Max for complete details refer authors popular book on

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online