In this lecture,
we are going to use the moment distribution method
to analyze a steel frame belonging to the skeleton of this glass house.
The house, known for its elegant simplicity,
is an architectural landmark designed and built by Phillip Johnson in 1949.
It is located near New York City.
This three-dimensional model is not a replica of the actual structure.
It has been modified to suit our purposes in this lecture.
The roof and the glass walls of the house are supported by a steel skeleton,
consisting of four beams,
and eight columns.
The three-dimensional frame measures 16.5 m in length,
10 m in width,
and 5.5 m in height.
For the purpose of our two- dimensional frame analysis,
we can view the skeleton as consisting of two long frames,
and two short ones.
Here, we are going to analyze one of the long frames under the dead roof load only.
For the sake of simplicity, let's assume there is no opening in the roof,
that we have a solid rectangular slab supported by the steel skeleton.
The roof is 208mm thick concrete slab weighing 24 kN/m^3.
This makes a uniform distributed load of 5 kN/m^2.
Since the roof rests on four beams, its weight is distributed among the beams.
We can determine the tributary area for each beam according to the yield line theory.
This is the correct approach for load distribution when we are dealing with two-way slabs
where the ratio of longer span to shorter span is less than two.
Here, the ratio is 1.65.
According to this theory, when bending moment in the slab reaches its ultimate value,
cracks develop along what are known as yield lines.
These are straight lines that run from the middle of the slab to the corners,
bisecting the 90-degree angles, like this:
So, we can consider each of these areas to be the tributary load area for its edge beam.
This means the long beams on the side of the roof are going to be subjected to a trapezoidal load,
and the short beams are going to be subjected to a triangular load.
Here, since we're focusing on this frame,
we only need to take into consideration this tributary area:
We need to transform the trapezoidal area load to a trapezoidal line load,
in order to place it on our long beam.
This transformation is rather straightforward.
All we have to do is multiply the magnitude of the area load, which is 5 kN/m^2,
by the height of the trapezoid.
Since the height is 5m, we end up with a load magnitude of 25 kN/m.
In essence, we are turning the flat trapezoidal area 90 degrees
and placing it on top of the beam, thereby turning it into a line load.
So, here is the frame that we need to analyze:
Since the columns are embedded in concrete blocks,
we consider them to be fixed at the base.
The rigid connection made between the steel member
and the concrete prevents rotation at the base of the column.
We are going to assume that all the members in the frame have the same section and material properties.
So, we have a constant EI throughout the structure.
Let's label the two ends of each structural member.
Since the frame has a constant EI,
we can calculate the distribution factor with relative ease.
At the fixed supports we get:
At joint B, we get 1/2 at each end.
At joints D and F, we get 1/3 at each of the three ends.
And, we get 1/2 at the ends of joint H.
Now, let's calculate the fixed-end moments.
For member DF, since it is subjected to a uniformly distributed load, we get:
For members DB in FH, the calculation is a bit more involved since the load is trapezoidal.
You can see the calculation details in video SA34-A.
Here are the results:
Now we can set-up the moment distribution table for calculating the member-end moments.
After going through the process of balancing the joints,
and determining the carry-over moments for a few iterations,
we get these results:
Knowing the moment values, we can easily determine the member-end shear forces:
Once the shear forces are determined,
we have all the support reactions. They are:
So, here are the results of our frame analysis using the moment distribution method: