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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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