PROBLEM 6.10 (Continued)
Free body: Joint B:
11
0: 7.07 0
22
yBD
FF
7.07 kN
BD
FT
11
0: 5 (7.07) 7.07 0
22
xBA
FF
PROBLEM 6.11
Determine the force in each member of the Gambrel roof
truss shown. State whether each member is in tension or
compression.
SOLUTION
Free body: Truss:
0: 0
xx
F H
Because of the symmetry of the truss and loading,
1total load
2
y
AH
1200 lb
y
AH
PROBLEM 6.11 (Continued)
Multiply Eq. (1) by 3, Eq. (2) by 4, and add:
100 120,000 lb
BD
F 1200 lb
BD
FC
Multiply Eq. (1) by 7, Eq. (2) by –24, and add:
500 30,000 lb
BE
F 60.0 lb
BE
FC
Free body: Joint D:
PROBLEM 6.12
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:
PROBLEM 6.12 (Continued)
Free body: Joint B:
444
0: (1500 lb) 0
555
xBDBC
FFF
or 1500 lb
BD BE
FF (1)
333
0: (1500 lb) 600 lb 0
555
yBDBE
FFF
PROBLEM 6.13
Using the method of joints, determine the force in each
member of the roof truss shown. State whether each
member is in tension or compression.
SOLUTION
Free body: Truss:
0: 0
xx
F A
From symmetry of loading:
PROBLEM 6.13 (Continued)
Multiply Eq. (1) by 2.5, Eq. (2) by 3, and add:
45 45 (6.24 kN) 7.2 kN 0, 4.16 kN,
13 13
BD BD
FF
4.16 kN
BD
FC
Multiply Eq. (1) by 5, Eq. (2) by –12, and add:
45 28.8 kN 0, 2.50 kN,
3.905
BC BC
FF
2.50 kN
BC
FC
Free body: Joint C:
PROBLEM 6.14
Determine the force in each member of the Fink roof truss
shown. State whether each member is in tension or
compression.
SOLUTION
Free body: Truss:
0: 0
xx
F A
Because of the symmetry of the truss and loading,
1 total load
2
y
AG
PROBLEM 6.14 (Continued)
Free body: Joint C: 44
0: (2.8125 kN) 0
55
yCD
FF
2.8125 kN,
CD
F 2.81 kN
CD
FT
33
0: 10.125 kN (2.8125 kN) (2.8125 kN) 0
55
xCE
FF
6.7500 kN
CE
F 6.75 kN
CE
FT
Because of the symmetry of the truss and loading, we deduce that
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: 0: 0
xx
FA
Due to symmetry of truss and loading,
1total load 6 kips
2
y
AG
Free body: Joint A: 6kips
AC
AB
F
F 7.50 kips
AB
FC
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: 0: 0
xx
FA
0: 6(36) (54) 0 4 kips
Gyy
MA A
0: 4 6 0 2 kips
y
FG G
PROBLEM 6.16 (Continued)
Free body: Joint D:
44
0: (2.5) 0
55
yDE
FF
2.5 kips
DE
F 2.50 kips
DE
FC
33
0: 6 (2.5) (2.5) 0
55
xDF
FF
PROBLEM 6.17
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:
0: 0
xx
FA
Due to symmetry of truss and load,
1 total load 21 kN
2
y
AH
PROBLEM 6.17 (Continued)
Free body: Joint C: 3
0: 10.5 0
5
yCD
FF
17.50 kN
CD
FT
4
0: (17.50) 44.625 0
5
xCE
FF
30.625 kN
CE
F 30.6 kN
CE
FT
PROBLEM 6.18
The truss shown is one of several supporting an advertising panel.
Determine the force in each member of the truss for a wind load
equivalent to the two forces shown. State whether each member is in
tension or compression.
SOLUTION
Free body: Entire truss:
0: (800 N)(7.5 m) (800 N)(3.75 m) (2 m) 0
F
MA
2250 NA
2250 NA
0: 2250 N 0
yy
FF
PROBLEM 6.18 (Continued)
Joint C: 8
0: 1200 N 1600 N 0
17
xCE
FF
850 N
CE
F 850 N
CE
FC
15
0: 0
17
yBCCE
FFF
PROBLEM 6.19
Determine the force in each member of the Pratt bridge
truss shown. State whether each member is in tension or
compression.
SOLUTION
Free body: Truss:
0: 0
zx
F A
0: (36 ft) (4 kips)(9 ft)
(4 kips)(18 ft) (4 kips)(27 ft) 0
A
MH
6kipsH
0: 6 kips 12 kips 0 6 kips
yy y
FA A
PROBLEM 6.19 (Continued)
Free body: Joint D:
We note that DE is a zero-force member: 0
DE
F
Also, 6.00 kips
DF
F
C
From symmetry:
F
F
F
F
F
F
F
F
F
F
GH
PROBLEM 6.20
Solve Problem 6.19 assuming that the load applied at G has
been removed.
PROBLEM 6.19 Determine the force in each member of
the Pratt bridge truss shown. State whether each member is
in tension or compression.
SOLUTION
Free body: Truss:
0: 0
xx
F A
0: (36 ft) (4 kips)(9 ft) (4 kips)(18 ft) 0
A
H M
3.00 kipsH
0: 5.00 kips
yy
F A
We note that DE and FG are zero-force members.
PROBLEM 6.20 (Continued)
Free body: Joint F:
We recall that 0,
FG
F
and from Eq. (1) that
D
FBD
F
F 4.50 kips
DF
F
C
4.50 kips
55 6
EF FH
FF
3.75 kips
EF
F
T
3.75 kips
FH
F
C
Free body: Joint H:
F
F
F