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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Concept explainers

Work and Heat Transfer

It is generally related to thermal engineering which tells about the use, creation, conversion and transfer of heat energy between the systems. It is also considered that to transfer heat from one source to another, transfer of masses is also required. There are various mechanisms by which, transfer of heat takes place i.e. thermal conduction, convection, radiation or by change of phase. All these mechanisms have their own specific features which sometimes occur simultaneously within the same system.

Question

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 in
Fig. 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 +
1200
dBE
1
8AD
1( FCF L`
FBEL 2( FADL
3 AE
AE
3
AE
F BE
FAD
FCF
(3)
Solving Eqs. (1), (2), and (3)
FCF
131 428.57 N
FBE
65 714.29 N F AD
32 857.14 N

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