Referring to the FBD of the rigid beam, Fig a, Referring to the geometry shown in Fig b, F AD F CF 2 F BE Solving Eqs (1), (2), and (3) yields F BE 43 33(10 3 ) N F AD 35 83(10 3 ) N F CF 50 83(10 3 ) N Thus, CF 113 MPa 12F, 0 FAD FBE FCF 50(10) 80(10) 0 3D ( YMp 0 Fge(2) FCr(4) 50(10)(1) 80(10)(3)...

Referring to the FBD of the rigid beam, Fig. a, Referring to the geometry shown in Fig.

Question:

Referring to the FBD of the rigid beam, Fig. a,

+12F, = 0; FAD + FBE + FCF - 50(10) – 80(10) = 0 %3D (+YMp = 0; Fge(2) + FCr(4) – 50(10)(1) – 80(10)(3) = 0

Referring to the geometry shown in Fig. b,

F_AD + F_CF = 2 F_BE

Solving Eqs. (1), (2), and (3) yields

F_BE = 43.33(10³) N F_AD = 35.83(10³) N F_CF = 50.83(10³) N

Thus,

Ïƒ_CF = 113 MPa

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Mechanics of Materials

ISBN: 978-0134319650

10th edition

Authors: Russell C. Hibbeler

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Question Posted: May 26, 2018 04:05 AM

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Referring to the FBD of the rigid beam, Fig. a, Referring to the geometry shown in Fig.

Question:

+12F, = 0; FAD + FBE + FCF - 50(10) – 80(10) = 0 %3D (+YMp = 0; Fge(2) + FCr(4) – 50(10)(1) – 80(10)(3) = 0

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Referring to the FBD of the rigid beam Fig a Referring ...View the full answer

Mechanics of Materials

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