Use the method of sections to determine the forces in members BD, CD, and CE.Note: The support reactions given below were determined in lecture last week (“Example:Method of Joints").[Ans.: Fsp = -4 kN (comp), FCD = 2.83 kN (tension), FCE = 1 kN (tension)]Cut bSome potential(3 unk.)ysection lines1 kNCut c(cuts a, b̟ and c):(3 unk.)Cut a(4 unk.),2 m2/n2 m2 th2 m→ F=-1 kNF = 2 kN5 kNA, = 3 kN%3D%3DNote: The length of each member is 2 m and hence 0 = tan-1(2 m/2 m) = 45°E.

Use the method of sections to determine the forces in members BD, CD, and CE. Note: The support reactions given below were determined in lecture last week ("Example: Method of Joints"). (Ans E. = -4 kN (comp), Fcp = 2.83 kN (tension), FcE = 1 kN (tension)]

Use the method of sections to determine the forces in members BD, CD, and CE. Note: The support reactions given below were determined in lecture last week ("Example: Method of Joints"). (Ans E. = -4 kN (comp), Fcp = 2.83 kN (tension), FcE = 1 kN (tension)]

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

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Problem 1.1MA

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Use the method of sections to determine the forces in members BD, CD, and CE.
Note: The support reactions given below were determined in lecture last week (“Example:
Method of Joints").
[Ans.: Fsp = -4 kN (comp), FCD = 2.83 kN (tension), FCE = 1 kN (tension)]
Cut b
Some potential
(3 unk.)
y
section lines
1 kN
Cut c
(cuts a, b̟ and c):
(3 unk.)
Cut a
(4 unk.),
2 m
2/n
2 m
2 th
2 m
→ F=-1 kN
F = 2 kN
5 kN
A, = 3 kN
%3D
%3D
Note: The length of each member is 2 m and hence 0 = tan-1(2 m/2 m) = 45°
E.

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