consent of McGraw–Hill Education.
SOLUTION Continued
Free body: Joint B:
( )
11
0: 7.07 0
22
y BD
FFΣ= − =
7.07 kN
BD
FT=
( )
11
0: 5 (7.07) 7.07 0
22
x BA
FFΣ= − − − =
5.00 kN
BA
FC=
By symmetry about the horizontal centerline:
7.07 kN , 5.00 kN
DG GF
F TF C= =
Joint D:
7.07 kN
DF DB DA DG
FFFF T= = = =
Free body: Joint A:
5.00 kN
AF
FC=
consent of McGraw–Hill Education.
PROBLEM 6.15
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:
0: 0
xx
F
Σ= =H
Because of the symmetry of the truss and loading,
1total load
2
y
AH
= =
1200 lb
y
= =AH
Free body: Joint A:
900 lb
54 3
AC
AB
F
F= =
1500 lb
AB C=F
1200 lb
AC T=F
Free body: Joint C:
BC is a zero–force member.
0
BC
=F
1200 lb
CE T=F
consent of McGraw–Hill Education.
SOLUTION Continued
Free body: Joint B:
444
0: (1500 lb) 0
555
x BD BC
F FFΣ= + + =
or
1500 lb
BD BE
FF+=−
(1)
333
0: (1500 lb) 600 lb 0
555
y BD BE
F FFΣ= − + − =
or
500 lb
BD BE
FF−=−
(2)
Add Eqs. (1) and (2):
2 2000 lb
BD
F= −
1000 lb
BD
FC=
Subtract Eq. (2) from Eq. (1):
2 1000 lb
BE
F= −
500 lb
BE
FC
=
Free Body: Joint D:
44
0: (1000 lb) 0
55
x DF
FFΣ= + =
1000 lb
DF
F= −
1000 lb
DF
FC=
33
0: (1000 lb) ( 1000 lb) 600 lb 0
55
y DE
FFΣ= −− − − =
600 lb
DE
F= +
600 lb
DE
FT=
Because of the symmetry of the truss and loading, we deduce that
EF BE
FF=
500 lb
EF
FC=
EG CE
FF=
1200 lb
EG
FT=
FG BC
FF=
0
FG
F=
FH AB
FF=
1500 lb
FH
FC=
GH AC
FF=
1200 lb
GH
FT=
consent of McGraw–Hill Education.
PROBLEM 6.16
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
FΣ= =H
Because of the symmetry of the truss and loading,
1total load
2
y
= =AH
1200 lb
y
= =AH
Free body: Joint A:
900 lb
AC
AB F
F= =
SOLUTION 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:
24 24
0: (1200 lb) 0
25 25
x DF
FFΣ= + =
1200 lb
DF
F= −
1200 lb
DF
FC
=
77
0: (1200 lb) ( 1200 lb) 600 lb 0
25 25
y DE
FFΣ= − − − − =
72.0 lb
DE
F=
72.0 lb
DE
FT
=
Because of the symmetry of the truss and loading, we deduce that
EF BE
FF=
60.0 lb
EF
FC=
EG CE
FF=
1200 lb
EG
FT=
FG BC
FF=
0
FG
F=
FH AB
FF
=
1500 lb
FH
FC=
GH AC
FF=
1200 lb
GH
FT=
consent of McGraw–Hill Education.
PROBLEM 6.17
Determine the force in each member of the truss shown.
SOLUTION
Joint D:
12.5 kN
2.5 6 6.5
CD DG
FF
= =
30 kN
CD
FT=
32.5 kN
DG
FC
=
Joint G:
0: 0
CG
FFΣ= =
0: 32.5 kN
FG
FF CΣ= =
Joint C:
1
2(2.5 m) 1.6667 m tan 39.81
32
BF
BF BCF
β
−
= = =∠= = °
0: 12.5 kN sin 0
y CF
FF
β
Σ= − − =
12.5 kN sin39.81 0
CF
F− − °=
19.526 kN
CF
F= −
19.53 kN
CF
FC=
0: 30 kN cos 0
x BC CF
F FF
β
Σ= − − =
30 kN ( 19.526 kN)cos39.81 0
BC
F− − − °=
45.0 kN
BC
F= +
45.0 kN
BC
FT=
Joint F:
66
0: (32.5 kN) cos 0
6.5 6.5
x EF CF
FF F
β
Σ= − − − =
6.5
32.5 kN (19.526 kN)cos39.81
6
EF
F
=−− °
48.75 kN
EF
F= −
48.8 kN
EF
FC=
2.5 2.5
0: (32.5 kN) (19.526 kN)sin39.81 0
6.5 6.5
y BF EF
FF FΣ = − − − °=
2.5 ( 48.75 kN) 12.5 kN 12.5 kN 0
6.5
BF
F−− − − =
6.25 kN
BF
F= +
6.25 kN
BF
FT=
consent of McGraw–Hill Education.
SOLUTION Continued
Joint B:
2.5 m
tan ; 51.34
2m
aγ
= = °
0: 12.5 kN 6.25 kN sin51.34 0
y BE
FFΣ = − − − °=
24.0 kN
BE
F= −
24.0 kN
BE
FC=
0: 45.0 kN (24.0 kN)cos51.34 0
x AB
FFΣ = − + °=
60 kN
AB
F= +
60.0 kN
AB
FT=
Joint E:
51.34
γ
= °
51.34
γ
= °
2.5
0: (24 kN)sin51.34 (48.75 kN) 0
6.5
y AE
FFΣ = − °− =
37.5 kN
AE
F= +
37.5 kN
AE
FT=
consent of McGraw–Hill Education.
PROBLEM 6.18
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
FΣ= =A
0: (36 ft) (4 kips)(9 ft)
(4 kips)(18 ft) (4 kips)(27 ft) 0
A
MHΣ= −
−− =
6 kips=H
0: 6 kips 12 kips 0 6 kips
yy y
FAΣ= + − = =A
Free body: Joint A:
6 kips
53 4
AC
AB
F
F= =
7.50 kips
AB
FC=
4.50 kips
AC
FT=
Free body: Joint C:
0:
x
FΣ=
4.50 kips
CE
FT=
0:
y
FΣ=
4.00 kips
BC
FT=
Free body: Joint B:
44
0: (7.50 kips) 4.00 kips 0
55
y BE
FFΣ= − + − =
2.50 kips
BE
FT=
83
0: (7.50 kips) (2.50 kips) 0
55
x BD
FFΣ= + + =
6.00 kips
BD
F= −
6.00 kips
BD
FC=
consent of McGraw–Hill Education.
SOLUTION Continued
Free body: Joint D:
SOLUTION Continued
Free body: Joint B:
( )
11
0: 7.07 0
22
y BD
FFΣ= − =
7.07 kN
BD
FT=
( )
11
0: 5 (7.07) 7.07 0
22
x BA
FFΣ= − − − =
5.00 kN
BA
FC=
By symmetry about the horizontal centerline:
7.07 kN , 5.00 kN
DG GF
F TF C= =
Joint D:
7.07 kN
DF DB DA DG
FFFF T= = = =
Free body: Joint A:
5.00 kN
AF
FC=
consent of McGraw–Hill Education.
PROBLEM 6.15
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:
0: 0
xx
F
Σ= =H
Because of the symmetry of the truss and loading,
1total load
2
y
AH
= =
1200 lb
y
= =AH
Free body: Joint A:
900 lb
54 3
AC
AB
F
F= =
1500 lb
AB C=F
1200 lb
AC T=F
Free body: Joint C:
BC is a zero–force member.
0
BC
=F
1200 lb
CE T=F
consent of McGraw–Hill Education.
SOLUTION Continued
Free body: Joint B:
444
0: (1500 lb) 0
555
x BD BC
F FFΣ= + + =
or
1500 lb
BD BE
FF+=−
(1)
333
0: (1500 lb) 600 lb 0
555
y BD BE
F FFΣ= − + − =
or
500 lb
BD BE
FF−=−
(2)
Add Eqs. (1) and (2):
2 2000 lb
BD
F= −
1000 lb
BD
FC=
Subtract Eq. (2) from Eq. (1):
2 1000 lb
BE
F= −
500 lb
BE
FC
=
Free Body: Joint D:
44
0: (1000 lb) 0
55
x DF
FFΣ= + =
1000 lb
DF
F= −
1000 lb
DF
FC=
33
0: (1000 lb) ( 1000 lb) 600 lb 0
55
y DE
FFΣ= −− − − =
600 lb
DE
F= +
600 lb
DE
FT=
Because of the symmetry of the truss and loading, we deduce that
EF BE
FF=
500 lb
EF
FC=
EG CE
FF=
1200 lb
EG
FT=
FG BC
FF=
0
FG
F=
FH AB
FF=
1500 lb
FH
FC=
GH AC
FF=
1200 lb
GH
FT=
consent of McGraw–Hill Education.
PROBLEM 6.16
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
FΣ= =H
Because of the symmetry of the truss and loading,
1total load
2
y
= =AH
1200 lb
y
= =AH
Free body: Joint A:
900 lb
AC
AB F
F= =
SOLUTION 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:
24 24
0: (1200 lb) 0
25 25
x DF
FFΣ= + =
1200 lb
DF
F= −
1200 lb
DF
FC
=
77
0: (1200 lb) ( 1200 lb) 600 lb 0
25 25
y DE
FFΣ= − − − − =
72.0 lb
DE
F=
72.0 lb
DE
FT
=
Because of the symmetry of the truss and loading, we deduce that
EF BE
FF=
60.0 lb
EF
FC=
EG CE
FF=
1200 lb
EG
FT=
FG BC
FF=
0
FG
F=
FH AB
FF
=
1500 lb
FH
FC=
GH AC
FF=
1200 lb
GH
FT=
consent of McGraw–Hill Education.
PROBLEM 6.17
Determine the force in each member of the truss shown.
SOLUTION
Joint D:
12.5 kN
2.5 6 6.5
CD DG
FF
= =
30 kN
CD
FT=
32.5 kN
DG
FC
=
Joint G:
0: 0
CG
FFΣ= =
0: 32.5 kN
FG
FF CΣ= =
Joint C:
1
2(2.5 m) 1.6667 m tan 39.81
32
BF
BF BCF
β
−
= = =∠= = °
0: 12.5 kN sin 0
y CF
FF
β
Σ= − − =
12.5 kN sin39.81 0
CF
F− − °=
19.526 kN
CF
F= −
19.53 kN
CF
FC=
0: 30 kN cos 0
x BC CF
F FF
β
Σ= − − =
30 kN ( 19.526 kN)cos39.81 0
BC
F− − − °=
45.0 kN
BC
F= +
45.0 kN
BC
FT=
Joint F:
66
0: (32.5 kN) cos 0
6.5 6.5
x EF CF
FF F
β
Σ= − − − =
6.5
32.5 kN (19.526 kN)cos39.81
6
EF
F
=−− °
48.75 kN
EF
F= −
48.8 kN
EF
FC=
2.5 2.5
0: (32.5 kN) (19.526 kN)sin39.81 0
6.5 6.5
y BF EF
FF FΣ = − − − °=
2.5 ( 48.75 kN) 12.5 kN 12.5 kN 0
6.5
BF
F−− − − =
6.25 kN
BF
F= +
6.25 kN
BF
FT=
consent of McGraw–Hill Education.
SOLUTION Continued
Joint B:
2.5 m
tan ; 51.34
2m
aγ
= = °
0: 12.5 kN 6.25 kN sin51.34 0
y BE
FFΣ = − − − °=
24.0 kN
BE
F= −
24.0 kN
BE
FC=
0: 45.0 kN (24.0 kN)cos51.34 0
x AB
FFΣ = − + °=
60 kN
AB
F= +
60.0 kN
AB
FT=
Joint E:
51.34
γ
= °
51.34
γ
= °
2.5
0: (24 kN)sin51.34 (48.75 kN) 0
6.5
y AE
FFΣ = − °− =
37.5 kN
AE
F= +
37.5 kN
AE
FT=
consent of McGraw–Hill Education.
PROBLEM 6.18
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
FΣ= =A
0: (36 ft) (4 kips)(9 ft)
(4 kips)(18 ft) (4 kips)(27 ft) 0
A
MHΣ= −
−− =
6 kips=H
0: 6 kips 12 kips 0 6 kips
yy y
FAΣ= + − = =A
Free body: Joint A:
6 kips
53 4
AC
AB
F
F= =
7.50 kips
AB
FC=
4.50 kips
AC
FT=
Free body: Joint C:
0:
x
FΣ=
4.50 kips
CE
FT=
0:
y
FΣ=
4.00 kips
BC
FT=
Free body: Joint B:
44
0: (7.50 kips) 4.00 kips 0
55
y BE
FFΣ= − + − =
2.50 kips
BE
FT=
83
0: (7.50 kips) (2.50 kips) 0
55
x BD
FFΣ= + + =
6.00 kips
BD
F= −
6.00 kips
BD
FC=
consent of McGraw–Hill Education.
SOLUTION Continued
Free body: Joint D: