Can someone slove the fx and fy equations in the below questions solution in steps to how we got f1 and f2 values. Thanks.

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 11.3.27P: A space truss is restrained at joints O, A,B, and C as shown in the figure. Load F is applied at...
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Can someone slove the fx and fy equations in the below questions solution in steps to how we got f1 and f2 values. Thanks.

An assembly consisting of tie rod (1) and pipe strut (2) is used to support an 80 kip load, which is applied to
joint B. Strut (2) is a pin connected structural A-36 alloy steel (E=29,000) pipe with an outside diameter of 8.625
in. and a wall thickness of 0.325 in. For the loading shown in figure, determine the facotr of safety with respect
to buckling for memeber (2)
12 ft
B
30 ft
80 kips
24 ft
Transcribed Image Text:An assembly consisting of tie rod (1) and pipe strut (2) is used to support an 80 kip load, which is applied to joint B. Strut (2) is a pin connected structural A-36 alloy steel (E=29,000) pipe with an outside diameter of 8.625 in. and a wall thickness of 0.325 in. For the loading shown in figure, determine the facotr of safety with respect to buckling for memeber (2) 12 ft B 30 ft 80 kips 24 ft
Step 1
Given
Young's Modulus, E = 29000 Pa
Diameter = 8.625 in
Wall thickness, t= 0.325 in
Find
Factor of safety, FS
Step 2
Solution
Equilibrium of joint B
12 ft
= 0.5 :. 0AB = 26.565°
= 1.25
tan AB
%3D
24 ft
30 ft
24 ft
%3D
tan eBC
. OBC = 51.340°
%3D
EF, = - F, cos (26.565°) - F2 cos (51.340°) = 0
EF, = F, sin (26.565°) – F2 sin (51.340°) – P = 0
%3D
%3D
Note: Tension assumed in each truss member.
Solve these equations simultaneously to obtain:
F = 51.110 kips F2 = -73.179 kips
%3D
Transcribed Image Text:Step 1 Given Young's Modulus, E = 29000 Pa Diameter = 8.625 in Wall thickness, t= 0.325 in Find Factor of safety, FS Step 2 Solution Equilibrium of joint B 12 ft = 0.5 :. 0AB = 26.565° = 1.25 tan AB %3D 24 ft 30 ft 24 ft %3D tan eBC . OBC = 51.340° %3D EF, = - F, cos (26.565°) - F2 cos (51.340°) = 0 EF, = F, sin (26.565°) – F2 sin (51.340°) – P = 0 %3D %3D Note: Tension assumed in each truss member. Solve these equations simultaneously to obtain: F = 51.110 kips F2 = -73.179 kips %3D
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