Determine minimum movement of the tool:

During turning operation, a shaft of diameter =100 mm is rotating at 300 revolutions per minute. The length of the shaft to be machined is 400 mm. The graduations on the hand wheel scale are in millimeters and the hand wheel is 50 divided into 100 divisions. Some of the feeds obtainable on the machine tool are 0.02, 0.05, 0.1, 0.15, . . . , 0.50. The feed specified in the present case is 0.05 mm/rev. The final diameter desired is 98.50 mm. The maximum allowed depth of cut is 0.50 mm. Determine

 I.            the minimum movement of the tool that can be made,

 II.            feed per second,

 III.            number of passes required, and

 IV.            total time needed to machine the shaft.

Solution

Data given are as follows:

D_i =100 mm, D_l = 98.50 mm, L = 400 mm, fr =0.05 mm/rev.,

D_l = 0.50 mm, RPM =300.

(i) From the problem it is obvious that the hand wheel has 100 divisions, and hence for one rotation of the hand wheel, the linear movement achieved is only 1 mm. Therefore, the resolution (or minimum movement that may be achieved) is = 1/100 mm = 0.01 mm.

(ii) Feed per second,         f_s = f_r × RPS    ---------------- (1)

RPS = RPM/60

        = 300/60 = 5

∴ f_s = 0.05 × 5

f_s = 0.25 mm/s

 (iii)      The diameter of the shaft to be compact

= D_i - D_l

= 100 - 98.50

=1.50

With permitted d = 0.50 mm, in one pass the diameter shall reduce to 99 mm (100 - 2 × 0.5 = 99 mm). Still the diameter is bigger by 0.5 mm. Therefore, the final pass may be taken along 0.25 mm depth of cut to bring the machined diameter of the shaft equal to 98.50 mm.

Therefore TWO passes are required.

 (iii) The equation to determine total machining time needed is given as

.-------------------(4(a))

Therefore, since y and Δ are not given hence they may be taken as zero (in actual practice it will have non-zero value). The value of p_n = 2 as discussed above. Consequently,

Cutting Conditions

       t_m = 53.33  min