# It opens and closes with constant acceleration and

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Unformatted text preview: (τ), 420 = K × 8 WC πd 2 = 1.184 × 8 × 346.6 × 8 πd 2 = 8360 d2 ...(Assuming τ = 420 MPa or N/mm2) ∴ d 2 = 8360 / 420 = 19.9 or d = 4.46 mm The standard size of the wire is SWG 7 having diameter ( d ) = 4.47 mm. Ans. (See Table 22.2). ∴ Mean diameter of the spring coil, D = C · d = 8 × 4.47 = 35.76 mm Ans. and outer diameter of the spring coil, Do = D + d = 35.76 + 4.47 = 40.23 mm Ans. (b) Number of turns of the coil Let n = Number of active turns of the coil. We know that maximum compression of the spring, δ= 8 W · C3 · n G·d or 8 C3 · n δ = W G ·d Internal Combustion Engine Parts n 1201 Power-brake mechanism of an automobile Since the stiffness of the springs, s = W / δ = 10 N/mm, therefore, δ / W = 1/10. Taking G = 84 × 103 MPa or N/mm2, we have 8 × 83 × n 10.9 n 1 = = 3 10 84 × 10 × 4.47 103 ∴ n = 103 / 10.9 × 10 = 9.17 say 10 For squared and ground ends, the total number of the turns, n′ = n + 2 = 10 + 2 = 12 Ans. (c) Free length of the spring Since the compression produced un...
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## This note was uploaded on 09/30/2012 for the course MECHANICAL 403 taught by Professor A.thirumoorthy during the Spring '11 term at Anna University Chennai - Regional Office, Coimbatore.

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