Determine the force in each member of the truss shown in Fig. 6-8a and indicate whether the members are in tension or compression. B 500 N 2 m SOLUTION 45 Since we should have no more than two unknown forces at the joint and at least one known force acting there, we will begin our analysis at joint B. 2 m (a) Joint B. The free-body diagram of the joint at B is shown in Fig. 6-8b. Applying the equations of equilibrium, we have 500 N FBC ΣF, - 0 500 N FBC Sin 45° = 0 FBC = 707.1 N (C) Ans. (b) +1 F, = 0; FBC cos 45° - FBA 0 FBA = 500 N (T) Ans. %3D 707.1 N
Determine the force in each member of the truss shown in Fig. 6-8a and indicate whether the members are in tension or compression. B 500 N 2 m SOLUTION 45 Since we should have no more than two unknown forces at the joint and at least one known force acting there, we will begin our analysis at joint B. 2 m (a) Joint B. The free-body diagram of the joint at B is shown in Fig. 6-8b. Applying the equations of equilibrium, we have 500 N FBC ΣF, - 0 500 N FBC Sin 45° = 0 FBC = 707.1 N (C) Ans. (b) +1 F, = 0; FBC cos 45° - FBA 0 FBA = 500 N (T) Ans. %3D 707.1 N
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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Transcribed Image Text:۱:V م الاثنين ۱4 شباط
trusses part1.pdf
their correct sense.
• Using the calculated results, continue to analyze each of the other
joints. Remember that a member in compression"pushes" on the
joint and a member in tension "pulls" on the joint. Also, be sure to
choose a joint having at most two unknowns and at least one
known force.
EXAMPLE 6.1
Determine the force in each member of the truss shown in Fig. 6-8a
and indicate whether the members are in tension or compression.
500 N
2 m
SOLUTION
45%
C
Since we should have no more than two unknown forces at the joint
and at least one known force acting there, we will begin our analysis at
joint B.
2 m
(a)
B
Joint B.
The free-body diagram of the joint at B is shown in Fig. 6-8b.
500 N
Applying the equations of equilibrium, we have
'FBC
EF, = 0;
500 N FBC sin 45° 0
FBC = 707.1 N (C) Ans.
(b)
+1EF, = 0;
FBC Cos 45° -
FBA = 0
FBA
= 500 N (T) Ans.
%3D
707.1 N
Since the force in member BC has been calculated, we can proceed to
anayzeoumEmoTrHNmmranrareran member CA and the sunnort
***
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