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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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