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Member CB has a square cross section of 25 mm on each side.

B 2m a a A Solution C 1.5 m Analyze the equilibrium of joint C using the FBD shown in Fig.

a, ΣFx = 0; S c ΣFy = 0; 3 FAC a b - FAB sin 45° = 0 5 (1) 4 FAC a b FAB cos 45° - P = 0 5 P (2) Solving Eqs. 2 Using these results, tallow = Va - a ; Aa - a FAB ; AAB 3(103) = 9(103) a th ei r sallow = 3t Ans.

(1) and (2) FAC = 0.7143P FAB = 0.6061P Average Normal Stress: Assuming failure of wire AB, sallow = FAB ; AAB 180 ( 106 ) = 0.6061P p ( 0.0052 ) 4 P = 5.831 ( 103 ) N = 5.83 k N Assume the failure of wire AC, sallow = FAC ; AAC 180 ( 106 ) = 0.7143P p ( 0.0062 ) 4 a th ei r P = 7.125 ( 103 ) N = 7.13 k N Choose the smaller of the two values of P, Ans. Determine the required minimum thickness t of member AB and edge distance b of the frame if P = 9 kip and the factor of safety against failure is 2. t A 30 b 30 C Solution Internal Loadings: The normal force developed in member AB can be determined by considering the equilibrium of joint A, Fig. ΣFx = 0; FAB cos 30° - FAC cos 30° = 0 FAC = FAB S c ΣFy = 0; 2FAB sin 30° - 9 = 0 FAB = 9 kip Subsequently, the horizontal component of the force acting on joint B can be determined by analyzing the equilibrium of member AB, Fig. ΣFx = 0; S (FB)x - 9 cos 30° = 0 (FB)x = 7.794 kip Referring to the free-body diagram shown in Fig.

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The wood has a normal failure stress of sfail = 6 ksi, and a shear failure stress of tfail = 1.5 ksi. c, the shear force developed on the shear plane a–a is ΣFx = 0; S Va - a - 7.794 = 0 Va - a = 7.794 kip Allowable Normal Stress: sallow = sfail 6 = = 3 ksi F.

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