GATE 2017

Question

A steel bar 20 mm in diameter is turned at a feed of 0.25 min/revolution with a depth of cut of 4 mm. The rotational speed of the work piece is 160 RPM. The material removal rate in cubic millimeter/second  is-

(1) 160

(2) 167.6

(3) 1600

(4) 1675.5

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Engineering Ishwarya Saro 8 months 1 Answer 106 views 0

Answer ( 1 )

  1. This comment is edited.

    We know that MRR (Material Removing Rate)(mm3) =  f * d * v
    Where we have f=0.25, d=4mm, v= (D w) /2 and  w =( 2 *22/7 * n ) / 60                                                                    So, v = ( 200/2 * 2*22/7 * 160 ) / 60 = 1675.52
    So, MRR =0.25* 4* 1675.52 = 1675.5 mm3 / s

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