PROBLEM 6.8
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: (3 m) (24 kN 8 kN)(1.5 m) (7 kN)(3 m) 0
Dy
MFΣ= − + − =
23.0 kN
y
=F
0: 0
xx
FΣ= =F
0: (7 24 8 7) 23 0
y
FDΣ = − + ++ + =
23.0 kN=D
Joint A:
0: 0
x AB
FFΣ= =
0
AB
F=
consent of McGraw–Hill Education.
SOLUTION Continued
15
0: ( 34.0) 0
17
x DE
FFΣ= − + =
30.0 kN
DE
F= +
30.0 kN
DE
FT=
Joint E:
0: 8 0
y BE
FFΣ = −=
8.00 kN
BE
F= +
8.00 kN
BE
FT=
Truss and loading symmetrical about
c.
L
consent of McGraw–Hill Education.
PROBLEM 6.9
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:
From the symmetry of the truss and loading, we find
600 lb= =CD
Free body: Joint B:
300 lb
21
5
BC
AB F
F= =
671lb
AB
FT=
600 lb
BC
FC=
Free body: Joint C:
3
0: 600 lb 0
5
y AC
FF
Σ= + =
1000 lb
AC
F= −
1000 lb
AC
FC=
4
0: ( 1000 lb) 600 lb 0
5
x CD
FFΣ= − + + =
200 lb
CD
FT=
From symmetry:
1000 lb , 671lb ,
AD AC AE AB
F F CF F T= = = =
600 lb
DE BC
FF C= =
consent of McGraw–Hill Education.
PROBLEM 6.10
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: 16 kN
Cx
MΣ= =A
0: 9 kN
yy
FΣ= =A
0: 16 kN
x
FΣ= =C
Joint E:
3 kN
54 3
BE DE
FF
= =
5.00 kN
BE
FT
=
4.00 kN
DE
FC
=
Joint B:
4
0: (5 kN) 0
5
x AB
FFΣ= − =
4 kN
AB
F= +
4.00 kN
AB
FT=
3
0: 6 kN (5 kN) 0
5
y BD
FFΣ= − − − =
9 kN
BD
F= −
9.00 kN
BD
FC=
Joint D:
3
0: 9 kN 0
5
y AD
FFΣ= − + =
15 kN
AD
F= +
15.00 kN
AD
FT=
4
0: 4 kN (15 kN) 0
5
x CD
FFΣ= − − − =
16 kN
CD
F= −
16.00 kN
CD
FC=
PROBLEM 6.11
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
5 12 13 ft
12 16 20 ft
AD
BCD
=+=
= +=
Reactions:
0: 0
xx
FD
Σ= =
..
165 lb
y
=D
0: 165 lb 693 lb 0
y
FEΣ= − +=
528 lb=E
Joint D:
54
0: 0
13 5
x AD DC
F FF
Σ= + =
(1)
12 3
0: 165 lb 0
13 5
y AD DC
F FFΣ= + + =
(2)
Solving Eqs. (1) and (2) simultaneously,
260 lb
AD
F= −
260 lb
AD
FC=
125 lb
DC
F= +
125.0 lb
DC
FT=
consent of McGraw–Hill Education.
SOLUTION Continued
Joint E:
54
0: 0
13 5
x BE CE
F FFΣ= + =
(3)
12 3
0: 528 lb 0
13 5
y BE CE
F FFΣ= + + =
(4)
Solving Eqs. (3) and (4) simultaneously,
832 lb
BE
F= −
832 lb
BE
FC=
400 lb
CE
F= +
400 lb
CE
FT=
Joint C:
Force polygon is a parallelogram (see Fig. 6.11, p. 290).
400 lb
AC
FT=
125.0 lb
BC
FT=
Joint A:
54
0: (260 lb) (400 lb) 0
13 5
x AB
FFΣ= + + =
420 lb
AB
F= −
420 lb
AB
FC=
12 3
0: (260 lb) (400 lb) 0
13 5 0 0 (Checks)
y
FΣ= − =
=
PROBLEM 6.12
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: (5 ft) (10 kips)(10 ft) (10 kips)(20 ft) 0
Bx
MAΣ= − − =
60.0 kips
x=A
0: 0
xx
F ABΣ= −=
60 kips=B
0: 10 kips 10 kips 0
yy
FAΣ= − − =
20.0 kips
y=A
Joint D:
10 kips
14
17
DC DA
FF
= =
41.2 kips
DA
FT=
40.0 kips
DC
FC=
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
SOLUTION Continued
Joint C:
1
0: 10 kips 0
5
y CA
FFΣ= − =
22.4 kips
CA
F=
22.4 kips
CA
FT=
2
0: (22.4 kips) 40 kips 0
5
x CB
FFΣ= − − =
60.0 kips
CB
F= −
60.0 kips
CB
FC
=
Joint B:
consent of McGraw–Hill Education.
PROBLEM 6.13
Determine the force in each member of the truss shown. State whether
each member is in tension or compression.
SOLUTION
Reactions for Entire truss:
By inspection: 4 kN 4 kN= ↑= ↑AE
Free body: Joint A:
4 kN
sin30
AB
F=
(4 kN)(cot 30 )
AF
F=
8.00 kN
AB
FC=
6.93 kN
AF
FT=
Free body: Joint F:
0:
x FG AF
F FFΣ= =
consent of McGraw–Hill Education.
6.93 kN
FG
FT=
0: 4.00 kN
y FB
FF TΣ= =
4.00 kN
FB
FT=
SOLUTION Continued
Free body: Joint B:
0: 8cos30 cos30 cos30 0 (1)
x BC BG
F FFΣ= + + =
0: 8sin30 4 sin30 sin30 0 (2)
y BC BG
F FFΣ = −+ − =
Solving (1) and (2) simultaneously yields:
4.00 kN
BC
FC=
4.00 kN
BG
FC=
Free body: Joint C:
By symmetry,
4.00 kN
CD CB
FF C= =
0: 4sin30 4sin30 0
y CG
FFΣ= − − =
4.00 kN
CG
FT=
4.00 kN , 6.93 kN , 4.00 kN , 8.00 kN , 6.93 kN
GD GH DH DE EH
F CF TF TF CF T= = = = =
SOLUTION Continued
15
0: ( 34.0) 0
17
x DE
FFΣ= − + =
30.0 kN
DE
F= +
30.0 kN
DE
FT=
Joint E:
0: 8 0
y BE
FFΣ = −=
8.00 kN
BE
F= +
8.00 kN
BE
FT=
Truss and loading symmetrical about
c.
L
consent of McGraw–Hill Education.
PROBLEM 6.9
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:
From the symmetry of the truss and loading, we find
600 lb= =CD
Free body: Joint B:
300 lb
21
5
BC
AB F
F= =
671lb
AB
FT=
600 lb
BC
FC=
Free body: Joint C:
3
0: 600 lb 0
5
y AC
FF
Σ= + =
1000 lb
AC
F= −
1000 lb
AC
FC=
4
0: ( 1000 lb) 600 lb 0
5
x CD
FFΣ= − + + =
200 lb
CD
FT=
From symmetry:
1000 lb , 671lb ,
AD AC AE AB
F F CF F T= = = =
600 lb
DE BC
FF C= =
consent of McGraw–Hill Education.
PROBLEM 6.10
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: 16 kN
Cx
MΣ= =A
0: 9 kN
yy
FΣ= =A
0: 16 kN
x
FΣ= =C
Joint E:
3 kN
54 3
BE DE
FF
= =
5.00 kN
BE
FT
=
4.00 kN
DE
FC
=
Joint B:
4
0: (5 kN) 0
5
x AB
FFΣ= − =
4 kN
AB
F= +
4.00 kN
AB
FT=
3
0: 6 kN (5 kN) 0
5
y BD
FFΣ= − − − =
9 kN
BD
F= −
9.00 kN
BD
FC=
Joint D:
3
0: 9 kN 0
5
y AD
FFΣ= − + =
15 kN
AD
F= +
15.00 kN
AD
FT=
4
0: 4 kN (15 kN) 0
5
x CD
FFΣ= − − − =
16 kN
CD
F= −
16.00 kN
CD
FC=
PROBLEM 6.11
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
5 12 13 ft
12 16 20 ft
AD
BCD
=+=
= +=
Reactions:
0: 0
xx
FD
Σ= =
..
165 lb
y
=D
0: 165 lb 693 lb 0
y
FEΣ= − +=
528 lb=E
Joint D:
54
0: 0
13 5
x AD DC
F FF
Σ= + =
(1)
12 3
0: 165 lb 0
13 5
y AD DC
F FFΣ= + + =
(2)
Solving Eqs. (1) and (2) simultaneously,
260 lb
AD
F= −
260 lb
AD
FC=
125 lb
DC
F= +
125.0 lb
DC
FT=
consent of McGraw–Hill Education.
SOLUTION Continued
Joint E:
54
0: 0
13 5
x BE CE
F FFΣ= + =
(3)
12 3
0: 528 lb 0
13 5
y BE CE
F FFΣ= + + =
(4)
Solving Eqs. (3) and (4) simultaneously,
832 lb
BE
F= −
832 lb
BE
FC=
400 lb
CE
F= +
400 lb
CE
FT=
Joint C:
Force polygon is a parallelogram (see Fig. 6.11, p. 290).
400 lb
AC
FT=
125.0 lb
BC
FT=
Joint A:
54
0: (260 lb) (400 lb) 0
13 5
x AB
FFΣ= + + =
420 lb
AB
F= −
420 lb
AB
FC=
12 3
0: (260 lb) (400 lb) 0
13 5 0 0 (Checks)
y
FΣ= − =
=
PROBLEM 6.12
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: (5 ft) (10 kips)(10 ft) (10 kips)(20 ft) 0
Bx
MAΣ= − − =
60.0 kips
x=A
0: 0
xx
F ABΣ= −=
60 kips=B
0: 10 kips 10 kips 0
yy
FAΣ= − − =
20.0 kips
y=A
Joint D:
10 kips
14
17
DC DA
FF
= =
41.2 kips
DA
FT=
40.0 kips
DC
FC=
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
SOLUTION Continued
Joint C:
1
0: 10 kips 0
5
y CA
FFΣ= − =
22.4 kips
CA
F=
22.4 kips
CA
FT=
2
0: (22.4 kips) 40 kips 0
5
x CB
FFΣ= − − =
60.0 kips
CB
F= −
60.0 kips
CB
FC
=
Joint B:
consent of McGraw–Hill Education.
PROBLEM 6.13
Determine the force in each member of the truss shown. State whether
each member is in tension or compression.
SOLUTION
Reactions for Entire truss:
By inspection: 4 kN 4 kN= ↑= ↑AE
Free body: Joint A:
4 kN
sin30
AB
F=
(4 kN)(cot 30 )
AF
F=
8.00 kN
AB
FC=
6.93 kN
AF
FT=
Free body: Joint F:
0:
x FG AF
F FFΣ= =
consent of McGraw–Hill Education.
6.93 kN
FG
FT=
0: 4.00 kN
y FB
FF TΣ= =
4.00 kN
FB
FT=
SOLUTION Continued
Free body: Joint B:
0: 8cos30 cos30 cos30 0 (1)
x BC BG
F FFΣ= + + =
0: 8sin30 4 sin30 sin30 0 (2)
y BC BG
F FFΣ = −+ − =
Solving (1) and (2) simultaneously yields:
4.00 kN
BC
FC=
4.00 kN
BG
FC=
Free body: Joint C:
By symmetry,
4.00 kN
CD CB
FF C= =
0: 4sin30 4sin30 0
y CG
FFΣ= − − =
4.00 kN
CG
FT=
4.00 kN , 6.93 kN , 4.00 kN , 8.00 kN , 6.93 kN
GD GH DH DE EH
F CF TF TF CF T= = = = =