978-0133915426 Chapter 6 Part 1

subject Type Homework Help
subject Pages 14
subject Words 3529
subject Authors Russell C. Hibbeler

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page-pf1
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
page-pf2
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
page-pf3
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
page-pf4
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
page-pf5
*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
page-pf6
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=
AC
D
3 kN
1.5 m
u
page-pf7
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
page-pf8
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 kNM
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
page-pf9
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.
page-pfa
*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
page-pfb
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
page-pfc
6–9. Continued
page-pfd
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
page-pfe
210 b
page-pff
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
page-pf10
*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
page-pf11
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
page-pf12
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
page-pf13
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
page-pf14
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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

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