5 603 363 12 17 640 mm3 z 60 2 induced bending

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Unformatted text preview: 2014.92 + 96.6 + 346.72 = 2458.24 say 2460 N Since the length of the two arms of the rocker are equal, therefore, the load at the two ends of the arm are equal, i.e. Fe = Fc = 2460 N. Internal Combustion Engine Parts n 1197 Front view of a racing car We know that reaction at the fulcrum pin F, RF = = ( Fe )2 + ( Fc ) 2 − 2 Fe × Fc × cos θ (2460) 2 + (2460)2 − 2 × 2460 × 2460 × cos 135º = 4545 N Let us now design the various parts of the rocker arm. 1. Design of fulcrum pin Let d1 = Diameter of the fulcrum pin, and ...(Assume) l1 = Length of the fulcrum pin = 1.25 d1 Considering the bearing of the fulcrum pin. We know that load on the fulcrum pin (RF), 4545 = d1 × l1 × pb = d1 × 1.25 d1 × 5 = 6.25 (d1)2 ...(For ordinary lubrication, pb is taken as 5 N/mm2) ∴ (d1)2 = 4545 / 6.25 = 727 or d1 = 26.97 say 30 mm Ans. and l1 = 1.25 d1 = 1.25 × 30 = 37.5 mm Ans. Now let us check the average shear stress induced in the pin. Since the pin is in double shear, therefore, load on the fulcrum pin (RF), π π (d )...
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