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 E Take: A= 1.8 m meter B= 2.8 m C= 2503 N B meter |0.5 m 1 m 0.9 m B -0.5 m| 1.5 m Solution: E E Z

International Edition---engineering Mechanics: Statics, 4th Edition
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ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
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Chapter6: Beams And Cables
Section: Chapter Questions
Problem 6.76P: The 50-ft measuring tape weighs 2.4 lb. Compute the span L of the tape to four significant figures.
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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| ||
Transcribed Image Text: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| ||
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