PROBLEM 6.20
Determine whether the trusses of Problems 6.22 and 6.23 are simple
trusses.
SOLUTION
Truss of Problem 6.22.
is not a simple truss
Truss of Problem 6.23.
but cannot continue, thus, this truss
is not a simple truss
consent of McGraw–Hill Education.
PROBLEM 6.21
For the given loading, determine the zero–force members in
the truss shown.
SOLUTION
Truss (a):
: Joint : 0
BC
FB B F =
: Joint : 0
CD
FB C F =
: Joint : 0
IJ
FB J F =
: Joint : 0
IL
FB I F =
: Joint : 0
MN
FB N F =
: Joint : 0
LM
FB M F =
The zero–force members, therefore, are
, ,,, ,BC CD IJ IL LM MN
(a)
consent of McGraw–Hill Education.
PROBLEM 6.22
For the given loading, determine the zero–force members in
the truss shown.
Truss (b):
: Joint : 0
BC
FB C F =
: Joint : 0
BE
FB B F =
: Joint : 0
FG
FB G F =
: Joint : 0
EF
FB F F =
: Joint : 0
DE
FB E F =
: Joint : 0
IJ
FB I F =
: Joint : 0
MN
FB M F =
: Joint : 0
KN
FB N F =
The zero–force members, therefore, are
, , , , ,, ,BC BE DE EF FG IJ KN MN
(b)
consent of McGraw–Hill Education.
PROBLEM 6.23
Truss (b)
: Joint : 0= =
AB AF
FB A F F
: Joint : 0=
CH
FB C F
: Joint : 0= =
DE EJ
FB E F F
: Joint : 0=
GL
FB L F
: Joint : 0=
IN
FB N F
(b)
PROBLEM 6.24
For the given loading, determine the zero-force members in the truss shown.
SOLUTION
Truss (a)
: Joint : 0=
BF
FB F F
: Joint : 0=
BG
FB B F
: Joint : 0=
GJ
FB G F
: Joint : 0=
DH
FB D F
: Joint : 0=
HJ
FB J F
: Joint : 0=
EH
FB H F
The zero–force members, therefore, are
,, , ,,BF BG DH EH GJ HJ
(a)
consent of McGraw–Hill Education.
PROBLEM 6.25
Determine the force in members BD and CD of the truss
shown.
SOLUTION
Reactions from Free body of entire truss
:
27 kips
y
= =AA
45 kips=H
We pass a section through members BD, CD, and CE and use the
free body shown.
0: (7.5 ft) (27 kips)(10 ft) 0
C BD
MFΣ= − − =
36.0 kips
BD
F= −
36.0 kips
BD
FC=
3
0: 27 kips+ 0
5
y CD
FF
Σ= =
45.0 kips
CD
F= −
45.0 kips
CD
FC=
PROBLEM 6.26
Determine the force in members DF and DG of the truss
shown.
SOLUTION
Reactions from Free body of entire truss
:
27 kips
y
= =AA
45 kips=H
We pass a section through members DF, DG, and EG and use the
free body shown.
0: (7.5 ft) (45 kips)(10 ft) 0
G FD
MFΣ= + =
60.0 kips
FD
F= −
60.0 kips
FD
FC=
3
0: 45 kips+ 36 kips 0
5
y GD
FF
Σ= − =
15.00 kips
GD
F= −
15.00 kips
GD
FC=
PROBLEM 6.27
Determine the force in members BD and DE of the truss shown.
SOLUTION
PROBLEM 6.28
Determine the force in members DG and EG of the truss shown.
SOLUTION
PROBLEM 6.20
Determine whether the trusses of Problems 6.22 and 6.23 are simple
trusses.
SOLUTION
Truss of Problem 6.22.
is not a simple truss
Truss of Problem 6.23.
but cannot continue, thus, this truss
is not a simple truss
consent of McGraw–Hill Education.
PROBLEM 6.21
For the given loading, determine the zero–force members in
the truss shown.
SOLUTION
Truss (a):
: Joint : 0
BC
FB B F =
: Joint : 0
CD
FB C F =
: Joint : 0
IJ
FB J F =
: Joint : 0
IL
FB I F =
: Joint : 0
MN
FB N F =
: Joint : 0
LM
FB M F =
The zero–force members, therefore, are
, ,,, ,BC CD IJ IL LM MN
(a)
consent of McGraw–Hill Education.
PROBLEM 6.22
For the given loading, determine the zero–force members in
the truss shown.
Truss (b):
: Joint : 0
BC
FB C F =
: Joint : 0
BE
FB B F =
: Joint : 0
FG
FB G F =
: Joint : 0
EF
FB F F =
: Joint : 0
DE
FB E F =
: Joint : 0
IJ
FB I F =
: Joint : 0
MN
FB M F =
: Joint : 0
KN
FB N F =
The zero–force members, therefore, are
, , , , ,, ,BC BE DE EF FG IJ KN MN
(b)
consent of McGraw–Hill Education.
PROBLEM 6.23
Truss (b)
: Joint : 0= =
AB AF
FB A F F
: Joint : 0=
CH
FB C F
: Joint : 0= =
DE EJ
FB E F F
: Joint : 0=
GL
FB L F
: Joint : 0=
IN
FB N F
(b)
PROBLEM 6.24
For the given loading, determine the zero-force members in the truss shown.
SOLUTION
Truss (a)
: Joint : 0=
BF
FB F F
: Joint : 0=
BG
FB B F
: Joint : 0=
GJ
FB G F
: Joint : 0=
DH
FB D F
: Joint : 0=
HJ
FB J F
: Joint : 0=
EH
FB H F
The zero–force members, therefore, are
,, , ,,BF BG DH EH GJ HJ
(a)
consent of McGraw–Hill Education.
PROBLEM 6.25
Determine the force in members BD and CD of the truss
shown.
SOLUTION
Reactions from Free body of entire truss
:
27 kips
y
= =AA
45 kips=H
We pass a section through members BD, CD, and CE and use the
free body shown.
0: (7.5 ft) (27 kips)(10 ft) 0
C BD
MFΣ= − − =
36.0 kips
BD
F= −
36.0 kips
BD
FC=
3
0: 27 kips+ 0
5
y CD
FF
Σ= =
45.0 kips
CD
F= −
45.0 kips
CD
FC=
PROBLEM 6.26
Determine the force in members DF and DG of the truss
shown.
SOLUTION
Reactions from Free body of entire truss
:
27 kips
y
= =AA
45 kips=H
We pass a section through members DF, DG, and EG and use the
free body shown.
0: (7.5 ft) (45 kips)(10 ft) 0
G FD
MFΣ= + =
60.0 kips
FD
F= −
60.0 kips
FD
FC=
3
0: 45 kips+ 36 kips 0
5
y GD
FF
Σ= − =
15.00 kips
GD
F= −
15.00 kips
GD
FC=
PROBLEM 6.27
Determine the force in members BD and DE of the truss shown.
SOLUTION
PROBLEM 6.28
Determine the force in members DG and EG of the truss shown.
SOLUTION