# We know that load on the fulcrum pin rf 1743 d1 l1 pb

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Unformatted text preview: e remains open = 33 + 180 + 1 = 214º Since the engine is a four stroke engine, therefore the camshaft angle for which the valve remains open = 214 / 2 = 107º Now, when the camshaft turns through 107 / 2 = 53.5°, the valve fifts by a distance of 16 mm. It may be noted that the half of this period is occupied by constant acceleration and half by constant decleration. The same process occurs when the value closes. Therefore, the period for constant acceleration is equal to camshaft rotation of 53.5 / 2 = 26.75 º and during this time, the valve lifts through a distance of 8 mm. We know that speed of camshaft N 475 = = 237.5 r.p.m. 2 2 ∴ Angle turned by the camshaft per second = 237.5 × 360 = 1425 deg / s 60 and time taken by the camshaft for constant aceleration, = 26.75 = 0.0188 s 1425 a = Acceleration of the valve. t= Let 1 a.t2 ... (Equation of motion) 2 1 8 = 0 × t + a (0.0188)2 = 1.767 × 10−4 a ...(∵ u = o) 2 ∴ a = 8 / 1.767 × 10–4 = 45 274 mm / s2 = 45.274 m / s2 and force due to valve acceleration, c...
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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 Chennai - Regional Office, Coimbatore.

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