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Structural Analysis Method of Joints / Method of Pins
I’m reading a book about an1gravity.
It’s impossible to put down.
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¢ What is today’s date?
Wednesday, October 17, 2012 Method of Joints 2
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Wednesday, October 17, 2012 Method of Joints 3
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
Bridge Truss
Wednesday, October 17, 2012 Method of Joints 4
3
Roof Truss
Wednesday, October 17, 2012 Method of Joints 5
Load Transfer
Wednesday, October 17, 2012 Method of Joints 6
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Wednesday, October 17, 2012 Method of Joints 7
Tools
¢ Equations of equilibrium ¢ Free Body Diagrams ¢ Trig and algebra ¢ Visualization
Wednesday, October 17, 2012 Method of Joints 8
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 are made at connections
¢ A simple truss analysis considers that all truss members are two-force members
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Wednesday, October 17, 2012 Method of Joints 9
Trusses
¢ This means that we can break a truss down into a collection of elements for analysis
Wednesday, October 17, 2012 Method of Joints 10
Trusses
¢ An example truss
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Wednesday, October 17, 2012 Method of Joints 11
Trusses
¢ The red arrows are the loads (applied externally)
Wednesday, October 17, 2012 Method of Joints 12
Trusses
¢ The dark black lines are the members
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Wednesday, October 17, 2012 Method of Joints 13
Trusses
¢ Each member goes from connection to connection
Wednesday, October 17, 2012 Method of Joints 14
Trusses
¢ The black line along the bottom of the truss represents three separate members
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Wednesday, October 17, 2012 Method of Joints 15
Trusses
¢ The support conditions in this example are a pin at the left and a roller at the right
Wednesday, October 17, 2012 Method of Joints 16
Trusses
¢ At each intersection of two or more members, we consider that the members are pinned together
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Wednesday, October 17, 2012 Method of Joints 17
Trusses
¢ To simplify our analysis, we assign each connection point with a letter
Wednesday, October 17, 2012 Method of Joints 18
Trusses ¢ I avoid using F and I as connection point
assignments but your text does not
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Wednesday, October 17, 2012 Method of Joints 19
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
GA
B C
DE
Wednesday, October 17, 2012 Method of Joints 20
Trusses ¢ At the point labeled G on the diagram, we have
four members coming to a point connected by a single pin
GA
B C
DE
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Wednesday, October 17, 2012 Method of Joints 21
Trusses ¢ Each of the members can now be labeled using
the points of connection
GA
B C
DE
Wednesday, October 17, 2012 Method of Joints 22
Trusses ¢ In this truss we have the following members: AB,
BC, CD, DE, CE, CG, EG, BG, and AG
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
12
Wednesday, October 17, 2012 Method of Joints 23
Trusses ¢ The order of the letters is not important, I choose
to put them in the order they appear in the alphabet
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
Wednesday, October 17, 2012 Method of Joints 24
Method of Joints ¢ To begin the analysis of the truss using this
method, we start by identifying all external forces acting on the truss
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
13
Wednesday, October 17, 2012 Method of Joints 25
Method of Joints ¢ To do the analysis, we don’t always have to
solve for the external reactions but it is important that we at least identify that they are there.
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
Wednesday, October 17, 2012 Method of Joints 26
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)
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
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Wednesday, October 17, 2012 Method of Joints 27
Method of Joints
¢ This is usually given in the problem
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
Wednesday, October 17, 2012 Method of Joints 28
Method of Joints ¢ Then we remove any supports and replace them
with reactions provided by the supports
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
15
Wednesday, October 17, 2012 Method of Joints 29
Method of Joints ¢ We have a pin at the left so we will have an x
and a y reaction
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200NAy
Ax
Wednesday, October 17, 2012 Method of Joints 30
Method of Joints ¢ And we have a roller at the right so we only have
a y reaction
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200NAy
Ax
Dy
16
Wednesday, October 17, 2012 Method of Joints 31
Method of Joints ¢ We are also usually given the dimensions and
geometry of the truss in the given
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200NAy
Ax
Dy
2 m
2 m 2 m 2 m
Wednesday, October 17, 2012 Method of Joints 32
Method of Joints ¢ Now we might solve for the reactions at the
supports using the methods developed previously
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200NAy
Ax
Dy
2 m
2 m 2 m 2 m
17
Wednesday, October 17, 2012 Method of Joints 33
Method of Joints ¢ If and which reactions we need to solve for will
be dictated by the problem.
BC
EG
CDC
EAB B
GCG
GA
B C
DE DEAG
500N 500N
200NAy
Ax
Dy
2 m
2 m 2 m 2 m
Wednesday, October 17, 2012 Method of Joints 34
Method of Joints ¢ We will use the entire truss as the FBD and
ignore the internal members (for the moment)
BC
EG
CDAB
GA
B C
DE DEAG
500N 500N
200NAy
Ax
Dy
2 m
2 m 2 m 2 m
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Wednesday, October 17, 2012 Method of Joints 35
Method of Joints ¢ Summing forces in the x-direction
BC
EG
CDAB
GA
B C
DE DEAG
500N 500N
200NAy
Ax
Dy
2 m
2 m 2 m 2 m
000
x
x
x
FAA
=
− ==
∑
Wednesday, October 17, 2012 Method of Joints 36
Method of Joints ¢ Summing moments about D
BC
EG
CDAB
GA
B C
DE DEAG
500N 500N
200NAy
Ax
Dy
2 m
2 m 2 m 2 m
( )( ) ( )( ) ( )( )( )( ) ( )( )
0
6 4 500 2 500 0
4 500 2 500500
6
D
y
y
M
m A m N m N
m N m NA N
m
=
− + + =
+= =
∑
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Wednesday, October 17, 2012 Method of Joints 37
Method of Joints ¢ Summing forces in the y-direction
BC
EG
CDAB
GA
B C
DE DEAG
500N 500N
200NAy
Ax
Dy
2 m
2 m 2 m 2 m
0
500 500 200 0500 500 500 200 0
700
y
y y
y
y
FA N N N DN N N N D
D N
=
− − − + =
− − − + =
=
∑
Wednesday, October 17, 2012 Method of Joints 38
Method of Joints
¢ Redrawing our system with the known reactions, we have
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
20
Wednesday, October 17, 2012 Method of Joints 39
Method of Joints ¢ Now for the problems that we had done earlier,
we would be finished
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
Wednesday, October 17, 2012 Method of Joints 40
Method of Joints ¢ In a truss analysis, we want to know what force
is being carried in each of the members
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
21
Wednesday, October 17, 2012 Method of Joints 41
Method of Joints
¢ Each member of the truss can be in one of three conditions: l It can be in COMPRESSION, that means that it
will be pushing on each end where it is connected
l It can be in TENSION, that means that it will be pulling on each end where it is connected
l It can have no force in the member BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
Wednesday, October 17, 2012 Method of Joints 42
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
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
22
Wednesday, October 17, 2012 Method of Joints 43
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
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
Wednesday, October 17, 2012 Method of Joints 44
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
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
23
Wednesday, October 17, 2012 Method of Joints 45
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
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
Wednesday, October 17, 2012 Method of Joints 46
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
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
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Wednesday, October 17, 2012 Method of Joints 47
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
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
Wednesday, October 17, 2012 Method of Joints 48
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!
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
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Wednesday, October 17, 2012 Method of Joints 49
Method of Joints
¢ At any joint, you can only solve for two unknowns!!!!
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
Wednesday, October 17, 2012 Method of Joints 50
Method of Joints
¢ Now to decide where to start the analysis, we need to find a joint where there are only two unknowns
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
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Wednesday, October 17, 2012 Method of Joints 51
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
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
Wednesday, October 17, 2012 Method of Joints 52
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
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
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Wednesday, October 17, 2012 Method of Joints 53
Method of Joints
¢ We choose A to start our analysis ¢ We now draw a FBD of the pin at A
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
Wednesday, October 17, 2012 Method of Joints 54
Method of Joints
¢ I always assume that the members are in compression so our FBD would look like
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
AB
GA
B
AG
500N1
1
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Wednesday, October 17, 2012 Method of Joints 55
Method of Joints
¢ Since I assume compression, the forces in the members are pushing on the connection (pin)
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
AB
GA
B
AG
500N1
1
Wednesday, October 17, 2012 Method of Joints 56
Method of Joints
¢ Now we can use our two equilibrium conditions to solve for the forces in the members
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
AB
GA
B
AG
500N1
1
01500 02
500 2
yF
N AB
AB N
=
− =
=
∑
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Wednesday, October 17, 2012 Method of Joints 57
Method of Joints
¢ We report all the magnitudes of the forces as positive numbers and the forces themselves as either compression or tension
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
AB
GA
B
AG
500N1
101500 02
500 2 C
yF
N AB
AB N
=
− =
=
∑
Wednesday, October 17, 2012 Method of Joints 58
Method of Joints
¢ Solving for AG
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
AB
GA
B
AG
500N1
10
1 02
1 500 2 5002
500 T
xF
AB AG
AG N N
AG N
=
− − =
= − = −
=
∑
30
Wednesday, October 17, 2012 Method of Joints 59
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
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
Wednesday, October 17, 2012 Method of Joints 60
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
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
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Wednesday, October 17, 2012 Method of Joints 61
Method of Joints
¢ At G, we have BG, CG, and EG as unknowns ¢ Three unknowns, so this pin won’t work
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
Wednesday, October 17, 2012 Method of Joints 62
Method of Joints
¢ At B, we have two unknowns, BC, and BG ¢ We can proceed to work at BG
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
32
Wednesday, October 17, 2012 Method of Joints 63
Method of Joints
¢ We draw a FBD of the pin at B
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
BC
AB BG
GA
B C
500N
Wednesday, October 17, 2012 Method of Joints 64
Method of Joints
¢ We know that AB is in compression, so we are not making an assumption about AB
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
BC
AB BG
GA
B C
500N
33
Wednesday, October 17, 2012 Method of Joints 65
Method of Joints
¢ We are assuming that BC and BG are in compression for this FBD
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
BC
AB BG
GA
B C
500N
Wednesday, October 17, 2012 Method of Joints 66
Method of Joints
¢ Now we use our conditions of equilibrium to solve for BC and BG
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
BC
AB BG
GA
B C
500N0
1 02
1 500 2 5002
500 C
xF
AB BC
BC N N
BC N
=
− =
= =
=
∑
34
Wednesday, October 17, 2012 Method of Joints 67
Method of Joints
¢ Now we use our conditions of equilibrium to solve for BC and BG
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
BC
AB BG
GA
B C
500N
01 500 02
1 500 2 5002
0
yF
AB BG N
BG N N
BG
=
+ − =
= − +
=
∑
Wednesday, October 17, 2012 Method of Joints 68
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
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
BC
AB BG
GA
B C
500N
35
Wednesday, October 17, 2012 Method of Joints 69
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
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
Wednesday, October 17, 2012 Method of Joints 70
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
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
EG
CG
GA
C
EAG
36
Wednesday, October 17, 2012 Method of Joints 71
Method of Joints
¢ We also have no longer included the force from BG, we know that it carries no force under there loading conditions
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
EG
CG
GA
C
EAG
Wednesday, October 17, 2012 Method of Joints 72
Method of Joints
¢ Using our equilibrium conditions
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
EG
CG
GA
C
EAG
01 020
yF
CG
CG
=
− =
=
∑
37
Wednesday, October 17, 2012 Method of Joints 73
Method of Joints
¢ Using our equilibrium conditions
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
EG
CG
GA
C
EAG
0500
500 500 T
xFAG EG N EG
EG N N
=
− − = − −
= − =
∑
Wednesday, October 17, 2012 Method of Joints 74
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
BC
EG
CD
CE
AB BG
CG
GA
B C
DE DEAG
500N 500N
200N
2 m
2 m 2 m 2 m
500N
700N
38
Homework
¢ Problem 6-2 ¢ Problem 6-5 ¢ Problem 6-6
Wednesday, October 17, 2012 Method of Joints 75
