# We know that the resultant force at the bearing 1

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ending moment on the shaft will be same as calculated eariler, i.e. MS = 10.08 × 106 N-mm and twisting moment on the shaft, TS = FT × r = 84 × 300 = 25 200 kN-mm = 25.2 × 106 N-mm ∴ Equivalent twisting moment on shaft, Te = = ( M S ) 2 + (TS ) 2 (10.08 × 106 ) 2 + (25.2 × 106 ) 2 = 27.14 × 106 N-mm We know that equivalent twisting moment (Te), π π ( d )3 τ = (135)3 τ = 483 156 τ 27.14 × 106 = 16 s 16 ∴ τ = 27.14 × 106 / 483 156 = 56.17 N/mm2 From above, we see that by taking the already calculated value of ds = 135 mm, the induced shear stress is more than the allowable shear stress of 31 to 42 MPa. Hence, the value of ds is calculated by taking τ = 35 MPa or N/mm2 in the above equation, i.e. π (d )3 35 = 6.873 (d s )3 16 s ∴ (ds)3 = 27.14 × 106 / 6.873 = 3.95 × 106 or ds = 158 say 160 mm Ans. (c) Design of shaft at the juncture of right hand crank arm Let ds1 = Diameter of the shaft at the juncture of the right hand crank arm. We know that the resultant force at the b...
View Full Document

Ask a homework question - tutors are online