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. Eз4-193 GPa E F G Take: A= 1.5 m meter B= 2.5 m C= 2500 N B meter Fo.5 m- 0.9 m B 1.5 m -0.5 m Solution: 0.5m Im FDE CF 0.5 m 1 m 1.5m 0.5m F, BG AH Page 1 1.5m 0.5 m Internal Forces in the wires : EM,= 0; 2F,G – C(1.5)=0 EF, = 0; F,+F,6-C =0 1875 N F 625 N AH EM, =0; 1.5F, -0.5F, =0 208.33 N CF AH EF = 0; F +Fg-FH = F, N 416.66 DE

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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Displacement :
FOLDE
0.21588
mm
DE
mm
0.108
Ac E
0.108
56
mm
1
1.5
8g = 6g + ốc =
mm
0.18
tan α
0.108000
Ans: a =
1500
mm
0.1943
AH
mm
0.26
mm
1.62
tan 6
1.619
Ans: B =
0.03
2000
Transcribed Image Text:Displacement : FOLDE 0.21588 mm DE mm 0.108 Ac E 0.108 56 mm 1 1.5 8g = 6g + ốc = mm 0.18 tan α 0.108000 Ans: a = 1500 mm 0.1943 AH mm 0.26 mm 1.62 tan 6 1.619 Ans: B = 0.03 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 mm?.
E304=193 GPa
Take:
A=
1.5
m
A meter
B=
2.5
m
C=
2500
N
B meter
1 1o.5 m
1 m
0.9 m
1.5 m
0.5 m
Solution:
0. 5m
Im
F
DE
F
CF
0.5 m
1 m
1.5m
F
0.5m
Page 1
1.5m
0.5 m
Internal Forces in the wires :
EM, = 0; 2Fac – C(1.5)= 0
EF, = 0; F+F -C = 0
F
F
1875
N
BG
N
625
AH
AH
Хм, - 0; 1.5F, -0.5F, -0
EF, = 0; F +F -FH
F
N
=
208.33
AH
= 0
416.66
N
DE
E E Z
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 Take: A= 1.5 m A meter B= 2.5 m C= 2500 N B meter 1 1o.5 m 1 m 0.9 m 1.5 m 0.5 m Solution: 0. 5m Im F DE F CF 0.5 m 1 m 1.5m F 0.5m Page 1 1.5m 0.5 m Internal Forces in the wires : EM, = 0; 2Fac – C(1.5)= 0 EF, = 0; F+F -C = 0 F F 1875 N BG N 625 AH AH Хм, - 0; 1.5F, -0.5F, -0 EF, = 0; F +F -FH F N = 208.33 AH = 0 416.66 N DE E E Z
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