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a, P 4 c ΣFy = 0; FBC a b - P = 0 FBC = 1.25P 5 Referring to the FBD of the cut segment of member BC Fig. ΣFx = 0; S c ΣFy = 0; 3 Na - a - 1.25P a b = 0 5 4 1.25P a b - Va - a = 0 5 The cross-sectional area of section 1.0417(10 - 3) m2. If the allowable tensile stress for wires AB and AC is sallow = 180 MPa, and wire AB has a diameter of 5 mm and AC has a diameter of 6 mm, determine the greatest force P that can be applied to the chain.

For Normal stress, sallow = Na - a = 0.75P Va - a = P a–a is Aa - a = (0.025)a 0.025 b = 3>5 Na - a 0.75P ; 150(106) = Aa - a 1.0417(10 - 3) P = 208.33(103) N = 208.33 k N For Shear Stress tallow = Va - a P ; 60(106) = Aa - a 1.0417(10 - 3) P = 62.5(103) N = 62.5 k N (Controls! C B 5 45 4 3 A Solution Normal Forces: Analyzing the equilibrium of joint A, Fig.

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That's why most of our educators have achieved an advanced degree in their field.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. We use cookies to offer you a better experience, personalize content, tailor advertising, provide social media features, and better understand the use of our services.To learn more or modify/prevent the use of cookies, see our Cookie Policy and Privacy Policy.Our faculty are passionate about the subjects they teach and bring this enthusiasm into their lessons.We are currently helping students , who need help with subjects like Mathematics , Physics , Statistics , Data Science , Data Analytics , Machine Learning , R Programming , Python , Economics , Operations Research , Mechanical Engineering , Production & Industrial Engineering and many more. The 2-Mg concrete pipe has a center of mass at point G.If it is suspended from cables AB and AC, determine the diameter of cable AB so that the average normal stress in this cable is the same as in the 10-mm-diameter cable AC.

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