We introduced the method of joints for analyzing statically determinate trusses in lecture SA04.
This lecture offers a review of this method in the context of a real-life structure.
Here, we have a one-story house, converted from an abandoned municipal garage,
which was located in a residential neighborhood.
The pitched roof of the garage was supported by six trusses.
For aesthetic and economic reasons,
the architect in charge of the renovation kept them in their original form.
In this lecture, we are going to analyze one of these trusses.
More specifically,
we want to use the method of joints to calculate member forces
in the exterior truss at the near end of the building,
this truss:
To begin, let's determine the loads being applied to our truss.
Generally, when designing such structures,
we need to consider several load types including dead load,
live load, snow load, wind load and seismic load.
For the sake of clarity and brevity however,
in this lecture, we are going to consider only one load type,
the dead load associated with the roof.
The key dimensions of the structure are:
Note that the left roof panel rests on three long beams,
which in turn are supported by the trusses.
So the roof's weight is going to be transferred to the trusses via these long beams.
Each beam is going to support a part of the roof's weight.
But how do we exactly distribute the weight among the beams?
To figure this out, we start by drawing parallel lines,
dividing the distance between each pair of consecutive beams in half, like this:
This way, we subdivide the root panel into three areas.
We call these tributary areas for the beams.
That is, the weight of the roof associated with this area is going to be supported by this beam:
The weight here is going to be supported by this beam:
And this beam supports the weight of this area:
Let's assume that the weight of the roof is 3 kN/m^2.
This represents the intensity of the load distributed uniformly over the roof panel.
We are going to convert this area load for each tributary area
into a line load applied to the beam supporting the area.
How?
We simply multiply the load intensity by the width of the tributary area.
For this area, we get:
So, this beam is going to be subjected to a uniformly distributed load of 5.52 kN/m, like this:
For this area, the uniformly distributed load on the beam becomes:
And here, we get:
Now, we need to transfer the beam loads to the trusses supporting the beams.
Here are the beams resting on the trusses:
Since each beam is supported by six trusses,
we can view it as a continuous beam with six pin supports, like this:
The supports are the points of contact between the beam and the trusses.
These are the points through which the beam load transfers to the trusses.
So, in order to determine the loads applied to our truss,
we need to calculate the beam's support reactions.
To do so, we can analyze this statically indeterminate beam.
If we did, we would get:
Alternatively, since the load is uniformly distributed over the entire length of the beam,
we can estimate the reactions by dividing each beam segment in half.
Then, we can consider each of these,
as a tributary segment for the reaction force supporting the distributed load above.
Therefore, the reaction forces can be estimated to be:
However, since we already have the more accurate values for the reaction forces,
let's use them instead of the estimated ones.
The upper roof beam exerts a force of 7.7 kN at this truss joint.
The middle beam exerts a force of 15.4 kN at this joint.
This joint also carries a force of 15.4 kN.
These are the forces applied to the left half of the truss.
To simplify the problem,
we can conservatively assume that the other half of the truss
is subjected to an identical set of forces, like this:
So, what are the applied loads for our truss analysis? Here they are:
Now we are ready to analyze the truss.
We take the left end of the truss as a pin support,
since the truss is connected to the supporting column using a few bolts.
The other end of the truss is resting on a large friction pad
that permits limited lateral movement,
so we treat it as a roller support.
Here is the line diagram of our truss:
Let's take a few seconds to examine the truss.
Do you see any zero-force members in the structure?
If there are such members, we can remove them now, before we start the analysis.
That would simplify our calculations.
I see eight such members.
If you're not sure how to identify zero force members,
please review lecture SA05.
Where are these members?
One is member BI,
another one is CK,
then there is DM,
and we have EO.
None of these members carry any force.
Why?
See lecture SA05.
So, let's remove them from the structure.
Now, take another look at the truss.
Do you see any other zero force members?
There are four more:
AG,
GH,
FQ,
and PQ are also zero force members.
Let's remove them from the truss.
the remaining truss has 11 joints.
Note that the truss geometry and loads are symmetrical about the center line of the structure.
That is, for every member in the left half of the truss,
there is a corresponding member in the right half of the truss.
Also, for every load applied to the left half,
there is a corresponding load applied to the right half.
So AH and FP carry the same axial force.
HJ and NP carry the same force,
so do JL and LN, AB and EF,
BC and DE and so on.
We can take advantage of this symmetry to reduce the amount of computation involved.
To do so however, we first need to calculate the support reactions.
There are three support reactions.
Applying the equilibrium equations to the entire system, we get:
Solving these equations for the unknown reactions, we get:
Now we are in a position to analyze the truss joint by joint.
We start at joint A.
Here is the free-body diagram of the joint:
Summing the forces in the X and Y directions, we can calculate the two unknown forces, like this:
Now we move to joint H.
Here is the free-body diagram of the joint:
Here are the joint equilibrium equations:
And here are the unknown forces:
Now we can draw this free-body diagram for joint B.
Then, we write the equilibrium equations and solve them for the two unknown forces, like this:
For joint J, we can write:
And for joint C, we have:
We are done with our analysis.
Let's summarize our findings by writing the member forces on the truss, like this:
Tensile forces and members are shown in blue.
And compressive forces and members are shown in red.
What about the rest of the truss members?
Since the structure is symmetrical,
we can easily determine the member forces for the right half of the truss. They are:
This member is the mirror image of this one
relative to the vertical axis passing through the center of the truss.
So the two members carry the same force.
Please keep in mind that we can do this only because:
one, the geometry is symmetrical,
and two, the applied load pattern is symmetrical about the vertical axis
passing through the center of the truss.
Otherwise, this shortcut would lead to erroneous results.
Using the same line of reasoning, we can conclude that these members carry the same force,
so do these members, and so on.
We can analyze the other roof trusses in a similar manner.
Give it a try, follow the procedure I just described to analyze one of the interior trusses in the house.