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
Free body: Entire truss
0: 0 0
x xx
FCΣ= = =C
0: (1.92 kN)(3 m) (4.5 m) 0
By
MCΣ= + =
1.28 kN 1.28 kN
yy
C=−=C
0: 1.92 kN 1.28 kN 0
y
FBΣ= − − =
3.20 kN=B
Free body: Joint B:
3.20 kN
53 4
BC
AB F
F= =
4.00 kN
AB
FC=
2.40 kN
BC
FC=
Free body: Joint C:
7.5
0: 2.40 kN 0
8.5
x AC
FFΣ= − + =
2.72 kN
AC
FT=
4(2.72 kN) 1.28 kN 0 (Checks)
8.5
y
FΣ= − =
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
22
22
3 1.25 3.25 m
3 4 5m
AB
BC
=+=
= +=
Reactions:
0: (84 kN)(3 m) (5.25 m) 0
A
MCΣ= − =
48 kN=C
0: 0
xx
F ACΣ= −=
48 kN
x=A
0: 84 kN 0
yy
FAΣ= = =
84 kN
y=A
Joint A:
12
0: 48 kN 0
13
x AB
FFΣ= − =
52 kN
AB
F= +
52.0 kN
AB
FT
=
5
0: 84 kN (52 kN) 0
13
y AC
FFΣ= − − =
64.0 kN
AC
FT=
Joint C:
48 kN
53
BC
F=
80.0 kN
BC
FC=
consent of McGraw–Hill Education.
PROBLEM 6.3
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:
0: 1260 lb
A
MΣ= =C
0: 0
xx
FΣ= =A
0: 960 lb
yy
FΣ= =A
Joint B:
300 lb
12 13 5
BC
AB F
F= =
720 lb
AB
FT=
780 lb
BC
FC=
Joint A:
4
0: 960 lb 0
5
y AC
FFΣ= − − =
1200 lb
AC
F=
1200 lb
AC
FC=
3
0: 720 lb (1200 lb) 0 (checks)
5
x
FΣ= − =
consent of McGraw–Hill Education.
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
Free body: Entire truss:
0: 0 0
yy y
FBΣ= = =B
0: (16 in.) (240 lb)(36 in.) 0
Cx
MBΣ= − − =
540 kN 540 lb
xx
B=−=
B
0: 540 lb 240 lb 0
x
FCΣ= − + =
300 lbC=
300 lb=
C
Free body: Joint B:
540 lb
54 3
BC
AB F
F= =
900 lb
AB
FT=
720 lb
BC
FT=
consent of McGraw–Hill Education.
SOLUTION Continued
Free body: Joint C:
300 lb
13 12 5
AC BC
FF
= =
780 lb
AC
FC=
720 lb (checks)
BC
F=
consent of McGraw–Hill Education.
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
0: 0
xx
FΣ= =A
0: (22.5) (10.8 kips)(22.5) (10.8 kips)(57.5) 0
A
MDΣ= − − =
38.4 kips=D
0: 16.8 kips
yy
FΣ= =A
Free body: Joint A:
16.8 kips
22.5 25.5 12
AB AD
FF
= =
31.5 kips
AB
FT
=
35.7 kips
AD
FC=
Free body: Joint B:
0:
x
FΣ=
31.5 kips
BC
FT=
0:
y
FΣ=
10.80 kips
BD
FC=
Free body: Joint C:
10.8 kips
37 35 12
CD BC
FF
= =
33.3 kips
CD
FC=
31.5 kips (Checks)
BC
FT=
consent of McGraw–Hill Education.
PROBLEM 6.6
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
0: 0
yy
FΣ= =B
0: (4.5 m) (8.4 kN)(4.5 m) 0
B
MDΣ= + =
8.4 kND= −
8.4 kN=D
0: 8.4 kN 8.4 kN 8.4 kN 0
xx
FBΣ=−−−=
25.2 kN 25.2 kN
xx
B=+=B
Free body: Joint A:
8.4 kN
5.3 4.5 2.8
AC
AB F
F= =
15.90 kN
AB
FC=
13.50 kN
AC
FT=
Free body: Joint C:
4.5
0: 13.50 kN 0
5.3
y CD
FFΣ= − =
15.90 kN
CD
F= +
15.90 kN
CD
FT=
2.8
0: 8.4 kN (15.90 kN) 0
5.3
x BC
FFΣ= − − − =
16.80 kN
BC
F= −
16.80 kN
BC
FC=
consent of McGraw–Hill Education.
SOLUTION Continued
Free body: Joint D:
8.4 kN
4.5 2.8
BD
F=
13.50 kN
BD
FC=
We can also write the proportion
15.90 kN
4.5 5.3
BD
F=
13.50 kN (Checks)
BD
FC=
consent of McGraw–Hill Education.
PROBLEM 6.7
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:
( ) ( )
22
6 m 3.2 m 6.80 mDE =+=
4
0: (4.5 m) (24 kN)(12 m) 0
5
B AC
MF

Σ= − =


80.0 kN
AC
T=F
( )
4
0: 80 kN 0
5
xx
FB

Σ= − =


64 kN
x
=B
( )
3
0: 80 kN 24 kN 0
5
yy
FB

Σ= + − =


24 kN
y=B
Joint E:
24 kN
6 6.8 3.2
CE DE
FF
= =
45.0 kN
CE
FT=
51.0 kN
DE
FC=
consent of McGraw–Hill Education.
SOLUTION Continued
Joint D:
( )
66
0: 51.0 kN 0
6.8 6.8
x BD
FF
 
Σ= − =
 
 
51.0 kN
BD
FC=
48.0 kN
CD
FT=
Joint C:
( ) ( )
4
0: 80 kN 45.0 kN 0
5
x BC
FF

Σ= − − + =


19.00 kN
BC
FC=
( )
3
0: 80 kN 48.0 kN 0 (checks)
5
y
F
Σ= − =


consent of McGraw–Hill Education.
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
22
22
3 1.25 3.25 m
3 4 5m
AB
BC
=+=
= +=
Reactions:
0: (84 kN)(3 m) (5.25 m) 0
A
MCΣ= − =
48 kN=C
0: 0
xx
F ACΣ= −=
48 kN
x=A
0: 84 kN 0
yy
FAΣ= = =
84 kN
y=A
Joint A:
12
0: 48 kN 0
13
x AB
FFΣ= − =
52 kN
AB
F= +
52.0 kN
AB
FT
=
5
0: 84 kN (52 kN) 0
13
y AC
FFΣ= − − =
64.0 kN
AC
FT=
Joint C:
48 kN
53
BC
F=
80.0 kN
BC
FC=
consent of McGraw–Hill Education.
PROBLEM 6.3
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:
0: 1260 lb
A
MΣ= =C
0: 0
xx
FΣ= =A
0: 960 lb
yy
FΣ= =A
Joint B:
300 lb
12 13 5
BC
AB F
F= =
720 lb
AB
FT=
780 lb
BC
FC=
Joint A:
4
0: 960 lb 0
5
y AC
FFΣ= − − =
1200 lb
AC
F=
1200 lb
AC
FC=
3
0: 720 lb (1200 lb) 0 (checks)
5
x
FΣ= − =
consent of McGraw–Hill Education.
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
Free body: Entire truss:
0: 0 0
yy y
FBΣ= = =B
0: (16 in.) (240 lb)(36 in.) 0
Cx
MBΣ= − − =
540 kN 540 lb
xx
B=−=
B
0: 540 lb 240 lb 0
x
FCΣ= − + =
300 lbC=
300 lb=
C
Free body: Joint B:
540 lb
54 3
BC
AB F
F= =
900 lb
AB
FT=
720 lb
BC
FT=
consent of McGraw–Hill Education.
SOLUTION Continued
Free body: Joint C:
300 lb
13 12 5
AC BC
FF
= =
780 lb
AC
FC=
720 lb (checks)
BC
F=
consent of McGraw–Hill Education.
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
0: 0
xx
FΣ= =A
0: (22.5) (10.8 kips)(22.5) (10.8 kips)(57.5) 0
A
MDΣ= − − =
38.4 kips=D
0: 16.8 kips
yy
FΣ= =A
Free body: Joint A:
16.8 kips
22.5 25.5 12
AB AD
FF
= =
31.5 kips
AB
FT
=
35.7 kips
AD
FC=
Free body: Joint B:
0:
x
FΣ=
31.5 kips
BC
FT=
0:
y
FΣ=
10.80 kips
BD
FC=
Free body: Joint C:
10.8 kips
37 35 12
CD BC
FF
= =
33.3 kips
CD
FC=
31.5 kips (Checks)
BC
FT=
consent of McGraw–Hill Education.
PROBLEM 6.6
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
0: 0
yy
FΣ= =B
0: (4.5 m) (8.4 kN)(4.5 m) 0
B
MDΣ= + =
8.4 kND= −
8.4 kN=D
0: 8.4 kN 8.4 kN 8.4 kN 0
xx
FBΣ=−−−=
25.2 kN 25.2 kN
xx
B=+=B
Free body: Joint A:
8.4 kN
5.3 4.5 2.8
AC
AB F
F= =
15.90 kN
AB
FC=
13.50 kN
AC
FT=
Free body: Joint C:
4.5
0: 13.50 kN 0
5.3
y CD
FFΣ= − =
15.90 kN
CD
F= +
15.90 kN
CD
FT=
2.8
0: 8.4 kN (15.90 kN) 0
5.3
x BC
FFΣ= − − − =
16.80 kN
BC
F= −
16.80 kN
BC
FC=
consent of McGraw–Hill Education.
SOLUTION Continued
Free body: Joint D:
8.4 kN
4.5 2.8
BD
F=
13.50 kN
BD
FC=
We can also write the proportion
15.90 kN
4.5 5.3
BD
F=
13.50 kN (Checks)
BD
FC=
consent of McGraw–Hill Education.
PROBLEM 6.7
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:
( ) ( )
22
6 m 3.2 m 6.80 mDE =+=
4
0: (4.5 m) (24 kN)(12 m) 0
5
B AC
MF

Σ= − =


80.0 kN
AC
T=F
( )
4
0: 80 kN 0
5
xx
FB

Σ= − =


64 kN
x
=B
( )
3
0: 80 kN 24 kN 0
5
yy
FB

Σ= + − =


24 kN
y=B
Joint E:
24 kN
6 6.8 3.2
CE DE
FF
= =
45.0 kN
CE
FT=
51.0 kN
DE
FC=
consent of McGraw–Hill Education.
SOLUTION Continued
Joint D:
( )
66
0: 51.0 kN 0
6.8 6.8
x BD
FF
 
Σ= − =
 
 
51.0 kN
BD
FC=
48.0 kN
CD
FT=
Joint C:
( ) ( )
4
0: 80 kN 45.0 kN 0
5
x BC
FF

Σ= − − + =


19.00 kN
BC
FC=
( )
3
0: 80 kN 48.0 kN 0 (checks)
5
y
F
Σ= − =


consent of McGraw–Hill Education.