The Conjugate Beam Method
How would you go about determining the deflection in a beam at an arbitrary point,
say deflection at point C in this beam?
One approach is to first derive an equation for deflection and then evaluate it at the point of interest.
The double integration method discussed in previous lectures is an example of this approach.
Another approach entails calculating deflection at the desired point,
without needing to write an explicit equation for deflection.
The conjugate beam method is an example of this approach.
This lecture presents and illustrates conjugate beam method
for calculating deflection in simply supported beams.
To start let's review the relationship between applied load and shear and moment in beams.
Shear can be defined in terms of the integral of the load function.
v(x) equals the integral of w(x),
where w(x) is the load function, and v(x) is the shear function.
The same relationship can be expressed graphically using load and shear diagrams.
For example, if w(x) represents a uniformly distributed load, then the load diagram looks like this:
And the shear diagram looks like this:
Similarly bending moment can be defined as:
m(x) equals the integral of v(x),
which equals to the double integral of w(x), since v(x) is defined as the integral of w(x).
So, if we express the applied load as a mathematical function,
its first integral describes shear in the beam,
and the function's second integral describes moment in the beam.
That is, this is a graphical representation of the integral of w(x),
and this is a graphical representation of the double integral of w(x).
But, what does all this have to do with calculating deflection?
We're going to use the relationship between load, shear and moment
as a template for calculating slope and deflection.
Let's see how.
Slope in a beam can be defined as the integral of M/EI, where M is bending moment,
E is modulus of elasticity,
and I is moment of inertia.
And deflection is defined as the integral of slope, or the double integral of M/EI.
See lecture SA12 if you would like to know why these equations hold true.
Note the similarity between these equations and these equations:
Here, if we view M/EI as a continuous load applied to a beam,
then theta represents shear in the beam, and Y represents moment in the beam.
Let me restate what I just said.
If we draw the moment diagram for the real beam,
divide the diagram by EI, to get what is known as M/EI diagram,
and treat this diagram as a distributed load placed on a beam similar to the real beam,
then shear in that beam corresponds to the slope of the elastic curve.
And bending moment in that beam corresponds to the real beam's deflection.
Here is a quick overview of these steps.
Suppose we wish to determine this beam's
deflection under the applied concentrated load at point C.
The beam has a constant EI.
We start by drawing the beam's moment diagram.
Divide the moment diagram by EI,
forming the M/EI diagram.
Now, draw what is called the conjugate beam.
I will explain how the conjugate beam may differ from the real beam in a minute.
For this problem the conjugate beam is the same as the real beam.
Place the M/EI diagram on the conjugate beam as applied load.
Then, determine the bending moment at point C in the conjugate beam.
This is the deflection in the real beam at point C.
Before we look at a few more examples, let's see how the conjugate beam could vary from the real beam.
The conjugate beam has the same length as the real beam, always.
The supports for the conjugate beam however,
could be different than those in the real beam.
If there is a fixed support at either end of the real beam,
we need to turn it into a free end in the conjugate beam.
Why?
Keeping in mind that shear in the conjugate beam represents
slope of the elastic curve in the real beam,
and moment in the conjugate beam represents deflection in the real beam,
then shear and moment at the boundaries of the conjugate beam
must match slope and deflection at the boundaries of the real beam.
Since slope and deflection at the fixed end of a real beam are zero,
then shear and moment at that end in the conjugate beam must also be zero.
The only end condition that produces zero shear and zero moment is a free end.
That is, there is no shear or moment at the free end of the beam.
Therefore the conjugate beam must have a free end where the real beam is fixed.
What if the real beam has a free end?
What would be the corresponding support in the conjugate beam?
The free end in the real beam becomes a fixed support in the conjugate beam. Why?
Because the real beam has a non-zero slope and a non-zero deflection at the free end.
Therefore, the conjugate beam must have a non-zero shear and a non-zero moment at the point.
The only support type that produces a non-zero shear and moment is a fixed support.
What if the real beam rests on a pin or a roller,
what would be the corresponding support type in the conjugate beam?
Well, that depends on the location of the support.
If the pin or the roller is an end support,
then the corresponding support in the conjugate beam is also going to be a pin or roller.
Why? Because a pin or a roller support allows rotation, but prevents deflection.
That is, the beam would have a non-zero slope, but a zero deflection at the support.
Therefore the conjugate beam must have a non-zero shear,
but a zero moment at that point.
A pin or a roller support placed at that end of the conjugate beam
indeed produces a non-zero shear and a zero moment.
If we're dealing with an interior support, if the roller or the pin is placed away from the beams ends,
then the conjugate beam is going to have an internal hinge at that point.
This is the case because such a hinge forces bending moment to vanish,
but permits shear to be developed at that point.
In summary, if this is the real beam,
then the conjugate beam should look like this:
Or, if this is the real beam,
then this would be the conjugate beam:
For this real beam,
we should use this conjugate beam:
So what if the real beam looks like this?
Can you draw the conjugate beam for it?
This turns into a free end,
this remains the same,
and this,
well here the beam has a deflection,
and the slope of the elastic curve is non-zero, and discontinuous.
These conditions need to be matched in the conjugate beam.
At C, there must be a non-zero moment, and a discontinuity in shear value.
A pin or roller at C produces the desired conditions.
So the conjugate beam looks like this:
Now let's look at a few examples: