Internal combustion engine parts n 1149 now the

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Unformatted text preview: n (i.e. big end of connecting rod), therefore, 3 the reaction at these two ends will be in the same proportion. i.e. 1 2 F , and RC = F RP = 3I 3I l ∫ 2 Honeycomb coated with metal catalysts To exhaust Metal casing Co Carbon monoxide CO2 Carbon dioxide NOx Nitrous oxides HC Hydrocarbons N2 Nitrogen H2O Water Emissions of an automobile. Internal Combustion Engine Parts n 1149 Now the bending moment acting on the rod at section X – X at a distance x from P, 1 x x × x× MX = R P × x − *m 1 × ω 2 r × 2 3 l 1 x3 m .l 2 1 = 3 FI × x − 2 × ω r × 3 2 l ...(Multiplying and dividing the latter expression by l) FI × x x F x3 − FI × 2 = I x − 2 ...(i) 3 3 3l l For maximum bending moment, differentiate MX with respect to x and equate to zero, i.e. 3 = dMX F 3x 2 = 0 or I 1 − 2 = 0 dx 3 l 2 3x l 1− ∴ = 0 or 3 x 2 = l 2 or x = 2 l 3 Maximum bending moment, Mmax 3 l = FI l − 3 33 l2 ...[From equation (i)] FI l l FI × l 2 2 FI × l − ×= = 3 = 3 33 93 3 3 3 m l l × ω2 r × = m × ω2 r...
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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.

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