Method of Joints
Let’s begin by calculating the forces in each member of the truss. We will assume
a load of 20N applied to the center of the bridge (this results in a 10N load applied
to each of the two trusses.) For our calculations we will be working with only one
of the parallel trusses. Since the bridge and loading are both symmetrical, the
reactions at both ends (A and E) of the truss are equal to half of the applied load.
10 N
B
C
D
12.5 cm
A
H
F
E
G
10 cm
10 cm
Starting with a free-body diagram at Joint A:
FAB
A
α
FAH
5N
Free-body Diagram at
Joint A
10 cm
10 cm
FAB
A
α
FAH
 Fx  F
 FAB cos( )  0
 Fx  F

 FAB 

AH
AH

0

5N
 Fy  5N  F
sin( )  0
 Fy  5N  F



AB
Free-body Diagram at
Joint A
AB

0

 FAB 
substituting FAB into the first equation
FAH 
What direction does the arrow point?
Let us look at the FAH arrow from the Joint at A:
A
FAH
This diagram shows the force applied to the pin (or joint) at A. Now let us add the
member AH to the diagram.
A
FAH
A
H
FAH
Since the pin A and the member AB are connected at Joint A, they must have
equal and opposite reaction forces. So the force FAH is the same magnitude at pin
A and on member AH but it is in opposite directions. These equal and opposite
reactions are necessary in order for the truss to stay in equilibrium.
Now that we have done the calculations, is member AH in tension or
compression? ___________________
What about member AB? __________________________________
Our final answer should indicate whether the members are in tension or
compression.
FAB = 6.4 N (C)
FAH = 4.0 N (T)
Now lets move on to Joint H and determine what direction the force FAH is
pointing.
A
FAH
H
H
FAH
From this, we can see that the force from member AH should be pointing to the
left at pin H.
The free-body diagram for Joint H is shown next.
 Fx 
FBH
FAH
H
FGH
 FGH  FAH 
Free-body Diagram at
Joint H
 Fy 
FGH =
 FBH 
FBH =
Is FGH in tension or compression? __________________
At joint H, we notice that FBH equals 0 N. This is called a zero-force member.
Zero-force members are important in trusses for several reasons.
What are some of the reasons you can think of?
Now let’s analyze Joint B. First, draw the free-body diagram in the space below.
 Fx  F
AB
FBC
α
FAB
6.4 N
(
α
FBH
0N
FBG
cos( )  FBG cos( )  FBC  0

)



  FBG 



  FBC  0

 Fy 
 FBG 
substitute FBG into the second equation above
 FBC 
Check for tension or compression
FBC =
FBG =
The only member left to solve for is CG. We can solve for it at joint C.
Draw the free-body diagram and solve for FCG.
FCG =
Now let’s calculate the expected failure load and location.
Consider member AB, is this member in tension or compression? _____________
From the data sheet, what is the expected failure load of member AB? _________
From the method of joints analysis, what was the internal member force in
member AB for the design load of 10N? ____________
In order to calculate the Factor of Safety (F.S.) for member AB use the equation
below.
F .S. 
Expected Failure Load
 ___________________
Internal Member Force
Member
Internal Member
Force (Newtons)
Expected Failure Load
(Newtons)
F.S.
AB, DE
6.4 (C)
56.8
8.9
BC, CD
8.0 (C)
AH, EF
4.0 (T)
23.5* x 2
11.8
BG, DG
6.4 (T)
GH, FG
4.0 (T)
CG
10 (C)
BH, DF
0
NA
NA
*Strength of tension members was experimentally determined to be 23.5N.
Where do you expect the truss to fail? ______________
Calculate the expected failure load for the bridge:
(F.S) x (number of parallel trusses) x (number of members at failure location) x (10 N) = _________
Maximum load actually held by the bridge: _________________
Member that failed: ______________