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Unformatted text preview: mm, therefore δ / W = 1 / 8. Taking G = 84 × 103 MPa or N/mm2, we have 8 × 63 × n 6.98 n 1 = = 3 8 84 × 10 × 2.946 103 ∴ n = 103 / 8 × 6.98 = 17.9 say 18 For squared and ground ends, the total number of turns, n′ = n + 2 = 18 + 2 = 20 Ans. (c) Free length of the spring Since the compression produced under W2 = 128 N is 16 mm, therefore, maximum compression produced under the maximum load of W = 176 N is 16 × 176 = 22 mm 128 We know that free length of the spring, LF = n’. d + δmax + 0.15 δmax = 20 × 2.946 + 22 + 0.15 × 22 = 84.22 say 85 mm Ans. (d) Pitch of the coil We know that pitch of the coil Free length 85 = = = 4.47 mm Ans. 20 − 1 n′ − 1 Design of cam The cam is forged as one piece with the camshaft. It is designed as discussed below : The diameter of camshaft (D′) is taken empirically as D′ = 0.16 × Cylinder bore + 12.7 mm = 0.16 × 140 + 12.7 = 35.1 say 36 mm The base circle diameter is about 3 mm greater than the camshaft diameter. ∴ Base circle diameter = 36 + 3 = 3...
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