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# Providing a clearance of 15 mm between the roller and

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Unformatted text preview: aking pb = 7 N / mm2) ∴ and (d 2 )2 = 2460 / 8.75 = 281 or d2 = 16.76 say 18 mm Ans. l2 = 1.25 d2 = 1.25 × 18 = 22.5 say 24 mm Ans. Let us now check the roller pin for induced shearing stress. Since the pin is in double shear, therefore, load on the roller pin (Fc), π π 2460 = 2 × (d 2 ) 2 τ = 2 × (18) 2 τ = 509 τ 4 4 ∴ τ = 2460 / 509 = 4.83 N/mm2 or MPa This induced shear stress is quite safe. The roller pin is fixed in the eye and thickenss of each eye is taken as one-half the length of the roller pin. ∴ Thickenss of each eye, l2 24 = = 12 mm 2 2 Let us now theck the induced bending stress in the roller pin. The pin is neither simply supported in fork nor rigidly fixed at the end. Therefore, the common practice is to assume the load distrubution as shown in Fig. 32.27. The maximum bending moment will occur at Y –Y . Neglecting the effect of clearance, we have Maximum bending moment at Y – Y, t2 = Fc l2 t2 Fc l2 M = 2 2 + 3− 2 × 4 Fc l2 l2 = 2 2 + 6 Fc l2 × − 4 2 5 5 × Fc × l2...
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