Force Method: Part 2
This is a continuation of lecture SA24
where we begun discussing the use of the force method
for analyzing statically indeterminate structures.
Please watch that video for an overview of the method
and how it can be used to analyze beams having a degree of indeterminacy of 1.
In this lecture we are going to focus on the use of the method
for analyzing beams with a degree of indeterminacy of 2.
Here is one such beam:
It rests on one pen and three rollers.
Hence, there are five support reaction forces.
Although the beam is going to deform under the applied load, like this,
but there is no vertical displacement at the supports.
Vertical displacement at A, B, C, and D remain 0.
Since the degree of indeterminacy of the beam is 2,
if we remove two of the vertical reactions, we get a statically determinate beam.
Let's pick B_y and C_y as the redundant forces.
If we remove them from the free-body diagram, we get:
Consequently, since points B and C are no longer restrained against vertical movement,
the beam is going to deflect downward like this:
Let's refer to the vertical displacement at B as Delta_B,
and call the displacement at C, Delta_C.
Okay, now visualize a pair of upward forces.
One placed at B, and the other placed at C.
They collectively push the displaced beam up such,
that the vertical displacement at B and C reverts back to zero, like this:
These forces, labeled B_y and C_y, represent the support reactions at B and C, respectively.
In order to come up with the correct equations for calculating B_y and C_y,
we need to examine them individually.
First, let's consider the effect of B_y on the beam.
If we place an upward unit load at B,
the simply supported beam without the applied distributed load is going to deform like this :
Note that there is a vertical displacement at B, and there is one at C.
Let's refer to the displacement at B, as delta_BB,
and denote the displacement at C, as delta_CB.
Why am I using a double subscript notation for these displacements?
Because I want the symbol to denote the location of the displacement,
as well as of the location of the unit load causing the displacement.
So this refers to the displacement at B, due to a unit load applied at B.
And this refers to the displacement at C, due to a unit load applied at B.
The first subscript points to the location of the displacement.
The second subscript indicates the location of the unit load that has caused the displacement.
Now if we replace the unit load with B_y, these displacements can be written as:
B_y delta_BB,
and B_y delta_CB.
Next, we want to consider the effect of C_y on the beam.
If we place an upward unit load at C,
the beam is going to deform like this:
We refer to this displacement as delta_BC.
And we call this one, delta_CC.
delta_BC is the displacement at B, due to the unit load placed at C.
and delta_CC is the displacement at C, due to the unit load at C.
Now if we replace the unit load with C_y, the displacements can be written as:
C_y delta_BC,
and C_y delta_CC.
Let's summarize what we have so far:
The simply supported beam subjected to the uniformly distributed load has downward displacements:
Delta_B at B,
and Delta_C at C.
The simply supported beam subjected to the upward reaction force B_y has displacements:
B_y delta_BB at B,
and B_y delta_CB at C.
And the simply supported beam subjected to the upward force C_y has displacements:
C_y delta_BC at B,
and C_y delta_CC at C.
We also know that the sum of these three diagrams gives us this diagram, meaning:
Delta_B = (B_y delta_BB) + (C_y delta_BC)
This equation is a mathematical way of stating that total displacement at B must be zero.
And, Delta_C = (B_y delta_CB) + (C_y delta_CC),
which means a beam's displacement at C must be zero.
So we end up with a system of linear equations.
We have two equations and two unknowns.
B_y and C_y are the unknowns.
And these displacements are coefficients, which we can calculate using, say, the virtual work method.
We call these compatibility equations.
The solution to the compatibility equations gives us two of the unknown reaction forces.
We can then use the static equilibrium equations to determine the remaining support reactions.
Let's illustrate the entire process using an example.
Suppose we have a beam subjected to a uniformly distributed load of 50 N/m.
Assuming a constant EI, we wish to analyze the beam using the force method.
We start by drawing the structure's free-body diagram.
Since the beam's degree of indeterminacy is 2,
we need to pick two of the unknown reactions as redundant forces.
Any pair of vertical forces will do, as long as the remaining structure is stable.
Let's pick B_y and C_y.
Now, remove them from the beam.
Here is the resulting simply supported beam.
The beam is going to deform like this:
Note the vertical displacement at B, and at C:
I can now write the compatibility equations.
This equation states that: at point B,
The upward displacement due to the redundant forces
must be equal to the downward displacement due to the applied load.
This equation makes the same statement, but about point C.
Upward displacement due to the redundant forces,
must be equal to the downward displacement due to the applied load.
There are six displacements here.
Can you identify them on these diagrams?
Pause the video if you want to give it a try.
This is Delta_B.
This is delta_BB, displacement at B, due to a unit load applied at B.
This is delta_BC, displacement at B, due to a unit load applied at C.
This is Delta_C.
This is delta_CB, displacement at C, due to a unit load applied at B.
And this is delta_CC, displacement at C, due to a unit load applied at C.
By the way, calculating these displacements
is the most time-consuming part of the force method.
I'm going to use the virtual work method for this.
But since the method was explained in previous lectures,
I'm going to go through the calculations rather quickly without any commentary.
If you're not sure how the virtual work method works, please see lecture SA22.
First, let's calculate Delta_B.
So, Delta_B equals:
Now let's compute Delta_C.
Delta_C equals:
Next we calculate delta_BB:
And here is delta_BC:
Finally, we determine delta_CC:
And, delta_CB:
Okay, now we're ready to solve the compatibility equations.
Here are the two equations in simplified form:
Solving them simultaneously we get:
B_y equals:
And C_y equals:
Now we are ready to calculate the remaining support reactions
using the static equilibrium equations.
Here is the result of the analysis:
We will continue our discussion on this topic in the next lecture.
Here are a few exercise problems involving statically indeterminate beams and frames,
all having a degree of indeterminacy of two.