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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. Gallery Items tagged Homework Assignment Show all Gallery Items Here we provide a selection of homework assignments templates and examples for school, college and university use.Students as well as working professionals can contact us for assignment help and other related services.

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A 30 45 C Solution B G Internal Loadings: The normal force in cables AB and AC can be determined by considering the equilibrium of the hook for which the free-body diagram is shown in Fig. ΣFx′ = 0; 2000(9.81) cos 45° - FAB cos 15° = 0 FAB = 14 362.83 N (T) ΣFy′ = 0; 2000(9.81) sin 45° - 14 362.83 sin 15° - FAC = 0 FAC = 10 156.06 N (T) Average Normal Stress: The cross-sectional areas of cables AB and AC are p p AAB = d AB2 and AAC = (0.012) = 78.540(10 - 6) m2.

4 4 Here, we require s AB = s AC FAC FAB = AAB AAC 14 362.83 10 156.06 = p 2 -6 d 78.540(10 ) 4 AB Ans.

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Assignment Solutions Middlebury Community Essay

To learn more or modify/prevent the use of cookies, see our Cookie Policy and Privacy Policy.You can contact Demenntor to find a solution to your research paper writing issues.As a highly trusted research paper writing service, we offer the best solutions that surpass your expectations. 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.You can contact us any time without worrying about any time constraints.See More Are you searching for a service that offers the best homework writing help online? As a reputed and reliable service, we offer the best homework solutions online with clear focus on timeliness and accuracy. Being a responsible assignment help company, we focus on delivering quality solutions at most competitive prices.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. d AB = 0.01189 m = 11.9 mm Ans: d AB = 11.9 mm 1 1–66.Determine the largest load P that can be applied to the frame without causing either the average normal stress or the average shear stress at section a–a to exceed s = 150 MPa and t = 60 MPa, respectively.

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