Solution for the forces FCF, FBE and FAD. Given the value just show how it get based from the given equation Equation of Equilibrium. Referring to the free-body diagram of the beam shown inFig. a,+1 ΣF, - 0;FAD + FBE + FcF - 230(10³) = 0(1)6+EMA = 0;FBE(400) + FcF(1200) – 230(10³)(800) = 0(2)FBE + 3FCF = 460(10³)Elastic Analysis. Referring to the deflection diagram of the beam shown in Fig. b,the compatibility equation can be written as( 8CF¯ d AD \(400)SBE = 8 AD +1200dBE18AD1( FCF L`FBEL 2( FADL3 AEAE3AEF BEFADFCF(3)Solving Eqs. (1), (2), and (3)FCF131 428.57 NFBE65 714.29 N F AD32 857.14 N

Answered: Image /qna-images/answer/4b71de09-da79-44ba-99b3-73bca59ed455.jpg

Equation of Equilibrium. Referring to the free-body diagram of the beam shown in Fig. a, +1£F, = 0; FAD + FBE + FcF - 230(10³) = 0 (1) 6+EMA- 0; FBE(400) + FcF(1200) – 230(10®)(800) = 0 FBE + 3FCF = 460(10³) (2) Elastic Analysis. Referring to the deflection diagram of the beam shown in Fig. b, the compatibility equation can be written as ( ScF - 8AD SBE = SAD + |(400) 1200 SCF FBEL 2(FADL) 3 AE (FCFL' 3 AE %3D AE FAD + Fer FBE FCF (3) Solving Eqs. (1), (2), and (3) FCF 131 428.57 N FBE 65 714.29 N F AD = 32 857.14 N %3D %3D %3D

Equation of Equilibrium. Referring to the free-body diagram of the beam shown in Fig. a, +1£F, = 0; FAD + FBE + FcF - 230(10³) = 0 (1) 6+EMA- 0; FBE(400) + FcF(1200) – 230(10®)(800) = 0 FBE + 3FCF = 460(10³) (2) Elastic Analysis. Referring to the deflection diagram of the beam shown in Fig. b, the compatibility equation can be written as ( ScF - 8AD SBE = SAD + |(400) 1200 SCF FBEL 2(FADL) 3 AE (FCFL' 3 AE %3D AE FAD + Fer FBE FCF (3) Solving Eqs. (1), (2), and (3) FCF 131 428.57 N FBE 65 714.29 N F AD = 32 857.14 N %3D %3D %3D

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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