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# We know that inertia force of the reciprocating parts

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Unformatted text preview: ing the big end cap Let dcb = Core diameter of the bolts, σt = Allowable tensile stress for the material of the bolts = 60 N/mm2 ...(Given) and nb = Number of bolts. Generally two bolts are used. We know that force on the bolts π π = (dcb )2 σt × nb = (dcb )2 60 × 2 = 94.26 (dcb )2 4 4 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 60 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 dcb = = 11.94 0.84 0.84 say 12 mm Ans. 4. Thickness of the big end cap Let 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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