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Unformatted text preview: ing the big end cap
dcb = Core diameter of the bolts, σt = Allowable tensile stress for the material of the bolts
= 60 N/mm2
nb = Number of bolts. Generally two bolts are used.
We know that force on the bolts
= (dcb )2 σt × nb = (dcb )2 60 × 2 = 94.26 (dcb )2
The bolts and the big end cap are subjected to tensile force which corresponds to the inertia
force of the reciprocating parts at the top dead centre on the exhaust stroke. We know that inertia
force of the reciprocating parts, cos 2θ FI = mR . ω2 . r cos θ +
l/r We also know that at top dead centre on the exhaust stroke, θ = 0.
∴ 2 2π ×1800 0.095 r FI = mR . ω2 . r 1 + = 2.25 0.095 1 + 0.38 N
l = 9490 N Equating the inertia force to the force on the bolts, we have
9490 = 94.26 (dcb)2 or (dcb)2 = 9490 / 94.26 = 100.7
∴ dcb = 10.03 mm and nominal diameter of the bolt, db = 10.03
say 12 mm Ans. 4. Thickness of the big end cap
tc = Thickness of the big end cap,
bc = Width of the big end cap. It is
taken equal to the length of the
crankpin or big end bearing (lc)
= 58 mm (calculated above)
σb = Allowable bending stress for the
material of the cap
= 80 N...
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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.
- Spring '11