# 84 d t tensile stress for the material of the stud

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: r of the stud in mm. It is usually taken as 0.84 d. σt = Tensile stress for the material of the stud which is usually nickel steel. ns = Number of studs. We know that the force acting on the cylinder head (or on the studs) π π × D 2 × p = (115) 2 3.15 = 32 702 N ...(i) = 4 4 The number of studs (ns ) are usually taken between 0.01 D + 4 (i.e. 0.01 × 115 + 4 = 5.15) and 0.02 D + 4 (i.e. 0.02 × 115 + 4 = 6.3). Let us take ns = 6. We know that resisting force offered by all the studs π π = ns × ( dc )2 σt = 6 × (0.84d ) 2 65 = 216 d 2 N ...(ii) 4 4 ...(Taking σt = 65 MPa = 65 N/mm2) From equations (i) and (ii), d 2 = 32 702 / 216 = 151 or d = 12.3 say 14 mm 1132 n A Textbook of Machine Design The pitch circle diameter of the studs (Dp) is taken D + 3d. ∴ Dp = 115 + 3 × 14 = 157 mm We know that pitch of the studs π × Dp π × 157 = = 82.2 mm = 6 ns We know that for a leak-proof joint, the pitch of the studs should lie between 19 d to 28.5 d , where d is the nominal diameter of the stud. ∴ Minim...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online