PROBLEM 6.40
Determine the force in members DG, FG, and FH of the
truss shown.
SOLUTION
Reactions:
0: 9 kips(8 ft) (45 ft) 5 kips(30 ft) 5 kips(15 ft) 0
k
MBΣ= − + + =
6.6 kips=B
0: 9 kips 0 9 kips
xxx
FKΣ= − + = =K
0: 6.6 kips 5 kips 5 kips 0 3.4 kips
y yy
FKΣ= −−+= =K
0: 9 kips(4 ft) 6.6 kips(22.5 ft) 5 kips(7.5 ft) (4 ft) 0
F DG
MFΣ= − + − =
18.75 kips
DG
F= −
18.75 kips
DG
FC=
8
0: 6.6 kips(15 ft) (15 ft) 0
17
D FG
MF
Σ=− + =
14.025 kips
FG
F= +
14.03 kips
FG
FT=
15
0: 5 kips(15 ft) 6.6 kips(30 ft) (8 ft) 0
17
G FH
MF
Σ= − + =
17.425 kips
FH
F= +
17.43 kips
FH
FT
=
consent of McGraw–Hill Education.
PROBLEM 6.41
Determine the force in members DG and FI of the truss shown. (Hint:
Use section aa.)
SOLUTION
0: (4 m) (5 kN)(3 m) 0
F DG
MF
Σ= − =
3.75 kN
DG
F= +
3.75 kN
DG
FT=
0: 3.75 kN 0
y FI
FFΣ= − − =
3.75 kN
FI
F= −
3.75 kN
FI
FC=
consent of McGraw–Hill Education.
PROBLEM 6.42
Determine the force in members GJ and IK of the truss shown. (Hint:
Use section bb.)
SOLUTION
0: (4 m) (5 kN)(6 m) (5 kN)(3 m) 0
I GJ
MFΣ= − − =
11.25 kN
GJ
F= +
11.25 kN
GJ
FT
=
0: 11.25 kN 0
y IK
FFΣ= − − =
11.25 kN
IK
F= −
11.25 kN
IK
FC
=
consent of McGraw–Hill Education.
PROBLEM 6.43
Determine the force in members AF and EJ of the truss shown when
1.2 kN.PQ= =
(Hint: Use section aa.)
SOLUTION
Free body: Entire truss
0: (8 m) (1.2 kN)(6 m) (1.2 kN)(12 m) 0
A
MKΣ= − − =
2.70 kNK= +
2.70 kN=K
Free body: Lower portion
0: (12 m) (2.70 kN)(4 m) (1.2 kN)(6 m) (1.2 kN)(12 m) 0
F EJ
MFΣ= + − − =
0.900 kN
EJ
F= +
0.900 kN
EJ
FT=
0: 0.9 kN 1.2 kN 1.2 kN 0
y AF
FFΣ= + − − =
1.500 kN
AF
F= +
1.500 kN
AF
FT=
consent of McGraw–Hill Education.
PROBLEM 6.44
The diagonal members in the center panels of the truss shown are
very slender and can act only in tension; such members are known
as counters. Determine the force in member DE and in the counters
that are acting under the given loading.
SOLUTION
Reactions from free body of entire truss:
0: 12 kips
H
MΣ= = ↑A
0: 15 kips
y
FΣ= = ↑H
Required vertical component of bar forces
and
BE CD
FF
is downward. We must have
tension and 0.
BE CD
FF
= =
3
0: 12 kips 6 kips 0
5
y BE
FF
Σ= − − =
10.00 kips
BE
FT=
SOLUTION Continued
3
0: 12 kips+15 kips 0
5
y EF
FF
Σ= − − =
5.00 kips
EF
F=
5.00 kips
EF
FT
=
Member DE: Free body of joint D
0: 0
y DE
FFΣ= =
Joint E: (check)
( ) ( )
33
0: 10 kips 5 kips 9 kips+ 0
55
y DE
FF
Σ= + − =
0 (checks)
DE
F=
consent of McGraw–Hill Education.
PROBLEM 6.45
Solve Prob. 6.44 assuming that the 9-kip load has been removed.
Problem 6.44 The diagonal members in the center panels of the
truss shown are very slender and can act only in tension; such
members are known as counters. Determine the force in member
DE and in the counters that are acting under the given loading.
SOLUTION
Reactions from free body of entire truss:
0: 7.5 kips
H
MΣ= = ↑A
0: 10.5 kips
y
FΣ= = ↑H
Portion to the left of vertical cut through panel BCDE:
Required vertical component of bar forces
and
BE CD
FF
is downward. We must have
tension and 0.
BE CD
FF= =
3
0: 7.5 kips 6 kips 0
5
y BE
FF
Σ= − − =
2.50 kips
BE
FT=
SOLUTION Continued
3
0: 12 kips+10.5 kips+ 0
5
y DG
FF
Σ= − =
2.5 kips
DG
F=
2.50 kips
DG
FT=
Member DE: Free body of joint D
( )
3
0: 2.5 kips 0
5
y DE
FFΣ= + =
1.500 kips
DE
F= −
1.500 kips
DE
FC=
consent of McGraw–Hill Education.
PROBLEM 6.46
The diagonal members in the center panels of the truss shown
are very slender and can act only in tension; such members are
known as counters. Determine the forces in the counters that
are acting under the given loading.
SOLUTION
Free body: Truss:
0: 0
xx
FFΣ= =
0: 4.8(2 ) 4.8 2.4 2.4(2 ) 0
Gy
M FaaaaaΣ=−+ +−− =
7.20
y
F=
7.20 kipsF=
Free body: ABF:
We assume that counter CF is acting.
9.6
0: 7.20 2(4.8) 0
14.6
y CF
FFΣ= + − =
3.65
CF
F= +
3.65 kips
CF
FT=
Since CF is in tension, O.K.
Free body: DEH:
We assume that counter CH is acting.
9.6
0: 2(2.4 kips) 0
14.6
y CH
FFΣ= − =
7.30
CH
F= +
7.30 kips
CH
FT=
Since CH is in tension, O.K.
consent of McGraw–Hill Education.
PROBLEM 6.40
Determine the force in members DG, FG, and FH of the
truss shown.
SOLUTION
Reactions:
0: 9 kips(8 ft) (45 ft) 5 kips(30 ft) 5 kips(15 ft) 0
k
MBΣ= − + + =
6.6 kips=B
0: 9 kips 0 9 kips
xxx
FKΣ= − + = =K
0: 6.6 kips 5 kips 5 kips 0 3.4 kips
y yy
FKΣ= −−+= =K
0: 9 kips(4 ft) 6.6 kips(22.5 ft) 5 kips(7.5 ft) (4 ft) 0
F DG
MFΣ= − + − =
18.75 kips
DG
F= −
18.75 kips
DG
FC=
8
0: 6.6 kips(15 ft) (15 ft) 0
17
D FG
MF
Σ=− + =
14.025 kips
FG
F= +
14.03 kips
FG
FT=
15
0: 5 kips(15 ft) 6.6 kips(30 ft) (8 ft) 0
17
G FH
MF
Σ= − + =
17.425 kips
FH
F= +
17.43 kips
FH
FT
=
consent of McGraw–Hill Education.
PROBLEM 6.41
Determine the force in members DG and FI of the truss shown. (Hint:
Use section aa.)
SOLUTION
0: (4 m) (5 kN)(3 m) 0
F DG
MF
Σ= − =
3.75 kN
DG
F= +
3.75 kN
DG
FT=
0: 3.75 kN 0
y FI
FFΣ= − − =
3.75 kN
FI
F= −
3.75 kN
FI
FC=
consent of McGraw–Hill Education.
PROBLEM 6.42
Determine the force in members GJ and IK of the truss shown. (Hint:
Use section bb.)
SOLUTION
0: (4 m) (5 kN)(6 m) (5 kN)(3 m) 0
I GJ
MFΣ= − − =
11.25 kN
GJ
F= +
11.25 kN
GJ
FT
=
0: 11.25 kN 0
y IK
FFΣ= − − =
11.25 kN
IK
F= −
11.25 kN
IK
FC
=
consent of McGraw–Hill Education.
PROBLEM 6.43
Determine the force in members AF and EJ of the truss shown when
1.2 kN.PQ= =
(Hint: Use section aa.)
SOLUTION
Free body: Entire truss
0: (8 m) (1.2 kN)(6 m) (1.2 kN)(12 m) 0
A
MKΣ= − − =
2.70 kNK= +
2.70 kN=K
Free body: Lower portion
0: (12 m) (2.70 kN)(4 m) (1.2 kN)(6 m) (1.2 kN)(12 m) 0
F EJ
MFΣ= + − − =
0.900 kN
EJ
F= +
0.900 kN
EJ
FT=
0: 0.9 kN 1.2 kN 1.2 kN 0
y AF
FFΣ= + − − =
1.500 kN
AF
F= +
1.500 kN
AF
FT=
consent of McGraw–Hill Education.
PROBLEM 6.44
The diagonal members in the center panels of the truss shown are
very slender and can act only in tension; such members are known
as counters. Determine the force in member DE and in the counters
that are acting under the given loading.
SOLUTION
Reactions from free body of entire truss:
0: 12 kips
H
MΣ= = ↑A
0: 15 kips
y
FΣ= = ↑H
Required vertical component of bar forces
and
BE CD
FF
is downward. We must have
tension and 0.
BE CD
FF
= =
3
0: 12 kips 6 kips 0
5
y BE
FF
Σ= − − =
10.00 kips
BE
FT=
SOLUTION Continued
3
0: 12 kips+15 kips 0
5
y EF
FF
Σ= − − =
5.00 kips
EF
F=
5.00 kips
EF
FT
=
Member DE: Free body of joint D
0: 0
y DE
FFΣ= =
Joint E: (check)
( ) ( )
33
0: 10 kips 5 kips 9 kips+ 0
55
y DE
FF
Σ= + − =
0 (checks)
DE
F=
consent of McGraw–Hill Education.
PROBLEM 6.45
Solve Prob. 6.44 assuming that the 9-kip load has been removed.
Problem 6.44 The diagonal members in the center panels of the
truss shown are very slender and can act only in tension; such
members are known as counters. Determine the force in member
DE and in the counters that are acting under the given loading.
SOLUTION
Reactions from free body of entire truss:
0: 7.5 kips
H
MΣ= = ↑A
0: 10.5 kips
y
FΣ= = ↑H
Portion to the left of vertical cut through panel BCDE:
Required vertical component of bar forces
and
BE CD
FF
is downward. We must have
tension and 0.
BE CD
FF= =
3
0: 7.5 kips 6 kips 0
5
y BE
FF
Σ= − − =
2.50 kips
BE
FT=
SOLUTION Continued
3
0: 12 kips+10.5 kips+ 0
5
y DG
FF
Σ= − =
2.5 kips
DG
F=
2.50 kips
DG
FT=
Member DE: Free body of joint D
( )
3
0: 2.5 kips 0
5
y DE
FFΣ= + =
1.500 kips
DE
F= −
1.500 kips
DE
FC=
consent of McGraw–Hill Education.
PROBLEM 6.46
The diagonal members in the center panels of the truss shown
are very slender and can act only in tension; such members are
known as counters. Determine the forces in the counters that
are acting under the given loading.
SOLUTION
Free body: Truss:
0: 0
xx
FFΣ= =
0: 4.8(2 ) 4.8 2.4 2.4(2 ) 0
Gy
M FaaaaaΣ=−+ +−− =
7.20
y
F=
7.20 kipsF=
Free body: ABF:
We assume that counter CF is acting.
9.6
0: 7.20 2(4.8) 0
14.6
y CF
FFΣ= + − =
3.65
CF
F= +
3.65 kips
CF
FT=
Since CF is in tension, O.K.
Free body: DEH:
We assume that counter CH is acting.
9.6
0: 2(2.4 kips) 0
14.6
y CH
FFΣ= − =
7.30
CH
F= +
7.30 kips
CH
FT=
Since CH is in tension, O.K.
consent of McGraw–Hill Education.