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
FBD Truss:
Joint FBDs:
Joint B:
N
Joint C:
N
()()()
0: 6.25 m 4 m 315 N 0 240 N
By y
MCΣ= − = =C
0: 315 N 0 75 N
yy y y
FB CΣ= − + = =B
0: 0
xx
FΣ= =B
75 N
54 3
AB BC
FF
== 125.0 N C
AB
F=W
100.0 N T
BC
F=W
By inspection: 260 N C
AC
F=W
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
FBD Truss:
Joint FBDs:
Joint C:
Joint A:
() ( )( )
0: 14 ft 7.5 ft 5.6 kips 0 3 kips
Ax x
MCΣ= − = =C
0: 0 3 kips
xxx x
FACΣ= − + = =A
0: 5.6 kips 0 5.6 kips
yy y
FAΣ= − = =A
3 kips
54 3
BC AC
FF
==
5.00 kips C
BC
F=W
4.00 kips T
AC
F=W
1.6 kips
8.5 4
AB
F=
3.40 kips T
AB
F=W
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
FBD Truss:
Joint FBDs:
Joint C:
Joint B:
()( )()
0: 6 ft 6 kips 9 ft 0 4 kips
Byy
MCΣ= − = =C
0: 6 kips 0 10 kips
yy y y
FB CΣ= − − = =B
0: 0
xx
FΣ= =C
4 kips
17 15 8
AC BC
FF
==
8.50 kips T
AC
F=W
7.50 kips C
BC
F=W
By inspection: 12.50 kips C
AB
F=W
10 kips
54
AB
F=
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
FBD Truss:
Joint FBDs:
Joint D:
Joint C:
Joint B:
()()( ) ( )
0: 1.5 m 2 m 1.8 kN 3.6 m 2.4 kN 0
By
MCΣ= + − =
3.36 kN
y=C
0: 3.36 kN 2.4 kN 0
yy
FBΣ= + − =
0.96 kN
y=B
2
0: 2.4 kN 0
2.9
yAD
FFΣ= − =
3.48 kN T
AD
F=W
2.1
0: 0
2.9
xCD AD
FF FΣ= − =
2.1 (3.48 kN)
2.9
CD
F= 2.52 kN C
CD
F=W
By inspection: 3.36 kN C
AC
F=W
2.52 kN C
BC
F=W
4
0: 0.9 kN 0
5
yAB
FFΣ= − =
1.200 kN T
AB
F=W
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
FBD Truss:
Joint FBDs:
Joint B:
Joint C:
Joint A:
0:
x
FΣ= 0
x=C
By symmetry: 6 kN
yy
==CD
1
0: 3 kN 0
5
yAB
FFΣ=− + =
3 5 6.71 kN T
AB
F== W
2
0: 0
5
xABBC
FFFΣ= − =
6.00 kN C
BC
F=W
3
0: 6 kN 0
5
yAC
FFΣ= − =
10.00 kN C
AC
F=W
4
0: 6 kN 0
5
xACCD
FFFΣ= − + =
2.00 kN T
CD
F=W
13
0: 2 3 5 kN 2 10 kN 6 kN 0 check
5
5
y
F
Σ=− + − =
By symmetry: 6.71 kN T
AE AB
FF== W
10.00 kN C
AD AC
FF== W
6.00 kN C
DE BC
FF== W
