SOLUTION
6–1.
Determine the force in each member of the truss and state
if the members are in tension or compression. Set
P1 = 20 kN, P2 = 10 kN.
CB
A
D
1.5 m
2 m
P1
P2
SOLUTION
5b
6–2.
Determine the force in each member of the truss and state
if the members are in tension or compression. Set
P1 = 45 kN, P2 = 30 kN.
CB
A
D
1.5 m
2 m
P1
P2
6–3.
SOLUTION
Ans.
Ans.
Joint B:
Ans.
Ans.
Joint C:
Ans.
Ans.
Joint D:
Ans.
:
+©Fx=0; FDF –140 –3
5(150) =0
FDE =120 lb (C)
+c©Fy=0; FDE –4
5(150) =0
FCE =180 lb (C)
:
+©Fx=0; –FCE +3
5(150) +3
5(150) =0
FCD =150 lb (T)
+c©Fy=0; a4
5bFCD –a4
5b150 =0
+c©Fy=0; FBC =0
FBD =140 lb (T)
:
+©Fx=0; FBD –140 =0
FAB =140 lb (T)
:
+©Fx=0; FAB –3
5(150) –5
13 (130) =0
FAC =150 lb (C)
Determine the force in each member of the truss. State if
the members are in tension or compression.
3ft3ft3ft
12
5
13
130 lb
AB
C
E
D
F
4 ft 4 ft
484
*6–4.
SOLUTION
Joint A:
Ans.
Ans.
Joint E:
Ans.
Ans.
Joint B:
Ans.
Ans.
Joint D:
Ans.
Ans.
Joint F:
Ans.
Ans.
Joint H:
Ans.
Ans.+c©Fy=0; 2.154 sin 21.80° –FHI =0FHI =0.8 kip (T)
FHG =2.154 kip =2.15 kip (C)
:
+©Fx=0; 2 –FHG cos 21.80° =0
FFG =8.078 kip =8.08 kip (C)
+R©Fx¿=0; FFG +3 sin 21.80° +4.039 sin 46.40° –12.12 =0
FFC =4.039 kip =4.04 kip (C)
+Q©Fy¿=0; FFC cos 46.40° –3 cos 21.80° =0
:
+©Fx=0; –FDC +7.75 =0FDC =7.75 kip (T)
+c©Fy=0; FDF =0
:
+©Fx=0; FBC –3.75 =0FBC =3.75 kip (T)
+c©Fy=0; FBI =0
FED =7.75 kip (T)
:
+©Fx=0; –FED –3.5 +12.12 cos 21.80° =0
FEF =12.12 kip =12.1 kip (C)
+c©Fy=0; 4.5 –FEF sin 21.80° =0
:
+©Fx=0; FAB –4.039 cos 21.80° =0FAB =3.75 kip (T)
FAl =4.039 kip =4.04 kip (C)
+c©Fy=0; 1.5 –FAl sin 21.80° =0
+c©Fy=0; Ey+1.5 –3–3=0Ey=4.5 kip
:
+©Fx=0; 1.5 +2–Ex=0Ex=3.5 kip
Determ
i
ne t
h
e force
i
n eac
h
mem
b
er of t
h
e truss an
d
state
if the members are in tension or compression. 2kip
1.5 kip
4ft
10 ft 10 ft 10 ft
3kip
3kip
10 ft
AB
I
H
G
F
CD
E
8ft
*6–4. Continued
Joint C:
Ans.
Ans.
J
oint G:
Ans.
+1.40 sin 21.80° –8.081 =0 (Check)
+R©Fx¿=0; 2.154 +3 sin 21.80° +5.924 sin 46.40°
FGI =5.924 kip =5.92 kip (C)
+Q©Fy¿=0; FGI cos 46.40° –3 cos 21.80° –1.40 cos 21.80° =0
FCG =1.40 kip (T)
+c©Fy=0; FCG +0.2692 sin 21.80° –4.039 sin 21.80° =0
FCI =0.2692 kip =0.269 kip (T)
:
+©Fx=0; –FCI cos 21.80° –4.039 cos 21.80° –3.75 +7.75 =0
6–5.
Determine the force in each member of the truss, and state
if the members are in tension or compression. Set .
SOLUTION
Method of Joints: We will use the above result to analyze the equilibrium of
joints Cand A, and then proceed to analyze of joint B.
Joint C:From the free-body diagram in Fig. b, we can write
Ans.
Ans.
Joint A:From the free-body diagram in Fig. c, we can write
Ans.
Ans.
Joint B:From the free-body diagram in Fig. d, we can write
Ans.
Note: The equilibrium analysis of joint Dcan be used to check the accuracy of the
solution obtained above.
4.167 –4.167 =0
(check!)©F
x=0;
:
+
F
BD =4 kN (T)
F
BD –4=0+c©F
y=0;
F
AB =4.167 kN =4.17 kN (T)
F
AB –3–1.458a4
5b=0©F
x=0;
:
+
F
AD =1.458 kN =1.46 kN (C)
0.875 –F
AD a3
5b=0+c©F
y=0;
F
CB =4.167 kN =4.17 kN (T)
5.208 a4
5b–F
CB =0©F
x=0;
:
+
F
CD =5.208 kN =5.21 kN (C)
3.125 –F
CD a3
5b=0+c©F
y=0;
Ay=0.875 kN
Ay+3.125 –4=0+c©F
y=0;
Ax=3 kN
3–Ax=0©F
x=0;
:
+
N
C =3.125 kN
u=0°
AC
D
3 kN
1.5 m
u
6–6.
Determine the force in each member of the truss, and state
if the members are in tension or compression. Set .
SOLUTION
Method of Joints: We will use the above result to analyze the equilibrium of
joints Cand A, and then proceed to analyze of joint B.
Joint C:From the free-body diagram in Fig. b, we can write
Ans.
Ans.
Joint A:From the free-body diagram in Fig. c, we can write
Ans.
Ans.
Joint B:From the free-body diagram in Fig. d, we can write
Ans.
Note: The equilibrium analysis of joint Dcan be used to check the accuracy of the
solution obtained above.
2.362 –2.362 =0
(check!)©F
x=0;
:
+
F
BD =4 kN (T)
F
BD –4=0+c©F
y=0;
F
AB =2.362 kN =2.36 kN (T)
F
AB –1.458a4
5b–1.196 =0©F
x=0;
:
+
F
AD =1.458 kN =1.46 kN (C)
0.875 –F
AD a3
5b=0+c©F
y=0;
F
CB =2.362 kN =2.36 kN (T)
5.208a4
5b–3.608 sin 30° –F
CB =0©F
x=0;
:
+
F
CD =5.208 kN =5.21 kN (C)
3.608 cos 30° –F
CD a3
5b=0+c©F
y=0;
Ay=0.875 kN
Ay+3.608 cos 30° –4=0+c©F
y=0;
Ax=1.196 kN
3–3.608 sin 30° –Ax=0©F
x=0;
:
+
N
C=3.608 kN
u=30°
AC
D
3 kN
1.5 m
u
488
6–7.
SOLUTION
Support Reactions:
a
Method of Joints:
Joint D:
Ans.
Ans.
Joint E:
Ans.
Ans.
Joint C:
Ans.
Ans.F
CB
=2.20 kN T
8.40 –8.768 cos 45° –F
CB
=0:
+©F
x
=0;
F
CF
=8.768 kN 1T2=8.77 kN 1T2
6.20 –F
CF
sin 45° =0+c©F
y
=0;
F
EC
=6.20 kN 1C2
23.0 –16.33
¢
5
234
≤
–8.854
¢
1
210
≤
–F
EC
=0+c©F
y
=0;
F
EA
=8.854 kN 1C2=8.85 kN 1C2
F
EA
¢
3
210
≤
–16.33
¢
3
234
≤
=0:
+©F
x
=0;
F
DC
=8.40 kN 1T2
16.33
¢
3
234
≤
–F
DC
=0:
+©F
x
=0;
F
DE
=16.33 kN 1C2=16.3 kN 1C2
F
DE
¢
5
234
≤
–14.0 =0+c©F
y
=0;
23.0 –4–5–D
y
=0D
y
=14.0 kN+c©F
y
=0;
4162+5192–E
y
132=0E
y
=23.0 kN+©M
D
=0;
Determine the force in each member of the truss and state
if the members are in tension or compression.
E
D
C
B
F
A5m
3m
5kN
4kN
3m 3m 3m
6–7. Continued
Joint B:
Ans.
Joint F:
(Check!)
Ans.F
FA
=6.20 kN 1T2
8.768 cos 45° –F
FA
=0:
+©F
x
=0;
8.768 sin 45° –6.20 =0+c©F
y
=0;
F
BF
=6.20 kN 1C2
F
BF
–4–3.111 sin 45° =0+c©F
y
=0;
F
BA
=3.111 kN 1T2=3.11 kN 1T2
2.20 –F
BA
cos 45° =0:
+©F
x
=0;
Ans.
*6–8.
Determine the force in each member of the truss, and state
if the members are in tension or compression.
SOLUTION
E
D
C
600 N
900 N
4 m
491
SOLUTION
22b
6–9.
Determine the force in each member of the truss and state
if the members are in tension or compression. Set
P1 = 3 kN, P2 = 6 kN.
AD
E
B C
P1P2
4 m4 m4 m
6 m
6–9. Continued
493
SOLUTION
22b
6–10.
Determine the force in each member of the truss and state
if the members are in tension or compression. Set P1 = 6 kN,
P2 = 9 kN.
AD
E
B C
P1P2
4 m4 m4 m
6 m
210 b
6–11.
Determine the force in each member of the Pratt truss, and
state if the members are in tension or compression.
SOLUTION
Joint A:
Joint B:
Joint L:
Joint C:
Joint K:
Joint J:
Ans.
Due to Symmetry
Ans.
F
AL =F
GH =F
LK =F
HI =28.3 kN (C)
F
JD =33.3 kN (T)
2 (23.57 cos 45°) –F
JD =0+c©F
y=0;
F
JI =23.57 kN (L)
23.57 sin 45° –F
JI sin 45° =0©F
x=0;
:
+
F
KJ =23.57 kN (C)
28.28 –10 cos 45° +7.454 sin (45° –26.57°) –F
KJ =0+Q©F
y=0;
F
KD =7.454 kN (L)
10 sin 45° –F
KD cos (45° –26.57°) =0R+©F
x–0;
F
CK =10 kN (T)
F
CK –10 =0+c©F
y=0;
F
CD =20 kN (T)
F
CD –20 =0©F
x=0;
:
+
F
LK =28.28 kN (C)
28.28 –F
LK =0+Q©F
y=0;
F
LC =0R+©F
x=0;
F
BL =0+c©F
y=0;
F
BC =20 kN (T)
F
BC –20 =0©F
x=0;
:
+
F
AB =20 kN (T)
F
AB –28.28 cos 45° =0©F
x=0;
:
+
20 –F
AL sin 45° =0+c©F
y=0;
A
BCDEF
G
H
I
J
K
L
2 m
10 kN 10 kN
20 kN
2 m
2 m
*6–12.
SOLUTION
Ans.
Joint C:
Ans.
Ans.
Joint E:
Ans.
Ans.
Joint B:
Ans.
Ans.
Joint B:
Ans.
Joint D:
FBF =155 lb (C)
+a©Fy=0; FBF sin 75° –150 =0
+Q©Fx¿=0; FBA –515.39 =0FBA =515 lb (C)
+a©Fy¿=0; FBF cos u=0FBF =0
+R©Fx¿=0; 515.39 –FEF =0FEF =515 lb (C)
+Q©Fy¿=0; FEB cos u=0FEB =0
FCB =515.39 lb =515 lb (C)
+Q©Fx¿=0; FCB –500 sin 75.96° –125 sin 14.04° =0
FCE =0
+a©Fy¿=0; FCE cos 39.09° +125 cos 14.04° –500 cos 75.96° =0
:
+©Fx=0; FCD –515.39 cos 75.96° =0FCD =125 lb (C)
Determine the force in each member of the truss and state
if the members are in tension or compression.
500 lb
3 ft
500 lb
C
B
E
D
9 ft
6 ft
3 ft 3 ft
6–13.
SOLUTION
Ans.
Ans.
Joint D:
Ans.FDB =1.33 P(T)
+c©Fy=0; FDB –0.687 P
¢
1
217
≤
–1
217 (0.687 P)–P=0
FCB =FAB =0.943 P(C)
FCD =FAD =0.687 P(T)
+c©Fy=0; P
2–1
22(FAB)+1
217 (FAD)=0
Determine the force in each member of the truss in terms of
the load Pand state if the members are in tension or
compression.
B
D
A
C
a a
a
a
3
—
4
1
—
4
498
6–14.
Members AB and BC can each support a maximum
compressive force of 800 lb,and members AD,DC,and BD
can support a maximum tensile force of 1500 lb.If ,
determine the greatest load Pthe truss can support.
a=10 ft
SOLUTION
Joint A:
OK
Joint D:
OK
Thus,
Ans.Pmax =849 lb
FBD =1131.3724 lb 61500 lb
+c©Fy=0; –848.5297 –583.0952(2)
¢
1
217
≤
+FDB =0
P=848.5297 lb
+c©Fy=0; P
2–1
22(800) +1
217 (583.0952) =0
FAD =583.0952 lb 61500 lb
:
+©Fx=0; –800
¢
1
22
≤
+FAD
¢
4
217
≤
=0
B
D
A
C
a
a
3
—
4
1
—
4
499
6–15.
Members AB and BC can each support a maximum
compressive force of 800 lb, and members AD, DC, and BD
can support a maximum tensile force of 2000 lb. If a = 6 ft,
determine the greatest load P the truss can support.
B
D
A
C
a a
a
a
3
—
4
1
—
4
SOLUTION
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*6–16.
SOLUTION
Determine the force in each member of the truss. State
whether the members are in tension or compression. Set
P=8 kN.
60••
60••
B
ED
C
A
4m