EGR 231 Engineering Statics:
Lecture 23: Truss Analysis: Method of Joints
Today:
Questions on Homework
Trusses:
Two force members
Method of Joints
New Homework
Next Class:
Exam 3
Homework 22:
Prob 6.3
Using the method of joints, determine
the force in each member of the truss
shown. State whether each member is
in tension or compression.
Fall 2014
FBD
FAB
FAC
FAC
FAD
FAB
FAD
48 kN
35 kN
Prob 6.12:
Using the method of joints, determine
the force in each member of the truss
shown. State whether each member is
in tension or compression.
24kN
Prob. 6.15:
Determine the zero force members in the truss shown for the given loading
Prob. 6.20:
The three dimensional truss is supported by the six reactions shown. If a vertical 5400 N
load is applied at A, determine a) the reactions b) the force in each member.
5400N
Structural Analysis:
Truss-- A structure composed of slender two force members, joined at their ends.
Members of a truss may only be subjected to tensile or compressive forces.
Two-force member: A two-force member is any body which has forces applied to it at
only two points of contact. In order for the body to remain in equilibrium, the line of the
force acting at each contact, must act along a line that passes through each point.
If the body is a straight member, then it is said to be in either tension or compression
Examples of Two Force Members:
Trusses are built out of straight, two-force members. Forces are only applied at the
connections between individual members.
A
Method of Joints:
In the method of joints, point equilibrium is set up for
forces that act on the pin at each joint of the body.
720 lb
B
64 in
Process: Assume all members are in tension
Identify zero-force members by inspection
Find the support reactions of the entire truss
or
Find a joint which has less than
three members and find member forces.
Example:
50 in
C
48 in
D
A
A
720 lb
720 lb
B
B
64 in
50 in
48 in
C
D
FCx
FD
FCy
C
Start by finding the reactions of the
entire body:
M
C
0
FCy
0  720lb(50in)  FD (48in)
(720lb)(50in)
FD 
 750lb
(48in)
F
x
0
D
FCx
F
y
0
0  FCx  720lb
0  FCy  FD
FCx  720lb
FCy   FD  750lb
FD
Next set up the equilibrium at point B because there are only two members
coming together there.
F
x
F
y
 0:
0:
48
FBA  720lb
50
FBA  750lb
50
0
FBA
14
14
0  FBA  FBD
50
14
14
FBD  FBA  (750)  210lb
50
50
Joint B
P=720 lb
48
FBD
Next move to Joint A and apply point equilibrium:
F
x
 0:
48
48
FAB  FAD
50
80
80
FAD   FAB
50
0

F
y
0:
80
(750)  1200lb
50
14
64
FAB  FAD  FAC
50
80
14
64
FAC   FAB  FAD
50
80
14
64
FAC   (750)  ( 1200)  750lb
50
80
FAB
FAC
FAD
48
80
48
50
14
64
0
Joint C
FCA
Finally, look at Joint C.
 Fx  0 :
0  FCx  FCD
FCD
FCx
FCD   FCx  (720lb)  720lb
FCy
Summary:
FBA  FAB  750lb (T )
FCD  FDC  720lb (T )
FBD  FDB  210lb (T )
FAC  FCA  750lb (T )
FAD  FDA  1200lb  1200lb (C)
Example 2:
Use method of joints to find the forces in each of the truss members.
4 kN
6 kN
3m
4 kN
3m
C
E
45o
45o
D
3m
3m
F
G
45o
45o
B
A
Solution:
Start by finding the reactions of the
entire body:
M
A
0
0  6kN (3m)  4kN (6m)  FB (6m)
6(3)  4(6)
FB 
 7 kN
6
F
x
0
F
y
0
0  FAx
0  FAy  FB  6kN  4kN  4kN
FAx  0
FAy  14  FB  14  7  7 kN
Next inspect for zero force members:
Identify EF and GC as ZFM
4kN
Start at Joint E:
FED
F
x
F
y
 0:
0  FED
FEA
0:
0  4  FEA
Joint E
FEA  4kN
Next note that this truss is symmetric, so FCB  4kN and FCD  0
Next Look at Joint D:
FDE
F
x
 0:
0   FDF cos 45o  FDG cos 45o
FDF  FDG
F
y
0:
0
o
45
FDF
6 kN
0
FDC
45o
FDG
0  6  FDF sin 45o  FDG sin 45o
2 FDF sin 45o  6
6
FDF  FDG 
 4.24
2sin 45o
Note that the force in members FA and GB will be the same as in DF and
DG, so FFA  4.24kN and FGB  4.24kN
Finally, look at Joint A to find force in AB.
F
x
 0:
0  FAx  FAB  FAF cos 45o
Joint A
FAE
FAF
FAB
FAx
FAB   FAF cos 45o  ( 4.24) cos 45 o  3kN
Summary:
FAB  3 kN (T )
FAF  FFD  FDG  FGB  4.24  4.24 kN (C)
FAE  FBC  4  4kN (C )
FED  FEF  FDC  FGC  0
FAy
6 kN
Example 3:
Use method of joints
to find the forces in
each of the truss members.
B
A
0.9 m
3 kN
C
E
1.2 m
Start at Joint E.
 Fx  0 :
D
1.2 m
 Fy  0 :
1.2
0
FEB  FED
1.5
0.9
0  3kN 
FEB
1.5
FEB  1.5(3kN ) / 0.9  5kN
1.5
0.9
1.2
Next move to Joint D:
 Fx  0 :
0   FDC  FDE 
1.2
FDA
1.5
FED
6kN
Next move to Joint B and apply point equilibrium:
 Fy  0
 Fx  0 :
1.2
FBE
1.5
FBA  1.2(5kN ) /1.5  4kN
P=3kN
FEB
FED  1.2(5kN ) /1.5  4kN
0   FBA 
Joint E
FBE
1.2
0.9
FBD
FBE  6kN
1.5
 0.9(5kN ) /1.5  6kN  9kN
0   FBD 
FBD
Joint B
FBA
0.9
1.5
Joint D
F
y
0
0.9
1.5
FDA
1.2
0.9
FDA
FDC
1.5
FDA  1.5( FDB ) / 0.9  1.5(9) / 0.9  15kN
0  FDB 
FDB
FDE
FDC  FDE  1.2FDA /1.5  (4)  1.2(15)  22kN
------------------------------------------------------------------------------------------Force in AB: 4 kN (T)
Force in AD: 15kN (T) Force in BD: 9kN (C)
Force in BE: 5 kN (T)
Force in CD: 22 kN (C)
Force in DE: 4kN (C)
Discussion Problem:
For the given loading, how many and where are the zero-force members in the truss
shown.
F1
F2
P
Q
O
M
L
N
A
B
R
K
S
H
E
D
C
Solution:
F1
F2
P
Q
O
M
L
N
A
B
C
Zero force members include:
R
K
S
D
E
H
BN, BM, QK, KR, SE, and ER
Extra Problems:
ExProb 1
Using the method of joints, determine the force in each member
of the truss shown. State whether each member is in tension or compression.
A
375 lb
500 lb
C
B
6 ft
8 ft
D
8.4 ft
ExProb 2:
Determine the force in each member of the Fink truss shown. State whether each
member is in tension or compression.
4.2 kN
D
2.8 kN
2.8 kN
E
B
C
3m
F
G
A
2m
2m
2m
2m
Ex Prob 3:
For the roof truss shown, determine the force in each of the members located to the left of
member GH. State whether each member is in tension or compression.
1.5 kN
1.5 kN
1.5 kN
1.5 kN
G
1.5 kN
D
1 kN
E
B
F
C
H
M
J
L
2.4 m
2.4 m
1.2 m
2.4 m
2.5 m
K
1m A
2.4 m
1 kN
I
1.2 m
Ex Prob 4:
For the given loading, determine the zero force members in the truss.
B
A
D
C
F
E
G
H
L
I
J
K