In this lecture,
we are going to use the slope-deflection method to analyze a frame structure subjected to live load.
Please review videos SA27 through SA33
for background information on the slope-deflection method before proceeding with this presentation.
This charming bungalow sits at the edge of a jungle,
with a clear view of the surrounding hills and valleys in the town of Matugama in Sri Lanka.
It was designed by architect Narein Perera and was constructed in the year 2010.
The three-bedroom house is made of steel, timber and bamboo,
in order to blend in with its surroundings, while maintaining a small footprint.
In this lecture, we focus on the analysis of the highlighted frame.
More specifically, we wish to determine the maximum positive bending moment
in the middle segment of the long horizontal beam under the design live load.
The floor rests on three long beams.
Given its rectangular shape, we can distribute the load to the three beams.
To do so, we divide the floor area between two adjacent beams in half, like this:
The load associated with each segment, there are four of them,
is to be carried by the beam closest to it.
This results in three tributary areas, one for each beam.
Here they are:
Since we want to analyze the near frame, we are only interested in this tributary area.
Suppose the floor carries a design live load of 1.94 kN/m^2.
Bending moment in the middle segment of the beam reaches its maximum positive value
when the segment itself is loaded
but there is no load applied to any of the adjacent segments.
So we are going to place the live load on the middle segment only, like this:
Since the floor is resting on a series of short beams,
we need to distribute the live load among the beams.
This is done, again, based on the tributary area associated with each beam.
Each beam, except for the one at the far end, has a tributary area equal to:
(3/4 m) by (3.33 m)
The tributary area for the far end beam is half as wide as the other ones.
So, that area equals to (3/8 m) by (3.33 m).
Each tributary area is subjected to a uniformly distributed load block.
Let us replace each block with a concentrated load
that would act directly on the long horizontal beam.
The magnitude of the concentrated load
equals to the tributary area, times the magnitude of the live load.
This gives a concentrated load of 4.8 kN for each interior beam,
and 2.4 kN for the far end beam.
So, we need to analyze our frame under this set of loads:
We consider the left end of the horizontal beam as a pin support.
The other two concrete blocks act as fixed supports.
Here are the key dimensions of the frame:
Let's assume a constant EI for the entire steel structure.
Note that since the cantilever beam segment to the right of joint C is not subjected to any loads,
it is not going to carry any internal forces either.
Therefore, to simplify our analysis, we are going to remove that segment from our diagram.
Since we are going to use the slope-deflection method, let's start by calculating the fixed-end moments.
Here, BC is the only member that has non-zero fixed-end moments.
Here are the fixed-end moments for a typical beam subjected to a concentrated load of P.
There are 15 concentrated loads applied to our member.
Let's calculate the left and right fixed-end moments for each load.
Now add up all the values in this column,
to get the total fixed-end moment for the left end of the beam.
And add up these values, to get the fixed-end moment for the right end of the beam.
We are now ready to use the slope-deflection method.
Note that the frame has three non-zero joint rotations at joints A, B and C.
Therefore, we need to write three joint equilibrium equations.
Let's begin by writing the slope- deflection equations for each member.
Here are the generic equations:
Applying them to member AB, we get:
For BC, we get:
The equations for BD are:
And for CE and CD, we have:
We can simplify the slope-deflection equations like this:
Now we are ready to write the joint equilibrium equations.
At joint A, there is only one internal moment, M_AB, so we can write:
At B, there are three member-end moments.
Here is the equilibrium equation for the joint.
There are also three member-end moments at joint C,
resulting in this equilibrium equation:
We are now going to expand these equations using the slope-deflection equations.
Solving these equilibrium equations simultaneously for the unknown rotations, we get:
Substituting these values in the slope-deflection equations, we get the member-end moments.
We are almost done.
To determine maximum positive moment in member BC,
let's draw the shear and moment diagrams for the member.
Knowing the member-end moments for the segment,
we can calculate its member-end shear forces using the equilibrium equations.
Now, first draw the shear diagram,
then, use it to construct the moment diagram.
If you are not sure how these diagrams are constructed, see Lectures SA06 through SA09.
So the maximum positive moment in the segment
occurs under the seventh concentrated load from the left end of a beam.
In conclusion, when considering floor live load only,
the maximum positive moment in the middle segment of our target frame
develops at this point under the floor.