84 d t tensile stress for the material of the stud

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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...
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