Structural analysis, statically determinant beams.
In this lecture, we are going to answer two questions:
What is a statically determinate beam? and how do we analyze one?
A beam is a properly supported structural member capable of carrying loads that are perpendicular to its longitudinal axis.
Such a member is called a beam because the loads cause it to bend, like this.
The analysis of such a beam involves calculating its support reactions
In this example, beam analysis entails calculating three support reactions: Ax, Ay, and By.
What is a statically determinate beam?
A beam is called statically determinate if its unknown reaction forces can be calculated solely using the equilibrium equations.
How do we analyze such a beam?
Before we can answer the question, let's review reaction forces associated with each type of support.
The three most common beam support types are: pin, roller and fixed supports.
There are two reaction forces associated with a pin support: A force in x direction and a force in y direction.
A roller support provides one reaction force only: a force that is perpendicular to the support surface.
A fixed support has three reactions: a force in the x direction, a force in the y direction, and a bending moment about the z-axis.
To clarify, let's look at two examples.
Suppose our beam has a pin support and a roller support.
Here, there are three reaction forces.
Two forces associated with the pin support and one force associated with the roller support.
In this example, the beam is fixed at one end, but free at the other end.
Therefore, there are three reactions involved here.
A force in the x direction, a force in the y direction, and a bending moment.
in order to analyze a beam, it is customary to draw a diagram called the free-body diagram,
showing the applied and reaction forces, and their locations in the Cartesian coordinate system.
Note how each unknown force in the diagram has been uniquely labeled.
The free-body diagram helps us to correct formulate of the equilibrium equations.
To determine the support reactions, that is, to analyze the beam, we use three equations:
One, sum of the forces in the x direction must be zero.
Two, sum of the forces in the y direction must be zero.
Three, sum of the moments about the z axis at, say, point A must be zero.
The first equation, in the expanded form gives us: Ax=0.
The second equation gives us: Ay + By - P = 0.
The third equation gives us: PL/2 – By L = 0.
Solving the third equation for By gives us: By = P/2.
Solving the second equation for Ay gives: Ay = P/2.
Before we end this session, let's look at two more examples.
Here, we have a cantilever beam subjected to a uniformly distributed load.
Let’s analyze the beam. First, we draw the free-body diagram for the beam.
For the purpose of finding the support reactions, we then replace the uniformly distributed load with its equivalent concentrated load.
Now, we can write the equilibrium equations.
Ax = 0
Ay - 20 = 0
Ma - 20 (5) = 0
Solving the last equation for Ma gives: Ma = 100.
From the second equation, we get: Ay = 20.
And the first equation gives us: Ax = 0.
The second example is a beam with a pin support at the left end, a roller support at the right end, and another roller support at the middle of the beam.
The beam is subjected to two concentrated loads.
To analyze the beam, we start by drawing its free-body diagram.
Note that the diagram shows four unknown forces: two forces for the pin support, and one reaction for each of the roller support.
Here, the equilibrium equations are:
Ax = 0
Ay + By + Cy - 2 P = 0
5P + 15P - 20Cy - 10By = 0.
Notice that we have four unknowns but only three equations.
Therefore, we cannot solve for the unknowns. The beam is not statically determinate.
We will discuss techniques for analyzing statically indeterminate beams in later lectures.