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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.

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

Elements Of Electromagnetics
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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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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