The force method is a classical technique for analyzing statically indeterminate structures.
Before we get into the details of the method,
let's spend a minute talking about statically indeterminate structures.
We call a structure statically indeterminate, if its number of unknown forces
is greater than the number of equilibrium equations.
For example, this is a statically determinate beam,
because the number of equilibrium equations that we can write for the beam,
equals to the number of unknown support reactions.
There are three equations and three unknowns.
This beam on the other hand, is statically indeterminate.
Why?
Because although we still have three equilibrium equations,
but the number of unknown forces has increased to 4.
So we have more unknowns than equations.
This means we cannot calculate the unknowns directly using the equations.
That is why the beam is called statically indeterminate.
In this example, the beam has a degree of indeterminacy of 1.
That is, if we remove one of the unknowns, the structure becomes statically determinate.
For example, if we remove the support reaction at B,
we get a simply supported beam.
So now let's talk about the force method.
The underlying idea behind this method is rather straightforward.
Here it is.
Suppose we have a statically indeterminate beam,
with a degree of indeterminacy of one.
say this beam:
Since the beam is resting on a pin at A,
and there is a roller at B and C,
no vertical displacement takes place at these points.
That is, while the beam is going to deform like this:
Vertical displacement at A, B and C remains zero.
Now suppose we remove one of the support reactions,
thereby turning the beam into a statically determinate one.
Say we remove the roller at B.
What do you think the deformed shape of the beam would look like?
It's going to look like this:
Let's refer to the vertical displacement at B as Delta_B.
So by removing a support reaction at B,
we've allowed the beam to displace downward by Delta_B at B.
Now let's place an upward vertical force at B,
and gradually increase its magnitude.
What would happen to the displacement of the beam at B?
It becomes smaller and smaller, as the force magnitude increases.
When the magnitude of the upward force reaches a specific value,
vertical displacement of the beam at point B becomes zero.
Let's refer to this force magnitude as B_y.
B_y indeed is the support reaction at B in the statically indeterminate beam.
Why, you ask?
Note the similarity between the deformed shape of a statically indeterminate beam,
and the statically determinate beam with the additional force at B.
Their displacement shapes are identical.
Since the two beams have identical geometric shape and properties,
and their load and displacement patterns are the same,
then they must have identical external and internal forces.
Therefore B_y, here, must be the support reaction at B, here:
So to determine the support reaction at B,
all we have to do is calculate Delta_B in this beam,
then determine the force magnitude that causes an upward deflection of Delta_B in this beam.
This upward force represents the vertical reaction at B in the statically indeterminate beam.
Put it differently, this, plus this, gives us this:
Let's write the equation that models this process.
We are going to denote the upward deflection of the beam
under a unit concentrated load at B, as delta_B.
Assuming linear elastic material,
then upward deflection of the beam due to a load with magnitude B_y,
can be written as: B_y delta_B
And since the vertical downward deflection at B,
under the distributed load is Delta_B, then we can write:
Or, B_y = Delta_B / delta_B.
To calculate B_y, we need to calculate the displacements in the numerator
and denominator of this ratio.
This can be done using the method of virtual work.
Let's go through the process of actually analyzing the beam.
Say, segments AB and BC, each has a length of L.
The intensity of the distributed load is w.
And the modulus of elasticity of the material,
and the beam's moment of inertia are constant.
Here's the free-body diagram of the beam:
Then we can write:
Where the numerator represents the vertical displacement at B,
under the uniformly distributed load,
and the denominator represents the vertical displacement at B,
under an upward unit load applied at B.
To calculate these displacements we are going to use the method of virtual work.
Here I'm going to use the method without explaining it.
For details of the virtual work method see lecture SA22.
We can calculate Delta_B using this equation,
where M represents bending moment equation in the beam, due to the distributed load,
and m* represents bending moment in the beam,
due to a unit load applied in the direction of the displacement at B.
Here is M:
And here's m*:
Then we can write:
Or, Delta_B equals:
To determine the other displacement, delta_B, we use this diagram:
Here delta_B can be written as,
where M represents bending moment in the beam due to the unit load,
and m* represents bending moment due to a virtual unit load.
Needless to say, we are going to have two identical moment equations in this case,
since the real and virtual loads are the same.
Solving for delta_B, we get:
Now we can determine B_y.
Once B_y is calculated,
the statically indeterminate beam becomes statically determinate.
This means we can now easily determine the remaining support reactions
using the static equilibrium equations.
We will examine the use of the force method for analyzing indeterminate beams
with degree of indeterminacy greater than 1 in the next lecture.
For now, see if you can analyze the following statically indeterminate beam:
The beam has four support reactions.
there are three reaction forces at A, and one reaction force at B.
You can use most of these reactions as the redundant force,
the force that should be removed from the beam,
in order to change it into a statically determinate one.
By the way, can you tell why A_x cannot be treated as a redundant force?