Joints

CHAPTER 3CHAPTER 7CHAPTER 7CHAPTER 3CHAPTER 6CHAPTER 6

3
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PROBLEM 6.1
Using the method of joints, determine the force in each member of the truss shown.
State whether each member is in tension or compression.
SOLUTION

AB
BC
= + =
= + =
3 1 25 3 25
3 4 5
2 2
2 2
. . m
m
Reactions:
+ SM CA = - =0 84 3 5 25 0: ( )( ) ( . )kN m m
C = 48 kN¨
+ SF A Cx x= - =0 0:
Ax = 48 kN Æ
+ SF Ay y= = =0 84 0: kN
Ay = 84 kN ≠
Joint A:
+
SF Fx AB= - =0 48
12
13
0: kN
FAB = +52 kN F TAB = 52 kN 

+
SF Fy AC= - - =0 84
5
13
52 0: ( )kN kN
FAC = +64 0. kN F TAC = 64 0. kN 
Joint C:

FBC
5
48
=
kN
3
F CBC = 80 0. kN 

4
PROBLEM 6.2
Using the method of joints, determine the force in each member of the truss
shown. State whether each member is in tension or compression.
SOLUTION
Free body: Entire truss

+
SF Bx x= =0 0:
+ SM CB = - =0 5 25 5 4 0: ( . ) ( )( )m kN m
C =
80
21
kN ≠
+ SF By y= + - =0
80
21
5 0: kN kN
By =
25
21
kN ≠
Free body: Joint B:

F F
F
AB BC
AB
5 4
25 21
3
25
63
125
68
= = =
=
/
kN
kN =1.9841 kN F CAB =1 984. kN 
FBC = =
100
63
1 5873kN kN. F TBC =1 587. kN 
Free body: Joint C:

F F
F
AC BC
AC
3 25 1 25
80 21
3
80
63
100
63
1 5873
. .
/
.
= =
= =
kN kN
kN kN
=
F CAC =1 587. kN 

F TBC =
100
63
kN (Checks)

5
PROBLEM 6.3
SOLUTION

+
SF Cx x x= = =0 0 0: C
+ SM CB y= + =0 1 92 3 4 5 0: ( . )( ) ( . )kN m m
Cy y= - =1 28 1 28. .kN kNC Ø
+ SF By = - - =0 1 92 1 28 0: . .kN kN
B = 3 20. kN ≠
Free body: Joint B:

F FAB BC
5 3
3 20
4
= =
. kN
F CAB = 4 00. kN 
F CBC = 2 40. kN 
Free body: Joint C:

+
SF Fx AC= - + =0
7 5
8 5
2 40 0:
.
.
. kN
FAC = +2 72. kN F TAC = 2 72. kN 

+ SFy = - =
4
8 5
2 72 1 28 0
.
( . ) . ( )kN kN Checks

6
PROBLEM 6.4
Using the method of joints, determine the force in each member
of the truss shown. State whether each member is in tension or
compression.
SOLUTION
Reactions:
+ SM FD y= - + - =0 6 20 12 3 5 6 0: ( ) ( )( ) ( )( )
Fy = 21 kN ≠
SFx x= =0 0: F
+ SF Dy = - + + + + =0 5 20 5 12 21 0: ( ) ( )
D = 21 kN ≠
Joint A:
SF Fx AB= =0 0: FAB = 0 
+ SF Fy AD= - - =0 5 0:
FAD = -5 kN F CAD = 5 00. kN 
Joint D: + SF Fy BD= - + + =0 5 21
8
17
0:
FBD = -34 kN F CBD = 34 0. kN 

+
SF Fx DE= - + =0
15
17
34 0: ( )
FDE = +30 kN F TDE = 30 0. kN 

7
PROBLEM 6.4 (Continued)
Joint E:
+ SF Fy BE= - =0 12 0:
FBE = +12 kN F TBE =12 00. kN 
Truss and loading symmetrical about Lc

8
PROBLEM 6.5
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.
SOLUTION
Free body: Truss

+
SFx x= =0 0: A
+ SM DA = - - =0 8 50 8 50 20 0: ( ) ( )( ) ( )( )kN kN
D =175 kN ≠
SFy = =0 75: Ay kNØ
Free body: Joint A:

F FAB AD
2 5
75
= =
kN
1
F TAB =150 0. kN 
F CAD =167 7. kN 
Free body: Joint B:
SFx = 0: F TBC =150 0. kN 
SFy = 0: F CBD = 50 kN 
Free body: Joint C:

F FCD BC
10 3
50
= =
kN
1
F CCD =158 1. kN 
F TBC =150 kN Checks( )

9
PROBLEM 6.6
SOLUTION
Free Body: Truss
+ SM FE = - - =0 3 900 2 25 900 4 5 0: ( ) ( )( . ) ( )( . )m N m N m
F = 2025 N ≠
+ SF Ex x= + + =0 900 900 0: N N
Ex x= - =1800 1800N NE ¨

+
SF Ey y= + =0 2025 0: N
Ey y= - =2025 2025N NE Ø
We note that AB and BD are zero-force members: F FAB BD= = 0 
Free body: Joint A:

F FAC AD
2 25 3 75
900
3. .
= =
N
F TAC = 675 N 
F CAD =1125 N 
Free body: Joint D:

F FCD DE
3 2 23
1125
3 75
= =
. .
N
F TCD = 900 N 
F CDF = 675 N 
Free body: Joint E:
+ SF Fx EF= - =0 1800 0: N F TEF =1800 N 

+
SF Fy CE= - =0 2025 0: N F TCE = 2025 N 

10
Free body: Joint F:
+
SF Fy CF= + - =0
2 25
3 75
2025 675 0:
.
.
N N
FCF = -2250 N F CCF = 2250 N 
+
SFx = - - - =
3
3 75
2250 1800 0
.
( )N N (Checks)

11
PROBLEM 6.7
Using the method of joints, determine the force in each member of the truss shown.
SOLUTION
Free body: Truss

+
SFy y= =0 0: B
+ SM DB = + =0 4 5 8 4 4 5 0: ( . ) ( . )( . )m kN m
D = -8 4. kN D = 8 4. kN ¨
+
SF Bx x= - - - =0 8 4 8 4 8 4 0: . . .kN kN kN
Bx x= + =25 2 25 2. .kN kNB Æ
Free body: Joint A:

F FAB AC
5 3 4 5
8 4
. .
.
= =
kN
2.8
F CAB =15 90. kN 
F TAC =13 50. kN 
Free body: Joint C:

+
SF Fy CD= - =0 13 50
4 5
5 3
0: .
.
.
kN
FCD = +15 90. kN F TCD =15 90. kN 
+ SF Fx BC= - - - =0 8 4
2 8
5 3
15 90 0: .
.
.
( . )kN kN
FBC = -16 80. kN F CBC =16 80. kN 

12
Free body: Joint D:

FBD
4 5
8 4
2 8.
.
.
=
kN
F CBD =13 50. kN 
We can also write the proportion

FBD
4 5
15 90
5 3.
.
.
=
kN
F CBD =13 50. kN (Checks)

13
PROBLEM 6.8
Using the method of joints, determine the force in each member
of the truss shown. State whether each member is in tension or
compression.
SOLUTION

AD
BCD
= + =
= + =
( . ) ( ) .
( ) ( )
2 5 6 6 5
6 8 10
2 2
2 2
m
m
Reactions: SF Dx x= =0 0:
+ SM D DE y y= - = ﬁ =0 10 5 3 2 5 0
5
7
: ( . ) ( )( . )m kN m kN Dy = 0 714. kN ≠

+
SF E
E
y = - + =
=
0
5
7
3 0
16
7
: kN kN
kN
E = 2 286. kN ≠
Joint D: + SF F Fx AD DC= + =0
5
13
4
5
0: (1)

+
SF F Fy AD DC= + + =0
12
13
3
5
5
7
0: kN (2)
Solving (1) and (2), simultaneously:
FAD = -
260
231
kN F CAD =1 126. kN 
FDC = +
125
231
kN F TDC = 0 541. kN 

14
Joint E:
+ SF F Fx BE CE= + =0
5
13
4
5
0: (3)

+
SF F Fy BE CE= + + =0
12
13
3
5
16
7
0: kN (4)
Solving (3) and (4), simultaneously:
FBE = -
832
231
kN F CBE = 3 60. kN 
FCE = +
400
231
kN F TCE =1 732. kN 
Joint C:
Force polygon is a parallelogram (see Fig. 6.11 p. 209)
F TAC =1 732. kN 
F TBC = 0 541. kN 
Joint A: + SF Fx AB=
Ê
ËÁ
ˆ
¯˜ +
Ê
ËÁ
ˆ
¯˜ + =0
5
13
260
231
4
5
400
231
0: kN kN
FAB = -
20
11
kN F CAB =1 818. kN 
+
SFy =
Ê
ËÁ
ˆ
¯˜ -
Ê
ËÁ
ˆ
¯˜ =
=
0
12
13
260
231
3
5
400
231
0
0 0
:
( )
kN kN
Checks

15
PROBLEM 6.9
Determine the force in each member of the Pratt roof
truss shown. State whether each member is in tension
or compression.
SOLUTION
Free body: Truss
SF Ax x= =0 0:
Due to symmetry of truss and load
�� A Hy = = =
1
2
21total load kN≠
Free body: Joint A:

F FAB AC
37 35
15 3
= =
. kN
12
F FAB AC= =47 175 44 625. .kN kN F CAB = 47 2. kN 
F TAC = 44 6. kN 
Free body: Joint B:
From force polygon: F FBD BC= =47 175 10 5. .kN, kN FBC = 10.50 kN C 
FBD = 47.2 kN C 

16
Free body: Joint C:
+
SF Fy CD= - =0
3
5
10 5 0: . F TCD =17 50. kN 
+ SF Fx CE= + - =0
4
5
17 50 44 625 0: ( . ) .
FCE = 30 625. kN F TCE = 30 6. kN 
Free body: Joint E: DE is a zero-force member FDE = 0 

17
PROBLEM 6.10
Determine the force in each member of the fan roof truss shown.
SOLUTION
Free body: Truss
SFx x= =0 0: A
From symmetry of truss and loading:
A Iy = =
1
2
Total load
A Iy = = 6 kN≠
Free body: Joint A:

F FAB AC
9 849 9
5
4.
= =
kN
F CAB =12 31. kN 
F TAC =11 25. kN 
Free body: Joint B:
+ SF F Fx BD DC= + + =
9
9 849
12 31 0
.
( . )kN
or F FBD BC+ = -12 31. kN (1)

+
SF F Fy BD BC= + - - =
4
9 849
12 31 2 0
.
( . )kN kN
or F FBD BC- = -7 386. kN (2)
Add (1) and (2): 2 19 70FBD = - . kN F CBD = 9 85. kN 
Subtract (2) from (1): 2 4 924FBC = - . kN F CBC = 2 46. kN 

18
Free body: Joint D:
From force polygon: F CCD = 2 00. kN 
F CDE = 9 85. kN 
Free body: Joint C:

+
SF Fy CE= - - =
4
5
4
9 849
2 46 2 0
.
( . )kN kN F TCE = 3 75. kN 
+ SF Fx CG= + + - =0
3
5
3 75
9
9 849
2 46 11 25 0: ( . )
.
( . ) .kN kN kN
FCG = +6 75. kN F TCG = 6 75. kN 
From the symmetry of the truss and loading:
F FEF DE= F CEF = 9 85. kN 
F FEG CE= F TEG = 3 75. kN 
F FFG CD= F CFG = 2 00. kN 
F FFH BD= F CFH = 9 85. kN 
F FGH BC= F CGH = 2 46. kN 
F FGI AC= F TGI =11 25. kN 
F FHI AB= F CHI =12 31. kN 

19
PROBLEM 6.11
Determine the force in each member of the Howe roof
truss shown. State whether each member is in tension or
compression.
SOLUTION
Free body: Truss
SFx x= =0 0: H
Because of the symmetry of the truss and loading:
A Hy= =
1
2
Total load
A H= =y 6 kN ≠
Free body: Joint A:

F FAB AC
5 4
4 5
3
= =
. kN
FAB C= 7 50. kN 
FAC T= 6 00. kN 
Free body: Joint C:
BC is a zero-force member
FBC = 0 FCE T= 6 00. kN 

20
Free body: Joint B:
+ SF F Fx BD BC= + + =0
4
5
4
5
4
5
7 5 0: ( . )kN
or F FBD BE+ = -7 5. kN (1)
+
SF F Fy BD BE= - + - =0
3
5
3
5
3
5
7 5 3 0: ( . )kN kN
or F FBD BE- = -2 5. kN (2)
Add Eqs. (1) and (2): 2 10FBD = - kN F CBD = 5 00. kN 
Subtract (2) from (1): 2 5FBE = - kN F CBE = 2 50. kN 
Free Body: Joint D:
+
SF Fx DF= + =0
4
5
5
4
5
0: ( )kN
FDF = -5 kN F CDF = 5 00. kN 

+
SF Fy DE= - - - - =0
3
5
5
3
5
5 3 0: ( ) ( )kN kN kN
FDE = +3 kN F TDE = 3 00. kN 
Because of the symmetry of the truss and loading, we deduce that
F FEF BE= F CEF = 2 50. kN 
F FFG BC= FFG = 0 
F FFH AB= F CFH = 7 50. kN 
F FGH AC= F TGH = 6 00. kN 

21
PROBLEM 6.12
Determine the force in each member of the Gambrel roof truss
SOLUTION
Free body: Truss
SFx x= =0 0: H
Because of the symmetry of the truss and loading
A H= =y
1
2
Total load
A H= =y 6 kN≠
Free body: Joint A:
F FAB AC
5 4
4 5
3
= =
. kN
FAB C= 7 50. kN 
FAC T= 6 00. kN 
Free body: Joint C:
BC is a zero-force member
FBC = 0 F TCE = 6 00. kN 
Free body: Joint B:
+
SF F Fx BD BE= + + =0
4
17
4
5
4
5
7 5 0: ( . )kN
or 0 9701 0 8 6. .F FBD BE+ = - kN (1)
+ SF F Fy BD BE= - + - =0
1
17
3
5
3
5
7 5 3 0: ( . )kN kN
or 0 2425 0 6 1 5. . .F FBD BE- = - kN (2)

22
Solving (1) and (2)
FBD = -6 185. kN F CBD = 6 19. kN 
FBE = ¥ -
1 9328 10 4
. kN F CBE ª 0 kN 
Free body: Joint D:
+
SF Fx DF= + =0
4
17
6 185
4
17
0: ( . )kN
FDF = -6 185. kN F CDF = 6 19. kN 
+
SF Fy DE= - - - - =0
1
17
6 185
1
17
6 185 3 0: ( . ) ( . )kN kN kN
FDE = ¥ -
1 6568 10 4
. kN F TDE ª 0 kN 
Because of the symmetry of the truss and loading, we deduce that
F FEF BE= F CEF = 0 kN 
F FFG BC= FFG = 0 
F FFH AB= F CFH = 7 50. kN 
F FGH AC= F TGH = 6 00. kN 
Note: Compare results with those of Problem 6.9.

23
PROBLEM 6.13
Determine the force in each member of the truss shown.
SOLUTION
Joint D:
12 5
6 6 5
.
.
kN
2.5
= =
F FCD DG F TCD = 30 kN 
F CDG = 32 5. kN 
Joint G: SF FCG= =0 0: 
SF F CFG= =0 32 5: . kN 
Joint C: BF BCF
BF
= = = – = = ∞-2
3
2 5 1 6667
2
39 811
( . ) . tan .m m b
+ SF Fy CF= - - =0 12 5 0: . sinkN b
- - ∞ =12 5 39 81 0. sin .kN FCF
FCF = -19 526. kN F CCF =19 53. kN 

+
SF F Fx BC CF= - - =0 30 0: coskN b
30 19 526 39 81 0kN kN- - - ∞ =FBC ( . )cos .
FBC = +45 0. kN F TBC = 45 0. kN 
Joint F:
+
SF F Fx EF CF= - - - =0
6
6 5
6
6 5
32 5 0:
. .
( . ) coskN b
FEF = - -
Ê
ËÁ
ˆ
¯˜ ∞32 5
6 5
6
19 526 39 81.
.
( . )cos .kN kN
FEF = -48 75. kN F CEF = 48 8. kN 
+ SF F Fy BF EF= - - - ∞ =0
2 5
6 5
2 5
6 5
32 5 19 526 39 81 0:
.
.
.
.
( . ) ( . )sin .kN kN
FBF - - - - =
2 5
6 5
48 75 12 5 12 5
.
.
( . ) . .kN kN kN 0
FBF = +6 25. kN F TBF = 6 25. kN 

24
Joint B: tan
.
; .a g= = ∞
2 5
2
51 34
m
m

+
SF Fy BE= - - - ∞ =0 12 5 6 25 51 34 0: . . sin .kN kN
FBE = -24 0. kN F CBE = 24 0. kN 
+ SF Fx AB= - + ∞ =0 45 0 24 0 51 34 0: . ( . )cos .kN kN
FAB = +60 kN F TAB = 60 0. kN 
Joint E:
g = ∞51 34.
+ SF Fy AE= - ∞ - =0 24 51 34 48 75
2 5
6 5
0: ( )sin . ( . )
.
.
kN kN
FAE = +37 5. kN F TAE = 37 5. kN 

25
SOLUTION
Free body: Truss
SFx x= =0 0: A
From symmetry of loading:
A Ey = =
1
2
Total load
A Ey = = 3 6. kN ≠
We note that DF is a zero-force member and that EF is aligned with the load. Thus
FDF = 0 
F CEF =1 2. kN 
Free body: Joint A:

F FAB AC
13 12
2 4
5
= =
. kN
F CAB = 6 24. kN 
F TAC = 2 76. kN 
Free body: Joint B:
+
SF F Fx BC BD= + + =0
3
3 905
12
13
12
13
6 24 0:
.
( . )kN (1)
+ SF F Fy BC BD= - + + - =0
2 5
3 905
5
13
5
13
6 24 2 4 0:
.
.
( . ) .kN kN (2)
PROBLEM 6.14
Determine the force in each member of the roof truss shown.

26
Multiply (1) by 2.5, (2) by 3, and add:

45
13
45
13
6 24 7 2 0 4 16F FBD BD+ - = = -( . ) . , .kN kN kN, F CBD = 4 16. kN 
Multiply (1) by 5, (2) by –12, and add:

45
3 905
28 8 0 2 50
.
. , .F FBC BC+ = = -kN kN, F CBC = 2 50. kN 
Free body: Joint C:

+
SF Fy CD= - =0
5
5 831
2 5
3 905
2 50 0:
.
.
.
( . )kN
F TCD =1 867. kN 
+
SF Fx CE= - + + =0 5 76 2 50
3
5 831
1 867 0: . ( . )
.
( . )kN
3
3.905
kN kN
F TCE = 2 88. kN
Free body: Joint E:

+
SF Fy DE= + - =0
5
7 81
3 6 1 2 0:
.
. .kN kN
FDE = -3 75. kN F CDE = 3 75. kN 
+ SF Fx CE= - - - =0 3 75 0: ( . )
6
7.81
kN
F F TCE CE= + =2 88 2 88. .kN kN (Checks)

27
SOLUTION
Free body: Truss SF Ax x= =0 0:
Due to symmetry of truss and loading
A Gy = = =
1
2
30Total load kN≠
Free body: Joint A:
F FAB AC
5 3
30
4
= = kN F CAB = 37 5. kN 
F TAC = 22 5. kN 
Free body: Joint B:
F FBC BD
5 6
37 5
= =
.
5
kN F TBC = 37 5. kN 
F CBD = 45 kN 
Free body: Joint C:

+
SF Fy CD= + - =0
4
5
37 5
4
5
30 0: ( . )
FCD = 0 
+ SF Fx CE= - - =0 22 5
3
5
37 5 0: . ( . )

+
FCE = +45 kN F TCE = 45 kN 
PROBLEM 6.15
Determine the force in each member of the Warren bridge truss

28
PROBLEM 6.16
Solve Problem 6.15 assuming that the load applied at E has
been removed.
PROBLEM 6.15 Determine the force in each member of the
Warren bridge truss shown. State whether each member is in
tension or compression.
SOLUTION
Free body: Truss SF Ax x= =0 0:
+ SM AG y y= - = =0 30 12 18 0 20: ( ) ( ) A kN ≠
+ SF Gy = - + = =0 20 30 0 10: G kN ≠
Free body: Joint A:
F FAB AC
5 3
20
4
= =
kN
F CAB = 25 0. kN 
F TAC =15 00. kN 
Free body Joint B:
F FBC BD
5 6
25
5
= =
kN
F TBC = 25 0. kN 
F CBD = 30 0. kN 
Free body Joint C:
+ SF Fy CD= + - =0
4
5
25
4
5
30 0: ( ) F TCD =12 50. kN 
+ SF Fx CE= + - - =0
3
5
12 5
3
5
25 15 0: ( . ) ( ) F TCE = 22 5. kN 

29
Free body: Joint D:
+
SF Fy DE= - - =0
4
5
12 5
4
5
0: ( . )
FDE = -12 5. kN F CDE =12 50. kN 
+
SF Fx DF= + - - =0 30
3
5
12 5
3
5
12 5 0: ( . ) ( . )
FDF = -15 kN F CDF =15 00. kN 
Free body: Joint F:
F FEF FG
5 5
15
= =
kN
6
F TEF =12 50. kN 
F CFG =12 50. kN 
Free body: Joint G:
FEG
3
10
=
kN
4
F TEG = 7 50. kN 
Also:
FFG
5
10
=
kN
4
F CFG =12 50. ( )kN Checks

30
PROBLEM 6.17
Determine the force in member DE and in each of the members
located to the left of DE for the inverted Howe roof truss shown.
SOLUTION
Free body: Truss:
SFx x= =0 0: A
+ SM d d d d A dH y= + + + - =0 2 4 4 3 4 2 4 4 0: ( )( ) ( )( ) ( )( ) ( ) ( )kN kN kN kN
Ay = 8 kN ≠
Angles: tan .
tan .
a a
b b
= = ∞
= = ∞
3
4
36 87
3
9
18 43
Free body: Joint A:

F FAB AC
sin . sin . sin .53 13 108 44
6
18 43∞
=
∞
=
∞
kN

F C
F T
AB
AC
=
=
15 2166
18 004
.
.
kN
kN F C F TAB AC= =15 22 18 00. .kN kN 

31
Free body: Joint B:

+
SF Fy BC= - - ∞ =0 4 18 43 0: ( )cos .kN
F F CBC BC= - =3 7948 3 79. .kN kN 

+
S ∞F Fx BD= + + =0 15 2166: . kN (4 kN) sin18.43 0
FBD = -16 4812. kN F CBD =16 48. kN 
Free body: Joint C:

+
SF Fy CE= - ∞ + ∞ -0 36 87 18 004 36 87 3 7948 18 43: sin . ( . )sin . ( . )cos .kN kN ∞∞ = 0
FCE =12 0037. kN F TCE =12 00. kN 
+ S ∞ ∞F Fx CD= - + -0 18 004 3 7948: ( . cos ( .kN) cos 36.87 (12.0037 kN) 36.87 kkN)sin .18 43 0∞ =
FCD = 5 9999. kN F TCD = 6 00. kN 
Free body: Joint E:

FDE
sin .
.
36 87
12 0037
∞
=
kN
sin71.57∞
F CDE = 7 59. kN 

32
PROBLEM 6.18
Determine the force in each of the members located to the right
of DE for the inverted Howe roof truss shown. State whether each
member is in tension or compression.
SOLUTION
Free body: Truss
+ SM A H d d d d d= - - - - =0 4 4 4 2 4 3 2 4 0: ( ) ( ) ( )( ) ( )( ) ( )( )kN kN kN kN
H = 8 kN ≠
Angles: tan .
tan .
a a
b b
= = ∞
= = ∞
3
4
36 87
3
9
18 43
Free body: Joint H:
FGH = ∞( )cot .6 18 43kN
F TGH =18 0052. kN F TGH =18 00. kN 
FFH =
∞
=
6
18 43
18 9786
kN
kN
sin .
. F CFH =18 98. kN 

33
Free body Joint F:
+
SF Fy FG= - - ∞ =0 4 18 43 0: ( )cos .kN
FFG = -3 7948. kN F CFG = 3 79. kN 

+
S ∞F Fx DF= - - + =0 18 9786 18: . kN (4 kN) sin .43 0
FDF = -17 7140. kN F CDF =17 71. kN 
Free body: Joint G:
+ SF F
F
y DG
DG
= ∞ - ∞ =
= +
0 36 87 3 7948 18 43 0
6 000
: sin . ( . )cos .
.
kN
kN
F TDG = 6 00. kN 
+
SF Fx EG= - + - ∞ - ∞0 18 0052 3 7948 18 43 6 00 36 87: . ( . )sin . ( . )cos .kN kN kN == 0
FEG = +12 0054. kN F TEG =12 01. kN 

34
PROBLEM 6.19
Determine the force in each of the members located to the left
of FG for the scissors roof truss shown. State whether each
SOLUTION
Free Body: Truss
SFx x= =0 0: A
+ SM AL y= + + + - =0 1 12 2 10 2 8 1 6 12 0: ( )( ) ( )( ) ( )( ) ( )( ) ( )kN m kN m kN m kN m m
Ay = 4 50. kN ≠
We note that BC is a zero-force member: FBC = 0 
Also: F FCE AC= (1)
Free body: Joint A: + SF F Fx AB AC= + =0
1
2
2
5
0: (2)

+
SF F Fy AB AC= + + =0
1
2
1
5
3 50 0: . kN (3)
Multiply (3) by –2 and add (2):
- - =
1
2
7FAB kN 0 F CAB = 9 90. kN 

35
Subtract (3) from (2):

1
5
3 50 0 7 826F FAC AC- = =. .kN kN F TAC = 7 83. kN 
From (1): F FCE AC= = 7 826. kN F TCE = 7 83. kN 
Free body: Joint B:

+
SF Fy BD= + - =0
1
2
1
2
9 90 2 0: ( . )kN kN
FBD = -7 071. kN F CBD = 7 07. kN 
+ SF Fx BE= + - =0
1
2
9 90 7 071 0: ( . . )kN
FBE = -2 000. kN F CBE = 2 00. kN 
Free body: Joint E: + SF Fx EG= - + =0
2
5
7 826 2 00 0: ( . ) .kN kN
FEG = 5 590. kN F TEG = 5 59. kN 

+
SF Fy DE= - - =0
1
5
7 826 5 590 0: ( . . )kN
FDE =1 000. kN F TDE =1 000. kN 
Free body: Joint D:
+ SF F Fx DF DG= + +0
2
5
1
2
7 071: ( ) ( . )kN
or F FDF DG+ = -5 590. kN (4)
+
SF F Fy DF DG= - + = - =0
1
5
1
2
7 071 2 1: ( ) ( . )kN kN kN 0
or F FDE DG- = -4 472. (5)
Add (4) and (5): 2 10 062FDF = - . kN F CDF = 5 03. kN 
Subtract (5) from (4): 2 1 1180FDG = - . kN F CDG = 0 559. kN 

36
PROBLEM 6.20
Determine the force in member FG and in each of the members
located to the right of FG for the scissors roof truss shown.
SOLUTION
Free body: Truss
+ SM LA = - - - =0 12 2 2 2 4 1 6 0: ( ) ( )( ) ( )( ) ( )( )m kN m kN m kN m
L =1 500. kN ≠
Angles: tan
tan .
a a
b b
= = ∞
= = ∞
1 45
1
2
26 57
Zero-force members:
Examining successively joints K, J, and I, we note that the following members to the right of FG are zero-force
members: JK, IJ, and HI.
Thus: F F FHI IJ JK= = = 0 
We also note that
F F FGI IK KL= = (1)
and F FHJ JL= (2)

37
Free body: Joint L:
F FJL KL
sin . sin
.
116 57 45
1 500
∞
=
∞
=
kN
sin18.43∞
FJL = 4 2436. kN F CJL = 4 24. kN 
F TKL = 3 35. kN 
From Eq. (1): F F FGI IK KL= = F F TGI IK= = 3 35. kN 
From Eq. (2): F FHJ JL= = 4 2436. kN F THJ = 4 24. kN 
Free body: Joint H:

F FFH GH
sin . sin .
.
sin .108 43 18 43
4 2436
53 14∞
=
∞
=
∞
F CFH = 5 03. kN 
F TGH =1 677. kN 
Free body: Joint F:
+
SF Fx DF= - ∞ - ∞ =0 26 57 5 03 26 57 0: cos . ( . )cos .kN
FDF = -5 03. kN

+
SF Fy FG= - - + ∞ - - ∞ =0 1 5 03 26 57 5 03 26 57 0: ( . )sin . ( . )sin .kN kN kN
FFG = 3 500. kN F TFG = 3 50. kN 

38
PROBLEM 6.21
Determine the force in each of the members located to the left
of line FGH for the studio roof truss shown. State whether each
SOLUTION
Free body: Truss SFx x= =0 0: A
Because of symmetry of loading:
A Ly = =
1
2
Total load
A Ly = = 6 kN≠
Zero-Force Members. Examining joints C and H, we conclude that
BC, EH, and GH are zero-force members. Thus
F FBC EH= = 0 
Also, F FCE AC= (1)
Free body: Joint A F F
F C
AB AC
AB
5 2
5
11 1803
= =
=
kN
1
kN.
F CAB =11 18. kN 
F TAC =10 00. kN 
From Eq. (1): F TCE =10 00. kN 
Free body: Joint B
+ SF F Fx BD BE= + + =0
2
5
2
5
2
5
11 1803 0: ( . )kN
or F FBD BE+ = -11 1803. kN (2)

+ SF F Fy BD BE= - + - =0
1
5
1
5
1
5
11 1803 2 0: ( . kN) kN
or F FBD BE- = -6 7082. kN (3)

39
Add (2) and (3): 2 17 8885
8 94425
F
F
BD
BD
= -
=
. kN
. kN F CBD = 8 94. kN 
Subtract (3) from (1): 2 4 4721
2 236
F
F
BE
BE
= -
= -
.
. kN
kN
F CBE = 2 24. kN 
Free body: Joint E
+ SF Fx EG= + - =0
2
5
2
5
2 236 10 0: ( . kN) kN
F TEG = 8 94. kN 
+ SF Fy DE= + - =0
1
5
8 944
1
5
2 236 0: ( . kN) ( . kN)
FDE = -3 00. kN F CDE = 3 00. kN 
Free body: Joint D
+ SF F Fx DF DG= + + =0
2
5
2
5
2
5
8 944 0: ( . kN)
or F FDF DG+ = -8 944. kN (4)
+ SF F Fy DF DG= - +
+ - =
0
1
5
1
5
1
5
8 944
3 2 0
: ( . kN)
kN kN
or F FDF DG- = -11 1803. kN (5)
Add (4) and (5): 2 20 1243FDF = - . kN F CDF =10 06. kN 
Subtract (5) from (4): 2 2 2363FDG = . kN F TDG =1 118. kN 

40
PROBLEM 6.22
Determine the force in member FG and in each of the members
located to the right of FG for the studio roof truss shown. State
whether each member is in tension or compression.
SOLUTION
Reaction at L: Because of the symmetry of the loading,
L =
1
2
Total load, L = 6 kN
(See F.B.D. to the left for more details)
Free body: Joint L

a
b
= = ∞
= = ∞
∞
=
∞
=
-
-
tan .
tan .
sin . sin .
1
1
3
6
26 57
1
6
9 46
63 43 99 46
5F FJL KL kkN
sin17.11
kN
∞
=F CJL 15 20. F TJL =15 20. kN 
F CKL =16 7637. kN F CKL =16 76. kN 
Free body: Joint K
+ SF F Fx IK JK= - - - =0
2
5
2
5
2
5
16 7637 0: ( . )kN
or: F FIK JK+ = -16 7637. kN (1)
+ SF F Fy IK JK= - + - =0
1
5
1
5
1
5
16 7637 2 0: ( . )
or: F FIK JK- = -12 2916. kN (2)
Add (1) and (2): 2 29 0553FIK = - .
FIK = -14 5277. kN F CIK =14 53. kN 
Subtract (2) from (1): 2 4 4721FJK = - .
FJK = -2 236. kN F CJK = 2 24. kN 

41
Free body: Joint J
+ SF F Fx IJ GJ= - - + - =0
2
13
6
37
6
37
15 20
2
5
2 236 0: ( . ) ( . )kN (3)
+ SF F Fy IJ GJ= + - - =0
3
13
1
37
1
37
15 20
1
5
2 236 0: ( . ) ( . )kN (4)
Solving (3) and (4) we obtain:
FIJ =1 8027. kN F TIJ =1 803. kN 
FGJ =12 1587. kN F TGJ =12 16. kN 
Free body: Joint I
+ SF F Fx FI GI= - - - + =0
2
5
2
5
2
5
14 5277
2
13
1 8027 0: ( . ) ( . )
or F FFI GI+ = -13 4097. kN (5)
+ SF F Fy FI GI= - + - - =0
1
5
1
5
1
5
14 5277
3
13
1 8027 2 0: ( . ) ( . )
or F FFI GI- = -6 7016. kN (6)
Add (5) and (6): 2 20 1113FFI = - .
FFI = -10 0557. kN F CFI =10 06. kN 
Subtract (6) from (5): 2 6 7081FGI = - . kN F CGI = 3 35. kN 
Free body: Joint F
From SF F F Cx DF FI= = =0 10 0557: . kN
+ SF Fy FG= - +
Ê
ËÁ
ˆ
¯˜ =0 2 2
1
5
0: kN
FFG = +7 00. kN F TFG = 7 00. kN 

42
PROBLEM 6.23
Determine the force in each of the members connecting joints A
through F of the vaulted roof truss shown. State whether each
SOLUTION
Free body: Truss
SFx x= =0 0: A
+ SM a a a aK = + + +0 1 2 6 2 4 5 2 4 4 1 2 3: ( . ) ( . ) ( . ( . )kN kN kN) kN
- = =A ay y( ) .6 0 5 40A kN≠
Free body: Joint A

F F
F
AB AC
AB
5 2
4 20
1
9 3915
= =
=
.
.
kN
kN F CAB = 9 39. kN 
F TAC = 8 40. kN 
Free body: Joint B
+ SF F Fx BD BC= + + =0
2
5
1
2
2
5
9 3915 0: ( . ) (1)
+ SF F Fy BD BC= - + - =0
1
5
1
2
1
5
9 3915 2 4 0: ( . ) . (2)
Add (1) and (2):

3
5
3
5
9 3915 2 4 0FBD + - =( . ) .kN kN
FBD = -7 6026. kN F CBD = 7 60. kN 

3
2
4 8 0FB + =. kN
FBC = -2 2627. kN F CBC = 2 26. kN 

43
Free body: Joint C
+ SF F Fx CD CE= + + - =0
1
5
4
17
1
2
2 2627 8 40 0: ( . ) . (3)
+ SF F Fy CD CE= + - =0
2
5
1
17
1
2
2 2627 0: ( . ) (4)
Multiply (4) by - 4 and add (1):
- + - =
7
5
5
2
2 2627 8 40 0FCD ( . ) .
FCD = -0 1278. kN F CCD = 0 128. kN 
Multiply (1) by 2 and subtract (2):

7
17
3
2
2 2627 2 8 40 0FCE + - =( . ) ( . )
FCE = 7 068. kN F TCE = 7 07. kN 
Free body: Joint D
+ SF F Fx DF DE= + +0
2
5
1
1 524
2
5
7 6026:
.
( . )
+ =
1
5
0 1278 0( . ) (5)
+ SF F Fy DF DE= - +0
1
5
1 15
1 524
1
5
7 6026:
.
.
( . )
+ - =
2
5
0 1278 2 4 0( . ) . (6)
Multiply (5) by 1.15 and add (6):

3 30
5
3 30
5
7 6026
3 15
5
0 1278 2 4 0
. .
( . )
.
( . ) .FDF + + - =
FDF = -6 098. kN F CDF = 6 10. kN 

3 30
1 524
3
5
0 1278 4 8 0
.
.
( . ) .FDE - + =
FDE = -2 138. kN F CDE = 2 14. kN 

44
Free body: Joint E
+ SF F F Fx EF EH CE= + - + =0
0 6
2 04
4
17
1
1 524
2 138 0:
.
.
( )
.
( . ) (7)
+ SF F F Fy EF EH CE= + - - =0
1 95
2 04
1
17
1 15
1 524
2 138 0:
.
.
( )
.
.
( . ) (8)
Multiply (8) by 4 and subtract (7):

7 2
2 04
7 856 0
.
.
.FEF - =kN F TEF = 2 23. kN 

45
PROBLEM 6.24
The portion of truss shown represents the upper part of a power
transmission line tower. For the given loading, determine the
force in each of the members located above HJ. State whether
each member is in tension or compression.
SOLUTION
Free body: Joint A

F FAB AC
2 29 2 29
1 2
. .
.
= =
kN
1.2
F TAB = 2 29. kN 
F CAC = 2 29. kN 
Free body: Joint F

F FDF EF
2 29 2 29
1 2
. .
.
= =
kN
2.1
F TDF = 2 29. kN 
F CEF = 2 29. kN 
Free body: Joint D

F FBD DE
2 21 0 6
2 29
. .
.
= =
kN
2.29
F TBD = 2 21. kN 
F CDE = 0 600. kN 
Free body: Joint B
+ SF Fx BE= + - =0
4
5
2 21
2 21
2 29
2 29 0: .
.
.
( . )kN kN
FBE = 0 
+ SF Fy BC= - - - =0
3
5
0
0 6
2 29
2 29 0: ( )
.
.
( . )kN
FBC = -0 600. kN F CBC = 0 600. kN 

46
Free body: Joint C
+ SF Fx CE= + =0
2 21
2 29
2 29 0:
.
.
( . )kN
FCE = -2 21. kN F CCE = 2 21. kN 
+ SF Fy CH= - - - =0 0 600
0 6
2 29
2 29 0: .
.
.
( . )kN kN
FCH = -1 200. kN F CCH =1 200. kN 
Free body: Joint E
+ SF Fx EH= - - =0 2 21
2 21
2 29
2 29
4
5
0: .
.
.
( . )kN kN
FEH = 0 
+ SF Fy EJ= - - - - =0 0 600
0 6
2 29
2 29 0 0: .
.
.
( . )kN kN
FEJ = -1 200. kN F CEJ =1 200. kN 

47
PROBLEM 6.25
For the tower and loading of Problem 6.24 and knowing that
FCH = FEJ = 1.2 kN C and FEH = 0, determine the force in member
HJ and in each of the members located between HJ and NO.
PROBLEM6.24Theportionoftrussshownrepresentstheupper
part of a power transmission line tower. For the given loading,
determine the force in each of the members located above HJ.
SOLUTION
Free body: Joint G

F FGH GI
3 03 3 03
1 2
. .
.
= =
kN
1.2
F TGH = 3 03. kN 
F CGI = 3 03. kN 
Free body: Joint L

F FJL KL
3 03 3 03
1 2
. .
.
= =
kN
1.2
F TJL = 3 03. kN 
F CKL = 3 03. kN 
Free body: Joint J
+ SF Fx HJ= - + =0
2 97
3 03
3 03 0:
.
.
( . )kN
F THJ = 2 97. kN 
+ F Fy JK= - - - =0 1 2
0 6
3 03
3 03 0: .
.
.
( . )kN kN
FJK = -1 800. kN F CJK =1 800. kN 
Free body: Joint H
+ SF Fx HK= + - =0
4
5
2 97
2 97
3 03
3 03 0: .
.
.
( . )kN kN
FHK = 0 
+ SF Fy HI= - - - - =0 1 2
0 6
3 03
3 03
3
5
0 0: .
.
.
( . ) ( )kN kN
FHI = -1 800. kN F CHI =1 800. kN 

48
Free body: Joint I
+ SF Fx IK= + =0
2 97
3 03
3 03 0:
.
.
( . )kN
FIK = -2 97. kN F CIK = 2 97. kN 
+ SF Fy IN= - - - =0 1 800
0 6
3 03
3 03 0: .
.
.
( . )kN kN
FIN = -2 40. kN F CIN = 2 40. kN 
Free body: Joint K
+ SF Fx KN= - + - =0
4
5
2 97
2 97
3 03
3 03 0: .
.
.
( . )kN kN
FKN = 0 
+ SF Fy KD= - - - - =0
0 6
3 03
3 03 1 800
3
5
0 0:
.
.
( . ) . ( )kN kN
FKD = -2 40. kN F CKD = 2 40. kN 

49
PROBLEM 6.26
Solve Problem 6.24 assuming that the cables hanging from the
right side of the tower have fallen to the ground.
PROBLEM6.24Theportionoftrussshownrepresentstheupper
part of a power transmission line tower. For the given loading,
determine the force in each of the members located above HJ.
SOLUTION
Zero-Force Members.
Considering joint F, we note that DF and EF are zero-force members:
F FDF EF= = 0 
Considering next joint D, we note that BD and DE are zero-force
members:
F FBD DE= = 0 
Free body: Joint A

F FAB AC
2 29 2 29
1 2
. .
.
= =
kN
1.2
F TAB = 2 29. kN 
F CAC = 2 29. kN 
Free body: Joint B
+ SF Fx BE= - =0
4
5
2 21
2 29
2 29 0:
.
.
( . )kN
FBE = 2 7625. kN F TBE = 2 76. kN 
+ SF Fy BC= - - - =0
0 6
2 29
2 29
3
5
2 7625 0:
.
.
( . ) ( . )kN kN
FBC = -2 2575. kN F CBC = 2 26. kN 

50
Free body: Joint C
+ SF Fx CE= + =0
2 21
2 29
2 29 0:
.
.
( . )kN
F CCE = 2 21. kN 
+ SF Fy CH= - - - =0 2 2575
0 6
2 29
2 29 0: .
.
.
( . )kN kN
FCH = -2 8575. kN F CCH = 2 86. kN 
Free body: joint E
+ SF Fx EH= - - + =0
4
5
4
5
2 7625 2 21 0: ( . ) .kN kN
FEH = 0 
+ SF Fy EJ= - + - =0
3
5
2 7625
3
5
0 0: ( . ) ( )kN
FEJ = +1 6575. kN F TEJ =1 658. kN 

51
SOLUTION
Free body: Truss
+ SM GF = - - =0 10 75 8 200 7 5 0: ( ) ( )( ) ( )( . )m kN m kN m
G = 210 kN≠
+ SF Fx x= + =0 75 0: kN
Fx = 75 kN¨
+ SF Fy y= - + =0 200 210 0: kN kN
Fy =10 kN Ø
Free body: Joint F
+ SF Fx DF= - =0
1
5
75 0: kN
FDF =167 705. kN F TDF =167 7. kN 
+ SF Fy BF= - + =0 10
2
5
167 705 0: ( . )kN kN
FBF = -140 00. kN F CBF =140 0. kN 
Free body: Joint B
+ SF F Fx AB BD= + + =0
5
29
5
61
75 0: kN (1)
+ SF F Fy AB BD= - + =0
2
29
6
61
140 0: kN (2)
PROBLEM 6.27
Determine the force in each member of the truss shown. State whether
each member is in tension or compression.

52
Solving (1) and (2)
FAB = -154 823. kN F CAB =154 8. kN 
FBD =107 391. kN F TBD =107 4. kN 
Free body: joint D
+ SF Fy AD= - + =0
2
5
2
5
167 705
6
61
107 391 0: ( . ) ( . )
F TAD = 75 467. kN 
+ SF Fx DE= + - - =0
1
5
75 467 167 705
5
61
107 391 0: ( . . ) ( . )
FDE =110 00. kN F TDE =110 0. kN 
Free body: joint A
+ SF F Fx AC AE= + + - =0
5
29
1
5
5
29
154 823
1
5
75 467 0: ( . ) ( . ) (3)
+ SF F Fy AC AE= - - + - =0
2
29
2
5
2
29
154 823
2
5
75 467 0: ( . ) ( . ) (4)
Multiply (3) by 2 and add (4):

8
29
12
29
154 823
4
5
75 467 0FAC + - =( . ) ( . )
FAC = -141 3602. kN, F CAC =141 4. kN 
Multiply (3) by 2 (4) by 5 and add:
- + - =
8
5
20
29
154 823
12
5
75 467 0FAE ( . ) ( . )
FAE = 47 5164. kN F TAE = 47 5. kN 

53
Free body: Joint C
From force triangle

F FCE CG
61 8
141 3602
29
= =
. kN
F TCE = 205 kN 
FCG = 210 0. kN F CCG = 210 kN 
Free body: Joint G
+ SFx = 0: FEG = 0 
+ SFy = - =0 210 210 0: ( )kN kN Checks

54
PROBLEM 6.28
Determine the force in each member of the truss shown. State
whether each member is in tension or compression.
SOLUTION
Free body: Truss
SFx x= =0 0: H
+ SM GH = - = =0 48 16 4 0 192: ( ) ( ) G kN ≠
+ SF Hy y= - + =0 192 48 0:
Hy y= - =144 144kN kNH Ø
Zero-Force Members:
Examining successively joints C, B, E, D, and F, we note that the following are zero-force members: BC, BE,
DE, DG, FG
Thus, F F F F FBC BE DE DG FG= = = = = 0
Also note: F F F FAB BD DF FH= = = (1)
F F FAC CE EG= = (2)
Free body: Joint A:
F FAB AC
8 73
48
= =
kN
3
FAB =128 kN F TAB =128 0. kN 
FAC =136 704. kN F CAC =136 7. kN 
From Eq. (1): F F F TBD DF FH= = =128 0. kN 
From Eq. (2): F F CCE EG= =136 7. kN 

55
Free body: Joint H
FGH
145
144
=
kN
9
F CGH =192 7. kN 
Also
FFH
8
144
=
kN
9
F TFH =128 0. kN (Checks)

56
PROBLEM 6.29
Determine whether the trusses of Problems 6.31a, 6.32a, and 6.33a
are simple trusses.
SOLUTION
Truss of Problem 6.31a
Starting with triangle ABC and adding two members at a time, we obtain
joints D, E, G, F, and H, but cannot go further thus, this truss
is not a simple truss 
Starting with triangle HDI and adding two members at a time, we obtain
successively joints A, E, J, and B, but cannot go further. Thus, this truss
Starting with triangle EHK and adding two members at a time, we obtain
successively joints D, J, C, G, I, B, F, and A, thus completing the truss.
Therefore, this truss
is a simple truss 

57
PROBLEM 6.30
Determine whether the trusses of Problems 6.31b, 6.32b, and 6.33b
are simple trusses.
SOLUTION
Truss of Problem 6.31b
Starting with triangle ABC and adding two members at a time, we obtain
successively joints E, D, F, G, and H, but cannot go further. Thus, this truss
Starting with triangle CGH and adding two members at a time, we obtain
successively joints B, L, F, A, K, J, then H, D, N, I, E, O, and P, thus completing
the truss.
Therefore, this truss is a simple truss 
Starting with triangle GLM and adding two members at a time, we obtain
joints K and I but cannot continue, starting instead with triangle BCH, we obtain
joint D but cannot continue, thus, this truss

58
PROBLEM 6.31
For the given loading, determine the zero-force members in each of
the two trusses shown.
SOLUTION
Truss (a) FB B FBC: Joint : = 0
FB C FCD: Joint : = 0
FB J FIJ: Joint : = 0
FB I FIL: Joint : = 0
FB N FMN: Joint : = 0
FB M FLM: Joint : = 0
The zero-force members, therefore, are BC CD IJ IL LM MN, , , , , 
Truss (b) FB C FBC: Joint : = 0
FB B FBE: Joint : = 0
FB G FFG: Joint : = 0
FB F FEF: Joint : = 0
FB E FDE: Joint : = 0
FB I FIJ: Joint : = 0
FB M FMN: Joint : = 0
FB N FKN: Joint : = 0
The zero-force members, therefore, are BC BE DE EF FG IJ KN MN, , , , , , , 

59
PROBLEM 6.32
For the given loading, determine the zero-force members in each of
the two trusses shown.
SOLUTION
Truss (a) FB B FBJ: Joint : = 0
FB D FDI: Joint : = 0
FB E FEI: Joint : = 0
FB I FAI: Joint : = 0
FB F FFK: Joint : = 0
FB G FGK: Joint : = 0
FB K FCK: Joint : = 0
The zero-force members, therefore, are AI BJ CK DI EI FK GK, , , , , , 
Truss (b) FB K FFK: Joint : = 0
FB O FIO: Joint : = 0
The zero-force members, therefore, are FK IOand 
All other members are either in tension or compression.

60
PROBLEM 6.33
For the given loading, determine the zero-force members in each of the two trusses shown.
SOLUTION
Truss (a) FB F FBF: Joint : = 0
FB B FBG: Joint : = 0
FB G FGJ: Joint : = 0
FB D FDH: Joint : = 0
FB J FHJ: Joint : = 0
FB H FEH: Joint : = 0
The zero-force members, therefore, are BF BG DH EH GJ HJ, , , , , 
Truss (b) FB A F FAB AF: Joint : = = 0
FB C FCH: Joint : = 0
FB E F FDE EJ: Joint : = = 0
FB L FGL: Joint : = 0
FB N FIN: Joint : = 0
The zero-force members, therefore, are AB AF CH DE EJ GL IN, , , , , , 

61
PROBLEM 6.34
Determine the zero-force members in the truss of (a) Problem 6.23, (b) Problem 6.28.
SOLUTION
(a) Truss of Problem 6.23
FB I FIJ: Joint : = 0
FB J FGJ: Joint : = 0
FB G FGH: Joint : = 0
The zero-force members, therefore, are GH GJ IJ, , 
(b) Truss of Problem 6.28
FB C FBC: Joint : = 0
FB B FBE: Joint : = 0
FB E FDE: Joint : = 0
FB D FDG: Joint : = 0
FB F FFG: Joint : = 0
The zero-force members, therefore, are BC BE DE DG FG, , , , 

62
PROBLEM 6.35*
The truss shown consists of six members and is supported by a short
link at A, two short links at B, and a ball and socket at D. Determine
the force in each of the members for the given loading.
SOLUTION
Free body: Truss
From symmetry:
D B D Bx x y y= =and
SM Az = - - =0 5 2 12 0: m kN m( ) ( )( )
A = -4 8. kN
SF B D Ax x x= + + =0 0:
2 4 8 0 2 4B Bx x- = =. , .kN kN
SF B Dy y y= + - =0 2 0: kN

2 2
1
B
B
y
y
=
= +
kN
kN
Thus B i j= +( . ) ( . )2 40 1 000kN kN 
Free body: C

F F
CA
CA
F
F F
CB
CB
F
CA AC
AC
CB BC
BC
= = - +
= = -
u ruu
u ruu
13
12 5
12 5
12
( )
.
(
i j
i ++
= = - -
3 5
12 5
12 3 5
. )
.
( . )
k
i kF F
CD
CD
F
CD CD
CD
u ruu
SF F F F j= + + - =0 2 0: kNCA CB CD ( )

63
PROBLEM 6.35* (Continued)
Substituting for F F FCA CB CD, , , and equating to zero the coefficients of i j k, , :
i: - - + =
12
13
12
12 5
0F F FAC BC CD
.
( ) (1)
j:
5
13
2 0FAC - =kN F TAC = 5 20. kN 
k:
3 5
12 5
0
.
.
( )F F F FBC CD CD BC- = =
Substitute for FAC and FCD in Eq. (1):
- - = = -
12
13
5 20
12
12 5
2 0 2 5( . )
.
( ) .kN kNF FBC BC F F CBC CD= = 2 50. kN 
Free body: B

F
CB
CB
F F
BA
BA
F
BC
BA AB
AB
= = - +
= =
( . ) ( . ) ( . )2 5 2 4 0 7
6
kN kN kN
u ruu
u ruu
i k
..
( . )
1033
5 3 5j k
k
-
= -F FBD BD
SF F F F i j= + + + + =0 2 4 1 0: ( . ) ( )BA BD BC kN kN
Substituting for F F FBA BD BC, , and equating to zero the coefficients of j and k:
j:
5
6 1033
1 0 1 22066
.
.F FAB AB+ = = -kN kN F CAB =1 221. kN 
k: - - + =
3 5
6 1033
0 7 0
.
.
.F FAB BD kN
FBD = - - + = +
3 5
6 1033
1 22066 0 7 1 400
.
.
( . ) . .kN kN kN F TBD =1 400. kN 
From symmetry: F FAD AB= F CAD =1 221. kN 

64
PROBLEM 6.36*
The truss shown consists of six members and is supported by a ball
and socket at B, a short link at C, and two short links at D. Determine
the force in each of the members for P = (-2184 N)j and Q = 0.
SOLUTION
Free body: Truss
From symmetry:
SF Bx x= =0 2 0:
B Dx x= = 0
SF Bz z= =0 0:
SM Bcz y= - + =0 2 2 8 2184 2 0: m N m( . ) ( )( )
By = 780 N
Thus B = ( )780 N j 
Free body: A

F F
AB
AB
F
F F
AC
AC
F
AB AB
AB
AC AC
A
= = - - +
= =
u ruu
u ruu
5 30
0 8 4 8 2 1
.
( . . . )i j k
CC
AD AD
AD
F F
AD
AD
F
5 20
2 4 8
5 30
0 8 4 8 2 1
.
( . )
.
( . . . )
i j
i j k
-
= = + - -
u ruu
SF F F F j= + + - =0 2184 0: NAB AC AD ( )

65
Substituting for F F FAB AC AD, , , and equating to zero the coefficients of i j k, , :
i: - + + =
0 8
5 30
2
5 20
0
.
.
( )
.
F F FAB AD AC (1)
j: - + - - =
4 8
5 30
4 8
5 20
2184 0
.
.
( )
.
.
F F FAB AD AC N (2)
k:
2 1
5 30
0
.
.
( )F FAB AD- = F FAD AB=
-
Ê
ËÁ
ˆ
¯˜ - = = -
16 8
5 20
2184 0 676
.
.
,F FAC ACN N F CAC = 676 N 
Substitute for FAC and FAD in (1):
-
Ê
ËÁ
ˆ
¯˜ +
Ê
ËÁ
ˆ
¯˜ - = = -
0 8
5 30
2
2
5 20
676 0 861 25
.
. .
( ) , .F FAB ABN N F F CAB AD= = 861 N 
Free body: B

F i j k
F
AB
BC BC
AB
AB
F
= = - - +
=
( . ) ( ) ( ) ( . )
.
861 25 130 780 341 25
2
N N N N
u ruu
88 2 1
3 5
0 8 0 6
i k
F
F
BC
BD BD
-Ê
ËÁ
ˆ
¯˜ = -
= -
.
.
( . . )i k
F k
SF F F F j= + + + =0 780 0: NAB BC BD ( )
Substituting for F F FAB BC BD, , and equating to zero the coefficients of i and k:
i: - + = = +130 0 8 0 162 5N N. .F FBC BC F TBC =162 5. N 
k: 341 25 0 6 0. .N - - =F FBC BD
FBD = - = +341 25 0 6 162 5 243 75. . ( . ) . N F TBD = 244 N 
From symmetry: F FCD BC= F TCD =162 5. N 

66
PROBLEM 6.37*
The truss shown consists of six members and is supported by a ball
and socket at B, a short link at C, and two short links at D. Determine
the force in each of the members for P = 0 and Q = (2968 N)i.
SOLUTION
Free body: Truss
From symmetry:
SF Bx x= + =0 2 2968 0: N
B Dx x= = -1484 N
SM Bcz y¢ = - - =0 2 2 8 2968 4 8 0: ( . ) ( )( . )m N m
By = -2544 N
Thus B i j= - -( ) ( )1484 2544N N 
Free body: A

F F
AB
AB
F
F F
AC
AC
F
AB AB
AB
AC AC
A
=
= - - +
= =
u ruu
u ruu
5 30
0 8 4 8 2 1
.
( . . . )i j k
CC
AD AD
AD
F F
AD
AD
F
5 20
2 4 8
5 30
0 8 4 8 2 1
.
( . )
.
( . . . )
i j
i j k
-
=
= - - -
u ruu
SF F F F i= + + + =0 2968 0: NAB AC AD ( )

67
Substituting for F F FAB AC AD, , , and equating to zero the coefficients of i j k, , :
i: - + + + =
0 8
5 30
2
5 20
2968 0
.
.
( )
.
F F FAB AD AC N (1)
j: - + - =
4 8
5 30
4 8
5 20
0
.
.
( )
.
.
F F FAB AD AC (2)
k:
2 1
5 30
0
.
.
( )F FAB AD- = F FAD AB=
-
Ê
ËÁ
ˆ
¯˜ - = = -
16 8
5 20
6 2968 0 5512
.
.
( ,F FAC ACN) N F CAC = 5510 N 
Substitute for FAC and FAD in (2):
-
Ê
ËÁ
ˆ
¯˜ -
Ê
ËÁ
ˆ
¯˜ - = = +
4 8
5 30
2
4 8
5 20
5512 0 2809
.
.
.
.
( ) ,F FAB ABN N
F F TAB AD= = 2810 N 
Free body: B

F i j k
F
i
AB
BC BC
BA
BA
F
= = + -
=
-
( ) ( ) ( ) ( )
.
2809 424 2544 1113
2 8 2
N N N N
u ruu
..
.
( . . )
1
3 5
0 8 0 6
k
i k
F k
Ê
ËÁ
ˆ
¯˜ = -
= -
F
F
BC
BD BD
SF F F F i j= + + - - =0 1484 2544 0: N NAB BC BD ( ) ( )
Substituting for F F FAB BC BD, , and equating to zero the coefficients of i and k:
i: + + - = = +24 0 8 1484 0 1325N N N. ,F FBC BC F TBC =1325 N 
k: - - - =1113 0 6 0N . F FBC BD
FBD = - - = -1113 0 6 1325 1908N N N,. ( ) F CBD =1908 N 
From symmetry: F FCD BC= F TCD =1325 N 

68
PROBLEM 6.38*
The truss shown consists of nine members and is supported
by a ball and socket at A, two short links at B, and a short
link at C. Determine the force in each of the members for
the given loading.
SOLUTION
Free body: Truss
From symmetry:
A Bz z= = 0
SF Ax x= =0 0:
SM ABC y= + =0 3 8 3 75 0: ( ) ( )( . )m kN m
Ay = -10 kN A j= -( .10 00 kN) 
From symmetry: B Cy =
SF By y= - - =0 2 10 8 0: kN kN
By = 9 kN B j= ( . )9 00 kN 
Free body: A SF F F F j= + + - =0 10 0: kNAB AC AD ( )
F F FAB AC AD
i k i k
i j j
+
+
-
+ + - =
2 2
0 6 0 8 10 0( . . ) ( )kN
Factoring i, j, k and equating their coefficient to zero:

1
2
1
2
0 6 0F F FAB AC AD+ + =. (1)
0 8 10 0. FAD - =kN F TAD =12 5. kN 

1
2
1
2
0F FAB AC- = F FAC AB=

69
Substitute for FAD and FAC into (1):

2
2
0 6 12 5 0 5 3033FAB AB+ = = -. ( . ) , . ,kN kNF F F CAB AC= = 5 30. kN 
Free body: B F
i k
i k
F k
F
BA AB
BC BC
BD
F
BA
BA
F
= = +
+
= +
= -
=
u ruu
( . ) ( . )( )5 3033
2
3 75kN kN
FF
F
BE
BE
F
BD
BE BE
BE
( . . )
.
( . )
0 8 0 6
6 25
3 75 4 3
j k
F i j k
-
= = + -
u ruu
SF F F F F j= + + + + =0 9 00 0: kNBA BC BD BE ( . )
Substituting for F F F FBA BC BD BE, , + and equate to zero the coefficients of i j k, , :
i: 3 75
3 75
6 25
0 6 25.
.
.
, .kN kN+
Ê
ËÁ
ˆ
¯˜ = = -F FBE BE F CBE = 6 25. kN 
j: 0 8
4
6 25
6 25 9 0 0.
.
( . ) .FBD +
Ê
ËÁ
ˆ
¯˜ - + =kN kN F CBD = 6 25. kN 
k: 3 75 0 6 6 25
3
6 25
6 25 0. . ( . )
.
( . )kN kN kN- - - - - =FBC
F TBC =10 5. kN 
From symmetry: F F CBD CD= = 6 25. kN 
Free body: D
SF F F F F i= + + + =0 0: DA DB DC DE
We now substitute for F F FDA DB DC, , and equate to zero the coefficient of i. Only FDA contains i and its
coefficient is
- = - = -0 6 0 6 12 5 7 5. . ( . ) .FAD kN kN
i: - + =7 5 0. kN FDE F TDE = 7 5. kN 

70
PROBLEM 6.39*
The truss shown consists of nine members and is supported by a ball
and socket at B, a short link at C, and two short links at D. (a) Check
that this truss is a simple truss, that it is completely constrained,
and that the reactions at its supports are statically determinate.
(b) Determine the force in each member for P = (-1200 N)j and Q = 0.
SOLUTION
Free body: Truss
SM i j i k j kB yC D D= ¥ + - ¥ +0 1 8 1 8 3: . ( . ) ( )
+ - ¥ - =( . . ) ( )0 6 0 75 1200 0i k j
- + -1 8 1 8 1 8. . .C D Dy zk k j
+ - - =3 720 900 0Dyi k i
Equate to zero the coefficients of i j k, , :
i: 3 900 0 300D Dy y- = =, N
j: Dz = 0, D j= ( )300 N 
k: 1 8 1 8 300 720 0. . ( )C + - = C j= ( )100 N 
SF B j j j= + + - =0 300 100 1200 0: B j= ( )800 N 
Free body: B
SF F F F j= + + + =0 800 0: NBA BC BE ( ) , with
F F i j kBA AB
ABBA
BA
F
= = + -
u ruu
3 15
0 6 3 0 75
.
( . . )
F iBC BCF= F kBE BEF= -
Substitute and equate to zero the coefficient of j i k, , :
j:
3
3 315
800 0 840
.
,
Ê
ËÁ
ˆ
¯˜ + = = -F FAB ABN N, F CAB = 840 N 
i:
0 6
3 15
840 0
.
.
( )
Ê
ËÁ
ˆ
¯˜ - + =N FBC F TBC =160 0. N 
k: -
Ê
ËÁ
ˆ
¯˜ - - =
0 75
3 15
840 0
.
.
( )N FBE F TBE = 200 N 

71
Free body C:
SF F F F F j= + + + + =0 100 0: NCA CB CD CE ( ) , with
F i j kCA AC
AC
F
CA
CA
F
= = - + -
u ruu
3 317
1 2 3 0 75
.
( . . )
FCB = -( )160 N i
F k F F i kCD CD CE CE
CE
F
CE
CE
F
= - = = - -
u ruu
3 499
1 8 3
,
( . )
j:
3
3 317
100 0 110 57
.
, .
Ê
ËÁ
ˆ
¯˜ + = = -F FAC ACN N F CAC =110 6. N 
i: - - - - = = -
1 2
3 317
110 57 160
1 8
3 499
0 233 3
.
.
( . )
.
.
, .F FCE CE F CCE = 233 N 
k: - - - - - =
0 75
3 317
110 57
3
3 499
233 3 0
.
.
( . )
.
( . )FCD F TCD = 225 N 
Free body: D
SF F F F j= + + + =0 300 0: NDA DC DE ( ) , with
F i j kDA AD
AD
F
DA
DA
F
= = - + +
u ruu
3 937
1 2 3 2 25
.
( . . )
F k kDC CDF= = ( )225 N F FDE DE= - i
j:
3
3 937
300 0
.
,
Ê
ËÁ
ˆ
¯˜ + =FAD N FAD = -393 7. ,N F CAD = 394 N 
i: -
Ê
ËÁ
ˆ
¯˜ - - =
1 2
3 937
393 7 0
.
.
( . )N FDE F TDE =1200 N 
k:
2 25
3 937
393 7 225 0
.
.
( . )
Ê
ËÁ
ˆ
¯˜ - + =N N (Checks)
Free body: E
Member AE is the only member at E which does not lie in the xz plane.
Therefore, it is a zero-force member.
FAE = 0 

72
PROBLEM 6.40*
Solve Problem 6.39 for P = 0 and Q = (-900 N)k.
PROBLEM 6.39* The truss shown consists of nine members
and is supported by a ball and socket at B, a short link at C, and
two short links at D. (a) Check that this truss is a simple truss,
that it is completely constrained, and that the reactions at its
supports are statically determinate. (b) Determine the force in
each member for P = (-1200 N)j and Q = 0.��
SOLUTION
Free body: Truss
SM i j i k j kB y zC D D= ¥ + - ¥ +0 1 8 1 8 3: . ( . ) ( )
+ + - ¥ - =( . . ) ( )0 6 3 0 75 900 0i j k kN
1 8 1 8 1 8. . .C D Dy zk k j+ -
+ + - =3 540 2700 0Dyi j i
Equate to zero the coefficient of i j k, , :
3 2700 0 900D Dy y- = = N
- + = =1 8 540 0 300. D Dz z N
1 8 1 8 0 900. .C D C Dy y+ = = - = - N
Thus: C j D j k= - = +( ) ( ) ( )900 900 300N N N 
SF B j j k k= - + + - =0 900 900 300 900 0: B k= ( )600 N 
Free body: B
SinceBis aligned with member BE:
F F F TAB BC BE= = =0 600, N 
Free body: C
SF F F F j= + + - =0 900 0: NCA CD CE ( ) , with

73
F i j kCA AC
AC
F
CA
CA
F
= = - + -
u ruu
3 317
1 2 3 0 75
.
( . . )
F k F i kCD CD CE CE
CE
F F
CE
CE
F
= - = = - -
u ruu
3 499
1 8 3
.
( . )
j:
3
3 317
900 0 995 1
.
, .
Ê
ËÁ
ˆ
¯˜ - = =F FAC ACN N F TAC = 995 N 
i: - - = = -
1 2
3 317
995 1
1 8
3 499
0 699 8
.
.
( . )
.
.
, .F FCE CE N F CCE = 700 N 
k: - - - - =
0 75
3 317
995 1
3
3 499
699 8 0
.
.
( . )
.
( . )FCD F TCD = 375 N 
Free body: D
SF F F k j k= + + + =0 375 300 0: ( (DA DE N) +(900 N) N)
with F F i j kDA AD
ADDA
DA
F
= = - + +
u ruu
3 937
1 2 3 2 25
.
( . . )
and F FDE DE= - i
Substitute and equate to zero the coefficient j i k, , :
j:
3
3 937
900 0 1181 1
.
, .
Ê
ËÁ
ˆ
¯˜ + = = -F FAD ADN N F CAD =1181N 
i: -
Ê
ËÁ
ˆ
¯˜ - - =
1 2
3 937
1181 1 0
.
.
( . )N FDE F TDE = 360 N 
k:
2 25
3 937
1181 1 375 300 0
.
.
( . )
Ê
ËÁ
ˆ
¯˜ - + + =N N N (Checks)
Free body: E
Member AE is the only member at E which
does not lie in the XZ plane. Therefore, it is
a zero-force member.
FAE = 0 

74
PROBLEM 6.41*
The truss shown consists of 18 members and is supported by a
ball and socket at A, two short links at B, and one short link at G.
(a) Check that this truss is a simple truss, that it is completely
constrained, and that the reactions at its supports are statically
determinate. (b) For the given loading, determine the force in each
of the six members joined at E.
SOLUTION
(a) Check simple truss.
(1) start with tetrahedron BEFG
(2) Add members BD, ED, GD joining at D.
(3) Add members BA, DA, EA joining at A.
(4) Add members DH, EH, GH joining at H.
(5) Add members BC, DC, GC joining at C.
Truss has been completed: It is a simple truss
Free body: Truss
Check constraints and reactions:
Six unknown reactions-ok-however supports at A and B
constrain truss to rotate about AB and support at G prevents
such a rotation. Thus,
Truss is completely constrained and reactions are statically
determinate
Determination of Reactions:

SM i j k i k j
j k i
A y zB B G= ¥ + + - ¥
+ - ¥ +
0 5 5 5 51 4 8
5 04 4 8 1375 1
: . ( ) ( . . )
( . . ) ( 2200 0k) =

5 5 5 5 5 5 1200 5 04 1375
5 04 1200 4
. . . ( . )( )
( . )( ) ( .
B B G Gy yk j k i k
i
- + + -
+ - 88 1375 0)( )j =
Equate to zero the coefficient of i, j, k:
i: . ( . )( )4 8 5 04 1200 0 1260G G+ = = - N G j= -( )1260 N 
j: . ( . )( )- - = = -5 5 4 8 1375 0 1200B Bz z N
k: . . ( ) ( . )( ) ,5 5 5 5 1260 5 04 1375 0 2520B By y+ - - = = N B j k= -( ) ( )2520 1200N N 

75

SF A j k j
i k
= + - -
+ + =
0 2520 1200 1260
1375 1200 0
: ( ) ( ) ( )
( ) ( )
N N N
N N
A i j= - -( ) ( )1375 1260N N 
Zero-force members
The determination of these members will facilitate our solution.
FB: C. Writing S S SF F Fx y z= = =0 0 0, , Yields F F FBC CD CG= = = 0 
FB: F. Writing S S SF F Fx y z= = =0 0 0, , Yields F F FBF EF FG= = = 0 
FB: A: Since Az = 0, writing SFz = 0 Yields FAD = 0 
FB: H: Writing SFy = 0 Yields FDH = 0 
FB: D: SinceF F FAD CD DH= = = 0, we need
consider only members DB, DE, and DG.
Since FDE is the only force not contained in plane BDG, it must be zero.
Simple reasonings show that the other two forces are also zero.
F F FBD DE DG= = = 0 
The results obtained for the reactions at the supports and for the zero-force members are shown on the
figure below. Zero-force members are indicated by a zero (“0”).
(b) Force in each of the members joined at E
We already found that F FDE EF= = 0 
Free body: A SFy = 0 Yields F TAE =1260 N 
Free body: H SFz = 0 Yields F CEH =1200 N 

76
Free body: E SF F F k j= + + - =0 1200 1260 0: ( ) ( )EB EG N N

F FBE EG
7 46
5 5 5 04
7 3
5 5 4 8 1200 1260 0
.
( . . )
.
( . . )i j i k k j- + - + - =
Equate to zero the coefficient of y and k:
j:
.
.
-
Ê
ËÁ
ˆ
¯˜ - =
5 04
7 46
1260 0FBE F CBE =1865 N 
k: -
Ê
ËÁ
ˆ
¯˜ + =
4 8
7 3
1200 0
.
.
FEG F TEG =1825 N 

77
PROBLEM 6.42*
The truss shown consists of 18 members and is supported by a
ball and socket at A, two short links at B, and one short link at G.
(a) Check that this truss is a simple truss, that it is completely
constrained, and that the reactions at its supports are statically
determinate. (b) For the given loading, determine the force in each
of the six members joined at G.
SOLUTION
See solution to Problem 6.41 for Part (a) and for reactions and zero-force members.
(b) Force in each of the members joined at G.
We already know that
F F FCG DG FG= = = 0 
Free body: H SFx = 0 Yields F CGH =1375 N 
Free body: G SF F F i j= + + - =0 1375 1260 0: ( ) ( )GB GE N N
F FBG EG
6 96
5 04 4 8
7 3
5 5 4 8 1375 1260 0
.
( . . )
.
( . . )- + + - + + - =j k i k i j
Equate to zero the coefficient of i, j, k:
i:
.
.
-
Ê
ËÁ
ˆ
¯˜ + =
5 5
7 3
1375 0FEG F TEG =1825 N 
j:
.
.
-
Ê
ËÁ
ˆ
¯˜ - =
5 04
6 96
1260 0FBG F CBG =1740 N 
k:
.
.
( )
.
.
( )
4 8
6 96
1740
4 8
7 3
1825 0
Ê
ËÁ
ˆ
¯˜ - +
Ê
ËÁ
ˆ
¯˜ = (Checks)

78
PROBLEM 6.43
A Warren bridge truss is loaded as shown. Determine the force in
members CE, DE, and DF.
SOLUTION
Free body: Truss
+ SFx x= =0 0: K
+ SMA y= -
- =
0 25 30 5
30 10 0
: ( ) ( )( )
( )( )
K m kN m
kN m
K K= =y 18 00. kN ≠ 
+ SF Ay = + - - =0 18 30 30 0: kN kN kN
A = 42 kN≠ 
We pass a section through members CE, DE, and DF and use the
free body shown.
+ SM FD CE= -
+ =
0 6 42 7 5
30 2 5 0
: ( ) ( )( . )
( )( . )
m kN m
kN m
FCE = +40 kN F TCE = 40 0. kN 
+ SF Fy DE= - - =0 42 30
6
6 5
0:
.
kN kN
FDE = +13 kN F TDE =13 00. kN 
+ SM
F
E
DF
= -
- =
0 30 5 42 10
6 0
: ( ) ( )( )
( )
kN m kN m
m
FDF = -45 kN F CDF = 45 0. kN 

79
PROBLEM 6.44
A Warren bridge truss is loaded as shown. Determine the force in
members EG, FG, and FH.
SOLUTION
See solution of Problem 6.43 for free-body diagram of truss and determination of reactions:
A = 42 kN K K= =y 18 kN 
We pass a section through members EG, FG, and FH, and use the free body shown.
+ SM FF EG= - =0 18 12 5 6 0: ( )( . ) ( )kN m m
FEG = +37 5. kN F TEG = 37 5. kN 
+ = + =SF Fy FG0
6
6 5
18 0:
.
kN
FFG = -19 5. kN F CFG =19 50. kN 
+ SM FG FH= + =0 6 18 10 0: ( ) ( )( )m kN m
FFH = -30 kN F CFH = 30 0. kN 

80
PROBLEM 6.45
Determine the force in members BD and DE of the truss shown.
SOLUTION
Member BD:
+ SM FE BD= - - =0 4 5 135 4 8 135 2 4 0: ( . ) ( )( . ) ( )( . )m kN m kN m
FBD = +216 kN F TBD = 216 kN 
Member DE:

+ SF Fx DE= + - =0 135 135 0: kN kN
FDE = +270 kN F TDE = 270 kN 

81
PROBLEM 6.46
Determine the force in members DG and EG of the truss shown.
SOLUTION
Member DG:
+ SF Fx DG= + + + =0 135 135 135
15
17
0: kN kN kN
FDG = -459 kN F CDG = 459 kN 
Member EG:
+ SM
F
D
EG
= +
+ =
0 135 4 8 135
4 5 0
: ( )( . ) ( )
( . )
kN m kN)(2.4 m
m
FEG = -216 kN F CEG = 216 kN 

82
SOLUTION
Free body: Truss
+ SF kx x= =0 0:
+ SM kA y= -
- -
0 4 8 4 0 8
4 1 6 3 2 4
: ( . ) ( )( . )
( )( . ) ( )( . )
m kN m
kN m kN m
- - - =( )( . ) ( )( ) ( )( . )2 3 2 2 4 1 4 8 0kN m kN m kN m
ky = 7 5. kN
Thus: k = 7 5. kN≠ 
+ SF A Ay = + - = =0 7 5 18 0 10 5: . .kN kN kN A =10 5. kN ≠ 
We pass a section through members CF, EF, and EG and use the free body shown.
+ SM FE CF= - +
+ =
0 0 4 10 5 1 6 2 1 6
4 0 8 0
: ( . ) ( . )( . ) ( )( . )
( )( . )
m kN m kN m
kN m
FCF = +26 0. kN F TCF = 26 0. kN 
+ SF Fy EF= - - - - =0 10 5 2 4 4
1
5
0: . kN kN kN kN
FEF = +1 118. kN F TEF =1 118. kN 
+ SM
F
F = + +
- -
0 2 2 4 4 1 6 4 0 8
10 5 2 4
: ( )( . ) ( )( . ) ( )( . )
( . )( . )
kN m kN m kN m
kN m EEG ( . )0 4 0m =
FEG = -27 0. kN F CEG = 27 0. kN 
PROBLEM 6.47
A floor truss is loaded as shown. Determine the force in
members CF, EF, and EG.

83
PROBLEM 6.48
A floor truss is loaded as shown. Determine the force in
members FI, HI, and HJ.
SOLUTION
See solution of Problem 6.47 for free-body diagram of truss and determination of reactions:
A =10 5. kN ≠, k = 7 5. kN≠ 
We pass a section through members FI, HI, and HJ, and use the free body shown.
+ SM
F
H
FI
= - -
- =
0 7 5 1 6 2 0 8 1 1 6
0 4 0
: ( . )( . ) ( )( . ) ( )( . )
( . )
kN m kN m kN m
m
FFI = +22 0. kN F TFI = 22 0. kN 
+ SF Fy HI= - - + =0
1
5
2 1 7 5 0: .kN kN kN
FHI = -10 06. kN F CHI =10 06. kN 
+ SM FI HJ= + - =0 0 4 7 5 0 8 1 0 8 0: ( . ) ( . )( . ) ( )( . )m kN m kN m
FHJ = -13 00. kN F CHJ =13 00. kN 

84
PROBLEM 6.49
A pitched flat roof truss is loaded as shown. Determine the force in
SOLUTION
Reactions at supports: Because of the symmetry of the loading,

A
A I
x
y
=
= = =
0
1
2
1
2
8( ) ( )Total load kN A I= = 4 kN≠ 
We pass a section through members CD, DE, and DF, and use the free body shown.
(We moved FDE to E and FDF to F)
Slope BJ = =
2 16
9 6
9
40
.
.
m
m
Slope DE
a
BJ
=
-
=
-
= = =
1
2 4
5
12
0 46 0 46
2 04449
40
m
m
m
Slope
m
m
.
. .
.
+ SM FD CE= + - =0 1 1 2 4 4 2 4 0: ( ) ( )( . ) ( )( . )m kN m kN m
FCE = +7 20. kN F TCE = 7 20. kN 
+ SM
F
K
DE
= -
- -
Ê
0 4 2 0444 1 2 0444
2 4 4444
5
13
: ( )( . ) ( )( . )
( )( . )
kN m kN m
kN m
ËËÁ
ˆ
¯˜ =( . )6 8444 0m
FDE = -1 047. kN F CDE =1 047. kN 
+ SM
F
E
DF
= + -
-
Ê
ËÁ
ˆ
¯˜
0 1 4 8 2 2 4 4 4 8
40
41
1
: ( )( . ) ( )( . ) ( )( . )
( .
kN m kN m kN m
554 0m) =
FDF = -6 39. kN F CDF = 6 39. kN 

85
PROBLEM 6.50
A pitched flat roof truss is loaded as shown. Determine the force in
members EG, GH, and HJ.
SOLUTION
Reactions at supports: Because of the symmetry of the loading,

A
A I
x
y
=
= = =
0
1
2
1
2
8( ) ( )Total load kN A I= = 4 kN ≠ 
We pass a section through members EG, GH, and HJ, and use the free body shown.
+ SM FH EG= - - =0 4 2 4 1 2 4 2 08 0: ( )( . ) ( )( . ) ( . )kN m kN m m
FEG = +3 4615. kN F TEG = 3 46. kN 
+ SM F FJ GH EG= - - =0 2 4 2 62 0: ( . ) ( . )m m
FGH = -
2 62
2 4
3 4615
.
.
( . )kN
FGH = -3 7788. kN F CGH = 3 78. kN 
+ SF F Fx EG HJ= - - =0
2 4
2 46
0:
.
.
F FHJ EG= - = -
2 46
2 4
2 46
2 4
3 4615
.
.
.
.
( . )kN
FHJ = -3 548. kN F CHJ = 3 55. kN 

86
PROBLEM 6.51
A Howe scissors roof truss is loaded as shown. Determine
the force in members DF, DG, and EG.
SOLUTION
Reactions at supports.
Because of symmetry of loading.
A A Lx y= = = = =0
1
2
1
2
48, ( (Total load) kN) 24 kN
A L= = 24 0. kN ≠ 
We pass a section through members DF, DG, and EG, and use the free body shown.
We slide FDF to apply it at F:
+ SM
F
G
DF
= + +
- -
0 8
24
2
: (4 kN)(6 m) N)(4 m) (8 kN)(2 m)
kN)(6 m)
(
(
22
2 5
3
1 5
2
2
+
Ê
ËÁ
ˆ
¯˜
=
.
( . m) 0
F F CDF DF= - =52 0 52 0. .kN, kN 

+ SM
F F
A
DG DG
= - -
-
+ ( )
-
0 8 8
2 3
2 2 3
4
2
22 2 2
: kN)(2 m) kN)(4 m)
m)
( (
/
/
(
++ ( )
Ê
ËÁ
ˆ
¯˜ =
2 3
5
32
/
m 0
F F CDG DG= - =16 8654 16 87. .kN, kN 

+ S - ( 4 )(4 ) -M
F
D
EG
= +
+
Ê
ËÁ
ˆ
¯˜
=0 4 4 8
2
2
3
1
2
2
: kN) m) kN)( ) m( ( ( ( )2 m 2 kN m
1
00
F F CEG EG= - =64 882 64 9. .kN, kN 

87
SOLUTION
Reactions at supports. Because of symmetry of loading:
Ax = 0, A Ly = =
=
1
2
1
2
48
( )
(
Total load
kN)
= 24 kN A L= = 24 0. kN≠ 
We pass a section through members GI, HI, and HJ, and use the free body shown.
+ SM
F
H
GI
=
+
Ê
ËÁ
ˆ
¯˜
+ - - =0
4
4
2
3
1 24 4 4 4 8 2 0
2
2
: ( ) ( )( ) ( )( ) ( )( )m kN m m m kN m
FGI = +64 883. kN F TGI = 64 9. kN 
+ SM F
F
L HI
HI
= - =
= +
0 8 2 4 0
4 0
: ( )( ) ( )
.
kN m m
kN
F THI = 4 00. kN 
We slide FHG to apply it at H.
+ SM
F
I
HJ
=
+
Ê
ËÁ
ˆ
¯˜
+ - -0
2
2
2 5
3
1 24 4 8 2 4 4
2
2
:
.
( ) ( )( ) ( )( ) ( )(m kN m kN m kN m)) = 0
FHJ = -69 333. kN F CHJ = 69 3. kN 
PROBLEM 6.52
A Howe scissors roof truss is loaded as shown. Determine
the force in members GI, HI, and HJ.

88
PROBLEM 6.53
A Pratt roof truss is loaded as shown. Determine the force in
SOLUTION
SF Ax x= =0 0:
Total load = +
=
5 3( kN) 2(1.5 kN)
18 kN
By symmetry:
A Ly = =
1
2
18( kN) A L= = 9 kN 
Free body: Portion ACD
Note: Slope of ABDF is

6 75
9 00
3
4
.
.
=
Force in CE:
+ SM FD CE= ¥
Ê
ËÁ
ˆ
¯˜ - + +0
2
3
6 75 9 1 5: . ( ( .m kN)(6 m) kN)(6 m) (3 kN)(33 m) 0=
FCE ( .4 5 36m) kN m 0- ◊ =
FCE = +8 kN F TCE = 8 kN 
Force in DE:

+
SM FA DE= + + =0 6 3 3: m) kN)(6 m) kN)(3 m) 0( ( (
FDE = -4 5. kN F CDE = 4 5. kN 

89
Force in DF:
Sum moments about E where FCE and FDE intersect.
+ SM FE CE= - + ¥0 9 3
2
3
6 75: (1.5 kN)(6 m) kN)(6 m) kN)(3 m) +
4
5
( ( . m
Ê
ËÁ
ˆ
¯˜ = 0

4
5
4 5 36FCE ( . m) kN m- ◊
FCE = -10 00. kN F CCE =10 00. kN 

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