In this session, I'm going to talk about stability of structures.
A structural system be it a beam, a truss, or a frame can be viewed as a rigid body
that is properly restrained by one or more supports.
Such a system is said to be externally stable,
if it does not undergo any rigid-body motion
regardless of the position of the applied loads.
Consider a simple frame structure supported by two rollers.
If we apply a horizontal load to the frame,
it starts rolling horizontally on the support surface,
the structure undergoes rigid-body motion.
Therefore, the frame is said to be externally unstable.
If, however, we replaced one of the rollers with a pin support,
the structure becomes externally stable,
as the pin prevents the rolling of the frame on the support surface.
Let's consider another example.
Here, we have a truss structure that rests on two supports: a pin and a roller.
Clearly, the supports prevent any rigid-body movement of the entire structure.
Yet, the truss is considered to be unstable.
Why?
Because, here we have localized rigid-body motion.
Some of the truss members are not properly restrained.
They can have limited free movement.
This is called internal instability.
A structure is said to be stable, if and only if, it is neither externally nor internally unstable.
To determine if a structure is stable, we need to ask three questions.
1) Does there exist a loading pattern that makes equation:
sum of the forces in the x-direction equals zero unsatisfiable.
2) Does there exist a loading pattern that makes equation:
sum of the forces in the y-direction equals zero unsatisfiable.
3) Does there exist a loading pattern that makes equation:
sum of the moments about the z-axis equals zero unsatisfiable.
If the answer to any of these questions is yes,
then the structure is considered unstable.
Consider a simple beam resting on two rollers.
Here, equation: sum of the forces in the x-direction = 0 cannot be satisfied,
if a nonzero force is placed on the beam.
Why?
Because summing the forces in the x-direction gives us P.
Since P is not zero, then the sum cannot be zero.
This implies that the structure is unstable.
Here is another example.
The structure is supported by a pin at A and a roller at B.
The free-body diagram for the structure shows three reaction forces, all passing through point A.
If a nonzero vertical force, P, is placed at the midpoint of member AB,
then the sum of the moments about point A becomes PL/2.
Since neither P nor L is zero, then the sum is not zero.
This means, the structure is unstable.
Let's look at one more beam example.
This beam has three reaction forces.
Note how all of them pass through a single point: point D.
If we subject the beam to a nonzero vertical force P at, say point A,
the moment equilibrium equation becomes P L.
Since this product is not zero, the equation is not satisfied.
Therefore, the structure is not stable.
Now, let's turn our attention to internal stability of structures.
The discussion generally comes up when dealing with trusses.
Assuming a truss is externally stable, we can determine its internal stability in one of two ways.
The simplest way is to compare the number of equilibrium equations
to the number of unknowns for the truss.
If the number of equations exceeds the number of unknowns, the structure is considered unstable.
Here is an example.
This truss has 6 joints and 8 members.
Since the number of equilibrium equations for a truss equals 2 times the number of its joints,
we get 2 times 6, or 12 equilibrium equations.
The number of unknown forces equals to the number of truss members plus the number of support reactions.
Here, we have 8 + 3, or 11 unknowns.
Since the number of equations turns out to be greater than the number of unknowns,
then we conclude that the truss is unstable.
This counting technique, however, does not work in all cases.
For example, consider this structure.
The truss has 8 joints and 13 members.
Therefore the number of equations equals 2 times 8, or 16.
And the number of unknowns equals 13 + 3, or 16.
Clearly here, the number of equations does not exceed the number of unknowns, yet the structure is unstable.
Why?
Because it can be shown that under certain loading pattern,
the static equilibrium of the structure cannot be maintained.
How?
Suppose we place a non-zero downward load of P at the midpoint of the bottom chord of the truss.
This makes the vertical reaction at the left and the right ends of the truss equal P/2.
Now suppose we conceptually cut the truss along a vertical plane as shown.
Then, we draw the free-body diagram of each segment.
We know that for the structure to be in equilibrium,
the equilibrium equations must be satisfied for each segment.
One of the equations, however, cannot be satisfied for either segment.
Sum of the forces in the y-direction is not zero.
For the left segment, we have: sum of the forces in the y-direction equals P/2.
For the right segment, we have: sum of the forces in the y-direction equals P/2 - P, or - P/2.
Since P is not zero, the segments are not in equilibrium.
This means the structure is unstable.
As you can see here, determining the internal stability of trusses may not be a simple task.
It may involve careful inspection of the geometry of the structure.
Generally speaking, if the geometry of a truss consists entirely of simple triangles, like this:
then the structure is considered stable.
Otherwise, the structure needs to be carefully inspected for geometric patterns that are inherently unstable.
Here is one such pattern.
Any truss structure that embodies this rectangular pattern is considered unstable.
see if you can figure out which of the following trusses are unstable.