Statics and Mechanics of Materials, Student Value Edition (5th Edition)
5th Edition
ISBN: 9780134382890
Author: Russell C. Hibbeler
Publisher: PEARSON
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Textbook Question
Chapter 9.5, Problem 39P
The load of 2000 lb is to be supported by the two vertical steel wires for which σY = 70 ksi. Originally wire AB is 60 in. long and wire AC is 60.04 in. long. Determine the cross-sectional area of AB if the load is to be shared equally between both wires. Wire AC has a cross-sectional area of 0.02 in2. Est = 29.0(103) ksi.
Prob. 9-38/39
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A steel column is of length 8m and diameter 600 mm with both ends hinged.Determine the crippling load by Euler’s formula. Take 5 E = 2.1x10 N/mm2
The load is supported by the four 304 stainless steel wires that are connected to
the rigid members AB and DC. Determine the angle of tilt of each member after the 500-lb
load is applied. The members were originally horizontal, and each wire has a cross-
sectional area of 0.025 in?.
E
F
G
3 ft
5 ft
Н
D
-1 ft- 2 ft
1.8 ft
I
A
В
-3 ft-
-1 ft-
500 lb
Question 1:
The load is supported by the four 304 stainless stell wires that are connected to the rigid members
AB and DC. Determine the angle of tilt of each member after the C (Newton) load is applied.
The members were originally horizontal, and each wire has a cross-sectional area of 15 mm?.
E304=193 Gpa
G
Take:
A=
1.8
m
A
meter
B=
2.8
m
C=
2503
N
B meter
H
D
|0.5 m
1 m
0.9 m
- 0.5 m|
1.5 m
C N
Solution:
0.5m
lm
FDE
FCF
0.5 m
1 m
Päge-1
1.5m
FAH
0.5m
to
0.5 m
Internal Forces in the wires :
EM, = 0; 2FRG -C(1.5)=0
УF, 3D 0; Fан + Fва — С%3D0
F,
FBG
N
A
FAH =
N
FCF
EM, = 0; 1.5Fcr -0.5FH = 0
EF, = 0; FDE + FeF – FAH = 0
FDE
N
Displacement :
8,
тт
FeLCF
8c =
тт
AcE
тт
1
1.5
8, = 8 +8c =
тm
tan a =
a
• Ans
1500
FAHLAH
A AE
O ALH =
тт
8 = 8µ+84H =
тm
FBGLBG
8, =
тт
ABGE
tan B =
B =
Ans
2000
I| ||
I| ||
Chapter 9 Solutions
Statics and Mechanics of Materials, Student Value Edition (5th Edition)
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