The slope-deflection method is built around a pair of equations,
called slope-deflection equations.
In lecture SA28, we derived the slope-deflection equations for a typical beam segment.
These equations define member-end moments in terms of member-end rotations.
This formulation however is limited in two ways.
One: the beam segment is assumed to be load-free.
And two: there is no relative vertical displacement between the ends of the segment.
That is, B does not move up or down relative to A.
Why do we need to talk about this relative end displacement, you ask?
I will explain shortly.
For now, just know that it is an important part of the slope-deflection method.
In lecture SA29, we expanded these slope-deflection equations
to account for the presence of external loads on the beam segment.
For example, we showed that when the segment is subjected to a uniformly distributed load, like this,
the slope-deflection equations become:
In this lecture, we're going to modify these equations
to account for relative vertical displacement at the ends of the beam segment.
But first, let's see why we need to have the differential displacement reflected in these equations.
As we have seen already,
the slope-deflection equations define member-end moments in terms of member-end rotations.
So if we have the end rotations, we can use the equations to determine the end moments.
But where do the end rotations come from?
What causes the ends of a beam segment to rotate?
There are three reasons for joint rotation in beams.
One: The load directly applied to this segment causes it to deform resulting in end rotations.
Two: Loads applied to adjacent segments force them to deform.
These deformations, when propagated throughout the structure, result in the end rotations in our member.
Three: a relative vertical displacement between the ends of the member produces end rotations.
It is this third reason that we want to investigate in this lecture.
Here is our beam segment.
Its ends are labeled A and B.
What happens to the shape of the segment when one of the ends
moves up or down in relation to the other end?
One of four scenarios would result.
If B moves up,
we get this configuration resembling a counterclockwise rotation of the member.
If B moves down, we end up with this configuration,
which involves a clockwise member rotation.
Further, an upward movement of A is equivalent to a clockwise rotation of the member.
And a downward movement of A,
corresponds to the member's counterclockwise rotation.
So when the beam segment experiences a differential end displacement,
member, and by extension, joint rotations result.
If the member displaces like this,
we could say,
tangent theta_A equals Delta over L.
And, tangent theta_B equals Delta over L.
But since our analysis is based on the assumption that joint rotations are very small, we could write:
And:
Therefore, when differential displacement delta is present,
the end rotations equal Delta divided by L.
Okay, but how do we incorporate this in the slope-deflection equations?
Let's begin by recalling a few diagrams and equations from lecture SA28.
This diagram shows member-end rotations due to a counterclockwise moment placed at A.
The equations that go with the diagram are:
Similarly, this diagram shows member-end rotations when a moment is placed at B.
And here are the corresponding equations.
Further, from lecture SA29, recall this diagram.
It shows member-end rotations due to a uniformly distributed load.
And here are the corresponding equations.
Now, let's add this diagram to the list.
The diagram shows the member-end rotations due to a differential displacement denoted by Delta,
where theta_A4 equals theta_B4 equals Delta over L.
What do we get if we superimpose these diagrams?
We get the total rotation at A and B,
that is, we get theta_A and theta_B,
where theta_A equals theta_A1 minus theta_A2, minus theta_A3 plus theta_A4.
And theta_B equals theta_B2 minus theta_B1, plus theta_B3 plus theta_B4.
Knowing the individual rotations, we can rewrite these equations as:
Treating M_AB and M_BA as unknowns,
let's rearrange the equations to get member-end moments
in terms of end rotations, load magnitude, and Delta.
As we mentioned in lecture SA29, these two terms are due to the applied load, 74 00:05:50,680 --> 00:05:53,320 we call them fixed-end moments.
When the load is uniformly distributed,
fixed-end moments are wL^2/12,
and negative wL^2/12.
But to write the slope-deflection equations in a more general form,
let's use FEM_AB and FEM_BA instead of these load specific terms.
So our general slope-deflection equations are:
Now let's see these equations in action.
Consider this beam.
It rests on one pin and two roller supports,
but the beam is free at the right end.
The entire beam is subjected to a uniformly distributed load.
We wish to analyze it using the slope-deflection method.
As usual, we assume a constant EI for the entire beam.
Before we proceed with the solution, let's make two important observations.
Observation one:
Since joint D is free to move up and down,
there is a differential vertical displacement in member CD.
This means Delta is not going to be zero for the member,
but Delta is zero for the other two members, since their ends are not free to move vertically.
Observation two:
There is a total of four unknown rotations, one at each joint of the structure.
There is also an unknown vertical displacement at joint D.
So we have a total of five unknowns.
But since the beam has four joints, we can write only four joint equilibrium equations,
therefore, we are one equation short for determining the unknowns.
This happens whenever we get a non-zero Delta in our equations.
But not to worry, as you will see shortly,
we can use our basic knowledge of static equilibrium to come up with an additional equation.
To begin, let's separate the joints from the members, like this:
Now, draw member-end moments and write the slope-deflection equations for each segment.
Here are the slope-deflection equations in their general form:
For member AB, we have a zero Delta,
and the fixed-end moments are wL^2/12 at the left end of the segment,
and negative wL^2/12 at the right end of the segment.
Therefore we get:
Similar to AB, for segment BC, we get:
And for segment CD, we have:
Note the non-zero Delta in these slope-deflection equations.
It is important to note the direction of Delta here.
It is assumed to be caused by a counterclockwise rotation of the member.
This means, Delta is assumed to be upward at D, like this:
Now let's place the balancing moments at the joints, like this:
And write the joint equilibrium equations like this:
When we expand these equations, we get:
Here we have four equations and five unknowns.
Now we need to come up with one more equation.
To do so, we isolate a segment of the structure and write the equilibrium equations for it.
In this problem, an ideal segment for this purpose is CD.
Let's cut the beam just to the right of C,
then draw the free-body diagram for segment CD, like this:
Since the conditions of equilibrium must be satisfied here, we can write:
Now we can combine these two equations to get one equation in terms of the member-end moments.
This can be done easily by solving the first equation for V_CD,
then substituting wL for V_CD in the second equation.
This gives us:
And since we can write M_CD in terms of the unknowns, we get this as our fifth equation:
Solving these equations simultaneously, we get:
Then we substitute these values back into the slope-deflection equations
to determine our member-end moments.
Now, draw the free-body diagram for each segment and determine member-end shear forces.
For segment AB, we get:
For BC, we have:
And for CD, we can write:
Finally, transfer the computed shear forces to the joints
and determine the support reactions, like this:
So here are the support reactions for the beam:
Now it's your turn. Analyze the following beams using the slope-deflection method.