Slope-Deflection Equations Derivation
This is a continuation of lecture SA27,
in which we introduced and illustrated the slope-deflection method for beam analysis,
but left the derivation of the slope- deflection equations for a later time,
a task that we are going to take up in this lecture.
Imagine a continuous beam having multiple segments.
A typical beam segment could be subjected to external loads,
like segments AB and CD,
or it could be external-load-free, like segment BC.
Let's consider each case separately.
First, we examine the simpler case,
where there are no external loads placed on the segment.
Although such a segment is not directly subjected to any loads,
it does deform, and carries internal forces.
For example, the concave up deflection of segment BC,
means that the member carries a counterclockwise moment at B,
and a clockwise moment at C.
Generally speaking, a typical beam segment has a bending moment at its left end,
and a bending moment at its right end.
If the ends are labeled B and C,
we refer to these moments as M_BC and M_CB, respectively.
Further, we denote the beam's rotation at B as theta_B,
and use theta_C for rotation at end C.
The slope-deflection equations define the mathematical relationship
between these rotations and bending moments.
More specifically, there are two slope- deflection equations for the segment:
One equation for defining M_BC in terms of rotations theta_B and theta_C,
and another equation for defining M_CB in a similar manner.
Before we start developing the slope-deflection equations,
let's pick a sign convention for member-end moments and rotations.
I'm going to use counterclockwise direction as positive.
So, I'm going to place M_BC like this, causing a positive counterclockwise rotation at B.
Similarly, we get a positive rotation at C when M_CB is placed in the positive direction.
This means that the entire beam segment is assumed to deflect like this:
Now we are ready to get to the essence of this lecture,
defining the member-end moments in terms of end rotations.
Consider a simply supported beam subjected to a positive moment at A, like this:
I'm going to label the moment M_AB.
The beam is going to deflect upward.
This means, end A rotates counterclockwise,
but end B rotates clockwise.
Let's refer to the rotation at A as theta_A1,
and call the rotation at B, theta_B1.
We can easily determine these joint rotations in terms of M_AB,
using the virtual work method.
The step-by-step details of these computations are presented elsewhere.
Here I'm going to give the results only:
theta_A1 equals: M_AB L / 3EI
And, theta_B1 equals: M_AB L / 6EI,
where L is the beam's length,
E is modulus of elasticity of the material,
and I is the beam's moment of inertia about the axis of bending.
Let's examine the beam one more time.
But this time, place the positive, counterclockwise moment at B like this:
I'm going to refer to this bending moment as M_BA.
The beam is going to deflect downward like this:
I am going to denote the end rotations as theta_A2 and theta_B2.
Similar to the first loading case,
we can determine the two end rotations in terms of M_BA,
using the virtual work method.
Here are the expressions for theta_B2 and theta_A2.
Okay, now what do we get if we superimpose these two diagrams?
We get this diagram,
where this rotation can be written as: theta_A1 minus theta_A2
And this rotation can be written as: theta_B2 minus theta_B1, or:
Let's refer to this overall rotation as theta_A, and call this one theta_B, so we can write:
And:
Here we have two equations in terms of the member-end moments and rotations.
Let's treat M_AB and M_BA as unknowns,
and write them in terms of theta_A and theta_B.
To do so, first rearrange the equations like this:
Summing these equations gives us:
Or:
Substituting this for M_BA in one of these equations we get:
Finally, here are the slope-deflection equations for a beam segment
that is not subjected to any external loads.
These equations follow our sign convention,
as the rotations and moments are shown in the positive,
counterclockwise direction in this diagram.
We also need to talk about the case in which the beam segment is subjected to external loads.
How would such loads affect the slope-deflection equations?
Before we delve into that discussion,
let's spend time analyzing a few simple indeterminate beams
to better digest what we have learned so far.
Consider the following indeterminate beam.
It is subjected to a concentrated moment at the left end.
We start by isolating the joints of the structure.
Here, we have two joints and one beam segment, so we get:
We denote the joint rotations as theta_A and theta_B.
And refer to the end moments as M_AB and M_BA.
The slope-deflection equations for segment AB can be written as:
Since the member-end moments also appear at the joints, we get:
Note that at B, in addition to M_BA, we have an unknown moment reaction labeled M_B.
Therefore, joint equilibrium equations become:
Substituting this expression for M_AB,
and this expression for M_BA,
in the equilibrium equations we get:
Here we have two equations but seemingly three unknowns, right?
So how do we solve for M_B, theta_A and theta_B using only two equations?
A close examination of the problem however,
reveals that the equations contain only two unknowns.
We actually know the numeric value for theta_B.
It is zero.
Why?
Because the right end of the beam is fixed,
meaning there is no rotation at joint B.
This allows us to rewrite the equilibrium equations as:
Solving these for the unknowns we get:
Substituting 25/EI for theta_A,
and zero for theta_B in the slope- deflection equations, we get:
Now that we have determined the member-end moments,
we can easily calculate member-end shear forces.
Here is the complete free-body diagram for segment AB.
So we can write:
Solving these equilibrium equations for the unknown shear forces, we get:
Now transfer these forces to the joints:
and determine the vertical support reactions:
Here is the analysis results:
We will discuss the formulation of slope-deflection equations
for the case of externally loaded members in the next lecture.
For now, see if you can analyze the following beams using what we have learned so far: