Post on 13-Aug-2020
transcript
1
Structural Analysis Method of Joints / Method of Pins
In the air duct installers union they have lots of opportunity to vent.
Wednesday, October 21, 2009 Method of Joints 2
Trusses
Trusses are common means of transferring loads from their points of application to the supports
They vary in design but their analysis always follows the same pattern
Wednesday, October 21, 2009 Method of Joints 3
Tools
Equations of equilibrium Free Body Diagrams Trig and algebra Visualization
2
Wednesday, October 21, 2009 Method of Joints 4
Trusses
A simple truss analysis, which we will do in this class, does not consider the weight of the members which make up the truss
A simple truss analysis considers that all loading is made at connections
A simple truss analysis considers that all truss members are two-force members
Wednesday, October 21, 2009 Method of Joints 5
Trusses
This means that we can break a truss down into a collection of elements for analysis
Wednesday, October 21, 2009 Method of Joints 6
Trusses
An example truss
3
Wednesday, October 21, 2009 Method of Joints 7
Trusses
The red arrows are the loads (applied externally)
Wednesday, October 21, 2009 Method of Joints 8
Trusses
The dark black lines are the members
Wednesday, October 21, 2009 Method of Joints 9
Trusses
Each member goes from connection to connection
4
Wednesday, October 21, 2009 Method of Joints 10
Trusses
The black line along the bottom of the truss represents three separate members
Wednesday, October 21, 2009 Method of Joints 11
Trusses
The support conditions are the pin at the left and the roller at the right
Wednesday, October 21, 2009 Method of Joints 12
Trusses
At each intersection of two or more members, we consider that the members are pinned together
5
Wednesday, October 21, 2009 Method of Joints 13
Trusses For example, at the point labeled G on the
diagram, we have four members coming to a point connected by a single pin
Wednesday, October 21, 2009 Method of Joints 14
Trusses
To simplify our analysis, we assign each connection point with a letter
Wednesday, October 21, 2009 Method of Joints 15
Trusses I avoid using F and I as connection point
assignments but your text does not
6
Wednesday, October 21, 2009 Method of Joints 16
Trusses Pick a starting point and label each connection, I
chose to start at the lower left and move across the top and then across the bottom
Wednesday, October 21, 2009 Method of Joints 17
Trusses Each of the members can now be labeled using
the points of connection
Wednesday, October 21, 2009 Method of Joints 18
Trusses In this truss we have the following members: AB,
BC, CD, DE, CE, CG, EG, BG, and AG
7
Wednesday, October 21, 2009 Method of Joints 19
Trusses The order of the letters is not important, I choose
to put them in the order they appear in the alphabet
Wednesday, October 21, 2009 Method of Joints 20
Method of Joints To begin the analysis of the truss using this
method, we start by identifying all external forces acting on the truss
Wednesday, October 21, 2009 Method of Joints 21
Method of Joints In this case it would be the forces represented by
the red lines The forces (loads) must always be applied at the
connections (joints)
8
Wednesday, October 21, 2009 Method of Joints 22
Method of Joints
This is usually given in the problem
Wednesday, October 21, 2009 Method of Joints 23
Method of Joints Then we remove any supports and replace them
with reactions provided by the supports
Wednesday, October 21, 2009 Method of Joints 24
Method of Joints We have a pin at the left so we will have an x
and a y reaction
9
Wednesday, October 21, 2009 Method of Joints 25
Method of Joints And we have a roller at the right so we only have
a y reaction
Wednesday, October 21, 2009 Method of Joints 26
Method of Joints We are also usually given the dimensions and
geometry of the truss in the given
Wednesday, October 21, 2009 Method of Joints 27
Method of Joints Now we can solve for the reactions at the
supports using the methods developed previously
10
Wednesday, October 21, 2009 Method of Joints 28
Method of Joints We will use the entire truss as the FBD and
ignore the internal members (for the moment)
Wednesday, October 21, 2009 Method of Joints 29
Method of Joints Summing forces in the x-direction
Wednesday, October 21, 2009 Method of Joints 30
Method of Joints Summing moments about D
11
Wednesday, October 21, 2009 Method of Joints 31
Method of Joints Summing forces in the y-direction
Wednesday, October 21, 2009 Method of Joints 32
Method of Joints
Redrawing our system with the known reactions, we have
Wednesday, October 21, 2009 Method of Joints 33
Method of Joints Now for the problems that we had done earlier,
we would be finished
12
Wednesday, October 21, 2009 Method of Joints 34
Method of Joints In a truss analysis, we want to know what force
is being carried in each of the members
Wednesday, October 21, 2009 Method of Joints 35
Method of Joints
Each member of the truss can be in one of three conditions: It can be in COMPRESSION, that means that it
will be pushing on each end where it is connected
It can be in TENSION, that means that it will be pulling on each end where it is connected
It can have no force in the member
Wednesday, October 21, 2009 Method of Joints 36
Method of Joints
In the method of joints, we are going to assume the condition of each of the members,
draw a free body diagram of the connection points, and
apply two of our three equilibrium conditions to solve for the force actually in the members
13
Wednesday, October 21, 2009 Method of Joints 37
Method of Joints
When we used a pin connection previously, we used an x and a y reaction to show what was being provided by the pin
Here we are actually drawing a FBD of the pin itself and showing the forces from the members onto the pin itself
Wednesday, October 21, 2009 Method of Joints 38
Method of Joints
You can assume any condition that you want for the members, I usually choose to assume that they are all in either tension or compression and then let the signs of the solutions dictate their actual condition
Wednesday, October 21, 2009 Method of Joints 39
Method of Joints
Since we can only use two of the equations of equilibrium, the sum of the forces in the x and the sum of the forces in the y direction, our analysis pattern will be dictated by the makeup of the truss itself
14
Wednesday, October 21, 2009 Method of Joints 40
Method of Joints
First, we can look at all the forces acting on each pin of the truss
It is critical to the analysis that you do not exclude either external forces or support reactions at each pin
Wednesday, October 21, 2009 Method of Joints 41
Method of Joints
If a member is pushing on a pin at one end, it is always pushing on the pin at its other end
Wednesday, October 21, 2009 Method of Joints 42
Method of Joints
Conversely, if a member is pulling at a pin on one end, it is pulling on the pin at its other end
Never push pull!
15
Wednesday, October 21, 2009 Method of Joints 43
Method of Joints
At any joint, you can only solve for two unknowns!!!!
Wednesday, October 21, 2009 Method of Joints 44
Method of Joints
Now to decide where to start the analysis, we need to find a joint where there are only two unknowns
Wednesday, October 21, 2009 Method of Joints 45
Method of Joints
In our case, the pins at A and at D both only have two unknowns, so we can start at either point
16
Wednesday, October 21, 2009 Method of Joints 46
Method of Joints
If you thought that there were three unknowns at A and D because of ABx and ABy, remember that the force in the member has a line of action along the member
Wednesday, October 21, 2009 Method of Joints 47
Method of Joints
We choose A to start our analysis We now draw a FBD of the pin at A
Wednesday, October 21, 2009 Method of Joints 48
Method of Joints
I always assume that the members are in compression so our FBD would look like
17
Wednesday, October 21, 2009 Method of Joints 49
Method of Joints
Since I assume compression, the forces in the members are pushing on the connection (pin)
Wednesday, October 21, 2009 Method of Joints 50
Method of Joints
Now we can use our two equilibrium conditions to solve for the forces in the members
Wednesday, October 21, 2009 Method of Joints 51
Method of Joints
We report all the magnitudes of the forces as positive numbers and the forces themselves as either compression or tension
18
Wednesday, October 21, 2009 Method of Joints 52
Method of Joints
Solving for AG
Wednesday, October 21, 2009 Method of Joints 53
Method of Joints
Now that AB and AB are known forces, we can revisit the FBD of the complete truss and find another joint/pin that has only two unknowns
Wednesday, October 21, 2009 Method of Joints 54
Method of Joints
Usually this will be adjoining the pin that we just solved for
For this truss it could be either pin B or G
19
Wednesday, October 21, 2009 Method of Joints 55
Method of Joints
At G, we have BG, CG, and EG as unknowns Three unknowns, so this pin won’t work
Wednesday, October 21, 2009 Method of Joints 56
Method of Joints
At B, we have two unknowns, BC, and BG We can proceed to work at BG
Wednesday, October 21, 2009 Method of Joints 57
Method of Joints
We draw a FBD of the pin at B
20
Wednesday, October 21, 2009 Method of Joints 58
Method of Joints
We know that AB is in compression, so we are not making an assumption about AB
Wednesday, October 21, 2009 Method of Joints 59
Method of Joints
We are assuming that BC and BG are in compression for this FBD
Wednesday, October 21, 2009 Method of Joints 60
Method of Joints
Now we use our conditions of equilibrium to solve for BC and BG
21
Wednesday, October 21, 2009 Method of Joints 61
Method of Joints
Now we use our conditions of equilibrium to solve for BC and BG
Wednesday, October 21, 2009 Method of Joints 62
Method of Joints
You might be asking yourself why BG is even included in the truss
This is a very preliminary analysis of one loading condition, there is actually much more to truss design than this
Wednesday, October 21, 2009 Method of Joints 63
Method of Joints
Now we could go to either joint C or joint G At C, we have CG, CE, and CD as unknowns At G, we have CG and EG Our next point is G
22
Wednesday, October 21, 2009 Method of Joints 64
Method of Joints
The FBD at G Notice that we have drawn AG as a tension
member, we know that from our earlier solution, it is no longer an assumption
Wednesday, October 21, 2009 Method of Joints 65
Method of Joints
We also have no longer included the force from BG, we know that it carries no force under there loading conditions
Wednesday, October 21, 2009 Method of Joints 66
Method of Joints
Using our equilibrium conditions
23
Wednesday, October 21, 2009 Method of Joints 67
Method of Joints
Using our equilibrium conditions
Wednesday, October 21, 2009 Method of Joints 68
Method of Joints
You can continue on with the pattern until all the members in the truss are solved for
You will always have one extra pin/joint available after all the members in the truss have been solved
Check the sum of the forces in the x and y direction at that point to see if the truss closes