Using 1 and 2, we can rewrite 3 and 4 as:
These equations define shear-end moment
at the two ends of the member in terms of end-slopes,
theta_i and theta_j, and relative displacement Delta.
Let's replace Delta with the end displacements for the member.
If we denote vertical displacement at end I as delta_i,
and refer to the displacement at end J as delta_j,
then we can write Delta = delta_j - delta_i.
Substituting delta_j - delta_i for Delta in equations 1 through 4, we get:
Now, I am going to reorder these equations before rewriting them in matrix form.
The first two equations give the shear and moment at joint I,
and the second pair of equations is for shear and moment at joint J.
In matrix notation, these equations can be written as:
This is called the member stiffness matrix.
This is the member-end force vector,
this is the member-end displacement vector
and this is the fixed-end load vector.
Now, let's apply this system of equations to each beam segment.
For member AB, we get:
Since the segment is not subjected to any loads,
the fixed-end load vector consists of zero elements only.
We can further simplify the equations by setting delta_A to theta_A to delta_B to 0.
That is, since A is fixed, displacement end rotation at that end of the beam would be 0.
And since there is a roller at B,
we know there would be no vertical displacement at that joint.
So the system of equations can be written as:
Here we have substituted zero for delta_A,
theta_A and delta_B in the member-end displacement vector.
It is important to note that in this lecture, we are ignoring axial deformation.
That is why axial forces are not present in our equations,
we will bring them into this formulation in future lectures.
We can shorten the member-end force vector by removing the three zeros,
and by adjusting the stiffness matrix accordingly.
Here are the equations for member AB.
And here they are in compact form:
Since the first column of the stiffness matrix represents the coefficients of delta_A,
the second column represents the coefficients of theta_A,
and the third column represents those for delta_B,
we can remove them from the matrix and rewrite the equations in a compact form.
And in case you were wondering, here are the exact same equations in algebraic form:
For member BC, we can write:
Since delta_B, vertical displacement at B, equals zero,
we can rewrite the equations as:
Here, we have replaced delta_B with zero.
Rewriting the equations in compact form, we get:
We achieve this by removing the zero from the force vector
and by removing the coefficients of delta_B from the stiffness matrix.
Let's summarize what we have done so far.
Given this beam, we have written an extended version of the slope-deflection equations in matrix form.
Here are the equations for member AB:
And here they are for member BC:
These equations give us member and shear forces as well as bending moments
in terms of slopes and displacements at the end joints.
Now, let's turn our attention to the joint equilibrium equations.
For this beam, we have a total of three unknowns,
theta_B, delta_C and theta_C.
We say that the beam has three degrees of freedom.
How many equations do we need for finding the unknowns?
We need three of them,
one equation per degree of freedom.
Since in this formulation we are only considering bending moment and shear force, but not axial force,
we can write two equilibrium equations per joint.
Sum of the forces in the Y-direction must be zero.
Sum of the moments at the joint must be zero.
The beam has three joints,
therefore we can write up to six equilibrium equations,
but we only need three of them.
Which ones do we need to use?
Simple. We use the equations associated with the degrees of freedom of the beam.
Here, rotation at joint B is one of the degrees of freedom,
so we use the moment equilibrium equation at B,
since bending moment and joint rotation go hand-in-hand,
they are cause and effect.
Another degree of freedom is the vertical displacement at C.
For this, we need to look into forces in the Y-direction.
So we write: sum of the forces in the Y-direction at C must be zero.
And, we use the moment equilibrium equation at C,
since theta_C is a degree of freedom.
Here are the three equilibrium equations in terms of member-end forces:
Let's expand them using member matrices.
In the first equation, we have M_BA and M_BC.
We find M_BA in the equations for member AB.
Here are the equations in matrix form:
And here is M_BA in matrix form:
For M_BC, we need the equations for member BC. Here they are:
And here is M_BC in matrix form:
So, the two terms in the first equilibrium equation are:
How do we add them together to get the first equilibrium equation?
Like this:
This is the sum of the coefficients for theta_B and M_BA and M_BC.
This is the sum of the coefficients for Delta_C.
And this is the sum of the coefficients for theta_C.
To better understand this addition operation, let's view the equations in algebraic form.
Here is the first equation:
Here is the second equation:
And here is their sum, it equals zero.
You may want to spend a few minutes here examining the relationship between the two notations.
Here is our second equilibrium equation.
We need V_CB in matrix form.
Here are the equations for member BC.
VC_B, the third equation in the set, can be easily extracted from the matrix. So we can write:
To write the third equilibrium equation, we need M_CB.
Here it is:
Therefore, this becomes our third equilibrium equation:
Given the three equilibrium equations in matrix form,
we can concatenate them to get our system of linear equations.
Let's write them this way:
We can find the unknown joint rotations and displacements by solving these equations.
This is called system stiffness matrix:
This is system displacement vector:
And this is system force vector:
There are various methods for solving such a system of equations.
If you are going to solve this by hand, Gaussian elimination method would be a good technique to use.
I am going to assume that you know how to solve this already and just present the results here.
The solution for this system is:
Now that we have determined the unknown slopes and deflection,
we can substitute them back in the member specific equations,
in order to find the member-end forces:
For AB, here are the equations for member forces:
Here is the solution vector:
Substituting this value for theta_B, we get these member forces:
Let's use them to draw the members free-body diagram.
For BC, we can write:
Substituting the solution vector here, we get:
Here are the member-end forces for segment BC.
And here is the segment's free-body diagram:
Now, let's determine the support reactions for the beam using the segment's free-body diagrams.
Here they are:
Finally, we draw the shear and moment diagrams for the entire beam:
See if you can analyze the following beams using the matrix method: