q1 Question 1:The load is supported by the four 304 stainless stell wires that are connected to the rigid membersAB 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 GPaEFTake:A=1.2mA meterB=2.2mB meterC=2497N10.5 m1 m0.9 m0.5 mt1.5 mSolution:1mFDE0.5m0.5 m1mFBGFAH1.5m0.5m1.5m0.5 mC Internal Forces in the wires:EM, = 0; 2F3G -C(1.5)= 0EF, = 0; FAH +F30 -C = 0F301877.25NFNAHEM, = 0; 1.5FCF -0.5F = 0EF, = 0; FDE +Fg - FAH = 0F =AHFDENDisplacement :FOELDE-mmApgEFeg LeFAcgE%3Dmmmm%3D1.58g = 8g. + dc =%3Dmmtan a =Ans: a =1500FLAE =mmAEHF.8, = 84 +84H =mmBGmmBGtan B =Ans: B=2000I|||

Answered: Image /qna-images/answer/beba08dd-52de-4031-8125-3af5282304d8.jpg

Juestion 1: he 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. he members were originally horizontal, and each wire has a cross-sectional area of 15 mm. E304=193 GPa E F Cake: A= 1.2 m meter B= 2.2 m C= 2497 N B meter 1 Fo.5 m 1 m 0.9 m B 1.5 m 0.5 m

Juestion 1: he 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. he members were originally horizontal, and each wire has a cross-sectional area of 15 mm. E304=193 GPa E F Cake: A= 1.2 m meter B= 2.2 m C= 2497 N B meter 1 Fo.5 m 1 m 0.9 m B 1.5 m 0.5 m

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Strain energy

The term strain indicates the variation in an object's length, area, and volume when the object is subjected to an external force. The external force may be of normal, shear, moment, torque, etc. The strain in an object depends on the types of external force and cross-section of the object. Mathematically, the evaluation of strain can be obtained with the help of elastic constants.

Question

Internal Forces in the wires:
EM, = 0; 2F3G -C(1.5)= 0
EF, = 0; FAH +F30 -C = 0
F30
1877.25
N
F
N
AH
EM, = 0; 1.5FCF -0.5F = 0
EF, = 0; FDE +Fg - FAH = 0
F =
AH
FDE
N
Displacement :
FOELDE-
mm
ApgE
Feg LeF
AcgE
%3D
mm
mm
%3D
1.5
8g = 8g. + dc =
%3D
mm
tan a =
Ans: a =
1500
FLAE =
mm
AE
HF.
8, = 84 +84H =
mm
BG
mm
BG
tan B =
Ans: B=
2000
I|||

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