Question 1: The load is supported by the four 304 stainless stell wires that are connected to the rigid members AB and DC. Detemine 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?. Eso4=193 Gpa E F Take: A= 1.5 m A meter B= 2.5 m C= 2500 N B meter 1 F0.5 m- 0.9 m o.5 m 1.5 m- Solution: 0.5m FDE Im Fa 0.5 m 1m F F, Fage 1.5m 0,5m AH 1.5m 0.5 m Intemal Forces in the wires : EM, =0: 2F30-C(1.5)=0 EF, =0; FAH+F30-C=0 FBG N BG F. %3D BG AH Ум, - 0: 1.5Fg -0.5F, EF, =0: FDE +Fcg -F4H =0 FCF FDE = 0 N AH N %3D

Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter11: Columns
Section: Chapter Questions
Problem 11.5.7P: Solve the preceding problem for a column with e = 0.20 in,, L = 12 ft, I = 2L7in4, and E = 30 ×...
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Displacement
Displacement :
FDELDE
ApgE
%3|
mm
Sc =
CF
mm
ACFE
mm
1
1.5
8 = 8p + 8c
mm
tan a =
α-
Ans
1500
AH
A/H
mm
8 = S +84H
mm
BG
BG
mm
tan B =
B =
Ans
2000
Transcribed Image Text:Displacement Displacement : FDELDE ApgE %3| mm Sc = CF mm ACFE mm 1 1.5 8 = 8p + 8c mm tan a = α- Ans 1500 AH A/H mm 8 = S +84H mm BG BG mm tan B = B = Ans 2000
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 mm².
E304=193 Gpa
E
Take:
A=
1.5
m
A meter
B=
2.5
m
B meter
C=
2500
N
1o.5 m
1 m
0.9 m
B
-0.5 m-
1.5 m
Solution:
0.5m
Im
FDE
CF
0.5 m
1 m
Fa
FAH
Fage
1.5m
0,5m
1.5m
0.5 m
C
Internal Forces in the wires :
N
EM, =0; 2FBG -C(1.5)=0
EF, = 0; FAH+F3G -C =0
FBG
F
AH
FCF
N
EM, = 0; 1.5FCF -0.5FH = 0
EF, = 0: FDE +FcF -FAH =0
FDE
N
%3D
||||
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 mm². E304=193 Gpa E Take: A= 1.5 m A meter B= 2.5 m B meter C= 2500 N 1o.5 m 1 m 0.9 m B -0.5 m- 1.5 m Solution: 0.5m Im FDE CF 0.5 m 1 m Fa FAH Fage 1.5m 0,5m 1.5m 0.5 m C Internal Forces in the wires : N EM, =0; 2FBG -C(1.5)=0 EF, = 0; FAH+F3G -C =0 FBG F AH FCF N EM, = 0; 1.5FCF -0.5FH = 0 EF, = 0: FDE +FcF -FAH =0 FDE N %3D ||||
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ISBN:
9781337093347
Author:
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Publisher:
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