6.2 Trusses: Method of Joints and Zero-Force Members
6.2 Trusses: Method of Joints and Zero-Force Members Example 1, page 1 of 3
1. Determine the force in each member of the truss and
state whether the force is tension or compression.
C
10 kN
3m
A
B
5m
Free-body diagram of entire truss. Calculating the reactions
is a good place to start because they are usually easy to
compute, and they can be used in the equilibrium equations
for the joints where the reactions act.
10 kN
C
2
Equilibrium equations for entire truss
+ Fx = 0: Ax + 10 kN = 0

(1)
Fy = 0: Ay + By = 0
(2)
MA = 0: (10 kN)(3 m) + By(5 m) = 0
(3)
+

1
+
3m
A
B
Ax
Ay
By
5m
Solving these equations simultaneously gives
Ax = 10 kN, Ay = 6 kN, and By = 6 kN
6.2 Trusses: Method of Joints and Zero-Force Members Example 1, page 2 of 3
3
5
Free-body diagram of joint C
FAC

C
10 kN
C
Geometry
 = tan-1( 5 ) = 59.04°
3

3m
FBC
A
B
5m
4
Equilibrium equations for joint C. It is a good idea
to assume all members in tension (forces point away
from the joint, not towards it). Then, after solving
the equilibrium equations, you will know
immediately that any member force found to be
negative must be compression.
Using  = 59.04° in Eqs. 4 and 5 and solving
simultaneously gives
FAC = 11.66 kN (T)
Ans.
FBC = 6.0 kN = 6.0 kN (C)
Ans.
and
+ F = 0: 10 kN F

x
AC sin  = 0
(4)
Fy = 0: FAC cos  FBC = 0
(5)
+

6
Writing "(T)" after the numerical value shows that
the member is in tension. We had arbitrarily
assumed member BC to be in tension. We then
found that the member force was negative, so we
know that our assumption was wrong. Member BC
is in compression, and we show this by writing a
positive "6.0" followed by "(C)".
6.2 Trusses: Method of Joints and Zero-Force Members Example 1, page 3 of 3
Free-body diagram of joint B
7
FBC = 6 kN
FAB
B
8 The force FBCis directed toward
the joint because member BC is
known to be in compression.
By = 6 kN
9
Equilibrium equation for joint B
+ Fx = 0: F

AB = 0
10 An "Answer diagram" summarizes the analysis
of the entire truss (All forces are in kN).
C
10 kN
Solving gives
FAB = 0
(T)
6
.6
11
Ans.
A
0
6.0 (C)
B
10
6
6
6.2 Trusses: Method of Joints and Zero-Force Members Example 2, page 1 of 7
2. Determine the force in each member of the truss and state
whether the force is tension or compression.
F
H
G
14 ft
A
E
10 ft
B
C
2 kip
4 kip
10 ft
D
2 kip
10 ft
10 ft
6.2 Trusses: Method of Joints and Zero-Force Members Example 2, page 2 of 7
Free-body diagram of entire truss.
Calculating the reactions is usually a
good way to start the analysis.
1
H
G
F
14 ft
A
E
Ax
B
D
C
Ey
Ay
4 kip
2 kip
Equilibrium equations for entire truss
2
10 ft
10 ft
10 ft
(1)
Fy = 0: Ay + Ey kip kipkip = 0
(2)
MA = 0: 2 kip)(10 ft)  (4 kip)(20 ft)  (2 kip)(30 ft) + Ey(40 ft) = 0
(3)
+
+

+ Fx = 0: Ax = 0

Solving simultaneously gives
Ax = 0, Ay = 4.0 kip, and Ey = 4.0 kip.
2 kip
10 ft
6.2 Trusses: Method of Joints and Zero-Force Members Example 2, page 3 of 7
Free-body diagram of joint E. This joint is
chosen because only two unknown forces are
present. Thus we know that we can solve for
these forces because two equations of
equilibrium are available for the joint. Note
also that we assume that both unknown forces
are in tension (directed away from the joint).
3
5
Geometry
 = tan-1( 14 ) = 54.46°
10
H
14 ft
FEH


E
D
FDE
Ey = 4 kip
6
4
Equilibrium equations for joint E
+ F = 0: F

x
DE  FEH cos  = 0
+

Fy = 0: FEH sin  4 kip = 0
(4)
(5)
7
E
10 ft
Using  = 54.46° in Eqs. 4 and 5 and solving
simultaneously gives
FDE = 2.857 kip (T)
Ans.
FEH = 4.916 kip = 4.916 kip (C)
Ans.
We arbitrarily assumed member EH to be in tension. We
then found that the member force was negative, so we
know that our assumption was wrong. Member EH is in
compression, and we show this by writing a positive
"4.916" followed by "(C)".
6.2 Trusses: Method of Joints and Zero-Force Members Example 2, page 4 of 7
9
8 Use a free-body diagram of joint H next because only
two member forces are unknown.
H
FEH = 4.916 kip (C)
G
F
Free-body diagram of joint H
H
FGH
FDH
A
E
2 kip
C
D
4 kip
2 kip
11 Equilibrium equations for joint H
FDE = 2.857 kip (T)
+ Fx = 0: F

GH  (4.916 kip) cos 54.46° = 0
(6)
Fy = 0: FDH + (4.916 kip) sin 54.46° = 0
(7)
+

B
10 As before, we assume
that the unknown
FEH = 4.916 kip (C)
member forces (GH and
DH in this instance) are
 = 54.46°
tension, so are directed

away from the joint.
The force in member
EH has already been
found to be 4.916 kip
compression, so it is
directed towards the
joint, not away from it.
Solving simultaneously gives
FGH = 2.858 kip = 2.858 kip (C)
Ans.
FDH = 4.0 kip (T)
Ans.
and
6.2 Trusses: Method of Joints and Zero-Force Members Example 2, page 5 of 7
FGH = 2.858 kip(C)
G
F
13 Free-body diagram of joint D
FEH = 4.916 kip (C)
H
FDG
 = 54.46° 
FCD
FDH = 4.0 kip
FDE = 2.857 kip
2 kip
A
E
B
C
D
FDE = 2.857 kip (T)
4 kip
2 kip
12 Use a free-body
diagram of joint
D because only
two member
forces are
unknown.
15 Equilibrium equations for joint D
+ F = 0: F

x
CD  FDG cos(54.46°) + 2.857 kip = 0
(8)
Fy = 0: FDG sin(54.46°) + 4.0 kip 2 kip = 0
(9)
+

2 kip
14 As before, we assume that the unknown
member forces are tension, so are directed away
from the joint. The forces in members DH and
DE have already been found to be tension and
so are directed away from the joint.
Solving simultaneously gives
FCD = 4.286 kip (T)
Ans.
FDG = 2.458 kip = 2.458 kip (C)
Ans.
and
6.2 Trusses: Method of Joints and Zero-Force Members Example 2, page 6 of 7
FGH = 2.857 kip(C)
FEH = 4.916 kip (C)
G
F
H
FDH = 4.0 kip (T)
FDG= 2.458 kip (C)
A
E
C
D
4 kip
2 kip
B
2 kip
FCD = 4.286 kip (T)
FDE = 2.857 kip (T)
19 Equilibrium equations for joint C.
16 Use a free-body diagram of
joint C because only two
member forces are
unknown.
17 Free-body diagram of joint C
FCG
FBC
C
4 kip
FCD = 4.286 kip (T)
+

+ Fx = 0: F

BC + 4.286 kip = 0
18 The unknown forces in
members CG and BC are
assumed to be tension, so
are directed away from
the joint. The force in
member CD has already
been found to be 4.286
kip (T).
Fy = 0: FCG 4 kip = 0
(10)
(11)
Solving gives
FBC = 4.286 kip (T)
Ans.
FCG = 4 kip (T)
Ans.
and
6.2 Trusses: Method of Joints and Zero-Force Members Example 2, page 7 of 7
20 All remaining bar forces follow from symmetry.
All forces in kips.
Answer diagram
H
G
F
2.86 (C)
2.86 (C)
4.0 (T)
4.00 (T)
C)
6(
2.4
4.0 (T)
C)
2(
4.9
C)
2
2.86 (T)
4.29 (T)
B
4
2(
4.29 (T)
2.86 (T)
4.9
C)
6(
2.4
A
D
C
4
E
2
4
6.2 Trusses: Method of Joints and Zero-Force Members Example 3, page 1 of 4
3. Determine the force in each member of
the truss and state whether the force is
tension or compression.
900 lb
30°
C
B
A
60°
12 ft
E
D
400 lb
10 ft
6.2 Trusses: Method of Joints and Zero-Force Members Example 3, page 2 of 4
1
Free body-diagram of entire truss
2
Because AB is a two-force member, the line of
action of FAB must pass through A and B.
900 lb
30°
B
C
FAB
60°
12 ft
3
Equilibrium equations for entire truss
E
D
Ex
Ey
400 lb
(2)
MC = 0: (400 lb)(10 ft + FAB cos 60°(10 ft) + Ex(12 ft) = 0
(3)
Solving simultaneously gives
10 ft
(1)
Fy = 0: FAB cos 60° + Ey + (900 lb) sin 30°  400 lb = 0
+
+

+ Fx = 0: F

AB sin 60° + Ex + (900 lb) cos 30° = 0
FAB = 347.8 lb
Ex = 478.2 lb,
Ey = 123.9 lb
6.2 Trusses: Method of Joints and Zero-Force Members Example 3, page 3 of 4
4
Free-body diagram of joint D. Joint D is chosen
because only two member forces are unknown there.
7
Free-body diagram of joint C. Joint C is chosen because
only two member forces are unknown there.
(900 lb) sin 30° = 450.0 lb
FBD
5
FDE
D
The unknown forces
have been assumed to
be tension.
400 lb
8
C
(900 lb) cos 30°= 779.4 lb
The unknown
forces have
been assumed
to be tension.
FCE
Equilibrium equations for joint D
+ Fx

9
= 0: FDE = 0
+

Fy = 0: FBD  400 lb = 0
(4)
(5)
Equilibrium equations for joint C
+ Fx = 0: F

BC +779.4 lb = 0
Fy = 0: FCE + 450 lb = 0
+

6
FBC
(6)
(7)
Solving gives
Solving gives
FBD = 400 lb (T)
FDE = 0
Ans.
Ans.
FBC = 779.4 lb (T)
Ans.
FCE = 450.0 lb (T)
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 3, page 4 of 4
10 Free-body diagram of joint B. Only one
member force is unknown at this joint.
B
FBC = 779.4 lb

FAB = 347.8 lb
60°
13 Using  = 50.19° in Eq. 8 and then solving gives
FBE
FBD = 400 lb
FBE = 746.9 lb = 747 lb (C)
14 Answer diagram (all forces in lb)
11 Equilibrium equations for joint B
+ Fx =0: F

BE cos  (347.8 lb) sin 60°
+ 779.4 lb = 0
900 lb
(8)
B
12 Geometry
 = tan-1( 12 )
10
= 50.19°
10 ft
B
Ans.
779 (T)
C
C
347.8 lb

60°
747 (C)
12 ft
450 (T)
400 (T)
D
0
E
478
E
124
400
30°
6.2 Trusses: Method of Joints and Zero-Force Members Example 4, page 1 of 9
4. Determine the force in each member of the truss and state whether
the force is tension or compression. The truss is symmetric.
6 ft
6 ft
6 ft
6 ft
6 ft
6 ft
2 kip
2 kip
2 kip
K
2 kip
2 kip
I
E
A
J
60°
F
60°
30°
H
G
D
B
C
6.2 Trusses: Method of Joints and Zero-Force Members Example 4, page 2 of 9
Free-body diagram of entire truss
1
6 ft
6 ft
6 ft
6 ft
6 ft
6 ft
2 kip
2 kip
2 kip
2 Equilibrium equations for entire truss
K
2 kip
I
A
Ax
60°
F
60°
30°
Fy = 0: Ay + Dy 2 kip
2 kip 2 kip 2 kip 2 kip = 0
+

E
+ Fx = 0: Ax = 0

J
H
G
D
Ay
B
C
Dy
MA = 0: 2(kip)(6 ft) 2 kip(12 ft)
2(kip)(18 ft) 2 kip(24 ft)
2(kip)(30 ft) + Dy(36 ft) = 0
+
2 kip
Solving simultaneously gives
Ax = 0
Ay = 5 kip
Dy = 5 kip
6.2 Trusses: Method of Joints and Zero-Force Members Example 4, page 3 of 9
2 kip
2 kip
2 kip
K
2 kip
2 kip
I
60°
E
3
F
60°
30°
A
Ax = 0
J
H
G
D
C
B
Ay = 5 kip
Dy = 5 kip
Use a free-body diagram of joint A because
only two unknown member forces are
present.
5 Equilibrium equations for joint A
+ Fx = 0: F

AB + FAE cos 30° = 0
Free-body diagram of joint A.
FAE
Fy = 0: FAE sin 30° + 5 kip = 0
+

4
30°
Solving simultaneously gives
A
FAB
Ay = 5 kip
FAB = 8.660 kip (T)
Ans.
FAE = 10 kip = 10 kip (C)
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 4, page 4 of 9
6
Three unknown member forces are present at joint I, but
two of them, FEI and FIK , are collinear, so summing
forces perpendicular to FEI and FIK would give an
equation with FFI as the only unknown.
2 kip
2 kip
2 kip
K
2 kip
2 kip
I
7
Free-body diagram of joint I
2 kip
y
A

FEI
I
60°
E
x
FIK
Ax = 0

J
F
60°
30°
H
G
D
C
B
Ay = 5 kip
Dy = 5 kip
FFI
Geometry of members at joint I
y
 = 60°
60° E
A
30°
I
60°
x
9
30°
60°
F
Equilibrium equations for joint I
Fy = 0: FFI sin 90° (2 kip) sin 60° = 0
+

8
 = 30° + 60°
= 90°
So member FI is
perpendicular to the x axis.
Thus the member force FFI
lies on the y axis.
Solving gives
FFI = 1.732 = 1.732 kip (C)
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 4, page 5 of 9
FFI = 1.732 kip (C)
(already known)
2 kip
2 kip
2 kip
K
2 kip
2 kip
I
E
60°
F
60°
30°
A
Ax = 0
J
Ay = 5 kip
B
12 Free-body diagram of joint F
H
G
D
FFI = 1.732 kip (C)
C
x
Dy = 5 kip
y
10 Use the same technique at
joint F as was used at joint I:
sum forces perpendicular to
collinear members BF and FK.
60°
FFK
F
FEF
60°
B
FBF
13 Equilibrium equations for joint F
Fy = 0: FEF sin 60° (1.732 kip) sin 60° = 0
+

11 Geometry of members at joint F
I
 = 180° (60° +60°)
60°
= 60°
E
F
60°
60°
Solving simultaneously gives
FEF = 1.732 kip (T)
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 4, page 6 of 9
FAE = 10.0 kip (C)
(already known)
2 kip
2 kip
2 kip
K
2 kip
2 kip
I
60°
E
H
G
F
60°
30°
A
D
C
B
Ay = 5 kip
14 At joint E, now only two member
forces, FEI and FEB, are unknown.
FEF = 1.732 kip (T)
(already known)
15 Free-body diagram of joint E
2 kip
60°
FEI
30°
FEF = 1.732 kip (T)
30°
FBE
FAE = 10 kip(C)
E
30°
Dy = 5 kip
16 Equilibrium equations for joint E
+ Fx = 0: (10 kip) cos 30° + F cos 30° + F cos 30° + 1.732 kip = 0

EI
BE
Fy = 0: (10 kip) sin 30° + FEI sin 30° FBE sin 30°  2 kip = 0
+

Ax = 0
J
Solving simultaneously gives
FEI = 9.0 kip = 9.0 kip (C)
Ans.
FBE = 3.0 kip = 3.0 kip (C)
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 4, page 7 of 9
2 kip
2 kip
FBE = 3 kip (C)
(already known)
2 kip
2 kip
K
2 kip
I
E
A
J
60°
F
60°
30°
D
C
B
Ax = 0
H
G
Dy = 5 kip
FAB = 8.660 kip (T)
Ay = 5 kip
(already known)
17 At joint B, now only two member forces, FBF and
FBC, are unknown.
18 Free-body diagram of joint B
FAB = 8.660 kip (T)
+

Solving simultaneously gives
60°
B
+ F = 0: 3 kip) cos 30°  8.660 kip + F

x
BF cos 60° + FBC = 0
Fy = 0: 3 kip) sin 30° + FBF sin 60° = 0
FBF
FBE = 3 kip(C)
30°
19 Equilibrium equations for joint B
FBC
FBF = 1.732 kip (T)
Ans.
FBC = 5.196 kip (T)
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 4, page 8 of 9
20 The remaining unknown member forces, FIK and FFK ,
can be found by re-using the free-body diagrams of
joints I and F.
23 Free-body diagram of joint F
FFI = 1.732 kip (C)
21 Free-body diagram of joint I
y
y
2 kip

x

FFK
F
FEF = 1.732 kip (T)
FIK
x
60°
FEI = 9.0 kip (C)
I
FBF = 1.732 kip (T)
FFI = 1.732 kip (C)
24 Equilibrium equation for joint F
22 Equilibrium equations for joint I
+
 Fx = 0: FFK  1.732 kip  (1.732 kip) cos 60°
 (1.732 kip) cos 60° = 0
+
Fx = 0: 9.0 kip  (2 kip) cos 60° + FIK = 0
Solving gives
FIK = 8.0 kip = 8.0 kip (C)
Solving gives
Ans.
FFK = 3.464 kip (T)
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 4, page 9 of 9
25 By symmetry, all forces on the right half of
the truss are also known.
Answer diagram
2
K
3.4
6
32
32
(T)
G
1.7
B
0 (C
)
1.732 (T)
(T)
C)
9.0
H
32
8.66 (T)
0(
F
2
J
1.7
3.0
1.7
1.732 (T)
E
A
5
(T)
0
32
0
10.
(C)
T)
1.7
C)
0(
0
.
9
6(
I
(T
)
8
2
8.0
0 (C
)
(C)
3.4
.00
2
(T)
2
All forces in kips
5.20 (T)
C
0
3.0
10.
0
0 (C
)
(C)
8.66 (T)
D
5
6.2 Trusses: Method of Joints and Zero-Force Members Example 5, page 1 of 8
5. Determine the force in each member of the truss and
state whether the force is tension or compression.
100 lb
D
B
1 ft
E
C
2.5 ft
A
F
4 ft
2 ft
4 ft
6.2 Trusses: Method of Joints and Zero-Force Members Example 5, page 2 of 8
Free-body diagram of entire truss.
B
100 lb
1
D
1 ft
E
C
2.5 ft
A
F
Ax
Ay
2
4 ft
2 ft
4 ft
Equilibrium equations for entire truss
+

+ Fx = 0: Ax + 100 lb = 0

Fy = 0: Ay + Fy = 0
+
MA = 0: 100 lb)(1 ft + 2.5 ft) + Fy(4 ft +2 ft + 4 ft) = 0
Solving these equations simultaneously gives
Ax = 100 lb
Ay = 35 lb
Fy = 35 lb
Fy
6.2 Trusses: Method of Joints and Zero-Force Members Example 5, page 3 of 8
Free-body diagram
of entire truss.
100 lb
3
D
B
1 ft
7
Geometry
E
C
2.5 ft
4
Ax = 100 lb
A
Only two unknown member
forces act at joint F.
4 ft
D
F
E
Fy = 35 lb
Ay = 35 lb
4 ft
2 ft
4 ft
3.5 ft

2.5 ft

Free-body diagram of joint F.
FDF

FEF
F

5
4 ft
2.5 ft
) = 32.01°
4 ft
 = tan-1( 4 ft ) = 48.81°
3.5 ft
 = tan-1(
Fy = 35 lb
6 Equilibrium equations for joint F
+ Fx = 0: F cos  F

EF
DF sin  = 0
+

Fy = 0: FEF sin + FDF cos  + 35 lb = 0
F
8
Solving the equilibrium equations with
 = 32.01° and  = 48.81° gives
FEF = 165.10 lb (T)
Ans.
FDF = 186.03 lb = 186.03 lb (C)
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 5, page 4 of 8
Free-body diagram
of entire truss.
100 lb
9
D
B
1 ft
E
C
2.5 ft
Only two unknown member
forces act at joint A.
Ay = 35 lb
4 ft
Fy = 35 lb
2 ft
10 Free-body diagram of joint A
FAB

FAC
Ax = 100 lb

A
Ay = 35 lb
By symmetry, the angles and at joint A are the
same as we calculated at joint F. Thus

F
4 ft
11 Equilibrium equations for joint A
+ Fx = 0: 100 lb + F

AC cos + FAB sin  = 0
+

Ax = 100 lb
A
Fy = 0: FAC sin + FAB cos   35 lb = 0
Solving the above equations with  = 32.01°
and  = 48.81° gives
= 32.01°
FAC = 247.70 lb (T)
Ans.
= 48.81°
FAB = 146.23 lb = 146.23 lb (C)
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 5, page 5 of 8
12 Free-body diagram of entire truss
B
D
C
E
13 Only two unknown member
forces act at joint D.
100 lb
FDF = 186.03 lb (C)
(already known)
48.81°
Ax = 100 lb
A
F
Ay = 35 lb
Fy = 35 lb
14 Free-body diagram of joint D
D
48.81°
+ Fx = 0: F  (186.03 lb) sin 48.81° = 0

BD
+

FBD
15 Equilibrium equations for joint D.
FDF = 186.03 lb (C)
FDE
Fy = 0: FDE + (186.03 lb) cos 48.81° = 0
Solving these equations gives
FBD = 139.99 lb = 139.99 lb (C)
Ans.
FDE = 122.51 lb (T)
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 5, page 6 of 8
16 Free-body diagram of entire truss
B
D
C
E
100 lb
FAC = 247.70 lb (T)
(already known)
Ax = 100 lb
A
Ay = 35 lb
32.01°
Fy = 35 lb
FCE
+

+ Fx = 0: (247.70 lb) cos 32.01° + F

CE = 0
FBC
FAC = 247.70 lb (T)
F
19 Equilibrium equations for joint C.
18 Free-body diagram of joint C.
C
17 Only two unknown member
forces act at joint C
Fy = 0: (247.70 lb) sin 32.01° + FBC = 0
Solving these equations gives
FCE = 210.04 lb (T)
Ans.
FBC = 131.23 lb (T)
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 5, page 7 of 8
20 Free-body diagram of entire truss
B
D
FDE = 122.51 lb (T)
(already known)
FEF = 165.10 lb (T)
(already known)
100 lb
E
C
Ax = 100 lb
FCE = 210.04 lb (T)
(already known)
A
Ay = 35 lb
32.01°
21 At joint E, member BE
is the only unknown
member force.

FCE = 210.04 lb (T)
Fy = 35 lb
24 Geometry
22 Free-body diagram of joint E.
FBE
F
B
FDE = 122.51 lb (T)

1 ft
E
32.01°
C
FEF = 165.10 lb (T)
23 Equilibrium equation for joint E.
+ Fx = 0: F

BE cos 210.04 lb + (165.10 lb) cos 32.01° = 0
 = tan-1( 1 ft ) = 26.57°
2 ft
2 ft
E
25 Substituting  = 26.57° in the equation for joint E
and solving gives
FBE = 78.31 lb = 78.31 lb (C)
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 5, page 8 of 8
26 Answer diagram
All forces in lb
100
B
131 (T)
6
14
C
)
(C
D
140 (C)
78 (
C)
210 (T)
123 (T)
E
18
6(
C)
T)
8(
24
16
5(
T)
100
A
35
F
35
6.2 Trusses: Method of Joints and Zero-Force Members Example 6, page 1 of 9
6. Determine the force in each member of the truss and state
whether the force is tension or compression.
2 kip
J
I
K
14 ft
L
H
G
A
B
C
16 ft
D
E
16 ft
F
6.2 Trusses: Method of Joints and Zero-Force Members Example 6, page 2 of 9
1
2 Equilibrium equations for entire truss
Free-body diagram of entire truss
+ Fx = 0: Ax = 0

+
+

2 kip
J
I
Fy = 0: Ay + Gy  2 kip = 0
MA = 0: 2 kip(16 ft) + Gy(16 ft + 16 ft) = 0
Solving simultaneously gives
K
Ax = 0
Ay = 1 kip
14 ft
H
Gy = 1 kip.
L
G
A
Ay
B
C
D
E
Ax
16 ft
16 ft
F
Gy
6.2 Trusses: Method of Joints and Zero-Force Members Example 6, page 3 of 9
2 kip
6 Geometry
J
 = tan-1( 14 ft ) = 41.19°
16 ft
J
I
K
14 ft
14 ft
H
L

A
G
B
Ay = 1 kip
C
D
16 ft
G
F
Gy = 1 kip
16 ft
Free-body
diagram of joint G
G
FFG
Gy = 1 kip
Only two unknown member
forces act at joint G.
3
7
Solving simultaneously gives
FFG = 1.143 kip (T)
5
FGL

16 ft
Equilibrium equations for joint G
+ Fx = 0: F

FG  FGL cos  = 0
Fy = 0: FGL sin  + 1 kip = 0
+

4
E
D
Ans.
FGL = 1.518 kip = 1.518 kip (C) Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 6, page 4 of 9
2 kip
J
I
K
14 ft
8
H
L
G
A
C
D
F
E
Gy = 1 kip
16 ft
16 ft
9
Free-body diagram of joint F
FFL

FEF
10 Equilibrium equation for joint F
+

B
Ay = 1 kip
At joint F, no external forces
act, three members meet, and
two of these members are
collinear. So FL is a
zero-force member, as will
now be shown.
Fy = 0: FFL sin  = 0
Since sin   0, it follows that
F
FFG = 1.143 kip
(already known)
FFL = 0
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 6, page 5 of 9
2 kip
J
I
12 At joint L, no external forces act, three
members meet, and two of these
members are collinear. So EL is a
zero-force member:
K
14 ft
FEL = 0
Ans.
H
L
A
G
B
Ay = 1 kip
C
D
E
11 Member LF has been omitted
because it is a zero-force member.
F
Gy = 1 kip
16 ft
16 ft
6.2 Trusses: Method of Joints and Zero-Force Members Example 6, page 6 of 9
2 kip
J
I
K
14 ft
H
L
G
A
B
Ay = 1 kip
C
D
E
F
Gy = 1 kip
16 ft
16 ft
14 Consideration of joint E shows that EK is a zero-force member:
FEK = 0
Ans.
But then consideration of joint K shows that DK is also a
zero-force member:
FDK = 0
Ans.
13 Members EL and FL have been omitted
because they are zero-force members.
6.2 Trusses: Method of Joints and Zero-Force Members Example 6, page 7 of 9
2 kip
17 You cannot conclude that member
DJ is a zero-force member by
looking at end J. (Instead, look at
end D.)
J
I
K
15 All zero-force members in the right
half of the truss have been omitted.
H
L
A
B
Ay = 1 kip
C
D
16 Because of symmetry, the members in
the left half of the truss must also be
zero-force and so can be omitted, too.
E
F
G
Gy = 1 kip
18 Consideration of joint D shows that DJ
must be a zero-force member:
FDJ = 0
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 6, page 8 of 9
20 Free-body diagram of joint L
2 kip
FKL
19 Zero-force member DJ
has been omitted.
L
J
FKL = FGL
I
FGL
K
21 We have previously shown that
FGL = 1.518 kip (C)
H
L
Consideration of free-body diagrams of K and L
show that
A
B
Ay = 1 kip
D
C
E
Ans.
F
G
FKL = FKJ = 1.518 kip (C)
Ans.
Gy = 1 kip
24 By symmetry,
FAH = 1.518 kip = FHI = FIJ
Ans
23 We have previously shown that
22 Free-body diagram joint F
F
FEF = FFG
FFG = 1.143 kip (T)
Ans.
Consideration of free-body diagrams of all joints in the lower
chord, B, C, D, E, and F, shows that all member forces there
must equal 1.143 kip (T).
6.2 Trusses: Method of Joints and Zero-Force Members Example 6, page 9 of 9
25 Answer diagram
2
All forces in kips
J
I
K
1.518 (C)
1.518 (C)
0
0
0
0
0
H
L
0
0
0
0
A
B
C
D
E
G
F
1
1
1.143 (T)
6.2 Trusses: Method of Joints and Zero-Force Members Example 7, page 1 of 7
7. Determine the force in each member and
state whether the force is tension or
compression.
1
Free-body diagram
of entire truss
4 kN
H
4 kN
J
I
60°
H
J
I
60°
F
E
D
D
G
Ax
A
Ay
A
C
B
2m
2m
Cy
C
B
2
+ Fx = 0: Ax = 0

Fy = 0: Ay + Cy  4 kN = 0
+

2m
Equilibrium equations for entire truss
MA = 0: (4 kN)(2 m) + Cy(2 m + 2 m) = 0
+
2m
G
F
E
Solving simultaneously gives
Ax = 0, Ay = 2 kN, and Cy = 2 kN.
6.2 Trusses: Method of Joints and Zero-Force Members Example 7, page 2 of 7
3
Free-body diagram of entire truss.
4 Two members meet at joint J, they are not collinear,
and no external force acts at the joint, so members IJ
and FJ must be zero-force members.
4 kN
H
J
I
Free-body diagram of joint J
60°
J
FIJ
60°
D
F
E
G
FFJ
5
A
Ay = 2 kN
B
C
+ Fx = 0: F F cos 60° = 0

IJ
FJ
+

Ax
Cy = 2 kN
Equilibrium equations for joint J
Fy = 0: FFJ sin 60° = 0
Solving simultaneously gives
FIJ = 0
Ans.
FFJ = 0
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 7, page 3 of 7
7 Two members meet at the joint, they are not
collinear, and no external forces act, so the
members carry zero force.
6
Members IJ and FJ have been omitted
because they are zero-force members.
4 kN
H
J
I
8 The same argument at G shows FG and
CG are zero-force members.
F
E
D
G
Ax
A
Ay = 2 kN
B
C
Cy = 2 kN
6.2 Trusses: Method of Joints and Zero-Force Members Example 7, page 4 of 7
4 kN
10 Free-body diagram of joint C
FCF
I
9 All members identified
as zero-force have been
omitted.
60°
FBC
F
E
C
Cy = 2 kN
11 Equilibrium equations for joint C
B
C
Cy = 2 kN
Ay = 2 kN
12 By symmetry,
FAE = 2.309 kN (C)
FAB = 1.155 kN (T)
+ F = 0: F

x
BC FCF cos 60° = 0
+

A
Fy = 0: FCF sin 60° + 2 kN = 0
Solving simultaneously gives
FCF = 2.309 kN = 2.309 kN (C)
Ans.
FBC = 1.155 kN (T)
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 7, page 5 of 7
4 kN
I
F
E
14 Free-body diagram of joint B
FBF
FBE
A
B
C
Cy = 2 kN
60°
FAB = 1.155 kN (T)
Ay = 2 kN
FAB = 1.155 kN (T)
(already known)
FBC = 1.155 kN (T)
(already known)
FBC = 1.155 kN (T)
B
Equilibrium equations for joint B
+ F = 0: F

x
BE cos 60° + FBF cos 60°  1.155 kN + 1.155 kN = 0
Fy = 0: FBE sin 60° + FBF sin 60° = 0
+

13
60°
Solving simultaneously gives
FBE = 0
Ans
FBF = 0
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 7, page 6 of 7
15 Zero-force members
BE and BF have
been omitted.
16 At joint F, no external force
acts, three members meet, and
two of these members are
4 kN
collinear, so member EF is a
zero-force member.
17 Free-body diagram of joint F
y
FFI
x
I
F
FEF = 0
(zero-force member)
F
E
FCF = 2.309 kN (C)
A
B
C
18 Equilibrium equations for joint F
+

Fy = 0: FFI + 2.309 kN (C) = 0
Ay = 2 kN
Cy = 2 kN
Solving gives
FFI = 2.309 kN = 2.309 kN (C)
Ans.
Then by symmetry
FEI = FFI = 2.309 kN (C)
Ans.
6.2 Trusses: Method of Joints and Zero-Force Members Example 7, page 7 of 7
19 Answer diagram
All forces in kN
4
H
I
C)
1(
2.3
1(
2.3
D
E
1(
2.3
0
A 1.155 (T)
2
0
G
0
C)
1(
0
F
0
2.3
C)
0
J
0
C)
0
0
C
B 1.155 (T)
2