Question: prove that in metal cutting. How velocity = cutting velocity x chip thickness coefficient ?
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Here,

$v_c$ = cutting velocity

$v_f$ = chip how velocity

$v_s$ = shear velocity

r = chip thickness coefficient

  • As in orthogonal cutting system

1) Every metal which is undergoing machining operation is considered as in compressible (i.e. no volumetric changes during machining)

$\therefore$ volume of material before machining = vol. after machining

$t_1$ $b_1$ $l_1$ = $t_2$ $b_2$ $l_2$ ........ $eq^n$ (01)

$t_1$ $b_1$ $v_c$ = $t_2$ $b_2$ $v_f$ ....... $eq^n$ (02)

2) width of the chip is assumed to be constant.

$\therefore$ $_b$1 = $b_2$ = b ......... $eq^n$ (03)

from $eq^n$ (01), (02) and (03)

As $_b$1 = $b_2$ = b

($\because$ r = $\frac{t1}{t2}$)

$\therefore$ $\frac{t1}{t2}$ = $\frac{l2}{l1}$ = r

similarly, $\frac{t1}{t2}$ = $\frac{vf}{vc}$ = r

$\therefore$ r = $\frac{t1}{t2}$ = $\frac{vf}{vc}$

r = $\frac{vf}{vc}$

$v_f = v_c \times r$ ..... Proved.

renu • 48 views
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modified 4 weeks ago  • written 6 months ago by gravatar for RB RB ♦♦ 110
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